TITLE: What's that shape? Inferring a 3D shape from random shadows
QUESTION [26 upvotes]: Let $P$ be a bounded, simply connected region of $\mathbb{R}^3$.
$P$ could be a polyhedron, or a smooth shape, or an arbitrary shape;
I'll assume below that $P$ is a (non-degenerate, perhaps non-convex) polyhedron of $n$ vertices.
Suppose you have available a number $k$ of shadows of $P$, projected orthogonally to
the $xy$-plane, of $P$ after a random 3D reorientation, uniformly random over
the sphere of orientations. You have only the outline of the opaque shadow;
no internal details are visible.
For example, here are $k=5$ random shadows of a 3D polyhedron:



You do not know the orientations of the object that produced the shadows,
only that they are chosen randomly.
Perhaps you know the number of vertices of the polyhedron; $n=20$ above.
My question is not the ideal sharp MO question, but still I think there is
a substantive issue here:

Q. Under what circumstances can one (approximately) reconstruct $P$
from the $k$ shadows?

It seems that as $k \to \infty$, there should be a convergence-to-$P$ result.
As this is a natural question,1
I expect it has been explored in the literature,
in which case, pointers would be appreciated.

Click inside the region below to see the polyhedron that produced the five shadows above:

 


One can view my question as a randomized version of the orthogonal
Gödel, Escher, Bach GEB projections:

(Of course, the same question can be asked for objects in $\mathbb{R}^d$
shadow-projected
to lower-dimensional hyperplanes.)

1

Watching the shadow of a hanging object twisting in the wind is a commonly encountered instance.


Addendum.
Several have pointed out that certain shape features are hidden from
influencing any shadow, and so cannot be reconstructed from shadows, even with
known orientations.
It turns out that the shape that can be reconstructed is known as the visual hull
(which I found by searching for "shadow hull"):

Laurentini, Aldo. "The visual hull concept for silhouette-based image understanding." IEEE Trans. Pattern Analysis Machine Intelligence,  16.2 (1994): 150-162.
(Journal link.)

This Laurentini paper points out that the visual hull of a polyhedron is not
always itself a polyhedron: Its surface can in general have curved surface patches.
There is work on reconstructing the visual hull when the orientations producing the shadows are known:

Matusik, Wojciech, Chris Buehler, and Leonard McMillan. Polyhedral visual hulls for real-time rendering. Springer Vienna, 2001. (PDF download.)

And Dustin Mixon identified literature addressing unknown orientations:

Singer, Amit, and Yoel Shkolnisky. "Three-dimensional structure determination from common lines in cryo-EM by eigenvectors and semidefinite programming." SIAM Journal Imaging Sciences. 4.2 (2011): 543-572. (Journal link.)

REPLY [3 votes]: In the case where you also have to deal with small signal-to-noise ratio, the problem of deducing the relative orientations the projection came from becomes a lot more challenging. There is a good series of paper by my PhD adviser Veit Elser and his former student Duane Loh on this problem, which they call cryptotomography. They derive information theoretic bounds on the limit of noise that allows faithful reconstruction. In their specific application they are actually interested in slices of Fourier-space data, since the data comes from X-ray diffraction, but their methods and theory also apply to projections of real-space data.
A sample:
Elser, "Noise Limits on Reconstructing Diffraction Signals From Random Tomographs"
Loh, Borgan, Elser, et al., "Cryptotomography: Reconstructing 3D Fourier Intensities from Randomly Oriented Single-Shot Diffraction Patterns"<|endoftext|>
TITLE: Intersection of connected components in $\mathbb{R}^n$
QUESTION [7 upvotes]: Let $n$ be a positive integer and let $K\subseteq \mathbb{R}^n$ be compact. Pick $x^* \in \mathbb{R}^n\setminus K$.
Let $E$ be the connected component of $\mathbb{R}^n\setminus K$ that contains $x^*$. Let ${\cal C}$ be the collection of connected components of $K$. For each $C\in {\cal C}$ let $E_C$ be the connected component of $\mathbb{R}^n\setminus C$ that contains $x^*$.
Is it true that $E=\bigcap\{E_C: C\in{\cal C}\}$?

REPLY [5 votes]: The answer to this problem is Yes.
Indeed, the inclusion $E\subset \bigcap_{C\in\mathcal C}E_C$ is trivial, so it remains to prove that for any point $x\in\mathbb R^n\setminus E$ there exists a connected component $C\in\mathcal C$ of $K$ such that $x\notin E_C$. By Zorn's Lemma, the compact set $K$ contains a minimal closed subset $S\subset K$ such that the points $x^*$ and $x$ belong to distinct components of $X\setminus S$. 
Such a minimal compact set $S$ is called an irreducible barrier between $x$ and $x^*$. The Claim below implies that the irreducible barrier $S$ is connected. Then for the connected component $C$ of $K$ containing the irreducible barrier $S$ we get $x\notin E_C$.
Claim. The irreducible barrier $S$ is connected.
Proof. The proof involves some tools of algebraic topology. I do not know if there is a more elementary proof. Assuming that $S$ is disconnected, we can write $S$ as the disjoint union $A\cup B$ of two non-empty compact sets. Since $S$ is a minimal barrier between $x$ and $x^*$ the points $x,x^*$ can be linked by a path in $\mathbb R^n\setminus A$ and  $\mathbb R^n\setminus B$. Since $x,x^*$ belong to different connected components, the 0-dimensional singular cycle $\alpha=x-x^*$  detemines a non-zero-element in the homology group $H_0(\mathbb R^n\setminus S)$.
Now observe that $(\mathbb R^n\setminus A)\cup(\mathbb R^n\setminus B)=\mathbb R^n\setminus(A\cap B)=\mathbb R^n$ and write the exact Mayer-Vietoris sequence 
$$H_0(\mathbb R^n)\to H_0(\mathbb R^n\setminus S)\to H_0(\mathbb R^n\setminus A)\oplus H_0(\mathbb R^n\setminus B).$$ By the minimality of $S$, the image of the singular cycle $\alpha$ in $H_0(\mathbb R^n\setminus A)\oplus H_0(\mathbb R^n\setminus B)$ is trivial. Now the exactness of the Mayer-Vietros sequence implies that $\alpha$ is zero in $H_0(\mathbb R^n\setminus S)$, which is a desired contradiction.<|endoftext|>
TITLE: Introductory article of knot Heegaard Floer Homology
QUESTION [10 upvotes]: I am looking for some article that gives an introduction to Heegaard Floer homology of knot.
I heard that it is very useful to determine the unknotting number of a knot, but I couldn't find any introductory article. (http://arxiv.org/abs/1411.4540 and http://arxiv.org/abs/1003.6041 are Greek to me!)
Can you point me to some introductory article? Could you also please give me some example of calculation of knot Floer homology, say for "small" knots like $3_1$ and $4_1$? 
If possible, can you also tell me why Heegaard Floer homology does not determine the unknotting number of 8_10?

REPLY [20 votes]: (Since this is just a string of references, I do not believe this constitutes a 'real answer' but it is too long for a comment, so I'm placing it in the answer field. Editors, please feel free to correct my etiquette.)
As for a general introduction or survey article, you might also look at these:

"An introduction to Heegaard Floer homology" by Ozsvath and Szabo. 
https://web.math.princeton.edu/~szabo/clay.pdf
"A introduction to knot Floer homology" by Manolescu. 
http://arxiv.org/abs/1401.7107

Regarding the second part of your reference request, about the calculation of the knot Floer groups of the trefoil or the figure eight; well, these are alternating knots, and so their Floer groups are completely determined by their signature and Alexander polynomials (see Theorem 1.3). However, I think what you are asking for is an explicit calculation from a Heegaard diagram. In the paper "Holomorphic disks and knot invariants" by Ozsvath and Szabo, you can find such a calculation for the trefoil in Section 6.1. However, this is not an introductory article --- it is full strength. You may also benefit from this expository article (in PDF form) written by Andrew Manion. His exposition also contains examples of explicit calculations, especially in Section 3. 
Sometimes the grid diagram approach to calculating knot Floer groups makes for a gentler introduction. For that, you might look at the paper "A combinatorial description of knot Floer homology" by Manolescu, Ozsvath and Sarkar. In Section 4 there are explicit calculations for the Hopf link and the trefoil.
Finally, for the third part of your reference request, I don't really understand what you mean by 'does not determine the unknotting number,' but I think you should look at the paper "Knots with unknotting number one and Heegaard Floer homology" by Ozsvath and Szabo, in particular Theorem 1.1 and Corollary 1.2. (The arXiv version is linked). They use the Heegaard Floer homology of the branched double cover of a knot to give an obstruction to that knot having unknotting number one. 
They apply their obstruction (the symmetry condition of Theorem 1.1) to show that the alternating knot $8_{10}$ does not have unknotting number one. I am under the impression this knot was already known to have $u(K)\leq2$, therefore they conclude it has unknotting number two.<|endoftext|>
TITLE: Linear projections of convex sets with unique preimages of boundary points
QUESTION [6 upvotes]: Fix a compact convex subset $C \subset \mathbb{R}^n$ with nonempty interior. For any subspace $S \subset \mathbb{R}^n$, let $P_S$ denote the orthogonal linear projection onto $S$. I'd like to claim that for almost every (in either a measure theory or topological sense) nontrivial subspace $S$ of a given dimension, the image of $C$ under $P_S$ has the following property: for every $y$ in the relative boundary of $P_S(C)$, the fiber $\{x \in C : P_S(x) = y\}$ is a singleton. 
In the case when $\dim S = 1$, my claim is equivalent to the statement that almost every linear functional attains its maximum at a single point in $C$. I'm particularly interested in the case when $\dim S = n-1$.  Has anyone ever seen anything like this? Note: I believe it is easy to verify the claim when $C$ is strictly convex or polyhedral, so the interesting situation is when $C$ is a general compact convex set.  

Thinking about this some more, it is not even obvious that the set of subspaces $S$ of a given dimension with this property (points on the relative boundary of $P_S(C)$ have unique preimages in $C$) is nonempty. Even a basic existence argument for general convex sets would be useful.

REPLY [2 votes]: I asked this question several years ago, and I recently found the answer in a paper by Ewald, Larman, and Rogers called "The directions of the line segments and of the r-dimensional balls on the boundary of a convex body in Euclidean space".  This was published in 1970 in Mathematika, and the main result in the paper is:
Theorem 1. If $K$ is a convex body in $\mathbb{R}^n$, the set $S$, of end-points of the vectors drawn from the origin in the directions of the line segments lying on the surface of $K$, is a set of $\sigma$-finite $(n-2)$-dimensional Hausdorff measure on the $(n-1)$-dimensional surface of the unit ball.<|endoftext|>
TITLE: Is this a rational function?
QUESTION [19 upvotes]: Is $$\sum_{n=1}^{\infty} \frac{z^n}{2^n-1} \in \mathbb{C}(z)\ ?$$
In a slightly different vein, given a sequence of real numbers $\{a_n\}_{n=0}^\infty$, what are some necessary and sufficient conditions for $\sum a_nz^n$ to be in $\mathbb{C}(z)$ with all poles simple?

REPLY [14 votes]: I would not argue with the irrationality of the function following from the functional equation relating $f(2z)$ to $f(z)$. In fact this function is a particular case of the so-called $q$-logarithm which received a quite special attention. Note that being rational over $\mathbb C(z)$ or over $\mathbb Q(z)$ for a power series with rational coefficients is equivalent.
If $f(z)$ where rational then $$f(1)=\sum_{n=1}^\infty \frac1{2^n-1}=\sum_{m=1}^\infty \tau(m)2^{-m}$$ would be a rational number, where $\tau(m)$ denotes the number of divisors of $m$. Erdős proved already in 1948 that this is not the case by showing that the base $2$ expansion of the number is not periodic. (This is still a nice exercise in analytic number theory!)
The irrationality of the values of $f(z)$ for rational $z\ne0$ was established by P. Borwein in 1992 using the Padé approximations to the function.<|endoftext|>
TITLE: What does "simplification of proofs as evaluation of programs" mean?
QUESTION [12 upvotes]: I am currently going through Philip Walder's "Proposition as Types" and a passage of the introduction has struck me:

for each way to simplify a proof
  there is a corresponding way to evaluate a program

What does simplify a proof means here? I have searched in my logic textbook and on the Internet but surprisingly enough I could not find any direct, informal explanation.

REPLY [9 votes]: Andrej's answer gives the technical explanation, but it might be useful to note, in addition, that "proof simplification" is a technical term that roughly corresponds to "lemma elimination", and is analogous to cut-elimination in the sequent calculus.
Informally, this operation is the opposite of a "proof simplification"! The things humans usually do to simplify proofs, e.g. introduce new notations, prove intermediate lemmas, generalize conclusions are the things that are eliminated by the operation you are referring to.<|endoftext|>
TITLE: Computation time of Smith normal form in Maple
QUESTION [8 upvotes]: I am using Maple to compute the Smith normal form (SNF) of a $120 \times 120$ matrix and it seems that I will never get an answer back. I have checked my code for small cases and I believe that it is correct. When I try to compute the SNF for a $24 \times 24$ matrix, the real time and CPU time are about 0.1~0.2 seconds. I don't think it will take more than 3 hours for a $100 \times 100$ matrix. I have also tried it on several operating systems and the results are similar. 
Anyway, what is the approximate time complexity of the computation time of SNF? Is there a limit for the matrix size in Maple to do SNF?
Thank you so much!

REPLY [6 votes]: (Dense) Smith Normal Form is theoretically computable in $O(\|A\| \log \|A\| N^4\log N)$ time (Arne Storjohann, 1996). Storjohann was at Waterloo at the time, so I would not be surprised if that is the algorithm Maple uses. Sparse SNF is much faster.<|endoftext|>
TITLE: Totally geodesic submanifold of a hyperbolic 3-manifold
QUESTION [6 upvotes]: If $M$ is a convex-cocompact hyperbolic 3-manifold, and $S$ is a closed surface with genus $\geq$ 2. Suppose $f:S\to M$ is a minimal immersion, and $f(S)$ is negatively curved. I know that all the closed geodesics in $f(S)$ are closed geodesics in $M$.  Can I conclude that $f(S)$ is totally geodesic in $M$?

REPLY [4 votes]: Reading the first paragraph of the introduction of the following paper:
http://homeweb.unifr.ch/parlierh/pub/BuserParlierOsaka.pdf
it seems to me that the set of unit tangent vectors $v_p$ to $S$ such that the $\gamma(t):=exp(tv_p)$ is a closed geodesic is dense. If I am not misunderstanding such a density then $\alpha(v_p,v_p) = 0$ for a dense set of unit vectors, where $\alpha$ is the second fundamental form of $f(S)$. Thus, $f(S)$ is totally geodesic in $M$ since $\alpha(v_p,v_p) \equiv 0 $ for all $v_p$ due to the density.<|endoftext|>
TITLE: Dominated convergence to characteristic function
QUESTION [7 upvotes]: Let $\phi_m(x):=\chi_{[0,1]} * \chi_{[0,1]} *...* \chi_{[0,1]}$ be the m -times convolution (so $m+1$ characteristic functions are involved).
Then the Fourier transform of this function is given by
$$\hat{\phi}_m(x):=\left( \frac{e^{-ix}-1}{-ix} \right)^{m+1}.$$
Now, I want to show that for
$$ \hat{f}_m(x):= \frac{\hat{\phi}_m(x)}{\sum_{l \in \mathbb{Z}} |\hat{\phi}_m(x+2\pi l)|^2}$$
we have $|| \hat{f}_m- \chi_{[-\pi,\pi]}||_{L^2} \rightarrow 0.$
I already showed that $| \hat{f}_m| \rightarrow \chi_{[-\pi,\pi]}$ pointwise and that there exists $|g| \in L^2$ such that $|\hat{f}_m| \le |g|.$
So the proposition would immdiately follow, if I would know that the imaginary part of this $\hat{f}_m$ would tend to zero for $m \rightarrow \infty$ (and I need this result only for $|x| \le \pi.$ Unfortunately, I don't see how this can be done. I tried expanding the product, but this was just messy.
Does anybody have an idea? Maybe there is more abstract reason, why the imaginary part has to vanish in the limit?
If you have any questions, please let me know in the comment section. In particular, any other possible proof of this statement is also highly appreciated.

REPLY [2 votes]: I think the sequence does not converge in $L^2.$
We have
$$\vert \hat f_{4m-1}(x)\vert= e^{i2xm}\hat f_{4m-1}(x)= {1\over{(2x\sin(x/2))^{4m}\sum_{l\in Z} {1\over{(x+2\pi l)^{8m}}}}}$$
Let us suppose $ \lim \hat f_m = f  ,\ \lim \vert \hat f_m\vert = \vert f \vert$ in $L^2$.
If we define $g_m(x)=e^{2imx}f(x),$
 we have 
$$\lim\Vert g_m-\vert \hat f_{4m-1}\vert\Vert_2 =\lim\Vert g_m- \vert f\vert\Vert_2 =0 \ \Rightarrow \ \  \lim\Vert g_{m}- g_{2m}\Vert_2 =0 .$$
If $\vert f\vert >\epsilon $  on some bounded interval $I$ we deduce
$$\lim_m\int_I \sin^2 (m x)  \vert f(x)\vert^2dx=\lim_m\int_I \sin^2 (m x)  dx=0$$
which is a contradiction.<|endoftext|>
TITLE: Ordinals in constructive mathematics ? (references)
QUESTION [13 upvotes]: I'm looking for references presenting a constructive treatment of the theory of ordinals. By constructive I mean valid in the internal logic of a topos (so no axiom of choice and no law of excluded middle).
I can easily guess that there is several non equivalent approaches to this question: for example it makes sense to define ordinal as being well ordered sets as such do exist internally in toposes (For example the Higgs object) but this would mean that the natural number object when it exist is not an ordinal.
For this reasons I will clarify a bit what kind of properties I want on ordinals:

The natural number object should be an ordinal.
I want to be able to do proof by induction over ordinals and induction over the natural number object should be a special case of ordinal indcution.
In the topos of sheaf over a topological space $X$ ordinals should be described the following way:
For every (classical) ordinal $\alpha$, one can define the sheaf $F_{\alpha}$ over $X$ of (local) continuous functions from $X$ to $\alpha$ (I guess with the order topology on $\alpha$, although I'm open to suggestion on that point).
If $\alpha' > \alpha$ there is a canonical inclusion $F_{\alpha} \hookrightarrow F_{\alpha'}$ and one can define $Ord_X$ as the (large) sheaf obtain as the union of all the $F_{\alpha}$.
Ordinals over $X$ should corresponds to sections of this large sheaf $Ord_{X}$.
One should be able to pullback ordinal along geometric morphisms (although it might not be exactly a pullback of sheaf), and if $f$ is a bounded geometric morphism there should be an operation of pushforward of ordinals, right adjoint to the pullback for the order relation.
Maybe ordinal of the effective topos are related to recursive ordinal ? (this is suggested by the look of the NNO of the effective topos, but this last one is just a guess)

I'm relatively convince that such a notion exists, and the third point can be used to answer any question that one might have about the properties ordinals should have (at least for the "geometric" properties): For example, it should not be expected that ordinals are totaly ordered, but any two ordinals should have a supremum (because the function max(a,b) is continuous in the order topology but $\{a \leqslant b \}$ is not open in $\alpha \times \alpha$). 
What I was wondering is if this kind of constructive theory of ordinals has been already developed and appears in the literature or not ?
Edit : It seems that the more restricitve notion of Ordinals among those that has been proposed in answer and comment is Paul Taylor notion of Plump ordinals (with an equivalent inductive-indutive definition in type theory given here ) but this definition seems already too weak for what I had in mind : one has the following Plump ordinals $0 = \emptyset$, $1 = \{ \emptyset \}$, $2 = \Omega$, $3 = \{$Initial segement of $\Omega \}$, $n =  \{ $ Initial segment of $ n -1 \}$. And as any element of an ordinal is again ordinal there is already way too many of them to be describe by a geometric theory as I mentioned in my third point (the elements of $3$ are already non geometric). SO there must be a more restrictive notion but it seems that no one has considered it yet...

REPLY [4 votes]: Perhaps the literature on W-types is what you are looking for? This is well-developed categorically and gives a good theory of inductive types.
If you are looking explicitly for ordinals, there's a recent discussion about ordinals in HoTT. @paul-taylor has worked on this. Since he is a regular here, I'll just provide the references. JSL paper and the section in his book.<|endoftext|>
TITLE: Symplectic reversing diffeomorphisms on a compact symplectic manifold
QUESTION [8 upvotes]: I Ask this question in MSE and I received interesting comments and ideas. I repeat the question here for more discussion:
Let  $(M,\omega)$ be  a compact symplectic manifold.
Is there always a diffeomorphism $f$ on M with $f^{*}\omega =-\omega$?

REPLY [6 votes]: I'd like to mention the work of Castaño-Bernard-Matessi-Solomon, who proved the existence of an anti-symplectic involution for symplectic manifolds carrying a Lagrangian torus fibration of a certain class. Such a class of Lagrangian fibration is constructed by gluing local models of Lagrangian fibrations to integral affine manifolds with singularities.
More precisely, let $(B,\mathscr{A},\Delta)$ be an integral affine manifold, whose affine structure $\mathscr{A}$ is singular along $\Delta\subset B$. In this case, we have a decomposition
$\Delta=\Delta_p\cup\Delta_g\cup\Delta_n$,
where the subscripts indicate the types of the singularity, namely positive, generic or negative. For simplicity, let's restricts to the case of a symplectic 6-manifold. After taking away the discriminant locus, $B\setminus\Delta$ carries the structure of an integral affine manifold, and we may form a Lagrangian torus bundle $T^\ast(B\setminus\Delta)/\Lambda^\ast$ in the obvious way, where $\Lambda\subset T(B\setminus\Delta)$ is a lattice bundle coming from the affine structure $\mathscr{A}$ on $B\setminus\Delta$ and $\Lambda^\ast$ denotes its dual. The aim is to glue local models of singular Lagrangian fibrations using fiber-preserving symplectomorphism to the torus bundle $T^\ast(B\setminus\Delta)/\Lambda^\ast$ so that we obtain a Lagrangian fibration $\pi:M\rightarrow B$ on the symplectic 6-manifold $M$ which serves as a compactification of the previous torus bundle.
For a small neighborhood $U_p\subset B$ so that $\Delta_p\subset U_p$, one considers the standard Lagrangian fibration $\pi_{std}:\mathbb{C}^3\setminus\{xyz=-1\}\rightarrow\mathbb{R}^3$ defined by
$(x,y,z)\rightarrow\left(\log|xyz+1|,\left(|x|^2-|y|^2\right)/2,\left(|x|^2-|z|^2\right)/2\right)$
Similarly one has local models of Lagrangian fibrations for $\Delta_g\subset U_g$ and $\Delta_n\subset U_n$. The outcome is a Lagrangian torus fibration $\pi:M\rightarrow B$ of preferred topology and singularities. In the paper of Castaño-Bernard-Matessi-Solomon, they call such Lagrangian fibrations belong to class $\mathscr{C}$.
For these Lagrangian fibrations, one can analyze their local models explicitly and glue local Lagrangian sections to a global one. One distinguished property of a Lagrangian fibration $\pi$ in $\mathscr{C}$ is that $\pi$ is not smooth near $\Delta_n$, so one has to impose some restrictions on the Lagrangian sections $\sigma$ of $\pi$ we work with, these restrictions specify a class $\mathfrak{C}$ of Lagrangian sections. With these preliminaries the main result of Castaño-Bernard-Matessi-Solomon can be stated as follows:
Theorem. Let $\pi:M\rightarrow B$ be a Lagrangian fibration of class $\mathscr{C}$. Given a Lagrangian section $\sigma$ of $\pi$ in class $\mathfrak{C}$, there is a unique antisymplectic involution $\phi_{\pi,\sigma}$ of $M$ such that $\phi_{\pi,\sigma}$ preserves the fibers of $\pi$ and fixes the Lagrangian section $\sigma$.
Notice that the class $\mathscr{C}$ is actually very general, it includes Lagrangian fibrations without singular fibers (cotangent bundles, tori), the so-called almost toric symplectic manifolds introduced by Leung-Symington (which plays an essential role in the recent deep work of Vianna in symplectic topology), and 6-dimensional symplectic Calabi-Yau manifolds homeomorphic to a quintic or Schoen's Calabi-Yau manifolds.
Speculations. From the point of view of mirror symmetry, the non-existence of an anti-symplectic involution roughly means the mirror of $M$ does not exist or at least carries a gerbe $\alpha\in H^2_{et}(M^\vee,\mathscr{O}^\ast)$. This is because the anti-symplectic involutions constructed by Castaño-Bernard-Matessi-Solomon (which are roughly reflections with respect to the Lagrangian section) are in some sense mirror to the functor $R\underline{Hom}(-,\mathscr{O}_{M^\vee})$ on the bounded derived category of coherent sheaves $D^b(M^\vee)$. It should be interesting to establish this speculation rigorously at least in some specific examples. But I don't know how to do this.<|endoftext|>
TITLE: An inequality for two independent identically distributed random vectors in a normed space
QUESTION [16 upvotes]: Suppose that $X$ and $Y$ are independent identically distributed random vectors in a separable Banach space $B$. Does it always follow that $E\|X-Y\|\le E\|X+Y\|$? 
Some background information on this question can be found at the end of the note posted on arXiv at additive decomposition of norms. In particular, the inequality in question holds if $B$ is two-dimensional or Euclidean.

REPLY [6 votes]: To get dimension 4, I think the norm 
\begin{equation} \|a\| =  \max_{\{i,j\}}  |a_i-a_j| \vee \|a\|_\infty,
\end{equation}
where $a=(a_1,a_2,a_3,a_4)$, 
works. Again $X$ samples the unit vector basis uniformly.
EDIT: It looks like a variation takes care of dimension 3.  Use again
\begin{equation} \|a\| =  \max_{\{i,j\}}  |a_i-a_j| \vee \|a\|_\infty,
\end{equation}
where $a=(a_1,a_2,a_3)$. This time, let $X$ sample $e_1, e_2, e_3 $ and $-(e_1+e_2+e_3)/2$ uniformly. The first three vectors have norm 1 and the last one norm $1/2$.  In the (unlikely) event that my arithmetic is correct, the expectation of $\|X+Y\|$ is $19/16$ and the the expectation of $\|X-Y\|$ is $21/16$.
If this is correct, the only remaining thing is whether the inequality is true for  random variables that take on three non zero values.<|endoftext|>
TITLE: $L^1$ norm of exponential sum of $n^2 x$
QUESTION [9 upvotes]: What is the asymptotic order of
$$
\int_0^1 \left| \sum_{n=1}^N e^{2 \pi i n^2 x} \right| ~dx
$$
as $N \to \infty$. This should be known, but I cannot find it in the literature.

REPLY [17 votes]: Jurkat and van Horne showed that the $L^1$ norm is asymptotic to a constant times $\sqrt{N}$ (see Theorems 4 and 5 in their paper which compute all moments).  For other related work see Jurkat and van Horne and Marklof.  Finally Vaughan and Wooley considered Weyl sums for powers larger than $2$, and formulated some conjectures -- the case for squares behaves differently from higher powers.<|endoftext|>
TITLE: A question involving e, floor, and all x > 0
QUESTION [9 upvotes]: Is $\lfloor(x+1/2)e\rfloor = \lfloor(x+1)(1+1/x)^x\rfloor$ for all $x > 0$?
The question occurred in connection with (nonhomogeneous) Beatty sequences, $\lfloor nr+h\rfloor$, where irrational $r>0$ and real $h$ are fixed, and $n = 1,2,\dots$.
Let
$$s_n = (n+1/2)e - (n+1)(1+1/n)^n$$ I checked that $(s_n)$ is strictly decreasing and $0 < s_n < 1$ for $n = 1,2,\dots, 10^6$.
Oops, thanks for noting that the leap from positive integers $n$ to real $x$ was blind.  So, the answer to the question as asked is "no" - leaving a subquestion, whether the proposed identity holds for positive integers $n$.  (Still, though, should anything else be said about the left side versus the right side for non-integer values of $x$.)

REPLY [6 votes]: The question is whether there is an integer $m$ with $(n+1/2)e < m \le (n+1)(1+1/n)^n$.  Note that $s_n \sim e/n$, so this would mean (approximately)
$$ 0 > e - \dfrac{2m}{2n+1} > \dfrac{2e}{n(2n+1)}$$
Now the continued fraction for $e$ is well-known:
$$ e = [2;1,2,1,1,4,1,1,6,1,1,8,\ldots]$$
The corresponding even-numbered convergents ($[2;1],\; [2;1,2,1],\; \ldots$) are greater than $e$, the odd-numbered ones are less than $e$.  But the even-numbered convergents all have odd numerators.  So we won't get a counterexample from those.  I suspect that with further work on the "not-quite-best" rational approximations of $e$ one can show that there are no counterexamples.<|endoftext|>
TITLE: Can we construct a Baas-Sullivan presentation of TMF?
QUESTION [12 upvotes]: Quick Review: 
The Baas-Sullivan construction cones off generators $\alpha_1, ..., \alpha_n \in \pi_*(MU)$ from $MU$ to get a new spectrum $MU/(\alpha_1, ..., \alpha_n)$, which is isomorphic to some spectrum $E$ we wish to present as a bordism theory with singularities. This is a "geometric" alternative to tensoring out generators as we do in the Landweber-exact functor construction.

To construct TMF, we're looking at $T: (Aff^{et}_{/M_{ell}})^{op} \to E_\infty$-$\text{Ring spectra}$
It's my understanding that to make sense of $TMF := \Gamma(T) := \text{holim } T(U_i)$, we need our target category to be enriched in spaces (where $U_i$ := an affine cover of $M_{ell}$, and $T(U_i)$ : cosimplicial diagrams).
Over a point, $TMF$ looks like either height 1 or height 2 Morava E-theory, both of which can certainly be presented via the Baas-Sullivan construction (Morava E-theories can be constructed by starting with BP and killing off generators).  
I've been told that we don't have a method to check that presentations of Morava E-theories (as bordism theories with singularities) are $E_\infty$-ring spectra. 
Can we construct a Baas-Sullivan presentation of TMF?
It doesn't quite make sense to construct a map from the (derived?) Landweber-exact theory to the Baas-Sullivan presentation to show that the presentation has $E_\infty$-structure, so I'm not sure what to do here.

REPLY [18 votes]: There are a couple of possible variant questions of this, and I'm not quite sure which is appropriate. The first question question is whether, without knowledge of the functor $T$, we could construct $TMF$ by Baas-Sullivan theory.
If you do not invert 6, then this is not possible. Any object constructed by Baas-Sullivan theory has a unit $S \to X$ which factors through $\pi_* S \to \pi_* MU \to \pi_* X$. This violates the fact that the 24-torsion element $\nu \in \pi_3 S$ is detected by any variant of $TMF$.
If you do invert 6, then some variants of this problem have positive solutions and some do not.

The connective spectrum $tmf$ has, away from 6, homotopy groups $\Bbb Z[c_4, c_6, 1/6]$, and can be obtained from $MU$ by inverting 6 and killing a regular sequence.
The version associated to the uncompactified moduli $M_{ell}$, often call $TMF$, has homotopy groups $\Bbb Z[c_4, c_6, \Delta^{-1}, 1/6]$. Because $\Delta^{-1}$ is in negative degrees, this cannot be constructed because any spectrum produced from the Baas-Sullivan method only has positive-degree homotopy groups. This is, however, only a small failure; localization is a relatively mild extra tool to add.
The version associated to the compactified moduli $\overline{M}_{ell}$, sometimes called $Tmf$, has (again, away from 6) the same homotopy groups as $tmf$ in positive degrees and a $\Bbb Z[1/6]$-dual set of groups in negative degrees. The negative homotopy groups obstruct using the Baas-Sullivan method, and it is also not possible to get it simply by localizing something constructed by those techniques. However, it is possible to construct it with a little more work: if we have already constructed $tmf$, then we can get a diagram of localizations
$$
c_4^{-1} tmf \to (c_4 \Delta)^{-1} tmf \leftarrow \Delta^{-1} tmf
$$
whose homotopy pullback is $Tmf$. (This last example illustrates the use of patching together constructions on a cover of $\overline{M}_{ell}$ to get a global construction. It works well because we can construct $tmf$ and then we can do localizations "in the category of $MU$-modules".)

The second question that you might ask is whether we might use Baas-Sullivan theory to construct a diagram (part of $T$) with $TMF$ or $Tmf$ as a homotopy limit. Even if Baas-Sullivan theory were completely functorial, this does not work either: the constructions of Baas-Sullivan theory naturally land in the category of $MU$-modules and so the primary objection about detecting $\nu$ still holds.
A third question that you might ask is whether we might construct objects using Baas-Sullivan theory, and maps between them by some other method, to give us a diagram whose homotopy limit is $[TMF\mid Tmf\mid tmf]$. That's a perfectly good idea (and using Landweber's theorem, after all, is how Morava got this started in "Forms of K-theory").
We run into a couple of problems with this program. One is that Baas-Sullivan theory classically produces objects in the homotopy category, and to construct a homotopy limit we need a lift to an honest diagram. There is no way around this; at some point we have to make sure that it is possible. Another is that, to my knowledge, Baas-Sullivan theory doesn't account for the possibility that we might need to relate the object constructed to another, isomorphic, formal group law constructed by a different regular sequence.
A third is that, as we saw above, to construct such the diagrams we typically want to patch together along open subobjects. That means localization, and localization is generally very difficult to carry out if we only have a multiplication on the homotopy groups rather than on the object proper. Above we were working in $MU$-modules and inverting elements in $MU_*$. Once we leave that situation and go back to situations where the primes 2 and 3 are in play, we no longer have localization available for free. Having a ring in the homotopy category is enough to do a localization if the ring of homotopy groups is commutative, but it's not easy to ensure that the result is still a ring in the homotopy category and allow us to do more localizations after that. Having a diagram of homotopy rings does not ensure a limit object, like a homotopy pullback or a fixed-point object for a group, also has a ring structure. There are also slightly delicate questions about whether localization is unique, or whether it has a universal property of any kind.
These are some reasons why one might move to trying to carry out a construction within a point-set theory of associative or commutative ring spectra. Unfortunately, once we are there Baas-Sullivan theory is no longer in our toolkit; it does not interact perfectly with point-set constructions and products (though Neil Strickland proved a lot about it in "Products on $MU$-modules"). I believe that Laures and McClure have work relating products in bordism theory to spectrum-level products, but I'm unsure of the degree to which one can make it run with singularity bordism.<|endoftext|>
TITLE: Does Peano's existence theorem admits a constructive proof?
QUESTION [11 upvotes]: $$y(t)=y_0+\int_0^t b(y(s))ds$$ $b\in C(R^d)\cap L^\infty(R^d)$
The classical proof for Peano's existence theorem in ODE need use the Ascoli's theorem, so it's not constructive. When $d=1$, in the paper of Walter "There is an Elementary Proof of Peano's Existence Theorem" the author showed there is an constructive proof for Peano's theorem. But the method can't transformed to $d>1$. I found some literature say there is no constructive proof for this theorem. 
I am not a logician and really confused about this. Can one give a rigorous meaning to "there is no constructive proof to the Peano's theorem"?

REPLY [11 votes]: I think that the heart of the question is "Can one give a rigorous meaning to 'there is no constructive proof to the Peano's theorem'?"   The answer to this is yes, but the answer is not as simple as one might naively hope. 
The way that one would give an affirmative answer is to first choose a constructive framework. The proof will not show that there is no constructive proof in any possible sense, only that there is no constructive proof within that framework. However, if the framework is sufficiently natural, then the negative answer in that framework will often be viewed as somewhat definitive. 
Once a particular framework has been chosen, if one can show that the theorem implies, within the framework, some principle that is known to be unprovable in the framework, then the theorem itself is certainly unprovable by any means that can be formalized within the framework. One such principle might be the law of the excluded middle, but there are many weaker but still nonconstructive principles that are also used in practice. 
There are several frameworks for constructive mathematics; an old but excellent introduction is Varieties of Constructive Mathematics by Bridges and Richman (1987). 
Two particular frameworks worth mentioning are:

Bishop's framework, which is presented in Bishop and Bridges, Constructive Analysis, 1985. Briefly put, this framework was intended to resemble ordinary informal mathematics (but with special care to avoid nonconstructive methods, and other issues of classical mathematics that Bishop felt obstructed constructive proofs). 
The "Markov-type" framework, developed by many constructivists but particularly associated with Russian constructivism. Briefly put, this framework focuses on algorithms, rather than more abstract mathematical objects. To have an object is to have an algorithm for it. 

It appears that the answer in both of these frameworks is that there is no constructive proof of Peano's theorem. 

Aberth (1971) showed that Peano's theorem fails in a Markov-type setting (full text available from the journal page). Reference: Oliver Aberth, The failure in computable analysis of a classical existence theorem for differential equations, Proc. Amer. Math. Soc. 30 (1971), 151-156. MR 302982
Bridges (2012) showed that, in a Bishop-type setting, Peano's theorem implies the principle LLPO, which is a well-known non-constructive principle used to gauge the non-constructivity of theorems. Reference: Bridges, Douglas S., 
Constructive solutions of ordinary differential equations. Logic, construction, computation, 67–77, Ontos Math. Log., 3, Ontos Verlag, Heusenstamm, 2012.  MR 3204972

It was pointed out in the comments that, in the context of classical Reverse Mathematics, Peano's theorem is known to imply the system $\mathsf{WKL}_0$. That is a clue that the theorem may not be constructive, but there are two issues that prevent it from being definitive. First, the implication proof may require non-constructive methods. Second, the system $\mathsf{WKL}_0$ is tightly tied to Bishop's program for constructive analysis, and some theorems that are equivalent to  $\mathsf{WKL}_0$ in the context of Reverse Mathematics are nonetheless considered constructive by Bishop, due to the difference between the Reverse Mathematics framework and Bishop's framework.  However, the two references above are both written in constructive frameworks, and I would view them as definitive.<|endoftext|>
TITLE: An inequality for eigenvalues of the Dirichlet problem
QUESTION [5 upvotes]: Is either of these inequalities true?
$$\lambda(tA + (1-t)B)\geq t\lambda(A) + (1-t)\lambda(B)$$
or
$$\lambda(tA + (1-t)B)\leq t\lambda(A) + (1-t)\lambda(B),$$
where $0\leq t \leq 1$, $A,B$ are bounded domains in $\mathbb{R}^n$ and $\lambda(\Omega)$ is an eigenvalue of the problem
$$-\Delta\,u=\lambda\,u\,\,\mbox{in}\,\,\, \Omega, \, u=0\,\,\,\mbox{on}\,\,\, \partial\Omega.$$

REPLY [3 votes]: The inequalities you state cannot hold. It is known that $\lambda(tA) = t^{-2}\lambda(A)$ so choosing $A = sB$ you would get
$$ \frac{1}{(ts+1-t)^2}\lambda(B) \leq \geq (t/s^2+1-t)\lambda(B)$$
Neither of the inequalities can hold for any $s$. Take $s=0.5$ and $s\to 0$.
However, a Brunn-Minkowski inequality holds for the $1$-homogeneous function $\lambda^{-1/2}$. In the paper https://core.ac.uk/download/pdf/81213094.pdf and the references therein you can see that the following inequality holds
$$ 
\lambda(tA+(1-t)B)^{-1/2} \geq t \lambda(A)^{-1/2}+(1-t)\lambda(B)^{-1/2}
$$
Therefore, an inequality of the type that you want holds for $\lambda^{-1/2}$. Equality cases are also described in the paper above, for the case when $A$ and $B$ are convex.<|endoftext|>
TITLE: The unpublished papers in reference to the published papers
QUESTION [29 upvotes]: Sometimes it happens that a published paper refers to an unpublished paper for a result used.
In this case, if we want to check this result by ourselves, we need to access to this unpublished paper.  
Question: What is the usual process for accessing to an unpublished paper?
 Is it the duty of the author (citing it in his published paper) to send it to anyone requiring it?
Or is it the duty of the editor, or of the author of the unpublished paper?   
What to do if we can not get it through these means?

REPLY [4 votes]: As the reader, you can also ask mathoverflow for help finding the paper.  There have been some very interesting MO questions along this line.  In particular, this record helps others to also track down the same paper (or discover it never existed in the first place).  Feel free to add to this list since this is CW:

Scott-Solovay unpublished paper on ``Boolean valued models of set theory'' (never existed)
Looking for a copy of Leo Harrington's unpublished notes on the first nonprojectible ordinal (now we have a version online)
Harrington's unpublished note "The constructible reals can be anything" (now we have a version online)<|endoftext|>
TITLE: Elementary proof of a triangular grid lemma
QUESTION [11 upvotes]: I am looking for an elementary proof of the following lemma, which concerns what Green and Tao call "triangular grids" (see arXiv:1208.4714).
Let $a_1$, $a_2$, $a_3$, $a_4$, $b_1$, $b_2$ be six arbitrary lines in the plane (in general position, for whatever the appropriate sense of "general position" is).
Let $c_1$ be the line through the intersection points $a_3b_1$ and $a_2b_2$.
Let $c_2$ be the line through $a_4b_1$ and $a_3b_2$.
Let $b_3$ be the line through $a_1c_1$ and $a_2c_2$.
Let $c_3$ be the line through $a_4b_2$ and $a_3b_3$.
Let $b_4$ be the line through $a_1c_2$ and $a_2c_3$.
Let $c_4$ be the line through $a_4b_3$ and $a_3b_4$.
Let $b_5$ be the line through $a_1c_3$ and $a_2c_4$.
Let $c_5$ be the line through $a_4b_4$ and $a_3b_5$.
Then the intersection points $b_1c_3$, $b_2c_4$, $b_3c_5$ are collinear.
As a consequence, this triangular grid can be indefinitely continued.
See figure below.
This can be proven using cubic curves, specifically the Cayley-Bacharach theorem: In the dual setting, all the points dual to the given lines lie on a common cubic curve. See the above-mentioned paper of Green and Tao.
But as I said, I was wondering whether an elementary proof exists.
I see there is previous literature on these triangular grids (also called with other names; see reference [6] in the above paper, and references cited there). I quickly glanced at the literature, and I didn't find what I'm looking for.

REPLY [9 votes]: We can prove this with a cross ratio chase, but first we need an easy lemma.

Lemma: If points $A,B,C,D$ are on one line and $E,F,G,H$ are on another line, then $(A,B;C,D) = (E,F;G,H)$ if and only if the three points $X = AF\cap BE, Y = BG\cap CF, Z = CH\cap DG$ lie on a line.
Proof: Let $P = AG\cap CE, Q = CG\cap XY$. By Pappus's Theorem, $P$ is on line $XY$. Projecting through $G$, we have $(A,B;C,D) = (P,Y;Q,DG\cap XY)$, and projecting through $C$, we have $(E,F;G,H) = (P,Y;Q,CH\cap XY)$. Thus $(A,B;C,D) = (E,F;G,H)$ if and only if $CH, DG$, and $XY$ meet at a point.

Now we need to name some points. Let $A = a_2\cap b_1, B = a_3\cap b_1\cap c_1, C = a_4\cap b_1\cap c_2, D = b_1\cap c_3, E = b_1\cap c_4$, let $F = a_1\cap b_3\cap c_1, G = a_2\cap b_3\cap c_2, H = a_3\cap b_3\cap c_3, I = a_4\cap b_3\cap c_4, J = b_3\cap c_5$, and let $K = b_5\cap c_2, L = a_1\cap b_5\cap c_3, M = a_2\cap b_5\cap c_4, N = a_3\cap b_5\cap c_5, O = a_4\cap b_5$. Finally, let $X = AN \cap DK$.
By two visually obvious applications of the Lemma we have $(A,B;C,D) = (F,G;H,I) = (K,L;M,N)$. Thus $(B,D;A,C) = (L,N;K,M)$, so by a slightly less obvious application of the Lemma $X$ must be on the line $GH = b_3$. Applying the Lemma again, since $X$ is on the line $GI = b_3$ we have $(E,C;A,D) = (O,M;K,N)$.
Putting $(A,B;C,D) = (K,L;M,N)$ and $(E,C;A,D) = (O,M;K,N)$ together, we see that $(B,C;D,E) = (L,M;N,O)$ (consider the projective transformation from line $b_1$ to line $b_5$ taking $A,C,D$ to $K,M,N$: the first equality says it takes $B$ to $L$, and the second says it takes $E$ to $O$). Applying the Lemma in another visually obvious way, we have $(L,M;N,O) = (G,H;I,J)$. Thus $(B,C;D,E) = (G,H;I,J)$, so by the Lemma the intersection $DJ\cap EI$ is on the line connecting $BH\cap CG, CI\cap DH$, which is $b_2$. Since $D = b_1\cap c_3$, $J = b_3\cap c_5$, $EI = c_4$, this means that $b_1\cap c_3, b_2\cap c_4, b_3\cap c_5$ are collinear.<|endoftext|>
TITLE: Lower density of {primes} times themselves
QUESTION [6 upvotes]: We say that a set $A\subseteq \mathbb{N}$ has lower density 0 if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 0.$$
Given $A,B\subseteq \mathbb{N}$ we set $A\cdot B = \{a\cdot b: a\in A, b\in B\}$. The set $A^n$ for $n\in\mathbb{N}$ is defined inductively in the obvious manner.
Let $P$ be the set of prime numbers in $\mathbb{N}$. What is the smallest positive $m_0\in\mathbb{N}$ such that the lower density of $P^{m_0}$ is $>0$? And what is the lower density of $P^{m_0}$?

REPLY [2 votes]: The following results are from the paper Ivan Niven, The asymptotic density of sequences. Bull. Amer. Math. Soc., 57(6):420-434, 1951. I changed the notation to denote upper asymptotic density by $\overline d(A)$ and asymptotic density by $d(A)$. (Which seems to be used more frequently nowadays than the notation $\delta_2(A)$ and $\delta(A)$ from that paper.)

For any prime $p$ let $A_p$ denote the subset $A_p=\{n\in A; p\mid n, p^2\nmid n\}$.
Theorem 1. If $\{p_i\}$ is a set of primes such that $\sum\frac1{p_i}=\infty$, then $\overline d(A)\le \sum\overline d(A_{p_i})$ for any $A$.
Corollary 1. If a set of primes $\{p_i\}$ we have $d(A_{p_i})=0$ for every $i$, and if $\sum p_i^{-1}=\infty$ then $d(A)=0$.
Corollary 2. For any fixed $k$, if $\{p_i\}$ is a set of primes such that $\sum\frac1{p_i}=\infty$, and if $A$ is any set whose members are divisible by at most $k$ of these primes to the first degree, then $d(A)=0$.

From Corollary 2 it follows that $d(P^{m_0})=0$ for each $m_0$. (In the other words, Corollary 2 implies that the set of all numbers having at most $m_0$ prime factors has density zero. Niven also mentions in connection with this weaker result the paper Willy Feller, Erhard Tornier: Mengentheoretische Untersuchung von Eigenschaften der Zahlenreihe, Mathematische Annalen 1933, Volume 107, Issue 1, pp 188-232.)<|endoftext|>
TITLE: What are algebras for the little n-balls/n-cubes/n-something operads exactly?
QUESTION [9 upvotes]: As a non expert in the theory of topological operads, I find it pretty hard, to understand what algebras for little balls/cubes/something operads are.
For all the other famous operads I know (like Lie, Com, Ass ect.) associated algebras are meanwhile given in terms of generators and relations, which makes them understandable without any reference to operads. However this seams to be different for the little something operads.
I know that their homology is sometimes a Poisson-n algebra for $n\geq 2$, therefore algebras for the homology are (homotopy) Poisson-n algebras in that case. But that's just the homology.
Then I heard, that in a category those algebras are just commutative algebras, 
as long as $n\geq 2$. Is that necessarily true?
So my questions are:
1.) Are there generator&relation style descriptions of the algebras for the little n-balls/n-cubes or related operads? 
2.) Do I need a higher category to get examples that are not just commutative for $n\geq 2$?
3.) Is there a standard reference for those algebras? (Not just the operads)
Edit:
4.) How could I start to get a good understanding of those algebras? Assuming I know just the basic about the underlying operads.

REPLY [12 votes]: An $E_n$ algebra (an algebra over the little $n$-cubes operad, etc.) is intuitively an object with $n$ compatible monoid structures. All of the subtlety in this theory lies in making "compatible" precise; in particular it is not a property but a structure. 
Here are some examples.

In $\text{Set}$, an $E_1$ algebra is a monoid. For $n \ge 2$ an $E_n$ algebra is a commutative monoid by the Eckmann-Hilton argument.
In $\text{Ab}$, an $E_1$ algebra is a ring. For $n \ge 2$ an $E_n$ algebra is a commutative ring by the Eckmann-Hilton argument.
In $\text{Top}$, a grouplike $E_n$ algebra (an $E_n$ algebra where $\pi_0$ is a group) is an $n$-fold loop space $\Omega^n X$ by the recognition principle. The $n$ monoid structures are the loop compositions in the $n$ loop directions. A grouplike $E_{\infty}$ algebra is an infinite loop space, or equivalently a connective spectrum. 
In $\text{Cat}$, an $E_1$ algebra is a monoidal category, an $E_2$ algebra is a braided monoidal category, and for $n \ge 3$ an $E_n$ algebra is a symmetric monoidal category. (This stabilization phenomenon is related to Freudenthal suspension and is part of the "periodic table of higher categories.")

In more detail, let's focus on $n = 2$. An $E_2$ algebra is intuitively an object with two compatible monoid structures. The Eckmann-Hilton argument shows that they're equivalent, but the way in which it shows that they're equivalent is itself interesting structure: along the way, it describes a map between $ab$ and $ba$ (for both monoidal structures), which is a braiding. This is where braided monoidal categories come into the picture. It might help to stare at the standard proof of Eckmann-Hilton where you move squares around and to explicitly think of the squares as describing binary operations in the little $2$-cubes operad. 
One way to make "compatible" precise is to write down a presentation of the $E_2$ operad, which is unique among the $E_n$ operads ($n \ge 2$) in that all of its spaces are $1$-truncated: that is, they are all groupoids, or equivalently have no higher homotopy $\pi_n, n \ge 2$. There is a particularly nice model of the $E_2$ operad as an operad in groupoids called the parenthesized braid operad, which has a "generators-and-relations" presentation, but where it's important to understand that when specifying an operad in groupoids there are three sorts of things you might want to write down, rather than two: 

Operations (which then generate other operations under operadic composition),
$1$-morphisms between operations (to describe the groupoid structure), and
Relations between $1$-morphisms between operations (to further describe the groupoid structure).

In general, higher category theory blurs the distinction between generators and relations: relations become generators one categorical level up. To avoid having to make the distinction you can just say "presentation."
The standard presentation of the parenthesized braid operad mimics exactly the standard axiomatization of braided monoidal categories: there is a generating binary operation (the monoidal structure), two generating $1$-morphisms (the associator and the braiding), and some relations between these (the pentagon and hexagon axioms). Here I'm ignoring units for simplicity. This presentation is important in discussions of Grothendieck-Teichmüller theory.
For $n \ge 3$ the spaces in the $E_n$ operads aren't truncated: for example the space of binary operations is the sphere $S^{n-1}$, which has nontrivial homotopy groups in arbitrarily high degrees. So I don't think there's any hope for a presentation along the above lines in general. You could ask for presentations of various truncations, but I imagine these are pretty horrible to work with in general. The $E_n$ operads exist precisely so that you can avoid having to do stuff like this. Of course it's a different story after taking rational homology. 
$E_n$ algebras show up in the story of factorization homology and topological field theory, so that's one place to go for some resources; see, for example, these notes by Ginot.<|endoftext|>
TITLE: Conjectured equivalent conditions on certain power-series
QUESTION [8 upvotes]: Let $P(x)=1+a_1x+a_2x^2+a_3x^3+...$ be a series such that every $a_i$ is an integer, $a_1<0$, and $a_i\ge 0$ for every $i\ge 2$. Are the following statements equivalent ?  

$P(y)=0$ for some $y>0$.  
Every coefficient of the series expansion of $\frac 1{P(x)}$ is positive.

REPLY [8 votes]: Yes. One direction is quite clear, the other one is carefully written in 
D. I. Piotkovskii, "On the growth of graded algebras with a small number of defining relations", Uspekhi Mat. Nauk, 48:3(291) (1993), 199–200.
(It is also mentioned without proof in Lemma 5.3 of
D. Anick, "Generic algebras and CW complexes", Algebraic Topology and Algebraic K-Theory (Proc. Conf. Princeton, NJ (USA)), Ann. Math. Stud., vol. 113 (1987), pp. 247–321)<|endoftext|>
TITLE: Geodesics on convex hypersufaces
QUESTION [6 upvotes]: Let $M^n$ be the boundary of a convex compact set in $\mathbb{R}^{n+1}$ with non-empty interior.
Question 1. Is $M$ geodesically complete, i.e. is it true that every geodesic (= locally shortest path) can be extended infinitely in both directions? Will it still be true if the convex set is not necessarily compact, but only closed.
Question 2. Is it true that every shortest path on $M$ has both left and right first derivatives at every point? 
UPDATE: The answer to Question 1 is NO, as explained by Igor Rivin below.
UPDATE: The answer to Question 2 is YES, as claimed by John Harvey below. Originally it was proved by I.M. Liberman in "Geodesic lines on convex surfaces", C. R. (Doklady) Acad. Sci. URSS (N.S.) 32, (1941). 310–313. 
The proof  for 2-dimensional hypersurfaces is also reproduced in the book "Intrinsic geometry of convex surfaces" by A.D. Alexandrov, see Ch. IV, $\S$ 6, Theorem 1.

REPLY [4 votes]: As for question 2, the tangent cone of $M$ can be defined as the cone on the space of directions, and the space of directions is the completion of the space of geodesic directions. Therefore every geodesic is represented in the cone.
On the other hand, the cone could be defined as a limit object by rescaling $M$ around the point, and this is the definition which we use when talking about derivatives.
$M$ is an Alexandrov space, and so by Theorem 7.8.1 of the Burago--Gromov--Perelman paper, these two definitions coincide. So this theorem shows that shortest paths have one-sided derivatives.
Probably this was already known for convex hypersurfaces without having to use the full generality of Alexandrov geometry, but I'm not sure where you would go to find that.<|endoftext|>
TITLE: Is this differential identity known?
QUESTION [64 upvotes]: Recently I discovered the differential identity
$$ \frac{d^{k+1}}{dx^{k+1}} (1+x^2)^{k/2} = \frac{(1 \times 3 \times \dots \times k)^2}{(1+x^2)^{(k+2)/2}}$$
valid for any odd natural number $k$; for instance $\frac{d^6}{dx^6} (1+x^2)^{5/2} = \frac{225}{(1+x^2)^{7/2}}$.  This identity was surprising at first, since usually the repeated application of the product rule and chain rule leads to far messier expressions than this, but there are now several proofs that adequately explain this identity (collected at this blog post of mine).  There is also the more general identity
$$ |\frac{d}{dx}|^{2s-1} (1+x^2)^{s-1} = \frac{2^{2s-1}\Gamma(s)}{\Gamma(1-s)} (1+x^2)^{-s}$$
valid for any complex $s$ (if everything is interpreted distributionally), which is related to the isomorphisms between principal series representations of $PGL_2({\bf R})$.
The purpose of my question here is not to ask for more proofs of this identity (but you are welcome to visit the above-mentioned blog post to contribute another proof, if you wish).  Instead, I am asking as to whether this identity (or something close to it) already appears in the literature - I find it hard to believe that such a simple identity has been missed for centuries, given that it could easily have been discovered and proven by (say) Euler.  The closest match that I know of so far are the Rodrigues formulae for the classical orthogonal polynomials, but I was not quite able to place the above identity as a special case of these formulae (the exponents don't quite match up).

REPLY [22 votes]: Here is a direct induction argument. Let $h_k(x):=(1+x^2)^{k/2}$, $g(x):=x$,
$p_k:=(1\times3\times\dots\times k)^2$. The identity in question is 
$$D^{k+1}h_k=p_k/h_{k+2}$$
for positive odd $k$, where $D$ is the differentiation operator. It is easy to check it for $k=1$. 
Then for odd $k\ge3$ 
$$D^{k+1}h_k=k D^k(gh_{k-2})=kg D^k h_{k-2}+k^2 D^{k-1}h_{k-2}$$
$$=kg D(p_{k-2}/h_k)+k^2 p_{k-2}/h_k=p_k/h_{k+2},$$
as desired; for the third equality in the above display we use the induction, and for the second we use the Leibniz rule.<|endoftext|>
TITLE: Zinn's "doubling" conjecture on weighted sums of independent Rademacher random variables
QUESTION [19 upvotes]: Let $a_1,\dots,a_n$ be real numbers such that $a_1^2+\dots+a_n^2=1$. Let $\eta_1,\dots,\eta_n$ be independent Rademacher random variables (r.v.'s), so that $P(\eta_i=\pm1)=\frac12$ for all $i$. Let $S:=a_1\eta_1+\dots+a_n\eta_n$, and let $T$ be an independent copy of $S$. 
Does then the inequality 
$$(*)\qquad E f_p(S)\le E f_p\Big(\frac{S+T}{\sqrt2}\Big)$$
hold for all real $p\ge2$, where $f_p(x):=|x|^p$?
This conjecture was communicated to me by Joel Zinn quite some time ago, and I think it deserves to be more broadly known. A motivation behind it was to obtain an alternative (and hopefully easier) proof of Haagerup's inequality 
$$E f_p(S)\le E f_p(Z)$$
Haagerup, which indeed easily follows from $(*)$ by the central limit theorem; here $Z$ is a standard normal r.v. For $p=2$, $(*)$ is trivial. For $p\ge3$, $(*)$ is easily proved; in particular, it follows immediately from Corollary 2.5 in T^2. 
So, actually the question is only about $p\in(2,3)$. In that case, in view of Lemma 2.2 in Figiel et al, it would be enough to prove $(*)$ with $g_t$ in place of $f_p$ (for all real $t>0$), where $g_t(x):=|x|^3-\max(0,|x|-t)^3$.

REPLY [3 votes]: Alas, the approach through the functions $g_t$ cannot possibly work. That is seen if one takes e.g. $n=2$, $a_1=4/5$, and $a_2=t=3/5$. 
Based on numerical evidence, the following approach promises to work. Without loss of generality (wlog) $n\ge2$ and $a_1\ge\dots\ge a_n\ge0$. Let $a:=a_1$, $b:=a_2$, and $Y:=|S-a_1\eta_1-a_2\eta_2|$. Then 
$$(1)\quad E Y^2=1-a^2-b^2$$ and 
$$(2)\quad 
EY^4=3\Big(\sum_3^n a_i^2\Big)^2-2\sum_3^n a_i^4\ge3(1-a^2-b^2)^2-2(1-a^2-b^2)\times\min(b^2,1-a^2-b^2),
$$
since $\max_3^n a_i^2\le\min(b^2,1-a^2-b^2)$. By induction, it is enough to show that 
$$ (3)\quad E|S_2+Y|^p\le E\Big|\frac{S_2+T_2}{\sqrt2}+Y\Big|^p$$
for all $p\in(2,3)$ and nonnegative r.v.'s $Y$ subject to conditions (1) and (2), where $S_2:= 
\eta_1 a+\eta_2 b$ and $T_2$ is an independent copy of $S_2$. At that, by well-known results  (see e.g. Hoeffding55), wlog the r.v. $Y$ takes at most $3$ values (say $0\le u\le v\le w$ with probabilities $r,s,1-r-s$). 

Thus, the problem is reduced to a calculus problem on proving (say) the nonnegativity of a function of $8$ variables $p,a,b,u,v,w,r,s$ subject to a finite number of restrictions on the values of these variables. So, in principle this problem is solvable, but seems very involved computationally.
Addendum: Unfortunately, other numerical evidence shows that conditions (1) and (2) on $Y$ are not enough for (3) to hold in general. For instance, if $a = b = 11/21$, 
$p = 93/46$, $u = 0$, $v = 11/95$, $w = 71/61$, and $r\approx0.642$ and $s\approx0.025$ are such that $E Y^2=1-a^2-b^2$ and $EY^4=3(1-a^2-b^2)^2$, then the difference between the right-hand and left-hand sides of (3) is $\approx-0.000163$.<|endoftext|>
TITLE: Is $\clubsuit_{\omega_1}$ enough to get Suslin tree?
QUESTION [10 upvotes]: This is problem 15.3 in Arnie Miller's problem list: 
(Juhasz) Suppose there exists $\langle A_{\alpha} : \alpha \in L \rangle$, where $L$ is the set of limit ordinals below $\omega_1$ and for each $\alpha \in L$, $A_{\alpha}$ is an unbounded subset of $\alpha$, satisfying: For every unbounded $A \subseteq \omega_1$, there exists $\alpha \in L$, $A_{\alpha} \subseteq A$. Must there exists a Suslin tree?
What is the current status of this problem? Thanks!

REPLY [10 votes]: The answer is negative apparently. It is consistent relative to ZFC that all Aronszajn trees are
special and that the club principle holds: 
http://home.mathematik.uni-freiburg.de/mildenberger/postings/paperspdf/988_2014_10_15no.pdf<|endoftext|>
TITLE: A cosmos where coproduct injections are not monic
QUESTION [11 upvotes]: The injections (coprojections) of a coproduct in a category are very often monomorphisms.  For instance, this happens in any extensive category (essentially by definition) and also in any category with zero morphisms (since in that case they are split monos).  However, there are examples of (complete and cocomplete) categories where this is not always the case, e.g. in the category of commutative rings the coproduct is the tensor product, and the injection $\mathbb{Z} \to \mathbb{Z} \otimes \mathbb{Z}/2 \cong \mathbb{Z}/2$ is not monic.
My question is, can you give an example of a complete and cocomplete closed monoidal category (a "Benabou cosmos") in which coproduct injections are not always monic?  (Or, I suppose, a proof that no such example exists, but that would surprise me.)

REPLY [7 votes]: Let me recall that injectivity of coproduct's injections follows from the distributivity of products over coproducts (rather than full extensivity of coproducts). Since every cartesian closed category is (obviously) distributive we cannot find counterexamples in cartesian closed categories.
However, the proof of injectivity of coproduct's injections highly relies on the cartesian structure of $\times$, and one should not expect to carry it to the context where product $\times$ is substituted by a general tensor $\otimes$.
One class of categories which cannot be cartesian closed (unless degenerated) are self-dual categories. I claim that very many of such categories do not have injective coproduct's injections, and this fact is (almost) unrelated to the existence of any closed monoidal structure. Here is an explicit example.
There are various notions of Chu spaces, but the underlying idea is common --- a Chu space is thought of as a "non-standard relation", and morphisms of Chu spaces are thought of as "adjoint pairs" between relations. Let me describe the category that is usually denoted by $\mathit{Chu}(\mathbf{Set}, \Omega)$. Its objects consist of typed binary relations: $$A = \langle A_!, A^*, A_! \times A^* \overset{A}\rightarrow \Omega \rangle$$ in $\mathbf{Set}$, and its morphisms $A \rightarrow B$ consist of pairs of functions (notice opposite directions!): $$f = \langle f_! \colon A_! \rightarrow B_!,  f^* \colon B^* \rightarrow A^* \rangle$$ in $\mathbf{Set}$ that satisfy the following adjoint-like condition:
$$B(b, f^*(a)) = A(f_!(b), a)$$
One may easily check that this category is self-dual, where the dualization swaps the domain with the codomain of a relation:
$$(A^\bot)_! = A^*$$
$$(A^\bot)^* = A_!$$
$$(A^\bot)(b, a) = A(a, b)$$
and complete (thus, also cocomplete) --- limits are constructed point-wise: the first component of a Chu space inherits limits from $\mathbf{Set}$, and the second from $\mathbf{Set}^{op}$.
Like in many self-dual categories, objects may have many (co)global coelements --- i.e. there are non-trivial morphisms to the initial object $0 = \langle \emptyset, \{ {*} \}, \emptyset \rangle$ (just pick any Chu space $A$ whose $A_! = \emptyset$ and whose $A^*$ is non-trivial). Therefore, the unique morphism $0 \rightarrow 1$ is not mono. So the canonical coproduct's injection $0 \rightarrow 0 \sqcup 1 \approx 1$ is not a monomorphism in $\mathit{Chu}(\mathbf{Set}, \Omega)$.
There is a closed monoidal structure on $\mathit{Chu}(\mathbf{Set}, \Omega)$ given by the following tensor:
$$(A \otimes B)_! = A_! \times B_!$$
$$(A \otimes B)^* = \{\langle h \colon A_! \rightarrow B^*, k \colon B_! \rightarrow A^* \rangle \colon B(b, h(a)) = A(a, k(b))\}$$
$$(A \otimes B)(a, b, h, k) = B(b, h(a)) = A(a, k(b))$$
In fact, $\mathit{Chu}(\mathbf{Set}, \Omega)$ is $\star$-autonomous (with linear implication $A \multimap B = (A \otimes B^\bot)^\bot$).
In particular, the construction is valid for any $\Omega$ (not necessarily the subobject classifier), and taking $\Omega = 1$ yields category $\mathbf{Set} \times \mathbf{Set}^{op}$ with linear exponentiation:
$$\langle A_!, A^*\rangle \multimap \langle B_!, B^*\rangle = \langle {B_!}^{A_!} \times {A^*}^{B^*}, A_! \times B^*\rangle$$<|endoftext|>
TITLE: Is there a motivic Cauchy integral formula?
QUESTION [22 upvotes]: Let $R$ be a complete dvr with fraction field $K$ and residue field $k$, and let $X, Y$ be two smooth projective $R$-schemes with isomorphic generic fibers.

Is it true that $[X_k]=[Y_k]$ in $K_0(\text{Var}_k)$?

Recall that $K_0(\text{Var}_k)$ is the so-called Grothendieck ring of varieties, namely, the free Abelian group on $k$-varieties, modulo the relation $[X]=[Y]+[X\setminus Y]$ for $Y\subset X$ closed.
That's the short version; let me say why I would call this a Cauchy integral formula and say what I know.  
First of all, why is this a Cauchy integral formula?  In this analogy, I'm thinking of a morphism $X\to Y$ as a function $Y(T)\to K_0(\text{Var}_T)$, sending a point $y$ to $X_y$.  You're supposed to think of $[X_K]$ as a function $\text{Spec}(K)$ (a puntured disk), and the question asks if we can recover the value at the central point $\text{Spec}(k)$ from this data.  The complex-analytic analogue is exactly the Cauchy integral formula.
Let me give a slightly non-trival example.  A trivial family of Hirzebruch surfaces $X_n$ may degenerate to either $X_n$ or $X_{n+2}$.  But of course $[X_n]=[X_{n+2}]=(\mathbb{L}+1)^2$.
Some other examples:  if $X_K, Y_K$ are isomorphic as polarized varieties and one of them is not ruled, all is good by Matsusaka-Mumford.  In particular, if $X, Y$ are canonically polarized, the answer to my question is affirmative.
Likewise suppose $X_K, Y_K$ have trivial canonical bundle.  Then $X_k, Y_k$ have trivial canonical bundle as well (at least if $K$ has characteristic zero) and are birationally equivalent, by spreading out the isomorphism $X_K\simeq Y_K$.  But birational Calabi-Yau's have the same class in (some mild localization of) $K_0(\text{Var}_k)$, by Kontsevich's motivic integration results.
A slightly better version of the question is:

Let $X, Y$ be smooth projective $R$-schemes with $[X_K]=[Y_K]$.  Does this imply $[X_k]=[Y_k]$?

This would give some kind of specialization map $K_0(\text{Var}_K)\to K_0(\text{Var}_k)$, at least if $\text{char}(K)=0$.
ADDED:  The result is also true for curves and surfaces.  For curves this is easy; for surfaces, observe that $X_k, Y_k$ are birationally equivalent.  Thus there is a surface $Z$ obtained by blowing up $X_k, Y_k$ at some number of points, so $$[X_k]+n\mathbb{L}=[Z]=[Y_k]+m\mathbb{L}.$$ But $n=m$ because $X_k, Y_k$ have equal Euler characteristic, so $[X_k]=[Y_k]$.

REPLY [9 votes]: I believe the answer is yes in the residue characteristic zero case. I will show it for a mild localisation of the Grothendieck ring (inverting $[L]$ and $[L]-1$). This may also follow from work of Denef and Loeser on the motivic nearby finer.
The real meat of the argument will be the weak factorisation theorem. 
First, by standard algebraization arguments + removal of singularities, we can reduce to the following problem:
Let $X$ and $Y$ be smooth varieties both mapping to a curve $C$, with the fibrations isomorphic away from a point $x$ in $C$, and both smooth over a neighbourhood of $x$. Show that $[X_x]=[Y_x]$.
To solve this, consider the involution of $K^0(Var)[L^{-1}]$ that sends $[X]$ to $[X] /[L]^{d}$ for $[X]$ a smooth projective variety of dimension $d$. We can check that this is well-defined using the blow-up relations of Franziska Bittner/Heinloth: For $Y \to X$ a blow-up with centre $Z$ of codimension $c$ and exceptional divisor $Z$, $$[X] -[L]^c [Z]= [Y]- [L][E]$$ follows from $$[X]-[Z]=[Y]-[E]$$ and $$([L]^c-1)[Z]) = ([L]-1)[Y],$$ both obvious.
Now applying the involution to the known relation $$[X]-[X_x]=[Y]-[Y_x],$$ we obtain $$[X]-[L][X_x]=[Y]-[L][Y_x].$$ Subtracting, we obtain $$ \left([L]-1\right) X_x = \left([L]-1\right) [Y_x]$$ and we win by dividing by $[L]-1$.
I claim the division by $[L]$ and $[L]-1$ in this argument may be removed if desired, to obtain an identity in the unlocalised Grothendieck ring. When working with varieties of dimension $\leq d$, rather than dividing $[X]$ by $[L]^{\dim X}$ we may multiply by $[L]^{d-\dim X}$ to achieve the same effect. This removes the need to adjoin $[L]^{-1}$. Furthermore, rather than working with the involution directly, observe that it only changes a class by a multiple of $[L]-1$ and work with this change divide by $[L]-1$, which is well-defined because we can remove an extra factor of $[L]-1$ from our argument for the relation. This removes the need to adjoin $([L]-1)^{-1}$.
One cool thing about this argument is it tells you something also in the case where $X_x$ is singular - something like "the integral over $X_x$ of the motivic nearby fiber of Denef and Loeser depends only on the $X_\eta$".<|endoftext|>
TITLE: Theorems proved using combinatorial nullstellensatz that have no other known proof
QUESTION [14 upvotes]: Alon's (or Alon and Tarsi's?) combinatorial nullstellensatz is a powerful algebraic tool with many applications in combinatorics and number theory. See this, this, this and this mathoverflow question. 
I am looking for good examples of results that were proved using combinatorial nullstellensatz (or its generalisation) but have no other known proof.

REPLY [5 votes]: Here is an example of a nice problem for which no combinatorial proof seems to be known. It is related to a problem studied in this paper https://arxiv.org/abs/1612.08698
Let $G$ be a $d$-degenerate graph (i.e. each subgraph contains a vertex of degree at most $d$). A classical result in graph theory is that if each vertex is given a list of $d+1$ colors, then every vertex can choose a color from its list such that the resulting coloring is proper.
Now, assume instead that each vertex of $G$ is given a list of size $d+1$ colors, except one vertex which has a list of size $d$. It can be proved with the combinatorial nullstellensatz that in this case again, each vertex can be colored with a color from its list, such that the resulting coloring of $G$ is proper (see the paper above, where a stronger result is proved when $d+1$ is prime, but the proof actually works in the weaker setting for general $d$).
I have worked on finding a combinatorial proof of this result, and I know several other researchers in the area who have studied this problem, without success.<|endoftext|>
TITLE: n-th root of unity in n-th division field of abelian variety?
QUESTION [5 upvotes]: Let $K$ be a number field and $A/K$ an abelian variety over it. 

Can it be that $K(A[n])$ does not contain a primitive $n$-th rooth of unity? 
If the answer is yes is it always possible to bound the bad $n$ uniformly in the degree of the number field and the dimension of $A$? 

Both references and arguments are well appreciated! Thanks in advance

REPLY [16 votes]: Actually, this is an exercise in Serre's Lectures on Mordell--Weil Theorem:
$K(A[n])$ always contains $\mu_n$ if $char(K)$ does not divide $n$ and $A$ is an abelian variety of positive dimension over $K$. (You don't need to assume that $K$ is a number field) 
Here is a solution. First, it suffices to check the case when $n=\ell^m$ is a power of a prime $\ell$.
Second, if $A^t$ is the dual of $A$ then let us take a $K$-polarization $\lambda: A \to A^{t}$ of smallest possible degree.  Then $\lambda$  is  not divisible by $\ell$, i.e., $\ker(\lambda)$ does not contain the whole $A[\ell]$. (Otherwise,  dividing $\lambda$ by $\ell$ we get a $K$-polarization of lesser degree.) 
Then the image $\lambda(A[\ell^m])\subset A^t[\ell^m]$ contains a point of exact order $\ell^m$, say $Q$. Otherwise, 
$$\lambda(A[\ell^m])\subset A^t[\ell^{m-1}]$$
 and therefore $A[\ell]=\ell^{m-1}A[\ell^m]$ lies in the kernel of $\lambda$, which is not the case. 
Since $A[\ell^m]\subset A[K]$ and $\lambda$ is defined over $K$, the image $\lambda(A[\ell^m])$ lies in $A^t(K)$. In particular, $Q$ is a $K$-rational point on $A^t$.
Third, there is a nondegenerate Galois-equivariant Weil pairing
$$e_n: A[\ell^m] \times A^t[\ell^m] \to \mu_{\ell^m}.$$
I claim that there is a point $P \in A[\ell^m]$ such that $e_n(P,Q)$ is a primitive $\ell^m$th root of unity. Indeed, otherwise
$$e_n(A[\ell^m],Q) \subset \mu_{\ell^{m-1}}$$
and therefore nonzero $\ell^{m-1}Q$ is orthogonal to the whole $A[\ell^m]$ with respect to $e_n$, which contradicts the nondegeneracy of $e_n$.
So,
$$\gamma:=e_n(P,Q)$$
is a primitive $\ell^m$th root of unity that lies in $K$, because both $P$ and $Q$ are $K$-points. Since cyclic $\mu_{\ell^m}$ is generated by $\gamma$,
$$\mu_{\ell^m}\subset K.$$<|endoftext|>
TITLE: First eigenvalue of the Laplacian on a regular polygon
QUESTION [12 upvotes]: Consider the Laplacian eigenvalue problem $-\Delta u = \lambda u$ on $\Omega$ with Dirichlet boundary conditions. Let $\lambda_1$ denote the first eigenvalue. The following theorem is well known: 
(Faber-Krahn) Let $c$ be a positive number and $B$ the ball of volume $c$. Then
$$\lambda_1(B) = \min\{\lambda_1(\Omega), \Omega\ \text{open subset of}\ \mathbb{R}^n, 
|\Omega| = c\}.$$
I am considering the question of minimizing $\lambda_1$ in the class of polygons with a given number $N$ as sides. If we denote by $\mathcal{P}_N$ the class of plane polygons with at most $N$ edges, then it is known that the problem
$$\min\{\lambda_1(\Omega), \Omega \in \mathcal{P}_N, |\Omega| = a\}$$
has a solution. This one has exactly $N$ edges.
For the case $N=3$, it has been proven the equilateral triangle minimizes $\lambda_1$. For $N=4$, the square minimizes $\lambda_1$. (Both proof uses the properties of the Steiner symmetrization of $\Omega$. The original proof was due to Pólya. Unfortuantely I could not find the original paper, but the proof can also be found in Extremum problems for Eigenvalues of Elliptic Operators by Henrot.)
Question: Is there a general result that the regular $N$-gon have the least first eigenvalue among all the $N$-gons of given area for $N \geq 5$?

REPLY [4 votes]: So few people are working on this problem. I have noticed that results are scattered far and wide. So, I am calculating and compiling data related to the eigenvalues of the Laplacian within polygons. For example, below is a list of the principal Dirichlet eigenvalues within regular polygons (with area Pi, not inscribe in a unit-radius circle), all correctly rounded to 27 decimal places. I actually bounded them to within a relative error of at most 1E-30; and the pentagon, I have to 1E-500. But, below is a good list. The first two are known in closed form, and the last entry is the square of the first root of the Bessel function J_0(x)=0. This last one is the number the sequence is approaching. One can use the formula L = j01^2*(1+4*zeta(3)/N^3+ O(1/N^5)) to estimate the eigenvalue, with improving results as N (the number of polygon sides) increases. I do not use that formula, and I challenge anyone to figure out the non-zero fifth order term. (Incidentally, each eigenvalue shown from 127 sides and up takes about a day of CPU time. I'm doing the 256 sided polygon now, but I can't get thirty digits. I'm at only about 20 digits now, 
5.7831876203689428759... where the trailing digits are yet uncertain.) -Bob Jones
   3    7.255197456936871402376313031 <--- 4*Pi/sqrt(3)
   4    6.283185307179586476925286767 <--- 2*Pi
   5    6.022137932042633878298008710
   6    5.917417831613661215688574577
   7    5.866449312655985857712474942
   8    5.838491433592442850516640380
   9    5.821826802270265731735546444
  10    5.811260359219116022788816469
  11    5.804230636717400721878394453
  12    5.799369804356500079315025311
  13    5.795900266856014709790771063
  14    5.793357005271194553273227079
  15    5.791450010651579975693848498
  16    5.789991899990208534349752214
  17    5.788857871981104698617196635
  18    5.787962591857846864212568380
  19    5.787246351381961243008036645
  20    5.786666514140372213530912962
  21    5.786192077596844273028203757
  22    5.785800129428365027574586044
  23    5.785473486454901632048264070
  24    5.785199089790024091834463613
  25    5.784966894130423501418670684
  26    5.784769086314842977992274718
  27    5.784599527236484640593222827
  28    5.784453347751719951196794842
  29    5.784326652365411207380293386
  30    5.784216299392264044119036734
  31    5.784119736080032703344528385
  32    5.784034873702444318330507487
  33    5.783959992040508812335032523
  34    5.783893665694809252033476569
  35    5.783834706770988202840700005
  36    5.783782119955880627699919966
  37    5.783735067049846291962440637
  38    5.783692838773292267706517922
  39    5.783654832210871143386207911
  40    5.783620532655973576951368559
  41    5.783589498912728857243541088
  42    5.783561351331960679963744950
  43    5.783535762021971971277446762
  44    5.783512446799268033474803358
  45    5.783491158538856302913858879
  46    5.783471681656168110105137225
  47    5.783453827508463297379645914
  48    5.783437430546865419250636065
  49    5.783422345083940177286945517
  50    5.783408442568212994780116196
  51    5.783395609277903442166398140
  52    5.783383744362702700019559015
  53    5.783372758175598244048049661
  54    5.783362570847292742897314144
  55    5.783353111064236385644810842
  56    5.783344315018129354638447816
  57    5.783336125500292103090369355
  58    5.783328491118809153913672275
  59    5.783321365620033853737939611
  60    5.783314707299059428210797800
  61    5.783308478486244250571058736
  62    5.783302645098928410170380598
  63    5.783297176249175699422050587
  64    5.783292043899785027461125829
  65    5.783287222561990216101379319
  66    5.783282689029249211147849046
  67    5.783278422142346980297970170
  68    5.783274402581728387648667337
  69    5.783270612683560625466506793
  70    5.783267036276517720104953505
  71    5.783263658536697267505357057
  72    5.783260465858434270390794564
  73    5.783257445739078946040870361
  74    5.783254586676063084321584559
  75    5.783251878074799951907284413
  76    5.783249310166151671634629193
  77    5.783246873932360298530618523
  78    5.783244561040478512305563119
  79    5.783242363782456338656120879
  80    5.783240275021144443460263290
  81    5.783238288141564708788818914
  82    5.783236397006877016631166830
  83    5.783234595918539141935169159
  84    5.783232879580215836619574007
  85    5.783231243065044797761869160
  86    5.783229681785912299987294190
  87    5.783228191468430723800997531
 127    5.783199538123680412174552014
 128    5.783199222432098956985238320
 129    5.783198916453726829015452454
 130    5.783198619817847494322697718
        5.783185962946784521175995758 <-- j_{0,1}^2 (J_0(j_{0,1})=0)<|endoftext|>
TITLE: Persistence barcodes and spectral sequences
QUESTION [29 upvotes]: Persistent homology is a well-developed tool which allows topological analysis of large data sets. From a topological perspective, the input is a filtered complex, and the output is a sequence of collections of intervals (one for each dimension) called a persistence barcode. The barcode gives information about homology classes which are born and die as you vary the scale (filtration parameter).
This is a very brief, non-expert summary. By now there are several good survey articles on the subject by experts in the field, for example by Gunnar Carlsson and Rob Ghrist.
On the other hand, given a filtered complex $X_\bullet$, one obtains a spectral sequence converging to the homology $H_\ast(X_\bullet)$ (or at least its associated graded object) . It is natural to ask how the persistence barcode relates to this spectral sequence. In a formative paper on the subject by Carlsson and Zomorodian, the authors ask exactly this question in section 1.4 of the introduction, claiming that a persistence interval of length $r$ in the barcode corresponds to a differential $d_{r+1}$. Thus, in principle, any algorithm for computing persistent homology should give an algorithmic way of computing the differentials in a spectral sequence. So persistent homology, which already has many applications outside of topology, becomes potentially applicable to topology itself.

Has anyone ever pursued this approach, and used algorithms for persistent homology to compute the differentials in a spectral sequence? Does this lead to any new theoretical insights?

I am imagining that by knowing the values of a differential in a given situation one might guess at a description of the differential (eg, in terms of cohomology operations) which applies more generally. 
Edit: The book Computational Topology: An Introduction by Edelsbrunner and Harer states a Spectral Sequence Theorem in Chapter VII.4, which says roughly that the total rank of the $E^r_{\ast,\ast}$ page of the spectral sequence equals the number of homology classes of persistence $r$ or larger. Here coefficients are taken mod 2. This makes precise the claim made by Carlsson and Zomorodian.

REPLY [7 votes]: there is actually a bit more to this story. The computational topology community has indeed wrestled with spectral sequences. The connection to spectral sequences has been discussed since as early as the Carlsson & Zomorodian paper mentioned by Vidit (although, not necessarily explicitly in that paper).
To make the connection clear, the $E_\infty$ page is a graded $k[t]$-module containing the "infinite" bars. the bars of length $p$ aren't exactly on the $p^{th}$ page, but, when you compute homology on page $p-1$, you can find the bars of length $p$ by hanging onto the images of the $p$-th differential. 
Speaking informally, in the language of numerical linear algebra, this is like looking at successively larger block diagonals of the boundary matrix written down in an appropriate way, reducing them, and then ``expanding." 
Infact, this algorithm (albeit not the direct connection to spectral sequences) is written down in the paper ``Clear and Compress" due to Bauer et. al 
In particular, there is also the question of whether other spectral sequences speed up computation in a practical sense. There are results that suggest asymptotic improvements in certain cases. In practice the jury is still out.<|endoftext|>
TITLE: Pull-back of a fibration along a homotopy equivalence and homotopy classes of sections
QUESTION [11 upvotes]: I previously asked this on Math.SE but didn't receive a satisfactory answer.
Let $p:E\rightarrow B$ be a fibration (i.e. have the homotopy lifting property with respect to all spaces), and $f: B'\rightarrow B$ and $g:B\rightarrow B'$ be homotopy inverses. Denote by $\pi_0\Gamma(B,E)$ the set of homotopy classes of sections of $p$, and likewise for other fibrations. I am interested in the following
Conjecture: There is a bijection $\beta:\pi_0\Gamma(B,E) \rightarrow \pi_0\Gamma(B',f^*E)$.
This would be a generalization of the elementary result $[B,X] \underset{\approx}{\xrightarrow{f^*}} [B',X]$, which is the case of trivial fibrations.
Some vague ideas:

It was pointed out by an author of [R. Brown and P.R. Heath, "Coglueing homotopy equivalences'', Math. Z. 113 (1970) 313-362] that the canonical projection $f':f^*E \rightarrow E$ is a homotopy equivalence (Corollary 1.4). Furthermore, there exists a map $g':E \rightarrow f^*E$, making the obvious diagram involving $g$ commute, such that $g'\circ f'$ and $f'\circ g'$ are homotopic to the identities via maps that factor through the bases (Theorem 3.4). This is an interesting result, but the issue is, unlike $f'$, that $g'$ doesn't seem to induce a map between sections in a natural way, so I don't know how this may be applied to my conjecture.
There's an induced map $f^*:\pi_0\Gamma(B,E) \rightarrow \pi_0\Gamma(B',f^*E)$ sending $\left[s:B\rightarrow E\right]$ to $\left[({\rm id}_{B'},s\circ f): B' \rightarrow f^*E\right]$, recalling that $f^*E=B'\times_{f,p}E$. One may try to prove $f^*$ is bijective. To do so, it would suffice to prove that the compositions $g^* \circ f^*$ and $f^* \circ g^*$ below are bijective:
\begin{equation}
\pi_0\Gamma(B,E) \xrightarrow{f^*} \pi_0\Gamma(B',f^*E) \xrightarrow{g^*} \pi_0\Gamma(B,g^*f^*E) \xrightarrow{f^*} \pi_0\Gamma(B',f^*g^*f^*E).
\end{equation}
Since $E\rightarrow B$ and $g^*f^*E\rightarrow B$ are pull-backs along homotopic maps, they are fiber homotopy equivalent (i.e. there eixst fiber-preserving maps between the total spaces, the compositions of which are homotopic to the identities via fiber-preserving maps), by e.g. Proposition 4.62 of Hatcher's "Algebraic Topology." It follows that
\begin{equation}
\pi_0\Gamma(B,E)\approx\pi_0\Gamma(B,g^*f^*E).
\end{equation}
Similarly,
\begin{equation}
\pi_0\Gamma(B',f^*E) \approx \pi_0\Gamma(B',f^*g^*f^*E).
\end{equation}
However, it is not known whether these bijections are given by $g^*\circ f^*$ and $f^* \circ g^*$.

Thank you in advance!

EDIT 6/5/2015: Upon encouragement by Dan Ramras, I made a renewed effort to carry my second idea further. I think the conjecture holds at least in "favorable cases," but I'm not sure how to conveniently characterize such cases, or if a more general proof is possible.
Our task boils down to the following. Let $p:E\rightarrow B$ be a fibration, and $F_t: B\rightarrow B$ be a homotopy such that $F_0={\rm id}_B$. I shall use $p_t$ to denote the pull-back fibration $F_t^*E\rightarrow B$. On the one hand, to each section $s\in \Gamma(B,E)$ of $p$ we can associate the section $({\rm id}_B, s \circ F_1)\in \Gamma(B, F_1^*E)$ of $p_1$. On the other hand, by the aforementioned Proposition 4.62 there is a fiber homotopy equivalence $\Phi:E\rightarrow F_1^*E$, so to each $s\in \Gamma(B,E)$ we can also associate the section $\Phi\circ s\in \Gamma(B, F_1^*E)$. The second way of associating sections is guaranteed to induce a bijection $\pi_0\Gamma(B,E)\xrightarrow{\approx} \pi_0\Gamma(B, F_1^*E)$, since $\Phi$ is a fiber homotopy equivalence. The task now is to show that the first way of associating sections induces the same map $\pi_0\Gamma(B,E)\rightarrow \pi_0\Gamma(B, F_1^*E)$. It suffices to show, given each $s\in \Gamma(B,E)$, that $\Phi\circ s$ and $({\rm id}_B, s \circ F_1)$ are in the same homotopy class of sections.
Let us recall the construction of $\Phi$. Regarding $F$ as a map $B\times I \rightarrow B$, there is the pull-back $\pi:F^*E\rightarrow B\times I$ of $p$ along $F$. Let
\begin{eqnarray}
L: E\times I &\rightarrow& B\times I \\
(e,t) &\mapsto& (p(e), t),
\end{eqnarray}
which can be thought of as a homotopy of maps $E\rightarrow B\times I$. Now consider the homotopy lifting problem
\begin{eqnarray}
E\times \{0\} &\xrightarrow{\widetilde L_0}&~ F^*E \\
\downarrow~~~~~~&   &~~~\downarrow\pi \\
E\times I ~~~& \xrightarrow{~L~} & B\times I
\end{eqnarray}
where $\widetilde L_0$ is the obvious injection $E\times\{0\} \xrightarrow{\approx} F_0^*E \hookrightarrow F^*E$. Let $\widetilde L: E\times I \rightarrow F^*B$ be the lift of $L$ extending $\widetilde L_0$. Then we define $\Phi$ as the restriction of $\widetilde L $ to $t=1$, i.e.
\begin{eqnarray}
\Phi: E &\rightarrow& F_1^*E \\
e &\mapsto& \widetilde L(e,1).
\end{eqnarray}
Remarkably, by the proof of Proposition 4.62, the homotopy class $[\Phi]\in\pi_0\Gamma(B,F_1^*E)$ of $\Phi$ is independent of the choice of the lift $\widetilde L$. Therefore, it suffices prove the following: given each $s\in \Gamma(B,E)$, there exists such a choice of $\widetilde L$ that $\widetilde L(s(-),1) = ({\rm id}_B, s\circ F_1) \in \Gamma(B, F_1^*B)$. The nice thing is this choice can depend on $s$.
Thus suppose $s$ is given, and we will construct an $\widetilde L$ in two steps. In the first step, define
\begin{eqnarray}
\psi: s(B) \times I &\rightarrow& F^*E \\
(s(b), t) &\mapsto& \left((b, (s\circ F_t)(b), t\right).
\end{eqnarray}
This is well-defined as one can verify $(b, (s\circ F_t)(b)$ is indeed in $F_t^*E$. Noting that $\psi(s(b),0) = ((b,s(b)), 0)$, we paste $\psi$ and $\widetilde L_0$ to obtain
\begin{eqnarray}
\widetilde L_0 \cup \psi: (E\times\{0\}) \cup (s(B)\times I) \rightarrow F^*E.
\end{eqnarray}
$\widetilde L_0 \cup \psi$ certainly extends $\widetilde L_0$, and it lifts $L$ because $\pi \left((b, (s\circ F_t)(b), t\right) = (b,t) = L(s(b),t)$. In the second step, we have to solve the following homotopy lifting problem for the pair $(E,s(B))$ (or "homotopy lifting extension problem"):
\begin{eqnarray}
(E\times\{0\}) \cup (s(B)\times I) &\xrightarrow{\widetilde L_0 \cup \psi}&~ F^*E \\
\downarrow~~&   &~~~\downarrow\pi \\
E\times I & \xrightarrow{~~~L~~~} & B\times I
\end{eqnarray}
This is where I had to make some favorable assumptions. Let us assume that every element of $\pi_0\Gamma(B,E)$ has a representative $s$ such that $(E,s(B))$ can be given a CW pair structure. By using a different representative if necessary we can assume that the given $s$ has this property. Now, a fibration is a Serre fibration, and a Serre fibration has the homotopy lifting property with respect to all CW pairs. Therefore the desired $\widetilde L:E\times I \rightarrow F^*E$ exists.
(To complete the proof of the original conjecture, of course, the same favorable assumptions should be made about $f^*E\rightarrow B'$ as well as $E\rightarrow B$.)

REPLY [5 votes]: Here is one way of proving the conjecture is true in general, using the modern method of weak factorization systems. 
A weak factorization system has at its core two classes of maps the left class and the right class and they satisfy lifting properties with respect to each other. Here the right class is the class of Hurewicz fibrations. The left class is the class of trivial Hurewicz cofibration (i.e. DR pair). These classes define each other. The left class consists of all maps with the left lifting property with respect to all maps in the right class; the right class consists of all maps with the right lifting property with respect to the left class. So trivial Hurewicz cofibrations have the left-lifting property with respect to the Hurewicz fibrations. They are exactly the Hurewicz cofibrations which are also homotopy equivalences. 
We won't really need the general notion, just a few special cases. You already know some trivial Hurewicz cofibrations. For example for any space $B$, the inclusion
$$B \times \{0\} \to B \times [0,1]$$
has the left lifting property with respect to any Hurewicz fibration (by definition), hence this is the first example of a trivial Hurewicz cofibration. 
Also the trivial Hurewicz cofibrations are closed under several operations: composition, taking retracts, and cobase change (aka pushouts). Using these we can form new examples of trivial Hurewicz cofibrations. Let $f: B' \to B$ be any map. Then the inclusion of $B$ in the mapping cylinder 
$$B \to B' \times [0,1] \cup^{B'\times \{1\}} B$$
is an example (a pushout of our previous example). We can also consider the inclusion on the other side
$$B' \times \{0\} \to B' \times [0,1] \cup^{B'\times \{1\}} B$$
If the map $f$ is a homotopy equivalence then this too is a trivial Hurewicz cofibration, though this takes a bit more work to see. The hard part is proven in prop 7 of these notes, as well as many textbooks.  
Now let us first consider the special case of your conjecture where $f: B' \to B$ is a trivial Hurewicz cofibration.  
Lemma: If $f: B' \to B$ is a trivial Hurewicz cofibration and $E \to B$ is a  Hurewicz fibration, then we get an induced bijection: 
$$ f^*: \pi_0 \Gamma(B, E) \to \pi_0 \Gamma(B', f^*E) $$
Proof: First let's show surjectivity. A section of $f^*E$ is the same as a map $s: B' \to E$ such that $ps = f$. This is a triangle which we can enlarge into a square where the left edge is $f$, the right edge is $p: E \to B$ and the bottom is the identity on $B$. This square is a "lifting problem". Now we use the property that trivial Hurewicz cofibrations have the left lifting property with respect to Hurewicz fibrations to solve the lifting problem. This solution is a map $B \to E$ which is exactly a section of $E$ which restricts on $B'$ to the original section. 
Injectivity is proven by the same argument but using the fact that
$$ B \times \{0,1\} \cup^{B' \times \{0,1\}} B' \times I \to B \times I$$
is also a trivial Hurewicz cofibration. [We leave this part as an exercise]. QED.
Now using this lemma we can prove the general conjecture as follows. We will construct a space Z with two trivial Hurewicz cofibrations
$$ i:B \to Z $$
$$ j:B' \to Z $$
and a map $k:Z \to B$ such that the composite
$$B \to Z \to B $$
is the identity map on $B$ and the composite
$$B' \to Z \to B $$
is our map $f$. Once we have a space with these maps, the previous lemma shows that the maps between homotopy classes of sections $i^*$ and $j^*$ are bijections. Since $k^* i^* = 1^*$ is also a bijection, this means that $k^*$ is a bijection too, and hence $f^* = k^* j^*$ is a bijection, which is what we wanted to show. 
Such a space $Z$ can be constructed as
$$ B' \times [0,1] \cup^{B' \times 1} B \times [1,2] $$ 
that is we glue $B' \times I$ to $B \times I$ at one end via the map $f$. The maps $i,j$ are given by including $B'$ and $B$ at 0 and 2, respectively (the ends of the "cylinder"). The map $k$ is given by projecting $B$ (by using $f$ for points on the first half of the cylinder).<|endoftext|>
TITLE: which norms can be realized as operator norms?
QUESTION [12 upvotes]: Assume $(V,∥∥_V),(W,∥∥_W)$ are both finite dimensional normed spaces. We have the induced operator norm on ${\rm Hom}(V,W)$.
It turns out that the operator norm is induced by an inner product iff both $V,W$ are inner product spaces and at least one of $V,W$ has dimension 1. (This is proved here).
Are there any other obstructions for a norm on ${\rm Hom}(V,W)$ to be realized as an operator norm for some suitable norms on $V,W$?
(The main interest is in the case where $\dim V>1,\dim W>1$. Take a norm on the Hom space which does not come from an inner product. Is it realizable?)

REPLY [11 votes]: Operator norms are the same thing as injective tensor norms or, equivalently, smallest dual cross norms on $\operatorname{Hom}(V, W) \simeq V^\ast \otimes W$. This means that the dual norm $\Vert \cdot \Vert^\ast$ on $V \otimes W^\ast$ has to satisfy the following:
$$\Vert x \Vert^\ast = \inf_{x = \sum_i y_i \otimes z_i} \sum_i \Vert y_i \Vert \Vert z_i \Vert$$
where the infimum is over all possible representations of a tensor as a combination of rank-ones. This follows immediately from the definition of the operator norm, i.e. that it's determined by testing against rank-ones.
In particular, the unit ball of $\Vert \cdot \Vert^\ast$ is the convex hull of its rank-one vectors. Thus, for example, it cannot be strictly convex (unless everything is of rank one, which means that $\min(\dim V, \dim W) = 1$). This generalizes the Euclidean case.<|endoftext|>
TITLE: Intuition for thinking about $R$-module of Kähler differentials, universal receptacles, derivations?
QUESTION [6 upvotes]: Suppose $k$ is a field of characteristic zero, and $R$ is a $k$-algebra. The $R$-module of Kähler differentials $\Omega_{R/k}$ of $R$ over $k$ with generators $\{dr\}_{r \in R}$ is the module subject to the following relations:

$dr = 0$ if $r \in k$.
$d(r+s) = dr + ds$.
$d(rs) = r\,ds + s\,dr$.

We have a natural map $d: R \to \Omega_{R/k}$ of vector spaces which sends $r \mapsto dr$.
$\Omega_{R/k}$ is the universal receptacle for derivations of $R$ into some $R$-module $M$. If $\delta: R \to M$ satisfies the axioms above, then there exists a map of $R$-modules $\Omega_{R/k} \to M$ such that $\delta$ factors through $d$.
What is the motivation/geometric intuition of these definitions? Is there a good way of thinking about all this in context of algebraic curves? Hopefully nothing fancy is needed?

REPLY [6 votes]: You want something that plays the role of cotangent bundle in nice cases, and be defined in general.
It may not be clear, a priori, that the universal derivation $d:R\to \Omega_{R/k}$ will give a reasonable object but it does. If you do a few  examples, you will see that it behaves quite well.
If $R=k[x_1,\ldots, x_n]$ which is the ring of functions on  affine space, then $\Omega_{R/k}$ is isomorphic to the module of polynomial $1$-forms $\bigoplus_i Rdx_i$, and 
$$df=\sum \frac{\partial f}{\partial x_i} dx_i$$
exactly as you might hope from calculus.  If $R$ is a dvr, i.e. the local ring of a regular affine curve, then $\Omega_{R/k} = Rdx$, where $x$ is a uniformizer.<|endoftext|>
TITLE: The size of Lindelof space
QUESTION [6 upvotes]: Question. Suppose that $X$ is a Lindelof space such
that every point of $X$ is a $G_{\delta}$-point. Then is it true that $|X| ≤ 2^{\omega}$?

REPLY [2 votes]: The consistency of this statement is Problem 1 in these slides by F. Tall, where the problem is attributed to Arhangelski. Some partial results already mentioned in Mohammad Golshani´s answer are given. The following is also mentioned:

Theorem (Tall-Usuba): Lévy-collapse a weak compact and then add $\aleph_3$ Cohen subsets of $\omega_1$. Then there are no counterexamples of size $\aleph_2$, even relaxing "points $G_\delta$" to "pseudocharacter $\leq \aleph_1$".<|endoftext|>
TITLE: Generalization of Krull dimension for commutative rings
QUESTION [7 upvotes]: In the paper How to construct huge chains of prime ideals in power series rings by B. Kang and P. Toan the Krull dimension of a commutative ring with $1$ is defined as follows: 
Let $R$ be a commutative ring and  $\mathfrak{C}$ be an arbitrary chain of prime ideals in $R$. Then length of $\mathfrak{C}$ is defined by $\left\vert{\mathfrak{C}}\right\vert-1$. The Krull dimension of $R$ is the largest cardinal number $\alpha$ (if any) such that there exists a chain of prime ideals in $R$ whose length is equal to $\alpha$. We write $\dim(R)\geq \alpha$ if there is a chain of prime ideals in $R$ whose length is $\geq \alpha$.
Note that this agrees with the usual definition of dimension when $\dim(R)$ is finite. However, there is no guarantee that the dimension of a ring exists in general. In the above paper, there are classes of rings with uncountable chains of prime ideals. Assuming that the Continuum hypothesis is true, there are a few examples where the Krull dimension is determined. This is done by noting that $\dim(R) \leq \left\vert{2^R}\right\vert$.

Does the dimension of a commutative ring exist in general? Is the existence of dimension for all commutative rings equivalent to the generalized continuum hypothesis?

REPLY [2 votes]: Here is one stupid obstruction.
Take any limit cardinal $\alpha = \bigvee_{i \in I} \alpha_i$, $\alpha_i < \alpha$, such that there is a ring $R_i$ with chains of any length $< \alpha_i$ but not of length $\alpha_i$. Then the ring $R$ defined as the unitalizatioon of $\bigoplus_i R_i$ doesn't have a dimension.
Indeed, any prime ideal is necessarily supported on some $i$, i.e. is pulled back from one of $R_i$, so every chain of prime ideals of $R$ is pulled back from a chain in some $R_i$. These don't have a maximal cardinality.<|endoftext|>
TITLE: For a ring $k$ and a set $X$, what are the $k$-algebra homomorphisms $k^X \to k$?
QUESTION [8 upvotes]: Let $k$ be a commutative ring.  Feel free to assume it's a field.
Let $X$ be a set.  This question is only interesting when $X$ is infinite.
Write $k^X$ for the $k$-algebra of functions $X \to k$, with the algebra operations defined pointwise.

What are the $k$-algebra homomorphisms $k^X \to k$?  

Trivially, for each $x \in X$ there is a projection/evaluation map $k^X \to k$.  

Under what circumstances are there any $k$-algebra homomorphisms $k^X \to k$ apart from the projections?

Here are some observations.  Observation 2 shows that the question is not entirely trivial, in the sense that there are sometimes nontrivial homomorphisms $k^X \to k$.

When $k$ an integral domain, any homomorphism $\Phi: k^X \to k$ gives rise to an ultrafilter $\mathcal{U}_\Phi$ on $X$.  To see this, write $\chi_S \in k^X$ for the characteristic function of a subset $S \subseteq X$.  Since $\chi_S$ is idempotent, $\Phi(\chi_S)$ is also idempotent, and is therefore either $0$ or $1$.  Write
$$
\mathcal{U}_\Phi
=
\{ S \subseteq X : \Phi(\chi_S) = 1 \}.
$$
It's easy to check that $\mathcal{U}_\Phi$ is an ultrafilter on $X$ —  in other words, that whenever we write $X = X_1 \amalg \cdots \amalg X_n$, there is precisely one $i$ for which $\Phi(\chi_{X_i}) = 1$.
When $k$ is a finite integral domain — that is, a finite field — the $k$-algebra homomorphisms $k^X \to k$ are in bijection with the ultrafilters on $X$.  One direction of this correspondence is given as in (1).  
For the other, start with an ultrafilter $\mathcal{U}$ on $X$.  We want to define a homomorphism $\Phi_{\mathcal{U}}: k^X \to k$, so take $\phi \in k^X$.  Since $k$ is finite, the fibres $(\phi^{-1}(c))_{c \in k}$ form a finite partition of $X$.  So there is precisely one element $c \in k$ such that $\phi^{-1}(c) \in \mathcal{U}$, and we put $\Phi_{\mathcal{U}}(\phi) = c$.  It's straightforward to check that $\Phi_{\mathcal{U}}$ is a homomorphism and that the processes $\mathcal{U} \mapsto \Phi_{\mathcal{U}}$ and $\Phi \mapsto \mathcal{U}_\Phi$ are mutually inverse.
When $k$ is an integral domain and $X$ (rather than $k$) is finite, the only homomorphisms $k^X \to k$ are the projections.  This follows e.g. from (1) and the fact that ultrafilters on a finite set are principal.
Denote by $X \cdot k$ the $k$-vector space with basis $X$.  Then $k^X$, as a $k$-vector space, is isomorphic to the space of linear maps $X \cdot k \to k$.  Now any $k$-algebra homomorphism $\Phi: k^X \to k$ is, in particular, a $k$-linear map, so $\Phi$ is an element of the double dual of $X \cdot k$.  Hence there can only be nontrivial homomorphisms $k^X \to k$ if there are nontrivial elements of the double dual of $X \cdot k$.  
So my question seems to be closely related to one that's come up a few times here before: how much Choice do we need in order to construct nontrivial elements of the double dual of an infinite-dimensional vector space?

REPLY [4 votes]: As explained in the comments, if $k$ is a field, then $k$-algebra homomorphisms $k^X\to k$ are in bijection with $|k|^+$-complete ultrafilters on $X$ (that is, ultrafilters closed under $|k|$-fold intersections, or equivalently, ultrafilters such that whenever you partition $X$ into $|k|$ pieces one of the pieces must be in the ultrafilter).  In particular, they are all projections unless (and only unless) there is a measurable cardinal $\kappa$ such that $|k|<\kappa\leq |X|$ (here I count $\aleph_0$ as measurable), since the least cardinal that supports a $|k|^+$-complete ultrafilter is measurable.  When $k$ is not a field, the question seems harder.<|endoftext|>
TITLE: Pontryagin Forms and Special Holonomy
QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian manifold. Recall that the $k^{th}$-Pontryagin class is a topological invariant which, by classical Chern-Weil theory, can be represented using the so-called Pontryagin forms. These forms are obtained by appropriate constant linear combinations and wedge multiplications of the closed $4p$-forms $\operatorname{Tr}(R^{p})$, where
$$\operatorname{Tr}(R^{p})(X_1,....,X_{4p}):=\sum_{\sigma}(-1)^{|\sigma|}\operatorname{Tr}(R(X_{\sigma(1)},X_{\sigma(2)})\circ\cdots\circ R(X_{\sigma(4k-1)},X_{\sigma(4p)}))$$ 
and $R$ is the curvature tensor of $(M,g)$. It is known that the Pontryagin forms depend only on the Weyl part of the curvature tensor (in particular they vanish for conformally flat manifolds).
In concrete cases, they can in principle be explicitly computed.
However I am interested in the case where $M$ has special holonomy (contained in) $SU(5)$, i.e. it is a not necessarily compact Calabi-Yau $5$-fold. My question is:

Are there any conditions under which the Pontryagin forms (or, equivalently, the above trace forms) are proportional to the appropriate powers of the Kahler form $\omega$?
More generally, are there any conditions under which the Pontryagin forms are parallel?

Note that the $k^{th}$-Pontryagin form of a Kahler manifold is always of type $(2k,2k)$ so that, when the value $h^{2k,2k}$ of the Hodge diamond is one, this form is necessarily a multiple of $\omega^{2k}$ up to exact terms.
Any example using algebraic geometry or any argument which might enlighten some properties (if any) of Pontryagin forms for manifolds with special holonomy in the sense of Berger's classification are appreciated.

REPLY [4 votes]: Here are a few remarks to show that your Questions 1 and 2 are almost equivalent.  (What the answers are is another matter, and I expect that to be somewhat difficult, as I'll explain below.)  Of course, Question 1 is a special case of Question 2 in the OP's situation because, if a closed form is a scalar multiple of $\omega^2$ or $\omega^4$ on $M^{10}$, then that scalar multiple must be constant, and hence the form is parallel.  What's interesting is that, in spite of appearances, having the Chern-Weil representatives of the Pontryagin forms be parallel when the holonomy is a subgroup of $\mathrm{SU}(5)$ is nearly the same as having these forms be scalar multiples of $\omega^{2k}$.
First, for Question 2, you can reduce to the case in which the holonomy acts irreducibly:  If $(M,g)$ is a (simply-connected) Riemannian manifold whose holonomy preserves a parallel splitting $TM = V_1\oplus V_2$, then, locally, $g$ is a product metric and the formula for the total Pontryagin class $p(TM) = p(V_1)p(V_2)$ and that fact that the holonomy is the product of the holonomies acting on the two subbundles $V_1$ and $V_2$ shows that the Chern-Weil representatives of the classes in $p(TM)$ are parallel with respect to the holonomy of the product if and only if each of the two factor metrics has the property that its Chern-Weil representatives of the classes $p(V_i)$ are also parallel.  By reduction, then, it suffices to classify the metrics $(M,g)$ for which the Pontryagin forms are parallel and the holonomy acts irreducibly.
Second, although you say you are mainly interested in the case of an $(M^{10},g)$ whose holonomy is (contained in) $\mathrm{SU}(5)$, you might as well generalize your question to considering the case of $(M^{2n},g)$ with holonomy (contained in) $\mathrm{SU}(n)$.  Since, by the first remark, we can assume the holonomy acts irreducibly, that means, by the Berger classification, that the holonomy has to be either $\mathrm{SU}(n)$ or, if $n$ is even, $\mathrm{Sp}(n/2)$.  
In the case the holonomy is $\mathrm{SU}(n)$, the only parallel forms of type $(2k,2k)$ are the (even) powers of the Kähler form $\omega$, so, in this case, having the Pontryagin forms be parallel implies having them be (constant) multiples of the  powers of $\omega$.  Meanwhile, for $k<n$ the only multiples of $\omega^k$ that are closed are the constant multiples.  In particular, when $n=5$, if the Chern-Weil representative of $p_1(M)$ (respectively,$p_2(M)$) is a multiple of $\omega^2$ (respectively, $\omega^4$), then it is a constant multiple, and hence parallel.
In the case the holonomy is $\mathrm{Sp}(n/2)$, things are a little more complicated:  In this case, the algebra of parallel forms is generated by $\omega$, $\phi$, and $\bar\phi$, where $\phi$ is a parallel holomorphic $(2,0)$-form.  Thus, the parallel $(2k,2k)$-forms are constant linear combinations of the forms $\omega^{2(k-j)}\wedge\phi^j\wedge\bar\phi^j$ for $0\le j\le k\le n/2$.  Since we only care about the cases $n\le 5$ for the OP's problem, we really only have to deal with $n=2$ and $n=4$ in this situation.  When $n=2$, since $\mathrm{Sp}(1)=\mathrm{SU}(2)$, this case is already handled. When $n=4$, the only case to deal with is $k=1$, where there are two parallel $(2,2)$-forms, $\omega^2$ and $\phi\wedge\bar\phi$.  However, in this case, because there is a $2$-sphere of parallel complex structures and because the Pontryagin form $p_1(M,g)$ will have to be of type $(2,2)$ with respect to all the complex structures of this family, one finds that, if the Chern-Weil Pontryagin form $p_1(M,g)$ is to be parallel, it must be a constant multiple of $\omega^2 + \phi\wedge\bar\phi$.  (Here, I am assuming that $\phi$ and $\omega$ have been suitably normalized.)  This multiple can't be zero because it is essentially the squared norm of the curvature; if it were zero, the metric would be flat, and the holonomy wouldn't be $\mathrm{Sp}(2)$.
Thus, the only case that you care about in which the two Questions essentially differ is the case of an $(M^{10},g)$ that is a product  $(N^8,h)\times (\mathbb{C},g_0)$ where $g_0$ is the flat metric and $(N^8,h)$ has holonomy $\mathrm{Sp}(2)$ with $p_1(N,h)$ being a (constant) multiple of $\omega^2 + \phi\wedge\bar\phi$ and $p_2(N,h)$ being a constant multiple of the volume form.
Finally, there is the question of whether any of these irreducible cases can actually occur.  Here, I don't have much to say.  I do not know of a single example of even a local, non-flat  Calabi-Yau metric for which the square norm of the curvature tensor is constant, even in complex dimension $2$, which is the very first nontrivial case.  The problem gets more and more overdetermined as the dimension goes up, so it seems unlikely that such metrics exist, even locally, but I think that actually writing down a proof might not be easy.<|endoftext|>
TITLE: Gersten complexes in Quillen's and Milnor's K-theories
QUESTION [10 upvotes]: Consider a good enough scheme $X$ (e.g. an algebraic variety over a field). Let $X_i$ be the set of points of dimension $i$ in $X$. Then we have the Gersten complex in Quillen's K-theory:
$$
\oplus_{x\in X_{i+1}}K_{n+1}(\kappa(x))\to \oplus_{x\in X_i}K_n(\kappa(x))\to \oplus_{x\in X_{i-1}}K_{n-1}(\kappa(x))
$$
Kato proved that there is a Gersten complex of the same type using Milnor's K-groups instead. As there are natural maps from Milnor K-groups to Quillen K-groups, I would like to know whether there is a compatibility between (the differential maps in) these two complexes. 
For the differentials from $K_2$ to $K_1$ or from $K_1$ to $K_0$, it is well-known that the two complexes give the same thing (up to a sign). But I'm not sure if someone has checked this for higher $K$-groups. Could anyone give some hints or references?

REPLY [5 votes]: Yes, the natural multiplication morphisms induce a morphism of Gersten complexes from Milnor to Quillen K-theory. The basic points are made in the paper 

M. Rost. "Chow groups with coefficients", Doc. Math. 1 (1996), pp. 319-393, link to DocMath page. 

That paper contains a definition of "cycle modules" and constructions of Gersten complexes for these. Milnor K-theory is almost by definition a cycle module, and Quillen K-theory is a cycle module by Remark 1.12. Remark 5.4 in Rost's paper states that the multiplication morphisms yield a morphism of cycle modules, and by the results and constructions in the paper, a morphism of cycle modules induces a morphism of corresponding Gersten complexes. (Note that Rost references the 1987 paper of Merkurjev and Suslin for an identification of the differentials for $K_3$.) 
Since Remark 5.4 doesn't contain a proof, we trace back a bit to Remark 1.12  in Rost's paper which discusses the structure of Quillen K-theory as cycle module. The notion of cycle module incorporates an action of Milnor K-theory which, by Remark 1.12, is given exactly by the multiplication morphism. In particular, the compatibility of the multiplication morphism with the differentials of the Gersten complexes is already present in the claim that Quillen K-theory yields a cycle module -- it is the compatibility of the action of Milnor K-theory (part D3 of the definition of cycle premodule with the differentials (part D4 of the definition of cycle premodule) via relations R3e resp. R3f. 
Now the claim that Quillen K-theory is a cycle module (and hence that all the compatibilities required above are satisfied) is also not done in Remark 1.12, but that remark contains references to the literature from which the verification of the rules "is a lengthy but straightforward exercise". The key point in the end is a K-linearity of the boundary map in the localization sequence for Quillen K-theory, which is discussed e.g. in Section V.6 of Weibel's K-book.<|endoftext|>
TITLE: The space of polynomials with all real roots
QUESTION [11 upvotes]: The question stems from an attempt to answer a question of David Speyer. Let $R \subseteq \mathbb{R}^n$ be the space of coefficients of all polynomials of degree $n$ whose all roots are real, i.e.
$R := \lbrace (a_0,\ldots, a_{n-1}): x^n + a_{n-1}x^{n-1} + \cdots + a_0$ has only real roots$\rbrace$
Let $R_+$ be the intersection of $R$ with the all-non-negative orthant. If $R_+$ has more than one connected component, the answer to David's question would be negative. 
As I describe below, for $n=3$, $R$ and $R_+$ are connected, and it 'looks' from the plot that David's question has affirmative answer (which, in this special case, can surely be proved in a more straightforward way). However, I would think the spaces $R$ and $R_+$ are interesting in their own right, and would like to know if they have been studied. The answers to even the simplest questions, e.g. if they are connected, seem not very obvious.
Edit: As Richard shows below, $R_+$ is connected. It follows by an even easier version of the same argument that $R$ is connected as well. 
Below follows a description of $R$ for the case $n=3$, following the excellent answer to this question.
The computation outlined in the above answer shows that the interior $R^0$ of $R$ is the region of positive definiteness of some symmetric matrix whose entries are polynomials in $a_j$'s. For $n=3$, using the characterization of positive definiteness in terms of minors, it follows that 
$R^0 = \lbrace (a_0, a_1, a_2): f_i(a_0, a_1,a_2) > 0,\ 1 \leq i \leq 2\rbrace$  
where $f_1 = a_2^2 - 3a_1$, $f_2 = -(27a_0^2 - a_0(18a_1a_2 - 4a_2^2) + 4a_1^3 - a_1^2a_2^2)$. The discriminant of $f_2$ (as a quadratic equation in $a_0$) is $16(a_2^2 -3a_1)^3 = 16f_1^3$ which is positive on $R$. For each $(a_1, a_2)$ such that $f_1(a_1, a_2) > 0$, the admissible $a_0$ values are therefore those belonging to the interval between the minimum $r_m(a_1, a_2)$ and the maximum $r_M(a_1, a_2)$ of the two roots of $f_2(\cdot,a_1,a_2)$. Setting $t := (a_2^2 - 3a_1)^{1/2}$ gives
$r_M(a_1, a_2) = \frac{1}{27}(a_2^3 - 3a_2t^2 + 2t^3) $
$r_m(a_1, a_2) = \frac{1}{27}(a_2^3 - 3a_2t^2 - 2t^3) $ 
It follows that 
$R^0 = \lbrace (a_0, a_1, a_2): a_2^2 - 3a_1 > 0,\ r_m(a_1, a_2) < a_0 < r_M(a_1, a_2) \rbrace$  
Here is a plot of the boundary of $R$ ($R$ is the region between the two surfaces):

And here is the boundary of $R_+$:

REPLY [4 votes]: Space $R$ of hyperbolic(i.e. real-rooted) polynomials of one variable, have actually been studied quite a lot, mostly in papers by V.I. Arnold and his students at the end of 80's-beginning of 90's
I would recommend a book by V.P. Kostov Topics in hyperbolic polynomials of one variable. Societe Mathematique de France, 2011, especially it's chapter 2.
It has a lot of references.
Moreover, Section 2. of a recent preprint http://arxiv.org/abs/1512.08645 has some review on this and some other close questions.
In particular, one can prove that $R$ is homeomorphic to ${\mathbb R}\times{\mathbb R_+^{n-1}}.$ 
This follows from the fact that the correspondence between roots and coefficients of polynomials is morphism of symmetric product of topological spaces (i.e. quotient by $S_n$ acting by permutations of roots -- see http://www.ams.org/journals/tran/1954-077-03/S0002-9947-1954-0065924-2/S0002-9947-1954-0065924-2.pdf P.538 or http://link.springer.com/article/10.1007%2FBF01094483) and $n$-th symmetric product of real line is exactly ${\mathbb R}\times{\mathbb R_+^{n-1}}$ (see e.g. Proposition 3 there: https://ncatlab.org/nlab/show/symmetric+product+of+circles).<|endoftext|>
TITLE: What can be said about the concentration of measure of product of Gaussian variables?
QUESTION [10 upvotes]: I have a set of random variables $X_1,\ldots,X_n$, all Gaussian with mean 0 and variance 1, indepedent. Let $p(x_1,\ldots,x_n)$ be some polynomial that takes products and sums of $x_1,\ldots,x_n$.
What can be said about the concentration of measure of $p(X_1,\ldots,X_n)$ around $E[p(X_1,\ldots,X_n)]$?
If there were only two-order interactions, I think I would look around for concentration of measure for chi-squared random variables, but unfortunately the interaction can be of a higher degree.

REPLY [3 votes]: Hypercontractivity implies that for a polynomial $P$ of total degree $d$ in Gaussian variables and $q \geq 2$, we have
$$ \|P\|_{L^q} \leq (q-1)^{d/2} \|P\|_{L^2} .$$
Applying Markov inequality for the optimal $q$ yields then for $t \geq C_d$
$$ \mathrm{Prob} \left(|P- \mathbf{E} P| \geq t \sqrt{\mathrm{Var}(P)} \right) \leq \exp(-c_d t^{2/d} ) $$
for some constants $C_d,c_d$.
Reference: Corollary 5.49 in Aubrun-Szarek, Alice and Bob meet Banach<|endoftext|>
TITLE: Packing obtuse vectors in $\mathbb{R}^d$
QUESTION [11 upvotes]: I came across this attractive theorem:

Theorem. In $\mathbb{R}^d$, there can be at most $d+1$ vectors that
form an obtuse angle with one another.

This was proved1 as a corollary of a lemma about irreducible matrices.
I am wondering if anyone knows of an alternative, more geometric proof that somehow
more directly captures the sense that one cannot "pack" more than
$d+1$ obtuse vectors in $\mathbb{R}^d$.

         


1Lipeng Ning, Tryphon T. Georgiou, Allen Tannenbaum, Stephen P. Boyd.
"Linear models based on noisy data and the Frisch scheme." 
SIAM Review. 57(2) 2015.
arXiv preprint.

REPLY [3 votes]: The proof I know of this result is essentially the same as Vladimir Dotsenko's, but in terms of Radon's lemma.:
Assume such a set exists with at least $d+2$ points.  By Radon's lemma, given any $d+2$ points in $\mathbb{R}^d$ there is a partition of them into two sets $A$, $B$ such that $conv(A) \cap conv(B) \neq \emptyset$.  Let $p$ be a point of this intersection.  If we write $p$ as a convex combination of $A$ and one of $B$, and using the fact that the dot product of any two points in the original set is negative, we immediately obtain $p \cdot p < 0$, a contradiction.<|endoftext|>
TITLE: Find all solution $a,b,c$ with $(1-a^2)(1-b^2)(1-c^2)=8abc$
QUESTION [14 upvotes]: Two years ago, I made a conjecture on stackexchange:
Today, I tried to find all solutions in integers $a,b,c$ to
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$  
I have found some solutions, such as
$$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$
$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$
Here is another simple solution:
$$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My question is: are there solutions of another form (or have we found all solutions)?

REPLY [3 votes]: I'm late for this party, but using math110's method employing Euler bricks, couldn't resist giving some simple rational solutions to,
$$(1-a^2)(1-b^2)(1-c^2) = 8abc$$
Solution 1:
$$a,\,b,\,c = \frac{-(x-z)(2x+z)}{(2x-z)y},\;\frac{z}{2x},\;\frac{-2y+z}{2y+z}\tag1$$
where $x^2+y^2=z^2.$
Solution 2:
$$a,\,b,\,c = \frac{2z^2}{xy},\;\frac{x-z}{x+z},\;-\frac{y+z}{y-z}\tag2$$
where $x^2+y^2=5z^2$, and which may also be solved as a Pell equation.<|endoftext|>
TITLE: Topological cobordisms between smooth manifolds
QUESTION [78 upvotes]: Wall has calculated enough about the cobordism ring of oriented smooth manifolds that we know that two oriented smooth manifolds are oriented cobordant if and only if they have the same Stiefel--Whitney and Pontrjagin numbers.
Novikov has shown that rational Pontrjagin classes may be defined for topological manifolds; thus smooth manifolds which are topologically cobordant have equal Pontrjagin numbers. It is also easy to see that they have the same Stiefel--Whitney numbers (for this they only need to be Poincare cobordant).
It follows that smooth manifolds which are topologically cobordant are in fact smoothly cobordant. Is there a direct geometric proof of this fact?

REPLY [13 votes]: Is there a direct geometric argument to show that two smooth manifolds that are  topologically bordant are actually smoothly bordant? This is the question to be addressed.
The response below is a guide to a more satisfying situation....
In dimension three there is Likorish's argument using Dehn twists that oriented three manifolds bound. There is also Rochlin's geometric proof that in dimension four the complete bordism invariant is the signature circa 1950 [Oleg Viro at Stonybrook, his student, has  these early references....] Rochlin was the student of Pontryagin. Thom suggested that he deserved the 1958 Fields Medal less than these predecessors.
Above that dimension, the only purely geometric step, to my knowledge, is the Pontryagin–Thom construction (1953 Thom, 1942 Pontryagin)reducing bordism questions to homotopy questions.
Serre's 1950 thesis opened the way to compute homotopy groups and the rest is the history alluded to in the question. Nevertheless.....
There is the theory of surgery or what was known as "constructive cobordisms" which one can learn from  Milnor–Kervaire (Annals 1963)
or Browder's book "Simply connected surgery theory" or Novikov's papers from early 60's, or Wall's book on "Non simply connected surgery"
There is also obstruction theory as explained in Steenrod's book "Topology of fibre bundles" circa late 50's.
With these  three  tools in hand one can try to do something step by step.
In my Princeton thesis January 1966 "Triangulating homotopy equivalences" I tried this for the problem of constructing a PL homotopy bordism of a homotopy equivalence to a PL homeomorphism when it was impossible to compute the torsion aspects in the smooth case and when the torsion aspects of the difference between PL bundles and smooth bundles was also unknown.
By a stroke of luck the two unknown torsion groups canceled each other and an obstruction theory for the question with  explicit coefficients in $0$ $\mathbb{Z}/2$ $0$ $\mathbb{Z}$ $0$ $\mathbb{Z}/2$ $0$ $\mathbb{Z}$.... emerged. [At the thesis defense Steenrod asked how these explicit obstructions obstructions could be calculated. The question was unexpected but a very good one. They were like any obstruction theory explicitly  ill-determined but in this case the effective obstructions could be determined using bordism theory itself.]
Takeaway, there may be a tweaked version of the challenge being discussed  which has a step by step  construction which is geometric and constructive  yet  still obstructed but now by more explicit groups related to the signature and Kervaire–Arf invariant of surgery & one might do something that is part geometric and part algebraic but still rather concrete. 
I would try to first lift the challenge to trying to lift a PL bordism to a smooth bordism because the relative homotopy group between topological and PL is one $\mathbb{Z}/2$ which is described by a signature divided by $8$ and reduced mod $2$.
Novikov's theorem/information expanded by et al comprises this information.
(BTW The sad news appeared today that Serge Novikov is very ill.)
Then  directly attack lifting a PL bordism to a smooth bordism. Now the elementary geometric argument of Thom from Colloque Topologie Mexico City 1957 suffices to understand the rational Pontryagin classes by geometry. And BTW Novikov's argument is built on this argument of Thom and both are geometric plus a dollop of Serre which is easy.
Conclusion: There is a chance to have a geometric understanding of this question using just basic algebraic topology/geometric  information about manifolds with a dash of serre algebraic topology/ algebra .
Dennis Sullivan   Friday December 13th, 2019.<|endoftext|>
TITLE: Derived functors - homotopical vs homological approach
QUESTION [24 upvotes]: This question is a crosspost of the second part of this MSE question.
In my first course in homological algebra, derived functors were defined in terms of universal $\delta$-functors. In the text Homotopy Limits Functors on Model Categories and Homotopical Categories, I learned a much more economic and conceptual definition as a Kan-extension along the localization functor.

Where can I find actual rigorous proof that in the abelian setting, the homologies of Kan extensions along localizations form universal $\delta$-functors?

In the MSE question, Zhen Lin proposed to simply calculate both in terms of acyclic resolutions, but I am looking for a proof using as little concrete calculations as possible, preferably employing only universal properties.
I'm a novice, so if you give a proof sketch, please be detailed.

Update: I know there are many great homotopy theorists here. That this question has remained unanswered but has not been downvoted makes me wonder - what's wrong with it?

REPLY [13 votes]: EDIT Corrected a couple of inaccuracies and mistakes, added some references.
For the sake of clarity, let me work with non-negatively graded cochain complexes, and analyze the case of a left exact covariant functor, to prove that its right derived functor is a universal (covariant) cohomological $\delta$-functor.
Let $\mathcal A$ and $\mathcal B$ be two abelian categories, and let $F : \mathcal A \to \mathcal B$ be a left exact functor. Let me denote by $\gamma_\mathcal{A}$ the localization functor $\mathrm{Ch}_+(\mathcal{A}) \to \mathrm{D}_+(\mathcal{A})$. Let $\tilde S_\mathcal{A}$ denote the full subcategory of $\mathrm{Ch}_+(\mathcal A)$ consisting of discrete complexes (i.e. of complexes whose differentials are all zeroes). Let $S_\mathcal{A}$ denote its essential image in the derived category.
First of all, observe that a $\delta$-functor is a particular case of what MacLane calls "connected sequences" (connected sequence is a $\delta$-functor iff the long sequence it induces is exact). Now, defining a connected sequence is equivalent to define a functor $S_\mathcal{A} \to \mathcal{B}$ (see Proposition XII.8.1 and Proposition XII.8.2 in [1]). Part of the equivalence works as follows: given any $T : S_\mathcal{A} \to \mathcal{B}$, define $T^n(A) := T(A[n])$, and check that, given any exact sequence $0\to A\to B\to C\to 0$, the image of the class corresponding to it in
$$S_\mathcal{A}(C[n],A[n+1]) \simeq \mathrm{D}_+(\mathcal A) (C[n], A[n+1]) \simeq \mathrm{Ext}^1_{\mathcal A}(C,A)$$
under $T$ gives you a morphism $\delta^n : T^n C \to T^{n+1} A$ giving $\{T^n\}$ the structure of a connected sequence. You can find the details of this construction in [1]. 
We have that
$$\mathbb{R}F := \mathrm{Lan}_{\gamma_\mathcal{A}(-[0])} (\gamma_\mathcal{B} F(-)[0]).$$
Now, we can postcompose the bottom corners of the square defining $\mathbb{R}F$, along the functors $$\gamma_{\mathcal A} \circ \bigoplus_{n\geq 0} H^n : \mathrm{D}_+(\mathcal{A}) \to S_\mathcal{A}$$ and
$$H^0 : \mathrm{D}_+(\mathcal{B}) \to \mathcal{B}$$ as follows
$$
\require{AMScd}
\begin{CD}
\mathcal{A} @>{F}>> \mathcal{B}\\
@VVV @VVV \\
\mathrm{D}_+(\mathcal{A}) @>{\mathbb{R}F}>> \mathrm{D}_+(\mathcal{B})\\
@VVV @VVV \\
S_\mathcal{A}@. \mathcal{B}
\end{CD}
$$
and further Kan extend, obtaining the connected sequence $RF : S_\mathcal{A} \to \mathcal{B}$ as
$$\mathrm{Lan}_{\gamma_\mathcal{A} \circ \oplus_n H^n \circ (-)[0]}(H^0 \circ (-)[0] \circ F) \simeq \\ \simeq \mathrm{Lan}_{\gamma_\mathcal{A} (-[0])}(F).$$
Now, given any other connected sequence (in particular, any $\delta$-functor) $T : S_\mathcal{A} \to \mathcal{B}$, by definition of left Kan extension we have
$$\mathrm{Nat}(RF,T) \simeq \\ \simeq \mathrm{Nat}(F, T \circ \gamma_{\mathrm{A}} (-[0])) \simeq \\ \simeq \mathrm{Nat}(F,T^0)$$
showing that the "universality" of $RF$ works not only for $\delta$-functors, but in general for connected sequences.
The fact that for $F$ left exact and $\mathcal A$ with enough injectives one has that $RF$ is not only a connected sequence, but really a $\delta$-functor, follows from XII.8.3, 4 and 5 in [1].

[1] MacLane - Homology<|endoftext|>
TITLE: When is $f(x^d)$ irreducible?
QUESTION [7 upvotes]: Let $f(x)$ be an irreducible polynomial of degree $n$ over a finite field $\mathbb F_p$. What can we say about $f(x^d)$? When is it irreducible ?

REPLY [6 votes]: Here’s a stab at a partial answer. Let $n$ be the degree of the original polynomial $f(x)$ over $\Bbb F_p$, so that any root $\alpha$ generates the field $\Bbb F_q$, $q=p^n$. The polynomial $f(x^d)$, of degree $nd$, is irreducible over $\Bbb F_p$ if and only if a root $\beta$ generates the field $\Bbb F_{q^d}$. The polynomial for $\beta$ over $\Bbb F_q$ is $x^d-\alpha$, of course, and we’re asking, precisely, whether this is irreducible over that field.
So I think the question boils down to this apparently simpler one: If $\alpha\in\Bbb F_q$, under what conditions is $x^d-\alpha$ irreducible over that field?
This certainly depends crucially on the multiplicative period of $\alpha$, as commenters have already noted. If $\alpha=i\in\Bbb F_9$, then $\sqrt\alpha$ is already in the same field, (because the multiplicative group is of order $8$). On the other hand, if $\alpha=1+i$ in that field, this is a generator of the multiplicative group (period eight), and $x^2-\alpha$ is irreducible, generating the field of cardinality $9^2=81$. Just to bring this back to $\Bbb F_3$, this is saying (since the polynomial for $i$ is $x^2+1$) that $x^4+1$ is not irreducible over the prime field, but (since the polynomial for $1+i$ is $x^2+x+2$) the polynomial $x^4+x^2+2$ is irreducible over $\Bbb F_3$.
So far so good. What values of $d$ are good, which are bad? Certainly anything divisible by the characteristic is bad, and more generally anything prime to $q-1$ is bad too, because a cyclic group of order $m$ automatically has $n$-th roots of all elements if $\gcd(m,n)=1$.
Indeed, that same argument shows that if $d=d'r$ where $r$ is relatively prime to $q-1$, adjoining the $d$-th root of an element is the same thing as adjoining the $d'$-th root. So we want $d$ to have among its prime divisors only the prime divisors of $q-1$.
And I think this answers the question completely if $\alpha$ is a generator of the cyclic multiplicative group of the field $\Bbb F_q$: $d$ must have for its prime divisors only primes that occur in $q-1$. There are a couple of gaps that I believe I know how to fill, but this posting is already too long.<|endoftext|>
TITLE: Geometric Mean of $L(1,\chi)$ for quadratic Dirichlet characters
QUESTION [14 upvotes]: Let $S = \{D_1, D_2, D_3, \ldots \}$ be the set of all prime discriminants 
(or positive prime discriminants) of quadratic number fields. For such a 
discriminant let $\chi_j(n) = (\frac{D_j}n)$ be the associated Dirichlet 
character and 
$$ L(1,\chi_j) = \sum_{n \ge 1} \frac1n \Big(\frac{D_j}{n}\Big) $$
the value of Dirichlet's L-series at $s = 1$. If we assume that
about half prime discriminants $D$ have $(D/p) = +1$ and the other half
satisfy $(D/p) = -1$, and if we interchange the limits, then the
geometric mean of the values of $L(1,\chi_j)$ is given by 
\begin{align*}
 \lim_{k \to \infty} \bigg( \prod_{j=1}^k L(1,\chi_j) \bigg)^{1/k} 
   & = \lim_k \prod_p \Big(\frac{p}{p-1}\Big)^{\frac{k}{2k}}
        \cdot \Big(\frac{p}{p+1}\Big)^{\frac{k}{2k}} \\
   &= \prod_p \Big(\frac{p^2}{p^2-1}\Big)^{1/2} = \sqrt{\zeta(2)}  
     = \frac{\pi}{\sqrt{6}}. 
\end{align*}
Has this result due to Scholz been studied anywhere?
Scholz also believed that if $S$ denotes the set of all fundamental
discriminants, then the corresponding limit is equal to
$$ \prod \Big( \frac{p^2}{p^2-1} \Big)^{\frac{p}{2p+2}}. $$

REPLY [15 votes]: The distribution of values of $L(1,\chi_d)$ as $d$ varies over fundamental discriminants has been extensively studied.  For example, see this paper of Granville and Soundararajan which gives uniform such results (and discusses other references and history).  The main result there shows that $L(1,\chi_d)$ is distributed like a random Euler product $L(1,X)= \prod_p (1-X(p)/p)^{-1}$ where $X(p)$ for primes $p$ denote independent random variables with $X(p)=1$ with probability $p/(2(p+1))$, $-1$ with probability $p/(2(p+1))$ and $0$ with probability $1/(p+1)$.   From this the last assertion you make about fundamental discriminants follows: you want to compute (for large $D$)
$$ 
\exp\Big( \frac{1}{|\{|d|\le D\}|} \sum_{|d|\le D} \log L(1,\chi_d) \Big)
\sim \exp\Big({\Bbb E} (\log L(1,X)) \Big) = \prod_p \Big(\frac{p^2}{p^2-1}\Big)^{\frac{p}{2p+2}}.
$$
Small modifications to the same techniques would allow you to study the family of prime discriminants that you mentioned -- the only difference is in adjusting the probabilistic model to reflect the fact that very few prime discriminants will be divisible a given prime $p$ (as opposed to all fundamental discriminants where this proportion is $1/(p+1)$).<|endoftext|>
TITLE: What is $\infty^6$?
QUESTION [23 upvotes]: The title of this question may make you want to close it immediately, but bear with me a moment.  In several older mathematics papers (early 20th century) I have seen statements such as

The motions of three-dimensional space are $\infty^6$.

I am curious what this means.  From context I guess that "being $\infty^6$" means something roughly like what we would nowadays call "being a 6-dimensional manifold".  But did it have a precise meaning?  Who defined it, where and when?  When did it fall out of use?

REPLY [19 votes]: Yes, indeed, this notation was used to state the dimension of the manifold. The idea of dimenson is very intuitive but it took long time and a lot of labor to formalize. Before the modern definitions of "dimension", "manifold" and "homeomorphism" were spread, people, especially geometers, expressed this fact by saying that something depends on $n$ parameters, or is an "n"-parametric family, and wrote $\infty^n$. This notation is out of date now, but they still speak of $n$-parametric families. When I was an undergraduate student (early 1970s) $\infty^n$ was still used in the lectures on projective geometry, for example. I was puzzled because I knew that Cantor proved that all these things have the same cardinaity, until I read about Brouwer's theory of dimension and "domain preservation property". 
(About a century before, Cantor and Poincare were similarly puzzled until Brouwer clarified the thing:-) I think Cantor was the first who tried to prove that manifolds of different dimensions are not homeomorphic, but he failed.
(Before Cantor it seemed evident to everyone that plane contains more points than the line, I do not know whether anyone cared to prove this, but those who did try, could not).<|endoftext|>
TITLE: Automorphism of genus 2 surface with 5 fixed points
QUESTION [7 upvotes]: Is there a self-homeomorphism of a genus 2 (closed, orientable) surface, which has finite order and exactly 5 fixed points?
Of course, the same question can be asked replacing 2 by $g$ and $5$ by any number $k$. An upper bound for possible values of $k$ is (generalized Lefschetz fixed point theorem) $2g+2$. 
For $g=0$, a sphere, only $k=0,2$ are possible.
For $g=1$, the torus, $k=0,1,2,3,4$ are all possible. For example, the map $(x,y) \mapsto (y,-x-y)$ is of order 3 and has exactly 3 fixed points.

REPLY [2 votes]: Although Jason and Dylan seemed to have answered this question in the comments, I decided to work out what the generalized version of this kind of statement is.
Let $\sigma$ be an automorphism of a Riemann surface. Any automorphism that is orientation-reversing with a fixed point has a fixed curve, so let's assume $\sigma$ is orientation-preserving.
By Rieman-Hurwitz, if $\sigma$ has order $k$ and $n$ fixed points than:
$$ 2k -n (k-1) \geq 2 - 2g$$
$$n \leq \frac{ 2k + 2g-2}{k-1}= 2+ \frac{2g}{k-1}$$
Moreover if $k$ has order $2$ then the number of fixed points is congruent to $2g+2$ modulo $4$. So this already rules out many possibilities.
In particular, for $g=2$ this rules out $5$. 
There are other congruence conditions. If $k$ is a power of a prime $p$ then all the terms in the Riemann-Hurwitz formula are multiples of $p$ except for the $-n(k-1)$ and $2-2g$ so we get $n \equiv 2-2g $ modulo $p$.
so the list of possibilities becomes something like:
$n \leq 2g+2$ and congruent to it mod $4$ (order $2$)
$n \leq g+2$ and congruent to it mod $3$ (order $3$)
$n \leq 2g/3+2$ and even (order $4$)
$n \leq g/2+2$ and congruent to $2-2g$ modulo $5$ (order $5$)
$n \leq g/5+2$ with no congruence condition (order $6$)
I think you can show that almost all of these are actually achieved, by solving the Riemann-Hurwitz equation, drawing the right orbifold, and using the formula for the orbifold fundamental group.<|endoftext|>
TITLE: Why is every variety of bands determined by a single identity?
QUESTION [17 upvotes]: A band is a semigroup where every element is idempotent: $a a = a$.  A collection of bands is a variety  if it is closed under taking subobjects (subsemigroups), quotient objects (images of homomorphisms) and products (including infinite products).  As usual in universal algebra, any variety of bands is defined by a collection of identities.  For example, there's a variety of semilattices, which are the bands obeying the identity
$$ a b = b a $$
for all $a,b$.  
You can define a variety of bands using more than one identity.  But amazingly --- to me, at least --- every variety of bands can in fact be defined using just one identity! 
(That is, one identity in addition to associativity $(a b) c = a (b c)$ and the idempotence law $a a = a$.)
This was shown here:

Charles Fennemore, All varieties of bands, Semigroup Forum 1 (1970), 172-179.

He even showed that there are exactly $8 + 10(n-2)$ varieties of bands that are determined by an identity involving $n$ variables, for $n \ge 2$... and he has a method for listing these identities explicitly.
The set of varieties of bands is partially ordered, based on whether one identity implies another.  It's actually a lattice, as usual in universal algebra, and it seems that Fennemore described the lattice operations explicitly in terms of his chosen identities.
A small portion of this lattice is shown on Wikipedia:

I'm wondering if there's a good explanation of 'why' any variety of bands can be defined using just one identity.  Fennemore's proof is not easy for me to follow.  Are there are other varieties, besides bands, that have this property?

REPLY [8 votes]: Here is a partial answer.
Claim. If $\mathcal V$ is an idempotent variety, then any subvariety of $\mathcal V$ that is finitely axiomatizable relative to $\mathcal V$  is axiomatizable by a single identity and the identities of $\mathcal V$.
Apply the claim to the variety of bands. This doesn't explain why all varieties of bands are finitely axiomatizable, but it does explain why only one identity is needed in the finitely axiomatizable cases.
Here is the idea behind the claim. 
Suppose that $s=s'$ and $t=t'$ are two identities, where $s, s'$ are terms in, say, two variables, and $t, t'$ are terms in three variables. If $\mathcal V$ is idempotent, then (modulo the identities of $\mathcal V$) the set $\{s=s', t=t'\}$ is equivalent to the single identity
$$
s(t(u,v,w),t(u',v',w'))=s'(t'(u,v,w),t'(u',v',w')).
$$
To explain this, let $M$ be the $2\times 3$ matrix
$$
M=\left[
\begin{matrix}
u&v&w\\
u'&v'&w'
\end{matrix}
\right]
$$
where the entries are distinct variables. Define a 6-ary term $s\diamond t(M)$
by applying $t$ to the rows of $M$ and then $s$ to the row results. I must explain why $\{s=s', t=t'\}$ is equivalent to $\{s\diamond t(M)=s'\diamond t'(M)\}$.
To derive the diamond identity from the original two,  argue from $\{s=s', t=t'\}$ that 
$$
s\diamond t(M) = s\diamond t'(M) = s'\diamond t'(M).
$$
To derive $s=s'$ from the diamond identity, just set variables in $M$ equal along rows, say equal to the first entry: from $s\diamond t(M)=s'\diamond t'(M)$ you obtain $s(u,u')=s'(u,u')$. To derive $t=t'$ from the diamind identity, set variables in $M$ equal along columns. (This paragraph is the part that needs idempotence.) 
[Note: lattices are defined with two band operations, and some varieties of lattices are not finitely axiomatizable, so there is still some kind of magic involved in the full result about bands.]<|endoftext|>
TITLE: q-Catalan numbers from Grassmannians
QUESTION [22 upvotes]: In this question by $q$-Catalan numbers I mean the $q$-analog given by the formula $\frac{1}{[n+1]_q}\left[{2n\atop n}\right]_q$. The polynomial $\left[{2n\atop n}\right]_q$ represents the class of the Grassmannian $G(n,2n)$ in the Grothendieck ring of varieties, and $[n+1]$ represents the class of $\mathbb P^n$. Is there a geometric reason why the fraction $[G(n,2n)]/[\mathbb P^n]$ is a polynomial in $[\mathbb A^1]$?
I guess one could ask more generally about why $\frac{[\mathbb P^r][G(k,2k+r)]}{[\mathbb P^k]}$ is a polynomial.

REPLY [12 votes]: There are a few nice answers to related questions. Unfortunately none of them quite answers the question you asked.

The $q$-Catalan number $\frac{1}{[n+1]_q}{ 2n \brack n}_q$ is the Hilbert series of a fairly natural graded representation of the symmetric group $S_n$ coming from an irreducible representation of a rational Cherednik algebra. This was originally proved by Berest-Etingof-Ginzburg and greatly generalized by Gordon-Griffeth:

http://arxiv.org/abs/0912.1578

Define the "other" $q$-Catalan number as the sum of $q^{|D|}$ where $D$ ranges over "Dyck paths" from $(0,0)$ to $(n,n)$ staying weakly above the diagonal and where $\binom{n}{2}-|D|$ is the number of unit squares between $D$ and the diagonal. Gorsky-Mazin proved that this other $q$-Catalan number evaluated at $t^2$ is the Poincare series of the "Jacobi factor" of the plane curve singularity $x^n=y^{n+1}$:

http://arxiv.org/abs/1105.1151

The whole story generalizes to the "rational $(q,t)$-Catalan numbers" $\mathrm{Cat}_{a,b}(q,t)$ where $a,b$ are positive coprime integers. The $q$-Catalan you mentioned comes from the formula $$q^{(a-1)(b-1)/2}\mathrm{Cat}_{a,b}(q,q^{-1})=\frac{1}{[a+b]_q}{ a+b \brack a}_q$$ by setting $(a,b)=(n,n+1)$ and the "other" $q$-Catalan number comes from setting $t=1$. The rational $(q,t)$-Catalan numbers are related to many things including the HOMFLY-PT polynomial of torus knots.

See here for some expositions:
http://www.math.miami.edu/~armstrong/Talks/RCC_AIM.pdf
http://thales.math.uqam.ca/~nwilliams/docs/AIM%202012/RCCAIMOutlineOnline.pdf
http://aimath.org/pastworkshops/rationalcatalanrep.pdf
https://www.math.ucdavis.edu/~egorskiy/Presentations/qtcat.pdf<|endoftext|>
TITLE: How much redundancy resides in an $n \times n$ orthogonal matrix?
QUESTION [19 upvotes]: Suppose one has an $n \times n$ orthogonal matrix $M$:
$$
\left(
\begin{array}{ccc}
 0.239326 & 0.846726 &
   0.475161 \\
 0.768893 & 0.13356 &
   -0.625272 \\
 0.592897 & -0.514992 &
   0.619077 \\
\end{array}
\right)
$$
Because it is orthogonal, $M^T M = I$.
Suppose one entry of $M$ is erased, say $M(2,2)$:
$$
\left(
\begin{array}{ccc}
 0.239326 & 0.846726 &
   0.475161 \\
 0.768893 & \color{red}{x} &
   -0.625272 \\
 0.592897 & -0.514992 &
   0.619077 \\
\end{array}
\right)
$$
It can be recovered from $M^TM=I$. For example, we must have the $(2,2)$ entry 
of $M^TM$ equal to $1$:
\begin{eqnarray}
0.982162 + x^2 &=& 1 \\
x &=& \pm 0.13356
\end{eqnarray}
and then, e.g., entry $(2,1)$ of $M^TM$ disambiguates (or determines on its own):
\begin{eqnarray}
-0.102694 + 0.768893 x &=& 0 \\
x &=& 0.13356
\end{eqnarray}
My question is:

Q. What is the maximum number $k$ of entries of an 
  $n \times n$ orthogonal matrix $M$
  that can be erased and then uniquely recovered, 
  knowing only that $M$ is orthogonal?

It could well be that $k$ depends on which entries are erased, which itself
could be interesting. But I am at the moment seeking the maximum of $k$ over
all possible entries that permits exact recovery.

REPLY [2 votes]: The fact that $\frac{n(n-1)}2$ well chosen coefficients determine the other ones is supported by the Schur parametrization : let $U$ be unitary (it works as well for real orthogonal matrices) and Hessenberg (that is $i\ge j+2$ implies $u_{ij}=0$). Up to multiplication by a diagonal unitary matrix, we may assume that the diagonal of $U$ is real non-negative. Then there are unique pairs $(a_k,b_k)$ with $b_k$ real and $|a_k|^2+b_k^2=1$, such that $U=G_1\cdots G_{n-1}$ with
$$G_k={\rm diag}\left(I_{k-1},\begin{pmatrix} -a_k & b_k \\ b_k & \overline{a_k} \end{pmatrix},I_{n-k-1}\right).$$
It is a case where one gives the sub-diagonal part of $U$ (with many zeroes).The pairs are determined by the entries $u_{i,i-1}$.
Notice that in the unitary case, the dimension of the tangent space $SK_n$ over $\mathbb R$ is $n^2$, and we are given $n^2$ real data, namely  $\frac{n(n-1)}2$ complex numbers (sub-diagonal entries) and $n$ real numbers (the imaginary parts of the diagonal entries).<|endoftext|>
TITLE: When is $(-1+\sqrt[3]{2})^n$ of the form $a+b\sqrt[3]{2}$?
QUESTION [6 upvotes]: When is $(-1+\sqrt[3]{2})^n$ of the form $a+b\sqrt[3]{2}$ ($n$ being an integer) , i .e.,  when does $(-1+\sqrt[3]{2})^n$ not have a non-zero term in $\sqrt[3]{4}$. As you might have noticed, I'm interested in solving the diophantine equation $x^3-2y^3=1$ using this specific method. Is there any way I can use Skolem's p-adic method here?

REPLY [13 votes]: In section 4 of my book on the Catalan conjecture, I present such a proof (for $p=3$).  I took it from Bill McCallum's 1977 honours project at the University of Sydney<|endoftext|>
TITLE: Anti-Mandelbrot set
QUESTION [24 upvotes]: I clearly remember seeing a paper where the dynamic of the anti-conformal map
$f(z)=\overline{z}^2+c$ was studied (the bar means complex conjugation). There was a picture of the analog of the Mandelbrot set, which looks very different from the usual one.
Can someone help me to locate this paper ?
EDIT. Thanks to Benjamin Dickman. Another common name is Mandelbar, which escaped my memory.
Once you know these two names, Tricorn and Mandelbar, the search becomes trivial.
EDIT 2. More information. Apparently this setting was introduced in the paper "On the structure of Mandelbar set by four authors, in Nonlinearity, 2 (1989), 541-553. I contacted one of the authors, and he told me the following story: someone wanted to draw a Mandelbrot set with a computer, but made a misprint in the program. The result looked interesting...
Tricorn was discovered by Milnor (independently of this) in his experiments with a holomorphic (cubic) family.
EDIT 3. If someone is curious why I asked this, see the Comment in the end of this preprint: http://arxiv.org/abs/1507.01704

REPLY [22 votes]: Perhaps the key term is tricorn? See Inou's Self-similarity for the tricorn (arXiv pdf) and its references.
Sample excerpt:<|endoftext|>
TITLE: discrete valuation ring and ring of witt vectors
QUESTION [5 upvotes]: Given a perfect field $F$ of prime characteristic the ring of Witt Vectors $W(F)$ is a discrete valuation ring. For example, $W(\mathbb{F}_p)$ is the ring of $p$-adic integers. Is it possible to embed an arbitrary unramified discrete valuation ring of mixed characteristic into  a Witt Vector ring of a perfect field?

REPLY [6 votes]: I think this is possible, if I understand your question correctly. If $R$ is a discrete valuation ring of mixed characteristic $(0,p)$ with residue field $k$ and maximal ideal $pR$, then $R$ is a Cohen ring for $k$. Cohen rings are unique up to (generally non-unique) isomorphism, and there is a construction of the Cohen ring for any field $k$ of characteristic $p$ which realizes the ring as a subring of $W(k)$. Now $W(k)$ is a subring of $W(k^{p^{-\infty}})$, where $k^{p^{-\infty}}$ is the perfect closure of $k$.<|endoftext|>
TITLE: About the prime divisors of values of polynomials
QUESTION [16 upvotes]: Let $P$ be a polynomial having integer coefficients (and degree $\geq 3$), and let $\mathscr P_P$ be the set of prime numbers dividing some value $P(n)$ with $n \in \mathbb Z$.

Is it true that  $\sum_{p \in \mathscr P_P} \frac1 p$ diverges?

The case of polynomials with degree $2$ can be studied by Dirichlet's theorem on arithmetic progressions (I suppose a quadratic-residue argument can be applied for every case), but I am not sure how to study the cases of greater degree.  
Any reference or proof (or disproof, which, by the way, would be very surprising) would be appreciated.

REPLY [11 votes]: I don't think one needs to invoke the Chebotareev Density Theorem here. 
Assume that the given sum converges, and let $K$ be a number field generated by any root of $P(x)$. Then, by assumption, the sum of $\mathrm{Norm}(\mathfrak{p})^{-1}$ over the degree $1$ prime ideals $\mathfrak{p}$ in $K$ converges, hence the same holds for the sum of $\mathrm{Norm}(\mathfrak{p})^{-1}$ over all prime ideals $\mathfrak{p}$ in $K$. This implies that the Dedekind zeta function of $K$,
$$ \zeta_K(s)=\prod_{\mathfrak{p}}\frac{1}{1-\mathrm{Norm}(\mathfrak{p})^{-s}},\qquad \Re(s)>1, $$ 
tends to a finite limit as $s\to 1+$, while it is known that $\zeta_K(s)$ has a simple pole at $s=1$.<|endoftext|>
TITLE: Do discrete groups with property $(T)$ have "modest" subgroup growth?
QUESTION [6 upvotes]: I saw it conjectured at http://www.mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0309.0317.ocr.pdf that "discrete subgroups with property $(T)$ may have modest subgroup growth." (Page 5, directly above the heading for section 4.)
According to the author, this conjecture was supported by some examples. 
I have three questions:

What are these examples?
Has any progress been made on this conjecture - are there nontrivial bounds on the subgroup growth of a discrete group with property $(T)$? (What are the trivial bounds?)
What do we expect "modest growth" to mean?

REPLY [13 votes]: This is now outdated: in 1994 there were very few known sources of Property T groups: lattices on the one hand (for which congruence subgroup property was true or unknown, and hyperbolic groups with no information on their finite index subgroups). This has dramatically changed. Restricting to cases for which we have information about finite quotients, we have, for instance

$\mathrm{SL}_3(\mathbf{Z}[X])$ is known to have Property T.
Also there are Golod-Shararevich groups with Property T (Ershov, see here
). See Jaikin-Zapirain's appendix to Ershov-Jaikin's paper showing that the subgroup of Golod-Shafarevich groups is large.<|endoftext|>
TITLE: Three-manifolds having a Reebless foliation but not a taut one
QUESTION [14 upvotes]: A straightforward argument reveals that a taut foliation is Reebless, and of course there are many examples of Reebless foliations that are not taut. I guess that there are many examples of three-manifolds having a Reebless foliation but not a taut one.
Looking for an example I found the following one. The 3-manifold obtained by doing $+37/2$-surgery along the $(-2, 3 ,7)$-pretzel knot contains a Reebless foliation (essentially because it contains an incompressible torus whose complement consists of two copies of the complement of the trefoil knot) but  does not contain a taut foliation since it is a Heegaard Floer L-space ($+18$-surgery along $K$ gives a lens space, and an L-space is forever!*).
So my main questions are: 
1) Is there a classical proof (i.e. that does not make use of Heegaard Floer homology) of the fact that the $+37/2$ surgery along $K$ does not support a taut foliation?
2) Can someone give me other (in some sense simpler) examples of three-manifolds having a Reebless foliation but not a taut one?
3) Are there examples in which the use of Heegaard Floer homology can't be apparently bypassed?
*i.e. if $S^3_n(K)$ is an L-space, then so is $S^3_q(K)$ for every rational number $q\ge 2g(K)-1$.

REPLY [5 votes]: I don't have definitive answers to your questions, but I'll try to make some comments. 
For 1), I haven't done a literature search to see if this was known. However, Boyer and Clay propose a related open question Problem 1.11. 
For 2), as I mentioned in the comments above, any irreducible manifold $M$ containing an incompressible torus admits a Reebless foliation. This follows from Theorem 5.1 of Gabai's paper. Cut $M$ along an incompressible torus to get manifolds $M_1, M_2$. Then let these be taut sutured manifolds with boundary being $R_+$. By Theorem 5.1, there is a foliation containing $\partial M_i$ as leaves, and which is taut in the interior (hence Reebless). Then glue these two foliations together to get a Reebless foliation of $M$. A caveat here is that the foliation might not be smooth. 
Hence, double branched covers of prime alternating knots which have a Conway sphere admit Reebless foliations but not taut ones. The preimage of a Conway sphere is an incompressible torus, but by Ozsvath-Szabo, the double branched cover is an L-space, so does not admit a (smoothe co-oriented) taut foliation. 
Simpler examples might come from manifolds admitting taut non-oriented foliations, but no taut oriented foliation, for example sol manifolds that semi-fiber (with trivial first betti number). 
For 3), I would suggest that the answer is yes, in the sense that before the work of Ozsvath and Szabo, we knew of very few classes of manifolds not admitting taut oriented foliations (hence, it appears that for most of their examples, Heegaard Floer homology cannot be bypassed). We knew that there was an algorithm to (in principle) detect if a given manifold admits a taut foliation, but it is not practical to implement. Something akin to this was implemented to find infinite collections of manifolds admitting no taut foliation (see also Calegari-Dunfield). The big open question though is whether the converse might hold: if (irreducible orientable) $M$ does not admit a taut (smooth co-oriented) foliation, then is it an L-space? I think Juhasz may have posed this as a question or conjecture.<|endoftext|>
TITLE: Distribution of Mordell–Weil ranks of higher genus curves
QUESTION [9 upvotes]: By "nice curve", I mean a smooth, projective, geometrically integral curve over $\newcommand{\Q}{\mathbb{Q}}\newcommand{\Jac}{\operatorname{Jac}}\Q$ with at least one $\Q$-rational point. The Mordell–Weil rank $r(C)$ of a nice curve $C$ is the rank (as an abelian group) of $\Jac(C)(\mathbb{Q})$, the group of $\Q$-rational points of the Jacobian of $C$.
Let $g$ be a positive integer, and consider the distribution of $r(C)$ among nice curves $C$ of genus $g$. For $g = 1$, the well-known "minimalist conjecture" says that $r(C) \leq 1$ for $100\%$ of elliptic curves, with $0$ and $1$ each having $50\%$ probability.
For fixed $g \geq 2$, what conjectures are there about the distribution of $r(C)$ among nice curves of genus $g$? What heuristics or evidence support such conjectures? (And, if it's not too speculative, what if we replace $\Q$ with some other number field?)

REPLY [5 votes]: General Katz-Sarnak heuristics suggest that the analogue of the minimalist conjecture should still be true. Let me sketch the reason why from two perspectives - the function field model, where we can establish a version of the conjecture, and the Sarnak-Shin-Templier conjectures on families of automorphic forms.
First note that it is not obvious what you mean by a random nice curve of genus $g$, because there is not always a clear height function on the moduli space of smooth curves of genus $g$. Moreover it is not always rational, so we do not always know how to count curves of bounded height!
It's safer to choose some parameterized family of curves, like hyperelliptic curves or plane curves, and ask for the distribution in that. However, my answers will not depend too much on the family.
In both cases, we will assume the truth of BSD and show that 50% of abelian varieties have analytic rank 0 and 50% have analytic rank 1 (under some heuristic.)
In a function field, an abelian variety $A$ over $\mathbb F_q(T)$ has analytic rank equal to the dimension of the Frobenius-invariant subspace of $H^1(\mathbb P^1_{\overline{\mathbb F_q}},  T_\ell(A))$ where $T_\ell(A)$ is the $\ell$-adic Tate module of $A$. Our family of curves will be parameterized by a space like $Maps(\mathbb P^1, \mathbb P^n)$ where $\mathbb P^n$ parameterizes our family of curves over $\mathbb F_q$. This cohomology group  $H^1(\mathbb P^1_{\overline{\mathbb F_q}},  T_\ell(A))$ forms a sheaf on this mapping space and we may compute its monodromy group.
If the monodromy group on a component is the full orthogonal group, the minimalist conjecture is true in the $q \to \infty$ limit for that component, because the distribution of Frobenius is as a random element in the monodromy group, and a random orthogonal matrix has a $0$-dimensional invariant subspace with probability $1/2$ and a $1$-dimensional invariant subspace with probability $1/2$. This argument is explained in Katz's book Twisted L-Functions and Monodromy.
For an explicit family it's not too hard to show that the monodromy group is orthogonal using the moment method. To show the monodromy group contains the special orthogonal group, it's sufficient to show that the first few moments agree with the moments of the orthogonal one. This can be done for sufficiently large degree maps from $\mathbb P^1$ to $\mathbb P^n$ using an independence argument, as long as $H^{2n-1}(\mathbb P^n_{\overline{\mathbb F_q}}, T_\ell(A))$ vanishes, which can be checked for the universal families of hyperelliptic and complete intersection curves by another independence argument. This sort of argument is explained in Katz's book Moments, Monodromy and Perversity.
In the note "Families of L-Functions and their Symmetry", Sarnak, Shin, and Templier conjecture equidistribution results for the L-functions of certain families of automorphic forms. Given any family of curves parameterized by an open subset of $\mathbb A^n$ (e.g. hyperelliptic, plane curves, complete intersection) the (conjectural) automorphic forms associated to their first etale cohomology groups are clearly a "geometric family" by the definition of the authors. In any of these families the local factors will be equidistributed in $SP_{2g}$, because a hyperelliptic/plane/complete intersection curve over a finite field has characteristic polynomial of Frobenius equidistributed in $SP_{2g}$.
Furthermore in their notation the rank of the family will be $0$ - this is actually the same $H^{2n-1}(\mathbb P^n_{\overline{\mathbb F_q}}, T_\ell(A))$ vanishing condition as before and can be checked in the same way.
Also, the $\epsilon$ factors should be equidistributed between $+1$ and $-1$ - I don't immediately see a reason for this but I don't see a reason why not.
Hence assuming Conjecture 2, the distribution of the low-lying zeroes follow the $SO_{even}(\infty)$ law half the time and the $SO_{odd}(\infty)$ law half the time. The first one has average analytic rank $0$ and the second has average analytic rank $1$.<|endoftext|>
TITLE: Homology class of a Lagrangian Klein bottle
QUESTION [9 upvotes]: Nemirovskii's 2008 paper, by the same title in this question, claims that any Lagrangian Klein bottle in a closed symplectic 4-manifold $M$ must realize a nontrivial homology class in $H_2(M; \mathbb Z/2\mathbb Z)$.  Unfortunately the paper is known to be flawed, as it would imply (as explained in the paper) that there are no Lagrangian Klein bottles in $\mathbb R^4$, and this predates the accepted proofs.  I haven't seen it claimed in another reference, but I am also not sure which references to trust.  So my question is: does anyone know if this theorem is true and where it is proven?
A secondary question to which I would love an answer just as much is: can anyone make more examples of Lagrangian Klein bottles in four dimensions?
I have just one: a construction of Lagrangian Klein bottle in $S^2\times D^2\subset S^2\times S^2$ with some area constraints, I briefly described it in an answer to Lagrangian Kleinian bottles and then realized I should ask this as a question.  This example does represent the nontrivial second homology class of $S^2$ with $\mathbb Z/2\mathbb Z$ coefficients.  I would love to know of more consructions.

REPLY [7 votes]: The homology class of a Lagrangian Klein bottle is non-zero in any ruled symplectic four-manifold, e.g. in $S^2\times S^2$ with a product symplectic form. This was first proved by Shevchishin (Izvestia Math., 2009, 73:4, 797-859) and then a shorter proof was given by Nemirovski (Izvestia Math., 2009, 73:4, 689-698). 
Nemirovski's earlier claim (Izvestia Math., 2002, 66:1, 151-164) that this holds for all closed symplectic manifolds is known to be false. An example of a Lagrangian Klein bottle with trivial homology class in a non-ruled symplectic four-manifold can be found in Section 1.6 of Shevchishin's paper.<|endoftext|>
TITLE: divisors of $p^4+1$ of the form $kp+1$
QUESTION [23 upvotes]: In group theory the number of Sylow $p$-subgroups of a finite group $G$, is of the form $kp+1$.
So it is interesting to discuss about the divisors of  this form. As I checked it seems that for an odd prime $p$, there is not any divisor $a$ of $p^4+1$, where $1<a<p^4+1$ and $a=kp+1$, for some $k>0$.
Could you help me about this question? If it is true how we can prove it?
Also when I checked the same fact for $p^8+1$, we get many counterexamples. What is the difference between $4$ and $8$?

REPLY [35 votes]: There's none indeed.
Lemma: if $1<m<n$ are coprime integers then $mn+1$ does not divide $n^4+1$.
First observe that for any $m,n$, of $mn+1$ divides $n^4+1$, then it divides $n^4m^4+m^4=((nm)^4-1)+1+m^4$, and since $mn+1$ clearly divides $((nm)^4-1)$, we deduce that it also divides $1+m^4$.
To prove the lemma, assume the contrary. As we have just seen, $mn+1$ divides $m^4+1$ as well. Write $(m^4+1)/(mn+1)=k$, and $k=(\ell m+r)$ with $0\le r\le m-1$. Then $m^4+1=(\ell m+r)((mn+1)=mN'+r$, so $r=1$. So $(\ell m+1)$ divides $m^4+1$. Then $m$ and $\ell$ are coprime: indeed, we have $(m\ell +1)(mn+1)=m^4+1$, so $\ell(mn+1)-m^3=-n$. If a prime $p$ were dividing both $m$ and $\ell$ then it would also divide $n$, contradicting that $m$ and $n$ are coprime, so $m$ and $\ell$ are indeed coprime. We have $\ell<n$ because otherwise
$$m^4+1=(mn+1)(m\ell+1)\ge (mn+1)^2> (m^2+1)^2>m^4+1.$$ So we found a new pair $(m,\ell)$ with $\max(m,\ell)<n$, with $m\ell+1$ dividing both $m^4+1$ and $\ell^4+1$. So, assuming that $n$ is minimal, we're done unless $\ell=1$. This happens if $m+1$ divides $m^4+1$, and since $m+1$ divides $m^4-1$ as well, if this occurs then $m+1=2$, contradicting $m>1$.
(Note: without the coprime assumption the conclusion fails, as $(m,n)=(m,m^3)$ for $m\ge 2$, e.g. $(m,n)=(2,8)$, satisfies $mn+1|n^4+1$.)
Proposition If $p$ is prime then $p^4+1$ has no divisor of the form $kp+1$ except $1$ and $p^4+1$.
Proof: write $p^4+1=(kp+1)(k'p+\ell)$ with $0\le\ell\le p-1$; then $\ell=1$. So exchanging $k$ and $k'$ if necessary we can suppose $k\le k'$. 
If by contradiction $k$ is divisible by $p$ then $k'\ge k\ge p$, so $p^4+1\ge (p^2+1)^2>p^4+1$, contradiction. So $k$ and $p$ are coprime, and the lemma yields a contradiction.

REPLY [2 votes]: Let $p$ be prime. Assume that $q \mid p^4+1$, where $q = ap+1$ for some
$a \in \mathbb{N}$.
Then $(p^4+1)/q \equiv 1 \!\! \mod p$, hence
$p^4+1 = (ap+1)(bp+1)$ for some $b \in \mathbb{N}$.
Now we have $p^4 = abp^2+(a+b)p$, respectively, $p^3-abp-a-b = 0$. Therefore, effectively
we are looking for solutions to the diophantine equation $x^3-xyz-y-z=0$ in positive
$x$, $y$, $z$, and prime $x$.
When allowing negative (or zero) $x, y, z$, we would get a lot of solutions,
and for positive $x, y, z \leq 1000$, we still find (up to switching of $y$ and $z$)
$$
  (x,y,z) \in \{(8,30,2),(27,240,3),(30,112,8),(112,418,30)\}.
$$
I don't know whether there is a solution with prime $x$ as required --
however we have a diophantine equation of total degree $3$ in $3$ variables,
and such are (at least potentially) hard.<|endoftext|>
TITLE: Exceptional values of real-valued functions on [0,1]
QUESTION [8 upvotes]: Given a continuous real-valued function $f$ from $[0,1]$ to itself with $f(0)=0$ and $f(1)=1$ such that $f^{-1}(c)$ is finite for all $c$ in $[0,1]$, let $E(f)$ be the set of $c$ in $[0,1]$ such that $|f^{-1}(c)|$ has (positive) even cardinality. Call a subset of $[0,1]$  an exceptional set if it is of the form $E(f)$ for some $f$ satisfying the stated conditions. How big can an exceptional set be? I intend "big" to be vague; you can interpret it in the sense of cardinality,  measure, category, or Hausdorff dimension. It is not hard to construct an exceptional set that is countably infinite (I believe this is an exercise in Spivak's calculus book).

REPLY [6 votes]: $E(f)$ is at most countable. An elementary argument based on the intermediate value theorem shows that any $c\in E(f)$ must be the value of a (strict) local extremum of $f$, at one of the points where this value is taken. However, if $c$ is a local maximum, say, then we can find $a,b\in\mathbb Q$ which let us recover $c$ as $c=\max_{a\le x\le b}f(x)$, so there are only countably many such values $c$.<|endoftext|>
TITLE: A finitely presented group with two simple relations
QUESTION [6 upvotes]: Is the group $G$ with the presentation $\langle x,y \;|\; x^7=1, y^2 x y=x^4\rangle$ solvable?  infinite?
I have computed by GAP the following fators of the derived series of $G$:
$G/G'\cong C_3 \times C_7$, $G'/G'' \cong C_2 \times C_2 \times C_2 \times C_7$, and $G''/G^{(3)}$ and $G^{(3)}/G^{(4)}$ are elementary abelian $2$-groups of rank $8$ and $16$, respectively. I couldn't go further, it apparently needs more time and ....
Maybe a simple trick needs here.

REPLY [7 votes]: The group $G$ is not solvable, since its quotient
$$
  \tilde{G} := \langle x, y \ | \ x^7 = 1, y^2xy = x^4, y^{15} = 1\rangle
$$
is a group of order $423360$ such that $\tilde{G}'' \cong {\rm PSL}(3,4)$.<|endoftext|>
TITLE: Is the set of multiplicatively even numbers thick?
QUESTION [13 upvotes]: A positive integer is multiplicatively even (odd) if, when decomposed into primes, the sum of the exponents is even (odd).
A subset of the integers is thick if it contains arbitrarily long intervals $\{n,n+1,\cdots,n+m\}$.
Is the set of multiplicatively even numbers thick?
It is easy to see that the set of multiplicatively even numbers is thick if and only if its complement (i.e. the set of multiplicatively odd numbers) is.

REPLY [17 votes]: You are asking about sign patterns of the Liouville function.  While conjecturally all finite sign patterns appear infinitely often, this has been surprisingly hard to establish.  One of the best results thus far is by Hildebrand, who showed that every sign pattern of length three occurs infinitely often, in particular the set of "multiplicatively even" or "multiplicatively odd" numbers contain infinitely many intervals of the form $\{n,n+1,n+2\}$.  The length four analogue of this remains open.
On the other hand, Szemeredi's theorem (and the prime number theorem) implies that both the set of multiplicatively even numbers and multiplicatively odd numbers contain arbitrarily long arithmetic progressions. (In fact, by using some rather complicated results of myself, Ben Green, and Tamar Ziegler, one can in fact show that all sign patterns occur for the Liouville function in arithmetic progressions, extending a previous result of Buttkewitz and Elsholtz, though this does not directly help with the current question.)
By the way, I don't see the "easy" argument you claim that the set of multiplicatively even numbers is thick if and only if the set of multiplicatively odd numbers is.  Could you elaborate?  (The only thing I can think of is that you are working with all integers and considering -1 to be a prime, but it appears that you are explicitly restricting attention to positive integers in your question.)  Never mind, it is obvious - every interval of length $2k$ contains the double of an interval of length $k$.<|endoftext|>
TITLE: The (Hecke) double coset von Neumann algebra
QUESTION [6 upvotes]: It it well-known in the von Neumann algebra theory that for $\Gamma$ a non-trivial countable group, the von Neumann algebra $L(\Gamma)$ generating by $\Gamma$ acting by left multiplication on $l^2(\Gamma)$, is a ${\rm II}_1$ factor iff $\Gamma$ is an ICC group.   
Now let $(G \subset \Gamma)$ be an inclusion of a finite group $G$ in a countable group $\Gamma$, Let $\mathbb{C}(G \backslash \Gamma /  G) $ be the (Hecke) double coset algebra: the subalgebra of $\mathbb{C}G $, generated by the elements $a_{\gamma} = \sum_{\alpha \in G \gamma G} \alpha$ (well-defined because $G$ finite) with $\gamma \in \Gamma$.  Let $L(\Gamma,G)$ be the von Neumann algebra generated by $\mathbb{C}(G \backslash \Gamma /  G) $ acting by left multiplication on $l^2(G \backslash \Gamma)$.
Question: What's the necessary and sufficient condition on $(G \subset \Gamma)$ for  $L(\Gamma,G)$ to be a ${\rm II}_1$ factor?    
In the case that there are inclusions $(G \subset \Gamma)$ with $\Gamma$ ICC, $G \neq \{ e \}$ and $L(\Gamma,G)$ ${\rm II}_1$ factor:
Optional question 1:   Is it true that $L(\Gamma,G) \simeq L(\Gamma)$?   
Optional question 2: How to generalize the construction above for $G$ infinite?

REPLY [5 votes]: $L(\Gamma,G)$ is the algebra of endomorphisms of the representation $l^2( \Gamma/G)$ (with $\Gamma$ acting by left multiplication). This answer your second optional question. Also, by classical results on $W^*$-categories, the category of normal representations of $L(\Gamma,G)$ will be equivalent to the category of unitary representations of $G$ that are retract of sums of copies of $l^2(\Gamma /G)$, which generally allow to determine the type in concrete situations but I don't know what a general criterion would be. In fact, in the general case this algebra can also be of type $III$ which make me think there is no simple criterion.
In the special case where $G$ is finite, $l^2(\Gamma /G)$ is a retract of $l^2(\Gamma)$ hence $L(\Gamma,G)$ is a corner of $L(\Gamma)$. If in addition $\Gamma$ is ICC, then $L(\Gamma)$ is a factor and hence any non trivial corner will be Morita equivalent to $L(\Gamma)$. So $L(\Gamma,G)$ will be Morita equivalent to $L(\Gamma)$ and hence of the same type.
Also note that in this situation $L(\Gamma)$ and $L(\Gamma,G)$ will often be isomorphic, but by a "non natural" isomorphism which will not going to be compatible with the natural Morita equivalence...<|endoftext|>
TITLE: What's the most simple proof of Kostant's version of Borel-Weil-Bott for Lie Algebra cohomology?
QUESTION [12 upvotes]: Besides Kostant's original proof (in http://www.math.tamu.edu/~jml/kostant61.pdf) of the above mentioned theorem (using the Lie Algebra Laplacian), there are a few other approaches:

Casselman-Osborne, using the center of the universal enveloping algebra (https://eudml.org/doc/89271)
Aribaud, using a Lie Algebraic analogue of Riemann-Roch (https://eudml.org/doc/87095)
BGG and Garland/Lepowski, using a tailored resolution (http://www.sciencedirect.com/science/article/pii/002186937790254X)

There is also an (algebro-) geometric proof by Demazure (https://eudml.org/doc/142384) that is amazingly short and simple, but it is not written in the language of Lie Algebras. 
Is there a paper where Demazure's ideas are written up solely in Lie Algebraic terms?
There is a paper by A. Joseph where this is very shortly remarked, but without explicit exposition. 
Also, Demazure treats Bott's theorem, i.e. the Borel subgroup case. Is there a generalization of this proof for the parabolic case?

REPLY [3 votes]: I'm not sure exactly what your header means, but maybe I can suggest partial answers to your questions.   First of all, there are by now many ways to approach the original Borel-Weil theorem, depending on what machinery you are inclined to use.   While it can be formulated in several related settings, this theorem basically provides a model for the finite dimensional highest weight representations of a semisimple algebraic or Lie group in characteristic 0: take global sections of a suitable line bundle on the full flag variety $G/B$.
Bott and Kostant both explored some of the same territory but in different ways. 
Bott's 1957 Annals paper extended the Borel-Weil theorem to other line bundles on $G/B$ and was both analytic and geometric in spirit, but as a consequence he observed that it was possible to derive the dimensions of certain Lie algebra cohomology groups.  Here the flag variety corresponds indirectly to the nilradical of an opposite Borel subalgebra in the corresponding semisimple Lie algebra.      In Kostant's 1961 Annals paper, he worked out more directly some of the related Lie algebra cohomology in the context of Borel subalgebras and their nilradicals (and more generally in terms of parabolic subalgebras).  
The BGG resolution of a simple finite dimensional representation of the Lie algebra later permitted a quick and transparent proof of the Bott-Kostant dimension theorem.   The resolution also has a parabolic version, as in Lepowsky's 1977 paper
here and in Rocha-Caridi's Rutgers thesis supervised by Wallach.   (Later work simplifies much of this, as recorded in my 2008 textbook on the BGG category: see 6.6 and 9.16.)    
As indicated in the question, the original geometric setting of Borel-Weil and Bott was greatly simplified by Demazure in his 1976 paper here.   This work relies on methods of algebraic geometry and only concerns the computations of line bundle cohomology on the flag variety, so it's not clear to me how to connect this directly with the Lie algebra cohomology.   As far as I know, the only relevant literature (due to people like S. Kumar and O. Mathieu in the late 1980s) deals much more generally with Kac-Moody groups and Lie algebras.  Maybe one can extract something from their work in the finite dimensional Lie algebra case which applies to the Bott-Kostant results.
[One side remark is that Henning Haahr Andersen, in his 1977 MIT thesis and in many papers thereafter, explored creatively in Demazure's style the more complicated cohomology of line bundles on flag varieties in prime characteristic.   Parabolic subgroups often play a role, as for example here.  But here one is much farther away from the setting in which Lie algebras play a prominent role.]<|endoftext|>
TITLE: Spaces that can't be embedded in the plane
QUESTION [6 upvotes]: If $X$ and $Y$ are topological spaces, let us write $X \preceq Y$ whenever $X$ embeds in $Y$.
Earlier today, I asked the question:

Is this a well-quasi-order on the completely metrizable spaces?

This was short-sighted, as Tom Goodwillie has pointed out in the comments that the closed surfaces give an easy counterexample.
Since I can't accept Tom's comment as an answer, I'd like to modify the question to make it more interesting (while still being very closely related to the original):

Is there a finite list $F$ of completely metrizable spaces such that, for any completely metrizable space $X$, $X \preceq \mathbb{R}^2$ if and only if $Y \not\preceq X$ for every $Y \in F$?

An affirmative answer would be something analogous to Wagner's Theorem, but with a more topological flavor.
[Considering this question was part of what led me to ask my other question: if embeddability were a wqo (which it isn't), then the answer to the present question would be yes.]
Candidate list: the topological graphs $K_5$ and $K_{3,3}$, the sphere $S^2$, and the subspace of $\mathbb{R}^3$ obtained by taking the X-Y plane and a sequence converging to the origin along the Z axis.
[Notice that every closed surface contains one of these.]

The following was a comment to the original question. It is not relevant to the modified question, but I am keeping it to explain the post of Nash-Williams below:
Embeddability is not a well-quasi-order for metric spaces generally. An easy way to get a counterexample is to build one by transfinite recursion: you can find infinitely many subsets of $\mathbb R$ that violate either/both of the conditions listed above. The examples you build will be very far from $G_\delta$, so not completely metrizable.
Completely ultrametrizable spaces are well-quasi-ordered by embedability. This follows (with a little bit of work) from a version of the Nash-Williams Tree Theorem (see Theorem 11 here), together with the fact that every completely ultrametrizable space can be represented as a tree.

REPLY [2 votes]: You get a positive answer if you restrict to compact, locally connected metric spaces.
See
https://www.semanticscholar.org/paper/On-planarity-of-compact%2C-locally-connected%2C-metric-Richter-Rooney/70090a524f2509408e5170f814ed1b33654bb585
and references therein.<|endoftext|>
TITLE: Embedding linear algebraic groups of a given dimension into a fixed $\mathrm{GL}_N$
QUESTION [10 upvotes]: Given $n$, can $n$-dimensional linear algebraic groups over $\mathbb{C}$ be embedded into $\mathrm{GL}(N,\mathbb{C})$ for a uniformly bounded $N$?
Thanks so much for your reply!

REPLY [9 votes]: As Dave points out, it's very risky to include finite (dimension 0) affine algebraic groups (in the traditional sense) in your question.   So I'm assuming $G$ is connected, in which case the answer to the question is YES.    
Even more than in the case of Cayley's theorem for embedding finite groups into symmetric groups, the old theorem of Chevalley (in arbitrary characteristic) embeds an arbitrary affine algebraic group into a general linear group without any control over the size of that linear group.    But if you appeal to the classication of semisimple groups along with the Levi decomposition in characteristic 0, you can control the size better.   To start with, there are only finitely many (connected) semisimple groups of a given dimension $n$, so that already bounds $N$ for them.   On the other hand, all $n$-dimensional algebraic tori are isomorphic to the direct product of $n$ copies of the multiplicative group, so they are also well-behaved with respect to embedding.    At least in characteristic 0, all $n$-dimensional unipotent groups are isomorphic to vector groups and have bounded embeddings.  Combined with the Levi decomposition over $\mathbb{C}$ (in the algebraic group setting), these individual results and the general Borel-Tits structure theory for reductive groups will provide $N$ indirectly.  
ADDED: My initial thinking about the question was not precise enough to handle unipotent radicals or semidirect product decompositions here.  So I would stick to reductive (connected) groups, where the structure and classification theorems easy to apply.   Aside from that it's worth asking why people don't work more often with affine algebraic groups in linear embeddings: probably it's because such embeddings (though possible by Chevalley's theorem) are too loosely structured and seldom of much help in applications.   Intrinsic properties such as Jordan-Chevalley decomposition and Bruhat decomposition are typically more useful in practice.<|endoftext|>
TITLE: Simple proof for property R conjecture
QUESTION [7 upvotes]: Gabai's property R theorem is:
If the 0-surgery manifold of a knot $K$ is homeomorphic to $S^1\times S^2$, then $K$ is the unknot.
Recently, 3-manifold topology has been developed rapidly by Agol, Wise and many other mathematicians.
Is there an another simple proof for property R conjecture?

REPLY [10 votes]: A new proof of Property R has appeared by Jamie Conway and Bülent Tosun. A slight shortcut of their argument may be summarized as:

If $K \subset S^3$ is a knot, and 0-framed surgery $S^3_0(K)\cong S^1\times S^2$, then (essentially) using the Floer exact triangle, one
may show that $S^3_1(K)$ is an L-space. An L-space is a rational homology 3-sphere $Y$ for which $HF^{red}(Y)=0$. $S^3$ and the Poincaré homology sphere are the only known irreducible L-spaces. This follows from the argument in Proposition 1.2 of Akbulut-Karakurt. 
Theorem 1.2 of 

Ozsváth, Peter; Szabó, Zoltán, On knot Floer homology and lens space surgeries, Topology 44, No. 6, 1281-1300 (2005). ZBL1077.57012.
implies that knot Floer homology of $K$ in each grading is rank 1.

Now by the main result of 

Ni, Yi, Knot Floer homology detects fibred knots, Invent. Math. 170, No. 3, 577-608 (2007); erratum ibid. 177, No. 1, 235-238 (2009). ZBL1138.57031.
$K$ is a fibered knot.

Hence 0-framed surgery is fibered. But $S^1\times S^2$ fibers in only one way, and hence the fibering must be genus 0, which implies that $K$ is the unknot. 

This proof still makes use of much of Gabai's theory, including his construction of taut foliations, which Ni's proof relies on. However, it eliminates the combinatorial topology technology of Scharlemann cycles which previous proofs relied heavily on. 
A simplified  version of Yi Ni's theorem was proved by Andras Juhasz using sutured Floer homology. 
Juhász, András, Floer homology and surface decompositions, Geom. Topol. 12, No. 1, 299-350 (2008). ZBL1167.57005.<|endoftext|>
TITLE: When is the convex hull of two space curves the union of lines?
QUESTION [10 upvotes]: I am interested in the convex hull of two space curves. Let $A,B\subset \mathbb R^3$ be two spaces curves. I am interested in when $\mathrm {con} (A \cup B)$ equals to $$\bigcup _{a\in A,b\in B} \{\lambda a + (1-\lambda) b | 0\le \lambda \le 1\}.$$ If I restrict this question to a more specific question, let $A=\{(\alpha(t),\beta(t),\gamma(t))|t\ge a\}$, $B=\{(\alpha'(t),\beta'(t),\gamma'(t))|t\ge b\}$ while $a,b\in \mathbb R$ and $\alpha,\beta,\gamma,\alpha',\beta',\gamma'$ be polynomials. Then when does the above statement hold?
I know that $\mathrm {con} (A \cup B)$ is the union of triangles; since this is in $\mathbb R^3$, and there are only two connected components, but I cannot figure out when the above statement holds. At first, I thought this was very, very special, but after I thought of many examples, I now think that this is not that special. 
I do not believe that this is a research-level question, but I didn't get any comments or answers in math.stackexchange.

REPLY [3 votes]: I will give partial answer in the following particular case: 
Assume you have only one closed space curve $\gamma(t)=(f_{1}(t),f_{2}(t),f_{3}(t)) \in C^{3}([0,1])$. Let $\tau$ be its torsion and let $k$ be its curvature. Assume that $k$ never vanishes, and $\tau$ is not identically zero on any subinterval of $[0,1]$. Assume also that the plane curve $(f_{1}(t),f_{2}(t))$ is convex. Let $n(\tau)$ be the number of sign changes of torsion $\tau$.  
Theorem
If $n(\tau)\leq  4$ then 
$$
 conv(\gamma)=\cup_{a,b\in \gamma}\{a\lambda+b(1-\lambda):0\leq \lambda \leq 1\}.
$$
Remark: It is known fact that $n(\gamma)\geq 4$, therefore in the theorem one should think that $n(\gamma)=4$.The proof can be extracted from this paper, see Section 3 and 4. In fact, what you can actually extract is that 
\begin{align*}
\partial[conv(\gamma)]=\cup_{a,b\in \gamma}\{a\lambda+b(1-\lambda):0\leq \lambda \leq 1\}.
\end{align*}
Here $\partial \Omega$ denotes boundary of the domain $\Omega$. Then it is not hard to show that this implies the theorem.  
I can sketch the idea: We are going to look to the boundary of  $conv(\gamma)$. You can think that it has two boundaries: upper one and lower one. What does it mean? The upper one is a graph of a minimal concave function $B^{min}(x,y)$ defined in the plane domain bounded by $(f_{1}(t),f_{2}(t))$,  and boundary of the graph $B$ is $\gamma$ i.e., $B^{min}(f_{1}(t),f_{2}(t))=f_{3}(t)$. Similarly the lower boundary is maximal convex function $B^{max}$ graph of which is attached to $\gamma$.
Now by Caratheodory's theorem it is enough to show that the graph of $B^{min}$ does not contain domains of linearity (triangles!), and this will mean that it consists only  by chords $\{a\lambda +b(1-\lambda), 0\leq \lambda\leq 1\}$ for some $a,b\in \gamma$. 
Suppose it contains triangles. Then let us consider any side of the triangle. Let it be the chord with endpoints $\gamma(a)$ and $\gamma(b)$ for some $a,b\in [0,1]$.
 Notice that this chord will be tangent to the curve $\gamma$ (this is not a difficult observation). In other words this means that there exists a plane containing the chord $[\gamma(a),\gamma(b)]$ and such that the curve $\gamma$ lies to one side of the plane. It is the same as to say that 
$$
\det(\gamma’(a), \gamma’(b), \gamma(b)-\gamma(a))=0 \quad (1)
$$
Now if you play with this equation for a while,  you will see that  the torsion must  change the sign from + to - on both of the side of its chord (moving counterclockwise). Since triangle has 3 sides in total you will have 3 times changing of signs of $\tau$ from + to - and this implies that $n(\tau)\geq 6$. 
In other words every time whenever you draw a such chord (bitangent line) torsion changes sign on its sides.  And this finishes the proof. 
Now the question remains: what happens in theorem if we assume that $n(\tau)>4$. Of course theorem is not true anymore, the quantity $n(\tau)$ does not give you any information about the structure of the graph $B^{min}$. There is another object (smooth transformation of torsion) which we call force function which gives the answer: there exists a source such that coming force have full tails,   but this is different story (some language was developed here)
Roughly peaking at the point where the torsion changes sign from + to -, you can (locally) construct tangent chords (a cup) $[\gamma(a),\gamma(b)]$. Now take one of them an try to extend them through out the curve (I mean foliate the domain, bounded by the curve $(f_{1}(t),f_{2}(t))$, by chords so that they will not intersect each other but will fill out the domain). 
Intuitively it means that you take this closed space curve (say closed wire), drop it on the ground, and try to roll it, so that in the beginning ground touches the wire exactly at one point (where torsion changes sign) and then it can be completely rolled over the ground (so that wire will never touch the ground at triangle) and eventually it will finish touching again at only one point (and again torsion will change the sing at that final point as well). This is possible if and only if you can extend equation (1) say by implicit function theorem throughout the curve $\gamma$, and there is one simple answer to the question when is it possible, it is possible if and only if there exists a force function with full tails. And this should be true for both: for $B^{min}$ and for $B^{max}$. 
Update: To illustrate what I wrote above I am attaching this picture. On this picture torsion of the curve changes sign $n(\tau)= 8$ times (4 times from + to - minus), and on each point where it changes sign from + to - you see the cup : family of chords coming from there. And you see that there are two triangular domains (domains of linearity: places without any thread): one close to you and one from another side which is not completely exposed on the picture. By the way this is how the upper boundary of the convex hull ($B^{min}$) look like for this space curve (closed wire).<|endoftext|>
TITLE: Generalisation of "tangent space" to not-necessarily connected sets
QUESTION [5 upvotes]: I vaguely recall having read somewhere a definition similar to (but probably not exactly the same as) the following. 

Definition (Blob) Let $S\subset \mathbb{R}^n$ be a set, and $p \in S$. The Blob of $S$ at $p$ is defined to be the subset
  $$ \left\{v \in \mathbb{S}^{n-1} \middle| \exists (x_n) \subset S, x_n \to p, \frac{x_n - p}{\|x_n - p\|} \to v \right\} $$
Definition (Widget-differentiable) Let $S\subset \mathbb{R}^n$ be a set, and $p\in S$. Let $v$ be in the Blob of $S$ at $p$. We say that a function $f:S \to\mathbb{R}$ is Widget-differentiable in the direction $v$ with derivative $\eta$ if for all sequences $(x_n)\subset S$ with $x_n \to p$ and $\frac{x_n - p}{\|x_n - p\|} \to v$, we have the limit 
  $$ \lim_{n \to \infty} \frac{f(x_n) - f(p)}{\|x_n - p\|} = \eta.$$

The above is, of course, an attempt to generalise the notion of differentiability and derivatives to functions defined on subsets of $\mathbb{R}^n$, which are not necessarily manifolds, but which have accumulation points. In particular, if $M\subset\mathbb{R}^n$ is a submanifold and $M \supset S$, then the Blob of $S$ is a subset of the unit-sphere bundle on $M$. And if a function $f$ on $M$ is differentiable at $p$, its restriction to $S$ is Widget-differentiable at $p$. 
My Question:
What are the actually names of "Blob" and "Widget"? Does anyone remember reading the same paper that I read which described this generalisation?

REPLY [3 votes]: Ah, I think I found the answer. (For future reference these notions apparently come up in non-smooth analysis.) The "Blob" I defined is closely (and rather trivially) related to the notion of the contingent cone or Bouligand tangent cones (the French Wikipedia has a better article than the English on the subject).<|endoftext|>
TITLE: Is there a big solvable subgroup in every finite group?
QUESTION [20 upvotes]: Definition: Let $G$ be a group, and let $H \leq G$ be a subgroup. We say that $H$ is big in $G$ if for every intermediate subgroup $H \leq L \leq G$ there exists some $x \in L$ such that $\langle H \cup \{x\} \rangle = L$.
Question: Is there a big solvable subgroup in every finite group?
Motivation: By Theorem A in M.Aschbacher and R.Guralnik, "Solvable generation of groups and Sylow subgroups of the lower central series", in every finite group $G$ there exists a pair of conjugate solvable subgroup $H_1,H_2 \leq G$ such that $\langle H_1 \cup H_2 \rangle = G$. So if $x \in G$ is such that $xH_1x^{-1} = H_2$ then $\langle H_1 \cup \{x\} \rangle = G$. In my question I am asking for a generalization of this conclusion.

REPLY [2 votes]: Michio Suzuki proved that every finite group is generated by a pair of conjugate solvable subgroups. See http://projecteuclid.org/download/pdf_1/euclid.hokmj/1381517825 (Open access). The subgroup $S$ which Suzuki exhibits is not a priori "big" in your sense (as far as I can see), but may suggest an approach to the question.
Later edit: Ah, I see that the existence part of Suzuki's result is no stronger than that of Aschbacher and Guralnick, it's just that his subgroup $S$ has more properties which may be useful from an inductive point of view.<|endoftext|>
TITLE: Differentiability of polytope shadow areas
QUESTION [5 upvotes]: Let $P$ be an opaque convex polyhedron containing the origin in $\mathbb{R}^3$,
and let $S$ be an origin-centered sphere strictly containing $P$: $S \supset P$.
For a point $x$ on $S$, let $\sigma(x)$ be the area of the shadow
of $P$ cast from a light at $x$ onto the plane tangent to $S$ at $-x$:

          


My question is:

Q. What is the differentiability class $C^k$ of $\sigma(x): S \to \mathbb{R}$?

I would be surprised if $\sigma$ is a smooth map, $C^\infty$, but it seems to be
at least $C^1$...
The question makes sense for a convex polytope in $\mathbb{R}^d$ for $d \ge 2$,
with $\sigma(x)$ the $(d{-}1)$-volume of the shadow cast on a $(d{-}1)$ hyperplane.

REPLY [4 votes]: Let me try to give a computation free (sketch of) proof that the shadow area is not $C^1$. The basic idea is the same as in the answer of Willie Wong: a problem happen when a corner shows up.
Consider a position $x_0$ where some corner $c$ of the polytope is projected right on a facet of the shadow. Then move $x_0$ along a smooth path $x_t$ such that for small negative $t$, $c$ is projected in the interior of the shadow, and for small positive $t$ it is projected "outside the shadow", i.e. it is projected to a vertex of the shadow. Then the area $\sigma(x_t)$ is given by the sum of a smooth function (corresponding to the shadow of the facets already contributing to the shadow at $x_0$) and a function which is zero for $t<0$ (when the facets of $c$ do not contribute to the shadow) and is linear for $t>0$ (when the facets of $c$ start contributing to the shadow). Thus $\sigma$ is not $C^1$ (but it is certainly Lipschitz).<|endoftext|>
TITLE: Status of Fontaine-Mazur conjecture
QUESTION [25 upvotes]: In the language of Richard Taylor's 2004 (extended) ICM article (''Galois Representations'', Annales de la faculté des sciences de Toulouse (2004) Tome XIII, no. 1, 73-119), the conjecture is the following

Conjecture: (Fontaine-Mazur) Suppose that
  $$R\colon G_{\mathbf{Q}}\rightarrow \mathrm{GL}(V),$$
  is an irreducible $\ell$-adic representation which is unramified at all but finitely many primes and with the $R|_{G_{\mathbf{Q}_\ell}}$ de Rham. Then there is a smooth projective variety $X/\mathbf{Q}$ and integers $i\ge 0$ and $j$ such that $V$ is a subquotient of $H^i(X(\mathbf{C}),\overline{\mathbf{Q}}_\ell(j))$. In particular $R$ is pure of some weight $w\in \mathbf{Z}$.

(here $G_K$ means absolute galois group of $K$ etc.) The notion of a de Rham representation is rather long - it may be found e.g. in the lecture notes of O. Brinon and B. Conrad here, and see loc. cit. for explanations of the other conditions. The references for the conjecture are Fontaine (J.M. Fontaine talk at ''Mathematische Arbeitstagung 1988'', Max-Planck-Institut für mathematik preprint no. 30 of 1988) and Fontaine-Mazur

Question: What is the current status of this conjecture? What results are known in its direction?

There seems to be a lot of research papers published in the last number of years on this topic. I would be grateful if anyone would be able to present some kind of rough panorama of results related to the conjecture.
--
Edit: In a Feb. 2015 survey article of C.M. Sorensen here on the Breuil-Schneider conjecture, there is a description given (in §1) of the contribution of M. Emerton to F.-M. conjecture together with a brief sketch of the method of proof.

REPLY [19 votes]: It is true if $V$ is of dimension 1, essentially by class field theory (as you are considering only representations of $G_{\mathbb Q}$). Otherwise, it is still largely open for the following reasons.
If $n\geq 2$, the best tool we have to study $G_{\mathbb Q}$-representations are so called modularity lifting theorems. These theorems take as hypotheses the hypotheses of the Fontaine-Mazur conjecture plus supplementary conditions (typically, assumptions on the residual irreducibility of $\rho$ and stronger conditions on $\rho|G_{\mathbb Q_{\ell}}$ than just being de Rham) and prove that $\rho$ then comes from an automorphic representation $\pi$ of $\mathbf{G}(\mathbb A_{\mathbb Q}^{(\infty)})$ (the finite adelic points of a reductive group $\mathbf{G}$). However, except in very rare cases, we don't know that such Galois representations (automorphic galois representations for short) occur in the cohomology of smooth projective varieties, so even when they apply, the full conclusion of the Fontaine-Mazur conjecture is largely out of reach.
Worse, notice how above I mentioned that modularity lifting theorems required some supplementary local assumptions at $\ell$. One of them is that the Hodge-Tate weights of $\rho|G_{\mathbb Q_{\ell}}$ are distinct. This happens for a very good reason: even when $n=2$, there are some automorphic Galois representations (conjecturally satisfying the hypotheses of the Fontaine-Mazur conjecture) we believe exist but don't know how to construct or study, namely the automorphic Galois representation attached to algebraic Maas forms (those with eigenvalue $1/4$). A proof of the full Fontaine-Mazur conjecture even for $n=2$ would need to deal with those, and this is currently seemingly out of reach.
A bit of good news after this gloom assessment. If $n=2$ and if you are fine with disregarding automorphic Galois representations attached to Maas forms (in particular if you assume that the Hodge-Tate weights are distinct), then the supplementary hypotheses required to establish the full Fontaine-Mazur conjecture are rather mild. I don't pretend to know the cutting edge, but the following theorem should certainly be true.

Assume $\rho:G_{\mathbb Q}\longrightarrow\operatorname{GL}_{2}(\bar{\mathbb Q}_{\ell})$ satisfies the following hypotheses. 1) $\rho|G_{\mathbb Q_{\ell}}$ is de Rham and its Hodge-Tate weights are distinct. 2) $\bar{\rho}$ is absolutely irreducible and odd. 3) $\bar{\rho}|G_{\mathbb Q_{\ell}}$ is neither the sum of two identical characters nor a twist of an extension of $\bar{\chi}_{cyc}$ by 1. Then $\rho$ occurs in the cohomology of smooth projective variety (namely the desingularization of some Kuga-Sato variety).

This is the culmination of the work of many people; the most recent being Pierre Colmez, Matthew Emerton and Mark Kisin.

REPLY [9 votes]: I'm far from an expert on this, and there are certainly users here more qualified to answer than me, but for now here's a short answer.
This is now known in the case of two-dimensional representations $R:G_Q\rightarrow GL_2(\bar{Q}_\ell)$, due to Emerton and Kisin.
Kisin proves the result under the additional hypothesis that $R$ is odd and has distinct Hodge--Tate weights at $\ell$ (the same $\ell$ as the residue characteristic of the coefficient ring of $R$). By "having distinct Hodge--Tate weights at $\ell$", I mean that the restriction of $R$ to the decomposition group at $\ell$ $D_\ell\simeq Gal(\bar{Q}_\ell/Q_\ell)$ has distinct Hodge--Tate weights. I'm not completely sure on this part, but I believe that it's since been proved (by Calegari if I recall correctly; maybe under some small, almost always satisfied hypotheses) that there are no even representations with distinct Hodge--Tate weights at $\ell$, removing the requirement that $R$ be odd in almost all cases.
I'm less familiar with Emerton's contribution. Certainly, Kisin's paper requires on the local-global compatibility result of Emerton and some subsequent result by Colmez; otherwise one is forced to require further that $R$ is potentially semistable over some abelian extension. I have a vague idea that perhaps (in the same local-global compatibility paper) Emerton gives an independent proof of the same 2-dimensional result, but I'm even less familiar with this.
I guess I should say that the two main papers that I've mentioned are The Fontaine--Mazur conjecture for $GL_2$ by Mark Kisin, and Local-global compatibility in the $p$-adic Langlands program for $GL_2$ by Matt Emerton.
To the best of my knowledge, there hasn't been any substantial progress on the result for higher-dimensional representations. This seems to me to be due to the fact that there are essentially no results towards a $p$-adic local Langlands correspondence for groups other than $GL_2$, but that isn't based on anything other than guesswork.<|endoftext|>
TITLE: Should we post on arXiv only papers in publishable shape (or very close)?
QUESTION [14 upvotes]: Question: Should we post on arXiv only papers in publishable shape (or very close)?   
This question should be distinguished from the following:
Should one post a paper on the arXiv if it is not intended to be published?
in the sense that a paper which has not a publishable "contents" can have a publishable "shape".
Moreover, a paper which is intended to be published can be not yet in a publishable shape.
Sometimes it happens that we start to write a paper, we develop some interesting new ideas, but then we do not continue the paper for several possible reasons:    

change of the research subject   
loss of motivation for this problem    
too difficult   
$\dots$  

Sometimes also we just want to share the current state of our work through a draft, even if all the proofs are not yet complete.
Whatever the reasons, the ideas contained in such drafts can be interesting and useful for the community, regardless of the state of advancement of the paper. So after this explanation:
Should we definitely not post such a paper on arXiv, even if its state is clearly specified at the beginning?

REPLY [24 votes]: The documentation of arXiv has this to say (my emphasis):

Inappropriate format. arXiv accepts only submissions in the form of an article that would be refereeable by a conventional publication venue. This excludes abstract-only submissions, submissions without references, book announcements or reviews, reports that do not contain original or substantive research, papers that contain inflammatory or fictitious content, papers that use highly dramatic and mis-representative titles/abstracts/introductions, or papers in need of significant review and revision. 

Source:  http://arxiv.org/help/moderation<|endoftext|>
TITLE: Can every genus $2$ curve be written as ramified cover of elliptic curve?
QUESTION [5 upvotes]: Suppose $C$ is a curve of genus $2$, does $C$ admit a surjective morphism onto some elliptic curve $E$?

REPLY [7 votes]: dhy's comment is correct, of course. You can also see that the answer is "no" by dimension counting. Fix a degree $d$. By a Riemann-Hurwitz computation, a degree $d$ cover of a genus $1$ curve by a genus $2$ curve is ramified over $2$ points (or doubly ramified over one point.) Given a positive integer $d$, a genus $1$ curve $E$ and two points $p$ and $q$, there are finitely many covers $X \to E$ of degree $d$ ramified only over $p$ and $q$. (Since each of them comes from one of finitely many maps $\pi_1(E \setminus \{ p,q \}) \to S_d$.) So, for fixed $d$, such covers are described by some finite covering of $M_{1,2}$. 
Now, $\dim M_{1,2} = 2$. So, for each $d$, we get a surface of such genus $2$ curves and, as $d$ varies, we get countably many such surfaces. Counteably many surfaces can't fill up the three-fold $M_2$.<|endoftext|>
TITLE: Algebraic spaces which are automatically schemes
QUESTION [5 upvotes]: Let $S$ be a scheme, and let $f:X\to S$ be a morphism of algebraic spaces.
If $f$ is smooth proper curve of genus at least two, then $X$ is a scheme. (Here I mean that $f$ is a smooth proper morphism whose geometric fibers are actual curves of genus at least two.)
Are there any other types of properties which one can impose on $f$ so that $X$ becomes a scheme?
My question is to see how far one can get without algebraic spaces. My guess is "not so far".
For instance, what if $f$ is a smooth proper morphism whose geometric fibres have ample or anti-ample canonical bundle?

REPLY [6 votes]: For a scheme $S$, every proper finitely presented map $f:X \rightarrow S$ from an algebraic space $X$ admitting a line bundle $L$ that is ample on each geometric fiber (which of course forces such fibers to be projective schemes, essentially by the very definition of ampleness for algebraic spaces) is automatically a scheme.  
Indeed, by standard limit methods we may assume $S= {\rm{Spec}}(R)$ is local and then we assume only that $L$ is ample on the special fiber. That case can be descended to $S= {\rm{Spec}}(R)$ local noetherian, and the aim is to prove that for some large $n$ the power $L^n$ is generated by global sections with the resulting map $f:X \rightarrow {\mathbf{P}}(\Gamma(X,L^n))$ quasi-finite (so $X$ is a scheme, being quasi-finite and separated over a scheme).  These are properties which are sufficient to check after faithfully flat base change to the completion of $R$, so we can assume $R$ is complete.  It suffices to prove here that $X$ is now a scheme, as then the cohomological theory of ample line bundles would then apply to provide the desired $n$, the crux being the remarkable EGA IV$_3$, 9.6.4 (which has no flatness hypotheses).
Every infinitesimal fiber is a scheme, so we can make sense of the formal completion along the special fiber as a proper formal scheme equipped with a residually ample line bundle.  By formal GAGA with the help of $L$, that formal scheme algebraizes to an actual proper scheme.  But we have formal GAGA for algebraic spaces too, so the algebraic space algebraization and the scheme algebraization coincide.
QED<|endoftext|>
TITLE: Counting fundamental units of real quadratic fields
QUESTION [6 upvotes]: For a given real quadratic field $K$, the group of units of its ring of integers is $\mathcal{O}_K^{\times}\cong(\pm1)\times \mathbb{Z}$ by the Dirichlet unit theorem. For each $\mathcal{O}_K$, pick the fundamental unit as $\epsilon >1$, then $\epsilon=m\sqrt{d}+n$, where $m,n>0$ are integers or half integers. Now for a large variable $x>>1$, define the counting function $$
  u(x):=\sum_{1<\epsilon<x} 1,
$$
where the sum is over all real quadratic fields. 
Firstly the sum is finite, since by the above token, there are only finitely many $\epsilon=m\sqrt{d}+n$ with bounded absolute value as $d\rightarrow\infty$.
What asymptotic information is known about $u(x)$?

REPLY [8 votes]: In Proposition 4.1 of Sarnak you'll find an asymptotic for the related quantity (take there $p=1$) when one considers all ring discriminants (not just the fundamental field discriminants that you want).  If restricted to field discriminants, the work of Raulf obtains such asymptotic formulae (see Lemma 5.1 in that paper).  In both papers, the interest is in obtaining averages for the class number $h(d)$ when the discriminants are arranged according to the size of $\epsilon_d$.<|endoftext|>
TITLE: "Family Tree" of Theorems
QUESTION [14 upvotes]: Is anyone aware of any attempt to describe the dependencies of theorems (perhaps in mathematics generally, perhaps in some limited areas) in the form of a "family tree"?  That is, each node on the tree might correspond to a theorem, and branches would indicate dependencies between theorems?  I realize that this would not constitute an actual tree -- as there can exist loops -- but this sort of meta-description of theorems might provide some insight not available in other manners.  
[Added:] Except in some very formalized proof systems, making the  notion of dependency precise is probably difficult, if not impossible.  But colloquially, people will often make statements such as "Theorem A is used / can be used to prove Theorem B".  I'm sure everyone here can think of many such statements to which few will object.  For those cases, it might be very nice to have some accessible database with this information.  Not only might the data in this "graph" be practically useful (e.g., I want to write an expository paper about five interesting consequences of the Borsuk-Ulam theorem), but perhaps even some "meta-data" might provide valuable insight.  Just a thought.

REPLY [2 votes]: There are fairly recent dependency graphs relating to Euclid's Book 1, Elements, in the literature.
See A Boxer's paper in Bridges Art/Math 2015
Also see M. Schiefsky (Harvard Classics) and Boman (Penn State Math) and Nyugen (Berkeley) for overviews and graphs.
The most interesting is that of Jesse Atkinson, Bridges Art/Math 2016, that shows a 3-D in-tree dependency graph of Euclid's proof of the Pythagorean Theorem (1.47) embedded in a discussion about the overall role of 1.45 and 1.47 in that book.  It is an elegant "little" paper, by an undergraduate math student, to say the least.<|endoftext|>
TITLE: Equivalence of Definitions of Quasiconformal Surfaces?
QUESTION [6 upvotes]: I have been reading John H. Hubbard's book Teichmüller Theory vol. 1 and I am a little bit concerned with his definition of Quasiconformal Surface.

Definition: A Quasiconformal surface $S$ is a topological surface with a Riemann-surface structure; two Riemann surface structures on $S$ define the same quasiconformal structure if the identity map between them is quasiconformal.
If $S_1,$ $S_2$, are two quasiconformal surfaces, a map $f:S_1 \to S_2$ is quasiconformal if it is a quasiconformal homeomorphism for one, hence all, analytic structures on each $S_1$ and $S_2$. This implies that all quasiconformal maps are isomorphisms.

I do understand the definition and one it's advantage is that it is easy to prove that if two compact quasiconformal surfaces $S_1$ and $S_2$ are homeomorphic, then they are isomorphic as quasiconformal surfaces.
However, I am use to define structure like that in a similar way as we define manifolds. Hence, if I had to give a definition of Quasiconformal Surface, I would say that it is a topological surface $S$ together with a maximal atlas that contains all the charts for which the transition maps are quasiconformal maps.
My question is are those 2 definitions equivalent (can I define a specific Riemann surface structure from a maximal atlas which is unique up to quasiconformal maps?), my first idea to solve this problem was to use the measurable Riemann mapping theorem which allow us to find local quasiconformal maps that will satisfy a Beltrami equation. The problem is that I am not sure if this argument will work for any surface, we might have to do the process on infinitely many charts (I want to consider surfaces with puncture and boundaries as well.)

REPLY [3 votes]: I believe the problem is exactly this. A composition of $K$-quasiconformal maps is not necessarily $K$-quasiconformal, which makes them difficult to work with. And a locally quasiconformal map is not necessarily globally quasiconformal. Normally when you define a type of manifold in terms of a class of permitted overlap maps the class of maps should be defined in terms of a local property and closed under composition. There's no way to do that so as to get structures that are then related by global quasiconformal maps.<|endoftext|>
TITLE: Is it true that any $3$-uniform hypergraph that is not $k$-colorable must have $\Omega(k^3)$ edges?
QUESTION [5 upvotes]: What is the best lower bound in terms of $k$ on the number of edges in a $3$-uniform hypergraph that is not $k$-colorable?
Thanks in advance.

REPLY [7 votes]: The following paper of Alon shows that the quantity you're after, $m(k)$, the minimum number of edges of a $3$-uniform hypergraph which is not $k$-colourable, is indeed $\asymp k^3$.
More precisely, he shows that
$$ 2\left\lceil \frac{k}{3}\right\rceil \left\lfloor \frac{2k}{3}\right\rfloor^2 < m(k) \leq \binom{2k+1}{3}$$
where the implied constants are absolute. The lower bound follows from a simple probabilistic argument -- colour all the vertices randomly, and then recolour a few necessary vertices to remove the small number of monochromatic edges which remain. The upper bound is just the number of edges of the complete 3-uniform hypergraph on $2k+1$ vertices, which is clearly not $k$-colourable
http://www.tau.ac.il/~nogaa/PDFS/Publications/Hypergraphs%20with%20high%20chromatic%20number.pdf<|endoftext|>
TITLE: Iwaniec-Kowalski Exponential Sum for Quadratic Function
QUESTION [12 upvotes]: I am reading about 'Exponential Sums' in the book 'Analytic Number Theory' by  Iwaniec and Kowalski. On page 199 they mention the bound:
$$|S_f(N)|^2 \le N +2N^2q^{-1}+4(N+q)\log q \tag{1}$$
where, $\displaystyle S_f(n) = \sum\limits_{n=1}^{N} e^{2\pi i(\alpha n^2 +\beta n)}$ and $1 \le q \le 2N$ satisfies $\displaystyle 2\alpha = \frac{a}{q} + \frac{\theta}{2Nq}$,with $|\theta| < 1$ and $(a,q) = 1$.
I am having trouble following why $(1)$ along with the trivial bound $|S_f(N)| \le N$ implies,
$$|S_f(N)| \le 2Nq^{-1/2}+q^{1/2}\log q\tag{2}$$
How do we see the claim that $(2)$ is best possible?

"Is it possible to deduce (2) more directly, without much numerical work at small values?"

Applying Vander Corput's Inequlity, we may derive:
$$|S_f(N)|^2 \le \frac{N^2}{H}+\frac{N(2N-H)}{q}+4N\left(1+\frac{q}{H}\right)\log q$$
where, $1\le q < 2H$, and $H$ is an integer less than $N$ we are free to choose. Is there a way to proceed to $(2)$ from here?

REPLY [9 votes]: If $\log(q)\ge4$, then the fact that $N\le2N^2q^{-1}$ and $4q\log(q)\le q\log(q)^2$ show that
$$
N+2N^2q^{-1}+4(N+q)\log(q)\le\overbrace{4N^2q^{-1}+q\log(q)^2+4N\log(q)}^{\left(2Nq^{-1/2}+q^{1/2}\log(q)\right)^2}
$$
This shows that $(1)\implies(2)$.

If $q\le4$, then
$$
N^2\le4N^2q^{-1}+q\log(q)^2+4N\log(q)
$$
This and $|S_f(N)|\le N$ also verifies $(2)$.

Therefore, we need to consider
$$
4\lt q\lt e^4
$$
Consider
$$
N^2\gt4N^2q^{-1}+q\log(q)^2+4N\log(q)
$$
which is only true for
$$
N\gt\frac{q\log(q)}{\sqrt{q}-2}
$$
and
$$
N+2N^2q^{-1}+4(N+q)\log(q)
\gt4N^2q^{-1}+q\log(q)^2+4N\log(q)
$$
which is only true for
$$
N\lt\frac q4\left(1+\sqrt{1+32\log(q)-8\log(q)^2}\right)
$$
Unfortunately, between $q\doteq12.592$ and $q\doteq48.802$ the bounds on $N$ above permit some common values of $N$.

The convex blue curve is $\frac{q\log(q)}{\sqrt{q}-2}$ and the concave red curve is $\frac q4\left(1+\sqrt{1+32\log(q)-8\log(q)^2}\right)$. The area above the blue curve and below the red curve is where the problem lies.
For example, at $q=30$ and $N=32$, we have
$$
\begin{align}
N^2&=1024\\
N+2N^2q^{-1}+4(N+q)\log(q)&=943.764\\
4N^2q^{-1}+q\log(q)^2+4N\log(q)&=918.931
\end{align}
$$
Thus, it doesn't seem that the third bound, $(2)$, can be derived from the first bound, (1), and the trivial bound. This doesn't mean that $(2)$ is not valid, just that is does not seem to follow from the other bounds.<|endoftext|>
TITLE: Extension-closed subcategories of triangulated categories as "almost exact" categories
QUESTION [5 upvotes]: Did anybody study those subcategories of triangulated categories that are closed with respect to "extensions" (in the sense of distinguished triangles; in particular, any such $B$ is additive)? If we consider these "extensions" as short exact sequences in $B$, we "almost" obtain the structure of a Quillen's exact category on it. The problem is that the corresponding admissible monomorphism are only "weak kernels" in $B$ in general (dually, admissible epimorphisms are only weak cokernels). Does some part of the theory of exact categories can be applied to this setting yet; can one associate an abelian category to $B$ (endowed with this "almost exact structure") in a reasonable way? Are there any results known for "almost exact" categories in this sense?
Upd. Possibly, it is better to consider a smaller set of exact sequences in this setting (so that the corresponding kernels and cokernels will be "strong").

REPLY [4 votes]: There is a beautiful paper by Iyama and Yoshino (http://arxiv.org/abs/math/0607736). The main result in Section 4 is closely related to your question. The authors consider some extension-closed subcategory of a triangulated category which behaves almost like a Frobenius exact category. They prove that the stable category is triangulated. The proof is similar to that of Happel, with the use of push-outs and pull-balcks replaced by that of the octaedron axiom. Their main motivation for this result was to show that cluster-tilting objects in Hom-finite 2-Calabi--Yau triangulated categories have a nice theory of mutation. 
This analogy with Happel's result was studied by Nakaoka in http://arxiv.org/abs/1006.1033, where he defines a common generalisation of exact categories and extension-closed subcategories of triangulated categories. However, I do not know if there exist an axiomatic definition along the lines of Quillen's.<|endoftext|>
TITLE: Double kissing problem
QUESTION [14 upvotes]: Consider two touching unit balls which will be called central balls. What is the maximum number $k$ of non-overlapping unit balls so that each ball touches as least one of two central balls?
An easy lower bound $k\geq 18$ is achieved by the face-centered cubic lattice. I conjecture that $k=18$.

REPLY [10 votes]: Using global nonlinear optimization one can obtain a configuration of $19$ spheres, that touch at least one of the central unit spheres and have almost no overlap. In fact, if one takes their radii to be $.99$ instead of $1$ they are non-overlapping.
Below are the coordinates; the two central spheres have radius $1$ and are centered around the origin and $(2,0,0)$. Here is a picture of the configuration: 

Maybe this is helpful as a starting point in the simulated annealing approach you mentioned in the comments, but I am not so sure, since it seems to be somewhat jammed already.
(1.30155675907051, 1.87408031823623, 0.000000000000000),
(3.30307693251716, -1.48756032724292, 0.298587978254738),
(3.77087392448039, -0.00565125397965555, -0.929501805767173),
(2.34028624585583, 1.21452857880052, -1.55213581949477),
(1.49324421375722, -1.89375136244257, -0.396111537780328),
(3.31658479709791, 0.0846873881998760, 1.50314088439273),
(1.46434039083497, 0.727511022954233, 1.78431961671711),
(1.82006459231285, -1.21042374283863, 1.58192844712867),
(2.17723615440802, -0.736686335972064, -1.85091344691338),
(3.24812939881743, 1.55286888817322, 0.175417273814499),
(-0.234051719598306, 1.57207701538230, 1.21399903223316),
(-0.182142247556194, -1.45688908753032, -1.35804947932633),
(-1.35706161319061, 0.134224681025498, -1.46299949180272),
(-1.82011109002214, 0.745105549605677, 0.363336400498419),
(0.560694305152308, 0.407540066521993, -1.87604184131108),
(-0.536544075078242, 1.78215924389985, -0.732139935326408),
(-1.60854779879003, -1.18847287273622, -0.0103058129362229),
(-0.850339283181967, -0.217696133065399, 1.79708972984823),
(0.0325533674370459, -1.76736000103794, 0.935616858014407)]<|endoftext|>
TITLE: Topology of algebraic varieties
QUESTION [6 upvotes]: Let $X$ be a projective variety (lets say normal and irreducible) with the topology coming from being a subspace of $\mathbb{P}^N$ (and not the Zariski topology). Surely one can then define the singular cohomology groups. My question is whether one can also make sense of the Cech cohomology groups $H^*(X,\mathbb{C})$ for the sheaf of locally constant $\mathbb{C}$-valued functions, and if the two cohomology groups agree, like would be the case if $X$ were a complex manifold. Is there also a notion of De Rahm cohomology where we look at forms in some suitable Sobolev space?
I apologize for the rather vague/soft question, but I was not able to find any references online. So it would also be great if someone could point out references for this sort of thing.

REPLY [8 votes]: Regarding your question on De Rham cohomology there are several approches to realize a De Rham complex that computes singular cohomology.
A. In Algebraic geometry.
You should look at R. Hartshorne "Algebraic De Rham cohomology" manuscripta  math. 7, 125-140 (1972). It is a research announcement and survey on the cohomology of algebraic De Rham forms on algebraic varieties, details are published in the Publications of IHES (1975).
In particular for a scheme $Y$ of finite type over a characteristic zero field $k$, he defines its algebraic De Rham cohomology. 
1) You embed $Y$ as a closed subscheme of a smooth scheme $X$. 
2) You consider $\Omega^*$ the complex of sheaves of regular differential forms on $X$ over $k$.
3) You take $\hat{X}$ the formal completion of $X$ along $Y$:
$$\hat{Y}=\bigcup_n Y(n)$$
where $Y(n)$ is the infinitesimal neighbourhood of order $n$ of $Y$ in $X$
and consider $\hat{\Omega}^*$ the completion of $\Omega^*$.
4) You define $H^*_{DR}(Y)$ as the hypercohomology of the complex $\hat{\Omega}^*$ on the formal scheme $\hat{X}$.
Then (theorem 1.6 of this paper) when $Y$ is a scheme of finite type over $k=\mathbb{C}$ we have a natural isomorphism
$$H^i_{DR}(Y)\cong H^i(Y^{an},\mathbb{C})$$
where $Y^{an}$ is the corresponding complex analytic space and $H^i(-,\mathbb{C})$ is the singular cohomology.
...............................
B. As stratified spaces
You can use the fact that a complex algebraic variety is stratified, for example it is a stratifold in the sense of M. Kreck, then you have a notion of De Rham complex that computes singular cohomology with real coefficients:

Christian-Oliver Ewald (PhD thesis): "Hochschild Homology and De Rham Cohomology of Stratifolds" http://www.him.uni-bonn.de/fileadmin/user_upload/kreck-phd10.pdf

Or you can use Whitney functions 

Bryce Chriestenson and Markus Pflaum: "Whitney functions determine the real homotopy type of a semi-analytic set" http://arxiv.org/pdf/1403.1627v1.pdf<|endoftext|>
TITLE: Space $X$ such that $X^\lambda\cong X$ for some $\lambda$
QUESTION [7 upvotes]: Which cardinals $\lambda > 2$ have the following property?
There is a space $(X,\tau)$ such that

for all cardinals $\kappa$ with $1<\kappa<\lambda$ we have $X\not\cong X^\kappa$, and
$X\cong X^\lambda$.

REPLY [9 votes]: As was conjectured by Adam in his answer, every finite $n > 2$ has the property you're looking for. Namely, there exists a topological space homeomorphic to its $n$th power and none of its $k$th powers, $1<k<n$. 
The relevant paper is "Homeomorphisms between finite powers of topological spaces" by Orsatti and Rodino. It can be found here. From the abstract:
"Let $\lambda$ be an infinite cardinal number. It is proved that, for each positive integer $r$, there exists a compact connected homogeneous topological space $X$ of weight $\lambda$ such that $X^n$ is homeomorphic to $X^m$ iff $n \equiv m$ (mod $r$)."
From glancing at the paper, the authors adapt a big theorem of Corner, that every countable reduced torsion-free ring is the endomorphism ring of some torsion-free Abelian group, to prove their result. (This is the same theorem Corner himself uses to prove the result cited in Adam's answer.) What they actually produce is a compact and connected topological Abelian group $X$ such that $X^n$ is topologically isomorphic $X^m$ iff $n \equiv m$ (mod $r$), and then use the fact that for such groups any homeomorphism between the underlying spaces must actually be a topological isomorphism.<|endoftext|>
TITLE: Gauss-Bonnet formula for 2-dimensional Alexandrov spaces
QUESTION [6 upvotes]: EDIT: Let $S$ be a closed orientable 2-dimensional surface equipped with a metric with curvature $\geq \kappa$ in the sense of Alexandrov. 
Questions 1. Can one define a measure $K$ on $S$ (thought to be an analogue of the Gauss curvature) satisfying the following properties:
(a) if the metric on $S$ is smooth then $K$ is the usual Gauss curvature times the Lebesgue measure.
(b) If a sequence of orientable surfaces $S_i$ with such metrics (with the same lower bound $\kappa$ on the curvatures) converges to $S$ in the Gromov-Hausdorff sense then $K_i\to K$ weakly (what is weak convergence of measures on different spaces should be made more precise, but I guess it is well known to experts). 
(c) Gauss-Bonnet formula: $\int_S K=2\pi \chi(S)$.
(It is very likely that (c) follows from (a)+(b).)
Question 2. If the answer to Question 1 is positive, it seems likely that if $S$ has non-negative curvature which in some open subset is $\geq \kappa>0$ (and $S$ is orientable) then $S$ is homeomorphic to the 2-sphere. Is this consequence known to be true in the context of Alexandrov spaces? (For smooth Riemannian metrics it is well known.)
UPDATE:. The answer to both questions is YES as follows from the answer below by Thomas Richard and the comment by Anton Petrunin.

REPLY [5 votes]: The answer to question 1 is almost yes. This more or less follows from the work of A. D. Alexandrov on convex surfaces. See "The intrinsic geometry of convex surfaces" by Alexandrov, and also "Intrinsic geometry of surfaces" by A. D. Alexandrov and V. Zalgaller (which is written in a more general context).
Note: the curvature measure is a signed measure.
The only thing missing from these reference is continuous dependence with respect to  GH convergence (which was not really in the at that time), however it is shown that the curvature measure is weakly continuous with respect to uniform convergence of the distance.
Another thing is that you need to assume a priori that your underlying topological space is a 2D surface without boundary (a closed disk with its usual metric is a perfectly legitimate alexandrov space).
The construction goes throught the following process (and actually shows why Gauss-Bonnet should hold). Let $abc$ be a geodesic (maybe convex or small enough, I don't remember the details) triangle in $S$ (considered as an simply connected region in $S$). Consider its three angles $\alpha,\beta,\gamma$, and define the excess of $abc$ to be $\alpha+\beta+\gamma-\pi$. On then sets the curvature of the interior region of $abc$ to be the excess (because one wants local Gauss-Bonnet to hold). 
Then the tedious part begins, where one shows that there exists a unique measure on borel subsets of $S$ which gives this result on the interior of geodesic triangles.
For your second question, once all of this has been set up, assume that the curvature measure (call it $\omega$) of $S$ is non-negative and not zero, this implies that $\omega(S)>0$. But Gauss-Bonnet tells you that $\omega(S)=2\pi\chi(S)$. This $S$ is an orientable surface with positive Euler characteristic.<|endoftext|>
TITLE: Is the Jordan decomposition of a self-adjoint functional constructive?
QUESTION [8 upvotes]: Let $A$ be an abstract C*-algebra, and let $\varphi\colon A \rightarrow \mathbb C$ be a bounded linear function. Assuming the axiom of choice, there exist unique positive bounded linear functions $\varphi_+$ and $\varphi_-$ such that $\varphi = \varphi_+ - \varphi_-$ and $\|\varphi\| = \|\varphi_+\| + \|\varphi_-\|$ (see section 3.2 of Pedersen's C*-algebras and their Automorphism Groups). Is the axiom of dependent choices sufficient to prove this result?
I believe that the axiom of dependent choices is sufficient to establish the result if $A$ is separable or commutative. If $A$ is separable, then the usual proof still works. If $A$ is commutative, then the decomposition can be obtained by lattice-theoretic methods, such as those in Schechter's Handbook of Analysis and its Foundations. In general, the self-adjoint operators of a noncommutative C*-algebra do not form a lattice (see examples II.3.3.3 in Blackadar's book).

REPLY [2 votes]: Let me expand a little my comment because it is more subbtle than what I suggested and it relies heavily on a recent result. I agree that this does not exactly answser your question but is I think relevant.
I will proove that this decomposition result hold in any boolean Grothendieck toposes (which are in particular place where no form of the axiom of choice holds, not even dependant choice... in this frame work completness is taken in the sense of Cauchy filter or Cauchy approximation but not in the sense of cauchy sequences...).
So let $T$ be a boolean Grothendeick topos, $C$ a $C^*$-algebra in $T$ and let $\phi$ be a linear form on $C$ whose norm is smaller than $1$.

In this paper, I proved that there is a classifying locale $Fn C$ for linear form of norm smaller than $1$ on $C$ (see section 3.5 and proposition 4.2.3). $\psi$ is a point of $Fn C$.
By Barr's covering theorem $T$ admit a covering $p:T' \rightarrow T$ such that $T'$ is boolean and the axiom of choice holds in $T'$. Moreover as $T$ is boolean $p$ is automatically an open surjection.
Pulling back everything to $T'$ ($C$ has to be completed after pull-back of course), one have a $C^*$-algebra $p^* C$ with a point of $Fn p^* C = p^* Fn C$. One can then apply the decomposition of $\psi$ in $\psi_+$ and $\psi_-$ in $T'$. They are again linear form of norm smaller than $1$, hence give rise (internally) in $T'$ to two new point of $p^* Fn C$.
By uniqueness of these two point, they actually lift to global section, i.e. to external morphisms from $T'$ to $p^* Fn C$ over $T'$.
$\psi_+$ and $\psi_-$ are characterised by properties which are pullback stable (the positivity and the condition one the norm) and are unique hence they are automatically compatible with the canonical descent data (for $p$) on the point and on $p^* Fn C$. Then as $p$ is an open surjection it is an effective descent map for locale and hence the two points $\psi_+$ and $\psi_-$ are pullback of two points $Fn C$ in $T$ that we also denote $\psi_+$ and $\psi_-$, these two points also satisfies the conditions on the norm and of positivity because their pullbacks along a surjection do. SO this prove the existence of the decomposition in $T$.
The uniqueness can also be proved by pulling back to $T'$, but I'm pretty sure that the uniqueness can be proved directly without the axiom of choice quite easily...

Let me mention the consequence of this for your question.

It is an extremely encouraging sign that the results can be proved without AC.
The result will holds in any model of ZF constructed using combination of forcing and permutation models (because these corresponds to certain boolean Grothendieck toposes).
It is going to be very hard to produce a counterexemple if it is false, first because of the previous observation and secondly because for any $C^*$-algebra "geometrically described" one should be able to use the previous argument in a booleanisation of a classifying topos to obtain a proof for this specific exemple. So the counter example has to be either non-explicit or has to rely on non geometric constrution (which in my opinion mean a very weird construction)
It might be possible to use a similar argument to proove that this result holds in the Solovay model (which if I remeber correctly is constructed by first taking a forcing model and then an inner model, but I'm not really familiar with inner model so I can't say anything clever about that...)<|endoftext|>
TITLE: Reference for a PL flat torus embedding in $\mathbb{R}^3$
QUESTION [9 upvotes]: A piecewise linear flat torus embedded in $\mathbb{R}^3$ is shown at http://www.mathcurve.com/polyedres/toreplat/toreplat.shtml.
It is flat in the sense that the angle defect at the vertices is zero. 
Here is a 3D printed hinged version:

Who came up with this construction? 
I asked Robert Ferréol who maintains the mathcurve.com site. He heard of it from Guy Valette, who remembers seeing a torus like this at Oberwolfach over 30 years ago.

REPLY [3 votes]: I believe the originator is Victor A. Zalgaller.
Permit me to quote myself from an earlier answer:

In the paper by
V. A. Zalgaller, "Some bendings of a long cylinder," Journal of Mathematical Sciences, 100(3):2228--2238, 2000 (translated from a 1997 article in the Russian journal
Zapiski Nauchnykh Seminarov POMI), he proves this theorem:

"Theorem 1. A direct flat torus can be isometrically embedded in $\mathbb{R}^3$ 'in the
origami style' if its development is a rectangle sufficiently large compared to its altitude."

He defines a direct flat torus as the result of identifying the opposite sides of a rectangle.<|endoftext|>
TITLE: An equivalence of derived categories by Happel-Reiten-Smalø
QUESTION [6 upvotes]: I have a problem in understanding the proof of a theorem by Happel-Reiten-Smalø. The original reference is this article
http://arxiv.org/abs/0911.4473
.
I write down the text of the theorem and a lemma which is needed in the proof.

Lemma 3.2.4 Let $k$ be a field and $\mathcal A$ a $k$-linear abelian category that is Ext-finite. Let $T\in\mathcal A$ be a tilting object. Then the functor $-\otimes_{\Lambda}T$ induces an isomorphism
  $$
\text{Ext}_{\Lambda}^n(M,N)\overset{\sim}{\longrightarrow}\text{Ext}_{\mathcal A}^n(M\otimes_{\Lambda}T,N\otimes_{\Lambda}T)
$$
  for all $M$, $N$ in $\mathcal Y(T)$ and $n\geq 0$. [$\mathcal Y(T) = \{M\in\text{mod }\Lambda\mid \text{Tor}_1^{\Lambda}(M,T) = 0\}$]
Theorem 3.2.5 (Happel-Reiten-Smalø).
  Let $\mathcal A$ be a $k$-linear abelian category that is Ext-finite. Let $T$ be a tilting object in $\mathcal A$ and $\Lambda=\text{End}_{\mathcal A}(T)$. Then the functor
  $$
-\otimes_{\Lambda}^{\mathbf{L}} T\colon\mathbf{D}^b(\text{mod } \Lambda) \to \mathbf{D}^b(\mathcal A)
$$ 
  is an equivalence of triangulated categories and its right adjoint $\mathbf{R}\text{Hom}_{\mathcal A}(T,-)$ is a quasi-inverse.
Proof. Set $F_T = -\otimes_{\Lambda}^{\mathbf L}T$ [the left derived functor of $\otimes_{\Lambda}$]. We identify objects in $\text{mod }\Lambda$ and $\mathcal A$ with complexes that are concentrated in degree $0$. For instance, $F_T M = M\otimes_{\Lambda} T$ for each $M$ in $\mathcal Y(T)$.
  We need to show that for each pair of complexes $X$, $Y$ in $\text{mod }\Lambda$, the induced map
  $$
\phi_{X,Y}\colon\text{Hom}_{\mathbf D^b(\text{mod }\Lambda)}(X,Y)\to\text{Hom}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY)
$$
  is bijective. Set
  $$
\ell(X) = \operatorname{card}\{n\in\mathbb Z\mid X_n\neq 0\}
$$
  and note that each bounded complex $X\neq 0$ fits into an exact triangle $X'\to X\to X''\to X'[1]$ such that $\ell(X') = \ell(X)-1$ and $\ell(X'') = 1$.
Using the five lemma and induction on $\ell(X)+\ell(Y)$, one shows that $\phi_{X,Y}$ is bijective. The case $\ell(X) = \ell(Y) = 1$ follows from lemma 3.2.4. To be precise, one uses that each $\Lambda$-module $M$ fits into an exact sequence $0\to M'\to P\to M \to 0$ with $M'$ and $P$ in $\mathcal Y(T)$, which yields an exact triangle $M'\to P\to M\to M'[1]$.
  [...]

Next they show that the given functor is dense, but I am stuck on this passage. I can't see how I could merge the given tools (induction, five lemma, every $\Lambda$-module fits into such an exact sequence) in order to get to the assertion.
I think this theorem is not difficult to find in the literature, for example in Happel-Reiten-Smalø, "Tilting in abelian categories and quasitilted algebras", theorem 4.6. Anyway, the proof always requires torsion pairs, and I have no knowledge about them, so I would like to understand a proof which avoids them.

REPLY [2 votes]: Here are some more details.
Lemma 3.2.4 written in the notation of Theorem 3.2.5 becomes the statement that $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M,N$ in $\mathcal Y(T)$ and $n \geq 0$. It is also an isomorphism when $n<0$ since then both sides are $0$.
For any $N$ in $\operatorname{mod} \Lambda$ there is a triangle $$N'\to P\to N\to N'[1]$$ with $N',P$ in $\mathcal Y(T)$, and for all $n \in \mathbb Z$ we get a commutative diagram
   $\require{AMScd}$
   \begin{CD}
\scriptsize
{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(M,N'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N'[n+1]) @>>>  \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n+1])\\
    @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\
    \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TP[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TP[n+1])
   \end{CD}
Since all the other downward arrows are isomorphisms, it follows from the five lemma that the middle downward arrow is an isomorphism. So $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M$ in $\mathcal Y(T)$, $N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$. A similar argument in the other variable gives that $\phi_{M,N[n]}$ is an isomorphism for all $M,N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$.
This is the basis for the induction. The inductive step uses triangles of the form $$Y' \to Y \to Y'' \to Y'[1]$$ with $\ell(Y') = \ell(Y)-1$ and $\ell(Y'') = 1$. We apply the five lemma to the commutative diagram
   $\require{AMScd}$
   \begin{CD}
\scriptsize
{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n]) @>>>  \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n+1])\\
    @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\
    \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n+1])
   \end{CD}
and a similar version for the other variable to conclude that $$\phi_{X,Y}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(X,Y) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY)$$ is an isomorphism for all $X,Y$ in ${\mathbf D^b(\operatorname{mod} \Lambda)}$.<|endoftext|>
TITLE: Continuity of length and area
QUESTION [10 upvotes]: Let $C_n$ be a sequence of rectifiable simple closed curves in $\mathbb{R}^2$ that converge to a rectifiable simple closed curve $D$ in the Hausdorff topology.  It is easy to construct examples where
$$\limsup_{n \mapsto \infty} \text{length}(C_n) \neq \text{length}(D).$$
Question 0: I actually do not know any examples where the limit on the LHS does not exist.  Can anyone provide one?
Question 1: In all the examples I know of, we have
$$\limsup_{n \mapsto \infty} \text{length}(C_n) \geq \text{length}(D).$$
Must this always hold?
Question 2: If the $C_n$ bound convex regions, must we have
$$\lim_{n \mapsto \infty} \text{length}(C_n) = \text{length}(D)?$$
Question 3: Let $R_n$ be the region bounded by $C_n$ and let $U$ be the region bounded by $D$.  Must we have
$$\lim_{n \mapsto \infty} \text{area}(R_n) = \text{area}(U)?$$

Edit: In the original version, I had the inequality in Question 1 backwards (thanks to Emil Jeřábek and kaleidoscop for pointing that out!).  This has now been corrected.  The examples I had in mind were a little easier than the ones in kaleidoscop's answer.  Namely, let $C_n$ consist of the union of the sets
$$\{\text{$(t,0)$ $|$ $0 \leq t \leq 1$}\}$$
and
$$\{\text{$(1,t)$ $|$ $0 \leq t \leq 1$}\}$$
with a "staircase" that starts at $(0,0)$, goes $1/n$ up, then goes $1/n$ to the right, then $1/n$ up, then $1/n$ to the right, etc, ending at $(1,1)$.  Each $C_n$ has length $4$.  However, the $C_n$ converge to the union of the sets
$$\{\text{$(t,0)$ $|$ $0 \leq t \leq 1$}\}$$
and
$$\{\text{$(1,t)$ $|$ $0 \leq t \leq 1$}\}$$
and
$$\{\text{$(t,t)$ $|$ $0 \leq t \leq 1$}\},$$
which has length $2+\sqrt{2}$.

REPLY [5 votes]: The answer of question 1 is positive. In fact, you can even replace the $\limsup$  by a $\liminf$: this is Golab's theorem that the $1$-dimensional Hausdorff measure is lower semi-continuous on the set of compact sets of the plane. I did not find a good reference to this theorem, but it is cited in a paper by Raphaël Cerf which seems to contain other relevant information.
I think the answer to question 2 is also positive. First, $D$ must bound a convex domain; for each $\varepsilon>0$ we can then find two convex approximations $D_i,D_o$ of $D$, one on the inside and one on the outside, which are disjoint from $D$ and at Hausdorff distance less than $\varepsilon$ from $D$, and of length $\varepsilon$-close to the length of $D$. Now, when $C_n$ is close enough to $D$, it must lie in the annulus bounded by $D_i$ and $D_o$. Using the projections $P_n$ to the domain bounded by $C_n$ and $P_i$ to the domain bounded by $D_i$, and recalling they are $1$-Lipschitz, we get (denoting lengths by $|\cdot|$):
$$ |D|-\varepsilon \le |D_i| = |P_i(C_n)| \le |C_n| = |P_n(D_o)| \le |D_o| \le |D|+\varepsilon $$
I also think the answer to question 3 is positive. I will not sketch a proof, but it should be doable using the regularity of Lebesgue measure and and the fact that the Minkowski content of a domain with rectifiable boundary is the length of its boundary. Maybe you will need to adapt the proof of that fact, but the key-word "Minkowski content" should get you going.<|endoftext|>
TITLE: Diameter of random segment intersection graph?
QUESTION [9 upvotes]: I have an even number of points $n$ randomly distributed (uniformly) in a disk.
Then the points are randomly connected to form $n/2$ segments, a perfect
matching.
Finally, I form the intersection graph $G$ whose nodes are the segments,
and each of whose edges represents a proper intersection between a pair of segments.
An example is shown below:

 


My question is:

Q. What is the diameter of $G$ as $n \to \infty$?

I have some evidence that the diameter
approaches a constant $> 1$,
perhaps $4$ or $5$, 
but to be honest,
the evidence is not robust.

Added.
This may not be helpful, but for $n=1188$, $n_{\mathop{segs}}=n/2=594$, one
simulation resulted in $G$ having $37,117$ edges, 
mean vertex degree $125$, one connected component, and diameter $4$:

REPLY [2 votes]: Here is a way of thinking.  It may turn out to be misleading, but it may also suggest a
way to think properly about this problem.
Lets take a random instance of a matching and reconstruct it edge by edge. We will
start by maximizing the diameter of the intersection graph. To me this means making
a large cycle of crossing sticks, so that the intersection graph is a path or a cycle.
Suppose we are lucky, and manage to use 2/3 (say) of the sticks to create a subarrangement
of (induced) maximal possible diameter.
Now we place the rest of the sticks, which can only decrease the induced diameter.
With high probability, the diameter is reduced by a minimum ratio with each
new stick or perhaps a small number of new sticks.  In any case, I see O(log G)
many sticks needed to bring the diameter back down.  For large G, this still
leaves many sticks left.
It still leaves the possibility that the diameter grows like log log G, but the idea
of finding the subconfiguration inducing the largest diameter and then reducing
it appeals to me as an inroad to the problem.<|endoftext|>
TITLE: Origin of group theory problem (bound on number of Sylow subgroups)
QUESTION [16 upvotes]: This problem (prove that the number of Sylow subgroups of a finite group $G$ is bounded by $\frac{2}{3}|G|$) posted on MSE proved rather difficult to solve. The OP has been silent about where the problem came from, even though he/she has been asked. Has anyone seen this result before? If so, where?

REPLY [8 votes]: Since this question has resurfaced (after 3 years) let me
say something about the origin of this Monthly problem.

A few years back a colleague gave me the task of making
up the algebra preliminary exam for our first-year
graduate students. Among the group theory problems that made
my initial list were:
(1) Show that there is no finite group with more
Sylow subgroups than elements.
(This later evolved to: Let $G$ be a finite group.
Show that the number of nontrivial Sylow subgroups of $G$
is at most $\frac{2}{3}|G|$.)
(2) Call a positive integer $c$
curious if there is a nontrivial finite group
$G$ such that, for every prime $p$ dividing $|G|$,
the number of Sylow $p$-subgroups is exactly $c$.
Find all curious integers.
(3) You are told that, for certain $n$ and $p$,
$S_n$ has exactly $n$ Sylow $p$-subgroups.
What are the possible pairs $(n, p)$?
(4) Prove or disprove: There is no group $G$ of
square-free order such that, for every prime $p$ dividing
$|G|$, the number of Sylow $p$-subgroups is
$p + 1$.
(5) Let $G$ be a finite group and let $S\subseteq G$
be a subset. Show that if $S$ has nonempty intersection
with each conjugacy class of $G$, then the
subgroup generated by $S$ is $G$.
(6) Show that if the conjugacy classes of $G$
have size at most $4$, then $G$ is solvable.
I ultimately used Problem (6) on the exam. My notes to myself
say about (6): Here $4$ can be replaced by any number up to $19$,
but the statement becomes false for $20$ --- $A_5$.

I decided Problem (1) was too hard for 
first-year grad students,
so I sent it to the problems section of the Monthly.
My solution to Problem (1) was similar to
the one by Richard Stong, which was published by the Monthly.
In particular, I also used Burnside's normal complement theorem
and the estimate $\sum_{p}\frac{1}{p^{2}} < \frac{1}{2}$.

Once the problem was type-set by the Monthly, I got a chance to
proofread it. I sent my final OK to the Monthly
on May 10, 2015. Strangely, the problem appeared on
the Art of Problem Solving website the next day, worded in exactly
the same way. The day after that it appeared at MSE, worded exactly
the same way. It did not appear in print in the Monthly for
another 3 months.

One last comment about this problem. When formulating the problem
I asked myself: is it possible to show that there is an injective
function $f:\textrm{Syl}(G)\to G$ such that $f(P)\in P$ for all $P$?
I don't know the answer to this version of the question.<|endoftext|>
TITLE: Avoiding countable subgroups of general uncountable groups
QUESTION [8 upvotes]: The following problem is a general form of another problem (motivation is available there). Initially, the problems were posted together, but the first one is solved below, a solution that does not apply to the more focused version of the problem, which is the needed scenario.
Problem 1. Let $G$ be an uncountable group, $H$ a countable subgroup of $G$, and $g\in G\smallsetminus H$.
Is there necessarily an element $x\in G\smallsetminus H$ such that $g\notin\langle H\cup \{x\}\rangle$?

REPLY [10 votes]: Here's another counterexample for Problem 1 not using the fancy (non-explicit) construction of Shelah.
Fix an odd prime $p$. Let $A=\mathbf{F}_p((t))$ be the ring of Laurent series over the field $\mathbf{F}_p$ on $p$ elements; we only view it as an abstract group, and let $A_0$ be the set of series $\sum a_nt^n$ in $A$ with $a_0=0$. Let $B$ be the non-unital subring $t^{-1}\mathbf{F}_p[t^{-1}]$. Define a symplectic form on $A_0$ by $\langle t^n,t^m\rangle=0$ if $n+m=0$ and $\langle t^n,t^{-n}\rangle=1$ for all $n\ge 0$, by extending it by "formal" linearity (namely
$$ \langle\sum a_n t^n,\sum b_nt^m\rangle=\sum_{n>0}a_nb_{-n}-a_{-n}b_n\qquad\qquad ).$$
That it is non-degenerate is straightforward. Observe that $B$ is a maximal isotropic subspace.
Define a (nilpotent) uncountable group $G$ as the set of pairs $(x,t)$ with $x\in A_0$, $t\in\mathbf{F}_p$, and group law $(x,t)(x',t')=(x+x',t+t'+\langle x,x'\rangle)$. Then $B$ is a countable subgroup and every group properly containing $B$ contains $(0,1)\notin B$.<|endoftext|>
TITLE: Why is it possible to normalize the Haar measure on the quotient?
QUESTION [5 upvotes]: I just asked a question which is related to the one I'm about to ask, but I realized my question can be reduced to the following: let $G$ be a locally compact abelian group with Haar measure $\mu$, and $H$ a discrete subgroup of $G$.  Then $G/H$ is locally compact with Haar measure $\bar{\mu}$.  
I believe it should be possible to normalize $\bar{\mu}$ to make it have the following property: if $W \subseteq G$ is any measurable set and the projection $\pi: G \rightarrow G/H$ maps $W$ injectively into the quotient, then $\mu(W) = \bar{\mu}(\pi(W))$.

REPLY [4 votes]: This is possible. Suppose that $\mu(W)<\infty$ and that $\pi(W)$ is measurable.
Then $\mu(W)$ is the supremum of all compact sets $K$ contained in $W$, as $\mu$ is a Radon measure.
The same holds for $\pi(W)$.
For a compact set $K$ the measure $\mu(K)$ is the infimum of all integrals $\int_Gf(x)d\mu(x)$ where $f\ge 0$ is continuous of compact support. (This is part of the proof of Riesz's representation theorem.) So the question boils down to the following: Can $\bar\mu$ be normalized such that
$$
\int_Gg(x)d\mu(x)=\int_{G/H}\sum_{h\in H}g(xh)d\bar\mu(x).
$$
Now the sum map $C_c(G)\to C_c(G/H)$ is surjective hence one can take the left hand side as a definition for a positive linear functional on $C_c(G)$ which by Riesz gives a unique Radon measure.<|endoftext|>
TITLE: Why there is a relation among the second-order minors of a symmetric $4\times 4$ matrix?
QUESTION [35 upvotes]: A $4\times 4$ symmetric matrix
$$
\left(
\begin{array}{cccc}
 a_{11} & a_{12} & a_{13} & a_{14} \\
 a_{12} & a_{22} & a_{23} & a_{24} \\
 a_{13} & a_{23} & a_{33} & a_{34} \\
 a_{14} & a_{24} & a_{34} & a_{44} \\
\end{array}
\right)
$$
contains exactly 21 minors of order 2, but there is a linear combination of them, namely
$$
-(a_{13} a_{24}-a_{14} a_{23})+(a_{12} a_{34}-a_{14} a_{23})-(a_{12} a_{34}-a_{13} a_{24})\, ,
$$
which vanishes, as opposed as to the case of symmetric $3\times 3$ symmetric matrices, whose 6 minors of order two are linearly independent.

BIG QUESTION: Why is that? I mean, what is the theory behind such a phenomenon?
PHILOSOPHICAL QUESTION: What is the "true number" of minors (from the example above), 20 or 21?

I already gave myself an explanation, but I still cannot see the big picture. I would be grateful if anyone pointed out a reference, sparing me the efforts of reinventing the wheel.
If an $n\times n$ matrix $A$ is regarded as an element of $V\otimes_{\mathbb{K}} V^\ast$, with $V\equiv \mathbb{K}^n$, then there is an obvious way to extend $A$ to a $\mathbb{K}$-linear map
$$
A^{(k)}:\bigwedge^kV\longrightarrow\bigwedge^kV^\ast\, .
$$
If $A$ is symmetric, then so is $A^{(k)}$, i.e., there is a (polynomial of degree $k$) map
$$
S^2(V^\ast)\ni A\stackrel{p}{\longmapsto} A^{(k)}\in W^k:=S^2\left( \bigwedge^kV ^\ast \right)\, .
$$
Observe that $W_k$ has dimension $\frac{{n\choose k}\left({n\choose k}+1\right)}{2}$, which is 21 for $n=4$ and $k=2$. So, I was led to identify $W^k$ with the space of "formal minors" of order $k$ of a $n\times n$ matrix (indeed, the entries of $A^{(k)}$ are precisely such minors).
How to explain now the dependency of three of them?
There is a linear map
$$
S^2\left( \bigwedge^2V ^\ast \right)\ni\rho\odot\eta\stackrel{\epsilon}{\longmapsto}\rho\wedge\eta\in \bigwedge^4V ^\ast\equiv\mathbb{K}\, ,
$$
and it can be proved that $\epsilon\circ p=0$, i.e., that $p$ takes its values in the subspace
$$
W^2_0:=\ker\epsilon\, ,
$$
which has dimension 20 (vanishing of the above linear combination corresponds precisely to the equation $\epsilon(p(A))=0$).

SIDE QUESTION: How to define this $W_0^k$ for arbitrary $n$ and $k$? It should be the "linear envelope" of the image of $p$: but how to exhibit it explicitly?

Probably, the theory of representations may answer this: $W^k$ is not always an irreducible $\mathfrak{gl}(n,\mathbb{K})$-module, and $W_0^k$ is the only irreducible component which contains the image of $p$. Provided this guess is true, it doesn't help me finding the expression of $W_0^k$ in terms of tensors on $V$.

REPLY [14 votes]: This is a comment regarding Robert Bryant's conjecture: It is indeed true and not hard if you know the representation theory of $GL(V)$ well enough. In fact, the following stronger statement is true:

The only common constituent of $(S^2 V)^{\otimes k}$ and $(\bigwedge^k V)^{\otimes 2}$ is the representation with highest weight $2 \alpha_k$.

One must then check that this constituent lies in the symmetric parts of the tensor powers, which is probably most easily done by pointing out that the $k \times k$ minors of a symmetric matrix give a nonzero $GL(V)$ equivariant map $(\bigwedge^k V)^{\otimes 2} \longrightarrow (S^2 V)^{\otimes k}$.
Write weights as partitions $(\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n)$, so $S^2 V$ has height weight $(2,0,0,\ldots,0)$ and $\bigwedge^k V$ has highest weight $(\overbrace{1,1,\ldots,1}^k,0,0,\ldots,0)$. Tensoring with $S^k V$ and with $\bigwedge^k V$ is described by the Pieri rule and its transpose. From the Pieri rule, any irreducible constituent $V_{\lambda}$ of $(S^2 V)^{\otimes k}$ has $\sum \lambda_i=2k$ and $\lambda_{k+1} = \lambda_{k+2} = \cdots = \lambda_n=0$. From the transpose of the Pieri rule, any $V_{\lambda}$ in $(\bigwedge^k V)^{\otimes 2}$ has all $\lambda_i \leq 2$. The only way to satisfy these conditions is $\lambda = (\overbrace{2,2,\ldots,2}^k,0,0,\ldots,0)$.
More generally, the only common constituent of $(S^{\ell} V)^{\otimes k}$ and $(\bigwedge^k V)^{\otimes \ell}$ is the representation of highest weight $\ell \alpha_k$. 
It is definitely in $S^{\ell} (\bigwedge^k V)$. If $\ell$ is even, then it is in $S^k(S^{\ell} V)$, otherwise it is in $\bigwedge^k (S^\ell V)$.<|endoftext|>
TITLE: Ref. request: Additive probability measure on $\mathcal P({\bf N})$ supplies subset of $\mathbf R$ without Baire property
QUESTION [6 upvotes]: ZFC proves, among the other things, the existence of a (finitely) additive probability measure $\theta: \mathcal P(\mathbf N) \to \mathbf R$ on the power set of $\mathbf N$ such that $\theta(X) = 0$ for every finite $X \subseteq \mathbf N$, which I will shortly refer to as an ADPM (where "D" stands for "diffuse", a term used e.g. by E. K. van Douwen in Finitely additive measures on $\mathbb{N}$, Topology Appl. 47 (1992), No. 3, 223-268): The proof is actually based on the Hanh-Banach theorem, which, just to recall something that is widely known, is implied by, but not equivalent to, the axiom of choice, see e.g. Section 23.19 in E. Schechter's Handbook of Analysis and its Foundations (Academic Press, 1996), so that, in principle, the existence of $\theta$ can even be established, for the record and those who care, in a weaker system than ZFC. On the other hand, ZF proves the following:

Proposition. The existence of an ADPM implies the existence of a subset of $\mathbf R$ without the Baire property.

But it follows from Theorem 1(3) in R. M. Solovay's celebrated paper A model of set theory in which every set of reals is Lebesgue measurable (Ann. of Math., 2nd
Ser. 92 (1970), No. 1, 1-56) that the existence of an uncountable transitive model of ZFC + I, where I is the statement: "There exists an inaccessible cardinal", supplies an uncountable transitive model of ZF in which every subset of $\mathbf R$ has the Baire property.
So, putting it all together, we see that the existence of an ADPM is provable in ZFC, but independent of ZF. With this said, my question is simply:

Do you know a reference where the proposition above is stated and proved?

To be clear: I am not looking for a proof, and am pretty sure the result is somewhere in Schechter's handbook, but couldn't find it.

REPLY [8 votes]: A very good reference for various forms of AC is the book Howard, Rubin: Consequences of the Axiom of Choice, AMS, 1998. (See AMS website or Google Books.)  This book contains a large database of various consequence of choice and tables showing relations between them.
There is also a website where you can search for relations between various forms mentioned in this book.
In this book we can find:

FORM 142. $\neg$PB: There is a set of reals without the property of Baire. Jech [1973b] p 7 and note 28.
FORM 222. There is a non-principal measure on $\mathcal P(\omega)$. Pincus/Solovay [1977] and note 147.

Then in the table which shows implications between various forms we find:

222 142 (1) Pincus [1972c]
  142 222 (3) Pincus [1972c]

(1) = "The implication is provable."
(3) = "The implication is not provable in ZF".
The reference given there is:

[1972c] D. Pincus: The strength of the Hahn-Banach theorem, Proc. of the Victoria Symp. on Nonstandard Analysis, Lecture Notes in Math. 369, Springer, Heidelberg, 1973, 203-248. DOI: 10.1007/BFb0066014.

(The correct year for this reference should probably be 1974 not 1973, see the comment below. I left it here in the same way it is written in the book I quoted from.)
D. Pincus writes there that this implication "is due to Solovay. It is stated without proof in [20]. Since it may interest readers of this paper we include our own proof at the end of §I." Where [20] is R. M. Solovay, A model of set theory in which every set of reals is Lebesgue measurable, Ann. of Math. 92 (1970), pp. 1-58; DOI: 10.2307/1970696.
A proof given there seems to be similar to the proof sketched (very briefly) in this comment.

EDIT: Since the OP also mentioned Schechter's Handbook of Analysis and Its Foundations, I will add that in this book we can find the result in Chapter 29.
In 29.37 the following three principles are listed and shown to be equivalent:

$(\ell_\infty)^* \supsetneqq \ell_1$; there is a bounded functional on $\ell_\infty$ that cannot be represented by a member of $\ell_1$.
There exists a measurable space $(\Omega,\mathcal S)$ and a bounded scalar-valued charge on $\mathcal S$ that is not a measure.
There exists a probability charge on $(\mathbb N,\mathcal P(\mathbb N))$ that vanishes on finite sets.

In 29.38 it is shown that they imply: (NBP) There exists a subset of $\{0,1\}^{\mathbb N}$ that lacks Baire property.
As far as the references are concerned, the author says that:

This implication was first stated without proof in Solovay [1970]; the first published proof apparently is that of Pincus [1974]. The slightly shorter proof below is essentially that of Taylor; it was published in Wagon [1985].

The first two references have already been mentioned. The third one is S. Wagon, The Banach-Tarski Paradox, Encyclopedia Math. Appl. 24, Cambridge Univ.
Press, Cambridge, 1985. The corresponding result is indeed given in as Theorem 13.5 in this book.<|endoftext|>
TITLE: Blow-up as polar coordinates?
QUESTION [5 upvotes]: While doing some explicit calculations involving a blow-up of the plane in a point, I realised what I was doing was basically writing things in polar coordinates. Somewhat astonished that I hadn't made the connections

tangent lines through point $\leftrightarrow$ lines $\theta=const$ in polar coordinates
exceptional divisor $\leftrightarrow$ the line $r=0$ in polar coordinates

before, I mentioned it to some other algebraic geometry people, and none of them had thought of it either. It's quite obvious when you see it, but somehow it's never mentioned anywhere, which may be because a) the geometric picture as usually presented is what you really want, who cares about coordinates anyway; or b) this isn't the actual motivation behind the construction, as originally conceived.
So, I propose the following conjectural origin story of the blow-up:
Look at picture of singular curve "Hmm, for no apparent reason I wonder what that looks like in polar coordinates." draw the picture "Hey, the curve isn't singular anymore!" work out how to express this in terms of polynomials, like a good algebraist -- and then you recover the usual text-book presentation of the blow-up.

Question: Is this story complete rubbish?

or if you will, I suppose I could just have asked

Question': what is the historical origin of the blow-up construction?

REPLY [6 votes]: Answer to the question: what is the historical origin of the blow-up construction?
Time ago I was reading Julian Coolidge's book "A history of geometrical methods". 
At page 197 he explain the idea of the great Isaac Newton in order to understand a singular point of a plane curve:
...a remarkable ingenious device called the analytic triangle. 
The starting point to construct such analytic triangle (that nowdays is called Newton's polygon) is (in Coolidge's words):  Suppose that the point in which we are interested is the origen. 
We put $y=vx^{\mu}$ and see out those terms ...
The above is exactly a blow-up.
So I should put Newton at the origen of the blow-up construction.<|endoftext|>
TITLE: Saturation of null ideal
QUESTION [10 upvotes]: In ZFC, can we find more than continuum many non null sets of reals whose pairwise intersections are null?

REPLY [6 votes]: The answer is no. Observe that it is enough to show that, consistently, the density of the boolean algebra $\mathcal{P}(\mathbb{R}) / Null$ is continuum (which means that there are continuum many non null sets such that every non null set contains one of them). For this, it suffices to show that, consistently, ($\star$) holds:
($\star$): $2^{\omega} = \kappa = \kappa^{< \kappa}$ and every non null set of reals has a non null subset of size less than $\kappa$.
Claim 1: If $\kappa = 2^{\omega}$ is real valued measurable, then ($\star$) holds.
Proof: That $\kappa^{< \kappa} = \kappa$, is a result of Prikry (Theorem 22.2 in Jech's book). Let $\langle x_i : i < \kappa \rangle$ be a one-one enumeration of a non null set of reals $X$. Force with the null ideal of a measure on $\mathcal{P}(\kappa)$. Let $j: V \to M$ be the generic elementary embedding with critical point $\kappa$. Note that, in $M$, $X$ is a non null initial segment of $j(X)$ - this is because forcing with a measure algebra preserves old non null sets; so this also holds in $V$.
Claim 2: ($\star$) holds in the random real model.
Proof: (Communicated by Arnold Miller) Let $V \models GCH$, and $P$ add $\omega_2$ random reals $\langle r_i : i < \omega_2 \rangle$. Suppose $X = \langle x_i : i < \omega_2 \rangle$ is non null. Define $X_i = V[\langle r_j : j < i \rangle] \cap X$. Choose $\alpha < \omega_2$ of cofinality $\omega_1$ such that no null set coded in $V[\langle r_j : j < \alpha \rangle]$ covers $X_{\alpha}$. So $X_{\alpha}$ is non null in $V[\langle r_j : j < \alpha \rangle]$ and it remains so in $V[\langle r_i : i < \omega_2 \rangle]$.<|endoftext|>
TITLE: Tensor products over operads and bar constructions
QUESTION [9 upvotes]: Let $O$ be an operad in spaces, $A$ an $O$-algebra and $R$ an right $O$-module. One can define $R \otimes_O A$ as the coequalizer of the two maps $ROA$ to $RA$. One can also define $B(R,O,A)$ (as in geometry of iterated loop spaces) to be the geometric realization of a simplicial space with space of $n$-simplicies given by $RO^nA$. Does anyone know a reference in the literature for sufficient conditions for when these two constructions yield weakly homotopy equivalent spaces?

REPLY [7 votes]: The space $B(R,O,A)$ is the tensor product $B(R,O,O)\otimes_OA$. It is a standard fact that the right module $B(R,O,O)$ is cofibrant in the projective model structure of right modules whenever $R$ is levelwise cofibrant.
By Theorem 15.1.A.(a) of Fresse's "Modules over operads and functors", if $A$ is cofibrant (as a space) and $O$ is levelwise cofibrant, then the functor $-\otimes_OA$ preserves weak equivalences between cofibrant (in the projective model structure) right modules. In particular the two constructions you want to consider coincide up to weak equivalence if $R$ is cofibrant in the projective model structure.
Using Theorem 15.1.A.(b) of the same book, you can also show that the two constructions coincide up to weak equivalence if $A$ is cofibrant as an algebra, $O$ is levelwise cofibrant and $R$ is levelwise cofibrant.<|endoftext|>
TITLE: How networks with high largest eigenvalues are more robust?
QUESTION [5 upvotes]: In the literature, it is sometimes indicated that network with high value of largest eigenvalue (either adjacency matrix or its Laplacian counterpart) are more robust against link/node removals. However, these statements are usually NOT accompanied with references. 
I am looking for some explanation on why is this so? Or pointers to some work that investigate/explain this.
I wonder if this statement is simply the result of difference between link density? I doubt its so simplistic. Is there any study on network robustness comparing networks with same link density but different largest eigenvalues (or spectral radius)?

REPLY [4 votes]: The following papers might contain relevant information to your question:
Robustness of networks against viruses: the role of the spectral radius, by
A. Jamakovic, R.E. Kooij, P. Van Mieghem, E.R. van Dam
https://www.nas.ewi.tudelft.nl/people/Rob/telecom/jamakovic.pdf
Epidemic Spreading in Real Networks: An Eigenvalue Viewpoint, by 
Yang Wang, Deepayan Chakrabarti, Chenxi Wang and Christos Faloutsos 
http://www-2.cs.cmu.edu/afs/cs.cmu.edu/user/christos/www/PUBLICATIONS/srds03-virus.pdf<|endoftext|>
TITLE: A new generalisation of Fermat's little theorem?
QUESTION [11 upvotes]: I would like to share a personal result, which I interpret as a "generalisation of Fermat's little theorem". I know that there exist already several generalisations (e.g. Euler's) but I would like to know whether the result I found is something original or not. Let me explain...
Some years ago, my researches on binary representation of numbers lead me to this sequence of naturals: 1, 2, 1, 2, 3, 6, 9, 18, 56, ... . After some researches, I found that this was already indexed as A001037 sequence in OEIS; this is known as number of "binary Lyndon words of length $n$" or "$n$-bead necklaces with beads of 2 colors". Such sequence can easily be generalised to other bases than 2. After naming $\lambda_a(n)$ the number of $n$-bead necklaces with beads of $a$ colors with $a \ge 2$, I found that
$$ a^n = \sum_{d|n}  d \lambda_a(d) $$
For example,
$$ 2^6 = 1\lambda_2(1) + 2\lambda_2(2) + 3\lambda_2(3) + 6\lambda_2(6) = 2 + 2 + 6 + 54 = 64 $$
This is NOT an original result since you can find this relationship on the afore-mentioned OEIS page, at least. 
QUESTION 1: Can someone provides references for a proof of this equation (papers, books, ...)?
Now, after defining
$$ \sigma_a(n) \triangleq \sum_{\substack{ d|n \\ 1<d<n}} d \lambda_a(d) $$
the previous equation can be rewritten
$$ a^n = a + n\lambda_a(n) + \sigma_a(n) $$
Then, using modular arithmetic, we obtain a generalisation of Fermat's little theorem:
$$ a^n \equiv a + \sigma_a(n) \pmod {n} $$
Note that this is indeed a generalisation since it applies on any natural $n$ and it boils down to 
$$ a^p \equiv a \pmod {p} $$
for any prime $p$, that is the Fermat's little theorem.
This relationship looks appealing to me, at least for studying Fermat pseudoprimes: these are the composites $n$ having the "rare" property that $\sigma_a(n) \equiv 0 \pmod {n}$.
QUESTION 2: Is this generalisation of Fermat's little theorem a known result? If yes, could you please provide references?
Thanks,
PS: Please, be indulgent: this is my very first post on MO!

REPLY [7 votes]: Here's a massive generalization, with references.
Consider a sequence $\{a_n \}_{n\ge 1}$ of integers. We'll say it satisfies the necklace congruences if $a_{n} \equiv a_{n/p} \mod n$ whenever $p \mid n$ ($p$ prime).
Here are some equivalent formulations:

$a_{p^{k+1} m} \equiv a_{p^k m} \mod {p^{k+1}}$ for all $p,m,k$
$\sum_{d \mid n} a_d \mu(n/d) \equiv 0 \mod n$ for all $n$
The Mobius function can be replaced by any arithmetic function $f$ satisfying $\sum_{d \mid n} f(d) =0 \mod n, f(1) = \pm 1$.

So any such sequence gives rise to an integer sequence $\lambda_{\tilde{a}}(n) := \frac{1}{n}\sum_{d \mid n} a_d \mu(n/d)$ which satisfies $a_n = \sum_{d \mid n} d \lambda_{\tilde{a}}(n)$ by Mobius inversion. One can then define $\sigma_{\tilde{a}}(n):=\sum_{d \mid n, 1<d<n} d\lambda_{\tilde{a}}(d)$ and obtain the following generalization of Fermat's little theorem:
$a_n \equiv a_1 + \sigma_{\tilde{a}}(n) \mod n$
Now, let's go back to necklace congruences. A less obvious equivalence is the following:
Theorem: $\{a_n\}_{n \ge 1}$ satisfies the necklace congruences iff $\zeta_{a}(x):=\exp(\sum_{n\ge1} \frac{a_n}{n}x^n)$ has integer coefficients.
Proof: Write, $\zeta_a$ formally as $\prod_{n\ge 1} (1-x^n)^{-\frac{b_n}{n}}$. The $b_n$'s are uniquely determined, and in fact (by taking logarithmic derivative) it can be seen that they are $b_n = \sum_{d \mid n} a_d \mu(n/d)$. Now one direction is immediate and the other requires an inductive argument. $\blacksquare$
(Reference: Exercise 5.2 (and its solution) in Richard Stanley's book "Enumerative Combinatorics, vol. 2")
As Sergei remarked, for any integer square matrix $A$, the sequence $a_n := \text{Tr}(A^n)$ satisfies the necklace congruences. This is immediate from the last theorem, as the corresponding "zeta" function $\zeta_a$ is just:
$\exp( \sum_{n \ge 1} \frac{\text{Tr}((Ax)^n)}{n}) = \exp (\text{Tr} (- \ln (I -Ax)))=\det(I-Ax)^{-1} \in \mathbb{Z}[x]$
In particular, by taking a 1 on 1 matrix, we recover your observation. 
Another generalization is $a_n = [x^n]f^n(x)$, where $f \in \mathbb{Z}[[x]]$. By taking the linear polynomial $f(x)=1+ax$ we recover your example, by taking $f(x)=(1+x)^m$ we get $a_n = \binom{nm}{n}$.
The really interesting feature is that this notion has a local version. Specifically, theorems of Dieudonne-Dwork and Hazewinkel give the following result:
Theorem: Given $\{ a_n \}_{n\ge 1} \subseteq \mathbb{Q}_{p}$ ($p$-adic numbers), the following are equivalent:

$\zeta_{a}(x) := \exp(\sum_{n\ge 1} \frac{a_n }{n}x^n)\in\mathbb{Z}_{p}[x]$
$\exp( \sum_{n \ge 1} p \frac{a_n - a_{n/p}}{n} x^n) \in 1+px\mathbb{Z}_{p}[x]$ (I defnite $a_{n/p}=0$ if $p\nmid n$)
$\sum_{n \ge 1}  \frac{a_n - a_{n/p}}{n} x^n) \in \mathbb{Z}_{p}[x]$
$p \mid n \implies a_n \equiv a_{n/p} \mod {n\mathbb{Z}_{p}}$

Applying this simultaneously for all $p$, we recover the previous theorem.
A reference for this theorem is "A Course in p-adic Analysis" by Alain M. Robert (the theorems are stated there in a much more general context).<|endoftext|>
TITLE: What is the analytic conductor of this Hecke L-function?
QUESTION [5 upvotes]: Following Iwaniek and Kowalski, S5.10, page 130 we consider an angle character $\xi_k$ on the Gaussian integers $\mathbb Z[i]$ defined by
$ \xi_k(\mathfrak a) = \left(\frac{\alpha}{|\alpha|}\right)^k $
where $k \equiv 0 \pmod 4$. 
This gives an $L$-function $L(s,\xi_k)$ and a functional equation described by the following data:
\begin{align*}
 L(s, \xi_k) &= \sum_{\mathfrak a} \xi_k(\mathfrak a) (N\mathfrak a)^{-s} \\
 \gamma(s, \xi_k) &= \pi^{-s} \Gamma\left( s + \frac{\left\lvert k \right\rvert}{2} \right) \\
 \Lambda(s, \xi_k) &= \pi^{-s} \Gamma\left( s + \frac{\left\lvert k \right\rvert}{2} \right) L(s, \xi_k) = \Lambda(1-s, \xi_k)
\end{align*}
Here $L(s,\xi_k)$ has degree $d=2$ and conductor $q=4$ (the discriminant of the Gaussian integers).
I&K had earlier written a prime number theorem $\psi(f,x) = rx - \frac{x^{\beta_f}}{\beta_f} + O\left(\sqrt{\mathfrak q(f)}x e^{ -\frac{c}{2d^4}\sqrt{\log x}} \right)$ which held for all $L$-functions, and apply this to deduce that
$$ \sum_{|\pi| \le x} \left(\frac{\pi}{|\pi|}\right)^k = 4\delta_{k=0} \operatorname{Li}(x) + O\left(|k|xe^{-c\sqrt{\log x}}\right). $$
What I'm confused about is how the analytic conductor $\mathfrak q(\xi_k)$ was computed; the calculation seems to suggest that it is $O(|q|^2)$. But the definition of the analytic conductor earlier was
$$ \mathfrak q(f) \overset{\text{def}}{=} q(f) \prod_{j=1}^d \left( |\kappa_j| + 3 \right)$$ but the $\gamma$ factor given above is not in the correct form to carry out this definition.
Is there something I'm missing? I'm guessing there might be some property of the $\Gamma$ function that transform $\gamma(s, \xi_k)$ to the right form but I'm not sure what it might be.

REPLY [2 votes]: Got it thanks to Lucia's comment. The main point is to use the duplication formula; we have
$$ \Gamma\left(s+\frac{|k|}{2}\right) \cdot \sqrt\pi
= 2^{s+\frac{|k|}{2}-1} \Gamma\left(\frac12s+\frac{|k|}{4}\right) \Gamma(\left(\frac12s+\frac{|k|}{4}+\frac12\right).$$
Since the conductor in our case is $q = 4$, this exactly knocks out the factor of $q^{-s/2}$ that is attached to the completed $L$-function. There is in fact an extra constant factor of $\pi^{-\frac12} 2^{\frac{|k|}{2}-1}$ floating around, but it has no effect on the results; the functional equation remains true and it goes away when we take the logarithmic derivative.<|endoftext|>
TITLE: How to prove this polynomial always has integer values at all integers?
QUESTION [47 upvotes]: Let $m$ be any positive integer.
$$
P_m(x)=\sum_{i=0}^{m}\sum_{j=0}^{m}{x+j\choose j}{x-1\choose j}{j\choose i}{m\choose i}{i\choose m-j}\frac{3}{(2i-1)(2j+1)(2m-2i-1)}.
$$
Question: $P_m(x)$ always has integer values at all integers.
Some remarks:
(1)  A polynomial is $\mathbb Z$-valued iff its unique expansion in basis $\left\{{x\choose k}\right\}$ has $\mathbb Z$-coefficients. This idea was Allen Knutson's.
For $m=1,2,3,4$, I list the polynomials in this basis, which is more clear that $P_m(x)$ is $\mathbb Z$-valued.
\begin{align*}
P_1(x)&=-4{x\choose 2}-2{x\choose 1}+2,\\
P_2(x)&=12{x\choose 4}+18{x\choose 3}+6{x\choose 2},\\
P_3(x)&=48{x\choose 6}+120{x\choose 5}+96{x\choose 4}+24{x\choose 3},\\
P_4(x)&=236{x\choose 8}+826{x\choose 7}+1070{x\choose 6}+610{x\choose
5}+134{x\choose 4}+4{x\choose 3}
\end{align*}
(2) Since ${-x+j\choose j}{-x-1\choose j}={x+j\choose j}{x-1\choose j}$, we have $P_m(-x)=P_m(x)$. Let $q_j(x)={x\choose 2j}+{-x\choose 2j}.$ Then
\begin{align*}
P_1(x)&=-2q_1(x)+2\\
P_2(x)&=6q_2(x)-6q_1(x)\\
P_3(x)&=24q_3(x)-72q_2(x)+48q_1(x)\\
P_4(x)&=118q_4(x)-704q_3(x)+1522q_2(x)-936q_1(x)\\
P_5(x)&=696q_5(x)-6900q_4(x)+30960q_3(x)-63252q_2(x)+38496q_1(x)\\
P_6(x)&=4824q_6(x)-71640q_5(x)+547572q_4(x)\\
&-2345904q_3(x)+4757916q_2(x)-2892768q_1(x).
\end{align*}
This idea was given by Wilberd van der Kallen.
(3) Let $S_m(x)=xP_m(x)$. Then
$$
S_m(x)=\sum_{i=0}^{m}\sum_{j=0}^{m}{x+j\choose 2j+1}{2j\choose j}{m\choose i}{j\choose i}{i\choose m-j}\frac{3}{(2i-1)(2m-2i-1)}.
$$
Note that ${2j\choose j}{m\choose i}{j\choose i}{i\choose m-j}\frac{1}{(2i-1)(2m-2i-1)}$ is an integer (I can prove this). So the question is equivalent to $$n|S_m(n)$$ for any positive integer $n,m$.
(4) It is easy to see that letting $|x|\le \left\lfloor \frac{m+1}{2} \right\rfloor$ be a integer, we have
$$
{x+j\choose j}{x-1\choose j}{j\choose i}{m\choose i}{i\choose m-j}=0.
$$
Then $P_m(n)=0$ for any integer $|n|\le \left\lfloor \frac{m+1}{2} \right\rfloor$.

REPLY [35 votes]: $$P_m(x)=\sum_{i=0}^{m}\sum_{j=0}^{m}\binom{x+j}{ j}\binom{x-1}{ j}\binom{j}{ i}\binom{m}{ i}\binom{i}{ m-j}\frac{3}{(2i-1)(2j+1)(2m-2i-1)}.$$
Our task is to show it takes integer values on integers.
Folowing Wadim Zudilin we put 
$$B_k(x)=\binom{x+k}{2k}+\binom{-x+k}{2k}.$$  
For $k\geq0$ the $B_k$ are even polynomials of degree $2k$ that take integer values on integers. One has $B_k(k)=1$ for $k\geq1$, but $B_0(0)=2$. Further $B_k(i)=0$ for $|i|<k$. 
So the matrix $$(B_k(i))_{0\leq k\leq m}^{0\leq i\leq  m}$$ is triangular.
Every even polynomial $f(x)$ of degree $2j$ is clearly a linear combination of $B_0,\dots,B_j$ and the coefficients are determined by $f(0),\dots,f(j)$.
When $f(0)=0$ it is actually a linear combination of  $B_1,\dots,B_j$. 
Rewrite $P_m(x)$ as $$P_m(x)=\sum_k d(m,k)B_k(x)$$ with $d(m,k)\in\mathbb{Q}$. As explained by Kevin, $P_m(k)$ vanishes if 
 $m>2|k|-2\geq0$ because all terms in the sum vanish. It can also be shown that  $P_m(0)=0$ for $m\geq2$, but that is more tricky. 
 Indeed we will show that $d(m,0)=0$ for $m\geq2$.
Note that $P_m(x)$ visibly lies in the local ring $\mathbb{Z}_{(2)}$ for integer $x$.
So it suffices to show that $d(m,k)$ lies in $\mathbb{Z}_{(p)}$ for any  odd prime $p$. 
In fact we will find that the $d(m,k)$ are integers for $m\geq1$. And $d(0,0)=3/2$ lies in $\mathbb{Z}_{(p)}$ for our odd prime $p$.
 For $m$ not too large one may simply compute all $d(m,k)$.
The matrix $$(d(m,k))_{0\leq m\leq10}^{0\leq k\leq10}$$ looks like this
$$\left(
\begin{array}{ccccccccccc}
 \frac{3}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 24 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 4 & 118 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 60 & 696 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 12 & 720 & 4824 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 336 & 8288 & 38240 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 60 & 6516 & 95928 & 336822 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 2520 & 109872 & 1131732 & 3215544 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 392 & 67904 & 1735320 & 13647840 & 32651544 \\
\end{array}
\right).$$
We will tacitly use it to deal with small values of $m$.
We will study the set $$V_p=\{ (m,k)\in \mathbb{Z}\times \mathbb{Z} \mid d(m,k)\in \mathbb{Z}_{(p)}\}.$$
Using a method of Zeilberger we will prove relations between the $d(m,k)$ that were first discovered experimentally.
 One relation allows us to rewrite  $m(m-1)(1 + 2 m) d(m, k)$ in such a manner that we can use
 the method of Floors described in 
question 26336.
With that method we show that $m(m-1)(1 + 2 m) d[m, k]$ is an integer multiple of $3m(m-1)$.
 Together with the relations this will allow us to show that $V_p$
 fills all of $ \mathbb{Z}\times \mathbb{Z}$ for odd primes $p$.
Our variables 
$i,j,k,m,n,q$ will 
take integer values only.
As in the A=B book we use the convention that $\binom{x}{j}$ is a polynomial in $x$ for fixed $j$. And it is the zero polynomial if $j<0$. So $\binom i j$ is defined for all integers $i$, $j$. It also vanishes  if $j>i\geq0$. Of course $\binom{i}{j}$  agrees with the usual binomial coefficient if $0\leq j\leq i$.
By inspecting the values at $x=0,\dots,j,$ we see that $$(-1)^j\binom{x+j}{j}\binom{x-1}{j}-(-1)^{j-1}\binom{x+j-1}{j-1}\binom{x-1}{j-1}$$ equals $(-1)^j\binom{2j}{j}B_j(x)/2$ for $j\geq0$.
Taking the telescoping sum over $j$ gives $$(-1)^j\binom{x+j}{j}\binom{x-1}{j}=\sum_{k=0}^j(-1)^k\binom{2j}{j}B_k(x)/2$$ for $j\geq0$. (Valid for all $j$, actually).
This allows us to conclude that
$$d(m,k)=\sum_{i=0}^m\sum_{j=k}^m\frac{ 3(-1)^{k+j}\binom{2k  }{ k }\binom{j  }{  i}\binom{ m }{ i }\binom{i  }{ m-j }}{2(2i-1)(2j+1)(2m-2i-1)}.$$
In particular $d(m,k)=0$ for $m<0$ and for $k>m$. We will see that $m(m-1)d(m,k)$ also vanishes for
$2k-2<m$. 
Let us use the notation $[$statement$]=
\begin{cases}1,&\text{if statement is true;}\\ 0,&\text{otherwise.}\end{cases}$
Then 
\begin{align}
d(m,k)=&\sum_{i,j}[j\geq k\geq0] \text{term}(m,k,i,j), \tag{$\Sigma ij$}
\end{align}
where $$\text{term}(m,k,i,j)=[m\geq0]\frac{3(-1)^{k+j}\binom{2k  }{ k }\binom{j  }{  i}\binom{ m }{ i }\binom{i  }{ m-j }}{2(2i-1)(2j+1)(2m-2i-1)}.$$
Put
\begin{align*}
\text{rel1}&(m,k)
 = \\
   -& 32 (3-2 k)^2 (-k+m+1)
   (-k+m+2) d(m,k-2)\\
 +&4 (-k+m+1) \left(2 k m^2-2 (k-1) (8 k-9) m+(2 k-3) (8
   (k-2) k+9)\right) d(m,k-1)\\
   + & k (-2 k+m+2) (-2 k+m+3) (-2 k+2
   m+1) d(m,k),
\end{align*}   
\begin{align*}
\text{rel2}&(m,k)
 = \\&-4 \left((m-1)^2-1\right) d(m-1,k-1)\\&-4 (2 (k-1)+m+1)
   (-k+m+1) d(m,k-1)\\&+k (2 k-m-2) d(m,k)
 \end{align*}   
Key results
$\bullet$ $\text{rel1}(m,k)$  vanishes.
$\bullet$ $m(m-1)d(m,k)$ vanishes for $2k-2<m$. 
$\bullet$ $\text{rel2}(m,k)$ vanishes.
$\bullet$ $m(m-1)( 2 m+1) d(m, k)$ is an integer multiple of $3m(m-1)$.
Before proving the Key results, let us draw conclusions from them. Let $m\geq2$. As $d(m,0)=0$, we have $P_m(0)=0$ and the $d(m,k)$ are 
determined by $P_m(1)\dots,P_m(m)$. 
Now the integral matrix $$(B_k(i))_{1\leq k\leq m}^{1\leq i\leq  m}$$ is triangular with ones on the diagonal.
We conclude that $d(m,k)\in \mathbb{Z}_{(2)}$ for $m\geq2$.
Let $p$ be a prime, $p\geq5$, and let $m\geq 2$.
If $p$ does not divide $2m+1$, then $d(m,k)\in \mathbb{Z}_{(p)}$ because $m(m-1)( 2 m+1) d(m, k)\in 3m(m-1)\mathbb{Z}_{(p)}$.
Now assume $p$ divides $2m+1$. Then it does not divide $2m+3$, so then $d(m+1,j)\in \mathbb{Z}_{(p)}$ for all $j$.
Also, $p$ does not divide $(m-1)(m+1)$, so it follows from $\text{rel2}(m+1,k+1)=0$ that $d(m,k)\in \mathbb{Z}_{(p)}$.
We have shown that  $d(m,k)\in \mathbb{Z}_{(p)}$ if  $p$ is prime, $p\geq5$, $m\geq2$. 
Remains $p=3$. Let $m\geq2$ again.
If $3$ does not divide $2m+1$, then $d(m,k)\in 3\mathbb{Z}_{(3)}$ because $m(m-1)( 2 m+1) d(m, k)\in 3m(m-1)\mathbb{Z}_{(3)}$.
If $m\equiv1$ mod 9, or $m\equiv7$ mod 9, then $(2m+1)/3$ is prime to $3$ and $d(m,k)\in \mathbb{Z}_{(3)}$ because $m(m-1)(( 2 m+1)/3 )d(m, k)\in m(m-1)\mathbb{Z}_{(3)}$.
If $m\equiv 4$ mod 9, then $(m-1)(m+1)/3$ is prime to 3 and $d(m,k)\in \mathbb{Z}_{(3)}$ because $\text{rel2}(m+1,k+1)=0$ shows 
$((m-1)(m+1)/3)d(m,k)$ is an integer linear combination of the integers $d(m+1,j)/3$.
We conclude that $d(m,k)\in \mathbb{Z}_{(3)}$ for $m\geq2$.
So the $d(m,k)$ are integers for $m\geq2$ and $P_m$ takes integer values on integers for $m\geq2$.
Recall that $P_0$, $P_1$ also take integer values. Done.
So we still have to prove the Key results.
First a technical issue.
If $x>0$ then $\binom{x}{j}=\frac{\Gamma(1+x)}{\Gamma(1+j)\Gamma(1+x-j)}$ and the bimeromorphic function $$f(x,y)=\frac{\Gamma(1+x)}{\Gamma(1+y)\Gamma(1+x-y)}$$ is continuous at $(x,j)$. However, if $i<0$ then $f$ has an indeterminate value at $(i,j)$. For example, $\binom{i}{i}$ equals $1$ if $i\geq0$, but it vanishes for $i<0$. At $(-1,-1)$ both $0$ and $1$ are values of $f$. Indeed Mathematica can be steered to give either answer.
$\mathtt {Binomial[i,j]~ /.~ i->-1~/.~j->-1}$ gives 1 
and 
$\mathtt {Binomial[i,j]~ /. ~j->-1~/.~i->-1}$ gives 0.
And  $\mathtt{FullSimplify[Binomial[i, i] == Binomial[i - 1, i - 1]]}$ yields $\mathtt {True}$.
This answer is correct, but it tells only that for generic complex numbers $i$ the
identity holds.
Thus we need to make case distinctions when using identities between multimeromorphic functions, explicitly or implicitly, to prove identities involving
the $\binom{i}{j}$.
We start proving that $\text{rel1}(m,k)$ vanishes.
As $[j\geq k+1]\big(2 (2 k+1)\text{term}(m,k,i,j)+(k+1)\text{term}(m,k+1,i,j)\big)=0$, we get from $(\Sigma ij)$ that
\begin{align}
2 (2 k+1) d(m,k)+(k+1) d(m,k+1)=&\sum_i \text{iterm}(m,k,i)\tag{$\Sigma i$}
\end{align}
where $$\text{iterm}(m,k,i)=2 (2 k+1)\text{term}(m,k,i,k).$$
Now we use the
Fast Zeilberger Package version 3.61
written by Peter Paule, Markus Schorn, and Axel Riese
Copyright 1995-2015, Research Institute for Symbolic Computation (RISC),
Johannes Kepler University, Linz, Austria.

It suggests to put
$$g(m,k,i)=\frac{3\times 2^{2 k+3} m (-2 i+m+1) \Gamma \left(k+\frac{3}{2}\right) \binom{k+1}{i-1}
   \binom{m-1}{k+1} \binom{k+1}{m-i}}{\Gamma \left(\frac{1}{2}\right) (k+1)!}$$
and show that
\begin{align*}
-32 (1 + 2 k) &(3 + 2 k) (k - m) (1 + k - m) \text{iterm}(m, k, i) \\
   -4 (1 + k - m) & (57 + 110 k + 72 k^2 + 16 k^3 - 34 m - 46 k m - 
   16 k^2 m + 4 m^2 + 2 k m^2)\\&\times \text{iterm}(m, k+1, i)\\
-(2 + k) (5 +  &2 k - 2 m) (3 + 2 k - m) (4 + 2 k - m)\text{iterm}(m, k+2, i)
 \\
  & -g(m, k, i + 1) +g(m, k, i)=0
\end{align*}   
for $m\geq0$, $k\geq0$.
So we do that and then sum over $i$, using $(\Sigma i)$. The $g$ terms drop out by telescoping and we get a relation 
\begin{align*}
-32 (1 + 2 k) &(3 + 2 k) (k - m) (1 + k - m) (2 (2 k+1) d(m,k)+(k+1) d(m,k+1) )\\
   -4 (1 + k - m) & (57 + 110 k + 72 k^2 + 16 k^3 - 34 m - 46 k m - 
   16 k^2 m + 4 m^2 + 2 k m^2)\\&\times  (2 (2 k+3) d(m,k+1)+(k+2) d(m,k+2) )\\
-(2 + k) (5 +  &2 k - 2 m) (3 + 2 k - m) (4 + 2 k - m) \\\times(2 (2 k+5) &d(m,k+2)+(k+3) d(m,k+3) )
 \\
  & =0
\end{align*}
valid for $k\geq0$, as it is obvious for $m<0$.
We may rewrite it as a recursion for $\text{ rel1}$:
$$2 (3 + 2 k) \text{rel1}(m, k + 2) + (2 + k)\text{ rel1}(m, k + 3)=0$$
for $k\geq0$.
As $d(m,k)$ vanishes for $k>m$, it follows from the recursion that $\text{rel1}(m, k )$ vanishes for $k\geq2$.
Put
$$\text{pterm}(m,x,i,j)=\binom{x+j}{ j}\binom{x-1}{ j}\binom{j}{ i}\binom{m}{ i}\binom{i}{ m-j}\frac{3}{(2i-1)(2j+1)(2m-2i-1)},$$
so that 
$$P_m(x)=\sum_{i,j}\text{pterm}(m,x,i,j).$$
If $k\geq1$ and $\text{pterm}[m,k,i,j]$ is nonzero, then 
$ k-1\geq j$ and $m\geq j \geq i \geq m-j $. We see that $$P_m(k)=0 \text{ if }0\leq 2k-2<m,$$ because all the $\text{pterm}(m,k,i,j)$ vanish.
In particular we get
$$0=P_m(1)=\sum_kd(m,k)B_k(1)=2d(m,0)+d(m,1),$$ and $$0=P_m(2)=\sum_kd(m,k)B_k(2)=2d(m,0)+4d(m,1)+d(m,2)$$ for $ m\geq 3$.
So then $d(m,1)=-2d(m,0)$ and $d(m,2)=6d(m,0)$. Substitute this into $\text{rel1}(m, 2 )=0$ and you find 
$$4 m (m - 1) (-2 m - 1) d(m, 0)=0.$$
This means that $d(m, 0)=0$ for $m\geq3$. As $d(2,0)$ also vanishes, we now have enough to conclude that
\begin{align}
m(m-1)d(m,k) \text{ vanishes for }&m>2k-2. \tag{SSE}
\end{align}
It is now easy to see that  $\text{rel1}(m, k )$ also vanishes for $k<2$. 
So we have established the vanishing of $\text{rel1}(m, k )$ and the vanishing of
$m(m-1)d(m,k)$ for $m>2k-2$.
Before turning to  $\text{rel2}(m, k )$ we compute $d(2k-2,k)$ and $d(2k-3,k)$ for $k\geq3$. These are the values that help to compute all $d(m,k)$ 
recursively with the recursion given by $\text{rel1}(m, k )$=0.
As $d(2k-2,j)$ vanishes for $j<k$, one has $$d(2k-2,k)=P_{2k-2}(k)=\text{pterm}(2k-2,k,k-1,k-1)$$
and similarly
$$d(2k-3,k)=P_{2k-3}(k)=\text{pterm}(2k-3,k,k-1,k-2)+\text{pterm}(2k-3,k,k-1,k-1).$$
So we know $d(m,k)$ for $m\geq 2k-3\geq3$. By (SSE) we also know  $d(m,k)$  for $k\leq1$ and any $m$.
Using these values we get $\text{rel2}(m,k)=0$ by inspection for $m\geq 2k-3$ or $k\leq1$.
Notice that $(-7+2 k)\text{rel2}(2k-4,k)-\text{rel1}(2k-4,k)$ is a combination of the known terms $ d(-5 + 2 k, -1 + k)$, 
$    d(-4 + 2 k, -2 + k)$,  $  d(-4 + 
  2 k, -1 + k)$. It also vanishes by inspection, so we now have that $\text{rel2}(m,k)=0$ for $m\geq 2k-4$ or $k\leq1$.
By substituting the definitions and expanding we check that
\begin{align*}
&(-1 + k) (-1 + 2 k - 2 m) (-3 + 2 k - m)(4 - 2 k + m) \text{rel2}(m, k)
\\& + 
 4 (-1 + (-1 + m)^2) \text{rel1}(m - 1, k - 1)
 \\& - 
 32 (5 - 2 k)^2 (-2 + k - m) (-1 + k - m) \text{rel2}(m, k - 2)
 \\& + 
 4 (1 - k + m)\\&\times (-99 + 16 k^3 - 2 m (32 + m) - 8 k^2 (11 + 2 m) + 
2 k (81 + m (31 + m))) \text{rel2}(m, k - 1)
   \\& - (-1 + k) (-4 + 2 k - m)\text{ rel1}(m, k)\\&  - 
 4 (-1 + k - m) (-5 + 2 k + m) \text{rel1}(m, k - 1)
 \\&=0
 \end{align*}   
As $\text{rel1}$ vanishes, this leads to the following recursion for $\text{rel2}$.
\begin{align*}
&(-1 + k) (-1 + 2 k - 2 m) (-3 + 2 k - m)(4 - 2 k + m) \text{rel2}(m, k) \\& - 
 32 (5 - 2 k)^2 (-2 + k - m) (-1 + k - m) \text{rel2}(m, k - 2)
 \\& + 
 4 (1 - k + m)\\&\times (-99 + 16 k^3 - 2 m (32 + m) - 8 k^2 (11 + 2 m) + 
2 k (81 + m (31 + m))) \text{rel2}(m, k - 1)
   \\&=0
 \end{align*}  
As $\text{rel2}(m,k)=0$ for $ 2k-4\leq m$ or $k\leq1$,  the recursion shows by induction on $k$
that $\text{rel2}(m,k)=0$ for all $m$, $k$.
So we have also established the vanishing of $\text{rel2}(m,k)$ and it is time to
show the Key result that $m(m-1)( 2 m+1) d(m, k)$ is an integer multiple of $3m(m-1)$. 
This is obvious for $m<2$, so we further assume $m\geq2$. Then we know that $d(m,0)=0$ and we have seen this implies
$d(m,k)\in \mathbb{Z}_{(2)}$. So it suffices to
show that $m(m-1)( 2 m+1) d(m, k)\in 3m(m-1)\mathbb{Z}[1/2]$.
Using relation $(\Sigma i)$ we may rewrite $\text{rel1}(m,k)=0$ as
\begin{align*}&2 (m-1 ) m ( 2 m+1) d(m, k-1 )\\&+
(2 - 2 k + m) (3 - 2 k + m) (1 - 2 k + 2 m) \sum_i \text{iterm}(m, k - 1,i)\\&+ 
 16 (3 - 2 k) (-2 + k - m) (-1 + k - m) \sum_i \text{iterm}(m, k - 2,i) \\&=0
 \end{align*}  
We claim that 
\begin{align*}&(2 - 2 k + m) (3 - 2 k + m) (1 - 2 k + 2 m)  \text{iterm}(m, k - 1,i)\\&+ 
 16 (3 - 2 k) (-2 + k - m) (-1 + k - m)  \text{iterm}(m, k - 2,i) 
 \end{align*} 
 lies in $3m(m-1)\mathbb{Z}[1/2]$. 
That will prove that the $ (m-1 ) m ( 2 m+1) d(m, k-1 )$ are  integer multiples of  $3m(m-1)$. 
Put 
$$\text{frac1}(m,k,i)=\frac{3 (m-1) m \binom{2 (k-1)}{k-1} (-2 k+2
   m+1) \binom{k-1}{i} \binom{m}{i} \binom{i}{-k+m+1}}{(2
   i-1) (2 m-2 i-1)}$$
   and 
   $$\text{frac2}(m,k,i)=6  (k-m-1)\binom{2 (k-1)}{k-1}  \binom{k-1}{i} \binom{m}{i} \binom{i}{-k+m+1}.$$
Then $\text{frac1}(m,k,i)+\text{frac2}(m,k,i)$ equals 
   \begin{align*}&(2 - 2 k + m) (3 - 2 k + m) (1 - 2 k + 2 m)  \text{iterm}(m, k - 1,i)\\&+ 
 16 (3 - 2 k) (-2 + k - m) (-1 + k - m)  \text{iterm}(m, k - 2,i) ,
 \end{align*} 
so it suffices to show that $\text{frac1}(m,k,i)/(6m(m-1))$ and $\text{frac2}(m,k,i)/(6m(m-1))$, which make sense for $m\geq2$,
  lie in $\mathbb{Z}[1/2]$ for $m\geq 2$.
 Recall that the Catalan numbers
 $$C(i)=\frac{\binom{2 i}{i}}{i+1}$$ are integers. See
A000108
We now look at $\text{frac1}(m,k,i)/(6m(m-1))$.
If $\text{frac1}(m,k,i)$ is nonzero then $ m\geq k-1\geq i\geq m+1-k\geq 0$. 
We distinguish two cases: $ m=k-1\geq i\geq 0$ and $ m> k-1\geq i\geq m+1-k\geq 0$. 
First let $ m=k-1\geq i\geq 0$. If $i=k-1$, then 
$$\text{frac1}(m,k,i)/(6m(m-1))=\text{frac1}(k-1,k,k-1)/(6(k-1)(k-2))=C(k-2).$$
Similarly $\text{frac1}(k-1,k,0)/(6(k-1)(k-2))=C(k-2).$
So we may assume $0<i<m=k-1$.
Then
$\text{frac1}(m,k,i)/(6m(m-1))=\text{frac1}(m,m+1,i)/(6m(m-1))$ equals
$$\small
\frac{-(2 i-2)! (2 m)! (-2 i+2 m-2)!}{2 (i!)^2 (2 i-1)!
   ((m-i)!)^2 (-2 i+2 m-1)!}$$ and we must show it takes values in $\mathbb{Z}[1/2]$.
This is the kind of expression to which one may apply the method of Floors explained in 
question 26336.
According to the method it suffices to check that $\text{test}(m,i,2n+1)\geq0$ for $n\geq1$, where
\begin{align*}
\text{test}(m,i,q)=&\\-2& \left\lfloor \frac{m-i}{q}\right\rfloor +\left\lfloor
   \frac{-2 i+2 m-2}{q}\right\rfloor -\left\lfloor \frac{-2
   i+2 m-1}{q}\right\rfloor\\ -2& \left\lfloor
   \frac{i}{q}\right\rfloor +\left\lfloor \frac{2
   i-2}{q}\right\rfloor -\left\lfloor \frac{2
   i-1}{q}\right\rfloor +\left\lfloor \frac{2
   m}{q}\right\rfloor
.\end{align*}
This is a tedious puzzle. For fixed $q$ the function $\text{test}(m,i,q)$ is periodic of period $q$ in both variables $i$ and $m$.
So for fixed $q$ one may simply compute all values. We do it for $3\leq q=2n+1<17$. The results are nonnegative.
But if $q$ is large we need to be more efficient. If both $q=2n+1$ and $m$ are
fixed, then $\text{test}(m,i,q)$ can only change value where at least one of the Floors jumps as a function of $i$. So it suffices to sample around the jumping
points (modulo $q$). We know where they are. More specifically, we only need to consider the 15 cases where one of $i-1$, $i$, $i+1$ lies in 
$\{0, 1, -1 + m, m,  -2 + m - n, -1 + m - n, 1 + n\}$.
So we can eliminate $i$ at the expense of having 15 cases. Similarly we can eliminate $m$ for each of those cases, ending up with
153 test functions that depend on $n$ only. Each test function is a linear combination of seven Floors.
Each of the Floors stabilises after $n$ has reached an easily computable bound.
For instance $\left\lfloor -\frac{8}{2 n+1}\right\rfloor$ is constant for $n\geq4$.
In fact the bound 5 suffices for all $7\times 153$ Floors. Compute the 153 stable values. They are nonnegative. 
 This solves the puzzle; the check for $3\leq q=2n+1<17$ was overkill.
So we now turn to the case $ m> k-1\geq i\geq m+1-k\geq 0$.
 Then 
 $$\text{frac1}(m,k,i)/(6m(m-1)C(i-1))$$ equals $$\small\frac{i! (2 k-2)! m! (-2 i+2 m-2)! (-2 k+2 m+1)!}{(2 i)! (k-1)! (-i+k-1)! (m-i)! (-2 i+2 m-1)!
   (-k+m+1)! (2 m-2 k)! (i+k-m-1)!}
 $$
We use  the method of Floors again to show that  $\text{frac1}(m,k,i)/(6m(m-1)C(i-1))\in \mathbb{Z}[1/2]$.
This time we eliminate $k$, $m$, $i$ in that order and take $n\geq6$ as bound where all  $13\times 3508$ Floors are stable.
So we have shown that  $\text{frac1}(m,k,i)/(6m(m-1))$ 
  lies in $\mathbb{Z}[1/2]$ for $m\geq 2$.
Remains showing that $\text{frac2}(m,k,i)/(6m(m-1))$ 
  lies in $\mathbb{Z}[1/2]$ for $m\geq 2$.
If $\text{frac2}(m,k,i)$ is nonzero then $m> k-1\geq i\geq m+1-k>0$ and
$\text{frac2}(m,k,i)/(6m(m-1))$ 
equals 
$$\frac{-(2 k-2)! (m-2)!}{i! (k-1)! (-i+k-1)! (m-i)! (m-k)!
   (i+k-m-1)!}$$
This can be treated like the previous case. We eliminate $k$, $m$, $i$ in that order and take $n\geq6$ as bound where all  $8\times 1278$ Floors are stable.
Done<|endoftext|>
TITLE: Z/p action on finite contractible complex
QUESTION [7 upvotes]: Let $p$ be a prime and $X$ a finite contractible CW-complex. Assume $\mathbb Z/p$ acts on $X$. Then it is easy to see that there has to be a fixed point. (E.g. use Lefschetz's fixed point theorem or group cohomology.) 
Assume there are only finitely many fixed points. 
Can there be more than one?
Remarks: (1) If X is a manifold, there can not be more than one, by the Lefschetz-Hopf fixed point theorem. But to generalize the proof to the general case, one would need to define the "index" of an isolated fixed point and I have no idea how to accomplish that, if one does not work with manifolds only.
(2) The statement is true for graphs.
(3) The statement is easily seen to be wrong for general contractible spaces, e.g. take a free action on $S^{\infty}$.

REPLY [13 votes]: If a finite $p$-group $P$ acts on a finite-dimensional $CW$-complex $X$ which is acyclic mod $p$, then the fixed point set $X^P$ is also acyclic mod $p$. This is a special case of "Smith theory" (see Theorem II of "Fixed-Point Theorems for Periodic Transformations", P. A. Smith, American Journal of Mathematics Vol. 63, No. 1 (Jan., 1941), pp. 1-8 for an early version of the theory where $P$ is cyclic).
For a cellular action of $P=\mathbb{Z}/p$ on a contractible finite $CW$ complex with only finitely many fixed points, the reduced mod $p$ cellular chain complex would be a bounded acyclic chain complex of $\mathbb{F}_pP$-modules, with all components free modules in non-zero degree, and the degree zero component the direct sum of a free module and a trivial module of dimension one less than the number of fixed points. But this is impossible by representation theory if that trivial module is non-zero, since a trivial $\mathbb{F}_pP$-module doesn't have finite projective dimension.<|endoftext|>
TITLE: Fourier expansion of Eisenstein Series
QUESTION [7 upvotes]: I have been reading a bit about the Fourier expansion of Eisenstein series (weight 1/2). I came across the fact that the coefficients contain Modified Bessel functions.
Further reading I found articles discussing the zeros of these Bessel functions to behave similar to that of the zeta function (Re(s) = 1/2).
My question, is this or why isn't this a popular way to study Riemann's zeta function? Or have i misunderstood an obvious vital key element?

REPLY [4 votes]: The relationship has actually motivated several studies:

Eisenstein series and the Riemann zeta function (1981) 
Moments of the Riemann zeta function and Eisenstein series part I and part II (2004)  
The Riemann hypothesis for
certain integrals of Eisenstein series (2004)

Some of its limitations are discussed in an answer to this MO question.<|endoftext|>
TITLE: Under what conditions a linear automorphism is an isometry of some norm?
QUESTION [9 upvotes]: Assume $V$ is a finite-dimensional vector space over $\mathbb{R}$, and $T: V \to V$ is a (linear) isomorphism. 

When is it possible to construct a norm on $V$ 
  making $T$ an isometry?

(Hopefully, I am looking for necessary & sufficient conditions $T$ should satisfy, i.e. a full characterization of the situation).
What I have so far:
A necessary condition: all the real eigenvalues of $T$ are of absolute value $1$. (Since $ T(v)=\lambda v \Rightarrow ||v||=||T(v)||=||\lambda v||=|\lambda| ||v||$ and an eigenvector $v$ must be nonzero).
This condition is certainly not sufficient:
For example look at $A$ = $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}: \mathbb{R}^2 \to \mathbb{R}^2$. It is an automorphism which has only one eigenvalue ($\lambda = 1$). However, $A\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+y \\ y \end{pmatrix}$, hence $A^n\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ny \\ y \end{pmatrix}$. Now assume there exist a norm $||$ on $\mathbb{R}^2$ making $A$ an isometry. 
In particular $A$ must map the open unit ball $B$ to itself. By properties of norms $B$ must be a bounded open set containing the origin. (Note that since all the norms on a finite dimensional vector space are equivalent, the notions of boundedness and opennes are independent of the norm).
$B$ is open $\Rightarrow$ $\exists y>0$ such that $(0,y)\in B \Rightarrow A^n\begin{pmatrix} 0 \\ y \end{pmatrix}= \begin{pmatrix} ny \\ y \end{pmatrix} \in B$ for every $n \in \mathbb{N}$. This contradicts the boundedness of $B$ (w.r.t to the standard Euclidean norm for instance).
Last remark:
If we want to be more restrictive and require $T$ to be an isometry of some inner product, then the answer is quite simple.
$T$ preserves some inner product on $V$ if and only if $V$ admits a basis for which the matrix of $T$ is orthogonal (in other words the matrix of $T$ on an arbitrary basis is similar to an orthogonal matrix). The occurs if and only if the complexification of T is diagonalisable, and all its (complex) eigenvalues have absolute value 1.
I am interested to know what additional potential isometries we can get when we allow more flexibility. (That is we allow arbitrary norms, not just those that come from inner products).

REPLY [6 votes]: For sake of completeness, I am writing a full answer based on the suggestion of
Pietro Majer.

The following are equivalent:
1) $A$ is an isometry w.r.to some norm.
2) $A$ is diagonalizable (over $\mathbb{C}$) , with all eigenvalues of modulus 1.
3) All orbits of $A$ are bounded ( $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$ for any $x\in \mathbb{R}^n$).
4) $A$ is an isometry w.r.to some inner product.

$(1)\iff (4)$ leads to an interesting point: The union of all isometries of all norms equals the union of all isometries of all inner products. 
Proof:
$(1) \Rightarrow (2):$
Assume $||$ is a norm preserved by $A$. Then the operator norm of $A$ w.r.t to $||$ equals 1. Also $||A^n||_{op}=1$. By the spectral radius formula: $\rho(A)=\lim_{n\rightarrow\infty}||A^n||^{1/n}=1 $. The same argument for $A^{-1}$ implies $\rho(A^{-1})=1$. This implies all the eigenvalues (including the complex ones) are of absolute value one.
Also, it is easy to see that If $A$ is an isometry of the norm $| |_1$ , $P∈GL(\mathbb{R^n})$, $P^{−1}AP$ is an isometry of the norm $||_2$ where $||x||_2=||Px||_1$. Thus, the property that a given matrix admit such a norm is invariant under similarity. 
So it is enough to focus upon matrices of Jordan form. (which is available to us since we work over $\mathbb{C}$).
Now assume $A$ is not diagonalizable. By looking at Jordan form of a non-diagonalizable matrix, we can see there is a vector $v∈\mathbb{C}^n$ such that $||A^kv||_{Euclidean}$ diverges. (Look at the example of $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}$ given in the question).
Since all the norms are equivalent This implies that $||A^kv||$ diverges. But since $v=x+iy$ with $x$ and $y$ in $\mathbb{R}^n$, and $A^kv=A^kx+iA^ky$, either $||A^kx||$ or $||A^ky||$ diverge. This is impossible if $A$ is an isometry of $||$.
$(2) \Rightarrow (4):
$ This is proved here (The basic idea is to look at each Jordan block separately).
$(4) \Rightarrow (1):$ Obvious.
It remains to prove $(1) \iff (3)$:
$(3) \Rightarrow (1):$ If all orbits of $A$ are bounded, then we can renorm $\mathbb{R}^n$ by $\|x\|_A:=\sup_{k\in \mathbb{Z}} \|A^k x\|$, which is an $A$-invariant equivalent norm. 
$(1) \Rightarrow (3):$ This follows immediately by the fact all norms on a finite dimensional vector space are equivalent. The orbits of $A$ are all of constant norm ($\|x\|$) w.r.t to the norm $A$ preserves, hence are bounded. (w.r.t any other norm).<|endoftext|>
TITLE: Do small prime (bounded) gaps imply larger (but still bounded) gaps?
QUESTION [6 upvotes]: The best result concerning bounded gaps between primes, whose existence was first proved by the seminal work of Yitang Zhang two years ago, are to my knowledge all of the form $\liminf_{n \rightarrow \infty} \left(p_{n+1} - p_n\right) \leq C$, where the record is currently $C = 246$, according to this source: http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes . These results imply that for at least one value of $k$, $k\leq 123$, that there are infinitely many primes separated by exactly $2k$. 
Suppose we can prove that for some positive integer $k$, that there are infinitely many consecutive primes $p_n, p_{n+1}$ such that $p_{n+1} - p_n = 2k$. Can we prove that there are infinitely many consecutive primes separated by exactly $2k+2, 2k+4, \cdots$ etc.? That is, does the existence of infinitely many twin primes for example imply that all even gaps must occur infinitely often? Certainly, whatever technique was used to prove the existence of infinitely many two primes should allow us to just as easily prove it for any even gap, but my question is a bit more specific: namely, can one directly prove the existence of infinitely many primes separated by $2k$ knowing only that there are infinitely many primes separated by $\max\{2, 2k-2\}$?

REPLY [5 votes]: The current technology works as follows. For any admissible finite set $\mathcal{H}$ of size at least $50$, one of the differences that occur within $\mathcal{H}$ also occurs as a difference of two (not necessarily consecutive) primes infinitely often. One can draw many interesting conclusions from this principle, but what you are looking for seems to be beyond this technology.
In particular, the set of numbers that occur as a difference of two (not necessarily consecutive) primes infinitely often has lower asymptotic density exceeding $\frac{1}{354}$. This is the current record in the sense that, to my knowledge, this lower bound has not been improved to $\frac{1}{353}$. The actual fraction here is not that important, but if we could prove what you are looking for, then we knew that the density is precisely $\frac{1}{2}$, because in that case all but finitely many even numbers would belong to the set of differences in question. Even assuming the generalized Elliott-Halberstam conjecture, we have no proof so far that the above lower asymptotic density is $\frac{1}{2}$.<|endoftext|>
TITLE: Open problems in hyperplane/subspace arrangements?
QUESTION [8 upvotes]: What are some open problems in hyperplane/subspace arrangements, preferably of the combinatorial algebraic topology kind, and where can one read about them? That is, where are they discussed, and where can one learn the necessary background knowledge to understand them properly and in context?
I have been working one one such problem for a while, but I'm ignorant about the big questions, and more generally, interesting problems in this field, and also current trends.
(I'm aware of a few unresolved questions in Richard P. Stanley's An Introduction to Hyperplane Arrangements (available online, the unresolved problems are signaled by [5]))

REPLY [5 votes]: The classic book ``Hyperplane Arrangements'' by Orlik and Terao gives many open problems (some listed there are solved). A more recent, but not published, book which gives many open problems is at http://www.math.uiuc.edu/~schenck/cxarr.pdf.
There are many other surveys that list open problems. Alex Suciu wrote two survery's, http://arxiv.org/pdf/math/0010105.pdf  and http://arxiv.org/pdf/1301.4851.pdf each with many problems. On the more algebraic side, Masahiko Yoshinaga wrote the survey http://arxiv.org/abs/1212.3523 which has many open problems. The survey by Sergey Yuzvinsky ``Resonance varieties of arrangement complements'' also has some open problems. The survey http://arxiv.org/abs/1101.0356 by Hal Schenck also has many open problems. One of the oldest open problems is Terao's conjecture: freeness depends on the intersection lattice (which is in the two books and in Yoshinaga's survey).<|endoftext|>
TITLE: Operator on a Banach space
QUESTION [5 upvotes]: Let $T$ be a continuous operator on a Banach space $V$. Assume there exist $T$-stable finite-dimensional subspaces $V_i$ such that $\bigoplus_{i=1}^\infty V_i$ is dense in $V$, on $V_i$ the operator $T$ has only one eigenvalue $\lambda_i$. One has $|\lambda_i|<1$ for each $i$ and the $\lambda_i$ tend to zero.
My question is: Is it true that $T^nv$ tends to zero for every $v\in V$?

REPLY [6 votes]: This is not necessarily true, and in fact you can get a counterexample where each $\lambda_i=0$.  Consider $V=c_0$ and let $N_i=\sum_{j=0}^i j$ be the $i$th triangular number.  Define $T:c_0\to c_0$ by  $$T(x)_k=\begin{cases} 0 & \text{if }k=N_i\text{ for some }i \\ 2x_{k-1} & \text{otherwise}\end{cases}$$
That is, $T$ is twice the right shift, except that the $N_i$th coordinates of $T(x)$ are declared to be $0$ for each $i$.  Clearly $T$ is bounded, with $\|T\|=2$.
If $V_i$ is the space of sequences supported on $[N_i,N_{i+1})$, then each $V_i$ is $T$-stable and $T$ is nilpotent on $V_i$.  As $\bigoplus V_i$ is dense in $c_0$, we conclude that $T$ satisfies your hypotheses with $\lambda_i=0$ for all $i$.  Now let $x\in c_0$ be the sequence such that $x_{N_i}=1/i$ for each $i$ and all other coordinates are $0$.  Then for any $n$, $T^n(x)$ will have $(N_i+n)$th coordinate $2^n/i$ whenever $i\geq n$.  In particular, $\|T^n(x)\|\to\infty$.<|endoftext|>
TITLE: $\Omega X$-action on spectral $X$-bundles
QUESTION [5 upvotes]: I can generalize the notion of a space over $X$ (where $X$ is a based and connected space) to the notion of a spectrum over $X$ by considering functors of quasicategories $X\to Mod_\mathbb{S}$. In the language of recent work of Ando, Blumberg and Gepner such functors can be thought of as bundles (i.e. locally constant sheaves) of $\mathbb{S}$-modules over $X$, or as $X$-parameterized spectra. In the case that we have a space parameterized by $X$, we know relatively obviously that $\Omega X$ acts on the fiber over the base point. This of course comes from just thinking about the fact that we can let the fiber travel along the path traced out by an loop in $X$. 
More generally however, it seems that it should be true that $\Omega X$ acts on the "fiber" of parameterized spectrum over $X$. How can I make a formal argument to this effect, which perhaps relies only on category-theoretic considerations? In particular, if I'm not mistaken, there is an equivalence between "spectra over $X$" and "spectra with an $\Omega X$-action" for which "taking the fiber" is one direction. 
In general I feel like this should be very generally true for the category of functors $X\to \mathcal{C}$ for any (presentable?) quasicategory $\mathcal{C}$.

REPLY [9 votes]: I hope I've understood your question.
Any connected based space $X$ is weak homotopy equivalent to the classifying space of a topological group $G$ in a functorial way (the group is the realization of the Kan loop group of the simplicial total singular complex). So we can assume $X = BG$. Then there is an equivalence of homotopy categories
$$
\text{ho(parametrized spectra}/X)  \qquad  \simeq   \quad \text{ho(naive $G$-spectra)}
$$
(This equivalence arises from a Quillen equivalence with respect to suitable choices of model structures.) 
The functors inducing the equivalences are given from right to left by taking the unreduced Borel construction and from left to right by base change along $EG \to X$ and then collapsing out the zero-section.
Actually the situation above arises from a Quillen equivalence between the model categories "Spaces over X" and "$G$-spaces." Then one can pass to categories spectra on both sides to get the result.<|endoftext|>
TITLE: Proof of a soft version of Moschovakis's lemma
QUESTION [6 upvotes]: The following fact, which I've heard being called "soft version of Moschovakis's lemma" (see top answer here) is the following:

Under AD, if there is a surjection $\Bbb R\rightarrow\alpha$, then there is a surjection $\Bbb R\rightarrow\mathcal{P}(\alpha)$.

I remember few days ago I have seen a really nice proof of this result, which directly used AD (and possibly DC, can't remember) and not some complex workaround using scales and things like that. However, when yesterday I wanted to show the proof to a friend of mine, I couldn't have find it anywhere, and I would be very thankful if anyone pointed out where I could have seen this proof.
The rough idea of the proof is the following:

Fix a surjection $f:\Bbb R\rightarrow\alpha$.
For each $X\subseteq\alpha$, using $f$, define a game $G(X)$ on $\omega$.
Show that, for any $X\neq Y$, no winning strategy for $G(X)$ is a winning strategy for $G(Y)$.
Thus, the function $F:\Bbb R\rightarrow\mathcal{P}(\alpha)$, defined as $F(r)=X$ if $r$ codes a winning strategy for $G(X)$, and $F(r)=\varnothing$ if $r$ codes no winning strategy, is well-defined.
By AD, every $G(X)$ has a winning strategy for some player, so $F$ is surjective.

Hope this helps to find the proof I meant. Thanks in advance.

REPLY [5 votes]: The argument you are looking for is given in Kanamori's book, see Theorem 28.15. 
For the more nuanced version of the lemma, see section 7D in Moschovakis's descriptive set theory book (particularly 7.D.5-8), or section 3.1 in the Koellner-Woodin chapter of the Handbook.<|endoftext|>
TITLE: The list of problems for Grothendieck's thesis
QUESTION [20 upvotes]: Is the list of open problems which were given by Dieudonne and Schwartz to Grothendieck for his thesis published somewhere? I know a quotation of Dieudonne that the problems concerned duality theory for general locally convex spaces.

REPLY [25 votes]: According to Chapter 3. From student to celebrity: 1949-1952 on Grothendieck Circle it was the 14 questions found at the end Dieudonne and Schwartz's article La dualite dans les espaces $\mathcal{F}$ et $\mathcal{LF}$ which can be found on EuDML.<|endoftext|>
TITLE: Non meager rectangle
QUESTION [11 upvotes]: Suppose $G \subseteq \mathbb{R}^2$ is dense $G_\delta$. Must there (in ZFC) exist non meager sets of reals $A, B$ such that $A \times B \subseteq G$?

REPLY [11 votes]: I pointed this question to Taras Banakh,
who pointed to me his joint paper with Lyubomyr Zdomskyy “Non-meager free sets for meager relations on Polish spaces” which contains an answer. 
Abstract. We prove that for each meager relation $E\subset X\times X$ on a Polish space $X$ there is a nowhere meager subspace $F\subset X$ which is $E$-free in the sense that $(x,y)\not\in E$ for any distinct points $x,y\in F$.
From myself I add a simple negative answer for descriptively good $A$ and $B$, that is when both $A$ and $B$ have the Baire Property. I recall, that a subset $B$ of a topological space $X$ has the Baire Property (BP) in $X$ if $B$ contains a $G_\delta$-subset $C$ of $X$ such that $B\setminus C$ is meager in $X$. By [Kech, 8.22] each Borel subset of a space $X$ has the Baire Property in $X$. Put $G=\mathbb{R}^2\setminus \{(x,y):x-y\in\Bbb Q\}.$ Assume that both $A$ and $B$ are non meager subsets of the space $\Bbb R$ with the Baire Property and $A\times B\subset G$. By [Kech, Prop. 8.22] both sets $A+\Bbb Q$ and $B+\Bbb Q$ have the Baire Property. By [Kech, Th. 8.46] both sets $A+\Bbb Q$ and $B+\Bbb Q$ are comeger. So $A+Q\cap B+Q\ne\varnothing$. That is, there exist points $a\in A$, $b\in B$, $q, q’\in\Bbb Q$ such that $a+q=b+q’$. Then $(a,b)\in A\times B\setminus G$, a contradiction. 
References
[Kech] A. Kechris. Classical Descriptive Set Theory, – Springer, 1995.<|endoftext|>
TITLE: Do relaxed Liouville functions violate Chowla's conjecture?
QUESTION [9 upvotes]: Let $\lambda$ be the Liouville function. One version of Chowla's conjecture says that for each set of distinct natural numbers $h_1 , \dots , h_k$,
$$\sum_{n\leq x} \lambda(n+h_1) \dots \lambda(n+h_k) = o(x)$$
Is this conjectural behavior stable under perturbations?
Say that a function $f: \mathbb N \to \pm 1$ is a relaxed Liouville function with error $\epsilon$ if for each $n$,  $f(nm)=\lambda(n)f(m)$ for a density $1-\epsilon$ set of $m$.
Note that this definition is meaningful for $\epsilon = 0$. Then the obvious questions are:

Does there exist a relaxed Liouville function $f$ with error $0$ and a set of numbers $h_1, \dots, h_k$ such that
$$\sum_{n\leq x} f(n+h_1) \dots f(n+h_k) \neq o(x)?$$

and more vaguely:

Does there exist a relaxed Liouville function $f$ with error $\epsilon$ for $\epsilon$ "small", a set of numbers $h_1, \dots, h_k$, and a "large" constant $\delta $ such that:
$$\left|\sum_{n\leq x} f(n+h_1) \dots f(n+h_k)\right| \not \leq ( \delta +o(1) ) x?$$
For instance, can we take $\delta/\epsilon$ to be arbitrarily large with fixed $h_1, \dots, h_k$?

Clearly negative answers to these questions would imply Chowla's conjecture. So I am hoping instead for positive examples.
My motivation is that many elementary approaches that can prove some weak partial results towards Chowla's conjecture are very stable and therefore apply even for relaxed Liouville functions. I want to understand the limits of these elementary methods by explicit counterexample.

Here is an alternate definition which I think captures the same spirit. Can we find, for some $h_1, \dots, h_k$, for each $N$ a function $f: \mathbb N \to \pm 1$ with $f(pn) = - f(n)$ for $p< N$ such that
$$\left|\sum_{n\leq x} f(n+h_1) \dots f(n+h_k)\right| \not \leq ( \delta +o(1) ) x$$
for some $\delta$ independent of $N$? Or even just some $\delta$ with $\delta \log N$ arbitrarily large?
The idea being that about $1/\log N$ numbers have no prime factors less than $N$ and hence are arbitrary.

REPLY [8 votes]: If you are willing to violate multiplicativity $f(nm)=f(n)f(m)$ about $\varepsilon$ of the time, then one can make $f$ more or less arbitrary on an interval of the form $[(1-\varepsilon) x, x]$, and this is enough to violate Chowla-type conjectures.  So the answer to your question as stated is "yes".
On the other hand, if one retains multiplicativity, then it was conjectured by Elliott that the analog of Chowla's conjecture should hold whenever $\sum_p \frac{1 - \hbox{Re} f(p) \chi(p) p^{it}}{p} = \infty$ for any Dirichlet character $\chi$ and real $t$, which would imply a negative answer to your question for any $\varepsilon < 1/2$ if one insists on perfect multiplicativity.  (Technically, Elliott's conjecture is false, see this recent paper of mine with Matomaki and Radziwill, but if one repairs it in the way indicated in that paper, the conclusion of the above argument still holds.)  The technical counterexample aside, all other known evidence points to the truth of the (corrected) Elliott conjecture, for instance our paper establishes an averaged form of this conjecture, and a different averaged form was recently established by Frantzikinakis and Host.<|endoftext|>
TITLE: Can we find two positive integers $n$ and $m$ ($n,m>1$) such that $n^\pi = m$?
QUESTION [33 upvotes]: I came across this apparent random question in some math questions website. At first, I thought it was easy to show that there are no non-trivial integer solutions to this equation, but then I realized that the question is far beyond what I can answer.
Irrationality and transcendence of $\pi$ play no role I think, because if we change $\pi$ with $\log_2(9)$ there are non-trivial solutions. My question is, Do you know some tool to attack this kind of problem?
I'll immediately delete this question if you think it's outside the scope of MathOverflow, but I know there are really knowledgeable people here who could say something about this, and my "innocent" curiosity for this question made me post it here. Thank you for your attention. 
PS: more accurate tags for the question are also welcomed.

REPLY [31 votes]: Unconditionally, this can essentially happen at most once.  It is a consequence of the Five Exponentials Theorem, which itself follows from a combination of the classical Six Exponentials Theorem and Baker's Theorem on linear forms in logarithms.
Five Exponentials Theorem: (See Waldschmidt, Diophantine Approximation on Linear Alebgraic Groups, 2000, Section 11.3.3)
Suppose that $\lambda_0, \lambda_1, \lambda_2 \in \mathbb{C}^{\times}$ are chosen so that $e^{\lambda_i} \in \overline{\mathbb{Q}}$ for each $i = 0, 1, 2$.  Suppose that $\lambda_1$ and $\lambda_2$ are linearly independent over $\mathbb{Q}$.  Then for any non-zero algebraic number $\beta$, at least one of the two numbers
$$
e^{\beta \lambda_0 \lambda_1}, \quad e^{\beta \lambda_0\lambda_2}
$$
is transcendental.
To apply to the present question, for positive integers $n_1$ and $n_2$ take
$$
\lambda_0 = \pi i, \quad \lambda_1 = \log n_1, \quad \lambda_2 = \log n_2,
\quad \beta = -i,
$$
and so
$$
e^{\beta \lambda_0 \lambda_1} = n_1^{\pi}, \quad e^{\beta \lambda_0 \lambda_2} = n_2^{\pi}.
$$
By the Five Exponentials Theorem, if both $n_1^{\pi}$ and $n_2^{\pi}$ are algebraic, then $n_1$ and $n_2$ must be multiplicatively dependent.  Thus up to multiplicative dependence there is at most one way for a solution to be found.<|endoftext|>
TITLE: Closed orientable 4-manifold with $H^1(M;\Bbb Z_2)=\Bbb Z_2$ and non-zero cup product $H^1\times H^1\to H^2$
QUESTION [8 upvotes]: I am looking for an example of a closed orientable 4-manifold $M$ with $H^1(M;\Bbb Z_2)=\Bbb Z_2$ and non-zero cup product $H^1(M;\Bbb Z_2)\times H^1(M;\Bbb Z_2)\to H^2(M;\Bbb Z_2)$.
A non-orientable example is $\Bbb RP^4$. An orientable example of dimension 3 is $\Bbb RP^3$.
I have asked at math stackexchange and the question was upvoted but no answers have been given.

REPLY [11 votes]: Take a closed oriented simply-connected fourfold $N$ with a fixed point free involution $\sigma $, and put $M=N/\sigma $ (for a typical example, take for $M$ an Enriques surface). Then $H^1(M,\mathbb{Z}_2)=$ $\mathrm{Hom}(\pi _1(M),\mathbb{Z}_2)=\mathbb{Z}_2$. Let $x$ be the nonzero element of $H^1(M,\mathbb{Z}_2)$; the square $x^2$ is the reduction (mod. 2) of $\beta x$, where  $\beta :H^1(M,\mathbb{Z}_2)\rightarrow H^2(M,\mathbb{Z})$ is the Bockstein homomorphism. If $x^2=0$, one  has $\beta x=2y$ for some class $y$ in $H^2(M,\mathbb{Z})$. But $\beta x\neq 0$ because $H^1(M,\mathbb{Z})=0$, hence $y$ is a 4-torsion class, which is impossible since $H^2(N,\mathbb{Z})$ is torsion free. Hence $x^2\neq 0$.<|endoftext|>
TITLE: Do locally convex topological vector spaces embed into diffeological spaces?
QUESTION [11 upvotes]: The nLab casually remarks that locally convex tvs embed into diffeological spaces by (discussion around) a corollary in Kriegl and Michor, namely 3.14, but this deals with Boman's theorem and results about smooth maps from a euclidean space to a locally convex spaces.
The definition of smoothness for maps $U \to V$ where $U$ is $c^\infty$-open in a lctvs $W$, and $V$ is a lctvs, is given so that smooth maps are automatically the same as between maps between the associated diffeological spaces.
However, there is a notion of smoothness for maps between lctvs (due to Michal [1,2] and Bastiani [3]) not relying on K&M's definition, but I don't how they relate (Keller [4] treats this, I think, but I don't have access to that book at the moment).

Is it still true that lctvs embed into diffeological spaces using Michal-Bastiani smoothness?

[1] Michal, A. D. Differential calculus in linear topological spaces, Proc. Nat. Acad. Sci. USA 24 (1938), 340-342 (pdf)
[2] Michal, A. D. Differential of  functions with arguments and values in  topological  abelian groups, Proc. Nat. Acad. Sci. USA 26 (1940), 356–359. (pdf)
[3] Bastiani, A., Applications différentiables et variétés différentiables de dimension infinie, J. Anal. Math. 13 (1964), 1–114. (article, paywalled)
[4] Keller, H. H., Differential Calculus in Locally Convex Spaces, Lecture Notes in Mathematics 417, Springer-Verlag, 1974 (Springerlink, paywalled)

REPLY [3 votes]: $\def\sp{\kern.4mm}\def\bbN{\mathbb N}\def\bbZ{\mathbb Z}\def\bbR{\mathbb R}$Here is the answer to David Roberts' residual query: is a convenient diffeomorphism necessarily MB-smooth? No. A counterexample follows:
Let $E$ be the vector space of all real (two-sided) sequences $x=\langle\sp x_i:i\in\bbZ\sp\rangle$ for which $x_i=0$ for sufficiently small $i$ , topologized 
so that we get a linear homeomorphism $E\to\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}$ defined by $x\mapsto(u,v)$ where $u_i=x_{i-1}$ and $v_i=x_{-i}$ for $i\in\bbN$ . Then define $f:E\to E$ by $x\mapsto y$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $y_0=x_0+\sum_{i\in\bbN}(x_i\cdot x_{-i})$ . The inverse is given by $y\mapsto x$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $x_0=y_0-\sum_{i\in\bbN}(y_i\cdot y_{-i})$ . Since the duality map $\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}\to\bbR$ is a discontinuous but bornological, and hence conveniently smooth bilinear map, the assertion follows.<|endoftext|>
TITLE: Exponential rule for Whitney-$\mathcal{C}^{\infty}$-topology
QUESTION [5 upvotes]: Let $M,N,X$ be smooth manifolds. Equip the space of smooth functions between two manifolds with the (strong) Whitney- $\mathcal{C}^\infty$-topology.
The evaluation map $$ev\colon N\times\mathcal{C}^\infty(N,X)\rightarrow X$$ is continuous. (This can e.g. be found in "Margalef-Roig/Dominguez - Differential Topology", Proposition 9.6.7)
Therefore one gets a map of sets $$\mathcal{C}(M,\mathcal{C}^\infty(N,X))\rightarrow \mathcal{C}(M\times N,X).$$
I searched the literature for a reference to the "other direction", but did not find anything. I.e. starting with a smooth map $M\times N\rightarrow X$ I want to know whether the adjoint $M\rightarrow \mathcal{C}^\infty(N,X)$ is continuous or if not, in which cases.

REPLY [4 votes]: The other direction is not true. The map
$$\mathcal{C}^\infty(M,\mathcal{C}^\infty(N,X))\rightarrow \mathcal{C}^\infty(M\times N,X)$$
is not surjective if $N$ is not compact. The image consists of functions $f$ such that

for each compact $K\subset M$ there is a compact $L\subset N$ such that $f(x,y)$ is constant in $x\in K$ for each $y\in N\setminus L$.

This is, because smooth (and continuous) curves in $C^\infty (N,X)$ can move only within a compact of $N$ in finite time. See 4.4.4 (page 34) of

Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp. (pdf)

for the very easy proof. There even the continuous case that you ask for is treated.
Added:
The following might give you feeling for the reason:
Consider $f\in C^\infty(N,\mathbb R)$, equip $C^\infty(N,\mathbb R)$ with the strong Whitney topology. If $t\mapsto t.f$ is continuous $\mathbb R\to C^\infty(N,\mathbb R)$, then $f$ has compact support. Thus $C^\infty_c(N,\mathbb R)$ is the maximal topological vector space in ($C^\infty(N,\mathbb R)$, Whitney topology).<|endoftext|>
TITLE: Non-algebraic K3 surfaces in characteristic $p$
QUESTION [11 upvotes]: I have a very naive question.
Recall that over the field of complex numbers, there exist non-algebraic K3 surfaces. Namely, smooth non-projective simply connected compact complex surfaces with trivial canonical bundle. Such surfaces often come in useful when studying the moduli and deformations of K3 surfaces over $\mathbb{C}$.

Is there a notion of "non-algebraic" K3 surfaces over fields of positive characteristic?

As I say above, I am well aware that this question is very naive. Possible guesses I had in mind were formal schemes / stacks / rigid analytic spaces / ...?

REPLY [15 votes]: Let me briefly expand on Jason's comment.
Actually, "formal scheme" is the right word here. 
For any K3 surface $X$ over an algebraically closed field $k$ of any characteristic one has $$h^0(X, T_X)=0, \quad  h^1(X, T_X)=20, \quad \, h^2(X, T_X)=0,$$
so the functor of Artin rings $$F \colon  (\textbf{Art}) \to (\textbf{Sets})$$  describing the first-order deformations of $X$ is pro-representable and unobstructed, hence it is represented by a complete, regular local ring $A$ of dimension $20$. More precisely, $$A \cong  k[[x_1, \ldots, x_{20}]],$$
where $x_1, \ldots, x_{20}$ are indeterminates. 
Thus there is a formal universal deformation $\widehat{\mathcal{X}} \to \textrm{Specf} (A)$ of $X$; however, such a deformation is not effective, in the sense that there exists no algebraic deformation $\mathcal{X} \to \textrm{Spec}(A)$, where $ \mathcal X$ is a scheme, such that $\widehat{\mathcal{X}}$ is the formal completion of $\mathcal{X}$ along the closed fibre $X$. In fact, a straightforward cup product computation shows that any algebraic family of K3 surfaces with injective Kodaira-Spencer map at any point has dimension $\leq 19$.
Deligne showed that any K3 surface over $k$ can be lifted to $\mathbb{C}$. So there is an analogy with the complex analytic theory, where the deformation space, as a complex manifold, has dimension $20$, but the algebraic K3 surfaces form a (countable union of) $19$-dimensional subfamilies. 
More details on this topic can be found in the books Deformations of algebraic schemes (Sernesi) and Deformation theory (Hartshorne).<|endoftext|>
TITLE: Categorical definition of infinite symmetric product
QUESTION [6 upvotes]: Let $(C,\otimes,I)$ be a symmetric monoidal category with coequalizers and directed colimits.
Fix some object $X$ and morphism $\tau\colon I\to X.$
Using $\tau$ one can construct a sequence of morphisms:
$$
I\rightarrow^{\tau}X\rightarrow^{\tau\circ\rho_X^{-1}} X\otimes X \rightarrow X^{\otimes 3} \ldots
$$
Let $(X,\tau)^{\infty}$ be a colimit of that diagram. 
Question: is it true(under which assumptions it is true?) that there exists a monomorphism $\eta\colon \Sigma_{\infty}\to Aut((X,\tau)^{\infty})$ induced by braidings on terms in the diagram? 
Here $\Sigma_{\infty}$ is a group of all permutations with finite support, we also assume that braiding $B_{X,X}$ is non-identity.
Why symmetric products?
Let us consider following diagram:
$$
\begin{array}{l}
I&\longrightarrow^{\tau}&X&\rightarrow^{\tau\circ\rho_X^{-1}} &X\otimes X& \longrightarrow& X^{\otimes 3}& \ldots\\
\downarrow_{id}&&\downarrow_{id}&&\downarrow_{Coeq(id,B_{X,X})}&&\downarrow_{Coeq(\Sigma_3)}\\
I&\longrightarrow^{\tau}&X&\rightarrow &X^{(2)}& \longrightarrow& X^{(3)}& \ldots\\
\end{array}
$$
Denote colimit of that diagram $(X,\tau)^{(\infty)}.$
Then we have canonical morphism $\zeta_{(X,\tau)}\colon (X,\tau)^{\infty} \to (X,\tau)^{(\infty)}.$
If $((X,\tau)^{(\infty)},\zeta_{(X,\tau)})$ is a coequalizer of $\eta(\Sigma_{\infty})$ then one can naturally call that pair an infinite symmetric product of $(X,\tau).$
UPDATE: Why it is non-trivial? Take a category with objects from the first diagram and terminal object. Then terminal object is colimit, but its automorphism group is trivial, therefore it does not contain $\Sigma_{\infty}.$

REPLY [6 votes]: This seems entirely straightforward unless I'm missing something.  For any $n\in\mathbb{N}$, $\Sigma_n$ acts on $X^{\otimes m}$ for any $m\geq n$ (on the first $n$ coordinates), and this action commutes with the maps in the colimit diagram.  Thus $\Sigma_n$ acts on the colimit of $X^{\otimes n}\to X^{\otimes (n+1)}\to\dots$, which is the same as $(X,\tau)^\infty$ because you've just taken a cofinal subdiagram.  Furthermore, these actions over different $n$ are compatible, so they glue together to give an action of $\Sigma_\infty$.
To put it another way, let $D$ be the category of diagrams in $C$ of the form $X_0\to X_1\to X_2\to\dots$, except that only the "tail" of the diagrams are required to be defined.  That is, for any $n\in\mathbb{N}$, a diagram of the form $X_n\to X_{n+1}\to X_{n+2}\to\dots$ also counts as an object of $D$, and a morphism between objects of $D$ need only be "eventually" defined (and parallel morphisms that eventually agree are identified).  It is straightforward to see that taking the colimit of the diagram gives a well-defined functor from $D$ to $C$.  Now observe that the diagram whose colimit is $(X,\tau)^\infty$ has an action of $\Sigma_\infty$ as an object of $D$, and hence so does its colimit.<|endoftext|>
TITLE: Kontsevich's flow on the space of Poisson structures
QUESTION [17 upvotes]: In §5.3 of Kontsevich's Formality Conjecture he writes:

This (...) gives a remarkable vector field on the space of bi-vector fields on $\mathbf{R}^d$. The evolution with respect to the time $t$ is described by the following non-linear partial differential equation: $$\frac{{\rm d}\alpha}{{\rm d}t} = \sum_{i,j,k,l,m,k',l',m'} \frac{\partial^3 \alpha_{ij}}{\partial x_k \partial x_l \partial x_m} \frac{\partial \alpha_{kk'}}{\partial x_{l'}} \frac{\partial \alpha_{ll'}}{\partial x_{m'}} \frac{\partial \alpha_{mm'}}{\partial x_{k'}} \left( \frac{\partial}{\partial x_i} \wedge \frac{\partial}{\partial x_j} \right),$$
  where $\alpha = \sum_{i,j} \alpha_{ij}(x) \frac{\partial}{\partial x_i} \wedge \frac{\partial}{\partial x_j}$ is a bi-vector field on $\mathbf{R}^d$.

What does ${\rm d}\alpha/{\rm d}t$ here mean exactly? We presume it gives a system of $d(d-1)/2 = {d \choose 2}$ partial differential equations, looking at the equation above component-wise (modulo anti-symmetry of the wedge), and a solution would be a flow $t \mapsto \alpha(t)$, where each $\alpha(t)$ is a bi-vector field.
Next he writes:

Also, in dimension $d = 2$ the direct calculation shows that the evolution operator gives a conjugation of bi-vector field $\alpha$ by a vector field whose coefficients are differential polynomials in coefficients of $\alpha$.

In this case we have one equation:
$$\frac{{\rm d}\alpha_{12}}{{\rm d}t} = \sum_{k,l,m,k',l',m'} \frac{\partial^3 \alpha_{12}}{\partial x_k \partial x_l \partial x_m} \frac{\partial \alpha_{kk'}}{\partial x_{l'}} \frac{\partial \alpha_{ll'}}{\partial x_{m'}} \frac{\partial \alpha_{mm'}}{\partial x_{k'}}, $$
where the $\alpha_{ij}$ in the nonzero terms are all $\pm \alpha_{12}$. Explicitly, writing $u = \alpha_{12}$, $x_1 = x, x_2= y$, we get $$u_t = u_{xxx}(u_y)^3 - u_{yyy}(u_x)^3 - 3u_{xxy}u_x(u_y)^2 + 3u_{xyy}(u_x)^2u_y.$$
How do we proceed to show that the solution is a conjugation by a vector field?

Edit (17/06/15): Or is it the right hand side of the equation that is a conjugation?
The Schouten bracket of a bivector $\alpha = \alpha^{12}(x) \partial_1 \wedge \partial_2$ and a vector $X = X^1\partial_1 + X^2\partial_2$ is a bivector: $$\begin{align*}[\![ \alpha, X ]\!] &= [\![\alpha, X^1\partial_1]\!] + [\![\alpha, X^2\partial_2]\!]\\ &=[\alpha^{12}(x)\partial_1, X^1\partial_1] \wedge \partial_2 - [\partial_2, X^1\partial_1] \wedge \alpha^{12}(x)\partial_1 \\ & \quad + [\alpha^{12}(x)\partial_1, X^2\partial_2] \wedge \partial_2 - [\partial_2, X^2\partial_2] \wedge \alpha^{12}(x)\partial_1\\ &=(\alpha^{12}(x)\partial_1X^1 - X^1\partial_1\alpha^{12}(x) - X^2\partial_2 \alpha^{12}(x) + \alpha^{12}(x)\partial_2X^2)\partial_1 \wedge \partial_2.\end{align*}$$
Putting $u = \alpha^{12}(x), f = X^1, g = X^2,$ this is: $$uf_x - fu_x - gu_y + ug_y = u(f_x + g_y) - fu_x - gu_y.$$
But this involves $u$'s without derivatives, so we can't match it directly to the RHS above.
The Jacobi identity is of no use either: every bivector field on $\mathbf{R}^2$ is Poisson.
(Strikethrough on 02/07/15)

REPLY [3 votes]: By looking at scaling, both $f$ and $g$ have to be cubic in $u$. Moreover, monomials in $f$ contain 3 derivatives in $y$ and 2 derivatives in $x$; similarly for $g$.
There are 12 possible monomials in each $f$ and $g$. E.g. the ones in $g$ are $u_{xxxyy}u^2, u_{xxxy}u_yu, u_{xxyy}u_xu, u_{xxx}u_y^2,u_{xxx}u_{yy}u,u_{xxy}u_xu_y,u_{xxy}u_{xy}u,u_{xyy}u_x^2,u_{xyy}u_{xx}u,u_{yy}u_{xx}u_x,u_{xy}u_{xx}u_y,u_{xy}^2u_x.$
The coefficients can be determined using a computer algebra system. For instance, the following works:
$$
f=u_{yyy}u_x^2+3u_{yy}u_xu_{xy}+u_yu_{xy}^2+u_yu_xu_{xyy}+2u_yu_{yy}u_{xx}+u_y^2u_{xxy}, \\
g=-u_xu_{xy}^2-u_x^2u_{xyy}-2u_{yy}u_xu_{xx}-3u_yu_{xy}u_{xx}-u_yu_xu_{xxy}-u_y^2u_{xxx}.
$$
Note that $f_x + g_y = 0$, so the terms containing $u$ are absent.<|endoftext|>
TITLE: Cheeger Numbers for 3-regular Graphs
QUESTION [8 upvotes]: A student wanted a challenging Graph Theory programming project and I had
him try to determine the maximum value of the Cheeger number (isoperimetric number) among all 3-regular graphs of order $n$, for small values of $n$.  The program we devised seems reasonably efficient, and I wonder if there is any similar data out there that we can use for comparison purposes?
(One amusing side note is that the Pappus graph seems to have an unusually large Cheeger number for it's order, larger than any order $16$ graph or any other order $18$ graph.)

REPLY [9 votes]: I did this calculation a few years ago (according to the timestamps on my programs).
Here is the summary of my results for $n=18$ (total of $41301$ graphs), with each line being the number of graphs followed by a particular Cheeger value.
190 0.111111
450 0.142857
795 0.200000
2002 0.250000
6280 0.333333
5542 0.428571
14909 0.500000
9793 0.555556
6 0.600000
69 0.666667
973 0.714286
291 0.750000
1 0.777778

I also conclude that the unique $18$-vertex graph with maximum Cheeger constant is the Pappus graph.
I can't actually remember why I calculated these numbers, but obviously whatever it was for did not lead to anything, and I haven't seen anything in the literature about Cheeger numbers of cubic graphs in particular. There are a few papers about isoperimetric numbers of families of graphs, but I am sure you can Google them as well as I can. 
ADDED: There are 11 cubic graphs on 20 vertices with the same extremal Cheeger constant as the Pappus graph. I expect that they are all minor perturbations of the Pappus graph, but do not know for sure.<|endoftext|>
TITLE: Is the unitary group of a pre Hilbert space contractible?
QUESTION [5 upvotes]: I already posted my question on mathstackexchange
For a separable Hilbert space $H$ it is known that the unitary group $U(H)$ is contractible, both for the norm topology (Kuiper's theorem) and for the strong operator topology (Dixmier and Douady, 1963). Note that the strong operator topology coincides with the compact open topology in this case.
Does somebody know whether contractibility of the unitary group of at least some special pre Hilbert spaces equipped with the strong operator topology or the compact open topology still holds?
The example I have in mind is the underlying algebraic Fock space $\bigwedge PH \oplus (PH^\perp)^*$ of the Fock space.
Thanks in advance,
StW

REPLY [7 votes]: The proof of contractibility in the strong(=weak=compact-open) topology is very easy:
Identify $H$ with $L^2([0,1])$.
The path $\{u_t\}$ that connects any unitary $u$ to the identity element is then given by:
$$u_t:L^2([0,1])= L^2([0,t])\oplus L^2([t,1])\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$ $$\qquad\qquad\qquad
\xrightarrow{\varphi_t u \varphi_t^{-1}\,\oplus\, id}
L^2([0,t])\oplus L^2([t,1])=L^2([0,1])
$$
were $\varphi_t:L^2([0,1])\to L^2([0,t])$ is the unitary given by
the obvious reparametrization.
To answer your question, I would say that it's a case-by-case analysis:
Examples:
For an infinite algebraic direct sum of Hilbert spaces, I can mimic the above argument and make it work: use $\bigoplus^\infty L^2([0,1])$ and do the same shrinking to $\bigoplus^\infty L^2([0,t])$.
For the $n$-th algebraic tensor power of a Hilbert space, I can also make it work.
Again, it's a matter of finding a 1-parameter family of embeddings of the pre-Hilbert space into itself that "shrinks it to zero". Think about $L^2([0,1]^n)$ shrinking to $L^2([0,t]^n)$ and note that it preserves the subspace given by the $n$-th algebraic tensor power of $L^2([0,1])$.
For exterior powers, it's the same as for tensor powers.
For your particular example of the algebraic Fock space, that same strategy also works, and so the unitary group is strong-contractible.
Think of $\mathcal F$ as given by by $\bigoplus_{n=0}^\infty\bigwedge^n (L^2([0,1]))$ and note that you can shrink it to $\bigoplus_{n=0}^\infty\bigwedge^n (L^2([0,t]))$. Once again, this shrinking procedure preserves the "algebraic" subspace of the Hilbert space $\mathcal F$.<|endoftext|>
TITLE: Is it possible to prove concentration bounds from optional stopping theorem?
QUESTION [5 upvotes]: It is known that the optional stopping theorem from martingale theory is a very powerful theorem in probability theory in statistics. 
I have heard of a probability course at Stanford where martingales and optional stopping is introduced at the beginning, then much of the course material is derived from this principle. 
Unfortunately there are no course notes. Does anyone know where I can find these kind of derivations, or for instance, whether it is possible to derive concentration bounds like Azuma's inequality from OST?

REPLY [6 votes]: Maybe you will be interested in David Williams's book Probability with Martingales, which is intended as a textbook for a first course.  It does take the approach of proving many classical results (strong law of large numbers, etc) using martingale techniques.  I don't believe that it gets as far as concentration bounds, however.
If you can find out who taught the course at Stanford, it may be worth asking them if they have any notes / references / materials they would be willing to share.<|endoftext|>
TITLE: Almost isometric embeddability implies isometric embeddability
QUESTION [7 upvotes]: Consider the following situation: Suppose $X$ is a Banach space such that for each finite metric space $M$ and each $\epsilon > 0$ for which $M$ bi-lipschitz embeds into $X$ with Lipschitz constant $1+\epsilon$, then $M$ isometrically embeds into $X$.
Is there anything known about the properties of such Banach spaces?
It follows from results of Ball that $\ell_p$ has this property for each $p$ (any $n$ point metric space embedding into $\ell_p$ embeds into $\ell_p^{n \choose 2}$, thus a compactness argument gives us that $M$ embeds into $\ell_p$ isometrically by passing to the limit.) Similarly the result is true in Hilbert spaces (much more easily.)
(Balls argument gives the result slightly more generally, but it's not particularly important.)
I am pretty sure the result isn't true in general, or at least, I can see no reason why it should be true: I think the space $(\oplus \ell^2_{p_n})_2$ with $p_n$ tending to 1 from above and the Hamming cube provide a counterexample (with some maybe slight modifications). If this doesn't work, in general there's no reason compactness should be applicable anywhere without really really nice properties (like the $\ell_p$ spaces where we can cut dimensions.)
Edit 1: It might be worth noticing (very, very trivially) that $C[0,1]$ and $\ell_\infty$ have this property, however apart from this I can't think of any other examples
I also can not show that even that if two Banach spaces $X,Y$ have the property that $X \oplus_p Y$ has the property for any $p$. I don't think such a statement should be true in general, but that's a real guess.

REPLY [7 votes]: The example can be generalized: if a Banach space $X$ is strictly convex (the unit sphere does not contain line segments), but is not uniformly convex (that is the unit sphere contains pieces which approach line segments), then it is another example for which $(1+\varepsilon)$-embeddability with an arbitrary $\varepsilon>0$ of a finite metric space does not imply its isometric embeddability. To see this one has to observe that the limiting space (like ultraproduct) will contain pairs of points with several different midpoints between them. Therefore some of finite metric spaces having several different midpoints for some pairs of points are $(1+\varepsilon)$-embeddable into $X$ with an arbitrary $\varepsilon>0$. On the other hand such metric spaces do not admit isometric embeddings into $X$. (I can make this argument more detailed, if needed.) This can be generalized to produce a general criterion in terms of ultraproducts (which does not seem very useful, since it just describes similar argument in more general terms).
Also, the last statement (about $Z=X\oplus_p Y$) can be proved using ultraproducts as follows: If $M$ is a finite metric space arbitrarily well embeddable into $Z$, then $M$ embeds isometrically into an ultrapower of $Z$. Write the corresponding sequences for $X$ and $Y$. They represent finite metric spaces which embed (by our assumption) isometrically into $X$ and $Y$, respectively. Combining these embeddings we get the desired isometric embedding of $M$ into $Z$.<|endoftext|>
TITLE: Is the affine closure of a quasi-affine variety again a variety?
QUESTION [5 upvotes]: Question: Let $X$ be a quasi-affine scheme of finite type over a finite field $k$ and let $A:=\Gamma(X,\mathcal{O}_X)$. Under which conditions is it true that $A$ is a finitely generated $k$-algebra?
As far as I understand, Amnon Neeman http://www.jstor.org/stable/2007052 gives in Remark 8.2 a counter example when $k=\mathbb{C}$ by using complex analytic techniques.
I am mainly interested in the case where $S=\mathbb{A}^1_k=\text{Spec }k[x]$, where $G$ is a closed subgroup scheme of the general linear group $GL_{n,S}$ over $S$, such that $G$ is flat over $S$ and such that $X:=GL_{n,S}\,/G$ is a quasi-affine scheme. Note that such an $X$ always is a scheme by Theorem 4.C of S. Anantharaman http://numdam.org/numdam-bin/fitem?id=MSMF_1973__33__5_0 and is smooth and of finite type over $S$ by [SGA3, VI$_{\rm B}$, Proposition 9.2(xii)].

REPLY [3 votes]: Here are some additional details for the example above.  First of all, it is homogeneous for a linear algebraic group scheme over $W = \text{Spec}(k[t,u]/\langle tu \rangle)$.  Since $W$ is itself finite and flat over $S=\text{Spec}(k[v])$, it is tempting to apply Weil restriction to try and produce an example over $\mathbb{A}^1_k$.  However, this does not work (at least not in the most obvious way).  This illustrates an interesting feature of Weil restriction: it can sometimes transform quasi-affine schemes to affine schemes.  
As above, let $Y$ be the affine scheme $\mathbb{A}^1_W = \text{Spec}(k[s,t,u]/\langle tu \rangle)$.  Let $Z$ be the closed subscheme associated to the ideal $\langle s,t \rangle$.  Let $X$ be the quasi-affine complement $Y\setminus Z$.  By working with the covering of $X$ by the basic open affines $D(s)\subset Y$ and $D(t)\subset Y$, it is straightforward to compute 
$$
\Gamma(X,\mathcal{O}_X) = (k[s,t,u]/\langle tu \rangle)[r_1,r_2,r_3,\dots,r_n,\dots]/\langle tr_n, s^nr_n - u, s^mr_{m+n} - r_n \rangle,
$$
where the restriction of $r_n$ to $D(s)$ is $s^{-n}u$ and where the restriction of $r_n$ to $D(t)$ is $0$.  In particular, $\Gamma(X,\mathcal{O}_X)$ is not a finite type $k$-algebra.
To see that $X$ is homogeneous, let $H_W$ be the $W$-group scheme $H\times_{\text{Spec}(k)} T$, where $H$ is the closed $k$-subgroup scheme $\text{Spec}(k[a,a^{-1},b])$ of $\mathbf{GL}_{2,k}$ of matrices of the form 
$$\left( \begin{array}{rr} a & b \\ 0 & 1 \end{array} \right). $$  A $W$-action of $H_W$ on $Y$ is the same as a $k$-action of $H$ on $Y$ such that projection from $Y$ to $W$ is $H$-invariant.  Define the action by
$$
\left( \begin{array}{rr} a & b \\ 0 & 1 \end{array} \right)\cdot (s,t,u) = (as+bt,t,u). 
$$
This action preserves the closed subscheme $Z$.  Thus it induces an action on $X$.  There is a section of the projection $X\to W$, namely $(t,u) \mapsto (1,t,u)$.  The stabilizer is the closed $W$-subgroup scheme $G_W\subset H_W$ that is the zero scheme of $a+bt-1$.  This is $W$-flat.  Thus, as a $W$-scheme, $X$ is the quotient $H_W/G_W$.  The scheme $X$ is finite type, quasi-affine and smooth over $W$, but the associated affine scheme $\text{Spec}\Gamma(X,\mathcal{O}_X)$ is not finite type over $W$.
We could try to produce such an example over $S=\text{Spec}(k[v])$ from this example by taking the Weil restriction of $X$ with respect to the finite, flat morphism 
$$
\phi:W\to S, \ \phi^*v = t + u.
$$ 
However, the Weil restriction of $X$ with respect to $\phi$ is actually affine.  The Weil restriction of $\mathbb{A}^1_W$ with respect to $\phi$ is isomorphic to $\mathbb{A}^2_S = \text{Spec}(k[s_0,s_1,v])$, where the associated universal $W$-morphism $\psi:\mathbb{A}^2_S\times_{S,\phi} W \to \mathbb{A}^1_W$ is
$$
\psi:\mathbb{A}^2_W \to \mathbb{A}^1_W, \ \ \psi(s_0,s_1,t,u) = (s_0+s_1t,t,u).
$$
With respect to this isomorphism, the Weil restriction of $X$ with respect to $\phi$ is the basic open affine subset $D(s_0)\subset \mathbb{A}^2_S$.  This illustrates one phenomenon of the Weil restriction: although it always transforms affine schemes to affine scheme, it also transforms some (non-affine) quasi-affine schemes to affine schemes.<|endoftext|>
TITLE: for which values of $\theta$ does this equation $x_{n+1}=\cos(\theta)x^2_{n}-\sin(\theta)x^2_{n-1}$ have bounded solutions?
QUESTION [6 upvotes]: I would like to investigate the global behavior of the following equation :
$$x_{n+1}=Ax^2_{n}-Bx^2_{n-1}$$
where $A(\theta)= \cos(\theta)$ and $B(\theta) =\sin(\theta)$  are nonnegative parameters and $x_{-1}$, $x_0$ are nonnegative with $x_{-1}+x_{0}>0$. Here is my question:

For which values of $\theta$ does the equation below have bounded solutions?
$$x_{n+1}=\cos(\theta)x^2_{n}-\sin(\theta)x^2_{n-1}$$ 

Note : $\theta$ $\in \mathbb{R}$, $n=0,1,2\cdots$.
Thank you for any help.

REPLY [9 votes]: I ran a small computation: The following plot is created as follows:
For each point $r(\cos \theta, \sin \theta)$, I use $x_{-1} = 0$, $x_0 = r$
as initial values, and $\theta$ as the parameter.
The white area is where the iterations is less than $2$ (bounded), the first 20 iterations. As we see, a small circle around the origin is white, meaning that there are small initial values that is bounded for every $\theta$. Now, this is not a proof, but the picture suggests this.
Changing the cut-off to 40 iterations does not change the picture much.

EDIT: I think I have a proof of the statement. 
Let $r=1/\sqrt{2}$. If $x_{-1},x_0 \leq r$, then the sequence is bounded for all $\theta$.
Proof: $|\cos(\theta)x_{n-1}^2 - \sin(\theta)x_{n}^2| \leq r^2(|\cos \theta|+|\sin \theta|) \leq r^2 \sqrt{2} \leq 1/\sqrt{2}$, and the statement follows from induction.
ValueFrom[x_, y_] := Module[{func, xf, xfp, iterTot, t, r, cc, sc},
   t = Arg[x + I y];
   r = Abs[x + I y];

   cc = Cos[t];
   sc = Sin[t];

   func[{xn_, xp_, i_}] := {xp, cc*xp^2 - sc*xn^2, i + 1}; 

   {xf, xfp, iterTot} = 
    NestWhile[func, {0, r, 0}, Abs[#[[2]]] < 2 && #[[3]] < 20 &];
   iterTot
   ];

DensityPlot[ValueFrom[x, y], {x, -2.0, 2.0}, {y, -2.0, 2.0},
 PlotPoints -> {150, 150}, PlotRange -> All, 
 ColorFunction -> "SunsetColors"]<|endoftext|>
TITLE: Style of mathematical writing vs. too many lemmas
QUESTION [21 upvotes]: I work in PDEs. I have now written 3 papers. I find my style is of the form: introduction, statement of results, paragraphs to introduce something, lemma, more text, lemma, more text, lemma, more text, theorem, concluding remarks (I missed the proofs).
I am getting sick of this type of writing. I want to write my next paper with more style and elegance than something which looks like the output from a workhouse. 
For example I have seen papers where they come up with some interesting result which they don't put into a lemma but instead put it in a paragraph. When I cite such a thing, I say something like "see the paragraph on page 10 of [1]" so I think it is a bad idea to put results outside environments. 
Does anyone have any ideas? Basically I want to write a more unpredictable paper instead of the usual routine which I wish to take myself out of. This is good because readers will also find it more interesting.

REPLY [9 votes]: The mathematical papers I find most enjoyable (as well as accessible) are those written in order of increasing technicality.  So, consider an order along the lines of Introduction and Conclusion (containing motivation, related work, main ideas, informal statement of results and applications, interesting open problems).  Essential definitions. Formal statement of results.  Applications/corollaries.  Further definitions. Elementary arguments (may be several sections).  Technical estimates (may be several sections).  Of course, the logical dependences require careful cross-referencing.<|endoftext|>
TITLE: Are maps homotopic with respect to a uniform number of local homotopies
QUESTION [7 upvotes]: I've encountered the following problem that I'm sure someone more topologically inclined can answer:
Say that a homotopy of maps $f:X\times[0,1)\to Y$ between two compact smooth manifolds $X$ and $Y$ is good (I don't know the exact term for it if there is one) if the homotopy is constant in time outside of some coordinate chart in $Y$, in which case we'd say the homotopy is supported in the chart.
Suppose you have two continuous maps $f_0$ and $f_1$ between compact smooth manifolds $X$ and $Y$ which are homotopic. Cover $Y$ with finitely many coordinate charts $U_\alpha$. My question is the following:
Are $f_0$ and $f_1$ homotopic with respect to a finite sequence of good homotopies supported in the elements $U_\alpha$ of the covering? Can the number of good homotopies required be uniformly bounded by topological data like the homotopy class of $f_0$ or the number of elements in the covering?
I am interested in this question because I have a functional which I can get a good estimate for if the above is true.

REPLY [3 votes]: Yes, you can always connect $f_0$ and $f_1$ by a finite sequence of good homotopies supported in any given open cover (and in fact, this sequence of good homotopies can be chosen to be homotopic to $f$ relative to $X\times \{0,1\}$).  By pulling back the $U_\alpha$ to an open cover of $X\times [0,1]$ via $f$, it suffices to show this in the universal case where $Y=X\times [0,1]$ and $f$ is the identity.  By refining the open cover, we may assume we have an open cover consisting of all the sets of the form $V_i\times (n\epsilon,(n+3)\epsilon)$ for some finite open cover $\{V_1,\dots,V_m\}$ of $X$ and some $\epsilon>0$.  Let $\{\varphi_i\}$ be a partition of unity subordinate to $\{V_i\}$.  Then you can connect $f_0$ to $f_1$ through good homotopies as follows.  First, homotope $f_0:x\mapsto (x,0)$ to the map $x\mapsto (x,\epsilon \varphi_1)$ through the homotopy $(x,t)\mapsto (x,t\epsilon\varphi_1)$.  Then homotope to $x\mapsto(x,\epsilon(\varphi_1+\varphi_2))$ similarly, and so on; after $m$ steps you will have reached $x\mapsto (x,\epsilon\sum \varphi_i)=(x,\epsilon)$.  Now repeat to get to $x\mapsto (x,\epsilon+2)$, and so on until you reach $x\mapsto (x,1)$.
On the other hand, you can't get any uniform bound on the number of steps needed using only $f_0$ or the open cover.  For instance, consider the case where $X=Y=S^1$, $f_0$ is a constant map, and $f_1$ winds around $n$ times in one direction and then $n$ times in the opposite direction.  By lifting to the universal cover, it is not hard to show that it takes at least $n$ good homotopies to get from $f_0$ to $f_1$ (for any covering of $Y$ by charts).<|endoftext|>
TITLE: Topological tameness beyond the Gandy-Harrington topology
QUESTION [7 upvotes]: The Gandy-Harrington topology on $\omega^\omega$ is the topology generated by all lightface $\Sigma^1_1$ sets; that is, all sets which are continuous-in-the-usual-sense images of $\omega^\omega$.
Although this topology is definitely less nice than the standard one - for example, it is non-metrizable - it satisfies the strong Choquet property: this is the statement that player I has a winning strategy in a certain topological game. From the strong Choquet property we can deduce many results whose proof would follow easily from metrizability, so in some sense the Gandy-Harrington topology is "close enough" to metrizable. One particularly nice result we can show is: if $A$ is a nonempty lightface $\Sigma^1_1$ subset of $\omega^\omega$, then $A$ has an element $x$ such that $\mathcal{O}^x\equiv_T\mathcal{O}\oplus x$.
A natural question now is to ask about topologies generated by lightface sets further up the projective hierarchy; e.g., let $\tau_k$ be the topology on $\omega^\omega$ generated by lightface $\Sigma^1_k$ sets. And similarly, we can define $\tau_\Gamma$ for any pointclass $\Gamma$ (although we're probably only interested in, say, Spector pointclasses). Unfortunately, for $k>1$ the strong Choquet property fails badly in $\tau_k$!
My question is:

Is there a weakening of the strong Choquet property that holds for $\tau_2$? (Or more generally for $\tau_\Gamma$ for nice enough pointclass $\Gamma$.)

Relatedly, for a real $x$ let $\mathcal{O}_2^x$ be the set of $x$-computable codes for nonempty $\Sigma^1_2$ sets, and let $\mathcal{O}_2=\mathcal{O}_2^\emptyset$.

Is it the case that every lightface $\Sigma^1_2$ subset of $\omega^\omega$ contains an $x$ with $\mathcal{O}_2^x\equiv_T\mathcal{O}_2\oplus x$?

I've tagged "set-theory" and "inner-model-theory" because it seems reasonable to me that answers might depend on structural hypotheses about the universe; feel free to remove either tag if this seems bonkers.

REPLY [5 votes]: As mentioned by Ted, the appropriate generalization is for odd levels only. In Hjorth "Variations of the Martin-Solovay tree" and "Some applications of coarse inner model theory", the following folklore result that generalizes Harrington's proof of Silver's dichotomy is stated:
Assume boldface $\bf\Delta^1_2$ determinacy. If $E$ is a thin $\Pi^1_3$ equivalence relation on $\mathbb{R}$, then for every $x$, there is a $\Delta^1_3(<u_\omega)$ set $A$ such that $x \in A \subseteq [x]_E$.
As defined in these papers, for a fixed $\alpha<u_\omega$, $A\subseteq \mathbb{R}$ is $\Sigma^1_3(\alpha)$ means there is $B\subseteq \mathbb{R}^2$ such that $x\in A$ iff  $(x,y)\in B$ for some sharp code $y$ for $\alpha$. $\Sigma^1_3(<u_\omega)$ means $\Sigma^1_3(\alpha)$ for some $\alpha<u_\omega$. $\Delta$-pointclasses are defined in the obvious way.
The correct generalization of the Gandy-Harrington topology appears to be generated by $\Sigma^1_3(<u_\omega)$ sets. Also as a corollary, every thin $\Pi^1_3$ equivalence relation is $\Delta^1_3$ reducible to a $\Pi^1_3$ equivalence relation on a $\Pi^1_3$ subset of $u_\omega$ (in the sharp codes); every thin $\Delta^1_3$ equivalence is $\Delta^1_3$ reducible to equality on $u_\omega$.
I haven't read the proof of this generalization. So it would be nice if someone can expand a bit more. I also don't know the proof of the basis theorem at lower level with this approach, even though it seems to be straightforward.<|endoftext|>
TITLE: Is $f(x,y)=\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{1}{n}e^{2\pi i(xn+yn^2)}$ essentially bounded?
QUESTION [8 upvotes]: Let $$f(x,y)=\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{1}{n}e^{2\pi i(xn+yn^2)}
$$
Is it true that $\|f\|_{L^{\infty}(\mathbb{R}^2)}<\infty$? i.e. is $f$ essentially bounded?

REPLY [5 votes]: Prof. Tao's answer is excellent. I also found two research papers answering the question so I list them below as complementary reference:

G.I.Arkhipov and K.I.Oskolkov, On a special trigonometric series and its applications, 1989 Math. USSR Sb. 62 145
Link to the article:
http://iopscience.iop.org/0025-5734/62/1/A10
See Theorem 1.
E.S.Stein and S.Wainger, Discrete analogues of singular Radon transform, Bulletin of AMS 1990
Link to the article: http://www.ams.org/journals/bull/1990-23-02/S0273-0979-1990-15973-7/S0273-0979-1990-15973-7.pdf
See the Lemma in Section 6

The key of their results is that the upper bound depends only on the degree (not the coefficients, i.e. $x$ and $y$ in the question) of the polynomial.<|endoftext|>
TITLE: How to embed $U(1)$ into a bigger group, using Dynkin diagrams
QUESTION [5 upvotes]: I am trying to find the embedding and the branching rules for some group decompositions. For example, I consider $E_7$ and its maximally compact subgroup $SU(8)$ and I want to "see" how the Dynkin diagram of $E_7$ is modified to get $A_7$. I tried to take a linear combination of roots to produce $A_7$. And also following other analog questions, with this I have (almost) no issues.
The problem is for an embedding like $SU(8)⊃SU(6)×U(1)$. How do I get this?
As all of you can imagine, my doubts arise when the embedding involves abelian subgroups, that I am not able to see from Dynkin diagrams.
All these issues comes from the calculus of centraliser groups. For instance, I want the centraliser group $\mathcal{C}_{E_7}(SU(3))$ of $SU(3)_{\mathrm{diag}} \subset SU(3) \times SU(3) \subset SU(6) \subset SU(8) \subset E_7$. I know it is
$$ \mathcal{C}_{E_7}(SU(3)) = SU(3) \times SU(2) $$
but I do not have a procedure to get something involving $U(1)$ factors.

REPLY [5 votes]: "I want to see how the Dynkin diagram of E7 is modified to get A7":
Take $E_7$:
$\bullet -\bullet -\bullet -\stackrel{\stackrel{\textstyle\bullet}{\textstyle |}}{\bullet} -\bullet -\bullet$
Add the lowest negative root (these roots are no longer linearly independent):
$\bullet -\bullet -\bullet -\stackrel{\stackrel{\textstyle\bullet}{\textstyle |}}{\bullet} -\bullet -\bullet -\bullet$
Remove the top root to get another basis of $\mathfrak t^*$:
$\bullet -\bullet -\bullet -\bullet -\bullet -\bullet -\bullet$<|endoftext|>
TITLE: Definitions of the module $R/(x_0^\infty,x_1^\infty,\ldots,x_{n-1}^\infty)$
QUESTION [6 upvotes]: There are several constructions of the Prüfer group $\mathbb{Z}/p^\infty$; here are two that are relevant for this question. 

It can be constructed via the short exact sequence
$$
0 \to \mathbb{Z} \to p^{-1} \mathbb{Z} \to \mathbb{Z}/p^\infty \to 0.
$$
It can be constructed as the colimit $$\mathbb{Z}/p^\infty = \operatorname*{colim}_k\mathbb{Z}/p^k\mathbb{Z}$$
where the homomorphisms are induced by multiplication by $p$. 

It is not too hard to see that these are isomorphic. 
There is a more complicated construction that often appears in topology. Now, let $R$ be a ring, and let $I = (x_0,\ldots,x_{n-1})$ be a finitely-generated ideal inside of $R$, generated by a regular sequence. Then, there is an $R$-module $R/(x_0^\infty,\ldots,x_{n-1}^\infty)$, defined in the following way, by first inverting $x_i$ and taking the cokernel, i.e., by the short exact sequences
$$
0 \to R \to x_0^{-1}R \to R/(x_0^\infty) \to 0 \\
0 \to R/(x_0^\infty) \to x_1^{-1}R/(x_0^\infty) \to R/(x_0^\infty,x_1^\infty) \to 0 \\
0 \to R/(x_0^\infty,x_1^\infty) \to x_2^{-1}R/(x_0^\infty,x_1^\infty) \to R/(x_0^\infty,x_1^\infty,x_2^\infty) \to 0 \\
 \cdots
$$
This is the analogue of construction (1) above. 
I would like to propose a different construction, which is the analogue of (2). Let $I^k$ denote the $k$-th power of the ideal $I$. There is a system
$$
\cdots \subset I^k \subset I^{k-1} \subset \cdots 
$$
which gives maps
$$
\cdots \to R/I^k \to R/I^{k-1} \to \cdots 
$$
Now we work in the derived category, and let $DM = \mathbb{R}\text{Hom}_R(M,R)$ be the (derived) dual. The first claim is that $DR/I^k = \Sigma^{-n}R/I^k$. To see this, start with the cofiber sequence
$$
R \xrightarrow{x_0} R \to R/(x_0);
$$
this is self dual, and $DR/x_0 = \Sigma^{-1}R/(x_1)$. Induction using the cofiber sequences
$$
R/(x_0,\ldots,x_{i-1}) \xrightarrow{x_i} R/(x_0,\ldots,x_{i-1}) \to R/(x_0,\ldots,x_i)
$$
proves the claim for $R/I$. For $R/I^k$ inductively use the cofiber sequences
$$
I^k/I^{k-1} \to R/I^k \to R/I^{k-1}
$$
and the fact that 
$$
\bigoplus_{k \ge 0} I^k/I^{k+1} = R/I[x_1,\ldots,x_n]
$$
to see inductively that $DR/I^k = \Sigma^{-n}R/I^k$. 
Thus (after desuspension) we can get a sequence
$$
\cdots \to R/I^{k-1} \to R/I^k \to \cdots.
$$
Define $R/I^\infty$ to be the colimit of this system. 
My question is the following:

Is $R/(x_0^\infty,\ldots,x_{n-1}^\infty) \simeq R/I^\infty$?

Here is an idea of how to prove this. Firstly, I believe one can show, by a cofinality argument, that
$$
R/I^\infty = \operatorname{colim}_i R/(x_0^i,\ldots,x_{n-1}^i).
$$
(Here I mean the colimit over the system of ideals  $I_t = (x_0^t,\ldots,x_{n-1}^t)$).
Then to prove the claim note that it is clear that $R/(x_1^\infty) = \operatorname{colim}_i R/(x_0^i)$. Then inductively we have
$$
R/(x_0^\infty,\ldots,x_{k}^\infty )= \operatorname{colim}_j(R/(x_0^\infty,\ldots,x_{k-1}^\infty)/x_k^j) \\
= \operatorname{colim}_j(\operatorname{colim}_i R/(x_0^i,\ldots,x_{k-1}^i)/x_k^j) \\
= \operatorname{colim}_j \operatorname{colim}_i R/(x_0^i,\ldots,x_{k-1}^i,x_k^j) \\
= \operatorname{colim}_i R/(x_0^i,\ldots,x_{k-1}^i,x_k^i)
$$
By induction this should give the result, I hope. 
Disclaimer: I originally asked this on math.se, but did not get a response.

REPLY [3 votes]: I strongly guess that, you mean,  the maps of your direct system, $$R/I_t\rightarrow R/I_{t+1},$$ whose colimit is, $$R/I^\infty,$$ are the multiplication by $x_0\ldots x_{n-1}$. So I think that your question has a positive answer. For a complete description I aim to firstly give some definitions. The following definitions  are given in the paper Sharp and Zakeri.
Let $R$ be a ring and $U$ be a non-empty subset of $R^n$ where $n\in \mathbb{N}$. Then $U $ is said to be a traingular subset of $R^n$ if it satisfies the following two conditions:
(1) for each $(u_1,\ldots,u_n)\in U$ and $\alpha_1,\ldots,\alpha_n\in \mathbb{N}$ we have $(u_1^{\alpha_1},\ldots,u_n^{\alpha_n})\in U$
(2) for each $(u_1,\ldots,u_n),(v_1,\ldots,v_n)\in U$ there exists $n\times n$ lower triangular matrices $H,K$ with entries in R and $(w_1,\ldots,w_n)\in U$ such that $H[u_1,\ldots,u_n]^\tau=[w_1,\ldots,w_n]^\tau=K[v_1,\ldots,v_n]^\tau$ (here the notation $\tau$ stands for the transpose of matrix).
Now fix a triangular subset $U$ of $R^n$. Let $M$ be an $R$-module. Then one can define an equivalance relation on $U\times M$ such that $((u_1,\ldots,u_n),m)$ is equivalent to $((v_1,\ldots v_n),n)$  if and only if there exists $(w_1,\ldots,w_n)\in U$ and lower triangular matrices $H,K$ such that, $$ H[u_1,\ldots,u_n]^\tau=[w_1,\ldots,w_n]^\tau=K[v_1,\ldots,v_n]^\tau $$ and $\text{det}(H)m-\text{det}(K)n\in (w_1,\ldots,w_{n-1})M$.
By this definition we get an $R$-module denote by $U^{-n}(M)$ which is called the module of generalized fractions. In fact one can verify that this definition generalizes the definition of the localization of an $R$-module at a multiplicative closed subset of $R$.
Now assume that $x_1,\ldots,x_n$ is sequence of elements of $R$. Extend this sequence to an infinite sequence, $x_1,\ldots,x_n,1,1,\ldots,1,\ldots$ ($x_i=1$ for $i\ge n+1$). Then for each $i$ we get the triangular subset, $$U_i:=\{(x_1^{\alpha_1},\ldots,x_i^{\alpha_i});\text{ for some } 0\le j\le i, \alpha_1,\ldots,\alpha_j\in \mathbb{N} \text{ and }  ,\alpha_{j+1}=\ldots=\alpha_i=0\},$$ of $R^i$. In particular for each $1\le i\le n$ we get the $R$-module $U_i^{-i}(R)$ of generalized fractions which form the complex,
  $$ 0\rightarrow R\rightarrow U_1^{-1}(R)\rightarrow \ldots \rightarrow U_i^{-i}(R) \overset{e^i}{\rightarrow}U_{i+1}^{-i-1}(R)\rightarrow\ldots,$$
in which the differential $e^i$ are defined by,
$$ e^i(r/(u_1,\ldots,u_i))=r/(u_1,\ldots,u_i,1).$$
In 2.2 Theorem of the paper On the Generalized Fractions of Sharp and Zakeri it is proved the above complex is isomorphic to the complex which is the same as your first construction, i.e. considering the  cokernel of the localization map at element $x_i$ at each step. 
Now in your situation we have the sequence $x_0,\ldots,x_{n-1}$ and the above mentioned Theorem states that $U_{n+1}^{-n-1}(R)\cong R/(x_0^\infty,\ldots,x_{n-1}^\infty)$ (with your notation).
  Now one can  prove that the module of generalized fractions $U^{-n-1}_{n+1}(R)$ is nothing but your direct limit $\lim_{\rightarrow} R/(x_0^i,\ldots,x_{n-1}^i)$ in which the maps $R/(x_0^i,\ldots,x_{n-1}^i)\rightarrow R/(x_0^j,\ldots,x_{n-1}^j)$ are multiplication maps by $x_0^{j-i}\ldots x_{n-1}^{j-i}$  (In order to prove this isomorphism you need to use Lemma 2.3 of the paper Modules of Generalized Fractions. Using this lemma you can reduce an arbitrary lower triangular matrix to a diagonal matrix whose diagonal is something like $x_0^t,\ldots,x_{n-1}^t,1$)). In fact your question has still an affirmative answer even if we drop the condition for the sequence to be a regular sequence. The modules you are interested are called the top local cohomology modules supported at the ideal $(x_0,\ldots,x_{n-1})$. 
By virtue of the paper Comparison of Certain Complexes of Modules of Generalized Fractions and Čech Complexes, over a Noetherian ring, it is always possible even to compute also the  other local cohomology modules by means of the above complex.<|endoftext|>
TITLE: Does the CGWH-fication change the (weak) homotopy type?
QUESTION [5 upvotes]: Let $Top$, $CG$, $WH$, $CGWH$ be the categories of topological spaces, compactly generated spaces, weak Hausdorff spaces and compactly generated weak Hausdorff spaces.
There is the CG-ification $X_{CG}$ of a space $X$ and the WH-ification $X_{WH}$ of a compactly generated space $X$. These provide adjoint pairs between $Top$ and $CG$ and also between $CG$ and $CGWH$. Combined, they give me the CGWH-fication $(X_{CG})_{WH}=X_{CGWH}$ of a space $X$. The latter one is the one used implicitly in most of today's algebraic topology.
I'm interested in the question which of these constructions can change the (weak) homotopy type and which of them preserves it. It seems to me that at least in many cases, the weak homotopy type must be preserved. I really don't want to have $\Omega X$ to have a different homotopy type depending on whether I used the compactly generated topology on the mapping space or not.

REPLY [11 votes]: It is very easy to see that CG-ification preserves weak homotopy type: it is a right adjoint, so for any CG space $K$, the maps $K\to X_{CG}$ are the same as the maps $K\to X$.  Letting $K$ be $S^n$ and $S^n\times I$ immediately gives that the map $X_{CG}\to X$ is a weak equivalence.
However, WH-ification does not preserve weak homotopy type.  For instance, if $X$ is a space with finitely many points, then $X$ is CG, and $X_{WH}$ will always be discrete (since it is a finite $T_1$ space).  However, finite spaces can have the weak homotopy type of any finite CW-complex!  So WH-ification can really destroy a lot of homotopy-theoretic information when applied to arbitrary spaces.  This rarely is a cause for concern though.  For instance, you mentioned being worried about the topology on loopspaces, but the CG-ification of the loopspace of a CGWH space is automatically already WH (see Proposition 2.24 of this excellent paper, for instance).<|endoftext|>
TITLE: Monte-Carlo computation of the Smith normal form
QUESTION [6 upvotes]: Quite some time ago I saw an article where a Monte-Carlo algorithm for computing the Smith normal form of an integer matrix was described. In this article the following problem was posed:
Suppose $P, Q\in\mathbb{Z}[X,Y]$ are polynomials without a common factor. Then $(P(n,m), Q(n,m))$ has a density, i.e. there exist real numbers $\delta_k$, such that
$$
\lim_{N\rightarrow\infty} \frac{1}{4N^2}\#\big\{(n,m):-N\leq n,m\leq N, (P(n,m), Q(n,m)) = k\big\} = \delta_k,
$$
and $\sum_k\delta_k = 1$.
Can someone please give me a reference to this paper?

REPLY [2 votes]: Typing "Monte-Carlo" and "Smith normal" into MathSciNet turned up a few possibilities. 
Mustafa Elsheikh, Mark Giesbrecht, Andy Novocin, B. David Saunders, Fast computation of Smith forms of sparse matrices over local rings, ISSAC 2012—Proceedings of the 37th International Symposium on Symbolic and Algebraic Computation, 146–153, ACM, New York, 2012, MR3206298. 
David Saunders, Zhendong Wan, Smith normal form of dense integer matrices, fast algorithms into practice. (English summary) ISSAC 2004, 274–281, ACM, New York, 2004, MR2126954 (2005k:15039). 
[The review mentions a "fast Monte Carlo algorithm of Eberly, Giesbrecht and Villard for computing the Smith form of an integer matrix," but gives no reference.]
Mark Giesbrecht, Fast computation of the Smith form of a sparse integer matrix, Comput. Complexity 10 (2001), no. 1, 41–69, MR1867308 (2003d:15014).<|endoftext|>
TITLE: Embedding of planar graphs
QUESTION [8 upvotes]: I've recently come across the following lemma. 
Lemma (Valiant): A planar graph $G$ with maximum degree $4$ can be embedded in the plane using $O(|V|)$ area in such a way that its vertices are at integer coordinates and its edges are drawn so that they are made up of line segments of the form $x=i$ or $y=j$, for integers $i$ and $j$.
I'd like to first point out that i'm not interested in the size of the area that the graph is embedded in, so we can for all intents and purposes assume it's possible to work in an area of unrestricted size. I'm interested to know (because they haven't explicitly stated) whether any graph in the class mentioned, can be embedded without having any 'bends' in the edges? If it's necessary to allow for 'bends' is there a restriction on the maximum number required? Secondly would i be able to create an embedding where the lengths of all the edges are multiples of $4$ (perhaps by applying some kind stretch to the original graph)?

REPLY [7 votes]: Bends are necessary, we have studied this problem in this paper: http://arxiv.org/abs/1009.1315.<|endoftext|>
TITLE: Countable group with uncountable number of subgroups $< 2^{\aleph_0}$
QUESTION [6 upvotes]: Is it consistent that there is a countable group $G$ such that the cardinality of the set of subgroups of $G$ is uncountable, but strictly less than $2^{\aleph_0}$?

REPLY [27 votes]: Subsets $H\subseteq G$ can be identified with their characteristic functions $\chi_H\colon G\to\{0,1\}$, which we can view as elements of the Cantor space $2^G$.
In this perspective, subgroups of $G$ form a closed subset of $2^G$: if $H\subseteq G$ is not a subgroup, then $1\notin H$, or $ab^{-1}\notin H$ for some $a,b\in H$, hence
$$\{f\in2^G:f(1)=0\}$$
or
$$\{f\in2^G:f(a)=f(b)=1,f(ab^{-1})=0\}$$
is a basic open neighbourhood of $\chi_H$ that excludes the characteristic functions of all subgroups.
Thus, $G$ can only have countably many or $2^\omega$ subgroups by the Cantor–Bendixson theorem.
There is nothing special about groups, the same argument applies to substructures of any countable algebraic structure $G$.<|endoftext|>
TITLE: Avoiding countable subgroups of a group homeomorphic to the Cantor space
QUESTION [13 upvotes]: Update: Further work with Adam (who answers below) and Piotr led to a rather satisfactory result about the problem that motivated the problem below, see our recent paper 
The Haar Measure Problem. In particular, we answer there a problem mentioned in the discussion below.
The following question is motivated by the paper 
[Brian, Mislove, Every compact group can have a non-measurable subgroup]. 
A positive solution to a variation of the following problem implies a positive
solution to the problem studied there, i.e., that every infinite compact group has a subgroup that is not Haar measurable. A more general form of this question was answered in the negative there.
Every infinite metric profinite group is, as a topological space, homeomorphic to 
the Cantor space [Hofmann, Morris, The Structure of Compact Groups, Theorem 10.40].
Problem.
Let $G$ be an infinite metric profinite group (or, more generally, one homeomorphic to the Cantor space), $H$ a countable subgroup of $G$, and $g\in G\smallsetminus H$.
Is there necessarily an element $x\in G\smallsetminus H$ such that $g\notin\langle H\cup \{x\}\rangle$?
Discussion. I find it intriguing that in this scenario, the problem is about a topological group structure on the Cantor space. A rather well-understood object from the topological point of view. In particular, this fact is used in the Brian--Mislove paper cited above to observe that the answer to the big question in the first paragraph is positive when there is a non-null set of reals of cardinality smaller than the continuum ($\mathrm{non}(\mathcal{N})<\mathfrak{c}$). Thus, it remains to consider the case $\mathrm{non}(\mathcal{N})=\mathfrak{c}$ and a transfinite induction can be used to construct the needed nonmeasurable subgroup. There, in each step, the intermediate subgroup is Lebesgue null. In our problem, we model this by "countable". 
Even assuming the Continuum Hypothesis, I do not know the answer to Problem 2 (or the big problem).
Comment: I have separated this question from the quetion there to make it possible to declare the other question as solved.

REPLY [8 votes]: Yes, there exists.
Since $G$ is profinite, we may write it as $G=\lim_{n\in\mathbb{N}} G_n$, where $G_n$ are finite and the projections $\pi_n:G\to G_n$ are onto.
Since $H$ is countable, we may write it as $H=\bigcup_{n\in\mathbb{N}} H_n$, where $H_n\subseteq H_{n+1}$ are finite, $H_0$ nonempty.
Let $\bar H_n = H_n\cup H^{-1}_n\cup\{e\}$ where $X^{-1}=\{x^{-1}\mid x\in X\}$. Let $X^r=\{x_1x_2\ldots x_r\mid x_i\in X\}$.
We construct inductively a sequence $A_{k}\subseteq G_{n_k}$ as follows. Pick any $h\in H_0$ and choose $n_0$ so that $\pi_{n_0}(h)\neq \pi_{n_0}(g)$. Let $A_0=\{\pi_{n_0}(h)\}\subseteq G_{n_0}$. Suppose $n_k$ and $A_k\subseteq G_{n_k}$ are defined. For every $a\in A_k$ we choose two points $h_a,h'_a\in H$ such that $\pi_{n_k}(h_a)=\pi_{n_k}(h'_a)=a$. This is always possible since $H$ is a group and the kernel of $\pi_{n_k}$ restricted to $H$ is infinite. Find $m_k$ such that every $h_a$ and $h'_a$ belongs to $H_{m_k}$. We may assume that the sequence $m_k$ is increasing. Then choose $n_{k+1}>n_k$ so that $\pi_{n_{k+1}}(g)\notin\pi_{n_{k+1}}(\bar H^{k}_{m_k})$. This is possible since $\bar H^{k}_{m_k}\subseteq H$ is finite and $g\notin H$. 
Define $A_{k+1}=\{\pi_{n_{k+1}}(h_a)\mid a\in A_k\}\cup\{\pi_{n_{k+1}}(h'_a)\mid a\in A_k\}\subseteq G_{n_{k+1}}$. Let $A=\lim_{k\in\mathbb{N}} A_{n_k}$. Then the cardinality of $A$ is the continuum and we are done once we prove the following.

Claim: For every $x\in A$ we have $g\notin\langle H,x\rangle$.

Suppose that, to the contrary, $g\in\langle H,x\rangle$ for some $x\in A$. Then we have elements $h_1,h_2,\ldots,h_s\in H$ and a word $w$ such that $w(x,h_1,h_2,\ldots,h_s)=g$. Let $r$ be the length of this word. There exists $k>r$ such that $h_1,h_2,\ldots,h_s\in H_{m_k}$. Then by construction of $A$ we have
$$\pi_{n_{k+1}}(x)\in\pi_{n_{k+1}}(A)=A_{k+1}\subseteq\pi_{n_{k+1}}(H_{m_k})$$
hence
$$\pi_{n_{k+1}}(g)\notin\pi_{n_{k+1}}(\bar H^{k}_{m_k})\ni\pi_{n_{k+1}}(w(x,h_1,h_2,\ldots,h_s))$$
which is a contradiction.<|endoftext|>
TITLE: Special case of Erdos Distance Problem in a plane
QUESTION [8 upvotes]: Erdos in his Distinct distance Problem in a plane conjectured that the minimal number of distinct distance determined by $n$ points in a plane be $g(n)$,
$$g(n) \sim \frac{cn}{\sqrt{\log n}}$$
But for the special case which asks the minimum number of distinct distances that must be determined by $n$ points in a plane such that no $3$ are collinear be $g_{\textrm{no}3}(n)$, it is known that,
$$g_{\textrm{no}3}(n) \ge \left\lceil \frac{n-1}{3} \right\rceil$$
and $\displaystyle g_{\textrm{gen}}(n) \ge \left\lceil \frac{n-1}{3} \right\rceil$, where, $g_{\textrm{gen}}(n)$ asks the same question but for points in general position in a plane (i.e., no $3$ collinear and no $4$ conclyclic.)
Has there been any improvements on these particular casees of Erdos's Distance Problem?

REPLY [2 votes]: There has been recent progress on this problem published in the annals of mathematics:
On the Erdős distinct distances problem in the plane
Pages 155-190 from Volume 181 (2015), Issue 1 by Larry Guth, Nets Hawk Katz
"In this paper, we prove that a set of N points in $\mathbb{R}^2$ has at least $c\frac{N}{\log N}$ distinct distances, thus obtaining the sharp exponent in a problem of Erdős."<|endoftext|>
TITLE: Measurement of "symmetry" of a convex body
QUESTION [6 upvotes]: I often hear that the regular simplex is "the least" symmetric convex body, and I've heard that there are some measures of symmetry of a body, that the simplex minimizes.
Could you please explain or refer me to what methods / measurements there are that measure a convex body's symmetry?
I'll give some context - I have some function on convex bodies and I want to run a computer simulation to find which polytopes with volume 1 and k vertices minimize this functional.
It is conjectured that of all bodies of volume 1 a ball will minimize, and I want to gather evidence for or against this conjecture.
I'll have to measure symmetry and/or maybe, how close to ellipsoid-like shape a convex polytope is.
Thanks

REPLY [4 votes]: The type of symmetry for which the simplex (not necessarily regular) is usually called "the least symmetric convex body" is the symmetry of reflection about a point (e.g. $x\mapsto-x$). There are a few measures of this symmetry that the simplex minimizes. For any convex body $K\subset\mathbf{R}^n$, consider the following symmetric bodies:
$$ D = \tfrac12 (K-K) $$
$$ C = \mathrm{conv}\{K\times \{0\}, -K\times\{1\}\}\subset\mathbf{R}^{n+1} $$
$$ R_a = \mathrm{conv}\{K, 2a-K\} $$
and let $R$ be the body of smallest volume among $\{R_a\}$.
Note that if $K$ is symmetric under reflection about a point, then $|D|=|C|=|R|=|K|$. When $K$ is not symmetric, $|D|,|C|,|R|>|K|$. And, for a fixed volume of $|K|$, the simplex maximizes $|D|$, $|C|$, and $|R|$.
See Convex Bodies Associated with a Given Convex Body, C. A. Rogers and G. C. Shephard (1957).
For your context it sounds like you want a measure of spherical symmetry. And since you want ellipsoids to be as "symmetric" as spheres, you probably want something affine invariant. I think you are looking for something like the Banach-Mazur distance to the sphere:
$$\delta(K,B) = \min\{t: B\subseteq TK+a\subseteq e^t B \text{ for some }T\in GL(n), a\in \mathbf{R}^n\}$$
The BM distance is usually defined for point-symmetric bodies, but there are versions out there without that assumptions (e.g. Macbeath AM (1951), "A compactness theorem for affine equivalence-classes of convex regions", Journal canadien de mathématiques. 3, 54–61).<|endoftext|>
TITLE: Does the Euler product for $L(s,\chi_4)$ also converge in the right half of the critical strip?
QUESTION [5 upvotes]: This question expands on this one from MSE.
In the literature about Dirichlet $L$-series, I found that their Euler products:
$$L(s, \chi) =\prod_p \bigg(\frac {1}{1-\frac{\chi(p)}{p^s}} \bigg)$$
are typically considered to be only converging for $\Re(s)>1$.
However, there seems to be an exception to this rule since Euler proved that:
$$L(1, \chi_4) =\prod_p \bigg(\frac {p}{p-\chi_4(p)} \bigg)=\prod_p \bigg(\frac {p}{p-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)=\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot\frac{13}{12}\dots=\beta(1)=\frac{\pi}{4}$$
does converge (albeit slowly).
I then decided to explore values for $\Re(s) \lt 1$ and numerical evidence suggests that the Euler-product:
$$\prod_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)$$
also (slowly) converges in the domain $\frac12 < \Re(s) \le 1$.
Questions:
1) Is the Euler product for $L(s,\chi_4)$ the only one known to converge for $s=1$?
2) Does this particular Euler product indeed converge in the right half of the strip?
Thanks.

REPLY [8 votes]: By taking the logarithm (or log derivative) of $L(s)$, you get a Dirichlet series whose convergence relates directly to the zeros. Your question about convergence in the right half of the critical strip is equivalent to the Riemann hypothesis for $\chi$. Similarly, (conditional) convergence of the Euler product (again as the "Euler sum" after taking the logs) on the 1-line is equivalent to the lack of zeros on this line.<|endoftext|>
TITLE: Book on the tetrahedron
QUESTION [11 upvotes]: Does anybody know of a book containing "all you want to know about the tetrahedron"? What you want to know should include basic geometry of the tetrahedron, study of orthocentric tetrahedra, the Monge point, various volume / edge length / face area formulae, volume via the Cayley-Menger determinant, the regular tetrahedron, etc.
The wikipedia page for "tetrahedron" is fairly interesting, but it will never treat things with the same level of detail as a textbook.

REPLY [8 votes]: Asking Math Reviews for books with "tetrahedr*" in the title turned up a couple of possibilities. 
Anđelko Marić, Tetrahedron. Definitions, theorems, formulas, problems. Translated by Juraj Šiftar. Publishing House ELEMENT, Zagreb, 2010. 176 pp. ISBN: 978-953-197-580-3, MR2963754. 
Kesiraju Satyanarayana, Angles and in- and ex-elements of triangles and tetrahedra. Studies in coordinate geometry with introductory results on determinants, linear equations, change of reference simplex, tetrahedra, etc. Bangalore Press, Bangalore City 1962 xiii+135 pp, MR0157268 (28 #504).
Also, this, but with no further information given: 
Dov Jarden, The tetrahedron: A collection of papers. Published by the author, Jerusalem 1963 41 pp, MR0149341 (26 #6831).<|endoftext|>
TITLE: Higher refinement of Seifert-van Kampen theorem on the language of hocolim
QUESTION [9 upvotes]: I like the following version of SvKT. If $\Pi_1$ is the functor of fundamental groupoid and $(X_i)_{i\in I}$ is a diagram of spaces then
$$\Pi_1({\sf hocolim}\: X_i)\simeq {\sf hocolim}\: \Pi_1(X_i).$$
Question: Is there a similar statement for higher homotopy? For example, if we replace $\Pi_1$ by some version of the infinity-groupoid $\Pi_\infty$. But it should be tricky because the homotopy pushout of the diagram $* \leftarrow S^1 \to *$ is $S^2$ whose homotopy groups are complicated.   
I know about Brown's version of this theorem but filtered spaces is not a convenient setting for me.

REPLY [2 votes]: This answer relates mainly to the term "Higher refinement of Seifert-van Kampen Theorem", rather than homotopy colimit. 
Last week I gave a talk to the 2015 Category Theory Meeting in Aveiro, entitled "A philosophy of modelling and computing homotopy types".  The Abstract and full and handout versions (slightly refined) are available on my preprint page. Homotopy colimits enter from the side in order to get a "nice" pushout of spaces, but the aim is 
explicit nonabelian pushout computations of certain **homotopy types. 
It is often hard work to compute from this homotopy groups and $k$-invariants! (There was also no time for indications in the talk.) 
I do not know of an account of a homotopy theory, and so of  "homotopy colimits",  for  the crossed squares used in this presentation for modelling, and some computations of, homotopy $3$-types. 
Later: I'll add that in 1965 I found the nice statement and proof for the fundamental groupoid $\pi_1(X)$ of a space which was a union of open sets. I thought: great! One can get rid of base points! 
Then I found this was not enough for my first aim,  getting as a Corollary the fundamental group of the circle. So then I discovered the applications of the fundamental groupoid  $\pi_1(X,A)$ on a set $A$ of base points, chosen according to the geometry of the given situation. In particular, the determination of the fundamental group of the circle required at least $2$ base points. These results were published in 1967. 
This simple extension of the theorem and proof of the usual van Kampen theorem is, to my knowledge, given in topology texts in English only in the 1968, 1988, 2006 editions of what is now titled "Topology and Groupoids". 
As for higher versions of the nonabelian fundamental group, these were sought by the topologists of the early 20th century topologists (Dehn, Cech, Hopf, ...) but I think did require for the proof a shift from groups to groupoids, and a 2-dim Seifert-van Kampen Theorem was published by Philip Higgins and I in 1978. That was a theorem on second relative homotopy groups. 
June 24: The updated Abstract of the Aveiro presentation is as follows, and I hope shows some special features of this approach to Higher Seifert-van Kampen Theorems: 
ABSTRACT
This philosophy involves  functors $\mathbb  H$  from (Topological Data) to (Algebraic Data), and conversely "classifying space" functors $ \mathbb B$ from (Algebraic Data) to (Topological Data). These should satisfy: 

$\mathbb H$ is homotopically defined. 
$\mathbb{ HB}$ is naturally equivalent to 1. 
The Topological Data has a notion of connected. 
For all Algebraic Data $A$, we have $\mathbb BA$ is connected. 
$\mathbb H$ preserves certain colimits of connected Topological Data. 

The Algebraic Data splits into several equivalent kinds, ranging from "broad" to "narrow", related by Dold-Kan type equivalences. The broad data is used for conjecturing and proving theorems; the narrow data is used for calculations and relating to classical methods. 
As examples of Algebraic Data we give groupoids, crossed modules and crossed squares. We give a sample computation, using crossed squares, of the homotopy 3-type of the mapping cone of the classifying space of a morphism of crossed modules.<|endoftext|>
TITLE: existence of totally geodesic hypersurfaces
QUESTION [5 upvotes]: Assume we are on a smooth, complete Riemannian manifold $(M,g), dim(M) \geq 3$.  What are the specific geometric/topological constraints for such a manifold to admit complete, totally geodesic hypersurfaces?  Please, I admit that I'm a rookie so any simple illustrations and sources are more than welcome, especially for the case of lowest dimension $(dim(M) = 3)$.

REPLY [21 votes]: You should be aware that, for $n\ge3$, the generic Riemannian metric $(M^n,g)$ has no totally geodesic hypersurfaces at all, even locally.  Typically, when they do exist, it is for some geometric reason that makes them obvious.  For example, if $(M^n,g)$ admits an isometry $\iota: M\to M$ of order $2$ (i.e., an involution) whose fixed point set is a hypersurface $H^{n-1}\subset M$, then $H$ will be totally geodesic (and it will be complete if $g$ is complete).
As for topological constraints, there really aren't any, because, for any closed, embedded hypersurface $H^{n-1}\subset M^n$, you can always construct a metric $g$ such that $H$ is totally geodesic in $(M,g)$; just take the metric to be a product metric on some tubular neighborhood of $H$.
As a practical matter, if someone hands you a Riemannian manifold $(M^n,g)$ that is described sufficiently explicitly that you can compute the various curvature tensors and functions to arbitrary orders and determine their vanishing loci, there is an algorithm to follow that will determine all of the totally geodesic hypersurfaces in $(M^n,g)$, and that is probably the best you can do in terms of giving explicit 'constraints' on $(M^n,g)$ for it to admit totally geodesic hypersurfaces (complete or not).  The algorithm basically, goes like this:  Let $\pi: \Sigma^{2n-1}\to M^n$ be the unit sphere bundle of $M$ with respect to the metric $g$.  Then, using the Levi-Civita connection, construct an $(n{-}1)$-plane field $D\subset T\Sigma$ such that a curve $\nu:\mathbb{R}\to \Sigma$ of the form $\nu(t) = \bigl (x(t), e(t)\bigr)$, where $e(t)\in T_{x(t)}M$ is a unit vector, is tangent to $D$ if and only if $x'(t)\cdot e(t)=0$ and $e(t)$ is parallel along $x(t)$, i.e., $\nabla_{x'(t)}e(t) = 0$.  Then a hypersurface $H\subset M$ is totally geodesic if and only if its 'unit normal graph' in $\Sigma$ is everywhere tangent to $D$.  (It is easy to show that this plane field $D$ is Frobenius if and only if $(M^n,g)$ has constant sectional curvature, which explains why only those manifolds have totally geodesic hypersurfaces through every point perpendicular to every direction.) Now the problem is reduced to finding the $(n{-}1)$-manifolds in $\Sigma$ that are tangent to $D$, which is a standard problem.
It is probably worth summarizing what this process says about Riemannian $3$-manifolds $(M^3,g)$.  When one goes through the above analysis, the first thing it tells one is that, if $H^2\subset M$ is a totally geodesic hypersurface with normal vector field $\nu:H\to \Sigma$, then, for all $x\in H^2$, the normal vector $\nu(x)\in T_xM$ must be an eigenvector of the Ricci curvature $\mathrm{Ric}(g)$.   In particular, if $\mathrm{Ric}(g)$ has distinct eigenvalues everywhere (or, what would be generic, on a dense open set $U\subset M$), then one can locally write
$$
g = {\omega_1}^2+{\omega_2}^2+{\omega_3}^2
\quad\text{and}\quad
\mathrm{Ric}(g) = \lambda_1\,{\omega_1}^2+ \lambda_2\,{\omega_2}^2+ \lambda_3\,{\omega_3}^2
$$
for some functions $\lambda_1 < \lambda_2 < \lambda_3$ and some coframing $\omega_i$, 
and one of the $\omega_i$ will have to vanish on $H$.  However, if $\omega_i$ vanishes on $H$, then $\mathrm{d}\omega_i$ must also vanish on $H$.  Defining the functions $f_{ij}$ that satisfy $\omega_i\wedge\mathrm{d}\omega_j = f_{ij}\,\omega_1\wedge\omega_2\wedge\omega_3$, one sees that $f_{ii}$ must also vanish on $H$.  
In the generic situation, each of the three equations $f_{ii} = 0$ will define a surface $H_i$ in $M$, and these $H_i$ are the only possible totally geodesic surfaces in $M$.  In fact, $H_i$ will be totally geodesic if and only if $\omega_i$ vanishes on $H_i$ along with the three functions $f_{jk}$, $f_{kj}$ and $f_{jj}{-}f_{kk}$, where $(i,j,k)$ is a permutation of $(1,2,3)$.  Thus, this provides the effective test in the case that the Ricci tensor has distinct eigenvalues.  
In the degenerate case of multiple eigenvalues (or, say, that $f_{ii}=0$ does not define a surface in $M$), more differentiation is required to formulate an explicit test of this nature, but I'll leave that to you.<|endoftext|>
TITLE: A weakening of the Littlewood conjecture
QUESTION [10 upvotes]: For real numbers $x$, let $\|x\|$ denote the distance from $x$ to the nearest integer. Define a function $\ell:\mathbb{R}^2\rightarrow\mathbb{R}$ by
$$\ell(\alpha,\beta)=\liminf_{n\rightarrow\infty}n\|n\alpha\|\|n\beta\|.$$
The Littlewood conjecture asserts that, for all $(\alpha,\beta)\in\mathbb{R}^2$, we have $\ell (\alpha,\beta)=0$.
Can anyone see how to prove the (seemingly much weaker) statement that, for all $(\alpha,\beta)\in\mathbb{R}$,
$$\inf_{A\in\mathrm{SL}_2(\mathbb{Z})}\ell\left(A\left(
\begin{array}{c}
\alpha\\
\beta\\
\end{array}
\right)\right)=0\quad ?$$
I wouldn't be surprised if this problem has an easy solution, but I haven't yet been able to find one. For the purposes of orientation note that, in the definition of $\ell$ if we replace $\liminf$ by $\inf$, then the problem becomes close to trivial.

REPLY [6 votes]: I believe so.  Let $K$ be a large positive integer, and $N$ an integer parameter going to infinity. By Dirichlet approximation we can find $n \in [N,2N]$ such that $\|n\alpha\|, \|n\beta\| = O(N^{-1/2})$; we can in fact assume that $\|n\alpha\|, \|n \beta\| \asymp N^{-1/2}$ since otherwise we are done.  For sake of notation let's suppose that the signed fractional parts $\{n\alpha\}, \{n\beta\} \in (-1/2,1/2]$ are positive (and thus comparable to $N^{-1/2}$).  Let $p_j/q_j$ be the last continued fraction approximant to $\{n\alpha\}/\{n\beta\}$ (a quantity comparable to 1) with $q_j \leq K$.  (Note we may assume this ratio is irrational, since Littlewood's conjecture is easy in the commensurate case.) Standard continued fraction theory tells us that
$$ p_{j-1} q_j - p_j q_{j-1} = \pm 1$$
and
$$ p_j - q_j \{n\alpha\}/\{n\beta\} = O(1/q_{j+1})$$
$$ p_{j-1} - q_{j-1} \{n\alpha\}/\{n\beta\} = O(1/q_{j})$$
and so $\| n (p_j \beta - q_j \alpha) \| = O( N^{-1/2}/q_{j+1} ) = O( N^{-1/2}/K)$ and $\| n (p_{j-1} \beta - q_{j-1} \alpha) \| = O( N^{-1/2}/q_j ) = O(N^{-1/2})$.  After pigeonholing we can find (for fixed $K$) an infinite sequence of $N$ such that the $p_j,q_j,p_{j-1},q_{j-1}$ are constant, and thus $\ell( p_j \beta - q_j \alpha, p_{j-1} \beta - q_{j-1} \alpha ) = O(1/K)$.  Letting $K$ to infinity we obtain the claim.<|endoftext|>
TITLE: Is there a universal $\omega$-limit set?
QUESTION [9 upvotes]: For the purposes of this question, a dynamical system means a compact metric space $X$ together with a continuous map $f: X \to X$.
For $x \in X$, the $\omega$-limit set of $x$, denoted $\omega(x)$, is the set of limit points of the orbit of $x$. That is,
$$\omega(x) = \bigcap_{n \in \omega} \overline{\{f^m(x) : m > n\}}.$$
It's easy to check that $(\omega(x),f)$ is a dynamical system (i.e., $\omega(x)$ is closed under $f$ and closed topologically). Let's say that $(X,f)$ is an abstract $\omega$-limit set if it is isomorphic to the $\omega$-limit set of some point in some dynamical system.

Is there a dynamical system $(X,f)$ such that every abstract $\omega$-limit set is a quotient of $(X,f)$?

[In this context, an isomorphism $(X,f) \rightarrow (Y,g)$ means a homeomorphism $h: X \rightarrow Y$ such that $h \circ f = g \circ h$. A quotient $(X,f) \rightarrow (Y,g)$ means a continuous surjection $h: X \rightarrow Y$ such that $h \circ f = g \circ h$.]
I suspect the answer might be negative, but I can't think of a way to get at a proof.

Motivation:
If we replace "compact metric" with "compact Hausdorff" (to broaden our notion of dynamical systems a bit), then there is a universal abstract $\omega$-limit set, namely $\mathbb{N}^* = \beta \mathbb{N} \setminus \mathbb{N}$ together with the shift map (the "shift map" being the map sending an ultrafilter $\mathcal F$ to the ultrafilter generated by $\{A+1 : A \in \mathcal F\}$). It seems to me that some smaller system should suffice to "capture" just the metric systems. Ideally, one would like a metric system to do this -- hence the question!
Potentially helpful information: The metrizable abstract $\omega$-limit sets have a very nice topological characterization due to Bowen. Namely, a metrizable system $(X,f)$ is an abstract $\omega$-limit set if and only if it is chain transitive. [$(X,f)$ is chain transitive if for some (equivalently, any) metric $d$ on $X$, for any two points $x,y \in X$, and for any $\varepsilon > 0$, there is a sequence $x = x_0, x_1, x_2, \dots, x_n = y$ such that $d(f(x_i),x_{i+1}) < \varepsilon$ for all $i < n$. The idea here is that this sequence is a "pseudo-orbit" (with error $\varepsilon$) from $x$ to $y$.]
Also possibly relevant is the well-known topological fact that every compact metric space is a continuous image of the Cantor space. So if $(X,f)$ is a system providing a positive answer to my question, I suspect $X$ should be the Cantor space.

REPLY [4 votes]: The answer is no, there is no universal $\omega$-limit set.
The same is true for the classes of metric minimal dynamical systems and metric dynamical systems in general.  I have written up proofs of these facts:
http://www.math.uni-hamburg.de/home/geschke/papers/NoUniversalMetricOmegaLimitSet.pdf
The basic idea of the proof is as follows:  as Will Brian already conjectured, it is enough to consider dynamical systems with 0-dimensional phase spaces.  Now, if $(X,f)$ is a dynamical system with a 0-dimensional metric phase space $X$ and there is a morphism from $(X,f)$ onto 
a dynamical system $(Y,g)$ with $Y$ 0-dimensional, then the Boolean algebra of clopen subsets of $Y$ embeds into the algebra of clopen subsets of $X$ by an embedding that respects the dynamical systems.  Since $X$ is metric, its Boolean algebra of clopen subsets is countable.
Using Sturmian subshifts we can cook up enough metric minimal dynamical systems $(Y,g)$ (that are automatically abstract $\omega$-limit sets) so that not all the corresponding Boolean algebras (with actions on them) embed into a single countable Boolean algebra (with action),
showing that there are no metric universal dynamical systems.
The most general result is this:  There is no metric dynamical system that maps onto all Sturmian subshifts.  The Sturmian subshifts are minimal and hence abstract $\omega$-limit sets.  Here it doesn't make a difference whether we consider $\mathbb Z$-actions, i.e., compact spaces with a homeomorphism, or $\mathbb N$-actions, i.e., compact spaces with a continuous map as in the original question.<|endoftext|>
TITLE: Left orthogonals to compact objects in triangulated categories: existence and "control"?
QUESTION [6 upvotes]: Let $C$ be a compactly generated triangulated category. Can it contain a non-zero object $M$ such that there are no non-zero morphisms FROM $M$ into compact objects? I would be grateful for any example; can one obtain a certain "description" for these "left phantom" objects? For example, what happens in the unbounded derived category of a ring of infinite cohomological dimension?
Note that the subcategory of objects satisfying this conditions is triangulated and closed with respect to coproducts. I am willing to localize by this category (if it is non-zero; I would actually prefer to impose certain extra "orthogonality" conditions on the subcategory I kill). Yet I do not know whether the hom-classes in the quotient are sets. The localization functor should respect coproducts; yet it does not seems to respect the compactness of objects.

REPLY [7 votes]: This happens in spectra.  By a theorem of Lin, there are no maps from the Eilenberg-MacLane spectra $H\mathbb{F}_p$ to finite spectra; the proof is an Adams spectral sequence computation using the fact that the Steenrod algebra is self-injective.
For a more algebraic example, let $A$ be ring containing an infinite regular sequence $(x_0,x_1,\dots)$ and let $M=A/(x_0,x_1,\dots)$.  We can resolve $M$ by an infinite Koszul complex and compute that $\operatorname{Ext}^*(M,A)=0$.  It follows that in the derived category of $A$, there are no maps from $M$ to compact objects.
As for getting some kind of control on these objects, I don't really know much, but I know Luke Wolcott has thought a lot about pathology in derived categories of non-Noetherian rings.  You might try taking a look at his work and seeing if you can find anything useful.<|endoftext|>
TITLE: Does this extension of Hodge structures split over $\mathbb{Q}$?
QUESTION [5 upvotes]: Let $X$ be a smooth projective curve of genus $\geq 1$ over $\mathbb{C}$, $H^\cdot=H^\cdot(X)$, and $K$ be the kernel of cup product $\cup: H^1\otimes H^1\rightarrow H^2$. Consider the extension of Hodge structures
\begin{equation}
0\longrightarrow K\longrightarrow H^1\otimes H^1\stackrel{\cup}{\longrightarrow} H^2\cong\mathbb{Z}(-1)\longrightarrow 0.
\end{equation}
This of course splits over $\mathbb{R}$. Does it split over $\mathbb{Q}$ as well? Equivalently, is there a Hodge class $\xi$ in $H^1\otimes H^1$ such that 
\begin{equation}
\int\limits_X\cup(\xi)\neq 0 ~?
\end{equation}
I tried taking $\xi$ to be the $H^1\otimes H^1$ Kunneth component of the class of the diagonal embedding $\Delta(X)$ of $X$ in $X^2$, but I don't seem to be able to show
$$\int\limits_{\Delta(X)}\xi\neq 0.$$

REPLY [5 votes]: Venkataramana gave an Hodge-theory argument. There is also a proof by intersection theory that your class $\xi$ gives a splitting.
Observe that the other two Kunneth components of the diagonal are a horizontal and vertical fiber. So $\xi$ is the diagonal minus a horizontal fiber and minus a vertical fiber. Now you're integrating that over the diagonal, which is the same as intersecting with the diagonal. $$\Delta(X) \cdot \Delta(X) =2-2g$$ The intersection of the diagonal with a horizontal or vertical line is $1$, from $1$ transverse point. So all in all $$\Delta(X) \cdot \xi = -2g$$$g \geq 1$, so that is nonzero.<|endoftext|>
TITLE: Which topological properties are preserved under taking box products?
QUESTION [8 upvotes]: Although the box topology is a topology worth studying and is similar to the strong topology in differential topology, the box topology is in many regards very badly behaved since the box product of even nice spaces has many undesirable properties. For instance, unlike ordinary products, non-trivial box products are never connected, metrizable, nor locally compact. I want to know what properties of topological spaces are preserved under taking box products. Let me list the more obvious properties.
$\textbf{Preserved under box products:}$
Separation axioms such as $T_{0},T_{1}$, Hausdorffness, regularity, complete regularity; zero-dimensionality, total disconnectedness(otherwise known as hereditary disconnectedness), total separation (two distinct points can always be separated by a clopen set), other disconnectedness properties, negative properties in general, $P$-spaces, discreteness, complete uniformizability.
$\textbf{Not preserved under box products:}$ Normality, paracompactness, ultraparacompactness, compactness, other compactness properties, connectedness, path connectedness, other connectedness properties, . . .
See chapter 4 in the Handbook of set-theoretic topology for more information on box products and a few other properties preserved under box products that I neglected to mention in this question.
For which other topological properties $P$ does it hold that if $X_{i}$ has property $P$ for all $i\in I$, then $\prod_{i\in I}^{\textrm{Box}}X_{i}$ also has property $P$? I am looking for specific examples of such properties since I do not believe that there is a nice general characterization of all such properties. I am looking for positive properties rather than the negation of certain well known properties.

REPLY [4 votes]: Some local network or base properties are preserved by countable box-products of topological spaces. 
In particular:
1) the existence of a countable $cs$-network at each point;
2) the existence of a countable $cs^*$-network at each point;
3) the existence of a countable $s^*$-network at each point;
4) the existence of an $\omega^\omega$-base at each point.
Now I recall the corresponding definitions.
Let $X$ be a topological space and $x$ be a point of $X$. A family $\mathcal F$ of subsets of $X$ is called a 
$\bullet$ a $cs$-network at $x$ if for any sequence $\{x_n\}_{n\in\omega}\subset X$ convergent to $x$  and any neighborhood $O_x\subset X$ of $x$ there exists a set $F\in\mathcal F$ such that $F\subset O_x$ and $F$ contains all but finitely many elements of the sequence $(x_n)$;
$\bullet$ a $cs^*$-network at $x$ if for any sequence $\{x_n\}_{n\in\omega}\subset X$ convergent to $x$  and any neighborhood $O_x\subset X$ of $x$ there exists a set $F\in\mathcal F$ such that $F\subset O_x$ and $F$ contains infinitely many elements of the sequence $(x_n)$;
$\bullet$ an $s^*$-network at $x$ if for any sequence $\{x_n\}_{n\in\omega}\subset X$ accumulating at $x$  and any neighborhood $O_x\subset X$ of $x$ there exists a set $F\in\mathcal F$ such that $F\subset O_x$ and $F$ contains infinitely many elements of the sequence $(x_n)$;
$\bullet$ an $\omega^\omega$-base at $x$ if each set $F\in\mathcal F$ is a neighborhood of $x$ and $\mathcal F$ can be written as $\mathcal F=\{F_\alpha\}_{\alpha\in\omega^\omega}$ so that $F_\alpha\subset F_\beta$ for any elements $\beta\le\alpha$ of $\omega^\omega$ (endowed with the ccordinatewise partial order).
Those notions are studied in this paper and often appear in Topological Algebra and Functional Analysis. 
For a topological space $X$ and a point $x\in X$ we have the implications:
($X$ has an $\omega^\omega$-base at $x$) $\Rightarrow$ ($X$ has a countable $s^*$-network at $x$) $\Rightarrow$ 
$\Rightarrow$ ($X$ has a countable $cs^*$-network at $x$) $\Leftrightarrow$ ($X$ has a countable $cs$-network at $x$).<|endoftext|>
TITLE: Sums of random variables mod p
QUESTION [6 upvotes]: Let $\varepsilon_1, \ldots, \varepsilon_n$ be independent random variables taking values $0,1$ each with probability $1/2$. It is well known that $R_n=\varepsilon_1+ \cdots+ \varepsilon_n$ modulo a prime $p$ tends to the uniform distribution on $\mathbb{Z}_p$ (say, in total variation distance, but also in a lot of other senses). 
Is it known what the speed of convergance is? Let us say I want to bound $\delta_n=\sup_{k}|\mathbb{P}(\varepsilon_1+ \cdots+ \varepsilon_n=k)-1/p|$ in terms of $n$. Is it true that $\delta_n$ converges to $0$ exponentially? If this is true, could you provide a decent reference?
Many thanks for the attention.

REPLY [10 votes]: Yes, and it's proportional to $\cos ( \pi/p)^n$. To see this, observe that adding one more independent random variable acts on the probability distribution as a linear operator, hence a $p \times p$ matrix. Using Fourier analysis mod $p$, compute the eigenvalues of this matrix. Note that one is $1$, corresponding to the uniform distribution, and the rest are at most $cos (\pi/p)$.
More generally, any Markov chain converges exponentially to a stable distribution.
Not a great reference but here's Wikipedia.<|endoftext|>
TITLE: Which degree does a motivic Galois representation show up in?
QUESTION [13 upvotes]: Consider a representation $\rho: \operatorname{Gal} (\overline{\mathbb Q} | \mathbb Q ) \to GL_n ( \overline{\mathbb Q}_\ell)$ that is a subrepresentation of $H^i(X, \overline{\mathbb Q}_\ell (j))$ for $X$ a smooth projective variety over $\mathbb Q$. This is a Galois representation of weight $w = i-2j$.
When do we expect a smooth projective variety $Y$ over $\mathbb Q$ such that $\rho$ shows up in $H^w(Y, \overline{\mathbb Q}_\ell)$?
I can see two obvious necessary conditions. One is that the coefficients of the characteristic polynomial of every Frobenius element had better be algebraic integers. Another is that the Hodge-Tate weights of $Y$ had better be in the interval $[0,w]$.

Do there exist any (conjectural) relations between the three conditions

Actually shows up in cohomology without Tate twisting.
Integral characteristic polynomial of Frobenius
Correct Hodge-Tate weights

other than 1 implying 2 and 3? Should they all be equivalent? Are there any counterexamples?

There are some partial results toward the claim that 2 and 3 imply 1.
Having an integral characteristic polynomial is a sufficient condition in the weight $0$ case. This would imply the eigenvalues of Frobenius are roots of unity. With compactness, they are all roots of unity in a bounded degree extension of $\mathbb Q_\ell$, hence of bounded order, so by Chebotarev all elements of the Zariski closure of the image of $\rho$ have eigenvalues roots of unity of bounded order, hence the image is finite, so it shows up in the cohomology of a $0$-dimensional scheme. I guess I'm counting non-geometrically connected schemes as varieties here.
Having the correct Hodge-Tate weights is a sufficient condition in the weight 0 and 1 case, at least assuming the Hodge and Tate conjectures. By the Tate conjecture we can associate a motive to the Galois representation, to which we can associate a Hodge structure. In the weight 0 case, this Hodge structure is supported in $H^{0,0}$ so all the classes are Hodge classes and hence are algebraic, hence the image of $\rho$ is finite again. In the weight 1 case, this Hodge structure is supported in $H^{1,0}$ and $H^{0,1}$, hence corresponds to an abelian variety.  $H^1$ of that abelian variety is a motive with the same Hodge structure as the motive of $\rho$. Because we assumed the Hodge conjecture, it must be the same motive and thus have the same Galois representation. A more detailed version of that argument is in this paper.
I guess the Langlands program will also give some cases of this.

REPLY [7 votes]: Do there exist any (conjectural) relations between the three conditions? Well Grothendieck essentially makes this conjecture, see $\S 2$ p.300 of the following paper:
http://boxen.math.washington.edu/home/wstein/www/sga/circle/HodgeConj.pdf
Especially the second paragraph of p. 301. Most people I have spoken to are fairly agnostic about this conjecture, but it at least suggests that finding a counterexample will not be so easy.
For two dimensional representations, it comes down to the question of whether any modular form $f$ has at least one ordinary prime (if not, then $\rho_{f}(-1)$ satisfies $2$). I think every one expects this to be true and nobody has any idea how to prove it (for weights $\ge 4$).  At least $3 \Rightarrow 1,2$ is OK here, and, as you noted, in some [but not all] other contexts where modularity is available)
One can make natural generalizations of the (infinitely many ordinary primes) condition to other motives as well, but since (trusting Grothendieck) we expect the conditions you listed to be equivalent, and since (considering ordinary primes for modular forms) we can't prove it even in the (almost) simplest case, that leaves us pretty much up the creek without a paddle with regard to proving anything.<|endoftext|>
TITLE: Preserving $\omega_1$ is Inaccessible to the reals
QUESTION [10 upvotes]: $\omega_1$ is inaccessible to the reals if and only if for all $x \in {}^\omega\omega$, $\omega_1^{L[x]} < \omega_1$. 
The question is if $\omega_1$ is inaccessible to the reals in $V$ and $\mathbb{P}$ is an $\aleph_1$-preserving forcing, in the generic $\mathbb{P}$-extension $V[G]$, is $\omega_1$ still inaccessible to the reals?
Being inaccessible to the reals is a $\mathbf{\Pi}_4^1$ statement. Under certain assumptions, any forcing will preserve $\omega_1$ is inaccessible to the reals.
Therefore, a more specific question would be if $\kappa$ is an inaccessible cardinal in $L$ and $G$ is generic for $\text{Coll}(\omega, <\kappa)$, is $\omega_1$ being inaccessible to the reals preserved in all $\aleph_1$-preserving forcings extension of $L[G]$?
In $L[G]$, the $\text{Coll}(\omega, \omega_1^{L[G]})$ extension of $L[G]$ is equal to a $\text{Coll}(\omega, \kappa)$ extension of $L$. So $L[G]$ is an example of a model where arbitrary forcings do not preserve $\omega_1$ being inaccessible to the reals.
Thanks for any information.

REPLY [2 votes]: If κ is the least inaccessible of L and g is Col(ω<κ) generic over L, then a.d. coding over L[g] is ccc and the extension is of the form L[x], x a real. A variant of the question is: which ω1 preserving forcings preserve e.g. analytic determinacy provably in ZFC? Consistently, again, there is a ccc counterexample (R. David), but Sacks forcing and others provably preserve analyt. determinacy (Castiblanco and Schlicht).<|endoftext|>
TITLE: Is "quotient" of projective variety projective?
QUESTION [7 upvotes]: Suppose $X$ is a projective variety, $f\colon X\to Y$ is a finite surjective morphism onto variety $Y$, must $Y$ be a projective variety?

REPLY [13 votes]: No, that is not true.  Let $X$ be $\mathbb{P}^3_k$.  Let $g:L\hookrightarrow X$ be a line in $X$. Let $h:C\hookrightarrow X$ be a plane conic in $X$ that is disjoint from $L$ and that contains a $k$-point.  Let $i:L\to C$ be an isomorphism of $k$-schemes.  Let $f:X\to Y$ be the coproduct of the two morphisms $g$ and $h\circ i$.  Then $Y$ is a proper $k$-variety, and $f$ is finite and surjective.
If $\mathcal{L}$ were an ample invertible sheaf on $Y$, then the pullback $f^*\mathcal{L}$ would be an ample invertible sheaf on $X$ whose degree on $L$ equals the degree on $C$.  Every invertible sheaf on $\mathbb{P}^3$ is of the form $\mathcal{O}(d)$ for some $d\in \mathbb{Z}$.  Only for $d=0$ is the degree on $L$ equals to the degree on $C$.  For $d=0$, this invertible sheaf is not ample.  Thus $Y$ is not projective.<|endoftext|>
TITLE: Does there exist a nonconstant, periodic, real analytic function with period 1 and rational Maclaurin coefficients?
QUESTION [8 upvotes]: Does there exist a nonconstant, real analytic function $f \colon \mathbb{R} \to \mathbb{R}$ such that $f$ is periodic with period 1 and whose Maclaurin coefficients are all rational?
(The function $\sin x$ satisfies all the conditions except that its period isn't 1.)
I can't recall exactly what I was thinking about when this question occurred to me and my primary motivation now is curiosity. Note that if the answer is no, a proof that doesn't use the irrationality of $\pi$ would be an alternative way to prove irrationality of $\pi$ because of the function $\sin(2\pi x)$.
Also, WLOG, we may assume $f(0)=0$ and if we define the vector $c = (c_k)$ where $f(x) = \sum\limits_{k = 1}^{\infty} \frac{c_k}{k!}x^k$ and let $v = (1, 1/2!, 1/3!, 1/4!, \dots)$, then by comparing the derivatives of $f$ at 0 and 1, we get that for all nonnegative integers $n$,
$$v \cdot C^n c = 0,$$
where $C$ shifts the components of a vector one space to the left.

REPLY [12 votes]: Yes, such $f$ exist, and can even be taken to be entire,
and with a preassigned finite initial segment $c_0,c_1,\ldots,c_d$
of the sequence of $x^k/k!$ coefficients.
Indeed if $f(x) = \sum_{k=0}^\infty a_k \sin^k (2\pi x)$
then $f$ is entire provided $a_k \rightarrow 0$ quickly enough,
say $a_k \ll 1/k!$; and the map from $\{a_k\}$ to $\{c_k\}$
is linear and invertible, with $a_k$ a linear combination of
$c_0,\ldots,c_k$ and $c_k$ a linear combination of $a_0,\ldots,a_k$
for each $k$.  So, solve for $a_0,\ldots,a_d$, and then inductively
choose small $a_{d+1}, a_{d+2}, a_{d+3}, \ldots$ to make each of
$c_{d+1}, c_{d+2}, c_{d+3}, \ldots$ rational.  (One way to
avoid invoking the Axiom of Choice here is to fix some
enumeration of ${\bf Q}$ and then at each step with $k>d$
use the least-indexed $c_k$ that makes $|a_k| < 1/k!$.)
[I see now that David Speyer gave much the same solution
in a comment (using $g_k$ for what I called $a_k$),
though without bothering to make $f$ entire
or preassign the start of its Taylor expansion.]<|endoftext|>
TITLE: Unicity of additive, $(-1)$-homogeneous, and shift invariant probability measures on $\mathbf N^+$
QUESTION [5 upvotes]: Let $\mathcal D$ be the set of all (finitely) additive probability measures $\mu^\ast: \mathcal P(\mathbf N^+) \to [0,\infty[$ such that $\mu^\ast(k \cdot X + h) = \frac{1}{k} \mu^\ast(X)$ for all $X \subseteq \mathbf N^+$ and $h,k \in \mathbf N^+$. It is proved in [1, Section 10] that $\mathcal D$ is nonempty, provided that we lay out the foundations of our mathematical believes, say, in ZFC (see here for some discussion on this point). However, [1] doesn't imply anything, insofar as I can tell, about the size of $\mathcal D$ (again, in ZFC). So my question is:

Do you know of any place in the literature where it's proved, say, that $\mathcal D$ has more than one element? And what about $\mathcal D$ being infinite?

I've also read through [2], but it's not clear to me whether the answer is there or not.
Bibliography.
[1] R. P. Agnew and A. P. Morse, Extensions of linear functionals, with applications to limits, integrals,
measures, and densities, Ann. of Math. (2) 39 (1938), No. 1, 20-30.
[2] E. K. van Douwen, Finitely additive measures on $\mathbb N$, Topology Appl. 47 (1992), No. 3, 223-268.

REPLY [4 votes]: This is basically Second construction 5.3 from Eric K. van Douwen. Finitely additive measures on $\mathbb N$. Topology Appl., 47 (3), (1992), 223–268. MR1192311 (94c:28004). This issue of Topology and Its Applications was a special issue dedicated to Eric K. van Douwen. 

EDIT: Only after making the construction I posted below I returned to the paper and noticed that it contains Appendix 5C: The number of elastic measures. It is shown here that the set of elastic measures has cardinality $2^{\mathfrak c}$. Every elastic measure has properties required in the above question. 
So this gives a better result than I obtained below. I have decided to keep the original version of my answer here. Just in case some of the information given there might be interesting for some people reading this question.


$\newcommand{\Flim}{\operatorname{\mathcal F-\lim}}\newcommand{\FF}{\mathcal F}$
The author uses diffuse mean in this construction. I will use $\FF$-limit, which is also a diffuse mean. (See that paper for terminology.) I will summarize which properties of limit along an ultrafilter I need. (Of course, you can use some other functional fulfilling these properties instead of $x\mapsto \Flim x_n$. I believe that such functionals can be obtained, for example, using Hahn-Banach theorem.)

For any filter $\FF$ we can define the $\FF$-limit of a sequence $(x_n)$ as
$$\Flim x_n=L \qquad\Leftrightarrow\qquad (\forall \varepsilon>0) \{n; |x_n-L|<\varepsilon\}\in\FF.$$
If $\FF$ is a free ultrafilter, then $\Flim x_n$ exists for each bounded sequence. Moreover, we know that for a free ultrafilter $\FF$ we have

If $(x_n)$ is convergent, then $\Flim x_n=\lim x_n$ (=extends limit);
$\Flim (x_n+y_n) = \Flim x_n + \Flim y_n$ and $\Flim (cx_n)=c\Flim x_n$ (=linearity);
$x_n\le y_n$ $\Rightarrow$ $\Flim x_n\le\Flim y_n$ (=is positive);
For a given sequence $(x_n)$ and any cluster point $c$ of $(x_n)$ there exists a free ultrafilter such that $\Flim x_n=c$.

Some references for $\FF$-limits can be found here or in the comments to this post. (There is also a Wikipedia article on ultralimit. However, this article discusses ultralimits of sequences only very briefly.)

For $A\subseteq\mathbb N$ we define
$$\mu_n(A) = \frac{\sum_{j\le n}\frac1j \chi_A(j)}{\ln n}.$$
Notice that $\overline\delta(A)=\limsup\limits_{n\to\infty} \mu_n(A)$ is precisely the upper logarithmic density. For limit inferior we get the lower logarithmic density $\underline\delta(A)=\liminf\limits_{n\to\infty} \mu_n(A)$.
Let $\FF$ be a free ultrafilter. We define
$$\mu(A)=\Flim \mu_n(A).$$
(Maybe $\mu_F$ would be a better notation, since it depends on the choice of $\FF$, but I will use $\mu$ for brevity.)
It is relatively easy to see that $\mu(A)$ is a finitely additive measure on $\mathbb N$.
We will show below that $\mu(kA+h)=\frac1k\mu(A)$.
Now if we fix some set $A$ such that $\underline\delta(A)=0$ and $\overline\delta(A)=1$, then every point in the interval $[0,1]$
is a cluster point of the sequence $(\mu_n(A))$. (Note that the set of all cluster points of this sequence is connected.)
So there are at least $\mathfrak c$ such measures, since $\mu(A)$ can attain all values between $0$ and $1$ (for various choices of the free ultrafilter $\FF$).

So it remains to show that the measures of the form described above fulfill the condition
$$\mu(kA+h)=\frac1k\mu(A).$$
In fact, this is shown van Douwen's paper. But since I changed notation a little bit I will include sketch of the proof.
$\mu$ is shift-invariant, i.e., $\mu(A+1)=\mu(A)$.
Let $B=A+1$. We have
$$|\mu_n(A)-\mu_n(B)| \le \frac{\frac1n+\sum_{k<n} \left(\frac1k-\frac1{k+1}\right)}{\ln n}\le\frac1{\ln n}.$$
This implies that $\lim\limits_{n\to\infty} (\mu_n(A)-\mu_n(B)) =0$, hence $\Flim (\mu_n(A)-\mu_n(B))=0$ which means that $\Flim\mu_n(A)=\Flim\mu_n(B)$ and $$\mu(B)=\mu(A).$$
$\mu$ is $\mathbb N$-scale invariant, i.e., $\mu(kA)=\frac1k\mu(A)$ for $k\in\mathbb N$.
Let $B=kA$. Now we have $$\mu_n(B)=\frac{\sum_{j\le n}\frac1j \chi_B(j)}{\ln n} =
\frac{\sum_{j\le n/k}\frac1{kj} \chi_A(j)}{\ln n} = \frac1k \frac{\sum_{j\le n/k}\frac1j \chi_A(j)}{\ln n}.$$
This yields
$$|\frac1k\mu_n(A)-\mu_n(B)| \le \frac1k \frac{\sum_{n/k<j\le n} \frac1j}{\ln n} \sim \frac1k \cdot \frac{\ln k}{\ln n}.$$
Again $\lim\limits_{n\to\infty} (\frac1k\mu_n(A)-\mu_n(B)) =0$ and $\mu(B)=\frac1k\mu(A)$.<|endoftext|>
TITLE: How slowly can a power of an ideal grow?
QUESTION [23 upvotes]: For a polynomial ideal $I\subset \mathbb{C}[x_1,x_2]$, let $D(I)$ be the smallest degree of any polynomial in $I$.

How slowly can $D(I^n)$ grow as a function of $n$? For example, if $D(I^n)\leq 1.01n$ for some $n$, does it imply that $I$ contains a linear polynomial?

Note that the single-variable case is trivial: $D(I^n)=n\cdot D(I)$.
I am interested in a more general situation than in the question, but the version is the simplest case where I am stuck.

REPLY [13 votes]: Consider $I = (y- x^k, x^{k+1})$.
For $k>1$ this does not contain any linear functions. It contains $xy$ so $D(I)=2$. But I claim $y^{k+1} \in I^k$, so $D(I^k) = k+1$.
By the binomial theorem
$$ y^{k+1} = (y-x^k + x^k)^{k+1}  = \sum_{i=0}^{k+1} \begin{pmatrix} k+1 \\ i \end{pmatrix} \left(y-x^k\right)^i x^{k (k+1-i) } $$
In the exponent:
$$k(k+1-i) =k^2 +k - ik =  (k+1)(k-i) + i$$
so this is 
$$(y-x^k)^{k+1} + \sum_{i=0}^{k} \begin{pmatrix} k+1 \\ i \end{pmatrix} x^i\left(y-x^k\right)^i \left(x^{k+1}\right)^{k-i} \in I^k$$
(Boris pointed out a flaw in my earlier argument, leading me to find this counterexample.)
In general, subadditivity shows $\lim_{n \to \infty} \frac{D(I^n)}{n}$ exists, and that any fixed value of $\frac{D(I^n)}{n}$ is at least this limit. So one version of this question is about how to compare $D(I)$ to this limit. Here we show the limit can go arbitrarily close to $1$ with $D(I)=2$. By adding random linear factors, the limit can get arbitrarily close to $D(I)-1$. But probably for larger $D(I)$ the limit can be less than $D(I)$ by even more than $1$.

Some lower bounds:
In the case where $I$ is radical, if $D(I) \geq 2$, then $D(I^n) \geq (3/2)n$ (and in fact $\lceil (3/2) n \rceil$ is achieved.) $V(I)$ must not be contained in any line, so it must contain $3$ noncolinear points, and we can assume that $I$ is the ideal of $3$ noncolinear points. Then $I^n$ is the ideal of functions vanishing of order $n$ at those $3$ points. This contains a function of degree $(3/2)n$, which is the product of powers of the lines through the points.
This is optimal, because given a polynomial $f$, which is the first line raised to the power $a$ times a polynomial of degree $d−a$, the polynomial of degree $d−a$ must intersect the two points on the first line with multiplicity $n−a$, so $d−a \geq 2(n−a)$ and if $d\leq (3/2)n$, $a \geq n/2$. Then the same is true for the multiplicity of the other $3$ lines, hence $d\geq 3n/2$.

Here's another interesting phenomenon. Take $I$ to be the ideal of $k (k+1) /2$ generic points. Then $D(I)= k$ by dimension counting. $I^n$ is the ideal of functions vanishing of order $n$ at $k(k+1)/2$ distinct points, which is an ideal of codimension $n (n+1)/2 \cdot k (k+1)/2$. This is less than $d (d+1)/2$ for $d$ approximately equal to $nk / \sqrt{2}$. So there is a degree $d$ polynomial in $I^n$, and $D(I^n)$ is asymptotically at most $nk/\sqrt{2}$.

I can show that if $D(I) \geq 2$, then $\lim_{n \to \infty} D(I^n)/ n> 1$. Take $I$ maximal with respect to the property $D(I) \geq 2$. Then each local factor of $I$ at a point of $V(I)$ either contains two linear functions, or is maximal with respect to the property of containing one linear function, and hence looks like $(y, x^2)$, or is maximal with respect to the property of containing no linear functions, and hence looks.
What do the last kind of ideals look like? There must be some length $1$ extension, which must contain some linear function $y$, and so it is of the form $(x^k,y)$ for some $n$. Length one extensions of that have the form $(x^{k+1}, xy, y^2, ax^k+ by)$ and we must have $a \neq 0$. If $b =0$, the ideal contains is contained in $(x^2, xy, y^2)$, which is one example of a maximal ideal with this property. Otherwise by scaling $y$, we may put it in the form of my example.
Case 1: $I= (x^2, xy, y^2)$. An element in $I^n$ vanishes to order $2n$ on $I$, hence has degree at least $2n$.
Case 2: $I = (y-x^k, x^{k+1})$. An element in $I^n$ intersects $y-x^k$ with multiplicity $n (k+1)$, hence has degree at least $n (k+1)/k$. Having $(y-x^k)$ divide the element doesn't help because it has degree $k$ but only removes $k+1$ of the intersection.
Case 3: $I$ contained in $(y, x^2)$. Then $I$ must also vanish somewhere else on the line $y=0$. If the degree is at most $(3/2)n$ the intersection multiplicity with the line $x=0$ is at least $2n$ so by the same logic as in the reduced case the polynomial contains a factor of $x^{n/2}$. The remainder of the polynomial must vanish to order $n$ at the other point on the line $y=0$ hence have degree at least $n$, so the minimum is $(3/2)n$.
Case 4: $I$ is contained in none of these and is maximal, hence reducd. We already did this case.

REPLY [4 votes]: Proof that $D(I^2)=2 \Rightarrow D(I)=1:$ Since $k=$ℂ is algebraically closed, every maximal ideal of $k[x,y]$ is of the form $(x-a,y-b)$, so by a translation of coordinates (which preserves degrees) we may assume $I \subset (x,y)$. Replacing generators of $I$ by $k$-linear combinations, we may assume that at most 2 generators (say, $u,v$) have degree=1 terms, and furthermore, that the highest terms of $u,v$ in the monomial ordering by (total degree, $y$-degree, $x$-degree) are different.  Now if $I^2$ contains a polynomial of degree=2 then it must contain a $k$-linear combination of $u^2,uv,v^2$ of degree=2, and the highest terms of $u^2,uv,v^2$ in the monomial ordering are different, so at least one of $u,v$ has degree=1.<|endoftext|>
TITLE: If a topological space has vanishing $n$th homology for every possible homology theory, does it have vanishing $n$th homotopy?
QUESTION [6 upvotes]: I don't have any strong preference as to whether or not the homology theories are required to be ordinary.
Also, if this does not hold in general, does it hold for some nice category of spaces, like CW-complexes?
Finally, in a more general context, does anyone have a reference for where to find information about things one can deduce from properties common to all homology theories evaluated on a given space?

REPLY [10 votes]: The first answer to your question is no. There are many acyclic spaces with nonvanishing $\pi_1$. Since generalized homology is a stable invariant, and the suspension of an acyclic space is contractible (exercise), this also means that any generalized cohomology theory vanishes on such a space. For an explicit example, take the classifying space of a perfect group. 
But now you'll say it's a $\pi_1$ issue, and that's sort of true. If $E_n(X) =0$ for every generalized homology theory, then that means that $E_*(X) = 0$ for any generalized homology theory, including ordinary homology. Indeed: if $E_*(-)$ is a homology theory, so is $E_{*+n}$. But if you're willing to stick to, say, connective cohomology theories (i.e. $E_k(pt) = 0$ for $k<0$), then there are simply connected counterexamples. Indeed:  $\pi_3S^2 = \mathbb{Z}$ but $H_3(S^2) = 0$. So let $X = S^2_{\mathbb{Q}}$ denote the rationalization of the 2-sphere, then for any homology theory $E$ with $E_{-1} \otimes \mathbb{Q} = 0$ we get $E_3(X) = 0$ but $\pi_3X = \mathbb{Q}$. Indeed, $\Sigma^{\infty}_+X$ is just $\Sigma^2H\mathbb{Q}$ so $E_3(X) = \pi_{-1}(H\mathbb{Q} \wedge E) = 0$.<|endoftext|>
TITLE: Motzkin polynomials and enumeration of chord diagrams
QUESTION [5 upvotes]: On page 12 of the paper Enumeration of chord diagrams on many intervals and their non-orientable analogs" by Alexeev, Andersen, Penner, and Zograf is a list of polynomials which are a refinement of the first few Motzkin polynomials listed in OEIS A055151. 
Can anyone provide a proof that this relation between the two sets of polynomials holds for all higher degrees in general?

REPLY [4 votes]: The bijection between planar partial chord diagrams and Motzkin paths is the following: the left end of a chord corresponds to U-step, the right end of a chord corresponds to D step, and "free" marked point corresponds to H step. For  example, the last diagram on Figure 3 in the paper is encoded by a UUHHDDUHHD Motzkin path. 
It is well-known, that the number of 2n-gon gluings into a sphere is n^th Catalan number, and so there is a bijection between planar polygon gluings and Dyck paths.
This is basically the same relation you mentioned.<|endoftext|>
TITLE: Semicircle law universality elsewhere
QUESTION [10 upvotes]: Wigner's semicircle distribution is:
$$f(x)=\frac{1}{2 \pi}\sqrt{4-x^2}, \ \ -2\leq x\leq 2.$$
Under reasonable conditions, the rescaled eigenvalue density of random symmetric matrices $M_n$ follows this distribution, from which we can infer both the number of rescaled eigenvalues in an interval $[a,b]$ and roughly where the $i$'th eigenvalue lies. Specifically, this concerns the statistics of the (unordered) vector $(\lambda_1(M),\lambda_2(M),\cdots,\lambda_n(M))$. 
My question is, where else does the Wigner semicircle law arise? To push the discussion in a particular direction, I am primarily interested in other random processes $(X_1,\cdots,X_n)$ which asymptotically have semicircle statistics similar to the above. Specifically, I am not interested in processes which can be somehow coupled to random matrices. I would love to see some examples where the semicircle law occurs more or less universally, that is, for a wide, reasonable class of different distributions on $X_i$'s.

REPLY [4 votes]: Following up on the answers in the context of free probability theory one should note that there is actually a quite important and canonical operator on a Hilbert space having the semicircle as distribution. In the answers above there were already mentioned concrete operators arising from the free group representations, but there one also has to take the limit $d\to\infty$ to get the semicircle. I want to highlight a situation without taking a limit: Let $l$ be the one-sided shift (acting on a Hilbert space
with orthonormal basis $(e_i)_{i\geq 0}$ via $le_i=e_{i+1}$), then the selfadjoint operator $l+l^*$ has with respect to the "vacuum expectation" $\langle e_0, \cdot e_0\rangle$ the semicirlce law as distribution.
The one-sided shift is arguably one of the most important operators in single operator theory, but it seems nobody before Voiculescu was interested in looking at the distribution of its real part.<|endoftext|>
TITLE: Is there a symmetric monoidal 2-category "SuperDuperVect"?
QUESTION [27 upvotes]: Recall that the category $\mathrm{SuperVect}$, as a category, consists of pairs of vector spaces, thought of as formal direct sums $V \oplus W\,\Pi$, where $\Pi$ is the "odd line".  (Called "$\Pi$" because tensoring with it is "parity reversal".)
Indeed, there is a good notion of "direct sum of categories" in which case, as categories, we have
$$ \mathrm{SuperVect} = \mathrm{Vect} \oplus \mathrm{Vect}\, \Pi$$
As a monoidal category, we declare that the one-dimensional vector space $\mathbf 1 \in \mathrm{Vect}$ is the monoidal unit and that $\Pi \otimes \Pi = \mathbf 1$, and give it the trivial associator (so that as a monoidal category $\mathrm{SuperVect}$ is the category of sheaves of vector spaces on $\mathbf Z/2$ with the convolution product; when $2$ is invertible, which I assume it is, this is also the category of representations of $\mathbf Z/2$).
The interesting part of $\mathrm{SuperVect}$ is its braiding/symmetry.  A braiding $\sigma$ on this monoidal category is uniquely determined by its value on
$$ \mathbf 1 = \Pi \otimes \Pi \overset \sigma \longrightarrow \Pi \otimes \Pi = \mathbf 1 $$
which is just a number $\sigma_{\Pi,\Pi}$.  The axioms of symmetric monoidal category force $\sigma_{\Pi,\Pi}$ to square to $1$, but do not force it to be $1$ itself.  The symmetric monoidal category $\mathrm{SuperVect}$ is determined by declaring that $\sigma_{\Pi,\Pi}$ is the other number that squares to $1$, namely $-1$.
At least over $\mathbb C$,
$\mathrm{SuperVect}$ has the following important property, due to Deligne.  Not every symmetric monoidal category is tannakian over $\mathrm{Vect}$ --- in particular, there does not exist a symmetric monoidal functor $\mathrm{SuperVect} \to \mathrm{Vect}$ --- but every symmetric monoidal category (satisfying some technical bounds on growth rates of objects) is tannakian over $\mathrm{SuperVect}$.
My question is whether this trick can be repeated one dimension higher.  Let $\mathrm{Vect}_{\mathrm{SuperVect}}$ denote the 2-category of all "supercategories", i.e. categories with an action by $\mathrm{SuperVect}$.  (I'd rather not make this precise.  Probably I should fill in some words like "abelian".  I'd be perfectly happy just working with the 2-category whose objects are the natural numbers and whose morphisms are matrices filled in with supervector spaces, just like one can model $\mathrm{Vect}$ as the category whose objects are natural numbers and whose morphisms are matrices of numbers.)
Then $\mathrm{Vect}_{\mathrm{SuperVect}}$ is symmetric monoidal (because $\mathrm{SuperVect}$ is) with unit object $\mathbf 1$, and with $\mathrm{End}(\mathbf 1)  = \mathrm{SuperVect}$.
Then there's a perfectly good 2-category
$$ \mathrm{SuperDuperVect} = \mathrm{Vect}_{\mathrm{SuperVect}} \oplus \mathrm{Vect}_{\mathrm{SuperVect}} \, \Xi$$
where the letter $\Xi$ is just a formal symbol playing the role of $\Pi$ above.  I would like to give this category a symmetric monoidal structure in which $\Xi \otimes \Xi = \mathbf 1$ but the braiding on $\Xi$ is the endomorphism
$$ \sigma_{\Xi,\Xi} = \Pi \in \mathrm{End}(\mathbf 1) = \mathrm{End}(\Xi \otimes \Xi). $$
The idea is that this is the other object that squares to $\mathbf 1 \in \mathrm{SuperVect}$.
Now, I should be a bit careful.  Symmetric monoidal 2-categories, when written out in full detail, consist of quite a lot of data and coherence conditions.  Perhaps there's some rule that says that $\sigma_{\Xi,\Xi}$ not only has to square to $\mathbf 1$ but also has to braid trivially with itself.  I don't know, and I'm not sure where to look up the axioms.

Does such a symmetric monoidal 2-category "$\mathrm{SuperDuperVect}$" exist?
Is it symmetric-monoidally equivalent to some more basic thing, say the 2-category of sheaves of supercategories on something with convolution product, or the 2-category of supercategorical representations of something?
Is there a symmetric monoidal 2-functor $\mathrm{SuperDuperVect} \to \mathrm{Vect}_{\mathrm{SuperVect}}$?

Of course, if the answer to the first question is "no", then the other two are moot.  If the answer to the first question is "yes", then I can't imagine positive answers to the other two, but maybe my imagination is faulty.

REPLY [15 votes]: Yes, the symmetric monoidal 2-category you are looking for does exist.
I think that there is a slightly different 2-category which is better, but yours embedds inside the one I will describe, which differs in that
there are interesting "cross-terms" from $\Xi$ to 1, i.e. morphisms between these objects.  This 2-category has a more familiar description. It is the Morita category of finite dimensional semisimple superalgebras (over $\mathbb{C}$). 
Here a superalgebra is just an algebra object in SuperVect. The theory of semisimple modules and algebras mirrors that for ordinary algebras, but with a few subtleties (it is super subtle). A very nice treatment which covers the case that the ground field is algebraically closed is this paper:
Semisimple Superalgebras Tadeusz Jozefiak Volume 1352 of the series Lecture Notes in Mathematics pp 96-113.
There you see that there is a classification of semisimple superalgebras over $\mathbb{C}$.  They are finite sums of simple superalgebras. The simple superalgebras are classified as either

$End(\mathbb{C}^{p|q})$ (which is super Morita equivalent to $\mathbb{C}$)
$Q(n) = M_n(\mathbb{C}) \otimes Cl_1$ where $Cl_1$ is the Clifford algebra on a one-dimensional complex vector space. 

They have multiplications (using the super-tensor product, of course):

$End(\mathbb{C}^{p|q}) \otimes End(\mathbb{C}^{m|n}) = End(\mathbb{C}^{pm + qn|pn + qm})$
$End(\mathbb{C}^{p|q}) \otimes Q(n) = Q(pn + qn)$
$Q(m) \otimes Q(n) = End(\mathbb{C}^{mn|mn})$

Now the 2-category I want to consider is the Morita 2category whose objects are fin. dim.  semisimple superalgebras and whose 1-morphisms are superbimodules between them. 
The isomorphism classes of objects in this 2-category are given by pairs of natural numbers which count the number of $End(\mathbb{C}^{p|q})$ and $Q(n)$ factors. 
The categories of morphisms are exactly what you describe, as long as you throw out the cross-term morphisms. For example the category of morphisms from $\mathbb{C}$ to $Q(n)$ is equivalent to the category of $Q(n)$-modules, where in your 2-category (if I understand your notation) you would have the zero category here. The 2-category you describe sits inside as the sub-2-category with only the zero cross term morphisms. 
Now for your final question. There is no symmetric monoidal functor 
$$ SuperDuperVect \to Vect_{SuperVect} $$ 
where SuperDuperVect is either the Morita category I describe or the subcategory you mention. You can see this by passing to maximal Picard sub-2-categories (i.e. the max 2-category in which all morphisms and objects are invertible). 
These Picard 2-categories are equivalent to spectra with 3 consecutive homotopy groups. The target gives a spectrum with 
$$\pi_0 = 0,  \;  \pi_1 = \mathbb{Z}/2, \; \pi_2 = \mathbb{C}^\times$$
while the source (in either case) has
$$\pi_0 = \mathbb{Z}/2, \; \pi_1 = \mathbb{Z}/2, \; \pi_2 = \mathbb{C}^\times$$.
Moreover the k-invariants of these are well-known. The latter looks like a truncated variant of the Brown-Comenetz dual of the sphere. It is closely related to real K-theory KO. The former (the target) looks like the connective cover. The k-invariant connecting the bottom $\mathbb{Z}/2$ to the other homotopy groups (eg. the next $\mathbb{Z}/2$) obstructs the existence of your map, and it is known to be non-zero.<|endoftext|>
TITLE: On combinatorial and cellular model categories and infinity categories
QUESTION [8 upvotes]: I am looking for a counterexample. Let me first give the set-up. When you work with model categories, it is extremely common to assume they are cofibrantly generated. For me, this means the definition in Hovey's or Hirschhorn's book, i.e. there are sets of maps I and J who detect the trivial fibrations and fibrations by lifting and whose domains are small relative to I-cell (resp. J-cell). There is a weaker notion used by Emily Riehl, which I will ask about at the very end.
When you want to do Bousfield localization it's common to assume $M$ is in addition either combinatorial or cellular. Combinatorial means cofibrantly generated and locally presentable as a category. Cellular is more mysterious. Morally, it's asking for good control over cell complexes (i.e. things built from the maps in I via gluing). Formally, it's asking for

Domains of maps in J are small relative to cofibrations
Cofibrations are effective monomorphisms, i.e. any cofibration $f:X\to Y$ is the equalizer of the two obvious maps $Y\to Y \coprod_X Y$
Domains and codomains of I are compact (in the sense of Hirschhorn, not Hovey) relative to I-cell.

Condition 1 is standard and usually easy to verify (for example, it comes for free if $M$ is locally presentable). Condition 2 is basically asking that the intersection of the two copies of $Y$ in $Y\coprod_X Y$ is equal to $X$, so it's also not that hard to check in categories where cofibrations are inclusions of some kind. Condition 3 is a pain to check, because it's really asking for a cardinal $\kappa$ such that for any presentation of a map $f:A\to B$ as a transfinite composition of pushouts along a chosen set of cells then any map from a (co)domain $X$ of a map in $I$ to $B$ factors through some subcomplex of size $\leq \kappa$. Again, if $M$ is locally presentable then this should be true automatically, at least if the collection of subcomplexes is filtered (since you know that all objects are small relative to filtered colimits).
It's clear that not every cellular model category is combinatorial. For example, Top is not (Hovey proves on page 49 of his book that the two-point Sierpinski space is not small). It's clear that not every combinatorial model category is cellular, because condition 2 can fail. For example, Hirschhorn provides such an example as 12.1.7. Another example is in Finnur Larusson's paper "The Homotopy Theory of Equivalence Relations." Another is the category of small categories Cat, as I learned today from a preprint of Amrani Ilias called "Stabilization of the category of simplicial objects in Cat." Similarly, if you take spectra valued in Cat it's not cellular (but it is combinatorial) as Deb Vicinsky has shown in her thesis. There are also examples of model categories which are not cofibrantly generated. My favorite is the Strom model structure on Top, and the proof is in Raptis's paper on Homotopy Theory for Posets. It seems to me that all the other conditions of cellularity are true here (suitably interpreted) except for cofibrantly generated, but perhaps I am wrong. Lastly, there is a model structure on a locally presentable category which is provably not cofibrantly generated, and it's given by Boris Chorny's example from the paper "The Model Category of Maps of Spaces is not Cofibrantly Generated." The category in question is simply the arrow category valued in sSet. When I tell these examples to people in the infinity category community I always get the same question, so now I'm putting it out to the MathOverflow world:

(1) Is there an example of a complete and cocomplete infinity category which does not admit any cofibrantly generated model?

This is probably a hard question. Chorny's example doesn't work, because every presentable infinity category gives rise to a combinatorial model category (which in this case will have the same weak equivalences but different cofibrations), obtained by embedding into simplicial presheaves then taking a Bousfield localization. A related question is

(2) Is there an example of a complete and cocomplete infinity category which does not admit a cellular model, but does admit a cofibrantly generated one?

This seems much more likely to be true to me, since from a model category perspective there really is a difference between cellularity and combinatoriality. But since all statements required for cellularity are about cofibrations, perhaps there's a sneaky way to always choose a nice set of cofibrations when you pass from a presentable infinity category to a model category. Perhaps you could shrink the cofibrations via Bousfield colocalizations, for example.

(3) Are there examples which have arisen in practice of non presentable infinity categories?

Lastly, I want to better understand where Riehl's definition of cofibrantly generated fits into this story. In a combinatorial model category the smallness hypotheses are automatic, so it seems Riehl's definition matches the usual one. Similarly, I suppose (1) and (3) in the definition of cellularity force Riehl's definition and the usual one to agree, though perhaps there is a way to weaken cellularity to algebraic cellularity in the sense of Riehl's algebraic wfs. Riehl also has notions of enriched weak factorization systems which allow you to do things like saying the Strom model structure is cofibrantly generated in her enriched sense. I am less certain about Chorny's examples. Anyway, for now my main question regarding Riehl's definition is:

(4) Is there an infinity category which admits the required colimits to form a factorization system but which does not admit any model that has a factorization system in the sense of Riehl (either simply an algebraic wfs or an enriched wfs)?

REPLY [8 votes]: If you believe Vopěnka's principle then any cofibrantly generated model category is Quillen equivalent to a combinatorial one and thus its underlying $\infty$-category is presentable. It follows that any non-presentable complete and cocomplete $\infty$-category would give a positive answer to (1). The pro-category of a small finitely complete $\infty$-category is not presentable but it is co-presentable (its opposite category is presentable). However, the pro category of a large cocomplete and finitely complete $\infty$-category is complete and cocomplete but neither presentable nor copresentable. In "Higher Topos Theory" Lurie considers the pro category of spaces (see Definition 7.1.6.1). It thus seems that this category gives a positive answer to (1) and (3). This example was also considered in the world of model categories: It is Isaksen's strict model structure on pro-simplicial sets. If you are interested in a cocombinatorial model category that arose naturally you have Morel's (resp. Quick's) model structure on simplicial pro-finite sets that models the $\infty$-category of p-pro-finite (resp. pro-finite) spaces.
Edit: It turns out that there is a mistake in Raptis's paper showing that under Vopenka's principle any cofibrantly generated model category is Quillen equivalent to a combinatorial one. Raptis and Rosicky posted a fixed proof, in which they had to assume that the domains of the generating (acyclic) cofibrations (in the cofibrantly generated model category) are small with respect to certain types of filtered colimits of cofibrations, more general than just chains of cofibrations. See also the remarks of Tim Campion below.<|endoftext|>
TITLE: Topology of hypersurface of sphere fixed by homeomorphic involution
QUESTION [5 upvotes]: I'm not an topologist, so I apologize in advance if this is a silly question.
I have the following situation, let $\mathbb{S}^n$ (for $n\geq 3$) be the standard (smooth if it matters) $n$-sphere and $\Sigma\subset \mathbb{S}^n$ a connected hypersurface.  Let $\Omega_\pm$ be the two components of of $\mathbb{S}^n\backslash \Sigma$ (there are exactly two by Jordan-Brouwer). Suppose moreover there is a non-trivial homeomorphic involution $\phi:\mathbb{S}^n\to \mathbb{S}^n$ which fixes $\Sigma$ in the strong sense that it restricts to the identity map and swaps $\Omega_\pm$.
My question is how much can one say about $\Sigma$?
So far what I can say (I only sketched this in my head so may have made a mistake):

By Mayer-Vietoris, the homology groups of the $\Omega_\pm$ vanish and $\Sigma$ is a homology sphere.
By Seifert-Van Kampen, $\pi_1(\Omega_\pm)=0$ and so by Hurewicz, all the homotopy groups of $\Omega_\pm$ vanish.

However, I'm now stuck as I don't see a reason for $\pi_1(\Sigma)$ to vanish and don't know enough examples to know if this can even be true. 
Can one go further?  Is $\Sigma$ a homotopy sphere?

REPLY [7 votes]: In "Smooth Homology Spheres and their Fundamental Groups" Kervaire proves that i) every 4-dimensional homology sphere bounds a contractible smooth manifold, and ii) for $d \geq 5$, every $d$-dimensional homology sphere bounds a contractible smooth manifold, after perhaps changing it by connect-sum with an exotic sphere.
Hence, let $\Sigma^d$ be any homology sphere of dimension $d \geq 4$, and modify it if necessary by connected-sum with a homotopy sphere so that it bounds a contractible manifold $\Delta^{d+1}$. Then
$$N := \Delta \cup_\Sigma \Delta$$
has an obvious (smooth) involution with fixed set $\Sigma$. Furthermore, this is easily seen to be a homotopy sphere (by Mayer--Vietoris and Seifert--van Kampen). It may not be diffeomorphic to $S^{d+1}$, but as it has an orientation-reversing involution it will have order $2$ in the group $\Theta_{d+1}$ of exotic spheres. When $d+1=5,6, 12, 13$ this group is trivial or $\mathbb{Z}/3$, and so $N$ must in fact be diffeomorphic to $S^{d+1}$ in these cases.
Thus $\Sigma$ need not be a homotopy sphere in general<|endoftext|>
TITLE: Can a homology $n-1$-sphere divide $\mathbb{S}^{n}$ into non-contractible components?
QUESTION [7 upvotes]: This is a follow-up to my earlier question.
Let $\Sigma\subset \mathbb{S}^{n}$ be a hypersurface -- here $\mathbb{S}^{n}$ is a smooth sphere (possibly exotic ... if this makes a difference).  If $\Sigma$ is a homology sphere, then is it possible for the components of $\mathbb{S}^{n+1}\backslash \Sigma$ to be non-contractible?

REPLY [15 votes]: Yes; this is not hard to arrange.  For instance, take any knot K in the 3-sphere with an n-fold branched cover, say M that is a homology sphere. A good example would be to take M to be the $k$-fold cover of the $(p,q)$-torus knot, where p, q, and $k$ are pairwise relatively prime. Then the amazing twist-spinning construction of Zeeman (Twisting Spun Knots, Trans. AMS 115 (1965), pp. 471-495) constructs a knot in the 4-sphere with fiber $M_0 = M - B^3$. (Zeeman also describes the monodromy as the generator of the covering group of the branched cover.)
Let $N = M_0 \cup -M_0$, embedded in $S^4$ as the boundary of a product neighborhood $M_0 \times I$.  Then both components of $S^4 - N$ are diffeomorphic to $M_0 \times I$, so neither is simply-connected, and hence not contractible.
A similar construction works in any dimension; you just need knots in $S^n$ with homology spheres for branched covers. You can get these by spinning.  Alternately, you can do the pairwise spin of the homology sphere $M$ above. By definition, for a manifold $M$, the spin of $M$, say $S(M)$, is gotten from $S^1 \times M$ by surgery on $S^1 \times x$ for some point $x \in M$. Two salient points are that $S(M)$ has the same fundamental group as $M$ and that $S(S^n) = S^{n+1}$. Moreover (I'm expecting the Spanish inquisition next) if $M$ is a homology sphere then so is $S(M)$. 
If $M \subset S^n$, then you can spin both simultaneously to get $S(M) \subset S^{n+1}$ and the complementary regions have the same fundamental groups as those of $M$.  This gives examples in all (ambient) dimensions $4$ and up.<|endoftext|>
TITLE: Ambidexterity and Quantization
QUESTION [6 upvotes]: Here the nlab says about Hopkins-Lurie's ambidexterity paper:

"The discussion in the article is apparently motivated as part of what it takes to make precise the discussion of quantization in sections 3 and 8 of [Freed-Hopkins-Lurie-Teleman]"

I don't understand this statement, and would appreciate if someone could shed more detail on it. I only found the following links that seem to be related: http://www3.nd.edu/~cmnd/graduate/abstracts/Lurie.pdf and $\S 1.2$ of http://arxiv.org/pdf/1409.0837.pdf. The latter has some commentary that is related to the nlab's statement, but I would appreciate a more detailed explanation. My question is therefore as follows:

Can anyone explain the nlab's statement above?

REPLY [3 votes]: Charles Rezk is right in quoting that part of [FHLT], which is probably the main motivation to ambidexterity; another one might come from Lurie's program to categorify Fourier Theory (however, it would be better to ask Lurie himself).
The obstacles in Lurie's abstract are probably related to the construction of the TQFT suggested in [FHLT] and the role played into it by the Nakayama isomorphism, as spelled out in [Morton, Two-Vector Spaces and Groupoids] where the factor $|G|^{-1}$ shows up.
In particular, the Nakayama isomorphism appears when you consider the quantization functor $Sum$ from $Fam_{1}(Vect)$ to $Vect$ (or its higher analogues). Here the source category is the one defined by Haugseng, in the case when $C=Vect$. In general the functor goes from $Fam_{n}(C)$ to $C$, and it seemed that an isomorphism between right and left Kan extensions along maps of spaces is needed when defining it on morphisms.  
As DamienC said, I've been working on this in my PhD dissertation. In dimension 1 it turns out that only cocompleteness and dualizability need to be imposed on our category $C$ (completeness comes for free). 
Under these hypothesis one can construct a canonical map going from the right to the left Kan extension which is NOT (a priori) needed to be an iso, and build up the quantization functor $Sum$. 
The interesting fact, is that once you have your quantization functor, you can show that the above canonical map is indeed an isomorphism. Ambidexterity is therefore a consequence, not an ingredient one needs. 
Ideally, the same procedure will work also for $(\infty,n)$-categories, where now the hypotheses will be cocompleteness and full dualizability, and I'm currently working on this.
I've uploaded the thesis on dropbox at this address 
https://www.dropbox.com/s/hci6utz2x0g0qmm/Trova-Quantization.pdf?dl=0
Edit: an incorrect guess on the possibility of the result to hold also in positive characteristic has been removed thanks to the remark by Yonatan Harpaz here below.<|endoftext|>
TITLE: Lurie's Endomorphism Space vs. Endomorphisms
QUESTION [9 upvotes]: In Jacob Lurie's book Higher Algebra, for an object $M$ of a monoidal $\infty$-category $\mathcal{C}$, he constructs a category $\mathcal{C}[M]$ which can be thought of as "maps in $\mathcal{C}$ of the form $A\otimes M\to M$ with associated coherence data". $\mathcal{C}[M]$ is shown to be monoidal, and algebra objects of this category are shown to be precisely the algebras which coherently act on $M$ (though one algebra that acts on $M$ in two different ways will be thought of as two different objects of $\mathcal{C}[M]$). Moreover, he shows that if $\mathcal{C}[M]$ has a final object, which we'll denote $End(M)$, this object is an algebra object of $\mathcal{C}$ and $M$ is an object of $LMod_{End(M)}$. Thus any algebra that acts on $M$ admits a morphism to $End(M)$. 
I'm in particular thinking about the quasicategory of $\infty$-groupoids, $\mathcal{C}=Top$, so $M$ and $End(M)$ and everything else will just be spaces for the rest of this. We have another more down-to-earth notion of "endomorphisms of $M$." That is, the space of morphisms $Map_{Top}(M,M)$. This is a monoid object of $Top$ whose monoid structure is given by composition. Moreover there is an evaluation map $ev:Map_{Top}(M,M)\times M\to M$ so there is a corresponding object of $Top[M]$. Is this object clearly final in $Top[M]$? 
In other words, if $M$ is an object of the quasicategory $LMod_A$ for $A$ some loop space in $Top$, is there induced a map $A\to Map_{Top}(M,M)$? Certainly given a map of topological spaces $A\times M\to M$ one can produce, by adjunction, a map $A\to Map_{Top}(M,M)$, but it's not clear to me that this is necessarily a map of algebras or that it carries all the necessary structure (though I suspect it does). In other words this would require a more complex kind of adjunction than just the one for tensor/hom.

REPLY [6 votes]: Suppose $\mathcal{C}$ is any closed monoidal $\infty$-category and $X$ is an object of $\mathcal{C}$. Then $\mathcal{C}[X]$ can be described as the $\infty$-category of pairs $(C, C \otimes X \to X)$, so it shouldn't be hard to see that the universal property of a final object in $\mathcal{C}[X]$ is precisely that of the internal Hom, say $X^{X}$, from $X$ to $X$. So what Lurie proves (4.7.2.40) is that there is a canonical associative algebra structure on $X^{X}$ - intuitively this is given by composing morphisms.
Now I think what you're asking about is the case where you've somehow constructed another algebra structure on $X^{X}$ (I suppose you could for instance do this if the $\infty$-category $\mathcal{C}$ came from a nice closed monoidal model category). Then by the uniqueness of this algebra, "all" you have to do to show it's the same as Lurie's endomorphism object is to promote it to an algebra object of $\mathcal{C}[X]$. That sounds hard, but it's actually not so bad since Theorem 4.7.2.34 tells you that giving an algebra in $\mathcal{C}[X]$ is the same as giving an algebra $A$ in $\mathcal{C}$ together with an $A$-module structure on the fixed object $X$. So all you need to do is check that your algebra structure on $X^{X}$ gives you a map $X^{X} \otimes X \to X$ compatible with composition, which you obviously ought to have if your algebra structure really was given by "composing maps"<|endoftext|>
TITLE: Berkovich stalk versus rigid analytic stalk
QUESTION [5 upvotes]: Let $A$ be a strictly affinoid algebra. Let $X^{Ber}$ bet its Berkovich spectrum and $X^{Tate} = \operatorname{Sp} A$ its affinoid variety in the sense of rigid analytic geometry. Let $\mathfrak{m} \subset A$ be a maximal ideal and $x$ the corresponding point in $X^{Ber}$, respectively in $X^{Tate}$. We know that for all $n \in \mathbb{N}$ there are isomorphisms
$$A / \mathfrak{m}^n \cong A_\mathfrak{m} / A_\mathfrak{m} \mathfrak{m}^n \cong \mathcal{O}_{X^{Tate},x} / \mathcal{O}_{X^{Tate},x} \mathfrak{m}^n.$$
Does $\mathcal{O}_{X^{Ber},x}$ also fit in? That is, is $\mathcal{O}_{X^{Ber},x} / \mathcal{O}_{X^{Ber},x} \mathfrak{m}^n$ also isomorphic to these rings?

REPLY [2 votes]: As Jérôme points out, the rings $\mathcal{O}_{X^{Ber},x}$ and $\mathcal{O}_{X^{Tate},x}$ are the same, thus the answer is "yes".<|endoftext|>
TITLE: Can a 3-ball divide a standard 4-ball into two exotic 4-balls?
QUESTION [7 upvotes]: Let $B^n$ denote the unit ball in $\mathbb{R}^n$ (wrt the standard euclidean metric) and $\bar{B}^n$ denote the unit closed ball.  Suppose that $\Sigma$ is a a smooth embedded hypersurface with boundary in $\bar{B}^4$ which is topologically $\bar{B}^3$ and so that $\Sigma$ meets $\partial \bar{B}^4$ transversely and $\partial \Sigma\subset \partial \bar{B}^4$. It should follow from known results that $B^4\backslash \Sigma=U_+\cup U_-$ where $U_\pm$ are open domains homeomorphic to $B^4$.  Is it also known whether they are diffeomorphic, or is this equivalent to the smooth 4D Schoenflies problem (it is clearly weaker)?
I'm not a topologist so I apologize if this is a silly question.

REPLY [5 votes]: I think this is equivalent to the smooth Schoenflies conjecture; the executive summary is that this is true because smooth balls (in any dimension) are isotopic to standard ones. 
Here are some details, starting with some preliminary remarks. Your $\bar{B}^3$ is diffeomorphic to a closed 3-ball, so I'll take that as given. Its boundary is a 2-sphere in $\partial \bar{B}^4$, and so is isotopic to the standard 2-sphere in $S^3$ by the 3D Schoenflies theorem (Alexander's theorem). Also, the standard result about balls in a manifold being standard applies to a codimension $0$ ball $\bar{B}^n$ embedded in an $n$-manifold in a non-proper (or neat) way. In other words the boundary of $\bar{B}^n$ is divided into $\bar{B}^{n-1}_\pm$ with $\bar{B}^n \cap \partial M = \bar{B}^{n-1}_-$ lying in the boundary of $M$, and $\bar{B}^{n-1}_+$ properly embedded. You really should worry a bit about corners here, but I'm going to ignore this (and in what follows).
Suppose first that we have $\bar{B}^4$ as described and that the Schoenflies conjecture holds.  By the remarks above, do a preliminary isotopy so the boundary is standard. Now add a 4-ball $B_0^4$, divided into balls $U^0_\pm$, where $\bar{U}^0_+ \cap \bar{U}^0_-$ is a standard $\bar{B}_0^3$.  Then $\bar{B}_0^3 \cup \bar{B}^3$ is an embedded sphere, and assuming the Schoenflies theorem, bounds balls on either side. Each of these balls is of the form $U^0_\pm \cup U_\pm$. But since we know that $U^0_\pm$ is a ball, its embedding is standard, so it follows that $U_\pm$ is a ball as well. 
The converse is also straightforward.  Given $S^3 \subset S^4$, dividing $S^4$ into regions $A_+ \cup A_-$, remove a small ball $B^4_0$ centered at a point of $S^3$. The complement of $B^4_0$ is a ball divided into regions $U_\pm$ as in your description, and by hypothesis each of $U_\pm$ is a ball. It follows that $A_\pm$ are balls as well.<|endoftext|>
TITLE: Why does the bitxor function appear in Nim?
QUESTION [15 upvotes]: I am conducting research in Combinatorial Game Theory (CGT). Although I have done a considerable amount of reading, I do not completely understand why the bit-xor function also known as the nim-sum appears in Nim. To be more clear, I will explain my current level of understanding. I completely grasp the mechanics of the proofs of both Bouton's Theorem and the Sprague-Grundy Theorem. However, I am looking for an intuitive explanation. 
I understand that the binary representations arise as a natural extension of the parity-mirroring arguments in two-heap Nim. However, I would like to know what particular properties of the bit-xor function allow it to work. For example, the bitwise-and would not work for Nim in the same way the bit-xor works; is there some functional equation that the bit-xor fulfills that makes it so ubiquitous?
Note that I am not asking for the motivation for the Sprague Grundy Theorem as in Motivation and Intuition for Sprague-Grundy Theorem. I am asking about the xor function in particular.
Thank you in advance.
By the way, this is my first question, so if anything is wrong, please critique me.
I feel that it it would be helpful to mention further context. I was reading A paper of Fraenkel and Kontorovich which used the bitwise and to explore Nim.

REPLY [3 votes]: Here's my understanding of the Nimbers and of your question.  Actually, you'll see that I do need to beg the question somewhere, but since you do know a proof of the bitxor rule, perhaps that's allowed --- then my answer can be understood as a proof that your proof implies your proof is natural.  (Said that way it sounds like an application of Lob's theorem, or perhaps a converse....)
By "Nimbers" I mean Nim games with Conway's game addition (put two games next two each other; on your turn, you choose one of the two boards to play on) modulo the second-player-win Games.  By definition, a "game" is one where you lose when you cannot make a turn.  The Nimbers are the classes of single-column Nim games.
The zeroth observation is that addition (henceforth "$+$") is commutative, and that for any game $g$, the game $-g$ in which the roles are reversed is its inverse.
The first observation, then, is that impartial games, and multi-column Nim games in particular, are 2-torsion: for any Nim game $g$, we have $g + g = 0$.  Thus the group generated by the Nimbers is a vector space over $\mathbb F_2$.
The next ingredient I don't really have an a priori reason for, which is that the sum of any two Nimbers is a Nimber. Actually, proving this is probably just about the same as finding the bitxor formula, so perhaps my whole story is question-begging.  But let's assume that this second ingredient is just an "observation".
The third observation is the following.  Let $G_k$ denote the group generated by the Nimbers $1,\dots,k$.  If you allow the second observation, then it is not hard to see that if $n \in G_k$, then for every $m < n$, $m\in G_k$.  Indeed, if $n\in G_k$, then I can write $n = \sum a_i$ for some sum of Nimbers with $a_i \leq k$.  Let's play the game $n + \sum a_i$, which is a second-player win by assumption.  Being magnanimous, I'll go first.  On my turn I turn $n$ into $m$.  Now you definitely have a move the return the sum to $0$.  It definitely doesn't involve the pile I touched, so it must involve dropping one of the $a_i$s to an $a_i' < a_i$.  But $a_i$ was one of our generators in $1,\dots,k$, and so $a_i'$ is also one of those generators.
Now we can put the observations together to describe the structure of the Nimbers.  We have $G_0 = \{0\}$ and $G_1 = \{0,1\}$ is the group of order $2$.  By induction, the set of Nimbers $G = \{0,1,\dots,2^k-1\}$ is closed under Nimber addition.  Consider $G_{2^k}$.  It is an $\mathbb F_2$-vector space generated by $G$, which has $2^k$ elements, and by one more element.  Thus $|G_{2^k}| = 2^{k+1}$.  Thus $G_{2^k} = \{0,\dots,2^{k+1}-1\}$.  The induction can then continue.
So the Nimbers are naturally organized as an $\mathbb N$-filtered $\mathbb F_2$-vector space: $$\{0\} \subset \{0,1\} \subset \{0,1,2,3\} \subset \{0,1,2,3,4,5,6,7\} \subset \dots \subset \{0,\dots,2^{k-1}\} \subset \dots.$$
This doesn't completely pin down the addition, but it makes bitxor seem very likely.  For example, it implies that if $m,n \in \{2^{k-1},\dots,2^k-1\}$, so that they have the same leading digit mod $2$, then their sum $m+n < 2^{k-1}$, and on the other hand if $m < 2^{k-1}$ and $n \in \{2^{k-1},\dots,2^k-1\}$, then $m+n \in \{2^{k-1},\dots,2^k-1\}$.  This gives the bitxor rule in the leading digit.
Of course, this analysis still allows lots of group structures on $\{0,\dots,2^k-1\}$.  The rule is only that the structure has to extend the one on $\{0,\dots,2^{k-1}-1\}$.  You can write down ad hoc group structures by twisting the given one by any permutation of $\{2^{k-1},\dots,2^k-1\}$.  To completely pin down the bitxor group law requires playing a bit more with the third observation, I think.  Let's see if we can do it.  We know that the bitxor rule applies to Nimbers $N < 2^k$, by induction.  We also know that $2^k + 2^k = 0$, so it applies to $N \leq 2^k$.  To prove the claim, it suffices to prove that the Nimber of height $2^k+2^j$ for $j<k$ is equal to the Nim addition $2^k + 2^j$.  Everything else will follow from linear algebra.  But now all I need to do is to tell you a second-player-win strategy for the three-column Nim game with heights $2^j$, $2^k$, and $2^k+2^j$.  Well, let's suppose you play on column $2^j$.  Then I have such a strategy by induction in $j$ (and linear algebra).  Suppose you play on column $2^k$.  Then you make it into something in $\{0,\dots,2^k-1\}$, and adding $2^j$ keeps me in that group, so I just need to drop the column of height $2^k+2^j$ to match.  Finally, suppose you play on column $2^k+2^j$.  If you leave it above $2^k$, I'll play on column $2^j$ to match, and use induction in $j$ to know that that works.  If you take it below height $2^k$, I'll drop the $2^k$-column to make the sum come out, using bitxor in the group $\{0,\dots,2^k-1\}$.
Perhaps this is the proof you already know.  It certainly is a follow-your-nose proof.<|endoftext|>
TITLE: Transcendence of $\Gamma(1/3), \Gamma(1/4)$
QUESTION [13 upvotes]: This is a re-post from MSE as I did not get even a single comment there.
Wikipedia mentions that the transcendence of $\Gamma(1/3), \Gamma(1/4)$ was proved by G. V. Chudnovsky. Does anyone have a reference to that proof? Or maybe some details on the essential ideas involved in the proof would be greatly helpful.
Also is there any simpler proof available when we restrict the conclusion to just "irrationality" instead of "transcendence"?

REPLY [15 votes]: Algebraic Independence of Values of Exponential and Elliptic Functions, G. V. Chudnovsky (1978)

(a.i. = algebraically independent)<|endoftext|>
TITLE: Do computational geometers use Lagrange multipliers?
QUESTION [6 upvotes]: Can anyone point me to an example of a problem that (more or less) originated in computational geometry whose solution requires the use of Lagrange multipliers (or Kuhn-Tucker conditions, or dual variables in a linear program, etc.)?  I have not been able to find such an example in any of the literature that I own.

REPLY [3 votes]: A convex optimization method for constructing a set of points in the plane with prescribed (combinatorial) Delaunay triangulation is given in 
Euclidean structures on simplicial surfaces and hyperbolic volume
I Rivin - Annals of Mathematics, 1994<|endoftext|>
TITLE: Understanding how to construct Bruhat-Tits buildings for non-split groups by Galois descent
QUESTION [7 upvotes]: Is there any way to get on top of the procedure for constructing Bruhat-Tits buildings for non-split groups over a non-archimedean local field $k$, by Galois descent, other than reading both the Bruhat-Tits articles "Groupes réductifs sur un corps local"? In particular is there any exposition which is written in English?

REPLY [8 votes]: There is a forthcoming book "Descent in buildings" by Bernhard Mühlherr, Holger Petersson and Richard Weiss, which should appear soon (published by Princeton University Press). See http://press.princeton.edu/titles/10649.html. It addresses exactly this issue, from a very building-theoretical point of view.

REPLY [6 votes]: Jiu-Kang Yu has an article "Bruhat-Tits Theory and Buildings" which has appeared in  "Ottawa Lectures on Admissible Representations of Reductive $p$-adic Groups". It doesn't directly address your question, instead it contains a guide to the literature on Bruhat-Tits buildings.
There is a 26 page pdf version of Yu's article. The published version contains a bit more information, but not a lot more (a few more references are added, plus a short introduction, plus the typesetting is updated). Unfortunately the printed version is  hard to find.<|endoftext|>
TITLE: Subsequence and integers as a sum of $\frac{1}{n}$
QUESTION [11 upvotes]: For all $M \in \mathbb{Z}$, is there a finite sequence of positive integers (not necessarily distinct) $(n_i)_{i \in I}$, s.t. $\sum_{i \in I} \frac{1}{n_i} = M$, and there is no subsequence $(n_i)_{i \in J}$ of $(n_i)_{i \in I}$ such that $\sum_{i \in J} \frac{1}{n_i}$ is an integer ?

REPLY [4 votes]: You have answered your question in the comments, just construct your sequence inductively.
Suppose you have a sequence $n_1,n_2,\ldots,n_k$ of pairwise relatively prime numbers. Put $q=n_1n_2\ldots n_k$. You put $q_i={q\over n_i}$ and find $d_i<n_i$ such that $d_iq_i+1=0$ mod $n_i$. This implies that $d_1q_1+d_2q_2+\ldots+d_nq_n+1=0\mathop{\rm mod}n_i$ for $i=1,2,\ldots,k$ hence $\mathop{\rm mod}q$. Dividing by $q$ we get some integer $$M={d_1\over n_1}+{d_2\over n_2}+\ldots+{d_k\over n_k}+{1\over q}$$.
So far this was your argument.
Now pick any $n_{k+1}$, $1<n_{k+1}<q$, relatively prime to every $n_i$ for $i\leq k$. Find $n_{k+2}<q$ such that $n_{k+1}n_{k+2}=1\mathop{\rm mod}q$. The numbers $n_{k+1}$ and $n_{k+2}$ may have a common factor but you may prevent that by choosing $n_{k+1}$ to be a prime bigger than $q/2$ (Edit: and such that $n_{k+1}^2\neq 1\mathop{\rm mod} q$, for example of the form $5r\pm 2$ if $5|q$, see comments by pallab1234 below). Put $q'=qn_{k+1}n_{k+2}=n_1n_2\ldots n_{k+2}$. When you find $d_i<n_i$ such that $d_iq'_i+1=0$ mod $n_i$ for $i=1,2,\ldots,{k+2}$ you see that those for $i\leq k$ didn't change. You obtain another integer 
$$
M'={d_1\over n_1}+\ldots+
{d_k\over n_k}+{d_{k+1}\over n_{k+1}}+{d_{k+2}\over n_{k+2}}+{1\over q'}=
M+\Delta
$$
where
$$
\Delta={d_{k+1}\over n_{k+1}}+{d_{k+2}\over n_{k+2}}+{1\over q'}-{1\over q}=
{d_{k+1}\over n_{k+1}}+{d_{k+2}\over n_{k+2}}+
{1\over q}\left({1\over n_{k+1}n_{k+2}}-1\right)
$$
must be an integer. Clearly $\Delta<2$. On the other hand
$$
\Delta>{1\over n_{k+1}}+{1\over n_{k+2}}-{1\over q}>{1\over n_{k+2}}-{1\over q}
$$
which is positive since $n_{k+2}<q$. Being an integer $\Delta=1$ hence $M'=M+1$.
Edit: deleted.
Edit2: Thus you may start with $n_1=3$ and $n_2=5$ for which you get $\displaystyle M={1\over 3}+{3\over 5}+{1\over 3\cdot 5}=1$ and go up to any positive integer you want.<|endoftext|>
TITLE: Gauss-Milgram formula for fermionic topological order?
QUESTION [7 upvotes]: For Bosonic topological order, a very useful formula was proved to be true:
$\sum_a d_a^2 \theta_a=\mathcal{D} \exp(\frac{c_-}{8}2\pi i) $
(for more detail: $d_a$ is the quantum dimension of anyon labeled by a, and $\theta_a$ is the topological spin.D is the total quantum dimension, $\mathcal{D}^2=\sum_a d_a^2$. And $c_-$ is the chiral central charge. If we assume bulk boundary correspondence, $c_-$ can be defined as $c_-=c_L-c_R$, the chiral combination of the central charge of boundary CFT. Alternatively, the chiral central charge is also well defined without referring to CFT, that is via the thermal Hall effect when we have an edge termination.)
So my question is straightforward: what's the fermionic version of this formula?
I also post this question in physics stackexchange: https://physics.stackexchange.com/questions/190902/fermion-version-of-gauss-milgram-sum

REPLY [3 votes]: We just posted a paper http://arxiv.org/abs/1507.04673 addressing this issue. For fermion topological orders, the fermionic version of this formula is 
$\Theta=\sum_a d_a^2 \theta_a=0$. See eq. 14 of the paper. So we cannot use eq. 14 to compute the chiral central charge of the fermionic topological orders. We have to use the bosonic extension of the fermionic topological orders to compute the chiral central charge of the fermionic topological orders.<|endoftext|>
TITLE: How do I evaluate this sum :$\sum_{n=0}^{\infty} \frac{\sin(n!)}{\cos(n!)}$ if it's not open problem?
QUESTION [6 upvotes]: I proposed this question on MSE but some comments affirmed that is  unsolved problem and no answer. I would like to see what MO say about it.
How do I evaluate this sum :$$\displaystyle\sum_{n=0}^{\infty} \frac{\sin(n!)}{\cos(n!)}$$
Note : I used many criterions of convergence to test whether it converges but i didn't succeed.
Thank you for any help .

REPLY [5 votes]: I computed $f(n)=\sum_{k=0}^n \tan(k!)$ for different $n$,
and got the following plot. It does not seem to have a limit.

Mathematica code:
Block[{$MaxExtraPrecision = 600},
 lst = Table[Tan[n!], {n, 0, 200}];
 ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
 ]<|endoftext|>
TITLE: Volume-minimizing submanifold implies calibrated?
QUESTION [13 upvotes]: Let $X$ be a smooth manifold of dimension $d$ and $M$ an oriented
submanifold of dimension $p < d$ so that the multiples k⋅M  are absolutely minimizing $p$-volume in their integral homology classes for all k∈Z .  Is $M$ calibrated by some
$p$-form w? (Thanks to Robert Bryant for correcting my initial question.)

Definitions: A $p$-form $w$ is called a calibration if it is closed and its evaluation on every geometric $p$-vector $v$ of norm 1 is at most 1 in norm, $|w(v)| \le 1$. One says a $p$-dimensional submanifold $M$ is calibrated (by a calibration $w$) if $w$ evaluates to 1 on each unit tangent $p$-vector to $M$. The word "geometric" above is used to distinguish primitive (or geometric) $p$-vectors from linear combinations of these; "geometric" means "rank one" in tensor language.
Note: It is elementary that if $M$ is calibrated by any $w$ then $M$ minimizes $p$-volume in its homology class, so I'm asking for a kind of converse. In the case $p+1=d$ the converse amounts to a continuum version of Max Flow/Min Cut which is discussed in John Sullivan's 1990 Princeton Ph.D. thesis. I tried asking a form of this question last year, but I did not use the term "calibration". I'm hoping with the correct language an expert will notice the question and be able to answer.
The question below is a good one. I not even know if one can "locally calibrate" some neighborhood of stable oriented minimal submanifold with oriented normal bundle. The local question may contain the important difficulties. A simple closed stable geodesics on a surface can run through areas of positive curvature, so even in that case, the local calibration is not found simply by pulling the length form of the geodesic back along normal coordinates.

REPLY [3 votes]: Just a related comment:
Not long ago Dima Burago and Sergei Ivanov showed that if one considers $\mathbb{R}^n$ with a translation-invariant $k$-area integrand (a $k$-area density), then the condition that all flat $k$-discs AND their multiples are area-minimizing is equivalent to the ellipticity (convexity) of the integrand (= all $k$-discs are calibrated by constant-coefficient forms). 
http://link.springer.com/article/10.1007%2Fs00039-004-0465-8?LI=true#page-1<|endoftext|>
TITLE: The stack of group algebraic spaces
QUESTION [5 upvotes]: The fibred category $\mathcal A$ of algebraic spaces over a scheme $S$ is a stack (over the category of affine schemes with the etale topology). This is proved in Laumon and Moret-Bailly's book (see (1.6.4) and (3.4.6)).
Let $\mathcal G$ be the fibred category of group algebraic spaces over a scheme $S$. Is $\mathcal G$ a stack? 
My guess is that the forgetful functor $\mathcal G\to \mathcal A$ is "representable" in some sense (as the stabilizers of $\mathcal G$ are smaller than those of $\mathcal A$). But I can't see how to make this rigorous.
Sidenote. Note that $\mathcal A$ is not an algebraic stack (see Claim 3.1 in http://arxiv.org/pdf/math/0602646v1.pdf).

REPLY [3 votes]: Here's a sketch proof. First, recall that inside the category of presheaves, sheaves are closed under taking limits. Second, inside the set of maps between the underlying algebraic spaces, the group homomorphisms are constructed by taking an equaliser:
$$
Hom_{Grp}(G,H) \to Hom_{Sp}(G,H) \rightrightarrows Hom_{Sp}(G\times G,H) \times Hom_{Sp}(\ast,H).
$$
Thus the condition that the hom-presheaf is a sheaf holds for the fibred category of group algebraic spaces, and so at the very least it is a prestack.
I claim that one can use the existence part of the universal property of descent (consider the map of descent data that is the multiplication map on the underlying groups) to show that given descent data for group algebraic spaces, the descended underlying algebraic space inherits a multiplication, and again by universality (uniqueness, this time), this makes the multiplication that of a group object. I can fill in details if needed, just not right now.<|endoftext|>
TITLE: Monge–Ampère with drift
QUESTION [6 upvotes]: Let $I\subseteq \mathbb{R}$ be an interval. 
Let smooth $M(x,y):I\times(0,\infty) \to \mathbb{R}$ satisfies PDE:
$$
M_{xx}M_{yy}-M_{xy}^{2}+\frac{M_{y}M_{yy}}{y}=0.
$$
My question is to describe/characterize solutions of this PDE. 
Here are few nontrivial examples which solve the PDE:
\begin{align}
&1)\quad M(x,y)=x\ln x-\frac{1}{2}\frac{y^{2}}{x};\\
&2)\quad M(x,y)=x^{2}-y^{2};\\
&3)\quad M(x,y)=\sqrt{[\Phi'(\Phi^{-1}(x))]^{2}+y^{2}} \quad \text{where}\quad \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{t^{2}}{2}}dt.
\end{align}
Update 1:
I have few more questions regarding Robert Bryant's answer:
1) (Philosophical question) Which PDE's $F(M, M_x, M_y, M_{xx}, M_{xy}, \ldots)=0$ can be "encoded" as an exterior differential system on $\mathbb{R}^{m}$? 
2)  (Incorrect question) After having  exterior differential system, how do you come up with "right" change of variables? Does it mean that you tried all possible (and reasonable) ones  and this was the best one? Maybe this question is not correct because I did not specify what is the meaning of ``right change of variables''. In this case goal was backward heat equation. 
3) (Technical question) What happens if $p<0$? I think nothing changes all you need is requirement that $p\neq 0$. Am I Right?
4) (The most important question) Does this "local" knowledge actually allow me to write down at least one global nontrivial solution? For example, can you somehow trace back to the solution 3) that I mentioned above?  You can assume some boundary conditions. 
Update 2:
1) Accepted. 
2) Accepted. 
3) Accepted. P.S. I had in my mind to consider $-M(x,y)$ instead of $M(x,y)$. 
4) ACCEPTED! By the way my first solution corresponds to the case $u(p,t)=\frac{t}{2}-\frac{p^{2}}{4}$. My second solution corresponds to the case $u(p,t)=-e^{-t-p-1}$.

REPLY [8 votes]: Perhaps the following observations will be of use to you:  First, for any (local) solution $M(x,y)$ of your equation on a simply-connected open domain $D\subset \mathbb{R}\times(0,\infty)$ in the $xy$-plane, consider its $1$-graph 
$$
(x,y,p,q)=\bigl(x,y,M_x(x,y),M_y(x,y)\bigr)
$$
in $xypq$-space.  This is a simply-connected surface $\Sigma$ in $4$-space on which $\Omega = \mathrm{d}x\wedge\mathrm{d}y$ is nonvanishing, but to which the two $2$-forms
$$
\Upsilon_1 = \mathrm{d}p\wedge\mathrm{d}x + \mathrm{d}q\wedge\mathrm{d}y
\qquad\text{and}\qquad
\Upsilon_2 = \bigl(y\,\mathrm{d}p+ q\,\mathrm{d}x\bigr)\wedge\mathrm{d}q
$$
pull back to be zero.  
Conversely, suppose given a simply connected surface $\Sigma$ in $xypq$-space (with $y>0$) on which $\Omega$ is nonvanishing but to which $\Upsilon_1$ and $\Upsilon_2$ pullback to be zero.  The $1$-form $p\,\mathrm{d}x + q\,\mathrm{d}y$ pulls back to $\Sigma$ to be closed (since $\Upsilon_1$ vanishes on $\Sigma$) and hence exact, and therefore there exists a function $m:\Sigma\to\mathbb{R}$ such that $\mathrm{d} m = p\,\mathrm{d}x + q\,\mathrm{d}y$ on $\Sigma$.  We then have (at least locally), $m = M(x,y)$ on $\Sigma$ and, by its definition, we have $p = M_x(x,y)$ and $q = M_y(x,y)$ on the surface.  Then the fact that $\Upsilon_2$ vanishes when pulled back to $\Sigma$ implies that $M(x,y)$ satisfies the desired equation.
Thus, we have encoded the given PDE as an exterior differential system on $\mathbb{R}^4$.  Note, though, that we can make a change of variables on the open set where $q>0$:  Set $y=qr$ and let $t=\tfrac12 q^2$.  Then, using these new coordinates on this domain, we have
$$
\Upsilon_1 = \mathrm{d}p\wedge\mathrm{d}x + \mathrm{d}t\wedge\mathrm{d}r
\qquad\text{and}\qquad
\Upsilon_2 = \bigl(r\,\mathrm{d}p+ \mathrm{d}x\bigr)\wedge\mathrm{d}t.
$$
Now, when we take an integral surface $\Sigma$ of these $2$-forms on which $\mathrm{d}p\wedge\mathrm{d}t$ is nonvanishing, it can be written locally as a graph of the form 
$$
(p,t,x,r) = \bigl(p,t,u_p(p,t),u_t(p,t)\bigr)
$$
(since $\Sigma$ is an integral of $\Upsilon_1$), where $u(p,t)$ satisfies $u_t + u_{pp} = 0$ (since $\Sigma$ is an integral of $\Upsilon_2$).  Thus, 'generically', your equation is equivalent to the backwards heat equation, up to a change of variables.   
(Note, by the way, that an integral surface of the $\Upsilon_i$ on which $\mathrm{d}p\wedge\mathrm{d}t$ vanishes identically must also have $\mathrm{d}t$ vanishing identically, since, otherwise, because $\mathrm{d}x\wedge\mathrm{d}t$ vanishes identically (since $\Upsilon_2$ vanishes), one would have $\mathrm{d}x$ and $\mathrm{d}p$ be multiples of $\mathrm{d}t$, which would then force $\mathrm{d}r\wedge\mathrm{d}t$ to vanish identically (since $\Upsilon_1$ vanishes), and hence each of $x,p,r$ would be functions of $t$ and this is impossible for a surface.  Thus, integral surfaces on which $\mathrm{d}p\wedge\mathrm{d}t$ vanishes identically are necessarily locally of the form
$$
(p,t,x,r) = \bigl(P(s),t_0,X(s),r\bigr)
$$
where $t_0$ is a constant and $r$ and $s$ are coordinates on the surface and $\bigl(P'(s),X'(s)\bigr)$ is never $(0,0)$.)
Thus, in some sense, your nonlinear Monge-Ampère equation is simply the (linear) backwards heat equation in disguise.  You should be able to use this to generate many nontrivial solutions and to solve the appropriate characteristic initial value problem.
Responses to updated question
1) In principle, any PDE can be encoded as an exterior differential system, often in more than one way.  There is a standard way to do it for Monge-Ampère equations, though, so that such an equation for one function of two variables can be written as an EDS on $\mathbb{R}^5$. For example, see the book Exterior Differential Systems by Bryant, et al for how this is usually done.
2) The first step is to determine the equivalence class of the EDS up to diffeomorphism (i.e., change of variables).  Not all Monge-Ampère equations are equivalent, but there are tests for when it is parabolic, when it is equivalent to a linear equation, etc.  In this particular case, I computed the conversation laws for the equation and realized that this space had infinite dimension, so I knew that it must be equivalent to a linear equation, in fact, the backwards heat equation.  For example, see my paper with Phillip Griffiths, Characteristic cohomology of differential systems II: Conservation laws for a class of parabolic equations, Duke Math. J. Volume 78, Number 3 (1995), 531-676.
3)  I think you mean $q<0$, not $p<0$.  In that case, yes, in fact, the mapping $(x,y,p,q)\mapsto(x,-y,p,-q)$ switches the domains $q>0$ and $q<0$ in $\mathbb{R}^4$ and preserves the EDS, so they are equivalent.
4) Yes.  Here are a couple of examples.  First,, if you follow the transformation through, you'll see that a solution $u(p,t)$ of $u_t+u_{pp}=0$ on some domain in $pt$-space corresponds to a solution of your original equation whose graph in $xyz$-space is the parametrized surface
$$
\begin{aligned}
x &= u_p(p,\tfrac12q^2)\\
y &=q u_t(p,\tfrac12q^2)\\
z &= p\,u_p(p,\tfrac12q^2) +q^2\,u_t(p,\tfrac12q^2) - u(p,\tfrac12q^2)
\end{aligned}
$$
You just need to choose your solution $u(p,t)$ on a $pt$-domain so that this surface is the graph of the form $z = M(x,y)$ over an appropriate domain and then $M$ will satisfy your
equation.  
For a simple example, take $u(p,t) = pt - \tfrac16 p^3$.  Running this through the above process yields in the solution
$$
M(x,y) = \tfrac23\bigl(\sqrt{x^2+y^2} + 2x\bigr)\bigl(\sqrt{x^2+y^2} - x\bigr)^{1/2}.
$$
As another example, take $u(p,t) = e^t\,\sin p$ on the open $pt$-strip $ \tfrac12\pi < p < \pi$ and $t>0$. Then the above surface in $xyz$-space is a graph $z = M(x,y)$ over the second quadrant in the $xy$-plane, so this gives a nontrivial solution $M(x,y)$ in the $xy$-domain where $x<0$ and $y>0$.  I believe that this solution is not the same as any of the solutions you have written down above.  In fact, I don't see how to get an 'explicit' formula for $M(x,y)$, i.e., to eliminate the parameter variables $p$ and $q$.<|endoftext|>
TITLE: Length of nearest neighbor path in travel salesman problem
QUESTION [7 upvotes]: Given $n$ nodes uniformly distributed in $[0,1]^2$, consider the nearest neighbor algorithm to solve traveling salesman problem, i.e., each time I select the nearest neighbor not visited so far as the next node to visit, my question is: what is the expected length of such TSP tour?

REPLY [6 votes]: A careful experimental analysis of the nearest-neighbor (NN) heuristic (among other heuristics) is described in this paper:

Johnson, David S., and Lyle A. McGeoch. "The traveling salesman problem: A case study in local optimization." Local search in combinatorial optimization 1 (1997): 215-310.
  (PDF download link.)

They find that the growth rate, in comparison to the Karp-Held lower bound, 
"appears to be proportional to $\log n$" for "random distance matrices."
This is the theoretical growth rate w.r.t. the optimum.
In contrast, for "random Euclidean instances" up to $n=10^6$, they find 
(Table 1, p.15) that the NN heuristic
leads to paths nearly constantly about 25% longer than the Karp-Held lower bound.
This is closer (but not identical) to the OP's "uniformly distributed within $[0,1]^2$."
The section, "Standard Test Instances" on pp.12-14 unpacks
"random Euclidean instances."
Incidentally, the observed running time grows subquadratically.
(See also the follow-up MO question, 
"Travelling Salesman Problem: Can the nearest neighbor algorithm be n times longer than the optimal solution?.")<|endoftext|>
TITLE: On the number of ends of a countable simple group
QUESTION [8 upvotes]: At the beginning I thought that the following statement could be an easy exercise after Stallings' theorem, but I found myself incapable of proving it:
Any countable f.g. simple group has one end.
It is obvious that a f.g. simple group cannot have two ends, as $\mathbb Z$ has many quotients. If it has infinitely many ends, 1) it can't be an HNN extension over a finite group, since it is a semi-direct product with a surjection on $\mathbb Z$, also 2) it can't be a free product $A*B$ since it surjects onto $A\times B$. So I'm left with the case of an amalgamated product over a non-trivial finite group.
Maybe I'm wrong, maybe there are example of f.g. groups with infinitely many ends... do you know some example?
EDIT
This is something that I can easily say studying the amalgamated product of two simple groups.
Let us consider the group $G=M_1*_{C}M_2$ be the amalgamated product of two simple groups. Let $\phi:G\to H$ be a nontrivial morphism which is not injective. Then the restriction $\phi_i=\phi\vert_{M_i}:M_i\to H$ is either trivial or injective, for $M_i$ is a simple group. It is not possible that $\phi_1$ and $\phi_2$ are both trivial, since $M_1$ and $M_2$ generate $G$. Also, if $\phi_1$ is trivial, then the copy of $C$ in $M_2$ is in the kernel of $\phi_2$, so $\phi_2$ must be trivial because $M_2$ is simple. As a consequence, both $\phi_1$ and $\phi_2$ are injective.

REPLY [10 votes]: Let $G$ be a group with infinitely many ends. According to Stallings' theorem, $G$ splits non trivially over a finite subgroup. Now, consider the action of $G$ on the associated Bass-Serre tree $T$. Because the edge stabilizers are finite, it is clear that the action $G \curvearrowright T$ is acylindrical. Furthermore, since there at most two orbits of vertices and that $T$ has necessarily infinitely many ends, we deduce that the limit set $\Lambda(G) \subset \partial T$ has infinitely many points, ie. the action $G \curvearrowright T$ is non elementary. Since $G$ has a non elementary acylindrical action on a hyperbolic space, $G$ is acylindrically hyperbolic (as defined by Osin), and is in particular SQ-universal (according to a result of Dahmani-Guirardel-Osin). This implies that $G$ has uncountably many normal subgroups, so that $G$ is far from being simple.
For more information on acylindrically hyperbolic groups, see Osin's article.
EDIT 1: As suggested by Yves Cornulier, these groups are in fact relatively hyperbolic (a stronger property than acylindrical hyperbolicity if the group is not virtually cyclic). It is essentially a consequence of the following criterion due to Bowditch:
Theorem: A group $G$ is hyperbolic relative to a collection $\mathcal{G}$ of infinite subgroups if it acts on a connected graph $\Gamma$ such that

$\Gamma$ is hyperbolic and fine (ie., for every $n \geq 1$, an edge belongs to finitely many simple cycles of length $n$),
there are finitely many orbits of edges, whose stabilizers are finite,
the elements of $\mathcal{G}$ are precisely the infinite vertex stabilizers of $\Gamma$,
every element of $\mathcal{G}$ is finitely generated. 

Notice that, if the vertex stabilizers are finite, then the action $G \curvearrowright \Gamma$ is properly discontinuous and cocompact, so that $G$ turns out to be hyperbolic.
Thus, because a finitely generated group $G$ with infinitely many ends splits non trivially over a finite subgroup, we deduce from the action on the associated Bass-Serre tree that $G$ is hyperbolic relatively to the factors of this splitting.
EDIT 2: In his article SQ-universality of free products with amalgamated finite subgoups, Lossov proves that the amalgamated product $A \underset{C}{\ast} B$ is SQ-universal provided that $[A:C] \geq 2$ and $[B:C] \geq 3$ (it corresponds to the case where the group has infinitely many ends). However, I did not find a similar reference for HNN extensions.<|endoftext|>
TITLE: Complex structure on $S^6$ gets published in Journ. Math. Phys
QUESTION [76 upvotes]: A paper by Gabor Etesi was published that purports to solve a major outstanding problem:
Complex structure on the six dimensional sphere from a spontaneous symmetry breaking
Journ. Math. Phys. 56, 043508-1-043508-21 (2015) journal version, current arXiv version.
Since this is obviously an important and groundbreaking 
result (if true), published in a physical journal, I 
am interested whether it is accepted by mathematicians.

REPLY [20 votes]: Here is an answer to the question of YangMills (thank you!) regarding the subbundle $H \subset TG_2$:
I did not claim in the text that the subbundle $H \subset TG_2$ should be integrable in any sense. What I only need is the formal fact that the functional in eq. (15) vanishes on the constructed triple $(\nabla_H,J_H,g_H)$ as can be verified by a direct calculation. However I acknowledge that YangMills was right and the Levi--Civita connection does not preserve $H\subset TG_2$ as it was written in the text. For a corrected version please visit:
http://www.math.bme.hu/~etesi/s6-spontan.pdf
Neverthekess after this observation the proof proceeds as follows: $J_H$ is Fourier expanded and the corresponding ground mode, denoted by $J$ in the text after eq. (21), descends to $S^6$. The very important subtlety however is this (explained carefully in Section III): in our situation (i.e., Fourier expanding general sections of general vector bundles), there is NO canonical way to perform a Fourier expansion. Instead there is "moduli space" of possible Fourier expansions resulting in inequivalent ground modes. This is because doing fiberwise integration does NOT commute with gauge transformations hence Fourier expanding a gauge transformed (on $H$) section is NOT the same as gauge transforming (on $TS^6$) the ground mode of a Fourier expanded section. I construct in Lemma 5.1 a distinguished  "$\alpha$-twisted" Fourier expansion of whose ground mode $J$ coincides with $J_H$ itself.   
I think that most of the concerns and uncertainty about the published version is related with the historical remark that the "relationship" between  Yang--Mills theory (mathematically invented in the 1980's) and classical complex manifold theory, more precisely the Kodaira--Spencer deformation theory (invented in the 1950-60's) is not fully clarified. By this I mean that apparently the action of the gauge group on an (almost) complex manifold $(M,J)$ is dubious: it can describe both just a symmetry transformation of $(M,J)$ or an effective deformation of $(M,J)$. But these certainly should be carefully distinguished. My suggestion is formulated in the "Principle" of Section II (but this point might require a more conceptional and less ad hoc work, I agree).<|endoftext|>
TITLE: Deformations of Ext rings
QUESTION [9 upvotes]: Let $k$ be a base ring and $k[x]$ the ring of polynomials in an indeterminate $x$ over $k$.  Consider a (not necessarily commutative) algebra $A$ over $k[x]$ and two $A$-modules $M$ and $N$.  Then for each element $q \in k$, we obtain specialisations
$$A_{q} := A\otimes_{k[x]} k, \; M_{q} := M\otimes_{k[x]} k,\; \mbox{ and } N_{q} := N\otimes_{k[x]} k$$ 
where $k$ becomes a $k[x]$-module by sending $x$ to $q$.  Then $A_q$ is a $k$-algebra, and $M_{q}$ and $N_{q}$ are $A_q$-modules.
I'm interested in how the groups $Ext_{A_q}^*(M_q, N_q)$ vary as $q$ ranges over $k$.  
My somewhat vague question is: if I know that a property holds for these Ext groups for a Zariski dense subset of elements $q \in U \subseteq \mathbb{A}^1(k)$ (which I suppose is simply fancy language for "all but finitely many elements of $k$"), can I conclude anything about these groups for $q \notin U$?  Properties that I'd be interested in include: finite cohomological dimension, finite generation, etc.
Perhaps less vague: are there any known techniques for getting at these groups for $q \notin U$?

REPLY [2 votes]: Let $A = k[x, y, z]/(yz - x^2)$ with augmentation $x \mapsto x$, $y \mapsto x$, and $z \mapsto x$ and with $M = N = k[x]$. Then $M_q = N_q = k$ and we are taking the ext groups over a nonsingular algebra when $q \not = 0$, but over a singular one if $q = 0$. Namely, for $q = 0$ we get $k[y, z]/(yz)$ and infinitely many of the Ext groups are nonzero, whereas for $q \not = 0$ you only have $2$ nonzero Ext groups.
This is not an answer but just an example showing that things can jump.<|endoftext|>
TITLE: Specialisation of rigid varieties
QUESTION [9 upvotes]: Recall that a variety $X$ over a field $k$ is called rigid if $H^1(X, T_X) = 0$. I am interested in understanding this property under specialisation.

Let $R$ be a discrete valuation ring and let $\pi: X \to \mathrm{Spec }R$ be a smooth proper morphism with geometrically irreducible generic fibre. Assume that the generic fibre of $\pi$ is rigid. Then is the special fibre of $\pi$ also rigid?

Note that basic facts in deformation theory show that if the special fibre is rigid then the generic fibre is also rigid. I am interested in the converse.

REPLY [2 votes]: Since in Jason Starr's answer (and comments) the Fano case has been mentioned, maybe this is worth noting:
A K-polystable Fano manifold X cannot specialise (in a $\mathbb{Q}$-Gorenstein family with general fiber isomorphic to X) to a klt K-polystable Fano variety Y different from X itself.
This follows by the fact that $\mathbb{Q}$-Gorenstein smoothable K-polystable Fano varieties over $\mathbb{C}$ form separated (and compact) moduli spaces. See the arXiv pre-prints by Li-Wang-Xu, Spotti-Sun-Yao and Odaka.  Note that the proof of such "algebraic" statement is, at present, highly transcendental, since it is based on the fact that smooth K-polystable Fanos admit (unique) Kahler-Einstein metrics, and vice versa.
An example (under the rigidity hypothesis): the klt $\mathbb{Q}$-Gorenstein degenerations of $\mathbb{P}^2$ (which is rigid and K-polystable since the Fubini-Study metrics is KE) have been classified by Hacking and Prokhorov http://arxiv.org/abs/math/0509529. Such degenerations are given by weighted projective planes $\mathbb{P}(a^2,b^2,c^2)$, where $a,b,c$ satisfies the Markov equation $a^2+b^2+c^2=3abc$ (and their partial $\mathbb{Q}$-Gorenstein smoothings). It is well-known that, beside the case of $\mathbb{P}^2$ itself, such (coarse) spaces do not admit singular (with isolated orbifold sing.) Kahler-Einstein metrics (hence they are not K-polystable).<|endoftext|>
TITLE: How to prove that this equation has only one solution?
QUESTION [7 upvotes]: I can't find a way to prove that the following equation has only one solution :
$$
X = \frac{2^Q - 1}{2^{P+Q} - 3^P}
$$
with $X,P,Q$ integers $> 0$.
One trivial solution is $X = 1, P = 1, Q = 1$.
Does anyone has an idea ?
Best regards

REPLY [7 votes]: I'm more used to the formulation in the following form:
$$ X(2^{P+Q} - 3^P)=2^Q-1 \\
 2^Q(2^Px -1) = 3^Px -1 $$
and then
$$ 2^Q = {3^P \cdot X - 1 \over 2^P \cdot X - 1} \tag 1
$$
and Ray Steiner has proved in 1976 in the context of the Collatz-problem (using Rhin's result given in the other answer), that there is only one solution (which you've already noticed). A bit more about this (and the references) can be found in the wikipedia-article on the Collatz-problem and also in a remark in Lagarias' survey about the research in the Collatz-problem.     
Footnote: The Waring-problem (which I mentioned in my comment just a minute ago) leads to a small modification; there the lhs is only required to be integer (instead of a power of 2) and even this conjecture (that there are no solutions for $P \gt 6$ ) seems to hold.

REPLY [4 votes]: Unless $3^P$ is very close to $2^{P+Q}$, the right hand side will be smaller than 1. Hence the linear form $(P+Q)\log 2 - P\log 3$ is exceptionally small, and you should be able to obtain effective upper bounds for $P$ and $Q$ by Baker's method. Looking at the continuous fraction of $\frac{\log 2}{\log 3}$ you can probably reduce the upper bound to a range where you can check everything using a computer.<|endoftext|>
TITLE: Can I find the gap between the two least eigenvalues of this special matrix A(t)?‎
QUESTION [7 upvotes]: I am interested in finding the gap between the two least eigenvalues of $A(t)$, a Hermitian $N\times N$ sparse ‎matrix whose diagonal elements are $a_it+b_i\,(1\leq i\leq N)$, and all off-diagonal non-zero elements ‎are the same and equal to $ct+d$. Furthermore, $A(t)$ is a banded matrix, that is, $a_{ij}=0$ for $( 1\leq i\leq N) \& (j>i+k)$ and $(1\leq j\leq N) \& (i> j+k)$, with $k<N$ fixed. Since real-scaled $N$ is larger than $10^6$, determining ‎the eigenvalues and then calculating the gap $(g)$ is not practical. Thus, even bounding the gap can ‎be satisfactory.‎ 
However, interlacing theorem and Courant-Weyl inequalities provide some bounds for the ‎eigenvalues, but it is not enough since they involve the most and the least eigenvalues, so the ‎bounds are to some extent wide. Another note I should mention is that while I need to calculate $g$ ‎for a range of values for $t$, so my main (and optimistic) goal is to derive $g$ in terms of $t$. However, ‎calculation of the gap (or its bounds) for some fixed $t_0 \neq 0$ can also do some good. 
Now my question is that, according to the structure of $A(t)$, can I calculate $g$ or derive any better bound ‎for it? Is there any sharp inequality that can help?‎
Any other useful comment would be appreciated. ‎
Edit: To be more illustrative, here is an example of $A(t)$ (with $N=8$): ‎
‎$$\begin{bmatrix}‎
‎ 3-t & t-1 & t-1 & 0 & t-1 & 0 & 0 & 0 \\‎
‎ t-1 & 3 (t+1) & 0 & t-1 & 0 & t-1 & 0 & 0 \\‎
‎ t-1 & 0 & 3 (t+1) & t-1 & 0 & 0 & t-1 & 0 \\‎
‎ 0 & t-1 & t-1 & 3-3 t & 0 & 0 & 0 & t-1 \\‎
‎ t-1 & 0 & 0 & 0 & 7 t+3 & t-1 & t-1 & 0 \\‎
‎ 0 & t-1 & 0 & 0 & t-1 & 3-t & 0 & t-1 \\‎
‎ 0 & 0 & t-1 & 0 & t-1 & 0 & t+3 & t-1 \\‎
‎ 0 & 0 & 0 & t-1 & 0 & t-1 & t-1 & 5 t+3 \\‎
‎\end{bmatrix}$$‎
and the plot for $g(t)$ for this example with $-1\leq t\leq 1 $ is:‎
           

Edit: Here is a long shot of an almost real-scaled example of $A(t)$ (with $N=32768$). While all off-‎diagonal non-zero elements are the same, Indeed, diagonal elements still vary from others (and ‎may be from each other) -- though this is not visible at this scale.‎

REPLY [5 votes]: You say that computing the gap for a fixed $t_0$ could also be good. I think I can do $t_0=0$ in a way that generalizes to examples of larger dimensions. (You haven't defined your matrix exactly for larger dimensions, but that is how I assume it works).
That matrix for $t=0$ is $A(0) = B\otimes I\otimes I + I\otimes B \otimes I + I\otimes I \otimes B$, where $I$ is the ($2\times 2$ in this example) identity matrix and $B=\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}$. Here $\otimes$ is the Kronecker product. By the properties of the Kronecker product, The eigenvalues of such a matrix are given by $\{\lambda_i+\lambda_j+\lambda_k: i,j,k \in \{1,2\}\}$, where $\lambda_1,\lambda_2$ are the eigenvalues of $B$. So in this case you get 0,2,2,2,4,4,4,6.
I imagine that what happens for larger dimensions is that $B$ is $I+L$, where $L$ is the tridiagonal matrix with (-1,0,-1), for which eigenvalues are known explicitly as $\lambda_j = 2\cos(j\pi/(n+1))+2$, $j=1,2,\dots,n$. So all eigenvalues are known and the gap is $\lambda_n-\lambda_{n-1}$.
This stuff is quite standard in numerical analysis (the key terms are "3d finite difference Laplace matrix").
I suspect that a similar trick may work for a generic $t$, but it looks tricky to make sense of the coefficients of $t$ on the diagonal. Do they come out of a formula?<|endoftext|>
TITLE: Number of $\mathbb F_p$ points constant mod $p$?
QUESTION [15 upvotes]: I have some affine varieties $X$ defined over $\mathbb Z$, and associated integers $c(X)$, with the property that $\# X_{\mathbb Z/p} \equiv c(X) \bmod p$ for all $p$. (In particular $c(X)$ is usually $1$ in my examples, but not always.) 

Is this property related to any other interesting properties of $X$, e.g. "ordinary", "unirational", etc.?

I am happy to entertain conjectures, necessary conditions, sufficient conditions, anything. I looked at the 33 occurrences of the word "constant" in Serre's Lectures on $N_X(p)$, but haven't delved deeper into that book.

REPLY [5 votes]: Nobody seems to have mentioned Fulton's (?) trace formula. It says that the number of points mod p equals the alternating trace of Frobenius on coherent cohomology $H^i(X, \mathcal{O})$.  So - and this is probably exactly the same as Sawin's point- the easiest reason that the number of points is congruent to $1$ modulo $p$ is that all the higher cohomology of the structure sheaf vanishes.<|endoftext|>
TITLE: Optimal strategy for game of 'online sorting' into a poset
QUESTION [9 upvotes]: Consider a single-player game played with an arbitrary finite poset, and a random number generator with a known distribution:
Each turn, the RNG produces a number, and the player must assign that number to an element of the poset as a label.  A new label may replace one already present, and each individual assignment need not respect the partial ordering. However, the goal of the player is to minimize the expected number of turns needed to label all elements of the poset such that the labels are consistent with the relation. 
Is there a demonstrably optimal strategy more elegantly expressible than a brute force game tree analysis?  (particularly in the case where the RNG has a continuous distribution?)  
Has this or a similar process already been considered in some other guise?  \
Edit to add:
Do posets with the same counts of elements and linear extensions necessarily share a common optimal expected time to completion?

REPLY [4 votes]: I worked out the case of a two-element chain and the uniform distribution on [0,1].  This was more complicated than I expected, and although the same sort of approach should allow one to compute chains of length 3, 4, 5, etc., it looks messy.  However, maybe someone else will see a clean way to argue the general case.
Let $r_1, r_2, \ldots$ denote the sequence of random numbers generated.
It is pretty clear that the optimal algorithm for labeling a two-element chain is the following:

If $r_1>1/2$ then label the larger element $r_1$; otherwise label the smaller element $r_1$.  On each subsequent turn, if labeling the unlabeled element with $r_i$ wins at once, then do so; otherwise, relabel the already-labeled element with $r_i$.

Let $N$ be the number of turns; then $N$ is a random variable, and we want to find the expected value of $N$.  So we need to compute $\Pr(N=k)$ for each integer $k$.  There are two ways that $N$ can equal $k$: Either
$$1/2 \le r_1 \le r_2 \le \cdots \le r_{k-1} > r_k $$
or
$$1/2 \ge r_1 \ge r_2 \ge \cdots \ge r_{k-1} < r_k.$$
By symmetry we may focus on the latter case and multiply by 2. The calculation splits into two cases depending on whether $r_k > 1/2$ or $r_k \le 1/2$.  The case $r_k > 1/2$ has probability
$${(1/2)^{k-1}\over (k-1)!} \cdot {1\over 2}$$
where $(1/2)^{k-1}$ is the probability that $r_1, \ldots, r_{k-1}$ are all less than $1/2$ and $1/(k-1)!$ is the probability that they are in decreasing order and the final $1/2$ is the probability that $r_k >1/2$.  The case $r_k \le 1/2$ has probability
$$ {1 \over 2^k} \cdot {k-1\over k!}$$
where $1/2^k$ is the probability that $r_1, \ldots, r_k$ are all less than $1/2$ and $(k-1)/k!$ is the probability that they are in decreasing order except that $r_{k-1} > r_k$.  So the expected value of $N$ is
$$E[N] = 2 \sum_{k\ge 2} k \cdot \left( {(1/2)^{k-1}\over (k-1)!} \cdot {1\over 2} + {1 \over 2^k} \cdot {k-1\over k!}\right) = 2\exp(1/2) - 1 \approx 2.29744.$$
To analyze the three-element chain, it seems that one will need to consider partially labeled posets.  The above analysis easily generalizes to show that the expected number of turns to finish labeling a two-element chain whose smaller element is already labeled with some number $x<1/2$ is
$$\sum_{k\ge 1} k \cdot \left( {x^{k-1}\over (k-1)!} \cdot {(1-x)} + x^k \cdot {k-1\over k!}\right) =  \exp(x). $$
However, as I said, extending the analysis to larger chains looks messy.<|endoftext|>
TITLE: Arctangents of odd powers of the golden ratio
QUESTION [40 upvotes]: While trying to answer this MSE question, I found that arctangents of many odd powers of the golden ratio $\varphi=\frac{1+\sqrt5}2$ are expressible as rational linear combinations of arctangents of positive integers:
$$\begin{align}
\arctan\varphi&=2\arctan1-\frac12\,\arctan2\\
\arctan\varphi^3&=\arctan1+\frac12\,\arctan2\\
\arctan\varphi^5&=4\arctan1-\frac32\,\arctan2\\
\arctan\varphi^7&=3\arctan1+\frac12\,\arctan2-\arctan5\\
\arctan\varphi^9&=\arctan1-\frac12\,\arctan2+\arctan4\\
\arctan\varphi^{11}&=5\arctan1+\frac12\,\arctan2-\arctan5-\arctan34\\
\arctan\varphi^{13}&=3\arctan1-\frac12\,\arctan2+\arctan4-\arctan89\\
\arctan\varphi^{15}&=-2\arctan1+\frac32\,\arctan2+\arctan11
\end{align}$$
I was not able to find such a representation for $\arctan\varphi^{17}$ though.

Question 1. Can we prove that it does not exist?


I also could not find such a representation for any positive even power.

Question 2. Can we prove that it does not exist for any positive even power?



Question 3. Is there a simple way to determine if such a representation exists for a given power?

REPLY [11 votes]: It may be helpful to transform this from a problem involving tangents to one involving complex exponentials.  Note that $\exp(2 i \arctan(x)) = \dfrac{i-x}{i+x}$.  Thus an identity $2 \arctan(x) = \sum k_j \arctan(x_j)$ for integers $k_j$ implies
$$ \left(\dfrac{i-x}{i+x}\right)^2 = \prod_j \left( \dfrac{i-x_j}{i+x_j} \right)^{k_j} \tag{1}$$
Conversely, (1) implies 
$2 \arctan(x) = \sum_{j} k_j \arctan(x_j) + n \pi$ for some integer $n$. 
Thus to have an identity for $2 \arctan(x)$ as an integer linear combination 
of $\arctan(x_j)$ (with $x_1 = 1$ to take care of the $n\pi$), we need $((i-x)/(i+x))^2$ to be in the group generated by $((i-x_j)/(i+x_j))$.
In the case at hand, for odd $n$ I get
$$ \left( \dfrac{i - \phi^n}{i+\phi^n}\right)^2 = \dfrac{L_{2n} - 6 - 4 i L_n}{L_{2n}+2}  = \dfrac{i+2}{i-2} \prod_{k=1}^{(n-1)/2} \left(\dfrac{i+L_{2k}}{i-L_{2k}}\right)^2$$
(I admit I might not have found these formulas without seeing Pakk's comment to ARupinski's answer).
 For even $n$, on the other hand, I get
$$\left(\dfrac{i-\phi^n}{i+\phi^n}\right)^2 = \dfrac{L_{2n} - 6 - 4 i \sqrt{5} F_n}{L_{2n}+2}$$
and clearly you need some factors involving $\sqrt{5}$ to have any hope.<|endoftext|>
TITLE: Computing an eigencuspform in $S_2(\Gamma_0(1776))$
QUESTION [6 upvotes]: Consider 
$$\bar{\rho}:G_{\mathbb Q}\longrightarrow\operatorname{GL}_2(\mathbb F_7)$$
the residual 7-adic Galois representation attached to the elliptic curve $y^2=x^3+x^2-4x-4$ of conductor 48. Then $\bar{\rho}$ is unramified exactly outside $\{2,3,7\}$ and the traces of $\operatorname{Fr}(\ell)$ for $\ell=5,11,13,17,\cdots,37$ are $-2,3,-2,2,\cdots,-1$.
If my computations (and my understanding) are correct, it follows from the fact that $37\cdot\operatorname{tr}(\operatorname{Fr}(37))^2\equiv 38^2$ modulo 7 that $\bar\rho$ satisfies the hypotheses of Ribet's theorem on level-raising and thus that there exists an eigencuspform $f\in S_2(1776)$ with residual representation equal to $\bar{\rho}$ ($1776=37\cdot 48$). I am interested in some properties of Kato's Euler system for $f$ (if $f$ indeed exists).

What are the first (say, 50) coefficients in the $q$-expansion of $f$?

I think the next question might be hard but let me try my luck nevertheless.

What is a system of equations defining the abelian variety $A_f$ attached to $f$?

I believe that $A_f$ is not an elliptic curve, as otherwise it would show up in the standard lists of elliptic curves, and I don't think it does.

REPLY [15 votes]: I did the computation in Sage, and there is no such form $f$. There are 21 Galois orbits of newforms of level $\Gamma_0(1776)$ and trivial character, of which the largest has size 3, and none of the reductions modulo any of the primes above 7 in the coefficient fields is congruent to the form associated to $\bar\rho$.
Looking again at your question, the problem is that you have misquoted Ribet's theorem; the condition for mod $p$ level-raising at $\ell$ should be that $a_\ell^2 = (1 + \ell)^2 \bmod p$, not $\ell a_\ell^2 = (1 + \ell)^2 \bmod p$. At least, that's what Theorem 1.1 of Ribet's paper says. So the smallest $\ell$ for which you can level-raise is not 37 but 53, leading to a newform of level 2544.
EDIT: Just for kicks, I did the computation for $\ell = 53$. The form $f$ is defined over $\mathbf{Q}(a)$ where $a$ is a root of $x^{3} - 10 x^{2} + 28 x - 23 = 0$, and it satisfies the required congruence modulo the prime $4 - a$ of norm 7. The $q$-expansion of $f$ is
$$   q + q^{3} + \left(-a + 2\right)q^{5} + \left(-2 a^{2} + 15 a - 21\right)q^{7} + q^{9} + \left(3 a^{2} - 23 a + 33\right)q^{11} + \left(-a^{2} + 10 a - 19\right)q^{13} + \left(-a + 2\right)q^{15} + \left(a^{2} - 7 a + 7\right)q^{17} + \left(2 a^{2} - 14 a + 14\right)q^{19} + \left(-2 a^{2} + 15 a - 21\right)q^{21} + \left(a^{2} - 8 a + 10\right)q^{23} + \left(a^{2} - 4 a - 1\right)q^{25} + q^{27} - 8q^{29} + \left(-3 a^{2} + 25 a - 39\right)q^{31} + \left(3 a^{2} - 23 a + 33\right)q^{33} + \left(a^{2} - 5 a + 4\right)q^{35} + \left(2 a^{2} - 17 a + 21\right)q^{37} + \left(-a^{2} + 10 a - 19\right)q^{39} + \left(2 a^{2} - 17 a + 23\right)q^{41} + \left(-2 a^{2} + 13 a - 10\right)q^{43} + \left(-a + 2\right)q^{45} + \left(-2 a + 8\right)q^{47} + \left(-3 a^{2} + 22 a - 26\right)q^{49} + O(q^{50}).$$
Email me if you want the original Sage code.<|endoftext|>
TITLE: Travelling Salesman Problem: Can the nearest neighbor algorithm be $n$ times longer than the optimal solution?
QUESTION [10 upvotes]: This is inspired by a recent question.
Given a positive integer $n\in\mathbb{N}$, is there a setting of finitely many points and a designated "starting point" $s$ in $\mathbb{R}^2$ such that the nearest-neighbor algorithm (described below) gives a tour that is $n$ times longer than the optimal solution starting at $s$?


Starting at $s$, pick the nearest neighbor not visited so far as the next node to visit.


EDIT: If the answer is no, what is the maximum value that the ratio $r$ of "nearest neighbor trip" vs "best trip" can take?

REPLY [15 votes]: The nearest-neighbor (NN) heuristic (among others) is analyzed in this paper:

Johnson, David S., and Lyle A. McGeoch. "The traveling salesman problem: A case study in local optimization." Local search in combinatorial optimization 1 (1997): 215-310.
  (PDF download link.)

They say:



In the specific case of "random Euclidean instances"
(in contrast to "random distance matrices"),
they observe experimentally a fixed ~25% longer path length, which
would exceed any fixed $n$ for large enough nodes $N$.
In terms of both theory and experiments with random distance matrices,
the growth rate is $\log N$.
Either way, any fixed $n$ could be exceeded with sufficiently large $N$,
so the answer to the OP's question is Yes.
The cited paper is:
Rosenkrantz, Daniel J., Richard E. Stearns, and Philip M. Lewis, II. "An analysis of several heuristics for the traveling salesman problem." SIAM Journal on Computing 6.3 (1977): 563-581.
(Journal link.)

Added (in response to question from Manfred).
Here is the essence of the Rosenkrantz et al. lower bound:

          


          

(Slide from David Johnson PowerPoint Lecture.)<|endoftext|>
TITLE: continuity of the Boltzmann entropy in the Wasserstein metric
QUESTION [5 upvotes]: For Lebesgue-absolutely continuous probability measures $\rho\ll \mathcal{L}^d$ in the whole space $\mathbb{R}^d$ with finite second moments (i-e $\rho\in \mathcal{P}^2_{ac}(\mathbb{R}^d)$), let
$$
\mathcal{H}(\rho)=\int_{\mathbb{R}^d}\rho\log \rho \,dx
$$
be the Boltzmann entropy.

Question: Is $\mathcal{H}$ continuous with respect to the quadratic Wasserstein distance $\mathcal{W}_2$? i-e is it true that
  $$
 \mathcal{W}_2(\rho_n,\rho)\to 0
 \quad\Rightarrow\quad
 \mathcal{H}(\rho_n)\to\mathcal{H}(\rho)\qquad ???
 $$

I guess this should be known and written down somewhere. I looked at the classical references (Villani's books [topics...] and [old and new...], also in [Ambrosio-Gigli-Savaré]) but surprisingly enough I couldn't find the precise statement. 
It is well known that $\mathcal{H}$ is $0$-displacement convex (in the sense of McCann). By analogy with the Euclidean setting, where any convex function $\phi:\mathbb{R}^d\to \mathbb{R}$ is continuous, I suspect that any displacement convex functional $\mathcal{F}:\mathcal{P}_2(\mathbb{R}^d)\to\mathbb{R}$ should be continuous with respect to the wasserstein distance $\mathcal{W}_2$. But maybe not...
This question arised in the context of my research: I am trying to construct weak solutions to some system of PDEs by means of the Jordan-Kinderlehrer-Otto/DeGiorgi's minimizing scheme, and I need the above statement in some technical step. Unfortunately the mere lower semi-continuity would not be enough for my purpose, I really need full continuity.

REPLY [7 votes]: Let $A_n=\bigcup_{i=0}^{n-1}[2i/(2n),(2i+1)/2n)$, $\rho_n=2\chi_{A_n}$, $A=[0,1]$ and $\rho=\chi_{[0,1]}$. 
Then it seems pretty clear that $\mathcal W_2(\rho_n,\rho)\to 0$. On the other hand $H(\rho_n)\not\to H(\rho)$, right?<|endoftext|>
TITLE: What is the easiest way to compute Ozsváth-Szabó tau invariant of a knot?
QUESTION [7 upvotes]: Suppose that we have a knot $K$ with 40 crossings which is not a cable knot or an alternating knot.
Then, what is the easiest way to compute Ozsváth-Szabó's invariant $\tau(K)$?
Are there any softwares which compute $\tau(K)$ if we insert Gauss code or PD code of $K$?
For Rasmussen's $s$-invariant, Knottheory package works very nicely for this purpose.

REPLY [7 votes]: This is to elaborate on the comment I made above about Livingston's paper. 
Livingston also proves (Theorem 4) that if a knot is null-homologous in a fiber surface of a torus knot and it bounds a quasipositive surface, then $\tau(K)=g(K)=g_4(K)$, where $g(K)$ is the three-genus and $g_4(K)$ is the four-genus. As pointed out by Hedden, this is equivalent to saying that if $K$ is strongly quasipositive, then $\tau=g(K)=g_4(K)$. Hedden also proves that for fibered knots the converse is true, i.e. $\tau(K)=g(K)=g_4(K)$ implies $K$ is strongly quasipositive.
So if you have some kind of Floer-homology-free way of figuring out whether your knot is fibered and (if it is) what its fiber surface is like (i.e. what's the three-genus? and is it strongly quasipositive?), that might be used to tell you $\tau(K)$.
As for figuring out the genera and fiberedness of your knot: in Gabai's paper ``Detecting fibered links in $S^3$" he gives a general strategy (using his sutured manifold theory) for figuring out whether any link in $S^3$ is fibered and what the fiber surface is like. Gabai classified the fiberedness and genera of pretzel knots and Hirasawa and Murasugi used Gabai's approach to do the same for Montesinos knots. I think you might be able to find similar info about arborescent knots. Apparently Stoimenow also has a paper classifying the strong-quasipositivity and fiberedness of closed 3-braids as well.
In that paper of Hedden's mentioned above he also connects the dots to contact topology --- in particular if your knot is fibered and the corresponding open book decomposition supports a tight contact structure, Hedden tells you that $\tau(K)=g(K)=g_4(K)$ as well. In all these approaches you need to be able to determine genus, which is not easy.<|endoftext|>
TITLE: Non-field example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\nsubseteq N$
QUESTION [13 upvotes]: I asked this question on math.stackexchange a month ago with no progress, even after a bounty. I hope to eliminate one if the other receives a satisfactory answer.

For an ideal $I\lhd R$ in a commutative ring $R$, let $ann(I)$ denote the annihilator of $\{x\in R\mid xI=\{0\}\}$. A commutative ring $R$ is said to be a dual ring if for every ideal $I$ of $R$, $ann(ann(I))=I$.

I am looking for an example (if one is possible) of a commutative, local, dual ring with Krull dimension greater than $0$ such that the nilradical $N$ satisfies $ann(N)\nsubseteq N$.

During my search in the literature, the only dual rings I found which weren't $0$-(Krull) dimensional are based on a construction which uses a valuation domain $D$, its field of fractions $Q$, and the $D$ module $M=Q/D$ in a trivial extension $R=D(+)M$ whose ideals are linearly ordered. The problem with this construction is that $0(+)M=N$ is the nilradical, $N$ is a faithful $D$ module, and $N^2=\{0\}$, which implies that $ann(N)=N$ (in $R$.)
I would be grateful for an example, or else some leads on easy methods to construct local dual rings that might lead to an example.

Additions:

it's worth noting that no example would be Noetherian: it's well known Noetherian dual rings are quasi-Frobenius, hence Artinian (and thus zero-dimensional).
UPDATE: as is apparent now, no such ring exists! See the accepted solution. It is a pleasant surprise...

REPLY [13 votes]: There is no such ring.
To lessen my typing, let me introduce some abbreviations.
C = commutative,
L = local,
D = dual,
K = Krull dimension greater than $0$,
$A(I)$ := $\textrm{Ann}(I)$ for $I\lhd R$,
N = $\textrm{Nil}(R)$.
Theorem. If $R$ is a CLDK ring, then $A(N)\subseteq N$.
The proof requires the following
Lemma. If $R$ is a CLD ring that is not a field
and $\mathfrak p\lhd R$ is a
prime ideal, then $A(\mathfrak p)\subseteq \mathfrak p$.
Proof of Lemma.
If $\mathfrak p$ is the maximal ideal, then $A(\mathfrak p)$
is a minimal ideal. Since $R$ is not a field,
this yields $R\neq A(\mathfrak p)$,
so $A(\mathfrak p)\subseteq \mathfrak p$ by locallness of $R$.
For the rest of the proof we consider only the case where
$\mathfrak p$ is nonmaximal.
As is well known,
if $\mathfrak p$ is a prime ideal, then it is $\cap$-irreducible.
(I.e., it is finitely meet irreducible.)
Reason: if $\mathfrak p = I\cap J$ and $\mathfrak p < I, J$,
then we contradict primeness by $IJ\subseteq I\cap J = \mathfrak p$.
Claim. If $\mathfrak p$ is nonmaximal, then it is not
$\bigcap$-irreducible. (I.e., $\mathfrak p$ is not infinitely meet irreducible.) Said another way, $\mathfrak p$ will equal the
complete intersection of all ideals that are properly above $\mathfrak p$.
Proof of Claim. Else $\mathfrak p^*:=\bigcap_{\mathfrak p < I} I$ is the smallest ideal
strictly above $\mathfrak p$. Then $\mathfrak p^*/\mathfrak p$ is the smallest nonzero ideal
in the domain $R/\mathfrak p$. Notice that this domain is not a field, since
$\mathfrak p$ was nonmaximal, so by its minimality $\mathfrak p^*/\mathfrak p$ is a proper ideal
of $R/\mathfrak p$.
$R/\mathfrak p$
is a domain and $0 < \mathfrak p^*/\mathfrak p\cdot \mathfrak p^*/\mathfrak p\leq \mathfrak p^*/\mathfrak p$,
forcing $\mathfrak p^*/\mathfrak p = (\mathfrak p^*/\mathfrak p)^2$. Now minimal idempotent ideals in commutative rings, like
$\mathfrak p^*/\mathfrak p$, are generated by idempotents, meaning
$\mathfrak p^*/\mathfrak p = (e)$ for some $e\in R/\mathfrak p$ satisfying $e^2=e$. The
element $e$ cannot be zero,
because $\mathfrak p^*/\mathfrak p$
is not zero, and it cannot be $1$, since $\mathfrak p^*/\mathfrak p$
is not $R/\mathfrak p$. Thus the domain $R/\mathfrak p$ has a proper idempotent $e$, which is absurd.
($e(1-e) = 0\neq e, (1-e)$ contradicts the definition of a domain.). This proves the claim. \\
Now back to the proof of the lemma. By the claim, if
$\mathfrak p$ is not maximal, then it equals the
intersection of all the ideals that properly contain it.
Applying the lattice anti-isomorphism $I\mapsto A(I)$, which must preserve
the complete lattice operations of the ideal lattice of $R$,
we obtain that $A(\mathfrak p)$ is the join $\bigvee_{\mathfrak p < I} A(I)$ of all ideals of the
form $A(I)$ where $\mathfrak p < I$. For any such $I$ we have
$$
I\cdot A(I) = 0\subseteq \mathfrak p,
$$
and $I \not\subseteq \mathfrak p$, so $A(I)\subseteq \mathfrak p$.
Hence the join $\bigvee_{\mathfrak p < I} A(I) = A(\mathfrak p)$
is also contained
in $\mathfrak p$. \\
Now
Proof of Theorem.
Suppose $R$ is a CLDK ring and that $\mathfrak p$ is a minimal prime
of $R$. If $\mathfrak q$ is a different minimal prime, then
$$
\mathfrak p \cdot A(\mathfrak p) = 0 \subseteq \mathfrak q,
$$
so either $\mathfrak p\subseteq \mathfrak q$ or
$A(\mathfrak p)\subseteq \mathfrak q$. The former cannot happen,
since $\mathfrak q$ is assumed to be minimal and different from $\mathfrak p$,
so we must have the latter. To reiterate: if $\mathfrak p$ is a minimal prime of $R$, then $A(\mathfrak p)$ is contained in every minimal prime different from $\mathfrak p$.
By the lemma, $A(\mathfrak p)\subseteq \mathfrak p$ as well,
so $A(\mathfrak p)$ is contained in all of the minimal primes. In other
words, if $\mathfrak p$ is a minimal prime, then
$$
A(\mathfrak p)\subseteq \bigcap_{\mathfrak q \textrm{min}} \mathfrak q
= N.
$$
Dualizing this yields $A(N)\subseteq \mathfrak p$ for every minimal prime $\mathfrak p$. It follows from this that
$$
A(N)\subseteq \bigcap_{\mathfrak p \textrm{min}} \mathfrak p
= N. 
$$
\\<|endoftext|>
TITLE: Maximum of the Vandermonde determinant / minimum of the logarithmic energy
QUESTION [16 upvotes]: The problem is to find the asymptotics (as $n\to\infty$) of the maximum (say $M_n$) of the Vandermonde determinant 
$$V_n:=\prod_{0\le i<j\le n-1}(a_j-a_i)
$$ 
over all $a_0,\dots,a_{n-1}$ such that $0=a_0<\dots<a_{n-1}=1$ (or, better, an upper bound on $M_n$ which is asymptotic to $M_n$; or, at least, the asymptotics of $\ln M_n$). It is clear that the maximum is attained. 
These questions can be obviously restated in terms of the minimization of the logarithmic energy
$$\sum_{0\le i<j\le n-1}\ln\frac1{a_j-a_i}. 
$$ 
I have encountered this problem working on a matter involving higher-order divided differences. It is known that the $n$-tuple $(a_0,\dots,a_{n-1})$ maximizing $V_n$ is given by the formula $a_i=(1+x_i)/2$, where the $x_i$'s are the roots of the polynomial $(1-x^2)P'_{n-1}(x)$ (taken in the ascending order) and $P_{n-1}$ is the Legendre polynomial of degree $n-1$; these points $x_0,\dots,x_{n-1}$ are also known as the Fekete points; see e.g. SE Mathematics. The picture below suggests that $\ln M_n\sim-(a+bn)^2$ as $n\to\infty$, for some real constants $a<0$ and $b>0$.

REPLY [5 votes]: I am amazed that nobody noticed that Iosif asks for the calculation of the transfinite diameter of the interval $[0,1]$. This notion applies to arbitrary compact domains $K$ in ${\mathbb R}^n$ :
$$d(K)=\lim_{k\rightarrow+\infty}\sup_{x_1,\ldots,x_k\in K}\left(\prod_{\alpha<\beta}|x_\alpha-x_\beta|\right)^{2/k(k-1)}.$$
In the particular case where $K\subset{\mathbb C}$ is simply connected, then the inverse of $d(K)$ is the conformal radius of ${\mathbb C}\setminus K$.<|endoftext|>
TITLE: Decidability of an Algebraic System in Real Numbers
QUESTION [8 upvotes]: Is there an algorithm to decide whether an algebraic system 
\begin{gathered}
  {f_1}({x_1}, \ldots ,{x_n}) = 0 \hfill \\
   \vdots  \hfill \\
  {f_m}({x_1}, \ldots ,{x_n}) = 0 \hfill \\ 
\end{gathered}
where $f_1,\ldots,f_m$ are polynomials with given rational coefficients, has a solution in real numbers?

REPLY [8 votes]: Yes, this follows from Tarski's theorem (1951) that the first order theory of real closed fields admits elimination of quantifiers. See also the Tarski-Seidenberg theorem.
P.S. A consequence of these theorems is that if the given polynomial system has a solution in real numbers, then it also has a solution in real algebraic numbers.<|endoftext|>
TITLE: A curious determinantal inequality
QUESTION [35 upvotes]: In my study, I come across the following curious inequality, which I do not know a proof yet (so I am asking it here). 
Let $A, B$ be $n\times n$ (Hermitian) positive definite matrices.  It is very likely true that
$$\det \left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \ge \det(A+B)^2. $$
Here $A^{\frac{1}{2}}$ is the unique positive definite square root of $A$.
I am able to confirm the $3\times 3$ case.
Comments: Only recently did I notice that the majorization $\lambda\left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \prec \lambda(A+B)^2$ follows immediately by THEOREM 2
of [R.B. Bapat, V.S. Sunder, On majorization and Schur products, Linear Algebra Appl. 72 (1985)
107–117.] http://www.sciencedirect.com/science/article/pii/0024379585901478

REPLY [5 votes]: Here is a complementary approach without using majorization. The answer is partial because it has an open "TODO". I am writing it down here already in case someone wishes to complete the argument.

Let $A, B, X, Y > 0$. It is easy to show using Schur complements that
\begin{equation*}
\tag{$*$}
 AX^{-1}A + BY^{-1}B \ge (A+B)(X+Y)^{-1}(A+B).
\end{equation*}
From $(*)$ it follows that $\det(AX^{-1}A + BY^{-1}B)\det(X+Y)\ge \det(A+B)^2$. 
Let $C=A^{1/2}(A+B)A^{1/2}$ and $D=B^{1/2}(A+B)B^{1/2}$. If we can find (TODO) $X$ and $Y$ such that 
\begin{equation*}
  X \gets A(X+Y)^{1/2}C^{-1}(X+Y)^{1/2}A,\quad Y \gets B(X+Y)^{1/2}D^{-1}(X+Y)^{1/2}B,
\end{equation*}
then we will obtain $$(X+Y)^{1/2}(AX^{-1}A + BY^{-1}B)(X+Y)^{1/2} = C+D = A^{1/2}(A+B)A^{1/2} + B^{1/2}(A+B)B^{1/2}.$$ Combining this identity with the above inequality immediately implies the desired inequality.
Notice that in particular, if $A$ and $B$ commute, then $X=A(A+B)^{-1}$ and $Y=B(A+B)^{-1}$ is a solution.<|endoftext|>
TITLE: Convex optimization with full subdifferential information
QUESTION [5 upvotes]: Can anyone direct me to any algorithms or theorems that describe the difficulty of solving a non-smooth convex optimization problem for the special case where the full subdifferential is available?  All of the results that I can find are designed for the case where one has an oracle that can generate a single subgradient.

REPLY [4 votes]: Regarding difficulty results (i.e., lower bounds) for iterative methods, worst-case complexity results in the oracle model hold regardless of whether you provide the full subgradient to the algorithm. What happens is that lower bounds for nonsmooth convex optimization are obtained by piecewise linear functions, and the adversary can always (consistently) adapt the instance so that at all query points the function is locally linear, thus the subdifferential is a singleton. You can see the proof at page 120 in the following notes http://www2.isye.gatech.edu/~nemirovs/Lect_EMCO.pdf
If you care about average-case analysis (where the adversary is not allowed to adapt) then the question is much more delicate, and I think there is no answer in the literature up to date.<|endoftext|>
TITLE: On an automatic translation of typed lambda calculus in untyped lambda calculus
QUESTION [6 upvotes]: I have a question regarding the "compilation" of typed lambda calculus in untyped lambda calculus.
Take for example the inductive definition of lists, with introduction rules:

and:

We can automatically derive the elimination rule and the computation rules (here
omitted as they are well known), as well as an interpretation in untyped lambda
calculus:
$[\![Nil]\!] = \lambda x.\lambda y.x$
$[\![Cons(a,l)]\!] = \lambda x.\lambda y.y \> [\![a]\!] \> ([\![l]\!] \> x \> y)$
$[\![ListCata(l,n,c)]\!] = [\![l]\!] \>[\![n]\!] \> [\![c]\!]$
This "automatic definition" can be done, up to my knowledge, if we are
defining an inductive type $I$ such that the occurrences of $I$ in the premises
of the introduction rules are strictly positive.
It seems to me that the power of this method
lies in his generality: I only have to state my introduction rules and I automatically get a translation in a language I know how to execute, in this case untyped lambda calculus.
Now, I'm interested in a better understanding of this type of translation, in different directions:

Where has this schema been first defined? Are there papers regarding this
translation?
What are the limits of this approach? For example, can I extend this
process to more inductive types, to co-inductive types, to a polymorphic
lambda calculus, to dependent types?
What if I'd like, given a definition of an inductive type, "compile" the
code in, for example, SKI calculus, or another system of symbolic computation?
Is there any work done in this direction?
Could one hope to give an inductive definition of the underlying "computational machine", and get automatically a translation?

REPLY [6 votes]: The translation process you describe is often referred to as the Church encoding, and it is rather well studied in the literature. Presumably it was first given by Church soon after the definition of the $\lambda$-calculus itself (some details appear here).
One way to see it is as a "control inversion", where an instance of pattern matching on an element of an inductive type simply becomes an application of each branch to the term.
This translation doesn't necessarily have to be to the untyped calculus: the usual Church encoding of (positive) data-types in System F results in well-typed terms! In general this extends to polymorphism and dependent types without difficulty, though the induction principles for those types can not be proven inside the type theories themselves! (See e.g. Geuvers for a proof).
The question about compiling to combinators seems unrelated. Of course it is possible to represent every $\lambda$-term using combinators, and using combinators as targets for compilation is another well studied subject (see e.g. Hudak & Kranz).<|endoftext|>
TITLE: How to explain the concentration-of-measure phenomenon intuitively?
QUESTION [31 upvotes]: One way to phrase the
"concentration-of-measure"
phenomenon is that, 
for a Euclidean sphere $S^d$ in $d$ dimensions, for large $d$,
"most of the mass is close to the equator, for any equator."1

Q. How could one explain/justify this intuitively—perhaps just verbally—to a mathematically literate but naive audience
  (say, advanced undergraduate math majors)?

That "most of the mass is close to the equator, for any equator"
seems almost contradictory (imagining orthogonal equatorial hyperplanes), or at the
least, superficially quite puzzling.
Can one only gain intuition via working
through details of the Brunn–Minkowski theorem
or the isoperimetric inequality?

1Boáz Klartag, in a book review in the AMS Bulletin, July 2015, p.540.
According to the Wikipedia article, the idea goes back to Paul Lévy.

REPLY [3 votes]: This answer is complementary to the other answers. I'll attempt to tackle the paradox

That "most of the mass is close to the equator, for any equator" seems almost contradictory (imagining orthogonal equatorial hyperplanes)...

head-on by leveraging some intuition about the 3-sphere.
Where the intuition struggles is in seeing how two $\epsilon$-width equatorial bands oriented in orthogonal directions can have any sort of substantial overlap, much less cover most of the sphere. And they must have substantial overlap: if an equatorial band covers 99% of the area of the sphere, and an orthogonal equatorial band also covers 99% of the sphere, then the overlap of the two bands has to cover at least 98% of the sphere.
The issue is that for the two spheres that are easy to visualize, the 1-sphere and the 2-sphere, the overlap of narrow equatorial bands in orthogonal directions is indeed insubstantial. For the 1-sphere it is, in fact, empty: one equatorial band consists of arcs on the west and east sides of the circe; the other consists of arcs on the south and north sides.

Of course if we increased $\epsilon$ so that each band covered 99% of the area (circumference in this case), then there would be an overlap, of four disconnected arcs, but as a schematic for the situation in higher dimensions this fails since the arcs disappear as $\epsilon$ shrinks.
The situation is hardly better for the 2-sphere. The overlap is non-empty this time, but it consists of two antipodal $\epsilon\times\epsilon$ patches (black regions) where the orthogonal belts overlap, and it is hard to imagine, even schematically, how the area of the patches can remain large as $\epsilon$ shrinks.

So how do things look for the 3-sphere? A picture of the 3-sphere is obtained by gluing together two 3-dimensional balls along their 2-dimensional surface. Let's redo the 1-sphere and 2-sphere by the analogous constructions. The 1-sphere is obtained by gluing together two 1-dimensional balls (line segments) along their 0-dimensional surfaces (endpoints). The west-east band originates with $\epsilon/2$-width segments at the surface (endpoints) of each line segment; the south-north band originates with an $\epsilon$-width segment spanning the equator (midpoint) of the line segments:

The 2-sphere is obtained by gluing together two 2-dimensional balls (disks) along their 1-dimensional surfaces (circles). One band originates with the $\epsilon/2$-width surface region (annulus) of each disk; the other originates with an $\epsilon$-width equatorial band of each disk:

The 3-sphere is obtained by gluing together two 3-dimensional balls along their spherical surface. One "equatorial band" will be the result of gluing the two spherical shells of thickness $\epsilon/2$ along their outer surfaces to form a spherical shell of thickness $\epsilon$. The other will be the result of gluing two thickened disks of thickness $\epsilon$ along their circumference to form a second spherical shell of thickness $\epsilon$. (One of the thickened disks is shown in the diagram with its center at $\infty$.) The two spherical shells intersect in a band of width $\epsilon$, thickness $\epsilon$, and circumference $2\pi R$, where $R$ is the radius of the balls. (The intersection is what looks like the rim of a bicycle wheel near the center of the last figure.)

As a schematic for the $n$-sphere, with $n$ large, this picture seems to me successful. As $\epsilon$ shrinks, the overlap remains global in scope, containing a circumference. Furthermore, the picture starts to satisfy some consistency checks: It is easy to believe that most of the volume of an $n$-dimensional ball lies in its outer layer. It's somewhat less obvious but still plausible that most of the volume of an $n$-dimensional ball lies in a thickened disk containing the equator. (Andreas Blass's answer may help convince you of this). For both to be true, most of the volume must lie in a thickened surface band around the equator. That this is so is again plausible: if $n$ is large, then $n-1$ is also large, so most of the volume of the disk lies near its outer edge rather than near its center, that is, in the part where it overlaps the spherical shell; at the same time, it isn't far fetched that in the spherical shell (when $n$ is large) one doesn't lose much by keeping the equatorial region and discarding the caps containing the poles. (This is explained in Andreas Blass's answer.) But this equatorial band containing most of the mass is precisely the overlap of the orthogonal equatorial regions in our 3-sphere construction.<|endoftext|>
TITLE: Automorphism group of a fiber bundle surjects onto diffeomorphism group?
QUESTION [6 upvotes]: This should surely be well-known by I have not been able to find a good reference to the following question: Given a smooth fiber bundle $\pi\colon P \longrightarrow M$ over a smooth manifold $M$ with typical fiber $F$, one has the group of fiber-preserving automorphisms of $P$: a diffeomorphism $\Phi\colon P \longrightarrow P$ is called fiber-preserving if $\pi \circ \Phi = \phi \circ \pi$ for some smooth map $\phi\colon M \longrightarrow M$, which then turns out to be a diffeomorphism of $M$. If $\phi = \mathrm{id}_M$ then one calls $\Phi$ a gauge transformation. Clearly they form a normal subgroup $\mathrm{Gau}(P) \subseteq \mathrm{Aut}(P)$, being the kernel of the group morphism $\Phi \mapsto \phi$. Hence we get a subgroup of the diffeomorphism group as the image of this quotient $\mathrm{Aut}(P) / \mathrm{Gau}(P) \subseteq \mathrm{Diffeo}(M)$. Of course, the case of principal fiber bundles is of particular interest here.
It is now well-known and not too hard to show that all the small diffeomorphisms of $M$ are contained in this image: this can be done by using a (complete) connection and it's parallel transport.
My question is about the large diffeomorphisms: are they also in the image, i.e. is the whole diffeomorphism group isomorphic to this quotient $\mathrm{Aut}(P) / \mathrm{Gau}(P)$? What conditions of $P$ would guarantee this (beside being trivial...)?

REPLY [8 votes]: Any fiber bundle $\pi:P \longrightarrow M$ with fiber $F$ is classified by a map $f_{\pi}: M \longrightarrow B Diff(F)$ where $B Diff(F)$ is the classifying space of the diffeomorphism group of $F$. If an automorphism $\phi: M \longrightarrow M$ lifts to $P$ then the bundle $\phi^*P$ is isomorphic to $P$ and hence the classifying map $f_{\pi}$ is homotopic to the classifying map $f_{\pi} \circ \phi$ of the bundle $\phi^*P$. This observation can be used to produce easy examples of diffeomorphisms $\phi: M \longrightarrow M$ which do not lift to diffeomorphisms of $P$. For example, take $F$ be the discrete manifold consisting of two points. Then $Diff(F) = \mathbb{Z}/2$ and fiber bundles with fiber $F$ are classified by maps $M \longrightarrow B\mathbb{Z}/2$, or equivalently, by elements in $H^1(M,\mathbb{Z}/2)$. Now take $M = \mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$ to be the $2$-dimensional torus and let us identify $H^1(\mathbb{T}^2,\mathbb{Z}/2)$ with $\mathbb{Z}/2 \oplus \mathbb{Z}/2$. Let $P \longrightarrow \mathbb{T}^2$ be the $F$-bundle corresponding to the class $\alpha = (1,0) \in H^1(\mathbb{T}^2,\mathbb{Z}/2)$. Let $\phi: \mathbb{T}^2 \longrightarrow \mathbb{T}^2$ be the automorphism of $\mathbb{T}^2$ induced by the matrix $\left(\begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right)$. Then $\phi^*(\alpha) \neq \alpha$ and hence the classifying map of $\phi^*P$ is not homotopic to the classifying map $P$. As a result the diffeomorphism $\phi$ does not lift to a diffeomoprhism of $P$.<|endoftext|>
TITLE: How to write an abstract for a math paper?
QUESTION [20 upvotes]: How would you go about writing an abstract for a Math paper? I know that an abstract is supposed to "advertise" the paper. However, I do not really know how to get started. Could someone tell me how they go about writing an abstract?

REPLY [50 votes]: Avoid notation if possible.  Notation makes it really hard to search electronically.
Put the subject in context, e.g., "In a recent paper, T. Lehrer introduced the concept of left-bifurcled rectangles.  He conjectured no such rectangles exist when the number of bifurcles $n$ is odd."
State your results, in non-technical language, if possible.  "In this paper we show the existence of left-bifurcled rectangles for all prime $n$."
Mention a technique, if there is a new one:  "Our methods involve analytic and algebraic topology of locally euclidean metrizations of infinitely differentiable Riemannian manifolds".
Never, ever, ever, cite papers in the bibliography by giving citation numbers; the abstract is an independent entity that should stand on its own.

REPLY [18 votes]: Jeffrey has made a good list.  I'll add one:
A major purpose of an abstract is to help interested people find your paper when they search for a topic. To that end, if there are multiple names in use for the concepts in the paper, I recommend that you try to mention them all, even if you have to write "also known as ...".

REPLY [15 votes]: One thing that I have been taught to do in the body of a paper, but which may also make sense in an abstract is to state an easily-understood interest-piquing corollary of the main result "As a special case of our results, we demonstrate the existence of infinitely many integer solutions to the equation $x^3-y^2=17xy$".<|endoftext|>
TITLE: Topology of ring of global sections of finite union of affinoid opens in a rigid analytic space
QUESTION [5 upvotes]: Let $X$ be a rigid analytic space over a non-Archimedean field $k$. If $U_1,\ldots,U_n\subseteq X$ are affinoid opens, then it's usually not clear whether or not the admissible open $U=U_1\cup\cdots\cup U_n$ is affinoid. But we have (due to the sheaf axioms, and the fact that the $U_i$ constitute an admissible covering of $U$) the canonical injection (arising from the restrictions $\mathscr{O}_X(U)\to\mathscr{O}_X(U_i)$) 
$$\mathscr{O}_X(U)\hookrightarrow\prod_{i=1}^n\mathscr{O}_X(U_i)\quad (*)$$
The target is a product of affinoid $k$-algebras, and in particular is a Banach algebra over $k$. One can endow $\mathscr{O}_X(U)$ (or $\mathscr{O}_X(V)$ for any admissible open $V$ of $X$) with a locally convex topology via the $k$-algebra isomorphism (again due to the sheaf axioms) $\mathscr{O}_X(U)\to\varprojlim_W\mathscr{O}_X(W)$, where $W$ runs over the affinoid opens of $X$ contained in $U$, by pulling back the projective limit topology on the target (using the canonical $k$-Banach topologies on each $\mathscr{O}_X(W)$). For this topology, the map (*) is certainly continuous. My question is as follows:
Is the map (*) a topological embedding?
This would follow if the sheaf of rings of a rigid space were in fact a sheaf of topological rings, but I don't think this is required in the definition of a $G$-ringed space in e.g. BGR like it is for adic spaces and formal schemes. I feel like this is the case for $X$ affinoid though...at least, I think this is true if it is the case that $k$-algebra maps between $k$-affinoid algebras, which are always continuous, are in fact strict (I don't know if this is true, though it's true for maps of finite modules over Noetherian $k$-Banach algebras, which affinoid algebras are by definition, but the base Tate algebra depends on the given affinoid algebra). 
A positive answer to my question would imply that the locally convex topology of $\mathscr{O}_X(U)$, while possibly not that of an affinoid $k$-algebra, is at least defined by a norm (the one got by restricting the max norm for a choice of norms on each $\mathscr{O}_X(U_i)$ along (*)). This consequence is what I really want. (I guess if we really have a sheaf of topological rings, then the map would be a closed topological embedding, since it's the first arrow in an equalizer sequence of Hausdorff topological rings, but I don't actually need this.)
EDIT: As user grghxy points out, I should assume that $X$ is quasi-separated to ensure that my set $U$ is in fact admissible open with admissible covering given by the $U_i$.

REPLY [2 votes]: So that this question doesn't remain unanswered, I will provide an elaboration of grghxy's answer in the comments. The quasi-separatedness of $X$ ensures that $U$ is admissible open with $U=\bigcup_{i=1}^n U_i$ an admissible covering (and in fact that any finite covering of $U$ by affinoid opens of $X$ is admissible). In my original formulation of the question I (again thinking about affinoid spaces) implicitly assumed each $U_i\cap U_j$ was affinoid, which would follow from separatedness of $X$, but is not needed. Quasi-separatedness ensures that each intersection $U_i\cap U_j$ is a finite union of admissible affinoid opens $V_{ijk}$, and we get an exact sequence
$$\mathscr{O}_X(U)\to\prod_{i=1}^n\mathscr{O}_X(U_i)\to\prod_{(i,j,k)}\mathscr{O}_X(V_{ijk})$$
where the second and third terms are products of finitely many affinoid $k$-algebras, hence are themselves naturally Banach algebras. Since the second arrow is continuous (it comes from maps between $k$-affinoid algebras in the factors, which are always continuous), the kernel of this map is closed, so by transport of structure we get a Banach topology on $\mathscr{O}_X(U)$ for which the injection $\mathscr{O}_X(U)\to\prod_{i=1}^n\mathscr{O}_X(U_i)$ is a closed topological embedding. If we use another finite covering of $X$ by affinoid opens (necessarily admissible), then the same reasoning shows that we get another Banach topology on $\mathscr{O}_X(U)$ as a closed subspace of the product of the affinoid algebras of the members of the covering. By taking a common refinement if necessary, in comparing the Banach topologies, we may assume one covering refines the other. Then the Banach topologies are comparable, and so coincide by the open mapping theorem. So there is an canonical Banach space topology on $\mathscr{O}_X(U)$ which arises in the manner above from any finite covering by affinoid opens. 
We can go slightly further using the above reasoning. If $W$ is any affinoid open contained in $U$, then the (redundant) covering $U=U_1\cup\cdots\cup U_n\cup W$ is admissible by quasi-separatedness, so, by the independence of the canonical Banach topology of the choice of finite affinoid covering, the map $\mathscr{O}_X(U)\to\mathscr{O}_X(W)\times\prod_{i=1}^n\mathscr{O}_X(U_i)$ is a closed topological embedding, and in particular, the restriction map $\mathscr{O}_X(U)\to\mathscr{O}_X(W)$ is continuous for the Banach topologies on source and target. 
It follows that the $k$-linear bijection $\mathscr{O}_X(U)\to\varprojlim_W\mathscr{O}_X(W)$ is continuous for the Banach topology on the source and the projective limit of Banach topologies on the target. Using the covering $U=\bigcup_{i=1}^n U_i$ to describe the Banach topology on $\mathscr{O}_X(U)$, a basic open set has the form $\{F\in\mathscr{O}_X(U):F\vert_{U_i}\in A_i,1\leq i\leq n\}$ for some open subsets $A_i\subseteq\mathscr{O}_X(U_i)$. But since the $U_i$ are among the $W$ in the projective limit, this is identified with a basic open subset of the projective limit, namely $\{(F_W)_W:F_{U_i}\in A_i,1\leq i\leq n\}$. The map $\mathscr{O}_X(U)\to\varprojlim_W\mathscr{O}_X(W)$ is therefore open, hence a topological isomorphism, so the Banach topology on $\mathscr{O}_X(U)$ coincides with its intrinsic topology as a projective limit. In particular, the answer to my question is ``yes."<|endoftext|>
TITLE: Generalization of Popoviciu's inequality
QUESTION [8 upvotes]: Popoviciu's inequality states that for convex $f$ and numbers $x_1,x_2,x_3$, we have
$f(x_1)+f(x_2) + f(x_3) + 3\cdot f(\frac{x_1+x_2+x_3}3) \geq 2\cdot f(\frac{x_1+x_2}2)+2\cdot f(\frac{x_1+x_3}2)+2\cdot f(\frac{x_2+x_3}2)$.
It can be proven just by using Karamata's inequality and arguing that the sequence of $6$ numbers on the LHS majorizes the sequence on the RHS.
Some modest generalizations were proposed, where we would have $n$ numbers, LHS would have still a sum of the function for each single argument + the mean and the RHS would have average of all possible $k$ of $n$ tuples.
Can we go full measure here with all possible combinations of numbers? That is,
$\sum f(x_i) - 2   \sum_{i_1 < i_2 } f(\frac{x_{i_1}+x_{i_2}}{2})+ 3 \sum_{i_1 < i_2 < i_3} f(\frac{x_{i_1}+x_{i_2}+x_{i_3}}{3}) + ... + (-1)^{n+1} nf(\frac{x_{i_1}+...+x_{i_n}}{n}) \geq 0$
I would like to tackle this with Karamata's inequality and argue that the sequence for positive numbers majorizes the sequence for negative numbers.
However already for $n=3$ the only way to show the majorization is with case work.
So I would like to know if there are some smarter ways of doing so.
Or maybe you know some obvious reasons for which the whole inequality just cannot be true.

REPLY [10 votes]: Let me elaborate on Darij's comment: Convex functions on a closed interval can be uniformly approximated by piecewise linear convex functions, and it is a theorem of Popoviciu that piecewise linear convex functions are of the form $$f(x)=ax+b+\sum c_i\left|x-r_i\right|,$$
where the $c_i$ are non-negative.
With this in mind, inequalities like Popoviciu's always can be checked by looking at the corresponding "abslute value" version. In this case it follows from the (more familiar to some people) Hlwaka inequality:
$$|x|+|y|+|z|+|x+y+z|\geq |x+y|+|y+z|+|x+z|.$$
Denis Serre asked a question equivalent to yours a few months ago, about an n-variable Hlawka inequality, but the same counterexample was pointed out.<|endoftext|>
TITLE: Four Dimensional Rook Domination
QUESTION [6 upvotes]: Let $\gamma(G)$ denote the domination number of a graph, and $G\,\square\,H$ denote the cartesian product of two graphs. Then $K_8\,\square\, K_8$ is the rook graph, whose vertices are the squares of a chessboard, with edges between squares a rook can move between. We can similarly define the rook graph of any square chessboard as $K_n \,\square \, K_n$, and define $d$-dimensional chessboards as $(K_n)^{\square \,d}$, where generalized rooks move by sliding parallel to one of the coordinate axes.
Clearly, $\gamma(K_n\,\square\, K_n)=n$. Put another way, it is possible to place $n$ rooks on an $n\times n$ chessboard so every square contains a rook or is attacked by one, while $n-1$ rooks are insufficient. Much less easily, you can prove $\gamma(K_n\,\square\,K_n\,\square\,K_n)=\lceil n^2/2\rceil$, which also has an interpretation with rooks on a 3D board. 
These nice answers for two and three dimensions gave me hope there was a nice one for 4, so my question is this:

Is anything known about $\gamma ((K_n)^{\square \,4})$? Or, how many rooks does it takes to dominate an $n\times n\times n \times n$ chessboard?

It's easy to show that $\gamma ((K_2)^{\square \,4})= 4$, and not hard to show $\gamma ((K_3)^{\square \,4})= 9$. But for a $4\times 4\times 4\times 4$ board, all I know is that the domination number is at least 23 and at most 32. Unfortunately, it is infeasible to brute force $\gamma((K_4)^{\square \,4})$.

REPLY [2 votes]: A dominating set of $(K_n)^{\square \,d}$ is equivalent to a $n$-ary covering code of length $d$ and covering radius $1$: just take the positions of the rooks to be the codewords. Known optimal sizes of covering codes are listed for example in Gerzson Kéri's website, although I don't know how up-to-date that information is.
The notation is such that $K_q(n,R)$ is the optimal size of a $q$-ary covering code of length $n$ and covering radius $R$. With the notation in this question, you are looking for $K_n(d,1)$. In particular, the answer for $(K_4)^{\square\,4}$ is $K_4(4,1)=24$, given in "Bounds for quaternary and quinary covering codes, listed for n <= 11, R <= 8 ".
(Yes, the $K_x$ part of the notation happens to be the same in both cases, which might be confusingly non-confusing; on the other hand the meaning of $n$ is different here and there.)<|endoftext|>
TITLE: Probability that random nonnegative integer matrix is singular
QUESTION [13 upvotes]: Q. What is the probability that an $n \times n$ matrix, whose elements
  are independent uniformly random integers in $\{0,1,\ldots,k\}$, is singular?

For example, for $n=3$ and $k=2$, the first matrix below is singular and
the second nonsingular:
$$
\left|
       \begin{array}{ccc}
        2 & 1 & 2 \\
        0 & 2 & 0 \\
        0 & 2 & 0 \\
       \end{array}
\right| = 0 \;,
$$
$$
\left|
       \begin{array}{ccc}
        0 & 2 & 0 \\
        0 & 0 & 1 \\
        2 & 2 & 2 \\
       \end{array}
\right| = 4 \;.
$$
For $n=3$ and $k=1,2,...,14$, 
the probabilities that the matrices are singular are $338/512 = 66.0\%,$ $6891/19683 = 35.0\%,$ $49246/(4^9) = 18.8\%,$ $228737/(5^9) = 11.7\%,$ $716214/(6^9) =7.11\%,$ $... 259500567/(15^9) = 0.675\%$. See A059976.
I find experimentally that for $n=5$ and $k=2$, about 20.8% are singular.
For $n=5$ and $k=10$, about .07% are singular.
Here is a graph for $n=3$:

          


Can results for these $(n,k)$-cases be derived from analogous results, e.g.,

Tao, Terence, and Van Vu. "On random $\pm 1$ matrices: Singularity and determinant." Random Structures & Algorithms 28.1 (2006): 1-23. 
  (arXiv abstract.)
Voigt, Thomas, and Günter M. Ziegler. "Singular 0/1-matrices, and the hyperplanes spanned by random 0/1-vectors." Combinatorics, Probability and Computing 15.03 (2006): 463-471.
  (Cambridge link.)
  (PDF download pre-publication version.)

?

REPLY [4 votes]: In the situation where $n$ is fixed, and $k$ is large, an asymptotic formula for this probability was found by Yonatan Katznelson.  From Theorem 1 of his paper (taking there ${\mathcal B}$ to be the box $[0,1]^{n^2}$) the chance of being singular is 
$$ 
\sim C \frac{\log k}{k^n}, 
$$
for a constant $C$.  Katznelson considers more generally singular matrices with integer entries lying in a large dilate of a fixed region ${\mathcal B}$ in ${\Bbb R}^{n^2}$ (the box $[0,1]^{n^2}$ being the example of this question).  His asymptotic is a little different from an earlier result (the difference is the $\log k$ factor) of Duke, Rudnick, and Sarnak who counted such matrices with a given non-zero determinant. Other references, dealing with variants of this problem, or in the regime where $k$ is fixed and $n$ is large may be found in the comments.<|endoftext|>
TITLE: Time averages and differentiability
QUESTION [23 upvotes]: Let $\varphi_t : M \rightarrow M$ be a smooth flow on a smooth manifold $M$. We may assume (although I'm not sure if this is important) that the flow preserves a smooth volume form on $M$. Given a continuous function $f : M \rightarrow \mathbb{R}$, is it true that if ALL the (partial) time averages
$$
A_T f(x) := {1 \over T} \int_0^T f(\varphi_t (x)) \, dt \hskip 1cm (T > 0)
$$
are smooth functions, then $f$ is also smooth? 
Remark. 
This is obviously true for the trivial action and it is non-trivially true for the case where $M = \mathbb{R}$ and the flow is the usual flow by translations. That's all I know for now. 
Motivation.
In the first volume of Dunford and Schwartz, where they motivate the Ergodic theorem (page 657), they state:
What is significant and measurable in the laboratory is not the quantity $f(\phi_t(x))$ but its average value
$$
 {1 \over T} \int_0^T f(\varphi_t (x)) \, dt
$$
computed over a certain time interval $0 \leq t \leq T$.
I'm asking whether knowing the regularity of all those average values says something about the regularity of the function, or "observable" $f$. It would be interesting to consider $f$ to be just locally integrable and then the question ties in a (very) little with Wiener's differentiation theorem.

REPLY [3 votes]: $\textbf{Update with some corrections}$:
Unfortunately, my original answer had a serious gap. Sorry for the mix-up. The argument I gave only works when we have estimates on $A_T f(x)$ that are uniform in $T$. However, all the hard work was not useless as it made clear the properties that a counterexample would have. Without further ado, here is an example for which $A_T f(x)$ is smooth while $f(x)$ is not.
Let $M= \mathbb{R^2} =(t,x)$ with the flow $\phi_s (t, x) := (t+s, x)$.
Let $\Psi$ be the following function
$$\Psi(x) = \begin{cases} 
      0 & x\leq 0 \\
      e^{\frac{-1}{x^2}} &  x > 0 \\ 
\end{cases}$$
In particular, we want $\Psi$ to be a smooth function which vanishes to all orders at $0$. Let $f$ be the following function:
$$f(t,x)= \begin{cases}
 e^{\frac{-t^2}{(\Psi(x))^2}} & x >0 \\
 0 & x \leq 0
\end{cases} $$
The function $f$ is not smooth. In fact, it is not even continuous at $(0,0)$. To show that $A_T f(x)$ is smooth for any $T$, we must only show that it depends smoothly on $x$ near $(0,0)$. For fixed $T$>0, we have the following:
$$A_T(f(t_1,x))= \frac{1}{T} \int_{t_1}^{t_1+T} f(t,x) dt <  \frac{\sqrt{ \pi}}{T} \Psi(x) = o(x^k) ~ \forall k >0 $$
From this, it follows that $\frac{\partial^k A_T f}{\partial x^k}(t,0) \equiv 0$, so the derivative of $A_T f$ exists to all orders at $(0,0)$ for $T>0$. The smoothness of $A_T f(x)$ elsewhere is immediate because $f$ is smooth away from the origin.
For an example where $f(x)$ is continuous but not smooth, we can consider the function $\hat f(t,x) = x \cdot f(t,x)$. Here, the same analysis holds except that $f$ is continuous but not differentiable at $(0,0)$.
$\textbf{Original answer with the some corrections}$
This is not a full answer, but I believe that your colleague's result does all the heavy lifting away from the fixed points of the flow, at least once you have good estimates on $A_T f(x)$. The case near a fixed point of a flow seems more difficult. I'm not sure how that works exactly, but if one can understand it in the $1$-dimensional case (i.e. when the flow has a limit point), this might give some good insight. I broke the answer into several sections to help make it more readable.
$\textbf{Finding the right coordinates}$
In this case, the smooth flow locally foliates your manifold, so around any point $p$ one can use the inverse function theorem to choose local coordinates $\{ x_0, x_1, \ldots, x_{n-1} \}$ for which the flow acts as $\phi_t(x_0) = x_0+t$ and $\phi_t(x_i) = x_i$ otherwise. By Professor Burnol's argument (which I'm black-boxing for this argument) $f(  x_0, x_1, \ldots, x_{n-1})$ depends smoothly on $x_0$ when all the other coordinates are fixed.
What remains to show is that it depends smoothly on $x_1, \ldots, x_n$. We start with the case where $x_0$ is some fixed value, say 0. In this case, the $A_T f(x)$ is exactly the following:
$$ \frac{1}{T}\int_0^T f(t, x_1, \ldots, x_n) ~dt$$
For the rest of this, I will use the parameter $t$ for $x_0$ and refer to the rest of the coordinates collectively as $\mathbf{x}$.
We know that $A_T f(x)$ depends smoothly on $\mathbf{x}$ and further that $f$ depends smoothly on $t$. What remains to show is that $f$ depends smoothly on $\mathbf{x}$. We can't take derivatives with respect to $\mathbf{x}$, so we need to estimate some difference quotients. I will be using the Euclidean norm on the $\mathbf{x}$ coordinates for my estimates.
$\textbf{The uniform Lipschitz estimate}$
Take two points $\mathbf{x}^1$ and $\mathbf{x}^2$. Since $A_T f(x)$ is smooth, it is locally Lipschitz, and so we have the following estimate:
$$ \frac{1}{T \| \mathbf{x}^1- \mathbf{x}^2 \| } | \int_0^T f(t, \mathbf{x}^1) - f(t, \mathbf{x}^2)  ~dt | < \mathcal{C} $$
In order for the rest of the argument to work, we must assume that this estimate is uniform in the choice of $T$, $\mathbf{x}^1$ and $\mathbf{x}^2$. This allows us to pick $T>0$ small enough so that that for our fixed $\mathbf{x}^1$ and $\mathbf{x}^2$, we have the following:
$$ \| f(t, \mathbf{x}^i) -f(s, \mathbf{x}^i)  \|_{C^2} < \epsilon$$
for $0<t,s < T$.
The choice of $T$ may depend on $\mathbf{x}^1$ and $\mathbf{x}^2$. In particular, we don't expect to be able to do this uniformly in $\mathbf{x}$, but that's okay because we are only doing it for two points.
Then, we have the following estimate.
\begin{eqnarray}
| \frac{1}{T} \int_0^T \frac{ f(t, \mathbf{x}^1) - f(t, \mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \| }  ~dt | & \geq & | \frac{1}{T} 
\frac{ T(f(0,\mathbf{x}^1) -  f(0,\mathbf{x}^2))}{\| \mathbf{x}^1- \mathbf{x}^2 \|} | \\
& & - \frac{1}{T} \left( \frac{\partial f}{\partial t} (0, \mathbf{x}^1)+ \frac{\partial f}{\partial t} (0, \mathbf{x}^2) + 2\epsilon \right) \frac{T^2}{2} \\
& \geq & | \frac{f(0,\mathbf{x}^1) -  f(0,\mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \|}| - \left( \frac{\partial f}{\partial t} (0, \mathbf{x}^1)+ \frac{\partial f}{\partial t} (0, \mathbf{x}^2) + 2\epsilon \right) T \\
\end{eqnarray}
Letting $T$ be go to zero using the uniformity of the earlier estimate while still fixing $\mathbf{x}^1$ and $\mathbf{x}^2$ for now, our work shows the following.
$$ | \frac{f(0,\mathbf{x}^1) -  f(0,\mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \|}| \leq \mathcal{C} $$
Notice that this $\mathcal{C}$ is the same constant as before in the estimate of $A_T f(x)$. Repeating this for arbitrary pairs of $\mathbf{x}$, this gives a uniform Lipschitz estimate on $f$ (at $t=0)$ in terms of the $\mathbf{x}$ coordinates. 
$\textbf{A sketch of how to use induction for higher order regularity}$
In fact, we can repeat essentially the same argument but with three points $\mathbf{x}^1$, $\mathbf{x}^2$ and $ \lambda \mathbf{x}^1 + (1- \lambda) \mathbf{x}^2$ where instead of using the difference quotient, we use the second order difference quotient. This should yield a uniform $C^{1,1}$ estimate on $f$ in terms of the smoothness of $A_T f(x)$. With a uniform $C^{1,1}$ estimate, this shows that the function $f$ was actually differentiable in terms of $\mathbf{x}$ all along. You can induct on all orders to get smoothness of $f$ in $\mathbf{x}$. This shows that $f$ is smooth in both $t$ and $\mathbf{x}$.<|endoftext|>
TITLE: When is a sequence the sum of two Beatty sequences?
QUESTION [7 upvotes]: In other words, given a sequence $(s_n)$, how can we tell if there exist irrationals $u>1$ and $v>1$ such that 
$$s_n = \lfloor un\rfloor +  \lfloor vn\rfloor$$
for every positive integer $n$?
A few thoughts:  Graham and Lin (Math. Mag. 51 (1978) 174-176) give a test for $(s_n)$ to be a single Beatty sequence $(\lfloor un\rfloor)$ (which they call the spectrum of $u$).  Perhaps someone knows a reference for a test for sums of two or more Beatty sequences?  A special case would be a test for a given sequence $(s_n)$ to be the sum of two complementary Beatty sequences (i.e., $1/u + 1/v = 1$).
In response to comments, the part of the question that says "for every positive integer n" indicates that the intended sequence is infinite.  It seems to me that the question, as stated above, is okay.  If it's undecidable - well, that's of interest. 
In any case, while it may be difficult to give a test that actually finds $u$ and $v$ when such numbers exist, there are some simple tests for deciding that $(s_n)$ is not a sum of two Beatty sequences:
(1) $\lim_{n\to\infty}s(n)/n$ must exist;
(2) if $u$ and $v$ exist, then $u$ + $v$ = $\lim_{n\to\infty}s(n)/n$; 
(3) $\lfloor(u+v)n\rfloor \in \{[un]+[vn],[un]+[vn]+1\}$ for every $n$;
(4) if $(s_n)$ is a sum of two Beatty sequences, then the difference sequence of $(s_n)$ consists of at most three terms; and if there are three, then they are consecutive integers.
It's easy to see how each of those generalizes to give a "negative test"; that is, a way to see that a given $(s_n)$ is not a sum of any prescribed number of Beatty sequences.  I hope that someone can find more "negative tests", or even better, a "positive test", perhaps similar to Graham and Lin's result.

REPLY [5 votes]: Let's use the notation $\{ x\}$ for the fractional part of a number $x$.
Assume $u, v$, and $u/v$ are all irrational.
Then, $\{un\}$ and $\{vn\}$ behave as independent uniform random variables. (This is proved by Fourier analysis, vindicating James Cranch's suggestion) $s_{n+1}-s_n$ depends on  $\{un\}$ and $\{vn\}$:
It is $\lfloor u \rfloor + \lfloor v \rfloor$ if $\{un\} < 1- \{u\}$ and $\{vn\} < 1 - \{v\}$
$\lceil u \rceil + \lceil v \rceil$ if $\{un\} \geq 1- \{u\}$ and $\{vn \} \geq 1- \{ v\}$
and the integer intermediate between those two otherwise.
So by measuring the frequencies with which these $s_{n+1}-s_n$ takes its maximal value, its minimal value, or the intermediate one, we can determine $\{u\} \{v\}$ and $(1-\{u\})(1-\{v\})$, hence by solving the equation $\{u\}$ and $\{v\}$. Then clearly how we distribute the integer part between $u$ and $v$ does not affect $\lfloor un \rfloor + \lfloor vn \rfloor$, so we can choose any values of $u$ and $v$ with the correct fractional part and sum and plug them in to see if they fit our sequence.
In fact the same thing works if just $u/v$ is irrational, because they are independent non-uniform random variables, and the probability that $\{un\}$ and $\{vn\}$ is in the range to increase by a larger amount is still $\{u\}$ or $\{v\}$ just because the average increase of $\lfloor un\rfloor$ or $\lfloor vn \rfloor$ must be $u$ and $v$ respectively.
Does this still work even if the ratio $u/v$ is rational?<|endoftext|>
TITLE: Convergence of Fixed-Point Iteration of a dependent map
QUESTION [6 upvotes]: Suppose that we have two mappings $T_1(\cdot): Y \mapsto Y$ and $T_2(\cdot,\cdot) : X \times Y \mapsto X$ where both $X$ and $Y$ are compact and convex subsets of the same Euclidean space. Furthermore $T_1(\cdot)$ is nonexpansive, i.e. has Lipschitz constant equal to $1$, $T_2(\cdot, y)$ is nonexpansive for all fixed $y \in Y$ and $T_2(x,\cdot)$ is uniformly Lipschitz continuous, i.e. there exists one Lipschitz constant for all $x \in X$ .
We know that for any $\lambda \in (0,1)$ and any $y_0 \in Y$ the iteration
$y_{n+1} = (1-\lambda) y_n + \lambda T_1(y_n)$ converges to a fixed point $\bar{y} = T_(\bar{y})$.
We also know that for any $\rho \in (0,1)$ and any $x_0 \in X$ and a fixed $y \in Y$ the iteration
$x_{n+1} = (1-\rho) x_n + \rho T_2(x_n, y)$ converges to a fixed point $\bar{x} = T_(\bar{x}, y)$.
My question is: Does the iteration
$x_{n+1} = (1-\rho) x_n + \rho T_2(x_n, y_n)$
converge to a fixed point $\bar{x} = T_2(\bar{x}, \bar{y})$?
Does anybody have a hint in proving or disproving it?

REPLY [2 votes]: Take $T_1(y)=y-y^2$ with $y\in[0,1]$ and $T_2(x,y)=e^{iy}x$, $x\in\mathbb C, |x|\le 1$. Now take $x_0=1$, $y_0=1/2$, say. Then all assumptions hold, but $y_n\approx c/n$, so the rotations in the iterations sum up to infinity like a harmonic series but the contractions of absolute value of $x$ multiply to a non-zero number like the product of $e^{-n^{-2}}$, and there is no convergence.
It looks like this is the only bad scenario in the sense that if you can somehow guarantee in addition that the sum $\sum_n|y_n-\bar y|$ is finite, or that the fixed point of $T_2(\cdot,\bar y)$ is unique, or something else that would prevent this ridiculous cycling over the set of the fixed points of the limiting mapping, then the desired conclusion should follow but, since I have no idea what exactly your setup is, I haven't tried to check the details, so I may be overly optimistic here.<|endoftext|>
TITLE: Is it possible to evaluate Connect 4 positions with Combinatorial Game Theory?
QUESTION [5 upvotes]: The surreal numbers in Combinatorial Game Theory only work for certain classes of games (e.g. they must satisfy normal play convention).  This rules out even reasonable games with fairly well-understood scoring (such as Go).  In that case, we approximate Go with a normal-play convention game where the last player to move wins and take the score of that (Mathematical Go: Chilling Gets the Last Point).  Dots and Boxes have been analyzed using Nimbers.
Can we analyze games of Connect-4?  Is it possible to quantify the degrees of advantage for one side or another using surreal numbers?  Here, whoever moves first loses so this game is a "number", if I recall, as the first move is not desirable for either side.
What could be the value of this position?  The reasoning could go:

$\color{#5256ED}{\textbf{blue}}$ has two potential connect 4's along the diagonal if $\color{#F76353}{\textbf{red}}$ moves first in the left or right column
$\color{#F76353}{\textbf{red}}$ must also avoid moving first in the 3rd column
if $\color{#5256ED}{\textbf{blue}}$ moves first in the 3rd column, $\color{#F76353}{\textbf{red}}$ must follow with the same, for the win
theoretially if $\color{#5256ED}{\textbf{blue}}$ moves twice in the right column, $\color{#F76353}{\textbf{red}}$ can place a chip to win
Columns 1 and 5 (from the left seem to have effect on the game leaving columns 1, 3 and 7
...

Very simplistically there is a $\color{#5256ED}{\textbf{3}}-\color{#F76353}{\textbf{1}}$ advantage, but the game tree should cover all this information and provide a "numerical" answer of sorts.

REPLY [4 votes]: There is a natural definition of a "connect-four monoid". I don’t know if it's mentioned in the literature, and I don’t know what it looks like, apart from some observations (below) that show that it is not a group. If there is a reasonably simple description of this monoid, it could lead to a “solution” of connect-four which is applicable to arbitrary board sizes. But it might be too complicated to be of any help. 

To begin with, one should think about how the game naturally splits into components, and what it means for a game position to be the “sum” of its components. 
As the game proceeds, the set of unoccupied sites can separate into components that do not interact with each other. In such a situation, a move consists in making a move in one of the components (this is like the classical theory), and winning in one component means winning the entire game (this differs from the so-called normal playing convention where the game ends when a player is unable to move). 
In classical combinatorial game theory, games are classified into four outcome classes: Positive (Left wins no matter who starts), Negative (Right wins), Fuzzy (Player to move wins) and Zero (Player not to move wins). In a game like connect-four that can end in a draw, there are nine outcome classes, namely the functions from {Red to move, Blue to move} to {Red wins, Draw, Blue wins}.
Just as in the classical theory, the set of positions is an abelian monoid under addition (position means what the board looks like, without information about who is to move), and one may define two positions $X$ and $Y$ to be equivalent if for every $Z$, the games $X+Z$ and $Y+Z$ belong to the same outcome class.
For instance, all positions with an even number of unoccupied sites, and where none of the players can win, are equivalent. These might be called “zero-positions”, since they include the empty position (neutral element of the monoid).
The additive structure carries over to addition of equivalence classes, but unlike the classical case, it doesn’t become a group. The zero class is a neutral element, but not all positions have inverses. For instance, if $X$ is a position is such that the player to move (whether Red or Blue) can win in one move, then there can be no other position $X’$ so that $X+X’$ becomes zero, because no matter what we add to $X$, the resulting game will still be a win for the player to move.
On the other hand some positions clearly do have inverses. There is a class that contains all positions with an odd number of free sites, and where nobody can win. We might call this class $\star$, since it is similar to the class "star" in the classical theory. These positions are clearly not equivalent to the zero positions, since adding them will turn a “zugzwang” into a first player win. But we do have the equation $\star + \star = 0$.
There are several games, in particular misère games and card games, where the analysis of a corresponding monoid leads to a solution, see for instance my paper The strange algebra of combinatorial games and its references. 
A couple of questions: Is the connect-four monoid infinite? What do the connect-two and connect-three monoids look like?
Finally, it seems worth taking a look at the Masters thesis of Victor Allis from 1988: A Knowledge-based Approach of Connect-Four.<|endoftext|>
TITLE: Can phase significantly concentrate a function's spectrum?
QUESTION [7 upvotes]: Let $F$ denote the Fourier transform over some group. What is known about the following quantity?
$$\gamma:=\inf_{x\neq 0}\frac{\|Fx\|_1}{\|F|x|\|_1}$$
Here, $|x|$ denotes the pointwise absolute value of $x$. We know $\gamma\leq1$ since $x$ can be pointwise nonnegative. In the case where $F$ is the DFT, we also know $\gamma>0$ by a compactness argument, and computer simulations give $\gamma\leq0.72$ by taking $x$ to be some random bandlimited even function.
EDIT: Following Josep's comment, we actually know $\gamma\geq1/\sqrt{n}$ in the DFT case by passing to the 2-norm and applying Parseval.
I would mostly like to know if $\gamma\gg0$ (i.e., there exists a constant $\epsilon>0$ independent of $n$ such that $\gamma\geq\epsilon$), since this would imply that $\|Fx\|_1$ is small (i.e., $Fx$ is concentrated) only if $\|F|x|\|_1$ is also small. This would confirm my intuition that phase doesn't play much of a role in the concentration of a function's spectrum. Presumably, this question is natural enough to have been studied already.

REPLY [5 votes]: Observe that if $n$ and $m$ are relatively prime,  $\gamma_{nm}\leq \gamma_n \gamma_m$ because you can combine functions on $\mathbb Z/n$ and $\mathbb Z/m$ using the Chinese remainder theorem.
So if we show $\gamma_n \leq 1- \epsilon$ for some $\epsilon$ and all sufficiently large $n$, then we can make $\gamma_n$ arbitrarily small.
Probably $x(t) =  \sin( 2\pi t/n)$ works for this. I think it's clear that in this case the ratio is asymptotic to
$$ \frac{ \sum_{k \in \mathbb Z}\left| \int_{0}^{2 \pi} \sin (x) e^{ i k x} \right|}{ \sum_{k \in \mathbb Z}\left| \int_{0}^{2 \pi} |\sin (x)| e^{ i k x} \right|}$$
where the numerator is clearly $2\pi$ and the denominator is
$$ \sum_{k \in 2 \mathbb Z} \left|\frac{-4}{k^2-1}\right|= 8 $$
This gives $\pi/4$, agreeing with Dustin's numerical experiments, which is bounded away from $1$.

The asymptotic statement is from the Poisson summation-like formula. If $f$ is a function on the unit circle, and $f_n$ is the function on $\mathbb Z/n$ you get by restricting it to the $n$ roots of unity, then
$$\hat{f_n} ( a) =\sum_{k \in \mathbb Z} \hat{f} ( a + nk)$$
so 
$$ \lim_{n \to \infty} ||\hat{f}_n||_1  = || \hat{f} ||_1$$
The integral I found the calculations too fiddly so I just did it on wolframalpha, but you can do it by breaking up $[0,2\pi]$ into $[0,\pi]$ and $[\pi, 2\pi]$. The sum is by telescoping from $\frac{1}{k-1} - \frac{1}{k+1} = \frac{2}{k^2-1}$.

Here's a method for arbitrary $\mathbb Z/n$. Let $a_1, \dots, a_k$ be elements of $\mathbb Z/n$ and consider
$$x(t) = \prod_{i=1}^k \sin (a_k x/n) $$
Supposing that $\sum_{i=1}^k \pm a_i$ is nonvanishing, the $L^1$ norm of $Fx$ is $1 $ if you normalize it properly.
Now I claim for any group homomorphism $G \to H$, given a function on $H$, the Fourier transform of its pullback to $G$ can be computed on a character of $G$ by summing the Fourier transform of $H$ over all characters of $H$ that pull back to that character of $G$. This is the same as the Poisson summation formula and is proved the same way. Here I'm normalizing by the measure of the group.
Apply this to the map $\mathbb Z/n \to (\mathbb R/\mathbb Z)^n$ that sends $t$ to $(a_1t/n, \dots, a_kt/n)$
Let $g(m)$ be the Fourier transform of $|\sin x|$. Then the Fourier transform of $\left| \prod_{i=1}^k sin x_i \right| $ is $\prod_{i=1}^k g(m_i)$.
$$F |x| ( \xi) = \sum_{(y_1, \dots, y_k) \in \mathbb Z^k } 1_{\sum_{i=1}^k a_i y_i \equiv \xi } \prod_{i=1}^k  g(y_i)$$
This is the sum of $\prod_{i=1}^k  g(y_i)$ over the lattice  in $\mathbb Z^k$ satisfying the condition $\sum_{i=1}^k a_i y_i \equiv \xi $. 
Now we know that $\prod_{i=1}^k  g(y_i)$ has $L_1$ norm $ (4/\pi)^k$, and at most $(2/3) (4/\pi)^k$ of its mass is in the box where $|y_i| < Ck$, because $1-O(1/k)$ of the mass of $g(y)$ is in the box where $|Y| < k $.
So as long as no equation of the form $\sum_{i=1}^k y_i a_i =0$ for $y_i$ not all $0$ but $|y_i| < 2 C k$ holds, then no two points in that box get sent to the same $\chi$, so the $L_2$ norm goes down by at most $1-2/3$ (in the worst case where everything outside the box cancels something inside the box.)
When can we find $a_1, \dots, a_k$ such that no such equation holds? If $n=p$ is a prime then by selecting $a_1, \dots, a_k$ randomly, the probability of any given equation holds is $1/p$, so as long as $p > (2 C k)^k$ then sometimes none of the equations hold.
If we're not a prime then we can just do the Chinese Remainder Theorem thing, or do the probabilistic argument more carefully.<|endoftext|>
TITLE: Equivalent Norms on Sobolev Spaces
QUESTION [6 upvotes]: When $k$ is a positive integer and $1<p<\infty$, we know that there is some
$C>0$ such that for all $u\in W^{k,p}\left(\mathbb{R}^{N}\right)  :$
$$
\left\Vert \left(  -I+\Delta\right)  ^{\frac{k}{2}}u\right\Vert _{p}\leq
C\left(  \left\Vert \left(  -\Delta\right)  ^{\frac{k}{2}}u\right\Vert
_{p}+\left\Vert u\right\Vert _{p}\right)  .
$$
(Stein, Singular Integrals and Differentiability Properties of Functions; etc)
The question is that can we have the following estimate: for $\varepsilon>0$,
there exists $C\left(  \varepsilon\right)  >0$ such that for all $u\in
W^{k,p}\left(\mathbb{R}^{N}\right)  :$
$$
\left\Vert \left(  -I+\Delta\right)  ^{\frac{k}{2}}u\right\Vert _{p}%
\leq\left(  1+\varepsilon\right)  \left\Vert \left(  -\Delta\right)
^{\frac{k}{2}}u\right\Vert _{p}+C\left(  \varepsilon\right)  \left\Vert
u\right\Vert _{p}?
$$
Note that when $p=2$, the answer is yes via Fourier transform. But I have no idea in the general case.

REPLY [4 votes]: If $k \in (0, 2]$, we define the multiplier 
$$
  m (\xi) =  (1 + \vert \xi \vert^2)^\frac{k}{2} - \vert \xi \vert^k.
$$
We observe that if $\vert \xi \vert \ge 2$, then by differentiability
$$
  \big \vert (1 + \vert \xi \vert^2)^\frac{k}{2} - \vert \xi \vert^k \big\vert
= \vert \xi \vert^k \Big \vert \Big(1 + \frac{1}{\vert \xi \vert^2}\Big)^\frac{k}{2} - 1 \Big\vert
\le C \vert \xi \vert^{k - 2},
$$
so that $m$ is bounded on $\mathbb{R}^N$.
Similarly,
$$
\vert   D^\ell m (\xi)\vert \le \frac{C_\ell}{\vert \xi \vert^\ell}.
$$
By the classical Mikhlin multiplier theorem, this implies that 
$$
 \big\Vert(-\Delta + I)^\frac{k}{2}u - (-\Delta)^\frac{k}{2}u \big\Vert_p
\le C \Vert u \Vert_p,
$$
from which the estimate follows.
If $k \in (2, \infty)$, we define the multiplier
$$
  m (\xi) =  \frac{(1 + \vert \xi \vert^2)^\frac{k}{2} - \vert \xi \vert^k}{1 + \vert \xi \vert^{k - 2}}.
$$
It can be checked that the multiplier satisfies also the conditions of the Mikhlin multiplier theorem. Therefore,
$$
\big \Vert \big((- \Delta + I)^\frac{k}{2} - (-\Delta)^\frac{k}{2}\bigr)\big((-\Delta)^{\frac{k}{2} - 1} + I\big)^{-1} v\Vert_p
\le C\Vert v \Vert_p.
$$
If we set $v = (-\Delta)^{\frac{k}{2} - 1}u + u$,
then 
$$\big \Vert \big((- \Delta + I)^\frac{k}{2} - (-\Delta)^\frac{k}{2}\bigr)u\Vert_p
\le \Vert (-\Delta)^{\frac{k}{2} - 1}u + u \Vert_p
\le \Vert (-\Delta)^{\frac{k}{2} - 1}u\Vert_p + \Vert u \Vert_p
$$
By a classical interpolation
$$ 
\Vert (-\Delta)^{\frac{k}{2} - 1}u\Vert_p
\le C \Vert (-\Delta)^\frac{k}{2}u\Vert_p^\frac{k - 2}{k}
\Vert u \Vert_p^\frac{2}{k}.
$$
The requested inequality follows then from Young’s inequality.<|endoftext|>
TITLE: What is the reverse mathematical strength of the fundamental theorem of algebra?
QUESTION [32 upvotes]: Reverse mathematics (RM) is that area that tries to pin down exactly which axioms are necessary to prove theorems, given some weak base theory. Harvey Friedman has pointed out several times (on the FOM mailing list) that $Con(PA)$ is equivalent to a variant of Bolzano-Weierstrass over the rationals between 0 and 1 inclusive (something like: every sequence has a subsequence $\{q_i\}$ which is Cauchy, in the that sense $\forall i,j \geq n, |q_i - q_j| < 1/n$). Apparently a very similar result is given in Simpson's book and "it is clear to the experts" how to get to Friedman's claim. (As an aside, I find this such an amazing result it should be written up for the average mathematician, and not buried in a vaguely equivalent form in a book that is hard to get one's hands on.)
The reason I bring up Bolzano-Weierstrass is that Todd Trimble, in a nice answer on Ways to prove the fundamental theorem of algebra, uses B-W to prove (as the key tool among other, elementary considerations) the fundamental theorem of algebra. Todd then had a look to see, at my behest, if the RM strength of FTA was known. He came up blank, so I ask here:

How close in reverse mathematical strength are the Bolzano-Weierstrass statement from Friedman's claim and the fundamental theorem of algebra?

If they are the same, then we find ourselves in the amazing situation that the consistency of PA is equivalent to a theorem that we all would use with no qualms whatsoever. However, I have a vague feeling that FTA is strictly weaker than BW (as used here), but cannot make this precise.

REPLY [10 votes]: It is perhaps instructive to see how Todd's proof of the FTA steps outside of $\mathsf{RCA}_0$ and how it could be modified to fit into $\mathsf{RCA}_0$.
First, let's review some aspects of compactness in subsystems of second-order arithmetic.

Totally bounded complete metric spaces can be formalized in $\mathsf{RCA}_0$ and it is straightforward to show in $\mathsf{RCA}_0$ that $[0,1]$, closed balls in $\mathbb{R}^n$ and closed discs in $\mathbb{C}$ are totally bounded complete metric spaces. [Simpson Definition III.2.3 and following examples. Somewhat confusingly, Simpson uses "compact metric space" for totally bounded complete metric spaces.]
The system $\mathsf{ACA}_0$ is equivalent over $\mathsf{RCA}_0$ to the statement that every totally bounded complete metric space is sequentially compact [Simpson Theorem III.2.7]. The Bolzano-Weierstrass Theorem is a special case of this and the further special case of $[0,1]$ is already equivalent to $\mathsf{ACA}_0$ over $\mathsf{RCA}_0$ [Simpson Theorem III.2.2].
The system $\mathsf{WKL}_0$ is equivalent over $\mathsf{RCA}_0$ to the statement that every totally bounded complete metric space is compact [Simpson Theorem IV.1.5]. The Heine-Borel Theorem is a special case of this and the further special case of $[0,1]$ is already equivalent to $\mathsf{WKL}_0$ over $\mathsf{RCA}_0$ [Simpson Theorem IV.1.2].

$\mathsf{WKL}_0$ is strictly stronger than $\mathsf{RCA}_0$ and $\mathsf{ACA}_0$ is strictly stronger than $\mathsf{WKL}_0$. The first-order fragment of $\mathsf{ACA}_0$ is $\mathsf{PA}$ and every model of $\mathsf{PA}$ can be expanded to a model of $\mathsf{ACA}_0$; the $\mathsf{RCA}_0$ and $\mathsf{WKL}_0$ have the same first-order fragment, $\mathsf{PA}$ with induction restricted to $\Sigma_1$-formulas, and likewise any first-order model of that theory can be expanded to a model of $\mathsf{WKL}_0$ (and hence of $\mathsf{RCA}_0$).
Now back to Todd's proof. As in Todd's proof, let $f(z)$ be a nonconstant polynomial over $\mathbb{C}$.
Todd first step uses the Bolzano-Weierstrass Theorem (BWT) for closed discs in $\mathbb{C}$. At first sight, this requires $\mathsf{ACA}_0$ but the point is to show that $|f(z)|$ attains a minimum value. The weaker subsystem $\mathsf{WKL}_0$ already proves that every continuous real-valued function on a totally bounded complete metric space attains a minimum value [Simpson Theorem IV.2.2]. The remaining parts of Todd's proof are direct computations, so now we have a proof in $\mathsf{WKL}_0$. Since the Extreme Value Theorem is equivalent to $\mathsf{WKL}_0$ over $\mathsf{RCA}_0$ [Simpson Theorem IV.2.3] it seems that we can't do much better. This would be the case if $f(z)$ were an arbitrary continuous function, but polynomials aren't arbitrary at all!
The first key fact about polynomials is that they have computable modulus of uniform continuity. In fact, for every closed disc $D$, we can use the coefficients of $f(z)$ to compute a constant $C_D$ such that $|f(x)-f(y)| \leq C_D|x-y|$ for all $x, y \in D$. This can be carried out in $\mathsf{RCA}_0$ and this is enough to show that $\inf\{|f(z)| : z \in D\}$ exists. However, the modulus of uniform continuity alone is not enough to show that this infimum is attained.
The second key fact about polynomials is that they can only have finitely many roots and we can calculate exactly how many there should be: the degree of $f(z)$ minus the degree of the gcd of $f(z)$ and $f'(z)$. Let's call this number $n$. This allows us to home in on a root in a deterministic fashion rather than using Bolzano-Weierstrass. Given disjoint closed  disks $D_1,\ldots,D_n$ such that $\inf\{|f(z)|: z \in D_i\} = 0$ for $i = 1,\dots,n$, we can effectively compute rapidly convergent Cauchy sequence for the unique roots $r_1 \in D_1,\ldots,r_n \in D_n$. To find an element of $D_i$ within $\varepsilon \gt 0$ of the purported $r_i$, find a small $\delta \gt 0$ and a cover of $D_i$ with open balls $B(z_1,\delta),\ldots,B(z_k,\delta)$ such that the elements of $\{z_j : |f(z_j)| \lt C_{D_i}\delta\}$ are all within $\varepsilon$ of each other and pick any element of that set.
I don't see how to adapt Todd's proof to show that $\inf\{|f(z)| : z \in \mathbb{C}\}$ must be $0$ without showing first that the infimum is attained. However, the two key facts above do explain how to circumvent Bolzano-Weierstrass when working with polynomials rather than arbitrary functions.<|endoftext|>
TITLE: Determinant of block tridiagonal matrices
QUESTION [5 upvotes]: Is there a formula to compute the determinant of block tridiagonal matrices when the determinants of the involved matrices are known?
In particular, I am interested in the case
$$A = \begin{pmatrix} J_n & I_n & 0 & \cdots & \cdots & 0 \\ I_n & J_n & I_n & 0 & \cdots & 0 \\ 0 & I_n & J_n  & I_n & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots &  \ddots & 0 \\ 0 & \cdots & \cdots & I_n & J_n & I_n \\ 0 & \cdots & \cdots & \cdots & I_n & J_n \end{pmatrix}$$
where $J_n$ is the $n \times n$ tridiagonal matrix whose entries on the sub-, super- and main diagonals are all equal to $1$ and $I_n$ is identity matrix of size $n$.
I have asked this question before on MathStackExchange, where a user came up with an algorithm. Nevertheless, I am interested if there is an explicit formula (or at least, if one can say in which cases the determinant is nonzero).

REPLY [5 votes]: The Kronecker product idea brought up in Algebraic Pavel's comment on the original maths stack exchange question seems like a good way to approach the particular case of interest to you. Specifically, assuming $A$ is $m n \times m n$, i.e., there are $m$ block rows and columns, then
$$A = J_m \otimes I_n + I_m \otimes J_n - I_{mn},$$
and the $mn$ eigenvalues of $A$ are given by
$$\lambda_{ij} = \Big(1+2 \cos \frac{i \pi}{m+1}\Big) + \Big(1+2 \cos \frac{j \pi}{n+1}\Big) - 1, \qquad 1 \le i \le m, 1 \le j \le n.$$
(I used the formula for the eigenvalues of the $J$ matrices from Denis Serre's answer here.) The determinant is then
$$\det A = \prod_{i=1}^m \prod_{j=1}^n \lambda_{ij}.$$
If you're only after characterizing when $A$ is singular, then you need only determine when any of the $\lambda_{ij}$ can be zero, which looks fairly straightforward.<|endoftext|>
TITLE: When is alternating sum $\sum_{i}f(a_i)-\sum_{i<j}f(a_i+a_j)+\ldots+(-1)^{n-1}f(a_1+\ldots+a_n)$ always positive?
QUESTION [13 upvotes]: Let $a_1,a_2,\ldots,a_n\geq 1$, and let $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$. Consider the sum
$$S(f)=\sum_{i}f(a_i)-\sum_{i<j}f(a_i+a_j)+\sum_{i<j<k}f(a_i+a_j+a_k)-\cdots+(-1)^{n-1}f(a_1+\cdots+a_n).$$
This question shows that if $f(x)=\frac1x$, then $S(f)>0$ for all $a_1,\ldots,a_n$. If we perturb $f$ a tiny bit, say $f(x)=\frac{1}{x}-\frac{1}{100x^{100}}$, I would imagine that $S(f)>0$ still always holds. But the proof method for $f(x)=\frac1x$ is hard to generalize to other functions. Can we prove it in some other way?
More generally, is there a theorem out there stating sufficient conditions under which $S(f)>0$ always holds?

REPLY [3 votes]: If $f$ is a polynomial with $\deg f< n$, then $S(f)=f(0)$.

ADDED. More generally, for an analytic function $f(x)$,
$$S(f) = f(0) - h(x),$$
where $h(x)$ is the Hadamard product of $f(x)$ and the function
$$g(x) := \int_0^{\infty} e^{-t} (1-e^{a_1tx})\cdots(1-e^{a_ntx})\,\mathrm{d}t.$$
It can be seen that $[x^k]\ g(x)=0$ for $k<n$, implying the above result for polynomials.<|endoftext|>
TITLE: Kullback Leibler "variance": does that divergence have a name?
QUESTION [6 upvotes]: If you consider two probability distributions $p$ and $q$, one way to measure the distance between the two is the Kullback-Leibler divergence:
$$KL(p,q)=\int p \log (p/q) = E_p(\log p/q)$$
and this has many good properties.
I'm currently writing an article in which I want to use what I call the KL variance:
$$KL_{var}(p,q) = var_p(\log p/q) = \int p \log^2  (p/q) - KL(p,q)^2$$
Which also has many good properties
I have searched around quite a bit for references to this divergence, and I haven't found anything. Does anybody have an existing reference to this divergence ? Are there any names which would be slightly more catchy than KL-variance?

REPLY [6 votes]: Too late to help you I guess but I'll leave this here for future reference. This quantity shows up a lot in Bayesian nonparametrics, when proving (frequentist) posterior contraction rates, so "posterior contraction rates" is a useful search. I don't know of any snappy names for it, but theorem 8.3 (and the comment afterwards) in Ghosal, Ghosh, van der Vaart, Convergence rates of posterior distributions, 2000, gives the following bound:
$$var_p \log(p/q) \leq 4h^2(p,q) \lVert{p/q}\rVert_\infty,$$ where $h$ is the Hellinger distance $h(p,q)^2=\int (\sqrt{p}-\sqrt{q})^2$:
https://projecteuclid.org/euclid.aos/1016218228<|endoftext|>
TITLE: Hyperbolic knot complement groups and relative dimension
QUESTION [5 upvotes]: Let $G$ be a the fundamental group of a hyperbolic knot complement. 
Then $G$ is hyperbolic relative to a subgroup $P\cong \mathbb Z \oplus \mathbb Z$. 
The knot complement has a $2$-dimensional spine with contractible universal cover illustrating that $G$ has geometric dimension $2$.
Question: Does there exist a $2$-dimensional $G$-complex $X$ with the property that the fixed point set $X^Q$ is contractible (in particular non-empty) for every subgroup $Q$ conjugate to a subgroup of $P$ (in particular $X$ is contractible)? 
EDIT: Add the condition that the $G$-stabilizer of every cell is a parabolic subgroup -- this is part of the definition of $E_{\mathcal F}G$. 
The question can be phrased as whether there exist a $2$-dimensional model for $E_{\mathcal F}G$ where $\mathcal F$ is the full family of parabolic subgroups of $G$.
One obtains a $3$-dimensional $G$-space with this property by coning-off the flats in the universal cover of the $2$-dimensional spine. The flats are isolated and it is conceivable that one can push them towards the cone points. 
The motivation of the question is that I have recently proved an extension for toral relatively hyperbolic groups of Gersten's result that finitely presented subgroups of hyperbolic groups of cohomological dimension $2$ are hyperbolic. If the answer to the question is known, then $G$ illustrates the result since finitely generated subgroups of $G$ are either a virtual fibers or geometrically finite.  ---Apologies, for the many references of important work that I am skipping here---

REPLY [7 votes]: Proposition 9.6 of M. Kapovich's 'Homological dimension and critical exponent of Kleinian groups' (it's Proposition 9.5 of the arXiv version) asserts that the cohomological dimension of the pair $(G,\mathcal{F})$ is equal to $\mathrm{dim}~\Lambda(G)+1$, where $\Lambda(G)$ is as usual the limit set of $G$. For cofinite $G$, $\Lambda(G)$ is the full 2-sphere and so $cd(G,\mathcal{F})=3$. In particular, no 2-dimensional model exists.<|endoftext|>
TITLE: Complements of unknotted tori (higher dimensions)
QUESTION [10 upvotes]: It is well-known that an unknotted 2-torus in $S^3$ provides the standard Heegaard splitting, in particular its complement consists of two solid tori.
It is also known that an unknotted 3-torus in $S^4$ bounds some $D^2\times T^2$, but that (somewhat surprisingly) the complement of that unknotted $T^2\times D^2$ is what is called the "Montesinos twin", which is constructed as the regular neighborhood of the union of two 2-spheres (intersecting transversely in 2 points with opposite intersection number) embedded in the 4-sphere.
Question: what is the higher-dimensional analogue of the Montesinos twin, i.e., has someone described in explicit terms (say, a handle decomposition) the topology of the 2 connected components of $$S^{n+1}\setminus T^n,$$
the complement of an unknotted n-torus in $S^{n+1}$?

REPLY [2 votes]: You are asking two quesions: embeddings of a torus ( I am assuming a prodcut of two spheres of some dimensions) in codimension 1 (one dimension higher) and in codimension 2.  I will answer both.
Use the join structure for $S^n$ so that $S^p * S^q = S^{p+q+1}$.
Think of the join as $S^p \times S^q \times D^1$ with identifications.
Then the standard unknotted torus in $S^p \times S^q \subseteq S^{p+q+1}$ is the "middle of the join" and corresponds to $S^p \times S^q \times \{0\}$. 
The subset of the join corresponding to $S^p \times S^q \times [-1,0]$ is a mapping cylinder of the projection map $S^p \times S^q \to S^p$ and is homeomorphic to $S^p \times D^{q+1}$.  
Similarly the subset of the join corresponding to $S^p \times S^q \times [0,1]$ is a mapping cylinder of the projection map $S^p \times S^q \to S^q$ and is homeomorphic to $S^q \times D^{p+1}$.
So the standard unknotted torus separates $S^{p+q+1}$ into two generalized solid tori. 
That is the codimension 1 case; the second example is codimension 2 and is more interesting.  Basically the complement of the unknotted $S^p \times S^q$ in $S^{p+q+2}$ is "a thickening of a $p+1$-sphere and a $q+1$-sphere which meet transversely in two points".
We take the standard unknotted  to be 
$S^p \times S^q \subseteq S^p \times S^{q+1} \subseteq S^p * S^{q+1} = S^{p+q+2}$.  
Suppose $S^0$ consists of two points $n$ and $s$. Now view $S^{q+1} = S^0 * S^q$---this is more commonly called the "suspension of$S^q$" and $n$ called the "north pole", $s$ the "south pole".
Now consider $X_n = S^p \times \{n\} \subseteq S^p \times S^{q+1} \subseteq S^p * S^{q+1}$ and $X_s = S^p \times \{s\} \subseteq S^p \times S^{q+1} \subseteq S^p * S^{q+1}$.  and finally let $X= X_n \cup X_s$. Restrict the projection $S^p \times S^q \to S^p$ to $X$ and the resulting mapping cylinder $M_1$ is homeomorphic to $S^p \times D^1$.  Restrict the projection $S^p \times S^q \to S^q$ to $X$ and the resulting mapping cylinder $M_2$ is homeomorphic to two $p+1$ balls. 
Let $A= M_1 \cup M_2$---then $A$ is homoemorphic to a $p+1$-sphere.  Let $B$ be the natural copy of $S^q$ in $S^p *S^q$.  Note that $A \cap B$ consists of two points.  If you carefully relate the join structure with the standard smooth structure of the sphere you can see that $A$ and $B$ will meet transversely. 
This union is the equivalent of the Montesinos twin in higher dimensions.  In particular the complement of the unknotted co-dimension 2 torus deforms in a very simple way to $A \cup B$.<|endoftext|>
TITLE: PSL(2,p) as quotient of triangle groups
QUESTION [8 upvotes]: As a by-product of some Magma computations, I've observed that, for each prime $p$
such that ${\rm PSL}(2,\mathbb{F}_p)$ can be a quotient of the $(3,3,4)$-triangle group
(i.e. $p \equiv \pm 1 (\!\!\!\mod 8)$), there seem to be exactly two normal subgroups
of the $(3,3,4)$-triangle group such that the quotient is isomorphic to
${\rm PSL}(2,\mathbb{F}_p)$.  Surely this is a known fact.  Can anyone please provide a proof or a reference?

REPLY [4 votes]: I think I can show that there are at most $2$ for all primes $p$ using rigid triples / rigid local systems.
We are looking at triples $x,y,z$ in $SL_2(\mathbb F_p)$ that generate the group and multiply up to $1$, with $x$ and $y$ of order $3$ and $z$ of order $4$, up to global conjugacy
Any such triple can be lifted to $SL_2(\mathbb F_p)$ in four different ways, and there is a unique way of doing it so that $x$ and $y$ remain order $3$. Then $z$ will be order $8$. It is sufficient to show that there are only two of these up to conjugacy.
There is one conjugacy class of order $3$ in $SL_2(\mathbb F_p)$ but there are two conjugacy classes of order $8$, because there are four primitive $8$th roots of unity in two pairs that each multiply to $1$.
I claim for each conjugacy class, there is a unique triple of two order $3$ elements, and an order $8$ element in that conjugacy class, that multiply to one up to global conjugacy. That is, it is a "rigid triple".
Given two such triples, view them as representations of the free group $F_2$ and hence as sheaves on $\mathbb P^1$ minus three points. Take the tensor product $V \otimes W^{\vee}$, and take the parabolic cohomology / cohomology of its middle extension to $\mathbb P^1$. This is a rank $4$ sheaf on a surface of Euler characteristic $2$, for an Euler characteristic of $8$, but it drops to a rank $2$ sheaf at the $3$ singular points, for a contribution of $-6$, so the total is $2$. 
Because the Euler characteristic is positive, $H^0$ or $H^2$ must be nonvanishing. But that can only happen if the sheaf has a global section, which would be an isomorphism between the two representations, showing that the two triples are conjugate.
Hence there are at most two pairs, one for each conjugacy class. By Ian Agol's argument we can show that they are both inhabited when $p \equiv \pm 1\mod 8$, so there are exactly two in that case.<|endoftext|>
TITLE: Generate Bernoulli vector with given covariance matrix
QUESTION [9 upvotes]: I am from different background, so please forgive me if the answer is so well known. 
Let $C=(c_{ij})$ be a given $n\times n$ matrix. Do we have a way to generate samples of random Bernoulli vectors with covariance matrix equal to $C$?
More specifically, are there functions that take in Bernoulli random vectors with i.i.d. entries, and output Bernoulli random vectors with covariance matrix equaling $C$.

REPLY [6 votes]: I'm going to provide two algorithms here:

A super-exponential-time solution that works in all cases.
A polynomial-time solution that applies if the mean values are known and if a certain matrix is positive semi-definite.

The polynomial-time solution essentially recycles the end of the proof of Goemans-Williamson's result for approximating MAX-CUT.  I include the super-exponential-time case just because (Dustin G. Mixon's clear sketch notwithstanding) the other solutions do not seem to specify their algorithms completely.
General setup
Let $x=(x_1,...,x_n)$ be the random variable with covariance matrix $C$.
Note that by a "random Bernoulli vector" we might mean $x_i\in\{0,1\}$ or $x_i\in\{-1,1\}$.  We can convert the covariance matrix from the former version to the latter by multiplying by 4, so we can adopt either convention without issue.  I'll consider $\{-1,1\}$.
Let $\overline{x_i}=\mathbb{E}[x_i]$.  We are given the covariances:
$$c_{i,j}=\mathbb{E}[(x_i-\overline{x_i})(x_j-\overline{x_j})]=\mathbb{E}[x_ix_j]-\overline{x_i}\,\overline{x_j}$$
Note that $c_{i,i}=\mathbb{E}[x_i^2]-\overline{x_i}^2=1-\overline{x_i}^2$, which implies that
$$\overline{x_i}=\pm\sqrt{1-c_{i,i}}$$
Super-exponential-time algorithm
With an outer loop of size $2^n$, we can try all possible sign combinations for the $\overline{x_i}$.
For the inner loop, then, we can assume that we know $\overline{x_i}$.  Suppose we are in the inner loop.
Let $\mathcal{B}$ be the domain of $x$ (so, $\mathcal{B}$ has $2^n$ elements).  Consider a linear program with variables $p_B$ for each $B\in \mathcal{B}$.  This represents the probability that $x=B$.  All the constraints can be expressed as linear combinations of these variables.
To see how, let $B=(b_1,...,b_n)$.  Then the constraint that we have a probability distribution is:
$$0\leq p_B \leq 1$$
$$\sum_{B\in\mathcal{B}} p_B=1$$
The value of $\overline{x_i}$ (given to us by the outer loop):
$$\left(\sum_{B\in\mathcal{B}, b_i=1}p_B\right)-\left(\sum_{B\in\mathcal{B}, b_i=-1}p_B\right)=\overline{x_i}$$
The covariances (for $i\neq j$):
$$\left(\sum_{B\in\mathcal{B}, b_i=b_j}p_B\right)-\left(\sum_{B\in\mathcal{B}, b_i\neq b_j}p_B\right)=c_{i,j}+\overline{x_i}\overline{x_j}$$
If there exists a distribution of Bernoulli vectors consistent with $C$, then the linear program will be feasible (which we can determine in time polynomial in $|\mathcal{B}|=2^n$), and we exit, returning the distribution.  On the other hand,
if all the linear programs are infeasible, then $C$ is not consistent with any random variable over Bernoulli vectors
Polynomial-time algorithm
We provide an algorithm that works under the assumptions that (1) the $\overline{x_i}$ are known, and (2) a certain matrix (specified below) is positive semi-definite.
Let $f(x)=\sin(\pi x/2)$, and note that $f:[-1,1]\rightarrow [-1,1]$.  We will abuse notation and apply $f$ element-wise to matrices, i.e. for any matrix $X$ with elements in $[-1,1]$, we write $f(X)=(f(x_{i,j}))$.
Consider, the second moments:
$$d_{i,j}=\mathbb{E}[x_ix_j]$$
Let $D=(d_{i,j})$ be the matrix of second moments.
Note that if we are given first moments $\overline{x_i}$, we can translate between $C$ and $D$:
$$d_{i,j}-\overline{x_i}\,\overline{x_j}=c_{i,j}$$
Note that $d_{i,i}=1$, so that in some sense $C$ contains more information than $D$.  To address that imbalance, consider the $(n+1)\times (n+1)$ matrix $G=(g_{i,j})$.  For $1\leq i,j \leq n$, we set $g_{i,j}=d_{i,j}$.  For $1\leq i \leq n$, we set $g_{i,n+1}=g_{n+1,i}=\overline{x_i}$.  Finally, $g_{n+1,n+1}=1$.
Note that all the entries of $G$ lie within $[-1,1]$.  Set
$$H=f(G)$$
If $H$ is a positive semidefinite matrix, then the following algorithm will produce a Bernoulli vector with covariance matrix $C$:

Produce a sample $(a_1,...,a_{n+1})$ from a Gaussian random variable with covariance matrix $H$.
Threshold the vector by setting $b_i=sign(a_i)$.
Set $c_i=b_i b_{n+1}$ for $i=1,...,n$.
Return $(c_1,...c_n)$.

Why does this work?  Using a Cholesky decomposition, if $H$ is rank $r$, we can write $H=J^TJ$, where $J$ is an $r\times n$ matrix.  Since the main diagonal of $H$ is all ones, each column of $J$ is a vector that lies on the unit sphere in $R^r$.
Consider two columns, $s,t$, lying on the unit sphere in $R^r$.  Select a random $v$ (as an i.i.d. normal random vector).  Let $\hat{s}=sign(v \cdot s)$, and $\hat{t}=sign(v \cdot t)$.  In other words, if we put a hyperplane perpendicular to $v$ through the origin and split the sphere into two halves, $\hat{s}=1$ if it lies in the same half as $v$, and -1 otherwise.
Consider the two-dimensional plane spanned by $s$ and $t$.  The vectors lie on the unit circle.  The hyperplane divides the circle into two halves.  If the angle between $s$ and $t$ is $\theta=\arccos(s \cdot t)$, then the covariance between $\hat{s}$ and $\hat{t}$ is (noting that $\mathbb{E}[\hat{s}]=\mathbb{E}[\hat{t}]=0)$:
$$\mathbb{E}[\hat{s}\hat{t}]$$
which is the probability that they lie on the same side of the circle $(1-\theta/\pi)$ minus the probability that they lie on opposite sides $(\theta/\pi)$, which is:
$$=1-\frac{2}{\pi}\theta$$
$$=\frac{2}{\pi}\arcsin(s \cdot t)=f^{-1}(s\cdot t)$$
Some steps of the algorithm may make more sense now:

The $f()$ function tells us what Gaussian covariance is necessary to map to the target covariance on the Bernoulli random variables.
With a little squinting, you can recognize that the business with $v$ and the hyperplane is the same as selecting a Gaussian random vector and thresholding it.

The algorithm above will produce $(b_1,...,b_{n+1})$ with the second moment matrix given by $G$.  Unfortunately, $\mathbb{E}[b_i]=0$ for all $i$.
To recover our target mean values $\overline{x_i}$,
we use $b_{n+1}$.  Letting $c_i=b_ib_{n+1}$, note that the covariance of $(c_i,c_j)$ is the same as $(b_i,b_j)$.  However, the mean of $c_i$ has shifted to $\overline{x_i}$.
Other notes
Because of paywalls, I haven't been able to follow V.C.'s references yet, but the presence of the arcsine above makes me suspicious that I'm reinventing those wheels.  The definition of Van Vleck's arcsine law given here seems related but a little different from what we need.  When I manage to track down more of V.C.'s references I'll update this post.
If we are given the second moment matrix $D$ directly, we do not need to make any assumptions about knowing the $\overline{x_i}$; we only use those values to convert from $C$ to $D$.
If the main diagonal of the covariance matrix is identically one, that implies that $\overline{x_i}=1$ or all $i$, so in that case we can deduce all the $\overline{x_i}$.
The extra row and column in $G$ is a bit of a hack; it seems like it should be possible to incorporate the mean more directly and perhaps apply to a wider class of covariance matrices.
Rather than guessing all $2^n$ possibilities for the signs of the $\overline{x}_i$, we could also express this problem as a semidefinite program with a rank 1 constraint, and remove the rank 1 constraint to make the problem tractable.  However, I don't know any guarantees for how well that method would work.
I suspect that a polynomial time solution for all $C$ would probably solve MAX-CUT (and hence imply that $P=NP$).  So such an algorithm is unlikely to exist.<|endoftext|>
TITLE: Is Every Holomorphic Near an Entire?
QUESTION [12 upvotes]: Let $K\subset \mathbb C$ be a closed subset of the complex plane, not necessarily bounded. Let $U$ be the interior of $K$.
Let $f:K\to \mathbb C$ be a continuous bounded function, whose restriction to $U$ is holomorphic.
Assume furthermore that for every closed curve $\gamma\subset K$, the integral $\int_\gamma f(z)dz$ vanishes. (this is only relevant if $K$ is not simply connected). If you prefer, you can assume that $K$ is simply connected.

Can $f$ be uniformly approximated by entire functions?

REPLY [9 votes]: This theorem does not follow in a straightforward way from Mergelyan's theorem, which on its face applies only to bounded domains.
There is, however, a related theorem called Arakelian's theorem which settles the matter.  For completeness:
Definition: A "hole" of a closed subset $E$ of $\mathbb{C}$ is any bounded component of the complement of $E$.
Definition A set $E$ is Arakelian if $E$ has no holes, and if for every closed disk $D$, the union of all holes of $E \cup D$ is bounded.
Theorem Let $E$ be Arakelian. If $f$ is continuous on $E$ and holomorphic on the interior of $E$, then $f$ can be uniformly approximated by entire functions.
A proof deriving this result from Mergelyan's theorem can be found here.

I would also like to remark that Mergelyan's theorem (while usually stated for polynomial approximation) actually applies to domains with holes as well, if you just allow rational approximation with "poles in the holes". 
I would also like to draw attention to how crazy Mergelyan's theorem is.  One consequence is that one can approximate an antiholomorphic function by entire functions on any set without interior.  For instance, one can approximate $\overline{z}$ on the "plus sign" consisting on the interval $[-1,1]$ on the real axis and "$[-i,i]$" on the imaginary axis.  If anyone can explicitly construct such approximating functions, I would be very happy to see them!<|endoftext|>
TITLE: Does projective imply flat?
QUESTION [21 upvotes]: Let $\mathcal C$ be an abelian category equipped with a closed symmetric monoidal structure.  This implies in particular that the monoidal structure $\otimes$ is right exact in each variable.  I care most about the situation where $\mathcal C$ is finite $\mathbb C$-linear in the sense of arXiv:1406.4204 (in which case the monoidal structure is closed iff it is right exact in each variable).  Note that, unlike in that paper, I specifically care about monoidal structures which are not rigid.  An example of the category I have in mind is the category $\mathrm{Mod}^f_A$ of finite-dimensional modules for any finite-dimensional commutative algebra $A$ (with $\otimes = \otimes_A$).
Recall that an object $P \in \mathcal C$ is projective if $\hom(P,-) : \mathcal C \to \mathrm{AbGp}$ is right exact.  (It is already left exact.)  Note that this has nothing to do with the monoidal structure.
An object $F \in \mathcal C$ is flat if $F \otimes : \mathcal C \to \mathcal C$ is left exact.  (It is already right exact.)  Note that this has everything to do with the monoidal structure.

Are projective objects necessarily flat?

Of course, in $\mathrm{Mod}_A^f$ they are.  The other examples I usually use of non-rigid monoidal categories are the representation theories of non-Hopf bialgebras, but there every object is flat.

REPLY [18 votes]: The paper:
When projective does not imply flat, and other homological anomalies, Theory and Applications of Categories, Vol 5, pp. 202-250, 1999, available here
by Gaunce Lewis shows that this behaviour is quite common in categories of Mackey functors for compact Lie groups. These categories arise very naturally in equivariant stable homotopy category. Here is a quote from the abstract: "These examples were not fabricated to illustrate the abstract possibility of misbehavior. Rather, they are drawn from the literature."<|endoftext|>
TITLE: What is the most transparent, rigorous definition of the Univalence Axiom?
QUESTION [20 upvotes]: I've been studying homotopy type theory and trying to grasp the Univalence Axiom.  I have yet to find a concise, accessible, rigorous definition of Univalence. I have several excellent survey papers that present the "flavor" of Univalence (i.e. "identity is equivalent to equivalence") but only informally.  Obviously the HoTT book has a rigorous definition but it is substantial work getting there, at least to someone new to type theory.
Can anyone give a reference to a paper that develops the minimal amount of machinery to provide a formal definition of the Univalence Axiom?

REPLY [21 votes]: I’ll give first a simplicial definition of univalence, and then a type-theoretic one, and discuss the equivalence between them as we go.
The first thing to know is that univalence is a property that can be defined for any family of types, or in models, for any fibration.  The Univalence Axiom then says that a particular family of types — a “universe” — is univalent.  I’ll come back to the universe at the end.
So: what is univalence for a fibration?  Roughly, a fibration is univalent just when the map from “paths in the base” to “equivalences between fibers” is itself an equivalence.
Precisely: given two fibrations $Y_1, Y_2$ over a common base $X$, write $\newcommand{\Hom}{\mathcal{Hom}}\Hom_X(Y_1,Y_2)$ for the fibred mapping space between them, i.e. the exponential in $\newcommand{\SSet}{\mathbf{SSet}}\SSet/X$, which represents maps between (pullbacks of) $Y_1$ and $Y_2$, and write $\newcommand{\Eqv}{\mathcal{Eqv}}\Eqv_X(Y_1,Y_2)$ for the subobject of $\Hom_X(Y_1,Y_2)$ that represents equivalences between them.  So an $n$-simplex of $\Eqv(Y_1,Y_2)$ over a simplex $x \in X_n$ corresponds to an equivalence of fibers $(Y_1)_x \to (Y_2)_x$ over $\Delta[n]$.
Now, given a single fibration $Y$ over $X$, we can consider maps between different fibers of $Y$ by first pulling it back to $X \times X$ along the two projections $\pi_1$, $\pi_2$.  So $\pi_1^*(Y)_{(x_1,x_2)} \cong Y_{x_1}$, and $\pi_2^*(Y)_{(x_1,x_2)} \cong Y_{x_2}$.  Now consider the fibred space of equivalences between these: $\Eqv_{X \times X}(\pi_1^*Y,\pi_2^*Y)$.  So an element of this over $(x_1,x_2)$ represents an equivalence $Y_{x_1} \to Y_{x_2}$ between fibers of $Y$.
There’s always a map $i : X \to \Eqv_{X \times X}(\pi_1^*Y,\pi_2^*Y)$, living over the diagonal $\Delta_X : X \to X \times X$, where $i(x)$ represents the identity map $Y_x \to Y_x$.  Then we define: the fibration $Y \to X$ is univalent if this map $i : X \to \Eqv_{X \times X}(\pi_1^*Y,\pi_2^*Y)$ is a weak equivalence.
Some easily equivalent forms:

(in $\SSet$, or anywhere else where cof = mono) $i$ is a trivial cofibration;
$i$ makes $\Eqv_{X \times X}(\pi_1^*Y,\pi_2^*Y)$ a path object for $X$;
the induced map $P(X) \to \Eqv_{X \times X}(\pi_1^*Y,\pi_2^*Y)$ over $X \times X$ is an equivalence (for any other path object $P(X)$).

A slightly-less-trivially equivalent form is the same statement, but using an alternative construction of the object of equivalences — call it $\Eqv'_X(Y_1,Y_2)$ — which represents not just “equivalences from $Y_1$ to $Y_2$”, but “maps $f$ from $Y_1$ to $Y_2$, equipped with a homotopy left inverse $(g_l, \alpha)$ (where $\alpha$ is the homotopy $g_l \cdot f \to 1_{Y_1}$) and a homotopy right inverse $(g_r,\beta)$”.  This is equivalent to the other versions just since the evident projection $\Eqv'(Y_1,Y_2) \to \Eqv(Y_1,Y_2)$ is always a trivial fibration.
Now let’s move to type theory.  First a couple of notes about language.  It’s familiar in homotopy theory to think of a fibration as a family of spaces varying over a base space.  In type theory, this is literally the case in the language — you work with a family of types indexed by a variable varying over a base type, just like in traditional settings you work with the family of sets $\newcommand{\R}{\mathbb{R}}\newcommand{\N}{\mathbb{N}}\R^n$ indexed by $n \in \N$ — but such a family will generally behave like a fibration, not just like a discretely indexed family of discrete sets.  And in the simplicial model (and other similar models), a family of types gets interpreted as a fibration.
Also, like the HoTT book, I’ll work in prose, not in formal symbolic type theory, just like how when one does maths over a traditional foundation, one writes in prose rather than the formal symbols of first-order set theory.  (In fact the gap between prose and formal language is significantly smaller in type theory than in set-theoretic foundations, for most mathematics.)
The only basic notions we need are path types, function types, and $\Sigma$-types (types of tuples of data).
Given two types $Y_1$, $Y_2$, we can define the type $\newcommand{\tyEqv}{\mathsf{Eqv}}\tyEqv(Y_1,Y_2)$ of equivalences between them as the type of tuples $(f,g_l,\alpha,g_r,\beta)$, where $f$ is a function $Y_1 \to Y_2$, $g_l$ is a function back the other way, $\alpha$ is a function giving for each $y \in Y_1$ a path from $g_l(f(y))$ to $y$ (so $(g_l,\alpha)$ together are a homotopy left inverse for $f$), and similarly $(g_r,\beta)$ is a homotopy right inverse.  Saying a function $f : Y_1 \to Y_2$ is an equivalence means equipping it with suitable $(g_l,\alpha,g_r,\beta)$.
In general, $Y_1$ and $Y_2$ may have been dependent all along on some variable(s) — say, $Y_1(x)$ and $Y_2(x)$, where $x$ ranges over some other type $X$.  Then their interpretations in the simplicial model will be fibrations $[Y_i] \to [X]$; and $\tyEqv(Y_1(x),Y_2(x))$ is itself dependent on $x : X$, so its interpretation is also a fibration $[\tyEqv(Y_1,Y_2)] \to [X]$.  And in fact this comes out to be precisely $\Eqv'([Y_1],[Y_2])$ as defined above, since path-types are modelled by (fibred) path-objects and function types by (fibred) mapping objects.
Now for a type $Y(x)$ depending on $x:X$, and for any $x_1,x_2 : X$, there’s a canonical map $\newcommand{\tyP}{\mathsf{P}}i : \tyP_X(x_1,x_2) \to \tyEqv(Y(x_1),Y(x_2))$, defined by giving the identity equivalence $Y(x) \to Y(x)$ on reflexivity paths.  (To construct something depending on a general path, it’s enough to construct it for reflexivity paths; this is the defining property of path-types, and is analogous to extending maps defined on $X$ to maps defined on a path-object $P(X)$ along the inclusion $X \to P(X)$.)
Now, say that a family of types $Y(x)$, indexed by $x : X$, is univalent if this map $i : \tyP_X(x_1,x_2) \to \tyEqv(Y(x_1),Y(x_2))$ is an equivalence for each $x_1,x_2 : X$.
I hope the remarks so far about interpretation make it reasonably plausible, if not quite watertight, why a family of types is univalent in this sense just if its interpretation as a fibration is univalent in the simplicial sense.
Now, in type theory (as in other foundations), one often wants to consider a universe — a family of types closed under common constructions (e.g. forming function types).  Indeed, one may consider multiple such universes.  The Univalence Axiom, for a given universe, says just that this universe, considered just as a family of types, is univalent.
I’ll stop here, because this is plenty long enough already!  But a full treatment of all the above, together with the construction of univalent universes in $\SSet$, can be found in The Simplicial Model of Univalent Foundations (Kapulkin, Lumsdaine, Voevodsky).  Specifically, it’s in Section 3, using the universe introduced in in Section 2; you can probably comfortably skip Section 1, which is about the technicalities of constructing models of type theory, and also most of Section 2 after the universe is constructed.<|endoftext|>
TITLE: E-infinity structure on singular cochains
QUESTION [13 upvotes]: Is there a transparent explanation of why the singular cochain complex of a topological space X is an $E_\infty$ algebra. There are combinatorial proofs using, say, the surjection operad, but is there a topological picture behind those? If we wanted to restrict,say, to the level of little (2-)discs, could we describe the structure using something like Eilenberg-Zilber contractions of $C^*(X^m)$ onto $C^*(X)^{\otimes m}$, various diagonal maps and so on?

REPLY [12 votes]: If you wrote $E(n)$ for the chain complex of natural transformations $C_*(-) \to C_*(-)^{\otimes n}$, the $E(n)$ collectively form an operad, parametrizing all natural "one-to-many" transformations on chains, called the Eilenberg-Zilber operad. The defining co-action on chains turns into a natural action on cochains, and so the Eilenberg-Zilber operad acts on the singular cochain functor $C^*(-)$. (I seem to recall that there is something to be careful with here regarding the order of composition and whether this is naturally an operad or the reverse of an operad, but the details escape me.)
One can show that the homology groups $H_* E(n)$ are just $\Bbb Z$, concentrated in degree zero, for all $n$, and that the composition operations make these homology groups into the commutative operad. The method of acyclic models is a nice way to prove this (but that seems to have fallen by the wayside as a standard part of the curriculum). It's not clear that Eilenberg-Zilber operad is an $E_\infty$ operad because the symmetric groups aren't guaranteed to act nicely enough, but this is enough to show you that it accepts a weak equivalence from an $E_\infty$ operad (e.g. a cofibrant replacement) without having to know a combinatorial construction.
It seems, from your question, like you would like a close connection to the little discs operads. Unfortunately, this point of view does not provide one at all, and I don't know of a "natural" way to show this that doesn't rely on either a specific combinatorial construction.<|endoftext|>
TITLE: What are a couple of examples of finite sized but interesting categories?
QUESTION [6 upvotes]: I'm studying category theory and, given that I don't have a background in topology, I'm struggling to think of some finite categories that interesting.
The main one I know of is finite preorders -- I find this category interesting because it has products, sums, etc. I'd love a couple of other categories like that that I could use to better understand (by graphing out the objects and morphisms) things like equalizers, pullbacks, etc.
I could of course do this by construction, but I'd much prefer some finite categories that include these ideas in a useful way.
Thank you!

REPLY [2 votes]: A well-known example is the equivalence between groups and
and one-object categories with all morphisms invertible
(i.e. all morphisms are isomorphisms).  The group is finite
iff the category is.
A more exotic example of a finite category with all morphisms invertible is
Conway's $M_{13}$ groupoid
(see also this
blog entry by John Baez, 
who is fond of most things categorical).
Conway calls it "$M_{13}$" because there are $13$ objects each of whose
automorphism groups is isomorphic with the Mathieu group $M_{12}$ 
(and any two objects are connected by $|M_{12}|$ isomorphisms).<|endoftext|>
TITLE: History of Geometric Analogies in Number Theory
QUESTION [23 upvotes]: My question, put simply, is: When did mathematicians/number theorists begin viewing questions in number theory through a geometric lens? 
For example, was it before Grothendieck introduced schemes to generalize the notion of covering spaces and algebraic curves to include primes in rings? Today we call p-adic fields local and number fields global, which suggests a very geometric interpretation of their relationship. I can certainly see the parallels more and more as I learn more of the theory, but I am curious as to when this "realization" that geometric ideas were all over number theory rose to prominence among number theorists. 
Does it go all the way back to classical number theory (i.e. Euler and Gauss), or perhaps does it begin with Hensel and Hasse? Any resources on the matter would also be appreciated.

REPLY [44 votes]: Treating number and function fields on the same footing or (for instance) the idea that ramification in algebraic number theory and in the theory of covering of Riemann or analytic surfaces are two incarnations of the same mathematical phenomenon are classical ideas of the German school of the second half of the 19th century.
It is very present in the research as well as expository material of Kronecker, Dedekind and Weber (see for instance the algebraic proof of Riemann-Roch by the last two). In fact, it is so ubiquitous in Kronecker's work that in some of his results on elliptic curves, it is often hard to ascertain if the elliptic curve he is studying is supposed to be defined over $\mathbb C$, $\bar{\mathbb Q}$, the ring of integers of a number field or over $\bar{\mathbb F}_p$ (or over all four depending on where you find yourself in the article). This is why the introduction of SGA1 says that the aim of the volume is to study the fundamental group in a "kroneckerian" way.
At any rate, the analogy was so well-known to Hilbert that Takagi actually says in his memoirs that Hilbert had a negative influence on his definition and study of ray class field: Hilbert always wanted Takagi's theorem to make sense for Riemann surfaces and so was asking Takagi to only consider extension of number fields unramified everywhere.
In the 1920s and 1930s (so to mathematicians like Artin, Hasse or Weil), this was thus very common knowledge. The revolutionary idea of Weil, in fact, is not at all the idea that arithmetic and geometry should be unified or satisfy deep analogies, it was the idea that they should be unified by topological means (at a low level, by systematically putting Zariski's topology to the forefront, at a high level, by introducing the idea that the rationality of the Zeta function and Riemann's hypothesis for varieties over function fields of positive characteristics were the consequences of the Lefschetz formula on a to be defined cohomology theory). A fortiori, the idea of viewing arithmetic through a geometric lens should certainly not be credited to Grothendieck, whose contribution (at least, the first and most relevant to the question) was the much more precise and technical insight that combining Serre's idea of studying varieties through the cohomology of coherent sheaves on the Zariski topology and Nagata's and Chevalley's generalization of affine varieties to spectrum of arbitrary rings, one would get the language required to carry over Weil's program.    
I cannot resist concluding with the following anecdote of Serre. In a talk he gave in Orsay in Autumn 2014 on group theory, he started by explaining that finite group theory should be of interest to many different kind of mathematicians, if only because the Galois group of an extension of number fields or the fundamental group of a topological space are examples of finite groups. In fact, he continued, these two kind of groups are the same thing and (quoting from memory and in my translation) "that they are the same thing is due to German mathematicians of the late 19th century, of course, except in Orsay where it is due to Grothendieck."<|endoftext|>
TITLE: Determining if a matrix is orthogonal
QUESTION [13 upvotes]: Let g be an element of $GL_n(\mathbb C)$. We know that there are orthogonal groups $O(\beta)=\{X\in GL_n(\mathbb C) \mid X^t\beta X=\beta\}$ for any $\beta$, invertible symmetric matrix. Though these groups are conjugate since all symmetric bilinear forms over $\mathbb C$ are equivalent. In literature one can find a way to determine when a matrix $X$ satisfies $X^tX=I$ thus an element of the orthogonal group $O(I)$. 
My question is as follows: Is there a way to determine if an invertible matrix X belongs to $O(\beta)$ for some $\beta$. To me it seems that I have to check for all conjugates. Sure enough there will be an easier way out!  
More precisely I want to know the following: Given $X$ an invertible matrix how to determine $\beta$, an invertible symmetric matrix, so that $X^t\beta X=\beta$?

REPLY [12 votes]: There is a complete characterization of matrices that belong to at least one
orthogonal group. It reads as follows over any arbitrary field $\mathbb{F}$
with characteristic different from $2$ (with algebraic closure denoted by $\overline{\mathbb{F}}$:

Given a matrix $M \in \mathrm{GL}_n(\mathbb{F})$, there exists an
  invertible symmetrix matrix $\beta$ such that $M^T \beta M=\beta$ if
  and only if, for every $\lambda \in \overline{\mathbb{F}} \setminus \{0,1,-1\}$ and 
  every positive integer $k$, one has
  $\mathrm{rk}(M-\lambda I_n)^k=\mathrm{rk} (M-\lambda^{-1} I_n)^k$ and,
  for each one of the (possibly absent) eigenvalues $1$ and $-1$ and
  every positive integer $k$, there is an even number of Jordan cells of
  size $2k$ in the Jordan reduction of $M$. 

Moreoever, if you have access to the Jordan reduction of $M$ and the above conditions
are satisfied, then coming up with an explicit solution $\beta$ is not difficult.
This characterization has been known for a very long time. See my preprint http://arxiv.org/abs/1008.4458 for a recent account using elementary methods.<|endoftext|>
TITLE: Algorithms for calculating R(5,5) and R(6,6)
QUESTION [23 upvotes]: Calculating the Ramsey numbers R(5,5) and R(6,6) is a notoriously difficult problem. Indeed Erdős once said:

Suppose aliens invade the earth and threaten to obliterate it in a year's time unless human beings can find the Ramsey number for red five and blue five. We could marshal the world's best minds and fastest computers, and within a year we could probably calculate the value. If the aliens demanded the Ramsey number for red six and blue six, however, we would have no choice but to launch a preemptive attack. 

I am curious what algorithm we would employ if such a situation were to occur. I know analytic results have been used to put bounds on R(5,5) and R(6,6), but I am mostly interested in the problem from a computational perspective. If we were to set a computer to the task and let it run for however long it might take, what algorithm would we use? How many operations might we expect it to take/what would it's time complexity be?
Edit: I should clarify that I am seeking the best classical algorithm. It was after reading the paper that Carlo Beenakker cites using quantum annealing that I became interested in finding the best classical alternative.

REPLY [6 votes]: Several years ago I found a method based on continuous optimizaton which can sometimes 
find lower bound. It also applies to the SAT problem but it may not be exactly the same as the SAT solver method.
Given the number of vertices $n$ and positive
integers $k,l$, we can code all $2$-colorings of the edges of $K_n$
by (the upper triangular part of) a binary matrix $x_{ij}, 1\le i <
j \le n$, and write down the energy function
$$\phi_{kl}^{(n)}(x)= \sum_{1 \le i_1 <...<i_k \le n} \prod_{1 \le s<
t \le k} x_{i_s i_t} + \sum_{1 \le j_1<...<j_l \le n} \prod_{1 \le s
< t \le l}(1- x_{j_s j_t}),$$
which counts the total number of monochromatic $1$-$K_k$ or
$0$-$K_l$. We now interpolate and think of $\phi(x)$ as a polynomial
function of the ${ n \choose 2}$ real variables $x_{ij}$ which is
linear in each, and must therefore satisfies a strong mean-value
property :
Lemma. Let $f(x_1,...,x_n)$ be a real function which is linear in
each variable. Then for any axes-parallel rectangular box of any
dimension, the value of $f$ at the center is the average value of
$f$ over the vertices of the box.
Proof. This is true for a $n=1$ (i.e. for a line), since $f$ is linear; for a $k$-dimensional box, we just have to take the average over the straight line joining the centers of two opposite $k-1$-dimensional faces which equals the
average over the whole box. Apply this to $\phi$, we see that if
$$\phi^{(n)}_{kl}(1/2,...,1/2)={ n \choose k }  2^{-{k \choose 2}}
+{n \choose l} 2^{-{l \choose 2}}<1,$$
we get a lower bound $R(k,l)>n$, which is the same bound as the original probabilistic counting method. We now observe that we can make the same conclusion
if we can find any $x$ inside the unit box $[0,1]^{n \choose 2}$,
with $\phi(x)<1$ because of the maximum principle: $\phi$ is
harmonic in any subset of the variables so any extremal value of
$\phi$ inside the box  must already occur at some vertex. So instead
of doing brute force searching over the vertices discretely, we can
try to minimize $\phi$ over the whole box continuously by some form
of gradient descent algorithm.
Finding the minimum of a polynomial inside a box is however an NP-hard problem since the energy function which counts the number of
falsified clauses of a truth assignment of a boolean formula in
conjunctive normal form has exactly the same sum-product form as
$\phi$ and SAT is NP-complete.
Evaluating $\phi$ and its gradient is compute intensive but is
polynomial in $n$ for fixed $k$ and $l$ and is certainly doable for
the range of $R(5,5)$ up to $n=49$.
Maybe someone with more computing power and better optimization
algorithm can try this. The idea is that we don't look at all the
vertices just the good ones following the gradient. The problem is
of course that we cannot be sure the best we have found is the true minimum
unless it is of zero energy.<|endoftext|>
TITLE: Idea of using etale site
QUESTION [13 upvotes]: I have just read an article which mentions that, when Grothendieck considered using etale morphism, he did borrow the idea from Riemann that multivalued function on an open subset of complex plane should live in Riemann surface covering it instead of the open subset itself.
The question is: how these things are related? Any detailed explanation is very welcome and appreciated. More generally, what are the good properties etale morphism has which make it essential in solving Weil conjecture.I understand that Grothendieck tried to search for a cohomology theory in algebraic geometry which is similar to classical cohomology theory on manifold theory but  I dont know what obstructions are there preventing a cohomology theory in AG behaves similar to one in manifold for which etale site was introduced to overcome them.

REPLY [6 votes]: With regards to the Weil conjectures:
I believe it was known to Weil that, to prove most of the Weil conjectures, a cohomology theory for varieties with coefficients in a field of characteristic zero would just have to satisfy a few properties that are familiar from singular cohomology:

The Lefschetz fixed point formula, applied to the endomorphism Frob_p, and finiteness of Betti numbers immediately give rationality of the zeta function.
Poincare duality, suitably twisted to account for the fact that Frob_p should act on $H^{2n} (X)$ for $X$ an $n$-dimensional smooth projective variety by multiplication by $p^n$, gives the functional equation.
The Riemann hypothesis is the only one I'm not sure about - Deligne's proof of this involves many clever steps that had no antecedent in singular cohomology.
The fact that Betti numbers are constant in smooth proper fibrations suffices for the Betti number conjecture, and a comparison theorem to singular cohomology over the complex numbers, (although Weil may not have had the full picture of an arithmetic family of varieties as being the same as a geometric family as clearly as Grothendieck did) 

So why does etale cohomology have these features? I think experience shows that any cohomology theory we define has these features as long as it satisfies some very basic tests - off the top of my head, I can't think of any cohomology theory that has the right Betti numbers for a curve of genus $g$ that doesn't have all these properties.
$\ell$-adic etale cohomology avoids the trivial failure of all the cohomology theories you can construct with just the Zariski topology - either cohomology of the constant sheaf $\mathbb Z$ in the Zariski topology, or cohomology of some coherent sheaf, probably powers of the cotangent bundle to mimic Dolbeaut cohomology in characteristic zero. The first one has the wrong Betti numbers, and the second one has coefficient field the base field, so has coefficients of characteristic $p$ in characteristic $p$, and so cannot prove the Weil conjectures.<|endoftext|>
TITLE: When is the boundary of an open planar set a Jordan curve?
QUESTION [6 upvotes]: Is the boundary of an open, regular, bounded, path-connected, and simply connected set a Jordan curve?

Trying to find weakest condition on an open bounded set to apply Carathéodory's theorem. 
My bounded open sets can be assumed to be pretty well-behaved, but I wonder if the above conditions are sufficient.

REPLY [3 votes]: To help clarify a well known characterization: If U is a connected open bounded simply connected planar set, then the boundary of U is a simple closed curve iff the boundary of U is locally path connected and contains no cut points.
Comments:
0) Definition. p is a cut point of the connected space X iff X\p is not connected. 
1) Definition. By `boundary' of an open bounded planar set U, we mean the difference U closure minus U.
2) For necessity, observe that the unit circle X is locally path connected and X\p is connected for all p in X.
3) For sufficiency, every orientation preserving conformal homeomorphism ( Riemann map) f: int(D) --->U extends continuously to the respective closures iff the boundary of U is locally path connected. (Caratheodory). If the extension is not 1-1, a straightforward geometric construction yields a non cutpoint on the boundary of U.
4) Mirko's nice sketch of the `Warsaw circle' shows a non locally path connected boundary can have no cut points, and pac-man with a closed mouth shows a locally path connected boundary can have cut points. Thus two the mentioned conditions are independent.<|endoftext|>
TITLE: Expected size of determinant of $AA^T$ for random circulant and Toeplitz matrices
QUESTION [10 upvotes]: If $A$ is chosen uniformly at random over all possible $n$ by $n$ Toeplitz (or circulant) (0,1)-matrices, can we give any bounds for the expected size  of the determinant of $AA^T$? All arithmetic is to be performed over the reals.
Richard Stanley gave a very nice and clean exact answer for the general non-square (0,1)-matrix case here.  
For small $n$ it seems that $\mathbb{E}(\det(AA^T))$ is much larger when $A$ is Toeplitz than in the general $(0,1)$ case and even larger still when $A$ is circulant. The means for $n =1,\dots, 12$ are:

Toeplitz:  0.5, 0.5, 0.875, 1.3125, 2.90625, 7.953125, 28.40625, 83.8125, 322.52734375, 1282.98730469, 6394.79248047
Circulant: 0.5, 0.5, 1.875, 2.5,    6.71875, 25.125,   185.8828125, 366.5,   2071.70507812, 9026.09375,   88453.690918
General $(0,1)$ matrices using the $(n+1)!/4^n$ formula: 0.5, 0.375, 0.375, 0.46875, 0.703125, 1.23046875, 2.4609375, 5.537109375, 13.8427734375, 38.0676269531, 114.202880859


Is it possible to prove that the general case is a lower bound for
  either the Toeplitz or the circulant case?


Answer accepted which gives the required result for circulant matrices with added interesting conjectures.

REPLY [2 votes]: I assume that in the question it is intended to ask for the expectation value $\mathbb{E}\{\det(AA^T)\}$, not $\mathbb{E}(AA^T)$. 
I have no good idea yet how to attack the general Toeplitz case, but for the special case of square circulant matrices one can use the fact that their eigenvalues can explicitly be expressed in terms of the matrix elements by a discrete Fourier transform. Below I will show that in this case the answer to the question is yes, and I will also present some evidence for the validity stronger statements concerning the growth rate of 
Denoting as $a_i$ (with $i=0,...,n-1$) the elements in the first row of a $n \times n$ circulant matrix $A$, the eigenvalues $\lambda_k$ of $A$ are given by: 
$$ 
\lambda_k = \sum_{i=0}^{n-1} a_i\, \omega^{ik} \quad (k=0,...,n-1) \ ,
\hspace{2cm} (1)
$$
with the primitive $n^{\rm th}$ root of unity $\omega={\rm e}^{\frac{2\pi{\rm i}}n}$. 
The determinant of $A$ can then be expressed as
$$ 
\det A = \prod_{k=0}^{n-1} \left( \sum_{i=0}^{n-1} a_i\, \omega^{ik} \right) \ , 
\hspace{2cm} (2)
$$
which I will use as starting point for this (partial) answer to your question. 
For later convenience, let us make a variable change to new random variables $\varepsilon_i = 2a_i-1 \in \{-1,1\}$. Using that $\sum_{i=0}^{n-1} \omega^{ik} = n\, \delta_{k0}$, and splitting off the $k=0$ factor in $(2)$, we obtain: 
$$ 
\det A = 2^{-n} \left( n + \sum_{i=0}^{n-1} \varepsilon_i \right) \prod_{k=1}^{n-1} \left( \sum_{i=0}^{n-1} \varepsilon_i\, \omega^{ik} \right) \ . 
\hspace{2cm} (3)
$$
Since  $A$ is a real matrix and thus $\det(AA^{\rm T}) = (\det A)^2$, the moments of the $\det(AA^{\rm T})$ distribution are equal to the even moments of the $\det A$ distribution. To calculate these moments, powers of $(3)$ have to be averaged over the i.u.d. variables $\varepsilon_i \in \{-1,1\}$. This may be done in a convenient way with help of a generating function $F$ defined as follows:
\begin{align}
F(x_0,...,x_{n-1}) 
&= \mathbb{E}\left\{ \exp \sum_{i,k=0}^{n-1} x_k\, \varepsilon_i\, \omega^{ik} \right\} 
= \mathbb{E}\left\{ \prod_{i=0}^{n-1} \exp\left( \varepsilon_i \sum_{k=0}^{n-1} x_k\, \omega^{ik} \right) \right\}  \\
&= \prod_{i=0}^{n-1} \cosh\left( \sum_{k=0}^{n-1} x_k\, \omega^{ik} \right) \ .
\hspace{2cm} (4)
\end{align}
Let us describe the $p^{\rm th}$ moment of the $\det A$ distribution, for our $n \times n$ circulant matrices, by $N_{\rm cir}(p,n) = 2^{pn}  {\big\langle} (\det A)^p {\big\rangle}$. With $(3)$ and $(4)$ we can express this quantity as follows by derivatives of the generating function (using the shorthand notation $\partial_k = \frac\partial{\partial x_k}$): 
$$ 
N_{\rm cir}(p,n) 
= \left[ \left( {\big(} n + \partial_0 {\big)} \prod_{k=1}^{n-1} \partial_k \right)^p F \right]_{x_0=x_1=...=x_{n-1}=0} \ . 
\hspace{1cm} (5)
$$
In order to further evaluate the right hand side of $(5)$, we first note that $(4)$ implies that any multiple derivative of $F$ with respect to the $x_k$ can, according to the product rule, be written as a sum over many terms, each of which is a product of (i) exactly $n$ factors of either $\cosh_i = $ $\cosh\left( \sum_{k=0}^{n-1} x_k\, \omega^{ik} \right)$ or $\sinh_i =$ $\sinh\left( \sum_{k=0}^{n-1} x_k\, \omega^{ik} \right)$ (one for each index $i=1,...,n$), and moreover (ii) some factors $\omega^{ik}$ (whose number is equal to the order of the derivative), each of which may have different values of $i$ and $k$. Each new derivative $\partial_k$ on one of these terms "pulls out" one new factor $\omega^{ik}$ together with a summation over $i$, and at the same time converts one of the $\cosh_i$ factors into a $\sinh_i$ or vice versa. In the end, terms having any $\sinh_i$ factors left will give a vanishing contribution this sum, since $\sinh(0)=0$; so in particular, the entire sum can be nonvanishing only if the total number of derivatives is even. This is certainly the case for any even $p$, and in particular for $p=2$ which is the case of interest here. 
Furthermore, carrying out the summations over $i$ leads to certain constraints between the different $k$ values in each term, since for instance $\sum_{i=0}^{n-1} \omega^{ik} \omega^{il} = n\delta_{kl}$ etc.. This again eliminates a large number of terms, and in the end only a certain narrow class of terms remains, each of which is real and contributes with a combinatorial factor and one factor of $n$ for each remaining $i$ sum. The combinatorial factors can be estimated using a diagrammatic approach. (I could explain it in some detail later, if there is enough interest and if I find the time). For $p=2$ this procedure leads to
$$
N_{\rm cir-}(2,n) < N_{\rm cir}(2,n) < N_{\rm cir+}(2,n) \ ,
\hspace{2cm} (6)
$$
with the following explicit bounds (here $\lfloor \frac{n-1}2 \rfloor$ denotes the integer part of $\frac{n-1}2$): 
\begin{align}
N_{\rm cir-}(2,n) &= 2^{\lfloor \frac{n-1}2 \rfloor} (n+1)! \ \ , \\ 
N_{\rm cir+}(2,n) &= 2^{\lfloor \frac{n-1}2 \rfloor} (n+1) \, n^n \ .
\hspace{2cm} (7)
\end{align}
For general $\{0,1\}$-matrices we have, as mentionend in the question, 
$$
N_{\rm gen}(2,n) = (n+1)! \ ,
\hspace{2cm} (8)
$$
so that $(6)$ and $(7)$ in particular imply that always $N_{\rm cir}(2,n) > N_{\rm gen}(2,n)$. This answers the question for circulant matrices. 
The results above can be further visualized by considering the following expression for what we may call the "asymptotic prefactor of $n$ in the exponent", or a little sloppy "relative growth rate": 
$$
f_{\alpha,c}(n) = \ln N_\alpha(2,n)/n - \ln(n{+}c) \ , 
$$
where $\alpha \in \{{\rm cir{-}}, {\rm cir{+}}, {\rm cir},{\rm gen}\}$, and $c$ is a positive constant of our choice (which won't affect the limit $n \to \infty$ of $f_{\alpha,c}(n)$ but can be adjusted to make the curves look nice). For the three  expressions in $(7)$ and $(8)$, 
$$
\lim_{n \to \infty} f_{\alpha,c}(n) =
\begin{cases} \frac1 2 \ln2 -1 \approx -0.65, \quad \alpha = {\rm cir{-}} \\ \quad \frac1 2 \ln2 \approx 0.35, \quad \alpha = {\rm cir{+}} \\ \qquad -1 , \qquad \alpha = {\rm gen} \end{cases} \ .
\hspace{2cm} (9)
$$
(Note that the Hadamard absolute upper limit (for arbitrary $n \times n$ matrices $A$ with all elements lying in the closed unit disk) corresponds to a $n \to \infty$ limit of $2\ln2 \approx 1.39$). 
A plot of numerical results for $f_{\alpha,c}(n)$ (with $c=4$ and $n = 1,...,20$) is shown below. It suggests that also $\lim_{n \to \infty} f_{{\rm cir},c}(n)$ exists, and thet it has a value ${\approx}{-0.3}$ (red curve). 

The following questions then arise naturally: 

Can the existence or non-existence of $\lim_{n \to \infty} f_{{\rm cir},c}(n)$ be proved? 
If  the limit exists, can its value be calculated, or can better bounds be provided than those in the first two lines of $(9)$?<|endoftext|>
TITLE: Prove that the Log-Euclidean distance is negative-definite
QUESTION [5 upvotes]: Let $\Bbb{S}_{++}^n$ be the $\frac{n(n+1)}{2}$-dimensional Riemannian manifold of the symmetric positive definite (SPD) $n\times n$ real matrices.
The Log-Euclidean distance between two points of $\Bbb{S}_{++}^n$, i.e. between two SPD matrices $A,B\in\Bbb{S}_{++}^n$, is given by
$$
d(A,B)=\lVert\log(A)-\log(B)\rVert_{F},
$$
where $\|\cdot\|_{F}$ denotes the regular Frobenius norm.
We want to prove that $d\colon\Bbb{S}_{++}^n\times\Bbb{S}_{++}^n\to\Bbb{R}$ is a conditionally negative definite function, as defined below. 
For a topological space $\mathcal{X}$, the function $f\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$ is called conditionally negative definite if for any $m\in\Bbb{N}$, $x_1,\ldots,x_m\in\mathcal{X}$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true
$$
\sum_{i,j=1}^{m}c_ic_jf(x_i,x_j)\leq0.
$$
In our case, we want to prove that for any $m\in\Bbb{N}$, $X_1,\ldots,X_m\in\Bbb{S}_{++}^n$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true
$$
\sum_{i,j=1}^{m}c_ic_j\lVert\log(X_i)-\log(X_j)\rVert_{F}\leq0.
$$
Thank you very much in advance!
Edit: In case of the squared Frobenius norm, I think it would be proven more easily, but I still need some help...

REPLY [3 votes]: First, recall that if $\psi$ is negative definite, then $\exp(-\gamma \psi)$ is positive definite for all $\gamma >0$. Now, from Corollary 2.10 of Harmonic analysis on Semigroups (Berg, Christensen, Ressel) we also know that if $\psi$ is negative definite, then $\psi^\alpha$ is also negative definite for $0 < \alpha < 1$ as long as $\psi(x,x) \ge 0$. 
Consider now some map $f: \mathcal{X} \to \mathbb{R}^n$, and set $\psi(x,y) = \|f(x)-f(y)\|^2$. Clearly, $\psi$ is negative definite. 
In particular, choosing $\mathcal{X} = \mathbb{S}^n_{++}$,  $f \equiv \log$, and $\alpha=1/2$, from the above corollary it follows that $\|\log X - \log Y\|_\text{F}$ is negative definite.<|endoftext|>
TITLE: Probability a random matrix contains a short integer vector in its kernel
QUESTION [6 upvotes]: Consider a random $m$ by $n$ matrix $M$ with $m \leq n$, chosen uniformly over all those whose elements are in $\{0,1\}$  (or $\{-1,1\}$ if it is any easier).    Is there any mathematical theory that can give bounds or estimates for the probability that the kernel of $M$ contains at least one short non-zero vector $v\in \mathbb{Z}^n $?
By short I mean with small $L_2$ norm.  Although I am interested in a general theory, my specific interest is in vectors no longer than $\sqrt{n}$.
Added I would also be interested in any ideas for the $L_1$ norm as well. Specifically if the $L_1$ norm is bounded by $n$ for example.

REPLY [3 votes]: For $m$ around $n/log n$ the probability goes to $0$, with $(0,1)$ or $(-1,1)$, it doesn't matter.
Combinatorial Lemma: Any set of subsets of $\{1,\dots, n\}$ which contains no pair of subsets $A, B$ with $A \subset B$ has size at most $\begin{pmatrix} n \\\lfloor n/2 \rfloor \end{pmatrix}$ elements.
Proof: Consider all paths from the empty set to the full subset that add one element at each time. Each one can intersect the set once, and there are $n!$ such paths. Each subset of size $k$ in the set must intersect $k! (n-k)!$ paths. This is minimized when $k = \lfloor n/2 \rfloor$, giving the upper bound.
Consequence: For a vector $v$ with $c$ nonzero entries, the probability that it is in the kernel of a random matrix is at most
$$ \left( \frac{ \begin{pmatrix} c \\\lfloor c/2 \rfloor \end{pmatrix}}{ 2^c}  \right)^m \approx \left( \frac{2}{\sqrt{\pi c} } \right)^m, \leq \left( \frac{1}{2} \right)^{m}$$
Proof: We may ignore the columns corresponding to the entries in the vector that are zero. So we may assume $n=c$. Each row is independent, so the probability is the probability that a single row is orthogonal to the vector, raised to the power $m$. Form a bijection between row vectors and subsets, where, if the entry in $v$ is positive, $1$ corresponds to being in the subset and $0$ or $-1$ does not, and if $v$ is negative, $0$ or $-1$ corresponds to being in the subset and $1$ does not. Then the row dot the vector is a monotonic function, so the set where it is zero contains no two subsets that dominate each other, so has the upper bound from the combinatorial lemma. Dividing it by $2^n$, we obtain the probability, and then raise it to the power $m$.
The asymptotic is well-known, though I may have the constant wrong, and the upper bound is obvious.
Now we split the short vectors into the ones with $<C$ entries and the ones with more than $C$ entries.
There are at most
$$\begin{pmatrix} n \\ C \end{pmatrix} \sqrt{n}^C \approx n^{ (3/2) C}$$
vectors with fewer than $C$ nonzero entries, and each has probability of at most $2^{-m}$ of being in the kernel, so we need:
$$ n^{ (3/2) C} < 2^m $$
There are at most $k^n$ vectors of length $<\sqrt{n}$ for a constant $k$, and each has a probability at most $(2/\sqrt{\pi C})^m$ of being in the kernel, so we need:
$$ k^n < C^{m/2} (\pi/2)^{m/2} $$
Taking $m$ a small constant multiple of $n/\log n$ and $C$ a small constant multiple of $n/(\log n)^2$ does the trick.
The improvement over Igor Rivin's bound comes only from the better estimate for the number of vectors of length $\sqrt{n}$. The proof of this is simple: Consider the sum over $v \in \mathbb Z^n$ vectors of $e^{- |v|^2}$. This is $\left( \sum_{x \in \mathbb Z} e^{-x^2} \right)^n$, and each vector of length at most $\sqrt{n}$ contributes at least $e^{-n}$, giving the desired bound. The complicated combinatorial calculations only serve to make the argument more rigorous.<|endoftext|>
TITLE: Is there a homomorphism between $\pi_8(S^5)$ and $\pi_8(SO(6))$?
QUESTION [7 upvotes]: I am currently thinking about a physics model related to framed bordism $\Omega_3^{fr}=\mathbb{Z}/24=\pi^s_3$, and the first stable example is $\pi_8(S^5)$, so I was curious about the generator, and happened to see $\pi_8(SO(6))=\mathbb{Z}/24$ as well.
So my question is: is there a homomorphism sending $\pi_8(SO(6))$ to $\pi_8(S^5)$? More generally, what's the relation between homotopy groups $\pi_n(S^m)$ and $\pi_n(SO(m+1))$ in general?

REPLY [22 votes]: There is a fibration $p : SO(m+1) \to S^m$ with fibre $SO(m)$ which induces a long exact sequence in homotopy
$$\dots \to \pi_n(SO(m)) \to \pi_n(SO(m+1)) \xrightarrow{p_*} \pi_n(S^m) \to \pi_{n-1}(SO(m)) \to \dots$$
For your particular question, we have the fibration $p : SO(6) \to S^5$ with fibre $SO(5)$ which gives 
$$\dots \to \pi_8(SO(5)) \to \pi_8(SO(6)) \to \pi_8(S^5) \to \pi_7(SO(5)) \to \pi_7(SO(6)) \to \pi_7(S^5) \to \pi_6(SO(5)) \to \dots$$
I don't know how to compute these groups myself, but they have been computed by others. From the bottom of page $3$ of this we see that $\pi_8(SO(5)) = 0$, $\pi_7(SO(5)) = \mathbb{Z}$, $\pi_7(SO(6)) = \mathbb{Z}$, and $\pi_6(SO(5)) = 0$. From here we see that $\pi_7(S^5) = \mathbb{Z}_2$ so we have
$$\dots \to 0 \to \pi_8(SO(6)) \xrightarrow{p_*} \pi_8(S^5) \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_2 \to 0 \to \dots$$
As the map $\mathbb{Z} \to \mathbb{Z}_2$ is surjective, it has kernel $2\mathbb{Z}$. So the map $\mathbb{Z} \to \mathbb{Z}$ must be multiplication by $2$ (or $-2$) which is injective. Therefore the map $\pi_8(S^5) \to \mathbb{Z}$ is the zero map, so $p_*$ is an isomorphism.
The fibration always gives rise to a homomorphism $p_* : \pi_n(SO(m+1)) \to \pi_n(S^m)$ but it is not necessarily an isomorphism as $p_* : \pi_7(SO(6)) \to \pi_7(S^5)$ demonstrates.<|endoftext|>
TITLE: Upper bound for number of prime numbers in a range
QUESTION [7 upvotes]: Theorem 3.2 in http://arxiv.org/pdf/1405.2593.pdf shows that for any $x$ there are $\gg x\exp(-\sqrt{\log x})$ integers $x_0 \in [x; 2x]$ such that $\pi(x_0 + \log x) - \pi(x_0) \gg \log\log x$. 
Is there an upper bound for number of such $x_0$? I think it must be $<x(\log x)^{-c}$ for any $C$.
UPD: It is interesting to find such upper bound for prime numbers $x_0$ such that $\pi(x_0 + \log x) - \pi(x_0) \gg \log\log x$

REPLY [5 votes]: If $x_0<x$ satisfies that $[x_0, x_0+\log x]$ contains $\log\log x$ primes, then for a parameter $r$ we have that this interval contains $\binom{\log\log x}{r}$ different $r$-tuples $p, p+d_1, p+d_2, \ldots, p+d_{r-1}$ of primes, such that $0<d_1<\dots<d_{r-1}\leq\log x$. The number of possible choices for $d_1, \ldots, d_{r-1}$ is $\binom{\log x}{r-1}$. Apply Selberg's sieve to each of them, and take the sum over all tuples. This will lead to some lengthy computation involving singular series, but on average the singular series will be of magnitude $\mathcal{O}(1)$. From this you obtain that the number of short $r$-tuples is bounded above by $$\ll C(r)\binom{\log x}{r-1} \frac{x}{\log^r x} \ll \frac{C(r)}{r!}\frac{x}{\log x},$$ where $C(r)$ is the coefficient of Selberg's sieve. Each tuple belongs to $\leq\log x$ values of $x_0$, hence 
comparing with the $\binom{\log\log x}{r}$ tuples produced by a single $x_0$ and restricting $x_0$ you obtain that there are $\ll C(r)\frac{x}{(\log\log x)^r}$ possibilities for $x_0$. Optimizing for $r$ should give an upper bound of magnitude $\frac{x}{(\log x)^c}$ for some $c>0$, which falls somewhat short of your expectation. 
Addendum: Since $r$ is large, it is better to use the large sieve in place of Selberg's sieve. The details will become more complicated, but the results should be better. For a model you can look at the proof of Lemma 2 in Elsholtz, On cluster primes, Acta Arith. 109, 281-284.<|endoftext|>
TITLE: How to prove the existence of the polytope in $\mathbb{R}^d$ with a given number of faces, minimizing the isoperimetric ratio?
QUESTION [7 upvotes]: This is the isoperimetric type question. We know that in $\mathbb{R}^d$, balls are the sets that minimize the isoperimetric ratio $\frac{S^{d}}{V^{d-1}}$, where $S$ is the surface area and $V$ is the volumn.
Now consider the polytopes with $f$ faces. Lindelof's theorem says, among all proper convex polytopes in $\mathbb{R}^d$ with given exterior normals of the facets, it is precisely the polytopes circumscribed to a ball that have minimum isoperimetric quotient. This theorem can be found in http://link.springer.com/book/10.1007%2F978-3-540-71133-9, page 308, Theorem 18.4. 
However, on Page 309, the author made a Corollary 18.2 that among all proper convex polytopes in $\mathbb{R}^d$ with a given number of facets, there are polytopes with minimum isoperimetric quotient and these polytopes are circumscribed to a ball.
Now my question is, I think the two claims above are different. To prove the Corollary 18.2, one has to prove the existence of the polytopes minimizing the isoperimetric constant among all polytopes circumscribed to a unit ball. I searched a lot of references, but I didn't find any proofs of such an existence. Is this an obvious result?

REPLY [3 votes]: This is an explanation of Anton's comment. For each set of $f$ unit vectors, one finds the polytope minimizing the isoperimetric quotient. The set of configurations of $f$ (not necessarily distinct) unit vectors is compact (it is homeomorphic to $\mathbb{S}^{(d-1) \times f}$), and the isoperimetric quotient is continuous thereon (this requires an argument), therefore it achieves its minimum for some set of $f$ unit vectors. These vectors might not be distinct a priori, but it is easy to see that counting a face with multiplicity (which is what "not distinct" means) does not help you. So, the faces are distinct, and by Lindelof the minimizer is circumscribed.<|endoftext|>
TITLE: Easiest way to see that $\zeta_{\mathbb{Z}[i]}(s) = \zeta(s) L(s, \chi)$?
QUESTION [8 upvotes]: As the question suggests, what is the easiest way to see that$$\zeta_{\mathbb{Z}[i]}(s) = \zeta(s)L(s, \chi)?$$Here, $\chi$ is the homomorphism $(\mathbb{Z}/4\mathbb{Z})^\times \to \mathbb{C}^\times$ which sends $3$ mod $4$ to $-1$ and $L(s, \chi)$ is the Dirichlet $L$-function of $\chi$.

REPLY [8 votes]: Kevin Yang posted a stellar solution to this problem while I was typing. Here is mine.
We have that$${{\mathbb{Z}[i]}\over{(p)}} \cong \begin{cases} {{\mathbb{Z}[i]}\over{\mathfrak{m}}}\text{ for maximal ideal }\mathfrak{m} \neq \mathfrak{m}_2 & \text{if $p \equiv 1 \text{ }(\text{mod }4)$} \\{{\mathbb{Z}[i]}\over{\mathfrak{m}_1}}{{\mathbb{Z}[i]}\over{\mathfrak{m}_2}} \text{ for maximal ideals }\mathfrak{m}_1 \neq \mathfrak{m}_2 & \text{if $p \equiv 3 \text{ }(\text{mod }4)$}\\ {{\mathbb{Z}[i]}\over{((1+i)^2)}} \text{ for maximal ideal }(1 + i) & \text{if $p = 2$.}\end{cases}$$Therefore,$$\#{{\mathbb{Z}[i]}\over{(p)}} \cong \begin{cases} \#{{\mathbb{Z}[i]}\over{\mathfrak{m}}}\text{ for maximal ideal }\mathfrak{m} \neq \mathfrak{m}_2 & \text{if $p \equiv 1 \text{ }(\text{mod }4)$} \\\#{{\mathbb{Z}[i]}\over{\mathfrak{m}_1}}{{\mathbb{Z}[i]}\over{\mathfrak{m}_2}} \text{ for maximal ideals }\mathfrak{m}_1 \neq \mathfrak{m}_2 & \text{if $p \equiv 3 \text{ }(\text{mod }4)$}\\ \#{{\mathbb{Z}[i]}\over{((1+i)^2)}} \text{ for maximal ideal }(1 + i) & \text{if $p = 2$.}\end{cases}$$So we have that$$\zeta(s)L(s, \chi) = \left(\prod_{p\, \equiv\, 1 \,(\text{mod}\,4)} {1\over{1 - p^{-s}}}{1\over{1 - p^{-s}}}\right)\left(\prod_{p\, \equiv\, 1 \,(\text{mod}\,4)} {1\over{1 - p^{-s}}}{1\over{1 + p^{-s}}}\right)\left({1\over{1 - 2^{-s}}}\right).$$Now, we have that$$\zeta_{\mathbb{Z}[i]}(s) = \prod_{\mathfrak{m} \in \text{max}(\mathbb{Z}[i])} {1\over{\left(1 - \#{{\mathbb{Z}[i]}\over{\mathfrak{m}}}\right)^{-s}}}$$$$= \left({1\over{1 - \#\left({{\mathbb{Z}[i]}\over{(1+i)}}\right)^{-s}}}\right)\left(\prod_{\mathfrak{m} \in \text{max}(\mathbb{Z}[i])} {1\over{\left(1 - \#{{\mathbb{Z}[i]}\over{\mathfrak{m}}}\right)^{-s}}}\right)$$$$= \left({1\over{1 - 2^{-s}}}\right)\left(\prod_{p \text{ is prime }\neq\,2} \prod_{\substack{\mathfrak{n} \in \text{max}(\mathbb{Z}[i]),\\ (p) \subseteq \mathfrak{n}}} {1\over{\left(1 - \#{{\mathbb{Z}[i]}\over{\mathfrak{n}}}\right)^{-s}}}\right)$$$$=\left({1\over{1 - 2^{-s}}}\right)\left(\prod_{p\,\equiv\,1\,(\text{mod}\,4)} \prod_{\mathfrak{n} = (p)} {1\over{1 - \#\left({{\mathbb{F}_p[T]}\over{(T^2 + 1)}}\right)^{-s}}}\right)\left(\prod_{p\,\equiv\,3\,(\text{mod}\,4)} \prod_{\mathfrak{n} \in \{\mathfrak{m}_1,\mathfrak{m}_2\}} {1\over{1 - \#\left({{\mathbb{Z}[i]}\over{\mathfrak{n}}}\right)^{-s}}}\right)$$$$=\left({1\over{1 - 2^{-s}}}\right)\left(\prod_{p\, \equiv\, 1\,(\text{mod}\,4)} {1\over{1 - p^{-2s}}}\right)\left(\prod_{p \,\equiv\,3\,(\text{mod}\,4)} {1\over{(1 - p^{-s})^{2}}}\right)$$$$=\zeta(s) L(s, \chi),$$as desired.

I want to augment the above with some philosophical reflection. We compare$$\zeta_{\mathbb{Z}[i](s)} = \zeta(s)L(s, \chi)\tag*{(*)}$$and$$\zeta_{\mathbb{Z}[\sqrt[3]{2}]}(s) = \zeta(s)L(s, f).\tag*{(**)}$$$(**)$ is a good example that demonstrates the Langlands correspondence.
We proved $(*)$ above. $\chi$ in $(*)$ is a Dirichlet character $(\mathbb{Z}/4\mathbb{Z})^\times \to \mathbb{C}^\times$ defined by $\chi(1) = 1$, $\chi(3) = -1$. This $(*)$ tells that a prime number $p \neq 2$ decomposes into two if $\mathbb{Z}[i]$ if and only if $p \equiv 1\,(\text{mod}\,4)$.
In $(**)$, we have that$$f(z) = \eta(6z)(18z) = q\prod_{n=1}^\infty (1 - q^{6n})(1 - q^{18n}),\text{ }q = e^{2\pi iz}$$and$$L(s, f) = \sum_{n=1}^\infty a_nn^{-s},$$with $a_n$ determined by$$q =\prod_{n=1}^\infty (1 - q^{6n})(1 - q^{18n}) = \sum_{n=1}^\infty a_nq^n$$$$=q - q^7 - q^{13} -q^{19} + q^{25} + 2q^{31} - q^{37} + 2q^{43} - q^{61} - q^{67} - q^{73} - q^{79} + q^{91} - q^{97} - q^{103} \dots$$(The proof of $(**)$ is quite difficult, so we will not include it here.) $(**)$ tells us that for a prime number $p \neq 2, 3$, we have that $(p)$ in $\mathbb{Z}[\sqrt[3]{2}]$ decomposes into a product of three maximal ideals if and only if $a_p = 2$. (It is not very hard to see that $(**)$ implies this, but we will not explain it here.) For example, $31$ decomposes into three in $\mathbb{Z}[\sqrt[3]{2}]$ as$$31 = (3 + 2^{1/3})(3 + 3(2^{1/3}) + 2^{2/3})(3 - 3(2^{1/3}) + 2^{2/3}).$$This $f$ is a modular form, and this is an example of the Langlands correspondence between arithmetic and modular forms. The Langlands correspondence tells us, roughly speaking, that Hasse zeta functions are expressed by zeta functions of modular forms. (The Dirichlet character is regarded as a special case of a modular form.)

REPLY [7 votes]: As mentioned in the comment, the trick is to look in the Euler product (which holds in the half plane of $Re(s)$ large enough and then extend by analytic continuation). Since it's an arithmetic identification, one should look at the behavior of primes in $\mathbb{Z}$ splitting in $\mathbb{Z}[i]$. 
It's not a hard fact to show that $2$ is the only ramified prime in $\mathbb{Z} \subset \mathbb{Z}[i]$ (the discriminant of this number field is $4$). It also follows via the following formula $\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}{2}}$ that an odd prime splits if and only if it reduces to $1 \pmod 4$.
Let's unfold the RHS, the product of (dimension $1$) $L$-functions, first looking at odd primes $p \equiv 1 \pmod 4$, so that $p = \mathfrak{p} \mathfrak{p}'$ in $\mathbb{Z}[i]$. The local factor at $p$ on the RHS is given by $\left(1 - p^{-s} \right)^{-2}$. The corresponding local factor on the LHS is given by $\left(1 - N(\mathfrak{p})^{-s}\right)^{-1} \left(1 - N(\mathfrak{p}')\right)^{-1}$, where $N$ is the norm map given by the extension $\mathbb{Q}(i)/\mathbb{Q}$. But $N(\mathfrak{p}) = N(\mathfrak{p}') = p$.
We unfold a similar argument for other odd primes $p$ that remain inert in $\mathbb{Z}[i]$. The local factor at $p$ on the RHS is given by $\left(1 - p^{-s}\right)^{-1} \left(1 + p^{-s} \right)^{-1} = \left(1 - p^{-2s}\right)^{-1}$. On the RHS, the local factor corresponding to $p$ is given by $\left(1 - N(p)^{-s}\right)^{-1}$. But $N(p) = p^2$ if $p$ is inert.
Lastly, for $p = 2$, we have $2 = \mathfrak{q}^2$ in $\mathbb{Z}[i]$. Because the character vanishes at the place $2$, the local factor on the RHS lives exclusively in $\zeta(s)$, and is thus $\left(1 - 2^{-s}\right)^{-1}$. But the LHS has a corresponding local factor of $\left(1 - N(\mathfrak{q})\right)^{-1}$, and $N(\mathfrak{q}) = 2$.
Just for completeness, convergence of our infinite Euler products is not an issue. 
Just as a brief remark (up to you how seriously to take this for now), although these functions are analytic manifestations of arithmetic gadgets, these factorization problems are, in nature, more algebraic/arithmetic than analytic. In particular, $L$-functions given by continuous, char. $0$ Galois representations illustrate this theme well (decomposing representations is given by factoring corresponding $L$-functions, which is crucial to showing such an $L$-function is given by a product of Hecke $L$-functions).<|endoftext|>
TITLE: Stability of the Solar System
QUESTION [27 upvotes]: Is the Solar System stable? 

You can see this Wikipedia page.
In May 2015 I was at the conference of Cedric Villani at Sharif university of technology with this title: "Of planets, stars and eternity (stabilization and long-time behavior in classical celestial mechanics)"
, at the end of this conference one of the students asked him this question and he laughed strangely(!) with no convincing answer!
Edit: The purpose of "long-time" is timescale more than Lyapunov time, hence billions of years.

REPLY [32 votes]: Due to chaotic behaviour of the Solar System, it is not possible to precisely predict the evolution of the Solar System over 5 Gyr and the question of its long-term stability can only be answered in a statistical sense. For example, in
http://www.nature.com/nature/journal/v459/n7248/full/nature08096.html (Existence of collisional trajectories of Mercury, Mars and Venus with the Earth, by 
J. Laskar and M. Gastineau) 2501 orbits with different initial conditions all consistent with our present knowledge of the parameters of the Solar System were
traced out in computer simulations up to 5 Gyr. The main finding of the paper is that one percent of the solutions lead to a large enough increase in Mercury's eccentricity to allow its collisions with Venus or the Sun.
Probably the most surprising result of the paper (see also http://arxiv.org/abs/1209.5996) is that in a pure Newtonian world the probability of collisions within 5 Gyr grows to 60 percent and therefore general relativity is crucial for long-term stability of the inner solar system.
Many questions remain, however, about reliability of the present day consensus that the odds for the catastrophic destabilization of the inner planets are in the order of a few percent. I do not know if the effects of galactic tidal perturbations or possible perturbations from passing stars are taken into account. Also different numerical algorithms lead to statistically different results (see, for example, http://arxiv.org/abs/1506.07602).
Some interesting historical background of solar system stability studies can be found in http://arxiv.org/abs/1411.4930 (Michel Henon and the Stability of the Solar System, by Jacques Laskar).<|endoftext|>
TITLE: Existence of Hecke operators with distinct eigenvalues?
QUESTION [6 upvotes]: Consider the space of modular forms $M_k(N)$.  Any modular form $f \in M_k(N)$ is determined by a finite number of Fourier coefficients (e.g., Sturm's bound), thus there is a finite set of Hecke operators that lets us distinguish eigenforms from each other.  In fact this is true for $T_p$'s rather than $T_n$'s
Question: Does there always exist a Hecke operator $T_p$ that distinguishes eigenforms?  I.e., is there always some $T_p$ acting on $M_k(N)$ with distinct eigenvalues?  
This is not true for all $p$ certainly (e.g., this question), but I want to know if you can have strange situations like $f_1, f_2, f_3$ are distinct eigenforms with $a_p(f_1) = a_p(f_2)$ for $p \equiv 1, 2$ mod $4$ and $a_p(f_1) = a_p(f_3)$ for $p \equiv 3$ mod $4$, say.
I would also be interested in partial results, e.g., cuspidal newforms in weight 2.
Edit: As pointed out in a comment and an answer, it's easy to come up with counterexamples using quadratic twists.  I would still like to know what happens if one restricts to "minimal" modular forms, say newforms of prime level.

REPLY [6 votes]: I think that this is false for Dirichlet characters (since you can take, say, $\chi_1, \chi_2, \chi_1 \chi_2$ when $\chi_1 \chi_2$ are quadratic  -- at least one will take the value $1$ on a prime) and then you can just twist your favorite modular form by these guys.<|endoftext|>
TITLE: Counting equivalence relations with marked classes
QUESTION [6 upvotes]: The number of equivalence relations on a set of $n$ elements is the Bell number $B_n$. 
If we wish to count the number of equivalence classes on a set of $n$ elements where one of the classes is marked, and the marked class is allowed to be empty, we get $B_{n+1}$.
Now, if two classes are marked (with identical markers), and as before, the markers can be attached to empty classes) we get $(B_{n+2} - B_{n+1} + B_n)/2$ partitions; the idea is to introduce two new elements (markers) to the set with $n$ elements, and then correct for the fact that we are using indistinguishable markers, and that the markers can not occur in the same equivalence class.
For equivalence relations with three marked parts (identical markers, marked classes can be empty) I got $(B_{n+3} - 3B_{n+2} + 5B_{n+1} + 2B_n)/6$ using similar techinques.
Is there a general theory to count the number of equivalence classes on a set of size $n$, with $k$ marked classes, the markings being identical and the marked classes being allowed to be empty? The leading term should be $B_{n+k}/k!$.

REPLY [4 votes]: If $B_{n,t}$ is the number of partitions of a set of size $n$, with $t$ parts marked (hopefully as desired, though I find the description unclear), then
$$ \sum_{n=0}^\infty \sum_{t=0}^\infty \frac{B_{n,t}}{n!} x^n y^t
   = \frac{\exp((1+y)(e^x-1))}{1-y}. $$
Now you can expand with respect to $y$ to look at a particular number of marks.  The value $y=0$ gives the Bell numbers.<|endoftext|>
TITLE: How small can the Mumford-Tate group of hypersurface be?
QUESTION [6 upvotes]: Is there some way  of giving a lower bound on the dimension of the Mumford-Tate group of a hypersurface? Let's say it's of general type, say, of degree $10$ inside $ \mathbb{P}^3$. 
(Edited from here onward, because I forgot about Fermat hypersurfaces):
I would expect small Mumford-Tate groups to be rare. 
For example, a Fermat hypersurface has Mumford-Tate group a torus. Is it possible to say, for example, that points with toral Mumford-Tate group are not Zariski-dense?

REPLY [4 votes]: For hypersurfaces in $\mathbb P^2$, i.e. curves, this follows from the Andre-Oort conjecture for $\mathcal A_g$, $g=(d-1)(d-2)/2$, as long as $d>4$. The moduli space of hyper surfaces is a sub variety of the moduli space of abelian varieties. By Andre-Oort, if the CM points are dense then it must be a special sub variety, that is, a Shimura variety. But the monodromy group of $H^1$ is full symplectic, so if it is a Shimura variety it is all of $\mathcal A_g$, which only happens for $d=3,4$ by dimension counting.
For higher dimension hyper surfaces I would imagine this follows from an Andre-Oort-like statement, although the moduli space of Hodge structures would not usually be a Shimura variety.<|endoftext|>
TITLE: Rectifying the definition of a closed category
QUESTION [14 upvotes]: The definition of a closed category I'm using is here. 
Suppose $V$ is a closed category and that for each object $b\in V$, $[b,-]$ has a left adjoint $- \otimes b$. The result is nearly a monoidal category, but the associator $\alpha \colon (a \otimes b) \otimes c \rightarrow a \otimes (b \otimes c)$ is not generally an isomorphism.
Question: Can we modify the definition of closed category directly so that whenever the adjoint above exists, result is always closed monoidal?
I know that we can fix this by requiring the adjunction to be "internal" in the sense that we have a natural isomorphism $\Phi \colon [a \otimes b, c] \rightarrow [a, [b, c]]$ but I'd rather there not be an asymmetry between closed and monoidal categories, since we can get from monoidal to closed by only demanding an ordinary adjuntion.
Edit: In light of my answer below, I've decided this question needed clarifying. What I want is some additional structure or property added to the axioms of a closed category such that
1) If $V$ is monoidal and $- \otimes b$ has a right adjoint $[b, -]$, then $[-,-]$ makes $V$ into this modified closed category
2) If $V$ is this modified closed category and $[b, -]$ has a left adjoint $-\otimes b$, then $-\otimes-$ makes $V$ into a monoidal category.
In the current state of affairs, 1 holds but 2 does not. With the additional property proposed in my answer below, 2 holds, but 1 does not.

REPLY [4 votes]: A symmetric closed category is a closed category together with isomorphisms
$$s:[A,[B,C]] \cong [B,[A,C]]$$ satisfying a few axioms: see Definition 1.1 of the paper ``On embedding closed categories" by Day and Laplaza that Buschi Sergio mentioned.
If you have one of these $(C,[-,-],I)$, together with adjoints $- \otimes A \dashv [A,-]$, then you get a symmetric monoidal category $(C,\otimes,I)$.  It is possible to give an elementary argument proving this, but it takes a bit of work to write down.  
It seems that you have realised this in your discussion with Buschi Sergio already but let me point it out anyway.  The result follows from Proposition 2.2 of Day and Laplaza's paper which asserts that a symmetric closed category gives rise to a symmetric promonoidal one with promonoidal structure $$P(A,B;C)=C(A,[B,C])$$: that is, a symmetric pseudomonoid in the symmetric monoidal bicategory Prof of profunctors.  Now if you have the adjoints you then have a representable symmetric promonoidal structure $$C(A \otimes B,C) \cong P(A,B;C)$$ and such amounts to a symmetric monoidal structure on $C$.  (Because the strong symmetric monoidal pseudofunctor Cat → Prof is essentially fully faithful - i.e. locally an equivalence).
The same paper gives examples showing that the canonical associator for the (almost) monoidal structure associated to a closed category needn't be invertible.  So there is a genuine asymmetry between the definitions of monoidal and closed category.  
The notions of skew monoidal and skew closed category rectify the asymmetry in a different way -- there is a perfect correspondence between skew monoidal structures $(C,\otimes,I)$ and skew closed structures $(C,[-,-],I)$ related by a natural isomorphism $C(A \otimes B,C) \cong C(A,[B,C])$.  See Proposition 18 of the paper "Skew closed categories" http://arxiv.org/abs/1205.6522 by Ross Street.<|endoftext|>
TITLE: Log smooth models for abelian varieties
QUESTION [6 upvotes]: Let $K$ be a field complete for a discrete valuation. Assume that the residue field has characteristic $p > 0$. Let $A$ be an abelian variety over $K$ having the property that (for some prime $\ell \neq p$) the action of the absolute Galois group of $K$ on the $\ell$-adic Tate module $T_\ell(A)$ is tamely ramified.
One could hope that under this condition, there is a projective model for $A$ over $\mathcal{O}_K$ which is log smooth; here the log structure is the one induced by the special fibre. I am not sure whether this is known; does anyone know a reference for this?
If one can find a projective, semistable, Galois-equivariant model for the abelian variety after a tamely ramified base change, then I know how to go from there - but my argument uses heavy machinery... There is a nice paper by Künnemann (Duke 1998) in which he proves that one can find a projective semistable model after a finite base change. But it is not clear to me whether a tame base change is sufficient to obtain such a model if one assumes moreover that the Galois action on the Tate module is tamely ramified.

REPLY [5 votes]: This question has been answered in the following preprint:
http://arxiv.org/abs/1512.02464<|endoftext|>
TITLE: Is the $\infty$-category of presentable $\infty$-categories presentable?
QUESTION [5 upvotes]: Let $\mathit{Pr}^L$ be the $\infty$-category of presentable $\infty$-categories and continuous functors in some universe.  Is it presentable itself a larger universe?

REPLY [7 votes]: Let us fix a universe and use the words "large" and "small" with respect to that universe. Presentable $\infty$-categories are typically large $\infty$-categories (since, as the previous answer mentioned, small $\infty$-categories are rarely presentable). One might then expect that $Pr^L$ would be a "very large" $\infty$-category (like the $\infty$-category $\widehat{Cat}_\infty$ of possibly large $\infty$-categories). In that case $Pr^L$ would turn from being very large to just large upon increasing the universe, and it would be natural to ask if it then becomes presentable. However, the $\infty$-category $Pr^L$ is actually not "very large", but just large. This is because presentable $\infty$-categories, though being large, are actually determined by a small amount of data. To formally prove this one might consider, for example, for each cardinal $\kappa$, the subcategory $Pr^L_{\kappa} \subseteq Pr^L$ consisting of $\kappa$-compactly generated presentable $\infty$-categories and functors preserving $\kappa$-compact objects between them (see section 5.5.7 of higher topos theory). Proposition 5.5.7.10 loc. cit. shows that when $\kappa > \omega$ the $\infty$-category $Pr^L_{\kappa}$ is equivalent to the $\infty$-category of small $\infty$-categories admitting $\kappa$-small colimits, and hence $Pr^L_{\kappa}$ is large (but not very large). Consequently, $Pr^L$ is a large colimit of large $\infty$-categories, and hence large (but again not very large). It follows that if we increase the universe $Pr^L$ will become small, and it will not be so natural to ask if it is presentable. On the other hand, since $Pr^L$ is just large you might ask if it is presentable without increasing the universe. In principle the answer would have to be no, because then $Pr^L$ would contain itself, and we know that such stories do not end well (although I admit I do not have a direct proof in mind, and would like to see one). Morally, $Pr^L$ should not be a presentable $\infty$-category, but some $(\infty,2)$-version of the notion, similarly to how the $\infty$-category of $\infty$-topoi should be something like an $(\infty,2)$-topos (but not an $(\infty,1)$-topos). Unfortunately, I am not aware of these ideas being made precise anywhere.    
Edit: a formal argument why $Pr^L$ is not presentable (in the current universe) is that it is not locally small, see comments below.<|endoftext|>
TITLE: Does the critical sequence for subalgebras of elementary embeddings with finitely many generators have order type $\omega$?
QUESTION [5 upvotes]: Suppose that $\lambda$ is a cardinal. Let $\mathcal{E}_{\lambda}$ be the set of all elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$. If $j,k\in\mathcal{E}_{\lambda}$, then define
$j[k]=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}})$.  Suppose that $j_{1},...,j_{n}\in\mathcal{E}_{\lambda}$. Let $\langle j_{1},...,j_{n}\rangle$  denote the smallest subset of $\mathcal{E}_{\lambda}$ that contains $\{j_{1},...,j_{n}\}$ and is closed under the operation $(j,k)\mapsto j[k]$. Let $K=\{\mathrm{crit}(j)|j\in\langle j_{1},...,j_{n}\rangle\}$. 

Then does $K$ necessarily have order type $\omega$?

If $\gamma<\lambda$ is a limit ordinal, then let $\equiv^{\gamma}$ to be the equivalence relation on $V_{\lambda}$ where $j\equiv^{\gamma}k$ if and only if $j(x)\cap V_{\gamma}=k(x)\cap V_{\gamma}$ whenever $x\in V_{\gamma}$. Then $\equiv^{\gamma}$ is a congruence on $\mathcal{E}_{\lambda}$.

Is the quotient algebra $\langle j_{1},...,j_{n}\rangle/\equiv^{\gamma}$ finite whenever $\gamma$ is a limit ordinal with $\gamma<\lambda$?

An affirmative answer to question 2 will give an affirmative answer to question 1 as well.

REPLY [3 votes]: I claim that the answer to both questions is yes
.
Suppose that $A$ is a finite set, $j_{a}\in\mathcal{E}_{\lambda}$ for each $a\in A$ and $\gamma<\lambda$ is a limit ordinal. Then I claim that $\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is finite and the sequence $\{\textrm{crit}(j)|j\in\langle\{j_{a}|a\in A\}\rangle\}$ has order type $\omega$.
Let $M$ be the set of all strings $a_{1}...a_{n}$ in $A^{+}$ so that either $n=1$ or if $m<n$, then $\textrm{crit}(j_{a_{1}}[j_{a_{2}}]...[j_{a_{m}}])<\gamma$. Then $M$ is finite; if $M$ were infinite, then the tree $M$ would have an infinite branch consisting of all substrings of $(a_{n})_{n=1}^{\infty}$, but for such an infinite branch, the sequence of elementary embeddings $(j_{a_{1}}[j_{a_{2}}]...[j_{a_{n}}])_{n=1}^{\infty}$ contradicts the Laver-Steel theorem.
Let $F$ be the set of all maximal strings in $M$ (in other words, $x\in F$ if and only if $xa\not\in F$ whenever $a\in A$) and let $L=M\setminus F$. Then define a binary operation $*$ on $M$ according to the following rules:
i. $x*y=y$ whenever $x\in F$
ii. $x*a=xa$ whenever $x\in L,a\in A$
iii. $x*y=(x*z)*xa$ whenever $y=za$ and $x\in L$.
One can show that $*$ produces a well-defined total operation on $M$ the same way that one can show without resorting to large cardinals that the Laver table operation operation on $\{1,...,2^{n}\}$ is well-defined. Indeed, the structure $(M,*)$ is a generalization of the notion of a Laver table (the structure $(M,*)$ is self-distributive although we shall not use the self-distributivity of $(M,*)$ in this answer).
Define a mapping $\Phi:(M,*)\rightarrow(\mathcal{E}_{\lambda}/\equiv^{\gamma})$ by letting
$\Phi(a_{1}...a_{n})=j_{a_{1}}[j_{a_{2}}]...[j_{a_{n}}]$ whenever $a_{1}...a_{n}\in M$. Then one can show that $\Phi(x)*\Phi(y)=\Phi(x*y)$ by a double induction which is descending on the length of $x$ and for each individual $x$ one uses an induction which is ascending on the length of $y$. In other words, $\Phi$ is a homomorphism. However, we have $\Phi[M]=\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ since the algebra $M$ is generated by $A$. Therefore the algebra $\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is a finite algebra.
Since each algebra $\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is finite, the set $\{\textrm{crit}(j)|j\in\langle\{j_{a}|a\in A\}\rangle\}\cap\gamma$ is finite. Therefore, the set
$\{\textrm{crit}(j)|j\in\langle\{j_{a}|a\in A\}\rangle\}$ has order type $\omega$.<|endoftext|>
TITLE: Minimum value of $|p(1)|^2+|p(2)|^2 +...+ |p(n+3)|^2$ over all monic polynomials $p$
QUESTION [20 upvotes]: Let $n$ be a positive integer. Determine the smallest possible value of $|p(1)|^2+|p(2)|^2 +...+ |p(n+3)|^2$ over all monic polynomials $p$ of degree $n$.

This question was proposed (problem A.611)
some time ago at KoMaL.
The minimal values for $n=0,1,2,3$ are $3,5,14,324/5$.
It was also discussed at math.se, where the polynomial minimising such sum was found using Gram-Schmidt algorithm.
However the minimum value was not determined.
Any suggestion is welcome.

REPLY [33 votes]: [edited to explain a few steps and connect with the Hahn polynomials]
The answer is
$$
\frac{(2n+1)(2n+3) n!^4}{(2n)!}
$$
assuming that I did the algebra right, which seems likely because this formula
agrees with the previously computed values $3,5,14,324/5$ for $n=0,1,2,3$.
Consider first the minimum of $\sum_{i=1}^{n+1} p(i)^2$ over monic $p$
of degree $n$.  Any vector $v = (a_1,a_2,\ldots,a_{n+1})$ is the list of
values at $1,2,\ldots,n+1$ of some polynomial of degree at most $n$;
the leading coefficient of this polynomial is $(u,v) / n!$ where
$u$ is the vector whose $(n+1-i)$-th coordinate is $(-1)^i {n \choose i}$
(each $i$ in $0 \leq i \leq n$).$\color{red}{\bf[1]}$
We thus seek the minimum of $(v,v)$ subject to $(u,v) = n!$, 
and by Cauchy-Schwarz the answer is $n!^2 / (u,u)$, attained iff
$v = n! u / (u,u)$.  The denominator $(u,u)$ is $\sum_{i=0}^n {n \choose i}^2$,
which is well-known to equal $2n \choose n$.$\color{red}{\bf[2]}$
Hence the answer is $n!^2 / {2n \choose n} = n!^4 / (2n)!$.
With a bit more work we can find for each $k$ the minimum of
$\sum_{i=1}^{n+k+1} p(i)^2$ over monic $p$ of degree $n$.
Here's how it goes for $k=2$.  There are now three linear conditions
on $v = (a_1,a_2,\ldots,a_{n+3})$ to be the list of values at
$1,2,\ldots,n+3$ of a monic polynomial of degree at most $n$.
We can write them as $(u_0,v)=n!$, $(u_1,v)=0$, $(u_2,v)=0$,
where $u_j$ is the vector whose $(n+3-i)$-th coordinate is
$(-1)^i {n+j \choose i}$ for each $i$ in $0 \leq i \leq n+2$.$\color{red}{\bf[1a]}$
As was the case for $k=0$,
the minimum of $(v,v)$ over all such $v$ is attained by
a linear combination of $u_0,u_1,u_2$.  So we need only
calculate the $3 \times 3$ Gram matrix of inner products
$(u_j,u_{j'})$ $(j,j'=0,1,2)$, and invert it to find the linear combination
$v$ such that $(u_j,v) = n! \delta_j$.  Each $(u_j,u_{j'})$ is
$\sum_{i \geq 0} {n+j \choose i} {n+j' \choose i} 
 = {2n+j+j' \choose n+j}$.$\color{red}{\bf[2a]}$
So write each of these entries of the Gram matrix as $2n \choose n$ times
some rational function of $n$, solve the resulting linear equations for
the coefficients of $v$ in $u_0,u_1,u_2$, and recover $(v,v)$.
This calculation yields the formula $(2n+1)(2n+3) n!^2 / {2n \choose n}$
displayed (in equivalent form) at the start of this answer.
For general $k$ the minimum seems to be
$$
\frac{n!^4}{(2n)!} {2n+1+k \, \choose k},
$$
which presumably can be proved from the above analysis
 and known identities.$\color{red}{\bf[3]}$

$\color{red}{\bf[1],[1a]}$ Taking the inner product with $u$ 
 amounts to evaluating an $n$-th finite difference.
Likewise, taking the inner product with $u_i$ amounts to evaluating
 an $(n+i)$-th finite difference.
$\color{red}{\bf[2],[2a]}$ The formula
$\sum_{i \geq 0} {m \choose i} {m' \choose i} = {m+m' \choose m}$
has at least two well-known proofs, one bijective and one
generatingfunctionological.
For the former, write ${m+m' \choose m}$ as the number of $(m+m')$-tuples
of $m$ 0's and $m'$ 1's, and let $i$ be the number of 1's among the
first $m$ coordinates.  For the latter, compute the $X^m$ coefficient 
of $(1+X)^m \, (1+X)^{m'} = (1+X)^{m+m'}$ in two ways.
$\color{red}{\bf[3]}$ Further corroboration is that this is also 
consistent with the extreme cases $n=0$ and (with a bit more work) $n=1$.
I later obtained a proof by transforming the relevant determinants into 
Vandermonde determinants.  The existence of such a formula for all $n,k$ 
suggested that the $p$'s (which are orthogonal polynomials for a 
discrete measure) must be known already, and after some Googling 
found that indeed they are the special case $\alpha=\beta=0$ of the 
Hahn polynomials
$Q_n$ evaluated at $x-1$ (with $N = n+k+1$).  The orthogonality relation,
together with the formula for the leading coefficient of $Q_n$,
soon yields the evaluation for all $n,k$ of the minimum of 
$\sum_{i=1}^{n+k+1} p(i)^2$ over monic polynomials $p$ of degree $n$.<|endoftext|>
TITLE: Uniformly small sums of roots of unity
QUESTION [12 upvotes]: I have considerable numerical evidence that
for all $0\leq k\leq{{n-1}\over 2}$ ($n$ odd) there exists a subset $
S_k$ of {1,2,...,n} of cardinality $k$
such that the modulus square of $g(z)=\sum_{j\in S_k}z^j$ is less than
$n/2$ for ALL $n^{th}$ roots of unity $z\neq 1.$ [By the way, I am not 
assuming that $n$ is prime.] Am I having a run of bad 
days, or is this a bummer to prove?
Greg

REPLY [5 votes]: This is a reworked version of my original answer, which now solves the problem for infinitely many pairs $(n,k)$ ( but certainly not all pairs).
Let $G:={\mathbb Z}/n{\mathbb Z}$. You want to show that for every $1<k<n/2$, there is a subset $S\subset G$ of size $|S|=k$ such that the indicator function $1_S$ has all its non-trivial Fourier coefficients $\hat1_S(\chi)$ smaller than $\sqrt{n/2}$ in absolute value. 
Denoting by $r_S(g)$ the number of representations of $g$ in the form $g=s'-s''$ with $s',s''\in S$, we have the easily-verified identity
  $$ |\hat 1_S(\chi)|^2 = \sum_{g\in G} r_S(g)\chi(g),\quad \chi\in\hat G. \tag{$\ast$} $$
Suppose now that $S$ is a $(n,k,\lambda)$-difference set in $G$; that is, a $k$-element subset of $G$ such that $r_S(g)=\lambda$ for each non-zero element $g\in G$. A simple double counting shows that a necessary condition for an $(n,k,\lambda)$-difference set in $G$ to exist is that $k(k-1)=\lambda(n-1)$. Consequently, it follows from ($\ast$) that for such a set $S$ and any non-principal character $\chi$, we have
  $$ |\hat1_S(\chi)|^2 = k-\lambda \le \frac{k(k-1)}{2\lambda} = \frac{n-1}2, $$
as wanted.
This answers your question in the situation where the group $G={\mathbb Z}/n{\mathbb Z}$ admits an $(n,k,\lambda)$-difference set, for an appropriate value for $\lambda$. Such difference set are known to exist for infinitely many pairs $(n,k)$, just google for "cyclic difference sets" (here "cyclic" indicates that we are interested in difference subsets of cyclic groups).<|endoftext|>
TITLE: Does the forgetful functor from presentable $\infty$-categories to $\infty$-categories preserve filtered colimits?
QUESTION [5 upvotes]: Marc's answer to my previous question gives a way to compute colimits in the category of presentable $\infty$-categories and continuous functors, using the (discontinuous) right adjoints to those functors.  But in particular it is not true that the forgetful functor from presentable $\infty$-categories and continuous functors, to all $\infty$-categories and functors, preserves all colimits.
Does it preserve all filtered colimits?  I am sorry if the answer is easy to find in Lurie's textbook.  I wasn't able to do so right away.

REPLY [14 votes]: It doesn't. For instance, the $\infty$-category of spectra is the colimit of the tower
$$ \mathcal{S}_* \stackrel\Sigma\to \mathcal{S}_* \stackrel\Sigma\to ... $$
in $Pr^L$, but its colimit in $Cat_\infty$ is the Spanier-Whitehead category whose objects are formal desuspensions $\Sigma^{-n}X$ of pointed spaces. The latter category is stable and (uncountably) accessible but lacks some infinite colimits, like the sum of $\Sigma^{-n}S^0$ for $n\geq 0$.<|endoftext|>
TITLE: Does the following $ C^{*} $-algebraic result have a purely algebraic proof?
QUESTION [8 upvotes]: While studying the proof of Bott periodicity for operator $ K $-theory in this set of notes, I learned this fact:

Theorem. Let $ A $ and $ B $ be $ C^{*} $-algebras. Let $ f,g: A \to B $ be $ * $-homomorphisms. Then $ f + g $ is also a $ * $-homomorphism if and only if the ranges of $ f $ and $ g $ are orthogonal, i.e.,
  $$
f[A] g[A] = g[A] f[A] = \{ 0_{B} \}.
$$

In general, $ f + g $ is only a $ * $-preserving linear map, and multiplication may not be preserved unless further conditions are imposed. I managed to prove the theorem, but my argument is not entirely algebraic in the sense that it uses topological facts about $ C^{*} $-algebras.
My proof
The backward implication is trivial enough, so let us prove the forward one only.
Suppose that $ f + g $ is a $ * $-homomorphism. Then for all $ a_{1},a_{2} \in A $, we have
\begin{align}
    (f + g)(a_{1} a_{2})
& = (f + g)(a_{1}) \cdot (f + g)(a_{2}) \\
& = [f(a_{1}) + g(a_{1})] [f(a_{2}) + g(a_{2})] \\
& = f(a_{1}) f(a_{2}) + f(a_{1}) g(a_{2}) + g(a_{1}) f(a_{2}) + g(a_{1}) g(a_{2}), \\
    (f + g)(a_{1} a_{2})
& = f(a_{1} a_{2}) + g(a_{1} a_{2}) \\
& = f(a_{1}) f(a_{2}) + g(a_{1}) g(a_{2}).
\end{align}
It follows immediately that $ (\star) ~ f(a_{1}) g(a_{2}) + g(a_{1}) f(a_{2}) = 0_{B} $ for all $ a_{1},a_{2} \in A $.
Next, let $ a \in A $ be any self-adjoint element. As $ (\star) $ implies that $ f(a) g(a) = - g(a) f(a) $, we get
$$
  f(a^{2}) g(a^{2})
= f(a) f(a) g(a) g(a)
= - f(a) g(a) f(a) g(a)
= f(a) g(a) g(a) f(a),
$$
and similarly,
$$
  g(a^{2}) f(a^{2})
= g(a) g(a) f(a) f(a)
= - g(a) f(a) g(a) f(a)
= f(a) g(a) g(a) f(a).
$$
We also know from $ (\star) $ that $ f(a^{2}) g(a^{2}) + g(a^{2}) f(a^{2}) = 0_{B} $, so $ f(a) g(a) g(a) f(a) = 0_{B} $. Hence, by the self-adjointness of $ a $, we have
$$
[f(a) g(a)] [f(a) g(a)]^{*} = f(a) g(a) g(a) f(a) = 0_{B}.
$$
Therefore, $ f(a) g(a) = 0_{B} $, and by interchanging $ f $ and $ g $, we also obtain $ g(a) f(a) = 0_{B} $. As our choice of $ a $ was arbitrary, the discussion in this paragraph applies to all self-adjoint elements of $ A $.
Finally, let $ (e_{i})_{i \in I} $ be any self-adjoint approximate identity in $ A $. Then for all $ x,y \in A $, we get
\begin{align}
    f(x) g(y)
& = \lim_{i \in I} f(x e_{i}) g(e_{i} y) \qquad
    (\text{$ C^{*} $-homomorphisms are automatically continuous.}) \\
& = \lim_{i \in I} f(x) f(e_{i}) g(e_{i}) g(y) \\
& = \lim_{i \in I} f(x) ~ 0_{B} ~ g(y) \qquad (\text{By the previous paragraph.}) \\
& = 0_{B}.
\end{align}
Similarly, $ g(x) f(y) = 0_{B} $ for all $ x,y \in A $. This concludes the proof. $ \quad \blacksquare $


Question. Can we obtain the same result if we merely assume that $ A $ and $ B $ are $ * $-algebras over $ \Bbb{C} $? For convenience, we may suppose that $ (a^{*} a = 0_{A}) \Rightarrow (a = 0_{A}) $ for all $ a \in A $ and likewise for $ B $.

REPLY [4 votes]: Here is a small extension of your idea. You have, for any $a,b\in A $,
$$
f (a)g (b)+g (a)f (b)=0.
$$
Then
$$
f (ab)g (ba)=f (a)f (b)g (b)g (a)=-f (a)g (b)f (b)g (a)=f (a)g (b)g (b)f( a)
$$
and
$$
g (ab)f (ba)=g (a)g (b)f (b)f (a)=-g (a)f (b)g (b)f (a)=f (a)g (b)g (b)f (a).
$$
Now
$$
0=f (ab)g (ba)+g (ab)f (ba)=2f (a)g (b)g (b)f (a).
$$
When $a,b $ are selfadjoint we get
$$
f (a)g (b)[f (a)g (b)]^*=f (a)g (b)g (b)f (a)=0,
$$
and we conclude that $f (a)g (b)=0$ for all selfadjoint  $a,b $. But then, as any $x,y\in A $ can be written $x=a+ib $, $y=c+id $,
$$
f (x)g (y)=f (a+ib)g (c+id)=f (a)g (c)-f (b)g (d)+i [f (b)g (c)+f (a)g (d)]=0.
$$<|endoftext|>
TITLE: A question of Erdős on entire functions
QUESTION [12 upvotes]: At the end of the following paper, Erdős asked if there is a family $F$ of entire functions of size continuum such that for every $z \in \mathbb{C}$, $\{f(z) : f \in F\}$ has size less than continuum. He also showed how to construct such a family under CH.
Did someone solve it?

REPLY [10 votes]: The following (negative answer to Erdos' question) will appear in a joint work with Shelah.
Claim: Suppose $V \models 2^{\aleph_0} = \lambda > \kappa = \aleph_1$. Let $P$ add $\kappa$ Cohen reals. Then in $V^{P}$, there is no such family.
Proof: Let $r \in {}^{\kappa}2$ be the Cohen generic sequence. Clearly $V[r] \models 2^{\aleph_0} = \lambda$. Suppose $\langle f_{\alpha} : \alpha < \lambda \rangle$ is a sequence of pairwise distinct analytic functions in $V[r]$. Choose $X \in [\lambda]^{\lambda}$, $\xi_{\star} < \kappa$ such that for each $\alpha \in X$, $f_{\alpha}$ is coded in $V[r \upharpoonright \xi_{\star}]$. Let $z_{\star} \in \mathbb{C}$ be Cohen over $V[r \upharpoonright \xi_{\star}]$. Since two distinct analytic functions only agree on a countable set, it follows that $\langle f_{\alpha}(z_{\star}) : \alpha \in X \rangle$ are pairwise distinct.
Update: Shelah and I showed that the answer to Erdos' question is independent of ZFC + not CH so the other direction also holds. 
An interesting question that remains open is: Is the following consistent: $2^{\aleph_0} = \aleph_2$ and there exists $U \in [\mathbb{C}]^{\aleph_1}$ such that for every $A \in [\mathbb{C}]^{\aleph_1}$, there is a non constant entire map that sends $A$ into $U$?<|endoftext|>
TITLE: Counterexamples for strengthening Whitehead's theorem?
QUESTION [14 upvotes]: Let $f:X\to Y$ be a pointed map of pointed connected $n$-dimensional CW complexes. Whitehead's theorem says that if $f_*:\pi_qX\to \pi_qY$ is an isomorphism for $q\le n$ and a surjection for $q=n+1$, then $f$ is a homotopy equivalence (e.g. Theorem (Whitehead) on p.75 of May's "Concise Course in Algebraic Topology"). 
I am interested in counterexamples to this when you drop the surjectivity condition for $q=n+1$. That is,

Question: What examples are there of a map $f:X\to Y$ of pointed connected $n$-dimensional CW complexes that induces isomorphism on $\pi_q$ for $q\le n$, but is not a homotopy equivalence?

I would also like to know what are the "minimal" examples of this. For example, it seems impossible for $n\le 2$ (the induced map on universal covers is a homology isomorphism by the Hurewicz theorem, and hence is a weak equivalence and thus induces isomorphism on all higher homotopy groups). I also wonder if there is an example with finite complexes.

REPLY [8 votes]: I think isos up to dimension $n$ is enough to deduce $f_*:[A,X]\to[A,Y]$ onto when $dim(A)\leq n$.  This gives a right inverse to $f$ which also induces isos up to $n$ and hence has a right inverse. So $f$ has a homotopy inverse.<|endoftext|>
TITLE: Geometric generic fibre
QUESTION [18 upvotes]: This is a pretty elementary question about schemes, but it came up in the course of research, so let's try it here rather than MSE.

Question 1: Are the fibres of a family of complex varieties isomorphic as schemes to the geometric generic fibre, outside of a union of countably many subfamilies?

That seems outlandish, so let me explain below the reasoning that led me to ask. If my argument is flawed I would be glad to know why. If, by contrast, this is well-known, I would be glad of a reference, so I ask:

Question 2: Does anyone know of a published reference for this fact, if it is correct? 


I fix the following simple setup for concreteness. 
Let $f: X \rightarrow \mathbf A^1$ be a family of varieties over $\mathbf C$. 
Let $k = \mathbf C(t)$, the function field of the base, and $K=\overline{k}$. Let $G$ be the geometric generic fibre of $f$, that is, the $K$-variety$ X \times_{\mathbf A^1} \operatorname{Spec K}$. 
Now $K$ is algebraically closed, has characteristic zero, and has the same cardinality as $\mathbf C$. So there is a field isomorphism $\alpha: \mathbf C \simeq K$. (As I understand it, this depends on the axiom of choice, but that's alright.) So base change by $\alpha$ turns $G$ into a variety $G_\alpha$ over $\mathbf C$, isomorphic to $G$ as a scheme. (OK so far?)
Now suppose for concreteness that $f$ is a family of hypersurfaces in projective space $\mathbf P^n$, so it is given by a form $F(x_0,\ldots,x_n;t)$ where the $x_i$ are coordinates coordinates on $\mathbf P^n$ and $t$ is the coordinate on $\mathbf A^1$. 
Now pick any number $z \in \mathbf C$ which is algebraically independent from all the coefficients of $F$. Then we can choose our field isomorphism so that $\alpha^{-1}$ fixes all the coefficients of $F$, and $\alpha^{-1}(t)=z$. Then base change by $\alpha$ just has the effect of substituting $z$ in place of $t$ in the form $F$: in other words, $G \simeq G_\alpha \simeq G_z$, the fibre of $f$ over $z \in \mathbf A^1$.

This argument has the disturbing (to me) consequence that all but countably many fibres of $f$ are isomorphic, albeit in a weird way, as schemes. (Of course, it doesn't claim that they are isomorphic as varieties over $\mathbf C$, which would be absurd.) But maybe this just shows that my intuition about scheme isomorphism is lacking. Either way, I would be glad to know!

REPLY [7 votes]: This statement is indeed pretty remarkable. A reference is Lemma 2.1 in C. Vial. Algebraic cycles and fibrations. Documenta Math. 18 (2013). The statement there is given for varieties, but it seems likely that it holds more generally.<|endoftext|>
TITLE: Which journals publish research announcements?
QUESTION [14 upvotes]: Can anybody advise mathematical journals that publish research announcements? (I mean little papers without proofs.) 
It sometimes happens that a proof is so long that it takes years to review and few journals are willing to accept it. I think in this case it would be useful (apart from posting your paper in arXiv) to  announce the result in a short communication (this is how this problem is resolved in Russia, but I am asking about the rest of the world).
I am working in Functional analysis and in Geometry.
EDIT. More generally, I am curious, 

how do people solve the problem of long texts? 

Suppose you write a long text, where the main results are just several propositions, say, 5 theorems, and the rest are various technical lemmas, many of them, say, 100. Your text is devoted to the explanation of one idea, those 5 theorems appear only in the end, and it is impossible to separate them so that some of them could be proved in the middle of the text. As a corollary, it is impossible to divide your text into several papers, on 40-50 pages, so that they could be sent to usual journals.
You have to send this long paper to a journal, that publishes long texts, they will spend some time on finding a reviwer, he will check everything in your text, after that they will put your paper into the queue, and this is a long process, you understand... It can take 2-3-5 years before your paper will be published.
Of course, it will be difficult to explain to your employer, what you were doing the prevoius 5 years, when you were writing this paper. 
So how do people resolve this problem? In Russia there is a possibility to write several  little papers, without proofs, but with the formulations of the main results (those 5 theorems), then send them to some journals (together with that long text with accurate proofs), then the reviewer checks everything, agrees, they publish these little papers (this is much quicker, since these papers are little, there is no need to put them into a long queue), and everything is OK. As far as I understand, in France the situation is more or less similar. What about the rest of the world?

REPLY [6 votes]: Tl;dr: I don't think the math publication ecosystem has a perfect solution to your problem.
First, note that even moderate-length papers can take years to be published (my personal record is of 5 years from submission to print, involving only one submission, for a 15-pages paper). If your employer only considers published papers, and not accepted or arXiv posted ones, you are basically out of good options (of course if any journal where to count, you could find a quick predatory journal, but I would not advise this).
The usual process is to post the complete paper on the arXiv and submit to a suitable journal (Memoires de la SMF is another good available option, similar to Memoirs AMS, and I think Documenta is able to publish long papers). Publishing an announcement has become rare, but is still possible (several venues for this have been mentioned already). However, when you say that proofs are thoroughly check, that is usually not true; and if it were, then you announcement would also take much time to be accepted, since the review time would be long.
So, it somehow seems that your main concern is about the accepted-to-published delay. I thus advise you to have a look at the AMS data on delays (published yearly in the Notices), you might find a good venue. Also, you could have a shot at an electronic only journal where this delay can be much shorter. Forum of Mathematics (Sigma or Pi) could be an option if your paper is very strong, but I think they now charge authors, and you will have the usual problem that you could spend a year or more waiting for a report which could turn out negative.
We lack a math megajournal which would publish any good paper (good meaning good enough to deserve publication) electronically, avoiding unecesary delays in resubmission along the prestige ladder and in finding room into issues of fixed length. That would be about the best option for you, where it to exist.
To finish on another note: once your large paper is written and waiting for a venue, you may consider working on smaller scale projects to get your beans to be counted. I do not like to advise this, but given the incentives you mention that might be the only solution. It is a bit like a painter working for hire in advertisement and buying himself time for his masterpiece; be careful that your smaller scale project have some interest instead of being noise in the publication record.<|endoftext|>
TITLE: What technical and/or theoretical challenges are involved in automatically extracting proofs from books and papers into Coq code?
QUESTION [23 upvotes]: Over the years, advances in machine learning has allowed us to communicate and interact, using the same natural language, more and more semantically with computers, e.g. Google, Siri, Watson, etc. On the other side, proof automation and formalization has gained more and more steam, culminating at, for example, recent efforts by Gonthier et al. to encode the Feit-Thompson theorem, and this trend will not stop as homotopy type theory has become a hot research topic. It then only seems natural to ask what open problems or technical challenges still lie ahead before we arrive at a Google Books style of automatically extracting human proofs written on paper (or just in latex) to formalized and checked proofs in something like Coq?
Ideally, answers regarding both the mathematical side and the more empirical machine learning side (or perhaps other viewpoints I've not considered here) are welcome, but I'm not quite sure if the latter is on topic on mathoverflow. And ideally, I should think we want to avoid babbling philosophy here, e.g. what's the point of mathematics, etc.

REPLY [22 votes]: I think Andrej's opinion is very accurate. Here are a few more comments based on my experience as a contributor to the Coq proof of the Feit-Thompson theorem (sorry it's a bit long).
This formal development comprises three natures of libraries: some are for infrastructure purposes and no
mathematician cares about the lemmas they prove (litanies of lemmas on
booleans, lists or natural numbers, seemingly cryptic boilerplate code...),
some contain constructions (permutation groups, rational numbers,...),
and some contain meaningful theories (finite
group theory, linear algebra, character theory,...). At least in this
last category, which is by far the largest in size, the length of both
statements and proof scripts are of the same order of size than the
LaTeX code of their paper counterpart. The statements are kept as
readable as possible by a choice of LaTeX-like notations which mimic
the corresponding type-setting source
code. Proof scripts on the other hand look rather different, but long
ones (for long proofs) are usually structured with many explicitly
stated intermediate lemmas, that are as readable as the main one.
Beside a (dumb and merciless) proof checker, interactive proof
assistants actually include
various tools that are there to help with the gap issue.
As already mentioned in this thread, comprehensive and well-organized
libraries of formalized mathematics are a mandatory ingredient. Also
it is possible to automate the description of computational parts in proofs
by implementing formal-proof-producing decision or normalization
procedures (like for instance computer-algebra-like routines for
algebraic expressions). But this is not enough --and by the way the
libraries of the Feit-Thompson theorem actually barely use this latter
form of automation.
One crucial point there was to be able to input mathematical
sentences, for
instance the formal statement of wannabe theorems, with as much
facilities as one would write them on paper. Otherwise very soon you
just cannot read them (nor are you able to input them in fact).
But in that case you need to have the proof
assistant automatically and silently decorate the objects with the
implicit information carried by the standard mathematical notations.
Once the appropriate infrastructure is set (which requires
in particular understanding the common notational practices of the
authors in the field), the routine arguments mentioned by Andy Putman
require close to zero input from the user. For instance if you know
that $K$ and $H$ are two subgroups of a same group, showing that
$0 \lt \#(H \cap N(K))$ is just trivial (i.e. a two character Coq proof
script), because the system knows that the
normalizer of a group and the intersection of two
groups are themselves groups, and that the cardinal of a group is always
positive for a group has at least one (neutral) element. Indeed,
these bureaucratic steps can be implemented by a program chaining some
identified lemmas for you behind the scene. This either closes some
branches in the proof or calculates the glue needed to make your
current goal match the known result you want to use at this stage.
Crafting this infrastructure was one of the fun and creative parts of
the formalization work (it is not just about tagging some lemmas) and
I do not really see how automated tools could significantly help
here. There is still much to be done: we do not have yet, in any proof
assistant on the market, a comprehensive set of compatible formal
mathematical libraries that would encompass a common socle of theories
for graduate level in algebra and geometry and analysis etc.
For instance the libraries we wrote do not address real and complex
analysis at all. Although I am near to ignorant in AI or machine learning,
this seems to me a prerequisite step to have automated tools produce
formal proofs from any kind (script, paper, voice,...) of input description.
Yet the advent of modern ways of interacting with proof assistants (to
describe formal proofs, and search existing content) is probably
one of the major challenges to overcome, in order for proof assistants
to become as popular as computer algebra systems (or even type-setting
environments) among researchers in mathematics.<|endoftext|>
TITLE: Commutative von Neumann algebras and localizable measure spaces
QUESTION [5 upvotes]: This is not my subject so I apologize if my question is too obvious or understood from other pages.
I read some pages such as
Reference for the Gelfand-Neumark theorem for commutative von Neumann algebras and
von neumann algebras and measurable spaces.
If I understand correctly, there is some correspondence between localizable measure spaces and commutative von Neumann algebras given by
$$(\Omega,\nu)\mapsto L^{\infty}(\Omega,\nu).$$ 
But I wanted to clarify:
(1) What is the correct notion of a morphism of commutative von Neumann algebras? Is it a normal *-homomorphism? What is the exact definition of normal? is it the same as being σ-weakly continuous?  
(2) Is it true that the opposite category of the category of commutative von Neumann algebras (with the appropriate class of morphisms) is equivalent to the category of localizable measure spaces and measurable maps? Or do we need to use another type of morphisms between localizable measure spaces?
Any good references on the above will be highly appreciated.

REPLY [3 votes]: The split interval $I^{\parallel} = \{t^+ : t \in [0,1]\} \cup \{t^- : t \in [0,1]\}$ yields the standard counterexample to the second question (details can be found in Fremlin Vol 3I, section 343, especially 343J, the exercises and the notes and comments at the end of the section). There usually are many maps of the measure space that induce the same morphisms of the measure algebra.
The identity homomorphism of the measure algebra of $I^{\parallel}$ is induced both by the identity map $f$ and by the map $g$ exchanging $t^+$ with $t^-$. Since the measure algebra uniquely determines the associated von Neumann algebra, both these maps induce the identity map of the algebra. Obviously $f(x) \neq g(x)$ for all $x \in I^{\parallel}$, so no identification modulo zero helps here.
Similarly, the measure algebra of $I^{\parallel}$ is canonically isomorphic to the measure algebra of the unit interval, and the two maps $t^{\pm} \mapsto t^-$ and $t^{\pm} \mapsto t^+$ both induce this isomorphism, but they are nowhere equal.<|endoftext|>
TITLE: Equivalence of "Weyl Algebra" and "Crystalline" definitions of rings of differential operators between modules?
QUESTION [18 upvotes]: Let $B$ be a commutative $A$-algebra, and let $M$, $N$ be two $B$-modules. We can talk about the set of $A$-linear module homomorphisms $M \to N$, i.e. the set $\text{Hom}_A(M, N)$. Differential operators of order zero should be the $B$-linear maps from $M$ to $N$, i.e. $\text{Hom}_B(M, N)$.
First, note that the commutator $[f, b]$ (where $b \in B$) is a well-defined morphism $M \to N$. Then we make our first definition, the "Weyl Algebra" one.

Definition 1 (Weyl Algebra). Let $\mathcal{D}_A^0(M, N) = \text{Hom}_B(M, N)$. Define $$\mathcal{D}_A^n(M, N) = \{f \in \text{Hom}_A(M, N) \text{ such that }[f, b] \in \mathcal{D}_A^{n-1}(M,N)\}.$$We set $\mathcal{D}_A(M, N) = \bigcup_{n \ge 0}\mathcal{D}_A^n(M, N)$.

In order to formulate the crystalline definition, we introduce some notation. Let $D: M \to N$ be an $A$-linear map. Then, $D$ induces the map $\overline{D}: \delta_{B/A} \otimes_B M\to N$. We now have our "Crystalline" definition.

Definition 2 (Crystalline). Let $I$ be the kernel of the diagonal map (i.e., the map $B \otimes_A B \to B, \  b \otimes b' \mapsto bb'$). Then $D: M \to N$ is said to be a differential operator of order $\le n$ if $\overline{D}$ annihilates $I^{n+1} \otimes_B M$. Let $\mathcal{D}_A^n(M, N)$ be the $B$-module of differential operators of order $\le n$. We define $\mathcal{D}_A(M, N) = \bigcup_{n \ge 0} \mathcal{D}_A^n(M, N)$.

My question is, what is the easiest way to see that/the intuition behind the definitions of rings of differential operators between modules given above are equivalent?

EDIT: In the comments, Michael Bächtold is asking me to spell out the definition of $\delta_{B/A}$ and $\overline{D}$.
So say we have $B$ a commutative $A$-algebra. We want to formalize the notion of an $A$-linear endomorphism of $B$ which is ``close" to being $B$-linear. Let $D: B \to B$ be an $A$-linear endomorphism of $B$. Using $D$, we obtain a map$$\tilde{D}: B \otimes_A B \to B$$defined by $\tilde{D}: b \otimes b' \mapsto bD(b')$, which can also be viewed as a map$$\overline{D}: B \otimes_A B \otimes_B B \to B,$$where we have identified $B$ and $B \otimes_B B$ and the map is defined by $\overline{D}: b \otimes b' \otimes b'' \mapsto bD(b'b'')$. Let us define $\delta_{B/A} = B \otimes_A B$. Then, we have a map:$$\overline{D}: \delta_{B/A} \otimes_B B \to B.$$

In order to formulate the crystalline definition, we introduce some notation. Let $D: M \to N$ be an $A$-linear map. Then, $D$ induces the map $\overline{D}: \delta_{B/A} \otimes_B M\to N$ defined by the same formula as above (that is, $\overline{D}: b \otimes b' \otimes b'' \mapsto bD(b'b'')$ for $b \in B$, $b' \in B$ and $b'' \in M$). We now have our "Crystalline" definition.

In the quoted text, the inducing is in perfect analogy to what I wrote above.

REPLY [3 votes]: Another way to catch the common footing of the two definitions is encoded in the following commutative diagram (I shall assume that both the modules $M$ and $N$ are $B$, just to make the point clearer):
$$
\require{AMScd}
\begin{CD}
B @>{D}>> B\\
@V{\mu}VV @| \\
\delta_{B/A} @>{\widetilde{D}}>> B\, ,
\end{CD}
$$
where $\mu(b):=1\otimes b$ and $D\in\mathrm{End}_A(B)$. Indeed, it is easy to check that $D\in\mathcal{D}_A^n(B)$ if and only if $\widetilde{D}$ vanishes when composed with all the $A$-homomorphisms of the form
$$
[b_1\cdots [b_{n},[b_{n+1},\mu]]\cdots]\, ,\quad b_1,\ldots,b_{n+1}\in B\, ,
$$
and that the images of the above homomorphisms span precisely the $\delta_{B/A}$-module $I^{n+1}$ (the proof is by induction, as in Michael Bächtold's answer).
In other words, if $D\in\mathcal{D}_A^n(B)$,   the above diagram descends to the commutative diagram 
$$
\require{AMScd}
\begin{CD}
B @>{D}>> B\\
@V{\widetilde{\mu}=:j_n}VV @| \\
J^nB:=\frac{\delta_{B/A}}{I^{n+1}} @>{\widetilde{\widetilde{D}}}>> B\, ,
\end{CD}
$$
where, by construction,  $j_n\in\mathcal{D}^n_A(B,J^nB)$. Then
$$
\mathcal{D}_A^n(B)\ni D\longrightarrow \widetilde{\widetilde{D}}\in\mathrm{Hom}_B(J^nB,B)
$$
is a $B$-module isomorphism, capturing the equivalence you were interested about (it all works for projective and finitely generated modules).<|endoftext|>
TITLE: Coarsest admissible topology on $\text{Cont}(X,Y)$
QUESTION [6 upvotes]: Let $X, Y$ be topological spaces and let $\text{Cont}(X,Y)$ be the collection of continuous functions $f:X\to Y$. We say that a topology $\tau$ on $\text{Cont}(X,Y)$ is admissible if the evaluation map $$e: \text{Cont}(X,Y)\times X\to Y; \ (f,x)\mapsto f(x)$$ is continuous. 
Is there an example of spaces $X,Y$ such that the intersection of all admissible topologies on $\text{Cont}(X,Y)$ is no longer admissible?

REPLY [10 votes]: This topic is pretty well-known in studies of function spaces and when they satisfy the appropriate adjointness condition (where the functor $- \times X: \mathbf{Top} \to \mathbf{Top}$ is left adjoint to $Cont(X, -)$; we call such spaces $X$ exponentiable). When one tries to apply the general adjoint functor theorem to construct a right adjoint $Cont(X, -)$, one is led pretty quickly to the condition that there should be a coarsest admissible topology, in order to get the correct topology on $Cont(X, Y)$). 
Stefan Geschke made a very pertinent comment that (for Hausdorff spaces at least) it is local compactness of $X$ which is the decisive factor for this condition; more exactly, when $X$ is Hausdorff, $X$ is exponentiable iff it is locally compact.  
This paper by Escardó and Heckmann gives a general analysis, showing that for more general (possibly non-Hausdorff) $X$, the decisive condition is something called core-compactness (if $X$ is Hausdorff, then core-compactness is equivalent to local compactness). Probably the most elegant way of formulating it is that $X$ is core-compact iff the topology of $X$ is a continuous lattice. 
To be more explicit: For open subsets $U$ and $V$ of a topological space $X$, let us write $V\ll U$ to mean that any open cover of $U$ admits a finite subcover of $V$; this is read as $V$ is relatively compact under $U$ or $V$ is way below $U$.  We say that $X$ is core-compact if for every open neighborhood $U$ of a point $x$, there exists an open neighborhood $V$ of $x$ with $V\ll U$.  In other words, $X$ is core-compact iff for all open subsets $V$, we have $V = \bigcup \{ U | U\ll V \}$. The theorem is that $X$ is exponentiable iff it is core-compact. 
For what it's worth, I did a little write-up at the $n$-Category Café some years back, giving a concrete counterexample much the same as Eric's but with $X$ the space of rationals $\mathbb{Q}$; it can be found here. This example was based on my reading of the paper mentioned above; it turns out that if $X$ is not core-compact, then a counterexample (to the assertion that the intersection of admissible topologies is still admissible) can be always be found by taking $Y$ to be Sierpinski space $\mathbf{2}$ (as in Eric's example).<|endoftext|>
TITLE: Does hypoellipticity imply the existence of a parametrix?
QUESTION [12 upvotes]: Let $M$ be a smooth manifold, like $\mathbb{R}^n$ for instance.  The existence of a parametrix for an operator $P$ on $C^\infty(M)$ in any reasonable pseudodifferential calculus implies that $P$ is hypoelliptic.  Is there a converse to this statement?
The converse would have to take place in some very general setting that encompasses all possible pseudodifferential calculi.  Here's how I'd like to phrase it...
An operator $P$ on $C^\infty(M)$ is "very regular" if its Schwartz kernel has the following two properties:

(1) $p$ is properly supported and semiregular in both variables, ie, $p \in (C^\infty(M) \hat\otimes \mathcal{E}'(M)) \cap (\mathcal{E}'(M) \hat\otimes C^\infty(M))$,
(2) $p$ is equal to a smooth function off the diagonal.

The first condition means $P$ maps each of $C_c^\infty(M)$, $C^\infty(M)$, $\mathcal{E}'(M)$ and $\mathcal{D}'(M)$ to itself.  The second means $P$ is pseudo-local.  
Suppose a very regular operator $P$ is hypoelliptic, in the sense that every preimage of a smooth function is smooth.  Does this mean that there is a very regular $Q$ which is a parametrix in the sense that $PQ-I$ and $QP-I$ are smoothing operators?

REPLY [3 votes]: In the case of hypoelliptic operators with constant coefficients, you have a characterization found by L.Hörmander, which implies that you do have a pseudo-differential parametrix with a symbol in a class $S^{-m}_{\rho,\delta}$.
For hypoelliptic operators with variable coefficients, this is a different story: although the hypoelliptic operator in two dimensions
$$
D_x^2+x^2 D_y^2,
$$
does have a pseudo-differential parametrix, it is not so easy to put it in a classical framework of pseudo-differential operators. Moreover, the operator in three dimensions
$$
D_x^2+x^4(D_y+x D_z)^2,
$$
is hypoelliptic, but is not likely to have a  parametrix with a symbol in any $\rho, \delta$ class and one can even doubt that a parametrix could live in a general class of pseudo-differential operators such as the ones created by R.Beals & C.Fefferman or L.Hörmander.<|endoftext|>
TITLE: About Abhyankar's conjecture
QUESTION [6 upvotes]: I just came to this conjecture (proved by M.Raynaud and D.Harbater in 1994) last weekend, in Fresnel and v.d.Put's book Rigid Geometry and Its Applications. It claims that all quasi $p$-group $G$ could be characterized as certain Galois group of a Galois covering $Y $to $\mathbf{P}_1$, only ramified at infinity (over algebraic closed field with positive character).
Since many grandmasters have researched this conjecture, what is the importance of Abhyankar's conjecture ? 
Well, of course it relates to the inverse Galois theory...

REPLY [6 votes]: This is not an answer to the question, "What is the deeper meaning of Abhyankar's conjecture?"  It is an answer to the question, "What is one application of Abhyankar's conjecture?"  Over an algebraically closed field $k$ of characteristic $p$, every $k$-action of a finite $p$-group on every proper, separably rationally connected $k$-variety (e.g., every rational $k$-variety) has a $k$-point fixed by the entire group.  In particular, every finite $p$-group in every semisimple group of adjoint type is contained in a Borel subgroup (maybe there is a direct proof of this, but I do not know it).  
The proof, following an argument introduced by Kollár and Debarre, combines Raynaud's solution to Abhyankar's conjecture with the theorem of de Jong and myself (the positive characteristic generalization of the theorem of Graber, Harris and myself).
Please confer the following answer of Chambert-Loir as well as my comment, Are rational varieties simply connected?.
Edit.  There are two remarks.  First, once the $p$-group is contained in a Borel subgroup, automatically it is contained in the unipotent radical of that Borel subgroup, since the multiplicative quotient has trivial $p$-torsion group (the $p$-torsion group scheme structure, of course, is nontrivial).  Second, this result completely fails in characteristic prime to $p$.  For every integer $n$ that is divisible by $p$, for every algebraically closed field $k$ of characteristic prime to $p$, there is a copy of $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$ in $\textbf{PGL}_{n,k}$ that is contained in no Borel subgroup.  There is a lift of this group to a $p$-group of order $p^3$ in $\textbf{SL}_{n,k}$.
Second edit. Poonen explained to me a counting argument for the $p$-group above: first reduce to the case of finite fields of characteristic $p$, and then observe that the number of rational points of the flag variety of $G$ is congruent to $1$ modulo $p$.  Thus there must be an orbit of size $1$.  However, the argument above also applies to groups of the form $Q\times \mathbb{Z}/\ell\mathbb{Z}$, where $P$ is a finite $p$-group and $\ell$ is an integer prime to $p$.  The counting argument does not imply this case, but Abhyankar's conjecture does.  Also, for flag varieties, Lang and Steinberg proved the existence of rational points in the 1960's.<|endoftext|>
TITLE: What is descent data (of higher categories), conceptually?
QUESTION [10 upvotes]: First consider a scheme $X$ with an open cover $\mathcal{U}=\{U_i\}$.  An object with descent data on $\mathcal{U}$ is a collection  $(\mathcal{E}_i,\phi_{ij})$ where $\mathcal{E}_i$ is a quasi-coherent sheaf on each $U_i$ and $\phi_{ij}$ is an isomorphism $pr_2^*\mathcal{E}_j\to pr_1^*\mathcal{E}_i$ in $Qcoh(U_{ij})$ which satisfies the cocycle condition
$$
pr_{13}^*\phi_{ik}=pr_{12}^*\phi_{ij}\circ pr_{23}^*\phi_{jk}.
$$
We can make the descent data into a category where the morphisms are those compatible with the $\phi_{ij}$'s. Moreover we have an equivalence of categories
$$
\text{Desc}(X,\mathcal{U})\simeq Qcoh(X).
$$
Of course the definition of descent data has various generalizations. For example instead of the category of quasi-coherent sheaves we can consider an arbitrary fibered category $\mathcal{F}$ over spaces. For any map between spaces $\pi: U\to X$ we have a cosimplicial diagram of categories
$$
\mathcal{F}(X)\to \mathcal{F}(U)\rightrightarrows \mathcal{F}(U\times_X U)\ldots
$$
 An object with descent data on is a collection  $(\mathcal{E},\phi)$ where $\mathcal{E}$ is an object in $\mathcal{F}(U)$ and $\phi$ is an isomorphism $pr_2^*\mathcal{E}\to pr_1^*\mathcal{E}$ which satisfies the cocycle condition
$$
pr_{13}^*\phi=pr_{12}^*\phi\circ pr_{23}^*\phi.
$$
We notice that the category $\text{Desc}(U\to X)$ is equivalent to the fiber product of categories $\mathcal{F}(U)\times_{\mathcal{F}(U\times_X U)} \mathcal{F}(U)$ (as Zhen Lin points out in the comment, we need to also consider $\mathcal{F}(U\times_X U\times_X U)$).
We also notice that if $\pi$ is flat then $\phi$ can be also expressed as the comodule structure $\mathcal{E}\to \pi^*\pi_*\mathcal{E}$ since $\pi^*\pi_*\mathcal{E}\cong (pr_2)_*pr_1^*\mathcal{E}$. Moreover in the flat case  (together with some condition on $F$ I guess) we have
$$
\text{Desc}(U\to X)\simeq \mathcal{F}(X).
$$
Now we consider the case that $\mathcal{F}$ contains some higher structure. In this case we have an augmented cosimplicial diagram of (higher) categories
$$
\mathcal{F}(X)\to \mathcal{F}(U)\rightrightarrows \mathcal{F}(U\times_X U)\ldots
$$
and the definition of descent data should be modified. For example in the recent version of Yekutieli's Twisted Deformation Quantization of Algebraic Varieties 
Section 5, an explicit definition of descent data of cosimplicial cross groupoid is given.
Before given the explicit construction, I would like to know what SHOULD the descent data be. Or more precisely what are the properties that descent data must satisfy?
One attempt is to say that descent data are extra structures (say comodule) on  $\mathcal{F}(U)$ such that we have the (weak) equivalence $\text{Desc}(U\to X)\simeq \mathcal{F}(X)$, but the equivalence only exists in good (say flat) cases while the descent data can be given in general cases. 
Therefore is the  alternative description better? The category of descent data should be the homotopy limit (or totalization) of the cosimplicial diagram
$$
\mathcal{F}(U)\rightrightarrows \mathcal{F}(U\times_X U)\ldots
$$

REPLY [10 votes]: Before given the explicit construction, I would like to know what SHOULD the descent data be. Or more precisely what are the properties that descent data must satisfy?

Zhen Lin's answer is, of course, correct, but it's a bit terse, so here is some elaboration.
Suppose $f : X \to Y$ is a map of sets and that you want to know a real-valued function $r : Y \to \mathbb{R}$. However, you aren't given $r$, but only its pullback $r \circ f : X \to \mathbb{R}$ along $f$. When can you always recover $r$ from this information, and how do you identify when such a function $s : X \to \mathbb{R}$ is a pullback along $f$? The answers, of course, are

Iff $f$ is surjective, and
Iff $s$ respects the equivalence relation on $X$ determined by $f$ (where $x_1 \sim x_2$ iff $f(x_1) = f(x_2)$).

This situation can be cast in more categorical language as follows. In a category with pullbacks, given a morphism $f : X \to Y$ we can take its kernel pair, which is the pullback $X \times_Y X$. In $\text{Set}$, this is
$$X \times_Y X = \{ (x_1, x_2) \in X^2 : f(x_1) = f(x_2) \}$$
and hence it, together with the two projection maps to $X$, exactly reproduces the equivalence relation on $X$ determined by $f$ alluded to above. The kernel pair is so named because it generalizes the kernel to nonabelian situations. In any case, a natural thing to try to do with the two projections $X \times_Y X \rightrightarrows X$ is to take their coequalizer. If $f : X \to Y$ is already this coequalizer, $f$ is said to be an effective epimorphism. Our discussion for $\text{Set}$ implicitly uses the fact that every epimorphism of sets is effective. 
The significance of effective epimorphisms is that the following version of descent holds for them: suppose $F$ is a contravariant functor from the ambient category to some other category which sends coequalizers to equalizers (in particular, any representable presheaf). Then $F(Y)$ is the equalizer of the pair of maps
$$F(X) \rightrightarrows F(X \times_Y X)$$
given by applying $F$ to the kernel pair projections. Our discussion for $\text{Set}$ says precisely this for the special case $F(-) = \text{Hom}(-, \mathbb{R})$.
Descent is the higher version of this story where the kernel pair is replaced by the Cech complex and $F$ takes values in a higher category. The reason it is not just a formal exercise involving functors which send homotopy colimits to homotopy limits is that one needs to identify, in a given higher category, which morphisms are effective in the relevant sense.<|endoftext|>
TITLE: Is an explicit $c$ known to lead to a noncomputable Julia set?
QUESTION [11 upvotes]: Braverman & Yampolsky have shown that there exist noncomputable Julia sets,
i.e., there exist $c \in \mathbb{C}$ such that the Julia set of $f(z) = c + z^2$
is not computable.
"A set is computable, if, roughly
speaking, its image can be generated by a computer with an arbitrary precision."

Braverman, Mark, and Michael Yampolsky. "Non-computable Julia sets." Journal of the American Mathematical Society (2006): 551-578.
  (PDF download.)

My questions are:

Q. Is an explicit such $c$ known? A computable $c$?

It seems likely these questions are answered, perhaps in the cited paper.
If anyone is familiar enough with this line of work to answer, I'd appreciate it.

Answered.
The question is answered in the paper Igor identified, particularly in its
full version:

Braverman, Mark, and Michael Yampolsky.
  "Computability of Julia sets." arXiv link. 2007.

They prove there exist computable $c \in \mathbb{C}$ such that the Julia set
of $c + z^2$ is not algorithmically computable, and provide an algorithm 
for computing such a $c$. Under the assumption of a complex dynamics conjecture
(due to Buff & Chéritat),
they obtain a polynomial-time algorithm for computing such a $c$,
i.e., $n$ bits of $c$ can be computed in time polynomial in $n$.
No explicit $c$ is known, as far as I can tell.
(Their algorithms would not be easy to implement.)

REPLY [7 votes]: Constructing such polynomials is the topic of a follow-up paper by Braverman and Yampolsky. (which, apparently, appeared in STOC '07). They don't give an explicit example, but give an algorithm to construct (arbitrarily close) approximations thereto.<|endoftext|>
TITLE: Moments of a random variable and of its conditional expectation
QUESTION [7 upvotes]: Let $X$ be a bounded random variable with $\mathbb{E}X=0$. Since $X$ is bounded, all its moments exist. Let $\mathcal{G}$ be any $\sigma$-field and let $Y:=\mathbb{E}[X|\mathcal{G}].$ I am interested in proving the following inequality relating even moments of $X$ with even moments of $Y:$ 
For non-negative integers $i<j,$ we have 
$$\mathbb{E}X^{2i}\cdot\mathbb{E}Y^{2j}\leq \mathbb{E}X^{2j}\cdot\mathbb{E}Y^{2i}.$$
Some special cases:
a) If $i=0,$ then this follows from Jensen’s inequality.
b) If $X$ is equiprobable on $\{-1,+1\},$ then this follows from the fact that $|Y|\leq 1$ almost surely.
Intuitively, I expect this to be true because $Y$ is a smoothed version of $X.$
Does the inequality follow easily from some known results? If not, what tools might help me prove it?

REPLY [4 votes]: So I think this is false. I want to consider the case $i=1$, $j=\infty$ first. Let $Y$ take values in $\{\pm 1,\pm\frac 12\}$ with equal probability of $\frac 14$ each. Now let $X$ take values $\pm 1$ with probability $\frac 14$ each and $\pm \frac 34,\pm\frac 14$ with probability $\frac 18$ each. The $\sigma$-algebra is $\mathcal G=\{\{X=1\},\{X=-1\},\{X\in \{\frac 14,\frac 34\},\{X\in -\frac 14,-\frac 34\}\}$ so that $\mathbb E(X|\mathcal G)=Y$.
Now we have $\mathbb EX^{2i}\mathbb EY^{2j}=\frac 12\mathbb EX^2$ while the right side is $\frac 12\mathbb EY^2$ so that the inequality goes in the opposite direction. However $\mathbb EX^{2j}$ and $\mathbb EY^{2j}$ converge to $\frac 12$ as $j\to\infty$, so that the wrong inequality persists for finite $j$.<|endoftext|>
TITLE: Infinite-dimensional admissible representations of SL(2,C)
QUESTION [7 upvotes]: I'm working in my research with the infinite dimensional (admissible) irreducible representations of $\mathrm{SL}(2,\mathbb{C})$ introduced by Harish-Chandra in his paper "Infinite Irreducible Representations of the Lorentz Group". I'm interested in particular in the non-unitary ones.
As the paper was written before Harish-Chandra switched to mathematics, it is not completely rigorous (as he later admitted himself). In particular, although one can show that these representations are admissible $(\mathfrak{g},K)$-modules, it is not clear that they integrate to representations of the group. Does anyone know of a reference on the treatment of these representations in the language of $(\mathfrak{g},K)$-modules (which Harish-Chandra hadn't introduced yet) which addresses my concerns?

REPLY [2 votes]: Though I'm not at all a specialist in this area, I'd certainly urge you to look into some of the relatively modern mathematical textbooks.   For example, David Vogan's 1981 book Representations of Real Reductive Lie Groups (Progress in Mathematics, Birkhauser, Boston) might still be a useful source even though he focuses mostly on the real groups and uses your group over $\mathbb{R}$ as an example in his first chapter.    At any rate, he does adopt the language of $(\mathfrak{g},K)$-modules throughout.    (The book looks much larger than it really is, due to being a photocopy of typescript in the pre-TeX manner.)
There are also useful books by Knapp, in particular his (also large) book Representation Theory of Semisimple Groups: An Overview  Based on Examples, Princeton, 1986.   He frequently uses the rank 1 groups over $\mathbb{R}$ and $\mathbb{C}$ as examples along the way.   
There is also a somewhat idiosyncratic book by Lang based on a course he gave and dealing with  $\mathrm{SL}_2(\mathbb{R})$ (1975, republished by Springer in 1985).    The experts don't seem to find Lang's approach attractive, however.  But one drawback to most of the textbook literature is the tendency to be very general, which might make it hard to extract your special case.<|endoftext|>
TITLE: Why is there a $\sqrt{5}$ in Hurwitz's Theorem?
QUESTION [24 upvotes]: Hurwitz's theorem is an extension of Minkowski's Theorem and deals with rational approximations to irrational numbers. The theorem states:
For every irrational number $\alpha$, there are infinitely many coprime integers $p$ and $q$ such that:
$$\left|\alpha - \frac{p}{q} \right| < \frac{1}{\sqrt{5}q^2} $$
It turns out that the $\sqrt{5}$ term is sharp. My question is why does the $\sqrt{5}$ term appear. What properties of $\sqrt{5}$ enable this to be a sharp bound? I am looking for an intuitive understanding.

REPLY [26 votes]: As Terry mentions in the comments, the reason for the $\sqrt{5}$ is that the limiting case, the golden ratio, forces it. There is a very neat explanation of all of this in the classic number theory book by Hardy and Wright, pages 209 to 212. I give a brief sketch of the ideas.

Why $\phi$ is the worst case.

As Hardy and Wright put it, "from the point of view of rational approximation, the simplest numbers are the worst. The "simplest" of all irrationals, from this point of view, is the number $\phi$."
The reason for this is that if we consider the best approximation for a given $\alpha$,
$$\left|\alpha - \frac{p_n}{q_n} \right| = \frac{1}{q_nq'_{n+1}} < \frac{1}{a_{n+1}q^2_n}$$
it is best when $a_{n+1}$ is large. But in the case of $\phi$, every $a_{n+1}$ is as small as possible.

Why it leads to $\sqrt{5}$

The idea is to simply see what happens when we approximate $\phi$. It roughly goes like this:
$$\left|\phi - \frac{p_n}{q_n} \right| = \frac{1}{q_nq'_{n+1}} \sim \frac{1}{q^2_n}\frac{1}{1+2\phi}=\frac{1}{q^2_n\sqrt{5}}$$

$\sqrt{5}$ is best possible

This follows easily by contradiction. There are no infinitely many $p$, $q$ such that
$$\alpha=\frac{p}{q}+\frac{\delta}{q^2}$$ and $$|\delta|<\frac{1}{\sqrt{5}}$$

Hurwitz's theorem

Now any proof of the theorem should look convincing enough, knowing where the $\sqrt{5}$ it presupposes comes from.
EDIT. I include for completeness a nice alternative proof brought up by Marty and Halbort in the comments.
L. R. Ford, "Fractions" (Amer. Math. Monthly, Vol 45, No 9 (Nov 1938))<|endoftext|>
TITLE: Recursive sequence of binomial random variables
QUESTION [7 upvotes]: Fix $p>0$ and define a recursive sequence of random variables with $X_1 =1$ and
$$X_{k+1} = X_k + \text{Bin}(X_k,p).$$
Thus, $\mathbf E [ X_k ] = (1+p)^k$. 
I would like a left tail bound. Perhaps, for some $a>0$ and $0 < \epsilon < p$,
$$\mathbf P[X_k < (1+ p - \epsilon)^k ] \leq e^{-a k}.$$

REPLY [4 votes]: I claim is that $X_k$ stochastically dominates the following process, $W_k$. To make things a bit simpler, assume $X_1 = 2 = W_1$. (This is harmless, since we can just wait some geometric time until $X_{k}=2$ then start the process.)  Let $$q = \inf_{k \geq 2} \{\mathbf{P}[\text{Bin}(k,p) > pk]\}.$$ Since this converges as $k \to \infty$ (by normal approximation) we know that $q>0$. Now, define $W_1 = 2$ and 
$$W_{k+1} = W_k + (1/2)W_k \text{Ber(q)}.$$
Essentially the $W_k$ increase whenever $X_k$ increases by at least $X_k/2$ and otherwise does not change. One could then write a formal coupling, so that $W_k \preceq X_k$. The $W_k$ are much simpler to analyze. Each successful Ber($q$) trial results in an increase by a factor of $(1+p)$. So we can write
$$W_k = 2*(1+p)^{ \text{Bin}{(k,q)}}.$$
Let $q_k = \mathbf P[ \text{Bin}(k,q) \leq kq/2]$. By Chernoff, $q_k \leq e^{-a k}$ for some $a>0$. It follows that 
$$\mathbf P [ W_k \leq ( (1+p)^{q/2})^k]\leq q_k \leq e^{-a k}. $$
As $W_k \preceq X_k$ we take $\epsilon = (1+p)^{q/2} -1$ and have
$$\mathbf P[X_k \leq (1 + \epsilon)^k ] \leq e^{-ak}.$$<|endoftext|>
TITLE: Does "$\forall Z(C(X,Z) \cong C(Y,Z))$" imply $X\cong Y$?
QUESTION [5 upvotes]: If $X, Y$ are topological spaces, let $C(X,Y)$ denote the collection of continuous maps $f: X\to Y$, endowed with the compact-open topology.
Assume that we are given topological spaces $X,Y$ such that for all spaces $Z$ we have $C(X,Z) \cong C(Y,Z)$. Does this imply that $X\cong Y$?

REPLY [13 votes]: Let $Z$ be the Sierpinski 2-point space.  Then the underlyying set of $C(X,Z)$ is naturally identified with the collection of open sets of $X$, and the specialization order from the compact-open topology is the inclusion order.  Thus we can recover the poset of open sets of $X$ from $C(X,Z)$.  So if you restrict your question to sober spaces, the answer is yes.
However, for arbitrary spaces the question seems more subtle.  For instance, let's consider the case where $X$ and $Y$ are indiscrete.  If $X$ is indiscrete, then for any $Z$, it is easy to see that $C(X,Z)$ looks like $Z$ except each maximal indiscrete subspace $A\subseteq Z$ has been replaced by $A^X$.  In particular, if $X$ and $Y$ are sets such that $|A^X|=|A^Y|$ for all sets $A$, then $C(X,Z)\cong C(Y,Z)$ for all $Z$, where $X$ and $Y$ are equipped with the indiscrete topology.  So the question reduces to the following: if $\kappa$ and $\lambda$ are cardinals such that $\mu^\kappa=\mu^\lambda$ for all cardinals $\mu$, must $\kappa=\lambda$?  The answer is yes: if $\kappa<\lambda$ and $\mu$ is a strong limit cardinal of cofinality $\kappa^+$, then $\mu^\kappa=\mu$ but $\mu^\lambda>\mu$.
So the answer is also yes if $X$ and $Y$ are both indiscrete.  However, it is noteworthy that this argument might require you to take a very large $Z$ to distinguish $X$ and $Y$.  For instance, when $|X|=\aleph_0$ and $|Y|=\aleph_1$, if $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$ and GCH holds above $\aleph_1$, then $C(X,Z)\cong C(Y,Z)$ for all spaces $Z$ of cardinality less than $\aleph_{\omega_1}$.  So if the answer is yes in general, this is a sign that proving it probably isn't going to be very easy.<|endoftext|>
TITLE: Another name for coin-flipping polynomials
QUESTION [7 upvotes]: In his paper Functions arising by coin flipping (section 4), Johan Wästlund coined the term "coin-flipping polynomial" for polynomials that arise in connection with observing a finite number of coin tosses. By "coin tosses" I mean i.i.d Bernoulli random variables with parameter $x$ (probability of head) not necessarily $1/2$.
Specifically, these polynomials represent the probability of an event defined from a finite number of tosses $N$. For example,

$x^2(1-x)$: $N=3$; probability of observing head, head, tail, in that order.
$x^2+(1-x)^2$: $N=2$; probability of the two outcomes being equal.
$3x(1 − x) = 3x(1-x)^2+3x^2(1-x)$: $N=3$; probability that not all three outcomes are the same.

Searching for more information about these polynomials, I haven't found any (other than the fact that they are a superset of Bernstein basis polynomials). This may be because they haven't received much study, but I feel they must have. So the other possibility is that they go by some other name that I don't know about.
So, my questions:

Are these polynomials known by some other name?
Do you know of any result related to these polynomials?

REPLY [2 votes]: These are all the polynomials $f(x) \in \mathbb Z[x]$ satisfying two inequalities:

They had better actually give you probabilities, so $f(x) \in [0,1]$ for $x \in [0,1]$.
They are either totally deterministic or indeterministic, so either $f=0$, or $f=1$, or $f(x) \in (0,1)$ for $x \in (0,1)$.

Ignoring the $f=0$, $f=1$ case, this inequality plus the fact that $f$ is a polynomial implies that $f(x) \geq \epsilon x(1-x)$ for some $\epsilon$ and $1-f(x) \geq \epsilon x(1-x)$ for some $\epsilon$. 
For the proof, note that for each $n$ greater than the degree of $f$, we may write $f$ uniquely as $$ \sum_{i=0}^n c_{i,n} x^i (1-x)^{n-i}$$ by choosing the $c_{i,n}$ in order to get the $i$th coefficient of $x$ right.
It is sufficient to show that for $n$ sufficiently large, $0 \leq c_{i,n} \leq \pmatrix { n \\ i}$.
In fact we will show that:
$$ c_{i,n} = \pmatrix { n \\ i} \left( f\left( \frac{i}{n}\right) + O\left( \frac{1}{n} \frac{i}{n} \left( 1- \frac{i}{n} \right) \right)\right)$$
or setting $x=i/n$, this looks a little more elegant as:
$$ c_{i,n} = \pmatrix { n \\ i} \left( f( x) + O\left( \frac{1}{n} x\left( 1- x \right) \right)\right)$$
Then choosing $n$ large enough that $O(1/n)< \epsilon$, we get the right bounds on $c_{i,n}$ and win.
Let $d$ be the degree of $f$. We can go from $c_{i,d}$ to $c_{i,n}$ by ignoring the last $n-d$ coin flips, which gives the formula
$$ c_{i,n} = \sum_{j=0}^d \pmatrix{ n-d \\ i-j}c_{j,d}$$
Now on the other hand we have:
$$ f(i/n) = \sum_{j=0}^d \left( \frac{i}{n} \right)^j \left( \frac{n-i}{n} \right)^{d-j} c_{j,d}$$
So it is sufficient to show that:
$$\pmatrix{ n-d \\ i-j} = \pmatrix { n \\ i} \left(\left( \frac{i}{n} \right)^j \left( \frac{n-i}{n} \right)^{d-j} + O\left( \frac{1}{n} \frac{i}{n} \left( 1- \frac{i}{n} \right) \right)\right)$$
or equivalently:
$$ \frac{ \pmatrix{ n-d \\ i-j} }{\pmatrix { n \\ i}} = \left( \frac{i}{n} \right)^j \left( \frac{n-i}{n} \right)^{d-j} + O\left( \frac{1}{n} \frac{i}{n} \left( 1- \frac{i}{n} \right) \right)$$
The left side is the probability of drawing $j$ white balls and then $d-j$ black balls from an urn with $i$ white balls and $n-i$ black balls, sampling without replacement. For the outcome in the two cases to be different, you have to draw the same ball twice, which happens $O(1/n)$ of the time, and then have it be a different color from the ball you would have gotten otherwise, which happens $O\left(\frac{i}{n} \left( 1- \frac{i}{n} \right) \right)$ of the time, which justifies the error term.<|endoftext|>
TITLE: Your favorite papers on geometric group theory
QUESTION [16 upvotes]: I would like to improve my "depth of understanding" in geometric group theory.  So I am interested in short and accessible papers on subjects related to this field but which are not always available in the classical references.
To make my question more precise: By short, I mean a research paper of at most twenty pages or a book containing at most one hundred pages. By accessible, I mean a(n almost) self-contained paper for postgraduate students.
Here are some examples which I think are suitable:

Topology of finite graphs, Stallings (15 pages). One of my favorite papers. Stallings shows how to apply covering spaces to finite graphs in order to prove several non-trivial properties of free groups.

Topological methods in group theory, Scott and Wall (about 60 pages). The authors proves several classical results of geometric group theory (Stallings' ends theorem, Grushko's and Kurosh's theorems, Bass-Serre theory) using the formalism of graphs of spaces.

Subgroups of surface groups are almost geometric, Scott (12 pages). Peter Scott proves that surface groups are LERF by using hyperbolic geometry.


I hope my question is precise enough to be of interest.
EDIT: I don't think my question is a duplicate of Introductory text on geometric group theory?. Rather, I see it as complementary: I ask about some interesting subjects which are typically not available in these classical references.

REPLY [3 votes]: The paper B. Krön. Cutting up graphs revisited - a short proof of Stallings' structure theorem gives a short and accessible proof of Stalling's End Theorem in full detail.<|endoftext|>
TITLE: Concrete examples of covering from the 3-torus to the 3-sphere
QUESTION [11 upvotes]: There is a two-fold branched covering from 2-torus to the 2-sphere, $T^2 \rightarrow S^2$, whose covering transformation group is  generated by the map $x \mapsto -x$ (Note that $T^2$ is an abelian group).
I heard that there is a three-fold branched covering from the 3-torus to the 3-sphere. Then what would be the covering transformation group of this case? 
Probably it is trivial for topologists but could anyone can help me out?

REPLY [8 votes]: There is an algorithm, due to Montesinos, for converting a surgery diagram of a 3-manifold $M$ into a description of $M$ as a 3-fold (irregular; as remarked above, there is no regular branched cyclic covering $T^3 \to S^3$) cover of $S^3$. It is described nicely in Rolfsen, Knots and Links, Chapter 10.G. You need to start with a surgery description where all of the framings are $\pm 1$. 
For the 3-torus, such a description is easily found. $T^3$ is surgery on the Borromean rings, with framings 0 on each component. Add a $+1$ framed meridianal circle to each component of the Borromean rings, changing the framing on that component to $1$. The picture below shows what I mean. If you follow the description in Rolfsen's book, you will have the branched covering. I haven't tried to draw it, though.<|endoftext|>
TITLE: Quantum fields and infinite tensor products
QUESTION [22 upvotes]: As I understand it, a naive interpretation of the state space of a quantum field theory is an infinite tensor product
$$\otimes_{x\in M} H_x,$$
where $x$ runs over the points of space. This corresponds to the fact that a field $\phi$ and the conjugate momentum $\pi$ can be viewed as a composite system of the array of $\phi(x)$ and $ \pi(x)$. Thus, again naively, the amplitude assigned by a quantum state $\Psi(\phi, \pi)$ to a classical initial condition $(\phi,  \pi)$  is a tensor product of the amplitudes $\Psi(\phi(x), \pi(x))\in H_x$.
Of course, this doesn't quite make sense, for many reasons including the fact that the infinite tensor product is quite badly behaved. Instead, the standard way of quantising, say a scalar field satisfying the Klein-Gordon equation, is to write it in terms of Fourier modes
$$\phi(x,t)=(2\pi)^{-3/2}\int [a(p)e^{i(px-\epsilon_pt)}+a^*(p)e^{-i(px-\epsilon_pt)}]\frac{d^3p}{2\epsilon_p}$$
with $\epsilon_p=\sqrt{p^2+m^2}$ (this being the KG equation). The canonical commutation relation for $\phi$ and $\pi$ become
$$[a(p), a^*(p')]=2\epsilon_p\delta(p-p'); \ [a(p), a(p')]=[a^*(p), a^*(p')]=0,$$
which can individually be quantised in a Segal-Bargmann fashion to act on a Hilbert space $H_p$. To quantise all these operators as we run over all momenta, we would again require an infinite tensor product
$$\otimes_p H_p.$$
This is avoided by imposing an additional condition, the existence of a vector $\Psi_0$ (interpreted as the vacuum), satisfying $$a(p)\Psi_0=0$$ for all $p$. After this, it all works out and we have a nicely quantised free field by putting the operators into the integral above. I think I sort of understand this procedure, with the level of uncertainty that I'm normally stuck with when thinking about physics.
However, I've come upon the following passage in the book by Streater and Wightman, page 86-87.

When, in fact, do non-separable Hilbert spaces appear in quantum mechanics? There are two cases which deserve mention. The first arises when one takes an infinite tensor product of Hilbert spaces... Infinite tensor products of Hilbert spaces are always non-separable. Since a Bose field can be thought of as a system composed of an infinitely of oscillators, one might think that such an infinite tensor product is the natural state space. However, it is characteristic of field theory that some of its observables involve all the oscillators at once, and it turns out that such observables can be naturally defined only on vectors belonging to a tiny separable subset of the infinite tensor product. It is the subspace spanned by such a subset that is the natural state space rather than the whole infinite tensor product itself. Thus, while it may be a matter of convenience to regard the state space as part of the infinite tensor product, it is not necessary.

My question is, how does one relate this passage to the usual quantisation procedure described above. In particular, what is the 'tiny separable subset' alluded to by Streater and Wightman?
Because the infinite tensor product picture is so intuitively compelling (this is emphasised, I think, by all authors on QFT), it would be nice to spell out the relation between it and standard quantisation with at least some level of mathematical clarity.

REPLY [3 votes]: There has been many answers, from many points of view on QFT: canonical quantization, algebraic QFT, etc. Let me add another perspective using the Euclidean path integral quantization, as emphasized in textbooks like "Quantum Field Theory and Critical Phenomena" by Jean Zinn-Justin.
For simplicity, I will take $M=\mathbb{R}^d$
In quantum mechanics, of say one particle in one dimension, the position is say $\phi\in\mathbb{R}$ and the momentum is say $\pi$. The physical Hilbert space is $L^2(\mathbb{R},d\phi)$ made of square integrable wave functions $\Psi(\phi)$, with respect to Lebsgue measure $d\phi$. When going to to QFT, the single number $\phi$, now becomes a classical field configuration $\phi=(\phi(x))_{x\in\mathbb{R}^d}$ and the physical Hilbert space should again be the $L^2$ space for a space of fields $\phi$ with respect to some measure $d\nu(\phi)$. Intuitively, one can think of this as a tensor product indexed by $x\in\mathbb{R}^d$ of $L^2$'s of $\mathbb{R}$, but trying to use this idea rigorously is probably counterproductive.
Suppose your theory is given by an honest Borel probability measure $\mu$ on $\mathscr{S}'(\mathbb{R}^{d+1})$, i.e., the Euclidean correlations or Schwinger functions
$$
S_n(x_1,t_1;\ldots;x_n,t_n)=\frac{1}{Z}\int 
\phi(x_1,t_1)\cdots\phi(x_n,t_n)\ e^{-S(\phi)}D\phi
$$
seen as distributions, namely, elements of $\mathscr{S}'(\mathbb{R}^{n(d+1)})$, are moments of a probability measure $\mu$. Suppose one can define a Borel-measurable restriction map $R:\mathscr{S}'(\mathbb{R}^{d+1})\rightarrow\mathscr{S}'(\mathbb{R}^{d})$ which sends the distribution $\phi(x,t)$ to $\phi(x,0)$. Then one can define a push-forward measure $\nu=R_{\ast}\mu$
and the physical Hilbert space should essentially be
$$
L^2(\mathscr{S}'(\mathbb{R}^d),d\nu)\ .
$$<|endoftext|>
TITLE: Which nice/deep elaborations on the (operators <-> sheaves) / (endomorphisms <-> objects) theme are there?
QUESTION [13 upvotes]: A linear operator $T:V\to V$ on a (say) vector space over a field $k$ is just a $k[T]$-module, and may be viewed as the sheaf $\mathscr F_T$ over $\mathbb A^1_k$, with fibre over $\lambda\in k$ equal to $V/\langle T(v)-\lambda v\mid v\in V\rangle$; going backwards one may reconstruct $V$ as global sections of $\mathscr F_T$, with $T$ acting via $T(s)(\lambda)=\lambda s(\lambda)$.
This is of course a commonplace observation nowadays, but for some reason I still stay fascinated by this fact, and I learned about it many years ago.
I remember being particularly struck, having shortly before that studied some category theory, by the fact that this construction establishes a connection between endomorphisms in one category and objects in another category, which seemed somewhat unforeseen by category theory.
Later I learned about categories of bordisms, where also an endobordism of an $n$-manifold produces an $n+1$-manifold, but I still have no idea whether this is related in any way to the fact I started with.
[LATER - Re: comment by Will Sawin - this could be in principle formulated in terms of (category-theoretic) traces; however I only know traces of endomorphisms with values in an object of the same category, whereas here we seemingly should have traces with values themselves being objects of another category] 
I've still got a feeling that there is something so deep hidden in this that its deepest roots still wait to be dug out.
In algebraic geometry, this is a simple particular case of the correspondence between modules over a ring and sheaves over the corresponding affine scheme;
The whole spectral theory may be viewed as built on it, things like direct integrals of linear spaces, etc. included;
Various dualities involving vector bundles and projective modules may be also viewed as generalizations of this...
Let me also mention certain "noncommutative" version of the spectrum of a $C^*$-algebra invented by Segal, which I briefly described in an answer to Why the Dold-Thom theorem?
But I will stop here and ask my questions.
Does this circle of ideas/facts have a common name?
Which constructions/theories/theorems may be viewed as stemming from it?
Is there a way to describe its place in mathematics?
Is there really a connection between it and the endobordism thing I mentioned?
Is there a category-theoretic or other abstract/axiomatic treatment of similar phenomena?

REPLY [8 votes]: What you describe in the opening of your question is most naturally an equivalence of categories, not some dimension-shifting thing between endomorphisms and objects as you suggest.  Indeed, let $\mathcal C$ be a category.  The category of endomorphisms in $\mathcal C$ is the functor category $\mathcal C^{\circlearrowleft}$, where $\circlearrowleft = \mathrm B \mathbb N$ denotes the category with one object and endomorphism monoid $\mathbb N$, i.e. the category freely generated by an object with an endomorphism.  In some corners of category theory, $\circlearrowleft$ would be called "the walking endomorphism".  In particular, it is a category, being a category of functors.
If $\mathcal C$ is a nice enough monoidal category ($\mathrm{Vect}$ is nice enough), then $\mathcal C^{\circlearrowleft}$ can also be described as the category of $\mathcal C$-linear functors from the free $\mathcal C$-linearization of $\circlearrowleft$.  This is the $\mathcal C$-enriched category with one object and endomorphism algebra $\mathbf 1_{\mathcal C}[x] \in \mathcal C$, the "polynomial algebra" in one variable with coefficients in the unit object $\mathbf 1_{\mathcal C} \in \mathcal C$.
But to move to your more geometric questions requires more than just this elementary category theory.  For example, it happens that $\mathbf 1_{\mathcal C}[x]$ is a commutative algebra, and so its spectrum is an affine variety.  This would not hold if $\circlearrowleft$ were replaced by some other more complicated walking object, say if you were asking about pairs of endomorphisms and not just endomorphisms.  (On the other hand, $\circlearrowleft \times \circlearrowleft$ is the walking pair of commuting endomorphisms.)
I don't think there's anything deeper to your observation about endobordisms except to observe that $\mathcal C^{\circlearrowleft}$, being a functor category, is naturally a category.  
Actually, there's something else to say about bordisms.  The category of bordisms is naturally a double category, i.e. a category object internal to categories.  Bordisms and their compositions form the "horizontal" morphisms but the "vertical" morphisms are smooth functions.  (Important variations: use just embeddings or just local diffeomorphisms.)  In any double category, if you fix an object, you get a category whose objects are the horizontal endomorphisms of that object.  In the case of $\emptyset \in \mathrm{Bord}$, you get the category of $n$-manifolds and smooth maps (or embeddings or local diffeomorphisms).<|endoftext|>
TITLE: Is there a topological Chevalley-Shephard-Todd Theorem?
QUESTION [6 upvotes]: Is the following true:
For a representation of a finite group $G$ on $\mathbb{C}^n$, the quotient $\mathbb{C}^n/G$ is a topological manifold if and only if $G$ is generated by pseudo-reflections.
( A pseudo-reflection is an element of $G$ whose fixed set is of codimension $\leq 1$)

REPLY [3 votes]: To elaborate on Jason's great answer.
The answer is no and the counter-example is the binary icosahedral group together with the sum of an irreducible representation of degree $2$ and a unit representation, over $\mathbb{C}$.
Let $\Gamma$ be the binary icosahedral group--the unique perfect and fixed-point-free finite group, and consider one of it irreducible representations of degree  $2$ over complex numbers (it is fixed-point free), for example the one resulting from embedding of $\Gamma$ in $SL_2(\mathbb{C})$. Then the quotient $\mathbb{C}^2/\Gamma$ is called the Poincare Homology Sphere.  Now by the affirmative answer to Milnor's double suspension problem, the double suspension of a homology sphere is a true sphere.  Now the image of the origin in $(\mathbb{C}^2\times \mathbb{C})/\Gamma$, where the second factor comes with a trivial $\Gamma$-action, looks locally like the vertex of the cone over double suspension of the Poincare Homology Sphere, hence the quotient space is locally Euclidean at the image of the origin.  But then, given the scaling action, it is locally Euclidean everywhere.
EDIT  That said, I claim the following sufficient condition:
Suppose no sub-quotient of a finite group $\Gamma$ is isomorphic to the binary icosahedral group.  Then $\Gamma$ satisfies the topological Chevalley-Shephard-Todd Theorem:  
Given a linear representation of $\Gamma$, finite degree over $\mathbb{C}$, the quotient is a topological manifold iff $\Gamma$ is generated by pseudo-reflections.<|endoftext|>
TITLE: Is there a tropical geometric proof for counting genus g curves in any n dimensional projective space?
QUESTION [11 upvotes]: Consider the following question: Let $X$ be a compact complex manifold 
and $\beta \in H_2(X, \mathbb{Z})$ a fixed homology class. Let 
$\mu_1, \mu_2, \ldots, \mu_k$ denote certain generic submanifolds (of the 
right dimensions) in $X$. 
What is $N_{g,\beta}(\mu_1, \ldots, \mu_k)$, the number of genus $g$ curves 
in $X$ representing the class $\beta$ and intersecting the submanifolds $\mu_i$? 
If $X:= \mathbb{P}^2$, then there is a complete formula for 
$N_{g,d}(p_1, \ldots, p_{3d-1+g})$, the number of genus $g$ degree $d$ curves 
through $3d-1+g$ generic points, using Tropical Geometry (and also using the 
Caporaso Harris formula). In fact I think if $X$ is any toric two dimensional 
variety there is a formula for $N_{g, \beta}$ using Tropical Geometry. 
$\textbf{Question:}$ Is there a $\textit{Tropical geometric}$ solution 
for computing $N_{g,d}$ for $\mathbb{P}^n$ where $n>2$? Kontsevich's derivation 
of $N_{0,d}$,
the number of degree $d$ rational curves in $\mathbb{P}^2$ generalizes for any
$\mathbb{P}^n$. More generally, is there a tropical geometric solution 
(or actually any solution) for $N_{g, \beta}$ on complex manifolds $X$ other 
than $\mathbb{P}^n$ (my question here primarily being for $g>0$, although I 
am interested in knowing what is known for $g=0$ as well.) 
$\textbf{Remark:}$ One of the reasons for asking this question is as follows 
(the reason/hope might be very naive): I am just wondering if using any of 
these enumerative results one can make some partial (but direct) verification 
of some predictions made by Mirror Symmetry (particularly higher genus mirror 
symmetry predictions). To take one example, it is predicted in this paper (page 34) 
http://arxiv.org/pdf/hep-th/0612125v1.pdf 
that the number of degree $4$, genus $2$ curves on the quintic three fold 
is $534750$. Is it conceivable that there could be a direct way to see that? 
$\textbf{Added Later:}$ My last comment (about degree $4$ genus $2$ curves on the quintic threefold) is incorrect as has been pointed out. I was very naively concluding that the numbers on page $34$ (at least the initial ones) were enumerative, which they are not (the authors don't claim that either). My hope that using enumerative formulas for genus g curves one can explain some of these 
BPS numbers is perhaps very naive. I am nevertheless interested to know if there 
are enumerative formulas for genus g curves into manifolds with dimension greater than two (irrespective of whether they have any use in explaining the numbers obtained from Mirror Symmetry).

REPLY [11 votes]: Here is an attempt at an overview of tropical curve counts by someone who has been involved in the story for a while but certainly hasn't followed everything that has happened. I look forward to being corrected by others. 
Toric surfaces are done: That's Mikhalkin's work and has since been redone in from many perspectives.
Genus zero curves in any toric variety are done: That's Siebert and Nishinou.

If you want to get beyond these cases, even for toric varieties, there are two major problems. I'll talk about them, and then move to the non-toric case. 
First, I want to talk about expected dimension. Fix a toric variety $X$ of dimension $n$ and a class $\beta \in H_2(X)$. We can think of $\beta$ as the list (possibly with multiplicities) of directions for the unbounded rays of the tropical curve. Call the length of this list $x$; in terms of $\beta$ and $X$, this is $\beta \cup c_1(X)$. The ``expected dimension" of the moduli space of curves is $D:=x-(n-3)g+3(n-1)$. 
Now, consider a corresponding tropical curve $\Gamma$ which could come from a limit of genus $g$ degree $\beta$ curves. So $\Gamma$ has $x$ unbounded rays, and $b_1(\Gamma) \leq g$. Let $e$ be the number of internal edges of $\Gamma$. Then $e \leq x+3(b_1(\Gamma)-1)$, with equality if and only if $\Gamma$ is trivalent, and the space of deformations of $\Gamma$ is of dimension $\geq e-n(b_1(\Gamma)-3)$. (See section 5.1 in my thesis.) In the nice cases which are done, we always wind up studying cases where $g=b_1(\Gamma)$, $\Gamma$ is trivalent and the space of deformations of $\Gamma$ has dimension $e-n(b_1(\Gamma)-3)$. When all these things hold, $\Gamma$ has deformations of dimension $e-n(b_1(\Gamma)-3)=x+3(b_1(\Gamma)-1)-n(b_1(\Gamma)-3) = x+3(g-1)-n(g-3)=D$.
In general, all of these things can go wrong:
Superabundant curves The space of classical curves doesn't have to be $D$ dimensional. When it isn't, we say that the curves are super-abundant, and we have to refer to virtual fundamental classes. But, even if the space of classical curves is $D$-dimensional, the space of tropical curves need not be. For example (see my thesis again; I learned this example from a talk of Mikhalkin) there are genus $1$ tropical cubic curves in $\mathbb{P}^3$ which vary in a $13$ dimensional family, whereas classical cubics only have $D=12$ dimensions of freedom.
Most of these extra tropical curves can not be limits of classical curves, so one must find additional conditions to exclude them before on can hope to do enumerative geometry, or even prove finiteness results. I did some of this for genus $1$ and Katz and Bogart-Katz did some more, but I think we are very far from a complete answer.
We cannot reduce to maximally degenerate curves A tropical curve encodes the dual graph of a nodal curve. I'll say that the tropical curve is maximally degenerate if the nodal curve has no moduli, so $\Gamma$ is trivalent and $b_1(\Gamma) = g$. 
In the cases $g=0$ or $n=2$, we can force this to be true by generic conditions to impose on the curve. Roughly speaking, in order for the tropical curve to obey $D$ many conditions, it must be able to vary in a $D$-dimensional family. Taking our curve not to be trivalent reduces $e$, and hence reduces the dimensionality with which our curve can vary. Taking $b_1(\Gamma)<g$, when $n=2$, increases $(n-3) b_1(\Gamma)$ and thus likewise reduces the variation.
However, when $g \geq 1$ and $n=3$, we can take $\Gamma$ trivalent but with $b_1(\Gamma) < g$, and still have $x+3(b_1(\Gamma)-1)-n(b_1(\Gamma)-3)$ come out the same. So there are $D$-dimensional families of tropical curves, corresponding to limits where there are moduli hidden in the stable curve. Life is even worse when $n \geq 4$: Then the $(n-3) b_1(\Gamma)$ term REWARDS taking $b_1(\Gamma) < g$.
These issues are why I got a bit depressed about tropical curve counting.

Moving beyond the toric case: This project is being driven almost entirely by Gross and Siebert. (Maybe I should be listing Konsevich and Soibelman as well? I'm not sure. And Hacking and Keel are also involved, but my ignorant impression is that they are not thinking about the actual question of lifting curves back out of the tropical realm.) I don't feel competent to summarize how far they have gotten. They are THE people to talk to about trying to do things beyond the toric case. (And beyond abelian varieties, which other people have also worked on.) But I don't see how leaving the toric case could possibly make the above issues easier.<|endoftext|>
TITLE: Maximal entropy distribution with given conditionals
QUESTION [7 upvotes]: It is well known that of all the joint distributions $p(x,y)$ with fixed marginals $p(x),p(y)$, the one with the highest entropy is:
$$
p(x,y)=p(x)p(y).
$$
Suppose instead that we have conditionals. Namely:

Which probability distribution $p(x,y,z)$, with fixed $p(x|y)$ and $p(x|z)$, has maximal entropy?

is there even an explicit formula?
(Note: one is tempted to take $p(x,y,z)$ such that $p(x|y,z)=p(x|y)p(x|z)$, but I think this is misleading, right?)
Feel free to edit question and tags appropriately.
Thanks!
(Crossposted from Cross Validated, where nobody could answer.)

REPLY [2 votes]: You can find a full formula in http://arxiv.org/pdf/1512.00752v3.pdf<|endoftext|>
TITLE: the true reason of the incompleteness of formal systems
QUESTION [21 upvotes]: A 3/4 year ago, I read Gödel's beautiful paper "Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme 1". There is one thing, I never understood.
In a footnote, Gödel says the following:
"Der wahre Grund für die Unvollständigkeit, welche allen formalen Systemen der Mathematik anhaftet, liegt, wie im zweiten Tell dieser Abhandlung gezeigt werden wird, darin, dass die Bildung immer höherer Typen sich ins Transfinite fortsetzen lässt, während in jedem formalen System höchstens abzählbar viele vorhanden sind. Man kann nämlich zeigen, dass die hier aufgestellten unentscheidbaren Sätze  durch Adjunktion passender höherer Typen (z. B. des Typus $\omega$ zum System $P$) immer entscheidbar werden. Analoges gilt auch für das Axiomensystem der Mengenlehre"
Meltzer's translation renders this in English as:
"The true source of the incompleteness attaching to all formal systems of mathematics, is to be found — as will be shown in Part II of this essay — in the fact that the formation of ever higher types can be continued into the transfinite (cf. D. Hilbert 'Über das Unendliche', Math. Ann. 95, p. 184), whereas in every formal system at most denumerably many types occur. It can be shown, that is, that the undecidable propositions here presented always become decidable by the adjunction of suitable higher types (e.g. of type $\omega$ for the system $P$). A similar result also holds for the axiom system of set theory."
Famously, Gödel never published part 2 of his paper.
Is the theorem which states that the undecidable propositions presented by Gödel become decidable by the adjunction of suitable higher types proved by someone? Has someone formulated Gödel's idea more precisely? Is there any research in this area?

REPLY [20 votes]: I think the main idea here is that, if you have a reasonably strong formal system $T$ (so that the incompleteness theorems apply to it) and you then strengthen it to a "higher type" system $T^+$ in which you can talk about (and quantify over) subsets of the universe that $T$ describes (so that second-order concepts from the point of view of $T$ are first-order from the point of view of $T^+$) and if $T^+$ has suitable comprehension axioms, then $T^+$ will be able to formalize a definition of truth for the language of $T$ and prove that the axioms of $T$ are true and that formal deduction preserves truth.  Thus, $T^+$ can prove the consistency of $T$ and, as a consequence, prove the usual Goedel sentence of $T$.
Typical examples are: (1) $T$ is Peano arithmetic and $T^+$ is second-order arithmetic.  (2) $T$ is Zermelo-Fraenkel set theory and $T^+$ is Morse-Kelley class theory. (3) (the example Goedel mentioned) $T$ is type theory with an $\omega$-indexed hierarchy of types and $T^+$ is type-theory with an $(\omega+1)$-indexed hierarchy of types.<|endoftext|>
TITLE: Why can't a nonabelian group be 75% abelian?
QUESTION [66 upvotes]: This question asks for intuition, not a proof.
An earlier question,
Measures of non-abelian-ness
was thoroughly answered by Arturo Magidin.
A paper by Gustafson1
proves that, for a nonabelian group,
the probability that two randomly selected
elements commute is at most $5/8$, a tight bound2
that also holds for a class of infinite topological groups.
One might say that a nonabelian group cannot be more than 62.5% abelian.
My question is:

Q. Is there some intuitive reason that a nonabelian
  group cannot be "nearly completely abelian" in the sense that
  the probability that two element commute approaches $1$?
  Why, intuitively, is there an upper bound less than $1$?

Gustafson's proof uses the conjugacy class equation,
and bounds on the terms of this equation.
I am seeking an underlying logic that can be conveyed without these calculations.
(Incidentally, there is no universal lowerbound on the probability, so
groups can be nearly completely nonabelian.)

1Gustafson, W. H.
"What is the probability that two group elements commute?"
American Mathematical Monthly (1973): 1031-1034.
(Jstor link.)
2"The reader may verify this bound is sharp,
by examining the nonabelian
groups of order $8$":
The dihedral group $D_8$,
and the quaternion group $Q_8$.

REPLY [65 votes]: (In response to suggestion of Johannes Hahn): As indicated in my comment, and developed further in j.p.'s comment, a fairly intuitive "explanation" is provided by the fact that Lagrange's Theorem tells us that an element $x$ of a finite group $G$ is already central in $G$ once it commutes with more than half the elements of $G$ (since $C_{G}(x)$ is a subgroup of $G$). In other words, once the probability that $x$ commutes with elements of $G$ gets above $\frac{1}{2}$, it has to be $1$.
Also, as I mentioned, there are many papers on this topic in the literature. At the other extreme to the question, in one paper, Bob Guralnick and I proved that the probability that two elements of  a finite group $G$ commute tends to $0$ as $[G:F(G)] \to \infty$, where $F(G)$ is the largest nilpotent normal subgroup
of $G$.
Later note: These arguments work not just with finite groups, but also whenever a group  $G$ admits a measure $\mu$  which is invariant under right (or left if preferred)  translation, making $
(G,\mu)$ a probability space. For then it is still true that if the probability that $x$ commutes with a group element gets above $\frac{1}{2}$, we must have $C_{G}(x) = G$, for otherwise $\mu(C_{G}(x)) \leq \frac{1}{2}$, as the $[G:C_{G}(x)]$ right cosets of $C_{G}(x)$ all have equal measure. Similarly, we have $[G:Z(G)] \ge 4$ if $G$ is non-Abelian, since otherwise $\mu(\langle x \rangle Z(G) ) > \frac{1}{2}$ for $x \in G \backslash Z(G)$.<|endoftext|>
TITLE: Homology spheres and fundamental group
QUESTION [16 upvotes]: I have a curiosity about homology spheres: I was wondering if they were uniquely characterized by their fundamental group. I.e. given two $n-$dimensional (integral) homology spheres with isomorphic fundamental groups, are they homeomorphic? If not, how many homeomorphism classes corresponds to a given fundamental group?

REPLY [20 votes]: See edit at bottom for further information answering the question in all dimensions.
In all odd dimensions $2k -1 > 3$, there are non-homeomorphic homology spheres with fundamental group G = the binary icosahedral group (fundamental group of the Poincaré homology sphere). This follows from basic surgery theory; the Wall group $L_{2k}(G)$ contains a subgroup of the form $\mathbb{Z}^n$ for some $n$ related to the number of irreducible complex representations of $G$. This subgroup is detected by the so-called multisignature, as described in Wall's book on surgery theory. 
Choose one homology sphere $M^{2k-1}$ with $\pi_1(M) = G$. For instance, you can repeatedly spin the Poincaré sphere, where spinning $P$ means doing surgery on the obvious $S^1$ in $S^1\times P$. For any $M$ with $\pi_1(M) = G$, there's an invariant originally described by Atiyah and Singer. A priori, it's a smooth invariant, but is known to be a homeomorphism invariant. Roughly speaking, one knows that for some $d \in \mathbb{N}$, the manifold $d\cdot M$ is the boundary of a $2k$-manifold $X$ with $\pi_1(X) = G$.  Then $X$ has a collection of equivariant signatures (associated to the action of $G$ on the universal cover of $X$), known collectively as the multisignature.  The multisignature (divided by $d$) is a topological invariant. (Technically, this is only well-defined up to a choice of isomorphism $\pi_1(X) \to G$ but this is readily dealt with.)
Now, the group $L_{2k}(G)$ acts on the structure set of $M$, as described in Wall's book. The effect of the action is to change the multisignature, and hence it changes the homeomorphism type of $M$. In this construction, you not only preserve the fundamental group, but also the (simple) homotopy type of $M$.
It's possible that acting on an even-dimensional homology sphere $M^{2k}$ with fundamental group $G$ by elements of $L_{2k+1}(G)$ could change the homeomorphism type. But odd dimensional $L$-groups are much harder, and you'd need some serious expertise to see what the effect should be. I rather suspect that you could do something simpler, and change the homology with local coefficients or something like that.
Addendum: There are two papers of Alex Suciu that answer this question in dimensions at least 4.  In "Homology 4-spheres with distinct k-invariants," Topology and its Applications 25 (1987) 103-110, he gives examples of homology 4-spheres with the same $\pi_1$ and $\pi_2$ that are not homotopy equivalent. In "Iterated spinning and homology spheres", Trans AMS 321 (1990) he constructs homology n-spheres in dimensions $n \geq 5$ with the same property. Combined with the remarks above about dimension 3, this answers your question.<|endoftext|>
TITLE: Is there a subset of the plane that meets every line in two open intervals?
QUESTION [30 upvotes]: Using the Axiom of Choice, it is possible to construct a subset of the plane that meets every line in two points (these are called "$2$-point sets"). What if, instead of points, we ask for two open intervals?
Some related ideas/constructions were explored in this paper (for example, it is shown that there is a subset of the plane meeting every line in a copy of the Cantor set).
If you want one open interval instead of two, then $\mathbb{R}^2$ itself suffices (and this is the only solution). If you want three open intervals, then just take the complement of a "$2$-point set" (and this idea can be modified to give you $n$ open intervals for every $n > 2$). Infinitely many open intervals is even easier (take the union of the interiors of the hexagons in this tiling, for example). The case of two intervals seems more stubborn.
UPDATE: This question has now been answered: there is no such subset of the plane.
Terry Tao posted some "partial progress" on the problem, which later turned out to be the first half of a complete proof. The second half can be found in my answer below. Neither post stands alone as a complete answer, but both taken together do the job.
I can't accept both answers, so I've accepted Terry's, since it came first and ended up being the first half of a correct proof. This seems to me to be in line with what the help center says about what it means to accept an answer.

REPLY [13 votes]: The answer to the question is no.
Rather than typing a new answer, I've just edited my old one (which contained some partial progress). I think this is OK since the main idea of the old answer (or at least its main "trick" -- fixing a special line, and then rotating it slightly about a point not in $E$) is still present in this one. The proof I'm about to give combines this trick with the results from Terry Tao's answer.

Theorem: There is no subset of the plane meeting every line in two open intervals.

Proof: Suppose $E$ is such a set.
Following the terminology in Terry Tao's answer, we will say that $\omega \in S^1$ is a limit direction of $E$ if $E$ does not avoid an infinite open sector containing the direction $\omega$ in the limit. Terry Tao proves:
Lemma: Not every direction is a limit direction.
(Actually he proves more, but this is all we will need.)
In other words, there is an infinite open sector $S$ that misses $E$. Translating and rotating $E$ if necessary, we may assume that $S$ contains the origin $O$ and the positive $X$-axis. Consider the expression (using polar coordinates)
$$(*) \ \ \ \ \ \ \ \ \ \ \ \{(r,\theta) : 0 < \theta < \phi, r \geq 0\} \cap E = \emptyset.$$
Clearly $\pi$ does not satisfy $(*)$ (otherwise $E$ does not meet the line $y = 1$). Let
$$\phi_0 = \inf \{\phi \in [0,\pi] : \phi \text{ does not satisfy }(*)\}.$$
(Since $S$ is open, we will have $0 < \phi_0 < \pi$, although we won't actually need this fact.) Let $L$ be the line through the origin in the direction $\phi_0$.
$E$ is open (this observation is also due to Terry Tao; see his first comment on Igor Rivin's answer). It follows that the ray $\{(r,\phi_0) : r \geq 0\}$ contains no points of $E$ (if it did, then a slightly smaller value of $\phi_0$ would also fail to satisfy $(*)$). Therefore $E$ meets $L$ in two open intervals that are both on the same side of the origin $O$.
Let $U$ and $V$ be these two open intervals, labeled in such a way that $V$ is between $U$ and $O$. Fix a point $x \in L \setminus E$ with $x$ strictly between $U$ and $V$. For $\rho \in [0,\pi)$, let $R_\rho^x$ denote the rotation of the plane about the point $x$ by the angle $\rho$. Fix $u \in U$ and $v \in V$.
We have $u,v \in E$ and $O \in S$, with $E$ and $S$ both open. By the continuity (in $\rho$) of $R^x_\rho$, there is some $\rho_0$ small enough that if $0 < \rho < \rho_0$ then $R_\rho^x(u), R_\rho^x(v) \in E$ and $R_\rho^x(O) \in S$. We may assume $\rho_0 < \frac{\pi}{2}$.
By the maximality of $\phi_0$, there is some $\varepsilon < \rho_0$ such that
$$\{(r,\phi_0+\varepsilon) : r \geq 0\} \cap E \neq \emptyset.$$
In other words, we may choose a point $e \in E$ such that, if we let $p = (r,\phi_0)$ for some $r > 0$, the angle $pOe$ is less than $\rho_0$. Since angle $pOe$ is acute (recall $\rho_0 < \frac{\pi}{2}$), and since $x$ is on the other side of $O$ from $p$, the angle $pxe$ is even smaller than the angle $pOe$. 
Let $\rho$ be the measure of the angle $pxe$. The line $L' = R^x_\rho(L)$ does not meet $E$ in two open intervals. To see this, observe that $L'$ passes through $E$ near $u$, through $x$ (not in $E$), through $E$ again near $v$, through $S$ near $O$ (not in $E$), and then through $e$ (in $E$). QED<|endoftext|>
TITLE: Who was/were the first to note that if $\sum_{x \in X} \frac{1}{x} < \infty$ then the natural density of $X$ is zero?
QUESTION [10 upvotes]: It is a result of folklore that the natural density of a set $X$ of positive integers such that $\sum_{x \in X} \frac{1}{x} < \infty$ is zero. This is reproved, e.g., in T. Šalát's paper: Convergence of subseries of the harmonic series and asymptotic densities of sets of integers (1987), which in turn redirects to Šalát's On subseries for a different proof. However, I am convinced that the result is (much?) older than that, and was told by G. Grekos that Šalát himself in his talks presented it as well-known. So my question is:

Do you have any clue about the first (explicit) occurrence of the result in the literature?

REPLY [11 votes]: I believe this is due to Kronecker. Namely, if you look at theorem 3 here, which is due to Kronecker, and says that if
$$\sum_{n=1}^\infty a_n$$ is convergent, and $(p_n)_{n\geq 1}$ is an increasing and unbounded sequence, then 
$$\lim_{n\rightarrow \infty}\frac{p_1 a_1 + p_2 a_2 + \dotsc + p_n a_n}{p_n} = 0.$$
Now, let your set be $X =\{x_1, \dotsc, x_k, \dotsc\},$ in order. Set $a_n = 1/ x_n,$ while $p_n = x_n,$ your assertion follows. 
I should note that in (one of) his papers, Salat attributes the result independently to Leo Moser (Monthly, 1958, DOI: 10.2307/2308884), and Krzyś (Prace Matem 1956) - I could not find the latter paper. Neither can I find the original Kronecker paper.<|endoftext|>
TITLE: Exotic $C^k$ manifolds
QUESTION [7 upvotes]: How much is known about exotic $C^k$-manifolds? For example,

Is it known whether there are $C^k$-differentiable manifolds that are homeomorphic, but not $C^k$ diffeomorphic?
More generally, is it known whether there are $C^k$-differentiable manifolds that are $C^l$ diffeomorphic, but not $C^k$ diffeomorphic for $l < k$?
Do any of the famous examples for $k=\infty$ (such as $\mathbb R^4$ or Milnor-Kervaire's exotic spheres) give examples for $k < \infty$?

REPLY [19 votes]: There are no such exotic manifolds.  Whitney proved in
Whitney, Hassler Differentiable manifolds. Ann. of Math. (2) 37 (1936), no. 3, 645–680.
that every $C^1$-differential structure can be uniquely smoothed to a $C^{\infty}$ differential structure.  This was improved by Morrey and Grauert to show that in fact it can be uniquely smoothed to a real analytic structure; see
H. Grauert, On Levi's problem and the imbedding of real-analytic
manifolds, Ann. of Math. 68 (1958), 460-472.
C. B. Morrey, The analytic embedding of abstract real-analytic manifolds,
Ann. of Math. 68 (1958), 159-201.<|endoftext|>
TITLE: Question on paper of Stewart and Top about ranks of elliptic curves over Q(t)
QUESTION [7 upvotes]: I'm reading "On Ranks of Twists of Elliptic Curves and Power-Free Values of Binary Forms" by Stewart and Top, and struggling to understand the argument on pg 962 which shows that the rank of a particular elliptic curve $E_{D(t)}/\mathbb{Q}(t)$ is exactly 2.  
Here are the relevant details: 
Start with the elliptic curve $$E/\mathbb{Q}: y^2 = x^3 + 1$$ and the polynomial $$D(t) =  2t(t - 1)(t + 1)(2t + 1)(t + 2) \in \mathbb{Z}[t].$$  Let $C/\mathbb{Q}$ be the curve given by $s^3 = D(t)$ and let $$E_D/\mathbb{Q}(t): y^2 = x^3 + D(t)^2.$$  For each point $P = (x(t), y(t))$ in $E_D(\mathbb{Q}(t))$, we define an element $\phi_P$ of $\text{Mor}_\mathbb{Q}(C, E)$ by $$\phi_P(t, s) = (x(t)/s^2, y(t)/s^3).$$ Then we have a map $$\lambda: E_D(\mathbb{Q}(t)) \to H^0(C, \Omega^1_{C/\mathbb{Q}})$$ given by $$\lambda(P) = \phi_P^\ast \omega_E$$ which is shown to be a homomorphism with finite kernel. 
We want to use this homomorphism to show that the rank of $E_D/\mathbb{Q}(t)$ is exactly two.  
First, we can find two points, and show they are independent by looking at their images under $\lambda$.  This is fine, and shows that the rank is at least 2. 
Next, we want to show that the rank is at most 2.  We know that the image lands in $H^0(C, \Omega^1_{C/\mathbb{Q}}(\zeta_3))$, the eigenspace on which the automorphism of $C$ given by $\zeta(t, s) = (t, \zeta_3 s)$ acts on differentials as multiplication by $\zeta_3$.  This constrains the image to the 3-dimensional space, say spanned by $\omega_1, \omega_2, \omega_3$ (this numbering is different from the paper).  
All of this makes sense to me.  What doesn't make sense is how the authors constrain the image to a 2-dimensional subspace.  
They define three involutions on $C$, called $\sigma_1, \sigma_2, \sigma_3$, and show that the space of $\sigma_i^\ast$-invariant holomorphic differentials is generated by $\omega_i$.  Hence the quotient of $C$ by $\sigma_i$ is an elliptic curve.  In two cases, the curve is isogenous to $E$ over $\mathbb{Q}$, and in the third case it is not.  From this, they somehow infer that the rank is 2.  I'm very confused about this, and would appreciate some more details or a reference.  
Thanks!

REPLY [2 votes]: Suppose that $\omega$ is the invariant differential of an elliptic quotient $C/\sigma$ of $C$. In particular, $\omega$ is non-zero. If $\omega$ lies in the image of $\lambda$, then there exists a point $P = (x(t), y(t))$ on $E_D(\mathbb{Q}(t))$ such that $\lambda(P) = \omega$. Let $\rho_1(P)$ denote the morphism in $\operatorname{Mor}_{\mathbb{Q}}(C/\sigma, E)$ corresponding to $\lambda_1(P)$. Then, as a morphism of curves, it must be either finite or constant. Since $\omega$ is non-constant, it must be the former; so $\rho_1(P)$ is an isogeny between $C/\sigma$ and $E$. Hence, this shows that any elliptic quotient corresponding to an element in the image of $\lambda$ must be isogenous to $E$.
Since, as you noted in the question, that the quotients $C/\sigma_1, C/\sigma_2$ are isogenous but $C/\sigma_3$ is not isogenous to the first two, $E$ can be isogenous to at most two of them. Since we have the lower bound of two, the rank is exactly two.<|endoftext|>
TITLE: Can one define "Ramanujan Summation" over algebraic number fields?
QUESTION [9 upvotes]: With some trepidation, I ask to "evaluate" badly divergent sums.  Generalizing $\sum n = -\tfrac{1}{12}$ what would be the value of this sum over $\mathbb{Z}[i]$?
$$\sum_{m,n \geq 0} (m+in) \hspace{0.25in}\text{and}\hspace{0.25in}\sum_{0 < m < n } (m+in)$$
This cursory and highly dangerous check shows that we would probably need higher order terms in the Euler McLarurin expansion.
$$  \sum_{m,n \geq 0} (m+in)  = \sum_{m,n \geq 0} m + i \sum_{m,n \geq 0}n 
=  \sum_{n \geq 0} -\tfrac{1}{12} + i \sum_{m \geq 0} -\tfrac{1}{12} = ??? $$
In a sense, all Ramanujan sums do is subtract the sum from an approximating integeral.  I tried:
$$ \sum_{m,n \geq 0}^{M,N} (m+in)  - \int_0^M \int_0^N (x + i y)\,dx \, dy
 = \tfrac{MN(M+N+2)}{2} - \tfrac{MN(M+N)}{2} = \tfrac{MN}{2}$$
This doesn't look right at all, but I did find something called the Khovanskii-Puhlikov theorem which looks like it might give the answer.

In general for any number field $K$ and lattice cone $V \subseteq \mathcal{O}_K$ in the ring of integers we could try to define:
$$ \sum_{x \in V \subset \mathcal{O}_K} x $$
So I wonder if someone has attempted to define such summations and what they evaluate to.

REPLY [6 votes]: I don't think this is exactly what you are looking for, but it seems too related not to mention. You can generalize zeta regularization to any number field $K$, although you run into an open problem.
Over $\mathbb{Q}$ we have the classic (Riemann) zeta regularization $\sum n=\zeta(-1)$, the right hand side of which can be defined by analytic continuation and equals the (in)famous $-1/12$.
Now, for an arbitrary number field $K$, take its Dedekind zeta function, which is also a meromorphic function with a simple pole on 1, and then define the sum over some (see Gerry's comment below) integral ideals of $K$:
$$\sum_{\mathfrak{a}' \in \mathcal{O}_K} \mathfrak{N(a)}=\zeta_K(-1)$$
The exact value of this sum is not know in general. The better understood case is for $K$ totally real, in which it is known to be an rational number (Klingen-Siegel) and conjectured to be related to algebraic K-groups (Birch-Tate).<|endoftext|>
TITLE: Why do we need filtered categories to index ind-objects?
QUESTION [6 upvotes]: I edited the question in view of several helpful replies (thanks). 
When we define ind-objects in a category, we use  in general filtered diagrams in a category, not just sequences $A_1 \rightarrow A_2 \rightarrow A_3 \dots$ indexed by the integers.   More generally, if one wants a "bigger" limit, we could look at diagrams indexed by an ordinal. 
Is there a simple example of an ind-object that can't be indexed by an ordinal?

REPLY [6 votes]: As far as I can tell, the idea to use filtered categories instead of just ordinal-indexed diagrams is due to Grothendieck. In fact, let me use this opportunity to advertise my favorite text on abstract category theory: Exposé I of SGA 4. It rocks.
Anyway: Proposition 8.1.6 (loc. cit.), which Grothendieck attributes to Deligne, says that every filtered category receives a cofinal map from an ordered set. 
Grothendieck remarks that, while this result says that the two points of view on filtered objects (general or ordered) are essentially equivalent, filtered objects arise more naturally. 
A nice example is Grothendieck's Theorem 8.3.3 on ind-representability. Aside from some set-theoretic issues, this basically boils down to the statement that exactness of a presheaf of sets $F$ on a category $C$ is equivalent to the category $C_{/F}$ being filtered.<|endoftext|>
TITLE: References for Forcing with Side Conditions
QUESTION [5 upvotes]: I'm looking for some good references about Forcing with Side Conditions, including expository papers that explain the main ideas with some details in order to give me a fairly clear insight of those set theoretic problems that are solvable using this approach. Any other reference is also welcome.

REPLY [5 votes]: S. Todorcevic, Notes on Forcing Axioms, Chapter 7.
Itay Neeman, Forcing with side conditions. Oberwolfach, 2011. http://www.math.ucla.edu/~ineeman/
B. Velickovic, G. Venturi, Proper forcing remastered.  http://www.math.cmu.edu/~eschimme/Appalachian/Pfr.pdf<|endoftext|>
TITLE: Homological vs. cohomological dimension of a group/space
QUESTION [8 upvotes]: I have several related questions regarding homological vs. cohomological dimension of a space/group (this is not a duplicate of this).
The standard definition of the cohomological dimension $cd(X)$ of a space $X$ is 
definition 1: $cd(X)=$ The minimum over $n\in \mathbb{N}$, such that $H^i(X,M)=0$ for all $i>n$ and all local coefficient systems $M$. 
One can similarly define the homological dimension $hd(X)$ (I am not sure it is a standard terminology)
definition 2: $hd(X)=$ The minimum over $n\in \mathbb{N}$, such that $H_i(X,M)=0$ for all $i>n$ and all local coefficient systems $M$. 
My main question is

Question1: Do we always have $cd(X)=hd(X)?$ What if one or both assumed to be finite? what if $X$ is a finite CW complex?

For $X = BG$, i.e a classifying space of a group $G$, the definitions reduce to the standard ones in group cohomology. I believe that the $cd(BG)$ coincides with the minimal length of a projective resolution of the trivial $G$-module $\mathbb{Z}$ over the group ring $\mathbb{Z}[G]$ (Namely, the projective dimension of $\mathbb{Z}$). 

Question 2: Is it true that the $hd(BG)$ coincides with the flat dimension of $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module?

Finally, if $X$ is a connected CW complex with $cd(X)=0$, then $X$ is contractible. One can similarly ask

Question 3: If $X$ is a CW complex with $hd(X)=0$, is it true that $X$ is contractible?

(Edit: The following argument has a gap is wrong. It is not clear true that if $hd(X)=0$ then $X$ is a filtered colimit of finite subcomplexes $X'$ with $hd(X')=0$. An exmaple is given here)
Regardless of the answer to question 1, I think the answer to this is yes by the following argument. Since the universal cover is acyclic and simply connected, it is contractible, so this reduces to a question about group cohomology. If $G$ is finitely generated (if $X$ is a finite CW for example), then $\mathbb{Z}$ is a finitely presented $\mathbb{Z}[G]$-module so if it is flat then it is projective and hence the cohomological dimension is zero as well. Can one reduce the general case to this one since every CW complex is a filtered colimit of its finite subcomplexes. Is there a more direct argument?

REPLY [4 votes]: As noted in the question, "Question 3" reduces to asking whether, if $H_i(G,M)=0$ for all $i>0$ and all $\mathbb{Z}G$-modules $M$, then $G$ must be trivial.
But this is true, since if $G$ is non-trivial, we can choose a non-trivial abelian subgroup $A$, and then, by Shapiro's Lemma,
$$H_1\left(G,\mathbb{Z}[G/A]\right)\cong H_1\left(A,\mathbb{Z}\right)\cong A\neq0.$$<|endoftext|>
TITLE: Roots of characteristic function of "reciprocal gamma measure"
QUESTION [5 upvotes]: Let us call a measure $\mu$ on the Borel $\sigma$-algebra $\mathfrak{B}_{(0,\infty)}$ of subsets of $(0,\infty)$ a reciprocal gamma measure if it is absolutely continuous with respect to the Lebesgue measure $m_L$ on $((0,\infty),\mathfrak{B}_{(0,\infty)})$ with the Radon-Nikodym derivative given by $$\frac{d\mu}{dm_L}(x) = \frac{1}{\Gamma(x)}\quad\text{for } x\in (0,\infty).$$
Let $\varphi$ be the characteristic transformation of the reciprocal gamma measure, that is, the complex valued function on $\mathbb{R}$ defined by
$$\varphi(u)=\int_{0}^{\infty}\!e^{ixu}\, \mu(dx)\quad\text{for } u \in \mathbb{R}.$$
Question: Is the collection of all zeros  $\{u\in\mathbb{R}\colon \varphi(u)=0\}$ an empty set? 
Remarks I: From numerical inspections, I am tempted to believe that the answer is in the affirmative. But I do not think there are any standard methods of deciding whether or not there are any zeros. 
If we set $c=\varphi(0)$ then $c^{-1}\frac{d\mu}{dm_L}$ is a density of a probability measure $\lambda$ on $\mathfrak{B}_{(0,\infty)}$ having moments of every order. It is well known that if $\lambda$ is infinitely divisible, then $\varphi(u)$ is non-zero for every real number $u\in\mathbb{R}$. I have not been able to verify this infinite divisibility for $\lambda$. I have scrolled through the pages of Steutel & van Harn's book on infinitely divisible probability distributions, but I have not found anything on a probability distribution where the density is given by the reciprocal gamma function normalized. I therefore do not think it is known whether or not the probability distribution $\lambda$ is infinitely divisible. The density is not likely to belong to the class Bondesson so it might be very difficult to determine whether or not it is infinitely divisible. 
Now the moments of the probability distribution $\lambda$ satisfy Carleman's condition so that $\lambda$ is "determined", that is, there are no other probability distributions on $((0,\infty),\mathfrak{B}_{(0,\infty)})$ having the same sequence of moments. As far as I can tell, the characteristic transformation can be extended to an entire holomorphic function and this too gives that the distribution is determined.
Remarks II: It is immediate, from the continuity of $\varphi$, that the set $\Lambda=\{u\in \mathbb{R}\colon \varphi(u)\neq 0\}$ is an open set in the Euclidean metric topology of $\mathbb{R}$. Moreover, since 0 is an element of $\Lambda$, it is a nonempty set. To prove that the collection of zeros of $\varphi$ is an empty set, according to the connectedness of $\mathbb{R}$, it suffices to show that $\Lambda$ is also a closed subset of $\mathbb{R}$. I have not been able to prove that $\Lambda$ is closed, but perhaps my patience was too short.

REPLY [3 votes]: This is not an answer. It's just a comment about the above integral representation.
If $H$ is a Hankel contour to the left ( a path that comes from $\infty$ with argument $-\pi$, turns around 0 - always at a distance from 0 bigger than 1) and then comes back to $\infty$ with argument $\pi,$ we have
$${1\over{\Gamma (z)}}={1\over {2\pi i}}\int_H {{ e^t}\over {t^z}}dt, \ \ z\in C,$$
where $t^z$ uses a branch of the logarithm $L,$ with $-\pi <Im L <\pi.$
When we compute now its Fourier transform, we can interchange integrals and write
$$\varphi (u)=\int_0^{\infty} {1\over{\Gamma (z)}}e^{iz u}dz={1\over {2\pi i}}\int_H \int_0^\infty e^te^{izu}e^{-z L t}dzdt={1\over{2\pi i}}\int_H {{ e^t}\over{ Lt-iu} }dt.$$
For $u\in (-\pi,\pi),$ we can deform the path to the negative real line plus a circle surrounding the singularity $e^{iu},$ and so  we get
$$\varphi(u)= e^{iu}e^{e^{iu}}+\int_0^\infty{{e^{-t}}\over{(\log t-iu)^2+\pi^2}}dt, \ \ \ \ u\in (-\pi,\pi).$$
Diferently, for $ u\in R-[-\pi,\pi],$ we write:
$$\varphi(u)= \int_0^\infty{{e^{-t}}\over{(\log t-iu)^2+\pi^2}}dt, \ \ \ \ u\not\in [-\pi,\pi].$$<|endoftext|>
TITLE: Best known zero-free region for Dirichlet $L$-functions in the $q$-aspect
QUESTION [5 upvotes]: It is classical that there is a $c > 0$ such that for all Dirichlet characters $\chi$ except for at most one exception, one has that $L(s,\chi)$ has no zeroes for $\sigma > 1 - \frac{c}{\log{q} + \log{(t+2)}}$, where $q$ is the conductor of $\chi$ and $s = \sigma + it$. What is the state of the art?
For instance, is it even known that one has $1 - \sigma > \omega(\log{(q(t+2))}^{-1})$? (That is, that one can change the "there exists $c > 0$" to a "for all $c > 0$". Of course if there is a sufficiently 'bad' Siegel zero then Deuring-Heilbronn says yes, but hopefully something more is known.)
I know there is a Vinogradov-Korobov-type estimate available as $t$ grows and $q$ is fixed, but I couldn't find anything better than the classical bound as $q$ grows. I only really care about the case of real zeroes and quadratic characters, but of course this is the hardest case.
Sorry for the spam of questions about Siegel zeroes, I am very much a nonexpert.

REPLY [7 votes]: No, such a result is not known unconditionally, even if one restricts to the low-lying case $t=O(1)$ and excludes real zeroes, and it would be a great breakthrough if one could do this.  If one had such an improved zero-free region, one could for instance resolve Vinogradov's conjecture on the least quadratic nonresidue (e.g. by applying the results of Granville and Soundararajan, see Theorem 1.6 of http://arxiv.org/abs/1501.01804, in the contrapositive; one can also use the older results of Rodosskii, http://www.ams.org/mathscinet-getitem?mr=82521), which is currently open.  Indeed the problem of enlarging the zero-free region in the q-aspect is very closely tied to the problem of estimating short character sums, in much the same way that the Vinogradov-Korobov type enlargements to the zero-free region in the t-aspect are tied to the problem of estimating short exponential sums.
On the other hand, we do have the Deuring-Heilbronn repulsion effect, which shows that the presence of an exceptional zero does allow one to enlarge the classical zero-free region in more or less the manner requested; see e.g. Chapter 18 of Iwaniec-Kowalski.  Of course this improvement is "illusory" in the sense that we don't actually believe that exceptional zeroes exist, but this result is still important for obtaining unconditional proofs of results such as Linnik's theorem on the least prime in an arithmetic progression.
ADDED LATER: it may also be possible to get better zero-free regions for those characters for which improved character sum estimates are known, such as when the modulus is very smooth or the character has a small odd order.  I don't know if this has been explicitly addressed in the literature though.<|endoftext|>
TITLE: Infinitely many irreducible polynomials of the form f(X^2) + X mod 3?
QUESTION [12 upvotes]: Are there infinitely many polynomials $f \in \mathbb{F}_3[X]$ for which $f(X^2) + X$ is irreducible?

REPLY [6 votes]: (Not an answer, just too long for a comment.)
Let $A_{2n, a,q} := \{ x^{2n} + a_1 x^{2n-2} + a_2 x^{2n-4} +\cdots + a_{n-1} x^{2} + ax + a_n \mid a_i \in \mathbb{F}_{q} \} \cap \{f \in \mathbb{F}_{q}[x], \text{ irreducible}\}$.
You are considering the infinitude of $A:=\cup_{n\ge 1} \cup_{a \neq 0} A_{2n,a,3}$ (actually, you force $a=1$ but don't require monic; I reversed the roles of these 2 properties).
There's extensive literature on counting irreducibles with coefficients coming from affine varieties, where your example is actually one of the simplest (you just prescribe certain coefficients). The first 3 pages of this paper offer a nice survey.
In short, algebraic geometry counting methods give asymptotic information on sizes of sets like $A_{2n,a,q}$, with a very accurate main term but a rather big error term, which allows one to extract information only for big $q$, i.e. bounded from below by a function of $n$. This means that for any $q$, you can bound the size of $\cup_{n\ge 1} \cup_{a \neq 0} A_{2n,a,q}$ from below, and that size will grow to infinity as $q$ does. 
(Hence, one reason that your problem is especially hard is that your $q$ is the smallest possible apart from $2$, and $2$ was at least studied w.r.t applications.)
It seems that the only known case where the error term is much better than described is when you count irreducibles in an arithmetic progression, as you can use the machinery of L-functions (theorem 1.2 in the paper). For instance, the results concerning your polynomials, and results concerning counting the sparser set of irreducible trinomials $x^{2n}+ax+b$ are basically the same, although the 2nd problem seems much harder (theorem 1.3 in the paper). But prescribing instead the $m$ consecutive lower coefficients instead of odd-indexed-coefficients gives asymptotic information as $q^{n-2m} \to \infty$, not just $q \to \infty$.
(Here we see another reason for your problem being very difficult: you prescribe half of the coefficients to be 0. Even the better error term of the arithmetic progression case gives information only when prescribing slightly less than half of the coefficients, otherwise you get the "worse" error-term of theorem 1.3.)
There are 3 standard routes you can take now: 

Use the above methods for your specific set, but perform a much more delicate calculation (by using, for example, the explicit formula for $\mathbb{F}_{q}$-points on a variety instead of Lang-Weil estimates). 
Try to find an ad-hoc construction of a family of the form you seek. You won't get them all, but there's a chance of finding infinitely many. A good starting point is page 60 in this handbook.
Relax some of the conditions, and try again :)

Good luck!<|endoftext|>
TITLE: Which subgroup order of the symmetric group is the most frequent?
QUESTION [5 upvotes]: Question: What is the most frequent order of subgroups of $S_n$? 
More precisely: Let $a_k$ be the number of subgroups of $S_n$ with order $k$. What is the maximum of $a_k$?
This question came up during a discussion of the open problem of determining the size of the largest antichain in the subgroup lattice of $S_n$.

REPLY [7 votes]: This is already sort of in the comments, but just to say it clearly:
In the paper Enumerating Finite Groups of Given Order
Author(s): L. Pyber
Source: Annals of Mathematics, Second Series, Vol. 137, No. 1 (Jan., 1993), pp. 203-220
Pyber proves that the number of subgroups of $S_n$ is bounded below by $2^{(1/16 + o(1))n^2}.$ He shows this by noting that $S_n$ contains $C_2^{[\frac{n}2]},$ and the latter has the specified number of subgroups. He then conjectures that the lower bound is, in fact, sharp. In other words, modulo this conjecture, at least morally, the maximally present order of a subgroup is, indeed, a power of two, and the actual power of two  is presumably $2^{[[\frac{n}2]/2]}.$ It seems that Pyber's conjecture is still open.<|endoftext|>
TITLE: Are there enough additive permutations?
QUESTION [20 upvotes]: I am hoping to learn enough about additive permutations to help with a number theory problem.  These permutations also have connections to difference sets, orthomorphisms, transversals, and other structures.  As a meta question, I would like to know more references or applications for additive permutations, especially applications involving combinatorial number theory.  First, the basic setup.
Let $l$ be a nonegative integer, and let $k$ stand for both the set $\{t : t$ is an integer and $ \mid t \mid \leq l \}$ and the cardinality of the set $k=2l+1$.  I will represent the set of permutations $S_k$
 on the set $k$ by vectors indexed by $k$ in increasing order.  So for $l=2$, the
identity permutation $e$ is written as $\langle -2, -1 ,0 , 1, 2 \rangle$.  Let me take
$\pi \in S_k$ and write it as a vector, and I will write vector addition as $++$. I can then write $e  ++ \pi$ as a vector in $\mathbb{R}^k$ with $i$th coordinate being $i + \pi(i)$.  This vector $e ++ \pi$ is a vector with integer coordinates, and may not look anything like a vector representation of an element of $S_k$.  However, sometimes it does, and when this happens, we call $\pi$ an additive permutation.
As an example when $l=1$, one has two additive permutations $\langle 1,-1, 0 \rangle$ and
$\langle 0 , 1, -1 \rangle$, each of which is the negative reverse of the other ($\pi(i) = -\rho(-i)$ for all $i \in k$).  Except when $l=0$, the identity permutation $e$ is not an additive permutation.  Also, the fact that $\langle 1,0, -1 \rangle$ is not an additive permutation shows that this definition depends on representation: it cannot be defined as a characteristic subset of the permutation group on $k$ members.
The OEIS entry A002047 contains some references to the literature, which I am slowly absorbing.  However, I don't see the answer to either of the following questions:

1) Given $l$, how many members of $S_k$ are additive permutations?  I have not found
an asymptotic formula, although the paper by Cavenagh and Wanless suggests an
exponential lower bound.  I have a weak upper bound which for most $l$ is slightly better than $l^{2l}$.
2) Using just the group operation of the symmetric group $S_k$ (so no inverse, but $k$ is finite so inverse is not needed), do the additive permutations generate $S_k$?  It seems to be so for $l=0,1,2$.  (It is less interesting to me but still valid to ask for $l$ large if $S_k$ is generated using $++$.)

In addition to the OEIS references, I am perusing work of D.G. Rogers, and am open to other suggestions for references.  I am also looking at a related paper, but the operation $++$ there is over a finite ring, and I don't think I can use those results yet.
Gerhard "First Question On This Account" Paseman, 2015.07.10

REPLY [2 votes]: I want to share a partial answer to question 1), and raise a few more questions.  I found what I think is a neat and likely unoriginal bijection; I'm hoping the combinatorialists can provide a reference and perhaps use it to help with my questions above.
I decided to try a $k$ by $k$ chessboard visualization of the enumeration problem, and succeeded, sort of.  I had to cut off triangular pieces of the chessboard and ask for a maximal nonattacking and covering queen placement, except that the queens were restricted in movement relative to actual queens in chess.  So I shifted that set up to a hexagonal board, and invented a (name for a likely unoriginal) piece called a "jack", which is like what a rook would be for a board of hexagonal cells (from a hexagonal cell, move in one of the 6 directions perpendicular to a side of the cell, along 3 lines which I call "diagonals"), but decided it needed to be partly royal.
Let's imagine a hexagonal array with $l+1$ cells on a side, with vertex cells at the even numbers on a clock, and for orientation label the six vertex cells A,B,C,D,E, and F in clockwise fashion.  (So A is at 12, B is at 2, and D is at 6 o'clock.)  Now, starting with the vertex cell E and going clockwise up to A, label (just outside the board) those cells with $-l, -l+1, \ldots$ all the way up to $l$, so that the $k$ "diagonals" from upper left to lower right are labeled with the $\pi(i)$ index. Cell F will have a $0$ label for $\pi(0)$ and cell A will have a $l$ label for $\pi(l)$.
In a similar fashion, we label the other diagonals from $-l$ at Cell A clockwise through $0$ for Cell B and ending with $l$ at cell C.  We have labeled these with $j$- values, which will represent values being placed in position $\pi(i)$.
If we look at a cell, it belongs to exactly two of the labeled diagonals, $\pi(i)$ and $j$
say.  If we were to decorate all the cells with the value $i+j$ when it is on diagonals
$\pi(i)$ and $j$, we would see constant values running up and down.  In particular, the
line of cells from A down to D would get the value $0$, $-l$ for the cells between (and including) cell E and F, and $l$ for cells B to C.
Once we have the observation of the values being constant on the "third" (vertical) diagonal, I can now assert the bijection.  Let G be a placement of $k$ jacks on this board
that do not attack each other and consequently cover all cells of the board.  Since I have
labeled the diagonals, I will call such a G a labeled placement.
Assertion:  The number of labeled placements on this $k$ board is the same as the number of additive permutations of $S_k$.
For an additive permutation $\rho \in S_k$ place a jack at cell on diagonals $\pi(\rho(j))$ and $j$ for each $j \in k$.  After I have verified the details, I will call the Assertion a Proposition.  However, I don't see what could go wrong, yet.
I am willing to shorten this description if someone will provide a graphic version.  Now for the payoff:  Referring to the third diagonal, place a jack on the 0 diagonal somewhere in one of the $k$ cells; this will leave usually $k-3$ and at most $k-2$ uncovered cells on the 1 diagonal to place another jack, so place one there.  We get an upper bound of $k!!$ placements of those $l+1$ mutually nonattacking jacks, and $l!$ trivially for the remaining $l$ jacks, giving an upper bound of $k!/2^l$ for the number of additive permutations in $S_k$.
I am still working on the lower bound, but have a feeling that $l!$ or even $(l+1)!$ might be achievable with the Assertion and this picture.  Now for the additional questions:

3) has anyone seen this bijection before, and will they please give me a reference?
4)  has labeled placement of nonattacking jacks on a hexagonal board appeared before, hopefully with enumeration?
5) assuming I did not screw up and the Assertion is soon to be a Proposition, can anyone
see a good lower bound (better than exponential, and hopefully factorial) for the number
of additive permutations with this or any other picture?
6) leaving additive permutations aside, it is tempting to view the board as a representation of a three-dimensional cubical array.  Is there a combinatorial advantage to such a perspective for this problem?  That is, picking $k$ cubes out of an array of $l+1$ cubes on a side, does this count or represent a nice entity in finite geometry or some other field?

I suspect that a factorial lower bound would imply improvements on the current literature.
I also would not mind improvements on the upper bound.
Gerhard "Likes Looking At Suggestive Pictures" Paseman, 2015.07.12<|endoftext|>
TITLE: Reconstruction Conjecture: are almost all digraphs reconstructible?
QUESTION [13 upvotes]: The Reconstruction Conjecture for simple graphs remains unresolved.  Most attempts I've seen at resolving the conjecture aim at proving it to be true (or partially true).  I don't believe there is a compelling reason for it to be true, so I'm brainstorming how to generate a counterexample.
Bollobás showed that almost all graphs have reconstruction number 3, which means that there exist three cards in their deck that uniquely determine the graph.  This essentially thwarts any obvious attempt at generating random graphs to find a counterexample.
On the other hand, the Reconstruction Conjecture for digraphs is false (there are known families of counterexamples).  It might also be similarly impractical to generate random counterexamples to the Reconstruction Conjecture for digraphs, despite knowing it is false.
Hence my question:
Question: Are almost all digraphs reconstructible?

REPLY [2 votes]: I'm thinking the answer is yes. In the undirected case, Bollobas proves that any 3 cards determine the graph. So given a random directed graph $D(n,p)$, ignore orientations and collapse the resulting multi-edges. This gives an undirected random graph $G(n,q)
$ with $q = 1 - (1-p)^2$. This graph can be reconstructed with any 3 cards, so revealing the orientations on the cards should give back the original directed graph.<|endoftext|>
TITLE: Quantitative approximation of invariant measures by periodic ones
QUESTION [7 upvotes]: It is well known that many strongly chaotic dynamical systems have the property that periodic measures are (weak-star) dense in the space of all invariant probability measures. Is there some knowledge on how accurate on can approximate a given good measure by a periodic one in terms of length of the periodic orbit supporting it?
I mean, is there a known statement of the following kind:
Let $M$ be a compact manifold, $f:M\to M$ is continuous, periodic measures are dense in the space of invariant probability measures. Let $f$ has a physical measure $\mu$ (i.e. for Lebesgue-almost every point $x$ one has $\frac{1}{n}\sum_{i=0}^{n-1} \delta_{f^i(x)} \to \mu$ ) that is not a finite sum of $\delta$-measures. Then under some additional condition on $f$ (for example, specification property) there exists an accuracy function $\gamma:\mathbb{N} \to \mathbb{R}_{\geq 0}$, $\gamma(n)\to 0, n\to \infty$ such that for every natural $N$ there exists a natural $n>N$ and an $n$-periodic orbit $p$ such that 
$$ \rho\left( \frac{1}{n}\sum_{i=0}^{n-1} \delta_{f^i(p)} , \mu \right) < \gamma(n) ,  $$
where $\rho$ is some metrisation of the weak-* topology.
If there is no general statement like this, are there examples (say, piecewise-expanding maps of the interval) where something like this is known to hold?

REPLY [2 votes]: For a result of a somewhat similar type, check the paper "Rate of approximation of minimizing measures" by Bressaud and Quas, 2007, especially Theorem 4. In their case the dynamics is a subshift of finite type. They consider an arbitrary invariant measure, and look for periodic orbits in an $\varepsilon$-neighborhood of the support of the measure. True, it's not the type of approximation you want (weak-star), but it may be something to start with. They prove that $\varepsilon$ can be took superpolynomially small as a function of the period of the orbit, i.e., for any $K>0$ and $n_0 \ge 1$ it's possible to find a periodic orbit of period $n \ge n_0$ in the $n^{-K}$-neighborhood of $\mathrm{supp}(\mu)$. They also show that these bounds are sharp, so for example exponential approximation is not always possible.<|endoftext|>
TITLE: Do models-and-homomorphisms always form an accessible category?
QUESTION [8 upvotes]: It's well-known that the category of models of any first-order theory $T$ form an accessible category if we take the elementary embeddings as morphisms. This is true in finitary first-order logic or infinitary first-order logic. This is hinted at by Makkai and Paré and shown by Adámek and Rosický, Thm 5.42. 
But what happens when we take our morphisms to be arbitrary homomorphisms? Is the resulting category still accessible? It turns out (Adámek and Rosický Thm 5.35, Makkai and Paré Thm 3.2.1) that a category is accessible if and only if it is equivalent to the category of models of a basic theory of infinitary first-order logic, with homomorphisms as morphisms. A basic theory is one which is axiomatized by a small set of sentences each of the form $\forall \vec x \, (\phi(\vec x) \to \psi(\vec x))$ where $\phi$ and $\psi$ are positive existential, i.e. built up from $\vee, \wedge, \exists$ and atomic formulas.
It seems too much to ask that every theory be "Morita equivalent" (in the sense of having equivalent categories of models-and-homomorphisms) to a theory of this restricted form, so perhaps the answer is no. On the other hand, every category of models-and-elementary-embeddings  is accessible, and hence (surprisingly, to me) equivalent to the category of models-and-homomorphisms of another theory with this restricted form -- so maybe an affirmative answer wouldn't be so surprising after all.
I don't expect the answer to depend on whether we use finitary or infinitary logic, but if it does, that would be very interesting.

REPLY [14 votes]: For a general theory, one does not need to have $\lambda$-directed colimits for any $\lambda$.
The simpliest example is formed by sets and the axiom $(\exists x,y)(x\neq y)$; idempotents do not split here.<|endoftext|>
TITLE: Which ordinals are proof-theoretic ordinals?
QUESTION [8 upvotes]: Few months ago I have posted this question on MO, but I must admit that at the time, admittedly, I had no idea on how technical proof-theoretic considerations can be. I have decided to revise this question, and I thought that asking a new question is a better option that rewriting the old one (especially given that the other one has a partial answer to old question).
Given theory $T$ and recursive relation $\prec$, we say that $T$ proves $\prec$ to be well-ordered if $T$ proves that $\prec$ is a linear order and that $\forall X:((\forall n\prec m:n\in X)\Rightarrow n\in X)\Rightarrow\forall n:n\in X$.
We define proof-theoretic ordinal (PTO) of $T$ to be supremum of order types of relations which $T$ proves to be well-order.

Which recursive ordinals are PTOs of some theory extending $RCA_0$? How about theories extending $ACA_0$?

Side question: What is the PTO of theory $ACA_0+$ "$\varepsilon_0$ is well-ordered", where with $\varepsilon_0$ I mean a canonical well-ordering with order type $\varepsilon_0$?

REPLY [8 votes]: If $\alpha$ is a reasonable presentation of a computable ordinal then the proof-theoretic ordinal of $ACA_0+\alpha$ is well-ordered is the smallest $\epsilon$ number $>\alpha$.  (There's a proof of the upper bound in Epsilon substitution method for $ID_1(\Pi^0_1\vee\Sigma^0_1)$ by Arai, though there are probably simpler proofs out there.  The lower bound follows by relativizing the usual argument to the representation of $\alpha$.)
It follows that every computable $\epsilon$-number is the proof-theoretic ordinal of an extension of $ACA_0$.  For weird theories, the notion of a proof-theoretic ordinal may not be well-defined (it may depend which definition you use), but if $T$ extends $ACA_0$ and proves the well-foundedness of an ordering of order-type $\alpha$, it also proves the well-foundedness of an ordering of order-type $\omega^\alpha$: there's apparently a proof in Proof theory and logical complexity by Girard, and some related exposition in the papers by Marcone-Montalbon and by Rathjen and his students where they prove similar theorems for order ordinal operaions.  That basically shows that the computable $\epsilon$-numbers are the only proof-theoretic ordinals over $ACA_0$.<|endoftext|>
TITLE: Approximating a real by a ratio of primes
QUESTION [7 upvotes]: Let $x$ and $y$ be positive reals in $(0,1)$ with $x < y$ and $y-x =\epsilon$.
I seek smallest primes $p$ and $q$ such that 
$$x \le \frac{p}{q} \le (x+\epsilon) = y \;.$$

Q. What upper bound $u(\epsilon)$ can be placed on $\max\{p,q\}=\max(q)$ as a function of $\epsilon$?

Examples.
$x=\sqrt{2}/2 \approx 0.707107$, $\epsilon=10^{-4}$, $y=(x+\epsilon)$, then
$$
x = \sqrt{2}/2 \approx 
0.\color{blue}{707}{\color{red}{1}}07 < \frac{p}{q}=\frac{1217}{1721}\approx 0.\color{blue}{707}147 < 0.\color{blue}{707}{\color{red}{2}}07
\approx (\sqrt{2}/2 + \epsilon) = y \;.
$$
For $\epsilon=10^{-5}$, 
$$
x = \sqrt{2}/2 \approx 
0.\color{blue}{7071}\color{red}{0}7 < \frac{p}{q}=\frac{3491}{4937}\approx 0.\color{blue}{7071}096 < 0.\color{blue}{7071}{\color{red}{1}}7
\approx (\sqrt{2}/2 + \epsilon) = y \;.
$$
I believe those are optimal prime ratios for those $x$ & $y$.
If so, $u(10^{-4}) \ge 1721$ and $u(10^{-5}) \ge 4973$;
and $u(10^{-6}) \ge 8597$ 
and $u(10^{-7}) \ge 38287$
(data not presented):
$$
\begin{array}{cccc}
\epsilon=10^{-4} & \epsilon=10^{-5} & \epsilon=10^{-6} & \epsilon=10^{-7}\\
u(\epsilon) \ge 1721 & u(\epsilon)\ge 4937 & u(\epsilon)\ge 8597 & u(\epsilon)\ge 38287
\end{array}
$$

(The above question is tangentially related to the earlier question,
"Visibility in a prime orchard.")

REPLY [2 votes]: Here is an idea which might help show that the smallest prime is not much larger than the smallest integer needed.
One of the processes I like to use is the mediant $\frac{a+c}{b+d}$ of two positive fractions $\frac ab$ and $\frac cd$ to approximate a real with a fraction of small denominator.  Suppose we use the process to approximate $x +\epsilon/2$, say.  At some point we will end with two ratios of integers of, say, $k$ decimal digits each, call them $\frac ab$ and $\frac cd$,
with $$x \lt \frac ab \lt x + \frac{\epsilon}{2} \lt \frac cd \lt x + \epsilon$$.
Then the mediant $\frac{a+c}{b+d}$ will have at most $k+1$ digits in numerator and denominator, and this can be tweaked to adjust the numerator and denominator to be primes without moving the ratio out of the interval of interest.  Probably one can tweak $\frac ab$ or $\frac cd$ similarly, but the point is that the prime representation won't need many more digits than the pure integer representation, and if $\epsilon$ is measured in negative powers of 10, I predict fewer decimal digits than that will be needed for the prime representation.
Gerhard "Caution: Much Vigorous Handwaving Nearby" Paseman, 2015.07.11<|endoftext|>
TITLE: Prove that expression is integer
QUESTION [17 upvotes]: Numerical experiments suggest that
$\binom{2m}{m + k}\cdot\frac{3m - 1 - 2k^2}{2m - 1}$
is integer for all $-m \le k\le +m$. It means that expression evaluation could be implemented very efficiently, only using integer addition and multiplication.
However, I've failed to derive computationally efficient expression so far.
The two ideas I have are linear combination of binomials, and recursion -- but more insight is needed to go further.

REPLY [42 votes]: It equals
$$
\binom{2m}{m+k}\frac{3m-1-2k^2}{2m-1}=-(m-1)\binom{2m}{m+k}+4m\binom{2m-2}{m+k-1}.
$$
I got it by expanding $3m-1-2k^2=2(m^2-k^2)-(2m^2-3m+1)=2(m-k)(m+k)-(2m-1)(m-1)$.<|endoftext|>
TITLE: Resource Constrained Routing with Refueling
QUESTION [5 upvotes]: What are good algorithms (resp. models) for calculating optimal or near optimal routes while taking into account fuel consumption, options for refueling and, limited tank capacity?
Especially modeling refueling gives me some headaches.  
The reason why I'm asking is because with the growing success of electric vehicles in presence of a still underdeveloped infrastructure for recharging stations, that kind of route calculation is important for either deciding, whether it is possible to come from $A$ to $B$ and if so, which route to take in order to optimize a secondary criterion such as length or time. 
So far I found the The Aircraft Routing Problem with Refueling, that addresses the subject, but apart from that, not much seems to be available.  
What I am looking for are articles with either formulations as a network-flow problem or as a breadth-first search in a Dijkstra fashion and a focus on practical applicability.

REPLY [4 votes]: Here are three sources. The 2nd paper is recent (2012) and its
literature review may be helpful. The 3rd is a 2013 PhD dissertation that incorporates traffic congestion.

(1) Lin, Shieu-Hong. "Finding optimal refueling policies in transportation networks." In Algorithmic aspects in information and management, pp. 280-291. Springer Berlin Heidelberg, 2008.
  (PDF download.)
"we can
  then (i) determine all-pairs optimal ($k$-stop-bounded) refueling policies given
  various initial fuel levels at the vertices and ending with a fuel level at least at
  the lower limit $L$ in $O(n^3)$ time, and (ii) determine an optimal refueling policy
  given a pair of vertices, an initial fuel level, and a required minimal final fuel
  level in $O(n^2)$ time."



(2) Sweda, Timothy M., and Diego Klabjan. "Finding minimum-cost paths for electric vehicles." In Electric Vehicle Conference (IEVC), 2012 IEEE International, pp. 1-4. IEEE, 2012.
  (PDF download.)
"In this paper, the problem
  of finding a minimum-cost path for an EV [Electric Vehicle] when the vehicle must
  recharge along the way is modeled as a dynamic program. It
  is proven that the optimal control and state space are discrete
  under mild assumptions, and two different solution methods are
  presented."



(3) Fontana, Matthew William. "Optimal routes for electric vehicles facing uncertainty, congestion, and energy constraints." PhD diss., Massachusetts Institute of Technology, 2013. (MIT link.)
"We develop an energy model, an optimization-based formulation using robust optimization, and algorithms to quickly find good routes for battery electric vehicles. The combination of using robust optimization, the A-Star algorithm to find shortest paths, and Lagrangian relaxation allows us to solve the problem in seconds or less."<|endoftext|>
TITLE: (A kind of) Irreducibiliy of regular open convex sets in the Cartesian space
QUESTION [5 upvotes]: I am looking for a proof of the fact which is formulated at the bottom of this post. The property of regular convex sets which the fact expresses seems to be true to me, yet I have not been able to demonstrate it in a rigorous way. Any hints will by very appreciated.
Let $\mathrm{r}\mathscr{O}$ be the family of open domains (regular open sets) of a topological Cartesian space $\langle\mathbf{R}^2,\mathscr{O}\rangle$ (with the standard topology), that is:
$$A\in\mathrm{r}\mathscr{O}\iff A=\mathrm{int(\mathrm{cl(A)})}.$$
It is well know that $\langle\mathrm{r}\mathscr{O},+,\cdot,-,0,1\rangle$ is a complete Boolean algebra, where: $A+B=\mathrm{int(\mathrm{cl(A\cup B)})}$, $A\cdot B=A\cap B$ and $-A=\mathrm{int}(\complement A)$ (with $\complement$ the standard set-theoretical complement operation).
In this algebra define the disjoint sum operation in the usual way:
$$A\oplus B=(A-B)+(B-A)\,.$$
By a convex set $A$ I understand, standardly, a regular open set for which it is the case that for any two points $x,y\in A$: $[x,y]\subseteq A$. The fact I would like to prove is (the order $\leqslant$ is standard) the following kind of irreducibility:

Let $A,B,C\in\mathrm{r}\mathscr{O}$ be convex:
$$A\leqslant B\oplus C\ \mbox{and}\ B\cdot C\neq 0\Longrightarrow A\leqslant B-C\ \mbox{or}\ A\leqslant C-B\,.$$

REPLY [4 votes]: Let $B$ and $C$ be convex regular open sets such that $B\cdot C\neq 0$, and write $D=B-C$ and $E=C-B$.  It suffices to show that the union $D\cup E$ is regular.  Indeed, if it is, then $B\oplus C=D\cup E$ and $D$ and $E$ are disjoint open sets, so any connected subset of $B\oplus C$ must be contained in either $D$ or $E$.
So suppose for a contradiction that $D\cup E$ is not regular.  Let $x$ be a point in $(D+E)\setminus(D\cup E)$; then there is some open ball $U$ around $x$ such that $D\cup E$ is dense in $U$.  If $x\in B$, then $E$ is disjoint from a neighborhood of $x$, and so $D$ would have to be dense in a neighborhood of $x$ and hence contain $x$ by regularity.  Thus $x\not\in B$, and by symmetry $x\not\in C$.  By convexity of $B$, we can find a line $L$ passing through $x$ such that $B$ is contained in one of the open half-planes $V$ formed by $L$; let $W$ be the other half-plane.  We thus have $D\cdot U\leq V\cdot U$, and so $E$ must be dense in and hence contain all of $W\cdot U$.  If $y\in V\cdot C$, then the line from $y$ to $x$ extends into $W\cdot U$, so by convexity $C$ would have to contain $x$.  Thus $V\cdot C$ must be empty.  But now $B\cdot C\leq V\cdot C=0$, a contradiction.<|endoftext|>
TITLE: Do the bounded isophase lines of a complex polynomial $f$ through the zeroes of $f’$ define a spanning tree?
QUESTION [5 upvotes]: Let $f: \mathbb{C} \to \mathbb{C}$ be a polynomial and let $\arg(f(z))$ be the phase of $f(z) = | f(z)| \exp(\mathrm{i} \arg(f(z)))$. The zeroes of $f'(z)$ are saddle points of  $\arg(f(z))$, i.e. intersections of isochromatic lines (if we color-encode the phase, cf. http://www.ams.org/notices/201106/rtx110600768p.pdf, page 772ff). 
In the following we assume $f$ to have only simple zeroes.
 Questions: 
 a1) Is it correct that through every (generic) saddle point  of $\arg(f)$  (= zero of $f’$) there are two isochromatic lines, one of which is unbounded and one of which connects two zeroes of $f$?  
 a2) If a1) is correct: Is there a simple (geometric) criterion determining which zeroes are directly connected via an isochromatic line?   
 b1)   Is it correct that the set of isochromatic lines which connect two zeroes define a spanning tree (of all zeroes)? 
 b2) If b1) is correct: Can the (graph theoretical, i.e. topolocial) structure of the spanning tree be solely defined by the position of the zeroes of $f$ (e.g. minimal weigth spanning tree where weight is square of Euclidean distance of zeroes)? 


The figure shows the color-encoded phase plot of an example of a polynomial of degree four (the thin black lines are lines of equal absolute value, they are orthogonal to the isochromatic lines).  The four zeroes of $f$ can be seen as the minima of the absolute value. The isochromatic lines through the zeroes of $f’$ are the thick black lines. The three chromatic saddle points (=zeroes of $f’$) are clearly visible.  In this example the spanning tree has one central vertex (the zero of $f$ below the middle) from which there are edges to the other three zeroes of $f$. 
An electrostatic interpretation could possibly be helpful. If we consider the zeroes of $f$ to be equal charges in 2d, the question amounts to how can one know which charges are connected via a field line of the electrostatic field.  
 Motivation: 
Every complex polynomial with $n$ distinct zeroes has $n-1$ critial points (zeroes of $f’$). For the location of the zeroes there is huge literature (for a starting point see e.g. Q.I. Rahman und G. Schmeisser, Analytic Theory of Polynomials,
Oxford University Press, 2002). 
From numerical studies it seems to be interesting to ask which spanning tree of the zeroes could be helpful in describing the location of the critical points (note that the spanning tree through $n$ vertices has $n-1$ edges). 
The question was inspired by Alth\“ofer/Voigt, Spiele, R\“atsel, Zahlen, Springer Spektrum 2014 and Kaffka, Zuordnungen zwischen Nullstellen und kritischen Punkten von Polynomen, Diploma Thesis supervised by Prof. Dr. I. Alth\"ofer, Jena, 2012, page 42, http://www.althofer.de/kaffka-diplom.pdf. 
In this context I have proven a slight strengthening of the theorem of Gauss-Lucas for polynomials of degree four (http://arxiv.org/abs/1405.0689).

REPLY [5 votes]: Edited
As @Alexandre Eremenko noted, in the generic case the answer is yes, and in fact it can be shown that any connected finite tree can appear as the isochromatic tree graph of some polynomial (see Theorem 6.1 here, this essentially has the parallel result for level curves, an induction argument will give you the desired application to isochromatic trees).
In the non-generic case the picture is even more interesting, and the isochromatic tree graphs must allow critical points as vertices as well, as I describe below.
The sets you are considering I will call "gradient lines" of $f$, namely components of the sets $\{z:\operatorname{Arg}(z)=\theta\}$ for various values of $\theta\in[0,2\pi)$.  First I need a bit of info about the orthogonal objects, namely the level curves of $f$ (sets where $|f|$ is constant).  (I call them "orthogonal" because at any point $z$ not a critical point of $f$, the level curve and gradient line of $f$ containing $z$ are orthogonal to each other at $z$.)
EDIT: I just realized that I don't think I need the second bullet point in the theorem below for this answer, but it is none the less true and I think interesting.

THEOREM Let $p$ be a polynomial, and let $\Lambda$ be a level curve of $p$ (ie. a component of the set $\{z:|p(z)|=\epsilon\}$ for some $\epsilon>0$).

$\Lambda$ consists of a connected finite graph whose vertices are critical point of $f$, and satisfying the following properties.
1.) There are evenly many (and more than two) edges of $\Lambda$ incident to each vertex of $\Lambda$.
2.) Each edge of $\Lambda$ is incident to both the unbounded face of $\Lambda$ and a bounded face of $\Lambda$.
That is, $\Lambda$ is a sort of complicated "figure eight" graph.
Let $D$ be a bounded face of $\Lambda$.  Then either $f$ has a single distinct zero of $f$ in $D$, or there is some unique level curve $\Lambda_D$ in $D$ which is maximal in the sense that every critical point of $f$ in $D$ is either in $\Lambda_D$ or contained in a bounded face of $D$, and every zero of $f$ in $D$ is contained in a bounded face of $D$.


Now we can define the non-generic version of the isochromatic graphs.  For any point $z\in\mathbb{C}$, let $\Lambda_z$ denote the level curve of $p$ containing $z$, and let $\Gamma_z$ denote the gradient line of $p$ containing $z$.
Let $z_o$ be a critical point of $p$.  Let $D_1,D_2,\ldots,D_k$ denote the bounded faces of $\Lambda_{z_o}$ which are incident to $z_o$.  For each $i\in\{1,2,\ldots,k\}$ a portion of the gradient line $\Gamma_{z_o}$ extends from $z_o$ into $D_i$, and ends at a zero or critical point of $p$ in $D_k$.  Let therefore $A_{z_o}$ denote the portion of $\Gamma_{z_o}$ which is contained in the bounded faces $D_1,D_2,\ldots,D_k$, stopping when a zero or other critical point of $p$ is reached (ie. if a gradient line of $p$ reaches down from $z_o$ to a zero of $p$ which has multiplicity greater than $1$, then other branches of that gradient line will emanate from that zero, and we want to disregard those other branches, similarly for other critical points).
If we now let $\mathcal{T}$ denote the union of the $A_z$ over all critical points $z$ of $p$, then $\mathcal{T}$ has the following nice properties.

$\mathcal{T}$ is a connected finite tree, the vertices of which are either critical points or zeros of $p$.
Every zero and critical point of $p$ appears as a vertex of $p$.
$\operatorname{Arg}(p(z))$ is constant on each edge of $\mathcal{T}$.
$\mathcal{T}$ is the minimal such tree that satisfies the above criteria (that is, it is the intersection of all such trees).

In my own work I have classified the way in which the critical level curves of the polynomial $p$ (level curves containing critical points) may be arranged.  This geometric property of of polynomials is a strong conformal invariant.  That is, for any two polynomials $p_1$ and $p_2$, they have (geometrically) the same configuration of critical level curves if and only if there is a degree $1$ polynomial $\phi:\mathbb{C}\to\mathbb{C}$ such that $p_2=p_1\circ\phi$ on $\mathbb{C}$.  A similar analysis might be done using the trees described above.  See papers 2, 3, and 4 here for this work.

REPLY [2 votes]: You are asking about the foliation (with some singular points) provided by the preimages of radial lines. Clearly you can say the following:
a1. Generically, every critical point is simple, and all critical values have different arguments. Hence clearly there are four lines meeting at each critical point, and analytic continuation of two of these will lead to a zero, while the other two will lead to infinity. Hence the answer is positive.
a2. I am not quite sure what you are looking for here. There are a number of ways to describe the mapping behaviour of the polynomial (classical "line complexes", "dessins d'enfants" in the case where there are only two critical values, various constructions connected with studies of Hurwitz numbers), and most of these should allow you to read off the desired information. But your construction is very similar to this already.
b1. To see that the set $A$ consisting of the lines that you are considering indeed forms a spanning tree of the zeros, we must only prove that it is connected with connected complement. Note that $A$ is the preimage of a "star" $S$ connecting zero to each of the critical values. The map from the complement of $A$ to the complement of $S$ is a covering map on each component. Since the complement of $S$ is topologically a punctured disc, and since your map is a polynomial and hence has no poles, it follows easily that there is only one complementary component, which is homeomorphic to a punctured disc. So the answer is again positive.
b2. Same answer as a2.
EDIT. Rereading your question, I now see that by "geometric", you mean, "in terms of the position of the zeros". I think that, in the way you mean the question, the answer is negative. For example, by work of Bishop ("True trees are dense"), your set can approximate any given compact set arbitrarily closely - even for polynomials with just two critical values, say $0$ and $1$, where your tree is the "dessin d'enfant" (and the polynomial is called a "Shabat polynomial"). In particular, you can make a polynomial of this type for which the tree contains some vertices that are very close together, but not connected by any short path - and some vertices relatively far together that are connected by an edge. [Basically, you start Bishop's construction with an arc of a circle that just leaves out some very small interval.] This seems to exclude any of the kind of answers that you are looking for.
Of course, this is non-generic, but that does not seem essential (and probably you can just make a small perturbation).<|endoftext|>
TITLE: Integral inequality similar to Hardy's inequality
QUESTION [7 upvotes]: I am not very sure if that's research level, I hope you don't find it too elementary for this place.
I am trying to solve the following puzzle:
We are given a real function $f$, where $f(x) \geq 0$ and $F := \int_0^x f(t) dt$ and some real $p>1$.
Does $\int_0^\infty f(x)^p e^{-x}dx < \infty$ imply $\int_0^\infty F(x)^pe^{-x}dx < \infty$ ? 
My current approach was modifying proofs of Hardy's theorem. This approach has not been very successful. I keep running into hard to compute integrals that I would need to bound. I hope somebody could maybe provide me with some direction in which I should look. If somebody was interested I could provide details of calculations so far.

REPLY [2 votes]: The question has been already solved by Paata Ivanisvili. I just want to share my insight.
I think I have found the solution. I will use generalized Hardy's inequality, which unfortunately is a bit like killing a fly with a cannon.
As a source I will cite http://www.encyclopediaofmath.org/index.php/Hardy_inequality where you can find references to the original source. 
Lemma - generalized Hardy's (cited from Encyclopedia of Math).
$$
\int_0^\infty \left| \phi(x) \int_0^x f(t) dt \right|^p dx \leq C \int_0^\infty \left| \psi(x)  f(x) \right|^p dx
$$
$\iff$
$$
\sup_{x > 0} \left[ \int_x^\infty  | \phi(t) |^p dt \right]^{\frac{1}{p}} \left[ \int_0^x | \psi(t) |^{-q} dt \right]^{\frac{1}{q}} < + \infty,
$$ where $\frac{1}{p} + \frac{1}{q} = 1 $. 
After using this inequality, the task comes down to verification of the assumption, which is pretty easy. Let me verify.
Given a function $f \geq 0$ and $F(x) = \int_0^{x} f(t)
dt$ I claim  $\int_0^\infty f(x)^p e^{-x}dx < \infty$ implies
$\int_0^\infty F(x)^pe^{-x}dx < \infty$. 
Proof: Assume $\psi(t) = \phi(t) = e^{- \frac{t}{p} }$. When
$\frac{1}{p} + \frac{1}{q} = 1 $, then $\frac{1}{q} = \frac{p-1}{p}$ and $-q = \frac{p}{1-p}$. We investigate
$$
\sup_{x > 0} \left[ \int_x^\infty  e^{-t} dt \right]^{\frac{1}{p}} \left[
\int_0^x e^{-t \frac{1}{1-p}} dt \right]^{\frac{p-1}{p}} = \sup_{x > 0} e^{- \frac{x}{p}} \left( e^{\frac{x}{p-1}} - 1 \right)^{\frac{p-1}{p}}
=
$$
$$
\sup_{x > 0} \left( e^{- \frac{x}{p-1}} e^{\frac{x}{p-1}} - e^{- \frac{x}{p-1}} \right)^{\frac{p-1}{p}}
=
\sup_{x > 0}
    \left(
        1 -
    e^{- \frac{x}{p-1}}
    \right)^{\frac{p-1}{p}},
$$
which is bounded as $p>1$. Now we apply Hardy and our theorem is verified.<|endoftext|>
TITLE: Is there a "free abelian group of rank 1" in the category of affine group schemes?
QUESTION [15 upvotes]: Let's fix an algebraically closed field $k$.
The group $\mathbb Z$, as a discrete group scheme, is not affine since it's not quasi-compact. Is there an affine algebraic scheme over $k$ whose representation category is isomorphic to that of $\mathbb Z$?
Here is the motivation for this question. Fix a point $x$ on the circle. The $k$-linear tensor category of local systems of finite dimensional $k$-vector spaces, equipped with the functor $\mathcal F \mapsto \mathcal F_x$, is a Tannakian category. By the Tannakian duality, the category is isomorphic to the representation category of an affine group scheme, possibly not of finite type. This category is isomorphic to the representation category of an affine group scheme $G$. I want to understand what $G$ looks like.

REPLY [4 votes]: Claim : if $k$ is perfect of characteristic $0$, then :
$$\mathbb Z^{k, alg}\simeq  \mathbb G_a\times D_k(k^*)$$
Here $\mathbb Z^{k, alg}$ stands for the pro-algebraic completion of $\mathbb Z$ (the group you are interested in), $\mathbb G_a$ is the additive group, and $D_k(M)$ denotes as usual the diagonalisable $k$-group associated to the abstract abelian group $M$.
Sketch of a proof.
We denote $G=\mathbb Z^{k, alg}$ to simplify. By definition, this is the Tannaka group of the Tannaka category ${\rm Rep}_k \mathbb Z$, that is canonically isomorphic to the category ${\rm Iso_k}$ whose objects are couples $(V,\phi)$, where $V$ is a finite dimensional $k$-vector space, and $\phi\in {\rm GL}(V)$. 
Since $\mathbb Z$ is abelian, so is  $G=\mathbb Z^{k, alg}$, hence since $k$ is perfect, it splits canonically into a product diagonalizable x unipotent :
$G=G^m\times G^u$.  
Moreover $G^m=D_k(\Gamma)$, where
$\Gamma={\rm Hom}_{gr}(G,\mathbb G_m)={\rm Pic}({\rm Rep}_k G)\simeq {\rm Pic}({\rm Iso}_k)\simeq k^*$.
On the other hand $G^u$ is the tannaka group of the category $({\rm Rep}_k G)^u\simeq({\rm Iso}_k)^u={\rm Uni}_k$, where ${\rm Uni}_k$ is the category whose objects are couples $(V,\phi)$, where $V$ is a finite dimensional $k$-vector space, and $\phi$ is a unipotent endomorphism.
Now since $k$ is of characteristic $0$, there is an equivalence
${\rm Rep}_k(\mathbb G_a)\simeq {\rm Uni}_k $ sending  $\rho :\mathbb G_a\rightarrow {\rm GL}(V)$ to $(V,\rho(1))$, compatible with fiber functors, so one deduces  $G^u\simeq \mathbb G_a$.
Complement, probably related to your original interest.
All this is implicitely contained in 
Saavedra Rivano, Neantro
Catégories Tannakiennes. 
Lecture Notes in Mathematics, Vol. 265. Springer-Verlag, Berlin-New York, 1972. 
http://link.springer.com/book/10.1007%2FBFb0059108
last chapter 
Exemples tirés de la géométrie algébrique
p 319 ex 1.2.6 a).
that states that the category of regular connections on $\mathbb P^1\backslash \{0,\infty\} $ admits $\mathbb G_a\times D_k(k/\mathbb Z)$ as Tannaka group, hence, if $k$ is algebraically closed of characteristic $0$, is non-canonically isomorphic to $\mathbb Z^{k, alg}$. This result also shows that the diagonalisable part is not compatible with base change. This explains why one is generally interested only by the unipotent fundamental group.<|endoftext|>
TITLE: Stabilization of a generic pointed model category
QUESTION [5 upvotes]: Let $\mathcal C$ be a pointed model category. It is well-known that its homotopy category $\mathrm{Ho}(\mathcal C)$ is naturally a $\mathrm{Ho}(\underline{\mathrm{sSet}}_*)$-category, where $\mathrm{sSet}_*$ is the Quillen model category of pointed simplicial sets.
With this structure one can define suspension and loop functors on $\mathrm{Ho}(\mathcal C)$. The suspension $\Sigma$ is the (derived) smash product with the simplicial $\mathbf S^1$ and the loop functor is its right adjoint. By definition, $\mathcal C$ is called stable iff $\Sigma$ is an autoequivalence of $\mathrm{Ho}(\mathcal C)$.
In case $\mathcal C$ is not necessarily stable, I'm looking for canonical ways to stabilize $\mathcal C$. In case $\mathcal C$ fulfills extra axioms, e.g. it is left proper and combinatorial, one can find by [1] a simplicial model structure on the simplicial objects $\mathrm s \mathcal C$ of $\mathcal C$ which is Quillen equivalent to $\mathcal C$ itself. So for left proper and combinatorial model categories $\mathcal C$, one may assume that they are already simplicial so that the suspension is already defined on the level of the model category. Then one can use [2] or [3] to consider (symmetric) spectrum objects in $\mathcal C$ with respect to the suspension and gets a canonical stable model category, where the model structure is the stable one.
Now my question is as follows: Is there a canonical way to stabilize a model category without going through simplicial enrichments (and which may work with any closed model category)? Bonus questions are: Are there better or more canonical references than the two I gave? Does it make sense just to stabilize the homotopy category $\mathrm{Ho}(\mathcal C)$, e.g. just to consider spectra on the homotopy level?
[1] Dugger: Replacing Model Categories with Simplicial Ones
[2] Hovey: Spectra and symmetric spectra in general model categories
[3] Schwede: Spectra in model categories and applications to the algebraic cotangent complex

REPLY [3 votes]: This question is near and dear to my heart. Since it is asked within the context of model categories I will try to answer in that context. I agree with Dmitri that you should not try to stabilize the homotopy category. Your main question seems to be about whether or not $C$ needs to be simplicial. The answer is no. Hovey's machine does not require a simplicial model category, but requires you to have the functor you want to stabilize. Via the machinery of framings (see chapter 5 of Hovey's book) you can write down that functor even when $C$ is only cofibrantly generated. This is the approach taken in chapter 6 of Hovey's book. Next, let's discuss the hypotheses that $C$ be left proper and cellular or combinatorial.
As far as I am aware all stabilization results involve Bousfield localization. However, the version of Bousfield localization which Schwede uses (improved upon in Bousfield's paper on Telescopic localization) does not require combinatoriality or cellularity. Those hypotheses are present in order to actually build the localization functor, which works via some transfinite process you need to know will eventually end. But if you already have the localization functor then you don't need those hypotheses. So I think what you propose can be done, but you will want to be careful to determine which model of spectra you want. As Hovey points out, Schwede's machine and Hovey's machine build different model categories of spectra, and only Hovey's appears to have the property you want. So basically you would want to take Hovey's proof and make it go with a localization functor built "by hand" so as to avoid the need for $C$ to be combinatorial or cellular. Note that without left properness you can still say something, and I have a preprint on that if you would like to email me. It's not yet ready to write about publicly here. I advise working out an example, e.g. take $C$ to be the stable module category and figure out what the Ho(sSet) action does. Use a framing to write down the suspension endofunctor and try to build the localization following Hovey and Schwede. If it works then you have a nice roadmap for what to do in general, and the material in chapters 5-6 of Hovey's book will help fill in the details.
There are also other ways to stabilize, e.g. Biedermann's paper L Stable Functors, which basically defines the stabilization you want, tries to build it, and has a comparison to Hovey's machine. The relevant results are in section 5, $M$ is what you called $C$, and $V$ can be simplicial sets or chain complexes or something else. This uses the version of Bousfield localization I mentioned above, but it (and also Proposition 2.2 of Motivic Functors by Dundas, Rondigs, Ostvaer) technically require locally presentable in order to embed into a presheaf category. Perhaps this can be avoided; I advise asking a category theorist.
Since Dmitri mentioned Higher Algebra, let me just say that I don't understand the proof of 1.4.2.24. It first does the case where $C$ is presentable, which we already knew how to do thanks to the references in the question, and then it says without loss of generality we can assume $C$ is small. I don't follow this step. Perhaps it's one of these arguments where you enlarge the Grothendieck universe but in this context it seems to be getting something for free. In any event, this step is probably not replicable in the context of model categories because once you bump up to a larger Grothendieck universe I don't know what happens. If the model category truly becomes small then it's uninteresting (e.g. just a poset), but I think what really happens is more in line with the discussion at Is the $\infty$-category of presentable $\infty$-categories presentable?, i.e. in this new universe $C$ won't have all limits and colimits, only those of some bounded size. In any event, I don't think you can reduce questions about model categories to dealing with small ones.<|endoftext|>
TITLE: Why is the set-theoretic principle $\diamondsuit$ called $\diamondsuit$?
QUESTION [9 upvotes]: A shallow answer would just point to theorem 6.2 in Jensen's 1972 paper "The fine structure of the constructible hierarchy", where Jensen introduces this property.  Or was this symbol used already earlier for the combinatorial principle? 
Is there some (deeper) reason explaining how $\diamondsuit$ got its name? Perhaps because $\Box$ was already established as a combinatorial principle?

REPLY [15 votes]: I once asked Jensen this question, when we were at a conference at Oberwolfach. 
I told him that I had always assumed that the diamond $\Diamond$ principle was called $\Diamond$ because it expresses that there is an object exhibiting an elaborate degree of internal reflection. After all, if $\langle A_\alpha\mid\alpha<\omega_1\rangle$ is a $\Diamond$-sequence, then it very often happens for $\alpha<\beta$ that $A_\beta\cap\alpha=A_\alpha$, which is an instance of $A_\beta$ reflecting to $A_\alpha$. 
But alas, this was not his reason. The actual reason was less interesting, having mainly to do, he said, with him simply needing a new symbol that had not yet been used, and that one was available.
Meanwhile, I believe that we should all adopt my explanation anyway! Henceforth, let it be known that a $\Diamond$ sequence is one exhibiting beautifully captivating internal reflections!<|endoftext|>
TITLE: Is every real a member of some CTM?
QUESTION [7 upvotes]: I read somewhere about the hyperuniverse of countable transitive models of ZFC (http://www1.maths.leeds.ac.uk/maloa/lecturenotes/RW3%20Munster/friedman.pdf). It states an assumption, namely that every real is a member of some countable transitive model of ZFC.
Is there some way to prove this assumption (assuming the hyperuniverse is not empty), or rather the (seemingly stronger) one: Every finite set of reals is a member of a ctm?

REPLY [16 votes]: It is (relatively) consistent with ZFC that this is false. For example, suppose that there are $\alpha<\beta$ with $L_\alpha$ and $L_\beta$ both being models of ZFC. (This is a very weak large cardinal axiom). Let us take $\alpha$ and $\beta$ to be least. So $L_\beta$ thinks that there is only one height $\alpha$ of a model of ZFC. It follows that $\alpha$ is countable inside $L_\beta$, and so inside $L_\beta$, the hyperverse is not empty---it contains $L_\alpha$ and many forcing extensions of $L_\alpha$ and so on. Let $x$ be a real in $L_\beta$ coding  a relation on $\omega$ of order-type $\alpha$. This real cannot be in any CTM of ZFC inside $L_\beta$, since that model would have to have height at least $\beta$ by the minimality of $\beta$, and so it could not be an element of $L_\beta$.
Another way to argue: let $M_0$ be any countable transitive model of ZFC, and let $M$ be any $\in$-minimal countable transitive model of ZFC with $M_0\in M$. So $M_0$ is countable in $M$, but by minimality, $M$ can have no CTM containing a real coding $M_0$. 
Meanwhile, if there is an inaccessible cardinal $\kappa$, or merely a worldly cardinal, then every real is contained in the corresponding $V_\kappa$, and by Löwenheim-Skolem, we may collapse a countable elementary substructure to place it into a countable transitive model of ZFC.<|endoftext|>
TITLE: When to postpone a proof?
QUESTION [44 upvotes]: One possible practice in writing mathematics is to prove every theorem and lemma right after stating it.
A long, technical proof — and sometimes even a short one — can interrupt the flow of the presentation, so postponing the proof can improve clarity.
But if too many proofs are postponed in a long(ish) paper, it can be difficult to keep track of what depends on unproven facts and what remains to be proven, not to mention the danger of circular proofs.
Are there good guidelines for deciding when to postpone a proof?
What should I take into account when making such decisions?
Below are some examples of situations where I might postpone a proof, and I would like to expand my understanding by more principles and examples.

Example 1:
Giving main theorems in the introduction as soon as the necessary notions are defined instead of giving them only when you are ready to prove them.
Example 2:
Pausing to explain to convince the reader that this is a natural thing to prove.

Lemma 1:
  Every strong polar bear is a weak penguin.
Before embarking on the proof, let us see why we should not hope for much more.
  A strong polar bear can fail to be a strong penguin and a weak polar bear can fail to be a penguin in the first place.
  Moreover, a penguin, even a strong one, need not be a polar bear of any kind.
  (For counterexamples, see the works of Euler and Gauss.)
Proof of lemma 1: Take a finite igloo containing the polar bear and a penguin…

Example 3:
Clear exposition of the main argument.
The following could be presented in the introduction or right after it, but the proofs of the lemmas would follow in subsequent sections.

Lemma 1: $A\implies B$.
Lemma 2: $B\implies C$.
Lemma 3: $C\implies D$.
Theorem 4: $A\implies D$.
Proof: Combining lemmas 1—3, we obtain $A\implies B\implies C\implies D$ as claimed. $\square$


Edit:
So far the discussion has mostly been around example 1.
The actual question has not been addressed so far.
An answer to "When to state the main theorem of a paper?" gives only a partial answer to my question.
Let me quote Tom Church's comment below (emphasis mine):
"Your Example 1 is dragging this discussion in the wrong direction, since everyone is going to agree with it: who would object to stating the main theorems in the introduction? The real question is about the structure of arguments within the body of the paper, and when it is appropriate to postpone a proof there. This is an important question, which too many authors neglect to think about when outlining their papers; I hope we'll get to hear a discussion on this point."

REPLY [6 votes]: I agree with most comments that more or less always the reader should know at least a good example of what you prove in the paper by reading the introduction. I think usually if you cannot state the main result in the introduction because it's too technical, probably the best thing is to have the section after the introduction contain the necessary definitions and accurate statements. It's not much fun for the reader to look through fifty pages of paper for the statement, and then discover that the technical conditions don't allow them to use it the way they wanted.
I quite often postpone proofs in the style of Example 3. Of course there is the danger of writing a circular argument, but this is something that both the author and reader should be easily able to check for: if not, then the paper isn't explained well enough. The advantage of this is that often in fifty-plus pages you have several proof strategies, maybe even a couple which use superficially similar methods but with an important difference. I find it much easier to understand a paper if I can take the lemmas as black boxes for understanding the main result, and then think about each lemma individually without having to keep my place mentally in the main result. This style of writing (I find) also discourages the author from opening a proof environment on page 3 which continues for the next forty pages, leaving the reader with the impression that they have to keep in memory all constant choices, definitions and equations made outside lemmas, which is usually false - but the important ones aren't indicated. It also discourages the author from stating a 'lemma' which is true in the context of the proof but false in general (in other words, with implicit assumptions) which I find very annoying.
Of course, with Example 3 one has to do more explanation: first an overview of the proof strategy, then a few sentences of why each lemma might be true or what it's used for, then an overview of the lemma proof (if it's long, which can easily be the case), which would probably be more compressed in the linear style (you probably don't explain why the lemma is true or how it's used if you are about to prove it). I don't think this is a good thing; most papers would be better for an extra five percent length in explaining what, why, how and perhaps why the obvious approach doesn't work.<|endoftext|>
TITLE: The infinity-type of automorphic representations in the Langlands correspondence
QUESTION [5 upvotes]: Let $K$ be a number field, $\rho\colon \mathrm{Gal}_K\to \mathrm{GL}_n(\overline{\mathbf{Q}_p})$ a geometric (i.e.: unramified a.e., de Rham above $p$) irreducible Galois representation. One piece of the global Langlands conjectures is that there exists a cuspidal automorphic representation $\pi$ of $\mathrm{GL}_n(\mathbf{A}_K)$ such that (up to a twist) the Satake parameters of $\pi$ correspond with the Frobenius eigenvalues of $\rho$ at the unramified places. 
Even better, for each $v\nmid p$, we can associate to the local Galois representation $\rho_v\colon \mathrm{Gal}_{K_v}\to \mathrm{GL}_n(\overline{\mathbf{Q}_p})$ a representation $\pi_v$ of $\mathrm{GL}_n(K_v)$ via the local Langlands correspondence. The restricted tensor product $\pi^p=\bigotimes_{v\nmid p}' \pi_v$ is a representation of $\mathrm{GL}_n(\mathbf{A}_{K,f}^p)$ (finite adeles away from $p$), and by multiplicity one theorems, is the away-from-$p$ part of at most one automorphic representation. Call it $\pi$, if it exists. 

Question: how does one "read off" the expected $\pi_\infty$ from $\rho$? In other words, how does one see the infinity-type of the expected automorphic representation on the Galois side? 

Surely $\rho_\infty\colon \mathrm{Gal}_{K_\infty}\to \mathrm{GL}_n(\overline{\mathbf{Q}_p})$ does not "know" $\pi_\infty$. For example, when $K=\mathbf{Q}$ and $n=2$, $\rho_\infty$ only knows the "parity" of $\pi_\infty$.

REPLY [5 votes]: Given $\pi$, Langlands' philosophy in its original form would predict the existence of a representation $\sigma$ from the (conjectural) global Langlands group of $\mathbf{Q}$ to $GL(n,\mathbf{C})$. For this representation there is "perfect local-global" -- the local behaviour of $\pi$ at a place $v$ (finite or infinite) should be related to the local behaviour of $\sigma$ at $v$ via the local Langlands correspondence.
As you've spotted though, things aren't quite the same with $\rho$ and $\pi$ because the symmetry is broken. $\pi$ is a representation over the complexes, and $\rho$ is a representation over the $p$-adics. So the first thing you have to do is to choose embeddings of $\overline{\mathbf{Q}}$ into $\mathbf{C}$ and $\overline{\mathbf{Q}}_p$. If you were working with a general number field you would now have set up some sort of weird correspondence between the places of the field above $p$ and the places above infinity, and there is some interplay in the local-global story with these places. 
You have base field $\mathbf{Q}$ so the interplay is between $p$ and $\infty$. Basically, the behaviour of $\pi$ at both $p$ and $\infty$ should be determined by the behaviour of $\rho$ at both $p$ and $\infty$, but as you have spotted it is not the case that the behaviour of one of these things at $p$ is determined by the other one, and the same for infinity; you have to treat both places at once.
The answer to your question is that the Langlands parameter for $\pi_\infty$ can be read off from the Hodge--Tate weights of $\rho$ plus the behaviour of complex conjugation. The infinitesimal character of $\pi_\infty$ is the Hodge--Tate cocharacter of $\rho$, for example. 
All the details can be found in
http://arxiv.org/abs/1009.0785<|endoftext|>
TITLE: Does a continuous map $f$ from the $n$-ball $B$ into $R^n$ such that $B\subset f(B)$ have a fixed point?
QUESTION [6 upvotes]: If $f$ is a continuous map from the $n$-ball $B$ into itself, the Brouwer fixed point theorem guarantees a fixed point. What if we assume that $f$ maps $B$ into all $R^n$, and $f(B)$ contains $B$? For the one-dimensional case it is still true that such $f$ has a fixed point. Does this property also hold for higher dimensions?

REPLY [5 votes]: (Really just a comment on Nik Weaver's answer, but too long for a comment)
The same idea with a different (more easily expressible) parametrization:  Use polar coordinates, and 
consider the "square"  $D:=\{(r,\psi):  \frac12 \le r \le 1, 0\le \psi \le \alpha\}$, with some small angle $\alpha$.

For $0\le\psi \le\frac14 \alpha$, map $(r,\psi)$ to $(r,\frac12\alpha + 2\psi)$, covering the region between $\frac12\alpha$ and $\alpha$. 
For $\frac34 \alpha\le\psi\le \alpha$, map $(r,\psi)$ to $(r,2*(\psi-\frac34\alpha))$, covering the region between $0$ and $\frac12\alpha$. 
For $\frac 14 \alpha\le \psi\le \frac34\alpha$, map $(r,\psi)$ to $(r, c(\psi-\frac14\alpha)+d)$, with $d:=\alpha$ and $c\cdot\frac12\alpha+d=2\pi$.<|endoftext|>
TITLE: Stinespring's dilation without $C^{\ast}$-algebras
QUESTION [5 upvotes]: Does Stinespring's dilation theorem hold if the algebra of interest is a topological $\ast$-algebra instead of the usual $C^{\ast}$-algebra?
I will now state the version of Stinespring's dilation theorem that I know, and am wondering what happens if I were to replace the $C^{\ast}$-algebra with a topological $\ast$-algebra.
Theorem. Let $\mathfrak{A}$ be a unital $C^{\ast}$-algebra, and let $\Phi : \mathfrak{A} \to \mathcal{B}(\mathcal{H})$ be a completely positive map. Then there exists a Hilbert space $\hat{\mathcal{H}}$, a unital $\ast$-homomorphism $\pi : \mathfrak{A} \to \mathcal{B}(\hat{\mathcal{H}})$ and a bounded operator $V : \mathcal{H} \to \hat{\mathcal{H}}$ with $||\Phi(1_{\mathfrak{A}})||=||V||^{2}$ such that 
\begin{equation}
\Phi (a)= V^{*} \pi (a) V, \; \; \; \; \;   a \in \mathfrak{A}.
\end{equation}
Notation. Let $1_{\mathfrak{A}}$ denote the unit in $\mathfrak{A}.$
References to literature on the subject would be greatly appreciated as well.

REPLY [8 votes]: The surprising fact is that the GNSStinespringKasparov theory is in fact completely algebraic, at least to a very large extend: the following results have been obtained by a PhD of mine but are, unfortunately, not published (yet?). 
So the setting is a *-algebra over a ring $C$ which is of the form $C = R(i)$ with $i^2 = -1$ and $R$ being an ordered ring. In this framework one can define positive functionals to be linear functionals $\omega\colon A \longrightarrow C$ satisfying $\omega(a^*a) \ge 0$. This gives also rise to the notion of a positive element by asking for $\omega(a) \ge 0$ for all positive $\omega$. One can then state what a positive linear map is (mapping positive elements to positive ones) and hence what $n$-positive and completely positive maps are. So far, the notion is entirely algebraic and, of course, reproduces the notions for a $C^*$-algebra you know.
Then one can proceed by constructiong pre-Hilbert modules over *-algebras. Without topology, we can of course not speak of completeness, but this "pre" is already interesting. The only catch is that one has to ask for completely positive inner products: for a Hilbert module over a $C^*$-algebra, this is a consequence of positivity, but in this algebraic setting, there seems to be no way to prove that (hint very welcome!).
From there on, one can proceed with many things like a strong Morita theory for *-algebras, etc. Among them is the algebraic part of the Stinespring construction: if you go through it, then we will see that most of it is algebraic and can be transfered to this setting.
The name of the PhD student is Florian Becher and you can find his thesis somewhere on the document server of Freiburg university.
On my homepage you find a lot of ref's on the algebraic *-representation theory of *-algebras, Morita theory and things, including a (not yet finished) lecture note on this.<|endoftext|>
TITLE: Posets isomorphic to their endomorphism poset
QUESTION [15 upvotes]: Let $(P,\leq)$ be a poset. We set $$\text{End}(P)=\{f: P\to P: f\text{ is order-preserving}\}$$ and order $\text{End}(P)$ pointwise.
Is there a poset with more than 1 point such that $P\cong \text{End}(P)$?

REPLY [5 votes]: The answer to 

Is there a poset with more than 1 point such that $P\cong\text{End}(P)$?

is No. This follows immediately from Theorem 3 in

Roy O. Davies, Allan Hayes and George Rousseau, Complete lattices and the generalized Cantor theorem, Proc. Amer. Math. Soc. 27 (1971), 253–258.

This reference was pointed out to me by user bof in a (now deleted) comment to this post. More precisely bof gave a link to this answer of them.<|endoftext|>
TITLE: Natural isomorphisms: what is the status now of "the Eilenberg/Mac Lane Thesis"?
QUESTION [13 upvotes]: I asked this some days ago over at math.se, and while the question got 10 upvotes, I didn't get too many answers. Although it is a "soft question", maybe the general issue is interesting enough to raise again here in the hope of more answers. 
The Church/Turing Thesis that we all know and love asserts that every algorithmically computable function (in an informally characterized sense) is in fact recursive/Turing computable/lambda computable. A certain intuitive concept, the Thesis claims, in fact picks out the same functions as certain (provably equivalent) sharply defined concepts. 
Evidence? Two sorts: (1) "quasi-empirical", i.e. no unarguable clear exceptions have been found, (2) conceptual, as in for example Turing's own efforts to show that when we reflect on what we mean by algorithmic computation we get down to the sort of operations that a Turing machine can emulate.
OK, now compare. The "Eilenberg/Mac Lane Thesis" in one version (but does anyone call it that??) is that if an isomorphism between widgets and wombats is intuitively "natural" (i.e. doesn't depend on arbitrary choices of co-ordinates, or the like) then it can be regimented as an natural isomorphism between suitable functors in the official category-theoretic sense. A certain intuitive concept, the Thesis claims, in fact picks out the same isomorphisms as a certain sharply defined concept.
Evidence? We'd expect two sorts. (1*) "quasi-empirical", i.e. no clear exceptions. (2*) conceptual ...
Two main questions:
(A) But are there no exceptions? Or (to take the perhaps more likely direction of failure) are there well known cases where we can say "Hey, this is the sort of isomorphism whose intuitive naturalness was surely of the kind that Eilenberg/Mac Lane were trying to chraracterize, back in the day: but actually, you can't shoehorn this case into the framework of their theory of natural isomorphisms."
(B) Assuming the Thesis isn't defeated by counter-example (or is true for some nice class of cases), what are the best efforts at trying to show that, conceptually, it "ought" to be true (when it is)?
And then I suppose that there is a supplementary question arising of category theoretic etiquette:
(C) Assuming again the Thesis isn't obviously defeated by counter-example, is it secure enough for it now to be acceptable to slide without further argument from showing that there is an isomorphism between widgets and wombats constructed in an intuitively "natural", unarbitrary, way to announcing that there is therefore a natural isomorphism here?

REPLY [5 votes]: There is probably no definite answer to this question. But let me propose a few ideas:
1) this kind of general "thesis" only makes sense for naturality with respect to isomorphisms:
There is plenty of examples of perfectly natural and canonical construction (in a naive sense) that are not functorial with respect to arbitrary morphisms:
The center of a group or an algebra, the group of automorphism of some object etc... are example of construction that are functorial only with respect to isomorphisms.
For example of a natural transformation which is only functorial on isomorphims not an isomorphism, consider the category $P$ of partially ordered set and monotone map between them. Consider the functors, $F:P \rightarrow sets$ the functor that forget the ordering, and $B: P \rightarrow Set$ that send any poset to $\{0,1\}$ you can then define a map from $F(p) \rightarrow B(p) = \{0,1\}$ for any posets $p$ as the map which send any element to $1$ if it is a maximal element and $0$ otherwise. It is not functorial with respect to arbitrary morphism but is functorial with respect to isomorphisms.
2) I think there is actually reasonable hopes to be able to give a formal statement and a prof of the isomorphisms part of the thesis: one can imagine purely categorical (and somehow tautological) argument for that of course, or just the philosophical stand that "if a construction does not depends on any choice it cannot distinguishes isomorphic structures", even better I like to take "functorial with respect to isomorphism" as the definition of "canonical".
But what I had in mind here is the Univalence axiom and the developement of category theory with homotopy type theory as it is developed in the Hott book, which I think goes a lot further along those ideas:
To sum things up, in this formalism every category is 'forced'* to satisfied a version of the univalence axiom (for any two object of the category the type of isomorphism between them is equal to the type of equality between two such object). For such categories one has the functoriality along isomorphism of any possible construction (because it boils down to functoriality along equality which is tautological...). So in some sense one can say that any construction that is natural enough to be performed within a version of type theory consistant with univalence should satisfies this Eilenberg-MacLane thesis with respect to isomorphism.
Of course it is not currently well understood to what extent one can develop ordinary mathematics in a way compatible to univalence but it is hoped that this is possible, and that would be in my opinion one of the best possible answer to this question.
*: by 'forced' I mean that they only call categories those which satisfies this univalence axiom. Classical construction on categories generally preserve this property and most usual categories satisfies it. Moreover any category can be replaced by an equivalent categories satisfying this axioms by looking at the essential image of its Yoneda embeddings.<|endoftext|>
TITLE: Measurability and Axiom of choice
QUESTION [38 upvotes]: In some situations, you need to show Lebesgue-measurability of some function on $\mathbb{R}^n$ and the verification is kind of lengthy and annoying, and even more so because measurability is "obvious" because "why would it not be".
In such a situation, I have heard the argument: The function is clearly measurable, because the axiom of choice was not used to define it.
This argument makes some sense, because (as far as I know, I am not an expert) there are models of ZF (no C), where every function on $\mathbb{R}^n$ is Lebesgue measurable. So, suppose that we have a function $f$ in our model of ZFC constructed without the axiom of choice, then it is also a function in the model of ZF constructed above. Hence it is measurable there, and "clearly" all functions that are measurable in the model above are also measurable in our model.
But the question is: The bold statement above is very "meta". So how rigorous is this argument? Can it be made rigorous?
/Edit: I changed "Borel"- to "Lebesgue"-Measurable.

REPLY [43 votes]: The bold statement is not true in the generality in which you
state it. Nevertheless, something very like it is true, if one
adopts the perspective and philosophy of large cardinal set theory
and restricts the kinds of definitions that are considered.
First, let's get a little more clear on what you mean. One does
not formally use axioms at all in a definition, but rather in a
proof. To define an object means to provide a statement
$\varphi(x)$ that one and only one object satisfies. What one
means by not using an axiom in a definition, is that one can
prove, without using that axiom, that there is such a unique
object fulfilling the definition. Perhaps one has in mind a
constructive procedure, but this is really just a sequence of such
definitions, and such a construction does not use the axiom of
choice, if at every step of the construction, the definition used
at that step is a definition in any model of ZF.
One can easily make a counterexample, now, by the definition: let
$f$ be the characteristic function of the least non-measurable set
of reals in the constructible universe $L$, using the canonical 
definable well-ordering of $L$.
This definition does not use the axiom of choice, since it is
sensible as a definition in any model of ZF, and picks out a
unique function on the reals in any model of ZF. But it is not
necessarily true in ZF that this function is measurable, since if
the axiom of constructibility holds, that is, if we are living in
$L$, then $f$ is definitely non-measurable. Meanwhile, it is
consistent with ZFC that the set of all reals in $L$ is countable
in $V$, and in this case, the function $f$ is the characteristic
of a countable set, and hence measurable in $V$. So the
definition, which did not use the axiom of choice, sometimes
defines a measurable function and sometimes does not, in the
various ZF worlds.
Let's give another concrete counterexample. The canonical
well-ordering of the reals in the constructible universe $L$, mentioned by Andres,
is a definable subset of the real plane $A\subset\mathbb{R}^2$,
which in $L$ has complexity $\Delta^1_2$ in the descriptive-set-theoretic
projective hierarchy. Thus, in our current universe $V$, the set $A$ has complexity at worst $\Sigma^1_2$, and so it arises from a certain definable
closed subset of $\mathbb{R}^4$ by projecting onto $\mathbb{R}^3$,
taking the complement, and then projecting to $\mathbb{R}^2$. So $A$ is definable in a highly concrete manner, without
making any use of the axiom of choice. Nevertheless, it is not
necessarily true that the resulting set is measurable, since
inside the constructible universe itself, the resulting set is not
measurable; in contrast, it is also consistent with ZF that there
are only countably many constructible reals, and in this case the
set $A$ would be countable and hence measurable. So the
measurability of the set $A$ is not determined, despite the simple
definition.
At the end of your post, you seem to suggest that, ("clearly") if
a definition defines a measurable set in some model of ZF, then it
defines a measurable set (in our current ZFC universe). But this
is not quite right. One can write down a definition $\varphi(x)$
that ZF proves defines a unique set of reals, but the set of reals
defined is measurable in an inner model and non-measurable in a
larger model.
Lastly, let me explain the sense in which your bold statement is
on the right track. One of the truly surprising and remarkable
discoveries of large cardinal set theory is that the existence of
large cardinals has effects on fundamental mathematical truth at
the level of descriptive set theory. In particular, the existence
of sufficient large cardinals implies that every projectively
definable set of reals is Lebesgue measurable. If there is a
supercompact cardinal, and much less suffices, as explained in the
article Saharon Shelah, Hugh Woodin, Large Cardinals Imply That
Every Reasonably Definable Set of Reals Is Lebesgue Measurable,
Isreal Journal of Mathematics, vol. 70, (1990) pp. 381-394 (reviewed by J. Bagaria in BSL 8:4(2002) pp. 543-545, as linked to by Andres in the comments), then every
set of reals in $L(\mathbb{R})$ is Lebesgue measurable. The universe $L(\mathbb{R})$ consists of those sets that are constructible relative to reals.
So, if you assume large cardinals, and you define a set of reals
by a definition that is absolute to $L(\mathbb{R})$ — and
this is very likely the case if your definition works in ZF and
does not involve set theory explicitly — then your set is
Lebesgue measurable.
In particular, assuming that there are sufficient large cardinals, then every projective set of reals is Lebesgue measurable, and this may provide a soft sufficient criterion. The projective statements are those that can be expressed using quantifiers only over the reals and the integers, with the usual algebraic and order structure. Alternatively, the projective sets are those that you get by closing the Borel sets under continuous images and complements. 
Let me point out that this kind of consequence of large cardinals
is often pointed to by large cardinal set theorists as evidence
that the large cardinal axioms themselves are on the right track,
since they provide a such a rich, coherent and desirable structure
theory for our everyday mathematics. We infinitely prefer the
smooth and elegant descriptive set theory of large cardinals to
the awkward land of counterexamples provided by the axiom of
constructibility $V=L$.

REPLY [6 votes]: Analytic sets that are not Borel can be explicitly defined.  AC is not used in the definition.  It is only the proof that they are not Borel that requires AC.
However: Lebesgue measurable is much more like your description.<|endoftext|>
TITLE: Upper density of the set of $n$'s such that $p(n)$ is prime, where $p$ is polynomial
QUESTION [7 upvotes]: The starting point for this question is the following (false) statement


$\forall n\in \mathbb{N} (n^2 + n + 41 \text{ is prime}).$


Given a polynomial function $p:\mathbb{N} \to \mathbb{N}$ we define $$\text{prime}(p) = \{n\in\mathbb{N}: p(n) \text{ is prime}\}.$$
For $A\subseteq \mathbb{N}$ we define the upper density by $$\operatorname{ud}(A)=\limsup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$
Is there a non-constant polynomial $p$ such that $\operatorname{ud}(\text{prime}(p)) > 0$? 
If yes, what is $\sup\{\operatorname{ud}(\text{prime}(p)): p:\mathbb{N}\to\mathbb{N} \text{ is a non-constant polynomial}\}$?

REPLY [8 votes]: This is a supplement to Eric Naslund excellent answer, and also an elaboration of Terry Tao's comment above. The crucial asymptotic formula
$$\frac{1}{\pi(x)}\sum_{p\leq x}v_{F}(p)=1+o(1)$$
also follows from two 19th century results: the Frobenius theorem on splitting types and the Cauchy-Frobenius orbit counting lemma (also known as Burnside's lemma). 
Indeed, consider the Galois group $G$ of the splitting field of $F$ over $\mathbb{Q}$ as a permutation group acting on the roots of $F$. The Frobenius theorem implies that the left hand side is asymtotically the average number of fixed points of $G$, which by the Cauchy-Frobenius formula equals the number of orbits of $G$. However, $G$ acts transitively, so the number of orbits equals $1$.
Added 1. As Vesselin Dimitrov pointed out, the above asymptotic formula also follows from Landau's prime ideal theorem (published in 1903).
Added 2. I just learned from the paper Stevenhagen-Lenstra: Chebotarev and his Density Theorem that the above asymptotic formula was originally shown by Kronecker (published in 1880), and this formed the basis of the quoted work of Frobenius of more group theoretic flavor.<|endoftext|>
TITLE: Metrics on Teichmüller spaces
QUESTION [5 upvotes]: I know that Teichmüller $\mathcal{T}_g$ spaces support different metrics. One of them is the Bergman metric; which is a particular case of the Bergman metric on any domain of holomorphy. On the other hand $\mathcal{T}_g$ is a Stein manifold by Bers' work. So $\mathcal{T}_g$ is a domain of holomorphy and support the Kähler structure induce by $\mathbb{C}^{3g-3}$ since there exists an holomorphic embedding of  $\mathcal{T}_g$ in $\mathbb{C}^{3g-3}$.
Now my question is: Is the Bergman metric equivalent to the Kähler metric induce by $\mathbb{C}^{3g-3}$ If so how can I prove it? Otherwise how can I prove the contrary?

REPLY [6 votes]: These two metrics are not equivalent to each other:

The Bergman metric $d_B$ on $T_g$ is complete: this result is essentially due to Earle (who proved that Caratheodori metric $d_C$ on $T_g$ is complete, while $d_C$ is bounded from above by $d_B$), a "better" proof is due to B.-Y. Chen (2004) who proved that $d_B$ is equivalent to the Teichmuller metric on $T_g$ and the latter is complete. 
This proof is better since it provides a bilipschitz model for $d_B$. 
If you use the Bers embedding $T_g\to {\mathbb C}^{3g-3}$ then the pull-back metric from ${\mathbb C}^{3g-3}$ is incomplete.<|endoftext|>
TITLE: positive not completely positive maps
QUESTION [16 upvotes]: In extension to this question 
Positive but not completely positive?
I'd like to know, for $k>1$, examples of $k$-positive linear maps of a matrix algebra into itself that are not $k+1$-positive. (I don't know a single one.) By M.D. Choi's theorem
the size of the matrices involved must grow with $k$, but how fast?

REPLY [16 votes]: The "canonical" example of a map that is $k$-positive but not $(k+1)$-positive is the map defined by
$$
\Phi_k(X) = k\cdot\mathrm{Tr}(X)I_n - X.
$$
Above, $n$ denotes the size of $X$ (i.e., $X \in M_n$) and $I_n$ is the $n \times n$ identity matrix. This map was introduced in "J. Tomiyama. On the geometry of positive maps in matrix algebras II. Linear Algebra Appl., 69:169–177, 1985." (though it had been considered by Choi earlier in the special case when $k = n-1$).
It's not difficult to prove that this map is $k$-positive just by elementary linear algebra. I won't post details here since it's a bit long and messy, but it's almost exactly the same as the proof that starts at the bottom of page 4 of these notes.
To show that this map is not $(k+1)$-positive (when $k < n$), simply let $\mathbf{v} = \sum_{i=1}^{k+1} \mathbf{e}_i \otimes \mathbf{e}_i \in \mathbb{C}^{k+1} \otimes \mathbb{C}^n$ (here $\{\mathbf{e}_i\}$ is the standard basis of $\mathbb{C}^{k+1}$ on the first subsystem, and it's just embedded into $\mathbb{C}^n$ in the natural way on the second subsystem). Then compute that
$$
(id_{k+1} \otimes \Phi_k)(\mathbf{v}\mathbf{v}^*) = k(I_{k+1} \otimes I_n) - \mathbf{v}\mathbf{v}^*,
$$
which has $-1$ as an eigenvalue and is thus not positive.

A couple of side notes:

This map shows that Choi's theorem on complete positivity is optimal in some sense: if $k \geq n$ then $k$-positivity implies complete positivity, but if $k < n$ then $k$-positivity and $(k+1)$-positivity are indeed different sets.
In the $k = 1$ case this map comes up frequently in quantum information theory under the name of the "reduction map".<|endoftext|>
TITLE: Representations of the unit group in a ring of integers
QUESTION [7 upvotes]: Let $K/\mathbb{Q}$ be a finite extension of degree $d > 1$. Suppose that $\omega_1, \cdots, \omega_d$ is a basis for $K$ over $\mathbb{Q}$. Further, we assume that $\omega_1, \cdots, \omega_d \in \mathcal{O}_K$, the ring of integers of $K$, and that it is an integral basis for $\mathcal{O}_K$. 
Recall that a unit in $\mathcal{O}_K$ is an invertible element in $\mathcal{O}_K$, or equivalently, an element of norm $\pm 1$ in $\mathcal{O}_K$. By the norm we mean the function
$$\displaystyle N_{K/\mathbb{Q}}(u) = \prod_{j=1}^d \sigma_j(u),$$
where $\sigma_1, \cdots, \sigma_d$ are the distinct embeddings of $K$ into $\mathbb{C}$.
It is then easy to see that if $\alpha \in \mathcal{O}_K^\ast$, then for all $u \in \mathcal{O}_K$ we have
$$\displaystyle N_{K/\mathbb{Q}}(\alpha u ) = N_{K/\mathbb{Q}}(u).$$
If we consider the polynomial
$$\displaystyle N(x_1, \cdots, x_n) = N_{K/\mathbb{Q}}(\omega_1 x_1 + \cdots + \omega_n x_n)$$
for $n \leq d$, then it is clear that on writing $\mathbf{x} = (x_1, \cdots, x_n)$ and $A$ for the linear operator induced by multiplication by $\alpha \in \mathcal{O}_K^\ast$ that
$$\displaystyle N(A \mathbf{x}) = N(\mathbf{x}).$$
This equation only makes sense in the vector space $K/\mathbb{Q}$, because any unit in $\mathcal{O}_K^\ast$ can be realized as a linear endomorphism of $K \cong \mathbb{Q}^d$. However, we want to make sure that $A$ is defined over the linear span of $\{\omega_1, \cdots, \omega_n\}$, say $S$. Further, if $n < d$, we do not want $\{\omega_1, \cdots, \omega_n\}$ to all lie in a proper subfield of $K$. Indeed, we want $K = \mathbb{Q}(\omega_1, \cdots, \omega_n)$. 
Hence, our goal is to find a subgroup of positive rank of the unit group $\mathcal{O}_K^\ast$ which can be realized as a subgroup of $\operatorname{GL}(S)$. Does anyone know how to go about this problem?

REPLY [3 votes]: Notation: I'll use $S$ to denote the $\mathbb{Z}$-linear span of $\lbrace \omega_1, \ldots, \omega_n\rbrace$. (In the question, the $\mathbb{Q}$-span may have been meant.)
When you allow yourself to change the representation, and look for arbitrary homomorphisms from the abstract finitely generated abelian group $\mathcal{O}_K^\ast$ to $\mathrm{GL}(S)$, not much can be said, because too much can happen. For example, pick one non-torsion generator and map it to any invertible matrix, and map the other generators to the identity. (Here, $\mathrm{GL}(S)$ would just mean the group of invertible endomorphisms of the free abelian group $S$, an additive group which isn't a ring in general - so we can't even exploit the multiplication in $K$ to constrain such homomorphisms.)
So let us limit ourselves to multiplication by units. For simplicity, let me also assume $\omega_1=1$, and let us let a subgroup $G \le \mathcal{O}_K^\ast$ act.
In order for this to work, $S$ must contain at least the image of $\omega_1$, which is just $G$ again, and thus it must contain the additive group spanned by $G$, which is a ring - an image of the group ring $\mathbb{Z}[G]$. Its quotient field $L$ is just the $\mathbb{Q}$-linear span of $G$ inside the additive group of $K$ (because denominators can be replaced by their norms to make them rational integers). If $S$ is of smaller rank than $\mathcal{O}_K$, then $L$, being contained in the $\mathbb{Q}$-span of $S$, must be a proper subfield of $K$, and then $G$ must be contained in the units $\mathcal{O}_L^\ast$ of $L$. This will now give you a lot of examples, by taking $S$ no larger than necessary, but they may not be very interesting.
When $G$ is all of $\mathcal{O}_K^\ast$, the above means that all the units in $K$ come from a proper subfield $L$, and by Dirichlet's unit theorem this can only happen when $L$ is totally real and $K$ is a quadratic totally complex extension (a CM-extension) of $L$ with no roots of unity other than $\pm1$.<|endoftext|>
TITLE: Euler characteristic of a curve
QUESTION [14 upvotes]: Is it a complete coincidence that the Euler characteristic of a curve is exactly twice the Euler characteristic of its structure sheaf?

REPLY [7 votes]: The answers that this follows from Hodge theory and HRR are of course spot on.  If what is meant is rather why those theorems are true, here is a short proof at least of HRR for curves:
If two smooth plane curves have the same degree, hence the same topology, they are linearly equivalent, so by the exact sheaf sequence (of a divisor on a surface) also have the same chi(O).  So chi(O) is a topological invariant for smooth plane curves.  But why does it equal 1-genus?
In degree one, both (1-genus) and chi(O) are equal to 1.  Moreover when you add a general line L to raise the degree of a curve C from n to n+1, the topological genus of the (smoothed) curve goes up by n-1, so (1-genus) goes down by n-1.  By the exact sequence relating the structure sheaf of the sum C+L of these divisors, to the sum of their structure sheaves, chi(O) also goes down by n-1.  
Thus both invariants are the same for degree one curves and change the same way when we raise degree, hence they agree.  But all curves are at least nodal plane curves, and we can modify the argument to allow for the nodes.
As to why there exist g holomorphic differentials, by taking real and imaginary parts this is equivalent to why there are 2g harmonic differentials.  I.e. we are trying to understand why Hodge’s theorem is true.  As mentioned, the maximum modulus principle implies a non zero harmonic form must have non zero period over some loop, and a harmonic form is uniquely determined by its periods over all homology cycles.  Thus existence of harmonic forms means finding ones with prescribed periods.  This is done in two steps.  The first step is to find a smooth form with given periods.  This is done by putting a collar around a non separating loop, forming a smooth function that climbs from 0 to 1 when crossing the collar, setting it equal to zero elsewhere, and taking d of it.  Thus we see that topology does tell us how many smooth closed forms we can construct that are cohomologically independent.  Then the deep part, i.e. the Hodge theory, says that each such smooth closed form can be written as a sum of a harmonic form and an exact form.  For details see Springer chapters 7 and 8 page 207, lemma 8-1, especially using Weyl’s smoothing lemma 7.3 p. 199.
To “see” why this occurs, one may view the flow diagrams in Springer’s chapter 1 (fig.1.27. p.27), taken from Klein’s lectures on Riemann surfaces.  I.e. it seems a differential form, or covector field, is determined by its flow lines, and when these flow lines are of minimal length in some sense, then they satisfy the Laplace equation and hence are harmonic.  So the intuitively plausible fact that these lines may be allowed to deform within a cohomology class until they are minimal (Dirichlet principle), suggests the existence of a harmonic form in each cohomology class (???  I just made this up, but maybe some expert will help out here.)
It seems Riemann reduced this problem to the “Dirichlet problem” of existence of a harmonic function with prescribed boundary conditions, on the polygon of 4g sides obtained by dissecting the genus g Riemann surface by 2g transverse cuts.  One constructs a harmonic function f which agrees on opposite pairs of sides except on one pair where it differs by a ≠ 0 constant on the two identified edges.  Then taking df gives a harmonic differential with a non trivial period.  Since there are 2g choices of such paired sides, and a harmonic form is determined by its periods, one can construct exactly 2g independent harmonic differentials this way, hence g holomorphic differentials.<|endoftext|>
TITLE: Properties of loop space functor from homotopy types to group objects in homotopy types
QUESTION [5 upvotes]: I am trying to understand some properties of categories enriched in homotopy types, and the following question has become important:
When we take the loop-space of a (connected) homotopy type, we get a homotopy type group object (since as a suspension object, the circle is canonically a cogroup object).  When we forget the group structure, we are left with the normal loop-space endofunctor, and when we restrict attention to the group of components of the loop-space, this is the fundamental group.  I am wondering about when we retain all of this information.  What properties does the resulting functor to the category of group objects in the homotopy category have (where morphisms are morphisms of homotopy types which respect the group structure)?
Is it full?  Is it faithful?  Is it essentially surjective (this is a very interesting question on its own for me -- does every homotopy type group object arise from a loop space)?  Is it a right adjoint (my intuition tells me yes)?
More context: there is an obvious composable pair of right adjoints $\text{Set}\rightarrow\text{Grpd}\rightarrow\text{Cat}$ whose left adjoints turn a category into a groupoid by freely inverting all arrows (quotients become necessary in the case of idempotent arrows; I think of this as the categorification of the Grothendieck group operation) and turn a groupoid into its set of components.  I would like to find an analogous sequence of functors when $\text{Set}$, $\text{Grpd}$, and $\text{Cat}$ are replaced by $\text{HoSSet}$, $\text{HoSSet-Grpd}$, and $\text{HoSSet-Cat}$ respectively.
One more point: the functor sending a groupoid to its components induces a functor $\text{Grpd-Cat}\rightarrow\text{Set-Cat}\cong\text{Cat}$.  When we use its closed structure to think of $\text{Grpd}$ as an object of $\text{Grpd-Cat}$, we see that this functor sends the $\text{Grpd}$-category $\text{Grpd}$ to a category equivalent to the category of sets (forgive me for playing fast and loose with size issues; insert "small" where appropriate).  I wonder if there is a functor which does something similar in the case of $\text{HoSSet}$-categories.

REPLY [12 votes]: Group objects in the homotopy category are the wrong thing to look at, and the reason is precisely that this is not enough structure to construct a classifying space. The things that have classifying spaces are topological groups or, more generally, $E_1$ / $A_{\infty}$ spaces. These are higher categorical and cannot be defined solely in terms of the homotopy category. 
With this modification, the loop space functor establishes an equivalence of higher categories between pointed connected spaces (by which I mean weak homotopy types) and grouplike $E_1$ spaces, with the inverse given by taking classifying spaces. In particular, every grouplike $E_1$ space is a loop space; this is the recognition principle for loop spaces. 
Regarding why group objects in the homotopy category are the wrong thing to look at, I don't have a counterexample off the top of my head on the level of objects, but here's a counterexample on the level of morphisms. Let $G$ be a group and let $A$ be an abelian group. Homotopy classes of pointed maps from $BG$ (the Eilenberg-MacLane space $K(G, 1)$) to $B^2 A$ (the Eilenberg-MacLane space $K(A, 2)$) are classified by second group cohomology $H^2(G, A)$. After taking loop spaces, whatever homotopy classes of maps from the group object $G$ to the group object $BA$ is, it's a subset of homotopy classes of maps from $G$ to $BA$. But since $G$ is a discrete space and $BA$ is connected, there is a unique such homotopy class. 
The problem is that the data of a map of $E_1$ / $A_{\infty}$ spaces from $G$ to $BA$ involves more information than the data of a map between their underlying spaces: that is, "preserving $E_1$ / $A_{\infty}$ structure" is itself a structure, not a property.<|endoftext|>
TITLE: Is there an $\infty$ version of the Wasserstein distance between two distributions?
QUESTION [8 upvotes]: If I have two probability distributions $\mu$ and $\nu$ defined on $X$ and $Y$ respectively, then the $p$-th Wasserstein distance between the two of them is defined as $$W_p(\mu,\nu) = \left(\inf_{\pi\in\Pi(\mu,\nu)}\int_{X\times Y}d(x,y)^{p}\,\mathrm{d}\pi(x,y)\right)^{1/p}
 $$
where $\Pi(\mu,\nu)$ is the set of all distributions on $X\times Y$ whose marginals are $\mu$ and $\nu$ respectively and $p\geq 1$.  My question is:  is the limiting behavior of this quantity as $p\to\infty$ defined?  Does it have a name?

REPLY [5 votes]: I see no reason why $W_\infty := \inf_{\pi\in\Pi}||d||_{L^\infty(\pi)}$ wouldn't fit. This is not to claim that $\lim_{p\to\infty} W_p=W_\infty$ (which I believe is true anyway), but at least it exists, although with properties different from the $W_p$s, $p<\infty$.<|endoftext|>
TITLE: Generalized geometries
QUESTION [9 upvotes]: Let $S$ be a non-empty set. A geometry of type $n$ for $n\geq 1$
on $S$ (consisting of at least $n$ elements) is a set ${\mathfrak P}\subseteq 
{\mathcal P}(S)$ such that

all members of $\mathfrak P$ have at least $n$ elements,
any $n$ elements of $S$ are contained in exactly one member of $\mathfrak P$, 
for $l_1\neq l_2 \in \mathfrak P$ we have $|l_1\cap l_2| = n-1$, and
there is $T\subseteq S$ with $|T|=n+1$ and $T\notin \mathfrak P$.

Geometries of type $1$ are "traditional" partitions -- they define
an equivalence relation on the set $S$.
A geometry of type $2$ is a projective plane.
Question: Is there for every $n\geq 1$ a geometry $\mathfrak P$ of type $n$ on $\omega$ such that $|\mathfrak P| \geq 2$?

REPLY [2 votes]: Can we construct a free such object? 
Step 1: Start with $n+2$ elements.
Step 2: For every set with $n$ elements, put it in $\mathfrak B$.
Step 3: For each pair of sets in $\mathfrak B$ that intersect in $n-k$ elements, add $k-1$ new elements and add those elements to those sets and those two sets.
Step 4: For each set of $n$ elements that is not already contained in a set in $\mathfrak B$, put it in $\mathfrak B$.
Repeat the last two steps infinitely. Axiom 1 is always satisfied, cause each set is created with $n$ elements. The uniqueness part of Axiom 2 is always satisfied, because we never add a set of $n$ elements to $\mathfrak B$ if they are already in a set in $\mathfrak B$. For Axiom 3, $|l_1 \cap l_2| \leq n-1$ is always satisfied, because new sets are created with $n$ elements that do not all intersect any other set, and because we only add new elements up to that limit. Axiom 4 is always satisfied by the orginal set of $n+2$ elements.
The existence part of Axiom 2 is always satisfied after even-numbered steps, and $|l_1 \cap l_2| \geq n-1$ is always satisfied after odd-numbered steps. So both are satisfied in the limit.
Combining these we get Axioms 1, 2, 3, and 4.<|endoftext|>
TITLE: References on quaternionic geometry
QUESTION [10 upvotes]: Is there any analog, in the quaternionic setting, of Kahler potentials?
In particular I'm interested in a similar construction of the Fubini-Study 2-form on $\mathbb{P}^1(\mathbb{C})$ over the quaternionic projective line $\mathbb{P}^1 (\mathbb{H}).$
In the complex setting it is well known that $\omega_{\textrm{FS}}=\frac{i}{2} \partial \overline{\partial} \log (|z|^2).$ Is there a natural extension of this form and of this potential over $\mathbb{P}^1 (\mathbb{H})$?
That is, can the standard 4-form $\Omega=\omega_i \wedge \omega_i + \omega_j \wedge \omega_j + \omega_k \wedge \omega_k$ on $\mathbb{P}^1(\mathbb{H})$ be written as some differential operator applied on some function (i.e. the potential)?
Answers, comments and references are welcome.

REPLY [4 votes]: Actually, a more direct answer to the original question explains why there is nothing like Kähler potential theory in the hypercomplex and quaternionic cases.
Recall what a Kähler potential does:  You start with a complex $n$-manifold, i.e., a $2n$-manifold $M^{2n}$ endowed with a torsion-free $\mathrm{GL}(n,\mathbb{C})$-structure $B$, represented by an integrable complex structure tensor $J:TM\to TM$.  Now, $\mathrm{U}(n)$ is a maximal compact in $\mathrm{GL}(n,\mathbb{C})$, and we can select a torsion-free $\mathrm{U}(n)$-structure $B'\subset B$ over an open set $U\subset M$ by choosing a smooth function $u\in C^\infty(U)$ such that $\omega = i\partial\bar\partial u$ is a positive $(1,1)$-form.  Locally, this is how all of the torsion-free (i.e., Kähler) structures subordinate to the $\mathrm{GL}(n,\mathbb{C})$-structure $B$ are described.  Of course, every torsion-free $\mathrm{U}(n)$-structure $B'$ on $M$ underlies a unique torsion-free $\mathrm{GL}(n,\mathbb{C})$-structure $B = B'{\cdot}\mathrm{GL}(n,\mathbb{C})$.  Now, the nice thing in this case is that, modulo diffeomorphism, locally at least, integrable $\mathrm{GL}(n,\mathbb{C})$-structures are all the same, i.e., they have no local geometry, whereas the torsion-free $\mathrm{U}(n)$-structures depend locally on $1$ function of $2n$-variables (which is represented exactly by the Kähler potential as above).
Consider what you would need to be true in either the hypercomplex case or the quaternionic case in order to have such an analogue:
First, consider the hypercomplex case:  Now, the group is $\mathrm{GL}(n,\mathbb{H})\subset \mathrm{GL}(4n,\mathbb{R})$, with maximal compact $\mathrm{Sp}(n)\subset \mathrm{GL}(n,\mathbb{H})$.  Thus, on a $4n$-manifold $M^{4n}$ you'd want a way to start with a torsion-free $\mathrm{GL}(n,\mathbb{H})$-structure $B$ on $M$ (i.e., a hypercomplex structure) and specify the torsion-free $\mathrm{Sp}(n)$-structures (i.e., hyperKähler structures) that it contains.  As in the complex/Kähler case, the reverse operation is automatic:  Each torsion-free $\mathrm{Sp}(n)$-structure $B'$ lies in a unique torsion-free 
$\mathrm{GL}(n,\mathbb{H})$-structure $B= B'{\cdot}\mathrm{GL}(n,\mathbb{H})$.  However, you immediately see the problem:  Modulo diffeomorphism, the torsion-free $\mathrm{GL}(n,\mathbb{H})$-structures in dimension $4n$ depend on $4n^2$ functions of $2n{+}1$ variables, while the the torsion-free $\mathrm{Sp}(n)$-structures in dimension $4n$ depend on only $2n$ functions of $2n{+}1$ variables, so the 'generic' torsion-free $\mathrm{GL}(n,\mathbb{H})$-structure in this dimension does not contain any torsion-free $\mathrm{Sp}(n)$-structures, so there can't be a potential theory like the complex/Kähler potential theory in this case.  You might think that you could get around this by considering the underlying $\mathrm{SL}(n,\mathbb{H})$-structures instead, but, modulo diffeomorphism, the torsion-free $\mathrm{SL}(n,\mathbb{H})$-structures in dimension $4n$ depend on $4n^2{-}2n$ functions of $2n{+}1$ variables, which is still too high (except when $n=1$, when $\mathrm{SL}(1,\mathbb{H}) = \mathrm{Sp}(1)$, so there is nothing to do).  (This is why they punt and go for HKT structures instead in this case.)
Second, in the quaternionic case, the group is the group is $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)\subset \mathrm{GL}(4n,\mathbb{R})$, with maximal compact $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)\subset \mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$.  Thus, on a $4n$-manifold $M^{4n}$ you'd want a way to start with a torsion-free $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structure $B$ on $M$ (i.e., a quaternionic structure) and specify the torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structures (i.e., quaternion-Kähler structures) that it contains.  As in the complex/Kähler case, the reverse operation is automatic:  Each torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structure $B'$ lies in a unique torsion-free 
$\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structure $B= B'{\cdot}\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$.  However, again, there's a problem:  Modulo diffeomorphism, the torsion-free $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structures in dimension $4n$ depend on $2n(2n{+}1)$ functions of $2n{+}1$ variables, while the the torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structures in dimension $4n$ depend on only $2n$ functions of $2n{+}1$ variables, so, again, the 'generic' torsion-free $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)$-structure in this dimension does not contain any torsion-free $\mathrm{Sp}(n){\cdot}\mathrm{Sp}(1)$-structures, so, again, there can't be a potential theory like the complex/Kähler potential theory in this case.
For these function counts for the generality of torsion-free structures modulo diffeomorphism, in case you are curious, you might consult my paper: R. Bryant - Classical, exceptional, and exotic holonomies: a status report, in Actes de la Table Ronde de Géométrie Différentielle en l’Honneur de Marcel Berger, Soc. Math. France, 1996, 93–166.<|endoftext|>
TITLE: Lifting torsors in characteristic $p$ to characteristic zero
QUESTION [6 upvotes]: Let $R$ be a local integral domain with residue field $k$ such that $R$ is of characteristic zero and $k$ is of characteristic $p>0$. Let $G$ be a smooth finite type affine group scheme with geometrically connected fibres over $R$. Suppose that $T$ is a $G_k$-torsor over $k$.
Does $T$ lift to $R$? 
That is,  does there exist a $G$-torsor $\mathcal T$ over $R$ such that $\mathcal T \otimes_R k$ is isomorphic to $T$ over $k$?
If so, can we then show the stronger assertion that the map on cohomology sets $H^1_{et}(R,G_R)\to H^1_{et}(k,G_k)$ is surjective for all complete local regular rings $R$ with residue field $k$?
Motivation: I'm trying to understand what it means for a variety over $k$ to lift to characteristic zero. I know curves, ppav's, K3 surfaces and hypersurfaces lift to characteristic zero. I also know examples of Fano varieties, Calabi-Yau threefolds and some surfaces of general type which don't lift to characteristic zero. Out of curiosity I wondered what one can say about liftability of torsors (under affine group schemes). I hope the comments and answers will shed some light on this matter for me.

REPLY [6 votes]: The answer is affirmative with $R$ any henselian local ring and $G$ any smooth $R$-group scheme; this is a special case of Lemma 11.4 in Grothendieck's article "Le Groupe de Brauer III: Examples et Complements" (whose proof simplifies quite a bit when $G$ is affine).  This is exactly the statement that $\rho_G:{\rm{H}}^1(R,G) \rightarrow {\rm{H}}^1(k, G)$ is surjective in such cases since such torsors split over finite etale covers of the base; strictly speaking that surjectivity result is for commutative $G$ (for which Grothendieck gives a result for higher cohomology too), but the method applies for H$^1$ without a commutativity assumption, as noted in Remark 11.8(3) of loc. cit.  (The key insight of Grothendieck is to prove smoothness for the differential from a "scheme of 0-cochains" to a "scheme of 1-cocycles".)  It also gives bijectivity.<|endoftext|>
TITLE: How to compute this $\mathrm{Ext}^1$?
QUESTION [5 upvotes]: Let $A$ be a regular local $\mathbb{C}$-algebra of dimension $2$, such as the localization of $\mathbb{C}[x,y]$ at $(x,y)$, and let $\nu=(\nu_1\geq\nu_2\geq\cdots\geq\nu_{\ell}\geq0)$, $\mu=(\mu_1\geq\mu_2\geq\cdots\geq\mu_m\geq0)$ be partitions of the same $n\in\mathbb{N}$.  Consider the ideals $I=(y^{\nu_1},xy^{\nu_2},x^2y^{\nu_3},\ldots,x^{\ell - 1}y^{\nu_\ell},x^\ell)$ and $J=(y^{\mu_1},xy^{\mu_2},x^2y^{\mu_3},\ldots,x^{m-1}y^{\mu_m},x^m)$ generated by elements in the maximal ideal (here $x,y$ are the uniformizers). So, $A/I$ and $A/J$ have the same length equal to $n$.

How to compute $\mathrm{Ext}^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

Edit: I would add that it's probably possible to make the computation using the theory of ADHM data developped e.g. in Nakajima's Lectures on Hilbert schemes of points on surfaces. But I would like to get a direct answer that does not use ADHM data and such.

REPLY [4 votes]: First off, Ext and localisation commute in situations like this (Weibel An introduction to homological algebra Proposition 3.3.10 p.76) so we can work with $A=k[x,y]$ if we like.  I will drop the assumption that the two partitions have the same size, not least because you get a nice interpretation of $\operatorname{Ext}^1$ when the first partition is (1): it has a basis indexed by addable nodes of the second partition.
Let 
$\lambda=\lambda_1^{a_1} \lambda_2^{a_2}\cdots$
be a partition of $n$, where
$\lambda_1>\lambda_2>\cdots> \lambda_l >0$.
A minimal $A$-projective resolution of the module you associate to $\lambda$ is
$$0 \to \bigoplus_{i=1}^l A \otimes (x^{a_i}\otimes x^{\sum_{j=0}^{i-1}a_j } y^{\lambda_i}-y^{\lambda_i-\lambda_{i+1}}  \otimes x^{\sum_{j=0}^i a_j}y^{\lambda_{i+1}})
\to \bigoplus _{i=0}^l A\otimes x^{\sum_{j=0}^i a_j}y^{\lambda_{i+1}}
\to A \to A/I_\lambda \to 0$$
in which you should interpret $\lambda_{l+1}$ as zero.
This leads to a direct but awkward description of $\operatorname{Ext}^1_A(A/I_\lambda, A/I_\mu)$ as follows.  Consider the Ferrers diagram of $\mu$ drawn in the French style: by this I mean the subset of $\mathbb{Z}\times \mathbb{Z}$ consisting of $(0,0), (0,1), \ldots, (0,m_1)$, $(1,0), (1,1), \ldots, (1,m_2)$ and so on, where $\mu=(m_1,m_2,\ldots)$.
Pick $0 \leq r \leq s \leq l$ and choose elements $\rho_s,\ldots,\rho_s$ of the Ferrers diagram of $\mu$ with the property that 
$\rho_r + (0,\lambda_r-\lambda_{r-1}) \notin \mu$, 
$\rho_j + (a_{j+1},0)=\rho_{j+1} + (0,\lambda_{j+1}-\lambda_j) \in \mu$ for each $r<j<s$, and
$\rho_s + (a_{s+1},0) \notin \mu$.
The first condition should be dropped if $r=0$ and the last if $s=l$.
Thus these elements form the concave corners in a fragment of the outer rim of the Ferrers diagram for $\lambda$.  The Ext-group is the free vector space on these "paths," modulo linear combintations of paths that arise from the entire outer rim of the Ferrers diagram for $\lambda$.
Things look a little nicer in the case $\lambda = (1)$.  Then a path with $r=s=0$ is an element of the Ferrers diagram of $\mu$ with nothing directly to its right, a path wtih $r=s=1$ is an element with nothing directly above it, and every path with $r=0,s=1$ is a coboundary.  Furthermore an $r=s=1$ path is a boundary if there's an element of the diagram to its left and a $r=s=0$ path is a boundary if there's an element of the diagram above it.  It follows the non-bounding paths sit next to the addable nodes of $\mu$.<|endoftext|>
TITLE: Torsors trivializing over a fixed finite etale cover
QUESTION [9 upvotes]: Let $S$ be an integral regular scheme and let $T\to S$ be a finite etale morphism. Let $G$ be a smooth affine finite type group scheme over $S$.
Is the set of $S$-isomorphism classes of $G$-torsors over $S$ which are trivial over $T$ finite?
I guess the set-up is ridiculously general. So what if 
1) $S$ is of finite type over $\mathbb C$, or
2) $S$ is of finite type $\mathbb F_p$, or
4) $S$ is of finite type over $\mathbb Z_p$, or
5) $S$ is of finite type over $\mathbb Q_p$, or
6) $S$ is of finite type over $\mathbb Z$?

REPLY [6 votes]: The answer to the general question is NO.
We take $S=\mathrm{Spec}\, K$, where $K$ is a number field.
Let $L/K$ be a finite Galois extension. We set $T=\mathrm{Spec}\, L$.
We consider the norm homomorphism
$$N_{L/K}\colon R_{L/K}\mathbb{G}_{m,L}\to \mathbb{G}_{m,K}\,,$$
where $R_{L/K}$ denotes the Weil restriction of scalars.
Set $G=\mathrm{ker}\, N_{L/K}$.
Then our torus $G$ splits over $L$, hence $H^1(L,G)=1$ by Theorem 90.
From the short exact sequence 
$$1\to G\to R_{L/K}\mathbb{G}_{m,L}\to \mathbb{G}_{m,K}\to 1$$
and the induced Galois cohomology exact sequence
$$ L^*\to K^* \to H^1(K,G)\to H^1(K,R_{L/K}\mathbb{G}_{m,L})=1$$
we obtain that $H^1(K,G)= K^*/ N_{L/K}(L^*)$, and this group is infinite.
Thus the kernel
$$\mathrm{ker}[H^1(K,G)\to H^1(L,G)]=H^1(K,G)= K^*/ N_{L/K}(L^*)$$
is infinite, as required.<|endoftext|>
TITLE: When the restriction of a derived functor to a subcategory is the derived functor of the restriction
QUESTION [9 upvotes]: Let $\mathcal{D},\mathcal{E}$ be abelian categories and $\mathcal{C}$ be a Serre subcategory of $\mathcal{D}$.
Let $D(\mathcal{C}), \, D(\mathcal{D})$ denote the derived categories of $\mathcal{C},\mathcal{D}$. 
Suppose that the natural functor between the derived category $D(\mathcal{C})$ and the category $D_C(\mathcal{D}):=\{X\in D(\mathcal{D}) \, | \, H^i(X) \in \mathcal{C}\}$ is an equivalence of categories. 
Let $F:\mathcal{D} \rightarrow \mathcal{E}$ be a right exact functor, that has derived functor $DF:D(\mathcal{D}) \to D(\mathcal{E})$. Does this imply that the restriction of $DF$ to $D(\mathcal{C})$ is the derived functor of the restriction of $F$ to $\mathcal{C}$?
The derived categories can bounded unbounded or semi-bounded, Whatever make the answer simpler.
A typical example is when $\mathcal D$  is the category of abelian groups and $\mathcal C$ is the category of finite abelian groups.

REPLY [15 votes]: In the example where $\mathcal{D}$ is the category of abelian groups and $\mathcal{C}$ is the category of finite abelian groups, take $F(X)=X\otimes_\mathbb{Z}\mathbb{Q}/\mathbb{Z}$. Then the restriction of $F$ to $\mathcal{C}$ is zero, and so has the zero functor as its derived functor. However, the derived functor of $F$ doesn't vanish on $\mathcal{C}$.<|endoftext|>
TITLE: Posets preserving stationary subsets of $\omega_1$ and no new $\text{cof}(\omega)$ ordinals, but without countable covering property
QUESTION [12 upvotes]: What are some examples of posets $\mathbb{P}$ which have the following properties?  It's OK if the definition uses large cardinals or some other hypothesis.

$\mathbb{P}$ preserves stationary subsets of $\omega_1$
$\mathbb{P}$ fails to have the $\omega$-covering property.  So in particular, $\mathbb{P}$ cannot be $\sigma$-distributive, or proper on any stationary set of countable models.
Whenever $\kappa$ has uncountable cofinality, then this remains true in $V^{\mathbb{P}}$.

Item (3) rules out Namba forcing,Prikry forcing, and the standard stationary tower forcings with critical point at least $\omega_2 $ (the latter preserve stationary subsets of $\omega_1 $ if the height of the tower is a Woodin cardinal, but change lots of regular cardinals to cof $\omega $).
Item (2)  rules out several forms of antichain sealing forcings (all variants of sealing forcings I know of are of the form "proper followed by shooting a club", which has the countable covering property).

REPLY [11 votes]: Let $(\kappa_n: n<\omega)$ be an increasing sequence of measurable cardinals with limit $\kappa,$ and for each $n$ let $U_n$ be a normal measure on $\kappa_n.$ Let $\mathbb{P}$ be the Prikry type forcing notion which adds one element Prikry to each $\kappa_n:$
$p\in \mathbb{P}$ iff $p=(s, A),$ where 
1)$s$ is a finite sequence,
2) $\forall i< |s|, s(i) < \kappa_i,$
3) $A=(A_n: |s| \leq n < \omega),$ and $A_n \in U_n$,
$p=(s,A)\leq q=(t, B)$ iff:
1) $s$ end extends $t$,
2) $\forall |t| \leq i < |s|, s(i)\in B_i,$
3) $\forall |s| \leq n < \omega, A_n \subseteq B_n.$
The forcing adds an $\omega$-sequence to $\kappa,$ and can not be covered by any ground model set of size $\omega$ (in fact of size $< \kappa$), it adds no bounded subsets to $\kappa,$ and does not change cofinalities. 
So the forcing is as requested.<|endoftext|>
TITLE: Is the set of the convolutions of two-point measures dense in the set of all measures?
QUESTION [15 upvotes]: A measure supported in two points is a measure of the form 
$$
\mu=\alpha\delta_a+(1-\alpha)\delta_b, 
$$
where $a<b$ and $\alpha\in (0,1)$. 
The question is: 
Given a finite non-negative measure $\sigma$ on the real line, do there exist, for every $N>0$, a real number $\lambda_N>0$ and a finite sequence of two-point measures $(\mu_{1,N},\dots,\mu_{N,N})$ such that the $\lambda_N$-rescaled convolution $(\mu_{1,N}*\dots*\mu_{N,N})(\lambda_N\cdot)$ converges in the weak* topology to $\sigma$?

REPLY [11 votes]: The answer is no. Without loss of generality (w.l.o.g.), the measures $\mu_{j,n}$ are probability measures (otherwise, multiply them by appropriate factors). 
Lemma 1 (probably well known). For each natural $n$, let $X_n$ and $Y_n$ be independent random variables (r.v.'s) such that $X_n+Y_n\to Z$ (in distribution), for some r.v. $Z$ with $|Z|\le1$. Let $y_n$ be a median of $Y_n$. 
Then $X_{n_k}+y_{n_k}\to X$, for some sequence $(n_k)$ of natural numbers and some r.v. $X$. 
Proof. W.l.o.g., $y_n=0$ for all $n$; otherwise, replace $Y_n$ by $Y_n-y_n$. Take any $b>1$ and any $\epsilon>0$. Then for all large enough $n$ one has 
$$\epsilon=\epsilon+P(Z\ge b)\ge P(X_n+Y_n\ge b)\ge P(X_n\ge b)P(Y_n\ge0)\ge P(X_n\ge b)/2, 
$$
whence $P(X_n\ge b)\le2\epsilon$. Similarly, $P(X_n\le a)\le2\epsilon$ for any $a<-1$ and all large enough $n$. So, the sequence of the distributions of the $X_n$'s is tight, and the lemma follows. 
Lemma 2. Let $X_n$, $Y_n$, $y_n$, and $Z$ be as in Lemma 1. Let $x_n$ be a median of $X_n$. Then, for some sequence $(n_k)$ of natural numbers and some independent r.v.'s $X$ and $Y$ such that $X+Y$ equals $Z$ in distribution, one has $X_{n_k}-x_{n_k}\to X$ and $Y_{n_k}+x_{n_k}\to Y$. 
Proof. By Lemma 1, w.l.o.g. $X_n+y_n\to\tilde X$ for some r.v. $\tilde X$ (or pass to a subsequence). Applying Lemma 1 again (now with $y_n$ in place of $Y_n$), w.l.o.g. one has $x_n+y_n\to z$ for some real $z$. It follows that $X_n-x_n\to\tilde X-z=:X$. Also, Lemma 1 implies that w.l.o.g. $Y_n+x_n\to Y$ for some r.v. $Y$. At that, w.l.o.g. the r.v.'s $X$ and $Y$ are independent. Hence, $X_n+Y_n=(X_n-x_n)+(Y_n+x_n)\to X+Y$. Comparing this with the condition $X_n+Y_n\to Z$, one concludes that $X+Y$ equals $Z$ in distribution. Thus, the proof of Lemma 2 is complete. 
Lemma 3. For any independent r.v.'s $X$ and $Y$, one has the following property of the support sets: $S_X+S_Y\subseteq S_{X+Y}$. 
Proof. Take any $z\in S_X+S_Y$, so that $z=x+y$ for some $x\in S_X$ and $y\in S_Y$. Take any real $\epsilon>0$. Then 
$$P(|(X+Y)-(x+y)|<\epsilon)\ge P(|X-x|<\epsilon/2,|Y-y|<\epsilon/2)
=P(|X-x|<\epsilon/2)P(|Y-y|<\epsilon/2)>0,$$
since $x\in S_X$ and $y\in S_Y$. So, $z=x+y\in S_{X+Y}$, for any $z\in S_X+S_Y$. Thus, Lemma 3 is proved. 
Lemma 4. The support set $S_\mu$ of any infinitely divisible distribution $\mu$ is either infinite or a singleton. 
Proof. I have not been able to find this statement in quite such a ready form in the literature (any hints?). The closest I have found is Theorem 2 (parts (1) and (2)) in https://projecteuclid.org/euclid.aop/1176995668 , from which Lemma 4 easily follows. However, it is even easier to prove Lemma 4 from scratch, based just on Lemma 3. Indeed, since $\mu$ is infinitely divisible, for each natural $n$ there is a unique probability measure $\mu_n$ such that $\mu_n^{*n}=\mu$. If $S_{\mu_n}$ is a singleton for some $n$, then $S_\mu$ is clearly a singleton. Otherwise, for each natural $n$ the set $S_{\mu_n}$ contains at least two points. On the other hand, by Lemma 3, 
$S_\mu\supseteq\underbrace{S_{\mu_n}+\dots+S_{\mu_n}}_n$. So (by induction and, say, Exercise 4 in http://goo.gl/lNPZnX), the cardinality of $S_\mu$ is no less than $n+1$, for any natural $n$. Thus, Lemma 4 is proved. 
Let now the support of $Z$ be the set $\{-1,1/2,1\}$. For each natural $n$, let $X_{n,1},\dots,X_{n,n}$ be independent r.v.'s such that $P(X_{n,j}=b_{n,j})=p_{n,j}:=1-q_{n,j}=1-P(X_{n,j}=a_{n,j})$, for some $a_{n,j}$ and $b_{n,j}$ such that $a_{n,j}<b_{n,j}$ and some $q_{n,j}\in(0,1)$. Suppose that $X_{n,1}+\dots+X_{n,n}\to Z$; that is, $\mu_{n,1}*\dots*\mu_{n,n}\to\sigma$, where $\mu_{n,1},\dots,\mu_{n,n},\sigma$ are the distributions of the r.v.'s $X_{n,1},\dots,X_{n,n},Z$, respectively. 
W.l.o.g., $p_{n,1}\wedge q_{n,1}\ge\dots\ge p_{n,n}\wedge q_{n,n}$ and zero is a median of $X_{n,1}$. W.l.o.g., $c:=\lim_n (p_{n,1}\wedge q_{n,1})$ exists (or pass to a subsequence). One of the following two cases must occur. 
Case 1: $c>0$. Then, by Lemma 2, w.l.o.g. $X_{n,1}\to U$ and $X_{n,2}+\dots+X_{n,n}\to V$ for some independent r.v.'s $U$ and $V$ such that $U$ has a two-point support $S_U$ and $U+V$ equals $Z$ in distribution. So, $\{-1,1/2,1\}=S_Z\supseteq S_U+S_V$, by Lemma 3. Moreover, $S_Z=S_U+S_V$ if $S_V$ is a singleton set, in which case $S_Z$ would be a two-point set, just as $S_U$ is, an so, the equality $\{-1,1/2,1\}=S_Z$ would be impossible. In the remaining case, when $S_V$ contains at least two distinct points, the containment $\{-1,1/2,1\}\supseteq S_U+S_V$ is impossible, and so, the equality $\{-1,1/2,1\}=S_Z$ is again impossible. 
Case 2: $c=0$. Let here $Y_{n,j}:=X_{n,j}-x_{n,j}$, where $x_{n,j}:=a_{n,j}$ if $p_{n,j}<q_{n,j}$ and $x_{n,j}:=b_{n,j}$ if $p_{n,j}\ge q_{n,j}$. Then the condition $c=0$ implies the uniform negligibility condition: $\max_j P(|Y_{n,j}|>\epsilon)\to0$ for each $\epsilon>0$. Indeed, $$\max_j P(|Y_{n,j}|>\epsilon)\le \max_j P(Y_{n,j}\ne0)=\max_j (p_{n,j}\wedge q_{n,j})=p_{n,1}\wedge q_{n,1}\to c=0.$$
Also, 
$\sum_j Y_{n,j}-y_n=\sum_j X_{n,j}\to Z$, where $y_n:=-\sum_j x_{n,j}$. It follows from well-known results that the r.v. $Z$ is infinitely divisible; see e.g. Theorem 2 in Chapter IV of Petrov '75. So, by Lemma 4, the support set $S_Z$ cannot be the set $\{-1,1/2,1\}$.<|endoftext|>
TITLE: Smallest strongly regular graph whose automorphism group is not vertex transitive?
QUESTION [12 upvotes]: I'm looking for a small strongly regular graph whose automorphism group is not vertex-transitive.
This answer to a different question shows that the Chang graphs on 28 vertices are such graphs.  Is there an example on less vertices?
Thanks for all replies.

REPLY [13 votes]: There is no smaller example.
Various places, including Andries Brouwer's list of parameters and existence for small SRGS (http://www.win.tue.nl/~aeb/graphs/srg/srgtab.html) show that there are only a handful of parameter sets to check.
Those with fewer than 25 vertices can almost be checked by hand as they fall into a few families (Paley graphs) or are well-known individual graphs (e.g Clebsch graph). 
For the parameter set  $(25, 12, 5, 6)$ there are exactly $15$ graphs and they have automorphism groups of orders $1$ (twice), $2$ (four times), $3$ (twice), $6$ (four times), $72$ (twice) and $600$.<|endoftext|>
TITLE: Comparing norms on tensor products of matrices
QUESTION [6 upvotes]: Given a Hilbert space $H$, let $S_1(H)$ denote the space of trace-class operators on $H$, with the trace-class norm or Schatten 1-norm. That is
$$ \Vert T \Vert_1 = \sum_{j\geq 1} |s_j| $$
where $(s_1,s_2,\dots)$ is the sequence of eigenvalues of the operator $|T|=(T^*T)^{1/2})$ written in any order.
Let $H_1$ and $H_2$ be Hilbert spaces of dimention $n$ and $m$ respectively. Let $H_1\otimes_2 H_2$ denote their Hilbert-space tensor product (hence, an inner product space of dimension $nm$).
Consider the map $$\varphi:S_1(H_1)\hat{\otimes}S_1(H_2)\to S_1(H_1\otimes_2 H_2),$$ which is defined by 
$$\varphi( A \otimes B) (\xi_1\otimes\xi_2) = (A\xi_1 )\otimes (B\xi_2) \quad(\xi_1\in H_1, \xi_2\in H_2) $$
Is it the case that $$\Vert\varphi^{-1}\Vert=\min\lbrace{m‎, ‎n}\rbrace ?$$
Here, $\hat{\otimes}$ denotes the projective tensor product of Banach spaces.

REPLY [4 votes]: Revised answer:
(Notation from p 61ff of the reference below).
Note that for trace class norm we have 
$\mathcal{B}(H_1) = H_1 \hat\otimes H_1' = H_1\hat \otimes H_1$. For operator norm we have $\mathcal{B}(H_1) = H_1\hat{\hat\otimes} H_1'= H_1\hat{\hat\otimes} H_1$ where the tensor product is now the inductive one (also called $\epsilon$-tensor product). Using this and finite dimensionality we then can write
$\varphi:\mathcal{B}(H_1)\hat{\otimes}\mathcal{B}(H_2)\to\mathcal{B}(H_1\hat{\otimes}H_2)$
as the natural isomorphism
$$
(H_1 \hat\otimes H_1)\, \hat\otimes\, (H_2 \hat\otimes H_2) \to
(H_1 \hat\otimes H_2)\hat\otimes (H_1 \hat\otimes H_2)' =
(H_1 \hat\otimes H_2)\hat\otimes (H_1 \hat{\hat\otimes} H_2)
$$
which is $Id\,\hat\otimes\,j$ where $j:H_1 \hat\otimes H_2\to H_1 \hat{\hat\otimes} H_2$
is the embedding (here iso) of trace class norm into operator norm. The norm of $j$ is indeed $min\{\dim(H_1,\dim H_2)$, and the norm of $j^{-1}$ is its inverse, which can be seen by fixing bases and then writing any operator $A:H_1\to H_2$ as $U_2\circ D\circ U_1$ for isometries $U_i$ and a diagonal operator $D$. 
Ah,
I see now that the question changed. Here is the version for the new question:
$$
(H_1 \hat\otimes H_1)\, \hat\otimes\, (H_2 \hat\otimes H_2) =
(H_1 \hat\otimes H_2)\, \hat\otimes\, (H_1 \hat\otimes H_2) 
\to
(H_1 \otimes_2 H_2)\hat\otimes (H_1 \otimes_2 H_2)' =
(H_1 \otimes_2 H_2)\hat\otimes (H_1 \otimes_2 H_2)
$$
which is $i\hat\otimes i$ where $i: H_1 \hat\otimes H_2 \to H_1 \otimes_2 H_2$ is the natural iso. Its norm is the norm of the embedding $\ell^1\to \ell^2$ for the s-numbers. 

Johann Cigler, Viktor Losert, Peter W. Michor: Banach modules and functors on categories of Banach spaces. Lecture Notes in Pure and Applied Mathematics 46, Marcel Dekker Inc., New York, Basel, (1979) (pdf)<|endoftext|>
TITLE: Isotropic subvarieties of $ V(x_1^2+\dots+x_n^2-t_1^2-\dots-t_n^2) $
QUESTION [5 upvotes]: Let $ K $  be a field, $ \operatorname{char} K = 0, $ let $ Q = x_1^2+\dots+x_n^2-t_1^2-\dots-t_n^2 $ be the totally isotropic form, then the maximal linear isotropic
subspaces of  $ V(Q) \subset K^{2n} $ are the translates of 
$ V(x_1-t_1,\dots,x_n-t_n) $ under the action of  $ O(n,n) $  and therefore  have dimension $ n. $
Question:  Do there exist isotropic homogeneous subvarieties of $ V(Q) $ 
of dimension $ n $ which are not linear subspaces ?

REPLY [7 votes]: There is a much stronger result:  Any irreducible isotropic subvariety of dimension $n$ in $K^{2n}$ (with its split inner product) is an isotropic $n$-plane.  No assumption of homogeneity is necessary.
The proof is as follows:  First, it simplifies things to make the inner product be the spit quadratic form $Q = 2\,u_iv^i = 2(u_1v^1+\cdots +u_nv^n)$.  As you have remarked, any $n$-dimensional $Q$-isotropic linear subspace is equivalent, up to $\mathrm{Iso}(Q)$, to the subspace defined by $u_1 = \cdots = u_n = 0$.  Now, if $W\subset K^{2n}$ is an $n$-dimensional variety that is isotropic (i.e., its tangent space at each smooth point is $Q$-isotropic), then, by translation, one can assume that the origin is a smooth point of $W$ and that its tangent space at the origin is $u_i = 0$.  
Suppose that $W$ were not equal to its tangent plane at the origin.  Then it osculates to some finite order $k-1\ge1$ to its tangent plane at the origin, so it osculates to order $k$ to a graph $\Gamma$ of the form
$$
u_i = F_{ij_1\cdots j_k} \,v^{j_1}\cdots v^{j_k} = f_i(v),
$$
where $F_{ij_1\cdots j_k}$ is symmetric in its last $k\ge2$ indices, and not all of these coefficients are zero (otherwise the osculation of $W$ to its tangent plane at the origin would be higher than $k-1$).  However, then the isotropic condition, which is that the quadratic differential form $2\,\mathrm{d}u_i\circ \mathrm{d}v^i$ must vanish on $TW$, implies that this quadratic differential form must also vanish on $\Gamma$ at the origin to order at least $k{-}1$.  However, on $\Gamma$, we have
$$
2\,\mathrm{d}u_i\circ \mathrm{d}v^i = \left(\frac{\partial f_i}{\partial v^j}+\frac{\partial f_j}{\partial v^i}\right)\, \mathrm{d}v^j\circ \mathrm{d}v^i,
$$ 
and this vanishes at the origin ($v=0$) to order $k{-}1$ if and only if
$$
F_{ij_1j_2\cdots j_k} + F_{j_1ij_2\cdots j_k} = 0.
$$
Thus, $F$ must be skew-symmetric in its first two indices but symmetric in its last $k$ indices.  Since $k\ge2$, it follows that $F_{ij_1\cdots j_k}$ vanishes identically, which contradicts the choice of $k$.  
Thus, $W$ must equal its tangent plane at the origin, i.e., $W$ is an isotropic $n$-plane.<|endoftext|>
TITLE: In modules over finite rings $Rv = Rw \iff R^\times v = R^\times w$
QUESTION [8 upvotes]: Let $R$ be a finite ring (with unit, possibly non-commutative), and $M$ a left module over $R$. Let $v,w\in M$. Then
  $$Rv = Rw \iff R^\times v = R^\times w.$$

This follows from Lemma 6.4 in Hyman Bass. K-theory and stable algebra. Publications Mathématiques de l'Institut des Hautes Études Scientifiques 22 (1964), 5–60.
The proof uses the Artin-Wedderburn classification theorem. My question is if there is a simpler proof. (The hard part is the direction $\Rightarrow$, of course.)
As a remark, the statement is not true any more for infinite rings. Counterexamples with $M={}_R R$ can be found on page 466 in Irving Kaplansky. Elementary divisors and modules. Transactions of the American Mathematical Society 66 (1949), 464–491. Counterexamle (b) is over a commutative Noetherian ring $R$.

REPLY [5 votes]: Let $r,s\in R$ with $rv=w$ and $sw=v$. Since $R$ is finite, there exists $n>0$ such that $f=(rs)^n$ and $e=(sr)^n$ are idempotent.  Note that $ev=v$ and $fw=w$.  Let $r'=fre$ and $s'=esf$.  Notice that $r'v=w$.  Also note that right multiplication by $r'$ gives an $R$-module homomorphism $Rf\to Re$.  In fact, it is surjective (as is easily verified using the definitions of $e,f,r'$).  Similarly right multiplication by $s'$ gives a surjective $R$-module homomorphism $Re\to Rf$.  By finiteness of $R$ we conclude both these homomorphisms are isomorphisms.  It follows that $R(1-f)\cong R(1-e)$ by the Krull-Schmidt theorem (I don't know if this counts as more or less elementary than Wedderburn-Artin).  Such an isomorphism is given by right multiplication by an element $x\in (1-f)R(1-e)$.  
Consider $u=r'+x$.  Then $u$ is a unit since right multiplication by $u$ gives an isomorphism from $R=Rf\oplus R(1-f)$ to $R=Re\oplus R(1-e)$ (as it is the direct sum of the two isos $Rf\to Re$ and $R(1-f)\to R(1-e)$) and in a finite ring an element with a one-sided inverse is invertible.    Also $uv=(r'+x)v=(r'+x)ev=r'ev=r'v=w$ because $x\in (1-f)R(1-e)$ implies $xe=0$.  Thus $w\in R^\times v$.<|endoftext|>
TITLE: Make mathematical sense of the Dirac well Potential Equation
QUESTION [5 upvotes]: A classical problem in quantum mechanics involving the Dirac Delta function is given by
$$
y''+(\delta(x)-\lambda^2)y=0.
$$ 
Then, to find ''bound states'', you solve on the right and find the converging solution as $x\rightarrow \infty$, then solve on the left similarly. Assume continuity of the solution $y$. The jump condition on $y'$ is found by integrating from $-\epsilon$ to $\epsilon$   and taking $\epsilon\rightarrow 0$. The only value of $\lambda$ giving a bound state is then found to be $1/2$. The corresponding state is continuous everywhere and differentiable everywhere except at zero.
My issue with this is that to make this solution mathematically correct, one has to make sense of the multiplication of a continuous non-differentiable function by the delta Dirac function. In a previous question of mine, Alan U. Kennington suggested to use Radon measures. However, I believe there must be something simpler that makes sense of the problem above by mathematically making sense of multiplying a continuous function by a Dirac delta function. I am not necessarily looking for a complete answer but rather a reference that discusses the problem.
The reason I am asking this question is because I am facing a third order equation with coefficients involving the Dirac delta function. I am able to find a solution but I would like to make my computations more mathematically sound. I already asked this question on Math Stack Exchange but did not get an answer.

REPLY [2 votes]: Probably the correct type of space for y is actually not a Hilbert space but a Rigged Hilbert space. Here's a paper about Rigged Hilbert spaces in quantum physics:
http://arxiv.org/pdf/quant-ph/0502053v1.pdf<|endoftext|>
TITLE: Modification of matching
QUESTION [5 upvotes]: Suppose i have an $n \times n$ random bipartite graph and suppose that i repeat the following process $n$ times. At the start (stage 1) each edge is selected independently with probability $p(n)$, and we output the largest matching $\mathcal{M}_{1}$. In stage 2 all edges are selected randomly again regardless of how they appeared before but this time we take the largest matching $\mathcal{M}_{2}$ which doesn't include any of the edges of $\mathcal{M}_{1}$ (alternatively we could assume the edges of $\mathcal{M}_{1}$ aren't present anymore and all others are selected randomly). Similarly in stage $i$ we select the largest matching $\mathcal{M}_{i}$ which doesn't include any of the edges in $\mathcal{M}_{1}\cup...\cup \mathcal{M}_{i-1}$. Is there a way we can gain some knowledge on the expected size of $\mathcal{M}_{1} \cup...\cup \mathcal{M}_{n}$?

REPLY [4 votes]: Sudakov and Vu proved that, for $p \gg \log n / n$, $G_{n,p}$ has a perfect matching with high probability even after adversarially deleting $(1 - o(1))pn/2$ edges at each vertex.  The same argument works in the bipartite case (and can in fact be slightly simplified).  So for this range of $p$ you expect to be pulling out perfect matchings for at least the first $pn/2$ steps.  In fact, this will still be the case until almost $n/2$ matchings have been removed, as until then it is unlikely that you choose a random graph in which more than half of the edges at any vertex have been used at any previous stage.
Following the comments, here's a slightly more detailed explanation of why we expect to get perfect matchings until around time $n/2$.
At some stage in the process, let $G \sim G_{n,n,p}$ and let $H$ be the subgraph of $K_{n,n}$ consisting of the edges that have already been used.  For each $x \in V(G)$, let $d(x)$ be the degree of $x$ in $G \cap H$.  Let $\epsilon > 0$ and suppose that $n$ is sufficiently large.  We make the following claims.

With probability at least $1-\epsilon$, $G$ is such that after deleting any subgraph of maximum degree at most $(1-\epsilon)pn/2$, the remainder $G'$ still has a perfect matching.
For each $x \in V(G)$, with probability at least $1-\epsilon/2n$, $d(x) \leq (1+\epsilon)p\Delta(H)$.

The first part is Sudakov and Vu's result.  The second is an easy application of Chernoff's inequality.  It follows by a union bound that, with probablity at least $1-2\epsilon$, $G$ will have a perfect matching even after deleting every edge of $G \cap H$ provided $\Delta(H) < \frac {1-\epsilon} {1+\epsilon} \frac n 2$.
This is enough to tell you that one stage of your process is very likely to succeed, and by iterating that in expectation you will cover close to half of your graph by time $n/2$.  By analysing the failure probabilities more carefully you should be able to show that, with probability tending to $1$ as $n$ tends to infinity, you in fact pull out $(1-\epsilon)n/2$ perfect matchings in the first $(1-\epsilon)n/2$ steps.<|endoftext|>
TITLE: What is $Sq^i(\alpha^j)$ for all $i$ and $j$?
QUESTION [5 upvotes]: Write $H^*(\mathbb{R}P^\infty; \mathbb{Z}_2) = \mathbb{Z}_2[\alpha]$, $\deg \alpha = 1$. What is $Sq^i(\alpha^j)$ for all $i$ and $j$? I am not an algebraic topologist by trade but need to know this result...

REPLY [8 votes]: The Cartan formula 
$$Sq^i(xy)=\Sigma _{j+k=i}Sq^j(x)Sq^k(y)$$
together with the instability condition 
$$Sq^d(\alpha)=\alpha ^2 \mbox{ if $d=deg(\alpha )$}, Sq^i(\alpha )=0 \mbox{ if $d>deg(\alpha )$}$$
and the property $Sq^0=id$ (see, e.g. https://en.wikipedia.org/wiki/Steenrod_algebra)
give $Sq^i (\alpha ^j)= \binom{j}{i}\alpha ^{i+j} $.  All of these can be found, for example, http://isites.harvard.edu/fs/docs/icb.topic1525101.files/steenrod-operations.pdf page 8.<|endoftext|>
TITLE: What is the (co-)homology of $K(\mathbb{R}_\delta,n)$?
QUESTION [5 upvotes]: To elaborate on the question from the title, $\mathbb{R}_\delta$ is the additive group of real numbers (without any topology) and $K(\mathbb{R}_\delta,n)$ is an Eilenberg-MacLane space. I would like to know what its integral (co-)homology is, but I'm also happy about any results in this direction.
Bill Thurston has stated in a comment (Nontrivial finite group with trivial group homologies?) that the homology of $B\mathbb{R}_\delta = K(\mathbb{R}_\delta,1)$ has rank $2^\omega$ in each degree, but I don't know why this holds and I would also like to have results about $n>1$.
The background behind this question is that I'd like to show that $K(\mathbb{R}_\delta,n)$ cannot be a retract of a (smooth) manifold, and I'm hoping that (co-)homology will help me to figure this out.

REPLY [10 votes]: Given any $\mathbb Q$-vector space $A$ (in your case, $A=\mathbb R_\delta$), the integral homology of $K(A,n)$, which is the same as its rational homology, is given by the cofree graded-cocommutative coalgebra cogenerated by $A$ in degree $n$.
In particular, for $n$ an even number, we have
$$
H_m(K(A,n),\mathbb Z)=H_m(K(A,n),\mathbb Q)= \begin{cases}Sym^k A &\text{if $m=kn$}
\\0 &\text{otherwise}\end {cases}
$$
The rational cohomology of $K(A,n)$ is obtained by the universal coefficient theorem (UCT):
$$
H^m(K(A,n),\mathbb Q)=\hom(H_m(K(A,n),\mathbb Q),\mathbb Q).
$$
The integral cohomology is more messy to describe: it involves a non-trivial $ext$-term in the UCT short exact sequence (see the wikipedia entry on the UCT for more details).<|endoftext|>
TITLE: Determining when combinatorial sums are zero
QUESTION [6 upvotes]: To keep things simple with a specific example, we ask: 
Prove that $\displaystyle\ a_n:=\frac{1}{n!}\sum_{k=0}^n \binom{n}{k} \frac{1}{k!} (-1)^{n-k}$ is zero if and only if $n=1$. (Or find a counter-example). 
In fact, determining when said sequence is positive and negative is even more challenging. We list the first few calculations, to show that it does not alternate in sign: 
$\{a_n\}_{n=0}^\infty=\{1, 0, -\frac{1}{4}, \frac{1}{9}, -\frac{5}{192}, \frac{7}{1800}, -\frac{37}{103680}, \frac{17}{2116800}, \frac{887}{232243200}, \ldots\}$
Posted a related question: https://mathoverflow.net/questions/211767/determing-signs-of-taylor-coefficients-in-entire-functions

REPLY [8 votes]: (Non-vanishing is established by multiplying by $n!^2$ and getting an expression which is 1 modulo $n$.)
Robert Israel's answer and the comments following it settle your problem completely, but here's a general strategy that often works:
Suppose you have a sum of the form $S(n) = \sum_{k=0}^{n} f(n,k)$, where $f$ is "hyper-geometric" in the sense that $f(n+1,k)/f(n,k), f(n,k+1)/f(n,k)$ are rational functions.

Find the maximum of $|f(n,k)|$ as a function of $k$ ($n$ fixed). This is rather easy because $|f(n,k+1)|/|f(n,k)|$ is the absolute value of a rational function. In your specific problem, it is $\frac{n-k}{(k+1)^2}$, so $|f(n,k)|$ increases for $k \lesssim \sqrt{n}$ and then it decreases. The sign of the largest term is usually the resulting sign of the sum.
Suppose the maximum is attained at $k=M(n)$. Focus on estimating the sum around $k=M(n)$ (the rest of the terms should be easier to bound, as we know they are smaller, the decay is exponential usually). Specifically, Stirling's approximation is usually enough for estimating $\sum_{k=-\Delta}^{\Delta} f(n,M(n)+k)$ - plugging Stirling and using some Taylor series estimates ($\Delta$ is chosen so that the Taylor series estimates are valid), we get a sum that can sometimes be evaluated even by comparing to a Riemann's integral. 

This is a rather elementary approach; Asymptotic methods such as steepest descent can yield much better results if you can express your sum as the coefficient of some analytic function. This can be done in your case as we have the relevant generating function: $e^{\frac{t}{1+t}-\ln(1+t)}$. See de Bruijn' "Asymptotic Methods in Analysis" (the example in section 4.7 is solved twice in the text - Laplace' method and Steepest descent, and it is reminiscent of your sum as it contains a sign-changing term).
The sign-changing property of your sum makes other nice methods obsolete. Hayman's method, for example, works in extracting the coefficients of $e^{P(t)}$ where $P$ is a function satisfying some positivity conditions ($\frac{t}{1+t}-\ln(1+t)$ is not admissible as its coefficients are sign-changing, but the method should succeed in approximating your sum if we remove the sign-changing term).<|endoftext|>
TITLE: Almost but not quite a Lie algebroid: what is it?
QUESTION [10 upvotes]: In some calculations, I have arrived at the following algebraic structure, reminiscent of a Lie algebroid, but not quite.
I have a real line bundle $E \to M$, on whose smooth sections $\Gamma(E)$ I have a Lie algebra structure.  I also have a Lie algebra homomorphism $\rho:\Gamma(E) \to \Gamma(TM)$, obeying
$$
[e_1,f e_2] = f [e_1,e_2] + \rho(e_1)(f) e_2
$$
for all sections $e_1,e_2 \in \Gamma(E)$ and functions $f \in C^\infty(M)$.  However I do not have a Lie algebroid because the map $\rho$ is not $C^\infty(M)$-linear, so it is not induced by a bundle map $E \to TM$.
Question: Does such a structure have a name?  Any references where such structure has been studied?
Thanks in advance.
Edit: I thought it would be instructive to mention that this is a situation which can only arise when $E$ is a line bundle.  If $E$ were of higher rank, then $C^\infty(M)$-linearity follows.  To see this simply evaluate $[f e_1, g e_2]$ in two different ways:
$$
[f e_1, g e_2] = g [f e_1, e_2] + \rho(f e_1)(g) e_2 = f g [e_1,e_2] - g \rho(e_2)(f) e_1 + \rho(f e_1)(g) e_2~,
$$
but also
$$
[f e_1,g e_2] = -[g e_2, f e_1] =  f g [e_1,e_2] + f \rho(e_1)(g) e_2 - \rho(g e_2)(f) e_1~.
$$
Equating the two,
$$
(\rho(f e_1)-f\rho(e_1))(g)e_2 + (\rho(g e_2)-g \rho(e_2))(f) e_1 = 0~.
$$
If $\text{rank} E>1$, then around any point there is a neighbourhood where we can choose  $e_1,e_2$ to be linearly independent, whence their coefficients must vanish.  Clearly this is not possible for a line bundle.

REPLY [9 votes]: José, after your last comment, I am pretty sure that you are simply in the presence of a Jacobi structure on a nontrivial line bundle. Most of what I write below is taken from this paper by Crainic and Salazar, called Jacobi structures and Spencer operators. 
Just for completeness, given a real line bundle $E \to M$, a Jacobi structure on its space of sections $\Gamma(E)$ is a local Lie bracket $[\cdot,\cdot]$ on $\Gamma(E)$, whereby local means that, for any (possibly locally defined) $e_1,e_2 \in \Gamma(E)$, the support of $[e_1,e_2]$ is contained in the intersection of the supports of $e_1$ and $e_2$. These were originally studied by Kirillov (under the name local Lie algebras) and by Lichnerowicz (first under the assumption that the line bundle be trivial). 
Informally speaking, Jacobi structures are to Poisson structures as contact manifolds are symplectic manifolds (this is a very useful analogy if you are to work with these objects). Just like a Poisson structure on a manifold $M$ induces a Lie algebroid structure on $T^*M$, so does a Jacobi structure on $E \to M$ induce a Lie algebroid structure on $J^1 E$, the first jet bundle of $E \to M$. If I am not mistaken, this was first proven by Dazord and is Proposition 3.4 in Crainic and Salazar's paper. 
The map $\rho$ that José mentions is simply what Crainic and Salazar call $\rho^1$ in their proof of Proposition 3.4, while the bivector $T_e$ in José's question is $\rho^2(- \otimes e)$ in Crainic and Salazar's language. The relation between the two (which is what José wrote in answer to David's comment) is equation (8) in the paper. The anchor of the Lie algebroid structure on $J^1 E$ is completely determined by $\rho^1$ and $\rho^2$ (this is the content of the first part of the proof of Proposition 3.4 in the paper), so I suspect that José may have just rediscovered this fact!<|endoftext|>
TITLE: A question on representation of graphs
QUESTION [14 upvotes]: Take a complete graph $K_n$. You want to assign a vectors from $\Bbb F_2^d$ to every edge such that sum of vectors in every simple cycle does not sum to $0$ vector. The question is what is minimum $d$ that is needed. Does this question have any connection to representation of symmetric group over $\Bbb F_2^d$?
Update: Fundamental roadblock here seems to be related to finding a nice way to provide labelling that encompasses every even alternating/reverse alternating cycle which prevents a better than $O(n\log{n})$ solution.

ADDED by David Speyer: I'd like to explain why I find this question interesting and challenging, and wish that people better at extremal combinatorics had turned some attention to it. 
If we instead ask that every disjoint union of simple cycles have nonzero sum, the problem is easy (at least to leading order). A winning strategy is to use $d=n \log_2 n$ bits, divided into $n$ blocks of length $\log_2 n$. For edge $ij$, put the binary label of $j$ in the $i$-th block and the binary label of $j$ in the $i$-th block. And it is easy to show you can't do much better. For simplicity, take $n=2m$ even. Look at the $(2m-1) (2m-3) \cdots 5 \cdot 3 \cdot 1$ perfect matchings of the graph. If any two of them get the same sum, then their symmetric difference is a disjoint union of cycles with weight $0$. So we need to have at least $(2m-1) (2m-3) \cdots 5 \cdot 3 \cdot 1$ different vectors, so we need to use at least $\log_2 (2m-1) (2m-3) \cdots 5 \cdot 3 \cdot 1 \approx \frac{n}{2} \log_2 n$ bits. We have found the answer up to a factor of $2$.
Now, out of the $n!$ disjoint unions of cycles, roughly $(e/n) n!$ are a single cycle. One would think that reducing the number of things to avoid by a factor of $n$ shouldn't be able to effect the number of bits very much. Yet it seems incredibly hard to show this!
How can we show that the cycles are basically randomly distributed among the disjoint unions of cycles?
Here is one way we could attempt to give a lower bound. Once again, take $n=2m$. Look at the $m!$ matchings from the first $m$ vertices to the second, which we can identify with the group $S_m$. Adding up the edges in a matching gives a coloring of $S_m$ with $2^d$ colors. If we make $S_m$ into a graph by joining $\sigma$ and $\tau$ when $\sigma^{-1} \tau$ is a single cycle, then this must be a proper coloring. So we are trying to find a lower bound for the chromatic number of the Cayley graph of $S_m$ with respect to the cycles. There is literature on the chromatic number of Cayley graphs, but I couldn't find anything that would help. I did find that a random graph on $m!$ vertices where an edge would appear with probability $e/m$ should be expected to have chromatic number $\frac{m! (e/m)}{2 \log m!} \approx \frac{e m!}{2 m^2 \log m}$. If that is the case here, we once again get $d \geq \log_2 m!$ (discarding lower terms). Is there some $S_m$ representation theory which would allow us to actually compute the chromatic number?
It really frustrates me that there is an exponential separation between by upper and lower bounds.

I will give the bounty for a construction which beats $(1-\delta) n \log_2 n$, or a lower bound which beats $(\log n)^{1+\delta}$, for any $\delta>0$.

Finally, I'd like to share a wild musing of mine. Consider two problems in graph theory and optimization. In the first problem, we are given an $n \times n$ matrix of weights $w_{ij}$, and we are trying to find a permuation $\sigma$ of $n$ which minimizes $\sum_i w_{i \sigma(i)}$. The second problem is the same, but we require that $\sigma$ is an $n$-cycle. The first problem is the assignment problem and there are good algorithms for it. The second problem is the traveling salesman problem and it is NP Hard. I am reminded of this problem: A large fraction of permutations are cycles, yet restricting myself to cycles makes thing incredibly harder.

REPLY [6 votes]: Here is a modest improvement on the upper bound that shows $d=O(n\log\log n)$.
The base of the construction is the following.
Order the vertices according to some permutation as $v_1,\ldots,v_n$.
Fix $2n$ independent vectors from $\mathbb F_2^d$ and assign two vectors, $\ell_i$ and $r_i$, to each vertex $v_i$.
Assign to the edge $v_iv_j$ the vector $r_i+\ell_j$ if $i<j$ and $\ell_i+r_j$ if $i>j$.
Notice that this usually gives a nonzero sum for long cycles.
Our strategy will be to take $t\approx \log \frac nk$ permutations (with $t\cdot 2n$ independent vectors) and take the sums on each edge to take care of all cycles longer than $k$.
The standard probabilistic approach with $\approx k\log n$ further independent vectors takes care of all cycles whose length is at most $k$.
This in total gives $d\approx t\cdot n + k\log n =O(n\log\log n)$.
Claim. $t\approx \log \frac nk$ permutations are enough to take care of all cycles longer than $k$.
Proof of the claim.
Divide $n$ into groups of size $k$.
Fix an order inside each group, this will be the same in each permutation, which thus practically act on $m=\frac nk$ elements.
For every cycle $C=c_1\ldots c_k$ of length $k$, either there are three different groups that contain three consecutive vertices from $C$ (thus $c_{i-1}\in G_1$, $c_{i}\in G_2$, $c_{i+1}\in G_3$), or two different groups, one of which contains the first two of three consecutive vertices and the other group the third one (thus $c_{i-1},c_{i}\in G_1$, $c_{i+1}\in G_2$)
In this latter case, all we need is a permutation where $G_2$ is after $G_1$, then $c_i$ will be between its neighbors and thus the respective $\ell$ and $r$ vectors won't be negated when taking the sum over the edges of $C$.
Similarly, in the first case, we need that $G_2$ is between $G_1$ and $G_3$.
Therefore, it is enough to show that for $m=\frac nk$ elements there are $O(\log m)$ permutations such that for any three elements $G_1,G_2,G_3$ there is a permutation in which $G_1<G_2<G_3$.
A standard probabilistic argument shows that $O(\log m)$ random permutations work.
In more detail, there are $m^3$ ordered triples, each permutation has probability $\frac 16$ to work for each triple, thus if we take $t$ independent permutations, the chance for any ordered tripe that none of the permutations work for it is $(1-\frac 16)^t$ and taking the union bound we need $m^3(1-\frac 16)^t<1$.<|endoftext|>
TITLE: A question related to Woodin's $HOD$ conjecture
QUESTION [9 upvotes]: Recall that an $\kappa$ is $\omega$-strongly measurable in $\text{HOD}$ if there exists 
$\lambda < \kappa$ such that $(2^{\lambda})^{\text{HOD}} < \kappa$ and such that
there is no partition of $S = \{\alpha < \kappa: \text{cf}(\alpha) = \omega\}$ into $\lambda$ many sets $\langle S_{\alpha}: \alpha < \lambda\rangle \in \text{HOD}$ such
each set $S_{\alpha}$ is stationary in $V$.
It is not known if the successor of a singular strong limit of uncountable cofinality can be $\omega$-strongly measurable in $\text{HOD}.$
Now my question is about countable cofinality:
Question 1.  Is it consistent that the successor of some singular strong limit cardinal of countable cofinality  is $\omega$-strongly measurable in $\text{HOD}$?
If consistent, would you please give some references for it.

REPLY [5 votes]: I asked the question from Prof. Woodin. The answer is yes,  assuming the consistency of Axiom I0. It has appeared as  Lemma 190 on page 298 of Woodin's paper ``Suitable Extender Models I''.<|endoftext|>
TITLE: Why there are only finitely many $\overline{\mathbb{Q}}$-isomorphism classes of elliptic curves with CM by $\mathcal{O}$?
QUESTION [9 upvotes]: For someone who does not have a very extensive knowledge of number theory, what is a good intuitive explanation as to why there are only finitely many $\overline{\mathbb{Q}}$ isomorphism classes of elliptic curves with complex multiplication by $\mathcal{O}$?

REPLY [14 votes]: Here is the standard argument. Obviously, there are details to fill in. I'm glad to fill in the ones you have trouble with, but I want to get the overview down first.
Isomorphism over $\overline{\mathbb{Q}}$ is the same as isomorphism over $\mathbb{C}$, by the usual nonsense that allows us to exchange any two algebraically closed fields of the same characteristic.
Over $\mathbb{C}$, every elliptic curve is of the form $\mathbb{C}/\Lambda$ for a discrete rank two sublattice $\Lambda$ of $\mathbb{C}$. A complex analytic homomorphism $\mathbb{C}/\Lambda_1 \to \mathbb{C}/\Lambda_2$ must lift to a map $\mathbb{C} \to \mathbb{C}$ of the form $z \mapsto a z$ for some $a \in \mathbb{C}$ such that $a \Lambda_1 \subseteq \Lambda_2$. 
So $\mathbb{C}/\Lambda$ has complex multiplication by $\mathcal{O}$ if and only if, for all $a \in \mathcal{O}$ we have $a \Lambda \subseteq \Lambda$. And two lattices $\Lambda_1$ and $\Lambda_2$ give isomorphic quotients if and only if there is a scalar $a$ with $a \Lambda_1 = \Lambda_2$.
So we need to prove the following result: There are only finitely many lattices $\Lambda$ such that $\{ a \Lambda \subseteq \Lambda : \forall a \in \mathcal{O} \}$, up to the relation that $\Lambda_1 \sim \Lambda_2$ if there is a scalar $a$ with $a \Lambda_1 = \Lambda_2$.
This is the result that the class semi-group of $\mathcal{O}$ is finite. I think the shortest proof is via Minkowski's theorem on lattice points in convex bodies.<|endoftext|>
TITLE: Intuition behind Kronecker's congruence?
QUESTION [6 upvotes]: The modular polynomial is defined by$$\Phi_n(X, \tau) = \prod_{\tau} (X - j(\tau)),$$where $j$ is the elliptic modular function and $\tau$ is running through classes of imaginary quadratic integers of discriminant $n$. What is an easy way to see that$$\Phi_p(X, Y) \equiv (X - Y^p)(X^p - Y) \text{ (mod }p \text{)},$$i.e. what is the intuition behind Kronecker's congruence?

REPLY [8 votes]: David Speyer posted his answer while I was typing, and it's great, but here's a slightly different take.
Consider a generic curve $E/\mathbb C_p$ having multiplicative reduction. Then $E/\mathbb C_p$ has a Tate model,
$$ E(\mathbb C_p) \cong \mathbb C_p^*/q^{\mathbb Z}, $$
where $q\in\mathbb C_p$ satisfies $0<|q|<1$. What are the curves $p$-isogenous to $E$. Letting $Q=q^{1/p}$ ($Q$ is any of the $p$'th roots of $q$), it's a nice exercise to check that they are the curves
$$ E_\zeta := \mathbb C_p^*/(\zeta Q)^{\mathbb Z} \quad\hbox{and}\quad E_0:=\mathbb C_p^*/(q^p)^{\mathbb Z},$$
where  $\zeta$ is any primitive $p$'th root of unity.
(These correspond to the subgroups $\boldsymbol\mu_p$ and $\langle\zeta Q\rangle$.)
From the $q$-expansion of the $j$-invariant, we have
$$ j(E) = q^{-1} + 744 + 196884q + \cdots, $$
So
$$ \Phi_p(X) = (X-j(q^p))\prod_\zeta (X-j(\zeta Q)) .$$
We have $j(q^p)\equiv j(q)^p\pmod{p}$, while the product looks like
$$
  \prod_\zeta (X-j(\zeta Q)) = \prod_\zeta (X-(\zeta Q)^{-1} - \cdots)
  = (X^p - Q^{-p} + \cdots) = (X^p-q^{-1} + \cdots) = (X^p-j(q) + \cdots).
$$
(I'm being a bit imprecise here, but we're looking at neighborhood of $q=0$, and
the dots represent lower order terms as $q\to0$.) Putting it together, we get
$$ \Phi_p(X) \equiv (X-j(q)^p)(X^p-j(q))\pmod{p}. $$
N.B. I'm not suggesting that this is a rigorous proof. But you asked for some intuition. For me, (1) it's always been writing $E/K$ as $K^*/q^{\mathbb Z}$ that's explained Kronecker; (2) you can turn this into a proof over $\mathbb C$ by being a bit more careful with the $q$ expansions; (3) you can also make the proof I've sketched over $\mathbb C_p$ rigorous with more work.
Of course, the fancy way to express it is that when you reduce the Hecke correspondence $T_p$ modulo $p$, you get $\Phi_p+\Phi_p^t$, the sum of the graph of Frobenius and it's transpose.<|endoftext|>
TITLE: Number of nonzero terms in polynomial expansion (lower bounds)
QUESTION [14 upvotes]: Let $f(x) = a_1x^{z_1} + a_2x^{z_2} + \cdots + a_kx^{z_k}$ be a polynomial with coefficients $(a_1, \ldots, a_k) \in \mathbb{F}_q^*$ and $z_i$ are distinct positive integers. If I need to compute the number of nonzero terms in the expansion of $f(x)^m$, the upper bound would be $\binom{m+k-1}{k-1}$ (from the multinomial theorem). Instead, I'd like to compute the lower bound, since sometimes there will be cancellations, for example:
Case 1: From $a_1a_3x^{z_1+z_3} + \cdots + a_2a_4x^{z_2+z_4} + \cdots$, if $z_1+z_3 = z_2+z_4$ then we have a collision $(a_1a_3 + a_2a_4)x^{z_1+z_3}$. From this, the number of nonzero terms decreases by 1. 
Case 2: If $a_1a_3 + a_2a_4 \equiv q\mbox{ (mod q)}$ then this term is completely cancelled and the number of terms is decreased by an additional 1 (2 in total). 
How can I compute this lower bound?
Also, is it possible to determine the best value for the exponents in $f(x)$ so that the lower bound is the lowest possible? I have a special case where $f(x) = a_1x^{z} + a_2x^{z + \alpha} + \cdots + a_kx^{z + (k-1)\alpha}$ seems to be the best case but I can't prove it. It seems that, if I use an arithmetic progression, the number of nonzero terms considering only exponents collisions (case 1) is $mk - (m-1)$. So for a fixed $k$, this is really not bad, but is it possible to prove that the number of collisions is optimal (maximum)?

REPLY [5 votes]: Existence of a lower bound that goes to infinity used to be a long-standing problem of Rényi and Erdős. It was finally resolved by Schinzel with a bound of about $\log \log k$. The bound was later improved by Schinzel and Zannier to about $\log k$. Also, Zannier proved a lower bound on the number of terms of $g(f(x))$ for any $g$.
Given an example with $K(f^2)=A$ and $K(f)=B$, one can obtain a polynomial $g$ with $K(g^2)=A^2$ and $K(g)=B^2$. Indeed, consider $g(x)=f(x)f(x^{BIG})$ So, in view of the example posted by Richard Stanley in another answer, it follows that there is an $\varepsilon>0$ and infinitely many polynomials $f$ such that $K(f^2)\leq K(f)^{1-\varepsilon}$.
The question of whether the correct bound is logarithmic or polynomial in $k$ (or is in between) is open.<|endoftext|>
TITLE: Dividing the edges and diagonals of a polygon among disjoint sub-polygons
QUESTION [9 upvotes]: Let $P$ be a convex $n$-gon ($n$ is odd and $n \geq 5$).
  Determine the smallest $m$ such that all edges and diagonals of $P$ can be covered by the edges 
  of $m$ convex sub-polygons of $P$ which do not share any edges.
($Q$ is a sub-polygon of $P$ if all vertices of $Q$ are vertices of $P$).

This question was proposed at a Chinese contest and maybe it's a known result in graph theory.
For $n=5$, for example, it's not hard to see that $m(5)=3$. If $P:=P_{1}P_{2}P_{3}P_{4}P_{5}$,
$m(5)=3$ can be achieved by 3 convex sub-polygons: 
$P_{1}P_{2}P_{4}P_{5}$; $P_{1}P_{3}P_{4}$; $P_{2}P_{3}P_{5}$.
I also found that $m(7) \leq 6$, but I couldn't go further.
Any suggestion would be appreciated. Thanks in advance.

REPLY [2 votes]: I think the following construction realizes the lower bound in Brian's answer. Let $n=2k+1$ and label the first $2k$ vertices with $\mathbb{Z}_{2k}=\{0,\ldots,2k-1\}$ and the last vertex with $\infty$. First we take all the $(n-1)(n-3)/8=2k(k-1)/4$ 4-gons of the form
\[(i,\,i+j,\,i+k,\,i+k+j)\]
with $i\in\{0,\ldots,2k-1\}$ and $j\in\{1,\ldots,k-1\}$. That covers everything not involving $\infty$ except the "long" diagonals $(i,i+k)$, so we add $(n-1)/2=k$ triangles $(i,i+k,\infty)$ for $i\in\{0,\ldots,k-1\}$.<|endoftext|>
TITLE: What sort of cardinal number is the Löwenheim–Skolem number for second-order logic?
QUESTION [18 upvotes]: In their paper “On Löwenheim–Skolem–Tarski numbers for extensions of first order logic”, Magidor and Väänänen make the following statement:
“For second order logic, $\mathrm{LS}(L^{2})$ [the Löwenheim–Skolem number for second order logic—my comment] is the supremum of $\Pi_{2}$-definable ordinals..., which means  that it exceeds the first measurable, the first $\kappa^{+}$-supercompact $\kappa$, and the first huge cardinal if they exist”.
[“The Löwenheim–Skolem number $\mathrm{LS}(L^{2})$ of second order logic $L^{2}$ is the smallest cardinal $\kappa$ such that if a theory $T\subset L^{2}$ has a model, it has a model of cardinality $\le\max(\kappa,|T|)$”, and “$L^{2}$ extends first order logic with quantifiers of the form $\exists R\,\phi(R,x_0,\dots,x_{n-1})$, where the second order variable $R$ ranges over $n$-ary relations on the universe for some fixed $n$”—my comment also but substantially quoting the authors.]
Assume that the first measurable, the first $\kappa^{+}$-supercompact $\kappa$, and the first huge cardinals exist.  What type of large cardinal, then, is $\mathrm{LS}(L^{2})$?   If the answer is known, please provide the reference.

REPLY [2 votes]: The Löwenheim number (compared to LS, this uses sentences rather than theories) for the second order logic $L^2$ is the least $κ$ such that $V_κ$ satisfies all true $Σ_2$ sentences.  This $κ$ has cofinality $ω$, and $V_κ$ does not satisfy ZFC, but $κ$ is a limit of large cardinals.  For example, if there is a proper class of huge cardinals, then $κ$ is a limit of huge cardinals, and same with other $Σ_2$ properties.
Assuming countable vocabulary, the Löwenheim-Skolem (LS) number for $L^2$ is the least $κ$ such that $V_κ$ satisfies all true $Σ_2$ sentences with real parameters.  $ω_1≤\operatorname{cf}(κ)≤c$, where $c$ is the cardinality of the continuum.  As before, $V_κ$ does not satisfy ZFC, but $κ$ is a limit of large cardinals.
If we allowed a proper class of constant symbols, then $L^2$ would not have a Löwenheim-Skolem number.  Still, even then, if $V_κ ≺_{Σ_2} V$, then every theory of cardinality $<κ$ with an $L^2$ model has such a model of cardinality $<κ$.  For reference, among $κ$ with $V_κ ≺_{Σ_2} V$, the following are in strictly increasing order of the least example (if any, if not assuming large cardinals):  $\operatorname{cf}(κ)=ω$, $\operatorname{cf}(κ)=ω_1$, $V_κ⊨\text{ZFC}$, $\operatorname{cf}(κ)=ω_1 ∧ V_κ⊨\text{ZFC}$, $κ$ is regular, $κ$ is Mahlo, $κ$ is weakly compact, $κ$ is measurable, $κ$ is strong.  Note that for every strong $κ$, $V_κ ≺_{Σ_2} V$.
Also, the Löwenheim-Skolem-Tarski (LST) number of a logic $L$ is the smallest cardinal $κ$ such that every structure for $L$ has an elementary substructure of size $≤κ$.  As noted in the paper in the question, the LST number for $L^2$ (with countable vocabulary) is the least supercompact cardinal, or none if none exists; the Vopěnka's principle holds iff every logic with set-sized vocabulary has an LST number.<|endoftext|>
TITLE: Image of an $F$-acyclic resolution homotopic to a projective resolution?
QUESTION [7 upvotes]: This is a crosspost of this MSE question according to the recommendation in the comments. I know this question is elementary, but I'm hoping the author of these notes could provide a more detailed explanation.

Here is an excerpt from some notes I stumbled upon online:


From what I understand, the "elementary proof" is just the fundamental lemma of homological algebra which says the homotopy type of chain maps out of projective resolutions is determined by maps between the objects being resolved.
It seems that the author says that the image of every $F$-acyclic resolution is homotopic to some injective/projective resolution, but I don't think this follows from the fundamental lemma.

Why is this true?
  How does it prove one may compute derived functors using $F$-acyclics?
  Does this approach circumvent dimension shifting?

REPLY [2 votes]: This is explained on the n-lab here. In particular in Theorem 3.15.<|endoftext|>
TITLE: Products of elliptic isometries
QUESTION [6 upvotes]: A well-known property on groups acting on trees is:

Theorem: Let $T$ be a tree and $g,h \in \mathrm{Isom}(T)$ two elliptic isometries. If $\mathrm{Fix}(g) \cap \mathrm{Fix}(h) = \emptyset$ then the product $gh$ is a loxodromic isometry.

I am pretty sure that a similar statement exists for $\delta$-hyperbolic spaces, where $\mathrm{Fix}$ is replaced with $\mathrm{Fix}_{\delta}$, but I did not succeed in remembering in which article I saw it...
Does somebody know a reference?

REPLY [3 votes]: Check Lemme 2.3 in chapter 9 of "Geometrie et theorie des groupes" by Coornaert, Delzant and Papadopoulos. It implies that your conclusion holds provided that the quasi-fixed point sets are sufficiently far apart.<|endoftext|>
TITLE: Field of definition of dominant morphisms
QUESTION [8 upvotes]: Let $k$ be an algebraically closed field and $k_0$ a sub-field. Let $X,Y$ be two projective varieties defined over $k_0$. Suppose that that there exists a dominant morphism $f$ between $X_k=X\otimes k$ and $Y_k=Y\otimes k$. 
Such a dominant morphism cannot always been defined over a finite extension of $k_0$. Take by exemple an elliptic curve $X=Y$ defined over $k_0=\mathbb{Q}$, $k=\mathbb{C}$ and $f$ a translation by a point not in the algebraic closure of $\mathbb{Q}$.
But it is still natural to ask the following:
With the above hypothesis on the existence of $f$, do there always exists a dominant morphism $g$ between $X$ and $Y$ defined over a finite extension $k_1$ of $k_0$ ?
According to the appendix 1 of Mumford’s book « Abelian Varieties », the answer is yes if $X$ and $Y$ are Abelian varieties. For the proof, using a translation, one reduces to the case of an homomorphism. Then the (dense set of)  torsion elements of $X$ are sent to torsion elements of $Y$, therefore the homomorphism is defined over a finite extension.
The last implication is a bit obscure for me, maybe the following can help:
field of definition of isogenies of abelian varieties
?
Anyway I would like to understand the general case, in which one can also replace the word "dominant" by "isomorphism" or "birational morphism"...

REPLY [9 votes]: The general technique goes as follows.  Let $\{A_i\}$ be a directed system of rings with limit $A$; e.g., $A=k$ and $\{A_i\}$ the set of finitely generated $k_0$-subalgebras of $k$.  Let $X$ and $Y$ be schemes of finite presentation over some $A_{i_0}$, and define $X_i = X \otimes_{A_{i_0}} A_i$ for $i \ge i_0$, $X_A = X \otimes_{A_{i_0}} A$, and similarly for $Y_i$ (with $i \ge i_0$) and $Y_A$.  There is a natural map of sets
$$\varinjlim {\rm{Hom}}_{A_i}(X_i,Y_i) \rightarrow {\rm{Hom}}_A(X_A,Y_A)$$
and it is bijective by EGA IV$_3$, 8.8.2(i). 
So given an $A$-morphism $f:X_A \rightarrow Y_A$, there exists some $i_1$ and an $A_{i_1}$-morphism $f_{i_1}:X_{i_1} \rightarrow Y_{i_1}$ which descends $f$.  But we want more: for various properties $\mathbf{P}$ of morphisms of schemes, if $f$ satisfies $\mathbf{P}$ then we want $f_i := f_{i_1} \otimes_{A_{i_1}} A_i$ to also satisfy $\mathbf{P}$ for some $i \ge i_1$.  The property "isomorphism" is immediate by applying the preceding formalism to the inverse $A$-morphism too (using bijectivity in the displayed map of sets above).  See IV$_3$ 8.10.5 for the tip of the iceberg on many possibilities for $\mathbf{P}$. By IV$_3$, 8.6.3, open subschemes of $X_A$ with finitely presented complement also descend to such open subschemes of some $X_i$. 
For $i \ge i_1$, the (set-theoretic) image 
$Z_i = f_i(X_i) \subset Y_i$ is constructible and if $j \ge i$ then $Z_j$ is the preimage of $Z_i$ under $Y_j \rightarrow Y_i$ and likewise $f(X_A)$ is the preimage of $Z_i$ in $Y_A$. Define $E_i$ to be the set of of $s \in S_i := {\rm{Spec}}(A_i)$ such that $(Z_i)_s$ is dense in $(X_i)_s$,
and likewise for an analogous subset $E \subset S := {\rm{Spec}}(A)$. The density or not of a constructible subset of a scheme of finite type over a field is insensitive to extension of the ground field, so  for $j \ge i$ the preimage of $E_i$ under $S_j \rightarrow S_i$ is $E_j$ (due to the analogue for $Z_i \subset Y_i$ and $Z_j \subset Y_i$ under $Y_i \rightarrow Y_i$ noted above) and likewise the preimage of $E_i$ under $S \rightarrow S_i$ is $E$. But $E_i$ is constructible in $S_i$ by IV$_3$ 9.5.3 and likewise for $E$ inside $S$. Thus, if $E=S$ then $E_i = S_i$ for all large $i$ due to IV$_3$ 8.3.5. 
Thus, in the initial setup of interest, if $X_k \rightarrow Y_k$ is dominant then we get a finitely generated $k_0$-subalgebra $R \subset k$ and an $R$-morphism $h:X_R \rightarrow Y_R$ which is dominant between fibers over all points of Spec($R$). 
In a similar manner, using the "spreading out" for open subschemes (with finitely presented complement) mentioned above, if there are dense open subschemes of $X_k$ and $Y_k$ between which $f$ restricts to an isomorphism, then by increasing $R$ further we can arrange that there exist fiberwise dense open subschemes $U \subset X_R$ and $V \subset Y_R$ between which $h$ restricts to an isomorphism. 
Finally, thanks to the Nullstellensatz over $k_0$, using the pullback of  $h$ between fibers over any closed point of such a Spec($R$) allows us to conclude for the properties of dominance, isomorphism, or birational morphism. 
This overall method is sometimes called the "Principle of finite extensions" (i.e., whatever happens after some extension of the ground field already happens over a finite extension), and complete proofs of virtually every such property you could ever imagine wanting is rigorously documented in remarkable and useful generality in EGA IV$_3$, $\S8$-$\S11$ (skip $\S10$) and IV$_4$, $\S17$-$\S18$.<|endoftext|>
TITLE: Natural topologies for the space of rational functions
QUESTION [7 upvotes]: I am looking for natural families of Hausdorff topologies (metrics, norms, if possible) for the space of rational functions of a single complex variable of arbitrary, unbounded denominator degree (and, if needed, a finite, or zero, limit for $z\to\infty$). Or spaces of meromorphic functions containing these functions.
The main requirement is that any sequence of functions $r_n$ defined by $r_n(z):=a_n/(b_n-z)$ for $z$-independent $a_n\to a\ne 0$ and $b_n\to b$ converges to to the function $r$ defined by $r(z):=a/(b-z)$.
If possible, I'd like to have a metric which gives a nice formula for the distance 
of $a/(b-z)$ to $a'/(b'-z)$.

REPLY [7 votes]: Karl-Goswin Grosse-Erdmann, The locally convex topology on the space of meromorphic functions, J. Austr. Math. Soc., 59 (1995) 287-303

REPLY [5 votes]: What happens with this way of doing it?  
A rational function is a map from the Riemann sphere 
$\mathbb C \cup \{\infty\}$ to itself.  The Riemann sphere is compact, so has a unique uniform structure.  (So choose a nice metric for it, say the one from an actual sphere.)  Use "uniform convergence".  
This topology is metric.

REPLY [2 votes]: $\hbox{Rat}_d$, the space of rational functions of degree $d$, is a $2d+1$ dimensional affine algebraic variety, so it embeds into $\mathbb A^N$ for some suffficiently large $N$. Since you're presumably interested in working over $\mathbb C$ (or maybe $\mathbb R)$, you can embed $\hbox{Rat}_d(\mathbb C)$ into $\mathbb C^N$ and then just use the usual topology on $\mathbb C^N$. 
To be more precise, if you view a degree $d$ rational function $f(z)=F(z)/G(z)$ as the $2d+2$-tuple of the coefficients of $F$ and $G$, it is defined as a point in $\mathbb P^{2d+2}$, and $\hbox{Rat}_d$ is the complement of the resultant locus $\mathcal R:\hbox{Res}(F,G)=0$. The complement of a hypersurface such as $\mathcal R$ is an affine variety. There's a brief description of how this embedding works in (see especially Proposition 4.27)
The Arithmetic of Dynamical Systems, Springer, Section 4.3 "The Space $\hbox{Rat}_d$ of Rational Functions"
Addendum in answer to comment that "I need a single space independent of $d$"
Ah, that's much more problematic. Now you're into the whole yoga of degenerations of rational maps of degree $d$ on the boundary of their natural moduli space. For those who work in dynamical systems, it is more natural to look at the quotient space $\mathcal M_d := \hbox{Rat}_d/\hbox{PGL}_2$, where the action is via conjugation, $f^\phi=\phi^{-1}\circ f\circ\phi$. These spaces were studied by Milnor (Experimental Math 2(1), 37-83, 1993), who proves that $\mathcal M_d(\mathbb C)$ is an orbifold and that it has a reasonable compactification. In particular, for $d=2$ he proves that $\mathcal M_2(\mathbb C)\cong\mathbb C^2$, and its natural compactification is $\overline{\mathcal M_2}(\mathbb C)\cong\mathbb P^2(\mathbb C)$. So then you can take any natural metric on projective space. I took this up from the perspective of algebraic geometry (geometric invariant theory) in Duke Math J. 94(1), 41-77, 1998.  Of course, this may not be what you need, but it provides one approach. The point is that if you want a nice compactification of a moduli space, you usually need to restrict what sort of degenerations are allowed. In GIT language, the magic word is semi-stability.<|endoftext|>
TITLE: Counting subspaces
QUESTION [8 upvotes]: We are given the finite vector space $V = V(n,p) = \mathbb{F}_p^n$ and two fixed subspaces $W_1, W_2 \subseteq V$ of dimensions $m_1$, $m_2$ respectively. Suppose 
that the dimension of $W_1 \cap W_2$ is $r$. 
How many $k$-dimensional subspaces of $V$ simultaneously intersect $W_1$ in dimension $n_1$, $W_2$
in dimension $n_2$, and $W_1 \cap W_2$ in dimension $r'$?

Further, is there a general argument for counting subspaces when more than two fixed subspaces are involved, given their dimensions and those of their $s$-fold intersections?

P.S. I would also appreciate any references to textbooks or papers where this
type of "$q$-combinatorics" is explored in depth.

REPLY [3 votes]: Ok, so the right way to think about this is to consider the maximal filtration generated by $W_1$ and $W_2$. This consists of $0,W_1\cap W_2,W_1,W_2, W_1+W_2,V$. Let these have dimension $0, a, a+b, a+c, a+b+c,a+b+c+e$ and let their intersection with the fixed subspace have dimension $0,a',a'+b',a'+c',a'+b'+c'+d', a'+b'+c'+d'+e'$
This is a relabeling of your parameters except i have added one new variable. Then such a subspace gives:
An $a'$-dimensional subspace of the $a$-dimensional $W_1 \cap W_2$
A $b'$-dimensional subspace of the $b$-dimensional $W_1/(W_1\cap W_2)$
A $c'$-dimensional subspace of the $c$-dimensional $W_1/(W_1\cap W_2)$
Two $d'$-dimensional subspaces, one of $W_1+W_2/W_2$ modulo its $b'$-dimensional subspace, and one of $W_1+W_2/W_1$ modulo its $c'$-dimensional subspace, and an isomorphism between them.
An $e'$-dimensional subspace of the $e$-dimensional $V/(W_1+W_2)$.
A bunch of extension classes.
Other than the extension classes we just get a $q$-analogue of
$$\pmatrix{ a \\ a'}\pmatrix{ b\\ b'}\pmatrix{ c\\ c'}\pmatrix{ b-b'\\ d'}\pmatrix{ c-c' \\ d'}\pmatrix{ e \\ e'} (d'!)$$
And the extension class is 
$$p^{(b'+c'+d')(a-a')+e'(a+b+c-a'-b'-c'-d')}$$
Then you can sum over $d'$ to get the answer to your problem.

Messy version:
Given such a subspace, we can choose a basis consisting of

$r'$ vectors in $W_1 \cap W_2$
$n_1-r'$ additional vectors in $W_1$ that remain linearly independent in $W_1 / (W_1 \cap W_2)$
$n_2-r'$ additional vectors in $W_2$ that remain linearly independent in $W_2 / (W_2 \cap W_2)$
$\alpha$ additional vectors in $W_1+W_2$ that, modulo the $n_1+n_2-r'$-dimensional subspace spanned by the previous vectors, do not intersect the $m-n_1$-dimensional image of $W_1$ or the $m-n_2$-dimensional image of $W_2$.
$\beta$ additional vectors in $V$ that remain linearly independent in $V/(W_1+W_2)$.

where $\alpha+ \beta=k-n_1-n_2+r'$.
It's easy to count the number of possible lists of vectors for the first three, and the number of choices in a fixed subspace. I think it's $$\left[\matrix{ r \\ r'} \right]_p \left[\matrix{ m_1-r \\ n_1-r'} \right]_p \left[\matrix{ m_2-r \\ n_2-r'}  \right]_p p^{(r-r') (n_1+n_2-2r')} $$ (A simpler argument along these lines gives a proof of the identity Richard Stanly mentions in the comments)
Similarly for the fifth one
$$\left[\matrix{n-m_1-m_2+r \\ \beta}\right]_p p^{\beta (m_1+m_2-r-k+\beta)}$$
So it remains to see how much the fourth kind of vectors increases the count. Each successive vector of the fourth type we choose must avoid two subspaces - $W_1$ plus the previous vectors and $W_2$ plus the previous vectors. We can count these using inclusion-exclusion because we know the intersection of each type of subspace with the others, as the new vectors we are adding at the fourth step increase the dimension of each by one and the dimension of the intersection by two.
$$\prod_{i=0}^{\alpha-1} ( p^{m_1+m_2-k} - p^{m_1+ n_2-k' + i} - p^{m_2+ n_1-k' + i}+ p^{k+n_1+n_2-2k' +2i})=\prod_{i=0}^{\alpha-1} (p^{m_2-n_2+k'-k-i}-1)(p^{m_1-n_1+k'-k-i}-1)p^{k+n_1+n_2-2k'+2i}$$
The denominator is given by a simpler formula - the number of choices for the remainder of the basis is just $$(\alpha!)_p p^{ \alpha (n_1+n_2-r')}$$
$$\left[\matrix{ r \\ r'} \right]_p \left[\matrix{ m_1-r \\ n_1-r'} \right]_p \left[\matrix{ m_2-r \\ n_2-r'}  \right]_p p^{(r-r')(n_1+n_2-2r')}\sum_{\alpha+\beta=n_1+n_2-k'}\left[\matrix{m_2-n_2+k'-k \\ \alpha} \right]_p \left[ \matrix {m_1-n_1+k'-k \\ \alpha}\right]_p (\alpha!)_p\left[\matrix{n-m_1-m_2+r \\ \beta}\right]_p p^{\alpha(k+n_1+n_2-2k'+\alpha-1) +\beta (m_1+m_2-r-k+\beta)}$$
Note that as $p$ goes to $1$ this becomes a perfectly ordinary product of binomial coefficients, because the $\alpha!$ vanishes for $\alpha$ nonzero.<|endoftext|>
TITLE: Permutable (Lie) subgroups
QUESTION [5 upvotes]: Let's recall that, a group $G$ being given, 
two subgroups $A,B\subset G$ are called 
permutable iff $AB=BA$ for the Minkowski 
law. It is straightforward to see that $(A,B)$ 
are permutable iff $AB$ is a subgroup of $G$.  

Let now $G$ be a finite dimensional Lie group 
  (real to begin with) and suppose that if 
  $A,B\subset G$ are Lie subgroups (then closed 
  by Cartan's theorem), one can 
  easily show that, providing $G=AB$ and if it 
  is a factorization (means that the decomposition 
  is unique or, equivalently, 
  $A\cap B=\{1_G\}$), then $Lie(A)\cap Lie(B)=\{0\}$

My questions are the following 


Q1) If $G=AB$ is a factorization ($A\cap B=\{1_G\}$),  
   do we have 
  $Lie(A)\oplus Lie(B)=Lie(G)$ ?
   Q2) Does the result hold if we just have 
  $G=AB$ (and still $Lie(A)\cap Lie(B)=\{0\}$)  
  without supposing that $A\cap B=\{1_G\}$ ?
   Q3) What are the references about these 
  questions ? 
  

What was (and is still) not clear to me is why the tangent map of the multiplication 
$$
A\times B\rightarrow G
$$
must be surjective. More precisely, the fact that it is a factorization provides a section $$s: G\rightarrow A\times B$$ (in fact an inverse bijection) but I am stuck in proving that it is C^1 (continuous would do, I think). Maybe the action 
of $A\times B$ on itself by $(a,b)*(x,y)=(ax,yb^{-1})$ could be exploited but I do not see how.

REPLY [3 votes]: As desired in the question, the action of $A \times B$ on $G$ is the key point.  This is a transitive action, and it is a general fact in the theory of actions of (separable) Lie groups on manifolds (see Bourbaki, Lie Groups & Lie Algebras, Ch. III, no. 1.7, Proposition 14) that if a Lie group $H$ with at most countably many connected components (equivalently, a countable base for its topology) acts transitively on a smooth manifold $X$ then for any $x_0 \in X$ with stabilizer $H_{x_0} \subset H$ the natural orbit map $H/H_{x_0} \rightarrow X$ is a diffeomorphism. (We need a countability hypothesis on $H$ to rule out the case where $X$ is a positive-dimensional connected Lie group and $H$ is the underlying discrete group acting by left-translation.)  In particular, if the action is simply transitive then the orbit map $H \rightarrow X$ is an $C^{\infty}$-isomorphism.
Now consider a Lie group $G$ with only countably many connected components and $A, B$ closed Lie subgroups of $G$ (so they each have only countably many connected components)  such that the multiplication map $A \times B \rightarrow G$ is surjective. For the Lie subgroup $H := A \cap B$ setting $X=G$ with the indicated transitive action, we get that the natural map $A \times B \rightarrow G$ defined by $(a,b) \mapsto ab^{-1}$ is a $C^{\infty}$-submersion (in fact, a quotient map modulo $A \cap B$), and a $C^{\infty}$-isomorphism when $A\cap B = 1$. In particular, in all such cases ${\rm{Lie}}(A) + {\rm{Lie}}(B) = {\rm{Lie}}(G)$, and the sum is direct if and only if $A \cap B$ is discrete.  (The generalization to Lie subgroups that might not have the subspace topology is left to the interested reader!)<|endoftext|>
TITLE: Remark 4.23.4 in Hartshorne
QUESTION [5 upvotes]: Crosspost from math.stackexchange, since it's quite possible I might not get a response there.
Remark 4.23.4 in Chapter IV of Hartshorne's Algebraic Geometry references a paper by Elkies that explains that$$\mathfrak{B} = \{p \text{ prime}: X_{(p)} \text{ is nonsingular over }k_{(p)}, \text{ and }X_{(p)}\text{ has Hasse invariant }0\}$$is infinite.

The existence of infinitely many superisngular primes for every elliptic curve over $\mathbb{Q}$, Invent. Math. 89 (1987) 561-567.

However, in the paper, the main theorem is stated as follows.

Let $E$ be any elliptic curve defined over $\mathbb{Q}$, and let $S$ be a finite set of primes. Then $E$ has a supersingular prime outside $S$.

Can anyone comment on the precise relationship between these two statements? Are they identical?

REPLY [8 votes]: The term "$X_{(p)}$ has Hasse invariant 0" means that "$X_{(p)}$ is supersingular". See the definition in Hartshorne's book 2 pages before the quoted remark. So the quoted remark says that there are infinitely many supersingular primes for an elliptic curve over $\mathbb{Q}$, i.e., for any finite set $S$ of primes there is a supersingular prime outside $S$.<|endoftext|>
TITLE: Homeomorphism of closed manifold
QUESTION [6 upvotes]: Suppose that we have two closed n-manifold $M$ and $N$ such that 
the topological group of homeomorphisms $Homeo(M)$ is homotopy equivalent to $Homeo(N)$ (maybe as topological groups if needed), can we deduce that $M$ is homotopy equivalent (or even homeomorphic ) to $N$ ? Is there an easy counterexample ?

REPLY [11 votes]: It is a result of Whittaker (1963), and Rubin (1990, in greater generality with a better proof) that $Homeo(M)$ (viewed as a group) determines $M$ up to homeomorphism (see this question). I doubt that the homotopy type is enough: Gabai had shown that for closed hyperbolic 3-manifolds, the inclusion of $Isom(M)$ into $Diff(M)$ is a homotopy equivalence. $Isom(M)$ is a finite group, so its homotopy type is given by its order. I don't think it is hard to find two hyperbolic manifolds with the same cardinality of isometry group (in fact, I assume that order is usually equal to $1.$)<|endoftext|>
TITLE: Incidence geometry and matrices
QUESTION [8 upvotes]: Supposing I have a $0/1$ or $\pm1$ matrix $A$ of size $m\times n$, is there a minimum $d$ (that works for every $m\times n$ $A$) such that there exists $m$ lines $r_1,\dots,r_m$, $n$ lines $s_1,\dots,s_n$ in $\Bbb R^d$ such that $r_i$ intersects $s_j$ iff there are some permutations $\Pi_1,\Pi_2$ such that $\widetilde{A}[i,j]=1$ where $\widetilde{A}=\Pi_1A\Pi_2$?
If so what is $d$ as function of $m,n,r$?
It is clear this is possible for special $A$ with $d=2$. Is it possible for every $A$ at some $d$?
Related: When is a 0-1 matrix a one-intersection incidence matrix?

REPLY [4 votes]: It seems that the matrix
$$
A = \left(\begin{array}{cccccc}
  1 & 1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 0 & 1 & 1 \\
  1 & 1 & 1 & 1 & 0 & 1 \\
  1 & 1 & 1 & 1 & 1 & 0 \\
\end{array}\right)
$$
is not the incidence matrix for any distinct lines $r_1,\ldots,r_6$ and
$s_1,\ldots,s_6$ for any dimension $d$, over $\bf R$ or any other field,
and thus that there is no such $d(m,n)$ once $m,n \geq 6$.
The $5\times 5$ matrix $A_0$ formed by the first five rows and columns of $A$
is not enough: let $R,S$ be points on a plane $\Pi$;
let $r_1,r_2,r_3 \subset \Pi$ be lines containing $R$ but not $S$,
and $s_1,s_2,s_3 \subset \Pi$ lines containing $S$ but not $R$;
and then choose points $P,Q$ off $\Pi$ and collinear with neither $R$ not $S$,
and set
$$
r_4 = \overline{PS}, \quad
r_5 = \overline{QS}, \quad
s_4 = \overline{QR}, \quad
s_5 = \overline{PR}.
$$
But this seems to be the only way to realize $A_0$, and it does not extend to
$6+6$ lines with the intersections demanded by $A$ itself.
The key point is suggested by Boris Bukh's comment:
of any three $r_i$, two must intersect.  Indeed if three pairwise skew lines
$l,l',l''$ have more than one transversal then they determine a quadric surface
$Q$ with two rulings, one of which includes $l,l',l''$ and the other consists
of all the transversals.  Then $s_1,s_2,s_3$ are in the other ruling,
and are thus pairwise skew.  Then $r_1,r_2,r_3$ are in the first ruling,
and it follows that all five $r_i$ are in one ruling and all five $s_j$
in the other.  Hence $r_i$ meets $s_j$ for every $i,j$ $-$ contradiction.
Likewise, of any three $s_j$, two must intersect.
On the other hand, if any three of the $r_i$ meet pairwise,
but not at the same point, then they span a plane $\Pi$ and $s_1,s_2,s_3$
must be on $\Pi$; and then they must also coincide, else every $r_i \subset \Pi$
and we soon reach a contradiction ($s_4$ meets $r_1,r_2,r_3$ but not $r_4$,
so $r_1,r_2,r_3$ meet at some point $R$ and $s_4$ contains $R$ but is not
in $\Pi$; likewise the same is true of $s_5$; but then $r_4,r_5$ also
go through $R$, which contradicts both the assumption at the start of
this paragraph and the $A_0$ condition).  Likewise if any three $r_i$
meet at a point $R$ but are not coplanar then $s_1,s_2,s_3$ go through $R$
and are coplanar, else every $r_i$ contains $R$ and again a contradiction
soon ensues.
From here it should be just a matter of hunting down which configurations
of skew/coincident/coplanar lines are possible among the $r_i$ and $s_j$
until only the one configuration described earlier survives…
Another candidate counterexample, with $m=n=7$, is a "double seven"
with $r_i \cap s_j = \emptyset$ iff $i=j$.  Even $(m,n)=(6,7)$
would suffice if the only way to get $(m,n)=(6,6)$ is the famous
"double six" configuration on a smooth cubic surface, on which
no two $r_i$ meet but the $r_i$ are not all on the same quadric,
and likewise for the $s_j$.<|endoftext|>
TITLE: Lie Algebras over DVRs and basechange to the completion
QUESTION [5 upvotes]: Let $R$ be a discrete valuation ring containing an algebraically closed field $K$ of characteristic zero and let $L$ be a Lie algebra over $R$ whose underlying $R$-module is finitely generated and free.
Let $\mathfrak{m}$ denote the maximal ideal of $R$ and $\hat{R}$ the $\mathfrak{m}$-adic completion of $R$. While thinking about something else I came up with the following question, which I cannot answer:

Is it true that if $L \otimes_R \hat{R}$ is isomorphic (as Lie algebras) to $\mathfrak{g} \otimes_K \hat{R}$ for some simple Lie algebra $\mathfrak{g}$ over $K$, then $L$ is already isomorphic to $\mathfrak{g} \otimes_K R$ ?

I have the feeling that this is either well-known or false (or both), but since I am no expert in Lie-theory, I thought I might just ask it here. Thanks in advance.

REPLY [5 votes]: One can also give explicit high-rank counterexamples using special orthogonal groups of quadratic lattices over $R$ with non-degenerate reduction, and we can also arrange that $R$ has $K$ as its residue field (likewise for Jason Starr's examples). 
To give wider context, let $K$ be any field of characteristic 0, $R$ any $K$-algebra, $\mathfrak{g}$ any semisimple Lie algebra over $K$ (we'll use $\mathfrak{so}_n$ in the end for any odd $n \ge 3$), and consider in place of $\widehat{R}$ any faithfully flat $R$-algebra $R'$.  
Now consider a projective $R$-module $\Lambda$ of constant odd rank $n \ge 3$ and a quadratic form $q:\Lambda \rightarrow R$ over $R$ that is non-degenerate at every point of ${\rm{Spec}}(R)$.   Let $H = {\rm{SO}}(q)$ be the associated special orthogonal group scheme; this is a smooth affine $R$-group whose geometric fibers over Spec($R$) are the classical special orthogonal groups of the corresponding fibers of $q$. In particular, since $n$ is odd it follows that such fibers are of adjoint type (and $\mu_2 \times H = {\rm{O}}(q)$ is the orthogonal group scheme of $q$). 
Let $L = \mathfrak{so}(q)$ be the Lie algebra of $H$, and let $\mathfrak{g}$ be the Lie algebra for the special orthogonal group ${\rm{SO}}_n := {\rm{SO}}(q_n)$ of the corresponding "split" rank-$n$ quadratic form
$$q_n := x_0^2 + x_1 x_2 + \dots + x_{2m-1}x_{2m}$$
where $2m+1=n$. Let $R'$ be a faithfully flat $R$-algebra such that $q$ becomes homothetic to $q_n$ over $R'$; such an $R'$ always exists (can even be chosen to be etale over $R$ since $R$ is a $\mathbf{Z}[1/2]$-algebra).  Thus, $L$ and $\mathfrak{so}_n$ become isomorphic over $R'$.  But I claim that if $L$ and $\mathfrak{so}_n$ are $R$-isomorphic then necessarily $q$ and $q_n$ are homothetic over $R$ (i.e., $q$ is isometric to $u q_n$ for some $u \in R^{\times}$). 
It is a general fact that for odd $n \ge 3$, ${\rm{SO}}_n$ represents the automorphism functor
of its Lie algebra on the category of all $K$-algebras, as may be checked
on $\overline{K}$-points since ${\rm{char}}(K)=0$. (This is a special case
of a general fact about connected semisimple groups of adjoint type whose Dynkin diagram has no nontrivial automorphisms.) Thus, $L$ as an fpqc $R$-form of $\mathfrak{so}_n$ corresponds to an element in the cohomology set
${\rm{H}}^1(R'/R, {\rm{SO}}_n)$.  
But since $n$ is odd,  ${\rm{SO}}_n$ represents the functor of conformal isometries of $q_n$ (i.e., pairs $(T,u)$ consisting of a unit $u$ and 
$T \in {\rm{GL}}_n$ such that $q_n \circ T = u q_n$). Unraveling everything, 
this says that $L := {\mathfrak{so}}(q)$ is $R$-isomorphic to $\mathfrak{so}_n$ if and only if $q$ is homothetic to $q_n$.
Now we can give explicit counterexamples over discrete valuation rings $R$ containing an algebraically closed field $K$ of characteristic 0 (and even with $K$ as its residue field). Letting $\mathfrak{m}$ be the maximal ideal of such an $R$, 
consider a quadratic lattice $(R^n, q)$ of odd rank $n \ge 3$ with $\overline{q} := q \bmod \mathfrak{m}$ non-degenerate over $R/\mathfrak{m}$ such that $\overline{q}$ is isometric to $q_n$. There is no infinitesimal obstruction to lifting an isomorphism between non-degenerate quadratic spaces $(M,Q)$ and $(M',Q')$ over a $\mathbf{Z}[1/2]$-algebra $A$ (indeed, the Isom-scheme ${\rm{Isom}}(Q',Q)$ is an ${\rm{O}}(Q)$-torsor and hence is smooth since $A$ is a $\mathbf{Z}[1/2]$-algebra), so $q$ becomes isometric to $q_n$ over the completion $R' := \widehat{R}$.  Thus, $L := \mathfrak{so}(q)$ is a counterexample
(using $\mathfrak{g} = \mathfrak{so}_n$) provided that $q$ is not homothetic to 
$q_n$ over $R$.
Now writing $n=2m+1$, observe that the ring
$$S := K[t_0, \dots, t_{n-1},v_1, \dots, v_{n-1}]/(v_i^2 + t_0 v_i - t_i)$$ 
satisfies $S/t_0 S = K[v_1, \dots, v_{n-1}]$ (in which 
$t_i = v_i^2$), so $t_0 S$ is prime in $S$ and 
$S$ is normal (e.g., by Serre's criterion). Thus, $R = S_{(t_0)}$ is a dvr
and 
$$q = x_0^2 + t_1 x_1^2 + \dots + t_{n-1} x_{n-1}^2$$
over $R$ has reduction isometric to $q_n$. To show that $q$ is not homothetic to $q_n$ over $R$ it suffices to prove that $q$ is even anisotropic over $F = {\rm{Frac}}(R) = {\rm{Frac}}(S) = K(t_0, v_1, \dots, v_{n-1})$.  This anisotropicity is "obvious", and is rigorously proved by induction on $n$ (allowed to have any parity) or perhaps in other ways too.
That $R$ does not have residue field $K$.  But consider the field extension
$K(\!(t_0)\!)$ of $K(t_0)$. This has countably infinite transcendence degree, so we can find elements $t_1, \dots, t_{n-1} \in K(\!(t_0)\!)$ algebraically independent over $K(t_0)$ that are all 1-units. Thus, $X^2 + t_0 X - t_i$ has a root
$v_i \in K[\![t_0]\!]$. Thus, the ring $S$ above can be found inside $K[\![t_0]\!]$, so the restriction to ${\rm{Frac}}(S)$ of the $t_0$-adic valuation on $K(\!(t_0)\!)$ is a discrete valuation whose valuation ring $R$ contains $S$ (it is not $S_{(t_0)}$!) and has the same fraction field as $S$, and has residue field $K$.  Clearly $q$ makes sense as a quadratic form over $R$, and as such it does the job (by the same anisotropicity argument over ${\rm{Frac}}(R) = {\rm{Frac}}(S)$!).  
Remark. Theorem 3.5 of http://arxiv.org/pdf/math/0406508v5.pdf provides a much more far-reaching result: over any ring whatsoever (no field of char. 0 in the picture), the groupoid of semisimple group schemes of adjoint type is equivalent (via the Lie functor) to the groupoid of vector-bundle Lie algebras whose Killing form is fiberwise non-degenerate!<|endoftext|>
TITLE: Does there exist a homotopy equivalence from $\mathbb{C}P^{2n}$ to itself that reverses orientation?
QUESTION [7 upvotes]: Perhaps this is a foolish question, forgive my lack of knowledge of topology. My question is, does there exist a homotopy equivalence from $\mathbb{C}P^{2n}$ to itself that reverses orientation?

REPLY [27 votes]: No, there is not.  The cohomology ring $H^*(\mathbb{C}P^{N})$ is a truncated polynomial ring $\mathbb{Z}[x]/(x^{N+1})$, where $x$ has degree $2$.  Given any map $f:\mathbb{C}P^{N}\to\mathbb{C}P^{N}$, there is some $d\in\mathbb{Z}$ such that $f^*(x)=dx$, and then $f^*(x^N)=d^Nx^N$.  If $f$ is a homotopy equivalence, then $d=\pm 1$, and if $N$ is even, this implies $f$ induces the identity on $H^{2N}(\mathbb{C}P^N)$ and thus preserves the orientation.  (Note that for $N$ odd, $[z_0,\dots,z_N]\mapsto [\bar{z}_0,\dots,\bar{z}_N]$ is an orientation-reversing diffeomorphism $\mathbb{C}P^N\to\mathbb{C}P^N$.)<|endoftext|>
TITLE: Generalized functions on a product of two manifolds
QUESTION [5 upvotes]: Let $X,Y$ be smooth compact manifolds. Let $C^\infty(X)$ and $C^{-\infty}(X)$ denote the spaces of smooth and generalized functions on $X$ respectively. We have the obvious canonical linear map $$T\colon C^{-\infty}(X\times Y)\to Bil(C^\infty(X),C^\infty(Y)),$$
where the target is the space of continuous bilinear functionals $C^\infty(X)\times C^\infty(Y)\to \mathbb{C}$. By definition $(T\Phi)(f,g)=\Phi(f\otimes g)$.
It is well known that $T$ is isomorphism of vector spaces. For which standard topologies on the source and the target $T$ is an isomorphism of topological vector spaces?
For example let us choose on $C^{-\infty}(X\times Y)$ the strong topology, and on the target the topology given by seminorms
$$||B||_{K,L}:=\sup_{k\in K,l\in L}|B(k,l))|,$$
where $K\subset C^\infty(X), L\subset C^\infty(Y)$ are arbitrary bounded subset. Then $T$ will be a continuous map for these topologies. But is $T$ a topological isomoprhism?

REPLY [3 votes]: $T$ is indeed a topological isomorphism. Indepently of the concrete situation, this can be shown by a suitable version of the closed graph theorem, e.g. the one of de Wilde: the domain has to be ultrabornological an the range should have a web (see, e.g., the book Introduction to Functional Analysis of Meise and Vogt, chapter 24). Here $T$ acts from the dual of the nuclear Frechet space $C^\infty(X\times Y)$ to $Bil(C^\infty(X),C^\infty(Y))\cong L(C^\infty(X),C^{-\infty}(Y))$. It follows e.g. from Grothendieck's work that both spaces are ultrabornological, and both spaces also have web. Therefore, one can apply the closed graph theorem fot $T$ and its inverse. (One can even show that the domain and range space are both isomorphic to $s'$, the space of slowly increasing sequences.)<|endoftext|>
TITLE: When is the image of an integral polynomial contained in the image of another?
QUESTION [11 upvotes]: Suppose $f$ and $g$ are polynomials with integral coefficients and $f(\mathbb Z)\subset g(\mathbb Z)$. Is there any relation between $f$ and $g$? 
For instance, this happens if $f=g\circ h$ for some third integral polynomial $h$. In the particular case that $f$ takes on only square values, we could deduce that the converse is true: $f$ is a square, so $f=g\circ h$ where $h$ is some polynomial and $g(x)=x^2$. Does something like this hold more generally?

REPLY [27 votes]: The suggestion from the comments to use Siegel's Theorem about integral points on algebraic curves is vast overkill. The by far easier Hilbert's Irreducibility Theorem is sufficiently strong here:
Suppose that $f(\mathbb Z)\subseteq g(\mathbb Z)$. Write $f(X)-g(Y)=A_1(X,Y)A_2(X,Y)\cdots A_r(X,Y)$ with polynomials $A_i(X,Y)\in\mathbb Q[X,Y]$ which are irreducible over $\mathbb Q$. By Hilbert's Irreducibility Theorem, there is an infinite set $H$ of integers such that $A_i(h,Y)$ is irreducible for each $h\in H$ and each $i$. On the other hand, for each integer $h$ there is an integer $u$ such that $f(h)=g(u)$. So $f(h)-g(Y)$ and therefore some $A_i(h,Y)$ has an integral root. Running through $h\in H$, we see that there is some $i$ such that $A_i(h,Y)$ has an integral root for infinitely many $h\in H$.
So for this index $i$ there are infinitely many $h\in H$ such that $A_i(h,Y)$ is irreducible and has a rational root. So $A_i(X,Y)$ has degree $1$ in $Y$.
We obtain $f(X)-g(Y)=A(X,Y)(b(X)-Yc(X))$ for $A\in\mathbb Q[X,Y]$ and $b,c\in\mathbb Q[X]$. Setting $Y=b(X)/c(X)$ yields $f(X)=g(b(X)/c(X))$. Looking at poles shows that $c(X)$ actually is a constant. So $f(X)=g(b(X))$ for a polynomial $b(X)$ with rational coefficients.<|endoftext|>
TITLE: Are there congruence subgroups other than $\operatorname{SL}_2(\mathbb Z)$ with exactly 1 cusp?
QUESTION [17 upvotes]: Are there any congruence subgroups other than $\operatorname{SL}_2(\mathbb Z)$ which have exactly 1 cusp? By congruence subgroup, I mean a subgroup of $\operatorname{SL}_2(\mathbb Z)$ containing $\Gamma(N)$ for some $N$.
This question was motivated by the comments on the answer to Generators of the graded ring of modular forms. In particular, if there are no such subgroups, then all congruence subgroups other than $\operatorname{SL}_2(\mathbb Z)$ are generated in degree at most 5, with relations in degree at most 10, as follows from the comments on the answer to Generators of the graded ring of modular forms.
I have a feeling the answer to this question may be well known, but I could not find it in Diamond and Shurman or by a Google search.

REPLY [6 votes]: $\def\SL{\mathrm{SL}}\def\GL{\mathrm{GL}}\def\PSL{\mathrm{PSL}}\def\ZZ{\mathbb{Z}}\def\FF{\mathbb{F}}\def\PP{\mathbb{P}}$I previously gave a wrong answer to this question. In fact, such groups exist for $N$ any prime which is $3 \bmod 4$ and in many other cases, as I will now spell out.
If we have $G_1 \subset \SL_2(\ZZ/n_1 \ZZ)$ and $G_2 \subset \SL_2(\ZZ/n_2 \ZZ)$ with $n_1$ and $n_2$ relatively prime, then $G_1 \times G_2 \subset \SL_2(\ZZ/n_1 \ZZ) \times \SL_2(\ZZ/n_2 \ZZ) \cong \SL_2(\ZZ/(n_1 n_2) \ZZ)$ also has this property. So I'll concentrate on prime powers.

I'll concentrate first on the case of $N$ a prime $p$. As in my wrong answer, if $\Gamma \subset \SL_2(\FF_p)$ has the required property, so does any larger group containing $\Gamma$, so it is enough to restrict our attention to the maximal proper subgroups of $\SL_2(\FF_p)$. These are classified; see Corollary 2.2 in this paper by King. Note that types (d) and (e) only occur for $p$ a prime power, not a prime, so they are irrelevant to us. Group (a) has a two element orbit and group (c) is $\Gamma_0(p)$, with two cusps. This leaves (b) and (f), (g), (h).
Type (b) This is the one I got wrong; I misread King's paper. The correct description of Group (b) is as follows: Let $q$ be an isotropic quadratic form on $\FF_p^2$. For example, if $p \equiv 3 \bmod 4$, then 
we can take $q(x,y) = x^2+y^2$. Then Group (b) is the group of matrices which either preserve the quadratic form or multiply it by $-1$. (I misread this as the index $2$ subgroup that preserve the form.) For example, when $q(x,y) = x^2+y^2$, this is matrices of the form $\left[ \begin{smallmatrix} a & b \\ -b & a \end{smallmatrix} \right]$ with $a^2+b^2 = 1$ or $\left[ \begin{smallmatrix} c & d \\ d & -c \end{smallmatrix} \right]$ with $c^2+d^2 = -1$.
Now, $q$ descends to a map $\mathbb{P}^1(\FF_p) \to \FF_p^{\times}/(\FF_p^{\times})^2$. The two classes in $\FF_p^{\times}/(\FF_p^{\times})^2$ thus divide $\PP^1(\FF_p)$ into two orbits for the action of the index two subgroup that preserves $q$. What about the elements that switch the sign of $q$? If $-1$ is a square, they preserve the split of  $\PP^1(\FF_p)$ into two orbits; but, if $-1$ is not square, then they switch the orbits, and we get just one cusp. Of course, $-1$ is not square if and only if $p \equiv 3 \bmod 4$. Conveniently, in this case, we can take $q(x,y) = x^2+y^2$, so all of my examples are relevant.
Lifting to $\ZZ/p^k \ZZ$ Take an isotropic quadratic form on $\FF_p^2$ and lift it to an quadratic form $Q: \ZZ_p^2 \to \ZZ_p$ where $\ZZ_p$ is the $p$-adics. Let $\Sigma \subset \SL_2(\ZZ_p)$ be the matrices which rescale $Q$ by $\pm 1$. Then $Q$ descends to a surjection $\PP^1(\ZZ_p) \to \ZZ_p^{\times} /  (\ZZ_p^{\times})^2$ (here $\ZZ_p^{\times}$ is the unit group of $\ZZ_p$ and $\PP^1(\ZZ_p)$ is pairs $[u:v]$ in $\ZZ_p^2$, not both in $p \ZZ_p$, modulo simultaneous rescaling by $\ZZ_p^{\times}$). The group $\Sigma$ acts transitively on $\PP^1(\ZZ_p)$;  if $-1$ generates the quotient $\ZZ_p^{\times} /  (\ZZ_p^{\times})^2$, which happens for $p \equiv 3 \bmod 4$. Descending modulo $p^k$ gives us examples in $\ZZ/p^k \ZZ$. 
Types (f), (g), (h) These types arise from the exceptional subgroups $A_4$, $S_4$ and $A_5$ of $\PSL_2(\FF_p)$ (for $p \geq 5$). They can only act transitively on $\PP^1(\FF_p)$ if $p+1$ divides the order of these groups, which happens for $p \in \{ 5,11 \}$, $p \in \{ 5,7,11,23 \}$ and $p \in \{ 5,11,19,29 \}$ rspectively. I believe all of those cases occur; many of them occur in the list of near fields. I am not sure which of these cases can be lifted modulo powers of primes.<|endoftext|>
TITLE: Symmetries of module categories over the category of representations of quantum $sl(2)$
QUESTION [10 upvotes]: The category $\mathcal{C}_l$ of tilting modules of the quantum group $U_q(sl_2)$ quotiented out by the modules of zero quantum dimension has a natural structure as a semisimple monoidal category when the quantum parameter $q$ is a root of unity.  The module categories of $\mathcal{C}_l$ (or equivalently, quantum subgroups of $U_q(sl_2)$) are classified by ADE diagrams. See for example, section 6 of Ostrik.
Let $K_0(\mathcal{C_l})$ be the Grothendieck ring of $\mathcal{C}_l$.  The action of $\mathcal{C}_l$ on a module category by tensor product on the objects defines a based module $M$ over $K_0(l)$.  The generating element of $K_0(\mathcal C_l)$ acts as a matrix $A$ on $M$, where $A$ is the adjacency matrix of the corresponding ADE Dynkin diagram.
The ADE Dynkin diagrams have graph symmetries: the A-type and E_6 graphs can have the order of the vertices reversed, the D-type graphs can switch the two short legs, and the D_4 graph exhibits a full $S(3)$ symmetry.  All of these symmetries induce based module symmetries on the corresponding based module $M$.
In what ways can the symmetries on the based modules be extended to symmetries of the corresponding module categories?  Can you construct module functors from the module category to itself that induce this symmetry on the based modules?  If not, is there some lesser structure than a module category but greater than a based module that can be preserved under these symmetries?

REPLY [4 votes]: Marcel's answer is great if you're familiar with the literature, but let me try to give a more direct argument.
Recall that for $A_n$, $D_{2n}$,$E_6$, and $E_8$ the module category actually is itself a tensor category with the action given by applying a tensor functor and then tensoring on the left.  So suppose we have $\mathcal{F}:\mathcal{C} \rightarrow \mathcal{D}$ a tensor functor making $\mathcal{D}$ into a module category for $\mathcal{C}$.  Clearly tensoring on the right by any object in $\mathcal{D}$ gives a module endofunctor, and tensoring on the right with an invertible object gives a module autoequivalence.  For the 2-fold symmetry of $A_n$, the 3-fold symmetry of $D_4$ and the $2$-fold symmetry of $E_6$ it's easy to see that the Dynkin diagram automorphism is just given by tensoring on the right with an invertible object.
The harder case to deal with is the 2-fold symmetries of $D_{2n}$ and of $D_{2n+1}$.  In the former case, there's no invertibles around to tensor with, and in the latter case the module category doesn't come from a tensor product.  
In the odd case there's a nice trick.  If $\mathcal{C}$ is a braided tensor category and $\mathcal{M}$ is a module category, then you can use the braiding to make tensoring with an object $X$ into a module endofunctor.  If you take $X$ to be the nontrivial invertible object in quantum SU(2), then a quick combinatorial check shows that tensoring with it implements the diagram involution.  In the even case, the above trick doesn't work since tensoring with the nontrivial invertible object in quantum SU(2) fixes the two ends of the fork.  
In both the odd and even cases for type D, everything is governed by $\mathbb{Z}/2\mathbb{Z}$ so you can just do the calculation directly.  Recall that $D_{2n}$ is realized as a module category as the category of right modules for the algebra $1+X$ (where X is the nontrivial invertible object).  It's easy to see that the two ends of the fork correspond to two different module structures on the middle object.  As Marcel said, any module endofunctor is then given by tensoring on the right with some A-A bimodule.  Typically the only invertible A-A bimodules have underlying object A itself ($D_4$ has extras, we'll just not use them), and then it's just a $\mathbb{Z}/2\mathbb{Z}$ fact that there are exactly two such bimodules.
We want to know if tensoring with the nontrivial one switches or leaves fixed the two mod-A structures on the middle object.  This can be rephrased just in terms of the group $G=\mathbb{Z}/2\mathbb{Z}$, and so you can see that yes it switches them.  I feel like there should be a cleaner way to see this though.<|endoftext|>
TITLE: Automorphisms of del Pezzo surfaces
QUESTION [9 upvotes]: Let $S$ be a del Pezzo surface of degree six over $\mathbb{C}$. Then $S$ is the blow-up of $\mathbb{P}^2$ in three general points $p_1,p_2,p_3$.
Is it true that its automorphism group is $((\mathbb{C}^{*})^{2}\rtimes S_2)\times S_3$?
Here $(\mathbb{C}^{*})^{2}$ are the automorphisms of $\mathbb{P}^2$ fixing $p_1,p_2,p_3$, $S_2$ is the group generated by the standard Cremona centered at $p_1,p_2,p_3$, and $S_3$ are the permutations of $p_1,p_2,p_3$.

REPLY [10 votes]: Not exactly. The quadratic transformation commutes with the action of $S_3$, and they both act on $(\mathbb{C}^*)^2$; so the automorphism group is $(\mathbb{C}^{*})^{2}\rtimes (S_3\times S_2)$. You'll find a detailed study of the automorphisms of del Pezzo surfaces in Chapter 8  of Dolgachev's book "Classical Algebraic Geometry: a modern view".<|endoftext|>
TITLE: Is $\liminf \frac{\sigma_{k}(n)}{n}$ finite for every $k$?
QUESTION [6 upvotes]: Can someone show me how to prove that $$\liminf_{n \to \infty} \frac{\sigma_{k}(n)}{n} < \infty$$ for every natural number $k$? Or is this problem open?
Here, $\sigma_{k}(n)=\sigma(\sigma(\sigma(\dots n)))$ is the $k$-th iterate of the sum of divisors function.
Note: I think for $k=2$ this had been proved by Makowski and Schinzel and the limit equals $1$.
Thank you for any help.

REPLY [8 votes]: It is an open problem, only known for $k=1,2$.
Both the conjecture and the known cases are due to Schinzel.
You can find a nice survey here: "On the third iterates of the φ- and σ-functions" H. Maier (1984)
It shouldn't be hard to find more recent references that mention it as open.<|endoftext|>
TITLE: When is the adjoint to a monoidal functor monoidal?
QUESTION [9 upvotes]: Let $\mathcal C,\mathcal D$ be monoidal categories.  Recall that a functor $F : \mathcal C \to \mathcal D$ is lax monoidal if it is equipped with maps $1_{\mathcal D} \to F(1_{\mathcal C})$ and $F(X) \otimes_{\mathcal D} F(Y) \to F(X \otimes_{\mathcal C} Y)$, the latter natural in $X,Y\in \mathcal C$, compatible with associators and unitors.  It is oplax monoidal if it is instead equipped with maps in the other direction, and strong monoidal if it is equipped with isomorphisms.  For each choice of lax/oplax/strong, there is a bicategory of monoidal categories, lax/oplax/strong monoidal functors, and monoidal natural transformations.
Suppose that $F : \mathcal C \to \mathcal D$ is one of lax/oplax/strong monoidal, and also admits a left adjoint $F^L$ in the bicategory of all functors.  Under what circumstances is $F^L$ naturally lax/oplax/strong monoidal?  Under what circumstances does this adjunction come from an adjunction in the bicategory of monoidal categories and lax/oplax/strong monoidal functors?

REPLY [9 votes]: If $L$ and $R$ are a left and right adjoint, then doctrinal adjunction asserts that $L$ is oplax monoidal iff $R$ is lax monoidal. (I'm being a bit imprecise here, treating monoidality as if it were a property instead of a structure, but hopefully the meaning is clear.)<|endoftext|>
TITLE: A problem in functional calculus
QUESTION [9 upvotes]: This is embarrassing, I think it must work, but I can't see how to prove it works. If anyone knows enough functional calculus of operators on a Hilbert space to tell me how to do it, I would be very grateful. Of course, you might also tell me that it is wrong...
Take positive bounded operators $x,y$, and increasing continuous functions $f,g:[0,\infty)\to [0,1]$ with the property that if $f(t)>0$ then $g(t)=1$. Then show that
$$
f(x)\,g(x+y)= g(x+y)\,f(x)=f(x)\ .
$$
Reason why I think it should work: If $f(t)>0$ then $g(t+\mathrm{positive})=1$, so it works... OK, this argument is not legal, but can the statement be proved in functional calculus?
Purpose: This is to be used to show that for a separable C* algebra there is an approximate identity $u_n$ with the property that $u_n u_{n+1}=u_{n+1}\ u_n=u_n$. Maybe someone actually looked at special approximate identities with this sort of property?

OK, the conjecture is false, as noted below. However there was a comment below about the property of approximate identities and whether this might still be true. After some thought I think that the problem is related to functional calculus of several commuting operators. Or rather to this modification: What happens to the functional calculus of several bounded operators which only almost commute, $f(T,S)$ where $\|[T,S]\|<\epsilon$? For an approximate identity we do have for fixed $n$, $\|[u_n,u_m]\|\to 0$ as $m\to\infty$, so it might still be possible to formulate the construction if a small commutator does not matter...
Yes, making this into a general note about functional calculus of two non-commuting operators is a but of an extrapolation.....

NOTE - I have just read the comment on the existence of these approximate identities below - this certainly solves my immediate problem. I now have to find a reference. Many thanks to all involved! I am still curious about the two non-commuting operators though.

REPLY [9 votes]: [EDITED 2015-07-23 to fix a small error pointed out in comments. EDITED 2015-11-07 to fix a mathematical typo.]
It seem that this can fail even when $x$ and $y$ are 2-by-2 matrices. We'll choose them such that $xy\neq yx$, choose $f$ such that $f(x)=x$ and $g$ such that $g(x+y)=x+y$.
Fix $a>0$ to be determined later. Let
$\newcommand{\twomat}[4]{\begin{bmatrix} #1 & #2 \cr #3 & #4 \end{bmatrix}}$
$$x_0= \twomat{2a}{0}{0}{0} \quad,\quad y_0=\twomat{1}{1}{1}{1} $$
which are both positive semi-definite.
Since $x_0+y_0$ is hermitian with trace $2a+2$ and determinant $2a$, it has eigenvalues
$$ r = (1+a)-\sqrt{1+a^2}\quad,\quad R= (1+a)+\sqrt{1+a^2} $$
both of which are strictly positive. Furthermore, one checks that $r< 2a<R$.
Now let $x=R^{-1}x_0$, $y=R^{-1}y_0$, so that $\sigma(x)=\{0, 2a/R\}$ and $\sigma(x+y)=\{r/R, 1\}$. Pick $t_0$ such that
$$ 0 < r/R < t_0 < 2a/R < 1. $$

Define $f$ to take the value $0$ on the interval $[0,t_0]$ and $2a/R$ on the interval $[2a/R,\infty)$ with the obvious linear interpolation on the interval $[t_0,2a/R]$.
Define $g(t)=t$ on the interval $[0,r/R]$ and $g(t)=1$ on the interval $[t_0,\infty)$ with the obvious linear interpolation on the interval $[r/R, t_0]$.

Then both $f$ and $g$ are continuous increasing functions $[0,\infty)\to [0,1]$ and by construction $fg=f$. Since
$f(t)=t$ for all $t\in \sigma(x)$ and $g(t)=t$ for all $t\in \sigma(x+y)$, we have
$$ f(x)g(x+y) = x(x+y) = R^{-2} x_0^2+x_0y_0 = \frac{1}{R^2} \twomat{8a^2+1}{2a}{0}{0} $$
$$ g(x+y)f(x) = (x+y)x = R^{-2} x_0^2 + y_0x_0 = \frac{1}{R^2} \twomat{8a^2+1}{0}{2a}{0} $$
$$ f(x) = x = R^{-1} \twomat{2a}{0}{0}{0} $$
so that none of the desired identities hold.
(PS if one desires an example with "nice" numbers then taking $a=3/4$ should give
$$ x = \twomat{1/2}{0}{0}{0}\quad,\quad y = \frac{1}{3}\twomat{1}{1}{1}{1}\quad,\quad x+y = \frac{1}{6} \twomat{5}{2}{2}{2} $$
with $\sigma(x)=\{0,1/2\}$ and $\sigma(x+y) = \{1/6, 1\}$.)<|endoftext|>
TITLE: What is the normal closure of $GL_2(\mathbb{Z})$ inside $GL_2(\mathbb{Z}_\ell)$?
QUESTION [7 upvotes]: This weird problem popped up in my research:
Let $\ell$ be a prime. Is there a description of the smallest normal subgroup of $GL_2(\mathbb{Z}_\ell)$ containing $GL_2(\mathbb{Z})$?
Is there a description of the corresponding quotient group?

REPLY [12 votes]: It's the subgroup of these elements in $\mathrm{GL}_2(\mathbf{Z}_\ell)$ with determinant in $\pm 1$. Thus the quotient is naturally isomorphic to the quotient of $\mathbf{Z}_\ell^\times$ by $\{\pm 1\}$. 
Indeed, denote by $e_{ij}(x)$ (for $i\neq j$) the matrix with entry $(i,j)$ equal to $x$, diagonal entries equal to 1, and other entries 0, and $d_i(x)$, for $x$ invertible, the diagonal matrix with $(i,i)$ entry equal to $x$, and other diagonal entries equal to $1$; write $s_{ij}(x)=d_i(x)d_j(x^{-1})$. Then, for $\{i,j\}=\{1,2\}$, and $x\in \mathbf{Z}_\ell^\times$, we have $s_{ij}(x)e_{ij}(1)s_{ij}(x)^{-1}=e_{ij}(x^2)$. Since any element in $\mathbf{Z}_\ell$ is a sum of squares (of 2 squares for $\ell$ odd, 4 squares for $\ell=2$, if I remember correctly), it follows that the normal closure of $\mathrm{SL}_2(\mathbf{Z})$ contains all matrices $e_{ij}(y)$ for any $y\in\mathbf{Z}_\ell$. Since such matrices generate $\mathrm{SL}_2(\mathbf{Z}_\ell)$ (because it's a Euclidean ring), we deduce that the normal closure of $\mathrm{SL}_2(\mathbf{Z})$ is all of $\mathrm{SL}_2(\mathbf{Z}_\ell)$. (The same simple argument work in $\mathrm{SL}_d$ for $d\ge 2$.) Now since $\det(\mathrm{GL}_2(\mathbf{Z})=\{\pm 1\}$, we get the result.
(Thanks to grghxy for a correction, I was initially essentially thinking of $\mathrm{SL}_2$ and $\mathrm{GL}_2$ makes an issue with the determinant.)<|endoftext|>
TITLE: Must a finitely generated projective module over a group ring with vanishing coinvariants be trivial?
QUESTION [8 upvotes]: Let $G$ be a (possibly infinite) group. Let $\mathbb{Z}[G]$ be its integral group ring and let $P$ be a finitely generated projective module over $\mathbb{Z}[G]$. Suppose that the coinvariants of $P$ vanish, i.e. 
$$
P_G=P/I_G P=0
$$
where $I_G$ is the kernel of the augmentation map $\mathbb{Z}[G]\to\mathbb{Z}$. Does it follow that $P=0$?

REPLY [4 votes]: This isn't an area that I'm expert on, and it's quite possible there's a much more elementary and/or more general answer.
But if the Bass Conjecture on Hattori-Stallings ranks for group rings is true for $G$ then I think the answer is yes. I don't think the conjecture has been proved for finitely presented groups in general, but there are large classes of groups for which it is known to be true.
If $A$ is a ring, and $P$ a finitely generated projective $A$-module, then $P$ is the image of an idempotent endomorphism of a finitely generated free module, given by a square matrix $M$ over $A$, and the Hattori-Stallings rank of $P$ is the image in $A/[A,A]$ of the trace of $M$, which depends only on $P$ and not the choice of $M$.
If $A=\mathbb{Z}[G]$ for a group $G$, then $A/[A,A]$ is a free abelian group on the images of a set of conjugacy class representatives $[g]$, and Bass conjectured that the Hattori-Stallings rank of any finitely generated projective is a multiple of $[1]$. Previously (I think), Kaplansky proved that the coefficient of $[1]$ in the rank is strictly positive if $P\not\cong0$, which, if we also assume the Bass conjecture for $G$, proves that the image of $M$ under the augmentation map must have non-zero trace, so that the image of $M$ under the augmentation map is non-zero, which is equivalent to $P$ having a non-zero map to the trivial module $\mathbb{Z}$.<|endoftext|>
TITLE: Intuition behind the Kodaira Vanishing Theorem?
QUESTION [17 upvotes]: As the question suggests, what is the intuition behind the Kodaira Vanishing Theorem? The Kodaira Vanishing Theorem says that the cohomology groups $H^q(M, L \otimes K_M)$ vanish for $q \ge 1$ when $L$ is a holomoprhic line bundle over a compact complex manifold $M$ and $L$ admits a Hermitian metric whose curvature form is positive definite. Here, $K_M$ denotes the canonical line bundle.

REPLY [10 votes]: It is well known that a cohomology class can be uniquely expressed by a harmonic form. To show that a certain cohomology group vanishes, it is equivalent to show that every harmonic form form that cohomology group is identically zero.
How does one show that a harmonic form vanishes? Let us look at the case of a real-valued harmonic function. We know that on a compact manifold, every real-valued harmonic function is constant. There are two ways of showing this. One is to use the maximum principle, and another way is to use integration by parts. Instead of considering a harmonic function, we consider a real-valued function $f$ which satisfies the equation$$\Delta f = cf,$$for some positive constant $c$. Such a function $f$ must vanish identically. Here, one can use either the maximum principle or integration by parts.
For the maximum principle, we consider a point where the maximum of $f$ is achieved. At that point, $\Delta f$ is nonpositive. Because of the positivity of $c$, it forces $f$ to be nonpositive. So $f$ is nonpositive everywhere. Applying the same argument to $-f$ instead of $f$, we conclude that $f$ is nonnegative everywhere. Hence, $f$ is identically zero.
For integration by parts, we simply multiply$$\Delta f = cf$$ by $f$, integrate over $M$, and get$$-(df, df)_M = c(f, f)_M,$$ which forces $f$ to be identically zero by the positivity of $c$. The argument of integration by parts is the same as saying the operator $-\Delta + c$ is positive definite.
Let us go back to the consideration of harmonic forms $\varphi$. Instead of $\Delta$, we have the operator$$\square = \overline{\partial}^* \overline{\partial}  + \overline{\partial}\overline{\partial}^*.$$When we compute $\square \varphi$, we will get$$\square\varphi = -(\text{Tr}\, \nabla \overline{\nabla} \varphi) + A\varphi,$$ where $A$ is a zero-order linear operator. The term $-(\text{Tr}\,\nabla \overline{\nabla}\varphi)$ corresponds to $-\Delta f$, and the term $A\varphi$ corresponds to $c \varphi$. When the curvature form of $L$ is positive, the operator $A$ is a positive operator, it corresponds to the positivity of $c$, and we get the vanishing of the harmonic form $\varphi$.
In order to rigorously prove the Kodaira Vanishing Theorem, we would derive the formula for $\square\varphi$.<|endoftext|>
TITLE: Are there discontinuities in the large cardinal hierarchy?
QUESTION [13 upvotes]: Suppose that $\Phi(x,y)$ is a formula in the language of set theory so that for each natural number $n$, the axiom $\exists x\Phi(n,x)$ is a large cardinal axiom (for example consider $n$-huge cardinals). Then does $\text{Con}(ZFC+\exists x\Phi(n,x))$ for all $n$ imply $\text{Con}(ZFC+\forall n\exists x\Phi(n,x))$ (i.e. informally is the large cardinal heirarchy continuous in this sense or are there jumps in the large cardinal heirarchy)? Could there be a large cardinal axiom  whose consistency strength is greater than each axiom $\exists x\Phi(n,x)$ for all natural numbers $n$ but whose consistency strength is strictly less than $\forall n\exists x\Phi(n,x)$?
Is it possible that there is a large cardinal axiom $\Psi$ so that $ZFC\models\text{Con}(\Psi)\leftrightarrow\forall n\in\omega\,\text{Con}(\Phi(n,x))$? What would the answer to these questions be if we replaced $\text{Con}(\Phi)$ with the existence of a transitive model that satisfies $\Phi$ or some other strengthening of mere consistency?
If one needs a formalization of what is meant by a large cardinal axiom, one can use this formalization due to Woodin mentioned in Mohammad Golshani's answer here or something similar to that.
I am also be interested in discontinuities in the large cardinal heirarchy by sequences of cofinality other than $\omega$ (for instance, for a sequence of cofinality $\aleph_{1},\mathfrak{c}$ or the first inaccessible cardinal), but I do not know of a nice way to formalize this notion.

REPLY [16 votes]: By compactness, if Con(ZFC$+\forall n\exists x\Phi(n,x)$) were a consequence of the infinite set of statements Con(ZFC$+\exists x\Phi(n,x)$), where $n$ ranges over genuine natural numbers, then it would already be a consequence of finitely many of those statements, and that's clearly not the case in non-trivial examples (like $n$-huge or $n$-inaccessible, provided of course that these are consistent).<|endoftext|>
TITLE: Analytical formula for topological degree
QUESTION [10 upvotes]: At the first page of the following article http://arxiv.org/pdf/1004.1018v1.pdf [edit: the formula on the arXiv differs from the formula in the published paper, and the formula displayed below is the published version], I read "The degree of a Hölder continuous function $f:\mathbb{S}^1 \to \mathbb{S}^1$ of exponent $\alpha$ is expressed by an analytic formula..." 
$$
\text{deg} (f)=\frac{1}{(2i \pi)^{2k}} \int \overline{f(z_0)} \frac{f(z_1)-f(z_0)}{z_1-z_0} \ldots \frac{f(z_0)-f(z_{2k})}{z_0-z_{2k}} dz_0 \ldots dz_{2k}
$$
whenever $\alpha(2k+1)>1$. 
Problem is, I don't know how to understand this formula to make it work. For example, taking $f=id$ and $k=1$ and integrating over $\mathbb{S}^1$,
$$
\text{deg} (f)=   \frac{1}{(2i \pi)^2} \int_{\mathbb{S}^1} \overline{z_0}  dz_0 \int_{\mathbb{S}^1}  dz_1  \int_{\mathbb{S}^1} dz_2=0
$$
which is obviously false. The formula comes from "Non-commutative geometry" http://www.alainconnes.org/docs/book94bigpdf.pdf by Connes, p 215 which seems really interesting but doesn't help me much.

REPLY [7 votes]: First, let me apologize for the confusion regarding the pre-factor. It should read $(-1)^{k}(2\pi i)^{-2k-1}$. This can be seen in several ways. One way is to compare the constant $c_k$ appearing in Proposition 3.b on page 215 of Connes' book (http://www.alainconnes.org/docs/book94bigpdf.pdf) with those coming from the pairing with $K$-theory in Proposition 3.a on page 230. Beware that there is a missing $2\pi i$ coming from that the cyclic $1$-cocycle $\tau$ on page 215 needs to be normalized by $(2\pi i)^{-1}$ to reproduce the winding number. 
Another way to see the pre-factor is in the origin of the formula. It comes from the following index computation. Really, it is what the odd index pairing in cyclic cohomology boils down to once all normalization constants are cancelled. We consider a projection $P$ and a unitary operator $u$ on a Hilbert space $\mathcal{H}$ such that the commutator $[P,u]$ belongs to the Schatten class $\mathcal{L}^{2p}(\mathcal{H})$. The operator $T:=PuP:P\mathcal{H}\to P\mathcal{H}$ is Fredholm, its inverse in the Calkin algebra is represented by $Pu^{-1}P$. To compute the index of $PuP$, one can use the computations in Proposition 4 on page 303 of Connes' bok. Following the proof of this proposition, 
$$\mathrm{ind}(PuP)=(-1)^p\mathrm{Tr}(\gamma e[F,e]^{2p}),$$
where $\gamma=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$, $e=\begin{pmatrix}P&0\\0&P\end{pmatrix}$ and $F=\begin{pmatrix}0&U^*\\U&0\end{pmatrix}$ are operators on $\mathcal{H}\oplus \mathcal{H}$. After some computations, we arrive at the expression
$$(-1)^p\mathrm{Tr}(\gamma e[F,e]^{2p})=(-1)^p\mathrm{Tr}((2P-1)([U^*,P][U,P])^p).$$
If $[U^*,P]\in \mathcal{L}^{2p-1}(\mathcal{H})$ one can proceed and write
$$(-1)^p\mathrm{Tr}((2P-1)([U^*,P][U,P])^p)=\frac{(-1)^{p}}{2}\mathrm{Tr}((2P-1)[(2P-1),U^*][P,U]([P,U^*][P,U])^{p-1})=(-1)^{k-1}\mathrm{Tr}(U^*[P,U]([P,U^*][P,U])^{k}),$$
where $k=p-1$.
The winding number of a mapping $f:S^1\to S^1$ can be computed from an index as above by taking $u$ to be point wise multiplication by $f$ and $P$ to be the Szegö projection. To be precise, the index formula for Toeplitz operators on the circle says that $\mathrm{deg}(f)=-\mathrm{ind}(PuP)$. The Szegö projection on the circle is defined by 
$$Pg(z):=\frac{1}{2\pi i}\int_{S^1} \frac{g(w)}{w-z}\mathrm{d}w=\frac{1}{2\pi}\int_{0}^{2\pi} \frac{g(\mathrm{e}^{i\theta})}{1-\mathrm{e}^{-i\theta} z}\mathrm{d}\theta.$$
The integral should be interpreted as a non tangential interior limit, i.e. $Pg(z):=\lim_{r\to 1^-} Pg(rz)$. For $a\in C^\alpha(S^1)$, the operator $[P,a]$ has an absolutely integrable integral kernel given by $\frac{a(w)-a(z)}{w-z}$. Note that if $f:C^\infty(S^1)$ is smooth, $[P,f]\in\mathcal{L}^{2p}(L^2(S^1))$ for any $p>0$ and all traces considered above are well defined. More generally, for $f\in C^\alpha(S^1)$, $[P,f]\in\mathcal{L}^{\frac{1}{\alpha},\infty}(L^2(S^1))\subseteq \cap_{p>\frac{1}{\alpha}} \mathcal{L}^p(L^2(S^1))$. Using the integral formula in Theorem 2.4 from http://arxiv.org/pdf/1004.1018v1.pdf, we arrive at the expression 
$$\mathrm{deg}(f)=(-1)^{k}\int_{(S^1)^{2k+1}} \overline{f(z_0)}\frac{f(z_1)-f(z_0)}{2\pi i(z_1-z_0)}\prod_{j=1}^{k}\frac{\overline{f(z_{2j})}-\overline{f(z_{2j-1})}}{2\pi i(z_{2j}-z_{2j-1})}\frac{f(z_{2j+1})-f(z_{2j})}{2\pi i(z_{2j+1}-z_{2j})}\mathrm{d}z_0\mathrm{d}z_1\cdots \mathrm{d}z_{2k},$$
where we identify $z_{2k+1}=z_0$. This gives the pre-factor $(-1)^k(2\pi i)^{-(2k+1)}$, where the number of $2\pi i$ correspond to the number of commutators with the Szegö projection. 
There is a tricky point to this formula. The Szegö projection is defined from an interior limit. However, for $a\in C^\alpha(S^1)$ the commutator $[P,a]$ is defined from an absolutely integral kernel. The latter fact makes it unnecessary to include the interior limit in the expression for the mapping degree. However, when computing it makes life easier. To consider an example of this, take $f(z)=z$. The case $k=0$ holds formally as the integrand $\overline{f(z_0)}\frac{f(z_1)-f(z_0)}{2\pi i(z_1-z_0)}$ has the limit $\frac{1}{2\pi i}\overline{f(z_0)}f'(z_0)$ as $z_1\to z_0$ (this can be made rigorous using for instance Theorem 3.9 in Barry Simon's book on trace ideals). For $k=1$, we have 
\begin{align*}
\mathrm{deg}(f)&=
\frac{-1}{(2\pi i)^3}\int_{(S^1)^3} \overline{z_0}\frac{z_1-z_0}{z_1-z_0}\frac{\overline{z_2}-\overline{z_1}}{z_{2}-z_{1}}\frac{z_0-z_{2}}{z_{0}-z_{2}}\mathrm{d}z_0\mathrm{d}z_1\mathrm{d}z_{2}\\
&=\frac{-1}{(2\pi i)^3}\int_{(S^1)^3} \overline{z_0}\frac{\overline{z_2}-\overline{z_1}}{z_{2}-z_{1}}\mathrm{d}z_0\mathrm{d}z_1\mathrm{d}z_{2}.
\end{align*}
If we perform the integral over $z_2$, keeping in mind that it is defined from an interior limit, we arrive at
$$\frac{-1}{(2\pi i)^3}\int_{(S^1)^3} \overline{z_0}\frac{\overline{z_2}-\overline{z_1}}{z_{2}-z_{1}}\mathrm{d}z_0\mathrm{d}z_1\mathrm{d}z_{2}=\frac{1}{(2\pi i)^2}\int_{(S^1)^2} \overline{z_0}\overline{z_1}\mathrm{d}z_0\mathrm{d}z_{1}.$$
Here we can apply the residue theorem arriving at the identity $\mathrm{deg}(f)=1$.<|endoftext|>
TITLE: Invariants of the maximal unipotent subgroup of GL(n) acting on the space of n by n matrices
QUESTION [5 upvotes]: Let $G=GL(n,\mathbb{C})$ and let $U\subset G$ be a maximal unipotent subgroup. (For example,assume that U is the set of upper triangular matrices with ones in the diagonal.) Now let $X=M_{n}(\mathbb{C})$ be the space of $n$ by $n$ complex matrices and consider the usual action of $U$ on $X$ given by left multiplication. Is it possible to describe the space $\mathbb{C}[X]^{U}$ explicitly using generators and relations? In other words, is there an explicit realization of the categorial quotient of $X$ by $U$ as an affine algebraic variety?

REPLY [2 votes]: Questions of this type can be approached from the direction of algebraic geometry (in a suitable generality) or from the direction of invariant theory involving algebraic group actions.   I'd point especially to Chapter 3 in the 1997 monograph by Frank D. Grosshans, Algebraic homogeneous spaces and invariant theory, Lect. Notes in Math. 1673, Springer.   He works in some generality throughout, taking $k$ to be any algebraically closed field of any characteristic and using "traditional" algebraic geometry language.  In particular, he looks at the case of general linear groups in $\S13$ of that chapter and describes (for either left or right action) the rings of invariants for $U$ on suitable algebras of polynomials.
Grosshans builds on older foundations for reductive group actions on varieties and resulting homogeneous spaces, going back to Hilbert, Mumford, and others.  His results rely a lot on combinatorics and are often more comprehensive than what you ask for.   On the other hand, he is fairly straightforward and careful about the details.  Though he works in arbitrary characteristic, his approach is not too different from that of the Russian school (Vinberg, Popov, Panyushev, ...) who typically work over an algebraically closed field of characteristic 0.  Along the way, Grosshans does provide a fairly detailed concrete picture of the quotients obtained, starting with the generators of rings of invariants.<|endoftext|>
TITLE: Do Iwahori-Hecke algebras come from cohomology classes?
QUESTION [7 upvotes]: Let $W$ be a Coxeter group. The Iwahori-Hecke algebra $H_q(W)$ is a deformation of $k W$. 

 Question:  is there some way to interpret the deformation $H_q(W)$ as a cohomology class? It doesn't seem like $kW$ twisted by a class in $H^2(W,k^{\times})$...

REPLY [10 votes]: Not in an interesting way.  Cohomology classes describe formal deformations (i.e. deformations over an Artinian ring, or more generally complete local rings), and the Iwahori-Hecke algebra is trivial on a formal neighborhood of $q=1$.  In fact, the only points where it is not trivial on a formal neighborhood is when $q$ is a root of unity of small order.<|endoftext|>
TITLE: Automorphisms of finite order in $Out(\widehat{F_2})$
QUESTION [12 upvotes]: Let $\widehat{F_2}$ be the pro-$\ell$ completion of the free group of rank 2, where $\ell$ is some prime. 
Every outer automorphism of $F_2$ induces an outer automorphism of $\widehat{F_2}$, hence an injection $Out(F_2)\rightarrow Out(\widehat{F_2})$.
Let $\alpha\in Out(\widehat{F_2})$ be an outer automorphism. Consider the three properties:
(A) $\alpha$ has finite order.
(B) $\alpha$ does NOT lie in the closure $\overline{Out(F_2)} $ in $Out(\widehat{F_2})$.
(C) $\alpha$ is of determinant 1 under the abelianization map $Out(\widehat{F_2})\rightarrow GL_2(\mathbb{Z}_\ell)$ (does this map admit a section?)
Are there any outer automorphisms which satisfy properties $(A),(B),(C)$?
Are there any outer automorphisms which satisfy $(A)$ and $(B)$?
I'm trying to get a grip on the outer automorphisms of $Out(\widehat{F_2})$ which don't come from "limits" of outer automorphisms of $F_2$. These automorphisms seem very mysterious to me.

REPLY [2 votes]: If an automorphism fails (C), then obviously it satisfies (B). An (A)+(B) example easily extends from rank 1 to ranks 2: take a root of unity $\zeta$ in $\mathbb Z_p^\times$ that is not $\pm1$ (requiring $p\geqslant 5$). If the generators of the free group are $x$ and $y$, then there is an automorphism of the $p$-adic completion given by $x\mapsto \zeta x$ and $y\mapsto y$. It obviously has finite order, but it has nontrivial determinant, so it is not in the closure of $Out(F)$.

Define $SOut$ as the subgroup of $Out$ with determinant $1$. I believe that $SOut(F)$ is dense in $SOut(\hat F)$, so there is nothing satisfying (B)+(C), let alone all three conditions. I will not attempt to show that, but only that all torsion is conjugate into the closure, which is awfully close. Specifically, I will show that an $\ell$-Sylow subgroup is contained in the closure; and I will suggest that there is no $p$-torsion ($p\geqslant 5$). (Added at last minute: that only covers prime-power torsion and does not show that torsion of order with multiple prime factors lifts from $SL_2(\mathbb Z_p)$ to $SOut(\hat F)$, let alone to the closure of $SOut(F)$.) The key point is that $\hat F$ is a pro-p-group and thus pro-nilpotent. That makes it much easier to analyze than if it were, say, the 2,3-completion and pro-soluble, let alone the full completion with simple composition factors. The length $n$ nilpotent quotient of a group is a characteristic quotient, so an automorphism of a group induces a automorphism of its length $n$ nilpotent quotient. The group of automorphisms of a pro-nilpotent group is the inverse limit of the automorphism groups of the length $n$ quotients. (The transition maps in this inverse limit need not be surjective, but are in the free case, as indicated below.)
The automorphism group of a nilpotent group is easy to understand. If $G_n$ is a length $n$ nilpotent group, $G_{n-1}$ its maximal length $n-1$ quotient and $Z$ the kernel, then the kernel of $Aut(G_n)\to Aut(G_{n-1})$ consists of automorphisms that differ from the identity by shearing into the kernel: $\phi$ so that $\phi(g)g^{-1}\in Z$. Using the centrality of $Z$, the map $g\mapsto \phi(g)g^{-1}$ is a homomorphism $G_n\to Z$. A useful observation is $\phi(g)g^{-1}=g^{-1}\phi(g)$. The kernel is isomorphic to to the group of homomorphisms $G_n^{ab}\to Z$. $Aut(G_n)$ need not surject to $Aut(G_{n-1})$, but a similar analysis shows that if an automorphism of $G_{n-1}$ lifts to an endomorphism of $G_n$, it has an inverse. By induction that yields a nice clean statement: an endomorphism of a nilpotent group is an automorphism if and only if the induced endomorphism of the abelianization is an automorphism.
Applied to $\hat F$, this shows that the kernel $Aut(\hat F)\to GL_2(\mathbb Z_p)$ is a torsion-free pro-$p$ group. Also, the map is surjective because freeness makes it easy to lift endomorphisms, which are then automorphisms. (Similarly, it is easy to lift automorphisms of the length $n$ nilpotent quotient to endomorphisms of the free group, but it takes work to lift them to automorphisms, which is why I do not do it. That would imply that $SOut(F)$ is dense in $SOut(\hat F)$. That would answer your question, but the points about Sylow subgroups would still be interesting.)
Thus the kernel $Aut(\hat F)\to GL_2(\mathbb Z_p)$ is built out of composition factors of the form $Hom(A,B)$, where $A$ and $B$ are composition factors of $\hat F$, so the kernel is a pro-$p$ group. More careful consideration shows that $A$ and $B$ are torsion free, hence so the kernel. The kernel $Out(\hat F)\to GL_2(\mathbb Z_p)$ is a quotient of the prior kernel by $\hat F$, so also a pro-$p$ group. I believe that directly applying an analogous stage-by-stage analysis shows that it is torsion-free. The kernel of $GL_2(\mathbb Z_p)\to GL_2(\mathbb F_p)$ is also a pro-$p$-group, torsion-free unless $p=2$. Extension by a $p$-group cannot change the prime-to-$p$ Sylow subgroups. For each $\ell$ prime to $p$, the $\ell$-Sylow subgroups of $Aut(\hat F)$, $Out(\hat F)$, $GL_2(\mathbb Z_p)$, and $GL_2(\mathbb F_p)$ are isomorphic by the quotient map. Similarly, the $\ell$-Sylow subgroups of $SAut(\hat F)$, $SOut(\hat F)$, $SL_2(\mathbb Z_p)$, and $SL_2(\mathbb F_p)$ are isomorphic. Since $SOut(F)=SL_2(\mathbb Z)$ surjects to $SL_2(\mathbb F_p)$, its closures in $SAut(\hat F)$, $SOut(\hat F)$, and $SL_2(\mathbb Z_p)$ must contain full $\ell$-Sylow subgroups. In particular, all $\ell$-power torsion is conjugate into Sylow subgroups and thus into the closure of $SOut(F)$. And I claim that there is no $p$-power torsion (except for $p=2,3$, where all the torsion in $GL_2(\mathbb Z_p)$ is conjugate into $GL_2(\mathbb Z)=Out(F)$).
All that applies to any rank free group ($Out(F)$ always surjects to $GL(\mathbb Z)$, but is no longer a isomorphic), except that there are more possibilities for $p$-power torsion in $GL_n(\mathbb Z_p)$. I think it is all conjugate into $GL_n(\mathbb Z)$ (though I’m more sure about the fields $\mathbb Q_p$ and $\mathbb Q$). I don’t know how to tell if it lifts to $Out(\hat F)$, but if it does, it probably lifts to $Out(F)$. Anyhow, I claim without proof that $SOut(F)$ is dense in $SOut(\hat F)$.<|endoftext|>
TITLE: What can we say about the boundary of the level set of a Sobolev function?
QUESTION [5 upvotes]: I'm a beginner of the area of free boundary problem. Let me first give some background: 
$\Omega \subset \mathbb{R}^n$ is an open connected set, and locally $\partial \Omega$ is a Lipschitz graph. 
Consider the convex set $$K:＝\{v \in L^1_{loc}(\Omega): \nabla v \in L^2(\Omega) \,, v=u^0 \mbox{on $\partial \Omega$}\},$$
where $u^0\ge0,u^0 \in L^1_{loc}(\Omega)$, and $\nabla u^0 \in L^2(\Omega)$. 
We are looking for the minimizer $u$ of the functional 
$$J(v):=\int_{\Omega}(|\nabla v|^2+\chi_{\{v>0\}})$$ in the class $K$.
It is proved that the minimizer $u$ exists and satisfying the following properties: $$u \ge 0 , \, \Delta u=0 \, \mbox{on the open set $\{u>0\}$}, \, \mbox{and $u$ is subharmonic},$$ see section 1-2 in the paper by Alt and Caffarelli here
It is also proved in section 3-5 that $\partial\{u>0\}$ has locally finite $\mathcal{H}^{n-1}$ measure. However, a lot of intermediate theorems such as Corollary 3.3 and Remark 4.2 are based on the fact that $|\partial\{u>0\}|=0$, that is, $|\partial\{u=0\}|=0$. This fact is not proved in the paper, and generally it is not true if $u$ is merely continuous.  
Now my question is, why is $|\partial\{u=0\}|=0$ true?  I've been stucked on it for a couple of days. 
Another related question is, what conditions on a general function $u$, which is not necessarily the minimum of the functional $J$, can guarantee that $|\partial\{u=0\}|=0$? Is the assumption that $u$ is a Sobolev function enough? How about $u$ is subharmonic?  
Any suggestions would be appreciated. Thanks!

REPLY [2 votes]: I figured out the problem later and until today I could have time to write it down.
The proof of the rectifiability of free boundary $\partial \{u>0\}$ requires the Lipchitz regularity of $u$ across the free boundary and the nondegeneracy of the function $u$, see theorem 3.2-theorem 4.5 in Alt and Caffarelli(1981).
In the proof of the Lipchitz regularity of $u$, the authors proves that if $u(x)>0$, then $|\nabla u(x)|$ is bounded, and then they claim that $u$ is Lipchitz. At first I thought it was a problem, because if $|\partial\{u>0\}|$>0, then how can one give a bound for $|\nabla u(x)|$? That is the reason I asked the question here.
Later I found there was no problem.  By Gilbarg and Trudinger, $Du^+=Du$ on $\{u>0\}$ and $Du^+=0$ otherwise. This means the weak derivative of $u$ is always $0$ on $\{u=0\}$ no matter how crazy $|\partial\{u>0\}|$ is. Then by a well known but nontrivial result that $C^{0,1}=W^{1,\infty}$, one can finally say $u$ is Lipchitz.
To tell the truth, I was reluctant to post the answer here, because I think the general question related to the boundary of a level set of a function is still meaningful, even just for the mere object of level set of a "nice" function. I have seen some references studying partial results of level sets of functions. For example, this question is related to the generalization of Morse-Sard Theorem for "nice" functions, coarea formula for $\mathcal{H}^s$ rectifiable sets and so on. My adviser told me the regularity of the free boundary of a solution of a general fully nonlinear equation is not known, and the problem is very hard.
By the way, one can easily construct a Lipschitz function $u$ such that $\partial\{u>0\}$ can be as crazy as possible.<|endoftext|>
TITLE: Loop space generalization
QUESTION [8 upvotes]: Let $X$ be a based connected space. The space of based continuous morphisms $Top_{\ast}(S^1,X)$ 
is the space of loops $\Omega X$. Since $S^1$ is homotopy equivalent to the Eilenberg-Mac Lane Space $K(\mathbf{Z},1)$ I was wandering if there is any "geometric interpretation" of the space of based continuous morphisms $Top_{\ast}(\mathbf{C}P^{\infty},X)$.  $\mathbf{C}P^{\infty}$ can be identified with the Eilenberg-Mac Lane Space $K(\mathbf{Z},2)$
More generally, what would be the "geometric interpretation" of the space of based continuous morphisms $Top_{\ast}(K(\mathbf{Z},n),X)$.

REPLY [2 votes]: [Edited after the comment below]
Well, I don't know exactly what you mean by "geometric interpretation", but 
for topological group $G$ the unpointed mapping spaces $Map(BG,X)$ are examples of homotopy fixed points. and the pointed mapping spaces $Map_*(BG,X)$ fits in the fibration
$$Map_*(BG,X) \rightarrow Map(BG,X) \rightarrow X$$ where the last map is
the evaluation map.
Now, $$Map(BG,X)\cong Map^G(EG,X)$$ where we provide $X$ with trivial action,
and this latter is nothing but the homotopy fixed points of $X$ with trivial $G$-action.
If $G$-action on $X$ is trivial, $X$ is just the fixed point set, so one can consider $Map_*(BG,X)$ as the space that measures the difference between the genuine fixed point set and the homotopy fixed point set.
As one can construct a model for $K(A,n-1)$ that has a structure of a topological group, you can take $G=K(A,n-1)$ in the above.<|endoftext|>
TITLE: A question of Erdős
QUESTION [13 upvotes]: In the following paper (pages 122-23), Erdős asks if there is a constant $c > 0$ such that every subset $A$ of plane of area more than $c$ contains the vertices of a triangle of unit area.
Is this still open? Has anyone discovered interesting lower bounds for $c$?
As a "motivational" puzzle you can show that if $c = \infty$ then $A$ contains vertices of triangles of all possible areas.
Addendum: Monroe has suggested looking at outer measure: First note that there is a subset of plane of full outer measure avoiding vertices of unit area triangles. Why? Just use transfinite induction to construct a unit area triangle free set X which meets every compact positive area subset of plane. This is doable because at any stage we just need to avoid less than continuum many "heights" corresponding to the base lengths that we have accumulated by that stage. Similarly for category. But what about the following variation?
Question: Let X be a bounded subset of plane. Must there exist a full outer measure subset Y of X that avoids vertices of triangles of unit area? Of course the answer is yes under CH or MA, but does this hold in ZFC?
To illustrate why this could be hard, let me state a related problem of P. Komjath.
Question: Let X be a bounded subset of plane. Must there exist a full outer measure subset Y of X that avoids rational distances? Can we even avoid unit distance?
There are uncountably many variations on these but they all currently seem very hard to me.

REPLY [3 votes]: The review by Marton Elekes of a slightly more recent paper, G Pantsulaia and N Rusiashvili, On a certain version of the Erdos problem, Georgian Int. J. Sci. Technol. 6 (2014), no. 3, 257-264, MR3236619, confirms that the question is still open. 
Pantsulaia has another paper on the topic, On maximal plane sets containing only the vertices of a triangle with area less than one, Georgian Int. J. Sci. Technol. 6 (2014), no. 2, 113-117, MR3235996.<|endoftext|>
TITLE: 1-dimensional representations of the affine Hecke algebra for $SL_2$
QUESTION [7 upvotes]: Kazhdan-Lusztig theory gives a correspondence between irreducibles of the affine Hecke algebra for a simply connected linear algebraic group $G$ and certain homological data extracted from the Steinberg variety (for $q$ not a root of unity).  I want to see this correspondence explicitly for $SL_2$ by (1) looking at generators and relations for the Hecke algebra $H$ and (2) looking at the geometric data.  My problem is that for the 1-dimensional representations I can't get them to match up.  I expect it's a somewhat silly error but I can't find it, in any case.
Edit: The sources I'm relying on Criss/Ginzburg (paper version here: http://arxiv.org/abs/math/9802004).  The affine Hecke algebra in this case I think is some deformation of the group algebra of the affine Weyl group.  On the geometric side it's $G \times \mathbb{C}^*$-equivariant K-theory on the Steinberg variety.
Method 1: The affine Hecke algebra for $SL_2$ is an algebra $H$ with generators $T, X, X^{-1}$ and relations
$$(T + 1)(T - q) = 0$$
$$TX^{-1} - X^1 T = (1-q)X^1$$
$$X X^{-1} = X^{-1}X = 1$$
It's not too hard to compute the one-dimensional representations.  For central character $\sqrt{q} + 1/\sqrt{q}$ (the scalar which $X + X^{-1}$, the generator for $\mathcal{O}(T//W)$, acts by), one has the representations
$$(T, X) \mapsto (-1, 1/\sqrt{q}), (q, \sqrt{q})$$
and for central character $-\sqrt{q} - 1/\sqrt{q}$
$$(T, X) \mapsto (-1, -1/\sqrt{q}), (q, -\sqrt{q})$$
In total, giving four representations, two for each central character.
Method 2: We should be able to read off the irreducibles by Kazhdan-Lusztig theory.  Namely, for generic $q$, fix $g$ with desired central character, and take the $g$-fixed locus of the Springer resolution $\mu^g: \tilde{\mathcal{N}}^g \rightarrow \mathcal{N}^g$.  The total Borel-Moore homology of $\mu^{g, -1}(x)$, choosing representatives $x$ in each $C(g)$-orbit of $\mathcal{N}^g$, is a representation of $H$ specialized at this central character.
For $g = \pm \text{diag}(\sqrt{q}, 1/\sqrt{q})$ one has $\mathcal{N}^g = \mathbb{A}^1$ the nilpotent radical of the usual Borel.  It has two $C(g)$ orbits: $x = 0$ and $x \in \mathbb{A}^1 - 0$.
(1) For $x = 0$, the fiber has two-dimensional homology.  The double centralizer $C(g, 0) = T$, which is connected, so we don't have to worry about isotypical components of $A(g, 0) = C(g, 0)/C(g, 0)^\circ$.
(2) For $x = \left(\begin{array}{cc}0&1\\0&0\end{array}\right)$, the fiber is a point, so we have a one-dimensional homology.  The component group isn't relevant here.
So here, we only find two 1-dim representations, one for each central character.  I must have misunderstood something.

REPLY [3 votes]: It turns out I was confused about the statement on the geometric side.  I'm using Ginzburg's paper as a reference.  One doesn't just take the (co)homology of the fibers for each orbit representative.  The statement (page 44) is really that the irreducibles (for a given central character) are the summands we get by applying the decomposition to the map $\mu$:
$$\mu_* \mathbb{C}_{\tilde{\mathcal{N}}^a} = \bigoplus_{\phi} L_\phi(k) \otimes IC(k)$$
In our case, the map is $\mu: \mathbb{A}^1 \cup \text{pt} \rightarrow \mathbb{A}^1$.  The cohomology upstairs is $\mathbb{C}^2$, and decomposes into two one-dimensional representations.  So for each central character has two one-dimensional representations as expected; one for each $C(g) = \mathbb{G}_m$ orbit in $\mathbb{A}^1$.
Anyway, thanks for your help, everyone!  The comments and answers were very informative.<|endoftext|>
TITLE: Counterexample to high dimensional Nielsen realization problem
QUESTION [5 upvotes]: Can someone give me an example of a closed smooth oriented manifold $M$ and an orientation preserving diffeomorphism $f:M\rightarrow M$ such that $f^k$ is isotopic to the identity for some $k\geq 1$, but $f$ is not isotopic to any finite order diffeomorphism?  Similarly for homeomorphisms, and also for homotopies.
Of course, Kerckhoff's famous theorem says that this is impossible in dimension $2$.

REPLY [5 votes]: This is discussed at great length (with copious references) in Jonathan Block's and Shmuel Weinberger's paper. The title of the paper is the suggestive "On the generalized Nielsen realization problem" (comm math helv, 2008)<|endoftext|>
TITLE: Survey papers on the role played by PDE in mathematics
QUESTION [7 upvotes]: There are already several questions on MathOverflow that inquire about the many diverse relationships between PDE and several other 'areas' of mathematics (e.g., algebraic and differential geometry and topology, number theory, harmonic analysis, probability theory, dynamical systems, etc.); however, most of the answers give only a few particular examples. 
The aim of this question is to collect a [big-list] of references (i.e., broad surveys or monographs) that specifically focus on the role played by PDE in various other areas of mathematics, or on methods "stemming from other topics" that are used in the analysis of PDE.

REPLY [4 votes]: Klainerman, Partial Differential Equations, in Princeton Companion to Mathematics; a longer version is freely available here;
Klainerman, PDE as a Unified Subject;
Brezis and Browder, Partial Differential Equations in the 20th Century;
Yau, The Role of Partial Differential Equations in Differential Geoemtry;
Evans, Weak KAM Theory and Partial Differential Equations; 
Stroock, Partial Differential Equations for Probabilists;
etc.<|endoftext|>
TITLE: Elementary Proof of Infinitely many primes $\mathfrak{p} \in \mathbb{Z}[i]$ in the sector $\theta < \arg \mathfrak{p} <\phi $
QUESTION [16 upvotes]: A quick look at the primes in $\mathbb{Z}[i]$ suggests they might be evenly distributed by angle if we zoom out on a coarse enough scale.

I would like ask about the much weaker statement forgetting questions about angular bias.  
Do all sectors $(\theta, \phi)$ have infinitely many primes $ \mathfrak{p} $? 
The idea being to find cases where it is possible to mimick Euclid's proof of the infinitude of primes in $\mathbb{Z}$.  I was not able to find such an elementary proof.  Certainly let $|\theta - \phi| \ll \frac{\pi}{8}$.
This may be like the cause of Dirichlet's theorem of primes in arithmetic progressions, Euclid's proof extends to some arithmetic progressions (like $4k+3,6k+5, 8n+5$) and not others [1]

REPLY [19 votes]: Yes, and in fact there will still be many primes even if the size of the sector decrease to zero quite quickly. 
In the 2001 paper "Gaussian primes in narrow sectors," Harman and Lewis use sieve methods to prove that there are Gaussian primes with $|p|^2\leq X$ in sectors where the angle goes to zero at the rate of $X^{-0.381}$. Specifically they prove that:

Theorem: Let $X>X_0$. Then the number of Gaussian primes $p$ with $|p|^2\leq X$ in the sector $\beta\leq \arg p\leq \beta +\gamma$ is at least $$\frac{cX\gamma}{\log X}$$ for an absolute constant $c>0$, where $\beta,\gamma$ satisfy $0\leq \beta \leq \frac{\pi}{2},$ $X^{-0.381}\leq \gamma \leq \frac{\pi}{2}$.

One interesting corollary of this result is that there are infinitely many primes $p$ satisfying $$\{\sqrt{p}\}<p^{-0.262},$$ where $\{x\}$ denotes the fractional part of $x$.<|endoftext|>
TITLE: Degrees of generators of radical ideals
QUESTION [10 upvotes]: Let $I \subseteq \mathbb{C}[x_1,\ldots,x_n]$ be an ideal generated by polynomials $f_1,\ldots,f_r$ of degree at most $d$. Is it possible to generate the radical $\sqrt{I}$ of this ideal with polynomials of degree at most $d$? If not, is there any other upper bound known in terms of $d$ for the degrees of a set of polynomials generating $\sqrt{I}$?

REPLY [7 votes]: In general, one can not bound the degree of the radical $\sqrt{I}$ using only $d$, even in the polynomial ring $R=\mathbb C[x,y,z,t]$. Consider the (complete intersection) ideal $I= (x^mt-y^mz, z^{n+2}-xt^{n+1})$, generated in degrees $m+1$ and $n+2$. Then the radical of $I$ has a generator of degree $mn+2$, as shown in Lemma 2.4 of this paper.<|endoftext|>
TITLE: Relationship between étale and topological $K(\pi,1)$s
QUESTION [12 upvotes]: I was trying to find a proof, or a counterexample to the claim that if $X/\mathbb{C}$ is connected smooth projective, then $X$ is a $K(\pi^{\mathrm{\acute{e}t}},1)$ if and only if $X^\mathrm{an}$ is a $K(\pi,1)$.
For me, $X/\mathbb{C}$ should be a $K(\pi^{\mathrm{\acute{e}t}},1)$ if for all LCC $\mathcal{F}$ the natural map
$$H^i(\pi_1^{\mathrm{\acute{e}t}}(X,\overline{x}),\mathcal{F}_{\overline{x}})\to H^i(X_{\mathrm{\acute{e}t}},\mathcal{F})$$
is an isomorphism. 
I believe this is equivalent to $X$ having vanishing higher étale homotopy groups. So, this question naturally led me to the following questions:

Does there exist $X/\mathbb{C}$ smooth projective and some $i>1$ such that $\pi_i^\mathrm{\acute{e}t}(X)=0$ but $\pi_i(X^\mathrm{an})\ne 0$?
Does there even exist $X/\mathbb{C}$ smooth projective with $\pi_1^{\mathrm{\acute{e}t}}(X)=0$ but $\pi_1(X^\mathrm{an})\ne 0$?
Does there even exist a connected compact Kähler manifold $X$ such that $\pi_1(X)\ne 0$ but $\widehat{\pi_1(X)}=0$ (profinite completion)?

Any help with these questions would be much appreciated!

REPLY [8 votes]: Let $G$ be a group, $\iota: G\to \hat G$ its pro-finite completion. We call a $G$ a good group (cf. J.-P. Serre Cohomologie galoisienne, 2.6) if for every finite $\hat G$-module $M$, the maps
$$ \iota^* : H^q(\hat G, M) \to H^q(G, M) $$ 
are isomorphisms for all $q\geq 0$.
Some examples of good groups:
(1) finite groups,
(2) finitely generated free groups and finitely generated free abelian groups,
(3) iterated extensions of groups of type (1) and (2),
(4) braid groups (a special case of (3)),
Arithmetic groups are not good in general, e.g., as Donu Arapura commented, ${\rm Sp}(2n,\mathbb{Z})$ is not a good group for $n > 1$. It is not known whether the mapping class groups $\Gamma_{g ,n}$ (the orbifold fundamental group of $\mathcal{M}_{g, n}$) are good groups (cf. Lochak–Schneps 2006, par. 3.4).
Now, say $X$ is a connected scheme of finite type over $\mathbb{C}$. Consider the following three conditions:

The scheme $X$ is a $K(\pi, 1)$ scheme.
The topological space $X^{\rm an}$ is a $K(\pi, 1)$ space.
The fundamental group $\pi_1(X^{\rm an})$ is a good group.

Then (1)+(2) implies (3) and (2)+(3) implies (1). To prove this, consider an lcc sheaf $\mathcal{F}$ on $X$ and look at the commutative squares
$$ \begin{array}{c} H^q(\pi_1(X^{\rm an}, x), \mathcal{F}_x) & \to & H^q(X^{\rm an}, \mathcal{F}) \\  \uparrow & & \uparrow & \\  H^q(\pi_1(X, x), \mathcal{F}_x) & \to & H^q(X, \mathcal{F}). \end{array} $$ 
Note that we do not get that (1)+(3) implies (2), as neither (1) nor (3) gives us any information
about the cohomology of $X^{\rm an}$ with non-torsion coefficients. Still, we see that (1)+(3) implies
that $X^{\rm an}$ is a "$K(\pi, 1)$ for local systems of finite groups”.
The same should hold for fundamental groups of Deligne–Mumford stacks. In particular, the open question
whether $\Gamma_{g ,n}$ is a good group would be equivalent to the question whether the
stack $\mathcal{M}_{g, n}$ is a $K(\pi, 1)$ in the algebraic sense. Similarly (as in Donu Arapura's answer), as the orbifold fundamental group
of $\mathcal{A}_g$ is ${\rm Sp}(2g ,\mathbb{Z})$, while the orbifold universal cover of $\mathcal{A}_g$ is the Siegel upper-half space
(which is contractible), the stack $\mathcal{A}_g$
 ($g > 1$) gives a probable example of a smooth Deligne–Mumford stack which is a $K(\pi, 1)$ in the analytic sense but not in the algebraic sense.<|endoftext|>
TITLE: Hahn Banach type extension of a Lipschitz map
QUESTION [5 upvotes]: The problem that I posted was a much generalized form of what I had in my mind. All I want to know the literature of Hahn-Banach type extension of Lipschitz map. I know only about the result by Kirszbrzun and very few of its development. The main problem is:

Let $X$ be a real Banach space, $Y$ a subspace of it and $f$ a $1$-Lipschitz map from $Y$ to $\mathbb{R}$. Is it possible to get an extension of $f$ from $X$ to $\mathbb{R}$ with Lipschitz constant $1$?

The case when $X$ is a metrizable tvs  and $Y$ a finite dimensional subspace follows from MR0737400(86a:46018). But can we say anything about a infinite dimensional $Y$? Let us assume $X=C(K)$, $K$ is compact, $T_2$ (assume metrizable too if you want) and $Y=\{f\in X:f|_D=0\}$ (i.e. an M-ideal) of $X$. Now can a real valued $1$-Lipschitz map from $Y$ necessarily have a similar extension? Will it be unique ?

REPLY [11 votes]: If $X$ is any metric space, $Y$ any subset of $X$, and $f: Y \to \mathbb{R}$ a Lipschitz function, then there is an extension $\tilde{f}: X \to \mathbb{R}$ of $f$ with the same Lipschitz constant. This easy result (just extend the function one point at a time) has been rediscovered many times. See Theorem 1.5.6 of my book Lipschitz Algebras.
The result does not follow from the paper of Johnson and Lindenstrauss that you mention. The case when $X$ is "metrizable" is meaningless as you need an actual metric to have a notion of Lipschitz map. The answer by MKO is basically unrelated to the question as far as I can tell.<|endoftext|>
TITLE: Computer calculations in a paper
QUESTION [32 upvotes]: I think I can improve the current upper bound concerning an open problem. The ideas are purely combinatorial, but in the end I have to calculate the maximum of a really ugly, non elementary function with four variables. I did this with a computer, and the output was good so that's why I think I can improve the bound. (In the most precise sense, as this is not a real proof, but clearly not nothing.)
My first question is: What kind of computer calculations can be included in a paper? 
To be a little bit more specific, my function includes the inverse of the binary entropy function (actually both inverses). This is not an elementary function and it is not implemented in Wolfram Mathematica (the program I am working in). So I had to calculate its values numerically. My function sometimes involves two of such functions nested in each other, but only at most 3 times. During the process I had to obtain upper bounds on complicated functions with two variables, I did this by plotting them and plotting the plane of the corresponding constant, and observing that the graph of the function is under the plane. 
To elaborate a bit further, I saw in other papers, that if you prove that the required constant is a solution of a certain equation, it is okay to stop there and say that "and this constant is approximately 2,32", as this can be calculated easily with arbitrarily small error. But I feel like my proof is not complete, as I do not know the error of the approximation of my function, and somehow I am also not satisfied by the "plotting and observing" type bounds. I believe them to be true, but this method lacks rigour. On the other hand, I feel that my real contributions to the subject are the combinatorial ideas. It is of course important to prove that they really improve the current bounds. But I feel that this should be easy. But non elementary functions are making it not that easy. My second question is: What would be the best solution, if my goal is to write a good paper?

REPLY [9 votes]: I suggest glancing at two papers:
Hagedorn, Thomas. "Computation of Jacobsthal’s function $h(n)$ for $n< 50$." Mathematics of Computation 78.266 (2009): 1073-1087.
Hajdu, L., and N. Saradha. "Disproof of a conjecture of Jacobsthal." Mathematics of Computation 81.280 (2012): 2461-2471.
Both of them describe somewhat massive projects which involve a lot of computation.  They give algorithmic descriptions as well as some sample computations.  Although I personally want more, they provide enough detail that I do not question the validity of their work.  You might use them as examples for how to write up your results.
Gerhard "The Computer Says So, QED" Paseman, 2015.07.25<|endoftext|>
TITLE: Baire Category Theorem for complete uniform spaces
QUESTION [11 upvotes]: The version of the Baire Category Theorem I have in mind is the statement that a countable intersection of dense open subsets of a complete metric space is dense.  The question is:  is it likewise true that the countable intersection of dense open subsets of a complete uniform space is dense?
A look at the proof suggests that the answer should be no---it makes crucial use of the fact that the canonical uniform structure on a metric space is generated by the $\varepsilon$-balls of radius $\frac{1}{m}$ for $m\ \in \mathbb{Z}^+$.  I thus started looking at complete topological groups that were not first-countable for a counter-example, with no luck (topological groups because they provide easy examples of uniform spaces, and not first-countable because first-countable topological groups have uniformities generated by a countable collection of covers (and also because (I think?) first-countable topological groups are metrizable).)
Surprisingly, a quick Google search only pulled up one source that seems to mention this at all---Joshi in his Introduction to General Topology states (pg. 363)

Unfortunately, there seems to be no analogue of the Baire category theorem for complete uniform spaces.

but yet does not provide a counter-example either.
Also, I doubt this will make too much of a difference, but I would prefer the counter-example to be $T_0$.  Finding an example of a highly pathological space that is, uhh, pathological, isn't exactly what I'm looking for.

REPLY [5 votes]: It is very common for a topological space to be a complete uniform space in some uniformity, but it is less common for a topological space to satisfy the Baire category theorem since the proof of the Baire category theorem makes an essential use of countability of the metric rather than the uniformity..
Every paracompact space has a compatible complete uniformity simply by taking the uniform covers to be open covers and the entourages to be the neighborhoods of the diagonal. In fact, the paracompact spaces are precisely the spaces that have a compatible uniformity that satisfies a stronger form of completeness. A uniform space is said to be supercomplete if its hyperspace is complete. A topological space has a compatible supercomplete uniformity if and only if it is paracompact.
A space whose cardinality is below the first measurable cardinal is completely uniformizable if and only if the space is realcompact.
Every countable completely regular space without any isolated points $X$ can be given a compatible complete uniformity, but $X$ is never a Baire space. For example, $\mathbb{Q}$ can be given a compatible complete uniformity, but $\mathbb{Q}$ is the countable union of nowhere dense subspaces.
Every good general topology text will tell you that an $F_{\sigma}$-set in a paracompact space is paracompact. The spaces mentioned by Priel in his answer are paracompact since they are countable unions of closed subspaces of a paracompact space (the space of all smooth functions with compact support is metrizable since its topology is induced by countably many seminorms, namely $Sup_{x\in\mathbb{R}} |f^{(n)}(x)|$ and every metrizable space is paracompact).
If $X$ is a completely regular space, then let $Ba(X)$ denote the $\sigma$-algebra of Baire sets. In other words, $Ba(X)$ is the $\sigma$-algebra on $X$ generated by the zero sets (Recall that a zero-set is a set of the form $f^{-1}[\{0\}]$ for some continuous $f:X\rightarrow\mathbb{R}$ and that the complement of a zero set is a cozero set).
$\mathbf{Lemma}$ A completely regular space $X$ is realcompact if and only if each $\sigma$-complete ultrafilter on $Ba(X)$ is of the form $\{R\in Ba(X)|x_{0}\in R\}$ for some
$\mathbf{Proposition}$ Every $F_{\sigma}$-set in a realcompact space is also realcompact.
$\mathbf{Proof}:$ Suppose that $X$ is a realcompact space and $C_{n}\subseteq X$ is a closed subspace for each natural number $n$. Let $D=\bigcup_{n\in\omega}C_{n}$. Let $\mathcal{M}\subseteq Ba(D)$ be a $\sigma$-complete ultrafilter. Then define $\mathcal{N}=\{R\in Ba(X)|R\cap D\in\mathcal{M}\}$. Then $\mathcal{N}$ is a $\sigma$-complete ultrafilter on $Ba(X)$. Therefore, $\mathcal{N}=\{R\in\mathcal{M}|x_{0}\in R\}$ for some $x_{0}\in X$. I claim that $x_{0}\in D$.
Suppose to the contrary that $x_{0}\not\in D$. Then since $x_{0}\in\overline{D}$ and $X$ is completely regular, for all $n$ there is some cozero set $U_{n}\subseteq X$ with $x_{0}\in U_{n}$ but $U_{n}\cap C_{n}=\emptyset$. Therefore, $U_{n}\in\mathcal{N}$ for each $n$, so $\bigcap_{n}U_{n}\in\mathcal{N}$ as well. However, since $\bigcap_{n}U_{n}\in\mathcal{N}$, we have $\emptyset=D\cap\bigcap_{n}U_{n}\in\mathcal{M}$ which is a contradiction. Therefore, $x_{0}\in D$.
I now claim that $\mathcal{M}=\{R\in Ba(D)|x_{0}\in R\}$. Suppose that $R\in Ba(D)$ and $x_{0}\in R$. Then there are open subsets $V_{n}\subseteq D$ such that $x_{0}\in V_{n}$ for all $n$ but where $\bigcap_{n} V_{n}\subseteq R$. Therefore, for all $n$, there is some open subset $V_{n}^{\sharp}\subseteq X$ such that $V_{n}=V_{n}^{\sharp}\cap D$. Therefore,  since $x_{0}\in V_{n}^{\sharp}$, for all $n$, there is a cozero set $W_{n}^{\sharp}\subseteq X$ such that $x_{0}\in W_{n}^{\sharp}\subseteq V_{n}^{\sharp}$. Let $W_{n}=W_{n}^{\sharp}\cap D$. Then $W_{n}^{\sharp}\in\mathcal{M}$ for all $n$, so $W_{n}=W_{n}^{\sharp}\cap D\in\mathcal{N}$. Therefore $\bigcap_{n\in\omega}W_{n}\in\mathcal{N}$. However, since $W_{n}\subseteq V_{n}$ and $\bigcap_{n\in\omega}W_{n}\subseteq\bigcap_{n}V_{n}\subseteq R$, we have $R\in\mathcal{N}$ as well. We therefore conclude $\mathcal{N}=\{R\in Ba(D)|x_{0}\in R\}$. Therefore $D$ is realcompact. $\mathbf{QED}$
Using the above proposition, one can obtain a realcompact space that does not satisfy the Baire category theorem from any completely regular space that does not satisfy the Baire category theorem: Suppose that $X$ is a completely regular space which is the union of countably many nowhere dense sets $(L_{n})_{n\in\omega}$. Let $Y$ be a realcompact space such that $X$ is a dense subspace of $Y$ (for example, $Y$ could be the Hewitt-realcompactification of $X$). 
Let $C_{n}=CL_{Y}(L_{n})$ for all $n$. Let $D=\bigcup_{n\in\omega}C_{n}$. Then $D$ is an $F_{\sigma}$-set in $Y$, so $D$ is realcompact. However, each $C_{n}$ is nowhere dense in $D$, so $D$ is not a Baire space. In fact, since every $F_{\sigma}$ subset of a paracompact space is paracompact, if the space $Y$ is paracompact, then the obtained space $D$ would be paracompact but a countable union of nowhere dense sets.
Let me close with a couple propositions that further show that non-Baire uniform spaces and non-Baire complete uniform spaces are quite common.
$\mathbf{Proposition}$ Suppose that $X$ is a regular space without any isolated points such that for each $x\in X$ there is a collection $\mathcal{U}$ of pairwise disjoint open sets and $x_{U}\in U$ for each $U\in\mathcal{U}$ (i.e. every point is the limit of a discrete set) such that $x\in\overline{\{x_{U}|U\in\mathcal{U}\}}\setminus\{x_{U}|U\in\mathcal{U}\}$. Then for each $x_{0}\in X$ there is a non-Baire subspace containing $x_{0}$.
$\mathbf{Proof}$ We shall construct a sequence of disjoint sets $(C_{n})_{n\in\omega}$ along with open neighborhoods $U_{x}$ of $x$ for each $x\in X$ by induction on $n$.
For $n=0$, let $x_{0}\in X$, let $C_{0}=\{x_{0}\}$ and let $U$ be an open neighborhood of $x_{0}$. 
Now assume that $n>0$ and $C_{n-1}$ along with $U_{x}$ for $x\in C_{n-1}$ has been constructed already. Then
for each $x\in C_{n-1}$, let $L_{x}\subseteq U_{x}$ be a subset and let $(U_{z})_{z\in L_{x}}$ be a collection of sets with $\overline{U_{z_{1}}}\cap\overline{U_{z_{2}}}=\emptyset$ and such that $z\in U_{z}\subseteq\overline{U_{z}}\subseteq U_{x}$ for each $z\in L_{x}$. Let $C_{n}=\bigcup_{x\in C_{n-1}}L_{x}$
Since each $x\in C_{n-1}$ is non-isolated in $C_{n-1}\cup C_{n}$ and $U_{x}\cap C_{n-1}=\{x\}$, the set $C_{n-1}$ is nowhere dense in $C_{n-1}\cup C_{n}$. Therefore if $D=\bigcup_{n\in\omega}C_{n}$, then each set $C_{n}$ is is nowhere dense in $D$, so $D$ is not a Baire space. $\mathbf{QED}$.
$\mathbf{Proposition}$ Suppose that $X$ is a normal space without any isolated points such that for each nowhere dense subspace $C\subseteq X$ there is a collection $\mathcal{U}$ of pairwise disjoint open subsets of $X$ along with closed nowhere dense subsets $C_{U}\subseteq U$ such that $C\subseteq\overline{\bigcup\{C_{U}|U\in\mathcal{U}\}}\setminus\bigcup\{C_{U}|U\in\mathcal{U}\}.$ Then for each nowhere dense subspace $C\subseteq X$ there is a $F_{\sigma}$ set $D\subseteq X$ with $C\subseteq D$ and where $D$ is a non-Baire space and $D$ is a $F_{\sigma}$ subset of $X$.<|endoftext|>
TITLE: Non strictly-singular operators and complemented subspaces
QUESTION [7 upvotes]: If $T$ is a bounded operator which is not strictly singular, acting on a separable Banach space $X$, can one always find an infinite dimensional, closed and complemented, subspace $Y$ such that $T$ restricted to $Y$ is an isomorphism on $Y$?

REPLY [5 votes]: I think that the correct Markus' question concerns operators in L(X).In this case the answer is positive for separable C(K). I do not know what happens in the case of spaces with an unconditional basis.
 Moreover the following seems interesting.
 Let X be a separable reflexive space and T in L(X). Does there exist indecomposable Y containing isomorphically X and S in L(Y) that extends T?
  I also do not know what is the answer if in the previous question if we replace the indecomposable
   Y by the space C[o,1].<|endoftext|>
TITLE: Which journals publish applied mathematics with mostly pure mathematics content?
QUESTION [14 upvotes]: In the spirit of Which journals publish expository work? please advise:

What consistently high quality journals (1) today publish results that would otherwise go to a pure mathematics journal were it not for (2) the included applied content, the motivation, and/or the examples?

PNAS (if the paper is not long), Lecture Notes in Mathematics (if it is long), Journal of Interdisciplinary Mathematics, SIAM subject journals, *Journal of Modern Dynamics, Journal of Mathematical Biology, Bulletin of Mathematical Biology, and International Journal of General Systems go without saying.
(But are the last three read by working mathematicians these days?)
What about others? 
To make the question answerable: 

Ideally suggest a journal per answer with comments. So that others can vote on it, and an answer can be accepted eventually.

1 - Many applied journals are neither read nor refereed by mathematicians, primarily working in mathematics. 
That makes it difficult to get appropriate referees to give advise on submitted papers, either at all or else in a timely manner. If only because of, for example, notation or terminology lag (APS journals for example).
Please only suggest journals that are not too obscure or seem like they going to be obscure in the future. Journals that, due to the founders or team are likely "up and comers" are OK. (Nothing wrong with obscurity, but as you probably know . . . arXiv by itself is often more widely read in comparison . . . )
2 - Content that is inappropriate in a pure mathematics journal, for it contains sufficiently detailed applications. Yet the applications by themselves cannot be coherently published separately from the majority of the paper, which begins with and discusses a problem of primarily mathematical interest and with conventional mathematical rigour.
Updated. After a couple years maybe some new journals were founded that are not well known yet but may be good quality, especially with application to computer science. If so, those would be good answers.

REPLY [4 votes]: Came across my own question and saw that most discussion was in the comments. Here are some answers. Question also got relatively many views. So here are some answers.

Here is one recently that I would like to see succeed: Compositionality

https://compositionality-journal.org/

Mathematical structures in computer science

https://www.cambridge.org/core/journals/mathematical-structures-in-computer-science

Mathematics of computation

https://www.ams.org/journals/mcom/all_issues.html<|endoftext|>
TITLE: Transcendental distance sets
QUESTION [5 upvotes]: Define a set $S \subset \mathbb{R}^d$ as a
transcendental distance set if the distance between any pair of
distinct points of $S$ is transcendental.
For example, $S = \{ k \, \pi \;\mid\; k=1,2,\ldots \}$ is such a set in $\mathbb{R}^1$.

Q1. How large can a transcendental distance set be in $\mathbb{R}^d$?

The above $S$ is countably infinite. What is an example of an uncountably infinite $S$ in $\mathbb{R}^1$, and more generally in $\mathbb{R}^d$?
It would seem that random sets should be transcendental distance sets.
Suppose one further filters, in $\mathbb{R}^d$ requiring that the area of all triangles
determined by $3$-point subsets be transcendental, the volume of all tetrahedra
determined by $4$-point subsets be transcendental, and so on up to 
the measure of $d+1$ simplices.
Call such a set a transcendental measure set. 

Q2. How large can a transcendental measure set be in $\mathbb{R}^d$?

I know there is work on "Avoiding rational distances"
(e.g., Ashutosh Kumar. Real Analysis Exchange 38.2 (2012): 493-498.)
that may answer my questions, but I have not yet seen direct implications.

REPLY [9 votes]: Answer to this question provides an algebraically independent set of real numbers of size continuum. Enumerate it as $\alpha_{i,r}$ where $i\in\{1,...,d\},r\in\Bbb R$. Now define set $S=\{(\alpha_{1,r},...,\alpha_{d,r}):r\in\Bbb R\}\subseteq\Bbb R^d$. Now, measure of any simplex with vertices on these points can be expressed using coordinates of the vertices and radicals. If any of these turned out to be an algebraic number, we would have a non-trivial algebraic relation between $\alpha_{i,r}$, which is not possible, as they are algebraically independent.
As $|S|=\frak{c}$, we see that transcendental measure set can have size $\frak{c}$. By taking subsets of $S$, we can get transcendental measure set of any size at most continuum, and, clearly, not of any greater cardinality.<|endoftext|>
TITLE: Are periodic billiard trajectories stable on a manifold with strictly convex boundary?
QUESTION [5 upvotes]: Let $(M,g)$ be a compact Riemannian manifold with strictly convex boundary.
Let $\gamma:S^1\to M$ be a periodic billiard trajectory (geodesic in the interior and reflects specularly at the boundary).
For some small $\epsilon>0$, let $g_s$, $s\in(-\epsilon,\epsilon)$, be a family of metrics on $M$ so that $g_0=g$.
Assume that all metrics and their dependence on the parameter $s$ is smooth, but I don't want to constrain the variation of the metric otherwise.
Is there a (smooth) family of closed curves $\gamma_s:S^1\to M$ (with some $\epsilon>0$) so that each $\gamma_s$ is a periodic billiard trajectory with respect to the metric $g_s$ and $\gamma_0=\gamma$?
(With a suitable notion of smoothness at reflection points.)
In some cases a perturbation of the metric can destroy the periodic trajectory (consider the 2-periodic trajectory on a square), but it seems to me that strict convexity of the boundary should add stability.
The most interesting case to me is when $M$ is three dimensional, in particular when $M$ is the unit ball in $\mathbb R^3$ and the metric $g_0$ is conformally Euclidean, but I would like to understand the problem in greater generality.
Section 7.1 in this chapter (for example) discusses this question for planar polygons, but that is quite far from my goal.
In principle one could glue two copies of $M$ together along the boundary and study the stability of periodic geodesics on the closed manifold.
The problem is that the metrics on the doubled manifold are not even $C^1$ (only Lipschitz) because $\partial M$ is strictly convex, so I fear that many tools are not applicable.
The normal derivative of the metric tensor is the second fundamental form, so the following are equivalent: (1) the metric is $C^1$ (2) the metric is $C^2$ (3) the second fundamental form vanishes.
Edit:
Carl's answer below tells that most periodic orbits are not stable if the unperturbed manifold $(M,g_0)$ is too symmetric.
This is a useful answer but only to a special case.
Are the orbits stable for generic metrics or under some additional assumption?

REPLY [4 votes]: The result in the paper Spectrum of the Poincare map for periodic reflecting rays in generic domains of Petkov--Stojanov may be related. 
Setting: Let $X\subset \mathbb{R}^n$ be a closed $(n-1)$-dimensional manifold (as the boundary of a connected $n$-dimensional subset $D_X$). Consider the set of the embeddings $f:X\to \mathbb{R}^n$ (denote the corresponding $n$-set by $D_f$). 
The result in the above paper is very general. A very special version states that for a generic embedding $f:X\to\mathbb{R}^n$, every periodic point of the induced billiard system on $D_f$ is nondegenerate, and hence persists under any types of small perturbations of the induced billiard map. In particular, each pre-fixed periodic orbit of on $D_f$ for a generic $f$ persists under any small perturbations of the Euclidean metric of $D_f$.<|endoftext|>
TITLE: Pure motives and compatible systems of $\ell$-adic representations
QUESTION [6 upvotes]: I am trying to understand the statement of the conjectures of Deligne on special values of certain $L$-functions, from his article titled, "Valuers de Fonctions L et periodes d'integrales" which appeared in Proc. of Symposia in Pure Math., 33, (1979), Part 2, 313-346. 
In section 1 he makes the conjectures for a motive $M$ defined over $\mathbb{Q}$. He begins by assuming that the $\ell$-adic realizations of $M$, $H_l(M)$ form a strictly compatible system of $\ell$-adic representations. The fact that he assumes this is the reason I am asking this question, since I would have thought that this would always be the case. 
If I understand correctly, the assumption that, the $\ell$-adic realizations of $M$, $H_l(M)$ form a strictly compatible system of $\ell$-adic representations, means the following: There is a finite set of primes in $\mathbb{Q}$, say $S$, such that 

for each $p\notin S$ and $\ell\neq p$ the representation $\rho_\ell:G_\mathbb{Q}\to GL(H_{\ell}(M))$ is unramified at $p$, 
Because of (1) we can consider the characteristic polynomial of the geometric Frobenius. This polynomial should have coefficients in $\mathbb{Q}$ (apriori, this is a polynomial with coefficients in $\mathbb{Q}_\ell$), 
For $p\notin S$, the polynomial we get in (2) should be independent of $\ell$.

Now suppose $M$ were the motive $h^i(A)=(A,\pi_i,0)$, where $A$ is an abelian variety defined over $\mathbb{Q}$, and the $\ell$-adic realization of $h^i(A)$ is $H^i_\ell(A;\mathbb{Q}_\ell)$. Let $S$ be the set of primes where $A$ does not have good reduction. Then the $\ell$-adic realizations of $M$ form a strictly compatible system of $\ell$-adic representations (condition (1) holds since good reduction implies unramified, condition (2) and (3) would hold because of the Weil conjectures, I hope I'm correct). I would have thought that for pure motives (as in Scholl's article titled, "Classical motives") the same conclusion would hold. Is this true?
Having written the above question I realize that Deligne uses the notion of absolute Hodge cycles to define motives in this article. Perhaps this is what is causing the problem. Can someone shed some light on this matter. Thanks in advance.
EDIT: From pondering over the comment by Eric and the answer by Will, I realize that it is not clear to me why the Frobenius should act on the cohomology $H_\ell(M)$. Let $M=(X,\pi,0)$ be a pure motive as in Scholl's article, and for simplicity let us also assume that it is homogeneous; I mean that the Hodge realization has only one weight. The situation I have in mind is one in which $M$ cuts out a piece from the cohomology $H^i_\ell(\bar{X},\mathbb{Q}_\ell)$. If $p$ is a prime at which $X$ has good reduction then the Galois representation $H^i_\ell(\bar{X},\mathbb{Q}_\ell)$ is unramified and so we can choose 'a' Frobenius endomorphism 
$$Frob_p:H^i_\ell(\bar{X},\mathbb{Q}_\ell)\to H^i_\ell(\bar{X},\mathbb{Q}_\ell)$$
If $H^i_\ell(M)$ denotes the image of the action of the projector $\pi$ on the cohomology, then I see no reason why our choice of $Frob_p$ should preserve this subspace. Is this clear? Yes it is, as pointed out by Will in the comments to his answer.

REPLY [5 votes]: Yes, the absolute Hodge cycles cause the difficulty.
Let's consider what happens when we take a weak form of motive. Take a correspondence $C \subset X \times X$ that is an idempotent up to singular homological equivalence, and define the motive whose $\ell$-adic realization is the subspace of $H^i(X, \mathbb Q_\ell)$ on which $C$ acts with eigenvalue $1$.
Assuming enough big conjectures this will be equivalent to all the other definitions of motives, but it is a priori pretty weak.
Now let's prove independence of $\ell$ of the characteristic polynomial of $\operatorname{Frob}_p$. To compute the polynomial, it is sufficient to compute the traces of powers of $\operatorname{Frob}_p$, so we must show that those are independent of $\ell$. We can pass to characteristic $p$, where $\operatorname{Frob}_p$ becomes an endomorphism, and then the trace of $\operatorname{Frob}_p^n$ on the motive is the trace of $\operatorname{Frob}_p^n C$ on $H^i(X, \mathbb Q_\ell)$. I guess the Lefschetz operator that projects onto the $i$th cohomology exists in characteristic $p$ because of the Weil conjecture, so this is just the trace of $\operatorname{Frob}_p^n C P_i$ on $X$, which is equal to the intersection number of $\operatorname{Frob}_p^n C P_i$ with the diagonal and so is clearly independent of $\ell$.
But if $C$ is an absolute Hodge cycle, then it is not necessarily an algebraic cycle, nor does it necessarily become an algebraic cycle when we reduce mod $p$, and there is no reason to expect independence of $\ell$ for a trace of it times a power of Frobenius.<|endoftext|>
TITLE: On the global structure of the Gromov-Hausdorff metric space
QUESTION [26 upvotes]: This is a purely idle question, which emerged during a conversation with a friend about what is (not) known about the space of compact metric spaces. I originally asked this question at math.stackexchange (https://math.stackexchange.com/questions/1356066/global-structure-of-the-gromov-hausdorff-space), but received no answers even after bountying.
Background. If $A, B$ are compact subsets of a metric space $M$, the Hausdorff distance between $A$ and $B$ is the worst worst best distance between their points: $$d_H(A, B)=\max\{\sup\{\inf\{d(a, b): b\in B\}: a\in A\}, \sup\{\inf\{d(a, b): a\in A\}: b\in B\}\}.$$ For two compact metric spcaes $X, Y$, their Gromov-Hausdorff distance is the infimum (in fact, minimum) over all isometric embeddings of $X, Y$ into $Z$ via $f, g$ of $d_H(f(X), g(Y))$. The Gromov-Hausdorff space $\mathcal{GH}$ is then the "set" of isometry classes of compact metric spaces, with the metric $d_{GH}$.)
Question. How homogeneous is $\mathcal{GH}$? For example: while distinct points in $\mathcal{GH}$ are in fact distinguishable in a broad sense, it's not clear that distinct points can always be distinguished just by the metric structure of $\mathcal{GH}$, so it's reasonable to ask:

Are there any nontrivial self-isometries of $\mathcal{GH}$?

There are of course lots of related questions one can ask (e.g., autohomeomorphisms instead of isometries); I'm happy for any information about the homogeneity of $\mathcal{GH}$.

Please feel free to add tags if more are appropriate.

REPLY [10 votes]: We have written a text with more details on the proof given by George Lowther. Here one can see the result:
https://arxiv.org/pdf/1806.02100.pdf
However, locally there are many self-isometries, see
https://arxiv.org/pdf/1611.04484.pdf<|endoftext|>
TITLE: Is there a generalization of homotopy groups to fractional dimensions
QUESTION [20 upvotes]: Does there exist a reasonable candidate for such an object as $\pi_{\frac12}(X)$?

REPLY [19 votes]: As requested, here's a little more detail in the form of an answer.
Disclaimer: Let me start by saying that it'll be easier for me to talk about stable homotopy groups instead of homotopy groups, just because that's all I know about. By that I mean, I'm not gonna define, say, $\pi_{1/2}S^3$ but rather $\pi_{N+1/2}S^{3+N}$ when $N$ is really big. Throughout I will be lazy and write $\pi_kX$ for what should really be the stable value of $\pi_{k+N}\Sigma^NX$. Feel free to ignore this. Someone who knows more about this than I do can come around and fill in how to translate this back into something about ordinary homotopy groups. Also, unless I say otherwise, my prime is odd. (Contrary to normal practice here at Northwestern.)
Motivation and (pseudo-)History Here is a 'just-so' story about how you could have invented $p$-adically interpolated homotopy groups. I think it's pretty close to the actual history, but I wasn't in Mike Hopkins's head at the time so I don't know. (But I really wish I had been... what a great sequel to Being John Malkovich that would be).
As usual, it starts with $K$ theory. Except I said we're gonna $p$-adically interpolate, so we actually want $p$-complete $K$ theory, $K_p$. The nice thing about $p$-adic $K$-theory is that, for any number $m$ coprime to $p$, there's a natural transformation of (multiplicative) cohomology theories, called the Adams operations, $\psi^m: K_p \rightarrow K_p$. This is supposed to act like a linearization of taking exterior powers of vector bundles, and here's all you need to know about it: 

If $v \in K(S^2)$ denotes the Bott periodicity element, then $\psi^m(v) = mv$
$\psi^m$ induces a graded ring endomorphism of $K^*(pt)$

Now, for various reasons, coming from manifold theory (if you're Sullivan) or just trying to understand some elements in the stable homotopy groups of spheres (if you're Adams), people got interested in understanding the cohomology theory you get by taking the 'fixed points' of $\psi^m$. That is, they wanted to understand the cohomology theory $J$ that has a natural transformation $J \rightarrow K_p$ and such that, for any space $X$, you get a long exact sequence 
$$J^n(X)\rightarrow K^n_p(X) \stackrel{\psi^m -1}{\longrightarrow} K^n_p(X) \rightarrow J^{n+1}(X) \rightarrow \cdots$$
When you choose $m$ correctly (I'll say how in a second), this cohomology theory deserves the name $S_{K(1)}$ which stands for '$K(1)$-local sphere' because it sees a very special sector of the rainbow that makes up the sphere (or more accurately, the 'stable' phenomena of the sphere). 
It is this object which we will now raise to the "1/2 th" power, and maps out of this will constitute elements of $\pi_{1/2}$. The key is to notice the following:

$K^{2n}_p(pt)= \mathbb{Z}_p$
The map $x \mapsto (m^{n}x - x)$ on the $p$-adics makes sense for (invertible) $p$-adic values of $m$. 

Now I can tell you how to choose $m$ to get the $K(1)$-local sphere: it has to be a topological generator of the $p$-adic units. (Those following along at home just saw why my prime was odd.) If we think about Bott periodicity and stare at the formula a bit, it becomes clear that the $\lambda$th suspension, for $\lambda \in \mathbb{Z}_p$, should be obtained by taking the fiber of the map which, on homotopy groups, does $x \mapsto (m^nx - \lambda x)$. [Okay, it takes a little fussing to see this is the right thing to do- it helps to interpret $\lambda$ as a unit which is 1 mod $p$]. 
This fiber behaves like the $\lambda$th suspension of the $K(1)$-local sphere, and with it we may probe $K(1)$-local spaces and spectra $p$-adically. When $p=2$ there's a little more stuff you can probe with, but you get $2$-adic things as well. 
 The general story 
All the stuff above is spelled out in Hopkins-Mahowald-Sadofsky's paper here. That's where they launched the program of computing Picard groups. As I said before, the Picard group of a symmetric monoidal category is the set of (isomorphism classes of) $\otimes$-invertible objects. These are good for indexing things like homotopy groups in all sorts of settings.
Example. The Picard group of the category of spectra (home to cohomology theories and to stable homotopy groups) is just $\mathbb{Z}$ and you get ordinary homotopy groups. Same for $p$-local spectra.

Example The Picard group of the category of motivic spectra over any base contains at least a copy of $\mathbb{Z}^2$ spanned by the simplicial sphere and $\mathbb{G}_m$. The exotic grading allows one to state and prove some pretty powerful statements, and is sort of related to the Grothendieck-Deligne yoga of weights. 
Example The Picard group of the category of (genuine) equivariant spectra contains the one-point compactifications of orthogonal representations (i.e. 'representation spheres') and you need this is extra bit of juice to get things like Poincaré duality in the equivariant setting. 
The most mysterious example, however, is the $K(n)$-local category (this is the thing that sees the $n$th layer of the rainbow of the sphere, we met $n=1$ earlier). We are still in the very early stages of understanding the Picard group of this category, but it's very desirable to do so. For one, there are mysterious duality phenomena in the $K(n)$-local category, analogous to the Poincaré duality example mentioned in the equivariant setting, which require exotic suspensions (see here). On the other hand, it can be useful to index the homotopy groups of something as a function of a $p$-adic number, say. Who knows- maybe the only way we'll ever get an answer to 'what are the ranks of the homotopy groups of spheres?' is to look for solutions like 'the special values of some horribly incomputable arithmetic-p-adic-y-L-type function'. 
But that's speculation. For some recent calculations, see some of the papers  here  by Paul Goerss and his collaborators (Henn, Mahowald, Rezk).<|endoftext|>
TITLE: Number of bases of a matroid
QUESTION [5 upvotes]: I would like to know the minimum number of bases of a matroid of rank $k$ and $n$ elements, knowing that each singleton is independent. At least for small ranks.

REPLY [2 votes]: With the problem as stated, the answer is $n-k+1$. Take the uniform matroid of rank $1$ on $n-k+1$ elements and direct sum with $k-1$ co-loops. (Geometrically, take the standard basis $e_1$, $e_2$, ..., $e_k$ of $\mathbb{R}^k$ and duplicate the first basis element $n-k+1$ times.)
To see that this is optimal, suppose for the sake of contradiction that we could achieve $n-k$ bases: $B_1$, $B_2$, ..., $B_{n-k}$. Consider the exchange graph, where $B_i$ and $B_j$ are connected by an edge if $\#(B_i \setminus B_j)= \#(B_j \setminus B_i) = 1$. It is known to be connected. Reorder the $B_i$ such that the induced subgraph on $B_1$, $B_2$, ..., $B_r$ is connected. Then, for $2 \leq r \leq n-k$, there is at most one element of $B_r$ not in $\bigcup_{1 \leq i<r} B_i$. We deduce that $\# \bigcup_{1 \leq i \leq n-k} B_i \leq \# B_1 + (n-k-1) = n-1$. So there is some element not in $\bigcup B_i$, and this element is not independent.
As I commented above, I think it would be more natural to impose that $M$ has neither loops nor co-loops. The best I can find for that problem $k(n-k)$, by using $U(k-1,k) \oplus U(1, n-k)$, where $U(r,m)$ is the uniform matroid of rank $r$ on $m$ elements. If you ask for the matroid to be connected, I can achieve $k(n-k)+1$ by starting with $U(k,k+1)$ and replacing one element with $n-k$ parallel copies. I would guess these are optimal.<|endoftext|>
TITLE: Relation between moduli spaces and classifying spaces
QUESTION [12 upvotes]: I hope this question is suitable to be posted here on MO. 
I wonder if there is a systematic relation between the notation of a classifying space, and the notion of a moduli space. I don't consider myself as someone with an expert level knowledge in any of these objects, but let me explain what I think.
The classifying space of a group/H-group, or a monoial category, most of the time is ought to be a representing object for a cohomology theory. The word classifying here, is mainly referring to the fact that $BG$ does parametrise certain ``bundles'' over paracompact spaces with a $G$-structure. The moduli space, is also ought to be a space which somehow parametrises the class of objects in a certain category - hope this analogy is not too vague. It somehow puts the totality of certain object in a `topological space'. Hence, both of these spaces are meant to parametrise certain objects. 
Of course, I have not declared which definitions I take for these object, and I hope there is a common understanding about these spaces when an example is given. My main question is this. If $C$ is a category for which I can associated a classifying space $BC$ as well as a moduli space to the class of its objects, say $\mathcal{M}_C$. Then, is there a map
$$BC\to\mathcal{M}_C$$
or the way around?
Let me add that the work of Madsen and Weiss, on the stable Mumford conjecture, provides an example of the relation that I have asked above. They show that the infinite loop space associated to a certain spectrum $\mathbb{C} P_{-1}$ is related to the classifying space of the cobordism category of oriented surfaces, say $BC_2$ (in fact there is a weak homotopy equivalence) which is meant to solve Mumford's conjecture as $\Omega^{\infty-1}\mathbb{C} P_{-1}$ and the moduli space of Riemannian surfaces are ought to be the same?! But, I still don't know where a map such as  $\Omega^\infty\mathbb{C}P_{-1}\to \mathcal{M}_\infty$ or a map  $BC_2\to \mathcal{M}_\infty$ is constructed where $\mathcal{M}_\infty$ is the moduli space of Riemannian surfaces of arbitrary genus. I am aware of the map Madsen-Tillmann map $\mathbb{Z}\times B\Gamma_\infty^+\to\Omega^\infty\mathbb{C}P_{-1}$ with $\Gamma_\infty$ is the stable mapping class group and $(-)^+$ is Qullien's plus construction. The work of Galatius-Madsen-Tillmann-Weiss also provides 
$\alpha_\infty:BC_2\to\Omega^{\infty-1}\mathbb{C}P_{-1}$. But, I don't think of the source nor the target as a moduli space.
I will be very grateful for any advise. Please let me know if I am mixing some objects. 
Thanks.

REPLY [10 votes]: There is a sense in which the relation between moduli stacks
and classifying spaces can be formalized, at least
when we use smooth manifolds as parametrizing objects.
(Topological manifolds and PL-manifolds also suffice.)
Start with a stack F of spaces on the site of smooth manifolds, which we think of as the moduli stack of some smoothly parametrized objects.
Define the concordification CF of F
as the prestack CF(X) := hocolim_{n∈Δ^op} F(Δ^n×X),
where Δ^n is the smooth extended n-simplex,
i.e., the smooth manifold R^n,
with the appropriate face and degeneracy maps.
(The name concordification comes from the fact
that F[X] := π_0(CF(X)) is the set of concordance classes
of sections of F over X, where two sections x and y
over X are concordant if there is a section z
over R×X whose restrictions to 0×X and 1×X
are isomorphic to x and y respectively.)
It is a nontrivial result that for any stack F of spaces
the prestack CF is actually a stack.
(For the much easier case of a stack (sheaf) of sets
this is shown in Proposition 2.17 and Proposition A.1
in the cited paper of Madsen and Weiss.)
Furthermore, CF is concordance-invariant:
for any manifold X the canonical pullback map
F(X)→F(R×X) is a weak equivalence.
For any stack (or prestack) of spaces G
one can construct a natural map G(X)→Map(X,CG(pt)).
Indeed, consider the functor C(F) := (CF)(pt)
from prestacks to spaces.
This functor is enriched over spaces,
so we have a map Map(X,G)→Map(CX,CG).
Here the left Map denotes the mapping space of stacks (which are enriched over spaces) and X denotes the representable stack of X.
The right Map is simply a mapping space of spaces.
By the (enriched) Yoneda lemma we have Map(X,G)=G(X).
Furthermore, CX is simply the (smooth) singular simplicial
set of X, i.e., the underlying homotopy type of X.
Abusing the notation, we write Map(CX,CG)=Map(X,CG)=Map(X,CG(pt)).
Altogether, we have a map G(X)→Map(X,CG(pt)).
An observation that goes back at least to Morel and Voevodsky
(see Proposition 3.3.3 in their paper)
says that concordance-invariant stacks of spaces
on the site of smooth manifolds
are precisely locally constant stacks,
i.e., the canonical map G(X)→Map(X,G(pt))
is a natural weak equivalence for any concordance-invariant
stack of spaces G.
In particular, in our case we get a natural
weak equivalence CF(X)→Map(X,CF(pt)).
Taking π_0 on both sides we get an isomorphism of sets F[X]→[X,CF(pt)].
In other words, for any (moduli) stack of spaces F
the set F[X] of concordance classes of sections of F
over X is naturally isomorphic to [X,CF(pt)],
where CF(pt) = hocolim_{n∈Δ^op} F(Δ^n).
It is natural to call CF(pt) the classifying space of F.
For example, for F(X)=Ω^n_cl(X), the set of
closed n-forms on X, we get CF(pt)=K(R,n),
the nth Eilenberg—MacLane space of R
(this is covered by the result of Madsen and Weiss
because F(X) is a set).
If we take F(X) to be (the nerve of) the groupoid
of real vector bundles of dimension n over X
equipped with a connection (and isomorphisms that
preserve connections), then CF(pt)=BO(n),
the classifying space of n-dimensional vector bundles
(or concordance classes of bundles with connection;
two bundles with connection are concordant
if and only if they are isomorphic as bundles without connection).
Note that we get the same answer if we take vector
bundles without connection;
the concordification construction does not see
any local data such as connections, forms, etc.
If F(X) is the space of bundle (n−1)-gerbes with
(or without) connection, then CF(pt)=B^n U(1),
the classifying space of bundle (n−1)-gerbes
(or concordance classes of bundle (n−1)-gerbes with connection).
One can also apply the concordification functor C
to morphisms of stacks, for example,
applying C to the Chern—Weil construction
recovers Chern classes, etc.<|endoftext|>
TITLE: Is there a structure theorem or group law for finite groups generated by two elements?
QUESTION [5 upvotes]: Say that $a, b \in G$ are two elements of a finite group $G$. Is there a structure theorem for the structure of $\langle a,b\rangle$? Is there a way to derive group laws for the group operation in the generated group?
I can think of special cases (the two elements commute, one of the elements is a power of the other, the commutator of the two elements commutes with them, etc.). I am wondering if a general classification results exists, still.

REPLY [15 votes]: It is a theorem of Graham Higman, Bernhard Neumann, and Hanna Neumann (Embedding theorems for groups, J. London Math. Society 24 (1949) 247-254) that every countable group can be embedded in a 2-generator group. This was later simplified by Bernhard and Hanna Neumann (Embedding theorems for groups, J. London Math. Soc. 34 (1959) 465-479). Fred Galvin has a paper in the Monthly (Embedding Countable Groups in 2-Generator Groups, Amer. Math. Monthly 100 no. 6 (1993), 578-580; available from JSTOR ) giving a simple proof, showing that in fact:

Theorem. Every countable group is embeddable in a 2-generator group, with one generator of order 11 and the other of order 2.

Given this, it seems hopeless to expect a structure theorem for 2-generator groups.<|endoftext|>
TITLE: Intuition behind the Morse inequalities?
QUESTION [8 upvotes]: Forgive me if this is sort of a vague question, but can someone supply me with their intuition behind the Morse inequalities?

REPLY [17 votes]: If you understand intuition behind the fact that the Euler characteristic is the alternating sum of the betti numbers, then I think you can grasp the Morse inequalities. A Morse function gives rise to a CW structure on the manifold, by considering the unstable manifolds of index $k$ critical points as giving $k$-cells attached to the $k-1$-skeleton. The Morse inequalities apply more generally to compact CW complexes.
The Euler characteristic of the $\gamma$-skeleton $M^{(\gamma)}$ of a compact CW complex $M$ is given by the alternating sum of the betti numbers of the $\gamma$-skeleton, and by the alternating sum of the number of $j$-cells $C^j$ of the $\gamma$-skeleton, $j\leq \gamma$, which is the same as the corresponding sum for $M$. The betti numbers of $M$ will agree with those of $M^{(\gamma)}$ except possibly in dimension $\gamma$, where one has $b_\gamma(M^{(\gamma)})\geq b_\gamma(M)$, because some $\gamma+1$-cells might kill cellular homology generated by $\gamma$-cells. This translates directly into the Morse inequality of index $\gamma$. Following the notation of the Wikipedia page on Morse theory, one has 
$$C^\gamma - C^{\gamma-1} + \cdots +(-1)^\gamma C^0 =(-1)^{\gamma} \chi(M^{(\gamma)}) = b_\gamma(M^{(\gamma)}) - b_{\gamma-1}(M^{(\gamma)}) +\cdots +(-1)^\gamma b_0(M^{(\gamma)}) \geq b_\gamma(M) -b_{\gamma-1}(M) + \cdots +(-1)^\gamma b_0(M).$$ 
Thus, one may recall the inequality $b_\gamma(M^{(\gamma)})\geq b_\gamma(M)$, which has a fairly intuitive explanation in terms of cellular homology, and derive the Morse inequalities as a consequence. There is also a relative version which may be easier to understand.<|endoftext|>
TITLE: Zeta-Determinant for shifted Laplacians on the circle
QUESTION [6 upvotes]: Consider on the circle $S^1$ the operator
$$L := - \frac{\partial^2}{\partial \theta^2} + c$$
for some constant $c \in \mathbb{R}$. 
What is its $\zeta$-regularized determinant?
This should be well-known, I suppose, but I didn't find a reference.
Some background: The eigenvalues of $L$ are $n^2+c$ for $n \in \mathbb{Z}$ (with multiplicity two each), and therefore the zeta-function for positive $c$ is given by
$$\zeta_c(s) = 2\sum_{n=0}^\infty \frac{1}{(n^2+c)^s}.$$
Now the $\zeta$-regularized determinant is defined by
$$\det(L) = e^{-\zeta^\prime(0)}.$$
How does one compute such a thing?
\Edit: The hint to look for "thermal zeta functions" was good: Here is a paper that computes the determinant in question: http://arxiv.org/pdf/hep-th/9505154v1.pdf

REPLY [2 votes]: Let me write a bit of what seems to me that natural/naive approach, and perhaps the questioner can comment on the direction...
From $\int_0^\infty y^s\,e^{-ty}\;{dy\over y}=t^{-s}\,\Gamma(s)$, writing $Z(s)$ for your zeta, 
$$
\pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s})
\;=\; \int_0^\infty y^{s/2}\,\sum_{n\in \mathbb Z} e^{-\pi (n^2+c)y}\;{dy\over y}
$$ 
Unlike Riemann's argument, the presence of $c>0$ will cause Poisson summation to produce an expression that behaves well, without breaking the integral into two pieces. Namely, the Fourier transform of $x\to e^{-\pi (x^2+c)y}$ is ${1\over \sqrt{y}} e^{-\pi (x^2/y +cy)}$, so this becomes
$$
\sum_{n\in \mathbb Z} \int_0^\infty y^{{s-1\over 2}} e^{-\pi (n^2/y+cy)}\;{dy\over y}
$$
The $n=0$ term should be taken out, and is elementary. For the rest, replace $y$ by $\sqrt{c} y/n$, to get
$$
\pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s})
\;=\; (n=0) + \sum_{n\not=0}(\sqrt{c}/n)^{(s-1)/2}  \int_0^\infty y^{{s-1}\over 2} e^{-\pi \sqrt{c}\,n\,(y+1/y)}\;{dy\over y}
$$
The right-hand side seems to be entire in $s$, so the factor of $\Gamma(s/2)$ on the left-hand side means that the factor $Z(s)-c^{-s}$ vanishes at $s=0$. Thus, the right-hand side is close to computing the derivative of $Z(s)-c^{-s}$ at $s=0$.
But/and this makes me wonder whether about that extra term $c^{-s}$. It is needed to make Poisson summation work, but then it doesn't seem to me that $Z(s)=0$, so $Z'(0)$ would not be the leading term, which I would have thought would have been the object of interest.
In any case, the right-hand side is a sum of values of Bessel functions, but not obviously (to me) further simplifiable at $s=0$. At $s=1$, the sum over $n$ admits summing as geometric series, so at least the outcome is just an integral  in $y$... though it seems not elementary, still.
Comment?<|endoftext|>
TITLE: Is every nonabelian finite simple group a quotient of a triangle group $(a,b,c)$ with $a,b,c$ coprime?
QUESTION [6 upvotes]: Here by triangle group $(a,b,c)$ I mean the group with presentation
$$\langle x,y \;|\; x^a = y^b = (xy)^c = 1\rangle$$
In other words, for every finite simple nonabelian group $G$, do there exist pairwise coprime integers $a,b,c$ such that $G$ is generated by $x,y$ with $|x| = a$, $|y| = b$, and $|xy| = c$?
I'd also be interested in any result where we relax the "pairwise-coprime" condition to the condition that $(|x|,|y|)\cdot (|x|,|xy|)\cdot (|y|,|xy|)$ is small.

REPLY [2 votes]: If my memory is correct, it is a consequence of Thompson's N-group paper that a finite group $G$ is solvable if and only it not possible to find three elements $(x,y,z)$ of pairwise coprime orders, not all $1$, with $xyz = 1_{G}.$ It follows from this that every minimal non-Abelian finite simple group $G$ is a quotient of the type of "triangle group" asked for.
For there must be such a triple of elements $(x,y,y^{-1}x^{-1})$ of pairwise coprime orders. Hence $\langle x,y \rangle$ is not a solvable group, so must be all of $G$, and $G$ is a quotient of a group of the required form.
As has been remarked in comments, there is an extensive literature of generating pairs for finite simple groups, and it appears extremely likely that the answer to this question is "yes"<|endoftext|>
TITLE: Does every Lawvere theory arise in this way?
QUESTION [10 upvotes]: By a Lawvere theory, I mean a finite-product category $\mathsf{T}$ equipped with a distinguished object, such that every object of $\mathsf{T}$ can be expressed as a finite product of the distinguished object.
Now let $X$ denote an object of a finite-product category $\mathbf{C}$. Then $X$ gives rise to a Lawvere theory $\mathrm{Lawv}(X)$ with distinguished object $X$, defined as the full subcategory of $\mathbf{C}$ induced by the objects $\{X^n \mid n:\mathbb{N}\}.$
Examples.

Viewing $2$ as an object of $\mathbf{Set}$, we see that the arrows of $\mathbf{Lawv}(2)$ are "truth tables." Hence $\mathbf{Lawv}(2)$ is the Lawvere theory of Boolean algebras.
Let $\mathbb{K}$ denote a field. Write $U_{\mathrm{Mod}}(\mathbb{K})$ for the underlying $\mathbb{K}$-module. Then $\mathbf{Lawv}(U_{\mathrm{Mod}}(\mathbb{K}))$ is the Lawvere theory of $\mathbb{K}$-modules.


Question. Is it true that for all Lawvere theories $\mathsf{T}$, there exists a Lawvere theory $\mathsf{S}$ and an $\mathsf{S}$-algebra $X$ such that $\mathrm{Lawv}(X) \cong \mathsf{T}$?

REPLY [3 votes]: I can elaborate on the second example a bit more:
Let $T$ be a Lawvere theory. Then $T$ is commutative iff $T \cong Lawv(\text{hom}_T(x,-))$, where $x$ is the generic object in $T$.
A Lawvere theory is commutative if all its $n$-ary operations commute with each each other. A morphism of $T$-algebras is a natural transformation between functors; the naturality condition is precisely the requirement to commute with all $n$-ary operations. Your example of modules over a field (more generally a commutative semi-ring) is the prime example of a commutative theory.
Now commutative Lawvere theories are extremely useful, but also very restrictive. I will think about the general case some more.<|endoftext|>
TITLE: Are piecewise linear functions dense in $W^{1,\infty}$?
QUESTION [5 upvotes]: Are piecewise linear functions dense in $W^{1,\infty}$ ?

REPLY [7 votes]: Piecewise linear functions are not dense in $W^{1,\infty}(\Omega;\mathbb R^n)$ for any open set $\Omega\subset\mathbb R^m$.
If it were true for $\Omega$, it would also be true for any open subset, in particular any open ball.
Recall that the Sobolev space $W^{1,\infty}(\Omega;\mathbb R^n)$ is the space of Lipschitz functions $\Omega\to\mathbb R^n$ with the same norm if $\Omega$ is quasiconvex — and balls are quasiconvex.
(This and many other basic facts of Lipschitz functions can be found in Lectures on Lipschitz analysis by Heinonen.)
The Lipschitz condition behaves well when restricting to subspaces, which is not that obvious from the point of view of Sobolev spaces.
If piecewise linear functions were dense, for any $f\in W^{1,\infty}$ there would be a sequence of piecewise linear functions $f_i$ tending to $f$.
Every Lipschitz function on a line can be extended to a Lipschitz function on the whole domain and the restriction of a piecewise linear function to a line is piecewise linear.
Therefore, if we restrict all functions to a line segment in $\Omega$ and evaluate only one component, we end up with the claim that every Lipschitz function $[a,b]\to\mathbb R$ can be approximated by a piecewise linear function in the Lipschitz norm.
Now, let $f:[a,b]\to\mathbb R$ be an arbitrary Lipschitz function and suppose there is a sequence of piecewise linear functions $f_i:[a,b]\to\mathbb R$ converging to it.
Then $f_i(a)\to f(a)$ and $f_i'\to f'$ in $L^\infty$.
Since $f'$ can be any $L^\infty$ function, this means that piecewise constant functions are dense in $L^\infty$.
But this is false.
The function $g:[0,1]\to\mathbb R$, $g(x)=\sum_{k=1}^\infty\chi_{[2^{-2k},2^{1-2k}]}(x)$ is in $L^\infty$ but any piecewise constant function has at least distance $\frac12$ from it.<|endoftext|>
TITLE: Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
QUESTION [5 upvotes]: Related to the n-conjecture.
We are looking for Weierstrass form and map from it of the genus one curve:
$$ y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0 $$
It is of degree $40$ and has only five monomials.

Added We suspect there are finite number of rational points (likely none).
So the Weierstrass form may be over number field with monic defining polynomial with integer coefficients. The rank must be positive in addition. This relaxation might help Magma (we don't have it).

Maple gave weird error.
In machine readable form:
y^10*z^30-8000*y^4*z^20+1600*z^10*y^2+12800000*z^20-64

REPLY [13 votes]: (a) The given curve is irreducible over $\mathbb Q$, but reducible over $\mathbb Q(\sqrt[5]{2})$. So it does not have a Weierstrass form, and each rational point has to be a singularity. The curve however does not have rational singularities (there are two on the projective completion though). 
(b) From the latest comments by the OP it has become clear that he wants a Weierstrass model for a component of the curve. Set $\alpha=\sqrt[5]{2}$. Then the curve given by $y^2z^6 + 20\alpha^2z^4 -2\alpha=0$ is one of the algebraically conjugate components. Over $\mathbb Q(\alpha,\sqrt[4]{20})$ this is easily brought into Weierstrassform by hand.
What follows was written before I realized that the curve isn't absolutely irreducible. So it may help in such situations, though it has little value (and lacking correctness) here.
The following works: Set $F(y,z)=y^{5} z^{3} - 8000 y^{2} z^{2} + 12800000 z^{2} + 1600 y z - 64$. Then your elliptic curve $C$ is the solution set of $F(y^2,z^{10})=0$. Let $C'$ and $X$ be the solution sets of $F(y^2,z)=0$ and $F(y,z)=0$. So there are natural covers $C\to C'\to X$. Note that $C'$ has genus $1$, and $X$ has genus $0$.
Maple easily finds a rational parametrizaton of $X$. After some simplifications we see that $y=s^3-20s$, $z=\frac{4}{s^5}$ is a parametrization of $F(y,z)=0$.
So a Weierstrass form of $C'$ is given by $E':y^2=s^3-20s$, and $\alpha':E'\to C'$, $(s,y)\mapsto(\frac{4}{s^5},y)$ is birational. Let $\alpha:E\to C$ be a birational map (defined over some number field) of a Weierstrass model $E$ of $C$, and $\psi:C\to C'$ be the natural map $(z,y)\mapsto (z^{10},y)$.
Then $\alpha'^{-1}\circ\psi\circ\alpha$ is an isogeny $E\to E'$ of degree $10$.
If a non-singular model of $C$ had a rational point, then $E$ and $\alpha$ could be defined over $\mathbb Q$, and $E'$ would have an isogeny of degree $10$ defined over $\mathbb Q$. However, the $5$-division polynomial of $E'$ does not have a quadratic factor over $\mathbb Q$, so $E'$ and then also $E$ has no rational isogeny of degree $5$, and therefore also none of degree $10$.
In particular, as expected by the OP, $C$ has only finitely many rational points.
How to explicitly find $E$ and $\alpha$? Using (for instance) Sage one can find $E$ and an isogeny $\mu:E\to E'$ of degree $10$. (First find the degree $10$ isogeny $E'\to E$, then take its dual.) Finding $\alpha$ amounts to expressing a generic point $(z,y)\in C$ in terms of the generic point $(u,v)\in E$. This should be doable via $(z^{10},y)=\psi((z,y))=\alpha'(\mu((u,v))$, because we can compute everything on the right hand side.<|endoftext|>
TITLE: Automorphism groups for free groups with action
QUESTION [9 upvotes]: Let $A$ be a (finitely generated) free group and $G$ be a  (finitely generated) free $A$-group - that is, a group with an action of $A$, which is free in the category of groups with an $A$-action. Equivalently, $G$ is isomorphic to the free product of copies of a (finitely generated) free group $H$ (with a free basis $(h_1,...h_n)$) indexed by the elements of $A$, the $A$-action being the obvious one.
I feel that the automorphism group of $G$ (as a group with an $A$-action) should be generated by the automorphisms of $H$ and the automorphisms of the following form:
$h_i\mapsto\negmedspace^ah_i$ (where $a\in A$ and $i\in\{1,\dots,n\}$ are fixed), $h_j\mapsto h_j$ for $j\neq i$.
Possibly this is well-known and written somewhere. Unfortunately, I did not find anything about this by searching on the web or in classical books of group theory.
Do you have references, or ideas to approach the problem in the most efficient way?
Edit: I began to look carefully at Nielsen reduction arguments to try and adapt them to this situation with action, unfortunately it does not seem to work, at least simply. Maybe we have to be quite more clever in the orderings involved in the induction... or the result that I suggested is false?
Many thanks in advance.

REPLY [4 votes]: Finally, the answer is positive. See §3.1 in this preprint: https://hal.archives-ouvertes.fr/hal-01214646
that I have just posted. This gives also my motivation to study this question.<|endoftext|>
TITLE: Collapsing the cardinals between two singular cardinals
QUESTION [6 upvotes]: Question 1: Is it consistent that there is a forcing notion collapsing $\aleph_{\omega\cdot 2}$ to $\aleph_\omega$ without collapsing $\aleph_\omega$ or $\aleph_{\omega\cdot 2 + 1}$? 

If the answer to question 1 is affirmative, it will probably involve a forcing construction. It is natural to ask whether a similar situation can occur "naturally":

Question 2: Does some large cardinals assumption imply that there is a pair of singular cardinals $\mu < \lambda$ and a forcing notion collapsing $\lambda$ to $\mu$ without collapsing $\mu$ or $\lambda^+$?

REPLY [5 votes]: For question 2, some large cardinals imply the existence of such a forcing.  Suppose $\kappa$ is 2-huge, with $j : V \to M$ an elementary embedding, $\lambda = j(\kappa)$, $\theta = j(\lambda)$, and $M^\theta \subseteq M$.  Then using the embedding, we see $S = \{ x \subseteq \lambda^{+\omega+1} : (\forall \alpha \leq \omega+1) \mathrm{ot}(x \cap \lambda^{+\alpha}) = \kappa^{+\alpha} \}$ is stationary.  A standard argument gives that the following set is also stationary:  $T = \{ x \subseteq \lambda^{+\omega+1} : \mathrm{ot}(x \cap \lambda^{+\omega+1}) = \kappa^{+\omega+1}$ and $\kappa^{+\omega} \subseteq x \}$.  Also, there is a Woodin cardinal $\delta > \lambda$.  Forcing with the stationary tower up to $\delta$ below $T$ will produce a generic embedding $i : V \to N \subseteq V[G]$ with critical point  $\kappa^{+\omega+1}$ and $i(\kappa^{+\omega+1}) = \lambda^{+\omega+1}$, and $N^\delta \subseteq N$.  This has the desired collapsing effects.<|endoftext|>
TITLE: Asymptotics of a Bivariate Generating Function
QUESTION [7 upvotes]: I have the following generating function,
$$G(x,y)=\sum_{n,k \geq 0}a(n,k)x^ny^k = \frac{(y^2-y)x+1}{(y-y^3)x^2-(y+1)x+1}$$
and I am interested in obtaining an asymptotic for the sequence $a(n,k)$ i.e. some nice (relatively) closed form $f(n,k)$ s.t.
$$\frac{a(n,k)}{f(n,k)} \rightarrow 1$$
as $(n,k) \rightarrow \infty$. From what I've been able to gather the leading person on this type of analysis in Robin Pemantle, so I have been trying to follow one of his survey papers (https://www.math.upenn.edu/~pemantle/papers/twenty.pdf) in order to compute this. All of the theorems presented in this paper, however, seem to make heavy use of algebraic geometry which I am not very familiar with, so I have been struggling to figure out exactly what I need to compute and how. If someone more familiar with the subject could give me a run down of the necessary computations and how they fit in with the theorems given by Pemantle, that'd be great. One small thing I forgot to mention: for my particular sequence, we can assume that we are in what Pemantle calls the combinatorial case i.e. a priori we know $a(n,k) \geq 0$. Thanks.

REPLY [2 votes]: I was able to figure out a partial answer and few people have upvoted this so I will go ahead and give it, and maybe someone will respond with a full answer. The answer I was able to obtain is for the center of the sequence i.e. an asymptotic for $a(n,n)$. I essentially just follow Section 8.2 of Pemantle's paper a link for which is given in the question. So we write 
$$G(x,y)=\frac{F(x,y)}{H(x,y)}=\frac{(y^2-y)x+1}{(y-y^3)x^2-(y+1)x+1}$$
First we need to verify that the variety $V=\{H(x,y)=0\}$ is smooth. This is simply done by making sure that the equations $H(x,y)=0$, $\nabla H(x,y)=0$ do not have simultaneous solutions. This computation can be done using Gröbner basis, mainly the Mathematica command (Pemantle likes using Maple for whatever reason) 
$$GroebnerBasis[\{H, D[H, x], D[H, y]\}, \{x, y\}]$$
should yield a basis for the trivial ideal. Now we need to find the set of contributing critical points which Proposition 3.11 tells is given by the solutions to the equations $H(x,y)=0$, $xH_x-yH_y=0$. This again can be computed using Gröbner basis, so the Mathematica command 
$$GroebnerBasis[\{H, x*D[H, x] - y*D[H, y]\}, \{x, y\}]$$
gives $\{-2 + y + y^3, -1 + 2 x\}$. The roots of the first polynomial are $y=1,\frac{1}{2}(-1+i\sqrt{7}),{2}(-1-i\sqrt{7})$ and the second $x=\frac{1}{2}$, so there is only one positive, real, simultaneous solution  and thus the set of contributing critical points is the singleton $\{(\frac{1}{2},1)\}$. Now we simply plug-in to the formula given by Corollary 3.21 to obatin
$$a(n,n) \sim \frac{2^n}{\sqrt{\pi n}}$$<|endoftext|>
TITLE: Distributing points evenly on a sphere
QUESTION [36 upvotes]: I am looking for an algorithm to put $n$-points on a sphere, so that the minimum distance between any two points is as large as possible.
I have found some related questions on stackoverflow but those algorithms are not an exact solution more random distributions. For special number of points (inscribed Platonic solids) it is clear but how about 5 points for example. I would be grateful for hints to the literature.
Thank you everyone for your time.

REPLY [5 votes]: There is a rather comprehensive list of research articles about points on spheres and manifolds at https://my.vanderbilt.edu/edsaff/spheres-manifolds/
A nice paper is E. B. Saff, A. B. J. Kuijlaars, "Distributing Many Points on a Sphere", The Mathematical Intelligencer, Winter 1997, Volume 19, Issue 1, pp 5-11.<|endoftext|>
TITLE: A generously vertex transitive graph which is not Cayley?
QUESTION [5 upvotes]: A graph is vertex transitive if $x \mapsto y$ by an automorphism.
A graph is generously vertex transitive if $x \mapsto y \mapsto x$ by an automorphism.
Simple facts:

GVT $\rightarrow$ unimodular. Just plug in the definition of unimodular to see why.
Cayley $\not \rightarrow$ GVT. Free product $\mathbb Z_2 * \mathbb Z_3$ is an example.
GVT $\not \rightarrow$ Cayley. Petersen graph is a finite example.

Question:
Is there an infinite generously vertex transitive connected graph with finite degree which is not a Cayley graph?
(Before edition, it was also asked: "Is there a common name for this property?", answered by Chris Godsil in the comments.)
Thanks!

REPLY [5 votes]: Take the graph product $G = P \times \mathbb{Z}$ of the Petersen graph with the infinite path graph. This is clearly infinite, finite-degree, and generously vertex-transitive.
Then we have two distinguishable types of edge, namely $P$-edges (ones which belong to $5$-cycles) and $\mathbb{Z}$-edges (ones which do not). This means that the automorphisms of $G$ must preserve the obvious product structure, and can uniquely be written as compositions of automorphisms of $P$ with automorphisms of $\mathbb{Z}$.
If $G$ were a Cayley graph of some group $H$, then there are three generators corresponding to $P$-edges and two corresponding to $\mathbb{Z}$-edges. The actions of the $P$-generators and $\mathbb{Z}$-generators must, respectively, preserve the $\mathbb{Z}$-coordinate and $P$-coordinate of each vertex.
Hence the $P$-generators and $\mathbb{Z}$-generators each generate disjoint normal subgroups of $H$ (and $H$ is their internal direct product). The normal subgroup generated by the $P$-generators is sharply vertex-transitive on $P$, implying that the Petersen graph is a Cayley graph.
Contradiction.
So $G$ has all the properties you requested.<|endoftext|>
TITLE: Frobenius $A_{\infty}$-bialgebras?
QUESTION [8 upvotes]: Recall that a finite dimensional associative algebra $A$ over a field $k$ is called a symmetric Frobenius algebra (sometimes called "closed" Frobenius algebra) if its equipped with a symmetric non degenerate bilinear form $< , >: A \otimes A \to k$ satisfying $<ab,c>=<ca,b>$.
We can obtain a coproduct $v: A \to A \otimes A$ through the following composition of maps:
$A \cong A^* \to (A \otimes A)^* \cong A^* \otimes A^* \cong A \otimes A$
where the first map is the isomorphism of $A$-bimodules between $A$ and its dual $A^*$ defined by the pairing, the second is the dual of the product in $A$, the third follows from the finite dimensionality of $A$, and the last is again obtained from isomorphism induced by the pairing. One can check that that the compatibility of the product in $A$ with $<, >$ implies that $v$ is a map of $A$-bimodules. 
This yields the notion of a "Frobenius bialgebra" (sometimes called an "open" Frobenius algebra), which is essentially an associative algebra $A$ equipped with a coassociative coproduct which is a map of $A$-bimodules. A closed Frobenius algebra is equivalent to an open Frobenius algebra with unit and counit. 
There is a homotopy notion of a closed Frobenius algebra - I think originally due to Kontsevich- known as a cyclic $A_{\infty}$-algebra, which is by definition a finite dimensional $A_{\infty}$-algebra $(A, m_k:A^{\otimes k} \to A)$ equipped a non degenerate bilinear form $< , >: A \otimes A \to k$ cyclically compatible with the $m_k$'s in the sense that $<m_k(a_1,...,a_k),a_{k+1}>= \pm <m_k(a_{k+1},a_1,...,a_{k-1}),a_k>$. In particular, this cyclic compatibility condition implies that the isomorphism $A \to A^*$ is a map of $A_{\infty}$-bimodules over $A$.
My question is the following: Is there an explicit notion of a Frobenius $A_{\infty}$-bialgebra? In particular, given an a cyclic $A_{\infty}$-algebra $A$ we can obtain an $A_{\infty}$-coalgebra structure on $A$ by dualizing the structure maps and using the isomorphism $A \cong A^*$ induced by the pairing as we did above for the strict case. What are the explicit compatibilities between the $A_{\infty}$-algebra and $A_{\infty}$-coalgebra structure maps characterizing such structure?

REPLY [7 votes]: Certainly one can ask for the data of a (unital or nonunital) $A_\infty$-algebra $A$ equipped with a "comultiplication" map $A \to A \otimes A$ of $A_\infty$ bimodules.  Of course, this is infinitely much data --- all the higher coherences explaining how the comultiplication intertwines with the left and right actions.  I will call these higher coherences "(higher) Frobeniators".
Such data will not, in and of itself, give the notion you are after, because it does not include the coassociativity data.  It also clearly does not suffice to ask that the underlying map of vector spaces $A \to A \otimes A$ be part of a co-$A_\infty$ structure, as then you may not have good coherence between the (higher) coassociators and the (higher) Frobeniators.  Recall that the coassociator is a homotopy between two different maps $A \to A \otimes A \otimes A$.  What I expect is that you want to ask that it be a homotopy of maps of $A_\infty$-bimodules.
Almost surely that will give you a good notion of "open $A_\infty$ Frobenius algebra", although I don't know if/where it's been studied in the literature.  If it has, almost surely it's in the context of Poincare duality.
Here's another option.  Following Gan, call by the name dioperad a collection of chain complexes $\{P(m,n)\}_{m,n\in \mathbb N}$ with symmetric group actions and associative ``composition'' maps indexed by directed, but not necessarily rooted, trees.  Dioperads are one possibly many-to-many generalization of operads which, unlike cyclic operads, admit a straightforward theory of Koszul duality.  The open and coopen Frobenius dioperad is the one for which $P(m,n)$ is empty if $mn=0$ and otherwise is the free bimodule for the symmetric groups on $m$ and $n$ letters.  It is generated by a class in $P(1,2)$ and a class in $P(2,1)$ subject to "associativity", "coassociativity", and "Frobenius" relations.
This dioperad is known to be Koszul (which follows e.g. from the Koszulity of the associative operad along with some results of Vallette), and its homotopy algebras are a reasonable notion of "homotopy Frobenius algebra".  I believe, but have not checked, that they are precisely the same as the "open $A_\infty$ Frobenius algebras" mentioned above.
Dioperads only involve trees, and so everything in, say, "homotopy Frobenius algebra" when defined in terms of dioperads only involves trees.  So let me actually call such things "tree-level homotopy Frobenius algebras".  There is another reasonable many-to-many generalization of operads known as PROPs, which involve graphs.  In particular, there is a PROP who describes Frobenius algebras.  PROPs do not admit as good a Koszul duality theory, making it hard to work out "homotopy algebra", but pretty much every PROP in nature is the universal enveloping PROP of a properad, a notion due to Vallette, and moreover Vallette has shown that the universal enveloping functor from properads to PROPs is exact.  In particular, there is a properad of Frobenius algebras (generated by the same presentation as the dioperadic version), and PROPic and properadic homotopy algebras thereof are equivalent.  I will call homotopy algebras for the properad/PROP of Frobenius algebras "graph-level homotopy Frobenius algebras", since PROPs and properads involve non-tree graphs.
An aside: It is very hard to prove that the properad of Frobenius algebras is Koszul, and I don't think a proof is in the literature.  Its commutative cousin is known to be Koszul by arXiv:1402.4048, but that paper was a long time coming and required some new techniques.  Without Koszulity, you can still work out what a large presentation of the notion of "homotopy ...", but not a small one.
Now for the warning: the universal enveloping functor from dioperads to properads is NOT EXACT.  So you should not expect the notions of graph- and tree-level homotopy Frobenius algebra to be equivalent.
I studied some related questions in arXiv:1412.4664 and arXiv:1307.5812, where I showed that tree-level homotopy "shifted commutative" Frobenius algebras can fail to extend to graph-level ones, or if they do extend then the extension can fail to be canonical.  Indeed, the extension from tree- to graph-level is a "perturbative quantization" problem closely related to Feynman diagrams in quantum field theory.
The punchline is just that you should pay attention to which version of "homotopy Frobenius algebra" you want — graph- or tree-level — and be careful not to assume you have one when you've built the other.<|endoftext|>
TITLE: Does stationary reflection imply Mahloness?
QUESTION [13 upvotes]: Suppose $\kappa$ is strongly inaccessible and every stationary subset of $\kappa$ reflects.  Must $\kappa$ be Mahlo?
Remarks:

It is possible for every stationary subset of $\kappa$ to reflect, but $\kappa$ is only weakly inaccessible (and not strongly inaccessible).
If $V=L$ then the answer is "yes", and in fact $\kappa$ must be weakly compact.

REPLY [14 votes]: Not necessarily.  Here's a counterexample, but I'm sure it is a ridiculous overkill in consistency strength:
Suppose $\kappa$ is the least inaccessible limit of supercompact cardinals.  Then $\kappa$ is not Mahlo.  If $S \subseteq \kappa$ is stationary, then there is a stationary $T \subseteq S$ such that $T$ concentrates on some cofinality $\theta < \kappa$.  Taking a supercompact $\delta \in (\theta,\kappa)$ gives the reflection of $T$ by the usual argument.<|endoftext|>
TITLE: Surgery along an arc connecting the components of a $2$-component link gives the unknot
QUESTION [6 upvotes]: Math Overflow seems to have a dearth of low dimensional topology, but this seems like an interesting question. Let $L$ be a $2$-component link in $S^3$. Suppose that there is a framed arc joining the two components such that surgery along that arc yields an unknot. (In other words, fatten the arc to a long thin rectangle where two of its edges are subarcs of the link components. Delete these arcs and replace them with the other two edges of the rectangle.) I wonder how restrictive this condition is. For example, can one prove that there are links which don't have this property? I don't really have a good feel for what is possible. 
One way to construct an example is to start with an unknotted surface with one boundary component. Now attach a band to the boundary of this surface that travels through all of the surfaces holes. This is now a 2-component link. Snipping the band can be realized by surgery along a transverse arc, and yields the boundary of the original surface, which is an unknot.

REPLY [3 votes]: A recentish paper of mine (which generalizes portions of Scharlemann's and Eudave-Munoz's work) also addresses this question. 
MR3192616 Taylor, Scott A.  Comparing 2-handle additions to a genus 2 boundary component.
 Trans. Amer. Math. Soc.  366  (2014),  no. 7, 3747--3769.<|endoftext|>
TITLE: Detecting positive endomaps of the formal reals
QUESTION [8 upvotes]: A locale is a sort of "formal topological space", which "may not have enough points to separate its open sets".  For instance, there is a "locale of all real numbers that are both rational and irrational", which of course has no points at all (since there are no such numbers), but as a locale it is a nontrivial structure.
Locales are often better-behaved than topological spaces in constructive mathematics (i.e. in the absence of the law of excluded middle).  For instance, there is a "locale of formal real numbers" $R_f$ whose sublocale $[0,1]_f$ is always compact, despite that the Heine-Borel theorem for the space $[0,1]$ may fail constructively.  Classically, $R_f$ has enough points, but constructively it may not — though unlike the example above it of course has lots of points, indeed its points are dense in it.  (See for instance P.T. Johnstone's book Stone spaces, or section D4.7 in Sketches of an Elephant.)
My question is: suppose $f:R_f \to R_f$ is a continuous map of locales, such that $f(x)>0$ for all points $x$ of $R_f$ (i.e. for all actual real numbers $x$); does it follow (constructively) that $f$ factors through the "locale of formal positive real numbers"?
The latter means the open sublocale of $R_f$ defined by the open subset $(0,\infty)$.  Note that denseness of the points of $R_f$ is insufficient to answer "yes", since a function can be positive on a dense subset but zero at some other points.

REPLY [4 votes]: I think the answer is no. Work in a recursive context; eg the effective topos.
Consider a (uniformly continuous) function $f:R\to R$ such that $f(x)>0$ for all $x$, but we don't have a positive infimum.  This is standard (use the Kleene tree).
Now consider the embedding of pointwise spaces into locales; the work by Palmgren A constructive and functorial embedding of locally compact metric spaces into locales. Then you have a function as requested, but if it factors through the positive formal reals, then we have a positive infimum. The result is proved here and recalled in the introduction of this paper<|endoftext|>
TITLE: Is this (funny) combinatorial optimization problem NP-hard ? (cutting numbers and placing them in urns)
QUESTION [10 upvotes]: The parameters of the problem are $m$ numbers which are integers (these numbers are denoted $b_i$), $n$ urns and in each urn, we can place $C$ numbers. We assume $nC \geq m$ so that the problem is feasible.
Let's take an example with $m=3$ numbers (with $b_1=1100$, $b_2=540$, $b_3=170$) and $n=2$ urns with capacity $C=4$.
The problem consists to cut the $m$ numbers in order to have $nC$ parts. The cut must preserve each number, for example we have $nC=8$ parts by doing:

$b_1=100+400+450+150$.
$b_2=100+40+400$.
$b_3=170$.

Then we must place these $nC$ numbers in the urns (knowing that each urn has a capacity of $C=4$), for example:

Urn 1: $40$ (2) - $170$ (3) - $100$ (2) - $450$ (1)
Urn 2: $100$ (1) - $150$ (1) - $400$ (2) - $400$ (1)

The cost $f$ of any solution is defined as the sum of the biggest numbers in each urn, in this case: $f=450+400=850$.
The goal of the problem is to decide how to cut the numbers and how to gather them in the urns in order to minimize the cost (actually, I have no idea what is the optimal solution of the previous instance).
I am not looking for an algorithm to solve this problem but I am wondering if this problem is NP-hard. Maybe this problem can be linked with another combinatorial optimization problem ?
I already searched and all that I could find is that the problem is easy if $n=1$ (and we still assume $C \geq m$) and that there is a polynomial algorithm in that case.
Thank you very much!

REPLY [2 votes]: Edit Using 3-PARTITION instead of PARTITION in the reduction below, we get that the problem is strongly NP-hard.
I think we get NP-hardness by the following reduction from PARTITION. Let a PARTITION instance be given by positive integers $a_1,\ldots,a_N$ with $a_1+\cdots+a_N=2B$ where the question is if we can select an index set $I\subseteq\{1,\ldots,N\}$ with $\sum_{i\in I}a_i=B$. We define an instance of the spaghetti cutting problem with the following data

number of jars: $n=N$
jar capacity: $C=N+2$
number of spaghetti pieces: $m=N(N+1)+2$
spaghetti lengths: For each $i\in\{1,\ldots,N\}$, there are $N+1$ pieces of length $a_i$, and in addition there are 2 pieces of length $B$.

The trivial lower bound for this instance is $(\sum_{i=1}^mb_i)/C=2B$, and I claim that it can be achieved if and only if the PARTITION instance is a YES-instance. The "if"-part is obvious: You can cut the two pieces of length $B$ into $N$ pieces with lengths $a_1,\ldots,a_N$, and then we have one jar full of length $a_i$ pieces for every $i$. For the "only if" part it helps to assume $a_1>a_2>\cdots>a_N$ which is no problem (see here). In order to achieve the objective value $2B$ we need one jar full of pieces of length $a_1$, so we have to cut a piece of length $a_1$ from one of the two long spaghetti pieces. Then we can assume w.l.o.g. that we don't cut the pieces of length $a_1$. Next we need a jar full of pieces of length $a_2$, which can be achieved only by cutting it from one of the long pieces. Continuing in this way we complete the argument.
This leaves open what happens for bounded capacity $C$. The first interesting case is $C=2$, for which the question if the lower bound $(\sum_{i=1}^mb_i)/2$ is achievable can be formulated as follows: Given $b_1,\ldots,b_m$ and $n> m/2$, can we cut the $b_i$'s into $2n$ pieces such that every length appears an even number of times?
For what it's worth, here is a mixed binary programming formulation for the general problem:
\begin{align*}
\text{Minimize }\sum_{j=1}^n&x_{(j-1)C+1}\qquad\text{subject to}\\
x_1\geqslant &x_2\geqslant\cdots\geqslant x_{nC},\\
\sum_{k=1}^{nC}y_{ik} &= b_i &&i\in[m],\\
\sum_{i=1}^my_{ik} &= x_k && k\in[nC],\\
y_{ik} &\leqslant b_iz_{ik} && i\in[m],\ k\in[nC],\\
\sum_{i=1}^mz_{ik} &\leqslant 1 &&k\in[nC],\\
y_{ik} &\geqslant 0,\ z_{ik}\in\{0,1\} && i\in[m],\ k\in[nC].
\end{align*}<|endoftext|>
TITLE: presentations of subalgebras
QUESTION [6 upvotes]: Assume that I have a finitely presented algebra $A$ over the complex numbers (by which I mean that $A$ is generated over $\mathbb{C}$ by finitely many elements $x_1,...,x_n$ subject to finitely many relations). Let now $y_1,\ldots, y_m$ be a finite subset of elements of $A$. Is there an algorithm which gives a presentation of the subalgebra $B$ of $A$ generated by $y_1,...y_m$ ? 
In particular, I am interested in the following example: Let $A = \mathbb{C}B_3 = \mathbb{C}\langle \sigma_1,\sigma_2 | \sigma_1\sigma_2\sigma_1 = \sigma_2\sigma_1\sigma_2\rangle$ be the group algebra of the Braid group on 3 braids. Is there any nice presentation of the subalgebra generated by $\sigma_1$ and $\sigma_2 + \sigma_2^{-1}$ ?

REPLY [4 votes]: If we modify your first question only slightly, then the answer to the question is no, there is no algorithm.  By Theorem 1 of the paper G. Baumslag, W. W. Boone and B. H. Neumann, Some unsolvable problems about elements and subgroups of groups, Math. Scand. 7, 191-201 (1959), let $G$ be a finitely presented group for which is it undecidable whether elements have finite order.  Let $A=\mathbb{C}G$.  For each element $g\in G$, we can take $B_g=\mathbb{C}[g]$ to be the subalgebra generated by $g$.  Any non-trivial relation in $B_g$ would tell us $g$ has finite order, and so there can be no algorithm (which is independent of $g$) to yield a presentation of $B_g$.<|endoftext|>
TITLE: Isotrivial families with non-zero Kodaira spencer map
QUESTION [7 upvotes]: Let $S$ be a smooth quasi-projective curve over the complex numbers. Let $P$ be a closed point in $S$. Let $f:\mathcal X \to S$ be a polarized family of smooth projective connected varieties. To this family and the point $P$ we can associate the Kodaira-Spencer map 
$$ Tan_S(P) \to H^1(X,T_X).$$
Here $X $ is the fibre of $f$ over  $P$.
Of course, if $f$ is a product/trivial family (so $\mathcal X = X \times S$) then the Kodaira-Spencer map is zero.
On the other hand, I can't prove that the same holds if all closed fibres of $f$  are isomorphic (i.e. $f$ is isotrivial). Therefore, I suspect that there might be examples of non-trivial isotrivial families with non-zero Kodaira-Spencer map.
Are there examples of such families of varieties? (Note that such a family if it exists is not trivial, locally for the fppf topology.)

REPLY [6 votes]: The most general (most useful) formulation of the Kodaira-Spencer map uses the triangle of transitivity of cotangent complexes and relative duality.  Even for such mildly singular varieties as nodal curves, the following is not sufficient (I recommend Section 6  of K. Behrend and B. Fantechi, "The Intrinsic Normal Cone", for a quick introduction to Kodaira-Spencer in the case of singular, projective varieties).  For a proper, smooth morphism of relative dimension $n$, $f:\mathcal{X}\to S$, with $S$ a finitely presented $k$-scheme, the natural exact sequence $$ f^*\Omega_{S/k} \to \Omega_{\mathcal{X}/k} \to \Omega_{\mathcal{X}/S} \to 0,$$ is left exact (thus a short exact sequence), and the cokernel $\Omega_{\mathcal{X}/S}$ is locally free of rank $n$.  Denote the dual locally free sheaf by $T_{\mathcal{X}/S}$.  Tensoring by this locally free $\mathcal{O}_{\mathcal{X}}$-module gives a short exact sequence, $$ 0 \to 
f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S} \to
\Omega_{\mathcal{X}/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S} \to
\Omega_{\mathcal{X}/S}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S} \to
0.$$  This short exact sequence gives rise to a long exact sequence of higher direct image sheaves with respect to $f$.  The relevant part here is the first connecting map, $$ \delta: f_*\left(\Omega_{\mathcal{X}/S}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right) \to
R^1f_* \left(f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right).$$ The identity map on $\Omega_{\mathcal{X}/S}$ gives a natural global section, $\text{Id}:\mathcal{O}_S \to f_*\left(\Omega_{\mathcal{X}/S}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right)$.  Thus there is a global section, which is a version of the Kodaira-Spencer map,
$$
\delta'_{\mathcal{X}/S}:\mathcal{O}_S \to R^1f_* \left(f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right)
$$  
If $S$ is smooth over $k$, then $\Omega_{S/k}$ is locally free.  Then, 
by the projection formula, the natural map $$\Omega_{S/k}\otimes_{\mathcal{O}_S} R^1 f_*\left( T_{\mathcal{X}/S} \right) \to R^1f_* \left(f^*\Omega_{S/k}\otimes_{\mathcal{O}_{\mathcal{X}}} T_{\mathcal{X}/S}\right),$$ is an isomorphism.  Thus, $\delta'_{\mathcal{X}/S}$ is naturally an $\mathcal{O}_S$-module homomorphism, $$ \delta'_{\mathcal{X}/S} : \mathcal{O}_S \to \Omega_{S/k}\otimes_{\mathcal{O}_S} R^1 f_*\left( T_{\mathcal{X}/S} \right).$$  Denoting by $T_{S/k}$ the dual locally free sheaf to $\Omega_{S/k}$, this is equivalent to another
$\mathcal{O}_S$-module homomorphism (closer to the usual definition),
$$ \delta_{\mathcal{X}/S} : T_{S/k} \to R^1 f_*\left( T_{\mathcal{X}/S} \right),$$ where $T_{S/k}$ is the dual of the locally free sheaf $\Omega_{S/k}$. 
By Grauert's theorem, Corollary III.12.9, p. 288, of R. Hartshorne, "Algebraic Geometry", $R^1f_*\left(T_{\mathcal{X}/S}\right)$ is a locally free $\mathcal{O}_S$-module compatible with base change to reduced schemes (we only need closed points) if for every point $s$ of $S$ (closed points suffice), $h^1(\mathcal{X}_s,T_{\mathcal{X}_s})$ is constant.  In particular, if all fibers of $f$ are isomorphic as $k$-schemes, this is true.  In that case, $\delta$ is a map of locally free $\mathcal{O}_S$-modules.  On a reduced scheme, a morphism of torsion-free quasi-coherent $\mathcal{O}_S$-modules is zero if and only if it is zero at the generic point if and only if it is zero on some dense open subscheme.  Thus, in what follows, it suffices to restrict to a dense open subscheme of $S$.
Let $s_0$ be a $k$-point of $S$.  Let $X_0$ denote the fiber of $f$ over $s_0$.  As a corollary of existence of the Hilbert scheme (really, algebraic space if $f$ is merely a proper, flat morphism of algebraic spaces), the Isom functor is representable, and there is an Isom scheme, $$\pi: \text{Isom}_S(X_0\times S,\mathcal{X}) \to S.$$ By hypothesis, this morphism is surjective on $k$-points.  Now assume that $k$ is characteristic $0$ and uncountable (under the hypotheses above, $\delta$ is compatible with base change, so we can base change to make $k$ uncountable).  
Since $\pi$ is surjective, and since $k$ is uncountable, there exists an irreducible component $I$ of $\text{Isom}_S(X_0\times S,\mathcal{X})$ such that the restion $\pi|_I$ is dominant.  Denote this restriction by $$\rho:I\to S.$$  Since $k$ is algebraically closed, there exists a dense open subscheme of $I_{\text{red}}$ that is a smooth $k$-scheme.  Replace $I$ by this dense open subscheme of $I_{\text{red}}$.  Since $\rho$ is a dominant morphism of smooth $k$-schemes, and since $k$ has characteristic $0$, by generic smoothness there exists a dense open subscheme of $S$ over which $\rho$ is smooth.  Replace $S$ by this open subscheme and assume that $\rho$ is smooth and surjective.  By construction of the Kodaira-Spencer maps, there is a commutative diagram of $\mathcal{O}_I$-modules,
$$ \begin{array}{ccc} \rho^*\mathcal{O}_S & \xrightarrow{\rho^*\delta_{\mathcal{X}/S}}  & \rho^*(\Omega_{S/k})\otimes_{\mathcal{O}_I} \rho^*R^1f_* T_{\mathcal{X}/S} \\
\downarrow & & \downarrow \\
\mathcal{O}_I & \xrightarrow{\delta_{\mathcal{X}_I/I}} & \Omega_{I/k}\otimes_{\mathcal{O}_I} R^1f_{I,*} T_{\mathcal{X}_I/I} \end{array}.$$
Since $\rho$ is faithfully flat, to prove that $\delta_{\mathcal{X}/S}$ is zero, it suffices to prove that $\rho^*\delta_{\mathcal{X}/S}$ is zero.
By Grauert's theorem, the natural map $\rho^* R^1f_* T_{\mathcal{X}/S}  \to R^1f_{I,*} T_{\mathcal{X}_I/I}$ is an isomorphism.  Since $\rho$ is smooth, the morphism $$d\rho^\dagger: \rho^*\Omega_{S/k} \to \Omega_{I/k}, $$ is injective with locally free cokernel.  Thus, the second vertical arrow above is injective (the first vertical arrow is an isomorphism).  So to prove that $\rho^*\delta_{\mathcal{X}/S}$ is zero, it suffices to prove that $\delta_{\mathcal{X}_I/I}$ is zero.
By construction, there is an isomorphism over $I$ of the base change $\mathcal{X}_I = I\times_S \mathcal{X}$ with the constant family $X_0\times I$.  That isomorphism of $I$-schemes identifies the Kodaira-Spencer map of $\mathcal{X}_I/I$ with the Kodaira-Spencer map of $X_0\times I/ I$.  Of course since $X_0\times I/I$ is the pullback of $X_0/\text{Spec}(k)$ by the constant $k$-morphism $p:X\to \text{Spec}(k)$, there is another commutative diagram, $$ \begin{array}{ccc} p^*\mathcal{O}_{s_0} & \xrightarrow{p^*\delta_{X_0/s_0}}  & \rho^*(\Omega_{k/k})\otimes_{k} H^1(X_0,T_{X_0/k}) \\
\downarrow & & \downarrow \\
\mathcal{O}_I & \xrightarrow{\delta_{X_0\times I/I}} & \Omega_{I/k}\otimes_{\mathcal{O}_I} R^1f_{I,*} T_{X_0\times I/I} \end{array}.$$  Since $\Omega_{k/k}$ is the zero sheaf, it follows that $\delta_{X_0\times I/I}$ is the zero homomorphism.  Therefore also $\delta_{\mathcal{X}/S}$ is zero.<|endoftext|>
TITLE: Matroids similar to the cycle matroid
QUESTION [7 upvotes]: Let $G=(V,E)$ be a graph (loops and multiple edges are permitted). Three following systems of dependent sets in $E$ define matroids:
1) Set $A\subset E$ is dependent if $A$ contains cycle. This is a cycle matroid, maybe the most popular matroid.
2) Set $A\subset E$ is dependent if $A$ contains two different cycles in the same connected component. In other words, independent set is characterized by the property that number of edges in any connected component does not exceed number of vertices.
3) Set $A\subset E$ is dependent if $A$ contains either even cycle or two odd cycles in the same connected component. In other words, rank of a connected subgraph $G_1=(V_1,E_1)$ equals $|V_1|-1$ if $G_1$ is bipartite and equals $V_1$ if $G_1$ has an odd cycle.
My question is whether these examples may be included in a more large picture, and may I read about those matroids.

REPLY [4 votes]: The first two examples fall neatly into the class of $(k,l)$-sparsity matroids (see the references in this Wikipedia article and references contained within).  The cycle matroid (1) is the (1,1)-sparsity matroid on $G$ and the second matroid you describe is the (1,0)-sparsity matroid.
The third example is an example of a frame matroid and has been called the "even-circle matroid".  It appears in a paper of Doob on the eigenspace of -2 of the adjacency matrix of a graph, and in Zaslavsky's paper "Biased Graphs I" (the first in the series linked by Gordon Royle) it appears as example 6.3.  (Note that what Zaslavsky calls bias matroids are now called frame matroids).
There are some common generalizations of sparsity matroids and frame matroids, see e.g. these papers by Tanigawa.<|endoftext|>
TITLE: The term $H^1(N,A)^{G/N}$ in the inflation-restriction exact sequence
QUESTION [9 upvotes]: [a repost from SE due to the lack of response]
Given a group $G$, let $A$ be a $G$-module and let $N\trianglelefteq G$.
If I understand it correctly, the superscript "G/N" in the third term of the standard inflation-restriction exact sequence
$$
0\to H^1(G/N,A^N)\to H^1(G,A)\to H^1(N,A)^{G/N}\to H^2(G/N,A^N)\to H^2(G,A)
$$
means the fixed points under the action of $G/N$ on the first cohomology group
$H^1(N,A)$. But how is this action defined? Is there an elementary definition (in terms of cocycles) that does not refer to the Lyndon–Hochschild–Serre spectral sequence?

REPLY [13 votes]: You do not need LHS spectral sequence for this action.
The functors $H^*(N,-)$ are the derived functors of $(-)^N:G\text{-mod}\rightarrow G/N\text{-mod}$,
so they will carry a structure of $G/N$-modules. 
Explicitly, on the level of cocycles, it can be described as follows:
if you have a one cocycle $f:N\rightarrow A$ and $g\in G$, then the action on $f$ will be given by $g\cdot(f)(n) = g\cdot f(g^{-1}ng)$<|endoftext|>
TITLE: From polynomial ideal over $\mathbb{Q}$ to polynomial ideal over $\mathbb{Z}$
QUESTION [16 upvotes]: Is there an algorithm to compute, given a polynomial ideal $I\subset \mathbb{Q}[x_1,\dotsc,x_n]$, the ideal $I\cap \mathbb{Z}[x_1,\dotsc,x_n]$ in $\mathbb{Z}[x_1,\dotsc,x_n]$?

The input and output can be given via a finite set of generators. From chapter 10 of book "Gröbner bases" by Becker and Weispfenning, I am familiar with Gröbner bases over PID, and in particular understand how to compute the intersection of polynomial ideals over PID (though it is not explicitly mentioned in the book, as far as I can see). However, $\mathbb{Z}[x_1,\dotsc,x_n]$ is not an ideal in $\mathbb{Q}[x_1,\dotsc,x_n]$. It is tempting to work in the ring $\frac{1}{D}\mathbb{Z}[x_1,\dotsc,x_n]$ for large very composite integer $D$, but how large is large enough?

REPLY [10 votes]: I am aware that I am replying to an old question that has been satisfactorily answered, but I think it might be worth while pointing out, for the completeness of MathOverflow, that the question is discussed (in slightly greater generality, and with an example) as corollary 4.4.5 in the book An Introduction to Gröbner Bases by William Adams and Philippe Loustaunau (AMS 1994, Graduate Studies in Mathematics 3).  The method is the same as in the approved answer, but formulated in a slightly different way.  Their statements (very slightly reworded) are:
Proposition 4.4.1 [computation of the saturation]: Let $R$ be an integral domain and $I$ a non-zero ideal of $A = R[x_1,\ldots,x_n]$.  Let $g\in A$, $g\neq 0$.  Let $w$ be a new variable.  Consider the ideal $\langle I,wg-1\rangle$ of $A[w]$.  Then $I A_g \cap A = \langle I, wg-1\rangle \cap A$ where $A_g = A[\frac{1}{g}]$.
(Note that the intersection $\langle I, wg-1\rangle \cap A$ in $A[w]$ can be computed using an elimination order on the variables: it is theorem 4.3.6 in that book.)
Proposition 4.4.4: Let $R$ be an integral domain with $k$ its quotient field.  Let $I$ be a non-zero ideal of $A = R[x_1,\ldots,x_n]$ and let $G = \{g_1,\ldots,g_t\}$ be a Gröbner basis for $I$ with respect to some term ordering.  Let $s$ be the product of the leading coefficients of $g_1,\ldots,g_t$.  Then
$$I k[x_1,\ldots,x_n] \cap R[x_1,\ldots,x_n] = IR_s[x_1,\ldots,x_n] \cap R[x_1,\ldots,x_n]$$
(This is essentially what the approved answer to this question explains.)
Corollary 4.4.5: Let $R$ be an integral domain in which linear equations are algorithmically solvable and let $k$ be its quotient field.  Let $I$ be an ideal of $A = R[x_1,\ldots,x_n]$.  Then we can compute generators for the ideal $I k[x_1,\ldots,x_n] \cap R[x_1,\ldots,x_n]$.
(This follows immediately from the above results.)<|endoftext|>
TITLE: Analysis of the Laplacian of a random bipartite graph
QUESTION [7 upvotes]: My analysis of an engineering problem reduced to analysis of the Laplacian of a (random) bipartite graph. There are a few particular questions I am interested in, but not sure which direction to take at the moment. Below, I am outlining the problem. Would really appreciate any advice.
The Graph
I have an undirected weighted complete bipartite graph $G = (V_1, V_2, E)$, with $\vert V_1 \vert = n_1$, $\vert V_2 \vert = n_2$, $n_1 + n_2 \equiv n$. Each vertex from $V_1$ is connected with each one from $V_2$. $G$ is obviously connected. Edge weights are positive i.i.d. random variables: $e_{ij} \in (e_\text{min}, e_\text{max})$, where $e_\text{min},e_\text{min} \in \mathbb{R}_+$ and $e_\text{min} \ll e_\text{max}$.
The edge weights of $G$ can be represented by a (random) matrix $S \in \mathbb{R}_+^{n_1 \times n_2}$: each element of this matrix, $s_{ij}$, represents a weight of the edge that connects $v_i \in V_1$ and $v_j \in V_2$. In this case, weighted adjacency matrix, $A \in \mathbb{R}_+^{n \times n}$, and Laplacian matrix, $L = D - A$, will have the following block structures:
$$
A =
\begin{pmatrix}
0 & S\\
S^\top & 0
\end{pmatrix},\quad
L =
\begin{pmatrix}
D_1 & -S\\
-S^\top & D_2
\end{pmatrix},
$$
where $D_1 = \text{diag}(\sum_{j=1}^{n_2} s_{1j}, \dots, \sum_{j=1}^{n_2} s_{n_1j})$ and $D_2 = \text{diag}(\sum_{i=1}^{n_1} s_{i1}, \dots, \sum_{i=1}^{n_1} s_{in_2})$.
Analysis and Questions
Given a graph instance, $G$, I am interested in inferring its edge weights from resistance distances I can "query" for each pair of the vertices. It can be shown that the resistance distance between $v_i$ and $v_j$ can be expressed as follows [1]:
$$
r_{ij} = L^{\dagger}_{ii} + L^{\dagger}_{jj} - 2L^{\dagger}_{ij},
$$
where $L^{\dagger}$ is the Moore-Penrose pseudoinverse of the Laplacian, which can be expressed as $L^{\dagger} = (L + \frac{1}{n}J)^{-1} - \frac{1}{n}J$, where $J$ is an $n \times n$ matrix of all ones [1]. The resistance distance can be also expressed as [2]:
$$
r_{ij} = \frac{\det L(i,j)}{\det L(i)} = \frac{\det L(i,j)}{\det L(j)},
$$
where $L(i)$ is denotes $L$ without the $i$-th row and column, and  $L(i,j)$ denotes $L$ without $i$-th and $j$-th both rows and columns.
Given a set $\{r_{ij}\}$ for $G$, I want to resolve the above system of equations for the unknown variables $\{s_{kl}\}$.
Q1: Is it possible to show that there exists a (non-)unique solution?
The problem is that $r_{ij} = \det L(i,j) / \det L(i)$ leads to a system of polynomial equations, and I have never dealt with. Any suggestions how to approach Q1?
Q2: Suppose, the system for $s_{kl}$ is underdetermined (i.e., we don't have enough $r_{ij}$ values). Is there a way to analyze the distribution of the random graphs over the solution space, i.e., $\mathbb{P}(\{s_{kl}\} \vert \{r_{ij}\})$?
Given that determinants, $\det L(i,j)$ and $\det L(i)$, can be represented as products of the corresponding eigenvalues, Laplacian spectral theory and random matrix theory can probably be the right thing to start with. However, I am only slightly familiar with the former and have no idea whether methods of the latter can help answer my question. Do I need to take this path in my analysis, or there are better ways I should consider?
References
[1] Bapat, R. B. (2010). Graphs and matrices. Chapter 10. New York (NY): Springer.
[2] Bapat, R. B., Gutmana, I., & Xiao, W. (2003). A simple method for computing resistance distance. Zeitschrift für Naturforschung A, 58(9-10), 494-498.

REPLY [2 votes]: By a strange coincidence, this circle of questions is studied in the extremely recent arXiv.org preprint by V. Vengerovsky.<|endoftext|>
TITLE: Connes Embedding Conjecture and Fusion Categories
QUESTION [6 upvotes]: I was recently introduced to Connes' Embedding Conjecture (CEC) which states:

Every separable type $II_{1}$ factor is embeddable into $R^{\omega}$. Where $\omega$ is a generic free ultrafilter on $\mathbb{N}$, $R$ is a hyperfinite type $II_{1}$ factor, and $R^{\omega}$ is the ultrapower of $R$ with respect to $\omega$.

I'm not well versed in the theory of factors and this conjecture is mysterious and difficult for me to understand. However, there do seem to be a lot of equivalent formulations of CEC. For some examples of this see: N. Ozawa, "About the Connes Embedding Conjecture---Algebraic Approaches---, arXiv:1212.1700 [math.QA]
My background is in fusion categories, and given the relationship between subfactors and fusion categories I was led to wonder:
Question: Are there statements in the theory of fusion categories that are equivalent to CEC, or does CEC have implications for fusion categories?
My understanding of the connections between fusion categories and subfactors, I'm sorry to say, is quite weak. So it very well could be that there is simply no relation between CEC and fusion categories, which would be a completely acceptable answer.

REPLY [3 votes]: For fusion categories, there is a hyperfinite embedding result which does not require passing to ultraproducts. 
If $\mathcal{C}$ is a unitary fusion category, then by results of Popa [MR1055708] (see also [MR3028581, Theorem 4.1]), there is a finite index, depth two subfactor, which we could reasonably denote $R\subset R\rtimes\mathcal{C}$, whose principal even half is $\mathcal{C}$. (Here, $R$ is the hyperfinite II$_1$-factor.) Moreover, by Popa's theorem, $R\subset R\rtimes \mathcal{C}$ is the unique hyperfinite subfactor up to subfactor isomorphism with this standard invariant. 
As a corollary, every unitary fusion category has an essentially unique realization as a concrete category of bifinite bimodules over $R$. (One has to be careful to get the correct uniqueness statement here.)
I'm not really sure how to introduce ultraproducts or Connes' embedding conjecture into this story. I'll think it over, and if I have something reasonable to say, I'll edit this answer.<|endoftext|>
TITLE: Consistency strength of being strong cardinal and indestructible under collapses
QUESTION [8 upvotes]: What is the consistency strength of the following statement:

$\kappa$ is a strong cardinals and it is indestructible under $Col(\kappa, <\theta),$  where $\theta> \kappa$ is some fixed inaccessible cardinal.

Remark 1. By Laver's result, the above statement is consistent relative to the existence of a supercompact cardinal (and an inaccessible above it).
Remark 2. By results of Mitchell, Schimmerling and Steel, ``the covering lemma up to a Woodin cardinal'', if $\kappa$ remains measurable after collapsing $\kappa^+$ to $\kappa,$
then there is an inner model with a Woodin cardinal.
Remark 3. Apter's paper ``Strong cardinals can be fully indestructible'' have some positive consistency results related to the question.

REPLY [4 votes]: If κ is weakly compact in VCol(κ,κ+), then there is a non-domestic premouse, see Theorem 0.4 of  https://ivv5hpp.uni-muenster.de/u/rds/stacking_mice.pdf<|endoftext|>
TITLE: Automorphic factorization of Dedekind zeta functions
QUESTION [15 upvotes]: It is well known that for abelian number fields, the factorization of its Dedekind zeta function goes like this:
$$\zeta_K(s)=\zeta(s)\prod_{\chi \neq 1} L(s,\chi)$$
with the Dirichlet characters distinct and primitive.
If $K$ is non-abelian but Galois, we instead have (by Aramata-Brauer):
$$\zeta_K(s)=\zeta(s)\prod_{\rho \neq 1} L(s,\rho)$$
with the representations non-trivial and irreducible, although in this case we don't know unconditionally that that's a factorization of irreducible L-functions. In the non-Galois case, factors might appear with arbitrary integer powers. But let's forget about that for a moment.
In this answer, Kevin Dong mentions an explicit factorization of the zeta of $\mathbb{Z}[\sqrt[3]{2}]$ in terms of modular forms. It is very nice but not quite suprising: Artin L-functions are expected (and known in some cases) to always be automorphic.

I'm interested on the proof for $\mathbb{Z}[\sqrt[3]{2}]$ (I haven't been able to find a reference for it), and any other reference for known case of such a factorization (this is, not a factorization on terms of Artin L-function, but of automorphic ones).
I'd also want to know what the conjectures are (on the automorphic side) for what the factorization looks like for an arbitrary non-Galois Dedekind zeta function.

Any other information around those issues might be of help, but not generally about Langlands or the Artin conjecture.

REPLY [2 votes]: if you know theta series, you write $L(s,\rho)= \sum_{n>0 } a_n n^{-s}$ and you have ; $\sum_{n>0} a_nq^n = 1/2(\Theta_{1,0,27}-\Theta_{4,2,7})$. Where : 
$$
\Theta_{a,b,c} = \sum_{(x,y) \in \mathbb{Z}^2} q^{ax^2+bxy+cy^2}
$$
For diedral odd representation  all is explicit.<|endoftext|>
TITLE: SO$(4)$ (& SO$(n)$) characterization?
QUESTION [6 upvotes]: I believe it is the case that
any finite subgroup of SO$(3)$ 
(the $3 \times 3$ orthogonal matrices of determinant $1$)
is either a cyclic group $C_n$,
or a dihedral group $D_n$, or one of the groups for one of the five Platonic solids.

Q, What is the generalization of this characterization to SO$(n)$,
  $n>3$,
  and in particular (my specific interest), to SO$(4)$?

This is certainly well-known, so this is just a reference request. Thanks!

          


          

(Image from John O'Connor: S3(Cube): red:face(cents); green: vertices; blue:edge(midpts) .)

REPLY [6 votes]: For finite irreducible linear groups in low dimension (at most 4) , the answer was already known and discussed in books by Blichfeldt, and Miller, Blichfeldt and Dickson in the early 20th century. However, Blichfeldt did miss at least one case. Dealing with real representations complicates the issue somewhat, but if the real representation splits into two complex conjugate representations when the field of scalars is extended to $\mathbb{C}$, we are looking at finite subgroups of ${\rm GL}(2,\mathbb{C})$. It is also possible that the real irreducible representation could split as the sum of two equivalent irreducible complex representations, but the $2$-dimensional question for complex representations is easy.
For general $n$, the classification of irreducible finite subgroups of ${\rm GL}(n,\mathbb{C})$ can be refined in various ways. First, consider only primitive linear groups (where the representation is not equivalent to one induced from a proper subgroup). Next, consider representations which (even after taking central extensions) can't be decomposed as the tensor product of two representations of smaller dimension.
Next, consider representations which can't be tensor induced from a representation of a proper subgroup (even after taking central extensions).
This leaves two residual configurations for $G$: in one case, there is a quasisimple irreducible normal subgroup of $G$. In the other case, there is an "almost extraspecial" irreducible normal $p$-subgroup $U$ for some prime $p$, where $n = p^{k}$, and $G/UZ(G)$ is isomorphic to a subgroup of ${\rm Sp}(2k,p).$
Prior to the classification of finite simple groups, the situation was fairly explicitly understood in dimension up to $11$. Using the classification of finite simple groups, B. Weisfeiler got a close to optimal bound for Jordan's Theorem for $n > 70$ or so.
Recently, M.J. Collins has extended Weisfeiler's work to determine the maximal possible index of an Abelian normal subgroup of a finite subgrouup of ${\rm GL}(n,\mathbb{C})$. For $n > 72$ or so, the bound is $(n+1)!$, which is attained by $S_{n+1}$in its irreducible $n$-dimensional representation.
(Later edit: Perhaps I should have made clear that passing between complex representations of finite groups and orthogonal real representations is straightforward, as has already been touched upon in comments: a complex irreducible representation of degree $n$ of a finite group $G$ is equivalent to a unitary representation. A unitary irreducible complex representation of degree $n$ of $G$ may be replced by a real orthogonal representation of degree $n$ in standard fashion, replacing the complex entry $a+bi$ by the real $2 \times 2$ matrix $\left(\begin{array}{clcr} a& b \\-b &a \end{array}\right)$. If the character of the original representation was not real, the resulting representation is irreducible as a real representation. If the original character was real, the resulting representation is equivalent to twice an irreducible if the original character had Frobenius-Schur indicator $+1$, and is irreducible as a real representation if the Frobenius-Schur indicator was $-1$).
(Later remark: In general, for finite subgroups of ${\rm GL}(n,\mathbb{C})$, we can take scalar multiples of the elements to ensure that all elements have determinant $1$ without affecting the structure of $G/Z(G)$. This trick may not be available for real representations.)<|endoftext|>
TITLE: "Most Similar Vector Problem" on an Integer Lattice?
QUESTION [9 upvotes]: I am currently working on problem that I think could be expressed as an integer lattice problem.
Given $u \in \mathbb{R}^n$ and a bounded integer lattice $L = \mathbb{Z}^n \cap [-M,M]^n$ I would like to find an integer vector $v \in L$ that minimizes the angle between $u$ and $v$. That is, I would like $$v \in \text{argmax}_{w \in L} \frac{u.w}{\|u\|\|w\|}$$
Here, the objective is maximizing the cosine of the angle between $u$ and $w$ (i.e. minimizing the angle between $u$ and $w$). The vectors $u$ and $w$ are said to be "similar" if this quantity is close to 1.
I am wondering:

Is this problem related to a well-known integer lattice problem (e.g. a closest vector problem)?
Could this problem be solved using existing lattice algorithms (e.g. the LLL algorithm?)

REPLY [6 votes]: Writing up the comment: You just need to "pixelate" the line by finding all lattice boxes that it crosses:
Then the answer vector $v$ must connect to one of the corners of the shaded boxes. Instead of first collecting the boxes, try the following zig-zag algorithm: start from the origin and poll the up and right nearest lattice points. One of them is closer to your vector $u$, so pick that one. Again, looking up and right, pick the closer vertex. Repeating this procedure will give you a list of distances for each point reached. Pick the minimum distance and that's your answer. This algorithm is $O(Mn)$, and perhaps can be sped up by someone more clever than me.<|endoftext|>
TITLE: What does it mean that the Hessian is proportional to the metric?
QUESTION [7 upvotes]: Let $(M,g)$ be a smooth manifold equipped with a metric tensor $g$, and $f\in C^\infty(M)$ a regular function (i.e., with nowhere vanishing differential).
Denote by $\mathrm{Hess}_g(f):=\nabla df$ the Hessian tensor of $f$ with respect to the metric $g$, and by $N_f:=f^{-1}(\{0\})$ the one-codimensional submanifold of $M$ determined by $f$.

QUESTION: how to characterise the functions $f$, with $\mathrm{Hess}_g(f)$ proportional to $g$? Do they belong to some well-known class? What about the submanifolds $N\subset M$, which can be written as $N=N_f$, for such an $f$?

By "proportional" I mean w.r.t. a conformal factor, and by "characterisation" I mean "what makes them special". The question may be restricted to the case when $g$ is flat.
Motivating example. Let $M=\mathbb{R}^2$ and $g$ the Euclidean metric. Then,
$$\mathrm{Hess}_g(f)=f_{xx}(dx)^2+f_{xy}dxdy+f_{yy}(dy)^2$$ 
is proportional to $g=(dx)^2+(dy)^2$ if and only if $f_{xy}=0$ and $f_{xx}=f_{yy}$, which means
$$
f=a(x^2+y^2)+bx+cy+d\, ,
$$
i.e., $f$ must be proportional to the square of the norm induced by $g$, plus some lower-order terms. I was wondering whether a similar result holds in general (of course, assuming $g$ to be flat).

REPLY [8 votes]: It is known (say, Y. Tashiro, Complete Riemannian manifolds and some vector fields, Trans.Amer.Math.Soc. 117(1965) 251–
275; I am not sure  that Tashiro is the first who proved it and there were many later papers which independently prove the same result later. The proof is pretty straightforward.)  that the existence of such a function for  a metric implies that the metric is a warped product metric, 
$$
g= dx_1^2 + \sigma(x_1) \sum_{i,j=2}^{n} h_{ij}(x_2,...,x_n)dx_i dx_j,    \ \ \ (*)
$$ 
and for such metrics a solution $f$ of your  equation    $\nabla\nabla f= \lambda(x) g$ is some  function of $\sigma$. 
The special case when the metric $g$ is flat is much more easy. In this case  the function $\sigma(x_1)$  from $(*)$
is automatically $\textrm{const_1} \ (x_1+ \textrm{const}_2)^2$ (this fact is also classically known), so in the nontrivial case, when $\textrm{const_1}\ne 0$, after the affine reparameterization  of $x_1$,   the form  $(*)$    is the cone form  $$ 
g= dx_1^2 + x_1^2 \sum_{i,j=2}^{n} h_{ij}(x_2,...,x_n)dx_i dx_j,   
$$ 
This implies that in the flat case  the function $\sigma(x_1)$, in  some euclidean  coordinate system $(y_1,...,y_n)$, is  (after addition of linear terms and constant which change nothing) simply the function 
$$f= y_1^2 +...+ y^2_n.$$ 
For flat metrics of other signatures the answer is essentially the same,  in this case $f= \varepsilon_1y_1^2 +...+ \varepsilon_ny^2_n,$ where
 $\varepsilon_i\in \{\pm 1\}$ are responcible for the signature<|endoftext|>
TITLE: Asymptotics of a recurrence relation
QUESTION [16 upvotes]: The sequence $(a_n)_{n \ge 0}$ satisfies, $a_0 = a_1 = 1$ and the recursion relation:
$$a_n = \sum\limits_{k=0}^{[n/2]} \frac{a_k}{(n-2k)!}$$
where, $[x]$ is the nearest integer to $x$ not exceeding it.
Alternatively define $a_n$'s as:
$$\sum\limits_{n=1}^{\infty} a_nx^n = \exp\left(\sum\limits_{n=0}^{\infty} x^{2^n}\right)$$
We need to show that: $$\liminf_{n \to \infty} \frac{\log a_n}{\log n} \le \frac{1}{\ln 2} - 1 \le \limsup_{n \to \infty} \frac{\log a_n}{\log n}$$
How do we investigate the asymptotics of this type of recursion relation?

REPLY [3 votes]: Here is a sketch of the standard approach to a problem of this type; its expansion will probably of Lucia's solution.
If in Jeff Shallit's response you introduce the related monotone increasing sequence $A_n=\sum_{k=0}^na_n$ then the difference relations can be used to show that $$C^kA_{\max\{0,[n/2^k]-5\}}\le A_n\le C^kA_{[n/2^k]+1}$$ for any $k$ with the choice $C=1+1/2+1/24+1+1/6+1/120=163/60$. An inductive argument then shows that $C_1n^\xi\le A_n\le C_2n^\xi$ with $\xi=1/\log(2)$ and some positive $C_1,C_2$ and the final passage from $A_n$ to $a_n$ can be performed with the help of the Stolz–Cesàro theorem.<|endoftext|>
TITLE: almost diagonal Positive semidefinite Matrix
QUESTION [5 upvotes]: Consider the set $\mathcal{D}_n$ of $n$-dimensional positive semidefinite matrices.
A matrix $M\in \mathcal{D}_n$ is called $\epsilon$ diagonal in trace distance if there is a diagonal matrix $D\in \mathcal{D}_n$ such that
$$|M-D|_{tr}\leq \epsilon.$$
For any $M\in \mathcal{D}_n$ , we can define $D(M)$ to be its diagonal part --- set all off diagonal term to 0 and leave the diagonal elements.  Let $\Lambda(M)$ denote the diagonal matrix whose diagonal entries are the sorted eigenvalues of M.
Now suppose $M\in \mathcal{D}_n$ satisfies 
$$|D(M)-\Lambda(M)|_{tr}\leq \epsilon.$$
Can we conclude $M$ is $2\epsilon$ diagonal?

REPLY [5 votes]: Half a year ago, John Wright and I were considering almost the same question, in connection with the quantum tomography problem; we even asked a few people, including Suvrit and fedja.  The only very slight difference is we were hoping to show that $M$ is close to its own diagonal, rather than to any diagonal matrix.  (Note that since we have the hypothesis that $M$'s diagonal is close to its spectrum, our question is equivalent to asking if $M$ is close to the diagonal matrix made from its spectrum.)
Now if you insist on this, that $M$ be close to its own diagonal, then I think the answer to your question would be "no".  I'm not 100% certain, but here's my reasoning. Let $A$ be any diagonal matrix of nonnegative reals (sorted, say) and let $H$ be any Hermitian matrix.  Now suppose $M = \exp(-i t H) A \exp(i t H)$, so $M$ is PSD with spectrum given by $A$'s diagonal.  Think of $t$ as a positive real tending to $0$. Now let's consider $\Delta := D(M) - A = D(M-A)$.  By Taylor expansion, 
\begin{align*}
\Delta &= D(A + i t (AH - HA) - \tfrac{t^2}{2}(AH - 2H^2 + HA) + O(t^3) - A) \\
 &= \tfrac{t^2}{2} D(AH - 2H^2 + HA) + O(t^3),
\end{align*}
where we used that $AH - HA$ has $0$ diagonal.  Thus, thinking of $t \to 0$, we will have $|D(M) - A|_{\mathrm{tr}}$ proportional to $t^2$ (or smaller).  
That means you're hoping that $M$ is $\Theta(t^2)$-close to diagonal.  Now if, like us, you are further hoping $M$ is $\Theta(t^2)$-close to its own diagonal then you're out of luck: it's equivalent to showing that $|M - A|_{\mathrm{tr}} = \Theta(t^2)$, but as we saw, $M - A = it(AH - HA) + O(t^2)$, and hence $|M - A|_{\mathrm{tr}} = \Theta(t)$ (unless $AH- HA = 0$, but this need not be the case).
If you restrict $M$ to have trace 1, then we felt the desired result was true but with a weaker conclusion of $O(\sqrt{\epsilon})$ rather than $2\epsilon$. 
--
In case this is related to our upcoming tomography papers, I'd be happy to continue the discussion over email :)
Best,
Ryan<|endoftext|>
TITLE: Do cotangent bundles have "bounded geometry"?
QUESTION [6 upvotes]: I have often heard the phrase "a manifold $M$ has bounded geometry" thrown around without ever seeing a precise definition of what this means. Apparent examples are compact manifolds and $\mathbb{R}^n$. What is the definition and can I therefore assume that cotangent bundles of compact manifolds have bounded geometry?

REPLY [3 votes]: Alexander Grigor'yan ("Heat Kernel and Analysis on Manifolds", 2009) gives the following definition:

a Riemannian manifold $(M, g)$ (of $\dim M = n$) has bounded geometry if and only if there exist the "bounds" $0<c<C$ and the radius $\varepsilon > 0$ such that for every $x \in M$ the geodesic ball $B(x, \varepsilon)$ is diffeomorphic to the standard Euclidean ball $B(0, \varepsilon) \subset \Bbb R^n$ under the diffeomorphism $\varphi _x : B(0, \varepsilon) \to B(x, \varepsilon)$ and $c g_0 \le \varphi _x ^* g \le C g_0$ (where $g_0$ is the usual Riemannian structure on $\Bbb R^n$).

(The definition is assembled from pieces found at pages 93 and 312.)
Notice that in general $T^*M$ has no natural Riemannian structure, therefore asking whether it has bounded geometry is meaningless, regardless of whether $M$ is compact or not.<|endoftext|>
TITLE: Why do the model structures on dg-algebras and on dg-categories are not compatible?
QUESTION [10 upvotes]: First we talk about dg-algebras. According to this n-lab page, we write $dgAlg$ for the category of cochain dg-algebras in non-negative degree over a field $k$ of characteristic $0$. Write $CdgAlg\subset dgAlg$ for the subcategory of (graded-)commutative dg-algebras. Then The projective model category structure on $CdgAlg$ and on $dgAlg$ is given by setting:

weak equivalences are the quasi-isomorphisms
fibrations are the degreewise surjections.

Then we look at dg-categories. According to this n-lab page, we write $dgCat$ for the category of small dg-categories over $k$. Then the Dwyer-Kan model structure on $dgCat$ is given by setting:

a dg-functor $F:A\to B$ is a weak equivalence if $(1)$ for all objects $x,y\in A$ the component $F_{x,y}:A(x,y)\to B(F(x),F(y))$ is a quasi-isomorphism of chain complexes $(2)$ the induced functor on homotopy categories $H^0(F)$ (obtained by taking degree 0 chain homology in each hom-object) is an equivalence of categories.
a dg-functor $F:A\to B$ is a fibration if $(1)$ for all objects $x,y\in A$ the component $F_{x,y}$ is a degreewise surjection of chain complexes; $(2)$ for each isomorphism $F(x)\to Z$ in $H^0(B)$ there is a lift to an isomorphism in $H^0(A)$.

Now we compare this two definitions. A dg-algebra can be considered as a dg-category which consists of one object hence we have a inclusion $dgAlg\subset dgCat$. However, although very similar at the first glance, the two model structures are not compatible. In particular there are dg-maps $F: A\to B$ between dg-algebras which is a fibration in the projective model structure on $dgAlg$ but not a fibration in the Dwyer-Kan model structure on $dgCat$.
For example let $A$ be the de Rham algebra of the closed interval $[0,1]$ and $B$ be the de Rham algebra of the two end points ${0,1}$ (hence $B$ is concentrated in degree $0$). Let $F: A\to B$ be the restriction map. It is clear that $F$ is degreewise surjective hence $F$ is a fibration in $dgAlg$. However if we take $f\in B$ to be $f(0)=1$, $f(1)=-1$ then $f$ is invertible in $H^0(B)$ but $f$ cannot be a restriction of a closed element in $A^0$ hence $F: A\to B$ is not a fibration when considered as a morphism in $dgCat$.
$\textbf{My question}$ is: what is the reason and consequence of this incompatibility? Does it mean that we can construct a new model structure on $dgAlg$ by restricting the Dwyer-Kan model structure to $dgAlg$?

REPLY [7 votes]: I will try to give a example in topology and then in dg-world. 
Suppose that $G$ and $H$ are topological  groups. There is a fiber sequence 
$$Map_{\ast}(BG,BH)\rightarrow Map(BG,BH)\rightarrow BH$$
where $B$ is the classifying space. Here is a homotopy categorical interpretation of the previous fiber sequence.
$$Map^{\otimes}(G,H)\rightarrow Map_{top-cat}(\mathbf{G},\mathbf{H})\rightarrow Map_{Top-cat}(\ast, \mathbf{H})$$
where $Map^{\otimes}  $ is the derived mapping space in the model category of topological mono ids. $Map_{Top-cat}$ is the derived mapping space in the model category of small topological categories. The notation $\mathbf{H}$ is for the topological category with one object such that the endomorphism monoid is $H$.   
So what happens in the dg-world. Well, there is a fiber sequence 
$$Map_{dg-algebras}(A,B)\rightarrow Map_{dg-cat}(\mathbf{A},\mathbf{B})\rightarrow Map_{dg-cat}(\mathbf{k}, \mathbf{B})$$
The notation $\mathbf{A}$ is for the $dg_{k}$-category with one object such that the endomorphism monoid is $A$. $k$ is the ground commutative ring.
Now we see why $Map_{dg-algebras}(A,B)$ is different from
$Map_{dg-cat}(\mathbf{A},\mathbf{B})$.<|endoftext|>
TITLE: Lower bounding the probability that $\gcd(t,N)≤B$, for a random $t$ and fixed (large) $N$
QUESTION [21 upvotes]: $\newcommand{\Prb}[1]{\mathcal{P}_{#1}}$
I have the following number theory problem, related to Odlyzko's improvement on Shor’s factoring algorithm (see this cstheory.sx question for details).

Let $N$ be a (large) integer and $1≤B≤N$ a bound. If I chose the integer $1≤t≤N$ uniformly at random, can I lower bound the probability $\Prb B$ that $\gcd(t,N)≤B$ ? How should I chose the least possible $B$ such that $\Prb B$ is bounded away from 0 ? Or asymptotically close to 1 ?

Of course 
$$\begin{align}\Prb 1&=\frac{ϕ(N)}N>\frac{1}{e^γ\ln\ln N+\frac3{\ln\ln N}};& \Prb N&=1 \end{align}$$ 
where $ϕ$ is Euler’s totient function and $γ$ is the Euler-Mascheroni constant.
More generally, it is easy to see that 
$$
\Prb B=\frac{\sum_{d\vert N}^{d≤B}ϕ(\frac{N}{d})}N
 =1-\frac{\sum_{d'\vert N}^{d'<\frac{N}{B}}ϕ(d')}N,
$$
but I did not manage to go further.
If I interpret the 1995 “private communication” from Odlyzko to Shor (referred to in arXiv:quant-ph/9508027 and the cstheory.sx question) correctly, $∀ε>0$, $\Prb{(\log N)^{1+ε}}$ is bounded away from 0. However, I did not manage to find a proof of this.

REPLY [13 votes]: In fact one can prove a stronger result -- namely the probability is bounded away from zero, so long as $B \ge (\log N)^\delta$ for some $\delta >0$.  This is best possible, by taking $N$ to be the product of the first few primes.  Since $\phi(N/d) \ge \phi(N)/d$, our probability is bounded below by 
$$ 
\frac{\phi(N)}{N} \sum_{d|N, d\le B} \frac{1}{d}\ge \frac{\phi(N)}{N}\sum_{d|N, d\le B} \frac{\mu(d)^2}{d}, 
$$ 
where in the last sum we have restricted attention to square-free $d$ for simplicity. 
Now I claim that for large enough $B$ (and any $N$, independent of $B$) one has 
$$ 
\sum_{d|N, d\le B} \frac{\mu(d)^2}{d} \ge \frac{1}{4} \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big). \tag{1} 
$$ 
Assuming the claim, we get a lower bound for our probability of 
$$ 
\ge \frac 14 \frac{\phi(N)}{N} \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big) 
\ge \frac{1}{4} \prod_{p} \Big(1-\frac 1{p^2}\Big) \prod_{p|N, p>\sqrt{B}} \Big(1-\frac 1p\Big)= \frac{3}{2\pi^2} \prod_{p|N, p>\sqrt{B}} \Big(1-\frac{1}{p}\Big).  
$$
If now $B\ge (\log N)^{\delta}$ then 
$$ 
\prod_{p|N, p>\sqrt{B}} \Big(1-\frac 1p\Big) \ge \prod_{(\log N)^{\delta/2}<p <(\log N)^2}\Big(1-\frac 1p\Big) \prod_{p>(\log N)^2, p|N} \Big(1-\frac 1p\Big),
$$ 
and by Mertens's theorem the first factor above is $\gg \delta$, and trivially the second factor is $1+o(1)$.  This completes the proof. 
It remains to settle the claim (1).  With $\alpha =1/\log B$ note that 
$$ 
\sum_{d|N, d\le B} \frac{\mu(d)^2}{d} \ge \sum_{\substack{d|N, d\le B \\ p|d \implies p\le \sqrt{B}}} \frac{\mu(d)^2}{d} \ge \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big) - B^{-\alpha} \sum_{\substack{d|N\\ p|d\implies p\le \sqrt{B} }} \frac{\mu(d)^2 d^{\alpha}}{d}.
$$
The second term above is 
$$ 
e^{-1} \prod_{p|N, p\le \sqrt{B}} \Big(1 + \frac{p^{\alpha}}{p}\Big) \le e^{-1} \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big) \exp\Big(\sum_{p|N, p\le \sqrt{B}} \frac{p^{\alpha}-1}{p}\Big).
$$
Now for large enough $B$, (since $(e^{t}-1)/t \le (\sqrt{e}-1)/(1/2)$ for $0\le t\le 1/2$)
$$ 
\sum_{p\le \sqrt{B}} \frac{p^{1/\log B}-1}{p} \le\sum_{p\le \sqrt{B}} \frac{\log p}{p\log B} \Big(\frac{\sqrt{e}-1}{1/2}\Big) = \sqrt{e}-1 +o(1),
$$
and using this above, we obtain 
$$ 
\sum_{d|N, d\le B} \frac{\mu(d)^2}{d} \ge \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big) \Big(1- e^{-1} (e^{\sqrt{e}-1}+o(1)) \Big) \ge \frac 14 \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big), 
$$
proving the claim (1).<|endoftext|>
TITLE: Existence and characterization of transitive matrices?
QUESTION [8 upvotes]: We call a matrix $M \in \mathbb{R}^{d \times d}$ transitive if it satisfies the following: 

For any three vectors $u, v, w$ in $\mathbb{R}^d$. If $u^T M v > 0$ and $v^T M w > 0$ then $u^TMw > 0$. 

EDIT: The case when $d = 1$ is simple. For $d > 1$ the following is true: 
a) M must be Positive Semi Definite, because if $M$ is transitive and $\exists u$ s.t. $u^TMu < 0$ then we can let $v = -u, w = u$ and arrive at the contradiction that $u^t M u < 0 \implies u^tMv > 0, v^tMw > 0 \implies u^TMw > 0 \implies u^TMu > 0$
b) $M \ne I$ because if $M = I$ then we just need to construct three vectors such that $u^Tv > 0, v^Tw > 0$ but $u^Tw < 0$. Intuitively, I know that sum of two acute angles can be obtuse.
c) More generally, $M$ can not be full rank positive definite because then the factors in the Cholesky decomposition $LL^T$ of $M$ would also be full rank and the problem could be reduced to case b) by change of basis using $L$.
Now I am stuck in the space of low rank positive semidefinite matrices. Is there a way to characterize such matrices further? Can such a matrix even exist, apart from zero matrix?

EDIT2: The deduction that $M$ is PSD in step a) implicitly assumed that $M$ is symmetric. Keith Kearnes' answer explicitly shows that $M$ must be rank 1 in that case. The case when $M$ is anti-symmetric is open but I doubt it would be possible to prove anything in that case. I will wait a few days to get more answers. Thanks.
EDIT3: Darij Grinberg's answer proved that every transitive matrix must be symmetric. This completed the characterization.

REPLY [5 votes]: Here is a proof of the fact that any transitive matrix $M\in\mathbb{R}
^{d\times d}$ is symmetric. Together with the argument in the answer by Keith
Kearnes, this proves that any transitive matrix $M\in\mathbb{R}^{d\times d}$
is a positive-semidefinite symmetric matrix of rank $\leq1$.
We first prove a lemma: If $p\in\mathbb{R}^{d}\setminus\left\{  0\right\}  $
and $q\in\mathbb{R}^{d}$ are two vectors such that every $u\in\mathbb{R}^{d}$
satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$, then
(1) there exists a nonnegative real $\lambda$ such that $q=\lambda p$.
Proof of (1). Let $p\in\mathbb{R}^{d}\setminus\left\{  0\right\}  $ and
$q\in\mathbb{R}^{d}$ be two vectors such that every $u\in\mathbb{R}^{d}$
satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$. We have $p\neq0$ and thus
$p^{T}p>0$.
If $p$ and $q$ were linearly independent, then there would be a vector
$v\in\mathbb{R}^{d}$ satisfying $p^{T}v=1$ and $q^{T}v=-1$; but this would
contradict the assumption that every $u\in\mathbb{R}^{d}$ satisfying
$p^{T}u>0$ satisfies $q^{T}u\geq0$. Hence, $p$ and $q$ must be linearly
dependent. Since $p\neq0$, this shows that there exists a $\lambda
\in\mathbb{R}$ such that $q=\lambda p$. It remains to prove that this
$\lambda$ is nonnegative. Indeed, recall that every $u\in\mathbb{R}^{d}$
satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$. Applying this to $u=p$, we get
$q^{T}p\geq0$ (since $p^{T}p>0$). Since $q=\lambda p$, this rewrites as
$\lambda p^{T}p\geq0$, and thus $\lambda\geq0$ (since $p^{T}p>0$). This
finishes the proof of (1).
Another lemma, which is really well-known: If $V$ is a finite-dimensional
vector space, and if $\phi$ is a linear endomorphism of $V$ such that $\phi$
sends every vector in $V$ to a scalar multiple of this vector, then
(2) the endomorphism $\phi$ is a scalar multiple of the identity.
(In order to prove (2), fix a basis of $V$ and see what $\phi$ does to the
basis vectors and their pairwise sums.)
Now, let $M\in\mathbb{R}^{d\times d}$ be any transitive matrix. We need to
prove that $M$ is symmetric.
If $M=0$, then this is obvious. Thus, WLOG assume that $M\neq0$.
Let $w\in\mathbb{R}^{d}$ be any vector such that $M^{T}w\neq0$. Then, $w\neq0$
(since $M^{T}w\neq0$).
Let $u\in\mathbb{R}^{d}$ be any vector such that $\left(  M^{T}w\right)
^{T}u>0$. Let us now show that $w^{T}M\left(  M^{T}u\right)  \geq0$. Indeed,
if $M^{T}u=0$, then this is clear; otherwise it follows from the transitivity
of $M$ (in fact, from $w^{T}Mu=\left(  M^{T}w\right)  ^{T}u>0$ and
$u^{T}M\left(  M^{T}u\right)  =\left(  M^{T}u\right)  ^{T}\left(
M^{T}u\right)  >0$ (since $M^{T}u\neq0$), we obtain $w^{T}M\left(
M^{T}u\right)  >0$ (since $M$ is transitive)). Thus, we have proven that
$w^{T}M\left(  M^{T}u\right)  \geq0$. Hence, $\left(  MM^{T}w\right)
^{T}u=w^{T}MM^{T}u=w^{T}M\left(  M^{T}u\right)  \geq0$.
Let us now forget that we fixed $u$. We thus have shown that every
$u\in\mathbb{R}^{d}$ satisfying $\left(  M^{T}w\right)  ^{T}u>0$ satisfies
$\left(  MM^{T}w\right)  ^{T}u\geq0$. Thus, (1) (applied to $p=M^{T}w$ and
$q=MM^{T}w$) yields that
(3) there exists a nonnegative real $\lambda$ such that $MM^{T}w=\lambda
M^{T}w$.
Now, let us forget that we fixed $w$. We thus have proven that for every
$w\in\mathbb{R}^{d}$ satisfying $M^{T}w\neq0$, we have (3). But (3)
also holds for every $w\in\mathbb{R}^{d}$ satisfying $M^{T}w=0$ (since we can
use $\lambda=0$). Hence, (3) holds for every $w\in\mathbb{R}^{d}$.
Let $V=M^{T}\left(  \mathbb{R}^{d}\right)  $ be the image of $M^T$. Then,
$M\left(  V\right)  \subseteq V$ (because (3) holds for every
$w\in\mathbb{R}^{d}$). Thus, $M$ restricts to an $\mathbb{R}$-linear
endomorphism $\phi$ of $V$. This endomorphism $\phi$ sends every vector in $M$
to a scalar multiple of this vector (because (3) holds for every
$w\in\mathbb{R}^{d}$). Thus, $\phi$ must be a scalar multiple of the identity
(due to (2)). In other words, there exists some $\mu\in\mathbb{R}$ such
that every $v\in V$ satisfies $\phi\left(  v\right)  =\mu v$. In other words,
there exists some $\mu\in\mathbb{R}$ such that every $w\in\mathbb{R}^{d}$
satisfies $MM^{T}v=\mu M^{T}v$ (because of what $V$ is and what $\phi$ is). In
other words, there exists some $\mu\in\mathbb{R}$ such that $MM^{T}=\mu M^{T}
$. Consider this $\mu$.
We are working over $\mathbb{R}$. Hence, a well-known fact says that
$\operatorname*{Ker}\left(  MM^{T}\right)  =\operatorname*{Ker}\left(
M^{T}\right)  $. Thus, from $M^{T}\neq0$, we obtain $MM^{T}\neq0$, so that
$\mu M^{T}=MM^{T}\neq0$ and therefore $\mu\neq0$. Thus, we can transform
$MM^{T}=\mu M^{T}$ into $M^{T}=\dfrac{1}{\mu}MM^{T}$. The matrix $M^{T}$ is
thus symmetric (since $MM^{T}$ is symmetric). In other words, the matrix $M$
is symmetric.<|endoftext|>
TITLE: Why is it so hard to prove Toeplitz' conjecture?
QUESTION [23 upvotes]: I'm a layman in mathematics, so please excuse me in advance for anything in this question that may be inappropriate :D. Well: Four years ago, I was reading (and working to solve the puzzles on) Winkler's Mathematical Puzzles: A Connoisseur's Collection, and in the section "Unsolved Puzzles", there was the problem "Squaring the Lake" at p. 143: "Prove that every simple closed curve in the plane contains four points forming the vertices of a square." Obviously, I was not able to prove it (the max I could get, after much effort, was to prove that it was true for triangles!!!), so I went to "Comments and Sources" at p. 148, where he mentioned a web-site, a journal paper and a book on the subject, and he also made the comment that "It's a little embarrassing that mathematicians cannot [...]" (bolds is mine. "[...]" is equivalently to "solve the puzzle", I don't want to make a full quote because if I quote too much I may infringe some copyright...).
Well, much time later I came across http://www.ams.org/journals/notices/201404/rnoti-p346.pdf and according to https://en.wikipedia.org/wiki/Inscribed_square_problem, the problem is still open. So, I ask, why this (innocent-looking :D) problem is actually so hard? (I probably won't understand why, but you guys don't need to give an "answer to layman", it can be a technical answer!)

REPLY [25 votes]: Let me elaborate on Sam Hopkins' comment.
The main reason that makes this and other problems on continuous curves so hard is that a "simple closed curve" or "Jordan curve", i.e. a non-self-intersecting continuous loop in the plane, can be a horrible object, for instance a nowhere differentiable curve such as the Koch snowflake and other fractal curves. There are also Jordan curves of positive area (first constructed by Osgood in 1903).
In fact, as it is also explained in the Wikipedia article that you linked, the problem is actually solved for "well-behaved" curves, such as convex curves or piecewise analytic curves, i.e. for objects that are close to our intuitive notion of "continuous closed loop".
A possible strategy to solve the problem in the general case is to try to  approximate your Jordan curve by using well-behaved curves, for which we know that the conjecture is true, and then pass to the limit. The technical difficulty with this approach is that a limit of squares is not necessarily a square, but it can be a single point (i.e., a square "of side length 0"), so it is not clear whether it is always possible to find an approximating sequences of curves that works.<|endoftext|>
TITLE: Integral formula for $\int_{0}^{\infty}e^{-3\pi x^{2}}((\sinh \pi x)/(\sinh 3\pi x))\,dx$ by Ramanujan
QUESTION [11 upvotes]: The following is a re-post from MSE because I did not get any answer even after offering a bounty.
Towards the end of G. N. Watson's (one of the joint authors of famous book "A Course of Modern Analysis") paper "The Final Problem: An Account of the Mock Theta Functions" the following formula of Ramanujan is mentioned:
\begin{align}&\int_{0}^{\infty}e^{-3\pi x^{2}}\frac{\sinh \pi x}{\sinh 3\pi x}\,dx\notag\\
&\,\,\,\,\,\,\,\,= \frac{1}{e^{2\pi/3}\sqrt{3}}\sum_{n = 0}^{\infty}\frac{e^{-2n(n + 1)\pi}}{(1 + e^{-\pi})^{2}(1 + e^{-3\pi})^{2}\dots(1 + e^{-(2n + 1)\pi})^{2}}\tag{1}
\end{align} where the term corresponding to $n = 0$ in the sum on the right is $1$.
Is there way to establish this exotic integral formula? Or a reference to any existing proof of $(1)$ would be of great help.

REPLY [21 votes]: This relation seems to be part of the story of Ramanujan's mock theta function.  The  identity you discuss is a specialization of transformation properties of a mock theta function evaluated at $\tau =i$.  See this paper of Zwegers for proofs of the general transformation identity (and he gives other references to earlier work of Watson):  in particular, look at section 3 and Lemma 3.2 of Zwegers's paper.  Other surveys that you may find of interest are Ono and Duke.<|endoftext|>
TITLE: soft: Reference/ Suggested Read: Homological Algebraic techniques in PDEs
QUESTION [5 upvotes]: I was reading this article on wikipiedia and was interested by the apparent link between Homological Algebra and PDEs.  What is an accessible reference which showcases the link between these topics?

REPLY [14 votes]: First, a critique. Both homological algebra and PDEs are far from being monolithic subjects. So, a well-posed question about the relation between the two should be quite specific. Though you don't explicitly say it, your question suggests that you are interested in the homological algebra methods that appeared in the work of Spencer and others influenced by him. So, I'll restrict my answer to that.
The early ideas appeared already in the work of Kodaira and Spencer on the deformation of complex structures. The equations describing a complex structure are non-linear over-determined PDEs. One way to study the deformations of a given complex structure is to linearize these equations and then try to complete a linear solution to a 1-parameter family of non-linear (true) deformations. As Kodaira and Spencer discovered, not all linear solutions can be continued to non-linear deformations. A problem can appear as follows. Solving the defining equations of a complex structure perturbatively, at the quadratic order one needs to solve a linear inhomogeneous equation with a source quadratic in the linear order solution. If this source lies outside the range of the linear operator in the quadratic order equation, there is obviously no solution and the linear order solution cannot be continued to a non-linear one (not even to second order). In other segments of the PDE literature, this phenomenon is called linearization instability. So, some obstructions to continuing the linear order deformation of a complex structure to a $1$-parameter family of non-linear deformations lies in the discrepancy between the range of the linear differential operator appearing at the quadratic perturbative order and the non-linear map that is used to construct its source term from the linear order solution. Kodaira and Spencer figured out that this discrepancy is absent if some cohomology of the compatibility complex of the linearized defining equations of a complex structure vanishes. The same cohomology could also be described as the cohomology of the sheaf of complex vector fields on the underlying complex manifold. All of this is described in Kodaira's book:

Kodaira, Kunihiko, Complex manifolds and deformation of complex structures (Springer, 1986).

Later Spencer encouraged people to further apply ideas from homological algebra to the theory of over-determined PDEs. The basic problem, given a PDE system, is to get a complete list of its integrability conditions (equations, other than the ones already given, of the same or lower differential order that automatically hold for any solution). Integrability conditions are usually identified by taking derivatives of the given PDE and trying to eliminate as many of the highest derivative terms as possible. Once all integrability conditions have been identified the system is said to have been completed to involution. It has been known since the early 20th century that completion to involution can be established by an algorithm in a definite number of steps. However, the existing methods required either the use of an explicit coordinate system (Janet, Riquier) or the conversion of the PDE system into an equivalent first order Pfaffian system (Cartan, Kuranishi). Making liberal use of some homological complexes constructed from the symbol of a given PDE system (Spencer cohomology) a geometric theory of the completion of over-determined PDEs to involution was then created (Spencer, Bott, Quillen, Goldschmidt). Seminal early papers on this subject include

Goldschmidt, H. Existence theorems for analytic linear partial differential equations. The Annals of Mathematics 86, 246-270 (1967).
Goldschmidt, H. Integrability criteria for systems of nonlinear partial differential equations. Journal of Differential Geometry 1, 269-307 (1967).
Spencer, D. C. Overdetermined systems of linear partial differential equations. Bulletin of the American Mathematical Society 75, 179-240 (1969).

A modern comprehensive treatment of this theory, invoking even stronger ties to homological algebra can be found in

Seiler, W. M. Involution: The Formal Theory of Differential Equations and its Applications in Computer Algebra (Springer, 2010).<|endoftext|>
TITLE: Does the ring generated by the odd power sum symmetric functions have a name?
QUESTION [5 upvotes]: Let $\Lambda$ be the ring of symmetric functions and recall the power sum symmetric function $p_i = \sum x_1^i + x_2^i + \dots$ generate this ring. Let $\tilde\Lambda$ be the ring generated by the odd power sums. Then $\tilde\Lambda$ has several interesting properties: it satisfies the $Q$-cancellation property
$$
f(x_1,-x_1,x_2,x_3,\dots) = f(x_2,x_3,\dots),
$$
has the $Q$-Schur and $P$-Schur functions as orthogonal bases and is closely related to the theory of projective representations for the symmetric group. 
As best I can tell (after having asked several experts), $\tilde\Lambda$ lacks a simple name. I have seen/heard people refer to it as the ring of symmetric functions generated by the odd power sums or the ring of symmetric functions satisfying the $Q$-cancellation property, but given its prominence I would have expected a name like the __________ symmetric functions. Is anyone aware of such a name? If someone can propose an informative name, I would be happy to start referring to $\tilde\Lambda$ as such in my future work.

REPLY [2 votes]: Darij gave a very good reference indeed. One other instance where I saw this before is the Maple package SF, where they talk about "signed class functions" (http://www.math.lsa.umich.edu/~jrs/software/QFhelp.html#char2qf), this might be also useful in choosing the name, if you are so inclined.<|endoftext|>
TITLE: Are there effective small intervals in which primes are dense?
QUESTION [7 upvotes]: As mentioned in Terry Tao's comment to this question, it is constructively known

that there are primes between sufficiently large cubes. $\:$ According to wikipedia,

"there exists a constant $\: \theta < 1 \:$ such that $\;\;\; \pi \hspace{-0.03 in}\left(x\hspace{-0.04 in}+\hspace{-0.04 in}x^{\hspace{.02 in}\theta}\right)-\pi \hspace{.03 in}(x) \: \sim \: \dfrac{x^{\hspace{.02 in}\theta}}{\log(x)} \;\;\;$ as $x$ tends to infinity".
Is there any somewhat-similar corresponding effective result for the density of primes in short intervals?
Motivation:
Such a result could yield an effective randomized reduction from subset sum to this variant of factoring.

REPLY [7 votes]: One can read off such an effective result from the work of Dudek. Indeed, examining his Section 3.3, we can get the following explicit lower bound for $x>\exp(8\times 10^{14})$ and $h=3x^{2/3}$:
$$ \sum_{x<p\leq x+h}\log p>h-\tfrac{1}{2}(1-10^{-3})h-\tfrac{1}{2}(1-10^{-3})h-0.0008h. $$
(Note that Dudek's definition of $E(x,h,k)$ should be multiplied by $-1$.)
That is,
$$ \sum_{x<p\leq x+3x^{2/3}}\log p>0.0006x^{2/3},\qquad x>\exp(8\times 10^{14}).$$
Under the Riemann Hypothesis, much better explicit bounds are known, see e.g. Theorem 1.1 in the recent preprint of Dudek-Grenié-Molteni.<|endoftext|>
TITLE: Deceptive linear algebra problem
QUESTION [6 upvotes]: Does anyone recognize this problem? There are $2p$ equations and $2p$ unknowns, and it feels like a classic, but I've never encountered it before:
Given $d_1, d_2, \ldots d_p$, find $a_1, a_2, \ldots a_p$ and $n_1, n_2, \ldots n_p$:
$$ \left[ \begin{array}{ccc}
d_1\\
d_2\\
d_3\\
\vdots\\
d_{2p}
\end{array} \right]
= \left[ \begin{array}{ccc}
a_1, ~a_2, \cdots ~a_p\\
a_1^2, ~a_2^2, \cdots ~a_p^2\\
a_1^3, ~a_2^3, \cdots ~a_p^3\\
\vdots\\
a_1^{2p}, ~a_2^{2p}, \cdots ~a_p^{2p}
\end{array} \right] \cdot
\left[ \begin{array}{c} 
n_1\\
n_2\\
\vdots\\
n_p
\end{array} \right] $$
(and where $a \succeq 0$ and $n \succeq 0$ and only real values are considered, so wherever there can be positive and negative roots for the values of $a_i$, only the positive roots need be considered). 
It seems straightforward to solve with elimination: From the first row, let $a_1 = \frac{d_1-\sum_{i>1} a_i n_i}{n_1}$. Then from the second row, solve for $a_2$ (which will now involve a quadratic, etc.). Because the polynomials get larger, solving it with elimination in this manner can be quite slow (I crashed sage doing this on a fairly small problem).
Has this family of problems been characterized well in the literature and can it be solved efficiently in practice? Even a numerical solution would suffice, as long as it will converge certainly (no multi-start Levenberg-Marquardt, etc.)...

REPLY [6 votes]: Actually, the problem precisely as you've stated it can't be solved: you can't write $(1,0,0,0)$ as a sum of multiples of two vectors $(a,a^2,a^3,a^4)$ and $(b,b^2,b^3,b^4)$, but this is easily fixed by starting with the zeroth power instead of the first. (Even more degenerately, $(1,0)$ isn't a multiple of $(a,a^2)$ for any $a$.)
Dongryul Kim's answer is basically correct, but let me expand a little bit on it: the span of the vectors $(1,a,\dots, a^{n-1})$ for $a\in A$ for some finite subset $A$ in any field has dimension equal to the size of $A$ if $\# A< n$, and spans the whole space otherwise (note how important it is that we start at the zeroth power to make this work).  If $\# A< n$, then the orthogonal to this space is spanned by vectors of the form $(0,\dots,0,e_{\# A},\dots, e_1,1,0\dots,0)$ where $\prod_{a\in A}(X-a)=X^{\# A}+e_1X^{\# A-1}+\cdots+e_{\# A}$ (these are obviously orthogonal to the vectors, and have the correct dimensional span).  Thus, if we want to write $(d_1,\dots, d_n)$ as a sum of these vectors, we have to find numbers $1,e_1,\dots,e_q$ such that $(d_1,\dots, d_n)$ is orthogonal to all of these vectors.  That is, we want to consider the equation 
\begin{equation} \begin{bmatrix} d_{q+1} & d_{q} & \cdots & d_1 \\ d_{q+2} & d_{q+1} & \cdots & d_2 \\ \vdots & \vdots & \ddots & \vdots \\ d_{n} & d_{n-1} & \cdots & d_{n-q} \end{bmatrix} \begin{bmatrix} 1 \\ e_1 \\ \vdots \\ e_q \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}. \end{equation}
For $q=1$, this will only have a non-trivial solution if all the $d_i$'s are 0.  Otherwise, there will be a unique smallest integer $q>1$ such that this equation has a non-trivial solution.  In that case, we have that $(d_1,\dots, d_n)$ is in the span of the vectors $(1,a,\dots, a^{n-1})$, for $a$ ranging over the roots of $X^{q}+e_1X^{q-1}+\cdots+e_q$, and one can easily solve for the $n_i$'s; by construction, we've seen that these roots are the unique minimal set such that $(d_1,\dots, d_n)$ is in the span.  By basic linear algebra, we must have  $q\leq \lceil n/2\rceil$, since at that point, the matrix has more columns than rows, and must have a non-trivial solution. Obviously, generically, we'll have equality there.<|endoftext|>
TITLE: Terminology in combinatorics
QUESTION [6 upvotes]: I met the following two combinatorial concepts during a study outside of combinatorics. I am wondering if there are common terminologies in combinatorics. 

A finite graph $G$ has the following property:


For any vertex $a$ and edge $\{b,c\}$ of $G$, there is an edge connecting them: there is at least one of $\{a,b\}$ or $\{a,c\}$ in $G$.


Let $P = \lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_k$ be an integer partition of $n$. Suppose that $P$ has the following property:


If there are two index sets $I$ and $J$ such that $\sum_{i \in I}\lambda_i = \sum_{j \in J}\lambda_j$, then the restricted partitions $\{\lambda_i\}_{i \in I}$ and $\{\lambda_j\}_{j \in J}$ are same. 

For instance, $5 \ge 3 \ge 2 \ge 2$ does not have the property because $2+3=5$, but $5 \ge 5 \ge 3 \ge 3$ has the property.
Are there common names for these two properties?

REPLY [2 votes]: The second property is almost exactly the concept of "distinct subset sums", about which there are many beautiful unanswered questions. See, eg, http://www.openproblemgarden.org/op/sets_with_distinct_subset_sums
The only difference between what you ask and the classical case is that you allow for repeated values.<|endoftext|>
TITLE: Random suborbits of a rotation
QUESTION [7 upvotes]: Let $u_n = x + n\alpha \pmod 1$ with $\alpha$ irrational. We know that $(u_n)_{n \geq 0}$ is dense in $\mathbb{R}/\mathbb{Z}$ (equivalently $(u_n)_{n \geq 0}$ visits every open interval infinitely many times).
Now, let $(\epsilon_n)_{n \geq 0}$ be a sequence of independent uniform Bernoulli random variables on $\{0,1\}$, and define the random integers $K_n=\sum_{i = 0}^n \epsilon_i2^i$. Is it true that the random sequence $(u_{K_n})_{n \geq 0}$ almost surely visits every open interval infinitely many times ? 
As a conditional and secondary question, if it is true, I would also like to know:

is it more generally true for random integers $K_n$ defined in the same way but using another representation of the integer numbers instead of the binary one ? 
is it more generally true for the orbits of a minimal homeomorphism ? (after replacing intervals with open sets)

REPLY [2 votes]: For your initial question I think I have an answer. WLOG assume $x=0$.
Assume that $\alpha$ has a universal binary expansion, i.e., the one which contains every finite 0-1 word as a factor. (This is a generic property in $\alpha$.) Then the sequence $2^n\alpha\bmod1$ is clearly dense in $(0,1)$. (Alternatively, you may use Weyl's criterion, as Vaughn suggested, but my condition is an explicit one.)
Now, we have $K_n=K_{n-1}+\epsilon_n2^n$, whence we have a map
$$
x\mapsto \begin{cases} x+2^n\alpha\bmod1,& 
\\ x,\end{cases}
$$
with equal probabilities. 
Fix $(a,b)\subset(0,1)$. Since $(\epsilon_n)$ is generic, for any $x=K_{n-1}\alpha\bmod1$ we will have $\epsilon_n=\epsilon_{n+1}=\dots=\epsilon_{n+k-1}=0, \epsilon_{n+k}=1$ with a positive probability, where $n+k$ is such that $x+2^{n+k}\bmod1\in(a,b)$. And this will happen infinitely many times by some general probability nonsense. ;) 
So, the claim appears to be true for any such $\alpha$. 
I think if a counterexample exists, one might want to look at a Liouville $\alpha$ with very sparse 1s in its binary expansion. 
EDIT. Come to think about it, there's no reason why we should stick to $2^n\alpha$. We have $K_{n+k}=K_n+2^nN$, where $0\le N\le 2^{k-1}$, and all of these have positive probability (where $k$ is chosen large enough to satisfy $b-a>>2^{-k}$). So, since the set $\{ N2^n\alpha\bmod1 \mid 0\le N\le 2^k-1\}$ intersects every sufficiently large interval, with probability 1 we will have $K_n\alpha\in (a,b)$. 
So, the answer is YES for all irrational $\alpha$.<|endoftext|>
TITLE: Proof of a Fourier pair with Bessel functions?
QUESTION [7 upvotes]: How can we prove that the Fourier transform of the function
$$
f(x)
=
\begin{cases}
(a^2-x^2)^{c/2}  BesselJ[c,b\sqrt{a^2-x^2}] & \text{for }x^2 < a^2\\
0 & \text{otherwise}
\end{cases}
$$
is 
$$
\hat f(y)
=
\sqrt{2\pi a} (a^c) (b^c) (b^2+y^2)^{-c/2-1/4} BesselJ[c+1/2,a\sqrt{b^2+y^2}]
?
$$
This Fourier transform pair is given in the book
Formeln und Satze fur die speziellen Funktionen der mathematischer Physik
(Julius Springer, Berlin, 1943) p. 119. http://www.pokyrek.cz/Nijboer/Magnus_Hettinger.pdf
Numerical computation suggests this is correct.
I need this formula for $c=1$.

REPLY [2 votes]: The proof is not very complicated, but even a sketch needs more space than a comment. So here is a sketch.
I want to prove that
$$
\int_{-a}^{a} dx \ e^{i y x} \  f(x) = \hat{f}(y)
$$
with $f$ and $\hat{f}$ defined as in the question.
First one observes that it suffices to prove the equality for $a=1$. Then because of $f(-x)=f(x)$ and the symmetric integration interval one only has to prove that
$$
2 \int_{0}^{1} dx \ cos(y x) \  f(x) = \hat{f}(y).
$$
Expanding the Bessel function under the integral (use e.g. http://dlmf.nist.gov/10.2.E2) and exchanging sum and integration leaves us with integrals of the form ($m$ is the summation index)
$$
\int_{0}^{1} d x \ cos (y \ x) \ (1-x^2)^{c + m}
$$
which can be calculated by the so called Poisson's integral formula (found e.g. here:  http://dlmf.nist.gov/10.9.E4) resulting in essentially another Bessel function, $J_{c+m+1/2}(y)$.
We are thus confronted with a sum over Bessel functions each with argument $y$ attached with some factors.
After a (trivial) change of sign of these Bessel functions' argument (use http://dlmf.nist.gov/10.11.E1) we can evaluate the sum using the Multiplication Theorem for Bessel functions (see http://dlmf.nist.gov/10.23.E1), which finishes the proof.<|endoftext|>
TITLE: Does "cardinal arithmetic is well-defined" imply axiom of choice?
QUESTION [15 upvotes]: Let me quickly explain what I mean with my question.
Let $(\kappa_i)_{i\in I}$ be a collection of cardinal numbers, indexed by elements of some set $I$. We can try to define $\sum\limits_{i\in I}\kappa_i$ as follows: take collection $(A_i)_{i\in I}$ of sets such that $|A_i|=\kappa_i$, and let $A=\{(a,i):i\in I,a\in A_i\}$. Then we say $\sum\limits_{i\in I}\kappa_i=|A|$. We can do similar trick with multiplication, by setting $B=\{f:f\text{ is a function from }I\text{ such that }f(i)\in A_i\}$.
If we assume AC, then these notions are well-defined: First of all, we can choose such a sequence $(A_i)_{i\in I}$, and if we take two collections $(A_i)_{i\in I}$, $(A_i')_{i\in I}$ such that $|A_i|=|A_i'|=\kappa_i$, then we can use choice to get bijections between $A_i$ and $A_i'$ for every $i\in I$ and then use there to create bijection between $A=\{(a,i):i\in I,a\in A_i\}$ and $A'=\{(a',i):i\in I,a'\in A_i'\}$, so this procedure defines only one cardinal. Similar trick for multiplication.
However, if we do not assume choice, then 1. we don't know a priori that we can always choose any sequence $(A_i)_{i\in I}$, and even if we can, we have no guarantee that there is only one possible size of $A$ we can get in this way. I don't know of scenario where we cannot find any sequence of sets with given cardinalities, but I know that uniqueness of size can fail quite spectacularly, even if we add $2$ countably many times (see here).
I have three questions:


Is it possible in abscence of choice that there is a collection of cardinals $(\kappa_i)_{i\in I}$, but there is no collection of sets $(A_i)_{i\in I}$ having respective cardinalities? (I suspect the answer is "yes")
If the answer to 1. is yes, then is it known that existence of such a collection implies axiom of choice? (I suspect answer to this to be "doesn't imply" or "not known")
Does existence of collections like above and "value of the sum and product of cardinals doesn't depend on choice of collection" imply axiom of choice? (no clue, but I'm hoping for "yes")


These are main questions I'm interested in, but we can ask more by asking which combinations of below three imply choice:

For any collection of cardinals, there is collection of sets of respective cardinalities.
  Sum of cardinals, if defined, is well-defined. (i.e. sum doesn't depend on choice of collection, but we allow possibility that the collection doesn't exist)
  Product of cardinals, if defined, is well-defined.

Thanks in advance for all feedback.

REPLY [17 votes]: First let me answer the first question. Yes. It is possible. You can find the proof in Jech "The Axiom of Choice" in Theorems 11.2 and 11.3. Also you might be interested in reading Pincus' paper,

David Pincus, Cardinal representatives, Israel J. Math. 18 (1974), 321--344.

Where he shows that it is also consistent that there are representatives to all cardinals uniformly, but the axiom of choice fails.
To the third question, which seems to mix two separate ideas, I don't know the answer, and I don't know if it was asked in explicit details. But even without any assumptions on cardinal representatives, just talking about disjoint unions (and products), let me give you the following answer.
If you allow talking about products, then trivially you get the axiom of choice back.
Suppose that $\{Y_i\mid i\in I\}$ is a family of non-empty sets. Pick a large enough non-empty set $Y$ such that each $Y_i$ satisfies $|Y\times Y_i|=|Y|$.1 Now consider $\prod_{i\in I} Y$ and $\prod_{i\in I}(Y\times Y_i)$, since the former is non-empty, the latter is non-empty as well. Now pick some $f$ such that $f(i)\in Y\times Y_i$ and consider the projection onto $Y_i$, as a choice function.
For addition it seems to be open. The last progress I am aware of can be found in Higasikawa's paper,

Masasi Higasikawa, Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434. 

Where you can find a proof that this assumption implies The Partition Principle (whose choice strength is quite open for quite some time now). You can find other proofs related to this principle, and that in the presence of some additional assumptions it will imply the axiom of choice.
Two remarks that might be useful, taken from Gregory Moore's "Zermelo's Axiom of Choice":

The assumption is in fact equivalent to the statement that $\sum_{i\in I}|A|=|A|\cdot|I|$.
If we also assume that "If $Y$ is a set and for every $y\in Y$, $|y|=|Y|$, then $|\bigcup Y|=|Y|$, then we can prove the axiom of choice.


Footnotes.

Such $Y$ always exists. For example $(\bigcup Y_i)^\omega$.<|endoftext|>
TITLE: How algebraic is the holonomy map?
QUESTION [8 upvotes]: Let $G$ be either a compact, simple, simply-connected Lie group, or a simply-connected complex reductive group (so either $SU(n)$ or $SL(n,\mathbb{C})$, for instance), or even complex affine. I don't know the correct setting for this question, so assume the case that makes it work. Please bear with the tentativeness of this question.
Consider the space $\mathcal{A}$ of polynomial or Laurent connections (again, take the option that works or makes sense) on the trivial $G$-bundle on $\mathbb{C}^\times$. This is something like an ind-scheme (perhaps only in the reductive case, and possibly only after picking a faithful representation). In the case of smooth connections on $G\times S^1$ (and probably for analytic too) we have a holonomy map $\mathcal{A}^{sm} \to G$, and this is a surjective submersion. I think there is also such a map for the case of $\mathcal{A}$, going around $S^1\subset \mathbb{C}^\times$. 

In the event this exists, how algebraic is this map? Is it surjective?

My thought is that this may be a cover in some generalised sense. Maps from ind-schemes to schemes aren't so common, though that I know how to find how they work or what they do. There are some notes by Gaitsgory that don't help me at all.

REPLY [3 votes]: Connections on the trivial $G$-bundle can be identified with maps from the base to $\mathrm{Lie}(G)$ via $x\frac{d}{dx} + f(x) \leftrightarrow f(x)$.  One way to present this as an ind-scheme is by choosing generators of $\mathcal{O}_{\mathrm{Lie}(G)}$, and considering the finite dimensional affine space of ring maps $\mathcal{O}_{\mathrm{Lie}(G)} \to \mathbb{C}[x,x^{-1}]$ for which the generators land in $\bigoplus_{n=-N}^N \mathbb{C} x^n$.
Following Ben Webster's example in the comments, consider the tangent field $x \frac{d}{dx} - a$ for $a \in \mathrm{Lie}(G)$.  This connection lies in the very first affine space of our sequence, with $N=0$.  If $a$ is semisimple, then solutions to the equation $x\frac{df}{dx} = af$ look like "$cx^a$".  The holonomy with respect to the basepoint $1$ is exponential in $a$ - in the one-dimensional case, identifying $\mathrm{Lie}(G)$ with $\mathbb{C}$, we get $e^{2 \pi i a}$.
In general, this example gives us an exponential map from $\mathrm{Lie}(G)$ to $G$.  This (and more generally, the holonomy map) exists in the category of (ind) complex analytic spaces, but outside the case of unipotent groups it is not algebraic.<|endoftext|>
TITLE: Forbidden coin flips
QUESTION [6 upvotes]: Suppose I have a (possibly infinite) bag of coins with various weights.  I select a coin  and flip it $n$ times.  Averaging over the choice of coins from the bag, there is some probability of seeing exactly $k$ heads, for $k=0,...,n$.  Let $r_k$ be the probability of seeing exactly $k$ heads.
More formally, let $D$ be a probability distribution on the unit interval $[0,1]$, and let $h$ be a binomial random variable with parameters $n$ and $p$, where $p$ is drawn from $D$.  Then marginalizing out $p$, the probability that $h=k$ is:
$$r_k:=\int_{0}^{1} \binom{n}{k}p^k(1-p)^{n-k} dD(p)$$
Given $D$, we can (in principle) compute $(r_0,...,r_n)$.  My question is: given $(r_0,...,r_n)$, does there exist some $D$ (i.e. bag of coins) that could have generated it?  Or is the set of probabilities forbidden?  More specifically, is there some finite procedure  that I could follow to determine whether or not such a $D$ exists?
A few comments:

Clearly we need $0\leq r_i\leq 1$ and $\sum_i r_i=1$.  
As a simple example of a forbidden configuration, take $n=2$ and $(r_0=0,r_1=1,r_2=0)$.
Note that when $D$ exists, it is usually not unique.
If we quantize the unit interval, e.g. if $D$ is supported on $\{0,\epsilon ,2\epsilon ,3\epsilon, ...,1\}$, we can express the quantized problem as a linear program.  However, the LP fails to solve the original problem and also feels (to me) like overkill.

REPLY [2 votes]: As fedja noted in the comments, I am essentially asking a classical moment problem.  I'm not sure if the $x=p/(1-p)$ change of variables reduction quite works (consider, e.g., the singleton probability distribution with all mass at $p=1$), but the general point is certainly correct.
For the benefit of future readers, I thought I'd summarize some relevant results.  First, note that 
$$r_n=\mathbb{E}_D p^n$$
Next, for $k<n$, 
$$r_k/\binom{n}{k}=\mathbb{E}_D p^k + \sum_{i=1}^k\binom{k}{i}\mathbb{E}_D p^{k+i}$$
so by (downward) induction, we can compute $\mathbb{E}_D p^k$ for all $k$.  Given the $\mathbb{E}_D p^k$, we can reverse the calculation to recover the $r_k$, so knowing the $r_k$ is exactly equivalent to knowing the first $k$ moments.
Second, determining whether a set of $k$ putative moments is consistent with any real distribution supported on $[0,1]$ is called the Hausdorff $k$-truncated moment problem.
To solve it, let $y=2p-1$, and let $\mu_i=\mathbb{E}_D y^i$ (i.e. we stretch $[0,1]$ to $[-1,1]$ and recompute the moments; we can compute these $\mu_i$ in terms of the $\mathbb{E}_D p^i$ moments).  Suppose we have $k$ moments and suppose $k=2d+1$ is odd.  Let
$$ A = \begin{bmatrix} \mu_0&\mu_1&\cdots&\mu_d \\ \mu_1&\mu_2&\cdots&\mu_{d+1} \\
\vdots & \vdots & \ddots & \vdots \\ \mu_d&\mu_{d+1}&\cdots&\mu_{2d}
\end{bmatrix}$$
$$ B = \begin{bmatrix} \mu_1&\mu_2&\cdots&\mu_{d+1} \\ \mu_2&\mu_3&\cdots&\mu_{d+2} \\
\vdots & \vdots & \ddots & \vdots \\ \mu_{d+1}&\mu_{d+2}&\cdots&\mu_{2d+1}
\end{bmatrix}$$
Then there exists a probability distribution with support on $[-1,1]$ with moments $\mu_0=1,\mu_1,...\mu_{2d+1}$ if and only if $A+B$ and $A-B$ are both semidefinite positive matrices.
I am quoting the last result from some class notes of Pablo Parrilo available here.  When I come across the "even $k$" version of this result I will update this answer.<|endoftext|>
TITLE: Action generated by geodesic flow is hamiltonian
QUESTION [5 upvotes]: I'm trying to understand why a certain action of a Lie Group is hamiltonian.
Let $(M,g)$ be a geodesically complete Riemannian manifold.
Then there exists a canonical one-form on the cotangentbundle $T^*M$ given by $\lambda_0 : T( T^*M) \to R, (v, \alpha) \mapsto \alpha(\pi_*v)$, with $\pi_*$ the derivative of the natural bundle projection $T^*M \to M$.
It is now possible to pullback this one-form to the tangentbundle using our riemannian metric, so $\lambda = (g^b)^*\lambda_0$. Using this one-form we define a symplectic structure on $TM$ by $\omega = -d\lambda$.
Now let $\phi:R\times  \text{TM} \to \text{TM}, V \mapsto \dot \gamma_V(t) $ be the action generated by the geodesic flow.
My question is, if $\phi$ acts by symplectomorphisms, that is, if $\phi_t^*\omega = \omega$ for all $t\in R$ or is it even exact symplectic, that is, if $\phi_t^*\lambda = \lambda$ for all $t\in R$? Because Im trying to show that this is an hamiltonian action, so that there exists a moment map for this action.
And it would be clear if we have an exact symplectic action.
Edit: To have a well-defined action of $R$, we assume that the manifold is geodesically complete.

REPLY [3 votes]: $\phi$ is the Hamiltonian flow corresponding to the ``Hamiltonian'' $E:TM\rightarrow\mathbb{R}$ given by $E(V) = \frac{1}{2}g(V,V)$. This is pretty standard, and you can read about it in most references on symplectic geometry, for example Foundations of Mechanics (in particular, Theorem 3.7.1). The corresponding momentum map of this flow is simply the Hamiltonian $E$ itself (with the canonical identification of $\mathbb{R}^*$ with $\mathbb{R}$).
The flow is not however exact symplectic, and in fact the generator of Hamiltonian flow $X_E\in\mathfrak{X}(TM)$ satisfies $\mathcal{L}_{X_E}\lambda = dL$, where $L$ is the Lagrangian, which equals $E$ in this case. To see this in coordinates, note that
$$
i_{X_E}\lambda = p_i dq^i \left(\dot{q}^i\frac{\partial}{\partial q^i} \right) = p_i\dot{q}^i.
$$
So using Cartan's Magic Formula
\begin{align*}
\mathcal{L}_{X_E} \lambda &= i_{X_E}d\lambda + d(i_{X_E}\lambda) \\
&= -i_{X_E}\omega + d(p_i\dot{q}^i) \\
&= d(p_i\dot{q}^i - E) \\
&= dL.
\end{align*}
For a more geometric treatment, see Foundations of Mechanics, sections 3.5-3.7.<|endoftext|>
TITLE: Is there a $q$-L'Hospital's Rule?
QUESTION [13 upvotes]: Let $\binom{n}{j}_q$ be a $q$-binomial coefficient and $(x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x).$ 
Consider the sum $$f(n,m,r,k)= \sum\limits_{j =  0}^{2n} {( - 1)}^{ j}q^{mj^2+rj}
\binom{2n}{j}_{q^k}$$ for integers $m,r$ and positive integers $k$.
Note that a famous result of Gauss says that $f(n,0,0,1)=(q;q^2)_n$.
For $n=1$ we get $f(1,m,r,k)=1-q^{m+r}-q^{k+m+r}+q^{4m+2r}$ 
and therefore
$$\frac{f(1,m,r,k)
}
{1-q}
={\frac{1-q^{k+m+r}}{1-q}}-q^{m+r}{\frac{1-q^{3m+r}}{1-q}}$$
which for ${q\to1}$ converges to ${(k+m+r)-(3m+r)}={k-2m}.$  
Computer experiments suggest that for each positive integer $n$ 
$$\lim_ {q\to1}\frac{f(n,m,r,k)
}
{{(q;q^2)_n}
}=(k-2m)^n.$$
Till now I did not find a method which leads to a proof of this result.
I asked the special case $ \lim_ {q\to1}\frac{f(n,m,m,1)}{f(n,1,1,1)}$
of this Problem already in Mathematics Stack Exchange 1359886.

REPLY [9 votes]: I think you can achieve this using the actual $q$-L'Hospital's Rule. Consider
$$\frac{ \partial^i f(n,m,r,k)}{\partial q^i} (1)= \sum_{i=0}^a \binom{a}{i} \sum\limits_{j =  0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{(m-k/2)j^2+rj}}{ \partial q^a}(1) 
\frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$
Now $\frac{\partial^a q^{(m-k/2) j^2+rj}}{ \partial q^a}(1)$ is a polynomial of degree $2a$ in $j$. So we can write this in terms of
$$ F(b,c) = \sum\limits_{j =  0}^{2n} {( - 1)}^{ j}j^b
\frac{ \partial ^{c} q^{(k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{c}}(1)$$ 
where $b+2c \leq 2i$.
We want to show that this vanishes for $i < n$ and evaluate it for $i=n$. For the first, it suffices to show that for all $b$ and $c$ with $b+2c < 2n$ we have:
$$F(b,c)=0$$ 
But these terms all show up in computing the partial derivatives for the sum 
$$\sum\limits_{j =  0}^{2n} {( - 1)}^{ j}
 q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}$$
$$  \frac{ \partial ^i \sum\limits_{j =  0}^{2n} {( - 1)}^{ j}
 q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}}{ \partial q^i} (1) =  \sum_{i=0}^a \binom{a}{i} \sum\limits_{j =  0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{rj}}{ \partial q^a}(1) 
\frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)  
$$ 
$$ = \sum_{i=0}^a \binom{a}{i} \sum\limits_{j =  0}^{2n} {( - 1)}^{ j} (rj)(rj-1) \dots (rj+1-a)
\frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$
$$ = \sum_{i=0}^a \binom{a}{i} \sum_{b=0}^a[j^b] (rj)(rj-1) \dots (rj+1-a) F(b,i-a) $$
Now suppose by induction that $F(b,c)=0$ for $b+c < i $ for some fixed $i \leq 2n$. Then this sum simplifies to only include the terms where $b=a$. Here
$$[j^a] (rj)(rj-1) \dots (rj+1-a) = r^a$$
so we obtain
$$= \sum_{i=0}^a \binom{a}{i} r^a F(a,i-a) $$
If $i<2n$ then this polynomial is identically $0$ by Max's computation, so all the coefficients vanish, so $F(a,i-a)=0$. This verifies the induction step.
Hence $F(b,c)=0$ for $b+c<2n$, which includes $b+2c<2n$, so the first $n-1$ partial derivatives vanish.
What about the $n$th partial derivative? Well the only $F(b,c)$ with $b + 2c \leq 2n$ that we don't know is $0$ is $F(2n,0)$. So dropping all the other terms we get
$$\frac{ \partial^n f(n,m,r,k)}{\partial q^n} (1)=   F(2n,0) [ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1) $$
Clearly
$$ [ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1) = [j^{2n}] ( (m-k/2)j^2+rj) ((m-k/2)j^2+rj-1) \dots ((m-k/2)j^2+rj+1-n) = (m-k/2)^{n}$$
So we get a constant depending only on $n$ times $(2k-m)^n$. To get the correct constant, you can compute it directly, or just note that it is sufficient to handle one special case for each $n$ and note that you did the $f(n,0,1,2)$ case in the comments to Max's answer.<|endoftext|>
TITLE: The conjugacy classes of diagonalizable $2 \times 2$ matrices can be identified with their eigenvalues, what about pairs?
QUESTION [14 upvotes]: For sake of simplicity, let's say that we live in $G = SL(2, \mathbb{C})$. Every conjugacy class of diagonalizable matrices $$[A] := \{gAg^{-1} \mid g \in G\}$$ can be identified with its set of eigenvalues $\{\lambda, \lambda^{-1}\}$. As such, the collection of diagonalizable matrices in $SL(2, \mathbb{C})$ can be identified with $\mathbb{C}^\times / \{z \sim z^{-1}\}$.
How about the conjugacy class of pair of matrices $$[A, B] = \{(gAg^{-1}, gBg^{-1}) \mid g \in G\}?$$ Can they similarly be identified with their eigenvalues in a natural way? What is the significance of this identification?

REPLY [2 votes]: Let me complete Ehud Meier's answer. One cannot distinguish one eigenvalue among $(\lambda,\lambda^{-1})$. In other words, one cannot distinguish between the projections $p_j$ and $q_j=1-p_j$. Therefore, $\alpha$ is only one among four relevant numbers. If $p_1q_2p_1=(1-\alpha)p_1$ is clear, it is less obvious that $q_1p_2q_1=(1-\alpha) q_1$.
Therefore one expect that the conjugacy classes are parametrized by the triples $(\alpha,\lambda,\mu)$, modulo the group generated by the involutions
$$(\alpha,\lambda,\mu)\mapsto(1-\alpha,\lambda^{-1},\mu)\quad\hbox{or}\quad(1-\alpha,\lambda,\mu^{-1}).$$<|endoftext|>
TITLE: Polynomials vanishing modulo some integer $n$
QUESTION [12 upvotes]: It is well-known that a polynomial $q \in \mathbb Z[t]$ vanishes modulo $p$ only if it lies in the ideal $J_p$ generated by $p$ and $t^p-t$. This means that either the degree is large (at least $p$) or the coefficients are large (divisible by $p$). 
Is there anything useful like this that one can say if a polynomial vanishes modulo $n$? For example, let $n=p_1 \cdots p_k$, where $p_1<\dots<p_k$ are different primes. (For me, this could be the list of all primes less than some number $x$ for example.) It is clear that if $q(t)$ vanishes modulo $n$, then
$$q(t) \in J_{p_1} \cap \cdots \cap J_{p_k} = J_{p_1} \cdots J_{p_k}.$$
Examples are $q(t)=t\prod_{i=1}^k (t^{p_i-1}-1)$ or $q(t)=p_1 \cdots p_l$ or anything in the ideal generated by polynomials divisible for each $1 \leq i \leq k$ by either $p_i$ or $t^{p_i}-t$.

Question: Is it true that again either the degree must necessarily be large or some coefficient (or let's say that sum of absolute values of coefficients) must be large? Here, large could mean for example comparable with $\sum_{i} p_i$ or $n$.

It is easy to see that any polynomial $q(t) \in \mathbb Z[t]$ that vanishes modulo $n$ is such that $q(t)/n$ maps $\mathbb Z$ to $\mathbb Z$, and hence 
$$q(t) = \sum_{i} n a_i \binom{t}{i},$$ for some $a_i \in \mathbb Z$ -
but I do not see how this helps. I also tried to apply Chebotarev density theorem (which together with the error analysis of Lagarias-Odlyzko gives a way to produce small primes modulo which a polynomial has to have a root), but the estimates are too coarse and do not seem to make efficient use of the assumptions on the polynomial. 
EDIT: Motivated by a discussion with David Speyer below, let me formulate a more precise question:

Question: Let $f \in \mathbb Z[t]$ be a monic polynomial of degree $d$ that vanishes modulo all primes $\leq P$. Is it true that $d \log \|f\|_{\infty}$ cannot be much smaller than $P^2$?

Here, $\|f\|_{\infty}$ denotes the maximum of the absolute value of the coefficients of $f$.

REPLY [4 votes]: As stated the answer is negative. Denote $N=lcm \{p_i-1|1\leq i \leq k\}$. Then $t^{N+1}-t$ is always divisible by $n$. But it may appear that $\sum (p_i-1)=N+N/2+\dots+N/k$, so it may be much greater than $N$. 
But some another reasonable bound on degree/norm may hold.<|endoftext|>
TITLE: Matrix from a homomorphism of simply connected groups
QUESTION [5 upvotes]: Let $H$ be a simple algebraic group of type $\mathbf{G}_2$ over $\mathbb{C}$.
Let $\rho$ be the standard 7-dimensional complex representation
$$ \rho\colon H=\mathbf{G}_2\to \mathrm{SO}_7.$$
We consider the corresponding homomorphism
$$H\to \mathrm{Spin}_7=G.$$
We regard $H$ as a subgroup of $G$.
Let $T_H\subset H$ and $T_G\subset G$ be compatible maximal tori.
We obtain a homomorphism of cocharacter groups
$$\rho_*\colon X_*(T_H)\to X_*(T_G).$$
We choose compatible Borel subgroups $B_H$ and $B_G$ such that $T_H\subset B_H\subset H$
and  $T_G\subset B_G\subset G$, then we obtain  bases consisting of simple coroots in $X_*(T_H)$ and $X_*(T_G)$.
Since we have bases, we can associate with $\rho_*$ an integral  $3\times 2$ matrix.

Question. How can one compute this matrix?

Note that I need  this matrix only modulo 2.

REPLY [4 votes]: Like Jason I'd be more inclined to look at this kind of embedding in a geometric or conceptual way.  But for your purposes a concrete description seems needed.  For this it might be simpler to work in the Lie algebra setting.    I don't see immediately what the group viewpoint does for you, since the centerless group $G_2$ embeds in either the special orthogonal group or its slightly elusive simply connected covering group Spin.   The basic roots-and-weights technology here depends just on the Lie algebra embedding.    For this to be made concrete, however, you'd have to line up the two Cartan subalgebras (Lie algebra of maximal tori) in a compatible way and relate the two root systems via simple roots.   
A description of the embedding here is given (in an adapted version going back to one of the Paris seminars) in the first part of section 19.3 of my 1972 Lie algebra textbook.    (Needless to say, I haven't spent a lot of time with this material since then.)    The main deficiency in the details written down is that a choice of simple roots for $G_2$ isn't made explicit (though it is in Bourbaki).   So you'd have to work that out further.   Since the weight lattice and root lattice of $G_2$ coincide, passage to the duals can be done over $\mathbb{Z}$.   
ADDED: With apologies for the delay in filling in some details, I still have some doubts about the basic setting here.    On the level of root lattices it seems fairly clear geometrically.   For the embedding $\mathfrak{g}_2 \hookrightarrow \mathfrak{so}_7$, one can work over $\mathbb{R}$ (since the Lie algebras are split).    Here the bigger root lattice lives in a 3-dimensional euclidean space with standard orthonormal basis $\varepsilon _1, \varepsilon_2, \varepsilon_3$.   
For $B_3$ there are 9 positive roots.   Two simple roots are long: $\alpha_1 = \varepsilon_1 - \varepsilon_2$ and $\alpha_2 = \varepsilon_2 - \varepsilon_3$.  A third simple root is short: $\alpha_3=\varepsilon_3$.   Then the root lattice for $G_2$ lies in the plane through 0 defined by the condition that coordinates sum to 0.  Here you can take a short simple root $\alpha = \alpha_1$ and a long simple root $\beta = \varepsilon_2 + \varepsilon_3 -2\varepsilon_1$.    Then you can compute your $2 \times 3$ or $3 \times 2$ matrix in terms of these bases, keeping in mind that for $G_2$ the root lattice equals the weight lattice and maps into the root lattice for $B_3$ (which however has index 2 in its weight lattice).
My quick arithmetic wasn't reliable, but while the matrix you get depends on the choice of bases, the relationship between the two root systems depends only on the Cartan matrices.<|endoftext|>
TITLE: Oracle queries asked in parallel
QUESTION [7 upvotes]: Definition: Assume that $\phi(q)$ is of the form $\exists y \leq 2^{p(n)} \varphi(q,y)$, where $p$ is a polynomial and $n = |q|$ (i.e. $n$ is the length of the binary representation of $q$). Then a witness query $q$ to $\phi$ returns a witness $y \leq 2^{p(n)}$ satisfying $\varphi(q,y)$, if such exists, and returns $2^{p(n)} + 1$, otherwise. 
Question: Let $f$ be a function in $FP^{\Sigma^p_1}[wit]$, i.e. $f$ is computed by a polynomial time Turing machine $M$ that makes polynomially many witness queries to a $\Sigma^p_1$ oracle $\phi(q) \equiv \exists y \leq 2^{p(n)} \varphi(q,y)$. 
My question is the following: Can we somehow construct a polynomial time Turing machine $M_1$ which makes polynomially many witness queries to $\phi(q)$
in parallel (i.e. all the queries to $\phi$ come first, then the sequence of replies  to the queries come after) and such that $M^{\phi}_1(x) = M^{\phi}(x)$.

REPLY [3 votes]: By a standard argument that also applies in this case, polynomially parallel many witness queries give the same power as $O(\log n)$ many sequential witness queries:
$$\mathrm{FP^{\|\Sigma^P_1[wit]}=FP^{\Sigma^P_1[wit,\log]}}.$$
In particular, the predicates (i.e., 0–1 valued functions) in $\mathrm{FP^{\Sigma^P_1[wit]}}$ are exactly those from $\Delta^P_2=\mathrm{P^{NP}}$, whereas predicates in $\mathrm{FP^{\|\Sigma^P_1[wit]}}$ are exactly those from the complexity class $\Theta^P_2=\mathrm{P^{NP[\log]}=P^{\|NP}=L^{NP}}$ (see zoo). It is widely expected that $\Theta^P_2\ne\Delta^P_2$, and this is supported by oracle separation results. Thus, the answer to your question should be no.
See also https://cstheory.stackexchange.com/q/31482 for more information about this class.<|endoftext|>
TITLE: Does there exist an algebraic space with large fundamental group but no finite etale covers by schemes
QUESTION [7 upvotes]: Does there exits a smooth proper algebraic space $X$ over $\mathbb C$ with "large" fundamental group such that no finite etale cover of $X$ is a scheme?
By "large" fundamental group I mean that $X$ has infinitely many non-trivial finite etale covers (of increasing degree).
Equivalently, does there exist a Moishezon space $X$ with large fundamental group such that no finite degree topological covering of $X$ is the analytification of a scheme?

REPLY [8 votes]: You can get one class of examples by modifying the Hironaka construction, cf. Appendix B, Examples 3.4.1 and 3.4.2 of Hartshorne's "Algebraic Geometry".  What is perhaps surprising is that Hartshorne begins with a scheme -- the scheme from Example 3.4.1 -- that has a fixed point free $\mathbb{Z}/2\mathbb{Z}$-action, and then he forms the algebraic space as a quotient of this.  So it appears that, by construction, there is a finite cover that is a scheme.  
However, you can choose the ambient scheme so that the $\mathbb{Z}/2\mathbb{Z}$-action is fixed point free on a Zariski open that contains the relevant "exceptional surface", yet which has fixed points elsewhere.  When you form the quotient and desingularize, the quotient will typically be simply connected.  For instance, if the ambient scheme is a blowing up of projective space, then the quotient is rationally connected, hence any desingularization is simply connected.  Finally, take the product of this simply connected algebraic space with any curve of genus $g\geq 1$.<|endoftext|>
TITLE: Generalization of a theorem of Burnside to non-compact groups
QUESTION [5 upvotes]: The following two theorems are often attributed to Burnside:
Theorem Let $G$ and $H$ be compact groups. Then the irreducible representations of $G\times H$ are precisely the representations $\pi\otimes \sigma$ where $\pi$ and $\sigma$ are irreducible representations of $G$ and $H$, respectively.
Corollary Let $(\pi, V)$ be an irreducible representation of the compact group $G$. Then every linear transformation of $V$ is of the form

$\sum_i a_i\pi(g_i)$ (finite sum) for some $a_i\in \mathbb{C}$ and $g_i\in G$. 
$\pi(f)$ for some continuous function $f$.

The Theorem is a fairly simple consequence of orthogonality relation for characters and the Corollary is an application of the Theorem to $\pi\otimes \pi^*$.
My questions are: 
(1) how does this beautiful result of Burnside generalize to non-compact (Lie) groups? (I can imagine that both results fail to hold although I haven't cooked up any counterexamples yet.)
(2) does the Corollary hold when $V$ is finite dimensional?

REPLY [4 votes]: (Note : all the groups are locally compact topological groups)
I will prove the following theorem, and then prove that at least for Hilbert space this result is pretty much optimal: 
Theorem: Let $G$ and $H$ be two postliminal groups, then every unitary irreducible representations of $G \times H$ over a Hilbert space is of the form $\rho \otimes \rho'$ where $\rho$ and $\rho'$ are unitary irreducible representations of $G$ and $H$ (over Hilbert spaces).
A post-liminal group (also called a type I group) is a group $G$ such that every for every unitary irreducible representation $\rho$ of $G$ on $H$, every compact operator on the Hilbert space $H$ can be approximated by linear combination of operator of the form $\rho(g)$. Equivalently, the group maximal $C^*$ algebra is type I/post liminal. Equivalently, all factor representation of the group generate a type I factor, etc... see Dixmier "C* algebra" for more information on post-liminal/type I $C*$ algebra and group...
All compact group are post-liminal, all commutative groups are post-liminal, all semi-simple lie group are post-liminal, nilpotent (lie ?) groups are post-liminal I, the class of postliminal group has some stability property (obvious exemple: a quotient of a post-liminal group is postliminal) etc...
But there exist an exemple (in dimension 5) of a solvable lie group which is not postliminal, free group are not postliminal, in fact all non-amenable group are not post-liminal etc...
The proof would go as follow:
Let $\rho$ be a irreducible unitary representation of $G \times H$ on a Hilbert space $T_0$. Let $A_G$ and $A_H$ the von Neuman algebra generated by $G$ and $H$ in $B(T_0)$.
$A_G$ and $A_H$ are factors: indeed if $u \in Z(A_G)$ then $u$ commutes to the action of $H$ because it is in $A_G$ and it commutes to the action of $G$ because it is in $A_H$ hence by Schur's lemma (which hold for unitary representation) it is a scalar.
Hence $A_G$ and $A_H$ are type I factor because $G$ and $H$ are type I algebras.
Then (this is the key step) using the decomposition theory of Dixmier's C* algebra chapter 5.4 one can decompose $T_0$ into $T_1 \otimes T_2$.
Indeed, $A_G$ is a type I factor acting on $T_0$ hence $T_0$ is isomorphic to $T_1 \otimes T_2$ with $A_G = B(T_1)$ and as the action of $A_H$ is in the commutant of $A_G$ it is in $B(T_2)$. One then easily see that $T_1$ and $T_2$ are irreducible representation of $G$ and $H$:
if $u$ is an endomorphism of $T_1$ commuting to the action of $G$ then it clearly also (when seen as acting on $T_0$) it clealry commutes to the action of $G \times H$ and hence is a scalar.
This concludes the proof of the theorem.
In fact one can state the following:
Theorem: let $G$ and $H$ be two groups and $V$ be a unitary irreducible representation of $G \times H$ then $V$ is of the form $\rho \otimes \rho'$ if and only if the actions of $G \times \{1\}$ and $\{1 \} \times H$ on $V$ (respectively) generates type I von Neumann algebras.
Indeed one direction is exactly the previous proof, and conversely if the representation can be decomposed, then (using the notation of the previous proof) $A_G$ is isomorphic to $B(T_1)$ and hence is a type I factor.
Corollary:
Let $G$ be a group, then the following conditions are equivalent:

$G$ is postliminal.
Every (unitary) irreducible representation of $G \times G$ decompose as a tensor product of two irreducible representations of $G$.

proof:
If $G$ is post-liminal then this is just the first theorem.
Conversely, assume $G$ is not post liminal, then there exists a Hilbert space $H$ with a unitary representation of $G$ that generate a type II or type III factor $A_G$.
Now considering the $l^2$ representation of $A_G$ one obtains a representation $H'$ of $A_G$ such that $End_{A_G}(H) = (A_G)^{op}$. (this refers to Tomita-Takesaki theory and has nothing to do with the regular representation of $G$)
One can make $G \times G$ acts on $H'$: the first component acts though $G \rightarrow A_G$ and the second component to $G \rightarrow A_G^{op}$ (sending $g$ to $g^{-1}$).
$H'$ is irreducible: indeed an endomorphism of $H'$ commuting with the action og $G \times G$ commute with the action of $A_G$, and of $(A_G)'$ (the commutant of $A_G$) hence by the double commutant theorem it belong to the center of $A_G$ which is reduce to the scalar as $A_G$ has been assume to be a factor.
By the second theorem $H'$ is a rep of $G \times G$ that cannot be decomposed: indeed the algebra generated by $G$ is $A_G$ and is of type II or type III by assumption.<|endoftext|>
TITLE: Intuition behind basic facts about homogeneous ideals?
QUESTION [5 upvotes]: What is the intuition (hopefully, geometric) behind these basic facts about homogeneous ideals? An ideal $I$ in $S$ is homogeneous if an element $f = \sum_{n \ge 0} f_n$ of $S$ lies in $I$ if and only if each homogeneous part $f_n$ lies in $I$. Here, we let $S = \bigoplus_{n \ge 0} S_n$ be an $\mathbf{N}$-graded ring.

An ideal generated by homogeneous elements is a homogeneous ideal.
A homogeneous ideal $I$ in $S$ is prime if and only if it is a proper ideal and $$fg \in I \implies f \in I \text{ or }g \in I$$for homogeneous $f, g \in S$.
The kernel of $\mathbf{N}$-graded rings is a homogeneous ideal.
For any homogeneous ideal $I$ of $S$ there is a natural $\mathbf{N}$-grading on $S/I$.

I am not an algebraic geometrer by trade, but I need to use these results.

REPLY [6 votes]: One thing to grok is that, in my opinion, graded rings are a hack. Occasionally graded rings as defined are of interest, but usually, at least in the fields I'm familiar with, the inhomogeneous elements don't play a role at all, and their presence is only a trick so that you can leverage the existing body of knowledge of commutative algebra.
The actual algebraic structure you care about is the essentially algebraic one made up out of the graded components, and the arithmetic with them (i.e. addition should only be defined for things of the same grading). This can be phrased in terms of preadditive categories, with one object per grading. A ring, for example, is a preadditive category with one object.
(also, I believe this foundation for graded rings wasn't known for a long time, so the version you cite is less of a trick and more of the only thing people had to work with)
The intuition behind the theorems you list is simply that when using this hack, the usual definitions of things have some relevance to what you want for graded rings (e.g. that a homogeneous prime ideal when represented as an ordinary ideal is actually prime). That is, the inclusion of the inhomogeneous things don't get in the way of the theory working in a simple way.<|endoftext|>
TITLE: Relationship of Weisfeiler-Lehman algorithm to weak isomorphism of coherent algebras
QUESTION [9 upvotes]: A coherent algebra is a matrix algebra (over $\mathbb{C}$) closed under conjugate transpose and Schur (entrywise) product, and that contains the identity matrix $I$ and all ones matrix $J$. Given coherent algebras $\mathcal{A}$ and $\mathcal{B}$, a weak isomorphism from $\mathcal{A}$ to $\mathcal{B}$ is a bijective linear map $\varphi$ satisfying:

$\varphi(MN) = \varphi(M)\varphi(N)$;
$\varphi(M \circ N) = \varphi(M) \circ \varphi(N)$ (where $\circ$ denotes the Schur product);
$\varphi(M^*) = \varphi(M)^*$ where $^*$ denotes conjugate transpose;
$\varphi(J) = J$.

Given a graph $G$ with adjacency matrix $A$, the coherent algebra of $G$ is defined as the intersection of all coherent algebras containing $A$. In this paper they define two graphs $G$ and $H$ to be equivalent if there exists a weak isomorphism between their corresponding coherent algebras that also maps the adjacency matrix of $G$ to the adjacency matrix of $H$. In the paper they say that there exists a polynomial time algorithm for testing whether two graphs are equivalent and cite Weisfeiler and Lehman's paper.
I gather that the Weisfeiler-Lehman algorithm is closely related to coherent algebras, but have so far only been able to find references which describe the algorithm in a purely combinatorial manner, i.e. in terms of graphs not coherent algebras. The paper of Weisfeiler and Lehman linked above is not available to me and is anyways written in Russian. I am wondering if there is a direct connection with the definition of graph equivalence given above, and the Weisfeiler-Lehman algorithm for distinguishing graphs. More specifically, is it true that two graphs are equivalent if and only if they cannot be distinguished by ($k$-dimensional) Weisfeiler-Lehman (for some [specific] $k$)?

REPLY [3 votes]: A standard reference is "On Construction and Identification of Graphs",
ed. by: B.Weisfeiler, Springer
Lect. Notes Math.
Vol. 558 (1976)
ISBN: 978-3-540-08051-0 
There you will find details on Weisfeiler-Lehman algorithm (called stabilization there, IIRC). Although it's not an easy read.
A modern reference with some details on various "generalised" isomorphisms is here.<|endoftext|>
TITLE: When do partial derivatives $p_x$, $p_y$ of a polynomial over $\mathbb{C}$ not have any common factor?
QUESTION [7 upvotes]: When do partial derivatives $p_x$, $p_y$ of a polynomial over $\mathbb{C}$ not have any common factor? Is there a general approach for any number of variables, aka when is the variety defined by the partial derivatives vanishing (the would-be singular points) a finite point set? I.e. when in $\mathbb{C}[x, y, \dots]$ is changing the constant term not enough to render a variety nonsingular? Which means, when does $\mathbb{C}[x, y, \dots]^{\text{sing}}$, polynomials defining singular varieties, which we know is a Zariski closed subspace of generic codimension $\ge$ the number of variables $-\,2$, contain an affine hyperplane in $\mathbb{C}[x, y, \dots]$ which allows the constant to vary?

REPLY [5 votes]: In the case of two variables, my example is the only one. Let $g$ be an irreducible polynomial that divides both $p_x,p_y$. On the curve $g=0$, we have $dp = p_xdx+p_ydy=0$ so $p$ is constant, say $p=c$ on this is curve, so $g$ divides $p-c$. Let's write $p-c = gh$ for some $h$. Then, again on $g=0$, $0=p_x = g_xh, 0=p_y=g_yh$. But $g_x,g_y$ cannot both vanish on $g=0$, so $h$ is divisible by $g$ and $p-c$ is divisible by $g^2$.<|endoftext|>
TITLE: Non-homeomorphic spaces such that taking away a point makes them homeomorphic
QUESTION [19 upvotes]: Are there topological spaces $X,Y$, each having more than $2$ points, satisfying the following two properties?

$X\not\cong Y$, and
there is a bijection $\varphi: X\to Y$ such that for all $x\in X$ the spaces $X\setminus \{x\}$ and $Y\setminus \{\varphi(x)\}$ are homeomorphic.

REPLY [2 votes]: For a minimal example consider the following two 3-element posets equipped with the Alexandroff topology (i.e. open sets are down-sets):
(a)   (b)                      (C)
  \   /                        / \
   \ /                        /   \
   (c)                      (A)   (B)

The bijection is given by: $a\mapsto A$, $b\mapsto B$, $c\mapsto C$.
The reconstruction conjecture for ordered sets (see e.g. Jean-Xavier Rampon, What is reconstruction for ordered sets?) - together with equivalence of finite posets and finite $T_0$ topological spaces (via the  specialization order/Alexandroff topology functors) - says the are no such examples among finite $T_0$ spaces with at least four elements. However, the conjecture is open.
(The related reconstruction conjecture for graphs is better known and also open.)<|endoftext|>
TITLE: Three and a half basic questions on the Weil restriction of scalars
QUESTION [5 upvotes]: (This is reposted from mathstackexchange, where it received no answer so far.)
I am currently trying to get familiar with the Weil Restriction functor. 
For a finite field extension $L|K$ it associates a variety over $K$ to every variety $X$ over $L$ as the representing object of the functor defined via $Res_{L|K}X(A)=X(A\otimes_K L)$ for every $K$-Algebra $A$. There is also an explicit description in terms of defining equations for the variety.
I have the following questions:
$1$) Is there a characterisation of the (essential) image of this functor? i.e. can we say which varieties over $K$ "come from" $L$?
$2$) Is Restriction of scalars "injective"  i.e. if a $K$-variety admits the structure of an $L$-variety is this structure unique? (up to isomorphism)
$3$) In the case of $L=\mathbb{C}$ and $K=\mathbb{R}$ there are also analytification functors and we can "restrict scalars" on the analytic side by regarding a complex manifold as a real analytic one. It seems to me, by the explicit construction of the Weil-Restriction, that analytification and restriction of scalars commute. Is this correct?
$3.5$) Can you recommend a reference to learn about this stuff which does not focus only on algebraic groups?

REPLY [9 votes]: It is best not to use the word (quasi-projective) "variety" in this context because (as alluded to in Scott Carnahan's comment to the question posed) Weil restriction does not preserve geometric integrality or even geometric irreducibility or geometric reducedness or non-emptiness in general (allowing $L/K$ that is possibly not separable), and even if $L/K$ is separable one can encounter some oddities such as ${\rm{R}}_{L/K}(X)$ being non-reduced with $X$ that is reduced (though such $X$ are not geometrically reduced, so the latter oddities are a bit anti-climactic after a moment's thought) . A reference for such counterexamples (giving $X$ geometrically integral over $L$ for which ${\rm{R}}_{L/K}(X)$ is disconnected with a connected component that is everywhere non-reduced) is provided at the end of this answer.  So one should speak in terms of "schemes" (as always!). 
The explicit description of Weil restriction in terms of equations is usually (though not always) useless:
Example: Assume $(X,e')$ is an $L$-group scheme, and consider the natural isomorphism of $K$-vector spaces $T_e({\rm{R}}_{L/K}(X)) \simeq T_{e'}(X)$ (via evaluating on the dual numbers). It is true that this is an isomorphism of Lie algebras, moreover respecting the $p$-operation in characteristic $p$, but I don't think that obvious by thinking in terms of "equations" (even if $X$ is smooth, which is not necessary to assume). 

The answers to #1 and #2 are certainly negative in general (if for #1 one intends for an interesting rather than tautologous answer); it is akin to analogues with induction of representations from a (possibly non-normal) subgroup of finite index.  For example, if $L/K$ is finite Galois with Galois group $G$ and $Y$ is a quasi-projective $L$-scheme that is not $L$-isomorphic to its twist $g^{\ast}(Y)$ by some $g \in G$ (elliptic curves provide plenty of examples) then nonetheless ${\rm{R}}_{L/K}(Y) \simeq {\rm{R}}_{L/K}(g^{\ast}(Y))$ due to the description of finite separable Weil restriction as mentioned in Fonarev's answer.
As for #3.5, section 7.6 of the book "Neron Models" gives a very elegant  introduction to the topic of Weil restriction (over general rings -- as one really ought to do, since $L \otimes_K K'$ is rarely a field when $K'/K$ is an extension field yet we do want to contemplate how Weil restriction interacts with scalar extension from $K$ to $K'$). That reference goes way beyond the setting of algebraic groups, and addresses good behavior of the Weil-restriction functor (on quasi-projective schemes) with respect to etale morphisms and open immersions and much else. It doesn't include the proof that the output is again quasi-projective in general, but that property -- which lies a bit deeper in the story -- is settled affirmatively in Proposition A.5.8 in the book "Pseudo-reductive groups"; Appendix A.5 of this latter book (recently published in a revised/expanded 2nd edition that I will refer to below) gives an alternative general development of the topic of Weil restriction (also going beyond algebraic groups).  
Remark: One input into the proof of quasi-projectivity of ${\rm{R}}_{L/K}(X)$ is knowing in advance that this $K$-scheme is quasi-compact. Such quasi-compactness for general (not necessarily separable) finite extensions $L/K$ is not part of the usual construction, which merely gives something locally of finite type over $K$ (due to failure of Weil restriction to carry Zariski covers to Zariski covers, a posteriori -- see Example A.5.3 of "Pseudo-reductive groups" for simple counterexamples with any nontrivial finite separable extension $L/K$, not such a surprise if you think in terms of pushforwards of sheaves of sets) and separated (by valuative criterion). Fortunately, there always does exist a finite Zariski affine open cover $\{U_i\}$ of $X$ for which the associated affine opens ${\rm{R}}_{L/K}(U_i)$ cover ${\rm{R}}_{L/K}(X)$: see page 504 of "Pseudo-reductive groups".

As for question #3, it seems you intend to ask, for a smooth quasi-projective $\mathbf{C}$-scheme $X$, if the real-analytic manifold associated to the smooth quasi-projective $\mathbf{R}$-scheme ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(X)$ is isomorphic naturally in $X$ to the real-analytic manifold underlying the complex manifold $X^{\rm{an}}$.  In other words, is the evident identification of sets 
$$\varphi_X: ({\rm{R}}_{\mathbf{C}/\mathbf{R}}(X))(\mathbf{R}) \simeq X(\mathbf{C})$$ a real-analytic isomorphism?  The case when $X$ is an affine space is clear, and for a clean argument it is tempting to try to reduce the general case to that case by using the Zariski-local descrption of smooth morphisms in terms of etale maps to affine spaces. This basically works, but there is a tiny wrinkle, as we will soon see.
We noted above that although Weil restriction generally does not carry Zariski covers to Zariski covers, there always exists some finite open affine 
cover $\{U_i\}$ of $X$ such that ${\rm{R}}_{L/K}(X)$ is covered by its open affine subschemes ${\rm{R}}_{L/K}(U_i)$. The construction of such $\{U_i\}$ does not permit us to choose such $\{U_i\}$ to be subordinate to a chosen cover of $X$, so that might seem to doom the attempt to reduce to working with $U_i$'s that are etale over affine spaces. 
But actually we don't need this: if $\{U_i\}$ is any Zariski-open cover of 
the smooth $\mathbf{C}$-scheme $X$ then the open submanifolds $({\rm{R}}_{\mathbf{C}/\mathbf{R}}(U_i))(\mathbf{R})$ of $({\rm{R}}_{\mathbf{C}/\mathbf{R}}(X))(\mathbf{R})$ visibly constitute a cover since $\{U_i(\mathbf{C})\}$ covers $X(\mathbf{C})$ (even though the ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(U_i)$'s may fail to cover ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(X)$, which is to say that $\{U_i(\mathbf{C} \otimes_{\mathbf{R}}\mathbf{C})\}$ may not cover $X(\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C})$). Likewise $\{U_i^{\rm{an}}\}$ is an open cover of $X^{\rm{an}}$.  Hence, the problem for $X$ does reduce to the same for each $U_i$ separately after all, so in this way we can reduce to the case when there is an etale map $f:X \rightarrow \mathbf{A}^n_{\mathbf{C}}$.  But then ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(f)$ is an etale map between smooth $\mathbf{R}$-schemes, so it induces a local real-analytic isomorphism on $\mathbf{R}$-points by the real-analytic inverse function theorem and the Zariski-local description of etale morphisms.  Likewise, $f^{\rm{an}}$ is a local complex-analytic isomorphism.  In this manner, the bijective map $\varphi_X$ is a real-analytic isomorphism because of the same for $\varphi_{\mathbf{A}^n_{\mathbf{C}}}$.  That settles #3 affirmatively.

Although some answers to #3.5 have been given above, I want to warn you again about the dangerous use of the word "variety" in these settings by providing a reference for the disorienting counterexamples alluded to at the start. To give some context, I first note that if the quasi-projective $L$-scheme $X$ is smooth and geometrically connected over $L$ then the $K$-scheme ${\rm{R}}_{L/K}(X)$ is smooth (easy, by infinitesimal criterion; not by staring at equations) and geometrically connected. One cannot "see" connectedness properties in terms of equations, so this geometric connectivity over $K$ is really not obvious in general: the case that $L/K$ is separable is very easy because one has the $K_s$-isomorphism 
$${\rm{R}}_{L/K}(X)_{K_s} \simeq  {\rm{R}}_{(L \otimes_K K_s)/K_s}(X_{L \otimes_K K_s}) = \prod_{\sigma} \sigma^{\ast}(X)$$
where $\sigma$ varies through the $K$-embeddings of $L$ into $K_s$ (so the smoothness hypothesis isn't needed in such cases), but an entirely different idea is needed to handle the possibility that $L/K$ is not separable. The general case (allowing any finite extension of fields $L/K$) is Proposition A.5.9 in "Pseudo-reductive groups".
To appreciate the optimality of the smoothness hypothesis when $L/K$ is not separable, we finally come to the mind-boggling counterexamples: for any imperfect field $K$ one can make examples of purely inseparable finite extensions $L/K$ and  geometrically integral affine plane curves $X$ over $L$ that are smooth away from one point (and even regular at that non-smooth point when $[K:K^p] > p$, with $p = {\rm{char}}(K)$) such that ${\rm{R}}_{L/K}(X)$ is disconnected and has a connected component that is everywhere non-reduced!  This is discussed immediately after the statement of Proposition A.5.9 in "Pseudo-reductive groups".  
[If we relax geometric integrality to geometric irreducibility then other weird things can happen. For instance, ${\rm{R}}_{L/K}(X)$ can be non-reduced with separable $L/K$ and reduced $X$ that is not geometrically reduced -- not so surprising if you think about it for a minute -- and this Weil restriction can be empty for $X = {\rm{Spec}}(F)$ with a purely inseparable finite extension field $F/L$ and suitable purely inseparable $L/K$.]<|endoftext|>
TITLE: Infimum of Gaussian process
QUESTION [5 upvotes]: Consider a Gaussian Process $g\sim GP(\mu,k)$ with mean zero $\mu\equiv0$ and continues covariance $k(t_1,t_2)=k(|t_1-t_2|)$ defined on the interval $A=[0,T]$. I'd like to make no assumptions about g(0).
What can be said on the infimum of $g(t)$? For example, what is the distribution of $m=\inf_A g(t)$? What is the mean? What is the median?
Related and perhaps simpler version would be to ask what is $Pr(\{g(t)>c\ :\ \forall t\in A\})=Pr(m>c)$? Or at least what is $Pr(m>0)$, the probability of no zero crossing in $[0,T]$?
Note that unlike related question I'm not interested in the asymptotic behavior for large $c$ but rather what is the probability of infimum near 0 (or how $Pr(m>c)$ behave for small $c$).
EDITED: Per Ylvisaker 1965 we know the expected number of zeros in $A$ is $\frac{T}{\pi}\sqrt{-k''(0)}$. But I couldn't see how it help get $Pr(m>0)$.

REPLY [2 votes]: I'll summarize what I've learned.
Denote a Gaussian Process $g\sim GP(\mu,k)$ with mean zero $\mu\equiv0$ and continues covariance $k(t_{1},t_{2})=k(|t_{1}-t_{2}|)$ defined on the interval $[0,T]$ and also $m=\inf_{t\in[0,T]}g(t)$. 
I could not find any result regarding the pdf (or cdf) of $m$, not even its mean. Also $P(m>c)$ yielded no results except for the case $c=0$, where from symmetry $P(m>0)=\frac{1}{2}p(0,T)$ with $p(n,T)$ is the probability of $g(t)$ having exactly $n$ zeros in $[0,T]$. This latter quantity is discussed extensively in the literature on “zero crossing intervals” and can be calculated numerically, as we shall now explain.
As already mensioned, the mean number of zeros in time unit is given by Rice' Formula $\beta=\sqrt{-k''(0)}/\pi$ (when $k$ is normalized, $k(0)=1$). Also denote $P_{n}(\tau)$ the probability density for of the event that the $n+1$th zero crossing occuring at interval $\tau$ after the first zero crossing. 
McFadden's paper from 1958, The axis-crossing intervals of random functions--II
show that $U(\tau)=\sum_{0}^{\infty}P_{n}(\tau)$, the probability of some zero crossing at interval $\tau$ after a zero crossing have a (complicated) closed-form expression $U(\tau)=\frac{2}{\pi}(\frac{1}{H(\tau)}-\tan^{-1}H(\tau))R(\tau)$ (eq. 20-21 there) with $R(\tau)=\frac{1}{2\beta\pi}\frac{k(\tau)k'(\tau)^{2}+(1-k(\tau)^{2})k''(\tau)}{(1-k(\tau)^{2})^{3/2}}$ and $H(\tau)=\frac{k(\tau)k'(\tau)^{2}+(1-k(\tau)^{2})k''(\tau)}{\sqrt{[k'(\tau)^{2}+(1-k(\tau)^{2})k''(0)]^{2}-[k(\tau)k'(\tau)^{2}+(1-k(\tau)^{2})k''(\tau)]^{2}}}$. Now $P_{0}(\tau)$ can be calculating numerically from the convolutional equation $P_{0}(\tau)=\frac{1}{2}[U(\tau)+R(\tau)]-\frac{1}{2}[U(\tau)-R(\tau)]*P_{0}(\tau)$ (eq. 47 there). Then $p(0,T)$ can be calculated numerically by integrating $p(0,T)=1+\beta[\int_{0}^{T}d\tau\int_{0}^{\tau}dlP_{0}(l)-\tau]$.<|endoftext|>
TITLE: Which finite groups have no irreducible representations other than characters?
QUESTION [14 upvotes]: A classical result states that all the irreducible representations of a finite group over $\mathbb{C}$ are characters if and only if $G$ is abelian. I would like to know what happens if we consider a field different from $\mathbb{C}$. Clearly as long as the field is algebraically closed and its characteristic does not divide $|G|$ the same conclusion holds.
Furthermore, I have proved the following claim: 
Let $G$ be a finite group, and let $F$ be a field of characteristic not dividing $|G|$. Then all the irreducible representations of $G$ over $F$ are $1$-dimensional if and only if $G$ is an abelian group and $F$ has a primitive root of unity of order $\mathrm{exp}(G)$.

Is this a known result? Is there a reference to it?

However, I am not sure what happens if the characteristic of $F$ divides $|G|$. For example, $S_3$ is a nonabelian group with all irreducible representations over $\mathbb{F}_3$ $1$-dimensional. 
One verifies that all the irreducible representations are characters if and only if the image of $G$ in $\mathrm{GL}_{|G|}(F)$ under the representation coming from the action of $G$ on itself can be simultaneously triangulated. This implies that again $F$ must have a primitive root of certain orders, and that $G$ must be solvable to say the least (it is embedded in the subgroup of upper triangular invertible matrices). 
In spite of that, I do not have a converse.

Is there a characterization (similar to those given above) of the
  cases in which a finite group has only 1-dimensional irreducible
  representations over a field $F$ of characteristic dividing $|G|$.

Hasn't this question been addressed before?
Both references and proofs are very much appreciated.

REPLY [12 votes]: A group satisfies this property if and only if it is an extension of a $p$ group by an abelian group. The reason is the following: Assume that indeed all the irreducible representations of $G$ are one dimensional. This means that all elements of the form $x-1$ where $x$ belongs to $[G,G]$ are in the Jacobson radical, because they act trivially on all irreducible representations. But since this is a finite dimensional algebra over a field, the Jacobson radical is nilpotent. This means that the element $x-1$ is nilpotent for every $x\in [G,G]$, and this can only happen if the order of $x$ is a power of $p$. Therefore, $[G,G]$ is a $p$-group, and the quotient $G/[G,G]$ is abelian. In particular, this implies that the group is solvable.<|endoftext|>
TITLE: Conditions for underlying space of an orbifold $\Bbb T^n/\Gamma$ to be a sphere?
QUESTION [8 upvotes]: Given a $n$-dimensional torus, is it always possible to find a discrete action to produce an orbifold such that its underlying space is the $n$-dimensional sphere? Or does it only happens for specific dimensions? Is there some particular property of the action $\Gamma$ -which I expect to be some realization of $\Bbb Z_n$- that I should check?
Bonus question: And to be a projective plane? I am thinking that according Arnold-Kuiper-Massey theorem one can see $\Bbb CP^2$ as a branched covering of $S^4$, so if $T^4$ could be a branched covering of $\Bbb CP^2$ we would have an interesting way to cover $S^4$
The personal motivation for this is, as usual, more fuzzy that the question itself but it could add some heat to the comments: I am interested on how to sequence manifolds in a way that increases some continuous symmetry. For manifolds with a metric the two extreme cases in a given dimension would be the n-Torus, where you can act with group $U(1)^n$ and the n-sphere, where you act with $SO(n+1)$. Of course the sphere that comes out from orbifolding is not exactly the sphere, one must consider the conical singularities and all the metric properties go amok, but still it is, to me, a hint of how do symmetries appear and disappear. For physics this is specially funny in dimension seven because you have also the the manifold $CP^2\times S^3$ where the isometry group has 14 generators and then you could try to produce a sequence 7 -> 14 -> 28
EDIT: It seems than for n=2 the physicists call "pillows" to the spheres, in order to stress the existence of singular points. So the question is, when is the orbifold a n-pillow?

REPLY [3 votes]: Let me clarify my comment. $S_n$ acts on $T^n$ in the obvious way, and preserves the multiplication map $T^n \to T$, so acts on the fiber $T^{n-1}$ in a natural way. I think the quotient by this action is a simplex, the Weyl alcove of type $A_{n-1}$.
The quotient by $A_n$ is a double cover of this. So it is two simplices glued together. I hope that they are glued together in the obvious way, to form a sphere. But I don't see immediately how to see this.<|endoftext|>
TITLE: C$^*$-algebras isomorphic after tensoring with $M_n(\mathbb C)$
QUESTION [20 upvotes]: In 1977, Joan Plastiras gave a striking example of two non $*$-isomorphic C$^*$-algebras $\mathcal A$ and $\mathcal B$ such that $$\mathcal A \otimes M_2(\mathbb C) \simeq \mathcal B\otimes M_2(\mathbb C)$$ (http://www.ams.org/journals/proc/1977-066-02/S0002-9939-1977-0461158-9/S0002-9939-1977-0461158-9.pdf).
My question is this: can you find two non $*$-isomorphic C$^*$-algebras $\mathcal A,\mathcal B$ such that $$\mathcal A\otimes M_n(\mathbb C) \simeq \mathcal B\otimes M_n(\mathbb C), \ n=2,3?$$ 
Note that Plastiras' example does not satisfy this condition. Furthermore, can you find non $*$-isomorphic $\mathcal A, \mathcal B$ such that
$$\mathcal A\otimes M_n(\mathbb C) \simeq \mathcal B\otimes M_n(\mathbb C), \ \forall n\geq 2$$
or does this condition imply that $\mathcal A\simeq \mathcal B$?

REPLY [16 votes]: Such examples do exist. In the paper "Stability of C*-algebras is not a stable property,  Doc. Math. J. DMV , 2, (1997), 375-386.", Rordam gives examples of simple C*-algebras $A$ such that $M_2(A)$ is a stable C*-algebra but $A$ is not. Now take the pair $A$ and $B=M_2(A)$. These C*-algebras are not isomorphic, since $A$ is not stable, but $B$ is. On the other hand, $M_n(A)$ is isomorphic to $M_n(B)$ which is isomorphic to $A\otimes \mathcal K$ for all $n\geq 2$. (Once $M_n(A)$ is stable for some $n$, it remains stable for all larger $n$. This is proven in the same paper.)
Part of the hard work in Rordam's paper is to find an example as above with $A$ simple. At the heart of Rordam's construction is the following non-simple example:   There exists a countable generated Hilbert C*-module $H$ over $C(X)$, with $X=\prod_{i=1}^\infty S^2$ (an infinite product of 2-spheres) such that $H$ is not isomorphic to $\ell^2(C(X))$
but $H\oplus H$ is. Then one can choose $A=K(H)$ and $B=K(H\oplus H)$. $A$ is not stable since this is equivalent to $H\cong \ell^2(C(X))$ but $M_2(A)\cong B$ is stable.
This is only a comment:
It seems to me that it should be possible to find (hermitian) complex vector bundles $\eta_1$ and $\eta_2$ over some space $X$ such that $\eta_1^{n\oplus}\cong \eta_2^{n\oplus}$ for all $n\geq 2$ but $\eta_1\ncong \eta_2$. Then $A=\mathrm{End}(\eta_1)$ and $B=\mathrm{End}(\eta_2)$ are candidate examples. From $\eta_1^{n\oplus}\cong \eta_2^{n\oplus}$ for all $n\geq 2$ we can already deduce that $M_n(A)\cong M_n(B)$ for all $n\geq 2$. If we choose them with some care (they can't be line bundles) then we should also have that $A\ncong B$.<|endoftext|>
TITLE: What were the main ideas and gaps in Yoichi Miyaoka's attempted proof (1988) of Fermat's Last Theorem?
QUESTION [60 upvotes]: Out of sheer curiosity I have been reading Stewert and Tall's "Algebraic Number Theory and Fermat's Last Theorem" (2001). As it contains various bits of history, I found out to my own shame that I was not even aware of the fact that there was another serious attack at FLT in the 80s - that of Y. Miyaoka. 
Now, we usually don't pay much attention to purported proofs from cranks and obvious amateurs of famous open conjectures, but since this came from a serious mathematician, the issues with his attempted proof can only be instructive in terms of learning value. It is a good thing to learn not only from one's own mistakes but from the mistakes of others too.
So I started searching for more information, but to my surprise (and perhaps understandably) I have not been able to find any details for the past day (except that the author used techniques from Differential Geometry, which is still way too generic).
Hence my question:

What were the main ideas and the respective gaps in Miyaoka's attempted proof of FLT?

Addendum: As can be seen from Timothy Chow's answer, the statement in Stewert and Tall's book that "Miyaoka had used a technique parallel to that of Wiles, by translating the number-theoretic problem into a different mathematical theory — in this case, differential geometry" is actually a bit misleading in that regard. For, in fact, he did the opposite - he tried to transfer notions from differential geometry (and related alg. topology) to the arithmetic world rather than give a differential-geometric (in a strict sense) proof. Sorry to get the hopes of our differential geometers too high!

REPLY [45 votes]: Here's some information from Barry Cipra's June 1988 article "Fermat's Theorem remains unproved" in Science magazine.

Parshin showed that the arithmetical version of a certain inequality involving geometric invariants of surfaces—an inequality that Miyaoka proved for the geometric case in 1974—would lead by a series of steps to a bound on the size of possible exponents for which Fermat's Last Theorem could be false. $\ldots$
Miyaoka's work is directed at proving the arithmetical inequality.  Miyaoka, who is an expert in algebraic geometry but a relative newcomer to the arithmetical theory, proceeded by analogy with the geometric case.  But according to Enrico Bombieri, a professor of mathematics at the Institute for Advanced Studies [sic] in Princeton, the translation is not straightforward. "Things go over, but with some qualifications," Bombieri says.  "The naïve extension doesn't go through."
The problem, according to Barry Mazur of Harvard University, is the lack of a good arithmetical analog of a crucial geometric object known as the tangent bundle.  Mazur, who helped Miyaoka analyze the proof, explains that Miyaoka had "a very interesting idea" to replace the tangent bundle with a "generic" bundle, with the assumption that the generic bundle can be chosen so as to have suitably nice properties.  This seems not to be the case.
The effort is not wasted, however, Mazur says that Miyaoka has carried the idea of substituting generic bundles for the tangent bundle back to the original geometric case.  "Given any choice of a bundle, you'll get some inequalities," Mazur says. "It's a perfectly reasonable and interesting geometric question to ask what's the structure of this whole complex set of inequalities."  Answering such questions will very possibly lead to a deeper understanding of Miyaoka's original geometric proof.

More information about how the inequality in question (known as the Bogomolov–Miyaoka–Yau inequality) relates to Fermat's Last Theorem can be found in the appendix (by Paul Vojta) to Serge Lang's book Introduction to Arakelov Theory.<|endoftext|>
TITLE: Rearrangements that never change the value of a sum
QUESTION [26 upvotes]: I posted this question on math.stackexchange.com and so far the only answer posted (also mentioned in the comments under the question) shows that one of my rash initial guesses about the bottom-line answer was wrong.
Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$,
$$
\lim_{n\to\infty} \sum_{k=1}^n a_k = \lim_{n\to\infty} \sum_{k=1}^n a_{f(k)},
$$
where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?
Here's another rash initial guess, different from the one I posted on stackexchange: It's the bijections whose every orbit is finite.
(That there are uncountably many such bijections can be seen as follows: For each odd number $n$, let $f$ either fix $n$ and $n+1$ or interchange them.  That bijection never changes the values of sums.  It's a countably infinite sequence of binary choices, so it's uncountable.)

REPLY [19 votes]: For the purpose of recording an answer rather than just a pile of links: Michael Hardy requires that

if either limit exists then so does the other and in that case then they are equal?

Let's call this set $G$.
Levi answered a slightly different question, namely characterizing the permutations for which

if the left hand limit exists then so does the right and in that case then they are equal?

I'll call that $P$. Clearly, $G = P \cap P^{-1}$.
Theorem A permutation $f$ is in $P$ if and only if there is a constant $M$ such that, for any $N$, the set $f([1,N])$ can be written as $\bigcup_{i=1}^M [a_i, b_i]$, where $[a,b] = \{ a,a+1, \ldots, b \}$.
Levi provided a different characterization than this one; Agnew provided this characterization; I learned about both from Schaefer who points out that they are fairly directly equivalent.
Pleasants shows that $P$ is not closed under inversion. I haven't (in an one hour skim) found any papers which give a simpler characterization of $P \cap P^{-1}$ than the defining formula.
Remark I would find it nicer to restate Levi/Agnew's characterization as follows: For $S \subseteq \mathbb{N}$, define the blockiness of $S$ to be the least integer $\beta(S)$ such that $S$ can be written as $\bigcup_{i=1}^{\beta(S)} [a_i, b_i]$. (We could have $\beta(S) = \infty$.) Then $f$ is in $P$ if and only if there is a constant $M$ such that $\beta(f(S)) \leq M \beta(S)$. This makes it more obvious that $P$ is closed under composition.<|endoftext|>
TITLE: Number of representations as sums of squares in rings of integers of number fields
QUESTION [6 upvotes]: Let $K$ be some number field, $\mathcal O_K$ denote its ring of integers, and let $n$ be a positive integer. Take $\alpha \in \mathcal O_K$, and consider the quantity $r_{n,K}(\alpha)$, which denotes the number of solutions to the equation
$$
\alpha = x_1^2 + x_2^2 + \ldots + x_n^2
$$
with $x_1, x_2, \ldots, x_n \in \mathcal O_K$.
When $K=\mathbb Q$ and $\mathcal O_K=\mathbb Z$, the precise formulas for the computation of $r_{n,K}(\alpha)$ were developed for all $n \leq 4$ (perhaps for other $n$'s as well, but I am not familiar with them). Are there any formulas for $r_{n,K}(\alpha)$ when $[K:\mathbb Q] > 1$? I am especially interested in the cases when $n = 2, 3, 4$ and $K$ is a real quadratic extension of $\mathbb Q$.

REPLY [2 votes]: For $n=4$, in addition to what Jeremy mentioned, there is some addition information in the answers to sum of squares in ring of integers.  John Goes has also done some work but it seems his preprint is not yet available.
There are also some results for $n=2$ and $n=3$.  
For $n=3$, see Donkar's 1977 American Journal paper.  He uses quaternion algebras over number fields like he did earlier over $\mathbb Q$.  He has fairly general results with just an assumption on the even primes of the number field, and he worked out some very explicit statements in several examples such as $\mathbb Q(\sqrt 5)$, $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 17)$.
For $n=2$, there has at least been work for $\mathbb Q(\sqrt 2)$.  See Michele Elia & Chirs Monico's paper On the Representation of Primes in $\mathbb Q(\sqrt 2)$ as Sums of Squares (JP Journal of Algebra, Number Theory, 2007).
(There has been other work over imaginary quadratic fields when $n=2$, but I'm not sure about other results over totally real fields right now.  There are also results for $n>4$.)<|endoftext|>
TITLE: A Collatz-like question about permutations
QUESTION [14 upvotes]: An answer to this question would provide an explicit counterexample to this question, but otherwise I don't know if it is interesting.
Consider all permutations $\pi$ on the natural numbers such that for each $i$, $\pi(3i)=2i$ and $\lbrace \pi(3i+1),\pi(3i+2)\rbrace = \lbrace 4i+1,4i+3\rbrace$.
The question: Is there such a permutation for which all the cycles have finite length?
By computer it is easy to find choices that make the cycles containing the first several thousand integers finite.

REPLY [4 votes]: This is something between an extended comment and an answer, and might contain nothing that you haven't known already, but I am sure that the answer to your question is yes.
The reason is that there are only countable many cycles, so we can take care of them one by one to make them finite.
Suppose up to some point all cycles containing numbers less than $n$ are finite and consider the cycle of $n$.
It has two ends and at both ends we can make choices by picking the yet undecided values of $\pi(3i+1)$ and $\pi(3i+2)$ as we like.
By making some choices, after a while the two ends will either meet, or both go to some very large values, which means that we can make many choices, as in that high range nothing was decided earlier.
Since for your parameters $\frac 32 \cdot \frac 34<2$, the graph of the choices will behave like an expander, so the two ends can be made to meet.<|endoftext|>
TITLE: Can hypercomplete objects be coreflective?
QUESTION [6 upvotes]: The subcategory of hypercomplete objects in an ∞-topos is a left-exact-reflective subcategory by the remarks after 6.5.2.8 of Higher topos theory.  Can it ever happen that this subcategory is also coreflective?

REPLY [12 votes]: Let $\mathcal{X}$ be the $\infty$-category of $1$-excisive functors from (pointed) spaces to (unpointed) spaces: equivalently, the $\infty$-category of pairs $(X, E)$ where $X$ is a space and $E$ is a local system of spectra on $X$. Then $\mathcal{X}$ is an $\infty$-topos (for example, it's a left exact localization of the $\infty$-category of functors from finite pointed spaces to spaces, via the Goodwillie calculus).
The homotopy groups of an object $(X,E) \in \mathcal{X}$ are just the homotopy groups of $X$, and the pair $(X,E)$ is hypercomplete if and only if $E = 0$. The construction $(X,E) \mapsto (X,0)$ is both left and right adjoint to the inclusion of hypercomplete objects into all of $\mathcal{X}$.<|endoftext|>
TITLE: Can the Laplace operator on $n-$ manifolds be represented as a sum of $n$ second order derivational operators
QUESTION [18 upvotes]: EDIT: According to some comments on this post I revise the title to remove the misunderestanding.
Assume that $M$  is a  Riemannian manifold of dimension $n$. The natural Laplace operator associated to the metric is denoted by $\Delta$.

Are there $n$ vector fields $X_{1},X_{2}, \ldots, X_{n}$  such that $\Delta=\sum \partial^{2}/\partial X _{i}^{2}$?
Are there some local obstructions?(However our question search for  global vector fields $X_{i}s$)

The obvious motivation for this question is the usual metric on $\mathbb{R}^{n}$

REPLY [5 votes]: The point of this answer is to flesh out my comment above: Any such $X_i$ must be everywhere linearly independent, so they only exist if the manifold $M$ is parallelizable. Suppose, for the sake of contradiction, that the $X_i$ become linearly dependent at some point $p$. Pass to local coordinates $(x_1, \ldots, x_n)$ with $p=(0,0,\ldots, 0)$ and write
$$X_i = \sum_j a_{ij}(x) \frac{\partial}{\partial x_j}.$$
Let $A(x)$ denote the matrix $(A_{ij}(x))$. Our hypothesis is that $A(0)$ is not of full rank.
We compute the principal symbol of $\sum (\partial/\partial X_i)^2$. We have 
$$\left( \sum_j a_{ij}(x) \frac{\partial}{\partial x_j} \right)^2  = \sum_{j_1, j_2} a_{i j_1}(x) a_{i j_2}(x) \frac{\partial^2}{(\partial x_{j_1}) (\partial x_{j_2})} + \mbox{first order operators}.$$
So the principal symbol is
$$\sum_i \sum_{j_1, j_2} a_{i j_1}(x) a_{i j_2}(x) y_{j_1} y_{j_2} = y^T A^T(x) A(x) y$$
where $y$ is the vector $(y_1, y_2, \ldots, y_n)^T$. If $A$ is not of full rank, neither is this quadratic form.
But the symbol of $\Delta$ is positive definite, a contradiction.<|endoftext|>
TITLE: Fredholm operators in $K$-theory?
QUESTION [5 upvotes]: Do Fredholm operators show up in K-theory? Why or why not? The idea of infinite Grassmannians classifying vector bundles is pretty straightforward, but why would adding in additive inverses and what not give you this? Is it a generalization of some nice finite-dimensional concept? Does it have a deep connection to (finite-dimensional) vector bundles?

REPLY [13 votes]: A classical connection is the Atiyah-Jänich Theorem, see 
Klaus Jänich: Vektorraumbündel und der Raum der Fredholm-Operatoren. Math. Ann. 161 (1965) 129–142. 
Let ${\mathcal{F}}$ be the space of Fredholm operators with the operator norm, and $X$ any compact space. To a map $F\colon X\to {\mathcal{F}}$ one can associate the virtual vector bundle $(ker(F(x)))_x-(coker(F(x)))_x$ from Qiaochu Yuan's comment. (In general, there might be points where dimension of kernel and cokernel jump, but upon homotopy of $F$ one can assume that kernel and cokernel are indeed vector bundles.) 
Jänich proves that this yields a bijection $$\left[X,{\mathcal{F}}\right]\cong K(X),$$
where $\left[.,.\right]$ means homotopy classes.<|endoftext|>
TITLE: Do binary symmetric channels maximize mutual information?
QUESTION [6 upvotes]: Consider the following setup: $(X, Y)$ is a doubly symmetric binary source with parameter $0 < p < 1/2$, i.e., $X \sim \text{Bernoulli}(1/2)$, $Z \sim \text{Bernoulli}(p)$ and $Y = X \oplus Z$. Let $(U, V)$ be another pair of binary variables, where $U$ is derived from $X$ and $V$ is derived from $Y$, i.e., the Markov chain $U - X - Y - V$ holds. My question is the following:
Can you find binary random variables $U'$ and $V'$ such that $U' - X - Y - V'$ form a Markov chain and the channels $X \rightarrow U'$ and $Y \rightarrow V'$ are binary symmetric channels with ($I(\cdot;\cdot)$ denotes  mutual information.)
$I(U';X) \le I(U;X)$, $I(V';Y) \le I(V;Y)$, and $I(U'; V') \ge I(U;V)$ ?
The question can also be phrased in terms of the probabilities of the binary symmetric channels: Are there values $0 \le r,q \le 1/2$ such that
$1 - h(r) \le I(U;X)$, $1 - h(q) \le I(V;Y)$, and $1 - h(r * p * q) \ge I(U;V)$ ?
($h(x) = -x \cdot \log_2(x) - (1-x) \cdot \log_2(1-x)$ is the binary entropy function and $p * q = p*(1-q) + (1-p)*q$ the binary convolution.)
I did some numerical evaluation which suggests a positive answer to that question: Choosing $r,q$ such that $1 - h(r) = I(U;X)$ and $1 - h(q) = I(V;Y)$, it seemed that $1 - h(r * p * q) \ge I(U;V)$ is always satisfied. But I don't know how to formally prove this statement. I wasn't lucky trying to apply Mrs. Gerber's Lemma in this context. I also tried splitting the problem in two stages: First fix $V'=V$ and try to show that a BSC $X \rightarrow U'$ with $I(U';X) \le I(U;X)$ maximizes $I(U'; V')$. But I have a counterexample to this claim, so it appears necessary to consider both channels simultaneously.

REPLY [2 votes]: No. You can get a higher $I(U;V)$ using asymmetric channels. Below I construct a counterexample, but first a more succinct restatement of the question.

Restatement
To summarize, there is an input $U$ distorted by three independent binary hops, each described by $2\times 2$ stochastic matrices $C_L,\ C,\ C_R.$ Labeling all your RV's, $$U\overset{C_L}{\to}X\overset{C}{\to}Y\overset{C_R}{\to}V.$$ 
We are interested in maximizing $I(U;V)$ 
 subject to constraints: 

$C$ is determined by nature.
$C_L$ is such that $I(U;X) \leq r_L$, 
$C_R$ is such that $I(Y;V) \leq r_R$,
$U$'s distribution is such that the left channel's output (i.e. $X$) is $B(1/2)$.


Counterexample
You claim the $C_L,C_R$ which produce the maximum are binary symmetric. 

If $C_L$ is a BSC with $B(1/2)$ output then its input must also be $B(1/2).$ For a given rate $r_L \leq 1$, then there is at most one 'positive' (i.e. can't be improved by relabeling the outputs) BSC $C_L$ whose output is $B(1/2).$  
You have assumed $C$ is a BSC, so with a symmetric input its output is also symmetric. 
For a rate $r_R\leq 1$ there is only one positive choice for $C_R.$

So to say they are binary symmetric is to determine all of $U,C_L$ and $C_R$.
Now take $C$ a perfect channel, 
$C= \left[\begin{smallmatrix}
    1 & 0 \\
    0 & 1
    \end{smallmatrix}\right]$ and $r_L=r_R=0.4.$ The associated positive BSC for this rate has crossover probability $\approx 0.15,$ and the end-to-end mutual information can be computed: 
$$I(U_{BSC}, V_{BSC})< 0.1895$$
However, trying randomly[1] you can find $U^\ast, C_L^\ast, C_R^\ast$ that satisfy all the mutual information constraints, but have greater $U$-to-$V$ mutual information. One I found happens to be quite close to a Z-channel:
\begin{equation}
   I(U^\ast; V^\ast) > 0.19,
\end{equation}
\begin{equation}
   C_L^\ast \approx \left[\begin{smallmatrix}0.2493 & 0.7507 \\ 0.9657 & 0.0343 \end{smallmatrix}\right], \qquad
    C_R^\ast \approx \left[\begin{smallmatrix} 0.9821 & 0.0179 \\ 0.3374 & 0.6626 \end{smallmatrix}\right], \qquad
   U^\ast \sim B(0.35)
\end{equation}

Discussion
This result is to be expected since there is a vague sense that uniform noise over a bounded space is the most degrading, even holding mutual information fixed. (by one heuristic "uniform noise means you can't precode to mitigate it")
A gentle introduction for a good, visualisable framework for studying binary symmetric channels is given in a short paper, Algebraic Information Theory for Binary Channels by Martin, Moskowitz and Allwein. Under this framework your maximization can be restated as a convex optimization problem for which I see no easy special cases. 
An easier-to-investigate (and arguably more interesting)  problem is one identical to yours that omits the fourth constraint that $X\sim B(1/2)$. But I could not find an easy path towards an answer for this either.  
For both of these there might be some magical connection to KL divergence which I am not seeing.

Code
[1]: Below is a crude counterexample finder. 
% Helper functions
    % Binary entropy
fn_h = @(p) -p.*log2(p) - (1-p).*log2(1-p); 
    % MI across mtx_bc when v_distn is input
fn_I = @(mtx_bc,v_distn) fn_h(v_distn(1)) + fn_h(v_distn*mtx_bc(:,1)) ...
    - nansum(nansum(-log2(diag(v_distn)*mtx_bc).*(diag(v_distn)*mtx_bc)));
    % Channel matrix when P(out=0|in=0)=pa, P(out=0|in=1)=pb
fn_mtxBC = @(pa,pb) [pa, 1-pa; pb, 1-pb];

% Set params
d_r_L = 0.4; 
d_r_R = 0.4;
d_xp = 0.146102; % solution to 1-H(p) = 0.4
mtxBSC = fn_mtxBC(d_xp, 1-d_xp);

% Search 
while true
    mtxL = fn_mtxBC(rand, rand);
    mtxR = fn_mtxBC(rand, rand);
    v_d = (mtxL'\[0.5, 0.5]')';
    if (abs(sum(v_d)-1) > 0.001 || ...
        min(v_d) < 0)
        continue
    end
    if(fn_I(mtxL, v_d)      > 0.4 || ...
       fn_I(mtxR, v_d*mtxL) > 0.4)
        continue;
    end
    fprintf('+\n');
    if fn_I(mtxL*mtxR, v_d) > fn_I(mtxBSC*mtxBSC, [0.5, 0.5])
        break;
    end
end<|endoftext|>
TITLE: What is the homomorphism between the third exterior and third symmetric power of the adjoint representation of a simple Lie algebra?
QUESTION [19 upvotes]: Let $\mathfrak{g}$ be the adjoint representation of a simple Lie algebra (which is not of type $A$). Then the space of intertwiners between the third exterior power of $\mathfrak{g}$ and the third symmetric power of $\mathfrak{g}$ has dimension one. I have checked on a case-by-case basis using LiE that these two representations have a single irreducible component in common.
I am asking for a natural non-zero element in this space; where by natural I mean constructed from the Lie bracket, permutations, the Killing form and its adjoint. If you can give this in diagrammatic notation that would be even better.
The question stands independently of the motivation. However my motivation is that I am trying to understand Drinfeld's "terrific relation" in the definition of the Yangian. This relation can be interpreted as saying that two actions of the common irreducible representation agree.
I am looking at "A guide to quantum groups" by Chari & Pressley where the "terrific relation" is (4) in Theorem 12.1.1 page 376 (Google Books).
A related question is Lie algebra cohomology.
The constructions work perfectly well in type $A$.
The reason type $A$ is exceptional (!) is that in type $A$ (and only in this case) $\mathfrak{g}$ appears as a composition factor of the symmetric square $S^2\mathfrak{g}$. The projection is given by
$$
x\otimes y \mapsto xy+yx -\frac{2}{n}\mathrm{trace}(xy).1
$$
where $x,y$ are $n\times n$ matrices with zero trace.
This muddies the water.
Update I was confused when I asked the original question. The dimension
of the space of intertwiners seems to 2 for the exceptional groups, 3 for classical groups and 4 for $\mathfrak{sl}(n)$.
However the representation that is relevant in this construction of the Yangian is the kernel of the Lie bracket. This is an irreducible representation of $\mathrm{Aut}(\mathfrak{g})$ and is nonzero except in type $A_1$. Since $H^2(\mathfrak{g})=0$ this representation also appears in the third exterior power.
Then the right hand side of the "terrific relation" shows that this representation also appears in the third symmetric power. This is explained in Robert Bryant's answer.

REPLY [25 votes]: There is a natural homomorphism that I am pretty sure is not zero in general that can be described most easily using  the dual language of the Lie algebra $\frak g$:  
Recall that there is a differential $\mathrm{d}:{\frak{g}}^\ast\to \Lambda^2({\frak{g}}^\ast)$ that extends uniquely to a degree-one, square-zero derivation of $\frak{g}^\ast$.  It is defined by $\mathrm{d}\alpha(x,y) = -\alpha\bigl([x,y]\bigr)$ for all $\alpha\in \frak{g}^\ast$ and $x,y\in \frak{g}$.
Define a mapping
$$
P:S^3(\frak{g}^\ast)\to \Lambda^6(\frak{g}^\ast)
$$
by the rule
$$
P(\alpha_1\circ\alpha_2\circ\alpha_3) = 
\mathrm{d}\alpha_1\wedge\mathrm{d}\alpha_2\wedge\mathrm{d}\alpha_3\,.
$$
Using the Lie bracket one defines the Cartan $3$-form $\phi\in \Lambda^3(\frak{g}^\ast)$ by
$$
\phi(x,y,z) = \kappa\bigl(x,[y,z]\bigr),
$$
where $\kappa$ is the Killing form of $\frak{g}$.
This can be used to define a mapping
$$
\Phi: \Lambda^3(\frak{g}^\ast)\to \Lambda^6(\frak{g}^\ast)
$$
by $\Phi(\psi) = \phi\wedge\psi$
whose image is a proper subspace of $\Lambda^6(\frak{g}^\ast)$, except for the cases $A_1$ and $A_2$.  
Now, using the natural inner product on $\Lambda^6(\frak{g}^\ast)$ induced by the Killing form, define an orthogonal projection $\pi:\Lambda^6(\frak{g}^\ast)\to \Phi\bigl(\Lambda^3(\frak{g}^\ast)\bigr)$.  Then $\pi\circ P\bigl(S^3(\frak{g}^\ast)\bigr) =\Phi\bigl(\Lambda^3(\frak{g}^\ast)\bigr)$, and, as long as these two spaces are not zero (and I think that they are not in general), this will define a map of the kind you seek.  
$\Phi$ is not injective since $\phi$ is in its kernel, so you will have to use the adjoint of $\Phi$ (using the inner product induced by the Killing form) to construct the map explicitly.
Alternatively, you could just define the pairing
$$
\beta: S^3(\frak{g}^\ast)\times \Lambda^3(\frak{g}^\ast)\longrightarrow \mathbb{C}
$$
by 
$$
\beta(p,\psi) = \kappa\bigl(P(p),\Phi(\psi)\bigr)
$$
where I have extended $\kappa$ in the usual way to the exterior powers of $\frak{g}^\ast$, and, noting that both spaces are, of course, self-dual, verify that this pairing is not identically zero, which will then give the map.
P.S.  I haven't checked all the details of this, but I believe that the following strategy will yield a (highly unsatisfactory) proof that the pairing $\beta$ is nonzero in all cases except $A_1$:  First, check, by hand, that the pairing is nonzero for $A_2$, $B_2=C_2$ and $G_2$.  (This is not hard to do.  $A_2$ is just a few lines, and $B_2$ is only a little longer.  I haven't checked $G_2$.)  Second, suppose that $\frak{g}$ is simple and that $\frak{h}\subset\frak{g}$ is a simple sub algebra such that the pair $(\frak{g},\frak{h})$ is an irreducible symmetric pair.  Using the fact that the Killing form of $\frak{g}$ restricts to be a non-zero multiple of the Killing form of $\frak{h}$ while the Cartan $3$-form of $\frak{g}$ restricts to be a nonzero multiple of the Cartan $3$-form of $\frak{h}$, one shows that the corresponding restriction of $\beta_{\frak{g}}$ is a nonzero multiple of $\beta_{\frak{h}}$.  Thus, if $\beta_{\frak{h}}$ is not zero, then $\beta_{\frak{g}}$ is also not zero.  Finally, any simple Lie algebra other than $A_1$ and the $C_n$ $(n>2)$ is the end of a chain of symmetric pairs that starts with either $A_2$, $B_2$, or $G_2$, so the result follows for all of these except the $C_n$.  However, $C_n$ contains $\mathbb{C}+A_{n-1}$ as a symmetric subalgebra, and I think that the above inductive argument can be extended to this case.<|endoftext|>
TITLE: Ranks of elliptic curves depend only on the field?
QUESTION [8 upvotes]: Let $K/\mathbb{Q}$ be an algebraic extension, and let $E_1,E_2/\mathbb{Q}$ be elliptic curves. Is it possible that the Mordell-Weil rank of $E_1(K)$ is finite while that of $E_2(K)$ is infinite?

REPLY [12 votes]: The answer is yes, at least if you assume that Tate-Shafarevich conjecture.
Let $E$ be an elliptic curve over a number field $k$. Under some mild hypothesis (see Corollary 1.10 of http://arxiv.org/abs/0904.3709) there exist quadratic twists of $E$ which have trivial Mordell-Weil rank. In other words, there exists quadratic extensions in which the rank of $E$ does not go up. Assuming in addition the Tate-Shafarevich conjecture there are also quadratic twists of $E$ whose Mordel-Weil rank is $1$ (see Corollary 1.12 loc. cit.), and hence quadratic extensions in which the rank of $E$ goes up. Generalizing the methods of loc. cit. along the line of http://arxiv.org/abs/1504.02343 one can show that (assuming the Tate-Shafaretich conjecture) if $E_1$ and $E_2$ are elliptic curves over $k$ satisfying certain independence conditions then one can find a quadratic extension in which the rank of $E_1$ stays the same and the rank of $E_2$ goes up. Applying this construction successively one can construct an infinite extension $K/k$ such that the rank of $E_1(K)$ is finite and the rank of $E_2(K)$ is infinite.

REPLY [4 votes]: If you're willing to replace $\mathbb Q$ by a finite (quadratic) extension $k$, then my recollection is that there are $\mathbb Z_p$ extensions $K$ of $k$ such that some elliptic curves over $k$ have finite Mordell-Weil group over $K$ and some have infinite rank. In any case, it would be worth looking up what's known about rank growth in $\mathbb Z_p$-extensions, including for $k=\mathbb Q$, starting with the work of Mazur.<|endoftext|>
TITLE: What pairs of sets have additive energy?
QUESTION [6 upvotes]: In an abelian group, the additive energy between two sets is $$E(A,B)=|\{(a_1,a_2,b_1,b_2)
\in A\times A\times B\times B:a_1+b_1=a_2+b_2\}|$$ which is ranges from $|A||B|$ to $(|A||B|)^{3/2}$. What I want to know is, given that this is large, should $A$ look like $B$? 
To keep things simple, let's say all sets in question consist of integers. From the symmetry bound $E(A,B)^2\leq E(A,A)E(B,B)$ we see that the "best" additive partner for $A$ is $A$ itself, or any translate of $A$. From Plunnecke-Ruzsa, if we have small numbers $m,n$ then when $E(A,A)$ is (very) large, $E(mA,nA)$ is large, so that $dA+t$ is a good partner for $A$ as long as $d$ is a rational of small height.
If $r(x)$ is the number of solutions to $x=a+b$ with $a\in A$ and $b\in B$ then the fact that $$E(A,B)=\sum_x r(x)^2$$ tells us that if $E(A,B)\geq K|A||B|$ then there is an $x$ so that $|A\cap x-B|\geq K$. So $B$ contains a subset which looks like (i.e. is a translate of) a subset of $A$. Can we do better?

REPLY [6 votes]: If $\lvert A\rvert \approx \lvert B\rvert$ and $E(A,B)\geq K^{-1}\lvert A\rvert\lvert B\rvert^2$, then the Balog-Szemeredi-Gowers theorem yields large subsets $A'\subset A$ and $B'\subset B$ such that $\lvert A'-B'\rvert\ll K^{O(1)}\lvert A\rvert$. This is essentially the best result possible; from this one can deduce using covering lemmas that $A$ and $B$ are both sets of small doubling, each contained in $K^{O(1)}$-many translates of the other, which is essentially best possible.
The case where $\lvert B\rvert$ is much smaller than $\lvert A\rvert$ is more difficult. Tao and Vu have an asymmetric version of the Balog-Szemeredi-Gowers theorem (their Theorem 2.35) to handle this case: 

Suppose $E(A,B)\geq K^{-1}\lvert A\rvert\lvert B\rvert^2$, and $L^{-1}\lvert A\rvert \leq \lvert B\rvert\leq \lvert A\rvert$. Then, for any $\epsilon>0$, there are sets $H$ and $X$ such that 

$\lvert H-H\rvert \ll L^\epsilon K^{O(1)}\lvert H\rvert$,
$\lvert X+H\rvert=\lvert X\rvert\lvert H\rvert \ll L^\epsilon K^{O(1)} \lvert A\rvert$, 
$\lvert A\cap (X+H)\rvert \gg L^{-\epsilon}K^{-O(1)}\lvert A\rvert$, and
there is some $x$ such that $\lvert (B-x)\cap H\rvert \gg L^{-\epsilon}K^{-O(1)}\lvert B\rvert$,

where all the implicit constants depend on $\epsilon$.

That is, up to polynomial losses in $K$, the set $B$ can be approximated by a set $H$ with small doubling constant, and $A$ is roughly a disjoint set of translates of $H$. This is a generalisation of your observation, which says (adjusting notation slightly to fit),

If $E(A,B)\geq K^{-1}\lvert A\rvert\lvert B\rvert^2$ then there is an $x$ such that $\lvert (B-x)\cap A\rvert\geq K^{-1}\lvert B\rvert$,

so that not only does $A$ contain a translate of $B$, but also that $A$ is essentially a union of translates of $B$.
The dependence of the constants on $\epsilon$ is a little delicate, but can be worked out from the proof in Tao and Vu. I believe that one can take the constants in the $\ll,\gg$ to be independent of $\epsilon$, and replace
$$ K^{O_\epsilon(1)}\quad\textrm{ by }\quad K^{O(\exp(\epsilon^{-1}))},$$
where the constant is now absolute, provided
$$ \epsilon \gg \log\log L/\log L;$$
this is only a back of the envelope calculation though, and for any particular application one can probably do better by unpacking the proof in Tao and Vu.<|endoftext|>
TITLE: Zhang's generalization of Gross-Zagier formula
QUESTION [9 upvotes]: I have some naive questions about Zhang's generalization of Gross-Zagier formula stated  in Theorem 1.2 of  Yuan-Zhang-Zhang book The Gross-Zagier formula on Shimura Curves. 
I am mostly interested by the cases  of  non split cartan modular curves and  Shimura curves $X_D(N)$. I understand that the "incoherent Shimura curves" setting of Zhang contains those cases but 
1) could you explain me what are concretely the formulas in those cases ?
2) does this formula holds directly on the curve $X_U$ instead of on an abelian variety parametrized by $X_U$? 
Thank you very much. 
[edit : I copy here the statement of Theorem 1.2 (but without detailing the notations) : 
Assume $\omega_A.\chi|_{A_F^\times}=1$ [which ensures that $P_\chi(f) \neq 0$]. For any $f_1\in\pi_A, $ and $f_2\in\pi_{\check A}$, 
$$ \langle P_\chi(f_1),P_\chi(f_2)\rangle _L= \frac{\zeta_F(2) L'(1/2,\pi_A,\chi)}{4L(1,\eta)^2L(1,\pi_A,ad)}\alpha(f_1,f_2)$$ 
as an identity in $L\otimes_{\mathbb Q} \mathbb C$. Here $\langle,\rangle_L$ is the $L$-linear Neron-Tate height pairing. ]

REPLY [4 votes]: Could you explain me what are concretely the formulas in those cases?

These formulas are equalities, so their content is that the left-hand side is equal to the right-hand; a tautology for sure, but an interesting one once the left-hand side and right-hand side are properly understood. So let's them examine them in turns.
The left-hand side is the height pairing of two Heegner points constructed from the choice of two vectors; one in the representation space $\pi_{A}$ and one in its contragredient (which is almost the same by essential self-duality). So the left-hand side is a bilinear form on the vectors of $\pi_{A}$, or more precisely on $\pi_A\otimes\check{\pi}_A$, globally constructed from algebraic geometry. Now the important term on the right-hand side is the bilinear form $\alpha(-,-)$, which is a product of local bilinear forms on $\pi_{v}\otimes\check{\pi}_{v}$ for all finite places $v$ of $F$. Hence, you a have an essentially global and geometric bilinear form (the intersection pairing between certain cycles of interest) and an essentially local and automorphic pairing (the convolution of local automorphic factors). The meaning of the Gross-Zagier formula in the style of Yuan-Zhang-Zhang is that these two bilinear forms are essentially the same, and insofar as they differ, it is only through the special value of some $L$-function (the most interesting term being $L'(\pi,1/2,\chi)$).
Hence, this result fits in the large family of results of the form: when two objects with the same properties are constructed for motives/galois representations/algebraic automorphic representations, they are in fact the same thing except their difference is a period/special value of $L$-function/etc.

Does this formula holds directly on the curve $X_U$ instead of on an abelian variety parametrized by $X_U$?

The formula you quoted depends on a choice of an automorphic representation $\pi$ of $\mathbf{G}(\mathbb A_{\mathbb Q}^{(\infty)})$ where $\mathbf{G}(\mathbb Q)$ is either $\operatorname{GL}_{2}$ or $B^\times$ where $B$ is a quaternion algebra isomorphic to $\operatorname{GL}_{2}(\mathbb R)$ at exactly one real place of $F$ (depending on the exact setting you are considering). This choice of $\pi$ corresponds in the usual way to a choice of an abelian variety in the Jacobian of the Shimura curve $X_{U}$ attached to $\mathbf{G}$, so the statement you quoted depends indeed of a choice of $A$, if you want to see it that way. You can formulate a version for the full Jacobian of $X_U$ (so before cutting out the part on which the Hecke operators act they way they have to). I'm not sure I understand what you mean by a statement "directly on $X_U$" itself, but as ABCDveve said, in that respect, there is no difference between the work of Yuan-Zhang-Zhang and the original work of Gross-Zagier.<|endoftext|>
TITLE: Connected but no path-connected components
QUESTION [11 upvotes]: Is there an infinite Borel subset of plane which is connected but whose only path connected components are singletons?
I know that a Bernstein set is a non-Borel example of such a set. Thanks!

REPLY [8 votes]: I think your assumption should include that the set contains at least two points, otherwise there is a trivial example ...
Eric already mentioned the pseudo-arc, which is of course a perfect answer. Allow me to add a historical discussion, and some more elementary examples.
In addition to the pseudo-arc, any hereditarily indecomposable continuum has your property, by definition. However, there are also many other plane continua (compact and connected subsets of the plane) with this property, including ones that are hereditarily decomposable. I believe Nadler's book on continuum theory has such an example in the exercises, but I do not have it to hand right now. 
The first construction of a continuum not containing any curves - a modification of the $\sin(1/x)$-continuum - seems to have been made by Janiczewski in 1912 ("Über die Begriffe Linie und Fläche", in the Proceedings of the 5th ICM). This continuum is hereditarily decomposable, and is probably the first example of a set with the properties you mention.
Since you are not asking for the set to be compact, easier examples are possible. Indeed, any space having an Explosion or dispersion point will trivially have the desired property. In particular, the Knaster-Kuratowski fan is a simple example. Another example is given by the set of endpoints of the Lelek fan, together with its top (which has an explosion point).<|endoftext|>
TITLE: Is a morphism whose all fibers are $\mathbf{P}^n$ a projective bundle?
QUESTION [12 upvotes]: Suppose $X\to Y$ is a morphism between varieties with all fibers isomorphic to $\mathbf{P}^n$, is $X$ a projective bundle over $Y$, i.e. $X=\mathbf{P}(E)$ for some vector bundle over $Y$?

REPLY [11 votes]: No that is not true, and I am confident somebody has written the following example previously on MO.  Let $\mathbb{A}^2$ have coordinates $s$ and $t$.  Let $Y\subset \mathbb{A}^2$ be the singular plane curve with defining equation $$f(s,t) = t^2-s^2(1-s) = 0.$$  Let $u$ be a coordinate on $\mathbb{A}^1$.  The normalization of $Y$ is $$\nu:\mathbb{A}^1 \to Y, \ \nu(u) = (1-u^2,(1-u^2)u).$$  Let $V\subset \mathbb{A}^1$ be the open complement of the closed point $p$ where $u=1$.  Let $X$ be $V\times \mathbb{P}^n$.  Denote by $$\text{pr}_V:V \times \mathbb{P}^n\to V,$$ the usual projection.  Then one counterexample is $\pi = \nu\circ \text{pr}_V$.  
Edit.  One reference for this problem is Exercise II.7.10 of Hartshorne's "Algebraic Geometry".  If $\pi$ is a proper, flat morphism $\pi$ as in the question, then Ariyan's answer is perfectly correct and gives étale local triviality. If, further, $Y$ is Noetherian and regular (the hypothesis in Exercise II.7.10), then generic triviality of $\pi$ (i.e., the OP's hypothesis at the generic point of $Y$) implies Zariski local triviality of $\pi$. There should be an easy direct argument for this, but this certainly follows from the known cases of the Grothendieck-Serre conjecture (there is a direct argument after all, see the hint in that exercise).  Finally, Exercise II.7.10 implies that $X$ is isomorphic to $\mathbb{P}(E)$.<|endoftext|>
TITLE: Reverse of a termspace forcing fact
QUESTION [6 upvotes]: Suppose $\kappa$ is an inaccessible cardinal.  Consider the termspace forcing for adding a Cohen subset of $\kappa$ after $\mathbb P= Col(\omega,<\kappa)$. Members are Levy names for countable partial functions from $\kappa$ to 2, ordered by $\tau \leq \sigma$ iff $1 \Vdash \tau \leq \sigma$.  Call this $T(\mathbb P, Add(\kappa))$.
It is well known that the identity map is a projection from $\mathbb P \times T(\mathbb P, Add(\kappa))$ to the iteration $\mathbb P *  \dot Add(\kappa)$.  Furthermore in this situation it is not hard to show that $T(\mathbb P, Add(\kappa)) \cong Add(\kappa)$. So if $G \times H$ is $\mathbb P \times Add(\kappa)$-generic, then in the extension there is $I$ such that $G * \dot I$ is $\mathbb P *  \dot Add(\kappa)$-generic.
My question is what about the opposite. In $V[G* \dot I]$ is there an $Add(\kappa)^V$-generic?

REPLY [6 votes]: The answer is no. Forcing with
$\mathbb{P}*\dot{\text{Add}}(\kappa,1)$ adds no fresh subsets to
$\kappa$, that is, a new subset of $\kappa$ all of whose initial
segments are in $V$. Thus, it cannot add a $V$-generic filter for
$\text{Add}(\kappa,1)^V$.
To see this, note that $\mathbb{P}$ is productively $\kappa$-c.c.,
meaning that $\mathbb{P}\times\mathbb{P}$ is $\kappa$-c.c., since
$\mathbb{P}$ remains $\kappa$-c.c. after forcing with
$\mathbb{P}$, since in the extension it amounts to
$\text{Add}(\omega,\kappa)$, which is c.c.c. there.
It now follows from Spencer Unger's improved version of my lemma
on the approximation and cover properties, which go back to results implicit in 
Mitchell's dissertation. Specifically, it follows from lemma 1.3
in his paper Spencer Unger, FRAGILITY
AND INDESTRUCTIBILITY II that $\mathbb{P}*\dot{\text{Add}}(\kappa,1)$ has
the $\kappa$-approximation property, meaning this forcing adds no
new sets of ordinals all of whose small approximations are in the
ground model. Thus, it can add no $V$-generic subsets of $\kappa$
using $\text{Add}(\kappa,1)^V$.<|endoftext|>
TITLE: Product of a Finite Number of Matrices Related to Roots of Unity
QUESTION [9 upvotes]: Does anyone have an idea how to prove the following identity? 
$$
\mathop{\mathrm{Tr}}\left(\prod_{j=0}^{n-1}\begin{pmatrix}  
x^{-2j} & -x^{2j+1} \\
1 & 0
\end{pmatrix}\right)=
\begin{cases}
2 & \text{if } n=0\pmod{6}\\
1 & \text{if } n=1,5\pmod{6}\\
-1 & \text{if } n=2,4\pmod{6}\\
4 & \text{if } n=3\pmod{6}
\end{cases},
$$
where $x=e^{\frac{\pi i}{n}}$ and the product sign means usual matrix multiplication.
I have tried induction but there are too many terms in all of four entries as $n$ grows. I think maybe using generating functions is the way?

REPLY [2 votes]: UPDATE: As pointed out by Peter Mueller in the comments below, this reduction to continuants does not quite work as stated. I still believe that there is some connection with (possibly generalized) continuants and leave this answer as an unsuccessful attempt to demonstrate it (in hope to fix it later).
Notice that
$$\begin{pmatrix}  
x^{-2j} & -x^{2j+1} \\
1 & 0
\end{pmatrix} = 
\begin{pmatrix}  
-x^{2j+1} & 0 \\
0 & 1
\end{pmatrix}
\cdot
\begin{pmatrix}  
-x^{-4j-1} & 1 \\
1 & 0
\end{pmatrix}.$$
Since diagonal matrices commute with others and $\prod_{j=0}^{n-1} (-x^{2j+1}) = (-1)^n x^{n^2} = 1$, we have
$$\prod_{j=0}^{n-1} \begin{pmatrix}  
x^{-2j} & -x^{2j+1} \\
1 & 0
\end{pmatrix} = 
\prod_{j=0}^{n-1} \begin{pmatrix}  
-x^{-4j-1} & 1 \\
1 & 0
\end{pmatrix}.$$
The last product can be expressed in terms of continuants $K_m(a_1,\dots,a_m)$. Namely, for $j=0,\dots,n-1$, let $a_{j+1} = -x^{-4j-1}$. Then
$$\prod_{j=0}^{n-1} \begin{pmatrix}  
-x^{-4j-1} & 1 \\
1 & 0
\end{pmatrix}=
\begin{pmatrix} K_n(a_1,\ldots,a_n) & K_{n-1}(a_1,\ldots,a_{n-1}) \\ 
K_{n-1}(a_2,\ldots,a_n) & K_{n-2}(a_2,\ldots,a_{n-1}) \end{pmatrix}.
$$
So, it remains to compute values of these continuats. 
P.S. Fedor's answer provides a way to compute the continuants, using Euler's rule.<|endoftext|>
TITLE: Mathematical software wish list
QUESTION [141 upvotes]: Like many other mathematicians I use mathematical software like SAGE, GAP, Polymake, and of course $\LaTeX$ extensively. When I chat with colleagues about such software tools, very often someone has an idea of how to extend an existing tool, what (non-existent) tool would be useful, or which piece of documentation should be (re)written. Due to lack of time & energy and often also programming expertise, these ideas rarely materialize.
On the other hand, every now and then I meet programmers with a strong interest in mathematics (who are often actually trained mathematicians), and who are looking for a software project to work on. However, normally they don't really know what's needed and end up doing a non-mathematical project.
This gave me the idea to ask the mathematical community to compile a wish list for mathematical software. Wishes can be very small or or something bigger. Just try to make sure that it's realistic and maybe also give an explanation why you consider your project as interesting.
And if you happen to be a programmer fulfilling one of the wishes, please leave a comment.
It would be great if you could also include an estimate on how complex your project and what the math/coding ratio is -- but this is optional.

tl;dr

What software tool would you like to see created?
What existent software tool would you like to see extended by what feature?
What piece of documentation is missing or should be updated/extended?


One suggestion per answer, please.

REPLY [3 votes]: I would like to have a feature in a (La)TeX IDE that would allow one to ``collapse" one's document tree into a single file.   I modularize my typesetting: I have separate files for definitions, lemmas, \newcommands, etc., and it is not convenient to share all of the separate files with others.
Uploading one's document tree to the arXiv--one file at a time--is a tedious chore.  Also,  some editors request that submissions be put into a single file for publication.   The ability to work modularly and then readily convert the corpus to a single file would be useful.<|endoftext|>
TITLE: Self-Similar Graphs
QUESTION [11 upvotes]: Many fractals can be generated using and infinite sequence of graphs. For example, Sierpinski's Gasket could be generated by the following sequence of graphs.

Many definitions of fractal dimensions (box-counting, correlation, capacity etc.) give the fractal dimension of Sierpinski's Gasket as $\log_23 \approx 1.58$. Is there a nice way to recover this number from the graph construction. More generally, given a sequence of self-similar graphs, can one assign a fractional dimension that makes sense with the typical definition from continuous fractals ? 
I was trying this by myself. However, all the proofs I has seen for the fractal dimension of Sierpinski's Gasket involved a definition of length. This length was defined to be of dimension $1$. How would this be extended to graphs?
My idea was simply to use the diameter of a graph as a base one-dimensional measure. This works wonderfully with the Sierpinski Gasket. However, consider the following sequence of graphs. Let $G_1$ just be $4$ points arranged in a cycle. Now to construct $G_n$ we have the following algorithm:

Arrange $4$ $G_{n - 1}$ graphs in a square.
Call two of the $G_{n - 1}$ graphs adjacent if they are not diagonally opposite.
Given a point $A$ connect it to the corresponding point in both of the adjacent $G_{n - 1}$ graphs.

In this case, using diameter does not work because the diameter of $G_n$ is only $2$ more than the diameter of $G_{n - 1}$. On the other hand, the number of vertices has quadrupled. This is different from the Sierpinski Gasket in that the number of points roughly triples between consecutive Sierpinski graphs. However, the diameter doubles. How should a one-dimensional measure be defined? Or is this the wrong approach entirely? 
Is there an intuitive meaning to the dimension of a self-similar graph?

REPLY [8 votes]: One way to think of dimension on graphs is to consider Ball growth. In fact there is some literature that refers to a graph $G = (V,E)$ as being "fractal" if there is a constant $C$ and an exponent $q$ such that for any vertex $x$, $$ V_t(x):= \#\{ \text{vertices no more than } t \text{ steps away from }x  \} \leq C t^q.$$
If one had the corresponding lower bound, then this would correspond to the graph being Ahlfors $q$-regular with the graph distance and the measure which counts verteces.
Checking the examples, the Sierpinski latices $V_{2^n} (x) \leq C 3^n$. This is essentially because, as you point out, the number of vertices in $S_n \approx 3^n$ while the diameter is $2^n$. 
On the other hand, the counterexample graphs you mention are not fractal because the diameter grows too slowly.

REPLY [5 votes]: This should be more like a comment but I haven't got enough reputation :)
Have you tried
B. Krön. Growth of self-similar graphs. Journal of Graph Theory, 45 no. 3, 224-239 (2004).
?
or
B. Krön. Green functions on self-similar graphs and bounds for the spectrum of the Laplacian. Ann. Inst. Fourier (Grenoble) 52 no. 6, 1875-1900 (2002).
B. Krön, E. Teufl. Asymptotics of the transition probabilities of the simple random walk on self-similar graphs. Trans. Amer. Math. Soc., 356 393-414 (2004).
Especially the first paper should have some answers for you.

REPLY [2 votes]: You ask for intuition, so I'll assume I'll get some wiggle room in mathematical content.
The typical box counting dimension measures the exponent in the relation
$$M(s)\sim s^d$$
Where $M(s)$ is some notion of 'mass' and $s$ is some notion of scale. 
In this case, you'd just use $M(s)=E(n)$ and $n$. Where $E(n)$ is the number of edges in iteration n. You'd then find an exponent that fits the relationship between the two variables using.
$$d \sim {{\ln(E(n))} \over {\ln(n)}}$$
Things get more interesting when the edges are given varying weights. In that case, an analogy involving multi-fractals might be drawn.
Here's a reference to another MO discussion on the subject.<|endoftext|>
TITLE: Tighter Caratheodory on the moment curve?
QUESTION [12 upvotes]: The moment curve is the set of points of the form 
$$(t,t^2,t^3,...,t^n) \in R^n$$
Let $M$ be the portion of the moment curve where $t\in [0,1]$, and let $\overline{M}$ be the convex hull of $M$.
Caratheodory's theorem tells us that any point in $\overline{M}$ can be expressed as a convex combination of $n+1$ points in $M$.  However, when $n=1$ and $n=2$, we can get by with only $n$ points instead of $n+1$ points.
Question: Is this true for all $n$?  I.e., can we express any point in $\overline{M}$ as a convex combination of at most $n$ points from $M$?
Edit: In the answer below, eins6180 points out that $n$ points always suffice, answering my question.  However, the reference he suggested actually mentions something much stronger.  The moment curve is a convex curve, that is, a curve in which no $n + 1$ points lie in a single affine hyperplane.  Barany & Karasev mention (through reference) that convex curves only require
$$\left\lfloor \frac{n+2}{2} \right\rfloor$$
points to represent any point in their convex hull.  The moment curve example is discussed a little bit in this paper by Holmsen and Roldan-Pensado.
I think a dimension-counting argument shows that this bound is tight.

REPLY [13 votes]: The answer is yes for all dimensions.
An old theorem by Fenchel states that for a compact set $K$ in $\mathbb{R}^n$ every point in the convex hull can be either written as a convex combination of at most $n$ points or $K$ can be separated by a hyperplane. Since $M$ is path-connected, such a separation is not possible.
But there are also more modern theorems of this kind available. A good starting point is recent work by Barany & Karasev. Their work also implies an affirmative answer to your question.<|endoftext|>
TITLE: Conway's subprime Fibonacci sequences
QUESTION [6 upvotes]: I want to be certain I have the latest information on
Conway's subprime Fibonacci sequences,
arXiv-posted a year ago; I am referencing the status in
a review. 
To wit, starting with $(0,1)$:1
$$
0, 1, 1, 2, 3, 5, 4, 3, 7, 5, 6, 11, 17, 14, 31, 15, 23, 19, \ldots
$$

Richard K. Guy, Tanya Khovanova, Julian Salazar:
  
  "however, it seems more likely here than in the $3x + 1$
  problem that sequences do not increase indefinitely. Here is an informal argument that
  supports such a conjecture,"

$\color{Red}{Q}$: Has it been established that some sequence increases indefinitely?
Or no sequence, regardless of starting conditions, does not?


1"Start with the Fibonacci sequence 0, 1, 1, 2, 3, 5, . . . , but before you write down a composite
term, divide it by its least prime factor so that this next term is not 8, but rather 8/2 = 4."

REPLY [4 votes]: A somewhat different (but related) sequence has been analyzed satisfactorily. Quite recently, too (A Note on Prime Fibonacci Sequences, Jeremy Alm, Taylor Herald
(Submitted on 17 Jul 2015)).<|endoftext|>
TITLE: reference for "Topological algebra of Grothendieck"
QUESTION [6 upvotes]: I would like to have some references for Grothendieck's theory of "Topological algebra": a synthesis of homotopical and homological algebra,
with special emphasis on topoi.

REPLY [7 votes]: There are 'obvious' places to follow this up. The mentions in PS (Pursuing Stacks) are worth looking into (but skim that as you can easily get bogged down in it and weighted down by its length) and also see the 'letters to Larry Breen' from 1975 (for which look at the Grothendieck circle pages . These latter give some of the wish-list for a theory of $n$-groupoids, including the corresponding higher Galois theory (and here you could skip to 2015 and Marc Hoyois's preprint: http://arxiv.org/pdf/1506.07155v3.pdf). For the development since then the work of Lurie, and Toen and Vezzosi provide sources for some of the ideas, but I am not sure what aspect you are interested in. ('Higher algebra' by Lurie would be a good place to browse.)
Look through Ronnie Brown's page that relates to PS ( pages.bangor.ac.uk/~mas010/pstacks.htm) and follow up some of the links and look at http://webusers.imj-prg.fr/~georges.maltsiniotis/ps/agrb_web.pdf. 
There is in PS, I seem to remember, some explicit mention of 'topological algebra' in the sense you want. (Edit:  In fact he mentions it early on in the 'letter to Quillen', but not after that if my search was working.)<|endoftext|>
TITLE: Hausdorff dimension of a Cantor-like set
QUESTION [6 upvotes]: Suppose $K$ is a subset of $[0,1]$ with the following property: for almost $x,y \in K$, we have
$$\frac{x+y}{2} \not\in K.$$
(Here, "almost in $K$" means "in $K$ except for a countable subset").
Such a set must have holes, and the Cantor tridiagonal set has this property. What can we say about the Hausdorff dimension of $K$? Is it true that it is less than $\ln 2 / \ln 3$?
Can we even say that the Hausdorff measure $H^{\ln 2/ \ln 3}(K)$ is finite?

REPLY [4 votes]: In Construction of 1-dimensional subsets of the reals not containing similar copies of given patterns Tamás Keleti constructs a compact set of Hausdorff dimension $1$ which contains no similar copy of any out a given countable number of $3$-element sets. In particular, there is a compact set of Hausdorff dimension $1$ which does not contain any arithmetic progressions.
His construction does not use Behrend's example at all, and it is intrinsic to the reals as it depends on a multi-scale argument, so it is substantially different from Adam Goucher's nice solution.
On the other hand, an easy application of the Lebesgue density theorem shows that any set of positive Lebesgue measure contains an arithmetic progression of length $3$ (and of any finite length).<|endoftext|>
TITLE: Example of Genus 7 Curve whose Conormal Sheaf isn't Locally Free
QUESTION [7 upvotes]: Let $C \rightarrow \mathbb P^6$ be a genus 7 canonically embedded (singular) Gorenstein generically reduced curve. Are there any examples of such a curve so that the conormal sheaf $N^\vee_{C/\mathbb P^6} \cong \mathscr I_C/ \mathscr I_C^2$ is not locally free, or do all genus 7 Gorenstein canonical curves have locally free conormal sheaves? 
For example, if C is a regular embedding, then it is not hard to show the conormal sheaf is locally free. See, for instance, Qing Liu's Algebraic Geometry and Arithmetic Curves, Section 6.3, Corollary 3.8. So, any regular embedding $C \rightarrow \mathbb P^6$ would yield a locally free conormal sheaf.
If an example isn't known in genus 7, I'd also be interested in seeing an example in another genus.

REPLY [5 votes]: There are examples already beginning in genus $5$, and thus in all higher genera (including genus 7). By Serre's Corollary, Corollary 21.20 of the following, you cannot find examples with genus $3$ or $4$ (the problem does not make sense for $g=2$, since the canonical map cannot be a closed immersion).  The construction below is based on Exercise 21.11, p. 553, of Eisenbud's book.
MR1322960 (97a:13001) 
Eisenbud, David(1-BRND) 
Commutative algebra. With a view toward algebraic geometry. 
Graduate Texts in Mathematics, 150. Springer-Verlag, New York, 1995. xvi+785 pp.  
ISBN: 0-387-94268-8; 0-387-94269-6 
13-01 (14A05) 
There exists a finite morphism of schemes $f:\mathbb{P}^1\to C$ that is a universal homeomorphism, that is an isomorphism on the dense open subset $\mathbb{P}^1\setminus \{0\}$, and such that the induced finite ring homomorphism,
$$ \widehat{f}^{\#}_0: \widehat{\mathcal{O}}_{C,0}\to \widehat{\mathcal{O}}_{\mathbb{P}^1,0} $$
is equivalent to the finite ring homomorphism
$$
k[[t^5,t^6,t^7,t^8]]\hookrightarrow k[[t]],
$$
where $t$ is a generator of the maximal ideal $\mathfrak{m}_{\mathbb{P}^1,0} \subset \widehat{\mathcal{O}}_{\mathbb{P}^1,0}$.  The scheme $C$ is Gorenstein by Exercise 21.11.  
The embedding dimension of $C$ at the origin is $4$ with one minimal set of generators of $\mathfrak{m}_{C,0}$ given by $$x_5 = t^5,\ x_6 = t^6,\ x_7 = t^7,\ x_8 = t^8.$$ For the induced surjection $$ \phi^*: k[[x_5,x_6,x_7,x_8]] \to k[[t^5,t^6,t^7,t^8]],\ \  x_n \mapsto t^n,$$ for the associated ideal $I=\text{Ker}(\phi)$, $I/\mathfrak{m}I$ contains the following $k$-linearly independent elements, $$ g_1 = x_5x_7-x_6^2,\ g_2 = x_5x_8-x_6x_7, \ g_3 = x_6x_8-x_7^2, \ g_4 = x_7x_8 - x_5^3,\ g_5 = x_8^2-x_5^2x_6.$$ (This might be a $k$-basis; as it is irrelevant to what follows, I did not compute this.)  Thus $C$ is not a local complete intersection scheme.  
Now choose $t$ to be an element of $k(\mathbb{P}^1)$ that is regular on $\mathbb{P}^1\setminus\{\infty\}$, that has a simple pole at $\infty$ and that has a simple zero at $0$, i.e., $t$ is a usual coordinate on $\mathbb{A}^1 = \mathbb{P}^1\setminus\{\infty\}$.  Let $\alpha$ be the rational section $dt/t^2$ of $\omega_{\mathbb{P}^1/k}=\Omega^1_{\mathbb{P}^1/k}$.  Thus $\alpha$ is a regular generator of $\omega_{\mathbb{P}^1/k}$ on the open subset $\mathbb{P}^1\setminus\{0\}$, and $\alpha$ has a pole of order $2$ at $0$.  By Exercise 21.11, $H^0(C,\omega_{C/k})$ is generated by $$ \beta_0 = \frac{1}{t^8}\alpha, \ \beta_1 = x_5\beta_0 = \frac{1}{t^3}\alpha, \ \beta_2 = x_6\beta_0 = \frac{1}{t^2}\alpha,\ \beta_3 = x_7\beta_0 = \frac{1}{t}\alpha,\ \beta_4 = x_8\beta_0 = \alpha.$$
Moreover, $\beta_0$ is a generator of $\omega_{C/k}$ on $C\setminus\{\infty\}$, and $\beta_4$ is a generator of $\omega_{C/k}$ on $C\setminus\{0\}$.  Thus the canonical morphism $$\phi:C \to \mathbb{P}H^0(C,\omega_{C/k}),$$ restricts on the open affine $C\setminus\{\infty\} \to D_+(\beta_0)$ to the morphism of schemes associated to the ring homomorphism $\phi^*$ above.  So $\mathcal{I}_C/\mathcal{I}_C^2$ is not locally free of rank $3$: the rank at $0$ is (at least) $5$.
This example has arithmetic genus $5$.  You can make an example with arithmetic genus $7$ in various ways, e.g., glue a copy of a curve of arithmetic genus $2$ at a point $p\neq 0$ of $C$ such that the total curve has an ordinary double point at $p$.   
P.S. For the one or two people who know what I am talking about, this example solves the extra credit problem from Hartshorne's 1995 commutative algebra final -- not quite completed by the exam deadline :)<|endoftext|>
TITLE: central charge and Calabi-Yau dimension
QUESTION [7 upvotes]: I would like to know if there is any setting where the two notions of

central charge of 2D conformal field theories,
Calabi-Yau dimension of fractionally Calabi-Yau categories

can be understood as "one and the same".
One motivation (not very serious) is that 

on the one hand there are CY categories of dimension $(h-2)/h$ where $h$ is the Coxeter number of a Dynkin diagram,
on the other hand, Coxeter numbers often appear in central charges.

REPLY [12 votes]: Given a $N=(2,2)$ two dimensional superconformal field theory (SCFT), one can construct two topological field theories called the $A$ and $B$ models. To each of these topological field theories, one should be able to associate a ($A_\infty$) triangulated category of boundary conditions, called category of branes. Thus corresponding to $A$ and $B$ models one obtains the categories of $A$- and $B$-branes. These categories are Calabi-Yau categories of dimension $c/3$ where $c$ is the central charge of the $N=(2,2)$ superconformal field theory. This Calabi-Yau property should be a consequence of the existence of the spectral flow in the $N=(2,2)$ superconformal algebra (and so I should make a technical hypothesis on the $N=(2,2)$ SCFT: the differences between left and right R-charges should be integers).
When I have written "should be", it means that they are rather standard facts for some physicists but that I don't necessarely know how to make and prove general rigorous statements. For more information on this topic, one might want to have a look at the book http://www.claymath.org/library/monographs/cmim04.pdf
But there are many examples where the $A$ and $B$ models and the corresponding categories of $A$ and $B$-branes are known and where the central charge of the maybe not yet rigorously constructed $N=(2,2)$ SCFT is also known.
Examples: 
1) If $(X, J, \omega)$ is a Calabi-Yau manifold of complex dimension $d$, with $J$
a complex structure and $\omega$ a Kähler form then one should be able to construct a $N=(2,2)$ SCFT (the two dimensional sigma model of target $(X,J,\omega)$), of central charge $c=3d$. The corresponding category of $A$-branes is the Fukaya catgeory of $(X,\omega)$ and the category of $B$-branes is the derived category of coherent sheaves on $(X,J)$, which are indeed Calabi-Yau categories of dimension $d$.
2) If $W(x_1,...,x_n)$ is a quasihomogeneous polynomial, i.e. satisfying 
$W(\lambda^{w_1}x_1,...,\lambda^{w_n}x_n)=\lambda^d W(x_1,...,x_n)$
for some $w_1,...,w_n,d$ positive integers then one should be able to construct a $N=(2,2)$ SCFT (the Landau-Ginzburg model of superpotential $W$), of central charge $c=3 \sum_{i=1}^n(1-2 \frac{w_i}{d})$. I don't know what is the category of $A$-branes in general but the category of $B$-branes is the category of matrix factorizations of $W$, which is Calabi-Yau of dimension $\sum_{i=1}^n(1-2 \frac{w_i}{d})$. In 1), the dimension of the Calabi-Yau categories was an integer but it is in general a rational number in 2).
If $W$ is an $ADE$ surface singularity, then $c=3(1-\frac{2}{h})$ where $h$ is the $ADE$ Coxeter number and the corresponding $N=(2,2)$ SCFT is believed to be the ADE $N=(2,2)$ minimal model. It is possible to show that the category of matrix factorizations of an $ADE$ surface singularity is equivalent to the derived category of representations of the $ADE$ quiver, which is a standard example of Calabi-Yau category of dimension $1-\frac{2}{h}$.
Remark: as the comment by Hugh Thomas shows, the terminology "central charge" can be confusing because any central element in an algebra could be called central charge and there are many different algebras and central elements... The question is about the central charge of a $2d$ CFT, i.e. the central element of the Virasoro algebra. The central charge to which Hugh Thomas refers to is different but also appears naturally in the story: it is part of the information which should be obtained from the $N=(2,2)$ SCFT and should determine which of the branes of the topological theories are in fact branes of the superconformal theory. Its name central charge refers to the fact that in some cases (when the $N=(2,2)$ SCFT is used as internal space for a string compactification), it is really a central charge in a $N=2$ four dimensional supersymmetry algebra (corresponding to the four non-compact dimensions of the string compactification).<|endoftext|>
TITLE: Varieties with an ample vector bundle mapping to their tangent bundle
QUESTION [17 upvotes]: A well-known result of Andreatta and Wisniewski says: Let $X$ be a projective complex manifold whose tangent bundle $T_X$ contains an ample sub-bundle $\mathscr{E}$. Then $X$ is isomorphic to projective space $\mathbb{P}^n$ for some $n$.  Furthermore, the bundle $\mathscr{E}$ is either $\mathscr{O}(1)^{\oplus r}$ or $T_{\mathbb{P}^n}$ itself.
I would like to know if a slightly stronger statement is known/true.  

Suppose there is an ample vector bundle $\mathscr{E}$ and a non-zero map $$f: \mathscr{E}\to T_X.$$  Is $X$ necessarily isomorphic to $\mathbb{P}^n$ for some $n$?

Of course if $\text{im}(f)$ is a vector bundle, the result follows by Andreatta-Wisniewski, as quotients of ample vector bundles are ample.
Remarks (Added later). The statement is true if $\mathscr{E}$ is a line bundle, or if $X$ is a curve or surface.  I can also show that the existence of a non-zero map $\mathscr{E}\to T_X$ with $\mathscr{E}$ ample implies $X$ is birational to an (etale) $\mathbb{P}^k$-bundle over a lower-dimensional variety (and in particular, is uniruled).  The answer of pgraf gives a reference showing the statement is true if $\text{Pic}(X)$ has rank $1$.
The paper "Galois coverings and endomorphisms of projective varieties" by Aprodu, Kebekus, and Peternell, which pgraf has referenced and which covers the case of Picard rank $1$ only uses the hypothesis in Corollaries 4.9 and 4.11, as far as I can tell.  In particular, many of the other arguments in Section 4 of that paper hold true; for example, one knows that if $T$ is a family of rational curves in $X$ of minimal degree, any split curves must land in the singular locus of $\text{im}(f)$.  Analyzing the arguments of that paper seem to give several other cases, e.g. if $f$ has generic rank equal to $\text{dim}(X)$.

REPLY [4 votes]: The answer to this question is now known to be yes -- it is Corollary 1.2 of this paper of Jie Liu.<|endoftext|>
TITLE: Nice sign-expansions of special surreal numbers
QUESTION [18 upvotes]: What is the "right" surreal generalization of the fact that a real number $r$ is rational if and only if its sign-expansion is eventually periodic?
I can think of more than one natural way to generalize the notion of a rational number, but "ratio of omnific integers" is not one of them, since every real number is a ratio of omnific integers. Maybe ratios of ordinals are the thing to look at (as a generalization of the non-negative rationals). Or maybe we should look at the Field-closure of the ordinals. Or perhaps we should consider the Class containing every surreal number whose normal form involves only rational numbers at all levels. 
I can also think of more than one way to generalize the notion of eventual periodicity to sign-sequences indexed by a general ordinal alpha. One of them is a variant of "Kaufman decimals" (see https://mchouza.wordpress.com/2013/08/25/kaufman-decimals/ and http://www.jefftk.com/p/decimal-inconsistency) in which the digit-set {0,...,9} is replaced by {$+$,$-$} and every over-bar is assigned an ordinal.
A seemingly different but possibly equivalent notion generalizing eventual periodicity involves a kind of symbolic dynamics I haven't seen before, where the Monoid of ordinals acts on ordinal-indexed sequences: if $s$ is a sequence indexed by some initial segment of the ordinals, and $\iota$ is some ordinal, define $T^{\iota}(s)$ to be the sequence obtained by omitting the first $\iota$ terms of $s$ (with $T^{\iota}(s)$ defined to be the empty sequence if $\iota$ is greater than or equal to the length of $s$). Then eventual periodicity (in the case where $s$ is indexed by the natural numbers) is seen to be a special case of the condition that the orbit of $s$ under the action of the Monoid of all ordinals is finite. (See Joel Hamkins' recent post, showing that constant sequences satisfy this finiteness condition: http://jdh.hamkins.org/every-ordinal-has-only-finitely-many-order-types-for-its-final-segments/.)
If my original question seems too vague (what does "right generalization" mean?), here are two very concrete ones that are relevant: does the surreal number with sign-expansion $+-^{\omega}++-^{\omega}+++-^{\omega}++++-^{\omega}\cdots$ (indexed by $\omega^2$) lie in the field generated by $\omega$? and, is it expressible as a ratio of ordinals? (Note that this sign-sequence does not satisfy the aforementioned finiteness property, although perhaps it satisfies some weaker regularity condition.) Here $-^{\omega}$ denotes a string of $\omega$-many $-$'s and $\cdots$ denotes that the pattern continues $\omega$ times.
I'd be interested in any implications that might hold between these various properties.

REPLY [5 votes]: While I do not know if there is a 

"right" surreal generalization of the fact that a real number $r$ is
  rational if and only if its sign-expansion is eventually periodic

it is perhaps worth noting that Conway has identified what he believes "are perhaps the closest analogue in No of the ordinary rational numbers" (ONAG, p.47). These surreal "fractional numbers" are the continued fractions that terminate at a finite stage. Instead of using the integer part to generate the continued fraction, one uses the omnific integer part. Conway observes that: "If  $x$ is fractional, so are $x+1$, $-x$ amd $1/x$ (if $x \neq 0$), but neither the sum nor the product of two fractional numbers need be fractional….” Moreover, it is easy to show that there are fractional numbers in Conway’s sense having sign-expansions that are not eventually periodic.<|endoftext|>
TITLE: Diameter of the modified bubble-sort graph
QUESTION [9 upvotes]: The modified bubble-sort graph is the Cayley graph $Cay(S_n,S)$ of $S_n$ generated by $n$ cyclically adjacent transpositions. Thus $S = \{ (1,2),(2,3),\ldots,(n,1)\}$.  I was wondering whether the diameter of this graph is known.
A computer simulation gives the conjecture that the diameter is $\lfloor n^2/4 \rfloor$.  In (Heydemann, "Cayley graphs and interconnection networks", in Graph Symmetry: Algebraic Methods and Applications, Eds. Hahn and Sabidussi), it is stated in p. 213 that the diameter of the modified bubble-sort graph is $\lfloor n^2/4 \rfloor$, and the result is attributed to (J.-F. Sacle,"Diameter of some Cayley graphs", private communication, 1997).  However, I couldn't find any proof of this result.  Is any proof for this formula given in the literature?  
This problem was studied earlier in a paper of Jerrum's ("The complexity of finding minimum-length generator sequences, TCS, 1985), but no closed-form expression for the diameter is given there either.

REPLY [2 votes]: The claim seems to be correct, here's an argument.

For all $n$, the diameter of the modified bubble sort graph is $\lfloor n/4 \rfloor$. One element with this word norm is $\pi(i) = i+\lfloor n/2 \rfloor$ (addition modulo $n$).

First, consider even $n = 2k$, so $\lfloor n^2/4 \rfloor = k^2$.
First, the lower bound.
Let our set to permute be $N = \mathbb{Z}/\mathbb{Z}_n$ which I mostly biject with $\{0,1,...,n-1\}$, and for the mental picture it is useful to think it's laid out on a circle. Define the huskiness of $\pi \in S_n$ to be $h(\pi) = \sum_{i \in N} |i - \pi(i)|'$ where $|a|' = \min \{|b| \;|\; b \in \mathbb{Z}, a = b + n\mathbb{Z}\}$ for $a \in N$. In other words this is just the sum of distances from $i$ to $\pi(i)$ over $n \in N$, in the cyclic graph structure of $N$. Clearly huskiness is $0$ iff the permutation is trivial.
Now, $S_n$ acts on itself by precomposition, and in this action the modified bubble sort generators decrease huskiness by at most $2$, since you move at most two elements of $N$, each by at most a distance of one. The permutation $\pi(i) = i+k$ has huskiness $h(\pi) = n k$, so you cannot generate $\pi$ with less than $nk/2 = 2k^2/2 = k^2$ generators, giving the desired lower bound. 
A possibly helpful note if you have some chirality confusion (or in case I do): Note that precomposition basically corresponds to thinking of $\pi(i)$ as the "goal" for $i$, and when you precompose by a swap involving $i$, you can think of $i$ as moving. (Postcomposition moves the goalposts.) I recommend thinking of the permutation $\pi$ as an arrangement of the values $N$ on $N$ where the value at $i$ is initially $\pi(i)$. The precomposition action moves these numbers around, and our goal is to move the value $\pi(i)$ to $\pi(i)$ for all $i \in N$.
Second, then the upper bound.
We show the stronger fact that the diameter is $k^2$ even if you drop the generator $(0, n-1)$. For the mental picture, it is useful to open up $N$ from a circle to a line. Define the friskiness of $\pi \in S_n$ to be $f(\pi) = \sum_{i \in N} d(i, \pi(i))$ where $d : N \times N \to \mathbb{N}$ is the graph metric in the linear graph structure of $N$, or the Euclidean metric on $\mathbb{R}$ applied to the representative in $\{0,1,...,n-1\}$.
The friskiness of a permutation is at most $2k^2$. Here's an argument for that, though I suppose there must be some easier way: Suppose $f(\pi)$ is maximal. I claim that we may assume $\pi(i) = n-1-i$, by sorting an arbitrary permutation into this $\pi$ without decreasing friskiness: to do this, bubble $0$ from $\pi(0)$ up to $n-1$ by pre-composing with swaps $(i, i+1)$, $i \neq n-1$, observing no individual step decreases friskiness. Thus, we may assume $\pi(0) = n-1$. But assuming $\pi(0) = n-1$, bubbling $1$ from $\pi(1)$ up to $n-2$ does not decrease friskiness: the first step may reduce the contribution of $1$ to friskiness if $\pi(1) = 0$, but the contribution of $\pi^{-1}(1)$ is increased, since $\pi^{-1}(1) \neq 0$ (unless $n = 2$, in which case we were done anyway). Continue inductively. It follows that indeed for $\pi(i) = n-i$ the friskiness
$$ h(\pi) = \sum_{i = 0}^n d(i,n-1-i) = kn = 2k^2 $$
is the maximal possible.
(Not that it matters, but the sorting above is not bubble sort, it's basically selection sort but by swaps, which I suppose has no name.)
Friskiness is $0$ iff the permutation is trivial, so it is enough to show that we can always decrease the friskiness of a non-trivial permutation by $2$ by applying a swap at a suitable position. For this, simply swap (as always by precomposition) at any position where $\pi(i) > i$ and $\pi(i+1) < i+1$ (which you can find by looking at $i \in N$ in increasing order). This concludes the proof of the upper bound.
Second, consider odd $n = 2k+1$, so $\lfloor n^2/4 \rfloor = k^2+k$.
First, the upper bound.
This reduces directly to the previous case: take $2k \in N$, and move it to its final position, in at most $k$ steps. Now, by the upper bound from the even case (for the smaller set of generators), we can move all the $n-1 = 2k$ to their final places with $k^2$ swaps, giving the upper bound $k^2+k$ as required.
Second, the lower bound.
Huskiness is not a sufficient in variant in this case, as its maximal value is $k$ too small. We now define a more complex invariant called muskiness.
For each $i \in N$, define $v_i = \pi(i) - i$ if $\pi(i) - i \in [0,k]$ and $v_i = i - \pi(i)$ otherwise. This is the "vector field" that at $i \in N$ indicates the (unique) geodesic from $i$ to $\pi(i)$. We define the oriented duskiness of $\pi \in S_n$ to be
$$ d(\pi) = \sum_{i \in N} v_i/n. $$
Define the unoriented duskiness of $\pi$ to be $|d(\pi)|$, and finally define its muskiness to be the sum of its huskiness (defined as in the even case) and unoriented duskiness, $m(\pi) = h(\pi) + |d(\pi)|$.
We again show that we cannot lower muskiness by more than $2$ by applying a generator $(i, i+1)$. There are a few cases to consider. The main technical issue when you swap is that when the distance from $i$ to $\pi(i)$ is precisely $k$, the contribution of $i$ to duskiness can change by $2k$ when you move $i$, in case the geodesic jumps on the other side. Let's call this a sign flip.
Suppose first that $c_i, c_{i+1} > 0$, i.e. both values want to move  in the same direction along the cycle (counterclockwise the way I picture it). Applying the generator moves $i$ in the "correct" direction, and $i+1$ in the "wrong" direction. Now, the contribution to huskiness from $i$ is $-1$, and the contribution to oriented duskiness from $i$ is $-1/n$ as well. The contribution of $i+1$ to huskiness is either $0$ or $1$. If it is $1$, then there is no sign flip for $i+1$, so the contribution of $i+1$ to oriented duskiness is $+1/n$. If the contribution of $i+1$ to huskiness is $0$, then there was a sign flip, and thus its contribution to oriented duskiness is $-2k$. The maximal possible contribution of the swap to muskiness happens in the latter case, in the case when additionally the orientation of duskiness is such that muskiness decreases, and in this case the contribution to muskiness is $-1+0-1/n-2k/n = -2$. The case $c_i, c_{i+1} < 0$ is exactly analogous (though the orientation of duskiness is flipped).
Suppose then $c_i > 0, c_{i+1} < 0$, i.e. both of the two move in the direction they want to move. Then huskiness clearly decreases by two after the swap. Since both move in the direction they want to, there is no sign flip, and the changes in oriented duskiness cancel out. Thus, muskiness changes by $-2$ in this case.
Suppose then $c_i < 0, c_{i+1} > 0$, i.e. both of the two move in the "wrong direction". Then huskiness does not decrease, and may increase or stay invariant (for $i$ and $i+1$ separately). Oriented duskiness, thus also duskiness, may change by at most by $2 \cdot 2k / n < 2n$. Thus, muskiness can decrease by at most $2$ in this case.
There's also the cases $(c_i = 0, c_{i+1} > 0)$ and $(c_i = 0, c_{i+1} < 0)$ and $(c_i = 0, c_{i+1} = 0)$ (and a few more that are symmetric). These cases are easy and similar to the ones above (maybe I could've directly incorporated them somewhere). The huskiness increases in each case and the change in duskiness is not sufficent to counteract that.
(Square.)<|endoftext|>
TITLE: Intuitive functional analysis book
QUESTION [5 upvotes]: I want to know functional analysis book like Terence tao's real analysis and measure theory book, full of intuition. I am aware of linear algebra, real analysis, measure theory, Probability theory.

REPLY [6 votes]: The book  A. Ya.  Helemskii, Lectures and exercises on functional analysis. AMS, Providence, RI, 2006 (translation from the Russian original) is IMHO well written and quite intuitive (check e.g. the MathSciNet review which is quite glowing).<|endoftext|>
TITLE: Groups where word problem is solvable, but not quickly?
QUESTION [25 upvotes]: Are there finitely generated groups whose word problem is solvable, but not quickly? It would be great to have specific examples, but existence results would also be helpful. 
All of the groups that I know with solvable word problem (linear groups, hyperbolic groups, branch groups) have a low-degree polynomial time algorithm for solving the problem. I am looking for something between those groups and the ones with unsolvable word problem.

REPLY [5 votes]: Here's a recipe to produce f.p. groups with solvable word problem at least as hard as the membership problem in any set $A$ of positive integers.
Consider the group
$$\Gamma_A=\langle t,x\mid [t^nxt^{-n},x]=1, \forall n\in A\rangle.$$ 
Then for $n>0$, $[t^nxt^{-n},x]=1$ in $\Gamma_A$ iff $n\in A$. In particular, ($\#$) any algorithm for the word problem in $\Gamma_A$ with complexity $f(n)$ (supremum of halting time for inputs in the $n$-ball) solves the membership problem in $A$ with halting time $\le f(4n+4)$. 
The group $\Gamma_A$ has solvable word problem iff $A$ is recursive. So for finitely generated group we are already done with no prerequisites (of course this boils down to the question of complexity of the membership problem in recursive subsets integers, but this is no longer group theory). 
Now an adaptation of Higman's embedding theorem by Clapham (C. R. J. Clapham. An embedding theorem for finitely generated groups”, Proc.
London. Math. Soc. (3), 17, 1967, 419-430.) yields that every f.g. group with solvable word problem embeds into a finitely presented group with solvable word problem. If we embed $\Gamma_A$ into such a finitely presented group $\Lambda$ and require that $t,x$ are among a generating subset, then ($\#$) above also holds in $\Lambda$. 
The case of residually finite finitely presented groups is considerably harder and is addressed by the Kharlampovich-Myasnikov-Sapir paper referenced in Andreas Thom's answer.<|endoftext|>
TITLE: Grothendieck - sheaves as meter sticks
QUESTION [14 upvotes]: I'm trying to read parts of McLarty's Grothendieck on Simplicity and Generality. In the article, I read Grothendieck thought of sheaves over some topological space as meter sticks measuring it.

What did Grothendieck mean? What property of a topological space do
  these meter sticks measure? Why was a meter stick the chosen metaphor
  and in what sense is it appropriate? Is the fact meter sticks only
  "measure one dimension" relevant in any way?

Just putting this out there: when I think of a sheaf over a topological space, I think of a big floating cloud containing information comprised of small clouds - one for each open set (look at connected stuff). Then the pasting axiom just says you can look at a proper bunch of smaller clouds as one big cloud. This is probably very naïve, but might help you help me.

REPLY [11 votes]: Here are two (related) interpretation of this quote I can think of:
A first interpretation is just that Grothendieck was attach to the idea that you can study a 'space' (whatever this mean) by studing certain family of objects indexed by it, for example sheaves over a topological space, vector bundle over a manifold, Vector bundle or coherent sheave over a scheme or an algebraic stacks etc... From this perspective it might seem natural to think that each sheaves is just "something that will bring you a piece of information about the space" (for example through cohomology, taking globale section etc...) so a "meter stick" seem to be a correct name for this.
The second interpretation is based on another image which is very important for topos theory, and I think which makes this a lot clearer. The idea is that sheaves are just "generalized open sets".
I think if one replace sheaves by open sets in the quote you will understand what is the idea behind ? So I just need to explain in what sense sheaves generalise open sets.
There is two way to think about a sheaf over a topological space $X$:

as a set of locale section which yields this image you are talking about of "cloud of thing that you can patch together".
as an etale space over $X$, i.e. a topological space with a map $p:Y \rightarrow X$ which is a local homomorphism.

This second way is where we see the "generalised open sets" aspect appears: clearly, any $U \subset X$ defines a sheaf $U \rightarrow X$ and morphism of such sheaves are just inclusion of open sets. Moreover a general sheaf over $X$, as it is locally homeomorphic to $X$, is just a gluing of these sheaves corresponding to open set.
So sheaves are a bit like open set, but more flexible you can add sheaves (disjoint union), a sheaf can "self intersect" (for example $(0,2) \rightarrow \mathbb{R}/\mathbb{Z}$ gives you such an example), etc...
This analogy can be pushed very far, for exemple topos theory can be explain as completely replacing open set by sheaves in the definition of a topological space: instead of specifying a base of open sets to define a topology you will specify a base of sheaves to define a (Grothendieck) topos and this is exactly what a site is.<|endoftext|>
TITLE: Fermat's Last Theorem "$\pm k$"
QUESTION [5 upvotes]: Let $\mathbb{N}=\{1,2,3,\ldots\}$ be the set of positive integers. For $n,k\in\mathbb{N}$ we define  $$\text{Sol}(n,k) = \{(a,b,c)\in \mathbb{N}^3: |a^n + b^n - c^n| \leq k\}.$$ (The set $\text{Sol}(n,k)$ denotes the solutions of the inequality $|a^n + b^n - c^n| \leq k$ for fixed $n,k$.)
Moreover, for $j\in\mathbb{N}$ set $$\text{Inf}(j) =\{n\in \mathbb{N}: \text{Sol}(n,j) \text{ is infinite}\}.$$
Clearly, we have $\{1,2\}\subseteq \text{Inf}(j)$ for all $j\in\mathbb{N}$.
Questions:

Is there $j\in\mathbb{N}$ such that $\text{Inf}(j)\neq \{1,2\}$?
Is there $j\in\mathbb{N}$ such that $\text{Inf}(j)$ is infinite, and what is the smallest such $j$?

REPLY [12 votes]: A 4-variable version of the infamous ABC Conjecture says the following: Let $a,b,c,d\in\mathbb{Z}$ be non-zero, satisfy $a+b+c+d=0$ and $\gcd(a,b,c,d)=1$, and no subsum of two or three of $a,b,c,d$ equal to $0$. Then for every $\epsilon>0$ there is a constant $K_\epsilon$ such that 
$$ \max\{|a|,|b|,|c|,|d|\} \le K_\epsilon \prod_{p\mid abcd} p^{1+\epsilon}. $$
Applying this to an expression of the form $a^n+b^n-c^n-k$ gives a very strong bound. Assuming that I haven't made an error (which is quite possible), I think that if $n\ge5$ (and assuming the ABCD conjecture), then for any $k$, the equation
$$ a^n + b^n - c^n = k $$
has only finitely many solutions $a,b,c\in\mathbb{Z}$ with $|a|,|b|,|c|$ distinct and non-zero.
Actually, I guess the same (more or less) should be true for $n=4$. The point is that the surface
$$ x^n+y^n-z^n=k $$
is of general type for $n\ge5$, so the Bombieri-Lang conjecture says that the solutions in rational numbers $(x,y,z)\in\mathbb{Q}^3$ are not Zariski dense (lie on a finite set of curves). This also follows from Vojta's conjecture. And for $n=4$, the equation defines an affine piece of a K3 surface, so Vojta's conjecture implies that the set of integer solutions likewise lies on a finite set of curves. 
So your problem fits into a general framework, and for example, these statements are known if you replace $\mathbb Z$ by the ring of polynomials $\mathbb C[t]$. And as July suggests, you might want to read about how such problems are normally written, since your notation is not at all standard (and somewhat hard to parse).<|endoftext|>
TITLE: Explicit Chebotarev and Langlands - irreducibility of X^5-X-1 mod primes
QUESTION [15 upvotes]: Is there an explicit infinite set of primes, modulo which $X^5 - X - 1$ is irreducible?

Since our polynomial's Galois group over $\mathbb{Q}$ is $S_5$, Chebotarev's density theorem implies that there are infinitely many such primes, but it is seemingly unclear how to find them.
If the Galois group of a polynomial is abelian, then I think primes modulo which the polynomial is irreducible are just the primes satisfying some congruences.
If we take $X^3-X-1$ instead, then such a set of primes can be constructed using the first answer given in Galoisian sets of prime numbers
As far as I am concerned, once a representation of $S_5$ is given, the Langlands correspondence should provide some analytic object (e.g modular form) such that the set of primes for which the polynomial is irreducible (the primes which are inert) could be read off this object.
Can recent work on the Langlands correspondence help finding such a set of primes?

Is there a specific automorphic form which (at least conjecturally) 
  gives us the required primes?

REPLY [12 votes]: The object (conjecturally) associated to an Artin representation by Langlands is not a classical modular form except in a very limited number of situations (odd two dimensional representations). Let $K$ be the splitting field of some polynomial over $\mathbf{Q}$ with Galois group $S_5$. If $V$ is any finite dimensional representation of dimension $n$ of $S_5$, then, associated to 
$$\rho: \mathrm{Gal}(K/\mathbf{Q}) \simeq S_5 \rightarrow \mathrm{GL}(V)$$
one can write down an Artin $L$-function $L(V,s)$. Then Langlands' extension of the Artin Conjecture predicts that $L(V,s)$ is equal to $L(\pi,s)$, where $\pi$ is an algebraic automorphic representation of $\mathrm{GL}(n)/\mathbf{Q}$ (of some very particular infinity type). For $n \ge 3,$ these objects are not so easy to write down very explicitly. For all practical matters, it is much easier to actually compute the split primes (or primes whose corresponding Frobenius element is any conjugacy class in $S_5$) directly from the polynomial rather than automorphically. 
In the particular case of the polynomial $X^5 - X - 1$, the Artin conjecture is known for any representation $V$, as follows from a theorem of J. Dwyer (Real zeros of Artin L-functions corresponding to five-dimensional $S_5$-representations, Bull. Lond. Math. Soc. 46 (2014), no. 1, 51–58).
Response To Comments:
For computational purposes (say which primes $p < 10000$ are inert) the best method to determine the inert primes is using pari/gp by directly factoring the polynomial modulo $p$. As to what information can be inferred from automorphy, that is hopelessly vague, and depends on what you want to do with it. Unfortunately the margins here are too small for me to explain the entire Langlands programme. The entire point of non-abelian class field theory is that the corresponding Galoisian sets are very complicated. There will be no more "explicit" description of the primes which are inert than "primes such that the Hecke eigenvalues of some appropriate automorphic form $\pi$ have some certain property."<|endoftext|>
TITLE: Example of a triangulable topological manifold which does not admit a PL structure
QUESTION [6 upvotes]: I know there are some examples of manifolds which don't admit a PL structure (combinatorial triangulation), and that it has been recently proven that in dimension $n\geq5$ there are manifold which are not triangulable (i.e. which are not homeomorphic to a simplicial complex).
As far as I understand, in dimension $4$ the two concept (triangulable and PL) should coincide, while in dimension $n\geq 5$ they are different.
The only examples of non-combinatorial triangulations I have encountered are double suspensions of homology spheres (which are homeomorphic to spheres): so in that case there is a topological manifold which admit also non-combinatorial triangulations.
I was wondering if there are examples of topological manifolds which admit triangulations, but none of them is combinatorial.

REPLY [7 votes]: The answer is yes, see Rudyak's paper Piecewise linear structures on topological manifolds, Examples 21.4: 

There are topological manifolds that can be triangulated as simplicial complexes but do not admit any PL structure.

Such examples exist in fact in any dimension $n \geq 5$, and are of the form $$M_k=V \times T^k, \quad k \geq 1,$$ where $V$ is the famous $E_8$-manifold constructed by Freedman. See Theorem 7.2 and Corollary 7.4 in the quoted paper.
It is worth remarking that the $4$-manifold $V$ is not triangulable as a simplicial complex. However, by the work of Siebenmann and others, it is known that every orientable topological 5-dimensional closed manifold can be triangulated as a simplicial complex, see Theorem 21.5. This implies that $M_1= V \times S^1$ is triangulable, so $M_k=M_1 \times T^{k-1}$ is triangulable for all $k \geq 1$.<|endoftext|>
TITLE: Is the space of vectorial functions that are Dunford integrable complete?
QUESTION [6 upvotes]: Let $X$ a Banach Space and $(\Omega, \Sigma, \mu)$ a measure space. A function $F:\Omega\rightarrow X$ is Dunford integrable if $x^\ast\circ F$ is $\mu$-integrable for every $x^\ast\in X^\ast$. The space of functions that are Dunford integrable, denoted by $\mathbb{D}(\mu,X)$, is a normed space with
$$
  \|F\|:=\sup\left\{\int_\Omega|x^\ast\circ F|d\mu:x^\ast\in X^\ast,\,\|x^\ast\|\leq1\right\}
$$
Does anybody knows if the space $\mathbb{D}(\mu,X)$ is complete?

REPLY [5 votes]: This a revised and expanded version of my post.
No, it is not complete. For simplicity suppose that $X$ is reflexive and separable in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite and non-atomic. In this case the space of Pettis integrable functions is complete if and only if $X$ is finite-dimensional as proved by Thomas

G.E.F. Thomas, Totally Summable Functions with Values in Locally Convex Spaces,
  Measure Theory (Oberwolfach 1975), Lecture Notes in Math. Vol. 541 (1976), 117–131.

This is based on a result o his which asserts that there exist an absolutely sequence summable sequence $(x_n)_{n=1}^\infty$ in $X$ (reflexivity is not needed) and a sequence $(f_n)_{n=1}^\infty$ in the unit ball of $L_1(\mu)$ such that the vector measure
$$\nu(A) = \sum_{n=1}^\infty \int\limits_A f_n(t)\,{\rm dt}\cdot x_n$$
does not have a Pettis intregrable density.

G.E.F. Thomas, The Lebesgue-Nikodym Theorem for Vector Valued Radon Measures, Memoirs. of AMS, 139, American Mathematical Society, Providence, 1974.

The sequence $(\sum_{k=1}^n f_k\cdot x_k)$ is Cauchy in $\mathcal{P}$ because $(x_n)_{n=1}^\infty$ is absolutely summable, yet it is not convergent as there is no function $F$ such that
$$\lim_{n\to \infty}\int\limits_A \sum_{k=1}^n f_k(t)\cdot x_k\,{\rm d}t\to \int_A F(t)\,{\rm d}t.$$
It seems to me that Pettis integrable functions form a closed subspace of the space of Dunford integrable functions, hence you may extend the above result as in the case where $X$ is infinite-dimensional, you have an incomplete, closed subspace of a normed space, so the space itself cannot be complete.
Addendum. Let me point out that if you want to do some functional analysis with the space of Pettis integrable functions, even though incomplete, it is barrelled.<|endoftext|>
TITLE: A fun game related to knot theory
QUESTION [15 upvotes]: I recently learned the following rather fun game: a group of people is standing up roughly in circle, facing each other. Then participants randomly join hands, in such a way that nobody holds its own hand, and that everybody hold hands with two distinct person.
The result is clearly a link, and the goal of the game is to untangle it, i.e. participants have to move while still holding hands until they can stand up in (possibly several !) circle(s) holding hands with their direct left and right neighboors. Obviously people are allowed to not face the center of the circle (or to be upside down, I guess) so that the first Reidemeister move is allowed, but I guess it's not really important.
Clearly it requires quite some physical and intellectual skills, a lot of collaboration between the players, and a rather long time, which makes it a fun activity, and I recommend trying this at home with friends or at work with fellow mathematicians.
The person who explained the game sweared that it "always works", although it's fairly easy for someone having heard of knot theory to come up with an example where it doesn't. For example the trefoil is easily obtained already with 3 players, but it's also clear that this correspond to a rather particular choice.
In fact, this one time we actually did not manage to untangle the link though we managed to simplify it a lot. Of course even if the link is mathematically trivial it might be physically challenging to actually untangle it, but let's ignore that. 
Basically, this game can be formalized as a process which generate links randomly, and I'm curious whether something interesting can be said about it. I must admit that since this question did not actually come up in my research I haven't given much thought about it.. Obvious questions are: can every link be obtained this way ? If not do they correspond to a known family of links ? Is there any way to support the organizer's claim, ie is there a way to estimate the probability of getting a trivial link ?

REPLY [8 votes]: Yes, every link can be obtained in this way.  Here's an inefficient way to do it.  Put the link in braid form (via Alexander's theorem); suppose that we have $b$ strands.  We'll achieve each braid generator as a certain pattern among $2b$ people on the circumference of the circle.
Here's a picture that explains better than my words can.  All the horizontal (red) strands pass over the vertical (blue) strands.  The two half-flips introduced cancel each other out (though they conjugate the braid generator) and we can clearly make all braid generators (and join them together!), hence all links.

With a little more thought we can arrange the half-flips and modify the generators accordingly to eliminate the factor of 2, giving a quantitative (though not at all sharp!) bound of $\mathrm{min}_{\hat{b} = K}\mathrm{length}(b)\mathrm{width}(b)$ for the number of people needed.<|endoftext|>
TITLE: Minimal possible cardinality of a $(a_1, ..., a_k)$-distributable multiset
QUESTION [12 upvotes]: Suppose we have a multiset $M$ of positive rational numbers. Sum of $M$ equals $1$. We'll call this multiset $n$-distributable for some $n\in \mathbb{N}$, if there exists a partition $M_1 \sqcup ... \sqcup M_n$ of the $M$ such that the sum of each (multi)subset $X_i$ equals $\frac{1}{n}$. If the multiset is $n$-distributable and $m$-distributable for some $n,m\in\mathbb{N}$, we will call it $(n,m)$-distributable, and so on. 
The problem is to find for some fixed $a_1, \ldots, a_k \in \mathbb{N}$ the minimal possible cardinality of a $(a_1, ..., a_k)$-distributable multiset.
Real-world analogy. You are organizing a party. You know that the number of guests to attend your party can be anything from $a_1, \ldots, a_k \in \mathbb{N}$.  In order to be prepared you cut the cake beforehand into smaller pieces, not necessarily of equal size. The requirement is that, no matter how many guests come, you will be able to give each of them some pieces of the cake without having to cut the cake any further so that everybody will get the same amount of cake.  What is the minimum number of pieces of your cake you will have to cut it into?  
The question. Formulated like this, is it a solved problem? If it's not, what specific cases are discussed and where could we read about it? If it is solved, well... basically the same, what mathematical branch is it and where to read about it? I tried to ask it on MSE here: https://math.stackexchange.com/questions/1381042/dividing-the-whole-into-a-minimal-amount-of-parts-to-equally-distribute-it-betwe. After that some other gentleman asked essentially the same question: https://math.stackexchange.com/questions/1383406/minimum-cake-cutting-for-a-party. The latter even has a bounty which is running out, but noone seems to know the answer (or maybe we chose wrong tags). I tried to dig into this problem by myself, but being an amateur, all I wind up with is a lot of specific made up terminology, some useless properties, couple of toy model cases solved and a constant feeling that I'm trying to invent a bicycle. Any insight would be greatly appreciated.

REPLY [2 votes]: Too long for a comment.
For $k=2$ the answer is $a_1+a_2-\mathrm{gcd}(a_1,a_2)$, this is well known. The text below is essentially a duplicate of math.stackexchange.com/a/1388683/87355 and other existed proof.
The example is to take a circular cake, inscribe a regular $a_1$-gon and a regular $a_2$-gon which have $\mathrm{gcd}(a_1,a_2)$ common vertices, and use for sector partitions.
The estimate may be done as follows: if the first party is for $a_1$ women and the second for $a_2$ men, draw the edge between, say, Ann and Bob if they share a virtual piece of cake. If a connected component contains $b_1$ women and $b_2$ men, we get $b_1/a_1=b_2/a_2$ by a double counting of the cake. That yields $b_1\geqslant a_1/\mathrm{gcd}(a_1,a_2)$ and the total number of components does not exceed $\mathrm{gcd}(a_1,a_2)$. Thus the number of edges is not less than $a_1+a_2-\mathrm{gcd}(a_1,a_2)$.<|endoftext|>
TITLE: A function with positive decreasing Taylor coefficients?
QUESTION [7 upvotes]: The function $\frac{x}{\ln(1-x)}$ has a Taylor series $-1+c_1x+c_2x^2+\cdots$ and I want to show $c_1>c_2>\cdots>0$. More generally, is there a result about how a function has positive Taylor coefficients or has decreasing Taylor coefficients?

REPLY [5 votes]: We have
$$-\int_{-1}^0 {1\over {(z+1)^t }}dt={ {-z}\over{\log(1+z)}}, \ \ \ \int_{-2}^{-1}{1\over {(z+1)^t }}dt={{z(z+1)}\over{\log(1+z)}}, \ \ z>0$$
So the k-derivative of these functions is positive when k is even and negative  when k is odd.
So the coefficients of the Taylor series of ${z\over{\log(1-z)}},{{z^2-z}\over{\log(1-z)}}$ are positive, as desired.<|endoftext|>
TITLE: Is Rasmussen's s-invariant of a knot an invariant of a 4-manifold?
QUESTION [15 upvotes]: Let $K$ be a knot in the 3-sphere $S^3$.
Here we  denote by $s(K)$  Rasmussen's s-invariant for $K$, 
and by $X_{K}(n)$ the 4-manifold obtained from the standard 4-ball $B^4$
by attaching a $2$-handle along $K$ with framing $n$.
My question is the following.

For two knots $K$ and $K'$ such that  $X_{K}(0)$ and  $X_{K'}(0)$ are diffeomorphic,
  is it true that $s(K)=s(K')$?

I am also interested in the same question for Ozsváth-Szabó's $\tau$-invariant. 
Note that it is known that there exist knots $J$ and $J'$ such that 
 $\partial X_{J}(0)$ and  $\partial X_{J'}(0)$ are diffeomorphic,
and   $s(J) \neq s(J')$, see Yasui's paper Corks, exotic 4-manifolds and knot concordance.

REPLY [8 votes]: Lisa Piccirillo just posted a few days ago this paper on the arXiv. Corollary 1.3 asserts exactly that $s$ is not a 0-trace invariant of the knot.<|endoftext|>
TITLE: Quasicategories for non-simplicial model categories
QUESTION [14 upvotes]: If I have a simplicially enriched model category, then I can take the coherent nerve of the full subcategory of bifibrant opjects to obtain a quasicategory.  If I have a model category that is not simplicially enriched, then I could take the hammock localisation first and then take the coherent nerve.  I have no technical objection to this, but I find it aesthetically displeasing.  I would prefer to have a construction that accepts an arbitrary model category $\mathcal{C}$ and produces a quasicategory $N'(\mathcal{C})$ with the same homotopy theory in a single step.  Ideally, all objects of $\mathcal{C}$ would appear as $0$-simplices in $N'(\mathcal{C})$, not just the bifibrant ones.  Does such a thing exist in the literature?
One idea (in the spirit of the thesis of James Cranch) is as follows.  Given a left proper model category $\mathcal{C}$, an $n$-simplex of $N'(\mathcal{C})$ would be an $n$-fold cospan diagram consisting of objects $X_{ij}$ for $0\leq i\leq j\leq n$.  There would be cofibrations  $X_{ij}\to X_{i,j+1}$ and weak equivalences $X_{i+1,j}\to X_{ij}$ and the resulting squares would be cocartesian.  I have not made any serious attempt to check whether this works.
Alternatively, it may be that an answer is essentially contained in the paper of Barwick and Kan on Relative Categories, but I have not properly digested that.

REPLY [7 votes]: You can consider the marked simplicial set $(N(\mathcal{C}),\mathcal{W})$, where $N$ is the usual nerve functor and $\mathcal{W}$ is the class of weak equivalences in your model category $\mathcal{C}$. Then take $N'(\mathcal{C})$ to be the underlying simplicial set of any fibrant replacement of $(N(\mathcal{C}),\mathcal{W})$ in the Cartesian model structure. This paper by Hinich shows that this procedure gives the "correct" $\infty$-category associated to $\mathcal{C}$. Perhaps some unwinding of definitions will give you a more explicit construction.<|endoftext|>
TITLE: Countable path-connected Hausdorff space
QUESTION [7 upvotes]: Is there a topology $\tau$ on $\omega$ such that $(\omega,\tau)$ is Hausdorff and path-connected?

REPLY [14 votes]: I realise this is quite an old question, but here is an alternative to the already existing elegant answers, which shows that Hausdorff can be replaced by $T_1$:

Proposition. No countable $T_1$ space containing at least two points is the continuous image of an interval.

Proof. Suppose $X$ is a $T_1$ topological space (so points are closed), $I:=[0,1]$, and $f\colon I\to X$ is continuous. Then
$$ \{f^{-1}(x)\colon x\in X\}$$
is a partition of $I$ into closed, hence compact, subsets. But $I$ (more generally, any non-trivial continuum) is not the countable disjoint union of compact proper subsets. So $X$ is uncountable. $\square$
Observe that the proof, in contrast to Joel's, does not show that the cardinality of $X$ is that of the continuum. On the other hand, there are non-Hausdorff compact $T_1$ spaces, so his proof does not apply to $T_1$ spaces. Likewise for Todd's argument, as there are $T_1$ spaces that are path-connected, but not arcwise connected. 
On the other hand, Sierpiński space (consisting of two points, only one of which is closed) is both $T_0$ and path-connected. So $T_1$ cannot be replaced by $T_0$.
EDIT. It seems this argument is also here: http://topospaces.subwiki.org/wiki/Path-connected_and_T1_with_at_least_two_points_implies_uncountable<|endoftext|>
TITLE: A question about $O(3,1)$
QUESTION [6 upvotes]: Recall that $O(3,1)$ is the collection of matrices $A\in M_4(\mathbb R)$ such that
$$A\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}A^T=\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}.$$
Let $\psi\in \mathfrak o(3,1)$ be an element in its Lie algebra. Can we find a $Q\in O(3,1)$ such that 
$$Q\psi Q^{-1}=\begin{pmatrix} 0&a&&\\-a&0&&\\&&0&b\\&&b&0\end{pmatrix}$$
for some $a,b\in \mathbb R$?
Formally, for any $\psi\in \mathfrak o(3,1)$ can we find an $\hat \psi \in \mathfrak o(2)\oplus \mathfrak o(1,1)$, such that $\psi, \hat \psi$ belong to the same $O(3,1)$-cojugate class?

REPLY [13 votes]: No. E.g. $\psi=\begin{pmatrix}0&0&-1&1\\0&0&0&0\\1&0&0&0\\1&0&0&0\end{pmatrix}$ is nilpotent: $\psi^3=0$. If your $\phi=\begin{pmatrix}0&a&0&0\\-a&0&0&0\\0&0&0&b\\0&0&b&0\end{pmatrix}$ was $Q\psi Q^{-1}$, we would have $\phi^3=\begin{pmatrix}0&-a^3&0&0\\a^3&0&0&0\\0&0&0&b^3\\0&0&b^3&0\end{pmatrix}=0$, whence $a=b=\phi=\psi=0$, contradiction.

Edit (Re: your amended question below): If all (or even some) eigenvalues of $\psi$ are nonzero then yes, $\psi$ is conjugate to one of your $\phi$'s. It is equivalent but easier to classify adjoint orbits of the covering group $\mathrm{SL}(2,\mathbf C)$, whose Lie algebra consists of all
$$
Z=J(\mathbf z) := \frac1{2i}\begin{pmatrix}z_3&z_1-iz_2\\z_1+iz_2&-z_3\end{pmatrix},
\qquad \mathbf z = \mathbf a + \mathbf bi\in\mathbf C^3.
$$
The Lie algebra isomorphism maps $Z$ to $\psi=\begin{pmatrix}j(\mathbf a)&\mathbf b\\{}^t\mathbf b&0\end{pmatrix}$ where $j(\mathbf a)=\left(\begin{smallmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{smallmatrix}\right)$.
Now $\det(\lambda\mathbf1-2iZ)=\lambda^2 - (\mathbf z,\mathbf z)=(\lambda+\zeta)(\lambda-\zeta)$ where $\zeta$ is a square root of $(\mathbf z,\mathbf z):=$ $z_1^2+z_2^2+z_3^2$ $=\|\mathbf a\|^2-\|\mathbf b\|^2+2i(\mathbf a,\mathbf b)$. So there are two cases:

if $(\mathbf z,\mathbf z)= 0$ then $Z$ is nilpotent, hence either 0 or (Jordan) conjugate to $\smash{\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)}$ $= J(i\mathbf e_1-\mathbf e_2)$ which maps to my nilpotent $\psi$ above (eigenvalues: 0).
if $(\mathbf z,\mathbf z)\ne 0$ then $Z$ is diagonalizable and hence conjugate to $\smash{\frac1{2i}\left(\begin{smallmatrix}\zeta&0\\0&-\zeta\end{smallmatrix}\right)}=J(\zeta\mathbf e_3)$ which maps to your $\phi$ if $\zeta = a + bi$ (eigenvalues: $\pm ai,\pm b$).<|endoftext|>
TITLE: Maximal Number of Pairs of Orthogonal vectors in a set of $n$ vectors in $\mathbb{R}^3$
QUESTION [12 upvotes]: Suppose you are given a set of $n$ non-zero vectors in $\mathbb{R}^3$. What is the maximum number of pairs of them that are orthogonal? The current guess is $\le 2n$.
EDIT: I forgot to add that no two vectors should be colinear.

REPLY [12 votes]: The maximum is $cn^{4/3}$ for some constant $c$.
We may as well assume all our points are on the unit sphere $S$. Let $P$ be some plane not containing the origin, which we might think of as being far away. For each point $x\in S$ in our collection let $p_x$ be the intersection of the line through $0$ and $x$ with $P$ and let $\ell_x$ be the intersection of the hyperplane orthogonal to $x$ with $P$. Unless $P$ was chosen by your enemies then all these things are well defined. Note that $p_x\in\ell_y$ iff $x$ and $y$ are orthogonal, so orthogonality in our original collection becomes incidence in our new collection, and by the Szemeredi-Trotter theorem there are at most $O(n^{4/3})$ incidences.
To see that there is a construction with this many orthogonal pairs, take some example which shows that Szemeredi-Trotter is tight, and such that there are about as many points as lines, and read the above paragraph backwards. If $P$ is far away then this example will consist of two collections of about $n/2$ points carefully clustered around some two orthogonal points of $S$.
I read this construction in the following paper of Erdos, Hickerson, and Pach, which also contains references to many other things. See Theorem 2(ii) for this problem.
http://www.renyi.hu/~p_erdos/1989-02.pdf<|endoftext|>
TITLE: On shifted symmetric power sums
QUESTION [12 upvotes]: The functions $p^*_k(x)=\sum_{i=1}^N ((x_i-i)^k-(-i)^k)$ are analogues of power sum symmetric functions, called shifted symmetric by Okounkov and Olshanski. Define $p^*_{(k_1,k_2,...)}=p^*_{k_1}p^*_{k_2}...$.
We know that the sum $\frac{1}{n!}\sum_{\mu\vdash n}C_\mu \chi_\lambda(\mu)p_\mu(x)=s_\lambda(x)$ returns Schur functions (where $C_\mu$ is the size of a conjugacy class in the permutation group and $\chi$ are the irreducible characters of this group).
My question is: what is known about $f_\lambda(x):=\frac{1}{n!}\sum_{\mu\vdash n}C_\mu \chi_\lambda(\mu)p_\mu^*(x)$? Is this related to shifted Schur functions? What properties does this function share with the usual Schur functions?

REPLY [5 votes]: The super Schur function $s_\lambda(x/y)$ is obtained by applying to
$s_\lambda(x,y)$ (a Schur function in two sets of variables
$x=(x_1,x_2,\dots)$ and $y=(y_1,y_2,\dots)$) the algebra homomorphism
$\omega_y\colon \Lambda(x,y)\to\Lambda(x)\otimes \Lambda(y)$, where
$\Lambda(z)$ is the ring of symmetric functions in the variables $z$,
defined by 
  $$ \omega_yp_n(x,y)=p_n(x)+(-1)^{n-1}p_n(y). $$
Equivalently,
   $$ s_\lambda(x/y) =
   \sum_{\mu\subseteq\lambda}s_\mu(x)s_{\lambda'/\mu'}(y). $$
Thus $f_\lambda(x)=s_\lambda(x_1-1,x_2-2,\dots/1,2,3,\dots)$.<|endoftext|>
TITLE: solid commutative ring spectra
QUESTION [5 upvotes]: Let $R$ be a discrete (i.e. an ordinary) commutative ring and let $HR\rightarrow T$ be a map of $E_{\infty}$-ring spectra where $HR$ is the associated Eilenberg-Mac Lane ring spectrum. We say that $T$ is $R$-solid (in the derived sense) if the induced map of $E_{\infty}$-ring spectra
$$T\wedge_{HR}^{\mathbb{L}}T\rightarrow T$$ is a weak equivalence. 
Are there some nontrivial examples of such solid ring spectra? By "nontrivial" I mean that $T$ is not discrete. The discrete case (over $\mathbf{Z}$) is classified here .

REPLY [4 votes]: Here is an example of such pair $T$ and $R$. 

$k$ is a finite field.
$G$ is the Lie group $SO(2)$ and $G^{\delta}$ the same group but with dicrete topology.
$R$ is the group algebra $k[G^{\delta}]$.
$T$ is $C_{\ast}(G,k)$ the singular chain complex associated to $G$.

Then the natural map $k[G^{\delta}]\rightarrow C_{\ast}(G,k)$ of $E_{\infty}$-algebras induces a quasi-isomorphism 
$$ C_{\ast}(G,k)\otimes_{k[G^{\delta}]}^{\mathbf{L}} C_{\ast}(G,k)\rightarrow C_{\ast}(G,k)$$ of $E_{\infty}$-algebras.<|endoftext|>
TITLE: If $C$ has all geometric realizations of simplicial objects, what other colimits does it have?
QUESTION [5 upvotes]: Let $C$ be an $\infty$-category.  Suppose that every diagram $\Delta^{\mathit{op}} \to C$ has a colimit.  Is there any characterization of small categories $I$ such that every diagram $I \to C$ has a colimit?

REPLY [5 votes]: This is not a definitive answer, but more an attempt to provide some background.
You're essentially asking for the saturation, in the $\infty$-categorical setting, of (the terminal weight on) $\Delta^{\mathrm{op}}$. The saturation has been studied in the abstract by Albert and Kelly in the enriched case. In this setting, it's natural to work with weighted colimits. Albert and Kelly observe that it's natural to consider a slightly stronger property: you should look at diagrams $D$ such that every $\infty$-category with geometric realizations has $D$-colimits and every geometric-realization-preserving functor between categories with geometric realizations preserves $D$-colimits. Such a $D$ is said to lie in the saturation of $\{\Delta^{\mathrm{op}}\}$. Albert and Kelly actually show that it doesn't make a difference which question you consider in the case of weighted colimits, but they leave this question open for conical colimits.
Albert and Kelly's main result (in the case of enriched categories) is that a weight $\phi$ lies in the saturation of $\Phi$ if and only if $\phi$ lies in the free cocompletion of (the opposite of) its domain under $\Phi$-colimits. Presumably something similar should hold in the $\infty$-categorical setting. (I don't know whether weighted limits appear in HTT, but they shouldn't be hard to define, and at any rate the definition should be derivable in a standard way from Riehl and Verity's modules.
Some well-known saturated classes include filtered colimits, sifted colimits, simply-connected limits, the L-finite limits, which are the saturation of finite limits in ordinary category theory, and absolute colimits, which are the saturation of the empty class of colimits (for any fixed enriching category).
In 1-category theory, it is "almost" the case that the saturation of filtered colimits + reflexive coequalizers is all sifted colimits. See, for example here (though better results are now available). And my understanding is that this "near equation" persists to $\infty$-category theory. In any event, the nearest 1-categorical analog of $\Delta^{\mathrm{op}}$ is the weight for reflexive coequalizers, so we can look at the saturation of this class in 1-category theory.
By Albert and Kelly's result, to compute the saturation in enriched category theory, it suffices to compute the free cocompletion under this class of colimits. An elementary description of the free completion under reflexive coequalizers is given, e.g. in Adámek et al's book Algebraic Theories, Ch. 17. I don't know whether the free completion under geometric realization is similarly complicated in $\infty$-category theory.<|endoftext|>
TITLE: The Gelfand duality for pro-$C^*$-algebras
QUESTION [20 upvotes]: The Gelfand duality says that 
$$X\to C(X)$$ 
is a contravariant equivalence between the category of compact Hausdorff spaces and continuous maps and the category of commutative unital $C^*$-algebras and continuous $*$-homomorphisms. A well known generalization of $C^*$-algebras are pro-$C^*$-algebras. Pro-$C^*$-algebras are topological $*$-algebras that are cofiltered limits of $C^*$-algebras (in the category of topological $*$-algebras). See for instance this paper by Phillips. If $X$ is a weakly Hausdorff compactly generated space, it is not hard to see that $C(X)$, the continuous functions from $X$ to $\mathbb{C}$, is a pro-$C^*$-algebra. Indeed, it is the cofiltered limits of $C(K)$, as $K$ ranges over all compact Hausdorff subspaces of $X$. My question is:

Is the functor 
  $$X\to C(X)$$ 
  a contravariant equivalence between the category of weakly Hausdorff compactly generated spaces and continuous maps and the category of commutative unital pro-$C^*$-algebras and continuous $*$-homomorphisms?

Few remarks:

In the paper by Phillips mentioned above, Theorem 2.7, it is shown that the functor above is a contravariant equivalence between the category of completely Hausdorff quasitopological spaces and the category of commutative unital pro-$C^*$-algebras. It is also shown, in Example 2.11, that when restricted to the full subcategory of completely Hausdorff compactly generated spaces, this functor is not essentially surjective. But this still doesn't answer my question, because maybe on weakly Hausdorff compactly generated spaces this functor is essentially surjective. 
Apparently, the reason Phillips insists on working with completely Hausdorff (quasi)topological spaces is that the way he proves this result is by constructing an inverse equivalence which assigns to any commutative unital pro-$C^*$-algebra its spectrum. The way he defines the spectrum, it can be shown that if $A$ is a commutative unital pro-C*-algebra, the spectrum of $A$ is completely Hausdorff. But, again, this doesn't answer my question, because maybe one can define an inverse equivalence in a different way, or maybe one can just show that this functor is fully-faithful and essentially surjective.  
The main reason I am asking is that the category of weakly Hausdorff compactly generated spaces is very important in algebraic topology and homotopy theory, and such a result would imply that pro-$C^*$-algebras are exactly the non-commutative version of it.

Remark: I used the terminology of the paper of Phillips that I referred to. Note that the category of pro-$C^*$-algebras is not the pro-category of the category of $C^*$-algebras. In particular, its objects are topological $*$-algebras and not cofiltered diagrams of $C^*$-algebras. Some authors call pro-$C^*$-algebras locally $C^*$-algebras. It can be shown that the pro-category of the category of $C^*$-algebras contains the category of pro-$C^*$-algebras, as a full coreflective subcategory. 
Edit: Due to Simon's answer the functor $X\to C(X)$ defined above is clearly not faithful. But the following question remains: Is this functor full and essentially surjective. If so then it would induce a contravariant equivalence of categories between  $\overline{CGWH}$ and the category of commutative unital pro-$C^*$-algebras, where $\overline{CGWH}$ is the category of weakly Hausdorff compactly generated spaces and equivalence classes of continuous maps, where two continuous map $X\rightrightarrows Y$ are called equivalent if the induced maps $C(Y)\rightrightarrows C(X)$ are the same.

REPLY [5 votes]: The answer is yes, provided you change the category on the topological side slightly to: compactly generated functionally Hausdorff topological spaces with a distinguished family of compact sets; with continuous maps that preserve the distinguished family. See Theorem 6 of 

Michael Forger, Daniel V. Paulino, Locally $C^*$ Algebras, $C^*$ Bundles and Noncommutative Spaces, arXiv:1307.4458v1<|endoftext|>
TITLE: Submersion theorem for smooth tame Frechet manifolds
QUESTION [7 upvotes]: If $M$ and $N$ are Banach manifolds, $f:M\rightarrow N$ is a smooth map, and $q\in N$ is a regular value, so $f$ is a submersion on $f^{-1}(q)$, it is well known that the level set $f^{-1}(q)$ is a regular submanifold of $M$.
Question:  Is there an analogous result for maps between manifolds modeled off locally convex spaces, in particular maps between smooth tame Frechet manifolds?
Question:  Let $U$ be an open set in a smooth tame Frechet space $E$ (hence $U$ is a smooth tame Frechet manifold), $F$ be a smooth tame Frechet space, and $f:U\rightarrow F$ a smooth tame map.  Is there a submersion or regular level set theorem in this context?

REPLY [7 votes]: There are a few works which study submersions in the locally convex setting. The most extensive (and recent) is by Helge Glöckner (http://arxiv.org/abs/1502.05795). Note that the basic results about submersions and immersions still hold true in this general setting, at least if you use the right notion of sub/immersion (i.e. a map being locally a projection or injection). The problem is essentially that you cannot easily go from infinitesimal results to local ones (as this step requires an inverse function theorem). Submanifolds of co-Banach type are discussed in http://arxiv.org/abs/1305.3145. Moreover, Hamilton's 1982 treatise of the Nash-Moser inverse function theorem also contains some results about submanifolds in the tame Fréchet context. However, these are rather hidden and only worked-out in specific examples (for example, his proof that the volume-preserving diffeomorphism group is a Lie subgroup of all diffeomorphisms is essentially a regular-value type argument).<|endoftext|>
TITLE: How simple does a $\mathbb{Q}$-simple group remain after base change to $\mathbb{Q}_{\ell}$?
QUESTION [6 upvotes]: Of course the general answer to the question in the title is: not very simple.
I could not think of a better title, so let me explain my question in more detail.
I have a number field $E/\mathbb{Q}$, and a simple group $H/E$.
Let $G$ denote the Weil restriction of scalars of $H$ from $E$ to $\mathbb{Q}$.
Then $G$ is a $\mathbb{Q}$-simple group. 
The group $G_{\ell} := G_{\mathbb{Q}_{\ell}}$ is not $\mathbb{Q}_{\ell}$-simple in general.
Short aside: However, if $n = [E : \mathbb{Q}] \le 3$ there is always an $\ell$ such that $G_{\ell}$ is $\mathbb{Q}_{\ell}$-simple. (For a short proof: consider the Galois action of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ on the $n$ factors of $G_{\bar{\mathbb{Q}}}$. This action is transitive. Since $n \le 3$, there must therefore be an $n$-cycle in the action, and by Chebotarev there will be (infinitely many) primes $\ell$, such that $\mathrm{Gal}(\bar{\mathbb{Q}_{\ell}}/\mathbb{Q}_{\ell})$ acts transitively on the factors of $G_{\bar{\mathbb{Q}_{\ell}}}$. [Proof copied from D. Lombardo, arXiv: 1402.1478])
My question is about what happens if $n > 3$. (In my particular case $n \le 10$.) For example, I would be very happy, if there is always a prime $\ell$, so that all the $\mathbb{Q}_{\ell}$-simple factors of $G_{\ell}$ are not absolutely simple. Is something like that true?

REPLY [3 votes]: I assume that the question is:  Is it true that there is  always a prime $\ell$ such that $G_\ell$ has no absolutely simple direct factors? 
The answer is YES. There are infinitely many such $\ell$.

Claim. Let $\Gamma$ be a finite group acting transitively on a finite set $S$ of cardinality $n>1$. 
  Then $\Gamma$ has a cyclic subgroup $\Delta\subset \Gamma$ acting on $S$ without fixed points.

Proof of the claim (due to Guntram). Assume the contrary. 
Then any element $\gamma$ of $\Gamma$ is contained in the stabilizer $\Gamma_s$ of some point $s\in S$,
hence $\Gamma=\bigcup_s\Gamma_s$.
Since $\Gamma$ acts transitively  on $S$, each subgroup $\Gamma_s$ is of index $n$ in $\Gamma$. 
Since $n>1$ and $1\in \Gamma_s$ for all $s\in S$, we obtain that
$$ \#\bigcup_s\Gamma_s\le 1+ n\cdot (\#\Gamma_s-1)<\#\Gamma, $$
contradiction.
Now let $S$ denote the set of embeddings of $E$ into $\overline{\mathbb{Q}}$. The Galois group
of $\overline{\mathbb{Q}}$ over ${\mathbb{Q}}$ acts on $S$ via a finite quotient group $\Gamma$.
Let $\Delta=\langle\gamma\rangle$ be as in the Claim. Then
by Chebotarev's density theorem there exist infinitely many primes $\ell$ unramified in $E$ 
such that the corresponding Frobenius is conjugate to $\gamma$.
For such an $\ell$, the tensor product $E\otimes_{\mathbb{Q}} {\mathbb{Q}}_\ell$ 
does not contain ${\mathbb{Q}}_\ell$ as a direct summand, and therefore, 
$G_\ell$ has no absolutely simple direct factors.<|endoftext|>
TITLE: When is the time one map of a suspension flow ergodic?
QUESTION [5 upvotes]: I'm sure the answer to the following question is well known but I couldn't find the answer I needed.
Let $(\Sigma,\sigma)$ denote the full shift on $k$ symbols and let $\mu$ be an invariant measure for which the system is ergodic. Let $f:\Sigma\to\mathbb R^+$ be a Holder continuous roof function. Let $$\Sigma_f:=\{(\underline a,t):\underline a\in\Sigma, 0\leq t\leq f(\underline a)\} $$ and let $\mu_f$ denote the measure $(\mu\times Leb)|_{\Sigma_f}$. Let $\phi$ denote the usual suspension flow on $\Sigma_F$, it is known that $(\Sigma_f, \phi,\mu_f)$ is ergodic. 
Let $T:\Sigma_f\to\Sigma_f$ denote the time one map associated to $\phi$, $T$ preserves measure $\mu_f$. Under what conditions on $f$ is the system $(\Sigma_f, T, \mu_f)$ ergodic?
My guess is that the system is ergodic unless $f$ is cohomologous to a rational constant,  but if anyone knows the correct statement and a reference for it that would be great.

REPLY [2 votes]: Rufus Bowen proved  a topological version in [1] for a basic set $X$ of an Axiom A flow $\phi_t:M\to M$. Under your setting, Theorem 3.2 in [1] gives a dichotomy of $X$:

either $\Phi$ on $X$ is the constant suspension of a basic set of Axiom A map,
or the (strong) unstable leaf $W^u(p)$ is dense in $X$ for each periodic point $p\in X$.

In the first case, it is clear that the time-1 map $T$ is ergodic if and only if the constant is irrational.
In the second case, the time-1 map $T:X\to X$ is topologically mixing. Under your setting, the system $(T,\mu_f)$ should be ergodic, but I didn't find a short argument. 

[1] R. Bowen. Periodic Orbits for Hyperbolic Flows. Amer. J. Math. 94, No. 1 (1972), 1--30.<|endoftext|>
TITLE: Listing ORCiD in LaTeX papers
QUESTION [20 upvotes]: The ORCiD unique author identifier, run by a non-profit organisation, has been around for a number of years now. Its stated goal is to become a de facto standard for uniquely identifying authors, even in cases where they have non-unique names, have changed their name, or have had their name appear in different variants on different publications. A large number of organisations, including the American Mathematical Society, have signed up to the system. (However, integration with MathSciNet has not happened so far.) 
At the moment, users maintain their own record manually, but there is a possibility that papers would eventually be added automatically to a record as they appear, as the functionality of publishers automatically updating records has recently been implemented. According to this link, the metadata provided by publishers to Crossref for assigning a DOI includes ORCiD identifiers, and Crossref will at some point start to automatically update corresponding the records of those users who have permitted this. 
However, this raises the question of how the ORCiD identifier is provided to the publisher in the first place. So far my experience has been that few journals have the option of including such an identifier either at submission or acceptance. (Of course, this may change with recent developments). At some point, I saw some advice - which unfortunately I am now unable to locate - to routinely list your ORCiD within the text of the paper. I wonder what the best way of doing so would be. 

Question. Is there by now an accepted way (in mathematics) of including an ORCiD as part of the author information, say when using LaTeX with the amsart article class? If so, what is it?

I would also be interested to hear whether this would actually be worthwhile: 

Is there any evidence that journals, as part of their processing, will pick up on ORCiD identifiers listed in papers and pass these on automatically as part of the metadata?

Nb. While my main question concerns LaTeX usage, I feel that it is not primarily a technical question, but rather one concerning academic practice, and specifically within mathematics. Hence mathoverflow seems an appropriate place for it, rather than tex.stackexchange or academiae.stackexchange. But of course if others disagree I am open to migrating.

(The lead-in to this question has been edited to be more factual, removing some of my personal opinions on the identifier system that may be irrelevant.)

REPLY [8 votes]: I recently had a paper accepted where I included the ORCiD in the address field of the LaTeX code, as suggested in my comment to Carlo's answer. The publisher did not ask me for any form-based entry as part of the publication process.
The publisher (Springer) did automatically recognise the ORCiD identifier from this and included it in the metadata. IIRC the paper was automatically added to my ORCiD record by Crossref upon publication as a result.
This may not be representative of other publishers, but I do conclude that it is a good idea to include an ORCiD in the address field, and that there is evidence that this is picked up as part of the publication process.<|endoftext|>
TITLE: Is every polynomial a factor of a trinomial?
QUESTION [6 upvotes]: We say that $T(X) \in \mathbb{Q}[X]$ is a trinomial if there exist $A,B,C \in \mathbb{Q}$ such that $T(X) = AX^n + BX^m + C$ for some $n \geq m \in \mathbb{N}$.

Is it true that for each irreducible $g(X) \in \mathbb{Q}[X]$ there exists some
  nonzero $h(X) \in \mathbb{Q}[X]$ such that $g(X)h(X)$ is a trinomial?

If no, then what are some necessary/sufficient conditions on $g$ for the existence of such $h$?

REPLY [8 votes]: To summarize some of the discussion in the comments: a trinomial can have at most four distinct real roots. A random polynomial of degree $d$ (random = all coefficients are iid, there are other models) has $\Omega(\log d)$ real roots. Which means that the probability that a random polynomial divides a trinomial goes to zero (exponentially fast with degree, though this is a little harder to prove).<|endoftext|>
TITLE: Kissing number and overlapping number
QUESTION [7 upvotes]: Let $S$ be a certain family of geometric objects (e.g, the family of unit squares).
The kissing number of $S$ is the maximum number of nonoverlapping elements of  $S$ that can touch one element of $S$. See here an example where $S$ is the family of unit-squares: The kissing number of a square, cube, hypercube? 
Define the overlapping number of $S$ as the the maximum number of nonoverlapping elements of  $S$ that can overlap one element of $S$. 
Obviously, the overlapping number is not larger than the kissing number since, if the central object has to be overlapped (not only touched), there is less room for the peripheral objects.
What else is known about the relation between the kissing number and the "overlapping number"? In particular, is there any bound on the ratio: kissing-number/overlapping-number?
EXAMPLE: When $S$ is the family of axis-parallel unit-squares, the overlapping number is 4 and the kissing number is 8. So the ratio is 2.

REPLY [7 votes]: Even for a set almost the same as the one you mention in your question, unit squares but with the four corners removed, the overlapping number is larger than the kissing number. Only six cornerless squares can kiss another cornerless square, because kissing can only happen along the sides of a square, and if the kissed square has two kissing squares on one of its sides then it must have only one each on the two adjacent sides. However, seven nonoverlapping cornerless squares can overlap a central cornerless square:<|endoftext|>
TITLE: Inverse Galois problem for $GL_2$ of a compact local ring
QUESTION [8 upvotes]: Let $A$ be complete noetherian local ring with maximal ideal $m$ and residue field $A/m$ a finite field (in other words, $A$ is a noetherian compact local ring)

For which $A$ as above is there a subgroup of $GL_2(A)$ containing $SL_2(A)$ which is the Galois group of a Galois extension of $\mathbb Q$
  unramified outside a finite set of primes $S=S(A)$?

(The italic part of the question above has been edited: the first, ill-formulated, version of this question asked if $GL_2(A)$ itself was a Galois group over $\mathbb Q$ 
 and has been answered by Will Savin below).
For example, we know since the work of Serre on the points of torsion of elliptic curves that when $A=\mathbb Z_p$, the answer is yes. On the other hand, the answer should not be always yes: when $A$ has Krull dimension $5$ or more for instance, such a group would define
a deformation to $A$ of the representation $G_{\mathbb Q,S} \rightarrow GL_2(A/m)$, which would have to be a quotient of the universal deformation ring of that representation, and these deformation ring are expected to have dimension at most $4$ (see Mazur's article in Galois Groups over A).
Obviously, I would be very very surprised if I received a complete answer to this question, but is it possible to give a conjectural answer, based on the most optimistic conjectures on the inverse Galois problems (e.g. Shafarevich's conjecture that $Gal(\mathbb Q/ \mathbb Q^{ab})$ is a free profinite group of countable rank), or at least, some reasonable guess?

REPLY [2 votes]: This isn't an answer, just some idle thoughts about the following closely related problem:
Which complete local Noetherian rings $A$ 
can occur as the quotient of the universal deformation ring $R$ of a representation
$\rho: G_{\mathbb{Q},S} \rightarrow \mathrm{GL}_2(\mathbb{F}_q)$? 
Such a universal ring $R$ should always have dimension $4$ and  be a complete intersection, at least conjecturally.   
So straight away  the following question occurs:
is the set $\mathcal{X}$ of   isomorphism classes of complete, local, Noetherian, integral,  $4$-dimensional  $W(\mathbb{F}_q)$-algebras actually uncountable?    (Conditions slightly edited)
I don't know the answer  but if indeed $\mathcal{X}$ is uncountable, that
seems to be a problem:  Only finitely many of the elements of $\mathcal{X}$ could occur as quotients of each of the countably many  universal $R$s (namely, one for each minimal prime ideal of $R$).  
On the other hand, I think one  might be able to verify   that
the set of universal deformation rings is dense in $\mathcal{X}$ (for  some reasonable topology).  For example,
I believe that any Artin local  $A$ is a quotient of an $R$ as above. This  should follow from the method of Taylor and Wiles: they show that you can get a formal power series ring, in as many variables as you want, as a limit of such deformation rings $R$.<|endoftext|>
TITLE: Colorings of triangulations as a generalization of the four-color problem
QUESTION [8 upvotes]: My question can be viewed as a generalization of the four-color problem. Instead of a planar graph, consider a triangulation of a d-sphere. One wants to color vertices with N colors so that no two vertices connected by a 1-simplex have the same color. For a fixed d, is there an N which suffices for any triangulation? What are known upper bounds on such N?

REPLY [3 votes]: Already for $d=3$ the boundary complex of the cyclic polytope with n vertices has a complete graph and hence requires $n$ colors. 
You may ask if restricting the class of triangulations can lead to interesting extension of the FCT. Some answers in Generalizations of the Four-Color theorem are relevant.<|endoftext|>
TITLE: Generalising the Penrose Twistor Fibration
QUESTION [8 upvotes]: As is well known, there exists a fibration $\mathbb{CP}^3 \to S^4$, of the four sphere by complex projective $3$-space, called the Penrose twistor fibration. Does this fibration admit a "canonical" generalisation to a fibration of $S^6$ by some projective $n$-space, or possibly flag manifold. What about general $2n$-spheres?

REPLY [19 votes]: Yes, there is such a twistor fibration over each $S^{2n}$, and the resulting manifold is a complex manifold endowed with a holomorphic $n$-plane field transverse to the fibers of the mapping. Namely, one writes $S^{2n} = \mathrm{SO}(2n{+}1)/\mathrm{SO}(2n)$ and then, using the inclusion $\mathrm{U}(n)\subset\mathrm{SO}(2n)$, one has the coset fibration
$$
Z_n = \mathrm{SO}(2n{+}1)/\mathrm{U}(n)\longrightarrow\mathrm{SO}(2n{+}1)/\mathrm{SO}(2n).
$$
The manifold $\mathrm{SO}(2n{+}1)/\mathrm{U}(n)$ canonically has the structure of a complex manifold and is now known as the twistor space of $S^{2n}$.  Interestingly, this space was already used in 1967 by Calabi in his study of minimal $2$-spheres in $S^{2n}$ (Minimal immersions of surfaces in Euclidean spheres, J. Differential Geom. 1 (1967), 111–125), where, using a quite different language, he showed that such immersions all have lifts to the twistor space as holomorphic curves that are tangent to a certain holomorphic plane field on $Z_n$.
Now, there are generalizations of this picture for each of the so-called 'inner' symmetric spaces $G/K$ where $K$ is the fixed subgroup of an involution that is an inner automorphism of $G$.  The twistor fibration is of the form $G/U\to G/K$ where $U\subset K$ is a subgroup such that $K/U$ (the typical fiber of the fibration) is an Hermitian symmetric space.  (There are also other kinds of twistor spaces over $G/K$ that are flag manifolds of the form $G/T$ where $T\subset K$ is a maximal torus.)
For more details on these topics, especially the generalization to the inner symmetric spaces and flag manifolds, one can, for example, have a look at my paper Lie groups and twistor spaces,  Duke Mathematical Journal 52 (1985), pp. 223–261.<|endoftext|>
TITLE: A Generalization of the Ahlfors function to have varying degrees?
QUESTION [8 upvotes]: It's a classical result of Ahlfors that, for any sufficiently nice n-connected domain $\Omega \subset \mathbb C$ there is a holomorphic branched covering $f: \Omega \rightarrow \mathbb D$ to the disk $\mathbb D$, which extends continuously to the boundary and maps the boundary curves of $\Omega$ monotonically onto the boundary of the disk.
Now, suppose we have $n$ positive integers, $d_1,...,d_n$, and denote the boundary curves of $\Omega$ by $\gamma_1,...\gamma_n$. Can one find a holomorphic map $f: \Omega \rightarrow \mathbb D$ which extends continuously to the boundary of $\Omega$ and maps each $\gamma_i$ onto $S^1$ with degree $d_i$? 
Even the case of $n=2$ seems difficult--one can readily achieve this with a real analytic function $f$ by a pretty basic interpolation. 
The proof of the Ahlfors result also seems to be less than helpful--it involves finding an extremal for the analytic capacity functional and then analyzing the regularity of the optimal function. See, for example:
Krantz's Geometric function theory: explorations in complex analysis, Theorem 4.5.9.

REPLY [4 votes]: The answer to your question is yes. Indeed, one can replace $\mathbb{D}$ with the right half plane, applying a Mobius transformation. Then the argument is quite simple if you are familiar with the following classical theorem of Bieberbach :
Theorem
Let $\Omega$ be a domain bounded by $n$ non-intersecting Jordan curves $\gamma_1,\dots,\gamma_n$. For $1\leq j \leq n$, fix one point $b_j$ in $\gamma_j$. Then there exists a degree $n$ holomorphic branched covering $F$ from $\Omega$ to the right half plane such that $F$ extends continuously to the boundary and $F(b_j)=\infty$ for each $j$. Moreover, $F$ is unique up to a positive multiplicative and an imaginary additive constant.
Using this, it is easy to prove the following, which implies the result that you want.
Theorem
Let $\Omega$ be a domain bounded by $n$ non-intersecting Jordan curves $\gamma_1,\dots,\gamma_n$. For  $1\leq j \leq n$, let $d_j$ be a positive integer. Then there exists a holomorphic branched covering $F$ from $\Omega$ to the right half plane which extends continuously to the boundary such that each restriction $F|_{\gamma_j}$ has degree $d_j$.
Proof Pick any points $a_1,\dots,a_{n-1}$ in $\gamma_1,\dots,\gamma_{n-1}$ respectively, and choose two distinct points $a_{n,1}$ and $a_{n,2}$ in $\gamma_n$. Let $F_1,F_2$ be branched coverings as in Bieberbach's theorem for the points $a_1,\dots,a_{n-1},a_{n,1}$ and $a_1,\dots,a_{n-1},a_{n,2}$ respectively. Then $F:=F_1+F_2$ is a holomorphic branched covering of $\Omega$ onto the right half-plane whose restriction to $\gamma_n$ has degree two and each other boundary restriction has degree one. We may continue this process and add other boundary points that map to infinity until we get the correct degrees. This completes the proof of the theorem.
All the details can be found in the paper The structure of the semi-group of proper holomorphic mappings of a planar domain to the unit disk by Bell and Kaleem.
See section 5, Proper holomorphic mappings of higher degree.<|endoftext|>
TITLE: Determine if an $n$-dimensional mesh of simplices is a non-manifold
QUESTION [7 upvotes]: In an $n$-dimensional space I have a set of simplices where each simplex consists of facets. Some of the simplices are 'connected' by sharing facets. Each facet is made up on edges, each consisting two points. In 3D each simplex is a tetrahedron consisting of triangular facets.
I would like to know, if the set of simplices I have is a 'non-manifold'.
In 3D I can check if my mesh is a 'non-manifold' as follows:
- select all facets having more or less than two neighboring simplices
- select all edges of these facets
- per edge, count the number of facets connected to the edge
- in case an edge has more of less than 2 neighboring facets, the edge is 'non-manifold'
This algorithm does only work in 3 dimension (3D). What about higher-dimensional sets of simplices? How can I determine, if an $n$-dimensional mesh of simplices is non-manifold?
Thank you!

REPLY [6 votes]: In all dimensions, something is a manifold if the link of every cell is a sphere. This, sadly, is undecidable if the dimension of your complex is at least five. It is decidable (but not quickly) for complexes of dimension 4 (google "thin position", "Jaco-Rubinstein-Thompson"). In low dimensions ($1, 2, 3$) it is easy.<|endoftext|>
TITLE: The formal p-adic numbers
QUESTION [45 upvotes]: The real numbers can be defined in two ways (well, more than two, but let's stick to these for now): as the Cauchy completion of the metric space $\mathbb{Q}$ with its usual absolute value, or as the Dedekind completion of the ordered set $\mathbb{Q}$ with its usual ordering.
When these two constructions are performed internally in a topos, they generally yield different results, and often it is the Dedekind construction that gives a more useful answer.  In particular, for any topological space $X$, the Dedekind real number object in $\mathrm{Sh}(X)$ is the sheaf of continuous $\mathbb{R}$-valued functions on $X$, where $\mathbb{R}$ has its usual topology.  (And this generalizes to some big toposes too.)  Roughly, this is because the definition of "a Dedekind real number" is constructively equivalent to "a point of the locale of formal real numbers", where the latter locale can be defined constructively, and classically turns out to be equivalent to the usual topological space $\mathbb{R}$.
However, the Cauchy definition of $\mathbb{R}$ has other generalizations: we can complete with respect to a non-Archimedean absolute value instead, obtaining the $p$-adic numbers $\mathbb{Q}_p$.  We can do this internally in a topos too, but as with $\mathbb{R}$, in general we will not get the "right" answer.  My question is, is there any way to construct "the $p$-adic numbers" constructively which, when interpreted in $\mathrm{Sh}(X)$, will yield the sheaf of continuous $\mathbb{Q}_p$-valued functions on $X$, where $\mathbb{Q}_p$ has its usual topology?  In particular, is there a "locale of formal $p$-adic numbers" that can be defined constructively and that classically is equivalent to the usual topological space $\mathbb{Q}_p$?

REPLY [8 votes]: Localic completion is awesome and is the way to answer your question as asked (which requests a locale).
However, a simpler answer for getting the correct (internal) set of $p$-adic numbers (the one whose points are, externally, the $p$-adic valued continuous maps) is simply to complete using all Cauchy nets (or filters) rather than only the sequences (or sequential filters).  Simon has stated as much already, but I wanted to extract that fact.
It is just not true in constructive mathematics that metric spaces are sequential spaces, so one must always work with all nets and not just restrict to sequences.  (Predicative constructivists may have some difficulty with this, of course, although fortunately they can still deal with the locale via a prosite for it.)<|endoftext|>
TITLE: Does every set $X$ have a topology for which the only continuous self-surjection is the identity map?
QUESTION [15 upvotes]: This question is a special case of Dominic van der Zypen's question Reconstructing relations with the image relation of a topology, as discussed in the comments, particularly the comment of Eric Wofsey, which explains that if the answer to this question is affirmative, then the answer to Dominic's question is also affirmative. 
Question. Does every set $X$ have a topology for which the only continuous surjection $f:X\to X$ is the identity map?
My answer to Dominic's question shows that for finite sets, the answer is affirmative, since one need only place an order on $X$, and then let the topology be the up-sets of the order. Every continuous surjection is a permutation of $X$ and order-preserving, and hence the identity. Although it is tempting to try to use the same method with well-orders on an infinite set, it doesn't quite work out in that generality, because the predecessor function on the finite height elements (otherwise fixed) will be a continuous surjection, but not the identity. 
For an extreme negative answer, I would ask: can one show that there is no topology on a countably infinite set for which the only continuous surjection is the identity map?

REPLY [2 votes]: In the article Constructions and Applications of Rigid Spaces, I (Advances in Mathematics 29, 89--130 (1978), V. Kannan and M. Rajagopalan show that if $(2^{\aleph_0})^+ < 2^{2^{\aleph_0}}$, there is a countable space $X$ such that the only non-constant continous self-map is the identity. (Of course that's a notion of rigidity that is "more rigid" than what is asked in the original post.)<|endoftext|>
TITLE: How many rearrangements must fail to alter the value of a sum before you conclude that none do?
QUESTION [41 upvotes]: This will not be altogether unrelated to this earlier question.
For which classes $C$ of bijections from $\{1,2,3,\ldots\}$ to itself is it the case that for all sequences $\{a_i\}_{i=1}^\infty$ of real numbers, if $\displaystyle f\mapsto \lim_{n \to\infty} \sum_{i=1}^n a_{f(i)}$ remains the same for all $f\in C$ then it remains the same for all bijections $f$ from $\{1,2,3,\ldots\}$ to itself?
Thus if you have a conditionally convergent series, then at least one of the rearrangements that change the value of the sum is within $C$.

REPLY [9 votes]: This answer is an extended comment on the answers of Joel David Hamkins and Paul Larson. Let's say that the rearrangment number $\kappa$ is the smallest cardinality of a family of bijections that can change the value of any conditionally convergent sum. Our goal is to find the value of the rearrangement number in terms of other small cardinals.
Using a nice argument involving "padding with zeroes", Joel shows that $\kappa \geq \mathfrak{p}$. (At least, this is one way of summarizing his answer. The inequality $\kappa \geq \mathfrak{p}$ implies $\kappa$ is uncountable and also that MA proves $\kappa = \mathfrak{c}$, which are the two assertions in Joel's answer; see also Andreas Blass's comment on Joel's answer).
In this answer, I want to show how to take the "padding with zeroes" argument a bit further to show that


$\kappa \geq \mathfrak{b}$
by itself, Joel's "padding with zeroes" argument cannot prove anything stronger than $\kappa \geq \mathfrak{b}$.


(if you've forgotten what $\mathfrak{b}$ is, I'll define it below; for now just know that $\mathfrak{p} \leq \mathfrak{b}$ is always true and $\mathfrak{p} < \mathfrak{b}$ is consistent, so this inequality improves the former one).
First a few definitions:

$A \subseteq \mathbb N$ is preserved by $f: \mathbb N \rightarrow \mathbb N$ if $f$ does not change the order of $A$, except possibly on a finite set. More precisely, $A$ is preserved by $f$ iff there is some cofinite subset $A'$ of $A$ such that
$$x < y \qquad \Leftrightarrow \qquad f(x) < f(y)$$
for all $x,y \in A'$.
If $A$ is not preserved by $f$, we say that $A$ is jumbled by $f$.
A family $\mathcal B$ of bijections is jumbling if every infinite set is jumbled by some member of $\mathcal B$.
The "jumbling number" $\mathfrak{j}$ is the smallest cardinality of a jumbling family.

This last definition is just a placeholder -- we'll show in a bit that $\mathfrak{j}$ is equal to another well-known cardinal, namely $\mathfrak{b}$. So what's the point of defining $\mathfrak{j}$ at all? Consider the following proposition and its proof:
Proposition: $\kappa \geq \mathfrak{j}$.
Proof: Suppose $\mathcal B$ is a family of bijections $\mathbb N \rightarrow \mathbb N$ and $|\mathcal B| < \mathfrak{j}$. By definition, there is an infinite set $A \subseteq \mathbb N$ that is preserved by every member of $\mathcal B$. Fix any conditionally convergent sequence $\langle b_n \rangle$. Define a new sequence $\langle a_n \rangle$ so that $a_n = 0$ if $n \notin A$, and otherwise $a_n = b_k$, where $n$ is the $k^{th}$ element of $A$. (In other words, pad the sequence $b_n$ with extra zeroes, putting the new zeroes precisely on the complement of $A$.)
Let $f \in \mathcal B$. Since no member of $\mathcal B$ jumbles $A$, the nonzero terms of $\langle b_n \rangle$, $\langle a_n \rangle$, and $\langle f(a_n) \rangle$ are all in the same order, except perhaps for finitely many terms. Thus
$$\sum_{n \in \mathbb N}b_n = \sum_{n \in \mathbb N}a_n = \sum_{n \in \mathbb N}a_{f(n)}.$$
Therefore no $f \in \mathcal B$ can rearrange $\langle a_n \rangle$ sufficiently to change the value of its sum (even though such a rearrangement is clear possible, because $\langle b_n \rangle$ is only conditionally convergent). Since $\mathcal B$ was an arbitrary family of bijections with $|\mathcal B| < \mathfrak{j}$, this finishes the proof.
QED
The proof of this proposition is nothing new: it is simply Joel's "padding with zeroes" argument in a slightly more abstract setting. The thing to notice is that $\mathfrak{j}$ is the largest cardinal number for which this argument succeeds: if $\lambda \geq \mathfrak{j}$, we cannot claim that $\lambda$ bijections will leave the nonzero terms of some sequence in the same order (modulo a finite set). Indeed, I defined the number $\mathfrak{j}$ simply to be the cardinal number at which the "padding with zeroes" argument stops working.
Now, as promised, we'll show that $\mathfrak{j}$ is just a more familiar number in disguise. Recall that $\mathfrak{b}$ is the smallest cardinality of an "unbounded" family $\mathcal F$ of functions $\mathbb N \rightarrow \mathbb N$. A "bound" for $\mathcal F$ means a function $h: \mathbb N \rightarrow \mathbb N$ such that, for every $f \in \mathcal F$, $h(n) > f(n)$ for all but finitely many $n$.
Theorem: $\mathfrak{j} = \mathfrak{b}$.
Proof:
Part I: $\mathfrak{b} \leq \mathfrak{j}$. Suppose $\lambda < \mathfrak{b}$; we will show that $\lambda < \mathfrak{j}$ as well. To this end, let $\mathcal B$ be a family of bijections $\mathbb N \rightarrow \mathbb N$ with $|\mathcal B| = \lambda$. We must show that $\mathcal B$ is not a jumbling family.
To each $b \in \mathcal B$, we associate a (recursively defined) function $f_b: \mathbb N \rightarrow \mathbb N$ as follows:
$$f_b^0(n) = \max\{b(m) : m \leq f_b(n-1)\}+1,$$
$$f_b(n) = \max\{b^{-1}(m) : m \leq f_b^0(n)\}+1.$$
Because $|\mathcal B| = \lambda < \mathfrak{b}$, there is some $h: \mathbb N \rightarrow \mathbb N$ such that, for every $b \in \mathcal B$, $h(n) > f_b(n)$ for all but finitely many $n$.
Let $a_0 = 0$, let
$$a_{n+1} = h(a_n)+n,$$
and let $A = \{a_n : n \in \mathbb N\}$. $A$ is clearly infinite, and I claim that $A$ is preserved by every element of $\mathcal B$ (which means that $\mathcal B$ is not a jumbling family).
Fix $b \in \mathcal B$ and fix $N$ such that $h(n) > f_b(n)$ for all $n \geq N$. We will show that for all $n \geq N$ we have $b(a_n) < b(a_{n+1})$, from which it follows that $A$ is preserved by $b$. If $n \geq N$, we have $a_n \geq n$ and
$$a_{n+1} = h(a_n) + n > h(a_n) > f_b(a_n).$$
From the definition of $f_b$, it is clear that $a_{n+1} > f_b(a_n)$ implies $b(a_{n+1}) > b(a_n)$. Thus $b$ does not jumble $A$.
Part II: $\mathfrak{j} \leq \mathfrak{b}$. Fix an unbounded family $\mathcal F$ of functions $\mathbb N \rightarrow \mathbb N$. We will find a jumbling family $\mathcal B$ with $|\mathcal B| = |\mathcal F|$.
We may assume that every member of $\mathcal F$ is strictly increasing and strictly positive (if not, replace each $f \in \mathcal F$ with the function $g$ defined by $g(n) = \max\{f(0),\dots,f(n)\} + n + 1$). Thus each member of $\mathcal F$ naturally partitions $\mathbb N$ into nonempty intervals, namely $[0,f(0))$, $[f(0),f(1))$, $[f(1),f(2))$, etc. Let $b_f$ be the function that flips each of these intervals upside-down. That is, (setting $f(-1) = 0$ for convenience), if $f(m-1) \leq n < f(m)$, then $b_f(n) = f(m-1) + (f(m)-1) - n$.
Clearly $b_f$ is a bijection for every $f \in \mathcal F$. Let $\mathcal B = \{b_f : f \in \mathcal F\}$. We want to show that every infinite set is jumbled by some $b_f$.
Let $A$ be an infinite subset of $\mathbb N$. Let $h$ be the unique increasing enumeration of $A$. Since $\mathcal F$ is unbounded, there is some $f \in \mathcal F$ such that, for infinitely many values of $n$, we have $f(n) > h(n)+n$. By the pigeonhole principle, if $f(n) > h(n) + n$, then there are intervals of the form $[f(m-1),f(m))$ containing more than one member of $A$; in fact, the number of members of $A$ that do not have an interval to themselves is at least $n$. Since $f(n) > h(n)+n$ infinitely often, it follows that there are infinitely many intervals of the form $[f(m-1),f(m))$ containing multiple elements of $A$. Because each of these intervals is flipped upside-down by $b_f$, we see that $b_f$ jumbles $A$.
QED
Putting this together with Paul Larson's observation, we have what seems to me a decent range for $\kappa$:
Theorem: $\mathfrak{b} \leq \kappa \leq \operatorname{non}(M)$.
For the record, I can see no real reason to think that $\mathfrak{j} = \kappa$. In fact, I'll leave it as an exercise to show that the family $\mathcal B$ defined in Part II of my proof has the following property: for every $f \in \mathcal B$, if $\sum_{n \in \mathbb N}a_n$ converges, then $\sum_{n \in \mathbb N}a_{f(n)}$ converges also, and to the same value. In other words, we have a jumbling family that doesn't change the value of any sums!<|endoftext|>
TITLE: Coxeter subgroups of Coxeter groups
QUESTION [7 upvotes]: Is there an algorithm to determine all the Coxeter subgroups of a given Coxeter group? If we only want the Coxeter subgroups of finite index does that make the question easier? If we only want a Coxeter subgroup of a given type does that make the question easier? Are there special cases of these questions which can be answered?

REPLY [3 votes]: The embedding problem among Coxeter groups (i.e. deciding, given two Coxeter groups, whether or not one is (abstractly) isomorphic to a subgroup of the other) is widely open, even in the right-angled case. And this problem seems much simpler than determining all the Coxeter groups that embed into a given Coxeter group.
Nevertheless, there are a few (very very) particular cases where an answer to your problem can be found. For instance, the following statement, dealing with right-angled Coxeter groups, can be found in Morphisms between right-angled Coxeter groups and the embedding problem in dimension two.
Proposition 1. Let $\Gamma$ be a finite simplicial graph and let $C_n$ denote a cycle of length $n \geq 5$. Then $C(\Gamma)$ is isomorphic to a subgroup of $C(C_n)$ if and only if $\Gamma$ is either a disjoint union of segments or a single cycle of length divisible by $n -4$.
Observe that a Coxeter subgroup of a right-angled Coxeter group is necessarily a right-angled Coxeter group, so the proposition provides an answer to your problem for a specific family of Coxeter groups.
Another result from the same article, and which might be of interest for the discusion, is:
Proposition 2. Let $R, S$ be two finite trees. Then $C(R)$ is isomorphic to a subgroup of $C(S)$ if and only if there exists a graph morphism $\varphi : R \to S$ that sends a vertex of degree $2$ to a vertex of degree $\geq 2$ and a vertex of degree $\geq 3$ to a vertex of degree $\geq 3$.

Fixing $R$ and $S$, the existence of such a graph morphism can be checked algorithmically, so, fixing $S$, the set of Coxeter subgroups in $C(S)$ is enumerable.<|endoftext|>
TITLE: Grothendieck toposes in (very) weak foundation
QUESTION [11 upvotes]: There is on the nLab page "Grothendieck topos" a part about the theory of Grothendieck toposes in weak foundation.
It claims that the equivalence for a category between the Giraud's axioms and being a category of sheaves over a site can be proved under very weak foundation: predicative (even without small set of functions), finitarist, constructive. 
I would really like to know if there is some references about this or if someone has thought about it and could explain some details about this that seems a bit obscure to me.
I am especially concern by the absence of sets of functions: without them Grothendieck toposes shouldn't be expected to be locally small, hence it does not seem possible to associate a sheave to an object $X$ being given a set of generators (because the sheave should be $Hom( \_ , X)$ which might not be a set). Maybe the theorem still holds by constructing a localization functor from a presheaves category to the "category of sheaves" but without a right adjoint, hence sheaves should not be set valued functors.
Also, it is not clear what a Grothendieck topology should be (more precisely what should be small ? this might explains the fact that sheafication don't preserve smallness )
I would also be interested in knowing if it is possible to weaken even further the foundations, for example by getting ride of quotient sets and still have a result of this kind.

REPLY [4 votes]: (Toby Bartels wrote back, explaining that he was having some trouble logging in to his account here, but asking if I could post some comments he had. I'm going to post them under my name as Community Wiki, and invite him to edit this answer further once he's back, if he'd like -- or he can post separately of course.) 
I believe that local smallness follows (in conventionally strong foundations) from the other axioms (this is what the Elephant seems to say in C.2.2.8.vii), so the right thing to do should just be to remove it from the list of axioms, which I have now done at the nLab.  However, it would be good to have the argument written out in a clearly predicative way, to be certain.  I no longer have access to my copy of the Elephant (I had to check the wording of C.2.2.8 in Google Books), which I believe was my guide the last time that I was thinking through this, but hopefully I can find it in the library and extract an explicitly predicative and constructive argument from its proofs.
More speculatively, the classical proof that Giraud's axioms imply local smallness might have a strongly predicative variation proving, say, that there is a small generating set (possibly smaller than the original one) $G$ such that $\hom(G,X)$ is small for each $X$.  That solves your problem of getting a Set-valued sheaf from an object, and it's trivial (but impredicative) to prove that a category with such a generating set must be locally small.  But I'm not sure that it's actually correct!<|endoftext|>
TITLE: Augmentation ideal is finitely generated if and only if $A$ is finitely generated as a $k$-algebra?
QUESTION [6 upvotes]: Let $A \subset k[x_1, \dots, x_n]$ be a subalgebra, which is also a graded subspace $A = \oplus_{i \ge 0} A_i$. One can write $A = A_0 \oplus A_{> 0}$ where we have $A_0 = k^0[x_1, \dots, x_n] = k$ and $A_{> 0} := \oplus_{i \ge 1} A_i$ is a graded ideal of $A$, known as the augmentation ideal.
My question is, is it necessary the case that the ideal $A_{> 0}$ is finitely generated (as an $A$-module) if and only if $A$ is finitely generated as a $k$-algebra?

REPLY [2 votes]: This amounts to an exercise, applying what are perhaps the first few theorems you encounter in the context of noetherian rings. Moreover, the assumption that $A$ is a graded subring of $k[x_1,\ldots,x_n]$ is stronger than necessary. It is enough to assume that $k$ is a commutative noetherian ring and that $A$ is a commutative augmented $k$-algebra (but even these assumptions are stronger than necessary).
I will write $A_+$ for the augmentation ideal of $k$. For one direction of the equivalence, if you know that $A_+$ is a finitely-generated $A$-module, then it follows more or less immediately that $A$ is generated as a $k$-algebra by your set of $A$-module generators for $A_+$ together with $1$.
For the other direction, assume that $A$ is finitely-generated as a $k$-algebra, so that $A$ is a quotient of a polynomial algebra of the form $k[x_1,\ldots,x_n]$. Then it is a consequence of Hilbert's Basis Theorem that $A$ is a noetherian ring. Now $A_+$ is an $A$-submodule of the left $A$-module $A$, so $A_+$ must be a noetherian $A$-module. Then by one of the first theorems you encounter for noetherian rings, $A_+$ must be finitely-generated as a left $A$-module.
Edit: As Eric pointed out in the comments, I made a careless error in the second paragraph of my answer. As he says, if you take $A$ to be the power series ring $k[[x]]$, then $A_+$ is generated as a left $A$-module by $\{x\}$, but the set $\{x,1\}$ does not generate $k[[x]]$ as an algebra (it generates only the subring $k[x]$). On the other hand, if you assume that $A$ is a positively graded algebra, then the inductive argument indicated by Todd in his comment shows that $A$ is generated as a $k$-algebra by the generating set for $A_+$ together with $1$.<|endoftext|>
TITLE: Maximal number of subsets in $\{1,\dots,n\}$ such that neither is contained in a union of two others
QUESTION [16 upvotes]: What are known estimates for maximal $M$ for which their exists subsets $A_1,\dots,A_M$ in $\{1,\dots,n\}$ such that there do not exist different indexes $i,j,k$ for which $A_i\subset A_j\cup A_k$?
It is not hard to prove some exponential bounds $c_1^n<M<c_2^n$ for some $1<c_1<c_2<2$, but maybe sharp exponent is known?

REPLY [6 votes]: To best of my knowledge, the only bounds on this problem in the literature are in the the old paper of Erdős, Frankl, and Füredi. See Theorem 2 there. If you look at the MathSciNet review of the said paper you will find papers that referenced this paper. Perhaps, some might be of use to you.
There is also a follow-up paper by the same authors on families not containing sets $A,B_1,\dotsc,B_r$ such that $A\subset B_1\cup\dotsb\cup B_r$.<|endoftext|>
TITLE: Has by well-foundedness every non-empty class an $R$-minimal element? Also if axiom REG is not assumed?
QUESTION [9 upvotes]: I asked this question here on Math.SE but uptil now it was not answered. So I decided to give it a try. Thank you in advance.
Working in $\mathbf{ZF}$ let $R$ be a proper class of ordered pairs that is well-founded. This means that for every non-empty set $a$ there is a set $b\in a$ such that $cRb\implies c\notin a$. Here $cRb$ is a notation for $\langle c,b\rangle\in R$ and $b$ is a so-called $R$-minimal element of $a$. If $R$ is local (i.e. collections $\{x\mid xRb\}$ are all sets) then it can be shown that also non-empty proper classes have $R$-minimal elements. 
I encountered a proof that made the condition of being local redundant. It made use of an operation on classes that adds to each class a set that is contained in it (Bottom-operation) but the definition of this operation relied on the regularity axiom. 
My question:

Is there a proof that every non-empty class has an $R$-minimal element that does not make use of the regularity axiom?

Another formulation:

If $R$ is a class of ordered pairs that is well-founded, then is it legal to apply $R$-induction if all axioms of $\mathbf{ZF}$ are accepted with exception of the axiom of regularity?

Edit:
Maybe a bounty will help. If the question will not be answered then I will start cherishing the fact that I asked a good question ($6$ upvotes) that 'nobody' could answer :).

REPLY [6 votes]: Let me mention another counterexample. In [1, Thm. 11], we construct a model of $\mathrm{ZFC}^-$ with the collection schema which contains a definable class relation $\langle A,<\rangle$ such that

$<$ is a dense linear order on $A$ with no least element;
every subset of $A$ is well-ordered by $<$.

The second condition says that $<$ is well-founded in the sense used in the question, nevertheless it has no minimal element by 1.
[1] A. S. Daghighi, M. Golshani, J. D. Hamkins, E. Jeřábek, The foundation axiom and elementary self-embeddings of the universe, in: Infinity, Computability, and Metamathematics: Festschrift celebrating the 60th birthdays of Peter Koepke and Philip Welch (S. Geschke, B. Löwe, and P. Schlicht, eds.), College Publications, London, 2014, pp. 89–112.<|endoftext|>
TITLE: Most computationally efficient Littlewood-Richardson rule
QUESTION [9 upvotes]: There are many, many different versions of the Littlewood-Richardson rule: the original characterization via Yamanouchi words, Remmel's version, a description via the Poirier-Reutenauer bialgebra, the Knutson-Tao hive model rule and many more. Any of these rules provides an effective algorithm for computing products of Schur functions. It is known that computing Littlewood-Richardson coefficients is, in general, #P complete.
I am interested in the most computationially efficient Littlewood-Richardson rule. Since any rule will be in #P, for which I am not aware of easy characterizations of run time (please correct me if I'm wrong on this), I suspect any answer to this question would have to be based on experimental evidence. No doubt the various computer algebra systems have coded the various rules, computed a bunch of examples and checked which one was fastest. In one of my graduate courses, we were told the Remmel rule was the easiest, though this was for by hand computations of small examples. At the end of Exercise 2.7.10 of book Symmetric Functions, Schubert Polynomials and Degeneracy Loci, Manivel asserts that the Lascoux-Schutzenberger transition equations (see here) for Schubert polynomials (and hence Stanley symmetric functions) provides the most computationally efficient means of computing the product of Schur functions.
1) What is the (experimentally) most efficient method of computing a single Littlewood-Richardson coefficient?
2) Can anyone confirm or refute Manivel's assertion that the Lascoux-Schutzenberger transition equations are the most efficient rule for computing the product of Schur functions?

REPLY [2 votes]: I can't answer your second question, but I have a pretty fast recurrence for computing a single Littlewood-Richardson coefficient for the complete flag variety. It also works for equivariant coefficients, but unlike other equivariant recurrences I've seen you can just set the equivariant part to 0 and get a recurrence for the ordinary coefficients. It does not restrict to a recurrence for Grassmannian coefficients only, so we have to expand our attention to the entire symmetric group. It is not published anywhere so I will prove it here.
The main result of my dissertation is a Leibniz formula for divided difference operators. If $p$ and $q$ are two polynomials in variables $x_1,x_2,\ldots,x_n$ and $w$ is a permutation, then
$$\partial_w(pq)=\sum_{u,v}{c_{u,v}^w(x_1,\ldots,x_n)\partial_u(p)\partial_v(q)}$$
where $c_{u,v}^w(x_1,\ldots,x_n)$ is the equivariant Littlewood-Richardson coefficient, which is the coefficient appearing in the expansion
$$S_u(x;y)S_v(x;y)=\sum_{w}{c_{u,v}^w(y_1,\ldots,y_n)S_w(x;y)}$$
of a product of two double Schubert polynomials. When $\ell(u)+\ell(v)=\ell(w)$, these coefficients are the ordinary Littlewood-Richardson coefficients (by which I mean for the complete flag variety).
For an integer $k$ I denote by $s_k$ the simple transposition $(k,k+1)$. Now, let $s_i$ be a right descent of $w$, so that
$$\partial_w=\partial_{ws_i}\partial_{s_i}$$
Then since
$$\partial_{s_i}(pq)=\partial_{s_i}(p)q+p\partial_{s_i}(q)-(x_i-x_{i+1})\partial_{s_i}(p)\partial_{s_i}(q)$$
we can write
$$\partial_w(pq)=\partial_{ws_i}(\partial_{s_i}(p)q+p\partial_{s_i}(q)-(x_i-x_{i+1})\partial_{s_i}(p)\partial_{s_i}(q))=\sum_{u,v}{c_{us_i,v}^{ws_i}\partial_{us_i}\partial_{s_i}(p)\partial_v(q)}+\sum_{u,v}{c_{u,vs_i}^{ws_i}\partial_u(p)\partial_{vs_i}\partial_{s_i}(q)}+\sum_{u',u,v}{c_{u',u,v}^{ws_i}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
For the last sum,
$$\sum_{u',u,v}{c_{u',u,v}^{ws_i}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}=\sum_{u',v',u,v}{c_{u',v'}^{ws_i}c_{us_i,vs_i}^{v'}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
If $v'=ws_i$, then we get a contribution of
$$\sum_{u',u,v}{c_{us_i,vs_i}^{ws_i}c_{u',ws_i}^{ws_i}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}=\sum_{u,v}{c_{us_i,vs_i}^{ws_i}(x_{w(i+1)}-x_{w(i)})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
(I snuck in that last equality, it's true because acting with $ws_i$ is equivalent to summing over all $u'$ what you get by acting with $c_{u',ws_i}^{ws_i}\partial_{u'}$.) If $v'<ws_i$ (in Bruhat order) and the term is nonzero, then $\ell(u')=1$, so $u'=s_j$ for some $j$. Indeed, either $j=i-1$, $j=i$, or $j=i+1$. Thus we get
$$\sum_{s_j,v',u,v}{c_{us_i,vs_i}^{v'}c_{s_j,v'}^{ws_i}\partial_{s_j}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
By Monk's formula, in order for this to be nonzero we must have $\ell(v')=\ell(ws_i)-1=\ell(w)-2$ and there exist $a\leq j<b$ such that
$$ws_i=v'(a,b)$$
where $(a,b)$ denotes a transposition.
Long story short, picking $i$ so that $w(i)>w(i+1)$ we have
(1) If $u(i)<u(i+1)$ and $v(i)<v(i+1)$, then
$$c_{u,v}^w=0$$
(2) If  $u(i)>u(i+1)$, and $v(i)<v(i+1)$, then
$$c_{u,v}^w=c_{us_i,v}^{ws_i}$$
(3) If $u(i)<u(i+1)$, and $v(i)>v(i+1)$, then
$$c_{u,v}^w=c_{u,vs_i}^{ws_i}$$
Now for the fun part.
(4) If $u(i)>u(i+1)$ and $v(i)>v(i+1)$, then
$$c_{u,v}^w=c_{us_i,v}^{ws_i}+c_{u,vs_i}^{ws_i}+\sum_{\stackrel{a\leq i-1<b}{\ell(ws_i(a,b))=\ell(w)-2}}{c_{us_i,vs_i}^{ws_i(a,b)}}+\sum_{\stackrel{a'\leq i+1<b'}{\ell(ws_i(a',b'))=\ell(w)-2}}{c_{us_i,vs_i}^{ws_i(a',b')}}-2\sum_{\stackrel{a''\leq i<b''}{\ell(ws_i(a'',b''))=\ell(w)-2}}{c_{us_i,vs_i}^{ws_i(a'',b'')}}-(x_{w(i+1)}-x_{w(i)})c_{us_i,vs_i}^{ws_i}$$
giving a recurrence for the equivariant coefficients in terms of $x_1,\ldots,x_n$. The formatting is a bit off, so note that the first two summations have nothing in front of them (so coefficient $+1$), and the third summation has a coefficient of $-2$. If you're not interested in the positive degree coefficients, ignore the last term (setting $x=0$). This is not so much fun for a human to use but on a computer it gets you the single coefficient fast (provided you either memoize or traverse the Bruhat graph intelligently).
The base case(s) for the recurrence is
$$c_{u,1}^u=c_{1,v}^v=1$$
or, if you're really a minimalist, it's sufficient to only assume
$$c_{1,1}^1=1$$<|endoftext|>
TITLE: Prime divisors of values of a polynomial on an infinite set
QUESTION [5 upvotes]: This may be a well known problem:
Let $f$ be a  polynomial with integer coefficients. Is it possible for the set of prime divisors of values of $f$ on an infinite set of integers, to be finite?  
I  guess such a set exists  only for polynomials $f$ of the form $f(x) = c(ax+b)^n, a,b,c\in \mathbb{Z}, c\neq 0$. But I can't give a proof.
I'm specially interested in the case $f(x) = x^2+1$.
Thanks!

REPLY [6 votes]: Another way would be to appeal to Faltings theorem. Indeed, if $f(n)$ has finitely many prime factors $p_1,\dots p_k$ for infinitely many $n,$ then the equation, say $$f(x)=Dy^3$$ would have infinitely many solutions for some fixed $D,$ which is a product of $\prod_{i=1}^kp_i^{\alpha_i}$ for $0\le \alpha_i\le 2.$ 
EDIT: since OP is interested in the case of quadratics, one can use the same reasoning as above to reduce everything to the equation
$$x^2+D=Ay^n$$
for fixed $A.$ This is Ramanujan-Nagel type equation. See https://en.wikipedia.org/wiki/Ramanujan–Nagell_equation for the description of solution.

REPLY [2 votes]: Assume that it holds for, say, $x^2+1$ with given primes $p_1,\dots,p_m$. Consider factorization of $x^2+1=(x+i)(x-i)$ in Gaussian integers. For inifinitely many $x$ we have $x+i=A y^7$ for the same $A$. That is, $2i=A y^7-\bar{A} {\bar y}^7$, factor RHS in 7 multiples $A^{1/7}y-\bar{A}^{1/7}\bar{y}$. All but 1 are about $|y|$ by absolute value, hence another is about $1/|y|^6$. It means that algebraic number $(A/\bar{A})^{1/7}$ is approximated by $\bar{y}/y$ with accuracy $|y|^{-7}$. Taken real part we get approximation with the same accuracy by rational with denominator $|y|^2$. This may happen only finite number of times by Roth's theorem.

REPLY [2 votes]: For $f(x)=x^2+1$, one can reach the desired conclusion by applying an elementary method of Stormer. "Elementary" means that nothing as deep as Siegel's theorem is used, just some theory of Pellian equations. Stormer's method is discussed on Wikipedia:
https://en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem
(However, there the method is explained in terms of a slightly different problem: Finding all $n$ with $n, n+1$ supported on a given finite set of primes.)
A nice reference is the paper of Lehmer cited there. You might also want to look at the following:
Najman, Filip(CT-ZAGR)
Smooth values of some quadratic polynomials. 
Glas. Mat. Ser. III 45(65) (2010), no. 2, 347–355.<|endoftext|>
TITLE: set of centers of sphere inscribed in tetrahedron
QUESTION [5 upvotes]: Having a sphere and three diffrent point $A,B,C$ on this sphere. Find set of all centers of spheres inscribed in a tetrahedron $ABCD$, where $D$ is some point on the given sphere. The problem reduced to 2-dimensions is trivial it's just sum of two arcs of some circle, but in 3-dimensions the set is not so simple. Checking in geogebra it's not sum of some parts of sphere. I don't know what this set looks and how it's described.

REPLY [4 votes]: Here is a depiction of Robert Bryant's surface defined by the $56$-term polynomial
he details:

          


Note one component is the "triangular tea bag" discernable in 
my empirical investigation.
Here is the polynomial:
$$
2 x^5+x^4 y+x^4 z-7 x^4+2 x^3
   y^2+4 x^3 y z-10 x^3 y+2
   x^3 z^2-10 x^3 z+8 x^3+2
   x^2 y^3-12 x^2 y^2 z-15 x^2
   y^2-12 x^2 y z^2+6 x^2 y
   z+15 x^2 y+2 x^2 z^3-15 x^2
   z^2+15 x^2 z-2 x^2+x y^4+4
   x y^3 z-10 x y^3-12 x y^2
   z^2+6 x y^2 z+15 x y^2+4 x
   y z^3+6 x y z^2-6 x y z-4 x
   y+x z^4-10 x z^3+15 x z^2-4
   x z-2 x+2 y^5+y^4 z-7 y^4+2
   y^3 z^2-10 y^3 z+8 y^3+2
   y^2 z^3-15 y^2 z^2+15 y^2
   z-2 y^2+y z^4-10 y z^3+15 y
   z^2-4 y z-2 y+2 z^5-7 z^4+8
   z^3-2 z^2-2 z+1$$
The plot above restricts $(x,y,z)$ to lie on or in the sphere.<|endoftext|>
TITLE: The p-adic valuation of a linear recurrence
QUESTION [9 upvotes]: Let $(u_n)_{n \geq 0}$ be an integer-valued linear recurrence of order $k \geq 1$. Precisely,
$$u_n = a_1 u_{n-1} + \cdots + a_k u_{n - k} \quad \forall n \geq k ,$$
for some $a_1, \ldots, a_k \in \mathbb{Z}$.
Given a prime number $p$, I wonder: How much is known about the $p$-adic valuation $\upsilon_p(u_n)$? Are there some "explicit" formulas?
$\bullet$ For $k=1$, we have simply a geometric progression and clearly it holds $$\upsilon_p(u_n) = \upsilon_p(a_1)^n + \upsilon_p(u_0).$$
$\bullet$ For $k=2$, the problem was studied (in the more general setting of linear recurrences in the field of $p$-adic numbers) by Ward [1], who gave formulas for $\upsilon_p(u_n)$. However, those formulas rely on the computation of a $p$-adic number $v$ (see Theorem 10.1) which is essential as much as difficult as the computation of $\upsilon_p(u_n)$, so they do not seem to give really useful information on $\upsilon_p(u_n)$ (see Vesselin Dimitrov comments).
The particular case of the Fibonacci sequence $(F_n)_{n \geq 0}$ was also studied by Lengyel [2], who gave practical closed expressions for $\upsilon_p(F_n)$, in terms of $\upsilon_p(n)$, $z(p) := \min\{n > 0 : p \mid F_n\}$, and $e(p) := \upsilon_p(F_{z(p)})$.
$\bullet$ For $k\geq 3$ it seems to me that nothing general is known. I found only another article of Lengyel [3] about the $2$-adic valuation of the Tribonacci numbers $(T_n)_{n \geq 0}$.
Surely, without loss of generality, it can be assumed that: $(u_n)_{n \geq 0}$ is not degenerate; it has at least one zero modulo $p$ (and that can be effectively checked since $(u_n)_{n \geq 0}$ is periodic modulo $p$ and the period length is less than $p^k$); $u_0 \equiv 0 \bmod p$, eventually by shifting $(u_n)_{n \geq 0}$.
Thank you in advance for possible ideas and references!
[1] M. Ward. The linear p-adic recurrence of order two. Illinois J. Math. (6) 40--52, 1962.
[2] T. Lengyel. The order of the Fibonacci and Lucas numbers. The
Fibonacci Quarterly, (33) 234--239, 1995.
[3] T. Lengyel. The 2-adic Order of the Tribonacci Numbers and the Equation $T_n = m!$. Journal of Integer Sequences Vol. 17, 2014.

REPLY [3 votes]: Regarding $p$-adic valuation of Lucas sequences, a quite precise result is given in [1].
Theorem. Let $(u_n)_{n \geq 0}$ be a nondegenerate Lucas sequence with $u_0 = 0$, $u_1 = 1$, and $u_{n+2} = a u_{n+1} + b u_n$ for all $n \geq 0$, where $a$ and $b$ are two integers. Furthermore, let $p$ be a prime number not dividing $b$.
Then for any positive integer $n$ we have 
$$v_p(u_n) = \begin{cases}
v_p(n) + v_p(u_p) - 1 & \text{ if } p \mid \Delta ,\; p \mid n, \\
0 & \text{ if } p \mid \Delta ,\; p \nmid n, \\
v_p(n) + v_p(u_{p\tau(p)}) - 1 & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n ,\; p \mid n, \\
v_p(u_{\tau(p)}) & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n ,\; p \nmid n, \\
0 & \text{ if } p \nmid \Delta ,\; \tau(p) \nmid n ,
\end{cases}$$
where $\Delta := a^2 + 4b$ and $\tau(p)$ is the rank of apparition of $p$ in $(u_n)_{n \geq 0}$, i.e., the least positive integer $m$ such that $p \mid u_m$. Moreover, if $p \geq 3$ then
$$v_p(u_n) = \begin{cases}
v_p(n) + v_p(u_p) - 1 & \text{ if } p \mid \Delta ,\; p \mid n, \\ 
0 & \text{ if } p \mid \Delta ,\; p \nmid n, \\ 
v_p(n) + v_p(u_{\tau(p)}) & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n , \\ 
0 & \text{ if } p \nmid \Delta ,\; \tau(p) \nmid n ,
\end{cases}$$
while if $p \geq 5$ then
$$v_p(u_n) = \begin{cases}
v_p(n) & \text{ if } p \mid \Delta , \\ 
v_p(n) + v_p(u_{\tau(p)}) & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n , \\ 
0 & \text{ if } p \nmid \Delta ,\; \tau(p) \nmid n .
\end{cases}$$
Actually, in [1] the theorem is stated for $a$ and $b$ relatively prime. 
However, as explained in [2], the result holds even if $a$ and $b$ are not coprime.
[1] C. Sanna, The $p$-Adic Valuation of Lucas Sequences, Fibonacci Quart. 54 (2016), no. 2, 118–124.
(Free preprint: https://www.researchgate.net/publication/304251918_The_p-adic_valuation_of_Lucas_sequences)
[2] N. Murru, C. Sanna, On the k-regularity of the k-adic valuation of Lucas sequences http://arxiv.org/abs/1603.09310<|endoftext|>
TITLE: Finding the inertia group
QUESTION [5 upvotes]: Set $h(x) = x^5+x^4+x^3+x^2+x-1$, let $L$ be the splitting field of $h$ over $\mathbb{Q}$, and let $p$ be a prime of $L$ lying over $2$.

What is the isomorphism class of the inertia group $I_p$, and how do I find it?

I would be more happy with having some procedure for finding inertia groups, rather than only the isomorphism class of this specific group.
I would also like to find the field $O_L/p$, and the decomposition group $D_p$ if possible.
Some useful information is: $h$ is irreducible over the rational numbers, and $h(x)(x-1) = x^6 - 2x + 1$ which allows for an easy irreducible factorization mod $2$: $h(x) = (x + 1)(x^2 + x + 1)^2$.

REPLY [3 votes]: This answer produces $I$ somewhat indirectly. So it actually does not what the OP asked for.
The decomposition group $D$, which is the Galois group of $h(x)$ over $\mathbb Q_2$, can be computed as follows: Using resultants, one sees that the minimal polynomial over $\mathbb Q$ of the difference of two distinct roots of $h(x)$ is \begin{equation}
H(y)=y^{20}+6y^{18}+21y^{16}+46y^{14}-116y^{12}+694y^{10}+1837y^8-1810y^6-1527y^4+8560y^2+9584.
\end{equation}
From the factorization of $h(x)$ over $\mathbb Q_2$ we know that $D$ is a transitive subgroup of $S_4$. Over $\mathbb Q_2$ the polynomial $H(y)$ factors into irreducibles of degrees $4,4,4,8$. The only transitive subgroup of $S_4$ which has these orbit lengths on the $20$ pairs of distinct elements of $\{1,2,3,4,5\}$ is the dihedral group of order $8$.
As Will Sawin already remarked, the inertia group $I$ is a subgroup of $S_2\times S_2$. On the other hand, $D/I$, as a Galois group of an extension of a finite field, is cyclic. This forces $|I|\ge4$, because a dihedral group of order $8$ modulo a normal subgroup of order $\le2$ isn't cyclic. Thus $I=S_2\times S_2$.
There is a paper by Sybilla Beckmann which describes a method to compute inertia groups under certain restrictions. Her theorem does not apply here. The paper finishes with some remarks about how to possibly extend the methods.<|endoftext|>
TITLE: Radon transform between affine grassmannians
QUESTION [6 upvotes]: Let $\overline{GR}(n,k)$ be the manifold of all affine k-dimensional subspaces in $R^n$, and let
$R:C^{\infty}_c(\overline{GR}(n,k))\to C^{\infty}_c(\overline{GR}(n,l))$, $0\le k<l\le n-1$, be the Radon transform defined in a standard way as follows:
$$
(Rf)(E)=\int\limits_{\{L\in \overline{GR}(n,k)):L\subset E\}}f(L)dL.
$$
Here the integration is with respect to the Haar measure on $\{L\in \overline{GR}(n,k)):L\subset E\}=\overline{GR}(k,l)$.
QUESTION. Is the inversion formula for the Radon transform $R$ known?

REPLY [6 votes]: Yes, it is known.
See Boris Rubin, Radon transforms on affine Grassmannians, 2004, in particular theorem 4.2.
A free full text pdf is available at the linked page.
The reconstruction formula is not very simple, so I will not reproduce it here.
Radon transforms on Grassmannians are also discussed in Helgason's book Radon Transform, section II.4.F.<|endoftext|>
TITLE: Additivity of upper densities with respect to arithmetic progressions of integers
QUESTION [5 upvotes]: Let $\mathsf{d}^\star$ be the asymptotic upper density, defined on the power set of positive integers $\mathbf{N}^+$, so that
$$
\mathsf{d}^\star\colon \mathcal{P}(\mathbf{N}^+) \to\mathbf{R}\colon X\mapsto \limsup_{n\to \infty} \frac{|X\cap [1,n]|}{n}.
$$
It is easy to verify that if $k\cdot \mathbf{N}^++h:=\{kx+h\colon x \in \mathbf{N}^+\}$ is an arithmetic progression of $\mathbf{N}^+$, and $X$ is a set of positive integers having no elements in common with $k\cdot \mathbf{N}^++h$, then
$$
\mathsf{d}^\star(X\cup (k\cdot \mathbf{N}^++h))=\mathsf{d}^\star(X)+\frac{1}{k}.
$$ 
The same reasoning can be extended to the upper analytic, upper logarithmic, upper Banach and upper Buck densities, at least. That's why one may ask if this property holds in general:
We say that a set function $\mu^\star\colon \mathcal{P}(\mathbf{N}^+)\to \mathbf{R}$ is an "upper density" whenever it satisfies the following axioms:
(F1) $\mu^\star(\mathbf{N}^+)=1$
(F2) $\mu^\star(X) \le \mu^\star(Y)$ if $X\subseteq Y$
(F3) $\mu^\star(X\cup Y) \le \mu^\star(X)+\mu^\star(Y)$
(F4) $\mu^\star(k\cdot X+h)=\mu^\star(X)/k$
for all positive integers $k,h$ and sets $X,Y \subseteq \mathbf{N}^+$. Then, is it true that if $X \cap (k\cdot \mathbf{N}^++h)=\emptyset$ then
$$
\mu^\star(X \cup (k\cdot \mathbf{N}^++h))=\mu^\star(X)+\frac{1}{k}\,\,?
$$

REPLY [2 votes]: The answer is in the negative. 
Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function 
$$h := (\alpha f^q + (1-\alpha) g^q)^{\frac{1}{q}}$$
is an upper density too (in particular, condition (F3) follows from Minkowski's inequality, which is why we need $q \ge 1$). 
Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, let $x := 2f(X)$, $y := 2g(X)$ and $Y := 2 \cdot \mathbf N^+ + 1$, and suppose to a contradiction that $h$ is ``weakly additive'' (that is, $h(A \cup B) = h(A) + h(B)$ for all disjoint $A, B \subseteq \mathbf N^+$ such that $B$ is an (infinite) arithmetic progression), regardless of the actual values of the parameters $\alpha$ and $q$. Then, also $f$ and $g$ are weakly additive, and using that $f(Y) = g(Y)=\frac{1}{2}$, we obtain 
$$
\begin{split}
2h(X \cup Y) & = 2(\alpha (f(X \cup Y))^q + (1-\alpha) (g(X \cup Y))^q)^{\frac{1}{q}} \\ 
& = 2(\alpha (f(X) + f(Y))^q + (1-\alpha) (g(X) + g(Y))^q)^{\frac{1}{q}} \\
& = (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}}
\end{split}
$$
and
$$
\begin{split}
h(X) + h(Y) & = (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}} + \frac{1}{2} \\ & = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}}+ \frac{1}{2}
\end{split}
$$
which, together with $h(X \cup Y) = h(X) + h(Y)$, yields
$$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1.
$$ 
On the other hand, an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any prescribed value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper asymptotic density of $S$ is $0$, (ii) the upper Banach density of $S$ is $\frac{1}{2}$, (iii) the upper asymptotic and upper Banach densities are upper densities, and (iv) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).
Accordingly, we should have
$$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$
for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []
Added later. If you assume $\alpha = \frac{1}{2}$  and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation 
$$\sqrt{1 + (y+1)^2} = y + \sqrt{2},$$
which has a unique solution for $y \in \bf R$ (namely, $y = 0$).<|endoftext|>
TITLE: Open questions in "Spin geometry"
QUESTION [16 upvotes]: This is a very naive question. I have the impression that the area of "Spin geometry" is not an active research field. Sure Spin geometry is used in many different branches of mathematics and physics as a tool, but I don't see papers published on the development of Spin geometry by itself. Is this true? Loosely speaking, by "Spin geometry" I mean here the area in mathematics that would have the "Spin geometry" book of Lawson and Michelsohn as the basic reference. 
Are there any open questions in Spin geometry?
Thanks.

REPLY [7 votes]: As @314159 has explained, Spin geometry on "spin manifolds" is a very active field of research, but the subject goes way beyond the domain spin manifolds. The key point to notice is that every pseudo-Riemannian spin manifold $(M,g)$ admits a bundle of irreducible Clifford modules over the bundle of Clifford algebras $Cl(M,g)$, which is usually called the "spinor bundle", but remarkably enough, the opposite is not true. In other words, there exist manifolds which admit "spinor bundles" but do not admit any spin structure. This is already clear just by looking at the literature on $Spin^{c}$ manifolds, but it turns out the possibilities go beyond $Spin^{c}$ manifolds as well. In fact, so you can see a very recent paper on the subject, the following paper by C. Lazaroiu and Carlos Shahbazi:
https://arxiv.org/abs/1606.07894
classifies all the spinorial structures associated to a bundle of irreducible real Clifford modules, and gives the topological obstructions for the existence of such bundles on any pseudo-Riemannian manifold. You can check that these obstructions are in general different from that required by a spin structure, and correspond to what they call a "Lipschitz structure". They even have some application of his formalism to physics. This proves that "spin geometry" is not always associated to a spin structure and indeed it is a very rich and subtle construction. I think spin geometry in this more general setting is very little explored, and I believe is going to attract a lot of attention in the future.<|endoftext|>
TITLE: Cohomology of $G_3(\mathbb{R}^5)$
QUESTION [8 upvotes]: This is in some sense a specialization of the question integral or rational cohomology of real grassmannians. Let $G_3(\mathbb{R}^5)$ denote the real Grassmannian of (unoriented) $3$-planes in $\mathbb{R}^5$, which is a non-orientable closed manifold of dimension $6$. I would like to know the integral cohomology ring
$$
H^*(G_3(\mathbb{R}^5);\mathbb{Z}).
$$
Does anyone know of a reference where this is worked out, or how to go about doing so?

REPLY [12 votes]: The cohomology groups of the Grassmann manifold are worked out in combinatorial terms in Luis Casian and Yuji Kodama's paper, http://arxiv.org/pdf/1309.5520v1.pdf; they make a conjecture at the end about the multiplicative structure.  The authors actually do the example you ask about; they compute the groups for $G_2(R^5)$ but that's homeomorphic to $G_3(R^5)$ by taking the perpendicular complement. The answer is simple enough that you could probably work out the ring structure by the requirements of Poincaré duality (plus some considerations about the universal coefficient theorem and Bocksteins).<|endoftext|>
TITLE: What's a good introduction to category theory for someone doing analysis?
QUESTION [12 upvotes]: I do functional analysis, and diagrams are popping all over the place. It is about time I learned me some category theory.
Any recommendations?

REPLY [3 votes]: I could suggest Tom Leinster's book: "Basic Category Theory", Cambridge studies in advanced mathematics 143. It is a modern introduction to category theory which covers the basic topics of the subject (I had to write an abstract for the Mathematical Reviews so I got it for free). It is not specifically written for people doing functional analysis. It is written for undergraduate students. It contains a lot of examples.<|endoftext|>
TITLE: Conjugation in associative algebras over finite fields
QUESTION [5 upvotes]: Let $A$ be a finite dimensional associative algebra (with unity) over a finite field $F$. Let $L$ be a field extension of $F$. Suppose that after extending scalars to $L$, two elements $a,b$ of $A$ are conjugate, i.e. there is an invertible element $u \in A\otimes_F L$ such that $a=ubu^{-1}$ . Does it follow that there is an invertible element $s \in A$ such that $a=sbs^{-1}$?

REPLY [5 votes]: This is true, and it follows from Serge Lang's proof of Serre's "Conjecture I" for finite fields.  The analogous result is true for any perfect field $F$ of cohomological dimension $1$ by Steinberg's solution of the general conjecture.
Denote by $E_A$ the $F$-group scheme that is an affine space of dimension $\text{dim}_F(A)$ and such that $E_A(R)$ is identified with $A\otimes_F R$ for every commutative, unital $F$-algebra $R$.  The subfunctor of the Yoneda functor of $E_A$ parameterizing invertible elements $u\in A\otimes_F R$ is represented by a Zariski open subset $\textbf{GL}_A$ of $E_A$, namely the inverse image of the Zariski open subset $\text{Aut}_F(E_A)\subset \text{End}_F(E_A)$ under the $F$-linear morphisms of affine $F$-spaces, $$\alpha: E_A \to \text{End}_F(E_A), \ u \mapsto (v\mapsto u\cdot v).$$
For a fixed element $b$ of $A$, consider the locally closed subscheme $C_b$ of $\textbf{GL}_A$ such that for every commutative, unital $F$-algebra $R$, $C_b(R)\subset \textbf{GL}_A(R)$ is the set of invertible elements $u\in A\otimes_F R$ satisfying $ub = bu$.  The point is, there is a linear subspace $L_b \subset E_A$ parameterizing $u$ (invertible or not) such that $ub=bu$.  Thus the intersection $C_b = \textbf{GL}_A\cap L_b$ is a nonempty Zariski open subset of an $F$-affine space, and thus $C_b$ is geometrically integral.  By construction, $C_b$ is a linear algebraic $F$-group.  Thus, $C_b$ is a smooth connected affine $F$-group.
For an element $a$ of $A$, there is a closed subscheme $T_{b,a}\subset \textbf{GL}_A$ parameterizing $v$ such that $bv = va$.  As above, since this is a linear condition, it is a Zariski closed subscheme of $\textbf{GL}_A$.  Assuming this is not the empty scheme, $T_{a,b}$ is a torsor for the smooth connected affine $F$-group $C_b$ via the action, $$ \mu: C_b \times_F T_{b,a} \to T_{b,a}, \ \ (u,v) \mapsto u\cdot v.$$  For $F$ a finite field, Serge Lang proved that every torsor over a finite field $F$ for every smooth connected affine $F$-group admits an $F$-point.
MR0086367 (19,174a) 
Lang, Serge
Algebraic groups over finite fields. 
Amer. J. Math. 78 (1956), 555–563. 
http://www.jstor.org/stable/2372673?origin=crossref&seq=1#page_scan_tab_contents
For $F$ an arbitrary perfect field of cohomological dimension $1$, this was proved by Robert Steinberg.
MR0180554 (31 #4788) 
Steinberg, Robert 
Regular elements of semisimple algebraic groups.  
Inst. Hautes Études Sci. Publ. Math. No. 25 1965 49–80. 
http://www.numdam.org/item?id=PMIHES_1965__25__49_0
This is also Appendix 1 of Serre's book on Galois cohomology.
Thus the $C_b$-torsor $T_{b,a}$ has an $F$-point $v \in T_{b,a}(F) \subset A$ such that $bv$ equals $va$.
Edit. Darij Grinberg is correct that this result holds over all fields.  For every infinite field $F$, the affine $F$-space $\overline{T}_{b,a} \subset E_A$ has a Zariski dense set of $F$-points.  Thus, the dense, Zariski open subset $T_{b,a}$ has an $F$-point.

REPLY [3 votes]: $\newcommand{\End}{\operatorname{End}}$
$\newcommand{\op}{{\operatorname{op}}}$
The finiteness of $F$ is not needed. I have learnt the idea of the following proof from Torsten Ekedahl.
In the following, all algebras are associative and with unity. The word "finite-dimensional" shall always mean "finite-dimensional as an $F$-vector space".
Theorem 1. Let $B$ be a finite-dimensional $F$-algebra. Let $M$ and $N$ be two finite-dimensional $B$-modules. Assume that $M\otimes_F L \cong N\otimes_F L$ as $B\otimes_F L$-modules. Then, $M \cong N$ as $B$-modules.
Theorem 1 is the Noether-Deuring theorem (19.25 in T. Y. Lam, A First Course in Noncommutative Rings, 2nd edition 2001).
Corollary 2. Let $B$ be an $F$-algebra. Let $M$ and $N$ be two finite-dimensional $B$-modules. Assume that $M\otimes_F L \cong N\otimes_F L$ as $B\otimes_F L$-modules. Then, $M \cong N$ as $B$-modules.
Proof of Corollary 2. Corollary 2 differs from Theorem 1 only in the lack of a finite-dimensionality requirement on $B$. We shall prove Corollary 2 by reducing it to Theorem 1.
The action of $B$ on the $B$-module $M \oplus N$ gives rise to an $F$-algebra homomorphism $B \to \End_F \left(M\oplus N\right)$. The kernel of this homomorphism is an ideal $I$ of $B$, and the $F$-algebra $B/I$ is finite-dimensional (since $B/I$ is isomorphic to the image of this homomorphism, and said image is clearly finite-dimensional). Every element of $I$ acts as $0$ on $M \oplus N$, and thus also on $M$ and $N$; therefore, both $B$-modules $M$ and $N$ can be regarded as $B/I$-modules. Moreover, the $B\otimes_F L$-module isomorphism $M\otimes_F L \cong N\otimes_F L$ (which exists by assumption) can be regarded as a $\left(B/I\right)\otimes_F L$-module isomorphism, whereas the $B$-module isomorphism $M \cong N$ that we are looking for boils down to a $B/I$-module isomorphism. Thus, Corollary 2 follows from Theorem 1 (applied to $B/I$ instead of $B$).
Corollary 3. Let $B$ and $C$ be two $F$-algebras. Let $M$ and $N$ be two finite-dimensional $\left(B,C\right)$-bimodules. Assume that $M\otimes_F L \cong N\otimes_F L$ as $\left(B\otimes_F L,C\otimes_F L\right)$-bimodules. Then, $M \cong N$ as $\left(B,C\right)$-bimodules.
Proof of Corollary 3. Let $C^\op$ be the opposite algebra of $C$. (This is the $F$-algebra which is defined as a "copy of $C$ with the multiplication turned around"; more formally, it is the $F$-vector space $C$ endowed with the multiplication $*$ defined by $a*b = ba$.) It is known that $\left(B,C\right)$-bimodules are "the same as" left $B\otimes_F C^\op$-modules. (More precisely, any $\left(B,C\right)$-bimodule structure on an $F$-vector space can be transformed into a left $B\otimes_F C^\op$-module structure on the same vector space by first turning the right $C$-module structure into a left $C^\op$-module structure, and then combining the left $B$-module structure with the left $C^\op$-module structure into a left $B\otimes_F C^\op$-module structure because they commute.) Similarly, $\left(B\otimes_F L,C\otimes_F L\right)$-bimodules are "the same as" left $\left(B\otimes_F L\right) \otimes_L \left(C\otimes_F L\right)^\op$-modules, which, conveniently, are the same as left $\left(B\otimes_F C^\op\right)\otimes_F L$-modules. Thus, Corollary 3 follows from Corollary 2 (applied to $B\otimes_F C^\op$ instead of $B$).
Let us now return to the problem.
For every $u \in A$, we let $L_u$ denote the map $A \to A, \ x \mapsto ux$ (known as "left multiplication by $u$"). The map $L_u$ is a homomorphism of right $A$-modules.
Now, we can define an action of the $F$-algebra $F\left[X\right]$ on $A$ by letting $X$ act as $L_a$. Combined with the canonical right $A$-module structure on $A$ (by right multiplication), this gives rise to an $\left(F\left[X\right],A\right)$-bimodule structure on $A$. Let us denote this $\left(F\left[X\right],A\right)$-bimodule $A$ by $A_a$.
Similarly, we can define an $\left(F\left[X\right],A\right)$-bimodule $A_b$, which differs from $A_a$ in that $X$ acts as $L_b$ rather than $L_a$.
It is easy to see that the right $A$-bimodule homomorphisms $A_b \to A_a$ are precisely the maps $L_u$ for $u \in A$. Thus, the $\left(F\left[X\right],A\right)$-bimodule homomorphisms $A_b \to A_a$ are precisely the maps $L_u$ for $u \in A$ which satisfy $ub = au$. Thus, the $\left(F\left[X\right],A\right)$-bimodule isomorphisms $A_b \to A_a$ are precisely the maps $L_u$ for invertible $u \in A$ which satisfy $ub = au$. Hence, an invertible $u \in A$ which satisfies $ub = au$ exists if and only if $A_a \cong A_b$ as $\left(F\left[X\right],A\right)$-bimodules. In other words,
(1) the elements $a$ and $b$ of $A$ are conjugate in $A$ if and only if $A_a \cong A_b$ as $\left(F\left[X\right],A\right)$-bimodules.
A similar argument (applied to the elements $a\otimes_F 1$ and $b\otimes_F 1$ of $A\otimes_F L$ over the field $L$ instead of the elements $a$ and $b$ of $A$ over the field $F$) shows that the elements $a\otimes_F 1$ and $b\otimes_F 1$ of $A \otimes_F L$ are conjugate in $A \otimes_F L$ if and only if $\left(A\otimes_F L\right)_{a\otimes_F 1} \cong \left(A\otimes_F L\right)_{b\otimes_F 1}$ as $\left(L\left[X\right],A\otimes_F L\right)$-bimodules (where $\left(A\otimes_F L\right)_{a\otimes_F 1}$ and $\left(A\otimes_F L\right)_{b\otimes_F 1}$ are defined similarly to $A_a$ and $A_b$). Since we know that the elements $a\otimes_F 1$ and $b\otimes_F 1$ of $A \otimes_F L$ are conjugate in $A \otimes_F L$, we thus conclude that $\left(A\otimes_F L\right)_{a\otimes_F 1} \cong \left(A\otimes_F L\right)_{b\otimes_F 1}$ as $\left(L\left[X\right],A\otimes_F L\right)$-bimodules. In other words, $A_a \otimes_F L \cong A_b \otimes_F L$ as $\left(F\left[X\right]\otimes_F L, A\otimes_F L\right)$-bimodules. Thus, Corollary 3 (applied to $B = F\left[X\right]$, $C = A$, $M = A_a$ and $N = A_b$) shows that $A_a \cong A_b$ as $\left(F\left[X\right], A\right)$-bimodules. By (1), this shows that the elements $a$ and $b$ of $A$ are conjugate in $A$. Qed.<|endoftext|>
TITLE: Smooth quadric hypersurface, Hilbert scheme is blowup of Grassmannian?
QUESTION [5 upvotes]: Let $Q \subset \mathbb{P}^n$ be a smooth quadric hypersurface. Where can I find a proof of/can anyone supply a proof of$$\text{Hilb}_{2m + 1}(Q) \cong \text{Bl}_{OG(3, n+1)}G(3, n+1)?$$Can we conclude that when $n > 5$, the Picard group of $\text{Hilb}_{2m+1}(Q)$ has rank two?
Apologies in advance, for I am not an algebraic geometrer...

REPLY [7 votes]: I am sure there are more direct references, but it is fairly easy to prove as well.  First of all, the Hilbert scheme $\text{Hilb}_{2m+1}(\mathbb{P}^n)$ is a $\mathbb{P}^5$-bundle over the Grassmannian $G(3,n+1)$.  There must be other sources, but one source is Theorem 3.4.1 of Alex Lee's senior thesis.
Alex Lee
The Hilbert Scheme of Curves in $\mathbb{P}^3$
http://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/ALee_Hilbertschemes.pdf
Of course there is an obvious morphism from the total space of that $\mathbb{P}^5$-bundle to $\text{Hilb}_{2m+1}(\mathbb{P}^n)$.  It requires just a little work to prove that this morphism is an isomorphism.
Given this identification, it is easy to see that the underlying closed subset of $\text{Hilb}_{2m+1}(Q^n)$ and the closed subset $\text{Bl}_{OG(3,n+1)}G(3,n+1)$ are equal as closed subsets of $\text{Hilb}_{2m+1}(\mathbb{P}^n)$.  It only remains to prove that $\text{Hilb}_{2m+1}(Q^n)$ is reduced as a scheme.
On $\text{Hilb}_{2m+1}(\mathbb{P}^n)$ there is a rank $5$, locally free sheaf obtained by pulling back $\mathcal{O}_{\mathbb{P}^n}(2)$ to the universal curve and then pushing forward to the Hilbert scheme.  It requires some computation to prove that there are no higher direct images and that the pushforward is locally free, but that is straightforward because the fibers are all degree $2$ Cartier divisors in $2$-planes (so just use standard results about cohomology of invertible sheaves on $\mathbb{P}^2$).  The defining equation of your quadric defines a global section of this locally free sheaf, and $\text{Hilb}_{2m+1}(Q^n)$ is the zero scheme of this global section.  Thus, wherever $\text{Hilb}_{2m+1}(Q^n)$ has codimension $5$ in $\text{Hilb}_{2m+1}(\mathbb{P}^n)$, it is a local complete intersection.  By the set-theoretic argument, $\text{Hilb}_{2m+1}(Q^n)$ is everywhere a local complete intersection.  In particular, this scheme is Cohen-Macaulay.
Finally, on the dense open subset of $\text{Hilb}_{2m+1}(Q^n)$ parameterizing smooth, genus $0$ curves, it is straightforward to compute that $\text{Hilb}_{2m+1}(Q^n)$ is smooth by checking vanishing of $H^1$ of the normal sheaf of the curve in $Q^n$: the tangent sheaf of $Q^n$ is globally generated, hence the normal sheaf is globally generated, and every globally generated coherent sheaf on $\mathbb{P}^1$ has vanishing $H^1$.  Since $\text{Hilb}_{2m+1}(Q^n)$ is generically smooth, it is generically reduced.  Every Cohen-Macaulay scheme that is generically reduced is everywhere reduced.  Since it is everywhere reduced, $\text{Hilb}_{2m+1}(Q^n)$ equals the reduced closed subscheme $\text{Bl}_{OG(3,n+1)}G(3,n+1)$ of $\text{Hilb}_{2m+1}(\mathbb{P}^n)$.<|endoftext|>
TITLE: Proof for the derivative of the determinant of a matrix
QUESTION [28 upvotes]: I was looking for theorems that might be helpful in order for some proofs that I have and I came across the following one:
$$\frac{d}{dt} [\det A(t)]=\det A(t) \cdot \operatorname*{tr}[A^{-1}(t)\cdot \frac{d}{dt} A(t)]$$
where $A(t)$ is a matrix with a variable $t$.
The problem is that I have neither a reliable source for this theorem nor am I able to prove it.
Did anyone come across the aforementioned equation or is able to prove it?

REPLY [31 votes]: Another way to obtain the formula is to first consider the derivative of the determinant at the identity:
$$
 \frac{d}{dt} \det (I + t M) = \operatorname{tr} M.
$$
Next, one has 
$$
\begin{split}
   \frac{d}{dt} \det A (t)
&=\lim_{h \to 0} \frac{\det \bigl(A (t + h)\bigr) - \det A (t)}{h}\\
&=\det A (t) \lim_{h \to 0}  \frac{\det \bigl(A (t)^{-1} A (t + h)\bigr) - 1}{h}\\
&=\det A (t) \operatorname{tr} \Bigl(A (t)^{-1}\frac{d A}{dt} (t) \Bigr).
\end{split}
$$<|endoftext|>
TITLE: Important formulas in combinatorics
QUESTION [138 upvotes]: Motivation:
The poster for the conference celebrating Noga Alon's 60th birthday, fifteen formulas describing some of Alon's work are presented. (See this post, for the poster, and cash prizes offered for identifying the formulas.) This demonstrates that sometimes (but certainly not always) a major research progress, even areas, can be represented by a single formula. Naturally, following Alon's poster, I thought about representing other people's works through formulas. (My own work, Doron Zeilberger's, etc. Maybe I will pursue this in some future posts.) But I think it will be very useful to collect major formulas representing major research in combinatorics.  
The Question
The question collects important formulas representing major progress in combinatorics.
The rules are:
Rules
1) one formula per answer
2) Present the formula explicitly (not just by name or by a link or reference), and briefly explain the formula and its importance, again not just link or reference. (But then you may add links and references.) 
3) Formulas should represent important research level mathematics. (So, say $\sum {{n} \choose {k}}^2 = {{2n} \choose {n}}$ is too elementary.) 
4) The formula should be explicit as possible, moving from the formula to the theory it represent should also be explicit, and explaining the formula and its importance at least in rough terms should be feasible.
5) I am a little hesitant if classic formulas like $V-E+F=2$ are qualified. 
An important distinction:
Most of the formulas represent definite results, namely these formulas will not become obsolete by new discoveries. (Although refined formulas are certainly possible.) A few of the formulas, that I also very much welcome, represent state of the art regarding important combinatorial parameters. For example, the best known upper and lower bounds for diagonal Ramsey's numbers. In the lists and pictures below an asterisk is added to those formulas.  
The Formulas (so far)
In order to make the question a more useful source, I list all the formulas in categories with links to answer (updated Feb. 6 '17). 

Basic enumeration: The exponential formula; inclusion exclusion; Burnside and Polya; Lagrange inversion; generating function for Fibonacci; generating function for Catalan; Stirling formula; Enumeration and algebraic combinatorics: The hook formula; Sums of tableaux numbers squared, Plane partitions;  MacMahon Master Theorem; Alternating sign matrices;  Erdos-Szekeres; Ramanujan-Hardy asymptotic formula for the number of partitions; $\zeta(3)$; Shuffles; umbral compositional identity; Jack polynomials; Roger-Ramanujan; Littlewood-Ricardson; 

Geometric combinatorics: Dehn-Somerville relations; Zaslavsky's formula; Erhard polynomials; Minkowski's theorem.  Graph theory: Tutte's golden identity; Chromatic number of Kneser's graph; (NEW)Tutte's formula for rooted planar maps ; matrix-tree formula; Hoffman bound; Expansion and eigenvalues; Shannon capacity of the pentagon; Probability: Self avoiding planar walks; longest monotone sequences (average); longest monotone sequences (distribution); Designs: Fisher inequality; Permanents: VanderWaerden conjecture; Coding theory: MacWilliams formula; Extremal combinatorics: Erdos-Sauer bound; Ramsey theory: Diagonal Ramsey numbers (); Infinitary combinatorics: Shelah's formula (); A formula in choiceless set theory. Additive combinatorics: sum-product estimates (*); Algorithms: QuickSort. 

(Larger formulas): Series multisection; Faà di Bruno's formula; Jacobi triple product formula; A formula related to Alon's Combinatorial Nullstellensatz; The combinatorics underlying iterated derivatives (infinitesimal Lie generators) for compositional inversion and flow maps for vector fields (also related to the enumerative geometry of associahedra); some mysterious identities involving mock modular forms and partial theta functions; 

Formulas added after October 2015:
Hall and Rota Mobius function formula; Kruskal-Katona theorem; Best known bounds for 3-AP free subsets of $[n]$ (*);

After October 2016: Abel's binomial identity; Upper and lower bounds for binary codes;

REPLY [4 votes]: The Durfee square formula for the generating function of the number of partitions doesn't seem to be in this list yet, I find it rather appealing:
$$
\frac{1}{(q)_\infty} = \sum_{n\geq 0} \frac{q^{n^2}}{(q)_n (q)_n} \ .
$$<|endoftext|>
TITLE: Connection between Stalling's end theorem and Seifert-van Kampen Theorem
QUESTION [6 upvotes]: Stalling`s end Theorem (a group has more than one end iff it splits over a finite subgroup) and the Seifert-van Kampen Theorem (the fundamental group of a 'decomposable' space is a free amalgamated free product w.r.t the 'pieces' of the space) are well-known theorems. However I often read that they are kind of analogues or that Stalling's Theorem is the 'combinatorial' version or something like that.
I was wondering if this could be made a bit more precise?
I know that there are some major differences. Indeed Stalling's Theorem deals with amalgamated free products and HNN-Extensions over finite subgroups, where the Seifert-van Kampen Theorem 'just' deals with amalgamated free products.
In this sense I was also wondering if there is some sort of 'Seifert-van Kampen Theorem' for HNN-Extensions?

REPLY [2 votes]: Here is an extract from p.335 of Topology and Groupoids (T&G):
 (source)
In this $\mathbf I$ is the groupoid with 2 objects $0,1$ and one arrow $\iota:0 \to 1$, and $\dot{\mathbf I}$ is the discrete groupoid on the set consisting of $0,1$.  Thus $\mathbf I$ is a "interval object" in the category of groupoids and can be used to model in the category of groupoids topological objects such as double mapping cylinders, as stated in the text. To mirror the topology you need of course a Seifert-van Kampen Theorem for the fundamental groupoid on a set of base points. Compare for example this mathoverflow discussion. 
I do not know about applications to the theory of ends! Perhaps old papers of Chris Houghton (see the bibliography of T&G) are relevant.<|endoftext|>
TITLE: A question about certain sets of permutations of the ordered pairs $(1,1),(1,2),\cdots,(1,n),\cdots,(n,1),(n,2),\cdots,(n,n)$
QUESTION [15 upvotes]: Let $n>1$ be a given positive integer. For any $0\leq k\leq n^2$, let $A_k$ be the set of permutations $((i_1,j_1),(i_2,j_2),\cdots,(i_{n^2},j_{n^2}))$ of the ordered pairs $(1,1),(1,2),\cdots,(1,n),\cdots,(n,1),(n,2),\cdots,(n,n)$ satisfying $i_1\leq i_2\leq \cdots\leq i_k$ and  $j_{k+1}\leq j_{k+2}\leq \cdots\leq j_{n^2}$.
For any $$((i_1,j_1),(i_2,j_2),\cdots,(i_k,j_k),(i_{k+1},j_{k+1}),\cdots,(i_{n^2},j_{n^2}))\in A_k,$$ we have $$((j_{k+1},i_{k+1}),(j_{k+2},i_{k+2}),\cdots,(j_{n^2},i_{n^2}),(j_1,i_1),(j_2,i_2),\cdots,(j_k,i_k))\in A_{n^2-k}.$$
So it is easy to see that $\vert A_k\vert=\vert A_{n^2-k}\vert$ for any $0\leq k\leq [n^2/2]$. 

Question: Do we have that $\vert A_0\vert<\vert A_1\vert<\cdots<\vert A_{[n^2/2]}\vert$?

REPLY [6 votes]: This is not an answer, but perhaps a relevant thought. The combinatorial description of $A_k$, for $0\le k\le n^2$, together with the Gamma integral $\Gamma(n+1)=\int_0^{\infty}x^ne^{-x}dx$ leads to the generating function
$$\sum_{k=0}^{n^2} A_kt^k=\int_0^\infty\cdots \int_0^{\infty}e^{-\sum_{i=1}^n(x_i+y_i)}\prod_{i,j=1}^n(x_i+ty_j)dx_1dy_1\cdots dx_ndy_n.$$
A polynomial whose coefficients form a unimodal (resp. log-concave) sequence is called a unimodal (resp. log-concave) polynomial. It is known that a polynomial with real roots is log-concave and thus unimodal. So unimodality holds for the polynomial being integrated above. Unfortunately log-concavity or unimodality are not preserved under linear combinations so one can't conclude the desired property just from the integral representation above. Maybe someone with knowledge in analysis will find this wording more useful.<|endoftext|>
TITLE: Relations among Young symmetrizers of non-standard tableaux
QUESTION [5 upvotes]: For any Young tableau, one can form the Young symmetrizer. I'm naturally interested in young symmetrizers coming from standard tableaux, but I'm forced to look at Young symmetrizers of non-standard tableaux. Is there a known way of writing such a Young symmetrizer as a linear combination of Young symmetrizers of standard tableaux, perhaps something akin to the Garnir relations?

REPLY [2 votes]: Let $T$ be any (not necessarily standard) tableau of shape $\lambda$. Define $c_\lambda(T)=a_\lambda(T) b_\lambda(T)$ as usual. Let $C_i(T)$ denote the ith column of $T$. I Equation 2.4 in
https://www3.nd.edu/~craicu/papers/yngsymm.pdf
says that for two columns $C_i(T)$, $C_j(T)$ with $|C_i(T)|\le |C_j(T)|$ and any $a\in C_i(T)$,
$$c_\lambda(T)\bigg(1-\sum_{x\in C_j(T)}{(a,x)}\bigg)=0,$$ where $(a,x)$ denotes the transposition switching $a$ and $x$. From here one can derive that 
$$c_\lambda(T)=\sum_{x\in C_j(T)}{(a,x)[(a,x)c_\lambda(T)(a,x)]}$$ where each $(a,x)c_\lambda(T)(a,x)$ is the symmetrizer of another tableaux. In fact, following the proof of Equation 2.4 shows that one can replace the sum of transpositions with any order two permutation. This gives a way to "straighten" a symmetrizer of a non-standard tableaux.<|endoftext|>
TITLE: Morita equivalence of $K$-algebras
QUESTION [5 upvotes]: Given $K$ a unital commutative ring and $A$ a $K$-algebra different from $K$. Can $K$ be Morita equivalent to $A \amalg A$, where $A \amalg A$ is the coproduct in the category of unital associative $K$-algebras?

REPLY [2 votes]: If V is a vector space such that $V\oplus  V$ is isomorphic to V then A=TV, the tensor álgebra, is isomorphic to $T(V\oplus V)=A\coprod A$, in particular Morita equivalent<|endoftext|>
TITLE: Maslov class as a relative cohomology class in $H^2(M, L)$
QUESTION [9 upvotes]: Let $(M, \omega)$ be a symplectic manifold and $L \subseteq M$ - a Lagrangian submanifold. I am trying to understand under what circumstances the Maslov homomorphism $I_{\mu, L} \colon \pi_2(M, L) \to \mathbb{Z}$ is in fact induced by an element $\mu_L \in H^2(M, L;\mathbb{Z})$. I am encountering a few related issues:

If I have a map $f \colon (S, \partial S) \to (M, L)$ from an oriented surface $S$ with non-empty boundary, I can symplectically trivialise $(f^*TM, \omega)$ and calculate the Maslov index of $f\vert_{\partial S}^*TL$ using this trivialisation and the natural orientations of the different boundary components of $S$. But since $Sp(2n)$ is not simply connected, this might depend on the choice of trivialisation if $S$ has non-zero genus. 
A related question is when is a class $a \in  H_2(M, L; \mathbb{Z})$ representable by a surface as above? 
Even if the Hurewicz homomorphism $h\colon \pi_2(M, L) \to H_2(M, L;\mathbb{Z})$ is surjective, so I can represent every relative homology class by a disc, do I have that the Maslov index of a disc $a \in \pi_2(M, L)$ depends in fact only on $h(a)$?

REPLY [7 votes]: I think it is always true, and one can explicitly define the class as follows, at least in the compact case (this construction is surely well-known to some people, but I'm not sure where to find it written down).
Let $M^{2n}$ be an almost complex manifold and $L^n \subseteq M$ a totally real submanifold (of course in the symplectic/Lagrangian case you just pick a compatible almost complex structure before starting).  Over $M$ we have the complex line bundle $$E := \Lambda^n_{\mathbb{C}} TM,$$ and over $L$ we have the real line bundle $$F := \Lambda^n_{\mathbb{R}} TL.$$  $F$ is a subbundle of $E|_L$ in the obvious sense.
If $L$ is oriented then we can define $\mu_L$ to be twice the Poincaré dual of the vanishing locus of a generic smooth section $s$ of $E$ such that $s|_L$ lies in $F$ and is nowhere-zero.  In the non-orientable case, $F$ doesn't have a nowhere-zero section, but we can replace $E$ and $F$ by their tensor squares and then omit the factor of two after taking Poincaré duals.  It is clear from these definitions that $\mu_L$ maps to $2c_1(X)$ in $H^2(X)$ and that when $L$ is orientable $\mu_L$ is divisible by two.  One can also check by hand that it does indeed induce the Maslov homomorphism.
Without assuming compactness of $M$ and $L$ one probably has to work a bit more; morally the definition has to be at least as hard as that of $c_1(X)$.  You may also needs to make some reasonable assumption on the niceness of the embedding of $L$ in $M$.  If the pair $(M, L)$ is a relative CW complex then $E$ (or its tensor square when $L$ is non-orientable) can be represented by a map $$M \rightarrow BU(1)$$ extended cell-by-cell from a constant map $L \rightarrow EU(1)$, by using $F$ (or its square) to give a canonical homotopy class of trivialisation over $L$.  You could then define $\mu_L$ to be twice (or not) the class of the pullback of a cocycle representing $c_1$ of the tautological bundle.  Here it's less obvious that this class is independent of choices and induces the Maslov homomorphism, but it can probably be checked.<|endoftext|>
TITLE: Does there exist an uncountable partition of a Polish space so that the union of any collection of blocks is Borel?
QUESTION [10 upvotes]: Is it consistent that there exists a partition $P$ of the real number line $\mathbb{R}$ such that $|P|>\aleph_{0}$ but where $\bigcup R$ is Borel whenever $R\subseteq P$?
If $2^{\aleph_{0}}<2^{\aleph_{1}}$, then the answer to this question is $\textbf{no}$ since there would be at least $2^{\aleph_{1}}$ subsets of each uncountable partition $P$ but there are only $2^{\aleph_{0}}$ Borel sets. We therefore need to use a model such that $2^{\aleph_{0}}=2^{\aleph_{1}}$ to construct our counterexample. 
Under $MA+\neg CH$ does there exist a partition $P$ of the real number line $\mathbb{R}$ such that $|P|>\aleph_{0}$ but where $\bigcup R$ is Borel whenever $R\subseteq P$?
Perhaps such a partition $P$ is impossible to construct by some fact obvious to descriptive set theorists that I probably have overlooked.
I call a partition $p$ of a Boolean algebra $B$ such that $\bigvee R$ exists whenever $R\subseteq p$ a subcomplete partition and subcomplete partitions came up all the time in my work on Boolean partition algebras, so I would be interested in which partitions of a given Boolean algebra are subcomplete and which ones are not.

REPLY [16 votes]: Suppose $\{P_i : i < \kappa\}$ is such a partition. Let $f: R \to R$ be a function satisfying $|f[P_i]| = 1$ and for all $i < j < \kappa$, $f[P_i] \cap f[P_j] = \phi$. Then $f$ is Borel so its image is an uncountable analytic set of size less than continuum: Contradiction.<|endoftext|>
TITLE: Representing classes in *relative* homology by submanifolds
QUESTION [6 upvotes]: There are nice results for representing homology classes by submanifolds, in particular for any class in $H_i(X)$ with $i\le 6$, see here. When $X$ is low-dimensional I can start getting explicit, but this uses Poincare duality and appeals to classifying maps and characteristic classes of bundles. What is the right way to approach the problem for relative homology? Does the formulation of (Thom) spectra extend here? While I do have Lefschetz duality, I don't think I can do much with bundles, and I am not comfortable with (Thom) spectra.
Given a submanifold $A\subset X$ of a closed (smooth?) manifold, when can I (not) represent a homology class in $H_i(X,A;\mathbb{Z})$ by some submanifold with appropriate boundary?
Take whatever restrictions on $i$ and $\dim X$, but assume we're in a setting where the Hurewicz map isn't an isomorphism. If $i>\dim A+1$ then $H_i(X)$ surjects onto $H_i(X,A)$ in the LES, so we can take a submanifold which represents a homology class of $X$ to be the submanifold that represents the relative homology class.
Perhaps there is an example of an unrepresentable class in $H_2(X,S^1;\mathbb{Z})$ with $\dim X =3$ and $S^1$ highly knotted. When $X=S^3$ this group is $\mathbb{Z}$ (as seen by the LES), and Seifert surfaces do the trick.
As a toy model, consider the equator in a sphere. Then $H_2(S^2,S^1)\cong\mathbb{Z}^2$, as seen by reduced homology $\widetilde{H}_2(S^2/S^1)\cong H_2(S^2\bigvee S^2)\cong H_2(S^2)\oplus H_2(S^2)\cong\mathbb{Z}^2$. It looks like the classes $(1,0)$ and $(0,1)$ come from the hemispheres, and the relative Hurewicz theorem applies I think.

REPLY [6 votes]: I think the arguments are all pretty much the same.  Let $X$ be a manifold, and $A$ a subset that's "tame enough" to admit a tubular / regular neighbourhood $V$, then
$$H_i(X,A) \simeq H_i(X,V)$$ 
by homotopy.  And $H_i(X,V) \simeq H_i(X \setminus int(V), \partial V)$ by excision. 
But this group is isomorphic to $H^{n-i}(X \setminus int(V))$ by Poincar\'e duality (if you assume orientability).  $n=dim(X)$
At this point the standard Serre-Thom representability constructions work very nicely, for example if $i=n-1$ you use that $H^1(X \setminus int(V)) \simeq [X \setminus int(V), S^1]$ and apply transversality.<|endoftext|>
TITLE: Training towards research on k3 surfaces
QUESTION [5 upvotes]: I am a graduate student learning basic algebraic geometry (from Hartshorne, Shafarevich). I'm planning to work in k3 surfaces (arithmetic and geometric properties, in my guide's words). I came to know that I can start learning algebraic surfaces and get the concepts when required from AG.
Can you suggest books, lecture notes having this view point?

REPLY [7 votes]: Likely this should only be a comment, but I don't have enough reputation for that...
J.C. Ottem has provided a wonderful reference about the basics of K3 surfaces in his comment. It's my personal experience though that when I'm working through notes such as Huybrecht's, it's instructive and motivating to have short and interesting papers to read once I've learned enough theory. If this is also the case for you, I recommend having a look at Deligne's ``Relèvement des surfaces K3 en caractéristique nulle''. The article clearly explains the degeneration of the Hodge to de Rham spectral sequence and is a nice way to start using some characteristic $p$ methods and deformation theory.   

I should also mention that there are some cool notes here:
http://math.northwestern.edu/~dwilson/K3.html
and point out that there's a list of references at the bottom of the page which you might like to browse. The notes also include a discussion of some uses of K3 surfaces in physics, if that interests you.  

One of the Arizona Winter School 2015 courses was ``Arithmetic of K3 surfaces'', given by Várilly-Alvarado. Not only are the notes nice, but there are also videos of the lectures. Here is the link...
http://swc.math.arizona.edu/index.html<|endoftext|>
TITLE: when does elementwise-log preserve positive-semidefiniteness?
QUESTION [6 upvotes]: Let $Z$ be a positive semidefinite matrix with nonnegative entries, and define $X=\log(1+Z)$, where the $\log$ is taken entrywise, i.e., $X_{ij}=\log(1+Z_{ij})$. Are there some simple sufficient conditions that guarantee that $X$ is positive semidefinite? 
Note 1
Intuitively, this should work when $Z$ is small enough, because then $X \simeq Z \succeq 0$.
Note 2 A simple counter-example is given by $Z=\left[ \begin{array}{cc}
3 & 5\\
5 & 9
\end{array}
\right]$.
Note 3
It is easy to verify that the converse relation allways holds, that is,
$$X\succeq 0 \Longrightarrow Z:=exp(X)-1\succeq 0,$$
where the above operations are elementwise: $Z_{ij} = e^{X_{ij}} -1$. Indeed, we have the relation $Z=\sum_{k=1}^\infty X^{\circ k}/k!$, where $X^{\circ k}$ is the $k$th power of $X$ with respect to the Hadamard (elementwise) product, and it is well-known that the Hadamard product of 2 positive semidefinite matrices is positive semidefinite.
Note 4 Some background on this question: I ask myself when a random variable $Y$ with mean vector $\mu\in\mathbb{R}^n$  and variance-covariance matrix $\Sigma \succeq 0 \in \mathbb{R}^{n\times n}$ can be fitted to a multivariate log-normal distribution using the method of moments. Applying the formulas, we find that the elements of variance-covariance matrix $V$ of $\log Y$ must satisfy
$$V_{ij}= \log(\frac{\Sigma_{ij}}{\mu_i \mu_j} +1),$$
but in general this matrix is not guaranteed to be positive semidefinite.

REPLY [4 votes]: It's not true that it works for $Z$ small enough.  Consider the $2 \times 2$ case
$$ Z = \pmatrix{t & 2t\cr 2t & 4t\cr} $$
which is positive semidefinite for $t \ge 0$.
Then $$\det(X) = \log(1+t)\log(1+4t) - \log(1+2t)^2 $$
which appears to be negative for all $t > 0$, and certainly is negative for small $t > 0$: its Maclaurin series is
$ \det(X) = -2 t^3 + O(t^4)$.
What is true is that if $Z$ is positive definite, $\log(1+tZ)$ will be positive definite for sufficiently small $t > 0$.  This can be obtained using the
Maclaurin series for $\log(1+z)$.
EDIT:
Let $g(z) = z - \log(1+z)$, and 
write $X = Z - G$, where $G_{ij} = g(Z_{ij})$.
Note that $g$ is increasing on $[0,1]$.  If all $|Z_{ij}| \le b$, then 
$|W_{ij}| \le g(b)$.  If your matrices are $N \times N$, 
for any vector $x$ we have by Cauchy-Schwarz $x^T G x \le  N g(b) \|x\|^2$.
Thus to have $X \succeq 0$ it suffices for $Z$ to be positive definite with least eigenvalue $\lambda$ and elementwise bound $b$ where
$$ \lambda \ge N g(b) $$<|endoftext|>
TITLE: can $H^*(\mathbb{C}P^n;\mathbb{Z})$ be the cohomology of some Eilenberg-Maclane space $K(\pi,1)$?
QUESTION [8 upvotes]: Recently I came across the following question:
can $H^*(\mathbb{C}P^n;\mathbb{Z})$ be the integral cohomology ring of some Eilenberg-Maclane space $K(\pi,1)$?
I guess (without strong evidences) that the answer is negative. If so, how to prove it?
Thanks in advance!

REPLY [21 votes]: The Kan-Thurston theorem says that every path-connected space is homology equivalent to an Eilenberg-MacLane space, so the answer is "yes".  See
Daniel Kan and William Thurston, Every connected space has the homology of a K(π,1), Topology Vol. 15. pp. 253–258, 1976.<|endoftext|>
TITLE: Immersed quasi-Fuchsian surfaces surviving Dehn fillings
QUESTION [6 upvotes]: In papers like, Cooper - Long - Some surface subgroups survive surgery or Li - Immersed essential surfaces in hyperbolic 3-manifolds the game is to find some quasi-Fuchsian immersed surface $Q \looparrowright M$ where $M$ is a hyperbolic 3-manifold with toroidal cusps and then to Dehn fill $M$ along some (multi) slope $\gamma$ that is sufficiently far from the immersed boundary slope(s) of $Q$. In the Dehn-filled quotient $M(\gamma)$ something clever is done to the image of $Q$ to produce a closed essential (= immersed +  $\pi_1$-injective) surface $F\looparrowright M(\gamma)$. Doing so was justifiably a source of pride, this was pre Kahn-Markovich.
I am curious about how the techniques of these papers, or others, could be used to answer a completely different question. Here's what I want to do: I have an atoroidal 3-manifold with torus boundary components (i.e. its interior is hyperbolizable) and a fixed immersed boundary slope $\gamma$, denote by $f:M \hookrightarrow M(\gamma)$ the inclusion into the Dehn filling. Can I find some essential surface with boundary $i:(S,\partial S) \looparrowright (M,\partial M)$ such that $f \circ i$ is still $\pi_1$-injective? I know that this composition is a map from a surface with boundary to a closed
 3-manifold (i.e. with empty boundary), which is a weird thing to do for a 3-manifold person, but I'm actually more interested by the group theory.
The literature is technically difficult for me and I am unclear about a few things.

Both linked papers have a Theorem 1.2 that seems to say that if $M$ is a hyperbolic 3-manifold and  $\alpha$ is the immersed boundary slope of some essential  $Q\looparrowright M$ then if a filling slope $\gamma$ is "far" enough from $\alpha$ we can cook up an essential closed surface in $M(\gamma)$. A hyperbolic knot complement has a Dehn filling to the 3-sphere, does this mean that such 3-manifolds only admit finitely many immersed boundary slopes, since they all need to be close to the meridian (the slope that fills to give a sphere)?
There is a result due to Baker, as well as many generalizations, that in some cases infinitely many immersed boundary slopes from essential surfaces may occur. This phenomenon seems opposite to knot complements.  Is there a characterization of the hyperbolic 3-manifolds that admit an infinity of immersed boundary slopes?
If the construction given in 1. above goes through (i.e. the immersed boundary slope is far from the filling slope) is it know whether $Q$ (which the authors stop caring about) also $\pi_1$-injects into $M(\gamma)$?

Again, just to be clear because it's weird, I have a fixed Dehn filling and I want to find essential surfaces with "far" immersed boundary slopes. Any help, even pointing out that I've got it all wrong, would be greatly appreciated. Might as well assume that $M$ has one boundary component.
EDIT: Please ignore  1. I misread Theorem 1.2 in both papers. The quasi-Fuchsian surface $Q$ is embedded, the mystery closed surface that is constructed is immersed. This is embarrassing: I got it all wrong :-(

REPLY [3 votes]: For question 2., one can prove that for a 1-cusped orientable hyperbolic 3-manifold $M$, all but finitely many slopes bound an immersed essential surface. This follows by combining the hyperbolic Dehn filling theorem (in fact, the $2\pi$ or 6 theorems suffice) and Kahn-Markovic's theorem. If the slope $\alpha$ has length $> 6$ in an embedded cusp, then the core of the Dehn filling $\gamma$ will be infinite order and the filling $M(\alpha)$ will be hyperbolic. By Kahn-Markovic, there is an immersed $\pi_1-$injective closed orientable surface $f:\Sigma \to M(\alpha)$. Assume that $f\pitchfork\gamma$, then pass to the covering space $ \pi: N\to M(\alpha)$ so that $f$ lifts to a map $\tilde{f}:\Sigma \to N$ which is a homotopy equivalence and an embedding. Then $\tilde{f} \cap \pi^{-1}(\gamma)$ consists of finitely many points. If $\tilde{f}\cap N-\pi^{-1}(\gamma)$ is not incompressible in $N-\pi^{-1}(\gamma)$, then one may perform compressions via an isotopy in $N$ (since $N$ is irreducible and $\tilde{f}$ incompressible) reducing the number of intersections until $\tilde{f}\cap N-\pi^{-1}(\gamma)$ is incompressible in the complement of $N-\pi^{-1}(\gamma)$. Then this surface will push down to $M$ to give a surface with immersed boundary slope $\alpha$. One might object that the surface might be homotoped in $M(\alpha)$ to completely miss $\gamma$. But one may use properties of the Kahn-Markovic surfaces and the fact that $\gamma$ is essential in $M(\alpha)$ to show that this doesn't happen if the surface is close enough to being geodesic (via an argument of Bergeron-Wise). 
The more recent paper of Cooper-Futer proves the existence of closed quasi-fuchsian surfaces in $M$ using the Kahn-Markovic surfaces of Dehn fillings as a starting point (so similar to this argument).<|endoftext|>
TITLE: How much of the ATLAS of finite groups is independently checked and/or computer verified?
QUESTION [52 upvotes]: In a recent talk Finite groups, yesterday and today Serre made some comments about proofs that rely on the classification of finite simple groups (CFSG) and on the ATLAS of Finite Groups. Namely, he said that a proof that relied on the CFSG and said so was ok, but a proof that relied on the ATLAS was not so ok, because the content has not been completely independently verified, and has as its basis old computer and other calculations that have only been done once.
Given that (numerous?) little errors have been found over time, it would be good to have a definitive record as to which bits have either been calculated or proved elsewhere, or formally verified in the case of the computer calculations, and if so where and by whom (with code for the latter case). Note that merely being able to do some calculations in GAP is not quite enough, since, as the documentation says

Part of the constructions have been documented in the literature on almost simple groups, or the results have been used in such publications, see for example the references in [CCNPW85] and [BN95].

where CCNPW85 is the ATLAS and BN95 is Breuer and Norton's Improvements to the Atlas (in an appendix of Atlas of Brauer Characters).

EDIT Since it may not have been clear, I was after statements modeled on the following:

"All results about classical groups of Lie type are well-known and documented elsewhere"
"All results about conjugacy classes of the sporadic groups except Janko 4 [say] are calculated afresh and contained in X computer package"
"The results on [blah] about [some group] are only contained in the ATLAS, and no papers or independently written software have reproved/recalculated them"

If it's easier to specify what is only in the ATLAS then that would be good, since clearly a lot of the classical material would be known and calculated long before.

EDIT May 2017 In a recent talk (50 years ago: a great time for number theory — the first few minutes only before the main talk) Serre mentions his comments discussed here, the fact people got worked up about it, and the paper Breuer, Malle, and O'Brien - Reliability and reproducibility of Atlas information in Farrokh Shirjian's answer, which he feels addresses his complaints.

REPLY [27 votes]: The Atlas was a work of scholarship, not research.  Our aim in those days was to collect information together for convenience, and the large character tables, and all the other data, was all proved.  The fact that there were so many errors is down to human error - both in our sources and our own work.
Nevertheless it would be possible to re-compute much of it now fairly easily, and in particular computing the character tables of explicitly given matrices or permutations would be less heroic now than 30 years ago.  The result would be that there exists a group with the proven properties.  If anyone wants to have a go at that, I'd love to help.
The "uniqueness" part is more difficult.  There is much published work on properties of simple groups, and the definition of the group studied would need to be somehow be connected to the (re)computed data.  This would require further work.  For example one would need to show for the first Conway group that the 24x24 matrices used to define the group fix an even unimodular lattice in 24 dimensions with minimum norm 4 and that there is an involution centralizer of the form 2.1+8.O8+(2) and so on.
There is no doubt that the project is feasible, and I had a short discussion with Serre on the subject.  Let's hope someone cares enough to actually do it!<|endoftext|>
TITLE: Questions on poincare homology spheres and branched covers
QUESTION [8 upvotes]: I have two questions:
Question 1. Suppose that $K$ is a knot in $S^3$. Let $\Sigma(K)$ be the double branched cover of $S^3$ branched along $K$. If $\Sigma(K)=\#_{i=1}^n\Sigma(2,3,5)$, then $K=\#_{i=1}^nT_{3,5}$? 
Question 2. There is a conjecture of Ozsvath-Szabo that the only $L$-spaces which are integral homology 3-spheres are connected sums of Poincare homology 3-spheres. If this conjecture and Question 1 is true, then can we (re)prove that Khovanov homology is an unknot-detector using Ozsavth-Szabo's spectral sequence?

REPLY [7 votes]: As Ian Agol points out the Orbifold Theorem will answer question 1 affirmatively. Although the original result is due to Thurston, the common references in the literature are: 
Boileau, Michel, Sylvain Maillot, and Joan Porti. Three-dimensional orbifolds and their geometric structures. Vol. 15. Paris: Société mathématique de France, 2003.
and
Cooper, Daryl, Craig David Hodgson, and Steve Kerckhoff. Three-dimensional orbifolds and cone-manifolds. Vol. 5. Mathematical Society of Japan, 2000.
However, both of these results would still require one to understand the quotients of $\Sigma(2,3,5)$ by its order 2 symmetries. Dunbar provides such a classification, which serves a translation of Threlfall and Seifert's work, see:
Dunbar, William D. "Geometric orbifolds." Revista Matematica de la Universidad Complutense de Madrid 1.1 (1988): 67-100. (available here)
and 
Threlfall, William, and Herbert Seifert. "Topologische Untersuchung der Diskontinuitätsbereiche endlicher Bewegungsgruppen des dreidimensionalen sphärischen Raumes I and II." Mathematische Annalen 104 and 107 (1931 and 1932): 1-70 and 543-586.
More specifically Dunbar gives a proof that if a knot $L$ has the property that $\Sigma(L)=\Sigma(2,3,5)$ then $L=T_{3,5}.$<|endoftext|>
TITLE: Hyperelliptic curves with fixed genus and many rational points
QUESTION [8 upvotes]: It is a famous theorem of Faltings, previously a conjecture by Mordell, that any algebraic curve of genus at least $2$ defined over the rational numbers have at most finitely many rational points. A hyperelliptic curve is a special algebraic curve of the form
$$\displaystyle z^2 = f(x,y),$$
where $f(x,y) \in \mathbb{Z}[x,y]$ is a binary form of degree $2n+2$. The genus of the curve is equal to $n$. For genus $1$ and $2$, all algebraic curves are hyperelliptic. 
What is unknown from Faltings' theorem is an effective bound for the height of rational points lying on a given curve. Therefore, even if one obtains an effective upper bound for the number of rational points lying on a given curve $C$, there is no way to check how close to being optimal the bound is.
A recent theorem of Bhargava asserts that most hyperelliptic curves, ordered by naive height of the coefficients of $f$, have no rational points as the genus $n$ tends to infinity. Indeed, there are $o(2^{-n})$ hyperelliptic curves of genus $n$ defined over $\mathbb{Q}$ with at least one rational point. 
My question is, what is the quantity
$$\mathfrak{S}(n) = \displaystyle \sup_{\substack{C \text{ hyperelliptic} \\ \text{genus of } C = n}}\#\{(x,y,z) \in C: (x,y,z) \text{ rational; inequivalent}\}.$$
Here two points $(x, y,z), (x',y',z')$ are inequivalent if there does not exist a non-zero rational number $\lambda$ such that $(x,y,z) = (\lambda x', \lambda y', \lambda^{n+1} z')$. In other words, the points $(x,y,z), (x',y',z')$ are not in the same equivalence class in the weight projective space $\mathbb{P}(1,1,n+1)$. 
Plainly, $\mathfrak{S}(n) \gg n$. Indeed, this can be seen by considering the curve
$$\displaystyle C: z^2 = x^{2n+2} + (y - x)(y - 2x) \cdots (y - (2n+2)x)$$
which has the pairwise inequivalent points $(1,1,1), (1,2,1), \cdots, (1,2n+2,1)$. It is not clear to me whether or not these are essentially all of the inequivalent rational points on $C$.
Is $\mathfrak{S}(n)$ bounded? If so, then can we give an upper bound in terms of $n$? If not, can one give an explicit family of curves $C_1, C_2, \cdots$ such that
$$\displaystyle \# C_1(\mathbb{Q}) < \#C_2(\mathbb{Q}) < \cdots ?$$

REPLY [9 votes]: This is an open problem. As Felipe points out in his comment to the question, the existence of a bound would follow from the weak Lang conjecture (rational points on varieties of general type are not Zariski dense), as was proved by Caporaso, Harris and Mazur. The weak Lang conjecture is wide open, though.
There are bounds when one also bounds the rank $r$ of the Mordell-Weil group (it has to be $\le n−3$), see this paper for the case of hyperelliptic curves (with a bound $\ll (r+1)n$) and this one by Katz, Rabinoff and Zureick-Brown for a generalization to arbitrary curves.
There are (in my view) convincing heuristic arguments in favor of a bound depending on $n$ and $r$ only. So your question may be related to the (equally open) question whether Mordell-Weil ranks of Jacobians of (hyperelliptic) curves of fixed genus are bounded or not. If they are, then it seems very likely that the number of points is bounded, too.
Another note: the current best lower bound for ${\mathfrak S}(2)$ is 642, given by this curve.<|endoftext|>
TITLE: Is every non-empty $\Delta_0$ set provably the range of some primitive recursive function?
QUESTION [7 upvotes]: Suppose $A(x)$ is a $\Delta_0$ formula defining a non-empty set of natural numbers. It's an easy theorem that there is a primitive recursive function $f:\mathbb{N} \rightarrow \mathbb{N}$ such that $Range(f) = \{n \in \mathbb{N} \mid A(n)\}$. I'm wondering if it's known whether either of the following strengthenings of this theorem are true:
Strengthening 1: Suppose $A(x)$ is a $\Delta_0$ formula defining a non-empty set of natural numbers. Can we find a $c \in \mathbb{N}$ such that $\varphi_c$ is primitive recursive, $\mathbb{N} \models \forall x (A(x) \rightarrow \exists y \varphi_c(y) {\downarrow} = x)$, and $PA \vdash \forall y \exists x (\varphi_c(y) {\downarrow} = x \wedge A(x))$?
Strengthening 2: Suppose $A(x)$ is a $\Delta_0$ formula defining a non-empty set of natural numbers. Can we find a $c \in \mathbb{N}$ such that $\varphi_c$ is primitive recursive, $PA \vdash \forall x (A(x) \rightarrow \exists y \varphi_c(y) {\downarrow} = x)$, and $PA \vdash \forall y \exists x (\varphi_c(y) {\downarrow} = x \wedge A(x))$?
(In the above, "$\varphi_c$" refers to the (partial) recursive function defined by the Turing machine whose Gödel code is c. "$\varphi_c(y) {\downarrow} = x$" is shorthand for a $\Sigma_1$ formula which says that with input y, the Turing machine with Gödel code c eventually halts and outputs x.)
Pedantic Clarification: Typically "$\phi_c(x){\downarrow} = y$" is an abbreviation for "$\exists t M(c,x,y,t)$", where (i) $M(c,x,y,t)$ is a $\Delta_0$ formula and (ii) $\mathbb{N} \models M(c,x,y,t)$ if and only if the Turing machine with Gödel code c halts after t steps on input x and produces output y. Let's call any formula $M$ satisying (i) and (ii) a "computation predicate". By this paper by H.B. Enderton, Strengthening 2 is false for certain choices of computation predicate.
It follows that any argument for Strengthening 2 which assumes that our choice of $\Delta_0$ computation predicate is arbitrary cannot suffice. We also need that PA proves certain facts about our chosen computation predicate. I suspect that any reasonable construction of a computation predicate will work (e.g., whatever construction is used in your favorite recursion theory book), so my question should be phrased more precisely as: is there a computation predicate such that Strengthenings 1 and 2 hold?

REPLY [3 votes]: I think both strengthening 1 and 2 are true and it is a sketch for proof:
You can construct an effective procedure to find an index $c$ (of a primitive recursive function) for any $\Delta_0$ formula
$A(x)$ such that $\mathbb{N}\models \forall x (A(x)\leftrightarrow \exists y \varphi_c(y)\downarrow =x)$ (because for writing a computer program to simulate $A(x)$ we do not need an unrestricted search loop, and all programs in which all search loops are restricted by a primitive recursive bound, computes a primitive recursive function). If we drop the part $\wedge A(x)$ from the condition $PA\vdash \forall y \exists x (\varphi_c(y)\downarrow =x \wedge A(x))$, it becomes the assertion that $\varphi_c$ is a total function and we know that all primitive recursive functions are provably total in $PA$ (in fact $I\Sigma_1$). I think that with a suitable choose of $c$ it would be provable in $PA$ that $\forall x (A(x)\leftrightarrow \exists y \varphi_c(y)\downarrow =x)$ (of course it need an exact
proof by induction on complexity of $A(x)$) and if it can be proved, srengthening1 and 2 both hold.

REPLY [3 votes]: I believe even strengthening 2 is a true statement, that can be proven by "simply" internalizing the proof of the original statement into $PA$.
Proof: Suppose $A(n_0)$ holds for some number $n_0\in\mathbb{N}$. In particular this is provable in $PA$:
$$ PA\vdash A(\overline{n_0})$$
where $\overline{n}$ is the numeral representing $n$. This is a simple consequence of completeness of $PA$ (and indeed, much much weaker systems) for $\Delta_0$ sentences.
One can then build the following recursive function $f$:

$f(0)=n_0$
$f(n+1)= \begin{cases} n \mbox{ if $A(n)$ holds}\\ n_0 \mbox{ otherwise}\end{cases}$

The first case in the definition of $f(n+1)$ is easily decidable even for a p.r. function. This function is easy to define in $PA$, and it is easy to prove, again in $PA$, that $A(m)$ holds for every $f(n)=m$, and conversely that for every $m$ such that $A(m)$ hold, there is a corresponding $n$ with $f(n)=m$ (take $n=0$ if $m=n_0$ and $m+1$ otherwise).
The rule of thumb is:

$PA$ can prove everything you can prove in an informal argument that doesn't specifically involve consistency of $PA$ (or something stronger).

This has a few exceptions (e.g. the Paris-Harrington argument).<|endoftext|>
TITLE: Steenrod operations on cohomology of grassmannians
QUESTION [9 upvotes]: Let $G_k(\mathbb{R}^n)$, $n\geq k$ and $G_k(\mathbb{R}^\infty)$ be the finite-dimensional and infinite-dimensional grassmannians respectively. Their cohomology rings are expressed in terms of universal Stiefel-Whitney classes
$$
H^*(G_k(\mathbb{R}^\infty);\mathbb{Z}_2)=\mathbb{Z}_2[w_1,w_2,\cdots,w_k],$$
$$
H^*(G_k(\mathbb{R}^n);\mathbb{Z}_2)=\mathbb{Z}_2[w_1,w_2,\cdots,w_k]/(\bar w_{n-k+1},\bar w_{n-k+2}\cdots,\bar w_n).$$
What are the Steenrod operations $Sq^i$ on these cohomology rings? I want to know the expressions of $Sq^i w_j$ for all possible $i,j$.

REPLY [11 votes]: As in Prasit's comment, the action of the Steenrod squares on the Stiefel-Whitney classes of any vector bundle are given by Wu's formula
$$
Sq^i(w_j) = \sum_{t=0}^i \binom{j+t-i-1}{t} w_{i-t} w_{j+t}.
$$
Here $w_k$ is of course $0$ if $k$ exceeds the dimension of the bundle. Together with the Cartan formula, this completely answers your question.<|endoftext|>
TITLE: Approximation of topological dynamical systems?
QUESTION [7 upvotes]: I'm trying to find references to approximations of topological dynamical systems in the following sense:
A topological dynamical system $(X, f)$ consists of a topological space (typically compact Hausdorff, or even metrisable) and a continuous $f : X \to X$. We can topologise $C(X, X)$ in various ways, for instance with pointwise / compact-open / uniform topologies. Given such a topology $\tau$, we denote the topological space by $C_\tau(X, X)$.
Are there theorems along the lines of "Given a sequence $(f_n) \subseteq C_\tau(X, X)$, if $(f_n) \to f$ then the dynamical systems $(X, f_n)$ 'approximate' $(X, f)$"? Perhaps $f_n \in C_\tau(X_n, X_n)$ for some subspace $X_n$ and we can consider 'approximations' from subsystems? Or $\{f \in C_\tau(X, X) : (X, f) \text{ has property } P\} \subseteq \overline{\{f \in C_\tau(X, X) : (X, f) \text{ has property } Q\}}$ for dynamical properties $P$ and $Q$?

REPLY [5 votes]: In general, dynamical properties of dynamical systems are often quite unstable. The Mandelbrot set makes one nice parameter: the base equation $z \mapsto z^2 + c$ is varies continuously with respect to any reasonable topology on continuous functions. The Mandelbrot set contains large open regions (for instance, the central cardoid, or all the disks touching it). In the interior of these, the dynamics is stable: nearby maps are actually topologically conjugate. But outside of these almost all properties of the dynamics will depend quite sensitively on where you are. For instance, look up plots of Julia sets of the function $z \mapsto z^2 + c$ for $c$ close to $1/4$.<|endoftext|>
TITLE: Can the Kan-Thurston theorem be turned into some kind of equivalence between groups and spaces?
QUESTION [8 upvotes]: I not really familiar with these subjects. I read  this question and I was really surprised by the answer. My question is probably vague (so please do bear with me). 
The cited question/answer suggests that any integral homology of a simply connected space can be realized as integral homology of some Eilenberg-MacLane space. Does it mean that connected spaces are the same thing (up to ?) as groups in some sense ? My question is vague, very vague. I will be happy to see a correct formulation of my question (if it makes sense).

REPLY [19 votes]: Here's something that might be considered an answer.
In Section 11 of Baumslag, Dyer, Heller, "The topology of discrete groups", the following theorem is proved, basically building on the ideas of the Kan-Thurston theorem.  
A perfect homomorphism of groups $G\to H$ is a surjective group homomorphism such that the kernel is a perfect group.  Let $\mathcal{C}$ be the category whose objects are perfect homomorphisms, and whose maps are commutative squares. 
The theorem is:

There exists a class $\mathcal{F}$ of morphisms in $\mathcal{C}$ such that the category of fractions $\mathcal{C}[\mathcal{F}^{-1}]$ is equivalent to the homotopy category of pointed connected CW complexes.  

The functor from $\mathcal{C}$ to spaces is defined by the Quillen plus construction to $BG$ with respect to $P=\mathrm{Ker}(G\to H)$, and the class $\mathcal{F}$ is just the maps that become equivalences after applying the plus construction; the class $\mathcal{F}$ can also be characterized as the maps $\phi\colon (G'\to H')\to (G\to H)$ such that $H'\to H$ is an isomorphism and the induced maps $H_*(G',\phi^*M)\to H_*(G,M)$ in group homology are isos for every $H$-module $M$.<|endoftext|>
TITLE: Is it possible for a metric on a smooth manifold to be smooth?
QUESTION [25 upvotes]: Are there any smooth manifolds $M$ with the following property:
There exist a realizing metric $d$ (i.e $d$ induces the topology on $M$), and $d$ is smooth on all of $M \times M$? 
If not, is it  possibe to guarantee smoothness of the function $x \mapsto d(x,y)$ (for a fixed $y$ ), or smoothness of $d^2$ even on a compact manifold?
(I am trying to see if we can achieve "improved smoothness" if we do not force the metric to be Riemannian.) 
Of course, such a metric cannot be induced by a Riemannian metric.
(see here and here).
Update and further questions:
(1) Joonas Ilmavirta showed $d$ cannot be smooth at a neighbourhood of points on the diagonal. Actually, the proof shows $d$ cannot even be twice continuously differentiable. (This is the regularity needed to bound from above the Taylor remainder*).
Now a natural quesion is whether this regularity can be achieved by some metric? (I suspect not, in fact I think the distance should not even be differentiable once at the diagonal, the intuition is based on the example of absolute value on $\mathbb{R}$).
(2) Is it also necessary for a singularity to exist at the diameter of the metric (for compact manifolds)?

*In fact the proof works even if we only assume $x \mapsto d(x,y)$ is continuously twice differentiable, and continuity of the partial derivatives (as functions of two variables).
$ \frac{d}{dt}f(t,t) = \lim_{\Delta \to 0} \frac {f(t+\Delta,t+\Delta)-f(t,t)}{\Delta} = \lim_{\Delta \to 0} \left( \frac {f(t+\Delta,t+\Delta)-f(t+\Delta,t)}{\Delta} + \frac {f(t+\Delta,t)-f(t,t)}{\Delta} \right) = \lim_{\Delta \to 0} \left( \frac {f(t+\Delta,t+\Delta)-f(t+\Delta,t)}{\Delta} + \frac {f(t+\Delta,t)-f(t,t)}{\Delta} \right) = \lim_{\Delta \to 0} \frac{\partial f}{\partial s}(t+\Delta,t+\alpha(\Delta) \cdot \Delta) + \lim_{\Delta \to 0} \frac{\partial f}{\partial t}(t+\beta(\Delta) \cdot \Delta,t) = \frac{\partial f}{\partial s} (t,t) + \frac{\partial f}{\partial t} (t,t)$
($0 \le \alpha(\Delta), \beta(\Delta) \le 1$ , Lagrange mean value theorem)

REPLY [49 votes]: In short:
Non-smoothness at the diagonal is inevitable but that is the only obstruction.
(I added the second part much later.)
Diagonal singularity
It is not possible to have a smooth metric.
Non-smoothness at the diagonal is inevitable.
In fact, any smooth semimetric is zero whenever the two points are in the same connected component.
Suppose you had such a metric on a manifold $M$.
Take a smooth curve $\gamma:(-1,1)\to M$ with $\gamma'\neq0$.
Consider the function $f(t,s)=d(\gamma(t),\gamma(s))$, which is now smooth.
Since
$$
\begin{split}
0
&=
\frac{d}{dt}0
\\&=
\frac{d}{dt}f(t,t)
\\&=
\partial_1f(t,t)+\partial_2f(t,t),
\end{split}
$$
we have $\partial_1f=-\partial_2f$ on the diagonal.
But since $f(t,s)=f(s,t)$, we also have $\partial_1f=\partial_2f$ on the diagonal.
Thus $\partial_1f(t,t)=\partial_2f(t,t)=0$.
This implies that $f(t,s)\leq C(t-s)^2$ for some constant $C$, for all $t,s\in[-1/2,1/2]$.
Pick any number $a\in[0,\frac12]$ and a large integer $N$ and observe that by triangle inequality
$$
\begin{split}
f(0,a)
&\leq
f(0,\frac{a}{N})+f(\frac{a}{N},\frac{2a}{N})+\dots+f(\frac{(N-1)a}{N},a)
\\&\leq
NC\left(\frac aN\right)^2
\\&=
\frac{Ca^2}{N}.
\end{split}
$$
This holds for any $N$, so in fact $f(0,a)=0$.
This implies that as long as $x,y\in M$ are in the same connected component, they must satisfy $d(x,y)=0$.
In fact, this argument works for any $C^{1,\alpha}$ semimetric for $\alpha>0$ (a similar calculation leads to an estimate $f(0,a)\lesssim N^{-\alpha}$), and I guess the claim is true for $C^1$, too.
Off-diagonal regularity
For any differentiable manifold $M$ there is a metric $d\colon M\times M\to\mathbb R$ which is smooth everywhere outside the diagonal and gives the usual topology of $M$.
This is based on two observations:

There is a Riemannian metric $g$ on $M$ with injectivity radius at least one.
(If I am mistaken, this is at least possible on compact manifolds. On non-compact manifolds one should be able to start with any metric and multiply it with a slowly varying conformal factor to force injectivity radius above one everywhere.)
There is a smooth concave function $\phi\colon[0,\infty)\to[0,\infty]$ so that $\phi(0)=0$ and $\phi(x)=1$ for $x\geq1$.

Now if $d_g$ is the distance function of $g$, the metric $d(x,y)=\phi(d_g(x,y))$ is smooth outside the diagonal.
The distance from $x$ to nearby points is smooth (except at $x$) within the injectivity radius.
Outside that radius $d_g$ may be singular, but $\phi$ is constant at that scale, removing all singularities from $d$.
The composition of a metric and a concave function is always a metric.
It follows from the assumptions that $\phi$ is a strictly increasing bi-Lipschitz diffeomorphism in some neighborhood of zero, and thus $d$ and $d_g$ give the same topologies.
Any Riemannian metric gives the original manifold topology.
The metric $d$ is non-smooth at the diagonal, but its square $d^2$ is smooth everywhere.<|endoftext|>
TITLE: Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
QUESTION [50 upvotes]: Set $f(x) = \exp(-x-x^{-1})$. An easy induction shows that
$$\frac{d^n}{(dx)^n} f(x) = \phi_n(x^{-1}) f(x)$$
for $\phi_n$ a polynomial of degree $2n$. Clearly, the roots of $\phi_n(x^{-1})$ are the same as the roots of $f^{(n)}(x)$ and, by repeated use of Rolle's theorem, $f^{(n)}(x)$ has at least $n$ roots in $(0, \infty)$. Since $\phi_n$ has degree $2n$, there is room for many more positive real roots than that.
However, for $n \leq 50$, computer computations show that $\phi_n$ has exactly $n$ positive real roots! Why?
Motivation: Nothing really, I was just thinking about this question and fooling around.
Data: Here are the first $10$ values of $\phi_n(y)$:
1, -1 + y^2, 1 - 2 y^2 - 2 y^3 + y^4, -1 + 3 y^2 + 6 y^3 + 3 y^4 - 6 y^5 + y^6, 1 - 4 y^2 - 12 y^3 - 18 y^4 + 32 y^6 - 12 y^7 + y^8, -1 + 5 y^2 + 20 y^3 + 50 y^4 + 60 y^5 - 50 y^6 - 180 y^7 + 115 y^8 - 20 y^9 + y^10, 1 - 6 y^2 - 30 y^3 - 105 y^4 - 240 y^5 - 200 y^6 + 540 y^7 + 1095 y^8 - 1080 y^9 + 294 y^10 - 30 y^11 + y^12, -1 + 7 y^2 + 42 y^3 + 189 y^4 + 630 y^5 + 1295 y^6 + 420 y^7 - 5075 y^8 - 7140 y^9 + 10521 y^10 - 3990 y^11 + 623 y^12 - 42 y^13 + y^14, 1 - 8 y^2 - 56 y^3 - 308 y^4 - 1344 y^5 - 4256 y^6 - 7560 y^7 + 3430 y^8 + 48160 y^9 + 48664 y^10 - 108360 y^11 + 53788 y^12 - 11424 y^13 + 1168 y^14 - 56 y^15 + y^16, -1 + 9 y^2 + 72 y^3 + 468 y^4 + 2520 y^5 + 10668 y^6 + 31752 y^7 + 45234 y^8 - 83160 y^9 - 478674 y^10 - 330120 y^11 + 1186836 y^12 - 742392 y^13 + 201132 y^14 - 27720 y^15 + 2007 y^16 - 72 y^17 + y^18, 1 - 10 y^2 - 90 y^3 - 675 y^4 - 4320 y^5 - 22800 y^6 - 93240 y^7 - 256830 y^8 - 246960 y^9 + 1272348 y^10 + 5033700 y^11 + 1965810 y^12 - 13829760 y^13 + 10636800 y^14 - 3530520 y^15 + 614925 y^16 - 59760 y^17 + 3230 y^18 - 90 y^19 + y^20

The other roots are shrinking towards $0$ like $1/n$ while accumulating on some sort of curve. Here is a picture for $n=50$:

REPLY [5 votes]: Another solution, inspired by fedja's. This one also establishes Peter Mueller's stronger statement that the coefficients of $\phi_n(x)$ have $n$ sign changes (in the sense of Descartes' Rule of Signs).
For any Laurent polynomial $f(x) \in \mathbb{R}[x, x^{-1}]$, let $D(f)$ denote the number of sign changes in the coefficients of $f$.
Claim 1 For any integers $0 \leq r \leq m$,
$$D \left( \left( \frac{d}{dx} \right)^r (1-x/m)^m (1-x^{-1}/m)^m \right) \leq 2m-r.$$
Proof The only monomials with nonzero coefficient are $x^{-m-r}$, $x^{-m-r+1}$, ..., $x^{-r-1}$, $1$, $x$, ..., $x^{m-r}$, a total of $2m-r+1$ monomials. $\square$
Claim 2 For any Laurent polynomial $h(x)$ and any $a>0$, we have $D((x-a) h(x)) \geq D(h(x))+1$.
Proof This is a standard Lemma established in the course of proving Descartes' Rule of Signs. See, for example, this proof. $\square$
Claim 3
$$D \left(\frac{ \left( \frac{d}{dx} \right)^r (1-x/m)^m (1-x^{-1}/m)^m }{(1-x/m)^{m-r} (1-x^{-1}/m)^{m-r}} \right) \leq r.$$
This follows by $2m-2r$ uses of Claim 2.
Now, sending $m \to \infty$, we obtain
$$D\left( \frac{\left( \frac{d}{dx} \right)^r \exp(-x-x^{-1})}{\exp(-x-x^{-1})}  \right)\leq r$$
as desired.<|endoftext|>
TITLE: A question about $(0,1]$-valued multiplicative functions
QUESTION [6 upvotes]: Suppose $f:\mathbb{N}\to [0,1]$ is a multiplicative function (i.e. $f(nm)=f(n)f(m)$ whenever $m$ and $n$ are coprime). Suppose $f$ has non-zero mean, which means
$$
\lim_{N\to\infty}\frac{1}{N} \sum_{n=1}^N f(n) >0.
$$
Also, lets assume $f(n)\neq 0$ for all $n$. Then, can we say something about the set of all $n$ such that $f(n)\leq \varepsilon$? What is the density (or upper density) of this set? In particular, does this density go to zero as $\varepsilon\to 0$?
Thank you very much for your help.

REPLY [8 votes]: Yes, the density goes to zero as $\epsilon \to 0$, but the convergence to zero can be arbitrarily slow (depending on the choice of the function $f$).  To see this, first note that the condition that the mean value of $f$ is strictly positive is equivalent to 
$$ 
\sum_p \frac{1-f(p)}{p}<\infty,   
$$ 
which is also the same as 
$$ 
\sum_{p^k} \frac{1-f(p^k)}{p^k} <\infty, 
$$ 
where the sum is over all prime powers $p^k$.  This is a simple case of results of Wirsing, Halasz, Delange, and follows from the 
estimate (for non-negative multiplicative functions bounded by $1$) 
$$ 
\sum_{n\le x} f(n) \ll \frac{x}{\log x} \sum_{n\le x} \frac{f(n)}{n} \ll \frac{x}{\log x} \exp\Big(\sum_{p\le x} \frac{f(p)}{p}\Big). 
$$ 
(See for example Theorem 2 of Halberstam and Richert). 
Next, for any $1> \delta >0$, put 
$$ 
{\mathcal F}(\delta)= \sum_{\substack{p^k \\ f(p^k) \le \delta} } \frac{1}{p^k},  
$$ 
the convergence of the sum for any $\delta <1$ being guaranteed by our previous observation.
We claim that ${\mathcal F}(\delta) \to 0$ as $\delta \to 0$; this step crucially uses the assumption that $f(n)$ is 
strictly positive for all $n$.  For each $\delta >0$ let $p(\delta)$ denote the smallest prime power $p^k$ with $f(p^k)<\delta$.  Then the assumption that $f(n)>0$ gives that $p(\delta) \to \infty$ as $\delta \to 0$.  Therefore, for $\delta \le 1/2$ say,
$$ 
{\mathcal F}(\delta) \le \sum_{\substack {p^k \\ p^k \ge p(\delta) \\ f(p)\le 1/2}} \frac{1}{p^{k}}, 
$$ 
 and the RHS is the tail of a convergent series, and therefore goes to zero as $p(\delta)\to \infty$, or in other words as $\delta \to 0$.  This proves our claim.
Now suppose $\epsilon >0$ is given (and suitably small), and we want to bound the density of $n$ with $f(n) \le \epsilon$.  Put $\delta=\exp(-(\log 1/\epsilon)^{\frac 12})$,  which is bigger than $\epsilon$ but still goes to zero with $\epsilon$.  Suppose first that $p^k \Vert n$ for some prime power $p^k$ with $f(p^k) <\delta$.   The density of such $n$ is clearly at most ${\mathcal F}(\delta)$.   Now suppose that if $p^k \Vert n$ then $f(p^k) \ge \delta$.  Note here that (using $x \ge \exp(2\log (1/\delta) (x-1))$ for $\delta <x\le 1$)
 $$ 
 \epsilon > f(n) = \prod_{p^k \Vert n} f(p^k) \ge \exp\Big( 2\log \frac{1}{\delta} \sum_{p^k \Vert n} (f(p^k)-1) \Big),  
 $$
 and so, for such $n$, 
 $$ 
 \sum_{p^k \Vert n} (1-f(p^k)) \ge \frac{1}{2} \Big(\log \frac {1}{\epsilon}\Big)^{\frac 12}. \tag{1}
 $$ 
Since 
$$ 
\sum_{n\le N} \sum_{p^k \Vert n} (1-f(p^k)) \ll N \sum_{p^k}\frac{1-f(p^k)}{p^k} \ll N,
$$ 
we conclude that  the density of $n$ satisfying (1) is $\ll (\log \frac {1}{\epsilon})^{-\frac 12}$.  Thus our desired 
density is 
$$ 
\ll {\mathcal F}(\delta) + \Big(\log \frac{1}{\epsilon}\Big)^{-\frac 12}, 
$$ 
which goes to $0$ as $\epsilon \to 0$.<|endoftext|>
TITLE: Smoothening a measure, II
QUESTION [5 upvotes]: There is an almost invisible, but significant difference between the question below and that recently answered by Boris Bukh.
Given a probability measure $\mu$ supported on a finite set $S\subset{\mathbb R}^2$, define 
  $$ f(z):=\max\left\{\mu(x)\colon \frac{x+y}2=z,\ x,y\in S \right\},
       \ z\in{\mathbb R}^2. $$
Now let $\bar\mu$ be the uniform probability measure supported on the same set $S$, and define
  $$ \bar f(z):=\max\left\{\bar\mu(x)\colon \frac{x+y}2=z,\ x,y\in S \right\},\ z\in{\mathbb R}^2. $$

Is it true that for any choice of the measure $\mu$, the total mass of $f$ is at least as large as the total mass of $\bar f$?

(Notice that $\bar f$ is actually a scaled indicator function of the set $\{(x+y)/2\colon x,y\in S\}$, and that the total mass of $\bar f$ is actually the doubling coefficient $|S+S|/|S|$.)

REPLY [4 votes]: This is also false.
Let $S=A\cup B$ where $A$ is $\{0,1,\dotsc,n-1\}$ and $B$ is a subset of $\{n,n+1,\dotsc,2n\}$ of size $m\leq 4\sqrt{n}$ such that $B+B$ contains all the integers from $2n$ to $4n$. (We can take $B$ to be a union of all multiples of $\sqrt{n}$ and two suitable arithmetic progressions of step $1$.) Note that the mass of $\bar{f}$ is $\approx 4n\cdot (1/n)=4$. Consider the probability distribution that is uniform on $A$, and assigns zero (or nearly zero) weight to $B$. For such a distribution, we compute the total weight of $f$ to be $\approx 3n\cdot(1/n)+n\cdot 0=3$.
I am not sure what exactly you are trying to do, but it is extremely unlikely that an inequality of the approximate form that you seek exists. The reason is that it must behave well under disjoint unions, tensor products, and be stable under local perturbations. That is a lot to ask!<|endoftext|>
TITLE: Is there a reference for "computing $\pi$" using external rays of the Mandelbrot set?
QUESTION [20 upvotes]: I was recently reminded of the following cute fact which I will state as a proposition to fix notation:

Proposition
Given $\epsilon > 0$, let $c = -3/4 + \epsilon i \in \mathbb{C}$ and $q_c(z) = z^2 + c$. Define the sequence of polynomials $q_c^n$ inductively by $q_c^0(z) = z$ and
$$
  q_c^{n+1}(z) = q_c(q_c^n(z)).
$$
Since $q_c^n(0) \to \infty$ as $n \to \infty$,
$$
  N(\epsilon) = \min\{n ~:~ |q_c^n(0)| > 2 \}
$$
is well-defined. The cute fact is that:
$$
  \epsilon N(\epsilon) \to \pi,
$$
as $\epsilon \to 0$.

The next-simplest example of this phenomenon uses $c = 1/4 + \epsilon$. If we define $N(\epsilon)$ similarly then this time it turns out that:
$$
  \sqrt{\epsilon} N(\epsilon) \to \pi,
$$
as $\epsilon \to 0$.
The points $-3/4$ and $1/4$ are the "neck" and "butt", respectively, of the Mandelbrot set and there are various other examples of these "$\pi$ paths": they are suitably-parameterised curves simultaneously tangent to two bulbs of the Mandelbrot set where they meet it. In fact I'd guess any part of the boundary where a bulb bubbles off works and so there are infinitely many such paths. External rays seem the obvious paths and they have a natural parameterisation. (Note that although $\epsilon \mapsto -3/4 + \epsilon i$ is not an external ray, it is asymptotic to one, as least set-wise, which is really all that matters.)
If true, I'm sure this is all in the literature (perhaps implicitly) but I don't know the field and after of searching through some lovely papers I couldn't find what I wanted. A naive guess is that something like the following might be true:

Half-serious conjecture
Let $M \subset \mathbb{C}$ be the Mandelbrot set, $\overline D \subset \mathbb{C}$ the closed unit disc and $\Phi : \mathbb{C} - M \to \mathbb{C} - \overline D$ the unique conformal isomorphism such that $\Phi(c) \sim c$ as $c \to \infty$. Let $e^{i\theta} \in S^1$ (for $\theta$ rational) and let:
$$
  \alpha : (1, \infty) \to \mathbb{C} - M\\
  r \mapsto \Phi^{-1}(re^{i\theta})
$$
Let $c = \alpha(r)$ and define $N(r)$ as above, then:
$$
  N(r) \sim \pi / f_\theta(r)
$$
for some function $f_\theta$, such that $f_\theta(r) \to 0$ as $r \to 1$ and in particular $f_0(r) \sim \sqrt{\alpha(r)-1/4}$ and $f_{1/3} \sim -i(\alpha(r)+3/4)$.

Here then, at last, is my question:

Question
Is some suitably-modified form of the above conjecture correct and is there a proof in the literature toward which somebody could direct me?

Aside from general curiosity my motivation stems from the fact that it's not too hard to prove the stated results for the $c = -3/4 + \epsilon i$ and $c = 1/4 + \epsilon$ paths separately but I'd like to have a unified proof. E.g., to prove the result for the (easier) $c = 1/4 + \epsilon$ path we consider:
$$
  z_{n+1} = z_n^2 + 1/4 + \epsilon\\
  \Rightarrow z_{n+1} - z_n = (z_n - 1/2)^2 + \epsilon
$$
Then approximate:
$$
  \frac{dz}{dn} \simeq (z-1/2)^2 + \epsilon\\
  \Rightarrow z(n) = 1/2 + \sqrt{\epsilon}\tan(\sqrt{\epsilon}n)\\
  \Rightarrow \sqrt{\epsilon}N(\epsilon) \simeq \tan^{-1}(\frac{3}{2\sqrt{\epsilon}} ) - \tan^{-1}(-\frac{1}{2\sqrt{\epsilon}})
$$
And so provided we can justify the approximation is accurate as $\epsilon \to 0$ (which is tricky but possible) the result follows.

REPLY [6 votes]: These results (which are indeed cute - I hadn't seen them before) are well-explained by the theory of parabolic explosion, which is by now classical. Indeed, for the Mandelbrot set, I think that the relevant statements were known to the experts already in the 1980s; they may already be contained implicitly in the Orsay notes by Douady and Hubbard. For an introduction, see Shishikura, Bifurcation of parabolic fixed points, in The Mandelbrot set, Theme and Variations.
Let me focus on the case of "primitive" parabolics, i.e. the "cusps" of Mandelbrot copies, with $c=1/4$ being the simplest case. 
Then (see Theorem 3.2.2 of Shishikura's article) - under perturbations as you describe - the number of iterates to escape through the "eggbeater" dynamics into a prescribed part of the basin of infinity is essentially
$$N(c) \approx \frac{1}{|\alpha(c)|},$$
where $\lambda(c) = e^{2\pi i \alpha(c)}$ is the multiplier of one of the orbits bifurcating from the parabolic fixed point. (This multiplier is a holomorphic function defined on a double cover around the parabolic; hence the square root in the expression.)
So, for perturbations $c=1/4+\epsilon$, the repelling fixed point has multiplier 
$$\lambda(c) = 1 + 2\sqrt{-\epsilon}$$
by an elementary and well-known calculation. So
$$N(c) \approx \frac{2\pi}{|\log \lambda(c)|}\approx \frac{2\pi}{2\sqrt{\epsilon}}=\frac{\pi}{\sqrt{\epsilon}}.$$
Hence indeed
$$ N(c)\cdot \sqrt{\epsilon}\to\pi.$$
For other parameters, you will get similar results, which will however depend on the derivative of the multiplier map with respect to the parameter in a suitable sense. Note that the above formulation (in terms of repelling orbits) is, in any case, the natural one, since it is independent of parameterisation.
For the satellite case, there are similar results that you can find in the literature. Again, for the bifurcation at -3/4, you get an explicit formula because you can compute the multipliers of the orbits in question directly.<|endoftext|>
TITLE: Transitive actions of $Aut(F_2)$ on surjections from $F_2\twoheadrightarrow G$
QUESTION [7 upvotes]: Here $F_2$ is the free group on two generators $x,y$. I'm interested in examples of finite groups $G$ such that $Aut(F_2)$ acts transitively on the set of surjections $F_2\rightarrow G$. (In particular I'm interested in $G$'s where automorphisms of $F_2$ of determinant 1 act transitive on surjections $F_2\rightarrow G$)
In this case the conjugacy class of the image of the commutator $xyx^{-1}y^{-1}$ is an invariant of any $Aut(F_2)$-orbit, so a necessary condition for transitivity of the action is that any two commutators of generating pairs of $G$ are conjugate in $G$
If $G$ is simple then one expects that "most" pairs of elements of $G$ will actually generate $G$. Thus, for alternating groups $A_n$ ($n > 5$), where every element is a commutator, one expects the action of $Aut(F_2)$ can't possibly be transitive.
For groups of the form $G = PSL_2(\mathbb{F}_q)$ it's proven in 
http://www2.math.ou.edu/~dmccullough/research/pdffiles/traces.pdf
that for $q > 11$ almost every element of $\mathbb{F}_q$ appears as the trace of a commutator of a generating pair, and so again $Aut(F_2)$ can't act transitively in this case.
Is this generally true for all finite nonabelian simple groups? More precisely..

Does there exist a nonabelian simple group $G$ on which $Aut(F_2)$ acts transitively on surjections $F_2\rightarrow G$?
What heuristics are there for describing the groups $G$ admitting such a transitive action (by $Aut(F_2)$? by automorphisms of determinant 1?)

REPLY [7 votes]: The answer to the first part of OP's question is "no", but possibly with a finite number of exceptions. 
Indeed [Theorem 1.8, 4] established that the number of $T_2$-systems of a non-abelian finite simple group $G$ tends to infinity as the cardinality of $G$ tends to infinity. A $T_2$-system in the sense of B. H. Neumann and H. Neumann (T-system stands for transitivity system) is an orbit of $\operatorname{Aut}(F_2) \times \operatorname{Aut}(G)$ acting on the set $\operatorname{Epi}(F_2, G)$ of the epimorphisms $\pi: F_2 \twoheadrightarrow G$ in the following way: $\pi \cdot (\psi, \phi) = \phi^{-1} \circ \pi \circ \psi$. 
An orbit of $\operatorname{Aut}(F_2)$ is called a Nielsen equivalence class and clearly the number of such classes is bounded from below by the number of $T_2$-systems.
This result was already proved by M. Evans, R. M. Guralnick and I. Pak for the sequence $(PSL_2(\mathbb{F}_q))_q$ with $q$ the power of a prime, and was refined in [5], as noted in OP's question. I. Pak also proved in [Proposition 2.5.2, 1] the result for the sequence $(A_n)_n$ of the finite alternating groups and he conjectured with Guralnick the aforementioned general result in [2]. Estimates for the number of $T_2$-systems of these groups, and for the Suzuki groups as well, are collected in [Theorem 2.1, 3].
I don't know of any non-abelian finite simple group with exactly one $T_2$-system. In contrast, Wiegold's conjecture asserts that the number of $T_3$-systems should always be $1$ for such groups.
Regarding the second part of OP's question, I don't have any heuristics at hand. Stating the obvious, I would definitely try to know more about the classification of finite simple groups.
If the second part of the question concerns also two-generated finite groups which are not necessarily simple, then the following observation may help: the class of the finite groups $G$ for which $\operatorname{Aut}(F_2)$ acts transitively (the same holds if we require the determinant to be $1$) on $\operatorname{Epi}(F_2, G)$ is stable under quotient. This follows from Gaschutz'lemma [Lemma 2.1.5, 1] and imposes for instance that the order of the first invariant factor of $G_{ab} = G/[G, G]$ lies in $\{2, 3, 4, 6\}$ when $G_{ab}$ is not cyclic.

[1] "What do we know about the product replacement algorithm?", I. Pak, 2000.
[2] "On a question of B.H. Neumann", R. M. Guralnick and I. Pak, 2002.
[3] "Nielsen equivalence and stability graph of finitely generated groups", M. Evans, 2008.
[4] "Commutator maps, measure preservation, and T-system", S. Garion and A. Shalev, 2009.
[5] "Nielsen equivalence of generating pairs of $SL_2(q)$", D. Mc Cullough and M. Wanderley, 2013.<|endoftext|>
TITLE: Why, in terms of quantum groups, does the knot determinant appear as an evaluation of both the Jones and Alexander polynomials?
QUESTION [21 upvotes]: The Jones polynomial can be computed from the representation theory of $\mathcal{U}_q(\mathfrak{sl}(2))$. The Alexander polynomial has an analogous description in terms of the representation theory of the superalgebra $\mathcal{U}_q(\mathfrak{gl}(1|1))$ (see, for example, Sartori http://arxiv.org/abs/1308.2047). 
From this perspective, is there a nice explanation of the fact that $V_K|_{q = i} = \Delta_K|_{t = -1} = \det K$? Here $K$ is a knot, $V_K(q)$ denotes the Jones polynomial of $K$, $\Delta_K(t)$ denotes the Alexander polynomial of $K$ ($t$ and $q$ are related by $t = q^2$), and $\det K$ is the knot determinant.

REPLY [2 votes]: You haven't specified what sorts of explanations you would find "nice," so the following collection of thoughts might not be helpful, but here goes (or it might be helpful, but contain only things you already know). [By the way, this answer only attempts to "explain" why $V_K(i) = \Delta(-1)$ -- it does not address the additional question of why they both yield the knot determinant. That's a great question; I guess one "explanation" is simply that the invariants are determined by their skein relation.]
Firstly, a reference I like, in addition to the one you mention by Sartori, is this paper by Kauffman and Saleur. In that paper, many (though not all) of the theorems and computations are for the general Lie superalgebra $\mathfrak{gl}(m|n)$, where $m,n \in \mathbb{Z} >= 0$ are not necessarily equal. The Jones and Alexander cases are each specializations of this general case, with $(m,n) = (2,0)$ or $(1,1)$, respectively, so therefore a general strategy for a certain kind of "explanation" would be: observe that for certain specializations of the quantum parameter $q$, the generally-defined $\mathfrak{gl}(m|n)$ quantum link invariants, and even much of the structure of the quantum group $U_q(\mathfrak{gl}(m|n))$, depends only on $m+n$.
Here are two specific instantiations of the strategy. For the first, see page 299 (= page 7) of the paper I linked to. There the authors point out that for $q = i$, the quantum $R$ matrices for $U_q(\mathfrak{gl}(m|n))$ and $U_q(\mathfrak{gl}(m+n))$ agree. This is the essential ingredient for showing the relevant polynomial invariants agree at $q=i$, but to be complete one would want to go through the details of how those invariants are defined, in terms of the $R$ matrix.
The second is along those lines: the first link invariant defined in the Kauffman and Saleur paper has this skein relation (see p. 302):
$$ q^{n-m}P^+ - q^{-(n-m)}P^- = (q-q^{-1})P_{||} $$
(hopefully my notation is clear enough). However, one can see, by passing from a given knot to a split (1,1) tangle, say, that this invariant $P$ is forced to be zero for all knots if the quantum dimension $D_q$ of $U_q(\mathfrak{gl}(m|n))$ is zero. However, one can define another invariant $P':= P/D_q,$ which will be non-zero. The quantum dimension is $$ D_q = \frac{q^{n-m} - q^{-(n-m)}}{q-q^{-1}},$$ which is zero for all $q$ if $n=m$. Anyway, I haven't done it carefully myself, but I think it should be fairly straightforward to go from the above skein relation to the skein relation satisfied by $P'$ at $q = i$, and see that this only depends on $m+n$. (A final step would be to relate these normalizations of the invariants to those in the literature).
Let me know if I've said anything confusing or wrong, and if this helps in any way; or if you were imagining something different in terms of an explanation.<|endoftext|>
TITLE: Springer GTM Reprints in China?
QUESTION [14 upvotes]: I apologise if this is not the sort of question appropriate for MO; it does however seem that mathematicians are the most likely to know the answer:  
Many of the Chinese mathematicians and graduate students at my university have softcover copies of some of the older Springer GTMs, usually printed on thin paper with Chinese characters on the front cover, but with the actual text in English. Someone said that they can be bought quite cheaply in China and are legitimate `Chinese Reprints' of the original books. Internet searches on both Google and the Springer website turn up nothing, however, so I presume the origin of these books is somewhat dubious:
Does someone know the actual status of these books and has anyone purchased one outside of China?

REPLY [15 votes]: They are legitimate editions, published by an agreement with China Science Press. They are for sale in China only. These reprints are done not only for GTMs but also many other maths books. See the enclosed photoes for an example. In this case the price is 98 yuan in China.<|endoftext|>
TITLE: What algebraic structures are related to the McGee graph?
QUESTION [20 upvotes]: Recall that an $(n,g)$-graph is a simple graph where each node has $n$ neighbors and the shortest cycle has length $g$, while an $(n,g)$-cage is $(n,g)$-graph with the minimum number of nodes.  
The McGee graph is the unique $(3,7)$-cage.  It looks like this:

Now, the unique $(3,6)$-cage is the Heawood graph, which has strong connections to the Fano plane and its symmetry group, $PGL(3,\mathbb{F}_2)$.  The unique $(3,8)$-cage is the Tutte–Coxeter graph, which has strong connections to the Cremona–Richmond configuration and its symmetry group, $S_6$.  These stories are fascinating and strongly parallel.
What about the McGee graph?  All I know is what I read:

McGee graph, Wikipedia.  

It has 24 nodes and 36 edges.  Its symmetry group has 32 elements — which 32-element group is it?  Its symmetry group doesn't act transitively on the nodes: there are two orbits, containing 16 and 8 nodes.
Is this graph associated to an interesting finite geometry, or not?

REPLY [3 votes]: certainly Sage(math) can easily compute the structure of this group:
sage: g=graphs.McGeeGraph()
sage: a=g.automorphism_group()    
sage: a.structure_description()
'(C2 x D4) : C2'

(Sage makes an unusual choice to denote the dihedral group $D_{2k}$ of order $2k$ as $D_k$...)<|endoftext|>
TITLE: Definition of p-adic modular forms
QUESTION [9 upvotes]: I have been reading Hida's book "p-Adic automorphism forms on Shimura varieties" and I don't understand a point. 
He first describes p-adic modular forms of tame level N as functions on the Igusa tower which can be interpreted as functions on triples $(E,\phi_{p^\alpha},\phi_N)$ where $\phi_{p^\alpha}: \mu_{p^\alpha}\hookrightarrow E$ is the level structure Cartier dual to the identification $E[p^\alpha]^{et}\cong \mathbb{Z}/p^\alpha\mathbb{Z}$. Then he claims that since $\mu_{p^\alpha}$ has a canonical differential $\frac{dt}{t}$ we can consider p-adic modular forms as functions on triples $(E,\phi_{p^\alpha},\phi_N, \phi_{p^\alpha *}\frac{dt}{t})$.
I guess the idea is that such a differential is supposed to trivialize the pushforward of the sheaf of differential of the elliptic curve over the base, when we think about modular forms as section of that sheaf for the universal elliptic curve. However, I cannot make sense of the pushforward of the differential as a differential of the elliptic curve.
Does anyone know how to understand all this?
Thank you in advance.

REPLY [3 votes]: It's not a pushforward of a differential (or maybe it can be described as one, but it doesn't have to be).
Rather there is a pullback map from differentials on the elliptic curve to differentials on $\mu_{p^\alpha}$. However, this map is an isomorphism mod $p^\alpha$, so you can apply the inverse to reverse the process.
The fact that it is an isomorphism follows from:
The differentials on an elliptic curve over a ring $R$ are a rank $1$ free module over $R$.
The differentials on $\mu_{p^\alpha}$ over $R$ are $R/ p^\alpha$, because it's giving by the equation $x^{p^\alpha}=1$, hence $p^\alpha x^{p^\alpha-1} dx =0$, and $x$ is invertible so $p^\alpha dx=0$.
The map is a unit mod $p$ - after specializing to a field of characteristic $p$, it's sufficient to check that it's nonzero. But if it's zero, then the map $\mu_p \to E$ is trivial on differentials, so by exponentiating (and stopping before $p$th powers) it is zero, which contradicts the fact that it comes from Cartier duality.<|endoftext|>
TITLE: Is there a non-smooth algebraic group scheme in char $p$, all of whose defining relations have degree less than $p$?
QUESTION [9 upvotes]: Let $k$ be an algebraically closed field of characteristic $p>0$.
All the examples of non-smooth algebraic group schemes over $k$ that
I have seen (apart from "artificial" examples; see below) have been given by presentations with at least one defining
relation of degree a positive power of $p$. Here are the examples
I have looked at:

Frobenius kernels. 
Some automorphism schemes of algebras.
The paper "Non-reduced automorphism schemes" by Geiss and Voigt has
an interesting example in Section 2, which is given by a presentation
including some relations of degree 2. The authors state that this group scheme is not reduced if and only if $p=2$. They also mention that if $G$ is a finite $p$-group,
then the group scheme $\mathrm{Aut}(k[G])$ is not reduced. 
Results by Sopkina, "Classification of all connected subgroup
schemes of a reductive group containing a split maximal torus",
imply, if I have understood things correctly, that every non-smooth
subgroup scheme of a reductive group over $k$ (or at least of
$\mathrm{GL}_{n}$) containing a maximal torus, has a presentation including a
$p$-power relation.  
Non-smooth centralisers in classical groups in not very good characteristic.

On the other hand, it is easy to produce "artificial" examples
of a non-smooth group scheme in, for example, char 3 with a presentation
involving only quadratic relations. Namely, take $k$ of char 3 and
$\alpha_{3}=\mathrm{Spec}\, k[x]/(x^{3})$. Since $k[x]/(x^{3})$ is isomorphic
to $k[x,y]/(xy,x-y^{2})$ as $k$-algebras, we can transport the Hopf
algebra structure from the former to the latter. 

Question. Is there a non-smooth algebraic group scheme $G$ over $k$, and an
  embedding $G\rightarrow\mathbb{A}^{n}$ of $G$ as a closed subscheme of affine
  $n$-space, with $n$ minimal, such that every defining relation
  of $G$ in this embedding is of degree strictly less than $p$? (Even
  the case $p=3$ would be interesting.) 

Using explicit equations and a minimal embedding into affine space may seem a bit unnatural, but results of Kollár and Jelonek (see this previous MO question) - which boil down to estimating degrees and Bézout's theorem - imply that if we take
$p>d^{n}$, where $d$ is the maximal degree of a relation, then $p$ does not divide the nilpotency index of any element in the coordinate algebra of $G$, so a proof of Cartier's theorem can be carried through in this case. Hence an example as in the above question must have $d<p<d^n$.

REPLY [7 votes]: No. Let $f$ be a relation of minimal degree $2 \leq d <p$. Apply the comultiplication. This must be zero in $R \otimes R$, where $R$ is the ring of functions. So if $x_1, \dots, x_n$ are the variables, then it is zero in $k[x_1, \dots, x_n] \otimes k[x_1,\dots x_n]$ modulo the various relations.
Write $f$ as a sum of monomials in the $x_i$. When we apply the comultiplication to a monomial, the leading term does not depend on which monomial we pick. It's just the sum over the ways of splitting that monomial into a product of two monomials. So each pair of monomials in $k[x_1, \dots, x_n] \otimes k[x_1,\dots x_n]$ comes from a unique monomial of $f$, so we may ignore cancellation among different monomials. Moreover, because $d \geq 2$, we may split it into two monomials, each of degree $<d$. Because $d$ was the minimal degree of relations, we may ignore the relations. So we end up with something nonzero unless the number of ways of splitting is nonzero. But the number of ways of splitting is a product of binomial coefficients. All numbers involved in the binomial coefficients are less than $p$, so the binomial coefficients are prime to $p$ and hence nonzer.<|endoftext|>
TITLE: Maximum size of a union of incomparable chains
QUESTION [8 upvotes]: The following question was asked on math.stackexchange, where it received no answers. 
https://math.stackexchange.com/questions/1392669/maximum-size-of-a-union-of-incomparable-chains
Let $\mathbb{N}^{<\mathbb{N}}$ denote the set of finite sequences of natural numbers (not including the empty sequence). Order this set by $E\preceq F$ if $E$ is an initial segment of $F$. Call a collection $(s_i)_{i=1}^n$ of subsets of $\mathbb{N}^{<\mathbb{N}}$ incomparable if for $1⩽i,j⩽n$ with $i≠j$, no member of $s_i$ is comparable to any member of $s_j$. Given a finite subset $A$ of $\mathbb{N}^{<\mathbb{N}}$, what is a lower estimate on the largest cardinality of a union of incomparable chains in $A$, as a function of the cardinality of $A$? In particular, is there a number $c>0$ so that any finite subset $A$ admits a subset $B$ which is a union of incomparable chains so that $|B|⩾c|A|$?

REPLY [2 votes]: I don't know if there is any interest in exact small numbers. Probably not, but I worked out a few small numbers, so I will post them as an answer.
The question is about finite trees. For our purposes, a "tree" is just a finite partially ordered set $T$ in which the set of predecessors of any element is a chain. (I see no advantage in representing it concretely as a tree of sequences.) To save typing I will write "size" for "cardinality", and I will call a set $S\subseteq T$ a "quasichain" if the comparability relation is transitive on $S,$ i.e., if $S$ is a "union of incomparable trees".
The question asks, given a number $t,$ what is the greatest number $q$ such that every tree of size $t$ contains a quasichain of size $q$?
It is convenient to look at the inverse problem: given a number $n,$ define $g(n)$ as the maximum possible size of a tree in which every quasichain has size $\le n.$
It is even more convenient to introduce a second variable and define $f(m,n)$ as the maximum possible size of a tree in which every chain has size $\le m$ and every quasichain has size $\le n.$
Clearly, $f(1,n)=n,$ and $m\ge n\implies f(m,n)=f(n,n)=g(n).$ Also, a little consideration will show that, when $2\le m\le n,$ we have
$$f(m,n)=\max\{1+f(m-1,n),\ f(m,1)+f(m,n-1),\ f(m,2)+f(m,n-2),\dots,f\left(m,\left\lfloor\frac n2\right\rfloor\right)+f\left(m,\left\lceil\frac n2\right\rceil\right)\}.$$
I used this recurrence to compute the first dozen values of $f(n,n)=g(n)$ by hand, and got:

$$1,3,5,8,10,13,16,20,22,25,28,32$$

The sequence is not in the OEIS.<|endoftext|>
TITLE: Identities and inequalities in analysis and probability
QUESTION [15 upvotes]: Usually, at the heart of a good limit theorem in probability theory is at least one good inequality – because, in applications, a topological neighborhood is usually defined by inequalities. Of course, an explicit inequality may be even more useful by itself than its application to a limit theorem, which latter is in fact a statement about the mere existence of a certain kind of inequality (recall the presence of the quantifier $\exists$ in standard definitions of the limit). 
In turn, a good inequality is oftentimes based on at least one good identity -- as something gets rewritten in some other, easier to analyze form. Identities used to prove inequalities are oftentimes simple and routine, such as $\int_a^c=\int_a^b+\int_b^c$. Among well-known examples of nontrivial and useful identities in analysis and probability are Cauchy's integral theorem, Plancherel's theorem, isometric imbeddings into $L^p$, duality identities of the $\max_x\min_y f(x,y)=\min_y\max_x f(x,y)$ type (under appropriate conditions on $f$), Spitzer's identity for random walks, and identities of Stein's method. 
There are of course a great many other known identities (some of them nontrivial), such as thousands of formulas for integrals as e.g. in the Gradshteyn and Ryzhik collection. However, it appears that not many of those formulas can lead to interesting inequalities. 
I think it would be useful for a number of people to learn about other, not widely known identities that can be/have been used to obtain interesting inequalities. A desirable answer to this question would state such an identity and indicate its possible/actual applications to inequalities, with appropriate references. Thank you!

REPLY [3 votes]: Another, very recent example of a proof of an inequality based on an apparently not-widely-known identity is my answer at An inequality concerning Lagrange's identity; cf. the answer by Fedor Petrov on this page concerning the Cauchy--Schwarz inequality.<|endoftext|>
TITLE: Some questions about the Malcev completion
QUESTION [10 upvotes]: Let $G$ be an abstract group. The Malcev completion $\widehat{G}$ of $G$  (over $\mathbb{Q}$) is the set of group-like elements in the complete Hopf algebra $\widehat{\mathbb{Q}[G]} = \lim_n \mathbb{Q}[G]/I^n, I = \langle g-1 \,|\, g\in G\rangle.$ The  nilpotent completion of $G$ is also defined by $G^{\mathrm{nil}}=\lim_n G/G^{(n)}, G^{(1)}=G, G^{(n+1)}= (G,G^{(n)}).$ These limit definitions induce two natural filtrations (both denoted by $F$) on Malcev and nilpotent completion. Also in both cases $\forall m,n, \,(F_n,F_m)\subset F_{n+m}$ and so there exists a natural structure of Lie algebras on the direct sum of graded quotients: $E\widehat{G} = \oplus F_n \widehat{G}/F_{n+1}\widehat{G}, EG^{\mathrm{nil}}= \oplus F_n G^{\mathrm{nil}}/F_{n+1}G^{\mathrm{nil}}$. 
I have three questions on these concepts:
1- What is the relation between these two completion processes? I know (By theorems in Chapter 8 of Fresse's book: Homotopy of operads and Grothendieck-Teichmuller groups) in the case of finitely generated free groups, $E\widehat{G} \simeq EG^{\mathrm{nil}}\otimes \mathbb{Q}.$ Does this hold in general? If not, does there exist simple sufficient conditions for this to be true? 
2-It's evident that $\widehat{G}^{(n)}\subset F_n(\widehat{G})$. Does there exist examples of $G$ for which this inclusion is strict for some $n$?
3-Is $\widehat{G}=\widehat{\widehat{G}}$ in general?
Thanks!

REPLY [7 votes]: A nice exposition of all of this can be found in section 12,1 of the book of Chmutov-Duzhin-Mostovoy (http://www.pdmi.ras.ru/~duzhin/papers/cdbook/).
Assume for the sake of safety that $G$ is finitely generated. The series $G^{(n)}$ is characterized by the property that $G/G^{(n)}$ is the largest quotient of $G$ which is nilpotent of class $n$. There is another series $G_n$ characterized by the fact that the successive quotient are the largest torsion free quotient of $G$ which are nilpotent of class $n$. More or less by definition:
$$G_n=\{x \in G\ |\ \exists k, x^k \in G^{(n)}\}$$
Now one can show that $G_n$ coincides with the filtration induced on $G$ by the augmentation ideal of $\mathbb{Q}[G]$, i.e. $G_n=(1+I^n)\cap G$.
The moral is that the difference between $G^{nil}$ and $\hat{G}$ is about torsion, hence disapears after tensoring with $\mathbb{Q}$. So:
1) is always true, it's a theorem of Quillen
I'm pretty confident that 2) is always an equality since you already completed everything
3) is definitely true: $\hat{G}$ is characterized by the property that the map $G\rightarrow \hat{G}$ is universal among maps from $G$ to pro-unipotent groups. Hence the pair $(\hat{G},id:\hat{G}\rightarrow \hat{G})$ realizes $\hat{G}$ as its own Malcev completion.<|endoftext|>
TITLE: Is there a geometric proof for the upper semicontinuity of fiber dimension in algebraic geometry?
QUESTION [8 upvotes]: One of the first theorems encountered in algebraic geometry is the upper semicontinuity of fiber dimension:
Let $ f : X \to Y $ be a surjective regular map between irreducible varieties with irreducible fibers. Then $ {\rm dim} \; f^{-1}(y) \geq {\rm dim} \; X - {\rm dim} \; Y $ and the equation ${\rm dim} \; f^{-1}(f(x)) \leq b $ defines a zariski open subset of $X$. 
This theorem is not true in the category of smooth manifolds. Indeed, consider the height function $ h : S^2 \to \mathbb{R}$. Moreover, the proof uses commutative algebra. 


 Question 1:  Is there a more geometric proof? Maybe one that works for holomorphic manifolds?


Here is my idea for how a proof might go: If $f : M \to N $ is a surjective map of smooth manifolds, then ${\rm Null} \; f'(m) \leq b $ defines an open subset of $M$. Somehow relate this nullspace to the fiber dimension in the complex manifold case. I have no idea if this will work...

REPLY [5 votes]: Let $X$ and $Y$ be complex manifolds and $f:X \to Y$ a holomorphic map. 
If $f$ is surjective then, by Sard's theorem, the generic fiber of $f$ has dimension $\dim X - \dim Y$. So, once we prove upper semicontinuity, we will know that all fibers have dimension at least $\dim X - \dim Y$.
To show semicontinuity, it is enough to show $\{ x : \dim f^{-1}(f(x))=0 \}$ is open. Once we have done this, suppose that $x$ lies on a $b$-dimensional component of $f^{-1}(f(x))$. Then we can choose a neighborhood $U$ of $x$ and a map $g: U \to \mathbb{C}^b$ such that the fiber of $f \times g : U \to Y \times \mathbb{C}^b$ though $x$ is just the singleton $\{ x \}$. There will then be some open neighborhood $V$ of $x$ where the fibers of $f \times g$ are finite, and hence the fibers of $f$ have dimension $\leq b$.
So, suppose that $f^{-1}(f(x))$ is finite. We must build an open neighborhood of $x$ where all fibers are finite. Shrinking $X$ around $x$, we may assume that $f^{-1}(f(x)) = \{ x \}$ and that $X$ is open in $\mathbb{C}^{d}$ for $d = \dim X$. Let $B$ be a closed ball in $X$ around $x$. Let $B^{\circ}$ be the interior  and  $\partial B$ the boundary. Write $y = f(x)$.
Now, $\partial B$ is compact, so $f(\partial B)$ is closed in $Y$. Since $f^{-1}(y) = \{ x \}$, we see that $y \not \in f(\partial B)$, so we can take an open set $V$ around $y$ disjoint from $f(\partial B)$. Take $U = f^{-1}(V) \cap B^{\circ}$. I claim that, for any $x' \in U$, the fiber $f^{-1}(f(x')) \cap U$ is finite.
Let $y' = f(x')$. Note that $y' \in V$ and hence $f^{-1}(y') \cap U = f^{-1}(y') \cap B^{\circ} = f^{-1}(y') \cap B$. The last is a closed subset of the compact set $B$, hence is compact. So far, we have not used complex geometry in any way.
Let $z$ be any holomorphic function on $U$. Then the restriction of $z$ to $f^{-1}(y') \cap U$ is a holomorphic function on a compact complex manifold, so it is locally constant by the maximum modulus principle. (I am glossing over the fact that $f^{-1}(y') \cap U$ could have singularities.) Applying this fact with $z_1$, $z_2$, ..., $z_d$ the coordinate functions on $U \subset \mathbb{C}^d$, we see that $f^{-1}(y') \cap U$ is finite.<|endoftext|>
TITLE: Vector Fields in a Riemannian Manifold
QUESTION [6 upvotes]: Suppose $(M,g)$ is a Riemannian manifold. 
Is there a way to classify manifolds where there exists a vector field that commutes with the laplace beltrami operator?
Thanks

REPLY [8 votes]: I give a geometric explanation of the calculations of Willie, which simultaneously elaborates the suggestion of Deane. 
The flow of a vector field commuting with Laplacian preserves the Laplacian  and therefore its symbol which is the inverse of the metric (i.e., $g^{ij}$). Then, the flow of the vector field preserves the metric and the vector field is Killing 
A related question is 
is $\nabla \cdot ( c^2 \nabla)$ a Laplace-Beltrami operator?
Added later: Just have seen that  Tobias Fritz  in his comment which appeared before my answer already had the same explanation.<|endoftext|>
TITLE: Three involutions on the set of 6-box Young diagrams
QUESTION [11 upvotes]: The set of $n$-box Young diagrams classifies both conjugacy classes in $S_n$ and equivalence classes of irreducible representations of $S_n$.  There is an outer automorphism of $S_6$, of order 2.  This acts on the set of conjugacy classes of $S_6$ in an obvious way.   It also acts on the set of equivalence classes of irreducible representations of $S_6$ in an obvious way.  
So, we get two actions of $\mathbb{Z}_2$ on the set of 6-box Young diagrams---or in other words, two involutions on the set of 6-box Young diagrams.
Question 1: are these the same?
Question 2: is there some easy way to look at a 6-box Young diagram and see what Young diagram you get when you apply either of these involutions?
There is also a pictorially obvious involution on the set of $n$-box Young diagrams, which reflects them across the diagonal.  There are 11 6-box Young diagrams, shown below, and the pictorially obvious involution sends the $i$th one to the $(11-i)$th one:
1.
☐☐☐☐☐☐
2.
☐☐☐☐☐ 
☐
3.
☐☐☐☐ 
☐☐
4.
☐☐☐☐ 
☐ 
☐
5.
☐☐☐ 
☐☐☐
6.
☐☐☐ 
☐☐ 
☐
7.
☐☐ 
☐☐ 
☐☐
8.
☐☐☐ 
☐ 
☐ 
☐
9.
☐☐ 
☐☐ 
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☐
10.
☐☐ 
☐ 
☐ 
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☐
11.
☐ 
☐ 
☐ 
☐ 
☐ 
☐ 
It is easy to see that this pictorially obvious involution is neither of the other two involutions.  Each column in the Young diagram corresponds to a cycle in the cycle decomposition of a permutation, so diagram 1 corresponds to the conjugacy class of the permutation 
$$ (1 2 3 4 5 6) $$
which has $6!/6 = 120$ elements---this is the largest conjugacy class in $S_6$.  Diagram 11 corresponds to the conjugacy class of the permutation
$$ (1) (2) (3) (4) (5) (6) $$
that is, the conjugacy class of the identity, with just one element---this is the smallest conjugacy class.  So, the outer automorphism of $S_6$ can't switch these.
Similarly, the irreducible representation corresponding to diagram 1 corresponds to the trivial representation of $S_6$, while diagram 11 corresponds to the sign representation.  Both these are 1-dimensional, but the outer automorphism of $S_6$ doesn't switch these: the trivial representation has to get mapped to the trivial representation.
Some extra hints:

The conjugacy classes of $S_6$ have cardinalities

$$ 1, 15, 15, 40, 45, 90, 90, 120, 120, 144 $$
which is consistent with
$$1 + 15 + 15 + 40 + 40 + 45 + 90 + 90 + 120 + 120 + 144 = 720 $$
Just for fun, I conjecture that all the pairs of equal cardinality get switched by the outer automorphism.  Clearly the cardinalities put strong limitations on how the outer automorphism can act, since a conjugacy class has to be mapped to an equal-sized conjugacy class.

The irreducible representations of $S_6$ have dimensions

$$ 1, 1, 5, 5, 5, 5, 9, 9, 10, 10, 16 $$
which is consistent with
$$ 1^2 + 1^2 + 5^2 + 5^2 + 5^2 + 5^2 + 9^2 + 9^2 + 10^2 + 10^2 + 16^2 = 720 $$
Here we know that not all pairs of equal dimension get switched by the outer automorphism, since the trivial representation is mapped to itself (and therefore so is the sign representation).  What happens to the 4 representations of dimension 5?
I apologize for being too lazy to match up the cardinalities and dimensions to the Young diagrams.

For a concrete description of the outer automorphism of $S_6$, see this chart drawn by Greg Egan:


Any permutation of $S =\{1,2,3,4,5,6\}$ induces a permutation of the 15 'synthemes' shown in blue ovals in this chart, and thus a permutation of the 6 'synthematic totals' shown as colored bands at the edge of the chart.  (The duads, shown in blue ovals, are the 15 2-element subsets of $S$.  Each syntheme is a way of partitioning $S$ into three duads, and each synthematic total is a choice of 5 synthemes that contain all 15 duads.)

REPLY [13 votes]: Let $t$ be an outer automorphism of $S_6$. It's known that $t$ sends a $6$-cycle to an element of cycle type $(3,2)$. This forces three pairs of conjugacy classes to be swapped:
$$ (6) \leftrightarrow (3,2), \ (3,3) \leftrightarrow (3),\ (2,2,2) \leftrightarrow (2). $$
The only other classes sharing a common size are $(4)$ and $(4,2)$ but these turn out to be fixed. (I've included a proof below.) From the character table and the equation $\chi^t(g^t) = \chi(g)$ one then gets the action on characters:
$$ (5,1) \leftrightarrow (2,2,2), \ (2,1,1,1,1) \leftrightarrow (3,3), \ (4,1,1) \leftrightarrow (3,1,1,1). $$
Incidentally, this gives a nice way to construct the 'exceptional' reflection representations labelled by $(3,3)$ and $(2,2,2)$: just apply $t$ to the natural permutation module, and take the non-trivial constituent. 
As expected from Brauer's Permutation Lemma, the actions of $t$ on conjugacy classes and characters have the same cycle type.
Construction of $t$. Let $H \cong S_5$ be the transitive subgroup of $S_6$ given by the action of $S_5$ on its $6$ Sylow $5$-subgroups. By identifying $S_6/H$ with $\{1,\ldots,6\}$ we can define $t$ to be the map $S_6 \rightarrow \mathrm{Sym}(S_6/H)$. Since $(1,2,3)(4,5) \in S_5$ acts as a $6$-cycle on the Sylow $5$-subgroups of $S_5$, $H$ contains a $6$-cycle. Hence a $6$-cycle acting on $S_6/H$ has a fixed point, so must have cycle type $(3,2)$. Similarly $H$ contains a $4$-cycle, so $t$ sends $4$-cycles to $4$-cycles.<|endoftext|>
TITLE: How does one identify flow lines on a vector bundle with those on the base in Morse theory?
QUESTION [7 upvotes]: In Chapter 4.2 of Schwarz's book on Morse homology there is a brief discussion of Morse theory on the total space of a smooth vector bundle $E \to M$. In particular, one can take the Morse function $f_E : E \to \mathbb{R}$ given by
\begin{equation*}
f_E(v_m)=f(m) + q(v_m) \ ,
\end{equation*}
where $f: M \to \mathbb{R}$ is a Morse function on the (closed) base and $q$ is the quadratic form associated to a Riemannian metric on $E$. Then one has $f_E|_M = f$ and there is a bijection
\begin{equation*}
\mbox{Crit}_*(f_E) \cong \mbox{Crit}_*(f) \ .
\end{equation*}
Given that the singular homology of $E$ is just that of $M$, I would assume there is some way of relating the moduli space of gradient flow lines of $f_E$ on $E$ with those of $f$ on $M$. However, this is not explained in Schwarz's book (there is a different objective) and I have not seen this done anywhere else. In particular, is it possible to get an identification at the chain level or only at the level of Morse homology?

REPLY [2 votes]: The function $q$ strictly decreases along the solutions of the gradient flow outside of the zero section. Hence any orbit that starts outside the zero section will not converge to a critical point in backwards time, and does not show up in some moduli space of orbits connecting critical points. The orbits on the zero section correspond bijectively to orbits on $M$ if you choose a metric which respects a horizontal-vertical splitting of $TE$ in the obvious way.
If one chooses orientations of all the unstable manifolds in $M$, this naturally also gives you orientations for all the unstable manifolds of the critical points in $E$, hence even the orientations of the moduli spaces on $M$ and $E$ agree. 
Much more interesting is to not take the function you give, but the function
$$
f^2_E(v_m)=f(m)-q(v_m)
$$
The critical points on $E$ correspond to the ones on $M$ shifted by the dimension of $E$, as there are this many more unstable directions. The moduli spaces on $M$ and $E$ again correspond bijectively to each other, but the orientations might not agree. However, if $E$ is an oriented vector bundle, we can orient the unstable manifolds of the critical points on $E$ taking into account this orientation on the fibers. Doing this correctly will show that the moduli spaces are isomorphic as oriented manifolds. This will give an isomorphism
$$
HM_*(M)\cong HM_{*+\dim E}(E,f_E^2)
$$
which is the Morse theoretic Thom isomorphism. This can be found for example in the appendix of this paper http://arxiv.org/abs/0810.1995 by Abbondandolo and Schwarz
To see that the orientation assumptions are necessary, you can compute this explicitly for the Mobius strip.<|endoftext|>
TITLE: Algebras for probability monad
QUESTION [5 upvotes]: What is the Eilenberg-Moore category for the non-finitary probability distribution monad is, that is, the monad $D \colon \mathbf{Set} \to \mathbf{Set}$ defined by
$$
DX = \left\{ p \in [0,1]^X \ \Big|\  \left|p^{-1}(0, 1]\right| \le \aleph_0,\ \sum_{x \in X} p(x) = 1 \right\}?
$$
I can only find information on the finitary version, e.g. here.

REPLY [4 votes]: The algebras of this monad and closely related ones have been introduced by Pumplün and Röhrl. For example the introduction of a paper by Börger and Kemper provides a good summary,

Pumplün and Röhrl [6] introduced the notion of totally convex space. These spaces are the Eilenberg-Moore-algebras for the monad induced by the unit-ball functor $\bigcirc:\mathbf{Ban}_1\to \mathbf{Set}$, where $\mathbf{Ban}_1$ is the category of Banach spaces and linear operations of norm $\leq 1$. They can be described as sets with formal infinitary linear combination maps $X^\mathbb{N}\to X,\: (x_i)_{i\in\mathbb{N}} \mapsto \sum_{i\in\mathbb{N}}\alpha_i x_i$, where $\alpha_i\in K$, $\sum_{i\in\mathbb{N}}|\alpha_i|\leq 1$, $K\in\{\mathbb{R},\mathbb{C}\}$, subject to two axioms.
Pumplün and Röhrl also considered the categories obtained by restricting the operations to positive coefficients (maybe even with $\sum_{i\in\mathbb{N}}\alpha_i=1$) and/or finite arity ([4,5,8]). [..]

The relevant reference here is [4], which points to Pumplün's paper "Banach spaces and superconvex modules". The latter seem to be the algebras of your monad, as are the "superconvex spaces" of the paper linked to above. (Although there is a cryptic remark on whether the empty set counts as superconvex or not.) I haven't been able to track down reference [4], but a web search for "superconvex module" and "superconvex space" reveals some other papers.
Finally, it may be worth emphasizing that your monad probably arises from a very natural adjunction similar to the one formed by the unit ball functor $\bigcirc:\mathbf{Ban}_1\to\mathbf{Set}$ and its left adjoint. I haven't checked the details, but I bet that it is the monad induced by the forgetful functor from the category of norm-complete base norm spaces with base-preserving positive linear maps as morphisms to $\mathbf{Set}$. Maybe you already know?

REPLY [3 votes]: The algebras for this monad can be described in essentially the same way: they are sets in which it makes sense to to take "convex combinations" of countably many elements. More precisely, an algebra is a set $X$ together with an  operation 
$$((a_0, a_1, a_2, \ldots), (x_0, x_1, x_2, \ldots)) \mapsto \sum_{i = 0}^\infty a_i x_i$$
defined for every sequence $(a_0, a_1, a_2, \ldots)$ of non-negative real numbers such that $\sum_{i = 0}^\infty a_i = 1$ and every sequence $(x_0, x_1, x_2, \ldots)$ of elements of $X$, such that
$$\sum_{i = 0}^\infty a_i \left( \sum_{j = 0}^\infty b_{i,j} x_{i,j} \right) = \sum_{k = 0}^\infty \left( a_{p(k)} b_{p(k), q(k)} \right) x_{p(k), q(k)}$$
where $k \mapsto (p(k), q(k))$ is your favourite bijection $\mathbb{N} \to \mathbb{N} \times \mathbb{N}$, and
$$\sum_{i = 0}^\infty a_i x_i = \sum_{j = 0}^\infty b_j y_j$$
whenever $\sum_{i : x_i = z} a_i = \sum_{j : y_j = z} b_j$ for all $z \in X$.
The most concise way of giving the complete definition is to use the monad in your question.<|endoftext|>
TITLE: Applications of Lubotzky's linearity theorem?
QUESTION [12 upvotes]: Lubotzky's theorem is a necessary and sufficient set of conditions for a finitely generated discrete group to be linear, i.e. isomorphic to a subgroup of $GL_n(K)$, where $K$ is a field of characteristic 0. Its proof relies on the (relatively) advanced theory of pro-$p$-groups. It can be found with its proof in the book, "Analytic pro-$p$-groups" of Dixon, Du Sautoy, Mann and Segal, Interlude B, or in the original paper "A Group-Theoretic characterization of linear groups" in Journal of Algebra 113.  
My question is:

Has this criterion being used to prove or disprove the linearity of discrete groups of independent interest?

By independent interest I mean some group interesting for mathematicians outside of the theory of pro-$p$-groups, for instance the Braid groups $B_n$, the group $Out(F_n)$, fundamental groups of interesting manifolds, etc...
not some group

REPLY [8 votes]: According to Alex himself, this theorem is practically useless. It does not mean that it can't be applied, for instance when you have a group with assumptions that it has many quotients in some suitable sense, it can be applied. But for explicit examples of groups (e.g., given by a presentation, or as groups of automorphisms of some reasonable structure) as in the examples you provide, you don't have much info about finite quotients so you it's not practical: in examples known to be linear, the easiest way to check Lubotzky's criterion is to show that they are linear...<|endoftext|>
TITLE: How should a mathematician approach the physics literature concerning percolation?
QUESTION [11 upvotes]: I would like to read some of the physics literature on two-dimensional percolation, however in attempting this I have run into two problems.  (1) Physics papers on percolation are (relatively) hard for me to locate, and when I do locate them I don't know which are the important ones I should read.  (2) Given my lack of substantial background in mathematical physics, the physics papers I find are hard for me to read.
I have become interested in two-dimensional percolation since reading some of the (relatively) recent been spectacular progress in the (mathematical) theory of two-dimensional percolation due to Smirnov, Schramm, Lawler, Werner, and others (which confirms many conjectures from the physics literature,though many open questions still remain).
The best I currently can do is to look at references to the physics literature in mathematics papers on percolation.  I am looking for:

Specific outstanding physics papers about two-dimensional percolation which I should read.
Useful tips for mathematicians trying to read such papers.

Thanks very much!

REPLY [2 votes]: This is just one reference; by no means a comprehensive answer to your broad question.
This paper,

Jacob J.H. Simmons. "Logarithmic operator intervals in the boundary theory of critical percolation." Journal of Physics A: Mathematical and Theoretical 46, no. 49 (2013): 494015. (Journal link.)
  (Earlier arXiv abstract.)

uses conformal field
theory (CFT) "to probe the problems of critical two dimensional
percolation."
The author
calculates "crossing probabilities and expectation values of crossing cluster numbers" in hexagons for continuum percolation at criticality.

          


In the figure caption, "$O(n)$" refers to the $O(n)$ loop model,
referencing 

Domany E, Mukamel D, Nienhuis B and Schwimmer A 1981 Nucl. Phys. B190 279–287.<|endoftext|>
TITLE: C$^*$-algebras isomorphic after tensoring
QUESTION [13 upvotes]: From the negative answer to this question we know that C$^*$-algebras that are isomorphic after tensoring with $M_n$ for all $n\geq 2$ need not be isomorphic. So what happens when we strengthen this? 
In the following, if $\mathcal A\subset B(\mathcal H)$ and $\mathcal B\subset B(\mathcal K)$ then the minimum tensor product $\mathcal A\otimes_\textrm{min} \mathcal B$ is the closure of the algebraic tensor product $\mathcal A\odot \mathcal B$ in the $B(\mathcal H\otimes \mathcal K)$ norm. 

Question 1: Let $\mathcal A,\mathcal B$ be C$^*$-algebras such that for all C$^*$-algebras $\mathcal C$ non-isomorphic to $\mathbb C$
  $$\mathcal A\otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C $$ Does this imply $\mathcal A\simeq \mathcal B$?

Secondly, can one accomplish this with a single C$^*$-algebra.

Question 2: Does there exist a C$^*$-algebra $\mathcal C$ not isomorphic to $\mathbb C$ such that for all C$^*$-algebras $A,B$ we have
  $$
\mathcal A \otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C \ \Rightarrow \ \mathcal A  \simeq\mathcal B\ ?
$$

One can also ask these questions with the max tensor product.

REPLY [4 votes]: The answer to question 1 is `yes': Let $A$ and $B$ be any $C^*$-algebras. Let $N$ be a simple $C^*$-algebra of such high cardinality that it does not embed into either $A$ or $B$. Then take $C:=N\oplus\mathbb{C}$.
We have
$$
A\otimes C \cong (A\otimes N) \oplus A, \quad
B\otimes C \cong (B\otimes N) \oplus B.
$$
Let $\varphi\colon (A\otimes N) \oplus A\to (B\otimes N) \oplus B$ be an isomorphism.
The choice of $N$ ensures that no quotient of $A\otimes N$ embeds into $B$.
Thus, $\varphi$ maps the summand $A\otimes N$ to $B\otimes N$.
Applying the same argument to the inverse of $\varphi$, we see that $\varphi$ is given by an isomorphism $A\otimes N\cong B\otimes N$ and an isomorphism $A\cong B$.
This also answers question 2 if one puts some restrictions on the cardinality (or better: density character) of the $C^*$-algebras. For instance, for separable $C^*$-algebras, one could use the test-algebra $C:=\mathcal{R}\oplus\mathbb{C}$, where $\mathcal{R}$ is the hyperfinite II-1 factor.<|endoftext|>
TITLE: Closed leaves of a foliation
QUESTION [10 upvotes]: Let $M$ be a differentiable manifold of dimension $n + k$, let $\Delta$ be an $n$-dimensional integrable distribution (à la Frobenius), let $N$ be an $n$-dimensional connected integral manifold of $\Delta$, and assume that $N$ is closed in the ambient manifold $M$. It seems that $N$ should be an embedded submanifold of $M$ (not only immersed), how do you prove this?

REPLY [9 votes]: Choose a point $p$ in $N$, and choose coordinates for $M$ near $p$ such that $N$ appears as $\mathbb R^n\times S$ for some $S\subset \mathbb R^k$. The set $S$ is closed, countable, and nonempty, so by the Baire category theorem it must have an isolated point. 
This implies that some nonempty open subset of $N$ is embedded in $M$. Since the group of diffeomorphisms of $M$ preserving the foliation and the leaf $N$ acts transitively on $N$, $N$ is embedded.<|endoftext|>
TITLE: Large Cardinal Principles that Imply $\Sigma_3^1$-Generic Absoluteness
QUESTION [6 upvotes]: It is known that (light-face) $\Sigma_3^1$ generic absoluteness is consistent with $\mathsf{ZFC}$: Friedman and Bagaria showed that it holds in the $\text{Coll}(\omega, < \kappa)$ extension of $V$ where $\kappa$ is a $\Sigma_2$-correct cardinal.
My questions are about what large cardinal principles can prove $\Sigma_3^1$-generic absoluteness. In particular:
1) If $0^\sharp$ exists (or even $x^\sharp$ exists for all reals $x$), does (light-face) $\Sigma_3^1$-generic absoluteness holds. 
2) What if there is a measurable cardinal, then does $\Sigma_3^1$-generic absoluteness hold?
Is there any large cardinal whose existence implies $\Sigma_3^1$ generic-absoluteness?

REPLY [7 votes]: Re a proof of this result: I once wrote up an elementary proof of this which avoids the Martin-Solovay tree: See Theorem 1 of  https://ivv5hpp.uni-muenster.de/u/rds/talks-barcelona.dvi<|endoftext|>
TITLE: Minimum size of the union of sets
QUESTION [9 upvotes]: I came accross this combinatorial problem in my computer science research. 
You are given a collection of k sets $S_1,...,S_k$ such that for any $i \neq j$, $ \vert S_i \setminus S_j \vert \geq p$ for some fixed integer $p$.  
Then what is the minimum size of the union of the sets $S_i$?

REPLY [3 votes]: This is lower estimate on the size $n$  of the union proven on the way of standard
proof of Sperner's lemma. Or, better to say, it gives an upper bound for $k$ with given $n$. In full generality it is the same problem, of course.
Let $U$ be union of our $k$ sets, $|U|=n$. 
Consider random permutation $(x_1,\dots,x_n)$ of $U$. Clearly $|S_i|\leq n-p$
for any $i$, else $k\leq 1$.
For any $i$ consider the following random
event: there exists $T$, $|T|\leq p-1$,
sich that $S_i\sqcup T=\{x_1,\dots,x_k\}$
for $k=|S_i|+(p-1)$ and some set $T\subset U\setminus S_i$. Note that
no two events may hold simultaneously.  Hence sum of probabilities 
of our events does not exceed
$1$. Denote $|S_i|=a$ for a moment (to calculate probability). Then $T$ may be chosen by
$\binom{n-a}{p-1}$ ways, hence probability equals 
$$\binom{n-a}{p-1}/\binom{n}{a+p-1}=\binom{n+p-1}{p-1}/\binom{n+p-1}{a+p-1}.$$
Maximizing by $a$ we see that it is not less than $\binom{n+p-1}{p-1}/\binom{n+p-1}
{\lfloor\frac{n+p-1}2\rfloor}$, therefore
$$
k\leqslant \frac{\binom{n+p-1}
{\lfloor\frac{n+p-1}2\rfloor}}{\binom{n+p-1}{p-1}}.
$$<|endoftext|>
TITLE: Matrix equation $XAXBXC=I$
QUESTION [15 upvotes]: Let $A,B,C$ be unitary matrices. Does there always exist a unitary matrix $X$ such that $$(XA)(XB)(XC)=I,$$ where $I$ is the identity matrix? The quadratic equation $(XA)(XB)=I$ has the solution $A^*(AB^*)^{1/2}$, and I am hoping that the cubic and higher dimensional versions are always solvable.

REPLY [30 votes]: Here is an argument showing that the answer is 'yes'.  I'll let you check the details and that this result generalizes to all higher degrees.  
Consider the map $f_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f_{ABC}(X) = XAXBXC.
$$
Since the image of this map is compact, if this map were not onto, it would have to have topological degree equal to zero.
Next, since $\mathrm{U}(n)$ is connected, the map $f_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f_{ABC}(X) = XAXBXC
$$
is homotopic to the map $f_{III}$, and hence has the same degree as $f_{III}$. Thus, it suffices to show that the map $f_{III}$ has nonzero topological degree to show that $f_{ABC}$ is surjective.
I claim that the map $f_{III}$ has topological degree $3^n$.  
To prove this, it suffices to compute the local degrees around the pre-images of a regular value.  Let $Y = \mathrm{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})$, where $0<\theta_1<\theta_2<\cdots<\theta_n<\pi$.  Then $Y$ has distinct eigenvalues.  Hence, any solution $X$ to $X^3 = Y$ has distinct eigenvalues and has the same eigendirections as $Y$. Thus, $X$, too, must be diagonal and must be of the form
$$
X = \mathrm{diag}(e^{i\tau_1},\ldots,e^{i\tau_n})
$$
where $3\tau_k \equiv \theta_k\ \mathrm{mod}\ 2\pi$ for $k=1,\ldots, n$.  Thus, there are $3^n$ solutions $X$ to $X^3=Y$.  
I claim that $Y$ is a regular value of the map $f_3:\mathrm{U}(n)\to \mathrm{U}(n)$ defined by $f_3(X)=X^3$ and that $f_3$ is orientation preserving at each solution $X$ of $X^3=Y$.  This follows from the following computation:  Consider the pullback under $f_3$ of the canonical left-invariant form $g^{-1}\mathrm{d} g$.
$$
f_3^*(g^{-1}\mathrm{d}g) = g^{-3}\mathrm{d}(g^3) 
= (I + \mathrm{Ad}(g^{-1})+\mathrm{Ad}(g^{-2}))(g^{-1}\mathrm{d}g)
$$
When one computes the determinant of $\bigl(I + \mathrm{Ad}(X^{-1})+\mathrm{Ad}(X^{-2})\bigr):{\frak{u}}(n)\to {\frak{u}}(n)$ for each solution $X$ of $X^3=Y$, one finds that, because $Y$ has distinct eigenvalues, this determinant is a positive number.  Thus, $Y$ is a regular value of $f_3$, and each of the $3^n$ solutions $X$ to $X^3=Y$ contributes a $+1$ to the topological degree in the usual degree formula.
(By a similar argument, the map $f_k:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by $f(X) = X^k$ has topological degree $k^n$, which is nonzero, so it is necessarily surjective. This answers the higher degree cases as well.)<|endoftext|>
TITLE: Differential geometry without the Hausdorff condition or the second axiom of countability
QUESTION [7 upvotes]: I would like to know how the standard differential geometry of manifolds would change if we didn't assume the Hausdorff  condition and/or the second axiom of countability. There are some simple things that can be said at a first glance:

Manifolds would not necessarily admit partitions of unity, so not every Manifold would admit a Riemannian metric. 
A Haussdorf manifold admitting a connection would be automatically second-countable.
A Haussdorf manifold admitting a Lorentzian metric would be automatically paracompact.
Manifolds without these two assumptions wouldn't necessarily have the homotopy type of a CW-complex.

Since the question is maybe too broad, we can focus for example on the following especific issue:
How Berger's classification of manifolds of special holonomy would change without these two assumptions?
One motivation to ask this question is that manifolds describing space-times needn't to be Haussdorf, or at least physicists have studied non-Haussdorf solutions to General Relativity, see for example "The large scale structure of spacetime" of Hawking and Ellis.
Thanks.

REPLY [2 votes]: Probably the simplest example of a non-second-countable manifold is the long line, and it already exhibits some interesting phenomena.  See for example "Various smoothings of the long line and their tangent bundles" by Peter Nyikos [Advances in Math. 93 (1992) 129-213].<|endoftext|>
TITLE: Consistency of the nonrigidity of $P(\omega_1)/NS$
QUESTION [6 upvotes]: Is it consistent with ZFC that there exists an automorphism of $P(\omega_1)/\mathrm{NS}_{\omega_1}$ which is not the identity?

REPLY [6 votes]: It is consistent relative to large cardinals that $P(\omega_1)/NS \cong \mathcal B(\mathrm{Col}(\omega,\omega_1))$ and many other homogeneous algebras.
It almost looks like the answer is yes in $L$, but I'm not sure.  In $L$ there are no precipitous ideals, and GCH holds.  A theorem of Balcar and Franek implies that therefore $\mathcal P(\omega_1)/NS$ is forcing-equivalent to $\mathrm{Col}(\omega,\omega_2)$.  Moreover, by general facts about collapsing, there is a dense embedding of $\mathrm{Col}(\omega,\omega_2)$ into $\mathcal P(\omega_1)/NS$.  $\mathrm{Col}(\omega,\omega_2)$ carries many nontrivial isomorphisms, but as Yair pointed out, we may not necessarily be able to lift this to an algebra strictly smaller than the completion.<|endoftext|>
TITLE: What are key $\Sigma^0_2$ or $\Pi^0_3$ theorems?
QUESTION [5 upvotes]: I am researching a logical system that is limited to $\Pi^0_2$ sentences and I am busy to prove that FOL + PA is a conservative extension of that system. Meaning that with $\Sigma^0_n$ sentences (that are not $\Pi^0_2$) you can express things, but they are not really necessary for proving a $\Pi^0_2$ theorem. 
As part of this research I like to know if there are interesting/notable theorems in literature that actually proves a $\Sigma^0_n$ or $\Pi^0_{n+1}$ with $n \geq 2$ sentence (which is not $\Pi^0_2$). Such example would contradict my idea, because if according to my idea those sentences are not necessary, they are likely not be interesting. So, if they exists I like to study it.
Note, that if you use the axiom scheme for induction (in a notable proof) you may temporarily have a $\Pi^0_3$ sentence, but the obtained implication is mostly directly used and than falls back to a $\Pi^0_2$ sentence. So, I am not looking for that.
Finally, the result is different than with allowing to quantify over predicates, which does add strength and can prove things about ordinals etc.
Thanks in advance

REPLY [7 votes]: I'm not clear whether you're asking if there are any interesting non-$\Pi_2$ theorems in the literature, or any proofs of $\Pi_2$ theorems with interesting non-$\Pi_2$ intermediate steps which cannot be removed.
If your question is whether there are interesting $\Sigma_2$ or higher theorems in the mathematical literature, the answer is a great many of them.  Off the top of my head, the Thue-Sigel-Roth and Skolem–Mahler–Lech theorems are good examples.  (I believe these are both $\Pi_3$.)
If your question is whether those theorems are necessary to prove $\Pi_2$ consequences, the answer is no.  In general, if $PA$ proves a $\Pi_2$ statement then there is a constructive proof consisting only of $\Pi_2$ statements.  The usual proof of this works in $PA^\omega$ ($PA$ with constructive higher types) so that high quantifier complexity statements can be replaced by statements with functionals using the functional interpretation, and so the $\Pi_2$ statements in question involve quantifiers over computable functionals.  If you prefer to stick to $PA$, all these quantifiers over computable functionals can be replaced by quantifiers over numbers viewed as indices for Turing machines computing the functional.  (The coding gets messy, since one can't computably tell whether one has a computable functional, but this can be resolved while staying in a $\Pi_2$ format.)
EDITED: I should clarify that last bit.  If one stays in PA, one gets a proof of $\exists y\phi(n,y)$ for each $n$ in a uniform(-ish) way using only $\Pi_2$ statements; if you actually want a purely $\Pi_2$ proof of the whole $\Pi_2$ statement, one does have to work in a somewhat expanded language.<|endoftext|>
TITLE: Parameterizing rotations of a cube
QUESTION [10 upvotes]: For $g\in\mathrm{SO}(3),S\subseteq \mathbb{R}^3,$ define $g\cdot S:=\{g\cdot p : p\in S\}.$  In words, if $g$ is a rotation of $\mathbb{R}^3$, $g\cdot S$ is the set of elements of $S$ rotated by $g$.  For two elements $g,g'\in\mathrm{SO}(3)$, define an equivalence relation $\sim$ via rotations of the axis-aligned cube:  $g\sim g'\iff (g\cdot [-1,1]^3 = g'\cdot [-1,1]^3)$.
What is the structure of $\mathrm{SO}(3)/\sim$?  Does this space admit a simple parameterization, measure, or metric?
More broadly, is this a well-studied space with a name?  Some interesting problems in discrete geometry seem to involve this structure.
Note:  $\mathrm{SO}(3)$ is a simple group, so I expect this to a quotient space rather than a quotient group.  This is a refined version of an earlier question.

REPLY [3 votes]: More useful than embedding it in some higher-dimensional space will be to give it as a manifold with identifications, I think. Topologically, $SO(3)/\!\sim$ is the same space as what you get by taking an octahedron and gluing the opposite faces with a $1/3$ turn. This is related to the 24-cell, one of the 6 different regular 4-dimensional polytopes (analogues of Platonic solids in 4 dimensions), using the fact that $SO(3)$ is topologically $RP^3$, the quotient of the 3-sphere $S^3$ by the antipodal map.
The space of course comes with a natural metric and measure (both inherited from the one on $SO(3)$. You can see this metric/measure on the octahedron by embedding it as a regular octahedron of an appropriate size subset of $SO(3)$. You can work out the size by looking at the volume: 12 of these octahedra tile $SO(3)$, or 24 tile the double-cover $S^3$. Alternatively, each corner of the octahedron in $SO(3)$ is a rotation by $\pi/4$ around one of the faces of the cube, and the center of each face of the octahedron is a rotation by $\pi/3$ around one of the vertices of the cube.<|endoftext|>
TITLE: A generalization of Chebyshev polynomials
QUESTION [10 upvotes]: What is the monic polynomial $p(x)$ of degree $n$ which minimizes $\max_{x \in [-1,1]} |p(x)|$?  The answer is the Chebyshev polynomial, and its largest value on $[-1,1]$ is $1/2^{n-1}$. 
Now suppose we ask the following question: what is the polynomial of degree $n$ of the form $$x^n + a_d x^d + a_{d-1} x^{d-1} + \cdots + a_1 x + a_0$$ which minimizes  $\max_{x \in [-1,1]} |p(x)|$? Here $d<n$; if we take $d = n-1$, we recover the Chebyshev polynomial. 
My main purpose in asking this is to  estimate how big $\max_{x \in [-1,1]} |p(x)|$ is as a function of $n$ and $d$ for such a polynomial. 
One can emulate the derivation of Chebyshev polynomials: we would solve the problem if we could construct a polynomial of this form which never exceeds $A$ in absolute value and oscillates between $A$ and $-A$ sufficiently many times in $[-1,1]$. However, its not immediately clear to me how to do this.

REPLY [4 votes]: A little computation is often a useful way to get insight into such questions.  Fortunately, available open-source solvers can approximately solve a given instance of this problem with a few lines of code and in a second or less.
Code
Here is a Python program that makes use of CVXOPT to find the polynomial of least deviation for a given $n$ and $d$.  There is one important approximation here: it minimizes $p(x)$ only over a discrete set of points $x$.  However, by taking a moderately fine grid of the desired interval, reasonably accurate values are obtained.
import numpy as np

def find_min_poly(x, d, n):
from cvxopt import matrix
from cvxopt.modeling import variable, op, max

X = matrix(x)
y = np.empty( (len(x), d+2) )
y[:,0] = x**n
for i in range(d, -1, -1):
    y[:, d+1-i] = x**i
Y = matrix(y)
c = variable(d+1)

op(max(abs(Y[:,0] + Y[:,1:]*c))).solve()

return c.value, max(abs((Y[:,0] + Y[:,1:]*c.value)))

Here is a nested loop that calls this for a range of $n,d$ values with
the interval $[-1,1]$:
x = np.linspace(-1,1,10000)
N = range(1,11)
maxabs = np.zeros( (len(N),len(N)) )

for n in N:
    for d in range(n-1,-1,-1):
        print n, d
        coeffs, M = find_min_poly(x, d, n)
        maxabs[d, n-1] = M
        print M

print maxabs

You can download the code here.

One warning: for large enough values of $n,d$, this approach will give inaccurate results due to effects of roundoff errors.

Results
Here is the resulting table of values (computed in a few seconds), truncated to 4 decimal places:
$$\begin{bmatrix}
 1.    &  0.5   &  1.    &  0.5   &  1.     & 0.5    & 1.     & 0.5    & 1.     & 0.5     \\
  -    &  0.5   &  0.25  &  0.5   &  0.32645& 0.5    & 0.36491& 0.5    & 0.38843& 0.5     \\
  -    &   -    &  0.25  &  0.125 &  0.32645& 0.19245& 0.36491& 0.23624& 0.38843& 0.2675  \\
  -    &   -    &   -    &  0.125 &  0.0625 & 0.19245& 0.11157& 0.23624& 0.14921& 0.2675  \\
  -    &   -    &   -    &   -    &  0.0625 & 0.03125& 0.11157& 0.06346& 0.14921& 0.09216 \\
  -    &   -    &   -    &   -    &   -     & 0.03125& 0.01562& 0.06346& 0.03559& 0.09216 \\
  -    &   -    &   -    &   -    &   -     &  -     & 0.01562& 0.00781& 0.03559& 0.01972 \\
  -    &   -    &   -    &   -    &   -     &  -     &  -     & 0.00781& 0.00391& 0.01972 \\
  -    &   -    &   -    &   -    &   -     &  -     &  -     &  -     & 0.00391& 0.00195 \\
  -    &   -    &   -    &   -    &   -     &  -     &  -     &  -     &  -     & 0.00195
\end{bmatrix}$$
The value of $n$ is 1 for the leftmost column and increases to the right (up to 10); the value of $d$ is 0 in the first row and increases downward.  You can see the $1/2^{n-1}$ values, corresponding to the Chebyshev polynomials, along the main diagonal.
The code above also provides the coefficients of the optimal polynomial, of course.
Conjectures
By experimenting and investigating these values, you may be able to conjecture the general behavior.  For instance, it is obvious that the first superdiagonal is identical to the main diagonal; the 2nd and 3rd superdiagonals are identical, etc., apparently because the optimal polynomial has $a_d=0$ whenever $n-d$ is odd.
One might hope that the other diagonals exhibit geometric progressions like that of the Chebyshev polynomials, but examining the other diagonals shows this to be false.  However, by computing a larger table of values with higher precision, one is led to conjecture that asymptotically, the ratio of consecutive values along any diagonal approaches two.<|endoftext|>
TITLE: Can a product of conjugates be a Pisot number again?
QUESTION [7 upvotes]: Let $p(X) \in \mathbb{Z}[X]$ be an irreducible polynomial, and let $\alpha_1 \dots, \alpha_n$ be its roots in $\mathbb{C}$. Suppose that $\alpha_1$ is a Pisot number, that is, $\alpha_1 \in \mathbb{R}, \ |\alpha_1| > 1$, and for any $1 < i \leq n$ we have $|\alpha_i| < 1$.

Suppose that $\alpha_1 < 2$. Is it possible that for some $1 <  k \leq n$ the number $\alpha_1 \cdots \alpha_k$ is Pisot?

Note that the assumption that $\alpha_1 < 2$ implies that $\alpha_1 \in \mathcal{O}_{\mathbb{Q}(\alpha_1)}^*$.

REPLY [3 votes]: Still, it seems to me that the difference in definitions is not that negligible. But if you are OK with negative numbers --- here is an example.
Set $g(x)=x^2-x$ and $f(x)=g(x)^2-g(x)-1=x^4-2x^3+x-1$. Its roots are the roots of
$$
  g(x)=\frac{1\pm \sqrt5}2.
$$
When the '$+$' sign is chosen, we get two real numbers $\alpha_1\in(1,2)$ and $\alpha_2\in(-1,0)$ since $g(-1)=g(2)>\phi$ but $g(0)=g(1)<\phi$. For the '$-$' sign we have two complex roots whose (equal) absolute values are less than 1 since their product's one is such. Thus $\alpha_1$ is a required example, because $\alpha_1\alpha_2=-\phi$. 
Remark. I cannot transfom this example into one where the product is a positive Pisot number, and it looks like much more difficult to make, at least with the same method.<|endoftext|>
TITLE: Theorems that tell if an explicit analytical solution is possible for nonlinear PDEs
QUESTION [5 upvotes]: Are there any theorems that tell if a particular nonlinear PDE can be solved explicitly by analytical methods?
Where analytical methods I refer to methods such as power series or any methods that use special and elementary functions in some form, as opposed to numerical methods which use iterations or difference schemes etc.
Thanks in advance.

REPLY [12 votes]: As far as (local) power series solutions go (i.e., in the analytic category) the main existence theorem is the Cauchy-Kowalewski Theorem (in the determined, non characteristic case) and its generalization, the Cartan-Kähler Theorem (in the (possibly overdetermined) involutive case).  There are further generalizations that weaken the involutivity hypothesis somewhat, such as $2$-acylcity (in the works of the Spencer school) or weighted involutivity (in the works of the Japanese school), that are useful.  In all of these cases, the power series solution can be computed recursively and effectively, so in that sense, the power series solution is 'explicit'.
However, most people don't think of power series solutions as 'explicit analytic solutions'.  Instead, what they usually mean is something like writing the classical solution of the wave equation $u_{xy}=0$ in the 'explicit' form $u(x,y) = f(x) + g(y)$, involving 2 arbitrary functions of one variable (each).  There are nonlinear equations that admit such representations of solutions, perhaps the most famous being that of the Liouville equation $u_{xy} = e^{2u}$, which can be written in the form
$$
u(x,y) = \frac12\log\left(\frac{f'(x)g'(y)}{(f(x)+g(y))^2}\right).
$$
However, as Lie showed, the equation $u_{xy} = F(u)$ admits a 'general' solution in the form
$$
u(x,y) = U\bigl(x,y,f(x),g(y),f'(x),g'(y),\ldots,f^{(n)}(x),g^{(n)}(y)\bigr)
$$
for some given function $U$ of $2n{+}4$ variables with $f$ and $g$ being 'arbitrary' functions of a single variable, only when $F(u) = ae^{bu}$ for some constants $a$ and $b$.  Note that, by Lie's theorem, even the linear equation $u_{xy} = u$ does not have such a representation.  In fact, equations that admit solutions with such representations are extremely rare, and this is why modern PDE does not attempt to rely on them.
In the 1870s, Gaston Darboux developed a general method for determining when equations of various classes have representations of the above form.  For example, Darboux' Method could, in theory (the computations can be formidable), determine when a given second order equation of the form
$$
F(x,y,u,u_x,u_y,u_{xx},u_{xy},u_{yy})=0
$$
admits an explicit solution in a form that considerably generalizes Lie's form as described above.  This is probably the most advanced result of this kind about explicit analytic representations, even today, and it is quite useful in geometric situations.  For example, when Darboux' Method is applied to the minimal surface equation, an elliptic nonlinear equation of the above form, it yields the Weierstrass representation for minimal surfaces in terms of a holomorphic function of a single complex variable.
One place where you can read about Darboux' Method is in a couple of papers by myself, Phillip Griffiths, and Lucas Hsu, Hyperbolic exterior differential systems and their conservation laws, I and II, Selecta Mathematica 1 (1995).  For further study of Darboux' Method, there are works by Anderson, Fel, and Vassiliou, among others, that can be profitably consulted.  As was suggested in the comments above, if you pursue this, you will eventually need to understand Cartan's Equivalence Method, which is the principal tool needed in the study of geometric invariants of PDE, i.e., properties of PDE that are invariant under all changes of coordinates.<|endoftext|>
TITLE: Serre duality over a non-algebraically closed field
QUESTION [6 upvotes]: Suppose $X$ is a projective smooth variety over a non-algebraically closed field , do we still have $Ext^i(F,\omega)\to H^{n-i}(X,F)^{\vee}$? (Hartshorne's proof Thm III 7.6 requires $k$ to be algebraically closed)

REPLY [9 votes]: Ok, here's a proof via specializing Grothendieck duality.   It is probably useful for people to see this worked out.  Say $f: X \to \text{Spec }k$ is the structural map and its proper and $X$ is just a scheme of finite type over $k$, say $F$ is a coherent sheaf on $X$ (for simplicity).  Then Grothendieck duality says that 
$$R f_* R \mathcal{H}\text{om}_{O_X}(F, f^! k) \simeq R\mathcal{H}\text{om}_{k}(R f_* F, k).$$
In particular, this is an isomorphism in the derived category.
Analyzing the Left side
If $X$ is Cohen-Macaulay, then $f^! k$ is a canonical module on $X$ with a shift (locally by the dimension, note that Cohen-Macaulay means it is locally equidimensional).  So lets work with one connected component at a time, where it is of some dimension $d$.  Then $f^! k \simeq \omega_X[d]$ (take it as a definition if you aren't comfortable with it).  We also notice that $\mathcal{H}\text{om}(F, \bullet)$ takes injectives to flasques and so we get the composition of derived functors: 
$$
Rf_* R\mathcal{H}\text{om}(F, \bullet) = R\text{Hom}(F, \bullet).
$$
(ie, global sections of sheafy hom are non-sheafy-hom)
We take the $(i-d)$th cohomology of the left side and get
$$
\begin{array}{rl}
& \mathcal{H}^{i-d}(R f_* R \mathcal{H}\text{om}_{O_X}(F, f^! k) ) \\
= & \mathcal{H}^{i-d} R \text{Hom}_{\mathcal{O}_X}(F, \omega_X[d]) \\
= & \mathcal{H}^{i} R \text{Hom}_{\mathcal{O}_X}(F, \omega_X) \\
= & \text{Ext}^i(F, \omega_X).
\end{array}
$$
Analyzing the right side
This is even easier, $R\mathcal{H}\text{om}_{k}(\bullet, k)$ is just the derived functor of $k$-vector space duality, which we identify with itself (its already exact).  So I'll just write it as $(\bullet)^{\vee}$.  Again we take the $(i-d)$th cohomology and get 
$$
\mathcal{H}^{i-d} R\mathcal{H}\text{om}_{k}(R f_* F, k) = \mathcal{H}^{i-d}(( R f_* F)^{\vee}) = (\mathcal{H}^{d-i} Rf_* F)^{\vee} = (H^{d-i}(X, F))^{\vee}.  
$$
In conclusion
Plugging these two things in we recover the desired Serre duality.  Notice we never used the fact that $k$ was algebraically closed.  We only used the fact that it was a field when noticing that $\mathcal{H}\text{om}_k(\bullet, k)$ was exact already.  You can still say similar things (that look a bit more complicated) when $k$ is a Gorenstein ring instead of a field.<|endoftext|>
TITLE: Can every permutation group be realized as the automorphism group of a graph (acting on a subset of the vertices)?
QUESTION [6 upvotes]: By Frucht's theorem, every finite group can be realized as the automorphism group of a finite undirected graph. Because a permutation group is a finite group, it is clear that every permutation group be realized as the automorphism group of a graph. However, a permutation group acts on its underlying set, and I also want to realize this group action on a subset of the vertices of the graph.
An equivalent description of what I want is that I'm given a set of vertices and a permutation group acting on those vertices, and I want to construct a graph around those vertices such that the automorphism group of the graph realizes the given permutation group on the given set of vertices.

REPLY [4 votes]: FYI, this result exists in the literature. The two citations I know of are: 

L. Babai. Representation of permutation groups by graphs. In Combinatorial theory and its applications, I (Proc. Colloq., Balatonfured, 1969), pages 55–80. North- Holland, Amsterdam, 1970. 

and 

I. Z. Bouwer. Section graphs for finite permutation groups. J. Combinatorial Theory, 6:378–386, 1969.<|endoftext|>
TITLE: Bounded gaps between primes in arithmetic progressions
QUESTION [8 upvotes]: Has Zhang's work on bounded gaps between primes been extended to the following theorem?

For any arithmetic progression $an+b,\gcd(a,b)=1$, there is a constant $H$ (depending only on $a$) such that there are infinitely many prime pairs $p_1<p_2$ in this progression with $p_2-p_1<H$.

I have tried to look for this result online, the closest hit I got is this paper, which proves similar sounding statement.
How about extending this question to sets of primes of positive relative density (this extension and question as the whole inspired by answer to this question) and quantitative bounds?
Thanks in advance.

REPLY [8 votes]: Yes, see Deniz Ali Kaptan's recent arXiv preprint. (Added: See Terry Tao's comment below for more references.)
Concerning your second question, even full relative density is not enough to produce bounded gaps. Using standard upper bounds on the number of solutions $p'-p=d$ (for a given even $d>0$) it is surely straightforward to show that a certain full relative density subset of primes avoids bounded gaps (I have not checked the details though).<|endoftext|>
TITLE: Fibered example of topologically slice knots
QUESTION [7 upvotes]: Is there any known example of fibered knot which is topologically slice but not (expected to be) smoothly slice?

REPLY [17 votes]: Such a knot would yield a counterexample to one of two important conjectures in the area.  A preliminary definition: a slice knot is homotopically ribbon if the inclusion of the knot into the slice disk complement induces a surjection on the fundamental group. It's easy to see that a ribbon knot is homotopically ribbon; note that homotopically ribbon doesn't require any smoothness (but you still want local flatness to avoid trivialities). This criterion is viewed as the right substitute for ribbon in the topological setting. 
The remarkable theorem of Casson-Gordon (A loop theorem for duality spaces and fibred ribbon knots; Invent. Math. 74, 119-137 (1983)) says that a fibered knot is homotopically ribbon if and only if its monodromy (filled in over a disk) extends over a handlebody.  The if part constructs a smooth slice smooth homotopy ball with boundary the 3-sphere (if your knot lay in the 3-sphere.)  So in particular, a fibered knot that is topologically homotopy ribbon is smoothly homotopically ribbon!!!  So an example of the sort that you ask for would either be slice but not homotopically ribbon (contradicting the conjecture that this doesn't happen) or would give a smooth slice  counterexample to the 4-dimensional Poincaré conjecture. 
So to summarize, no such example is known.
PS If there were an MO question asking for "your favorite paper that people don't think about enough", the Casson-Gordon paper mentioned above might top my list.

REPLY [3 votes]: A common source of topologically slice knots are those with Alexander polynomial $1$. However these are not fibered. This follows from a classical result, that $2\mathrm{genus}(K) = \mathrm{deg}(\Delta_K(t))$ for fibered knots, which I learned in a paper of Stefan Friedl and Taehee Kim ("The Thurston Norm, Fibered Manifolds and Twisted Alexander Polynomials"). Of course there are other sources of topologically slice knots, so this doesn't settle the question.<|endoftext|>
TITLE: Is this an instance of any existing convex pentagonal tilings?
QUESTION [10 upvotes]: Inspired by Wikipedia's article on pentagonal tiling, I made my own attempt. 

I believe this belongs to the 4-tile lattice category, because it's composed of pentagons pointing towards 4 different directions. According to Wikipedia, 3 out of the 15 currently discovered types of pentagonal tiling belongs to the 4-tile lattice category (type 2, 4 & 6).
https://en.wikipedia.org/wiki/Pentagonal_tiling
I don't think my attempt is an instance of type 4 or 6, since in both type 4 & 6, any side of all pentagons overlaps with only one side of another pentagon, while in my attempt, half of the long sides overlap with two shorter sides.
At the same time, I can't figure out how my attempt is an instance of type 2... Please kindly offer your insight.

REPLY [8 votes]: The pentagon tiling problem is just settled by Michaël Rao: 
The list of $15$ types is
complete. See the Natalie Wolchover article in Quanta.

"Exhaustive search of convex pentagons which tile the plane": link.<|endoftext|>
TITLE: An equivalence relation on the space of polynomials in one complex variable
QUESTION [6 upvotes]: Let $P(z)$ be a  polynomial with complex variable $z$.  We consider the following distribution for the roots of  $P(z)=0$: the distribution is a triple $(n_{1},n_{2},n_{3})$
where these integers are the number of roots in the interior of unite circle, on the unit circle and out of unit disc, respectively.
$\ell^{2}$ is the Hilbert space of all square sumable sequence of  complex numbers. $S_1$ is the shift operator on $\ell^{2}.$
The equivalence relation: We say two polynomials $P,Q$  are equivalent if $P(S_1)$ is  conjugate to $Q(S_1)$  via  an invertible operator in $B(\ell^{2})$.
Assume that $P,Q$ are two equivalent polynomials:

1.Must they have the same degree?
2.Must they have the same root distribution?

The motivation for the second question is that when $P,Q$ have the same degree and $P$ has $(n_{1},0,n_{3})$ distribution, then $Q$ has the same distribution as $P$. The reason is that $P(S_1)$ is  a  Fredhom operator of index $-n_1$. Obviously this property is invariant under conjucacy. So $Q$ has the same root  distribution.

REPLY [5 votes]: Much more is true. According to the answer by Alexandre Eremenko (I do not have this paper, but I completely trust him), from $P(\mathbb{S}^1)=Q(\mathbb{S}^1)$ (which are essential spectra of $P(T)$, $Q(T)$ as noted in the comments) it follows that $P=f(z^n)$, $Q=f(wz^m)$ for some number $w$, $|w|=1$, positive integers $n,m$ and polynomial $f$. Further, I claim that $n=m$, it would mean just $Q(z)=P(wz)$. Indeed, choose complex number $a$, $|a|<1$, so that $f(a)\notin P(\mathbb{S}^1)$, and consider polynomials $P(z)-f(a)$, $Q(z)-f(a)$. They do not have roots on $\mathbb{S}^1$, thus the numbers of their roots inside the unit disc must be the same (as noted in your post, these numbers are just Fredholm indices of $P(T)-f(a)$, $Q(t)-f(a)$). But these numbers equal $n\cdot k$, $m\cdot k$ resp., where $k>0$ is the number of roots of $f(z)-f(a)$ in the open unit disc.  Thus $n=m$ as desired.<|endoftext|>
TITLE: Does the Galois group of a Pisot polynomial contain the alternating group?
QUESTION [9 upvotes]: Let $n \in \mathbb{N}$, and let $p(X) \in \mathbb{Z}[X]$ be a monic polynomial of degree $n$. Suppose that exactly one complex root of $p$ is of modulus $> 1$, and that the remaining $n-1$ roots of $p$ belong to the open unit disk in the complex plane (such a polynomial is necessarily irreducible, and is sometimes called a Pisot polynomial).

Is $\mathrm{Gal}(p(X)/\mathbb{Q}) \in \{A_n,S_n\}$ ?

Pisot polynomials are quite common, for instance Perron has shown that if $p(X) = X^n + a_{n-1}X^{n-1} + \dots + a_1X + a_0$ and $|a_{n-1}| > 1 + a_0 + a_1 + \dots + a_{n-2}$ then $p(X)$ is a Pisot polynomial.

REPLY [18 votes]: Any number field $K$ which has a real place is generated by a root of a Pisot polynomial. So the answer is no. For example, we can take $p$ a prime $\geq 11$, and use the real subfield of $\mathbb{Q}(\zeta_p)$, to get a Pisot polynomial whose splitting field is cyclic of degree $(p-1)/2$.
Suppose $K$ has $r$ real places, and $2s$ complex places. Let $V_i$ be $\mathbb{R}$ for $1 \leq i \leq r$, and be $\mathbb{C}$ for $r+1 \leq i \leq r+s$, and let $\sigma_i : K \to V_i$ be the $r+s$ embeddings of $K$. Then recall that
$$\bigoplus \sigma_i : K \to \bigoplus V_i$$
embeds $\mathcal{O}_K$ as a discrete full rank lattice in $\bigoplus V_i$. Projecting further onto $\bigoplus_{i \neq 1} V_i$, we get a dense additive subgroup of $\bigoplus_{i \neq 1} V_i$. In particular, there is some nonzero $\theta \in \mathcal{O}_K$ such that $|\sigma_i(\theta)|<1$ for all $i \neq 1$. And, the norm of any algebraic integer is at least $1$, we have $|\sigma_1(\theta)|>1$.

REPLY [14 votes]: David Speyer has beaten me to it, but for what it's worth, a simple explicit example is
$$p(X)=X^4-2X^3-5X^2-4X-1$$
which is a Pisot polynomial, and has Galois group the dihedral group of order 8.
How did I find it? I started from observing that if $\vartheta$ is the positive root of $X^2-X-1$, then $\alpha=1+\sqrt{\vartheta}$ is an algebraic unit with only one conjugate inside the closed unit disk, and thus $-1/\alpha$ is a Pisot-Vijayaraghavan number, with this minimal polynomial.
Addendum:  The Pisot polynomial
$$p(X)=X^4-2X^3+X-1$$
again has dihedral Galois group of order 8, corresponding to another quadratic extension of $\mathbb{Q}(\vartheta)$, and produces a root less than $2$ (which also happens to be the fourth smallest limit point of the set of PV numbers).  This number is mentioned on page 1251 of David W. Boyd, Pisot and Salem numbers in intervals of the real line, Math. Comp. 32 (1978), 1244-1260.<|endoftext|>
TITLE: On the Riesz representation theorem
QUESTION [5 upvotes]: Let $V$ be a vector space with inner product $(\phi,\psi)$ antilinear in the second argument - not necessarily a Hilbert space. Let $\Phi$ be an antilinear functional on $V$. 
What are the precise (necessary and sufficient) conditions that must be imposed on $\Phi$ that will guarantee the existence of a sequence or net of vectors $\phi_k\in V$ such that $$\Phi(\psi)=\lim_{k} (\phi_k,\psi) ~~~\forall~ \psi\in V?$$
The answer is given by the Riesz representation theorem when $V$ is a Hilbert space, but I am interested in the general case.

REPLY [9 votes]: Such a net exists for any $\Phi$ (in fact, there is a canonical such net).  First, note that if $F\subseteq V$ is a finite-dimensional subspace, then there is a unique $\phi_F\in F$ such that $\Phi(\psi)=(\phi_F,\psi)$ for all $\psi\in F$.  The collection of such $F$ form a directed set under inclusion, and the net $(\phi_F)$ will have the desired property.<|endoftext|>
TITLE: Numerically double-checking formula with L-values
QUESTION [7 upvotes]: I'm working with a special case of Ichino's triple product formula, which for classical holomorphic newforms $f$, $g$ ,$h$ of weights $k$, $m-k$, $m$ (and central characters satisfying $\chi_f \chi_g = \chi_h$) gives me an identity of the form 
$$ |\langle fg,h \rangle|^2 = C \cdot L(f\times g\times \bar{h},m-1) $$
(where $m-1$ is the central value for this L-function). This is great in principle, but if you're familiar with Ichino's formula you probably know it requires a bunch of work to actually nail down the constant $C$ correctly (there's lots of normalizations you have to get right, and then you have to deal with some unpleasant local integrals). So I want to be able to computationally check the formula I have in a bunch of cases, to convince myself it's actually right. I'm particularly interested in cases where $g$ and $h$ are CM newforms from Hecke characters of an imaginary quadratic field, and the triple-product $L$-value splits up as a product of two Rankin-Selberg $L$-values (twists of $f$ by Hecke characters).
So, my question is: what's the best way to work with these sort of quantities computationally? I haven't used computer algebra systems for number theory of this sort, and I'm not entirely sure where to start. Googling around I found that Pari and Sage both have algorithms that should (?) be able to compute my $L$-value, and another MO question giving some ways to compute Petersson inner products numerically. 
My thought is just to boot up Sage and start messing around with the things suggested in those links, but I figured I'd ask here first to see if those are actually the right tools for the job, or if I'm making things way harder on myself than I need to somehow. (I don't really need a ton of precision - just enough to convince me that my constant isn't off by powers of 2 or by Euler factors or something - but I do want to be able to try it for a bunch of modular forms).

REPLY [6 votes]: This is a good question around an important issue. The first thing that came to my mind is Popa's article "Central values of Rankin $L$-series over real quadratic fields", which is about a related problem. In Remark 6.3.3 he writes:

Using the PeriodMapping algorithm implemented by William Stein in MAGMA, and an algorithm implemented by Dokchitser [Dok04] in PARI for computing the special values, we have checked that the formula in Theorem 6.3.1 is exact up to more than 10 decimal places for a range of forms $f$ of weight 2, 4, 6, and 8, taking for $χ$ the trivial character.

In short, I recommend that you contact Popa and Stein (and perhaps also Rubinstein, Farmer, Booker) who are familiar with these kinds of calculations.<|endoftext|>
TITLE: What can we learn from the newly discovered monohedral convex pentagonal tiling?
QUESTION [11 upvotes]: Wikipedia: https://en.wikipedia.org/wiki/Pentagonal_tiling#Stein_.281985.29_and_Mann.2FMcLoud.2FVon_Derau_.282015.29
Media coverage: http://www.theguardian.com/science/alexs-adventures-in-numberland/2015/aug/10/attack-on-the-pentagon-results-in-discovery-of-new-mathematical-tile
It seems that the detail is not yet published.
For those who do not see any mathematics in the question, here are two possibilities for an answer:

Based on the published information, is there any property that is not present in the previously known tilings?
Does this tiling bring any new insight on the pentagonal tilings? How does it help the full classification?

REPLY [4 votes]: What we can learn from (or be reminded of by) the new tiling, is just how hard it is to know whether we have all the tilings. According to Wikipedia, Kershner claimed to have found all pentagonal tilings in 1968, but then James found one in 1975, and Rice found three in 1977, and Stein found one in 1985, and now this new one.<|endoftext|>
TITLE: Square of the distance function on a Riemannian manifold
QUESTION [16 upvotes]: Let $(M^n,g)$ be a smooth Riemannian manifold. Consider the square of the distance function
$$dist^2\colon M\times M\to \mathbb{R}$$
given by $(x,y)\mapsto dist^2(x,y)$. It is easy to see that this function is infinitely smooth near the diagonal.
Now fix a point $a\in M$. Consider the Taylor series of $dist^2$ at $(a,a)$ in some coordinate system (say normal). Can one compute explicitly its coefficients up to the third order?

REPLY [10 votes]: The function $dist^2$ satisfies   a Hamilton-Jacobi equation. Using this  you can  find its Taylor expansion  at $(a,a)$ up to any order. In Appendix A of this paper I  use this approach  to find  the degree 4 Taylor expansion  of this function; see Eq. (A.17) in the paper. The procedure  used in this appendix can be used to find the degree $6$ expansion as well, though the formulas become increasingly more complicated.  I should mention that this approach is also used in
B. DeWitt: The Global Approach to Quantum Field Theory, Oxford University Press, 2003,
pages 281-282, though the physicists' notation may be a bit hard to follow.<|endoftext|>
TITLE: Which sequential colimits commute with pullbacks in the category of topological spaces?
QUESTION [13 upvotes]: This question was asked on math.stackexchange.com without a reaction.

Given diagrams of topological spaces
$$X_0\rightarrow X_1\rightarrow\ldots$$
$$Y_0\rightarrow Y_1\rightarrow\ldots$$
$$Z_0\rightarrow Z_1\rightarrow\ldots$$
and maps $X_i\rightarrow Z_i$, $Y_i\rightarrow Z_i$, such that everything commutes. Taking the colimit yields a canonical map $$colim(X_i\times_{Z_i}Y_i)\rightarrow colim(X_i)\times_{colim(Z_i)}colim(Y_i)$$ from the colimit of the pullbacks to the pullback of the colimits. This map is a continuous bijection, since filtered colimits commute with finite limits in the category of sets and the forgetful functor from spaces to sets preserves all limits and colimits.
This answer of @StefanHamcke serves as a counterexample where the map is not a homeomorphism even in the case where the first three diagrams consist of closed inclusions.
I'm interested in mild point set topological restrictions on the spaces and maps, such that the canonical map is a homeomorphism. The answer linked suggests, that separation conditions will be necessary, so what e.g. about closed inclusions of T1 spaces, Hausdorff spaces, etc.?
Side question: Is this true in any convenient category of topological spaces like CGWH?

REPLY [4 votes]: Fiber products do commute with sequential colimits of closed embedding in CGWH, but one must remember that neither limits nor colimits in CGWH are the same as the corresponding limits and colimit in the category TOP of topological spaces. If you want to keep the fiber product and sequential colimits of TOP, here is a setting where this whould work:
Claim: Fiber products commute with sequential colimits of closed embeddings if the $X_i$'s and $Y_i$'s are Hausdorff locally compact and the $Z_i$'s are Hausdorff.
Proof:
As mentioned in the question we have a continuous bijection
$$ f:colim_i(X_i \times_{Z_i} Y_i) \longrightarrow colim_i(X_i) \times_{colim_i(Z_i)} colim_i(Y_i) $$
Our first step is to reduce to the case where $Z_i = \ast$ for every $i$. To do so consider the continuous bijection
$$ g:colim_i(X_i \times Y_i) \longrightarrow colim_i(X_i) \times colim_i(Y_i) $$
By the universal property we may identify $colim_i(X_i) \times_{colim_i(Z_i)} colim_i(Y_i)$ with a subspace of $colim_i(X_i) \times colim_i(Y_i)$ (equipped with the subspace topology). On the other hand, since each $Z_i$ is Hausdorff the maps $h_i:X_i \times_{Z_i} Y_i \longrightarrow X_i \times Y_i$ are closed embeddings. We claim that the induced map $h:colim_i(X_i \times_{Z_i} Y_i) \longrightarrow colim_i(X_i \times Y_i)$ is a closed embedding as well. First since sequential colimits of sets preserve injective maps it follows that $h$ is injective. Now let $A \subseteq colim_i(X_i \times_{Z_i} Y_i)$ be a closed set. Then the intersection of $h(A)$ with the $X_i \times Y_i$ coincides with $h_i(A \cap (X_i \times_{Z_i} Y_i))$, which is closed. Hence $h(A)$ is closed as desired. We may now conclude that the topology on $colim_i(X_i \times_{Z_i} Y_i)$ is the one induced from $colim_i(X_i \times Y_i)$. It will hence suffice to proce that the map $g$ is a homeomorphism. In other words, we may assume that $Z_i = \ast$.
Let $W$ be an open subset of $colim_i(X_i \times Y_i)$. We wish to show that $g(W)$ is open. For each $i$ let us denote by $W_i = W \cap (X_i \times Y_i)$, considered as an open subset of $X_i \times Y_i$. Let $(x,y) \in colim_i(X_i) \times colim_i(Y_i)$ be a point which belongs to $g(W)$. Then there exists an $i$ such that $x \in X_i$ and $y \in Y_i$ and we may find open neighborhood $x \in U_i \subseteq X_i$ and $y \in V_i \subseteq Y_i$ such that $U_i \times V_i \subseteq W_i$. Furthermore, since $X_i$ and $Y_i$ are locally compact we may assume in addition that the respective closures $\overline{U}_i$ and $\overline{V}_i$ are compact and that $\overline{U}_i \times \overline{V}_i \subseteq W$. Our next goal is to find open subsets $U_{i+1} \subseteq X_{i+1}$ and $V_{i+1} \subseteq Y_{i+1}$ such that $U_{i+1} \cap X_i = U_i$, $V_{i+1} \cap Y_i = V_i$, the closures $\overline{U}_{i+1}, \overline{V}_{i+1}$ are both compact, and $\overline{U}_{i+1} \times \overline{V}_{i+1} \subseteq W$. To construct $U_{i+1}$ and $V_{i+1}$ we begin by choosing for every $(u,v) \in \overline{U}_i \times \overline{V}_i$ open neighborhoods $u \in A_{u,v} \subseteq X_{i+1}$ and $v \in B_{u,v} \subseteq Y_{i+1}$ such that $\overline{A}_{u,v}$ and $\overline{B}_{u,v}$ are compact and $\overline{A}_{u,v} \times \overline{B}_{u,v} \subseteq W_{i+1}$. For each $u \in \overline{U}_i$ we may then find a finite collection $B_{u,v_1}, ...,B_{u,v_n}$ which covers $\overline{V}_i$. We now set $C_u = A_{u,v_1} \cap ... \cap A_{u,v_n}$ and $D_u = B_{u,v_1} \cup ... \cup B_{u, v_n}$. We then observe that $C_u$ is an open neighborhood of $u$ in $X_{i+1}$, $D_u$ is an open neighborhood of $\overline{V}_i$ in $Y_{i+1}$, and $\overline{C}_u \times \overline{D}_u \subseteq W$. Next we may find a finite collection $C_{u_1},...,C_{u_m}$ which covers $\overline{U}_i$. We may now set $E_{i+1} = C_{u_1} \cup ... \cup C_{u_m}$ and $F_{i+1} = D_{u_1} \cap ... \cap D_{u_m}$. Then $E_{i+1}$ is an open neighborhood of $\overline{U}_i$ in $X_{i+1}$, $F_{i+1}$ is an open neighborhood of $\overline{V}_i$ in $Y_{i+1}$, $\overline{E}_{i+1}$ and $\overline{F}_{i+1}$ are both compact and $\overline{U}_i \times \overline{V}_i \subseteq \overline{E}_{i+1} \times \overline{F}_{i+1} \subseteq W_{i+1}$. Finally, since $X_i \longrightarrow X_{i+1}$ is a closed embedding we may let $U_{i+1}$ be the intersection of $E_{i+1}$ with some open subset of $X_{i+1}$ whose intersection with $X_i$ is $U_i$. Similarly we define $V_{i+1} \subseteq Y_{i+1}$. Then $U_{i+1} \cap X_i = U_i$ and $V_{i+1} \cap X_i = V_i$. Furthermore, $\overline{U}_{i+1}$ and $\overline{V}_{i+1}$ are compact and $\overline{U}_{i+1} \times \overline{V}_{i+1} \subseteq W_{i+1}$, as desired.
To finish the proof we apply the procedure described repeatedly so that for every $j=i,i+1,...$ we have open subsets $U_j \subseteq X_j, V_j \subseteq Y_j$ such that $\overline{U}_j$ and $\overline{V}_j$ are compact and $\overline{U}_j \times \overline{V}_j \subseteq W_j$. Furthermore, by construction we have $U_{j'} \cap X_j = U_j$ and $V_{j'} \cap Y_j = V_j$ for every $j < j'$. Let $U = \cup_{j \geq i} U_j$ and $V = \cup_{j \geq i} V_j$. Then $U$ is an open subset of $colim_i X_i$, $V$ is an open subset of $colim_i Y_i$ and $(x,y) \in U \times V \subseteq g(W)$. This show that $g(W)$ is open, as desired.<|endoftext|>
TITLE: How big is the lattice of all functions?
QUESTION [14 upvotes]: Define the lattice $(\mathcal{L},\prec)$ as the set of all function $f:\mathbb{N}\rightarrow\mathbb{N}$ satisfying $f(n)\leq f(n+1)\leq f(n)+1$, where two functions are considered equal if they differ at finitely many points, and $f\prec g$ is defined as $f(n)\leq g(n)$ with finitely many exceptions. 
This lattice is large, e.g. you can construct chains and antichains of size continuum, chains of antichains of the same cardinality and so on. Although these constructions are structurally simple, defining them in a precise way gets quite complicated. Therefore I am interested into references dealing with this lattice or sublattices, making the intuitive "largeness" precise. 
Background: Y. Barnea and I constructed a set of groups parametrized by elements of $\mathcal{L}$, and want a way to convince group theorists that we really have "many" groups, and don't want to waste a lot of space on constructions which yield results far inferior to what every lattice theorist would immediately see.
Edit: As every function in $\mathcal{L}$ is equivalent to either a function $f$ satisfying $f(0)=0$ or a function $f(n)=n+c$, the interesting part is the sublattice of functions satisfying $f(0)=0$. Also I always assume AC and would not mind too much about assuming CH.

REPLY [2 votes]: Here are a few tricks to play with.
For every $A\subseteq\omega$ define $f_A$ by $f_A(0)=0$, $f_A(n+1)=f_A(n)+1$ if $n\in A$, and $f_A(n+1)=f_A(n)$ if $n\notin A$.
If $A\subset^*B$ (so $B\setminus A$ is infinite) then $f_A\prec f_B$: fix $m$ such that $A\setminus B\subseteq m$; in fact, $f_B(n)-f_A(n)$ will diverge to infinity.
Now take a tower $\langle T_\alpha:\alpha<\mathfrak{t}\rangle$ such that $T_{\alpha+1}\setminus T_\alpha$ is always infinite. Then $\langle f_{T_\alpha}:\alpha<\mathfrak{t}\rangle$ is a $\mathfrak{t}$-chain in $\mathcal{L}$.
One can also create a copy of the unit interval in $\mathcal{L}$ in this way.
If $\mathcal{A}$ is an almost disjoint family then we can modify it as follows: for every $A$ let $X_A=\bigcup_{n\in\omega}[3^n,3^{n+1})$. Then the set $\{f_{X_A}:A\in\mathcal{A}\}$ is an antichain in $\mathcal{L}$.
To see this consider two infinite sets $A$ and $B$ with finite intersection, say $A\cap B\subseteq m$, and write $g_A=f_{X_A}$ and $g_B=g_{X_B}$. For every $n$ we have $g_A(3^n), g_B(3^n)\le \sum_{k<n}3^k=\frac12(3^n-1)$.
Now, take $l\ge m$ and let $n\le l$ be the first member of $(A\cup B)\setminus l$; now if $n\in A$ then $g_A(3^{n+1})=g_A(3^n)+2\cdot 3^n$ and $g_B(3^{n+1})=g_B(3^n)$, so that $g_A(3^{n+1})>g_B(3^{n+1})$. If $n\in B$ then this reverses. It follows that $g_A$ and $g_B$ are $\prec$-incomparable.<|endoftext|>
TITLE: If all balls around fixed basepoints are isometric, are the spaces as well (length spaces)?
QUESTION [10 upvotes]: Let $X$ and $Y$ be two complete proper length spaces, $x \in X$ and $y \in Y$. Assume for every $r>0$ the closed balls $\overline{B_r(x)}$ and $\overline{B_r(y)}$ are isometric. 
Does there exist an isometry $X \to Y$? 
Remark: If you add that the isometries between the balls all map $x$ to $y$, you get a pointed isometry between $(X,x)$ and $(Y,y)$.
Definitions: A metric space is called proper if all closed balls are compact. A metric space is called length space if the distance of two points equals the infimum of the length of curves connecting these points, i.e. $d(x,y) = \inf \{ L(c) \mid c(0) = x, c(1) = y\}$. 
In fact, for a complete length space, this infimum is a minimum.

REPLY [2 votes]: The answer is NO. 
To get an idea look at the following picture. 
Imagine it is extended down infinitely and each segment exchanged to a thin rectangle with $\ell_\infty$-metric.

The center of the ball has to be at one of the ends.
There are ways to extend it down to get non-isometric spaces.
At one step down you have two choices --- extend it to the right $(+1)$ or to the left $(-1)$. The space can be encoded by an infinite sequence of $\pm1$. Two such spaces are isometric if the sequences coincide starting from some moment or they have opposite sign starting from some moment. In particular the space for 
$$(+1,+1,+1,+1,\dots)\ \ \text{and}\ \ (+1,-1,+1,-1,\dots)$$ 
are not isometric.<|endoftext|>
TITLE: Which algebraic relations are possible between algebraic conjugates?
QUESTION [19 upvotes]: For which non-constant rational functions $f(x)$ in $\mathbb{Q}(x)$ is there $\alpha$, algebraic over $\mathbb{Q}$, such that $\alpha$ and $f(\alpha) \neq \alpha$ are algebraic conjugates? More generally, can one describe the set of such $\alpha$ (empty/non-empty, finite/infinite etc.) if one is given $f$?
Examples:

$f(x)=-x$: These are precisely the square roots of algebraic numbers $\beta$ such that there is no square root of $\beta$ in $\mathbb{Q}(\beta)$. There are infinitely many $\alpha$ and even infinitely many of degree $2$.
$f(x)=x^2$: These are precisely the roots of unity of odd order, since $H(\alpha)=H(\alpha^2)=H(\alpha)^2$ implies $H(\alpha)=1$, so $\alpha$ a root of unity. Here $H(\alpha)$ is the absolute multiplicative Weil height of $\alpha$. There are infinitely many $\alpha$, but only finitely many of degree $\leq D$ for any $D$.
$f(x)=x+1$: There is no $\alpha$. If there was and $P(x)$ was its minimal polynomial, then $P(x+1)$ would be another irreducible polynomial, vanishing at $\alpha$, with the same leading coefficient and hence $P(x+1)=P(x)$. Looking at the coefficients of the second highest power of $x$ now leads to a contradiction. Analogously for $f(x)=x+a$ if $a$ is any non-zero rational number.

So the existence of such an $\alpha$ and the set of all possible $\alpha$ seem to depend rather intricately on $f(x)$, which seems interesting to me. As I found no discussion of this question in the literature, I post it here.
UPDATE: Firstly thanks to all who have contributed so far! As Eric Wofsey pointed out, any solution $\alpha$ will satisfy $f^n(\alpha)=\alpha$ for some $n>1$, where $f^n$ is the $n$-th iterate of $f$. So one should consider solutions of the equation $f^n(x)-x=0$ or $f^p(x)-x=0$ for $p$ prime.
If the degree of $f$ is at least 2, one can always find irrational such solutions $\alpha$ with $f(\alpha) \neq \alpha$ by the answer of Joe Silverman. However, for his proof to work, we'd need to know that $f^k(\alpha)$ and $\alpha$ are conjugate for some $k$ with $0 < k <p$. I'm not enough of an expert to follow through with his hints for proving this, but if someone does, I'd be very happy about an answer!
If the degree of $f$ is 1, then $f$ is a Möbius transformation and all $f^n$ will have the same fixed points as $f$ (so there's no solution) unless $f$ is of finite order. In that case, if $f(x) \neq x$, the order is 2, 3, 4 or 6 (see http://home.wlu.edu/~dresdeng/papers/nine.pdf). By the same reference, in the latter three cases, $f$ is conjugate to $\frac{-1}{x+1}$, $\frac{x-1}{x+1}$ or $\frac{2x-1}{x+1}$, so it suffices to consider these $f$, which give rise to the minimal polynomials $x^3-nx^2-(n+3)x-1$ (closely related to the polynomial in GNiklasch's answer), $x^4+nx^3-6x^2-nx+1$ and $x^6-2nx^5+5(n-3)x^4+20x^3-5nx^2+2(n-3)x+1$, if my calculations are correct. If the order is 2, the map is of the form $\frac{-x+a}{bx+1}$ or $\frac{a}{x}$, which leads to $x^2+(bn-a)x+n$ or $x^2+nx+a$ respectively. So this case is somewhat degenerate, which explains the unusual behavior of $f(x)=-x$ and $f(x)=x+1$ above.

REPLY [16 votes]: A first observation is that the field $\mathbb{Q}(\alpha)$ must admit a non-identity automorphism $\sigma$, since applying a rational function to $\alpha$ will not take us outside this field. Given such $\alpha$ and $\sigma$, you can immediately read off an $f$ by expressing $\sigma(\alpha)$ as a $\mathbb{Q}$-linear combination of $1, \alpha, \alpha^2,\ldots$, and there'll be additional non-polynomial (rational) expressions in terms of $\alpha$. This leads to plenty of examples, but now we want to turn this around and start from a given $f$.
Next, the application of $f$ can be iterated. If $g$ is the minimal polynomial of $\alpha$ over the rationals, $g \circ f$ will be a rational function whose numerator vanishes at $\alpha$, hence is divisible in $\mathbb{Q}[x]$ by $g$, hence vanishes at $f(\alpha)$ as well, so $g(f(f(\alpha))) = (g\circ f)(f(\alpha))=0$, so $f(f(\alpha))$ is another conjugate of $\alpha$, and so forth.
Since there are only finitely many conjugates to choose from, some iterate $f^{\circ n}(\alpha)$ must return to $\alpha$: in other words, $\alpha$ will be a root of (the numerator of) $f^{\circ n}(x)-x$ for some integer $n > 1$. (This leads to another argument ruling out $f(x)=x+1$ and friends, as has been noted in the comments.)
So in many cases you'll be able, given $f$, to find the $\alpha$ by picking an $n$ and factoring the numerator of $f^{\circ n}(x)-x$ into irreducible polynomials.
However, the $n$-th iterate of $f$ might be the identity function $x\in \mathbb{Q}(x)$ already, and the numerator to be factored might thus be zero! This happens in your first example $f(x)=-x$ with $n=2$, or with $$f(x)=1/x$$ again with $n=2$ which leads to reciprocal minimal polynomials $g(x)=x^{\mathrm{deg}(g)}g(1/x)$, or with (e.g.) $$f(x)=1-1/x$$ with $n=3$ which gives rise (among other things) to what D. Shanks in 1974 called The Simplest Cubic Fields (Math.Comp. 28, 1137-1152), with $g(x)=x^3+ax^2-(a+3)x+1$. Here, you can proceed by prescribing a degree for $g$, spelling out the condition that $g$ should divide the numerator of $g\circ f$, and comparing coefficients.

REPLY [9 votes]: The answer to the question you ask, namely "for which $f(x)$...?", is that there exists such an $\alpha$ for every $f(x)$ of degree at least $2$. Proof: If $p$ is a sufficiently large prime, then there is a $p$-periodic point $\alpha$ of $f(x)$ that is not in $\mathbb Q$. (This follows, e.g., by computing heights.) So $f^p(\alpha)=\alpha$ 
and $f(\alpha)\ne\alpha$. The orbit $\{f^i(\alpha):0\le i<p\}$ is Galois invariant [not necessarily true, see below] and contains an element $\alpha$ not in $\mathbb Q$, so there is some $\sigma\in\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$ such that $\sigma(\alpha)=f^k(\alpha)$ for some $1\le k<p$. Now let $j$ satisfy $jk\equiv1\pmod{p}$. Then
$$ f(\alpha) = f^{jk}(\alpha)
   = \underbrace{f^k\circ\cdots\circ f^k }_{\text{$j$ iterates}}(\alpha)
  = \sigma^j(\alpha).
$$
This shows that $f(\alpha)$ is a Galois conjugate of $\alpha$.
For your second question, as has been noted, if $f(\alpha)=\sigma(\alpha)$, then since $\sigma$ has finite order, one must have that $\alpha$ is a periodic point for iteration of $f$. So your question seems to come down to more-or-less classifying all irrational numbers that are periodic (but not fixed) for a given $f(x)\in\mathbb{Q}(x)$. There's been a lot of study of the  Galois properties of fields generated by periodic points of rational functions. See Section 3.9 of The Arithmetic of Dynamical Systems (Springer GTM 241) for an introduction and some further references. 
--- ADDITION ---
As Gabriel Dill pointed out, it is the set of points of period $p$ that is invariant under Galois, not the individual orbits. In general, the Galois group $G_p$ of the points of period $p$ is a subgroup of a wreath product $W_p$ of a cyclic group $C_p$ and a symmetric group $S_N$, where $N_p=(d^p-d)/p$. [This is assuming that the $p$-periodic points don't appear with multiplicities.] It is conjectured (I believe) that $G_p=W_p$ for all but finitely many $p$ (except maybe for $x^d$, Chebyshev, and Lattes maps?). This would certainly be enough. Alternatively, if one could find primes with $p\nmid N_p$, then it would suffice to know that $p\mid\#G_p$. (Of course, even for $d=2$ we don't know how to find infinitely many such primes.) Clearly one can get away with making various weaker assumptions about $G_p$, and possibly they could be proven.<|endoftext|>
TITLE: When are direct products exact in the category of quasi-coherent sheaves?
QUESTION [6 upvotes]: (This question is crossposted from MSE, since there the question did not recieve any attention whatsoever.)
I would like to know if there is a description (or at least some sufficient condition known) of (Noetherian) schemes $X$ such that the category $\mathrm{QCoh}_X$ does have exact direct products. Of course, I do not mean affine schemes since for affine schemes, direct products are exact (kind of) trivially.
Seems to me that viable candidates would be projective schemes, since direct products in the category of graded modules are exact, and the well-known functor $M \mapsto \tilde{M}$ is not far from being an equivalence (at least in my mind) of the category of graded modules over a graded ring and the category of quasi-coherent sheaves over the projective scheme. However, I was told this is not true, i.e. that the direct products are not exact even on simple projective schemes.
I would also like to know some example when this fails, i.e. when product of epimorphisms of quasi-coherent sheaves is not an epimorphism (especially if the scheme is projective).
I only stumbled upon the fact that direct products of quasi-coherent sheaves are not exact in general in the introduction to this paper by L. Positselski (first paragraph of the introduction, in fact). However, I cannot find any further information, so any reference would be greatly appreciated.
A reference to some text treating these question I would find most helpful.
Thanks in advance for any help.

REPLY [8 votes]: A counterexample showing that direct products in the category of quasi-coherent sheaves over the projective line $\mathbb P_k^1$ over a field $k$ are not exact functors can be found in the paper "The stable derived category of a Noetherian scheme", by H. Krause, http://arxiv.org/abs/math.AG/0403526 , Example 4.9 (attributed to B. Keller).
The point is that the functor $M\longmapsto\widetilde M$ from the category of graded modules over the projective coordinate ring $R$ to the category of quasi-coherent sheaves on the projective spectrum $X$ establishes an equivalence between the category of quasi-coherent sheaves on $X$ and the quotient category of the category of graded $R$-modules by the Serre/localizing subcategory of torsion modules (i.e., modules in which every element is annihilated by all the elements of high enough degree in $R$).
It is important that the full subcategory of torsion modules is not preserved by direct products in the category of graded $R$-modules (so the functor $M\longmapsto\widetilde M$ cannot possibly preserve direct products).  One proceeds from this observation in order to construct the counterexample in the paper under the link, which likely can be generalized to show that direct products in the category of quasi-coherent sheaves on a projective scheme $X$ over the spectrum of a Noetherian ring $A$ are not exact unless $X$ is finite over $\operatorname{Spec}A$ (and consequently affine), or something close to that.
It would be interesting to know whether direct products in the category of quasi-coherent sheaves over the affine plane without point $X=\mathbb A_k^2\setminus(0,0)$ are exact.  My guess would be that they are not.<|endoftext|>
TITLE: Are all separable algebras Frobenius algebras?
QUESTION [11 upvotes]: Let $\mathcal C$ be a [added later: semi-simple] tensor category, and let $A=(A,m:A\otimes A\to A,i:1\to A)$ be an algebra object in $\mathcal C$.
The algebra is...
Separable if there is an $A$-$A$-bimodule map $\Delta:A\to A\otimes A$ such that $m\circ\Delta=\mathrm{id}_A$
Frobenius if there is an $A$-$A$-bimodule map $\Delta:A\to A\otimes A$ that is coassociative and counital.

I'm wondering whether separable implies Frobenius.

I can show that separable implies coassociative but I suspect that separable does not imply counital. I'm having a hard time finding counterexamples.

PS: by a "tensor category", I mean a category that is monoidal (not necessarily symmetric) and linear over some field $k$.
PS2: In a previous version of this question, I had added the condition that $\mathcal C$ be rigid. I'd be actually more interested to have counterexamples where the ambient category $\mathcal C$ is semi-simple. I understand that there are many things one might mean by "semisimple" (e.g., is the category of all vector spaces over $k$ semisimple?), I'm deliberately keeping things vague to allow for more counterexamples.

REPLY [9 votes]: You should be able to get a (perhaps unnatural) counterexample by constructing a tensor category (or algebra object) with no maps to $1$. Let's try to make the most trivial such counterexample possible.
Let $C$ have two objects $A$ and $1$ with one non-identity morphism $i:1\to A$. Equip this with a structure where $1$ is the unit and $A\otimes A=A$. I believe these make $C$ unambiguously into a tensor category.
There are then unique maps $m:A\otimes A\to A$ and $\Delta:A\to A\otimes A$, both of which are the identity on $A$. Then $m$ and $i$ seem to make $A$ an algebra object in $C$ which trivially satisfies the separability condition. But this can't be counital because there are no maps $A\to 1$.
This is obviously a bit unnatural but the only "natural" tensor categories I know offhand where objects don't map to the unit are ones with disjoint union... But they can't supply any counterexamples because multiplication is forced to be the fold map but then you can't make a bimodule map $A\to A\sqcup A$ unless $A$ is empty.<|endoftext|>
TITLE: Can each non-open analytic subgroup of a Polish abelian group be covered by countably many closed Haar null subsets?
QUESTION [6 upvotes]: By a result of Laczkovich ('Analytic subgroups of the reals' Proc AMS Vol 126 (1998)), any non-open analytic subgroup of a Polish locally compact group can be covered by countably many closed Haar null sets. 
Is this result of Laczkovich true for any Polish (not necessarily locally compact) groups?
More precisely:
Problem 1: Does each non-open analytic subgroup $H$ of a Polish abelian group $G$ belong to the $\sigma$-ideal $\mathcal E$ generated by closed Haar null subsets of $G$?
A Borel subset $A$ of a Polish group $G$ is Haar null of there exists a Borel probability measure $\mu$ on $G$ such that $\mu(xAy)=0$ for all $x,y\in G$. 
A related weaker question also seems to be open.
Problem 2: Does each non-open analytic subgroup $H$ of a Polish abelian group $G$ belong to the $\sigma$-ideal $\mathcal M_h$ generated by closed Haar-meager subsets of $G$?
A Borel subset $A$ of a topological group $G$ is called Haar-meager if there exists a continuous map $f:K\to G$ defined on a compact metrizable space $K$ such that $f^{-1}(xAy)$ is meager in $G$ for all $x,y\in G$. It can be shown that a closed subset $A$ of a topological group $G$ is Haar-meager if and only if there exists a compact subset $K\subset G$ such that for every $x,y\in G$ the intersection $K\cap xAy$ is nowhere dense in $K$. It is easy to see that each closed Haar-null set is Haar-meager. More information on Haar meager sets can be found in the paper "On Haar meager sets" by Darji.

REPLY [2 votes]: The answer to both problems (1 and 2) is negative: the Polish group $G=\mathbb Z^\omega$ contains a dense meager Borel subgroup $H$ (which can be written as the difference $H=A\setminus B$ of two $F_\sigma$-sets in $G$) which cannot be covered by countably many closed Haar-meager subsets of $G$. Such subgroup $H$ is constructed in Example 6.3 of this paper.
The subgroup $H$ also gives a (partial) counterexample to the question Meager subgroups of compact groups .
Indeed, for any countable cover $\{A_n\}_{n\in\omega}$ of the group $H$ the closure $\bar A_n$ of some set $A_n$ is not Haar-meager (and hence not Haar-null) in $G$ and hence $\bar A_n\bar A_n^{-1}$ is a neighborhood of zero, which implies that $A_nA_n^{-1}$ is not nowhere dense in $H$.<|endoftext|>
TITLE: 2-dimensional sublattices with all vectors having very big square (in absolute value)
QUESTION [12 upvotes]: QUESTION: Let $\Lambda\times\Lambda\rightarrow {\Bbb Z}$ be a lattice,
that is, ${\Bbb Z}^n$ with a non-degenerate integer quadratic form, not 
definite, not necessarily unimodular, $n>2$. I want to show that for each $N>0$
there exists a 2-dimensional primitive sublattice $\Lambda_0\subset \Lambda$, 
also not definite, such that all nonzero vectors $x\in \Lambda_0$ satisfy $|(x,x)| > N$.
This question is motivated by considering a K3 surface 
(or a hyperkahler manifold) with 2-dimensional Picard
lattice. It is known that the squares of minimal rational
curves are bounded, and we want to find a manifold with
Picard number 2 having no minimal rational curves.

Bounty added This problem has kept me up late the last two nights, so I'm hoping some of the quadratic form experts will work on it. 
I tried to find a strategy working with rational quadratic forms rather than integers, but failed for the following reason:
Equip $\mathbb{Q}^3$ with the quadratic form $x^2+y^2-z^2$. Then I claim that any rank two subspace $L$ on which this form is non-degenerate contains a vector of norm $1$. 
Proof: Let $L^{\perp} = \mathbb{Q} v$. Since our form is nondegenerate on $L$, we have $\langle v,v \rangle \neq 0$, say $\langle v,v \rangle = N$, and $\mathbb{Q}^3 = L \oplus \mathbb{Q} v$. Now, $x^2+y^2-z^2$ is equivalent to $N (x')^2 + (y')^2 - N (z')^2$, by the change of variables $(x,y,z) = (\tfrac{N+1}{2} x' + \tfrac{N-1}{2} z', y', \tfrac{N-1}{2} x' + \tfrac{N+1}{2} z')$. So our form is equivalent to $L' \oplus \mathbb{Q} v$, where the form on $L'$ is $(y')^2 - N (z')^2$. By Witt cancellation, the forms on $L$ and $L'$ are equivalent. Since $(y')^2 - N (z')^2$ represents $1$, so does our form on $L$. $\square$
So, when we are trying to construct rank two sub-lattices with no vectors of norm $1$, we have to do so using lattices which do represent $1$ rationally. This seems hard to me...

REPLY [3 votes]: For the record: OP Misha Verbitsky writes that
"[$\Lambda$ of] rank $\geq 6$ or $\geq 7$ is a usual assumption
in these kind of applications", in which case 
Ekaterina Amerik's suggestion of using
$N\cdot H$ (for some fixed indefinite rank-$2$ form $H$)
surely yields the easiest construction;
but there are also primitive indefinite forms $\Lambda_0$ of rank $2$ 
that represent no small nonzero integers of either sign,
and those should work even for $\Lambda$ of rank as small as $3$ 
(as in the lattice that troubled the OP in the first place).
One reasonably simple construction is to take $M>N$ odd and set 
$$
Q(x,y) = \frac1M ((x+My)^2 - (M^4+1)x^2)
= -M^3 x^2 + 2xy + M y^2,
$$
which represents $\pm M$ (as for $(x,y)=(0,1)$ and $(M,M^2-1)$)
but no integer of smaller absolute value other than zero.
This can be checked using the fact that the associated
real quadratic field ${\bf Q}(\sqrt{M^4+1})$ has fundamental unit
$M^2 + \sqrt{M^4+1}$ (of norm $-1$) as small as possible.
Moreover, unlike the forms $N \cdot H$, this $Q$ is primitive,
so any embedding of the into $\Lambda$ of the indefinite lattice 
$\Lambda_0$ associated to $Q$ is automatically primitive as well.<|endoftext|>
TITLE: Derivatives of theta functions at zero
QUESTION [6 upvotes]: Let $L$ be a line bundle over complex elliptic curve, $\deg L = k>0$. Theta functions 
$$
\theta_s(z;\tau)_k=\sum_{r\in \mathbb{Z}} e^{\pi i [(\frac{s}{k} + r)^2 k \tau + 2kz(\frac{s}{k}+r)]}, \hspace{20pt} s=0,...,k-1$$
form a basis of $H^0(E, L)$. I'm especially interested in the case $k=3$ when theta functions embed the elliptic curve as Hesse cubic in $\mathbb{P}^2$. 
I have the following basic question: are there formulas for derivatives of such theta functions at zero $\left.\frac{d\theta_s(z,\tau)_3}{dz}\right|_{z=0}$?
The closest result I know is Jacobi formula for Jacobi theta functions
$$
\theta'_{\frac{1}{2} \frac{1}{2}}=-\pi \theta_{0 0} \theta _{\frac{1}{2} 0} \theta _{0 \frac{1}{2}},
$$
but (as far as I understand) it corresponds to the case $k=4$, so it does not answer my question.

REPLY [6 votes]: An eta-product identity for $k=3$, $s=1$ similar to that given in my comment above may be found in "Some eta-identities arising from theta series" by Günter Köhler (Math. Scand. 66, 1990, p. 146, identity (3)):
$$
\frac{\eta^2(\tau)\eta^2(4\tau)}{\eta(2\tau)}=\sum_{n=1}^\infty\left(\frac n3\right)ne^\frac{2\pi in^2\tau}3,
$$
which is equivalent to
$$
\prod_{n=1}^\infty\frac{(1-q^n)^2(1-q^{4n})^2}{1-q^{2n}}=\sum_{n=1}^\infty\left(\frac n3\right)nq^{\frac{n^2-1}3}=1-2q+4q^5-5q^8+7q^{16}-8q^{21}+...
$$
The rhs is the same as $q^{-\frac13}\sum_{n\in\mathbb Z}(3n+1)q^{\frac13{(3n+1)^2}}$, i. e., up to rescalings, $\frac\partial{\partial z}$ of your $\theta_1(z;\tau)_3$ at $z=0$.
I don't know whether such identities for other $k$ are systematically treated anywhere. Köhler also has a book "Eta Products and Theta Series Identities" but I have never looked inside. From "About this book":

The author brings to the public the large number of identities that have been discovered over the past 20 years, the majority of which have not been published elsewhere.

(Second part)
This might be a separate answer but I decided to just add it here. In "Eta-quotients and theta functions" by Robert J. Lemke Oliver (free online version available), a classification of all those theta-functions of the form $\theta_\psi(z)=\sum_n\psi(n)n^\delta q^{n^2}$ which admit an eta-product decomposition is given. Here $\psi$ is a Dirichlet character, and $\delta=0$ or $1$ according to the parity of $\psi$. There turns out to be very few of this kind - just eight in the even case and five in the odd case. The case of your $k=3$, $s=1$ is the third identity of Theorem 1.1 (2) in that paper; I am not reproducing it here as it is essentially the same as the Köhler's case above. If one allows some twists of the characters, there are a little bit more such products.<|endoftext|>
TITLE: What is the maximum size of a set system where the intersection of any two incomparable members is not in the set?
QUESTION [12 upvotes]: Let the set $\mathcal{F}$ consist of subsets of $[n]$.  Suppose that for any incomparable $A$ and $B$ in $\mathcal{F}$, we have $A \cap B \notin \mathcal{F}$.  What is the largest possible size of $\mathcal{F}$?

REPLY [4 votes]: It was proved by Kleitman that $|\mathcal F|\le {n\choose n/2}+2^n/n$, see http://www.sciencedirect.com/science/article/pii/0097316576900376.<|endoftext|>
TITLE: The coupon collector's earworm
QUESTION [36 upvotes]: [EDITED mostly to report on the answer by Kevin Costello
(and to improve the gp code at the end)]
I thank Nicolas Dupont for the following question
(and for permission to disseminate it further):

I have a playlist with, say, $N$ pieces of music.
  While using the shuffle option (each such piece is played randomly
  at each step), I realized that, generally speaking, I have to hear
  quite a lot of times the same piece before the last one appears.
  It makes me think of the following question:
At the moment the last non already heard piece is played, what is the
  max, in average, of number of times the same piece has already been played?

I have not previously enountered this variant of the
coupon
collector's problem.  If it is new, then (thinking of the original
real-world context origin of the problem) I propose to call it the "coupon
collector's earworm".

Is this in fact a new question?
  If not, what is known about it already?

Let us call this expected value $M_N$.  Nicolas observes that $M_1=1$
and $M_2=2$ (see below), and asks:

I doubt there is such an easy formula for general $N$ -
  would you be able to find some information on it,
  e.g. its asymptotic behavior?

[Nicely answered by Kevin Costello:
$M_N$ is asymptotic to $e \log N$ as $N \rightarrow \infty$.
Moreover, the maximal multiplicity is within $o(\log N)$ of $e \log N$
with probability $\rightarrow 1$.   I don't recall any other instance
of a naturally-arising asymptotic growth rate of $e \log N$...]
Recall that the standard coupon collector's problem asks for the expected value
of the total number of shuffles until each piece has appeared at least once.
It is well known that the answer is $N H_N$ where $H_N = \sum_{i=1}^N 1/i$
is the $N$-th harmonic number.  Hence the expected value of the average 
number of plays per track is $H_N$, which grows as $\log N + O(1)$.
The expected maximum value $M_N$ must be at least as large $-$ in fact
larger once $N>1$, because one track is heard only once, so the average
over the others is $(N H_N-1) / (N-1)$.  One might guess that $M_N$
is not that much larger, because typically it's only the last few tracks
that take most of shuffles to appear.  But it doesn't seem that easy to get
at the difference between the expected maximum and the expected average,
even asymptotically.  Unlike the standard coupon collector's problem,
here an exact answer seems hopeless once $N$ is at all large (see below),
so I ask:

How does the difference $M_N - H_N$ behave
  asymptotically as $N \rightarrow \infty$?

[Looks like this was a red herring: "One might guess" plausibly that
$M_N - H_N$ is dwarfed by $H_N$ for large $N$, but by Kevin Costello's
answer $M_N - H_N$ asymptotically exceeds $H_N$ by a factor $e - 1$,
and that factor is more complicated than
$\lim_{N\rightarrow\infty} M_N/H_N = e$,
so analyzing the difference $M_N-H_N$ is likely not a fruitful approach.]
Here are the few other things I know about this:
@ For each $N>1$, the expected value of the maximum count
is given by the convergent $(N-1)$-fold sum
$$
M_N = \sum_{a_1,\ldots,a_{N-1} \geq 1}
  N^{-\!\sum_i \! a_i} \Bigl(\sum_i a_i\Bigr)! \frac{\max_i a_i}{\prod_i a_i!}.
$$
Indeed, we may assume without loss of generality that the $N$-th track
is heard last; conditional on this assumption, the probability that
the $i$-th track will be heard $a_i$ times for each $i<N$ is
$N^{-\!\sum_i \! a_i}$ times the multinomial coefficient
$(\sum_i a_i)! \big/ \prod_i a_i!$.  Numerically, these values are
$$
2.00000, \quad
2.84610+, \quad
3.49914-, \quad
4.02595\!-
$$
for $N=2,3,4,5$.
@ A closed form for $M_N$ is available for $N \leq 3$ and probably
not beyond.  Trivially $M_1 = 1$; and N.Dupont already obtained
the value $M_2 = 2$ by evaluating $M_2 = \sum_{a \geq 1} a/2^a$.
But for $N=1$ and $N=2$ the problem reduces to the classical
coupon collector's problem.  Already for $N=3$ we have a surprise:
$M_3 = 3/2 + (3/\sqrt{5})$, which has an elementary but somewhat tricky proof.
For $N=4$, I get
$$
M_4 = \frac73 - \sqrt{3} + \frac4\pi \int_{x_0}^\infty
  \frac{(2x+1) \, dx}{(x-1) \sqrt{4x^3-(4x-1)^2}}
$$
where $x_0 = 3.43968\!+$ is the largest of the roots (all real) of
the cubic $4x^3-(4x-1)^2$.  I don't expect this to simplify further:
the integral is the period over an elliptic curve of a differential with two
simple poles that do not differ by a torsion point.  In general
one can reduce the $(N-1)$-fold sum to an $(N-2)$-fold one
(which is one route to the value of $M_3$ and $M_4$), or to an
$(N-3)$-fold integral, but probably not beyond.
@ It's not too hard to simulate this process even for considerably larger $N$.
In GP one can get a single sample of the distribution from the code
try(N) = v=vector(N); while(!vecmin(v),v[random(N)+1]++); vecmax(v)

[turns out that one doesn't need to call vecmin each turn:
try(N, m,i)= v=vector(N); m=N; while(m, i=random(N)+1; v[i]++; if(v[i]==1,m--)); vecmax(v)

does the same thing in $\rho+O(1)$ operations per shuffle rather then
$\rho+O(N)$, where $\rho$ is the cost of one random(N) call.]
So for example
sum(n=1,10^4,try(100)) / 10000.

averages 1000 samples for $M_{100}$; this takes a few seconds, and
seems to give about $11.7$.

REPLY [22 votes]: For the asymptotic case: Let $t_1=n \log n - Cn$ and $t_2 = n \log n + Cn$, where $C$ is slowly tending to infinity.  It is a classic result that as $C$ tends to infinity the probability all coupons are collected at time $t_1$ tends to $0$, and the probability all coupons are collected at time $t_2$ tends to $1$.  
So the number being asked for in the original post is with high probability sandwiched between the most collected coupon at time $t_1$ and the most collected coupon at time $t_2$.  This has been studied a fair amount, though often in language of load-balancing and balls-in-bins ("if you toss $m$ balls in $n$ bins, what is the typical number of balls in the bin with the most balls").  In particular, there's a nice analysis of Raab and Steger that uses the second moment method to give a tight concentration on this.  It follows from Theorem $1$ in their paper that at time $c n \log n$, the most common coupon has almost surely been collected $(d_c+o(1)) \log n$ times, where $d_c$ is the larger real root of 
$$1+x(\log c - \log x +1)-c=0$$
In our case, we have $d_1=e$, so the most common song will have been heard $(e+o(1))\log n = (e+o(1))H_n$ times.  

In response to the comment: The first moment part of the calculation also comes with come concentration estimates for free.  At time $t_1$, the probability that we've seen some coupon at least $C \log n$ times can be bounded by 
\begin{eqnarray*}
& & \sum_{j=C \log n}^{t_1} n \binom{t_1}{j} n^{-j} (1-\frac{1}{n})^{t_1-j} \\
&\leq& \sum_{j=C \log n}^{t_1} n \left(\frac{e t_1}{ nj}\right)^j (1-\frac{1}{n})^{n \log n + o(\log n)} (1-\frac{1}{n})^{-j} \\
&=& n^{o(1)} \sum_{j=C \log n}^{t_1} \left(\frac{e \log n(1+o(1))}{j}\right)^j \\
&\leq& n^{o(1)} \sum_{j=C \log n}^{t_1} \left(\frac{e}{C}+o(1)\right)^j
\end{eqnarray*} 
Which should decay rapidly enough in $C$ to guarantee the mean lies close to the concentration.  
The linked bounds for the probability of the completion time lying outside $[t_1, t_2]$ also decay quite rapidly (e.g. cardinal's answer, and David's subsequent comments on it).<|endoftext|>
TITLE: An extension of $K$-theory to topological $^*$-algebras
QUESTION [7 upvotes]: What I have in mind is the following:  a (sequence of) functor(s) $K_\bullet$ on the category of topological $^*$-algebras (with values in the category of commutative groups) that satisfies (among other properties one would wish a $K$-theory functor to possess) (i) $K_\bullet (A)=K_\bullet ^{\text{op}}(A)$ when $A$ is a $C^*$-algebra and (ii) $K_\bullet (A)=K_\bullet ^{\text{alg}}(A)$ when $A$ has involution the identity and the discrete topology.
As a matter of fact, a quick skim of the construction of $K_0^{\text{op}}(A)$ (as the grothendieck group of unitary equivalence classes of projections in $M_\infty (A)$) seems as if would work verbatim for an arbitrary topological $^*$-algebra, and furthermore, would give the same definition in the case of involution the identity and the discrete topology (this uses the fact that if self-adjoint elements are conjugate, then they are in fact unitarily equivalent).  (I suppose one possible issue with extending this to the higher $K$-groups in the most obvious way is determining the 'right' topology to put on the tensor product of topological $^*$-algebras in order to define the suspension.)
Does there exist such an extension?  If not, why not?  If so, is there a reference?

REPLY [5 votes]: Attempts to construct a unifying theory for arbitrary complete metric algebras were undertaken as early as in 70's by Karoubi and Villamayor, You may check this http://webusers.imj-prg.fr/~max.karoubi/Publications/11.pdf for the original paper, and the standard reference is the paper of Cortinas http://arxiv.org/abs/0903.3983 . The major issue one faces when constructing such theories is whether You
a) want them to be homotopy stable, and if yes, what kind of homotopy do You choose;
b) want them to be homology theories.
Since You mention $C^\ast$-algebras - let's even better stick to Banach algebras as it was suggested by Branimir Cacic above - as a theory standing on one side, we may assume the theory in question is supposed to be homotopy-stable. Here, You face two problems on both discrete and Banach sides. (BTW, $K$-theory for Banach algebras was defined by Vincent Lafforgue in http://www.researchgate.net/publication/267481248_Bivariant_K-theory_for_Banach_algebras_and_the_Baum-Connes_conjecture )
From the discrete side, the problem is that $K_\bullet$ is not homotopy-invariant. This may be cured by introducing polynomial homotopy, as did Karoubi and Villamayor. However, although they can construct all positive K-Theory groups, you cannon have negative ones such that they pass into a long exact sequence. This is cured by Weibel, who constructs his homotopy K-Theory $KH$. The problem with it is that in general neither $KH_0$ coincides with $K_0$, nor $KH_1$ with $KV_1$ (i.e. the Karoubi-Villamayor group), and that was one of the reasons Weibel did not publish his theory for about a decade.. But at least for $KH$ you may have definitions via spectra, as $KH_n(R)=\pi_n(BGl_+(R^\Delta))$, where $R^\Delta$ is a simplicial algebra based on the simplices $R^{\Delta^n}:=R[t_0,\dots,t_n]/<1-t_0-\dots-t_n>$. And $KH$ is perfectly a homology theory. 
However, if you go up to the the Banach case, another problem occurs: what is the general notion of homotopy. Karoubi and Villamayor propose the homotopy given by convergent formal power series. However, the elementary homotopies provided by convergent power series are not symmetric. You could of course symmetrize them, but still you cannot have a good definition of $A^\Delta$ like Weibel, and with that a definition via spectra. In the case when your ground ring is, say, $\mathbb{C}$, the notion of convergent series homotopy may be proved to coincide with the usual homotopy of Banach algebras, yet it's tricky to generalize this for arbitrary metric rings. It's doable with another notion of homotopy, something like probability measures taking values in the ground ring (shameless self-advertisement here), but it is not known (at least to me) whether this other notion of homotopy may yield any kind of interesting K-Theory (although you could do spectra once again).
By the way, all the problems become much more simple if You're doing ultrametric Banach algebras. A Karoubi-Villamayor-type theory is now being developed by Bunke and Tamme, see e.g. http://arxiv.org/pdf/1111.4109v4.pdf .<|endoftext|>
TITLE: Relationship between the syntomic cohomology of Kato and of Fontaine-Messing
QUESTION [10 upvotes]: Fix a prime $p$ and let $X$ be a $\mathbb{Z}_{p}$-scheme. Write $X_{n}:=X\otimes\mathbb{Z}/p^{n}$ and $\phi:X_{1}\rightarrow X_{1}$ for the absolute Frobenius. Let $X\hookrightarrow Z$ be a (suitable) closed immersion. Let $D_{n}$ be the PD-envelope of $X_{n}$ in $Z_{n}$ and set $J_{D_{n}}=\ker(\mathcal{O}_{D_{n}}\rightarrow\mathcal{O}_{X_{n}})$. Then for $0\leq r\leq p-1$ the syntomic complex $\mathscr{S}_{n}(r)_{X,Z}$ is the complex of étale sheaves on $X_{1}$ given by the mapping fibre of 
\begin{equation*}
1-\frac{\phi}{p^{r}}:\mathbb{J}^{[r]}_{n,X,Z}\rightarrow\mathbb{J}^{[0]}_{n,X,Z}
\end{equation*}
where $\mathbb{J}^{[r]}_{n,X,Z}$ is is the complex
\begin{equation*}
J_{D_{n}}^{[r]}\xrightarrow{d}J_{D_{n}}^{[r-1]}\otimes_{\mathcal{O}_{Z_{n}}}\Omega_{Z_{n}}^{1}\xrightarrow{d}\cdots
\end{equation*} 
As Kato explains in [1], the image of $\mathscr{S}_{n}(r)_{X,Z}$ in the derived category is independent of $Z$. Kato then defines
\begin{equation*}
H_{\text{syn}}^{i}(X,\mathscr{S}_{n}(r)):=\mathbb{H}_{\text{ét}}^{i}(X,\mathscr{S}_{n}(r)_{X,Z})
\end{equation*}
In the same paper, Kato says that this computes the syntomic cohomology defined by Fontaine-Messing using the syntomic site, but says that the details are given in [2]. But here I can only find Remark (1.1) which is the same statement, again without proof. 
Does anybody have a reference that explains how these two cohomologies are the same? Or perhaps somebody can sketch how this works? 
[1] The Explicit Reciprocity Law and the Cohomology of Fontaine-Messing, Bull. Soc. math. France,
119, 1991, p. 397–441.
[2] On $p$-adic Vanishing Cycles (Application of ideas of Fontaine-Messing), Adv. Studies in Pure Math. 10, 1987, pp207-251. 

Further remark:- Niziol has many (wonderful) papers studying generalisations of Kato's mapping fibre approach.  I haven't been able to find an answer in any of these articles either, but in Niziol's 2006 ICM article $p$-adic motivic cohomology in arithmetic, there is a remark on page 7 claiming that the Fontaine-Messing construction is "philosophically" the same as the Kato construction. This only confuses me further...

REPLY [5 votes]: Ok, maybe I've figured this out. Hopefully somebody can correct me if this is wrong. Also, I'd still like to know a reference that writes this out in detail, if anybody has one.
I'll change the notation from the question slightly: $k$ is a perfect field of characteristic $p>0$, and $W=W(k)$. First let's recall the definition of Fontaine-Messing's syntomic sheaf:- we are working on the site $(\text{Spf }W)_{\text{NILSYN}}$. This is the site whose objects are $W$-schemes on which $p$ is locally nilpotent, and the topology is the syntomic topology. For each $n\in\mathbb{N}$, we have some sheaves on this site:-
\begin{equation*}
\mathcal{O}_{n}:X\mapsto\mathcal{O}_{X_{n}}(X_{n})
\\
\mathcal{O}_{n}^{\text{cris}}:X\mapsto H_{\text{cris}}^{0}(X_{1}/W_{n})
\\
J_{n}:=\ker\left(\mathcal{O}_{n}^{\text{cris}}\rightarrow\mathcal{O}_{n}\right)
\end{equation*} 
Define $\tilde{J}_{n}^{[r]}:=\text{im}\left(J_{n+r}^{[r]}\rightarrow J_{n}^{[r]}\right)$. For $r<p$, Fontaine and Messing (2.3 on page 191) show that $\phi(\tilde{J}_{n}^{[r]})\subset p^{r}\mathcal{O}_{n}^{\text{cris}}$ where $\phi:\mathcal{O}_{n}^{\text{cris}}\rightarrow \mathcal{O}_{n}^{\text{cris}}$ is the Frobenius. So we have a map $\frac{\phi}{p^{r}}:\tilde{J}_{n}^{[r]}\rightarrow\mathcal{O}_{n}^{\text{cris}}$. The syntomic sheaf is then defined to be
\begin{equation*}
s_{n}(r)_{X}:=\ker\left(1-\frac{\phi}{p^{r}}:\tilde{J}_{n}^{[r]}\rightarrow\mathcal{O}_{n}^{\text{cris}}\right).
\end{equation*}
Now, for $X$ a $W$-scheme, let $u:X_{n,\text{SYN}}\rightarrow X_{n,\text{ET}}$ be the morphism of sites. The claim is that we have an isomorphism
\begin{equation*}
Ru_{\ast}s_{n}(r)_{X}\simeq\mathscr{S}_{n}(r)_{X}[-1]
\end{equation*} 
By Remark 1.8 on page 213 of Kato's ``On p-adic Vanishing Cycles (Application of ideas of Fontaine-Messing)'', it suffices to prove the claim when $X$ is quasi-projective. But then we can find a closed immersion $X\hookrightarrow Z$ into a smooth $W$-scheme endowed with a Frobenius (e.g. $Z$ could be projective space). In the notation of the question, we are writing $D_{n}=D_{X_{n}}(Z_{n})$ for the PD-envelope of $X_{n}$ in $Z_{n}$ (wrt the PD-structure on $pW_{n}$). In this notation then, Theorem 7.2 of Berthelot-Ogus gives an isomorphism
\begin{equation*}
\mathbb{R}u_{\ast}J_{n}^{[r]}\simeq\mathbb{J}_{X_{n},Z_{n}}^{[r]}
\end{equation*}
(I think we also need the result of this question
Crystalline cohomology via the syntomic site
to make this part legit).
Now we simply use the definition of $s_{n}(r)_{X}$ as $\ker\left(1-\frac{\phi}{p^{r}}\right)$ and the triangle associated to mapping fibres to see the claim. (Note, this is where the shift by $-1$ comes in).<|endoftext|>
TITLE: Geodesics on SO(3)
QUESTION [11 upvotes]: I have two 3D rotations about the origin, represented as
$3 \times 3$ orthogonal matrices $M_1$ and $M_2$
(specified by numerical entries),
and I would like to interpolate (and compute)
a continuous sequence of rotations
between $M_1$ and $M_2$.
Ideally this interpolation would follow a geodesic on SO($3$).
Assuming this is well-known, 
(Q1): I would appreciate a pointer to geodesics on SO($3$),
especially suited to my computational task.
(Q2) [added]. I wonder if the geodesics identified by 
Benjamin,
Paul Siegel,
and Francois Ziegler,
which effectively answer Q1,
can be viewed as equivalent to following a great-circle arc on $S^3$ between
unit quaternions representing the two rotations?
Update (25Jan2019): A paper was just published on this
topic: 

Novelia, Alyssa, and Oliver M. O’Reilly. "On geodesics of the rotation group $SO(3)$." Regular and Chaotic Dynamics 20, no. 6 (2015): 729-738. Springer link.

Update (26Jan2019): The paper above raises the interesting question
whether saccadic motions of the eye follow $SO(3)$ geodesics, 
an issue possibly dating back
to Helmholtz's investigations in the late $19^{\mathrm{th}}$ century:

von Helmholtz, H., "Über die normalen Bewegungen des menschlichen Auges," Archiv für Ophthalmologie, 1863, Vol. 9, No. 2, pp. 153–214; DOI: 10.1007/BF02720895.

REPLY [9 votes]: If you are assuming the use of the bi-invariant metric, then the geodesics are right/left translations of the one parameter sub-groups $O(t)=O(0) \exp(tG)$ where $G \in \mathfrak{so}(3)$ (i.e. it is anti-symmetric and traceless) and $O(0)\in SO(3)$.
The curve you want will have $O(0)=M_1$ and thus be of the form $O(t) = M_1 \exp(tG)$ with $G=\log(M_1^{-1}M_2)$ (the matrix log). You need to choose a specific matrix log as generically there will be many, you many wish to read about principal matrix logs if you are going to do any numerical computations. This is what will likely come out of Matlab etc. The curve is parameterised by $t \in [0,1]$, it is smooth, is a geodesic of the aforementioned metric and has the end points you hoped for.

REPLY [9 votes]: The geodesics (of the unique connection invariant under left and right translations and inversion) are the translates of one-parameter subgroups: see Helgason, exerc. 6(iii), p. 148. So given your $M_1$ and $M_2$, find (by surjectiveness of $\exp$) a $Z\in\mathfrak{so}(3)$ such that $\exp(Z)=M_1^{-1}M_2$, and put $g(t) = M_1\exp(tZ)$.<|endoftext|>
TITLE: Kernel of the natural map between group $C^*$-algebras
QUESTION [5 upvotes]: Let $\Gamma$ be a discrete group. We can form two $C^*$-algebras: the universal (or full) and reduced, to be denoted by $C^*_u(\Gamma)$ and $C^*_r(\Gamma)$ (respectively). Both of them are completions of the group algebra $\mathbb{C}\Gamma$ but with respect to different norms: universal norm is defined as the supremum of $\| \pi(\cdot) \|$ over all $*$-representations of $\mathbb{C}\Gamma$: the reduced norm is the ordinary operator norm where $\Gamma$ acts on $\ell^2(\Gamma)$ by the left regular representation (and then we extend this action linearly).
One can consider the identity mapping $id:(\mathbb{C}\Gamma,\| \cdot \|_u) \to (\mathbb{C}\Gamma,\| \cdot \|_r)$ and extend it to the whole $C^*_u(\Gamma)$ (call $\theta$ this extension). Since this is morphism beetween $C^*$-algebras its range is closed but it this dense. Therefore it is surjective: but there is no reason for this map to be injective (it is injective iff it is isomorphism iff the group is amenable). 

What is the kernel of $\theta$?

My guess is that its kernel should be the set $\{x \in C^*_u(\Gamma): \tau(x^*x)=0\}$ where $\tau$ is the canonical trace on $C^*_u(\Gamma)$ defined on $\mathbb{C}\Gamma$ by $\tau(x)=\langle x \delta_e,\delta_e \rangle$ where $e$ denotes the neutral element.
EDIT: I corrected the typo, as pointed in the comment below.

REPLY [2 votes]: (Caleb Eckhardt already answered, but this was too long for a comment)
This is indeed true for general $C^*$-algebra but use the fact that $\tau$ is a trace and not just a state.
An element of the maximal algebra is zero in the reduced algebra if it acts trivially on the regular representation, which is the GNS represenation assciated to the canonical trace $\tau$.
An element $h$ is in the kernel if for any $v$ in the regular representation one has $\Vert hv \Vert =0$. Up to an approximation, the element $v$ can be written $t \delta_0$ for $h$ in the $t$ in the maximal algebra (those are dense in the regular representation). Hence $h$ is in the kernel if and only if for all $t$: $0=\Vert ht \delta_0 \Vert^2 = \langle ht \delta_0| ht \delta_0 \rangle = \langle \delta_0 | t^* h^* h t \delta_0 \rangle = \tau( t^* h^* h t)$
In the general framework of a GNS representation this is the best you can obtain: $h$ is in the kernel if and only if for all $t$, $\tau( t^* h^* h t)=0$.
But because $\tau$ is not just a state but a trace one can improve:
$\tau( t^* h^* h t) = \tau(h^* h t t^*) = \tau(h tt^* h^*) \leqslant \Vert t \Vert^2 \tau(hh^*) = \Vert t \Vert^2 \tau(h^* h )$
(indeed $tt^* \leqslant \Vert t \Vert^2$ hence $htt^* h^* \leqslant \Vert t \Vert^2 hh^*$, because if $x$ is positive then $h x h^*$ is also positive)
Hence $h$ is in the kernel if and only if $\tau(h^* h ) =0$.<|endoftext|>
TITLE: Is an irreducible ideal in $R$ also irreducible in $R[x]$?
QUESTION [12 upvotes]: Let $R$ be a commutative Noetherian ring and $I\subset R$ an ideal that is irreducible in the sense that if $I = J_1 \cap J_2$, then $I=J_1$ or $I=J_2$. Is (the ideal generated by) $I$ irreducible in the polynomial ring $R[x]$?
The answer to the question is "yes" since the number of irreducible components has a homological characterization which can be found in Computational methods in commutative algebra and algebraic geometry by W. V. Vasconcelos.
Proposition 3.1.7. Let $(R,\mathfrak{m})$ be a local commutative Noetherian ring. Let $\mathfrak{p}$ be an associated prime of an $R$-module $M$ and denote $\Delta_\mathfrak{p}(M)$ the submodule of $M$ whose elements are annihilated by $\mathfrak{p}$. The number of irreducible $\mathfrak{p}$-primary components in an irredundant irreducible decomposition of $0\subset M$ is $\dim_{k(\mathfrak{p})}(\Delta_\mathfrak{p}(M))_\mathfrak{p}$.
The minimal number of irreducible intersectands of $I$ equals the $R_{P}/P_P$-vector space dimension of the socle of $R[x]_{P}/I_{P}$. Now when adjoining indeterminates the field $R[x]_P/PR[x]_P$ grows, but the socle dimension is the same.
Now the real question is: Is there a simpler proof (without localization(?)) or do we have to use homological invariants of ideal decompositions?
Background: This comes from Exercise 3.6 in Eisenbud's book on commutative algebra which asks for a characterization of irreducible monomial ideals.  I'm wondering if a reader who has only read Chapters 1,2,3 would be able to do that exercise.  This is a repost of my math.se question which can't be migrated because it is too old.

REPLY [9 votes]: I prove your question for (not necessarily Noetherian) commutative ring. Irreducible ideals in non-Noetherian ring are complicate (see this question). For Noetherian ring, see our paper for a study of the index of reducibility. The proof is elementary but quite long.
Edit: I realise that the proof also work for another question. If$(0)$ is an irreducible ideal of $R$ then it is a graded irreducible ideal of the graded ring $R[X]$. 
We can assume that $I = 0$. Suppose $ 0 \in R[X]$ is reducible as the intersection of two proper ideals. We can assume these two ideals are principal, so $0 = (f) \cap (g)$ with
$$f = X^r (a_0 + a_1X + \cdots + a_nX^n), a_0 \neq 0,$$
$$g = X^s (b_0 + b_1X + \cdots + b_mX^m), b_0 \neq 0.$$
Choose $f$ and $g$ so that $m+n$ is minimal.
Fact: Let $f = X^rf'$ and $g = X^sg'$. Then $(f) \cap (g) = 0$ if and only if $(f') \cap (g') = 0$.
Proof. The "if part" is clear since $(f) \subseteq (f')$ and $(g) \subseteq (g')$.
For the "only if" part, suppose $(f') \cap (g') \neq 0$. We have $(X^{r+s} f') \cap (X^{r+s}g') = X^{r+s}((f') \cap (g')) \neq 0$. Thus $(f)\cap (g) \neq 0$ since $(X^{r+s}f') \subseteq (f)$ and $(X^{r+s}g') \subseteq (g)$. 
Using the above fact we can assume 
$$f = a_0 + a_1X + \cdots + a_nX^n, a_0 \neq 0$$
$$g = b_0 + b_1X + \cdots + b_mX^m, b_0 \neq 0.$$
If $m+n = 0$ then $f, g \in R$ and this contradicts the assumption that $(0)$ is irreducible there. 
So $m+n> 0$.  Without loss of generality we assume that $m \ge n$ and thus $m>0$. Choose $0 \neq c \in (a_0) \cap (b_0)$ in $R$.  Then $c = da_0 = eb_0$ for some $d, e \in R$. Replacing $f$ and $g$ by $df$ and $eg$, respectively, we can assume henceforth that $a_0 = b_0$.
By the minimality of $m+n$ we have the following.
Claim 1: Let $r$ be an element of $R$ such that $ra_0 = 0$. Then $rf = 0$ and $rg = 0$.
Proof. Assume $ra_0 = 0$ and $rf \neq 0$. Since $(f) \cap (g) = 0$ also $(rf) \cap (g) = 0$. However, since $ra_0 = 0$, $rf = X^tf'$ with $\mathrm{deg}(f')<n$. By the Fact, $(f') \cap (g) = 0$ in contradiction to  minimality of $m+n$.  The same argument applies to $g$.
Applying Claim 1 inductively one can prove:
Claim 2: If a polynomial $h = c_0 + c_1X + \cdots + c_k X^k$ satisfies $hf = 0$ (resp. $hg = 0$), then each coefficient $c_i$ satisfies $c_if = 0$ (resp. $c_ig = 0$).
Combining Claims 1 and 2 we have
Claim 3: $hf = 0$ if and only if $hg = 0$.
We continue the proof. Let $g' = g-f$. Since $a_0 = b_0$, we get $g' = X^{m-m'}g' '$ for some $m' < m$ and polynomial $g' '$ of degree $m'$.
By minimality of $m+n$ and the Fact, we have $(f) \cap (g' ') \neq 0$. Thus there are polynomials $u,v,w$ such that $0 \neq w = uf = vg' '$, and thus $X^{m-m'}uf = vg' = v(g-f)$.  Now, if $vg \neq 0$ then $vg = (u+v)f \in (f) \cap (g)$, a contradiction. Therefore $vg = 0$. By Claim 3 we have $vf = 0$ so $w = 0$. This is also a contradiction. The proof is complete.<|endoftext|>
TITLE: A "universally non Hypercomplete" $\infty$-topos via Goodwillie calculus?
QUESTION [18 upvotes]: My question is :
Is there a classifying $\infty$-topos for $\infty$-connected objects ? Does this $\infty$-topos has a nice description (as an $\infty$-category ) ?
What I mean by $\infty$-connected object is an object $X$ such that all the $\pi_n(X)$ (seen as truncaturated objects of the $\infty$-topos) are isomorphic to the terminal object.
And by classifying $\infty$-topos I mean that for any other $\infty$-topos $\mathcal{E}$ there is a (natural) equivalence of $\infty$-category between $\infty$-connected object in $\mathcal{E}$ and geometric morphisms from $\mathcal{E}$ to our topos $\mathcal{T}$, this (inverse of this) equivalence being induced by $f \mapsto f^* (X)$ where $X$ is a given ("universal") $\infty$-connected object of $\mathcal{T}$.
I am relatively convinced that such an $\infty$-topos exists, so I'm more interested in knowing if it has an interesting description (something simpler than than a localization of the category of simplicial presheaves over the category of finite simplicial set would be great ! ).
This topos should have a unique point which should be its hypercompletion, I know one $\infty$-topos with this property (mentioned for example in this answer ). I guess it would be way too beautiful if this was the answer, but there is at least be a geometric morphism between them...

REPLY [12 votes]: As discussed in the comments, I'm writing here the proof of the following fact:

Let $\mathscr{X}_∞$ be the ∞-topos of sheaves on $\mathrm{FinTop}^{op}$ under the atomic topology (the topology where a sieve is covering iff it is nonempty). Then for any other ∞-topos $\mathscr{X}$, there is a natural equivalence between the ∞-category of geometric morphisms $\mathscr{X}\to \mathscr{X}_∞$ and the ∞-category of ∞-connective objects in $\mathscr{X}$.

To do so, we will need a lemma.
Lemma: Let $\mathscr{X}$ be an ∞-topos and let $X\in\mathscr{X}$ be an object. Then all the following statements are equivalent

The object $X$ is ∞-connected. That is $\pi_nX\to X$ is an equivalence for each $n\ge 0$ and $X\to *$ is an effective epimorphism.
For every $n\ge -1$ the diagonal map $X\to X^{S^n}$ is an effective epimorphism (where $S^{-1}=\varnothing$)
For every finite space $T$ the diagonal map $X\to X^T$ is an effective epimorphism.
For every nonempty collection of maps of finite spaces $\{T\to T'_i\}_{i\in I}$ the map $\coprod_{i\in I} X^{T_i'}\to X^T$ is an effective epimorphism.

The lemma immediately implies the fact we were looking for thanks to proposition 6.2.3.20 in Higher Topos Theory. In fact our condition 4 is exactly the condition that the left exact functor $\mathrm{FinTop}^{op}\to \mathscr{X}$
$$T\mapsto X^T$$
sends covering families to effective epimorphisms.
Proof: $4 \Rightarrow 3\Rightarrow 2$ is obvious. Let us prove $1\Leftrightarrow 2$. We know that $\pi_nX$ is the 0-truncation of $X^{S^n}\to X$ given by the evaluation at the basepoint. So $\pi_nX\to X$ is an equivalence iff $X^{S^n}\to X$ is a 0-connected. But by HTT.6.5.1.20, this is true iff the diagonal map $X\to X^{S^n}$ (which is a section of the evaluation) is -1-connected, i.e an effective epimorphism. 
In the following we will often use HTT.6.2.3.12, that is if $gf$ is an effective epimorphism, so is $g$.
$2\Rightarrow 3$ This follows from an induction on the number of cells of $T$. Let us assume that it is true for $T$ and we will show it is true for $T'=T\amalg_{S^{n-1}}D^n$. There is a cofiber sequence
$$T\to T'\to T'/T=S^n$$
and so we obtain a pullback square
$$\require{AMScd}
\begin{CD} 
X^{S^n} @>>> X^{T'}\\
 @VVV  @VVV\\
 X @>>> X^T
\end{CD}
$$
Since effective epimorphisms are stable under pullbacks (HTT.6.2.3.15), it follows that $X^{S^n}\to X^{T'}$ is an effective epimorphism. Finally, since $X\to X^{S^n}$ is an effective epimorphism by hypothesis and by HTT.6.2.3.12 effective epimorphisms are stable under composition, we are concluded.
$3 \Rightarrow 4$ This follows from HTT.6.2.3.12 applied to the compositions
$$ X\to X^{T_j'}\to \amalg_{i\in I}X^{T_i'}\to X^T$$
for some $j\in I$. $\square$

REPLY [9 votes]: To summarize the discussion in the comments:

$Fun(\mathsf{FinTop},\mathsf{Top})$ classifies objects
(Proof: since a left adjoint functor $Fun(\mathsf{FinTop},\mathsf{Top}) \to \mathcal E$ corresponds to an arbitrary functor $\mathsf{FinTop}^\mathrm{op} \to \mathcal E$, and if that functor is left exact, it just corresponds to a functor from the terminal category to $\mathcal E$.)
$\infty$-connected objects are objects with a certain property, so they are classified by a localization $\mathcal T$ of $Fun(\mathsf{FinTop},\mathsf{Top})$,  which is precisely the full subcategory of $\infty$-connected objects  [$Fun(\mathsf{FinTop},\mathsf{Top})$ is a presheaf topos, hence hypercomplete. So it's a bit more subtle.].
Jacob Lurie's comment describes this localization as being at the Grothendieck topology generated by all maps ( but I don't know why this is so  This is explained in Denis Nardin's answer).
This localization can be computed using the fact that every object $X$ has a finest cover in this topology, the map in $\mathsf{FinTop}^\mathrm{op}$ corresponding to the map $X \to \ast$ in $\mathsf{FinTop}$. We take a limit over the Cech nerve of this singleton cover in $\mathsf{FinTop}^\mathrm{op}$, whose $n$th level consists of an "$n$-pointed cone on $X$".
These cones also show up in Goodwillie calculus, so we see that an $n$-excisive functor $F$ lies in $\mathcal T$ because the descent object stabilizes at the $n$th stage. Moreover, the localization is closed under limits, so it contains all limits of $n$-excisive functors for varying $n$.
This leaves the obvious question: does $\mathcal T$ consist just of the limits of polynomial functors?<|endoftext|>
TITLE: Asymptotic $\int_M \mathrm{exp}[\mathbf{e}\left(n -\frac{t}{2\pi i}\right)] \left( 1 + \frac{5}{6} \mathbf{e}^2 \right)^{1/2} $ on quintic Calabi-Yau
QUESTION [5 upvotes]: Let $M = \{ G(x) = 0 \} \subseteq \mathbb{P}^4$ be a quintic Calabi-Yau and $\mathbf{e} \in H^2(M, \mathbb{Z})$ such that $\int_M \mathbf{e}^3 = 5$.  Then as $t \gg 1$:
$$
\int_M 
e^{n \mathbf{e}} 
e^{-\frac{t\mathbf{e}}{2\pi i}} 
\left( 1 + \frac{5}{6} \mathbf{e}^2 \right)^{1/2} = 
- \frac{5}{12}
\left( \frac{t}{2\pi i} - n\right)
\left( 2\left( \frac{t}{2\pi i} - n\right)^2 - 5 \right) \tag{$\ast$}$$
How does a high-frequency integral like this get computed?  This could be steepest descent or stationary phase.  In that case what are the Leftschetz thimble and Morse flow?  Perhaps set $G$ bo the the Fermat quintic or something for an example.

Question Because there are so many new elements for me here: the surface in $\mathbb{P}^4$, digesting what it means $\int_M \mathbf{e}^3 =5$.  In a deep way, this is hardly more than an Laplace transform. I would like help seeing how steepest descent / stationary phase is implemented and computed here.
The original paper talks about the pullback of the line bundle $\mathcal{O}_{\mathbb{P}^4}(n)$ to $\mathcal{O}_M(n)$ and details of $\hat{A}$-genus that I am relegating to a separate question.  
Withholding the physics context, the left side looks like particular case of a more general result, but this is already an involved computation.  I am struggling to see even this much and connections to more classical mathematics.

Tongue-in-cheek remark if the moments on the left side could be computed, then we get some analogue of Stirling formula $\log n! \approx n \log n$.

Clarifications
The notation $(1 + \frac{5}{6} \mathbf{e}^2)^{1/2}$ is taken from an arXiv.hep-th paper from 2013 which in turn cites others.  One possible interpretation is to use the Taylor series:
$$(1 + \frac{5}{6} \mathbf{e}^2)^{1/2}
= 1 + \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^1 \mathbf{e}^2
+  \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^2 \mathbf{e}^4
+  \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^3 \mathbf{e}^6
+ \dots $$
The remaining terms are $0$ since $M$ is 6-dimensional.  At least that's the gist of one of the comments.

Another comment suggests the integral $\ast$ is exact: an equality for all values of $t,n$.
Also $\mathbb{e}$ is just a 2-form like $x \, dx \vee dy$ restricted to the manifold $M$.  How are we assured these exists one with $\int_M \mathbb{e}^3 = 5$?

REPLY [10 votes]: OK, let's see if I can put my money where my commenting mouth is. Let me say at the outset that I have no idea where such an integral comes from, but I claim that it doesn't matter to answer the question.
First of all, $\mathbf e$ is nothing mysterious: it is the 2-form dual to a hyperplane section of $M$. The equation $\int_M \mathbf e^3=5$ is just a fancy way of writing that the intersection of 3 hypeplanes in $\mathbf P^4$ (i.e. a line) intersects $M$ in 5 points.
Next, the coefficient $\left(n -\frac{t}{2\pi i}\right)$ never gets modified in any way, so let's just write it as $A$ for simplicity.
Finally, everything in the integrand is supposed to be an element of $H^*(M)$, so we should expand things as power series in $\mathbf e$. Since $\mathbf e \in H^2(M)$ we already have $\mathbf e^4=0$ (contrary to what was written in the edit). So we get
$$ (1+\frac{5}{6}\mathbf e^2)^{\frac12} = 1 + \frac{5}{12} \mathbf e^2 \\
 \operatorname{exp} (A \mathbf e) = 1 + A \mathbf e + \frac{A^2}{2} \mathbf e^2 + \frac{A^3}{6} \mathbf e^3$$.
Now multiply these together: the result is 
$$ \left( \frac{A^3}{6} + \frac{5A}{12} \right) \mathbf e^3 + \text{lower order terms}. $$ 
Applying $\int_M$ kills the lower order terms and turns the $\mathbf e^3$ into a 5, so we end up with 
$$\frac{5A^3}{6}+\frac{25A}{12}$$
The sign of the last term is opposite to what you wrote above, but agrees with the formula on p.37 of the paper you linked. 
By the way, I think you misinterpreted what the paper is saying about taking $t \rightarrow + \infty$. In the paper, they are saying that as $t \rightarrow +\infty$, some other integral expression (higher up the page) reduces to the left-hand side of your equation. They are then saying that the left-hand side equals the right-hand side, but that equality is valid for all $t$, as we just saw.<|endoftext|>
TITLE: Quantum Hamiltonian for an Inverse Cube Force Law
QUESTION [16 upvotes]: If you have a nonrelativistic quantum particle in $\mathbb{R}^3$ in an attractive inverse cube force, its Hamiltonian is 
$$ H = -\nabla^2 - \frac{c}{r^2}  $$
where I'm keeping things simple by setting various constants to 1.  It's a notorious fact that if the force is too strong — if $c > 1/4$ — this operator $H$ is not essentially self-adjoint on $C_0^\infty(\mathbb{R}^3 - \{0\})$, and it's also not bounded below, so we can't use the Friedrichs extension, which is a canonical choice of bounded-below self-adjoint extension for a bounded-below symmetric densely defined operator.  However, because $H$ commutes with complex conjugation, its deficiency indices are equal, so it does have self-adjoint extensions.  
So, my question is: have these self-adjoint extensions been classified, and what are they like?  What are the deficiency indices of $H$?   
Typically the various self-adjoint extensions of a differential operator correspond to different choices of boundary conditions.  They often have a clear physical meanings: for example, for the Laplacian on a cube, they represent different choices of what happens when a particle hits a wall.   Here I think they should correspond to different choices of what a particle does after it hits the singularity at $r = 0$.  But I haven't seen anyone address this issue.
For the 'notorious fact', try:

Michael Reed and Barry Simon, Methods of Modern Mathematical Physics, Vol. 2: Fourier Analysis, Self-Adjointness, Academic Press, New York, 1975.

and look near the discussion of the KLMN theorem.
(Actually, now that I think about it, this borderline case may be too tricky; I'd also be happy to hear about the case
$$ H = -\nabla^2 - \frac{c}{r^p}  $$
with $p > 2$.  In this case $H$ fails to be essentially self-adjoint or bounded below on $C_0^\infty(\mathbb{R}^3 - \{0\})$ for any $c> 0$ and its deficiency indices are still equal, so the same issues arise.  One difference is that in this case there are functions $\psi \in C_0^\infty(\mathbb{R}^3)$ with 
$$ \int \frac{|\psi|^2}{r^p} d^3 x = \infty $$
whereas when $p = 2$, even if $\psi \in C_0^\infty(\mathbb{R}^3)$ doesn't vanish at the origin, this integral converges.)

REPLY [9 votes]: Here is some material on this question.  For related classical issues, see

John Baez, The inverse cube force law.

In quantum mechanics, a particle moving in an inverse cube force law has a Hamiltonian like this:
$$ H = -\nabla^2 + c r^{-2}$$
The first term describes the kinetic energy, while the second describes the potential energy.   I'm setting $ \hbar = 1$ and $ 2m = 1$ to remove some clutter that doesn't really affect the key issues.  
To see how strange this Hamiltonian is, let me compare an easier case.  If $ p \lt 2,$ the Hamiltonian
$$ H = -\nabla^2 + c r^{-p}$$
is essentially self-adjoint on $ C_0^\infty(\mathbb{R}^3 - \{0\}),$ which is the space of compactly supported smooth functions on 3d Euclidean space minus the origin.  What this means is that first of all, $ H$ is defined on this domain: it maps functions in this domain to functions in $ L^2(\mathbb{R}^3)$.  But more importantly, it means we can uniquely extend $ H$ from this domain to a self-adjoint operator on some larger domain.  In quantum physics, we want our Hamiltonians to be self-adjoint.  So, this fact is good.
Proving this fact is fairly hard!  It uses something called the Kato–Lax–Milgram–Nelson theorem together with this beautiful inequality:
$$ \displaystyle{ \int_{\mathbb{R}^3} \frac{1}{4r^2} |\psi(x)|^2 \,d^3 x \le \int_{\mathbb{R}^3} |\nabla \psi(x)|^2 \,d^3 x } $$
for any $ \psi\in C_0^\infty(\mathbb{R}^3).$   
If you think hard, you can see this inequality is actually a fact about the quantum mechanics of the inverse cube law!  It says that if $ c \ge -1/4,$ the energy of a quantum particle in the potential $ c r^{-2}$ is bounded below.  And in a sense, this inequality is optimal: if $ c \lt -1/4$, the energy is not bounded below.  This is the quantum version of how a classical particle can spiral in to its doom in an attractive inverse cube law, if it doesn't have enough angular momentum.  But it's subtly and mysteriously different.  
You may wonder how this inequality is used to prove good things about potentials that are 'less singular' than the $ c r^{-2}$ potential: that is, potentials $ c r^{-p}$ with $ p \lt 2.$  For that, you have to use some tricks that I don't want to explain here.  I also don't want to prove this inequality, or explain why its optimal!  You can find most of this in some old course notes of mine:

John Baez, Quantum Theory and Analysis, 1989.

See especially section 15.
But it's pretty easy to see how this inequality implies things about the expected energy of a quantum particle in the potential $ c r^{-2}$.  So let's do that.
In this potential, the expected energy of a state $ \psi$ is:
$$  \displaystyle{  \langle \psi, H \psi \rangle = 
\int_{\mathbb{R}^3} \overline\psi(x)\, (-\nabla^2 + c r^{-2})\psi(x) \, d^3 x }$$
Doing an integration by parts, this gives:
$$ \displaystyle{  \langle \psi, H \psi \rangle = \int_{\mathbb{R}^3} |\nabla \psi(x)|^2 + cr^{-2} |\psi(x)|^2 \,d^3 x } $$
The inequality I showed you says precisely that when $ c = -1/4,$ this is greater than or equal to zero.  So, the expected energy is actually nonnegative in this case!  And making $ c$ greater than $ -1/4$ only makes the expected energy bigger. 
Note that in classical mechanics, the energy of a particle in this potential ceases to be bounded below as soon as $ c \lt 0.$  Quantum mechanics is different because of the uncertainty principle!  To get a lot of negative potential energy, the particle's wavefunction must be squished near the origin, but that gives it kinetic energy.  
It turns out that the Hamiltonian for a quantum particle in an inverse cube force law has exquisitely subtle and tricky behavior.  Many people have written about it, running into 'paradoxes' when they weren't careful enough.  Only rather recently have things been straightened out.  
For starters, the Hamiltonian for this kind of particle
$$ H = -\nabla^2 + c r^{-2}$$
has different behaviors depending on $ c.$  Obviously the force is attractive when $c \gt 0$ and repulsive when $ c \lt 0,$ but that's not the only thing that matters!  Here's a summary:

$ c \ge 3/4.$  In this case $ H$ is essentially self-adjoint on $ C_0^\infty(\mathbb{R}^3 - \{0\}).$  So, it admits a unique self-adjoint extension and there's no ambiguity about this case.
$ c \lt 3/4.$  In this case $ H$ is not essentially self-adjoint on $ C_0^\infty(\mathbb{R}^3 - \{0\}).$  In fact, it admits more than one self-adjoint extension!  This means that we need extra input from physics to choose the Hamiltonian in this case.  It turns out that we need to say what happens when the particle hits the singularity at $ r = 0.$  
$ c \ge -1/4.$  In this case the expected energy $ \langle \psi, H \psi \rangle $ is bounded below for $ \psi \in C_0^\infty(\mathbb{R}^3 - \{0\}).$   It turns out that whenever we have a Hamiltonian that is bounded below, even if there is not a unique self-adjoint extension, there exists a canonical 'best choice' of self-adjoint extension, called the Friedrichs extension.  I explain this in my course notes.
$ c \lt -1/4.$  In this case the expected energy is not bounded below, so we don't have the Friedrichs extension to help us choose which self-adjoint extension is 'best'.

To go all the way down this rabbit hole, I recommend these two papers:

Sarang Gopalakrishnan, Self-Adjointness and the Renormalization of Singular Potentials, B.A. Thesis, Amherst College.
D. M. Gitman, I. V. Tyutin and B. L. Voronov, Self-adjoint extensions and spectral analysis in the Calogero problem,  Jour. Phys. A 43 (2010), 145205.

The first is good for a broad overview of problems associated to singular potentials such as the inverse cube force law; there is attention to mathematical rigor the focus is on physical insight.  The second is good if you want—as I wanted—to really get to the bottom of the inverse cube force law in quantum mechanics.  Both have lots of references.
Also, both point out a crucial fact I haven't mentioned yet: in quantum mechanics the inverse cube force law is special because, naively, at least it has a kind of symmetry under rescaling!  You can see this from the formula
$$ H = -\nabla^2 + cr^{-2} $$
by noting that both the Laplacian and $ r^{-2}$ have units of length-2.  So, they both transform in the same way under rescaling: if you take any smooth function $ \psi$, apply $ H$ and then expand the result by a factor of $ k,$, you get $ k^2$ times what you get if you do those operations in the other order.  
In particular, this means that if you have a smooth eigenfunction of $ H$ with eigenvalue $ \lambda,$ you will also have one with eigenfunction $ k^2 \lambda$ for any $ k \gt 0.$  And if your original eigenfunction was normalizable, so will be the new one!
With some calculation you can show that when $ c \le -1/4,$ the Hamiltonian $ H$ has a smooth normalizable eigenfunction with a negative eigenvalue.  In fact it's spherically symmetric, so finding it is not so terribly hard.  But this instantly implies that $ H$ has smooth normalizable eigenfunctions with any negative eigenvalue.  
This implies various things, some terrifying.  First of all, it means that $ H$ is not bounded below, at least not on the space of smooth normalizable functions.  A similar but more delicate scaling argument shows that it's also not bounded below on $ C_0^\infty(\mathbb{R}^3 - \{0\}),$ as I claimed earlier.
This is scary but not terrifying: it simply means that when $ c \le -1/4,$ the potential is too strongly negative for the Hamiltonian to be bounded below.  
The terrifying part is this: we're getting uncountably many normalizable eigenfunctions, all with different eigenvalues, one for each choice of $ k.$  A self-adjoint operator on a countable-dimensional Hilbert space like $ L^2(\mathbb{R}^3)$ can't have uncountably many normalizable eigenvectors with different eigenvalues, since then they'd all be orthogonal to each other, and that's too many orthogonal vectors to fit in a Hilbert space of countable dimension!
This sounds like a paradox, but it's not.  These functions are not all orthogonal, and they're not all eigenfunctions of a self-adjoint operator.  You see, the operator $ H$ is not self-adjoint on the domain we've chosen, the space of all smooth functions in $ L^2(\mathbb{R}^3).$  We can carefully choose a domain to get a self-adjoint operator... but it turns out there are many ways to do it.
Intriguingly, in most cases this choice breaks the naive dilation symmetry.  So, we're getting what physicists call an 'anomaly': a symmetry of a classical system that fails to give a symmetry of the corresponding quantum system.  
Of course, if you've made it this far, you probably want to understand what the different choices of Hamiltonian for a particle in an inverse cube force law actually mean, physically.  The idea seems to be that they say how the particle changes phase when it hits the singularity at $ r = 0$ and bounces back out.  
(Why does it bounce back out?  Well, if it didn't, time evolution would not be unitary, so it would not be described by a self-adjoint Hamiltonian!  We could try to describe the physics of a quantum particle that does not come back out when it hits the singularity, and I believe people have tried, but this requires a different set of mathematical tools.)
For a detailed analysis of this, it seems one should take Schrödinger's equation and do a separation of variables into the angular part and the radial part:
$$ \psi(r,\theta,\phi) = \Psi(r) \Phi(\theta,\phi) $$
For each choice of $ \ell = 0,1,2,\dots$ one gets a space of spherical harmonics that one can use for the angular part $ \Phi.$  The interesting part is the radial part, $ \Psi.$  Here it is helpful to make a change of variables
$$ u(r) = \Psi(r)/r $$
At least naively, Schrödinger's equation for the particle in the $ cr^{-2}$ potential then becomes
$$ \displaystyle{ \frac{d}{dt} u = -iH u }$$
where
$$ \displaystyle{ H = -\frac{d^2}{dr^2} + \frac{c + \ell(\ell+1)}{r^2} }$$
Beware: here I'm using $H$ for a new Hamiltonian, one that describes only the radial part of the dynamics of a quantum particle in an inverse cube force.  As in classical mechanics, the centrifugal force and the inverse cube force join forces in an 'effective potential'
$$ \displaystyle{ U(r) = kr^{-2} } $$
where
$$ k = c + \ell(\ell+1)$$
So, we have reduced the problem to that of a particle on the open half-line $ (0,\infty)$ moving in the potential $ kr^{-2}.$ The Hamiltonian for this problem:
$$ \displaystyle{ H = -\frac{d^2}{dr^2} + \frac{k}{r^2} }$$
is called the Calogero Hamiltonian.  Needless to say, it has fascinating and somewhat scary properties, since to make it into a bona fide self-adjoint operator, we must make some choice about what happens when the particle hits $ r = 0.$  The formula above does not really specify the Hamiltonian.
This is more or less where Gitman, Tyutin and Voronov begin their analysis, after a long and pleasant review of the problem.  They describe all the possible choices of self-adjoint operator that are allowed.  The answer depends on the values of $ k,$ but very crudely, the choice says something like how the phase of your particle changes when it bounces off the singularity.  Most choices break the dilation invariance of the problem.  But intriguingly, some choices retain invariance under a discrete subgroup of dilations!  
Finally, some remarks on the same problem in 4 dimensions. It's a bit odd to study the inverse cube force law in 3-dimensonal space, since Newtonian gravity and the electrostatic force would actually obey an inverse cube law in 4-dimensional space.  For the classical 2-body problem it doesn't matter much whether you're in 3d or 4d space, since the motion stays on the plane.  But for quantum 2-body problem it makes more of a difference!  
Just for the record, let me say how the quantum 2-body problem works in 4 dimensions.  As before, we can work in the center of mass frame and consider this Hamiltonian:
$$ H = -\nabla^2 + c r^{-2}$$
And as before, the behavior of this Hamiltonian depends on $ c.$  Here's the story this time:

$ c \ge 0.$  In this case $ H$ is essentially self-adjoint on $ C_0^\infty(\mathbb{R}^4 - \{0\}).$  So, it admits a unique self-adjoint extension and there's no ambiguity about this case.
$ c \lt 0.$  In this case $ H$ is not essentially self-adjoint on $ C_0^\infty(\mathbb{R}^4 - \{0\}).$  
$ c \ge -1.$  In this case the expected energy $ \langle \psi, H \psi \rangle $ is bounded below for $ \psi \in C_0^\infty(\mathbb{R}^3 - \{0\}).$   So, there is exists a canonical 'best choice' of self-adjoint extension, called the Friedrichs extension.  
$ c \lt -1.$  In this case the expected energy is not bounded below, so we don't have the Friedrichs extension to help us choose which self-adjoint extension is 'best'.

I've been assured these are correct by Barry Simon, and a lot of this material will appear in Section 7.4 of his book:

Barry Simon, A Comprehensive Course in Analysis, Part 4: Operator Theory, American Mathematical Society, Providence, RI, 2015.

See also:

Barry Simon, Essential self-adjointness of Schrödinger operators with singular potentials, Arch. Rational Mech. Analysis 52 (1973), 44-48.<|endoftext|>
TITLE: Construction of coherent sheaf such that $\text{Proj}\,\text{Sym}\,(\mathcal{F}) = \text{Sym}^n X$
QUESTION [6 upvotes]: Let $X$ be a smooth projective curve. How do I construct a coherent sheaf $\mathcal{F}$ on $\text{Pic}^n X$ (i.e., the component of the Picard scheme of $X$ parametrizing line bundles of degree $n$) such that$$\mathbb{P}(\mathcal{F}) := \text{Proj}\,\text{Sym}(\mathcal{F})$$equals the $n$th symmetric power $\text{Sym}^nX$?
(Here, "equals" means ``canonically isomorphic as a scheme over $\text{Pic}^nX$"; notice that $\text{Sym}^nX$ is equipped with a canonical map to $\text{Pic}^nX$ because $X$ is a smooth curve.)
Ideally, the construction of $\mathcal{F}$ should be as canonical as possible.

REPLY [7 votes]: There is an obstruction, as alluded to in my comments.  If $X$ has genus $0$, for instance (perhaps not the case you are most interested in), then there exists such a sheaf when $n$ is odd if and only if $X$ is isomorphic to $\mathbb{P}^1_k$, where $k$ is your field. So for a conic over $\mathbb{R}$ having no real points, you cannot find such a sheaf when $n$ is odd. For $n$ even, there always exists such a sheaf.  
Assume that there exists a Poincare invertible sheaf $\mathcal{P}_n$ on $\text{Pic}^n X\times X$.  The sheaf you want on $\text{Pic}^n X$ is $$\mathcal{F} = R\textit{Hom}^0_{ \mathcal{O}_{\text{Pic}^n X}}(R\text{pr}_{1,*}(\mathcal{P}_n),\mathcal{O}_{\text{Pic}^n_X}). $$  But what does that mean?  Following is a (somewhat) more explicit construction (used in Eisenbud-Harris, for instance) that depends on some choices.
Edit. I just found that the construction below is in Arbarello, Cornalba, Griffiths, Harris. Geometry of Algebraic Curves, Grundlehren der mathematischen Wissenschaften 267, Springer, 1985, in Section IV.3, pp. 176--178.  I do believe that the construction is also in the limit linear series paper of Eisenbud-Harris (but I might be misremembering).
Assume now that the genus is positive. Let $d$ be a positive integer such that $d(2g-2) + n > 2g-2$, e.g., $d=1$ always works.  Let $s$ be a nonzero section of $H^0(X,\omega_{X/k}^{\otimes d})$.  Let $D \subset X$ be the zero scheme of $s$.  This is an effective Cartier divisor of degree $d(2g-2)$.  The short exact sequence on $X$, $$ 0 \to \mathcal{O}_X \xrightarrow{s} \omega_{X/k}^{\otimes d} \xrightarrow{q} \omega_{X/k}^{\otimes d}|_D \to 0,$$ induces a short exact sequence on $\text{Pic}^n X\times X$, $$ 0 \to \mathcal{P}_n \xrightarrow{1\otimes \text{pr}_2^*s} \mathcal{P}_n\otimes_{\mathcal{O}} \text{pr}_2^*\omega_{X/k}^{\otimes d} \xrightarrow{1\otimes \text{pr}_2^*q} \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d}|_{\text{Pic}^n X\times D} \to 0.$$  Because of the hypothesis on $d$, the pushforward $$\mathcal{E}^0 := \text{pr}_{1,*}\left( \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d} \right)$$ is a locally free sheaf of rank $(2d-1)(g-1)+n$, and for every $i>0$, the higher direct image sheaf $R^i\text{pr}_{1,*}\left( \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d} \right)$ is the zero sheaf.  Similarly, because $\text{Pic}^n X \times D$ is finite and flat over $\text{Pic}^n X$ of degree $d(2g-2)$, also the pushforward
$$\mathcal{E}^1 := \text{pr}_{1,*}\left( \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d}|_{\text{Pic}^n X \times X} \right)$$ is locally free of rank $d(2g-2)$, and again all the higher direct image sheaves vanish.  The pushforward of the map $1\otimes q$ is a morphism of locally free sheaves, $$ d^0 : \mathcal{E}^0 \to \mathcal{E}^1.$$  Thus, one model of $R\text{pr}_{1,*}(\mathcal{P}_n)$ is the two-term complex $\mathcal{E}^\bullet$.  
Now to form $R\textit{Hom}_{ \mathcal{O}_{\text{Pic}^n X}}(R\text{pr}_{1,*}(\mathcal{P}_n),\mathcal{O}_{\text{Pic}^n_X})$, you just take duals and transposes.  Define $\mathcal{E}_0 = \left( \mathcal{E}^0 \right)^\vee$, define $\mathcal{E}_1 = \left( \mathcal{E}^1 \right)^\vee$, and define $d_1$ to be the transpose of $d^0$.  Then your sheaf $\mathcal{F}$ is the cokernel of $$d_1: \mathcal{E}_1 \to \mathcal{E}_0.$$
Of course that description of $\mathcal{F}$ is not so canonical, e.g., if your curve has automorphisms, none of $\mathcal{E}_0$, $\mathcal{E}_1$ nor $d_1$ is equivariant for the induced automorphisms of $\text{Pic}^n X$.  There are models of these total derived functors that are truly canonical.  But if you want an explicit model, typically you have to make choices that remove some of the functoriality.<|endoftext|>
TITLE: Finding combinatorial models / statistics
QUESTION [13 upvotes]: In many cases in combinatorics and especially algebraic combinatorics with some representation theory, the main problem is about finding the correct statistic on a mathematical object.
For example, finding the combinatorial description of symmetric functions, the invention of the inv and maj statistic to describe modified Macdonald polynomials,
or the rather tricky weight on certain fillings that yield the combinatorial formula for Jack polynomials by Knop and Sahi.
Another example is the complicated charge statistic defined on permutations, words, and semi-standard Young tableaux.
A lot of hard, open problems in combinatorics boils down to finding (guessing) a combinatorial statistic, and once the correct guess is done, the rest is usually proof by induction (not always easy, but at least you know what to prove).
What techniques to people use to guess such combinatorial statistics in the first place?
This is usually not explained in the papers, only the final conjecture or proof is presented.
I am currently battling such a problem myself, and here are some techniques I have used so far:

If the statistic is a generalization of an already existing one, see which properties might generalize. Is it linear, in some sense? Are there equalities of the form $\sigma(T) = \sigma(T')$ where $T'$ is simpler than $T$ in some sense?
Use computer algebra to test if additional assumptions lead to a contradiction or not.
Check if Findstat already has it in the database.
Check natural sequences of objects against OEIS.
Look for alternative underlying combinatorial objects, where the statistic might be more evident. For example, Young tableaux and Gelfand-Tsetlin patterns are in bijection, and some statistics are easier to define on one of these, rather than the other.

I would love to hear stories of how people guessed combinatorial statistics (or better underlying combinatorial objects) in stories like above.

REPLY [4 votes]: One additional test I usually do is to check if I can extract information about its generating function given by
$$\sum_{x \in \mathcal{C}_n} q^{\operatorname{stat}(x)}$$
for some decomposition $\mathcal{C} = \bigcup_{n} \mathcal{C}_n$ with $\big|\mathcal{C}_n \big| < \infty$ of your combinatoral collection.
In the best case, you find a known equidistributed statistic $\operatorname{stat'}$ to which you can possibly biject.
I guess an archetypical example would be to have the number of inversions of a permutation, so the decomposition is $Perm = \bigcup_n Perm_n$, the statistic for $n=3$ is
$$
inv(123) = 0, inv(132) = 1, inv(213) = 1, inv(231) = 2, inv(312) = 2, inv(321 = 3
$$
and its generating function $f_3 = 1 + 2q + 2q^2 + q^3$.
You then might realize that this is also the generating function of the major index on permutations.
Finally, let me remark: FindStat shows you the generating function (such as in http://www.findstat.org/St000018), and you can then automatically search the OEIS for the coefficients of the statistic distribution.<|endoftext|>
TITLE: Compensated compactness for system of conservation laws?
QUESTION [6 upvotes]: As far as I knew, the method of compensated compactness can be used only for one-dimensional scalar and $2\times 2$ systems of conservation laws, i.e. $u_t+f(u)_x=0$. But if I understood correctly there are some works which use it for $N\times N$ systems; for instance
1- Hyperbolic to Parabolic Relaxation Theory for Quasilinear First Order Systems, Pierangelo Marcati and Bruno Rubino.
2- Convergence of a relaxation scheme for hyperbolic systems of conservation laws, Corrado Lattanzio, Denis Serre.
3- Existence and Uniqueness of Solutions for some Hyperbolic Systems of Conservation Laws, Arnaud Heibig. 
Does Div-Curl lemma work in such cases? Could anyone give me any explanation about it?

REPLY [4 votes]: So far, only the div-curl Lemma could be used when applying compensated compactness to systems of conservation laws. But it has two flaws.

Because we only have divergences but no curls in general, we cannot handle multi-dimensional systems. Only one-dimensional systems can be considered by this method.
The idea behind Tartar's and DiPerna's use of compensated compactness is that, because of nonlinearity, a system displays some ellipticity. This feature could in principle be used to establish compactness of (approximate) solutions. But since the systems under consideration are not elliptic (they are hyperbolic), the amount of ellipticity is marginal. To carry out the method, we therefore need much more than just one information. Actually, we need infinitely many. This is why the CC has been successful only for systems having infinitely many independent entropy-entropy flux pairs. This covers the scalar case as well as the $2\times2$ case. Generic systems of large size usually have only finitely many independent entropies and CC cannot be fruitful. A notable exception is the  class of rich systems, to which CC can be efficiently applied. This is the class of systems of conservation laws which admits a diagonalisation in terms of Riemann invariants. The rich class contains the class of Temple systems, which is the object of Heibig's article.<|endoftext|>
TITLE: For a 3-manifold $Y$, when does $Y\times S^{1}$ admits a Riemannian metric with positive scalar curvature?
QUESTION [14 upvotes]: Let $Y$ be an orientable, smooth 3-manifold and let $X=Y\times S^{1}$. My question is that: when does $X$ admits a Riemannian metric with positive scalar curvature? 
An obvious case is when $Y$ itself admits a psc metric. Are there any other case?
 It was proved by Gromov and Lawson that any 3-manifold which contains a $K(\pi,1)$ prime factor does not admits a psc metric (another result in the paper implies that $Y$ should not be hyperbolic if $Y\times S^{1}$ admits psc), which, together with the Geometrization conjecture classifies all the 3-manifold admits psc metric. Does their proof works for $Y\times S^{1}$?
(I am most interested in the special case that $Y$ is an irreducible integer homology sphere)

REPLY [19 votes]: It follows from Theorem 2 of the following paper and the geometrization theorem that $Y^3\times S^1$ admits a metric of positive scalar curvature iff $Y$ does. 
Schoen, Richard; Yau, Shing-Tung, On the structure of manifolds with positive scalar curvature, Manuscr. Math. 28, 159-183 (1979). ZBL0423.53032.
On p. 9 of the paper, they define $C_3'$ to be the class of closed three dimensional manifolds which do not
admit any non-zero degree map to a compact three dimensional manifold
$M$ such that $\pi_2(M) = 0$ and $M$ contains a two-sided incompressible
surface with genus $\geq 1$ (i.e. a Haken 3-manifold).
It follows from the geometrization theorem and virtual Haken theorem that $Y$ admits a metric of positive scalar curvature iff every finite-sheeted cover $Y'\to Y$ has  $Y'\in C_3'$. 
Let $C_4'$ be the class of closed 4-manifolds $M$ such that every codimension one homology
class of $M$ can be represented, up to some non-zero integer, by a map
from a manifold of class $C'_3$ (a homology class $\alpha\in H_k(M)$ is represented by an oriented $k$-manifold $N$ if there is a map $f:N\to M$ such that $f_\ast([N])=\alpha$, where $[N]\in H_k(N)$ is the fundamental class determined by the orientation of $N$). 
Then Theorem 2 implies that if $M^4$ admits a metric of positive scalar curvature,
then $M$ is of class $C'_4$. 
Now, consider the 4-manifold $Y^3\times S^1$ and let $Y'\to Y$ be a finite-sheeted
cover. By Theorem 2, there exists an $N\subset Y'\times S^1$ such that $[N]=[Y'\times \ast]\in H_3(Y'\times S^1)$ and $N$ is of class $C_3'$. Moreover, there is a non-zero degree map $N\to Y'$ (by composing the embedding with the projection $Y'\times S^1\to Y'$), hence $Y'\in C_3'$ (the class $C_3'$ is clearly closed under taking non-zero degree maps). Since every finite-sheeted cover of $Y$ is in $C_3'$, we conclude that $Y$ admits a metric of positive scalar curvature.<|endoftext|>
TITLE: Brouwer's theorem for the Cauchy reals
QUESTION [17 upvotes]: Brouwer famously proved, using principles motivated by intuitionistic choice sequences, that every function $\mathbb{R}\to \mathbb{R}$ is continuous.  In Sheaves in geometry and logic (section VI.9), MacLane and Moerdijk exhibit a topos (the topos of sheaves on any sufficiently nice small full subcategory $\mathbf{T}\subseteq\mathrm{Top}$) in which this theorem holds in the internal logic.  However, their proof is about the Dedekind real numbers.  Can anyone point to a topos in which Brouwer's theorem holds for the Cauchy real numbers?
Of course it would suffice to give a topos in which Brouwer's theorem holds and in which the Cauchy and Dedekind real numbers coincide (e.g. if countable choice holds).  This is not the case for MacLane and Moerdijk's example, in which $\mathbf{R}_D$ is the sheaf of continuous $\mathbb{R}$-valued functions while (at least if all spaces in $\mathbf{T}$ are locally connected) $\mathbf{R}_C$ is the sheaf of locally constant $\mathbb{R}$-valued functions.  Moreover, it is easy to see that in this case there do exist discontinuous functions $\mathbf{R}_C \to \mathbf{R}_C$; so a different example is needed.

REPLY [3 votes]: I received the following answer from an anonymous contributor:
Historically the first model with validity of the Brouwer 'theorem' was the topos $\mathrm{Sh}(N^N)$ of sheaves on Baire space. This goes apparently back to Dana Scott here:
Scott, Extending the topological interpretation to intuitionistic analysis II , pp.235-255 in Myhill,Kino,Vesley (eds.) Intuitionism and proof theory, North-Holland 1970. (pay-walled here)
The first installment of Scott is probably also of interest in the wider context and available from numdam
According to an exercise in Moerdijk- Mac Lane (p.345) $\mathrm{Sh}(N^N)$ has dependent choice as well, so it should fit your bill.
The question of validity of the Brouwer theorem was studied by Martin Hyland in the early seventies, relevant references are the following in the 'applications of sheaves' LNM, in particular the second:
Fourman, Hyland, Sheaf models of analyis, pp.280-301 LNM 753 Springer Heidelberg 1979. (PDF)
Hyland, Continuity in spatial toposes, pp.442-465 LNM 753 Springer Heidelberg 1979.  (pay-walled from springer page)
It seems the question then is, can $R_C$ satify the Brouwer thm with coinciding with $R_D$ ?<|endoftext|>
TITLE: $Ext$-algebra generated by $Hom$ and $Ext^1$ as $A_\infty$-algebra?
QUESTION [6 upvotes]: In [Keller: A-infinity algebras in representation theory, Proposition 1(b)], Keller states that for an associative algebra the $Ext$-algebra of the simples is generated by $Ext^1(S,S)$ as an $A_\infty$-algebra. 
Are there any other conditions on a module $M$ known to prove that $Ext^*(M,M)$ is generated by $Hom(M,M)$ and $Ext^1(M,M)$ as an $A_\infty$-algebra  module $M$ over a finite dimensional associative algebra? E.g. conditions like $M$ generator of the derived category or something similar.

REPLY [6 votes]: Suppose $M=M_1 \oplus \dots \oplus M_r$ with $M_i$ indecomposable for $1 \leq i \leq r$. Let $\mathcal F (M_1, \dots, M_r)$ denote the category of modules which admit a filtration with subquotients in the $M_i$. The crucial assumption in Keller's proof is that $\mathcal F(M_1, \dots, M_r)$ is closed under syzygies.
With this assumption, any $n$-fold extension ($n>1$) $$\xi \in \operatorname{Ext}^n(M_u,M_v) \simeq \operatorname{Ext}^1(\Omega^{n-1}(M_u),M_v)$$ with $1 \leq u,v \leq r$ is the Yoneda splice of $n$ short-exact sequences $$0 \to X_{i+1} \to E_i \to X_i \to 0 \quad (0 \leq i \leq n-1)$$ where $X_0=M_u$, $X_n=M_v$, and each $X_i$ is an object in $\mathcal F(M_1, \dots, M_r)$ for $0 \leq i \leq n$.
As explained in the article you mentioned, the category $\mathcal F(M_1, \dots, M_r)$ is equivalent to a category of twisted stalks. A twisted stalk can be described as a pair $(B,\delta)$, where $B$ is a sequence of subfactors from $\{M_1, \dots, M_r\}$ and $\delta$ is an upper triangular matrix with entries from $\operatorname{Ext}^1(M,M)$.
If $X_i$ corresponds to a twisted stalk $(B,\delta)$ and $X_{i+1}$ corresponds to a twisted stalk $(B',\delta')$, then the short-exact sequence $0 \to X_{i+1} \to E_i \to X_i \to 0$ corresponds to a morphism of twisted complexes $(B,\delta) \to (B',\delta')[1]$. Such a morphism is a matrix with entries from $\operatorname{Ext}^1(M,M)$.
So each short-exact sequence $0 \to X_{i+1} \to E_i \to X_i \to 0$ can be described using only the $\operatorname{Ext}^1$ part of $\operatorname{Ext}^\ast(M,M)$. The higher multiplications enter the picture when we splice the short-exact sequences together. The rule for composing morphisms of twisted complexes involves $m_2$ and higher multiplications of the matrices involved. When splicing all the short-exact sequences together, we end up with an expression for $\xi \in \operatorname{Ext}^n(M_u,M_v)$ using only elements in $\operatorname{Ext}^1(M,M)$ and $m_2$ and higher multiplications. This shows that $\operatorname{Ext}^\ast(M,M)$ is generated in degree $0$ and $1$ as an $A_\infty$-algebra.
A class of examples where the assumption is satisfied is when $\{M_1, \dots, M_r\}$ are the standard modules over a quasi-hereditary algebra.<|endoftext|>
TITLE: Tangent space of Hilbert scheme
QUESTION [19 upvotes]: We have the following theorem:
Let $X$ be a projective scheme over an algebraically closed field $k$, and $Y \subset X$ a closed subscheme with Hilbert polynomial $P$. Then$$T_{[Y]}\text{Hilb}_P (X) = \text{Hom}_Y (I_Y/I_Y^2, \mathcal{O}_Y).$$In particular, if both $X, Y$ are both smooth, then$$T_{[Y]}\text{Hilb}_P (X) = H^0(Y, N_{Y/X}).$$My question is, what is the intuition behind this theorem? What are some examples to keep in mind when thinking of this theorem? Thanks.

REPLY [6 votes]: I'll write $H$ for the Hilbert scheme and $[Y]$ for the point of $H$ corresponding to a subscheme $Y \subseteq X$. When $Y$ and $X$ are both smooth, the claim is that the tangent space to $H$ at $[Y]$ is the global sections of $(T_X)|_Y/T_Y$.
Intuition 1 (This is basically Scott's comment.) Given a holomorphic vector field $\theta$ on $X$ near $Y$, we can flow $Y$ along $\theta$ to get a family $Y(t)$ of complex submanifolds of $X$, and thus a path $[Y(t)]$ in $H$. It is plausible (and true!) that the first order variation $\frac{d}{dt} [Y(t)]$, which is a tangent vector to $H$, only depends on $\theta|_Y$. Moreover, if $\theta$ is tangent to $Y$, then $Y$ flows to itself, so it makes sense that only the image of $\theta$ in the quotient $T_X/T_Y$ matters.
Intuition 2 Let's go the other way: Let $[Y(t)]$ be a path through $H$ with $Y(0) = Y$. For $t$ in some open disc $D$ around $0$, we can smoothly (not holomorphically!) trivialize the family and thus get a map $\phi: Y \times D \to X$. For every point $y \in Y$, we have a path $\phi(y,t)$ in $X$, whose derivative at $0$ lies in $T_X(y)$. But changing the trivialization can change this derivative by vectors in $T_Y(y)$, so the intrinsic thing is a vector in $T_X(y)/T_Y(y)$ for each $y \in Y$ -- in other words, a section of the normal bundle.
Bonus We have a long exact sequence $0 \to T_Y \to (T_X)|Y \to N_{Y/X} \to 0$ of vector bundles on $Y$. So we have a long exact sequence $H^0(Y, T_X|Y) \to H^0(Y, N_{Y/X}) \to H^1(Y, T_Y)$. The final vector space, $H^1(Y, T_Y)$, describes deformations of the complex structure of $Y$ and, indeed, the map $H^0(Y, N_{Y/X}) \to H^1(Y, T_Y)$ says, when you wiggle $Y$ inside $X$, how the complex structure on $Y$ changes. The tangent vectors coming from $H^0(Y, T_X|Y)$ are the ones that don't change complex structure. Intuitively, if we have an actual section of $T_X$, then we can flow the embedding $Y \to X$ along it to get a holomorphic map $Y \times D \to X$.
Of course, all of this is intuition only in the case that $X$ and $Y$ are smooth. In order to see that $\mathrm{Hom}(I_Y/I_Y^2, \mathcal{O}_Y)$ is the version which is still right when you put singularities in, I recommend Count Dracula's method.<|endoftext|>
TITLE: Learning the exponents in a sum of two modular roots of unity
QUESTION [12 upvotes]: $\newcommand{\Z}{\mathbb{Z}}$
Suppose that $n$ is a large and known integer (say, with 100 digits) and that you are given access to a function
$$f(x) = x^a + x^b$$
with unknown exponents $a,b \in \Z/n$.  You are allowed to evaluate $f$ on any cyclic ring $\Z/q$ with a solution to $x^n = 1$, where $x$ and $q$ are of your choosing as long as $\gcd(n,q) = 1$.  You are allowed several evaluations with distinct $x$ and $q$.  For simplicity you can assume that $q$ is a product of primes (which may not be distinct) that are all 1 mod $n$ and that $x$ is an $n$th root of unity, since for instance $n$ could be prime.  Linnik's effective version of Dirichlet's theorem says that there is a ready supply of values of $q$.
My ultimate question:  What algorithms in number theory are available to find the exponents $a$ and $b$?   Of course you can find them in principle with one enormous value of $q$.  The question is what is known about efficiency as a function of $d$, the number of digits of $n$.  The problem is like discrete logarithm, but more complicated because there are two terms.
I am also interested in this more tangible question: Can you find a moderate value of $q$ such that $f$ is injective?  Heuristically, $O(d)$ digits should be enough.  I am thinking that GRH implies that $f$ is strictly injective for most values of $q$ with $O(d)$ digits --- is this true?  Can you prove unconditionally $f$ that is usually injective in this range, or usually mostly injective?

REPLY [9 votes]: If you can compute discrete logarithms, there's an easy solution:
$$ f(\zeta) = \zeta^a + \zeta^b $$
$$ f(\zeta^2) = (\zeta^a)^2 + (\zeta^b)^2 $$
is a system of two equations in the quantities $\zeta^a$ and $\zeta^b$, allowing you to solve for $(\zeta^a, \zeta^b)$. There will be 2 solutions, but that just reflects the symmetry between $a$ and $b$.<|endoftext|>
TITLE: Can integration spoil real-analyticity?
QUESTION [14 upvotes]: Is there an example of a function $f:(a,b)\times(c,d)\to\mathbb{R}$, which is real analytic in its domain, integrable in the second variable, and such that the function 
$$ g:(a,b)\to\mathbb{R},\qquad g(x) = \int_c^d f(x,y) dy$$
is not real-analytic on $(a,b)$?
Edit: What about an example of bounded $f$ satisfying the above?

REPLY [6 votes]: Maybe this is too obvious to be interesting, but if we have bounds $\frac{1}{n!} \left| \frac{\partial^n}{(\partial x)^n} f(x,y) \right| < C_n$ on all of $(a,b) \times (c,d)$ such that $\sum C_n r^n$ converges for some $r>0$, then we get the conclusion. 
This is because the Taylor series with respect to $x$ then converges on circles of radius $r$ around any point of $(a,b)$. Write $U$ for the open subset of points in $\mathbb{C}$ which are within distance $r$ of some point of $(a,b)$. These Taylor series give an extension of $f$ to $U \times (c,d)$, holomorphic in the first variable. (They agree on overlaps, since they agree on the real interval where they overlap.) Moreover, $|f| \leq \sum C_n r^n$ everywhere on $U \times (c,d)$.
We want to claim that $\int_c^d f(x,y) dy$ is holmorphic on $U$. By Morera's theorem, it is enough to check that $\int_{x \in \gamma} \int_{y=c}^d f(x,y) dy dx=0$ for any closed curve $\gamma \subset U$. By Fubini, we may switch the integral to $\int_{y=c}^d \int_{x \in \gamma} f(x,y) dx dy$, and the inner integral is zero since $f$ is holomorphic. (We may apply Fubini because we have a uniform bound for $f$ throughout $U \times (c,d)$.)<|endoftext|>
TITLE: The structure map of topological K-theory
QUESTION [5 upvotes]: This may be a silly question but I don't know the answer.
I know the construction of (equivariant) K-spectrum $KU_G$ and the periodicity of (equivariant) K-theory. But I don't know its structure maps and how they are constructed.
Can anybody show me the answer or some reference on this? Thanks very much.

REPLY [5 votes]: That is a very reasonable question.  By a $G$-spectrum $KU_G$, $G$ a compact Lie group, one should mean a genuine $G$_spectrum, so suitably indexed on representations of $G$. One gets the structure maps by use of equivariant Bott periodicity.  A good sketch of how this goes, without full details, is given on pages 146-148 of Equivariant Homotopy and Cohomology Theory (available here: http://www.math.uchicago.edu/~may/BOOKS/alaska.pdf).  The relevant chapter was written by John Greenlees.<|endoftext|>
TITLE: A question on Ramanujan's $1/\pi$ formulas
QUESTION [10 upvotes]: It is known that Ramanujan discovered a number of formulas for $1/\pi$. All of these formulas are of the form $$\frac{1}{\pi}=\sum_{n=0}^{\infty}\frac{(1/2)_n(s)_n(1-s)_n}{(1)_n^3}(a+bn)z^n,$$where $(a)_n$ is Pochhammer symbol and $-1\leq z< 1$$, a, b$ are all algebraic numbers.
Question: Fix $s$, for example, $s=1/4$. Suppose that $z$ is rational number. Are there finite many triples $(z,a,b)$ that satisfy Ramanujan's identity?

REPLY [5 votes]: For a particular general formula, yes, there are only finitely many triples $(z,a,b)$ with rational $z$. For $\color{blue}{p = 2:}$
$$\begin{aligned}
\frac{1}{\pi} &= \sum_{n=0}^\infty \frac{256^n \big(\tfrac{1}{2})_n \big(\tfrac{1}{4})_n \big(\tfrac{3}{4})_n}{n!^3} \frac{An+B}{C^{n+1/2}}\\
\small{\color{brown}{where,}}\;&\\
A &= \sqrt{C}\big(1-2\beta)\tfrac{1}{p}\sqrt{d}\\
B &= \frac{\sqrt{C}}{\pi}\Big(\,_2F_1\big(\tfrac{1}{4},\tfrac{3}{4};1;\beta\big)\Big)^{-2}-\tfrac{1}{2}\cdot\tfrac{1}{4}\cdot\tfrac{3}{4}\cdot\tfrac{1}{p}\cdot\frac{256\sqrt{d}}{\sqrt{C}}\,\frac{_2F_1\big(\tfrac{5}{4},\tfrac{7}{4};2;\beta\big)}{_2F_1\big(\tfrac{1}{4},\tfrac{3}{4};1;\beta\big)}\\
C &= r_2(\tau)\\
\beta &= \frac{1-\sqrt{1-\frac{256}{r_2(\tau)}}}{2}\\
\small{\color{brown}{and,}}\;&\\
r_2(\tau) &= \left( \Big(\tfrac{\eta(\tau)}{\eta(2\tau)}\Big)^{12}+ \Big(\tfrac{\sqrt{2}\,\eta(2\tau)}{\eta(\tau)}\Big)^{12}\right)^2\\
\small{\color{brown}{with,}}\;&\\
\tau &= \frac{1}{4}\sqrt{-d},\quad \text{or}\quad \tau = \frac{1}{4}\big(2+\sqrt{-d}\big)
\end{aligned}$$
This formula generalizes the well-known one by Ramanujan and depends only on a single parameter, $\tau$, and ultimately on the discriminant $d$. Thus, it is easy to test various $d$ to see which yields rational $C$.
Examples:. Define,
$$h_2(n) = \frac{(4n)!}{n!^4} = \frac{256^n \big(\tfrac{1}{2})_n \big(\tfrac{1}{4})_n \big(\tfrac{3}{4})_n}{n!^3} = 1, 24, 2520, 369600, 63063000,\dots\tag1$$   
which is sequence A008977.

Let $d = 148 = 4\times37$, with class number $h(-d) = 2$, and $\tau = \frac{2+\sqrt{-148}}{4}$. Note that given the prime-generating polynomial $F(n) = 2n^2-2n+19$, then $F(\tau) = 0$. Then,
$$A = 85840i,\quad  B = 4492i,\quad  C = r_2(\tau) = -14112^2$$
$$\frac{1}{\pi} = 4\sum_{n=0}^\infty h_2(n) (-1)^n \frac{37\times580n+1123}{(14112^2)^{n+1/2}}$$
Let $d = 232 = 4\times58$, with class number $h(-d) = 2$, and $\tau = \frac{\sqrt{-232}}{4}$. Note that given the prime-generating polynomial $F(n) = 2n^2+29$, then $F(\tau) = 0$. Then,
$$A = 844480\sqrt{2},\quad  B = 35296\sqrt{2},\quad  C = r_2(\tau) = 396^4$$
$$\frac{1}{\pi} = 32\sqrt{2}\sum_{n=0}^\infty h_2(n) \frac{58\times455n+1103}{(396^4)^{n+1/2}}$$

which is Ramanujan's famous formula alluded to earlier.
Remarks:

The formulas for $p=1,2,3,4$ are given in this article Ramanujan's pi formulas and the hypergeometric function. They are my own formulation, but you can re-derive them from the more complicated versions by the Borweins or by Guillera.
Using $p=2$, I count only $12$ formulas with rational $C$, but some of the $A,B$ involve a square root. They're in Pi Formulas and the Monster. This is an older work, so is not as stream-lined as the previous article.
Ramanujan's formulas have been generalized for $p>4$ by using other well-defined integer sequences similar to $(1)$. See Ramanujan-Sato series.<|endoftext|>
TITLE: cup product and Steenrod operations in Serre spectral sequence
QUESTION [10 upvotes]: Let $F\to E\to B$ be a fibration with $B$ simply-connected. Suppose all differentials in the cohomology Serre spectral sequence (corresponding to the above fibration) are zero maps. Then as a graded module, 
$$
H^*(E)\cong H^*(F)\otimes H^*(B).
$$
Question 1: as cohomology rings with cup products,  do we still have the isomorphism
$$
H^*(E)\cong H^*(F)\otimes H^*(B)?
$$
My idea of Question 1: there is a product in the $E_r$-page
$$
E^{p,q}_r\times E^{s,t}_r\to E^{p+q,s+t}_r
$$
which induces a product in the $E_{r+1}$-page
$$
E^{p,q}_{r+1}\times E^{s,t}_{r+1}\to E^{p+q,s+t}_{r+1}.
$$
Since all differentials are zero, the products on  $E_r$-pages are same for each $r$. Hence there is a product on $E_\infty$-page which is the same as the product on $E_2$-page,  given by the product of the tensor algebra $H^*(F)\otimes H^*(B)$. However, these products are in the pages of the Serre spectral sequence, but not in $H^*(E)$.
Question 2: suppose the cohomology coefficient is taken in $\mathbb{Z}_2$ and the Steenrod operations on $H^*(F;\mathbb{Z}_2)$, $H^*(B;\mathbb{Z}_2)$ are already known. Is it possible to determine the Steenrod operations on $H^*(E;\mathbb{Z}_2)$?

REPLY [14 votes]: The behavior of the Steenrod squaring operations in the Serre spectral sequence was determined by Araki and independently by Vázquez (whose article I cannot locate online). However, it's a little work to extract the statement from Araki's paper. Roughly, there are $Sq^n$ operations, they are compatible with the differentials (though this is a little sensitive), and on the $E_\infty$ page they give the Steenrod operations on the associated graded of the cohomology of the total space (i.e. they give about the same amount of information as the cup product in the Serre spectral sequence).
One of the big differences is that the target of the Steenrod operations $Sq^n$ moves. For example, we have:
$$
\begin{align*}
Sq^n:&E_\infty^{p,q} \to E_\infty^{p,q+n} & \text{if }&n \leq q,\\
Sq^n:&E_\infty^{p,q} \to E_\infty^{p+(n-q),2q} & \text{if }&p+q \geq n \geq q.\\
\end{align*}
$$
There's actually a good reason for this. Roughly, if you imagine that your terms in the spectral sequence are $\alpha \otimes \beta$ for $\alpha$ in $H^p(B)$ and $\beta \in H^q(F)$, this corresponds to applying the Cartan formula for $Sq^n(\alpha \otimes \beta)$, eliminating the terms $Sq^a(\alpha) \otimes Sq^b(\beta)$ which are forced zero for degree reasons (if $a > p$ or $b > q$), and picking the term with maximal filtration degree $q+b$.<|endoftext|>
TITLE: A Point-free probability theory?
QUESTION [10 upvotes]: I must confess I hardly know anything about probability theory. Still, I'm interested in the following: Much like point-free topology, where one basically replaces topological spaces by their locales of open sets, I figured there is a way to do something similar with $\sigma$-algebras and with probability spaces.

Any thoughts on that? Does somebody know, whether this has been studied
  before?

Here are some more thoughts: I suppose a problem is how to recover the sample space $\Omega$ from a point-free probability space, as there is a no guarantee that there is an injection $\Omega \to \sigma$ from the sample space to the $\sigma$-algebra of a probability space. I wonder, how important it is to have a sample space at all. I (think I) know, that probability theory is actually about random variables, but do we really need a sample space to talk about those? Also, considering that there is no obvious notion of a morphism between probability spaces, maybe there are other objects we should look at?
(I asked this question on MSE and a user suggested to ask this on MO. I studied mathematics for about a year in university, so my background is not actually that sophisticated. I only know a little bit category theory, analysis and linear algebra)

REPLY [5 votes]: Possibly related... Caratheodory's book on measure and integration without points ...
Carathéodory, C. Mass und Integral und ihre Algebraisierung. (1956)
translated
Carathéodory, C. Algebraic theory of measure and integration. (1963) 
Translated from the German by F. E. J. Linton<|endoftext|>
TITLE: Does an element in the center of universal enveloping algebra becomes a scalar in irreducible representations?
QUESTION [6 upvotes]: I'm asking a question about Lie group representation. 
Let $G$ be a Lie group, not necessarily connected. Let $\Omega$ be an element in the center of the universal enveloping algebra $U(\mathfrak{g})$ of the Lie algebra $\mathfrak{g}$ of $G$, so $\Omega$ commutes with every vector $X\in\mathfrak{g}$. 
Let $\rho$ be a unitary representation of $G$ on a complex Hilbert space $H$. Then $\rho$ induces an algebra homomorphism (which will be denoted as just $\rho$) from $U(\mathfrak{g})$ into the linear endomorphism algebra of some dense subspace of $H$, the space of smooth vectors. If $\rho$ is irreducible, meaning that there is no proper nontrivial closed subspace of $H$ that is invariant under all of $\rho(g)$'s, then can we say that $\rho(\Omega)$ is a scalar multiple of the identity map on $H$?
I've been convinced that the Schur's lemma is still true for unbounded operators; more precisely, if $T$ is a densely-defined closable unbounded operator on $H$ commuting with every $\rho(g)$ (so in particular the domain $D(T)$ of $T$ is invariant under $\rho(g)$'s) then $T$ should be a scalar. But it is not clear to me that whether or not $\rho(\Omega)$ commutes with $\rho(g)$'s. Perhaps the answer to my question is true when $\Omega$ is invariant under the adjoint action of $G$, but this is also not clear. 
So my question is:

Is it true that $\rho(\Omega)$ should be a scalar?
Should $\Omega$ be invariant under $\mathrm{Ad}_{g}$?
If not, when those are true?
(1) What if $H$ is finite-dimensional?
(2) What if $G$ is connected, or simply-connected?

Thank you.

REPLY [8 votes]: No.  Let $G$ be $O_2(\mathbb{R})$, so the Lie algebra is one dimensional, and the center of the universal enveloping algebra is the symmetric algebra of the Lie algebra.  Then the usual 2-dimensional representation (which is irreducible) does not have elements of the Lie algebra acting by scalars.
You need $\Omega$ to be $Ad$ invariant.<|endoftext|>
TITLE: Roller's problem on median groups
QUESTION [8 upvotes]: At the end of his dissertation Poc Sets, Median Algebras and Group Actions, Martin Roller asks

A group $G$ is called median if it acts freely and transitively on a median algebra. This is equivalent to saying that the Cayley graph with respect to a certain set of generators is a cubing. [...] Characterize all presentations of median groups. 

Do you know any work on the subset? The only examples I am able to imagine are of the form
$$\langle x_1, \ldots, x_n \mid [x_i,x_j]=1 \ \text{and} \ x_k^2=1 \ \text{for some} \ 1 \leq i,j,k \leq n \rangle.$$
Are there other simple examples?

REPLY [4 votes]: Below are a few examples of groups admitting a median graph as a Cayley graph. (About the terminology "median groups", notice the possible confusion with groups acting geometrically on median spaces.)

Right-angled Artin groups. Given a (simplicial) graph $\Gamma$, define $$A(\Gamma)= \langle \text{vertices of $\Gamma$} \mid [u,v]=1, \ (u,v) \in E(\Gamma) \rangle.$$ Then the Cayley graph of $A(\Gamma)$ with respect to the generating set given by the previous presentation is median graph, or if you prefer, it is naturally the one-skeleton of a CAT(0) cube complex.
Right-angled Coxeter groups. The same comment holds for the group $$\langle \text{vertices of $\Gamma$} \mid u^2=1 \ (\text{$u$ vertex}), \ [u,v]=1 \ (\text{$u,v$ adjacent}) \rangle$$
LOG groups. Given an oriented graph $\Gamma$ whose edges are labelled by its vertices (such that any vertex labels at most one edge), one can define the group $$G(\Gamma)= \langle \text{vertices of $\Gamma$} \mid cac^{-1}=b, \ (a \mid c \mid b) \in E(\Gamma) \rangle$$ where $(a \mid c \mid b) \in E(\Gamma)$ means that there exists an oriented edge from $a$ to $b$ which is labelled by $c$. In his paper On the realization of Wirtinger presentations as knot groups, Rosebrock determines precisely when the square complex associated to the above presentation is nonpositively curved.
C(4)-T(4) presentations. For any C(4)-T(4) presentation whose relations have length four, the corresponding two-complex is a nonpositively curved square complex.
Right-angled mock reflection groups. In his paper Right-angled mock reflection and mock Artin groups, Scott classifies explicitely all the groups acting vertex-simply-transitively on CAT(0) cube complexes with $\mathbb{Z}_2$ edge-stabilisers.
Right-angled mock Artin groups. In the same way that a right-angled Artin group may be associated to a right-angled Coxeter group, Scott introduced right-angled mock Artin groups from right-angled mock reflection groups.
BMW-groups. In addition to Mark's answer, I would like to mention Caprace's survey Finite and infinite quotients of discrete and indiscrete groups (Section 4, Lattices in products of trees after M. Burger, S. Mozes and D. Wise).<|endoftext|>
TITLE: Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?
QUESTION [50 upvotes]: Similarly is the complement of any countable set in $\mathbb R^3$ simply connected?
Reading around I found plenty of articles discussing the path connectedness $\mathbb R^2 \setminus \mathbb Q^2$ and even an approach using cofiltered limits to approach that problem, but I am not read enough in that literature to see if this could be applied here.

REPLY [67 votes]: Yes, the complement of any countable set in $\mathbb{R}^3$ is simply connected, by the Baire category theorem.
Say your set is $X = \{x_1, x_2, ... \}$, and let $y$ be any point in  $\mathbb{R}^3 \setminus X.$
Let $f:S^1 \rightarrow \mathbb{R}^3 \setminus X$, and consider the space of homotopies $h:S^1 \times [0,1] \rightarrow \mathbb{R}^3$, where $h(x, 0) = f(x)$ and $h(x, 1) = y$. With the natural topology the space of homotopies is a Baire space, and for each $n$, the set of homotopies that avoid the points ${x_1, ... , x_n}$ is open and dense. So the set of homotopies that miss all of $X$ is nonempty.<|endoftext|>
TITLE: Interchanging the tensor product with infinite product
QUESTION [6 upvotes]: Let $R$ be a $k$-algebra (not necessary commutative) and let $\mathbf{D}(R)$ be its derived category (right modules). I'm interested in the class of objects $V$ of $\mathbf{D}(R^{op})$ having the following property (P):
For any (infinite) product $\prod_{i\in I} M_{i}\in \mathbf{D}(R)$ the natural map
$$ [\prod_{i\in I} M_{i}]\otimes^{L}_{R} V\longrightarrow \prod_{i\in I}[ M_{i}\otimes^{L}_{R} V]$$ is an isomorphism in  $\mathbf{D}(k)$.
I have two questions.


Does every perfect complex $V$ satisfy the property (P) ?

If the answer to the first question is positive, what is the biggest class of objects of $\mathbf{D}(R)$ having the property (P)? Does it have a name ?



PS: $R^{op}$ is the ring $R$ but with opposite multiplication.
Thank you.
Ed.

REPLY [5 votes]: The class of objects with property (P) is a thick subcategory of $\mathbf{D}(R^{op})$ (i.e., a triangulated subcategory closed under taking direct summands), and contains $R$, so it contains all perfect complexes, since the category of perfect complexes is the thick subcategory generated by $R$.
I think that the class of objects with property (P) is the thick subcategory generated by finitely generated flat left $R$-modules. In particular, if $R$ is left noetherian, so that finitely generated flat modules are projective, then this is just the perfect complexes.
Proof: If $N$ is a finitely generated flat module then, by flatness, $N$ has property (P) if
$$\left[\prod_{i\in I}M_i\right]\otimes_RN\to\prod_{i\in I}\left[M_i\otimes_RN\right]$$ 
for every collection of right modules $\{M_i\}$, and this is true since $N$ is finitely generated. So the class of objects with property (P) certainly contains the thick subcategory generated by finitely generated flat modules.
Now suppose that $V$ has property (P). For every collection $\{M_i\vert i\in \mathbb{Z}\}$ of right modules, the direct product $\prod_{i\in\mathbb{Z}}M_i[i]$ coincides with the direct sum $\bigoplus_{i\in\mathbb{Z}}M_i[i]$, and so by property (P) the map
$$\bigoplus_{i\in\mathbb{Z}}\left[M_i[i]\otimes^{\mathbf{L}}_RV\right]\to
\prod_{i\in\mathbb{Z}}\left[M_i[i]\otimes^{\mathbf{L}}_RV\right]$$
is an isomorphism. In particular (taking $M_i=R$ for all $i$) $V$ has bounded cohomology, and (taking $M_i$ where possible with $H^0\!\!\left(M_i[i]\otimes^{\mathbf{L}}_RV\right)\neq0$) $V$ has finite flat dimension.
So up to a shift $V$ is isomorphic to a complex
$$\dots\to0\to V^0\to V^1\to\dots\to V^{n-1}\to V^n\to0\to\dots\tag{$*$}$$
with $V^0$ flat and $V^i$ projective for $i>0$.
If $n=0$, so that $V=V^0$ is just a flat module, then property (P) for a countable collection of copies of $R$ gives that
$$\left[\prod_{i\in\mathbb{Z}}R\right]\otimes_RV^0\to\prod_{i\in\mathbb{Z}}V^0$$
is an isomorphism, which is only true when $V^0$ is finitely generated.
If $n>0$, the same argument, together with right exactness of tensor product, shows that the cokernel of $V^{n-1}\to V^n$ must be finitely generated. Taking $P$ to be a finitely generated projective module mapping onto this cokernel, and lifting to a map $P\to V^n$, we get a map of complexes $P[-n]\to V$ whose mapping cone also has property (P). But the mapping cone is isomorphic to a shorter complex of the form $(*)$, so we are done by induction on $n$.<|endoftext|>
TITLE: Advanced Differential Geometry Textbook
QUESTION [59 upvotes]: I tried this post on StackExchange with no luck. Hopefully the experts at MathOverflow can help.
In algebraic topology there are two canonical "advanced" textbooks that go quite far beyond the usual graduate courses.
They are Switzer Algebraic Topology: Homology and Homotopy and Whitehead Elements of Homotopy Theory. These are both excellent books that (theoretically) give you overviews and introduction to most of the main topics that you need for becoming a modern researcher in algebraic topology.
Differential Geometry seems replete with excellent introductory textbooks. From Lee to do Carmo to so many others.
Now you might be thinking that Kobayashi/Nomizu seems natural. But the age of those books is showing in terms of what people are really doing today compared to what you learn from using those books. They just aren't the most efficient way to learn modern differential geometry (or so I've heard).
I am looking for a book that covers topics like Characteristic Classes, Index Theory, the analytic side of manifold theory, Lie groups, Hodge theory, Kahler manifolds and complex geometry, symplectic and Poisson geometry, Riemmanian Geometry and geometric analysis, and perhaps some relations to algebraic geometry and mathematical physics. But none of these topics completely, just as Switzer does with a unifying perspective and proofs of legitimate results done at an advanced level, but really as an introduction to each of the topics (Switzer does this with K-theory, spectral sequences, cohomology operations, Spectra...).
The only book I have found that is sort of along these lines is Nicolaescu's Lectures on the Geometry of Manifolds, but this book misses many topics.
This was inspired by page viii of Lee's excellent book: link where he lists some of these other topics and almost implies that they would take another volume. I'm wondering whether that advanced volume exists.
Any recommendations for great textbooks/monographs would be much appreciated!
Edit: there are many excellent recommendations (I particularly like the Index theory text mentioned by Gordon Craig in the comments as it doesn't shy away from analysis, and does so many things in geometry plus has extensive references) below. 
One other reference that I found which people may find interesting is the following: link and link2 where Prof. Greene and Yau say: "It is our hope that the three volumes of these proceedings, taken as a whole, will provide a broad overview of geometry and its relationship to mathematics
in toto, with one obvious exception; the geometry of complex manifolds...Thus the reader seeking a complete view of geometry would do well to add
the second volume on complex geometry from the 1989 Proceedings to the
present three volumes". However most of the articles are research level articles and lack the coherence and unified vision of a textbook/monograph.

REPLY [3 votes]: A nice recent (2017) book that covers some of items listed in the question is Differential Geometry: Connections, Curvature, and Characteristic Classes by Loring W. Tu.<|endoftext|>
TITLE: Zero divisors with support of size 3 in group algebras of finite groups
QUESTION [6 upvotes]: Are there a finite group $G$ and a field $\mathbb{F}$ such that $\gcd(3,|G|)=1$ and the group algebra $\mathbb{F}[G]$ contains a zero divisor whose support is of size $3$?
Recall that the support of an element $\alpha$ of $\mathbb{F}[G]$ is  the set $\{x\in G \;|\; \alpha(x)\neq 0\}$.
This question is related to the following one:
Zero divisors of the form $1+x+y$ in the rational group algebra
One motivation to propose the question is the following well-known observation: if $a$ is an element of order $3$ in a group $G$ then $1+a+a^2$ is a zero divisor over any group algebra of $G$ (whose support is of size $3$). So the hypothesis $\gcd(3,|G|)=1$.

REPLY [6 votes]: It is not really an answer to the question, but Ilya Bogdanov's answer puts quite strong restrictions on zero divisors of $\mathbb{F}G$ when $G$ is finite. Let $I(G)$ denote the augmentation ideal of $\mathbb{F}G$, that is $\{ \sum_{g \in G} \lambda_{g} g : \sum_{g \in G} \lambda_{g} = 0 \}$, which is indeed a two-sided ideal of $\mathbb{F}G$. Then every element  of $I(G)$ is annihilated by $\sum_{ g \in G} g$, as Ilya observed, so $I(G)$ consists of zero divisors. 
In the opposite direction, we can draw the conclusion that whenever $ab = 0$ for $a,b \in \mathbb{F}G$, at least one of $a$ or $b$ is in $I(G)$. For if $a$ is not in $
I(G)$, we can write 
$ a = \lambda 1_{G} + i$ where $0 \neq \lambda \in \mathbb{F}$ and $i \in I(G)$.
Then $0 = ab = \lambda b +ib$. However, $ib \in I(G)$, so that $\lambda b \in I(G)$, and hence $b \in I(G)$. Similarly if $b$ is outside $I(G)$, we find that $a \in I(G)$, since $I(G)$ is a two-sided ideal. In particular, note that all nilpotent elements of $\mathbb{F}G$ lie in $I(G)$ by repeated applications of this.
(Later edit: In fact, note that a zero divisor $b \in \mathbb{F}G$ which annihilates
an element $a$ lying outside $I(G)$ must itself lie in $\cap_{n = 1}^{\infty}I(G)^{n}$, since we see above that there is some $i \in I(G)$ and non-zero $\lambda \in \mathbb{F}$ such that $ib = -\lambda b$ (and a similar argument if we had $ba = 0$).<|endoftext|>
TITLE: Remark on Fermat's Last Theorem by Darmon, Diamond and Taylor
QUESTION [8 upvotes]: In their paper, Darmon, Diamond and Taylor remarked the following :
(the previous paragraph of Section 2.2 (p. 55), https://www.math.wisc.edu/~boston/ddt.pdf)
If $\rho : G \rightarrow GL_2(\mathbb{C})$ is irreducible, then $\text{ad}^0 \rho$ is either irreducible or the direct sum of two representations (one is a non-trivial character of order 2 and the other is an induced representation.) For more detail, see the file linked above. Can someone help me to understand this remark?

REPLY [12 votes]: The assocated $\mathbb{C}G$-module, say $M$, is $2$-dimensional, and 
${\rm End}_{\mathbb{C}}(M)$ is $4$-dimensional. The action of $g \in G$ on 
${\rm  End}_{\mathbb{C}}(M)$ is via conjugation by $\rho(g)$. Then ${\rm End}_{\mathbb{C}}(M)$ decomposes as a direct sum of two modules under this this action as $S \oplus {\rm End}^{0}(M)$, where $S$ is the set of scalar matrices, and ${\rm End}^{0}(M)$ is the set of matrices of trace $0$. The latter module is $3$-dimensional. 
The representation $\rho$ may be primitive or imprimitive. It is primitive if it can't be induced from a representation of a proper subgroup, and imprimitive if it can be so induced.
In the latter case, it must be induced from a (non-trivial) $1$-dimensional representation of a subgroup $H$ of index $2$. This means that there are two $H$-invariant $1$-dimensional spaces $U$ and $V$ which are interchanged by $G$, and such that the actions of $H$ on $U$ and $V$ are different ( otherwise the induced representation of $G$ would not be irreducible).
In that case, with respect to the right basis, $G$ is acting by monomial matrices. The action of $G$ on ${\rm End}_{\mathbb{C}}(M)$ is also by monomial matrices with respect to a suitable basis, since the Kronecker product of monomial matrices is monomial.
Note that, in this case, the fixed point subspace of $H$ on ${\rm End}_{\mathbb{C}}(M)$ is $2$-dimensional. This fixed point space is $G$-invariant, and completely reducible since it is really a module for the finite group $G/H$. It therefore decomposes as the direct sum of the trivial module (from $S$, which is the whole space of $G$-fixed points by Schur's Lemma) and a $1$-dimensional module on which $H$ acts trivially and every element of $G \backslash H$ acts as $-1$.
In the monomial action of $G$ on ${\rm End}_{\mathbb{C}}(M)$, one can see that there remains a $2$-dimensional summand on which $H$ acts as non-trivial diagonal matrices, and $G$ interchanges the corresponding $1$-dimensional subspaces. This is the remaining induced $2$-dimensional summand.
There  is one scenario where that $2$-dimensional module is not irreducible for $G$ (and the authors did not specify whether that two dimensional module was irreducible or not so this remark is tangential). Let $\alpha$ and $\beta$ be the respective characters of $H$ afforded by the two $H$-invariant one-dimensional subspaces of $M$. Then the linear characters of $H$ in its diagonal action on ${\rm End}_{\mathbb{C}}(M)$ are $1,1,\alpha\beta^{-1}$ and $\alpha^{-1}\beta$. Hence if $\alpha(h)^{2} = \beta(h)^{2}$
for all $h \in H$, then $H$ has the same action on the two remaining spaces, and $G$ does not act irreducibly on that space. This can happen if $H$ itself has a normal subgroup $K$ of index $2$, in which case $H$ has a linear character $\lambda$ with kernel $K$.
Then if we have $\beta = \lambda \alpha$, the corresponding induced 2-dimensional submodule of ${\rm End}_{\mathbb{C}}(M)$ module is not irreducible for $G$, and decomposes as a direct sum of two one-dimensional modules. The dihedral group or order $8$ has a $2$-dimensional irreducible representation of this type, induced from a representation of a Klein $4$-subgroup.
Hence the second scenario specified by the authors occurs if the representation is imprimitive. If the representation is primitive, then it is easy to see (using Clifford's Theorem) that every normal subgroup of $G$ which does not act as scalars on $M$ acts irreducibly. If ${\rm End}_{\mathbb{C}}^{0}$ were reducible, then it would have a non-trivial $1$-dimensional irreducible $G$-submodule whose kernel $H$ would be a proper normal subgroup of $G$. Then $H$ does not acts irreducibly on $M$ by Schur's Lemma, so by primitivity, $H$ acts as scalars on $M$  and $H \leq S(G)$. Now the derived group $G^{\prime}$ is contained in $H$ since $G/H$ is Abelian. Hence $G^{\prime}$ acts as scalars on $M$. In fact every element of $G^{\prime}$ acts as $\pm I$ on $M$, since an element of $G^{\prime}$ must act a matrix of determinant $1$. Now for any element $x \in G \backslash S(G)$ we see that $\langle \rho(x), -\rho(x) \rangle$ is an Abelian (proper) normal subgroup of ${\rm Im}(\rho)$ ( since it contains the derived group $[\rho(G),\rho(G)]$) which does not consist of scalars, in contradiction to the primitivity of $\rho$.<|endoftext|>
TITLE: Injective model structure on sheaves of bounded complexes of $A$-modules
QUESTION [7 upvotes]: The following might be very well known for people who works with model categories, but I do not find the answer.
Let $A$-be a ring. Denote $\mathbf{Ch}_+(A)$ the category of positive degree cochain complexes of $A$-modules (complexes where differential rises the degree and which are zero in negative degree). There is the injective model structure on $\mathbf{Ch}_+(A)$ which has quasi-isomorphisms as weak equivalences, monomorphisms as cofibrations and epimorphisms with injective kernel as fibrations. Let $K^\bullet$ be a complex, then its injective resolution $I^\bullet (K^\bullet)$ turn out to be its fibrant replacement.
Let $X$ be a topological space and consider $\mathbf{Ch}_+(\mathbf{Shv}_X(A))$, the category of positive complexes of sheaves of $A$-modules on $X$. My question is the following:
Question: Is there an analogue model structure in $\mathbf{Ch}_+(\mathbf{Shv}_X(A))$ to the injective model structure in  $\mathbf{Ch}_+(A)$? More concretely, if we consider quasi-isomorphisms as weak equivalences, monomorphisms as cofibrations, and morphisms having the right lifting property with respect to injective quasi-isomorphisms, do they define a model structure? If $\mathcal{K^\bullet}$ is a complex of sheaves, will its injective resolution $I^\bullet (\mathcal{K}^\bullet)$ be its fibrant replacement?
Thank you very much

REPLY [4 votes]: David White's comment led me to adequate references to answer this question. Thank you very much
Question: Is there an analogue model structure in $\mathbf{Ch}_+(\mathbf{Shv}_X(A))$ to the injective model structure on $\mathbf{Ch}_+(A)$?  More concretely, if we consider quasi-isomorphisms as weak equivalences, monomorphisms as cofibrations, and morphisms having the right lifting property with respect to injective quasi-isomorphisms, do they define a model structure?
Answer: Yes, there is such model structure. The above mentioned three clasess do form a model structure. Find the result in a letter to A. Grothendieck from Joyal in Théorème 2 (p.10) 
Question: If $\mathcal{K}^\bullet$ is a complex of sheaves, will its injective resolution $I^\bullet(\mathcal{K}^\bullet)$ be its fibrant replacement? 
Answer: Yes, it will. The same argument that shows that for a chain complex of modules its fibrant replacement is a complex made of injective modules holds for sheaves. Find the argument for modules (although the case of projective resolutins) in this paper,<|endoftext|>
TITLE: Derived global functions on (derived) stacks $BG$ and $G/G$
QUESTION [7 upvotes]: In Toen's Affine Stacks, he computes that $\mathcal{O}(B\mathbb{G}_a) = k[\epsilon]$ with $|\epsilon| = 1$ and trivial differential (where here $\mathcal{O}$ is computed in a derived sense, and we choose dg algebras as a model for derived rings).  To prove this, he explicitly constructs a cosimplicial ring whose $\text{Spec}$ is $B\mathbb{G}_a$, verifying this by showing they satisfy the same universal property.
I'm wondering if there's a more explicit way to do this calculation.  In particular, I would like to compute $\mathcal{O}(BG)$ for various linear algebraic $G$ (unipotent, solvable, semisimple...).  I would also like to compute $\mathcal{O}(G/G)$ (the adjoint action).
Here's an idea for an approach, which might be totally wrong.  These stacks are Artin stacks, so in particular one can write $BG$ as a colimit of a simplicial diagram in schemes.  Then by definition, $\mathcal{O}(BG)$ is a limit of cosimplicial diagram in algebras.  Here I'm not exactly sure what I'm doing, but I want to say I can apply some kind of (dual) monoidal Dold-Kan to realize this limit in dg algebras.  I think (not sure) one only gets a commutative algebra structure up to homotopy here, but at least taking cohomology, this ends up just being the Hochschild complex for computing the ("rational") cohomology of the (rational) trivial representation of $G$.
Using this, I think one computes that $\mathcal{O}(B\mathbb{G}_m) = k$ (but haven't verified the details), and Jantzen's book (Representations of Algebraic Groups) verifies Toen's result for $\mathbb{G}_a$.  Very quickly though, this computation becomes difficult.  Are there general results/computations known for unipotent, solvable, or reductive $G$, and can one compute $\mathcal{O}(G/G)$ in a similar way (e.g. using the rational cohomology of $k[G]$ under the adjoint action)?  Do you have a preferred way to think about this?

REPLY [4 votes]: For a smooth groupoid $G_1\rightrightarrows G_0$ , here is a way to compute derived global sections $\mathcal O(X)$ of $\mathcal O_X$, where $X=[G_0/G_1]$. 
Observe that $X$ is the homotopy colimit of the nerve $G_\bullet$ of the groupoid. As you said then $\mathcal O(X)$ is the homotopy limit of $\mathcal O(G_\bullet)$. This homotopy limit can indeed be computed via cosimplicial Dold-Kan. 
This is particular tells you that, as Pavel wrote in his comment, $\mathcal O(BG)=C^*(G,k)$. 
Considering the action groupoid $G\times G\rightrightarrows G$ for the conjugation action you get (as suggested in your question) that $\mathcal O(G/G)=C^*(G,\mathcal O(G))$, where the action of $G$ on $\mathcal O(G)$ comes from the conjugation on $G$. 
For a reductive $G$, we know that $C^*(G,V)=V^G$ for any $G$-module $V$. Hence $\mathcal O(BG)=k$ and $\mathcal O(G/G)=\mathcal O(G)^G$ is the algebra of class functions. 
As Pavel said in his comment, $\mathcal O(B\mathbb{G}_a)=C^*(\mathbb{G}_a,k)=C^*(Lie(\mathbb{G}_a),k)$. Then $Lie(\mathbb{G}_a)=Free(k)$ is the free Lie algebra generated by $k$. Now for any perfect complex $V$, $C^*(Free(V),k)$ is quasi-isomorphic to the square-zero extension $k\oplus V^*[-1]$. One thus gets back that $\mathcal O(B\mathbb{G}_a)=k[\epsilon]$.<|endoftext|>
TITLE: The growth of a subset of a group
QUESTION [6 upvotes]: Let $S$ be a symmetric subset of a group $G$ containing the identity, and let $S^n$ be the set of all products of $n$ elements of $S$.  If $S^3\subset gS$ for some translate $gS$ of $S$ then it follows that $S^2=S^3$.  My question is, if $S^2\subset gS$ does it follow that $S=S^2$?
My question is about the growth of a group in geometric group theory.  For a finitely generated group $G$ the growth $G$ is the function that assigns to each natural number $n$ the cardinality $|S^n|$ for some finite generating set $S$.  Up to an equivalence of growth functions this does not depend on the generating set chosen.  Further, the function $(S^n:S)$ that assigns to each $n$ the smallest number of translates of $S$ that cover $S^n$ is in the equivalence class of $|S^n|$.  So my question is asking in a general group $G$ does $(S^2:S)=1$ imply $S=S^2$.  For $G$ finitely generated by $S$ the answer is yes but trivially because $gS$ and $S$ have the same cardinality.
In general I'm interested in the situation where $G$ is a topological group equipped with a coarse structure and $S$ is a small subset that generates $G$.  So for example, it is relevant to assume $G$ is connected and Haar measurable and $S$ is a precompact open set.
Thank you for considering my question.

REPLY [5 votes]: The answer is yes. In fact, it can be proved in a totally elementary way. 
Because $1 \in S^2 \subset gS$, we have $g^{-1} \in S$ and thus $g \in S$. Then since $S^2 \subset gS \subset S^2$, it follows that $$S^2 = gS.$$ By symmetry of $S$, we obtain $S^2 = (S^2)^{-1} =  (gS)^{-1} = S g^{-1}$. Hence $S \subset S^2 = gS = Sg^{-1}$, and thus $$g^{-1} S, Sg \subset S.$$
$ $
Let $x \in S$ be any element. By symmetry, $x^{-1} \in S$. Because $g^{-1} S, Sg \subset S$, both $$x^{-1} g, \; xg^2 \in S.$$
Since $S^2 = gS$, it follows that $g^{-1} S^2 = S$ and that $$g^{-1} (x g^2) (x) = g^{-1} x g^2 x \in S.$$
Also, $$g^{-1} (x^{-1} g) (g^{-1} x g^2 x) = gx \in S.$$ Thus for every $x \in S$, we have $gx \in S$. This means that $gS \subset S$, and as we have shown $S^2 = gS$, we get $$S \subset S^2 = gS \subset S. $$ Therefore $S^2 = S.$<|endoftext|>
TITLE: Non trivial rank 2 holomorphic vector bundles in complex dimensions greater than or equal 2
QUESTION [7 upvotes]: Does every compact complex manifold of complex dimension greater than or equal two 
possess a nontrivial rank 2 holomorphic vector bundle?

REPLY [4 votes]: Some partial answer:
On non-Kahler surfaces we can give stronger version of your question and we have allways the following thorem
Theorem: Let $S$ be non-Kahler surfaces, For any $n > 0$ there exists a Mumford stable rank-2 vector bundle $E$ with trivial determinant, ($det(E) = \mathcal O_S$ ) and $c_2(E)=n$.
We have also the following theorem due to Banica - Le Potier
Theorem: If $S$ is a non-projective surface and $N$ is enough big
, then there exists rank-2 vector bundles on $S$ with $c_2(E) = N$
which are non-filtrable.
We have following theorem due to Ballico and Schwarzenberger
Theorem: If $S$ be a projective surface and $E$ is any topological rank-2
vector bundle with $det(E) \in Pic(X)$, then $E$ has a holomorphic
structure by applying elementary transformations starting from the
trivial bundle.
We have Cartan-Serre theorem for construction of rank 2 vector bundles on projetive varieties.
Theorem: Let $X$ be a complex manifold, $L_1, L_2 ∈ Pic(X)$ are line bundles,
$Z ⊂ X$ with $codim_X(Z) = 2$. Under some cohomological conditions, the
sheaf $E$ sitting in 
$$0 \to L_1 \to E \to L_2 \otimes J_Z → o$$
is locally free. Moreover, If $X$ is projective, any rank-2 holomorphic vector bundle can be constructed this way<|endoftext|>
TITLE: Integral cohomology ring of K(Z,3)
QUESTION [17 upvotes]: Computing the cohomology of Eilenberg Maclane spaces is a feasible but difficult problem in algebraic topology. The general answer is quite complicated (see the MO answer and the reference therein https://mathoverflow.net/a/24759/184 )
However the cohomology of K(Z,2) is quite simple, it is just a polynomial algebra on a generator in degree two. I am wondering about the next easiest case.


What is the integral cohomology ring of K(Z,3)? 


You can get quite far with spectral sequence calculations, but if this is worked out in detail somewhere, why reinvent the wheel?

REPLY [13 votes]: Given a complex oriented cohomology theory $E$, one can define a formal scheme of Weil pairings on the associated formal group, as explained in the paper "Weil pairings and Morava $K$-theory" by Matthew Ando and me.  If we let $R_E$ denote the ring of functions on this scheme, then there is a natural map $R_E\to E^*K(\mathbb{Z},3)$.  This is an isomorphism if $E$ is Morava $K$-theory or Morava $E$-theory.  I think that it is also an isomorphism for $E=MU$ or $E=kU$ but not for $E=H$.  However, there is a natural short exact sequence
$$ kU^*K(\mathbb{Z},3)/v \to H^*K(\mathbb{Z},3) \to \text{ann}(v,kU^*K(\mathbb{Z},3)) $$
(where $v$ is the standard generator of $\pi_2kU=kU^{-2}(\text{point})$).  I think that this is probably an effective way to understand $H^*K(\mathbb{Z},3)$.
Some other things that are going on in the background here:

There is a fibration $K(\mathbb{Q}/\mathbb{Z},2)\to K(\mathbb{Z},3)\to K(\mathbb{Q},3)$.  Here $K(\mathbb{Q},3)$ is the rationalisation of $S^3$ and is not so hard to understand.
$K(\mathbb{Q}/\mathbb{Z},2)$ is the colimit of the spaces $K(\mathbb{Z}/n,2)$.
One can understand $K(\mathbb{Z}/n,2)$ using the multiplication map $K(\mathbb{Z}/n,1)\times K(\mathbb{Z}/n,1)\to K(\mathbb{Z}/n,2)$.<|endoftext|>
TITLE: Irreducible cubics modulo primes
QUESTION [7 upvotes]: Is there a small finite (perhaps of cardinality two or three) collection of cubic polynomials $p_1, \dotsc, p_k \in \mathbb{Z}[x]$ such that for every prime $p$ at least one of these is irreducible?

REPLY [11 votes]: No. It follows from Chebotarev's density theorem that for any polynomials $p_1,\dots,p_k\in\mathbb{Z}[x]$ there are infinitely many primes that split all these polynomials. Simply, apply this theorem for the number field $K$ generated by the roots of $p_1,\dots, p_k$, and note that if an unramified prime completely splits in $K$ then each $p_i$ splits into linear factors modulo that prime.<|endoftext|>
TITLE: cohomology of iterated loop space on spheres
QUESTION [8 upvotes]: In the book The homology of iterated loop spaces, the homology Hopf algebra
(1)
$$
H_*(\Omega^n \Sigma^n X;\mathbb{Z}_p)
$$ 
for primes $p\geq 2$ is obtained on p. 226, Thm. 3.2. In particular, the homology Hopf algebra
(2)
$$
H_*(\Omega^nS^n;\mathbb{Z}_2)
$$
is known. However, when I use the Cartan formula and Adem relations to compute the coproduct of (2), I find it is quite complicated when modulo the Adem relations,  and do not know how to get the explicit expression of the coproduct. 
Question: Are there any references where I can find the explicit expression of the cohomology algebra
$$
H^*(\Omega^nS^n;\mathbb{Z}_2)?
$$
I obtain that $\Omega S^1=\text{Map}_*(S^1,S^1)\simeq [S^1;S^1]_*=\pi_1(S^1)=\mathbb{Z}$. Hence $H^*(\Omega S^1;\mathbb{Z}_2)=\oplus_{k\in\mathbb{Z}} \mathbb{Z}_2a_k$, $|a_k|=0$. How to compute the following examples
$$
H^*(\Omega^2S^2;\mathbb{Z}_2)
$$
and
$$
H^*(\Omega^3S^3;\mathbb{Z}_2)?
$$

REPLY [6 votes]: As is known from J.W. Milnor, J.C. Moore, "On the structure of Hopf algebras" Ann. of Math. (2), 81 : 2 (1965) pp. 211–264, the algebra structure of an underlying algebra of a Hopf algebra is quite limited, so that it suffices to study $p$-th powers to determine the algebra structure.  Dually, it suffices to 
study the Verschiebung, and not all the coproduct.  This has been worked out by 
Wellington, in "The unstable Adams spectral sequence for free iterated loop spaces" Memoirs of the American Mathematical Society, Vol.36 Number 258 (1982).<|endoftext|>
TITLE: A problem in elementary geometry
QUESTION [17 upvotes]: Let us have a triangle ABC  in the Cartesian plane and consider the following transformation of this triangle: 
On  the ray  AB starting at A, select a point B' so that  so that |AB'|=|AC|. Likewise,
 on the ray BC starting at B, select a point C' so that  BC' so that |BC'|=|AB|,
 and on the ray CA starting at C select a point A' so that |BA'|=|BC|.
 As a result, a new triangle A'B'C' is obtained.
Prove (or disprove)  that the perimeter of a new triangle A'B'C'  does not exceed the perimeter of ABC.
Remark. The question has been stated by Ahmet Yaşar Özban, Atilim University in regard with his research on iterative methods.

REPLY [16 votes]: Let O be the point of intersection of the angle bisectors of the initial triangle ABC (the incenter). Then the length of OB' is equal to the length of OC, |OC'|=|OA| and |OA'|=|OB|. Now the statement of the problem will follow from the following fact: given point O and three length x,y,z, consider all triangles EDF that satisfy |OE|=x, |OD|=y, |OF|=z; among those the triangle with the greatest perimeter is unique (up to congruence) and it is the only triangle in the family for which O is the incenter. 
Consider the maximum point, clearly it exists, fix two of the rays, say, OE and OD, and let the third point F move along the circle centered at O. If the triangle EDF maximizes the perimeter, then this circle is tangent to the ellipse with foci at E and D, then by the property of the tangent of an ellipse, OF is the bisector of the angle EFD. This implies that for the triangle of maximal perimeter O is the incenter. 
The last remark is that there is only one triangle with fixed distances from the incenter to the vertices, it follows for example by applying the sine theorem to small triangles EOF, FOD and DOE.<|endoftext|>
TITLE: Commuting ODE's implies existence of nonzero vanishing two variable polynomial?
QUESTION [9 upvotes]: Write $\partial := d/dt$, fix $m, n > 0$, and let$$F = \partial^n + f_1(t)\partial^{n-1} + \dots + f_{n-1}\partial + f_0,\text{ }G:= \partial^m + g_1(t)\partial^{m-1} + \dots + g_{m-1}\partial + g_0$$be a pair of ordinary differential operators with rational coefficients (i.e. with coefficients in $\mathbb{C}(t)$). Assume that $F$ and $G$ commute (as operators). Then, for any polynomial$$p = \sum_{i, j} c_{i, j} x^i y^j \in \mathbb{C}[x, y],$$there is a well-defined differential operator$$p(F, G) = \sum_{i, j} c_{i, j} F^i \circ G^j,$$where$$F^i := F \circ \dots \circ F\text{ }(i \text{ times}),\text{ }G^j := G \circ \dots \circ G\text{ }(j \text{ times}).$$My question is, if $F$ and $G$ commute, does there exist a nonzero polynomial $p \in \mathbb{C}[x, y]$ such that one has $p(F, G) = 0$?
EDIT: $n = 2$ is not hard. For each $\lambda \in \mathbb{C}$, let $V_\lambda$ be the space of solutions $\psi \in C^\infty(a, b)$ of the differential equation $F(\psi) = \lambda \cdot \psi$, on a small enough segment $[a, b] \subset \mathbb{R}$. The operator $G$ acts on $V_\lambda$. We use that there exists a polynomial $p_\lambda \in \mathbb{C}[y]$ such that $p_\lambda(G|_{V_\lambda}) = 0$. Then, we analyze the dependence of solutions $\psi$ on the initial condition at the point $a$ to show that we can find $p_\lambda$ in a way that the function $(\lambda, \mu) \mapsto p_\lambda(\mu)$ gives a polynomial $p \in \mathbb{C}[x, y]$ such that $p(F, G) = 0$.

REPLY [4 votes]: See Mulase's notes. The result whereof you speak is the Theorem of Burchnall-Chaundy.<|endoftext|>
TITLE: Probability of random geodesics on the half-sphere intersecting
QUESTION [6 upvotes]: 4 end points (a,b,c,d say) are chosen uniformly randomly and connected a to b and c to d by two geodesics on the 2-dim half-sphere. Here, uniform means that, probability that a point lies on a surface area of size $da$ is proportional to $da$. What is the probability that the two curves cross?

REPLY [2 votes]: Alternatively to Igor Rivin's method, here is what I think could lead to the result.
To get four independent uniformly distributed random points on the hemisphere, we first chose two (independent, uniformly distributed) diameters of its boundary in the equatorial plane, say AOB and COD, with an angle $\theta$ between them, distributed uniformly on $[0,\pi/2]$. Then we chose two planes containing AOB and COD respectively, making with the equatorial plane an angle of $\xi$ (or $\eta$, respectively) uniformly distributed on $[0,\pi]$ ; they intersect the hemisphere on half-circles. Each point M on the half-circle is defined by the angle $\alpha$ (or $\gamma$) from OA (or OC) to OM, and the half-circle is equipped with the probability measure $\sin\alpha\ d\alpha/2$ (or $\sin\gamma\ d\gamma/2$). With this choice of conditional distribution, if M is chosen randomly it will be uniformly distributed on the half-sphere, I (strongly) believe.
Clearly then, $ab\subset$ AB and $cd\subset$ CD intersect iff both contain the point M$_0$ where the two half-circles intersect (at angles $\alpha_0$ on AB, $\gamma_0$ on CD): this (given $\theta$, $\xi$ and $\eta$) has probability $\frac12 \sin^2\alpha_0\times\frac12 \sin^2\gamma_0$.
These angles $\alpha_0$ and $\gamma_0$ of M$_0$ can be determined using spherical trigonometry (considering the spherical triangle with $\theta$ as one side and $\xi,\eta$ as adjacent angles). You get$$\tan\alpha_0=\frac{2\sin\eta}{\cot(\theta/2)\sin(\xi+\eta)+\tan(\theta/2)\sin(\xi-\eta)}$$and$$\tan\gamma_0=\frac{2\sin\xi}{\cot(\theta/2)\sin(\xi+\eta)-\tan(\theta/2)\sin(\xi-\eta)}$$
Integrating in $\theta$, $\xi$ and $\eta$ should then give the desired probability that $ab$ and $cd$ intersect.<|endoftext|>
TITLE: Computer algebra system for Weyl algebra computations
QUESTION [7 upvotes]: Does anyone have a suggestion for the best computer program to perform calculations in the 2nd Weyl algebra?

REPLY [4 votes]: I use GAP (http://www.gap-system.org/) with the GBNP package (http://www.win.tue.nl/~amc/pub/grobner/chap0.html).
You define Weyl algebra as a quotient of free associative algebra by the ideal
$<x_1 x_2 - x_2 x_1 - 1>$.
Easy to generalize to arbitrary dimension and can handle Clifford algebras
(of arbitrary signatures) in the same setting.<|endoftext|>
TITLE: About the relation between the categories $\text{Sch}$, $\text{LRS}$ and $\text{RS}$
QUESTION [10 upvotes]: I've asked this question https://math.stackexchange.com/questions/1407451/about-the-relation-between-the-categories-textsch-textlrs-and-text on math.stackexchange , however I don't think I will receive any response there (because of the current activity in my post). Therefore I'm asking it here. If the question seems inconvenient because of the excess of questions, I can split this question into other ones (let me know).
I've been trying to find some useful categorical facts about the category of schemes, locally ringed spaces and ringed spaces (that I shall denote by $\text{Sch}$, $\text{LRS}$ and $\text{RS}$ respectively). The motivation is that I'm trying to compute some (co)limits explicitly in the category of schemes.
There's this excellent answer here https://math.stackexchange.com/questions/102973/on-limits-schemes-and-spec-functor , but I still have some doubts.
More precisely, I want to know about the following assertions:
1) Is the category of locally ringed spaces (co)complete? The answer is yes by prop 1.6 in Demazure and Gabriel's "Groupes Algébriques" (I didn't notice that they proved the general case and not just the case of filtered colimits when I posted this question, sorry)
In the answer cited above, the references implies the existence of cofiltered limits and filtered colimits, however as I understand the notion of filtered in these cases is restricted to the case where the index category is a poset.
2)Is the category of ringed spaces (co)complete?
3)What can be said about the underlying topological space of the (co)limit of locally ringed spaces? (Is it the (co)limit of the topological spaces?)
4)What can be said about the underlying topological space of the (co)limit of  ringed spaces? (Is it the (co)limit of the topological spaces)
5)What can be said about the underlying topological space of the colimit of  schemes? (Is it the (co)limit of the topological spaces)
Obviously, the underlying topological space of the pullback of schemes is not the pullback of the topological spaces (for instance, $\text{Spec} (\mathbb{C}) \times_{\text{Spec} (\mathbb{R})}\text{Spec} (\mathbb{C}) \cong \text{Spec} (\mathbb{C}\times\mathbb{C})$ by $a \otimes z \mapsto (az, a\overline{z})$), but the case of push outs seems to be true.
6) Are (co)limits preserved under the inclusions $\text{Sch} \hookrightarrow\text{LRS} \hookrightarrow \text{RS}$?
The inclusion $\text{LRS} \hookrightarrow \text{RS}$ preserves colimits since it's a left adjoint (see below)
7) For each forgetful functor $U : \mathcal{C} \rightarrow \mathcal{D}$, where $\mathcal{C}$ and $\mathcal{D}$ are equal to $\text{Sch}$, $\text{LRS}$ and $\text{RS}$ (for all possible coherent substitutions), are there adjoint functors?
8) For each inclusion $\mathcal{C} \hookrightarrow \mathcal{D}$, where $\mathcal{C}$ and $\mathcal{D}$ are equal to $\text{Sch}$, $\text{LRS}$ and $\text{RS}$ (for all possible coherent substitutions), are there adjoint functors?
According to http://arxiv.org/abs/1103.2139 [Cor. 6], the inclusion $\text{LRS} \hookrightarrow \text{RS}$ have a right adjoint given by localization of the terminal prime system.
Thanks in advance.

REPLY [6 votes]: Edited to incorporate Marc's comments below:
1) The cocompleteness of LRS is Prop 1.6 in Demazure-Gabriel's Groupes algebriques. Completeness is proved in http://arxiv.org/abs/1103.2139, Corollary 5.
2) Yes. The category of ringed spaces is (co)fibered over the category of topological spaces (which is (co)complete) and has (co)complete fibers. EDIT: To answer the OPs question, I can't think of a reference offhand but it's not too bad to prove straight from the definitions. Take your diagram upstairs, form the (co)limit downstairs, choose (co)cartesian lifts of the projection/inclusion maps from/to your (co)limit, form the (co)limit in the fiber, and you're good. 
4) The forgetful functor RS → Top preserves limits and colimits. See (2), in this situation (co)limits are constructed via their projections to the underlying topological space.
3) The proof of Demazure-Gabriel's Prop 1.6 shows that the inclusion $LRS\subset RS$ preserves colimits (even better: creates them).
5) See (6).
6) Colimits are not preserved by the inclusion $Sch \subset LRS$, see Emerton's comment here: Colimits of schemes. Colimits are preserved by $LRS \subset RS$ as stated above in (3). The inclusion $Sch \subset LRS$ preserves finite limits by results in section 5.1 of Demazure-Gabriel.<|endoftext|>
TITLE: Optimisation of betting strategy
QUESTION [5 upvotes]: Consider integers $r \geq 1$ and $k \geq 1$ and consider the following game:
We start with $r$ tokens and at each round we choose $i \in \{1,...,r\}$ tokens to bet (if we have $N<r$ tokens we can't bet more than $N$). If we chose to bet $i$ tokens then we flip a coin $X_i$ with $P(X_i=0) = (1-i/r)^k$ and $P(X_i=1) = 1-P(X_i=0)$. The flip is independent of everything.
If $X_i=1$ then we get $r-i$ tokens, otherwise we lose $i$ tokens. (If we have $M$ tokens at the beginning of the round then at the end we have $M+r-i$ or $M-i$ tokens)
We stop playing the game when we have 0 tokens. It is more or less clear that we can play forever (if we bet $r$ tokens per round), but I want the opposite. What is the best strategy to maximise the probability to end the game?
I think that independently of $k$ and $r$, the best strategy to maximise such probability is to bet 1 token per round (such probability is the extinction probability of a branching process with binomial offspring distribution with parameters $r$ and $1-(1-1/r)^k$).
Can you suggest a strategy to prove that or to find  a counterexample? I was looking on the internet, but it was very hard to find similar problems, so if you can suggest some lectures It will be amazing.

REPLY [2 votes]: You can reduce the problem from optimizing on the space of strategies to analyzing the single-step deviations from what you believe is the optimal strategy. This is a common reduction.
Let $p_k(n)$ be the probability that the game ends if you bet one token at a time, starting with $n$ tokens. At the boundary, $p_k(0)=1$. If you bet one token at a time, then $p_k$ is a martingale. You can do better if and only if there exists $(i,n)$ so that betting $i$ from $n$ increases $p_k$ on average. If $p_k$ can't increase on average on a single step, then $p_k$ is a nonincreasing semimartingale under any strategy, it starts at $p_k(r)$, and so the probability of ending at $1$ is at most $p_k(r)$. If $p_k$ can increase on average on a step from $(i,n)$, then you can do better than betting one unit at a time, simply by deviating only the first time you hit $n$ tokens, which happens with positive probability.<|endoftext|>
TITLE: Number of distinct factors
QUESTION [13 upvotes]: Denote $\omega(m)$ to be number of distinct factors of $m$ as defined in http://mathworld.wolfram.com/DistinctPrimeFactors.html.
At every $c>0$, given $n\in\Bbb N$ define $$S(n,c)=\big\{m\in\Bbb N:\mbox{ }m<n,\mbox{ }\omega(m)<(\log\log m)^c\big\}.$$
What is known about cardinality $|S(n,c)|$?
What probabilistic model has a good fit for $\omega(m)$?
I am looking for finer information such as:
How does $\frac{\large\partial\big(\big|S(n,c)\big|\big)}{\large\partial c}$ behave? 
For a given $a>1$, is there a $c$ as a function of $m$ where $1-\frac 1a$ fraction of numbers less than $m$ lie in?

REPLY [10 votes]: Let me supplement the answer of i707107 for $c<1$, i.e. when we count integers with very few prime factors. Writing 
$$ \pi_k(n):=|\{m\in\Bbb N:\mbox{ }m\leq n,\mbox{ }\omega(m)=k\}|,$$
the cardinality $|S(n,c)|$ can be well approximated with the sum of $\pi_k(n)$ for $k<(\log\log n)^c$. Now, we have very precise information for $\pi_k(n)$ in this range, in fact even in the wider range $1\leq k\leq A\log\log n$ with $A>0$ fixed. Namely, Selberg (1954) proved that
$$\pi_k(n)=\frac{n}{\log n}\frac{(\log\log n)^{k-1}}{(k-1)!}
\left\{\lambda\left(\frac{k-1}{\log\log n}\right)+O_A\left(\frac{k}{(\log\log n)^2}\right)\right\}, $$
where
$$ \lambda(z):=\frac{1}{\Gamma(z+1)}\prod_p\left(1+\frac{z}{p-1}\right)\left(1-\frac{1}{p}\right)^z.$$
You can read the proof in Chapter II.6 of Tenenbaum: Introduction to analytic and probabilistic number theory. The notes to this chapter contain further information, e.g. an extension of the above result valid for $k\ll(\log\log n)^2$. This extension, with a slightly different looking asymptotic, is due to Hildebrand-Tenenbaum (1988), and it corresponds to $c\leq 2$ in your notation.<|endoftext|>
TITLE: Generate polyhedra by collapsing vertices of a polyhedron
QUESTION [9 upvotes]: I am looking for basic information about the following idea:
(I) Consider a square. By collapsing two adjacent vertices, we obtain a triangle.
(II) Consider a three-dimensional cube. By collapsing a single face to a line, we obtain a prism with triangular base. Next, by collapsing a line longitudinally, we obtain a pyramix with square base. Next, by contracting two adjacent vertices of the square base, we obtain a tetrahedron.
How can this can be generalized to higher dimensions? It is obvious that we can transform a hypercube to a simplex by a sequence of contractions. But not all contractions of faces produce a polyhedron (with flat faces), and many contractions produce the same object up to congruence. Since the idea is so basic, I hope somebody else has already written about it.
Hence I look for information, in particular about the following questions:

Which contractions actually produce polyhedra, and which class of polyhedra is actually produced?
Since different contractions produce congruent polyhedra, how can we relate (sequences of contractions) and congruence relations?
Do these operations have an analogon on the symmetry groups of the polyhedra?

Of course, it would be best to have a book reference on this topic.

REPLY [4 votes]: Perhaps this paper addresses some of your questions?

Nevo, Eran. "Higher minors and Van Kampen's obstruction." 
  Math. Scand. 101 (2007), no. 2, 161–176.
  (arXiv preprint.)

Nevo proves this:

This generalizes an $\mathbb{R}^3$ result of Dey et al., 
"Topology preserving edge contractions"
(PDF download.)<|endoftext|>
TITLE: What are some useful invariants for distinguishing between random graph models?
QUESTION [15 upvotes]: Quite a few probabilistic algorithms for generating random graphs exist in the literature, such as:

The Erdős-Rényi model
The Stochastic Block model
The Watts-Strogatz model
The Barabasi-Albert model

and many others. The general sort of question that I am trying to ask is: how do you do Bayesian statistics with random graphs?  In other words: what can you infer about the underlying random graph model by looking at data?  A natural strategy is to look at statistical properties of ordinary graph invariants, such as:

The degree distribution
The distribution of path lengths
The clustering coefficient

These specific invariants are nice for statistics because they are relatively predictable for many graph models (e.g. it is known that degrees are asymptotically Poisson for Erdős-Rényi graphs and asymptotically power law for preferential attachment graphs) and they aren't too hard to compute from data.

What are some other "nice for statistics" graph invariants?

I'm curious about spectral invariants in particular, though this might be no easier than the well-known hard problem of estimating eigenvalues of random matrices.
For this question let's assume that the graphs are undirected, simple, and large but finite.

REPLY [2 votes]: It really depends on your applications/goals, and the graphs you consider.
Do you want to compare graph sequences or graphs of a fixed size? 
First let's consider sequences.
The problem with property testing invariants is that they are trivial for any non-dense class of graphs (including the Watts-Strogatz and Barabasi-Albert models).
In other words, $\mathcal{G}_i \rightarrow W_0$ for any graph sequences with $o(n^2)$ edges, where $W_0$ is the empty graphon.
The close equivalent of such properties for sparse-graphs is the distribution of neighborhoods, which gives the framework of local weak convergence.
You can also look at the degree-degree correlations(Joint-Degree Distribution), or correlations for more vertices, but it becomes computationally expensive.
There are some invariants that work for any class of graph, such as the class of first-order definable properties, or connectivity, hamiltonicity, etc, but either they are a bit too simple (FO-properties) or they lack a proper way to be treated uniformly.
There are many other less known invariants, either in the social network community (Between-ness, Elite/Rich club or community analysis) or in the theoretical world (a good place to start is [1]). Here the keyword is "graph metrics".
If you are not interested in asymptotic analysis, you can pretty much use anything, including a mix of algorithms for sparse graphs and algorithms for dense graphs. With some applications in mind, you can also use the result/statistics of some algorithms as invariants (e.g. some routing algorithms). In that case rigorous mathematical may be out of the question, but it may be much more useful in practice [2][3].
[1] Bollobás, Riordan "Sparse graphs: metrics and random models."
[2] Sala, Cao, Wilson, Zablit, Zheng, "Measurement-calibrated graph models for social network experiments"
[3] Li, Alderson, Willinger, Doyle, "A first-principles approach to understanding the internet's router-level topology"<|endoftext|>
TITLE: Repeats of all binary strings of length k
QUESTION [8 upvotes]: The question seems like it should be known, but I was not able to find it anywhere. 
How many binary strings of length $n$ are required so that for every $k$ positions in these strings, all $2^k$ possible subsequences occur? 
For example, suppose $n=3$, and $k=2$. We want a set of binary strings of length $3$ so that if you look at the first and third symbols (or first two, or last two), you will see all $4$ patterns $00$, $01$, $10$, and $11$. For example, the strings with an even number of $1$s $\{ 000, 011, 110, 101\}$ induce all subsequences on each set of two positions.  
Let $f(n,k)$ be the minimum number. Trivially, $f(n,k) \ge 2^k$ so $f(3,2)=4$. 
I am interested in precise upper and lower bounds for $f(n,k)$. Bounds that are within a constant of each other (independent of $n$ and $k$) suffice for my purposes.

REPLY [7 votes]: Orthogonal arrays were mentioned in another answer, but you are not requiring that each $k$-tuple occurs exactly once in every set of $k$ columns, but rather that each $k$-tuple occurs at least once in every set of $k$ columns.
What you are looking for is called a covering array. In the usual notation for covering and orthogonal arrays, $v$ is the size of the alphabet ($2$ in this case), $k$ corresponds to your $n$, and $t$ corresponds to your $k$. Some known upper bounds for the number of rows for small values of $t,k,v$ are listed in http://www.public.asu.edu/~ccolbou/src/tabby/catable.html - you are interested in the numbers for $(t,k,2)$.<|endoftext|>
TITLE: What is the fewest number of points you must delete from $\mathbb{R}^3$ to make it not simply connected?
QUESTION [35 upvotes]: This question concerns a set-theoretic aspect that I found interesting in the recent question asked by user Nick R., namely, Is
$\mathbb{R}^3\setminus\mathbb{Q}^3$ simply connected? He had asked whether $\mathbb{R}^3$ remains simply connected after deleting a countable set of points, such as the collection of rational points $\mathbb{Q}^3$.
That question was answered affirmatively by Martin M. W. My question is, can we do better? Specifically, I want to understand, in general context where the continuum may be very large, exactly how many points we
may freely delete from $\mathbb{R}^3$, whilst remaining simply
connected. What is the fewest number of points that we must delete from $\mathbb{R}^3$ in order to make it no longer simply
connected?
Let us define the simply
connected deletion number, $\delta$, to be the smallest cardinality of a subset
$A\subset\mathbb{R}^3$, such that the complement
$\mathbb{R}^3\setminus A$ is no longer simply connected.
Martin's answer to the earlier question shows that deleting any
countable number of points preserves the simply connected
property, and so the simply connected deletion number is
definitely uncountable, at least $\omega_1$. And since it is
clearly at most the continuum, the question is settled if the
continuum hypothesis holds. Like all other cardinal
characteristics of the continuum, this number is more interesting
when the continuum hypothesis fails.
In a comment,
I had suggested that Martin's argument suggested that the simply
connected deletion number should be at least as large as
cov$(\cal{M})$, the covering number of the meager
ideal, which is the fewest number of meager sets whose union is
the whole space. My reason for suggesting this was that as far as I understand
Martin's answer (which I admit is imperfectly), he is proposing that for
any one point $x$, there is a comeager set of homotopies that
avoids $x$. So in order to avoid all the points in a set $P$, we need to
know that the intersection of $|P|$ many comeager sets in his space of homotopies is
nonempty. This is the same as knowing that the unions of $|P|$
many meager sets (the complements) is not the whole space of homotopies, in order that there is at
least one desired homotopy that avoids every point in $P$.
If this is right, then we would deduce that the simply connected
deletion number is at least cov$(\cal{M})$, provided that the covering number for meager sets in his space was the same as for our other more familiar spaces. (If someone
could explain and confirm this inequality in greater detail, please post
an answer! I would want to see more details than Martin had provided about the space of homotopies.)
Question. What is the simply connected deletion number
exactly? Is it consistent that this number is strictly less than
the continuum? Is it necessarily the continuum? How does it relate to the other standard cardinal
characteristics of the continuum? What is the value under Martin's
axiom? Is it equal to cov($\cal{M}$)? Can it be
strictly larger than cov$(\cal{M})$?

REPLY [13 votes]: The simply connected deletion number equals the continuum. This follows from the fact that for any dense subset $A$ in the real line the subset $A_3=(A\times \mathbb R\times \mathbb R)\cup (\mathbb R\times A\times\mathbb R)\cup (\mathbb R\times\mathbb R\times A)$ is 2-dense in the following sense: for any continuous ap $f:[0,1]^2\to\mathbb R^3$ and any $\varepsilon>0$ there is a continuous map $g:[0,1]^2\to \mathbb R^3$ such that $g$ coincides with $f$ on the boundary of $f$ and $g((0,1)^2)\subset A_3$, and $g$ is $\varepsilon$-near to $f$. The proof of this fact is a bit more complicated than I wrote in the preceding version of this answer, so we can proceed differently.
Given a subset $S\subset\mathbb R^3$ of cardinality $|S|<\mathfrak c$, construct by induction a dense countable set $C$ in $\mathbb R^3$ such that  any 3-element subset $B$ of $С$ is affinely independent and its affine hull $aff(B)$ does not intersect $S$. (To construct such a set $C$ fix any countable base $(U_n)$ of the  topology and in $n$th set $U_n$ choose a point $c_n$ such that any 3-element subset $B\subset\{c_0,\dots,c_n\}$ is affinely independent and its affine hull misses $S$).
Then the union $\Delta=\bigcup\{aff(B):B\subset C,\;|B|=3\}$ is 2-dense in $\mathbb R^3$. To prove the 2-density of $\Delta$, fix any function $f:[0,1]^2\to\mathbb R^3$. We need to construct a map $g:[0,1]^2\to\mathbb R^2$ such that $g$ coincides with $f$ on the boundary of $[0,1]^2$, $g((0,1)^2)\subset\Delta$ and $g$ is near to $f$. On 
$(0,1)^2$ the map $g$ can be defined by the formula $g(x)=\sum_{U\in\mathcal U}\lambda_U(x)c_U$ where

$\mathcal U$ is a cover of $(0,1)^2$ by open sets whose diameters tend to zero as $U$ tends to the boundary of the square,
$\mathcal U$ has order $\le 3$ in the sense that each point of $(0,1)^2$ is contained in at most 3 sets of the cover $\mathcal U$ (such choice of $\mathcal U$ is possible as $dim(0,1)^2=2$);
$(\lambda_U)_{U\in\mathcal U}$ is a partition of the unity subordinated to the cover $\mathcal U$, and
for every $U\in\mathcal U$ the point $c_U$ belongs to $C$  and is 
sufficiently near to a point in the set $f(U)$.<|endoftext|>
TITLE: Sums of unique squares
QUESTION [6 upvotes]: Let $\mathbb{N}$ denote the positive integers and let $S = \{n^2: n\in \mathbb{N}\}$. For any positive integer $k$ we define $$\text{sq}(k) = |\{F\subseteq S: F\neq \emptyset, F\text{ is finite and } k = \sum_{n\in F} n\}|.$$
Questions:

Is the function $\text{sq}:\mathbb{N}\to \mathbb{N}\cup\{0\}$ surjective?
Is there $m_0>1$ such that $\text{sq}^{-1}(\{m\})$ is infinite for all $m\geq m_0$?

REPLY [14 votes]: The generating function is 
$$
\sum_{n\geq 0} \text{sq}(n) z^n = \prod_{k\geq 1} (1+z^{k^2}).
$$
Using complex integration you can use this to get an asymptotic formula for $\text{sq}(n)$. This involves quite some work, but the path is well described in Andrews, The theory of partitions, chapter 6. You will arrive at something like $\text{sq}(n)\sim e^{n^{1/3}}$ times some minor terms, hence $\text{sq}$ is ultimately increasing at a pretty fast rate. In particular, $\text{sq}(n)$ is not surjective. For $\text{sq}^{-1}$ you can either derive an asymptotic series or compute the proportion of all partitions not containing the summand $1^2$ to find that $\text{sq}$ is strictly increasing from some point onwards. Hence you will most likely obtain that $\text{sq}^{-1}(\{m\})$ is infinite if and only if $m=1$.<|endoftext|>
TITLE: convert a special case of nonlinear fractional programming into a convex problem
QUESTION [5 upvotes]: Is it possible to convert a fractional problem (maximization) with objective function equal to the ratio of a concave function and convex function ? This question sound impossible but I have read this claim in wikipedia without any reference !
https://en.wikipedia.org/wiki/Nonlinear_programming#Methods_for_solving_the_problem
"If the objective function is a ratio of a concave and a convex function (in the maximization case) and the constraints are convex, then the problem can be transformed to a convex optimization problem using fractional programming techniques"

REPLY [2 votes]: Here is some insight into Jean's answer, showing what that final problem means.  Suppose you transform the problem to finding a vector $(y,t) \in \mathbb{R}^{n+1}$ to solve: 
\begin{align*}
&\mbox{Max: } & tf(y/t) \\
&\mbox{Subject to: } & tg(y/t) \leq 1  \\
& & y/t \in S  \\
& &  t>0 \\
& & y \in \mathbb{R}^n 
\end{align*}
where $S$ is a convex subset of $\mathbb{R}^n$, the functions $f(\cdot)$ and $g(\cdot)$ are defined over $S$, and $f$ is concave while $g$ is convex.   To show this is a convex optimization problem, it suffices to show: 
i) The constraints $y/t \in S$, $y \in \mathbb{R}^n$, $t >0$ together specify a convex set. 
ii) The function $tf(y/t)$ is concave in the joint variables $(y,t)$. 
iii) The function $tg(y/t)$ is convex in $(y,t)$. 
To this end, define $A$ as the set of all $(y,t)$ such that $t>0, y \in \mathbb{R}^n, y/t \in S$. 

Claim 1: The set $A$ is a convex subset of $\mathbb{R}^{n+1}$. 
Proof: Let $(y_1,t_1)$ and $(y_2,t_2)$ be elements if $A$.  Let $\theta \in [0,1]$ and define $\overline{\theta} = 1-\theta$. We want to show that
$$ (\theta y_1 + \overline{\theta}y_2, \theta t_1 + \overline{\theta}t_2) \in A$$
Since $t_1>0$ and $t_2>0$, it follows that $\theta t_1 + \overline{\theta}t_2>0$. It remains to check the ratio condition: 
$$ \frac{\theta y_1 + \overline{\theta} y_2}{\theta t_1 + \overline{\theta}t_2}  = \left(\frac{\theta t_1}{\theta t_1 + \overline{\theta}t_2}\right)(y_1/t_1)  + \left(\frac{\overline{\theta} t_2}{\theta t_1 + \overline{\theta}t_2}\right)(y_2/t_2) \in S$$
since $S$ is convex and $y_1/t_1 \in S$, $y_2/t_2 \in S$, and this is just a convex combination of $y_1/t_1$ and $y_2/t_2$. $\Box$.

Claim 2: The function $tf(y/t)$ is concave over $(y,t) \in A$. 
Proof: Fix $(y_1, t_1)$ and $(y_2, t_2)$ in $A$.  Fix $\theta \in [0,1]$ and define $\overline{\theta} = 1-\theta$. Then: 
\begin{align} 
\theta t_1f(y_1/t_1) + \overline{\theta}t_2f(y_2/t_2) &= \left(\theta t_1 + \overline{\theta}t_2\right)\left(\frac{\theta t_1f(y_1/t_1) + \overline{\theta}t_2f(y_2/t_2)}{\theta t_1 + \overline{\theta}t_2}\right) \\
&\leq \left(\theta t_1 + \overline{\theta}t_2\right)f\left( \frac{\theta t_1 (y_1/t_1) + \overline{\theta}t_2 (y_2/t_2)}{\theta t_1 + \overline{\theta}t_2}\right)\\
&= \left(\theta t_1 + \overline{\theta}t_2\right)f\left( \frac{\theta y_1 + \overline{\theta}y_2}{\theta t_1 + \overline{\theta}t_2} \right)
\end{align} 
where the inequality holds because the function $f()$ is concave. $\Box$

That the function $tg(y/t)$ is convex over $(y,t) \in A$ follows by a proof similar to that of Claim 2.<|endoftext|>
TITLE: Is there a nonabelian free group inside a group of positive rank gradient?
QUESTION [5 upvotes]: Let $G$ be a finitely generated residually finite group with positive
  rank gradient, and let $F_2$ be the free group on $2$ elements. Must
  there be an embedding $i \colon F_2 \to G$ ?

A group $G$ is called residually finite if the intersection of all of its finite index subgroups is trivial.
A group $G$ is said to have positive rank gradient if the rank of finite index subgroups grows linearly with the index. More formally, the Rank Gradient (RG) of a finitely generated group $G$ is defined to be:
$$\mathrm{RG}(G) = \inf_{H} \frac{\mathrm{rank}(H) - 1}{[G : H]}$$ where $H$ runs over all subgroups of finite index in $G$, and the rank of a group is the smallest cardinality of a generating set for the group.
The notion of rank gradient is also important in the theory of $3$-manifolds.

REPLY [4 votes]: Lackenby ( http://people.maths.ox.ac.uk/lackenby/lg070105.ps ) proved that a finitely presented group, which has a pro-$p$ completion of positive rank gradient, is large, i.e. it contains a subgroup of finite index that projects onto a non-abelian free group. So while the answer to the original question is no, if you add some extra conditions it becomes yes in a very strong sense.<|endoftext|>
TITLE: Parametrized Atiyah-Singer index theorem
QUESTION [7 upvotes]: Let $M$ be any smooth manifold (could be unorientable - I think). Let $E,F \to M$ be two complex vector bundles. Let $S$ be any compact space, and let $D_s:E\to F,s\in S$ be a continuous family elliptical pseudo-differential operator. Since $D_s$ is a Fredholm operator (choosing the right Sobolev spaces to represent sections) for each $s$ we must be able to associated an index bundle over $S$.
The Atiyah-Singer index theorem basically states that the dimension of this bundle at a point $s\in S$ can be computed using the formula explained here.
Question: How do you describe and prove what the entire map $S\to \mathbb{Z}\times BU$ which classifies the index bundle is?
Notes:
1) I tried googling different things, but whenever I tried a link it was always something different than the answer to this question, but maybe I missed an obvious link. It seems like this should be explained somewhere.
2) In this question Paul Siegel gives a K-theoretic description of the index formula. This leads me to believe that the answer might be something a long the line of: The symbol of $D_s$ defines an element of $K((S\times M)^{TM})$ (Thom space of pull back bundle to the product). We can use the Thom isomorphism to identify this with a $K$ class of the spectrum $(S\times M)^{-TM}$ (we subtract the complex vector bundle $\mathbb{C}\otimes TM = TM\oplus TM$ from the Thom construction). We then use that $M^{-TM}$ has a unit map from the sphere spectrum $\mathbb{S} \to M^{-TM}$ to compose:
$$S_+\wedge \mathbb{S} \to S_+\wedge M^{-TM} \cong (S\times M)^{-TM}$$
and then pullback the class. If this is correct I would like to known, and also get a reference to where this is proved.

REPLY [5 votes]: Since Thomas is satisfied with my comment, I will post it as an answer to close: this result is in the 4th Atiyah-Singer paper: jstor.org/stable/1970756<|endoftext|>
TITLE: Laplace-Beltrami and averaging
QUESTION [8 upvotes]: For a Riemannian manifold $M$ with metric $g$ and Laplace-Beltrami operator $-\Delta_{g}$, what conditions on $M$ guarantee that $-\Delta_{g} u(x)$ measures the difference between $u(x)$ and the average of $u$ over a geodesic ball (or sphere) centered at $x$? More precisely, what conditions on $M$ guarantee that
$$-\Delta_{g} u(x) ~ \propto ~ \lim_{h \to 0}{ \frac{2}{h^2} \left( u(x) - \frac{1}{|B(x,h)|} \int_{B(x,h)}{ u(y) dy } \right) } \ \ ? $$
Here, $B(x,h)$ is the geodesic ball with center $x$, radius $h$, and measure $|B(x,h)|$. My guess is that this holds on harmonic manifolds (where harmonic functions can be characterized by the mean value property), but I haven't found the result anywhere.

REPLY [8 votes]: This holds in complete generality. I sketch the case in which, instead of the ball $B(x,h)$, you consider the metric sphere $S(x,h)$. In particular (here $\dim M = n$ and the Laplacian is $\Delta = \mathrm{div}\circ \mathrm{grad}$):
$$ (\Delta u)(x) = \lim_{h\to 0} \frac{2n}{h^2}\frac{1}{|S(x,h)|}\int_{S(x,h)} [u(y)-u(x)] dy .$$
The case of the ball is similar with a different constant. Choose normal coordinates in a neighbhorhood of $x$, such that
$$u(y)-u(x) = \sum_{i=1}^k (\partial_i u)(x) y_i + \frac{1}{2} \sum_{i,j=1}^n (\partial_{ij}^2 u)(x) y_i y_j + O(|y|^3). $$
The Riemannian metric in these coordinates is $g_{ij}= \delta_{ij}+O(|y|^2)$, hence up to higher orders the measure is the Euclidean one. Also the metric sphere, in these coordinates, for small $h$, coincides with the Euclidean sphere $\mathbb{S}^{n-1}$.
When you take the average over the sphere of each term, the linear part averages to zero (as the integral of a linear function over the sphere). The integral of the quadratic part averages to the sum of second derivatives. To see this, use the fact that for any symmetric matrix $M$
$$\int_{\mathbb{S}^{n-1}} x^* Q x \,d\mu_{\mathbb{S}^{n-1}} = \frac{|\mathbb{S}^{n-1}|}{n} \mathrm{Tr}(Q).$$
Hence, at your point $x$, in normal coordinates, you have
$$ \lim_{h\to 0} \frac{2n}{h^2}\frac{1}{|S(x,h)|}\int_{S(x,h)} [u(y)-u(x)] dy = \sum_{i=1}^n (\partial_i^2 u)(x).$$
At the point $x$ this coincides with $(\Delta u)(x)$ written in normal coordinates centered in $x$. Since your averaging expression is coordinate invariant, the equality holds in general.<|endoftext|>
TITLE: Sums of reciprocals involving divisor sums
QUESTION [8 upvotes]: This question was asked at MSE but never received an answer.
Let $A\subset\mathbb{N}$ be a subset of the natural numbers, and let $\sigma(n)$
denote the sum of divisors of $n$.  Recall that we have the
bound  $\sigma(n) = O(n\log\log n)$. Consider the sums $\sum_{a\in A} 1/a$ and $\sum_{a\in A} 1/\sigma(a)$.
Question. Is there on $A\subset\mathbb{N}$ so that the first sum diverges but the second 
           sum converges ?
If the answer is already known, a reference would more than suffice.
Thank you

REPLY [11 votes]: Yes, this is possible.  Let $k\ge 3$, say and let $P(k)$ denote the product of the first $k$ primes, and let $A_k$ denote a set of integers that are all multiples of $P(k)$ and with 
$$ 
\frac{1}{k \log k} \le \sum_{a\in A_k} \frac{1}{a}  \le \frac{2}{k\log k}. 
$$ 
Since the harmonic sum diverges, we can clearly choose such $A_k$, and moreover we may arrange for the sets $A_3$, $A_4$, $\ldots$ all to be disjoint (just pick $A_k$ to be the multiples of $P(k)$ in suitable disjoint intervals).   Take $A$ to be the union of all the $A_k$ (with $k\ge 3$).  Clearly $\sum_{a\in A} 1/a$ diverges.  
Now for $a\in A_k$ we have $\sigma(a)/a \ge \sigma(P(k))/P(k) \gg \log k$ by Mertens.  Therefore 
$$ 
\sum_{a\in A_k} \frac{1}{\sigma(a)} \ll \frac{1}{\log k} \sum_{a\in A_k} \frac 1a \ll \frac{1}{k (\log k)^2}. 
$$ 
Therefore 
$$ 
\sum_{a\in A} \frac{1}{\sigma(a)}
$$ 
converges.<|endoftext|>
TITLE: A road to inter-universal Teichmuller theory
QUESTION [29 upvotes]: What would be a study path for someone in the level of Hartshorne's Algebraic Geometry to understand and study inter-universal Teichmuller (IUT) theory? I know that it heavily relies on anabelian geometry and earlier works of Mochizuki, but what's the order to study those material? I think I had seen somewhere a complete list of papers to read from beginning to end in order to come to a level of understanding to tackle the original four papers about IUT theory, but I can't find it.

REPLY [6 votes]: I'll try to cover the "considerable gap" mentioned in Myshkin's answer, from algebraic geometry on the level of Hartshorne to Mochizuki's work. As a disclaimer, I'll mention that my understanding of Mochizuki's approach is very, very far from satisfactory; however, I believe that it's still possible to figure out some of the more basic prerequisites to Mochizuki's work (which at the moment I'm also still trying to learn about) without entirely understanding Mochizuki's work itself.
"0". Algebraic Number Theory - Aside from algebraic geometry, some basic ideas of modern algebraic number theory are certainly important, such as the local-global principle, and why we might want to develop analogues for p-adic fields of things we are already familiar with in the real and complex case.
Suggested references: Algebraic Number Theory by Jurgen Neukirch or Algebraic Number Theory by J. W. S. Cassels and A. Frohlich.
I. Elliptic Curves - Elliptic curves play a big part in Mochizuki's approach. The abc conjecture is equivalent, via the Frey curve, to the Szpiro conjecture, which concerns elliptic curves. Mochizuki's approach is often said to be (including by Mochizuki himself) analogous to proofs of the Szpiro conjecture for the function field case; first, in his earlier "Hodge-Arakelov Theory", he tries to follow the proof of Szpiro himself (see Minhyong Kim's answer to this question), and in his "Inter-Universal Teichmuller Theory", the proof of Bogomolov and Zhang (see Zhang's paper, and Mochizuki's discussion of the analogy).
In addition, the Tate parametrization of elliptic curves plays an important role in Mochizuki's Inter-Universal Teichmuller Theory. See, for example, Kiran Kedlaya's lecture at the 2015 conference on Inter-Universal Teichmuller Theory at Oxford.
Suggested References: The Arithmetic of Elliptic Curves and Advanced Topics in the Arithmetic of Elliptic Curves by Joseph Silverman
II. Anabelian Geometry - Anabelian geometry is the idea that we can "reconstruct" certain algebraic varieties (called anabelian varieties) from their etale fundamental groups. Before developing Inter-Universal Teichmuller Theory, Mochizuki became well-known for proving that hyperbolic curves (which include, for example, elliptic curves with one point removed, and the projective line with three points removed) are anabelian varieties. Ideas of "reconstruction" figure heavily into Mochizuki's approach; although in what way exactly, I have not yet been able to figure out.
Suggested References: The Grothendieck Conjecture on the Fundamental Groups of Algebraic Curves by Hiroaki Nakamura, Akio Tamagawa, and Shinichi Mochizuki
Additionally, I second the recommendation of Fesenko's notes mentioned in Myshkin's answer, as well as notes from this seminar on Kummer Classes and Anabelian Geometry at Vermont.<|endoftext|>
TITLE: Are all transversely oriented, transversely measured foliations given by closed forms?
QUESTION [6 upvotes]: This paper, "AF-algebras and topology of 3-manifolds" seems to claim on page 5 that a [transversely] oriented measured foliation on a compact surface is given by a closed form. On the other hand, the book "Differential Geometry of Foliations: The Fundamental Integrability Problem" by Reinhart places a restriction on the transversely oriented foliation to have trivial holonomy. On the other hand, the latter result pertains to $C^k$-foliations, while the first one refers to the smooth case.
Is the claim in the paper false?
EDIT: "Smooth" is a bit of misnomer. What I had in mind is foliations suited to  train tracks (so the singular leaves only have cusps as singularities.)

REPLY [2 votes]: The claim is true and you find it on page 319 of Farb-Margalit's book.
The point is that the transition maps between foliation charts (with the leaves as y-level sets) are of the form $(x,y)\to (f(x,y),\pm y+c)$. (This is because the transverse measure has to be preserved. So the transverse measure seems essential for the claim to be true.)
Clearly $dy$ or $-dy$ define the singular foliation in a chart, and by coorientability of the foliation you can then consistently choose one of the two for each chart.<|endoftext|>
TITLE: Current status of the linearity of mapping class group
QUESTION [11 upvotes]: In the paper A faithful linear-categorical action of the mapping class group of a surface with boundary it is claimed that it was (as of 2014) still unknown that mapping class group is linear, but the authors provide a faithful linear-categorical action of the mapping class group of a surface with necessarily non-empty boundary using knot Floer homology that does not seem decategorifiable in a non-trivial way. 
On the other hand, there exists the following paper, Mapping class groups are linear, in which the author uses the non-commutative machinery to prove that they are. As I have asked in this question, there might perhaps be a mistake in one of the proof ingredients.
Hence the question: is it known that a mapping class group of some surface (possibly with non-empty boundary or non-zero number of punctures) is linear?

REPLY [6 votes]: Yes. There is the paper by Bigelow, and our own Ryan Budney, who do this for genus 2. I believe that is still the last word.<|endoftext|>
TITLE: Segal's 1999 Stanford lecture notes on TQFT, where to find them?
QUESTION [10 upvotes]: I am trying to track down a copy of Graeme Segal's 1999 lecture notes on topological field theory. These are sometimes referred to as the "Stanford lectures" or something similar. 
For many years these unpublished notes were available at Duke's mathematics website via the link: http://www.cgtp.duke.edu/ITP99/segal/
However this link is now dead. Every reference I have found by googling seems to lead back to this dead link. It would be a tragedy if these lecture notes were lost to the mathematics community. I am hoping that somebody out there has retained a copy.

REPLY [2 votes]: The page you're looking for is a mirror of http://web.math.ucsb.edu/~drm/conferences/ITP99/segal/. Unfortunately, lectures 4 and 6 are missing, as are all the diagrams in lecture 1.<|endoftext|>
TITLE: bar construction and loop space cohomology
QUESTION [9 upvotes]: Let $X$ be a topological space. Then its chain complex $C_{*}(X)$ is naturally a coalgebra (as explained here Does homology have a coproduct?). In particular if $X$ is simply connected  we have that the homology of the cobar construction $\Omega C_{*}(X)$ of $C_{*}(X)$ is isomorphic to the homology of the pointed path space $\Omega X$ (Adam's theorem).
I'm looking for reference about similar statements using the bar construction: 
a) Consider the cochain complex $C^{*}(X)$ as a dg algebra equipped with the cup product. Assume that $X$ is simply connected. Then under which conditions the cohomology of the bar construction $BC^{*}(X)$ is isomorphic to the cohomology of the path space?
b) $C_{*}(X)$ is a coalgebra where the coproduct is the composition of the Alexander-Whitney map with the diagonal map. By taking the dual we get a dg algebra $C^{*}(X)$ with a product $\mu$. Let $B'C^{*}(X)$ be the bar construction of ($C^{*}(X)$, $\mu$). What is the relation between $BC^{*}(X)$ and $B'C^{*}(X)$?

REPLY [13 votes]: Over the rationals or the reals, K.T. Chen described a concrete way of relating the bar construction of the CDGA algebra of differential forms on a manifold to the smooth singular cochains on the based loop space. It is dual to Adams' cobar construction and can be thought as a De Rham type theorem for the based loop space. The main statement is the following. Let $M$ be a connected smooth manifold and let $(\mathcal{A}(M), d, \wedge)$ be the CDGA of differential forms on $M$. Let $A$ be a sub CDGA of $\mathcal{A}(M)$ such that  $A^0=0$, $A^i=\mathcal{A}^i(M)$ for $i>1$, and $A^1$ is a complement of $d\mathcal{A}^0(M)$, i.e. $\mathcal{A}^1(M)=d\mathcal{A}^0(M) \oplus A^1$.
Consider the bar construction $(B(A), D)$. This is a DG commutative coassociative Hopf algebra with underlying vector space $T(sA)$, the tensor coalgebra on the shifted $A$, product given by shuffling monomials, coproduct given by deconcatenation of monomials, and differential given by extending $d + \wedge$ as a coderivation, as usual. Chen constructed a chain map
\begin{eqnarray}
\int: B(A) \to C^*(\Omega M)
\end{eqnarray}
inducing a map of Hopf algebras on cohomology, where $C^*(\Omega M)$ denotes the real smooth singular cochains on the based loop space of $M$.
Moreover, if $M$ is simply connected, the above map induces an isomorphism on cohomology. The map is constructed by a process of iterated integration (or integration over a simplex) on a given monomial of differential forms on $M$. More precisely, given $[w_1|...|w_m] \in B(A)$, $\int w_1 ... w_m$ is the cochain that sends any smooth simplex $\sigma: \Delta^n \to \Omega M$ to the integral 
\begin{eqnarray}
\int_{\Delta^m \times \Delta^n} \sigma_1^*(w_1)...\sigma_m^*(w_m)
\end{eqnarray}
where $\sigma_i: \Delta^m \times \Delta^n \to M$ is defined by $\sigma_i(t_1,...,t_m,s)=\sigma(s)(t_i)$. It is a beautiful theory exposed in several papers of Chen from the 70's and there are still paths to be explored.<|endoftext|>
TITLE: Category theory for Algebraic Geometry
QUESTION [10 upvotes]: How much of category theory should I know to view schemes, sheaves and cohomology concepts as concrete cases of abstract categorical concepts? Is there a textbook of category theory for AG people?

REPLY [5 votes]: The  answer to your first question "how much of category theory..." is partly that it is a matter of taste. Some people use a lot, while others don't. But, if you plan to study schemes, it is necessary to know some. For example, the product of varieties can be treated naively, but for schemes it definitely needs to be understood in a categorical sense. For existence, it helps to the know that the category of affine schemes is equivalent to the opposite of the category of commutative rings. Many schemes can be understood best in terms of the functor it represents etc. etc.<|endoftext|>
TITLE: Can I find Fermat's complete works anywhere?
QUESTION [12 upvotes]: I admire the mathematician very much and want to look at his writings.  Is there anywhere in book or web form that has a collection of his writings?

REPLY [2 votes]: You can find the letters  here.<|endoftext|>
TITLE: Curvature of a finite metric space
QUESTION [18 upvotes]: I am sorry to ask a very vague question, but:

What are good ways to define the curvature of a finite metric space?

The best way I can think of is: the curvature of a finite metric space $M$
is the infimum of the real $k$ such that there is a geodesic metric space $X$ which is $Cat(k)$ and $M$ embeds isometrically in $X$. However, I don't know how to compute this curvature form the distance matrice... Is there a way? Moreover
is this notion defined somewhere, and has its properties being studied? 
I would also be open to any other sensible definition, or reference discussing this question. Thanks.
PS: I ask this question for experimental purposes. I have a metric set of data I would like to define and compute the curvature of.

REPLY [5 votes]: If something more analogous to Ricci curvature than to sectional curvature would interest you, then there has been some work done on the Ricci curvature of discrete spaces. I don't know much about it, but you can find some articles on Yann Olivier's website, and Jürgen Jost and Christian Leonard gave a minicourse on it at the IHS last spring. These might be some starting points for you.<|endoftext|>
TITLE: Free $k[x_1, \dots, x_n]^{S_n}$-module?
QUESTION [7 upvotes]: Let $\text{char}\,k = 0$ and $n \ge 2$. What is the easiest way to see that $k[x_1, \dots, x_n]$ is a free $k[x_1, \dots, x_n]^{S_n}$-module with basis$$x_2^{m_2}x_3^{m_3} \dots x_{n-1}^{m_{n-1}} x_n^{m_n},\text{ }m_2 \in [0, 1],\text{ }m_3 \in [0, 2], \dots,\text{ }m_n \in [0, n-1]?$$

REPLY [4 votes]: It is enough to show that the given generators span $k[x_1, \ldots, x_n]$ as a $k[x_1, \ldots, x_n]^{S_n}$ module. Once we've shown this, it is easy to see that $k(x_1, \ldots, x_n)$ is dimension $n!$ as a $k(x_1, \ldots, x_n)^{S_n}$ vector space, so $n!$ vector which span must be linearly independent.
Notation: For every $r \in \left\{0,1,\ldots,n\right\}$, let $e^r_k$ be the $k$-th elementary symmetric polynomial in $x_1$, ..., $x_r$. Let $R = k[x_1, \ldots, x_n]$ and let $A = R^{S_n}$.
Lemma 1 The polynomial $e^r_k$ is in the ring $A[x_{r+1}, \ldots, x_n]$. 
Proof Induction on $n-r$. The base case, $r=n$, is that $e_k(x_1, \ldots, x_n) \in A$; this is true. Now, $e_k^r = e_k^{r+1} - x_{r+1} e_{k-1}^{r
+1}$, so the result follows by induction. $\square$.
Lemma 2 The monomial $x_r^r$ is in the $A[x_{r+1}, \ldots, x_n]$-linear span of monomials of the form $x_r^s$ with $0 \leq s < r$.
Proof Expand $(x_r-x_1) (x_r-x_2) \cdots (x_r-x_r) = 0$ to get $x_r^r - e^r_1 x_r^{r-1} + e^r_2 x_r^{r-2} - \cdots = 0$ or, in other words, $x_r^r = \sum_{s=0}^{r-1} (-1)^{r-s+1} x_r^s e^r_{r-s}$. Now apply Lemma 1. $\square$
We now prove the following result by induction on $r$: 

The ring $R$ is spanned as an $A[x_{r+1}, \ldots, x_n]$-module by monomials of the form $x_1^{d_1} x_2^{d_2} \cdots x_r^{d_r}$ with $0 \leq d_j < j$.

The base case $r=0$ is trivial (it says $R = A \cdot 1$); the case $r=n$ is the claim.
We now do the inductive step. Since $R$ is spanned as an $A[x_{r}, \ldots, x_n]$-module by monomials of the form $x_1^{d_1} x_2^{d_2} \cdots x_{r-1}^{d_{r-1}}$ with $0 \leq d_j < j$, we know that $R$ is spanned as an $A[x_{r+1}, \ldots, x_n]$-module by monomials of the form $x_1^{d_1} x_2^{d_2} \cdots x_{r-1}^{d_{r-1}} x_r^t$, with $0 \leq d_j < j$, and no bound on $t$. But Lemma 2 allows us to replace $x_r^t$ by lower powers of $x_r$ if $t \geq r$. QED<|endoftext|>
TITLE: vector spaces with uncountable dimension and a nice basis
QUESTION [6 upvotes]: Unlike the reals considered as a vector space over the rationals, I know of a number of nice examples of vector spaces with uncountable dimension that have a nice basis. 
For example, the space of rational functions on the real or complex numbers with complex coefficients whose numerator degree does not exceeds the denominator degree has as basis the fractional functions $1/(1+sx)^k$ with arbitrary complex $s$ and positive integers $k$.
Another example is the space of linear combinations of exponential functions $e^{sx}$ on the real or complex numbers.
Is there a common generalization? Or even some sort of general theory behind these examples?
Edit: Given the discussion so far, let me make the question more precise: Is there a fairly general way of proving that certain nice uncountable families of nice functions on some nice domain have the property that any finite subset is linearly independent? One kind of niceness criterion (but not the only relevant one) could be that the space spanned by the functions from the family can be characterizd in a basis-free way. Another niceness criterion is that the proof of linear independence should be not completely trivial.

REPLY [7 votes]: The linearly independent set $\{ e^{sx} \}$ is generated by a simple mechanism: namely, it consists of eigenvectors for an operator $\frac{d}{dx}$ acting on a vector space all of whose eigenspaces are $1$-dimensional. The rational functions, I think, don't naturally appear in this way but they are all annihilated by some polynomial differential operator in the Weyl algebra $k \left[ x, \frac{d}{dx} \right]$.
In general if you have an algebra $A$ acting on a vector space $V$ it's interesting to look at the vectors $v \in V$ such that the $A$-submodule $Av$ generated by them is simple. (This is one of a few possible natural generalizations of being an eigenvector.) Then if $v_i, i \in I$ are vectors such that the corresponding $A$-submodules $A v_i$ are both simple and nonisomorphic, the $v_i$ are linearly independent.<|endoftext|>
TITLE: Asymptotic expression for $j$ which satisfies $\binom{n}{j}/j! \sim k$ as $n\to\infty$
QUESTION [6 upvotes]: Suppose $k>0$ is some fixed constant, and $n$ is a positive integer tending to infinity.  Find $j\equiv j(n,k)$ such that 
$$ \frac{\binom{n}{j}}{j!} \sim k. $$
The asymptotic expression for $(n!)^{-1}$ (https://stackoverflow.com/questions/3084937/how-to-calculate-the-inverse-factorial-of-a-real-number) was initially helpful, but upon rearranging we have equivalently to solve for $j$ in
$$ k\ j!^2 = n(n-1)\cdots(n-j+1),$$
and in this expression both sides depend on $j$.
I've tried Stirling's approximation, and assuming I did the calculations correct, $e \sqrt{n}$ is interesting since $j \ge e \sqrt{n}$ implies $\binom{n}{j}/j! \to 0$ whereas $j < e \sqrt{n}$ implies $\binom{n}{j}/j! \to \infty$.
In de Bruijn's book there are implicit methods, but I wasn't able to successfully apply any of the approaches.

REPLY [4 votes]: This is not quite an answer to the question (Brendan McKay already gave quite a satisfactory one), but it's worth adding that 

$\binom{n}{j}/j!$ has an interesting probabilistic interpretation, namely it is the expected number of increasing subsequences of length $j$ in a (uniformly) random permutation of order $n$, and
the fact noted by the OP that this quantity goes to $0$ when $n\to\infty$ with $j>(e+\epsilon)\sqrt{n}$ can be used to give an easy proof that the expected length of a longest increasing subsequence in a random permutation of order $n$ is bounded from above by $(e+o(1))\sqrt{n}$. (And semi-conversely, the fact that $\binom{n}{j}/j!\to\infty$ for $j<(e-\epsilon)\sqrt{n}$ implies that this is the best bound that can be proved using this method. This bound is not sharp however, the correct answer for the expected length of a longest increasing subsequent is $(2+o(1))\sqrt{n}$.) See page 9 of this book.<|endoftext|>
TITLE: Does $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ ever support an almost complex structure?
QUESTION [72 upvotes]: This question has been crossposted from Math.SE in the hopes that it reaches a larger audience here.
$\Bbb{CP}^{2n+1} \# \Bbb{CP}^{2n+1}$ supports a complex structure: $\Bbb{CP}^{2n+1}$ has an orientation-reversing diffeomorphism (complex conjugation!), so this is diffeomorphic to the blowup of $\Bbb{CP}^{2n+1}$ at one point.
On the other hand, $\Bbb{CP}^2 \# \Bbb{CP}^2$ does not even support an almost complex structure: Noether's formula demands that its first Chern class $c_1^2 = 2\chi + 3\sigma = 14$, but if $c_1 = ax_1 + bx_2$ (where $x_1, x_2$ generate $H^2$, $x_1^2 = x_2^2$ is the positive generator of $H^4$, and $x_1x_2 = 0$), then $c_1^2 = a^2 + b^2$, and you cannot write $14$ as a sum of two squares.
Using a higher-dimensional facsimile of the same proof, I wrote down a proof here that $\Bbb{CP}^4 \# \Bbb{CP}^4$ does not admit an almost complex structure. The computations using any similar argument would, no doubt, become absurd if I increased the dimension any further.
Can any $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ support an almost complex structure?

REPLY [40 votes]: The $m$-fold connected sum $m\# {\mathbb{CP}}^{2n}$ admits an almost complex structure if and only if $m$ is odd, as we show in our recent preprint. (By the way, thanks to Mike for this interesting question, which motivated us to write the paper!)
Here's a brief summary of the proof's idea. Our main tool is a result by Sutherland resp. Thomas from the 60s which tells us when a stable almost complex structure is induced by an honest almost complex structure: this is the case iff its top Chern class equals the Euler class of the manifold.
As the connected sum of manifolds admitting a stable almost complex structure admits one as well, we certainly have stable almost complex structures on $m\# {\mathbb{CP}}^{2n}$ at our disposal, and we can understand the full set of all such structures by explicitly determining the kernel of the reduction map from complex to real K-theory. We then compute the top Chern class of all these structures: luckily for us, it turns out that in order to show the non-existence of almost complex structures for even $m$, it suffices to compute its value modulo 4 and compare it to the Euler characteristic of $m\# {\mathbb{CP}}^{2n}$. For odd $m$, we explicitly find a stable almost complex structure for which the criterion above is satisfied. 
Edit: The paper Connected sums of almost complex manifolds from Huijun Yang
generalizes our theorem to the following beautiful fact:

Let $M_i$ $i=1,\ldots,\alpha$ be $4n$-dimensional almost complex manifolds. Then the connected sum $$\#_{i=1}^\alpha M_i \#(\alpha-1) \mathbb C\mathbb P^{2n}$$ admits an almost complex structure. Moreover if $M$ is an almost complex manifold of dimension $2n$
  then $M\#\overline{\mathbb C\mathbb P^n}$ admits an almost complex structure. 

Thus the last statement means that the "blow-up" of almost complex manifolds is again almost complex in analogy to blow-ups of points in complex manifolds.<|endoftext|>
TITLE: Arithmetic Points are Dense on a Hida Family
QUESTION [9 upvotes]: I am reading the paper "Constancy of the Adjoint L-invariant" by H. Hida (http://www.math.ucla.edu/~hida/ConstP.pdf). 
Correct me if I'm wrong, but I've read/heard that the arithmetic points $p \in Spec(\mathbb{I})$ are Zariski dense (Where $Spec(\mathbb{I})$ is a connected component of $Spec(\mathbf{h})$, the spectrum of the Universal p-ordinary Hecke Algebra). Once we have defined the L-invariant on the dense points, there is some interpolation trick to define it on all points.
How can I understand these two facts? 
Also,
REFERENCE REQUEST:
I've been reading through Hida's papers on $\mathbf{h}$ between '81-'86, as well as his course notes, textbooks, etc. This paper I linked at the top (Constancy of L-Invariant) is one of the most understandable expositions I have read (Due to the Alg. Geom. style). Are there any modern expositions of Hida Theory floating around somewhere that one can learn from? I am not familiar with Motives, a language I have seen appear in Hida's work/books, so something that avoids extra-technical definitions would be great.

REPLY [6 votes]: Hida theory is a vast domain of research. I am assuming that that you are in the simplest and oldest setting: Hida theory for ordinary eigencuspforms for the group $\operatorname{GL}_2$ over $\mathbb Q$ (the arguments I am giving would apply with little modification to $\operatorname{GL}_2$ over a totally real field $F$ after an application of the Jacquet-Langlands correspondence).
In that setting the Hida-Hecke algebra $\mathbf T$ injects in the endomorphism ring of a module which is free of finite rank over $\Lambda$ (the usual Hida algebra of diamond operators), namely the ordinary part of the inverse limit on the level at $p$ of the first cohomology group of the modular curves. So $\mathbf T$ is finite, torsion-free as $\Lambda$-module and so the ring-extension $\mathbf T/P$ of $\Lambda$ for $P$ a minimal prime of $\mathbf T$ is integral. 
Hence the Zariski-density of arithmetic primes in $\mathbf T/P$ reduces to the density of primes below them in $\Lambda$ which are, by definition, the principal ideals $(\gamma-\chi(\gamma)\gamma^{k-2})$ for $\chi$ a finite order character and $\gamma$ a topological generator of $\Gamma=1+p\mathbb Z_p$. These are dense in $\Lambda=\mathbb Z_{p}[[\Gamma]]$.
These arguments carry over for many reductive groups (symplectic groups, unitary groups, units in quaternion algebras...) over totally real fields but it is typically not the case that arithmetic points are dense in the spectrum of the Hida-Hecke algebra for a reductive group over a number field which is not totally real. This is already false, for instance, for $\operatorname{GL}_2$ over a quadratic imaginary field $K$.
Regarding your question on the interpolation of the $\mathcal L$ invariant of the adjoint representation, this is a very subtle result, especially in the generality you consider here. It requires first deep modularity results of Wiles, Taylor-Wiles, Skinner-Wiles and Kisin on deformations of Galois representations with coefficients in a finite extension of $\mathbb Q_p$ (so with infinite residue field). These results establish the vanishing of some Selmer group for the adjoint representation and this in turns lead to an explicit computation of the $\mathcal L$ invariant which shows that it extends to an analytic function on the set of height-one primes of the universal Hecke algebra.
Finally, regarding your reference request, I like Control of nearly ordinary Hecke algebras (H.Hida) and Residually reducible representations and modular forms (C.Skinner, A.Wiles) Inst. Hautes Études Sci. Publ. Math. No. 89 (1999), 5--126 (2000) myself, but there are dozens of excellent references (none of which make essential use of the notion of motives, so don't get scared by that word and just mentally translate it to geometric Galois representations or compatible systems of geometric Galois representations whenever it appears).<|endoftext|>
TITLE: Generating (or availability of) large strongly regular graphs
QUESTION [8 upvotes]: Are there collections of already generated large strongly regular graphs available to download?  By large I mean $n \geq 200$ where $n$ is the number of vertices. I found  Ted Spence's page on srgs, which contains a very nice list of srgs up to $n \leq 64$, but they are kind of on the small side. I also searched for software that can generate them but wasn't able to find one. (Although Brendan McKay's nauty software  is extremely useful but doesn't seem to have an option to generate srgs).
Note that all I'm looking for is few samples (not an exhaustive list which is probably not practical). 
  Are there such sites or software available? Apologies if this does not qualify as a pure research question. Thank you!

REPLY [7 votes]: Sage(math) aims to have graphs from A.Brouwer's web pages on s.r.g.'s available. Namely, at least one example for each case for which the existence is known. E.g. it will be possible to do things like
sage: t=graphs.strongly_regular_graph(126, 60, 33); t
complement(Taylor two-graph SRG): Graph on 126 vertices

The current beta version (6.9.beta4) has about 20% missing; after the work in progress here merged (hopefully in the forthcoming Sagemath 6.9) there will be at most 192 graphs missing.
Many of these graphs are actually available already in the current stable release of Sage, but there they aren't easy to find just from parameters.

REPLY [4 votes]: I am not aware of any such site.
To me, it even seems highly unlikely that anyone would try to produce such a list.
For a start it would have be extremely selective - for example, there are
11,084,874,829 srgs on 57 vertices coming from Steiner triple systems
on 19 points. 
I (and collaborators) have used sage to construct srgs on up to something like 1000 vertices. This is not necessarily trivial (point graphs of flock generalized quadrangles say), but is certainly feasible.
If you gave some idea of what you want to do with these graphs, you might get
a more useful response.

REPLY [4 votes]: I have a program that makes SRG block-intersection graphs of Steiner Triple Systems, using the randomised hill-climbing method of Stinson and Mathon (?). It works in a fraction of a second on any size that is an integer of the form (v-1)v/6. I don't know if Sage has it (if not, it should).<|endoftext|>
TITLE: Conditions for existence of Penrose diagrams
QUESTION [5 upvotes]: A Penrose diagram (also known as a conformal diagram or Carter-Penrose diagram) is a technique for visualizing the causal (light-cone) structure of a 3+1-dimensional manifold. Usually the diagram is two-dimensional, but it is also possible to have more dimensions. My question is whether there are definite conditions for the existence of a Penrose diagram for a given spacetime, and if so, what they are.
It's not clear to me that this question has a definite answer, partly because the construction of these diagrams seems to be more like a body of techniques than a systematic theory. There does not seem to be any widely recognized definition of exactly what is meant by a Penrose diagram. Penrose,[Penrose 2010] decades after originating the idea, says he wants to distinguish "strict" from "schematic" diagrams. To explain the distinction he gives a reference to Carter 1966, which, unfortunately, does not use those terms and does not seem to make any such distinction.
My understanding of the process is that it's something like this. We start with some given spacetime, then:

Make an $n$-dimensional section or projection, where usually, but not always, $n=2$.
Do a conformal transformation to reduce the resulting manifold to a flat one of finite size.
Adjoin idealized surfaces and points at infinity.

At step 1, we want to take advantage of any symmetries, such as rotational symmetry, so that the final result will be informative, be representative of the whole spacetime, and accurately depict causal relationships in the original spacetime. At this step we also need to make sure that lightlike geodesics in the original space correspond properly to lightlike geodesics in the submanifold.
Step 2 is only possible if our $n$-dimensional space is conformally flat, but this is automatic in the usual case where $n=2$. In many cases of interest, the original 3+1-dimensional manifold is also conformally flat (which in four dimensions is equivalent to vanishing of the Weyl tensor).
Can anyone shed any light on the distinction between strict and conformal Penrose diagrams, and the conditions for their existence? From the sketch above, it seems to me that $n=2$ diagrams would always exist, but they might not be useful and informative. Penrose 2010 says that exact spherical symmetry is needed for strict conformal diagrams but not for schematic ones, but it's not clear to me why this is a requirement, or what would be so special about spherical symmetry as opposed to any other symmetry.
Carter 1966 - "Complete Analytic Extension of the Symmetry Axis of Kerr's Solution of Einstein's Equations," Phys. Rev. 141, 1242, http://luth2.obspm.fr/~luthier/carter/trav/Carter66.pdf
Penrose 2010 - Cycles of time: an extraordinary new view of the universe, Bodley Head, p. 106

REPLY [4 votes]: There is no definitive answer to your specific question, so I'm going to talk around the topic and hope that it's informative.
As you've noted the classical examples all basically look like special cases. This isn't helped by, pretty much all, books and articles using the same examples. There are a few common threads (which you've noted) and those threads are exemplified in Carter's paper. 
My guess, and this really is a guess, is that Penrose is pointing to those threads. My experience of Penrose's writing, particularly on these kinds of "meta-techniques", is that he does not provide strict definitions and is quite happy to point towards something while giving details in specific cases. If I was him (which I'm not) I'd do this because I thought something was interesting but had other things I wanted to study.
The "strict" vs "schematic" differentiation might very well be an example of this. I haven't found that terminology anywhere else in the literature on this kind of thing... and I did my PhD on it.
My best advice is to shrug your shoulders, note that it is interesting that complicated global conformal structure can be accurately represented in two-dimensional drawings and then to move on.
Why? Because this type of question is what birthed a subfield of General Relativity called "Boundary Constructions". This is a subfield that divides the community. Some think it's great, others not so much. I once had a paper recommended for rejection on the sole basis that (para-phrasing slightly) "the field should be allowed to die in this generation as it is an exercise in academic abstraction." (At the time I thought this reviewer probably also objects to category theory for the same reason.)
It is no overreach to say that most, if not all modern studies in GR have been influenced by Penrose diagrams and the even more important Penrose embeddings. For example, AdS/CFT is based on the conformal boundary. The conformal boundary is an example of the boundary of a Penrose embedding. Using Penrose embeddings Penrose gave, I think, the first geometric interpretation of the BMS group and super-translations. Even now one of the fundamental areas of study in numerical relativity is to do with propagation "at infinity" defined, you guessed it, by Penrose embeddings. If you look closely enough at a research area you'll see Penrose diagrams and embeddings influencing it.
But Penrose embeddings (which have a definition unlike Penrose diagrams) are only defined for weakly asymptotically flat space-times. Moreover, if you want certain fields to be nicely behaved on the boundary you'll also need the space-time to be weakly asymptotically empty. These are strict conditions that apply only to isolated gravitating systems.
Researchers wanted a tool that replicated the success of Penrose's embeddings that could be applied to a wider class of space-times... and thus was born the field of "Boundary Constructions". 
The field has had some success, most notably in the importance attached the causal relations (e.g. the causal sets quantum gravity thing), Krolak's weak cosmic censorship results and in application to the study of singularities. And there is a perennial interest in it (e.g. most recently, by people in condensed matter physics looking to replicate the AdS/CFT correspondence for aniostropic fields). But these are few and far between, particularly for the level of sophistication that has been reached in the most recent constructions. 
So Boundary Constructions are a dangerous road to tread. You'll immediately get offside with a large segment of the community, but there is something there. Something that a lot of people have chased and not quite got a handle on. This is very very hard stuff with little to no idea of what the pay off will be. You'd be much better served by working in a more active area.
Now, what if you want to completely ignore my advice? Take a look at my own most recent (and probably last) contribution http://arxiv.org/abs/1401.1287. And have a look at the references. Ashley's PhD thesis has a good review of the state of the art from around the turn of the millennium (https://digitalcollections.anu.edu.au/handle/1885/46055). You should also take a look at Senovilla's work on the iso-causal boundary, http://iopscience.iop.org/article/10.1088/0264-9381/22/9/R01, as well as the "controversy" around it, http://arxiv.org/abs/1103.2083. The guys working on the causal boundary, e.g. http://arxiv.org/abs/1001.3270 and the papers that cite it, have also staked a claim to generalising Penrose's constructions. These are all good places to start.
To summarise: It certainly feels like there is something deep about Penrose diagrams, that feeling has been enough to produce its own subfield and absorb substantial amounts of research. But, that subfield hasn't done anything truly astounding and that "deep" feeling might just be the same kind of thing as discovering that taking taylor expansions about infinity can be really useful.<|endoftext|>
TITLE: The sum of a series
QUESTION [8 upvotes]: Let $0< \alpha <1$ and $q>1.$ 
Consider the (alternating) series: $$
\sum_{k=1}^\infty 
(-1)^k \frac{q^k (q^k-1)^\alpha}{(q^k-1)\dots (q-1)}.$$
Denote its sum by  $f(q,\alpha).$  
Prove (or disprove)  that $f(q,\alpha)\neq 0$ for all
$\alpha\in(0,1)$ and all $q> 1.$
Computer computation confirms that there are no zeros for $ 1<q<2$ and $0<\alpha<1.$
For $\alpha\in N_0,$ we obtain  0 by Euler's formula. (Added on September 5 after the discussion, thanks to all who participated).
Remark. For $0<\alpha <1/2,$ and $q\geq 2$ the proof is rather simple, but, for $\alpha$ close to 1,I have not succeeded. The problem is related to the study of the convergence for the $q$-Bernstein polynomials.

REPLY [5 votes]: I have no doubt that you have figured this exercise out by now but let me post the solution just for the fun of it.
Denote $U_k=\prod_{m=1}^k\frac 1{q^m-1}$ with the usual convention $U_0=1$.
We have $\prod_{k\ge 1}(1-q^{-k}y)=\sum_{k\ge 0}(-1)^kU_ky^k$. In particular, it implies that
$$
\sum_{k\ge 1}(-1)^k U_{k-1}y^k<0\text{ for all } 0<y<q
$$
The sum you are interested in is 
$$
\sum_{k\ge 1}(-1)^k U_{k-1}\frac{q^k}{(q^k-1)^\beta}
$$
where $\beta=1-\alpha$.
Now just write 
$$
\frac{q^k}{(q^k-1)^\beta}=(q^{1-\beta})^k(1-q^{-k})^{-\beta}=(q^{1-\beta})^k\sum_{m\ge 0}c_{m}(\beta)q^{-mk}=\sum_{m\ge 0}c_m(\beta)(q^{1-m-\beta})^k
$$ 
and observe that all $c_m(\beta)$ (the Taylor coefficients of $x\mapsto (1-x)^{-\beta}$ at $x=0$) are positive.<|endoftext|>
TITLE: Negative coefficient in an almost cyclotomic polynomial
QUESTION [6 upvotes]: Let $a,b,c,d$ be four prime numbers. We set the polynomial :
$$P(X)=\frac{(1-X^{abc})(1-X^{abd})(1-X^{acd})(1-X^{bcd})(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})(1-X^{ac})(1-X^{ad})(1-X^{bc})(1-X^{bd})(1-X^{cd})}$$
By numerical tests, i see that $P(X)$ always has at least one negative coefficient, how can i prove it?

REPLY [6 votes]: Suppose that $a<b<c<d$. We show that the coefficient of $X^c$ or of $X^{b+c-1}$ of $P(X)$ is negative. In order to do so, it suffices to work in the power series ring $\mathbb Q[[X]]$ modulo $X^{b+c}$. Note that $ac>b+c$ and so on, hence
\begin{equation}
P(X)\equiv\frac{(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})}\pmod{X^{b+c}}.
\end{equation}
Set
\begin{equation}
F(X)=\frac{1-X^a}{1-X}\cdot\frac{1-X^b}{1-X}\cdot\frac{1}{1-X^{ab}}=(1+\dots+X^{a-1})(1+\dots+X^{b-1})(1+X^{ab}+\dots),
\end{equation}
so $P(X)\equiv F(X)(1-X^c-X^d)$ (recall that $c+d>b+c$).
For a power series $G$ let $G[k]$ be the coefficient of $X^k$.
So $P[k]=F[k]-F[k-c]-F[k-d]$.
The coefficients of $F$ lie between $0$ and $a$, and $F[k]=a$ if an only if the remainder of $k\pmod{ab}$ lies between $a-1$ and $b-1$. Furthermore, $F[k]=0$ if the remainder of $k\pmod{ab}$ lies between $a+b-1$ and $ab-1$.
Now assume that $P[b+c-1]\ge0$. From $P[b+c-1]=F[b+c-1]-F[b-1]-F[b+c-1-d]$ and $F[b-1]=a$ we infer that $F[b+c-1]=a$. Thus the remainder of $(b+c-1)\pmod{ab}$ lies between $a-1$ and $b-1$. This implies that the remainder of $c\pmod{ab}$, call it $r$, fulfills $ab+a-b\le r\le ab-1$ or $r=0$. The latter cannot happen, because then $a$ would divide the prime $c$. Thus the former holds. But $a+b-1\le ab+a-b$, hence $F[c]=0$.
But then $P[c]=F[c]-F[0]=0-1=-1$, and we are done.<|endoftext|>
TITLE: Fuzzy logic of Godel
QUESTION [7 upvotes]: In Gödel logic, is conjunction definable from implication, negation , and disjunction?
We know that conjunction in that logic is not definable from negation and implication.

REPLY [5 votes]: I assume that by "definable" you mean "expressible by a term".  In that case the answer is "no". 
Let $t(x,y)$ be any term (involving the variables $x$, $y$, and connectives $\to$, $\lnot$, $\vee$ -- possibly not all of them).   

If we interpret $t$ on the set $\Delta:= \{(x,y)\in (0,1)^2: x>y\}$ (under the Gödel interpretation, so in particular $p\to q$ has value $1$ for $p\le q$, and value $q$ otherwise), then $t$ must be constant with value $1$ or $0$, or equivalent to one of the terms $x$ or $y$.   
Similarly for $\nabla:= \{(x,y)\in (0,1)^2: x< y\}$. 

We will show that there is no term which is equivalent to $y$ on $\Delta$, and equivalent to $x$ on $\nabla$. 
So let $t$ be such a term of minimal length. Obviously $t$ is not of the form $\lnot t'$, and not equal to $x$ or $y$. 
So $t$ must be an implication $t_1\to t_2$ (or a disjunction $t_3\vee t_4$; this case is left as an exercise). 
Now consider all the possible behaviours of $t_1$ and $t_2$ on $\Delta$.
$t_1\restriction \Delta$ can be $x$, $y$ or $1$, similarly $t_2\restriction \Delta$.
Considering the 9 possible cases, we see that we get the desired result $y$ ($= x\wedge y$ on $\Delta$) in only 2 cases, as $x\to y$ and as $1\to y$. In both cases we must have $t_2=y$ (on $\Delta$). 
Similarly we must have  $t_2=x$ on $\nabla$ to get $t_1\to t_2$ equivalent to $x$ on $\nabla$. But then $t$ was not minimal. 
If you think this proof is ugly, consider my second proof, which is worse, but has the advantage that it can be outsourced to a computer: Consider binary functions on the interval $[0,1]$.   Start with the projections $x$ and $y$, close off under disjunction, negation, Gödel implication. This stops after finitely many steps. (19, if I counted correctly; at least if you consider only the behavior on the open square.)  Check that you have not generated the conjunction.<|endoftext|>
TITLE: Which polynomial's roots are its coefficients?
QUESTION [41 upvotes]: Start with any polynomial of degree $n$ with complex coefficients, e.g.,
$$z^3+z^2+2 z+3 \;.$$
Find its $n$ roots, and list them in order of their modulus:
$$-1.28, (0.14\pm 1.53 i)$$
Now form a new polynomial with leading coefficient $1$ and the other
coefficients the $n$ roots:
$$
z^3
-1.28\, z^2
+(0.14 -1.53 i)\, z
+(0.14 +1.53 i) \;.
$$
Here I used the sorting convention that $(a-b i) < (a + b i)$.
A bit more formally, let $P$ be
$$
P \;:\; z^n + a_{n-1}z^{n-1} + \cdots + a_0 \;,
$$
with roots
$$r_{n-1},  \ldots ,r_0 \;,$$
sorted so that $|r_i| \le |r_{i-1}|$.
Define $P_1$ as
$$
P_1 \;:\; z^n + r_{n-1}z^{n-1} + \cdots + r_0 \;.
$$
My question is:

Q1. Which are the polynomials whose roots are its coefficients,
  in the sense that $r_i=a_i$, $i=0,\ldots,n-1$, i.e., $P_1 = P$?

Here is one fixed point:
$$
z^3
-(0.782599 +0.521714 i)\, z^2
+(0.884646 -0.589743 i)\, z
+(0.680552 +1.63317 i)
$$
When this process is iterated, it appears to fall into cycles, not always of length $1$.

Q2. Does iteration of this process always fall into a cycle?

(Added. 3Sep15.) The question suggested by Igor Rivin is also interesting,
especially in light of Richard Stanley's remark concerning real polynomials:

Q3. Which are the polynomials whose roots are its coefficients, in any order? (This is an unordered version of Q1.)

(Added. 4Sep15.)
Concerning Q2, which I realize is less interesting than the other questions, here is
an example of a cycle of length $11$. I plot the magnitude of the vector
of $3$ roots versus iteration #, starting from a random cubic polynomial. The last
vector of roots is
$$
(-1.52+0.88 i,-0.48-0.90 i,1.00 +0.02 i)
$$
with magnitude $2.27$:

REPLY [23 votes]: For the unordered version, the solutions for degree $3$ are
$$ \eqalign{x^3&\cr
x^3& {}+x^2-2\,x\cr
x^3&{}+x^2-x-1\cr
x^3&{}+ \left( r^2/2-1 \right) x^2+rx-r-r^2+2 \ \text{where}\ r^3 - 2 r + 2 = 0\cr}
$$
The solutions for degree $4$ are
$$ \eqalign{
x^4\cr
x^4&{}+x^3-x^2-x\cr
x^4&{}+x^3-2\; x^2 \cr
x^4&{}+ \left( \frac{r^2}{2}-1 \right) x^3+rx^2+(-r^2-r+2) x \ \text{where}\ r^3-2\,r+2 = 0\cr
x^4&{}-r^2 x+rx^2+x^3+r^2-2\,rx+r-x-1 \ \text{where}\
r^3+2\,r^2+r+1 = 0\cr
x^4&{}+ \left( \frac{r^{12}}{4}+{\frac {53\,r^{11}}{116}}+{\frac {19\,
r^{10}}{116}}-{\frac {43\,r^9}{116}}-{\frac {59\,r^8}{116}}-
{\frac {9\,{r}^{7}}{29}}+{\frac {11\,{r}^{6}}{29}}+{\frac {31\,{r}^{5}
}{29}}+{\frac {35\,{r}^{4}}{58}}\right.\cr &\left.{}-\frac{{r}^{3}}{29}+{\frac {2\,{r}^{2}}{29}
}+{\frac {5\,r^{13}}{116}}+\frac{r}{29}-{\frac {12}{29}} \right) {x}^{3}+r{x
}^{2}\cr &{}+ \left( -{\frac {49\,r^{11}}{29}}+{\frac {123\,{r}^{10}}{58}}+
{\frac {207\,{r}^{9}}{58}}+{\frac {195\,{r}^{8}}{58}}-{\frac {27\,{r}^
{7}}{58}}-{\frac {373\,{r}^{6}}{58}}-{\frac {113\,{r}^{5}}{29}}+{
\frac {77\,{r}^{4}}{29}}\right. \cr &\left.{}+{\frac {13\,{r}^{3}}{29}}-{\frac {47\,{r}^{13
}}{58}}-\frac{5\,{r}^{12}}{2}+{\frac {3\,{r}^{2}}{29}}-{\frac {100\,r}{29}}+{
\frac {98}{29}} \right) x+2\,{r}^{12}+{\frac {45\,{r}^{11}}{58}}-{
\frac {71\,{r}^{10}}{29}}\cr &{}-{\frac {82\,{r}^{9}}{29}}-{\frac {68\,{r}^{8
}}{29}}+{\frac {63\,{r}^{7}}{58}}+{\frac {329\,{r}^{6}}{58}}+{\frac {
51\,{r}^{5}}{29}}-{\frac {112\,{r}^{4}}{29}}+{\frac {21\,{r}^{13}}{29}
}-{\frac {74}{29}}-{\frac {11\,{r}^{3}}{29}}\cr&{}-{\frac {7\,{r}^{2}}{29}}+
{\frac {69\,r}{29}}\ \text{where}\cr
 {r}^{14}&{}+2\,{r}^{13}-{r}^{12}-4\,{r}^{11}-{r}^{10}+4\,{r}^{8}+6\,{r}^{
7}-4\,{r}^{6}-6\,{r}^{5}+4\,{r}^{4}+4\,{r}^{2}-8\,r+4
=0
}$$
Note that you can always multiply a solution by $x^k$ for positive integer $k$, obtaining $k$ more coefficients of $0$ to match $k$ more roots of $0$.<|endoftext|>
TITLE: Relaxed Collatz 3x+1 conjecture
QUESTION [44 upvotes]: The Collatz $3x+1$ conjecture claims that any positive integer can eventually be reduced to $1$ by iterative application of the maps $x \mapsto 3x+1$ whenever $x$ is odd and $x \mapsto x/2$ whenever $x$ is even.
While the Collatz conjecture is still open, I wonder if the following relaxed version is any simpler. In this relaxed version we are allowed to apply maps in any order keeping the numbers integer. That is, if $x$ is odd, we still have to apply the $x \mapsto 3x+1$ map; but for even $x$, we have the freedom to choose between $x \mapsto 3x+1$ and $x \mapsto x/2$. The conjecture is that for any positive integer, we can reduce it to $1$ with some iterative sequence of maps.
Clearly, the Collatz conjecture would imply the relaxed version. But it may happen that the relaxed version is much simpler. Is it?
The question is inspired by discussion of the sequence http://oeis.org/A109732 which is a permutation of the odd positive integers iff the relaxed version of the $3x-1$ variant of the Collatz conjecture holds.
UPDATE. The minimum number of iterations to reach $1$ in this relaxed version is given by http://oeis.org/A127885 and this number is often smaller than that for the Collatz conjecture given by http://oeis.org/A006577 .

REPLY [6 votes]: I'll write $\longrightarrow$ for some standard iterations and $\implies$ for some possibly non-standard iterations.  The relaxed Collatz conjecture is that for all $x$, $x \implies 1$.  I'll call counterexamples to the conjecture escapees.
First of all, since it's a good example of the notation, a lemma: for all $a$, $4 a \implies 9 a + 1$.
Proof: $4a \implies 12 a + 1 \longrightarrow 36a + 4 \longrightarrow 18a + 2 \longrightarrow 9 a + 1$.
$\square$
Now here is what I will try to show:
Theorem: If $x$ is the smallest escapee, then $4x \implies 1$.
Proof: By elementary considerations, $x \equiv 3 \pmod{4}$.  If $x \equiv 2 \pmod{3}$, then $\frac{2 x - 1}{3} \longrightarrow 2 x \longrightarrow x$.  So in this case, $2 x$ can't be an escapee, and neither can $4 x$.
The next case is $x \equiv 0 \pmod{3}$.  Write $x = 12k+3$.  Using the lemma, $4x \implies 27k+7$.  Observe that $x > 9k+2 \rightarrow 27k+7$ if $k$ is odd.  Now if $x \equiv 15 \pmod{24}$, we also have that $4x \implies 1$.
It's also possible that $x \equiv 3 \pmod{24}$.  In that case, we have $x = 24 j + 3$ and $x \longrightarrow 27j+4$.  This means $j$ cannot be even.  So we drill down again, with $x = 48i + 27 \longrightarrow 81i+49$.  This means $i$ cannot be odd.  But also, $4x = 4(48i+27) \implies 81i + 92$, so if $i$ is even, the next step brings us under $x$.  This covers all the $x \equiv 0 \pmod{3}$ cases.
So far, our result is that if $x \implies 4x$, then $x \equiv 7 \pmod{12}$.  Now I'll handle that case.  Suppose $x \implies 4x$ and $x = 24k + 7$.  Then we have:
$4x = 4(24k+7) \implies 9(24k+7)+1 = 216k+64 \longrightarrow 108k+32 \longrightarrow 54k+16$.
But also, $18k+5 \longrightarrow 54k+16$.  This means $18k+5$ is an escapee too, but it can't be, since it's less than $x$.  So the hypothesis is false and we have $x = 24k + 19$ instead.
Now, consider:
$4x = 4(24k+19) \implies 216k + 172 \longrightarrow 108k + 86 \longrightarrow 54k + 43 \longrightarrow 162k + 130 \longrightarrow 81k+65$.
By similar reasoning, $k \neq 3 \pmod{4}$, or else we can continue the chain two more steps to get an escapee less than $x$.  Finally,
$x = 24k+19 \longrightarrow 72k+58 \longrightarrow  36k + 29 \longrightarrow 108k + 88 \longrightarrow 54k + 44 \longrightarrow 27k + 22$.
shows that $k$ can't be even, and substituting $k = 4 j + 1$:
$27k + 22 = 108j + 49 \longrightarrow 324j + 148 \longrightarrow 162j + 74 \longrightarrow 81j + 37 < x$
completes the proof, since that's all the cases. 
$\square$
Corollary: if the Collatz conjecture is false and the smallest counterexample leads back to itself then that number is not a counterexample to the relaxed Collatz conjecture.  In other words, if $y$ is the smallest number satisfying $\neg(y \longrightarrow 1)$, then $y \longrightarrow y$ implies $y \implies 1$.  This is a weaker version of what I originally thought I had proved.
Corollary: If $4x$ is an escapee, then there is an escapee smaller than $x$. This suggests to me a recursive descent strategy where we attempt to show $z \implies 4k < 4x$. But it's not clear if number-crunching will get us anywhere further.<|endoftext|>
TITLE: Positive Elements of a $\ast$-Algebra
QUESTION [6 upvotes]: In a $C^*$-algebra ${\cal A}$, a positive element is a one of the form $aa^*$, for some $a \in {\cal A}$. It is known that the set of positive elements is a cone, and that for $a,b$ two non-zero elements, and $\lambda \in {\bf R}_{>0}$, we have $a + \lambda b \neq 0$. 
If we generalize to $\ast$-algebras, do these facts still hold: is the set of positive elements still a cone, and are sums of the form $a + \lambda b$ still non-zero.

REPLY [8 votes]: The answer to the second part is also no. That is, $aa^*+bb^*$ could be zero for nonzero $a$ and $b$. Look at the $*$-algebra $\mathbb{C}^2$ with component-wise addition and multiplcation and $*$ given by $(x,y)^* = (\overline{y},\overline{x})$. It's straightforward to calculate that anything of the form $aa^*$ will look like $(z,\overline{z})$ for some $z\in\mathbb{C}$. Conversely, $(z,\overline{z})=(z,1)(z,1)^*$. We see $(z,\overline{z})$ and $(w,\overline{w})$ sum to $(z+w, \overline{z+w})$, so $aa^*+bb^*$ is still of the form $cc^*$. But taking $w=-z$ gives a sum of zero.

REPLY [4 votes]: The answer to the first part of the question, in general, is no. Consider the following counter-example: let $\mathscr{D}(\mathbb{R}^{kd})$ be the complex vector space of (complex-valued) smooth functions with compact support on $\mathbb{R}^{kd}$ (seen as functions $f(x_1,\ldots,x_k)$ with $k>0$ arguments in $\mathbb{R}^d$), and define $$ \mathfrak{A}=\mathbb{C}\oplus\bigoplus^\infty_{k=1}\mathscr{D}(\mathbb{R}^{kd})\ ,$$ whose elements have the form $$\underline{f}=(f_0,f_1,f_2,\ldots)\ ,\quad f_0\in\mathbb{C}\ ,\,f_k\in\mathscr{D}(\mathbb{R}^{kd})\ (k>0)$$ with $f_k=0$ for all but finitely many $k$. This vector space over $\mathbb{C}$ becomes a $*$-algebra if we endow it with the product $$\underline{f}\underline{g}=\left(f_0g_0,\,\ldots,\sum^k_{j=0}f_j\otimes g_{k-j},\,\ldots\right)$$ and the involution $$\underline{f}^*=(\bar{f_0},\,\ldots,\,f_k^*,\,)\ ,$$ where $$(f_j\otimes g_l)(x_1,\ldots,x_{j+l})=\begin{cases} f_0g_l(x_1,\ldots,x_l) & (j=0) \\ g_0f_j(x_1,\ldots,x_j) & (l=0) \\ f_j(x_1,\ldots,x_j)g_l(x_{j+1},\ldots,x_{j+l}) & (j,l>0) \end{cases}$$ and $$f_k^*(x_1,\ldots,x_k)=\overline{f_k(x_k,\ldots,x_1)}\ .$$ In this case, the element
$(0,0,f\otimes g+f'\otimes g',0,\ldots)$, where $f,g,f',g'\in\mathscr{D}(\mathbb{R}^d)$ are linearly independent, cannot be written in the form $\underline{h}^*\underline{h}$ for any $\underline{h}\in\mathfrak{A}$, for the same reason the sum of the tensor products of linearly independent vectors does not factorize. Of course, $\mathfrak{A}$ is not a C$*$-algebra, although it is a topological $*$-algebra. In general, one considers the cone generated by the positive elements (or the latter's closure in the topological case). This contains only positive elements in the case of C$*$-algebras thanks to the spectral theorem for self-adjoint elements, which does not hold in the above example.<|endoftext|>
TITLE: Upper bound on length of addition chain
QUESTION [8 upvotes]: An addition chain for $n$ is a finite sequence of integers starting at 1 and ending at $n$, such that each element is a sum of two previous elements. A short addition chain for $n$ can be used, for example, to calculate $x^n$ quickly by multiplying previously-calculated values.
Let $l(n)$ be the length of the shortest addition chain for $n$. I am looking for an upper bound on $l(n)$ of the form:
$$l(n) \leq C \cdot \log_2(n)$$
where $C$ is a constant. 
The Wikipedia page cites the following bound from 1975:
$$l(n) \leq \log_2(n) + \log_2(n)(1+o(1))/\log_2(\log_2(n)) < 2\log_2(n)$$
The OEIS page states several bounds, but without a clear reference:
$$l(n) \leq (5/2) \log(n) = 1.73 \log_2(n)$$
$$l(n) \leq (4/3) \lfloor\log_2(n)\rfloor + 2$$
What is a reference for the best known upper bound of the form $l(n) \leq C \cdot \log_2(n)$ (the smallest value of $C$)?

REPLY [2 votes]: In this paper:
Wattel, E. & Jensen, G. "Efficient calculation of powers in a semigroup"
Stichting Mathematisch Centrum. Zuivere Wiskunde, 1968, 1-18
they prove that  $l(n)/log_2(n)\le1.463$ and takes on this value for $n=71$.
This paper is interesting since it's an earlier usage of the Thurber sliding window method than the Thurber paper. The only reason I know about this paper is that it was referenced by a student doing a term paper and he was given the paper by one of the authors.
I don't understand the interest in a bound like this mind you.<|endoftext|>
TITLE: Dynamics of the distribution of prime factorization types in increasing intervals
QUESTION [5 upvotes]: I've tagged this as reference request as surely this question must be very well investigated, I just don't know how to look for it. Most likely the perfect answer will be in form of a keyword for googling :)
I would like to know about dynamics of relative frequencies of prime factorizations. More precisely, let $[k_1,k_2,...]$ be any multiset of natural numbers, and for $x>0$ let $f_x[k_1,k_2,...]$ be the number of those natural $n\le x$ with prime factorization of the form $p_1^{k_1}p_2^{k_2}\cdots$, divided by $x$. How does the "champion" (i. e. $[k_1,k_2,...]$ with largest $f_x[k_1,k_2,...]$) change with $x$? Do all "champions" have some common features? (Say, are they all of the form $[1,1,...]$?) At which $x$es do "champion changes" occur?
More generally, how does the list of all multisets $[k_1,k_2,...]$ ordered according to $f_x[k_1,k_2,...]$ vary with $x$?
Here is the plot for $x$ from $100000$ to up to ten million of the $f_x[k_1,k_2,...]$ for five most frequent (at ten million) $[k_1,k_2,...]$, namely, for $[1,1,1]$, $[1,1]$, $[1,1,1,1]$, $[1,1,2]$ and $[1]$.

REPLY [8 votes]: Indeed such problems have been studied before.  I claim that the champion multisets will just be a string of $r$ ones (for a suitable $r$), and in this case one is counting square-free integers up to $x$ with exactly $r$ prime factors.  This case  has been widely investigated.  For example, Balazard establishing a conjecture of Erdős showed that for each large $x$ this sequence is unimodal in $r$ (see Theorem E on page 24), and (as one can see by Hardy-Ramanujan/Erdős-Kac) attains its maximum for $r$ close to $\log \log x$.  Balazard's paper will have more references -- in particular to work of Hildebrand and Tenenebaum giving asymptotics in wide ranges.  In particular, it follows that the maximum number of integers up to $x$ having a factorization of type $p_1 \cdots p_r$ is 
$$ 
\sim \frac{6}{\pi^2} x \frac{1}{\sqrt{2\pi \log \log x}}, \tag{1}
$$ 
which is attained for $r$ around $\log \log x$.  (The $6/\pi^2$ comes from counting square-free numbers.) 
Now consider a general factorization type $[k_1,\ldots, k_s, 1, 1,\ldots ,1]$ where $k_1$, $\ldots$, $k_s$ are all at least $2$ (and the number of $1$'s in the multiset could possibly be zero).  We'll show that these factorization types will contribute an amount smaller than the largest possibility for all ones given above.  Write every number of this factorization type as $n=ab$ where $a$ is square-full and $b$ is square-free and coprime to $a$ (thus $a$ has type $[k_1,\ldots, k_s]$ and $b$ is of type $[1,\ldots, 1]$).   Thus the number of integers of this factorization type is (with $a$ and $b$ having the above meaning) 
$$ 
\sum_{1< a=p_1^{k_1}\cdots p_s^{k_s}} \sum_{b \le x/a} 1 
\le \sum_{a >\sqrt{x}} \frac{x}{a} + \sum_{a\le \sqrt{x}} \Big(\frac{6}{\pi^2} +o(1)\Big) \frac{x}{a \sqrt{2\pi \log \log x}}, \tag{2}
$$
where we bounded the sum over $b$ trivially in the first sum, and using (1) in the second sum (note $\log \log (x/a) \sim \log \log x$ there, and we also ignored the condition that $b$ is coprime to $a$).  Since the number of square-full integers up to $x$ is $\ll \sqrt{x}$, the first sum in (2) may be bounded by $\ll x^{\frac 34}$.  We may of course evaluate the second term exactly, for any given choice of $[k_1,\ldots,k_s]$.  Alternatively, we could bound that term by letting $a$ run over all square-full integers $>1$: since 
$$ 
\sum_{1<a, a\text{ square-full}} \frac 1a =\prod_{p} \Big(1+\frac{1}{p^2}+\frac{1}{p^3}+\ldots \Big) -1 = \frac{\zeta(2)\zeta(3)}{\zeta(6)}- 1 =0.943\ldots .
$$
This completes our proof that factorization types having at least one number strictly larger than $1$ contribute an amount that is strictly smaller than the maximum attained by all ones.
One can be extract more information than given above.  For example, after types involving all $1$'s the next champion will be $[2,1,1,1,\ldots,1]$, and here one can give an asymptotic like (1) for the maximum such number etc.<|endoftext|>
TITLE: Del Pezzo surfaces and homotopy groups of spheres
QUESTION [23 upvotes]: A (complex) del Pezzo surface is a smooth projective complex surface with ample anticanonical line bundle. Such surface has a degree defined as the self intersection of the canonical divisor. It is well known that the possible degrees run between $d=1$ and $d=9$ and that topologically del Pezzo surfaces are determined by their degree except for $d=8$. If $d$ is different from $8$, then a del Pezzo surface of degree $d$ has to be a generic blow up of the projective plane $\mathbb{P}^2$ in $9-d$ points. But if $d=8$, there are two choices: $\mathbb{P}^2$ blown up in one point and $\mathbb{P}^1 \times \mathbb{P}^1$.
On the other hand, the homotopy group $\pi_4(S^3)$ (which is also $\pi_4(S^2))$ is cyclic of order 2.
My question is: is there a natural relation between the fact that there are two del Pezzo surfaces of degree $8$ and the fact that $\pi_4(S^3)$ is of order $2$ ?
On the face of it, this question might sound vague: it seems that I am looking for an explanation of the existence of a bijection between two randomly choosen sets with two elements. But it is not the case: in fact, there exists a natural relation between these two sets with two elements but it is in the realm of theoretical physics (consider M theory compactified on a Calabi-Yau 3-fold containing a shrinking del Pezzo surface, decouple gravity, the effective low energy five dimensional theory is, for $d=8$, a $SU(2)$ gauge theory, and the various possible choices are parametrized from this point of view by a discrete theta angle, i.e. by the topological choice of a principal $SU(2)$-bundle on $S^5$, i.e. by $\pi_4(SU(2))$) and I would like to know if there exists a more geometric/topological way to understand this relation.

REPLY [8 votes]: Borrowing from 4 different comments to make a complete answer (and making it community wiki):
We can construct a bijection between the set of del Pezzo surfaces and $\pi_4(S_3)$ using topology. However, the proof that this is a bijection depends on the classification of del Pezzo surfaces and does not give an explanation for why there are exactly two.
Both del Pezzo surfaces of degree $8$ are $S^2$-bundles over $S^2$, by the projection map $\mathbb P^1 \times \mathbb P^1 \to \mathbb P^1$, or by projection from the blown-up point in $\mathbb P^2$ onto a line.
$S^2$-bundles over a manifold $M$ are classified by $$Map(M, BDiff(S^2))= Map(M, BO(3))$$ Thus $S^2$ bundles over $S^2$ are classified by $$Map(S^2, BO(3))= \pi_2 (BO(3)) = \pi_1(O(3)) = \pi_1(SO(3)) = \mathbb Z/2$$
Furthermore the map from del Pezzo surfaces to $S^2$-bundles on $S^2$ is bijective, because the two surfaces are topologically distinct. We can check this by computing the intersection form. By the Kunneth formula $H^2(\mathbb P^1 \times \mathbb P^1)$ has generators $H$ and $V$ with $H \cdot H = 0$, $H \cdot V = 1$, $V \cdot V=0$. On the other-hand by the blow-up formula $H^2(Bl_x( \mathbb P^2))$ has generators $H$ and $E$ with $H \cdot H = 1$, $H \cdot E = 0$, $E \cdot E=-1$. The first intersection form is even, while the second is odd, so the two spaces are not homoeomorphic.
Finally $\pi_4(S^3)$ is the first stable homotopy group of spheres, which by the Pontrjagin construction is the same as the group of framed $1$-cobordisms $\Omega_1^{fr}$. $\pi_1( SO(3))$. A framing on a circle is just an element of $\pi_1(SO) = \pi_1(SO(3))= \mathbb Z/2$. In fact this gives an isomorphism between the stable homotopy group and $\pi_1(SO(3))$.
Composing these bijections, we get a bijection between degree 8 del Pezzo surfaces and $\pi_1(SO(3))$.<|endoftext|>
TITLE: Wavelet-like Schauder basis for standard spaces of test functions?
QUESTION [12 upvotes]: Edit: A more precise formulation of my question follows the separation line.
The Schwartz space of test functions $\mathcal{S}(\mathbb{R})$ is isomorphic to $\mathfrak{s}$ the space of sequences of real numbers with faster than power-like decay. Likewise, the space $\mathcal{D}(\mathbb{R})$ of compactly supported test functions is isomorphic to $\oplus_{\mathbb{N}} \mathfrak{s}$.
My question: is it possible to find a Schauder basis which realizes such isomorphisms and has a wavelet-like structure, namely, the elements of this basis are scaled copies of one mother function?
Of course, this is related to this other MO question.
The only explicit Schauder basis for such Valdivia-Vogt isomorphisms that
I have seen is in this article by Bargetz. But it does not look wavelet-like to me.

A more detailed formulation: (Aug 2017 edit)
Let $f(x)$ be a function on $\mathbb{R}$. For $(m,k)\in\mathbb{Z}^2$ let me define its dyadic translates $f_{m,k}$ by
$$
f_{m,k}(x)=2^{\frac{m}{2}}f(2^m x-k)\ .
$$
For a Schwartz distribution $T$ in $\mathcal{D}'(\mathbb{R})$ let me also define its translates by the same formula " $T_{m,k}(x)=2^{\frac{m}{2}}T(2^m x-k)$ ", except that it has to be interpreted, as usual, in the sense of distributions. Namely, for every test function $f\in\mathcal{D}(\mathbb{R})$, the space of smooth compactly supported functions, I define
$$
\langle T_{m,k},f\rangle=\langle T, g\rangle
$$
with
$$
g(x)=2^{-\frac{m}{2}}f(2^{-m}(x+k))\ .
$$
Let me call a pair $(f,T)\in\mathcal{D}(\mathbb{R})\times\mathcal{D}'(\mathbb{R})$ an unobtainium biorthogonal pair if for all $k,l,n,m$ in $\mathbb{Z}$, one has (using Kronecker deltas)
$$
\langle T_{m,k},f_{n,l}\rangle=\delta_{m,n}\delta_{l,k}\ .
$$

My question is: does there exist unobtainium biorthogonal pairs?

This is an "extreme sport" version of biorthogonal wavelets.

REPLY [6 votes]: This is not at all a complete answer, but rather an expanded update on my above comment. I shall start with a few general considerations:

If $\{f_n\ |\ n\in\mathbb{N}\}$ is any Schauder basis for either $\mathscr{S}(\mathbb{R})$ or $\mathscr{D}(\mathbb{R})$, at least one $f_n$ must have a non-vanishing moment of order zero, otherwise we run into a contradiction with the fact that $u(f)=\int^{+\infty}_{-\infty}f(x)\mathrm{d}x$ is a continuous linear functional on both spaces. This means that one must employ inhomogeneous wavelet bases (i.e. starting from both a "mother wavelet" and a "father wavelet" in Meyer's terminology), unless one works in "homogeneous" subspaces defined by a certain number of vanishing moments (e.g. $\mathscr{S}_0(\mathbb{R})\subset\mathscr{S}(\mathbb{R})$, discussed in more detail below).

Secondly, as I remarked in my comment, Battle and Meyer have shown that there is no countably infinite orthonormal set of smooth wavelets with exponential decay, let alone of compact support, so any wavelet Schauder basis of $\mathscr{D}(\mathbb{R})$ must be non-orthogonal.

Thirdly, since the Fourier transform of any $f\in\mathscr{D}(\mathbb{R})$ is real analytic, no such $f$ can have all its moments vanishing unless it is identicaly zero, therefore any would-be wavelet basis of $\mathscr{D}(\mathbb{R})$ can have vanishing moments of positive order only up to a finite order $m\in\mathbb{N}$. This is not a problem for $\mathscr{S}(\mathbb{R})$, though (e.g. any $f\in\mathscr{S}(\mathbb{R})$ whose Fourier transform vanishes in a neighborhood of the origin has vanishing moments at all orders).

Finally, as for the form of the wavelet coefficients one should expect from members of $\mathscr{D}(\mathbb{R})$, the compact support restriction together with the group action of $\mathbb{Z}$ on any wavelet Schauder basis by translations should entail that "many" of these coefficients would be zero altogether. This scenario is encouragingly consistent with the Valdivia-Vogt characterization of $\mathscr{D}(\mathbb{R})$. Unfortunately, non-orthogonal wavelets are far more unwieldy than their orthonormal counterparts, so I do not have any more ideas to go on at the moment regarding this case.


That being said, it was shown by K. Saneva and J. Vindas (Wavelet Expansions and Asymptotic Behavior of Distributions, J. Math. Anal. Appl. 370 (2010) 543-554) that the (one-dimensional) homogeneous Littlewood-Paley wavelet basis is an orthonormal wavelet Schauder basis for the "homogeneous" version of $\mathscr{S}(\mathbb{R})$ - namely, the closed subspace $\mathscr{S}_0(\mathbb{R})\subset\mathscr{S}(\mathbb{R})$ consisting of all tempered test functions on $\mathbb{R}$ all of whose moments are vanishing: $$\mathscr{S}_0(\mathbb{R})=\bigg\{\phi\in\mathscr{S}(\mathbb{R})\ \bigg|\ \int^{+\infty}_{-\infty}x^n\phi(x)\mathrm{d}x=0\ ,\,\forall n\in\mathbb{N}\cup\{0\}\bigg\}\ .$$
Now, to go a bit on the details: recall as in e.g. Chapter 3 of the book of Y. Meyer, Wavelets and Operators, Cambridge University Press, 1993, that the homogeneous Littlewood-Paley wavelet basis is the countable orthonormal subset of $\mathscr{S}_0(\mathbb{R})$ given by $$\mathrm{LP}_0(\mathbb{R})=\big\{\psi_{m,n}(x)=2^{\frac{m}{2}}\psi(2^m x-n)\ \big|\ m,n\in\mathbb{Z}\big\}\ ,$$ where the "mother wavelet" $\psi\in\mathscr{S}_0(\mathbb{R})$ is given by its Fourier transform $\hat{\psi}$ as $$\hat{\psi}(\xi)=e^{-i\xi/2}\sqrt{\hat{\phi}^2(\xi/2)-\hat{\phi}^2(\xi)}=e^{-i\xi/2}\theta_1(\xi)$$ and the "father wavelet" $\phi\in\mathscr{S}(\mathbb{R})$ has a Fourier transform $\hat{\phi}\in\mathscr{D}(\mathbb{R})$ with the following properties:

$\hat{\phi}$ is real-valued and even;

$\hat{\phi}(\xi)=1$ if $0\leq\xi\leq 2\pi/3$ and $\hat{\phi}(\xi)=0$ if $\xi\geq 4\pi/3$;

$\hat{\phi}(\xi)$ is strictly decreasing for $2\pi/3\leq\xi\leq4\pi/3$;

$\hat{\phi}^2(\xi)+\hat{\phi}^2(2\pi-\xi)=1$ for all $\xi\in[0,2\pi]$.


Conversely, it is possible to recover $\phi$ from $\psi$ by noticing that $$\hat{\phi}(\xi)=\begin{cases} 1 & (|\xi|\leq 2\pi/3) \\ \sqrt{1-\theta_1^2(\xi)} & (2\pi/3\leq|\xi|\leq 4\pi/3) \\ 0 & (|\xi|\geq 4\pi/3) \end{cases}\ .$$ It is not difficult to find $\hat{\phi}$ with the above properties: for instance, set $$\hat{\phi}(\xi)=\hat{\phi}(-\xi)=\sqrt{\frac{\chi(4\pi/3-\xi)}{\chi(\xi-2\pi/3)+\chi(4\pi/3-\xi)}}\ ,\quad\xi\geq 0\ ,$$ where $$\chi(\xi)=\begin{cases} e^{-\frac{1}{\xi}} & (\xi>0) \\ 0 & (\xi\leq 0) \end{cases}\ .$$
It is shown in the aforementioned paper that the expansion of $f\in\mathscr{S}_0(\mathbb{R})$ in this wavelet basis $$f=\sum_{m,n\in\mathbb{Z}}c^\psi_{m,n}(f)\psi_{m,n}\ ,\,c^\psi_{m,n}(f)=\int^{+\infty}_{-\infty}\overline{\psi_{m,n}(x)}f(x)\mathrm{d}x$$ converges in the induced topology from $\mathscr{S}(\mathbb{R})$ and yields a topological isomorphism $f\mapsto(c^\psi_{m,n}(f))_{m,n\in\mathbb{Z}}$ of $\mathscr{S}_0(\mathbb{R})$ with the so-called space of dyadic rapidly decreasing (double) sequences $$\mathscr{W}=\big\{(c_{m,n})_{m,n\in\mathbb{Z}}\ \big|\ \|(c_{m,n})\|^{\mathscr{W}}_l<+\infty\ ,\,\forall l\in\mathbb{N}\cup\{0\}\big\}\ , \\ \|(c_{m,n})\|^{\mathscr{W}}_l\doteq\sup_{m,n\in\mathbb{Z}}|c_{m,n}|(1+|n|)^l(2^m+2^{-m})^l\ .$$ The above rapid decay of the wavelet coefficients easily follows from Parseval's formula and the support properties of $\hat{\psi}_{m,n}$. Particularly, $\mathrm{LP}_0(\mathbb{R})$ is an absolute (hence Schauder) basis of $\mathscr{S}_0(\mathbb{R})$.
Presumably, the argument of Saneva and Vindas can be adapted to the "inhomogeneous" space $\mathscr{S}(\mathbb{R})$ if we use instead the inhomogeneous Littlewood-Paley wavelet basis $$\mathrm{LP}(\mathbb{R})=\big\{\phi_n(x)=\phi(x-n)\ \big|\ n\in\mathbb{Z}\big\}\cup\big\{\psi_{m,n}(x)=2^{\frac{m}{2}}\psi(2^m x-n)\ \big|\ m\in\mathbb{N}\cup\{0\}\ ,\,n\in\mathbb{Z}\big\}\ ,$$ which is also orthonormal. It can be seen from the form of the seminorms $\|\cdot\|^{\mathscr{W}}_l$ that the outcome should be a Valdivia-Vogt-like isomorphism for $\mathscr{S}(\mathbb{R})$ after a suitable rearrangement of the wavelet coefficients into a single sequence.<|endoftext|>
TITLE: Physical interpretation of the mellin transform variable?
QUESTION [9 upvotes]: I shall keep this to the point: Given a time domain signal say microphone recording of a conversation:

Laplace tranfrom of x is some function X(s) say defined in the complex plane. I like to think of this variable s as a measure of frequency and damping inherrent in the signal x(t) (If anyone has a better interpretation of this I'd love to hear it).
Fourier transform of x is some function x(f) defined in a real domain f that splits the function x into it's constituent frequency components, like only listeneing to the drums in a jazz band.
Mellin transform? I cannot find any such explanation for the melling transform of a function?

Also it seems the Mellin tranform has limited applications in engineering, physics has always been ahead of engineers in adopting new techniques. Is there any reason for this or any new research?

REPLY [7 votes]: Physical interpretation: To develop a physical intuition, this article might be informative:

The power spectrum of the Mellin transformation with applications to scaling of physical quantities

The Mellin transform is used to diagonalize the dilation operator in a
  manner analogous to the use of the Fourier transform to diagonalize
  the translation operator. A power spectrum is also introduced for the
  Mellin transform which is analogous to that used for the Fourier
  transform. Unlike the case for the power spectrum of the Fourier
  transform where sharp peaks correspond to periodicities in
  translation, the peaks in the power spectrum of the Mellin transform
  correspond to periodicities in magnification.

For this reason the Mellin transform is also referred to as the "scale transform" (just as the Fourier transform is called a frequency transform). Just as the Fourier transform is insensitive (in absolute value) to a translation, the Mellin transform is insensitive (in absolute value) to a magnification. 

Applications in radar engineering: A classic application of the Mellin transform is in radar technology. Recall that the resolution in time of a signal is the reciprocal of the spread of its Fourier transform. In radar (or sonar) a signal is reflected from a moving target and you would like to accurately determine its velocity. The velocity resolution of the signal is the reciprocal of the spread of its Mellin transform. This application is worked out in:

Cramer Rao bound computation for velocity estimation in the broad-band case using the Mellin transform

Applications in pattern recognition: The invariance properties of the Mellin and Fourier transforms can be combined in the socalled Fourier-Mellin transform to detect for rotation and scale invariant patterns. This combination is realized by a logarithmic mapping followed by a Fourier transformation, and it is believed that the visual cortex operates in this way.

A biologically plausible transform for visual recognition that is invariant to translation, scale, and rotation<|endoftext|>
TITLE: The sum of a series, continued
QUESTION [7 upvotes]: In this question the OP asks whether the sum
$$
f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k}
$$
is ever zero. An experiment with Mathematica indicates, to any reasonable precision that
$$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$
but I, for one, can't actually prove it. Is it true?

REPLY [6 votes]: We can use the Euler identity
$\sum_{k=0}^\infty \frac{x^k}{q^{k(k-1)/2}(1/q;1/q)_k}=\prod_{j=0}^\infty(1+q^{-j}x)$ for $q>1$. Taking $x=-1$, we obtain the series $f(q,1)$ in the original post, and by the infinite product representation it equals 0.<|endoftext|>
TITLE: Complete regularity in C*-algebras
QUESTION [8 upvotes]: It is clear that commutative C*-algebras correspond to locally compact Hausdorff spaces. And locally compact Hausdorff spaces are completely regular. Now, does the complete regularity statement have some kind of generalization to general C*-algebras. In particular, I am looking for a statement using pure states of the C*-algebra. So something of the form:
If $\tau$ is a pure state on the C*-algebra and if $T$ is a collection of pure states satisfying some conditions (this should correspond to the closed set condition in commutative cases), then there exists an element $a$ of the $C^*$-algebra such that $\tau(a) = 1$ and $\sigma(a)=0$ for each $\sigma\in T$. 
Is is a statement like this true? Are there other interesting generalizations of complete regularity?

REPLY [5 votes]: Nik's answer seems perfectly suited to what the OP was looking for, but there are indeed other generalizations of (complete) regularity to C*-algebras, as shown in Rosický's "Multiplicative lattices and C*-algebras" Cahiers Top. Geom. Diff. Cat. tome 30 no 2 (1989), 95-110.  To paraphrase part of that paper, note first that regularity can be stated in terms of open and closed sets, without reference to points.  Specifically, a topological space $X$ is regular iff, for every open $O\subseteq X$,
$$O=\bigcup\{N:N\text{ is open and }\overline{N}\subseteq O\}.$$
So a natural question to ask is whether, for any C*-algebra $A$ and open projection $q\in A^{**}$,
$$q=\bigvee\{p\in A^{**}:p\text{ is an open projection and }\overline{p}\leq q\}?$$
The answer is yes.  To see this, take any $a\in A^1_+$ with $a\leq q$ and note that $\overline{a_{(\epsilon,1]}}\leq a_{[\epsilon,1]}\leq a_{(0,1]}\leq q$, for any $\epsilon>0$, where $a_S$ denotes the spectral projection of $a$ in $A^{**}$ corresponding to $S\subseteq\mathbb{R}$.  The result now follows because $q$ being open means
$$q=\bigvee\{a_{(0,1]}:a\in A^1_+\text{ and }a\leq q\}.\\$$
As $\overline{N\cup M}=\overline{N}\cup\overline{M}$, for open subsets $N$ and $M$, above we actually have a directed union.  On the other hand, we can have $\overline{p\vee r}\neq\overline{p}\vee\overline{r}$ for open projections $p$ and $r$, as in Example II.6 in Akemann's paper referred to by Nik, which also shows that the supremum above is not always directed.  However, we might ask:

Can the open projections above be replaced by a directed subset?

In general I do not know.  If $A$ is separable then the answer is yes, as then every open $p$ will be of the form $a_{(0,1]}$, for some $a\in A^1_+$, and $a_{(\epsilon,1]}$ is directed, for $\epsilon>0$.  More generally, the answer is yes if $A$ has an "almost idempotent" approximate unit $(a_\lambda)\subseteq A^1_+$, i.e. such that $a_\lambda a_\gamma=a_\lambda$ whenever $\lambda<\gamma$.  Wondering if, in fact, all C*-algebras have such an approximate unit led me to post this question.
Regardless, in general we can still get a weaker iterated form of directedness.  Specifically, note that $(a+b)_{(0,1]}=a_{(0,1]}\vee b_{(0,1]}$, for all $a,b\in A^1_+$, so every open projection $r$ is a directed supremum of open projections $q$, which are all themselves directed supremums of open projections $p$ with $\overline{p}\leq q$.
Also note the relation $\overline{N}\subseteq O$ can be expressed purely in terms of open sets as
$$\text{there exists open $M$ disjoint from $N$ such that }X=M\cup O.$$
Wondering if, for open projections $p,q\in A^{**}$, $\overline{p}\leq q$ can also be expressed similarly in terms of open projections just led me to post this question.  In fact, it is really relation 2. of that question, used with iterated directness as above, that Rosický considers as the appropriate notion of regularity for C*-algebras in the paper mentioned above.
For complete regularity one replaces $\overline{N}\subseteq O$ with the statement that there exist open $(O_r)$ for $r\in[0,1]$ (or $[0,1]\cap\mathbb{Q}$, as in Rosický's paper) such that $N\subseteq O_0$, $O_1\subseteq O$ and $\overline{O}_s\subseteq O_t$ whenever $s<t$.  Again, the analogous statement for complete regularity holds for open projections in C*-algebras although again it is not clear if the supremum can always be replaced with a directed supremum.<|endoftext|>
TITLE: Jordan-Hölder-like statements for modules with $\Delta$-filtrations over a quasihereditary algebra
QUESTION [8 upvotes]: Definitions
Let $A$ be an Artin algebra (for instance, take $A$ to be a finite dimensional algebra over some field) and label the isomorphism classes of simple $A$-modules by the elements of a partially ordered set $(\Lambda, \leq)$. 
Denote the simple $A$-modules (up to isomorphism) by $L_{\lambda}$, $\lambda \in \Lambda$, and let $P_{\lambda}$ be the projective cover of $L_{\lambda}$. 
For every $\lambda \in \Lambda$, denote by $\Delta(\lambda)$ the largest factor module of $P_{\lambda}$ all of whose composition factors are of the form $L_{\mu}$, $\mu \leq \lambda$. Call this module the standard module with weight $\lambda$.
We say that $A$ is quasihereditary with respect to the poset $(\Lambda, \leq)$ if:

The poset $(\Lambda, \leq)$ is adapted to $B$: for every finitely generated module $M$ with simple socle $L_{\lambda}$ and with simple top $L_{\mu}$, where $\mu$ and $\lambda$ uncomparable in $(\Lambda, \leq)$, there is $\nu \in \Lambda$, $\nu > \mu$ or $\nu > \lambda$, such that $L_{\nu}$ is a composition factor of $M$,
$L_{\lambda}$ has multiplicity one in $\Delta(\lambda)$, for all $\lambda \in \Lambda$;
$P_{\lambda}$ has a $\Delta$-filtration (i.e. a filtration whose factors are standard modules), for all $\lambda \in \Lambda$.

Finally, call a module $\Delta$-semisimple if it is a direct sum of standard modules.
Questions
Let $A$ be quasihereditary and let $M$ be some finitely genereted $A$-module with a $\Delta$-filtration. I can convince myself that the $\Delta$-filtrations of $M$ are essentially unique (à la Jordan-Holder), up to possible permutation of the factors (I could not find this in literature though -- does anyone know a reference?).
Let $N$ be a $\Delta$-semisimple submodule of $M$ such that $M/N$ still has a $\Delta$-filtration -- I have the feeling that $M$ has a unique maximal submodule $N$ with this property. I have not been able to prove this nor found a reference. Does anyone know if this is true? And would this be "real" uniqueness or just uniqueness up to isomorphism?

REPLY [2 votes]: Let $M$ be a module with a $\Delta$-filtration. Let $N$ be a $\Delta$-semisimple submodule of M such that $M/N$ still has a $\Delta$-filtration -- I prove that there is a unique maximal module with this property (and hope that my justification is not immensely stupid).
This will be a consequence of the following Lemma.
Lemma Let $M_1$ and $M_2$ be submodules of $M$, with $M_i$ isomorphic to a standard module and such that $M/M_i$ has a $\Delta$-filtration, $i=1,2$. Then either $M_1=M_2$ or $M_1 \cap M_2 =0$.
Proof Suppose $M_1 \cap M_2 \neq 0$ and let $M_i \cong \Delta(x_i)$, for some $x_i \in \Phi$, $i=1,2$. Let $\iota_i$ be the inclusion map of $M_1 \cap M_2$ on $M_i$, and let $\tau_i$ be the inclusion of $M_i$ on $M$,
$$
\iota_i : M_1 \cap M_2 \longrightarrow M_i,
$$
$$
\tau_i : M_i \longrightarrow M.
$$
Consider the (right exact) functor $F:=D\circ \operatorname{Hom}_A(-,T)$, where $D$ is the standard duality and $T$ the characteristic $A$-module. Denote $\operatorname{End}_A(T)^{op}$ by $\mathcal{R}(A)$ (the Ringel dual of $A$) -- the functor $F$ induces an equivalence between $\mathcal{F}(\Delta)$ and $\mathcal{F}(\nabla ')$ (where $\nabla '$ is the set of costandard modules over $\mathcal{R}(A)$). 
The functor $F$ maps the monics $\tau_i$ in $\operatorname{mod}A$ to monics $F(\tau_i)$ in $\operatorname{mod} \mathcal{R}(A)$ -- this is because the cokernel of $\tau_i$ is in $\mathcal{F}(\Delta)$. I claim that the functor $F$ maps the monics $\iota_i$ in $\operatorname{mod}A$ to nonzero morphisms $F(\iota_i)$ in $\operatorname{mod} \mathcal{R}(A)$. Let $\varepsilon$ be the projective cover of $M_1 \cap M_2$. The morphisms $\varepsilon \circ \iota_i$ are nonzero, with domain and codomain in $\mathcal{F}(\Delta)$, thus $F(\varepsilon) \circ F(\iota_i)$ is nonzero, and hence $F(\iota_i) \neq 0$. 
Now the image of $F(\iota_i)$ is contained in a module isomorphic to $\nabla '(x_i)$. Because $F(\tau_i)$ is monic, the image of $F(\tau_i) \circ F(\iota_i)$ has simple socle $L_{x_i}$. As
$$F(\tau_1) \circ F(\iota_1)=F(\tau_1 \circ \iota_1)=F(\tau_2 \circ \iota_2) =F(\tau_2) \circ F(\iota_2),$$
we must have $x_1=x_2=:x$.
We want to show that $M_1$ and $M_2$ coincide. Suppose (by contradiction) that $M_1\neq M_2$. Then the (simple) top $L_x$ of $M_1$ is a composition factor of $M/ M_2$. Indeed, there must be some subquotient of $M/ M_2$, say $M_3$, satisfying: $M_3$ is part of a $\Delta$-filtration of $M/M_2$; $M_3$ is isomorphic to $\Delta(y)$ (for some $y \in \Phi$); $M_3$ contains the top of $M_1$ as a composition factor. Note that we must have $y \geq x$ in $\Phi$. In fact, by the definition of standard module and by the structure of the module $M$, the module $M_1$ (which is isomorphic to $\Delta(x)$) would have to be contained in $M_3\cong \Delta(y)$. On the other hand, we are assuming that $M_3$ is a subquotient of $M/ M_2$ and that $M_1\cap  M_2\neq 0$.<|endoftext|>
TITLE: Is the locus of points which have irreducible fibers constructible?
QUESTION [5 upvotes]: Suppose $X \rightarrow Y$ is a map of projective schemes over a field $k$. Is $\{y \in Y: \pi^{-1}(y) \text{ is irreducible}\}$ a constructible subset of $Y$?
Note: One cannot hope to do "better" than constructible. That is, if we let H be the hilbert scheme of conics in $\mathbb P^2$ and let $\mathscr C$ be the universal curve over $H$, then $\mathscr C \rightarrow H$ has irreducible fibers consisting of smooth conics and double lines, which is constructible but not locally closed.

REPLY [4 votes]: I double-checked.  This is Théorème 4.10, p. 36 of the following.
MR0725671 (86b:13007) Reviewed 
Jouanolou, Jean-Pierre 
Théorèmes de Bertini et applications. (French) 
Progress in Mathematics, 42. Birkhäuser Boston, Inc., Boston, MA, 1983. ii+127 pp. 
ISBN: 0-8176-3164-X 
As mentioned above, you need to replace "irreducible" by "geometrically irreducible".  You can also prove this from Lemma 3.2 as I sketched above.  But Lemma 3.2 is really designed for a different purpose.<|endoftext|>
TITLE: Stiefel-Whitney class of unordered configuration space
QUESTION [6 upvotes]: Let $S^m$ be the $m$-sphere and $$F(S^m,2)/\mathbb{Z}_2=\{(a,b)\mid a,b\in S^m, a\neq b\}/(a,b)\sim (b,a)$$ be the $2$-nd unordered configuration space on $S^m$.  How to compute the total Stiefel-Whitney class of the tangent bundle of $F(S^m,2)/\mathbb{Z}_2$
$$
w(TF(S^m,2)/\mathbb{Z}_2)?
$$
Is $F(S^m,2)/\mathbb{Z}_2$ homeomorphic to $\mathbb{R}P^m\times \mathbb{R}^m$ or not?
Moreover, if we change $S^m$ to other manifolds, for example, projective spaces, Grassmannians, then how to compute the corresponding Stiefel-Whitney class? The question is on characteristic classes of tangent bundle of 2-nd unordered configuration space

REPLY [4 votes]: The configuration space $F(S^m, 2)/\mathbb Z_2$ is diffeomorphic to the total space of the $m$-dimensional vector bundle $E=\gamma^\perp$ over $\mathbb RP^m$, where $\gamma\subset \mathbb R^{m+1}$ is the tautological 1-dimensional line bundle and $\gamma^\perp$ is its orthogonal complement.
To define a diffeomorphism $\phi\colon Tot(E)\to F(S^m, 2)/\mathbb Z_2$ let us pick a point $([v], w)\in Tot(E)$ (where $[v]\in\mathbb R P^m$ and $w\in v^\perp$),
such that a vector $v\in \mathbb R^{m+1}$ has the unit norm $||v||_2=1$. 
Let $\phi \colon ([v],w)\mapsto\bigl(\frac{w+v}{||w+v||_2}, \frac{w-v}{||w-v||_2}\bigr)\in F(S^m, 2)/\mathbb Z_2$. You can easily verify, that $\phi$ is diffeomorphism.
Computation of the total Stiefel-Whitney class is now straighforward. Let $\pi\colon Tot(E)\to \mathbb R P^m$ be the natural projection, then one has the long exact sequence:
$$
0\to \pi^* E\to TTot(E)\to\pi^*T\mathbb RP^m\to 0,
$$
hence 
$$
w(TF(S^m, 2)/\mathbb Z_2)=w(\pi^*E)w(\pi^*T\mathbb RP^m)=\pi^*(1+a)^{-1}\cdot\pi^*(1+a)^{m+1}=(1+\pi^*a)^m,$$ where $a\in H^1(\mathbb RP^m,\mathbb Z_2)$ is the generator.<|endoftext|>
TITLE: Proof of Pinelis (1992) - Banach space inequalities
QUESTION [5 upvotes]: I am reading Pinelis "An approach to inequalities for the distributions of infinite -dimensional martingales" and cannot follow his proof of Theorem 3:
Let $(f_n)$ be a martingale in a separable Banach space $(\mathcal{X},||~||)$, $\mathcal{X} = L^p, p \geq 2$ and $f^*=\sup\{||f_n||\}$.
Theorem 3 says

\begin{align} P(f^*>r) \leq 2\exp\big(-r^2/2(p-1)\big), \quad r \geq 0
\end{align}

and Pinelis writes 

One can compare the last inequality with
  \begin{align}
P(f^*>r) \leq (r+1)\exp(-r^2/2)
\end{align}
  in Kallenberg ans Sztencel (1991) proved for $\mathcal{X} = L^2$.

Does he refer to
\begin{align}
P(f^*>r) \leq \frac{1+r}{1+rc}\exp\big(-\frac{r}{2c} \ln(1+rc)\big), \quad r \geq 0
\end{align}
from Kallenberg and Sztencel (1991)? If so, I still cannot see the path he takes to prove his theorem.
Kallenberg and Sztencel (1991): Some dimension-free features of vector-valued martingales

REPLY [14 votes]: As written in my paper [1], the inequality 
$$P(f^*>r) \le 2\exp\big(-r^2/2(p-1)\big)
$$
in Theorem 3 in [1] 
for martingales in $\mathcal{X}=L^p$ can be compared with the inequality 
$$
P(f^*>r) \le C (r+1)\exp(-r^2/2)
$$
with an unidentified absolute constant $C>0$, given 
(as noted in the comment by Nate Eldredge) in Theorem 5.3 of [Kallenberg and Sztencel], proved for $\mathcal{X} = L^2$.
As for the inequality 
$$P(f^*>r) \le C \frac{1+r}{1+rc}\exp\big(-\frac{r}{2c} \ln(1+rc)\big) 
$$
with an absolute constant $C>0$, 
which is inequality (5.2) of [Kallenberg and Sztencel], it can be compared with the inequality in Theorem 2 of [1].  
More general results were given in [3]. 
Also, you wrote: "I still cannot see the path he takes to prove his theorem." Can you specify the steps that seem unclear?<|endoftext|>
TITLE: $|\exp_p(x)\exp_q(T(x))|$ controlled by $|pq|?$ $T$ is parallel transportation in Alexandrov space
QUESTION [9 upvotes]: Let $M$ be an Alexandrov space with $sec \geqslant 0$. Let $p$ and $q$ be points of shortest path $\gamma$ in $M$, that are not end point. Then the tangent cone can split as $C_p=L_p \times \mathbb{R}$. Where $L_p=\{x\in C_p;x \perp \gamma\}$. 
The second variation says: "For any sequence $\varepsilon_n \to 0$, there is a subsequence $\{\epsilon_n\}\subset \{\varepsilon_n\}$ and the isometry (parallel transportation) $T: L_p \to L_q$ such that 
$$
|\exp_p(\epsilon_nx)\exp_q(\epsilon_n T(x))|\leqslant |pq|+o(\epsilon^2_n)
$$ 
Where $x\in L_p$ is any vector.
My question: is $|\exp_p(x)\exp_q(T(x))|$ can be controlled by $|pq|?$
I think this is wrong, but I can't think of a counter example.

REPLY [4 votes]: First note that in positively curved Riemannian manifolds you do have theestimate 
$$
|\exp_p(\varepsilon\cdot x)\exp_q(\varepsilon\cdot T(x))|\leqslant |pq|.
$$ 
for small $\varepsilon>0$ ONLY.
Consider Berger sphere and geodesic along Hopf fiber.
I do not know example of Alexandrov space where this inequality fails for all small $\varepsilon$ but I would not be surprized if one can build it based on the example above.
Yet closely related technical question. In principle the parallel transportation $T$ may depend on the sequence $\varepsilon_n$, but there are no known examples when it does happen. 
If such examples exist then for fixed $T$ the function 
$\varepsilon\mapsto|\exp_p(\varepsilon\cdot x)\exp_q(\varepsilon\cdot T(x))|$
might behave quite bad...<|endoftext|>
TITLE: A problem of potential theory arising in biology
QUESTION [27 upvotes]: Let $K_0$ and $K_1$ be two bounded, disjoint convex sets in $R^n,n\geq 3$,
and
$u$ the equibrium potential, that is the
harmonic functon in $R^n\backslash\{ K_0\cup K_1\}$ such that $u$
has boundary values $1$ on $K_1\cup K_2$ and $u(\infty)= 0$.
Denote
$$r_j=r_j(K_0,K_1)=
\int_{\partial K_j}\frac{\partial u}{\partial n}ds,\quad j=0,1,$$
where $n$ is the inner normal and $ds$ is the surface area element.

Is it true that each $r_j$ decreases if we move $K_0$ and $K_1$ closer
  to each other?

More precisely: Let $f:R^n\to R^n$ be a distance decreasing homeomorphism,
whose restriction on each $K_j$ is an isometry onto $f(K_j)$.
Is it true that $r_j(f(K_0),f(K_1))\leq r_j(K_0,K_1)$ for each $j$?
Same question can be asked with more than two sets, but there are counterexamples
with many balls $K_j$, even of equal radii: this is called a "Faraday cage".
Comments. This problem originates in the attempts of biologists to explain
why certain animals (like armadillos) group close together when they sleep.
In this model, $K_j$ are the animals with body temperature $1$, while the medum has temperature $0$. Then $u$ is the time independent temperature near the bodies, and $r_j$ is the rate of heat loss. Presumably moving closer together
desceases the rates of heat loss $r_j$. It is easy to prove that the TOTAL heat loss of the group
$r_1+r_2$ decreases when we move animals $K_1$ and $K_2$ closer together,
see http://www.math.purdue.edu/~eremenko/dvi/armadillo.pdf
and references there.
But each individual animal $K_j$ feels only his own $r_j$, not the sum.
Also some condition of convexity type is indeed needed here; think of a kangaroo
putting her child in the bag. This decreases the heat loss by the child but (slightly) increases the cooling rate of the mother.
The problem is unsolved even when $K_0$ and $K_1$ are two balls of different
radii. In this case one can obtain $r_k$ as infinite series of functions of the distance, but these
series are alternate and it is unclear how to show that they are monotone.
EDIT. By "Faraday cage" I mean the following. Take a spherical shell of the form $1-r\leq |x|\leq 1$, $r$ is small, and break it into little convex pieces of size approximately $r$, with some space between pieces. These pieces are $n$ animals. The temperature they
create is approximately the same as the shell itself would create, which is almost constant $1$ inside, and almost $1/|x|$ outside. Consider a new animal of the same size
outside. It slightly decreases the heat loss by some animals on the shell close to it.
Now move the same animal inside. It will have no effect at all on the shell animals.
This is the same principle as kangaroo bag, I just broke the bag into many small convex pieces. More careful argument shows that the animals can be balls of the same radius here.
(Faraday's own cage does the same with electric field which satisfies the same equation).

REPLY [8 votes]: The answer is negative. 
Assume that $K_{0}$ and $K_{1}$ are two disjoint balls. If $K_{0}$ is at least twice as big as $K_{1}$, then $K_{0}$ should keep some nonzero distance in order to minimize his heat loss while $K_{1}$ should try to be as close as possible to $K_{0}$. 
To be more precise let $r_{0}$ and $r_{1}$denote the radii of the  balls $K_{0}$ and $K_{1}$. Let 
\begin{align*}
Q_{j}(d) = \int_{\partial K_{j}} \frac{\partial u}{\partial n} ds\quad \text{for} \quad j=0,1
\end{align*}
denote the heat loss as a function of distance between the balls. Then $Q_{0}(d)$ is not monotonically increasing provided that $\frac{r_{0}}{r_{1}} > \ell$ where $\ell\approx 1.95$ is the positive solution of the equation
\begin{align*}
2(1+x^{3})\left(\gamma+\psi\left(\frac{1}{1+x}\right) \right)+x^{2}+x+2(x^{2}-x)\psi'\left(\frac{1}{1+x}\right)=0.
\end{align*}
In the above notation $\gamma$ is the Euler--Mascheroni constant, and $\psi$ is the digamma function. I have uploaded a preprint on arXiv  about this problem.
In the following figure the graph of $Q_{0}(d)$ is given when $r_{0}=20$, $r_{1}=1$, $0\leq d \leq 80$ and the temperature of the balls equals $1$. 

Actually numerical computations show that if the sizes of the balls are comparable, namely,  $\frac{1}{\ell} \leq \frac{r_{0}}{r_{1}} \leq \ell$ then the individual heat loss $Q_{j}$, $j=0,1$ is monotonically increasing, however, I was not able to prove this.
Also based on numerical computations if   $\frac{r_{0}}{r_{1}} > \ell$ then the nonzero distance that the big ball has to keep approximately equal to $r_{0}$. The heat loss of the small ball is always monotonically increasing.<|endoftext|>
TITLE: Partition relation at successor cardinal
QUESTION [11 upvotes]: Is it consistent that there are regular cardinals $\kappa < \lambda$, such that $\lambda$ is a successor cardinal and for every coloring $d\colon[\lambda]^2\to\kappa$ there is some $A\subseteq\lambda$ of cardinality $\lambda$ and $\eta < \kappa$ such that $\forall x,y\in A$, $d(x,y)< \eta$?

REPLY [11 votes]: If $\lambda\nrightarrow[\lambda]^2_\lambda$ then there is no such $\kappa<\lambda$:  modify the coloring that witnesses this by defining it to be $0$ on pairs of ordinals sent to values above $\kappa$. (The original coloring takes on all values on any unbounded subset of $\lambda$, hence the modified coloring takes on all values below $\kappa$ on such a set.)
This rules out $\lambda$ being successor of regular (in fact, it rules out $\lambda$ having a non-reflecting stationary set by a result of Todorcevic).  
For successor of singular, the question of whether $\lambda\nrightarrow[\lambda]^2_\lambda$  can fail is open.
Edit #1:
What you ask follows if $\lambda$ is a Jonsson successor of singular, as the failure of your condition at successor of singular gives a Jonsson algebra.
Edit #2: 
If you demand that your condition applies to colorings of finite subsets of $\lambda$, then you end up with something equivalent to $\lambda$ being Jonsson.<|endoftext|>
TITLE: generalized universal coefficient sequence
QUESTION [6 upvotes]: Take the familiar Universal Coefficient Theorem for ordinary homology with $\mathbb{Z}$-coefficients and ordinary cohomology with coefficients in some abelian group $A$:$$0\rightarrow \text{Ext}_\mathbb{Z}^1(H_{i-1}(X;\mathbb{Z}),A)\rightarrow H^i(X;A)\rightarrow \text{Hom}_\mathbb{Z}(H_i(X;\mathbb{Z}),A)\rightarrow 0$$This exact sequence can be rewritten in the language of topological spectra:$$0\rightarrow\text{ext}_{H\mathbb{Z}}^1([\Sigma^{i-1}S,H\mathbb{Z}\wedge X],HA)\rightarrow [X,\Sigma^iHA]\rightarrow [[\Sigma^iS,H\mathbb{Z}\wedge X],HA]\rightarrow 0$$where $[-,-]$ denotes the internal hom $\text{hom}_{H\mathbb{Z}}(-,-)$ of $H\mathbb{Z}$-module spectra. (I am confused about some basic points [see my question], but I presume that $\text{ext}$ makes sense in this context.)
How can one generalize this sequence to generalized (co)homology theories? One guess is to replace $H\mathbb{Z}$ with some commutative ring spectrum $E$ and $HA$ with an $E$-module spectrum $F$. I imagine this guess is far too ambitious and a more conservative theorem is the correct generalization.

REPLY [11 votes]: There is more than one possible generalization. The most common is the universal coefficient spectral sequence. Given a (homotopy) commutative ring spectrum $E$ and a spectrum $X$, there is under certain conditions a spectral sequence
$$ Ext^{p,q}_{E_*}(E_*(X), E_*) \Rightarrow E^{q-p}(X) $$
This is true for example if

$E$ is an $A_\infty$-ring spectrum (e.g. if $E =ko, ku, KO, KU, TMF, MO, MSO, MSpin \dots$) [EKMM, IV.4]
$E$ is even and Landweber exact (e.g. if $E = MU, E(n), E_n, \dots$) [Adams' lectures on generalized cohomology, which Peter alluded to, and Devinatz: Morava Modules and Brown-Comenetz Duality, Prop 1.3, and the discussion thereafter]

Sometimes, this spectral sequence is not extremely useful though. For example, take $E = KO$. In general, $KO_*X$ might have infinite cohomological dimension over $KO_*$ so that the spectral sequence might be difficult to control. In this case another perspective is more useful: Anderson duality. 
Consider the functor $E \mapsto Hom(\pi_*E, \mathbb{Q}/\mathbb{Z})$ from the homotopy category of spectra to graded abelian groups. As $\mathbb{Q}/\mathbb{Z}$ is injective, this is a cohomological functor, by Brown representability represented by a spectrum $I_{\mathbb{Q}/\mathbb{Z}}$. There is an evident map $H\mathbb{Q} \to I_{\mathbb{Q}/\mathbb{Z}}$ whose fiber we denote by $I$. For a spectrum $E$, we define its Anderson dual $IE$. to be the function spectrum $F(E, I)$. It is easy to show that we get for every spectrum $X$ a short exact sequence
$$ 0 \to Ext^1_{\mathbb{Z}}(E_{k-1}X, \mathbb{Z}) \to (IE)^kX \to Hom_{\mathbb{Z}}(E_kX, \mathbb{Z}) \to 0.$$
That means that there exists always a short exact sequence computing from $E$-homology the $IE$-cohomology. This is, of course, only useful if we can identify $IE$. Luckily, this has been done for a few spectra: 

$IH\mathbb{Z} \simeq H\mathbb{Z}$
$IKU \simeq KU$
$IKO \simeq \Sigma^4 KO$ [see e.g. Heard, Stojanoska]
$ITmf \simeq \Sigma^{21}Tmf$ (at least at primes $>2$) [see Stojanoska]

For example, for $KO$ this means that we get a universal coefficient sequence
$$ 0 \to Ext^1_{\mathbb{Z}}(KO_{k-1}X, \mathbb{Z}) \to KO^{k+4}X \to Hom_{\mathbb{Z}}(KO_kX, \mathbb{Z}) \to 0.$$<|endoftext|>
TITLE: Schematization of a topological space
QUESTION [7 upvotes]: I wanted to understand or at least to know if what follows make sense.
Given a connected toplogical space $X$, I want to associate a scheme. In the following way. 
For a space $X$ and $A(X)$ the Sullivan minimal model, first I associate the differential graded commutative ring $A(X)$. Then I associate to $A(X)$ the corresponding symmetric monoidal triangulated category of differential graded $A(X)$-modules. Finaly, we associate to symmetric monoidal triangulated category of differential graded $A(X)$-modules the scheme $\mathbf{X}$ using the $spec$ functor defined by paul Balmer.
My question is the following. What can we say about $X$ using  $\mathbf{X}$ ?
Can we reconstruct (partially)  $X$ from  $\mathbf{X}$ ?
Thank you very much for any help. 
Edit: I replaced the rational cohomology ring $H^{\ast}(X, \mathbf{Q})$ by $A(X)$.

REPLY [8 votes]: If $X$ has mild finiteness conditions (its even-dimensional cohomology is noetherian and its odd-dimensional cohomology is finitely generated as a module over it), then it is possible to completely determine the spectrum (as defined by Balmer) of the category of (perfect) modules over the cochain algebra $C^*(X; \mathbb{Q})$. The points of the spectrum correspond to homogeneous prime ideals of the even-dimensional cohomology in the natural sense, so perfect modules over $C^*(X; \mathbb{Q})$ are "stratified" by their cohomology as a module over $H^*(X; \mathbb{Q})$ (and the associated support). For example, if $X$ is  a finite CW complex and is connected, then the spectrum is just a point. 
This result holds more generally for $E_\infty$-ring spectra over $\mathbb{Q}$ that satisfy this type of noetherianness condition, and is contained in my preprint "Residue fields for a class of rational $E_\infty$-rings and applications." As indicated in the title, the idea is to construct "residue fields" via a transfinite iterated cell attachment procedure and then to prove an analog of the Devinatz-Hopkins-Smith nilpotence theorem (from which these types of thick subcategory theorems can be derived). Incidentally, these methods do not (at least so far as I am aware) lead to a classification of localizing subcategories, which has been carried in some other instances (notably for stable module categories by Benson-Iyengar-Krause).  
In any event, the conclusion is that one won't see very much of $X$ that one didn't have before.<|endoftext|>
TITLE: Questions about $\mathbb{C}[G/U^-]$ and $\mathbb{C}[B]$
QUESTION [5 upvotes]: Let $G = GL_n$. By algebraic Peter-Weyl theorem, we have 
$$ 
\mathbb{C}[G] = \bigoplus_{\lambda} V_{\lambda} \otimes V_{\lambda}^*,
$$
where $\lambda$'s are dominant weights. Let $U^-$ be the unipotent subgroup of $G$ consisting of all unipotent lower triangular matrices in $G$. Let $U^-$ act on $G$ by left multiplication. Then we have 
$$
\mathbb{C}[G/U^-] = \bigoplus_{\lambda} V_{\lambda}^*.
$$ 
Let $B$ be the Borel subgroup of $G$ consisting of all upper triangular matrices in $G$. Do we have
$$
\mathbb{C}[B] = \bigoplus_{\lambda \in \mathfrak{h}^*} V_{\lambda}^*?
$$
Here $\lambda \in \mathfrak{h}^*$ can be non-dominant. Thank you very much.

REPLY [4 votes]: This type of question is considered in the paper Longest weight vectors and excellent filtrations by Wilberd van der Kallen (Math. Z. 201, 19-31 (1989); MR). van der Kallen works over an arbitrary algebraically closed field $k$, and states his results for the Borel subgroup $B$ of a connected simply-connected semisimple algebraic group $G$ defined over $k$ (e.g., $G = SL_n$). In this context, he shows that the coordinate algebra $k[B]$ admits a filtration as a $B \times B$-module with sections of the form $P(-\lambda) \otimes Q(\lambda)$ for $\lambda$ an integral weight, where $P(-\lambda)$ is a "dual Joseph module," and $Q(\lambda)$ is "minimal relative Schubert module."
I am not well-versed on all of the terminology and conventions in van der Kallen's paper, so will leave it to the reader to consult van der Kallen's paper for the precise definitions of the relevant modules appearing in the filtration. I'm also not sure how this description fits with Ben's description when $k = \mathbb{C}$.<|endoftext|>
TITLE: On bounds for idoneal integer
QUESTION [9 upvotes]: What is the best known lower bound and upper bound known for such a number if it exists and have there been any attempts (computational including) to eliminate the existence of such a number in known range?
As mentioned below http://arxiv.org/pdf/1502.07953.pdf provides some computation which gets a lower bound from exhaustive checking.
Is there a theoretical way to get upper bound?
A new post https://mathoverflow.net/questions/217689/some-issues-on-idoneal-numbers-in-available-research-literature has been created moving some portion of modification of post originally made here.

REPLY [6 votes]: Regarding upper bounds, there are none.  In more detail:
By genus theory we know that for discriminants with one class per genus, the class group satisfies
$$
\mathcal C(-d)\cong \left(\mathbb Z/2\right)^{g-1},
$$
where $g$ is the number of prime divisors of $d$.  Obviously $d$ is bigger than the absolute value of the smallest fundamental discriminant with $g$ prime divisors,
$$
d_g\overset{\text{def.}}=3\cdot4\cdot5\cdot7\dots \cdot p_g.
$$
From lower bounds on the size of $p_g$, the $g$th prime and on $\theta(x)=\sum_{p<x}\log(p)$, one can show that
$$
d_g>g^g.
$$
Since $
2^{g-1} \ll \sqrt{g^g}, 
$
good lower bounds for the class number rule out the possibility for one class per genus for large $g$.  Chowla used this idea to show that the number of classes per genus tends to infinity with $d$.  
In 1973, Peter Weinberger showed that on GRH, no fundamental discriminant $-d<-5460$ has one class per genus, and unconditionally there is at most one more such $d$.  Weinberger used Tatuzawa's version of Siegel's Theorem to deduce there is at most one such $d$ bigger than $d_{11}=401120980260\approx 4\times 10^{11}$, and sieving to eliminate the $d<d_{11}$.  
In contrast, Oesterle explicitly observed that the lower bound due to Goldfeld-Gross-Zagier  is not strong enough to finish the classification of discriminants with one class per genus: 
$
\log(g^g) $ is $\ll 2^{g-1}
$.
Iwaniec and Kowalski observed that even the full strength of the Birch Swinnerton-Dyer conjecture, 'the best effective lower bounds which current technology allows us to hope for'  would not suffice, as $ \log(g^g)^r$ is $\ll 2^{g-1}$  for any $r$.  In fact, the outlook is still more bleak:  Watkins observes that if the discriminant $-d$ is divisible by all the primes up to $(\log\log d)^3$ (as $d_g$ certainly is), the product over primes dividing $d$ in the Goldfeld-Gross-Zagier lower bound is so small the resulting bound is worse than the trivial bound.
Regarding lower bounds, sieving can be used, as Weinberger did and as has been done with all class number problems going back to the work of D.H. Lehmer.  In resolving class number problems, beginning with the work of Stark on class number $1$ all the way up to the work of Watkins resolving all class numbers $\le 100$, there is always an intermediate range beyond where sieving is feasible and below where the deep theorems (e.g. Gross-Zagier-Goldfeld) take effect.  In this intermediate range one uses instead the Deuring-Heilbronn phenomenon, in which a counter-example to GRH (real a zero of a quadratic Dirichlet $L$-function close to $s=1$) will influence the location of zeros (on the critical line) of other $L$-functions.  Practical application of the Deuring-Heilbronn phenomenon was initiated by Stark in his PhD thesis, who used it to show a $10$th discriminant with class number  $1$ would have to have $d>\exp(2.2\cdot 10^7)$.
These ideas can be adapted to the study of discriminations with one class per genus (unpublished).  One can show that there is no such discriminant in the range $10^{67}\le d\le 10^{28000}$.  Extending this interval upwards requires finding new quadratic Dirichlet $L$ functions with very low-lying zeros.  What one really desires is to extend the interval downward to meet up with the range eliminated by sieving.  This is technically very difficult (see Watkins' thesis for the analogous problem for class number $\le 100$) and is the main reason the work is unpublished.<|endoftext|>
TITLE: An introduction to Macdonald polynomials other (better?!) than SFHP
QUESTION [15 upvotes]: Long story short, I personally find Macdonald's celebrated book Symmetric Functions and Hall Polynomials somewhat difficult to read for various reasons. I also know for a fact that I'm not the only one to hold that opinion =)
On the other hand, Macdonald polynomials have been extremely popular objects of study for quite some time now. Nonetheless, an expert in the field told me recently that he does not know of any other books that can be used as an alternative introduction to the subject. So here are my two questions.
1) What is your favorite introductory text on Macdonald polynomials, whether in the form of a book or anything else?
2) Are there any other books that focus on Macdonald's ring of symmetric functions, its generalizations and various bases therein? How do they compare to SFHP?

REPLY [10 votes]: I recommend reading Macdonald's volume University Lecture Series Vol 12 Symmetric functions and orthogonal polynomials.
It is a rather short introduction to Macdonald polynomials for the symmetric group and for general finite Weyl groups, but everything is very well explained, organized, and self-contained. After reading this, I found myself much more comfortable getting back to Symmetric functions and Hall polynomials for futher background and details.<|endoftext|>
TITLE: Hodge de Rham operator and orientability
QUESTION [12 upvotes]: Let $(M,g)$ be a Riemannian manifold. One can consider the exterior algebra bundle $\Lambda(T^*M)$. The sections of this bundle are differential forms, to be noted by $\Omega^k(M)$. One can consider de Rham differential $d:\Omega^*(M) \to \Omega^{*+1}(M)$. In order to define the so called Hodge-de Rham operator one needs the adjoint of this differential: this requiers some choice of scalar product on the space of all forms. If $M$ is orientable there is natural notion of integration which is defined via volume form (nonvanishing top form)-this notion is absent in the nonorientable situation. However there is also the notion of density: as far as I know, with this notion one can define scalar products on forms as well and so we are able to define the adjoint of $d$ and therefore the Hodge de Rham operator. My question is: 

Is there a flaw in above argument? If not, why we consider Hodge-de Rham operator only for orientable manifolds?

REPLY [8 votes]: The comments above almost say it all. Let me just add one small point of motivation. There are several notions of integration on manifolds. If you have a measure, you can integrate functions. A Riemannian metric is one way of getting a measure. If you have a top degree form, you need an orientation.
To combine both, consider odd top forms, that is, elements of $$\Omega^k(M;o(TM))=\Lambda^kT^*M\otimes o(TM)\;,$$ where $n=\dim M$ and  $o(TM)\cong\Lambda^nTM$ denote the bundle of local orientations. The integral over forms in $\Omega^n(M;o(TM))$ is well-defined because they "carry an orientation with them". These are the densities in the question, and they correspond to (signed) measures on $M$. A typical example is the Euler form of the Levi-Civita connection on $M$.
The Hodge star relates integration of functions with integration of forms. For example $$\int_M\|\alpha\|^2\,d\mathrm{vol}_g=\int_M\alpha\wedge *\alpha$$ on oriented Riemannian manifolds $(M,g)$. If you drop the orientation, you can still define $*\colon\Omega^k(M)\to\Omega^{n-k}(M,o(TM))$ and the formula above still works. You can also define $D=d+*^{-1}d*$ if you regard the second $d$ as acting on $\Omega^\bullet(M,o(TM))$. Because the local formula for both $d$s is the same, you won't even notice a difference. And with the formula above and Stokes theorem (for odd forms), you can check that $*^{-1}d*$ is the formal adjoint of $d$.
Edit To address the last question: the only reason I see why one would regard the Hodge-de Rham operator on oriented manifolds only is related to the decomposition of $H^{2k}(M)$ of an oriented $4k$-dimensional manifold into positive and negative forms. This is needed for example to define the signature, or to talk about antiselfdual metrics in dimension 4.<|endoftext|>
TITLE: Are rationally connected varieties robustly simply connected?
QUESTION [17 upvotes]: Let $X$ be a smooth projective rationally connected variety over $\mathbb C$. Let $C$ be a curve class in $X$ such that rational curves equivalent to $C$ connect every two points.
Let $f: Y \to X$ be a finite morphism from a normal variety $Y$ to $X$, generically of degree $d$. Let $D$ be the branch divisor of $f$. (This is the pushforward of the ramification divisor, which is well-defined because the ramification divisor depends only on the codimension 1 points, which are regular.)

Do we have the inequality
  $$ C \cdot D \geq 2 d - o(d) ?$$

Here the $o(1)$ goes to $0$ as $d$ goes to $\infty$ for any fixed $X, C$. 
Of course we know that rationally connected varieties are simply connected, so $D \neq 0$, and $C$ is ample, so $C \cdot D >0$. This question asks whether we can show a much stronger inequality, so that $X$ isn't even close to being simply connected.
I can show a positive answer in some special cases, like projective space: Pull back any cover to the space of pairs of a line and a point on that line. It remains connected because the space of pairs of a line through a point is simply connected. Stein factorize the map from the pullback to the Grassmanian of lines in $\mathbb P^n$. We obtain a ramified cover of the Grassmanian, each point of which corresponds to a connected cover of a line. But the ramified cover of the Grassmanian is ramified only at the lines contained in $D$, which is a codimension $2$ subset, so it is unramified by purity, so it is a trivial cover as the Grassmanian is simply connected. Thus a general line has a connected inverse image is $Y$, so by Riemann-Hurwitz its intersection with $D$ must be at least $2d-2$.
But this argument uses in a couple places special facts about the Grassmanian of lines that I don't know how to generalize to an arbitrary rationally connected variety.
Of course for projective spaces with the line class, this lower bound is optimal, as can be seen by the cover adjoining the $d$th root of the ratio of two coordinates.

REPLY [4 votes]: For my own edification I'm rewriting Jason's answer so as to excise all mention of loops (as he suggests), and trying to get the best possible bound from this technique.
Let $M$ be a variety with a $\mathbb P^1$-bundles $U \to M$ and two sections $s_1, s_2$ with a map $f: U \to X$.  Let $Y \to X$ be a finite covering and let $x$ be a general unramified point. We want to bound the degree of the finite part of the Stein factorization of that map $U \times_X Y \to M$ by a constant $n$.
Once we do that, we get that a general fiber of that map has at most $n$ connected components and hence has branch divisor of degree at least $2d-2n$.
It is sufficient to bound it restricted to the fiber of $f \circ s_1$ over a general point $x_0$. But then because $s_1$ has a constant map to $X$, the covering $U_{x_0} \times_X Y$ restricted to $M_{x_0} \times_X Y$ by $s_1: M_{x_0} \to U_{x_0}$ is a finite union of copies of $M_{x_0}$. So we are taking the Stein factorization of a map with many sections - in fact one section in each connected component. Hence the finite part will have a section in each connected component, so because it is finite each connected component will actually be a section (maybe after pasing to an open set).  But now the fiber $U_{x_0} \times_X Y$ restricted to $M_{x_0} \times_X Y$ by $s_2$ has one connected component meeting each connected component of $U_{x_0} \times_X Y$.  So we may take $n$ to be an upper bound for the connected components, which we can take to be the degree of the Stein factorization of the map $f \circ s_2 : M_{x_0} \to X$.
So if $M$ is the space of all rational curves of class $\beta$ with two marked points, we may take $n$ to be the number of connected components of the space of curves of class $\beta$ between two general points. In fact we can do a little better - it's the index of $\operatorname{Gal}(M_{0,2}(X, \beta))$ in the image of $\operatorname{Gal}(X)\times \operatorname{Gal}(X)$, where by the Galois group of a variety I mean the Galois group of its function field.<|endoftext|>
TITLE: primorial puzzlement
QUESTION [14 upvotes]: Let $x_n$ be the smallest positive integer which is not a quadratic residue modulo any of the first $n$ odd primes. The question is: is there any bound on how quickly $x_n$ grows as a function of $n?$ Now, since the probability of being a residue modulo any given prime is very close to $1/2,$ by an independence heuristic, $x_n$ should grow roughly like $2^n.$ The problem is that independence clearly does not hold here: if you ask the "opposite" question: how big is $y_n,$ the smallest integer which is a square modulo all of the first $n$ odd primes, the answer is $y_n=1.$ When in doubt, experiment, and computing $\log x_n$ for the $n=2, \dotsc, 20$ gives the graph: In case you wonder, the slope is around $0.8$ (so a bit bigger than $\log 2,$ but $20$ is pretty small, so $x_n \sim 2^n$ seems like a reasonable guess. Does anyone have any nontrivial bounds, even mod "standard conjectures"?
EDIT Noam Elkies points out that a few more terms have been computed (I have gotten up to 24 in the meantime, but 27 is even better :)) The slope of the log is down to $0.76,$ so the conjecture that $\log(x_n) \sim n \log 2$ seems quite reasonable.
Further Edit In fact, if you ponder Noam's comment, you will note that the heuristic will give something like $\log n 2^n$ for the conjectured growth rate...
More Values Using some of joro's ideas, and a little more, we can push the list of values to 40. 
{2, 2, 17, 17, 83, 167, 227, 398, 398, 5297, 64382, 69647, 116387,
214037, 214037, 430022, 5472953, 5472953, 8062073, 8062073, 8062073,
41941577, 86374763, 163520117, 163520117, 231912722, 231912722,
231912722, 545559467, 1728061733, 2832363203, 5638787822, 5638787822, 
6154772762, 6154772762, 7012246247, 7012246247, 7012246247, 6571091781638,    7218195919667}

REPLY [7 votes]: I'll try to unify all the previous answers (of GH from MO, Will Sawin and Hurkyl) and also indicate unconditional results on this problem.   It turns out 
 that one can get a surprisingly decent unconditional upper bound here, which is something I hadn't 
 appreciated fully previously.  As noted  in Hurkyl's answer above, a unifying problem is the following:  Given a non-trivial subgroup $H$ of $({\Bbb Z}/q{\Bbb Z})^*$ with index $h$, bound (i) the least prime $p$ lying in a given coset of $H$, or (ii) the least integer $n$ lying in a given coset of $H$.  Obviously the second problem is easier than the first, and as noted in Hurkyl's answer Theorem 1.4 of Lamzouri, Li and Soundararajan gives a good bound on GRH for the first problem: the least prime is essentially bounded by $((h-1) \log q)^2$  (see the precise form) and this result unifies both the least quadratic non-residue problem and also the least prime in an arithmetic progression problem.   
Unconditionally, for problem (i), by Linnik's theorem on the least prime in an arithmetic progression we always have a bound of $\ll q^{L}$, for an absolute constant $L$, and the current record here is that $L=5$ is permissible by Xylouris.  When $h$ is small, then a somewhat better bound can be given:  for $h=2$ (the case of the least prime that splits, or the least prime that is inert in a quadratic field) this is classical and well known, the case of split primes being harder and the bound here is $q^{\frac 14+\epsilon}$ (using Burgess for character sums and Siegel's ineffective bound for $L(1,\chi)$).   For larger, but still bounded $h$, the matter was treated by Elliott  in the context of the least prime $h$-th power residue, and he obtained the bound $q^{(h-1)/4+\epsilon}$.  Of course for large $h$ (i.e. $h\ge 22$) this becomes inferior to Linnik's theorem.  A recent paper of Pollack addresses a somewhat more general version of this question, and obtains the bound $\ll q^{(h-1)/4+\epsilon}$.  
Now for unconditional results towards the easier problem (ii).  Assume that $h \le q^{\epsilon}$ and express the 
condition of belonging to a coset of $H$ in terms of the characters that are trivial on $H$ (see the paper of Lamzouri, Li and Soundararajan for details).   Then use the Burgess bounds on the character sums that arise.  This argument gives that the least integer $n$ in any given coset is $\ll q^{\frac 14+\epsilon}$.  This is a (modest) improvement over the trivial bound of $q$.   For a general modulus $q$ the Burgess bound gives the best known result on this problem.  However, for moduli $q$ that are very smooth (and this will be important for the present problem) one can obtain a better result by a $q$-van der Corput argument due to Graham and Ringrose (following Heath-Brown).   From the Graham-Ringrose argument (see either the original paper, or Corollary 12.14 of Iwaniec and Kowalski, or estimate (4.1) and Lemma 4.2 from Granville and Soundararajan), one can show that if all the prime factors of $q$ are bounded by $q^{o(1)}$ then the least integer in any prescribed coset is $\ll q^{\epsilon}$ 
for any fixed $\epsilon >0$. Naturally there is a connection between how small $\epsilon$ can be and the smoothness of $q$, and one can find uniform bounds in the results referenced above. 
Returning to the problem at hand, if $p_1, \ldots, p_n$ are the first $n$ odd primes, and we 
 set $q=p_1\cdots p_n = n^{(1+o(1))n}$ and $H$ to be the subgroup of squares mod $q$ (which has index $2^n$) 
 then on GRH we can find a prime $p \ll 4^n (n\log n)^2$ lying in any given coset of $H$.  That is given any choice of signs $\epsilon_j = \pm 1$, there is a prime $p\ll 4^n (n\log n)^2$ with $(\frac{p}{p_j}) = \epsilon_j$ (in particular we can take all $\epsilon_j=-1$ as asked for in the problem).   This was noted in Hurkyl's answer.  Note now 
 that unconditionally by the Burgess estimate we may find a number $N \ll n^{n(1/4+\epsilon)}$ with $(\frac{N}{p_j}) =\epsilon_j$.  But note that we are dealing now with a modulus $q$ of size $n^n$ all of whose prime factors are 
 of size $\ll n\log n$; in other words, $q$ is very smooth indeed and Graham-Ringrose applies and gives an unconditional upper bound of $n^{\epsilon n}$.  In fact, by keeping track of uniformity, one gets an unconditional upper bound of $n^{c n/\log \log n}$ for some positive constant $c$. 
Now consider lower bounds in the problem.  Here we must make use of the fact that we are looking for the smallest number $N$ with $(\frac{N}{p_j})=-1$ for all $j\le n$ -- for a random choice of signs $\epsilon_j$ there may of course 
 be a very small number $N$ (just pick the number $N$ and see what the signs are!).  Note that $N$ or $4N$ would be a discriminant, and we are asking for a quadratic character $\pmod {N}$ or $\pmod {4N}$ taking value $-1$ on the first $n$ small primes.  This fits our framework by taking $q=4N$, and $H$ to be the subgroup of index $2$ on which this character  (say $\chi$) is trivial.  On GRH we know that there is an odd prime $\ell \le (1+o(1)) (\log N)^2$ with $\chi(\ell) =1$, and so we must have $(\log N)^2 \ge (1+o(1)) n\log n$ or 
 $$ 
 N\ge \exp((1+o(1)) \sqrt{n\log n}).
 $$ 
 This is the conditional bound mentioned in Will Sawin's answer, and what I indicated in my comment there.
 Unconditionally, the Burgess bound and Siegel's theorem argument (for problem (i)) gives that there is a prime $\ell \le N^{\frac 14+\epsilon}$ with $\chi(\ell)=1$, and  so $N\ge n^{4+o(1)}$.  
To summarize, on GRH we know that the answer lies between 
 $$ 
 \exp((1+o(1)) \sqrt{n\log n} ) \text{  and  } \ll 4^n (n\log n)^2 
 $$ 
 and unconditionally we know that it is between 
 $$ 
 n^{4+o(1)} \text{   and   } \ll n^{cn/\log \log n}.
 $$ 
 The probabilistic conjecture that the answer should be somewhere near $2^n$ seems reasonable.<|endoftext|>
TITLE: Windows into new mathematical worlds
QUESTION [7 upvotes]: Yitang Zhang's Annals of Mathematics primes-gap result
opened a new window, which 
Polymath's reduction from $70\times 10^6$ to $246$ attests.
Perhaps
Harald Helfgott's
celebrated proof of the odd Goldbach's conjecture
has not similarly open new avenues—or at least not yet.
Certainly the Green-Tao theorem has opened new windows.
Perhaps the
Guth-Katz breakthrough on the Erdős distance problem
has opened new windows.
My question is, 

Q. Which results in the recent past (~last decade+) have opened 
  significant windows into new mathematics?

I realize this is quite subjective,
but it requires a high-level view of fields of mathematics
to notice this while it is happening, in a way that others (like me) without that
expertise cannot discern.
It would be educational to learn of expert opinions, without diminishing
the significance of any particular result.
Rather I am hoping for a celebration of those results which seem not to be
the end of a line of investigation, but rather a new beginning.

REPLY [7 votes]: The work of Lurie, and others (for instance, Rezk, Joyal, and also older work of Toen, Vezzosi, Simpson, ...) on Higher Topos Theory and Higher Algebra, leading to e.g. a proof of the cobordism hypothesis (see also Freed's article) and the proof of the Weil conjecture on Tamagawa numbers for function fields over finite fields. 
I'm making this community wiki so others can add things.<|endoftext|>
TITLE: To what extent are modular parametrizations expected to generalize?
QUESTION [18 upvotes]: By the Modularity Theorem (a.k.a. the Shimura--Taniyama--Weil Conjecture), if $E$ is an elliptic curve over $\textbf{Q}$ with conductor $N$, then there exists a “modular parametrization” $\psi: X_0(N) \to E$, a non-constant surjective morphism defined over $\textbf{Q}$. Let me call this “geometric” modularity.
It is known that this is equivalent to the existence of a holomorphic modular form of weight 2 and level $\Gamma_0(N)$ such that $L(E, \chi, s) = L(f, \chi, s)$ for all Dirichlet characters $\chi$. Indeed, over $\textbf{Q}$, such a result follows from the weaker result that $L(E, s) = L(f,s)$. Let me call this “analytic” modularity.
According to the Langlands philosophy, one expects all motivic L-functions to be automorphic. In particular, for Galois representations coming from the etale cohomology of any variety over a number field, one expects to be able to attach an automorphic representation (or some set of them, depending on the context); so we expect “analytic” modularity in some general sense. It seems like most modularity/automorphy results aim to prove this kind of statement.
My question is the following:

Does one also expect a general form of “geometric” modularity to hold?

I have not seen any such parametrizations outside of the well-known case of modular parametrizations of elliptic curves over $\textbf{Q}$, and occasionally by Shimura curves (but am less familiar with the implications, compared to the modular curve case). Since many of the nice “coincidences” in this classical modularity setting seem to break down as one passes to more general settings, and since the connection between geometric and analytic modularity seems weaker in the function field setting (e.g. the relevant automorphic forms are not functions or sections of line bundles on the Drinfeld modular curve), does one expect a general "geometric" modularity statement to hold? For example, given a higher genus curve over $\textbf{Q}$, or an elliptic curve over a number field, or a surface (say, abelian) over $\textbf{Q}$, do you expect some form of modularity to correspond to the existence of a map from some space (like a Shimura variety) to the variety in question?
I am aware that one can sometimes, say, find the appropriate motives in the cohomology of Shimura varieties (e.g. if V is a variety such that $L(V,s) = L(\pi, s)$ for some automorphic representation $\pi$, and then you may have a Galois representation attached to $\pi$ lying in the cohomology of some Shimura variety), but let me call this a case of "analytic" modularity instead of a weak form of "geometric" modularity, for the sake of the question. In other words, by "geometric" modularity, I am specifically referring to the existence of something like a morphism from, say, a Shimura variety to a variety over a number field. When I asked this question (IRL), some suggested that such a statement may be better formulated in the language of correspondences (or purely in the language of motives), but then that such an answer may be essentially the case of the variety $V$ above. I would be interested in any such answers, or in any corrections to the formulation of the question.

REPLY [12 votes]: A natural generalization of the geometric modularity conjecture which is compatible with your formulation

Do you expect some form of modularity to correspond to the existence of a map from some space (like a Shimura variety) to the variety in question?

is to ask whether a variety always appears as a quotient of the Picard variety of (the smooth compactification of) a Shimura variety.
The answer to this question is negative, and in fact is already so for abelian varieties for the following reasons.
Let $X$ be a Shimura variety and let $X^*$ be a smooth compactification. The weights (as Hodge decomposition or as Galois representation) attached to an abelian variety are $(0,1)$ and $(1,0)$ so if an abelian variety appears as a quotient (up to isogeny) of the Picard variety of $X$, these weights have to appear in the weight decomposition of the Hodge structure attached to $X^*$. There, they can only appear inside the first graded piece of the $H^1$, which thus has to be non-trivial. However, this turns out to be a very restrictive condition. Unraveling its meaning in $L_2$-cohomology yields the following result which says that the non-CM abelian varieties which are geometrically modular in the sense of the question are the one you already know about.

Theorem Every non-CM abelian variety which occurs as a quotient of the Picard variety of any smooth compactification of a Shimura variety is isogenous to a base change of a factor of the Jacobian of a quaternionic Shimura curve.

This theorem is deemed folklore in Elliptic curves, Hilbert modular forms, and the Hodge conjecture (D.Blasius, 2004) but this could have been modesty on Don Blasius's part.
For general varieties, I have heard L.Clozel explain why the situation is (of course) still much worse but the afferent preprint is not available as far as I know. I think it is fair to say to sum up that one should not expect much geometric modularity beyond the already known cases.<|endoftext|>
TITLE: Uniform definition of $S(\mathbb{R})$ and $S(\mathbb{Q}_p)$
QUESTION [6 upvotes]: Let $\mathcal{P}=\{\infty, 2,3,5,7,11,\ldots\}$ be the set of primes of $\mathbb{Q}$ and let $\mathbb{Q}_p$ denote the corresponding completions, so in particular $\mathbb{Q}_{\infty}=\mathbb{R}$.
Is it possible to give a definition of the spaces of test functions $S(\mathbb{Q}_p)$ together with their standard topologies (finest locally convex for $p$ finite and usual Frechet in the real case) in a completely uniform way with respect to $p\in\mathcal{P}$?
It is not too hard to uniformly define $L^2(\mathbb{Q}_p, dx)$ as well as the Fourier transform as a unitary transformation on this space. The kind of definition I am looking for might look like:
$S(\mathbb{Q}_p)$ is the smallest subspace of $L^2$ which is invariant by Fourier transform and... insert mystery property here...

REPLY [3 votes]: François Bruhat in his paper of 1961 (see page 61) gave a definition (and proved some properties) of the space ${\mathcal S}(G)$ of the Schwartz test functions on an arbitrary abelian locally compact group $G$.<|endoftext|>
TITLE: Can a group be a union of finitely many subgroups of infinite index?
QUESTION [8 upvotes]: Is there a group $G$ and subgroups $H_1, \dots, H_n \leq G$ for some $n \in \mathbb{N}$, such that $[G : H_i] = \infty$ for each $1 \leq i \leq n$, and $$G = \bigcup_{i=1}^n H_i \ \ ?$$

REPLY [12 votes]: It may well be a lot easier than this, but it follows from the answer to the weaker question Can a group be a finite union of (left) cosets of infinite-index subgroups? that it's not possible.<|endoftext|>
TITLE: Modern comprehensive account of the Barnes G-Function
QUESTION [6 upvotes]: I previously put this forward on the math.stackexchange community with little luck: 
I am looking for a comprehensive account of the properties and applications of the Barnes G-Function. Everything from recurrence relations, proof of the infinite product representation, functional relations, e.t.c. Ideally with proofs?
I am primarily interested only on the detailed proof of it's infinite product representation and physical applications. Other proofs are a welcomed bonus for this beauty.

REPLY [5 votes]: Perhaps one of the most recent overviews of the Barnes G-function (with some new results) is Contributions to the Theory of the Barnes Function (2003). It is not a self-contained treatise, but it does contain many pointers to the literature.<|endoftext|>
TITLE: Many representations as a sum of three squares
QUESTION [16 upvotes]: Let $r_3(n) = \left|\{(a,b,c)\in {\mathbb Z}^3 :\, a^2+b^2+c^2=n \}\right|$. I am looking for the maximum asymptotic size of $r_3(n)$. That is, the maximum number of representations that a number can have as a sum of three squares.
Gauss proved that $r_3(n) = \frac{A\sqrt{n}}{\pi}\sum_{m-1}^\infty\left(\frac{-n}{m}\right) \frac{1}{m}$ (where $\left(\frac{a}{b}\right)$ is the Jacobi-Legendre symbol). Another similar-looking formula can be found, for example, in the bottom of the first page of this paper
A simple and non-constructive argument shows that there are infinitely many $n$'s for which $r_3(n) \ge c\sqrt{n}$ (for some constant $c$). Moreover, in an old paper of Erdős he mentions that $r_3(n) \ge c\sqrt{n}\log \log n$ can also be obtained. Unfortunately, Erdős does not provide any details and only states "By deep number theoretic results".
More specifically, my questions are (1) how can one obtain Erdős' bound? (2) What $n$'s achieve the above bounds? (3) Can one use the above formula for $r_3(n)$ to show this?
Many thanks,
Adam

REPLY [19 votes]: The formula for $r_3(n)$ essentially connects this with a class number of an imaginary quadratic field, or (apart from the $\sqrt{n}$ scaling) with the value of an $L$-function at $1$.  So your question may be reformulated as asking how large can $L(1,\chi_{d})$ be as $d$ runs over negative fundamental discriminants ($d$ is essentially $-n$).  The distribution of these values has been extensively investigated (see for example Granville and Soundararajan where you'll find more references.  To create large values of $L(1,\chi_d)$ you should find a $d$ with $\chi_d(p)=1$ for all small primes $p$ up to some point $y$.  One can do this with $d$ of size about $\exp(y)$.  Then for most such $d$ one will have
$$ 
L(1,\chi_d) \approx \prod_{p\le y} \Big(1-\frac{\chi_d(p)}{p} \Big)^{-1} \asymp \log \log |d|
$$ 
by Mertens.  Arguments of this type go back to Littlewood and  Chowla (see the linked paper for references).

REPLY [18 votes]: Let me restrict to the number of primitive representations
$$r_3^\ast(n) = \left|\{(a,b,c)\in {\mathbb Z}^3 :\, a^2+b^2+c^2=n\ \text{and}\ \gcd(a,b,c)=1 \}\right|.$$
Note that $r_3(n)$ can be easily expressed from this quantity as
$$r_3(n)=\sum_{d^2\mid n}r_3^\ast(n/d^2).$$
Clearly $r_3^*(n)=0$ when $n\equiv 0,4,7\pmod{8}$. For the remaining cases, it follows from the work of Gauss (1801) and Dirichlet (1839) on the class number that
$$r_3^\ast(n) = \frac{24}{\pi}\,\sqrt{n}\,L\left(1,\left(\frac{D}{\cdot}\right)\right),$$
where
$$D=\begin{cases}
-4n,&n\equiv 1,2,5,6\pmod{8},\\
-n,&n\equiv 3\pmod{8}.
\end{cases}$$
This tells us that the maximal (resp. minimal) order of $r_3^*(n)$ is determined by the maximal (resp. minimal) order of $L\left(1,\left(\frac{D}{\cdot}\right)\right)$, where $D$ depends on $n$ as above. Concerning the latter quantity, Littlewood (On the class-number of the corpus $P(\sqrt{-k})$, Proc. Lond. Math. Soc. (2) 27 (1928), 358-372) proved under GRH that 
$$ (\log\log|D|)^{-1}\ll L\left(1,\left(\frac{D}{\cdot}\right)\right)\ll \log\log|D|,$$
and he also proved in the same article that $L\left(1,\left(\frac{D}{\cdot}\right)\right)\gg\log\log|D|$ holds for infinitely many $D$'s. This last bound was proved unconditionally by Walfisz (On the class-number of binary quadratic forms, Trav. Inst. Math. Tbilissi 11 (1942), 57-71), and was further explicated by Granville-Soundararajan (The distribution of values of $L(1,\chi_d)$, Geom. Funct. Anal. 13 (2003), 992-1028). In particular, see Theorem 5b on page 998 in Granville-Soundararajan's paper, which is also available as an arXiv preprint.
In short, there are infinitely many $n$'s such that $r_3(n)\gg\sqrt{n}\log\log n$, and this is sharp under GRH.<|endoftext|>
TITLE: Littlewood-Richardson-Type Rule for Restriction from $S_{2n}$ to $S_{2(n-t)} \times (S_2 \wr S_t)$
QUESTION [8 upvotes]: It is well-known that the Littlewood-Richardson coefficient $c^{\nu}_{\lambda \mu}$ is the number of times the irreducible representation $V_\lambda \bigotimes V_\mu$ of the product of symmetric groups $S_{|\lambda|} × S_{|\mu|}$ appears in the restriction of the representation $V_{\nu}$ of $S_{|\nu|}$ to $S_{|\lambda|} × S_{|\mu|}$. 
It is also known that the irreducible representations of the hyperoctahedral group $H_t := (S_2 \wr S_t) \leq S_{2t}$ are indexed by pairs of partitions $(\alpha,\beta)$ such that $|\alpha| + |\beta| = t$.
Let $\nu \vdash 2n$, $\lambda \vdash 2(n-t)$, and let $c^{\nu}_{\lambda(\alpha,\beta)}$ be the number of times the irreducible representation $V_\lambda \bigotimes V_{(\alpha,\beta)}$ of the $S_{|\lambda|} × H_t$ appears in the restriction of the representation $V_{\nu}$ of $S_{2n}$ to $S_{|\lambda|} \times H_t$. My question is the following:
Is there a branching/Littlewood-Richardson-type rule for computing $c^{\nu}_{\lambda(\alpha,\beta)}$ in the literature?
By the Littlewood-Richardson rule, transitivity of induction, and Frobenius reciprocity, a branching rule for restricting from $S_{2(n-t)} \times S_{2t}$ to $S_{2(n-t)} \times H_t$ would also do the trick.  This is quite similar to an older post of Steven Sam, but unfortunately I do not have access to the reference that answers his question.  
I might add that I am really only interested in whether the trivial representation of 
$S_{|\lambda|} \times H_t$ appears when an irreducible representation $V_{\nu}$ of $S_{2n}$ is restricted to $S_{|\lambda|} \times H_t$. I'm not a rep. theorist/algebraist, so if there is some easier way of determining this, I would be happy to know.

REPLY [6 votes]: For the trivial representation the problem is much simplified. Let $1_G$ denote the trivial character of a group $G$. It is well known that $1_{S_2 \wr S_t}\!\!\uparrow^{S_{2t}} = \sum_{\mu \vdash t} \chi^{2\mu}$ where $2\mu$ is the partition obtained from $\mu$ by doubling the length of each part. 
By two applications of Frobenius reciprocity, we have
$$ \begin{align*} \bigl\langle \chi^\nu \!\!\downarrow_{S_{2(n-t)} \times S_2 \wr S_t}, 1_{S_{2(n-t)}} \times 1_{S_2 \wr S_t} \bigr\rangle &= \bigl\langle \chi^\nu \!\!\downarrow_{S_{2(n-t)} \times S_{2t}}, 1_{S_{2(n-t)}} \times 1_{S_2 \wr S_t}\!\!\uparrow^{S_{2t}} \bigr\rangle \\
&= \bigl\langle \chi^\nu, \sum_{\mu \vdash t} \bigl( 1_{S_{2(n-t)}} \times \chi^{2\mu} \bigr)\!\!\uparrow^{S_{2n}} \bigr\rangle \end{align*}
$$
Hence by Young's rule (or the more general Littlewood—Richardson rule), the multiplicity of the trivial character in $\chi^\nu \downarrow_{S_{2(n-t)} \times S_2 \wr S_t}$ is equal to the number of partitions $\mu$ of $t$ such that the Young diagram of $\nu$ can be obtained from the Young diagram of $2\mu$ by adding $2(n-t)$ boxes, no two in the same column.<|endoftext|>
TITLE: Curious inequality
QUESTION [24 upvotes]: Set 
$$
g(x)=\sum_{k=0}^{\infty}\frac{1}{x^{2k+1}+1}  \quad \text{for}  \quad  x>1. 
$$
Is it true that 
$$
\frac{x^{2}+1}{x(x^{2}-1)}+\frac{g'(x)}{g(x)}>0 \quad \text{for}\quad x>1? 
$$
The answer seems to be positive. I spent several hours in proving this statement but I did not come up with anything reasonable. Maybe somebody else has (or will have) any bright idea?  
Motivation? Nothing important, I was just playing around this question:
A problem of potential theory arising in biology

REPLY [9 votes]: Writing $k=\sum_{j=1}^k 1$ and then summing by parts, one has 
$$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2}
=\sum_{j\ge 1}(-1)^{j-1}s_j
=\sum_{k\ge0}[s_{2k+1}-s_{2k+2}],
$$
where 
$$s_j:=\sum_{k=j}^\infty(-1)^{k-j} a(k)=\sum_{m\ge0}(-1)^m a(m+j)
=\sum_{k\ge0}[a(2k+j)-a(2k+1+j)] 
$$
and
$a(q):=\dfrac1{(x^q+1)^2}$. 
So, 
$$s_j-s_{j+1}=\sum_{k\ge0}[a(2k+j)-2a(2k+1+j)+a(2k+2+j)]>0,
$$
since $a''>0$ and hence $a$ is strictly convex. 
So, $\sum_{k\geq 1} \dfrac{(-1)^{k-1}k}{(x^{k}+1)^2}>0$, and the result follows by Max Alekseyev's answer.<|endoftext|>
TITLE: Characterization of right properness using slice categories
QUESTION [5 upvotes]: I would like to know how to cite this theorem (which has a quite surprising consequence):

A model category $\mathcal{M}$ is right proper if and only if for any
  weak equivalence $f:A\to B$, the Quillen adjunction
  $\Sigma_f:\mathcal{M}/ A \leftrightarrows \mathcal{M} / B:f^*$ is a Quillen equivalence.

I had never heard of it (I never read that in any textbook I believe) before I browse the nLab site which refers to this blog comment from C. Rezk.

REPLY [4 votes]: This is Proposition 2.7 in Rezk's Every Homotopy Theory of Simplicial Algebras Admits a Proper Model<|endoftext|>
TITLE: Classifiying sphere eversions
QUESTION [34 upvotes]: For a year I have been giving lectures on a (probalby) new way to present an explicit sphere eversion. These lectures include a review of many other explicit eversions that have been described, as text, drawings or even computer generated movies, since Smale proved his theorem.
During these lectures, I have been frequently asked the following question: are these eversions all equivalent?
I suppose the meaning of this question is: can you deform one eversion into another?
A more general question is:
What is the $\pi_1$ of the space $I$ of immersions of $S^2$ in $\mathbb{R}^3$?
Recall that Smale proved that $I$ is connected so that there is a path from the canonical embedding to the antipodal embedding, which is the definition of a sphere eversion. The set of all eversions up to homotopy is in (non canonical) bijection with the $\pi_1$ above. If the latter were trivial, then all eversions would be equivalent, and conversely. But as far as I have understood, it is not the case.

REPLY [40 votes]: Answer Summary
The fundamental group of the space of immersions of $S^2$ into $\mathbb{R}^3$ is
$$ \pi_1 Im(S^2, \mathbb{R}^3) \cong \mathbb{Z}/2 \times \mathbb{Z}$$
This means that there are infinitely many different sphere eversions (where "different" mean not homotopic in the path space of immersions). 
Given two sphere eversions their "difference" (formed by running the first eversion, and then the reverse of the second eversion) is a loop in the space of immersions of $S^2$ into $\mathbb{R}^3$ based at the standard embedding. As such it determines an element in $\pi_1 Im(S^2, \mathbb{R}^3) \cong \mathbb{Z}/2 \times \mathbb{Z}$. Projecting to these different factors gives a pair of invariants (one $\mathbb{Z}/2$ valued, one integer valued), and below I will describe how to compute these two invariants. These are complete invariants: if both invariants are zero for the difference of two eversions, then these eversion can be deformed into eachother in the space of eversions. In this way we can completely classify eversions using these invaraints. 
The Space of Immersions, in general
Smale's theorem fits into a larger family of results known as $h$-principle results. I think it is fair to say that Smale's result was one of the major precursors of this field, and these techniques allow you to completely determine the homotopy type of the space of immersions, not just of spheres into Euclidean space but much more generally.  
In a general h-principle problem, at least formulated in modern language, you compare two sheaves of spaces on the site of $d$-manifolds via a map of sheaves:
$$ \mathcal{F} \to \mathcal{F}^h $$
What this means is that $\mathcal{F}$ is an assignment that associates to every $n$-manifold $U$ a topological space $\mathcal{F}(U)$ and to every open embedding $V \subseteq U$ a continuous restriction map:
$$ \mathcal{F}(U) \to \mathcal{F}(V). $$
A good case to keep in mind is $\mathcal{F}(U) = Im(U,Y)$, the space of immersions into a fixed manifold $Y$. For non-compact manifolds $U$ we have to be careful about which topology we are using on the space of immersions in order for the restriction map to be continuous. This is important for the general theory and the proof of the h-principle theorem, but since we will be applying this to a compact manifold $U = S^2$, I am going to skip over that subtlety. 
In a general h-principle set-up you have two such sheaves: $\mathcal{F}$ is a sheaf that you are interested in, and $\mathcal{F}^h$ is some approximation to $\mathcal{F}$ that you can compute.  Usually this easier one is sections of some fiber bundle, so it is also a "homotopy sheaf", and can be computed using methods from homotopy theory. Under some assumptions on $\mathcal{F}$ there is actually a universal such $\mathcal{F}^h$, but I don't really want to get too far into the general theory here. 
We say that $\mathcal{F}$ satisfies the $h$-principle if this comparison map is a weak homotopy equivalence. Maybe this is only true for a certain class of manifolds (eg. non-compact, etc), and then we say $\mathcal{F}$ satisfies the h-principle for that class of manifolds.  
In the case at hand we have $\mathcal{F} = Im(-, Y)$ the sheaf of immersions into an ambient manifold $Y$. Thus 
$$ \mathcal{F}(M) = Im(M, Y) $$
is the space of immersions of $M$ into $Y$. 
The sheaf $\mathcal{F}^h$ is the space of pairs $(f, \phi)$ where
$$ f: M \to Y$$
is a map of topological spaces, and
$$ \phi: TM \hookrightarrow f^*TY$$
is a bundle map which is a linear injection on fibers. 
These are maps which "formally look like immersions". If you have an actual immersion, $f: M \to Y$ then you get such a pair via $(f, df)$, where $df: TM \to f^*TY$ is the differential. By the definition of immersion $df$ is a linear injection on fibers.  This defines a comparison map between the space of immersions and the space of "formal immersions".
This latter space of formal immersions is much easier to understand since for a pair $(f,\phi)$ we do not require that the map $\phi$ is given as the differential of the map $f$. For example the space of $\phi$ compatible with a given $f$ only depends on the homotopy class of $f$ and not on the particular map $f$ itself. This space is much more amenable to standard homotopy theoretic techniques. 
The theorem here about immersions is that $\mathcal{F}$ satisfies the h-principle on $d$-manifolds $M$ provided that the dimension of $Y$ is strictly larger than $d$. (It is also satisfied for non-compact $M$ even when $\dim Y = \dim M$). In otherwords the comparison map from the space of immersions to the space of "formal" immersions is a weak homotopy equivalence. This later space is a much easier space to consider as it can be attacked by purely homotopy theoretical means. 
If you want to understand the proof of this theorem and more general h-principle type proofs, I highly recommend Micheal Wiess' preprint Immersion Theory for Homotopy Theorists.
The Space of Immersions, $S^2$ into $\mathbb{R}^3$
Now let's specialize to the specific case asked about in this question: immersions of $S^2$ into $\mathbb{R}^3$. The space of maps from $S^2$ to $\mathbb{R}^3$ is contractible, and since all maps of $S^2$ to $\mathbb{R}^3$ are null-homotopic, $f^*T\mathbb{R}^3$ is always a trivial bundle. Thus we conclude:
Theorem [Smale] The space of immersions of $S^2$ into $\mathbb{R}^3$ is homotopy equivalent to the space of bundle injections of $TS^2$ into a trivial rank three bundle, via the map which takes an immersion $f$ to its differential $df$. 
We can actually do better at identifying this space. There are several essentially equivalent ways we can proceed here, and I will describe one. Depending on your background it is more or less possible to translate between the different approaches. One thing to remember is that the following spaces are all just names for the same space thought of in different ways:
$$ O(3) / O(1) \cong V_2 \mathbb{R}^3 \cong SO(3) $$ 
In any case, the space of bundle injections of $TS^2$ into a trivial rank three bundle can, up to homotopy, be identified with the coset space of the space of vector bundle isomorphisms:
$$ Isom( TS^2 \oplus \varepsilon, \varepsilon^3)/O(1) $$
where $\varepsilon$ is a trivial(ized) line bundle on $S^2$. Here the $O(1)$ action is on the source copy of $\varepsilon$.  This means that 
$$ Isom( TS^2 \oplus \varepsilon, \varepsilon^3) $$
is a double cover of the space that we want to know about. As we will see in a moment, this is a disconnected space and in fact the space we want is really just the connected component of the above space. 
This latter space can be thought of as a space of framings of $TS^2 \oplus \varepsilon$ and is (non-canoncially) isomorphic to:
$$ Maps(S^2, O(3)) $$
where this is the free mapping space. This identification comes by picking our favorite framing of $TS^2 \oplus \varepsilon$ and measuring the difference. The space of framings is a torsor for the topological group $ Maps(S^2, O(3))$.
Next, we have a short exact sequence of topological groups:
$$ 1 \to Maps_*(S^2, O(3)) \to Maps(S^2, O(3)) \to O(3) \to 1 $$
coming from evaluation at our favorite basepoint of $S^2$. 
The evaluation map is split by the topological group homomorphism sending $g \in O(3)$ to the constant map with value $g$. This means we have an isomorphism as spaces:
$$ Maps(S^2, O(3)) \cong Maps_*(S^2, O(3)) \times O(3)$$
(this is only as spaces, as groups it is a semidirect product). For simplicity below I will write $\Omega^2X$ in place of $Maps_*(S^2, X)$.
To get back to the space of immersions, we need to realize this as the double cover of the space we want, which is a quotient of the above space by $O(1)$. The key is with our choice of standard framing. with respect to this the $O(1)$-action acts be a constant element in $O(3)$. This means the $O(1)$ factor really comes from the $O(3)$ factor in the above topological group. More importantly the action of $O(1)$ is to switch the two components of the above space. 
We conclude:
Proposition There is a non-canoncial homtopy equivalence:
$$ Im(S^2, \mathbb{R}^3) \simeq \Omega^2(O(3)) \times O(3)/O(1) \cong \Omega^2(SO(3)) \times SO(3).$$
What is more, we have a prescription for how to pass from the left-hand-side to the right-hand-side. 

First send our immersion to a "formal immersion", a certain bundle injection.    
Then pick an orientation of $S^2$. This allows us to frame the normal bundle of our immersion so that $TS^2 \oplus \varepsilon \cong \varepsilon^3$ is orientation preserving. 
Pick a framing of $TS^2 \oplus \varepsilon$ compatible with the orientation (this is the non-canonical part). 
This then identifies our space with $Maps(S^2, SO(3))$, which splits into the above two factors. 

Corallary 
$$ \pi_1  Im(S^2, \mathbb{R}^3) \cong \pi_1 \Omega^2 SO(3) \times \pi_1 SO(3) \cong \pi_3 SO(3) \times \pi_1 SO(3) \cong \mathbb{Z} \times  \mathbb{Z}/ 2 .$$
Identifying the Invariants
The mod 2 invariant
The $\mathbb{Z}/ 2 $ invariant comes from the fundamental group of the factor $SO(3)$ in the space of immersions, and the map to $SO(3)\cong V_2 \mathbb{R}^3$ was given by evaluating at a point. This means that this invariant only depends on the restriction of the loop of immersions to a neighborhood of this fixed favorite point in the sphere. 
Here is one way to compute it:  for $t \in [0,1]$, let $f_t: S^2 \to \mathbb{R}^3$ be an immersion, which starts and ends at our preferred basepoint immersion.  
Pick our favorite point $p \in S^2$ with local coordinates $(x,y)$ and look at the map $f_t \mapsto (\partial_x f_t, \partial_y f_t) \in V_2 \mathbb{R}^3$. As written this as a map to 2-frames in $\mathbb{R}^3$, but to be consistent with the above I should really be using orthonormal 2-frames. To get an orthonormal 2-frame I simply apply the Gram-Schmidt process. This is more relevant in the construction below. 
In any case this is the projection to the $V_2 \mathbb{R}^3 \simeq SO(3)$ factor. Our loop of immersions gives us an element in $\pi_1 SO(3) \cong \mathbb{Z}/2$. 
Example This description makes it clear that this invariant is non-trivial on the path of immersions (even embeddings) which rotates the sphere 360 degrees. (This can also be see by tracing the $SO(3)$ action on $\mathbb{R}^3$ through our identification of $Im(S^2, \mathbb{R}^3)$.)
Another consequence of this description: suppose that we have a loop of immersions such that this mod 2 invariant vanishes? What does this mean? Well it means that the induced loop in $SO(3)$ can be deformed to the constant loop. But more is true. We can actually deform the loop of immersions to an equivalent loop which is constant on a disk neighborhood $D$ of our chosen basepoint $p$.
The space of immersions which are fixed to be standard on $D$ is sometimes considered in the literature and denoted $Im_D(S^2, \mathbb{R}^3)$.  
The integral invariant
For example the integral invariant of loops of immersion, but restricted to loops in the space $Im_D(S^2, \mathbb{R}^3)$ was considered by Tahl Nowik. In that paper the invariant is constructed in a similar fashion to our mod 2 invariant, by considering what happens on the (closure of the) compliment of $D$ in $S^2$, which we will call $U$. 
We can fix coordinates $(x,y)$ on $U$ (not to be confused with our earlier $(x,y)$) and then the loop of immersions (fixed on $D$) induced a map
 $$U \times I \to SO(3) = V_2 \mathbb{R}^3 $$
$$(u, t) \mapsto (\partial_x f_t(u), \partial_y f_t(u) ). $$
Again I mean to apply the Gram-Schmidt process so that I get an orthonormal 2-frame here. I am suppressing this from the notation. This map is subject to certain boundary constraints, and so factors through a quotient of $U \times I$ homeomorphic to $S^3$. The pointed map
$S^3 \to SO(3)$ then represents an element in $\pi_3SO(3) = \mathbb{Z}$, which is the invariant considered by Nowik. This also agrees with the integral invariant we want, but there is an easier way to get at this invariant using degrees of maps. 
Let's suppose that we are going to use the invariant as described by Nowik, but that we would like to actually compute the integer itself? How can we do this from just the map $S^3 \to SO(3)$? How do we tell which homotopy class we have?
Since this is a map between equal dimensional oriented manifolds, one thing we can do is pass to homology and compute the degree of the map. The degree is actually quite readily computable. We simply make the map generic near a point in $q \in SO(3)$ and the we count with signs the number of points in $S^3$ which are inverse images of $q$. 
Computing the degree is relatively easy, but how does that translate to the homotopy groups? It gives us a homomorphism
$$ \mathbb{Z} = \pi_3SO(3) = [S^3, SO(3)]_* \stackrel{deg}{\to} \mathbb{Z}$$ 
It is not the zero homomorphism, and so it is injective. Which means the degree detects our desired integral invariant.
The generator of $\pi_3SO(3)$ is represented by the double cover map $S^3 = Spin(3) \to SO(3)$, which has degree two. Thus we can be more precise. Our desired integral invariant is precisely half of the induced degree map. 
The thing I like about the degree map is that it is locally computable. It depends on the local behavior near the inverse images of $q \in SO(3)$. This means that we don't actually have to pass through Nowik's particular construction. 
Instead, here is something else we can do. We follow the four-step instructions above, and so pick a reference framing of $TS^2 \oplus \varepsilon$. An immersion into $\mathbb{R}^3$ gives a different framing and by measuring the difference we get a map $S^2 \to SO(3)$. For a loop of immersions we get a map:
$$S^2 \times S^1 \to SO(3) $$
And we can just compute the degree (divided-by-two) of this map of 3-manifolds. This agrees with Nowik's invariant for immersions constant on the disk $D$. 
example Apparently  Nowik's invariant was considered earlier in a paper by Max-Banchoff. I can't seem to track down a copy of their paper, so I don't know for sure, but according to the summary in Nowik's paper in the course of their work they construct a specific loop of immersions (fixed on $D$) and prove it is a generator of $\pi_1 Im_D(S^2, \mathbb{R}^3) \cong \mathbb{Z}$.  
[Note: This answer was edited from the original to address some of the comments below, expand on some of the details of the arguments, and generally improve the exposition]<|endoftext|>
TITLE: Linear occurrences of finite simple groups
QUESTION [13 upvotes]: Let $S$ be a finite simple group. All representations below are over the complex numbers. 
Let 

$d_0(S)$ be the smallest dimension of a faithful representation of $S$,
$d_1(S)$ be the smallest dimension of a faithful representation of some central extension of $S$, and
$d_2(S)$ be the smallest dimension of a faithful representation of some finite group admitting $S$ as quotient (or, equivalently, of some finite group admitting $S$ as Jordan-Hölder factor).

In a sense, $d_2(S)$ can be thought of the smallest dimension in which $S$ occurs linearly. On the other hand, $d_0$ and $d_1$ are well-documented, but, as far as I know, not $d_2$.
Trivially $d_0\ge d_1\ge d_2$, and the left-hand inequality can be strict: $d_0(\mathrm{Alt}_5)=3>2=d_1(\mathrm{Alt}_5)$. 

Does there exist $S$ with $d_1(S)>d_2(S)$?

I'd be interested as well by any information about $d_2$, including values for small non-abelian simple groups, or some particular families.
Note: in a naive attempt to show $d_1=d_2$, we could wonder if any finite group having $S$ as quotient admits a subgroup isomorphic to some central extension of $S$; a counterexample is pointed out by Derek Holt here.

REPLY [11 votes]: Searching and starting from Noam's answer, I found some extra information:
the phenomenon pointed out by Noam on automorphism groups of extraspecial 2-groups seems to appear in: R.L. Griess, Jr. Automorphisms of extra special groups and nonvanishing degree 2 cohomology. Pacific J. Math 48 (73) 403-422. (Link, MR link)
The discussion about $d_1$ and $d_2$ seems to appear in:
W. Feit and J. Tits, Projective representations of minimum degree of group extensions, Canad. J. Math. 30 (1978) 1092-1102, and continues in P. Kleidman and M. Liebeck. On a theorem of Feit and Tits, Proc. AMS 107(2), 1989, 315-322. (Link, MR link)
Feit-Tits establish that if $d_2(S)<d_1(S)$, then $S$ is of Lie type over a finite field of characteristic 2, and in this case $d_2$ is a power of 2. Kleidman and Liebeck establish the precise list of those such groups of Lie type that indeed satisfy $d_2<d_1$, including the computation of $d_2$. 
In particular, it follows from this classification that the smallest dimension in which some finite simple groups ``occur" but not through a projective representation is 16, namely for the groups $\Omega_8^-(2)$, $\mathrm{Sp}_8(2)$, $\mathrm{Sp}_4(4)$.  
Besides, if we consider representations over an algebraically closed field of characteristic $p\ge 5$, defining $d_i^{(p)}$ in the same fashion: the picture is exactly the same. In char. 3 it's essentially the same up to a minor difference in the classification among those $S$ of Lie type over a field of characteristic 2. On the other hand in characteristic 2 it's dramatically different: $d_1^{(2)}(S)=d_2^{(2)}(S)$ for all $S$.<|endoftext|>
TITLE: "Separated" version of Sauer's lemma on VC classes
QUESTION [7 upvotes]: Sauer's lemma, a well-known result in computational complexity theory, learning theory, and combinatorics, states the following:
Let $\Phi$ be a collection of subsets of a set $U$, and assume that for all $d$-element subsets $x_1, \dots, x_d$ of $U$ we have
$$
\Big| S \cap \{x_1, \dots, x_d\}:~~ S \in \Phi\} \Big| < 2^d.
$$
(That means, $\Phi$ is of Vapnik-Chervonenkis dimension at most $d$.)
Then for all $x_1, \dots, x_n \in U$ we have
$$
\Big| S \cap \{x_1, \dots, x_n\}:~~S \in \Phi \Big| \leq \sum_{i=0}^d \binom{n}{i} \leq \left( \frac{en}{d} \right)^d.
$$
Question: Let $m$ be a fixed positive integer. Assume that $\Phi$ is of VC dimension at most $d$. Let $x_1, \dots, x_n \in U$. Furthermore, let $\mathcal{A}$ denote a family of sets which has the property that

Every set $A \in\mathcal{A}$ is of the form $S \cap \{x_1, \dots, x_n\}$ for some $S \in \Phi$.
For two sets $A_1,A_2$ we have $\big| A_1 \triangle A_2 \big| \geq m.$

What is a good upper bound for the maximal cardinality of $\mathcal{A}$?
In words, what I am looking for is a version of Sauer's lemma under the additional assumption that not all possible sets obtained by intersection are counted, but only those which are "separated" in the sense of the symmetric difference being not too small. (Sauer's lemma is the special case $m=1$.) The application which I have in mind requires $m \approx \varepsilon n$ for some small $\varepsilon$, but I think it is in general an interesting question.

REPLY [2 votes]: The following paper by Haussler: http://www.sciencedirect.com/science/article/pii/0097316595900527.
Gives an appropriate upper bound for the case $\mathcal{A} \subset \Phi$.
Which results in $|\mathcal{A}| = O((\frac{n}{m+d})^d)$
If i'm not mistaken, you may add to $\Phi$ the set $\{S| \exists A \in \Phi, S \subset A, |S| = d\}$ and it's VC dimensional will not change. If so this gives a bound to your case as well.
For a more modern approach to this packing lemma you can read 5.3 in Matousek, Geometric Discrepancy (http://www.springer.com/us/book/9783540655282).<|endoftext|>
TITLE: Abelianization of characteristic quotients of $F_2$
QUESTION [5 upvotes]: Let $F_2$ be the free group on two generators.
Let $U\le F_2$ be a characteristic subgroup of finite index, and let $f : F_2\rightarrow\mathbb{Z}^2$ be the abelianization map.
It's easy to check that $f(U)$ is characteristic in $\mathbb{Z}^2$, so it must be of the form $N\mathbb{Z}\times N\mathbb{Z}$ for some $N\ge 1$, and hence the induced quotient $\mathbb{Z}^2/f(U)$ must be of the form $(\mathbb{Z}/N\mathbb{Z})^2$.
I apologize if this is a basic question, but my general question is - Is there a way of determining the $N$ from the characteristic subgroup $U$?
Of course this is too vague - I suppose an underlying question was - what are the characteristic subgroups of $F_2$? (There seems to be very little literature on this)
A more precise question might be:
Let $G$ be a finite group, and let $H\lhd F_2$ with $F_2/H \cong G$, and let $\{H_i\}$ be the $Aut(F_2)$-orbit $H$, then $U_H := \bigcap_i H_i$ is characteristic in $F_2$. Then we know that $f(U) = N\mathbb{Z}\times N\mathbb{Z}$. Is there a way of determining $N$ from $H$?
If we let $U_G$ be the intersection of all $G$-defining subgroups of $F_2$, then is there a way of determinine the $N$ associated to $f(U_G)$ from $G$?

REPLY [2 votes]: Note that $N$ is clearly the exponent of $F_2/[F_2,F_2]U$.
For any normal subgroup $U$ of $F_2$, the exponent of $F_2/[F_2,F_2]U$ is the same as the exponent of $K/[K,K]$, where $K=F_2/U$. Thus, $N$ is the exponent of $K/[K,K]$.
Now, $K$ is a subgroup of the Cartesian product $G\times G\times\dots$ such that the projection of $K$ to each factor is surjective. This implies that $K/[K,K]$ is a subgroup of
$G/[G,G]\times G/[G,G]\times\dots$ and the projection of $K/[K,K]$ to each factor is surjective.  
Thus, the exponent of $K/[K,K]$ (that is $N$) is equal to the exponent of $G/[G,G]$. 
So, Derek is right.<|endoftext|>
TITLE: Enriched Cauchy completions and underlying categories
QUESTION [5 upvotes]: The ordinary Cauchy completion $\overline{C}$ of a small category $C$ satisfies a number of conditions: Every idempotent in $\overline{C}$ splits, there's an equivalence of categories $[C^{op}, Set] \simeq [\overline{C}^{op}, Set]$, etc...
There's also a notion of Cauchy completion for enriched categories, my questions are about it:
1 - Let $X$ be a $V$-enriched category (where $V$ is a closed symmetric monoidal category with all limits and colimits), what properties does its enriched Cauchy completion $\overline{X}$ satisfy? Like is there an equivalence $[X^{op}, V] \simeq [\overline{X}^{op}, V]$, etc?
2 - What can be said about the underlying categories of $X$ and $\overline{X}$ ($X_0$ and $\overline{X}_0$)?? Is $\overline{X}_0$ the ordinary Cauchy completion of $X$? Do we have $[X_0^{op}, Set] \simeq [\overline{X}_0^{op}, Set]$, etc?
I just wanted to ask before I go about trying to answer this myself.

REPLY [3 votes]: It works the other way around --- you should have tried to answer your question by yourself before you posted it here :-)

Yes, there is an equivalence $[X^{op}, V] \simeq [\overline{X}^{op}, V]$. You may find more details in "Basic Concepts of Enriched Category Theory" by M. Kelly (Chapter 5.5).
No. The name "Cauchy completion" has been chosen to suggest that it is a  generalization of the usual concept of Cauchy completion for metric spaces. Indeed, if you take a generalized Lawvere metric space $X$, then the usual completion of $X$ under Cauchy sequences coincide with the Cauchy completion of $X$, when $X$ is thought of as a category enriched over poset $[0, \infty]$ with monoidal structure $\langle 0, {+}\rangle$. On the other hand, the underlying category of $X$ is always Cauchy complete in the sense of ordinary categories.<|endoftext|>
TITLE: Derived algebraic geometry: how to reach research level math?
QUESTION [61 upvotes]: I know the question "how to study math" has been asked dozens of times before in many  variations, but (I hope) this one is different.
My goal is to study derived algebraic geometry, where derived schemes are built out of simplicial commutative rings rather than ordinary commutative rings as in algebraic geometry (there's also a variant using commutative ring spectra, which I don't know anything about). Anyways, since the category of simplicial rings form a model category, we can apply homotopy theoretic methods to study derived schemes. 
I thought the first thing I should do is study simplicial homotopy theory, in order to learn about model categories and simplicial objects. So I started reading Simplicial Homotopy Theory by Goerss and Jardine. How should I study this book? There are very few exercises, unlike standard graduate textbooks like Hartshorne, and a lot of the proofs are simplex/diagram chasing, so I decided to skip a lot of the proofs and read the book casually. 
A big disadvantage to this method is that I don't understand anything at a deep level and I'm only familiar with a few buzzwords. But I feel overwhelmed by the amount of prerequisite material I need to understand to learn DAG, because most of it is written in the language of $\infty$-categories. So what should I do? How can I get to "research level mathematics"?
EDIT: I'm a senior math major and I've taken the graduate algebraic geometry and algebraic topology sequences. I've also studied some deformation theory.

REPLY [7 votes]: I'm going to take a dissenting view, here. I think the best way to assimilate concepts in derived algebraic geometry (for finite fields, $\mathbb{R}$ or $\mathbb{C}$), is to understand where and why they are used. Then, work backwards when the need arises. Personally, I found it formidable to read through any section of Toen-Vezzosi's homotopical algebraic geometry series straight through. I'd first recommend reading and understanding the content of Vezzosi's AMS notice, here: https://www.ams.org/notices/201107/rtx110700955p.pdf. Once you begin digesting the need for replacing the source category for Grothendieck's functor of points approach to algebraic geometry with derived commutative algebras, browse through the literature and find instances where this becomes necessary. From my perspective, the most striking application is here: https://arxiv.org/pdf/1102.1150.pdf, where one sees (sloppily speaking here), that even replacing the source category with truncated derived objects goes a very long way in recovering classical results. Feel free to let me know if you'd like me to explicate further.<|endoftext|>
TITLE: Can a class be represented by both a $(p,q)$-form and a $(p',q')$-form?
QUESTION [11 upvotes]: Suppose $X$ is a complex manifold. 
If $X$ is Kähler, the cohomology groups decompose into subgroups represented by $(p,q)$-forms. 
If $X$ is not Kähler, I think the decomposition may not hold? 
Is there an example where we have a nonzero class be represented by both a $(p,q)$-form and a $(p',q')$-form with $(p, q) \neq (p',q')$?

REPLY [6 votes]: This is meant as a supplement to Michael's answer above. Here is an example of a compact complex threefold $X$ and two forms $\gamma_1 \in \mathcal{E}^{p,q}$ and $\gamma_2 \in \mathcal{E}^{p',q'}$ such that $(p,q) \neq (p',q')$ and $[\gamma_1] = [\gamma_2] \in H_{dR}^2(X)$, but also $[\gamma_1] \neq 0$. This will make use of results found in Cordero, Fernández, Gray, The Frölicher spectral sequence for compact nilmanifolds.
Applying Theorem 6 in this paper gives us that there exists a simply connected 6-dimensional real Lie group with a left-invariant complex structure whose left-invariant $(1,0)$-forms are spanned by $\omega_1, \omega_2, \omega_3$ and whose left-invariant $(0,1)$-forms are spanned by $\bar \omega_1, \bar \omega_2, \bar \omega_3$, with structure equations given by \begin{align*} d\omega_1 &= 0, \ d\bar \omega_1 = 0, \\ d\omega_2 &= 0, \ d\bar \omega_2 = 0, \\ d\omega_3 &= \omega_1\omega_2 - \bar \omega_1 \omega_2, \ d\bar \omega_3 = \bar \omega_1 \bar \omega_2 - \omega_1 \bar \omega_2. \end{align*} Since the (dual) structure constants are rational, a theorem of Malcev (see Corollary 8 in the paper) tells us that there exists a discrete subgroup $\Gamma$ in $G$ such that $\Gamma \backslash G$ is a closed 6-manifold. The forms $\omega_i$ and $\bar \omega_i$ descend to $\Gamma \backslash G$ along with the complex structure on $G$. A theorem of Nomizu (referenced in this context in Hasegawa, Minimal models of nilmanifolds  at the end of Section 1) tells us that the de Rham cohomology of $\Gamma \backslash G$ is isomorphic to to the cohomology of the differential graded algebra generated in degree one by $\{\omega_1, \omega_2, \omega_3, \bar \omega_1, \bar \omega_2, \bar \omega_3\}$ with the differential $d$ prescribed above. 
Now, take the $(2,0)$-form $\omega_1\omega_2$ and the $(1,1)$-form $\bar \omega_1 \omega_2$. Note that they are both closed, and their difference is $d$-exact (since $d\omega_3 = \omega_1\omega_2 - \bar \omega_1 \omega_2$). So, $[\omega_1\omega_2] = [\bar \omega_1 \omega_2]$ in de Rham cohomology. We see that $\omega_1\omega_2$ is not in the image of $d$, that is $[\omega_1\omega_2] \neq 0$.<|endoftext|>
TITLE: Interval arithmetic with different definitions of intervals
QUESTION [12 upvotes]: Interval arithmetic normally deals with intervals defined as $[a,b]$ with rules like $$[a,b]+[c,d]=[a+c,b+d]$$ I am interested in interval arithmetic with different interval definitions such as $$\{a\}_\delta=[a(1-\delta);a(1+\delta)]$$ or $$[a]_\epsilon := \left[ae^{-\epsilon};ae^\epsilon\right]$$ which base on some kind of relative error around a number $a$. For example for the later definition one has the rule $$[a]_\epsilon\cdot[b]_\delta=[ab]_{\epsilon+\delta}$$
My Question: Can you point me to textbooks/papers where interval arithmetic with different interval definitions such as the two above is discussed? I am looking for those interval definitions which base on some kind of relative error so that multiplication with those intervals is easy.
Note: It may help, that $$\begin{align}
[a]_\epsilon &= \left\{ y \in \mathbb R^{+}: \left\|\frac{y}{a}\right\|_{\log} = |\log(y)-\log(a)| < \epsilon \right\} \\
&= \left[ae^{-\epsilon};ae^{\epsilon}\right] \end{align}$$ whereby $\left\|\frac{y}{a}\right\|_{\log}$ is the so called log ratio distance.

REPLY [3 votes]: I know of a variant of interval arithmetic called ball arithmetic.  It seems as though it may be like what you're looking for.  Ball arithmetic is currently implemented in the C library, Arb.
Here is a link to a paper about this sort of arithmetic.  Also, it discusses the differences between ball arithmetic and interval arithmetic.
http://www.texmacs.org/joris/ball/ball-abs.html
Also, you may be able to find what you're looking for at http://www.cs.utep.edu/interval-comp/books.html.  Which contains a list of books on interval arithmetic.
Also, take a look at http://fredrikj.net/blog/2012/04/high-precision-ball-arithmetic/.  This is extremely similar to what you wrote.
I can't find any other explicit definitions, but it shouldn't be hard to formulate ball arithmetic over general normed vector spaces.<|endoftext|>
TITLE: Is being Noetherian a quasi-isometry invariant for f.g. groups?
QUESTION [5 upvotes]: Recall that a group $G$ satisfies max (or is said to be Noetherian) if all its proper subgroups are finitely generated. Similarly $G$ satisfies max-n if all its normal subgroups are normal closures of finite subsets. Note that property max is "closed with respect to extension", i.e if $N \unlhd G$ and $G \backslash N$ have this property than so does $G$ (in particular this implies that polycyclic groups satisfy max). However max-n is not inherited by subgroups. Nonetheless we have the following Theorem:
If a group $G$ satisfies min-n (resp. max-n) and $H$ is a subgroup of $G$ with finite index, then $H$ satisfies min-n (resp. max-n).
My question is, if these properties max, min, max-n and min-n are preserved under quasi-isometries (of f.g. groups).
Edit 1: as YCor pointed out in a comment the answer is negative for max-n and min-n. The question for max and min seems still open (in general), however we may have a lack of examples.
In this sense we may collect examples of groups satisfying max resp. min.

REPLY [8 votes]: For finitely presented groups, you're close to two well known open questions/conjectures.
Question 1: If a finitely presented group is Noetherian, is it virtually polycyclic?
(This question is FP11 here, where it's attributed to S. Ivanov.) 
Question 2: The class of virtually polycyclic groups is closed under quasi-isometry.
(Alex Eskin conjectures this here.)
Putting these together, I think it's fair to say that, for finitely presented groups, it's unknown whether the property of being Noetherian is a qi-invariant.<|endoftext|>
TITLE: Complex Geometry Consequences of Serre's Kähler-Zeta Function
QUESTION [8 upvotes]: Serre's famous paper Analogues Kählériens de Certaines Conjectures de Weil proves an analogue of the Weil conjectures for compact Kähler manifolds. It would go on to inspire the line of attack that eventually solved the Weil conjectures. 
I would like to know is whether the result had any applications or consequences in complex geometry. Since I guess this is a little broad of a question for this site, let me write: Can anyone name a substantial result in complex geometry that uses Serre's zeta function in a non-trivial way. (Apologies for fluffy words like substantial and non-trivial.)

REPLY [7 votes]: This is a purity result for a polarized Kähler dynamical system. The precise statement is that if $(X,\omega)$ is compact Kähler and $\phi : X \to X$ an endomorphism having the class $[\omega] \in H^2(X,\mathbb{R})$ as eigenvector of eigenvalue $q > 1$ (a polarized Kähler dynamical system of positive entropy), then all eigenvalues for the action of $\phi^*$ on $H^i(X,\mathbb{R})$ have modulus $\sqrt{q}^i$.
A number of consequences of this result for algebraic and Kähler dynamics are derived by Shou-wu Zhang in [Distributions in algebraic dynamics, Surveys in Differential Geometry, vol. X, 2006], which also reproduces the proof of Serre's theorem in mildly generalized form.
As for the zeta function, not explicitly considered in Serre's paper, it is a special case of a dynamical zeta, which is a vast subject. 
Incidentally, there is something else going by the name of Kähler zeta function, discussed in Anton Deitmar's article A panorama of zeta functions from the volume [Erich Kähler. Mathematische Werke., De Gruyter, 2003], also available on ArXiv. Kähler's zeta function is actually arithmetical; it is a different idea of generalizing Riemann's zeta to finitely generated fields than the much better known (and better behaved) Hasse-Weil zeta function. Kähler's interest in finitely generated fields is well known (suffice it to recall the notion of the Kähler differential), but I was amused to learn from J.-B. Bost's article in that volume (A neglected aspect of Kähler's work in arithmetic geometry) that the modern subject of Arithmetic Geometry apparently takes its name from the title of a paper (Geometria Aritmetica) that Kähler wrote in Italian, and that Kähler's writings contain the germs of ideas and notions that later found their proper place in Arakelov theory, such as the Faltings height of an abelian variety.<|endoftext|>
TITLE: Must uncountable compact Hausdorff spaces have large discrete subsets?
QUESTION [5 upvotes]: The situation is this.  I have a space $X$ which is second countable, compact, and Hausdorff (it's a modified form of a type space, though I don't think that matters here).  It has size continuum.  It may or may not have isolated points.
Must $X$ have a discrete subset of size continuum?
The obvious inductive constructions yield only countable discrete sets, and there isn't obviously a Zorn's Lemma argument, since the ascending union of discrete sets is not necessarily discrete.  But this is not my specialty; is this known one way or the other?
The purpose of this is to produce a many-models theorem, and seemingly has nothing to do with topology.

REPLY [11 votes]: Without even using compactness (or metrisability), a subspace of a second-countable space is second-countable. There is no discrete, uncountable, second-countable space.<|endoftext|>
TITLE: Asymptotics of the least common multiple of the first natural numbers
QUESTION [7 upvotes]: What is $$ \limsup_{n \to \infty} \frac{\log(\mathrm{lcm}(1,2, \dots, n))}{n} \ \ ?$$

REPLY [16 votes]: It is well-known that $\operatorname{lcm}(1,\ldots,n) = e^{\psi(n)}$, where $\psi$ is the Chebyshev's function. Since $\psi(x) = x + o(x)$, as $x \to +\infty$, (a form of the Prime Number Theorem) it follows that actually $\lim_{n \to +\infty} \frac{\log(\operatorname{lcm}(1,\ldots,n))}{n} = 1$. (See Part 1 of G. Tenenbaum - Introduction to Analytic and Probabilistic Number Theory).
Not really a MO question, in my opinion.

REPLY [3 votes]: Well, certainly $L=\text{lcm}(1,...,n)$ is divisible by $p^k\leq n$. We can take $k$ up to $[\log_p n]\geq \frac{\log n}{\log p}-1$. So $$\log L\geq \sum_{p\leq n}\left(\frac{\log n}{\log p}-1\right)\log p\approx n$$ by the Prime Number Theorem. So the lim sup should be 1.<|endoftext|>
TITLE: Solve SDE $dX_t=(c+\sigma_\zeta W'_t)X_tdt + \sigma_\epsilon dW_t$
QUESTION [7 upvotes]: I am trying to solve the following SDE
$$dX_t=(c+\sigma_\zeta W'_tX_t)dt + \sigma_\epsilon dW_t$$
$c\in \mathbb{R}$ is a constant, $X_t$ is a stochastic process, $\sigma_\zeta,\sigma_\epsilon \in \mathbb{R}^+$ are constants and $W'_t,W_t$ are independent Wiener processes. Let $x_0\in \mathbb{R}$ be the starting value of $X_t$.
I tried using mathematica (see https://mathematica.stackexchange.com/questions/90088/solving-sde-fracdytdt-c-sigma-wwtyt-epsilont-in-mathematica). This failed. I guess because $X_t$ is not an Ito process. I think it is not an Ito process because there is a random process in the $dt$ term. Sadly, I have no knowledge about solving non Ito process SDEs.
If a solution is to hard to obtain, it would also be sufficient to show that $X_t$ is not a Gaussian Process.

REPLY [5 votes]: Using a (random) integrating factor, the solution is explicitly: $$
X_t =\exp\left( \sigma_{\zeta} \int_0^t W'_s ds \right) \times \left[ x_0+\int_0^t \exp\left(- \sigma_{\zeta} \int_0^s W'_r dr \right) c ds + \sigma_{\epsilon} \int_0^t \exp\left(-\sigma_{\zeta} \int_0^s W'_r dr \right) dW_s \right]
$$ For example, using the fact that $\int_s^t W'_r dr \sim \mathcal{N}(0,(s-t)^2 (2 s +t)/3)$, its expected value is: 
$$
\mathbb{E} \{ X_t \} = \exp\left( \frac{\sigma_{\zeta}^2}{6}  t^3 \right) x_0 + c \int_0^t \exp\left( \frac{\sigma_{\zeta}^2}{6}  (t-s)^2 (2 s + t) \right) ds
$$<|endoftext|>
TITLE: Special rational numbers that appear as answers to natural questions
QUESTION [102 upvotes]: Motivation:
Many interesting irrational numbers (or numbers believed to be irrational) appear as answers to natural questions in mathematics. Famous examples are $e$, $\pi$, $\log 2$, $\zeta(3)$ etc. Many more such numbers are described for example in the wonderful book "Mathematical Constants" by Steven R. Finch.
The question:
I am interested in theorems where a "special" rational number makes a surprising appearance as an answer to a natural question. By a special rational number I mean one with a large denominator (and preferably also a large numerator, to rule out numbers which are simply the reciprocals of large integers, but I'll consider exceptions to this rule). Please provide examples.
For illustration, here are a couple of nice examples I'm aware of:

The average geodesic distance between two random points of the Sierpinski gasket of unit side lengths is equal to $\frac{466}{885}$. This is also equivalent to a natural discrete math fact about the analysis of algorithms, namely that the average number of moves in the Tower of Hanoi game with $n$ disks connecting a randomly chosen initial state to a randomly chosen terminal state with a shortest number of moves, is asymptotically equal to $\frac{466}{885}\times 2^n$. See here and here for more information.
The answer to the title question of the recent paper ""The density of primes dividing a term in the Somos-5 sequence" by Davis, Kotsonis and Rouse is $\frac{5087}{10752}$.

Rules:
1) I won't try to define how large the denominator and numerator need to be to for the rational number to qualify as "special". A good answer will maximize the ratio of the number's information theoretic content to the information theoretic content of the statement of the question it answers. (E.g., a number like 34/57 may qualify if the question it answers is simple enough.) Really simple fractions like $3/4$, $22/7$ obviously do not qualify.
2) The question the number answers needs to be natural. Again, it's impossible to define what this means, but avoid answers in the style of "what is the rational number with smallest denominator solving the Diophantine equation [some arbitrary-sounding, unmotivated equation]".
Edit: a lot of great answers so far, thanks everyone. To clarify my question a bit, while all the answers posted so far represent very beautiful mathematics and some (like Richard Stanley's and Max Alekseyev's answers) are truly astonishing, my favorite type of answers involve questions that are conceptual in nature (e.g., longest increasing subsequences, tower of Hanoi, Markov spectrum, critical exponents in percolation) rather than purely computational (e.g., compute some integral or infinite series) and to which the answer is an exotic rational number. (Note that someone edited my original question changing "exotic" to "special"; that is fine, but "exotic" does a better job of signaling that numbers like 1/4 and 2 are not really what I had in mind. That is, 2 is indeed quite a special number, but I doubt anyone would consider it exotic.)

REPLY [3 votes]: $5\over 8$ths of (propositional) classical tautologies are intuitionistically valid.
This is surprising to me on multiple levels. First, it's fairly robust: a couple different reasonable approaches yield the same answer. Second, of course the fact that it's rational strikes me as deeply weird, I'd have definitely expected something quite odd. Finally, I'm quite surprised that it's greater than $1\over 2$ - my naive take would definitely be that a randomly-picked classical tautology probably isn't intuitionistically valid.<|endoftext|>
TITLE: Theorems of the Galois groups of quintics appears not to work for the ${F}_{20}$ group determination
QUESTION [7 upvotes]: I am computing the Galois groups of quintics using the theorems from Ryan Kavanagh paper "On Irreducible Rational Quintics" using the decic resolvent ${P}_{10} \left({x}\right) = \prod\limits_{1 \le i < j \le 5} \left({x - \left({{\alpha}_{i} + {\alpha}_{j}}\right)}\right)$.  The relevant part is the second part of Theorem 2 where given the discriminant ${\Delta}_{5} \not\in \mathbb{Q}^{2}$ of the quintic then ${P}_{10} \left({x}\right)$ is irreducible over $\mathbb{Q}$ if and only if $\text{Gal} \left({f/\mathbb{Q}}\right) \cong {S}_{5}$ and otherwise, ${P}_{10} \left({x}\right)$ is the product of two quintics irreducible over $\mathbb{Q}$ and $\text{Gal} \left({f/\mathbb{Q}}\right) \cong {F}_{20}$.
Now I have computed ${P}_{10} \left({x}\right)$ both symbolically and numerically for a given test polynomial that I know has Galois group ${F}_{20}$ such as ${x}^{5} + a$ or the two cases of ${x}^{5} + a\, {x}^{2} + b$.  The problem is that the decic resolvent ${P}_{10} \left({x}\right)$ does not factor as two irreducible quintics over $\mathbb{Q}$.
I am using Mathematica for the calculations.  My computation of ${P}_{10} \left({x}\right)$ compares correctly to the examples given.  I am using two different methods which all agree.  For the ${F}_{20}$ Galois group cases the factoring is not correct.  Is there an error in the statement of the theorem?

REPLY [7 votes]: Alas, I have insufficient reputation to comment, so I'll submit my comment as an answer.
Thanks to everybody for the feedback; I'll look at working it into the paper and trying to salvage it. I wrote it during my third year of undergraduate studies for my Galois theory class, and to be honest, I never expected anybody else to ever look at it.
Here's a direct link to the paper: http://ryanak.ca/files/papers/galois.pdf
Best wishes,
Ryan<|endoftext|>
TITLE: About the generating structure of Borel field
QUESTION [5 upvotes]: This is a graduate-level measure theory problem. I have thought throught it and asked on math.SE but received no satisfying answer. 
On P.32 of [P.Billingsley] Probability and Measure, 3ed, 1993, the author wrote:

...and there are Borel sets that cannot be arrived at from the
  intervals by any finite sequence of set-theoretic operations, each
  operation being finite or countable. It can even be shown that there
  are Borel sets that cannot be arrived at by any countable sequence of
  these operations. On the other hand, every Borel set can be arrived at
  by a countable ordered set of these operations if it is not required
  that they be performed in a simple sequence.

Then the author referred me to the following exercise which I cannot figure it out:
For any class $\mathcal{H}$ of sets in $\Omega$ let $\mathcal{H}^{*}$ consist of sets in $\mathcal{H}$, the complements of sets in $\mathcal{H}$ and the finite and countable unions of sets in $\mathcal{H}$. Given a class $\mathcal{A}$, put $\mathcal{A}_{0}=\mathcal{A}$ and define $\mathcal{A}_{1},\mathcal{A}_{2},\cdots$ inductively by:
$$\mathcal{A}_{n}=\mathcal{A}_{n-1}^{*}$$
That each $\mathcal{A}_{n}$ is contained in $\sigma(\mathcal{A})$ follows by induction.
(1)Prove that if $\Omega=(0,1]$, $\mathcal{A}_{0}=\emptyset$ and the intervals with rational endpoints in $\Omega$,$\cup^{\infty}_{n=0}\mathcal{A}_n$ is strictly smaller than $\sigma(\mathcal{A}_{0})$
(2)Extend the above construction to infinite ordinals $\alpha$ be defining 
$$\mathcal{A}_{\alpha}=(\cup_{\beta<\alpha}\mathcal{A}_{\beta})^{*}$$
Show that if $\Omega$ is the first uncountable ordinal, then
$$\cup_{\beta<\Omega}\mathcal{A}_{\beta}=\sigma(\mathcal{A})$$
Moreover, show that if the cardinality of $\mathcal{A}$ does not exceed that of the continuum, then the same is true of $\sigma(\mathcal{A})$, Thus the Borel field has the power of the continuum as its own cardinality.
(i)I cannot figure out how to prove the second problem altough the author provided a hint:

Use the fact that, if $\alpha_{1},\alpha_{2},\cdots$is a sequence of
  ordinals satisfying $\alpha_{n}<\Omega$, then there exists an ordinal
  $\alpha$ such that $\alpha<\Omega$ and $\alpha_{n}<\alpha$ for all
  $n$.

(ii)In particular, I do not quite understand what is the significance of "countable ordered set" here? What role does the "order" play in this construction?
(iii)"It can even be shown that there are Borel sets that cannot be arrived at by any countable sequence of these operations." Where can I find a reference for this issue?

REPLY [2 votes]: Hint Use the fact that, if $\alpha_{1},\alpha_{2},\cdots$is a sequence of
  ordinals satisfying $\alpha_{n}<\Omega$, then there exists an ordinal
  $\alpha$ such that $\alpha<\Omega$ and $\alpha_{n}<\alpha$ for all
  $n$.

Proof By the $\mathbb{R^{1}}$ ordered topology ([H&S] (4.42)), the first uncountable $\Omega$ ([H&S] (4.49) proved its existence) is often written as $[0,\Omega)$ Every increasing sequence ${\alpha_{i}}$ of elements of $[0,\Omega)$ converges to a limit in $[0,\Omega)$ since the supremum ($\cup_{i=1}^{\infty}\alpha_{i}$) of every countable set of countable ordinals is another countable ordinal.
This proves the hint above. The $\alpha$ in that hint is actually the limit of the sequence ${\alpha_{i}}$.#
Since for any class $H$ of sets in $\Omega$ the class $H^{∗}$ consist of sets in $H$, the complements of sets in $H$ and the finite and countable unions of sets in $H$, we have:

Prop 1 If the cardinality of  $H$ is countable, the cardinality of $H^{*}$
  cannot exceed countable cardinality.

Proof All the complements of this class is another class with the same cardinality, and the countable unions of countably many elements in $H$ is still countable.#

Prop 2 $|A_{\alpha}|\neq \omega$(countable cardinal)

Proof If there exists some $\alpha_{0}$ such that $|A_{\alpha_{0}}|=\omega$, then choose the least such $\alpha_{0}$ ([H&S] (4.47)). $|A_{\alpha_{0}}|=|(\cup_{\beta<\alpha}A_{\beta})^{*}|=\omega$, hence by Prop 1 $|\cup_{\beta<\alpha}A_{\beta}|=\omega$ and hence $|A_{\beta}|=\omega,\forall \beta < \alpha_{0}$. Now each such $\beta$ violates the minimality of $\alpha_{0}$#
And since $c\geq |A_{\alpha}|\neq \omega$, it has to be $|A_{\alpha}|=c $ due to the continuum hypothesis.

Prop 3$\cup_{\beta<\Omega}A_{\beta}=\sigma(A)$

Proof $\cup_{\beta<\Omega}A_{\beta}\supset \sigma(A)$ is clear since every element in this sigma field should be obtained via a sequence(countable or not) of these operations. $\cup_{\beta<\Omega}A_{\beta}\subset \sigma(A)$ is also clear by definition of $\sigma(A)$ that it is the smallest sigma field containing these operations' result.#
Then by Prop 2 and the formula $\cup_{\beta<\Omega}A_{\beta}=\sigma(A)$, $|\sigma(A)|=c^{\omega}=c$
I still feel a bit uncertain about Prop 3, can anyone make it clearer to me? I am deeply appreciate all your help.
Reference
[H&S] Hewitt & Stromberg, Real and Abstract Analysis.
Comments by Dave L. Renfro:
What follows is a proof of "Prop 3", which your comments indicated a concern with. My notation varies slightly from your notation because I wrote it earlier this morning where I didn't have access to the internet.
If $\mathcal{C}$ is a collection of subsets of a given set $X,$ let ${\mathcal{C}}^{*}$ be the collection of sets, each of which is an at most countable union of sets, each of which either belongs to $\mathcal{C}$ or its complement belongs to $\mathcal{C}.$ We now define the possibly larger collections ${\mathcal{C}}_{\alpha}$ for each ordinal $\alpha$ by letting ${\mathcal{C}}_0 = \mathcal{C} \cup \{X\}$ and ${\mathcal{C}}_{\alpha} = \left(\cup_{0 \leq \beta < \alpha} {\mathcal{C}}_{\beta} \right)^{*}$ for each ordinal $\alpha > 0.$ I will show that ${\mathcal{C}}_{\omega_1} = \sigma(\mathcal{C})$ by showing ${\mathcal{C}}_{\omega_1} \subseteq \sigma(\mathcal{C})$ and $\sigma(\mathcal{C}) \subseteq {\mathcal{C}}_{\omega_1}.$
First, note that ${\mathcal{C}}_{\alpha} \subseteq \sigma(\mathcal{C})$ for each ordinal (at most countable or not, although we only need this for at most countable ordinals). This is either obvious or it can be proved by transfinite induction, depending on how formal you want to make things. It follows that ${\mathcal{C}}_{\omega_1} \subseteq \sigma(\mathcal{C}).$ To show the opposite inclusion, I'll prove that ${\mathcal{C}}_{\omega_1}$ is a $\sigma$-algebra containing each of the sets in $\mathcal{C},$ and from this the result is immediate by using the fact that $\sigma(\mathcal{C})$ is the minimal (with respect to inclusion) $\sigma$-algebra that contains each of the sets in $\mathcal{C}.$ For closure under complements, if $A \in {\mathcal{C}}_{\omega_1},$ then we have $A \in {\mathcal{C}}_{\alpha}$ for some countable ordinal $\alpha,$ and hence $A^c$ (the complement of $A$ relative to $X$) belongs to ${\mathcal{C}}_{\alpha + 1} \subseteq {\mathcal{C}}_{\omega_1}.$ For closure under at most countable unions, if each of $A_{1},$ $A_{2},$ $A_{3},$ $\ldots$ belongs to ${\mathcal{C}}_{\omega_1},$ then there exist at most countable ordinals $\alpha_{1},$ $\alpha_{2},$ $\alpha_{3},$ $\ldots$ such that $A_1 \in {\mathcal{C}}_{\alpha_1},$ $A_2 \in {\mathcal{C}}_{\alpha_2},$ $A_3 \in {\mathcal{C}}_{\alpha_3},$ $\ldots$ Let $\lambda$ be an at most countable ordinal such that $\alpha_n < \lambda$ for each positive integer $n.$ Then $\cup_{n=1}^{\infty} A_n$ is an element of ${\mathcal{C}}_{\lambda},$ and hence $\cup_{n=1}^{\infty} A_n$ is an element of ${\mathcal{C}}_{\omega_1}$ since ${\mathcal{C}}_{\lambda} \subseteq {\mathcal{C}}_{\omega_1}.$ The proof for closure under at most countable intersections is exactly analogous to that for at most countable unions, but you can also appeal to the fact that any nonempty collection of sets closed under complements and at most countable unions is also closed under at most countable intersections (use De Morgan's Laws).
Some Useful References:
Robert B. Ash, Measure, Integration, and Functional Analysis (1972).

Section 1.2 (pp. 3-13 & 244-245) is useful. In particular, Exercise 9 (p. 12) gives two constructions (as induction over the positive integers) for the algebra generated by a collection of subsets from a given set, and Exercise 11 (p. 13) gives a construction (as transfinite induction over the countable ordinals) for the $\sigma$-algebra generated by a collection of subsets from a given set. Note that fairly detailed solutions to both of these exercises are given on pp. 244-245.

Billingsley, Probability and Measure, 3rd edition (1995).

See Problems 2.22, 2.23 on p. 36 and hints for their solution on pp. 554-555.

Inder K. Rana, An Introduction to Measure and Integration, 2nd edition (2002).

Theorem 4.5.1 (p. 111) states and proves a construction of the algebra generated by a collection of sets (as induction over the positive integers) and Theorem 4.5.2 (pp. 111-112) states and proves a construction of the $\sigma$-algebra generated by a collection of sets (as transfinite induction over the countable ordinals).

Halmos, Measure Theory (1950)

See Chapter I (pp. 9-29).

Felix Hausdorff, Set Theory (1937+ English translation of 3rd edition by John R. Aumann)

See §17, §18, §30, §32, §33.

Kuratowski/Mostowski, Set Theory, 2nd English edition (1976).<|endoftext|>
TITLE: The finiteness of the associated primes of $Ext^i_R(R/I, M)$
QUESTION [5 upvotes]: Let $(R,m)$ be a Noetherian local ring and $M$ an $R$-module. If $M$ is a finite $R$-module, then we know that $Ext^i_R(R/I,M)$ is a finite $R$-module for all $i\geq 0$; Now suppose $supp(M)\subseteq V(I)$ and $Ass_R(M)$ is a finite set, then $Ass_R(Hom(R/I,M))=Ass_R(M)$ is finite. For $i>0$, is $Ass_R(Ext_R^i(R/I,M))$ still a finite set?

REPLY [2 votes]: Take $R = \mathbb{Z}[X]$, so $\mathbb{Z} = R/(X)$. Let $I = (X)$. For each prime number $p$ we consider the module $M_p := (p,X)/(X^2)$. We have $\mathrm{Supp}M_p = V(I)$ and $\mathrm{Ass}_R(M_p) = (X)$. Consider the short exact sequence
$$0 \to M_p \to R/(X^2) \to R/M_p \cong \mathbb{Z}/p\mathbb{Z} \to 0.$$
Applying the $\mathrm{Ext}^i_R(R/(X),-)$ to the above sequence with note that
$$\mathrm{Hom}_R(R/(X), M_p) \cong (X)/(X^2) \cong \mathrm{Hom}_R(R/(X), R/(X^2)$$
we have
$$0 \to \mathrm{Hom}_R(R/(X), \mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z} \to \mathrm{Ext}^1_R(R/(X),M_p) \to \cdots$$
Thus $(p,X) \in \mathrm{Ass}_R(\mathrm{Ext}^1_R(R/(X),M_p))$. Now set
$$M = \bigoplus_{p: prime} M_p$$
 we have $\mathrm{Supp}(M) = V(I)$, $\mathrm{Ass}_R(M) = (X)$ but 
$$\mathrm{Ass}_R(\mathrm{Ext}^1_R(R/(X),M)) = \{(p,X): p \text{ is prime}\}$$
is an infinite set.<|endoftext|>
TITLE: Smoothness of Hecke algebras
QUESTION [5 upvotes]: First I will introduce some notation and definitions.
Fix a level $N$ (take $N=1$ if it makes things easier) and a prime $p$. Let $k$ be a finite field of characteristic $p$ and let $\mathcal{C}$ be the category of complete noetherian local rings $(R, \mathfrak{m}_R)$ together with an isomorphism $R/\mathfrak{m}_R \cong k$. Let $\Lambda$ be an artinian object of $\mathcal{C}$ (I'm interested in cases where $\Lambda = \mathcal{O}_K / \pi^n$ where $K$ is a finite extension of $\mathbb{Q}_p$ with uniformiser $\pi$). Let $S_k(\mathbb{Z}_p)$ denote the set of $q$-expansions of cuspforms of level $N$ and coefficients in $\mathbb{Z}_p$, and for a $\mathbb{Z}_p$-algebra $R$ let $S_k(R) := S_k(\mathbb{Z}_p)\otimes_{\mathbb{Z}_p}R$. Let $S(R) = \sum_k S_k(R)$, and let $\mathbb{T}(R)$ denote the $R$-subalgebra of $\mbox{End}(S(R))$ generated by the Hecke operators $T_n$ for all $n$ such that $(n,p)=1$.
Now if I understand things correctly, it follows from deformation theory of Galois (pseudo-)representations that $\mathbb{T}(\Lambda)$ is a complete noetherian semilocal ring. The maximal ideals correspond to mod $p$ modular representations. Let $\mathfrak{m}$ be a maximal ideal. 

My question is the following: can we under some conditions assert that
  $\mathbb{T}(\Lambda)_\mathfrak{m}$ is smooth or $\mathfrak{m}$-smooth
  as a $\Lambda$-algebra for all $\Lambda$, or at least for all the
  rings $\Lambda$ I'm interested in (described above) ?

What I know:
Please feel free to correct me if I'm mixing things up in the following. 
Let $\mathbb{T} = \mathbb{T}(W(k))$. Let $\mathcal{R}$ be the universal deformation ring attached to the residual representation corresponding to $\mathfrak{m}$. If the deformation problem is unobstructed, then I know that $\mathcal{R}$ will be a power series ring and therefore smooth. Moreover the restriction of the deformation functor to the category of $\Lambda$-algebras is represented by $R_\Lambda = \mathcal{R}\mathbin{\hat\otimes} \Lambda$ and is therefore smooth again. Generally we have a surjective homomorphism $\mathcal{R}\twoheadrightarrow\mathbb{T}$ but under some conditions we have $\mathcal{R} \cong\mathbb{T}$.
However it seems to me that even if we have $\mathcal{R} \cong\mathbb{T}$, we might not have that $\mathcal{R}_\Lambda \cong \mathbb{T}(\Lambda)$, where $\mathbb{T}_\Lambda$ is as defined above. For an example there is the following paper by Bellaiche and Khare 
http://people.brandeis.edu/~jbellaic/preprint/Heckealgebra4.pdf
In the cases they consider, $\mathbb{T}(k)_\mathfrak{m}$ (which they call $A_{\overline{\rho}}$) is not the reduction of $\mathbb{T}$ modulo $p$, but a quotient of that modulo some element in the maximal ideal. Actually, under some conditions (e.g. the deformation problem is unobstructed) they identity $\mathbb{T}(k)_\mathfrak{m}$ with the universal ring of pseudo-deformations to $k$-algebras with constant determinant. In these cases, we still can deduced that $\mathbb{T}(k)_\mathfrak{m}$ is smooth.
I would appreciate any hints  or references on this question. Thanks.

REPLY [2 votes]: It follows from deformation theory of Galois (pseudo-)representations that $\mathbb T(\Lambda)$ is a complete noetherian semilocal ring. The maximal ideals correspond to mod $p$ modular representations.

The fact that $\mathbb T(\Lambda)$ is a semi-local ring follows from the existence of the natural surjective map
$$\mathbb T(\mathcal O_K)/\pi^{n}\longrightarrow \mathbb T(\Lambda)$$
and the fact that system of Hecke eigenvalues (otherwise known as eigenforms) are in finite numbers (not a priori from arguments with pseudo-representations).
Moreover, for stupid reasons of compatibility of coefficients, the maximal ideals of $\mathbb T(\Lambda)$ cannot correspond to mod $p$ modular representations. Now maybe you meant to write "correspond to modular representations with coefficients in $\Lambda$" however, as in your setting the coefficient ring is artinian but not a field, it is not obviously true that such a statement hold: maybe the pseudo-representation $t:G_{\mathbb Q}\longrightarrow \mathbb T(\Lambda)_{\mathfrak m}$ sending $\operatorname{Fr}(\ell)$ to $T(\ell)$ does not lift to an actual representation $\rho_{\mathfrak m}$ with coefficients in $\Lambda_{\mathfrak m}$. Conversely, a single residual representation $\bar{\rho}$ can correspond to several congruent pseudo-representations with coefficients in $\mathbb T(\Lambda)$.
Regarding your actual question, I think the only known line of attack of such problems at present is the one you describe, that is to say the one of Nicolas-Serre-Bellaïche-Khare: either find an explicit description (totally hopeless in your case) or assume that $R(\bar{\rho})=\mathbb T_{\bar{\rho}}$ is smooth and try to relate $\mathbb T_{\bar{\rho}}\otimes_{\mathcal O}\mathcal O/\pi^n$ to $\mathbb T(\Lambda)_{\mathfrak m}$. As I indicated above, this seems to require much clarification as $\mathfrak m$ and $\bar{\rho}$ are not in obvious correspondence anymore (and in particular, I would certainly not bet that unobstructed implies smooth in your setting; I don't see how to rule out the possibility of many congruences between pseudo-representations destroying the smoothness when one passe from $\mathbb T(\mathcal O/\pi)$ to $\mathbb T(\mathcal O/\pi^n)$).<|endoftext|>
TITLE: What is wrong with this deterministic algorithm efficiently generating large primes?
QUESTION [7 upvotes]: According to PolyMath

(Strong) conjecture. There exists deterministic algorithm which, when given an integer k, is guaranteed to find a prime of at least k digits in length of time polynomial in k. You may assume as many standard conjectures in number theory (e.g. the generalised Riemann hypothesis) as necessary, but avoid powerful conjectures in complexity theory (e.g. P=BPP) if possible.

According to answers in this question.
Let $p$ be prime. Under GRH there exists prime $p' \equiv 1 \pmod{p}$ satisfying
$$p' \leq 70 p (\log p)^2 \qquad (1)$$
Doubling lemma Let $p$ be odd prime. In time polynomial in $\log{p}$
we can find prime $p' > 2p$.
From (1), the interval $[p+1,70 p (\log p)^2]$ contains
about $70(\log{p})^2$ integers congruent to $1$ modulo $p$ 
of form $mp+1$ and
at least one $p'$ is prime. Since $p+1$ is even, $p' \ge 2p+1$.
To find $p'$, enumerate the candidates and check them for primality.
The complexity is polynomial in $\log{p}$.
Start from $p=3$ and repeat Doubling lemma.
Each iteration produces prime at least twice larger than the previous step.
So this appears to give algorithm under GRH to find $p>k$ in time
polynomial in $\log{k}$, which is equivalent to the Strong conjecture.

What is wrong with this?

REPLY [11 votes]: I have a somewhat different answer. The bound (1) is not known, but it is expected by some optimist paper. See for example this paper which says that one expects $p' = O(p (\log p)^2)$ and gives (page 1718) three references to papers discussing this. This estimate is way better that anything one can prove with GRH. If it holds, then there is nothing wrong with your algorithm, it works fine (and is nice, by the way). 
However that estimate is so strong it is very doubtful. Actually it is related to Montgomery's conjecture, that $\psi(x,p,a) - x/(p-1) = O_\epsilon( (x/p)^{1/2+\epsilon} \log(x))$, where $\psi(x,p,a)$ counts the (weighted by their log) primes congruent to $a$ modulo $p$ up to $x$. If I'm not mistaken, this implies that the first such prime $p'$ is $O_\epsilon(p (\log p)^{2+\epsilon})$ which is very close to your estimate and enough for your algorithm. The problem is: Montgomery's conjecture 
has been proved false: see this article by Friedlander and Granville.
As a replacement, the same authors proposes (conjecture 1)  $\psi(x,p,a) - x/(p-1) = O_\epsilon( (x/p)^{1/2} x^\epsilon)$. This gives an estimate of $p'$ in $O(x^{1+\epsilon})$ and kills your algorithm. 
So in conclusion, your algorithm works assume your estimate, but this estimate, while it has been conjectured, is at best very doubtful.<|endoftext|>
TITLE: The number of solution of $x_1^2 + \cdots + x_k^2 \equiv \lambda \bmod q$
QUESTION [7 upvotes]: I'm playing with exponential sums...
If $q$ is an odd prime and $a$ an integer such that $q \nmid a$, then the following formula for the Gaussian sum is known
$$\sum_{x=0}^{q-1} e_q(ax^2) = \left(\frac{a}{p}\right) i^{\left(\frac{q-1}{2}\right)^2} \sqrt{q} ,$$
where $e_q(z) := \exp\left(\frac{2\pi i z}{q}\right)$ and $\left(\frac{a}{q}\right)$ is the Legendre symbol. Using the orthogonality identity
$$\delta_q(b) := \frac1{q}\sum_{y = 0}^{q-1} e_q(by) = \begin{cases}1 & \text{ if } q \mid b; \\ 0 & \text{ if } q \nmid b;\end{cases}$$
for integer $b$, and standard manipulation of exponential sums (I can give more details, but, as I said, it is all standard stuff), I obtained that for all integers $k \geq 1$ and $\lambda$, the number of $x_1, \ldots, x_k \in \{0, \ldots, q-1\}$ such that
$$x_1^2 + \cdots + x_k^2 \equiv \lambda \bmod q$$
if equal to
$$q^{k-1} + \begin{cases}(-1)^{\frac{k}{2}\left(\frac{q-1}{2}\right)^2} q^{\frac{k}{2}-1} (q \delta_q(\lambda)-1) & \text{ if } 2 \mid k; \\
\left(\frac{-\lambda}{q}\right)(-1)^{\frac{k+1}{2}\left(\frac{q-1}{2}\right)^2}q^{\frac{k-1}{2}} & \text{ if } 2 \nmid k \text{ and } q \nmid \lambda;\\ 0 & \text{ if } 2 \nmid k \text{ and } q \nmid \lambda .\end{cases} $$
I'm sure that this result should be already known, but I failed to find a reference.
Thank you in advance for any suggestion!

REPLY [6 votes]: I just want to share a character-free proof - very accessible and too long for a comment (also - probably not original). Of course, it's just a restatement of a character-full proof.
Let $\vec{v} \in \mathbb{R}^{\mathbb{F}_q}$ be the vector whose $i$'th coordinate is the number of solutions to $x^2 \equiv i \mod q$. Then the distribution of $\sum_{i=1}^{k} x_i^2$ is given by the convolution (coming from the additive structure of $\mathbb{F}_q$) of $\vec{v}$ with itself $k$ times, i.e. $\underbrace{\vec{v} * \cdots * \vec{v}}_{k \text{ times}}$.
The main observation is that the convolution of $\vec{v}$ with itself is almost constant, in the following sense: there are only 2 (possibly equal) values appearing in it, one corresponding to the number of solutions to $x_1^2+x_2^2=0 \mod q$ and the other for all the rest:

If $q \equiv 1 \mod 4$, there is a root of $-1$ in $\mathbb{F}_q$, denote it by $i$. The equation $x_1^2+x_2^2 =a\mod q$ factors as $(x_1+ix_2)(x_1-ix_2)=a$, which is equivalent (after an $\mathbb{F}_q$-linear transformation) to $y_1 y_2 = a \mod q$. In this form the result is evident.
If $q \equiv -1 \mod 4$, $i:=\sqrt{-1}$ still exists but in $\mathbb{F}_{q^2}$. The equation still factors as $(x_1 +ix_2)(x_1-ix_2) = a$ in $\mathbb{F}_{q^2}$. Raising to $q$'th power is a (Frobenius) automorphism acting as $(x_1+ix_2)^q=x_1-ix_2$, so we can rewrite this as $z^{q+1} = a$ where $z \in \mathbb{F}_{q^2}$. Because of the cyclic structure of $\mathbb{F}_{q^2}^{\times}$ and because $\mathbb{F}_{q}^{\times}$ is a subgroup of index $q+1$, the number of solutions is $q+1$ when $0 \neq a \in \mathbb{F}_{q}$ (there are $q+1$ roots of unity of order dividing $q+1$ in $\mathbb{F}_{q^2}$).

We can summarize both cases as $(\vec{v}*\vec{v})_a = \begin{cases} q+(\frac{-1}{q})(q-1)& a =0 \\ q-(\frac{-1}{q}) & a\neq 0 \end{cases}$.
The second obserivation is that if $\{ \vec{u_i}\}_{i=1}^{2}$ are two vectors of the form $\vec{v_i} = (a_i, \underbrace{b_i, \cdots, b_i}_{q-1 \text{ times}})$, then their convolution is of the form $(a, \underbrace{b, \cdots, b}_{q-1 \text{ times}})$, where $a,b$ satisfy $a+(q-1)b=(a_1+(q-1)b_1)(a_2+(q-1)b_2)$ and $a-b=(a_1-b_1)(a_2-b_2)$. In other words, the functional $(a, \underbrace{b, \cdots, b}_{q-1 \text{ times}}) \to \begin{cases} a+(q-1)b\\ a-b \end{cases}$ respect convolution and turn it into multiplication.
Now it is only a computational matter: if $k$ is even, we need to convolve $\vec{v} * \vec{v}$ with itself $\frac{k}{2}$ times using the second observation. We get the vector $(a, \underbrace{b, \cdots, b}_{q-1 \text{ times}})$ where $a,b$ are given as a solution of the following linear system: $a+(q-1)b=q^{k/2}, a-b=(q(\frac{-1}{q}))^{k/2}$.
If $k$ is odd, there's an additional complication - we first compute $\underbrace{v * \cdots * v}_{k-1 \text{ times}}=(a, \underbrace{b, \cdots, b}_{q-1 \text{ times}})$, and then we need to convolve it once again with $\vec{v}$ - the $j$'th coordinate will be $av_j + (q-v_j)b= qb + (a-b)(1+(\frac{j}{q}))$.<|endoftext|>
TITLE: What does it mean to take the diagonal of the group $SU(2) \times SU(2) $?
QUESTION [8 upvotes]: I am reading Witten's paper on topological field theories, in specific the topological twist in page 359. In order to perform the twist he takes the diagonal subgroup of $K = SU(2)_{\text{Right}} \times SU(2)_{\text{Isospin}}$ where the first comes from the $SU(2)_{\text{Left}} \times SU(2)_{\text{Right}} \simeq SO(4) $ while the latter is part of the $\mathcal{R}$-symmetry group $SU(2)_{\text{Isospin}} \times U(1)_{\mathcal{R}}$.
What does it mean to take the diagonal of $K$ and how can I understand how the representations of the sections/fields change after the twist? I mean, he says it is "easy to see" how the fields transform under the new global group but I cannot see it.

REPLY [9 votes]: As noted in Vit's post, the diagonal group $G_D \subset G \times G$ consists of the elements $\{(g,g) : g \in G \}$ with multiplication defined to be $(g,g) \cdot (h,h) = (gh,gh)$. A representation $(R_1,R_2)$ of $G \times G$ thus transforms as the representation $R_1 \otimes R_2$ under $G_D$, where $\otimes$ denotes the tensor product.
Let me illustrate through an example in the paper you are reading. In equation (2.16), it is stated that the fermions are in the representation
\begin{equation}
(1/2,0,1/2)^1 \oplus (0,1/2,1/2)^{-1}
\end{equation}
under $SU(2)_L\times SU(2)_R \times SU(2)_I \times U(1)$. Now under
\begin{equation}
SU(2)_L\times SU(2)_{R'} \times U(1) \subset SU(2)_L\times SU(2)_R \times SU(2)_I \times U(1) \,,
\end{equation}
where $SU(2)_{R'}$ is the diagonal subgroup of $SU(2)_R \times SU(2)_I$, this representation transforms as
\begin{equation}
(1/2,0 \otimes 1/2)^1 \oplus (0,1/2 \otimes 1/2)^{-1} \,.
\end{equation}
We know the tensor product decomposition rule for $SU(2)$ pretty well:
\begin{equation}
0 \otimes 1/2 = 1/2 \,, \quad
1/2 \otimes 1/2 = 0 \oplus 1 \,.
\end{equation}
Recall that the numbers here denote the "spin" of the representation. A spin $j$ representation has dimension $(2j+1)$. Note that the representation ``0" is the trivial (one-dimensional) representation. Thus we recover equation (2.18):
\begin{equation}
(1/2,1/2)^1 \oplus (0,0)^{-1} \oplus (0,1)^{-1} \,.
\end{equation}
I hope this makes everything clear to you now.<|endoftext|>
TITLE: Why is there a duality between spaces and commutative algebras?
QUESTION [46 upvotes]: 1) The category of affine varieties over $\mathbb{C}$ is equivalent to the opposite category of finitely generated reduced algebras over $\mathbb{C}$. The equivalence associates to an affine variety its algebra of regular functions to $\mathbb{A}^1$, and to each finitely generated reduced algebra its (ringed) space of algebra homomorphisms to $\mathbb{C}$. This is almost a tautology, but in some sense a deep one.
2) The category of compact Hausdorff spaces is equivalent to the opposite category of $\mathrm{C}^*$-algebras. The equivalence associates to a compact Hausdorff space its algebra of continuous functions to $\mathbb{C}$ (equipped with the sup norm), and to each $\mathrm{C}^*$-algebra its space of $\mathrm{C}^*$-algebra homomorphisms to $\mathbb{C}$. This is due to Gelfand.
3) The category of totally-disconnected compact Hausdorff spaces is equivalent to the opposite category of Boolean algebras. The equivalence associates to a totally-disconnected compact Hausdorf space its algebra of continuous functions to $\{0,1\}$, and to each Boolean algebra its space of Boolean algebra homomorphisms to $\{0,1\}$. This is due to Stone.
4) The $\infty$-category of simply-connected rational spaces of finite type is equivalent to the opposite $\infty$-category of simply-coconnected coconncetive $\mathbb{E}_\infty$-algebras of finite type over $\mathbb{Q}$. The equivalence associates to a rational space $X$ its cochain complex $C^*(X,\mathbb{Q})$, and to each $\mathbb{E}_\infty$-algebra over $\mathbb{Q}$ its space of $\mathbb{Q}$-linear $\mathbb{E}_\infty$-algebra maps to $\mathbb{Q}$. This is an $\infty$-categorical reformulation due to Lurie of a classical theorem of Sullivan. 
5) The $\infty$-category of pro-$p$-finite spaces is equivalent to the opposite category of solvable $\mathbb{E}_\infty$-algebras over $\overline{\mathbb{F}}_p$. The equivalence associates to a pro-$p$-finite space $\{X_i\}_{i \in I}$ the corresponding colimit of cochain complexes $colim_i C^*(X_i,\overline{\mathbb{F}}_p)$. If $X$ is a $p$-finite space then $X$ can be reconstructed as the space of $\overline{\mathbb{F}}_p$-linear $\mathbb{E}_\infty$-algebra maps from $C^*(X,\overline{\mathbb{F}}_p)$ to $\overline{\mathbb{F}}_p$. Here $\overline{\mathbb{F}}_p$ is the algebraic closure of $\mathbb{F}_p$. The theorem in this form is due to Lurie, but a the original version of the theorem (working with $p$-complete spaces) is due to Mandell.

Indeed, if $R$ is a commutative-ring type object in some category of geometric objects, and $X$ is a geometric object carrying a sufficient supply of maps to $R$, then it is worthwhile to look at the ring of functions from $X$ to $R$, as this object will carry a lot of information on $X$. In the other direction, it is often useful to study commutative-ring objects by trying to realize them as rings of functions on some geometric object, their "spectrum". However, even with this in mind, it is still quite striking that this approach yields so many anti-equivalences between full subcategories of geometric objects and full subcategories of commutative-algebraic objects. It is like the geometer/topologist/homotopy-theorist is walking on one continent, and the commutative-algebraist is walking on another, and yet they keep stumbling upon the exact same species of categories (only in reverse).
What is going on here? Is there any conceptual explanation to this phenomenon? Is there any general argument which suggests, even heuristically, that the above equivalences are to be expected?

REPLY [5 votes]: I have not yet considered the new answer in full detail, so apologies for not addressing it. I'm coming back to this question after a while, so I thought I'd share some observations which came up during that time, and which might interest someone (at least, those someones who thought the original question was of any interest).
1) Example (3) above is actually a particular case of Example (5) with $p=2$, in the sense that it is obtained by restricting from 2-pro-finite spaces to pro-finite sets. Indeed, recall the notion of "solvable" which appears in Lurie's $p$-adic homotopy theory paper. In the discrete case, the data of a solvable $\overline{\mathbb{F}}_2$-algebra is essentially equivalent to the data of a Boolean algebra. In fact, example (3) admits a version for every prime $p$, replacing Boolean algebras with algebras which satisfy the relation $a^p=a$ for every $a$. These categories of "$p$-Boolian algebras" are in fact all equivalent to the opposide category of pro-finite sets (or totally disconnected compact Hausdorff spaces).
2) In every context where there is an interesting notion of geometric objects and commutative algebra objects which are connected by some duality, there is usually also a related category of "abelian group objects" underlying the situation. In particular, the commutative algebras are commutative algebra objects in this category, and there is typically an adjunction between geomtric objects and abelian group objects which associates to an abelian group its "underlying space" and to a geometric obejct the free abelian group generated from it. This later abelian group should be dual in some since to the underlying abelian group of the algebra of functions on the space. The adjunction between geometric objects and abelian group objects then induces a comonad on the category of abelian group objects. In that case one can ask two types of questions: (i) to what extent is this adjunction comonadic? in other words, how close is it to exhibiting spaces as coalgebras over this comonad? and (ii) how close is this comonad to being the cocommutative coalgebras comonad? When both answers are sufficiently positive we get that spaces look a lot like cocommutative coalgebras in abelian groups. Since abelian groups often have something that is close to self duality, this makes spaces close to the opposite of commutative algebras.
3) If one is working in the $\infty$-categorical setting, and $\mathcal{C}$ is some $\infty$-category of geometric objects, then one has a natural choice for the $\infty$-category $\mathcal{A}$ of abelian group objects, namely, the stabilization $Sp(\mathcal{C})$ of $\mathcal{C}$. Alternatively, one can tensor the stabilization with a field, such as $\mathbb{Q}$ or $\overline{\mathbb{F}}_p$, the choices underlying examples (4) and (5) above. If $\mathcal{C}$ is furthermore presentable then we will have an adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$ where we can consider the left functor as an analogue of the free abelian group functor. If $\mathcal{C}$ carries a symmetric monoidal structure which preserves colimits in each variable separately then $\mathcal{A}$ will inherit the same type of structure and the left functor $\mathcal{C} \to \mathcal{A}$ will be monoidal. 
4) Let $K: \mathcal{A} \to \mathcal{A}$ be the comonad corresponding to the adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$. Since $\mathcal{A}$ is monoidal it makes since to ask if $K$ is the coalgebra comonad of some operad enriched in $\mathcal{A}$. This question can be tackled using Goodwillie calculus and obstruction theory, see work of Gijs Heuts (e.g., https://arxiv.org/abs/1510.03304). In particular, these obstructions vanish when working over a field of characteristic $0$, so if we take $\mathcal{A}$ to be the stabilization tensored with $\mathbb{Q}$, then the comonad $K$ will always come from some operad in $\mathcal{A}$. The spaces of $n$-ary operations of this operad (considered as $\Sigma_n$-objects in $\mathcal{A}$) are given by the Goodwillie derivatives of $K$ of $\mathcal{A}$. For this operad to be the commutative operad we need that the derivatives are all the unit object. This is likely to happen for completely abstract reasons when the monoidal structure on $\mathcal{C}$ is the Cartesian one, i.e., when $\mathcal{C}$ is Cartesian closed (a typical property of categories of geometric objects). Here is a cool heuristic explanation of this phenomenon that I learned from Tomer Schlank: the right adjoint $\mathcal{A} \to \mathcal{C}$ preserves Cartesian products and under our assumptions the left adjoint sends Cartesian products to tensor products in $\mathcal{A}$. Since $\mathcal{A}$ is stable products and coproducts in $\mathcal{A}$ coincide, and are often known simply as direct sums. One may then conclude that the comonad $K: \mathcal{A} \to \mathcal{A}$ sends direct sums to tensor products. In other words, in the analogy between functors and functions underlying the Goodwille calculus, $K$ is analogous to a function $\mathbb{R} \to \mathbb{R}$ which sends addition to multiplication. In other words, $K$ is analogous to an exponential function. As such, its derivatives at $0$ are the sequence of powers of some number $a$. There is a natural family of operads which have the same property. If $A \in \mathcal{A}$ is a cocommutative bialgebra object then the theory of commutaive $A$-algebras is controlled by an operad whose $\Sigma_n$-object of $n$-ary operations is $A^{\otimes n}$. It seems likely (though I don't have a proof of this), then for every Cartesian closed $\infty$-category $\mathcal{C}$ there is a cocommutative bialgebra object in $\mathcal{A} = Sp(\mathcal{C}) \otimes \mathbb{Q}$ such that the comonad associated to the adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$ is the comonad of cocommutative $A$-coalgebras. For example, when $\mathcal{C}$ is the $\infty$-category of spaces this $A$ is the unit object of $\mathcal{A}$ and one gets the commutative operad. 
Edit: A closer look at 1 shows that this $A$ is actually always the unit, see Lemma B.3. On the other hand, the situation is more complicated than I thought, so remark (4) should really be taken loosely.<|endoftext|>
TITLE: Is every locally compactly generated space compactly generated?
QUESTION [9 upvotes]: [Parse it as (locally compact)ly generated.]
I stumbled across this one whilst supervising an undergraduate thesis. Convenient categories for homotopy theory (e.g. CGWH) have been discussed here before. As an alternative to CGWH, Rainer Vogt proposed the category of locally compactly generated spaces; see also the recent German-language point-set topology textbook Grundkurs Topologie by Gerd Laures and Markus Szymik.
If (as is now usual) one means (compact Hausdorff)ly generated when one says compactly generated, then the category of compactly generated spaces is a full subcategory of the category of locally compactly generated spaces. Back in 1971, Vogt asked whether this inclusion is strict or not. Do we know the answer yet?

REPLY [8 votes]: The paper "A distinguishing example in k-spaces" by John Isbell constructs an example of a locally compact space $X$ which is not compact-Hausdorffly generated.<|endoftext|>
TITLE: Bicoloring of $\mathbb{N}^2$, avoiding set of patterns, is the maximal limit density rational?
QUESTION [15 upvotes]: Consider a bi-coloring of $\mathbb{N}^2$, (black and white), where we wish to maximize the limit (limsup) of the density of black squares in $[n] \times [n]$ as $n \to \infty$. Here, we identify each point with a square in the obvious way.
However, we have specified a finite set of black sub-patterns that are not allowed.
That is, a pattern is a subset of $[k]\times [k]$, and any translation of this subset is not allowed in the coloring as all black squares.
For example, perhaps we do not allow two adjacent black squares.
Then, a checkerboard coloring has limit density $1/2$.
Updated after answer
Q1: For any finite set of forbidden patterns, will the coloring(s) maximizing the density always be periodic? NO
Q2: Is there a finite set of forbidden patterns, such that the maximal limit density can only be obtained by a non-periodic coloring? YES
Q2b: Is there a finite set of forbidden patterns, such that the maximal limit density is an irrational number? Note that this requires a non-periodic pattern.
Q3: Is there a limit density that is not realizable, that is, there is a sequence of colorings, $c_1,c_2,\dots$ with increasing limit density, but the limit density is not realizable by any coloring?
Q4: Since this feels closely related to tiling problems and permutation patterns, might it be that the question "Does the set $F$ allow a coloring with limit density $\geq d$?" is undecidable?  YES 
According to bijection with Wang tiles, where all tiles having equal density.
Deciding if a finite set of Wang tiles, tile the plane, is undecidable.
*Note that for every set of forbidden patterns, coloring all squares white is a valid coloring. *
Note: We only consider a finite set of forbidden patterns in all questions.

REPLY [3 votes]: I believe the answer to Q2b is yes, there does exist setups where the limit density is irrational.
Consider the Kari-Culik (Wang) tilings, and let $\Phi$ be projection onto the top label of each tile (treating $0'$ as $0$).  Durand, Gamard, and Grandjean in http://arxiv.org/abs/1312.4126 show that for any Kari-Culik tiling $x$, the row averages of $\Phi(x)$ converge.  Further, if $(\gamma_i)_{i\in\mathbb Z}$ is a vector of row averages, 
$\gamma_i=f(\gamma_{i-1})$ where $f(t) = \begin{cases}2t&t\in[1/3,1)\\ t/3 & t\in[1,2]\end{cases}$.
$f$ is conjugate to an irrational rotation via the map $\varphi:[1/3,2]\to [0,1]$ given by $\varphi(t) = \frac{\log t+\log 3}{\log 6}$, and so since an irrational rotation gives a uniform orbit density, pushing forward through $\varphi$, we can compute the average
$\mathrm{Ave}(\Phi(x)) = \frac{5}{\log 216}$
for any Kari-Culik tiling $x$.  I didn't think about the correspondence of Wang tilings and your problem in detail, but after translating the Kari-Culik tiles into the framework of your bi-coloring, an irrational average should be preserved.<|endoftext|>
TITLE: How stable is the top cell of a Lie group?
QUESTION [13 upvotes]: It is well known that the fundamental class of a compact Lie group $G$ is stably spherical (see "H-Spaces and Duality" by Browder and Spanier, or "Thom Complexes" by Atiyah), and there is a stable equivalence $G\simeq A\vee S^n$ for some subcomplex $A$, with $n$ being the dimension of $G$.
My question relates to exactly when the attaching map of the top cell becomes trivial? How many suspensions are required for the top cell to split off? Is there an exact answer, or bounds for some special cases perhaps?
As an example, Mimura shows in "On the Number of Multiplications on $SU(3)$ and $Sp(2)$" that this occurs for $Sp(2)$ after exactly two suspensions, $S^2\wedge Sp(2)\simeq S^5\cup e_9\vee S^{12}$, and for $SU(3)$ after exactly three suspensions, $S^3\wedge SU(3)\simeq S^6\cup e_8\vee S^{11}$. These are very special cases. Are more general results or estimates available?

REPLY [4 votes]: Let me just give the naive answer. The homotopy category of $n$-connected $(n+k)$-dimensional CW-complexes is stable for $k<n-1$, by the Freudenthal suspension theorem etc. So the $(\dim G+3)$-fold suspension splits in general, if $G$ is connected the $(\dim G+2)$-fold suspension suffices, and if $G$ is simply connected suspending $\dim G$ times is OK ($\pi_2 G=0$).
This naive bound gives $n^2+2$ for $U(n)$. As Matthias says $n^2$ is sufficient, which is pretty close. For $O(n)$ the bound is $\frac{n^2-n}{2}+3$, which for $n\geq 6$ improves John's estimate. For $SU(n)$ it is $n^2-1$, which is far from optimal as you indicate. The same happens for $Sp(n)$.<|endoftext|>
TITLE: tree properties on $\omega_1$ and $\omega_2$
QUESTION [12 upvotes]: Are the following mutually consistent (relative to large cardinals)?
(1) There are no $\omega_2$-Aronszajn trees.
(2) There is an $\omega_1$-Kurepa tree.
In the models I know of the tree property at $\omega_2$, it also holds that there are no weak Kurepa trees on $\omega_1$ (also called Canadian trees).

REPLY [12 votes]: I wrote a short note with the consistency proof, which can be found at http://www.math.cmu.edu/users/jcumming/papers/kurepa/kurepa.pdf. It is pretty rough, please tell me if there are problems.<|endoftext|>
TITLE: Update for 2015: least prime of form nq+1, with q prime?
QUESTION [11 upvotes]: I have received a complaint about my 2011 answer 
least prime in a arithmetic progression 
which, indeed, gives conflicting reports about this:

given a prime $q,$ what can we say about an upper bound for the smallest prime $p$ in the arithmetic progression $n q + 1$?

Note that I do not see anything using the number $70$ in THIS.
I guess there are about three parts:
(A) What is the most optimistic upper bound, i.e. numerical computations? I had a short computer run in my answer.
(B) What is the strongest result one gets assuming a generalized/extended Riemann Hypothesis?
(C) What is the strongest unconditional  result?

REPLY [2 votes]: The $70$ comes from the claimed result of Oesterle mentioned here (which is supposed to use GRH only).<|endoftext|>
TITLE: Constructive Mathematics and Termination
QUESTION [15 upvotes]: In the 1988 book The Universal Turing Machine A Half-Century Survey
there is the paper "The Confluence of Ideas in 1936" by Robin Gandy. In section 4.2, Gandy writes:

"If one accepts, on whatever grounds, that a process terminates after a finite number of steps, then one should also accept, on the same grounds, that the number of steps can, in principle, be computed - one only needs a clock![8]"

Footnote [8] states:

"A constructive mathematician may have good grounds for believing that the process doesn't fail to terminate, without believing that it does terminate. Markov used his principle to jump over this gap."

My question is, can there exist a constructive logic and a Turing machine such that 
1) the logic proves that the Turing machine doesn't fail to always terminate and
2) the logic does not prove that the Turing machine does always terminate?

REPLY [8 votes]: Ingo Blechschmidt already explained in the comments why we should expect a negative answer to the question (for most readings of "constructive logic"). Namely, if classical arithmetic proves $\forall n \in \mathbb{N} . \exists k \in \mathbb{N} . \phi(n, k)$, where $\phi(n,k)$ is quantifier-free, then so does intuitionistic arithmetic. So then, if intuitionstic arithmetic proves $\lnot\lnot \forall n \in \mathbb{N} . \exists k \in \mathbb{N} . \phi(n, k)$, then so does classical arithmetic, but classically we may remove the $\lnot\lnot$, and then go back to intuitionistic logic to get $\forall n \in \mathbb{N} . \exists k \in \mathbb{N} . \phi(n, k)$. The statement "Turing machine $M$ halts on every input" is of this form, namely
$$\forall n \in \mathbb{N} . \exists k \in \mathbb{N} . T(m, n, k),$$
where $m$ is a code of $M$ and $T$ is Kleene's predicate $T$.
What I would really like to explain is that in a sense this is the wrong question to ask. Let $H(m)$ be the statement that the Turing machine encoded by $m$ always halts, i.e., $$H(m) \iff \forall n \in \mathbb{N} . \exists k \in \mathbb{N} . T(m, n, k).$$ In terms of $H$, the question is: "Is there a number $m$ such that intuitionistic logic proves $\lnot\lnot H(m)$ but does not prove $H(m)$?" This seems to indicate to me that the author of the question is trying to imagine how $$\forall m \in \mathbb{N} . (\lnot\lnot H(m) \Rightarrow H(m))$$ might fail, and he expects to be able to find an instance of $m$ in which the implication fails. But in intuitionistic logic this is not the right way to think of quantification and implication!
The classical reading of $\forall x \in A . \psi(x)$ is "$\psi(a)$ holds for every element $a \in A$", whereas the intuitionistic reading of the same statement is "there is a procedure which takes as input any $a \in A$ and outputs evidence of $\phi(a)$. Here the word "procedure" is not fixed: it cold mean a computable map, or a continuous map, or computable with respect to an oracle, etc. But the point is this: in intuitionistic logic $\forall x \in A . \psi(x)$ may fail because there is no procedure, and not because there is a specific $b \in A$ for which $\lnot \psi(b)$ holds.
Applying the last paragraph to Markov's principle, we see that the "correct" question to ask was:

Is there a procedure which takes as input (the code of) a Turing machine $M$ that never runs forever, and a number $n \in \mathbb{N}$, and halts and outputs the running time of $M(n)$?<|endoftext|>
TITLE: Homotopy types of schemes
QUESTION [16 upvotes]: Let $X$ be a scheme over $\mathbb{C}$. 

When does the topological space $X\left(\mathbb{C}\right)$ of $\mathbb{C}$-points have the homotopy type of a finite CW-complex?
When does the topological space $X\left(\mathbb{C}\right)$ of $\mathbb{C}$-points have the weak homotopy type of a finite CW-complex, (i.e. when is it a finite space)?

By "when", I mean what adjectives do I have to add to make this true, e.g. finite type, separated, smooth...

I'm also interested in question 1.) for when $X$ is affine.

If you happen to know a reference also, that would be fantastic. Thanks!
P.S. I'm aware of this mathoverflow question:
How to prove that a projective variety is a finite CW complex?
However, it addresses only the case of varieties, unless I am missing something..

REPLY [21 votes]: Any scheme which is separated of finite type, has at least a triangulation, hence is, in particular, a CW-complex. In fact, by a theorem of Lojasiewicz, this is true for any semi-algebraic set (one can even get this for subanalytic sets, by a result of Hironaka, in Triangulation 
of algebraic sets, Proc. Amer. Math. Soc. Inst. Algebra Geom. Arcata(1974)); however, the case of (possibly singular) algebraic varieties goes back to the early times of Algebraic Topology: e.g. these papers of van der Waerden and of Lefschetz and Whitehead). 
If you only are interested in weak homotopy types, it follows from Lurie's proper base change theorem that considering complex points satisfies proper (hyper)descent (this is Prop. 3.21 in this paper of A. Blanc, which is now published in Compositio Math.). Using Hironaka's resolution of singularities theorem, this implies that, for any scheme of finite type $X$, the space $X(\mathbf{C})$ is a finite homotopy colimit of spaces of the form $Y(\mathbf{C})$ with $Y$ affine and smooth (using Mayer-Vietoris-like homotopy pushouts associated to blow-ups and to coverings by Zariski open subschemes). A smooth affine algebraic variety has the homotopy type of a finite CW-complex: this follows from Morse theory, as can be seen from (the proof of) Theorem 7.2, page 39 in Milnor's book Morse Theory.
From all this, we get that a sufficient condition for $X(\mathbf{C})$ to have the weak homotopy type of a finite CW-complex is to be of finite type, while a sufficient condition to get the homotopy type of a finite CW-complex is to be separated of finite type. A sufficient condition to get an actual finite triangulated space is to be proper.
If we drop the assumption that the scheme is of finite type, I don't see how we can control/define what happens unless we work with pro-homotopy types of some sort.<|endoftext|>
TITLE: Obstructions to Picard-graded groups of maps
QUESTION [8 upvotes]: Suppose $(C,\odot,\Bbb I)$ is an additive category with a compatible symmetric monoidal structure and $Pic(C)$ is the group of isomorphism classes of objects which have an inverse under $\odot$. For $\alpha \in Pic(C)$, we would like define the $\alpha$'th coefficient functor by choosing a representative $P^\alpha \in C$ for the isomorphism class, and letting $\pi_\alpha X = Hom(P^\alpha,X)$.
In Section 14 of Morava K-theories and localisation, Hovey and Strickland discuss an obstruction to extending this to something compatible with the monoidal structure (possessing appropriate maps $\pi_\alpha X \otimes \pi_\beta Y \to \pi_{\alpha \odot \beta}(X \odot Y)$ satisfying all the desired naturality properties). They comment that there is an obstruction cocycle in $H^3(Pic(C), Aut(\Bbb I))$, and that if this obstruction vanishes then the possible choices of this natural isomorphism are parametrized by $H^2(Pic(C), Aut(\Bbb I))$. They did not know of an explicit reference for this fact.

Are now there references where this obstruction is worked out in detail?
This cohomology group is a group of Postnikov invariants, parametrizing spaces whose only two homotopy groups are $\pi_1$ and $\pi_2$. This is equivalent to the theory of (nonsymmetric) Picard groupoids: monoidal categories where all morphisms are invertible and where every object has an inverse under the monoidal structure. There is a natural candidate monoidal category: the category of tensor-invertible objects in $C$ and isomorphisms between them. What is the relationship between these two classes in $H^3$?
If the answer to (2) is roughly "these are the same", then because $C$ is symmetric monoidal this Postnikov invariant deloops, at least as far as an element of $H^5(K(Pic(C),3),Aut(\Bbb I))$. A back-of-the-envelope calculation seems to indicate that the obstruction must vanish because Postnikov invariants won't deloop at this low stage. Is this the case?
If we suspect that this is true, then: Is there an explicit proof using the symmetry isomorphism that Hovey-Strickland's obstruction cocycle must vanish? (If this does vanish, then the $H^5$-cocycle is still an obstruction to making this compatible with the symmetry isomorphism.)

REPLY [3 votes]: Let $\mathbf{Pic}(C)$ be the monoidal subgroupoid of tensor invertible objects and isomorphisms between them, so $\pi_0\mathbf{Pic}(C)=Pic(C)$. For the definition of those appropriate maps, it is clearly necessary and sufficient to choose representatives in such a way that we get a monodical splitting $Pic(C)\to \mathbf{Pic}(C)$ of the quotient map $\mathbf{Pic}(C)\to Pic(C)$. The obstruction to the definition of such a splitting is precisely the cohomology class $H^3(Pic(C),Aut(\mathbb I))$ you mention in 2, compare Joyal-Street. In the braided case, this class comes from $H^4$, and in the symmetric case from $H^5$, which is stable, as you say in 3. Therefore the class in $H^3$ vanishes in the symmetric case, compare this paper for a proof. You can in principle obtain the splitting from the symmetry using proofs therein, but the paper is written in terms of stricter but equivalent structures, so you'd have to strictify and go back and forth. It's also remarkable that the trivialisation seems to be unnatural in $C$.<|endoftext|>
TITLE: Fibrant-cofibrant models of Eilenberg-MacLane spectra
QUESTION [13 upvotes]: There are many models for spectra, by which I mean a model category whose homotopy category is triangulated-equivalent to the stable homotopy category. In each model, there are ways to construct Eilenberg-MacLane spectra $HA$, where $A$ is an abelian group. In $S$-modules, this is described in Section IV.2 of this version of EKMM. In symmetric spectra, this is described in Example 1.14 of Stefan Schwede's book project (version 3.0).

Question: Are there models of Eilenberg-MacLane spectra that are
  fibrant, cofibrant, and (strict) abelian group objects with respect to
  the addition map $+ \colon HA \times HA \to HA$?

My first candidate was symmetric spectra because there, the construction of $HA$ follows directly from a standard construction of Eilenberg-MacLane spaces $K(A,n)$ as topological abelian groups (or simplicial abelian groups, if working in simplicial sets). In particular, $HA$ is an abelian group object. Moreover, $HA$ is an $\Omega$-spectrum, and those are the fibrant objects in, for instance, the absolute projective stable model structure (Theorem III.4.11 in Schwede's book, or Theorem 3.4.4 in Hovey—Shipley—Smith). However, I'm still missing cofibrancy, and I suspect that a cofibrant replacement would mess up the abelian group object structure.
Another idea would be to use the various stable model structures on symmetric spectra. One could also try in $S$-modules, where every object is fibrant.
For the record, an associative smash product is not crucial to my purposes, though of course it would be nice.

REPLY [12 votes]: The following four categories are models for spectra with Eilenberg-MacLane spectra of the desired form.

Kan's category of semisimplicial spectra [1]
The category $\mathbf{Sp}^\mathbb{N}(\mathbf{\Sigma})$ of sequential spectra of pointed simplicial sets together with the Kan suspension $\Sigma$
The category $\mathbf{Sp}^\mathbb{N}(S^1\wedge -)$ of Bousfield-Friedlander spectra 
The category $\mathbf{Sp}^\mathbb{N}(\mathcal{T})$ of sequential spectra of pointed weak Hausdorff $k$-spaces $\mathcal{T}$

These four models are connected by Quillen equivalences
$$\mathbf{Sp}^\mathbb{N}(S^1\wedge -) \rightleftarrows  \mathbf{Sp}^\mathbb{N}(\mathcal{T}) \leftrightarrows \mathbf{Sp}^\mathbb{N}(\mathbf{\Sigma)} \rightleftarrows\{\text{semisimplicial spectra}\}$$
as described by Bousfield and Friedlander in [2]. (Strictly speaking, Bousfield and Friedlander work with the category of sequential spectra of pointed topological spaces instead of  $\mathbf{Sp}^\mathbb{N}(\mathcal{T})$, but the stable model structure and the corresponding Quillen equivalences also exist for $\mathbf{Sp}^\mathbb{N}(\mathcal{T})$.)
We elaborate on each model.

Semisimplicial spectra: Ken Brown equipped the category of semisimplicial spectra with a model structure in which every object is cofibrant and group objects are fibrant in [3]. Kan's stable Dold-Kan correspondence asserts that the category of abelian group objects of semisimplicial spectra is equivalent to the category of unbounded chain complexes of abelian groups. Considering an abelian group $A$ as an unbounded chain complex concentrated in degree zero yields under this correspondence an abelian group object $HA$ in semisimplicial spectra that models the Eilenberg-MacLane spectrum and that is both cofibrant and fibrant.
$\mathbf{Sp}^\mathbb{N}(\mathbf{\Sigma})$: There is a right Quillen equivalence $\mathrm{Ps}$ from the category of semisimplicial spectra to the category of sequential spectra $\mathbf{Sp}^\mathbb{N}(\mathbf{\Sigma})$ with the stable model structure. For any semisimplicial spectrum $X$, the structure maps of $\mathrm{Ps}(X)$ are monomorphism. Thus $\mathrm{Ps}(X)$ is cofibrant. Hence if $X$ is a group object (and therefore in particular fibrant), then $\mathrm{Ps}(X)$ is cofibrant, fibrant and a group object as well, since $\mathrm{Ps}$ is a right Quillen functor. For the semisimplicial spectrum $HA$ above, the sequential spectrum $\mathrm{Ps}(HA)$ is an Eilenberg-MacLane spectrum of the desired form and is explicitly given by the sequence of pointed simplicial sets 
$$A, \overline{W}A, \overline{W}\overline{W}A, \ldots, \overline{W}^n A,\ldots$$
where $A$ is considered as a constant simplicial abelian group and $\overline{W}$ is "dual" to the right adjoint of Kan's loop group functor.
Bousfield-Friedlander spectra: A model for an Eilenberg-MacLane spectrum of the desired form is given by the sequence
$$A, BA, BBA, \ldots, B^nA,\ldots$$
where $B$ is the classifying space functor given by the diagonal of the bar construction. The structure map $S^1\wedge B^n A\to B^{n+1}A$ in level $k$ is just the inclusion of the $k$-fold wedge of $(B^n A)_k$ into the $k$-fold product of $(B^n A)_k$. In particular, this model is cofibrant. One way to show that this model is fibrant is to note that it is precisely the Bousfield-Friedlander spectrum construction of the $\Gamma$-space associated to $A$.
$\mathbf{Sp}^\mathbb{N}(\mathcal{T})$: The left Quillen functor from any of the two categories of sequential spectra of pointed simplicial sets to $\mathbf{Sp}^\mathbb{N}(\mathcal{T})$ is induced by geometric realization. In particular, it preserves finite products and thus group objects. As a left Quillen functor, it preserves cofibrant objects. It preserves fibrant objects as well. Thus the left Quillen functor applied to any model of an Eilenberg-MacLane spectrum of the desired form in $\mathbf{Sp}^\mathbb{N}(\mathbf{\Sigma})$ or $\mathbf{Sp}^\mathbb{N}(S^1\wedge -)$ yields a model of the desired form in $\mathbf{Sp}^\mathbb{N}(\mathcal{T})$.

References:
[1] Kan, Semisimplicial spectra
[2] Bousfield and Friedlander, Homotopy theory of $\Gamma $-spaces, spectra, and bisimplicial sets
[3] Brown, Abstract homotopy theory and generalized sheaf cohomology<|endoftext|>
TITLE: Why does $GL_n(F)\backslash GL_n(\mathbb{A}_F)/\prod_xGL_n(O_x)$ classify vector bundles over $X$?
QUESTION [6 upvotes]: Suppose $X$ is an algebraic curve, $F$ its function field, $\mathbb{A}_F$ its adele, $O_x$ the ring of integers at local field of $x$.  Why does $GL_n(F)\backslash GL_n(\mathbb{A}_F)/\prod_xGL_n(O_x)$ classify vector bundles over $X$?

REPLY [5 votes]: This is a good exercise. Here's a big hint for one direction. Given a rank $n$ vector bundle, write down an isomorphism between it and the trivial rank $n$ vector bundle on the generic point of the curve (note: different choices will differ by an element of $GL_n(F)$). Now write down an isomorphism between it and trivial rank $n$ vector bundle at the completed local ring at every closed point~$x$ of the curve (note: different choices will differ by an element of $GL_n(O_x)$). Now at the generic point of each completed local ring we have an isomorphism between a trivialised bundle and itself, which is represented by an element of $GL_n(Frac(O_x))$ and now just put everything together to get the element of $GL_n(A_F)$.<|endoftext|>
TITLE: The Schwartz space is not normable
QUESTION [5 upvotes]: The Schwartz space of rapidly decreasing function (as well as their derivatives) on $\mathbb  R^n$ is a Fréchet space, whose (metric complete) topology is given by the usual countable family of semi-norms $(p_k)_{k\in \mathbb N}$
$$
p_k(\phi)=\max_{\vert \alpha\vert, \vert \beta\vert\le k}\Vert x^\alpha\partial_x^\beta
\phi\Vert_{L^\infty(\mathbb R^n)}.
$$
Is there a simple proof of the fact that this topological space is not normable?

REPLY [9 votes]: More elementary than Ascoli:
If it was normable, it would mean that there exists a norm $n$ which is continuous, hence for some $k$, $n(\phi) \leqslant C\ p_k(\phi)$ and which defines the topology, i.e. such that for all $k$, $p_k(\phi) \leqslant C_k\ n(\phi)$.
This would imply that all the norms $p_k$ are equivalent for $k$ large enough, which is not the case: for example $\exp(-|x|^2) \sin(N x_1) /N^k$ tends to $0$ for the first $k$ norms but not for the following.<|endoftext|>
TITLE: Parahoric Group Scheme
QUESTION [9 upvotes]: I am looking for the definition of a parahoric group scheme in the sense of Bruhat and Tits? I couldn't find a reference for that? at least a "clear" reference!
thanks

REPLY [4 votes]: What George McNinch has written is most helpful, but maybe I can point out an obstacle to locating the sort of precise reference you ask for.    The most relevant work of Bruhat-Tits, apart from surveys such as the one by Tits in English which George refers to, is found online at numdam.org, starting with their IHES papers I and II.   In each paper there are many separate occurrences of the terms schema and sous-groupes parahoriques but apparently no explicit formulation of the definition you want.  Another paper by them here may also be useful.    Their style however is quite formal, as you realize.   (I guess it's clear that the artificial term parahorique combines the notions of parabolic subgroup and Iwahori subgroup in the case of local fields.)<|endoftext|>
TITLE: Twists of cubic threefolds
QUESTION [12 upvotes]: Let k be a field of characteristic $0$.


Let $X$ be a variety over k which is isomorphic to a smooth cubic threefold over $\bar{k}$. Then is $X$ isomorphic to a smooth cubic threefold over $k$?


For motivation, let's consider some other cases. 

For cubic curves, the analogous question has a a negative answer in general, as elliptic curves can have quite complicated twists. 
For cubic surfaces however the answer is yes. This is because cubic surfaces are embedded by the anticanonical divisor, which is already defined over $k$.

This latter argument does not work however for cubic threefolds, as for a cubic threefold, the hyperplane section is half the anticanonical divisor. So the question is whether the anticanonical divisor is always divisible by $2$ in the Picard group of $X$.
Note that the Hochschild-Serre spectral sequence yields the exact sequence
$$0 \to \mathrm{Pic}(X) \to \mathrm{Pic}(X_\bar{k})^{\mathrm{Gal}(\bar{k}/k)}=\mathbb{Z} \to \mathrm{Br}(k),$$
so the obstruction to $X$ being isomorphic to a cubic threefold over $k$ is given by an element of $\mathrm{Br}(k)[2]$.

REPLY [7 votes]: As pointed out by Noam Elkies, my comment above can be turned into a positive answer to the question.
In the exact sequence

$\DeclareMathOperator{\Pic}{Pic}\DeclareMathOperator{\Br}{Br}0 \to \Pic X \to (\Pic \bar{X})^{G_k} \to \Br k$,

the arrow $\delta \colon (\Pic \bar{X})^{G_k} \to \Br k$ sends the class of an effective Cartier divisor $D$ to the class of the Severi--Brauer variety whose $\bar{k}$-points are isomorphic to the projectivisation of $\mathrm{H}^0(\bar{X},\mathcal{O}(-D))$.  See Serre, Local Fields, X.6 for a passing mention of this fact, and Grothendieck, Le groupe de Brauer, III, 5.4 for more details (though he doesn't actually prove it, saying "on (plus précisément, J. Giraud) vérifie que la classe de cet élément dans $\Br(Y)$ est bien $\delta(\xi)$.")
In the particular example here, $\Pic X$ is of index $2$ in $(\Pic \bar{X})^{G_k}$, so the image of $\delta$ is killed by $2$.  On the other hand, taking $D=-K_X/2$, the vector space $\mathrm{H}^0(\bar{X},\mathcal{O}(-D))$ has dimension $5$, since we are assuming that $\bar{X}$ is isomorphic to a cubic threefold, on which $D$ is the class of a hyperplane section.  It follows (for example Serre, loc. cit.) that $\delta(D) \in \Br k$ is killed by $5$.  So $\delta(D)=0$, and $D$ lies in $\Pic X$.<|endoftext|>
TITLE: Homotopies with prescribed regular values
QUESTION [12 upvotes]: Let $M_1$ and $M_2$ be connected smooth manifolds and let $f_0,f_1:M_1 \rightarrow M_2$ be homotopic smooth maps such that some fixed point $p \in M_2$ is a regular value for both $f_0$ and $f_1$.  Question: Can I always find a smooth homotopy $F:M_1 \times I \rightarrow M_2$ such that $p$ is a regular value for $F$?
This question came up in the smooth manifolds course I'm teaching right now.  If it had a positive (and relatively elementary) answer, then it would allow me to greatly simplify some of the proofs I'm getting ready to present.

REPLY [6 votes]: The answer is "yes" if $M_1$ is compact.  Here's a sketch of a proof.
Consider any smooth homotopy $G:M_1 \times I \rightarrow M_2$ between $f_0$ and $f_1$.  Since $M_1$ is compact, the set of regular values of $f_0$ and $f_1$ are both open.  Moreover, the condition of being a regular value is stable.  We can thus find some $\epsilon>0$ and some tiny open ball $U$ around $p \in M_2$ such that every point of $U$ is a regular value of the function $g_t:M_1 \rightarrow M_2$ defined via $g_t(x) = G(x,t)$ for all $t \in [0,\epsilon] \cup [1-\epsilon,1]$.
Using Sard's theorem, we can find some $q \in U$ such that $q$ is a regular value of $G$.  Choose some smooth function $H:M_2 \times I \rightarrow M_2$ with the following properties:

For $0 \leq t \leq 1$, the function $h_t:M_2 \rightarrow M_2$ defined via $h_t(x) = H(x,t)$ is a diffeomorphism supported on $U$.
$h_0 = \text{id}$ and $h_1(q)=p$.
$h_{t} = h_{t'}$ for all $t,t' \in I$ that are sufficiently close to $1$.

Now define a smooth function $F:M_1 \times I \rightarrow M_2$ via the following formula:
$$F(x,t) = \begin{cases}
G(h_{t/\epsilon}(x),t) & \text{if $t \in [0,\epsilon]$},\\
G(h_1(x),t) & \text{if $t \in (\epsilon,1-\epsilon)$},\\
G(h_{1-(t-1+\epsilon)/\epsilon}(x),t) & \text{if $t \in [1-\epsilon,1]$}.
\end{cases}$$
Thus $F$ is a smooth homotopy from $f_0$ to $f_1$.  The third condition above is needed to ensure that $F$ is smooth.  I claim that $p$ is a regular value of $F$.  Indeed, consider $(x,t) \in F^{-1}(p)$.  There are three cases:

If $t \in (\epsilon,1-\epsilon)$, then $(x,t) \in G^{-1}(q)$ and $F$ agrees with $G \circ h_1$ in a neighborhood of $(x,t)$.  Since $q$ is a regular value of $G$, it follows that $G$ is a submersion at $(x,t)$, and thus since $h_1$ is a diffeomorphism it follows that $F$ is also a submersion at $(x,t)$.
If $t \in [0,\epsilon]$, then setting $f_t(x) = F(x,t)$ we have that $f_t = g_t \circ h_{t/\epsilon}$.  Since $h_{t/\epsilon}$ is a diffeomorphism and $h_{t/\epsilon}^{-1}(p) \in U$ and every point of $U$ is a regular value for $g_t$, it follows that $f_t$ is a submersion at $x$.  This implies that $F$ is a submersion at $(x,t)$.
If $t \in [1-\epsilon,1]$, then an argument similar to the previous one shows that $F$ is a submersion at $(x,t)$.<|endoftext|>
TITLE: Fundamental group of the space of immersions of the 2-sphere in 3-space modulo diffeomorphisms of the first
QUESTION [13 upvotes]: In a previous Mathoverflow question, we saw that the fundamental group of the space $Imm(S^2,\mathbb{R}^3)$ of immersions the 2-sphere in ordinary 3-space is isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}$.
We also saw that rotating the sphere 360 degrees on itself gives an element of order $2$: what I mean, and more generally, if $f_0\in Imm(S^2,\mathbb{R}^3)$ is your basepoint and if $R_t\in SO(3)$ is a path which represents the non-trivial element in $\pi_1(SO(3))\equiv \mathbb{Z}/2$, then $f_0\circ R_t$ is a path which is of order $2$ in $\pi_1(Imm(S^2,\mathbb{R}^3))$.
Of course pre-composing an immersion by a diffeo does not change the image of the immersion. So the path above is not significant in terms of the image. It is interesting, instead of considering $Imm(S^2,\mathbb{R}^3)$, to look at the quotient space $Imm(S^2,\mathbb{R}^3)/Diff(S^2)$ where $Diff(S^2)$ is the space of self-diffeomorphisms of the 2-sphere. In fact we wish to retain the orientation of the sphere (to be able to label each face of the immersed spheres in a coherent way), hence I would rather look at
$$Imm(S^2,\mathbb{R}^3)/Diff^+(S^2)$$
where $Diff^+(S^2)$ denotes the set of orientation preserving diffeomorphisms of $S^2$.
I think the action of $Diff^+(S^2)$ on $Imm(S^2,\mathbb{R}^3)$ is free, i.e. that there is no immersion $f$ that is invariant by a non-trivial orientation preserving diffeo of $S^2$.
Is it true that we get a fibre bundle $Imm(S^2,\mathbb{R}^3) \mapsto Imm(S^2,\mathbb{R}^3)/Diff^+(S^2)$ with fiber $Diff^+(S^2)$?
Is it true that this leads to a short exact sequence $$1\to \pi_1 Diff^+(S^2)\to \pi_1 Imm(S^2,\mathbb{R}^3) \mapsto \pi_1 (Imm(S^2,\mathbb{R}^3)/Diff^+(S^2)) \mapsto 1$$
and that the latter is equivalent to
$$0\to \mathbb{Z}/2 \to \mathbb{Z}/2\times \mathbb{Z} \to \mathbb{Z} \to 0\ ?$$
Now if you insist on looking at $Imm(S^2,\mathbb{R}^3)/Diff(S^2)$, I would also be happy to know what is its fundamental group. Note that the double cover of Boy's surface is fixed by the antipodal map so the action is non-free, so we do not get a fibre bundle to begin with. However, the map
$$Imm(S^2,\mathbb{R}^3)/Diff^+(S^2) \to Imm(S^2,\mathbb{R}^3)/Diff(S^2)$$ exists (it is continuous but not a cover) and we may wonder about its action at the level of the respective fundamental groups. It is interesting to note that eversions are loops in the space on the right but not on the space on the left.

REPLY [10 votes]: The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free. 
The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.
It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration. 
Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective homomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)
e: $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ is the quotient of $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$ by the involution given by precomposing with the antipodal map. The fixed point set is of infinite codimension. Its complement, then, by transversality arguments, has the same fundamental group; and away from the fixed point set this is a covering map. So we obtain that $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ has fundamental group $\Bbb Z$, and the quotient map induces multiplication by two on the fundamental group.<|endoftext|>
TITLE: Infinite topological direct sum of amenable Banach algebra
QUESTION [6 upvotes]: Is an infinite topological direct sum of amenable Banach algebras amenable again?
Can you give me a good reference about this notion?
Thanks

REPLY [8 votes]: The OP has now clarified that he or she is asking about $c_0$-direct sums. To answer the question we need the notion of the amenability constant of a Banach algebra. Everything that follows is probably "folklore", in the sense that
(i) Johnson knew how to do it in 1972 but, due to the different culture at the time of mathematical publication, did not bother formally writing it down in his 1972 monograph Cohomology in Banach Algebras or his Amer J. Math article of the same year;
(ii) successive generations of researchers, who take seriously the study of amenability of Banach algebras, will almost surely have worked these results out for themselves during their own studies.
Actually, this result might also be somewhere in Runde's book, but I haven't checked. (UPDATE 2015-11-10: Tomasz Kania has pointed out that this is Corollary 2.3.19 in Runde's Lectures on Amenability.)
Definition. Given an amenable Banach algebra $B$, the amenability constant of $B$ is defined to be 
$${\rm AM}(B):=\inf \{ \Vert M \Vert_{(B\widehat\otimes B)^{**}}\;\; \colon \;\;\hbox{$M$ is a virtual diagonal for $B$}\}. $$
(By weak-star compactness, the infimum is actually attained.) We adopt the convention that if $B$ is a non-amenable Banach algebra, ${\rm AM}(B)=+\infty$.
Remark 1.
It is an easy exercise to show that if $q:A\to B$ is a surjective homomorphism of Banach algebras then ${\rm AM}(B)\leq \Vert q\Vert^2{\rm AM}(A)$.
Lemma 2. Suppose $A$ is a Banach algebra and $(B_i)_{i\in \Lambda}$ is a family of closed subalgebras such that $\bigcup_{i\in\Lambda} B_i$ is dense in $A$. Then ${\rm AM}(A)\leq \sup_{i\in\Lambda} {\rm AM}(B_i)$.
Note that since closed subalgebras of ${\bf M}_2({\mathbb C})$ can be non-amenable, the inequality in this lemma is usually not an equality.
Outline of the proof of Lemma 2. We may assume the RHS of the inequality is finite, bounded above by $C$ say; otherwise there is nothing to prove. Consider $(M_i)_{i\in\Lambda}$ where $M_i$ is a virtual diagonal for $B_i$ such that ${\rm AM}(B_i)=\Vert M_i \Vert$. Then $(M_i)$ is a net in $(A\widehat{\otimes} A)^{**}$, bounded by $C$; take any weak-star cluster point $M$. Then $\Vert M\Vert\leq C$ and one can check $M$ is indeed a virtual diagonal for $A$. QED.
Lemma 3.
Let $A_1$ and $A_2$ be Banach algebras, and let $B=A_1\oplus_\infty A_2$ i.e. with the norm $\Vert (a_1,a_2)\Vert := \max (\Vert a_1\Vert, \Vert a_2\Vert)$.
Then ${\rm AM}(B)= \max \{{\rm AM}(A_1), {\rm AM}(A_2)\}$.
Proof of lemma 3. Since there are contractive, surjective algebra homomorphisms $q_j:B\to A_j$ ($j=1,2$) we have ${\rm AM}(B)\geq \max \{{\rm AM}(A_1), {\rm AM}(A_2)\}$ by Remark 1. Conversely, using the contractive embedding
$$ (A_1\widehat\otimes A_1) \oplus_\infty (A_2\widehat\otimes A_2) \to (A_1\oplus_\infty A_2)\widehat\otimes(A_1\oplus_\infty A_2) $$
and the corresponding embedding at the level of biduals, we can combine a virtual diagonal $M_1$ for $A_1$ and a virtual diagonal $M_2$ for $A_2$ to get a virtual diagonal $M$ for $A$, which satisfies $\vert M \Vert \leq \max( \Vert M_1\Vert, \Vert M_2\Vert)$. QED.
Proposition. Let $(A_n)_{n=1}^\infty$ be a sequence of Banach algebras, and let $A=c_0{\rm-}\bigoplus_{n=1}^\infty A_n$. Then 
${\rm AM}(A) = \sup_n {\rm AM}(A_n) <\infty$.
Proof that LHS ≥ RHS. This is like the proof of Lemma 2, using the contractive surjective homomorphisms $A\to A_n$ for $1\leq n\leq\infty$, and I omit the details.
Proof that LHS ≤ RHS. Let $C=\sup_n {\rm AM}(A_n)$; we may assume $C<\infty$, otherwise there is nothing to prove. Now let $B_n = A_1 \oplus \dots \oplus A_n$. By Lemma 3 and induction we know that ${\rm AM}(B_n)\leq C$, for each $n$. Since $\bigcup_{n\geq 1} B_n$ is dense in $A$, applying Lemma 2 completes the proof. QED.

OLD answer, before the question was clarified
Tomek Kania has already pointed out the key "counterexample" (assuming that by infinite topological direct sum you mean what I would call the $\ell^\infty$-direct sum).
I should also point out that there are commutative counterexamples. Indeed, let $A={\ell^\infty}{\rm-}\bigoplus \ell^1({\bf T}_d)$ where ${\bf T}_d$ denotes the unit circle equipped with the discrete topology, and the multiplication on $\ell^1({\bf T}_d)$ is given by convolution. Then, as observed in e.g. Proposition 5.8 of http://arxiv.org/abs/0708.4195 $A$ quotients onto $M(G_0)$ where $G_0$ is the Bohr compactificaton of ${\bf T}_d$.
If $A$ were weakly amenable, then $M(G_0)$ would be weakly amenable; but by results of Brown and Moran $M(G_0)$ has non-zero point derivations, hence is not weakly amenable. Therefore $A$ is not weakly amenable, and in particular cannot be amenable.<|endoftext|>
TITLE: stable homotopy groups and zeta function
QUESTION [18 upvotes]: I have heard during a discussion that there is a well known relation between the stable homotopy groups of a sphere (more precisely the order of stable homotopy groups of localized sphere spectrum with respect to some homology theory $E$) and the values of the zeta function at some integers.
$$ |\pi_{i}^{s}L_{E}\mathbb{S}| =^{?} \zeta(-n)$$
I'm not sure that I understood well, I will be glad if someone can explain this relation.

REPLY [12 votes]: Here is a slightly more fleshed out version of my comment. Let $K(1)$ be the first Morava $K$-theory. When $p$ is odd one can calculate the homotopy groups of the $K(1)$-localised sphere spectrum to be 
$$
\pi_nL_{K(1)}\mathbb{S} = \begin{cases}
\mathbb{Z}_p, &n=0,1\\
\mathbb{Z}/p^{\nu_p(t')+1} & n=2(p-1)t'-1, t' \in \mathbb{Z}.
\end{cases}
$$
Here $\nu_p(x)$ is the $p$-adic valuation of $x$. 
Following Adams define a function $m(l)$ by 
$$
\nu_p(m(l)) = \begin{cases}
0 & l \not \equiv 0 \mod (2(p-1)) \\
1+ \nu_p(l) & l \equiv 0 \mod (2(p-1)).
\end{cases}
$$
Adams shows (following Milnor and Kervaire) that $m(2s)$ is the denominator of $\beta_{2s}/4s$, where $\beta_s$ is the $s$-th Bernoulli number, and the fraction is expressed in the lowest possible form. 
Using standard properties of $\nu_p(x)$ there is an equivalence $\nu_p(t')+1  = \nu_p((n+1)/2)+1$. Since $(n+1)/2 \equiv 0 \mod (2(p-1))$ we see
$$
\nu_p(m\left(2\cdot \frac{n+1}{4}\right)) = \nu_p((n+1)/2)+1 
$$
and so the order of $\pi_nL_{K(1)}S^0$ is the denominator of $\beta_{(n+1)/2}/(n+1)$. 
Edit: Let me try and say something about the image of $J$ then. This is a homomoprhism $J:\pi_nSO \to \pi_n\mathbb{S}$. When $n=4k-1$ the order of the image of $J$ is cyclic of order the denominator of $\beta_{2k}/4k$.  Let $\text{Im}(J_n)_p$ denote the image of the composite  $\pi_nSO \to \pi_n\mathbb{S} \to \pi_n \mathbb{S}_{(p)}$. I believe this is meant to be isomorphic to $\pi_nL_{K(1)}\mathbb{S}$ for $n>1$.<|endoftext|>
TITLE: Diameter of hyperbolic 3-manifolds
QUESTION [10 upvotes]: Is there a good method to estimate the diameter of a closed hyperbolic 3-manifold?
I am particularly interested in know the diameter of the Weeks manifold.

REPLY [7 votes]: On page 356 of his paper The ortho-length spectrum for hyperbolic 3-manifolds, Meyerhoff shows that if $M$ is a closed hyperbolic $3$-manifold of volume $V$ then its diameter $\mathrm{diam}(M)$ satisfies $$\mathrm{diam}(M) < \frac{V}{\pi\sinh^2(\ell/4)},$$
where $\ell$ is the real length of the shortest geodesic of $M$. 
In their paper Symmetries, Isometries and Length Spectra of Closed Hyperbolic Three-Manifolds, Hodgson and Weeks compute the initial part of the length spectra of the ten smallest closed hyperbolic $3$-manifolds. If I am reading their Table 1 correctly, for the Weeks manifold you should have $\ell=0.58460369$. The volume of the Weeks manifold is $V=0.94270736$, hence we conclude from Meyerhoff's inequality that $$\mathrm{diam}(M)<13.9487057150346
.$$<|endoftext|>
TITLE: Local cohomology groups and linearity
QUESTION [7 upvotes]: I am reading local cohomology and am confused on a silly point. Let $U$ be an affine, non-singular variety and $Z \subset U$ a hypersurface section on $U$ (i.e., complete intersection in $U$ of codimension $1$). We know that we have an exact sequence, $$H^0(\mathcal{O}_U) \to H^0(U-Z,\mathcal{O}_U|_{U-Z}) \xrightarrow{\delta} H^1_Z(\mathcal{O}_U)$$
My question is: Is $\delta$ a $\mathcal{O}_U(U)$-linear morphism? If so, then it seems that $H^1_Z(\mathcal{O}_U) \otimes_{\mathcal{O}_U(U)} \mathcal{O}_Z(Z)$ should be zero. Is this correct? Am I missing something?

REPLY [4 votes]: Please compute through an example before asking! What happens if $U = \text{Spec}(k[x])$ and $Z = V(x)$?
Later edit. The answer to your question is that if $A$ is a ring and $f \in A$ an element, and $M$ an $A$-module, then we can look at the sequence
$$
0 \to M[f^\infty] \to M \to M_f \to M_f/M \to 0 \to 0 \to \ldots
$$
This sequence will be isomorphic, as a sequence of $A$-modules, to the sequence
$$
0 \to H^0_Z(X, \mathcal{F}) \to H^0(X, \mathcal{F}) \to H^0(U, \mathcal{F}) \to
H^1_Z(X, \mathcal{F}) \to H^1(X, \mathcal{F}) \to \ldots
$$
where $X = \text{Spec}(A)$, $Z = V(f)$, $U = X \setminus Z$, and $\mathcal{F}$ is the quasi-coherent $\mathcal{O}_X$-module associated to $M$. Since the functor $\otimes_A A/fA$ is right exact and since $M_f \otimes_A A/fA = 0$ we find that the tensor product you wrote down is zero.
In the above I am using cohomology with supports $H^i_Z$. This can be defined for an arbitrary closed subset $Z$ of a topological space $X$. This then gives me long exact sequences relating $H^i_Z(X, -)$, $H^i(X, -)$, $H^i(U, -)$. If $-$ is an abelian sheaf, then these are long exact sequences of abelian groups. If $-$ is a sheaf of $\mathcal{O}_X$-modules, then these are long exact sequences of $\Gamma(X, \mathcal{O}_X)$-modules. You can prove this by taking a resolution by injectives in the category of $\mathcal{O}_X$-modules and using general theorems saying that such resolutions compute the right thing (i.e., the same as what you get for abelian sheaves). Or you can use the functoriality of the construction (because multiplying by an global section of the structure sheaf commutes with everything in sight).<|endoftext|>
TITLE: Lattice random walk under gravity
QUESTION [12 upvotes]: Suppose a random walk on $\mathbb{Z}^2$ takes a step left or right with
probability $\frac{1}{4}$, but up with probability $\frac{1}{2} p$
and down with probability $\frac{1}{2} (1-p)$, where $p \in [0,1]$ indicates
a constant bias against north when $p < \frac{1}{2}$.
(So we have $\frac{1}{4}+\frac{1}{4}+ p \frac{1}{2}+ (1-p)\frac{1}{2}=1$.)
For example, here are two $10^5$-step random walks with $p=0.49$:

          


          


          

The red dot marks the starting origin.



What is the probability of returning to the origin for $p < \frac{1}{2}$?

It is not $0$, and I don't think Polya's recurrence theorem
holds unless $p = \frac{1}{2}$.
Added. Here is a distribution of path endpoints,
again using $p=0.49$, and paths of $10^5$ steps:

          


          

$500$ random walk endpoints, with $p=0.49$ bias against north.

REPLY [20 votes]: Let $E_n$ denote the probability of the first return after $n$ steps, and let $P_n$ denote the probability of return after $n$ steps (but not necessarily the first return).  Note that $E_n=P_n=0$ if $n$ is odd.
 We are interested in $E= E(p)= \sum_{n=1}^{\infty} E_n$.   Now note that for $n\ge 1$
 $$
 P_n = E_n + \sum_{k=1}^{n-1}  E_k P_{n-k},
 $$ 
 from which it follows that 
 $$ 
 P = \sum_{n=1}^{\infty} P_n = \sum_{n=1}^{\infty} E_n + \sum_{k=1}^{\infty} E_k \sum_{j=1}^{\infty} P_j,
 $$ 
 or in other words $P= E(1+P)$, or 
 $$ 
 E= \frac{P}{1+P}. 
 $$ 
 All this is due to Polya, of course. 
Now $P_n$ is quite easy to calculate:  for odd $n$ it is zero, and for even $n$ it is simply (counting $a$ steps to the right and left, and $b$ steps up and down)
 $$ 
 P_n  = \sum_{a+b=n/2} \frac{n!}{a!^2 b!^2} \Big(\frac 14\Big)^a \Big(\frac 14\Big)^a \Big(\frac 12p \Big)^b \Big(\frac 12 (1-p)\Big)^b.
 $$ 
 We can also express $P_n$ as a double integral: 
 $$ 
\frac{1}{(2\pi )^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \Big( \frac 14 e^{i\alpha} + \frac 14 e^{-i\alpha} + \frac 12p e^{i\beta} + \frac 12 (1-p) e^{-i\beta} \Big)^n d\alpha d\beta.
$$
From this we can see that 
$$ 
1+P= \frac{1}{(2\pi)^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \frac{2}{2-(\cos \alpha + \cos \beta + i(2p-1)\sin \beta)} d\alpha d\beta. 
$$ 
Note that when $p=\tfrac 12$ this integral diverges, and we recover Polya's theorem that $E=1$.  For any given $p<\tfrac 12$ we may clearly calculate the integral.  Moreover, it is not hard to get asymptotics from the above, and show that  $P$ behaves like $\log (1-2p)^{-1}$ for $p$ near $\tfrac 12$ (forgetting constants here).<|endoftext|>
TITLE: The use of modules in control theory
QUESTION [5 upvotes]: So far I have seen the use of vector spaces in control theory and other notions from linear algebra; So I wonder if there's a use of this abstraction of modules over rings in control theory? any literature you can suggest?
Thanks.

REPLY [7 votes]: Modules theory over the rings of principal ideals is the main tool of control theory.
Linear control theory mainly deals with matrices whose entries are polynomials.
Polynomials is the ring of principal ideals. Matrices over this ring require modules theory.
This fact is usually hidden in the books of control theory written for engineers,
because modules (unfortunately) are not a part of the standard engineering math curriculum.
These textbooks (on my opinion) only obscure the matter.<|endoftext|>
TITLE: $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$
QUESTION [90 upvotes]: Are there abelian groups $A$ with $A \cong A \oplus \mathbb{Z}^2$ and $A \not\cong A \oplus \mathbb{Z}$?

REPLY [2 votes]: Also not an answer, but the following Boolean algebra gives a similar example of William Hanf.
Let B be the two element boolean algebra, and we will use an uncountable subuniverse of a countable power of B.  Let A be the subalgebra with subuniverse $\{\sigma \in \{0,1\}^{\omega}:$ there is $n$ so that for all $k \geq n, \sigma(2k)=\sigma(2k+1)\}$.  Then A Is isomorphic to A $\oplus$ B $\oplus$ B , but not to A $\oplus$ B .
This suggests putting $Z$ in for B and seeing what comes of it.
For more details, consult Algebras, Lattices, Varieties Volume I of McKenzie, McNulty, and Taylor, chapter 5 section 5.1 exercise 9 (p. 267 in my copy).  I do not know what happens if the signature is restricted to that of Abelian groups.
Gerhard "Not Feeling That Ambitious Today" Paseman, 2016.01.02<|endoftext|>
TITLE: Proof for new deterministic primality test
QUESTION [25 upvotes]: Claim:

Let $p$ be a positive prime. Let $n \in \left\{1, 2, 3, ...\right\}$. Then $N =
p\cdot 2^n+1$ is prime, if and only if it holds the congruence $3^{(N-1)/2} \equiv \pm1\ ($mod $N)$.

If the claim is true, we would have a fast deterministic test for numbers of the form $p\cdot2^n + 1$. That means, with small $p$ and large $n$, we could generate huge prime numbers, similar to Mersenne primes or Fermat primes. 
A proof is needed. Thanks for Your attention.

REPLY [6 votes]: I thought I could prove it in the case of a negative sign but I can only show that in this case, for fixed $n$, there can be only finitely many counterexamples. Nothing magical about $3$ by the way.
Let $p$ be prime and $n$ an integer such that $N=2^np + 1$ is such that, for some integer $a$ we have $a^{(N-1)/2} \equiv -1 \pmod N$. Then $N$ is prime or $p \le a^{2^{n-1}}/2^n$.
Proof: Let $m$ be the order of $a$ modulo $N$, then $m | 2^np$, so $m = 2^k$ or $2^kp$ for some $k \le n$. Since we have $-1$ in the congruence in the hypothesis, we conclude that $k=n$. If $m = 2^np$, then $N$ is prime ($\phi(N)=N-1$ iff $N$ is prime). The only other possibility is $m=2^n$. Assume that's the case. Then $2^n | \phi(N)$ but we cannot have $p|\phi(N)$ as that would force $\phi(N) \ge N-1$. So $(p,\phi(N))=1$ and the congruence $a^{(N-1)/2} \equiv -1 \pmod N$, then implies that $a^{2^{n-1}} \equiv -1 \pmod N$. So $N \le a^{2^{n-1}} + 1$ giving the result.<|endoftext|>
TITLE: Is the free product $\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}$ linear over $\mathbb{Z}$?
QUESTION [9 upvotes]: Let $H:=\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}=\langle p,q| q^n=1\rangle.$ I want to know if $H$ is a ($\mathbb{Z}$)linear group that is to say is there an injective homomorphism $f: H\to GL_m(\mathbb{Z})$ for $m\geq n.$
I asked the question on Math Stack Exchange (https://math.stackexchange.com/questions/1430677/is-the-free-product-mathbbz-mathbbz-n-mathbbz-a-linear-group) and Derek Holt suggested me to also ask it here.
By advance thank you.

REPLY [14 votes]: The group $Z*Z/n$ is virtually free: the kernel $K$ of the projection to $Z/n$ is free of index $n$ (say by the Kurosh theorem and since each finite order element is conjugate to an element of $Z/n$). Since a free group has a faithful representation over $Z$ of degree 2, the induced representation of this representation gives a faithful representation over $Z$ of your group of degree $2n$.<|endoftext|>
TITLE: How do the direct and inverse image sheaf functors interact with homotopy?
QUESTION [10 upvotes]: This is a crosspost of this MSE question.
The direct image sheaf functor $f_\ast$ and inverse image sheaf functor $f^\ast$ (here I mean the usual inverse image sheaf functor often denoted by $f^{-1}$) form a well-known adjunction for $\mathsf{Set}$-valued sheaves (I think also for sheaves valued in algebraic categories).

Is there anything interesting to be said about $f^\ast,g^\ast$ and
  $f_\ast,g_\ast$ when $f\simeq g$? What about $f_!,g_!$ and $f^!,g^!$?

REPLY [2 votes]: By naturality, it suffices to consider the projection map $p : X \times I \to X$.  If $X$ and $I$ are locally conctractible Hausdorff spaces and $\mathcal{F}$ is a sheaf of abelian groups on $X$, then the Vietoris-Begle theorem (Bredon, Sheaf theory, II.13) states that the unit $\mathcal{F} \to p_* p^* \mathcal{F}$ is an isomorphism. There are also cohomological and Ext-functor versions of this result. See here in section 2.d.<|endoftext|>
TITLE: Hopf Galois extensions and conditional expectations for C* algebras
QUESTION [7 upvotes]: Suppose that $H$ is a Hopf algebra with normalised invariant integral (appropriate side) $\int:H\to \mathbb{C}$. The $H$ right comodule algebra $P$ is a Hopf Galois extension, so the canonical map $P\otimes_A P\to P\otimes H$ is a 1-1 correspondence, where $A$ is the $H$ invariant part of $P$. 
There is an averaging map $E:P\to A$ given by $E(p)=p_{(0)} \ \int(p_{(1)})$ which gives a projection to the subalgebra $A$ and is an $A$ bimodule map.
Now let us also suppose that $P$ is a $C^*$ algebra, and that $H$ is a Hopf star algebra with whatever other properties we need, and that the integral preserves positivity. Just when is $E$ a conditional expectation for $C^*$ algebras, i.e. when does it preserve positivity? Any nice sufficient conditions would be interesting. 
Apologies if I have missed something obvious in the literature...

REPLY [2 votes]: First off, as you can see, the definition of $E$ doesn't require the Hopf-Galois condition. The positivity is a consequence of that for slicing by states. If $\phi$ is a state on a C$^*$-algebra $C$, the map $\iota\otimes\phi\colon B \otimes_{\rm min} C \to B$ characterized by $b \otimes c \mapsto \phi(c) b$ is (completely) positive, as the GNS representation for $\phi$ provides a Stinespring factorization of $\iota\otimes\phi$. Now, with $\phi = \int$ being the Haar state of a compact quantum group, $E$ can be presented as the composition of $(\iota \otimes \phi)$ and the coaction $*$-homomorphism $P \to P \otimes_{\rm min} H$ which is (completely) positive. So $E$ is (completely) positive as expected for a conditional expectation.<|endoftext|>
TITLE: Class of hypergraphs that are always the neighborhood hypergraph of some simple graph
QUESTION [9 upvotes]: Let $G=(V,E)$ be a graph. Its (open) neighborhood hypergraph $\mathcal{H}(G)$ has the same vertex set $V$ with a hyperedge for the (open) neighborhood of every vertex $v \in V$. 
It seems that not every hypergraph is the (open) neighborhood hypergraph of some graph. Indeed, it is NP-complete to decide if a given hypergraph is the (open) neighborhood hypergraph of some graph.
This makes me wonder: what kind of hypergraphs are always the (open) neighborhood hypergraphs of some graph? It would be particularly interesting to know the weakest possible structural requirements imposed on the class of hypergraphs such that each hypergraph belonging to the class is still a neighborhood hypergraph of some simple graph. Are there some examples at least?

REPLY [5 votes]: Here is another way of thinking about the problem. Suppose for simplicity that your hypergraph $\mathcal{H}$ has exactly $|V(\mathcal{H})|$ hyperedges (as was mentioned by Dominic, we can immediately exclude any hypergraph with more hyperedges). 
Let $G'$ be a bipartite graph where both parts are of size $|V(\mathcal{H})|$. We associate the left side of $G'$ with the vertices of $\mathcal{H}$ and, for each edge of $\mathcal{H}$, we make a vertex on the right side whose neighbourhood into the left side is exactly that hyperedge.
Now, if there is an automorphism $\phi$ of $G'$ such that $\phi^2$ is the identity, the vertices on the right are mapped to those on the left, and no vertex is mapped to a neighbour of itself, then identifying each vertex $v$ on the left with its image $\phi(v)$ gives you the desired graph $G$. In fact, when $\mathcal{H}$ has exactly $|V(\mathcal{H})|$ hyperedges, this is an "if and only if."
Thinking of if this way leads to some obvious conditions. For example, the degree sequence of the right side must be the same as the left side, etc.
If $\mathcal{H}$ has fewer than $|V(\mathcal{H})|$ hyperedges, then one can still construct $G'$, but will have to duplicate some vertices on the right side before trying to find the isomorphism. This seems to make things more complicated. 
This brings up another question. You mention that the decision problem is NP-complete in general. However, is it possible that the problem becomes tractable for hypergraphs $\mathcal{H}$ with exactly $|V(\mathcal{H})|$ hyperedges? This seems to be related to the Graph Isomorphism Problem, but perhaps easier. If it is tractable for this class, then the difficulty in general must come down to finding the right way of duplicating vertices on the right side of $G'$.
By the way, this type of problem can have some very nice applications. For example, there is a paper of Labeznik, Ustimenko and Woldar (1999) where they obtain a new lower bound on the Turán number of the $6$-cycle by showing that a special class of hypergraphs, known as "generalised polygons," can be represented by the neighbourhood hypergraph of a graph (perhaps with loops). Their approach is pretty much exactly what I described above, but is presented in different language.<|endoftext|>
TITLE: Do one-relator groups satisfy Haagerup property?
QUESTION [6 upvotes]: The question is in the title:

Do one-relator groups satisfy Haagerup property?

I think the answer is known at least in some specific cases, but is the problem completely solved?

REPLY [2 votes]: As far as I know, the problem is not solved, but much is known. The canonical reference is
the suggestively named Groups with Haagerup property: Gromov's a-T-menability, by Cherix, Cowling, Jolissant, Julig, Valette. See, in particular, Theorem 6.3.1 and Chapter 7<|endoftext|>
TITLE: Are compact, complex, affinely flat manifolds geodesically complete?
QUESTION [7 upvotes]: Let $M$ be a real, even dimensional, compact manifold endowed with a symplectic form  $\omega$ and a flat, torsionless connection $\nabla$ compatible with $\omega$, that is $$\nabla \omega=0.$$
Under this assumptions, taking into account a simplified version of the  Markus conjecture, it should follow that $M$ is geodesically complete.
My question is: Under what additional geometric assumptions is the result known to be true?
One concrete example is: If in addition we assume the existence of a parallel almost complex structure $J$ on $M,$ then the result is true for $2$ dimensional and $4$ dimensional manifolds as observed by Misha.
What if instead of the the parallel almost complex structure we assume that there is a nonzero parallel vector field $V$ on $M?$ Does it follow that the manifold is geodesically complete?

REPLY [4 votes]: This is an answer to the last question of the revised version of your question which is 
"What if instead of the the parallel almost complex structure we assume that there is a nonzero parallel vector field V on M? Does it follow that the manifold is geodesically complete?"
The answer is the same as to the   question without the assumption. Indeed, having an incomplete manifold with $\omega$ parallel with respect to a flat torsionsfree connection, we can directly multiply it by $R^2$ with the standard connection and the standard symplectic structure.  The resulting  manifold remains   incomplete but now it has even two parallel vector fields.<|endoftext|>
TITLE: Is there an easier proof to show that the closed convex hull of a normalized weakly null sequence is weakly compact?
QUESTION [6 upvotes]: In a paper that I am reading there is a following step:

Let $X$ be a Banach space and let $(x_k) \subset X$ be a normalized sequence that converges weakly to $0$.
  Then $\overline{co}(x_k)$ is a weakly compact set.

(notice that $\overline{co}(x_k)$ denotes the norm-closure of the convex hull of $(x_k)$.)
I think that I managed to prove the claim, but I had to do a lot of manual checking. My line of thought is given below.
My question is: Can this be proved more directly than I did (assuming that my proof is without error)? For example, is closed convex hull of a weakly compact set always weakly compact? (similar question was asked here, but in a bit different context, so it does not seem to apply to this situation: Convex hulls of compact sets).
I reasoned as follows:

$\{x_k \mid k \in \mathbb{N} \}$ is an weakly compact set.
I checked that given a family $(y_{\alpha}) \subset co(x_k)$, it contains an weakly convergent subfamily, which weakly converges to an element in $\{ \sum_{k=1}^{\infty} {\alpha_k x_k \mid (\alpha_k) \in B_{\ell_1} }\}$.
I observed that given any family $(y_{\alpha})_{\alpha \in I} \subset \overline{co}(x_k)$, I can construct a family $(z_{\alpha, \epsilon})_{\alpha \in I, \epsilon > 0} \subset {co}(x_k)$ such that $\forall \alpha \in I, \epsilon > 0$ we have that $\lVert y_{\alpha} - z_{\alpha, \epsilon} \rVert < \epsilon$. By defining order for family $z_{\alpha, \epsilon}$ in such a way that $(\alpha_1, \epsilon_1) \leq (\alpha_2, \epsilon_2) \Leftrightarrow \alpha_1 \leq \alpha_2$ and $\epsilon_1 \geq \epsilon_2$, I can verify that (assuming that I did not make a mistake):
\begin{equation*}
(y_{\alpha}) \text { converges weakly to } w \Leftrightarrow (z_{\alpha, \epsilon}) \text{ converges weakly to } w
\end{equation*}
Since $(y_{\alpha}) \subset \overline{co}(x_k)$, then $(z_{\alpha, \epsilon})_{\alpha \in I, \epsilon > 0} \subset {co}(x_k)$. The latter family contains an weakly convergent subfamily $(z_{\beta}')$, which converges to an element $c \in \{ \sum_{k=1}^{\infty} {\alpha_k x_k \mid (\alpha_k) \in B_{\ell_1} }\} \subset \overline{co}(x_k)$. Therefore the original family $(y_{\alpha}) \subset \overline{co}(x_k)$ can also be shown to have a subfamily $(y_{\gamma}')$ which converges weakly to the same element $c$.
Therefore $\overline{co}(x_k)$ is a weakly compact set.

REPLY [5 votes]: Rauni’s link to the Freeman et al paper perhaps provides sufficient excuse for me to give here a simple proof (which, however, uses some theory) of the theorem in that paper. The background can more or less be found in Diestel’s book “Sequences and Series in Banach Spaces”, Springer GTM 92.   The theorem says that if every weakly compact subset of the Banach space $X$ is in the closed convex hull of a weakly null sequence, then $X$ is a Schur space (which means that every weakly convergent sequence in $X$ is norm convergent).  If $X$ has the property but is not Schur, then there is a normalized weakly null sequence $(y_n)$ in $X$ that, by passing to a subsequence, can assumed to be basic.  As in my other answer to this MO problem, let $T:\ell_1 \to X$ be the weak$^*$ to weak continuous bounded linear operator that maps the $n$th unit vector in $\ell_1$ to $y_n$. The operator $T$ is weakly compact but not compact, and $T$ is injective because $(y_n)$ is basic.  By the  factorization theorem for weakly compact operators, $T= JS$ where $S:\ell_1 \to R$, $J:R\to X$ are injective bounded linear operators and $R$ is reflexive. We can now forget about $T$ and focus on $J$. Necessarily $J$ is not compact. But the image $JB_R$ of the closed unit ball $B_R$ of $R$ is weakly compact, so by hypothesis  is in the closed convex hull of a weakly null sequence $(x_n)$ in $X$.  Again, the sequence $(x_n)$ is the image under a weak$^*$ to weak continuous bounded linear operator $U: \ell_1 \to X$ that maps the $n$th unit vector in $\ell_1$ to $x_n$. If $U$ were injective we would be done since $U^{-1}J$ would be a necessarily non compact bounded linear operator from a reflexive space into $\ell_1$, which is impossible because $\ell_1$ has the Schur property.  The same contradiction is available as long as $\ell_1/U^{-1}(0)$ has the Schur property.  But since $U$ is weak$^*$ to weak continuous, the kernel $U^{-1}(0)$ of $U$ is weak$^*$ closed in $\ell_1$, which is to say that $\ell_1/U^{-1}(0)$ is the dual of the subspace $U^{-1}(0)_\perp$ of $c_0$. But the dual of every subspace of $c_0$ has the Schur property (proofs that $\ell_1$ has the Schur property also give this).<|endoftext|>
TITLE: "Lebesgue-measurable" cardinals and real-closed fields
QUESTION [6 upvotes]: I understand the motivation behind measurable cardinals is to ask the question: "is there any set large enough to admit a non-trivial measure on all of its subsets?"
Hence, it's also worthwhile to ask the question "is there any real-closed field large enough to admit a non-trivial 'Lebesgue' measure on all of its subsets?" Here's one way to formulate this question precisely:
First, let $\Bbb R^*$ be our real-closed field, noting that it can be non-Archimedean.
Second, let's generalize our notion of "measure." Given some $\Bbb R^*$, let $\Bbb R^{**}$ be a strictly larger real-closed field. Then we will allow our measure on $\Bbb R^*$ to take values in $\Bbb R^{**}$.
Now, we want to find real-closed fields $\Bbb R^*$ and $\Bbb R^{**}$, and a function $\mu: 2^{\Bbb R^*} \to \Bbb R^{**}$ which obeys the following:

For all $S \subseteq \Bbb R^*$, $\mu(S) \geq 0$.
$\mu(\emptyset) = 0$.
Countable additivity of pairwise disjoint sets.
$\mu([0,1]^*) = 1^{**}$.
$\mu$ is a complete measure.
$\mu$ is translation-invariant.

Now, here are the questions:

Can this measure space exist?
If so, then what relationship does this measure space have with the concept of a measurable cardinal?

REPLY [3 votes]: This is NOT an answer but a question related to the original question. 
Take $R^*=R$, the usual reals, then $R^{**}$ is $R^*$, a non-Archimedean extension. By your 3 and 4, the measure of the whole of $R$ is a sum of infinitely many ($\omega$) values $1$. How this can be a value in $R^*$?<|endoftext|>
TITLE: Kronecker polynomial or not?
QUESTION [10 upvotes]: Given a polynomial $f$ of degree $d$ with integer coefficients, I am interested in an effective algorithm to determine whether or not $f$ has all its roots on the unit circle.
(So the output should be a YES or NO, no further information required.)
In this context the maximum integer $m$ such that $\phi(m)$ (Euler totient) does
not exceed $d$ is of some importance. One could take greatest common divisors of $f$ and the consecutive cyclotomic polynomials $\Phi_d(x)$ up to $\Phi_m(x)$.

REPLY [2 votes]: As explained in the answers, applying a Möbius transform and then using Sturm's theorem on the number of real roots gives an algorithm which solves the problem.
A natural question is what is an efficient implementation of this algorithm. There are some improvements on implementing the above method literally. For example, if $P$ is of degree $d$ and $P(0)\neq 0$, then every root $x$ of $P$ on the unit circle is also a root of $P^*(x) = x^d P(1/x)$ (this is because $\overline{x} = 1/x$). So the condition of $P$ having all the roots on the unit circle implies that $P$ is reciprocal or anti-reciprocal, that is, $P^* = \pm P$. It may be good to check that first.
Furthermore, there are several variants of the above method where one applies Sturm's theorem to other polynomials. One reference is Cerlienco, Mignotte, Piras, Computing the Measure of a Polynomial, J. Symbolic Computation (1987) 4, 21-33 (the article is old and I would be grateful if someone knows more recent references). The last section of this article also investigates the cost of these algorithms.<|endoftext|>
TITLE: What are the applications of the depth 2 reduction to the subfactors theory?
QUESTION [5 upvotes]: Let $(N \subset M)$ be an irreducible finite depth ($>2$) finite index unital inclusion of hyperfinite ${\rm II}_1$ factors, then for $n$ sufficiently large the subfactor $(N \subset M_n)$ is depth $2$ (reducible) and is isomorphic to $(R^H \subset R)$ with $H$ a weak Kac algebra (also called quantum groupoid, see prop. 9.1.1 p37 here).  Then the initial subfactor is isomorphic to an intermediate of $(R^H \subset R)$ and by Galois correspondence  (see here), it is completely given by the inclusion $(J \subset H)$ with $J$ a left coideal subalgebra of $H$.
Question: what are the applications of this result to the subfactors theory?
(In what this help to better understand the subfactors?)

REPLY [3 votes]: Two answers of Dmitri Nikshych: 

Dmitri
  This reduces the study of subfactors to that of weak Hopf algebras.
  Equivalently (and more importantly)  it relates the theory of finite
  index finite depth subfactors to the theory of fusion categories and
  module categories over them.
There are many papers by Snyder, Morrison and others in this
  direction.
A concrete example: it follows from the theory of fusion categories
  that the index of a subfactor is a cyclotomic integer.

Sebastien
Yes the fusion category framework helps a lot!
My question is more about the weak Kac algebra framework.
An irreducible finite index finite depth subfactor is completely given by an inclusion of a left coideal subalgebra in a weak Kac algebra,
i.e. an inclusion of finite dimensional ${\rm C}^{\star}$-algebras with an additional "quantum group"-like structure.
In what this help to better understand the subfactors?

Dmitri
  To me weak Hopf algebras are a bridge between subfactors and fusion
  categories.
The importance of weak Hopf algebra description of finite depth
  subfactors is that  it reduces a seemingly analytic problem (inclusion
  of von Neumann algebras)  to purely algebraic theory.  It also folows
  that weak Hopf algebras are the correct  symmetries of subfactors just
  like groups are symmetries of field extensions.<|endoftext|>
TITLE: Current Research in Numeric Mathematics
QUESTION [16 upvotes]: To me, as an non-expert in the field, it seems as if numeric mathematics should have lost its importance because nowadays symbolic calculations or calculations with unlimited precision are generally available.
So, just out of curiosity, I would like to know, whether my impression is wrong and what current hot research-topics of practical relevance in numeric mathematics are.

REPLY [9 votes]: Most of the problems to which mathematics is applied nowadays cannot be solved symbolically; think of flow in an oil reservoir or air flow past a vehicle.  These problems are modelled by complicated systems of nonlinear PDEs and have highly irregular boundaries, all of which make analytic solution impossible.  Numerical approximations are the only reasonable approach, and they require billions, trillions (or more) arithmetic operations.  If those operations were performed in "unlimited precision" (i.e., rational arithmetic) the size of the denominators required would be similarly large and it would quickly become impractical to store the solution.  It would also be millions of times slower (compared to doing it in double precision floating point).  Even in floating point, there are plenty of problems (e.g. turbulence modeling) that we cannot solve to acceptable accuracy in reasonable time on the largest supercomputers.
One other note: rounding errors are only one source of error in numerical computations; discretization error is frequently more significant.  Even if you could use infinite precision for these computations, the stability of the numerical discretizations would continue to be an important area of research.<|endoftext|>
TITLE: Does a rational function extend to a point if all its 1-parameter extensions at that point agree?
QUESTION [5 upvotes]: Let $f: X \dashrightarrow Y$ be a rational map where $X$ and $Y$ are varieties over $\mathbb{C}$ with $X$ smooth and $Y$ proper (or projective if necessary). Choose an open set $U \subset X$ on which $f$ is defined.
Pick a closed point $x \in X$ and an algebraic curve $C$ passing through $x$ such that $C \cap U \neq \emptyset$. Since the restriction of $f$ to $C$ can be uniquely extended to all of $C$ we may define $f|_C(x)$.
Let us say that all 1-parameter limits of $f$ at $x$ agree if there exists a point $y \in Y$ such that for any curve $C$ as above $f|_C(x) = y$. 
Could anyone tell me if some variation of the following is true?

If all 1-parameter limits of $f$ agree at $x \in X$ can $f$ be extended to an open set containing $x$?

One suggested alternative in case the first fails:

Does $f$ extend to an open set $\tilde U \subset X$ if for all points in $\tilde U$ all 1-parameter limits of $f$ agree?

REPLY [2 votes]: Thanks to Jason Starr who answered my question in the comments. I would like to beef up his answer here for future reference.

The answer to the first question--and thus for the second--is in the affirmative. In fact, there are no restrictions on the singularities of $X$; for instance it need not be normal let alone smooth.

Here is the proof:
Step 1
Let $U \subset X$ be any open subset where $f$ is defined, $\Gamma_U \subset U \times Y$ be the graph of $f$ and $\Gamma \subset X\times Y$ be the closure of $\Gamma_U$.
The restricted projection map $\pi: \Gamma \to X$ is proper, moreover it is an isomorphism over $U$. The goal is to show that $\pi$ is in fact an isomorphism over an open set containing $x$. By precomposing with this isomorphism we can extend $f$ around $x$. 
To that end we will show that $\pi$ is quasi-finite of degree 1 around $x$.
Step 2
Let $\Gamma_x$ be the fiber of $\Gamma$ over $x$. Using the valuative criterion of closedness and the universal property of the fiber product it is easy to show that $\Gamma_x$ is in bijection with the set of values that appear as 1-parameter limits of $f$ at $x$. 
The hypothesis states that this latter set consists of a single element. Using semi-continuity we conclude that in a neighbourhood of $x$, the fibers of $\Gamma$ consist of a single element. 
Thus $\pi$ is invertible in a Zariski neighbourhood of $x$ and $f$ extends to this neighbourhood.<|endoftext|>
TITLE: Why study the p-completions of a space?
QUESTION [27 upvotes]: Given a nice topological space $X$ there are various notions of a 'completion' at a set of primes. Some of the most common constructions may be found in Bousfield-Kan's, May's, Neisendorfer's or Sullivan's classic textbooks - that last three of which I have read. 
But what information about a space is contained in its p-completions? Whilst I have a good handle on the information that may be gleamed from study of the p-localizations, the slightly more abstract nature of the completion leaves me unsure of how to interpret its properties. I am aware of the various arithmetic squares and have no trouble with the concepts or algebra, I just lack any concrete examples of the theory of completions yielding any useful information about the homotopy type of a given space.

REPLY [21 votes]: I'll give an answer from the point of view of an algebraic topologist,
somebody who cares about examples and computations.  This all goes way back 
and has nothing to do with modern generalities.  First off, calculationally, algebraic 
topologists can compute, like to compute, and need to compute mod p homology.  It is 
far more accessible than p-local homology, and it is the focus of completion, not localization, 
at p.  
An elementary concrete example of 
homotopy information invisible to p-localization is that $K(\mathbb Z/p^{\infty},n)$ 
becomes equivalent to $K(\mathbb Z_p,n+1)$ after $p$-completion.  This jacks up to Quillen's
equivalence between the $p$-completions of the algebraic $K$-theory space $K(\bar{\mathbb F}_q)$ and $BU$ where $p\neq q$.  The homotopy groups of $K(\bar{\mathbb F}_q)$ are 
$\oplus_{p\neq q} \mathbb Z/p^{\infty}$ in odd degrees and $0$ in even degrees.
The homotopy groups of $BU$ are $0$ in odd degrees and $\mathbb {Z}$ in even degrees.
Obviously $p$-localization knows nothing whatsoever about this equivalence.   
Calculationally,
the classical Adams spectral sequence converges to the stable homotopy groups of
the $p$-completion, not the $p$-localization, of the space (or spectrum) one 
starts with; that is also true unstably. The Atiyah-Segal completion theorem says that
the topological $K$-theory of $BG$ is the completion of $R(G)$ at its augmentation
ideal.  If $G$ is a $p$-group, this is just $p$-adic completion (on the reduced level). 
That is a place where use of pro-groups helps calculation, but in most of the 
calculational applications use of pro-objects is not especially helpful.  
There are tons of places where $p$-completion is essential to the proof or statement 
of results: the Sullivan conjecture, the Segal conjecture, the classification of 
$p$-compact groups, the study of atomic spaces, etc.  It is $p$-completion that is 
relevant to all of these and many more.  
A technical reason $p$-completion is so convenient is that there are no phantom maps 
to worry about if one restricts to nilpotent spaces of finite type.  That is also 
true for rationalization but not for localization at $p$.
A more sophisticated example is Mandell's algebraization of $p$-adic homotopy theory, the closest analogue we have of the classical algebraization of rational homotopy theory.
(The last part of Harpaz's answer concerns this.) 
I could go on for many pages.  The more algebraic topology one learns, 
the more frequently $p$-completion appears and the more natural it seems. 
It is worth adding that we do not understand $p$-completion very well for
non-nilpotent spaces, where there are several very different notions,
none well understood calculationally.  This deserves more study.<|endoftext|>
TITLE: Computations in modular cohomology of finite groups
QUESTION [8 upvotes]: Let $k$ be an algebraically closed field of characteristic $p$, let $G$ be a finite group whose order is divisible by $p$, and let $H(G)$ be the commutative cohomology algebra of $G$ with coefficients in $k$ (viewed as a trivial module), i.e.,
$$H(G):=\begin{cases}H^*(G,k)&p=2\\H^{ev}(G,k)&p>2\end{cases}$$
For what sorts of groups have we explicitly computed $H(G)$?  Or if not an explicit computation, what can we say about $H(G)$ (or at least $|\operatorname{Spec}H(G)|$)?  Below is what I'm aware of:

If $G$ is elementary abelian of rank $r$, then $H(G)$ is known by using a particularly nice projective resolution of a cyclic group of order $p$ and using the Kunneth formula.
Thanks to Quillen, the Krull dimension of $H(G)$ is the $p$-rank of $G$, that is the length of the largest chain of prime ideals in $H(G)$ is equal to the largest rank of an elementary abelian $p$-subgroup of $G$.
Jon Carlson has used Magma to compute $H(G)$ for some $2$-groups here.

If you find the question too broad, at the moment I'm particularly interested in finite groups of Lie type, and to be even more specific, $\operatorname{GL}_n(\mathbb{F}_q)$.  I recall reading a paper that computed $H(\operatorname{GL}_2(\mathbb{F}_q))$, and I also recall reading another paper showing certain vanishing results of $H(\operatorname{GL}_n(\mathbb{F}_q))$ as one allows $n$ or $q$ to grow independently, but I can't seem to find either of these papers and I don't remember the authors.


What do we know about the commutative algebra $H(\operatorname{GL}_n(\mathbb{F}_q))$?  Have any computations been made for certain $n$ and $q$?


Any references are appreciated, even those that address the question for general $G$, and not just $\operatorname{GL}_n(\mathbb{F}_q)$.

REPLY [6 votes]: There is a comprehensive set of calculations of the homology of classical groups over finite fields in Z. Fiedorowicz and S. Priddy. Homology of classical groups over finite fields and their associated infinite loop spaces.  Springer Lecture Notes in Mathematics Volume 674, 1978. I'm a little surprised this has not yet been mentioned, since it seems to give more complete answers than some that are cited above.  It is written with a topological slant, following up Quillen's original work that used such homological calculations to determine the K-theory of finite fields, by comparison of Frobenius in algebra with Adams operations in topology.<|endoftext|>
TITLE: When is Chevalley Warning's bound best possible?
QUESTION [6 upvotes]: Chevalley Warning's theorem (a form of) states that any homogeneous form over a finite field of degree $d$ in more than $d$ variables has a nontrivial zero in the field. However, for diagonal forms, this result is often much weaker than it needs to be. I am wondering if anyone knows of a particularly good reference on improving the bound for certain families of diagonal forms? I apologize if this is a question better fit for MSE.

REPLY [5 votes]: Consider a degree $d$ homogeeous diagonal form in $n$ variables over $\mathbb F_q$.
I think that for $d$ quite large this is a problem in additive combinatorics. Essentially you have sets $A_1, \dots, A_n$, each a translate of the set of $d$th powers, and you're trying to show that $0 \in A_1+ \dots + A_n$ nontrivially.
One possible method is Fourier analytic. Without loss of generality $d$ divides $q-1$.  We may write the number of solutions of $\sum_{i=1}^n a_i x_i^{d}=0$ as
$$\frac{1}{q} \sum_{\psi: \mathbb F_q \to \mathbb C^{\times}}\prod_{i=1}^n \left( \sum_{x_i \in \mathbb F_q }\psi( a_i x_i^d) \right)$$
The main term $\psi=1$ is $q^{n-1}$, so to have the other terms smaller and thus guarantee nontrivial solutions we need an estimate for $\psi$ nontrivial:
$$ \left|\sum_{x \in \mathbb F_q }\psi( a_i x_i^d)\right| \leq q^{1-1/n}$$
The Weil bound gives the estimate:
$$\left|\sum_{x \in \mathbb F_q }\psi( a_i x_i^d)\right| \leq (d-1) \sqrt{q}$$
so for $d  \leq q^{1/2-1/n}$ we win.
This can be improved when $q$ is prime using better estimates for character sums, e.g. Estimates for the number of sums and products and for exponential sums in fields of prime order by Bourgain, Gibichuk, and Konyagin.
However I suspect that the Gauss sum method is not optimal. For $q$ a perfect square and $d= \sqrt{q}+1$, say, there is no cancellation in the Gauss sums, but still for $n=3$ we get a nontrivial solution by linear algebra over $\mathbb F_{q^{1/2}}$.<|endoftext|>
TITLE: Shortest vector problem over polynomials
QUESTION [6 upvotes]: In shortest vector problem, given a lattice in $\Bbb Z^n$, we seek the shortest non-zero vector in the lattice. This problem is computationally difficult.
Answer in Evidence for integer factorization is in $P$ seems to suggest some connections between polynomial analog of this problem that is easy? What is the precise polynomial analog of this problem and is there any connection to discrete logarithms?

REPLY [3 votes]: Not sure about the first, but for the second, yes:
Claus-Peter Schnorr. Factoring Integers and Computing Discrete Logarithms via Diophantine
Approximations. In Eurocrypt 1991, volume 547 of LNCS, pages 281–293.
Springer, 1991
Alexander May. Using LLL-Reduction for Solving RSA and Factorization Problems. In
Nguyen and Vallee [NV10], pages 315–348.<|endoftext|>
TITLE: Well-understood bases for Grothendieck groups of modular representation categories
QUESTION [6 upvotes]: Let $\mathfrak{g}$ be a semi-simple Lie algebra. 
So in characteristic $0$, the Grothendieck group of a block of category $\mathcal{O}$ is given by the classes of the Verma modules. Unlike the simples (or projectives), Verma modules are very easy to understand explicitly, and one can easily perform computations with them (eg. involving translation functors). 
In characteristic $p$, the analogous object is the category of modules with a fixed central character: one must specify both a Harish-Chandra character, and a character of the Frobenius (or p-) center. Now each simple is a quotient of a baby Verma module; however the classes of the baby Verma's no longer give a basis in the Grothendieck group -- in fact, they all have the same class! 
I'm looking for a basis for the Grothendieck group of this category, where the corresponding objects are easy to understand (unlike the simples or projectives). In fact, I'd be happy with a collection of objects whose images span the Grothendieck group. I'm most interested in the case where the Frobenius character is 0, but the Harish-Chandra character is singular.
I've heard that the Weyl modules (coming from the char $p$ version of Borel-Weil theory) might be useful here; though I'm not entirely sure if they can be defined in this generality. This link might be relevant: Weyl modules and reduction modulo $p$.

REPLY [2 votes]: I'm not sure what sources you are mainly relying on, but there are several points to be made:
1) When you say the Lie algebra is "semisimple", I suspect you mean (as people sometimes do when using shorthand) the Lie algebra of a semisimple algebraic group.   There are lots of other simple Lie algebras (mostly classified for $p>3$ in the work of Premet-Strade and others) which don't come from algebraic groups, and their representations are less well studied.    Anyway, it's probably best to limit to simple algebraic groups and their Lie algebras, the latter usually but not always being simple.    (For example, the special linear Lie algebra of trace zero $n \times n$ matrices over an algebraically closed field of characteristic $p>0$ has a center consisting of scalar matrices when $p|n$.)
2) The algebraic group framework provides some useful tools for studying the Lie algebra representations, as seen in Jantzen's book Representations of Algebraic Groups: you can construct "mixed" group schemes, using a torus action to restore some ordering to "highest weights", but at the cost of creating an infinite rank Grothendieck group.  Anyway, the simple modules for the Lie algebra always come from simple modules for the group (where the highest weights have to be "restricted"); then Steinberg's tensor product theorem produces all simple modules for the group.  
3) The blocks in these various situations have been worked out by now, but for the Lie algebras alone there seems to be no realistic candidate for an easy basis of the corresponding Grothendieck group.  As you say, the (baby) Verma modules for such a block all have the same composition factor multiplicities and define a single element of the Grothendieck group.
4) The Lie algebras here admit infinitely many simple modules, mostly not coming directly from the group and depending on a linear functional $\chi$ in the dual Lie algebra.   For a short survey with references (not up-to-date), see my 1998 paper here.    The conjecture there on blocks of the finite dimensional reduced enveloping algebra for $\chi$ was proved by Brown-Gordon here.  
5) The methods of geometric representation theory have been used by Bezrukavnikov and others to get better information about the Lie algebra representations.   But there still seems to be no easy way to approach this in terms of a Grothendieck group basis coming from modules that are easy to construct.   The simple modules are mostly still elusive.  (But those with singular "highest weights" are usually obtainable using Jantzen's translation functors from those with regular weights.)<|endoftext|>
TITLE: Can one twist fibred knots and still get fibred knots?
QUESTION [6 upvotes]: Suppose we have a fibred knot $K$ with a fiber surface $F$ and let $c$ be an unknot disjoint from $F$ (but not homotopically trivial in the complement of $F$). Is it possible that every twist along $c$ leaves $K$ fibred with $F$ still being the fiber surface?

REPLY [10 votes]: Such examples were constructed by Morton. He showed that one can find unknotted curves lying on fiber surfaces with zero framing. Twisting about them preserves fiberedness and the fact that it is a knot in $S^3$. Also, curves on the fiber can be pushed disjoint from the fiber meeting your requirement.
Edit: As Danny points out in the comments John Stallings originally came up with this construction.<|endoftext|>
TITLE: Advice for PhD Supervisors
QUESTION [53 upvotes]: My first PhD student is having his viva tomorrow. Hence, I began contemplating a bit about the whole process of supervising. One thing I realized is that while there seems to be plenty of advice for PhD students I cannot recall ever seeing advice for supervisors. So 

What resources are available for PhD supervisors?
If you are experienced supervisor, then what would be your advice?

REPLY [9 votes]: A new blog post came out yesterday from the AMS, with some nice advice:
https://blogs.ams.org/matheducation/2018/01/08/advice-for-new-doctoral-advisors/<|endoftext|>
TITLE: Asymptotic behaviour of an integral
QUESTION [6 upvotes]: For $k\in\mathbb{N}_{0}$ and $x\in\mathbb{R}$, define
$$I_{k}(x):=\int_{0}^{\pi/2}\cos(xg(\theta))\sin^{2k}\theta\,\mathrm{d}\theta$$
where
$$g(\theta)=\int_{\sin\theta}^{1}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-\alpha^{2}t^{2})}}$$
and $0<\alpha<1$ is a fixed parameter.
I arrived at the expression $I_{k}(x)$ while working on certain asymptotic analysis of Jacobi elliptic functions. I doubt that the integral $I_{k}(x)$ could be evaluated somehow (even in terms of some known special functions). Nevertheless, for my purpose, it would be sufficient to know the asymptotic behavior of $I_{k}(x)$, as $k\to\infty$. It seems, however, that for my calculations, I need to know not only the leading term of the asymptotic expansion, but the second term as well.
Does anybody know how to derive the asymptotic expansion for $I_{k}(x)$, as $k\to\infty$? Many thanks.

REPLY [7 votes]: With the aid of Mathematica I found that 
$$g(\theta)=\textrm{EllipticK}(a^2)-\textrm{EllipticF}(t,a^2).$$
I get the first terms of the asymptotic expansion
$$\frac{\sqrt{\pi}}{2\sqrt{k}}-\frac{\sqrt{\pi} x^2}{8(1-a^2)k^{3/2}}+
\Bigl(-\frac{k}{6}+\frac{a^2 x^2}{6(1-a^2)^2}+\frac{x^4}{24(1-a^2)^4}\Bigr)
\frac{3\sqrt{\pi}}{8k^{5/2}}$$
For example with  $a=2/3$, $x=1/5$ and $k=1000000$ Mathematica evaluates 
$$I_k(x)=0.00088622679872235827901$$
The first, second and third approximations are equal to 
$$0.00088622692545275801365,\quad 0.00088622690950067335550, \quad
0.00088622679872231419831.$$
I used the standard Laplace method. I expect my computations are correct. I do not make adequate checks.<|endoftext|>
TITLE: Structure of Hopf algebras - trouble understanding an old paper
QUESTION [26 upvotes]: UPDATE: I am grateful to Peter May for the accepted answer, which makes most of the details below irrelevant.  However, I will leave them in place for the record.
I am trying to understand the proof of Theorem 2 in the 1969 paper Remarks on the Structure of Hopf Algebras by Peter May (because I would like to prove a similar result with different hypotheses).  The statement is given in the discussion below, and the proof contains the following claim:

Since $\tilde{f}$ is the identity on $RA$, $\tilde{f}(F_sRA)=RA\cap F_sA$

I do not see why this should be true; I hope that someone will enlighten me (or suggest a workaround, or an alternative proof or reference for the same theorem).
The context is as follows:

$A$ is a connected, graded, commutative algebra over a perfect field $K$ of characteristic $p>0$.  You can assume if you like that it is concentrated in even degrees.  Later, we will assume that $A$ is a Hopf algebra, but we do not need that for the moment.
$IA$ is the augmentation ideal, and $QA=IA/IA^2$.
$f\colon QA\to IA$ is a section of the natural projection.
$\xi\colon A\to A$ is the $p$'th power map.  I am interested in the case where this is injective, so you can assume that if you like.
$RA=\sum_{i\geq 0}\xi^i(f(QA))$.
$B=V(RA)$ is the commutative graded $K$-algebra freely generated by $RA$ subject to the relation $\xi(r)=r^p$ for all $r\in RA$.  (In the case where $\xi$ is injective, this is just the free commutative algebra on $f(QA)$.)
$\tilde{f}\colon B\to A$ is the unique homomorphism of $K$-algebras that acts as the identity on $RA$.  This is easily seen to be surjective.  
The statement of the theorem is that if $A$ is a Hopf algebra, then $\tilde{f}$ is an isomorphism.
We will write $RB$ for the obvious copy of $RA$ inside $B$, so $\tilde{f}\colon RB\to RA$ is an isomorphism.
The statement that I am worried about says that $\tilde{f}(RB\cap IB^t)=RA\cap IA^t$ for all $t\geq 0$.
If $r\in RB$ and $r$ can be written as a sum of terms like $u_1\dotsb u_t$ with $u_i\in IB$, then we can apply $\tilde{f}$ to express $\tilde{f}(r)$ as a sum of terms in $IA^t$.  Thus $\tilde{f}(RB\cap  IB^t)\subseteq RA\cap IA^t$.
Suppose instead that $s\in RA$ and $s$ can be written as a sum of terms like $v_1\dotsb v_t$ with $v_i\in IA$.  As $\tilde{f}\colon RB\to RA$ is an isomorphism, there is a unique $r\in RB$ with $\tilde{f}(r)=s$; we want to show that this lies in $IB^t$.  As $\tilde{f}\colon IB\to IA$ is surjective, we can choose $u_i\in IB$ with $\tilde{f}(u_i)=v_i$.  This gives an element $r'\in IB^t$ with $\tilde{f}(r')=\tilde{f}(r)$, so $\tilde{f}(r-r')=0$.  However, we only know that $\tilde{f}$ is injective on $RB$, and we do not know that $r'\in RB$, so we cannot conclude that $r=r'$.  Thus, we do not seem to have a proof that $f(RB\cap IB^t)\supseteq RA\cap IA^t$.
Suppose that the element $r$ above does not lie in $IB^t$, but only in $IB^{t-1}$.  Then $r$ represents a nonzero element in the associated graded group $E^0RB$ in filtration degree $t-1$, whose image in $E^0RA$ is zero.  Thus, the map $\tilde{f}\colon E^0RB\to E^0RA$ is not injective, so we cannot proceed with the next step.

The proof in the paper does not seem to use the coproduct in any essential way when reaching the statement that I have questioned.  Thus, we should ask whether it is true in the absence of a coproduct.  Consider the following case (where $p$ is an odd prime):

$A=\mathbb{F}_p[x,y]/(x^{p+1}-y^p)$, with $|x|=2p$ and $|y|=2p+2$
$QA=\mathbb{F}_p\{x,y\}$
$RA=\mathbb{F}_p\{x^{p^k},y^{p^k}:k\geq 0\}$, with $y^{p^k}=x^{p^k(p+1)}$ for $k>0$
$B=\mathbb{F}_p[x,y]$
$RB=\mathbb{F}_p\{x^{p^k},y^{p^k}:k\geq 0\}$

If we consider $y^p$ as an element of $RB$, then it is equal to $x^{p+1}$ and so lies in $RA\cap IA^{p+1}$.  However, the corresponding element of $RB$ does not lie in $IB^{p+1}$, so $\tilde{f}(RB\cap IB^{p+1})\neq RA\cap IA^{p+1}$.

REPLY [20 votes]: Ah, Neil, my apologies, sins of a forgetful old man.  You are quite 
right to object to Theorem 3 in that 1969 paper, because it is 
false as stated. In fact, Paul Goerss gave me a counterexample ages ago.  That appears as Example 23.4.2 in More Concise
Algebraic Topology, by Kate Ponto and myself, which is available
here: http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf
(and at booksellers everywhere).  Advertising that book, it reworks
Milnor and Moore, among other things, in Chapters 20-23.  In response 
to Paul's counterexample, it also reworks the relevant part of the
1969 paper, proving a correct variant of the theorem you object to,
namely Theorem 23.4.1.  Work over a perfect field $K$.  
Theorem. Let $A$ be a connected, commutative, and associative
quasi Hopf algebra. For a morphism of $K$-modules $\sigma: QA
\rightarrow IA$ such that $\pi\sigma = id$, where $\pi: IA \rightarrow
QA$ is the quotient map, let $R(A;\sigma)$ be the sub abelian restricted 
Lie algebra of $A$ generated by the image of $\sigma$. For a suitable 
choice of $\sigma$, the morphism of algebras $f: V(R(A;\sigma))
\rightarrow A$ induced by the inclusion of $R(A;\sigma)$ in $A$ is
an isomorphism.
Quasi here just means not necessarily coassociative. The proof is an adaptation 
of the proof of Theorem 4 in the 1969 paper, which is correct.  The point 
is that you must choose the section carefully to make it go.<|endoftext|>
TITLE: Zeta-Determinant Theorem
QUESTION [5 upvotes]: Recently, someone asked on MO about lecture notes from Graeme Segal's "Stanford lectures" on TQFT, and the answer was to check here. 
When scrolling over the notes, I stumpled of Prop. 2.8.2 in lecture 2. It says (modulo typos) that if $P$ is a Dirac operator and $S$ is an invertible operator such that $S-\mathrm{id}$ is trace-class, then
$$\det\nolimits_\zeta(PS) = \det(S)\det\nolimits_\zeta(P)),$$
where $\det\nolimits_\zeta$ denotes the zeta-regularized determinant and $\det$ denotes the usual determinant.
However, there are neither references nor proofs there. Does anyone know of a proof?

REPLY [2 votes]: The mathematics here is definitely out of the realm of my understanding, so hopefully someone who knows this stuff will post a better answer. 
I believe that Prop 2.8.2 in Segal's Lecture 2 is more or less Proposition 6.4 in "Determinants of elliptic pseudo-differential operators" by Kontsevich and Vishik.
There seems to be an ample literature on zeta determinants of the Laplacian and Dirac operators so you probably can find other sources as well. For example, I actually first found the paper "Zeta Determinants on Manifolds with Boundary" by Scott, which includes a stronger statement as Prop. 2.19.<|endoftext|>
TITLE: Smoothed exponential sums: bounds and sources?
QUESTION [17 upvotes]: Let $f:\mathbb{R}\to\mathbb{C}$ be differentiable $k$ times, with $f, f',\dotsc,f^{(k)}\in L^1$. Let $\alpha\in \mathbb{R}/\mathbb{Z}$, $\alpha\ne 0$. In "Every odd number..." (Math. Comp. 83, 2014), Lemma 3.1, Tao shows that
$$\left|\sum_{n\in \mathbb{Z}} f(n) e(\alpha n)\right|\leq \frac{1}{|2 \sin(\pi \alpha)|^k} |f^{(k)}|_1,$$
where $e(t) = e^{2\pi i t}$. 
The proof goes essentially by summation by parts.
(a) Are there older sources for this? Somewhat confusingly, Tao credits Gallagher ("The large sieve") and Lemma 1.1 in Montgomery's Topics in Multiplicative Number Theory, but they give only equation (3.1) in Tao's papers, not the inequality above.
(b) For $k=2$, this is not in general optimal: one can show 
$$\left|\sum_{n\in \mathbb{Z}} f(n) e(\alpha n)\right|\leq \frac{1}{|2 \sin(\pi \alpha)|^2} |\widehat{f''}|_\infty,$$
which is no weaker and often strictly stronger, since $|\widehat{f''}|_\infty\leq |f''|_1$.
This is Lemma 2.1 in my three-prime book draft on the arxiv; the proof I give goes by the Poisson summation formula, plus Euler's formula for the cotangent.
Are similar bounds true for general $k$? (Is $\left|\sum_{n\in \mathbb{Z}} f(n) e(\alpha n)\right| \leq |\widehat{f'}|_\infty/|2 \sin \pi \alpha|$, for instance?) Again, can such results be found in older sources?

REPLY [7 votes]: By Poisson:
$$ \sum_{n \in \mathbb Z} f(n) e(\alpha n)= \sum_{m \in \mathbb Z} \hat{f} ( 2 \pi m + \alpha) $$
By the formula for the derivative of the Fourier transform:
$$= \sum_{m \in \mathbb Z} \frac{1}{ (2\pi i m + \alpha i)^k}\widehat{f^{(k)}} ( 2 \pi m + \alpha) \leq \sum_{m \in \mathbb Z} \frac{1} {\left| 2 \pi m + \alpha \right|^k}\left|\widehat{f^{(k)}}\right|_{\infty}$$
if $k >1$ then this sum converges and we may replace your constant by $\sum_{m \in \mathbb Z} \frac{1} {\left| 2 \pi m + \alpha \right|^k}$. I'm not sure if there's an analytic formula for this for various values of $k$, we can certainly estimate.
If $k=1$ then this sum diverges and this method fails to prove a bound. Does this mean that no bound is posssible? Yes. Simply take a Schwartz function $g$ that does something nice and smooth on $[-1,1]$, behaves like $1/|\xi|$ on $[1,n]$ and $[-n,-1]$, and then declines rapidly to $0$ outside $[-n,n]$.  Then let $f$ be the inverse Fourier transform, also a Schwartz function, so $\hat{f}=g$. Then the Poisson summation formula is justified but:
$$\sum_{m \in \mathbb Z} \hat{f} ( 2 \pi m + \alpha)  \approx \log n$$
$$|\widehat{ f'} |_\infty \approx 1$$
So no bound in terms of $|\widehat{ f'} |_\infty$ is possible.<|endoftext|>
TITLE: Is there an octonionic analog of the K3 surface, with implications for stable homotopy groups of spheres?
QUESTION [64 upvotes]: The infamous K3 surface has many constructions in many fields ranging from algebraic geometry to algebraic topology. Its many properties are well known. For this question I am really interested in the K3 surface from a algebraic topology perspective (hence I view it as a particular smooth 4-manifold).
As we have seen in this MO question, and also this one, the K3 surface  plays an important role in the third stable homotopy group of spheres. It can be viewed as the source of the 24 in this group $\pi_3^{st} = \mathbb{Z}/24$. Here is a brief review of how that goes: the stable homotopy groups (in degree n) of spheres are the same as cobordism classes of stably framed manifolds (of dimension n). In dimension three the generator is given by $\nu = (S^3, \text{Lie})$, the 3-sphere with its Lie group framing (where we think of $S^3 \subseteq \mathbb{H}$ as the group of unit quaterions). 
The K3 surface is responsible for the relation $24 \nu = 0$ which comes from the existence of a certain framed 4-manifold whose boundary is a disjoint union of 24 copies of the framed 3-sphere $\nu$. The two important properties of the K3 surface which give rise to this are:

The K3 surface is an example of an almost hyperkahler manifold, which means that its tangent bundle admits a reduction to an $S^3 = SU(2)$ vector bundle. In fact it is integrable (so a hyperkahler manifold), but that is not important for this. What is important is that it admits three different complex structures $I,J$ and $K$ which generate a quaterionic algebra. This means that given a vector field $v$ on the K3 surface we get a quadruple of vector fields 
$$(v, Iv, Jv, Kv)$$ 
which form a frame whereever $v$ is non-zero. 
The Euler characteristic of the K3 surface, the obstruction to the existence of a non-vanishing vector field, is 24.

This means that if we cut out 24 small balls from the K3 surface, we get a manifold $X$ with boundary $\partial X = \sqcup_{24} S^3$, 24 copies of the 3-sphere. $\chi(X)= 0$, and so $X$ admits a non-vanishing vector field, which restricts to the inward pointing normal vector field at each boundary component. 
Using the hyperkahler structure this gives us an (unstable) framing of $X$ which restricts to the Lie group framing at each of the boundary 3-spheres. Hence $24 \nu = 0$. 
What I would like to understand is if there is an analog of this story which replaces the quaternions with the octonions?  
The $\pi_7^{st} = \mathbb{Z}/240$ and is generated by the framed 7-sphere $\sigma$ which is $S^7$ with its "Lie group framing" coming from its embedding as the unit octonions $S^7 \subseteq \mathbb{O}$. It is not actually a Lie group, because the multiplication is non-associative, but there is still enough structure to obtain a framing (by, say, left-invariant vectors), and it is the generator. 
I am wondering if there is an octonionic analog of the K3 surface which, in a similar way, leads us to the conclusion $240 \sigma = 0$? In other words I am looking for a smooth 8-manifold such that:

Its tangent bundle admits a multiplication by the octonions, in the same way K3's tangent bundle admits a multiplication by the quaternions. In particular given a vector field $v$ we should get an octuple:
$$(v, Iv, Jv, Kv, Ev, IEv, JEv, KEv)$$
which is a frame wherever $v$ is non-zero; and
The Euler characteristic should be 240.

Is there such a manifold? and how can we construct it? Does this sort of octonionic multiplication on the tangent bundle have a name? If there isn't such a manifold is there perhaps some replacement which still allows us to see, geometrically, that $240 \sigma = 0$?

REPLY [44 votes]: Yes, such M exists.  The boundary connected sum of 28 copies of the Milnor plumbing has boundary diffeomorphic to $S^7$ so it can be closed off with $D^8$ and you can let $M$ be the connected sum of the resulting closed manifold and 8 copies 7 copies of $S^4 \times S^4$.
Why does that work?  Let $f: M \to \mathbb{O}P^1 = S^8$ have degree 240 and let $L$ denote the canonical bundle over $\mathbb{O}P^1$ whose euler class generates $H^8(S^8;\mathbb{Z})$.  Since $M$ has signature $28 \cdot 8 = 224$ and since $$\mathcal{L}_2 = \frac{7}{45}p_2 = \frac{42}{45}e \in H^8(\mathbb{O}P^1;\mathbb{Q})$$ we see from the signature formula that $p_2(f^*(L)) = p_2(TM)$ from which we deduce that $TM$ and $f^*(L)$ are stably isomorphic (they are both stably trivial away from the top cell of $M$ so $p_2$ is the only obstruction).  Since the two bundle have the same euler class we deduce that they are also unstably isomorphic, i.e. $$f^*(L) \cong TM,$$ which should give the bundle condition you're looking for.
Remark: Maybe it follows from the results in Kreck's paper "Surgery and duality" that up to diffeomorphism this is the only possible 3-connected $M$ which is parallelizable away from a point.  (EDIT: No, see this comment.)<|endoftext|>
TITLE: What is the mean value of a pair of Ramanujan Sums when summed over squares?
QUESTION [6 upvotes]: Does anyone know of the mean value of two Ramanujan Sums when summed over the square of integers?
In my research on the Landau problem regarding nearly square primes, I have run into the mean value of a product of Ramanujan Sums:
\begin{equation*}
\lim\limits_{N \to \infty}
\frac{1}{N}
\sum\limits_{n =1}^{N} 
c_r\left( n^2 \right)
c_s\left( n^2 \right)
\end{equation*}
Does anyone know of a listing of properties of Ramanujan Sums similar to the this one, but with summations taken over the squares of positive numbers instead summation over the integers?  
The closest I can find is the function, $E_G$ found in Sums of products of Ramanujan sums by Toth. I am interested in the particular case where $g_1(n) = n^2$ though and not general case case represented by $E_G$ for and arbitrary set of functions $g_i(n)$

REPLY [9 votes]: Here is a partial answer (Added: completed below). Assuming $(r,s)=1$ and denoting $q=rs$, we have
$$ \sum_{n=1}^N c_r\left( n^2 \right) c_s\left( n^2 \right) 
= \sum_{n=1}^N c_q\left( n^2 \right) = \sum_{n=1}^N \sum_{d\mid(q,n^2)}\mu\left(\frac{q}{d}\right)d = \sum_{d\mid q}\mu\left(\frac{q}{d}\right)d\sum_{\substack{{1\leq n\leq N}\\{d\mid n^2}}} 1. $$
Now let $f(d)$ be the number of residue classes modulo $d$ whose square is zero modulo $d$. Then 
$$ \sum_{n=1}^N c_r\left( n^2 \right) c_s\left( n^2 \right) 
= \sum_{d\mid q}\mu\left(\frac{q}{d}\right)d f(d)\left(\frac{N}{d}+O(1)\right)
= N \sum_{d\mid q}\mu\left(\frac{q}{d}\right) f(d) + O_q(1),$$
whence
$$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N c_r\left( n^2 \right) c_s\left( n^2 \right) 
= \sum_{d\mid q}\mu\left(\frac{q}{d}\right) f(d).$$
The right hand side is the multiplicative convolution $g:=\mu\ast f$ evaluated at $q$, in particular it is multiplicative in $q$. Hence it suffices to determine $g$ at prime powers $p^k$, which is straightforward:
$$ g(p^k) = f(p^k)-f(p^{k-1}) = p^{\lfloor\frac{k}{2}\rfloor}-p^{\lfloor\frac{k-1}{2}\rfloor}.$$
We infer that the sought mean value equals
$$ g(q) = \begin{cases}\phi(\sqrt{q}),&q=\square\\0,&q\neq\square\end{cases} \tag{$\ast$}$$
Note that under our initial assumptions, $q$ is a square if and only if both $r$ and $s$ are squares.
Added. In the general case, i.e. without the coprimality condition on $r$ and $s$, we can proceed as follows. The function $n\mapsto c_r(n^2)c_s(n^2)$ is periodic mod $[r,s]$, hence the mean value exists and equals
$$ h(r,s):=\frac{1}{[r,s]}\sum_{n=1}^{[r,s]}c_r(n^2)c_s(n^2). $$
This function satisfies the multiplicativity relation
$$ h(rr',ss') = h(r,s)h(r',s'),\qquad (rs,r's')=1, $$
as follows from the Chinese Remainder Theorem and the invariance property
$$ c_q(m)=c_q(km),\qquad (k,q)=1. $$
Therefore,
$$ h(r,s) = \prod_{p\mid rs}h(p^{v_p(r)},p^{v_p(s)}), $$
so that it suffices to evaluate $h(p^k,p^l)$ for any prime $p$ and any exponents $k,l\geq 0$. This is straightforward, given the explicit formulae for $c_{p^k}(m)$ here. In particular, for $l>k\geq 0$ we obtain
$$ h(p^k,p^l)=\phi(p^k)h(1,p^l)
=\begin{cases}\phi(p^k)\phi(p^{l/2}),&\text{$l$ even}\\0,&\text{$l$ odd}\end{cases}$$
which generalizes the explicit formula ($\ast$) above.<|endoftext|>
TITLE: Can you functorially "reconstruct" a branched cover of curves from its etale locus?
QUESTION [5 upvotes]: I'm sure this must be covered somewhere, but all the references I have only treat this in very special cases (mostly when working over fields).
Suppose $f : X\rightarrow S$ is smooth of finite presentation with geometrically connected fibers of dimension 1, and let $g_i : S\rightarrow X$ be finitely many sections. Let $X^\circ := X - \sqcup_i g_i(S)$ and let $p^\circ : Y^\circ\rightarrow X^\circ$ be finite etale.
I would like a procedure to find a branched cover of $X$ which agrees with $p^\circ$ over $X^\circ$, ideally a construction that commutes with (arbitrary) base change. 
One way is to consider the normalization $X'$ of $X$ in $Y^\circ$ (though this only commutes with smooth base change), so we get a commutative diagram, which I don't know how to draw here, but the point is that the following two compositions are equal:
$$Y^\circ\stackrel{p^\circ}{\rightarrow} X^\circ\hookrightarrow X$$
and
$$Y^\circ\stackrel{h}{\rightarrow} X'\stackrel{\nu}{\rightarrow} X$$
Under the mild assumption that $S$ is Nagata, we know that $\nu$ is finite (http://stacks.math.columbia.edu/tag/03GR), so lets assume $S$ is Nagata.
Further, by Zariski's main theorem (http://stacks.math.columbia.edu/tag/02LQ) , we know that $h$ is an open immersion.
My questions are:

Intuitively, what does this relative normalization do? I really don't have any intuition for how relative normalization will behave over the $g_i$'s... I guess some more specific questions are:
If it's possible to extend $p^\circ : Y^\circ\rightarrow X^\circ$ to an etale cover of $X$, will the normalization $\nu$ be etale? (ie, does the normalization "prefer" "unramified completions" of the cover?)
When will $X'$ be smooth over $S$?
Is this the "best" way to recover a branched cover from its etale locus? Unfortunately, as Jason Starr pointed out, it does not commute with general base change. Is there a way to do this that does commute with base change?

References would be appreciated.
EDIT: Perhaps I shouldn't have used "reconstruct" in the title, since that implies that there exists a unique/canonical/best candidate for an extension of $p^\circ$ to a finite map over $X$. Basically I'm asking if there are decent alternatives to the normalization, and also for some intuition about how the normalization behaves.

REPLY [3 votes]: 2) Yes, if the cover is finite etale. (Maybe we need to be over a Noetherian base as well?) All the functions on the cover are going to be integral over the base and contained in the field of fractions of the open subset. Furthermore, there will be no additional functions of this type in the field of fractions of the cover (check this etale-locally, where it becomes obvious.)
3) In the case of wild ramification, where the $g_i$ don't intersect, smoothness implies some additional conditions on the ramification that you can see by looking at transverse curves. If your etale cover extends smoothly, then the extension by zero of the pushforward of the constant sheaf from that etale cover has no vanishing cycles, which implies (Deligne's semicontinuity theorem) that the Swan conductor is constant.
Also when it's smooth the standard argument that gives you the structure of the Galois group of local fields applies and says something about the structure of the cover - you can split it into an unramified part, a tame part, and a wild part. When it's not smooth, this argument fails, and the result can fail as well.
The conditions I describe on monodromy are invariant under inseparable base change, because it preserves the etale topology. But smoothness is not invariant under inseparable base change. Take a finite etale cover $Y^\circ \to X^\circ$ that extends to $Y \to X$, with $Y$ a family of nontrivial smooth projective curves over $S$. Take some inseparable morphism $S \to T$ which $X$ descends down but $Y$ does not (Maybe $X$ is $\mathbb P^1$, $Y$ is an elliptic curve with nonconstant $j$ invariant, and $T$ is the ring generated by the $p$th power of the $j$ invariant). Finite etale covers always descend, so $Y^{\circ}$ descends, but it no longer has a smooth compactification. (Smooth compactifications of curves are unique, so we also know there isn't some other compactification that does the job.)
There may be analogues of the first two phenomena in characteristic zero when $g_i$ intersect.
4) The normalization is optimal in a pretty clear sense: Given any finite cover that extends your cover, all the function son it are integral, and all lie in the ring of functions on the open subset that is your original cover, so they are all functions on the normalization. Hence any other branched cover $Y$ that extends $X^{circ}$ has a map $X' \to Y$.
When you look at an individual fiber, so this is a map of curves, note that in a birational map of proper curves, the target always has worse singularities (by some reasonable measurement) than the source. So in this sense, it's optimal.
Using this, we can show that no base change-invariant construction exists. Simply consider families which contain the same etale cover, but which have arbitrarily bad singularities. If you consider the cover $y^2 = (x - t)(x-2t)(x-3t) \dots (x-nt)$, at $t=0$ you get the singularity $y^2 =x^n$, which gets worse and worse as $n$ goes to $\infty$.  However on the open set where $x\neq t, 2t, 3t, \dots, nt$, restricted to the fiber $t=0$, the singularity only depends on $n$ mod $2$.
1) Intuitively, to me, the relative normalization makes each fiber as smooth as it can be considering the nearby fibers. So if the nearby fibers are really quite ramified (in number of ramification points or in Swan conductor), but a single fiber is not so ramified, then that difference will show up in a singularity of the special fiber. I don't know how to make this precise.<|endoftext|>
TITLE: Hodge to de Rham spectal sequence with twisted coefficients
QUESTION [12 upvotes]: Let $M$ be a smooth compact Kahler manifold and let $\mathcal{F}$ be a local system on $M$.
Question 1: I assume that there exists a twisted Hodge to de Rham spectral sequence converging to $H^{p+q}(M;\mathcal{F})$ whose $E_1$-page is of the form
$$E_1^{p,q} = H^p(M;\mathcal{F} \otimes \Omega^q).$$
Is this correct, and if so does anyone know a good reference for it?
Question 2: What conditions can I put on $\mathcal{F}$ to ensure that the above spectral sequence degenerates at the $E_1$-page?  For example, what if there exists some finite unbranched cover $\widetilde{M} \rightarrow M$ such that the pullback of $\mathcal{F}$ to $\widetilde{M}$ is just the sheaf of locally constant functions?

REPLY [9 votes]: (Edit: I answered Q2 initially, ignoring Q1.)
Q1: The spectral sequence is right except that it starts at $E_1$.
Q2: It is true if $\mathcal{F}$ is unitary in the sense that the underlying representation of $\pi_1(M)$ is unitary, then the spectral sequence degenerates at the $E_1$ page. This seems like a folklore fact, so I would need to think to find a good reference. But one that comes to mind is Timmerscheidt's appendix to a paper of Esnault and Viehweg, Inventiones, vol 86, pp 189-194. (He proves something more general, so it might be hard to follow.) By the way, this should be enough to conclude that condition of the last sentence of your Q2 is sufficient.
I don't think degeneration is true for non unitary local systems.
The best result in general is due to Simpson that, at least when $M$ is projective, $H^*(M,\mathcal{F})$ is isomorphic to the hypercohomology of the Higgs complex of the associated Higgs bundle. But perhaps this going too far afield.
Added Actually why don't just give you the proof. The key analytic fact that
makes the usual degeneration work is the Kähler identity 
$$d d^*+d^*d= 2(\bar \partial \bar\partial^* +\bar\partial^*\bar\partial)$$
This continues to hold for a unitary local system, where $d$ is now the connection. The point is that the  arguments needed to establish the usual identity carry over almost word for word when you work with a unitary local frame.<|endoftext|>
TITLE: Cohomology of configuration space as a representation of the symmetric group
QUESTION [15 upvotes]: Let $X_n$ be the space of $n$ distinct labeled points in $\mathbb{R}^3$, which is equipped with an action of the symmetric group $S_n$.  It is well known that the total cohomology of $X_n$ is isomorphic to the regular representation of $S_n$, but I would like to know what representation one gets in each degree.
When $\mathbb{R}^3$ is replaced with $\mathbb{R}^2$, this question has been answered by Lehrer and Solomon, and Getzler expresses this answer in a beautiful form in Equation (2.5) of http://arxiv.org/pdf/alg-geom/9510018.pdf.  If we let $f_{n,i}$ be the degree $n$ symmetric function associated with the action of $S_n$ on the degree $i$ cohomology of the configuration space, then Getzler gives a nice infinite product expansion for the power series $\sum_{n\geq 0}\sum_{i\geq 0}(-x)^if_{n,i}$.  If there were a formula similar to this one for the configuration space of points in $\mathbb{R}^3$, that would be great.

REPLY [18 votes]: One can indeed modify the formulas in the paper of Getzler to get an answer for any $\mathbf R^d$, by judiciously inserting minus signs and making substitutions $x \mapsto x^{d-1}$ in various places, but if I try I'll probably get it wrong. So let me explain how it works instead, and hopefully there are no sign errors and everything ends up in the right cohomological degree.
The cohomology ring of $F(\mathbf R^d,k)$ was computed by Cohen. It is the graded commutative algebra generated by variables $\omega_{ij}$ of degree $(d-1)$ for distinct $i,j \in \{1,\ldots,k\}$, modulo the relations

$\omega_{ij}^2 =0$,
$\omega_{ij} = (-1)^d \omega_{ji}$,
$\omega_{ij}\omega_{ik} + \omega_{jk}\omega_{ji} + \omega_{ki}\omega_{kj} = 0$.

One can associate a graph to a monomial in these generators, with vertices $\{1,\ldots,k\}$ and an edge between $i$ and $j$ if $\omega_{ij}$ occurs in the monomial. The relations imply that the monomial is zero unless this graph is a forest. 
The top degree $H^{k(d-1)}(F(\mathbf R^d,k))$ is spanned by monomials such that the corresponding graph is a tree. One can interpret such a monomial as a bracketing of $\{1,\ldots,k\}$: if $d$ is odd then the second and third relation correspond to anticommutativity and Jacobi identity for this bracketing, so the $S_k$-representation is given by  the space $\mathrm{Lie}(k)$ spanned by Lie words on $k$ letters, and if $d$ is even we get instead $\mathrm{Lie}(k) \otimes \mathrm{sgn}_k$. 
In lower degrees we get instead a sum over partitions of $\{1,\ldots,k\}$, and for each block $B$ in the partition a monomial such that the corresponding graph is a tree on the vertex set $B$. This tells us that we are considering a plethysm of symmetric functions.
Let $\widehat\Lambda$ be the degree completion of the ring of symmetric functions in infinitely many variables and let $\widehat\Lambda[[t]]$ be the power series ring in one variable over it. If $V$ is a representation of $S_k$, let $\mathrm{ch}_k(V)$ be the corresponding degree $k$ symmetric function. Define 
$$ C = \sum_{k \geq 0} h_k \in \widehat\Lambda,$$
 the sum of the trivial representation in each arity. Let 
$$ L_d = \begin{cases} \sum_{k \geq 0} \mathrm{ch}_k \mathrm{Lie}(k) \otimes \mathrm{sgn}_k t^{k(d-1)} \in \widehat \Lambda [[t]] & d \text{ even} \\ \sum_{k \geq 0} \mathrm{ch}_k \mathrm{Lie}(k) t^{k(d-1)} \in \widehat \Lambda [[t]] & d \text{ odd}  \end{cases}$$
Then
$$ \sum_{k \geq 0} \sum_{i \geq 0} t^i \mathrm{ch}_k H^i(F(\mathbf R^d,k)) = C \circ L_d.$$
The plethysm here satisfies $p_n \circ f(t,p_1,p_2,\ldots) = f(t^n,p_n,p_{2n},\ldots)$. You can also write this plethysm in terms of an infinite product or the "plethystic exponential", like Getzler does. One can describe $\mathrm{ch}_k\mathrm{Lie}(k)$ explicitly in terms of symmetric functions, and Getzler is using this in his paper. Namely, there is an isomorphism $\mathrm{Lie}(k) = \mathrm{Ind}_{C_k}^{S_k} \chi$ where $\chi$ is a primitive character of the cyclic group on $k$ elements, and 
$$ \mathrm{Ind}_{C_k}^{S_k} \chi = \frac 1 k \sum_{{d \mid k}} \mu(d)p_k^{k/d}.$$ 
Lurking in the background is an operadic perspective. The operad $e_d$ is given by the homology of the little $d$-dimensional disks and what we are trying to do is calculate the characteristic of this operad. The operad $e_d$ can be constructed as the product of the commutative operad and a $(d-1)$-fold suspension of the Lie operad by a distributive law. This tells us that the characteristic of $e_d$ is the plethysm of the characteristic of $\mathrm{Com}$ (which is $C$) and the characteristic of this $(d-1)$-fold suspension, which is $L_d$.<|endoftext|>
TITLE: "Database" of simplicial polytopes/spheres
QUESTION [11 upvotes]: Reading through various papers on polytopes I have come across really interesting examples of simplical polytopes and non-shellable (or non-PL) simplicial spheres but sometimes it is hard to keep track of their provenance. The nice thing about the simplicial case is that in order to get the whole combinatorial structure all one needs is the list of the facets. My question:
Are there any "comprehensive" resources for examples of simplicial polytopes or spheres?
Here by comprehensive I mean either examples obtained via some construction method or an aggregate of various results in the field. These examples could come in any form I guess, but a simple list of facets would probably be the easiest to work with.
I've been searching around a bit and found the following sources:

John Palmieri gives an implementation in Sage of a few interesting examples of nonshellable spheres and manifolds here.
Frank Lutz gives examples of small non-constructible spheres here and he also has a few papers enumerating $3$-manifolds with $10$ vertices here and vertex-transitive triangulations (with E. Köhler) here. Although in the latter $2$ papers the facet-lists are not given.

I am sure there are many other papers that give examples of various interesting triangulations that I am not aware of so even if you do not know of a database of such examples, answering with specific papers would be useful, I think.
Thank you!

REPLY [2 votes]: I want to mention a couple of resources that I haven't seen anyone else put up:

The simpcomp GAP package includes a simplicial complex library.  As I understand it, it includes Frank Lutz's list (and some more stuff).  See the simpcomp documentation.
Masahiro Hachimori has a small library of simplicial complexes (really more of an encyclopedia) with various properties.  It's well-curated (if not updated since 2001) and a good place to start looking for some standard-ish examples.  I've found it a useful place to start looking on several occasions.
On a different note, Frank Lutz started a journal-type collection of Electronic Geometry Models.  The emphasis is more on visualization than facet lists.  If you want to e.g. visualize Rudin's non-shellable ball, though, then this is the place to go!  Unfortunately, it's looking a bit dated:  the last submission was 2013, and the visualization requires java for best results.<|endoftext|>
TITLE: The law of large numbers for diverging moments
QUESTION [5 upvotes]: I have a random variable $X$ whose first and second moments are given as
$$
E[X] \propto C_n^{1-a},\quad E[X^2] \propto C_n^{2-a}\quad (0 < a < 1),
$$
where $C_n$ satisfies
$$
\lim_{n\to\infty} C_n = \infty, \quad \lim_{n\to\infty} \frac{C_n^a}{n} = 0.
$$
I want to see the limit of a ratio, 
$$
R_n = \frac{\sum_{i=1}^n X_i^2}{(\sum_{i=1}^n X_i)^2},
$$
where $X_1,\dots, X_n$ is i.i.d. samples of $X$. If the first and second moment is finite, one can easily see that
$$
\lim_{n\to\infty} R_n = \lim_{n\to\infty} \frac{1}{n} \frac{\sum_{i=1}^n X_i^2/n}{(\sum_{i=1}^n X_i)^2/n^2} = 0,
$$
by the law of large numbers. But in this case, $C_n$ also goes to infinity and as far as I know, the law of large numbers does not hold for infinite moments. At first, I forgot this fact and just thought as follows,
$$
\lim_{n\to\infty} R_n \propto \lim_{n\to\infty} \frac{1}{n} \frac{C_n^{2-a}}{C_n^{2-2a}}
= \lim_{n\to\infty} \frac{C_n^a}{n} = 0,
$$
by the condition given to $C_n$. I empirically checked whether $R_n$ converges to zero as $n\to\infty$, and it indeed goes to zero. Is there any clear proof, or if it is wrong, can anybody explain why it is wrong?
Thanks in advance for the answers.

Sorry for the confusion. What Iosif Pinelis interpreted is right. To be more specific, I have a r.v. $X_n$ constrained on an interval $[0, C_n]$, and its first and second moments
are given as
$$
EX_n = \frac{C_n^{1-a}}{1-a} - \gamma(1-a, C_n),\quad EX_n^2 = \frac{C_n^{2-a}}{2-a} - \gamma(2-a, C_n),
$$
where $\gamma(\cdot,\cdot)$ is a lower incomplete gamma function. I compute the ratio $R_n$ by drawing $n$ i.i.d. samples $X_{n,1}, \dots X_{n,n}$ of $X_n$. For a fixed $n$ (and thus $C_n$), there is no problem, but I want to know how $R_n$ behaves as $n\to\infty$, so $C_n\to\infty$.

REPLY [4 votes]: I interpret the question as follows (cf. the comment by Nate Eldredge).
For each natural $n$, let $X_n,X_{n,1},\dots,X_{n,n}$ be independent identically distributed (i.i.d.) random variables (r.v.'s) such that 
$$
EX_n \bowtie C_n^{1-a},\quad EX_n^2\bowtie C_n^{2-a}\quad (0 < a < 1),
$$
where $C_n$ satisfies
$$(1)\qquad
\lim_{n\to\infty} C_n = \infty, \quad \lim_{n\to\infty} \frac{C_n^a}{n} = 0, 
$$
and $a\bowtie b$ means that $a\triangleleft b\triangleleft a$ and $a\triangleleft b$ means $a=O(b)$. 
Let
$$
R_n := \frac{\sum_{i=1}^n X_{n,i}^2}{(\sum_{i=1}^n X_{n,i})^2}. 
$$
Show that $R_n\to0$; here and elsewhere, the convergence of r.v.'s is in probability, as $n\to\infty$.

The conclusion $R_n\to0$ is indeed true and can be proved as follows, actually under the more general condition
$$(2)\qquad 
EX_n \triangleright C_n^{1-a},\quad EX_n^2 \triangleleft C_n^{2-a}\quad (0 < a < 1),
$$
where $a\triangleright b$ means $b\triangleleft a$. 
For each natural $n$, let $V_n,V_{n,1},\dots,V_{n,n}$ be i.i.d. r.v.'s. 
By the well-known necessary and sufficient condition for the (weak) law of large numbers (LLN) (see e.g. Theorem 3 of Ch. IX of Petrov), for $\sum_{i=1}^n V_{n,i}\to0$ it is sufficient that for each real $\tau>0$ 
(i) $nP(|V_n|\ge\tau)\to0$, 
(ii) $nEV_n^2\,I\{|V_n|<\tau\}\to0$, and 
(iii) $nEV_n\,I\{|V_n|<\tau\}\to0$, where $I$ denotes the indicator.  
Let now $Y_{n,i}:=X_{n,i}^2/B_n$ and $Y_n:=X_n^2/B_n$, where $B_n:=\rho_n n C_n^{2-a}$ and the $\rho_n$'s are any positive real numbers such that $\rho_n\to\infty$ and $\rho_n C_n^a/n\to0$; by (1), such $\rho_n$'s exist. Then, by (2), for each real $\tau>0$,
$$nP(|Y_n|\ge\tau)\le nE|Y_n|/\tau=nEX_n^2/(\tau B_n)\triangleleft nC_n^{2-a}/B_n=1/\rho_n\to0,$$ 
$$nEY_n^2\,I\{|Y_n|<\tau\}\le\tau nE|Y_n|\triangleleft1/\rho_n\to0\ \text{(cf. the previous line)},$$
$$\big|nEY_n\,I\{|Y_n|<\tau\}\big|\le nE|Y_n|\triangleleft1/\rho_n\to0.$$ 
So, by the above sufficient condition for the LLN, 
$$(3)\qquad \sum_{i=1}^n X_{n,i}^2/B_n=\sum_{i=1}^n Y_{n,i}\to0.
$$
Next, let $Z_{n,i}:=(X_{n,i}-EX_{n,i})/E_n$ and $Z_n:=(X_n-EX_n)/E_n$, where $E_n:=nC_n^{1-a}$. Then, by (2) and (1), for each real $\tau>0$,
$$nP(|Z_n|\ge\tau)\le nEZ_n^2/\tau^2\triangleleft nEX_n^2/E_n^2\triangleleft nC_n^{2-a}/E_n^2=C_n^a/n\to0,
$$ 
$$nEZ_n^2\,I\{|Z_n|<\tau\}\le nEZ_n^2\to0\ \text{(cf. the previous line)},$$
$$\big|nEZ_n\,I\{|Z_n|<\tau\}\big|=\big|nEZ_n\,I\{|Z_n|\ge\tau\}\big|\le nEZ_n^2/\tau\to0.$$ 
So, by the same sufficient condition for the LLN, $\sum_{i=1}^n Z_{n,i}\to0$; that is, 
$\sum_{i=1}^n X_{n,i}/E_n-nEX_n/E_n\to0.
$
On the other hand, by (2), $nEX_n/E_n\triangleright1$. So, $\sum_{i=1}^n X_{n,i}/E_n\triangleright1$ and hence 
$$\Big(\sum_{i=1}^n X_{n,i}\Big)^2\triangleright E_n^2=n^2C_n^{2-2a}. 
$$
Comparing this with (3), one concludes that 
$$R_n<<\frac{B_n}{E_n^2}=\rho_n C_n^a/n\to0, 
$$
by the choice of $\rho_n$. 
This completes the proof.<|endoftext|>
TITLE: Is there a Gelfand-Naimark-like characterization of group algebras $L_1(G)$?
QUESTION [12 upvotes]: The Gelfand-Naimark theorem establishes that a complex commutative Banach algebra $A$ with an identity and an involution $x\to x^*$ satisfying $\|x x^*\|=\|x\|^2$ is (isometrically isomorphic to) a $C(K)$ space where $K$ is the compact space consisting of the maximal ideals in $A$. 
Is there an analogous characterization of the complex commutative Banach algebras with an identity and an involution which are a convolution algebra $L_1(G)$ with $G$ a locally compact abelian group?

REPLY [5 votes]: Rieffel has given in his Ph.D. thesis a fairly satisfactory characterisation of commutative $L_1(G)$-algebras. This requires some terminology.
Let $A$ be a commutative (possibly non-unital) Banach algebra and let $f\in A^*$ be a (norm-one) character. Then $f$ is termed $L^\prime$-inducing, when $A$ is an abstract $L$-space under the ordering given by the cone
$$\{x\in A\colon \|x\| = f(x)\}$$
and $| x\cdot y | \leqslant |x| \cdot |y|$ in the sense of the above-introduced ordering (here $|x|$ stands for the Banach-lattice modulus). Rieffel then offers the following characterisation:
Theorem. Let $A$ be a commutative, semi-simple Banach algebra. Suppose that every norm-one character on $A$ is $L^\prime$-inducing and $A$ is Tauberian. Then $A$ is Banach-algebra isometrically isomorphic to $L_1(G)$ for some locally compact group $G$.
It is to be noted that there exist characterisations of $L_1(S)$-algebras, where $S$ is a locally compact semigroup, that are in a spirit similar to the above result. There is a recent (2016) result by Lau and Hung extending this further to general Fourier-Stieltjes algebras.<|endoftext|>
TITLE: Mathematical writing : using an "out-of-date" notation
QUESTION [12 upvotes]: When I wrote my master's thesis, a professor who read it said that I should not use the phrase "A function of class $k$." but instead "A function of class $C^k$". I am not an expert about mathematical history of notations, but I read that in Geometric Measure Theory, H. Federer  actually uses the first one, and it seems logical for me: I think that $C^k$ is the abbreviation for "of class $k$". Therefore, employing "class $C^k$" seems like a repetition. Or maybe the other notation is just not used any more and should simply be prohibited?

REPLY [54 votes]: Federer was not exactly known, even to his contemporaries, for employing standard notation.  Here is a quote from Steenrod's 1948 Math Review of some mimeographed notes of Federer for a course on differential geometry.

The most striking feature of the book to the casual reader is the notation. The author adopts the view that certain familiar notations are misleading, and obscure the meanings of definitions and theorems. He replaces them by more elaborate notations based on the roots in set theory of the concepts represented (e.g., the polynomial x becomes the sequence of its coefficients [0,1]). A few such changes would not be worthy of comment; but he has carried out the prodigious task of applying the same stern standards to every phase of the work. The result can be described by saying that a resemblance to any notation, living or dead, is purely coincidental.<|endoftext|>
TITLE: local quasi geodesics in hyperbolic spaces
QUESTION [6 upvotes]: I asked this question on math stackexchange (see here) but didn't get any answer so I thought I post it here too.
We have the following two well-known Theorems:

T1) For all $\delta > 0, \lambda \geq 1, \varepsilon \geq 0$ there exists a constant $R = R(\delta, \lambda, \varepsilon)$ with the following property:
  If $X$ is a $\delta$-hyperbolic geodesic space, $c$ is a $(\lambda, \varepsilon)$-quasi-geodesic in $X$ and $[p,q]$ is some geodesic segment joining the endpoints of $c$, then the Hausdorff distance between $[p,q]$ and the image of $c$ is less than $R$. (Hence there is some constant such that $[p,q]$ is contained in the neighbourhood of $c$ and vice versa)
T2) Let $X$ be a $\delta$-hyperbolic geodesic space and let $c: [a,b] \to X$ be a $k$-local geodesic, where $k > 8\delta$. Then:
(i) im(c) is contained in the $2 \delta$-neighbourhood of any geodesic segment connecting the endpoints of $c$.
(ii) $[c(a),c(b)]$ is contained the $3 \delta$-neighbourhood of im(c)
(iii) $c$ is a $(\lambda, \varepsilon)$-quasi-geodesic, where $\varepsilon = 2 \delta$ and $\lambda = (k + 4 \delta)/(k - 4 \delta)$.

I am wondering if Theorem 2 (T2) is also true (in an apropriate way) for $k$-local-$(\lambda, \varepsilon)$-quasi-geodesic, i.e. that such local quasi geodesics are actually quasi-geodesics.
In the book of Bridson and Haefliger (Metric spaces of non-positive curvature) this should follow in the 'obvious' way of Theorem 1 (T1) and Theorem 2 (T2) above. However I have sme troubles writing this down explicitly.
More precisely in the above mentioned book the following is written:

"By combining (1.7) [our T1] and (1.13) [our T2] in the obvious way one gets a criterion for seeing that paths which are locally quasi-geodesic (with suitable parameters) are actually quasi-geodesics."
Question: This quote is exactly the content of my question, i.e. how to combine these two theorems and in particular what is the exact statement we get (e.g. for the constant $k$ in T2)?

My heuristic would be as follows: A local quasi geodesic is somehow close to a local geodesic by T1 which in turn is a quasi geodesic by T2. Thus the local quasi geodesic we were starting with is close to a quasi geodesic and hence one may deduce that it is itself a quasi geodesic.

REPLY [6 votes]: I don't have the book in front of me but in the book of Coornaert, Delzant and Papadopoulos you can find the fact that local quasi-geodesics are quasigeodics provided that the locality Parameter is sufficiently large.
PS: I checked, it is Théorème 1.4 of Chapitre 3<|endoftext|>
TITLE: How far wrong could the Continuum Hypothesis be?
QUESTION [24 upvotes]: I hear it's consistent with ZFC to have
$$ 2^{\aleph_0} = \aleph_n $$
for any $n = 1, 2, 3, \dots $.  How much worse can it get?
More precisely: are there models of ZFC with $2^{\aleph_0} \gt \aleph_n$ for all $n$?   What's the 'world record' when it comes to finding models where $2^{\aleph_0}$ is 'very large' in some sense?  And on the other hand, are there theorems putting interesting bounds on how large $2^{\aleph_0}$ can be?
(I count $2^{\aleph_0} \lt 2^{2^{\aleph_0}}$ as an uninteresting bound.)
According to Wikipedia:

A recent result of Carmi Merimovich shows that, for each $n \ge 1$, it is consistent with ZFC that for each $\kappa$, $2^\kappa$ is the $n$th successor of κ.  On the other hand, László Patai (1930) proved, that if $\gamma$ is an ordinal and for each infinite cardinal $\kappa$, $2^\kappa$ is the $\gamma$th successor of $\kappa$, then $\gamma$ is finite.

However, I'm interested in failures of the Continuum Hypothesis, not the Generalized Continuum Hypothesis.

REPLY [29 votes]: Solovay proved shortly after Cohen's result on the independence of CH that in any model of set theory $V$, if $\kappa^\omega=\kappa$, then there is a forcing extension in which $2^\omega=\kappa$. The forcing is simply $\text{Add}(\omega,\kappa)$, the forcing to add $\kappa$ many Cohen reals. Thus, the continuum $2^\omega$ can be $\aleph_{\omega+1}$ or $\aleph_{\omega_1}$ or $\aleph_{\omega_1+\omega^2+17}$ or what have you, even weakly inaccessible, if you had such a cardinal available in the ground model. But it cannot be $\aleph_\omega$ or $\aleph_{\omega_1+\omega}$ or any cardinal with cofinality $\omega$, because $(2^\omega)^\omega=2^\omega$ but König's theorem shows $\kappa^{\text{cof}(\kappa)}>\kappa$ for any infinite cardinal $\kappa$. 
In particular, if you start in a model of the GCH, then the continuum can be made to be equal to $\delta^+$ for any infinite successor cardinal in the ground model, and this is cardinal-preserving forcing, so the meaning of the successor function is the same in the ground model as in the extension.
Easton vastly generalized this result to show that if you start in a model of the GCH, and if $E$ is any function defined on the infinite regular cardinals and obeying the following requirements (now called an Easton function)

$E(\kappa)>\kappa$
Further, $\text{cof}(E(\kappa))>\kappa$
$\kappa\leq\lambda\to E(\kappa)\leq E(\lambda)$

then $E$ can become the continuum function $\kappa\mapsto 2^\kappa=E(\kappa)$ as computed in a cardinal-preserving forcing extension of the universe. See Easton's theorem. 
The Easton requirements amount to the trivial requirements on the continuum function imposed by the facts that $\kappa\leq\lambda\to 2^\kappa\leq 2^\lambda$ and König's theorem $\text{cof}(2^\kappa)>\kappa$. 
As a result, we can have $2^{\aleph_n}=\aleph_{\omega^2+\omega+p_n}$ where $p_n$ is the $n^{th}$ prime, if we like. There is infinite flexibility, and basically anything is allowed subject to the Easton requirements, which are trivial limitations.
Easton's theorem does not apply at all to singular cardinals, however, and controlling the continuum function at singular cardinals is an extremely subtle issue.<|endoftext|>
TITLE: Large cardinal consistency strength and size
QUESTION [12 upvotes]: My understanding is that large cardinals are ordered by "consistency strength", but how does this correlate with their size (cardinality)?
More specifically, are there any systematic results on the lines of:
If A and B are two types of large cardinals such that 
Cons(ZFC + Type A exists) => Cons( ZFC + Type B exists)
THEN
Cardinality of smallest Type A cardinal >= Cardinality of smallest Type B cardinal

REPLY [5 votes]: Since the relevant kind of large cardinal was just named on Joel Hamkins's blog, let me point to it as an interesting dual to some of the existing answer.
A cardinal $\kappa$ is otherworldly if there is some $\lambda$ such that $V_\kappa\prec V_\lambda$. (Exercise 4.1.7 in Drake's large cardinal book shows that otherworldly implies worldly, hence the name). We say $\kappa$ is totally otherworldly if there are arbitrarily large $\lambda$ such that $V_\kappa\prec V_\lambda$.
Now totally otherworldly cardinals are very low in consistency strength: a single inaccessible cardinal suffices to show the consistency of a proper class of totally otherworldlies, via the usual Löwenheim–Skolem argument. But every totally otherworldly cardinal is $\Sigma_2$-correct, which makes it at least larger than the least measurable, the least superstrong, and the least huge, if consistent.
So this is an example of very low consistency strength but very large size.<|endoftext|>
TITLE: Is there a one relator group with property (T)?
QUESTION [8 upvotes]: Is there a one-relator group with property (T)?
That is, is there an $n > 2$, and some $x \in F_n$ (the free group on $n$ generators) such that the quotient of $F_n$ by the normal subgroup generated by $x$ has Kazhdan's property $\mathrm{(T)}$ ?

REPLY [3 votes]: Property (T) implies Property FA: every action on a tree has a global fixpoint. The Magnus–Moldavanskii hierarchy expresses every one-relator group as (a subgroup of) an HNN extension of a "simpler" one-relator group (over a free subgroup, but that doesn't matter here). The hierarchy terminates in finitely many steps with a free group or finite cyclic group. Putting this together, any group with (T) that embeds in a one-relator group fixes a vertex of the Bass–Serre tree, so is a subgroup of a simpler one-relator group, and acts on the Bass–Serre tree of that one-relator group, etc until we conclude that, since it can't be a non-trivial subgroup of a free group (infinite abelianization), it must be finite.<|endoftext|>
TITLE: What are Reinert's reproaches to the Ricardo theory?
QUESTION [11 upvotes]: Economists accuse me in vulgarization of their science, so I'll edit the text from the very beginning to remove the inaccuracies.  
Main question
I have just read the book by a norwegian economist, Erik Reinert, "How Rich Countries Got Rich… and Why Poor Countries Stay Poor". I think it is very interesting for people who want to understand how modern economics works, and, in particular, where poverty comes from. And, what is unexpected, the problem, as Reinert it presents, seems to be not so complicated from the point of view of a mathematician.  
Reinert blames in everything the Ricardo theory of comparative advantages.  He writes that it is exactly this theory that justifies poverty in the modern world. He gives numerous examples, when arguments based on the Ricardo theory made some countries extremely poor, without hope for correcting the situation in future. He paints horrible pictures of the situation in Mongolia, Peru, Equador, Haiti, Russia, etc. Simultaneously, from what he writes it becomes clear why there are so many migrants in the modern world, so many jobless people, so many refugees, etc. -- so I recommend his book to everyone.
My main question is what I wrote in the title:

What are Reinert's reproaches to the Ricardo theory?

Ricardo examples
This question was in the previous version of the text, and since nobody answered yet, I think that this must be difficut. That is why I want to add another question that I believe is much easier and can clarify the situation: 

In which extent the Ricardo examples (without further theory, just examples) explain the problems?

This needs explanation. First, I foresee new accusations in vulgarization from economists. So I think it will be useful to remind that vulgarization is a standard trick in mathematics: when people are trying to understand a problem, they often remove something from the theory (or replace this detail with something else), and look what happens then. A classic example is the Parallel postulate: in attempts to understand what is wrong Lobachevsky replaced this axiom in the theory and looked what happens after that. In modern mathematics this trick is widely used, since all axiomatic theories are based on this idea. 
Second, I have to explain to mathematicians what is meant by the Ricardo examples. 
The classic Ricardo example (I change the numbers for simplicity). Suppose we consider two countries, $A$ and $B$, and each of them produces two goods, $G_1$ and $G_2$, and the table of expenses is as follows:
$$
\begin{matrix}
 & G_1 & G_2 \\
A: & 1 & 1 \\
B: & 2 & 4 \\
\end{matrix}
$$
(this means that in the country $A$ one unit of $G_1$ costs 1 man-hour, and the same for $G_2$, and in the country $B$ one unit of $G_1$ costs 2 man-hours, while one unit of $G_2$ costs 4 man-hours). 
Both goods, $G_1$ and $G_2$, are cheaper in the coutry $A$, so it seems evident, that it is  more profitable to produce both $G_1$ and $G_2$ in the country $A$. But Ricardo notices that the difference in the comparative expences changes the situation: if the country $A$ conveys  the production of $G_1$ to the country $B$, while $B$ conveys $G_2$ to $A$, and they begin to trade, this becomes more profitable for both countries, since

in $A$ each man-hour still gives one unit of $G_2$, but when selling it to $B$, $A$ takes two units of $G_1$ (instead of one unit of $G_1$, as it was when they did not trade),
in $B$ each man-hour still gives $1/2$ unit of $G_1$, but when selling it to $A$, $B$ takes $1/2$ unit of $G_2$ (instead of $1/4$ of $G_2$, as it was when they did not trade).

This justifies the following observation

If in a given country $A$ the expences in the production of a good $G_2$ in comparison with another good, $G_1$, is less than in a country $B$, then it is more profitable (for both countries) to produce $G_2$ in the country $A$ (and to trade).  

The examples like this do not exhaust the Ricardo theory, of course, but I suspect that the essential effects can be explained with their help. As illustrations I suggest to consider the following two situations. 
Example #1. Consider three countries with two goods and with the following table of expences: 
$$
\begin{matrix}
 & G_1 & G_2 \\
A: & 1 & 1 \\
B: & 2 & 4 \\
C: & 4 & 2 \\
\end{matrix}
$$
Here, according to Ricardo, 

the least comparative expenses for producing the good $G_1$ are in the country $B$, so the good $G_1$ must be produced in the country $B$,
at the same time the least comparative expenses for producing the good $G_2$ are in the country $C$, so the good $G_2$ must be produced in the country $C$.

And the Ricardo trick gives the conclusion that 

the country $A$ must abandon the production of both $G_1$ and $G_2$. 

Example #2. In the Ricardo example nothing prevents us to consider workforce as another, third commodity in this situation, and to look at the comparative expences in its production. We can just change the unit of measure, and use the good $G_1$ instead of "man-hours", then the table of expenses becomes the following:
$$
\begin{matrix}
 & \text{workforce} & G_2 \\
A: & 1 & 1 \\
B: & 1/2 & 2 \\
\end{matrix}
$$
(this means that in the country $A$ one man-hour costs one unit of $G_1$ and the same for one unit of $G_2$, and in the country $B$ one man-hour costs $1/2$ unit of $G_1$, while one unit of $G_2$ costs 2 units of $G_1$). 
And the Ricardo trick gives the conclusion that

the country $A$ must abandon the production of its workforce (i.e. make all its citizens jobless, and import the workforce from $B$).

Qualitative differences
This trick with the Ricardo examples seems to be an easiest way to explain the problem to an outlooker. That's why I ask specialists what I wrote above: in which extent these examples explain the situation? My question is, if it is possible that the general theory prescribes qualitatively the same conclusions as in these examples (i.e. the calculations just change the quantitative answer from $0$ to a little $\varepsilon>0$, and that's all)?  In particular,

Is it possible that in Example #1 the countries $B$ and $C$ take off all (or most of) the production from the country $A$, even if in $A$  it is more effective?
Is it possible that in Example #2 the country $A$ comes to a situation when it must discharge all (or most of) its citizens, even if the production there is more effective than in the other countries?

Reinert's criticism inspires this suspicion. 
Experiments
One more question: 

Do economists use experiments in this field? 

(I asked this in economic forum without success.)

REPLY [3 votes]: There is an extensive literature on Ricardian models with many goods and countries. For a survey which also answers some of the questions you make in your later edit, see:
Eaton, Jonathan, and Samuel Kortum. 2012. "Putting Ricardo to Work." Journal of Economic Perspectives, 26(2): 65-90.
https://www.aeaweb.org/articles.php?doi=10.1257/jep.26.2.65 (freely accessible full-text)
In a Ricardian world there is no unemployment by assumption. What ensures full employment is that wages need not be equal in all countries. The level of wages adjusts in response to productivity differences until a point is reached where each country is the cheapest supplier of some commodities to some destinations (which might include itself). Which commodities these are for each country is determined by the Ricardian principle of comparative advantage.
Original research publications:

Two countries, continuum of goods: Dornbusch, Rudiger, Stanley Fischer, and Paul Anthony Samuelson. "Comparative advantage, trade, and payments in a Ricardian model with a continuum of goods." The American Economic Review (1977): 823-839.  http://www.jstor.org/stable/1828066 
Many countries, many goods: Eaton, Jonathan, and Samuel Kortum. "Technology, geography, and trade." Econometrica (2002): 1741-1779. http://www.jstor.org/stable/3082019 

Ricardian theory has many simplifications: only one input in production, constant returns to scale, perfect competition, full employment, no dynamic effects on technology and resources … Weakening these assumptions may reverse the conclusions that Ricardo reaches and economists have come up with many scenarios where free trade may not be beneficial. 
But the papers cited above show that under Ricardo's assumptions it is possible to give a fully rigorous mathematical account of the theory with no logical problems while preserving many of the qualitative features of the 2-country, 2-goods case.<|endoftext|>
TITLE: Lebesgue differentiation theorem holds on locally doubling space?
QUESTION [7 upvotes]: It's known that for a metric space with doubling measure $(X,\mu)$, the Lebesgue differentiation theorem holds , i.e. If $f:X\to \mathbb{R}$ is a locally integrable function, then $\mu$-a.e. points are Lebesgue points. 
Can we relax the doubling condition to local doubling or local uniformly doubling?

REPLY [8 votes]: Yes, one can relax the doubling condition to 
$$\limsup_{r \to 0} \frac{\mu(B(x,2r))}{\mu(B(x,r))}<\infty \ \text{for}\ \mu-\text{a.e.}\ x \in X$$
This is done in Section 3.4 of the book ``Sobolev Spaces on Metric Measure Spaces" by Heinonen, Koskela, Shanmugalingam, and Tyson. 
The proof is essentially the same as in the doubling case (i.e., it still boils down to the Vitali covering property), but somewhat more techincal.<|endoftext|>
TITLE: Moduli space of boundary maps with prescribed chain and homology groups?
QUESTION [16 upvotes]: Let $R$ be a reasonable ring (maybe I mean a PID, or $\mathbb{Z}$, and when sufficiently desperate, a field). Now consider fixed sequences $C_n$ and $H_n$ of $R$-modules, which are tame in every possible sense -- finite rank, zero for large enough $n$, etc. Here's the problem:

What is the set (space? category?) of all $R$-linear maps $d_n:C_n \to C_{n-1}$ so that $(C_\bullet,d_\bullet)$ forms a chain complex with homology groups $H_n$?

I've tried to figure things out in baby cases without much luck in terms of seeing the bigger picture. You can easily create situations where no $d_\bullet$s will suffice, but most fascinating are the cases where several different sequences of boundary operators will work!
A very general thing to do is fix a basis of all chain groups in sight and examine conditions that entries of various $d_\bullet$-representing matrices have to satisfy. Just requiring the $d_\bullet$s to produce a chain complex yields an algebraic variety generated by a system of multivariate quadratic equations (which sounds nightmarish). So unless forcing the homology to be prescribed by $H_n$s greatly simplifies things, I should not expect a very explicit answer. However, I'd like to know what the obstructions to getting such an answer might be.
In the cases of interest, all my $d_\bullet$s can be assumed to have an upper-triangular form as $R$-matrices, although I'd also be interested in a solution that ignores such structural constraints.
Update: Thanks to David Speyer's answer, which provides a wealth of information on the problem when $R$ is a field. I'm still seeking answers for more general choices of $R$.

REPLY [7 votes]: Over a field $k$, this is a special case of quiver loci. The general framework is that you are given a triangular array $(r_{ij})_{1 \leq i \leq j \leq n}$ of nonnegative integers. Consider a list of $n-1$ matrices $M_1$, $M_2$, …, $M_{n-1}$ of size $r_{ii} \times r_{(i+1)(i+1)}$. The quiver locus is the set where $\mathrm{rank}(M_i M_{i+1} \cdots M_{j-1}) \leq r_{ij}$. Your case is where $r_{ij}=0$ for $j-i \geq 2$. 
See Knutson, Miller and Shimozono, and the many citations therein.
To be a bit more explicit, if you want $\dim V_i = d_i$ and $\dim H_i(V_{\bullet})=h_i$, then take $r_{ii} = d_i$, compute $r_{i(i+1)}$ by solving the linear equations $r_{(i-1)i} + r_{i(i+1)} + h_i= d_i$, and put all other $r_{ij}=0$.<|endoftext|>
TITLE: Morava $K$-theory of $K( \mathbb{Z}/p^2)$
QUESTION [10 upvotes]: The $p$-adic completion of $K( \mathbb{F}_p)$ is known (by Quillen's calculation) to be $H \mathbb{Z}_p$; in particular, $K(\mathbb{F}_p)$ is acyclic with respect to all Morava $K$-theories $K(n), 0 < n < \infty$ at the implicit prime $p$.
Does this hold for $\mathbb{Z}/p^2$? That is, is $L_{K(1)} K( \mathbb{Z}/p^2)$ (with the implicit prime for $K(1)$ equal to $p$) nontrivial? (A theorem of Mitchell implies that all the higher $K(i)$-localizations are trivial.)

REPLY [11 votes]: $L_{K(1)}K({\Bbb Z}/p^n)$ is always trivial. This is Proposition 2.14 in the recent preprint by Bhatt, Clausen and yourself. Alternatively, this also appears in recent work by Land, Meier and Tamme.<|endoftext|>
TITLE: $C^k$ one-parameter family of metrics
QUESTION [8 upvotes]: Consider a smooth Riemannian manifold $M$ and a $C^k$ one-parameter family of Riemannian metrics $g_t$ on $M$. Here $k$ could be any integer, $k$ could be infinity, when the one-parameter family $g_t$ is smooth in time, or $k = \omega$, when the one-parameter family $g_t$ is real analytic in time. Now, as the metric varies, the Laplacian associated to the metric varies, and hence its spectrum also varies in time. My question is, if $g_t$ is $C^k$ in time, are the eigenvalues of the Laplacian also $C^k$ in time for $k$ integer, $k = \infty$ or $k = \omega$? In case such results are well-known (which I am assuming they are), what is a good reference to learn about such results? Thanks for any guidance.
Edit after Umberto Lupo's answer: I was somewhat more interested in the compact case, which Umberto's answer and the paper he cites cover very nicely. I would, however, be highly interested in learning about possible results in the non-compact setting as well.

REPLY [6 votes]: The references given in this answer to a related issue provide at least a partial answer to your question. 
On a complete Riemannian manifold $(M,g)$, the Laplacian associated with $g$ is essentially self-adjoint on $C_0^\infty(M) \subset L^2(M)$. If $M$ is also compact, then its resolvent is also a compact operator.  Therefore, under these two additional assumptions you will satisfy the hypotheses of the main theorem on page 1 in the second article linked there:

Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. (pdf).

Cases (A)-(F) and (N) answer your question at a selection of regularity scales. I hope this helps.<|endoftext|>
TITLE: Minimal number of algebra generators of a group ring
QUESTION [5 upvotes]: Let $k$ be a commutative ring with unit and let $(A,\varepsilon)$ be a (not necessarily commutative) augmented, finitely generated $k$-algebra with augmentation ideal $I$. 
If $\mu(A)$ denotes the minimal number of algebra generators of $A$ and $\mu(I)$ the minimal number of generators of $I$ as ideal, we have $\mu(I) \le \mu(A)$. 
(For, if $A$ is generated as algebra by $x_1,...,x_n$, then it's also generated by $y_i := x_i - \varepsilon(x_i)\cdot 1\in I$. If $y$ is any element in $I$, write $y = a + f\langle y_1,...,y_n\rangle$ where $a \in k$ and $f$ is a polynomial in the (noncommuting) vaiables $y_1,..,y_n$ having no constant term. Hence $0=\varepsilon(y)=\varepsilon(a)=a$, i.e. $y = f\langle y_1,...,y_n\rangle$ is in the ideal generated by $(y_1,...,y_n)$. This shows $I=(y_1,...,y_n)$ and hence $\mu(I) \le n$)
I wonder what can be said in the case of group algebras: 
Question: Does for finitely generated groups the equality 
$\mu(\mathbb{Z}[G]) = \mu(I)$ hold ? 
Note: If $\mu(G)$ denotes the minimal number of generators of the group $G$, we have for finite $G$: 
$$\mu(I) \le \mu(\mathbb{Z}[G])  \le \mu(G).$$
So the equality would follow from $\mu(I)=\mu(G)$ but I don't know if the latter is true.

REPLY [6 votes]: The paper The presentation rank of a direct product of finite groups by Cossey, Gruenberg and Kovacs shows that the difference between the minimal number of generators of $G$ and  $I$ can be as big as you like for finite groups.<|endoftext|>
TITLE: Almost complex manifold fibered by holomorphic sub-manifolds
QUESTION [6 upvotes]: Suppose $p:(M,J)\rightarrow (N,I)$ is a submersion between smooth manifolds M and N such that:

$(M,J)$ is an almost-complex manifold.
$(N,I)$ is a complex manifold where $I$ is the integrable almost-complex structure.
$p$ is almost-holomorphic, that is $dp\circ J=I\circ dp.$
The fibers of $p,$ which are $J$-holomorphic by the previous condition, are holomorphic in the usual sense.  That is, the almost complex structure $J$ restricted to any fiber $p^{-1}(x)$ for $x\in M$ is integrable.

Is $J$ integrable?  
I believe that the answer to this question is likely to be no, though I am having trouble cooking up a counter-example.  The only immediate consequence I see is that the Nijenhuis tensor on $M$ must take values in vertical vector fields.  I am primarily interested in the case where $M$ and $N$ are not compact so standard techniques of deformation theory of compact complex manifold don't easily apply, though I don't know the answer in the case when $p$ is proper either.  My google searches haven't turned up anything useful yet, so I'd be very interested to hear about any additional conditions one could add to this list, namely additional structure on $M,$ which guarantees that $J$ is integrable if, as I suspect, the answer to my original question is negative.  Thanks in advance for any comments.

REPLY [2 votes]: If I am not reading something wrong, or did some silly calculation mistake,  than the answer to your question is no. Let us take $M=\mathbb R^4$ and $N=\mathbb R^2=\mathbb C$ and the map $p$  just the projection $(x_1,y_1,x_2,y_2)\mapsto (x_1,y_1)$. We can choose $J=\left[\begin{matrix} I & A\\0  & I \end{matrix}\right]$, where $I$ is the standard complex structure on $\mathbb R^2$, just as long as $A$ is of the form $\left[\begin{matrix} a & b\\b  & -a \end{matrix}\right]$. Almost any nonconstant $A$ should do the trick. For example, for  $A=\left[\begin{matrix} 0 & x_1\\x_1  & 0 \end{matrix}\right]$, you can check that $N_J(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2})=-\frac{\partial}{\partial x_1}$.<|endoftext|>
TITLE: electron configuration on manifolds
QUESTION [6 upvotes]: Let $M$ be a Riemannian manifold. For $k\geq 2$, suppose there are $k$ particles whose mass and volume can be regarded as zero and negatively charged with electricity equally. These $k$ particles move on $M$ freely without frictions and mutually repulse from each  other. When these $k$ particles stop at $(x_1,\cdots, x_k)$ and be stable under small disturbance, we just call $(x_1,\cdots,x_k)$ a "electron configuration". The collection of all "electron configuration"s  form the "electron configuration space". 
Question: Are there any reference about the "electron configuration space"?  What formal names of these spaces should I search online?
Question: Any references about the cohomology ring of the "electron configuration space"?

REPLY [6 votes]: There seems to be a considerable body of work on this:
Minimal Riesz energy point configurations for rectifiable d-dimensional manifolds
D.P. Hardin, , E.B. Saff1, 
More by Hardin and Saff. 
http://personales.unican.es/beltranc/archivos/FoCMBeltran2011volume.pdf (Refers to work by Noam Elkies, who will probably have a much more in depth response)
Perhaps most relevant: Papers by Burton Randol and coworkers:
Burton Randol. Stable configurations of repelling points on manifolds. Proc.
Amer. Math. Soc., 142:2769–2773, 2014.
Nechaeva and Randol<|endoftext|>
TITLE: Serre's surjective theorem importance
QUESTION [7 upvotes]: I'm studying Serre's paper in wich he shows the following theorem:
Let K be a number field, $E$ an elliptic curve over K without CM. Then the representation $$\rho_{\ell}:\mathrm{Gal}(\bar K/K)\longrightarrow\mathrm{Aut}(E[\ell])$$ is surjective for all but finitely many prime numbers $\ell$.
I see the beauty of this theorem, however what consequence has it? What is its importance?
For example I know that for a non-CM semi-stable elliptic curve $E$ over $\mathbb{Q}$, if the $\ell$-adic representation is surjective then $E[\ell](\mathbb{Q})$ is trivial.

REPLY [3 votes]: There is an obvious application to the inverse Galois problem. And, some other instances of this problem have been extensively studied along the same lines (that is by proving surjectivity of some Galois reps attached to cuspidal modular forms for instance).<|endoftext|>
TITLE: Is this integral representation of $\zeta(2n+1)$ known?
QUESTION [43 upvotes]: Background: I'm an undergraduate at an institution with no researchers in analytic number theory, and no ties to the analytic number theory community.  I believe I have found what is, as far as I can tell after some googling, a new family of integral representations of $\zeta(2n+1)$.  I was told I should post this here by an algebraic number theorist at my university, to see if it was a known result or not.
Feel free to look at Equations (4), (5), and (7), to see if you recognize them before reading the entire text.  Most of this text is for the special case of $\zeta(3)$, but all results generalize quite readily to the case of $\zeta(2n+1)$, which I discuss at the end.
Beginning of setup
As a preliminary step, we will split the sum $\zeta(3)$ into even and odd parts. Observe that $$
\sum_{n=1}^\infty \frac{1}{(2n)^3} = \frac{1}{8} \sum_{n=1}^\infty \frac{1}{n^3} = \frac{1}{8}\zeta(3)
$$
Therefore $$
\sum_{n=1}^\infty \frac{1}{(2n+1)^3} = \frac{7}{8}\zeta(3) \tag{1}
$$
We now write $$
\frac{1}{(2n+1)^3} = \left(\int_0^1 x^{2n} dx\right)^3 = \int_0^1 \int_0^1 \int_0^1 x^{2n} y^{2n} z^{2n} \tag{2} dx dy dz
$$
and plug this triple integral into Equation (1), obtaining $$
\zeta(3) = \frac{8}{7} \sum_{n=0}^\infty \left[ \int_0^1 \int_0^1 \int_0^1 x^{2n} y^{2n} z^{2n} dxdydz \right]
$$
Using the absolute convergence of the geometric series on $(0,1)$, we can interchange the sum and integrals, obtaining $$
\zeta(3) = \frac{8}{7} \int_0^1 \int_0^1 \int_0^1 \frac{1}{1-x^2y^2z^2} dxdydz \tag{3}
$$
End of setup
With this part out of the way, I will now describe the integral representations that form the meat of my question.
Consider the $u$-substitution for the integral given by Equation (3) $$
x = \frac{\sinh{u}}{\cosh{v}}, \qquad y = \frac{\sinh{v}}{\cosh{w}}, \qquad z = \frac{\sinh{w}}{\cosh{u}}
$$
Some computation shows that $$
dxdydz = [1-(\tanh{u}\tanh{v}\tanh{w})^2]dudvdw
$$
But dividing both sides by $1-(\tanh{u}\tanh{v}\tanh{w})^2$ and rewriting the left-hand side in terms of $(x,y,z)$ gives $$
\frac{1}{1-x^2y^2z^2}dxdydz = dudvdw
$$
After changing limits, the transformed integral then reads $$
\zeta(3) = \frac{8}{7} \int_0^\infty \int_0^{\sinh^{-1}(\cosh(w))} \int_0^{\sinh^{-1}(\cosh(v))} dudvdw \\
= \frac{8}{7} \int_0^\infty \int_0^{\sinh^{-1}(\cosh(w))} \sinh^{-1}(\cosh(v)) dv dw \tag{4}
$$
But the most interesting part of the above is the following generalization. Let $f : (0,\infty) \to (0,\infty)$ be a function satisfying the following three properties.

$f$ is surjective, with $\lim\limits_{x\to 0} f(x) = 0$ and $\lim\limits_{x \to \infty} f(x) = \infty$
$f$ is invertible
$f$ is differentiable

Now consider the $u$-substitution $$
x = \frac{\sinh{f(u)}}{\cosh{f(v)}}, \qquad y = \frac{\sinh{f(v)}}{\cosh{f(w)}}, \qquad z = \frac{\sinh{f(w)}}{\cosh{f(u)}}
$$
We then find that $$
dxdydz = f'(u)f'(v)f'(w)[1-(\tanh{f(u)}\tanh{f(v)}\tanh{f(w)})^2]dudvdw
$$
and hence $$
\zeta(3) = \frac{8}{7} \int_0^\infty \int_0^{g(w)} \int_0^{g(v)} f'(u)f'(v)f'(w) dudvdw \tag{5}
$$
where $g(x) = f^{-1}(\sinh^{-1}(\cosh(f(x))))$.
All of the above can be generalized to $\zeta(2n+1)$.  Equation (3) becomes $$
\zeta(2n+1) = \frac{2^{2n+1}}{2^{2n+1}-1} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{1}{1-x_1^2 x_2^2 \cdots x_{2n+1}^2} dx_1 dx_2 \cdots dx_{2n+1} \tag{6}
$$
Our $u$-substitution becomes (where $u_{(2n+1)+1} = u_1$) $$
x_i = \frac{\sinh(f(u_i))}{\cosh(f(u_{i+1}))}
$$
which transforms Equation (6) to $$
\zeta(2n+1) = \frac{2^{2n+1}}{2^{2n+1}-1} \int_0^\infty \int_0^{g(u_{2n})} \int_0^{g(u_{2n-1})} \cdots \int_0^{g(u_1)} f'(u_1)f'(u_2)\cdots f'(u_{2n+1}) du_1 du_2 \cdots du_{2n+1} \tag{7}
$$
My questions are then: 

Are Equations (4), (5), and (7) known results?
Equation (4) is a volume integral. Exploring the region of integration in Mathematica numerically, it looks like an octant of a hyperbolic cube with vertices at infinity. Is this in fact the case?

REPLY [23 votes]: As indicated in my comment, some of these integrals are essentially known, and involve the hyperbolic "Beukers-Kolk-Calabi" change of variables.
In particular, in this paper, Z. Silagadze shows in (27) that
\begin{equation*}
 \zeta(n) = \frac{2^n}{2^n-1}\int_0^1\cdots\int_0^1 \frac{dx_1\cdots dx_n}{1-x_1^2\cdots x_n^2},
\end{equation*}
which clearly yields (7) above. 
The hyperbolic change of variables is used later in that paper to obtain a related formula over the $2n$-simplex $\Delta_{2n}$, namely
\begin{equation*}
  \zeta(2n+1) = -\frac{1}{2n}\frac{2^{2n+1}}{2^{2n+1}-1}\int_{\Delta_{2n}} \log(\tan x_1 \tan x_2\ldots \tan x_{2n})dx_1\cdots dx_{2n}.
\end{equation*}
This integral is amenable to further interesting modifications.
EDIT:

The arXiv paper cited above has been published in the journal "Resonance"
Another paper by the same author (Z. Silagadze) comments on these integrals.
The "Beukers-Kolk-Calabi" change of variables was also considered in this paper.
This paper (cited below by Benjamin Dickman) is also relevant here.
The relation of these representations to Amoebas is quite interesting, and worthy of further exploration.

REPLY [8 votes]: I recommend this recent preprint "Sums of generalized harmonic series for kids from five to fifteen" by Z.K. Silagadze. The author explains identities for $\zeta(n)$ that come from your equation (6) by using hyperbolic function substitutions. It is pointed out that the Beukers-Kolk-Calabi substitution and its hyperbolic analogue can be explained by relating them to the amoeba's of a certain Laurent polynomial (the regions you have been computing in mathematica). Unfortunately I can't retell that part of the story beyond $\zeta(2)$, but this is somewhat discussed in the last section.<|endoftext|>
TITLE: Can the conformal structure on the projective light-cone detect hyperplane sections?
QUESTION [5 upvotes]: Let $(V,\langle\,\cdot\,,\,\cdot\,\rangle)$ be an $(n+1)$-dimensional real vector space, equipped with a nondegenerate symmetric bilinear form of indefinite signature, and denote by $\nu(v):=\langle v,v\rangle$ the corresponding quadratic form. 
As a homogeneous polynomial, $\nu$ cuts out the projective ``light cone'' (or null space) $\mathcal{N}:=\{[v]\in\mathbb{P}V\mid \nu(v)=0\}$, which is a smooth hypersurface in $\mathbb{P}V\equiv\mathbb{RP}^n$.

PRELIMINARY QUESTION: what is the ``remnant'' of the scalar product $\langle\,\cdot\,,\,\cdot\,\rangle$ on $\mathcal{N}$? I mean: does it induce a metric, a conformal structure, or something even weaker on $\mathcal{N}$? 
BIG EDIT: such a ``remnant" is a conformal structure $[g]$, according to Vít Tuček's answer below.

By a hyperplane section of $\mathcal{N}$ I simply mean the intersection $\mathcal{N}\cap\pi$ of $\mathcal{N}$ with a projective hyperplane $\pi\in \mathbb{RP}^{n\,\ast}$, and I'm interested in the possibility of characterising hyperplane sections by means of $[g]$.

MAIN QUESTION: Given a smooth hypersurface $S\subset \mathcal{N}$, is it possible to tell whether or not $S$ is an hyperplane section, just by using the conformal structure $[g]$ on $\mathcal{N}$?

Any reference/hint concerning the ``instrinsic geometry'' of this projective light cone will be warmly appreciated!
- - - COMMENTS & MOTIVATIONS - - -
This post is a major edit of a previous question, whose "no" answer was given in a very convincing way by Willie Wong (see below). Thanks to his geneorous explanation, I was able to rearrange my thoughts ane reformulate the question in a hopefully more robust way. The original question, which is a sort of "affine version" of the current one, is attached below, to keep sustaining Wong's answer and to provide some motivations as well. 
In my previous post, the light cone $\mathcal{N}$ was not projectivised, and I made the mistake of believing that $\langle\,\cdot\,,\,\cdot\,\rangle$ induced a metric $g$ on $\mathcal{N}$. Hence, I considered the Hessiian $\mathrm{hess}^V$ on $V$ (i.e., the one defined by means of the flat metric induced by $\langle\,\cdot\,,\,\cdot\,\rangle$) and the Hessian $\mathrm{hess}^\mathcal{N}$ on $\mathcal{N}$ (which in fact is ill-defined, as $g$ is degenerate). Finally, by ``hyperplane section'' I just meant the intersection of $\mathcal{N}$ with an affine hyperplane of $V$. Denoting smooth functions on $\mathcal{N}$ (resp., $V$) by $f$ (resp., $F$), I formulated the following question, which (if Wille Wong hadn't proved it wrong) would have answered the main question of the present post.

PREVIOUS QUESTION: is it true that  $\{f=0\}\subset \mathcal{N}$ is an hyperplane section if and only if $$\mathrm{hess}^\mathcal{N}f\textrm{ is proportional to }g\,\textrm{?} \quad\quad{(*)}$$

The reasoning below is what originally led me to suspect that hyperplane sections can be indeed characterised by means of formula $(*)$ above.
The hypersurface $\{F=0\}\subset V$ is an affine hyperplane if and only if $$\mathrm{hess}^VF=0\, ,\quad\quad{(**)}$$
so that it all comes down to show that $(*)$ is the ``restricted version'' of $(**)$.
Allow me to regard $\nu$ as a local coordinate function (normal to the light cone) and to decompose the Hessian of $F$ as follows:
$$
\mathrm{hess}^VF=\mathrm{hess}^{\mathcal{N}}(F|_{\mathcal{N}})+F_{\nu}\odot d\nu+F_{\nu\nu}\mathrm{hess}^V\nu\, .
$$
By restricting the last expression to the light cone, and observing that $\mathrm{hess}^V\nu=\langle\,\cdot\,,\,\cdot\,\rangle$, one gets $(*)$ for $f=F|_{\mathcal{N}}$.
Conversely, if $(*)$ is true for a scalar factor $\lambda$, then I can extend $f$ to a function $F$ on $V$, such that $(**)$ is valid: the idea is to set
$$
F:=\pi_{\mathcal{N}}^*(f)-\frac{1}{\lambda}\nu\, ,
$$
where $=\pi_{\mathcal{N}}$ is the projection of $V\stackrel{\textrm{loc.}}{\equiv}\mathcal{N}\times\mathbb{R}$ onto its first factor.
(The present question is linked to my previous one).

REPLY [5 votes]: For your new question, the answer is at least in part yes. I'll outline the proof here for the case your projective hyperplane is originally not a null hyperplane.  
First let us work in the ambient $\mathbb{R}^{n+1}$. Let $\Pi$ be a non-null hyperplane through the origin, this is our projective hyperplane. Let $\pi \in \Pi$ be a non-null point, and let $P$ be the orthogonal hyperplane through $\pi$, relative to the indefinite ambient metric. 
$P$ can be identified with $\mathbb{R}^n$ with $\pi$ sent to the origin, as a (plane) model of the projective space $\mathbb{R}P^n$ once you add the point at infinity. Since $P$ is non-null we have that the ambient metric induces a metric $\nu'$ on $P$ that is again pseudo-Riemannian; considering $P$ as a indefinite inner product space with $\pi$ at the origin you have that $\mathcal{N}$ is a hyperquadric in this picture. (It is the set of points $\nu'(x) = \pm c$ for some constant $c$.) In particular $\mathcal{N}$ is a pseudo-Riemannian manfiold. 
The projective hyperplane $\Pi$ now becomes a hyperplane through the origin in $P$. And it is well-known (see, e.g. O'Neill's Semi-Riemannian Geometry, pps 108 et seq) that $\Pi\cap \mathbb{N}$ is totally geodesic in $\mathbb{N}$. This is a characterisation in that a hypersurface in $\mathbb{N}$ is totally geodesic iff it is formed by intersecting with a plane through the origin. 

Next we use the mean-curvature change formula under conformal change of metrics. 
Let $(M,g)$ be a pseudo-Riemannian manifold, $(N,h)$ a non-null hypersurface. Let $\eta$ be a choice of unit normal along $N$. Let $f$ be a real function on $M$. 
Let $S$ be the second fundamental form of $N$ in $M$. 
Now let $\tilde{g} = e^{2f} g$ be a conformal metric and $\tilde{h} = e^{2f} h$ the induced conformal metric on $N$. And let $\tilde{S}$ be the second fundamental form of $N$ in $M$ relative to the conformal metric, we have that
$$ \tilde{S} = e^f S + \eta(e^f) h $$
where $\eta(e^f)$ is the derivative of $e^f$ in the direction of $\eta$. 
So we have the following statement:

Under a conformal change of metric $g \mapsto e^{2f} g$, the trace free part of the second fundamental form $A$ changes like $A \mapsto e^f A$. 

So it makes sense to speak of the conformal class of the trace free part of the second fundamental form. 

So from the above we have 

A necessary condition for $S \subset \mathcal{N}$ to be a hyperplane section is that the traceless part of the second fundamental form for the embedding vanishes. This condition is conformally invariant so can be checked using any realisation of the conformal class of $[g]$. 

Unfortunately I can't quite see whether this is a sufficient condition; my gut feeling is that probably it is not, but there probably is a slightly modified version of the statement for which the vanishing of the traceless part of the second fundamental form is both necessary and sufficient. 
At the very least you can now tell when $S$ is not a hyperplane section. :)<|endoftext|>
TITLE: actions of the hyperoctahedral group
QUESTION [14 upvotes]: I am looking for actions (i.e., permutation representations) of the hyperoctahedral group $H_n$ (also known as the group of signed permutations) studied in the literature, i.e., homomorphisms from $H_n$ to $\mathfrak S_X$, the set of bijections from a finite set $X$ to itself.  I am not really interested in general representations.
More generally, I am also interested in actions of the wreath product of a symmetric group $\mathfrak S_n$ with a cyclic group $\mathbb Z_r$, also known as the complex reflection group $G(r, 1, n)$.
One example of what I have in mind are the papers by Bill Chen and Richard Stanley Derangements on the n-cube and Bill Chen, 'Induced cycle structures of the hyperoctahedral group'.  I am also aware of
Anthony Henderson's 'Representations of wreath products on cohomology of De Concini-Procesi compactifications'.  There is also a natural action on signed graphs, which I found in a preprint of Brian Davis 'Unlabeled Signed Graph Coloring'. 
Are there any other examples?  If not so, is there a good reason that permutation representations of the symmetric group are ubiquitous in the literature, but permutation representations of the hyperoctahedral group are rare?

REPLY [4 votes]: Tymoczko has defined something called the "dot action" on the equivariant cohomology of regular semisimple Hessenberg varieties.  I think she explains all the details only in Type A but her definition generalizes straightforwardly to other types.
For example, in Type C, you start with an indifference graph $H$ on $2n$ vertices that is palindromic (i.e., the associated unit interval order $P$  is isomorphic to the poset you get by reversing all the inequalities of $P$).  Label the vertices $-n, -n+1, \ldots, -1,1,\ldots, n-1,n$ from left to right.  Edges in $H$ may be identified with elements of the hyperoctahedral group: An edge connecting $i$ with $j$ represents the element that swaps $i$ with $j$ and $-i$ with $-j$.
Now define the associated moment graph $G$, whose vertices are the elements of the hyperoctahedral group.  Two vertices of $G$ are adjacent if some edge of $H$ (thought of as an element of the hyperoctahedral group as explained above) takes one to the other.
For example, if $n=2$, we could take $H$ to be a path on four vertices:
$$\bar{2} - \bar{1} - 1 - 2$$
The edges of $H$ define two elements of the hyperoctahedral group—the element that swaps $1$ with $-1$, and the element that swaps $1$ with $2$ and simultaneously swaps $-1$ with $-2$.  Thinking of the group as acting on places (rather than letters), this means that the edges in $G$ are given by
$$12 - 21 - \bar{2}1 - 1\bar{2} - \bar{1}\bar{2} - \bar{2}\bar{1} - 2\bar{1} - \bar{1}2 - 12$$
where of course we wrap around the end (i.e., the two appearances of $12$ above are really the same vertex, so $G$ is abstractly isomorphic to a cycle on eight vertices).  Each vertex of $G$ is adjacent to two others, one of which is obtained by swapping the two (signed) letters and one of which is obtained by negating the letter in the first position.
The edges of $G$ are labeled with elements of the polynomial ring $R := \mathbb{C}[x_1, \ldots, x_n]$; specifically, if the letters that are swapped are $i$ and $j$ then we label the edge in $G$ with $x_i - x_j$, where we interpret $x_{-i}$ to mean $-x_i$.  (This only defines the edge label of $G$ up to sign because I have not said which letter is $i$ and which letter is $j$, but for the purposes of defining the dot action, this will not matter.)
In our running example, the corresponding edge labels (up to sign) are
$$x_1 - x_2, 2x_2, -x_2 - x_1, 2x_1, -x_1 + x_2, -2x_2, x_2 + x_1, -2x_1.$$
An element of the equivariant cohomology ring $\mathscr{H}$ is given by specifying an element of $R$ for each vertex of $G$, subject to the condition that if $p(u)$ is the polynomial at vertex $u$ and $p(v)$ is the polynomial at vertex $v$, and $u$ and $v$ are adjacent, then $p(u) - p(v)$ must be divisible by the edge label of the edge $uv$.
Finally, the dot action  on $\mathscr{H}$ is defined as follows. Let $w$ be an element of the hyperoctahedral group and let $p\in\mathscr{H}$, with $p(v)$ denoting the element of $R$ at the vertex $v$.  To specify $wp$, we need to specify the value of $wp(v)$ for every vertex $v$.  This is done via the formula
$$(wp)(v) = w p(w^{-1}v),$$
meaning that we first apply $w^{-1}$ to the letters of $v$ to obtain a new vertex $w^{-1} v$ of $G$, and then finally letting $w$ act naturally on the variables $x_i$ (again, interpreting $x_{-i}$ as $-x_i$).<|endoftext|>
TITLE: An elementary lower bound on the number of primes
QUESTION [7 upvotes]: Recall the second Chebyshev function: $$\psi(x) = \sum_{p \leq x} \lfloor \log_p x \rfloor \log p$$ where $x$ is a positive integer, and $p$ runs over all primes $\leq x$.
In a hunt for an "elementary" lower bound on the number of primes up to some $x \in \mathbb{N}$, it is sufficient to find a lower bound for $\psi(x)$ in view of its connection with the prime counting function $\pi(x)$. We can work instead with $$e^{\psi(n)} = \mathrm{lcm} (1, \dots, n)$$ for $n \in \mathbb{N}$, and try to bound it from below. This can be done by observing that for any $c_1, \dots, c_n \in \mathbb{Z}$ we have $$e^{\psi(n)}\sum_{i=1}^{n}\frac{c_i}{i} = \mathrm{lcm} (1, \dots, n)\sum_{i=1}^{n}\frac{c_i}{i} \in \mathbb{Z}$$ from which we conclude, in case that the sum does not vanish, that $$e^{\psi(n)} \geq \frac{1}{|\sum_{i=1}^{n}\frac{c_i}{i}|}$$.
Therefore, my question is:

Given $n \in \mathbb{N}$, how small can we make $|\sum_{i=1}^{n}\frac{c_i}{i}|$ (as a function of $n$) using $c_1, \dots, c_n \in \mathbb{Z}$ without making it zero?

I want to see choices of $\{c_i\}_{i=1}^n$ which are not necessarily explicit, but preferably elementary. I understand that bounds on this sum can be obtained using the prime number theorem, but this is not what I am looking for. Note also that $$\sum_{i=1}^{n}\frac{c_i}{i} = \int_{0}^1 \sum_{i=1}^{n}c_ix^{i-1}$$ so in some sense we are minimizing nonvanishing integrals of integral polynomials of degree at most $n$. Here is an example:
Take some $m \in \mathbb{N}$ and set $n = 2m+1$. Note that  $$0 < \int_{0}^1 x^m(1-x)^m \leq \frac{1}{4^m} = \frac{1}{4^{\frac{n-1}{2}}}$$ so indeed one way to tackle this is by working with positive valued polynomials. The latter (simple) bound already shows that $\pi(x) \geq \frac{x}{3 \log x}$.

REPLY [5 votes]: This question (with copious references, summarizing the progress to date, with best constants) is dealt with in this 2013 paper by D. Bazzanella.<|endoftext|>
TITLE: Above/below directed graph on cells of arrangement of lines
QUESTION [7 upvotes]: This question concerns the structure of a directed graph
built on the cells of an arrangement of lines.
My basic question is whether this graph has been
studied before, perhaps in another guise. I also ask a specific
question.

Let $\cal A$ be an arrangement of $n$ lines in $\mathbb{R}^2$,
with no line vertical, and no two lines parallel.
I associate to $\cal A$ a directed graph $G$,
with each node a cell of $\cal A$,
and $a \to b$ iff cell $a$ is adjacently above $b$
in the sense that (1) $a$ and $b$ share an edge $e$,
and (2) for a point $p$ in the interior of $e$,
points vertically just below $p$ lie in $b$, and points
vertically just above $p$ lie in $a$.
Thus $e$ is an upper edge of $b$ and a lower edge of $a$.
Let $L^+$ be the line of $\cal A$ with the largest positive slope,
and $L^-$ the line with the steepest negative slope.
(From the assumptions, $L^+$ and $L^-$ are unique and distinct.)
Then there is a unique unbounded cell $A$ of $\cal A$ that is
bounded by $L^+$ and $L^-$ and has no cell adjacently above it,
and similarly there is a unique unbounded cell $B$, also
bounded by $L^+$ and $L^-$, which has no cell adjacently below.
So $A$ is the unique source of $G$, and $B$ the unique sink.
Now I assign numbers/weights to the nodes of $G$ as follows.
$A$ is assigned $1$.
For any node $x$ of $G$, $x$ is assigned the sum of the
weights of the nodes adjacently above $x$,
i.e., with incoming arcs to $x$.
Every node but $A$ has at least one incoming arc, and so
this is well-defined.
Two $4$-line arrangements are shown below.

         


         

$L^+$ and $L^-$ are colored green.


Every node but $B$ has at least one outgoing arc, so the
weight of a node is always propagated downward.
Thus the sink node $B$ receives the highest weight.
In (a), $B$ has weight $8$, whereas in $B$
it has weight $9$.
A basic question is:

Q1. What is the largest possible weight assigned to the sink node $B$,
  over all arrangements of $n$ lines?

Here is a more complicated arrangement of $8$ lines, whose
sink node has weight $111$.

         


(This was calculated by hand, so apologies if there are errors.
Thanks to Will Brian for catching one error.)
The more general question is:

Q2. Has this graph and node-weighting scheme been studied
  before, perhaps in another context?

REPLY [2 votes]: We can always achieve exactly $2^{n-1}$ for the weight of the sink. This gives us a lower bound for the answer to Q1.
Essentially, the idea is that we draw our $n$ lines all tangent to the upper half of a circle. We then obtain a weight pattern identical to the first $n$ rows of Pascal's triangle, and the weight of the sink is gotten by adding together all the entries in the bottom row. I'm giving only a crude explanation, because a picture explains it much better. Here is $n = 5$ (the circle is shown in gray, and the faded red lines are showing you the rows of Pascal's triangle):<|endoftext|>
TITLE: Determinant of symmetric Latin square
QUESTION [8 upvotes]: Let $n=2m$ be an even number. Let us construct $n\times n$ symmetric matrices $S_n$ in the following way. The entries are indeterminates $X_1,\ldots,X_{n-1}$. We choose a $1$-factorization of the complete graph over $[\![1,n]\!]$ (which exists because $n$ is even). The $1$-factors (these are perfect matchings) are labelled $F_1,\ldots,F_{n-1}$. The indeterminate $X_r$ is attributed to the $(ij)$-entry whenever the edge $(ij)$ belongs to $F_r$. On the diagonal, we put zeros. In other words, every line and every column of $S$ contains a zero and all the indeterminates exactly once. You may call this a symmetric Latin square. For instance
$$S_2=\begin{pmatrix} 0 & X \\ X & 0 \end{pmatrix},\qquad S_4=\begin{pmatrix} 0 & X & Y & Z \\ X & 0 & Z & Y \\ Y & Z & 0 & X \\ Z & Y & X & 0 \end{pmatrix}.$$
It seems that the determinant $D_n$ of $S_n$, a homogeneous polynomial in the indeterminates, is symmetric.
One verifies easily that for $n=2$ or $4$, $D_n$ splits:
$$D_2=-X^2,\qquad D_4=(X+Y+Z)(X-Y-Z)(Y-Z-X)(Z-X-Y).$$ 
Actually, the characteristic polynomials $P_n(T,X_1,\ldots,X_{n-1})$ split for $n=2$ or $4$ :
$$P_2=(T-X)(T+X),\qquad P_4=(T-X-Y-Z)(T+X+Y-Z)(T+Y+Z-X)(T+Z+X-Y).$$ 

Is it true that $D_n$ or $P_n$ always split into linear factors ?

Notice that $P_n$ has always the factor $T-X_1-\cdots-X_{n-1}$.
For $n\ge6$, the $1$-factorization is not unique. There are already 6240 of them for $n=8$. Does the answer depend on the choice of the $1$-factorization ?
Edit. Patrik's negative answer raises a far-reaching question. The original question ressembles vaguely that of the splitting of the determinant associated with the Cayley graph of a finite group $G$. In the latter case, we know that the determinant splits into linear factors if and only if $G$ is abelian. For a general group, the nature of the splitting is given by Frobenius' Theorem about the irreducible characters of $G$. In the present situation, I suspect that another algebraic theory will tell us what are the degrees of the irreducible factors of $P_n$. When $n$ is small, these degrees are small, because the underlying algebra must be trivial ($n=2$ or $4$) or very simple ($n=8$).

REPLY [10 votes]: No it's not true in general. I claim that the following matrix
\begin{bmatrix} 
t&x&y&z&w&u&v&r\\
x&t&z&y&r&w&u&v\\
y&z&t&x&v&r&w&u\\
z&y&x&t&u&v&r&w\\
w&r&v&u&t&x&y&z\\
u&w&r&v&x&t&z&y\\
v&u&w&r&y&z&t&x\\
r&v&u&w&z&y&x&t
\end{bmatrix}
is of the desired form (when $t=0$) and has determinant
$[(r-u)^2-(t-y)^2+(v - w + x - z)^2]\cdot $
$[(r-u)^2-(t-y)^2+(v - w - x + z)^2]\cdot$ 
$(r + u + t + y + v + w + x + z)
(r + u  -t - y + v + w - x - z)$ 
$(r + u + t + y - v - w - x - z)(r + u -t - y - v - w + x + z)$.
Realize $K_8$ as two copies of $K_4$ and connect all vertices (one with vertex set $\{1,2,3,4\}$ and one with vertex set $\{1',2',3',4'\}$). The 1-factors $x,y,z$ come from uniting corresponding matchings of $K_4$ and $K_4'$, and the edges between $K_4$ and $K_4'$ are decomposed into the 1-factors $w,u,v,r$ given by $\{11',22',33',44'\},\{12',23',34',41'\},\{13',24',31',42'\},\{14',21',32',43'\}$ respectively, completing the 1-factorization. The determinant computation was done using a computer, so unfortunately I have no particular understanding of what is going on.<|endoftext|>
TITLE: Why are most coefficients of these minimal polynomials divisible by $p$?
QUESTION [6 upvotes]: For an odd prime $p$, let $\zeta:=e^{\frac{\pi i}p}$ and choose odd $1<n<p$. Further let $q(x)$ and $r(x)$ be integer polynomials such that $r(x)$ has no common factor with $x^n+1$, and $\xi$ any root of unity (not necessarily related to $\zeta$). 

I have observed that the minimal polynomial of $$(1+\zeta^n)\frac{q(\xi)}{r(\xi)}$$ always seems to have all its coefficients divisible by $p$, except either the first or the last one. 

Likewise for the minimal polynomial of $(1+\zeta^n)a$, where $a$ is any real algebraic number. 
How to prove this?

REPLY [5 votes]: The second statement is false as stated. Let $p=3$, $n=1$, $a=1/\sqrt{3}$. The minimal polynomial of $(1+\zeta)a$ is $x^4-x^2+1$. Joe Silverman's argument is inapplicable since the $p$-adic value of $(1+\zeta)a$ is 0.<|endoftext|>
TITLE: Finding the octonionic analog of the K3 surface, via (almost) hyperkahler geometry?
QUESTION [35 upvotes]: The K3 manifold is an amazing object in mathematics which plays an important role in several fields ranging from the study of smooth 4-manifolds to algebraic geometry to differential geometry and beyond.
It also has important implications for the 3rd stable homotopy group of spheres as shown in this MO question and this one. Quickly, the fact that the K3 surface (1) is hyperkahler and (2) has Euler characteristic 24 can be viewed as the source of the 24 in $\pi_3^{st} = \Omega^{fr}_3 = \mathbb{Z}/24$. (This is the 3rd stable homotopy group of spheres, also known as stably framed 3-manifolds up to cobordism).
Motivated by this I asked in this previous MO question if there was an octonionic analog of the K3 surface which could likewise account for the 240 in $\pi_7^{st} = \Omega^{fr}_7 = \mathbb{Z}/240$. This group is generated by the octonionic hopf map after all. This octonionic version of a K3 surface would be a smooth 8-manifold with Euler characteristic 240 and possessing an octonionic analog of a hyperkahler structure (more on this later).
My question was answered in the positive! Indeed there is such a manifold. However the construction given there used the celebrated result of Kervaire and Milnor that the group of exotic 7-spheres is $\Theta_7 = bP_8 = \mathbb{Z}/28$ and is generated by the boundary of "Milnor's pluming". This is all well and good, as it answers the question I posed, however Kervaire and Milnor's computation that $bP_8 = \mathbb{Z}/28$ depends crucially on the computation (by Adams and Quillen) that the image of the J-homomorphism in this dimension is $\mathbb{Z}/240$.
One reason I was interested in this manifold is that I wanted a purely geometric reason for the 240 to appear in this dimension, just as the 24 appears for the K3 surface. However as it stands the construction of this octonionic analog of a K3 uses the 240 as input.
The K3 surface has many different constructions and I still hold out hope that this octonionic analog of the the K3 might also admit many constructions. Hopefully it admits a construction which doesn't implicitly use the fact that $\pi_7^{st} = \mathbb{Z}/240$ or that $\Theta_7 = \mathbb{Z}/28$. In fact if such a manifold can be constructed by other means, then we can probably reverse the argument and use this manifold to give a new geometric identification of these groups (or more precisely identifications of $Im\;J$ in degree 7 and the group $bP_8$).
The manifold constructed in answer to my previous question is 3-connected, and, as was suggest by the OP of that answer, I think that it is likely that a modified form of Kreck's "Surgery and duality" paper will show that it is the only 3-connected 8-manifold with the desired properties (explained below). This is in analogy with the 4-dimensional case: K3 is the only simply connected hyperkahler 4-manifold.
The K3 manifold $N$ is uniquely specified by the following properties:

It is simply connected (i.e. connected and $\pi_1 = 0$);
It is a complex manifold;
The first Chern class  vanishes $c_1(\tau_N) = 0$.

Using the connectivity of the manifold, this last condition can be expressed in a different way. Since the manifold is complex, the classifying map of the tangent bundle factors as $N \to BU(2) \to BO(4)$. Using obstruction theory this last condition is equivalent to asking that it factors further as
$$N \stackrel{f}{\to} \mathbb{HP}^1 \to BU(2)$$
where the bundle on $\mathbb{HP}^1$ is the quaternionic projective line. This means that $\tau_N \cong f^*(\gamma_\mathbb{H})$ where $\tau_N$ is the tangent bundle of $N$ and $\gamma_\mathbb{H}$ is the tautological quaternionic line bundle. In particular the tangent bundle of $M$ inherits a quternionic action.
We can try to construct a higher dimensional analog of this by replacing the complex numbers with the quaternions. In particular instead of looking for a complex surface we will be looking for a hyperkahler manifold.

Question: Is there a construction of a smooth 8-manifold $M$ (an "octonionic K3") such that

$M$ is 3-connected (i.e. connected and $\pi_1 = \pi_2 = \pi_3=0$);
$M$ is hyperkahler almost hyperkahler;
The first pontryagin class vanishes, $p_1(\tau_M) = 0$.

and which does not make use of the computations of exotic 7-spheres $\Theta_7 = bP_8 = \mathbb{Z}/28$ or the stable 7-stem $\pi_7^{st} = (Im \; J)_7 = \mathbb{Z}/240$?

As with the usual K3 surface, this last condition can be reformulated as saying that the classifying map of the tangent bundle factors as
$$ M \stackrel{f}{\to} \mathbb{OP}^1 \to BHU(2) = BSp(2)$$
i.e. that there exists an isomorphism $\tau_M \cong f^*(\gamma_\mathbb{O})$ of the tangent bundle of $M$ with the pullback of the octonionic line bundle. In particular this gives an octonionic "action" on the tangent bundle of $M$.

Side question: Has this octonionic analog of hyperkahler, without the connectivity condition, been studied before? Does it have a name? To me it seems like a very natural generalization of the notion of (almost) hyperkahler.

If, as I suspect, there is indeed only one of these manifolds up to diffeomorphism then it will agree with the construction given here. In that case we can compute some other things about this manifold:

The Euler characteristic is $240$;
The integral cohomology is torsion free;
The Betti numbers are $b^0 = b^8 = 1$, $b^4_+ = 8 * 28 + 7 = 231$, $b^4_- = 7$, and all other Betti numbers vanish;
The signature is $\sigma = 8 * 28 = 224$;
The second Pontryagin number is $p_2 = 6 * 240 = 1440$.

I know there has been a lot of study of hyperkahler manifolds using methods from differential geometry and algebraic geometry. Can you help me? have you seen my manifold or know how to find it?

REPLY [5 votes]: The following shows that such a manifold exists, but unfortunately it uses $bP_8 \cong \mathbb Z_{28}$. 
According to Corollary 5.5 of this article a highly connected $8$-manifold $M$ with $p_1(M)=0$ admits an $Sp(2)$-structure (i.e. an almost hyperkähler structure) if and only if 

$p_2(M) -2 \chi(M)$=0
$\chi(M) \equiv 0 \mod 4$.

(I am identifying the Pontryagin class $p_2$ with the Pontryagin number in the obvious way)
Now let $W$ be the union of an closed $8$-disc $D^8$ with the Milnor plumbing mentioned
here and consider the manifold $M= 247\#(S^4\times S^4)\#W$.
We have that $p_1(M)=0$ since the tangent bundle of $M$ is stably equal to the Whitney sum
$TS^4\oplus\ldots\oplus TS^4\oplus TW$ (where we pull back the tangent bundles of $TS^4$ and $TW$ the the obvious collapsing maps $M \to S^4$ and $M \to W$) and $p_1(S^4)=p_1(W)=0$.
Furthermore we have $\chi(M) = b_4(W)+246\cdot b_4(S^4\times S^4)+2=224 +247\cdot 2+2=720(\equiv 0\mod 4)$
as well as $p_2(M) = \frac{45}{7}\sigma(M) = \frac{45}{7}\cdot 224 =1440$.
Edit: It seems to me that the example of the other post will not have an $Sp(2)$-structure if Corollary 5.5 is true. Moreover it will not be possible to construct a highly connected manifold with $Sp(2)$-structure, $p_1(M)=0$ and Euler characteristic equal to $240$ since the signature would not be an integer.<|endoftext|>
TITLE: Do there exist "topologically significant" (and not "algebraic") triangulated categories killed by the multiplication by $p$?
QUESTION [11 upvotes]: I have a somewhat vague question: does there exist a prime $p$ and a triangulated category killed by the multiplication by $p$ that would be "interesting for topologists"? This category would probably correspond to the study of $p$-torsion of (co)homology. As far as I remember, $S^0/p$ is not a ring spectrum by a result of Schwede (so, one cannot consider modules over it); yet does there exist any method for "overcoming" this difficulty (and still obtaining a "non-algebraic" triangulated category) somehow? Here non-algebraic means that the category should not be "closely related" to the derived category of any abelian category. I suspect that there should exist certain categories of this sort since certain cohomology theories (including $K$-theory) cannot be defined in terms of complexes.

REPLY [15 votes]: Depending on exactly what you mean by "killed by $p$", the answer may be no.
Let $\mathcal{C}$ be a stable $\infty$-category and let $\iota_{\mathcal{C}}$ be the identity functor from $\mathcal{C}$ to itself. If the ``multiplication by $p$'' map
$p: \iota_{\mathcal{C}} \rightarrow \iota_{\mathcal{C}}$ is nullhomotopic, then
$\mathcal{C}$ is $\mathbf{F}_p$-linear (and can therefore be obtained from a pretriangulated differential graded category over $\mathbf{F}_p$).<|endoftext|>
TITLE: Forcing Extension of Countable Linearly Iterable Structures
QUESTION [7 upvotes]: Let $V$ satisfy there exists a measurable cardinal. Let $\kappa$ be a measurable cardinal and $U$ be the normal measure on $\kappa$ witnessing this. Let $\mathbb{P}$ be a forcing of size less than $\kappa$. 
Let $M$ be a countable transitive structure such that there exists an elementary embedding $\pi : M \rightarrow V$ (or a large $V_\Theta$). Suppose $U' \in M$ is such that $\pi(U') = U$. Suppose $\mathbb{Q} \in M$ is such that $\pi(\mathbb{Q}) = \mathbb{P}$. 
It is known that $M$ can be iterated using $U'$ throughout the ordinals of $V$. In particular, $M$ can be iterated $\omega_1^V$ many times. 
Since $M$ is countable, there is a $g \subseteq \mathbb{Q}$ which is $\mathbb{Q}$-generic over $M$ and $g \in V$. 
My question is: Is $M[g]$ also $\omega_1^V$-iterable? 

My ideas were: Since $|\mathbb{P}| < \kappa$, $\kappa$ would still be measurable in any forcing extension of $V$ by a generic of $\mathbb{P}$. So if there exists $G \subseteq \mathbb{P}$ which is $V$-generic and $\pi'' g \subseteq G$, then if one could lift $\pi$ to a map $M[g] \rightarrow V[G]$ which takes some measure $\tilde U' \in M$ to some measure $\tilde U \in V[G]$, then this should show $M[g]$ is iterable. 
However, I am not sure such a $G$ can exists. Since $g \in V$, I do not think it can be used make $G$ in the ways that usual liftings are done.
Thanks for any information about this problem.

REPLY [7 votes]: Yes, in general, the model $M[g]$ will have the same amount of iterability as $M$. To see this, we need the following lemma giving the necessary and sufficient condition for when an ultrapower of a generic extension is the lift of an ultrapower of the ground model. 
Lemma: Suppose $M$ is a transitive model of ${\rm ZFC}^-$, $\mathbb P\in M$ is a poset and $G\subseteq \mathbb P$ is $M$-generic. Suppose further that $U\in M$ is an ultrafilter on a cardinal $\delta$ and $U^*\in M[G]$ is an ultrafilter on $\delta$ extending $U$, both with well-founded ultrapowers. Then the ultrapower by $U^*$ is a lift of the ultrapower by $U$ if and only if every $f:\delta\to M$ in $M[G]$ is $U^*$-equivalent to some $g:\delta\to M$ in $M$. 
Let $\delta$ be the preimage of $\kappa$ under $\pi$ and let $M=M_0$. Let $j_{\xi\mu}:M_\xi\to M_\mu$ be the iterated ultrapowers of $M_0$ by $U'$. We can lift the entire iteration to the directed system consisting of $j_{\xi\mu}:M_\xi[g]\to M_\mu[g]$. So it suffices to argue that this is precisely the iteration of $M_0[g]$ by $U^*$, where $U^*$ is the ultrafilter extending $U'$ generated by using  $\delta$ as a seed from the lift $j_{01}:M_0[g]\to M_1[g]$ (for $A\subseteq\delta$ in $M_0[g]$, we have $A\in U^*$ whenever $\delta\in j_{01}(A)$)  . The embedding $j_{01}:M_0[g]\to M_1[g]$ is precisely the ultrapower by $U^*$. So, the conclusion for $j_{01}$ follows by the definition of $U^*$. Thus, $M_0[g]$ satisfies that "every new function from $\delta$ into $M$ is $U^*$-equivalent to an old function" by the lemma. But, then by elementarity $M_1[g]$ satisfies this for $j_{01}(U^*)$ and $M_1$, which means that the ultrapower of $M_1[g]$ by $j_{01}(U^*)$ is the lift of $j_{12}:M_1\to M_2$, namely $j_{12}:M_1[g]\to M_2[g]$ (lifts of small forcing extensions are unique). By elementarity, we can continue to propagate this statement along the entire iteration. 
Indeed, the result is much more general. Even if $\mathbb Q$ is not small relative to $\delta$, but we are able to lift the first ultrapower $j_{01}:M_0\to M_1$ to $M_0[g]$, then every iterate by $U^*$ (obtained as before) will satisfy the "no new functions" statement and therefore it will be a lift of a ground model embedding and therefore well-founded. Thus, to lift the entire iteration, it suffices to lift the just the first step! Moreover, the lift of the iteration is the iteration of the lift. A more involved argument (and some more restrictions on the forcing) is needed in the case where the ultrafilter $U'$ is an $M$-ultrafilter that is external to $M$, because the statement whose elementarity we are using is not necessarily expressible in $M$ in that situation. This argument is given in my paper (with Arthur Apter and Joel Hamkins) Inner models with large cardinal features usually obtained by forcing.<|endoftext|>
TITLE: Probability that planar Brownian motion doesn't "encircle" 0
QUESTION [7 upvotes]: Suppose $B_t$ is a standard Brownian motion in $\mathbb{R}^2$ and $T = \text{inf}\{t : |B_t| = 1\}$. Let $E$ denote the event that $0$ is contained in the unbounded component of $\mathbb{R}^2 \setminus B[0, T]$. Do there exist $\lambda < \infty$ and $\alpha > 0$ such that for all $|x| < 1$, we have$$\mathbb{P}^x(E) \le \lambda |x|^\alpha?$$

REPLY [10 votes]: The probability is asymptotic to $\lambda |x|^{1/4}$.  This was proved by Schramm, Werner and myself using the Schramm-Loewner evolution. The exponent is called the disconnection exponent for Brownian motion.<|endoftext|>
TITLE: Convexity of Isoperimetric Domains
QUESTION [6 upvotes]: I am interested in what is known about the convexity of isoperimetric domains in compact Cartan-Hadamard manifolds (Riemannian manifolds that are complete and simply-connected and have non-positive sectional curvature).
In particular, we know that for a compact Riemannian manifold $M$ with smooth boundary, isoperimetric domains exist. That is, for each $V \in (0,vol(M))$ there exists $E\subset M$ such that $Vol(E) = V$ and $Area(\partial E) = \inf\{area(\partial E) \, | \, E\subset M, vol(E)=V\}$. We also know that isoperimetric domains are $C^{1,1}$ (loosely speaking) and the mean curvature is constant almost everywhere (the mean curvature is defined almost everywhere for $C^{1,1}$ surfaces). 
It is an open problem to show that Cartan-Hadmard manifolds satisfy a Euclidean isoperimetric inequality, so I'm wondering if these manifolds similarly have the property that isoperimetric domains are convex. So, if $M$ is a compact, convex Cartan-Hadamard manifold, is it known if isoperimetric domains in $M$ are convex? 
A less restrictive question would be: if $M$ is a compact, convex Cartan-Hadamard manifold, do isoperimetric domains $E_0 \subset M$ have the property that $$Area(\partial \,\overline{Conv(E_0)}) \leq Area(\partial E_0)$$ where $\partial \,\overline{Conv(E_0)}$ is the boundary of the closure of the convex hull of $E_0$? Note that in dimensions greater than 2, the surface area of the convex hull of a domain may be greater than the surface area of the domain, even in the Euclidean case.

REPLY [3 votes]: Using the definition of a Cartan Hadamard as a complete, simply connected manifold having non-positive curvature, a region in a Cartan Hadamard manifold realizing the optimal isoperimetric ratio need not be convex.  In fact it does not even need to be connected, even in dimension two.
An explicit example is constructed in 
Isoperimetric Regions in Nonpositively Curved Manifolds
arXiv:1604.02768<|endoftext|>
TITLE: Is the Number of Carries in Integer-Addition Associative?
QUESTION [10 upvotes]: Is it true that the number of carries, when calculating the sum of a finite set of finite positive integers,  is constant (i.e. independent of their permutation and the order in which the additions are carried out)? Carries are computed in base 2, so that 1+3 is $01_2+11_2=100_2$, which involves 2 carries: the least significant digits resulted in a carry, which then causes a second carry of the 2's digit. 
In case the assumption is wrong, I would also like some ideas for determining the optimal permutation and order of additions.

REPLY [10 votes]: In base $b$, 
let $G_n$  be the set of $n$-digit integers, thought of as the integers mod $b^n$.  Then we have an exact sequence
$$0\rightarrow G_1\rightarrow G_n\rightarrow G_{n-1}\rightarrow 0$$
For $n-1$ digit numbers, the leftmost 
carry digit is the two-cocycle associated to this group extension and therefore satisfies
the cocycle condition
$$c(x,y)+c(x+y,z)=c(y,z)+c(x,y+z)$$ 
In other words, the sum of the leftmost carry digits does not depend on the order of the summands.  
After reducing mod $b^k$, the same argument holds for any other carry digit.<|endoftext|>
TITLE: On continuous perturbations of functions of the first Baire class on the Cantor set
QUESTION [6 upvotes]: Is it true that for any function of the first Baire class $f:X\to\mathbb R$ on the Cantor cube $X=2^\omega$ there is a continuous function $g:X\to[0,1]$ such that the image $(f+g)(X)$ is disjoint with the set $\mathbb Z$ of integers? 
We recall that a function $f:X\to\mathbb R$ is of the first Baire class if it is a pointwise limit of a sequence of continuous functions. By a classical theorem of Baire, a function $f:X\to\mathbb R$ on a Polish space $X$ is of the first Baire class if and only if for any non-empty closed subset $A\subset X$ the restriction $f|A$ has a point of continuity.
It is also well-known that each separately continuous function $f:X\times Y\to\mathbb R$ on the product of two Polish spaces is of the first Baire class.
That is why the affirmative answer to the above problem will imply the affirmative answer to the problem
Uniform approximation of separately continuous functions on zero-dimensional spaces .

REPLY [6 votes]: After some thinking I realized that the answer to this question is negative. A counterexample can be constructed by a standard diagonal method of killing all possible candidatures.
We shall construct a function $f:X\to\mathbb R$ of the first Baire class on the Cantor cube $X=\{0,1\}^\omega$ such that for any continuous function $g:X\to\mathbb R$ the function $f+g$ contains zero in its image. 
Observe that the Banach space $C(X)$ of all continuous real-valued functions on $X$ is Polish and hence is the continuous image of the space $\omega^\omega$. Since $\omega^\omega$ is the image of the Cantor cube $X$ under a function of the first Baire class, there exists a surjective function $\xi:X\to C(X)$ of the first Baire class. Now define a function $f:X\to \mathbb R$ by $f(x)=-\xi(x)(x)$ and observe that it is of the first Baire class. For every continuous function $g\in C(X)$ we can find $x\in X$ with $\xi(x)=g$ and conclude that $f(x)+g(x)=-\xi(x)(x)+g(x)=-g(x)+g(x)=0$.  
On the other hand, by the proof of the Lebesgue-Hausdorff-Banach Theorem 24.10  in [A.Kechris, Classical Descriptive Set Theory, Springer, 1995], for each bounded function $f:X\to \mathbb R$ of the first Baire class and each $\varepsilon>0$ there is a function $g:X\to\mathbb R$ of the first Baire class such that $\|f-g\|<\varepsilon$ and $g(X)$ is finite.<|endoftext|>
TITLE: Are $E_n$-operads not formal in characteristic not equal to zero?
QUESTION [11 upvotes]: This is a short question:
Is it just unproven folklore (yet), or is it definitively known that $E_n$-operads are not formal, if the characteristic of the underlying field is not equal to zero?

REPLY [3 votes]: Paolo Salvatore put a paper on the arXiv last week (1807.11671) proving that $E_2$ is not formal over $\mathbb{F}_2$ as a non-symmetric operad (what he calls planar operad), i.e. you cannot find a zigzag of quasi-isomorphisms of operads (not necessarily $\Sigma$-equivariant) between $H_*(E_2; \mathbb{F}_2)$ and $C_*(E_2; \mathbb{F}_2)$. He uses obstruction theory for that.<|endoftext|>
TITLE: What are examples when the equality of some invariants is good enough in algebraic topology?
QUESTION [16 upvotes]: As far as my understanding goes, most of the tools of algebraic topology (homotopy groups, homology groups, cup product, cohomology operations, Hopf invariant, signature, characteristic classes, knot invariants...) are mostly designed with "discrimination" in mind: if two spaces/maps/bundles/... do not have the same invariant, then they are not "equivalent" for the correct notion of equivalence. And of course two things having the same invariants in some arbitrary big class is not enough for the things to be equivalent (e.g. there are spaces that have the same homotopy and homology groups that are not homotopy equivalent).
But sometimes computing only a few of those invariants is enough:

If a knot has the same crossing number as the unknot (zero), then it's the unknot. (Maybe a silly example.)
If a space has the same homotopy groups as an Eilenberg–MacLane space (i.e. they all vanish except one), then it has the same homotopy type as an Eilenberg–MacLane space.
If a closed (connected) 3-manifold has the same fundamental group as the 3-sphere, then it's actually homeomorphic to the 3-sphere (the famous Poincaré conjecture / Perelman theorem).
The result that made me think of this question: under some technical conditions, if a simplicial operad has the same rational homology as the little $n$-disks operad (as Hopf $\Lambda$-operads), then it is rationally equivalent to the little $n$-disks operad (Fresse–Willwacher).

The above results are the ones I could think of (and are probably very skewed towards my interests).

What are some other examples of situations where an algebro-topological invariant, that in general is not enough to characterize an object, is enough to actually completely characterize the object?

REPLY [7 votes]: Here are three results where the conclusion is not exactly an equality in the topological category, but is something quite close to that.  All three are of fundamental importance.

If $f\colon X\to Y$ is a map of simply connected CW complexes or of connective CW spectra, and $H_*(f)$ is an isomorphism, then $f$ is a homotopy equivalence.
If $f\colon X\to Y$ is a map of finite spectra, and $H_*(f;\mathbb{Q})=0$, then $nf=0\colon X\to X$ for some $n>0$.
If $f\colon\Sigma^d X\to X$ is a map of finite spectra, and $MU_*(f)=0$, then $f^n=0\colon\Sigma^{nd}X\to X$ for some $n>0$.

REPLY [6 votes]: Here is the easiest generalization of the fact you cite about Eilenberg-MacLane spaces. Spaces $X$ with exactly two nontrivial homotopy groups $\pi_n(X), \pi_m(X), 2 \le n < m$ are classified (up to (weak) homotopy equivalence) by these two homotopy groups together with one additional Postnikov invariant, which is a cohomology class in $H^{m+1}(B^n \pi_n(X), \pi_m(X))$, where $B^n A$ denotes the $n$-fold delooping $K(A, n)$ of a discrete abelian group $A$. This class classifies the fibration
$$B^m \pi_m(X) \to X \to B^n \pi_n(X).$$
If $n = 1$ then we need the additional data of the action of $\pi_1(X)$ on $\pi_m(X)$, and then the cohomology above is group cohomology with nontrivial / local coefficients.  
Example. The $3$-truncation of $S^2$ has two nontrivial homotopy groups $\pi_2(X) \cong \pi_3(X) \cong \mathbb{Z}$, and all other homotopy groups vanish. The Postnikov invariant is a class in $H^4(B^2 \mathbb{Z}, \mathbb{Z}) \cong H^4(\mathbb{CP}^{\infty}, \mathbb{Z}) \cong \mathbb{Z}$, and I believe it turns out to be a generator. 
The natural generalization of this fact is the theory of Postnikov towers, although there the relevant Postnikov invariants are defined in terms of spaces defined in terms of other Postnikov invariants, so it's trickier to tell whether the're different or the same.<|endoftext|>
TITLE: A natural Lascoux-Schützenberger involutions on plane partitions
QUESTION [6 upvotes]: The Lascoux-Schützenberger involutions, $s_i$, that permute the weight of semi-standard Young tableaux are fairly known.
They satisfy some nice Coxeter relations, for example, if $v$ and $w$ are reduced words of the same permutation, then $s_{v_1} \dots s_{v_l} = s_{w_1} \dots s_{w_l}$.
Each $s_i$ only acts on the entries $i$ and $i+1$ in the tableau, 
and columns that contain one entry of both $i$ and  $i+1$ is fixed.
Now, is there some generalization of these involutions that 

Generalize to plane partitions (or arbitrary fillings), where the same element can appear more than once in both rows and columns,
satisfy the independence of reduced word relation
specialize to the classical involutions on semi-standard Young tableaux?

Note that this generalization does not need to preserve the position of other entries if the filling is not a SSYT: What I am looking for will not have this property.

REPLY [2 votes]: There is a modern viewpoint on combinatorics Young tableaux  in Danilov, V. I.; Koshevoy, G. A. Massifs and the combinatorics of Young tableaux. (Russian) Uspekhi Mat. Nauk 60 (2005), no. 2(362), 79--142; translation in Russian Math. Surveys 60 (2005), no. 2, 269–334 (the LS-involutions are considered in Appendix A).
This paper is online here
http://www.mathnet.ru/links/b2797cbfa2d63906937371dbe7b7b519/rm1402.pdf<|endoftext|>
TITLE: Converse to Modularity I: weight 2 newforms
QUESTION [11 upvotes]: Since 2008 we have the following remarkable correspondence:

Odd irreducible 2-dim Galois repn $\longleftrightarrow$ weight 1
  newforms

note: all Galois representations in this question are ment to be continuous complex linear representations.
That this is a bijection is the consequence of three deep results.

Serre-Deligne: for each weight 1 newform there is an irreducible 2-dim Galois representation.
Weil-Langlands: the converse is true if the Artin conjecture holds for those representations.
Khare-Wintenberger: the Artin conjecture holds for odd irreducible 2-dim Galois representation (from Serre's modularity conjecture).

Now, we also famously have (Wiles-Breuil-Conrad-Diamond-Taylor):

Elliptic curves over $\mathbb{Q}$ $\longrightarrow$ weight 2
  newforms

My question is, what is known/conjectured on the converse direction? This is, does every newform of weight 2 arises from an elliptic curve over $\mathbb{Q}$? (Note that this correspondence can only hold up to isogeny) Are there any other source for newforms of this weight? How can we tell apart those that came from elliptic curves?

REPLY [11 votes]: This should be a comment, but is getting too long.
First, unless you are using the strange convention that Galois representations have by definition complex coefficients, odd, irreducible 2-dimensional Galois representations do not correspond to weight 1 newforms. Only a very small subset of the former set corresponds to the latter one, namely the one with values in $\operatorname{GL}_2(\mathbb C)$ or with finite image.
Second, you exchanged the contributions of Weil-Langlands and Deligne-Serre.
Third, the newforms whose $L$-function is the $L$-function of an elliptic curve are exactly those  with integral coefficients. This is elementary (at least if one disregards Euler factors at places of bad reduction, in order to conclude for all Euler factors, one needs Faltings's theorem): just take the quotient of the Jacobian of the modular curve given by the Hecke action on $f$. The other newforms of weight 2 are attached to modular abelian varieties (the dimension of the abelian variety being the degree of the field generated by the coefficients of $f$ over $\mathbb Q$). This is a classical result of G.Shimura.<|endoftext|>
TITLE: Can we find CH in the analytical hierarchy?
QUESTION [19 upvotes]: Recently I was talking to my friend and I have mentioned to him that it was proven that CH is not provably (over ZFC) equivalent to any statement in second-order arithmetic. However, today I found out that the result I was thinking about is the one mentioned in this answer, which says that GCH is independent of all sentences in analytical hierarchy. This now made me think:

Is there a sentence in the analytical hierarchy which is provably equivalent to continuum hypothesis?

I am expecting the answer to this question to be no, because CH concerns arbitrary sets of reals, but this is only an intuitive reason.
Thanks in advance.

REPLY [24 votes]: The collapsing forcing by countable partial functions from $\omega_1$ to $2^\omega$ is $\omega_1$-closed, hence it preserves $H_{\omega_1}$, and a fortiori the truth of all formulas in the analytical hierarchy; it also makes CH hold. Thus CH is not equivalent to any statement in the analytical hierarchy (assuming ZFC is consistent).
In fact, the argument implies that ZFC+CH is $\Pi^2_1$-conservative over ZFC, hence CH is not equivalent to a $\Pi^2_1$ statement. On the other hand, it is easy to see that it can be formulated as a $\Sigma^2_1$ statement.<|endoftext|>
TITLE: Euler Characteristic of Real Algebraic Surfaces
QUESTION [11 upvotes]: Given a (compact if needed) real smooth surface $V(f)$ defined by  $f\in \mathbb{R}[X,Y,Z]$, in particular it is oriented. Is there a formula which gives the Euler character of $V(f)$ ?
Thanks.

REPLY [8 votes]: A “formula” is a lot to ask for, but there are algorithms based on Morse theory.
E.g. §5 of Basu (1999), or §3 of Fortuna-Gianni-Luminati (2004).<|endoftext|>
TITLE: What does taking the graded algebra do to the Grothendieck group, and its relation to the Chow ring?
QUESTION [16 upvotes]: Let $X$ be a nonsingular variety. (Perhaps some/all of this works over more general smooth schemes, but let's stick to the simple case.)
In, e.g., Fulton's Intersection Theory chapter 15, and Soule's Lectures on Arakelov Geometry chapter 2, there are brief expositions on the associated graded $\mathbb{Z}$-algebra $\text{Gr }K(X)$ of the Grothendieck group under its filtration by dimension of support, and its relationship to the Chow ring $A(X)$. These treatments are pretty much just glancing mentions, so I'm left somewhat confused about the situation. Specifically:
We have a surjective graded-group homomorphism $\phi:A(X)\to \text{Gr }K(X)$ given by $[V]\mapsto [\mathcal{O}_V]$. Further, some geometry shows that the Chern character takes $[\mathcal{O}_V]$ to $[V] +\alpha$ for $\alpha$ of lower-dimensional support, hence induces a graded-group homomorphism in the opposite direction $\text{ch}':\text{Gr }K(X)\to \text{Gr }A(X)\otimes \mathbb{Q}$, so that $\text{ch}'\circ \phi$ is the natural inclusion. So when we pass to rational coefficients everywhere, everything becomes an isomorphism, and hence $\text{ch}$ is also an isomorphism of graded groups since the associated graded module is a faithful functor, and hence it is also an isomorphism of rings. All of this is fine. (Unless I made a mistake!)
But now since $A(X)$ is already a graded algebra, it is naturally isomorphic to $\text{Gr }A(X)$. So we have that $K(X)\otimes \mathbb{Q}$ is also isomorphic to $\text{Gr }K(X)\otimes \mathbb{Q}$ as graded groups, which makes sense - after killing torsion, $K(X)$ already seems to be a graded group since direct sum of sheaves preserves dimension so long as we don't get killed off by one of the exactness relations. I am pretty sure this is all right.
So my clarifying questions are:
1) How exactly does taking the associated graded algebra of $K(X)$ by dimension of support change the multiplicative structure? And - does it? $K(X)\otimes \mathbb{Q}$ is already a graded $\mathbb{Q}$-algebra by the Chern character isomorphism, since the Chow ring is. But it seems that the induced gradation by this isomorphism is very weird since, e.g. $[\mathcal{O}_V]$ corresponds not to $[V]$ but $[V]+\alpha$ as mentioned above, so it can't be the same gradation.
2) On a related note, $\phi,\text{ch'}$ can't be ring homomorphisms, right? Since then that would imply the impossible thing from #1. But this seems odd, because in the case of $\phi$ at least, $\phi([V]\cdot [W])=\phi([V])\cdot\phi([W])$ seems to be exactly the (true) Serre intersection multiplicity formula - or would be, if all the nonvanishing Tors had support on $V\cap W$, which I'm not sure is the case. What exactly is going on here? I'm particularly interested in whether there is a quick way to prove the Serre intersection multiplicity formula here quickly in full rigor.

REPLY [9 votes]: Does this help?
$K(X) \otimes \mathbb{Q}$ is a graded ring in the sense that it is isomorphic (by the Chern character map) to $A(X) \otimes \mathbb{Q}$, which is graded. I would perhaps prefer to say that it is "gradeable", since the grading isn't very obvious in terms of $K$-theory. The most $K$-theoretic way I know to describe it is that the Adams operators $\psi^k$ act by $k^j$ on the $j$-th graded piece. The corresponding descending filtration is the filtration by codimension. 
For example, let $L$ be a line bundle with $c_1(L) = D$. Then $ch(L) = e^D$. So $ch(\log L) 
=D$ and $\log [L]$, defined as the class $([L]-1) - ([L]-1)^2/2 + ([L]-1)^3/3 - \cdots$ in $K(X)$, is pure of degree $1$. Indeed, $\psi^k [L] = [L^k]$, so $\psi^k \log [L] = \log [L^k] = k \log [L]$. Let's abbreviate $\log [L]$ by $z$. The structure sheaf of the vanishing locus of a section of $L$ has $K$-class $1-L^{-1} = 1-e^z = z-z^2/2+z^3/6-z^4/24+\cdots$. So this class is in the filtered part that has degree $\geq 1$, but is not of pure degree.
A natural question, to which I don't know the answer, is whether the integer $K$-ring has a natural grading $K(X) \cong \bigoplus K_i(X)$, which turns into this grading when we tensor by $\mathbb{Q}$.<|endoftext|>
TITLE: A conjecture associated with n-gon cut curve of degree m
QUESTION [5 upvotes]: This conjecture is a generalization of A theorem for cubic-A generalization of Carnot theorem, in MSE question.  I'm an electrical engineer, I am not a mathematician. I don't know how to prove this result.
Let $A_1A_2....A_n$ be a n-gons, Let $m$ points $B_{ij}$  lie on $A_iA_{i+1}$ for j=1, 2,...., m. Then $m.n$ points  $B_{ij}$ for $i=1, 2, 3,...., n$ and $j=1, 2,...., m$ lie on a curve of degree $m$ if only if:
$$\prod_{i=1}^n \prod_{j=1}^m \frac{\overline{B_{ij}A_i}}{\overline{B_{ij}A_{i+1}}}=1$$
Note that: $A_{m+1}=A_1$

REPLY [6 votes]: I believe you meant to type "curve of degree m". Let's prove the "only if" direction first. If you can prove it for triangles then you can triangulate your $n$-gon and multiply together the expressions for each triangle. So let's assume $n=3$.
Now, pick coordinates $(t_0,t_1,t_2)$ on $\mathbb P^2$ so that the lines $t_i=0$ correspond to the three lines of your triangle, and let $at_0+bt_1+bt_2$ be a generic line that doesn't pass through any of your points. Consider the three functions $$f_i=\frac{t_i}{at_0+bt_1+ct_2}$$
on your degree $m$ curve. Weil reciprocity then tells you that $\prod_{P} (f_i,f_j)_P=1$ so you can write $$\prod_P (f_0,f_1)_P(f_1,f_2)_P(f_2,f_0)_P=1$$
and by the definition of the Weil symbol this ends up being exactly the product $\prod_{i=1}^n \prod_{j=1}^m \frac{\overline{B_{ij}A_i}}{\overline{B_{ij}A_{i+1}}}$ in your statement. I learned this proof here. Note that the result is true even if you let points coincide and count them with multiplicities. 
To got the other way, omit one of the points and pick a degree $m$ curve through the remaining $3m-1$, by the calculation above the remaining point will be uniquely determined.<|endoftext|>
TITLE: Converse to Modularity II: Maass cusp forms
QUESTION [6 upvotes]: (This comes from this other question. You can find more details there)
The following bijection is now a theorem:

Odd irreducible 2-dim Galois repn $\longleftrightarrow$ weight 1
  newforms

note: Galois representations are taken to be continuous, complex and linear.
It is natural to ask whether the equivalent correspondence for even representations holds:

Even irreducible 2-dim Galois repn $\longleftrightarrow$ Maass cusp forms with $\lambda = 1/4$

Every non-icosahedral even irr. 2-dim Galois representation is known to arise from a Maass cusp form with eigenvalue $1/4$, from the known cases of the strong Artin conjecture, so that

Even irreducible 2-dim Galois repn $\longrightarrow$ Maass cusp forms with $\lambda = 1/4$

is almost a theorem. You can assume that it is, for the purpose of this question: how far is this from being known to be a bijection, as in the odd case? Is there a converse theorem, analogous to the Serre-Deligne result?

REPLY [7 votes]: This is a great question. I am no expert, but I share what I know.
We conjecture that this is a bijection. The most relevant publication that I am aware of is due to Blasius-Ramakrishnan (MR1012167) who provided a strategy to associate $2$-dimensional even complex Galois representations to Maass forms with Laplace eigenvalue $1/4$. Gelbart wrote in his AMS review to this paper that subsequently the authors together with Clozel and Harris proved the algebraicity of Hecke eigenvalues of such Maass forms unconditionally. Henniart gave a Bourbaki seminar on this topic (MR1040577), and then he published an erratum (MR1157812) whose AMS review by Zink reads:

The theorem of Blasius, Clozel and Ramakrishnan that the eigenvalues of Hecke operators for Maass forms of Galois type are algebraic numbers, which the author had discussed in the original paper, has to be considered unproved up to now. The problem is that the transfer $\Pi$ of a Maass representation $\pi$ from $\mathrm{GL}_2(\mathbf{A}_\mathbf{Q})$ to $\mathrm{GSp}_4(\mathbf{A}_\mathbf{Q})$ is unlike what has been stated before, namely its infinite component $\Pi_\infty$ occurs in an $L$-packet which does not contain limits of the discrete series. 

The erratum is based on a letter by Blasius and Ramakrishnan, dated 9 April 1991.
Added. A similar question was asked at MO already, see here.<|endoftext|>
TITLE: q-Integer-valued polynomials
QUESTION [8 upvotes]: For $n \in \mathbb{Z}_{\geq 0}$, let $[n]_q := (1-q^n)/(1-q) = (1+q+...+q^{n-1})$ as is customary, with $[0]_q=0$.
Let $R$ be the subring of $\mathbb{Q}(q)[x]$ consisting of all $f$ such that $f([n]_q) \in \mathbb{Z}[q]$ for all $n \in \mathbb{Z}_{\geq 0}$.
Define $f_0(x) = 1$ and $f_k(x) = f_{k-1}(x)\cdot \frac{x-[k-1]_q}{q^{k-1}[k]_q}$ for $k \geq 1$. Note $f_k([n]_q) = \frac{[n]_q!}{[n-k]_q! [k]_q!}=\binom{n}{k}_q \in \mathbb{Z}[q]$, so indeed $f_k \in R$ for all $k \geq 0$.
Is it the case that $R$ is spanned as a $\mathbb{Z}[q]$-module by the $f_k$? If so, what is known about the structure constants $c_{ij}^{k}(q)$?
EDIT: More information about this ring, such as the structure constants and a classification of all maps to fields, can be found in this preprint with Nate Harman: http://arxiv.org/abs/1601.06110v1.

REPLY [7 votes]: (Below is the proof that module $R$ is generated by $f_i(x)$, without calculation of structure constants.)
Polynomial $f(x)$ of degree $n$ may be interpolated in points $[0],[1],\dots,[n]$ (I omit index $q$): $f(x)=\sum_{i=0}^n f([i])\prod_{j\ne i} \frac{x-[j]}{[j]-[i]}$. For any summand leading term $f([i])\prod_{j\ne i} \frac1{[j]-[i]}$ is divisible in $\mathbb{Z}[q]$ by the leading term of $f_n(x)$. Hence $f(x)-Af_n(x)$ has degree less than $n$ for some $A\in \mathbb{Z}[q]$. This is induction step in the proof that $f$ is $\mathbb{Z}[q]$-linear combination of $f_0,f_1,\dots,f_n$.<|endoftext|>
TITLE: For a simplicial set $X$, is the category of non-degenerate simplices of $X$ a full subcategory of the category of simplices of $X$?
QUESTION [7 upvotes]: I'm slightly confused, but I think you can help me. Let $X$ be a simplicial set. The category of simplices of $X$ and its subcategory of non-degenerate simplices are defined at http://ncatlab.org/nlab/show/category+of+simplices. Is this subcategory a full subcategory? I think it is, because I think that I can prove that it is, but as I don't really trust myself, I'd like to have your confirmation. What irritates me is that on the referred nLab page, and also in Hovey's "Model categories" (just before Lemma 3.1.4, which, by the way, had to be repaired slightly: see http://hopf.math.purdue.edu/Hovey/model-err.pdf), it is specifically emphasized that the morphisms be monomorphisms (or injective, respectively).

REPLY [6 votes]: As far as I can tell you are correct.  In fact, any map in the category of simplices coming out of a nondegenerate simplex must be injective.  If $a:\Delta^n\to X$ is a nondegenerate simplex and $f:\Delta^n\to\Delta^m$ is a map to some other simplex $b:\Delta^m\to X$, factor $f=hg$ as a surjection followed by an injection.  If $g$ is not the identity, then it makes $a$ a degeneracy of the simplex $bh$ in $X$.  Thus $g=1$ and $f=h$ is injective.<|endoftext|>
TITLE: formula for Eta invariant
QUESTION [18 upvotes]: Hirzebruch's signature formula is not valid for manifolds with boundary.
An error term is introduced by Atiyah-Patodi-Singer to fix it.More precisely:
  $$sign (M)=L(M)[M]+\eta(\partial M)$$
Yet better,in some situations,this invariant could be used to distinguish homeo types within a simple homotopy type.
My question is about basic properties of eta invariant:Is there a formula for 
$\bullet$ $\eta(M\times N)$ where $M$ and $N$ are closed manifolds
$\bullet$ $\eta(\widetilde{M})$ where $\widetilde{M}$ is a regular covering space of closed manifold $M$ 
$\bullet$ $\eta(2M)$ where $M$ is a manifold with boundary and $2M$ is the double $M\cup_{\partial M} M$.

REPLY [5 votes]: You may find some information from the following paper:
Donnelly, Harold
Eta invariant of a fibered manifold. 
Topology 15 (1976), no. 3, 247–252.
Review from Mathscinet<|endoftext|>
TITLE: How is the free modular lattice on 3 generators related to 8-dimensional space?
QUESTION [34 upvotes]: Here are three facts which sound potentially related.  What are the actual relationships?

In 1900, Dedekind constructed the free modular lattice on 3 generators as a sublattice of the lattice of subspaces of an 8-dimensional vector space.  If the basis of the vector space is $e_1, \dots, e_8$, he looked at the subspaces
$$ X = \langle e_2, e_4, e_5, e_8 \rangle , \; Y = \langle e_2, e_3, e_6, e_7 \rangle, \;
Z = \langle e_1, e_4, e_6, e_7 + e_8 \rangle $$
By repeatedly taking intersections and unions, you can build up 28 subspaces starting from these three.  This proves the free modular lattice on 3 generators has at least 28 elements.  In fact it has exactly 28 elements.  I think Dedekind showed this by working out the free modular lattice 'by hand' and noting that it, too, has 28 elements.
The dimension of $\mathrm{SO}(8)$ is 28.  Its Lie algebra $\mathfrak{so}(8)$, also called $D_4$, has 12 positive roots, and its Cartan algebra has dimension 4.  As usual, the Lie algebra is spanned by positive roots, an equal number of negative roots, and the Cartan subalgebra, so we get
$$   28 = 12 + 12 + 4 $$
The 3 subspace problem asks us to classify triples of subspaces of a finite-dimensional vector space $V$, up to invertible linear transformations of $V$.  There are finitely many possibilities, unlike the situation for the 4 subspace problem.  One way to see this is to note that 3 subspaces $X, Y, Z \subseteq V$ give a representation of the $D_4$ quiver.  This is nothing profound: a representation of the $D_4$ quiver is just 3 linear maps $X \to V$, $Y \to V$, $Z \to V$, and here we are taking those to be inclusions.  The nontrivial part is that indecomposable representations of any Dynkin quiver correspond in a natural one-to-one way with positive roots of the corresponding Lie algebra.  So, in particular, the $D_4$ quiver has 12 indecomposable representations.  The representation coming from  $X, Y, Z \subseteq V$ must be a direct sum of indecomposable representations, so we can classify the possibilities and solve the 3 subspace problem.

Here is a way of making my question more concrete.  However, it may be on the wrong track:

Is there a natural 1-1 correspondence between the 28 elements of the free modular lattice on 3 generators and some basis of $\mathfrak{so}(8)$?  

Here is another more cautious way:

What is known about the relation between the free modular lattice on 3 generators and the 3 subspace problem?

It would help if one could somehow directly relate the free modular lattice on 3 generators to something built from $D_4$.  The free modular lattice on 3 generators looks like this:

For more background information, see:

John Baez, The free modular lattice on 3 generators, The n-Category Café, September 19, 2015.

REPLY [6 votes]: Here's a nice clean statement that emerged from discussions on The n-Category Café.
A representation of the $D_4$ quiver consists of three linear maps $f_i : L_i \to L$ ($i =1,2,3$) between finite-dimensional vector spaces over your favorite field.   We can take direct sums of these representations, and define an
indecomposable representation to be one that’s not a direct sum
of two others.
Given a representation of the $D_4$ quiver, the images of the maps $f_i$ are subspaces of $L$.  These generate a sublattice $\mathcal{L}$ of the lattice of all subspaces of $L$.  Clearly $\mathcal{L}$ is a modular lattice with 3 generators.
Theorem. If we take a direct sum of indecomposable representations
of the $D_4$ quiver, one from each isomorphism class, we obtain a
representation of the $D_4$ quiver whose corresponding modular lattice is the free modular lattice on 3 generators.  In this representation the spaces $\mathrm{im} (f_i)$ have dimension 5 and the space $L$ has dimension 10.  10 is the smallest possible dimension for a vector space $L$ containing subspaces that generate a copy of the free modular lattice on 3 generators.
(Here I should emphasize that I'm using lattice to mean a poset for which every finite subset has a least upper bound and greatest lower bound.  Such a thing has a top and bottom as well as binary operations $\vee$ and $\wedge$.  So, the free modular lattice on 3 generators, in this sense, has 30 elements, including a freely adjoined top and bottom element.  This gives a cleaner statement of the result than working with Dedekind's original definition of lattice.)
The proof is lurking in discussions here:

John Baez, The free modular lattice on 3 generators, The n-Category Café, September 19, 2015.

but I will put a cleaned-up version onto Visual Insight on January 1, 2016.<|endoftext|>
TITLE: Flag varieties and orbit of highest weight vector
QUESTION [7 upvotes]: I asked this here https://math.stackexchange.com/questions/1441408/orbit-under-lie-group-and-projective-variety but did not get any answers, so I am asking here. I apologize if this is not appropriate for this site. 
On https://en.wikipedia.org/wiki/Generalized_flag_variety#Highest_weight_orbits_and_homogeneous_projective_varieties there is a section which says

If G is a semisimple algebraic group (or Lie group) and V is a (finite dimensional) highest weight representation of G, then the highest weight space is a point in the projective space P(V) and its orbit under the action of G is a projective algebraic variety. This variety is a (generalized) flag variety, and furthermore, every (generalized) flag variety for G arises in this way.

I am interested in the case of Lie groups. Does anyone have a reference for this, which includes a proof? The bold parts are what really interest me.
I am also interested in whether the action of the real form $G_0 \subset G$ on the highest weight vector is also a projective algebraic variety.
Thanks.

REPLY [9 votes]: This is part of the Borel-Weil theorem, more precisely it's Theorem 3 in Serre's 1954 exposition. As he notes there, the algebraicity you ask about is a consequence of Chow's theorem. Also, the orbit of the highest weight space under the compact real form $G_0$ is open in the $G$-orbit by dimension and closed by compactness, so it coincides with the $G$-orbit.<|endoftext|>
TITLE: Every finitely generated flat module over a ring with finitely many minimal primes is projective
QUESTION [16 upvotes]: Over a commutative ring $R$, a finite type locally free (weak sense) module for which the rank function is locally constant is projective.
If we notice that for each minimal prime $p$ of the ring, the rank function is constant on $V(p)$, the adherence of $p$ in the Zariski topology (because if $p\subset q$ the rank at $p$ is equals to the rank at $q$) on finite locally free (weak sense) modules, then if the ring has only a finite number of minimal primes, the rank function is always locally constant on finite flat modules (Reasoning: the image of the rank function is finite in $\mathbb{N}$. Every reciprocal image of an integer $n$ is a finite union of closed sets like the $V(p)$. So it is a closed set. The disjoint union of these closed sets is the whole spectrum. Each one of them is therefore open right?). Am I right ?
I am asking this simple question because I read this great answer here
https://mathoverflow.net/a/33574/3333
where an important paper of Raynaud-Gruson is mentioned. It gives amongst a lot of generalizations the simple criteria that if $R$ has a finite number of associated primes then without any other hypothesis on $R$ every f.g. flat modules is actually projective. My reasoning above seems quite simple and gets a slightly more general result, but perhaps I am wrong ?
Edit2: I dug around but I did not find any reference to this simple criteria. I only found the criteria about the finiteness of the number of associated primes. Is it equivalent ? Is there a counterexample with an infinite number of embedded primes but a finite number of minimal (isolated) primes ?

REPLY [5 votes]: There are two questions here. First of all, yes, the argument is fine; secondly, yes, there are rings with finitely many minimal primes, but infinitely many associated primes. So all together, the criterion is slightly more general than the one by Raynaud-Gruson mentioned in the question, but the proof is much easier. It also admits a further generalisation (very slightly) as shown below.
First, the construction of a ring with finitely many minimal primes but infinitely many associated primes. It is obviously inspired by the ring $k[t,x]/(t^2,tx)$ which corresponds to the line with an embedded point in the origin, just that we take infinitely many embedded points. Of course, we have to leave world of polynomials for that.
Let $H$ be the ring of holomorphic functions on the complex line $\mathbb{C}$ (with coordinate $z$) and $f\in H$ a non-trivial holomorphic function with infinitely many zeros.
Then $A = H[t]/(t^2,tf)$ is such an example.
The nilradical of $A$ is the prime ideal $(t)$. Thus, $(t)$ is the unique minimal prime.
However, for each root $\alpha\in\mathbb{C}$ of $f$, we easily see that $(t,z-\alpha) = \mathrm{ann}_{A}(t(z-\alpha)^{-1}f)$ is an associated prime of $A$. By assumption, there are infinitely many roots, hence infinitely many associated primes.
To give a complement to the arguments given in the question and the comments on MSE, here is the generalisation to schemes (using the same arguments).

Let $X$ be a scheme with open connected components, each of which is a finite union of maximal irreducible components.
  Then every flat quasi-coherent $\mathcal{O}_X$-module of pointwise finite rank is Zariski-locally trivial.

This is a local question and we assumed the connected components to be open, so we may just assume that $X$ is connected and show that the rank is constant then.
Let $F$ be a quasi-coherent flat $\mathcal{O}_X$-module with $F_x$ finitely generated for each $x\in X$.
Let $\eta_1,\eta_2,\dots \eta_s\in X$ be the generic points of the maximal irreducible components $X_i := \{\eta_i\}^{-}$, $i = 1,2,\dots s$.
We first have to show that the rank of $F$ is constant on each $X_i$.
The basic thing to note is: if $y\leadsto x$ is a specialisation, i.e., $x\in\{y\}^{-}$, then $\mathcal{O}_{X,y}$ is a localisation of $\mathcal{O}_{X,x}$ and with respect to this structure we have $F_{y} = F_{x}\otimes_{\mathcal{O}_{X,x}}\mathcal{O}_{X,y}$.
Therefore, if $F_x$ is a free $\mathcal{O}_{X,x}$-module, then $F_y$ is a free $\mathcal{O}_{X,y}$-module of the same rank.
By definition, each $x\in X_i$ is a specialisation of $\eta_i$ and so the rank is constant on each maximal irreducible component.
To show that the rank is constant on our connected scheme $X$ we consider the subsets
$V_k := \{x\in X\mid \mathrm{rk}_{\mathcal{O}_{X,x}}(F_x) = k\}$
disjointly covering $X$ as $k\geq 0$. 
Since the rank is constant on maximal irreducible components, each $V_k$ is a union of such.
But there are only finitely many maximal irreducible components, so there are only finitely many nonempty $V_k$ and all of them are closed, hence, also open.
Since $X$ is connected, the only possibility for this to happen is when $X = V_k$ for some $k\geq 0$, i.e., if the rank is $k$ at every point.<|endoftext|>
TITLE: Is $T^{**}$ unconditionally $p$-summing whenever $T$ is unconditionally $p$-summing?
QUESTION [6 upvotes]: A sequence $(x_{n})_{n}$ in a Banach space $X$ is said to be unconditionally $p$-summable if $$\sup_{x^{*}\in B_{X^{*}}}\Bigl(\sum_{n=m}^{\infty}\lvert\langle x^{*},x_{n}\rangle\rvert^{p}\Bigr)^{1/p}\rightarrow 0\qquad(m\rightarrow \infty).$$ We say that an operator $T:X\rightarrow Y$ is unconditionally $p$-summing if $(Tx_{n})_{n}$ is unconditionally $p$-summable in $Y$ whenever $(x_{n})_{n}$ is weakly $p$-summable in $X$. We can prove that $T$ is unconditionally $p$-summing whenever $T^{**}$ is unconditionally $p$-summing. Is the converse true?

REPLY [3 votes]: The answer is no.  Bourgain and Delbaen constructed a Banach space $X$ that has the Schur property and $X^{**}$ is isomorphically universal for separable Banach space (it is even isomorphic to the the second dual of $C[0,1]$).  Every operator from $\ell_p$, $1<p<\infty$, and from $c_0$ into $X$ is thus compact, so that every operator with domain $X$ is unconditionally $p$-summing for all $1\le p < \infty$. But the second adjoint of the identity operator on $X$ is not unconditionally $p$-summing for any $1\le p < \infty$ because it is an isomorphism on copies of $\ell_p$ for all $p$.  
Bourgain, J.; Delbaen, F. A class of special $L_\infty$ spaces. Acta Math. 145 (1980), no. 3-4, 155–176.<|endoftext|>
TITLE: When is the tangent bundle of a manifold naturally a complex manifold?
QUESTION [8 upvotes]: It is well-known that the cotangent bundle of a manifold is naturally a symplectic manifold. Inspired by mirror symmetry, when is the tangent bundle $TM$ of a manifold $M$ naturally a complex manifold? There is always canonical almost complex structure as the tangent bundle of $TM$ is point wise $TM\times TM$, i,e. $J(x,y)=J(-y,x)$. When is this integrable?

REPLY [12 votes]: $TM$ is not ever `naturally' a complex manifold, in the sense that there is no complex structure on $TM$ that is preserved by all of the diffeomorphisms $f':TM\to TM$ where $f:M\to M$ ranges over all diffeomorphisms of $M$ with itself.
In spite of what the OP wrote, there is no canonical (i.e., natural) almost complex structure on $TM$ either.  This is a misunderstanding about $TTM$, which is not $TM\times TM$ 'pointwise' is any natural sense.<|endoftext|>
TITLE: Elementary proof of a special case of Chebotarev's density theorem
QUESTION [11 upvotes]: A special case of Cheboratev's density theorem states that, for $K/\mathbb{Q}$ a Galois number field of degree $n$, then the rational primes that split completely in $K$ have density $1/n$.
Is there an elementary proof of this fact?
Actually I'm asking this only to apply it to $K=\mathbb{Q}(\zeta_n)$ and get that the density is nonzero, and there is an elementary proof of this. However I'm also interested in the more general result.

REPLY [20 votes]: There does exist an elementary proof, and is given in many books on algebraic number theory. I think it is in Lang's book. I reproduce the proof below.
Consider $ \zeta _K(s)=\prod _{\mathfrak p}(1-\frac{1}{(N\mathfrak p)^{s}})^{-1}$ where $\mathfrak p$ runs over all prime ideals in the ring of integers in $K$. Therefore, 
$$\log (\zeta _K(s))=\sum _{\mathfrak p}\sum _ {m\geq 1} \frac{1}{m(N\mathfrak p)^{ms}}$$
Since $\zeta _K(s)=\frac{1}{s-1}(a_0+a_1(s-1)+\cdots)$ with $a_0\neq 0$, it follows that $\frac{\log (\zeta _K(s))}{\log \frac{1}{s-1}}$ tends to $1$ as $s$ tends to $1$ from the right. On the other hand, in this limit, only the term $m=1$ and $\mathfrak p$ of degree one over $\mathbb Q$ need be considered. Hence we get 
$$1= \lim _{s \rightarrow 1} \frac{\sum _{\mathfrak p} \frac{1}{(N\mathfrak p) ^s}}{\log \frac{1}{s-1}}.$$ Now, degree $1$ primes $\mathfrak p$ lie over rational primes $p$ which split completely in $K$, and over each $p$, there are $n=\deg(K/{\mathbb Q})$ primes $\mathfrak p$ of degree $1$. Hence we get 
$$1=\lim _{s\rightarrow 1} n\left( \frac{\sum _p \frac{1}{p^s}}{\log \frac{1}{s-1}}\right)$$ where the sum is over primes $p$ which split completely. This gives what you want.<|endoftext|>
TITLE: fixed points of permutation groups
QUESTION [10 upvotes]: As is well-known (see, for example, a nice exposition by our own Qiaochu: https://qchu.wordpress.com/2012/11/07/fixed-points-of-random-permutations/) that the distribution of the number of fixed points of random permutations (so, uniformly chosen elements of $S_n$) is Poisson, for largish $n.$ The question is: what is known for proper subgroups of $S_n?$

REPLY [2 votes]: Given $G\leq S_n$ let $X_G$ be the random variable on $\mathbf{N} = \{0,1,\dots,\}$ defined by the number of fixed points of a random $g \in G$.

Claim: For every random variable $X$ on $\overline{\mathbf{N}} = \mathbf{N} \cup \{\infty\}$ there is a sequence of groups $G_n$ such that $X_{G_n} \to X$ in distribution.

In other words, the set $\{\mu_{X_G}: G\leq S_n\}$ is dense in the space of all probability measures on $\overline{\mathbf{N}}$. As established in the comments, there are multiple ways of interpretting the question. This answers one particularly concrete interpretation.
Lemma 1: If both $X$ and $Y$ can be obtained, then so can their (independent) sum $X+Y$ and product $XY$.
(In the event of $0\cdot\infty$, our convention is $0\cdot\infty = 0$.)
Proof: Suppose $X_G \approx X$ and $X_H \approx Y$, where $G \leq \text{Sym}(\Omega_1)$ and $H \leq \text{Sym}(\Omega_2)$. Then $G\times H$ acts on $\Omega_1 \sqcup \Omega_2$ via
$$(g,h) \omega = \begin{cases} g \omega : \omega \in \Omega_1, \\ h \omega : \omega\in \Omega_2.\end{cases}$$
For this action we have
$$\text{fix}_{\Omega_1 \sqcup \Omega_2}((g,h)) = \text{fix}_{\Omega_1}(g) \cup \text{fix}_{\Omega_2}(h),$$
so the random variable for this action is $X_G + X_H \approx X + Y$.
Alternatively, $G \times H$ acts on $\Omega_1 \times \Omega_2$ via
$$(g,h)(\omega_1,\omega_2) = (g\omega_1, h \omega_2).$$
For this action we have
$$\text{fix}_{\Omega_1 \times \Omega_2}((g,h)) = \text{fix}_{\Omega_1}(g) \times \text{fix}_{\Omega_2}(h),$$
so the random variable for this action is $X_G X_H \approx X Y$.
Lemma 2: For each $t\in[0,1]$ we can obtain a random variable $X_t$ such that $X_t = 0$ with probability $t$ and $X_t = 1$ with probability $1-t$.
Proof: For prime $p$ consider the "$ax+b$ group" $\mathbf{F}_p^\times \ltimes \mathbf{F}_p$ acting on $\mathbf{F}_p$ via $(a,b) x = ax+b$. Evidently
$$|\text{fix}_{\mathbf{F}_p}(a,b))| = \begin{cases} 1 & \text{if}~a\neq 1,\\ 0&\text{if}~a=1, b \neq 0,\\ p & \text{if}~a=1,b=0.\end{cases}$$
So we can obtain a random variable $Y$ such that $Y = 1$ with probability $1-1/(p-1)$, $Y=0$ with probability $1/(p-1) - 1/p(p-1)$, and $Y=p$ with probability $1/p(p-1)$. By Lemma 1 then we can obtain $Y^* = \prod_{i=1}^k Y_i$, where each $Y_i$ is an independent copy of $Y_i$. Note that $Y^*=1$ with probability $(1-1/(p-1))^k$ and $Y^*\geq 2$ with probability $\leq k/p(p-1)$. Put $k = \lfloor \log(1/(1-t)) p \rfloor$ and send $p\to\infty$.
Proof of main claim: It suffices to show that we can obtain any random variable supported on $\{0,\dots,N-1\}\cup\{\infty\}$ for each $N$. We will prove this by induction on $N$. For the case $N=0$ (i.e., $X=\infty$ with probability $1$), consider $X_{S_n}$ as $n\to\infty$. Now let $N\geq 1$ and let $X$ be an arbitrary random variable supported on $\{0,\dots,N-1\}\cup\{\infty\}$. Let $p_i = \mathbf{P}(X = i)$. Let $Y$ be the random variable supported on $\{0,\dots,N-2\}\cup\{\infty\}$ defined by
$$\mathbf{P}(Y = i) = p_{i+1} / (1 - p_0).$$
By induction we can obtain $Y$. Thus by the lemmas we can obtain
$$(X_0 + Y) X_{p_0} \sim X.$$<|endoftext|>
TITLE: Isometric imbedding of a sphere with positively curved metric
QUESTION [11 upvotes]: QUESTION. Given a Riemannian metric on the sphere $S^n$ with positive sectional survature. Can it be isometrically imbedded into $\mathbb{R}^{n+1}$ (of any class of regularity) as a boundary of a convex set?
This question was motivated by the following three well known facts on isometric imbeddings (please correct me if references are not precise).
1) Given a smooth Riemannian metric with strictly positive Gauss curvature on the 2-sphere $S^2$, it can be isometrically imbedded into $\mathbb{R}^3$ such that its image bounds a convex set (Nirenberg, Pogorelov).
2) On $S^n$ with $n>2$ there exist smooth Riemannian metrics (even with positive sectional curvature) which cannot be smoothly (!) imbedded into $\mathbb{R}^{n+1}$. (Common knowledge.)
3) Any Riemannian metric on $S^n$ can be $C^1$-regularly isometrically imbedded into $\mathbb{R}^{n+1}$. (Special case of Nash's theorem.)

REPLY [5 votes]: Igor is essentially right. If $n > 2$ and the embedding is convex, then the second fundamental form exists almost everywhere. Using the Gauss equations per Robert's answer, it extends uniquely and smoothly to everywhere. The rest is straightforward. I'm omitting it because I'm typing this on an iPhone.
The $C^1$ but non-convex case is due to Nash (who did it in codimension 2) and Kuiper (who did it in codimension 1). It is not a special case of smooth Nash embedding theorem.
ADDED: The argument can be made rigorous by taking a family of smooth convex hypersurfaces converging uniformly to the original convex hypersurface and using the inverse to the map from positive definite second fundamental forms to curvature tenors defined by the Gauss equations to show that the second fundamental form has to converge uniformly to a continuos limit that also weakly solves the Codazzi equations. This can then be integrated to show that the embedding is in fact smooth. Uniqueness then follows by the Cohn-Vossen theorem.<|endoftext|>
TITLE: What are the possible automorphism groups of Riemann surfaces of low genus?
QUESTION [16 upvotes]: Skipping the easy cases of genus 0 and 1, what groups can arise as the group of conformal transformations of a Riemann surfaces of genus, say, 2 or 3?
I'm frustrated because there are papers that supposedly answer this question (which are even open-access):

Izumi Kuribayashi and Akikazu Kuribayashi, Automorphism groups of compact Riemann surfaces of genera three and four, Journal of Pure and Applied Algebra 65 (1990), 277-292
Izumi Kuribayashi, On an algebraization of the Riemann-Hurwitz relation, Kodai Math. J. 7 (1984), 222-237.

Unfortunately, I don't understand the notation for groups used in this paper.  There are lots of groups with names like $G(60,120)$ and $H(5 \times 40)$.  These are actually specific subgroups of $\mathrm{GL(g,\mathbb{C})}$ where $g$ is the genus of the surface, obtained by looking at how automorphisms of a Riemann surface act on its space of holomorphic 1-forms.  But I don't understand how they are defined, so I don't know how to answer questions like this:

Which 32-element groups show up as automorphism groups of Riemann surfaces of genus 3?

This last question is the one I really want answered right now, but in general I would like to know more about automorphism groups of low-genus Riemann surfaces.  I get the feeling that when such a group is reasonably big, it preserves a regular tiling of the surface by regular polygons, making the Riemann surface into the quotient of the hyperbolic plane by some Fuchsian group.  
The classic example is of course Klein's quartic curve, the genus-3 surface tiled by 24 regular heptagons, whose automorphism group is $PSL(2,7)$, the largest allowed by the Hurwitz automorphism theorem.

REPLY [5 votes]: To answer your specific question, there is a unique curve of genus three with a 32-element automorphism group, the hyperelliptic curve
$$ y^2 = x^8 - 1.$$
As for any hyperelliptic curve, the automorphism group fits in a central extension
$$ 0 \to C_2 \to G \to D_{16} \to 1 $$
where $C_2$ is the subgroup generated by the hyperelliptic involution and $D_{16}$ permutes the branch points of the hyperelliptic map, i.e. the solutions of $x^8-1=0$ on $\mathbf P^1$.
The group can be given a presentation $$ \langle s, t \mid  s^4, t^4, (st)^2, (s^{-1}t)^2\rangle.$$
I just learned this from a paper of Shaska and Wijesiri (a decidedly noncanonical reference).<|endoftext|>
TITLE: Examples of Stiefel-Whitney classes of manifolds
QUESTION [9 upvotes]: Let $M$ by an compact, connected $n$-dimensional manifold without boundary. 
Are there any other computable examples of the Stiefel-Whitney class $w(M)$ except for $M=S^m, \mathbb{R}P^m,\mathbb{C}P^m, \mathbb{H}P^m$?

REPLY [3 votes]: The manifold $P(m,n) := (S^m\times \mathbb{CP}^n)/(\mathbb Z/2)$, where $\mathbb Z/2$ acts by the antipodal map on $S^m$ and by complex conjugation on $\mathbb{CP}^n$, is called a Dold manifold. In "Erzeugende der Thomschen Algebra $\mathfrak N$", Dold computes the mod 2 cohomology, Steenrod module structure, and Stiefel-Whitney classes of $P(m,n)$. The calculations are fairly straightforward, and I found them to be a good example when I was learning Stiefel-Whitney classes and wanted more examples than projective spaces.
The paper is in German, but it's nonetheless quite readable, especially with a dictionary.<|endoftext|>
TITLE: canonical action of symmetric groups on orthogonal groups
QUESTION [5 upvotes]: There is a canonical faithful orthogonal representation of the symmetric group $S_{n+1}$, for $n\geq 1$:
$$
S_{n+1}\to O(n)
$$
given as follows.
(1). I regard $O(n)$ as the isometry group of the unit sphere $S^{n-1}$ in $\mathbb{R}^{n}$.
(2). Let $\Delta^n$ be a regular $n$-simplex embedded in $\mathbb{R}^{n}$ such that
(i). all its edges are of the same length;
(ii). all its $(n+1)$-vertices are in $S^n$;
(iii). its center is the 0.

(3). I observe that any permutation on the $(n+1)$-vertices of $\Delta^n$ can be uniquely extended to an isometry of $S^n$. Hence I get an embedding of $S_{n+1}$ into $O(n)$.
Regarding $O(n)$ as a manifold, we have a canonical action of $S_{n+1}$ on $O(n)$. Hence we have a covering map
$$
O(n)\to O(n)/S_{n+1}.
$$
Let the vector bundle (the action of $S_{n+1}$ on $\mathbb{R}^{n+1}$ is given by permutation of coordinates of $\mathbb{R}^{n+1}$)
$$
\eta: \mathbb{R}^{n+1}\to O(n)\times _{S_{n+1}}\mathbb{R}^{n+1}\to O(n)/S_{n+1}.
$$
Question: I want to know the Stiefel-Whitney class of $w(\eta)$. How to compute it?

REPLY [2 votes]: I think the bundle you consider is trivial, because the representation of $S_{n+1}$ on $\mathbb R^{n+1}$ by permutation of coordinates extends to a representation of $O(n)$. I am not completely sure about the first part of the argument, however. 
I think the map $S_{n+1}\to O(n)$ you describe actually comes from the $n$-dimensional standard representation of $S_{n+1}$. Indeed, permutation of coordinates gives a homomorphism $S_{n+1}\to O(n+1)$, but this has actually values in the stabilizer of the vector $v:=(1,\dots,1)$ and that stabilizer is isomorphic to $O(n)$, via the action on $v^\perp$. Now I think that if you orthogonally project the unit vectors to this hyperplane, you get (up to a multiplication by a constant factor depending on $n$) the $n+1$ points in the sphere that you describe. 
If this is correct, then the representation of $S_{n+1}$ on $\mathbb R^{n+1}$ indeed extends to $O(n)$ as the direct sum of the defining representation of $O(n)$ on $v^\perp$ and the trivial representation on the line spanned by $v$. But then it follows from general principles that the bundle $O(n)\times_{S_{n+1}}\mathbb R^{n+1}$ can be trivialized. To do this, consider the map $O(n)\times\mathbb R^{n+1}\to (O(n)/S_{n+1})\times\mathbb R^{n+1}$ defined by $(g,v)\mapsto (gS_{n+1},g\cdot v)$. This evidently factorizes to an isomorphism on the induced bundle.<|endoftext|>
TITLE: Relative null-ness
QUESTION [21 upvotes]: Here, "measure" always means Lebesgue measure on $\mathbb{R}$. This question is partly motivated by my answer https://math.stackexchange.com/questions/1444498/is-there-a-categorizaiton-system-for-null-quantities/1444524#1444524. 
Some measure zero sets are more measure zero than others:

A set $X\subseteq\mathbb{R}$ is strong measure zero if for any sequence $(\epsilon_i)_{i\in\mathbb{N}}$ of positive reals, there is a cover $\{I_n: n\in\mathbb{N}\}$ of $X$ by intervals with $\mu(I_n)<\epsilon_n$.
A set $X\subseteq\mathbb{R}$ is microscopic if for any sequence $(\epsilon_i)_{i\in\mathbb{N}}$ of positive reals where for some positive real $\delta$ we have $\epsilon_i=\delta^{i+1}$, there is a cover $\{I_n: n\in\mathbb{N}\}$ of $X$ by intervals with $\mu(I_n)<\epsilon_n$.
Etc. (See e.g. http://www.sav.sk/journals/uploads/0721132912Horbac.pdf.)

More generally, if $F$ is a family of functions from $\mathbb{N}$ to $\mathbb{R}_{>0}$, say $X\subseteq\mathbb{R}$ is $F$-microscopic if for every $f\in F$, there is a cover $\{I_n: n\in\mathbb{N}\}$ of $X$ by open intervals such that $\mu(I_n)<f(n)$. To measure how null a given null set is, we can look at its scope (I'm not sure what this is actually called, I couldn't find a reference to it): $scope(X)=\{f: \mathbb{N}\rightarrow\mathbb{R}_{>0}: X\text{ is $\{f\}$-microscopic}\}.$ This leads to a natural preorder on the set of null subsets of $\mathbb{R}$: $$X\le_{null}Y\iff scope(X)\subseteq scope(Y).$$
My question is: 

What does the resulting degree structure $\mathfrak{N}=(Null/\equiv_{null}, \le_{null})$ look like? 

I'm particularly interested in the extent to which set-theoretic hypotheses such as large cardinals or forcing axioms are relevant; based on the fact that even when just studying strong measure zero sets, set-theoretic hypotheses become important, I suspect there is in fact some relevance.

Comment 1: Arguably my definition of "scope" is wrong, and we should instead look at something slightly more well-behaved, like $\{f: X\text{ is $\{\alpha f: \alpha\in\mathbb{R}_{>0}\}$-microscopic}\}$ or similarly. If tweaking the definition of scope would lead to a better result, feel free to do so.
Comment 2: Of course we can work in much more generality than $\mathbb{R}$ with Lebesgue measure, but already this case seems really interesting.

EDIT: It seems like a good first step would be to try to understand forcings which add sets of a prescribed scope. Ideally, this could be used to prove e.g. that it's consistent that there are null sets with incomparable scopes.
The following forcing notion might be a good first try. For $F$ a set of functions from $\mathbb{N}$ to $\mathbb{R}_{>0}$, let $\mathbb{P}_F$ be the set of ordered pairs $(D, C)$ where 

$D$ is a countable set of reals, and
$C$ is a finite set of infinite families of intervals $\mathcal{C}_{f_0}, . . . ,\mathcal{C}_{f_n}$ where $f_i\in F$ and each $\mathcal{C}_{f_i}=\{I_n^{f_i}: n\in\mathbb{N}\}$ satisfies $\mu(I_n^{f_i})<f_i(n)$, such that
$D\subseteq \bigcap_{\mathcal{C}_{f_i}\in C} (\bigcup_{n\in\mathbb{N}} I_n^{f_i})$, and
the set $\bigcap_{\mathcal{C}_{f_i}\in C} (\bigcup_{n\in\mathbb{N}} I_n^{f_i})$ has positive measure (to prevent the forcing from being trivial).

ordered by $(D, C)\le (D', C')$ if $D\supseteq D'$ and $C\supseteq C'$. Forcing with $\mathbb{P}_F$ yields a set of reals which is uncountable, and is $F$-microscopic. It also, unfortunately, does a fair bit of damage to the ground reals. A countably closed forcing would be nicer, but it's not clear to me how to make that work without accidentally building a set of strong measure zero.

REPLY [3 votes]: $\newcommand{\scope}{\mathrm{Scope}}$
$\newcommand{\res}{\upharpoonright}$
The edit makes it seem like already the question of whether incompatible null sets can exists is of interest to you, I hope that is correct. I will construct $\subseteq_{\mathrm{Null}}$-incompatible null sets (without any additional assumptions). (I use the definition of $\subseteq_{\mathrm{Null}}$ in the question, not the one in Comment 1).
All the null sets here will be "Cantor-like" sets. As a first step, we will analyse (to some extend) for which $f$ the Cantor set is $f$-microscopic. Recall the usual "take away the middle third" construction of the Cantor set: Let $I_\emptyset=[0, 1]$ and if $I_s$ is defined let $I_{s^\frown 0}$ be the (closed) "first third" of $I_s$ and $I_{s^\frown 1}$ the (closed) "last third" of $I_s$. The Cantor set is
$$\mathcal C=\bigcup_{x:\mathbb N\rightarrow\{0, 1\}}\bigcap_{n\in\mathbb N}I_{x\res n}$$
(with $n=\{0,\dots, n-1\}$). Set $\mathcal C^\ast=\mathcal C\setminus\{1\}$. Via proof by picture, there is a (red) interval cover $\langle J_n\mid n<\omega\rangle$ of $\mathcal C^\ast$ with $\mu(J_n)=3^{-(n+1)}$:

Curiously, there is no such cover for $\mathcal C$ itself, but there is one after any one  point is removed from $\mathcal C$.

Lemma: If $f(n)\leq\frac{2}{3^{n+1}}$ for all $n$, then $\mathcal C^\ast$ is $f$-microscopic iff $f(n)>\frac{1}{3^{n+1}}$ for all $n\in\mathbb N$.

Proof: We already know that the "if" part holds. For the other direction, let us first assume that for some $\epsilon>0$ and all $n$, $f(n)\leq\frac{2-\epsilon}{3^{n+1}}$. Find $N\geq 1$ large enough so that $3^{-N}\leq\epsilon$ for all $n$. Exemplarily, we deal with the case $N=2$ in the pictures. Let $\langle J_n\mid n\in\mathbb N\rangle$ witness that $\mathcal C^\ast$ is $f$-microscopic. By induction we will find $A_n\subseteq {}^{N+n}\{0, 1\}$ so that

$\vert A_n\vert\geq 2^{N-1}$
If $s\in A_n$ and $m<n$ then $s\res N+m\in A_m$
If $s\in A_n$ then $I_s\cap
    J_k=\emptyset$ for all $k\leq n$.

We start with $n=0$. Since $\mu(I_0)<\frac{2}{3}-\frac{1}{3^N}$, $J_0$ must be disjoint from at least $2^{N-1}$ of the $2^N$-many intervals $I_s$ with $s$ of length $N$. We proof this by picture and imagination: Imagine one slides the red interval $J_0$ to the left so that its lower endpoint is $0$. If it is now slid to the right, then it disjoint from the first interval on the third level before it meets the next interval:

And this pattern continues. Thus $A_0=\{s:N\rightarrow 2\mid I_s\cap J_0=\emptyset\}$ has size at least  $2^{N-1}$. Suppose we have managed to get to stage $n$. Since every interval $I_s$, $s\in A_n$ splits into two smaller intervals on the next level, there are at least $2^N$-many of those. By a similar argument as above, $J_{n+1}$ can cover at most half of those. Hence
$$A_{n+1}=\{s:n+1\rightarrow \{0, 1\}\mid s\res n\in A_n\wedge I_{s}\cap J_{n+1}=\emptyset\}$$
works.
Now $A=\bigcup_{n\in\mathbb N} A_n$ forms a tree via end-extension with all levels non-empty and finite. By König's Lemma, there is a branch $x:\mathbb N\rightarrow \{0, 1\}$ through $A$ (i.e. $x\res N+n\in A_n$ for all $n$) and the unique real $y\in \bigcap_{n\in\mathbb N} I_{x\res n}$ is in $\mathcal C$ and avoids every interval $J_n$. Thus $y$ must equal $1$ and $x\equiv 1$. This also means that $\vert A_n\vert = 2^{N-1}$ for any $n$: If $A_n$ is larger, after taking $x\res N+n$ out of $A_n$, we can still continue the argument indefinitely and get a branch through $A$ different from $x$, contradiction.
Finally assume, for a contradiction,  that for some $m$, $f(m)\leq 3^{-(m+1)}$ so that $\mu(J_m)<3^{-(m+1)}$. Again by a sliding argument, one can be convinced that $J_m$ cannot completely cover $2^{N-1}$-many of the intervals $I_{s^\frown i}$, $s\in A_{m-1}$, $i<2$. Thus there must be some such $t=s^\frown i$ so that $I_t\cap J_m\neq\emptyset (\Rightarrow t\notin{A_m})$ and $I_t\not\subseteq J_m$. In particular, one of the endpoints $z$ of the interval $I_t$ is not covered by $J_m$ (and any earlier $J_n$, $n<m$). (If we are in the nasty case of $z=1$, replace $z$ by a slightly smaller real in $\mathcal C\cap I_t\setminus J_m$) This real $z$ must be accounted for at a larger stage: There is $M>m$ with $z\in J_M$. But this is a big problem: This $z$ is so far away from from the other intervals still in the game that $J_M$ cannot both cover $z$ and enough of them:

$J_m$ misses $z$ and the later $J_M$ covers $z$ but cannot reach the $4$ blue intervals of which it should cover $2$.
Thus $\vert A_{M}\vert>2^{N-1}$, contradiction.
The final trick to remove the $\epsilon$ is to diagonalise the above argument, i.e. one increases $N$ along the construction of the $A_n$'s if necessary.$\Box$
To construct incompatible null sets, we relativise the construction of the Cantor set:
Let $g:\mathbb N\rightarrow(0, \infty)$ so that

$g(0)\leq \frac{1}{3}$
$\forall n\ g(n+1)\leq\frac{g(n)}{3}$

Then let $I^g_\emptyset=[0, 1]$ and if $I^g_s$ is constructed $\mathrm{len}(s)=n$, $I^g_{s^\frown 0}$ is the (closed) initial segment of $I^g_s$ of length $g(n+1)$ and $I^g_{s^\frown 1}$ is the (closed) end-segment of $I^g_s$ of length $g(n+1)$. Then
$$\mathcal C_g=\bigcup_{x:\mathbb N\rightarrow 2}\bigcap_{n\in\mathbb N} I^g_{x\res n}$$
is an uncountable closed null set. If we set $\mathcal C^\ast_g=\mathcal C_g\setminus\{1\}$, then by the same argument as above we get:

If $f\leq 2g$ then $\mathcal C_g^\ast$ is $f$-microscopic iff $f>g$.

From this we can construct $\subseteq_{\mathrm{Null}}$-incompatible null sets: Take two functions $g_i$, $i<2$ for which $\mathcal C_{g_i}^\ast$ is defined so that

For all $n$ there is some $i$ with $g_i(n)\leq g_{1-i}(n)<2g_i(n)$
For both $i$ there is $n_i$ with $g_i(n_i)<g_{1-i}(n_i)$

If we let $f_i$ so that $f_i(n_i)\in (g_i(n_i), g_{1-i}(n_i)]$ and for $n\neq n_i$
$$f_i(n)=\mathrm{min}\{2g_0(n), 2g_1(n)\}$$
then
$$f_i\in\scope(\mathcal C_{g_i}^\ast)\setminus\scope(\mathcal C_{g_{1-i}}^\ast)$$
as the single mishap at $n_i$ excludes $f_i$ from the scope of $\mathcal C_{g_{1-i}}^\ast$ but not from the scope of $\mathcal C_{g_i}^\ast$. Observe that we only used two integers for "mishaps", but we have countably infinite many possibilities for them to occur. So in a similar manner one can construct a countably infinite antichain in $\mathcal N=(\mathrm{Null}/\equiv_{\mathrm{Null}},\subseteq_{\mathrm{Null}})$.
In my opinion, this construction can either be seen optimistically, in the sense that it gives some insight into the structure of $\mathcal N$, or pesimistically in the sense that it is evidence for the definition of $\scope$ being incorrect (in the same vein as your Comment 1).
I hope there is still some interest in this question after all these years!<|endoftext|>
TITLE: Bounding Schur symmetric polynomials on the unit circle
QUESTION [20 upvotes]: Recall the Schur polynomial in $n$ variables, indexed by the partition $\lambda$, with $\ell(\lambda) \leq n$, is given by 
\begin{equation}
s_\lambda(x_1,\ldots, x_n) = a_{\lambda + \delta}(x_1, \ldots, x_n) / a_\delta (x_1, \ldots, x_n),
\end{equation}
 where $\delta = (n-1,n-2,\ldots, 0)$ and $a_\lambda = \det (x_i^{\lambda_j})$ is the Vandermonde determinant of the $n \times n$ matrix whose $(i,j)$ element is $x_i^{\lambda_j}$. My notation here is consistent with Ian Macdonald's Hall polynomial book chapter 1.
Also let $e_j$ be the $j$th elementary symmetric polynomial in $n$ variables, $0 \le j \le n$. These are defined by 
\begin{equation}
\prod_{i=1}^n (1 + x_i t) = \sum_{j=0}^n e_j(x_1,\ldots, x_n) t^j.
\end{equation}
By the Jacobi-Trudi formula we know that 
\begin{equation}
s_\lambda = \det(e_{\lambda^t_i -i + j}),
\end{equation}
 where $\lambda^t$ is the transpose of the partition $\lambda$, that is, $\lambda^t_i = \mid\{j: \lambda_j \geq i\}\mid$. Thus $s_{1^j} = e_j$, for $j \le n$. 
Now we specialize to $x_i \in \mathbb{T}:= \{z \in \mathbb{C}: \mid z \mid = 1\}$. Given that
\begin{equation}
e_1(x_1, \ldots, x_n) = 0,
\end{equation}
 that is, $\sum x_i = 0$, I am interested in bounding $s_\lambda$. 
I have a conjecture for $\lambda = 1^{n/2}$, and $n = 4m$, $m \in \mathbb{N}$, namely, 
\begin{equation}
\mid e_{n/2}(x_1, \ldots, x_n) \mid \le \binom{n/2}{n/4}.
\end{equation}
This is attained when $x_i = (-1)^i$, that is, when half of them equal $1$ and the other half equal $-1$, since 
\begin{equation}
\prod_i (1  +x_i t) = (1-t)^{n/2} (1+t)^{n/2} = (1-t^2)^{n/2}.
\end{equation} 
I don't know if this conjecture is true for all qualifying $x_i$'s. Full credit will be given to solve this special case. However, I am also interested in a general conjecture for arbitrary $\lambda$.

REPLY [14 votes]: In the special case mentioned in the problem, I'll show the bound 
$$ 
|e_{n/2}(x_1,\ldots, x_n)| \le 2^{n/2}. 
$$ 
Let 
$$ 
F(z) = \prod_{j=1}^{n} (1+zx_j) = C \prod_{j=1}^{n} (z + \overline{x_j}), 
$$ 
where $C= \prod_{j} x_j$ has magnitude $1$.  (The roots of $F$ are $-\overline{x_j}$.)
Now by Cauchy's theorem 
$$ 
e_{n/2}(x_1,\ldots,x_n) = \frac{1}{2\pi i} \int_{|z|=1} \frac{F(z)}{z^{n/2+1}} dz 
$$ 
and so in magnitude this is bounded by 
$$ 
\le \sup_{|z|=1} |F(z)|. 
$$ 
Now we use a very nice Theorem of Carneiro and Vaaler (see Theorem 8.1 there), which gives a bound for the maximum of a polynomial whose roots are on the unit circle in terms of the first few symmetric power sums of the roots.  Let me quote their result fully:  Suppose $G(z) = \prod_{j=1}^{N}(z-\alpha_j)$ is a polynomial with $|\alpha_j| \le 1$ for all $j$.  Then for any natural number $K$ we have 
$$ 
\sup_{|z|\le 1} \log |G(z)| \le \frac{N}{K+1} \log 2 + \sum_{k=1}^{K} \frac{1}{k} \Big| \sum_{j=1}^{N} \alpha_j^k \Big|.
$$ 
This is Theorem 8.1, display (8.6) of their paper.   
Apply this to our polynomial $F$, with $K=1$.  Since the sum of the $x_j$ is zero, by assumption, we deduce that 
$$ 
\max_{|z|=1} \log |F(z)| \le \frac{N}{2} \log 2, 
$$ 
which proves the claimed estimate. 
Maybe there is an easier proof of this particular bound, rather than appealing to the Carneiro-Vaaler work.  I didn't see one immediately, and would be very interested if someone found an alternative approach.  
Edit:  In the particular case $K=1$, which is what is needed for this question, zeb kindly showed me the following one line proof: assuming $|\alpha_j|\le 1$ and $|z|\le 1$ we have 
$$
|G(z)|^2 =\prod_{j} |z-\alpha_j|^2 \le \prod_{j} (2-2\text{Re }\overline{z}\alpha_j) \le 2^N \prod_j \exp(-\text{Re }\overline{z}\alpha_j)\le 2^N \exp\Big(\Big| \sum_j \alpha_j\Big|\Big).
$$<|endoftext|>
TITLE: What's the difference between a PL simplicial sphere and a shellable simplicial sphere?
QUESTION [8 upvotes]: Shellability of a simplicial sphere tells us that we can build up the complex one facet at a time such that at each step (except the last step) the complex is a PL-ball. At the last step it is of course a PL-sphere.
Lickorish showed that there exist non-shellable d-spheres for $d \geq 3$.
My question, perhaps a cynical one, is 'Are there any practical implications for why someone would care to have a shellable simplicial sphere, rather than just a PL-sphere?

REPLY [2 votes]: I think the question is slightly missing the point, in a way that it may be useful to explain. Frequently, the reason you would want to show something is a shellable simplicial sphere is in order to be able to deduce that it is a sphere at all. Given some crazy simplicial complex, it is not obvious whether it is a sphere or not, but for a shellable complex, it's much easier to tell.<|endoftext|>
TITLE: Seifert--van Kampen for the loop space dga
QUESTION [6 upvotes]: I recently noticed that I could mostly prove a special case of the following statement. I think it's true in general, though perhaps only for nice spaces.

Let $X$ be a topological space, and suppose $X=X_0\cup X_1$, where $X_0$, $X_1$, and $Y:=X_0\cap X_1$ are all nonempty and path-connected. Then
  $$
C_*(\Omega X)\cong C_*(\Omega X_0)\otimes_{C_*(\Omega Y)}C_*(\Omega X_1)
$$
  where $\otimes$ is the homotopy pushout of dg-algebras and $\cong$ is $A_\infty$ quasi-isomorphism.

For disconnected spaces, it should still be true, where one replaces $C_*(\Omega X)$ with the dg-category of chains on the path space.
This seems like a natural mostly-extension of the classical Seifert--van Kampen theorem, which makes me think it exists somewhere in the literature, but I couldn't find it. Does anyone know where to find this statement?
Thanks!

REPLY [5 votes]: Sketch of proof. I will use the following ingredients
0) the category of spaces will be the category of simplicial sets. All the computation are in the derived sense. 
1) use the Quillen adjunction $$F: sSet^{\otimes}\longleftrightarrow sMod_{k}^{\otimes}: U$$ 
between the category of monoids in simplicial sets and  monoids in simplicial $k$-modules .
2) the category of simplicial $k$-algebras is Quillen equivalent to the category of differential graded $k$-algebras (i.e. monoids in the category of chain complexes of $k$-modules in positive degree) 
$$N: sMod_{k}^{\otimes}\longleftrightarrow Ch^{\otimes}_{k}: S$$ 
Now suppose that $T$ is  a homotopy push out $X\leftarrow Y\rightarrow Z$ of connected pointed simplicial sets (Kan complexes), then $\Omega T$ is the homotopy pushout of  $\Omega  X\leftarrow \Omega Y\rightarrow \Omega Z$ in the category in $sSet^{\otimes}$ (notice that the model that I'm using for loop space gives me a honest simplicial monoid). The functor $F$ commutes with homotopy push out, it means that that $F(\Omega T)$ is a homotopy push out of $F(\Omega  X)\leftarrow F(\Omega Y)\rightarrow F(\Omega Z)$ in $sMod_{k}^{\otimes}$, therefor the $NF(\Omega  X)\leftarrow NF(\Omega Y)\rightarrow NF(\Omega Z)$ is a homotopy push out in the category $Ch^{\otimes}_{k}$ and the functor $NF$ can be identified with $C_{\ast}(-,k)$.
It means that $$C_{\ast}(\Omega T)\simeq C_{\ast}(\Omega X)\sqcup_{C_{\ast}(\Omega Y)}^{h}C_{\ast}(\Omega Z)$$ as differential graded algebras, where $\sqcup$ is the coproduct in the category of $DGA$. This statement is true for any pointed connected spaces $X, Y$ and $Z$.<|endoftext|>
TITLE: An effective version of Kronecker's approximation theorem and its variations
QUESTION [10 upvotes]: Let $\theta_1,\dots,\theta_n$ be algebraic numbers of degree $\leq D$ and Weil height $\leq H$ such that $1,\theta_1,\dots,\theta_n$ are $\mathbb Q$-linearly independent. An effective version of Kronecker's approximation theorem would then assert that, given $\epsilon > 0$ and $\alpha \in \mathbb R^n$, there exist integers $q,p_1,\dots,p_n$ such that
$$|q\theta_i - \alpha_i - p_i| \leq \epsilon$$
for all $1 \leq i \leq n$, where $q$ is bounded appropriately in terms of $D$, $H$, and $\epsilon$. I was told that a result like this was originally proved by Erdos and Turan, but I cannot find a reference. Would anyone know of such a reference?
Additionally, I need a following variation of this theorem: let $s,t$ be fixed integers, then given $\epsilon > 0$ and $\alpha \in \mathbb R^n$, there exist integers $q,p_1,\dots,p_n$ such that
$$|(sq+t)\theta_i - \alpha_i - p_i| \leq \epsilon$$
for all $1 \leq i \leq n$, where $q$ is bounded appropriately in terms of $D$, $H$, and $\epsilon$. Is something like this known? What are the best known bounds?
I would appreciate any information on this!

REPLY [8 votes]: The numbers $\theta_i$ here are real of course. Also there is only one question really; upon replacing $\boldsymbol{\alpha}$ with $(\boldsymbol{\alpha} - t \boldsymbol{\theta})/s$ the second problem reduces to the case that $s = 1$ and $t = 0$.
That being said, there is nothing inherently ineffective in neither of the two best known proofs of Kronecker's approximation theorem: the diophantine approximations one based on the Dirichlet approximation theorem with an inductive scheme on $n$, and the harmonic analysis one based on Weyl's equidistribution criterion. Into either of these proofs you may input Liouville's lower bound on non-vanishing algebraic quantities, in the form $\mathrm{dist}(\mathbf{k} \cdot \boldsymbol{\theta}, \mathbb{Z}) \geq (2n \cdot \| \mathbf{k} \|_{\infty} \cdot H)^{-D^n}$ for non-zero $\mathbf{k} \in \mathbb{Z}^n$ and $\| \mathbf{k} \|_{\infty} := \max_{i=1}^n |k_i|$.
Let me give some indication of how the latter route goes. It is Weyl's theorem that Erdos and Turan made completely explicit; their inequality is where you were referred to. You will however need the extension of the Erdos-Turan estimate to the higher dimensional torus $\mathbb{T}^n = (\mathbb{R}/\mathbb{Z})^n$, which is due to Koksma. A good reference for this is Theorem 1.21 of the book Sequences, Discrepancies and Application by Michael Drmota and Robert Tichy (LNM 1651, 1997). This bounds the discrepancy $D_N := \sup_{I \subset \mathbb{T}^n} \big| \#\{ \mathbf{x}_j \in I, \, j \leq N\}  - N \cdot \mathrm{vol}(I) \big|$ of a sequence $\mathbf{x}_j \in \mathbb{T}^n$ by an explicit linear form of its Fourier coefficients:
Theorem. (Erdos, Turan, Koksma). For any sequence $\mathbf{x}_j \in \mathbb{T}^n$ and every $K \in \mathbb{N}$ it holds
$$
D_N = D_N(\mathbf{x}) \leq N \cdot (3/2)^n \cdot \Big( \frac{2}{K+1} + \sum_{0 < \|\mathbf{k}\| \leq K} \frac{1}{r(\mathbf{k})} \Big| \frac{1}{N} \sum_{j=1}^N e( \mathbf{k} \cdot \mathbf{x}_j ) \Big| \Big),
$$
with $e(z) := e^{2\pi i z}$ and $r(\mathbf{k}) := \prod_{i=1}^n \max (1,|k_i|)$.
Apply this theorem taking $\mathbf{x}_j := j \, \boldsymbol{\theta}$. The exponential sums in this case are geometric series with quotients of the form $e( \mathbf{k} \cdot \boldsymbol{\theta} )$ and $\mathbf{k} \neq \mathbf{0}$. They are bounded in magnitude by $(2 \, \mathrm{dist}(\mathbf{k} \cdot \boldsymbol{\theta}, \mathbb{Z}))^{-1}$, which in turn is bounded by the Liouville diophantine estimate above. This is the same point as in the Polya-Vinogradov inequality on character sums.<|endoftext|>
TITLE: New base of matroid from old
QUESTION [7 upvotes]: Let $M$ be a matroid of rank 3 and $E_1, E_2, E_3$ 3 basis of $M.$ Let $e_{i,j}$ be the $i$-th element of base $E_j$.

Is it true that you can always find a permutation $s: \{1,2,3\} \to \{1,2,3\}$ such that $e_{s(1),1}, e_{s(2),2}, e_{s(3),3}$ is also a base? 

I could not find a counter example.
I am also interested in arbitrary rank (it is obviously false for a matroid of rank 2).

REPLY [4 votes]: Let $M$ be the cycle matroid of the graph shown below.

Consider this grid:
1 2 3
2 3 4
1 2 5

Each row gives a basis for $M$.  However, no transversal gives a basis.<|endoftext|>
TITLE: Iterated Homotopy Quotient
QUESTION [6 upvotes]: If one has a normal Lie group inclusion $H\to G$, with quotient $G/H$, and a $G$-manifold $X$, one can take the quotient $X/H$. Then the $G$-action on $X/H$ factors thru a $G/H$-action, so one can take the quotient again to obtain $(X/H)/(G/H)$. It is known classically that $(X/H)/(G/H)\cong X/G$ (see, e.g. Bourbaki's book on Lie Groups and Lie Algebras, Section 1.6 of Chapter III). 
In a less structured setting, say, May's "Classifying Spaces and Fibrations" where the $H$ and $G$ are not Lie groups, but only topological monoids, and $X$ is just an object of the category of compactly generated Hausdorff spaces, we can still produce the quotient monoid $G/H$ and quotient space $X/H$ by a bar construction.
One could take this one step further and work just with a spectrum $X$ with an action of a topological monoid $G$. One can still take the quotient by $H$, and can still try to then take the quotient again by $G/H$. This all seems like it should be some purely formal bar construction thing.
My question is: where (if anywhere) can I find a proof that for a general model or quasi- category $C$, and a $G$-action on an object $X\in C$, the iterated quotient $(X/H)/(G/H)$ is equivalent to $X/G$? 
It'd also be nice to see a general proof for spaces or spectra.

REPLY [7 votes]: Here's a sketch. The sequence of maps to look at is actually
$$BH \to BG \to B(G/H).$$
Recall that an object $X$ with a $G$-action in an $\infty$-category $C$ is the same thing as a functor $BG \to C$. I claim that the left Kan extension of this to a functor $B(G/H) \to C$ has the effect of quotienting by $H$ and then remembering the $G/H$-action; this follows from the fact that the above is a fiber sequence, and that left Kan extensions take "fiberwise colimits." Then the further quotient by the $G/H$ action is just taking the further left Kan extension along the unique map $B(G/H) \to \bullet$. 
On the other hand, taking the left Kan extension along $BG \to \bullet$ has the effect of quotienting by $G$. So the desired result follows from the observation that Kan extensions compose.<|endoftext|>
TITLE: Some "axiom of choice" and "dependent choice" issues
QUESTION [10 upvotes]: I am probably about to ask some fairly basic questions, and yet I have found it quite hard to find the answers to these.
If I understand correctly, mathematicians tend to be quite happy working with ZF+DC, but other forms of choice that are not implied by DC can be more controversial.
[Therefore it seems natural that people should give higher priority to discussing the differences in provable theorems between ZFC and ZF+DC -- or at least, the differences in provable theorems between ZFC and ZF+(countable choice) -- than to discussing the differences in provable theorems between ZFC and ZF. (Indeed, you basically can't do any analysis in just ZF.)]
My questions are:

Is it consistent with ZF+DC that every subset of $\mathbb{R}$ is Borel-measurable?
If the answer to Q1 is no: Is it consistent with ZF+DC that a countably generated $\sigma$-algebra can have a cardinality strictly larger than that of the continuum?
Is it a theorem of ZF+DC that there exists an injective map from the set $\omega_1$ of well-orderings of $\mathbb{N}$ into $\mathbb{R}$?

Thanks.

REPLY [3 votes]: Let me add that the answer to (3) is negative, if you are willing to assume that inaccessible cardinals are consistent.
We say that $\omega_1$ is inaccessible to reals if for every real $x$, $\omega_1^{L[x]}<\omega_1$. This implies that $\omega_1$ is a limit cardinal in $L$. But if we also assume $\sf DC$, then $\omega_1$ is regular, in which case it is also regular in $L$. The conjunction of the two means that it is inaccessible there.
It is consistent that there is no injection from $\omega_1$ into the reals, e.g. in Solovay's model, where $\sf DC$ holds. And on the other hand, if $\omega_1$ is accessible to reals, then for some real number $x$, there is an injection from $\omega_1$ into the reals constructible from $x$, and in particular into $\Bbb R$.
So we see that it is consistent that $\omega_1$ is inaccessible to reals, and that there is no injection from $\omega_1$ into $\Bbb R$. (Of course it is consistent that $\omega_1$ is inaccessible to reals and there is such an injection. Simply look at the model obtained by collapsing an inaccessible to be $\omega_1$.)<|endoftext|>
TITLE: Non null Turing antichain
QUESTION [9 upvotes]: This interesting question resulted from a query of Mushfeq: In ZFC, can we find a non null set of pairwise Turing incomparable reals?

REPLY [12 votes]: This is a $ZFC$-theorem. Actually for any $\Sigma^1_1$-locally countable partial ordering, there is a non-null antichain.
The proof is not quite simple. See my paper
http://ims.nju.edu.cn/~yuliang/lcpfinal
I feel that it might be helpful to give more details.
The idea is as following: 
By Harrison's theorem, the question concerning  $\Sigma^1_1$-locally countable partial ordering can be reduced to the question about the measure of antichains of hyperarithmetic degrees. So it is sufficient to construct a non-null antichain of hyperarithmetic degree. By an application of some algorithmic randomness results due to Kucera, Miller and me, any such antichain must be nonmeasurable.
So we just need to construct a nonmeasureable antichain of hyperdegrees. By a generlaization of the results due to Miller and me, it can be proved that sufficient randomness (in the paper "sufficient randomness" means $\Delta^1_2$-randomness. Now we know that $\Pi^1_1$-randomness is sufficient after the development of higher randomness theory) is $\leq_h$-downward closed. 
Now take a maximal set $X$ of reals so that any two different reals in $X$ bounds disjoint "sufficiently random" reals. If $X$ is not null, then we are done. Otherwise, by the randomness result above, the $\leq_h$-upward closure of $X$ must be null.  For each $e$, let $X_e$ be the collection of the $e$-th "sufficiently random" real hyperarithmetically below some $x\in X$ (such an $\omega$-type well ordering is "natural" for each $x$, i.e. the $e$-th hyperarithmetic reduction.) It cannot be true that for every $e$, $X_e$ is  null (Otherwise, by the randomness result and maximality of $X$, the set of "sufficiently random" reals would be null). By the property of $X$, $X_e$ is an antichain for every $e$. So there must be some $e_0$ so that $X_{e_0}$ is an antichain and non-null.
Note that even the hyperarithmetic closure of $X_{e_0}$ does not have a positive measure.
Under  $MA+\neg CH$, any locally countable partial ordering has a non-null antichain. However, under $V=L$, it fails for the $\Delta^1_2$ well-ordering $\leq_L$. I am not sure about $\Pi^1_1$-locally countable partial orderings.

REPLY [6 votes]: This is a great question. 
Let me get things started by showing that the answer is yes under
the continuum hypothesis. Assume CH. Let $\langle
U_\alpha\mid\alpha<\omega_1\rangle$ enumerate all the open sets
with less than full measure. I shall build an antichain $\langle
a_\alpha\mid\alpha<\omega_1\rangle$ of Turing incomparable reals,
which are not contained entirely in any $U_\alpha$. Thus, the set
$\{a_\alpha\mid\alpha<\omega_1\}$ will have full outer measure. At
any stage of the construction, we will have already specified $a_\beta$
for $\beta<\alpha$, which is countably many reals. Since
$U_\alpha$ is an open set with less than full measure, I claim
that we may find a real $a_\alpha$ that is not in $U_\alpha$ and
which is Turing incomparable with all $a_\beta$ for
$\beta<\alpha$. To see this, observe first that if we force to add
a random real $r$ not in $U_\alpha$, then $r$ will fit the bill in
the forcing extension $V[r]$, since clearly $r$ is not computable from any ground model real, and conversely I believe that no non-computable ground model real is computable from a random real. But next, observe that the existence
of such a real is a $\Sigma^1_1$ assertion, and hence it was
already true in $V$ by Shoenfield absoluteness. So we may find $a_\alpha\notin U_\alpha$ that
is incomparable with all earlier $a_\beta$. Thus, we have built
the antichain of Turing incomparable reals, which is not contained
in any open set of less than full measure. And so it is not
measure zero. QED<|endoftext|>
TITLE: Lattice points on the boundary of an ellipse
QUESTION [8 upvotes]: How many points of the integer lattice ${\mathbb Z}^2$ can an axis-parallel ellipse of radius $r$ contain on its boundary? (that is, we consider ${\mathbb Z}^2$ as lying in ${\mathbb R}^2$). Alternatively, this appears to be very similar to the question of how many points can the boundary of any axis-parallel ellipse contain from an $r\times r$ section of the integer lattice.
Let us say that the ellipse is defined as $(ax-b)^2+(cy-d)^2=r^2$. If $a,b,c,d$ are integers, we can reduce the problem to asking for the number of integer solutions to of an equation of the form $ax^2+by^2=r^2$ with integers $a,b$. This is known to have at most $r^{\frac{c}{\lg \lg r}}$ solutions, for some constant $c$. Does a similar bound exist for every axis-parallel ellipse? 
In case it helps, I mainly care about the case where $a,b,c,d$ are not much larger than $r$.

REPLY [3 votes]: For general ellipses I doubt you can do much better than Bombieri-Pila sort of bounds:
MR1016893 (90j:11099) Reviewed 
Bombieri, E.; Pila, J.
The number of integral points on arcs and ovals. 
Duke Math. J. 59 (1989), no. 2, 337–357. 
11P21 (11D99) 

(you should check out the many papers that cite this, as well, but most of them seem to be interested in higher degrees/dimensions). 
For integer coefficients, the best seems to be 
Lattice points on Ellipses
Cilleruelo and Cordoba, DMJ 1994<|endoftext|>
TITLE: Approximating integers with prime quotients
QUESTION [5 upvotes]: Is this statement true for all positive integers $n\in\mathbb{N}$?


For all $\varepsilon >0$ there are prime numbers $p,q$ such that $|\frac{p}{q} - n| < \varepsilon$.

REPLY [6 votes]: It is known that there exists a prime between $n$ and $n+n^{\epsilon}$ for $n$ large enough, where the best known $\epsilon$ is 0.525. Therefore, given a large enough prime $q$, choose a prime $p$ between $q\alpha$ and $q\alpha+(q\alpha)^{\epsilon}$ to get 
$$|\alpha-p/q|<Cq^{\epsilon-1},$$
where $C$ depends only on $\alpha$. Assuming the Reimann Hypothesis, this bound can be improved to 
$$|\alpha-p/q|<Cq^{-1/2}\log q,$$
for infinitely many pairs of primes $(p,q)$. With recent results on prime gaps, this bound is extremely better when $\alpha=1$. One can then ask this question:
${\bf Question:}$ Let $\alpha$ be a positive real number. Does there always exist infinitely many pairs of primes $(p,q)$ such that
$$|\alpha-p/q|<C/q,$$
where $C$ is a universal constant?<|endoftext|>
TITLE: The category of elements, enrichment, and weighted limits
QUESTION [13 upvotes]: This is a crosspost of this MSE question.

Every so often, when reading notes online or skimming through books, the category of elements and the Grothendieck construction pop up. I don't know anything about the Grothendieck construction, and I don't understand the significance of its special case - the category of elements. This is probably going to be the first in a series of questions about the Grothendieck construction (I'll interlink with later ones if/when I ask them).
My main motivation for asking this question is the excerpt below from Borceux vol II, sec 6.6. I have tried to go into detail about how I don't understand the significance of the category of elements in this context.

The first observation for enriching the notion of limit is the fact in many fundamental results of the previous chapters, we had to deal with a situation like in diagram 6.3.6, where $F,G$ are functors and $\mathsf{Elts}(G)$ is the category of elements of $G$; the limit considered was that of $F\circ \phi _G$ [...]. Restricting one's attention to limits of the form $\varprojlim (F\circ \phi _G)$ is not at all a restriction since, choosing for $G$ the constant functor on the singleton, the category of elements of $G$ is just $\mathscr A$ itself and $\phi _G =1_\mathscr{A}$, so that we recapture the limit of $F$.

But the key observation for enriching the notion of limit is the following fact
Lemma 6.6.1 In the situation which has just been described, there exist bijections natural in $B\in \mathscr B$,
  $$\mathsf{Nat}(\varDelta_B,F\circ \phi _G)\cong \mathsf{Nat}(G,\mathscr B(B,F-))$$ where $\varDelta _B:\mathsf{Elts}(G)\longrightarrow \mathscr B$ is the constant functor on the object $B\in \mathscr B$.

From this lemma Borceux concludes that $\varprojlim (F\circ \phi_G)$ exists iff $\exists L\in \mathscr B$ with bijections natural in $B$:
$$\mathsf{Nat}(G,\mathscr B(B,F-))\cong \mathscr B(B,L)$$
And from this, he almost has to write the definition of a weighted limit.

Alternatively, here's an approach I find somewhat more intuitive, but one that does not explicitly mention the category of elements at all.
In classical CT, we define limits using $\varDelta$, which is not available in a general enriched category $\mathsf C$ since $\mathsf C(A,B)$ has no elements to work with. In the unenriched case, one could also try to define $\varDelta$ via the cartesian closed structure of $\mathsf{Cat}$ as a right adjoint of the projection $\pi:\mathsf C\times \mathsf J\longrightarrow \mathsf C$, but this doesn't generalize either since most enriched categories are not cartesian closed. So we're left with trying the representability approach - a limit of $F$ is just a representation of $\mathsf{Nat}(\varDelta-,F)$. Here we can escape the "problems of enrichment" because I think the following holds:
Proposition There's a bijection natural in $X$: $\mathsf{Nat}(\varDelta \ast,\mathsf{Hom}(X,F-))\cong \mathsf{Nat}(\varDelta X,F)$, where $\ast$ is the singleton.
The point is to now only mention the functor $\varDelta \ast$ instead of the general $\varDelta$.
Digression - is the above bijection also natural in $F$? (I think it is.)
Now since $\varprojlim F$ is a representation of $\mathsf{Nat}(\varDelta -,F)$, we can define it as an object with bijections natural in $X$$$\mathsf{Hom}(X,\varprojlim F) \cong \mathsf{Nat}(\varDelta \ast,\mathsf{Hom}(X,F-))$$
Now we have reduced the usage of $\varDelta$ to the functor $\varDelta \ast$ alone. The role of $\varDelta \ast$ as a weight is also easily seen, as each component of natural transformation $\eta _A :\varDelta \ast \Rightarrow \mathsf{Hom}(X,F-)$ picks out an arrow $X\rightarrow FA$, and so replacing 
$\varDelta \ast$ by say, $\varDelta (\ast\coprod \ast)$ would give a "double cone". Replacing $\varDelta \ast$ by more general functors $G$ would produce more intricate cones.
This alternative approach is (I think) completely ignorant of the category of elements, and yet manages to give a nice motivation for the definition of a weighted limit. On the other hand, Borceux's approach just poops it out by category-of-elements-magic.

What exactly is happening here? What is the role of the category of elements in this story and why does it effortlessly yield the definition of a weighted limit?

REPLY [6 votes]: The question is a bit imprecise, so I'm not quite sure whether this will really answer it, but let me try to paint the reduction of weighted colimits to conical colimits in a more "conceptual" light.
First of all, your identity $[\mathcal{A},\mathsf{Set}](\Delta \ast, \mathcal{B}(X,F)) \cong [\mathcal{A},\mathcal{B}](\Delta X, F)$ is correct; we can write it out in terms of ends:
$\int_{A \in \mathcal{A}} [\Delta \ast(A), \mathcal{B}(X,F(A))] = \int_{A \in \mathcal{A}} \mathcal{B}(X,F(A)) = \int_{A \in \mathcal{A}} \mathcal{B}(\Delta X(A), F(A))$
Now, let me argue that the key identity is $G = \operatorname{Lan}_{\phi_G} (\Delta \ast)$. This identity implies that the $G$-weighted limit of $F$ is the $\Delta \ast$-weighted limit of $F \phi_G$, because we have
$[\mathcal{A},\mathsf{Set}](G, \mathcal{B}(X,F)) = [\mathcal{A},\mathsf{Set}](\operatorname{Lan}_{\phi_G} (\Delta \ast), \mathcal{B}(X,F)) = [\mathcal{A},\mathsf{Set}](\Delta \ast, \mathcal{B}(X,F\phi_G))$
That is, the defining property of a left Kan extension relates the functor represented by a weighted limit, on the left, to the functor represented by the conical limit, on the right.
So from this perspective, the key property of the category of elements is that it allows us to express our copresheaf as the left Kan extension of a constant-at-1 presheaf. I'm not sure whether this is a universal property, but it is reminiscent of the universal property of a subobject classifier, and of a universal property the category of elements does possess -- it is the classifying discrete opfibration. We have a diagram:
$\require{AMScd}$
\begin{CD}
    \mathsf{el}G @>>> \ast \\
    @V \phi_G V V @VV \ast V\\
    \mathcal{A} @>>G> \mathsf{Set}
    \end{CD}
The property I'm pointing out is that the bottom is a left Kan extension, while the usual universal property says that this is a pullback comma square. I'm not quite sure of the relationship.
EDIT
Just an additional miscellaneous thought: if this were also true in the enriched setting, then this would tell us for example that every group representation was the induced representation of a trivial representation of some algebra (or rather, algebroid -- and of course, "trivial representation" doesn't even make sense in this generality). This isn't true, but it's still a good strategy to look for representations as subrepresentations of induced representations of trivial representations.<|endoftext|>
TITLE: Can the standard map $\Sigma \Omega X \to X$ be a homotopy equivalence?
QUESTION [6 upvotes]: The question is in the title : are there spaces X such that the adjoint of the identity on the loop space $\Omega X$, i.e. $\Sigma\Omega X \to X$, is a homotopy equivalence ?

REPLY [15 votes]: If $X$ is such a space (a CW complex, say), then it must be a suspension $\Sigma Y$ (as it is the suspension of $y :=\Omega X$). The James splitting gives
$$\Sigma \Omega \Sigma Y \simeq \bigvee_{n=1}^\infty \Sigma Y^{\wedge n},$$
the wedge of suspensions of smash products, so under your assumption $\Sigma \Omega \Sigma Y \overset{\sim}\to \Sigma Y$ it follows that $\Sigma Y^{\wedge n} \simeq *$ for all $n > 1$. In particular, by the Kunneth theorem $Y$ must have the homology of a point, and in particular be path-connected. But then $X=\Sigma Y$ must be simply-connected, and also have the homology of a point: therefore it is contractible.<|endoftext|>
TITLE: A system of non-linear equations that is decomposable as a product -- uniqueness of solution?
QUESTION [7 upvotes]: I have a system of non-linear equations 
$ a_1=f_0 g_1$
$a_2=f_1 g_1 + f_0 g_2$
$a_3=f_2 g_1 + f_6 g_2 + f_0 g_3 $
$a_4=f_3 g_1 + f_7 g_2 + f_6 g_3 + f_0 g_4 $
$a_5=f_4 g_1 + f_8 g_2 + f_7 g_3 + f_6 g_4 + f_0 g_5$
$a_6=f_5 g_1 + f_9 g_2 + f_8 g_3 + f_7 g4 + f_6 g_5 + f_0 g_6$
$a_7=f_5 g_3 + f_9 g_4 + f_8 g_5 + f_2 g_6 + f_0 g_7 $
$a_8=f_5 g_5 + f_4 g_6 + f_7 g_7 + f_0 g_8$
$a_9=f_5 g_7 + f_8 g_8 + f_0 g_9$
$a_{10}=f_0 g_{10} + f_5 g_9$
here both $f=(f_0,..,f_5,,..f_9)$ and $g=(g_1,…,g_{10})$ are unknown.
My problem is to figure out when this system has no solution, a unique or multiple solutions.
What is known is that this system of equations allows to decompose itself as
$\left(\begin{array}{c}  a_1\\ a_2 \\ \vdots \\a_9 \\a_{10} \end{array} \right)
 = 
\begin{bmatrix} 
f_0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 
f_1 & f_0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
f_2 & f_6& f_0&  0& 0& 0& 0& 0 &0 & 0 \\
f_3 & f_7 & f_6 & f_0 & 0 & 0 & 0 & 0 & 0&  0 \\
f_4 & f_8 & f_7 & f_6 & f_0&  0& 0& 0& 0& 0 \\ 
f_5 & f_9 & f_8 & f_7 & f_6&  f_0& 0& 0& 0& 0 \\ 
0 & 0 & f_5 & f_9 & f_8 & f_2&  f_0&  0 & 0&  0 \\ 
0 & 0 & 0 & 0 & f_5 & f_4 & f_7 & f_0 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 0 & 0 & f_5& f_8 & f_0 & 0 \\ 
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & f_5 & f_0 
\end{bmatrix} 
\times 
\left( \begin{array}{c} g_1 \\ g_2 \\ \vdots \\ g_9 \\ g_{10} \end{array} \right) $
Then whenever I take only vector of $f$ as unknowns (and so $g$ are the parameters), or vice versa, there is a unique solution for a generic vector $a$, similarly to how it works for linear systems of equations. While when I am letting one extra element of $f$ or of $g$ be "a parameter", i.e. there are more equations than there are unknown $f$ or unknown $g$, there is no solution.
But I have no idea what happens when both $f$ and $g$ are unknown. I have then 10 equations and 20 unknowns, but it is not clear to me that it automatically should lead to a continuum of solutions. (And I cannot solve this system explicitly..) 
If there is a continuum of solutions (for a "generic" $a$), it seems to be an interesting case: a system of $N$ non-linear equations and $M(>N)$ unknowns does not have a solution (for generic vector $a$) whenever the size of the smallest of two sets of unknowns is less than $N$. Or it has a continuum of solutions in the complementary case, but there is no case in-between -- with a unique solution. 
Are there any known results for such system of non-linear equations that are "decomposable"?
I would appreciate any opinions on this!

REPLY [8 votes]: The linear-in-$g$ system is uniquely solvable whenever $f_0\ne0$. therefore the solution set is parametrized by $f$ in the form 
$$g=f_0^{-10}P(f;a),$$
where $P$ is linear in $a$, polynomial in $f$ and homogenenous of degree $9$. Not only it is a continuum, but it has dimension $10$.
When $f_0=0$, there is no solution in general, say for $a_1\ne0$.<|endoftext|>
TITLE: Different ways of making $HOD$ far from $V$
QUESTION [7 upvotes]: There are different criteria for building a model $V$ of $ZFC$ which is far from its 
$HOD$, for example:
$(A)$ Cardinality criteria: For this in a joint work with James Cummings and Sy Friedman, we produced a model $V$ in which for all infinite cardinals $\alpha$ we have $\alpha^+ > (\alpha^+)^{HOD}.$
$(B)$ Having large cardinals: The idea is to build models which contain some very large cardinals but such that their $HOD$ does not contain them. For this see the nice paper 
Large cardinals need not be large in HOD by Yong Cheng, Sy Friedman and Hamkins.
$(C)$ The continuum function criteria: To idea is to build models in which $2^\kappa$
is large in $V$ but is small in $HOD$ (for this I recently showed that we can have a model in which $GCH$ fails everywhere but its $HOD$ satisfies the $GCH$).
$(D)$ The cofinality criteria: To build models which contain many singular cardinals which are regular in $HOD$. For this see for example Shoshana Friedman's talk given at the  5th European Set Theory Conference (5ESTC). 

I wonder to know what other natural criteria one can consider? In other words what other properties one can consider to measure as a witness for having $HOD$ far from $V$?

References for similar works are appreciated.

REPLY [7 votes]: Another criterion may be to make HOD far from $V$ with respect to forcing. For example, in our paper 

G. Fuchs, J.D. Hamkins, J. Reitz, Set-theoretic geology, Annals of Pure and Applied Logic, vol. 166, iss. 4, pp. 464-501, 2015

we prove that any given model $W$ of ZFC can be made into the Mantle and generic Mantle and HOD and generic HOD of another model $V$, which is a class forcing extension of $W$. The HOD of our model $V$ is not a ground model, so this is a sense in which HOD is far from $V$.
We have other models where the HOD of $V$ is the original model $V_0$, but the mantle is large.<|endoftext|>
TITLE: How bad can a circle domain get?
QUESTION [13 upvotes]: Let $X$ be a domain in the Riemann sphere $\widehat{\mathbb{C}}$. We say that $X$ is a circle domain if every connected component of its boundary is either a circle or a point.
It was conjectured by Koebe in 1909 that circle domains represent all planar domains up to conformal equivalence. This was proved by Koebe himself in the case of finitely many boundary components, and by Zheng-Xu He and Oded Schramm in 1993 in the countable case.
I'm trying to find examples of "bad" circle domains. In particular :
Question : Is there a circle domain whose boundary is not the union of countably many circles, countably many Cantor sets and countably many points?
I'm mostly interested in the case where the boundary has zero area.
Of course, the boundary of a circle domain contains at most countably many circles. The question is whether the set of point components can be written as a countable union of closed totally disconnected sets. The problem seems to be the set of point components that are accumulation points of circles...
Another question :
Question 2 : If $X$ is a circle domain, must there exist an element $z \in \partial X$ which is isolated from circles (in other words there is no sequence of circles in $\partial X$ converging to $z$)?
The answer is yes if the set of point components in closed, by Baire Category considerations...
Thank you,
Malik

REPLY [6 votes]: Let $K$ be the boundary of your circle domain $\Omega$. 
Let us suppose that every point of $K$ is accumulated on by a sequence of (pairwise different) circle components. Such an example is easy to construct (see e.g. the already existing answer to your Question 2, or simply add the circles inductively - see below for the details).

Claim. A countable union of closed and totally disconnected subsets of $K$ cannot contain all non-circle components of $K$. 

Remark. This shows that the answer to your first question is negative.
Proof. First, suppose that $A$ is a totally disconnected closed subset of $K$, and that $U$ is some open set that intersects one of the circle components, say $C$. Then for any point $z$ on any arc of $C\setminus A$, there are arbitrarily small circle components accumulating on $z$. If they are small enough, then these components are themselves in the complement of $A$, and then there is also a small clopen (in $K$) neighbourhood of each of these components that is contained in the complement of $A$.
So, in summary: $U$ contains a clopen subset $X$ of $K$, which contains a circle component. Furthermore, the diameter of $X$ can be chosen as small as we wish. 
Now if $(A_i)$ is a sequence of totally disconnected closed subsets of $K$, then we can proceed inductively: Find a nonempty clopen subset $X_1$ in the complement of $A_1$ as above, say having diameter less than $1$. Then find a nonempty clopen subset $X_2\subset X_1 \setminus A_1$ of $K$, having diameter less than $1/2$, etc. 
Since each $X_k$ is clopen, and the diameters tend to zero, their (non-empty) intersection is disjoint from all circle components, and belongs to the complement of the union of all $A_i$. This completes the proof.
The proof in fact never used that the non-trivial sets in question are circles. Hence it shows:
Proposition. Let $K\subset\mathbb{C}$ be a compact set such that, for every $\newcommand{\eps}{\varepsilon}\eps>0$, there are only finitely many connected components of $K$ of diameter at least $\eps$. Assume furthermore that every point $z\in K$ is accumulated on by non-point components not containing $z$. Then any countable collection of closed and totally disconnected subsets of $K$ must omit some point components of $K$.
Remark. A slight adaptation of the proof shows furthermore that the set of omitted point components has the cardinality of the continuum.
EDIT. For completeness, let me outline the elementary construction of the domain in question, which provides more details about Misha's "Sierpinski carpet" suggestion. At each stage of the inductive construction, we have a collection $\mathcal{C}_k$ of finitely many pairwise disjoint circles, with $\mathcal{C}_{k+1}\supset \mathcal{C}_k$. 
For each circle $C\in \mathcal{C}_k$, we also pick an open annulus $A_k(C)$ that surrounds $C$, separates $C$ from all other elements of $\mathcal{C}_k$, and whic becomes closer and closer to $C$ as $k\to\infty$. 
Start with $\mathcal{C}_0$ having one circle $C$ in it, with some annulus $A(C)$ picked around it. Then, inductively, for every circle $C$ in $\mathcal{C}_k$, add a whole bunch of small circles to $\mathcal{C}_{k+1}$, within the inner curve of $A_k(C)$, such that every point of $C$ is close to one of these new circles. Then pick the annuli with the desired property, and so that they are pairwise disjoint. 
Let $K$ be the closure of the union of all these circles. The construction ensures that this set is totally disconnected (any two points will be separated by one of the annuli), and clearly $K$ has all the desired properties).<|endoftext|>
TITLE: Computations with conetypes of hyperbolic groups
QUESTION [5 upvotes]: I'd like to know if there exists (and, in this case, where I can find it) some computer program/programming language/any kind of software that can find explicitly the conetypes of a hyperbolic group on which I am working (a presentation of which is known).

REPLY [9 votes]: My KBMAG package can compute a finite state automaton that accepts the language of geodesic words in a hyperbolic group, and I think the states of that automaton correspond exactly to the conetypes.
It is not particularly easy to use. If you write down the presentation of the group you are interested in, then I can try out the computation for you.<|endoftext|>
TITLE: Can exponential sums be small on a whole interval?
QUESTION [8 upvotes]: This is almost certainly routine to an analyst, so forgive me in advance.
Let $\alpha_i\in \mathbb{R}$. Consider the functional $$\varphi: L^1[0.9A,A]\to \mathbb{C}$$ via $$f\mapsto \sum_i \hat{f}(\alpha_i),$$ where by '$\hat{f}$' I mean the Fourier transform of $f$ regarded as a function on $\mathbb{R}$ (via extension by zero).
What can one say about the norm of $\varphi$?
I am wondering about the case where all $|\alpha_i| = O(1)$, $A$ is large, and the number of $\alpha_i$ is also large (comparable to $A$). As the title suggests, I'm hoping for a lower bound on $||\varphi||$. The thing that seemed most natural to me was to write this as integrating against a sum of exponentials (say $g(t)$, so that $||g||_\infty = ||\varphi||$) and then calculate the $L^2$ norm of $g$, but I can't bound the off-diagonal terms well enough.
The idea is that if, for some $t$ in the interval, $g(t)$ is very small, then some small-order derivative of $g$ should be bounded below so that $g((1\pm \delta)t)\gg_\delta A^{-O(1)}$ maybe. One could get a lower bound on some derivative via the nonvanishing of a Vandermonde determinant, but I think that would require too many derivatives.
Perhaps I am mistaken and there is an example with $||\varphi||\ll \exp(-c A)$?
Anyway, thanks much!

REPLY [6 votes]: I think I can prove a polynomial bound using complex analysis. This really seems like it shouldn't work but it seems to.
The bound is that for all $\alpha_1, \dots, \alpha_n \in \mathbb R$ 
$$ \sup_{t \in [A,B]}\left| \sum_{i=1}^n e( \alpha_i t) \right| \geq n^{1-\frac{\pi}{4 \arctan {\sqrt{\frac{B}{A}}}-\pi}}$$
So in the special case $[.9A,A]$ this is $n^{-28.8313 \dots}$.
We may freely translate the $\alpha_i$, therefore, we assume $\min \alpha_i=0$. Form the holomorphic function
$$f(z) = \sum_{i=1}^n e(\alpha_i z)$$
We have the following estimates:
In the whole upper half plane, $|f(z)| \leq n$, as all the $\alpha_i$ are nonnegative so $e(\alpha_i z) \leq 1$.
On the imaginary line, $|f(z)| \geq 1$, as some $\alpha_i$ is zero giving $e(\alpha_i z) =1$ and the rest of the terms are positive reals.
Assume that for $z$ on the real line in the interval $[A,B]$, $|f(z)|<C$. (Then the same holds for $[-B,-A]$, which helps to get a better constant but isn't essential).
$f$ is holomorphic so $\log | f(z)|$ is subharmonic. Hence by the mean value theorem for subharmonic functions and the various bounds we stated on $f$:
$$ 0 \leq \log |f(it)| \leq\frac{t}{2\pi} \int_{-\infty}^{\infty} \frac{\log |f(s)|}{t^2+s^2}ds \leq \frac{t}{2\pi} \int_{-\infty}^{\infty}  \frac{\log n}{t^2+s^2}ds+ 2\frac{t}{2\pi} \int_{A}^{B}  \frac{\log C - \log n}{t^2+s^2}ds$$
$$ = \log n + \frac{\theta}{\pi}  (\log C - \log n)$$
where $\theta$ is the hyperbolic angle between $A$ and $B$ from $it$. So we have:
$$0 \leq \log n + \frac{\theta}{\pi} (\log C - \log n)$$
$$ \frac{\theta}{\pi} (\log n - \log C) \leq \log n$$
$$ \log C \geq (1- \frac{\pi}{\theta}) \log  n$$
$$C \geq n^{1-\pi/\theta}$$
as desired.
The optimal value of $\theta$ is a hyperbolic trigonometry problem which I may or may not have gotten the right answer to. Anyways it's some explicit formula.
Tao's example starting with, $\alpha_1=0$, $\alpha_2= \frac{1}{A+B}$ has an upper  bound of the form $n^{\frac{\log \left|1- e\left(\frac{A}{A+B}\right)\right|}{\log 2}}$. My exponent is far from this as $A$ goes to $B$ so sharpness is an interesting question.<|endoftext|>
TITLE: SnapPea for the uninitiated
QUESTION [24 upvotes]: SnapPea (http://www.math.uic.edu/~t3m/SnapPy/) is a program with extensive facilities for doing various kinds of calculations with hyperbolic 3-manifolds.  The official documentation assumes that the reader is intimately familiar with all the relevant mathematical background.  Does there exist any other document which explains some of the theory in parallel with explaining the code?  Failing that, is there a recommended source which just explains the theory, but in a way which meshes nicely with the code?

REPLY [8 votes]: First as pointed out in the comments, the documentation of the SnapPea kernel (now maintained as SnapPy) is extensive. It contains theorems and proofs as well as a fairly thorough treatment many of the functions of SnapPea/SnapPy.
Also, in addition to Thurston's notes, one might also consider Section E.6 of Benedetti and Petronio's Lectures on Hyperbolic Geometry as it provides a good explanation of the polynomial equations that one must solve to find a hyperbolic structure associated to the triangulation.
However, the question asks for theory that is most closely connected to the code, on that note, there is also Weeks' article in the Handbook of Knot Theory:
Weeks, Jeff. "Computation of hyperbolic structures in knot theory." Handbook of Knot Theory (2005): 461-480.
http://arxiv.org/abs/math/0309407 
This article mainly focuses on the algorithm to triangulate the complement of a knot or link and then solve a relevant system of equations (coming from Thurston), before finishing with a discussion of how the program considers Dehn filling. As Carlo Beenakker points out, other aspects of the program, such as the computation isometry group (via a computation of the canonical triangulation) are covered elsewhere in Jeff Weeks' works.<|endoftext|>
TITLE: Characterisations of closed embeddings in $Top_1$?
QUESTION [6 upvotes]: Let $Top_1$ be the category of topological spaces which are $T_1.$   
I am curious as to whether there is a categorical definition of what a closed embedding is in this environment. With a categorical definition, I mean something along the lines that in the category $Top_2$ , consisting of topological spaces which are $T_2,$ the closed embeddings are precisely the extremal monomorphisms.  
In $Top$ (and $Top_1$) the extremal monomorphisms are precisely the embeddings. In $Top$ we can use the Sierpinski space to single out the ones with closed image, but this is not possible in $Top_1$ since the Sierpinski object is not $T_1.$  
Any comments or thoughts would be welcome.

REPLY [9 votes]: As partial motivation, let me start with the observation that in $\mathrm{Set}$, if $i: A \to B$ is a monomorphism, then $i$ is retrieved as the pullback of the monomorphism $\ast \cong A/A \hookrightarrow B/A$ along the canonical map $q: B \to B/A$ (where $B/A$ is the pushout or cofiber product of $i$ and the unique map $A \to 1$). This is one of the exactness properties singled out by Freyd in a categories-list discussion that are common to pretoposes and abelian categories. The square 
$$\begin{array}{ccc}
A & \to & 1 \\
i \downarrow & & \downarrow \\
B & \stackrel{q}{\to} & B/A
\end{array}$$ 
which is simultaneously a pushout and pullback is sometimes called, affectionately, a "dolittle square" (after the pushmi-pullyu in the Doctor Dolittle story). So not knowing quite what to call such monomorphisms in general, I'll call it a dolittle monomorphism. 
In other words, a monomorphism $i$ in a category with finite limits and pushouts will be called dolittle if it is the pullback of the inclusion $1 \to B/A$ in the pushout square above. Then in $\mathrm{Top}_1$, closed embeddings $i: A \to B$ are precisely dolittle monomorphisms. (Notice that the pushout construction $B/A$ in $\mathrm{Top}_1$ is constructed as the pushout $B/\bar{A}$ in $\mathrm{Top}$.)<|endoftext|>
TITLE: Examples of n+1D TQFT with 1 dimensional Hilbert spaces on n-torus and n-sphere but higher dimensional Hilbert spaces on other n-manifolds
QUESTION [15 upvotes]: Are there simple examples of $n+1$D TQFT that assign 1-dimensional Hilbert spaces to both $n$-torus and $n$-sphere but higher dimensional Hilbert spaces to some other $n$-manifolds? Here I am assuming that the manifolds under consideration are all orientable.
For $n=2$, the requirement of 1-dimensional Hilbert spaces on both 2-torus and 2-sphere implies that the corresponding modular tensor category associated to the TQFT has only 1 simple object. Such TQFT have 1-dimensional Hilbert spaces on all oriented 2-manifold. Therefore, examples I am looking for should not exist with $n=2$. For $n>2$, I cannot think of any reason why such example cannot exist.

REPLY [7 votes]: If your topological field theory is at least once-extended, by which I mean it assigns values to $(n+1)$-manifolds, $n$-manifolds, and also to $(n-1)$-manifolds, than this cannot happen. 
More precisely we have the following result:

Theorem: Suppose that $ Z: Bord_{n+1} \to C$ is an $(n+1)$-dimensional once-extended (oriented) topological field theory valued in the symmetric monoidal 2-category $C$. Then the value $Z(T^n)$ assigned to the $n$-torus is invertible if and only if the whole field theory is invertible (assigns invertible values to all manifolds in all dimnesions). 

So in particular suppose $C$ is any 2-category that deloops the category of vector spaces (like linear categories, functors, transformations or algebras, bimodules, maps, etc). Then if the vector space assigned to the n-torus is 1-dimensional (hence invertible) than every $n$-manifold is assigned an invertible (i.e. one-dimensional) vector space. This also holds if you replace vector spaces with super vector spaces. 
A version of this theorem also holds for bordism equipped with arbitrary tangential structures. You can read about it here:1511.01772.
The main ingredients in the proof are dimensional reduction and surgery. Dimensional reduction lets you relate theories of different dimensions and so you can attack the problem inductively. This already lets you prove that many bordisms take invertible values. Then next idea is that handle decompositions use handles which have at most codimension two corners. This means that if you are at least once-extended, then you can implement handle decompositions of (n+1)-manifolds, and hence surgery for n-manifolds, entirely in the bordism 2-category in categorical terms. 
That is not the complete argument, but those are two of the main ideas. 
I don't know what happens if your field theory is not extended. Even in dimension n=2  (n+1 =3) the OP's argument used that we assign categories to 1-manifolds. So if the theory is not extended, perhaps it is possible to assign an a 1-dimensional vector space to the 2-torus, but higher dimensional spaces to other surfaces? András Juhász has a classification of non-extended (2+1)-TQFTs, but it is a bit complicated. Perhaps it is possible to use his work to prove or disprove this in dimension (2+1)? I think that is an interesting question to explore.<|endoftext|>
TITLE: Connection between Bernoulli numbers and Riemann-Siegel theta function?
QUESTION [16 upvotes]: I have come across a strange approximation for the Riemann-Siegel theta function involving the Bernoulli numbers - namely that
$$\frac{1}{2} \log \left| B_{2 n}\right|\approx \vartheta (2n)\ ,\quad n \in \mathbb{N}$$
where $B_n$ is the $n$th Bernoulli number, and $\vartheta,$ the Riemann-Siegel theta function. More accurately, it seems that
$$\frac{1}{2} \left(\log \left(\frac{\left| B_{2 n}\right| }{\sqrt{2 n}}\right)-\left(\frac{2}{3}\right)^{2/3} \pi \right)$$
is closer to the actual value of $\vartheta (2n),$ and it may be true to say that 
$$\lim_{n\rightarrow\infty}\frac{1}{2} \left(\log \left(\frac{\left| \sqrt{2}n \zeta(1-2n)\right| }{\sqrt{ n}}\right)-\left(\frac{2}{3}\right)^{2/3} \pi \right)=\vartheta (2n),$$
and it seems likely that large values of $|B_{2n}|$ can be estimated with
$$\sqrt{2 n} \exp \left(2 \vartheta (2 n)+\pi  \left(\frac{2}{3}\right)^{2/3}\right).$$
I haven't seen any references anywhere alluding to any connection between the absolute values of the Bernoulli numbers at even $n,$ and the Riemann-Siegel theta function previously. Is the above statement correct, and if it is, what is the connection?

REPLY [16 votes]: Summary: Your approximation is very good, but not quite perfect. The absolute error does not go to $0$ as $n\to\infty$. But it is astonishingly close: the error is approximately $1.3\times 10^{-8}+o(1)$.

Detailed answer:
we can easily check your proposed approximation for $\vartheta(2n)$, since both $|B_{2n}|$ and $\vartheta(t)$ have well-known asymptotic expansions. In the case of $|B_{2n}|$, we have
$$
B_{2n} = (-1)^{n-1} \zeta(2n) \frac{2(2n)!}{(2\pi)^{2n}} = (1+O(2^{-2n})) (-1)^{n-1} \frac{2(2n)!}{(2\pi)^{2n}},
$$
so, using this together with Stirling's approximation in the form
$$
\log(m!) = m\log m-m+\frac12 \log(2\pi m)+O(m^{-1})
$$
we have that
\begin{align}
\frac12\log \frac{|B_{2n}|}{\sqrt{2n}} &= 
\frac12 \Big(\log 2+
2n\log(2n)-2n+\frac12\log(4\pi n)+O(n^{-1}) - (2n)\log(2\pi) \\ & \qquad - \frac12\log(2n)
+O(2^{-2n}) \Big)
\\ &= n \log\frac{n}{\pi}-n + \frac14 \log(8\pi) + O(n^{-1}).
\end{align}
On the other hand, for $\vartheta(t)$ we have the asymptotic formula
$$
\vartheta(t) = \frac{t}{2}\log\frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8} + O(t^{-1}).
$$
which for $t=2n$ gives
$$
\vartheta(2n) = n\log\frac{n}{\pi} - n - \frac{\pi}{8} + O(n^{-1}).
$$
It follows that the difference $\frac12\log \frac{|B_{2n}|}{\sqrt{2n}}-\vartheta(2n)$ satisfies
$$
\lim_{n\to\infty}\left( \frac12\log \frac{|B_{2n}|}{\sqrt{2n}}-\vartheta(2n)\right)
= \frac{\pi}{8} + \frac14 \log(8\pi) \neq \frac12\left(\frac23\right)^{2/3}\pi.
$$
Now note that, amazingly,
$\frac{\pi}{8} + \frac14 \log(8\pi)\approx 1.19874193858$ and $\frac12\left(\frac23\right)^{2/3}\pi \approx 1.19874195162$. The difference between these two constants is $\approx 1.304\times 10^{-8}$. 
To summarize, there is certainly something interesting going on here, and I am very curious to know how you obtained your approximation. Was it just a lucky guess? If not, there seems to be some mysterious mechanism at play causing the constant
$$
\frac12\left(\frac23\right)^{2/3}\pi - \frac{\pi}{8} - \frac14 \log(8\pi)
$$
to assume a surprisingly small numerical value.
Note however that my approach of using the standard asymptotic expansions for $|B_{2n}|$ and $\vartheta(t)$ says nothing about a possible connection between the Bernoulli numbers and the Riemann-Siegel theta function, other than maybe the trivial fact that the asymptotics of the Bernoulli numbers is governed by factorials (aka, the gamma function), and the Riemann-Siegel $\vartheta(t)$ is also defined in terms of the gamma function.<|endoftext|>
TITLE: SVD vs Fourier analysis for data.
QUESTION [9 upvotes]: Fourier analysis is useful for analysis in the frequency domain. SVD on the other hand is useful for analysis of data, and expressing noise in the data. I have a problem that needs extensive data analysis, it is in the area medicine. This could be generalized to other problems.  
The problem is that of gene expression, in case of long term gene mutation. Using Fourier analysis we can get a time series analysis of the genes(and thereby get noisy gene expression), and as time progresses, the changes in a particular organ. On the other hand, we could use Singular Value Decomposition, and the noisy gene expresses itself. This, is just an outline of the problem. Both SVD, and Fourier lend themselves to solve the problem that of expressing noisy genes. Is there any comparison of the two techniques, why one would be preferred over another qualitatively, or references that one can use for the problem of gene expression, thanks in anticipation.

REPLY [3 votes]: Assuming your data is vector-valued time series, SVD and Fourier analysis give different information.
(Singular vectors obtained by) SVD will essentially give you the "dominant" noise components  without any useful time information.
On the other hand, doing Fourier analysis on a bunch of different scalar time series will give you dominant temporal frequencies of the noise components, but will not give you the "vector" noise components associated with those frequencies.<|endoftext|>
TITLE: Motivic cohomology and pushforward maps
QUESTION [9 upvotes]: I have a question about pushforward maps for the motivic cohomology groups $H^p(X, \mathbf{Q}(q))$, for $X$ a smooth variety over a characteristic 0 field.
According to Mazza--Voevodsky--Weibel "Lectures on motivic cohomology", these groups have the following covariant functoriality properties (in addition to their usual contravariant functoriality)

if $f: X \to Y$ is a proper flat morphism, with fibres of dimension $d$, there is a pushforward map $f_*: H^{p + 2d}(X, \mathbf{Q}(q + d)) \to H^{p}(Y, \mathbf{Q}(q))$;
if $f: X \to Y$ is the inclusion of a closed subvariety of codimension $c$, then there is a pushforward map $f_*: H^{p}(X, \mathbf{Q}(q)) \to H^{p + 2c}(Y, \mathbf{Q}(q + c))$.

(Note that the first map lowers the cohomological degree, while the second map raises it.)
There are also corresponding maps in $\ell$-adic étale cohomology (for any prime $\ell$), or in Beilinson's absolute Hodge cohomology when the base field is $\mathbf{R}$ or $\mathbf{C}$; and there are "realisation" maps from motivic cohomology to these other two cohomology theories.

Are the pushforward maps (1) and (2) known to be compatible with their étale and Hodge analogues under the realisation maps?

I guess this must be known, but I've been unable to find a reference which states this directly. For étale cohomology, the work of Ciskinski--Deglise "Etale motives" apparently gives a realisation functor (on some category of motivic complexes) that is "compatible with Grothendieck's six operations", which sounds promising, but I'm not sufficiently expert to know if the concrete compatibility statements I want follow from this. And I've been unable to find any reference treating the case of absolute Hodge cohomology.

REPLY [5 votes]: The answer is yes, they are compatible. This is consequence of the machinery developped for the Riemann-Roch theorem. 
Recall that motivic cohomology, $l$-adic cohomology, and absolute Hodge cohomology have additive Chern classes (that is to say, $c_1(L\otimes L')=c_1(L)+c_1(L')$) and that the realization maps Chern classes to Chern classes. You should take a look to Déglise's preprint. The theorem you use is 4.3.2. The spectra $\mathbb{E}=\mathrm{H}_B$, Beilinson's motivic cohomology spectrum, and $\mathbb{F}$ the spectrum representing either $l$-adic cohomology (done here) or absolute Hodge cohomology (done in Brad Drew's thesis). The $\varphi$ from Déglise would be your realization and the $\mathrm{Td}_\varphi$ from Déglise would be 1 because the realization preserves Chern classes.<|endoftext|>
TITLE: History of set-class distinction
QUESTION [18 upvotes]: I have two questions concerning the history of set theory, both related to the distinction between the notion of a set and the notion of a class:

Who was the first mathematician to make this distinction explicit?
Did the terminology for this distinction immediately settle to the current terms "set" and "class", or was there some variance in the terminology used to make this distinction in the beginning?

Question 2 can also be extended to other languages, e.g. whether the terminology immediately settled to "Menge" and "Klasse" in German.

REPLY [6 votes]: A detailed discussion of the early history of set theory is in :

José Ferreiros, Labyrinth of thought : A history of set theory and its role in modern mathematics (2ed 2007).

The starting point are Dedekind and Cantor.
Dedekind, in his exposition of the theory of fields (1871) writes [§159] :

By a field [Körper] we mean an infinite system [System] of real or complex numbers [...].

In a draft for Zahlen, written between 1872 and 1878, he introduces the concepts of system and of mapping ["Die Begriffe des Systems, der Abbildung"].
In  Was sind und was sollen die Zahlen? (1888) we find also : "zusammengesetze System" and "Gemeinheit der Systeme" ["compounded system" (union - Def.8), and "community of systems" (intersection - Def.17)].

Cantor in the years 1863-64 introduced his theory od point-sets, and in 1872 speaks of set of points [Punktmenge].

Up to now, we have "sets of" something : numbers or points.
The "pure" notion of set has been introduced by Cantor in 1879 :

I say that a manifold (a collection, a set) of elements that belong to any conceptual sphere is well-defined ["Eine Mannichfaltigkeit (ein Inbegriff, eine Menge) von Elementen, die irgend welcher Begriffsphare angehoren"]

followed in 1883 by the "mature" concept :

By a manifold or a set I understand in general every Many that can be thought of as One, i.e., every collection of determinate elements which can be bound up into a whole through a law, and with this I believe to define something that is akin to the Platonic εἶδος or ἰδέα. ["Unter einer Mannichfaltigkeit oder Menge ...].


We can find class into the claculus of class of The Algebra of Logic Tradition : from Boole to Schröder.
Peano in Arithmetices principia (1889, page x) uses classis, followed by Russell (Principles, 1903). Of course, also Frege did [Appendix to vol.2 of Grundgesetze, discussing Russell's paradox, also if class is a "derived" notion for Frege, "concept" [Begriffe] being the primitive one] :

I cannot see how arithmetic could be given a scientific foundation, how numbers could be
  conceived as logical objects and introduced, if it is not allowed - at least conditionally - to go from a concept over to its extension. Can I always speak of the extension of a concept, of a class? 


Zermelo (1908) starts from Cantor's definition of set (1895) as "a collection, gathered into a whole, of certain well-distinguished objects of our perception or our thought" [Unter einer 'Menge' verstehen wir jede Zusammenfassung ...]
He introduce also the notion of "Klassenaussage" ('class-statement') meaning a logical expression with one variable, what Russell called a propositional function.
Finally, von Neumann (1925) speaks of : set [Menge] and class [Bereich] :

"sets" are the sets that (in the earlier terminology) are "not too big", and "classes" are all totalities [Gesamtheiten] irrespective of their "size". 



See also The Early Development of Set Theory.<|endoftext|>
TITLE: Is the axiom schema of replacement used in algebraic number theory (or more generally outside logic)
QUESTION [10 upvotes]: Here's a precise question. Does Wiles' proof of FLT run just fine in the set theory that logicians would perhaps call "Zermelo + choice" -- i.e. drop the axiom schema of replacement but assume the axiom of choice to make analysis work sensibly? [Wiles' proof needs some cyclic base change, which involves a lot of functional analysis, and one would imagine that one needs enough AC to make this stuff go through, but my question is not about AC. I am also aware that there are theorems in logic of the form "if a statement has some properties and it is provable in ZFC then it's provable in certain other systems", but my question is not about the truth or provability of FLT in Z+AC, it's about whether Wiles' proof works].
My understanding of some justifications for including replacement: without it you can't construct the ordinal $\omega+\omega$, or the cardinal $\aleph_\omega$. But I am not sure a number theorist would need such things. If you really need something with order-type $\omega+\omega$ you could just use $\{0,1\}\times\omega$ with some lexicographic ordering, and $\aleph_\omega$ itself is not something I've ever seen in a number theory paper.
Of course Wiles' proof uses a lot of commutative algebra, directly or indirectly, and I do remember a proof in Matsumura's commutative algebra book that uses induction over ordinals, but if I remember correctly the induction was only needed in situations where some module was massively infinitely-generated over some ring.
Maybe Wiles' proof of FLT is asking too much -- what about something less technical, like Gross-Zagier or some other random thing? I'm sure people realise what I'm trying to ask -- take a "standard" result in "standard" number theory -- does one need Replacement? I'm not interested in statements about things that logicians might have constructed (for example Goodstein sequences) that they may perhaps argue are "number theoretic" -- I know something about these things and they are interesting, but this question is not about them. I want to focus on standard well-known theorems in mainstream number theory from the last 50 years or whatever. Perhaps another reason to shy away from FLT is that on the face of it it uses some category theory and there was a certain amount of fuss in the 1990s about whether the proof actually was a ZFC proof [conclusion: it was, although it undoubtedly used references which talked about subtle constructions in very large categories so some rewriting would be necessary to stuff it in.]
The reason I ask is that I have it in my head (perhaps I read it in a paper by Adrian Mathias) that Bourbaki don't assume replacement in their treatise, and I remember at the time thinking "oh wow that's crazy, what an omission" but now having actually used Bourbaki as a reference for various facts in e.g. representation theory and analysis, and having begun to comprehend just how far they get and how much further they could have got, I am now wondering how important replacement is outside logic and neighboring areas.

REPLY [7 votes]: At eric's request, I am expanding my comment into an answer.
If we set aside specific theorems in algebraic number theory for the moment, the question of whether Replacement is used in "ordinary mathematics" has come up on MO before, e.g.,
here, here, and here, and it also surfaces on the Foundations of Mathematics mailing list from time to time.  There is a relatively short list of known examples outside of set theory where Replacement is provably necessary.  These include some theorems about Borel sets due to Martin, Friedman, and others as well as some more algebraic examples due to Adrian Mathias.  Replacement is also needed to show that various axioms of infinity are equivalent.
Now, if we lower the bar and declare that an argument "uses" Replacement if its most natural and direct interpretation invokes Replacement, then it can be argued that there are many such examples in ordinary mathematics.  However, many of these arguments can in principle be reformulated using a combination of powerset and separation.  So they do not "use" Replacement in the strongest sense that the theorems are unprovable without it.
Returning to number theory, the only way to be absolutely sure that Replacement is not needed in some argument is to check each step of the argument.  However, given that provable uses of Replacement outside of set theory are so rare, it would be surprising if any "ordinary" theorem such as the ones you mentioned requires Replacement in a strong sense.<|endoftext|>
TITLE: Refinement of Dirac's theorem on Hamiltonian graphs
QUESTION [7 upvotes]: Dirac's theorem states that if degree of each vertex of a graph $G=(V,E)$ is not less than $|V|/2$, then it has Hamiltonian cycle. It is less known, but still known and not so hard to prove (though I do not know the reference other than competition problem) that if all degrees are exactly $(|V|-1)/2$, the graph still have Hamiltonian cycle. So the question.
Let $G$ be a graph with $2n+1$ vertices and all degrees at least $n$ but without Hamiltonian cycle. What is minimal possible number of edges in $G$?  There are two examples with $n(n+1)$ edges: $K_{n,n+1}$ and two $K_{n+1}$'s glued by a vertex. Maybe, this is least possible actually?
UPDATE. It looks that what actually is proved in the result to which I refer is that the graph on $2n+1$ vertices without Hamiltonian cycle and all degrees at least $n$ either contains $K_{n,n+1}$ or coincides with two $K_{n+1}$'s glued by a vertex. This answers the question, if anybody is interested I may leave a proof here, if not, just remove the question.

REPLY [3 votes]: Let $G$ be a graph with $2n+1$ vertices and all degrees at least $n$ but without Hamiltonian cycle. Consider two cases.
1) There is a cycle $C=x_1\dots x_{2n}x_1$ of length $2n$. Let $y$ be the vertex not in this cycle. If it is joined with two consecutive vertices in $C$, then we have a Hamiltonian cycle. Thus $y$ is joined, say, with $x_1,x_3,\dots,x_{2n-1}$. Replace $C$ to a new cycle, replacing fragment $x_{2k-1}x_{2k}x_{2k+1}$ to $x_{2k-1}yx_{2k+1}$. Apply the same argument, now $x_{2k}$ plays role of $y$. We see that $x_{2k}$ must be joined with $x_{2i-1}$ for all possible indices $2i-1,2k$. It follows that $G$ contains $K_{n,n+1}$ ($x_1,x_3,\dots,x_{2n-1}$ is first part and $y,x_2,x_4,\dots,x_{2n}$ the second part. It may contain also arbitrary set of edges between vertices of the first part.
2) There is no such a cycle. Consider Hamiltonian path $P=x_1\dots x_{2n+1}$ in $G$. Then $x_1$ is joined with some (at least) $n$ vertices in $P$. If $x_1$ is joined with $x_k$, then $x_{2n+1}$ can not be joined with $x_{k-1}$, nor with $x_{k-2}$. If for some $k$ it appears that $x_1$ is joined with $x_k$, $k>2$, but not with $x_{k-1}$, we get already $n+1$ forbidden vertices for $x_{2n+1}$. Thus $x_1$ must be joined with $x_2,\dots,x_{n+1}$ and $x_{2n+1}$ with $x_{n+1},\dots,x_{2n}$. Now for any $k=2,\dots,n$ change path to $x_kx_{k-1}\dots x_1x_{k+1}\dots x_{2n+1}$. Repeat the same argument. We see that $x_1,\dots,x_{n+1}$ form a clique, as well as $x_{n+1},\dots,x_{2n+1}$. Thus $G$ coincides with two cliques of size $n+1$ glued by their common vertex. Adding any edge produces a Hamiltonian cycle.<|endoftext|>
TITLE: Area of square to wrap a torus
QUESTION [9 upvotes]: The Nash-Kuiper
$C^1$ isometric embedding of flat torus into $\mathbb{R}^3$
has recently been spectacularly visualized by the
Hevea Project.
This suggests two questions.

Q1. What is the area of the smallest square that realizes the
  corrugated Nash-Kuiper embedding?

The Hevea project shows this image of how the square
maps to the torus after the 4th corrugation "wave":



And here is the 3D embedded torus, overview (left) and closeup (right),
which appears to be approximately $R$-major / $r$-minor axes, $R=2$ / $r=1$:


Independent of the above, one can "gift-wrap" non-flat surfaces
with paper or foil, as was studied for wrapping a ball
by Demaine et al.1

Q2. What is the area of the smallest square that
  can gift-wrap a torus, in the sense of completely covering
  its surface?
  For specificity, let the major axis (center of tube to center of torus)
  be $R=2$ and the minor axis (radius of the tube) be $r=1$.
  Thus the aspect ratio $R/r=2$.
  (When $R>r$, it is a ring torus.
  As j.c. says, 
  $R=2$ / $r=1$ is known as the "unit ring torus.")

If this torus were a present, one might wrap its convex hull,
but here I mean truly covering all of the surface
with paper, so that
the wrapped result looks like a torus (with many crinkling creases).
From the Demaine et al. paper,
here is a definition of what constitutes a wrapping:

Definition of wrapping.
  A wrapping is a noncrossing contractive mapping of a piece of paper
  into Euclidean 3-space.
  The contractive constraint means that every distance either decreases or
  stays the same, as measured by shortest paths on the piece of paper before and after mapping via
  the folding.
  Contractive mappings are called short or contracting by Burago and Zalgaller [BZ96]
  and submetry maps by Pak [Pak06].


1
Demaine, Erik D., Martin L. Demaine, John Iacono, and Stefan Langerman. "Wrapping spheres with flat paper."
Computational Geometry 42, No. 8 (2009): 748-757.
(PDF download.)

REPLY [4 votes]: This answer addresses Q2, as Q1 was more or less answered in the comments. Edit of Oct 21, 2017, an upper bound and a big correction.
You can get a lower bound on the area by the fact that the diameter (in the sense of metric spaces) of the starting square must be greater than or equal to the diameter of the final embedded torus with its intrinsic metric.
I think the diameter of the standard embedded torus is realized by the distance between two points on the outer equator that are opposite each other.
I was able to extract the following from the note Geodesics on the Torus
and other Surfaces of Revolution
Clarified Using Undergraduate Physics Tricks
with Bonus:  Nonrelativistic and Relativistic
Kepler Problems by Robert Jantzen.
For a standard torus with major radius 2 and minor radius 1 (what Jantzen calls the "unit ring torus" and what I'll denote $T_{emb}$), Jantzen computes that the shortest path connecting two antipodal points on the outer equator is of length approximately 15.26/2 = 7.63.  Below is his picture (Fig. 11) of the "[1,1;0]" geodesic that connects them:

The path is depicted in blue.  Somewhat confusingly, a meridian is also colored in blue in this image; that should be disregarded.
The diameter of an open square of area $a^2$ is $\sqrt{2}a$, realized by a diagonal joining two corners.

The square with diameter 7.63 has side length $a=5.39$ and area 29.1, which gives a lower bound on the area of a wrapping square.
This is about 1.47 times the area of the unit ring torus $T_{emb}$ $Rr\pi^2=2\pi^2\approx19.739$ (formula from MathWorld).

Now, this is a pretty awful bound since it's intuitively obvious that if we try to wrap $T_{emb}$ with a square so that the diagonal of the square lies along the above-depicted diameter, there's a lot of the torus that won't be covered (more than half, it seems). This does beat the trivial "area" bound though.

I now consider a different problem: that of finding a map from the flat square torus with "side length" $a$, $T_{flat}(a)$, to $T_{emb}$ which is a wrapping in the sense of the question.
Note that every wrapping from $T_{flat}(a)$ to $T_{emb}$ induces wrapping maps from a square with side length $a$ to $T_{emb}$, so it is possible to get upper bounds for the area of the smallest wrapping square by constructing a wrapping from a flat torus.
For instance, we can get an upper bound by the following obvious map from $T_{flat}(a)$ to $T_{emb}$: let the square fundamental domain of $T_{flat}(a)$ be parametrized by $(x,y)\in[0,a]^2$. Then we map this to the embedded torus so that constant $x$ circles go to longitudinal circles and constant $y$ circles go to meridians.
For what values of $a$ is this map a wrapping? All meridians have the same length $2\pi$ which means that we must take $a\geq 2\pi$ for a wrapping.  The longest longitudinal circle is the outer equator of the torus, which has length $6\pi$, which means that in fact we must take $a\geq 6\pi$. One can check that this is also sufficient for the map to be a wrapping (by computing the Jacobian of the above map and seeing that its operator norm $\leq1$ precisely when $a\geq 6\pi$). Therefore:

The above map is a wrapping from $T_{flat}(6\pi)$ to $T_{emb}$ and so an upper bound for the area you're looking for is $36\pi^2\approx 355.3$, which is 18 times the area of $T_{emb}$.

Unfortunately, results relating to this problem don't obviously give lower bounds to Q2.  This is because wrappings from a square to $T_{emb}$ don't have to "glue up" to maps from $T_{flat}(a)$ to $T_{emb}$. [In earlier versions of this answer, I did not realize this!]
Nonetheless, I think this is an independently interesting problem (maybe a worthy "Q3"?) and perhaps the results below inspire some other ideas.  In particular, I find it unlikely that the best wrapping surjection from a square has a smaller area than the lower bound given for $T_{flat}(a)$ below, though I don't know how to prove it.  Indeed, I would be surprised to see any wrapping that does better than the one given as an upper bound above!
In what follows I'll discuss a lower bound for wrappings from $T_{flat}(a)$ to $T_{emb}$ using the idea described in the first part of this answer. The diameter $T_{flat}(a)$ is $a/\sqrt{2}$, realized by a point in the "center" of the square fundamental domain and a vertex.

The flat square torus with the same diameter as $T_{emb}$ has $a=10.79$ and area 116.4, which gives a lower bound on the area of a wrapping flat torus.
This is about 5.90 times the area of $T_{emb}$, $Rr\pi^2=2\pi^2\approx19.739$.

This bound is very unlikely to be sharp for this problem. Note that the diameter of $T_{flat}(a)$ is realized by 4 paths (the segments of the "crossed diagonals").  See the green paths in this figure, which is an edit of this file from wikipedia.

The diameter of $T_{emb}$ is also realized by 4 paths (the "top" and "bottom" of the [1,1;0] geodesic depicted above, as well as the reflections of these through the $xy$ plane).  A crude cartoon is depicted below, where the green paths represent the geodesics (but were not actually computed as such).

However, the graphs formed by the union of these 4 paths in the two spaces have topologically inequivalent embeddings in their respective tori!  One way to see this is to consider the (topological) circles formed by the edges in the graphs.  For the graph in $T_{flat}(a)$, such a circle never bounds a disk, but it can (in multiple ways) for the graph in $T_{emb}$.
In the following figure, the left panel shows the previous sketch of the diameters on $T_{emb}$ with a red longitudinal circle and blue meridional circle highlighted. If we imagine cutting $T_{emb}$ along those circles and unwrapping it, the graph of diameters will be isotopic to the one in the figure in the right panel.

I don't know how to show that this precludes the existence of a wrapping of this size but it is certainly boggling my mind.<|endoftext|>
TITLE: Are there non-contractible spaces $A$ and $B$ such that $A \wedge B$ is contractible?
QUESTION [5 upvotes]: The question is in the title : Can we find spaces $A$ and $B$, each non contractible, such that their smash product $A \wedge B$, i.e. the homotopy cofibre of $A \vee B \to A \times B$, is a contractible space ?

REPLY [6 votes]: I just saw this in Ravenel's "orange book"! Let $X$ be any simply connected CW complex whose reduced homology is all torsion. Let $X_{(p)}$ be the p-localization for a prime $p$. Then for primes $p\neq q$, $X_{(p)}\wedge X_{(q)}$ is contractible. But neither $X_{(p)}$ nor $X_{(q)}$ is contractible if $H_*(X)$ has $p$ and $q$ torsion.<|endoftext|>
TITLE: Non-realizability of $\mathbb{Q}$ as a cohomology group
QUESTION [12 upvotes]: (This is a re-post of 1)
In the paper "On the realizability of singular cohomology groups" by Kan and Whitehead, it is shown that there is no space $X$ and integer $n\geq 1$ such that $H^{n−1}(X)=0$ and $H^n(X)=\mathbb{Q}$ (cohomology with integral coefficients).
At the very end of the article there is a remark where it is stated that, at the time of writing (around 1960, I suppose), it was not known whether $\mathbb{Q}$ could be a (integral) singular cohomology group at all.
My question is: is this still not known?

REPLY [15 votes]: This may depend on your axioms, see
S. Shelah "The consistency of Ext(G,Z)=Q", Israel J. Math. 39 (1981), no. 1-2, 74–82. 
There it is shown that it is consistent with the generalised continuum hypothesis that there exists a group $G$ having $Ext(G, \mathbb{Z})=\mathbb{Q}$. Then a Moore space $M(G,n-1)$ has the required property.<|endoftext|>
TITLE: almost disjoint ladder system on $\omega_2$
QUESTION [5 upvotes]: Suppose $\langle s_\alpha : \alpha \in \omega_2 \cap \mathrm{cof}(\omega_1) \rangle$ is a sequence such that each $s_\alpha$ is an increasing cofinal map from $\omega_1$ to $\alpha$.  Is it possible that for all $\alpha < \beta$, $\mathrm{ran}(s_\alpha) \cap \mathrm{ran}(s_\beta)$ is finite?  Note that a pressing-down plus pigeonhole argument shows that CH must fail.

REPLY [7 votes]: Following the suggestion, here is the solution, from Baumgartner: Almost-disjoint sets, the dense set problem and the partition calculus, Annals of Math. Logic, 10(1976), p. 424, part 6.
Let $\{A_\xi:\xi<\omega_2\}$ be sets of size $\aleph_1$ with pairwise countable intersection. 
Set $p\in P$ iff $p$ is a function, $Dom(p)\in [\omega_2]^{<\omega}$, $p(\xi)\in [A_\xi]^{<\omega}$ ($\xi\in Dom(p)$). $p'\leq p$ if $Dom(p')\supseteq Dom(p)$, $p'\supseteq p(\xi)$ and $p(\xi)\cap p(\eta)=p'(\xi)\cap p'(\eta)$ ($\xi\neq\eta\in Dom(p)$). 
If $G$ is generic, we set $B_\xi=\bigcup\{p(\xi):\xi\in Dom(p),p\in  G\}$. Density arguments give $|B_\xi|=\aleph_1$, $|B_\xi\cap B_\eta|<\omega$, essentially the only nontrivial point is to show that $(P,\leq)$ is ccc. 
For that, assume that $p_\alpha\in P$ ($\alpha<\omega_1$). By the $\Delta$-system lemma, we can assume that $Dom(p_\alpha)=s\cup s_\alpha$ with $\{s,s_\alpha:\alpha<\omega\}$ disjoint. The common extension of $p_\alpha$, $p_\beta$ will clearly be the coordinatewise union, the only trouble can be if $p_\alpha(\xi)\cap p_\alpha(\eta)\neq p_\beta(\xi)\cap p_\beta(\eta)$ for some pair $\{\xi,\eta\}\in[s]^2$. However, as $|B_\xi\cap B_\eta|\leq\omega$ for $\xi\neq\eta$, there are just countably many possibilities for $p_\alpha(\xi)\cap p_\alpha(\eta)$ for each pair $(\xi,\eta)$, so we can find $\alpha$ and $\beta$ that $p_\alpha(\xi)\cap p_\alpha(\eta)=p_\beta(\xi)\cap p_\beta(\eta)$ for all pairs $\{\xi,\eta\}$ in $s$, and then $p_\alpha$, $p_\beta$ will be compatible. 
The nice thing in the argument is that it is iterable, i.e., $|A_\alpha|=\aleph_2$, $|A_\alpha\cap A_\beta|\leq\aleph_1$ for $\alpha<\beta<\omega_3$, then one forcing shrinks each $A_\alpha$ to $A'_\alpha$ so that $|A'_\alpha|=\aleph_2$, $|A'_\alpha\cap A'_\beta|\leq\aleph_0$, then the next forcing thins out $A'_\alpha$ to $A''_\alpha$ such that $A''_\alpha\cap A''_\beta$ is finite, etc.<|endoftext|>
TITLE: Universal covering and double cover functors
QUESTION [6 upvotes]: Initially posted on MSE
Let $\mathsf{CW}$ be the category of CW-complexes and $\mathsf{CW}_*$ that of pointed CW-complexes (possibly disconnected, one basepoint in each component). I would like to know whether there is a functor $\mathsf{CW} \to \mathsf{CW}$ mapping a CW-complex to its universal cover. In the pointed case this works, i.e. there is a functor $\mathsf{CW}_* \to \mathsf{CW}_*$ mapping a pointed CW-complex $(X,x_0)$ to its universal cover 
$$ \widetilde X(x_0) \colon = \{[\alpha \colon I \to X] \mid \alpha(0)= x_0 \} $$
where square brackets denote homotopy relative endpoints, and the covering projection is $[\alpha] \mapsto \alpha(1)$. The basepoint of $\widetilde X(x_0)$ is the class of the constant map, and induced maps are given by requiring that basepoints map to basepoints, using lift uniqueness. But without the basepoint I am unable to define this as a functor.
A related question (one that I am more interested in) is: define the category $\mathsf{CW}^\text{tw}$ to be the category of pairs $(X,w)$ with $X$ a CW-complex and $w \in H^1(X;\mathbb{Z}_2)$, and maps preserving cohomology classes. Is there a functor $\mathsf{CW}^\text{tw} \to \mathsf{CW}$ which maps a $(X,w)$ to a two sheeted covering space $\overline X \to X$ whose first Stiefel-Whitney class is $w$? Again, this is possible in the pointed setting, and a positive answer to the first question would imply a positive answer here as well. The second functor actually exists on the subcategory $\mathsf{MFD} < \mathsf{CW}^\text{tw}$ of smooth (actually also topological) manifolds with their first Stiefel-Whitney classes: take the sphere bundle of the top exterior algebra of the tangent bundle. This makes me think it would not be an easy task to show that they don't exist over all CW-complexes, i.e. the statement is not that implausible after all.
I would be very grateful for any help, including references and hints. Thank you for reading :)

REPLY [5 votes]: I am supposing that you want to take all continuous maps between CW-complexes in $\mathsf{CW}$, and I am also supposing for simplicity that you want the purported universal cover functor $U : \mathsf{CW} \to \mathsf{CW}$ to be continuous, and that you want it to be augmented $\eta: U \Rightarrow Id_{\mathsf{CW}}$ so that $\eta_X : U(X) \to X$ is the universal covering map.
Then this is impossible. This is because for each CW-complex $X$ one can then form the continuous map
$$X = \mathrm{Hom}_{\mathsf{CW}}(*, X) \to \mathrm{Hom}_{\mathsf{CW}}(U(*), U(X)) = \mathrm{Hom}_{\mathsf{CW}}(*, U(X)) = U(X)$$
a canonical map from each space to its universal cover, and the natural transformation $\eta$ shows that this is a section of the covering map $\eta_X : U(X) \to X$.<|endoftext|>
TITLE: Why does McMahon formula look like the inclusion-exclusion principle?
QUESTION [31 upvotes]: The McMahon formula for the number of tilings of an $a \times b \times c$ hexagon by lozenges:
$$ \Big[H(a)H(b)H(c)\Big] \Big[H(a+b)H(b+c)H(c+a)\Big]^{-1} \Big[H(a+b+c)\Big]$$
looks oddly like the inclusion-exclusion formula:
$$ |A \cup B \cup C| = |A|+|B|+|C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|$$
Here $H(a) = 1! 2! \dots a!$ is the hyperfactorial.  
Perhaps there is a more general explanation via Gelfand-Tsetlins or something?

(source: microsoft.com)

REPLY [2 votes]: This is not an answer, but seeing as this is a fresh post, I would like to add:

These are the $20\; 2\times2\times2$ plane partitions. Note there are $4$ of each color when viewed as hexagons.
If we let $a=1$, we get:
$$\dbinom{b+c}{c}=\prod_{i=1}^b\prod_{j=1}^c \dfrac{i+j}{i+j-1}=\prod_{i=1}^c \dfrac{i+b}{i}$$
A few more links:

Mathworld
Wikipedia
DLMF

Note the I-E principle is divisive and not subtractive.<|endoftext|>
TITLE: Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $Gal(T(f))$ $\simeq$ The commutator subgroup of $Gal(f)$?
QUESTION [5 upvotes]: I  asked this  question at MSE but I did not receive an answer. So I ask it at MO:
We denote the field of rational numbers by $\mathbb{Q}$. The Galois group of a polynomial $f$ is  denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:
Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$

REPLY [17 votes]: The answer is no. First we see that $T$ is injective: Let $0\ne g\in\mathbb Q[X]$ with $T(g)=0$. Then $\textrm{Gal}(f-\alpha g)'=\textrm{Gal}(f)'$ for all $f\in\mathbb Q[X]$ and $\alpha\in\mathbb Q$. Pick $\beta\in\mathbb Q$ with $g(\beta)\ne0$, and $f\in\mathbb Q[X]$ of degree $n$ larger than $\deg g$ and $2$ such that $\textrm{Gal}(f)=S_n$. Then $\textrm{Gal}(f)'=S_n'=A_n$. But the polynomial $f(X)-\frac{f(\beta)}{g(\beta)}g(X)$ has the root $\beta$, so its Galois group is a subgroup of $S_{n-1}$. But no subgroup of $A_{n-1}=S_{n-1}'$ is isomorphic to $A_n$.
Thus we find $f\in\mathbb Q[X]$ of degree $2$ such that $F=T(f)$ has degree at least $2$. Since $\textrm{Gal}(F)=\textrm{Gal}(f)'=1$, $F$ splits into linear factors. Next pick $\gamma\in\mathbb Q$ such that $T(X+\gamma)=T(X)+\gamma T(1)$ has no common root with $F$. (Recall that $T(1)\ne0$.) Then $F$ and $G=T(X+\gamma)$ are relatively prime. As $f+\alpha(X+\gamma)$ has degree at most $2$ for all rational $\alpha$, we see that $F(X)+\alpha G(X)$ splits into linear factors for all rational $\alpha$. However, $F(X)+YG(X)$ is irreducible in $\mathbb Q[Y][X]$, so by Hilbert's irreducibility theorem there are infinitely many $\alpha\in\mathbb Q$ such that $F(X)+\alpha G(X)$ is irreducible. On the other hand, we saw that this polynomial factors into linear factors, so $\deg F\le 1$, contrary to $\deg F\ge2$.<|endoftext|>
TITLE: Pontryagin dual of the surreal numbers?
QUESTION [9 upvotes]: Has any work been done on the Pontryagin dual of the surreal numbers (suitably topologized)? I have not been able to find anything and am not sure if this is still unknown.
Alternatively, has this been worked out for the various hyperreal fields, or real-closed fields in general?

REPLY [9 votes]: For any infinite cardinal $\kappa$, let $S_\kappa$ be the surreal numbers of rank $<\kappa$, considered as a group under addition and topologized with the order topology (if you want to consider all the surreal numbers, suppose $\kappa$ is inaccessible).  Suppose now that $\kappa$ has uncountable cofinality and $f:S_\kappa\to U(1)$ is a continuous homomorphism.  Since $\kappa$ has uncountable cofinality, $S_\kappa$ is countably saturated as an ordered set, and it follows that a countable intersection of open sets in $S_\kappa$ is still open.  In particular, this implies that $\ker(f)$ is an open subgroup of $S_\kappa$.
That is, every continuous homomorphism $f:S_\kappa\to U(1)$ factors through a discrete group.  The open subgroups $K_\alpha=\bigcup_{n\in \mathbb{N}}(-n\omega^{-\alpha},n\omega^{-\alpha})$ for ordinals $\alpha<\kappa$ are a neighborhood base at $0$, so the group $G$ of continuous homomorphisms $f:S_\kappa\to U(1)$ can be considered as the direct limit of the Pontryagin duals of the discrete groups $S_\kappa/K_\alpha$.  It is easy to see that for each $\alpha<\kappa$, $K_\alpha/\bigcup_{\beta<\alpha} K_\beta$ is a $\mathbb{Q}$-vector space of dimension $\kappa$, as is $S_\kappa/K_0$.  Choosing bases for all these vector spaces, we obtain the following description of the group $G$.  Consider $\mathbb{Q}$ as a discrete group, let $B$ be its Pontryagin dual, and let $A=B^\kappa$.  Then $G$ is isomorphic to the group of functions $\kappa\to A$ which are eventually $0$.  Note that in this description, the group of all (possibly discontinuous) homomorphisms $S_\kappa\to U(1)$ can be identified with the full product $A^\kappa$.
I don't know whether there is any particularly natural topology to put on $G$, but it is not hard to check that the compact-open topology is just the product topology on $G$ as a subgroup of $A^\kappa$ (to show this, first show that any compact subset of $S_\kappa$ is finite).<|endoftext|>
TITLE: Do there exist elliptic curves over schemes which have all primes as residue characteristics?
QUESTION [6 upvotes]: It's well known that there are no elliptic curves over Spec $\mathbb{Z}$, but it's unclear (to me at least) if the proof generalizes.
My question is: If $S$ is a connected scheme such that has every prime as a residue characteristic, then can there exist an elliptic curve over $S$?

REPLY [7 votes]: Another perspective: Let $K$ be any number field and let $E$ be any elliptic curve over $K$ with $j$-invariant in $\mathcal{O}_K$. Then there is an extension $L$ of $K$ such that $E \times_K L$ extends to an elliptic curve over $\mathcal{O}_L$. 
Proof: By properness of the stack $\overline{M}_{1,1}$, there is some $L$ so that we get a stable genus one curve with one marked point over $\mathrm{Spec}\ \mathcal{O}_L$. We just need to rule out the possibility that some fibers are nodal. If the fiber over a prime $\pi$ of $\mathcal{O}_L$ is nodal, than a computation with Tate curves shows that $v_{\pi}(j)<0$, contradicting that $j \in \mathcal{O}_L$. $\square$
I think it is worth bringing this up, as well as the CM solutions above, because there are many more integral $j$-invariants than there are $j$-invariants with CM.<|endoftext|>
TITLE: Zeta zeros standard normal distribution about $\vartheta (\gamma_n)$
QUESTION [6 upvotes]: Asked at MSE here without response.
I realise that this resembles Odlyzko's famous nearest neighbours plot, and was wondering whether this is simply a manifestation of the same phenomenon.
That said, below is a partially scaled histogam plot of $\vartheta (\gamma_n) - \pi (n - 3/2) ,$ where $\gamma_n$ is the imaginary part of the $n$th zeta zero, and $\vartheta $ is the Riemann-Siegel theta function, for the first $2$m zeros:

It certainly looks to have standard normal distribution, is this indeed the case? Does it depend on the RH, or is it independent of it?

REPLY [8 votes]: What you're observing is a remarkable theorem of Selberg.  The usual notation is to let $N(T)$ denote the number of zeros of $\zeta(s)$ with ordinates between $0$ and $T$.  Then the argument principle (with Stirling's formula) gives 
$$
N(T) = \frac{T}{2\pi} \log \frac{T}{2\pi e} + \frac{7}{8} + S(T) + O\Big(\frac 1T\Big),  
$$ 
where $S(T) = \frac {1}{\pi} \text{arg} \zeta(\tfrac 12+iT)$ (and the argument is 
defined by continuous variation along line segments from $2$ to $2+iT$ and then from $2+iT$ to $1/2+iT$).  Then Selberg showed that $\pi S(T)=\text{arg}\zeta(\tfrac 12+iT)$ has a normal distribution with mean $\sim 0$ and variance $\sim \frac{1}2 \log \log T$.  What you're looking at is the distribution of this remainder term in the asymptotic for the number of zeros when evaluated at the ordinates of the $n$-th zero, and from Selberg's work one can deduce that this is Gaussian.  You'll find a discussion of Selberg's theorem in the books of Edwards or Titchmarsh.<|endoftext|>
TITLE: Explicit eigenvalues of the Laplacian
QUESTION [36 upvotes]: Let $(M,g)$ be a compact manifold without boundary.

Question: For which $(M,g)$ are the eigenvalues of the Laplace operator on functions explicitly known?

An important example is the $n$-sphere with its standard metric. To find eigenvalues, we embed $S^n$ inside $\mathbb{R}^{n+1}-\{0\}$ in the usual way, consider a positive homogeneous function $f\in C^\infty(\mathbb{R}^{n+1}-\{0\})$ of degree $s$, and then take the restriction to the sphere of the Laplacian $\Delta$ on $\mathbb{R}^{n+1}-\{0\}$ applied to the function $|x|^{-s} f$. The result is that if $f$ is harmonic relative to the Laplacian on $\mathbb{R}^{n+1}-\{0\}$, then the restriction to $S^n$ of $\Delta(|x|^{-s} f)$ is a scalar multiple of the restriction of $f$ to $S^n$, with the scalar being $s(s+n-2)$.
One sees very quickly that for more complicated manifolds, such a method does not apply. Various authors comment that the spectrum of the Laplacian is not easy to determine explicitly, and much of the literature seems to be consumed only with estimates for certain eigenvalues of the Laplacian given various constraints on the geometry of $(M,g)$.
Are there other interesting manifolds for which the spectrum of the Laplacian is known? In particular, are they known for ellipsoids?

REPLY [7 votes]: Generalizing the case of flat tori, one can compute explicitely the spectrum of many compact flat manifolds. See for instance 

spectrum on $p$-forms Miatello and Rossetti or the survey on isospectral compact flat manifolds in Contemp. Math. 491 AMS, 83--113.

There are also more progress for lens spaces than the one by Sakai mentioned by Ziegler.
See for instance

the description of the associated generating functions in Thm. 3.2 given by Ikeda and Yamamoto,
the same for the spectrum on $p$-forms in Thm. 2.3 by Ikeda (``Riemannian manifolds $p$-isospectral but not $p+1$-isospectral'', in Geometry of manifolds, Perspect. Math. 8, 1989),
the multiplicity of each eigenvalue in Thm. 3.8 of this article,
the associated generating function written as a rational function in Thm. 3.6 here,
the Dirac spectrum in Thm. 4.3 here. 

Other contributions on compact homogeneous spaces:

spectra on $p$-forms on $S^n$ and $P^n(\mathbb{C})$ Ikeda & Taniguchi, 
Grassmann Manifolds Tsukamoto, Fida El Chami, Tsagas & Kalogeridis,
Quaternionic Grassmann manifolds Fida El Chami,
$Sp(n)/U(n)$ Hong & Chen, Tsagas & Kalogeridis.<|endoftext|>
TITLE: Amount of math research published in other languages?
QUESTION [27 upvotes]: I'm curious what languages contribute the largest fraction of published research mathematics. That is, for a given language the percent of new research being published in that language. I'm especially curious to see how you come up with such numbers.
From some googling I managed to get a very crude estimate of total publication volume: 28643 arXiv articles in 2014. Looks like arXiv sometimes has articles in other languages but I can't search by language.
Edit: Thanks S. Carnahan for pointing out that I could search the full text by common words from different languages!

REPLY [14 votes]: I researched the language distribution using the Web of Science. Since your interest was in new content, I restricted the search to the period 1985-2015. The total number of mathematics articles in that database is 882565 for this period. Here is the breakdown by language:

[1985-2015]

The Web of Science goes back to 1945, so I also made the search for the entire period 1945-2015. That gives a total of 1120170 mathematics articles, with this breakdown by language:

[1945-2015]

Finally, for the last ten-year period 2005-2015 the breakdown (for a total of 459213 mathematics articles) is 

[2005-2015]

It is clear that the coverage of Web of Science is not representative for Asia, but for the European languages I would expect it to give a reliable breakdown.<|endoftext|>
TITLE: Ordering of large cardinals by cardinality
QUESTION [16 upvotes]: Let Type A and Type B be two types of large cardinals from, say, Cantor's Attic (http://cantorsattic.info/Upper_attic)
Now assuming that ZFC + Type A + Type B is consistent (ie, both Type A and Type B cardinals can coexist), I define:
*Type A > Type B if smallest Type A cardinal has higher cardinality than smallest Type B
*Type A = Type B if smallest Type A and Type B have same cardinalities
*Type A $\perp$ Type B if the ordering in the sense above is undecidable 
So, for instance, Inaccesssible < Hyper Inaccessible < Mahlo
What is known about the ordering of large cardinals from (http://cantorsattic.info/Upper_attic) in this sense ?
I am particularly interested in the "large" large cardinals from measurable upwards.
For eg: How would one order measurable, extendible, huge and rank-into-rank ?
Motivation: I understand researchers are mostly focused on consistency strength, but I am interested in the intuitive notion of getting "much bigger infinities" from each successive large cardinal axiom.

REPLY [11 votes]: Let me add one extra example that might be interesting. 
Let $\pi_n^m$ and $\sigma_n^m$ denote respectively the least $\Pi_n^m$-indescribable and the least $\Sigma_n^m$-indescribable cardinal  (if they exist). Then:
Fact 1. If $V=L,$ then  $\sigma_n^m < \pi_n^m,$ for all $n, m \geq 1,$
Fact 2 (Hauser). If the existence of a $Σ^m_n$ indescribable above a $Π^m_n$ indescribable is consistent with $ZFC$, then the theory $ZFC+GCH+σ^m_n>π^m_n$ is consistent.<|endoftext|>
TITLE: Is SL(n,Z[x]) generated by transvections?
QUESTION [15 upvotes]: Is $\mathrm{SL}(n,\mathbb{Z}[x])$ equal to $E(n,\mathbb{Z}[x])$, the subgroup generated by transvections?

REPLY [7 votes]: First, for $n=2$, the result is certainly false by the counterexample of Cohn quoted in Oblomov's comment.
For higher $n$, start with the fact that the Bass Stable Rank of a commutative ring is at most 1 more than the Krull dimension.  (I'm sure you can find this hidden in Chapter V of Bass's book on K-theory, though extracting it might require some work to get familiar with his notation).  Therefore ${\mathbb Z}[X]$ has stable rank at most 3.  In fact, it's almost surely exactly 3.  (There is some discussion here about whether this stable rank might in fact be 2, but there's something close to a proof there that this is not the case.)
Now from Bass, Chapter V, 3.3, it follows that for all $n\ge 4$, we have
$$GL_n({\bf Z}[x])=GL_3({\mathbb Z}[X])E_n({\mathbb Z}[X])$$
In other words, for any matrix of size $4x4$ or greater, you can, by applying elementary operations, reduce to a matrix of the form 
$$\pmatrix{A&0\cr 0&I\cr}$$
where $A$ is 3 by 3 and $I$ is the $(n-3)\times (n-3)$ identity matrix.
It's not immediately clear to me how much better you can do, but I bet either that Wilberd van der Kallen (who shows up here occasionally) or Vaserstein (who I think does not) could answer this in his sleep.  You might want to email one or both of them.<|endoftext|>
TITLE: Trivialisation of vector bundles on Stein spaces
QUESTION [10 upvotes]: Does every vector bundle on a Stein space have a finite local trivialisation?
Definitions:

Stein space means either a complex analytic Stein space or a nonarchimedean Stein space in the sense of Kiehl.
Vector bundle means either holomorphic vector bundle or rigid analytic vector bundle (= locally free sheaf). Edit: Say, of constant rank $r$.
finite local trivialisation = local trivialisation where the covering consists of finitely many sets.

In the nonarchimedean case, assume moreover that the base field is spherically complete or that the Stein space is unbounded.
My thoughts so far: Usually, the space is noncompact, so the answer is probably "no". On the other hand, I could not come up with an example yet. If the Stein space is the analytification of an algebraic variety, then at least those vector bundles that are analytifications of algebraic bundles will have a finite trivialisation. Every vector bundle on a Stein space is determined by finitely many global sections. Could that help? For an example, one should probably use the non-finitely generated ideals in the algebra of global functions of the Stein space. I cannot write down any of them either.
As said, I suspect the answer to be "no". It would be great if somebody knew an example or a reference.

REPLY [4 votes]: I think that it is also possible to give a positive answer to the question using purely topological arguments, namely Ostrand's theorem: for every paracompact space $X$ of dimension $n$ and every open covering $\mathscr U$ of $X$, there exists $n+1$ families $\mathscr V_1,\dotsc, \mathscr V_{n+1}$ of disjoint open sets whose union is a covering that refines $\mathscr U$. 
Apply this to a covering trivializing your vector bundle $E$. Then, for every $i$, $E$ is trivial on the open set that is the union of the elements of $\mathscr V_i$, and you are done.
Note that the argument does use the fact that the space is finite-dimensional, but not so much the Stein hypothesis (maybe for paracompactness).<|endoftext|>
TITLE: Embedding planar graphs into the grid
QUESTION [9 upvotes]: I've seen the following lemma in a paper. The result is by Valiant.
A planar graph $G$ with maximum degree $4$ can be embedded in the plane using $O(|V|)$ area in such a way that its vertices are at integer coordinates and its edges are drawn so that they are made up of line segments of the form $x=i$ or $y=j$, for integers $i$ and $j$.
I have two questions. 

Although not explicitly stated i assume the embedding is planar too?
Is there anything regarding the shape of the area that the graph is embedded into other than the fact it's area is $O(|V|)$ ? More specifically can we for example ensure that the graph can be embedded in a $|V|\times |V|$ grid?

REPLY [3 votes]: There is a huge literature on this topic. Search for "orthogonal graph drawing". The best possible area bound is $O(n^2)$.<|endoftext|>
TITLE: Surreal compactness
QUESTION [9 upvotes]: In a comment here, Joel David Hamkins said:

...I think perhaps every set-sized open cover of a bounded interval in the surreals has a finite subcover, but there are proper class open covers with no set-sized subcover. (But I have to think more about this to be sure.) –  Joel David Hamkins 15 hours ago 

What are references (or proofs) for those two things?
I'll provide the second one, modulo the first one.
So: assume any set-size open cover of a closed interval admits a finite subcover.
Let $\omega$ be as usual, in particular
$\omega > n$ for all $n \in \mathbb N$.  Consider the closed interval
$$
[0,1] := \{x \in \mathbf{No}\;:\; 0 \le x \le 1\} .
$$
A proper-class open cover can be made up of all open intervals
$$
\left]a-\frac{1}{\omega}, a+\frac{1}{\omega}\right[
$$
where $a$ ranges over the proper class $[0,1]$.  But of course no finite set of
those covers $[0,1]$, since each of them can contain at most one rational number and $[0,1]$ has infinitely many rationals in it.
Thus we are left with:
Question:

Let $[0,1]$ be the closed interval in $\mathbf{No}$ defined above.  Is it true that any cover of $[0,1]$ by a set of open intervals admits a finite subcover?

REPLY [7 votes]: Suppose $\mathcal{A}$ is a set-size open cover of $[0,1]$ with open intervals. Without loss of generality, $\mathcal{A}$ is closed under overlapping unions (i.e. the union of any two elements of $\mathcal{A}$ that happens to be an open interval is again in $\mathcal{A}$). Let $B$ be the set of right endpoints $b$ of all intervals $(a,b) \in \mathcal{A}$ such that $a < 0$. We must show that $B$ contains a number bigger than $1$.
Let $\gamma$ be a sufficiently large ordinal, at least larger than all the birthdays of endpoints of intervals from $\mathcal{A}$ and larger than $\omega$. Let $c$ be the simplest number such that $b \leq c$ for every $b \in B$ and $c < d$ for every upper bound of $B$ born before time $\gamma$. If $B$ doesn't contain a number bigger than $1$, then certainly $c \leq 1$ since $1$ was born before time $\gamma$. So $c$ must be covered by an interval $(a,b)$ from $\mathcal{A}$. Since $a$ was born before time $\gamma$, it must be that $a < b'$ for some $b' \in B$. But then if $(a',b')$ is an interval from $\mathcal{A}$ with $a' < 0$ then $(a',b') \cup (a,b) = (a',b)$ shows that $b \in B$, which is impossible since $c < b$. From this contradiction, we conclude that $B$ must contain an element greater than $1$.

REPLY [5 votes]: $\newcommand\No{\text{No}}$
Here is a proof of the second claim, that there is a proper class-sized open
cover of the surreal init interval with no set-sized subcover.
In particular, there is no finite subcover. (Note that your open cover does admit a set-sized subcover.)  The point is that $\No$ is not order-complete with respect to
proper class subsets. To see this, build two sequences
$$0=a_0<a_1<\cdots<a_\alpha <\cdots\cdots<b_\alpha<\cdots <b_1<b_0=1$$ of
length Ord, with no surreal number filling cut between them. Given
$a_\alpha< b_\alpha$, we can let $a_{\alpha+1}$ be the average of
$a_\alpha$ and $b_\alpha$, and $b_{\alpha+1}$ the average of
$a_{\alpha+1}$ and $b_\alpha$. At limits, the limits are not
equal, since every set-sized cut in $\No$ is filled with many
distinct surreal numbers, and so we can find $a_\lambda<b_\lambda$
in between, specifically $a_\lambda=\{a_\alpha\mid b_\alpha
\}_{\alpha<\lambda}$ and $b_\lambda=\{a_\lambda\mid b_\alpha\}_{\alpha<\lambda}$. Note that the birthdays of
these points $a_\alpha$ and $b_\alpha$ are strictly increasing as
$\alpha$ increases, and furthermore, the interval
$(a_\alpha,b_\alpha)$ contains no surreal numbers with earlier
birthdays, since $b_\alpha$ is the successor of $a_\alpha$ among
the surreals that are born on that day or earlier. It follows that
there can be no surreal number $z$ with $a_\alpha<z<b_\alpha$ for
all $\alpha$, because it would have to be born at some ordinal
stage $\alpha$, and then it would contradict the claim that
$(a_{\alpha+1},b_{\alpha+1})$ has no surreals with earlier
birthday.
Thus, let $U_\alpha=[0,a_\alpha)\cup(b_\alpha,1]$. This is a
proper-class open cover of $[0,1]$, but it has no set-sized
subcover, since any set of $U_\alpha$'s would miss out on the
later $a_\beta$'s.
Update. Regarding the open cover property, there are some subtle set/class issues here to be found. 
François has explained that every set-sized open cover of $[0,1]$
consisting of intervals admits a finite subcover, and it follows
immediately from this that every set-sized open cover consisting
of set-sized unions of intervals also admits a finite subcover,
since one may simply take the set of all intervals appearing in
any such union as another open cover and reduce to the finite
subcover.
But let me show that there are countably many open proper classes
$U_n$ for $n\in\omega$, such that $[0,1]\subset\bigcup_n U_n$, but
no finitely many $U_n$ cover $[0,1]$. In fact, we will arrange
that the $U_n$ are disjoint open classes, and so we will have an
open partition into open proper classes.
To see this, we use the fact that cuts of the type I constructed
above appear densely in the surreal numbers, that is, proper class
cuts $Z$ for which there is a lower sequence $a_\alpha$ converging
up to $Z$ and an upper sequence $b_\alpha$ converging down to $Z$,
with no surreal number filling this cut. Specifically, the reader
may construct countably many such cuts $Z_n$, each with a lower
sequence $a^n_\alpha$ converging up to it, and an upper sequence
$b^n_\alpha$ converging down to it, so that there is no surreal
number $z$ filling the cut, with $a^n_\alpha<z<b^n_\alpha$ for all
$\alpha$. We may construct these cuts so that $0<Z_0<Z_1<\cdots
<Z_n<\cdots<1$, meaning that the sequences $a^n_\alpha, b^n_\alpha$ are
all separated within the surreal unit interval $[0,1]$, and we may
assume furthermore that the upper and lower sequences for each
such cut do not overlap or interfere with one another.
Now, let $U_n=(Z_n,Z_{n+1})$, meaning the proper class open union $U_n=\bigcup_\alpha (b^n_\alpha,a^{n+1}_\alpha)$, which is a convex class in the surreals, but the endpoints are not realized in $\No$. Adding the bottom and top, let $U_{-1}=(-2,Z_0)\cup(\sup
Z_n,2)$, where by this latter interval, we mean the surreal
numbers above all the $b^n_0$'s and below $2$. This is an open
interval, since no increasing $\omega$-sequence converges.
Since the cuts $Z_n$ are not filled by any surreal number, it
follows that $[0,1]\subset \bigcup_n U_n$, and so we have a
countable open cover of the unit interval consisting of open
proper classes. But there is no finite subcover, since in fact
these open classes form a partition of $[0,1]$ into open classes.
So this is a sense in which the surreal numbers are connected,
with respect to separations into disjoint open sets, but it is
disconnected, if one allows the separating open classes to be
proper classes.<|endoftext|>
TITLE: Smash product of spheres in $\mathbf{SH}$ and product in cohomology
QUESTION [6 upvotes]: I have two very concrete and simple question. Just in case I write downwards what led me into this.
My questions: Let $\mathbf{SH}(X)$ be the stable homotopy category of Voevodsky. Denote $S^n$ the simplicial $n$-sphere and also the spectrum it defines by infinite suspension.
1.- Is the following diagram commutative?
$$
\begin{matrix}
S^a\wedge S^{b}&\buildrel{\sim}\over{\to} & S^{a+b}\\
\sigma \downarrow  & &\downarrow ^{(-1)^{a+b}}\\
S^b\wedge S^a &\buildrel{\sim}\over{\to} & S^{a+b}
\end{matrix}
$$
where $\sigma$ denotes the permutation.
2.- If so, can you give a proof or a concrete reference?
Thanks in advance.

What led me to this question? Recall from Cisinki and Deglise that the algebraic de Rham cohomology is represented by a commutative ring spectrum $E_{\mathrm{dR}}$, in other words:
$$
H^p(X,\Omega_X^\bullet)=\mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p]).
$$
Note that commutative spectrum means that there is a product $\mu \colon E\wedge E\to E$ satisfying the commutative diagram if we switch the source. We know that the algebraic de Rham cohomology is anticommutative or graded commutative, (because the product comes from the wedge product on differential forms) In other words, $\alpha \in H^p(X,\Omega_X^\bullet)$ and $\beta \in H^q(X,\Omega_X^\bullet)$ then $\alpha \cdot \beta =(-1)^{p+q}\beta\cdot \alpha$. 
Let me recall the definition of product in the $E$-cohomology. Let $\alpha\in \mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p])$ and $\beta \in \mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p'])$, then 
$$
\alpha\cdot \beta \colon 1_X\to E_{\mathrm{dR}}[p]\wedge E_{\mathrm{dR}}[p']\simeq E_{\mathrm{dR}}\wedge E_{\mathrm{dR}}\wedge 1_X[p+p']\buildrel{\mu}\over{\longrightarrow}E_{\mathrm{dR}}\wedge 1_X[p+p']\simeq E_{\mathrm{dR}}[p+p']
$$
Recall that $1_X[p]=S^p\wedge 1_X$, in order for this product to be anticommutative or graded commutative the diagram 
$$
\begin{matrix}
1_X[p]\wedge 1_X[p']&\buildrel{\sim}\over{\to} & 1_X[p+p']\\
\sigma \downarrow  & &\downarrow (-1)^{p+p'}\\
1_X[p']\wedge 1_X[p]&\buildrel{\sim}\over{\to} & 1_X[p+p'],
\end{matrix}
$$
where $\sigma$ denotes the permutation, must commute. Right?

REPLY [6 votes]: Thanks to Marc Hoyois which pointed the answer in a comment above. I am just writting that with more details: 
1.- Yes, the diagram is commutative. 
2.- This already happens in classic stable homotopy category. One may find the concrete statement in Adam's "Stable homotopy and generalised homology" at the beggining of section III.4   It is not anything particular from the product of spheres $S^n$ but it is a general fact from the smash product of spectra. Since this reference is hard to find in electronic format I copy the page here 
(One obtains our diagram of 1 for classic spectra out of the lower right diagram of Adam's page. Note that there is a missprint on the upper right corner, it should be $Y\wedge X$ and not $X\wedge Y$). Thanks everyone for the comments and help.<|endoftext|>
TITLE: Pseudo-holomorphic curves in the six-sphere
QUESTION [22 upvotes]: Equip $S^6$ with the almost complex structure coming from the cross product on $\mathbb R^7$ (i.e. the product on the pure imaginary octonions).  What is known about the psudo-holomorphic curves in this $S^6$?
There are certainly a bunch of holomorphic $S^2$'s coming from the inclusions $\mathbb R^3\to\mathbb R^7$ compatible with the cross product.  If these $S^2$'s are cut out transversally (seems likely), then one can start gluing them together at intersection points to get lots of other examples.  Are there any other natural constructions?
Note that this gluing construction of lots of $S^2$'s (which are all null-homologous) shows that the moduli space of (even connected) pseudo-holomorphic curves is very non-compact (glue together arbitrarily many spheres).  Of course, this does not contradict Gromov's compactness theorem since this almost complex structure $S^6$ is not tamed by a symplectic form (since $H^2(S^6;\mathbb R)=0$).

REPLY [29 votes]: It might be immodest of me to mention my own work, but there is quite a lot known about the pseudo-holomorphic curves in $S^6$.  For example, it is known that the pseudoholomorphic rational curves are all algebraic and have area a multiple of $4\pi$ (and, yes, there are many of them). (It's true that the moduli space is not compact; however, because the area is quantized, the rational curves of a given area, can, in fact, be compactified to yield a compact moduli space, even an algebraic one.)  As another example, every compact Riemann surface occurs as a pseudoholomorphic curve (possibly ramified) in $S^6$.  (A student of Kevin Corlette, some years ago, proved that you can actually get every Riemann surface as an unramified pseudoholomorphic curve in $S^6$.)
For example, see my paper Submanifolds and special structures on the octonians, Journal of Differential Geometry 17 (1982), 184–232.  (I regret very much the misspelling 'octonians' that permeates that paper.)
Added references:  (January 2020) Recently, I was asked some questions about pseudo-holomorphic curves in $S^6$, and that inspired me to look at some of the literature that has been generated in the nearly 40 years since I wrote the paper above.  A fair amount has been worked out, and anyone who wants to know more about the subject would benefit from consulting a few of the following papers:

J. Bolton, L. Vrancken, and L. Woodward, On almost complex curves in the nearly Kähler $6$-sphere, Q. J. Math. Oxford (2) 45 (1994), 407–427.
H. Hashimoto, Deformations of super-minimal $J$-holomorphic  curves of a $6$-dimensional sphere,  Tokyo J. Math. 27 (2004), 285–298.
J. K. Martins, Superminimal surfaces in the $6$-sphere, Bull. Braz. Math. Soc. (N.S.) 44 (2013), 25–48.
L. Fernández, The space of almost complex $2$-spheres in the $6$-sphere,
Trans. Amer. Math. Soc. 367 (2015), 2437–2458.
M. Dajczer and T. Vlachos, A representation for pseudoholomorphic surfaces in spheres, Proc. Amer. Math. Soc. 144  (2016), 3105–3113.

This is by no means a complete list of relevant articles, but these do cover a range of interesting developments of the results in my 1982 JDG paper.<|endoftext|>
TITLE: On determinants formed by binomial coefficients
QUESTION [23 upvotes]: Let $q$ be a number. Let us consider the $q^2-1$-th line of the Pascal triangle (i.e. numbers ${{q^2-1} \choose i}$, $i=0,1,...q^2-1$). We have $q^2$ numbers. 
Let us form naively a $q \times q$ matrix from them. Example for $q=2$:
\[\begin{pmatrix} 1 & 3 \\
3 & 1 \end{pmatrix}\]
And let us take its determinant: $-8$ if $q=2$. 
How to prove that if $q$ is prime then $q$ enters in this determinant with 
$(q+4)(q-1)/2 $-th exponent?
Example: $q=2$, $(q+4)(q-1)/2 =3$. Really, $8=2^3$. 
Case $q=3$: the binomial coefficients are $1,8,28,56,70,56,28,8,1$.
The matrix: 
\[\begin{pmatrix} 1 & 8 & 28 \\
56 & 70 & 56 \\
28 & 8 & 1 \end{pmatrix}\]
The determinant is $2 \cdot 3^7 \cdot 7$. Really, $(q+4)(q-1)/2 =7$.

REPLY [13 votes]: For the specific determinant in the question there is a slightly more direct proof than Lemma 3 in Krattenthaler. We can evaluate it using the Jacobi-Trudi identity. Let's denote  $\lambda _i=(q-i+1)(q-1)$. The determinant in question, $\det\left(\binom{q^2-1}{qi-q+j}\right)$, is equal (up to sign) to the determinant $\det\left(e_{\lambda_i-i+j}\right)$ when you put all variables equal to $1$. But this is equal to the Schur polynomial of the conjugate partition $\lambda^{\star}=(q^{q-1}(q-1)^{q-1}\cdots 1^{q-1}0^{q-1})$ evaluated at $1,1,\dots,1$. (I added a bunch of zeros to the partition so that the number of parts matches the number of variables.)
There is a simple product formula for Schur polynomials evaluated at $1,1,\dots,1$, and in this case it gives
$$s_{\lambda^{\star}}(1,1,\dots,1)=\prod_{i<j}\frac{\lambda^{\star}_i-\lambda^{\star}_j+j-i}{j-i}.$$
The denominator has the form $1!2!\cdots (q^2-2)!$ so the highest power of $q$ dividing it is $q^{\frac{(q-1)(q^2-2)}{2}}$. For the numerator, notice that $\lambda^{\star}_i-\lambda^{\star}_j+j-i$ is divisible by $q$ iff $j-i$ is divisible by $q-1$. This means that exactly $(q-1)\binom{q+1}{2}$ factors in the numerator are divisible by $q$. Out of these the pairs $(i,j)=(r,q^2-q+r)$ for $r\in\{1,2,\dots,q-1\}$ are divisible by $q^2$. So the power of $q$ dividing your determinant is exactly
$$(q-1)\binom{q+1}{2}+(q-1)-\frac{(q-1)(q^2-2)}{2}=\frac{(q-1)(q+4)}{2}$$
as desired.<|endoftext|>
TITLE: Are the Sasaki metrics on tangent and cotangent bundle isomorphic?
QUESTION [6 upvotes]: Let $(M,g)$ be a Riemannian manifold. Then there is the well-known
Sasaki metric that makes $(TM,\hat{g})$ a Riemannian manifold. In a
similar way, one can construct a Sasaki metric $\bar{g}$ on the
cotangent bundle. My question is, is the map
$g\colon (TM,\hat{g}) \to (T^*M,\bar{g})$ an isometry?
For completeness, the Sasaki metrics are given as follows (not 100%
sure about the cotangent one). Let $X,Y$ be vector fields on $M$ and
$\alpha,\beta$ be one-forms on $M$, let $\pi$ be the projection from
both bundles onto $M$ and let superscripts $h$ and $v$ denote
horizontal and vertical lifts. Then these Sasaki metrics are uniquely
defined by the following conditions:
$$
\hat{g}(X^h,Y^h) = \hat{g}(X^v,Y^v) = g(X,Y) \circ \pi, \quad \hat{g}(X^h,Y^v) = 0
$$
and
$$
\bar{g}(X^h,Y^h) = g(X,Y) \circ \pi, \quad
\bar{g}(\alpha^v,\beta^v) = g^{-1}(\alpha,\beta) \circ \pi, \quad
\bar{g}(X^h,\alpha^v) = 0.
$$

REPLY [6 votes]: The answer is yes, and you can see this by realizing that the vertical (resp. the horizontal) subbundle of $TTM$ is mapped to the vertical (resp. the horizontal) subbundle of $TT^*M.$ For the vertical part, this follows from the fact that $g\colon TM\to T^*M$ is a bundle homomorphism, and for the horizontal bundle this follows from the fact that $g\colon TM\to T^*M$ is parallel (with respect to the natural connections defined by $g$ which are used to define the horizontal bundles).<|endoftext|>
TITLE: Stiefel-Whitney class of fibre bundles
QUESTION [12 upvotes]: Let $F,E,B$ be manifolds (we may assume them to be compact, without boundary if necessary) and $$F\to E\to B$$ be a fibre bundle. Suppose $$w(F), w(B),$$ the Stiefel-Whitney class of $F$ and $B$, are known. I notice that in particular, if the bundle is trivial, then $E=B\times F$ and $w(E)=w(B)w(F)$. 
Question: In general, are there any formulas to compute the Stiefel-Whitney class of the total space $E$
$$
w(E)
$$ 
in terms of $w(B)$ and $w(F)$? Or even in terms of the cohomology ring 
$$
H^*(B;\mathbb{Z}_2), H^*(F;\mathbb{Z}_2)
$$
And other factors?

REPLY [8 votes]: Just to get started (a bit long for a comment):
For a fiber bundle $F\rightarrow E\overset{\pi}{\rightarrow}B$ one has an exact sequence of bundles over $E$
$$
0\rightarrow T_vE\rightarrow TE\rightarrow \pi^{*}TB\rightarrow 0.
$$
Here $T_vE:=\ker T\pi$ is the vertical part of the tangent space to $E$, i.e. these are the tangent vectors to the fibers. The choice of a complementary bundle amounts to the choice of a connection, but can be identified with $\pi^*TB$. So one sees that 
$$
w(TE)=w(T_vE)\pi^*(w(TB)).
$$
I think one needs more information if one wants to continue the computation. For example one can go on if $E$ is the projectivization of a vector bundle over $B$.<|endoftext|>
TITLE: Is the Brauer group functor a Zariski sheaf?
QUESTION [13 upvotes]: For any scheme $X$, let $\operatorname{Br}X$ denote the (Azumaya) Brauer group of $X$, namely the Morita equivalence classes of Azumaya $\mathcal{O}_{X}$-algebras.

Is the functor $$\operatorname{Br} : \operatorname{Sch}^{\operatorname{op}} \to \operatorname{Ab}$$ sending $X \mapsto \operatorname{Br}X$ a sheaf for the Zariski topology on $\operatorname{Sch}$?

In other words, if $X = \bigcup_{i \in I} U_{i}$ is a Zariski open covering of a scheme $X$, is the sequence $$0 \to \operatorname{Br}X \to \prod_{i \in I} \operatorname{Br}U_{i} \to \prod_{i_{1},i_{2} \in I} \operatorname{Br}(U_{i_{1}} \cap U_{i_{2}}) $$ exact?
Thoughts: My naive guess is "no" since $\operatorname{Br}$ is not an etale sheaf and $\operatorname{Pic}$ is not a Zariski sheaf. Exercise 8(f) of http://www-personal.umich.edu/~bhattb/teaching/mat731fall2011/ex6.pdf shows that the functor $U \mapsto \mathrm{H}_{et}^{2}(U,\mathbb{G}_{m})$ is a Zariski sheaf if $X$ is regular Noetherian. See also the discussion on Milne's "Etale Cohomology", IV, Remark 2.10.

REPLY [17 votes]: No, $\mathrm{Br}$ is not a Zariski sheaf: it is possible for a non-trivial Azumaya algebra on a variety to become trivial when restricted to a Zariski cover.  This can happen even for a normal surface with rational singularities.  Some references:

M. Ojanguren, "A non-trivial locally trivial algebra", J. Algebra 29, 1974
F. DeMeyer and T. Ford, "Nontrivial, locally trivial Azumaya algebras", Contemporary Mathematics 124, 1992
My answer to this question and more generally this article.<|endoftext|>
TITLE: Basic questions about simplicial commutative rings
QUESTION [5 upvotes]: I am trying to learn about simplicial commutative rings, and would be grateful if one can help with some basic facts about them. Basically, I would like to understand how to do homological algebra over a simplicial ring.

Let $A$ be a simplicial commutative ring. 
Is the category of simplicial modules over $A$ abelian?
If so, does its derived category exist?
Does any simplicial module have a projective resolution? a flat resolution? an injective resolution?
Consider the associated DG-algebra $N(A)$. Is there any relation between the derived 
category of DG-modules and the derived category I was hoping exists in 3 above?
Any references that discuss these basic issues?

REPLY [6 votes]: Yes, there are many references that discuss this. The first is Quillen's Homotopical Algebra. Chapter II contains much of what you're asking about, especially II.4 and II.6
For a more modern version, see Schwede-Shipley Algebras and Modules in Monoidal Model Categories. Section 5 is all examples and contains precise references in Quillen's work above.
This covers your (2), (3), and (6). I agree with Zhen Lin that (4) is the wrong question to ask, since the structure as an abelian category is less important than the model structure if you are trying to study the homotopy category from (3). For (5), check out the Dold-Kan theorem. A nice write-up is in Goerss-Schemmerhorn Model Categories and Simplicial Methods 4.1. This also includes the model structure on simplicial R modules and what resolutions look like for simplicial R modules, answering your (4) in the correct homotopically meaningful sense. It is clear from 4.1 that $N$ is both an equivalence of categories (if by DG we mean non-negatively graded) and a Quillen equivalence, so the homotopy theories agree, and this answers your (5)<|endoftext|>
TITLE: Statistical independence and the Fourier transform
QUESTION [7 upvotes]: Consider the random variable $S=(s_0, \dots ,s_{N-1})$, a sequence of signs uniformly distributed on the hypercube $\{-1,1\}^N$. With the Fourier transform we can define $N$ random walk variables
$$
\hat{s}_q=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} \zeta^{q k} s_k, \qquad q=0, \ldots, N-1,
$$
where $\zeta=e^{2\pi i/N}$ and to keep things simple we take $N$ to be prime. As we take $N\to\infty$ all of these have normal distributions: $\hat{s}_0$ on the real line and the others in the complex plane. 
I would like to be able to claim that these random variables become independent as $N\to\infty$ in a certain qualified sense. First, since $\hat{s}_q$ and $\hat{s}_{-q}$ are complex conjugates we only consider $q\in\{1,\ldots,(N-1)/2\}$. Second, I am only interested in "fixed-information" counterparts of these. We define these random variables as $f(\hat{s}_q)$, where $f$ maps the complex plane to a finite set. For example (in fact the case that generated this question), $f$ might be take the magnitude of the complex number and round it to 7 significant digits, or to $\infty$ if the magnitude exceeds a certain bound (in order to keep the cardinality of the range of $f$ fixed as $N\to\infty$).
Is it true, say for particular kinds of fixed-information maps, that $f(\hat{s}_q)$ and $f(\hat{s}_{q'})$, $q\ne q'$, are independent as $N\to\infty$?

REPLY [2 votes]: With help, offline, from P. Diaconis:
I believe my "fixed information" map is really just a license to take the limit $N\to \infty$ before asking if the random variables are independent. Now, if I exercise this license, I notice that $\hat{s}_q$ and $\hat{s}_{q'}$ have a bivariate normal distribution (in the limit), and when I work out the covariance matrix, which I can do for finite $N$, I see that indeed the two variables are independent for $q\ne q'$ (taking the $N\to \infty$ limit of the covariance matrix).<|endoftext|>
TITLE: Mochizuki's "phenomena in number theory" outside the scope of Langlands
QUESTION [28 upvotes]: (Crossposted from math.stackexchange by suggestion)
On page 12 of Shinichi Mochizuki's "On the Verification of Inter-universal Teichmuller Theory: A Progress Repor", he writes

"The representation-theoretic approach exemplified by the Langlands
  program does indeed constitute one major current of research in modern
  number theory. On the other hand, my understanding is that the idea
  that every essential phenomenon in number theory may in fact be
  incorporated into, or somehow regarded as a special case of, this
  representation-theoretic approach is simply not consistent with the
  actual content of various important phenomena in number theory."

Does anyone know what these 'various important phenomena' are that he's referring to?

REPLY [30 votes]: As noted by Lucia, large parts of number theory are completely "beyond the scope of the Langlands program": most of analytic number theory, obviously, is, but also many important, active and beautiful subfields of algebraic number theory -- for a list of examples, see for instance the list of publications of Bjorn Poonen, which cover a large scope of subjects in algebraic number theory, but touch the Langlands program only tangentially.
To make sense of the proposed quote of Mochizuki, it is therefore necessary to give a much more restrictive interpretation of "number theory",
which I imagine would be in this context the set of number-theoretic questions that can be solved or at least attacked by understanding the Galois groups of number fields, together with all its attached tractors of decomposition groups, inertia subgroups, and Frobenius elements. This set contains all the reciprocity laws, proved or conjectured, and many diophantine equations and problems, from Fermat's last theorem to Mordell's conjecture and its generalizations, through Birch and Swinnerton-Dyer. 
My guess is that Mochizuki means that even in this relatively restricted sense, many phenomenons of number theory lie beyond the scope of Langlands. In saying so, he is following an idea that Grothendieck tried to disseminate in the 80's ("avec force", as Deligne said), which I will try to summarize as follows. The Langlands program tries to understand Galois groups by looking at their representations, that is, their action over vector spaces over fields or slightly more generally over modules over commutative rings. At any rate, these objects form an abelian category (even Tannakian), a linear object. Grothendieck's theory of Motives
should be attached to this general program, because the motives also
are  "linear objects", and as Langlands himself famously argued in his famous 1979 paper "Automorphic Representations, Shimura Varieties, and Motives. Ein Märchen".
But, Grothendieck argues, there are many other kind of objects on which we may let Galois group act in order to study them in a different direction than with representations: sets, for instance, as in the version of Galois theory developed in SGA 1 (1960), or non-abelian (profinite) groups, as the étale fundamental groups of various variety over $\mathbb Q$: to this aspect belong all the theory of Grothendieck-Teichmüller, the Dessins d'enfants, the anabelian geometry. That is of course a subject in which Mochizuki has himself proved some remarkable results, and which he claims to have enormously developed up to the point of deducing the ABC conjecture.
An example of a number theoretic (in the restricted sense mentioned above) problem that arguably lies beyond the Langlands program but that 
the practicians of anabelian geometry plans to study successfully is Mordell's conjecture (proved by Faltings, it is true, with methods not so far from the Langlands program, at least in that they study "linear objects", namely abelian varieties) and all its generalizations (still open). For a very interesting, if speculative, discussion of these questions, see "Galois Theory and Diophantine Geometry" by Minhyong Kim.

REPLY [9 votes]: The basic idea is that sometimes arithmetic objects aren't "nice enough" for the tools that support the Langlands program (representation theory, noncommutative harmonic analysis, etc) to work.
They work fine, pretty much by definition, as long as the arithemtic object is essentially linear ($GL_n(\mathbb{A})$, cohomologies, Galois groups...).
But when you consider the kind of highly non-linear arithmetic objects (étale fundamental groups, moduli spaces...) that anabelian geometry studies, those theories just aren't meant to apply.
I would recomend section 1.3.3 of "Survey of the Hodge-Arakelov Theory of Elliptic Curves I" (pages 17 and 18), called "The Meaning of Nonlinearity", for a concrete (yet philosophical) example of this kind of reasoning in Mochizuki's own words.
For a pre-anabelian example of the perceived limitations of the Langlands tools and techniques, notice that after all these years, no natural proof (from the point of view of the theory of automorphic representations) exists of the Ramanujan conjecture for classical modular forms.<|endoftext|>
TITLE: Who needs RCS iterations?
QUESTION [15 upvotes]: According to this paper of Chaz Schlindwein, any countable support iteration of semi-proper forcings is semi-proper.  This seems like a breakthrough simplification, and I wonder why it is not more well-known.  Before diving into the arguments I want to ask the forcing community whether this work is widely viewed as correct?  Thanks for your help.

REPLY [13 votes]: Following up on Martin's answer:
Definition 10 of the paper defines the concept "hemi-properness", shows that semi-proper forcings are hemi-proper, and then claims (Theorem 16) that hemi-properness is preserved under countable support iterations.  This last is demonstrably false, as we will show that "hemi-properness" is equivalent to "preserves $\omega_1$", and it is well known that this property is not preserved by countable support iteration.
Definition [Definition 10 from the paper]
A poset $P$ is hemi-proper if and only whenever $\lambda$ is a sufficiently large regular cardinal, $M$ and $N$ are countably elementary submodels of $H(\lambda)$, $P\in M\in N$, and $q\in P\cap M$, then 
$q\nVdash``\omega_1\cap M[G_P]>\omega_1\cap N$''.
Lemma 11 in his paper shows that hemi-proper forcings do not collapse $\omega_1$.
Claim:  If forcing with $P$ preserves $\omega_1$, then $P$ is hemi-proper.
Let $\lambda$, $M$, $N$, and $q$ be given.  Any condition forces that $M[G_P]$ is countable (it is just the interpretations of the countably many $P$-names living in $M$).  In particular, any condition forces that $M[G_P]\cap\omega_1$ is an ordinal $\alpha<\omega_1$ since $\omega_1$ is preserved.  The model $N$ can see $M$, and so $N$ will contain a name $\dot\alpha$ for the countable ordinal $M[G_P]\cap\omega_1$ from $M[G_P]$. In $N$, we can extend $q$ to a condition $r$ deciding a particular value $\alpha$ for $\dot\alpha$.  Notice that since $r$ and $\dot\alpha$ are in $N$, the countable ordinal $\alpha$ is in $N$ as well.  Necessarily we have $\alpha<N\cap\omega_1$.  Since $r$ extends $q$, it follows that $q$ did not force $\omega_1\cap M[G_P]$ to be greater than $N\cap \omega_1$.  This establishes that $P$ is hemi-proper.
Schlindwein lectured on this at a few meetings in the 1990s. The paragraph at the top of page 9 was added in an attempt to meet my objections that his notion was the same as "preserves $\omega_1$'', but there are no such $M$ and $N$ as he describes there.
Chaz was a nice guy. I'm sad to hear the news that he has passed away.<|endoftext|>
TITLE: Intertwiners and Clebsch-Gordan coefficients
QUESTION [8 upvotes]: Consider two unitary irreducible representations on vector spaces $V_1$ and $V_2$ of a Lie group $G$. For $G$ is compact and $V_1$ and $V_2$ finite dimensional there is a unique decomposition of $V_1 \otimes V_2$ into a set of irreducible representations $V_k$, $k \in K$. The Clebsch-Gordan coefficients can then be viewed as intertwiners, so $G$-equivariant maps from $V_1 \otimes V_2 \to V_k$. Conversely, I think that nontrivial intertwiners $V_1 \otimes V_2 \to V_m$ can only be nonzero if $m \in K$ and then they are essentially the Clebsch-Gordan coefficients.
Now if $G$ is non-compact (but $V_1$ and $V_2$ still unitary) then it seems that I can define intertwiners from $V_1 \otimes V_2$ to another $V_m$ which does not appear in the decomposition of $V_1 \otimes V_2$. For example, if $V_1$ and $V_2$ are in the principal series representation of $SL(2,\mathbb R)$ then I can obtain such intertwiners by analytic continuation of the Clebsch-Gordan coefficients.
I was first of all wondering if the above story is true. If so, is there a way to understand the domain of these intertwiners? Do we understand for which triplet of representations they exist? Is there a general theory of their properties? If not, maybe just for $SL(2,\mathbb R)$?

REPLY [2 votes]: I have some trouble understanding notation (what is $K$?), so here is a very general answer. 
The case of ${\rm SL}(2,\Bbb{R})$ was studied in the old work of Repka and, if I am not mistaken, described in Lang's eponymous book and Lyon-Vergne's book. Much more is known, in particular, through the theory of theta correspondence in the case of classical groups. You may want to focus your question more.<|endoftext|>
TITLE: Logarithmic weights on number theoretic sums
QUESTION [6 upvotes]: Suppose we are interested in the sum
$\sum _{n\leq x}a_n.$
The study of the sum
$\sum _{n\leq x}a_n\log (x/n)$
may be easier.
What can one say about the first sum from knowing the behaviour of the second?  
In the case I have in mind, I have
$\sum _{n\leq x}a_n\log (x/n)=x+\mathcal O\left (x^{1/2}\right )$
and would like a similar result for the sum without weights.  
How feasible is this?  I'm quite sure an asymptotic formula holds for the sum without weights, but I don't know if I should expect to lose a power saving.  (Logarithms and epsilons in the error are not of concern.)
I'm not sure which properties exactly of the $a_n$ are important.  It may be useful that they are all non-negative.  
Any pointers as to what one should expect in such a situation would be very much appreciated.  
If it seems that one can't really say much without knowing at least something more about the $a_n$ I can give some more details.  But I think my problem has more to do with the fact that I'm lacking some general principles in dealing with weights, so I'll leave the $a_n$ arbitrary for now.
Thanks very much in advance.

REPLY [7 votes]: Put $A(x) =\sum_{n\le x} a_n$, and $B(x) =\sum_{n\le x} a_n\log x/n$.  Then 
$$ 
B(x) = \int_1^x A(t)\frac{dt}{t}. 
$$ 
So information about $A(x)$ readily translates to information about $B(x)$ and there is no loss since integration (which makes things smoother) is involved.  But to pass from $B(x)$ to $A(x)$ we need to differentiate, and based on the situation this could be either impossible, or could involve some loss.  
Suppose first that the $a_n$ are bounded.  Then 
$$ 
B(x+h) - B(x) = \int_{x}^{x+h} A(t) \frac{dt}{t} = (A(x)+O(h)) \log\frac{x+h}{x}, 
$$ 
and choosing $h=x^{3/4}$ we obtain $A(x)=x+O(x^{\frac 34})$.  This was noted in Alpoge's comment above, and also holds if the $a_n$ are given to be non-negative rather than bounded.  However there is a loss in this argument and the $x^{3/4}$ error term cannot be improved, even for bounded non-negative $a_n$.  Here is an example:  Take $a_n=2$ if $n \in [m^4,(m^4+(m+1)^4)/2)$ for some integer $m$, and $a_n=0$ if $n\in ((m^4+(m+1)^4)/2,(m+1)^4)$.  Then you can check that $B(x) = x +O(x^{\frac 12})$ holds, but no estimate sharper than $A(x) = x +O(x^{\frac 34})$ can hold.  
Finally without some restrictions on $a_n$, one can say nothing at all.  For example, take $a_n=1+ (-1)^n n$.  Then you can check that $A(n)$ alternates between about $n/2$ and about $3n/2$, whereas $B(n)$ is very close to $n$.<|endoftext|>
TITLE: Braid group on 4 strands
QUESTION [15 upvotes]: Consider the braid group $B_4$ with generators
$a=\ $, 
$b=\ $
and  $c=\ $
Assume 
$$\alpha, \alpha'
\in
\langle a,c\rangle{\smallsetminus}(\langle c\rangle{\cdot}\langle a\rangle{\cdot}\langle c\rangle),$$
where $\langle c\rangle{\cdot}\langle a\rangle{\cdot}\langle c\rangle=\{\,c^k\cdot  a^l\cdot c^m\mid k,l,m\in\mathbb{Z}\,\}$ and
$$\beta, \beta'
\in
\langle b,c\rangle{\smallsetminus}(\langle c\rangle{\cdot}\langle b\rangle{\cdot}\langle c\rangle).$$
Is it true that
$$\alpha{\cdot}\beta{\cdot}\alpha'{\cdot}\beta'\ne 1?$$
P.S. I am also interested in algorithms for solving this and similar problems. (There are some issues with the suggestion of Lee Mosher.)

REPLY [8 votes]: This may be resolved by identifying $B_4$ with the mapping class group of the 4-punctured disk (fixing the boundary). I.e. $B_4\cong Mod(D^2-\{p_1,p_2,p_3,p_4\})$. I will consider isotopy classes of arcs in $D^2-\{p_1,p_2,p_3,p_4\}$ with endpoints on $\partial D^2$ and how they intersect each other, and the action of $B_4$ on the isotopy classes of arcs. 
In your pictures defining the generators of $B_4$, I'm assuming the points $p_i$ are numbered in order from top to bottom.  Choices of isotopy classes of arcs may be made by choosing a complete hyperbolic metric on the punctured disk so that the boundary is totally geodesic, and making the arcs geodesics. Then the intersection number between pairs of arcs will be minimized by these representatives. Given an arc $w$ and $\sigma \in B_4$, I'll abuse notation $\sigma(w)$ to mean the arc made geodesic after applying $\sigma$ to the arc $w$. 
Consider the arcs $x, y, z$: 

Then we have that $stab(y)=\langle a,c\rangle$, $stab(x)=\langle b,c\rangle$, $stab(x\cup y)=\langle c\rangle$, which may be seen again by regarding the braid group
as a mapping class group. 
What you would like to know is the intersection of the double
cosets $$U=( \langle a,c\rangle\cdot\langle b,c\rangle ) \cap ( \langle b,c\rangle\cdot\langle a,c\rangle ).$$
Suppose that we have $\gamma \in U$. Then one may see that $\gamma(x)\cap y =\emptyset$ and $\gamma(y)\cap x = \emptyset$. Moreover $\gamma(x)\cap \gamma(y)=\emptyset$ since $x\cap y=\emptyset$. Write $\gamma=\alpha \beta, \alpha\in \langle a,c\rangle, \beta \in \langle b,c\rangle$. Then $\beta(x)=x$ and $\alpha(y)=y$, so we see that  $\gamma(x)\cap y =\alpha\beta(x)\cap \alpha(y) =\alpha(x)\cap \alpha(y)=\emptyset$, and similar for the other intersection.  
Consider the subsurface $P\subset D^2$ obtained by cutting $D^2$ along $x \cup y$, and keeping the middle piece containing $p_2, p_3$. Then $P'=P-\{p_2,p_3\}$ is a twice-punctured disk, and there is only one isotopy class of essential arc with endpoints on $\partial P'$. Then each component of $\gamma(x)\cap P', \gamma(y)\cap P'$ must be isotopic to this arc (the case when one of the arcs is boundary parallel may be dealt with in a similar fashion). However, all but one component of $\gamma(x)\cap P'$ must have endpoints on $x\subset \partial P'$, and all but one component of $\gamma(y) \cap P'$ must have endpoints on $y\subset P'$. Up to composing $\gamma$ with an element of $\langle c\rangle = Mod(P')$, we must see a picture like this: 

Thus replace $\gamma$ by $c^{-i}\gamma$ if necessary to
get this configuration. We see now that $\gamma(x)\cap z=\emptyset$,
$\gamma(y)\cap z=\emptyset$. Now focus on the subsurface
$L\subset D^2$ obtained by cutting along $z$ and taking the
left piece containing $p_1, p_2$. There is only one isotopy
class of arc $x$ and $\gamma(x)$ in $L'=L-\{p_1,p_2\}$. But
the endpoints of these arcs lie in $\partial L' -z$, and hence
we see a picture like this: 

Thus we see that up to composing with a power of $a$ (which
generates $Mod(L')$), we may assume that $\gamma(x)=x$. 
Similarly, up to composing with a power of $b$, we may assume
that $\gamma(y)=y$. 
But now $stab(x \cup y) =\langle c\rangle$ in $B_4$. So
we've shown that $\gamma = c^i a^j b^k c^l$. 
Now we have $$\gamma = c^ia^jb^kc^l = \alpha \beta, \alpha\in\langle a,c\rangle, \beta\in \langle b,c\rangle.$$
Then $$\alpha^{-1} c^i a^j = \beta c^{-l} b^{-k} \in \langle a,c\rangle\cap \langle b,c\rangle = stab(x\cup y)=\langle c\rangle.$$
Thus $\alpha = c^i a^j c^{-m}, \beta =c^m b^k c^l$, as desired.<|endoftext|>
TITLE: Behaviour of $\zeta(1-it)/\zeta(1+it)$?
QUESTION [8 upvotes]: I am trying to understand the behaviour of
$$\int^\infty_{-\infty}\frac{\xi(1-it)}{\xi(1+it)}h(t)\frac{dt}{t}$$
where $h$ is a Schwartz function on $\mathbb R$, and $\xi(s)$ the completed Riemann zeta function. Clearly it is the quotient of zeta functions that is the most difficult to study.
One knows certain things about $\zeta(1+it)$, for example the pole at $t=0$ and the nonvanishing for all $t$. (See this question, Also the paper referenced in the answer to this question.) But what can we say about this quotient, at least on average, say?

REPLY [7 votes]: For zeta itself, there is a clear result, that eventually the argument of zeta on the line $\Re(s)=1$ becomes very regular. Some more sophisticated things around this are in 
D. Hejhal, "On a result of G. Polya concerning the
Riemann $\xi$-function", J. D'analyse Math. 55 (1990), 60-95.
Also, (5.14) of that paper recalls the consequence
$$
\log \zeta(1+iu) - \log \zeta(1+it) \;=\; O\Big({\log t \over \log\log
t}\Big)\cdot (u-t)
$$
for $u\ge t$ from [Titchmarsh~1986]. In an earlier edition, this was on page 98. In
the edition revised by Heath-Brown, this is (5,17.4) on page 112.
For more generaly automorphic $L$-functions, somewhere in Iwaniec-Kowalski they show that similar results follow for self-adjoint (data for) $L$-functions assuming sufficient GRH's ... but/and it is not clear to me whether/how the delicate-but-elementary arguments for zeta and maybe Dirichlet $L$-functions over $\mathbb Q$ would generalize...<|endoftext|>
TITLE: Essential Klein bottle in simply connected symplectic 4 manifolds
QUESTION [9 upvotes]: Consider the following question:
Let $X$ be a simply connected, symplectic 4-manifold. Does there exists a smoothly embedded Klein bottle $K\subset X$ such that the following conditions are both satisfied?
(1) The self intersection number of $K$ is 0 (which means that the normal bundle has a non-where vanishing section.)
(2) $[K]\neq 0\in H_{2}(X;Z_{2})$.
I am not aware of any kind of adjunction inequalities ruling out this possibility. But I also don't have a example of such embedded Klein bottle.

REPLY [8 votes]: Let $X$ be the algebraic variety over $\mathbb{R}$ obtained from the projective plain $\mathbb{P}^2$ by blowing up one point $P \in \mathbb{P}^2(\mathbb{R})$. Since $X$ is a smooth projective algebraic variety the space of complex points $X(\mathbb{C})$ admits a Kahler structure and hence the underlying $4$-dimensional real manifold admits a symplectic structure. Furthermore, $X(\mathbb{C}) \cong \mathbb{CP^2} \sharp \mathbb{CP}^2$ (Correction: $X(\mathbb{C}) \cong \mathbb{CP^2} \sharp \overline{\mathbb{CP}^2}$) is simply connected. Now the space $K = X(\mathbb{R})$ of real points can be identified with $\mathbb{RP}^2 \sharp \mathbb{RP}^2$, and is hence isomorphic to the Klein bottle. Let us check that $K$ satisfies Properties (1) and (2) of the question. Let $F$ be the restriction of the complex vector bundle $TX$ to $K$. Then the tangent bundle $TK$ can be identified with the real sub-bundle of $F$ and we have a decomposition $F \cong TK \oplus iTK$, identifying the normal bundle of $K$ with the imaginary sub-bundle of $F$. Since $TK$ admits a nowhere vanishing section it follows that $iTK$ admits a nowhere vanishing section. This shows (1). For (2), let $f: X \longrightarrow \mathbb{P}^2$ be the natural map. Then it will be enough to show that $f_*[K] \in H_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z}/2)$ is non-trivial. Now note that the image of $K$ in $\mathbb{P}^2(\mathbb{C}) = \mathbb{CP}^2$ is given by $\mathbb{P}^2(\mathbb{R}) = \mathbb{RP}^2$ and that the corresponding map $K \longrightarrow \mathbb{RP}^2$ induces an isomorphism on $H_2(-,\mathbb{Z}/2)$. It will now be enough to show that $[\mathbb{RP}^2] \in H_2(\mathbb{CP}^2,\mathbb{Z}/2)$ is non-trivial. To see this, observe that if $\mathbb{CP}^1 \cong L \subseteq \mathbb{CP}^2$ is any complex line that is not defined over $\mathbb{R}$ then $L$ meets $\mathbb{RP}^2$ transversely at exactly one point (the intersection point of $L$ and $\overline{L}$), and that $[L] \in H_2(\mathbb{CP}^2,\mathbb{Z}/2) \cong \mathbb{Z}/2$ is a generator.<|endoftext|>
TITLE: Two strengthenings of "strong measure zero"
QUESTION [18 upvotes]: A set $X\subseteq\mathbb{R}$ is strong measure zero if, for every sequence $(\epsilon_i)_{i\in\mathbb{N}}$ of positive reals, there is a sequence $(I_i)_{i\in\mathbb{N}}$ of open intervals covering $X$ such that $\mu(I_n)<\epsilon_n$.
Looking at various computability theoretic questions around strong measure zero (see e.g. https://math.stackexchange.com/questions/1446602/anti-random-reals), I've run into two possible strengthenings; I'm curious what is known about them.
For the first, we demand that the cover depend on the sequence one bit by bit:

$X\subseteq\mathbb{R}$ is strategically strong measure zero if player II has a winning strategy in the following game: on move $n$, player I plays a positive real $\epsilon_n$, and player II plays an interval $I_n$ with $\mu(I_n)<\epsilon_n$; and player II wins if the $I_n$ form a cover of $X$.

For the second, we merely ask that the cover depend on the sequence continuously. Let $Eps$ be the set of infinite sequences of positive reals, and let $Int$ be the set of infinite sequences of open intervals in $\mathbb{R}$, each topologized as usual.

$X\subseteq\mathbb{R}$ is continuously strong measure zero if there is a continuous $F$ from $Eps$ to $Int$ such that, for every $f\in Eps$, $F(f)$ is a cover of $X$ and $\mu(F(f)(n))<f(n)$.

Clearly strategically strong measure zero implies continuously strong measure zero implies strong measure zero; consistently all strong measure zero sets are countable (this is Borel's conjecture), so no nonimplications can be proved over ZFC. My question is whether we can say anything else; in particular,

Are any other implications provable over ZFC+A, where A is some reasonable axiom which does not imply Borel's conjecture?


Note that ZFC proves that there are continuum many continuously strong measure zero sets $S_r$ such that every continuously strong measure zero set is contained in one of the $S_r$. This seems likely to not be the case in general for strong measure zero sets in ZFC, so I suspect that the situation is not entirely trivial.
EDIT: To clarify, this informal argument merely suggests that not every strong measure zero set is continuously strong measure zero; as Andreas' answer below shows, it's quite likely that every continuously strong measure zero set is countable, which would be triviality in the other direction.
FURTHER EDIT: See "Nicely" strong measure zero sets for a continuation of this question.

REPLY [20 votes]: Strategically strong measure zero is equivalent to countable. To prove the nontrivial direction, suppose $X$ is strategically strong measure zero and $s$ is a winning strategy for player II. Consider the tree of finite plays in which player I plays only rational numbers and player II follows $s$. For any partial play $p$ in which player I is to move next (i.e., the length of $p$ is even), say that an element $x\in X$ is killed at $p$ if, no matter what (rational) move player I makes next, player II's response (using $s$) is an interval that covers $x$.  
I claim first that, at any $p$, at most one $x$ is killed.  The reason is that, if $x\neq x'$, then player I could play an $\epsilon<| x-x'|$ and then player II's response would be an interval too short to cover both $x$ and $x'$.
Since the tree of partial plays is countable (that's why I restricted player I to rational $\epsilon$'s), only countably many $x$'s are killed anywhere in the tree.
To finish the proof, I claim that every element of $X$ is killed somewhere in the tree.  Suppose, toward a contradiction, that $x$ is a counterexample.  Then, at the initial position, player I can move so that II, using $s$, doesn't immediately cover $x$ (because $x$ isn't killed at the initial position).  Then, player I can again move so that II doesn't immediately cover $x$ (because $x$ isn't killed at the position after II's first move).  continuing in this way, player I can ensure that II never covers $x$.  Since $x\in X$, this contradicts the assumption that $s$ is a winning strategy for II.<|endoftext|>
TITLE: Inducing up the group homomorphism between mapping class groups
QUESTION [10 upvotes]: There are many ways to embed the braid group into the mapping class group of a surface. To describe one of them, let ${C}_{2g+2}(\mathbb{D}^2)$ be the configuration of unordered $2g+2$ points in the unit disk and let $\mathcal{M}_{g,2}$ be the moduli
space of connected Riemann surfaces of genus $g$ with two ordered and parametrized boundary components. There is a map
$$
\psi: {C}_{2g+2}(\mathbb{D}^2)\to \mathcal{M}_{g,2}
$$
 which sends $2g+2$ points ${\bf z}= \{z_1,z_2,\dots, z_{2g+2}\}$ to the Riemann surface $\Sigma_{{\bf z}}$ given by 
$$
 f_{{\bf z}}(z)^2 =\prod_{i=1}^{2g+2} (z-z_i)
$$
 which gives a branch cover over the unit disk with ${\bf z}$ as branched points. We consider the map that is induced on the fundamental groups
$$
 \phi: \mathrm{Br}_{2g+2}\to \mathrm{Mod}(\Sigma_{g,2}).
$$
 Tillmann and Song proved that $\phi$ induces the trivial map in the stable homology.
Motivated by the Lie theoretic version of Margulis Superrigidity which asserts that homomorphisms between lattices (virtually) extend to homomorphisms of the ambient groups, Aramayona and Suoto in  asked if the same is true for group homomorphisms between mapping class groups. In our case, the question becomes 
Question:
 Is it known whether the group homomorphism $\phi$ is induced by a homomorphism $\Phi$ between diffeomorphism groups, in other words does there exist  a homomorphim $\Phi$ that makes the following diagram commutative
$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$
$$ \begin{array}{c} \mathrm{Diff}^{\delta}(\mathbb{D}^2- (2g+2) \text{ points},\partial\mathbb{D}^2) & \ra{\Phi} & \mathrm{Diff}^{\delta}(\Sigma_{g,2},\partial\Sigma_{g,2})  \\\da{} &  & \da{}  \\ \mathrm{Br}_{2g+2} & \ra{\phi} & \mathrm{Mod}_{g,2} \end{array}, $$
where $\mathrm{Diff}^{\delta}(\mathbb{D}^2- (2g+2) \text{ points},\partial\mathbb{D}^2)$ denotes the discrete group of diffeomorphisms of the punctured disk which are identity near $\partial\mathbb{D}^2$ and $\mathrm{Diff}^{\delta}(\Sigma_{g,2},\partial\Sigma_{g,2}) $ denotes the discrete group of diffeomorphisms of the surface $\Sigma_{g,2}$ of genus $g$ and $2$ boundary components?

REPLY [3 votes]: The real question is: does every (virtual) homomorphism of mapping class groups have a “geometric meaning”? Your example is defined by geometry, so the answer is yes. Aramayona and Souto make “geometric meaning” precise by defining it as (virtually) extending to the diffeomorphism group. So they are implying that your example should easily extend. And it does.
Such ramified space constructions work just as well for the diffeomorphism group as for the mapping class group. They all give virtual homomorphisms, which is all that Aramayona and Souto ask. It takes some work to figure out in which cases they give actual homomorphisms, but the answer is the same for the mapping class group and the diffeomorphism group. Your construction does not seem to include that extra work, which may be fooling you into thinking that the mapping class group case is easier than the diffeomorphism case.
Your construction is just taking the double cover ramified in the specified points. This is a coordinate-free construction, so it ought to work just as well in the smooth category as in the holomorphic category. That is, if it gave you a map of holomorphic moduli stacks, it should give you a map of stacks of $C^\infty$ surfaces. But that is just a functor between groupoids, which is pretty much the desired homomorphism. However, maps between moduli stacks are not quite that easy to write down. You cannot associate to a surface “the” double cover because the cover has automorphisms. In your example, there is a real homomorphism of mapping class groups, but in other examples, it is only a virtual homomorphism. For example, the hyperelliptic mapping class group is not a genus 0 mapping class group, but a nontrivial central extension thereof: $M_{0,2g+2}\leftarrow H_g\to M_g$.
It is difficult to construct maps between moduli spaces, but fairly easy to construct such zig-zags or correspondences. For any finite map of surfaces $\Sigma’\to\Sigma$, there is a moduli of such maps $M_{\Sigma’\to\Sigma}$, a diffeomorphism group $Diff(\Sigma’\to\Sigma)$, and a mapping class group. The forgetful map from the relative object to the downstairs object is almost an isomorphism. For the diffeomorphism group and the mapping class group, they are virtual isomorphisms, while in the case of the moduli space, it is a finite covering space. Thus they give virtual homomorphisms from the groups associated to $\Sigma$ to the groups associated to $\Sigma’$.
More precisely, let $\Sigma’\to\Sigma$ be a finite map of surfaces, in some category: algebraic, holomorphic, or just smooth. They could have decorations, like a tangent vector or the boundary you ask, but I will suppress that in the notation. Let $D\subset \Sigma$ be the branch locus. We will produce zig-zags $M_{\Sigma,D}\leftarrow M_{\Sigma’\to\Sigma}\to M_{\Sigma’}$ and $Diff(\Sigma,D)\leftarrow Diff(\Sigma’\to\Sigma)\to Diff(\Sigma’)$. Here $M_{\Sigma,D}$ is the moduli of curves of the same genus as $\Sigma$ marked by $D$ and anything else suppressed in the notation (such as the boundary in your case). If $\Sigma’$ is unmarked and $n$ is the size of $D$, then $M_{\Sigma,D}=M_{g,n}$. The upstairs object $M_{\Sigma’}$ could be the moduli of curves of the genus of $\Sigma’$ with or without a marking of size the preimage of $D$. The most structure is to include a marking, but people usually prefer to compose with the forgetful map.
The key ingredient is $M_{\Sigma’\to\Sigma}$. This is just what is has to be to fit in the middle: it is the moduli of pairs of surfaces with a map and appropriate branching structure. The maps are simply forgetting $\Sigma$ or $\Sigma’$. Similarly, $Diff(\Sigma’\to\Sigma)$ is the group of pairs of a diffeomorphism of $\Sigma’$ covering a diffeomorphism of $\Sigma$. Thus the diffeomorphism of $\Sigma$ must preserve $D$ and the diffeomorphism of $\Sigma’$ must preserve the inverse image of $D$. 
The key theorem is that $Diff(\Sigma’\to\Sigma)\to Diff(\Sigma,D)$ is a virtual isomorphism. That is, its kernel and index are both finite. This divides into three parts: that the kernel is finite; that the map on the connected component is surjective; and that the map on mapping class groups is a virtually surjective. The kernel consists of diffeomorphisms of $\Sigma’$ that cover the identity of $\Sigma$. They are nontrivial because they act on the fibers of the map. The generic fibers are all “the same” so the kernel acts faithfully by permutations on a single fiber, which is a finite set, so the kernel is itself finite. The connected component of the diffeomorphism group is a Lie group with Lie algebra consisting of vector fields that vanish at the marked points. That is, it is generated by short flows along such vector fields. Vector fields on $\Sigma$ that vanish at $D$ lift to vector fields on $\Sigma’$, showing surjectivity (thence isomorphism). 
This reduces to understanding the corresponding map on mapping class groups. After deleting the branch locus $D$ and its preimage $D’$, the map $\Sigma’-D’\to\Sigma-D$ becomes a finite covering space. The basic idea is that this corresponds to a subgroup of $\pi_1(\Sigma-D)$ of finite index $i$. Since the fundamental group is finitely generated, there are only finitely many subgroups of index $i$. A mapping class of $(\Sigma,D)$ cannot lift to a relative mapping class if it does not preserve the subgroup of the covering space, but this is the only obstruction. This is not quite right because we need a basepoint to talk about fundamental groups. I included the step about surjectivity of the connected component diffeomorphism group just to make it easier to talk about basepoints; the alternative is to work with groupoids and multiple basepoints. Now, more rigorously: take a mapping class of $(\Sigma,D)$. Compose with a diffeomorphism in the connected component so that it preserves the basepoint. It now induces a self-map on the fundamental group $\pi(\Sigma,D)$. If that does not preserve the conjugacy class of the subgroup corresponding to the cover $\Sigma’-D’$, then it cannot lift. If it does, find a path of diffeomorphism of $\Sigma-D$ that moves the basepoint around the conjugating path. The path exhibits the final diffeomorphism as being in the connected component of $Diff(\Sigma-D)$ (but not $Diff(\Sigma-D,*)$). Compose with the final diffeomorphism to get a diffeomorphism of $\Sigma-D$ that is in the same mapping class as the original, but now preserves the subgroup corresponding to the cover (and not just its conjugacy class). By the standard theorems about covering spaces, the two covering spaces are isomorphic, exhibiting the covering diffeomorphism of $\Sigma’-D’$, thence of $\Sigma’$.
Since $Diff(\Sigma’\to\Sigma)\to Diff(\Sigma,D)$ is a virtual isomorphism, $Diff(\Sigma,D)\leftarrow Diff(\Sigma’\to\Sigma)\to Diff(\Sigma’)$ is a virtual homomorphism corresponding to the virtual homomorphism on mapping class groups. That is the most that is true in general. 
In your case, this can be promoted to an actual homomorphism, both for diffeomorphism group and thence mapping class group. First, $Diff(\Sigma’\to\Sigma)\to Diff(\Sigma,D)$ is surjective. As in the discussion of surjectivity above, this must be because there is a unique subgroup of $\pi_1(\Sigma-D)$ of index 2 that does not contain the loop about any puncture. To put it another way, there is a unique double cover of the plane with specified ramification. This is not true in higher genus, because the fundamental group is not generated by the punctures. Specifically, the set of covers is a torsor over $H^1(\Sigma;\mathbb Z/2)$. So in the genus 0 case $Diff(\Sigma’\to\Sigma)\to Diff(\Sigma,D)$ is surjective, but it is not an isomorphism; its kernel is $\mathbb Z/2$, the diffeomorphism that swaps the two sheets. In some examples, such as the hyperelliptic group, the extension is nontrivial, but in this example it is split by the homomorphism $Diff(\Sigma’\to\Sigma)\to \mathbb Z/2$ that detects whether the diffeomorphism permutes the two boundary components of $\Sigma’$. Thus the kernel of this homomorphism is isomorphic to $Diff(\Sigma,D)$, but it is a subgroup of $Diff(\Sigma’\to\Sigma)$ and thus of $Diff(\Sigma’)$, so it gives a homomorphism $Diff(\Sigma,D)\to Diff(\Sigma’)$.<|endoftext|>
TITLE: Order between two completely monotone functions?
QUESTION [5 upvotes]: I am wondering if the following assertion is true: 
Let $f,g:\mathbb{R}_+\rightarrow [0,1]$ be completely monotone functions on $\mathbb{R}_+^*$, that is, $(-1)^n f^{(n)}(x)\geq 0$ and $(-1)^n g^{(n)}(x)\geq 0$ for any $x>0$ and any $n\in \mathbb{N}$. Assume that $f(0)=g(0)$ and $\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow \infty} g(x)=0$, and that there exist $0<a<A$ such that $f\leq g$ on $[0,a]$ and $f\leq g$ on $[A,\infty)$.
Then $f\leq g$ on $\mathbb{R}_+$ (?).
If not, does anyone have a counter-example?
N.B.: 1) one can think of $f$ and $g$ as two Laplace transforms of positive measures.
2) Removing any of the assumptions seems to lead to a counter-example.
Thank you for your help!

REPLY [6 votes]: No. Let's take $g(x)=\epsilon e^{-\epsilon x}+(1-\epsilon)e^{-\alpha x}$,
$$
f(x) = \int_{\epsilon}^{1+\epsilon} e^{-tx}\, dt = \frac{1}{x}e^{-\epsilon x}(1-e^{-x}) .
$$
Then clearly $f\le g$ near infinity, and near zero, $f(x)\simeq 1-(1/2+\epsilon)x$, $g(x)\simeq 1-(\epsilon^2+(1-\epsilon)\alpha)x$. I also want $f\le g$ near zero, and this gives a condition on $\alpha$ (given $\epsilon$). For small $\epsilon$, I can take $\alpha\approx 1/2$.
On the other hand, as $\epsilon\to 0$, $\alpha\to 1/2$, we have that $f(1)\to 1-e^{-1}$, $g(1)\to e^{-1/2}$, and now a calculator will tell us that $f(1)>g(1)$ for sufficiently small $\epsilon$. (There might be easier counterexamples.)<|endoftext|>
TITLE: A case of nested central limits
QUESTION [7 upvotes]: Consider the random variable $S=(s_0, \dots ,s_{N-1})$, a sequence of signs uniformly distributed on the hypercube $\{-1,1\}^N$. We are interested in $N$ large and prime. The Fourier transform $\hat{S}=(\hat{s}_0, \dots ,\hat{s}_{N-1})$, where 
$$
\hat{s}_q=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} \zeta^{q k} s_k, \qquad \zeta=e^{2\pi i/N},
$$
defines a new multivariate random variable. Finally, consider the univariate random variable
$$
h=\left(\prod_{q=1}^{N-1}\hat{s}_q\right)^\frac{1}{N-1}=\left(\prod_{q=1}^{(N-1)/2}|\hat{s}_q|^2\right)^\frac{1}{N-1}.
$$
This variable is the scaled algebraic norm of the cyclotomic integer whose coefficients are $S$.
I would like to prove that $h$ has the central limit property for $N\to \infty$, that is, its distribution becomes increasingly concentrated at one value. The plausibility argument goes like this. Consider a "baby" estimator of the variable given by the geometric mean of a fixed number $M$ of the $|\hat{s}_q|^2$, instead of all $(N-1)/2$. For this fixed number of Fourier transform components we have a multivariate central limit theorem, for $N\to \infty$. When we apply CLT and compute the covariance matrix, we find that these $|\hat{s}_q|^2$ are independent and identically distributed random variables. The baby estimators therefore have the central limit property for $M\to\infty$. Unfortunately, the $M\to\infty$ limit of the baby estimators may be a different random variable than the $N\to \infty$ limit of $h$.

REPLY [3 votes]: The central limit property of $h$ follows from the main result of:
David Freedman and David Lane, "The Empirical Distribution of Fourier Coefficients", Ann. Statist. Volume 8, Number 6 (1980), 1244-1251.
They show that all the Fourier coefficients (not trivially related as complex conjugates) are asymptotically independent.<|endoftext|>
TITLE: Gerstenhaber conjecture for free loop space
QUESTION [12 upvotes]: I- Is the following statement still a conjecture see this article ?
Conjecture (?)
Let $M$ be a simply connected compact oriented $d$-manifold (smooth), then $HH^{\ast}(C^{\ast}(M))$ the Hochschild cohomology of cochain complexes associated to $M$ is isomorphic as a Gerstenhaber algebra to $H_{\ast+d}(LM)$ the $d$-shifted homology of the free loop space on $M$.
II- What are the recent advances in solving the conjecture ? 
III- Is there a short proof of the following fact: $HH^{\ast}(C^{\ast}(M))$ is isomorphic to $H_{\ast+d}(LM)$ as a graded vector space ?

REPLY [4 votes]: For III:
First show that the Hochschild homology of the DGA of cochains on $M$ is quasi-isomorphic to cochains on $LM$. To do this you can use Jones argument using a cosimplicial model for $LM$ or over the rationals one can construct an explicit map through Chen iterated integrals. 
Now the "Poincaré duality step": relate Hochschild homology to Hochschild cohomology.  To do this one may construct a lift of the iterated integrals map mentioned above by choosing a Thom form supported near the diagonal in $M \times M$. The details of this approach can be found on Merkulov's De Rham string topology. 
There are other ways of achieving this last Poincare duality step by using finite dimensional models for the cochains on $M$: either strict Frobenius algebras  or cyclic $A_{\infty}$ algebras.<|endoftext|>
TITLE: Atiyah Singer index theorem and Hodge de Rham operator
QUESTION [10 upvotes]: When I read about Atiyah Singer index theorem I met the following example: let $M$ is (orientable closed smooth) Riemannian manifold and consider Hodge-de Rham Dirac operator defined by $d+d^*$ (adjoint takes into account the metric). This operator is self adjoint and interchanges odd and even degree forms: therefore is of antidiagonal form. The index of $(d+d^*)^+$ is equal to the Euler characteristic of $M$:
$$ind(d+d^*)^+=\chi(M)$$ 


Can we derive this formula from the right hand side of the Atiyah Singer index formula (involving Todd genus and Chern character)?  


Moreover there is so called generalized Gauss Bonnet theorem which relates Euler characteristic to the geometric quantity $\int_{M}Pf(\Omega)$ (Pfaffian of the curvature form): $$\chi(M)=(2\pi)^{-n}\int_MPf(\Omega).$$ 
So we have at all three quantities: the index, Euler characteristic and the integral of Pfaffian. 


Is it possible to obtain the equality 
  $$ind(d+d^*)^+=(2\pi)^{-n}\int_MPf(\Omega)$$
  without invoking Gauss-Bonnet (which as far as I know requires some work).  


The first question is something like "first step" in order to get into index theory and to get some better understanding of the background of Atiyah Singer index theorem. Second question is inspired by the following: $\chi(M)$ is a topological quantity: from the other hand $Pf(\Omega)$ is of geometric flavour. Gauss-Bonnet formula relates these two quantities: but as $d^*$ requires choice of the metric, $ind(d+d^*)^+$ should also be viewed as geometric (or maybe better, analytical quantity) so maybe it would be easier to relate this two geometric quantities and then refer to the Euler characteristic.

REPLY [10 votes]: The Euler operator $d+d^*$ of a spin manifold is the spin Dirac operator
twisted by the $\mathbb Z/2$-graded spinor bundle $W=S^+\ominus S^-$, as explained in Heat Kernels and Dirac Operators by Berline, Getzler, Vergne [BGV, chapter 4.1]. The Chern character $\mathrm{ch}(W)$ equals $e(TM)\hat A(TM)^{-1}$, see [BGV] or Nicolaescu's notes. The cohomological index theorem therefore gives
$$\mathrm{ind}(d+d^*)^+=(\hat A(TM)\wedge\mathrm{ch}(W))[M]=e(TM)[M]$$
The de Rham representative of $e(TM)$ is given by the Pfaffian, and you get your second formula [BGV, Theorem 4.6].
The formulas above are pairings of "characteristic cohomology classes" with "fundamental homology classes". If you regard the Dirac operator as a fundamental $K$-homology class for the $K$-orientation of the  tangent bundle given by the chosen spin structure, then the left hand side also constitutes a pairing of the $K$-theory class of the spinor bundle with the corresponding fundamental class.
To avoid the spin condition, you define the ``twist Chern character'' $\mathrm{tr}(\mathrm{exp}(-F^{W/S}))$ as explained in [BGV, chapter 4.1].
The formulas above still hold.
You can even get rid of orientability by regarding $e(TM)\in H^{\dim M}(M,o(TM))$ as a cohomology class twisted by the orientation bundle. Similarly, the Pfaffian is in fact a differential form twisted by the orientation bundle. Then all formulas above make sense without an orientation of $TM$.
On the other hand, the Euler number is the "trace" of id$_M$ in a sense explained nicely here (Def 2.2) - no matter if you count cells or dimensions of (co-) homology groups.
To relate it to the pairings above, you may use either Hodge theory, or the Gauss-Bonnet-Chern theorem, or deduce $e(TM)[M]=\chi(M)$ from the Poincar'e-Hopf theorem, e.g. using Morse theory. It seems that because $\chi(M)$ and $e(TM)[M]$ are different kinds of objects, some work is needed for this step.<|endoftext|>
TITLE: Relationship between two universal properties of the category of elements?
QUESTION [10 upvotes]: Let $G: \mathcal{A} \to \mathsf{Set}$ be a functor. Recall that the category of elements $\mathsf{el}G$ is defined by the comma square
$\require{AMScd}$
\begin{CD}
    \mathsf{el}G @>!>> \ast \\
    @V U_G VV \Downarrow @VV \Delta_\ast V\\
    \mathcal{A} @>>G> \mathsf{Set}
    \end{CD}
where $\ast$ is used to label the terminal category and $\Delta_\ast$ is the functor that is constant at the terminal set.
But there is another key universal property of this diagram: it exhibits the left Kan extension $ G = \operatorname{Lan}_{U_G} (\Delta_\ast !)$. This fact is key in the reduction of weighted colimits to conical ones (as explained here).
It's easy to verify that this is a Kan extension directly, but it's a bit mysterious, because arbitrary comma squares are not Kan extensions (for example, the same square in $\mathsf{Cat}^{\mathrm{co}}$ doesn't seem to be a left Kan extension). So what is going on here? Why, conceptually (by which I suppose I mean: 2-categorically) is this particular comma square also a Kan extension triangle?

REPLY [7 votes]: The conceptual explanation is that a comma square exhibits the bottom morphhism as a left Kan extension if the right morphism is dense.
There is, of course, a problem what we mean by a "dense morphism" in an arbitrary 2-category. If we take the usual definition and say that a morphism is dense if its pointwise left Kan extension along itself exists and is the identity, then the above follows almost by the definition of density and the definition of "pointwiseness" in the sense of R. Street (i.e. stability under comma squares). 
In your example $\Delta_* \colon 1 \rightarrow \mathbf{Set}$ is obviously dense (in the classical sense) --- therefore its left Kan extension along itself is the identity:
$$\begin{CD}
    \ast @>\Delta_\ast>> \mathsf{Set} \\
    @V \Delta_\ast VV \Downarrow @| \\
    \mathsf{Set} @>>\mathit{id} = \mathit{Lan}_{\Delta_\ast}\Delta_\ast> \mathsf{Set}
\end{CD}$$
and when you draw the above left extension diagram on the right side of your comma square, then the fact that the bottom morphism is the left extension follows directly from the definition of pointwiseness.<|endoftext|>
TITLE: vanishing theorem in algebraic geometry
QUESTION [5 upvotes]: This is a general question: As we know there are a lot of vanishing theorems like Fujita vanishing, kodaira Nakano vanishing, vanishing for big nef line bundle, Kollár vanishing, etc. Those just for projective guys.
My question is: are there some important vanishing problems for general complete algebraic manifold? Although we don't have nice line bundle like $\mathcal{O}_X(1)$, are there still some vanishing expectation for cohomology of coherent sheaves in some sense.
More specific question, if we have a proper surjective morphism $f:X\rightarrow Y$, $X$ complete, $Y$ projective, do we expect that $R^jf_*\mathcal{O}_X(K_X)=0$ for $j>dim X-dim Y$?

REPLY [5 votes]: I guess I could have just told you in person, but  anyway, yes your specific question has a positive answer. To see this, use Chow's theorem and resolution of singularities to find a birational map $\pi:\tilde X\to X$ with $\tilde X$ smooth and projective. Now apply Kollár vanishing twice (or really Grauert-Riemenschneider) to get $R^if_*K_X= R^i(f\circ \pi)_* K_{\tilde X} = 0$ for $i>\dim X-\dim Y$.<|endoftext|>
TITLE: How can I prove that these two graph coloring problems are polynomial time equivalent?
QUESTION [6 upvotes]: Given a graph $G(V,E)$. The standard $k$-coloring problem consists in finding a feasible coloring (no two adjacent nodes share the same color) of the nodes with $k$ colors. Let this problem be $P_1$.
A $\theta$-improper coloring of $G$ is a coloring where each node can have at most $\theta$ neighbors with the same coloring.
Now, given a weighted digraph $G(V,A,\omega)$ with weight function $\omega(u,v), (u,v)\in A$, a weighted $\theta$-improper $k$-coloring of $G$ is a coloring of the nodes such that for every node $v$:
$$
\sum_{u \in N(v)|c(u)=c(v)}\omega(u,v)\le \theta,
$$
where $N(v)$ denotes the adjacent vertices of $v$, and $c(v)$ denotes the color of vertex $v$. Note that this definition generalizes $\theta$-improprer colorings (chose weights equal to 1), and standard colorings (…and chose $\theta=0)$. Let this problem be $P_2$.
It is fairly easy to prove that finding a weighted $\theta$-improper coloring is an $\mathcal{NP}$ complete problem by proving that $P_1\propto P_2$ (by choosing an appropriate weight function), but I am struggling to prove that $P_2 \propto P_1$ (this is necessarily true, since both problems are $\mathcal{NP}$ complete.
In other words, given a weighted digraph $G(V,A,\omega)$, how can one find a 
weighted $\theta$-improper $k$-coloring of $G$ with a procedure that solves the standard $k$-coloring problem?

REPLY [3 votes]: It's interesting to notice that weighted improper $k$-coloring seems  $NP$-complete even for $\theta = 1, k = 2$; so unless $P=NP$ there is not a polynomial time reduction to $k$-coloring ($k = 2$); because $2$-coloring of graphs is in $P$.
A sketch of the reduction from NAE 3-SAT to unweighted $\theta=1$ improper 2-coloring of undirected graphs is the following [NOTE: I quickly made it for exercise, so it can be wrong or probably the result is already known].

represent each variable $x_i$ with two $K_4$ gadgets, one node of the first $K_4$ represents $+x_i$, one node of the second $K_4$ represents $-x_i$; the nodes $+x_i$ and $-x_i$ are linked together and they cannot have the same color;
represent each clause $Cj = (\ell_{j,1} \lor \ell_{j,2}  \lor \ell_{j,3})$ with a $K_3$ gadget; the three nodes $\ell_{j,1}, \ell_{j,2}, \ell_{j,3}$ of a clause gadget cannot have the same color; 
if $\ell_{j,p} = x_i$ then add an edge between $\ell_{j,p}$ and $+x_i$; if $\ell_{j,p} = \neg x_i$ then add an edge between $\ell_{j,p}$ and $-x_i$.


It's not hard to prove that the resulting graph is 1-improper 2-colorable if and only if it exists an assignment of the $x_i$s such  that in each $C_j$ at least one literal is true and at least one literal is false (Not-All-Equal). 
The NP-completeness of weighted $\theta = 1$ improper 2-coloring of directed graphs follows from:

weighted $\theta = 1$ improper 2-coloring of directed graphs is simply a generalization of unweighted $\theta = 1$ improper 2-coloring of directed graphs (build an NPC reduction setting weight = 1);
$\theta = 1$ improper 2-coloring of directed graphs $\theta = 1$ improper 2-coloring of directed graphs is simply a generalization $\theta = 1$ improper 2-coloring of undirected graphs $\theta = 1$  (build an NPC reduction setting $w(u,v) = w(v,u)$)<|endoftext|>
TITLE: Approximation via finite rank Cameron-Martin projections
QUESTION [10 upvotes]: Let $(W, \|\cdot\|_W)$ be a real separable Banach space equipped with
a non-degenerate Gaussian Borel measure $\mu$.  Let $H \subset W$ be
the corresponding Cameron-Martin Hilbert space (also known as the
reproducing kernel Hilbert space or RKHS), which is dense in $W$; let
$\langle \cdot, \cdot \rangle_H$ denote the inner product on $H$.
Then the inclusion map $i : H \to W$ is continuous, injective and has
dense range, so the same is true of its adjoint $i^* : W^* \to H$.
Let $W_* \subset H \subset W$ be the image of $i^*$.
Suppose $\{e_j\}$ is an orthonormal basis for $H$ which is contained
in $W_*$, so that $e_j = i^* f_j$ for some $\{f_j\} \subset W^*$.
Define the finite rank projection $P_n : W \to W_*$ by $P_n x =
\sum_{j=1}^n f_j(x) e_j$.
Note that if $h \in H$ then $P_n h = \sum_{j=1}^n \langle e_j, h
\rangle e_j$.  So the restriction $P_n|_H$ is orthogonal projection
and we have $P_n h \to h$ in $H$-norm.  In other words, $P_n|_H \to
I_H$ strongly on $H$, where $I_H$ is the identity map on $H$.
Moreover, it can be shown that $P_n \to I_W$ in
$L^2(W, \mu; W)$.  In other words, $\int_W \|P_n x - x\|_{W}^2 \mu(dx)
\to 0$.  In particular, there is a subsequence $P_{n_k}$ such that for
$\mu$-almost every $x \in W$, we have $P_{n_k} x \to x$ in $W$-norm.  (This is essentially the fact that $\|\cdot\|_W$ is a measurable norm  on $H$ in the sense of Gross.  I believe this result is due to Dudley, Feldman and Le Cam.  A short proof due to Daniel Stroock can be found in these lecture notes by Bruce Driver; see Theorem 44.8.) 
Now in some cases much more is true.  For instance, consider the
standard Wiener measure on $W = C_0((0,1])$ (i.e. all continuous paths
starting at 0).  Then $H$ is the Sobolev space $H^1_0((0,1])$.  If we
take $e_j$ to be the Schauder (aka Faber-Schauder) basis (i.e. the
integrals of the Haar wavelets), then $P_n x$ is a piecewise linear
function agreeing with $f$ at its vertices, which include all the
dyadic rationals as $n \to \infty$.  So $P_n x \to x$ uniformly,
i.e. in $W$-norm, for every $x$ (not just almost every).  In other
words, $P_n \to I$ strongly on $W$.
As a related, not-quite example, let $W = \mathbb{R}^\infty$, viewed
as the space of all real sequences with the product topology, and let
$\mu$ be standard product Gaussian measure (I know this is not a Banach
space).  Then $H = \ell^2$.  If we choose the standard basis $\{e_j\}$
for $H$, then $P_n x$ is merely the sequence that agrees with $x$ in
its first $n$ terms and is 0 thereafter.  Clearly $P_n x \to x$ in the
$W$ topology, for every $x$.
I suppose this can't always happen, so I am looking for a
counterexample.

What is an example of $W$, $\mu$ and $\{e_n\}$ such that it is not
    the case that $P_n \to I$ strongly on $W$?  Better, is there an
    example where there is no subsequence $P_{n_k}$ with $P_{n_k} \to I$
    strongly on $W$?

REPLY [3 votes]: For an example, take for $W$ any separable Banach space that fails the approximation property, or even just does not have a Schauder basis.  Every $\ell_p$ for $p\not= 2$ has such a subspace; see, for example, volume 1 of Classical Banach Spaces by Lindenstrauss and Tzafriri.<|endoftext|>
TITLE: Antirandom reals
QUESTION [21 upvotes]: This is a crossposting of https://math.stackexchange.com/questions/1446602/anti-random-reals, which has not gotten any answers; after thinking about the problem, I've become more convinced that it belongs here instead.
For a function $f: \mathbb{N}\rightarrow\mathbb{R}_{>0}$ and a set $X\subseteq \mathbb{R}$, an $f$-cover of $X$ is a sequence $(I_n)_{n\in\mathbb{N}}$ of open intervals with rational endpoints such that 

$X\subseteq\bigcup I_n$, and
$\mu(I_n)<f(n)$.

Say that a set $X\subseteq\mathbb{R}$ is $f$-small if $X$ has an $f$-cover.
Talking about $f$-covers provides us with many different refinements of the notion of "measure zero": e.g., a set is strong measure zero if it has an $f$-cover for every function $f$.
Say that a set of reals $X$ is computably strong measure zero (csmz) if there is an $e$ such that $\Phi_e^f$ is an $f$-cover of $X$ whenever $f$ is a function from $\mathbb{N}$ to $\mathbb{R}_{>0}$. (This is sadly not the same as effective strong measure zero, a notion introduced by Kihara in his thesis; see also http://www.sciencedirect.com/science/article/pii/S016800721400044X.) 
In computability theory, a real is said (informally) to be "random" if it is not in any "simple" measure-zero set; there are of course many ways to formalize this, but this is the basic theme. Csmz sets provide a very strong notion of non-randomness: say that an individual real $r$ is antirandom if $\{r\}$ is csmz - equivalently, if $r$ is contained in some csmz set. My question is:

What are the antirandom reals?

I strongly suspect that every antirandom real is computable, but I can't prove it. EDIT: As Joe Miller's answer below shows, my guess was really wrong!

Comment 1. The set of antirandom reals is countable, but the proof is surprisingly nontrivial. There is a countable set $\{A_i: i\in\omega\}$ of csmz sets such that every csmz set is contained in one of the $A_i$s; specifically, take $A_i$ to be the largest csmz set witnessed to be csmz by $\Phi_i$. Now, it can be forced that every strong measure zero set is countable (this is Borel's conjecture) - in such a forcing extension, the antirandoms are countable. Meanwhile, antirandomness is a $\Pi^1_1$ property. This means the statement "there are countably many antirandom reals" is $\Sigma^1_3$, and so absolute assuming large cardinals.
This argument is probably overkill, but (1) we do seem to need Borel's conjecture for coanalytic sets, which is independent of ZFC, and (2) we do seem to need $\Sigma^1_3$-absoluteness for proper forcing, which has nontrivial large cardinal strength (see http://www.logic.univie.ac.at/~sdf/papers/bagfried.pdf).
Comment 2. Thomas Andrews, in the comments section to the original question, proposed looking at "c-csmz" sets - these are sets which are csmz, but where we only allow computable sequences of epsilons. Say that a real $r$ is weakly antirandom if $\{r\}$ is a c-csmz set. I believe that there are noncomputable weakly antirandom reals, but I haven't been able to prove that yet; I would be interested in these reals as well. Note that the countability proof breaks down: a c-csmz set is not strong measure zero.

REPLY [13 votes]: Let A be a c.e. set with the property that there are infinitely many stages $s$ such that $A_s\upharpoonright s = A\upharpoonright s$. Call such a stage "strongly true". Noncomputable $A$ with infinitely many strongly true stages can be produced using a movable marker construction.
I claim that any such $A$ is antirandom.
Define $\Phi_e^f$ as follows: First, we can compute from $f$ an increasing function $g$ such that $2^{-g(n)}<f(n)$ for all $n$. The $n$th interval in our cover should be $I_n = \left[\;A_{g(n+1)}\upharpoonright g(n)\;\right]$.
To verify that this works, just note that if $s\in[g(n),g(n+1)]$ is a strongly true stage, then $I_n$ covers $A$.
Improved Example: I claim that if $A$ is a rank 1 point in a $\Pi^0_1$ class, then $A$ is antirandom. It can be shown that this subsumes the previous answer, and furthermore, there is a rank 1 point $A\equiv_T\emptyset''$, so not all antirandom sets are $\Delta^0_2$.
To see the claim, let $P$ be a $\Pi^0_1$ class with exactly one limit point $A$. First, note that it is true that there is an $f$-cover of $P$ for every $f$, and in fact, one that uses only finitely many intervals. To see this, fix $f$ and start by covering $A$ with an interval $I_0$ of length $f(0)$. Now there are only finitely many points of $P$ not covered, and they can be covered by finitely many of the remaining $I_n$'s.
Now note that by compactness, we will eventually see that an $f$-cover of $P$ using only finitely many $I_n$'s is correct, so relative to $f$, we can search for such a cover.
Even Better: This argument can be improved to any member of any countable $\Pi^0_1$ class. This is because countable classes are strongly measure zero and, by compactness, only finitely many of the $I_n$'s are needed for any fixed $f$. So a search (relative to the given $f$) will find an $f$-cover.
This means that antirandom sets can have complexity as high as we would like in the hyperarithmetical hierarchy.
Suggestion: Maybe there is a nice relationship between antirandom and never continuously random (NCR)? After all, Kjos-Hanssen and Montalbán showed that members of countable $\Pi^0_1$ classes are NCR. I'm just copying their idea.
My Final Edit: Actually, it is easy to see that every antirandom $A$ is NCR. Given a name $m$ for a continuous measure $\mu$, we can find a function $f$ such that every interval of length $f(n)$ has $\mu$-measure at most $2^{-n}$. Now using the fact that $A$ is antirandom, we can build a $\mu$-ML test relative to $m$ covering $A$.
Note that Reimann and Slaman showed that every NCR is hyperarithmetic (which gives another proof that the collection of antirandom sets is countable).
So we are left with the following:
Question: If $A$ is NCR, must it be antirandom?<|endoftext|>
TITLE: Tangent fields spanning the distribution of principal directions on a surface
QUESTION [8 upvotes]: Suppose $S$ is an orientable regular surface in $\mathbb R^3$ without umbilical points (not necessarily compact, and with no boundary). There are two well-defined smooth $1$-dimensional tangent distributions on $S$ corresponding to the directions of curvature.

Can each of these be generated by a single unitary vector field?

Of course this is possible locally and a standard argument using Čech cocycles gets one from there to simply connected surfaces, which is not very much. This argument is generic —does not use the fact that these are the directions of curvature of the surface— so this is to be expected. It may well be the case that the cocycles one gets from this are special in some way which I cannot see, so that one does not need the vanishing of $\pi_1$.

REPLY [4 votes]: Let me add a small comment to the answer of Professor Bryant: There is the notion of constant mean curvature (CMC) surfaces which are defined as critical points of the area functional with the constrained of fixed enclosed volume. These surfaces are characterized (in space-forms) by the fact that the complex bilinear part (the so-called Hopf differential $Q$) of the second fundamental form is a holomorphic quadratic differential with respect to the induced Riemann surface structure. There do exists CMC surfaces whose Hopf differentials have simple zeros (e.g. the Lawson surface $\xi_{2,2}$ or its conjugate cousin (here is an image showing the curvature lines)), i.e., with respect to an appopriate holomorphic coordinate $z$ the Hopf differential  is given by $Q=z(dz)^2,$ and after removing the zeros you cannot find a vector field $X$ such that $Q(X,X)>0$ everywhere. But this inequality would hold for a vector field which generates a principal curvature distribution.
Globally on a CMC surface whose Hopf differentials have simple zeros the obstruction to the existence of a vector field $X$ which generates a principal curvature distribution is that the Hopf differential $Q$ admits a square root $\omega$ (which is a holomorphic 1-form).<|endoftext|>
TITLE: Why should the map $-\Delta^{-1}$ be continuous?
QUESTION [9 upvotes]: I'm reading an article by Wei-Ming Ni about the existence of solutions for the elliptic problem $$\Delta u +|x|^\lambda |u|^\tau =0,$$ 
in the unit ball $\Omega$ in dimension $>2$. I'm looking for solutions in $E=\tilde{H}^1_0$, which means this functions are also radially symmetric and in $H^1_0$. The scalar product is $$\langle u,v \rangle_E = \int_\Omega \nabla u \cdot \nabla v $$
The last step of the proof is based on the definition of a functional $T:E\to E$ such that 
$$\langle Tu,v \rangle_E = \int_\Omega |x|^\lambda |u|^\tau v.$$
How do I know such an operator is well-defined? 
Moreover, it says it can be written in the form $Tu = -\Delta^{-1} (|x|^\lambda |u|^\tau),$ and it goes on saying that the map $T_1 : H^{-1} \to E$ such that $|x|^\lambda|u|^\tau \mapsto  -\Delta^{-1} (|x|^\lambda |u|^\tau)$ is continuous. I can't get why this should be well-defined (and continuous).

REPLY [4 votes]: First of all, there should be a constraint on the exponent $\tau$ to guarantee certain integrability  conditions. Define
$$n^*:=\frac{2(n+\lambda)}{n-2}, $$
where $n$ denotes the dimension of the ball. Then $E$ embeds continuously in the Banach space $ X:=L^{n^*}(0,1; r^{\lambda+n-1} dr)$. (For a proof of this I refer to this old paper of mine.)
Define $\tau^*:=n^*-1$. (This is the critical Sobolev exponent.) Assume
$$\tau \leq \tau^*. $$
Set $r:=|x|$, $x\in \mathbb{R}^n$. Use  Holder's inequality  for the conjugate exponents $n^*$ and $\frac{n^*}{\tau^*}$  and the Sobolev embedding $E\to X$ to verify that
$$\left\vert\int_\Omega r^\lambda |u|^\tau v dx\right\vert \leq C\Vert u\Vert_E\cdot \Vert v\Vert_E. $$
This proves that $Tu$ defines a linear functional on $E$. The dual of $E$ is $H^{-1}$.
The Laplacian  defines an isomorphism $\Delta E\to H^{-1}$ with inverse  $\Delta^{-1}$. The continuity statement   is proved by observing that the map
$$ E\ni u\mapsto |u|^\tau \in L^{\frac{n^*}{\tau^*}}(0,1, r^{\lambda+n-1} dr). $$
is continuous.<|endoftext|>
TITLE: Geometric meaning of unimodular matrix
QUESTION [10 upvotes]: Rotations are given by unitary matrices.
What is the geometric meaning of unimodular matrices that are not unitary?

REPLY [8 votes]: If the matrix has integer elements, then the geometric meaning is that a unimodular transformation maps the integer lattice onto itself:

Consider a basis $B$ of an $m$-dimensional lattice
  $L(B)=\{Bx:x\in\mathbb{Z}^m$}, and another basis $C$, then $L(B)=L(C)$
  if and only if there exists a unimodular matrix $M$ (an $m\times m$
  matrix with integer entries and determinant $\pm 1$) such that $B=CM$.<|endoftext|>
TITLE: Existence of closed operators with arbitrary dense domain of a given Banach space
QUESTION [12 upvotes]: Consider any Banach space $X$, and let $Y$ be any dense subspace, then does it necessarily exist a closed linear operator $T$ defined on $X$, such that the domain of $T$ is exactly $Y$, i.e., $D(T)=Y$? Perhaps this looks not like a problem of research level, however, I can't solve it myself and can't find any reference on google.
Furthermore, does it exist a closed linear operator $T$ on $X$, such that $D(T)=Y$ and $T$  is also the generator of some $C_0$-semigroup on $X$ ? 
Added (from a comment Sep. 30 '15): If $X$ is a Hilbert space, and the dense subspace $Y$ is also a Hilbert space under some norm, then is the existence of such closed operator true?

REPLY [5 votes]: You remarked in comments that $T$ should be an unbounded operator from $X$ to $X$.
If $X$ is separable then $D(T)$ has to be a Borel set in $X$.
Note that $X \times X$ is a separable Banach space under a norm such as $\|(x_1, x_2)\| = \|x_1\| + \|x_2\|$.  The graph $\Gamma(T) = \{(x, Tx) : x \in D(T)\}$ is by assumption a closed linear subspace of $X \times X$, hence it too is a separable Banach space under the same norm.  In particular it is a Polish space.  Moreover the map $S : \Gamma(T) \to X$ defined by $S(x, Tx) = x$ is continuous and injective, and its image is $D(T)$.  So by a theorem of Lusin and Souslin, $D(T)$ is a Borel subset of $X$.  This appears as Theorem 15.1 in Kechris's Classical Descriptive Set Theory and probably any other descriptive set theory book.
In particular, if $D(T) \ne X$ then $D(T)$ has to be meager.  This follows from the fact that every Borel set has the property of Baire, together with the Pettis lemma.  Every infinite dimensional Banach space has proper linear subspaces which are not meager, so they are not the domain of any closed operator.  See Are proper linear subspaces of Banach spaces always meager? for more details on this.
Note that if we allow $T$ to be an unbounded operator from $X$ to some other Banach space $Z$ then this argument doesn't work.  Consider for example $X = L^2([0,1])$, let $D(T) = L^\infty([0,1]) \subset X$, and let $T : D(T) \to L^\infty([0,1])$ be the identity map.  Then the map $\Gamma(T) \ni (f,f) \mapsto f \in L^\infty$ is bi-Lipschitz so $\Gamma(T)$ is not separable.  But in this case $D(T)$ is still Borel (even $F_\sigma$, since the unit ball of $L^\infty$ is closed in $L^2$), so maybe there is another way to make it work in general.
Edit. You asked in a comment:

If X is a Hilbert space, and the dense subspace Y is also a Hilbert space under some norm, then is the existence of such closed operator true? 

No, you need more than that.  For instance suppose $H$ is a separable Hilbert space, whose Hamel dimension must be $\mathfrak{c}$.  If $f$ is a discontinuous linear functional on $H$ whose kernel $Y=\ker f$ is nonmeager, as constructed here, then $Y$ is not Borel, so there is no closed operator $T$ from $H$ to $H$ with domain $Y$.  But $Y$ has codimension 1 so the Hamel dimension of $Y$ is again $\mathfrak{c}$.  Since the dimensions match, there is a linear isomorphism $S$ from $Y$ to any separable Hilbert space, say $H$ itself.  Then $\|y\|_Y := \|Sy\|_H$ makes $Y$ into a separable Hilbert space too.<|endoftext|>
TITLE: A question on compact operators with domain $l_{p}$
QUESTION [5 upvotes]: Suppose that $T$ is an operator from a Banach space $X$ to a Banach space $Y$. Let $1<p<q<\infty$. If $TS$ is compact for any operator $S:l_{p}\rightarrow X$, is $TR$ compact for any operator $R:l_{q}\rightarrow X$?

REPLY [4 votes]: Yes.  If $R:\ell_q \to X$ and $TR$ is not compact then there is a normalized block basis $u_n$ of the unit vector basis for $\ell_q$ s.t. $TRu_n$ is bounded away from zero.  Let $V:\ell_p \to \ell_q$ be the bounded linear operator that maps $e_n$ to $u_n$ and set $S=RV$.  Then $TS$ is not compact.<|endoftext|>
TITLE: Embedding Z into Z^2 with large distortion
QUESTION [12 upvotes]: Is it possible to find a 2-way infinite (self-avoiding) path $\{x_i\}_{i\in \mathbb Z}$ in the standard Cayley graph of $\mathbb Z^2$, i.e. the square grid, such that the distance between $x_i$ and $x_{i+n}$ is of order $o(n)$? If yes, how small can this distance be? Here I'm asking for upper bounds $f(n)$ that are independent of $i$. Let me make this more precise:
Is there a 2-way infinite (self-avoiding) path $\{x_i\}_{i\in \mathbb Z}$ in $\mathbb Z^2$, and a number M, such that for every i and every $n>M$, we have $d(x_i,x_{i+n}) < f(n)$ where $f(n)$ is $o(n)$?
Here $d$ denotes the graph-distance on $\mathbb Z^2$. 
If the answer is yes, I would like to know what is the smallest $f(n)$ for which this is possible. Easily, $f(n)= \Omega(\sqrt{n})$.

REPLY [9 votes]: This question is answered, in greater generality, in a paper by Richard Stong. For any dimensions $r \leq s$ he constructs an embedding ${\phi}: {\mathbb Z}^r\longrightarrow {\mathbb Z}^s$ such that for all $x,y\in {\mathbb Z}^r$, $$\parallel {\phi}(x)-{\phi}(y)\parallel \, <\, C \parallel x-y \parallel^{r/s}$$ for some constant $C$. (Conversely, there is a constant $K$  such that for any embedding ${\mathbb Z}^r\longrightarrow {\mathbb Z}^s$ there exist infinitely many pairs $x,y\in {\mathbb Z}^r$ with  $\parallel {\phi}(x)-{\phi}(y)\parallel \, >\, K \parallel x-y \parallel^{r/s}$.)
In terms of the original question $( {\mathbb Z}\longrightarrow {\mathbb Z}^2)$ this gives $f=\Theta(\sqrt{n})$. To follow up on the comment in the answer above, note that Stong gives a single embedding that works for all $n$.<|endoftext|>
TITLE: Does the equation $241+2^{2s+1}=m^2$ have a solution?
QUESTION [17 upvotes]: Let $p$ be a prime congruent to $1$ mod. 8. 
If $p= 17$ one has : $p+ 8 = 5 ^2$.
If $p= 41$ one has : $p+ 8 = 7 ^2$.
If $p= 73$ one has : $p+ 8 = 9 ^2$.
If $p= 89$ one has : $p+ 32 = 11 ^2$.
If $p= 97$ one has : $p+ 128 = 15 ^2$.
If $p= 113$ one has : $p+ 8 = 11 ^2$.
If $p= 137$ one has : $p+ 32 = 13 ^2$.
If $p= 193$ one has : $p+ 32 = 15 ^2$.
If $p= 233$ one has : $p+ 128 = 19 ^2$.
For $p=241$, there is no value $s\leq 4000$ for which $p+2^{2s+1}$ is a square. 
Question 1 : Is there a simple way to prove that no such $s$ can exist ?
Question 2 : Can one estimate the density of those primes $p\equiv 1$ mod. 8 that can be written under the form
$$p=m^2-2^{2s+1}\ \ \ ?$$

REPLY [38 votes]: To answer your first question: there is indeed no $s$ such that
$241+2^{2s+1}$ is a perfect square. -- Proof: $2^{2s+1}$ is always
congruent to either $2$, $8$ or $32$ modulo $63$, which makes
$241+2^{2s+1}$ congruent to either $21$, $54$ or $60$ modulo $63$.
However none of these values is a quadratic residue modulo $63$,
and thus $241+2^{2s+1}$ cannot be a perfect square, as claimed.

REPLY [5 votes]: I was able to allow the power of $2$ to go up to $2^{29} = 536870912,$ no higher....
Stefan's argument works word for word for $p=337.$ 
For $p=569,$ we need to expand to numbers $\pmod{255},$ where odd powers of $2$ are $2,8,32,128,$ $569 \equiv 59 \pmod {255},$ and $569 + 2^{\mbox{odd}} \equiv 61,67,91,187 \pmod {255},$ while the quadratic residues are
   0       1       4       9      15      16      19      21      25      30
  34      36      49      51      55      60      64      66      69      70
  76      81      84      85      94     100     106     111     115     120
 121     135     136     144     145     151     154     166     169     171
 174     186     189     195     196     204     205     219     220     225
 229     234     240     246




      17          8         25      5
      41          8         49      7
      73          8         81      9
      89         32        121     11
      97        128        225     15
     113          8        121     11
     137         32        169     13
     193         32        225     15
     233        128        361     19
     241     FAIL  
     257         32        289     17
     281          8        289     17
     313        128        441     21
     337     FAIL  
     353          8        361     19
     401        128        529     23
     409         32        441     21
     433          8        441     21
     449        512        961     31
     457       8192       8649     93
     521          8        529     23
     569     FAIL  
     577        512       1089     33
     593         32        625     25
     601        128        729     27
     617          8        625     25
     641     FAIL  
     673     FAIL  
     761       2048       2809     53
     769     FAIL  
     809         32        841     29
     857        512       1369     37
     881     FAIL  
     929         32        961     31
     937     FAIL  
     953          8        961     31
     977       2048       3025     55
    1009        512       1521     39
    1033     FAIL  
    1049     FAIL  
    1097        128       1225     35
    1129     FAIL  
    1153     FAIL  
    1193         32       1225     35
    1201       2048       3249     57
    1217          8       1225     35
    1249     FAIL  
    1289     FAIL  
    1297     FAIL  
    1321     FAIL  
    1361          8       1369     37
    1409     FAIL  
    1433       2048       3481     59
    1481     FAIL  
    1489         32       1521     39
    1553        128       1681     41
    1601     FAIL  
    1609       8192       9801     99
    1657     FAIL  
    1697        512       2209     47
    1721        128       1849     43
    1753     FAIL  
    1777     FAIL  
    1801     FAIL  
    1873     FAIL  
    1889        512       2401     49
    1913     FAIL  
    1993         32       2025     45
    2017          8       2025     45
    2081        128       2209     47
    2089        512       2601     51
    2113     FAIL  
    2129     FAIL  
    2137     FAIL  
    2153     131072     133225    365
    2161     FAIL  
    2273        128       2401     49
    2281     FAIL  
    2297        512       2809     53
    2377     FAIL  
    2393          8       2401     49
    2417       8192      10609    103
    2441       2048       4489     67
    2473        128       2601     51
    2521     FAIL  
    2593          8       2601     51
    2609     FAIL  
    2617     FAIL  
    2633     FAIL  
    2657     FAIL  
    2689     FAIL  
    2713       2048       4761     69
    2729     FAIL  
    2753     FAIL  
    2777         32       2809     53
    2801          8       2809     53
    2833       8192      11025    105
    2857     FAIL  
    2897        128       3025     55
    2953      32768      35721    189
    2969        512       3481     59
    3001     FAIL  
    3041     FAIL  
    3049     FAIL  
    3089     FAIL  
    3121        128       3249     57
    3137     FAIL  
    3169     FAIL  
    3209        512       3721     61
    3217         32       3249     57
    3257       8192      11449    107
    3313     FAIL  
    3329     FAIL  
    3361     FAIL  
    3433     FAIL  
    3449         32       3481     59
    3457        512       3969     63
    3529     FAIL  
    3593        128       3721     61
    3617     131072     134689    367
    3673     FAIL  
    3697     FAIL  
    3761     FAIL  
    3769     FAIL  
    3793     FAIL  
    3833     FAIL  
    3881       2048       5929     77
    3889     FAIL  
    3929     FAIL  
    4001    8388608    8392609   2897
    4049     FAIL  
    4057     FAIL  
    4073     FAIL  
    4129       8192      12321    111
    4153     FAIL  
    4177     FAIL  
    4201     FAIL  
    4217          8       4225     65
    4241     524288     528529    727
    4273     FAIL  
    4289     FAIL  
    4297     FAIL  
    4337     FAIL  
    4409     FAIL  
    4441     FAIL  
    4457         32       4489     67
    4481          8       4489     67
    4513       2048       6561     81
    4561     FAIL  
    4649     FAIL  
    4657     FAIL  
    4673     FAIL  
    4721     FAIL  
    4729         32       4761     69
    4793     FAIL  
    4801     FAIL  
    4817        512       5329     73
    4889     FAIL  
    4937     FAIL  
    4969     FAIL  
    4993     FAIL  
    5009         32       5041     71
    5081     FAIL  
    5113        512       5625     75
    5153     FAIL  
    5209     FAIL  
    5233     FAIL  
    5273     FAIL  
    5281     FAIL  
    5297         32       5329     73
    5393     FAIL  
    5417        512       5929     77
    5441     FAIL  
    5449     FAIL  
    5521       2048       7569     87
    5569     FAIL  
    5641     FAIL  
    5657     FAIL  
    5689     FAIL  
    5737     FAIL  
    5801        128       5929     77
    5849     FAIL  
    5857     FAIL  
    5881     FAIL  
    5897         32       5929     77
    5953     FAIL  
    6073     FAIL  
    6089     FAIL  
    6113        128       6241     79
    6121     FAIL  
    6217     FAIL  
    6257     FAIL  
    6329     FAIL  
    6337     FAIL  
    6353     FAIL  
    6361     FAIL  
    6449       8192      14641    121
    6473     FAIL  
    6481     FAIL  
    6521     FAIL  
    6529         32       6561     81
    6553          8       6561     81
    6569     131072     137641    371
    6577     FAIL  
    6673     FAIL  
    6689     FAIL  
    6737     FAIL  
    6761        128       6889     83
    6793     FAIL  
    6833      32768      39601    199
    6841     FAIL  
    6857         32       6889     83
    6961     FAIL  
    6977       2048       9025     95
    7001     FAIL  
    7057        512       7569     87
    7121     FAIL  
    7129     FAIL  
    7177     FAIL  
    7193         32       7225     85
    7297     FAIL  
    7321     FAIL  
    7369     FAIL  
    7393     FAIL  
    7417     FAIL  
    7433       8192      15625    125
    7457     FAIL  
    7481     FAIL  
    7489     FAIL  
    7529     FAIL  
    7537         32       7569     87
    7561          8       7569     87
    7577     FAIL  
    7649     FAIL  
    7673     FAIL  
    7681     FAIL  
    7753       2048       9801     99
    7793        128       7921     89
    7817     FAIL  
    7841     FAIL  
    7873     FAIL  
    7937       8192      16129    127
    7993     FAIL  
    8009     FAIL  
    8017     FAIL  
    8081     FAIL  
    8089     FAIL  
    8161     FAIL  
    8209     FAIL  
    8233     FAIL  
    8273          8       8281     91
    8297     FAIL  
    8329     FAIL  
    8353     FAIL  
    8369     FAIL  
    8377     FAIL  
    8513        512       9025     95
    8521        128       8649     93
    8537     FAIL  
    8609     FAIL  
    8641          8       8649     93
    8681     FAIL  
    8689     FAIL  
    8713     FAIL  
    8737     FAIL  
    8753     FAIL  
    8761     FAIL  
    8849     FAIL  
    8929     FAIL  
    8969       8192      17161    131
    9001     FAIL  
    9041     FAIL  
    9049     FAIL  
    9137     FAIL  
    9161     FAIL  
    9209     FAIL  
    9241     FAIL  
    9257      32768      42025    205
    9281        128       9409     97
    9337     FAIL  
    9377         32       9409     97
    9433     FAIL  
    9473     FAIL  
    9497       8192      17689    133
    9521     FAIL  
    9601     FAIL  
    9649     FAIL  
    9689        512      10201    101
    9697     FAIL  
    9721     FAIL  
    9769         32       9801     99
    9817     FAIL  
    9833       2048      11881    109
    9857     FAIL  
    9929     FAIL  
   10009     FAIL  
   10169         32      10201    101
   10177     FAIL  
   10193          8      10201    101
   10273       2048      12321    111
   10289     FAIL  
   10313     FAIL  
   10321     FAIL  
   10337     FAIL  
   10369     FAIL  
   10433     FAIL  
   10457     FAIL

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
I told it to just print the values $\pmod {4^t - 1}$ that could not be expressed in the prescribed manner. The first few primes that say "FAIL" can be proved to fail in Stefan's manner, as $241 \equiv 52 \pmod {63},$
$337 \equiv 22 \pmod {63},$ $569 \equiv 59 \pmod {255},$ $641 \equiv 641 \pmod {1023},$ $673 \equiv 43 \pmod {63}.$
Bad mod 15
       0       3       5       6       9      10      12


 Bad mod 63
       0       3       6       9      12      13      15      18      19      21
      22      24      25      27      30      33      36      37      39      42
      43      45      46      48      51      52      54      57      58      60


 Bad mod 255
       0       3       5       6       9      10      12      15      18      20
      21      24      25      27      29      30      31      33      35      36
      39      40      42      45      48      50      51      54      55      57
      59      60      63      65      66      69      70      71      72      75
      78      80      81      84      85      87      90      93      94      95
      96      99     100     102     105     108     110     111     114     115
     116     117     120     121     123     124     125     126     129     130
     132     135     138     140     141     144     145     147     150     151
     153     155     156     159     160     162     165     168     170     171
     174     175     177     179     180     183     185     186     189     190
     192     195     198     199     200     201     204     205     206     207
     209     210     213     215     216     219     220     222     225     228
     229     230     231     234     235     236     237     240     241     243
     245     246     249     250     252



 Bad mod 1023
       0       3       6       9      10      11      12      15      18      21
      22      24      26      27      30      33      36      39      40      42
      44      45      48      51      54      55      57      60      63      66
      69      72      75      77      78      81      83      84      87      88
      90      93      96      99     102     104     105     106     108     110
     111     114     117     120     121     123     126     129     132     135
     138     141     143     144     147     150     153     154     156     159
     160     162     165     168     171     173     174     176     177     180
     181     183     186     187     189     192     195     197     198     201
     204     205     207     209     210     211     213     216     219     220
     222     225     228     231     234     237     240     242     243     246
     249     252     253     255     258     261     264     267     270     273
     275     276     279     282     285     286     288     291     294     297
     299     300     301     303     305     306     307     308     309     312
     315     318     319     321     324     327     330     331     332     333
     336     339     341     342     345     348     351     352     354     357
     360     362     363     366     367     369     372     374     375     378
     381     383     384     385     387     390     393     396     399     402
     405     407     408     411     414     416     417     418     420     423
     424     425     426     429     432     435     438     440     441     444
     445     447     450     451     453     456     459     462     465     468
     471     473     474     477     479     480     483     484     486     489
     492     495     498     501     503     504     506     507     509     510
     513     514     516     517     518     519     522     525     528     531
     534     537     538     539     540     543     546     549     550     552
     555     558     561     564     567     570     572     573     576     579
     582     583     585     588     591     594     597     600     602     603
     605     606     609     612     615     616     618     621     624     627
     630     633     636     638     639     640     641     642     645     646
     648     649     651     654     657     660     662     663     666     669
     671     672     673     675     677     678     681     682     684     687
     690     692     693     696     699     702     703     704     705     708
     711     714     715     717     720     722     723     724     726     729
     732     735     737     738     741     744     747     748     750     753
     756     757     759     762     765     766     768     770     771     774
     777     780     781     783     786     788     789     792     795     798
     801     803     804     807     810     813     814     816     819     820
     822     825     828     831     834     836     837     840     842     843
     844     846     847     849     850     852     855     858     859     861
     863     864     867     869     870     873     876     879     880     882
     885     887     888     891     893     894     897     900     902     903
     906     909     912     913     915     918     921     924     927     930
     933     935     936     939     942     943     945     946     948     951
     954     957     960     963     966     968     969     972     975     978
     979     981     982     983     984     987     989     990     993     996
     999    1001    1002    1003    1005    1008    1011    1012    1013    1014
    1017    1018    1020

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= 
I could have saved some of the printing just above. The original question was about primes in any case, so any of the numbers such as $3,5 \pmod {15}$ are not useful for the original problem. At the same time, expanding to all integer values, we are asking when $n = m^2 - 2 \cdot 4^k = m^2 - 2 w^2$ where $w$ is a power of $2.$ That is, such $w$ is not divisible by any odd prime. By a simple argument involving Legendre symbol $(2 |q),$ we find that such $n$ cannot be divisible by $3,5,11,13, $ or by any prime $q$ with  $(2 |q)= -1.$ That is, such $n$ cannot be divisible by $3,5,11,13, $ or by any prime $q$ with  $ q \equiv 3,5 \pmod 8.$<|endoftext|>
TITLE: What is $\hat{A}=\{[\pi]:\pi$ is a irreducible representation of $A$} ( $A$ is a $C^*$-algebra)?
QUESTION [6 upvotes]: Let $A=\{f:[0,1]\to M_2(\mathbb{C}): $f continuous and $ f(0)=\begin{pmatrix} f_{11}(0) & 0 \\ 0 & f_{22}(0) \end{pmatrix}\}$ be a $C^*$-algebra with pointwise multiplication, involutions and norm $\|f\|=\sup\limits_{t\in [0,1]}\|f(t)\|_{op}$. I want to determine $\hat{A}=\{[\pi]:\pi$ is a irreducible representation of $A$}, (Here is $\pi\sim \rho :\iff$ there is an unitary operator $V:H_{\pi}\to H_{\rho}$ such that $V\pi(a)=\rho(a)V$ for all $a\in A$). 
I want to use the following fact: If $I$ is a closed ideal in a $C^*$-algebra $A$, it is $$\hat{A}=\hat{A}_I\coprod \hat{A}_{A/I}=\widehat{I}\coprod  \widehat{A/I},$$ where $\hat{A}_I=\{[\pi]\in \hat{A}: \pi(I)\neq 0\}$ and $\hat{A}_{A/I}=\{[\pi]\in \hat{A}: \pi(I)=0\}$. 
Consider $I=\{f\in A: f(0)=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\}$. $I$ is a closed Ideal in $A$, because $I$ is the kernel of the evaluation-homomorphism $\epsilon_0:A\to M_2(\mathbb{C}), f\mapsto f(0)$. Therefore $I$ is a $C^*$-subalgebra of $A$. Then  it follows that $I$ is isomorphic to $C_0((0,1],M_2(\mathbb{C}))$ as $C^*$-algebra. Maybe we get $\widehat{A/I}$ now, because we know what $A/I$ is, it is $A/I=A/ \ker(\epsilon_0)\cong Im(\epsilon_0)\cong \mathbb{C}^2$. But here I'm stuck, I'm not sure what $\widehat{A/I}$ should be and I really don't know what $\hat{I}$ is, I assume $\hat{I}$ is homeomorphic to $(0,1]$, but I can't prove it. 
I need that because I want to describe the hull-kernel-topology on $\hat{A}$.
I'm not sure if the question is ok for mathoverflow, I'm sorry if not.
But I appreciate your help. 
edit: is $\widehat{A/I}$ a discrete set with 2 elements?

REPLY [3 votes]: Let $X$ be a compact Hausdorff space, $A$ a unital $C^*$-algebra, first you can prove that $$\gamma:X\times \hat{A}\to \widehat{C_(X,A)},\; (x,[\pi])\mapsto  [\pi\circ ev_x]$$ 
is a bijection, where $ev_x: C(X,A)\to A,\; f\mapsto ev_x(f):=f(x)$ is the evaluation map.
Then you can deduce the same for $\widehat{C_0(X,A)}$ and  $X\times \hat{A}$, if $X$ is local compact, Hausdorff and if $A$ is an arbitrary $C^*$-algebra. 
[Note: $\widehat{C_0(X,A)}$ and $\hat{A}$ are homeomorphic if $\widehat{C_0(X,A)}$ and $\hat{A}$ are endowed with the kern-hull-topology and $X\times \hat{A}$ takes the product topology. ]
Therefore it is $\hat{I}=\widehat{C_0( (0,1], M_2(\mathbb{C}) )}=(0,1]\times \widehat{M_2(\mathbb{C}) }$. 
Next, note that all irreducible $*$-representations $\pi:M_2(\mathbb{C})\to L(H_{\pi})$ are equivalent to the irreducible $\ast$-representation $id:M_2(\mathbb{C})\to  M_2(\mathbb{C})\cong L(\mathbb{C}^2)$, i.e. $ \widehat{M_2(\mathbb{C}) }=\{[id]\}$. So $\hat{I}$ is basically $(0,1]$.
To obtain $\hat{A}$ we also need $\widehat{A/I}$, because as sets we have the equality $$\hat{A}=\widehat{I}\coprod  \widehat{A/I}.$$ It is $\widehat{A/I} =\widehat{ \mathbb{C}\oplus \mathbb{C} }=\hat{\mathbb{C}}\coprod \hat{\mathbb{C}}$, a discrete set with 2 elements.
In summary: $\hat{A}$ has the form $\{\pi_t|t\in (0,1]\}\cup \{\chi_1,\chi_2\}$ with $\pi_t(f)=f(t)$ and $\chi_i(f)=f_{ii}(0)$.
Thanks to Rasmus Bentmann for the help.<|endoftext|>
TITLE: ACC (DCC) implies upper (lower) sets are upper (lower) closure of antichains?
QUESTION [5 upvotes]: I have read around (e.g. in Wikipedia) that if the ascending (descending) chain condition holds, all upper (lower) sets are the upper (lower) closure of an antichain, but I cannot find a proof. 
More precisely, let $(P, \leq)$ be a poset, $A(P)$ be the set of antichains on $P$, $U(P)$ be the set of upper sets on $P$ and $L(P)$ be the lower sets. Let $\uparrow$ ($\downarrow$) be the upper (lower) closure:
$$
  \uparrow S = \{p \in P: \exists s \in S: s\leq p  \}
$$
$$
  \downarrow S = \{p \in P: \exists s \in S: s \geq p\}
$$
Then I'm looking for a proof of the following:
Lemma (?): If DCC holds, then for all $u\in U(P)$, there exists $a\in A(P)$ such that $u=\,\uparrow a$.
Dually:
Lemma (?): If ACC holds, then for all $l\in L(P)$, there exists $a\in A(P)$ such that $l=\,\downarrow a$.
(I can see how this is true for finite posets but not more generally.)

REPLY [3 votes]: $\newcommand\P{\mathbb{P}}$Let $U$ be any upper set in a partial
order $\langle\P,\leq\rangle$, which satisfies the descending
chain condition, which asserts that every descending sequence of
points terminates. In other words, the order is a well-founded relation,
which is equivalent to saying that every nonempty subset of $\P$
has a minimal element. (Note that one needs the principle of dependent choice to prove that these characterizations are equivalent in all relations.) 
Let $A$ consist of the minimal elements of $U$. These form an
antichain (in the sense of pairwise incomparability). Since $A\subset U$ and $U$ is upward-closed, we clearly have 
$A{\uparrow}\subset U$. But the converse inclusion is also true, because if
$y\in U$, then consider the collection of $\{u\in U\mid u\leq
y\}$, which is a nonempty subset of $\P$. Thus, it has a minimal
element $a$, which is also minimal in $U$. So $a\in A$ and $a\leq
y$, which puts $y\in A{\uparrow}$. So $A{\uparrow}=U$, and thus $U$ is
the up-set of an antichain.
The dual situation for down-sets, using the ACC, is analogous.
Update. Finally, let me add that without the axiom of dependent choice, the lemmas are no longer necessarily true. For example, it is consistent with ZF that there is a tree of finite sequences, such that every sequence has an extension in the tree, but there is no infinite branch through the tree (basically, one is not able to choose successors successively in order to build the branch). If we orient this tree growing downwards, then the relation has the DCC, precisely because the tree has no infinite branch. But meanwhile, the whole tree is upward-closed, but it is not the up-set of an antichain, because every node in the tree has extensions below.
Since these kind of trees are intimately connected to the failure of dependent choice, this shows that the lemmas are actually each equivalent over ZF to the principle of dependent choice.<|endoftext|>
TITLE: Why would one "attempt" to define points of a motive as $\operatorname{Ext}^1(\mathbb{Q}(0),M)$?
QUESTION [20 upvotes]: I'm a novice when it comes to motives. (I've read multiple introductory texts.)
I'm attempting to read Galois Theory and Diophantine geometry by Minhyong Kim. In it, he says that "One might attempt, for example, to define the points of a motive M over
Q using a formula like
$$\operatorname{Ext}^1(\mathbb{Q}(0),M)$$
or even
$$\operatorname{RHom}(\mathbb{Q}(0),M)$$,
hoping it eventually to be adequate in a large number of situations."
I'm stumped. Why would one attempt to define it as $\operatorname{Ext}^1(\mathbb{Q}(0),M)$? Why the word "even" when describing $\operatorname{RHom}(\mathbb{Q}(0),M)$? Why only "in a large number of situations"? I'm confused as to the intuition here.
A hint comes in the next sentence: "However, even in the best of all
worlds, this formula will never provide direct access to the points of a scheme, except in very special
situations like $M = H_1(A)$ with $A$ an abelian variety."
My guess that the intuition lies in the abelian variety case. Can you help me understand what the author meant by all of this? I'm sure it's trivial for experts, but it's currently a mystery to me.

REPLY [13 votes]: In the spirit of You Could Have Invented Spectral Sequences by T.Chow, I claim you could have invented $\operatorname{Ext}^{1}(\mathbb Q(0),M)$ as group of "rational points" of a motive. Here is how.
The motivation comes from conjectures on special values of $L$-function. Before the general conjectures were formulated by S.Bloch and K.Kato, there were two settings in which the statement of the conjecture was known and which thus provided the main inspiration: the special value of the Dedekind zeta function of a number field $K$ at $s=0$ and the conjecture of Birch and Swinnerton-Dyer for the $L$-functions of an abelian variety $A$ at $s=1$.
In the first case, the special value of the $L$-function is a rational multiple of a regulator, which is equal to the covolume of the lattice $\mathcal O_{K}^\times$ inside $\mathbb R^{r_1+r_2-1}$ and the denominator of this rational number is the cardinality of the torsion group of $\mathcal O_{K}^{\times}$. In the second case, the regulator involves the lattice of rational points (more precisely, their heights) and the rational value equals the cardinality of the torsion group of rational points of $A$ and of its dual. In both cases, then, the transcendental and rational part involve some lattice.
The works of P.Deligne, A.Beilinson, S.Bloch and K.Kato on special values of $L$-function of motives aimed at generalizing this lattice to general (putative) motives.
But for the Tate motive $X=\mathbb Q(1)$, we have  that $H^1(X)$ is $\operatorname{Ext}^1(\mathbb Q(0),\mathbb Q(1))$ (for any realization, that is extending the scalar to $\mathbb R$ and considering Deligne cohomology or to $\mathbb Q_{p}$ for some prime $p$ and considering étale or Galois cohomology). Likewise, Bloch showed in A Note on Height Pairings, Tamagawa Numbers, and the Birch and Swinnerton-Dyer Conjecture (Invent. math. 1980) that all the terms of the BSD conjecture could be expressed in terms of real and $p$-adic realizations of the pairing between $\operatorname{Ext}^1(A,\mathbb Q(1))=\operatorname{Ext}^1(\mathbb Q(0),A^*(1))$ and $A$.
Based on these evidence, it seems reasonable to conjecture that $\operatorname{Ext}^1(\mathbb Q(0),M)$ is the group playing for any motive the role rational points play for abelian varieties (extension in the category of mixed motives, whatever that means). This is further confirmed by the fact that Beilinson constructed (conjecturally) a regulator map with source in this space after extension of scalars to $\mathbb R$. Finally, a complete formalism including $p$-adic places was developed by Jean-Marc Fontaine and Bernadette Perrin-Riou (see for instance Valeurs spéciales des fonctions $L$ des motifs Fontaine (1992)).
Note that it would be more accurate to say that this group actually corresponds to the rational points of the dual motive (as this is what happens for abelian varieties if I didn't mess up my normalizations) and that this group is an analogue of rational points only in the limited sense of conjectures on special values of $L$-functions (you could equally validly call it the group of units of a motive, adapting the vocabulary of number fields rather than of abelian varieties). For other properties of rational points, especially diophantine ones, there is no reason to expect that this group will be interesting, which I guess explains M.Kim's remark. To see clearly the discrepancy, you can remark that the Ext group is supposed to be a Chow group when $M$ is pure of weight $-1$ so involves rational cycles, not points, when the motive comes from a variety of large dimension. 
I am a bit puzzled by the "even" part of the remark as the category of mixed motives, if it exists and if all the conjectures about it are true (a rather bold requirement, admittedly), is of cohomological dimension 1, so no higher non-trivial $\operatorname{Ext}$ groups are believed to exist.<|endoftext|>
TITLE: Is there a notion of "flat vector bundle over a topological space"?
QUESTION [13 upvotes]: I am reading this paper and at the top of page 5 the author makes reference to categories consisting of flat complex vector bundles over $X$ where $X$ is an arbitrary topological space. However, the only notion of flat vector bundle which I have seen uses a connection on the vector bundle, which in turn requires that $X$ have a cotangent bundle, hence $X$ must be a smooth manifold.
My questions are therefore as follows:

What is the definition of a flat vector bundle over a topological space?
What is the action of $\pi_1(X)$ on such a vector bundle? (I need to understand $H^\bullet(X;\text{Hom}_X(V,W))$ for flat vector bundles $V,W$ over $X$.)

REPLY [21 votes]: A flat vector bundle over a topological space is a bundle whose transition functions can be taken to be locally constant; equivalently, over a path-connected space, it's the same data as a principal $G$-bundle ($G = GL_n(\mathbb{R})$ or $GL_n(\mathbb{C})$ as appropriate) where $G$ is given the discrete topology. Over a reasonable space $X$ this is the same thing as a functor from the fundamental groupoid $\Pi_1(X)$ to vector spaces.

REPLY [10 votes]: A way to rephrase QY's answer is to say that a vector bundle E is flat if there is a local system L such that $L \otimes_{\mathbb{C}} C_X \simeq E$ where $C_X$ is the trivial vector bundle (or, said differently, $C_X$ is the bundle whose sections on an open U are given by $C_X(U)$ - the continuous functions $f\colon U \to \mathbb C$).<|endoftext|>
TITLE: Gaps between descending order statistics
QUESTION [7 upvotes]: Let $\{X_{1},X_{2},\cdots,X_{n}\}$ be a random sample of size $n$. Denote $(X_{(1)},X_{(2)},\cdots,X_{(n)})$ to be its descending order statistics. Define gap $g_{i}(n)$ to be $g_{i}(n)=X_{(i)}-X_{(i-1)},1\leq i\leq n$. My question is what is the limiting distribution of $g_{i}(n)$ as $n\to\infty$ or after some scaling if necessary. Are there any results on this problem?

REPLY [2 votes]: If the $X_i$ are independent and uniformly distributed on $[0,1]$, the (finite) point process $\sum_1^n \delta_{nX_i}$ converges in law to a Poisson point process on $\mathbb R^+$ with unit intensity, so the scaled gaps $ng_i(n)$ are exponentially distributed (density $e^{-x}$) in the limit.
More generally, if $F(y)=Pr\{X_i>y\}$ , the distribution of $ng_i(n)$ should converge to $\exp[-x/\ell_i]\ dx/\ell_i$ where $\ell_i=F'(F^{-1}(i/n))$, at least if $F$ is smooth enough (and $F'(F^{-1}(i/n))>0$). There are probably published results of this kind (Kolmogorov?...)<|endoftext|>
TITLE: Maximal subgroups of special linear groups over finite fields
QUESTION [5 upvotes]: Let $p$ be a prime number, and denote by $\mathbb{F}_p$ the field with $p$ elements. 
Is there a classification of the maximal subgroups of $G = \mathrm{SL}_3(\mathbb{F}_p)$ ? 
I am interested in the indices of these subgroups in $G$ and in their ranks (minimal cardinality of a generating set).

REPLY [9 votes]: The original references are:
H.H. Mitchell.
Determination of the ordinary and modular ternary linear groups.
Trans. Amer. Math. Soc. 12 (1911), 207-242.
R.W. Hartley.
Determination of the ternary collineation groups whose coefficients
lie in the $\mathrm{GF}(2^n)$.
Ann. of Math. 27 (1925/6), 140-158.
Our book is
The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, John N. Bray, Derek F. Holt, Colva M. Roney-Dougal, London Mathematical Society Lecture Notes 407, CUP, 2013.
I hope you won't find the complete book online, because that might adversely affect its sales! If you send me an e-mail, I can send you the table for ${\rm SL}(3,q)$ from the book.<|endoftext|>
TITLE: Another formula for Bell numbers
QUESTION [7 upvotes]: Here is an observation (thanks to OEIS):
$$\sum_{i=0}^\infty \frac{i^k}{i!}= B_k e,$$ where $B_k$ is the $k$-th Bell number. I might be having reading comprehension issues, but I don't see this formula in the OEIS notes. I assume this is very well known - can someone point me at a reference or a simple proof?

REPLY [12 votes]: That is essentially Dobinski's formula.<|endoftext|>
TITLE: Is each closed convex set a manifold with corners?
QUESTION [10 upvotes]: Assume that $C$ is a convex set in $\mathbb{R}^{n}$ with non empty interior.
Then consider its closure, is it a smooth manifold with corners?
Edit: 
1) The closure of $C$ should be a smooth manifold with corners equipped with the induced smooth structure from $\mathbb{R}^{n}$.
2) With non empty interior I mean that $C$ contains a non empty open set.

REPLY [14 votes]: See de Rham's "cutting corners" curve.  



Picture from an MO answer by Bill Thurston; also see a description there.
The limiting curve is $C^1$ but not $C^2$.  It has a tangent everywhere, but curvature zero almost everywhere.  When you say "manifold with boundary" what do you require?
Plug: English translation of de Rham's paper is in my book Classics on Fractals

REPLY [14 votes]: Let $C \subset S^1 \subset \mathbb{R}^2$ be a Cantor set. Let $H_C$ be its convex hull, the smallest closed subset of $\mathbb{R}^2$ containing $C$. Then $H_C$ is not a manifold with corners. Its boundary $\partial H_C$ is $C^1$ at every point except for the countably many points which are endpoints of components of $S^1-C$, these points forming a dense subset of $C$. Also, $\partial H_C$ fails to be $C^2$ at every point of $C$. But $H_C$ is homeomorphic to the closed 2-disc, indeed there is an ambient homeomorphism $\mathbb{R}^2 \mapsto \mathbb{R}^2$ taking $H_C$ to the unit closed disc.<|endoftext|>
TITLE: "Additive version" of Kronecker product
QUESTION [7 upvotes]: Let $A$ and $B$ be two square matrices with complex entries.
Let $\lambda_1, \ldots, ,\lambda_n$ be the Eigenvalues of $A$ and 
$\mu_1, \ldots, ,\mu_m$ be the Eigenvalues of $B$.
 Then the Eigenvalues of the Kronecker product are exactly the products $\lambda_i \cdot \mu_j$.
Is there an analogue for the sums of Eigenvalues? My precise question is the following:
For given natural numbers $m$ and $n$ are there polynomials $f_{rs} \in \mathbb{C}[x_{ij},y_{kl}: \, 1 \leq i,j \leq m, \, 1 \leq k,l \leq n]$ such that for every $n \times n$ matrix  $A$ and every $m \times m$ matrix $B$ the Eigenvalues of the matrix $C=(f_{rs}(A,B))_{1 \leq r,s \leq mn}$ are exactly the sums of an Eigenvalue of $A$ and an Eigenvalue of $B$? Here $f_{rs}(A,B)$ stands for the complex number obtained by substituting $x_{ij}$ by the $(i,j)$th entry of $A$ and $y_{ij}$ by the $(i,j)$th entry of $B$.
I am aware of some similar construction where the matrix $C$ has the desired Eigenvalues among others. But for me it is important that they are no other Eigenvalues.

REPLY [7 votes]: Federico already mentioned the keyword. The precise answer may be found among others as Theorem 13.16, of this book. (That theorem makes a restriction to real matrices, but that is not necessary).<|endoftext|>
TITLE: Loci in the moduli space of K3 surfaces associated to lattices
QUESTION [6 upvotes]: The moduli space of K3 surfaces forms a 20-dimensional family with countably many 19-dimensional components $M_d$ corresponding to the polarized K3s $(X,L)$ with $L^2=d$. The moduli space $M_d$ has a natural locus $M_d^W$ for each lattice $W$ with an embedding $W\subset H^{1,1}(X)\cap H^2(X,\mathbb Z)$. Is it known how these loci intersect? 
For instance, is it true that $M_{d}^W\cap M_{e}\neq \emptyset$ for each $W,d,e$ provided $M_{d}^W\neq \emptyset$?

REPLY [3 votes]: I won't completely answer your question, but will try to just rephrase it in a certain way. You are asking when two given moduli spaces of lattice-polarized K3 surfaces $M_L$ and $M_{L'}$ intersect. This is equivalent to the existence of a lattice with $L,L'\hookrightarrow N$ such that $M_N$ is non-empty. Necessary is the existence of such an $N$ which is a sub-lattice of the K3 lattice of rank less than or equal to $20$ and hyperbolic signature.
One approach to the second question would be to consider a hyperbolic lattice $L$ whose quadratic form represents the integer $d$, and enlarge it to a hyperbolic lattice $N$ which also represents the integer $e$. For instance by adding in vectors of negative square in $L^{\perp}$. This may help resolve your question in the low-rank case. The only obstructions to a lattice of rank $5$ or more representing an integer are congruence obstructions. I think one can eliminate any such obstructions by increasing the rank of $L$ by two. So maybe $rk(L)\leq 18$ is fine?
The Picard lattice of any Kummer surface will represent every integer because it contains a rank $16$ lattice with intersection form $(-2)^{16}$. When  the rank of $L$ is $19$ or $20$, your K3 is necessarily a Kummer surface, so perhaps that resolves your second question.<|endoftext|>
TITLE: What is the maximal possible rank of a subgroup of a special linear group mod a prime?
QUESTION [7 upvotes]: Let $p$ be a prime number, and let $\mathbb{F}_p$ be the unique field of cardinality $p$. 
What is $\max \{d(H) : H \leq \mathrm{SL}_3(\mathbb{F}_p)\}$?
Here we denote by $d(G)$ the smallest cardinality of a generating set of a group $G$.

REPLY [5 votes]: I am very confident that, for $q=p^e$ with $p$ prime,  the answer is $2e$ when $q$ is even, and $2e+1$ when $q$ is odd. There is an elementary abelian subgroup $P$ of rank $2e$ with
$$P=\left( \begin{array}{ccc}1&a&b\\0&1&0\\0&0&1\end{array} \right).$$
When $q$ is odd, all elements of $P$ are inverted by the diagonal matrix $t$ with entries $(1,-1,-1)$, and $\langle P,t \rangle$ has rank $2e+1$. Note that this maximal rank is at least $3$ except in the case $q=2$, when you can check directly that it is $2$.
I have not attempted to write down a complete formal proof, and I don't plan to. If you need a formal proof, then I would happy to help! Here is a sketch proof.
Let $G \le {\rm SL}(3,q)$. I proved a result in a recent paper with Roney-Dougal that any finite primitive irreducible subgroup of ${\rm GL}(n,K)$ for any field $K$ has rank at most $1 + \lceil(2\log_3 2)\log_2 n \rceil$, which gives rank at most $3$ for $n=3$.
An imprimitive irreducible subgroup is a subgroup of $(q-1)^2:S_3$ and it is not hard to show that any such group has rank at most $3$.
So suppose that $G$ is reducible. By replacing $G$ by its dual if necessary, we can assume that it fixes a $2$-dimensional subspace, and so it is a subgroup of the maximal parabolic subgroup $N:{\rm GL}(2,q)$, where $N$ is elementary abelian of order $q^2$ with the natural induced action of ${\rm GL}(2,q)$ on $N$. Let $H$ be the image of the projection of $G$ onto ${\rm GL}(2,q)$.
If $H$ acts irreducibly on $N$, then the rank of $G$ is equal to the rank of $H$ (that is a known result). In a 1991 paper, Kovacs and Robinson proved that any finite completely reducible subgroup of ${\rm GL}(n,K)$ (for any field $K$) has rank at most $3n/2$. So $H$ and hence $G$ has rank at most $3$ in this case.
So we are left with the situation when $H$ does not act irreducibly on $N$, when $G$ is conjugate to a group $U$ of upper triangular matrices. Then $U$ has the strucure $Q:D$, where $Q$ is a $p$-group of class $2$ and rank $2e$, and $D \cong C_{q-1}^2$. I haven't tried to write out a formal proof in this case. The maximal ranks of subgroups of $Q$ and $D$ are $2e$ and 2, so we easily get a bound of $2e+2$. But in fact generators projecting nontrivially onto $D$ generally reduce the numbers of generators in $Q$ that you need. The only exception is the involution $t$ above, giving the maximal rank $2e+1$ for odd $q$.<|endoftext|>
TITLE: Cardinality of definable sets of reals
QUESTION [11 upvotes]: Throughout this question we assume ZFC.
If CH holds, then the following is obvious:

(S) Every definable infinite subset of $\mathbb R$ has size either $\aleph_0$ or $2^{\aleph_0}$.

(It's true because every subset of reals satisfies this, in particular so does every definable)
Recently my friend has asked me whether the same is true if we assume CH fails. My answer was that this is independent of ZFC, because I'm fairly sure that results about pointwise definable models will gives us such model of ZFC + not CH (correct me if I'm wrong!) in which a set of reals of size $\aleph_1$ is definable, and I can recall seeing a result that (S) can hold, possibly under some large cardinal assumption. However, I failed to find a reference for that.
Hence my question is for a reference of the following result (if my memory isn't failing me and it's actually true):

It is consistent (relatively to large cardinals) that CH fails and (S) holds.

I also wanted to ask if this is true if CH fails badly:

Is it consistent for every cardinal $\kappa$ that (S) holds and $2^{\aleph_0}>\kappa$?

Thanks in advance.

REPLY [3 votes]: EDIT: This isn't really an answer - see Andres' comment - but you might find it interesting, and it is too long for a comment.
Your second question is vague, but has an affirmative answer in the following sense:

Let $V\models ZFC+V=L$, and let $\kappa$ be a cardinal in $V$. Then there is a forcing extension $V[G]$ of $V$ in which $2^{\aleph_0}>\kappa$ but there is a definable set of reals of cardinality $\aleph_1$.

Proof: Add $\kappa^+$-many Cohen reals; this bumps the continuum up to $\kappa^+$. Meanwhile, it does not collapse $\aleph_1$, so the set $V\cap\mathbb{R}$ of ground reals is still of size $\aleph_1$. And $V\cap\mathbb{R}=(\mathbb{R}^L)^V=(\mathbb{R}^L)^{V[G]}$, so is definable in $V[G]$ without parameters.

The first question seems harder. Any countable model of ZFC has a class forcing extension which is pointwise definable, but this class forcing extension doesn't seem to preserve large cardinal properties. Meanwhile, "big enough" large cardinals kill off all the obvious ways to try to define sets of reals of intermediate cardinality.<|endoftext|>
TITLE: A question on characterizing a Banach space containing no copy of $l_{1}$
QUESTION [5 upvotes]: Let $X$ be a Banach space. My question is: $X$ contains no copy of $l_{1}$ if and only if any operator from $X$ to $l_{1}$ is compact? I guess that the necessary part may be true. But is the sufficient part true? At least, the sufficient part is true for $X=c_{0},l_{p}(1<p<\infty)$.

REPLY [3 votes]: Since weakly compact operators into $\ell_1$ are compact, and since by a result of Kadec and Pelczynski every non-weakly compact operator into $\ell_1$ fixes a copy of $\ell_1$, we have that if $X$ contains no copy of $\ell_1$ then every operator from $X$ to $\ell_1$ is compact. 
However, the converse is not true in general. Since $ C [0,1] $ is universal for separable Banach spaces, it contains a copy of $\ell_1$, but every operator from $ C [0,1] $ is compact; indeed, Pelczynski showed that non-weakly compact operators from a $ C (K) $ space fix a copy of $ c_0$, but there is no copy of $ c_0$ in $\ell_1$.

Typed painfully slowly on my Samsung Galaxy S3<|endoftext|>
TITLE: Pullback of line bundles and existence of divisors representing line bundles
QUESTION [7 upvotes]: I am studying the proof of Lemma 7.2. on page 108 in Dolgachev's "Lectures on invariant theory". It states (everything is done over the field $k = \overline{k}$):

Let $X$ be a normal affine variety (for example, nonsingluar) and $G$ is a connected affine algebraic group. Let $L$ be a line bundle on $G \times X$. Then there exist line bundles $L_1 \in Pic(G)$, $L_2 \in Pic(X)$ such that $L \cong pr_1^*(L_1) \otimes pr_2^*(L_2)$ (where $pr_i$ is the $i$-th projection of $G \times X$).

The proof uses the following facts about algebraic groups:

$G$ containes a Zariski open set $U$, which is isomorphic to $(\mathbb{A}^1 \setminus \{0\})^N$.
$pr_2^* \colon Pic(X) \to Pic((\mathbb{A}^1 \setminus \{0\}) \times X)$ is an isomorphism.

Then it is clear that $L$ restricted to $U \times X$ is isomorphic to $pr_2^*(L_2)$ for some $L_2 \in Pic(X)$.
The following is said afterwards:
Let $D$ be a Cartier divisor on $G \times X$ representing $L$.
Question 1: Why does such a $D$ exist? $X$ is not projective, so I don't quite see why $D$ should exist.
The proof continues as follows:

Then the preceding isomorphism implies that there exists a Cartier Divisor $D_2$ on $X$  such that $D' = D-pr_2^*(D_2)|_{U \times X} = 0$.

Question 2: Why exactly does $D_2$ exist?
Now, take any irreducibly component $D_i'$ of $D'$ and consider the projection $D_i' \to G$. The image is contained in $G \setminus U =: Z$. By the theorem on the dimension of fibres, the fibres of this projection have dimension equal to $\dim X$. Then the last part which I don't understand follows:

This easily implies that $D_i' = pr_1^*(D_i)$, where $D_i \subset  Z$.

Question 3: Why?
Thanks in advance for any help.

REPLY [2 votes]: Question 1: Maybe there's some subtlety I'm missing here, but isn't there a Cartier divisor attached to any line bundle?  You just pick an affine where it's trivial, and let the trivialization be your rational section (using the definition here).
Question 2: Since $pr_2^*$ is an isomorphism to $\operatorname{Pic}(U\times X)$, it's surjective.
Question 3: Because $Z\times X$ is already a divisor.  Thus, the only divisor that could be supported on it is a sum of the components of $Z\times X$.<|endoftext|>
TITLE: Is dgCat a category or a 2-category?
QUESTION [8 upvotes]: Let us consider dgCat, the "collection" of all small dg-categories. In On differential graded categories and Lectures on dg categories the authors state that they form a category, i.e.  dgCat has small dg categories as objects and the dg functors as morphisms. They then give a model structure on dgCat.
However, as pointed in What do DG-categories form?, it is more likely that dgCat is a 2-category: it does not make too much sense to say whether two dg-functors $F$ and $G$ are equal.  Instead we can talk about morphisms between dg-functors. 
How to integrate these two viewpoint? In particular, could we still have a model category structure on dgCat if we treat it as a 2-category?

REPLY [11 votes]: Adeel answer is perfect, I will be more basic. 
There is many notions of "2-category" structure here around. 
1) the category of small dg-categories $\mathbf{dgCat}$ is symmetric monoidal closed category. Closed means that there is an internal $HOM$ in the sense that for any two small dg-categories $A$ and $B$ there is $HOM(A,B)\in \mathbf{dgCat} $ such that there is a natural isomorphism of sets $\mathbf{dgCat}(X,HOM(A,B))\cong \mathbf{dgCat}(X\otimes A,B)$. That means $\mathbf{dgCat}$ is enriched over it self. 
There is a functor from $H^{0}:\mathbf{dgCat}\rightarrow \mathbf{Cat}$ which gives you an enrichment of the category $\mathbf{dgCat}$ over $\mathbf{Cat}$, hence you can see $\mathbf{dgCat}$ as a 2-category. 
On an other hand, the internal $HOM(-,-)$ described before has the wrong homotopy type, you can not derive it since it does not take Dwyer-Kan equivalences (between fibrant-cofibrant objects) to  Dwyer-Kan equivalences (the $\mathbf{dgCat}$ is not symmetric monoidal model category in the sense of Hovey). Bertand Toen constructed, for the model category $\mathbf{dgCat}$, the right notion of the derived internal hom denoted by $RHOM(A,B)\in \mathbf{dgCat}$ (using bimodules, I will not write the details). Moreover this new derived $RHOM(A,B)$ induces the derived Mapping space $Map
_{\mathbf{dgCat}}(A,B)$ via the nerve functor of some well choosen subcategory of $RHOM(A,B)$). This new derived internal allows you to see the category $\mathbf{dgCat}$ as $(2,\infty)$-category and in the same time as symmetric monoidal $(1,\infty)$-category.  
An important consequence is the following isomorphism in $Ho(\mathbf{sSet})$:
$$Map_{\mathbf{dgCat}}(A\otimes^{L}B,C)\cong  Map_{\mathbf{dgCat}}(A,RHOM(B,C))$$<|endoftext|>
TITLE: On the zero set of a $C^2$ function on $[0,1]^2$
QUESTION [6 upvotes]: Let $f:[0,1]^2\rightarrow \mathbb{R}$ be a twice continuously differentiable function with the property that for all $x\in [0,1]$, there is an interval $I_x\subset [0,1]$ such that $f(x,y)=0$ for all $y\in I_x$. Does it follow that there must exist an open ball in $[0, 1]^2$ where $f$ is identically $0$?

REPLY [12 votes]: Yes. Each $I_x$ contains some segment $[p,q]$ with rational endpoints. Let $A(p,q)$ be a set of corresponding $x$. It is impossible by Baire category theorem that each $A(p,q)$ is nowhere dense. Thus there exist $p,q$ and a segment $\Delta$ of positive length such that $A(p,q)$ is dense in $\Delta$. Hence $f$ does vanish on $\Delta\times [p,q]$, continuity of $f$ is enough here.<|endoftext|>
TITLE: How (if at all) does category theory deal with situations where the usual notion of isomorphism isn't right?
QUESTION [7 upvotes]: This question will potentially rub some people the wrong way; I can't do much about this, except state right here at the outset that this question is motivated by a genuine desire to understand, and to conceptualize things better and more clearly. I realize that this isn't the "usual" way of conceptualizing things under category theory; all I can say is that maybe there are people out there who have thought about this point of view already and that, if so, I want to understand what they already understand.
There's a category-like object that can be described (in vague terms) as follows.
Objects. Well-ordered sets.
Arrows. Functions.
"Special" arrows. Monotone functions.
Monoidal product. Given well-ordered sets $X_0$ and $X_1$, we define that their monoidal product $X_0 \otimes X_1$ is the result of taking the coproduct of these objects viewed as posets, and then declaring that each element of $X_1$ is greater than each element of $X_0$. Given functions $f_0 : X_0 \rightarrow Y_0$ and $f_1 : X_1 \rightarrow Y_1$, there is a corresponding function $f_0 \otimes f_1 : X_0 \otimes X_1 \rightarrow Y_0 \otimes Y_1$, defined in the obvious way. If $f_0$ and $f_1$ happen to be "special" (i.e. they're monotone), then so too is $f_0 \otimes f_1$.
Call this objects $\mathbf{Wos}$ (for "well-ordered sets"). Then we have appear to have a "bifunctor" $$\otimes : \mathbf{Wos}, \mathbf{Wos} \rightarrow \mathbf{Wos}.$$
The weird thing about this situation is that under the usual definition of "isomorphic," $\mathbf{Wos}$ has isomorphic objects that aren't the same. For example, the ordinal $\omega$ and $\omega 2$ are "isomorphic" as objects of $\mathbf{Wos}$, in the sense that there exists an arrow $\omega \rightarrow \omega 2$ that has a two-sided inverse, despite that $\omega$ and $\omega 2$ aren't isomorphic as ordered sets.
The standard solution, of course, is to just "throw out" all the morphisms that aren't monotone. This seems a bit drastic though; after all, the arrow $f_0 \otimes f_1$ makes sense even if $f_0$ and $f_1$ aren't monotone. In other words, the bifunctor $\otimes$ is defined on the whole of $\mathbf{Wos}$, not just the wide subcategory of special morphisms. Therefore, let us instead change our notion of isomorphism; let us simply define that two objects of $\mathbf{Wos}$ are isomorphic iff there exists a special arrow $X \rightarrow Y$ that has a special two-sided inverse.
That's fine, but I want a more systematic viewpoint here. As I see it, two objects $X$ and $Y$ of a category-like structure ought to be "the same" iff $\mathrm{Hom}(X,-)$ and $\mathrm{Hom}(Y,-)$ are naturally isomorphic. However, and here's the rub, I wish to view $\mathrm{Hom}(X,-)$ as a functor defined on the whole of $\mathbf{Wos}$, not just the subcategory of special arrows. And I want $\mathrm{Hom}(X,-)$ and $\mathrm{Hom}(Y,-)$ to be naturally isomorphic iff there is a special arrow $X \rightarrow Y$ with a special inverse.
My question is as follows.

Question. How (if at all) does category theory deal with (or think about, or conceptualize) these kinds of situations, where the usual notion of isomorphism isn't right?

Having motivated the question, let me finish up by offering a "minimal example." There is a category-like object given as follows.
Objects. Sets equipped with a distinguished subset.
Arrows. Functions.
"Special" arrows. Functions that preserve membership in the distinguished subset.
Denote this $\mathbf{Sds}.$ The neat thing about this is that its "self-enriched," in the sense that given objects $X$ and $Y$ of $\mathbf{Sds}$, the collection of functions $X \rightarrow Y$ forms an object of $\mathbf{Sds}$ in a natural way, by taking the distinguished subset of $\mathrm{Hom}(X,Y)$ to consist of precisely those functions that preserve membership in the distinguished subset. In fact, $\mathbf{Sds}$ seems to have "hom-tensor adjunction," in the sense that the object of functions $X \times Y \rightarrow Z$ is the "same as" the object of functions $X \rightarrow \mathrm{Hom}(Y,Z).$

REPLY [6 votes]: One standard way of looking at your first example would be to see it as two categories: the usual category $\newcommand{WOS}{\mathbf{WOS}}\WOS$ of well-ordered sets and monotone maps, and the category $\newcommand{\fn}{\mathrm{fn}}\WOS^\fn$ with arbitrary functions as morphisms.  Then $\omega$ and $\omega^2$ are not isomorphic as objects of $\WOS$, but they are isomorphic as objects of $\WOS^\fn$ (equivalently, their underlying sets are isomorphic).
But if you really want to see the pair $\WOS \subseteq \WOS^\fn$ as a single category (and you reasonably might) then as Finn Lawler suggests in comments, you could consider it as a category enriched over the cartesian closed category $\newcommand{\SDS}{\mathbf{SDS}}\SDS$ you describe.  Giving such an enrichment is equivalent to giving a subcategory containing all objects (sometimes called a lluf subcategory).
One well-studied case like this is that of dagger categories and relatives.  These generalise examples like the category of finite-dimensional Hilbert spaces, where one wants to consider all linear maps, but wants to consider two such spaces the same only if they are unitarily isomorphic, not just linearly isomorphic.  This has been viewed in various ways in the literature, including both the ways above: a distinguished lluf subcategory of unitary (or self-adjoint) maps, or equivalently as a distinguished subset of each hom-set.

Edit: I originally missed the part where you say you want “special isomorphisms” to correspond to “natural isomorphisms $\newcommand{\Hom}{\mathrm{Hom}} \Hom(X,-) \to \Hom(Y,-)$, where $\Hom(X,-)$ is viewed as a functor on the whole category, not just the special morphisms”. This is a bit ambiguous, so let’s take some notation: call the larger category $\newcommand{\C}{\mathcal{C}}\newcommand{\D}{\mathcal{D}}\newcommand{\sp}{\mathrm{sp}}\C$, and the special subcategory $\C^\sp$.
If by $\Hom$ you mean $\Hom_\C$, then the Yoneda lemma says that natural isomorphisms $\Hom_\C(X,-) \to \hom_\C(Y,-)$, considered as functors on $\C$, will always correspond to isomorphisms in $\C$ (not generally special ones as you want).  Similarly, natural isos $\Hom_\sp(X,-) \to \hom_\sp(Y,-)$, as functors on $\C^\sp$, always correspond to isomorphisms in $\C^\sp$.
We can restrict $\Hom_\C(X,-)$ to a functor on just $\C^\sp$.  Considered as such, there may be more natural isomorphisms $\Hom_\C(X,-) \to \Hom_\C(Y,-)$ (natural just w.r.t. special maps) that don’t correspond to any isos in $\C$, special or otherwise.  (Consider the case where $\C$ is a group, viewed as a one-object category, and only the identity is designated as special; then any automorphism of the underlying set will give a natural isomorphism of this form.)
It sounds most like what you want is to consider $\Hom_\sp(X,-)$ as a functor on the whole of $\C$.  But this is not possible: the functorial action doesn’t extent, because when you compose a special arrow with an arbitrary map, it doesn’t generally stay special (except in the trivial case where all maps are special).<|endoftext|>
TITLE: $(L, \nabla)$ comes from a $G$-bundle with connection for some abelian algebraic subgroup $G \subset GL(n)$?
QUESTION [6 upvotes]: Let $A$ be an abelian variety over a field $k$ of characteristic $0$. How do I prove, without using transcendental methods, that if $\nabla$ is an integrable connection on a vector bundle $L$ on $A$ then $(L, \nabla)$ comes from a $G$-bundle with connection for some abelian algebraic subgroup $G \subset GL(n)$? And the corollary there exists a $\nabla$-stable flag$$L = L_n \supset L_{n-1} \supset \dots \supset L_1 \supset 0$$with $\text{rank }L_i = i$? It is easy to prove for $k = \mathbb{C}$ by looking at the monodromy of $(L, \nabla)$, and the general case follows by the Lefschetz principle, but I want to find a different proof which does not use $\mathbb{C}$ at all.
Any help would be greatly appreciated!

REPLY [2 votes]: Here is an outline of a different approach from that of t3suji.
First, define an algebraic group structure on the group $G$ of pairs $(a, \tau)$, where $a \in A$ and $\tau$ is a lift of the transformation automorphism $T_a: A \to A$ to an automorphism of $L$, then interpret $\nabla$ as an abelian Lie subalgebra of $\text{Lie}\,G$, and show that for any algebraic group $G$, an abelian Lie subalgebra of $\text{Lie}\,G$ is contained in the Lie algebra of an abelian algebraic subgroup of $G$.<|endoftext|>
TITLE: Why should we believe in the axiom of regularity?
QUESTION [52 upvotes]: Today I started reading Maddy's Believing the axioms. As I knew beforehand, it includes some discussion of ZFC axioms. However, I really hoped for a more extensive discussion of axiom of foundation/regularity. 
Apparently, the reason why we usually take it is because it makes sets well-founded and makes $\in$-induction work, or because it puts all sets into a hierarchy (namely $V$). However, these reasons sound to me more like "we take this, because it's convenient". Another reason commonly given is "It's difficult to think of a set which is an element of itself". This is not a good reason, because many things are difficult to think of, and also one could argue that a set represented by $\{\{\{...\}\}\}$ should do the trick.
That brings me to my question:

Are there any "philosophical" reasons to believe that the axiom of regularity holds?

I understand that this question is quite vague and maybe too broad, but I will be thankful for any responses.
${}{}$

REPLY [4 votes]: I know that this question is pretty old, but since it has reappeared I take the opportunity to give a justification of this axiom due to Dana Scott (I think).
First, some general remarks. I understand that to be justified is to be justified from a list of principles characterizing the notion of set-formation set theory is supposed to be about. The old cantorian notion of set-formation as unlimited/undisciplined/unrestricted gathering leads to paradoxes and has long been replaced by an ordered/organized/disciplined set-formation which became the standard conceptual basis of set theory. This new notion can be characterized by the following principles:
1) Sets ought to be determined by their elements and produced in ordered stages without upper bound. If a set is produced at some stage, then each of its elements ought to be produced at an earlier stage.
2) At each stage any plurality of sets produced in previous stages ought to determine a set.
3) Given a set and stages functionally connected to its elements, there ought to be a stage after all those given stages.    
There is no need to demand that the stages are well-ordered, contrary to a common belief.
Now, Dana Scott's proof:
First, a set $x$ is said to be grounded if and only if every set containing $x$ as an element has a minimal element. The relation between sets and stages, that $x$ is produced at stage $s$, is denoted by $x R s$, and the ordering of stages is denoted by $\triangleleft$.
If all elements of a set $x$ are grounded, then $x$ is also grounded. For, suppose $x$ is an element of $y$. Then, either $y$ and $x$ have no common member, in which case $x$ is minimal, or there is a $z$ such that $z \in y$ and $z \in x$. From $z \in x$, it follows that $z$ is grounded, and from $z \in y$, it follows that $y$ has a minimal element.  
If $s$ is a stage, then, from the second principle, there is a unique set $G_s$ which is produced at stage $s$ and whose elements are exactly the grounded individuals related to some stage below $s$. The set $G_s$ is grounded. Moreover, if $t \triangleleft s$, then $G_t \in G_s$.
Now, let $x$ be a non-empty set. Suppose that $x R s$. Let $y$ be the set of all $G_t$, for $t \triangleleft s$, such that there is a $z \in x$ such that $z R t$. Since any such $G_t$ is grounded, there is a stage $u$ such that $G_u \in y$ and $G_u$ is minimal. Since $G_u \in y$, there is a corresponding $z \in x$, such that $z R u$. So, $z$ must be minimal. Otherwise, there are $w \in z\cap x$ and a stage $v \triangleleft u$ such that $w R v$, hence $G_v \in y$. This contradicts the minimality of $G_u$, for $G_v \in G_u$.<|endoftext|>
TITLE: Characters of cuspidal representations
QUESTION [8 upvotes]: Let $\pi$ be an irreducible cuspidal representation of a semi-simple $p$-adic group $G$. It is well-known that the character of $\pi$ is concentrated in the set of compact elements in $G$.
What is the best reference for this result?
Also, if there is a really simple proof of this result, I would be glad to learn it!

REPLY [4 votes]: This is an old result due to Deligne:
Le support du caractère d'une représentation supercuspidale. 
C. R. Acad. Sci. Paris Sér. A-B 283 (1976), no. 4, Aii, A155–A157.<|endoftext|>
TITLE: What's the minimal embedding of orthogonal Grassmannian?
QUESTION [5 upvotes]: Suppose $X$ is the orthogonal Grassmanian.  We know the Plücker embedding does not span the whole background $\mathbb{CP}^n,$ just a subspace $\mathbb{CP}^m.$  

My question:  is there an expression of the isometric embedding of $X$ into $\mathbb{CP}^m$?  

By expression, I mean a way to write down explicitly the coordinates such as Plücker coordinates for the Grassmannian?

REPLY [3 votes]: I stumbled over here looking for something else. 
I think the short answer is to construct the orthogonal Grassmannian of isotropic n-planes in an 2n-dimensional space, take a list of all the principal pfaffians of a skew-symmetric n by n matrix. Since odd-pfaffians automatically vanish, the construction is slightly different in the even and odd cases. 
In general there is a construction of the minimal rational embedding of each compact hermitian symmetric space. In these cases the variety is the unique closed orbit of a highest weight vector and the minimal rational embedding is into the span of the orbit of the highest weight vector. For a complete picture see  Landsberg and Manivel's work https://arxiv.org/pdf/math/9902102.pdf<|endoftext|>
TITLE: asymptotic for the number of involutions in GL(n,2)
QUESTION [7 upvotes]: Is it known how the number of involutions in $GL_n(2)$, the group of $n\times n$ matrices over $\mathbb{Z}/2\mathbb{Z}$, behaves as $n\to\infty$ ?
Equivalently, one may ask this for the number of $n\times n$ matrices $A$ over $\mathbb{Z}/2\mathbb{Z}$ satisfying $A^2=0$, as $(A+I)^2=I \mod 2$.
Needless to say, there are $\lfloor n/2\rfloor$ conjugacy classes of involutions in $GL_n(2)$, and one can write down formula for the centraliser order for each class, 
$$\frac{\left|GL_n(2)\right|}{2^{k^2+2k(n-2k)}\left|GL_k(2)\right|\left|GL_{n-2k}(2)\right|},\quad 1\leq k\leq \left\lfloor \frac{n}{2} \right\rfloor,$$
but it's quite a mess to just sum them up.

REPLY [3 votes]: We may write $|{\rm GL}(n,2)| = 2^{n^{2}} \prod_{j=1}^{n}( 1- \frac{1}{2^{j}}).$
As $n \to \infty$, the rightmost factor tends to $\left( \sum_{r=0
}^{\infty} \frac{p(r)}{2^{r}} \right)^{-1}$, where $p(r)$ is the number of partitions of $r$. Let us write $|{\rm GL}(n,2)| = 2^{n^{2}}f(n)$.
Then we see that the number of involutions of ${\rm GL}(n,2)$ is given by 
$\sum_{k = 1}^{\lfloor \frac{n}{2} \rfloor} 2^{2k(n-k)} \frac{f(n)}{f(k)f(n-2k)}$ using the formula given in the question ( if $k = \frac{n}{2}$, we should interpret $|{\rm GL}(n-2k,2)|$ as $1$).
Hence the number of involutions in ${\rm GL}(n,2)$ may be expressed as 
$2^{\frac{n^{2}}{2}} \left(\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} 2^{- 2\left(\frac{n}{2}-k\right)^{2}}\left( \frac{\prod_{j=k+1}^{n}( 1- \frac{1}{2^{j}})}{ \prod_{m=1}^{n-2k}( 1- \frac{1}{2^{m}})}\right)\right)$ if $n$ is odd,
and $2^{\frac{n^{2}}{2}} \left( \prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}})+ \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor-1} 2^{- 2\left(\frac{n}{2}-k\right)^{2}}\left( \frac{\prod_{j=k+1}^{n}( 1- \frac{1}{2^{j}})}{ \prod_{m=1}^{n-2k}( 1- \frac{1}{2^{m}})}\right)\right)$ if $n$ is even.
Hence when $n$ is even, the number of involutions is at most
$2^{\frac{n^{2}}{2}} 
\prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}}) 
\left( 1 +  \frac{1}{\prod_{m=1}^{n-2}( 1- \frac{1}{2^{m}})} 
\left( \sum_{k=1}^{\frac{n}{2}} 2^{- 2(\frac{n}{2}-k)^{2}}\right)\right)$ and is at least $2^{\frac{n^{2}}{2}} \prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}})$.
As (even) $n \to \infty$, the first product appearing approaches $1$ from below
and (the reciprocal of) the second product appearing tends to $\sum_{r=0}^{\infty} \frac{p(r)}{2^{r}}$, while the inner sum is at most $\sum_{j=0}^{\infty} \left(\frac{1}{4} \right)^{j^{2}}$. 
I omit the similar analysis when $n$ is odd.<|endoftext|>
TITLE: Do representations of real semisimple algebraic group have to be algebraic?
QUESTION [6 upvotes]: If $G$ is the real points of a semisimple algebraic group and $\rho:G\to GL(n,\mathbb R)$ is continuous representation. Is $\rho$ an algebraic morphism?

REPLY [7 votes]: If $G$ is a real simple algebraic group and $\rho$ is a finite dimensional continuous representation, and if $G$ is not the group of complex points and $G(\mathbb C)$ is simply connected, then $\rho$ is indeed algebraic. This follows, for example, by going to the Lie algebra level and using classification of lie algebra representations. 
If $G=SL_2(\mathbb C)$, for example, you can have $\rho\otimes {\overline \rho}$ which is not quite algebraic, but can be viewed to be algebraic, if we view the complex group $SL_2(\mathbb C)$ as the group of real points of the Weil restriction of scalars $R_{\mathbb C/\mathbb R}(SL_2)$.   
The same sort of thing holds for any simply connected complex simple algebraic group, viewed as the group of real points by restriction of scalars. 
If $G(\mathbb C)$ is not simply connected, then there is some (small) ambiguity. Take 
$$G_0=SL_3(\mathbb R)\subset G_0(\mathbb C)= SL_3(\mathbb C)/\text{scalars}.$$ We can view $SL_3(\mathbb R)$ as an algebraic group in $G_0(\mathbb C)$; then the standard representation of $SL_3(\mathbb R)$ is not algebraic with this algebraic structure, but algebraic, if we view $SL_3(\mathbb R)$ as real points of $SL_3(\mathbb C)$. 
[Edit] Since @jerry has asked for details, I give them; I am sure that all this is explained in some textbook, but not having one at hand, I cannot give references. 
The first point is that given a complex semi-simple Lie algebra $\mathfrak g$ , there is a simply connected complex algebraic group  $G$ with Lie Algebra $\mathfrak g$ (this is perhaps due to Hermann Weyl). So any representation of the Lie algebra $\mathfrak g$ integrates to a holomorphic representation of the complex group $G$. That holomorphic finite dimensional representations of $G$ are algebraic is also well known, perhaps by using the Borel Weil Theorem. 
Secondly, if $G_0$ is an algebraic semisimple group defined over $\mathbb R$ with $G(\mathbb C)$ simply connected, and $\rho:G(\mathbb R) \rightarrow SL_n(\mathbb C)$ is a representation, it is known (as Dimitrov has remarked) that $\rho $ is smooth and hence yields a complex representation of the Lie algebra $\mathfrak g_0$, and therefore of its complexification $\mathfrak g$. Since $G$ is simply connected, this gives an algebraic representation of $G$. Since $G= G_0(\mathbb C)$ it follows that this complex representation is an algebraic  representation of $G_0(\mathbb C)$. 
Incidentally, these kinds of algebraicity results are proved in great generality by Borel and Tits in an Annals paper "abstract homomorphisms of algebraic groups" (it is in French).<|endoftext|>
TITLE: Exponentiation of vector spaces?
QUESTION [17 upvotes]: This question occurred to me while thinking on another one here, Name for an operation on matrices?
Can one define in an invariant way a binary operation on finite-dimensional vector spaces - let us denote it somehow suggestively by $(V,W)\mapsto V^{\otimes W}$ - with the property$$\dim(V^{\otimes W})=\dim(V)^{\dim(W)}?$$
Added later (and I should do it from the beginning as it is important): the construction from the linked question suggests that this operation seemingly should act on linear operators by assigning to $f:V\to X$ and $g:Y\to W$  certain operator (explicitly given there in terms of chosen bases)
$$
f\dagger g:V^{\otimes W}\otimes Y\to X\otimes W,
$$
with $\operatorname{rank}(f\dagger g)\geqslant\operatorname{rank}(f)\operatorname{rank}(g)$.
I'm aware that this looks even more impossible, but anyway - with bases it is done in that question.

REPLY [14 votes]: Okay, so you can get pretty close as follows: I still don't think $V^{\otimes W}$ makes sense, but riffing off of your comment, we can make sense of $(1 \oplus V)^{\otimes W}$ (where $1$ denotes the $1$-dimensional vector space). The guiding intuition is the binomial expansion
$$(1 + V)^W = 1 + {W \choose 1} V + {W \choose 2} V^2 + \dots $$
which can be upgraded to the functor
$$(1 \oplus V)^{\otimes W} = 1 \oplus (W \otimes V) \oplus (\Lambda^2 W \otimes V^{\otimes 2}) \oplus \dots $$
When $V = 1$ we reproduce the exterior algebra functor. Note that there's no hope to upgrade the $GL(V)$ action on this to a $GL(1 \oplus V)$ action: each component of this direct sum contains as a factor a different irrep of $GL(W)$, so any such upgrade must continue to respect this direct sum decomposition, but this is impossible already for the summand $W \otimes V$.<|endoftext|>
TITLE: Scheme of irreducible components
QUESTION [10 upvotes]: Let $\pi:X \to S$ be a morphism of schemes (I can assume that $\pi$ is sufficiently nice, e.g. proper and flat, but certainly not smooth).

Does there exist a scheme $I_{X/S}$ which parametrises the irreducible components of the fibres of $\pi$?

In the case where $S = \mathrm{Spec}(k)$ for some field $k$ and $X$ is reduced, it is possible to construct such a scheme by hand. One takes $I_{X/k}$ to be the product of the spectrum of the algebraic closure of $k$ in the function field of each irreducible component of $X$. This has the property that for any field extension $k \subset K$, the points $I_{X/k}(K)$ are in bijection with those irreducible components of $X$ over $\bar{k}$ which are actually defined over $K$. 
I am looking for such a construction which works in greater generality. I'm naively hoping that this might be buried in EGA somewhere...
As a closely related example, note that it is quite easy to construct a scheme of connected components, at least when $X/S$ is proper. Namely, one simply takes the Stein factorisation of $\pi$ to obtain a finite scheme $S'/S$, whose fibres parametrise the connected components of the fibres of $\pi$ in a natural way.

REPLY [6 votes]: There is a definition of a functor of irreducible components in that paper. If $\pi$ is finitely presented with geometrically reduced fibres, then the functor is representable by an étale algebraic space. As mentioned by Martin Bright, it is not separated in general; if it is, then it's a scheme (like all quasi-finite separated algebraic spaces).<|endoftext|>
TITLE: q-Virasoro and q-Heisenberg algebras
QUESTION [12 upvotes]: The literature has definitions (seemingly plural, though they might be linked) of a $q$-deformed Virasoro algebra. But is there any link of these to a $q$-deformed Heisenberg algebra? (Classically there is an expression for $L_n$ in terms of a normal ordered quadratic in the generators of the Heisenberg algebra.) I would be grateful for a reference to somewhere that discusses this in terms understandable to a non-expert in conformal field theory. However anything to shed light on the current status of the q-Virasoro and q-Heisenberg algebras, or relates them to the undeformed theory, would be of interest.
Apologies for my ignorance on this subject, as I pointed out I am certainly not an expert in CFT. I am currently trying to sort this out as an example of the action of vector fields in noncommutative geometry...

REPLY [6 votes]: The main sources are Awata et al or Frenkel-Reshetikhin. In http://arxiv.org/pdf/q-alg/9507034v5.pdf section 4, you can see the q,t case. You can also look at http://arxiv.org/pdf/q-alg/9505025v1.pdf where the introduction gives more references to how this relates to the undeformed case. This gives the classical Virasoro as functions on opers on the punctured disc.<|endoftext|>
TITLE: Two vector spaces with homeomorphic open subsets are isomorphic?
QUESTION [6 upvotes]: Is it true that if $ E,F$ are two topological vector spaces (or say Banach spaces) over $\mathbb{R}$ such that they have nonempty open subsets $U\subset E, V\subset F$ which are homeomorphic, then the two vector spaces are isomorphic? If false, then what can we say if the two open subsets are $\mathcal{C}^1$-diffeomorphic?

REPLY [9 votes]: This is false.  All separable Banach spaces, for example, are homeomorphic.Indeed, there is a considerable body of work on when topological vector spaces are homeomorphic (see Bessaga and Pelczynski "Selected Topics in infinite-dimensional Topology" for starters).<|endoftext|>
TITLE: Embedding of a proper scheme into a smooth one
QUESTION [7 upvotes]: Let $\mathbb{K}$ be any field and let $X$ be a proper $\mathbb{K}$-scheme. Does it exist a smooth, proper $\mathbb{K}$-scheme $Y$ and a closed immersion from $X$ to $Y$? This is tautological if $X$ is projective, but what about the general case?

REPLY [14 votes]: For a counterexample over $\mathbb{C}$, it sufficies to take a proper scheme $X$ with trivial Picard group (there are examples among toric varieties).  
Such a scheme cannot be embedded in any smooth variety $Y$; in particular, $X$ is necessarily singular. 
In fact, assume the contrary and take a dense affine subset $U$ of $Y$. Then we can choose an effective divisor $D_U$ on $U$, whose closure would give a non-trivial Cartier (since $Y$ is smooth) divisor $D_Y$ on $Y$, hence a non-trivial Cartier divisor $D_X$ on $X$, against the assumption $\textrm{Pic}\,X=\{\mathcal{O}_X \}$.<|endoftext|>
TITLE: Type of a modular form
QUESTION [8 upvotes]: Let $f$ be an arbitrary weight 1 newform. We know by Serre-Deligne that there is an odd 2-dimensional irreducible Artin representation $\rho$ such that $L_f(s)=L(\rho,s)$.
I was wondering how much can we tell about $\rho$ from $f$ alone, without computing $\rho$.
(For instance, one could make complete use of not enough Fourier coefficients to make the representation completely determined)
Of course, Serre-Deligne gives us automatically the conductor and the determinant of the representation. I'm particularly interested in:


Can we say what type of representation (dihedral, symmetric, alternating) $\rho$ is?


Unless I am missing something, Serre-Deligne's construction doesn't give any information in that direction. But on the survey "Modular forms of weight one and Galois representations", section 9, Serre does some very interesing calculations for prime conductor/level, for example breaking a basis of $S(\Gamma_0(p),1,\omega)$ into a concrete number of dihedral, $S_4$ and $A_5$-type forms, or ruling out $A_4$-type modular forms.

REPLY [5 votes]: EDIT : I wanted to add details after the OP's comment, and I realized I forgot one case (the dihedral case $D_{2n}$ with $n$ even). Below is a slightly corrected version.

You can determine the type of the representation by looking at the density of primes $p$ such that $a_p = 0$. If that density is $\geq 1/2$, then the form is dihedral, and the converse is true. In the case of density between $1/2$ and $5/8$, the set of $p$ contains the set of primes that are inert in a quadratic field $K$, and this means your form is CM or RM by $K$ (depending of whether $K$ is imaginary or real).
In the rare cases the density is $>5/8$, it is actually $3/4$, and the projective image of the Galois representation $\rho$
is the group $\mathbb Z/2 \mathbb Z\times \mathbb Z/2\mathbb Z$, which is dihedral in three possible ways, and accordingly your form has CM by two imaginary fields and RM by a quadratic field (all three subfields of a bi-quadratic extension of the rationals).
If the density is less than $1/2$ it is either $3/8$ or $1/4$, and your in one of the so-called exceptional cases. If $\rho$ is of type $A_4$, $S_4$, $A_5$ instead, the density will be $1/4$, $3/8$ and $1/4$, so you can use this to recognize the forms that have a $\rho$ octahedral or icosahedral from the ones that have a $\rho$ tetrahedral. 
How do you determine that density? It depends of what you want. If you're happy with guessing what type is your form, just count the coefficients $a_p$ non zero up to a certain $x$ with a computer and believe what it looks like. If you want something rigorous, you can use the error bound in the effective Chebotarev's density theorem, that will give you explcit values of $\epsilon(x)$ such that if the proportion of $p$ up to $x$ such that $a_p = 0$ is in an $\epsilon(x)$-range of $\geq 1/2$, $3/8$ or $1/4$, then the density is that number.

EDIT: I add some details. The proof of the stated results is easy. The representation satisfies, by Deligne-Serre, $tr \rho(Frob_p) = a_p$ for almost all $p$. The density of primes $p$ such that $a_p$ is $0$ is therefore, by Chebotarev's density theorem, the proportion of elements of trace $0$ in the subgroup $G:=$ Im $\rho$ of $GL_2(\mathbb C)$. Let $G'$ be the image of $G$ in $PGL_2(\mathbb C)$. An element $g$ of $G$ has trace $0$ if and only if $g$ has order $2$ in $G'$. We must therefore count the proportion of element of order $2$ in $G'$. For a group dihedral $D_{2n}$, this proportion is $1/2+1/(2n)$ if $n$ is even, $1/2$ if $n$ is odd. In $A_4$ there are $3$ elements of order $2$, giving a proportion of $3/12=1/4$. In $S_4$ the are
$9$ elements of order $2$, so the proportion is $9/24=3/8$. In $A_5$ there are 15 elements of order $2$ (the products of two disjoint transpositions, which form a single conjugacy class), and the proportion $15/60$ is again $1/4$. That justifies most of the assertion of the first paragraph. 
To complete the second paragraph, you need a statement of Effective Chebotarev (initially due to Lagraias-Odlyzko), for which a very good reference is Serre's paper at IHES in 1981 (Quelques applications du théorème de densité de Chebotarev), available on numdam.org<|endoftext|>
TITLE: Is this a functor on the category of $C^{*}$ algebras?
QUESTION [10 upvotes]: The category of $C^{*}$ algebras is denoted by $\mathcal{A}$.
Is there a  functor $\mathcal{F}$ on $\mathcal{A}$ which send each object $A\in \mathcal{A}$ to its center $Z(A)$. In the other words, can we extend  the maping $A\mapsto Z(A)$ on objects to a functor on this category?

REPLY [10 votes]: Here is another attempt at proving no such functor exists — I apologize to Chris and to Manny if something like this is already in the papers which they cite.$\newcommand{\Mat}{{\bf M}}\newcommand{\Cplx}{{\bf C}}\newcommand{\Cst}{{\rm C}^*}$
Let $\Mat_2$ denote the algebra of $2\times 2$ complex-valued matrices, and let $D_2$ be the algebra $\Cplx\oplus\Cplx$ regarded as the subalgebra of $\Mat_2$ consisting of diagonal matrices. Let
$$ A = \{ f\in C([0,1], \Mat_2) \mid f(1)\in D_2\}. $$
Since $Z(\Mat_2)=\Cplx$, one shows (with a small appeal to continuity) that
$$ Z(A)= \{ f\cdot I_2 \mid f\in C([0,1])\}\cong C([0,1]).$$
Let $\phi: D_2 \to A$ be the $*$-homomorphic embedding $\phi(d)(t) = d$ for all $t\in [0,1]$ and all $d\in D_2$. Let ${\rm ev}_1 : A\to D_2$ be the homomorphism "evaluate at $1$".
Now suppose $\mathcal F$ is a functor on the category of $\Cst$-algebras and $*$-homomorphisms, such that ${\mathcal F}(B)=Z(B)$ for every $\Cst$-algebra $B$. Then ${\mathcal F}({\rm ev}_1) \circ {\mathcal F}(\phi)$ must be the identity homomorphism on ${\mathcal F}(D_2)=D_2$, and thus the identity homomorphism on $D_2$ must factor somehow through the algebra $Z(A)\cong C([0,1])$. But this is impossible, since there is no injective algebra homomorphism from $D_2$ into $C([0,1])$. One way to see this last claim is to note that $D_2$ contains two non-trivial projections $e_1$ and $e_2$ which sum to $1$, while the only projections in $C([0,1])$ are $0$ and $1$.

Update 7th Oct. 2015
In response to a comment: we can find examples of a retract $D\to B \to D$ (that is, the composition of these two maps is the identity map on $D$) where $D$ is commutative and infinite-dimensional and $B$ has trivial centre.
Here is one way. Let $A$ be an infinite, countable group, and $H$ a countable amenable group, equipped with an action $\alpha: A\to {\rm Aut}(H)$. Form the semidirect product $G=H\rtimes_\alpha A$, which can be defined as the set $H\times A$ equipped with the product
$$
(h,a)(k,b) = (h\cdot\alpha_a(k), ab).
$$
The subset $\{e_H\} \times A$ is a subgroup of $G$; there is a quotient map of $G$ onto $A$, defined by "projection in the second variable"; and doing the inclusion $A \to G$ followed by the quotient $G\to A$ gives the identity map on $A$. (In other words, we have a retract.)
Note, for later reference, that 
$$(h,a)^{-1} = ({\alpha_a}^{-1}(h^{-1}), a^{-1} )$$
and so, using the fact that $A$ is abelian, a short calculation gives
$$ 
(h,a) (k,b) (h,a)^{-1} = ( h\alpha_a(k)\alpha_b(h^{-1}), b).
$$
I now want to impose conditions on $\alpha$ which will ensure that $G$ is an ICC group, i.e. all conjugacy classes except $\{e\}$ are infinite. It is well known that if $G$ is countable and ICC then its group von Neumann algebra ${\rm VN}(G)$ has trivial centre: since the centre of $\Cst_r(G)$ is contained in the centre of ${\rm VN}(G)$, we deduce that $\Cst_r(G)$ has trivial centre.
To find such conditions: note that $(0,a)(k,b)(0,a)^{-1} = (\alpha_a(k), b)$
and $(h,0)(e,b)(h,0)^{-1} = (h\alpha_b(h^{-1}), b)$.
So if we assume that $\alpha$ satisfies the following conditions
1) for each $k \in H \setminus\{e_H\}$, the orbit $\{\alpha_a(k) : a\in A\}$ is infinite
2) for each $b\in A\setminus\{e_A\}$, the set $\{h \alpha_b(h^{-1}): h\in H\}$ is infinite
then $G$ will be ICC, as desired, and so $\Cst_r(G)$ will have trivial centre.
$\newcommand{\Cst}{{\rm C}^*}$
Recall that on the category ${\sf Grp}$ of (discrete) groups and homomorphisms, the full group $\Cst$-algebra defines a functor ${\sf Grp}\to {\sf Cstar}$ is functorial. Since $A \to G \to A$ is a retract in ${\sf Grp}$, we get a retract $\Cst(A)\to \Cst(G)\to\Cst(A)$.
Clearly $\Cst(A)$ is commutative and infinite-dimensional. Now since $A$ and $H$ are amenable, so is $G$; hence $\Cst(G)=\Cst_r(G)$, which has trivial centre by the remarks above.
To get an actual, concrete example: let $A=\{2^n : n \in {\mathbb Z}\}$ be the multiplicative group formed by all integer powers of $2$, and let $H$ be the set of dyadic rationals, regarded as an additive group. The action of $A$ on $H$ is just by scaling, i.e. for $a\in A$ and $h\in H$ we have $\alpha_a(h)=ah$. It is easily checked that both conditions 1) and 2) are satisfied; note that $e_H=0$, $e_A=1$, and $h\alpha_b(h^{-1})$ is just $h-bh$.
[I learned of the existence of solvable countable ICC groups in a seminar many years ago, which is why I knew there should be some way to manufacture examples that did what you want, once one has the idea of trying to build an example in ${\sf Grp}$ and transport it functorially to ${\sf Cstar}$.]<|endoftext|>
TITLE: Average of Short Character Sum over All Dirichlet Characters Mod n
QUESTION [5 upvotes]: Cross-posted from M.SE.
Given $a,n$ coprime positive integers, let $L = \{(x,y)\in \mathbb{Z}^2, ax=y(n)\}$ be the lattice of all points satisfying $ax=y\pmod{n}$.
I want to find an order-of-magnitude bound on the shortest (minimizing say $\max(|x|,|y|)$) vector in $L$ with $x,y$ coprime. If you restrict to merely finding a nonzero vector an easy pigeonhole argument gives a bound of the form $|x|,|y|\ll\sqrt{n}$. Is this still true if we require additionally that $x,y$ are coprime? In particular note that just because $(x,y)$ is in the lattice doesn't mean $(x/g,y/g)$ is, where $g$ is their greatest common divisor - problems can occur if $g$ has common factors with $n$.
Explicitly can anyone prove the existence of a primitive point $|x|,|y|\ll n^{1/2+\varepsilon}$?
Note that if $a\ll \sqrt{n}$ then the problem is trivial by picking $(x,y)=(1,a)$. I am interested in bounds for generic $a\in (\mathbb{Z}/n\mathbb{Z})^\times$.

Here are my ideas so far to reduce to a character sums bound.
We want to show that the number $S(a)=\#\{ax\equiv y (n): |x|, |y| \le N, x,y$ coprime to n$\}$ is nonzero; then $N$ is an upper bound for the primitive point's size. By discrete Fourier analysis against the Dirichlet characters mod $n$,
$$S(a)=\frac{\phi(N,n)^2}{\phi(n)} + \frac{1}{\phi(n)}\sum_{\chi \ne \chi_1} \Big |\sum_{|y| \le N} \chi(y)\Big|^2 \chi(a)^{-1}.$$
Here $\phi(N,n)$ is the number of $|y|\le N$ coprime to $n$.
Applying Polya-Vinogradov to the inner sums just misses a nontrivial bound - is there any bound on these short character sums on average, twisted by $\chi(a)^{-1}$? This method seems doomed to fail because the choice of $a=1$ gives a large (positive) error term.

REPLY [7 votes]: Suppose $n=2m$ and $a=m+1$. Clearly $(a,m)=1$. Then there are no $x,y$ odd with $ax \equiv y$ mod $n$, $x<m$, $y<m$. Indeed in this case
$$y \equiv ax= mx+x \equiv m+x \mod 2m$$
Clearly this cannot be satisfied for $0<x<m$, $0<y<m$.
You can try to write the sum instead using additive characters instead and you will see this kind of obstruction as a large term coming from a single additive character.<|endoftext|>
TITLE: $\kappa$-support iterations of $<\kappa$-strategically closed forcing
QUESTION [5 upvotes]: Let $\kappa$ be an uncountable regular cardinal, and suppose that $\langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\delta\rangle$ is a $\kappa$-support iteration of $<\kappa$-strategically closed notions of forcing, i.e.,
a) $\Vdash_{\mathbb{P}_\alpha}``\dot{\mathbb{Q}}_\alpha\text{ is $<\kappa$-strategically complete''}$, and
b) inverse limits are taken at every limit stage of cofinality $\leq\kappa$, and direct limits elsewhere.
It is well-known (even folklore) that the limit forcing $\mathbb{P}_\delta$ is also $<\kappa$-strategically closed (see, e.g., Proposition 7.9 of Cummings chapter in the Handbook of Set Theory).
Is there a place where a proof of this appears in the literature?
I am asking because I need to use that the obvious strategy has some additional properties, and if I can find an appropriate reference then I can avoid taking a rather technical detour in the middle of another proof.

REPLY [2 votes]: To finish this question off:
Proposition A.1.6 of the paper "Sheva-Sheva-Sheva: Large Creatures" by Roslanowski and Shelah gives the required proof and explicitly points out much of what I was looking for. The rest is obvious from what is there.
The paper appears in Israel Journal of Mathematics 159(2007), 109-174.
It is number 777 in Shelah's numbering of papers (hence the name!)<|endoftext|>
TITLE: a generalization of gamma matrices
QUESTION [9 upvotes]: Is it possible to find matrix solutions to the following :
$$\left(\sum_1^m M_k x_k\right)^n=\left(\sum_1^m x_k^n\right)I_d$$
where $M_k$ are the desired $d \times d$ matrices (no restriction on $d$) and $x_i$ are indeterminate variables;
For n=2 the gamma matrices satisfing $M_i M_j + M_j M_i = 2\delta_{ij}$ work; so in a way this is a generalization of these to larger $n$.

REPLY [3 votes]: Yes, these are called Generalized Clifford Algebras. The earliest reference I could find was an article by Yamazaki from 1964.
An explicit construction is given by Morris
For example, with $m=2$ and $n=3$, we find as a solution $M_1 = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\end{pmatrix}$ and $M_2 = \begin{pmatrix} 1 & 0 & 0\\ 0 & e^{2 \pi i/3} & 0\\ 0 & 0 & e^{4\pi i /3}\end{pmatrix}$<|endoftext|>
TITLE: What is the role of projective spaces in GAGA?
QUESTION [8 upvotes]: The GAGA theorem is a celebrated elaboration of the idea that complex analytic and complex algebraic geometry are equivalent, at least for smooth projective varieties/manifolds.
I am aware why this is a theorem about projective varieties; historically the two classes of varieties people cared about were projective and affine varieties. I am also aware that GAGA fails to hold for affine varieties and Stein manifolds.
I wonder if there is any deeper/conceptual reason why projective spaces in particular ought to appear though?
For instance I feel like I understand why projective spaces appear when you work with bundles, say when working with Chern classes since they classify complex vector bundles. But GAGA is a theorem about categories of coherent sheaves and not just vector bundles ...
Maybe if one does "relative" GAGA in the sense of Grothendieck, can we work over bases other than projective spaces or more generally projective varieties? And even if we can, is there some significance in choosing projective spaces anyway?
Thanks a lot and sorry if the question is too vague and philosophical!

REPLY [13 votes]: Short version of the proof of GAGA is this:
it's an immediate consequence of Serre's twisting theorem that every analytic coherent sheaf $\mathcal{F}$ is a quotient of $\mathcal{O}^{\mathrm{an}}(-m)^{\oplus n}$ by the image of a map from $\mathcal{O}^{\mathrm{an}}(-m')^{\oplus n'}$.  That map must be algebraic (that's where properness is important!); in fact it's just a $n\times n'$ matrix of polynomials of degree $m'-m$. So we can get the desired algebraic coherent sheaf we want by taking the cokernel of the "same" map $\mathcal{O}(-m')^{\oplus n'}\to \mathcal{O}(-m)^{\oplus n}$.  
Morally, I would say the important thing about properness is that analytifying is fully faithful.  The essence of the proof is that once you know this, you just need a set of algebraic coherent sheaves whose analytifications generate the derived category of analytic coherent sheaves and you'll be done.  In the projective case, this is easy: you just use $\mathcal{O}(-m)$; in the general proper case, this is harder, but possible; that's what Grothendieck is doing on the last page of the article cited above.<|endoftext|>
TITLE: Large deviation for Brownian path on $[0,\infty)$
QUESTION [6 upvotes]: It seems strange to me that all we can find about Schilder's theorem in the literature is on a finite interval of Brownian path. 
If we equip the space of continuous function starting from $0$, defined on $\mathbb{R}_+$ with the topology of uniform convergence on compacts. Can we have the similar large deviation principle? With of course the rate function
$$I(f) = \frac{1}{2}\int_0^{\infty} \dot{f}(t)^2 dt.$$
I don't see any objection so far but I don't have any reference to confirm my guess. If it is wrong can you tell me why?
The same question for the Cameron-Martin formula. If we write $\mu$ the Wiener measure on $C_0([0,\infty))$, and $\mu^h$ the measure of $B+h$ where $B \sim \mu$. Can we have, for $h$ of finite functional value $I(h) < \infty$, that $\mu^h$ is absolutely continuous with respect to $\mu$,  of density
$$\frac{d\mu^h}{d\mu} = \exp(\int_0^{\infty} \dot{h}_t dx_t - \frac{1}{2}\int_0^{\infty} \dot{h}_t^2 dt),$$
where $\int_0^{\infty} \dot{h}_t dx_t$ is defined in $L^2(\mu)$ by Wiener Integral. 
Does that make any sense?
Thank you in advance.

REPLY [4 votes]: It is not in the uniform topology but with a topology tapered off at infinity it is correct. It is done that way in the book of Deuschel and Stroock on large deviations.<|endoftext|>
TITLE: Is the boundary of Alexandrov space again an Alexandrov space?
QUESTION [5 upvotes]: Let $X$ be a finite dimensional (possibly compact) Alexandrov space with curvature $\geq K$. Is it true that its boundary is again Alexandrov space with curvature bounded from below? If yes, is the curvature at least $K$?

REPLY [5 votes]: This is an open problem. 
It is a special case of the following question: 

Is it true that every extremal subset is again an Alexandrov space?

The answer to this question is "No". Petrunin has constructed a counterexample in codimension three, here.<|endoftext|>
TITLE: Mapping class groups in high dimension
QUESTION [23 upvotes]: $\DeclareMathOperator\MCG{MCG}\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\Homeo{Homeo}$Let $M$ be a $1$-connected, closed, smooth manifold with $\dim(M)>4$ and let us set $\MCG(M)=\pi_0(\Diff(M))$. Dennis Sullivan proved that $\MCG(M)$ is commensurable to an arithmetic group.
Edit: regarding A. Kupers' remark on commensurability, here is a nice note
https://comptes-rendus.academie-sciences.fr/mathematique/item/CRMATH_2020__358_4_469_0/

I was wondering if the same type of result holds for topological manifolds and $\pi_0(\Homeo)$.
Let us consider the canonical morphism $i:\Diff(M)\rightarrow \Homeo(M)$, what do we know about the morphism induced on $\pi_0$?
And more generally about the homotopy groups of the homotopy fiber?
For which classical manifolds this homotopy fiber has been determined?
For spheres, this should be well known and related to exotic spheres, and what do we know about complex projective spaces?

REPLY [11 votes]: Let me assume that M is at least 5-dimensional.
Sullivan's proof only uses surgery theory and properties of O(n) that also hold for Top(n), so the answer to your first question is yes.
Regarding your other questions: the homotopy fiber Homeo(M)/Diff(M) of the map BDiff(M) -> BTop(M) is subject of smoothing theory. There is a map from Homeo(M)/Diff(M) to the space of sections of the bundle over the d-manifold M with fiber Top(d)/O(d) which is associated to the frame bundle of M using the action of O(d) on Top(d)/O(d). This map is injective on path components and an isomorphism on all homotopy groups. This can be found for instance in the book of Kirby and Siebenmann.
Up to a question about components, this reduces the study of the homotopy fiber you asked about to understanding the homotopy type of Top(d)/O(d) (and the twist of the bundle), which is hard. The homotopy groups of Top(d)/O(d) are object of study in geometric topology for a long time and can be divided into three ranges:

For $i\lesssim d$, we have $\pi_i(Top(d)/O(d))\cong \Theta_i$, where $\Theta_i$ is the group of homotopy spheres which is understood in terms of the stable homotopy groups of spheres by Kervaire--Milnor.
For $d \lesssim i\lesssim \frac{4}{3}d$, the groups $\pi_i(Top(d)/O(d))$ can in principle be understood in terms of Waldhausen's algebraic $K$-theory of spaces. Rationally, this can be used to compute that in this range $\pi_*(Top(d)/O(d))\otimes \mathbb{Q}$ vanishes for $d$ even and is isomorphism to $K_{*-d+1}(\mathbb{Z})\otimes \mathbb{Q}$ if $d$ is odd (that's a calculation of Farrell--Hsiang). The rational $K$-groups of the integers were computed by Borel: $$ K_{*}(\mathbb{Z})\otimes \mathbb{Q}\cong \begin{cases}\mathbb{Q}&\text{if }*\equiv 1 (4)\\0&\text{else }\end{cases}$$ in positive degrees.
In the range $i \gtrsim \frac{4}{3}d$ very little is known, but some progress has been made in recent years. For instance, work of Watanabe (building on ideas of Kontsevich) and Weiss shows that there are plenty of nontrivial rational classes.

Later edit: The identification of $\pi_i(Top(d)/O(d))\otimes\mathbb{Q}$ in the range $i\lesssim \frac{4}{3}d$ mentioned above holds in fact for $i\lesssim 2d$ by Corollary 4.2 of Randal-Williams and Corollary C of Krannich. Beyond this range, progress in even dimensions has been made by Kupers--Randal-Williams and in odd dimensions by Krannich--Randal-Williams.<|endoftext|>
TITLE: Graph homomorphisms and line graph
QUESTION [8 upvotes]: If $G,H$ are simple, undirected graphs, we say that they are in a hom(omorphism)-relation if there is a graph homomorphism from $G$ to $H$ or from $H$ to $G$.
For any graph $G$ let $L(G)$ denote its line graph.
If $G,H$ each have at least one edge, and they are in a hom-relation, does the same hold for $L(G), L(H)$?

REPLY [7 votes]: Things are even worse than you imagine.
Tony's answer shows that the homomorphism order of line graphs can be partitioned into intervals whose endpoints are the complete graphs. I made use of this fact in my thesis. Then the line graphs in the interval $[K_d, K_{d+1})$ are the line graphs of graphs with maximum degree $d$ (ignoring exceptions caused by $K_3$). For $d = 1$, this interval contains only $K_1$, for $d = 2$ it contains only $K_2$ and the odd cycles of length at least 5 (up to homomorphic equivalence). After that however things get more complicated. In fact, it is shown here that every other such interval is universal, meaning that ANY countable partial order can be embedded order-preservingly into it. Finally, the graphs in that paper whose line graphs are used to prove universality of the interval $[K_d, K_{d+1})$ are all homomorphically equivalent to $K_d$.
All together this means that not only are there two homomorphically equivalent graphs whose line graphs have no homomorphisms between them in any direction, as Gordon's answer shows. But for any $d \ge 3$, there are an infinite number of graphs all homomorphically equivalent to $K_d$, but whose line graphs form a partial order that any countable partial order can be embedded in.<|endoftext|>
TITLE: Quantitative and elementary proofs of the Prime Number Theorem
QUESTION [10 upvotes]: I would like to know two things: one, whether the best quantative bounds in the Prime Number Theorem are still basically those given by the Vinogradov-Korobov zero-free region? and two, whether there are any elementary proofs substantially different from the Erdős/Selberg proofs?
I realise this is probably trivial to answer for experts, but it seems to be hard to find clear statements in the literature for non-experts.

REPLY [4 votes]: The proof that is broadly speaking elementary, in the sense that it does use the analytic continuation of Dirichlet L-fucntions, which gives the same error term as Korobov-Vinogradov zero free region was given a few years ago by Koukouloupolos, see http://www.dms.umontreal.ca/~koukoulo/documents/publications/pnt.pdf<|endoftext|>
TITLE: Does non-stablity imply that there is a difference between non-forking and coheir extension
QUESTION [8 upvotes]: Fix some theory $T$.
Let $p$ be a type over some Model M and let $q$ be some global extension of $p$. 
Note:
The number of global coheirs of $p$ is bounded by the number of ultrafilters on $M$.
Also if $T$ is stable, it follows that $q$ is a non-forking extension of $p$ if and only if $q$ is a coheir of $p$.
If $T$ is simple and non-stable, then there is a type (say $p$) with a unbounded number of non-forking extension. Hence if $q$ is a non-forking global extension, then it does not need to be a coheir of $p$.
Is this true in general non-stable theories or for example NTP2-non-stable? Is there always a type $p$ over some model and some non-forking extension $q$ which is no coheir of $p$

REPLY [4 votes]: In general there are TP2 theories in which every global type non-forking over a small model is finitely satisfiable in it. An example of this is constructed in "On non-forking spectra", Artem Chernikov, Itay Kaplan, Saharon Shelah, http://arxiv.org/abs/1205.3101, Section 5.3.
Restricting to NTP2 theories, bounded number of non-forking extensions characterizes NIP, so in particular if every non-forking type is a coheir and T is NTP2, then it is NIP. See Section 4 in the paper above for a proof.
Finally, if T is a pseudofinite NIP theory, then a global type does not fork over a model M if and only if it is finitely satisfiable in it, see "A guide to NIP theories", Pierre Simon, Exercise 6.19. There are pseudofinite NIP theories that are not stable (e.g. the theory of a discrete linear order with the first and the last element - in this example it is easy to check this directly).<|endoftext|>
TITLE: Quotients of posets
QUESTION [6 upvotes]: Let $\mathbf{Poset}$ denote the category of partially ordered sets and order-preserving maps. Does $\mathbf{Poset}$ have quotients?

REPLY [9 votes]: (There may be some users who think this would have been better asked at Mathematics StackExchange, but I'll go ahead and answer because there are several ways of looking at it.) 
The answer is "of course". See The Joy of Cats, p. 119. The coequalizer of two maps $f, g: X \rightrightarrows Y$ in $\mathbf{Poset}$ is computed in two steps: first take the coequalizer of $f, g$ as if in the category $\mathbf{Preord}$ of preordered sets (sets with reflexive transitive relations). This is done by taking the coequalizer in $\mathbf{Set}$, say $q: Y \to Q$, and then endowing $Q$ with the smallest reflexive transitive relation that makes $q$ an order-preserving map. Second, take the "posetal reflection of $Q$", i.e., the quotient $R$ where $x, y \in Q$ are identified if $x \leq y$ and $y \leq x$ in $Q$; the order relation on $R$ inherited from $Q$ makes $R$ the coequalizer of $f, g$ in $\mathbf{Pos}$. 
In other language: $\mathbf{Preord}$ is topological over $\mathbf{Set}$ (see The Joy of Cats for details), and therefore is cocomplete (and much more). The full inclusion $\mathbf{Pos} \hookrightarrow \mathbf{Preord}$ is a reflective subcategory, and therefore $\mathbf{Pos}$ is also cocomplete, although not topological over $\mathbf{Set}$. 
In still other language: it may be shown that $\mathbf{Preord}$ and $\mathbf{Pos}$ are locally finitely presentable, i.e., they are categories of structures defined entirely in terms of suitable finite limit diagrams in $\mathbf{Set}$, and on general abstract grounds such categories are cocomplete. See Adámek & Rosický, Locally presentable and accessible categories, Cambridge University Press 1994.<|endoftext|>
TITLE: A vanishing condition for cup products in Galois cohomology
QUESTION [7 upvotes]: Let $k$ be a field of characteristic $\neq 2$. For a non-zero element $a \in k^*$, let us write $[a] \in H^1(k,\mathbb{Z}/2)$ for the Galois cohomology class corresponding to the quadratic extension $k(\sqrt{a})/k$. If $a,b \in k^*$ are non-zero elements then the class $[a] \cup [b] \in H^2(k,\mathbb{Z}/2)$ is trivial if and only if $b$ is a norm from $k(\sqrt{a})/k$. In particular, if $ab = -1$ then $b=-a$ is a norm from $k(\sqrt{a})/k$ and hence $[a] \cup [b] = 0$. I have a reason to believe the following generalization is true:
Claim: Let $a_1,...,a_n \in k^*$ be such that $\prod_i a_i = -1$. Then $[a_1] \cup [a_2] \cup ... \cup [a_n] = 0$.
Is this claim true? 
One way to approach this problem is via quadratic forms. Let us denote by $\left<b_1,...,b_n\right>$ the isomorphism class of the quadratic form $\sum_i b_i x_i^2$. The Witt ring $W(k)$ of $k$ is the ring generated by isomorphism classes of non-degenerate quadratic forms over $k$, modulu the relation $\left<1,-1\right> \sim 0$ (more precisely, the addition in $W(k)$ is determined by the direct sum operation and multiplication by tensor product). Let $I \subseteq W(k)$ be the ideal generated by forms of even rank. A deep result in the theory of quadratic forms (closely related to Milnor's conjecture), is that there is a natural isomorphism of graded rings 
$$ \oplus_n I^n/I^{n+1} \stackrel{\cong}{\longrightarrow} H^*(k,\mathbb{Z}/2) .$$ 
In particular, the class of $\left<1,-a\right> \in I$ (mod $I^2$) is sent to the class $[a] \in H^1(k,\mathbb{Z}/2)$, and hence the class of $\left<1,-a_1\right> \otimes ... \otimes \left<1,-a_n\right>$ is sent to $[a_1] \cup ... \cup [a_n]$. A possible strategy is then to show that if $\prod_i a_i = -1$ then the quadratic form $\left<1,-a_1\right> \otimes ... \otimes \left<1,-a_n\right>$ (which has rank $2^n$) is trivial (in the sense that it is isomorphic to a direct sum of $2^n$ copies of $\left<1,-1\right>$).

REPLY [2 votes]: After realizing that the claim is wrong (thanks to the answers above), I managed to find a weaker statement, that turned out to be what I needed anyway. So just in case someone else happens to be interested in this question, here is the idea.
For each $l$, let $\mu_l$ denote the Galois module of $l$-roots of unity. In particular, let us use $\mu_2 = \{1,-1\}$ instead of $\mathbb{Z}/2$. For each $n$ we have an inclusion of Galois modules $\mu_l \longrightarrow \mathbb{G}_m$. Now indeed it is not true that if $\prod_i a_i = -1$ then $[a_i] \cup ... \cup [a_i] = 0$. However, it is true that in this case the image of $[a_1] \cup ... \cup [a_n]$ in $H^n(k,\mathbb{G}_m)$ is $0$ (which happens to be sufficient for my original application). 
To see this, consider the short exact sequence of Galois modules
$$ 0 \longrightarrow \mu_2 \longrightarrow \mu_4 \longrightarrow \mu_2 \longrightarrow 0 .$$
We then obtain a long exact sequence in homology groups, and in particular a boundary map
$$ H^{n-1}(k,\mu_2) \longrightarrow H^n(k,\mu_2) .$$
Unwinding the definitions, one can check that this boundary map sends a class $\alpha \in H^{n-1}(k,\mu_2)$ to the class $\alpha \cup [-1] \in H^n(k,\mu_2)$. Now, as explained in the answer above, it is sufficient to consider the case where $n > 1$ is odd. Let $a_1,...,a_n$ be such that $\prod_i a_i = -1$. Then, again as explained above, we have
$$ [a_1] \cup ... \cup [a_n] = [a_1] \cup ... \cup [a_{n-1}] \cup [-1] $$
and so $[a_1] \cup ... \cup [a_n]$ is in the image of the boundary map $H^{n-1}(k,\mu_2) \longrightarrow H^n(k,\mu_2)$. It follows that the image of $[a_1] \cup ... \cup [a_n]$ in $H^n(k,\mu_4)$ is $0$, and hence the image in $H^n(k,\mathbb{G}_m)$ is $0$ as well.<|endoftext|>
TITLE: Existence of a measurable map between metric spaces
QUESTION [6 upvotes]: Let $X$ and $Y$ be separable complete metric spaces (if necessary, they may be assumed to be compact). Let $R\subset X\times Y$ be a closed subset such that the projection of $R$ to $X$ is onto. 
Is it true that there exists a Borel measurable map $f\colon X\to Y$ such that $(x,f(x))\in R$ for every $x\in X$?
If this is not true in general, are there known sufficient conditions for its existence?

REPLY [7 votes]: For a compact space $Y$ the answer is affirmative, but in general case of Polish space $Y$ it is negative. 
Results yielding nice selections of relations $R$ are known in Descriptive Set Theory as Uniformization Theorems, see Section 18 of the standard textbook [A.Kechris, Classical Descriptive Set Theory, Springer, 1995]. 
According to Theorem 18.18 of Arsenin, Kunugui (in this section), for Polish spaces $X,Y$ and a Borel set $R\subset X\times Y$ with all sections $R_x=\{y\in Y:(x,y)\in R\}$, $x\in X$,  $\sigma$-compact and non-empty, there exists a Borel function $f:X\to Y$ whose graph is contained in $R$. 
On the other hand, by Exersice 18.9 in the book of Kechris, for the Polish spaces $X=Y=\omega^\omega$ there exists a closed subset $R\subset X\times Y$ such that for every $x\in Y$ the section $R_x$ is uncountable, but $R$ admits no Borel uniformization (i.e., no Borel function $f:X\to Y$ whose graph is contained in $R$). Yet by Jankov, von Neumann Uniformization Theorem, $R$ always has a $\sigma(\Sigma_1^1)$-measurable uniformization.<|endoftext|>
TITLE: Is $\beta \mathbb{D}\setminus \mathbb{D}$ a group?
QUESTION [5 upvotes]: NB the original question asked about $\beta\mathbb{D}$ rather than the corona, hence some of the initial comments.

Is there a group operation on $\beta \mathbb{D} \setminus \mathbb{D}$ extending complex multiplication on $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$?

REPLY [9 votes]: No, because $\beta\mathbb D$ and $\beta\mathbb D\setminus \mathbb D$ are not Dugundji compact (and not supercompact) whereas all compact topological groups are Dugundji compact according to a result of Uspenskii
https://www.researchgate.net/publication/266010994_Topological_groups_and_Dugundji_compacta
and supercompact (by a result of Mills reproved by Kubis and Turek in http://link.springer.com/article/10.2478/s11533-011-0019-x).
Another reason why $\beta \mathbb D$ or $\beta\mathbb D\setminus\mathbb D$ cannot possess a group structure with continuous left (or right) shifts is that those spaces are not topologically homogeneous (because of the existence of weak P-points, for example).<|endoftext|>
TITLE: Hashimoto Matrix (Non-backtracking operator) and the Graph Laplacian
QUESTION [6 upvotes]: The question is: how can we recover the graph Laplacian or its spectrum from the Hashimoto Matrix (also commonly called the Non-backtracking operator)?
To make the question as self-contained as possible, here I give the main definitions:
Let $W$ be a given symmetric adjacency matrix of graph $G = (V,E)$.
Let $D$ be a diagonal matrix such that $D_{ii} = d_i$, 
where $d_i$ is the degree of vertex $i$, i.e. $d_i = \sum_i^n w_{ij}$ 
Then, 

The Laplacian matrix $L$ is defined as 
$$ L=D-W $$
The Hashimoto matrix $B$ is defined as 
$$
B_{(i \rightarrow j), (k \rightarrow l)} = 
\begin{cases} 
   1 & j=k \text{ and } i\neq l \\
   0 & \text{ otherwise }
\end{cases}
$$
Observation: See that the Hashimoto Matrix is a kind of directed linear graph of $G$.

Is there an operator such that we can recover $L$ from $B$, i.e. an operator $\mathcal{M}$ such that $\mathcal{M}(B) = L$?
Is there any relation between the spectrum of $B$ and $L$?
Answers with references are highly appreciated.

REPLY [2 votes]: Is there any relation between the spectrum of B and L?

One can say some things, but it follows from computations on small graphs that one spectrum does not determine the other.  For instance, compare Tables 2.1 and 5.1 in my paper with C. Durfee.  The T column in the first table counts graphs of order n not determined by their Hashimoto spectrum, and the L column in the latter table does the same for the Laplacian.
In this paper, we talk about some properties the Hashimoto spectrum determines.  Since it determines the Ihara zeta function, it at least determines some mild things about the signed Laplacian $|L|$, e.g., the product of the eigenvalues of $|L|$.  See Sec 2.1.<|endoftext|>
TITLE: Bounding exponential sum with square roots
QUESTION [10 upvotes]: It is well known that for each $m\in\mathbb{N}$
$$\lim_{N\to\infty}\frac1N\sum_{n=1}^Ne^{2\pi i\sqrt{nm}}=0$$
My question is whether there is some uniformity in the variable $m$.
More precisely, is it true that
$$\lim_{N\to\infty}\frac1N\sum_{n=1}^Ne^{2\pi i\sqrt{nN}}=0?$$
I would be particularly interested in a power saving of the form
$$\lim_{N\to\infty}\frac1{N^{1-\epsilon}}\sum_{n=1}^Ne^{2\pi i\sqrt{nN}}=0$$
for some $\epsilon>0$.

REPLY [6 votes]: The answer given by Xiaoyu He uses the same arguments as usual van der Corput's theorem (see Theorem 2.2 from Graham, S. W. & Kolesnik, G. "van der Corput's method of exponential sums"):

Suppose that $f$ is a real valued function
  with two continuous derivatives on $I$. Suppose also that there is some $\lambda > 0 $ and some $a > 1$ such that
  $\lambda  < |f''(x)| < a\lambda$
  on $I$. Then
  $$\sum_{n\in I} e(f(n))\ll a|I|\lambda^{1/2} + \lambda^{-1/2}.$$

So we can try to apply this theorem directly. For $f(x)=\sqrt{ N x}$ on the interval $I(X)=[X,2X)$ we have $f''(x)\asymp\lambda =N^{1/2}X^{-3/2}$. Van der Corput's theorem gives the estimate
$$\sum_{n\in I(X)} e(f(n))\ll N^{1/4}X^{1/4} + N^{-1/4}X^{3/4}.$$
Summation over $X=1,2,4,8,\ldots,2^k\asymp N$ leads to the upper bound 
$$\sum_{n=1}^N e(f(n))\ll N^{1/2}.$$<|endoftext|>
TITLE: Near model completeness
QUESTION [12 upvotes]: A theory $T$ is called near model complete if every formula is equivalent to a Boolean combination of existential formulas mod $T$. I wonder whether there is an equivalent "semantic" definition of this notion like model completeness. (A theory is model complete if every formula is equivalent to an existential formula in that theory or, equivalently, if any embedding of its models is elementary.)
Also, I wonder whether near model completeness of a theory implies its inductiveness or not. Recall that a theory is inductive if the union of any chain (or directed family) of its models is again a model. Equivalently, a theory is inductive iff it is $\forall \exists$-axiomatisable.

REPLY [5 votes]: I'm still thinking about your main question. Meanwhile here is my answer to your additional question.
A sandwich criterion by Kueker and Turnquist
You may have seen this before, but it's still worth stating here as it may well be the best answer possible. Proposition 2.3 in Nearly Model Complete Theories by Kueker and Turnquist, MLQ 45 (1999), says that near model completeness is equivalent to the following sandwich condition:
Whenever $M\subseteq N\subseteq M'$ are all models of $T$, $\bar a\in M$ and $(M,\bar a)\equiv (M',\bar a)$, then in fact $(M,\bar a)\equiv(N,\bar a)\equiv(M',\bar a)$.
It's formula-free, but it's not a category theoretic definition.
A nearly model complete theory that is not inductive
Let $T$ be the theory of infinite discrete linear orders, i.e. of non-empty linear orders in which every element has a successor and a predecessor. Define binary existential formulas $<_n$ such that $a<_nb \iff \exists c_1\ldots c_n(a < c_1 < \ldots < c_n < b)$. $T$ clearly has quantifier elimination down to boolean combinations of these formulas. So $T$ is nearly model complete.
For every $n$, $M_n=\frac 1{2^n}\mathbb Z\subset\mathbb Q$ is a model of $T$. However, the union of the chain $(M_n)_{n<\omega}$ consists of the dyadic rationals and is dense, hence not a model of $T$. So $T$ is not inductive.<|endoftext|>
TITLE: An equivalent definition for the adiamond lattices
QUESTION [5 upvotes]: A lattice is called adiamond if it admits no sublattice equivalent to the diamond lattice $M_3$ below:

The top interval of a lattice is the interval between the meet of all the maximal elements and the greatest element. A lattice is called selftop is it is equal to its top interval.  
Let $\mathcal{L}$ be a selftop finite lattice, $M$ the set of its maximal elements, and $e$ the smallest element.
Question: Is  $\mathcal{L}$ adiamond iff $\forall S \subsetneq M$, $\bigwedge_{b \in S}b >e$?

REPLY [6 votes]: No, neither implication holds.
If you take $\mathcal L$ to be the 8-element Boolean lattice, then remove one atom, the resulting poset is a selftop adiamond lattice which fails your condition. That is, the three coatoms meet to $0$, but one subset of two coatoms also meets to $0$.
On the other hand, start with any nonempty finite lattice $\mathcal K$ and form the poset that is the parallel sum of $\mathcal K$ and a singleton $z$. Now adjoin a top element $1$ and a bottom element $0$ to this poset. This is a selftop lattice with two maximal elements ($z$ and the top of $\mathcal K$), and no one of the coatoms is $0$. Hence this lattice satisfies your condition that no proper subset of coatoms meets to $0$. But there is no reason for this lattice to be adiamond, since $\mathcal K$ is an interval of arbitrary structure.<|endoftext|>
TITLE: Veronese embeddings of elliptic curves in weighted projective space
QUESTION [5 upvotes]: Let $E$ be an elliptic curve and $D_k=kp$ a divisor on $E$, where $p\in E$, for $k\in\mathbb{N}$.
Then we can reconstruct $E$ from the graded ring $R(D_k)=\bigoplus_{n\geqslant0}\mathcal{L}({nD_k})$: $$E\cong \mathrm{Proj}\bigoplus_{n\geqslant0}\mathcal{L}(nkp).$$
Below is a table of what this results in for the first few values of $k$.

(A really useful resource here is this problem sheet by Miles Reid.)
I have two questions for which I can't seem to find any reference:

What happens if we carry on -- can we get an embedding for every value of $k\in\mathbb{N}$? If so, are they all different enough to be interesting, or is there a reason that most sources seem to stop after $5$?
Why do some embeddings never occur in the above sequence (if this is even true)? For example, $C_7\subset\mathbb{P}(1,2,3)$.

Edit:
The answer to 1. is apparently that we can keep on going, obtaining embeddings in $\mathbb{P}^{k-1}$ (since $D_k$ is very ample for $k\geqslant3$), but they do get increasingly complicated.

REPLY [3 votes]: You are just considering the map defined by a linear system $|D_k|$ of degree $k$ (that $D_k=kp$ is irrelevant, any degree $k$ divisor on $E$ is linearly equivalent to $kp$ for some $p$). Any introductory book on curves will tell you that this is an embedding in $\mathbb{P}^{k-1}$ for $k\geq 3$.
This gives only one embedding type for each $k$, so other types do not appear in this way.<|endoftext|>
TITLE: Are graphs with sparse $r$-balls necessarily sparse?
QUESTION [17 upvotes]: Let $G$ be an unweighted undirected graph with the following property:
For some integer $r$, for all nodes $v$, we have
$$\frac{\sum \limits_{u \in B(v, r)} \deg(u)}{|B(v, r)|} \le D$$
where $B(v, r) = \{u \, | \, \mathop{dist}(u, v) \le r\}$.
Does this imply that $|E(G)| = O(nD)$?
The claim is trivially true for $r=0$ or $r = \mathop{rad}(G)$.  It is also true for $r=1$ (see Brendan McKay's nice comment below).

REPLY [4 votes]: I've decided to post my argument anyway but, since nobody (including the OP) expressed any interest, I'll just sketch the proofs. I shall also assume that $r\ge 2$.
Part 1. The upper bound The condition immediately implies that 
$$
\sum_{y\in B(x,r)}d(y)\le D|B(x,r)|\le D\sum_{y\in B(x,r-1)}d(y)
$$
Let now $\sigma(x)=\sum_{y\in B(x,r)}d(y)$. If $x$ and $z$ are adjacent, then 
$$
\sigma(x)\le D\sum_{y\in B(x,r-1)}d(y)\le D\sigma(y)
$$
just because $B(x,r-1)\subset B(y,r)$. It follows immediately that for any $y\in B(x,r)$, we have 
$$
D^{-r}\sigma(x)\le \sigma(y)\le D^r\sigma(x)\,.
$$
Now multiply the inequalities $\sum_{y\in B(x,r)}d(y)\le D|B(x,r)|$ by $d(x)/\sigma(x)$ add them up and change the order of summation. We get
$$
\sum_y d(y)\sum_{x\in B(y,r)}\frac{d(x)}{\sigma(x)}\le D\sum_y\sum_{x\in B(y,r)}\frac{d(x)}{\sigma(x)}
$$
Replacing $\sigma(x)$ by $\sigma(y)$ can only create a multiplicative error $D^r$ on each side, so we finally get
$$
2|E|=\sum_y d(y)\le D^{2r+1}\sum_y 1=D^{2r+1}|V|
$$
Part 2. An example of power growth. Consider the road network consisting of a cycle $ABCDEFA$ and an extra road $BF$ where all $7$ roads have length $1$ each. Make the roads $AF$ and $AB$ toll roads with uniform toll $7/3$ per unit length, the roads $DC$ and $DE$ toll roads with uniform toll $4/3$ per unit length, and the remaining roads free. Then for every ball of radius $2$ in the metric space just constructed, the total toll associated with that ball does not exceed the total road length associated with it (it is enough to check it for balls centered at the towns because both functions are linear on each road) but the total toll is $22/3$, which is strictly larger than the total road length. Taking a large $r$, putting about $r/2$ extra towns along each road, and dropping the tolls just a tiny bit, we shall get a graph and a function $g$ on vertices with the properties $\sum_{y\in B(x,r)g(y)}\le |B(x,r-1)|$ for every $x$ but $\sum_y g(y)>(1+\delta)|V|$. Observe also that we have $|B(x,r)|\le (1+Cr^{-1})|B(x,r-1)|$. Now take the $n$-th power of that graph choosing $n$ so that $(1+Cr^{-1})^n\approx D$, choose a huge $U$ and attach to each vertex $x=(x_1,\dots,x_n)$ $U$ vertices of degree $1$ and a complete graph with $\sqrt{g(x_1)\dots g(x_n)U}$ vertices.<|endoftext|>
TITLE: The density of integers represented by a binary form
QUESTION [17 upvotes]: Suppose that $F(x,y)$ is a binary form of degree $d \geq 3$ with integral coefficients, and non-zero discriminant. It is known (from a paper due to Erdős and Mahler from 1938) that the density of integers in $[1,B]$ which can be represented by $F$ is $\Theta(B^{2/d})$ as $B \rightarrow \infty$, provided that $F(1,0) > 0$. 
It is not too much of a leap to guess that we can replace $\Theta(B^{2/d})$ with an asymptotic formula, i.e. there should exist a positive constant $C_F$ such that
$$\displaystyle \#\{t \in \mathbb{N} \cap [1,B]: \exists (x,y) \in \mathbb{Z}^2 \text{ s.t. } F(x,y) = t \} \sim C_F B^{2/d}.$$
This fact has been confirmed by C. Hooley for the case $d = 3$ and 
$$F(x,y) = ax^3 + 3bx^2 y + 3cxy^2 + dy^3$$
irreducible in two separate papers, the first in 1967 and the second in 2000. In the latter, he obtains the constant
$$\displaystyle C_F = \left(1 - \frac{3}{2 \Delta'}\right) \frac{\sqrt[3]{3}\Gamma(1/3)^2}{2 \sqrt[3]{\Delta} \Gamma(2/3)}, $$
where $\Delta$ is an integer such that $D(F) = 81 \Delta^2$, where $D(F)$ is the discriminant of $F$, and $\Delta' = \Delta/G$, where $G = \gcd(b^2 - ac, bc - ad, c^2 - bd)$. 
Hooley's work seems to indicate that $C_F$ should exist in general, but probably not very easy to compute. I have not seen any later works where $C_F$ (if it exists) is worked out explicitly. 
Has there been further work on investigating what $C_F$ is in other cases, in particular cases with $d \geq 4$? If not, are there any heuristics as to what it should be in general?

REPLY [4 votes]: This question was open in general for degree $d \geq 5$ at the time this question was posted and in addition to the irreducible cubic case Hooley also dealt with the special case of bi-quadratic quartic forms in a paper in 1986 (bi-quadratic as in forms of the shape $Ax^4 + Bx^2y^2 + Cy^4$); see http://www.degruyter.com/view/j/crll.1986.issue-366/crll.1986.366.32/crll.1986.366.32.xml?format=INT . 
The determination of the existence of the constant $C_F$ in general is done in the following joint paper of myself and Cam Stewart: http://arxiv.org/abs/1605.03427 . We showed that $C_F$ is equal to $W_F A_F$, where $A_F$ is the area of the region 
$$\displaystyle \{(x,y) \in \mathbb{R}^2 : |F(x,y)| \leq 1 \}$$
and $W_F$ is a positive rational number which depends on the $\text{GL}_2(\mathbb{Q})$-automorphism group $\text{Aut}(F)$ of $F$. 
However, in this paper we did not generalize Hooley's work entirely since we were not able to determine the value of $W_F$ as an explicit function of the coefficients of $F$. This is likely impossible to do when $d \geq 5$, in a similar way that the general quintic (and beyond) is unsolvable from a Galois theory perspective. However, much like the fact that degree 3 and 4 polynomials are always solvable, it is possible to determine $\text{Aut}(F)$ and therefore $W_F$ explicitly when $F$ is a binary cubic or quartic form. This is contained in an upcoming paper of mine.<|endoftext|>
TITLE: Does every (co)homology functor (in particular, stable homotopy) factor through chain complexes?
QUESTION [12 upvotes]: Ordinary homology and cohomology factor through chain complexes via singular homology and cohomology. What about other (co)homology theories?
That is, for each spectrum $E$, do we have a lift in the following diagram?
$\begin{array}[ccc]
& \mathsf{HoTop} & \overset{E}{\to} & \mathsf{GrAb} \\
& \underset{?}{\searrow} & \uparrow \\
& & \mathcal{D}(\mathsf{Ch}_{\mathbb{Z}})
\end{array}$
Where $\mathsf{HoTop}$ is the homotopy category, $E$ is $E$-homology or $E$-cohomology (in which case it's contravariant, of course), $\mathsf{GrAb}$ is graded abelian groups, $\mathcal{D}(\mathsf{Ch}_{\mathbb{Z}})$ is the derived category of chain complexes in abelian groups, the functor $\mathcal{D}(\mathsf{Ch}_{\mathbb{Z}}) \to \mathsf{GrAb}$ is homology, and the functor labeled "?" is the desired lifting.
It would just about suffice to do this in the universal example of stable homotopy ("just about" only because we have to do this for all spectra now), because we have a factorization:
$\begin{array}[ccc]
& \mathsf{HoTop} & \overset{E}{\to} & \mathsf{GrAb} \\
\downarrow & \underset{\pi^S}{\nearrow} \\
\mathsf{HoSp}
\end{array}$
where $\mathsf{HoSp}$ is the stable homotopy category, $\pi^S$ takes stable homotopy groups, and $\mathsf{HoTop} \to \mathsf{HoSp}$ is either $E\wedge(\Sigma^\infty-)$ (for homology) or $\operatorname{Fun}(\Sigma^\infty -,E)$ (for cohomology).
So in some sense I'm asking about the relationship between two sorts of graded-abelian-group-valued decategorification: taking homotopy groups of a spectrum, and taking homology groups of a chain complex.

REPLY [15 votes]: As written by Fernando, the answer is no if you suppose that you have exactness fo pairs at the chain level. In fact for a functor 
$$L_*:CW^{pairs}\rightarrow Ch(Ab)$$
if you have a short exact sequence of chain complexes
$$0\rightarrow L_*(A,\emptyset)\rightarrow L_*(X,\emptyset)\rightarrow L_*(X,A)\rightarrow 0$$
and if the homology $\mathcal{L_*}=H_*(L_*(-))$ is a generalized homology theory then we have:
Theorem (Burdick, Conner, Floyd 1968).To each finite CW-pair (X,A) we have a natural isomorphism:
$$\mathcal{L}_n(X,A)\cong \bigoplus_{p+q=n} H_p(X,A;\mathcal{L}_q(pt)).$$
Ref: "Chain Theories and their Derived Homologies", Proc . Amer. Math. Soc. l9 (l968) Proceedings of the AMS (1968)<|endoftext|>
TITLE: On Bailey–Borwein–Plouffe formula for irrational numbers
QUESTION [12 upvotes]: A BBP-type formula for an irrational number $\alpha$ in the integer base $b\geq 2$ is a formula in the form $\alpha=\Sigma_{k=0}^{\infty}\frac{1}{b^k}\frac{p(k)}{q(k)}$ ($p, q$ are polynomials in integer coefficients) that allows computing the $n$th digit of $\alpha$ in the basis $b$ directly and without any need to computing the preceding $n-1$ digits of $\alpha$. It is also connected to the problem of normality of irrational numbers. 
1- For which irrational numbers do we have a known Bailey–Borwein–Plouffe formula? For which of them do we have such a formula in the base $10$? (Please add a reference to a list if there is any). 
2- Is there any known BBP formula for $e$ in an integer base $b$? 
3- For which irrational numbers, the normality of the number in a given base is proved using a BBP formula?

REPLY [4 votes]: Concerning (1), there is a large compendium of BBP formulas in this paper by Bailey. Regarding (2), it is strongly suspected (but not proved) that there is no BBP formula for $e$.  Here is a quote from the paper by Bailey. 

Powers of $e$ are specified here because, as far as anyone can tell (although this has not
  been rigorously proven), $e$ is not a polylogarithmic constant in the sense of this paper,
  and thus it and its powers are not expected to satisfy BBP-type linear relations (this
  assumption is confirmed by extensive experience using the author’s PSLQ programs).<|endoftext|>
TITLE: Schur Multiplier of Tarski Monsters
QUESTION [6 upvotes]: Is it known whether the Schur Multiplier of the Tarski monsters are finitely generated?

REPLY [5 votes]: There is no such thing as "the Tarski monster". There exists an uncountable set of Tarski monsters, as was first proved by Olshanskii. His Tarski monsters have infinitely generated Schur multipliers as shown in  Section 5 of "Lacunary hyperbolic groups". See also the earlier paper by Ashmanov and Olshanskii MR0829100 where a similar statement was proved for the free Burnside groups (the proof is essentially the same).<|endoftext|>
TITLE: Grothendieck's "La longue Marche à travers la théorie de Galois"
QUESTION [30 upvotes]: It seems that Grothendieck's familly has given permission for the distribution of his unpublished works, so I hope it is ok to ask this.

Is there any way to obtain a copy (online or not) of "La longue Marche à travers la
  théorie de Galois"?

Note: I am aware that sections 26 to 37 and section 49 are avaible online at the Grothendieck Circle.
Note: a very informative article by Leila Schneps on this particular work can be found here: Grothendieck’s "Long March through Galois theory".

REPLY [18 votes]: All the manuscripts have been finally made avaible by Montpellier university.
You can find all of them here, avaible in pdf.
The items related to “La longue Marche" are:
La "Longue Marche" à travers la théorie de Galois

"Longue Marche" [brouillon] [36 pages]
La "Longue Marche" à travers la théorie de Galois... [574 pages]
Deuxième partie: variations sur l'équation des lacets [695 pages]
La "Longue Marche" à travers la théorie de Galoi [279 pages]

Autour de La "Longue Marche" à travers la théorie de Galois

Théorie de Kummer ? + note Alexandre (bi tenseur...) [15 pages]
Action de Galois sur Teichmüller [121 pages]
Paradigmes des courbes algébriques (version provisoire) [33 pages]
Teichmülleries, 1981-1982 [183 pages]
Printemps 81 [37 pages]
Groupoïdes de Teichmuller et les S0 Tg,!v, S0 TG [133 pages]
Structure à l'infini des Mg,v [189 'ages]
Teichmüller (1983) [71 pages]
Autour de La "Longue Marche" à travers la théorie de Galois [135 pages]<|endoftext|>
TITLE: The proof that a vertex algebra can lead to a Wightman QFT
QUESTION [9 upvotes]: On p. 13 of "Vertex Algebras for Beginners", 2nd edition, Kac writes:
"Under certain assumptions and with certain additional data one may reconstruct the whole QFT from these chiral algebras, but we shall not discuss this problem here".
However, he does not give any references to support this claim (at least not next to the claim) and I was not able to find any on my own.
Does anybody know of any references?

REPLY [2 votes]: I found Nikolov's paper "Vertex Algebras in Higher Dimensions and Globally Conformal Invariant Quantum Field Theory" arXiv:hep-th/0307235. The abstract:

We propose an extension of the definition of vertex algebras in arbitrary space-time dimensions together with their basic structure theory. An one-to-one correspondence between these vertex algebras and axiomatic quantum field theory (QFT) with global conformal invariance (GCI) is constructed. 

Now he did not explicitly show how are these generalized vertex algebras related to the usual chiral vertex algebras. So in my master's thesis     arXiv:1607.05078 I restricted myself to unitary (quasi-)vertex operator algebras and reversed Kac's proof.<|endoftext|>
TITLE: Generalized density functions on the natural numbers
QUESTION [6 upvotes]: If $a_1,a_2,\dots$ are IID random bits (correction as per Anthony Quas: these "bits" are $+1$ and $-1$ with equal probability), then with probability 1, the set of natural numbers $n$ such that $a_1+a_2+\dots+a_n \leq 0$ has lower density 0 and upper density 1, so it has no density in the ordinary sense. Still, I wonder if there is a principled way to generalize the manner in which we assign "densities" to subsets of the natural numbers in such a fashion that, with probability 1, the aforementioned set has generalized density 1/2 -- and, more generally, for every real $t$, the set of $n$ such that $(a_1+a_2+\dots+a_n)/\sqrt{n} \leq t$ has generalized density equal to the probability that the relevant Gaussian random variable has value less than $t$.

REPLY [4 votes]: EDIT: As pointed out by Anthony Quas below, this approach suffers from what looks to be a quite serious measurability issue.
A more abstract approach: 
Let $\theta$ be a shift-invariant probability mean   on $\mathbb{N}$  (i.e., a finitely additive but not necessarily $\sigma$-additive set function with total weight 1 such that $\theta(\{n : n+1 \in A\}=\theta(A)$ for every subset $A$ of $\mathbb{N}$). (Such a mean can be obtained e.g. by taking a subsequential limit of the functions sending $A$ to $\frac{1}{n}\sum_{x\in[0,n]}1(x\in A)$.)
Let's define $A_+=\{n : a_1 + \cdots + a_n >0\}$, $A_0=\{n : a_1 + \cdots + a_n =0\}$ and $A_-=\{n : a_1 + \cdots + a_n <0\}$.

The values of $\theta(A_+)$ and $\theta(A_-)$ are non-random by e.g. the Hewitt-Savage 0-1 law.
By symmetry, $\theta(A_+)=\theta(A_-)$.
$\theta(A_0)=0$ for every choice of $\theta$ since the upper density of $A_0$ is zero .

It follows that $\theta(A_+)=\frac{1}{2}$ almost surely.
(here's a reference from google for the third item above https://books.google.ca/books?id=eFFyBgAAQBAJ&lpg=PA16&ots=srDM7KjW76&dq=translation%20invariant%20means%20and%20upper%20density&pg=PA16#v=onepage&q=translation%20invariant%20means%20and%20upper%20density&f=false)<|endoftext|>
TITLE: Angular distribution of zero sets of sparse polynomials
QUESTION [8 upvotes]: Consider a sequence of complex polynomials $f \in \mathbb{C}[z]$, $f(0) \neq 0$, that are composed of a negligible fraction $o(\deg{f})$ of monomials. Are the zeros of such polynomials necessarily equidistributed in angle, for the uniform measure $d\theta/2\pi$ on $S^1 = \mathbb{C}^{\times} / \mathbb{R}^{> 0}$?
Certainly the zeros are equidistributed in angle when the number of monomials is bounded while the degree goes to infinity. This follows from much more general results of A. G. Khovanskii exposed in his book Fewnomials (Transl. Math. Monographs, vol. 88). But growing like $o(\deg{f})$?
Not knowing the answer, it might make more sense to look at this question contrapositively. Clearly, for any $\varepsilon > 0$ we can construct an $\varepsilon$-sparse sequence with $\deg{f} \to \infty$ and not equidistributed in angle; just consider $h(z^n)$ as $n$ is fixed subject to $1/n < \varepsilon$ while $h$ ranges over the $\mathbb{R}$-split real polynomials. Can we do better than this?

REPLY [4 votes]: A lot depends on how exactly to understand this question. The crudest form (for an $m$-nomial of degree $n$ with non-zero free term, the number of all zeroes in a sector of aperture $2\pi\theta$ is $\theta n$ with an error at most $m$) is an exercise in elementary complex analysis. Indeed, consider a sector of aperture $2\pi\theta$ with the vertex at $0$ and of huge radius $R$.
The increment of the argument of $p$ over the arc is then $2\pi\theta n$ for all practical purposes. On the other hand, each loop around the origin on the interval $\{zt:0\le t\le R]$ creates $2$ real zeroes of $\Re p(zt)$, so the "side" increment of the argument is at most $\pi$ times the total number of real roots of $2$ real $m$-nomials (one for each side). However, by the Descartes rule of signs, an $m$-nomial can have at most $m$ real roots, and the story is over.<|endoftext|>
TITLE: positions of a methane molecule with carbon atom at the origin
QUESTION [10 upvotes]: Let $\text{CH}_4$ be the molecule of Methane:

The four hydrogen atoms form vertices of a regular tetrahedron with the carbon atom in the center of the regular tetrahedron.
Here we regard all atoms to be points in $\mathbb{R}^3$ with volume zero (or we can consider the centers of these atoms, which are points with volume zero). We want to embed the $\text{CH}_4$ molecule into $\mathbb{R}^3$ such that the $\text{C}$-atom is at the origin $0$.

The collection of all such embeddings of $\text{CH}_4$ molecule into $\mathbb{R}^3$ form a manifold $G$. In fact, 
$$
G=O(3)/{\frak S}_4=SO(3)/{\frak A}_4,
$$
where ${\frak S}_4$ is the symmetric group and ${\frak A}_4$ the alternating group. Here we regard $O(3)$ as the isometric embedding Lie group of the $\text{CH}_4$ molecule with the four hydrogen atoms distinct, labelled by $1,2,3,4$. Then we quotient the permutation action of $S_4$ on the four hydrogen atoms, regarding all the four hydrogen atoms the same and do not distinguish them.
Question: as a manifold, what is the cohomology ring (with cup product)
$$
H^*(G;\mathbb{Z}_2)
$$
and the Steenrod square $Sq$'s acting on the cohomology ring?

A contradiction with the following answer by W. Schlieper:
We consider a vector bundle
$$
\xi: \mathbb{R}^4\to O(3)\times_{S_4}\mathbb{R}^4\to O(3)/S_4
$$
where $S_4$ permutes the order of coordinates in $\mathbb{R}^4$. The action of $S_4$ on $\mathbb{R}^4$ may have determinant $-1$, hence it is not contained in $SO(4)$, but only in $O(4)$. Therefore, the first Stiefel-Whitney class $w_1(\xi)\neq 0$. However, $w_1(\xi)\in H^1(O(3)/S_4;\mathbb{Z}/2)$. This contradicts your result
$$
H^1(O(3)/S_4,\mathbb{Z}/3) \simeq H^2(O(3)/S_4,\mathbb{Z}/3) \simeq \mathbb{Z}/3.
$$
Why we have such a contradiction?

REPLY [8 votes]: Over $\mathbb{Z}/2$ it turns out to be pretty boring.  Your space $G$ (which you really shouldn't call that, as it's merely a manifold, not a group), which you define as a quotient of $O(3)$ or equivalently as one of $SO(3)$, can be thought of as the quotient of this group's double cover $SU(2) \cong S^3$ by the binary tetrahedral group $2T$, which turns out to be your fundamental group as $S^3$ is simply connected.
The first homology group over $\mathbb{Z}$ will be $2T/[2T,2T]$, but $[2T,2T]$ is the quaternion group $Q_8$, and $2T/Q_8 \simeq \mathbb{Z}/3$, so $H_1(G,\mathbb{Z}/2)=H^1(G,\mathbb{Z}/2)=0$ and by Poincaré duality, $H^2(G,\mathbb{Z}/2)=0$ will have the same values as that of the $3$-sphere.
Over $\mathbb{Z}/3$ you'll get $H^1(G,\mathbb{Z}/3) \simeq H^2(G,\mathbb{Z}/3) \simeq \mathbb{Z}/3$ and a nontrivial Bockstein, though.<|endoftext|>
TITLE: Geometric interpretation of the conjugation operation in the dual Steenrod algebra
QUESTION [6 upvotes]: As the dual mod 2 Steenrod algebra, $A$, is a Hopf algebra, it has the conjugation operation, $\chi:A\to A$. Milnor also gives a formula for this.
I wonder if there is any source telling about a geometric interpretation of this operation.

REPLY [13 votes]: I do not know if this is what you are searching for, but for the Steenrod algebra itself $\chi$ is closely related to Spanier-Whitehead duality. Let $X$ be a spectrum and $DX=F(X,S)$ its Spanier-Whitehead dual we have an isomorphism
$$Hom(H^*(X,\mathbb{F}_2),\mathbb{F}_2)\rightarrow H^{-*}(DX,\mathbb{F}_2)$$
the contragredient action on the left of a Steenrod operation $\Theta$ corresponds to the action on the right of $\chi \Theta$.
When $X=M$ is a smooth manifold, Atiyah duality identifies $DM$ with the Thom spectrum of $Th(\nu)$ where $\nu=-TM$ is the stable normal bundle of $M$. And we get an isomorphism
$$H_i(M,\mathbb{F}_2)\cong Hom(H^i(M,\mathbb{F}_2),\mathbb{F}_2)\rightarrow H^{-i}(DM,\mathbb{F}_2)\cong H^{dim(M)-i}(M,\mathbb{F}_2).$$
This explains the Wu formula, that relates the total Stiefel-Whitney classe $w(M)$ of $M$ with the Wu class $v(M)$:
$$v(M)=\chi(Sq)(w(M)).$$<|endoftext|>
TITLE: A matching that covers vertices with maximum degree
QUESTION [8 upvotes]: We have a graph G with maximum degree $\Delta$. The induced subgraph on vertices with degree equal to $\Delta$ is a bipartite graph (while the original graph is not).
Prove that G has a matching that covers all vertices with degree $\Delta$.
For example consider $K_{3,3}$ and add a vertex on one edge. So the graph has 7 vertices and 10 edges. The graph is not bipartite, but the induced subgraph on vertices with degree $\Delta(G) = 3$ is a bipartite subgraph. You can easily find a matching that covers all of the 6 vertices with degree 3.

REPLY [8 votes]: We use Tutte's theorem: if a graph $G$ with even number of vertices does not have a perfect matching, then there exists a set $S\subset V(G)$ such that the graph $G\setminus S$ has at least $|S|+2$ odd components.
Let $V=M\sqcup U$ be a vertex set, where $M$ is the set of vertices with maximal degree $\Delta$. We prove that if the induced subgraph $G(M)$ is bipartite, then
there exists a matching covering $M$. Add new set $W$,
$|W|=|M|+|U|$, of vertices to our graph, join 
them with each other and with all vertices of $U$. We have to prove that
in the new graph there exists a perfect matching. Assume the contrary,
then by the Tutte theorem there exists a set $S$ of vertices such that $G\setminus S$ has at least $|S|+2$ odd components. 
If $W\subset S$, it is a clear nonsense. Thus in $G\setminus S$ all vertices
of $(U\cup W)\setminus S$ are in the same component, hence there
exist at least $|S|+1$ odd components containing only vertices of
$M$. Consider each such odd component $K$. It is bipartite graph having, say, 
$k$ vertices in larger part and $\leq k-1$ vertices in a smaller part. 
Totally $\Delta k$ edges of $G$ go from the larger part of $K$. Between them, at most
$\Delta(k-1)$ edges go to the smaller part of $K$, hence at least $\Delta$ edges go outside $K$. 
They all go to $S$. Summing up by all odd components of $G\setminus S$ which belong to $M$ we see that at least $\Delta(|S|+1)$ edges from them come to $S$, it is impossible.<|endoftext|>
TITLE: Minimal immersions of the 2-sphere
QUESTION [9 upvotes]: Following the ideas of S.-S. Chern, J. L. M. Barbosa associated a holomorphic curve in $\mathbb{C}P^m$ to a minimal immersion of the 2-sphere into the $2m$-sphere in his 1972 paper. However, his construction seems to be very unclear because of heavy use of coördinates. Are there any more coördinate-free expositions of this topic?

REPLY [10 votes]: I don't know exactly where it appears in the literature, but a coordinate-free interpretation is certainly known, based on the idea of Penrose's twistor construction.  The main points can be summarized as follows:
First, everything is equivariant under the group $\mathrm{SO}(2m{+}1)$:  One has $S^{2m} = \mathrm{SO}(2m{+}1)/\mathrm{SO}(2m)$ as the space of unit vectors $\mathbb{R}^{2m+1}$, and $\mathrm{Gr}^+_2(\mathbb{R}^{2m+1})$, the Grassmannian of oriented $2$-planes in $\mathbb{R}^{2m+1}$ is $\mathrm{SO}(2m{+}1)/\bigl(\mathrm{SO}(2){\times}\mathrm{SO}(2m-1)\bigr)$.  This latter space can be realized as the complex quadric $Q_{2m-1}\subset\mathbb{CP}_{2m}$ by identifying a unit bi-vector $e_1\wedge e_2$ (where $(e_1,e_2)$ is an oriented orthonormal basis of the $2$-plane $P\in\mathrm{Gr}^+_2(\mathbb{R}^{2m+1})$) with the 'isotropic' complex line $[e_1{-}ie_2]\in\mathbb{CP}_{2m}$. (It is called 'isotropic' because it is spanned by the complex vector $v = e_1{-}ie_2$ that satisfies $v\cdot v = 0$, i.e., $v$ is 'null', aka 'isotropic'.)  It is easy to see that the correspondence $P\leftrightarrow [e_1{-}ie_2]$ is one-to-one and onto.
It's also worthwhile to note that a maximal torus $T^m$ in $\mathrm{SO}(2m)$ is also a maximal torus in $\mathrm{SO}(2m{+}1)$ and can be understood as the stabilizer of a splitting of $\mathbb{R}^{2m}$ as an orthogonal direct sum of $m$ oriented $2$-planes.  In particular, the space $\mathrm{SO}(2m{+}1)/T^m$ can be thought of as the space of orthogonal splittings $\mathbb{R}^{2m+1} = L \oplus P_1\oplus\cdots\oplus P_m$, where $L$ is an oriented line (i.e., an element of $S^{2m}$) and each $P_i$ is an oriented $2$-plane (i.e., an element of $\mathrm{Gr}^+_2(\mathbb{R}^{2m+1})$).  Note that, since $\mathrm{SO}(2m{+}1)/T^m$ is a compact simple group modulo a maximal torus, it has a natural complex structure.  In fact, it has more than one, but there is a particularly nice one to choose for applications to this problem, one such that the mapping $p: \mathrm{SO}(2m{+}1)/T^m\to Q_{2m-1}\subset\mathbb{CP}_{2m}$ given by 
$$
p\bigl(L \oplus P_1\oplus\cdots\oplus P_m) = \overline{P_m}
$$
is holomorphic. 
Next, back in the 60s, Calabi had shown that, if $x:S^2\to S^{2m}$ is a (branched) minimal immersion of the $2$-sphere that is non-degenerate (i.e., its image does not lie in a great $(2m{-}1)$-sphere), then $x$ has a canonical lifting $\hat x:S^2\to\mathrm{SO}(2m{+}1)/T^m$ as a holomorphic rational curve of the form
$$
\hat x =  \mathbb{R}\,x \oplus \pi_1(x)\oplus \cdots \oplus \pi_m(x)
$$
Basically, the idea is that one chooses a splitting of the $x$-pullback of the tangent bundle of $S^{2m}$ into oriented $2$-planes corresponding to the osculating spaces of the immersion.  For example, $\pi_1(x)$ represents the (oriented) tangent space to the image, $\pi_2(x)$ represents the first normal space to the image, etc.
What Barbosa observed is that the map 
$$
p(\hat x):S^2\to \mathrm{Gr}^+_2(\mathbb{R}^{2m+1}) = Q_{2m-1}\subset \mathbb{CP}_{2m}
$$
is holomorphic, and he showed that there is a way to recover $x$ from $p(\hat x)$.  
Moreover, he showed that, if one starts with a holomorphic rational curve $\xi:S^2\to Q_{2m-1}\subset \mathbb{CP}_{2m}$ that is nondegenerate in the sense that its image does not lie in a hyperplane in $\mathbb{CP}_{2m}$, then it is of the form $\xi = p(\hat x)$ for some unique branched minimal immersion $x:S^2\to S^{2m}$.<|endoftext|>
TITLE: Equivariant Fredholm operators classify equivariant K-theory
QUESTION [15 upvotes]: Let $\mathcal{F}$ be the space of Fredholm operators on a separable Hilbert space $H$ with the topology induced by the operator norm. 
If $X$ is compact,
Atiyah-Jänich proved that
$$[X,\mathcal{F}]\simeq K^0(X). $$
Here $[\cdot,\cdot]$ denotes the set of homotopy classes.
For the equivariant version, for a compact Lie group $G$, let $\mathcal{F}(G)$ be the space of Fredholm operators on $L^2(G,H)$. The group $G$ acts on $\mathcal{F}(G)$ in a natural way. For compact $G$-space $X$, one has 
$$[X,\mathcal{F}(G)]_G\simeq K_G^0(X). $$
Here $[\cdot,\cdot]_G$ denotes the set of $G$-homotopy classes of $G$-maps.
(Matumoto, T.,
Equivariant K-theory and Fredholm operators.
J. Fac. Sci. Univ. Tokyo Sect. I A Math. 18 1971 109–125.) 
On the other hand, let $\hat{\mathcal{F}}_*$ be the space of self-adjoint Fredholm operators on $H$ such that its elements have infinite positive as well as infinite negative spectrum. Atiyah proved that
$$[X,\hat{\mathcal{F}}_*]\simeq K^1(X). $$
My question:
is there a equvariant isomorphism for $K_G^1$ as in $K_G^0$?
That is,
$$[X,\hat{\mathcal{F}}(G)_*]_G\simeq K_G^1(X)? $$
As the proof of Atiyah, I think it turns to prove $\hat{\mathcal{F}}(G)_*\rightarrow \Omega \mathcal{F}(G)$ is a $G$-homotopy equivalence.

REPLY [5 votes]: The isomorphism for $K_G^1$ in the question is right.
Freed, Daniel S.; Hopkins, Michael J.; Teleman, Constantin,  Loop groups and twisted K-theory I. J. Topol. 4 (2011), no. 4, 737–798. 
Section 3.5.4 and Appendix A.5<|endoftext|>
TITLE: Inverse Galois problem for simple Lie type groups
QUESTION [6 upvotes]: Progress towards the Inverse Galois problem over $\mathbb{Q}$ is very well documented for sporadic groups ($M_{23}$ is the only case open) and for $PSL_n(q)$ (a lot of cases known, but wide open in general).

I'm interested in progress for families of simple Lie type groups other than
  $PSL_n(q)$.

Important note (edit). I'm looking for results for complete families, not individual groups.
I haven't been able to find almost any information, and that suggests that the answer is "nothing is known". But it would be nice to have a reference for that, if it is the case.
On a side note, as an example of progress for those groups (so that they are not completely intractable), Belyi proved that the 6 families of classical simple Lie groups are realizable over $\mathbb{Q}^{ab}$.

REPLY [7 votes]: There is Thompson and Volklein, who prove that the symplectic groups are Galois groups:
Thompson, J. G.(1-FL); Völklein, H.(1-FL)
Symplectic groups as Galois groups. 
J. Group Theory 1 (1998), no. 1, 1–58. 
12F12 

More information on realization of simple groups can be found in Volklein's book:
MR1405612 (98b:12003) Reviewed 
Völklein, Helmut(1-FL)
Groups as Galois groups. (English summary) 
An introduction. Cambridge Studies in Advanced Mathematics, 53. 
Cambridge University Press, Cambridge, 1996. xviii+248 pp. ISBN: 0-521-56280-5 
12F12 

Finally, a zoo of low-order (not that low) is discussed by David Zywina.

REPLY [5 votes]: It's hard to keep track of all relevant literature on the inverse Galois problem for finite groups of Lie type, but at this point many special cases have been worked out while others remain open.   One rather recent paper here indicates how subtle the approaches have become.   (For an arXiv preprint, see here.)
Work by Gunter Malle, including his joint book with B.H. Matzat, is an important source; Malle might also have advice to give by email.    (One obvious question is what kind of search tools you are using and what kind of library access you have?)  
As you realize, finite special linear groups and their relatives pose extra problems---especially in low ranks.   The study of finite simple or almost-simple groups is natural but has so far been successful only in special cases rather than through some unified argument.<|endoftext|>
TITLE: Erdös-Turán via Hardy-Littlewood circle method?
QUESTION [14 upvotes]: For any set $B\subseteq \mathbb{N}$ one can associate the formal series
$$f_B(z) = \sum_{b\in B}z^b$$
and obtain
$$f_B(z)^k = \sum_{n\geqslant 0} r_{B,k}(n)z^n,$$
where $r_{B,k}(n) = |\{(x_1,\cdots,x_k)\in B^k : x_1+\cdots+x_k=n\}|$.  As it is characteristic of the Circle Method, one them have
$$ r_{B,k}(n) = \frac{1}{2\pi i}\int_{|z|=\rho} \frac{f_B(z)^k}{z^{n+1}}~\mathrm{d}z = \int_0^1 f_B(e(t))^k e(-nt)~\mathrm{d}t,$$
where $e(z) = e^{2\pi iz}$.
I am starting to study the Circle Method and the first thing I noticed is that, although it is a very powerful way to deal with both Waring's and weak Goldbach's problems, it seems that it relies heavily on the number-theoretical properties of the primes and Waring bases to work properly, and doesn't have strong "combinatorial powers" per se.
However, this first step of the setup seems to be a very very natural starting point to deal with problems involving (asymptotic) bases in general, but I found way less discussions about this in the literature than I expected. More especifically, my question is:

Is there any approach of the Erdős–Turán conjecture on additive bases via the Hardy-Littlewood circle method? Or it is known to be an ineffective approach?

As far as I know, the best results up to date has been made in the closely related problem of finding thin bases, namely

Erdös-Tetali theorem (1990) on the existence of asymptotic $h$-bases $B$ with $r_{B,h}(n) = \Theta(\log(n))$, using, basically, only probabilistic methods.
Vu's thin subbases of Waring bases (2000), showing the existence of asymptotic $h_k$-bases $B^{(k)} \subseteq P_k = \{n^k : n\geqslant 0\}$ with $r_{B^{(k)},h_k}(n) = \Theta(\log(n))$ for every $k$. He uses both probabilistic methods AND the Circle Method, but as far as I understood the last one is used "only" in dealing with the $P_{k}$'s. (By "only" I mean that it is not in the sense that I am asking in this question.)

Other approaches like the Borwein-Choi-Chu result (2006) that asserts that if $B$ is a $2$-basis then $\liminf\limits_{n\to\infty} r_{B,2}(n) > 7$ seems fruitful too, but I didn't spent enough time reading it to be able to make a proper remark. On the other hand, the current probabilistic methods apparently have a "technical limitation" involving the $\Theta(\log(n))$, as it seems to be the best one can get in this direction.
After reading Xiao's master's dissertation and a few books (Alon & Spencer's Probabilistic Method, Nathanson's Additive NT: Classical Bases, Tao & Vu's Additive Combinatorics), I had a feeling that the Circle Method is a possible way to understand better this limitation of the probabilistic method.
In any case, this is more a raw idea than anything. I'd appreciate any references, thoughts and thinghs to think about. Thanks in advance!

REPLY [6 votes]: Your question probably doesn't have a definitive answer, as it is unlikely that one can prove that the `circle method' (it has evolved in many directions taking in tools from a very broad area of mathematics, so it is likely difficult to define exactly what the circle method entails) can never be used to prove the Erdös-Turán conjecture for additive bases (colloquially, the 'Erdös-Turán conjecture' is usually reserved for this: https://en.wikipedia.org/wiki/Erd%C5%91s_conjecture_on_arithmetic_progressions). That said, I would argue that the current knowledge of the circle method is unlikely to produce significant results on the conjecture.
The reason is that we expect a basis $B$ such that $r_{B,k}(n)$ has as small an order of magnitude (which is conjectured by Erdös to be $\log n$) as possible is very close to the random sets constructed by Erdös and Erdös-Tetali. To date, we have yet to produce a single concrete example of such a set, so it is unclear what properties, if any, such a set should have. However, as you mentioned the circle method is heavily dependent on certain key ingredients to work. Indeed, it mostly works in settings somewhat related to algebraic varieties because then you can expect that your exponential sums turn into Weyl sums, where you can obtain savings via Weyl differencing (this explanation is due to Sam Chow). Even then, the circle method seems to be fairly inefficient in the sense that many more variables are needed than expected to produce acceptable savings.
One way to think about the weakness in the circle method to tackle problems of this type is in the proof of Vinogradov's theorem. The set of primes are in some sense extremely ripe to be a candidate for an additive basis: for every admissible congruence class they are equidistributed, they are very dense (has density exceeding $\Omega(n^{1-\epsilon})$ for any $\epsilon > 0$), and there are many heuristics which point to them being `random'. Even for the set of primes, the circle method can only succeed in proving that they are an additive basis of order 4 (i.e. every (sufficiently large) positive integer is the sum of four primes, a fact implied by Vinogradov's theorem which asserts that every sufficiently large odd number is the sum of three primes. Recent work of Helfgott seems to have removed the 'sufficiently large' requirement, though I believe this work is still being refereed), while the expected truth is that they are an additive basis of order 3 (Goldbach's conjecture). If you examine why the circle method has (to date) failed to produce a proof of the Goldbach conjecture, I think you will gain a better understanding as to why the circle method likely cannot be applied to a set whose only known property is 'thin additive basis'. 
To further clarify, my point is the following: the additive bases which are as sparse as possible are much sparser than the primes, so if the circle method has failed to prove the Goldbach conjecture, then it is unlikely that it can produce results for a set which is in almost all ways much harder to work with (i.e. has less arithmetical structure and much sparser).<|endoftext|>
TITLE: What are retracts of polynomial rings?
QUESTION [27 upvotes]: Is there a known example of a ring endomorphism $f: \mathbb{Z}[x_1, \ldots, x_n] \to \mathbb{Z}[x_1, \ldots, x_n]$ such that $f \circ f = f$ but whose image is not isomorphic to a polynomial ring? 
My own interest in this has to do with better understanding the (categorical) Cauchy completions of Lawvere theories for some familiar types of algebraic objects; here we are dealing with the Lawvere theory of commutative rings. But apparently this type of problem is of interest to algebraists in the context of hard problems like the Jacobian conjecture and the cancellation problem, so there is a certain body of work out there already on related material. 
From the literature I've scanned so far (articles by Costa, Shpilrain, Picavet, Gupta, and others), a lot of attention is paid to retracts of polynomial algebras over fields, but I'm having quite a hard time finding a clear statement for the case of polynomial algebras over $\mathbb{Z}$. One tantalizing lead was a statement by Picavet here: "We were motivated by an unsolved conjecture: a projective algebra of finite type over a field $A$ is a polynomial ring. An example by Costa shows that the statement is false if $A$ is not a field." I couldn't find a statement by Costa which treated every non-field $A$ (including $\mathbb{Z}$ in particular); I suspect Picavet meant that there exist non-fields $A$ for which the statement is false. An interesting example of such $A$ can be extracted from Gupta's later negative solution to the cancellation problem, as mentioned by Jeremy Rickard here at the MO discussion Is a retract of a free object free?. (Actually, Gupta's constructions more significantly show that the statement is false for any field of positive characteristic, but this work can be used to derive some non-field examples as well.) 
Akhil Mathew asked essentially the same question at Math.SE here, and got pointers to literature from Mariano Suárez-Alvarez, but I'm hoping to get something more definitive now.

REPLY [13 votes]: Existence of such an example follows from the same result of Asanuma that is crucial for Gupta's work, see the article 
Teruo Asanuma, "Polynomial fibre rings of algebras over noetherian rings", Inventiones mathematicae 87 (1987), 101–127 (DOI link). 
It follows from Corollary 5.3 of that article that if
$$
A=\mathbb{Z}[x,y,z]/(-x^{p^e}+y+y^{sp}+pz),
 $$
where $p$ is a prime number and $e,s$ are positive integers such that $p^e\not\mid sp$, $sp\not\mid p^e$, then $A$ is not isomorphic to a polynomial ring, but $B=A[t]$ is isomorphic to a polynomial ring in three variables. Of course, $A$ is a retract of $B$ via the evaluation at $t=0$.
EDIT:
Moreover, consider 
 $$
A_n:=\mathbb{Z}[x,y,z,x_1,\ldots,x_n]/(-x^{p^e}+y+y^{sp}+px_1\cdots x_nz),
 $$
then the proof of the same Corollary 5.3 (in the case $R=\mathbb{Z}[x_1,\ldots,x_n]$) can be adapted to show that $A_n[t]$ is isomorphic to the polynomial ring in $n+3$ variables, but $A_n$ is not isomorphic to the polynomial ring in $n+2$ variables.<|endoftext|>
TITLE: Young-Fibonacci lattice and purely periodic continued fractions
QUESTION [6 upvotes]: The Fibonacci lattice $\mathcal{F}$ is the poset of all finite words consisting of 1's and 2's where a word $v$ covers a word $u$ if $v$ is obtained from $v$ by either (a) inserting a 1 in $u$ prior to its leftmost 1 or (b) changing the leftmost 1 occurring in $u$ into a 2. Let $\mathcal{L}$ be the partially ordered set obtained by "vetting" the Young-Fibonacci lattice of all "constant" words --- i.e. all words of length at least two of the form $11 \dots 1$ and $22 \dots 2$. Interpret
the remaining words as purely periodic continued fractions; e.g.
the word $1121$ would correspond to the quadratic surd $\tau$ defined by
\begin{equation} \tau \ = \ {1 \over {1 + {\displaystyle 1 \over {\displaystyle 1 + {\displaystyle 1 \over {\displaystyle 2 + {\displaystyle 1 \over {\displaystyle 1 + \tau} }}}}}  }} \end{equation}
Recall that the Lagrange number $L(\tau)$ of a positive real number $\tau$ is defined as
\begin{equation} \text{sup} \, \Bigg\{ L \ : \ \Big| \tau - {p \over q} \Big| < {1 \over {Lq^2}} \ \, \begin{array}{l} \text{for infinitely may reduced} \\ \text{postive rational numbers ${p \over q}$} \end{array} \Bigg\} \end{equation}
Finite words consisting of 1's and 2's correspond, under the recipe described above, to congruence classes of quadratic surds $\tau$ whose
Lagrange numbers $L(\tau)$ are strictly less than $3$, where two real numbers $\alpha$ and $\beta$ are understood to be congruent if there exists 
$g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{GL}_2 \big( \Bbb{Z} \big)$ with $\det(g) = \pm 1$ such that 
\begin{equation} \alpha \ = \ {a \beta + b \over {c \beta +d}} \end{equation}
Question: Can the order relation on $\mathcal{L}$ be meanfully interpreted as a kind of closure relation on the congruence classes of the Lagrange spectrum beneath 3?
regards, Ines Institoris.

REPLY [2 votes]: It seems there is some confusion in what you wrote. There are purely periodic continued fractions consisting only of 1's and 2's whose Lagrange numbers are greater than 3 --- for example the Lagrange number of $\Big[ \overline{121} \Big]$ is $\sqrt{10}$. See Martin Aigner's book "Markov's Theorem and 100 Years of the Uniqueness Conjecture" page 27 for this computation. However the converse is true,
namely: If $\tau$ is a real number whose Lagrange number is strictly 
less than 3 then $\tau$ is congruent (in the sense you explain) to a 
continued fraction consisting only of 1's and 2's. So it seems to me that
the Young-Fibonacci lattice ought to be further "vetted" in order to exclude
surds like Aigner's example. 
The question about $G$-orbit "closures" and covering relations (presumably in the spirit of how one defines the Bruhat order ?) where $G$ is the group of 
all $2 \times 2$ integer matrices with determinant $\pm 1$ is compelling. 
My apologies for not adding this to the comment section; I don't have enough "points" to exercise this privilege yet :P
best, 
A. Leverkühn<|endoftext|>
TITLE: Is a manifold generically real analytic (with generic real analytic metric)?
QUESTION [8 upvotes]: I have heard it said in some differential geometry talks that "the generic situation in such and such case is real analytic". My question is, is the generic smooth manifold also real analytic in some sense of genericity? I do not have a specific purpose in asking this question, just curious. Thanks!
Edit after Robert Bryant's comment: As an extension to my question, consider a Riemannian manifold with a smooth metric. Is this metric generically real analytic in a certain sense? That is, can we equip a smooth manifold with a real analytic metric  generically (in a certain sense)?

REPLY [20 votes]: Your two final questions are not the same, in spite of the "That is,..." that starts the second one, but, in the interpretations that make the most sense to me, the answer to the first is 'no' and the answer to the second is 'yes'.
For the second question, you know that a smooth manifold $M$ has a compatible real-analytic structure and, by the theorem of Grauert and Morrey, it can be real-analytically embedded into an Euclidean space of sufficiently high dimension.  Then the induced metric from such an embedding is real-analytic, so every smooth manifold does have a metric that is real-analytic with respect to some compatible real-analytic structure. 
Moreover, when a smooth $M$ is compact (and maybe even in general), every smooth Riemannian metric can be approximated arbitrarily closely (in practically any sense) by a metric that is real-analytic with respect to some compatible real-analytic structure on $M$.  You just use the given metric as the initial metric for the Ricci flow (which exists for short time $0\le t < T$), and the resulting metrics $g(t)$ for any positive time will be real-analytic with respect to some real-analytic structure (that is independent of $t$).  Thus, each smooth metric $g$ on a compact $M$ determines a canonical real-analytic structure, and $g$ can be approximated arbitrarily closely by metrics $g(t)$ (for $t$ small and positive) that are real-analytic in that real-analytic structure.
Now, my interpretation of the OP's first question is this:  Given a smooth Riemannian manifold~$(M^n,g)$, is there an analytic structure on $M^n$ in which $g$ is real-analytic?  If one doesn't always exist, does one exist for a 'generic' smooth metric $g$ on $M$?
Well, it's certainly not true always.  For example, if $n=2$ and $g$ is a metric whose Gauss curvature vanishes to infinite order at a point of $M$ but is not identically zero, then $g$ clearly cannot be real-analytic for any real-analytic structure subordinate to the given smooth structure.  It's easy to construct such examples.  For example, take $g = e^{2h(x,y)}(dx^2+dy^2)$ where $h$ is a smooth function on $\mathbb{R}^2$  such that $\Delta h$ vanishes to infinite order at $(x,y)=(0,0)$ but isn't identically zero.
Moreover, it's also not true 'generically', in the following sense:  Again, in dimension $2$, we know that any smooth metric is conformally flat, i.e., each point lies in the domain of a smooth coordinate chart $(x,y)$ such that $g = e^{2h(x,y)}(dx^2+dy^2)$.  Moreover, if $g$ is analytic in some local coordinate system $(u,v)$, then any conformal coordinate chart $(x,y)$ in that domain must be analytic as functions of $(u,v)$.  Thus, $g$ is analytic in some coordinate system if and only if it is analytic in conformal coordinates.  However, when you write the 'generic' $g$ in conformal coordinates $(x,y)$, the function $h(x,y)$ will not, in general, be an analytic function of $(x,y)$.  Thus, the generic smooth metric in dimension $2$ is not real-analytic in any coordinate system.
Similar remarks hold in all higher dimensions.  A 1981 result of DeTurck and Kazdan  shows that a metric $g$ has its maximum regularity in $g$-harmonic coordinates, i.e., $g$ is real-analytic if and only if it has analytic coefficients when expressed in a coordinate system $x=(x^i)$ where $\Delta_g x^i = 0$ and where $\Delta_g$ is the Laplace operator associated to $g$.  In particular, if $g$ is analytic in any real-analytic atlas, its system of $g$-harmonic local coordinates must have real-analytic transitions functions, and these belong to the only possible real-analytic atlas in which $g$ could be real-analytic.
This 'test', though, may not be entirely satisfactory because it requires one to find (local) solutions of a PDE to make a (local) coordinate system and then recognize whether the coefficients of $g$ in that coordinate system are real-analytic functions of the coordinates.  There's no way to avoid needing a way to recognize when some function is an analytic function of some other functions, but one can usually avoid the step of solving a PDE.  
For example, again, in dimension $2$, given a metric $g$, one can compute its Gauss curvature $K$ and the squared norm of its gradient $L = |\nabla K|^2_g$.  Suppose that these are independent functions (i.e., local coordinates) on some open set $U\subset M$.  Now compute $\Delta_g K$.  If $g$ is real-analytic in some coordinate system with domain $U$, then $\Delta_gK$ will be an analytic function of $K$ and $L$.  One can show that, conversely, if $\Delta_gK$ is an analytic function of $K$ and $L$, then $g$ has real-analytic coefficients in the $(K,L)$ coordinate system.<|endoftext|>
TITLE: Are there irreducible polynomials with all zeros on two concentric circles?
QUESTION [27 upvotes]: This is somewhat similar to this recent question, but extending in a different direction.
Let $f(x)$ be an irreducible polynomial of degree $n$ with integer coefficients. Call such $f$ a bicycle polynomial if all its roots are located on two circles around $O$, i.e. all roots have one of two moduli. (Of course we'll exclude polynomials of cyclotomic type like $\Phi_n(mx)$ for $m\in\mathbb Z$, which have all their roots fit on one circle.)
For $n=2k$ or $n=3k$, examples of bicycle polynomials are $f(x)=g(x^k-1)$, where $g$ is irreducible of degree $2$ or $3$. Taking here a $g$ of degree $4$ with two pairs of complex roots yields still other bicycle polynomials for $n=4k$.   
Now: except replacing the "$-1$" by any other nonzero integer, those types of constructions already seem to be about all of it... 


Do bicycle polynomials of degree $n$ exist if $n$ has only prime factors $>3$?


Assuming the existence of such a polynomial, it appears (?) to boil down to the existence of a degree $m$ polynomial ($m>3$ odd) with $m-1$ roots on one circle. This circle must presumably have an irrational radius because of what is known about Salem polynomials, but I'm stuck here.  
Also related: How to best distribute points on two concentric circles?
Another question:  


Does anything change if we allow complex integers as coefficients?

REPLY [10 votes]: I couldn't resist to include a picture showing how asymmetric the distribution of roots can be for a bicycle polynomial. Here are the roots of $$z^{16}+2 z^{15}+z^{14}-2 z^{13}-4 z^{12}-2 z^{11}+3 z^{10}\\ +5 z^9+3 z^8+z^7-z^5-z^4-z^3+z^2+z+1,$$ which is an example of an extended Salem polynomial given by Dubickas and Smyth in their paper.<|endoftext|>
TITLE: Traces of operators in nuclear spaces
QUESTION [9 upvotes]: I am currently reading up on nuclear spaces in Jarchow, "Locally Convex Spaces", but I got confused and don't seem to find my mistake. In said book, theorem 21.5.9 states:
Let $F$ be a nuclear Frechet space. Then, $F'_\beta \otimes_{\pi} F = L_\beta (F,F)$.
i.e. the projective tensor product of $F$ with its strong dual is homeomorphic to the operators on $F$ with the strong topology.
Now, that means every operator on $F$ has a trace, right? 
So take the the space $X=\Pi_\mathbb{N} \mathbb{R}$. This is a nuclear Frechet space as it is a countable infinite product of those. Let $id_k : \mathbb{R} \rightarrow \mathbb{R}$ denote the identity on the k'th component of $X$. We have $id_X = \Sigma_k id_k$. Now, by above theorem $id_X$ and all $id_k$ must be trace class, but clearly, the sum can not converge to $id_X$ as otherwise the trace must be infinite. 
If I read correctly, the strong topology has a 0-basis of $L(B,U)$, maps from some bounded set into an open set. The partial sums $f_n=\Sigma_{1≤k≤n}
id_k$ however satisfy that $f_n-id_X$ is zero on the first $n$ components and thus maps any set into a chosen open set for n big enough. Thus, the partial sum converges and there is a contradiction. 
I simply can't spot my mistake, perhaps you can help me?

REPLY [3 votes]: Your answer is wrong (unfortunately, I can't comment, so here is another answer instead). 
Now, I'm quoting from "Notes on locally convex vector spaces" by J.L. Taylor. 
Since $F$ is nuclear, the (completed, you are right, I meant that) injective and projective tensor products agree, so "is smaller than" doesn't apply here.
Furthermore, theorem 21.5.9 in Yarchow explicitely states that the maps are homeomorphisms, and not just an isomorphism.
We have:
$F'_\beta \hat{\otimes}_\pi F=F'_\beta \hat{\otimes}_\epsilon 
F=L_\beta(F,F)$ 
(corrolary 5.19 in the aforemention lecture notes). Thus the injective tensor product is just as big, "is smaller than" does not make much sense here.
Also, we have (proposition 3.5):
Any  bilinear continuous map $\phi : F'_\beta \times F \rightarrow \mathbb{R} $ extends to a map $\phi' : F'_\beta \hat{\otimes}_\pi F \rightarrow \mathbb{R}$. 
Which brings me to the real answer to the problem, which I just found myself: 
Now, this would be fine if the trace was indeed continuous. But: Let $U$ be an open set in $\mathbb{R}^\mathbb{N}$, $V$ be an open set in $\mathbb{R}_\mathbb{N}$. Then, we always have: $U(V)=\mathbb{R}$, as can always be seen by choosing a sufficiently small open subset. Hence, the trace is not a bilinear continous map, hence can not be extended! Please correct me if you see any mistake, but I think that is the answer, no?<|endoftext|>
TITLE: Do the complex zeros of the sum/difference of these series all reside on the line $\Re(s)=\frac12$?
QUESTION [9 upvotes]: The following series seems convergent for all $s\in \mathbb{C}$:
$$\displaystyle f(s):=\sum_{n=1}^\infty \frac{(-1)^n}{(n+s)^{n+s}}$$
The function itself does not appear to have any real or complex zeros, however numerical evidence suggests that all zeros of:
$$f(s) \pm f(1-s)$$
reside on the line $\Re(s)=\frac12$. Their density apparently increases in a quite regular manner. 

1) Could this be proven? When using the finite series $\sum_{n=1}^N$, the claim seems to hold for each $N$.
2) Could there exist a functional equation between $f(s)$ and $f(1-s)$? The only relation I found so far is that $f(-1)=f(1)$.
EDIT: A partial result on the second question. Based on Joro's findings below and realising that for integers and half-integers an infinite number of terms could be cancelled out when adding or substracting $f(s)$ and $f(1-s)$, the following reflective formulae hold:
For all integers $s \in \mathbb{Z}$:
$$f(s)+f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$ 
For all half-integers $s+\dfrac{1}{2}$ with $s \in \mathbb{Z}$:
$$f(s)-f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$
The non-alternating series $\displaystyle g(s):=\sum_{n=1}^\infty \frac{1}{(n+s)^{n+s}}$ has one formula for integers and half-integers:
$$g(s)-g(1-s) +\sum_{n=0}^{2(s-1)} \frac{1}{(s-n)^{s-n}} =0$$

REPLY [4 votes]: EDIT Due to confusing XRay and programming mistake, I erroneously claimed real zeros. Root finding didn't found any zeros off the line.

Very partial answer for functional equation.
For integer $s$, $\{1,f(s),f(1-s)\}$ appear linearly dependent
over the rationals with high precision.
Let $K=f(s),K_2=f(1-s),K_3=1/s^s,K_4=(K-K_2+K_3)^2$.
$$s=2, 1+4K+4K_2=0$$
$$s=3, 131-108K-108K_2=0$$
$$s=4, 36059+ 6912K+ 6912K_2=0$$
$$s=5, -695877463+ 21600000K+ 21600000K_2=0$$
$$s=6, 168087904001+ 583200000K+ 583200000K_2=0$$
$$s=7, -1639334733641495543+ 480290277600000K+ 480290277600000K_2=0$$
And in addition for $s=3/2, K^{2} - 2 \, K K_{2} + K_{2}^{2} + 2 \, K K_{3} - 2 \, K_{2} K_{3} + K_{3}^{2} - 2=0$.
...And with the help of Agno's observation, $s=5/2,8503056K_4^4 - 61096032K_4^3 + 202411224K_4^2 - 171663192K_4 + 252460321$
I suspect for $s$ half integer there is non-trivial algebraic dependency.<|endoftext|>
TITLE: What is the Euler characteristic of a mapping space?
QUESTION [5 upvotes]: Suppose that $A$ and $B$ are topological spaces homotopy equivalent to finite cell complexes, and let $B^A = \mathrm{maps}(A,B)$ denote the space of maps from $A$ to $B$.  Is it there a formula for the Euler characteristic $\chi(B^A)$ of this mapping space in terms of, say, the Euler characteristics of $A$ and $B$ (or similar not-too-difficult data, like the Betti numbers)?  I think that if $\chi(B^A)$ were $\chi(B)^{\chi(A)}$, then Wikipedia would have said so, and I would have known about it.
I care most about the case when $B$ is a manifold and $A$ is a surface, which is fairly specific, so I will accept answers that demand good behavior of $A$ and/or $B$.

REPLY [25 votes]: This Euler characteristic usually won't be well-defined. For example, take $A = S^1$ and $B = S^3$. Then the mapping space $[A, B]$ is the free loop space $L S^3$, which decomposes as a product
$$L S^3 \cong S^3 \times \Omega S^3$$
because $S^3$ has a Lie group structure. This makes the rational cohomology of $L S^3$ easy to compute: it's the tensor product of the rational cohomology of $S^3$ and the rational cohomology of $\Omega S^3$, and the latter is free on a generator of degree $2$. Hence the Poincare series of $L S^3$ is
$$\frac{1 + t^3}{1 - t^2}$$
and in particular $LS^3$ has nonvanishing rational cohomology in arbitrarily high degrees (which is usually the case with free loop spaces). Removing the common factor of $1 + t$ and plugging in $t = -1$ for kicks gets us $\frac{3}{2}$, which doesn't have any obvious relationship to $\chi(S^1) = 0$ or $\chi(S^3) = 0$.<|endoftext|>
TITLE: Is Frac $\mathbb{Z}((x))$ Hilbertian?
QUESTION [9 upvotes]: Note that Frac $\mathbb{Z}((x)) \ne\mathbb{Q}((x))$.
As a result of Some questions about the ring Z((x)), we know that it is a Dedekind domain with uncountably many primes, each of which is of the form
$$p^n + a_1q + a_2q^2 + \cdots$$
By the Weierstrass preparation theorem, the residue fields of $\mathbb{Z}((x))$ are precisely fields of the form $\mathbb{Q}_p(\alpha)$, where $\alpha$ has positive $p$-adic valuation.
Another question is:
Does the Tate elliptic curve over Frac $\mathbb{Z}((x))$ satisfy weak weak approximation? (this would imply that Frac $\mathbb{Z}((x))$ is Hilbertian) Does it have the Hilbert property (does it satisfy Hilbert's irreducibility theorem?)

REPLY [5 votes]: This does not answer the question about the Tate curve, but the question in the title: The field ${\rm Frac}(\mathbb{Z}((x)))={\rm Frac}(\mathbb{Z}[[x]])$ is Hilbertian.
Namely, by a result of Weissauer (Satz 7.2 in [1]), the quotient field of a (generalized) Krull domain of dimension at least 2 is Hilbertian. The PID $\mathbb{Z}$ is a Krull domain, and if $R$ is a Krull domain, then so is $R[[x]]$. For a more direct and more general statement see for example Corollary B in [2].
[1]: Rainer Weissauer. Der Hilbertsche Irreduzibilitätssatz. J. reine angew. Math. 334 (1982), pp. 203–220. https://eudml.org/doc/152457
[2]: Arno Fehm and Elad Paran. Klein approximation and Hilbertian fields. J. reine angew. Math. 676 (2013), pp. 213—225. https://www.degruyter.com/view/journals/crll/2013/676/article-p213.xml<|endoftext|>
TITLE: Extensions of $\Bbb Z_3$ by $PGL(2,q)$ where $q$ is odd
QUESTION [5 upvotes]: Let $q$ be odd. If $G$ is a finite group such that $G$ has a normal subgroup $H$ of order $3$ such that $G/H\cong {\rm PGL}(2,q)$, what can we say about $G$. Is it true in general that $G\cong {\Bbb Z}_3\times {\rm PGL}(2,q)$?
The motivation for this question: If we change ${\rm PGL}(2,q)$ to ${\rm PSL}(2,q)$, then the result is true.

REPLY [6 votes]: Since the Schur Multiplier of the perfect group ${\rm PSL}(2,q)$ has order $2$, $G$ must have a normal subgroup $N$ of index $2$ isomorphic to $C_3 \times {\rm PSL}(2,q)$. Since ${\rm PGL}(2,q) \setminus {\rm PSL}(2,q)$ contains an element of order $2$, we have $G = \langle N,t \rangle$, with $t^2=1$. The direct factors of $N$ are both characteristic in $N$ and hence normalized by $t$, and clearly $\langle t, {\rm PSL}(2,q) \rangle = {\rm PGL}(2,q)$, so the group $G$ is determined by the conjugation action of $t$ on $H=C_3$.
There are two possibilities for that, $t$ can either centralize of invert the elements of $H$, giving two isomorphism classes of groups.
Added later: As pointed out by Geoff Robinson, the case $q=9$ is exceptional, because the Schur Multiplier of ${\rm PSL}(2,9)$ has order $6$. There is a third isomorphism class of groups $G$ in this case, which is a group with a normal series $1 < H < K < G$ with $|H|=3$, $K$ isomorphic to the perfect triple cover $3.{\rm PSL}(2,9)$, and $G/H \cong {\rm PGL}(2,9)$.
In this example, an element outside of $K$ inverts the elements of $H$.<|endoftext|>
TITLE: Generating function of the Thue-Morse sequence
QUESTION [14 upvotes]: Let $T$ be the generating function of the Thue-Morse sequence; thus,
$T(x)=x+x^2+x^4+x^7+\dotsb$. It is known that $T$ satisfies the nice
congruence
  $$ (1+x)^3 T^2(x) + (1+x)^2 T(x) + x \equiv 0 \pmod 2 $$
(the congruence is actually modulo the principal ideal generated by $2$ in
the ring of formal power series ${\mathbb Z}[[x]]$). Does $T$ satisfy any
identity of this or some other sort? Is there some product representation
for $T$? In brief, I am interested in the properties of $T$ in the ring
${\mathbb Z}[[x]]$ itself rather than in its quotient rings.
Thanks in advance for any pointers!

REPLY [14 votes]: Let 
$$F(x)=1-x-x^2+x^3-x^4+x^5+\ldots=(1-x)(1-x^2)(1-x^4)\ldots.$$
Then $F(x)=(1-x)F(x^2)$ and
$$F(x)=1+x+x^2+x^3+\ldots-2T(x)=\frac{1}{1-x}-2T(x).$$
So
$$T(x)-(1-x)T(x^2)=\frac{x}{1-x^2}.$$<|endoftext|>
TITLE: Getting out a system of linear ODEs by knowing the Magnus expansion
QUESTION [5 upvotes]: Assume we are given for a transition between two time points $t_0 = 0$ and $t_1$ a matrix relationship, eventually describing the solution of a system of linear with non-constant coefficients,
$$Y(t_1) = \exp(\Omega(t_1,0))Y(0),$$
or, in a more general setting,
$$Y(t_1) = \exp(\Omega(t_1,0))Y(0)+c(t_1,0),$$
where the $\Omega(t_1,0)$ is the Magnus expansion just as in https://en.wikipedia.org/wiki/Magnus_expansion and http://arxiv.org/pdf/0810.5488.pdf .
If I assume that the system of linear ODEs is given by,
$$\frac{dY}{dt} = A(t)Y(t)$$
or, more generally,
$$\frac{dY}{dt} = A(t)Y(t)+b(t),$$
is there any way to get out some information about $A(t)$ or $\int_0^{t_1} A(t) dt$ or $b(t)$ by just knowing $\Omega(t_1,0))$ (and $c(t_1,0)$).

REPLY [4 votes]: You can directly get the trace of $A$ from the identity
$${\rm det}\,\left[\exp\bigl(\Omega(t,0)\bigr)\right]=\exp\left[\int_0^t {\rm tr}\,A(s)ds\right]$$
The full matrix $A$ is determined by $\Omega$ via the inverse Magnus expansion [see equation 4.2 from this thesis]:
$$A=\sum_{k=0}^{\infty}\frac{(-1)^k}{(k+1)!}\left[\sum_{q=0}^{k}(-1)^q{k\choose q}\Omega^{q}\frac{d\Omega}{dt}\Omega^{k-q}\right]$$
This is useful if $\Omega$ is proportional to some small parameter $\epsilon$, so that the expansion gives you $A$ as a power series in $\epsilon$.<|endoftext|>
TITLE: In what sense is $\Omega_p$ universal?
QUESTION [6 upvotes]: In Chapter 3 of the book ''A Course in $p$-adic Analysis'' A.M Robert defines the field $\Omega_p$. He calls the field ''universal'' but doesn't show a universal property. I would have guessed that it's the universal spherical complete and algebraically complete extension of ${\mathbb Q}_p$, but the book doesn't claim anything of the sort, at least I didn't find it. The construction of the field depends on the choice of an ultrafilter and there's not even a remark on the (in)dependence of that choice.
What I mean by universal property is this: I expect something of the following form: Let $Z$ be a field extension of ${\mathbb Q}_p$ with an absolute value extending the one of ${\mathbb Q}_p$ such that $Z$ is algebraically closed and spherically complete. Then there exists an isometric field homomorphism $\Omega_p\to Z$.
Now my question is, whether $\Omega_p$ is indeed universal in the described sense and where I would find a proof of that fact.

REPLY [2 votes]: You need the concept of maximal completeness.
In a word, $\Omega_p$ is the minimal spherically complete field containing $\mathbb{C}_p$ and is the maximal mixed-characteristic valued field with redisue field $\bar{\mathbb{F}}_p$ and valuation group $\mathbb{Q}$.
For details, see this survey.<|endoftext|>
TITLE: Normal approximation of tail probability in binomial distribution
QUESTION [16 upvotes]: My problem: From the Berry--Esseen theorem I know, that $$\sup_{x\in\mathbb R}|P(B_n \le x)-\Phi(x)|=O\left(\frac 1{\sqrt n}\right),$$ where $B_n$ has the standardized binomial distribution and $\Phi$ is the standard normal distribution function. I can prove this for $x\approx 0$ with Stirling's formula and in a way similar to this proof.
Unfortunately Stirling's approximation becomes worse the bigger $|x|$ is, so that I can only show that
$$\sup_{|x| \le c}|P(B_n \le x)-\Phi(x)|=O\left(\frac 1{\sqrt n}\right)$$ for a fixed $c > 0$.
My question: Is there a direct proof for $\sup_{|x| > c}|P(B_n \le x)-\Phi(x)|=O\left(\frac 1{\sqrt n}\right)$, based on Stirling's formula? How can I estimate the error of the difference in the tail probability of the binomial distribution to the normal distribution?
Note: Chebyshev's inequality does not provide the right convergence speed. As zhoraster shows in his answer to Finding an error estimation for the De Moivre–Laplace theorem, also Chernoff's inequality does not provide the right convergence speed, too. That's why I am looking for another method. I also want to use more easy and direct approximations than Stein's method which is used in the proof of the Berry--Esseen theorem.
I already asked this question on MSE (see this thread), but I haven't got an answer there (even after adding a bounty). That's why I want to ask this question here.

REPLY [7 votes]: The needed observation here is that, whereas the relative error of the Stirling approximation is worse for large $|x|$, it gets multiplied by a fast decreasing normal density function, and then the product gets integrated to produce the desired result. Here are the details. 
Let $X$ be a binomial random variable with parameters $n\in\mathbb{N}$ and $p\in(0,1)$, where $n\to\infty$ and $p$ is fixed. For any integer $k$, let 
$$z_k:=\frac{k-np}{\sqrt{npq}},\quad q:=1-p. 
$$
Let 
$$K:=\{k=0,\dots,n\colon|z_k|\le n^{1/4}\},$$
so that $k/n\to p$ uniformly over $k\in K$. 
Let 
$$\hat p:=k/n,\quad\hat q:=1-\hat p=(n-k)/n,$$
$$\epsilon:=\epsilon_k:=z_k/\sqrt n,\quad\lambda:=\sqrt{q/p}, 
$$
whence 
$$\frac{\hat p}{p}=1+\epsilon\lambda,\quad\frac{\hat q}{q}=1-\epsilon/\lambda.$$
Uniformly over $k\in K$ one has $k\sim np$ and $n-k\sim nq$, so that $\frac1k=O(\frac1n)$ and $\frac1{n-k}=O(\frac1n)$. 
Here and in what follows, the constant in $O(\cdot)$ depends only on $p$. So, by Stirling's formula, over all $k\in K$ 
$$P(X=k)=\frac{n!}{k!(n-k)!} p^kq^{n-k}
=\frac1{\sqrt{2\pi n\hat{p}\hat{q}}}\,\frac1R\,\Big(1+O\Big(\frac1n\Big)\Big)$$
$$
=\frac1{\sqrt{2\pi npq}}\,\frac1R\,\Big(1+O\Big(|\epsilon|+\frac1n\Big)\Big)
=\frac\delta{\sqrt{2\pi}}\,e^{-nH}\Big(1+O\Big(|\epsilon|+\frac1n\Big)\Big)
$$
where
$$R:=\Big(\frac{\hat p}{p}\Big)^k \Big(\frac{\hat q}{q}\Big)^{n-k},  
$$
$$\delta:=\frac1{\sqrt{npq}},\quad H:=H_k:=\ln(R^{1/n})=\hat p\ln\frac{\hat p}p+\hat q\ln\frac{\hat q}q
:=p\psi(\epsilon\lambda)+q\psi(-\epsilon/\lambda),$$
$$\psi(s):=(1+s)\ln(1+s).
$$ 
Since $\psi(s)=s + s^2/2 +O(|s|^3)$ for $|s|\le1/2$ and 
$\frac{|z_k|^3}{\sqrt n}+|\epsilon|=\frac{|z_k|^3}{\sqrt n}+\frac{|z_k|}{\sqrt n}=O\big(\frac{|z_k|^3+1}{\sqrt n}\big)$, we find that
$$(1)\qquad P(X=k)=\delta\,\phi(z_k)\Big(1+O\Big(\frac{|z_k|^3}{\sqrt n}\Big)\Big)\Big(1+O\Big(|\epsilon|+\frac1n\Big)\Big)$$
$$
=\delta\,\phi(z_k)\Big(1+O\Big(\frac{|z_k|^3+1}{\sqrt n}\Big)\Big)  
$$
over all $k\in K$, 
where $\phi$ is the standard normal density. 
Next, 
for $u=O(n^{1/4})$ and $|h|\le\delta/2$, 
$$\frac{\phi(u+h)+\phi(u-h)}{2}=\phi(u)\cosh(hu)e^{-h^2/2}
=\phi(u)(1+O(h^2u^2+h^2))=\phi(u)\Big(1+O\Big(\frac1{\sqrt n}\Big)\Big). 
$$
So, for $k\in K$
$$\delta\,\phi(z_k)=\frac12\,\int_{-\delta/2}^{\delta/2}[\phi(z_k+h)+\phi(z_k-h)]\,dh\,\Big(1+O\Big(\frac1{\sqrt n}\Big)\Big)$$
$$
=\int_{z_k-\delta/2}^{z_k+\delta/2}\phi(z)\,dz\,\Big(1+O\Big(\frac1{\sqrt n}\Big)\Big). 
$$
Also, 
for any $k\in K$ and any $z\in[z_k-\delta/2,z_k+\delta/2]$ one has 
$|z_k|^3-|z|^3\le3\delta(|z_k|+\delta/2)^2=O(1)$, whence $|z_k|^3/\sqrt n=O((|z|^3+1)/\sqrt n)$. 
So, by $(1)$, for some positive real $c$ depending only on $p$ and for all $k\in K$ 
$${J_k\!}^-\le P(X=k)\le J_k^+,$$
where
$$J_k^\pm:=\int_{z_k-\delta/2}^{z_k+\delta/2}\phi(z)
\Big(1\pm\frac{c(|z|^3+1)}{\sqrt n}\Big)\,dz.  
$$
Hence, for any $x\in[0,n^{1/4}]$, letting $K_x:=\{k\in\mathbb{Z}\colon0<z_k\le x\}$, one has 
$$P(0<B_n\le x)=\sum_{k\in K_x}P(X=k)
\le\sum_{k\in K_x}J_k^+$$
$$
\le\int_{-\delta/2}^{x+\delta/2}\phi(z)\Big(1+\frac{c(|z|^3+1)}{\sqrt n}\Big)\,dz$$
$$
\le\int_0^x\phi(z)\,dz+A\delta+\frac B{\sqrt n}
\le\int_0^x\phi(z)\,dz+\frac C{\sqrt n}, 
$$ 
where $A,B,C$ depend only on $p$. Quite similarly one obtains the corresponding lower bound on $P(0<B_n\le x)$, so that 
$$(2)\qquad |P(0<B_n\le x)-(\Phi(x)-1/2)|\le\frac C{\sqrt n}  
$$
for $x\in[0,n^{1/4}]$, where $\Phi$ is the standard normal cumulative distribution function. 
On the other hand, for real $x>n^{1/4}$ one has $1-\Phi(x)=O(1/\sqrt n)$ and, 
by Bernstein's inequality, 
$$P(B_n>x)\le\exp\Big(-\frac{x^2}{2+x/\sqrt{npq}}\Big)
\le e^{-x^2/4}+e^{-x\sqrt{npq}/2}=O(1/\sqrt n). 
$$
It follows that $(2)$ holds for all real $x\ge0$ (perhaps with a greater constant $C$). Similarly, $|P(B_n\le0)-1/2)|\le\frac C{\sqrt n}$. 
Thus, for all real $x$
$$|P(B_n\le x)-\Phi(x)|\le\frac{2C}{\sqrt n},   
$$
as desired.<|endoftext|>
TITLE: Are dagger categories truly evil?
QUESTION [60 upvotes]: Recall that a dagger category is a category equipped with an involution $*:Hom(x,y)\to Hom(y,x)$ that satisfies $f^{**}=f$ and $f^* g^*=(gf)^*$. A prominent example of a dagger category is the category of Hilbert spaces and continuous linear maps.

Now, dagger categories are evil!
For example, here's a quote from the Nlab:

"Note that regarded as an extra structure on categories, a †-structure is evil, since it imposes equations on objects."

This is not just a little thing. It's a fundamental philosophical problem with category theory: category theory does not seem to be able to cope with this rather important category-theory-related notion which are dagger categories.
Also, because of this buit-in evilness, there's a whole bunch of very familiar concepts that one cannot apply to dagger categories without rethinking everything very, very carefully from the beginning: equivalences, limits, colimits, adjoint functors, duals, algebra objects, etc...
Here's another quote:

"It is possible that this problem will force a change in thinking in either the concept of the principle of equivalence or our thinking in quantum theory."

and yet another one:

"Often concepts violating the principle of equivalence (like the concept of “strict monoidal category”) have equivalence-invariant counterparts (like the concept of “monoidal category”). But in this particular case there appears to be no known way to express the idea without equations between objects."

Now here's my question:
Question: Can one improve the above "there appears to be no known way to express the idea without equations between objects" to "there is no way to express the idea without equations between objects"?
I expect dagger categories to be truly evil (in some yet-to-be-defined technical sense of "evil"). Having a proof of that fact might be very informative, and would at least force us to define the term "evil" at a mathematical level of precision.

REPLY [9 votes]: Here's a notion of undirected category, which is equivalent to the notion of dagger category. It's a bit cumbersome to work with, but the main point is to underscore Qiaochu Yuan and Manuel Bärenz's points that daggers are not really a categorical structure, and dagger categories are their own thing.
An undirected category consists of a pair of maps $s,t:C_1 \to C_0$, with constant $\mathrm{id} : C_0 \to C_1$ and instead of one composition operation for composing arrows tip to tail, there are four composition operations for composing in any direction (composition is in diagrammatic order):
${}_t;_s : Hom(A,B) \times Hom(B,C) \to Hom(A,C)$
${}_s;_s : Hom(B,A) \times Hom(B,C) \to Hom(A,C)$
${}_t;_t : Hom(A,B) \times Hom(C,B) \to Hom(A,C)$
${}_s;_t : Hom(B,A) \times Hom(C,B) \to Hom(A,C)$
Satisfying the following equations ($a, b$ and $c$ are variables in $s$ and $t$, and $-s = t$, $-t = s$)
$f = f {}_t;_a \mathrm{id} = \mathrm{id} {}_a;_s f$ (usual identity)
$f {}_s;_a \mathrm{id} = \mathrm{id} {}_a;_t f =: f^\dagger$
$(f {}_a;_b g) {}_s;_c h = (g {}_b;_a f) {}_t;_c h = g {}_b;_s  (f {}_{-a};_c h) = g {}_b;_t  (h {}_{c};_{-a} f)$ (the middle one includes usual associativity).
It's not hard to show that given a dagger category, you can get an undirected category by defining the compositions as $f {}_s;_t g := f^\dagger ; g^\dagger$ and the like, with the same identity, and given an undirected category, define a dagger as in the second line of equations above, and to show that these are inverses of each other.
Then there is also the fact that algebras of the path monad on the category of undirected graphs is exactly an undirected/dagger category, just as an algebra of the path monad on the category of directed graphs is a category.<|endoftext|>
TITLE: What can we say about tropical maps $\mathbb{P}^1 \to A$ for an Abelian variety $A$?
QUESTION [10 upvotes]: It's well known that maps $\mathbb{P}^1_\mathbb{C} \to A$ are constant for any Abelian variety $A$ (in fact, for any complex torus). 
Is there any similar statement in the tropical case? Naively, the answer seems to be no. We can see this in the following way: a tropical $\mathbb{P}^1$, $\mathbb{TP}^1$ is just a metric tree, and (for example) a circle (with a view vertices for good measure) produces an example of a one-dimensional tropical abelian variety. We can include $\mathbb{TP}^1 \hookrightarrow C$ as one of the edges quite easily which is a tropical map (unless I'm making a silly mistake).
So naively, the answer seems to be that there are non-constant tropical maps. Is there some way to eliminate these in a meaningful way? That is, it would be nice to have a statement of the form
Theorem: All (possibly restricted) tropical maps $\mathbb{TP}^1 \to A$ are constant (up to some appropriate notion of equivalence).
Edit: As pointed out in the comments, there is a balancing condition to be considered. For the case of maps to a tropical elliptic curve, this is given as the following.
For each point in $C$, the target, and each pre-image $p$ of $C$, the sum of all weights of the edges incident to $p$ must sum to zero (paying attention to outward orientation of edges).
One sees immediately that this implies that no $\mathbb{TP}^1$s can map to $C$, since the image of a leaf of a tree can never satisfy the balancing condition (unless the weight of every edge is zero).
However, this doesn't address maps to higher dimensional abelian varieties, but one can hope that a similar balancing condition would be true.

REPLY [4 votes]: The answer should be that there are no non-constant maps from tropical curves of genus zero to tropical abelian varieties. By "should" I mean that there are a lot of definitions of these things in the literature and it would be frustrating to write an answer which is correct for every definition, but any definition which doesn't have this consequence is a "bad" definition.
Here is why. A tropical curve of genus zero is a metric tree $\Gamma$. In particular, it is simply connected. So any continuous map $\phi: \Gamma \to \mathbb{R}^g/\Lambda$ lifts to a continuous map $\Gamma \to \mathbb{R}^g$, which can't obey the balancing condition at a leaf.
One caveat: I could imagine someone making a definition (though I haven't see it) which would allow $\mathbb{R}$, with its standard metric, as a tropical genus zero curve with two punctures; $\mathbb{R}/\mathbb{Z}$ with the quotient metric as a tropical genus $1$ curve, and the covering map as a tropical map. This is precisely analogous to the analytic map $\mathbb{C}^{\ast} \to \mathbb{C}^{\ast}/q^{\mathbb{Z}} \cong E$ for some $q \in \mathbb{C}$ with $0 < |q|<1$.<|endoftext|>
TITLE: Distribution of the number of prime factors
QUESTION [8 upvotes]: Count the number of prime factors of a number $n$
to include multiplicity,
so that
$$n=24=2^3 \cdot 3 = 2 \cdot 2 \cdot 2 \cdot 3$$
has $4$ prime factors, and
$$n =
6500 =
2^2 \cdot 5^3 \cdot 13 =
2 \cdot 2 \cdot 5 \cdot 5 \cdot 5 \cdot 13
$$
has $6$ prime factors.
The distribution is quite regular.
Here it is for $n \le n_\max$, $n_\max=10^7$:

         


         

About a quarter of $n \le 10^7$ have $3$ prime factors.



Q. What is this distribution explicitly?
  Where is its peak, for $n \le n_\max$?

REPLY [14 votes]: this variable $\Omega(n)$, the number of prime factors of $n$ counting multiplicity, has for large $n$ a normal distribution with mean [*] $1+\log(\log n)$ and standard deviation $[\log(\log n)]^{1/2}$; see, for example, Prime Numbers and Computer Methods for Factorization, page 167 [first edition], page 159 [second edition].
[*] more precisely, this additive constant 1 should be replaced by $1.03465\ldots$ as calculated by Knuth and Trabb-Pardo (appendix A); incidentally, if we don't count multiplicities the normal distribution has mean $0.26+\log(\log n)$ with the same standard deviation $[\log(\log n)]^{1/2}$, so the only difference is a slight displacement of the whole curve.

--- update 2020, in response to query:   
the "0.26" number is defined as
$$c_1= \gamma+\sum_{p\;\text{prime}}\biggl(\log(1-1/p)+\frac{1}{p}\biggr)= 0.261497212847643$$
while the "1.03" number is defined as
$$c_2=\gamma+\sum_{p\;\text{prime}}\biggl(\log(1-1/p)+\frac{1}{p-1}\biggr)= 1.034653881897438$$
The number $c_1$ is known as the Meissel-Mertens constant.
Both $c_1$ and $c_2$ are referred to as Hadamard-de la Vallée-Poussin constants (see also this MathWorld entry).<|endoftext|>
TITLE: Differential structures and K-homology groups
QUESTION [6 upvotes]: What is an example of a (compact) manifold, which has two non-equivalent differential structures such that the K-homology groups are non-isomorphic? If no such example exists, i.e. "K-homology does not see the differential structure", then can someone give a heuristic explanation of why an object (originally at least) defined in terms of pseudo-differential operators does not depend on a choice of differential structure.

REPLY [3 votes]: This is more or less an addendum to AlexE's answer: 
Even though the $K$-homology groups themselves do not depend on the smooth structure there are classes in $K$-theory, which can tell apart some of the smooth structures: 
Let $\mathcal{S}(M)$ denote the group of isotopy classes of smooth structures on $M$ and $M_{\alpha}$ denote $M$ equipped with the smooth structure $\alpha$. The identity map induces a map on the corresponding sphere bundles $\varphi \colon SM_{\alpha} \to SM$. There is a unit $u \in K^0(SM)$, such that $\varphi_*([SM_{\alpha}]) = u \cap [SM]$, where $[SM] \in K_1(SM)$ denotes the fundamental class of the spin$^c$-manifold $SM$ in $K$-homology. The class $u$ descends to a unit $\theta(\alpha) \in K^0(M)$, such that $\pi^*(\theta(\alpha)) = u$, where $\pi \colon SM \to M$ is the bundle projection. It was proven by Jerome Kaminker that $\theta \colon \mathcal{S}(M) \to K^0(M)$ is a homomorphism and that there are smooth manifolds $M$ for which $\theta$ is nontrivial (see "Pseudo-differential Operators and differential structures"). 
This implies for example that the algebra of order $0$ pseudodifferential operators $\mathcal{P}_M$ depends on the smooth structure. To see this, note that it fits into an exact sequence
$$
0 \to \mathcal{K} \to \mathcal{P}_M \to C(SM) \to 0
$$
(where $\mathcal{K}$ are the compact operators) giving an element in $K_1(SM)$ by BDF-theory. This class agrees with the fundamental class $[SM] \in K_1(SM)$.<|endoftext|>
TITLE: Essays and thoughts on mathematics
QUESTION [74 upvotes]: Many distinguished mathematicians, at some point of their career,
  collected their thoughts on mathematics (its aesthetic, purposes,
  methods, etc.) and on the work of a mathematician in written form.

For instance:

W. Thurston wrote the lovely essay On proof and progress in mathematics in response to an article by Jaffe and Quinn; some points made there are also presented in an answer given on MathOverflow (What's a mathematician to do?). 
More recently, T. Tao shared some personal thoughts and opinions on what makes "good quality mathematics" in What is good mathematics?.
G. Hardy wrote the famous little book A Mathematician's Apology, which influenced, at least to some extent, several generations of mathematicians.

Personally, I've been greatly inspired by the two writings listed under (1.) --  they are one of the main reasons why I started studying mathematics -- and, considering that one of them appeared on MathOverflow, I'd like to propose here -- if it is appropriate -- to create a "big-list" of the kind of works described in the above blockquote. 

I'd suggest (again, if it is appropriate) to give one title (or link) per answer with a short summary.


A related question, which I've found very interesting, is 
Good papers/books/essays about the thought process behind mathematical research.
Only slightly related (but surely interesting): Which mathematicians have influenced you the most?
A single paper everyone should read? is not quite related, but still somewhat relevant (especially the most up-voted answer).

REPLY [2 votes]: A really nice article by Andrei Toom about mathematical education, especially in the US, got recently mentioned in a comment to this question.<|endoftext|>
TITLE: What is the etale sheafification of the (unramified) Milnor-Witt $K$-theory
QUESTION [9 upvotes]: I would like a reference/argument for the truth/falsity of the following statement: 
The etale sheafification of the unramified Milnor-Witt K-theory (Nisnevich) sheaves are the (etale sheafification of?) unramified Milnor K-theory sheaves. 
Thanks!

REPLY [7 votes]: We have a short exact sequence of sheaves of abelian groups (see e.g. Morel's book on $\mathbb{A}^1$-algebraic topology): 
$$
0\to\mathbf{I}^{n+1}\to\mathbf{K}^{\rm MW}_n\to \mathbf{K}^{\rm M}_n\to 0,
$$
where $\mathbf{I}^{n+1}$ is the sheaf of $n+1$-th powers of the fundamental ideal in the Witt ring. Over a strictly henselian local ring (essentially smooth), this yields a short exact sequence of global sections (by exactness of the Gersten complex). So the question is equivalent to asking if the etale sheafification of $\mathbf{I}^{n+1}$ is trivial. (Alternatively, this can be deduced from the new presentation of Milnor-Witt K-theory of local rings by Gille-Scully-Zhong.)
So let $R$ be an essentially smooth strictly henselian local ring with $1/2\in R$. Its 2-cohomological dimension will be at most the Krull dimension of $R$. This implies that the restriction of $\mathbf{I}^{n+1}$ to the small Nisnevich site of $R$ vanishes whenever $n>\dim R$ (well, actually we only need the global sections). This follows as in Proposition 5.1 of

A. Asok and J. Fasel. A cohomological classification of vector bundles on smooth affine threefolds. Duke Math. J. 163 (2014), 2561-2601.

Now we want to show that $\mathbf{I}^j(R)=0$ for all $j$. By descending induction, it suffices to show that $I^n(R)/I^{n+1}(R)=0$. By Voevodsky's solution of the Milnor conjecture, we know that we have an identification of Nisnevich sheaves $\mathbf{I}^j/\mathbf{I}^{j+1}\cong \mathbf{K}^{\rm M}_j/2$. Now if $(R,\mathfrak{m})$ is a local henselian ring with residue characteristic $\neq 2$, then $K^{\rm M}_j(R)/2\cong K^{\rm M}_j(R/\mathfrak{m})/2$ by rigidity. Since $(R,\mathfrak{m})$ is in fact a strictly henselian local ring, the residue field $R/\mathfrak{m}$ is algebraically closed and therefore has trivial mod 2 Milnor K-theory. This shows that $I^j(R)=0$ for all $j$, proving that the natural projection $\mathbf{K}^{\rm MW}_n\to\mathbf{K}^{\rm M}_n$ becomes an isomorphism after etale sheafification. 
Some more remarks on rigidity: one possibility is to apply Hornbostel-Yagunov rigidity for orientable $\mathbb{A}^1$-representable theories which works for Milnor K-theory. Another possibility is to use Kerz's identification of Milnor K-theory of local rings with motivic cohomology. Yet another possibility is to use the other part of the Milnor conjecture, the identification of $\mathbf{I}^n/\mathbf{I}^{n+1}$ with the Nisnevich sheafification of ${\rm H}^n_{\rm et}(-,\mu_2^{\otimes n})$. This would also satisfy rigidity by results of Gabber. Essentially, the rigidity argument requires $\mathbb{A}^1$-invariance, existence of suitable transfers and mod $p$ coefficients (prime to the characteristic). These things are satisfied for either Milnor K-theory or etale cohomology.<|endoftext|>
TITLE: Groups without property (T) but all finite quotients are expanders
QUESTION [11 upvotes]: What is an example of a group $G$ which 
1- is finitely generated by $S$,
2- does not have property (T), 
3- admits infinitely many finite quotients which do not factor through an homomorphism $G \to H$ for some [fixed & infinite] property (T) group $H$
4- all the Cayley graphs (w.r.t. $S$) of those finite quotients are $\epsilon$-expanders (for some fixed $\epsilon >0$)
The question is motivated by the false assertion: "a group has property (T) if and only all its finite quotients are $\epsilon$-expanders".
There are "simple" answers if one does not put the factor condition in 3. For example, one could consider $A \times H$ where $A$ is a finitely generated simple amenable group  (as proven/constructed by Juschenko-Monod/Matui) and $H$ is some residually finite property (T) group ($SL_3(\mathbb{Z})$ being the canonical choice). Because $A$ is amenable, this does not have (T) and its finite quotients (which all come from $H$) are $\epsilon$ expanders. 
Actually, to enlarge the list of examples of this form, a side question would be: what are the finitely generated groups which do not have property (T) and do not have finite non-trivial quotients?

REPLY [7 votes]: A bit late, but let me draw your attention to my first paper:
http://arxiv.org/abs/1005.4566
It shows that property (T) is not determined by the finite quotients, and gives in particular an answer to your question.

Later edit -- Yves' comment calls for more details:
The aim of the above mentioned note is to show that property (T) is not profinite, that is, it is not determined by the set of the finite quotient of the group (a nice exercise is to show that considering the set of isomorphism classes of finite quotient or considering the set of finite quotients counting multiplicity amount to the same - so there is no ambiguity in the above definition). For finitely generated groups, having the same set finite quotients is equivalent to having the same profinite completion. 
Now I can explain the simple idea behind the construction. The Congruence subgroup Property (CSP) essentially transforms this problem into a local-to-global problem:
Essentially, under CSP, the profinite completion of $G(\mathbb Z)$ is 
$$\widehat{G(\mathbb Z)}=\prod_p G(\mathbb Z_p).$$
So in order to have two groups with the same profinite completion we look for two groups $G_1(\mathbb Z)$ and $G_2(\mathbb Z)$ (with CSP) such that they are isomorphic locally.
If, moreover, we can arrange that  $G_1(\mathbb Z)$ (resp. $G_2(\mathbb Z)$) is a lattice in a product of rank 1 (resp. high rank $\geq 2$) groups, we win, as this should imply that the first does not have $(T)$ while the second has $(T)$.
From this, one is led to the construction naturally. One uses the fact that the quadratic forms $$\sum^n x_i^2-\sum^m y_i^2,\quad \sum^{n-4} x_i^2-\sum^{m+4} y_i^2$$ agrees everywhere locally, but not globally. This leads to the Spin groups construction that appear in Yves' comments.
Another sentence of advertising: This construction is one of three types of constructions that I consider in a different paper, in order to show that the set of isomorphism classes of arithmetic groups in high rank which have the same profinite completion is finite.<|endoftext|>
TITLE: Simple argument regarding sums of two units in a number field?
QUESTION [33 upvotes]: I wonder if it is possible to show, without using the Schmidt subspace/Roth theorem/Baker's bounds on linear forms in logarithms or other very deep results, that, in a number field, not all integral elements are sums of two units/most of them are not.
The known results that state that there are few elements that are sums of two units in number fields (see e.g. Fuchs, Tichy, Ziegler, "On quantitative aspects of the unit sum number problem", Arch. Math. 93 (2009)) seem to use either Thue-Siegel-Roth theorem or Baker's bounds on linear forms in logarithms. Both I find very difficult to reach. Since I rely on a similar result in a statement I have made, the motivation behind the question is a wish to find a simpler argument (or to convince oneself that existence of (many) elements that are not sums of two units is a claim that is itself of strength that is seemingly out of reach to significantly easier means than Thue-Siegel-Roth/Baker theorems).

REPLY [16 votes]: $\newcommand\p{\mathfrak{p}}$
$\newcommand\OL{\mathcal{O}}$
$\newcommand\P{\mathfrak{P}}$
Here is a solution which is essentially an elaboration on Felipe's answer.
Instead of working with squares, consider working with $m$th powers instead. 
Lemma: If $(1 - v u^m)$ is exactly divisible by a prime $\p$ of $\OL_K$, then, assuming $K(v^{1/m}) \ne K$, the prime $\p$ is not inert in $L = K(v^{1/m})$.
Proof: In $\OL_L$, the ideal $\P = (\p,1 - v^{1/m} u)$ has norm $\p$.
By Cebotarev, the density of primes $\p$ which remain inert in $\OL_L$ is non-zero. Hence by the analytic
 arguments Felipe alluded to, the set of principal ideals of the form $(1 - v u^m)$ with $v$ ranging over a the (finite) set of non-zero representatives in $\OL^{\times}_K/\OL^{\times m}_K$ has density zero. So it remains to deal with ideals of the form $(1 - u^m)$, where we now have flexibility in choosing $m$.
Lemma: Let $\ell$ be any prime. Suppose that $m = |(\OL_K/\ell)^{\times}|$. Then the density of principal ideals of the form $(1 - u^m)$ is at most $1/\ell$.
Proof: Since $u^m \equiv 1 \mod \ell$, this is the same as saying that the density of principal ideals divisible by $\ell$ is at most $1/\ell$.
Taken together, it follows that the density of principal ideals of the form $(1 - u)$ has density at most $1/\ell$ for any prime $\ell$, and hence has density zero.<|endoftext|>
TITLE: Weil's paper under a pseudonym on deforming singular varieties
QUESTION [8 upvotes]: I am looking for a paper of Weil that is published under a pseudonym, in which he proves a statement along the lines of: a singular algebraic variety cannot be deformed into a nonsingular one.
Thanks in advance.

REPLY [13 votes]: this pseudonomous letter mentioned by Jim Humphreys is too amusing not to summarize here:
R. Lipschitz (Ann. of Math. 69, 1959, 247-251)
reprinted in A. Weil, Collected Papers, volume II (Springer, 1979):


and for the record, here is a translation of the 1957 anonymous note in Italian:

A correspondent, who wishes to remain anonymous, writes as follows:
--- A very well known conjecture by F. Severi (Rend. Pal. 28 (1909), p. 45) asserts that "every variety with multiplicities in some points
  can be considered as the limit of one without singularities, belonging
  to the same space."
According to a note distributed by Nancago from "United Press" agency,
  this hypothesis from that famous author would have been refuted by the
  respectable French geometer René Thom, based on the example of cone of
  3d order in the space $S_6$ and following quite delicate topological
  considerations.
It might not displease the readers of your esteemed journal to find
  here an elementary geometric treatment of Thom's example. In fact, we
  will determine all the varieties of 3d order in any space $S_n$. From
  this we will obtain, as an immediate consequence, the falseness of the
  hypothesis in question.
Let $V_r$ be a variety of 3d order in the space $S_n$, of dimension $r<n-1$, 
  not contained in a hyperplane. Let $C$ be the cone that projects $V_r$
  from an arbitrary simple point $M$ of $V_r$; let $M'$ be another point
  in $V_r$, simple in the cone $C$. The cone $C$ is of second order;
  hence its projection from $M'$ is a linear variety of dimension $r+1$,
  so that $V_r$ is contained in a linear variety of dimension $r+2$. It
  follows that $r=n-2$ and $C$ has dimension $n-1$. The same holds for
  cone $C'$, that projects $V_r$ from $M'$. The cones $C$ and $C'$ are
  distinct, hence $M'$ is simple in cone $C$; their intersection is
  therefore a reducible variety of 4th order, broken in $V_r$ and a
  linear variety. Let $A=0,B=0$ be the equation of this variety. Then
  the equations for the cones $C$, $C'$ can be written in the form
  $$(1)\qquad\qquad AP=BQ,\;\;AP'=BQ',$$ where we denote by $P,Q,P',Q'$ four linear
  forms homogeneous in the space coordinates. Let $s$ be the number of
  independent forms among the six forms $A,B,P,Q,P',Q'$; this number is
  at least 4, because, if it where 2, the cones $C,C'$ would not be
  reducible; if it where 3, then their intersection would be broken in
  four distinct or non-distinct varieties. The possible values for $s$
  are therefore 4,5 and 6.
Let $L$ be the linear variety of dimension $n-2$, defined by the
  equations $A=B=0$, $P=Q=P'=Q'=0$. Each linear variety that projects
  from $L$ a point in the intersection of $C$ and $C'$ is contained in
  it, as one can see immediately from equations (1). Projecting $V_r$
  from $L$ one therefore obtains a variety $W$ of 3d order
  and dimension $s-3$ in a space $S_{s-1}$; and $V_r$ is nothing else
  than the cone projecting $W$ from $L$. In the space $S_{s-1}$ one can
  take as homogeneous coordinates the $s$ independent forms out of the
  forms $A,B,P,Q,P',Q'$; therefore the equations (1) define a variety of
  4th order, broken in $W$ and a linear variety. Now we can distinguish
  three cases:
(a) $s=6$: $W$ is the variety $W_3$ of Segre embedded in $S_5$, the
  bi-rational image without exceptions of the product of a plane and a
  line; as it is known, this has no points with multiplicities.
(b) $s=5$: $W$ is the hyperplane section of $W_3$. It is easy to see
  that all the irreducible hyperplane sections of $W_3$ are projectively
  equivalent among each other; denoting by $W_2$ one of these, it is
  also a rational variety without points having multiplicities.
(c) $s=4$: $W$ is the well known cubic rational $W_1$ embedded in
  $S_3$.
In this way we have demonstrated that every variety of 3d order
  belongs to one of the following types:

a variety of dimension $r$, contained in a linear variety of dimension $r+1$;
the three rational varieties $W_3$, $W_2$, $W_1$ enumerated above;
the projective cone of one of these three varieties from a linear variety of arbitrary dimension.

Evidently, a variety of type two or three can not be the limit of a
  variety of type one, and hence a variety of type three, of dimension
  greater than 3, can not be the limit of a variety without
  multiplicities in some points.<|endoftext|>
TITLE: separating points in $\mathbb{R}^d$ by minimal number of planes
QUESTION [15 upvotes]: Given $n$ points of general position in $\mathbb{R}^d$ (say, $n>d$ and no $d+1$ lie in a hyperplane.) We want to draw $k$ hyperplanes not passing through those points so that they all are in different open regions of the complement of the drawn planes. What is the minimal value $k(n,d)$ for which it may be always done? Case $d=2$ is solved by D. Gerbner, G. Tóth, `Separating families of convex sets' (arxiv:1211.2982). The answer is $\lceil n/2\rceil$, the upper estimate holds for points in convex position, the lower is proved by rotating the line which halves the number of points. For $d$, I may only show lower estimate like $(n-1)/d$ by considering points on the moment curve $(t,t^2,\dots,t^d)$ (and, of course, $k(n,d)\leq k(n,d-1)$ by projection argument).

REPLY [6 votes]: $\lceil n/d\rceil+d-2$ hyperplanes always suffice. To achieve this, we first choose $H_1,\dots,H_{d-1}$ in a manner that $H_1$ separates $\lceil n/d\rceil$ points from the rest, $H_2$ separates $\lceil n/d\rceil$ of those rest ones, and so on. Thus we get $d$ sets $S_1,\dots,S_d$ separated from each other, each consisting of at most $k=\lceil n/d\rceil$ points. 
We prove by induction on $k$ that such $d$ sets can be separated completely by $k-1$ hyperplanes. To perform the induction step, we implement the ham-sandwich theorem to bisect att $S_1,\dots,S_d$. We may shift this bisecting hyperplane a bit, so that on one side we get a collection of sets $S_1^1,\dots,S_d^1$ with at most  $\lceil k/2\rceil$ points in each, and on the other side we have a collection of $S_1^2,\dots,S_d^2$ with at most $\lfloor k/2\rfloor$ points in each. Now apply the induction hypothesis to each of these collections separately. 
Perhaps, the `$+d$' term can be improved by choosing the initial hyperplanes in a smarter way?<|endoftext|>
TITLE: Can we recover a topological space from the collection of Borel probability measures living on it?
QUESTION [8 upvotes]: Let $(X, \tau)$ be a topological space, and $\mathcal{P}(X, \tau)$ be the Borel probability measures living on $X$. Can we recover $(X, \tau)$ from $\mathcal{P}(X, \tau)$?

REPLY [9 votes]: I write this as an answer due mainly to space constraints.    I want to   elaborate a bit on Nate Eldredge's useful comments. 
Think of   Gelfand's classical  theorem stating that a compact space $X$ is determined by the  space $C(X)$ of continuous complex valued functions   on $X$.  This  statement is more   precise than this.  
The space  $C(X)$  is   given an algebraic-topologic structure  (commutative $C^*$-algebra). This  structure alone completely determines the space $X$ as the spectrum of this algebra equipped with an appropriate topology. Continuous maps between two compact spaces $X_1$ and $X_2$ induce continuous morphisms between the corresponding $C^*$-algebras,  and the isomorphisms of $C^*$-algebras induces homeomorphisms between the spectra. $\newcommand{\bR}{\mathbb{R}}$
It may help to assume first that your topological space $(X,\tau)$ is compact, for then the space of   finite Borel measures can be identified with the space of   bounded continuous linear maps $C(X)\to\bR$. We denote $\newcommand{\eP}{\mathscr{P}}$ by $\eP(X)$ the space of  Borel probability measures on $X$. Then $\eP(X)$ is a closed convex subset of  the dual $C(X)^*$.   The map  
$$ X\ni x\mapsto \delta_x \in \eP(X) $$
maps $X$ bijectively  onto the set of extremal points of $\eP(X)$.  (Above $\delta_x$ denotes the Dirac $\delta$-measure supported at $x$.)  This map is continuous  (w.r.t. to the weak topology on $C(X)^*$) and thus establishes a homeomorphism between $X$ and the set of extremal points of $\eP(X)$. In this  sense $\eP(X)$ determines $X$ but this answer is  far less satisfactory   than Gelfand's theorem mentioned above.<|endoftext|>
TITLE: Does $WKL_0$ provide more comprehension than $RCA_0$?
QUESTION [6 upvotes]: $WKL_0$ extends $RCA_0$ with the statement that any infinite subset of the infinite binary tree has an infinite branch. Does $WKL_0$ Prove that there are sets which are not proven to exist by the $\Delta_0$-comprehension schema of $RCA_0$? If so, to what level of comprehension do such sets belong?

REPLY [6 votes]: If I understand your question correctly, the answer is 
"no" in the following sense:
Fix a model $M=(\omega_M, \mathbb{R}_M)$ of $RCA_0$, and let $\mathcal{C}_M$ be the set of structures $N=(\omega_N, \mathbb{R}_N)$ such that

$N\models WKL_0$,
$\omega_N=\omega_M$ (same first-order parts), and
$\mathbb{R}_N\supseteq\mathbb{R}_M$ ($N$ is gotten by adding sets to $M$).

Then we have:

$\mathcal{C}_M$ is nonempty. (This is due to Harrington, and is how he shows that $WKL_0$ is $\Pi^1_1$-conservative over $RCA_0$.)
$\bigcap_{N\in\mathcal{C}_M} \mathbb{R}_N=\mathbb{R}_M$. That is, there is no specific set which must be added when we expand $M$ to a model of $WKL_0$. (This is a direct consequence of the proof of Jockusch and Soare's Low Basis Theorem, but I don't know who first explicitly observed this.)

In particular, the intersection of all the Scott sets is precisely $REC$, the set of recursive sets.
Does this answer your question?<|endoftext|>
TITLE: Existense of semi-stable vector bundles on smooth curves in positive characteristic
QUESTION [5 upvotes]: Let $k$ be an algebraically closed field of positive characteristic and $X$ be a smooth projective curve over $k$ of genus $g \ge 2$. Fix a polarization $L$ on $X$. Does there exist a semi-stable vector bundle on $X$ of rank $r$ and degree $d$ with gcd$(r,d)=1$? 
I know that this result is true in characteristic zero. I have heard this to be true in positive characteristic but have not been able to find a good reference for this fact or a counterexample.
Any reference/hint would be very helpful.
EDIT Assume further that $k$ is countable.

REPLY [5 votes]: It seems relatively easy to construct such a bundle by induction. Namely:
Step 0 If $r=1$, there are clearly line bundles of given degree $d$ on $X$, since $k$ is assumed to be algebraically closed.
Step 1 Choose $(r',d')$ such that $0<r'<r$, $d'/r'>d/r$, and there are no integral points within (or on the edges) of the triangle with vertices $(0,0)$, $(r',d')$, and $(r,d)$. For instance, can take $(r',d')$ satisfying the first two inequalities whose distance from the line through $(0,0)$ and $(r,d)$ is minimal. 
Step 2 The condition that there are no integral points on the boundary of the triangle implies that $gcd(r',d')=gcd(r-r',d-d')=1$. Let $E_1$ and $E_2$ be stable bundles such that $rk(E_1)=r'$, $deg(E_1)=d'$, $rk(E_2)=r-r'$, $deg(E_2)=d-d'$, which exist by the induction hypothesis. Choose a non-trivial extension
$$0\to E_2\to E\to E_1\to 0,$$
which exists by the Riemann-Roch Theorem (here we need that the genus of $X$ is at least one).
Step 3 It is easy to see that $E$ is stable. Indeed, assume it is not, and $F\subset E$ is destabilizing. Then the slope of $F$ is at most $d'/r'$, and dually the slope of $E/F$ is at least $(d-d')/(r-r')$. From the condition that there are no integral points in the triangle, we see that $rk(F)=r'$ and $deg(F)=d'$. Because of stability of $E_1$ and $E_2$, this implies that the map $F\to E\to E_1$ is an isomorphism, which would split the exact sequence for $E$. Contradiction.
Remark The only place where we need the field to be algebraically closed is Step 0. So as long as you know that line bundles of every degree exist on $X$, the statement holds. For instance, it works if $X$ has a $k$-point, or if the base field is finite (see this question), but fails if the field is $\mathbb{R}$.<|endoftext|>
TITLE: Measure of chords from a cantor set
QUESTION [7 upvotes]: The following problem is inspired by a problem in Pugh's Mathematical Analysis book. (Chapter 2 Problem 42).
In the problem he asks one to consider the standard Cantor set on the unit interval, and then to map it to the unit circle using the standard mapping $z \mapsto e^{2 \pi i z}$. Then given this set $D \subset S^1$ one can consider the subset of chords between all points in $D$. Namely $Chords(D) = \{t x + (1-t)y ~|~ x, y \in D, t \in [0,1]\}$. He asks if this set is compact and/or convex, to which the answers are yes and no respectively. 
A further question can be asked: Does this set have measure zero? I believe naive bounds would show that if one constructs a cantor set by removing over $1/2$ of the interval at each iteration we indeed have the chords have measure 0. I am unsure of the result in the standard case of removing middle thirds.

REPLY [6 votes]: I agree with questioner that the area measure is positive for a nice cantor set if and only if the hausdorf dimension of the cantor set is at least 1/2.There are nice examples of such in the theory of convex cocompact Fuchsian groups. The set in question is diffeomorphic to the set of full  non euclidean geodesics contained in the convex hull of the limit set of the group and has been studied a lot.
(Dennis Sullivan)<|endoftext|>
TITLE: first Chern class of complex vector bundles and first Pontrjagin class of quaternionic vector bundles
QUESTION [11 upvotes]: Let $\xi$ be a (real) vector bundle of dimension $n$. Then the first Stiefel-Whitney class 
$$
w_1(\xi)=0
$$
if and only if $\xi$ is orientable, i.e. the structure group of $\xi$ can be reduced to $SO(n)$.
Question 1: Let $\xi^\mathbb{C}$ be a complex vector bundle of dimension $n$. Then the first Chern class 
$$
c_1(\xi^\mathbb{C})=0
$$
if and only if  the structure group of $\xi^\mathbb{C}$ can be reduced to $SU(n)$? Is it true or false? Any references?
Question 2: Let $\xi^\mathbb{H}$ be a quaternion vector bundle of dimension $n$. Then the first Pontrjagin class 
$$
p_1(\xi^\mathbb{C})=0
$$
if and only if  the structure group of $\xi^\mathbb{H}$ can be reduced to $SSp(\mathbb{H},n):=\{A\in Sp(n)\mid \text{Det}(A)=1\}$? Is it true or false? Any references?
I have a further question characteristic classes of a covering space with symmetric group action

REPLY [4 votes]: I think haohaizi was looking for the following standard facts:
$BSO(n)$ is the fiber of $Bdet : BO(n) \longrightarrow BO(1)$
and $Bdet = w_1$.  This uses $BO(1) = K(Z/2,1)$.
$BSU(n)$ is the fiber of $Bdet : BU(n) \longrightarrow BU(1)$
and $Bdet = c_1$.  This uses $BU(1) = K(Z,2)$.
There is no determinant function for quaternions;  they are not commutative.
As Sullivan suggests, $Sp(n)$ is already `special', i.e., oriented.  Further,
$Sp(1) = S^3$ is not a $K(Z,3)$, so the obstruction to orientability here is not a cohomology class.<|endoftext|>
TITLE: Why does strong convergence of the EMSS imply that Tot commutes with suspension spectrum?
QUESTION [10 upvotes]: Given a fiber square of simplicial sets
$$\begin{array}{cc}
             & \hspace{-7mm} E  \\
&\hspace{-7mm}\downarrow \\
           \ast\longrightarrow &\hspace{-7mm} B 
\end{array}$$
and a homology theory $h(-)$, there is an associated Eilenberg-Moore spectral sequence converging to the homology of the homotopy fiber $F$. This is just the spectral sequence for a cosimplicial space, specifically a cobar construction $C^\bullet(E,B,\ast)$. It is claimed in this paper of Hopkins and Ravenel, in the middle of page 7, that if this Eilenberg-Moore spectral sequence converges strongly then $\Sigma_+^\infty Tot(C^\bullet(E,B,\ast))\simeq Tot(C^\bullet(\Sigma^\infty_+E,\Sigma^\infty_+ B,\mathbb{S}))$.
Why is this true? I know from an old paper of Bousfield that the EMSS for stable homotopy converges strongly in certain nice cases (e.g. if $B$ is simply connected), but does that give me an actual equivalence of objects? That only seems to tell me nice things about the algebra. For instance that there exists a finite filtration of $\pi_n^S(F)$ whose filtration quotients are the $E_{i,j}^\infty$ for $i+j=n$. That seems several steps away from making a general statement about the suspension spectrum functor commuting with totalization.
More generally, are there other functors $Top\to Spectra$ which commute in this fashion? For instance, it seems like the Thom spectrum functor (a sort of twisted suspension spectrum functor) for suitable fibrations should also commute with totalization is suspension spectrum does.  
EDIT----------------------------
I've also discovered this mysterious correspondence between Tom Goodwillie and Mike Hopkins (this is how people talked about this stuff before MO I guess!) that shows that suspension commutes with Tot in certain cases.
http://www.lehigh.edu/~dmd1/tg7192

REPLY [10 votes]: I'm going to avoid the question and answer the edit.
Hopkins has some results on this in his Oxford thesis which were announced without proof in his Asterisque paper of 1984. 
For a quite thorough analysis of a different, but related, spectral sequence, there is also the nice paper of Goerss (1993), Barratt's Desuspension Spectral Sequence and the Lie Ring Analyzer. See § 1. Barratt's spectral sequence, with a generalization by Hopkins. 
See Theorem 1.1.
There is, however, a (potential?) error in Hopkins' work involving connectivity assumptions, which is addressed in Comultiplication and suspension J. Klein, R. Schwaenzl, R.M. Vogt.(1997)
Their analogous result is Theorem 1.4 of that paper.
I don't see how your claim follows from the above results, however.<|endoftext|>
TITLE: Confusion about Subcategories of Category $\mathcal{O}$
QUESTION [8 upvotes]: So, in learning about category $\mathcal{O}$ representations of a semisimple Lie algebra $\mathfrak{g}$, I've come across two natural kinds of subcategories, and I think I'm confused about their structure and the relationship between them. (In particular, if they are equivalent or not).
Fix $\lambda \in \mathfrak{h}^{*}$ and let $\chi_{\lambda}:Z(\mathfrak{g}) \rightarrow \mathbb{C}$ denote the corresponding central character ($Z(\mathfrak{g})$ denoting the center of $U\mathfrak{g}$)
Type 1: Modules $M$ such that $(z - \chi_{\lambda}(z))\cdot m = 0$ for all $m \in M$, $z \in Z(\mathfrak{g})$. Let's call this $\text{Mod}_{\mathcal{O}}(\mathfrak{g},\lambda)$
Type 2: Modules $M$ such that $(z - \chi_{\lambda}(z))^{n} \cdot m = 0$ for all $m \in M$, $z \in Z(\mathfrak{g})$, where $n$ is an integer possibly depending on $z$ and $m$. Let's call this $\mathcal{O}_{\lambda}$.
Now one clear reason to prefer the $\mathcal{O}_{\lambda}$ is that category $\mathcal{O}$ is a direct sum of these subcategories. Additionally, if $L \in \mathcal{O}_{\lambda}$ is simple, it's projective cover will also be in $\mathcal{O}_{\lambda}$, but not necessarily in $\text{Mod}_{\mathcal{O}}(\mathfrak{g},\lambda)$.
But it seems like, in some cases, these categories should be equivalent?
For simplicity, let $\lambda = 0$. If I'm understanding correctly:
1. Beilinson-Bernstein says that there is an equivalence of categories: $\text{Mod}_{\mathcal{O}}(\mathfrak{g},0) \cong \text{Mod}_{B}(D_{X})$, where the second thing is weakly $B$-equivariant $D$-modules on the flag variety $X$.
2. Riemann-Hilbert says that there is an equivalence $\text{Mod}_{B}(D_X) \cong \mathcal{P}_{B}(X)$, the second category being perverse sheaves on $X$ whose cohomology is constant along $B$-orbits.
3. In the BGS paper on Koszul duality, it is claimed that $D^{b}(\mathcal{O}_{0}) \cong D^{b}_{B}(X)$, the latter being the bounded derived category of constructible sheaves on $X$ whose cohomology is constant along $B$-orbits, and this congruence respects the $t-$structures giving $\mathcal{O}_{0} \cong \mathcal{P}_{B}(X)$.
Is this all correct? If it is, it would seem to establish an equivalence $\mathcal{O}_{0} \cong \text{Mod}_{\mathcal{O}}(\mathfrak{g},0)$. Which seems plausible, but also somewhat strange, since one is a strict subcategory of the other (not that that by itself would imply non-equivalence..). And even trying the simplest case, $\mathfrak{sl}_2$, I am unable to realize this equivalence, for example, the category $\text{Mod}_{\mathcal{O}}(\mathfrak{sl}_2,0)$ only seems to have four indecomposables (since it doesn't have the projective cover of $L(-2)$).

REPLY [10 votes]: The source of your confusion is the difference between B-equivariance and being smooth along B-orbits.  Since in step 3, you want only smoothness along B-orbits, you have to do the same thing in step 1.  In order to do that, you need to take some modules not in category $\mathcal{O}$, and instead consider modules where the Cartan subalgebra $\mathfrak{h}$ acts locally finitely (not necessarily semi-simply!), and the subalgebra $\mathfrak{n}$ nilpotently.  
The important point here is that now there's no containment of Type 1 or Type 2 modules: there are some modules on which the center acts semi-simply and the Cartan subalgebra does not, and some where it's vice versa.  These categories are equivalent by an old theorem of Soergel.  This theorem is secretly hiding in the result cited in BGS.<|endoftext|>
TITLE: Closeness graph of a topological space
QUESTION [5 upvotes]: Let $(X,\tau)$ be a topological space. We say that $x, y \in X$ are close if for every neighborhood $U$ of $x$ and $V$ of $y$ we have $U\cap V \neq \emptyset$. Let $E$ be the set of $\{x,y\}$ where $x,y\in X$ with $x\neq y$ and $x,y$ are close. We say $G(X,\tau) = (X,E)$ is the closeness graph of $(X,\tau)$.
Given any simple undirected graph $G$, is there a topological space $(X,\tau)$ such that $G\cong G(X,\tau)$?

REPLY [7 votes]: There is no topology on 4 elements whose closeness graph is the $4$-cycle.
(Proof: Let's call the elements $\{A,B,C,D\}$. Define a directed graph with these elements as vertices and a directed edge $u\to v$ whenever every open set containing $u$, also contains $v$, this has to be transitive and contain all self loops, lets call this directed graph $H(X,\tau)$. Two vertices $u,v$ in the closeness graph are connected iff there is a $w$ such that $u\to w$ and $v\to w$ in $H$. Now, we see that we can't have edges between $A$ and $C$ or between $B$ and $D$, however any orientation of the $4$-cycle $ABCD$ either includes a diagonal in the transitive closure, or it has edges $A\to B$, $C\to B$, or it has edges $B\to C$, $D\to C$. In each case the closeness graph must contain one of the diagonals.)
Update: Every undirected graph $G$ can be realized as a connected component of the closeness graph of some topological space. To prove this, let $Y=V(G)\sqcup E(G)$, and let $\tau$ be the topology generated by the open sets $\{e\}$ for all edges $e$, and $\{v,e\}$ whenever $v$ is an endpoint of $e$. Then $G$ is a connected component of $G(Y,\tau)$.<|endoftext|>
TITLE: Convex embedding with a positivity condition
QUESTION [5 upvotes]: We have a $n$-dimensional hypersurface $\Sigma$ embedded in the Euclidean $(n+1)$-space $\mathbb{R}^{n+1}$. We know that $\Sigma$ is compact without boundary, convex (not necessarly strictly convex), and that the $k$-th symmetric polynomial in the principal curvatures is constant.

Can I deduce that $\Sigma$ is a sphere?

Note: in the case $k=1$ (maybe in the case $k=n$?) the answer is "yes".

REPLY [5 votes]: The answer is yes. Even without convexity, a version of Alexandrov's theorem holds true for symmetric functions of the principal curvatures, i.e. if the $k$-th symmetric polynomial of the principal curvatures of a compact embedded hypersurface in $\mathbb{R}^{n+1}$ is constant, then the surface is a sphere. 
See the following references:
A. Ros "Compact hypersurfaces with constant higher order mean curvatures." Rev. Mat. Iberoamericana 3 (1987), no. 3-4, 447–453.
S. Montiel and A. Ros "Compact hypersurfaces: the Alexandrov theorem for higher order mean curvatures." Pitman Monogr. Surveys Pure Appl. Math., 52, Longman Sci. Tech., Harlow, 1991.<|endoftext|>
TITLE: Power's Theorem for irreducible representations
QUESTION [5 upvotes]: Let $A_{\alpha}\subset B(H)$ be a bunch of unital C*-algebras acting on a Hilbert space $H$ given together with their character spaces $M(A_{\alpha})$'s. A very nice theorem of Stephen C. Power identifies the character space of $C=C^{\ast}(\cup_{\alpha}A_{\alpha})$ the C*-algebra generated by the union of $A_{\alpha}$'s as a certain subset of the cartesian product of $M(A_{\alpha})$'s. Is there a similar characterisation for the spectrum i.e. the space of all irreducible representations instead of characters?

REPLY [3 votes]: An important point to remember about Power's result is that if $\varphi$ is a character of $C$ then $\varphi|_{A_\alpha}$ will be a character of $A_\alpha$. This does not happen for irreducible representations. 
In particular, consider $C = \overline{\cup M_{2^n}}$, the CAR algebra. Any non-zero irreducible representation $\varphi$ of $C$ will be faithful, since $C$ is simple, and thus $\varphi|_{M_{2^n}}$ is nonzero and so is unitarily equivalent to an infinite direct sum of the identity representation because $M_{2^n}$ has a one point spectrum. However, it is known that the CAR algebra has an uncountable number of non-equivalent irreducible representations (Gårding and Wightman, Representations of the anticommutation relations. 
Proc. Nat. Acad. Sci. 40, (1954). 617–621.). 
Therefore, there cannot be such a characterization.
You may find On representations of finite type by Kadison of some interest as well.<|endoftext|>
TITLE: Topological structure of SO(n) as a product
QUESTION [7 upvotes]: I’m interested in the question for which $n$ the special orthogonal group is homeomorphic to the product
$$ \mathrm{SO}(n) \approx S^{n-1} \times \mathrm{SO}(n-1). $$
Allen Hatcher [1, p. 293 f.] claims (?) that this is true for $n \in \{ 2, 4, 8 \}$ and wrong for all other values (although I’m not sure what is meant by “twisted product”).
Does anyone have a reference where this is done in more detail? Maybe I just didn’t have the right keywords for a proper search.
[1] Algebraic Topology, downloadable at https://www.math.cornell.edu/~hatcher/AT/ATpage.html

REPLY [18 votes]: Another line of reasoning: If $SO(n)$ were homeomorphic to the product $S^{n-1}\times SO(n-1)$ then $S^{n-1}$ would be a retract (not deformation retract!) of $SO(n)$, hence $S^{n-1}$ would be an H-space since a retract of an H-space is an H-space, assuming the retract contains the identity element, which we can arrange here by a suitable choice of retraction.  Then Adams' theorem restricts $n$ to $2,4,8$.<|endoftext|>
TITLE: Smooth complete intersections and sharpness of the Chevalley-Warning theorem
QUESTION [6 upvotes]: Let $X$ be a complete intersection in $\mathbb{P}^n$ of multidegree $(d_1,\ldots,d_r)$. If we're working over a finite field $\mathbb{F}_q$, the Ax-Chevalley-Warning theorem says that if $X$ is in the Fano range, i.e., $$ \sum d_i \leq n,$$ then $|X(\mathbb{F}_q)| \equiv 1 \pmod q.$
In the case that $X$ is not in the Fano range, one can cook up examples of such $X$ with no $\mathbb{F}_q$-points at all using the norm form, but these are not smooth.
In SGA 7 II Exposè XXI, Katz shows that for a general complete intersection in the Fano range, this congruence is not satisfied, possibly after extending the ground field.
Given a multidegree outside the Fano range, are there explicit examples of smooth complete intersections which don't satisfy this congruence?
Edit: Sorry, I wasn't sufficiently explicit about what I was asking. Restricting to hypersurfaces, what I would like is for each $d$, $p$, and $n$ with $d \geq n+1$, an explicit example of a smooth hypersurface of degree $d$ in $\mathbb{P}^n$ which doesn't satisfy the Chevalley-Warning congruence.
Thanks for the examples, though; I knew some explicit examples before but no infinite families.

REPLY [6 votes]: Extension of Daniel Loughan's example (which is also known, but not as
well-known as it should(?) be):  if a prime $p$ is of the form $dn+1$ then
the Fermat hypersurface $\sum_{i=1}^d x_i^d = 0$ in ${\bf P}^{d-1}({\bf F}_q)$
is smooth and its number of rational points is not congruent to $1 \bmod p$. 
Indeed the usual argument for Chevalley(-Warning) shows that
the number of rational points is congruent mod $p$ to $1 \pm t$
where $t$ is the $(x_1 x_2 \cdots x_d)^{p-1}$ coefficient of
$\left(\sum_{i=1}^d x_i^d\right)^{p-1}$, and when $p = dn+1$
this coefficient is $(p-1)!/n!^d$ which is clearly not $0 \bmod p$.<|endoftext|>
TITLE: How to get convinced that there are a lot of 3-manifolds?
QUESTION [25 upvotes]: My question is rather philosophical : without using advanced tools as Perlman-Thurston's geometrisation, how can we get convinced that the class of closed oriented $3$-manifolds is large and that simple invariants as Betti number are not even close to classify ?
For example i would start with :

If $S_g$ is the closed oriented surface of genus $g$, the family $S_g \times S^1$ gives an infinite number of non pairwise homeomorphic $3$-manifolds.
Mapping tori of fiber $S_g$ gives as much as non-diffeomorphic $3$-manifolds as conjugacy classes in the mapping class group of $S_g$ which can be shown to be large using the symplectic representation for instance.

I think that I would like also say that Heegaard splittings give rise to a lot of different $3$-manifolds which are essentially different, but I don't know any way to do this. 
So if you know a nice construction which would help understanding the combinatorial complexity of three manifolds, please share it :)

REPLY [5 votes]: Read Chapter 4 of Thurston's notes http://library.msri.org/books/gt3m/. He produces infinitely many closed hyperbolic 3-manifold of different volumes, and hence non-homeomorphic, just by doing Dehn surgery on the figure 8 knot in $S^3$. This does not used advanced geometrisation theorems, just some basic (but clever) hyperbolic geometry, combined with the Mostow rigidity theorem.<|endoftext|>
TITLE: Fundamental class in $KO[1/2]$
QUESTION [6 upvotes]: Let $M^m$ be an oriented Riemannian manifold.
The signature operator associates to $M$ a class $\Delta_M\in KO_m(M)[1/2]$.
I have two questions about this class $\Delta_M$:

Rationally, $\Delta_M$ is computed by the Chern character, i.e.
 $$\text{ch}(\Delta_M)=L(M)\cap [M].$$
 Where could I find details of the proof of this formula? 
Here $\text{ch}$ denotes the Chern character, $L$ denotes the 
 $L$-class and $[M]$ the fundamental class in the ordinary
 homology theory.
Is there a formula for $\Delta_{M\times N}$ in terms of
 $\Delta_M$ and $\Delta_N$?

REPLY [6 votes]: Concerning your first question, I think a good place to learn about it is Markus Banagl's book "Topological invariants of stratified spaces" (chapter 6) or have a look at Paul Siegel's paper "$KO$-homology at odd primes" Amer. J. Math. 105 (1983), 1067-1105 (in particular the appendix where Sullivan's approach is explained).
In Siegel's paper you have a bordism theory $\Omega^{Witt}_*(-)$ build from cobordisms of singular spaces called Witt spaces. 
A Witt space $X$ carries rational homology L-classes $l_i\in H_{dim(X)-4i}(X,\mathbb{Q})$. This homology $L$-classes were defined by Goresky and McPherson in this singular context extending the homology $L$-classes of a manifold $M$ which are poincaré dual to Hirzebruch $L$-classes $L_i\in H^{4i}(M,\mathbb{Q})$:
$$l_i(M)=L_i(M)\cap [M]$$ 

In Witt bordism any Witt space has a fundamental class $[X]\in \Omega^{Witt}_{dim(X)}(X)$ represented by $id:X\rightarrow X$. 
Moreover we have a natural transformation 
$$\Phi:\Omega^{Witt}_k(X)\rightarrow \oplus_{i} H_{k-4i}(X,\mathbb{Q})$$such that for a Witt space we have $\Phi([X])=l_0+l_1+\ldots$, for a manifold we get that $\Phi([M])=L(M)\cap [M]$.
And this bordism theory when tensored by $\mathbb{Z}[1/2]$ is naturally isomorphic to $ko_*(-)\otimes \mathbb{Z}[1/2]$. Siegel used Sullivan's construction of $ko_*(-)\otimes \mathbb{Z}[1/2]$ to get a natural transformation
$$\mu: \Omega^{Witt}_*(-)\rightarrow ko_*(-)\otimes \mathbb{Z}[1/2].$$
such that $\mu([M])=\Delta_M.$

Concerning your second question you can have a look at theorem 11.1 "Multiplicativity of the L-theory fundamental class" of this paper:
"The  L-homology fundamental class for IP-spaces and the stratified  Novikov conjecture"
by Markus Banagl, Gerd Laures and Jim McClure (available on Banagl's homepage).<|endoftext|>
TITLE: Weighted Permutation Sum
QUESTION [7 upvotes]: I am trying to find out a closed-form formula (or a generating function at least) for the number of permutations $\sigma$ that satisfy $$ S = \sum_{i = 1}^{n} i\sigma(i)$$ for a given value of $S$. We can find a maximum and minimum possible sum for a particular $n$ and I have observed that the intermediate sums (between max and min) will always exist for $n>3$. The frequency of the occurrence of each sum is approximately a bell-shaped curve although it is not unimodal.
I have tried various approaches to solve this problem including using a partitions formulation but I have not been able to derive much from it. Any clue as to how to proceed with this problem? Has a similar problem been solved elsewhere?

REPLY [6 votes]: You can search this statistic (normalized so that the smallest value is 0) in www.FindStat.org and you will find that this is the rank of the permutation inside the lattice of alternating sign matrices. You find further information at http://www.findstat.org/St000055 and the references there:
Sack, J., Úlfarsson, H. Refined inversion statistics on permutations MathSciNet:2880660
Striker, J. A unifying poset perspective on alternating sign matrices, plane partitions, Catalan objects, tournaments, and tableaux MathSciNet:2794039<|endoftext|>
TITLE: Is there a way to simplify the following trace expression?
QUESTION [6 upvotes]: I'd like to simplify the following expression:
$$\text{tr}\{\mathbf{A}^HE(\mathbf{C}^H \begin{bmatrix} \mathbf{0}_{M\times M} & \mathbf{0}_{M\times N} \\ \mathbf{0}_{N\times M} & \mathbf{I}_{N\times N} \end{bmatrix} \mathbf{C})\mathbf{A}\}$$, where the matrix $\mathbf{C}$ is Toeplitz and is constructed by shifting the vector $[c_0,\cdots, c_{M-1}]$ through the rows while filling the rest of elements in every row with $N$ zeros, $M<N$.
$$\mathbf{C}  = \begin{bmatrix} c_0 & 0 & \cdots & 0 & c_{M-1} & \cdots & c_1\\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots & c_{M-1}\\
c_{M-1} & \cdots & \cdots & c_0 & 0 & \cdots & 0\\
0 & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\
0 & \ddots & 0 & c_{M-1} & \cdots & \cdots & c_0 
\end{bmatrix}~,$$ $\mathbf{A} \in \mathbb{C}^{(N+M)\times N}$, $\mathbf{C} \in \mathbb{C}^{(N+M)\times (N+M)}$ and $E(c_ic_i^*)=\frac{\sigma^2}{2}$,  $E(c_ic_j^*)=0$. $\text{tr}(\cdot)$ is the trace, $E(\cdot)$ is the expectation, $(\cdot)^H$ is the hermitian and $(\cdot)^*$ is the conjugate. $\mathbf{I}$ and $\mathbf{0}$ are the identity and zero matrices respectively.
In particular, I want to get rid of the expectation operator, any suggestions?
Cross posted at "https://math.stackexchange.com/questions/1466196/is-there-a-way-to-simplify-the-following-trace-expression"

REPLY [8 votes]: After a cyclic permutation of the trace, the expression you need is
$$Y=\text{tr}\left\{\mathbf{A}^HE(\mathbf{C}^H \begin{bmatrix} \mathbf{0}_{M\times M} & \mathbf{0}_{M\times N} \\ \mathbf{0}_{N\times M} & \mathbf{I}_{N\times N} \end{bmatrix} \mathbf{C})\mathbf{A}\right\}=\text{tr}\left\{E(\mathbf{C}\mathbf{A}\mathbf{A}^H \mathbf{C}^H) \begin{bmatrix} \mathbf{0}_{M\times M} & \mathbf{0}_{M\times N} \\ \mathbf{0}_{N\times M} & \mathbf{I}_{N\times N} \end{bmatrix} \right\}$$
Let me abbreviate $\mathbf{H}=\mathbf{AA}^H$, $\mathbf{X}=\mathbf{C}\mathbf{A}\mathbf{A}^H \mathbf{C}^H$, and let me denote by $\mathbf{P}$ the projection matrix in square brackets, so $Y={\rm tr}\{E(\mathbf{X})\mathbf{P}\}$. The expectation value can be evaluated using the Toeplitz property $C_{ij}=c_{i-j}$. 
From your construction I gather that
$$E(c_i c^\ast_j)=\tfrac{1}{2}\sigma^2\delta'_{ij}$$
where $\delta'_{ij}=1$ if $i=j\in\{0,1,2,\ldots M-1\}$ $\text{modulo}\,(N+M)$, while $\delta'_{ij}=0$ otherwise.
Carrying out the average,
$$E(X_{ij})=\sum_{kl}E(C_{ik}H_{kl}C^\ast_{jl})=\sum_{kl}E(c_{i-k}H_{kl}c^\ast_{j-l})=\tfrac{1}{2}\sigma^2\sum_{kl}H_{kl}\delta'_{i-k,j-l}$$
The projector $\mathbf{P}$ identifies $j=i=M+1,M+2,\ldots M+N$, so we arrive at
$$Y={\rm tr}\{E\mathbf{(X)P}\}=\tfrac{1}{2}\sigma^2 \sum_{i=M+1}^{N+M}\sum_{k,l=1}^{N+M}H_{kl}\delta'_{i-k,i-l}=\tfrac{1}{2}\sigma^2 \sum_{k=1}^{N+M}H_{kk}N_k$$
with the definition $N_k=\sum_{i=M+1}^{N+M}\delta'_{i-k,i-k}$. This number is a bit tedious to evaluate [*], but I presume you can easily take it from here.
[*] If I have not made a mistake, I find:
$$N_k=
\begin{cases}
k-1& \text{if}\quad 1\leq k\leq M\\
M& \text{if}\quad M+1\leq k\leq N+1\\
N+M-k+1& \text{if}\quad N+2\leq k\leq M+N\\
0&\text{otherwise}
\end{cases}$$<|endoftext|>
TITLE: Maximal cyclic quotient of a $p$-group
QUESTION [6 upvotes]: Let $G$ be a finite abelian $p$-group, $p$ a prime. I say that a pair $(G',\varphi)$ is a maximal cyclic quotient (please excuse me if this definition already exists and refers to a different concept) of $G$ if $G'$ is a cyclic group and $\varphi\colon G\to G'$ is a surjective map with the following property: if $H\leq \ker\varphi$ is a subgroup such that $G/H$ is cyclic, then $H=\ker\varphi$.
My first question is the following: is it true that if $(G',\varphi)$ is a maximal cyclic quotient of $G$ then $G\simeq G'\times\ker\varphi$?
Let now $G=\prod_iG_i$ be the decomposition of $G$ into cyclic $p$-groups and let $H$ be a subgroup of $G$. Does there exist an automorphism of $G$  which maps $H$ to a product $\prod_iH_i$ with $H_i\leq G_i$ for all $i$?
An affirmative answer to this question would imply an affirmative answer to the first one, but I have no clue about its truth; it sounds as a pretty strong statement indeed...

REPLY [8 votes]: Let $G = C_p \times C_{p^3} = \langle x \rangle \times \langle y \rangle$, and let $H = \langle z\rangle$, with $z = xy^p$. Then no automorphism of $G$ can map $H$ into one of the direct factors of $G$. Since $|H|=p^2$, it would have to map it into the $C_{p^3}$ factor, and hence to $\langle y^p \rangle$. But $z$ has order $p^2$ and is not the $p$-th power of an element of $G$ of order $p^3$, so $z$ cannot be mapped to $y^p$, or to any other element of $\langle y \rangle$ of order $p^2$, because these elements are all $p$-th powers. That answers the second question in the negative.
Also $G/H$ is a maximal cyclic quotient of $G$: it is cyclic of order $p^2$, but $H/\langle z^p \rangle = G/\langle y^{p^2} \rangle$ is not cyclic. But $H$ is clearly not a direct factor of $G$. So the answer to the first question is also no.<|endoftext|>
TITLE: Historical developement of analysis and partial differential equations (especially in the 20th century)
QUESTION [10 upvotes]: Q: Is there a set of some comprehensive surveys or monographs describing (in
  technical detail) the historical development of the various
  subareas of analysis and partial differential equations?

I'm especially (but not only) interested in expositions of the most recent ($20^{\text{th}}$ century onwards) developments.
For instance, I already know the following works: 

History of Functional Analysis, by J. Dieudonné;
Partial Differential Equations in the 20th Century, by H. Brezis and F. Browder;
A History of Analysis, edited by H. N. Jahnke.


A  slight variation on the theme of this question is:

Q': Which "texbooks"/monographs (dealing with subareas of analysis and PDE) strongly embrace an historical point of view?

For example, I've some good memories of the following book (which deals with topics in basic calculus): 

Analysis by Its History by E. Hairer and G. Wanner.


A quite related question is Motivation for and history of pseudo-differential operators.

REPLY [2 votes]: J. Lutzen, The prehistory of the theory of distributions, a companion book of which can be considered the autobiography of Laurent Schwartz, which contains quite nontrivial remarks on the math side.<|endoftext|>
TITLE: Is there an integrable complex structure on $\mathrm{SU}(3)$?
QUESTION [17 upvotes]: Is there a complex manifold diffeomorphic to  $\mathrm{SU}(3)$?
This question arises in a StackExchange discussion by HK Lee, Ted Shifrin and Jason DeVito:
https://math.stackexchange.com/questions/488959/way-distinguishing-whether-or-not-complex-manifold 
I believe it is also related to Etesi's work on a complex structure for $S^6$ where he makes use of the fibration  $\mathrm{SU}(3) \to G_2 \to S^6$.   So as an aside question, I would like to know whether Etesi's work implies the existence of a complex manifold dffeomorphic to  $\mathrm{SU}(3)$?

REPLY [33 votes]: It is an old theorem of Samelson that any compact Lie group $G$ of even rank has an integrable complex structure, which, in particular applies to the case of $\mathrm{SU}(3)$.  Basically, one chooses a Cartan subalgebra $\frak{t}\subset\frak{g}$, which gives a splitting of the Lie algebra into (complex) root spaces, choose a base for the root system and then consider the positive root spaces, use those to define the $(1,0)$-vector fields and then define the complex structure however one likes on the Cartan sub-algebra $\frak{t}$ (which one can do because it has even dimension, by assumption), and then one shows that the left-invariant almost complex structure that this defines on $G$ is always integrable.  (It follows almost immediately from the structure equations.)
Here is an explicit construction in the case of $\mathrm{SU}(3)$:  Let $g:\mathrm{SU}(3)\to \mathrm{GL}(3,\mathbb{C})\subset M_{3,3}(\mathbb{C})$ be the standard inclusion homomorphism of $\mathrm{SU}(3)$ into the group of invertible $3$-by-$3$ complex matrices.  Then $g$ satisfies $g^*g = I_3$ and $\det(g)=1$ (where $g^*$ means the conjugate transpose of $g$), so the canonical left invariant form $\gamma = g^{-1}\,\mathrm{d}g$ satisfies $\gamma^*+\gamma=0$ and $\mathrm{tr}(\gamma)=0$, so it can be written in the form
$$
\gamma = \begin{pmatrix}
i\theta_1 & \omega_3 & \omega_2\\
-\overline{\omega_3} & i\theta_2 & \omega_1\\
-\overline{\omega_2} & -\overline{\omega_1} & i\theta_3\end{pmatrix},
$$
where the $\omega_i$ are $\mathbb{C}$-valued $1$-forms and the $\theta_i$ are $\mathbb{R}$-valued $1$-forms satisfying $\theta_1+\theta_2+\theta_3 = 0$.  By the definition of $\gamma$, we have the structure equation $\mathrm{d}\gamma = - \gamma\wedge\gamma$. 
Now let $\lambda\in \mathbb{C}\setminus\mathbb{R}$ be any constant that has nonzero imaginary part and set $\omega_0 = \theta_1 + \lambda\,\theta_2$.  Then the $\mathbb{C}^4$-valued $1$-form $\omega = (\omega_0,\omega_1,\omega_2,\omega_3)$ defines a left-invariant $\mathbb{C}^4$-valued coframing on $\mathrm{SU}(3)$, i.e., it provides a linear identification of each tangent space of $\mathrm{SU}(3)$ with $\mathbb{C}^4$, and hence defines a unique almost complex structure $J_\lambda$ on $\mathrm{SU}(3)$ for which the $1$-forms $\omega_i$ are of $J_\lambda$-type $(1,0)$.  (This $J_\lambda$ really does depend on the choice of $\lambda$ above.)
Finally, the structure equation $\mathrm{d}\gamma = -\gamma\wedge\gamma$ implies
$$
\mathrm{d}\omega_i \equiv 0 \mod \omega_0,\omega_1,\omega_2,\omega_3
$$
for $0\le i\le 3$.  (For example, 
$$
\mathrm{d}\omega_1 = i(\theta_3-\theta_2)\wedge\omega_1 +\overline{\omega_3}\wedge\omega_2
\equiv 0 \mod \omega_0,\omega_1,\omega_2,\omega_3\,,
$$
the other cases are similar.)  Thus, the Nijnhuis tensor of $J_\lambda$ vanishes, and so there is a unique complex structure on $\mathrm{SU}(3)$ for which the forms $\omega_i$ are of type $(1,0)$.  (Because this almost complex structure $J_\lambda$ is left-invariant, it is a fortiori real-analytic, so the Newlander-Nirenberg theorem is not needed, the integrability follows from earlier results.)<|endoftext|>
TITLE: Equality of codimension under Lusztig-Spaltenstein induction
QUESTION [7 upvotes]: Pardon if this is well known. Suppose I have a (say complex) connected reductive group $G^{\vee}$ with the $\tilde{\Delta}=\Delta\cup\{\alpha_0\}$ being the simple roots plus the negative highest root of $G^{\vee}$. Any proper subset of $\tilde{\Delta}$ defines a connected reductive subgroup $H^{\vee}\subset G^{\vee}$. Let $H$ and $G$ be the corresponding dual groups. For any special nilpotent orbit $u_H$ in $H$ we can define the Lusztig-Spaltenstein induction $u$ as a nilpotent orbit in $G$.
(that is, $(u_H,triv)$ corresponds to some representation of $W_H$ by Springer correspondence, we do the $j$-induction on this to get a representation of $W$, which correspond to $(u,triv)$ for some $u$)

Question: Is it true that the codimension of $u$ in the nilpotent cone of $G$ is
  the same is the codimension of $u_H$ in the nilpotent cone of $H$?

Thanks!

Edited: Just to say what I understand / guess so far: For any $u_H$, say the Springer fiber $\mathcal{B}_{e_H}$ at $u_H$ has dimension $e_H$. Then the corresponding Weyl group representation appears as the component of $H^{2e_H}(\mathcal{B}_{e_H})$ on which $A_{e_H}$ acts trivially. This component receives a surjection from $H^{2e_H}(H/B_H)$, which is a quotient of $\text{Sym}^{2e_H}(std_H)$, where $std_H$ is the standard representation of $W_H$. The Lusztig-Spaltenstein $j$-induction then, I think (I cannot find a reference for why this works for $H^{2e_H}(H/B_H)$ instead of $\text{Sym}^{2e_H}(std_H)$), gives a $W$-representation in $H^{2e_H}(G/B)$, which then corresponds to some orbit in $G$ whose Springer fiber has dimension $e_H$.

REPLY [4 votes]: What you're asking for can be phrased entirely in terms of nilpotent orbits. If $\mathcal{O}$ is the nilpotent orbit of $H$ then the nilpotent orbit of $G$ you obtain via your process is the induced nilpotent orbit $\mathrm{Ind}_H^G(\mathcal{O})$. It is well known that the codimensions of these orbits coincide. See Proposition 7.1.4 of the book "Nilpotent Orbits in Semisimple Lie Algebras" by Collingwood and McGovern.
EDIT: Sorry, I just saw that you want this for a more general class of subgroups. The statement is still true. See 13.3 in Lusztig's book "Characters of reductive Groups over a finite field" and also Lusztig's paper "Unipotent classes and special Weyl group representations", J. Algebra (321), no. 11, 3418-3449, where more details are provided concerning some of the statements in the book.<|endoftext|>
TITLE: Can $L$ be thin?
QUESTION [21 upvotes]: I have recently been wondering if the following is consistent with ZFC:

For every infinite ordinal $\alpha$: $|V_\alpha\cap L|=|\alpha|$.

Intuitively, this states that for $L$ is very "thin", in that it doesn't branch off too much from the set of ordinals as we climb cumulative hierarchy.
I have heard that it is possible to have it for $V_{\omega+1}$, which is essentially having $\Bbb R^L$ countable. I suspect that this is consistent, and we can construct a model of this starting from $V=L$ and adding a proper class of generic sets giving bijections between parts of $L$ and ordinals, but I have no real background in forcing to see if that makes any sense.
Thanks in advance.

REPLY [29 votes]: Claim: $|V_\alpha \cap L| = |\alpha|$ for every $\alpha \geq \omega$ implies that $0^\#$ exists. 
Proof: Let's assume, toward contradiction, that $0^\#$ doesn't exist that $|V_\alpha \cap L| = |\alpha|$ for every $\alpha \geq \omega$.
Let $\mu$ be a singular cardinal in $V$. Let's apply the assumption $|V_\alpha \cap L| = |\alpha|$ for $\alpha = \mu + 1$. In $V$, $\mu = |\mu + 1|=|V_{\mu + 1} \cap L| \geq |\mathcal{P}^L(\mu)| \geq |(\mu^{+})^L|$.
In particular, $\mu^{+} > (\mu^{+})^L$. Therefore, the cofinality of $(\mu^{+})^L$ in $V$ is strictly below $\mu$. Applying Jensen's covering theorem, we conclude that the cofinality of $(\mu^{+})^L$ in $L$ is strictly below $\mu$ - a contradiction to the regularity of $(\mu^{+})^L$ in $L$. 
By the way, if we restrict the values of $\alpha$ for which we want $|V_\alpha \cap L| = |\alpha|$ to be $\leq \aleph_{\omega}$, the consistency strength drops considerably:
Claim: $|V_\alpha \cap L| = |\alpha|$ for every $\omega \leq \alpha \leq \aleph_\omega$ is equiconsistent with the existence of $\omega$ inaccessible cardinals. 
Proof: For the first direction, force over $L$ with an iteration of Levi collapses in order to make the $n$-th inaccessible in $L$ equal to the $\aleph_n$ of the generic extension. For the second direction, note that the assumption implies that for all $1\leq n < \omega$, $\aleph_n$ is a $\beth$-fixed point in $L$, since for every infinite $\beta < \aleph_n$, $|V_\beta \cap L| = |\beth_\beta^L| < \aleph_n$. In particular, $\aleph_n$ is a limit cardinal in $L$. It is regular in $V$ and thus also in $L$ and therefore it is inaccessible.<|endoftext|>
TITLE: Modified mean value property
QUESTION [5 upvotes]: Let $L=\Delta + c$ in 3 dimensions, where $c$ is a positive constant.
I met this modified mean value property of a solution $u$ of $Lu=0$ as 
$$u(\xi)=\frac{\sqrt{c}\rho}{sin(\sqrt{c}\rho)}\frac{1}{4\pi \rho^2}\int_{\partial B(\xi,\rho)} u(x)d\sigma(x)$$
where $sin(\sqrt{c}\rho) \ne 0$, and $\sigma$ denotes the surface measure.
I'm trying to prove it. The following is what I did.
First I figured out that $$K(x,\xi)=-\frac{cos(\sqrt{c}r)}{4\pi r}, r=|x-\xi|$$ is a fundamental solution of $L$ with pole $\xi$. Then I tried to apply the second Green identity (which is also true for $L$ for sure)
$$\int_{\Omega}(LG) u-\int_{\Omega}G (Lu)=\int_{\partial \Omega} \frac{\partial G}{\partial \nu} u- G \frac{\partial u}{\partial \nu}$$ For fixed $\rho$, let $G=K+\frac{cos(\sqrt{c}\rho)}{4\pi \rho}$ and $\Omega=B_{\rho}(\xi)$, so that $G$ vanishes on the boundary of $\Omega$, and $LG=\delta_{\xi}+c\frac{cos(\sqrt{c}\rho)}{4\pi \rho}$, where $\delta_{\xi}$ denotes the Dirac mass at $\xi$. Then the identity above becomes $$u(\xi)+c\frac{cos(\sqrt{c}\rho)}{4\pi \rho}\int_{B_{\rho}(\xi)}u(x)dx=\frac{\sqrt{c}\rho sin(\sqrt{c}\rho)+cos(\sqrt{c}\rho)}{4 \pi \rho^2} \int_{\partial B(\xi,\rho)} u(x)d\sigma(x)$$
Then no matter how hard I tried, I just could not get the desired form. 
I also tried to take the derivative with respect to $\rho$ of the desired form, and I got a complicated form which doesn't seem to be zero. Then I took the second derivative, and things got worse...
Can anyone either give the right reference about the solution of this operator $L$, or help me figure out how to obtain the desired form? Thanks!

REPLY [2 votes]: Ah, it is one of these cases when you better know your formula in all dimensions. 
    It is 
    $$
  \int_{|x-y|=\rho} u(x) d\sigma(x) = (2\pi)^{n/2} \cdot \frac{J_{n/2-1}(\sqrt{c}\rho)}{(\sqrt{c}\rho)^{n/2-1}} \cdot u(y)
 $$
    One of the ways to prove it:
    $$
 \Delta u + c u = 0,
 $$
    $$
 \int_{|x-y|\leq R} \Delta u(x) + c u(x)  dx = 0.
 $$
    Apply Green's formula to the first integrand, 
    $$
 \int_{|x-y|= R} \frac{\partial u}{\partial r}(x) d\sigma(x) + \int_{|x-y|\leq R} c u(x)  dx = 0,
 $$
    (here $\frac{\partial u}{\partial r}(x) = \nabla u(x)\cdot x/|x|$).
    $$
 R^{n-1}\int_{|x|=1} \frac{\partial u}{\partial r}(y-Rx) d\sigma + c \int_{|x-y|\leq R} u(x) dx = 0.
 $$
    Differentiate w.r. to $R$ 
    $$
 R^{n-1}\int_{|x|=1} \frac{\partial^2 u}{\partial r^2}(y-Rx) d\sigma +(n-1)R^{n-2} \int_{|x|=1} \frac{\partial u}{\partial r}(y-Rx) d\sigma + c R^{n-1} \int_{|x|=1} u(y-Rx) d\sigma = 0,
 $$
    $$
 \int_{|x|=1} \frac{\partial^2 u}{\partial r^2}(y-Rx) d\sigma +\frac{n-1}{R}\int_{|x|=1} \frac{\partial u}{\partial r}(y-Rx) d\sigma + c \int_{|x|=1} u(y-Rx) d\sigma = 0.
 $$
    $$
 \frac{d^2}{dR^2}\int_{|x|=1} u (y-Rx) d\sigma +\frac{n-1}{R} \frac{d}{dR}\int_{|x|=1} u(y-Rx) d\sigma + c \int_{|x|=1} u(y-Rx) d\sigma = 0.
 $$
    So $v(R) = \int_{|x|=1} u (y-Rx) d\sigma $ is a solution of 
    $$
 v''(R) + \frac{n-1}{R} v'(R) + c v(R)=0.
 $$
    This equation is one of the forms of Bessel's equation 
    $$
 x^2 y''+ xy' + (x^2-\alpha^2) y = 0. 
 $$
    In particular, it is not difficult to show that 
    $$
 v(R) = C_1 \frac{J_{n/2-1}(R\sqrt{c})}{R^{n/2-1}},
 $$
    here $J_\nu$ is the Bessel function of the first kind.
    To get the right constant just send $R\to 0$: on the one hand 
    $$
 \lim_{R\to 0} v(R)  = \omega_{n-1} \cdot u(y),
 $$ 
    on the other 
    $$
 \lim_{R\to 0} v(R) = C_1\lim_{R\to 0} \frac{J_{n/2-1}(R\sqrt{c})}{R^{n/2-1}} = C_1 \frac{(\sqrt{c}/2)^{n/2-1}}{2\pi^{n/2}} \omega_{n-1}. 
 $$
    So $C_1 = (2\pi)^{n/2} c^{1/2-n/4} u(y)$.<|endoftext|>
TITLE: What is the point of pointwise Kan extensions?
QUESTION [34 upvotes]: Recall that a Kan extension is called pointwise if it can be computed by the usual (co)limit formula, or equivalently if it is preserved by (co)representable functors.
I have seen pointwise Kan extensions defined in many texts, such as Mac Lane's book or Borceux's book, but these texts don't typically prove any theorems about them. I've also heard it claimed that all mathematically important Kan extensions are pointwise. And apparently it should be easier to compute a pointwise Kan extension since there is a formula for it. But I'd like to know: what are some interesting categorical facts that are true about pointwise Kan extensions that are not true about general Kan extensions? I'm happy with properties whose proofs don't generalize to the non-pointwise case, but I'd also be interested to see actual counterexamples to such generalizations.
In Basic Concepts of Enriched Category Theory, Kelly chooses to reserve the term "Kan extension" for the pointwise version. In explaining this choice, he says "Our present choice of nomenclature is based on
our failure to find a single instance where a weak Kan extension plays any mathematical role whatsoever." If I comb through his book for uses of Kan extensions, will I find that "most" facts used about Kan extensions fail in the non-pointwise case?
One fact which I have seen stated is that in a Kan extension
$\require{AMScd}$
\begin{CD}
A @>G>> C\\
@VFVV \nearrow \mathrm{Lan}_F G \\
B 
\end{CD}
If $G$ is fully faithful and $\mathrm{Lan}_F G$ is pointwise, then the comparison 2-cell is an isomorphism. Is this false when the pointwiseness hypothesis is dropped?
Part of the reason I ask is that there are several definitions in the literature generalizing pointwiseness of Kan extensions to more general categorical frameworks (enriched categories, 2-categories, equipments, double categories...) but I'm confused because I don't know what the theory is that these theories are attempting to generalize. Pointwiseness also shows up in the theory of exact squares, but again I'm confused because I don't know the point of pointwise Kan extensions.
EDIT
I'm starting to compile a list of these properties. Here's what I have so far:

As above, a pointwise extension along a fully faithful functor has an isomorphism for a comparison cell.
A functor $F$ is dense if and only if its left Kan extension along itself $\mathrm{Lan}_F F$ exists, is pointwise, and is isomorphic to the identity functor.
To show that taking total derived functors is functorial, one needs that they are pointwise extensions.
To show that the total derived functors of a Quillen adjunction form an adjunction between homotopy categories, one needs to know that they are pointwise extensions.

(3) and (4) were mentioned by Emily Riehl in introductory remarks to a lecture at the Young Topologists' Meeting 2015 (link).

Pointwise Kan extension along a fully faithful functor is itself a fully faithful functor. More precisely, if $F: A \to B$ is fully faithful, and $\mathrm{Lan}_F G$ and $\mathrm{Lan}_F G'$ exist and and are pointwise, then $\mathrm{Nat}(G,G') \to \mathrm{Nat}(\mathrm{Lan}_F G, \mathrm{Lan}_F G')$ is an isomorphism.

So far the only counterexamples I have to dropping the pointwise hypothesis in these statements are in the cases (1) and (2), which I've detailed in CW responses.

REPLY [6 votes]: In a Kan extension $\mathrm{Lan}_F G$,
$\require{AMScd}$
\begin{CD}
A @>G>> C\\
@VFVV \nearrow \mathrm{Lan}_F G \\
B 
\end{CD}
If $F$ is fully faithful and the extension is pointwise, then the comparison 2-cell is an isomorphism. This is because, if the extension is pointwise, we have
$C(\mathrm{Lan}_G F(Fa),c) \cong [A^\mathrm{op},\mathsf{Set}]( B(F-,Fa),C(G-,c))$
$\qquad \qquad \qquad \, \cong [A^\mathrm{op},\mathsf{Set}]( A(-,a),C(G-,c))$
$\qquad \qquad \qquad \, \cong C(Ga,c)$
and we can take $c = Fa'$.
In order to get an example of a fully faithful $F$ such that $\mathrm{Lan}_F G$ does not have an isomorphism for a comparison cell, we need to move beyond monoids, because a fully faithful functor between monoids is an isomorphism. So the simplest sort of fully faithful functor which is not an equivalence would be the inclusion $F: \mathbf{B}M \to \mathbf{B}M_+$ of a monoid into the category which freely adjoins either an initial or terminal object to it. If we adjoin a terminal object, then left Kan extensions will all be pointwise (with the terminal object being sent to the colimit of the original diagram), so we adjoin an initial object instead. And we might as well take $M$ to be the simplest possible monoid $\mathbb{N}$:
$\require{AMScd}$
\begin{CD}
\mathbf{B}\mathbb{N} @>G>> C\\
@VFVV \nearrow L \\
\mathbf{B}\mathbb{N}_+ 
\end{CD}
It turns out there is a "minimal" $C$ admitting such a left Kan extension $L$, which looks like this:
$G\bullet \overset{(\eta_n)_{n \in \mathbb{N}}}{\overset{\to}{\to}} L\bullet \overset{L!}{\leftarrow} L\emptyset$
Here $G\bullet$ is the fully faithful image of $G$ (so it's a copy of $\mathbf{B}\mathbb{N}$), and $L\emptyset \overset{L!}{\to} L\bullet$ is the fully faithful image of $L$ (so it's a copy of $\mathbf{B}\mathbb{N}_+$ where $L\emptyset$ is the initial object). $C(G\bullet, L\bullet)$ is generated by an arrow $\eta = \eta_0$ under the equation $\eta_n = \eta \circ Gn = Ln \circ \eta$. So $\eta$ constitutes a comparison natural transformation $G \implies LF$.
An exhaustive analysis of the functors $H: \mathbf{B}\mathbb{N}_+ \to C$ (there are 5 families of them, depending on where the objects are sent) reveals that the only one which admits a natural transformation $G \implies HF$ or $L \implies H$ is $L$ itself, and this diagram is in fact a left Kan extension $L = \mathrm{Lan}_F G$ with $F$ fully faithful, but the comparison 2-cell $\eta$ is not invertible.<|endoftext|>
TITLE: "Interesting" projective varieties being quotients of $\mathbb{A}^n\setminus \{0\}$ by an action of an algebraic group?
QUESTION [6 upvotes]: The algebraic (multiplicative) group $G^m$ acts on $\mathbb{A}^n$ (diagonally) and the quotient of $\mathbb{A}^n\setminus \{0\}$ by $G_m$ is $\mathbb{P}^{n-1}$ (which is a proper variety). I would like to find a "more interesting" example of an action of an  algebraic groups $G$ on  $\mathbb{A}^n$ such that the quotient of $\mathbb{A}^n\setminus A$ by $G$ is proper, where $A$ is a certain "small" subvariety of $\mathbb{A}^n$ (say, over the field of complex numbers). Under these conditions does the quotient have to be toric (and does $G$ have to be a torus)?
I would be deeply grateful for any hints or references (and I know very little on algebraic groups and toric varieties)!

REPLY [4 votes]: As Daniel Loughran pointed out, this situation has been largely investigated (and generalized) in recent years. I don't know if the theory of Cox Rings is overkill for your situation, but let me refer you to this book that is treating the subject systematically and not assuming much prior knowledge. The first chapter is available here.
(I would've written this as a comment as it's not really an answer to your question, but apparently you have to have asked or answered a few questions before being able to do that.)<|endoftext|>
TITLE: Mapping class group of a punctured genus 0 surface
QUESTION [5 upvotes]: Let $T_{0,n}$ be the Teichmuller space of $n$-punctured genus $0$ Riemann surface, and $M_{0,n}$ the Moduli space (assume $n\geq 3$ and the punctures are numbered). What is the correct notion of the mapping class group $\Gamma_{0,n}$, so that $M_{0,n} \cong T_{0,n} / \Gamma_{0,n}$? 
Now $\pi_1(M_{0,n}) \cong \Gamma_{0,n}$ since the Teichmuller space is contractible and the action of $\Gamma_{0,n}$ is free.
What does the Dehn-Nielsen-Baer theorem say in this case? Let 
$S_{0,n}$ be the $n$ punctured complex projective line then which subgroup of $Out(\pi_1(S_{0,n}))$ does $\Gamma_{0,n}$ correspond to? Also how does $\Gamma_{0,n}$ act on $H_1(S_{0,n})$?

REPLY [3 votes]: Let $S_g$ be a compact Riemann surface of genus $g$ with $n$ marked points $x_1, \ldots, x_n$, and set $S_{g, \, n}:=S_g - \{x_0, \ldots, x_n\}$. Also, denote by $\pi_{g, \, n}$ the fundamental group of $S_{g, \, n}$ (we omit the decoration $n$ when $n=0$).
Then the mapping class group $\Gamma_{g,\, n}$ is defined as $$\Gamma_{g, \, n} := \pi_0 \, \textrm{Diff}^{+}(S_{g, \, n}),$$
where $\textrm{Diff}^{+}(S_{g, \, n})$ is the group of orientation-preserving diffeomorphisms of $S_g$ that fix each $x_i$.
The group $\Gamma_{g, \, n}$ naturally acts on the Teichmueller space $T_{g, \, n}$. Such an action is properly discontinuous and there is a subgroup of finite index acting freely, in such a way that, as an orbifold, $$M_{g, \, n}= T_{g, \, n}/ \Gamma_{g, \, n}.$$
In particular, $\pi^{\textrm orb}_1(M_{g, \, n}) = \Gamma_{g, \, n}$.
By a result of Baer and Nielsen, with this representation $\Gamma_{g}$ is identified with the subgroup of $\textrm{Out}(\pi_{g})$ acting trivially on $H_2(\pi_{g})$. When $n \geq 1$, there are similar characterizations. 
For (many) more details, see the paper
R. Hain, E. Looijenga: Mapping class groups and moduli spaces of curves,  Algebraic Geometry-Santa Cruz 1995, 97–142, Proc. Sympos. Pure Math., 62, Part 2, Amer. Math. Soc., Providence, RI (1997),
also available on the arXiv.<|endoftext|>
TITLE: Does a monotone subadditive $f: \mathcal{P}(\bf N)\to [0,1]$ admit a finite partition with values in $(0,1)$?
QUESTION [10 upvotes]: A function $f\colon \mathcal{P}(\mathbf{N})\to [0,1]$ is said to have the Darboux property whenever for all $X \subseteq \mathbf{N}$ and $y \in [0,f(X)]$, there exists $Y \subseteq X$ such that $f(Y)=y$. 
Moreover, $f$ is said to be monotone if $f(X)\le f(Y)$ whenever $X\subseteq Y$ and subadditive if $f(X\cup Y) \le f(X)+f(Y)$ for all $X,Y \subseteq \mathbf{N}$.

Question. Let $f\colon \mathcal{P}(\mathbf{N})\to [0,1]$ be a monotone subadditive function with the Darboux property such that $f(\mathbf{N})=1$. Does there exist necessarily a finite partition $\{A_1,\ldots,A_k\}$ of $\mathbf{N}$ for which
  $$
f(A_1), \ldots, f(A_k) \in (0,1)?
$$


A simpler version of this question has been posted on MSE one week ago here, showing that in general the answer is negative if $k$ is bounded. 
The question comes from the study of properties of a class of functions satisfying certain axioms, which can be called as "upper densities" (with this respect, you can see this MO question). Here, instead, there is MO question related to the definition of Darboux property. 
A related construction (which does not ask for subadditivity) by Salvo Tringali has been provided here.

REPLY [7 votes]: $\let\eps\varepsilon$It seems that the following function fits:
$$
  f(A)=\inf\left\{\alpha\colon \quad \sum_{x\in A}x^{-\alpha}<\infty\right\}.
$$
(I assume that $\mathbb N$ starts with 1.)
Clearly, it is monotone, and $f(\mathbb N)=1$. Moreover, it is more than subadditive: we have $f(A\cup B)=\max\{f(A),f(B)\}$. This also implies that if $A_1\cup A_2\cup\dots\cup A_k=\mathbb N$, then $f(A_i)=1$ for some $i$.
It remains to show that $f$ has the Darboux property. Let $f(A)=\alpha$ and $\beta\in[0,\alpha)$; we need to find $B\subseteq A$ with $f(B)=\beta$. We have $\sum_{x\in A}x^{-\beta}=\infty$. Set $X_n=[2^n,2^{n+1})\cap \mathbb N$, $A_n=A\cap X_n$, and $a_n=\sum_{x\in A_n}x^{-\beta}$.  Then the series $\sum_n a_n$ diverges.
Set $b_n=\min\{a_n,1\}$. The series $\sum_n b_n$ diverges as well (this is trivial if $b_i=1$ infinitely many times, as well as if it happens only finitely many times). Now, choose $B_n$ to be the subset of  $A_n$ such that $S_n=\sum_{x\in B_n}x^{-\beta}\leq b_n$, and $S_n$ is maximal subject to this property. Then $S_n\geq b_n/2$ for all $n\geq 1$. 
Set $B=\cup_n B_n$; then $\sum_{x\in B}x^{-\beta}\geq \sum_n b_n/2=\infty$. On the other hand, for every $\eps>0$ we have
$$
  \sum_{x\in B}=
  \sum_n\sum_{x\in B_n}x^{-\beta-\eps}
  \leq \sum_n 2^{-\eps n}\sum_{x\in B_n}x^{-\beta}
  \leq \sum_n2^{-\eps n}<\infty.
$$
Thus $f(B)=\beta$.<|endoftext|>
TITLE: Does pullback in the category of smooth manifolds always exists?
QUESTION [10 upvotes]: I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist. 
Remarks:
1) A pullback in a certain category is defined as a space satisfying a universal property, not as a fiber product. (Which is just the usual form of many pullbacks...). See a clarification at the bottom.
2) I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category). 
This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.
3) In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other) 
4) In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

Clarification of the definition of pullback:
A space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$ which complete $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}Y'$ into a commutative square) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$ (which commutes with $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$), there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).

REPLY [11 votes]: If a pullback exists in the category of smooth manifold then, its underlying set of points has to be what you described simply by looking at morphism from the point. Moreover a map into the pullback is smooth if and only if the map to the product is smooth (because it is smooth if and only if each component is smooth by the universal properties of the pullback).
So the only things surprising things that can happen if the categorical pullback exists, is that the fiber product has a smooth structure which is not induced by the differentiable structure on the product, but still have the same smooth map into it, like the example you gave a link to... But this situation is rather weird, and it is a lot more common to simply have no pullback:
So take for example the pullback of $\{0\}$ along the map $(x,y) \rightarrow xy$. If this pullback existed it would be a smooth structure on the union of the horizontal and the vertical line in $\mathbb{R^2}$ such that a map into it is smooth if and only if it is smooth when seen as a map into $\mathbb{R^2}$.<|endoftext|>
TITLE: second dual of minimal tensor products of $C^*$-algebras
QUESTION [5 upvotes]: Let $A$ be a unital $C^*$-algebras and $K(H)$ is $C^*$-algebras of compact operators on separable Hilbert space $H$. Is it true that $(A \otimes K(H))^{**}= A^{**} \overline{\otimes}B(H)$?

REPLY [4 votes]: Yes (but maybe there is a more direct argument ? ):
$A$ and $K(A) = A \otimes K(H)$ are Morita equivalent so they have equivalente categories of representations, moreover this equivalence is implemented as following: any representation of $K(A)$ is of the form $H \otimes R$ with $R$ a representation of $A$.
$A^{**}$ can be constructed as the weak closure of $A$ into some "universal" representation of $A$, i.e. a representation $V$ such that any other representation is a direct sum of subrepresentations of $V$.
A universal representation of $A$ and of $K(A)$ corresponds to each other under the above equivalence (because being universal is a purely categorical property)
SO this mean that there is a representation $V$ of $A$ such that:
$A^{**}$ is the weak closure (double commutant) of $A$ into $B(V)$.
$K(A)^{**}$ is the weak closure of $A \otimes K(H)$ in $B(V \otimes H)$.
Because the commutant of $A \otimes K(H)$ is $A' \otimes \mathbb{C}$ (still because of the above equivalence of catégores: the commutant is the set of endomorphism of representations) and the bi-comutant of $A \otimes K(H)$ is indeed $A'' \otimes B(H)$ (for the spatial tensor product).<|endoftext|>
TITLE: Tori in three-space
QUESTION [11 upvotes]: Recently I was talking to an alien who does not know complex function theory. I was trying to convince her that the set of conformal equivalence classes of smooth embedded tori in $R^3$ is two parametric. Is there a nice two-parametric family of tori in $R^3$ which are pairwise conformally non-equivalent? 
I do not have an exact definition of ``nice''. Maximally symmetric, canonical in some sense. Or such that it is intuitively clear that they are all in different conformal classes. The standard model (a circle rotated around a line) gives only
one-parametric family up to conformal equivalence.
EDIT. I understand that all conformal tori EXIST in $R^3$. But is it possible to
show (describe, draw, make a 3-d model) a torus in $R^3$ which is not conformally equivalent to a ``rectangular'' torus ? And this can be proved just by looking at its shape in $R^3$.

REPLY [8 votes]: It is a 1961 result by Adriano Garsia that every conformal class can be represented by an embedded surface. His proof seems to be reasonably constructive, see this question (and answers thereto).
EDIT Pinkall, in the paper cited by j.c. in his comment, explicitly solves the question for the case of genus 1 (as asked by the OP).<|endoftext|>
TITLE: Structure of Deligne-Lusztig representations $R_{T,\theta}$ for ministropic $T$ and cuspidal representations
QUESTION [7 upvotes]: Let $G$ be a reductive group over a finite field $k$, let $F$ be a Frobenius morphism on $G$.
I'll start with a somewhat vague question and make my question more specific further down:
How do cuspidal representations fit into Deligne-Lusztig characters $R_{T,\theta}$?
A little about what is known classically: if $\rho$ is a cuspidal representation of $G^F$, then $\rho$ is a subrepresentation of $\pm R_{T,\theta}$ for some pair $(T,\theta)$, where $T$ is a minisotropic torus (so it is not a subtorus of any proper Levi subgroup of $G$).  Moreover, for such a $T$, if $\theta: T^F \to \mathbb{C}^\times$ is in general position (so not fixed by any nontrivial element of the Weyl group), then $\pm R_{T,\theta}$ is irreducible cuspidal.
In the case of $G = GL_n$, it turns out that these are all of the irreducible cuspidal representations; i.e. every irreducible cuspidal representation is isomorphic to some $\pm R_{T,\theta}$, where $T$ is ministropic, and $R_{T,\theta} \cong R_{T',\theta'}$ if and only if $(T,\theta)$ is $G^F$-conjugate to $(T',\theta')$.
Of course, it's too much to hope that this should happen in every case.  Let $T$ be an anisotropic torus in $SL_2$. Then there is a character $\theta: T^F \to \mathbb{C}^\times$ that is not in general position, where $\pm R_{T,\theta}$ splits as a sum of two irreducible cuspidal representations (perhaps you need the base field to have odd characteristic).
Now to specify my question.  Throughout, $T$ is a minisotropic torus of $G$, and $\theta:T^F \to \mathbb{C}^\times$.
1). Are there groups $G$ with pairs $(T,\theta)$ and cuspidal representations $\rho,\, \rho'$ such that $\rho$ occurs in $R_{T,\theta}$ with positive multiplicity, and $\rho'$ occurs with negative multiplicity?
2). Are there groups $G$ with pairs $(T,\theta)$ such that $R_{T,\theta}$ such that $R_{T,\theta}$ contains both cuspidal and non-cuspidal representations with nonzero multiplicity?

REPLY [5 votes]: EDIT: It's easy to answer Question 2 affirmatively by pointing to the groups $G=\mathrm{Sp}(4,q)$ of Lie type $B_2= C_2$ in odd characteristic.  Bhama Srinivasan first worked out the ordinary irreducible characters of $G$ using ad hoc methods in her thesis work at Manchester: the resulting paper is here and can be downloaded freely from the AMS site.  She was advised by J.A. Green, whose 1955 paper on characters of finite general linear groups was the inspiration for this approach.  After the landmark Annals of Mathematics (1976) paper by Deligne and Lusztig, their generalized characters $R_T^\theta$ could be compared with her work (and then reduced modulo $p$ by Jantzen in his 1987 exposition for the Arcata Conference on Finite Groups).   Some of this comparison is not formally published but was written up in her own notes.
A striking feature of the family of groups of Lie type is the existence of just one cuspidal "unipotent" character (in the later terminology): this was called $\theta_{10}$ by Srinivasan.   She later computed (unpublished) the explicit decomposition of various unipotent characters $R_T^\theta$ (where $\theta$ is the trivial character but $T$ ranges over all types of maximal tori).  This shows that $\theta_{10}$ occurs with coefficient $\pm 1, \pm 2$ in two cases where $T$ is anisotropic, along with other non-cuspidal irreducible unipotent characters.
The answer to Question 1 is also yes, but it's most conveniently seen in type $G_2$ (again using unipotent characters). Following Srinivasan's work, but still prior to 1976, B. Chang and R. Ree (1974) worked out the ordinary irreducible characters of these groups (again assuming odd characteristic) in the spirit of Macdonald's conjectures which were later proved by Deligne-Lusztig.  The four irreducible characters denoted by $X_{17}, X_{18}, X_{19}, X'_{19}$ in the paper by Chang and Ree were later checked by Lusztig to be precisely the irreducible cuspidal unipotent characters. 
If I'm reading the computations correctly in the 1984 Bonn Diplomarbeit by D. Mertens (student of Jantzen, one decomposition of a unipotent Deligne-Lusztig (generalized) character involves six irreducible characters including $X_{18}$ and $X_{19}+X'_{19}$ with opposite signs (plus three other non-cuspidal unipotent characters with coefficients $\pm 1$).   It seems likely that this kind of mixed decomposition will occur frequently as the rank increases, though it's unclear whether it has significance.
An obstacle to seeing this much detail is the sheer complexity of writing down explicit decompositions of Deligne-Lusztig characters as $\mathbb{Z}$-linear combinations of ordinary irreducible characters.  The
papers of Lusztig (including his 1984 monograph) and the 1985 book by Roger Carter contain the machinery needed for this purpose, but the case-by-case computations are nontrivial.   Most of the published work of Lusztig and others has emphasized more the theoretical algorithms involved.    But the rank 2 groups do serve as a useful laboratory for experimental work.  

A couple of notes about terminology: The reference to a "Frobenius element" $F$ of $G_k$ is out of focus, since the finite group consists of fixed points of a Frobenius morphism $F$ of the ambient algebraic group (such as raising all matrix entries to the $q$th power for $q$ a power of $p$).         
Concerning "minisotropic" tori (terminology, like "cuspidal", which I guess comes from the original Harish-Chandra program for representations over real and $p$-adic fields), these are relevant only in reductive groups which are not semisimple---such as general linear groups.   Otherwise one can just refer to "anisotropic" tori over $k$.<|endoftext|>
TITLE: In a noetherian commutative ring with only one associated prime, is the nilradical locally free?
QUESTION [5 upvotes]: The title says it all.
I suspect that the answer in general is no, although my intuition tells me that a jump in the dimension of the fibre of the nilradical at some point of Spec(A) can occur only when one meets an associated prime cycle.
Note the easiest case: when the ring is generically reduced then it has to be reduced hence its nilradical is 0 hence free. This follows easily from the fact that Ass={pt} implies that the ring is the union of its nilpotents and its nonzerodivisors.

REPLY [6 votes]: Actually, this never happens unless the nilradical is $0$ (i.e., unless the ring is a domain).  Let $A$ be a Noetherian ring with only one associated prime whose nilradical $N$ is locally free.  Then $A$ has only one minimal prime, so $N$ is prime.  Now localize at $N$.  We now have a $0$-dimensional Noetherian local ring $A_N$, so it is artinian.  In particular, $A_N$ has finite length over itself, and the length of the maximal ideal $NA_N$ is strictly smaller than the length of $A_N$.  But we are assuming $N$ is locally free, so $NA_N$ is free over $A_N$; the only way this can happen is if $NA_N=0$.  But, as you observe, every element of $A\setminus N$ is a nonzerodivisor, so the canonical map $A\to A_N$ is injective.  Thus $NA_N=0$ implies $N=0$.<|endoftext|>
TITLE: Does the hyperdeterminant calculate a quantity akin to the volume of a parallelepiped?
QUESTION [10 upvotes]: If $M$ is an $n \times n$ matrix, $|\det(M)|$ is the volume of the $n$-dimensional
parallelepiped spanned by the column vectors of $M$.

          


          

(Image from Wikipedia's Determinant article.)


In the 19th-century, Cayley defined the hyperdeterminant of a hypermatrix $H$.
(A hypermatrix 
can be viewed as a representation of a
tensor.)
My question is:

Q.
  Does the hyperdeterminant have a geometric interpretation somehow analogous
  to the parallelepiped-volume interpretation of the determinant?

Perhaps an answer resides in the
book by Gelfand, Kapranov, and Zelevinsky entitled Discriminants, Resultants and Multidimensional Determinants (Birkhäuser, Boston, 1994;
MAA link),
which I have yet to
examine.

REPLY [12 votes]: At least, in the classical case treated by Cayley, the $2{\times}2{\times}2$ hyperdeterminant, there is an interpretation in terms of volumes generalizing the classical determinant case.  It goes like this:
First, recall that, when two vector spaces $U$ and $V$ have the same dimension, say, $r$, there is a polynomial map of degree $r$
$$
\mathbf{det}: U\otimes V \to \Lambda^r(U)\otimes\Lambda^r(V)
$$
defined, relative to any basis $u_i$ of $U$ and $v_j$ of $V$, by 
$$
\mathbf{det}( a^{ij} u_i\otimes v_j) = \det(a^{ij})
\,(u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes (v_1\wedge v_2\wedge\cdots\wedge v_r)
$$
where $\det$ is the usual determinant of a matrix.  It is easy to see that this definition of $\mathbf{det}$ is independent of the choice of basis $u_i$ of $U$ and $v_\rho$ of $V$.  Note that, since $\Lambda^r(U)$ and $\Lambda^r(V)$ are the top exterior powers of $U$ and $V$, their elements represent 'volume elements' (sometimes called 'mass elements') in $U$ and $V$ respectively, so a tensor $\alpha\in U\otimes V$ gives rise, via $\mathbf{det}(\alpha)$ a way to 'multiply' volume elements in the two vector spaces.  
(Of course, in the usual linear algebra interpretation, we consider $\alpha\in U^*\otimes V$ as a linear map from $U$ to $V$, and then 
$$
\mathbf{det}(\alpha)\in \Lambda^r(U^*)\otimes \Lambda^r(V)\simeq
\Lambda^r(V)\otimes \Lambda^r(U)^*\simeq\  "\Lambda^r(V)/ \Lambda^r(U)"
$$
can be thought of as a ratio of volume forms on the two vector spaces.  Even more specially, we can take $U=V^*$, and then, because $\Lambda^r(V)\otimes \Lambda^r(V)^*$ is canonically isomorphic to $\mathbb{R}$, $\mathbf{det}(\alpha)$ becomes just a scalar.)
Now, when you have three vector spaces $U$, $V$, and $W$ of the same dimension $r$, you can expand a tensor $\alpha\in U\otimes V\otimes W$ in terms of a basis $u_i$ of $U$, $v_j$ of $V$, and $w_k$ of $W$ as
$$
\alpha = a^{ijk}\,u_i\otimes v_j\otimes w_k
$$
and you can just regard the $w_k$ as 'indeterminates' and define
$$
\mathbf{det}_{UV}(\alpha) = \det(a^{ijk}w_k)\otimes
(u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes(v_1\wedge v_2\wedge\cdots\wedge v_r),
$$
where, now,
$$
\mathbf{det}_{UV}(\alpha) \in S^r(W)\otimes \Lambda^r(U) \otimes \Lambda^r(V).
$$
To go further, you need to look at particular values of $r$.  What Cayley did was consider the fact that, for quadratic forms in $r$ variables, there is a well-defined discriminant mapping $\mathrm{discr}_W:S^2(W)\to S^2(\Lambda^r(W))$ that is a homogeneous polynomial of degree $r$.  It is defined by 
$$
\mathrm{discr}_W(q^{ij}w_iw_j) = \det(q^{ij})\,
(w_1\wedge w_2\wedge\cdots\wedge w_r)^2. 
$$
In particular, when $r=2$, we can compose to get an element
$$
\mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha))
\in S^2(\Lambda^2(U))\otimes S^2(\Lambda^2(V)) \otimes S^2(\Lambda^2(W)),
$$
which is a polynomial of degree $4$ in the coefficients of $\alpha$.
This is (the negative of) Cayley's hyperdeterminant in the $2{\times}2{\times}2$
case. Note that it is, indeed, expressed in terms of volume forms on the three vector spaces $U$, $V$, and $W$.  It's just that it is now the product of squares of volume forms.  By the way, it is not hard to show that, if we had, instead, computed $\mathbf{hdet}_{VWU}(\alpha)$, we would have got the same result.  
I think that, at least in the $r=2$ case, this is probably the best interpretation of the hyperdeterminant in terms of volumes.
In the case when $(\dim U, \dim V, \dim W) = (2,2,s)$ where $s > 2$, this gives an expression
$$
\mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha))
\in S^s(\Lambda^2(U))\otimes S^s(\Lambda^2(V)) \otimes S^2(\Lambda^2(W))
$$
of degree $2s$, but, of course, this vanishes identically when $s > 4$.
When you go to higher values of $r$, it's not so clear.  For example, when $r=3$ (again, with all vector spaces of the same dimension $r$), there are the two Aronhold relative invariants:  $Q^4: S^3(W)\to S^4(\Lambda^3W)$ (of degree $4$) and $Q^6:S^3(W)\to S^6(\Lambda^3W)$ (of degree $6$), and so we can define two expressions
$$
\mathbf{Q}^4_{UVW}(\alpha) = Q^4_W(\mathbf{det}_{UV}(\alpha))
\in S^4(\Lambda^3(U))\otimes S^4(\Lambda^3(V)) \otimes S^4(\Lambda^3(W))
$$
of degree $12$ and 
$$
\mathbf{Q}^6_{UVW}(\alpha) = Q^6_W(\mathbf{det}_{UV}(\alpha))
\in S^6(\Lambda^3(U))\otimes S^6(\Lambda^3(V)) \otimes S^6(\Lambda^3(W))
$$
of degree $18$.  Thus, these clearly relate powers of volume elements of the underlying vector spaces.  According to the Wikipedia page, though, the hyperdeterminant of $\alpha$ in this case must have degree $36$; presumably it is a polynomial in the above two invariants.<|endoftext|>
TITLE: Are free positive operators equivalent to almost-commuting operators?
QUESTION [11 upvotes]: Set $A:=C_0((0,1]) * C_0((0,1])$ (the free product C*-algebra), with canonical generators $a,b$ (positive contractions). Does there exists some $\gamma>0$ such that, for any $x,y \in A$ if $x^*x=a$ and $y^*y=b$ then
$$ \|[xx^*,yy^*]\| > \gamma? $$

REPLY [7 votes]: YES. (However, the answer for the same question for von Neumann algebras is NO.) 
I take $$A:=\lbrace f\in C([0,1],M_2) : f(0), f(1) \in D_2\rbrace.$$ 
Here $D_2\cong\ell_\infty^2$ is the diagonal. 
Let $$Q:=\mathrm{ev}_0\oplus\mathrm{ev}_1\colon A\to\ell_\infty^2\oplus\ell_\infty^2.$$
Let $$a=\left(\begin{matrix} 1 & 0\\ 0 & \frac{1}{2}\end{matrix}\right)\mbox{ (constant) and }
b=\left(\begin{matrix} t & \sqrt{t(1-t)}\\ \sqrt{t(1-t)} & 1-t\end{matrix}\right)\mbox{ (projection)}.$$
Suppose $a=x^*x$ and $b=y^*y$. Then, $x=u|x|$ and $u\in A$ (unitary). One has
$$\|[xx^*,yy^*]\|=\|[a,u^*yy^*u]\|.$$
Since $Q(A)$ is commutative, $u^*yy^*u$ is a {\bf projection} such that 
$Q(u^*yy^*u)=Q(b)=\mathrm{diag}(0,1)\oplus\mathrm{diag}(1,0)$.
This implies $\|[a,u^*yy^*u]\|\geq 1/4$.<|endoftext|>
TITLE: Reduction formula for Schubert polynomials
QUESTION [10 upvotes]: In my endless fiddling with formulas I discovered one that fills in the blanks in a generic formula I saw in a paper, but I'm wondering if maybe it's already known and the paper was just mentioning the form of it casually. The formula I saw, which has undetermined constants, expresses a Schubert polynomial in $n$ variables as a sum of products of Schubert polynomials in a smaller number of variables with nonnegative coefficients. It looks like
$$S_w(x_1,x_2,\ldots,x_n)=\sum_{u,v}{d_{u,v}^wS_u(x_1,x_2,\ldots,x_k)S_v(x_{k+1},x_{k+2},\ldots,x_n)}$$
where $d_{u,v}^w$ is mentioned to be nonnegative. I discovered that if you let $w_0$ be the longest element of $S_{n+1}$ and let $w_0'$ be the longest element of $S_{n+1-k}$ (identified with the parabolic subgroup of $S_{n+1}$ corresponding to the elements in the first $n+1-k$ positions) then in fact
$$S_w(x_1,x_2,\ldots,x_n)=\sum_{a,b}{c_{a,b}^{ww_0}S_{aw_0'w_0}(x_1,x_2,\ldots,x_k)S_{bw_0'}(x_{k+1},x_{k+2},\ldots,x_n)}$$
where $\ell(aw_0'w_0)+\ell(a)=\ell(w_0'w_0)$, $\ell(bw_0')+\ell(b)=\ell(w_0')$, and $c_{a,b}^{ww_0}$ is the corresponding structure constant for multiplying Schubert polynomials, which of course is known to be nonnegative. Is the formula known to this degree of specificity?

REPLY [2 votes]: This is essentially Theorem 4.6.1 in (http://arxiv.org/pdf/alg-geom/9703001v1.pdf) Schubert polynomials, the Bruhat order, and the geometry of flag manifolds by Bergeron and Sottile.<|endoftext|>
TITLE: Weak convergence of random variables in $L^2$ and vague convergence
QUESTION [5 upvotes]: Dumb question: Let $X_n:\Omega \to \mathbf{R}$ be a sequence of $L^2(\Omega,\Sigma,\mathbf{P})$ random variables that has a weak limit $X$ in $L^2$.
Suppose also that $\mu_n$, the distributions of $X_n$ on $\mathbf{R}$ converge vaguely to $\mu$. 
Does the distribution of $X$ have to be $\mu$? If the $X_n$ are uniformly bounded by constants, does this have to be true?

REPLY [7 votes]: No, it is not true. Take $\Omega = [0,1]$ equipped with the Lebesgue measure and let $(X_n)_{n \in \mathbb{N}}$ be the sequence of Rademacher functions -- all of these have the same distribution (uniform on $\{-1,1\}$) but they converge weakly in $L^{2}[0,1]$ to $0$, since they form an orthonormal system. They are also uniformly bounded.<|endoftext|>
TITLE: vector bundles associated to a covering space
QUESTION [7 upvotes]: Let $M$ be a $m$-dimensional manifold whose cohomology ring and cell structure are well-understood, such that there is a free action of the symmetric group $S_n$ on $M$. Then we have a $n!$-sheeted covering space
$$
\pi:M\to M/S_n.
$$
Let $S_n$ act on the Euclidean spaces $\mathbb{R}^n$ and $\mathbb{C}^n$ by permuting the order of coordinates.
We obtain vector bundles by attaching Euclidean spaces as fibres to the covering $\pi$:
$$
\xi:\mathbb{R}^n\to M\times_{S_n}\mathbb{R}^n\to M/S_n,
$$
$$
\xi^\mathbb{C}: \mathbb{C}^n\to M\times_{S_n}\mathbb{C}^n\to M/S_n.
$$
Question: (1). Any  methods to know the Stiefel-Whitney classes $$
w(\xi)?
$$
Question: (2). Any  methods to know the Chern classes $$
c(\xi^\mathbb{C})?
$$
Note: if we take coefficients to be $\mathbb{Q}$, then all the Chern class vanish. Hence we take the coefficients to be $\mathbb{Z}/p$ for $p$ prime.

REPLY [11 votes]: When you have an Euclidean vector space of type:
$$\mathbb{R}^n\rightarrow M\times_{S_n}\mathbb{R}^n\rightarrow M/S_n$$
Then its classifying map $\phi:M/S_n\rightarrow BO(n)$ factors as:
$$M/S_n\rightarrow BS_n\stackrel{\rho_n}{\rightarrow} BO(n)$$
where $\rho_n$ is induced by the regular reprsentation $S_n\rightarrow O(n)$.
Thus what you want to compute is the morphism 
$$\rho_n^*:H^*(BO(n),\mathbb{F}_2)\rightarrow H^*(BS_n,\mathbb{F}_2)$$
A nice way to formulate such a result is to put all $n$ together. Then you have a map of Hopf rings 
$$\rho:H^*(BO(\bullet),\mathbb{F}_2)\rightarrow H^*(BS_{\bullet},\mathbb{F}_2).$$
Then set $w_{i,n}=\rho^*_n(w_i)\in H^i(BS_n,\mathbb{F}_2)$ where $w_i$ is the i-th Stiefel-whitney class and $w(k,l)=w_{2^k(2^l-1),2^{k+l}}$. Then Giusti, Salvatore and Sinha proved in their paper "Mod-2 cohomology of symmetric groups as a Hopf ring" (theorem $10.7$):
"The classes $w(k,l)$ generate $H^*(BS_{\bullet},\mathbb{F}_2)$ as a Hopf ring."<|endoftext|>
TITLE: Average digit sum in different bases
QUESTION [12 upvotes]: Given two natural numbers $n\geq 1$ and $b\geq 2$, denote by $S_b(n)$ the sum of the digit of $n$ in its representation in base $b$. Clearly $S_b(n)$ varies from 1 (when $n$ is a power of $b$) to $(b-1)\lfloor\ln{n}/\ln{b}\rfloor$  (when $n=b^k-1$ for some integer $k$). However, for a fixed $n$, the value of $S_b(n)$ for a generic base $b$ should be of order $(b-1)/2\cdot\ln{n}/\ln{b}$. 
This lead to consider the average of $S_b(n)$ for $b\leq n$
$$S(n):=\sum_{b\leq n}S_b(n)$$
or (maybe better) the wieghted average
$$S_w(n):=\sum_{b\leq n}\frac{S_b(n)}{b}$$
There exists some asymptotics or non trivial bounds for $S(n)$ and $S_w(n)$?

REPLY [10 votes]: The observation by Greg Martin is indeed correct. I have worked with these expressions in my bachelor thesis, which can be accessed for free here: Digit Sums. Proposition 2.10, page 12 states that $$\sum_{2\leq b\leq n}S_b(n)\sim (1-\frac{\pi^2}{12})n^2$$
as $n \to \infty$.
Relationship between the two sums
We have
$$\sum_{2\leq b\leq n}\frac{S_b(n)}{b}=\frac{1}{n}\sum_{2\leq b\leq n}S_b(n)+\int_{1}^{n}\frac{\sum \limits_{2 \leq b \leq x}S_b(n)}{x^2}dx$$
Proof:
Set $n_0$ equal to a fixed positive integer, and consider the sum $\sum \limits_{2\leq b\leq n}\frac{S_b(n_0)}{b}$. If we let $(a_k)_{k=1,2,\ldots}$ be the sequence defined by $a_1=0$ and $a_t=S_t(n_0)$ for $t\geq 2$, and $\phi:t\mapsto\frac{1}{t}$. Then the sum becomes $\sum \limits_{k=1}^{n}a_k\phi(k)$, which by Abel's summation formula equals
$$\phi(n)\sum_{k=1}^na_k -\int_{1}^{n}\phi'(x)\sum_{k\leq x}a_k \;\;dx$$
or equivalently
$$\frac{1}{n}\sum_{2\leq b\leq n}S_b(n_0)+\int_{1}^{n}\frac{\sum \limits_{2 \leq b \leq x}S_b(n_0)}{x^2}dx$$
Setting $n_0$ equal to $n$ completes the proof.
Asymptotic formula for the second sum
Proposition 4.4, page 27 states that $$\int_{1}^{n}\frac{\sum \limits_{p \leq n}S_p(n)}{x^2}dx \sim (\frac{\pi^2}{12}-\gamma)\frac{n}{\log(n)},$$ as $n \to \infty$, where $\gamma$ is the Euler- Mascheroni constant. Here, the summation is taken over the primes $p$. If you modify the deduction of this formula, you could conclude that
$$\int_{1}^{n}\frac{\sum \limits_{2 \leq b \leq x}S_b(n)}{x^2}dx \sim (\frac{\pi^2}{12}-\gamma)n$$
Using the relationship we obtained for the two summands, along with these two asymptotic formulae, we get:
\begin{align}
\sum_{2\leq b\leq n}\frac{S_b(n)}{b}\;=&\;\frac{1}{n}\sum_{2\leq b\leq n}S_b(n)+\int_{1}^{n}\frac{\sum \limits_{2 \leq b \leq x}S_b(n)}{x^2}dx 
\\ \sim \; &(1-\frac{\pi^2}{12})n+(\frac{\pi^2}{12}-\gamma)n \\[2mm]
\sim \; &(1-\gamma)n & \\[2mm]
=&\; n \cdot 0.42278433\ldots
\end{align}
Which perfectly matches the the numerical heuristics as pointed out by Greg Martin.
What if we sum over only primes p?
Proposition 2.12(p.14), Propostion 4.4(p.27) and the forumlae on page 25 show that
\begin{align}\sum_{p \leq n}S_p(n)\; =\; &(1-\frac{\pi^2}{12})\frac{n^2}{\log(n)}+C\frac{n^2}{\log^2(n)}+o(\frac{n^2}{\log^2(n)}) \\
\sum_{p\leq n}\frac{S_p(n)}{p}\sim& \;(1-\gamma)\frac{n}{\log(n)},
\end{align}
where $C=0.119\ldots$<|endoftext|>
TITLE: Simplest case of Langlands-Shahidi method
QUESTION [6 upvotes]: I would like to read the simplest examples of Langlands-Shahidi method carried out to prove the functional equation of $L$-function. 
Could the constant term of $\mathrm{GL}(2)$-Eisenstein series be used to prove the functional equation of $L$-functions? (standard on $\mathrm{GL}(2)$ or Rankin-Selberg on $\mathrm{GL}(2) \times \mathrm{GL}(2)$ or adjoint/symmetric square on $\mathrm{GL}(2)$)

REPLY [4 votes]: I think that the best introduction to the Langlands-Shahidi method still is the following book by Shahidi and Gelbart:

Gelbart & Shahidi, "Analytic Properties of Automorphic L-Functions" (1988)

It includes a detailed exposition of the $\mathrm{GL}(2)×\mathrm{GL}(2)$ case, from both the Langlands-Shahidi and the Rankin-Selberg point of view (and the relation between them). Of course, general $\mathrm{GL}(n)×\mathrm{GL}(m)$ and symmetric powers are also treated.
As for your other questions, yes, those examples of L-functions are well covered by this method, see the list here.<|endoftext|>
TITLE: $\pi_1$ of the moduli of G-bundles on elliptic curves and the double affine braid group
QUESTION [11 upvotes]: For a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$, the fundamental group of $\mathfrak{h}_\text{reg}/W$ (where $\mathfrak{h}$ is the Cartan subalgebra, $\mathfrak{h}_\text{reg}$ is the subset of its regular elements, and $W$ the Weyl group) is the Artin braid group of type $\mathfrak{g}$.
Similarly, if we take a simple and  simply-connected Lie group $G$ over $\mathbb{C}$ and consider the orbifold fundamental group of $T_\text{reg}/W$ (where $T$ is the Cartan torus of $G$), it is the affine Artin braid group of the corresponding type. 
The references to the proofs to these statements can be found e.g. in Charney-Davis.
Now, take the moduli space of $G$-bundles on an elliptic curve $E$, which is known to be given by $(E\otimes_{\mathbb{Z}}\Lambda) /W$ (cf. Laszlo ) where $\Lambda$ is the coweight lattice of $T$.  I believe the fundamental group of this space (minus 'non-regular' elements) is more or less given by the double affine braid group of the corresponding type, but I don't find any explicit reference. Could you give me one?

REPLY [4 votes]: After a day of hard literature searching myself, apparently the statement goes back to Cherednik's original paper on DAHA in 1992, and also to van den Lek's article "Extended Artin Groups" in 1983. 
A modern reference with all the detailed proofs can be found in this paper by B. Ion. If I understand the paper correctly, the $\pi_1$ of the regular points of the moduli space of flat G-bundles on $T^2$ is the quotient of the double affine Artin group by the central subgroup generated by an element usually denoted by  $X_\delta$.<|endoftext|>
TITLE: Operads and the Stable Module Category
QUESTION [10 upvotes]: I am seeking references to places where operads and their algebras have been studied for the stable module category. Colored operads are fine too.
Let $k$ be a field and $R$ a $k$-algebra. The stable module category StMod(R) is the category of R-modules where we've modded out by all morphisms that factor through a projective. This triangulated category is useful in representation theory (see Happel's book) and for the study of Tate Cohomology (see work of Benson, Carlson, Rickard, Iyengar, Christensen, Minac, Beligiannis, Gillespie, Bravo, Stovicek, Krause, Pevtsova). The stable module category is also the homotopy category of a stable model structure on $R$-mod (shown originally by Pirashvili then by Hovey in his book). This is a monoidal model category when $R = k[G]$ for a finite group $G$ (see section 9 of Hovey's "Cotorsion pairs, model category structures, and representation theory), and this is the setting I am interested in.
Right now the only examples of operad-algebras in this category I can think of are:

Algebras over $Ass$ are associative $R$-algebras
Algebras over $Com$ are commutative $R$-algebras
Algebras over $Lie$ should work here and be $R$-algebras that additionally have a Lie structure. 

My reason is that, motivated by applications in Top and Ch(k), I recently did some model categorical work on the interplay between localization and colored operads. I've realized the model structure on $R$-mod from above also satisfies all my hypotheses, so there might be applications of my work in this setting. Obviously, I need to read up first on what has been done and what sorts of questions people care about, hence the reference request. The first two examples above seem intrinsically interesting, but I don't know if their homotopy theory has ever been studied elsewhere. I know almost nothing about the third example. References where these three examples or others have been studied in StMod(R) would help me figure out what is currently known and what sorts of questions are the right questions to work on in this area.
Thanks!

REPLY [3 votes]: The answer to this question is now yes. I asked this question while working on the following paper with Donald Yau, about preservation of operad-algebra structure under right Bousfield localization:
http://arxiv.org/abs/1512.07570
After getting no answers here (as well as Dan Christensen sharing that he'd never heard of this being done), we went ahead and did it in Section 12, i.e. proved the stable module category satisfied our conditions and got results about preservation of operad algebras. If anyone comes across this thread later and thinks of another paper that looked at operads in the stable module category, please let me know so I can cite that paper.<|endoftext|>
TITLE: Algebraic proof without using comparison theorem for étale cohomology
QUESTION [7 upvotes]: Let $X$ be some smooth scheme over $\mathbf C$ equipped with an action of $\mu_n$ (the group of $n$th roots of unity). 
The étale cohomology groups of X are therefore equipped with an action of $\mu_n$.
Now, let's suppose that the action of $\mu_n$ on $X$ extends to an action of $\mathbf G_m$.
Then, the analytic space $X(\mathbf C)$ is equipped with an action of $\mu_n$ which extends to an action og $\mathbf C^\times$. As $\mathbf C^\times$ is connected, one concludes that the action of $\mu_n$ on the singular cohomology of $X(\mathbf C)$ is trivial (the multiplication by each element $\xi\in \mu_n$ is homotopic to the identity).
Going backwards to étale cohomology, using a comparison theorem between étale and singular cohomology, one concludes that the action of $\mu_n$ on the étale cohomology of $X$ is trivial.
Is there a direct proof  that the action on the étale cohomology groups is trivial without having to refer to the topological case?

REPLY [4 votes]: Here's an alternative to grghxy's answer. Let me stick to the original assumptions for simplicity, but they can certainly be relaxed. Let $G$ be the image of the $\mu_n$ action on étale cohomology with finite coefficients. In particular, $G$ is finite. Let $g\in G$. Since the action is assumed to extend to a $\mathbb{C}^*$-action, we can solve $g=h^N$ for arbitrary $N>0$. Now choose $N=|G|$. 
NB  See some comments by grghxy & Yonatan Harpaz below.<|endoftext|>
TITLE: Plethysm of $S^3(S^2V)$ as $\mathfrak{sl}_3(\mathbb{C})$-module
QUESTION [5 upvotes]: I have asked this question in MSE before, but have not got any answer. So here I am asking it again with some more detail.
I believe that the following sequence of $\mathfrak{sl}_3(\mathbb{C})$-modules is exact:
$$0\to\mathbb{C}\to V\otimes\wedge^2V\stackrel\phi\to S^2 V\otimes S^2(\wedge^2 V)\stackrel\psi\to S^3(S^2V)\stackrel{N}\to S^6V\to 0$$
where $V=\mathbb{C}^3$ and $S^2$ denotes the symmetric square. The exactness of this sequence can be shown easily using eigen value diagram. But I want to show the exactness explicitly by constructing the maps and showing exactness of those maps.
Let's call the third and fourth map $\phi$ and $\psi$ respectively, the fifth map is the natural map (call it $N$). I also know what $\psi$ is, it is given by the followig $$\psi ((s\cdot t)\otimes (u\wedge v)\cdot(w\wedge z))=((u\cdot w)\cdot(v\cdot z)-(u\cdot z)\cdot(v\cdot w))\cdot (s\cdot t)$$
and I have checked that $\operatorname{im}\psi=\ker N$. But I am not sure what $\phi$ is. 
So my question is, how to define $\phi$ so that $\ker\phi$ is isomorphic to $\mathbb{C}$ and $\psi\circ\phi$ is zero i.e. $\operatorname{im}\phi\subset\ker\psi$ (the other direction $\ker\psi\subset\operatorname{im}\phi$ follows by computing the dimensions which turns out to be equal) and of course $\phi$ should be $\mathfrak{sl}_3(\mathbb{C})$-linear.

REPLY [6 votes]: Under the action of $\mathfrak{sl}_3(\mathbb{C})$, $\Lambda^2V\simeq V^\ast$. This isomorphism can be written explicitly in terms of the invariant volume form $\omega\in\Lambda^3V\simeq \mathbb{C} $, by $(v_1,v_2\wedge v_3)=v_1\wedge v_2\wedge v_3=\lambda \omega$ (in other words, for $\theta\in V^\ast$, $\theta\mapsto \iota_\theta\omega\in\Lambda^2V$).
Thus $V\otimes\Lambda^2V\simeq V\otimes V^\ast = \mathfrak{gl}_3(\mathbb{C}) = \mathfrak{sl}_3(\mathbb{C})\oplus \mathbb{C}$. Moreover, $S^2V\otimes S^2(\Lambda^2 V)\simeq S^2V\otimes S^2 V^\ast = \mathfrak{gl}(S^2V)$. The latter space contains a subalgebra isomorphic to $\mathfrak{sl}_3(\mathbb{C})$, which is spanned by the representation matrices of the induced action on $S^2V$. 
Your map $\phi$ is then given by composing the following operations: first use the isomorphism $\Lambda^2V\simeq V^\ast$, then remove trace of the resulting operator by $A\mapsto A-\text{Tr}(A)/3 I$, then send the resulting matrix to its induced matrix on $S^2V$, and finally use the isomorphism $\Lambda^2V\simeq V^\ast$ to obtain an element of $S^2V\otimes S^2(\Lambda^2 V)$.
The kernel property is clear, because $S^3S^2V$ contains no submodule isomorpic to $\mathfrak{sl}_3(\mathbb{C})$. (This follows from highest weight calculations).<|endoftext|>
TITLE: $n$ groups of $n$ queens on a toroidal chessboard
QUESTION [5 upvotes]: An interesting question came up in the Puzzling Stack Exchange a few days ago about "queen-connected sets". When trying to solve this problem, I came across an arrangement of five colours of queens that would not attack each other on a toroidal 5×5 chessboard, and discovered that 4 queens could be placed into two non-attacking groups but not four individual ones.

Let a queen number be a number $n$ where there exists an arrangement of $n$ colours of $n$ queens that are placed on an $n \times n$ chessboard, such that no queen can attack a queen of the same colour, including through broken diagonals.
5 is the smallest queen number, and the arrangement I found is the one where all of them are a knight's move away from each other:
1 2 3 4 5
3 4 5 1 2
5 1 2 3 4
2 3 4 5 1
4 5 1 2 3

I haven't come across a way to prove that 6 is not a queen number, although I suspect that's true. I also tried making an arrangement for 10 just now using the same method that I used for 5, but it didn't work:
1234567890
4567890123
7890123456
0123456789
3456789012
6789012345
9012345678
2345678901
5678901234
8901234567

If you notice, every queen attacks the next queen 5 squares right and 5 squares down. But I don't know if there is another arrangement that does work.

Is there any way to determine the set of queen numbers without going through every arrangement to verify that it solves the problem?

REPLY [8 votes]: The "queen numbers" are precisely those numbers whose smallest prime factor is at least 5.
It is a theorem of Polya that

You can place n queens on an nxn toroidal board such that no two queens can attack each other if and only if the smallest prime factor of n is at least 5.

I couldn't find Polya's paper in English, but this paper of Chandra states the theorem above, and also solves some interesting multi-dimensional generalizations of your question.
To get from Polya's theorem to answering your question:
If n is a queen number, then looking at just one color I see that I have place n queens on an nxn toroidal board so that no two can attack each other. Conversely, if I have placed n queens on an nxn toroidal board in this way, then by shifting my solution n-1 times, I can find a way to place the other n-1 colors (this idea was given in a comment by user76105)<|endoftext|>
TITLE: Non-negative polynomials on $[0,1]$ with small integral
QUESTION [17 upvotes]: Let $P_n$ be the set of degree $n$ polynomials that pass through $(0,1)$ and $(1,1)$ and are non-negative on the interval $[0,1]$ (but may be negative elsewhere).
Let $a_n = \min_{p\in P_n} \int_0^1 p(x)\,\mathrm{d}x$ and let $p_n$ be the polynomial that attains this minimum.
Are $a_n$ or $p_n$ known sequences?  Is there some clever way to rephrase this question in terms of linear algebra and a known basis of polynomials?

REPLY [20 votes]: Following Robert Israel's answer, we also scale everything to $[-1,1]$ (thus multiplying the result by 2). As he mentions, the optimal polynomial is always a square of some other polynomial, $p_{2n}=p_{2n+1}=q_n^2$, and $q_n$ is either even or odd (see Lemma below). So we are left to find the minimal $L_2[-1,1]$-norm of an odd/even polynomial $q_n$ such that $\deg q_n\leq n$ and $q_n(\pm1)=(\pm1)^n$. In other words (recall the division by 2 in the first line!), $a_{2n}=a_{2n+1}=d_n^2/2$, where $d_n$ is the distance from $0$ to the hyperplane  defined by $q(1)=1$ in the space of all odd/even polynomials of degree at most $n$ with $L_2[-1,1]$-norm.
Now, this hyperplane is the affine hull of the Legendre polynomials $ P_i(x)=\frac1{2^ii!}\frac{d^i}{dx^i}(x^2-1)^i$, where $i$ has the same parity as $n$ (since we know that $ P_i(1)=1$ and $P_i(-1)=(-1)^i$). Next, by $\| P_i\|^2=\frac{2}{2i+1}$, our distance is
$$
  \left(\sum_{j\leq n/2}\| P_{n-2j}\|^{-2}\right)^{-1/2}
  =\left(\sum_{j\leq n/2}\frac{2(n-2j)+1}{2}\right)^{-1/2}=\sqrt{\frac4{(n+1)(n+2)}},
$$
attained at
$$
  q_n(x)=\left(\sum_{i\leq n/2}\frac{2}{2(n-2i)+1}\right)^{-1}
    \sum_{i\leq n/2}\frac{2P_{n-2i}(x)}{2(n-2i)+1}.
$$
Thus the answer for the initial question is $a_{2n}=a_{2n+1}=\frac2{(n+1)(n+2)}$ and $p_{2n}=p_{2n+1}=q_n^2$.
Lemma. For every $n$, one of optimal polynomials on $[-1,1]$ is a square of an odd or even polynomial.
Proof. Let $r(x)$ be any polynomial.which is nonnegative on $[-1,1]$ with $r(\pm 1)=1$. We will replace it by some other polynomial which has the required form, has the same (or less) degree and the same values at $\pm 1$, is also nonnegative on $[-1,1]$, and is not worse in the integral sense. Due to the compactness argument, this yields the required result.
Firstly, replacing $r(x)$ by $\frac12(r(x)+r(-x))$, we may assume that $r$ is even (and thus has an even degree $2n$).  Let $\pm c_1,\dots,\pm c_{n}$ be all complex roots of $r$ (regarding multiplicities); then
$$
  r(x)=\prod_{j=1}^n\frac{x^2-c_j^2}{1-c_j^2},
$$
due to $r(\pm 1)=1$.
Now, for all $c_j\notin[-1,1]$ we simultaneously perform the following procedure.
(a) If $c_j$ is real, then we replace $\pm c_j$ by $\pm x_j=0$. Notice that 
$$
  \frac{|x^2-c_j^2|}{|1-c_j^2|}\geq 1\geq \frac{|x^2-0^2|}{|1-0^2|}.
$$
for all $x\in[-1,1]$.
(b) If $c_j$ is non-real, then we choose $x_j\in[-1,1]$ such that $\frac{|c_j-1|}{|c_j+1|}=\frac{|x_j-1|}{|x_j+1|}$. Notice that all complex $z$ with $\frac{|c_j-z|}{|x_j-z|}=\frac{|c_j-1|}{|x_j-1|}$ form a circle passing through $-1$ and $1$, and the segment $[-1,1]$ is inside this circle. Therefore, for every $x\in[-1,1]$ we have
$$
  \frac{|c_j-x|}{|x_j-x|}\geq\frac{|c_j-1|}{|x_j-1|},
$$
thus
$$
  \frac{|x^2-c_j^2|}{|1-c_j^2|}
  =\frac{|c_j-x|}{|c_j-1|}\cdot\frac{|c_j+x|}{|c_j+1|}
  \geq \frac{|x_j-x|}{|x_j-1|}\cdot\frac{|x_j+x|}{|x_j+1|}
  =\frac{|x^2-x_j^2|}{|1-x_j^2|}.
$$
So, we replace $\pm c_j$ and $\pm\bar c_j$ by $\pm x_j$ and $\pm x_j$ (or simply $\pm c_j$ by $\pm x_j$ if $c_j$ is purely imaginary).
After this procedure has been applied, we obtain a new polynomial whose roots are in $[-1,1]$ and have even multiplicities, and its values at $\pm1$ are equal to $1$. So it is a square of some polynomial which is even/odd (since the roots are still split into pairs of opposite numbers). On the other hand, its values at every $x\in[-1,1]$ do not exceed the values of $r$ at the same points, as was showed above. So the obtained polynomial is not worse in the integral sense, as required. The lemma is proved.<|endoftext|>
TITLE: Bounding function of norms in constrained vector space
QUESTION [5 upvotes]: $v$ is a vector of length $n$, where $v_1 = 1$ and every element $v_i \in [0,1]$
$w = \| v \|_1^1 = \sum_i |v_i| = \sum_i v_i$
$x = \| v \|_2^2 = \sum_i |v_i|^2 = \sum_i v_i^2$
$y = \| v \|_3^3 = \sum_i |v_i|^3 = \sum_i v_i^3$
$z = \| v \|_4^4 = \sum_i |v_i|^4 = \sum_i v_i^4$
Can you recommend a strategy for achieving a bound on
$$f = \frac{w z - x y + \sqrt{w^2 z^2 - 6 w x y z + 4 w y^3 + 4 x^3 z - 3 x^2 y ^2 }}{2 (w y - x^2)}$$
for all possible $v$ described above and where the denominator is nonzero: $(w y - x^2) \neq 0$ ? The nonzero denominator implies that the vector $v$ has an element $v_{i\neq 1} \in (0,1)$, which can be seen by assuming all elements are $\in \{0,1\}$ and yielding a contradiction. Thus it can be shown that when the denominator equals zero, the numerator also equals zero (because $w = x = y = z$). When I've simulated numerically (choosing values in $v_{2 \ldots n}$ uniformly in $[0,1]$ a million times), and have observed that $f$ is generally quite close to $1$.
Can you help me find a bound (even a loose bound) $f \in [lower(n), upper(n)]$? I've tried starting from the KKT criteria to define $v$ that maximizes (or, alternatively, minimizes) $f$, but the expression was so messy that Mathematica choked even when using a small vector length $n$.
Optimizing over $w, x, y, z$, which should each be $\in [0, n]$ and where $w \geq x \geq y \geq z$ (because, e.g. $v_i \geq v_i^2 \rightarrow w \geq x$); however, that is still not constrained enough, because it allows that the denominator can be close to zero when the numerator is nonzero (e.g., $w \approx x \approx y << z$. But there is no vector $v$ where $w \approx x \approx y << z$: $\| v \|_p^p$ is decreasing, concave up, meaning that $z - y < y - x$, and thus $z$ cannot jump quickly if $w \approx x \approx y$. But even with this additional constraint on $w,x,y,z$, I have not been able to find a bound on $f$. 
I've also tried letting $\alpha$ denote an element of $v_{i\neq 1} \in (0,1)$ so that $w = 1 + \alpha + \| u \|_1^1$, $y=1 + \alpha^2 + \| u \|_2^2$, etc., where $u$ is a vector of length $n-2$ and all elements in $[0,1]$, and then maximizing (or minimizing) $f$ with respect to $w',x',y',z',\alpha$, where $w' = \| u \|_1^1, x' = \| u \|_2^2, \ldots$, but have also not yet found success.
Do you see appropriate additional constraint(s) that will adequately describe that $w,x,y,z$ behave as norms (without letting $v$ creep back into the problem)? Do you see any place where I can massage $f$ into some larger (or smaller for the lower bound) expression that permits a loose bound? Or, more generally, can you recommend a plan of attack?
In case you're interested, I've come to this problem through statistics; $f$ is an estimate that should ideally be 1, and I'm trying to bound the error of my estimate. Note that this problem is equivalent to
$$ max( roots( nullSpace \left[
\begin{array}{ccc}
\|v\|_1^1 & \|v\|_2^2 & \|v\|_3^3 \\
\|v\|_2^2 & \|v\|_3^3 & \|v\|_4^4 \\
\end{array} \right]
) ) )$$
Thank you for any advice or ideas you have.

REPLY [3 votes]: Experimentally,
$$0.7 <\ f \ \le\ 1.$$
The upper bound is rigorous.  First we show $wy-x^2=\sum_{i<j}v_i v_j (v_i-v_j)^2 \ge 0$.  Then we can use this to simplify the upper bound to
$(wy-x^2)+(xz-y^2)-(wz-xy) = \sum_{i<j} v_i v_j(1-v_i)(1-v_j)(v_i-v_j)^2 \ge 0$.  The extreme case is $v$ of any length of the form
$$v=(1,1/2,\ldots,1/2),\ f=1.$$
The lower bound is more conjectural.  Experimentation suggests looking at $v$'s of the form $(1,a,b,\ldots,b)$.  For these $v$, we can show that the value of $f$ approaches $b$ as $v$ grows longer, so long as $(1-b)^3 \le a(b-a)^3$.  We minimize $b$ subject to that constraint with $a=0.176$, $b=0.704$.  Those values give the apparently extreme case where $v$ is long and of the form
$$v=(1,\,0.176,\,0.704,\,\ldots,\,0.704),\ f=0.704.$$<|endoftext|>
TITLE: Are the following identities well known?
QUESTION [19 upvotes]: $$
x \cdot y = \frac{1}{2 \cdot 2 !} \left( (x + y)^2 - (x - y)^2 \right)
$$
$$
\begin{eqnarray}
x \cdot y \cdot z &=& \frac{1}{2^2 \cdot 3 !} ((x + y + z)^3 - (x + y - z)^3 \nonumber \\
&-& (x - y + z)^3 + (x - y - z)^3 ), 
\end{eqnarray}
$$
$$
\begin{eqnarray}
x \cdot y \cdot z \cdot w &=& \frac{1}{2^3 \cdot 4 !} ( (x + y + z + w)^4 \nonumber \\
&-& (x + y + z - w)^4 - (x + y - z + w)^4 \nonumber \\
&+& (x + y - z - w)^4 - (x - y + z + w)^4 \nonumber \\
&+& (x - y + z - w)^4 + (x - y - z + w)^4 \nonumber \\
&-& (x - y - z - w)^4 ). 
\end{eqnarray}
$$
The identity that rewrites a product of $n$ variables $( n \ge 2$, $n \in \boldsymbol{\mathbb{Z}_+})$ as additions of $n$ th power functions is as given below:
$$
\begin{eqnarray}
& &x_0 \cdots x_1 \cdot x_{n-1} = \frac{1}{2^{n - 1} \cdot n !} \cdot \sum_{j = 0}^{2^{ n - 1} -1} ( - 1 )^{\sum_{m = 1}^{n - 1} \sigma_m(j)} \times ( x_0 + (-1)^{\sigma_1(j)} x_1 + \cdots + (- 1)^{\sigma_{n - 1}(j)} x_{n - 1})^n, \\
& &\sigma_m(j) = r(\left\lfloor \frac{j}{2^{m - 1}}\right\rfloor, 2), m (\ge 1), j (\ge 0) \in \mathbb{Z}_+, \; \left\lfloor x \right\rfloor = \max \{ n \in \mathbb{Z}_+ ; n \le x, x \in \mathbb{R} \}
\end{eqnarray}
$$
where $r(\alpha, \beta)$,  $\alpha, \beta (\ge 1) \in \mathbb{Z}_+$ means the remainder of the division of $\alpha$ by $\beta$ such that $r(\alpha, \beta)$ $=$ $\alpha - \beta \left\lfloor \frac{\alpha}{\beta}\right\rfloor$

REPLY [33 votes]: Although not the exactly the same due to $2^{n-1}$ instead of $2^n$ terms, the OP's formula seems to be essentially the well-known polarization formula for homogeneous polynomials, which is stated as following:

Any polynomial $f$, homogeneous of degree $n$ can be written as $f(x)=H(x,\ldots,x)$ for a specific multilinear form $H$. One has the following polarization formula for $H$ (see also this MO post):
  \begin{equation*}
 H(x_1,\ldots,x_n) = \frac{1}{2^n n!}\sum_{s \in \{\pm 1\}^n}s_1\ldots s_n f\Bigl(\sum\nolimits_{j=1}^n s_jx_j\Bigr)
\end{equation*}

In your case, $f=x^n$, so $H(x_1,\ldots,x_n) = x_1\cdots x_n$ (please note off by 1 indexing).<|endoftext|>
TITLE: Is an elliptic curve that is isomorphic to its Frobenius conjugate defined over $\mathbb{F}_p$?
QUESTION [7 upvotes]: Let $p$ be prime and $q = p^n$. Let $E$ be an elliptic curve over $\mathbb{F}_q$, and let $E^{(p)}$ be the pullback of $E$ by the $p$-power Frobenius of $\mathbb{F}_q$. If $E$ is isomorphic (over $\mathbb{F}_q$) to its Galois conjugate $E^{(p)}$, then does it follow that $E$ is the base change of an elliptic curve over $\mathbb{F}_p$? If so, what is the argument?
Note that similar statements are not true over infinite fields: for instance, $\mathbb{Q}$-curves are isomorphic to all their Galois conjugates, yet need not descend to actual elliptic curves over $\mathbb{Q}$.

REPLY [9 votes]: Almost, but not quite. If $j(E)$ is the $j$-invariant of $E$ then, under your hypothesis, $j(E)=j(E^{(p)})=j(E)^p$, so $j(E) \in \mathbb{F}_p$. Hence $E$ is either defined over $\mathbb{F}_p$ or is a twist of such a curve. If you take the quadratic twist in $\mathbb{F}_{p^2}$ of an elliptic curve defined over $\mathbb{F}_p$ you get a counterexample.<|endoftext|>
TITLE: Conformal changes of metric and geodesics
QUESTION [9 upvotes]: Suppose $(M,g)$ is a Riemannian manifold. Let us assume that $X$ denotes a vector field in this manifold and consider the integral curves of this vector field.
Does there exist a conformal factor $c$ such that locally these integral curves will be geodesics with respect to $ \hat{g} = c g$ ?
Thanks

REPLY [8 votes]: In "A foliation of geodesics is characterized by having no “tangent homologies”, J. of Pure and Applied Algebra Volume 13, Issue 1, 1978, Pages 101–104 D. Sullivan explains a necessary and sufficient condition for a flow to be "taut", i.e all leafs are geodesics. The example of the so-called Reeb flow on the annulus (Gluck example Figure 2.A in Sullivan's paper) is not. In dimension $2$, a vector field $X$ is taut iff there exists a closed $1$-form $w$ such that the product $\langle X , w \rangle$ is everywhere strictly positive.<|endoftext|>
TITLE: Schur polynomial, change of variable
QUESTION [7 upvotes]: Let $k=(k_1,k_2,k_3,k_4)\in \mathbb{N}^4$ and let $s_k(x_1,x_2,x_3,x_4)$ be the Schur polynomial on $GL_4$.
Question 1: If I replace $x_3$ with $x_1$ and $x_4$ with $x_2$, can $s_k(x_1,x_2,x_1,x_2)$ be written as Schur polynomials on $GL_2$ in variables $x_1$ and $x_2$? Is there any clean formula for such reduction?
Question 2: Can $s_k(x_1,x_1^{-1},x_3,x_3^{-1})$ be written in Schur polynomial of $(x_1,x_1^{-1})$ and Schur polynomial of $(x_3,x_3^{-1})$?

REPLY [7 votes]: To expand a bit on the answer of Matt Samuel, let
$x$ and $y$ be two sets of variables (of any
length) and $\lambda$ any partition. Then
  \begin{eqnarray*} s_\lambda(x,y) & = & \sum_{\mu\subseteq\lambda}
  s_\mu(x)s_{\lambda/\mu}(y)\\ & = &
   \sum_{\mu,\nu} c_{\mu\nu}^\lambda s_\mu(x)s_\nu(y).
   \end{eqnarray*}
See for instance equation (7.66) of Enumerative Combinatorics,
vol. 2. Now put $y=x$ to get
  $$ s_\lambda(x,x) = \sum_{\mu,\nu,\rho}c_{\mu\nu}^\lambda
           c_{\mu\nu}^\rho s_\rho(x). $$<|endoftext|>
TITLE: references for faithful orthogonal (or unitary) representation of symmetric groups
QUESTION [6 upvotes]: Let $S_n$ be the symmetric group of $n$ points. I want to find references (or proofs) for the following statement (1).
(1). There does not exist any faithful orthogonal representation 
$$
S_n\longrightarrow O(n-2).
$$
In order to prove (1), I want to use the result that any unitary representation of $S_n$ is over $\mathbb{Q}$ and find references or proofs for the following statement (2).
(2). There does not exist any faithful unitary representation 
$$
S_n\longrightarrow U(n-2).
$$
Where to find references for (1)?
(if cannot, then find references for (2)? )

REPLY [5 votes]: The smallest degree faithful representation of $S_n$ is $n-1$ in characteristic does not divide $n$. It is Theorem 22 of Chapter 19, Section 8 of Y. G. Berkovich, E. M. Zhmud; Characters of finite groups. Part 2.
Translated from the Russian manuscript by P. Shumyatsky, V. Zobina and Berkovich. Translations of Mathematical Monographs, 181. American Mathematical Society, Providence, RI, 1999.
The result is due to Burnside. The proof is inductive.
Dickson proved of the characteristic divides n and n is at least 5, then n-2 is the smallest degree faithful representation.<|endoftext|>
TITLE: What is the algebraic fundamental groups of $SO(n)$ and $Sp(2n)$?
QUESTION [5 upvotes]: Let $k$ be an algebraically closed field of characteristic zero. and let $$\sigma: SL_n(k)\rightarrow SL_n(k)$$
be an involution.
My questions are: 

How could one calculate the fundamental group of $SL_n(k)^\sigma$ ? (the invariant subgroup)
In particular, what is $\pi_1(SO_n(k))$ and $\pi_1(Sp_{2n}(k))$? 

thanks

REPLY [7 votes]: Let $k$ be an algebraically closed field of characteristic 0.
Let $G$ be a connected reductive group over $k$.
The notion of the algebraic fundamental group of $\pi_1(G)$
was introduced in 
my memoir here
and generalized to arbitrary characteristic 
here
and to reductive group schemes here.
Let $G^{\rm ss}=[G,G]$ denote the commutator subgroup of $G$ (which is semisimple).
Let $G^{\rm sc}\twoheadrightarrow G^{\rm ss}$ denote the universal covering of $G^{\rm ss}$ (then $G^{\rm sc}$ is simply connected).
We consider the composite homomorphism
$$ \rho\colon G^{\rm sc} \twoheadrightarrow G^{\rm ss} \hookrightarrow G.$$
Let $T\subset G$ be a maximal torus.
By abuse of notation, we write $T^{\rm sc}$ for the preimage of $T$ in $G^{\rm sc}$.
We have a homomorphism
$$\rho\colon T^{\rm sc}\to T,$$
which in general is neither surjective nor injective.
Let $X_*(T)=\{\chi\colon \mathbf{G}_{m,k}\to T\}$
denote the cocharacter group of $T$.
We obtain a homomorphism
$$ \rho_*\colon X_*(T^{\rm sc})\to X_*(T). $$

Definition. $\pi_1(G)=X_*(T)/\rho_* X_*(T^{\rm sc}). $

This algebraic fundamental group $\pi_1(G)$
does not depend on the choice of $T$ (up to a canonical isomorphism).
Further, if $K$ is an algebraically closed field extension of $k$,
then clearly
$$ X_*(T)=X_*(T\times_k K)$$
and 
$$ \pi_1(G)=\pi_1(G\times_k K).$$
Let $k={\mathbb{C}}$. A cocharacter $\chi\colon \mathbf{G}_{m,{\mathbb{C}}}\to T$
induces a continuous homomorphism ${\mathbb{C}}^\times\to T({\mathbb{C}})$
and a homomorphism of topological fundamental groups
$$ \pi_1^{\mathrm{top}}({\mathbb{C}}^\times)\to\pi_1^{\mathrm{top}}(T({\mathbb{C}}))\to\pi_1^{\mathrm{top}}(G({\mathbb{C}})).$$
By Proposition 11.1 of the memoir, in this way we obtain
a canonical isomorphism
$$ 
\pi_1(G)\overset{\sim}{\to}\mathrm{Hom}\left[\pi_1^{\mathrm{top}}({\mathbb{C}}^\times)
\to\pi_1^{\mathrm{top}}(G({\mathbb{C}}))\right]. 
$$
After we choose one of the two generators of $\pi_*^{\mathrm{top}}({\mathbb{C}}^\times)$,
we obtain a noncanonical isomorphism 
$$ \pi_1(G)\overset{\sim}{\to}\pi_1^{\mathrm{top}}(G({\mathbb{C}})). $$
Now reducing to the case $k={\mathbb{C}}$, one can easily see that Igor Rivin's comment
works over any algebraically closed field $k$ of characteristic 0.
I  show below how to see this without reducing to ${\mathbb{C}}$,
by elaborating on the comment of Matthias Wendt.
First assume that $G$ is a simply connected semisimple group.
Then $G^{\rm sc}=G^{\rm ss}=G$, hence $T^{\rm sc}=T$ and $\pi_1(G)=0$ (as one should expect!).
Since $\mathrm{Sp}_{2n}$ is simply connected,
we conclude that $\pi_1(\mathrm{Sp}_{2n})=0$.
Then assume that $G$ is a torus.
Then $G^{\rm sc}=1$, $T=G$, $T^{\rm sc}=1$, hence $\pi_1(G)=X_*(G)$.
Since $\mathrm{SO}_2$ is a 1-dimensional torus, 
we conclude that $\pi_1(\mathrm{SO}_2)\simeq\mathbb{Z}$.
Now let $G$ be a semisimple group over $k$.
Note that $\ker\rho$ is always a finite abelian group.
By Example 1.6(3) in the memoir, we have a canonical isomorphism
$$\pi_1(G)\cong \mathrm{Hom}(\mathrm{Hom}_k(\ker\rho,\mathbf{G}_{m,k}),\mathbb{Q}/\mathbb{Z}).$$
It is well known that for $\mathrm{SO}_n$ for $n>2$ we have $\ker\rho\simeq\mu_2$,
hence $\pi_1(G)\simeq\mathbb{Z}/2\mathbb{Z}$ (again, as one should expect).
Unfortunately,  Example 1.6(3) was given without proof.
For a proof see, e.g., this preprint, Lemma 15.2.
The point of introducing the algebraic fundamental groups was as follows. 
If $G$ is actually defined over a
nonclosed field $k_0$ such that $k$ is an algebraic closure of $k_0$, 
then the Galois group $\mathrm{Gal}(k/k_0)$ acts on $\pi_1(G)$. 
Then, following an idea of Kottwitz, from the Galois module $\pi_1(G)$ one can compute arithmetic invariants of $G$ over $k_0$,
such as the Galois cohomology $H^1(k_0,G)$ when $k_0$ is a $p$-adic field,
and the Tate-Shafarevich kernel and the defect of weak approximation for $G$ when $k_0$ is a number field.<|endoftext|>
TITLE: Which weighted projective spaces (and their finite quotients) are local complete intersections?
QUESTION [9 upvotes]: Let $G$ be a finite subgroup of $\textrm{Gl}_{n+1}(k)$ (where $k$ is an algebraically closed field). My question is: do there exist examples of $G$ such that the corresponding quotient $P$ of $\mathbb{P}^n$ by $G$ is not isomorphic to  $\mathbb{P}^n$ but yet is locally a set-theoretic complete intersection (if we embed it into  $\mathbb{P}^m$ for some $m>n$)? It seems that the most "popular" quotients of this type are the so-called weighted projective spaces; so what is known for them? Can $P$ be smooth, and still not isomorphic to $\mathbb{P}^n$? 
I am also interested in cases when $P$ can locally be "described by not so many extra equations" (say, by $m-3/4 n$ for $P\subset \mathbb{P}^m$) and when the "non-LSTCI locus" (i.e., the set of points where $m-n$ equations are not sufficient) is of dimension less than $n/4$. 
Upd. I am deeply grateful to Francesco Polizzi for his answer (including examples). So I would like to ask the following: can one describe weighted projective spaces (or more general $P$ as above) that are locally (set-theoretic) complete interestions? Is it possible to calculate the number of "extra local equations" for weighted projective spaces? Any references (and electronic versions of these text) would also be very welcome!

REPLY [4 votes]: Weighted projective spaces (WPS) are toric varieties of Picard number one. In general, any complete $\mathbb{Q}$-factorial toric variety of Picard number one is a "Fake weighted projective space" and it is obtained as a finite abelian quotient of a WPS as follows :
Let $X$ be a $\mathbb{Q}$-factorial complete toric variety of Picard number one and dimension $n$. Then, its fan is composed by exactly $n+1$ simplicial cones of maximal dimension. These cones are generated by $n+1$ primitive lattice vectors $v_0,\ldots,v_n\in N \subseteq N_\mathbb{R}\cong \mathbb{R}^n$. As they are linearly dependent, they satisfy the relation
$$\sum_{i=0}^n a_i v_i = 0. $$
Let $N'\subseteq N$ be the sublattice generated by the vectors $v_0,\ldots, v_n$. Then, $N/N'$ is a finite abelian group (isomorphic to the torsion group of $\operatorname{Cl}(X)$, which is also isomorphic to the fondamental group in codimension 1 of $X$) inducing a quotient map
$$ \mathbb{P}(a_0,\ldots,a_n) \to X $$
Now, if you want to know which of them are local complete intersections you might refer to Nakajima's classification.
I've never seen the computation Nakajima's conditions in the context of WPS (as far as I know), but it should be doable (?).
Some references:

W. Buczynska, "Fake weighted projective spaces". http://arxiv.org/abs/0805.1211
M. Rossi, L. Terracini, "A Q-factorial complete toric variety is a quotient of a poly weighted space". http://arxiv.org/abs/1502.00879v1
H. Nakajima, "Affine torus embeddings which are complete intersections". http://projecteuclid.org/euclid.tmj/1178228538
Lemma 2.7 in D. Dais, C. Haase, G. Zeigler, "All toric local complete intersection singularities admit projective crepant resolutions". http://projecteuclid.org/euclid.tmj/1178207533<|endoftext|>
TITLE: Vector bundles on open (affine) curves
QUESTION [7 upvotes]: It is well-known by Grothendieck (or earlier by Dedekind-Weber) that every vector bundle on $\mathbb{P}^1_k$ for $k$ a field decomposes into a sum of the line bundles $\mathcal{O}(k)$.
As investigated by Hübl and Sun, this fails if we replace $k$ by a discrete valuation ring $D$: There are vector bundles on $\mathbb{P}^1_D$ that do not decompose into line bundles and the whole situation is considerably more complicated than over a field. But if we pass to the open subset $\mathbb{A}^1_D$, life is easy again and every vector bundle on $\mathbb{A}^1_D$ is even trivial. A possible argument is the following: By Horrock's theorem, it is enough to extend the vector bundle on $\mathbb{A}^1_D$ to $\mathbb{P}^1_D$. In general, we can extend a vector bundle defined on an open subset only to a reflexive sheaf on the whole scheme; but on a regular scheme of dimension $2$ all reflexive sheaves are vector bundles.
If we pass to complete curves of genus $\geq 1$, the situation is complicated even over an (algebraically closed) field. Atiyah could classify vector bundles on elliptic curves, but on higher genus curves the classification problem is wild. Replacing the field by a discrete valuation ring certainly makes the situation even worse.
If we pass to an open subset of a complete smooth curve, the situation certainly cannot be worse than that: We still can extend every vector bundle defined on an open subset of a complete smooth curve (defined over a field or a DVR) to the whole curve; on the other hand, many vector bundles on the complete curve might be isomorphic over the given open subset. My question is:

Are there known simplification of the problem of classifying vector bundles on curves if we pass to open (affine) subsets? Is there maybe even a number of points I can delete from a complete smooth curve so that every vector bundle decomposes into line bundles?
Again, I am interested not only in the case of a base field, but also in the case of a base DVR.

REPLY [7 votes]: By a theorem of Serre, see Theorem 1 in this paper, vector bundles over smooth affine curves over fields are direct sum of a line bundle and a trivial bundle. Then the classification problem is given by the line bundle classification. In particular, both questions have positive answers over fields.
For smooth affine curves over a discrete valuation ring, the same result allows to reduce to the classification of rank two bundles. However, it will probably not be possible to remove a fixed number of points to split all vector bundles; only the other way round will work. Obstruction theory in  $\mathbb{A}^1$-homotopy theory will probably allow to get a vector bundle classification in terms of the first two Chern classes, but I would have to check the details (I only know this for surfaces over fields...)<|endoftext|>
TITLE: triviality of Whitney sums of a vector bundle
QUESTION [5 upvotes]: Let a $3$-dimensional subspace $V$ of $\mathbb{R}^4$ be $$V=\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4\mid\sum_{i=1}^4x_i=0\}.$$ The alternating group $A_4$ acts on $V$ by 
$$\sigma(x_1,x_2,x_3,x_4)=(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)} ,x_{\sigma(4)})$$ for any $\sigma\in A_4$.
Since $V$ is linearly isometric to $\mathbb{R}^3$, $SO(3)$ acts on $V$ canonically. Moreover, $A_4$ acts on $V$ as a subgroup of $SO(3)$. Hence we have a covering map
$$
A_4\to SO(3)\to SO(3)/A_4.
$$
Letting $A_4$ act on $\mathbb{R}^4$ by permuting basis and attaching  $\mathbb{R}^4$ as fibres, we have a vector bundle associated to the covering map
$$
\xi: \mathbb{R}^4\to SO(3)\times _{A_4}\mathbb{R}^4\to SO(3)/A_4.
$$
Question: (1). Is $\xi$ a trivial vector bundle? 
(2). Is $\xi^{\oplus 2}$ (2-fold Whitney sum) a trivial vector bundle?
(3). What is the smallest integer $n$ such that $\xi^{\oplus n}$ is a trivial vector bundle?
My attempt: I plan to compute the Stiefel-Whitney class. But I am not sure whether all Stiefel-Whitney classes of $\xi$ are trivial or not?

REPLY [5 votes]: There is a simple general argument showing that $\xi$ is trivial: Take any homogeneous space $G/H$ and any representation $V$ of $G$ and restrict the representation to $H$. Then the homogeneous vector bundle $G\times_H V\to G/P$ is a trivial as a vector bundle. A trivialization can be written down explicitly. It is induced by the map $G\times V\to (G/H)\times V$ defined by $(g,v)\mapsto (gH,g\cdot v)$. This is evidently $H$-invariant and factorizes to an isomorphism $G\times_H V\to (G/H)\times V$ of vector bundles.   
Edit (following the comments by @Sebastian_Goette and @Asghar_Ghorbanpour): This argument can be applied both to the three dimensional representation of $A_4$ called $V$ in the question and to the representation on $\mathbb R^4$ by permutation of coordinates. The latter representation also extends to $SO(3)$ as the direct sum of the representation on $V$ and a trivial representation.<|endoftext|>
TITLE: unwinding the definition of $H_i(KU)$ as a map of spectra $\mathbb{S}^i \to HZ \wedge KU$
QUESTION [5 upvotes]: I asked this on mathstackexchange but didn't get any response (or many views) so I'm asking it here, although clearly it belongs over there.
In the answer to this question on mathoverflow, it says:
"The integral homology group $H_i(KU)$ is the direct limit of 
$$\dots \to H_{2n+i}(BU)\to H_{2n+2+i}(BU)\to\dots,"$$
My question is why? Here is my understanding: Given two spectra $E, X$, I believe the definition of the homology group $E_i(X)$ is 
$$E_i(X):=[\Sigma^i \mathbb{S}, E \wedge X]_{stable}=colim_j[S^{i+j}, (E \wedge X)_j]$$ 
where the square brackets denote homotopy classes of maps. 
In the case at hand, $E=H\mathbb{Z}$ (or maybe $E=\Sigma H\mathbb{Z}$?) is the spectrum whose $n$th space is the Eilenberg-MacLane space $K(\mathbb{Z}, n)$, and $X=KU$ is the spectrum whose $n$ space is either $BU \times \mathbb{Z}$ or $BU$, depending on if $n$ is even or odd, respectively. I have no idea what the $n$th space of $HZ \wedge KU$ is, since I am  confused by smash product of spectra. I'm still learning how to best think of maps between spectra.

REPLY [7 votes]: I'll define a naive prespectrum to be a system of spaces $X_i$ with maps $\Sigma X_i\to X_{i+1}$.  That seems to be what you want to work with, but it is not technically very satisfactory.  However, if $\mathcal{C}$ is one of the fancier categories of spectra, then one can define $FX$ to be the homotopy colimit in $\mathcal{C}$ of the objects $\Sigma^{-i}\Sigma^\infty X_i$, and that will give an object of $\mathcal{C}$ that has the homotopy type that you want.  From this perspective, the real point is that $FX \wedge FY$ is the homotopy colimit of the diagram of objects $\Sigma^{-i-j}\Sigma^\infty X_i\wedge Y_j$, and that colimit can be broken down in three different ways:

Take the colimit over $i$ first, then the colimit over $j$
Take the colimit over $j$ first, then the colimit over $i$
Take the colimit of the diagonal terms $\Sigma^{-2i}\Sigma^\infty X_i\wedge Y_i$.

It is a fairly formal argument to show that these are all the same, and so 
$$ X_nY = \text{colim}_{i,j}\pi_{n+i+j}(X_i\wedge Y_j) = 
 \text{colim}_i \pi_{n+2i}(X_i\wedge Y_i) = 
 \text{colim}_i X_{n+i}Y_i = \text{colim}_i Y_{n+i}X_i.
$$
If you want to work exclusively with naive prespectra, you can use the "handicrafted smash product" 
$$ (X\wedge Y)_{2i} = X_i\wedge Y_i $$
$$ (X\wedge Y)_{2i+1} = X_i\wedge Y_{i+1} $$
It is unpleasant to prove any properties based on this definition, but it is adequate for the calculation that you asked about.<|endoftext|>
TITLE: Lefschetz Principle for semisimplicity
QUESTION [5 upvotes]: I think I can prove the following using the compactness of first order logic and I am wondering what a purely algebraic proof would look like. 


Let $R$ be a unital ring (not necessarily commutative but definitely associative) whose additive group is finitely generated. Suppose that $K\otimes R$ is semisimple for some algebraically closed field of characteristic 0. Then for any algebraically closed field $F$ of characteristic 0 or of characteristic outside of a finite set of primes, $F\otimes R$ is semisimple. 


Of course, if $G$ is a finite group and $R=\mathbb ZG$, then this is true by Maschke's theorem. I realized this is true for finite monoids the other day. I believe the the above statement is true because the semisimplicity of $K\otimes R$ can be written as a statement in the first order theory of algebraically closed fields of characteristic 0 and hence what I want can be obtained from standard compactness arguments in the model theory of algebraically closed fields.  
My question is what does a purely algebraic proof look like?

REPLY [3 votes]: There are really two separate assertions here: one concerns the generic fiber $A = \mathbf{Q} \otimes_{\mathbf{Z}} R$, and the other concerns orders in semisimple $\mathbf{Q}$-algebras.  To clarify the framework for such arguments (e.g., to convey how "characteristic zero" is a red herring and one can make arguments that work in all characteristics uniformly), we pose the assertions more generally:
(i) If $A$ is a nonzero finite-dimensional associative algebra over a field $k$ (e.g., $k = \mathbf{Q}$) and $A_K := K \otimes_{k} A$ is semisimple for one algebraically closed extension $K/k$ then the same holds for all algebraically closed extensions of $k$. (This is called "absolute semisimplicity'' over $k$.)
(ii) If $B$ is a noetherian domain (e.g., $\mathbf{Z}$) with fraction field $k$, $A$ is a nonzero finite-dimensional absolutely semisimple $k$-algebra, and $R$ is a $B$-order in $A$ (i.e., a $B$-subalgebra finitely generated as a $B$-module that spans $A$ as a $k$-vector space) then there exists a nonzero $b \in B$ such that $R[1/b]$ is an Azumaya algebra over $B[1/b]$ (i.e., $R[1/b]$ is finite flat as a $B[1/b]$-module and $R[1/b] \otimes_{B[1/b]} \kappa(\mathfrak{p})$ is absolutely semisimple over $\kappa(\mathfrak{p}) := {\rm{Frac}}(B[1/b]/\mathfrak{p})$ for all primes $\mathfrak{p} \in {\rm{Spec}}(B[1/b])$.
Each of these is a consequence of standard "direct limit" methods that are exhaustively developed and used in a vast array of situations in EGA IV$_3$--IV$_4$, combined with knowledge of the structure of nonzero finite-dimensional semisimple algebras over algebraically closed fields. 
At a basic level this amounts to apt use of the Nullstellensatz and denominator-chasing, but is often much deeper than that because for properties such as flatness and properness and conditions on geometric fibers that are not expressed in terms of "equations" one has to come up with some substantial ideas.
Proof of (i): To save notation, we'll write $V_K$ to denote $V \otimes_k K$ for any $k$-vector space $V$. The center $Z$ of $A$ is the kernel of the system of linear conditions $a \mapsto ae_i - e_i a$ for a $k$-basis $\{e_i\}$ of $A$, so it is clear that $Z_K$ is the center of $A_K$ for any extension field $K/k$.  By hypothesis there is some such algebraically closed $K$ for which $A_K$ is semisimple, and we want to deduce the same for all such $K/k$.  
Fix an algebraic closure $\overline{k}$ of $k$, so there is a $k$-embedding of $\overline{k}$ into any such $K/k$.  Hence, by replacing $k$ with $\overline{k}$ we are reduced to showing that if $k$ is algebraically closed and $K/k$ is an algebraically closed extension then $A_K$ is semisimple if and only if $A$ is semisimple.  (This is an instance of what EGA called "the principle of algebraically closed fields", essentially a glorified version of the Nullstellensatz.) 
We know that a (nonzero) finite-dimensional associative algebra over an algebraically closed field is semisimple if and only if it is a product of matrix algebras, in which case its center is a product of copies of the ground field.  But over an algebraically closed field, the products of finitely many copies of the field are precisely the etale commutative algebras, and the property of being etale for a commutative algebra with finite linear dimension over a field is insensitive to extension of the ground field (consider the non-vanishing or not of the discriminant attached to a choice of basis). Hence, if either $A$ or $A_K$ is semisimple then each of $Z$ and $Z_K$ is etale (over $k$ and $K$ respectively), so we may now assume this etaleness holds. 
The $k$-algebra $Z$ is now a direct product of copies of $k$. Since a direct product of two $k$-algebras is semisimple if and only if each factor is semisimple (and likewise over $K$), we may decompose $A$ as a direct product according to the primitive idempotents of $Z$ to reduce to the case $Z=k$. Then $Z_K=K$, so our task has become proving that $A$ is a matrix algebra if and only if $A_K$ is a matrix algebra.  In either case $\dim_k A = n^2$ is a perfect square, so we may assume this dimension expression holds for some $n$.
Now comes the part with algebro-geometric content. 
Consider the finite type affine $k$-scheme
$I$ representing the functor on $k$-algebras $R \rightsquigarrow {\rm{Isom}}(A_R, {\rm{Mat}}_n(R))$; it is easy to build $I$ by choosing a $k$-basis of $A$ containing 1 and working with "structure constants" for the $k$-algebra structure on $A$ relative to that basis. 
Our task is to show that $I(k)$ is non-empty if and only if $I(K)$ is non-empty.  Note that $I(K) = I_K(K)$. By the Nullstellensatz, since $k$ and $K$ are algebraically closed and $I$ is a $k$-scheme of finite type, this is equivalent to saying that $I$ is non-empty if and only if $I_K$ is non-empty. But this final equivalence is clear (as it expresses the non-vanishing of a coordinate ring, and non-vanishing of an algebra over a field is insensitive to extension of the ground field). 
[The use of $I$ could be expressed entirely ring-theoretically, so its true power only becomes  apparent in a wider context; e.g. one can use deformation theory to show that $I$ is actually $k$-smooth, so it yields that any absolutely semisimple algebra over a field becomes a matrix algebra over a finite etale extension of the field, a classical fact which is nonetheless understood in a more conceptual manner via the smoothness of $I$ and thereby adapts to a useful technique in the relative setting over rings and schemes.]
Proof of (ii): This will be all about "denominator-chasing" (the Nullstellensatz hiding in the way we will invoke (i) below). 
Since $k = {\rm{Frac}}(B)$ and $A = R_k$ is a free $k$-module, by denominator-chasing with a choice of $k$-basis we can find a nonzero $b_0 \in B$ such that the subset $R[1/b_0] \subset R_k$ is spanned by a $B[1/b_0]$-linearly independent set, which is to say $R[1/b_0]$ is $B[1/b_0]$-free. We may replace $B$ with $B[1/b_0]$ and $R$ with $R[1/b_0]$ to reduce to the case that $R$ is $B$-free. 
By (i), it now suffices to find a nonzero $b \in B$ and a $B[1/b]$-algebra $C$ such that ${\rm{Spec}}(C) \rightarrow {\rm{Spec}}(B[1/b])$ is surjective and $R_{C} := R \otimes_B C$ is a product of matrix algebras over $B'$. There is certainly a finite extension $k'/k$ such that $A_{k'}$ is a product of matrix algebras. 
Let $B' \subset k'$ be a module-finite $B$-algebra such that $B'_k = k'$ (easy to build by denominator-chasing with a $k$-basis of $k'$). Since $B'_k$ is $k$-free, there is a nonzero $b_1 \in B$ such that $B'[1/b_1]$ is $B[1/b_1]$-free.  By localizing at $b_1$ we may then assume $B'$ if $B$-free, so ${\rm{Spec}}(B') \rightarrow {\rm{Spec}}(B)$ is open.  Hence, we may replace $(k, B, R)$ with $(k', B', R_{B'})$ to reduce to the case
that $A=R_k$ is a direct product of matrix algebras and $R$ is $B$-free. 
Fix an isomorphism of $k$-algebras 
$$k \otimes_B R= A  \simeq \prod {\rm{Mat}}_{n_i}(k) =
k \otimes_B \prod {\rm{Mat}}_{n_i}(B).$$
Since $k = {\rm{Frac}}(B)$ and both $R$ and $\prod  {\rm{Mat}}_{n_i}(B)$ are free as $B$-modules, by expressing the above $k$-algebra isomorphism
relative to $B$-bases of each  we can denominator-chase to find a nonzero $b \in B$ such that the isomprphism arises from a $B[1/b]$-algebra isomorphism
$$R[1/b] \simeq \prod {\rm{Mat}}_{n_i}(B[1/b]).$$<|endoftext|>
TITLE: Forcings that are not equivalent to Levy collapse
QUESTION [12 upvotes]: Assume GCH and that $\kappa$ is a regular uncountable cardinal.  Let $\mathbb{P}$ be a separative, $<\kappa$-directed closed, nowhere trivial, $\kappa^+$-cc poset of size $\kappa^+$.  Must $\mathbb{P}$ be forcing equivalent to $\text{Col}(\kappa, < \kappa^+)$?  (which I believe is equivalent to adding $\kappa^+$ many Cohen subsets of $\kappa$).  If not, then what if we also assume that the $<\kappa$ directed closure of $\mathbb{P}$ is witnessed by greatest lower bounds?  (i.e. that $\mathbb{P}$ satisfies a strong version of the ``well-met" requirement).
This question is somewhat related to the following earlier question on MO:"Resembling the Levy collapse".  There it was shown that for $\delta > \kappa$ where $\delta$ is inaccessible, the Silver collapse to turn $\delta$ into $\kappa^+$ has all the relevant properties but is not equivalent to Levy collapse $\text{Col}(\kappa, < \delta)$.  I don't think the analogue of the Silver collapse between $\kappa$ and $\kappa^+$ is $\kappa^+$-cc, however, so I don't think that provides a counterexample to this question.

REPLY [8 votes]: First note that in the special case $\kappa = \omega_1$, countably closed implies countably directed-closed.
Answer 1:  No.  In any model of CH, there is a countably closed, $\omega_2$-c.c. forcing that is inequivalent to $\mathbb P = \text{Add}(\omega_1,\omega_2)$.
Proof: Let $\mathbb Q$ be Jensen's partial order for adding a Kurepa tree.  Conditions are $(t,b)$, where $t$ is a countable tree of successor height $\alpha$, $b$ is a countable subset of $\mathcal P(\omega_1)$, and for all $x \in b$, $x \cap \alpha \subseteq t$.  Ordering is $(t_1,b_1) \leq (t_0,b_0)$ when $t_1$ end-extends $t_0$ and $b_0 \subseteq b_1$.  It is not hard to show that $\mathbb Q$ is countably closed and $\omega_2$-c.c. under CH.  Any two conditions with the same first coordinate are compatible.  The union of the first coordinates in any generic is a Kurepa tree $T$ by a density argument.
Note that every subset of $\omega_1$ added by $\mathbb P$ is added by some $\omega_1$-sized regular suborder.  If $\mathbb Q$ were forcing equivalent to $\mathbb P$, then there would be some $\omega_1$-sized regular subalgebra $\mathbb R$ of $ro(\mathbb Q)$ that adds the generic Kurepa tree $T$.  $\mathbb Q$ forces an $\omega_2$-size subset of $\mathcal P(\omega_1)^V$ to be contained in the set of branches of $T$.  Thus suppose $G * H$ is $\mathbb R * \mathbb Q / \mathbb R$-generic. But in $V[G]$ we can compute whether $(p,q) \in H$, since we just determine whether $p \subseteq T$ and whether all $x \in q$ are branches of $T$.  Thus $V[G * H] = V[G]$, and $\mathbb Q$ must have a dense set of size $\omega_1$.  But this is false:  If $\{ (t_i,b_i) : i < \omega_1 \} \subseteq \mathbb Q$, then pick $x \notin \bigcup b_i$.  No $(t_i,b_i)$ can force $x$ to be a branch of $T$ since we can extend $t$ to rule out $x$ as a branch.
Note that $\mathbb Q$ doesn't have infima to countable chains since we have a lot of choices for the "top level" of the countable tree.  But...
Answer 2:  Not necessarily, even for $\omega_1$-closed with infima.  Let $V$ be a model of GCH, and let $G \subseteq \text{Col}(\omega_1,\omega_2)$ be generic.  $\omega_3^V = \omega_2^{V[G]}$.  I claim that in $V[G]$, $\mathbb P = \text{Col}(\omega_1,<\omega_2)^{V[G]}$ and $\mathbb Q = \text{Col}(\omega_2,<\omega_3)^{V}$ are both countably closed with infima and both $\omega_2$-c.c., but are not forcing equivalent.
The closure follows from the fact that $V$ and $V[G]$ have the same countable sequences.  The $\omega_2$-c.c. for $\mathbb Q$ follows from the fact that $\mathbb Q \times \text{Col}(\omega_1,\omega_2)$ is $\omega_3$-c.c. in $V$.
Next, one can prove a similar fact to Mohammad's answer.  In $V[G]$,
(1) $\Vdash_{\mathbb P} \exists A \in [\omega_2]^{\omega_2}$ such that $A$ does not contain any $\omega_1$-sized set from $V[G]$.
(2) $\Vdash_{\mathbb Q} \exists A \in [\omega_2]^{\omega_2}$ such that for all $X \in [\omega_2]^{\omega_2} \cap V[G]$, there is $y \in [X]^{\omega_1} \cap V[G]$ such that $y \cap A = \emptyset$.
The justification for (1) is the same as for Mohammad's answer.  To see (2), let $\dot A$ be a name for a code of the generic $H$ for $\mathbb Q$.  In $V$, let $\dot B$ be a name for a code for $G \times H$.  If $X \in ([\omega_2]^{\omega_2})^{V[G]}$, let $\dot X$ be a name for it in $V$.  Let $p \in \text{Col}(\omega_1,\omega_2)$ be arbitrary, and let $\{ p_i : i < \omega_3 \} \subseteq \text{Col}(\omega_1,\omega_2) \restriction p$ be such that $p_i$ decides the $i^{th}$ member of $\dot X$.  Let $p'$ be such that $p_i = p'$ for $\omega_3$ many $i$, giving a large set $Y$ such that $p' \Vdash \check Y \subseteq \dot X$.  Let $q \in \mathbb Q$ be arbitrary.  Interpreting $Y$ as a subset of $\mathbb Q$, find $Y' \subseteq Y$ of size $\omega_3$ such that $q \nleq y$ for all $y \in Y'$.  For each such $y$, choose $q_y \leq q$ such that $q_y \perp y$.  Now form a $\Delta$-system among the $q_y$, and let $z$ be an infimum of a set $S$ of $\omega_1$ many of them.  Then $z$ forces that the generic $H$ misses $y$ for $q_y \in S$.  As $p$ and $q$ were arbitrary, we are done.
Likewise, we can show that $\mathbb P$ forces the negation of the the property under the forcing sign in (2), and vice versa.  Some details are in section 2.1.2 of my thesis.  I may post more later.<|endoftext|>
TITLE: Spin Structures for Quaternionic-Kaehler and Hyper-Kaehler Manifolds
QUESTION [7 upvotes]: As is well-known (see Friedrich's book for example) every Kähler manifold is spin (or at least spin$^c$) and the Dirac is given (up to a twist) by $\partial + \partial^*$. What happens in the quaternionic-Kaehler and hyper-Kähler cases? Are they spin$^c$, does the Dirac admit a nice description? I'm particularly interested in the case of Wolf spaces (Quaternion-Kähler symmetric space).

REPLY [2 votes]: It is a result of Salamon that $8n$-dimensional quaternion-Kähler manifolds are spin.<|endoftext|>
TITLE: Questions on topologies on space of Radon measures
QUESTION [7 upvotes]: Consider the space $C_c(\mathbb{R})$ of continuous real-valued functions on $\mathbb{R}$ equipped with the inductive limit topology by $C_c(\mathbb{R}) = \bigcup_{n \in \mathbb{N}} C_c(\mathbb{R}, K_n)$ where $K_n$ is some compact exhaustion of $\mathbb{R}$ by compact sets and $C_c(\mathbb{R}, K_n)$ is the Banach space with uniform norm. In particular, $C_c(\mathbb{R})$ is an LB-space. Let $M := M(\mathbb{R})$ denote the topological dual $C_c'(\mathbb{R})$. Then $M$ can be identified with the space of all Radon measures (bounded and unbounded, e.g. the Lebesgue measure belongs to $M$) - for details, see Bourbaki: Integration, Chapter III. Every measure $\mu \in M$ has the Jordan decomposition $\mu = \mu^+ - \mu^-$ into two positive measures $\mu^+$ and $\mu^-$. Let $M^+ \subseteq M$ denote the subset of all positive measures. Then we have a surjective map $s : M^+ \times M^+ \to M$, $(\mu, \nu) \mapsto \mu - \nu$.
One can equip $M$ with the weak-* topology $\tau_v$ which Bourbaki calls the vague topology.
Some properties of $(M, \tau_v)$:

$M$ is by duality the projective limit of the weak-* duals $C_c'(\mathbb{R}, K_n)$.
$M$ is Hausdorff locally convex and in particular, addition and scalar multiplication are continuous
$M$ is not first-countable and thus not metrizable
$M$ is quasi-complete but not complete
The map $s$ is continuous ($M^+$ carries the subspace topology (which is Polish))
$M$ is Souslin (as the image of the Polish space $M^+ \times M^+$ under the continuous map $s$)
$M$ is separable (since Souslin)

I don't know of the following:

Is $M$ sequential or even Fréchet-Urysohn? (This is a specialization of the question here.)
Is $M$ Lusin? (For an uncertain proof idea see below.)

Bourbaki studies also other topologies like compact convergence or strictly compact convergence.
But one has also the map $s$ from above and can equip $M$ with the final topology $\tau_f$ induced by $s$
which is then finer than $\tau_v$
and turns $M$ into a quotient space. In particular, $(M, \tau_f)$ is sequential.
Does anyone know whether this topology on $M$ was already studied before and has some nice properties
(or missing some good properties)?
In particular, is $\tau_f$ a vector space topology (addition and scalar multiplication continuous)?

Proof for $(M, \tau_v)$ is Lusin:
I'm trying to transfer the ideas of Trèves, "Topological Vector Spaces", p. 556 (in particular Proposition A.9)
which is based on the Borel Graph Theorem.
$(M, \tau_v)$ is the projective limit of the countable family $C_c'(\mathbb{R}, K_n)$ equipped with the weak-* topology $\tau_{w*}$.
The space $C_c'(\mathbb{R}, K_n)$ equipped with strong norm topology is Banach but it is not separable, thus not Polish
and thus we can't directly say whether $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ is Lusin.
Now fix $n$ and set $E := C_c(\mathbb{R}, K_n)$.
Since $E$ is separable Banach we have a countable basis $U_k$ of open nbhds. of $0$ and
the dual $E'$ is the union of all the polars $U_k^0$.
Each $U_k^0$ is weakly compact (Banach-Alaoglu) and thus weakly closed equicontinuous.
By Trèves Exc. 32.9 it follows that $U_k^0$ equipped with the weak-* topology is metrizable compact, thus Polish
and therefore Lusin. A countable union of Lusin subsets of a Hausdorff space is Lusin and thus $E'$ with the weak-* topology is Lusin.
In other words, $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ is Lusin.
Moreover, the product of the countable family $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ of Lusin spaces is Lusin
and the projective limit $M$ of all the $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ is a closed subset of the product
thus a Borel subset of the product and thus Lusin.
The only transfer of the proof is that I replaced "Souslin space" by "Lusin space". So is there some major gap which I did oversee?

REPLY [3 votes]: This is a comment rather than an answer but it will be too long.  It mentions some tools that might be of use.  To begin with, there are advantages in using the context of a general locally compact space $X$ rather than the special case of the real line, at least in the beginning.

In the case of a compact space $K$, there are three natural topologies on $M(K)$, the weak topology induced by $C(K)$, the norm topology and a third less familiar one, the bounded weak star topology which is the finest locally convex topology which agres with the weak topology on the unit ball.  This has several advantages over the other two.  In particular, it is complete as the weak topology is not.  However, it has the same convergent sequences and coincides with it on bounded sets, importantly on the set of probability measures.
As you point out, the dual space in the general case is a projective limit of the duals of analogues of those above, namely of Banach spaces of those functions with supports in a fixed compact subset.  It is natural to consider it as a (complete) locally convex space with the corresponding limit structure.  This will not be identical to the weak topology you use but will be closely related.  
The space of measure will then be naturally isomorphic to a closed subspace of a product of spaces of measures with compact suport, as mentioned above.  In the case of a $\sigma$-compact space this will be a countable product.
Hence it will inherit topological properties of the space of measures on compacta (in your case, even compact metric spaces) which are stable under (countable) products and closed subspaces.
How ever, in the case where $X$ is paracompct (and so, in particular, $\sigma$-compact), it is even a complemented subspace and thus a continuous image of the product.  This follows from results of de Wilde and Keim on projective limits with partitions of unity which can be found in Manuscripta Mathematica, vols. 5 and 10.

I have no expertise in descriptive topology and can't say whether this helps in your problem but it might be worth your while to have a look.<|endoftext|>
TITLE: Doob Martingale: Where is the catch?
QUESTION [9 upvotes]: I am working on a research problem in uncertainty propagation that involves sums of possibly dependent random variables with bounded sets of support. 
I am  attempting to use the method of bounded differences as an amateur. In general, most presentations of this method promise to handle arbitrary dependences between random variables but all worked out examples involve independent random variables. 
In particular, let $$X_0, \ldots, X_n$$ be a sequence of random variables wherein 
$$ X_0 = \left\{ \begin{array}{cc} 1 & \mbox{with probability}\ \frac{1}{2} \\
-1 & \mbox{with probability}\ \frac{1}{2} \\
\end{array}\right. \,,$$
and $X_{i+1} = X_i$ for $i \geq 0$. In other words, $X_i$s for $i \geq 1$ are "very highly correlated" with $X_0$. :-)
We are looking at $\sum_{i=0}^n X_i $ and with probability $\frac{1}{2}$ each it can be either $\pm (n+1)$.
Let us apply the method of bounded differences to the summation  with the Doob martingale sequence (I am blindly following the presentation in [1] here :-)
$$ Y_i = \mathbb{E}\left( \sum_{j=0}^n X_j\ |\ X_0, \ldots, X_{i-1} \right)\,.$$
We can show that $$|Y_{i+1} - Y_i| \leq 1\,.$$
Applying Azuma's theorem gives us the "concentration result"
$$ \mathbb{E}( Y_n  \geq t ) \leq \exp\left( \frac{ -2t^2}{ n } \right) $$
Clearly I am wrong! This is the same as the Chernoff Hoeffding bound for the iid case! I am wondering where I went wrong.
Very grateful for your help!
[1] Dubhashi and Panconesi, Concentration of measure for the analysis of randomized algorithms.

Resolution: it seems to be resolved thanks to the comments below from @alpoge. The mistake here is $Y_0 = \mathbb{E}(\sum X_i) = 0$. But $Y_1 = Y_2= \cdots = Y_n = (n+1) X_0$ and therefore, when applying Azuma's theorem, the denominator will be something like 
$$\mathbb{P}( Y_n  \geq t) \leq \exp\left( \frac{-t^2}{2(n+1)^2} \right) $$
Now the bounds obtained are much weaker and consistent with having $Y_n$ being $n+1$ with probability $\frac{1}{2}$ and $-(n+1)$ with probability $\frac{1}{2}$.
The lesson is then when the method of bounded difference is applied to sums of dependent random variables, care should be taken to calculate $\mathbb{E}(X_j | X_0,\ldots, X_{j-1})$ carefully when $X_{j}$ could be dependent on $X_0,\ldots, X_{j-1}$. It is tempting to equate this to  $\mathbb{E}(X_j)$.

REPLY [3 votes]: What if we work it out?  All of $X_j$ are equal to $X_0$, in particular measurable with respect to $X_0$.  So
$$
\sum_{i=0}^n X_i = (n+1)X_0,
\\
Y_i = (n+1)X_0\qquad\text{for all } i
\\
Y_{i+1}-Y_i = 0
$$
Certainly bounded differences...
What other hypotheses are there?<|endoftext|>
TITLE: History of spectral methods to the study of real analytic $GL_2$-Eisenstein series
QUESTION [6 upvotes]: I'm trying to sort out the history of spectral methods in the study of real analytic $GL_2$-Eisenstein series. From what I read so far, I would say that the subject was really kicked off by the seminal work of Selberg who developped a a very broad theory which applies to any weakly symmetric Riemannian spaces and not just to symmetric spaces coming from $GL_2$.
A very good reference that I found on the subject is the book of Tomio Kubota entitled "Elementary theory of Eisenstein series". I found that book extremely well written and very accessible for somebody who wants to learn the subject. I have noticed that this book is not so often cited (as I think it should be) as compared for example to the more recent book of Iwaniec entitled "Spectral methods of Automorphic forms". In fact, I find it very curious that Kubota's book is not even cited once in Iwaniec's book, since after reading the two books, I found a lot of overlaps, especially on the topic surrounding $GL_2$ real analytic Eisenstein series.
For a paper that I'm currently writting, I'm planning to cite Kubota's work and give him the proper credits, but before doing so I would like to make sure that I'm not missing an earlier reference on $GL_2$-real analytic Eisenstein series (besides of course the paramount work of Selberg, Langlands, Harish-Chandra, Gelfand, Pyatetskii-Shapiro who worked in much greater generality).

REPLY [10 votes]: R. Rankin's 1939 paper giving a non-trivial estimate on Ramanujan's $\tau$ function used the "real-analytic Eisenstein series" for $SL_2(\mathbb Z)$, at least. Selberg's related paper just-slightly later seemed to express the same awareness for such cases, as opposed to the more general situations treated in the 1950's. In some informal remarks I saw elsewhere, Rankin credited his advisor, Ingham, with knowing how to meromorphically continue such Eisenstein series.
The idea for congruence subgroups of $SL_2(\mathbb Z)$ was recapitulated (at least) in one of R. Godement's articles in the 1965/66 Boulder Conference proceedings (AMS Proc. Symp. IX), which was where I saw it first.
The appendix in Langlands' SLN 544 in which he stitches together several $GL_2$ cases to get minimal-parabolic Eisenstein series for $GL_n$ and $Sp_n$ certainly presumes that the $GL_2$ case is well-known. Apparently these notes were written up by about 1967 even though they were not public until 1976.
My own (then-naive) impression by the mid-1970s was that the meromorphic continuation of $GL_2$ things was a cliche. Selberg's ideas from the 1950's seemed to suffice for rank-one situations without subtle data on the (possibly non-abelian) Levi components. Lectures of G. Shimura at Princeton in the mid-1970's proved things, including meromorphic continuation of Eisenstein series, by looking at Fourier expansions of all sorts of $GL_2$ Eisenstein series, related to applications of Maass-Shimura operators and "nearly holomorphic" automorphic forms on $GL_2$ over totally real fields.
(So my own 1989 book on Hilbert modular forms gave such an argument over totally real fields, maybe mentioning Rankin and Selberg from 1939...)
So, although it might have been reasonable for (e.g.) Iwaniec to mention Kubota, I think it is not accurate to worry that a deserved citation was inappropriately denied. Rather, already by the time of Kubota's book those ideas were perhaps considered "general knowledge", so by the 1990's (and Iwaniec' book) the ideas would have been even more "classical".
Edit (in response to Hugo Chapdelaine's clarifying comments): Selberg's spectral approach seems/seemed limited, if one insists on a full spectral decomposition, to situations where no cuspidal-data or other subtleties enter. My colleague D. Hejhal confirmed (from Selberg's unpublished papers) my suspicion that Selberg had not considered (to whatever purpose) cuspidal-data Eisenstein series, nor corresponding parts of a spectral decomposition for higher-rank groups. Oppositely, so far as I can tell Harish-Chandra's relatively early results on automorphic forms on general semi-simple groups (e.g., 1959) were aimed primarily at the cuspidal part of the spectrum. Especially the residual part of the spectrum, not in the functional analysis sense but in the sense of being a residue of Eisenstein series, is the most ... obscure... part of Langlands' 544.
Yes, the Lax-Phillips 1976 argument that not only do spaces of cuspforms discretely decompose (e.g., for a relevant Laplacian), but even "pseudocuspforms", was a striking point, although I think their literal result was considerably obscured by its context, and a certain unclearness about how much was a physically-based heuristic and how much was really proven/provable. Nevertheless, yes, Colin de Verdiere's 1981 "new proof" of meromorphic continuation of Eisenstein series showed the peculiar potential in that fact, even though (to my recollection) many people were confused about the precise nature of "Friedrichs extensions", and how it could be that certain truncated Eisenstein series (palpably not smooth) could become eigenfunctions for some variant of an elliptic operator. (I recall wrangling between P. Cohen and Selberg at Stanford over such points c. 1980, and at the time I was certainly baffled.)
My perception at the time of the reception of Colin de Verdiere's result was that, mostly, since it reproved a true, known result, was just an "announcement" (rather than detailed paper), and used a tricky feature of unbounded self-adjoint operators, people did not engage with it as much as they might have. Nevertheless, W. Mueller showed in the 1990s that this approach certainly immediately applies to rank-one cases, and H. Jacquet at least sufficiently assured C. Moeglin and J.-L. Waldspurger that the analogous argument would succeed generally, so that they outline a similar argument in their book.
Of course, in all these cases the crux of the matter is that some artfully arranged operator has compact resolvent, therefore discrete spectrum, therefore various things have meromorphic continuations... Thus, in contrast to Roelcke's meromorphic continuation up to $\Re(s)=1/2$, which follows from very general spectral considerations, getting beyond that line when it parametrizes the continuous spectrum is non-trivial.
Yes, in fact, by now, I like the Colin de Verdiere argument better than other forms, since it gives more meaning to poles. In slight contrast, the J. Bernstein reformulation of Selberg's "third" proof (the latter from Hejhal's trace formula volumes) is a general-enough idea that, ironically, it seems not to obviously give sharp information about the case-at-hand, namely, automorphic forms. 
One should also mention some experiments done by A. Venkov not only about spectral expansions for automorphic forms, in the usual sense, but also some Schroedinger operators or Dirac operators for automorphic forms...
I think the most serious subtlety is still correct identification of residual (=residue of Eisenstein series) spectrum. Harish-Chandra's SLN 62 from 1968 explicitly doesn't attempt this, and somewhere there's a quote from Langlands that he doesn't remember how that part of his SLN 544 went. For $GL_n$, Jacquet's conjecture from about 1983 (Maryland conference) proven by Moeglin-Waldspurger (1989) is the only (to my knowledge, at this date) systematic result for any family of groups. Of course, various methods can prove that some Eisenstein series have no poles in various cones/half-planes, but that's not the question.
For that matter, residues of maximal-proper-parabolic Eisenstein series are relatively tractable, via Maass-Selberg relations and their implications (as Borel already did in his little $SL_2(\mathbb R)$ book), but iterated residues entail complications that are not trivially dispatched.<|endoftext|>
TITLE: Conditions on the fusion data of symmetric fusion category
QUESTION [7 upvotes]: We know that every symmetric fusion category (SFC) gives rise to data
$N^{ij}_k$ that describe the fusion of simple objects:
$i\times j = N^{ij}_k k$, and the data $\theta_i =\pm 1$ that describe the twist of simple objects.
My question is what are the conditions on the data $N^{ij}_k,\theta_i$ such that the data correspond to a SFC?
Any conditions beyond those for fusion category are welcome. (One may try to classify finite groups via a classification of SFCs.)
One also can ask a related (and more general) question: what are the conditions on the data $N^{ij}_k,\theta_i$ such that the data correspond to a pre-modular braided fusion category?

REPLY [5 votes]: The information in the $\theta$'s is very weak, for example if a FC admits a symmetric structure with $\theta_i=-1$ for at least one $i$, it also admits a symmetric structure with $\theta\equiv 1$. Even, it is possible to have several non-equivalent symmetric structures with $\theta\equiv 1$ over a fixed fusion category. 
On the other hand, since every SFC is equivalent to the representation category of a (super)-group a necessary condition over the fusion algebra $N_k^{ij}$ is that Frobenius-Perron dimension of every simple object is an integer. 
Let us call a fusion algebra without non-trivial fusion subalgebra a simple fusion algebra. Since quotient groups of a finite group $G$ are in biyective correspondence with fusion subalgebras of $K_0(Rep(G))$, a condition on the fusion data of symmetric fusion category is that  every  simple fusion subalgebra have to be realized (in a unique way) as the fusion algebra of the SFC of representation of (a unique) simple group (the fusion data can be read from the character table). Also, you can read from your fusion data if it corresponds to a nilpotent group, see http://arxiv.org/abs/math/0610726 in this case using group cohomology is possible to find obstructions to the existence of a SFC that realize your fusion data.<|endoftext|>
TITLE: Enriching categories and equivalences
QUESTION [10 upvotes]: Let $\mathcal{C}$ and $\mathcal{D}$ be  two equivalent categories. Furthermore, assume  $\mathcal{C}$ is enriched over a monoidal category $(\mathcal{M}, \otimes)$. Can one use the equivalence to enrich $\mathcal{D}$  over $(\mathcal{M}, \otimes)$?

REPLY [9 votes]: I think most category theorists would answer "yes, obviously", and not bother to write down a proof.  But presumably that isn't sufficiently convincing, since you ask the question, so let me try to make it a bit more explicit with some big words.  (-:
An $M$-enriched category with set of objects $A$ is equivalently a lax functor $A_{ch}\to B M$, where $A_{ch}$ is the chaotic category with $A$ as objects (exactly one morphism between any two objects) and $B M$ is the 1-object bicategory associated to $M$.  A function $f:A\to B$ induces a functor $f_{ch}: A_{ch} \to B_{ch}$, hence by precomposition a map $f^*$ from $M$-categories with object set $B$ to $M$-categories with object set $A$, such that $f^*\underline{B}(a_1,a_2) = \underline{B}(f(a_1),f(a_2))$ for any $M$-category $\underline{B}$ with object set $B$.
Now suppose that $C$ is an $M$-category with object set $B$, that $D$ is an ordinary category with object set $A$, and $f:D \to C_0$ is a fully faithful and essentially surjective functor, where $C_0$ is the underlying ordinary category of $C$.  Then $f$ induces a function $A\to B$ and thereby an induced $M$-category $f^*C$ with object set $A$.  Moreover, the action of $f$ on homs consists of bijections $D(x,y) \cong C_0(f(x),f(y))$ that respect composition and identities; but by construction of $f^*C$ these are equivalently $D(x,y) \cong (f^*C)_0(x,y)$, giving an isomorphism of categories $D\cong (f^*C)_0$, i.e. an enrichment of $D$ over $M$.
Note that nowhere did we use the essential surjectivity of $f$, so actually we've proven a bit more.<|endoftext|>
TITLE: Another question on Heath-Brown's "Prime twins and Siegel zeros"
QUESTION [5 upvotes]: With a graduate student, I'm going through the paper (Proc. London Math. Soc. (3) 47 (1983), no. 2, 193–224.)  
Here's the background and notation.  
We have a quadratic character $\chi$ modulo $q$, with Siegel zero $\beta_0$.  $\eta=((1-\beta_0)\log q)^{-1}$, so $3\le \eta\ll q$ is known.  Let $L=\log q$.  We take
$$
q^{250}\le x\le q^{500}.
$$
Lemma 3 (stated on p. 198) says
$$
\sum_{p<x,\chi(p)=1}\frac{\log(p)}{p}\ll L\log(\eta)^{-1/2}
$$
Heath-Brown says "This is not necessarily the best bound of its type, but suffices for our purposes."  I think the proof he gives is flawed.  I know results like this are found elsewhere.  My question is
Can this proof be fixed?  If not, can you provide a reference for a more robust proof?
(Side note:  In a recent blog post on this same paper, Tao proves his Lemma 5 which is similar, summing instead $1/p$, up to a $o(1)$ error.  "For more precise estimates on the $o(1)$ error, see the paper of Heath-Brown (particularly Lemma 3).")
The proof starts in Section 4 on p. 206.  The overall structure is to write 
$$
\frac{L^\prime}{L}(s,\chi)-\frac{L^\prime}{L}(s^\prime,\chi)
$$
as a both sum over zeros and as a sum over primes.  Here $s=1+L^{-1}$ and $s^\prime=1+aL^{-1}$, where $a$ is to be chosen later.  The zeros side comes down to estimating
$$
aL^{-1}\sum_{\rho\ne\beta_0}|\rho-1|^{-2}.
$$
The zeros $\rho$ at height $\ge 1$ give no trouble.  Heath-Brown quotes Prachar to estimate the number of zeros in the disk $|s-1|\le r$ as
$$
\ll 1+r\log q
$$ 
for $r\le 2$.  To count zeros below height $1$ it would make sense to take $r=\sqrt{2}$ here.
With $r_0$ the closest non-Siegel zero to $1$, he quotes the Deuring-Heilbronn phenomenon to say $r_0\gg L^{-1}\log\eta$.
One would expect then the bound to involve the $aL^{-1}$ term, the number of zeros at height below $1$, and the worst case for the term being summed, namely $r_0^{-2}\ll L^2\log(\eta)^{-2}$, i.e. a bound of
$$
aL(1+\sqrt{2}L)\log(\eta)^{-2}\ll aL^2\log(\eta)^{-2}.
$$
Heath-Brown uses instead the count of zeros inside the circle of radius $r_0$ (which makes no sense, by definition of $r_0$) and gets a better estimate $\ll aL/\log\eta$.  The correct (I think) estimate does not suffice for the error bound the lemma claims.

REPLY [4 votes]: "Those oft are stratagems which errors seem, 
Nor is it Homer nods, but we that Dream."
Heath-Brown's proof is fine.  It needs just one more line of explanation.
Note that 
$$
\sum_{\substack{\rho \neq \beta \\ |\gamma| \le 1}} \frac{1}{|\rho -1|^2} \ll 
\int_{r_0}^{2} \# \{ \rho \neq \beta: |\rho -1| \le x \} \frac{dx}{x^3}, 
$$
and by the quoted line from Prachar this is 
$$ 
\ll \int_{r_0}^2 (1+x \log q) \frac{dx}{x^3} \ll r_0^{-2} + (\log q) r_0^{-1}, 
$$ 
which is what Heath-Brown writes.<|endoftext|>
TITLE: Conditions on the hierarchy for Thurston's hyperbolization theorem
QUESTION [8 upvotes]: From my understanding the proof of Thurston's hyperbolization theorem for Haken $3$--manifolds consists of cutting the manifold along a hierarchy (collection of incompressible, $\partial$-incompressible surfaces) to obtain a collection of $3$--balls. A hyperbolic structure is put on the $3$--balls, and then a bootstrapping argument results in re-gluing the hierarchy components so as to get a hyperbolic structure on the original manifold.
I certainly do not know the details for the above argument, but was wondering if there are any conditions on the hierarchy surfaces needed for the above argument to go through? In other words, are we required to choose a particular hierarchy to ensure that the re-gluing can be done consistently (or can one assume that any hierarchy will do ie- only the existence of a hierarchy is needed)? If a hierarchy with certain properties is needed, what are the conditions on the hierarchy?

REPLY [5 votes]: Yes, there are conditions; basically you want to maintain the hypothesis of not having incompressible annuli at each stage. A really good reference for this is Morgan's essay, "On Thurston's uniformization theorem for three-dimensional manifolds", in Morgan, John W.; Bass, Hyman, The Smith conjecture (New York, 1979), Pure Appl. Math. 112, Boston, MA: Academic Press, pp. 37–125.  I'd suggest looking there.
(I edited out the "tori"; this will be automatic if the original manifold is atoroidal as it should be.)<|endoftext|>
TITLE: Descent theorems for fundamental groups and groupoids?
QUESTION [11 upvotes]: Grothendieck in his 1984 "Esquisse d'un programme" (Section 2) wrote (English translation):
" ..,people still obstinately persist,  when calculating with fundamental groups, in fixing a single base point, instead of cleverly choosing a whole packet of points which is invariant under  the symmetries of the situation, which thus get lost on the way.  In certain situations (such as descent theorems for fundamental groups `a la van  Kampen) it is much more elegant,  even indispensable for understanding  something, to work with fundamental groupoids with respect to a suitable    packet of base points,.."
Question: Does anyone have any reference to relevant work on "descent theorems for fundamental groups"? 
Relevant to this question is this mathoverflow  discussion on several base points.

REPLY [6 votes]: I think I can now reasonably point to the book 
Bourbaki, Topologie Algébrique: Chapitres 1 à 4 (Elements de Mathématique) 7 Apr 2016
which does use ideas of descent and the fundamental groupoid both for the van Kampen theorem and for discrete groups acting properly on spaces. They call the fundamental groupoid the "Poincaré groupoid",  though I think its topological use goes back only to Reidemeister's 1932 book. However, despite the comment of Grothendieck, they do not use the fundamental groupoid on a set of base points. 
Their Theorem 3 on p. 419 seems to overlap with the work on orbit groupoids of Chapter 11 of Topology and Groupoids, which gives the application to  the symmetric square of a space.  
Comments welcome!   
Aug 21, 2016  I should mention the paper 
"Van  Kampen   theorems   for  categories   of  covering 
morphisms   in  lextensive   categories"   R. Brown,  G.   Janelidze, J. Pure  Applied  Algebra  I 19 (1997)  255-283. (pdf)
This considers the whole fundamental groupoid, not "many base points", but does use descent  notions in general situations.<|endoftext|>
TITLE: Tiling a square with rectangles
QUESTION [31 upvotes]: Is it possible to completely tile a square with different rectangles of integer sides but all with the same area?
The original problem, not requiring integer sides for rectangles, was proposed by Joe Fendel in Gary Antonick´s New York Times Monday puzzle column (June 13, 2013). A tiling with seven rectangles was subsequently provided.
Here is an image from Nick Baxter's solution (see Fendel's article link above), but with lengths normalized such that the outer square has side $1$. (Each "v" represents $\sqrt{19}$). It is  easy to check that each rectangle has area $\dfrac17$.

REPLY [13 votes]: The Nick Baxter solution is actually Blanche's Dissection, published in 1971. 
I've outlined a general solution method at my Commuunity post Blanche Dissections. As a proof of concept, here are 16 dissections of a square into equal-area non-congruent rectangles:  

So far, none of these gives a rational solution.  But many polyhedral graphs will give a Blanche-style solution.  Since we have an infinite number of polyhedral graphs to choose from, there is very likely an integral solution. I just need to devote a bit more programming and computer power. If a solution doesn't pop out, then at least we'll have a few million more non-integral solutions.
Relax the problem to allow non-congruent integer-sided rectangles of different areas, but minimize the difference between the largest and smallest rectangle. This is the Mondrian Art Puzzle.  Here are best solutions for 10x10 to 17x17. 

We could also attack this from the numerical side.  Some good candidate squares are the following:  
Side 2520 into 35 rects of area 181440 (35 available).
Side 4080 into 34 rects of area 489600 (34 available).
Side 3960 into 33 rects of area 475200 (40 available).
Side 3780 into 30 rects of area 476280 (32 available).
Side 2520 into 30 rects of area 211680 (35 available).
Side 3120 into 26 rects of area 374400 (32 available).
Side 2340 into 26 rects of area 210600 (27 available).
Side 1560 into 26 rects of area 93600 (26 available).
Side 2760 into 23 rects of area 331200 (26 available).
Side 3300 into 22 rects of area 495000 (22 available).
Side 3080 into 22 rects of area 431200 (23 available).
Side 2772 into 22 rects of area 349272 (25 available).
Side 2640 into 22 rects of area 316800 (31 available).
Side 2310 into 22 rects of area 242550 (22 available).
Side 1980 into 22 rects of area 178200 (26 available).
Side 1848 into 22 rects of area 155232 (24 available).
Side 1320 into 22 rects of area 79200 (25 available).
Side 3150 into 21 rects of area 472500 (21 available).
Side 3024 into 21 rects of area 435456 (21 available).
Side 2940 into 21 rects of area 411600 (23 available).
Side 2772 into 21 rects of area 365904 (23 available).
Side 2730 into 21 rects of area 354900 (21 available).
Side 2520 into 21 rects of area 302400 (37 available).
Side 2310 into 21 rects of area 254100 (21 available).
Side 1890 into 21 rects of area 170100 (22 available).
Side 1680 into 21 rects of area 134400 (23 available).
Side 1260 into 21 rects of area 75600 (28 available).
Side 2280 into 19 rects of area 273600 (25 available).
Side 1140 into 19 rects of area 68400 (19 available).
Side 2856 into 17 rects of area 479808 (23 available).  
The 3960x3960 square seems especially promising.
For 9 rectangles, square sizes {360, 504, 540, 630, 720, 756, 840, 990, 1008, 1080, 1170, 1188, 1260, 1386, 1404, 1440, 1512, 1530, 1584, 1620, 1638, 1680, 1800, 1872, 1890, 1980, 2016, 2100} are promising.
For 10 rectangles, square sizes {360, 420, 600, 660, 720, 840, 900, 1050, 1080, 1200, 1260, 1320, 1400, 1440, 1560, 1680, 1800, 1890, 1980, 2040, 2100, 2160} are promising.
For 11 rectangles, square sizes {330, 660, 792, 924, 990, 1320, 1386, 1540, 1584, 1650, 1716, 1848, 1980, 2244, 2310} are promising.<|endoftext|>
TITLE: 6j symbols of SU(4) at level 4
QUESTION [7 upvotes]: Does anybody know of a reference that gives the (quantum) 6j symbols of SU(4) at level 4?
Alternatively, I know the S-matrix and the fusion rules, in the form
$a \times b = \sum_i N^{ab}_{c_i} c_i$
Is there a simple way to compute all quantum 6j symbols from these?

REPLY [3 votes]: In answer to your second question, no, there is not a simple way to compute $6j$-symbols from the fusion rules and the $S$-matrix.<|endoftext|>
TITLE: Which journals publish short notes in discrete mathematics?
QUESTION [12 upvotes]: The journal Discrete Mathematics contains a lot of short notes (i.e., less than 7 journal pages). What are some other journals that publish short notes in discrete mathematics? I've looked at other journals, but most of them seem to contain primarily long papers.

REPLY [4 votes]: Discrete Mathematics Letters http://www.dmlett.com/ publishes only short notes (that is, the papers consisting of maximum 8 published pages).<|endoftext|>
TITLE: Fundamental solution of Discrete Laplace in the plane
QUESTION [6 upvotes]: We consider a discretization of the Laplace operator on $\mathbb Z^2$, https://en.wikipedia.org/wiki/Discrete_Laplace_operator
Then, it is natural to consider its fundamental solution $u$, i.e. $|u(x)|\leq C \ln|x|,\Delta u = 1$ at $(0,0)$ and $\Delta u=0$ elsewhere. I am sure that somewhere it is proven that the discrete derivatives of $u$ behave exactly as we expect them : $|u'(x)|\leq \frac{1}{|x|}, |u''(x)|\leq c\frac{1}{|x|^2}$.
But I can not find anything like that in the Internet. Could you provide me with a reference, please? Any information about $c$?

REPLY [6 votes]: For nearest neighbor Laplacian, you can find a short self-contained proof of the formula $$u(x)=\frac1{2\pi}\log |x|+c+O\left(\frac1{|x|^2}\right)$$ at these lecture notes, pages 6-7. It is a simple exercise in Laplace's method to work out further coefficients of the asymptotic expansion, if necessary. The proof there is not the shortest possible; the easiest way would be to compare the double integral formula defining $G_0$ on page 6 with its counterpart for continuous Laplacian. But then you do not get the beautiful formula (4.4) along the way.
"Random Walk: a modern introduction" by Lawler and Limic does the same in much greater generality (not necessarily nearest neighbour), using local CLT with error bounds.
Historically, the first sources usually cited for the above formula are
Stohr, A., [50]: "Uber einige lineare partielle Differenzengleichungen mit konstanten Koefizienten," I, II, and Ill, Mathematische Nachrichten 3 (1950),
or even
W. H. McCrea and F. J. W. Whipple, Random paths in two and three dimensions, Proc. Roy. Soc. Edinburgh 60 (1940),
although the latter apparently proves a slightly weaker result (I have no access to it at the moment to check).<|endoftext|>
TITLE: State of the art in the expected length of the Longest Increasing Subsequence of a random permutation
QUESTION [9 upvotes]: I have been reading about the topic motivated by a problem I read that asked for the first three digits of the sum of the LIS lengths in all permutations of length $n$. It is easy to see that we are really interested in the expected value of the LIS length in a random permutation of the first $n$ positive integers. A paper by Mike Phulsuksombati states
$$l_n=2\sqrt{n}+cn^{1/6}+o(n^{1/6})$$
Where $c\approx-1.771088$ as a known asymptotic.
Questions: Is there a better asymptotic known? Is there a polynomial or logartihmic(in terms of $n$) time algorithm for finding $l_n$ or approximating it to a desired precision?

REPLY [13 votes]: @user61318 and @JosephORourke, thanks for the advertisement for my book.
@chubakueono, as far as I know the answer to your questions are no and no. Pages 148-149 in my book have the state of the art (essentially the formula you wrote, along with some additional background) and the relevant references. However, in a broader sense much more is understood about the asymptotic formula for $l_n$ and where it comes from. The main important points as relates to your question are:

The asymptotic formula for $l_n$ is closely related to a more detailed distributional limit for the random variable $L(\sigma_n)$ defined as the maximal length of an increasing subsequence in a uniformly random permutation $\sigma_n$ of order $n$. The result, known as the Baik-Deift-Johansson theorem, says that for any $t\in\mathbb{R}$,
$$ \mathbb{P}\left(
\frac{L(\sigma_n)-2\sqrt{n}}{n^{1/6}} \le t 
\right) \to F_2(t) \quad \textrm{as }n\to\infty,
$$
where $F_2(t)$ is a certain probability distribution known as the Tracy-Widom distribution. $F_2(t)$ can be expressed explicitly as
$$F_2(t) = \exp\left(
-\int_x^\infty (x-t)q(x)^2\,dx
\right),
$$
where $q(x)$ is the unique solution to the Painleve II ODE
$$y''(x)=2y(x)^3+xy(x)$$
with certain asymptotic boundary conditions.
The constant $c$ in the asymptotic formula for $l_n=\mathbb{E}(L(\sigma_n))$ is simply the expected value of this distribution, that is
$$ c = \int_{-\infty}^{\infty} x F_2'(x)\,dx. $$
Proving a rate of convergence estimate in the asymptotic formula for $l_n$ would be extremely interesting (and, I suspect, difficult) and is an open problem as far as I know.<|endoftext|>
TITLE: Frobenius elements in infinite extensions
QUESTION [6 upvotes]: Let $K$ be a number field, $\bar K$ an algebraic closure and $G$ the associated absolute Galois group. How can I define the Frobenius elements of $G$ or at least their conjugacy class?
I know how these elements arise in finite Galois extensions, and I know that $G$ is the inverse limit among this finite extensions, however I don't know how to put together these fact to obtain an answer.

REPLY [11 votes]: [This answer is an elaboration on what KConrad mentions in his comments below, posted while I was first writing this.]
At finite level one has a perfectly good theory of Frobenius elements attached to maximal ideals upstairs with ambiguity of inertia subgroups (so no ambiguity in the unramified case, but don't get too attached to that).  This goes similarly at "infinite level" and one can in fact think about Frobenius elements in a "naive way" for $\overline{K}/K$ by working directly with the integral closure $O_{\overline{K}}$ of $O_K$ in $\overline{K}$. 
That integral closure is quite gigantic (and not noetherian) but is really not as frightening as it may seem because it is the directed union of the rings of integers $O_{K'}$ of the subextensions $K'/K$ of finite degree, with those $O_{K'}$ quite concrete things.  If you think about it, a prime ideal of $O_{\overline{K}}$ "is" nothing more or less than a compatible system of prime ideals $P_{K'} \subset O_{K'}$ (meaning $P_{K''} \cap O_{K'} = P_{K'}$ whenever $K' \subset K''$). 
Let's see how this plays out with maximal ideals in $O_{\overline{K}}$ and $O_K$, and then with that experience under our belts we'll see that a "naive" approach to Frobenius elements works very well. 
For a maximal ideal $M$ of $O_K$, the set $S_{K'}$ of maximal ideals in $O_{K'}$ over $M$ is a non-empty finite set, and we have the natural maps $S_{K''} \rightarrow S_{K'}$ defined by $\mathfrak{m} \mapsto \mathfrak{m} \cap O_{K'}$ (even surjective, but doesn't matter). 
In this way $\{S_{K'}\}$ is an inverse system of non-empty finite sets, so the inverse limit $\lim S_{K'}$ is non-empty (no need for Zorn stuff here since these are all countable inverse systems).  
But this inverse limit is precisely "the same" as a prime ideal $P$ of $O_{\overline{K}}$ satisfying $P \cap O_K = M$. For any such $P$, each prime ideal $P \cap O_{K'}$ is maximal (i.e., a nonzero prime ideal). Thus, $O_{\overline{K}}/P$ is a field, as it is the direct limit of the quotients $O_{K'}/(P \cap O_{K'})$ that are all fields, so $P$ is maximal.  The upshot is that every maximal ideal $M$ in $O_K$ lifts to a maximal ideal $\overline{M}$ of $O_{\overline{K}}$.  
Conversely, every maximal ideal of $O_{\overline{K}}$ is nonzero and hence its prime ideal intersection with some $O_{K'}$ is nonzero and thus maximal, whence its intersection with $O_K$ is maximal.  So we conclude that $\overline{M} \mapsto \overline{M} \cap O_K$ is a surjection from the set of maximal ideals of $O_{\overline{K}}$ onto the set of maximal ideals of $O_K$.  There is an evident action of ${\rm{Gal}}(\overline{K}/K)$ on the set of such $\overline{M}$ which permutes those over a common $M$. Let's next see that this Galois action is transitive.
If $\overline{M}$ and $\overline{N}$ are two maximal ideals of $O_{\overline{K}}$ over the same $M$ then the same goes for
the maximal ideals $\overline{M} \cap O_{K'}$ and $\overline{N} \cap O_{K'}$ for each finite $K'/K$, especially those which are Galois. But the usual story for number fields assures that the finite set $T_{K'}$ of elements of ${\rm{Gal}}(K'/K)$ carrying $\overline{M} \cap O_{K'}$ over to $\overline{N} \cap O_{K'}$ is non-empty, and for $K''/K'$ with $K''/K$ also Galois the natural map ${\rm{Gal}}(K''/K) \twoheadrightarrow {\rm{Gal}}(K'/K)$ clearly carries $T_{K''}$ into $T_{K'}$ (surjectively, but doesn't matter).  So once again the inverse limit $$\lim T_{K'} \subset \lim {\rm{Gal}}(K'/K) = {\rm{Gal}}(\overline{K}/K)$$
is non-empty.  But if you think about it, this inverse limit is precisely the set of $g \in {\rm{Gal}}(\overline{K}/K)$ carrying $\overline{M}$ over to $\overline{N}$.  Hence, the fibers of the map $\overline{M} \mapsto \overline{M} \cap O_K$ of sets of maximal ideals consist precisely of ${\rm{Gal}}(\overline{K}/K)$-orbits on maximal ideals of $O_{\overline{K}}$.  In that sense, all $\overline{M}$ over a given $M$ are "created equal" from the perspective of the Galois-action.
Now finally (!) we come to your question.  For each maximal ideal $M$ of $O_K$ we choose a maximal ideal $\overline{M}$ of $O_{\overline{K}}$ over it, and define the maximal ideal $\overline{M}_{K'} := \overline{M} \cap O_{K'}$ of $O_{K'}$ over $M$ for each $K'/K$. Consider the stabilizer subgroup ("decomposition group")
$$D(\overline{M}|M) = \{g \in {\rm{Gal}}(\overline{K}/K)\,|\,g(\overline{M}) = \overline{M}\}.$$
The above transitivity shows that each of the natural surjections
${\rm{Gal}}(\overline{K}/K) \twoheadrightarrow {\rm{Gal}}(K'/K)$ 
with finite Galois $K'/K$ carries $D(\overline{M}|M)$ onto the decomposition group $D(\overline{M}_{K'}|M)$. But for each $K''/K'$ that is Galois over $K$ the natural map $D(\overline{M}_{K''}|M) \rightarrow D(\overline{M}_{K'}|M)$ is surjective by the usual theory, so in particular for the non-empty finite subset $\phi_{K'} \subset D(\overline{M}_{K'}|M)$ of "Frobenius elements" we get a natural surjection $\phi_{K''} \rightarrow \phi_{K'}$ induced by the restriction map ${\rm{Gal}}(K''/K) \twoheadrightarrow {\rm{Gal}}(K'/K)$.
Finally, consider the inverse limit $\lim \phi_{K'} \subset {\rm{Gal}}(\overline{K}/K)$.  This is non-empty and consists of precisely those elements $g \in {\rm{Gal}}(\overline{K}/K)$ such that $g(\overline{M}) = \overline{M}$ and the induced map on the field $O_{\overline{K}}/\overline{M}$ has restriction to each finite subfield $O_{K'}/\overline{M}_{K'}$ is the $q$-power map for $q = \#(O_K/M)$.  But that says $g$ induces the $q$-power map on $O_{\overline{K}}/\overline{M}$, so such $g$ certainly deserve to be called "Frobenius elements" (at $\overline{M}$).  Note that given one such $g$, all others are related to it through multiplication against the subgroup $I(\overline{M}|M) = \lim I(\overline{M}_{K'}|M)$ ("inertia subgroup") of elements of $D(\overline{M}|M)$ that induces the identity on the residue field $O_{\overline{K}}/\overline{M}$. 
So that finally is the definition of a Frobenius element in ${\rm{Gal}}(\overline{K}/K)$ relative to $\overline{M}$ over $M$: it preserves $\overline{M}$ and induces the map $t \mapsto t^{\#(O_K/M)}$ on the residue field $O_{\overline{K}}/\overline{M}$. The above shows that such elements exist, and for any $\overline{N}$ over $M$ conjugation against any choice (which we saw does exist!) of element of ${\rm{Gal}}(\overline{K}/K)$ carrying $\overline{M}$ to $\overline{N}$ clearly carries $D(\overline{M}|M)$ over to $D(\overline{N}|M)$ and carries Frobenius elements over to Frobenius elements. (With a tiny bit of work one sees that $O_{\overline{K}}/\overline{M}$ is algebraically closed, so it is even an algebraic closure of $O_K/M$, but this was
not needed above; indeed, $\overline{K}$ everywhere above could have been
an arbitrary Galois algebraic extension of $K$.) What could be better than that?<|endoftext|>
TITLE: Proof-theoretic ordinals after liberalizing induction to $RCA_0$
QUESTION [9 upvotes]: This is kind of a follow-up to this question.
For a class $\Gamma$ of second-order formulas (here either $\Sigma_n^0$ or $\Sigma_n^1$), let $X\Gamma$ be a formal theory consisting of $RCA_0$ together with induction schema for formulas in $\Gamma$. In particular, $X\Sigma_0^1$ adds arithmetical induction, and let $RCA$ be $RCA_0$ with induction schema for all formulas.

What are the proof-theoretic ordinals of $X\Gamma$ and $RCA$?

The question linked at the top asks whether these theories have any additional strength comprehension-wise (OP there considers sets $\Sigma_n$, which I suspect meant $\Sigma_n^0$ there). I have asked under one answer what the PTOs of these theories are, and I was told that $X\Sigma_n^0$ has a PTO at least $\omega$ stacked $n+1$ times (which makes sense, because we are allowing "stacked quantifies" in provable part of PA), and I was advised to ask this as a separate question, so here it is.
I highly suspect that this was considered before, but it's hard to find any results.
Thanks in advance.

REPLY [2 votes]: I don't have complete answer but I think that my remarks still may be useful.
Let me consider theory $\mathsf{NT}$ which extends $\mathsf{PRA}$ by one unary predicate $X$ and axiom scheme of induction in extended language and weaker theories $\mathsf{NT}_n$ in the same language but with induction scheme restricted to for $\Sigma^0_n(X)$-formulas (here $\Sigma^0_0(X)$-formulas are all the formulas from the language of $\mathsf{NT}$, where all quantifiers are bounded). The proof-theoretic ordinal $|\mathsf{NT}|$of $\mathsf{NT}$ (i.e. the limit of well-ordered order types of primitive-recursive binary relation for which $\mathsf{NT}$ proves transfinite induction for $X$) is equal to $\varepsilon_0$ (essentially it is ordinal analysis of $\mathsf{PA}$ in the terms used by W. Pohlers in his "Proof Theory. The First Step into Impredicativity", Chapter 7; Pohlers considered $\mathsf{NT}$ in the language with countable family of free unary predicates but it can't influence proof-theoretic ordinal).
Clearly,
$$|\mathsf{X}\Sigma^0_n|\ge |\mathsf{NT}_n|,$$
$$|\mathsf{RCA}|\ge |\mathsf{X}\Sigma^1_n|\ge |\mathsf{NT}|.$$
Let me consider the interpretation of $\mathsf{RCA}_0$ in $\mathsf{NT}_1$ that encodes $\mathsf{RCA}_0$-sets by $\Delta^0_1(X)$-sets. This interpretation consists of $\Pi^0_2(X)$ definition of set of codes for $\mathsf{RCA}_0$-sets and $\Delta^0_1(X)$-interpretation of $\in$. Clearly, the same syntactical translation gives an interpretation of $\mathsf{X}\Sigma^0_n$ in $\mathsf{NT}_n$ and an interpretation of $\mathsf{RCA}$ in $\mathsf{NT}$.
This interpretations gives the following bounds:
$$|\mathsf{X}\Sigma^0_n|\le |\mathsf{NT}_n|,$$
$$|\mathsf{RCA}|\le |\mathsf{NT}|.$$
Thus $|\mathsf{RCA}|=|\mathsf{X}\Sigma^1_n|=\varepsilon_0$. Also, $|\mathsf{X}\Sigma^0_n|= |\mathsf{NT}_n|$. There is natural conjecture that $|\mathsf{NT}_n|=\omega_{n+1}$, where $\omega_m$ is the tower of $\omega$-exponentiation of the height $m$. But I don't know whether it have been proved somewhere; Pohlers in Exercise 7.3.19(f) of the mentioned book asked the question about the upper bound for $|\mathsf{NT}_n|$ without definite conjecture about the answer.<|endoftext|>
TITLE: What polytope is this? Bounded sums with choice of coefficients
QUESTION [7 upvotes]: Let $b\gt a\gt 0$ be constants.  Define $P_n(a,b)$ to be the set of all $(x_1,\ldots,x_n)\in\mathbb{R}^n$ satisfying
$$ |c_1 x_1 + \cdots + c_n x_n| \le 1$$
for every choice of $c_1,\ldots,c_n\in\lbrace a,b\rbrace$.
This is a finite convex polytope and I wonder if it has been studied and perhaps has a name.
The vertices are rather few, $n(n+1)$ in total, consisting of all permutations of the points
$$ \Bigl(-\frac{1}{b}, 0,\ldots,0\Bigr),~~
  \Bigl(\frac{1}{b}, 0,\ldots,0\Bigr),~~
  \Bigl(-\frac{1}{b-a},\frac{1}{b-a},0,\ldots,0\Bigr). $$

REPLY [8 votes]: I've seen some papers where the convex hull of the roots in a root lattice is called the root polytope of that root system. In type $A_n$ this is the convex hull in $\mathbb R^{n+1}$ of the vectors $e_i-e_j, 1\le i,j\le n+1$. These lie in the hyperplane $\sum_{i=1}^{n+1} x_i=0$. If you project to $\mathbb R^n$ by $x_i'=x_i+\frac{a}{b}x_{n+1}$ you get your polytope.
For example, the f-vector and some other stuff about the root polytope of type $A_n$ was done in "Root polytopes and growth series of root lattices" by Ardila, Beck, Hosten, Pfeifle, and Seashore.<|endoftext|>
TITLE: A condition for a sequence defined by a recurrence relationship to all be integers
QUESTION [11 upvotes]: I am interested in a specific sequence $\{a_n \}$ defined by a simple recurrence relationship: $$a_n = \frac {a_{n-1} ^2 +c} {b} $$ where $b,c\in \mathbb{Z}$. I want to find all $b,c$ such that there exists some integer $a_1$ such that $\{a_n\} $ are all integers.
I think that this would be a well-known problem, but I have trouble finding it. Even if this is not well-known, I cannot find a way to solve this question.
For example, if $b=3$ and $c=1$, it is fairly trivial that there exists no such $a_1 $. if $b=3$ and $c=2$, we can let $a_1 = 1 $ to make all $\{a_n \}$ be integers. However, I cannot easily work this out even in a single case, such as $b=3, c=17$. Furthermore, how can I find all $(b,c)$ such that there exists some $a_1$ such that $a_n \in \mathbb Z$ for all $n\in \mathbb{N}$?
*I posted this on SEMath, but I did not get any answers nor references. https://math.stackexchange.com/questions/1470398/a-condition-for-a-sequence-defined-by-a-recurrence-relationship-to-all-be-intege

REPLY [3 votes]: $\let\eps\varepsilon$This is not an answer, but just an attempt to summarize flaming (at least inspired by myself, sorry) appeared in the comments to the answer by @Joe Silverman. I would kindly ask Joe's permission to delete those comments, if all of them are correctly reflected here. It is not that convenient to walk through all of them... 
On the other hand, I may easily delete this if it disturbes someone.
So, following Joe's approach, we fix an odd $p\mid b$ (assuming $b$ is odd and square-free, and $b\nmid c$) and search for $p$-adic integers $a$ such that for $a_1=a$ the whole sequence consists of $p$-adic integers. Surely, we assume that $-c\equiv \mu^2\pmod{p}$ for some integer $\mu$. Then it is easy to see that if $(a_n)$ is as desired, then $a_n\equiv\pm \mu\pmod p$ for all $n$.
We claim that for every sequence $\eps=(\eps_1,\dots,\eps_n)$ with $\eps_i\in\{-1,1\}$, there exists a unique residue $r_{\eps}$ such that for $a_1\equiv r\pmod{p^n}$ all $a_1,\dots,a_n$ are ($p$-adic) integers with $a_i\equiv\eps_i\mu\pmod p$. The base $n=1$ is clear. For the step, apply the inductive hypothesis to $a_2,\dots,a_{n+1}$; we find that $a_2\equiv r_{\bar\eps}\pmod{p^n}$, where $\bar\eps=(\eps_2,\dots,\eps_{n+1})$. Now, from $a_1^2=ba_2-c$ we obtain two possible residues for $a_1\pmod {p^{n+1}}$ corresponding to $a_1\equiv \pm \mu\pmod p$, as required. The claim is proved.
Thus, for every infinite sequence $\eps_1,\eps_2,\dots$ there exists a sequence $r_{\eps_1},r_{\eps_1,\eps_2},\dots$ converging to some $p$-adic integer $R_{\eps_1,\eps_2,\dots}$ (obviously, these residues agree).
Now again comes the hard part. We need to find out whether one of these $p$-adic numbers is actually an integer. 
An easy case is when the sequence $\eps_1,\eps_2,\dots$ is periodic. Then, due to the uniqueness, the sequence starting from $R_\eps$ is periodic as well.  This sequence may indeed be constant, e.g. 
$$
  a_n=\frac{a_{n-1}^2+b-1}b, \quad a_1=1.
$$ 
It may also be cyclic with longer period, e.g., 
$$
  a_n=\frac{a_{n-1}^2-(1+b+b^2)}b, \quad a_1=1, \quad a_2=-(b+1).
$$ 
I am almost sure that arbitrarily long periods may appear; but for every fixed choice for $b$ and $c$ they can be found easily: since $\frac{x^2+c}b>x$ for all sufficiently large $x$, it suffices to check only the sequences consistinc of small numbers.
The case when $\eps$ is an aperiodic sequence seems to be, eh, much harder (and, perhaps, not approachable by such elementary methods). But still, some heuristics: all the $R_\eps$ form some Cantor set in $p$-adic numbers, and it has zero measure, so it is quite improbable that one of these numbers is integer. Nevertheless, it is not clear why it cannot happen (if it really cannot).
Notice also that in any case, some of the $R_\eps$ are algebraic (two of them satisfies $x^2-bx+c=0$, and some generate longer cycles), while some are transcendent (since there are continuum many of them).<|endoftext|>
TITLE: How does a mathematician choose on which problem to work?
QUESTION [50 upvotes]: Main question:

How does a mathematician choose on which problem to work?

An example approach to framing one's answer:

What is a mathematical problem - big or small - that you solved or are working on, how did you choose this problem, and why did/do you continue to work on it? 

An earlier incarnation of this question asked also for problem sources, and about finding mathematics questions (and engaging in mathematical research) more generally; hopefully this revised version focuses it a bit more (though already posted answers should be understood as responses to the previous version).
As one example of what a retrospective look at a problem (problems) might look like, see the note (How the Upper Bound Conjecture Was Proved) provided by Richard Stanley in a comment below.

REPLY [21 votes]: I've found that "problem creation by analogy" can be very helpful. In grad school I learned about elliptic curves from many great sources (courses of Mazur and Serre, grad student friends too numerous to enumerate, survey articles by Cassels and by Tate, books by Lang,...) and started working on an elliptic curve problem posed by Lang. And every few weeks I'd go to the library and skim the titles and abstracts of lots of journals. (Nowadays, the ArXiv can serve as a similar source.) And I noticed an article with a new improvement on something called Lehmer's conjecture, which I'd never heard of, but it had something to do with heights of algebraic numbers. So I thought, well, algebraic numbers (more properly, the multiplicative group $\bar{\mathbb Q}^*$) are analogous to points on elliptic curves. So I translated Lehmer's conjecture to elliptic curves and proved a result. (Admittedly, it was rather weak, and Masser and other people had stronger results via different methods; but over the years, I've returned to the problem and have papers with Marc Hindry and with Matt Baker.) Fast-forward a few years, I was at a conference at Union College, where the inimitable John Milnor gave a beautiful colloquium-level talk on complex dynamics. I knew nothing about the subject, but for the first half, which he devoted to a survey of the classical theory (Fatou, Julia, etc.), almost every concept that he mentioned seemed to have an elliptic curve (or arithmetic geometry) analog. Thus orbits of points via iteration of rational maps looked analogous to the Mordell-Weil group of an elliptic curve, points with finite orbits were the torsion points, one could look at integer points in orbits as being analogous to integer points on curves (Siegel's theorem), etc. Pursuing that analogy has lead me, and many other people, to a host of fascinating problems, including 10 PhD theses that my students have written in this relatively new field of arithmetic dynamics.<|endoftext|>
TITLE: Is every $C_0$ semigroup on a Hilbert space automatically a $C_0$ group on a larger space?
QUESTION [7 upvotes]: Let $\{T(t),t\ge 0\}$ be a $C_0$ semigroup on a Hilbert space $X$, does that exist a larger Hilbert space $Y$ such that $X\subset Y$, and $T(t)$ extend to a  $C_0$ group $T'(t)$ (so $t<0$ make sense now) on $Y$?  
Edit: Here "extend" means that $(T(t)x, y)=(T'(t)x, y)$ for all $x, y\in X$, which is weaker than its usual meaning. If the semigroup $T(t)$ is a contraction, then the conclusion is true, see Theorem 8.1 (p.29) in "Harmonic analysis of operators on Hilbert spaces" written by  Sz. Nagy, C. Foias,  H. Bercovici and L. Kérchy.

REPLY [5 votes]: An operator $A$ is a group generator if and only if both $A$ and $-A$ are semigroup generators. So under your assumptions, if we denote by $\tilde{A}$ the generator of the extension $\{\tilde{T}(t),t\ge 0\}$ of $\{T(t),t\ge 0\}$ to $Y$, then $-\tilde{A}$ would be a semigroup generator, too.
If by "extend" you mean the usual thing (viz, that $X$ is left invariant under $\{\tilde{T}(t),t\ge 0\}$ and the restriction to $X$ of $\{\tilde{T}(t),t\ge 0\}$ is $\{T(t),t\ge 0\}$ again), then $X$ would be left invariant under $\{T(t),t\le 0\}$, too, and the generator of this semigroup would necessarily be the part in $X$ of $-\tilde{A}$. So, the analytic semigroup $\{T(t),t\ge 0\}$ would actually imbed into a group $\{T(t),t\in \mathbb R\}$ on $X$ as well.
Now, an analytic semigroup on a Banach space $X$ can imbed into a group if and only if $X$ is finite dimensional its generator is bounded.<|endoftext|>
TITLE: Are the algebraic numbers dense everywhere on the boundary of the Mandelbrot set?
QUESTION [16 upvotes]: Let $\mathcal{B}$ denote the boundary of the Mandelbrot set, and let
$\overline{\mathbb{Q}}$ denote the algebraic closure of the rationals.
Further put $\mathcal{B}_{\overline{\mathbb{Q}}} := \mathcal{B} \cap 
\overline{\mathbb{Q}}$.
Questions:

Is $\mathcal{B}_{\overline{\mathbb{Q}}}$ dense in $\mathcal{B}$
in the sense that for every $z \in \mathcal{B}$ and every
$\varepsilon > 0$, the $\varepsilon$-neighborhood of $z$ has nontrivial 
intersection with $\mathcal{B}_{\overline{\mathbb{Q}}}$?
Given $z \in \mathbb{C}$, let
$\mathcal{B} + z := \{b + z \ | \ b \in \mathcal{B}\}$
denote the translate of $\mathcal{B}$ by $z$.
Is there a $z \in \mathbb{C}$ such that
$(\mathcal{B} + z) \cap \overline{\mathbb{Q}} = \emptyset$?
What are the answers to (1.) and (2.) if we replace
$\overline{\mathbb{Q}}$ by the Gaussian rationals $\mathbb{Q}[i]$
or by the cyclotomic integers $\mathbb{Z}[e^{\frac{2\pi i}{5}}]$?

REPLY [27 votes]: Post-critically pre-periodic quadratic polynomials, i.e. those for which the orbit of the critical point $0$ is pre-periodic, are well-known to be dense in the boundary of the Mandelbrot set. (This is essentially a normality argument.)
Each of these is determined by an algebraic equation. This answers your first question (the question in the title).
EDIT. As pointed out by Malik, post-critically pre-periodic parameters are also called Misiurewicz points. (Unfortunately, the term Misiurewicz is sometime also used, particularly in real dynamics I think, to refer to a larger class of systems.)<|endoftext|>
TITLE: Why should curves be two-dimensional?
QUESTION [21 upvotes]: In Weil cohomology, a nice curve has cohomology up to degree 2, or more generally a nice $n$-dimensional variety has cohomology up to degree $2n$.
I know that this was motivated at least in part by a desire to extend cohomology of complex manifolds to algebraic varieties over other fields, and since complex curves are two-dimensional manifolds, curves should have cohomology up to degree 2.
My question is if this is just an artifact of the motivation, or if there is a more 'intrinsic' reason why cohomology should work out that way? Or put differently, should one expect there to be cohomology theories where, say, curves act like they're five-dimensional?

REPLY [23 votes]: This is a very subtle question, I think.  First of all, the sense in which an $n$-dimensional algebraic variety $X$ acts as if it is "cohomologically $2n$-dimensional" is quite complicated--for example, unless one uses some notion of cohomology with compact support, an affine $n$-dimensional variety looks $n$-dimensional, not $2n$-dimensional (just as over $\mathbb{C}$, affine varieties have the homotopy types of $n$-dimensional CW complexes).  So this is really something about proper varieties, or about the more complicated notion of cohomology with compact support.
Second, this is just a fact about algebraically closed fields.  For example, if $k$ is a finite field and $X$ is an $n$-dimensional smooth projective $k$-variety, $X$ "looks" like it is $2n+1$-dimensional (indeed, it looks a lot like a $2n+1$-manifold fibered over a circle).  For example, there is an arithmetic Poincare duality between $H^i$ and $H^{2n+1-i}$.  (Of course the reason for this is that $k$ has cohomological dimension $1$ and has reasonable duality properties.)
Nonetheless, let me try my hand at some kind of explanation.  I think the proof of Grothendieck's vanishing theorem gives a reasonable explanation of why Zariski cohomology of $X$ vanishes in degree greater than $\dim(X)$.  Typically when one defines a fancier cohomology theory (e.g. etale or crystalline cohomology) one chooses a topology on some category of schemes which is finer than the Zariski topology.  Call such a topology $\tau$.  As $\tau$ is finer than the Zariski topology, there is a forgetful functor $f_*$ from sheaves on the topology $\tau$ to Zariski sheaves.  Grothendieck teaches us to think of this as a morphism $$f: X_\tau\to X_\text{Zar}.$$
(Maybe a good analogy is that there is a natural continuous map of spaces $$g: X(\mathbb{C})^{\text{an}}\to X_{\text{Zar}}$$ if $X$ is a finite-type $\mathbb{C}$-scheme.)
For all of the topologies $\tau$ which we like to use to define Weil cohomology theories, the morphism $f$ has cohomological dimension $n$, in the sense that the right-derived functor of $f_*$, denoted $Rf_*$, sends certain sheaves to complexes with cohomology concentrated in degrees $[0, n]$.  For example, if $\tau$ is the etale topology, and we are working over an algebraically closed field, $Rf_*\mathcal{F}$ is concentrated in degrees $[0,n]$ if $\mathcal{F}$ is constructible.  Or if $\tau$ is the crystalline site (over a field of characteristic zero), $$Rf_*\mathcal{O}_X\simeq \Omega^\bullet_{X, dR}$$
which is concentrated in degrees $[0,n]$ (and there is a similar formula with any flat vector bundle in place of $\mathcal{O}_X$).  So the fact that varieties in these topologies are $2n$-dimensional comes from the formula $$2n=n+n,$$ where the first $n$ comes from $Rf_*$ and the second comes from the cohomological dimension of the Zariski topological space.  Likewise, the honest continuous map $g: X(\mathbb{C})^{\text{an}}\to X_{\text{Zar}}$ has cohomological dimension $n$.
Of course (1) proving that $f$ has the claimed cohomological dimension is usually quite involved, and (2) we like these cohomology theories because they behave in ways which match our intuitions.  So it's unclear to me whether this is really a non-anthropological answer to your question.  But I think the pattern (of finding a site over the Zariski site which has relative cohomological dimension $n$) is ubiquitous enough to be worth commenting on.<|endoftext|>
TITLE: Is every ordinal potentially definable?
QUESTION [7 upvotes]: It is easy to see that, if $V\models\alpha>\omega_1^{CK}$, then $\alpha$ is not recursive in any forcing extension of $V$. The argument goes as follows:

The relation "$\Phi_e=r$" is $\Pi^0_2$.
The predicate "$r$ codes an ill-founded order" is $\Sigma^1_1$.
So if $V[G]\models $"$\Phi_e$ is a well-ordering," then so did $V$; and moreover, $V[G]$ and $V$ interpret $\Phi_e$ in the same way.

In the presence of generic absoluteness, we can extend this. For instance, it follows from Projective Determinacy that the set of ordinals which have projective copies in some set-generic extension of $V$ is precisely the set of ordinals which have projective copies already in $V$.
My question is about what happens when we don't have generic absoluteness.

Is it consistent with $ZFC$ that: "For every ordinal $\alpha$, there is a set-generic extension of the universe in which $\alpha$ has a projectively definable copy?"

If so, what is the least $n$ such that it is consistent with ZFC that every ordinal is "potentially $\Pi^1_n$"?
I am especially interested in what we know if $V=L$. In particular, I don't know that "there is a non-potentially-$\Pi^1_3$ ordinal" is even consistent with $ZFC+V=L$!
Conversely, I would be delighted (and amazed) if "Every ordinal is potentially $\Pi^1_3$" had nonzero consistency strength over $ZFC$, despite contradicting large cardinals.

REPLY [4 votes]: Great question!
The answer is yes. Start with $V=L$, and fix any ordinal $\alpha$.
Now go to the forcing extension $V[G]$ obtained by collapsing
$\aleph_\alpha^L$ to $\omega$, so that
$(2^\omega)^{V[G]}=\aleph_{\alpha+1}^L=\omega_1^{V[G]}$. In
$V[G]$, the ordinal $\alpha$ is definable in the structure
$H_{\omega_1}^{V[G]}$, since this structure sees all countable
objects, including all $L_\beta[G]$ for
$\beta<\omega_1^{V[G]}=\aleph_{\alpha+1}^L$, and therefore
$H_{\omega_1}^{V[G]}$ can identify the unique $\alpha$ for which
$\omega_1=\aleph_{\alpha+1}^L$. This definition is projective, since
we need to quantify only over countable objects.
I'll have to think a bit more to identify the exact complexity of
the definition.<|endoftext|>
TITLE: Lie algebra and base change
QUESTION [5 upvotes]: I am very confused by the following and would appreciate any help.
Let $\mu_p \subset \mathbb{G}_m$ be the $p$-torsion subgroup scheme of the multiplicative group over $\mathbb{Z}_p$. I would like to compute the Lie algebra of $\mu_p$ (at the identity section) to make sure that I understand Lie algebras well. I have heard that "the formation of the Lie algebra at the identity section of a group scheme commutes with base change" and I know that the Lie algebra of $\mu_p$ has to be killed by $p$, so by looking at the residue field I get that this Lie algebra is $\mathbb{Z}/p\mathbb{Z}$. 
On the other hand, from the $\epsilon$-points definition, I get that the Lie algebra of $\mu_p$ should be a subfunctor of the Lie algebra of $\mathbb{G}_m$. The Lie algebra of $\mathbb{G}_m$ is just $\mathbb{Z}_p$ because $\mathbb{G}_m$ is smooth of relative dimension $1$. But $\mathbb{Z}/p\mathbb{Z}$ cannot sit inside $\mathbb{Z}_p$, so where have I made a mistake? Does the $\epsilon$-points definition maybe not apply for some reason?

REPLY [10 votes]: The quotation you have heard is false because over an affine base $S = {\rm{Spec}}(k)$ for a commutative ring $k$ and a $k$-group scheme $G$, the Lie algebra is the linear dual ${\rm{Hom}}_k(e^{\ast}(\Omega^1_{G/k}),k)$ and that generally does not commute with non-flat base change when $G$ is not $k$-smooth.  You have already noticed yourself that this is false for $\mu_p$ over $\mathbf{Z}_p$.
There is always a "base change morphism" ${\rm{Lie}}(G) \otimes_{k} k' \rightarrow {\rm{Lie}}(G_{k'})$ for a $k$-algebra $k'$, but one cannot say more when $k'$ is not $k$-flat or $G$ is not $k$-smooth.  Read section A.7 (through A.7.6) for a discussion of Lie algebras of group schemes locally of finite type over rings in the book "Pseudo-reductive groups" (2nd edition) for a discussion of several of the key points from SGA3 (including that ${\rm{Lie}}(G)$ admits a natural Lie bracket compatibly with the base change morphism, even when $G$ isn't $k$-smooth).
The functor $k' \rightsquigarrow {\rm{Lie}}(G_{k'})$ on $k$-algebras is generally not representable if $e^{\ast}(\Omega^1_{G/k})$ is not a vector bundle over $k$, as usually fails if $G$ is not $k$-smooth and $k$ is not a field.<|endoftext|>
TITLE: Does the degeneracy of the Frölicher spectral sequence vary in families?
QUESTION [6 upvotes]: I would like to know if there are any known examples of families of complex manifolds for which the Frölicher spectral sequence of one fibre degenerates on the $E_m$ page and the spectral sequence of a different fibre degenerates on the $E_n$ page for $m \neq n$.
Since the spectral sequence degenerates on the $E_1$ page for compact Kähler manifolds and since small deformations of compact Kähler manifolds are compact and Kähler, one would have to look at either non-compact manifolds or non-Kähler manifolds (or "large" deformations).

REPLY [6 votes]: For example, in Corollary 4.7 in "Invariant complex structures on 6-nilmanifolds: classification, Frölicher spectral sequence and special Hermitian metrics" by Manuel Ceballos, Antonio Otal, Luis Ugarte, Raquel Villacampa (JGA), a family of complex non-Kähler structures such that the Frölicher spectral sequence degenerates for any fibre except $X_0$ is provided on the Iwasawa manifold. For other examples, see the references therein and other works by the authors.<|endoftext|>
TITLE: References for Stiefel-Whitney class of Stiefel manifolds and Grassmannians
QUESTION [13 upvotes]: Let $M$ be a manifold. The total Stiefel-Whitney class of $M$ is defined to be the Stiefel-Whitney class of the tangent bundle $TM$
$$
w(M)=1+w_1(TM)+w_2(TM)+\cdots
$$
I want to find references for
$$
w(SO(n)/SO(k)),(i.e. w(V_{n-k}(\mathbb{R}^n))), \\
w(U(n)/U(k)),(i.e. w(V_{n-k}(\mathbb{C}^n))), \\
w(Sp(n)/Sp(k)),(i.e. w(V_{n-k}(\mathbb{H}^n))),\\
w(SO(n)/(SO(k)\times SO(n-k))),(i.e. w(G_{n-k}(\mathbb{R}^n))), \\
w(U(n)/(U(k)\times U(n-k))),(i.e. w(G_{n-k}(\mathbb{C}^n))), \\
w(Sp(n)/(Sp(k)\times Sp(n-k))),(i.e. w(G_{n-k}(\mathbb{H}^n))).
$$
Which ones of the above are known? Where could I find these formulas?

REPLY [5 votes]: In this note, the first two Stiefel-Whitney classes of unoriented, oriented, and complex grassmannians are determined in terms of the Stiefel-Whitney classes of their tautological bundles. This is achieved via the method mentioned at the end of Mark Grant's answer.
For the unoriented grassmannian $\operatorname{Gr}(m, m+n) = O(m+n)/(O(m)\times O(n))$, we have 
\begin{align*}
w_1(\operatorname{Gr}(m, m+n)) &= (m + n)w_1(\gamma)\\
w_2(\operatorname{Gr}(m, m + n)) &= \left[\binom{m}{2} + \binom{n}{2} + m^2 + mn - 1\right]w_1(\gamma)^2 + (m^2 + n^2)w_2(\gamma)\\
&= \begin{cases}
0 & m - n \equiv 2 \bmod 4\\
w_2(\gamma) & m - n \equiv 1 \bmod 4\\
w_1(\gamma)^2 & m - n \equiv 0 \bmod 4\\
w_2(\gamma) + w_1(\gamma)^2 & m - n \equiv 3 \bmod 4.
\end{cases}
\end{align*}
For the oriented grassmannian $\operatorname{Gr}^+(m, m+n) = SO(m+n)/(SO(m)\times SO(n))$, we have 
\begin{align*}
w_1(\operatorname{Gr}^+(m, m + n)) &= 0\\
w_2(\operatorname{Gr}^+(m, m + n)) &= \begin{cases}
0 & m - n \equiv 0 \bmod 2\\
w_2(\gamma_+) & m - n \equiv 1 \bmod 2.
\end{cases}
\end{align*}
For the complex grassmannian $\operatorname{Gr}^{\mathbb{C}}(m, m + n) = U(m + n)/(U(m)\times U(n))$, we have
\begin{align*}
w_1(\operatorname{Gr}^{\mathbb{C}}(m, m + n)) &= 0\\
w_2(\operatorname{Gr}^{\mathbb{C}}(m, m + n)) &= (m + n)w_2(\gamma_{\mathbb{C}}).
\end{align*}
As for the quaternionic grassmanian $\operatorname{Gr}^{\mathbb{H}}(m, m + n) = Sp(m + n)/(Sp(m)\times Sp(n))$, it is $2$-connected, so $w_1(\operatorname{Gr}^{\mathbb{H}}(m, m + n)) = 0$ and $w_2(\operatorname{Gr}^{\mathbb{H}}(m, m + n)) = 0$.
In principle, you can calculate all of the Stiefel-Whitney classes using the splitting principle argument, but the calculations get more and more complicated. A short cut for the third Stiefel-Whitney class is to use the fact that $\operatorname{Sq}^1(w_2) = w_3 + w_1w_2$. It follows that we have
\begin{align*}
w_3(\operatorname{Gr}(m, m + n)) &= \begin{cases}
0 & m - n \equiv 0, 2 \bmod 4\\
w_3(\gamma) & m - n \equiv 1 \bmod 4\\
w_3(\gamma) + w_1(\gamma)^3 & m - n \equiv 3 \bmod 4
\end{cases}\\
& \\
w_3(\operatorname{Gr}^+(m, m + n)) &= \begin{cases}
0 & m - n \equiv 0 \bmod 2\\
w_3(\gamma_+) & m - n \equiv 1 \bmod 2
\end{cases}\\
& \\
w_3(\operatorname{Gr}^{\mathbb{C}}) &= 0\\
& \\
w_3(\operatorname{Gr}^{\mathbb{H}}) &= 0.
\end{align*}<|endoftext|>
TITLE: Conformal compactification of Kerr spacetime
QUESTION [8 upvotes]: I'm looking for a book/paper where the conformal compactification of Kerr spacetime is calculated. I've seen plenty of reference for the Minkowski, but none (explicitly calculated) for Kerr.
Thank you.
PS: prior to writing this post, I was already referred to "Large scale structure of spacetime" by Hawking and Ellis, but there they only discuss the maximal extension of Kerr (and other solutions), not conformal compactification calculation.
EDIT on what I meant by "compactification for Minkowski": The Minkowski metric is
$ds^{2}=-dt^{2}+dr^{2}+r^{2}d\Omega^{2}$
where $-\infty<t<\infty$ and $0\leq r<\infty$. (We now want to get the coordinates with finite ranges)
First go to the double-null coordinates $u=t-r$ and $v=t+r$, and then change to $T=\arctan u+\arctan v$, $R=\arctan v-\arctan u$. We then obtain
$ds^{2}=\omega^{-2}(T,R)(-dT^{2}+dR^{2}+\sin^{2}Rd\Omega^{2})$
with $\omega(T,R)=2\cos U \cos V=\cos T+\cos R$.
Therefore, the original metric is related to the new one by (explicitly given) conformal transf. $\omega^{2}$ as
$d\tilde{s}^{2}=\omega^{2}(T,R)ds^{2}=-dT^{2}+dR^{2}+\sin^{2}R d\Omega^{2}$
where $R,T$ have the finite ranges $0\leq R<\pi$ and $R-\pi <T <\pi -R$.

REPLY [3 votes]: Although the focus of the original question was on conformal compactification, a necessary step along the way is an introduction of double-null coordinates that are regular on the horizons and bifurcation spheres of Kerr (where the standard Boyer-Lindquist coordinates become singular). An alternative set of such coordinates, which are obtained in a briefer and more elementary way than the classic Pretorius-Israel references cited in Willie's answer can be found in

Hayward, Sean A., Kerr black holes in horizon-generating form, Phys. Rev. Lett. 92, No. 19, Article ID191101, 4 p. (2004). [arXiv:gr-qc/0401111] ZBL1267.83059.<|endoftext|>
TITLE: Existence of a non-null-homotopic simple closed curve
QUESTION [6 upvotes]: Assume that $X$ is a path-wise connected Hausdorff space, and assume that its fundamental group is non-trivial. Does it always exist a simple curve in $X$ which is non-null-homotopic?
Such curve does exist if we further assume that $X$ possesses a universal cover.

REPLY [12 votes]: No. The harmonic archipelago is a compact $2$-dimensional counterexample, embedded in $\mathbb R^3$.
Let $S_n$ denote the planar circle with center on the $x$-axis, and whose intersection with the $x$-axis is 
$$
\left\{\Big(\frac{1}{n+1},0\Big), \Big(\frac{1}{n},0\Big)\right\}.
$$ Take the closure of the union of the circles. Now make each circle the base of a cone, raised in the $z$-direction, of height $1$, and take the closure of the union of the cones. Every simple closed curve is inessential, but the space is not simply connected. (A curve running once around each base circle is essential).<|endoftext|>
TITLE: Invariant regular cones in Lie group representation
QUESTION [6 upvotes]: I am following Analysis and Geometry on Complex Homogeneous Spaces by Faraut et al. I'll set up all of what I need and then ask my questions.
Let $G$ be a connected semi-simple non-compact real Lie group. Let $K \subset G$ be a chosen maximal compact subgroup. We say an irreducible representation $(\pi,V)$ is spherical the space of $K$-fixed vectors $V^K$ is exactly $1$-dimensional. 
We know that $(\pi,V)$ is spherical if and only if $V$ contains a cone $C$ invariant under the action of $G$. Moreover if $(\pi,V)$ is spherical then we can let $u_0 \in V$ be a $K$-fixed vector and $v_0 \in V$ a highest weight vector. We then have minimal and maximal cones $C_{min} = \overline{\mathbb{R}^+ conv(Gu_0 \cup \{0\})} = conv(G v_0)$ and $C_{max} = C_{min}^*$, the dual cone.
Here are my questions:
(1) In the general case, when we're not looking at the adjoint representation, are the cones interesting? Can the properties of the cone (a semi-algebraic set) tell us anything about the representation?
(2) Occasionally one gets $C_{min}=C_{max}^*$. Why is this interesting? Or what is the significance of it not happening?
I've worked out that $C_{min} \neq C_{max}$ for spherical representations of $SL(2,\mathbb{R})$. Is there some significance to this?

REPLY [4 votes]: Regarding your second question, the cases where $C_{min} = C_{max}$ for cases where $G$ is a real form a complex semisimple Lie group have been classified by Misyureva. Basically, you get that $V$ is a Euclidean Jordan algebra and the cone is the closure of the self dual symmetric cone in $V$. According to Hilgert and Neeb this question in its general form was open in 1998.
As for your first question, let me quote from the latter reference:

Problems related to invariant cones in representations occur in various contexts. They seem to have been studied first by Vinberg in the context of homogeneous cones (cf.[Vi63]). Apart from the study of the adjoint and coadjoint representations (cf. Example 1.2) most of the work done more recently is connected to isotropy representations of symmetric spaces, where invariant cones show up in connection with orderings, Wiener-Hopf operators and Hardy spaces (cf. Example 1.4, [KN96] [HO96], and the references given there).<|endoftext|>
TITLE: Polynomial inequalities in ordered fields
QUESTION [6 upvotes]: Let $p(x)$ be a polynomial over an ordered field. If $p'(x)\ge 0$ for all $x$ in an interval over that field, is it true that $p(x)$ is increasing over that interval?

REPLY [3 votes]: It is remarked in A Course in Model Theory: An Introduction to Contemporary Mathematical Logic by Bruno Poizat, page 101 (Google Books link), that the assertion is true for all ordered fields, but that its proof is difficult. He gives the proof for real closed fields just prior to that remark.<|endoftext|>
TITLE: Complex scalar field, generating functional?
QUESTION [6 upvotes]: The Lagrangian for the complex scalar field is$$\mathcal{L} = \partial_\mu \phi^* \partial^\mu \phi - m^2 \phi^* \phi.$$Can anyone work out or provide me a reference to the computation of the generating functional$$Z[J, J^*] = \int \mathcal{D}\phi \,\exp\left\{i \int d^4x\,[\mathcal{L} + J(x)\phi^*(x) + J^*(x)\phi(x)]\right\}?$$I need this result for my research, but unfortunately, I am not a physicist by training. Thank you very much!

REPLY [3 votes]: you can find this calculation, for example, in chapter 7 of Kleinert's path integral book, I reproduce the relevant equations:<|endoftext|>
TITLE: Closed curve whose neighborhood is as large as possible
QUESTION [7 upvotes]: Let $C$ be a closed curve in the plane and let $N_\epsilon(C)$ be an $\epsilon$-neighborhood of $C$, like this: 
(ignore the fact that the "curve" is polygonal in this picture, I drew it in MATLAB)
My question is:  given fixed $\epsilon$, if we search among all closed curves $C$ of fixed length $\ell$, what is the optimal curve that makes the area of $N_\epsilon(C)$ as large as possible?  It seems very intuitive to me that the curve should just be a circle, which gives an area of $2\epsilon\ell$ provided that $\epsilon\leq \ell/(2\pi)$.

REPLY [6 votes]: Just to emphasize Thomas Richard's remark about smoothness, unless I've miscalculated, a $\frac{1}{4} L$-square leads to area
$$2 \epsilon L - \epsilon^2 (4-\pi) <  2 \epsilon L \;.$$

      

Added. This is to illustrate my "hypothesis" in the comments: Any smooth curve (not necessarily convex) such that its radius of curvature exceeds $\epsilon$ (at every point) leads to area $2 \epsilon L$.
In the right figure, too-sharp curvature leads to gaps between
the $\epsilon$-disk and the tube boundaries.<|endoftext|>
TITLE: confounding riddle about fine moduli schemes and twists of elliptic curves
QUESTION [7 upvotes]: I've encountered a strange situation while thinking about modular curves... Consider the modular curve $Y(3)$ parametrizing elliptic curves with a symplectic basis for their 3-torsion. This curve has degree 12 over the $j$-line $Y(1)$. Let $y\in Y(1)$ be a $\mathbb{Q}$-rational point, then its fiber in $Y(3)$ has degree 12, and hence there exists a number field $K$ of degree dividing 12 over $\mathbb{Q}$ such that over $K$, the fiber above $y$ in $Y(3)_K$ is completely decomposed. 
Now for any elliptic curve $E$ over $\mathbb{Q}$ with $j$-invariant $y$, $E[3]$ is defined over some number field $L$. Thus, over $L$, by picking a suitable basis for $E_L[3]$ we get an $L$-point of $Y(3)$ over $y$. Thus, by our construction of $K$ we may as well have taken $L = K$. 
My question is this: Fix an elliptic curve $E$ over $K$ with $j$-invariant y, then there are infinitely many nonisomorphic twists $E^d$, parametrized by $d$ in $K^*/(K^*)^2$. Now fix a set of representatives $D_K$ for $K^*/(K^*)^2$, and consider the set:
$$\{E^d : d\in D_K\}$$
Now each of these twists $E^d$ is an elliptic curve over $K$ with $j$-invariant $y$, so by the above discussion, for each such twist $E^d$ with $d\in D_K$ we may choose a suitable basis $(P(d),Q(d))$ for $E^d[3]$ so that $(E^d/K,P(d),Q(d))$ corresponds to a $K$-point of $Y(3)$ above $y$. However, there are only 12 $K$-points of $Y(3)$ above $y$, but infinitely many nonisomorphic triples:
$$\{(E^d/K,P(d),Q(d)) : d \in D_K\}$$
Where have I gone wrong?
$\newcommand{\QQ}{\mathbb{Q}}$
EDIT: The above question was mostly answered by Ari in his comment below, but I feel like it doesn't resolve my confusion. Here's another way of articulating my confusion:
Let $\mathcal{Y}(3)$ be the stacky version of $Y(3)$. Fix an elliptic curve $E$ over $\QQ$, corresponding to a $\QQ$-point of $\mathcal{M}_{1,1}$. The fiber of $\mathcal{Y}(3)$ over $E/\QQ$ is a representable stack, finite etale over $\QQ$, whose corresponding scheme is a degree 12 etale $\QQ$-algebra $F$. For any other nonisomorphic twist $E^d$ over $\QQ$, we may play the same game, and we find that the fiber of $\mathcal{Y}(3)$ over $E^d/\QQ$ is also a degree 12 etale $\QQ$-algebra which I will call $F^d$. Now, passing to coarse moduli schemes, the $F$'s and $F^d$'s are all somehow related to the etale algebra corresponding to the fiber $Y(3)_y$. In fact, by suitably picking a $y$, by Hilbert's irreducibility theorem we may assume that $Y(3)_y$ is connected (ie a field), so lets call it $K$. What is the relation between $F^d$ and $K$?
If there is no relation, then what is $K$? Surely it must have some moduli-theoretic meaning?
Okay, I'm beginning to suspect that there is a particular special twist of $E/\QQ$ such that $K$ is just $F^d$, where Spec $F^d$ is the scheme of $\Gamma(3)$-structures on $E^d$. This special $E^d$ might just be the twist representing the $K$-isomorphism class of the fiber of the universal elliptic curve $\mathcal{E}(3)_K/Y(3)_K$ at a $K$-point $x\in Y(3)$ lying above $y$. If this were true, it would seem to "imply" that all fibers of $\mathcal{E}(3)_K$ over points $x\in Y(3)$ lying above $y$ must be $K$-isomorphic?

REPLY [2 votes]: $\newcommand{\QQ}{\mathbb{Q}}$
$\newcommand{\ZZ}{\mathbb{Z}}$
Okay, so the solution appears to be this (Thanks to Ari Shnidman, Joseph Silverman, nfdc23, and eric for their comments)
Fix an $N\ge 3$. Let $y\in Y(1)$ be a $\QQ$-point, then the fiber $Y(N)_y$ of $Y(N)$ above $y$ is a $\QQ$-algebra $A$ of degree $d_N := |PSL(2,\ZZ/N)|$. Since $Y(N)/Y(1)$ is galois, we find that $A = \prod_{i=1}^{k_y} K[x]/(x^e)$ is a product of $k_y$-many connected $\QQ$-algebras, where $e = 1$ if $y\ne 0, 1728$, otherwise $e = 3,2$ if $y = 0,1728$. Here we have $k_y\cdot e\cdot[K:\QQ] = d_N$. Note that by the Weil Pairing $K$ must contain a primitive $N$th root of unity $\zeta_N$.
Thus, we find that for any such $y$, there exists a field $K$ of degree over $\QQ$ dividing $d_N$ such that there exists a $K$-point on $Y(3)$ lying above $y$. This means, that there exists an elliptic curve $E/K$ with a pair of points $P,Q\in E(K)[N]$ which generate $E[N]$ and such that $e_N(P,Q) = \zeta_N$. In particular, $E[N]$ is defined over $K$. Note that given one such pair $(P,Q)$ there exist $SL_2(\ZZ/N)$-many such pairs with Weil pairing $\zeta_N$, whose equivalence classes modulo $Aut(E)$ occupy all of the points in the fiber $Y(3)_y$ (ie, over $y\ne 0,1728$ there are $d_N$ such equivalence classes. Otherwise there are $\frac{d_N}{3}$ or $\frac{d_N}{2}$ if $y = 0$ or $1728$).
The fact that all the points of $Y(3)_y$ is taken up by this single curve $E$ means that there is (up to $K$-isomorphism), precisely one elliptic curve $E$ with $E[N]$ defined over $K$ (this is the special twist I was referring to in the OP). This elliptic curve is precisely the fiber of the universal elliptic curve $\mathcal{E}(N)$ over $Y(N)$ at the $K$-point, which indeed implies that all fibers of $\mathcal{E}(N)$ with a particular $j$-invariant are isomorphic.
Lastly, in the OP I had said that the algebras $F^d$ are degree $d_N$ over $\QQ$. This is incorrect. They are actually degree $2d_N$ (the map from $\mathcal{M}_{1,1}$ to $Y(1)$ is degree "one half").
If we do a little relabeling and say that the scheme of full level-$N$ structures on $E^d/\QQ$ is a galois $\QQ$-algebra $B^d$ of degree $2d_N$, then $B = \prod_{i=1}^{k_y} F^d[x]/(x^e)$, then again we find that since there must be a unique elliptic curve $E'$ over $F^d$ (up to $F^d$-isomorphism) whose full level-$N$ structures occupy all the $F^d$ points of $Y(N)_y$, that this elliptic curve must be $E^d$, and thus $F^d = K(\sqrt{d})$.<|endoftext|>
TITLE: Closed form for $\sum_{n=1}^{\infty} \frac{1}{1+n+n^2+\cdots+n^a}$
QUESTION [10 upvotes]: I am looking at a certain sequence, and consequently I am wondering if anyone knows if there happens to exist a closed form solution for this sum:
$$\sum_{n=1}^{\infty} \frac{1}{1+n+n^2+\cdots+n^a}$$
or, an alternate form:
$$\sum_{n=1}^{\infty} \frac{n-1}{n^{a+1}-1}$$
For low values of $a$, Wolfram Alpha gives a closed form in terms of the polygamma function function of order $0$, so I am wondering if there is a general closed form in terms of $a$.
I asked this question on MSE, but have not received answers yet. I realize I did ask it recently there, but I also wanted to ask it here just to see if anyone had seen this sum before.

REPLY [15 votes]: We use expansion $$\frac{z-1}{z^{a+1}-1}=\frac1{a+1}\sum_{w^{a+1}=1} \frac{w(w-1)}{z-w}.$$
Now use Abel-Poisson regularization of your sum, multiplying $n$-th term by $t^n$ for $t<1$:
$$
\sum \frac{n-1}{n^{a+1}-1}=\frac1{a+1}\lim_{t\rightarrow 1-0} \sum_{w^{a+1}=1} w(w-1)\sum_n \frac{t^n}{n-w}.
$$
We have for $w\ne 1$ 
$$\sum \frac{t^n}{n-w}=\int_0^1 \sum_{n=1}^{\infty} t^nz^{n-w-1}dz=t\int_0^1 \frac{z^{-w}dz}{1-tz}.$$
Hence our sum equals
$$
\frac1{a+1}\int_0^1 \frac{\sum_{w^{a+1}=1} w(w-1)z^{-w}}{1-z}dz.
$$
This may be further rewritten in several ways. For example, we may replace $(1-z)^{-1}$ to $(1-z)^{\varepsilon-1}$ and then tend $\varepsilon$ to $+0$. For each single $w$ we get 
$$\int z^{-w}(1-z)^{\varepsilon-1}dz=B(1-w,\varepsilon)=\frac{\Gamma(1-w)\Gamma(\varepsilon)}{\Gamma(1-w+\varepsilon)}=\Gamma(\varepsilon)(1-\varepsilon \psi(1-w)+O(\varepsilon^2)).$$ 
Multiply this by $w(w-1)$ and sum up. Singularity $\Gamma(\varepsilon)$ goes out as expected, we get a sum like $$
\frac1{a+1}\sum_{w^{n+1}=1,w\ne 1} w(1-w)\psi(1-w).$$
Or just use the formula (proved by the same way) $\int_0^1 \frac{1-z^{-w}}{1-z}=\gamma+\psi(1-w).$

REPLY [7 votes]: This is not an answer but a conjecture. It based on a first examples examples calculated by Mathematica.
$$S(a)=\frac1{a+1}-\sum_{i=1}^a\frac{\psi(-\alpha_i)}{f_a'(\alpha_i)},$$
where $f_0(x)=1$, $f_1(x)=x+3$, $f_2(x)=x^2+5x+7$,...
$$f_{n+1}(x)=(x+2)f_n(x)+1\qquad(n\ge 0),$$
(see A193844) and $\alpha_i$ are the roots of $f_a$.
UPD: From Fedor's formula $f_{n}(x)=\frac{(x+2)^{n+1}-1}{x+1}$ follows that for $\alpha_k=w_k-2$ we have $$f_n'(\alpha_k)=\frac{n+1}{w_k(w_k-1)}.$$<|endoftext|>
TITLE: Critical with respect to chromatic, but not Hadwiger number
QUESTION [7 upvotes]: For any simple, undirected graph $G=(V,E)$ where $V$ is finite, we define the Hadwiger number $\eta(G)$ to be the maximum $n$ such that $K_n$ is a minor of $G$.
Is there a graph $G$ on such that removing a vertex or contracting an edge reduces the chromatic number, but has no effect on the Hadwiger number?
To be more precise, I'm looking for a graph $G$ such that

$G$ is vertex-critical (for any $v\in V(G)$ we have $\chi(G\setminus\{v\}) = \chi(G) - 1$);
$G$ is edge-contraction-critical (for any $e\in E(G)$ contracting $e$ reduces the chromatic number);
for any $v\in V(G)$ we have $\eta(G) = \eta(G\setminus \{v\})$;
for any $e\in E(G)$ contracting $e$ does not change the Hadwiger number.

REPLY [3 votes]: Here's an example that does not require a computer or exercises left to the reader.  Let $G$ be the graph obtained by taking two $7$-cycles $C_1$ and $C_2$ and joining each vertex of $C_1$ to each vertex of $C_2$.  The chromatic number of this graph is $6$ since $\chi(C_1)=\chi(C_2)=3$.  The Hadwiger number of this graph is $7$ (we get a $K_7$ by contracting a perfect matching between $V(C_1)$ and $V(C_2))$.  Deleting a vertex yields a path on one side, so the chromatic number goes down to $5$.  Contracting an edge of $C_1$ or $C_2$ gives a $6$-cycle on one side, so the chromatic number goes down to $5$.  Contracting an edge between $V(C_1)$ and $V(C_2)$ yields two $7$-cycles $C_1'$ and $C_2'$ meeting at a vertex $v$, together with all edges between $V(C_1') \setminus v$ and $V(C_2') \setminus v$. This graph is obviously $5$-colourable, since there is a $3$-colouring of $C_1'$ where only $v$ is coloured red.  Finally, deleting a vertex or contracting an edge does not change the Hadwiger number, since in either case the resulting graph still contains $K_{6,7}$.  Contracting a matching of size $6$ in $K_{6,7}$ yields a $K_7$-minor. 
Edit. Having read my answer again after a few years, I see a mistake in the above argument.  The Hadwiger number of $G$ is actually $8$ (not $7$ as claimed), since contracting the edges of a matching of size $6$ between $V(C_1)$ and $V(C_2)$ yields a $K_8$-minor.  Thus, the Hadwiger number of $G$ appears to go down when contracting an edge of $C_1$ or $C_2$.<|endoftext|>
TITLE: "Pythagoras number" for integral matrices
QUESTION [11 upvotes]: It is classically known that every positive integer is a sum of at most four squares of integers, i.e. every sum of squares of integers is a sum of four squares of integers. Now consider a symmetric $n\times n$ matrix $M$ with integer entries which can be written as $M= Q^{\rm T} Q$ for an $m \times n$ matrix $Q$ with integer entries. Can we bound $m$ in some way? I.e. is there a constant $c(n)\in \mathbb{Z}$ depending only on $n$ such that in this situation we can always find a decomposition $M= \widetilde{Q}^{\rm T} \widetilde{Q}$ for a $c(n) \times n$ matrix $\widetilde{Q}$ with integer entries? What is the smallest possible such constant (perhaps even linear growth?)?
If we look at the same situation over the polynomial ring $\mathbb{R}[T]$ in one variable, we have the following: Every sum of squares in $\mathbb{R}[T]$ is a sum of two squares and  one can show that $c(n)$ grows linearly in $n$: $c(n)=2n$. Thus, I hope that a similar result might hold over the integers too. Perhaps $c(n)$ even grows linearly in $n$?
Is someone aware of something in that direction?

REPLY [4 votes]: In Maria Icaza's Ohio State PhD thesis (1992?) she established the following result.  For any positive integer $n$, let $S(n)$ be the set of $\mathbb Z$-lattices (or integral quadratic forms, if one prefers polynomials) of rank $n$ that can be represented by some sums of squares.  Then there exists an integer $g(n)$ such that all lattices in $S(n)$ are represented by $g(n)$ number of squares.  She also obtained an explicit upper bound on $g(n)$, which is certainly not optimal.  All of these were published later in a couple of papers which you can find in MathSciNet.  
Precise values of $g(n)$ are known up to $n = 6$.  For $n \leq 5$, $g(n) = n + 3$.  However, $g(6) = 10$ which was proved by M-H Kim and B-K Oh in the 90's (in Journal of Number Theory).  Later they obtained much better upper bounds for $g(n)$ in a series of papers.<|endoftext|>
TITLE: Density of Gaussian measures on Banach spaces
QUESTION [5 upvotes]: I am trying to get my head around this question and was reading (1) which states the same a little bit more general:

Let $X$ be a separable Banach space and $X^*$ the dual space. The mean
  value $m \in X$ and covariance $C\colon X^*\to X$ are defined by
  \begin{align} x^*(m) &= \int_X x^*(x) \mu(dx),\\ y^*(Cx^*) &= \int_X
x^*(x)y^*(x) \mu(dx) - x^*(m)y^*(m) \end{align} $x^*,y*\in X^*$ and
  measure $\mu$.

Moreover

A measure $\mu$ is called Gaussian if all the functionals $x^*\in
X^*$, regarded as random variables on $(X,\mu)$ are Gaussian random
  variables.

I considered $X=\mathbb{R}^2$ since this is easier to visualize and I can check the results with my experience of $\mathbb{R}^2$ random variables.
I generated samples $Z$ from a Gaussian distribution with mean $m = (1,1)^T$ and covariance $C = \bigl(\begin{smallmatrix}
1&0.5 \\ 0.5&0.4
\end{smallmatrix} \bigr)$.

Choosing a $x^* = (2,3)$ (or any other value) then $x^*(Z)$ seems to be Gaussian with
$x^*(m) = (2,3)\cdot(1,1)^T = 5$ and $x^*(Cx^*) = (2,3)\cdot (C(2,3)\cdot (2,3)) = 13.6$ as in the histogram:

and everything seems to be fine.
My question now is, can I somehow access the density of the (multivariate Gaussian) distribution at $x^*$ if $X$ is a Banach space, to get the value of
\begin{align}
f(x^*) = \frac{1}{\sqrt{2\pi^2|C|}} \exp\big(-0.5(x^*-m)^TC^{-1}(x^*-m)  \big)
\end{align}
It would be great to get an example assuming that $X$ and the measure are sufficiently "nice" (eg. the measure is not degenerate etc, X is separable or a Hilbert space, etc.).
Edit: As Nate Eldredge pointed out there is no Lebesgue measure. Let me give a possible application: Failure detection on an engine. There is some sensor delivering a curve (function) which carries information about the engine properties. So each test of an enigne delivers a curve (function) and therefore we model $X$ to be a separable Hilbert space of square-integrable functions. A malfunction is now detection if a test yields a curve outside two standard deviations.
(1) Rudnìcki, Ryszard. "Gaussian measure-preserving linear transformations." Univ. Iagel. Acta Math 30 (1993): 105-112.

REPLY [3 votes]: First of all, please note that I am not an expert for this. Nevertheless, I think what you want (in order to transfer your intuition from $\mathbb{R}^2$ to the Banach space case) is to write the integral of a suitable function $f$ over the Gaussian measure $\mu$ as
$$ \int_X f(x)\,\mu(d x) \,\,\,= \,\,\, \int_X f(x)\,\exp\big(q(x)\big)\,\lambda(dx) $$
where $\lambda$ is analogous to the standard Lesbegue-meassure on $\mathbb{R}^2$ and $q$ some quadratic function on $X$. The problem then is that in order to define the standard Lesbegue measure on $\mathbb{R}^2$ one makes use of the standard basis of $\mathbb{R}^2$ in order to define the measure of a rectangle (therefor, if you transform to a different basis via a linear transformation $T$ you get a factor $\det(T)$).
Because of this, there is no analogoue of the standard Lesbegue measure on an abstract Banach space. Nevertheless, on a Hilbert space $H$ you would have a natural notion of a "length" and one might try to find an analogue in that case. But even then there are more problems arising because integration on infinite dimensional spaces is way more complicated than on finite dimensional ones. Heuristically you could say that
$$ \int_H \exp\big(-\pi<x\,|\,x>\big)\,\lambda(dx) \,\,\,=\,\,\,1$$
because it is the infinite product of $1$-dimensional integrals that evaluate to $1$. But then
$$ \int_H \exp\big(-2\pi<x\,|\,x>\big)\,\lambda(dx) \,\,\,=\,\,\,\infty$$
because it is the infinite product of $1$-dimensional integrals that evaluate to something larger than $1$. So you could only give a meaning to
$$ \int_H c\,\exp\big(-\pi<(x-x_m)\,|\,V\,(x-x_m)>\big)\,\lambda(dx)$$
where $c\in \mathbb{R}$ is for normalisation, $x_m\in H$ is the mean and $V$ is a bounded operator on $H$ that is sufficiently close to the identity (maybe the exponential of an operator of Hilbert Schmidt type) and is of course related to the Variance. I think with some work one can get at least a formula for this heuristic -- but not necessarily a measure on $H$.<|endoftext|>
TITLE: Can you have many independent reals?
QUESTION [11 upvotes]: Working in $\sf ZFC$, is it provable, or at least consistent (say, over $L$), that you have $\aleph_1$ forcings, $\Bbb P_\alpha$ such that:

$\Bbb P_\alpha$ is c.c.c.
$\Bbb P_\alpha$ adds a real which determines the generic.
For every countable $A\subseteq\omega_1$, and $\alpha\notin A$ the finite support product $\prod_{\beta\in A}\Bbb P_\beta$ does not add a $V$-generic real for $\Bbb P_\alpha$?

REPLY [7 votes]: First add $\omega_1$ Cohen reals, then partition this set of Cohen reals into $\omega_1$ disjoint sets $A_i$ each of size $\omega_1$. Let $P_i$ be a sigma centered forcing whose generic is a real coding a meager set covering $A_i$. Then, it is easy to check that the family $\{P_i : i < \omega_1\}$ is as required.
Some details: Let $p \in Q_i$ iff $p = (F, \overline{n} = \langle n_k : k \leq N \rangle, \overline{\sigma} = \langle \sigma_k : k < N \rangle)$ where $F$ is a finite subset of $A_i$, $n_0 = 0, n_k \in \omega$ are increasing and $\sigma_k \in 2^{[n_k, n_{k+1})}$, $q \leq p$ iff $F_p \subseteq F_q$, $\overline{\sigma}_q, \overline{n}_q$ extend $\overline{\sigma}_p, \overline{n}_p$, and for every $x \in F_p$, every new string $\sigma$ from $\overline{\sigma}_q$, $x$ disagrees with $\sigma$ somewhere. $Q_i$ is sigma centered and adds a real, namely $\bigcup \{\overline{\sigma}_p : p \in G_{Q_i}\}$, coding a meager set covering $A_i$. Let $P_i$ be the complete subalgebra of $Q_i$ generated by this real. Now note that if $y$ is any real in $\prod \{P_i : i \neq i_0\}$, then the meager set coded by $y$ cannot cover $A_{i_0}$ - in fact none of the reals in $A_{i_0}$ is covered. Clauses 1 and 2 are obvious.<|endoftext|>
TITLE: What are the usual deadlines in paper submission procedure?
QUESTION [7 upvotes]: I've submitted a paper to a journal 10 days ago, and I did not yet get any news from the handling editor.
Of course, 10 days is quite short, but I hope I will not wait one year without any news for then getting a rejection, because then it would be a big waste of time.  
Question: What are the usual deadlines in paper submission procedure?

REPLY [10 votes]: I think that your experience may vary quite a bit depending on what type of journal you submit to and what type of submission system they use.
For journals that instruct you to send your manuscript to one of the editors directly via email, my experience has been that within a day or two you usually receive a terse reply thanking you for your submission and informing you that they will get the paper to a referee right away and will get back to you once they've heard back from the referee. The entire email is usually a sentence or two and is often the only thing you will hear from them for the next six months or so. If you sent your paper to an especially selective journal though, the editor may want to get a quick opinion about whether or not your paper is at the right level. In this case it may take up to a week to ten days to hear back from the editor.
For journals that have you submit via an online submission system, you should receive some sort of automatically generated confirmation of your submission more or less immediately. In my experience this is often the only email that I will receive until the refereeing process has ended. Oftentimes you'll be provided with a link to a website where you can track the status of your paper. In theory this should let you see that the paper is being assigned to an editor, has been sent to a referee, that the referee's report has been received, etc. I have found that more often or not you are simply told that the status of your paper is that it is "Under Review".
Once your paper makes it this far there really isn't anything you can do but wait. I have found that six months is about the average time it takes to receive a referee's report, though it might take a couple of months less if the paper is very short or a few months more if the paper is especially long or technical. 
I think that once you've waited six months (perhaps a few months more or less depending on the length of your submission) it is reasonable to send a brief email to the editor inquiring about the status of your submission and asking when you might expect a report. Probably the editor will just reply by saying that your paper is still with the referee, though your email may result in the editor asking the referee for an update about when the report might be ready.
As for your fear that you'll have to wait a long time only to receive a rejection, unfortunately this happens all the time. For starters, it is not always that case that your paper will be sent to a referee immediately. It could take several weeks for this to happen. If your paper is especially technical or draws on techniques from a lot of different areas then it could take months for the editor to even find someone willing to referee the paper. You also have to remember that the person refereeing your paper is not being paid to do so. He has his own papers to write, classes to teach, etc. So for instance if you submit your paper right before a very busy time of year (say, when classes are just beginning) it could very well take a month or two for the referee to even bother printing out your paper and begin looking at it. And even then there is no guarantee that your long awaited report is going to provide you with useful feedback. I have certainly had multiple experiences in which I have waited six or eight months only to receive a three or four sentence report which made me question whether the referee even read past the introduction. In these situations I think you just have to have a thick skin and submit the paper again. On the other hand your paper may be rejected with a report that provides a lot of insightful feedback that helps you strengthen your paper. In this case I would hardly call the wait "a big waste of time".<|endoftext|>
TITLE: What are some examples of total derived functors that can't be computed from a functorial replacement?
QUESTION [7 upvotes]: (Or more generally, what are some examples of Kan extensions which are not pointwise?)
By a total derived functor of a functor $F: C \to D$, I simply mean a (left or right) Kan extension of $F$ along some localization $C \to C[W^{-1}]$ at a class of arrows $W$. Usually when such a Kan extension exists, it comes from a functor $L: C \to C$ and a natural transformation $\eta: 1_C \implies L$ (or the other way around depending on handedness) such that on the image of $L$, $F$ already takes arrows in $W$ to isomorphisms, and the components of $\eta$ are in $W$ (Riehl calls this a deformation of the functor $F$, but I'm not sure how standard this terminology is). When such an $L$ exists, the total derived functor is simply $F \circ L$.
My question is: does a total derived functor ever exist without such a deformation / replacement functor $L$ existing? I presume the answer is yes, and I'd like to see some examples.
My motivation for this question actually comes from the fact that when a deformation exists, the Kan extension involved is pointwise, and in fact absolute (preserved by all functors). I'm actually just trying to understand pointwiseness of Kan extensions better by looking at examples of non-pointwise Kan extensions. There is an example of a non-pointwise Kan extension in Borceux's Handbook of Categorical Algebra, and it generalizes a bit, but I'd like to see more examples. Since the codomain category of a total derived functor rarely has many (co)limits, it's a good candidate to admit non-pointwise Kan extensions into it, and it seems one just has to get around the fact that these Kan extension are pointwise if they are computed as deformations.

REPLY [3 votes]: Yes. In fact, one such example comes from homotopical algebra:

Proposition. Let $\mathcal{C}$ be a small homotopical category and let $\gamma : \mathcal{C} \to \operatorname{Ho} \mathcal{C}$ be the localising functor. Then $\mathcal{C} (-, -) : \mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathbf{Set}$ admits a left Kan extension along $\gamma^\mathrm{op} \times \gamma : \mathcal{C}^\mathrm{op} \times \mathcal{C} \to \operatorname{Ho} \mathcal{C}^\mathrm{op} \times \operatorname{Ho} \mathcal{C}$, namely $\operatorname{Ho} \mathcal{C} (-, -) : \operatorname{Ho} \mathcal{C}^\mathrm{op} \times \operatorname{Ho} \mathcal{C} \to \mathbf{Set}$. 

Bizarrely enough, this is a pointwise left Kan extension – so the proof is a direct calculation. If $\gamma : \mathcal{C} \to \operatorname{Ho} \mathcal{C}$ has a (necessarily fully faithful) right adjoint, then this can also be calculated using a right deformation. I think the converse is also true, but at any rate, right deformations usually do not exist.<|endoftext|>
TITLE: Motivating Lubin-Tate theory
QUESTION [39 upvotes]: The Lubin-Tate theory gives an amazingly clean and streamlined way of constructing the subfield (usually denoted) $F_\pi\subset F^\mathrm{ab}$ for a local field $F$ fixed by the Artin map associated to the prime element $\pi$ (i.e. such that $F^{\mathrm{ab}}=F_\pi\cdot F^{\mathrm{un}}$ with the usual notations). The idea to consider 1-dim. formal groups over the ring of integers $\mathcal{O}_F$ is a deus ex machina for me, and I wonder if anyone can explain Lubin-Tate's motivation to consider such a thing?
Related, on page 50 of J.S. Milne's online notes on the class field theory, he offers the speculation that the motivation comes from complex multiplication of elliptic curves and how one might try to get an analogue of the theory for local fields. But this requires again that it is somehow natural to consider formal groups as an analogue which I think still needs a motivation.

What is the motivation to consider formal groups a la Lubin-Tate theory? Is there a way to motivate their construction?

REPLY [82 votes]: Sorry I didn’t see this earlier. My memory is vague, and probably colored by subsequent events and results, but here’s how I recall things happening.
Since I had read and enjoyed Lazard’s paper on one-dimensional formal group (laws), which dealt with the case of a base field of characteristic $p$, I decided to look at formal groups over $p$-adic rings. For whatever reason, I wanted to know about the endomorphism rings of these things, and gradually recognized the similarity between, on the one hand, the case of elliptic curves and their supersingular reduction mod $p$, when that phenomenon did occur, and, on the other hand, formal groups over $p$-adic integer rings of higher height than $1$.
I had taken, or sat in on, Tate’s first course on Arithmetic on Elliptic Curves, and was primed for all of this. In addition, I was aware of Weierstrass Preparation, and the power it gave to anyone who wielded it. And in the attempt to prove a certain result for my thesis, I had thought of looking at the torsion points on a formal group, and I suppose it was clear to me that they formed a module over the endomorphism ring. Please note that it was not my idea at all to use them as a representation module for the Galois group.
But Tate was looking over my shoulder at all times, and no doubt he saw all sorts of things that I was not considering. At the time of submission of my thesis, I did not have a construction of formal groups of height $h$ with endomorphism ring $\mathfrak o$ equal to the integers of a local field $k$ of degree $h$ over $\Bbb Q_p$. Only for the unramified case, and I used extremely tiresome degree-by-degree methods based on the techniques of Lazard. Some while after my thesis, I was on a bus from Brunswick to Boston, and found not only that I could construct formal groups in all cases that had this maximal endomorphism structure, but that one of them could take the polynomial form $\pi x+x^q$. Tate told me that when he saw this, Everything Fell Into Place. The result was the wonderful and beautiful first Lemma in our paper, for which I can claim absolutely no responsibility. My recollection, always undependable, is that the rest of the paper came together fairly rapidly. Remember that Tate was already a master of all aspects of Class Field Theory. But if the endomorphism ring of your formal group is $\mathfrak o$ and the Tate module of the formal group is a rank-one module over this endomorphism ring, can the isomorphism between the Galois group of $k(F[p^\infty]])$ over $k$ and the subgroup $\mathfrak o^*\subset k^*$ fail to make you think of the reciprocity map?<|endoftext|>
TITLE: Invertible combinations of linear maps on infinite-dimensional vector spaces
QUESTION [6 upvotes]: Let $V$ be a real infinite-dimensional vector space of cardinality $\kappa$. Does there exist a set $\Omega$ of cardinality $\kappa$ of linear maps from $V$ to $V$ such that for every $n\geq 1$, every nonzero vector $(x_1,\ldots, x_n) \in \mathbb{R}^n$, and distinct $A_1,\ldots, A_n \in \Omega$, the map 
$$x_1A_1+\cdots+x_nA_n$$
is an isomorphism from $V$ to $V$?
When $V$ is an infinite-dimensional separable Hilbert space, there exists a countable set $\Omega$ such that 
$$U^2=-I,~UV+VU=0,~~~~~~~~~~~~~~~(1)$$
for all distinct $U,V \in \Omega$. Then every nontrivial linear combination of elements in $\Omega$ is invertible. The question is that if generally one can forego the conditions (1) but increase the cardinality of the set and still have invertible nontrivial linear combinations.

REPLY [3 votes]: I think this is true for any infinite $\kappa$, which is presumably the case you are interested in. (For finite $\kappa$, the situation is more interesting but well understood: see for instance here.) 
If $\kappa$ is infinite, you can construct such a family essentially by using the Clifford algebra. Namely, take a real vector space $W$ of dimension $\kappa$. Equip it with a positive definite quadratic form, and generate the Clifford algebra $V:=Cliff(W)$, which will be the union of Clifford algebras corresponding to finite-dimensional subspaces of $W$. It is easy to see that $\dim V=\dim W$, and that (say, left) action of $W$ on $V$ is invertible. Now let your family of operators be the action of some basis vectors of $W$. 
Remark. I find it interesting that if you ask the same question over the field $\mathbb{C}$, you must assume that $\kappa\ge c$ instead of $\kappa$ being infinite (and then you can use fields of rational functions instead of Clifford algebras).<|endoftext|>
TITLE: 'Updated' book in the same spirit as Dieudonné's Panorama des mathématiques pures
QUESTION [11 upvotes]: Today a colleague of mine asked me if I knew of any "more modern version" of J. Dieudonné's Panorama des mathématiques pures. Le choix bourbachique. 
The very first thing that instantly came to my mind was The Princeton Companion to Mathematics; however, upon careful consideration, as I see it, the spirit of these two works is different (for the context and a summary of Dieudonné's work, you may refer to this review by P. Halmos, for example).
Being myself interested in possible answers to the question, I'll pose it here: 

Does there exist an 'updated' book in a similar spirit as Dieudonné's
  Panorama des mathématiques pures?


I've just found out that a similar question was asked on Mathematics Stack Exchange, and the only answer proposed was indeed The Princeton Companion to Mathematics: An updated alternative to “A Panorama of Pure Mathematics”.

REPLY [2 votes]: you might consider Oddifreddi's The Mathematical Century: The 30 Greatest Problems of the Last 100 Years

The twentieth century was a time of unprecedented development in
  mathematics: more theorems were proved and results found in a hundred
  years than in all of previous history. Here this development is told
  through thirty highlights of pure and applied mathematics. Each tells
  the story of an exciting problem, from its historical origins to its
  modern solution, in lively prose free of technical details. The book
  opens by discussing the four main philosophical foundations of
  mathematics of the nineteenth century and ends by describing the four
  most important open mathematical problems of the twenty-first century.<|endoftext|>
TITLE: Characteristic Variety of the Principal Symbol solves PDE system?
QUESTION [8 upvotes]: In the study of partial differential equations, it is often considered enough to analyze the principal symbols and their characteristic variety (see for example, http://www.sciencedirect.com/science/article/pii/S0001870802000993, and https://en.wikipedia.org/wiki/Symbol_of_a_differential_operator).  In the first link, only the principal symbol is used to count the number of solutions to a holonomic system.  But there are easy examples for which this information does not appear to be enough.  Consider the two-dimensional system:
$$
(\partial+1)\phi(z,z')=0
$$
$$
(\partial+\partial')\phi(z,z')=0
$$
and,
$$
(\partial+z)\psi(z,z')=0
$$
$$
(\partial+\partial')\psi(z,z')=0.
$$
Both have the same principal symbols, namely, $p$ and $p+p'$, however, the first has the solution $\phi(z,z')=e^{z'-z}$, while the second has no solution (other than $\psi(z,z')=0$).  
It seems to be suggested in the first link that the second has a solution on an integral curve, perhaps, but it is not clear to me if that is the right context and if so, how does one construct such a curve?

REPLY [3 votes]: Let me elaborate a little about what Robert wrote. When analyzing a system of PDE's, there are two separate issues that need to be considered. Below, I'll assume that you want to study the space of smooth solutions to a homogeneous linear system of PDE's.
Before you even look at the characteristic variety (i.e., study the system microlocally), you have to look at the formal properties of the system. In other words, are there any formal power series solutions (without considering convergence) to the system? If there aren't any, then there's no point in going any further. If there are, the next important question is how "many" formal solutions are there?
Figuring this out is not always easy, especially with an overdetermined system like yours. Note that each of your systems has 2 unknown real-valued functions but 4 real-valued equations. As Robert says, the two systems have completely different formal properties. To analyze them properly, you often have to do what's known as prolongation which involves adding more unknown functions and equations obtained by differentiating the original equations. To determine whether you have formal solutions and "how many", you have to keep doing this until the system satisfies a condition known as involutivity.
Only after you have an involutive system does the characteristic variety (or ideal or sheaf) of the involutive system (and not the original system) tell you anything meaningful about the system of PDE's.
In your examples, the symbol and therefore the characteristic (variety, ideal, sheaf) are the same for both systems. However, since the second one is not involutive, the characteristic object tells you nothing about the second system. However, if you prolong the second system until it becomes involutive, then the characteristic object of the prolonged system will tell you something about the second system of PDE's and its solutions. In particular, they will be different from those of the first system.
The theory for this is quite involved. I'll let others provide more specific references for all of this.<|endoftext|>
TITLE: Woodin on Posner-Robinson for the hyperjump and sharp
QUESTION [13 upvotes]: The Posner-Robinson theorem states that, if $X$ is noncomputable, there is some $G$ such that $X\oplus G=G'$; that is, even though genuine jump inversion only works above $0'$, every (nontrivial) $X$ is "almost" the jump of something. There are a number of extensions and variations of the Posner-Robinson theorem; I'm interested in two, due to unpublished work of Hugh Woodin.
In multiple places - e.g. page 208 of "Proceedings of the 13th Asian Logic Conference" - we find the claim that Woodin proved the following higher-order analogues of the Posner-Robinson theorem:

If $X$ is not hyperarithmetic, then there is some $G$ such that $X\oplus G\equiv_T \mathcal{O}^G$.
(Assume $X^\#$ exists for every real $X$.) If $X$ is not constructible, then there is some $G$ such that $X\oplus G\equiv_T G^\#$.

My question is: are these proofs available anywhere?

REPLY [6 votes]: This result by Woodin is stated as "unpublished" in a paper published by his former Ph.D. student Xianghui Shi in July 2015, so that seems to be an authoritative source:



Axiom $I_0$ and higher degree theory
The Journal of Symbolic Logic 80, 970-1021 (2015).<|endoftext|>
TITLE: Is the heat kernel for the hyperbolic plane uniformly continuous in $t\in(0,\infty)$?
QUESTION [5 upvotes]: let $\mathbb{H}$ be the hyperbolic plane and let $k(t,x,y)$ be the associated heat kernel.
I am wondering, if for any fixed $y\in M$ and $\epsilon >0$ the function $u_t(x):=k(t,x,y)$ is continuous in $S:=\lbrace x\in M : d(x,y)>\epsilon \rbrace$ uniformly in $t\in(0,\infty)$?
For instance in Euclidean space $\mathbb{R}^n$ the heat kernel is given by $k(t,x,y)=\frac{1}{(4\pi t)^{n/2}}e^{-\frac{\Vert x-y \Vert^2}{4t}}$. Hence for $\mathbb{R}^n$ the claim is true because the function $u_t(x)$ has bounded derivative on S, i.e. $\sup_{t>0} \vert  \nabla u_t(x) \vert\leq C$.
There is a formula for hyperbolic plane as well, given by
$$k(t,x,y)=\frac{\sqrt{2}}{(4\pi t)^{3/2}}e^{-\frac{t}{4}}\int\limits_{d(x,y)}^{\infty}\frac{s e^{-\frac{s^2}{4t}}}{\sqrt{\cosh{s}-\cosh{d(x,y)}}}ds$$
but I do not know how to estimate the derivative for this function.
Q: How can I prove/disprove that for any $x^{*}\in\mathbb{H}^2, x^{*}\neq y, \forall \epsilon>0$ $\exists\delta>0,$ s.t. $\vert u_t(x)-u_t(x^{*}) \vert\leq \epsilon$ for all $x\in \mathbb{H^2}$ with $d(x,x^{*})<\delta$? If the above statement is wrong, What happens if we restrict the domain to $t\in (0,T)$ for $T>0$ arbitrary and fixed?
Any help will be very appreciated!
Best wishes

REPLY [4 votes]: It follows from the interior  Schauder estimates for parabolic equations. Namely, put for simplicity's sake $y=0$, $k(t,x)=k(t,x,0)$ and for negative values of $t$ continue $k$ by zero: $k(x,t)=0$, $t<0.$ For $r>0$ denote $B_r(x_0)=\{x\in \mathbb{H}^2|d(x,x_0)<r\}$ a ball on the plane and $B_r=B_r(0)$. Then $k(x,t)$ is bounded on a cylindrical domain  $H_r=\mathbb R\times (\mathbb{H}^2\backslash B_r)$. Now fixing $\alpha\in(0,1)$ and applying the interior Schauder estimate in the cylinder $G_r(x_0,t_0)=(t_0-r^2,t_0)\times B_r(x_0)$, $d(x_0,0)\ge 2r$, we have
$$
\|k\|_{C^{2,\alpha}(G_{r/2}(x_0,t_0))}\le C_r\sup_{G_{r}(x_0,t_0)}|k|\le C_r\sup_{H_r}|k|.
$$
It follows that 
$$
\sup_{H_{2r}}|\nabla_x k|\le C_r\sup_{H_r}|k|.
$$<|endoftext|>
TITLE: A question about simple closed plane polygons
QUESTION [6 upvotes]: For any positive integer $n$ greater than $3$, let $P(1),P(2),...,P(n)$ be a set of $n$ pairwise distinct points in the Euclidean plane, no three of which are collinear. Let $H(P(1),P(2),...,P(n))$ be the set of simple closed (not necessarily convex) polygons whose set of vertices is exactly the set $P(1),P(2),...,P(n)$. Partition $H(P(1),P(2),...,P(n))$ into equivalence classes, with two polygons in the same class just in case they are congruent. Let $N(P(1),P(2),...,P(n))$ be the number of these equivalence classes. Is $H(P(1),P(2),...,P(n))$ always non-empty? For which positive integers $n$ greater than 3, has the maximum possible value of $N(P(1),P(2),...,P(n))$ been computed-where the maximum is to be taken over all sets of n pairwise distinct co-planar points, no three of which are collinear?

REPLY [7 votes]: What you call set $H$ is the set of simple polygonalizations of the $n$ points $P$.
$H$ is only empty if all $n$ points are collinear.
For one may form a star-shaped polygon by selecting a point $x$
in the convex hull, not in $P$ and not collinear with any two points of $P$, and connecting $P$ in angular order around $x$:

         


The maximum number (your $N$) of polygonizations of a $n$-point set $P$
is $c^n$ for $c$ somewhere between about $4.6$ and $56$.

          


          

(Image from Erik Demaine's webpage.)<|endoftext|>
TITLE: Is there a class of functions acting on a set of projected points that remain invariant under changes in projection parameters?
QUESTION [5 upvotes]: Suppose I have a set of $k$ points $\{x_1,x_2,\ldots,x_k\}$ in $\mathbb{R}^n$ that I can project into $\mathbb{R}^m$ with the linear operator $\mathcal{P}$, with $\alpha, \beta, \ldots$ parameters of the projection operator. Is there a well known best method for determining the class of functions $f:(\mathbb{R}^m)^k \rightarrow \mathbb{R}$ such that $f(\mathcal{P}x_1,\mathcal{P}x_2,\ldots,\mathcal{P}x_k)$ is invariant under changes in $\alpha, \beta, \ldots$?
As an example, a situation I'm interested in has $\mathcal{P}: \mathbb{R}^3 \rightarrow \mathbb{R}^2, x \mapsto A[\phi, \theta, \eta]x$, where:
$A[\phi, \theta, \eta] = \eta \left(
\begin{matrix}
    \cos\phi\cos\theta & \sin\phi\cos & \sin\theta \\
    -\sin\phi & \cos\phi & 0
\end{matrix}\right)$
i.e. the projection of points in 3-D space onto the plane with unit normal $(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$, with an additional scaling parameter of $\eta$.
Setting derivatives of $f$ with respect to the parameters equal to zero yields a set of equations with no obvious general solution, and a feeling that I'm missing a certain set of tools.

REPLY [3 votes]: I can think of (at least) two ways of interpreting this question.
First: You are given some specific list of $k$ points $x_1$, $x_2$, ..., $x_k$ in $\mathbb{R}^n$, and you want to detect whether $k$ points $y_1$, ..., $y_k$ in $(\mathbb{R}^m)^k$ could be a linear projection of the original $k$ points. 
Solution: Compute the vector space of all $(c_1,\ldots, c_k)$ such that $\sum c_i x_i=0$. For a basis of this space, whether or not $\sum c_i y_i=0$. If so, the $y_i$ are a projection, if not, they are not.
Second: You want to construct functions $f$ on $(\mathbb{R}^m)^k$ such that, if $(y_1, \ldots, y_k)$ and $(y'_1, \ldots, y'_k)$ are both projections of the same points $(x_1, \ldots, x_k)$, then $f(y_1, \ldots, y_k) = f(y'_1, \ldots, y'_k)$.
Solution: There are no nonconstant examples. I will show that we must have $f(y'_1, \ldots, y'_k) = f(y_1,\ldots,y_k)$ for any $y$ and $y'$ in $(\mathbb{R}^m)^k$.
For notational simplicity, I'll take $n=m+1$. Let $y^j_i$ be the vector in $\mathbb{R}^m$ whose first $j$ coordinates are that first $j$ coordinates of $y_i$ and whose last $m-j$ coordinates are the last $m-j$ coordinates of $y'_i$. So $y^0_i = y'_i$ and $y^m_i = y_i$.
Let $x^j_i = \mathbb{R}^{m+1}$ be the vector whose first $j$ coordinates are those of $y_i$ and whose last $m+1-j$ coordinates are those of $y'_i$. Then $(y_1^j, \ldots, y_k^j)$ and $(y_1^{j+1}, \ldots, y_k^{j+1})$ are both projections of $(x^j_1, \ldots, x^j_k)$. So
$$f(y_1^0,\ldots,y_k^0) = f(y_1^1, \ldots, y_k^1) = \cdots = f(y_1^m, \ldots, y_k^m)$$
and $f(y'_1, \ldots, y'_k) = f(y_1,\ldots,y_k)$.
I can also think of other possibilities, like that you have a data bank of a few billion options for $x$, and you want to find the best match, but this seems like enough for now.<|endoftext|>
TITLE: Growth of dimension of fixed spaces in $GL_n(\mathbb{Q}_p)$-representations
QUESTION [12 upvotes]: Let $\pi$ be a generic irreducible admissible representation of $GL_n(L)$, where $L$ is a $p$-adic field, $R$ is its ring of integers, and $\mathfrak{p}$ is its prime ideal.  The conductor of $\pi$ has the following definition (follwoing Jacquet, Pietetski-Shapiro, and Shalika): fix a non-negative integer $r$, and let $K(r)$ be the subgroup of matrices:
\begin{pmatrix} x & y \\ u & v \end{pmatrix}
where $x\in GL_{n-1}(R)$, $y\in R^{n-1}$, $u$ is a vector of elements of $\mathfrak{p}^r$ and $v\in 1 + \mathfrak{p}^r$.  The conductor of $\pi$ is the minimal $r$ such that $\pi$ has a $K(r)$-fixed vector.  In this case, $\pi$ has a unique $K(r)$-fixed vector (up to scalar multiplication).
For $n = 1$ or $2$, the growth of the $K(r)$-fixed vectors is known.  Clearly, if $\pi$ is an irreducible $GL_1(L)$-representation of conductor $r$, then $\dim(\pi^{K(r')}) = 1$ if $r' \geq r$ and $0$ otherwise.  If $\pi$ is an irreducible $GL_2(K)$-representation and $r' \geq r$ then $\dim(\pi^{K(r')}) = r' - r + 1$.
I have two questions, the first of which I assume is known:
1). Let $n \geq 3$ and let $\pi$ be a $GL_n(L)$-representation of conductor $r$.  Is there a formula for computing $\dim(\pi^{K_(r')})$, when $r' \geq r$?
2). If $G$ is a reductive group defined over $L$, is there a sequence of open-compact subgroups $K'(r)$ that mimic the behavior of the $K(r)$ in $GL_n$, in the sense that for every irreducible admissible representation $\pi$ of $G(L)$,  there is an $r$ such that $\pi$ has a unique $K'(r)$-fixed vector?

REPLY [11 votes]: For question 1, you are asking about the theory of oldforms (or oldvectors). I think the canonical reference for $\mathrm{GL}_n$ is "Oldforms on $\mathrm{GL}_n$" by Mark Reeder. In particular, he discusses how to find a basis of $\pi^{K(r')}$ if one already has a basis of $\pi^{K(r)}$ for all $r < r'$.
For question 2, I don't believe there is a particularly satisfactory answer, though this is an active area of research. For some groups of small rank this has been studied; for example, on $\mathrm{GSp}_4$ (or rather $\mathrm{PGSp}_4$), Brooks Roberts and Ralf Schmidt have studied this question extensively, with most of the results appearing in their book "Local Newforms for $\mathrm{GSp}_4$ ". For split special orthogonal groups, the theory has been developed in Pei-Yu Tsai's recent thesis, "On Newforms for Split Special Odd Orthogonal Groups".<|endoftext|>
TITLE: Can there be computable non-standard models of PA in a weaker sense?
QUESTION [16 upvotes]: By Tennenbaum's theorem, in the usual sense of computability for models,
neither addition nor multiplication can be computable in a countable non-standard model of PA.
Weak version:

Can addition or multiplication become computable if non-equality
only needs to be computably enumerable, rather than computable?

In other words, can there be a countable structure on which [either addition is computable or multiplication is computable] and some quotient of that structure is a non-standard model of PA?
If I understand the proof of the addition version of Tennenbaum's theorem correctly, then for addition to be computable there would need to be a representative $r_0$ of a non-standard number such that $\{r : r\text{ represents the same element as }r_0\}$ ​ is not computably enumerable.
I'm mainly after an answer to either version, rather than both, so the following will be quite strong.
Super-Strong version:

Are there a computable function ​ $d : \{0,1,2,3,\ldots\}^2 \to \mathbb{Q}$ and
binary operations $+_M$ and $\times_M$ on $\{0,1,2,3,\ldots\}$ ​ (by Tennenbaum, they can't be computable) such that

composing d with the inclusion from $\mathbb{Q}$ to the real numbers gives a metric and
the induced metric space is complete and
there is an algorithm that approximates $+_M$ and $\times_M$ to arbitrary accuracy and
$+_M$ and $\times$ make $\{0,1,2,3,\ldots\}$ into the non-standard part of a model of PA


The standard part could just be put in with distance 1 from everything, so requiring it to be a non-standard part is stronger than requiring it be a full non-standard model.
Furthermore, we can get a non-standard model in the sense of the weak version from any witnesses to the truth of the Super-Strong version, by taking the set of well formed expression using ${0, 1,{+},{\times}, {(\; , )}}$ and elements of the metric space, and noting that both operations are computable.

REPLY [8 votes]: This is a great question!
Let me give a meager partial answer, for the case where we are talking
about nonstandard models of true arithmetic. 
Theorem. No nonstandard model of true arithmetic arises as the
quotient of a structure $\langle N,+,\cdot\rangle$ by an equivalence relation $\approx$, if $+$ is computable and $\approx$ is co-c.e. Indeed, there is no model of true arithmetic that is 
the quotient of a structure $\langle
N,+,\cdot\rangle$ by an equivalence
relation $\approx$, if both $+$ and $\approx$ are arithmetic.
Proof. Suppose that $\mathcal{N}=\langle
N,+,\cdot\rangle/\approx$ is a model of true arithmetic, and
$+$ and $\approx$ are each arithmetically definable. Let
$Z\subset\mathbb{N}$ be an arithmetic set that is not computable
from $+$ and $\approx$, such as the jump of the join of this operation and relation. Let $[s]_\approx>\mathbb{N}$ be a
nonstandard element in the quotient structure, and consider
$Z^{\mathcal{N}}\cap [s]$, using the arithmetic definition of $Z$.
Since $\mathcal{N}$ is a model of true arithmetic, this set agrees
with $Z$ on the standard numbers. Since $Z^{\mathcal{N}}\cap [s]$
is a finite set in $\mathcal{N}$, it is coded in the model. So
there is some $r$ such that $i\in Z\iff \mathcal{N}\models p_i$
divides $[r]$, where $p_i$ is the $i^{\rm th}$ prime. Using the
operation $+$ and the relation $\approx$ as oracles, we
can compute whether $\mathcal{N}$ thinks that $p_i$ divides $[r]$. Specifically, we search for a solution to $r\approx p_i\cdot x+y$, where $y$ is equivalent to one of the corresponding numbers less than $p_i$, and then check whether $y\approx 0$. Note that since $p_i$ is standard finite, we can compute $p_i\cdot x$ by repeated addition of $x$, and so we need only $+$. In this way, we
can compute $Z$ from those oracles, contrary to the choice of $Z$.
QED
Another way to think about the argument is this: there is no arithmetically definable nonstandard model of true arithmetic. This is essentially the same as Tennenbaum, and the point now is that if you had an arithmetic quotient structure, then you could produce an actual arithmetic structure, which is impossible. 
By paying a little closer attention to complexity, the same argument shows:
Theorem. If $\langle N,+,\cdot\rangle/\approx$ is a nonstandard model of arithmetic that is $\Sigma_2$-sound, then it cannot be that $+$ and $\approx$ are computable from $0'$. 
Proof. We simply take $Z=0''$ in the previous argument. If the model is $\Sigma_2$ sound, then it will agree on the standard elements of $0''$, and so using $0'$ as an oracle, we would be able to compute both $+$ and $\approx$, and hence the coded version of $0''$, giving a contradiction. QED
Here is a slightly improved version:
Theorem. If $\mathcal{N}=\langle N,+,\cdot\rangle/\approx$ is a nonstandard model of PA, then every real in the standard system of $\mathcal{N}$ is computable from $+$ and $\approx$. 
Proof. Suppose that $Z$ is in the standard system of $\mathcal{N}$, so that it is coded in $\mathcal{N}$, in the sense that there is $r\in N$ such that $i\in Z\iff p_i$ divides $r$ in $\mathcal{N}$. Using $+$ and $\approx$ as oracles, we can find $x$ and $y<p_i$ for which $r\approx p_i\cdot x+y$, and then check whether $y\approx 0$. So $Z$ is computable from $+\oplus\approx$. QED
So the models of PA that do arise in the way you describe must have comparatively low standard systems. If there is any real in the standard system of the quotient $\langle N,+,\cdot\rangle/\approx$ that is not computable from $0'$, then it cannot be that $+$ is computable and $\approx$ is co-c.e.<|endoftext|>
TITLE: What is the chromatic number of the "conic hypergraph" on a non-singular plane cubic?
QUESTION [11 upvotes]: Can we color the points of a complex non-singular plane cubic curve with finitely many colors so that no conic intersects the curve at 6 distinct points of the same color?
If so, what is the smallest number of such colors?

REPLY [9 votes]: You mean the six points to be distinct, of course (or not all six points to be the same point).
Fixing the analytic identification $(\wp(z),\wp'(z))$ with $T = \mathbb{C}/\mathbb{\Lambda}$, the Abel-Jacobi theorem implies that the six points of intersection of $E \subset \mathbb{CP}^2$ with any conic sum to zero on the torus. Now take the fundamental parallelogram $P$ of $\Lambda$, subdivide it into $36$ equal subparallelograms, and use $72$ colors so that monochromatic subsets of $P$ are either interior points of one of the subparallelograms or of its left or down boundary edges, or else a single lattice point from the subdivision of $P$ (a $6$-division point).<|endoftext|>
TITLE: When is the image of a 2-dim l-adic representation associated to a modular form open
QUESTION [15 upvotes]: I know the following theorems by Serre:
1, The 2-dim l-adic representation associated to a non-CM elliptic curve is open.
2, The  2-dim l-adic representation associated the weight-12 cusp form $\Delta$ has open image (even before Deligne's construction of 2-dim l-adic representations).
So is there any general theorem about When the image of a 2-dim l-adic representation associated to a modular form is open?

REPLY [18 votes]: This is more subtle than it looks. I asked exactly the same question some years back (see here); but I'm not going to flag this question as duplicate, because the answer that was given to my question at the time, which was identical to @eric's comment above, was wrong.
The correct answer is this. Let $f$ be a normalized non-CM newform. Associated to $f$ there is a number field $L = \mathbf{Q}(a_n(f) : n \ge 1)$, and for each prime $\mathfrak{P}$ of $L$, we get a Galois representation
$$ \rho_{f, \mathfrak{P}}: G_{\mathbf{Q}} \to \operatorname{GL}_2(L_{\mathfrak{P}}). $$
It is not, however, generally the case that this has open image. There is a trivial obstruction coming from the fact that the determinant of $\rho_{f, \mathfrak{P}}$ is the product of a finite-order character and a power of the cyclotomic character, so there is a finite-index subgroup of $G_{\mathbf{Q}}$ whose determinant lands in $\mathbf{Q}_p^\times$, which is generally much smaller than $L_{\mathfrak{P}}^\times$.
One can still hope that the image contains an open subgroup of $\operatorname{SL}_2(L_{\mathfrak{P}})$, but this also turns out to be false, for a much more subtle reason. Consider the subgroup $\Gamma \subseteq \operatorname{Aut}(L / \mathbf{Q})$ consisting of those automorphisms $\sigma$ for which $f^\sigma$ is equal to the twist of $f$ by some Dirichlet character ("inner twists"). The fixed field of $\Gamma$ is a subfield $F \subseteq L$, and Momose showed that there's a finite-index subgroup of $G_{\mathbf{Q}}$ such that for any $\tau$ in this subgroup $\rho_{f, \mathfrak{P}}(\tau)$ has trace in $F_{\mathfrak{q}}$, where $\mathfrak{q}$ is the prime of $F$ below $\mathfrak{P}$. So if $F_\mathfrak{q} \ne L_{\mathfrak{P}}$, which occurs for infinitely many $\mathfrak{P}$ if $F \ne L$, then the image of $\rho_{f, \mathfrak{P}}$ cannot contain an open subgroup of $\operatorname{SL}_2(L_{\mathfrak{P}})$.
One might still hope that the image at least contains an open subgroup of $\operatorname{SL}_2(F_{\mathfrak{q}})$ but this isn't generally true either -- there is an obstruction coming from a Brauer group, which can genuinely be nontrivial! But this weaker statement does, at least, hold for all but finitely many $\mathfrak{P}$, as was shown by Momose.
If $f$ has level 1, then the group $\Gamma$ of inner twists is always trivial, so $F = L$ and the image contains an open subgroup of $\operatorname{SL}_2(L_\mathfrak{P})$ for all $\mathfrak{P}$. More generally, if $f$ has trivial character it's "usually" the case that there are no nontrivial inner twists. But every non-CM form of non-trivial character has at least one nontrivial inner twist, coming from the action of complex conjugation, so this bad behaviour is unavoidable.
All of this is very nicely explained in Ribet's paper "On l-adic representations attached to modular forms. II" (Glasgow Math. J. 27, 1985, 185--194), which you can find on Ribet's website here.<|endoftext|>
TITLE: How and why did mathematicians develop spin-manifolds in differential geometry?
QUESTION [27 upvotes]: First of all, I am neither a physicist nor a mathematician. And I am afraid that mathoverflow is not a suitable place for my question, but having asked similar questions on math SE it is obvious that this question is not appropriate for math.SE.
As far as I have searched in mathematical physics literature, historically, physicists such as Pauli and Dirac pioneered the concept of spin for a particle. Dirac developed his theory for spin-1/2 electrons by factorizing the Klein-Gordon equation to find a linear relativistic equation that is compatible with Schrodinger's wave equation but doesn't give negative probability density. He factorized the Klein-Gordon equation and found algebraic constraints that gave the Clifford algebra $Cl(1,3)$.
Now that's the algebraic part of the story. What I don't understand is how the concepts in Spin manifolds and Spin geometry were developed from the point of view of differential geometry and topology. 
Apparently, Elie Cartan was one of the pioneers and he has written a book  about it (I have read its first few chapters). But his language is very different from the language of differential geometry that we use today. He doesn't talk about covering spaces, vector bundles or connections the way that has become common in today's math literature. So, I find it very difficult to trace the chain of thoughts that has led physicists and mathematicians to develop spin geometry in its current language.
I'd like to know how our today's mathematical physics literature has been developed historically and what is the aim of using and further developing this sophisticated language.

REPLY [5 votes]: The concept of spinor is a fascinating one, not only from the mathematical point of view but also from the physical point of view: I find extremely remarkable that some of the fundamental particles of nature are actually described in terms of spinors, a fact that has been extensively verified experimentally. Therefore, spin geometry is not just an abstract mathematical theory, but the language in which describe at least some aspects of the universe.
Spinors were invented, or rather discovered, by E. Cartan 1913 in his work on the classification of complex irreducible representations of simple Lie groups, where he discovered the "spinorial" representations of orthogonal groups. Fifteen years later, P. Dirac independently invented spinors in his attempt to find a "relativistic wave equation". 
E. Cartan made the connection between both constructions much later, in 1937. However, he was dissatisfied with the status of the theory of spinors, as he was fearing that it would not be possible to define spinors globally on a manifold. Then, spin structures on manifolds came along (I am not sure exactly when and how) and it was clear that one can define spinors on a manifold through the associated bundle construction of a spin structure.
What I want to stress in this answer is the following: it has been widely accepted that the right mathematical set up to deal with spinors is that of spin structures on manifolds. However this is some times not the most convenient formalism: in particular it is too restrictive for many physical applications, especially in the context of String Theory (I can elaborate on this if needed). In order to obtain a general theory of spinors on a manifold one has to get rid of the concept of spin structure. Instead, one defines a spinor bundle $S$ as a bundle of (say irreducible) Clifford modules over the bundle of Clifford algebras $Cl(M,g)$ of your pseudo-Riemannian manifold $(M,g)$. This approach was already advocated by E. Schrodinger (in the language of that time), but it was soon forgotten when the description in terms of spin groups gained favor. Note that $Cl(M,g)$ is not obstructed, but, depending on the representation used, $S$ may be. Clearly, if $(M,g)$ admits a spin structure $P$, it admits a bundle of irreducible Clifford modules over $Cl(M,g)$ which can be constructed as an associated vector bundle to $P$. However, what about the other direction? The manifold $(M,g)$ needs to be spin in order to admit a bundle of irreducible Clifford modules over $(M,g)$? The answer to the latter question is big no. This can already be seen from the theory of spin$^c$ structures. There are manifolds that admit spin$^c$ structures that do not admit any spin structure, and such spin$^c$ structure can be used to construct bundles of irreducible complex Clifford modules over $Cl(M,g)$. Obtaining the general topological obstruction to the existence of a bundle of irreducible/faithful Clifford modules over $(M,g)$ is a subtle mathematical problem addressed only relatively recently in the literature in the work of T. Friedrich, A. Trautman, C. Lazaroiu and C. S. Shahbazi.<|endoftext|>
TITLE: Smoothness of the fourth power of the geodesic distance in a Finsler geometry
QUESTION [10 upvotes]: The simplest form of Finsler metric is:
$ds (X, dx)=\sqrt[4]{g_{\alpha, \beta, \gamma, \delta}(X).dx^{\alpha}dx^{\beta}dx^{\gamma}dx^{\delta}}$, where $g_{\alpha, \beta, \gamma, \delta}$ is a fourth degree polynomial that smoothly depends on $X$.
My question is: is it always true that $s^4(X)$, the fourth power of the geodesic distance from an origin point $O$ to $X$, constructed from the Finsler metric $ds$, is everywhere smooth on a neighboorhood of $O$?
My question arises from the fact that, when we take a Riemmanian metric $ds (X, dx)=\sqrt[2]{g_{\alpha, \beta}(X).dx^{\alpha}dx^{\beta}}$, then we know that the squarred geodesic distance $s^2(X)$ is everywhere smooth on a neighboord of $O$. Smoothness is easy to prove outside the origin $O$. At the origin, we need to use this argument: the exponential map is smooth and radially isometric.
Unfortunately, this argument seems not to be decisive in the Finslerian case, where the exponential map is not anymore smooth at the origin. I don't find any answer to my question in classical books on Finsler geometry.
Thanks.

REPLY [9 votes]: It turns out that the answer is 'No, the fourth power of the geodesic distance from a point $p$ in a Finsler space $M$ whose norm, raised to the fourth power, is a smooth convex quartic form is not generally a smooth function on a neighborhood of $p$'.  
While the desired smoothness does hold in the Minkowski case (i.e., when, in some local coordinate system, the functions $g_{\alpha\beta\gamma\delta}(X)$ are constant) and (of course) in the case that the norm is the square root of a positive definite quadratic form, it does not hold in general.  (In fact, at least in the $2$-dimensional case, it appears that the above two cases are the only cases in which, for all points $p$, the fourth power of the geodesic distance from $p$ is $C^6$ near $p$. — RLB)
For example, let $M$ be the upper half-plane $y>0$ in the $xy$-plane, and let
$$
ds^4 = \frac{\mathrm{d}x^4 + 2A\,\mathrm{d}x^2\mathrm{d}y^2 + \mathrm{d}y^4}{y^4},
$$
where $A$ is a constant satisfying $0<A<1$. It is easy to show that this does define a strictly convex Finsler metric on $M$ that is, in fact, complete.  Moreover, it can also be shown that the geodesic distance $s_p:M\to [0,\infty)$ from any given point $p = (x_0,y_0)\in M$ has the property that $({s_p})^4$ is $C^5$ at $p$ but it is not $C^6$ there.  (It is smooth everywhere else, of course.)
It does not seem to be so easy to explain why, however.  Even in this example, which is homogeneous under translation in $x$ and simultaneous scaling in $x$ and $y$, simply computing the geodesic flow and using this explicitly to compute $s_p$ for some specific $p$ turns out to be surprisingly tricky.  Ultimately, I had to use a different method.
Here is a sketch of the method I used; details can be supplied upon request:  
In a Finsler space $(M,\mathrm{d}s)$ that is smooth (i.e., $\mathrm{d}s:TM\to[0,\infty)$ is strictly convex fiber wise and smooth away from the zero section in $TM$), the geodesic distance function from either a point or a submanifold satisfies the so-called eikonal equation wherever it is $C^1$.  Here, the eikonal equation is a first order PDE for functions $f:M\to\mathbb{R}$ that is characterized by the condition that, for each $x\in M$, the $1$-form $\mathrm{d}f_x:T_xM\to \mathbb{R}$ lies in the Legendre transform $\Sigma^*\subset T^*M$ of the tangent indicatrix $\Sigma\subset TM$ consisting of 'unit vectors' in $TM$, i.e., $\Sigma = \{ u\in TM\ |\ \mathrm{d}s(u)=1 \}$.   (For $u\in \Sigma_x = \Sigma\cap T_xM$, its Legendre transform is the element $u^*\in T^*_xM$ that satisfies $u^*(u)=1$ and $u^*(v)\le 1$ for all $v\in \Sigma_x$.)
When $\mathrm{d}s^4$ is a smooth function on $TM$ that is a homogeneous quartic on each fiber $T_pM$, one can write down the eikonal equation for $f:M\to\mathbb{R}$ as an explicit polynomial PDE on $F=f^4$.  In the above example, the eikonal equation takes the form $P\bigl(F,\,\tfrac14yF_x,\,\tfrac14yF_y\bigr)=0$, where $P$ is an irreducible polynomial in its three arguments.  In fact, one finds, after some calculation,
$$
P(r_0,r_1,r_2) =  \left(1-{A}^{2}\right)^{2}{r_{{0}}}^{9}+ \left(1-{A}^{2}\right)  
\left( \left( {A}^{2}-3 \right) {r_{{1}}}^{4}+12\,A{r_{{1}}
}^{2}{r_{{2}}}^{2}+ \left( {A}^{2}-3 \right) {r_{{2}}}^{4} \right){r_
{{0}}}^{6} - \left(  \left( -2\,{A}^{2}+3 \right) {r_{{1}}}^{8}+
 \left( 10\,{A}^{3}-6\,A \right) {r_{{1}}}^{6}{r_{{2}}}^{2}+ \left( {A
}^{4}+26\,{A}^{2}-21 \right) {r_{{1}}}^{4}{r_{{2}}}^{4}+ \left( 10\,{A
}^{3}-6\,A \right) {r_{{1}}}^{2}{r_{{2}}}^{6}+ \left( -2\,{A}^{2}+3
 \right) {r_{{2}}}^{8} \right) {r_{{0}}}^{3}- \left({r_{{1}}}^{4}+ 2\,A{r_{{1}}}^{2}{r_{{2}}}^{2}+{r_{{2}}}^{4} \right)^{3}
$$ 
Now, if $f = s_{(0,1)}$ had the property that $F = f^4$ were smooth, then it is not hard to show that
$$
F = x^4 + 2Ax^2(y{-}1)^2 + (y{-}1)^4 + R_5(x,y{-}1),
$$
where $R_5(u,v)$ vanishes to order $5$ at $(u,v)=(0,0)$. Using the explicit form of the eikonal equation above, one can then show that
$$
F = x^4 + 2A\,x^2(y{-}1)^2 + (y{-}1)^4 - 2\,x^4(y{-}1) - 2A\,x^2(y{-}1)^3 + R_6(x,y{-}1) 
$$
where $R_6(u,v)$ vanishes to order $6$ at $(u,v)=(0,0)$. However, now going back and substituting this into the eikonal equation and examining the lowest order terms (which have degree $38$), one finds that there is no solution to these equations for any term of order $6$.  Hence $F=f^4$ cannot be $C^6$.<|endoftext|>
TITLE: Is every ideal part of an operator ideal?
QUESTION [6 upvotes]: An operator ideal $\mathfrak J$ is a class of continuous operators. Namely, for every pair of complex Banach spaces, $\mathfrak X,\mathfrak Y$, we have that $\mathfrak J(\mathfrak X,\mathfrak Y) \subseteq \mathfrak L(\mathfrak X,\mathfrak Y)$ is a closed two-sided ideal, which means 
\begin{align*}
1.& \ \ \ A,B\in \mathfrak J(\mathfrak X,\mathfrak Y) \Rightarrow A+B \in \mathfrak J(\mathfrak X,\mathfrak Y), \\
2.& \ \ \  \mathfrak L(\mathfrak W,\mathfrak X)\mathfrak J(\mathfrak X,\mathfrak Y)\mathfrak L(\mathfrak Y,\mathfrak Z) \subseteq \mathfrak J(\mathfrak W,\mathfrak Z),\ \textrm{and} \\
3.& \ \ \ \mathfrak J(\mathfrak X,\mathfrak Y) \supseteq \mathfrak F(\mathfrak X,\mathfrak Y), \ \textrm{the finite-rank operators}.
\end{align*}
Now for my question: Let $\mathfrak X$ be any complex Banach space and suppose $J$ is a non-trivial closed two-sided ideal of $\mathfrak L(\mathfrak X)$. Can you always find an operator ideal $\mathfrak J$ such that $\mathfrak J(\mathfrak X,\mathfrak X) = J$?

REPLY [5 votes]: The answer is yes; take 
$$ \mathfrak{I}(\mathfrak{Y},\mathfrak{Z} ) = {\rm span}\{ T \in \mathfrak{L}(\mathfrak{Y},\mathfrak{Z}) \mid \exists U \in \mathfrak{L}(\mathfrak{Y},\mathfrak{X}) , \exists  V \in \mathfrak{L}(\mathfrak{X},\mathfrak{Z}) ,  \exists  S  \in  \mathfrak{I}(\mathfrak{X}) , T= VSU \} $$
More generally, a 2003 survey article by Laustsen and Loy on closed ideals in $\mathfrak{L}(\mathfrak{X})$ defines the operator ideal generated by a set of operators; they comment that their definition has its roots in wotk of Porta.<|endoftext|>
TITLE: Why does $\sum_{p=1}^n \exp\left(\frac{i\pi p l}{2m}\right)/\prod_{k=1,k\neq p}^n\sin\left(\frac{\pi (k-p)}{2m}\right)$ vanish?
QUESTION [15 upvotes]: While trying to prove some identities for generating functions, I ended up needing to show that 
$$\sum_{p=1}^n \exp\left(\frac{i\pi p l}{2m}\right)\prod_{\substack{k=1\\k\neq p}}^n\frac{1}{\sin\left(\frac{\pi (k-p)}{2m}\right)} \stackrel{?}{=} 0$$
for integers $m \geq 1$, $2\leq n\leq 2m$, and $l = -n+2,-n+4,\ldots n-2$. Is this identity known? I have checked it to be valid for small values, but so far I have been unable to prove the general case.

REPLY [16 votes]: Let us consider the case when $n$ is odd. Let $$P(x):=\exp\left(\frac{ixl}{2}\right)\left(\exp\left(\frac{ix}{2}\right)-\exp\left(\frac{-ix}{2}\right)\right),$$
and notice that the degree of $P$ as an exponential polynomial equals $$\max\left(\left|\frac{l+1}{2}\right|,\left|\frac{l-1}{2}\right|\right)\leq\frac{n-1}{2}.$$ Therefore this is the unique trigonometric polynomial of degree at most $\frac{n-1}{2}$ passing through any given $n$ distinct points. Let these given points be $x_p=\frac{\pi p}{m}$ for $p\in \{1,\dots,n\}$. Then, the trigonometric version of Lagrange's interpolation formula tells us that $P(x)$ can be written as 
$$P(x)=\sum _{p=1}^n P(x_p)\prod_{\substack{k=1\\k\neq p}}^n\frac{\sin\frac{1}{2}(x_k-x)}{\sin\frac{1}{2}(x_k-x_p)}.$$
By plugging $x=0$ we obtain
$$0=\sum _{p=1}^n  \exp\left(\frac{i\pi p l}{2m}\right)2i\prod_{k=1}^n\sin\left(\frac{\pi k}{2m}\right)\prod_{\substack{k=1\\k\neq p}}^n\frac{1}{\sin\left(\frac{\pi (k-p)}{2m}\right)},$$
and by dividing by the middle factors the OP's identity follows.
The case of even $n$ can be done similarly.<|endoftext|>
TITLE: Vector field on a K3 surface with 24 zeroes
QUESTION [19 upvotes]: In https://mathoverflow.net/a/44885/4177, Tilman points out that one can use a $K3$ surface minus the zeroes of a generic vector field to build a nullcobordism for $24[SU(2)]$. Given that a) this is a purely topological notion and b) we can write down defining equations for smooth projective varieties that are $K3$s, it may well be easy for experts to write down an explicit and, ideally, highly symmetric vector field on eg the Fermat quartic with 24 zeroes. 
More generally: what's an example, algebraic or otherwise, of such a vector field on an explicitly given $K3$ surface?

REPLY [7 votes]: This is probably not as explicit as what you requested, but there are recipes for how to build a Morse function on any smooth hypersurface in $CP^3$ with no index 1 or 3 critical points. There will be exactly 24 critical points, and any gradient vector field for this Morse function will be of the sort you asked about.
Such Morse functions are due to Rudolph (Topology 14,301-303. 1975), Mandlebaum-Moishezon (Topology 14, 23-40. 1976) and Harer (Math Annalen 238, 51-158, 1978).  The first two use a Morse function coming from the distance (squared) of the hypersurface to a point in the complement, in the fashion of the Morse theory proof of the Lefschetz hyperplane theorem (cf. Andreotti-Frankel,  Annals of Math 69 713–717, 1959, Bott, Michigan Math Journal 6 (3): 211–216, 1959, or Milnor's book on Morse theory). The latter uses a Lefschetz pencil.  In principle, all of these could be converted to explicit formulas, but I don't know how, or if anyone has written this down.
You can find pictures of the resulting handlebody structure, with only index 0,2 and 4 handles (so exactly 24 of them) in Harer-Kas-Kirby (Mem. AMS 1986). You can find similar pictures for all elliptic surfaces in the book of Gompf-Stipsicz.<|endoftext|>
TITLE: Congruences between modular forms and the eigencurve construction
QUESTION [5 upvotes]: This question might be too conceptual.
Congruences between modular forms (due to Shimura, Hida, etc) are  really amazing. I know that the  eigencurve construction are closely related to these relations. The basic reference is "The Eigencurve" by Coleman and Mazur. Besides, I think  "A brief introduction to the work of Haruzo Hida" by Mazur is a good introduction. 
It seems that we at first prove the congruence and then interpolate them into a family, like the Eisenstein family, and the property of the eigencurve are deduced from properties of modular forms. 
So I wonder  conversely, can  the  eigencurve construction  explain (prove) more congruences between modular forms (not just these used in building the eigencurve)? Or other interesting facts about modular forms? For example, see section 5 of "A brief introduction to the work of Haruzo Hida".
I type slowly, sorry...

REPLY [5 votes]: Here's a theorem about congruences between modular forms:
Theorem. Let $f$ be a (normalised) eigenform of weight $k$ and level $\Gamma = \Gamma_1(N) \cap \Gamma_0(p)$. Then for any $r$, and any $k'$ sufficiently close (*) to $k$, there exists an eigenform of weight $k'$ and level $\Gamma$ that is congruent to $f$ modulo $p^r$.
This is an incredibly powerful theorem; it's virtually self-evident once you know the eigencurve exists; and I don't think I know of any way of proving it without constructing the eigencurve in the process.
Does that answer your question?
(*) Here "close" means that $k'$ has to be congruent to $k$ modulo $(p-1) p^j$, for some $j$ depending on $f$, $p$ and $r$.<|endoftext|>
TITLE: Hilbert's (cancelled) 24th problem
QUESTION [41 upvotes]: Hilbert's 23 problems, ten of which were presented at the 1900 ICM in Paris, are too famous for any mathematician to not know. If one reads the descriptions of the problems in Hilbert's paper, one realizes that some questions are concrete whereas the others are stated somewhat vaguely. The 24th problem that I will quote below definitely falls into the latter category.
It seems that there was a 24th problem which was "cancelled". The following is from an article that appeared in American Mathematical Monthly in 2003.

Let me begin by presenting the problem itself. The twenty-fourth
  problem belongs to the realm of foundations of mathematics. In a
  nutshell, it asks for the simplest proof of any theorem. In his
  mathematical notebooks [38:3, pp. 25-26], Hilbert formulated it as
  follows (author's translation):
The 24th problem in my Paris lecture
  was to be: Criteria of simplicity, or proof of the greatest simplicity
  of certain proofs. Develop a theory of the method of proof in
  mathematics in general. Under a given set of conditions there can be
  but one simplest proof. Quite generally, if there are two proofs for a
  theorem, you must keep going until you have derived each from the
  other, or until it becomes quite evident what variant conditions (and
  aids) have been used in the two proofs. Given two routes, it is not
  right to take either of these two or to look for a third; it is
  necessary to investigate the area lying between the two routes.
  Attempts at judging the simplicity of a proof are in my examination of
  syzygies and syzygies [Hilbert misspelled the word syzygies] between
  syzygies [see Hilbert [42, lectures XXXII-XXXIX]]. The use or the
  knowledge of a syzygy simplifies in an essential way a proof that a
  certain identity is true. Because any process of addition [is] an
  application of the commutative law of addition etc. [and because] this
  always corresponds to geometric theorems or logical conclusions, one
  can count these [processes], and, for instance, in proving certain
  theorems of elementary geometry (the Pythagoras theorem, [theorems] on
  remarkable points of triangles), one can very well decide which of the
  proofs is the simplest. [Author's note: Part of the last sentence is
  not only barely legible in Hilbert's notebook but also grammatically
  incorrect. Corrections and insertions that Hilbert made in this entry
  show that he wrote down the problem in haste.]

The paper I linked above discusses the history and the role of Hilbert's problems and I think is worth reading. Most of mathematical logic, as we know it right now, did not exist when this question was asked and you can simply disregard the question by saying "this is not a mathematical question". On the other hand, the same could be said about the second problem on the consistency of arithmetic today, if mathematicians did not develop the necessary tools to deal with this problem.
My point is that one might be able to answer Hilbert's 24th problem if one finds the "correct" statement of the problem. With our current knowledge and understanding of mathematical logic, can we define a criteria for a proof to be "simple"? Have there been any attempts to define such a notion? Should "simple" merely mean "short"?
In Thiele's article, you can find some quotations in Section 5 but they do not really give any useful information about how Hilbert perceived the word "simple". Having stumbled upon this article only today, I admit that I have not searched for other articles yet. So I would also appreciate being directed to other books and articles on this cancelled 24th problem.

REPLY [4 votes]: Since this question has been bumped to the first page, it seems safe to mention a paper that I wrote in 2002, developing a notion of simplicity in first-order logic and a better-behaved one in propositional logic (so in a much more restricted context than what Hilbert had in mind). The MathSciNet data are 
MR2023260 (2004j:03014) Reviewed 
Blass, Andreas(1-MI)
Resource consciousness in classical logic. (English summary) 
Games, logic, and constructive sets (Stanford, CA, 2000), 61–74, 
CSLI Lecture Notes, 161, CSLI Publ., Stanford, CA, 2003. 
A pre-publication version is available at http://www.math.lsa.umich.edu/~ablass/llc9.pdf .<|endoftext|>
TITLE: What is this construction using iterated face maps of semisimplicial sets?
QUESTION [6 upvotes]: Let $X$ be a semisimplicial set (face maps but no degeneracy maps). Fix a positive integer $k$. Let $Y_n$ be $X_{(n+1)k}$ and then define $\partial^Y_i:Y_n\to Y_{n-1}$ by 
$$\partial^Y_i = (\partial^X_{ik})^k.$$
These maps give $Y$ the structure of a semisimplicial set.
There are many variations possible on this theme, changing the indexing numbers around. I'm not only interested in this particular example, but in any similar semisimplicial set created using iterated face maps of a semisimplicial set. 
Has this construction or similar ones been studied? Does it have a name? Any obvious properties? Is there a variation with a straightforward or canonical extension to simplicial sets? Are there interesting variations where $\partial_0$ and $\partial_n$ are perturbed or different in some way? If there is something interesting that can only be said in a linear category that would also be of interest.

REPLY [4 votes]: I don't know if this is exactly what you're looking for, but there is the edgewise subdivision of simplicial sets. It seems to be very similar to the construction you're introducing. As far as I know it's a variant of a construction introduced by Segal, and it was introduced by Bökstedt–Hsiang–Madsen (and it was later used by McCarthy in the context of Hochschild homology, and probably also other people). What's next basically comes from [BHM] below.
The construction is as follows. Fix a positive integer $r$. There's an endofunctor $$\DeclareMathOperator{\sd}{sd}\sd_r : \Delta \to \Delta$$ given by $\sd_r[m-1] = [mr-1]$ (where $[n] = \{0 < \dots < n \}$, and on nondecreasing maps is given by $\sd_r(f)(am+b) = an + f(b)$ (where $f : [m-1] \to [n-1]$, and the Euclidean division $0 \le b < m$). Then the edgewise subdivision of a simplicial set $X : \Delta^{op} \to \mathsf{Set}$ is given by $\sd_r X = X \circ \sd_r$. More precisely, $\sd_r X_n = X_{r(n+1) - 1}$, and the face and degeneracy maps are given by:
$$\bar{d}_i = d_i \circ d_{i+(n+1)} \circ \dots \circ d_{i+(r-1)(n+1)}, \\
\bar{s}_j = s_{j+(r-1)(n+2)} \circ \dots \circ s_{j + (n+2)} \circ s_j.$$
There's a canonical homeomorphism $|\sd_r X| \to |X|$, they have very a nice picture in [BHM]:

I think it's practical to write these down as matrices to see what's happening, see [McC] below. If $X$ is actually a cyclic set it's even possible to make everything compatible with the $S^1$-action. McCarthy uses these subdivisions to define explicit Adams operations on Hochschild homology (pinch the circle using the subdivision and fold). I know Ginot also uses them to define Adams operations on higher Hochschild homology. It's probably been used in other contexts too.

[BHM] M. Bökstedt, W. C. Hsiang, and I. Madsen. “The cyclotomic trace and algebraic K-theory of spaces”. In: Invent. Math. 111.3 (1993), pp. 465–539. ISSN: 0020-9910. DOI: 10.1007/BF01231296. MR1202133.
[Seg] G. Segal. Configuration-spaces and iterated loop-spaces. Invent. Math. 21 (1973), 213–221. DOI: 10.1007/BF01390197 MR331377
[McC] R. McCarthy. “On operations for Hochschild homology”. In: Comm. Algebra 21.8 (1993), pp. 2947–2965. ISSN: 0092-7872. DOI: 10.1080/00927879308824712. MR1222750.
[Gin] G. Ginot. “Higher order Hochschild cohomology”. In: C. R. Math. Acad. Sci. Paris 346.1-2 (2008), pp. 5–10. ISSN: 1631-073X. DOI: 10.1016/j.crma.2007.11.010. MR2383113.<|endoftext|>
TITLE: Chromatic numbers of nowhere dense graphs
QUESTION [7 upvotes]: Given $c\in (0,1)$ and a graph $G=(V,E)$ such that any subset $U\subset V$ contains an independent subset of cardinality at least $c|U|$. Does it allow to bound the chromatic number $\chi(G)$ by the constant depending only on $c$? If yes, what is the best constant?
UPDATE. For $c\geq 1/2$ we get that $G$ is bipartite. For $c<1/2$ it is proved in Gjergji's answer (by considering Kneser graph) that no uniform bound does exist.

REPLY [8 votes]: Take the Kneser graph $K(2k+r,k)$, defined as the graph where the vertices are the $k$-element subsets of $\{1,2,\dots,2k+r\}$, and there is an edge between two vertices if the corresponding sets are disjoint. It is not hard to prove that we can take any $c < \frac{1}{2+\frac{r}{k}}$, but the chromatic number is $r+2$. So the answer to your question is negative.
Remark: The quantity $\rho(G)=\max_{H\subseteq G}\left(\frac{|H|}{\alpha(H)}\right)$ is called the Hall ratio of $G$. It is known that $\rho(G)\le \chi_f(G)\le \chi(G)$, where $\chi_f$ is the fractional chromatic number. The reason why Kneser graphs are natural to bring up in this context is because they are known as a natural example of graphs where $\chi$ is not bounded as a function of $\chi_f$.<|endoftext|>
TITLE: Higher-dimensional Artin L-functions
QUESTION [7 upvotes]: I begin by clarifying that the "higher-dimensional" in my question refers to analogues of Artin L-functions over higher dimensional base schemes than $\mathrm{Spec}(\mathbb{Z})$.
Now for the set-up. This will be very similar to the set-up of Chapter 9 of Serre's book "Lectures on N_X(p)".
Let $R$ be a finite type flat $\mathbb{Z}$-algebra of dimension $n$ with fraction field $K$ and let $X = \mathrm{Spec}(R)$. Let $L/K$ be a finite Galois extension with Galois group $G$ and let $Y$ be the normalisation of $X$ in $L$. At any point I am quite happy to shrink $X$. For example, I may assume that $X$ is regular and that $Y/X$ is finite étale.
Now for the analogue of Artin L-functions. For a complex representation $\rho$ of $G$ we define
$$L(s,\rho) = \prod_{\substack{\text{closed points} \\ x \in X}}\mathrm{det}\left(I - \frac{\rho(\mathrm{Frob}_x)}{N(x)^s}\right)^{-1}, \quad \text{re}(s) > n.$$
See the above book of Serre for the definition of Frobenius elements in this setting. That the above Euler product converges in the range $\text{re}(s) > n$ follows from the Weil conjectures.
My first question is the following.


Have such L-functions been studied before?


It seems that Deligne's Weil II paper should be relevant here, but alas I don't understand it well enough.
But what I would really like to know is the following.


Assume that $\rho$ is irreducible and not the trivial representation. Then does $L(s,\rho)$ admit a holomorphic continuation to the line $\text{re}(s) = n$ without zeros?


This latter fact is proved in the classical case where $n = 1$ by using Brauer's induction theorem to reduce to the case where $G$ is abelian. Here one essentially obtains a dirichlet L-function, where the result is well-known (being part of the proof of Dirichlet's theorem on primes in arithmetic progressions). I imagine a similar reduction to the abelian case can be made in my setting, but of course the point is that I don't know how to deal with the abelian case itself.

REPLY [3 votes]: According to Serre in "Zeta and L-functions" (MR0194396), these L-functions have been defined by Artin himself. Moreover, it seems that in the formula
$$
L(s,\rho) = \prod_{\substack{\text{closed points} \\ x \in X}}\mathrm{det}\left(I - \frac{\rho(\mathrm{Frob}_x)}{N(x)^s}\right)^{-1}
$$
one should define $\rho(\mathrm{Frob}_x)$ as the average of the $\rho(g)$ for $g \in G$ mapping to $\mathrm{Frob}_x$.
Serre further says that $L(s,\rho)$ can be continued to a meromorphic function in the half-plane $\mathrm{Re}(s)>n-\frac12$, and that the singularities « can be determined, or rather reduced to the classical case $n=1$ » (there is an interesting lemma in the case $Y \to Y'$ is a fibration of curves, enabling to do induction on the dimension of $Y$). I don't know whether this answers your Question 2.
Artin's paper is Zur Theorie der L-Reihen mit allgemeinen Gruppencharakteren. Abh. Math. Sem. Univ. Hamburg 8 (1931), no. 1, 292--306 (MR3069563).<|endoftext|>
TITLE: Do "most" modular forms have no extra twists?
QUESTION [10 upvotes]: Let $f$ be a modular form -- more specifically, a normalized new eigenform which is not of CM type.
We say $f$ has extra twists if there exists some $\sigma \in \operatorname{Aut}( \mathbf{C})$ such that the Galois conjugate $f^\sigma$ is equal to the twist of $f$ by some non-trivial Dirichlet character, so $a_n(f)^\sigma = \chi(n) a_n(f)$ for all $n$ coprime to the level.
Any newform of nontrivial character has at least one extra twist (because the complex conjugate $\bar{f}$ is a twist of $f$.) But browsing some tables suggests that extra twists are pretty unusual for newforms of trivial character.


Do "most" non-CM newforms of trivial character (in some precisely quantifiable sense) have no extra twists?
Do "most" newforms of non-trivial character have no extra twists other than the obvious one?


(This question is prompted by this earlier question, which is subtle precisely when extra twists exist.)

REPLY [11 votes]: If a newform $f \in \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N),\varepsilon)$ has an inner twist by some $\sigma \in \operatorname{Aut}(\mathbb{C})$, then $f^{\sigma}$ is a newform of the same level as $f$. Moreover, if $\varepsilon$ is trivial, then so is the nebentypus of $f^{\sigma}$ (see (3.8) of Ribet's paper), and so any inner twist must arise from a quadratic Dirichlet character.
Ribet's paper actually shows (see (3.9)) that if $N$ is squarefree, then there are no nontrivial inner twists. More precisely, he shows that if $N$ is squarefree and $\chi$ is a quadratic Dirichlet character, then the twist $f \otimes \chi$ cannot have level $N$ and trivial nebentypus.
This is not hard to see by a local argument: if $N$ is squarefree, then for any $p \mid N$, the local component $\pi_p$ of the associated automorphic representation must be isomorphic to $\mathrm{St}$, the Steinberg representation of conductor $p$ associated to the trivial character. Any nontrivial twist of $f$ leaving the nebentypus unchanged must necessarily be quadratic, but any twist of $\mathrm{St}$ by a ramified quadratic character has conductor at least $p^2$, not $p$ (and in fact exactly $p^2$ if $p$ is odd). Similarly, at any unramified place, the twist of an unramified principal series representation by a ramified quadratic character has conductor at least $p^2$. So the level of $f \otimes \chi$ is strictly greater than $N$.
But if $N$ is not squarefree, then this is no longer the case! Indeed, in a recent paper (Theorem 6.4), I proved the following result (well, technically I proved it for Maaß cusp forms, but it generalises in an obvious way to holomorphic cusp forms).

Fix a nonsquarefree odd integer $N$, and let $N' > 1$ be squarefree and such that $N'^2 \mid N$. Let $\mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon_{\mathrm{quad}(N')})}$ denote the vector space spanned by newforms $f$ of weight $k$, level $N$, and trivial nebentypus such that $f$ does not have CM by the quadratic Dirichlet character $\varepsilon_{\mathrm{quad}(N')}$ modulo $N'$, but that the twist $f \otimes \varepsilon_{\mathrm{quad}(N')}$ is also a newform of level $N$ and trivial nebentypus. Then
  \[\frac{\dim  \mathcal{S}_{k}^{\mathrm{new}} (\Gamma_0(N))_{\mathrm{nonmon} (\varepsilon _{\mathrm{quad}(N')} )} }{\dim \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))} \sim \prod_{\substack{p \mid N' \\ p^2 \parallel N}} \left(1 - \frac{p}{p^2 - p - 1}\right)\]
  as $k$ tends to infinity over the even integers.
Furthermore, this still holds if we replace $\mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon_{\mathrm{quad}(N')})}$ with
  \[\bigcap_{\substack{N^* \mid N' \\ N^* > 1}} \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon _{\mathrm{quad}(N^*)})}.\]

A similar result holds even when $N$ is squarefree if $\varepsilon$ is nontrivial; see Proposition 6.5 of my paper.
That inner twists occur in abundance when $N$ is not squarefree was observed as far back as 1977 by Ribet; at the bottom of page 48 of this paper, Ribet writes 

It would be of interest to give an a priori construction of forms
  with extra twists. If the level $N$ is divisible by a high power of a
  prime, these forms seem to be more the rule than the exception.<|endoftext|>
TITLE: Is the moduli of formal groups smooth?
QUESTION [21 upvotes]: There is a notion of smooth stack in "homotopical algebraic geometry 2" and a notion of a cotangent complex for certain kinds of stacks (including representable stacks). I have two questions. 

Is the moduli stack of one dimensional commutative formal groups of height less than or equal to $n$ smooth in this sense?
Is the formal moduli stack of deformations of a one dimensional commutative formal group over a perfect field of characteristic $p$ of height equal to $n$ smooth in this sense? Here I mean the quotient stack of deformations by the action of the stabilizer group.

It seems to me that (2) should be true as this is the "ind-étale" quotient of Lubin-Tate. Question (1) should be true because of locality for smoothness, but a reference to the result (or at least the relevant tools in the concrete from) would be nice.

REPLY [4 votes]: Q1: I did not find the definition of a smooth stack in HAG2 (it would be nice if you provide a concrete citation!), but the definition of smoothness I know is: A morphism $X \to Y$ of algebraic stacks is smooth if there is a commutative diagram 
$\require{AMScd}$
\begin{CD}
    U @>f>> V\\
    @V g V V @VV h V\\
    X @>>> Y
    \end{CD}
where $U$ and $V$ are schemes (or algebraic spaces) and $f,h,g$ are smooth and $g$ is surjective. (see http://stacks.math.columbia.edu/tag/075U)
In particular, we can take in our example $X = \mathcal{M}_{FG}^{\leq n}$, $V = Y = Spec \mathbb{Z}_{(p)}$ and $U = Spec \mathbb{Z}_{(p)}[v_1,\dots, v_n, v_n^{-1}]$. So at least in this sense of smooth, the answer is yes: $\mathcal{M}_{FG}^{\leq n}$ is smooth. 
Edit: As explained by Jacob Lurie in the commments, the map $g$ is not smooth and my argument fails. I am sorry for being careless.<|endoftext|>
TITLE: Asking for Advices for Choosing a Ph.D thesis problem (in PDE area)
QUESTION [8 upvotes]: I'm a first year phd student in Germany. I've started my phd study one year ago and I'm currently confused about the topic I've chosen. The program is in the area of PDEs, and actually I didn't learn much about PDEs in my master study, only for instance the solution of the four fundamental types of PDEs and some Sobolev space theory. It was a challenge for me to learn the PDE theory at the beginning, since one doesn't only need to know a lot of things involving mathematics, but should also learn something about the physical modelling. This was a little bit hard for me since my minor subject was economics. But after the one year learning I think studying physical models is also very attractive and interesting. But as more as I learned, I know one can generally not conclude a very central theory for all the models, which causes a problem to my phd topic. My topic is regularity theory for a general setting of physical models. As one knows, the general trick is to consider the Nirenberg-difference quotient in such regularity theory, supposing the difference quotient is still in the space of test function. But now, if we give some first order differential conditions to the test functions, for instance $\nabla\cdot\phi=0$ or $\nabla\times\phi=0$ for a test function $\phi$, then one needs not to have that the difference quotient is still divergence free. My task is to find a trick to solve these problems. But as more as I learned, I have more and more confusion:

I can not find any models which use restricted test functions. The restrictions here are mostly given by divergence or curl free conditions. But these conditions are usually to solutions but not to test functions. Examples are the Maxwell's equations and piezoelectric model. 
My advisor asked me to give a generalization which concludes a class of such problems, but as you know, PDE questions are usually formulated very differently. Indeed, I can give some sort of generalization to the conditions, but it seems that these generalization does not make any sense in physics. I can not see that a generalization makes a contribution to physical models, since in general only the gradient, divergence or curl appear in the most models and have a clear physical meaning.

I will have a talk to my advisor in next week, eventually I will ask her to give me some real world models that relate to my topic, which seems not actually possible since I have asked her once earlier and got no answer… If this is the situation, I would ask her to give me a new topic which actually comes from physics and for which one can actually work on any unknown issues.
But before I meet her, I have a question to my confusion for which I might find a solution here: Does anyone know some real world models which indeed use restricted test functions like first order linear differential restrictions as described above?
Any kind of help is welcome!

EDIT: Thanks for the comments! Ok, let me explain more what I have researched. Let us consider for instance the following equations: Find a solution $u\in M$
\begin{equation}
\int_\Omega u:v=\int_\Omega f:v 
\end{equation}
for all $v\in M$ and a given $f\in L_2$, where $M$ is defined by $M:=\{v\in L_2:\nabla\cdot v=0\}$, and the divergence is understood as distribution. The existence and uniqueness of solution is guaranteed by Lax-Milgram. Now the question is, can $u$ be as regular as $f$ if $f$ is for instance in $H^1_{loc}$? Since the test functions do not include all smooth functions with compact support, we can not say that $u=f$ almost everywhere. But in fact, one can still prove $u$ is in $H^1_{loc}$ if $f$ is in $H^1_{loc}$. The suggestion from my adviser was to use Piola transformation, but I found another approach which is easier to handle this problem and can be used for distribution like $\nabla\times v=0$ or $\nabla v=0$. Using my approach one can even derive a criterium for the general distributional restrction
\begin{equation}
(\sum_{ij}a_{ijk}\partial_i v_j)_k=0,
\end{equation}
where $a_{ijk}$ are constants and fullfill certain properties. Eventually one can do the same trick to an elliptic equation or a Maxwell's system (but I haven't seen such models, namely with restricted test functions, in any references.) Now to be honest, I dont have any physical models related to this. But in general, this could relate to constrained optimal PDE problems (they have a similar formulation, but still, I can not find any concrete models so far). My advisor didn't tell me anything about models, she just said one may find some... At that time I wasn't really familiar with the models, so I accepted this topic, but for this moment, after having some basic modeling knowledge, I still can't find any appropriate models.

REPLY [2 votes]: Well, I'm not sure if I got your question correctly. If I understood apropriately, you are asking for real life models where the test functions have restrictions as being divergence/curl free. Fluid dynamics is one of the fields where this restriction appear, not only on the solution, but also on the test function. Check the book "Navier-Stokes Equations and Turbulence" by Foias, Manly, Rosa and Temam for instance. Let me also copy the link to the appropriate page (the page is in google books) 
https://books.google.com/books?id=f4zaccq9KREC&pg=PA39&lpg=PA39&dq=divergence+free+test+function&source=bl&ots=ZboqaQ8IOo&sig=wzvufxa7QsBZHK14JUUuNnaa8HE&hl=es&sa=X&ved=0CDYQ6AEwA2oVChMI5-STvsXuyAIVQcFjCh03yghh#v=onepage&q=divergence%20free%20test%20function&f=false<|endoftext|>
TITLE: Are coarse spaces of 1-dimensional smooth proper Artin stacks smooth?
QUESTION [12 upvotes]: Let $\mathcal{X}$ be a regular proper 1-dimensional Artin stack with finite diagonal, with coarse space morphism $\mathcal{X} \to X$. 
Question: Is $X$ regular?
Some comments:

I'm happy to assume that $\mathcal{X} \to X$ is birational.
I'm happy to assume that $\mathcal{X}$ is over a field and to replace regular with smooth. (But I am also interested in the case where $X$ is Spec of a Dedekind domain.
If $\mathcal{X}$ is additionally a Deligne-Mumford stack, or tame, then $X$ is smooth. 
Jack Hall pointed out to me that we can weaken tameness a bit: it is enough to assume that $\mathcal{X} \to X$ is an adequate moduli space, in the sense of Alper.

REPLY [3 votes]: Jack Hall pointed out to me over email that a stack with a coarse space is adequate, so this does it.<|endoftext|>
TITLE: Is conditional expectation with respect to two sigma algebra exchangeable?
QUESTION [5 upvotes]: $(\Omega, \mathcal{F}, P)$ is a probability space. $X$ is a r.v. defined on it, and $\mathcal{G}_1, \mathcal{G}_2$ are two $\sigma$-algebra, can we claim the following:
$$
\mathbb{E}\{\mathbb{E}[X|\mathcal{G}_1]|\mathcal{G}_2\}=\mathbb{E}\{\mathbb{E}[X|\mathcal{G}_2]|\mathcal{G}_1\}=\mathbb{E}[X|\mathcal{G}_1\cap\mathcal{G}_2].
$$
If this is not true, why the following is true:
$X_1,X_2,\ldots,X_n$ are independent r.v. on the same probability space, $Z=f(X_1,X_2,\ldots,X_n)$ is another r.v. And we have:
$$
\mathbb{E}\{\mathbb{E}[Z|X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n]|X_1,\ldots,X_i\}=\mathbb{E}[Z|X_1,\ldots,X_{i-1}].
$$

REPLY [3 votes]: It's not true.  For a counterexample, take $\Omega = \{a,b,c\}$ to be a sample space with 3 points, $\mathcal{F} = 2^{\Omega}$, and $P(A) = \frac{1}{3} |A|$ to be the uniform probability measure assigning probability 1/3 to each outcome.  Let's represent a random variable $X : \Omega \to \mathbb{R}$ as the ordered triple $(X(a), X(b), X(c))$.
Set $\mathcal{G}_1 = \{\Omega, \emptyset, \{a\}, \{b,c\}\}$, and $\mathcal{G}_2 = \{\Omega, \emptyset, \{b\}, \{a,c\}\}$.  Let $X$ be the random variable $(1,2,3)$.  Then one can directly compute
$$\begin{align*} 
E[X \mid \mathcal{G}_1] &= (1, 2.5, 2.5) \\
E[X \mid \mathcal{G}_2] &= (2,2,2) \\
E[E[X \mid \mathcal{G}_1] \mid \mathcal{G}_2] &= (1.75,2.5,1.75) \\
E[E[X \mid \mathcal{G}_2] \mid \mathcal{G}_1] &= (2,2,2).
\end{align*}$$
To prove the second claim, I assert the following: suppose $U,V,W$ are mutually independent random variables or vectors, and $Y$ is $\sigma(U,V)$-measurable and integrable.  Then $E[Y \mid V,W] = E[Y \mid V]$.
Clearly $E[Y \mid V]$ is $\sigma(V,W)$ measurable, so it suffices to show $E[Y 1_A] = E[E[Z \mid V] 1_A]$ for any $A \in \sigma(V,W)$.  Suppose first that $A$ is of the form $A = B \cap C$ for $B \in \sigma(V)$, $C \in \sigma(W)$. Then
$$\begin{align*} 
E[E[Y \mid V] 1_A] &= E[E[Y\mid V] 1_B 1_C] \\
&= E[E[Y 1_B \mid V] 1_C] && \text{since $B \in \sigma(V)$} \\ &= E[E[Y 1_B \mid V]] E[1_C] && \text{since $1_C \in \sigma(W)$ is independent of $\sigma(V)$} \\ &= E[Y 1_B] E[1_C]
\end{align*}$$
while $E[Y 1_A] = E[(Y 1_B) 1_C] = E[Y 1_B] E[1_C]$ since $Y 1_B \in \sigma(U,V)$ is independent of $\sigma(W)$.  The general case follows by a monotone class or $\pi$-$\lambda$ argument (the collection of all $B \cap C$ is a $\pi$-system that generates $\sigma(V,W)$).
Now apply this taking $U = (X_{i+1}, \dots, X_{n})$, $V = (X_{1}, \dots, X_{i-1})$ and $W = X_i$, and $Y = E[Z \mid U,V]$.<|endoftext|>
TITLE: Graph with group structure?
QUESTION [7 upvotes]: Is there any established theory of graphs which themselves are groups? I don't mean Cayley graphs or "graphs of groups". I mean a graph whose set of vertices forms a group, where the group operation is compatible with the graph structure in some way. Like for example any two adjacent vertices have adjacent inverses, and some similar adjacency condition for the multiplication operation. 
Simple cycles would have this property, though I'm not sure what else. Is there a name for this? Hard to google due to many similar concepts.

REPLY [7 votes]: I can at least propose a definition. To my mind, the categorically best behaved category of graphs is the category of presheaves on $\{ \bullet \rightrightarrows \bullet \}$ ("directed multigraphs"); explicitly, it's the category of tuples $(E, V, s, t)$ of an edge set $E$, a vertex set $V$, and two functions $s, t : E \to V$ (the source and target of an edge). Among other things, this is the category over which the category of small categories is monadic. So:

Proposal: A "group graph" is a group object in the category of graphs (in the above sense). 

Explicitly, a group graph is a tuple $(E, V, s, t)$ of two groups $E, V$ (so edges admit a multiplication too, not just vertices) and two group homomorphisms $s, t : E \to V$. Note that this is the same thing as a graph object in the category of groups, and that another way to say "graph object" is "coequalizer diagram." 
Said another way, first we need a group operation $\otimes$ on vertices, and then given two edges $e_1 : s_1 \to t_1, e_2 : s_2 \to t_2$, they admit a product which takes the form $e_1 \otimes e_2 : s_1 \otimes s_2 \to t_1 \otimes t_2$, and this product also satisfies the group axioms.
Example. Given any homomorphism $f : G \to H$ of groups, taking its kernel pair gives a diagram
$$G \times_H G \rightrightarrows G \to H$$
and the first part $G \times_H G \rightrightarrows G$ of this diagram (where the two morphisms are the two projections) is naturally a graph object. This construction generalizes to any category with pullbacks.
A related but more familiar thing you could ask for is a "group object" in categories rather than graphs; that is, a monoidal category in which every object is invertible (this is the motivation for the notation $\otimes$). This notion and various related notions are important in algebraic topology and other fields which make use of higher category theory; compare the notion of 2-group and "Picard groupoid." 
Another related but more familiar thing you could ask for is a simplicial group (thinking of graphs as being like simplicial sets, but over a truncated version of the simplex category).<|endoftext|>
TITLE: Vietoris-Begle theorem for simplicial sets
QUESTION [9 upvotes]: I've learned the theorem when reading a comment by Vidit Nanda to my question see here.
Here is the (simplified) version of the theorem for topological spaces:
Vietoris-Begle Theorem
Let $f:X\rightarrow Y$ be a surjective closed continuous map between paracompact Hausdorff spaces. Assume that for any $y\in Y$, $f^{-1}(y)$ is weakly contractible, then the induced map
$$H^{\ast}(f): H^{\ast}(Y,G)\rightarrow H^{\ast}(X,G)$$ 
is an isomorphism (reduced cohomology) for any abelian group $G$. 
It seems that the assumption that $f$ is closed is essential. 
My question is the following: Do we have a simplicial version of Vietoris-Begle Theorem ? i.e., 
simplicial Vietoris-Begle Theorem ?
Let $f:X\rightarrow Y$ be a surjective morphism of simplicial sets (Kan complexes). Assume that for any $y\in Y_{0}$, $f^{-1}(y)$ is weakly contractible simplicial sets, then the induced map
$$H^{\ast}(f): H^{\ast}(Y,G)\rightarrow H^{\ast}(X,G)$$ 
is an isomorphism (reduced cohomology) for any abelian group $G$.
reference for the topological case: reference
Edit: Please feel free to add more conditions on $f,X ,Y$ but not the trivial ones such as $f$ is a fibration or something similar...
There is for example a simplicial (strong) version here (lemma 26, case (3))

REPLY [6 votes]: Here is a theorem due (I think) to E. Dror Farjoun (it is Theorem 9.A.11 in his book on homotopy localization).
Theorem: Let  $f:X\to Y$ be a surjective map of connected pointed simplicial sets.  Then the homotopy fiber $F_f$ of $f$ is in the smallest closed class containing $f^{-1}(\sigma)$ for every simplex $\sigma$ in $Y$.
A closed class is a class $\mathcal{C}$ of spaces which is closed under weak equivalence and (pointed) homotopy colimit.  For pointed spaces use pointed homotopy colimit; for unpointed ones, restrict to homotopy colimits of diagrams whose nerve is contractible.
It follows, in particular, if every $f^{-1}(\sigma)$ is weakly contractible, then $f$ will be a weak homotopy equivalence (Corollary 9.B.3.1).<|endoftext|>
TITLE: In Gromov-Witten theory, why is the string coupling constant weighted by $2g-2$?
QUESTION [11 upvotes]: Let $X$ be a Calabi-Yau threefold and let us fix a homology class $\beta\in H_2(X,\mathbb Z)$, just for simplicity. The generating series of Gromov-Witten invariants of $X$ in class $\beta$, $$\mathsf Z=\sum_{g}\textrm{GW}_{g,\beta}(X)\color{red}{u}^{2g-2},$$ has the variable $\color{red}{u}$, taking care of the varying genera, weighted by $\color{red}{2g-2}$. Some time ago, I remember having read this variable is called the string coupling constant, sometimes denoted $g_s$.

Question. Why is this variable weighted by $2g-2$, instead of, say, $g$? Is there a reason coming from Physics, maybe from String Theory?

I feel the reason why I cannot answer my question is that I do not really understand the meaning of $u$ (or $g_s$), or its role inside $\mathsf Z$. It seems not to be just a variable, or indeterminate, which one uses to write down $\mathsf Z$. It seems, for instance, that it makes sense to talk about "small values" of the coupling constant (which confuses me even more, since it is called a constant). I hope I made clear my confusion. And, my apologies for the non-research-level of the question.
Thanks you all.

REPLY [6 votes]: Here are two ways how physicists think about the string coupling constant:
1) In usual quantum field theory defined by quantization of a classical field theory, the partition function is defined by a path integral of the form
$Z= \int D\phi e^{\frac{S(\phi)}{\hbar}}$ where $S$ is the classical action. In particular, in the classical limit $\hbar \rightarrow 0$, we expect a localization around the classical solution $\phi_{cl}$ to classical equation of motion, and so a behaviour like $Z \sim e^{\frac{S(\phi_{cl})}{\hbar}}$, i.e.
$F=log(Z) \sim \frac{S(\phi_{cl})}{\hbar}$. Perturbative expansion in $\hbar$ will give a pertubative expansion of $F$. Standard Feynman diagram tecnhiques show that this expansion is of the form 
$F \sim \sum_{L} F_L \hbar^{L-1}$ where $F_L$ receives contribution from (connected) Feynman diagrams with $L$ loops.
We expect an open string theory to reduce at low energy to an usual quantum field theory with an expansion as above. But the closed string coupling $g_s$ is related to the open string coupling $\hbar$ by $g_s \sim \hbar^2$: this follows by unitarity from the annulus diagram which can be interpreted either as a one loop open string or a tree level closed string.
2) The point 1) is a spacetime point of view. There is also a worldsheet point of view. The non-linear sigma model is a two dimensional theory of maps from a two dimensional $\Sigma$ to a space $X$. The action of such theory has to be local and one natural term with this property is given by the integral of the curvature of $\Sigma$, which by Gauss-Bonnet is given by the Euler characteristic $2-2g$.
This origin of the appearance of the Euler characteristic is essentially the one mentioned in the answer given by Jim Bryan: additivity gives a way to reconstruct a global quantity from a local one.<|endoftext|>
TITLE: If $X∼F_1$, $Y∼F_2$, under what conditions on $F_1$, $F_2$ can we construct $Y=E(X\mid\mathscr{G})$ for some $\mathscr{G}$?
QUESTION [8 upvotes]: Suppose that we have distributions $F_1 $ and $F_2$. Under what conditions on $F_1,F_2$ is it possible to construct random variables $X\sim F_1,Y\sim F_2$ such that $Y=E(X|\mathscr{G})$, that is, $Y$ is the conditional expectation of $X$ with respect to some $\sigma$-field $\mathscr{G}$?

REPLY [7 votes]: The condition proposed by Nate Eldredge is not only necessary, but also sufficient. This is classical Strassen's theorem:

Theorem For integrable random variables $X$ and $Y$ the following is equivalent:

There exist $\hat X \overset{d}{=} X$ and $\hat Y \overset{d}{=} Y$ such that $E[\hat X | \hat Y] = \hat Y$.

$Y\prec X$ in convex stochastic order, i.e. $E[f(Y)]\le E[f(X)]$ for any convex function $f$.



Note that the formulation in Strassen's paper differs from this one (and Strassen attributed it himself to Hardy-Littlewood-Polya-Blackwell-Stein-Sherman-Cartier-Fell-Meyer). In the present form it is given e.g. in Ruschendorf (Theorem 6 for $i=2$).<|endoftext|>
TITLE: Synthetic projective lines
QUESTION [15 upvotes]: The classical synthetic notion of projective plane consists of a set of points, a set of lines, and a relation of incidence between the two, such that any two distinct points lie on a unique line and any two distinct lines intersect in a unique point (plus some nondegeneracy assumptions).  There are similar notions of projective 3-space, $n$-space, and so on — but 1-dimensional projective space seems harder to capture synthetically, since there is no "room", dimensionally, for subspaces in between the points and the entire space.
Has anyone attempted to define a synthetic notion of "projective line"?  Ideally such a definition would have properties like the following:

The space $P^1(k)$ is naturally a projective line for any division ring $k$, and from any projective line $L$ satisfying enough axioms we can construct a skew field $c(L)$ such that $c(P^1(k)) \cong k$ and $P^1(c(L))\cong L$ (unnaturally).  The corresponding facts for Desarguesian projective planes are classical.
Any line in a projective plane is a projective line, and any projective line satisfying enough axioms can be embedded as a line in some projective plane.  This would be analogous to how any plane in a projective 3-space is a Desarguesian projective plane, while any Desarguesian projective plane can be embedded in a projective 3-space.

I have an idea for how one might do this, by axiomatizing the "quadrangular hexad" relation on a line in a projective plane; but before I try very hard, I'm looking for references where something like it has been tried before.

REPLY [5 votes]: You might benefit from reading Section 5.3 of John Faulkner's book The role of nonassociative algebra in projective geometry AMS 2014. The results there may have been what you had in mind by 'axiomatizing the "quadrangular hexad" relation'. These are (almost) all very old theorems, appearing in (and mostly predating) Pickert's Projektive Ebenen book from 1955, one going back to Veblen and Young from 1910 and another to von Staudt from 1860. The basic underlying idea of using quadrangular sets goes back to Desargues in 1639. (It was Desargues who introduced the term involution into mathematics, in this context, often referred to as an involution of six points.) But Desargues' work is an elaboration and extension of results of Pappus from c.340.<|endoftext|>
TITLE: Motivation for Hirzebruch-Jung Modified Euclidean Algorithm
QUESTION [6 upvotes]: Let $a,b \in \mathbb{N} \ \ s.t. \ \ a > b$ have $\gcd(a,b) =1$. We can define the Hirzebruch-Jung modified euclidean algorithm as follows:
Let $e_i \in \mathbb{N} >2$, and $ r_k \in \mathbb{N}$ where $0\le r_k<r_{k-1}$. Then we can do the following procedure (analogous to the standard Euclidean algorithm):
$$ \begin{array}{rcl} 
a &= & e_1b -r_1\\
b &= & e_2r_1 - r_2\\
r_1 &= &e_3r_2 -r_3\\
&...\\
r_{k-2} &= &e_ir_{k-1} = e_i
\end{array}$$
That last line is true because the $\gcd(a,b) =1$, by hypothesis.
Considering that the greatest common divisor is known by hypothesis, what is the utility of this modified Euclidean algorithm, and what was the motivation for its development? I have heard that it has applications to toric geometry, but I am not familiar with those.

REPLY [4 votes]: I think the motivation for the Hirzebruch-Jung algorithm is not the algorithm itself, but the fact the it yields directly a very interesting continued fraction expansion. It is those, now called Hirzebruch-Jung continued fractions, that have a wide number of applications. A bit more on that later.
Using your notation (but changing $b$ with $r_o$ for clarity).
$$ \begin{array}{rcl} 
a &= & e_1r_0 -r_1\\
r_0 &= & e_2r_1 - r_2\\
r_1 &= &e_3r_2 -r_3\\
&...\\
r_{k-2} &= &e_ir_{k-1} = e_i
\end{array}$$
By simple division you get:
$$ \begin{array}{rcl} 
a/r_0 &= & e_1 -r_1/r_0\\
r_0/r_1 &= & e_2 - r_2/r_1\\
r_1/r_2 &= &e_3 -r_3/r_2\\
&...\\
r_{k-2}/r_{k-2} &= &e_i
\end{array}$$
And from that you can simply read off:
$$a/r_0 = e_1 -\frac{1}{e_1-\frac{1}{...-\frac{1}{e_i}}}$$
It is those finite continued fractions $a/r_0$ that have, among many others, applications to toric varieties. To quote from D. I. Dais' "Geometric Combinatorics in the Study of Compact Toric Surfaces" (2000):

"Examining two-dimensional toric singularities "under the microscope"
  one discovers a peculiar algebro-geometric world endowed with a rich
  combinatorial structure. Viewed historically, everything begins with
  Hirzebruch-Jung continued fractions."

For some more motivation and applications, I quote form a very recent (2015) paper:

"Hirzebruch-Jung (H-J) continued fractions are widely used in various
  branches of mathematics as well as in theoretical physics. First of
  all, HJ-continued fractions arise naturally in the minimal resolution
  of cyclic quotient (that is, Hirzebruch-Jung) surface singularities of
  the type $\mathbb{C}/\mathbb{Z}_p$, which is also known as
  HJ-resolution" [...]
HJ-continued fractions are used also to describe the plumbing
  decomposition of the other type of link of surface singularities,
  namely, Seifert fibered homology spheres (Sfh-spheres), particularly
  Brieskorn homology spheres (Bh-spheres) [...]
In condensed matter theory, the HJ-continued fractions are used to
  describe fractional quantum Hall (FQH) systems with k levels of
  hierarchy [...]
Also, in topological string theory, the structure of internal space
  (Calabi-Yau threefold) can be encoded in terms of HJ-continued
  fraction expansion of positive integers, which are the topological
  invariants of the internal space and define the mode of interactions
  between D-branes"

REPLY [2 votes]: This algorithm gives best possible left (or right, it depends on the first step) convergents to a given number, while usual continued fractions give left and right convergents. 
From gemetrical point of view convergents to ususal continued fractions for $a/N$ correspond to the verteces of two sails (in 1st and 2nd quarter) for the lattice generated by vectors $(a,1)$  and $(N,0)$ while "negative-regular" continued fractions (aka reduced regular continued fractions) give the verteces of only one sail (in 1st or 2nd quarter), see Geometric proof of Rödseth's formula for Frobenius numbers for the picture.<|endoftext|>
TITLE: Evaluating elliptic integrals
QUESTION [5 upvotes]: I am interested in evaluating some elliptic integrals, and I have not been able to secure a reference to do exactly what I need. Most of the references I've found seem to focus on reducing more general elliptic integrals to Legendre form, but leave out the part about actually dealing with complete elliptic integrals. In particular, I am interested in the following toy problem, which is to show the following:
$$\displaystyle \int_0^\infty \frac{dx}{\sqrt{x(x+2)(x+3)}} = \sqrt{2}K(-1/2),$$
where $K(k)$ is the complete elliptic integral of the first kind. The above result is due to Wolfram Alpha.
I tried the obvious substitution which is $x = \tan \theta$, and after some labour we obtain the integral
$$\displaystyle \int_0^{\pi/2} \frac{2 d \theta}{\sqrt{7 \sin 2\theta + 5 \sin^2 2 \theta + 5 \sin 2 \theta \cos 2 \theta}},$$
which again can be checked to evaluate to $\sqrt{2}K(-1/2)$, although in this case Wolfram only gave the numerical value and not the closed form. Further, this last one does not look like what the 'correct' form should be, which is
$$\displaystyle \int_0^{\pi/2} \frac{\sqrt{2} d \theta}{\sqrt{1 - (1/4)\sin^2 \theta}}.$$
Based on some data I got from playing around with Wolfram, I suspect that the following is true: Suppose that $0 < a < b$. Then
$$\displaystyle \int_0^\infty \frac{dx}{\sqrt{x(x+a)(x+b)}} = \frac{2 K(1 - b/a)}{\sqrt{a}}.$$
Any help would be much appreciated, I apologize if this problem is in fact trivial or well-known.

REPLY [10 votes]: This question has been out for a while, and I think it deserves a thorough answer, even tho I came quite late to this party.
As both the classical Legendre-Jacobi theory and the Carlson theory have been mentioned by other users, I'll treat the OP's integral from both viewpoints.

Legendre-Jacobi
The OP came pretty close to using the correct substitution. One thing that could have been done instead is to recall the Pythagorean identity $1+\tan^2 u=\sec^2 u$, so that the substitution that should have been used is $x=a \tan^2 u$, where $a=2$ or $a=3$. Taking the smaller value of $a$, and after some amount of algebra, we obtain
$$\begin{align*}\int_0^\infty\frac{\mathrm dx}{\sqrt{x(x+2)(x+3)}}&=\int_0^{\pi/2}\frac{2}{\sqrt{3-\sin^2u}}\mathrm du\\
&=\frac2{\sqrt{3}}\int_0^{\pi/2}\frac{\mathrm du}{\sqrt{1-\frac13\sin^2u}}\\&=\frac2{\sqrt{3}}K\left(\frac13\right)\end{align*}$$
where I use the parameter convention for elliptic integrals. (This is the same convention used in Abramowitz and Stegun. Relatedly, see here for an extended discussion on the notational confusion surrounding elliptic integrals.)
Had we chosen the substitution with $a=3$ instead, we would have instead obtained the result $\sqrt{2}K\left(-\frac12\right)$, which is equivalent through the imaginary modulus transformation
$$K(-m)=\frac1{\sqrt{1+m}}K\left(\frac{m}{m+1}\right),\quad m>0$$

Carlson
In Carlson's theory, there is the general hypergeometric function
$$R_{-a}(b_1,\dots,b_k;z_1,\dots,z_k)=\frac1{\mathbf B\left(a,-a+\sum_j b_j\right)}\int_0^\infty u^{-a-1+\sum_j b_j}\prod_j \left(u+z_j\right)^{-b_j}\mathrm du$$
where $\mathbf B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is the usual Euler beta function.
This multivariate hypergeometric function is a homogenized/symmetrized version of the classical Lauricella $F_D$ function (see e.g. this paper where Carlson introduced his function, tho that reference uses an opposite sign convention for $a$).
With this consideration, the OP's original integral can be expressed either as a three-variable Carlson integral, or a two-variable Carlson integral:
$$\begin{align*}\int_0^\infty\frac{\mathrm dx}{\sqrt{x(x+2)(x+3)}}&=2\,R_{-\frac12}\left(\frac12,\frac12,\frac12;0,2,3\right)\\&=\pi\,R_{-\frac12}\left(\frac12,\frac12;2,3\right)\end{align*}$$
The three-variable (incomplete) case occurs often enough that it is given the notation
$$\begin{align*}R_F(x,y,z)&=\frac12\int_0^\infty\frac{\mathrm du}{\sqrt{(u+x)(u+y)(u+z)}}\\&=R_{-\frac12}\left(\frac12,\frac12,\frac12;x,y,z\right)\end{align*}$$
and the two-variable form corresponds to the so-called "complete case",
$$\begin{align*}R_K(x,y)&=R_{-\frac12}\left(\frac12,\frac12;x,y\right)\\&=\frac2{\pi}R_F(0,x,y)\end{align*}$$
In fact, the two-variable form has the integral representation
$$R_K(x,y)=\frac2{\pi}\int_0^{\pi/2}\frac{\mathrm du}{\sqrt{x\cos^2 u+y\sin^2 u}}$$
which one recognizes to be related to the integral representation of Gauss's arithmetic-geometric mean (AGM):
$$R_K(x,y)=\frac1{\operatorname{agm}(\sqrt{x},\sqrt{y})}$$
Thus, the OP's integral is $\pi/\operatorname{agm}(\sqrt{2},\sqrt{3})$.
Additionally, the two-variable Carlson function is also related to the Gauss hypergeometric function, being a homogeneous version of it:
$$R_{-a}(b_1,b_2;x,y)=y^{-a}{}_2 F_1\left({{a,b_1}\atop{b_1+b_2}}\middle|1-\frac{x}{y}\right)$$
so one has
$$\begin{align*}\pi\,R_{-\frac12}\left(\frac12,\frac12;2,3\right)&=\frac{\pi}{\sqrt{3}}{}_2 F_1\left({{\frac12,\frac12}\atop{1}}\middle|\frac13\right)\\&=\frac2{\sqrt{3}}K\left(\frac13\right)\end{align*}$$
where we have used the hypergeometric representation of the complete elliptic integral of the first kind.<|endoftext|>
TITLE: Graphs in which any two odd cycles have a common vertex
QUESTION [7 upvotes]: Let $G$ be a graph in which any two odd cycles have a common vertex. It is easy to see that $\chi(G)\leq 5$ (choose minimal odd cycle $C$, use two colors for $G\setminus C$ and three colors for $C$). And this is sharp as $K_5$ shows.
May we improve this bound to $\chi(G)\leq 4$ assuming additionally that $G$ does not contain $K_5$? Or to $\chi(G)\leq 3$ assuming that $G$ does not contain $K_4$? I can not prove even $\chi(G)\leq 4$ for $G$ without $K_3$.

REPLY [14 votes]: The claim on graphs without $K_5$ is a particular case of the (still open in general) Erdos--Lovasz Tihany conjecture. (Tihany is not a surname, but the name of a peninsula on Balaton lake in Hungary.)
This particular case has been proved in W.G. Brown and H.A. Jung, On odd circuits in chromatic graphs, Acta Math. Acad. Sci. Hungarica, V. 20(1), pp. 129-134. In fact, this paper contains more general results.
As I already mentioned in the comments, this bound cannot be lowered to 3 even if the graph contains no $K_3$.  An example is provided by the Mycielski graph of $C_5$. To see that it does not contain two disjoint odd cycles, notice that each odd cycle either contains the apex and an edge of $C_5$, or contains at least one of any two neighbors in $C_5$.<|endoftext|>
TITLE: harmonic extension of a curve by different parametrization
QUESTION [5 upvotes]: Let us consider a curve $\gamma :S^1 \rightarrow \mathbb{R}^3$ (or even a planar convex one if it simplifies). Then I look to the harmonic extension to the disc $h:\mathbb{D}\rightarrow \mathbb{R}^3$ (i.e $\Delta h=0$ and $h=\gamma$ on $\partial\mathbb{D}$). 
If I change the parametrization of $\gamma$ into $\gamma\circ \phi$, where $\phi$ is a diffeomorphism of $S^1$ :
1)is that true that the new harmonic extension can be write $h\circ \psi$ where $\psi$ is a diffeomorphism of $\mathbb{D}$. (it seems to be true in the planar convex case thanks to Rado-Kneser-Choquet theorem).
2) is there any chance to know how to "compute" $\psi$ with respect to $\phi$?
Last but not least, in fact it is the point I reallly want to understand, even in the planar case. it is a kind a Scharwz theorem: I would like to maximize the the norm of $\vert h_x \wedge h_y \vert(0)$ (the Jacobian in the planar case) with respect to the parametrization of $\gamma$. More precisely if the area bounded by the curve is equal to $\pi$ in the planar case, else we consider the area of the unique minimal surface when the curve is not planar but nice enough to insure the uniqueness.... it is just a normalization for the following question:
3) is there (always) a parametrization of the curve such that $\vert h_x \wedge h_y \vert(0) \geq 1$ (I am sure but I get no proof). is there even an optimal one? is there is many?
For the circle, the only one is the $\theta \mapsto e^{i\theta}$, up to rotation, but I can't say anything for another curve.
I am open to any suggestion, I have read quickly the book of Peter Duren, Harmonic maps into the plane, but i seems to be not enough...
Thanks in advance.
Edit (10/22/15):
My problem is indeed link to minimal surfaces, but not to the Palteau problem directly, but to the exterior Plateau problem. A simple by product is the following question: 
let $\Gamma$ a planar curve, is there $h:\mathbb{D} \rightarrow \mathbb{C}$ such that $\frac{dh}{dz}=1$ and $h_{\vert \partial \mathbb{D}}$ is a monotone parametrisation of $\Gamma$. The answer is probably (clearly?) yes. But the real question is how many (really different)? 
For $\Gamma$ the unit circle, the answer is one, you can prove it using Fourier series. But for instance, for the image of $e^{i\theta}+\frac{1}{2}  e^{i2\theta}$, I don't know how to proceed.

REPLY [3 votes]: The answer to (1) in the space case is a definite 'no'.  The actual image 'surface' $h(\Delta)$ in $\mathbb{R}^3$ can actually change when you reparametrize with $\phi:S^1\to S^1$, in which case, there can't be a reparametrization of the disk to match.  
It sounds as though you are not aware of the work of Douglas and Rado on minimal surfaces using harmonic parametrization, which definitely addresses your questions in the space case.  I'd recommend that you consult a source such as the discussion of this topic in H. Blaine Lawson's Lectures on minimal submanifolds for the details.
I think that the answer to (2) is also 'no' in any realistic sense, but that may depend on your ideal of 'realistic'.<|endoftext|>
TITLE: When is curvature given by a connection?
QUESTION [20 upvotes]: Suppose we have a curvature-like tensor $R\in \wedge^2 T^*_p M \otimes T^*_p M \otimes T_p M$ on a manifold $M$, that is $R(X,Y)Z = - R(Y,X)Z$. How does one determine whether or not this is a curvature tensor for some torsion-free affine connection?
One obvious condition is that the first Bianchi identity $R(X,Y)Z + R(Y,Z)X + R(Z,Y)X = 0$ must be satisfied. Are there other conditions that can be checked?

REPLY [29 votes]: Yes, there are typically many further conditions.  In dimension $n$, the space of curvature-like tensors that satisfy the first Bianchi identity are the sections of a bundle of rank $\tfrac13n^2(n^2{-}1)$, while the space of torsion-free connections is the space of sections of an affine bundle of rank $\tfrac12n^2(n{+}1)$, so, when $n>2$, the partial differential equation for the connection $\theta$ given the curvature-like tensor $\Omega$, i.e., $\mathrm{d}\theta + \theta\wedge\theta = \Omega$, is typically over-determined and has no solutions.  [Added comment:  For a discussion about why over-determined systems (such as this one when $n>2$) generally have no solution, see Rigorous justification that overdetermined systems do not have a solution.]  A trivial example of such an extra condition (when $n>2$) is that one must have $\mathrm{d}\bigl(\mathrm{tr}(\Omega)\bigr) = \mathrm{d}\bigl(\mathrm{tr}(\mathrm{d}\theta)\bigr) = 0$, but there are many more.
In fact, right away, one can see that the second Bianchi identity, i.e., $\mathrm{d}\Omega = \Omega\wedge\theta - \theta\wedge\Omega$ constitutes a number of inhomogeneous linear algebraic equations on $\theta$, and, when $n$ is sufficiently large ($n>3$ will do), these equations typically will have no solution, so such $\Omega$, even though they satisfy first Bianchi, cannot be the curvature of any connection, much less a torsion-free one.
Addendum: In my original answer, I wrote "This problem has been well-studied in the intermediate cases when $n$ is not too large (so that there are some interesting PDE problems to discuss).", but when the OP asked for references, I searched for a little while but couldn't find any that treated this particular problem.  I am sure that I have seen this problem treated somewhere, but I couldn't turn anything up.  Sorry about that.  However, I can sketch the analysis via exterior differential systems, as it is straightforward.  Perhaps this will be useful to the OP.  Here is how it goes:
To understand the local solvability, suppose that a curvature-like tensor $R$ satisfying the first Bianchi identity has been specified on $M^n$.  Choose a coframing $\eta = (\eta^i):TU\to \mathbb{R}^n$ on an open set $U\subset M$.  (One could restrict to the case of a coordinate coframing, i.e., take $\eta^i = \mathrm{d}x^i$ for some local coordinate system $x = (x^i):U\to\mathbb{R}^n$, but the extra freedom of using a general coframing is sometimes useful.) Then $R$ will be represented by an $n$-by-$n$ matrix $\Omega = (\Omega^i_j)$ of $2$-forms on $U$, say $\Omega^i_j = \tfrac12 R^i_{jkl}\,\eta^k\wedge\eta^l$, where $R^i_{jkl}=-R^i_{ilk}$.  The condition that $R$ satisfy the first Bianchi identity is simply that $\Omega\wedge\eta = 0$,
i.e., that $R^i_{jkl}+R^i_{klj}+R^i_{ljk}=0$.  (Note that the $R^i_{jkl}$ are $\tfrac13n^2(n^2{-}1)$ in number.)  By Cartan's Lemma, the equations $\Omega\wedge\eta = 0$ imply that there exist $1$-forms $\rho^i_{jk}=\rho^i_{kj}$ on $U$ such that $\Omega^i_j = \rho^i_{jk}\wedge\eta^k$.
A torsion-free connection on $U$ will be represented, relative to the coframing $\eta$, by an $n$-by-$n$ matrix $\theta = (\theta^i_j)$ of $1$-forms on $U$ that satisfies the first structure equation $\mathrm{d}\eta = -\theta\wedge\eta$.  We want to know when it is possible to choose $\theta$ so that it satisfies the second structure equation $\mathrm{d}\theta + \theta\wedge\theta = \Omega$.  
Now, if $\mathrm{d}\eta^i = -\tfrac12 T^i_{jk}\eta^j\wedge\eta^k$, then we must have
$$
\theta^i_j = (T^i_{jk} + p^i_{jk})\eta^k
$$
for some (as yet unknown) functions $p^i_{jk}=p^i_{kj}$.  (Note that the $p^i_{jk}$ are $N=\tfrac12n^2(n{+}1)$ in number.)  Let us regard the $p^i_{jk}$ as fiber coordinates on the (trivial) bundle $V = U\times \mathbb{R}^N$ over $V$. We now work on $V$, where we have $\mathrm{d}\eta = -\theta\wedge\eta$. 
Differentiating the first structure equation yields that the matrix $\Theta = \mathrm{d}\theta + \theta \wedge\theta $ satisfies $\Theta\wedge\eta = 0$, so it follows (again, by Cartan's Lemma) that there exist 
$$
\pi^i_{jk}=\pi^i_{kj} = \mathrm{d}p^i_{jk} + (\text{terms in $\eta$}).
$$
so that $\Theta^i_j = \pi^i_{jk}\wedge\eta^k$. 
Thus, we can write
$$
\Upsilon = \mathrm{d}\theta+\theta\wedge\theta - \Omega 
= \bigl((\pi^i_{jk}-\rho^i_{jk})\wedge\eta^k\bigr) 
= \bigl(\tilde\pi^i_{jk}\wedge\eta^k\bigr),
$$
and we see that the algebraic ideal generated by the components of $\Upsilon$ is generated by the $2$-forms $\Upsilon^i_j = \tilde\pi^i_{jk}\wedge\eta^k$.  A solution to our problem will be a section $u:U\to V = U\times \mathbb{R}^n$, that pulls back $\Upsilon$ to be zero.
Now, when $n=2$, the algebraic ideal suffices because we are only looking for a $2$-dimensional integral manifold, and the exterior derivative of $\Upsilon$ is a $3$-form, which will necessarily vanish when pulled back via any section $u$.  Now, it is easy to see, from the above description that, in this case, the Cartan characters of the system are $(s_1,s_2) = (4,2)$ and the space of integral elements at each point has dimension $S = 8 = s_1 + 2s_2$, so the system is involutive, and local solutions exist and depend on $s_2 = 2$ functions of $2$ variables (at least in the analytic case).  (It is not hard to show that, in fact, local solutions always exist, even in the smooth case, when $n=2$, but let me skip over that discussion now. Basically, one can impose two conditions, such as $p^i_{jj}=0$, and then the restricted system becomes elliptic.)
However, when $n>2$, the ideal generated by the components of $\Upsilon$ is not differentially closed.  In fact, one has
$$
\mathrm{d}\Upsilon = \mathrm{d}(\Theta - \Omega) 
= \Theta\wedge\theta-\theta\wedge\Theta - \mathrm{d}\Omega
= (\Omega\wedge\theta-\theta\wedge\Omega - \mathrm{d}\Omega)
  + (\Upsilon\wedge\theta-\theta\wedge\Upsilon),
$$
so, to get a differentially closed ideal, one must add the components of the $3$-form $\Psi = (\Omega\wedge\theta-\theta\wedge\Omega - \mathrm{d}\Omega)$.  Note that the $3$-form $\Psi$ does not involve any derivatives of $\theta$, and, in fact, is linear in the 'unknowns' $p^i_{jk}$.  In fact,
$$
\Psi^i_j = \tfrac12\bigl(R^i_{qkl}p^q_{jm}-R^q_{jkl}p^i_{qm} - S^i_{jklm}
\bigr)\,\eta^k\wedge\eta^l\wedge\eta^m
$$
for some functions $S^i_{jklm}$ on $U$, so it follows that the graph of any solution to the problem must lie in the affine `subbundle' of $V = U\times\mathbb{R}^N$ defined by the vanishing of these coefficients.  Most of the time, when $n>3$, this is more equations than unknowns $p^i_{jk}$, and this locus will be empty.  It is a nontrivial set of conditions on $R$ (and its derivative) that this locus be nonempty.  (Of course, a (small) part of these conditions is that $\mathrm{d}\bigl(\mathrm{tr}(\Omega)\bigr) = 0$.)  
When $n=3$, things are a little more interesting.  Typically, as long as $\mathrm{d}\bigl(\mathrm{tr}(\Omega)\bigr) = 0$ and the tensor $R$ is 'algebraically generic' in the appropriate sense, the equations above will define an affine bundle $W\subset V$ of rank $9$ over $U$, and generically, the Cartan characters will be $(s_1,s_2,s_3) = (9,0,0)$.  However, there will be nontrivial torsion, and the symbol will not be involutive, so you will get more obstructions at the next level.  A detailed analysis is somewhat messy, but there are interesting special cases in which the curvature takes values in a simple subalgebra, such as ${\frak{so}}(3)$ or ${\frak{so}}(2,1)$.  
Addendum 2:  The OP asked for 'conditions that can be checked', and the above discussion does not really address that question.  I had one further thought about this in the dimensions above $n=3$ that I thought that OP might find useful, so here it is:
When $n\ge 4$, there is an algorithmic way to proceed that works in the 'generic' case.  What I mean by 'generic' is this:  Say that a curvature-like tensor $R$ that satisfies the first Bianchi identity is algebraically generic if the kernel of the mapping $L(\phi) = \Omega\wedge\phi - \phi\wedge\Omega$, from $1$-forms with values in $n$-by-$n$ matrices to $3$-forms with values in $n$-by-$n$ matrices, consists of the elements of the form $\phi = \alpha\,I_n$, where $\alpha$ is a $1$-form (such elements are always in the kernel of $L$).  It is not hard to show that, when $n\ge 4$, the generic curvature-like tensor $R$ that satisfies the first Bianchi identity is algebraically generic.
Suppose that one is given an algebraically generic $R$ and one wants to check whether it is the curvature of a torsion-free connection.  Here are the steps:

Test whether $\mathrm{d}\Omega$ is in the image of $L$, i.e., whether there exists a traceless matrix with $1$-form entries $\phi$ such that $\mathrm{d}\Omega = L(\phi) = \Omega\wedge\phi-\phi\wedge\Omega$.  Because of the algebraically generic hypothesis, if $\phi$ exists (which is a matter of linear algebra), it will be unique.  If no such $\phi$ exists, then there is no solution.
Supposing that $\phi$ exists, consider the $2$-form $\mathrm{d}\eta + \phi\wedge\eta$.  Either this can be written in the form $-\alpha\wedge\eta$ where $\alpha$ is a (scalar-valued) $1$-form, or it cannot. (When it can be written in this form, $\alpha$ is necessarily unique.)  If it can, then $\theta = \phi + \alpha\,I_n$ is a torsion-free connection form that satisfies $\mathrm{d}\Omega = \Omega\wedge\theta-\theta\wedge\Omega$ (and it is the only such torsion-free connection form).  If it cannot, there is no solution.
Finally, check whether $\mathrm{d}\theta + \theta\wedge\theta = \Omega$.  If this equation holds, then you have found the (unique) torsion-free connection that solves the problem.  If this $\theta$ doesn't work, then there is no solution. 

Thus, you can usually check whether your problem has a solution in dimensions greater than $3$.  The only interesting cases that remain are those in dimension $3$ (where the problem can be subtle, since $L$ always has a large kernel) or algebraically special curvature-like tensors in higher dimensions.<|endoftext|>
TITLE: Ordinal-indexed transitive antichain of sets with urelements
QUESTION [7 upvotes]: Operate in ZFC. Can we find a function-class $\phi$ whose domain is the class of ordinals such that the following properties hold?

If $x \in \phi(\alpha)$, then either $x \in \mathbb{N}$ or there exists some ordinal $\beta < \alpha$ with $\phi(\beta) = x$;
If $\phi(\alpha) \subseteq \phi(\beta)$, then $\alpha = \beta$.

The first of these conditions equivalently states that the image of $\phi$ is a transitive set, except that the natural numbers are treated as urelements, and with the constraint that $\phi(\beta) \in \phi(\alpha) \implies \beta \in \alpha$.
The second condition means that $\phi$ is injective and its image is an antichain under inclusion.

We can construct such a function-class if we assume additional axioms, such as the existence of no inaccessible cardinals. In particular, the following construction will work:

$\phi(0) := \{ 1 \}$
$\phi(\alpha + 1) := \{ 2, \phi(\alpha) \}$ for every $\alpha$
$\phi(\omega) := \{ 3 \}$
$\phi(\beta) := \{ 4, \phi(\textrm{cf}(\beta))\} \cup \{ \phi(\gamma) : \gamma \in C_{\beta} \}$ (where $C_{\beta}$ is a cofinal subset of $\beta$ with order-type $\textrm{cf}(\beta)$ such that no element of $C_{\beta}$ is less than or equal to $\textrm{cf}(\beta)$), for singular limit ordinals $\beta$
$\phi(\omega_{\alpha+1}) := \{ 5, \phi(\alpha) \}$ for every $\alpha$

The first of these rules deals with the zero ordinal, and the second deals with successor ordinals. The third of these handles the special case $\omega$. The fourth handles limit ordinals which are not regular ordinals. The fifth handles initial ordinals of successor cardinals. This construction, however, does not define an image if there are inaccessible cardinals (whose initial ordinals do not fit the form for either the fourth or fifth rules).
Proof: We can verify transitivity since every element of $\phi(\alpha)$ is, by definition, either a natural number or a $\phi(\beta)$ for some earlier $\beta$. Verifying the antichain property is more complicated, but it boils down to the following:

Clearly $\phi(\alpha) \subseteq \phi(\beta)$ means that their unique urelements must agree, which means $\alpha$ and $\beta$ were processed by the same rule (out of the five provided).
The first and third rules each only process one ordinal, and the second and fifth are clearly injective as well.
For the fourth rule, we can recover $\textrm{cf}(\alpha)$ and $C_{\alpha}$ from $\phi(\alpha)$ by removing the urelement, inverting $\phi$, and separating the preimage into its first element and the remaining elements. If $\phi(\alpha) \subseteq \phi(\beta)$, it thus follows that $C_{\alpha} \subseteq C_{\beta}$, and that they have the same order-type (which is a regular ordinal), which implies they have the same limit, so $\alpha = \beta$.


This works when there are no inaccessible cardinals, and one can introduce further rules to provide a construction which works under the weaker hypothesis that there are no hyper-inaccessible cardinals. Is there a construction which does not depend on additional axioms beyond ZFC?

REPLY [7 votes]: If there is a subtle cardinal then such $\phi$ doesn't exist. 
An subtle cardinal is a cardinal $\kappa$ such that for every sequence of sets $\langle S_\alpha \mid \alpha < \kappa\rangle$ such that $S_\alpha \subseteq \alpha$ and a club $C\subseteq \kappa$ there are $\alpha < \beta$ in $C$ such that $S_\beta\cap \alpha = S_\alpha$.
In our case, let us fix some bijection $f$ between $On \times \{0\} \cup \omega \times \{1\}$ and $On$ such that $f(n, 1) < \omega$ for all $n < \omega$ and $f(\beta, 0) = \beta$ for all $\beta \geq \omega$. Define:
$$S_\alpha = \{f(\beta, 0) \mid \phi(\beta)\in\phi(\alpha)\}\cup \{f(n, 1) \mid n \in \phi(\alpha)\cap\omega\}$$
Note that $S_\alpha \subseteq \alpha$ for all $\alpha\geq \omega$. Clearly, $S_\alpha \subseteq S_\beta$ implies $\phi(\alpha) \subseteq \phi(\beta)$. Assume that $\kappa$ is subtle. Then there are $\omega\leq \alpha < \beta$ such that $S_\alpha = S_\beta\cap \alpha$ and in particular, $S_\alpha \subseteq S_\beta$. 
Remark: subtle cardinals are consistent with $V=L$ (assuming that they are consistent with ZFC) and if $0^\#$ exists then every Silver indiscernible is subtle.<|endoftext|>
TITLE: Determining the stretch of a cluster of points
QUESTION [5 upvotes]: I am trying to determine a metric for measuring cluster stretch. Let $C$ be a cluster of points $P_0, P_1,...,P_n$ in a two dimensional space with the same units.
I need a metric that will allow me to differentiate clusters that are long and thin from other clusters. 
I imagine it like a function that would be close to 0 when the cluster is shaped like a circle or a square and would be close to 1 when the cluster is shaped like a long and thin line or curve.
I thought of finding the length of the cluster by finding the shortest path between extreme points of the cluster, then finding the shortest distance from each point to the center line, then finding the mean distance between any point and the center line and then getting the ratio.
The ratio between the mean width of the cluster and its length would give a good estimation of the cluster stretch.
Unfortunately, this approach is quite difficult to implement in reality, so I thought I would ask for ideas. 
Maybe I should calculate cluster boundaries and then go from there? How about noise problems?
Thanks a lot for your help!

REPLY [8 votes]: There are many roundness measures that have been explored for different applications, which may give you ideas.
A good source for roundness in image processing is this paper, which analyzes
and compares several different measures:

Ritter, Nicola, and James Cooper. "New resolution independent measures of circularity." Journal of Mathematical Imaging and Vision 35.2 (2009): 117-127.
  (Springer link.)

The Wikipedia article on roundness could also be useful to you.
Much depends on your application. I doubt there is one "best" measure.<|endoftext|>
TITLE: Sum of inverse of multinomial coefficients
QUESTION [7 upvotes]: Find an asymptotically tight estimate for the sum
$$
A_n^{k}(\lambda)= \sum_{
\substack{a_i\geq \lambda_i
\\
a_1+a_2+\dots a_k=n
}} \prod_{i=1}^k a_i!
$$
Is the leading term going to  be
$$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i!
$$
Edit: As of right now there is some discrepency as to weather this conjecture is correct or not. 
This question has been asked before in the binomial situation Here

REPLY [2 votes]: I add two hopefully useful remarks.
I consider the general situation. In the sequel $k\geq 1$ and $\lambda=(\lambda_1,\ldots,\lambda_{k+1})$ are fixed,  $s:=\sum_{i=1}^{k+1}\lambda_i$ and $n\geq s$.
Let $A_n^{(k+1)}(\lambda):=\sum\limits_{\stackrel{a_i\geq \lambda_i, i=1,\ldots,k+1}{a_1+\ldots +a_{k+1}=n}} \prod_{i=1}^{k+1} a_i!$ 
and let  $\mathcal{S}_{k}:=\{ (x_1,\ldots,x_{k+1})\;:\;x_i\geq 0, \sum_{i=1}^{k+1} x_i=1\}$ denote the $k$-dimensional standard simplex.
(i)  $A_n^{(k+1)}(\mathbf{\lambda})$ has a nice geometric representation:
$$ A_n^{(k+1)}(\lambda)=\frac{(n-s+k)!\,(n+k)!}{(n-s)!}  \int_{\mathcal{S}_{k}} \int_{\mathcal{S}_{k}} \left(x_1y_1+\ldots+x_{k+1}y_{k+1}\right)^{n-s}\,\prod_{i=1}^{k+1} x_i^{\lambda_i}d\mathbf{x}\,\,d\mathbf{y} $$
Proof: recall
 Dirichlet's integral: $$((\sum_{i=1}^{k+1}a_i)+k)!\, \int_{\mathcal{S}_{k}} \prod_{i=1}^{k+1} x_i^{a_i}\,d\mathbf{x}=\prod_{i=1}^{k+1} a_i!$$
Now expand the integrand using the multinomial theorem and use Dirichlet's integral repeatedly.
EDIT:  I have replaced the previous (unnecessarily complicated) derivation and again corrected the factor (apologies). (The conclusions remain unchanged).
(ii) Thus the large $n$ asymptotic of $A_n^{(k+1)}(\lambda)$ can be investigated along the lines of the Laplace method (explained e.g. in chapter 4 of de Bruijn's book).
Remark: I have done that, and (unfortunately?) the result confirmed your original conjecture. Thus either my analysis or the accepted answer (or both) are incorrect (no offence). 
EDIT:
The Laplace analysis follows the usual scheme:
 (1) identify maxima
(2) cutoff tails, approximate integrand (3) complete tails
I sketch the main steps. Let $I$ denote the integral above.
Basic intuition:
(1) Consider the scalar product $\langle \mathbf{x},\mathbf{y}\rangle>=\sum_{i=1}^{k+1}x_iy_i\leq 1$. On the domain of integration $\frac{1}{k+1}\leq \langle\mathbf{x},\mathbf{y}\rangle\leq 1$, and $\langle\mathbf{x},\mathbf{y}\rangle$ can be 1 only in a "corner" $x_iy_i=1$ for a certain  $i$.
(2) For any $0<\epsilon<1$ the part of $I$ where $\langle\mathbf{x},\mathbf{y}\rangle<1-\epsilon$ is asymptotically negligible (compared to its complement). Asymptotically thus only the parts around the corners (i.e. $x_iy_i\approx 1$) need to be considered.
With this in mind:
(3) choose $\epsilon$ so small that the corners $C_i:=\{ x\in \mathcal{S}_k\;:\; x_i\geq 1-\epsilon, y_i\geq 1-\epsilon\}$ are disjoint, and discuss their influence individually.
E.g. for corner $k+1$ choose $x_1,\ldots,x_k$ resp. $y_1,\ldots,y_k$ as independent coordinates. 
Then  $x_{k+1}=1-\sum_{i=1}^k x_i=:1-u$, $y_{k+1}=1-\sum_{i=1}^k y_i=:1-t$.
We  have $\langle\mathbf{x},\mathbf{y}\rangle=(1-u)(1-t)+\sum_{i=1}x_iy_i\leq (1-u)(1-t)+ut$, thus $(\langle\mathbf{x},\mathbf{y}\rangle)^2\leq (1-2u(1-u))(1-2t(1-t))$. 
The parts of the integral where  $u\geq 1/n^{2/3}$ or $t\geq 1/n^{2/3}$
are negligible because $\langle\mathbf{x},\mathbf{y}\rangle^n=\mathcal{O}(\exp(-n^{1/3}))$ there.
On the remaining part $\langle\mathbf{x},\mathbf{y}\rangle^n =\exp(-n(u+t))(1+o(1))$ uniformly.
Therefore introduce new coordinates $x_1,\ldots,x_{k-1},u$, $y_1,\ldots,y_{k-1},t$, then
\begin{align*}
&\int_{{C}_{k+1}}\int \left(x_1y_1+\ldots+x_{k+1}y_{k+1}\right)^n\,\prod_{i=1}^{k+1} x_i^{\lambda_i}d\mathbf{x}\,\,d\mathbf{y} \\
&=\int\limits_{{x_i\geq 0, x_1+\ldots +x_k=u}\atop{{y_i\geq 0, y_1+\ldots +y_k=t}\atop {0\leq u,t\leq n^{-2/3}}}} e^{-n(u+t)}\prod_{i=1}^k x_i^{\lambda_i}(1-u)^{\lambda_{k+1}} dx_1\ldots,dx_k dy_1\ldots dy_k\,du\,dt\big(1+o(1)\big)\\
&= \int_0^{n^{-2/3}}\int_0^{n^{-2/3}} e^{-n(u+t)} (\frac{u}{n})^{s-\lambda_{k+1}+k-1} \frac{\prod_{i=1}^k\lambda_i!}{(s-\lambda_{k+1}+k-1)!} (1-u)^{\lambda_{k+1}} 
(\frac{t}{n})^{k-1}\frac{1}{(k-1)!}\,du\,dt\big(1+o(1)\big)\\
&=\frac{\prod_{i=1}^k \lambda_i!}{n^{2k+s-\lambda_{k+1}}}\left(1+o(1)\right)
\end{align*}
where last line follows after substituting $z=nu,w=nt$ and completing the tails.
Summing over all corners one now gets your formula (after applying Stirling's formula to it, and up to the additional factors).<|endoftext|>
TITLE: On Hamkins' answer to a problem by Michael Hardy
QUESTION [14 upvotes]: Based on a post by Michael Hardy and Hamkins' answer to it Andreas Blass, Will Brian, Joel Hamkins, Michael Hardy and Paul Larson introduced a new cardinal characteristic of the continuum $\mathfrak{rr}$ and as the main result they proved $\mathfrak{b}\leq\mathfrak{rr}\leq\mathop{\bf non}(\mathcal{M})$.(See the post by Hamkins on his homepage.)
The following question is motivated by a discussion I had with my friends Shahram Biglari and Mohammad Golshani.
Let $\mathfrak{i}\mathfrak{i}$ be the smallest cardinal such that there exists a family of injective functions $i:\omega\longrightarrow\omega$ of size $\mathfrak{i}\mathfrak{i}$ that guarantees the absolute convergence of a convergent real series $\sum\limits_{n=1}^{\infty} a_{n}$. Thus it's clear that $\mathfrak{r}\mathfrak{r}\geq\mathfrak{i}\mathfrak{i}$.

Question) What happens if one considers the set of all injective functions $i:
\omega\longrightarrow\omega$ instead of permutations. More precisely is it possible to have $\mathfrak{i}\mathfrak{i}<\mathfrak{r}\mathfrak{r}<\mathfrak{c}$?

Is it possible also to have $\mathfrak{i}\mathfrak{i}=\aleph_0$? Note that if $CH$ holds then $\mathfrak{r}\mathfrak{r}=\mathfrak{c}$.

REPLY [6 votes]: Let me show something about the subseries number, which I
suggested in the comments, and which I find very natural to
consider, in light of the following:
Fact. A series $\sum_n a_n$ is absolutely convergent just in
case every subseries $\sum_{n\in A} a_n$ is convergent, for every
$A\subset\mathbb{N}$.
So let us define the subseries number, denoted
$\newcommand\ss{\bf ß}\ss$ (German sharp s), to be the size of the
smallest family $\mathcal{A}$ of sets of natural numbers, such
that if a series $\sum_n a_n$ has the property that $\sum_{n\in
A}a_n$ converges for all $A\in\mathcal{A}$, then $\sum_n a_n$ is
absolutely convergent.
Clearly, $\frak{i\!i}\leq\ss$, since considering subseries amounts
to insisting on increasing injections in the context of the OP.
The splitting number $\newcommand\s{\frak{s}}\s$, in contrast,
is the smallest size of a splitting family $S$, which is a
family $\mathcal{S}$ of infinite sets of natural numbers, such
that if $B\subset\mathbb{N}$ is infinite, then there is some
$A\in\mathcal{S}$ such that $B\cap A$ and $B-A$ are both infinite,
and in this case we say that $A$ splits $B$ and that
$\mathcal{S}$ is a splitting family.
Meanwhile, ${\bf\text{non}}(\mathcal{M})$ is the size of the
smallest non-meager set in the space (it will be convenient to
consider) of all subsets of $\mathbb{N}$.
Theorem. $\s\leq\ss\leq{\bf\text{non}}(\mathcal{M})$.
Proof. For the first inequality, I shall show that every
subseries family $\mathcal{A}$, witnessing the defining property
of $\ss$, is also a splitting family. To see this (just as Will
had argued), consider any infinite set $B\subset\mathbb{N}$. Let
$\sum_n a_n$ be any conditionally convergent series whose nonzero
terms $a_n$ arise solely for $n\in B$. For example, $\sum_n a_n$
could be the alternating harmonic series, padded with $0$'s so
that the nonzero terms arise only on indices in $B$. Thus, $\sum_n
a_n$ converges, but only conditionally. So there must be some
$A\in\mathcal{A}$ such that $\sum_{n\in A} a_n$ diverges. Such a
set $A$ must have $B\cap A$ infinite, in order to have infinitely
many non-zero terms, and it must have $B-A$ infinite, in order to
differ infinitely from $\sum_n a_n$. So $A$ splits $B$, and
therefore $\mathcal{A}$ is a splitting family. So
$\s\leq|\mathcal{A}|$, as desired.
For the second inequality, I shall show that every nonmeager
family of subsets of $\mathbb{N}$ has the subseries property.
Suppose that $\mathcal{A}$ is such a nonmeager family, and we have
a conditionally convergent series $\sum_n a_n$. Since the positive
and negative terms each sum to $\infty$, it follows that any
finitely many terms of this series can be extended in such a way
to cause an additional huge oscillation; from this, it follows
that every conditionally convergent series has a comeager set of
subsets $A\subset\mathbb{N}$ for which $\sum_{n\in A}a_n$ does not
converge. Since $\mathcal{A}$ is non-meager, it must intersect
this set, and so there must be some $A\in\mathcal{A}$ for which
$\sum_{n\in A}a_n$ does not converge. So $\mathcal{A}$ has the
subseries property, as desired. QED<|endoftext|>
TITLE: Two questions about the "grasp" cardinal function
QUESTION [10 upvotes]: For a topology $\mathcal{T}$ on a set $S$, where $\mathcal{T}$ does not have a finite base, I define the grasp $g(\mathcal{T})$ to be the least infinite cardinal $\kappa$ such that $\mathcal{T}$ has a base $\mathcal{B}$ with $|\mathcal{B}|= w(\mathcal{T})$ , ($w$ is the usual weight function), satisfying $$\mathcal{T}=\{\bigcup V : V\in [\mathcal{B}]^{\leq \kappa}\}.$$ 
That is, every open set is the union of at most $\kappa$ members of $\mathcal{B}$, and $|\mathcal{B}|$ is minimum among all bases.
The following can be shown by elementary means: 

$g(\mathcal{T})\leq w(\mathcal{T})$. (Obvious).
$|\mathcal{T}|\leq w(\mathcal{T})^{g(\mathcal{T})}$. (Obvious).
$g(\mathcal{T})\leq g(D(w(\mathcal{T})))$  where  $D(\lambda)$ is discrete space of cardinality $\lambda$.
If $Y$ is a subspace of $X$ and $w(Y)=w(X)$ then $g(Y)\leq g(X)$.
When $\kappa$ is an infinite cardinal with the discrete or the order topology then 

$\operatorname{cf}(\kappa)\leq g(\kappa)\leq \kappa$.
If $\kappa$ is a singular strong limit then $g(\kappa)=\operatorname{cf}(\kappa)$.

(This last one helps to distinguish $g$ from other topological cardinal functions.)

I have two questions.

Question 1: Referring to (2) above, $g(T)$ is not necessarily the least cardinal $\lambda$ such that $|\mathcal{T}|\leq w(\mathcal{T})^\lambda$ because if we assume $2^{\omega}=2^{\omega_1}$ and let $\mathcal{T}$ be the discrete topology on $\omega_1$ then $|\mathcal{T}|=\omega_1^{\omega}$ but by (5)(i) $g(\mathcal{T})=\omega_1$. So is there an example like this in $\mathsf{ZFC}$?  
Question 2: By (5)(i) we have $\operatorname{cf}(2^{\omega})\leq g(D(2^{\omega}))\leq 2^{\omega}$. What values for $g(D(2^{\omega}))$ other than $2^{\omega}$ are consistent?

REPLY [2 votes]: Proposition (with Z. Szentmiklóssy): It is consistent that 
$\omega_1=cf (2^{\omega})<g(D(2^{\omega}))=2^{\omega}$.
Proof:
Assume GCH in the ground model. 
For ${\alpha}<{\omega}_1$ let 
$$
P({\alpha})=Fn({\omega}_{{\alpha}+1}\times {\omega}_{{\omega}_1+1},2;{\omega}_{{\alpha}+1}),
$$
and 
$$
P=\prod_{{\alpha}<{\omega}_1}P({\alpha}),$$
where the product is  the product with full support.
So   $P$ is just an Easton-forcing, hence it preserves all the cardinals,
and $2^{{\omega}_{{\alpha}+1}}={\omega}_{{\omega}_1+1}$ for all ${\alpha}<{\omega}_1$
in $V^P$.
Claim: For each  ${\beta}<{\omega}_1$
there is an almost disjoint family 
$\mathcal F_{\beta}\subset [{\omega}_{{\beta}+1}]^{{\omega}_{{\beta}+1}}$ 
with $|\mathcal F_{\beta}|={\omega}_{{\omega}_1+1}$ in $V^P$.
Proof of the Claim:
Write $P^{\beta}=\prod_{{\beta}\le {\alpha}<{\omega}_1}P({\alpha})$ and
$P_{\beta}=\prod_{{\alpha}<{\beta}}P({\alpha})$.
Since $P^{\beta}$ is ${\omega}_{{\beta}+1}$-complete, and  forcing 
with $P({\beta})$ introduces ${\omega}_{{\omega}_1+1}$ new subsets of ${\omega}_{{\beta}+1}$ , 
we have
$$
V^{P^{\beta}}\models 2^{{\omega}_{\beta}}={\omega}_{{\beta}+1}\land 
2^{{\omega}_{{\beta}+1}}\ge {\omega}_{{\omega}_1+1},
$$
hence in the model $V^{P^{\beta}}$ we have 
an almost disjoint family 
$\mathcal F_{\beta}\subset [{\omega}_{{\beta}+1}]^{{\omega}_{{\beta}+1}}$ 
with $|\mathcal F_{\beta}|={\omega}_{{\omega}_1+1}$.
Since $P=P^{{\beta}}*P_{\beta}$, we have such an $\mathcal F_{\beta}$ even in  $V^P$.
So we proved the Claim. Q.E.D
Let $Q=P*Fn ({\omega}_{{\omega}_1},2)$.
Since $P$ is $\sigma$-complete, we have 
$$
V^P\models 2^{\omega}={\omega}_1\land  ({\omega}_{{\omega}_1})^{\omega}={\omega}_{{\omega}_1},
$$
and so 
$$
|2^{\omega}|^{V^Q}\le ((|Q|^{\omega})^{\omega})^{V^P}={\omega}_{{\omega}_1}.
$$
So $2^{\omega}={\omega}_{{\omega}_1}$ in $V^Q$.
Assume on the contrary that 
$g(D({\omega}_{{\omega}_1}))={\omega}_{\alpha}<{\omega}_{{\omega}_1}$ in $V^Q$.
Let $\mathcal  B$  be a base witnessing $g(D({\omega}_{{\omega}_1}))={\omega}_{\alpha}$.
Since every $F\in \mathcal F_{\alpha}\subset [{\omega}_{{\alpha}+1}]^{{\omega}_{{\alpha}+1}}$
is the union of at most ${\omega}_{\alpha}$-many elements of $\mathcal B$, so 
 $F$ contains at least one element  $B_F\in \mathcal B$ of cardinality ${\omega}_{{\alpha}+1}$.
But the elements of $\mathcal F_{\alpha}$ are almost disjoint  (i.e.
the cardinalities of the intersections $\le {\omega}_{\alpha}$), so 
$B_F\ne B_{F'}$ for $F\ne F'$. 
Thus $|\mathcal B|\ge |\mathcal F_{\alpha}|>{\omega}_{{\omega}_1}$. Contradiction.<|endoftext|>
TITLE: Number of intervals needed to cross, Brownian motion
QUESTION [6 upvotes]: Let $B_t$ be a standard Brownian motion. Let $E_{j, n}$ denote the event$$\left\{B_t = 0 \text{ for some }{{j-1}\over{2^n}} \le t \le {j\over{2^n}}\right\},$$and let$$K_n = \sum_{j = 2^n + 1}^{2^{2n}} 1_{E_{j,n}},$$where $1$ denotes indicator function. I have three questions.

What is $\lim_{n \to \infty} \mathbb{P}\{K_n = 0\}$?
What is $\lim_{n \to \infty} 2^{-n} \mathbb{E}[K_n]$?
Does there exist $\rho > 0$ such that for $\mathbb{P}\{K_n \ge \rho2^{n}\} \ge \rho$ for all $n$?

REPLY [7 votes]: I'll address the second question on the expected value of the sum $K_n$.
Let $\phi(x)$ and $\Phi(x)$ be the probability density function and cumulative distribution functions for a standard normal distribution.
Let $h(a,b)$ be the probability that a Brownian motion without drift returns to $0$ at some time in $[a,b]$. Let $h(b)=h(1,b)$. Then by rescaling, $h(a,b)=h(1,b/a)=h(b/a)$. We can calculate this exactly.
For $x \gt 0$ let $f(x,t)$ be the probability that a Brownian motion released at position $x$ will hit $0$ by time $t$. Let $f(x) = f(x,1)$. By rescaling, $f(x,t)= f(\frac{x}{\sqrt{t}},1) = f(\frac{x}{\sqrt{t}})$. By reflection, $f(x) = 2 \Phi(-x) = 2-2\Phi(x)$. 
$$\begin{eqnarray}h(b) &=& 2 \int_0^\infty \phi(x) f(x,b-1)~dx \newline &=&2 \int_0^\infty \phi(x)\cdot 2 \Phi\left(\frac{-x}{\sqrt{b-1}}\right)~dx\end{eqnarray}$$ 
We can use differentiation under the integral sign. $h(1)=0$ and $h(b) = \int_1^b h'(t) dt$. 
$$\begin{eqnarray} h'(b) &=& 4 \int_0^\infty \phi(x) \phi\left(\frac{x}{\sqrt{b-1}}\right) \left(\frac{1}{2} \frac{x}{(b-1)^{3/2}}\right) dx \newline &=&\frac{2}{(b-1)^{3/2}}\int_0^\infty x \phi(x) \phi\left( \frac{x}{\sqrt{b-1}}\right) dx \newline &=& \frac{2}{(b-1)^{3/2}} \int_0^\infty \frac{x}{2 \pi} e^{-x^2 \cdot \left(\frac{1}{2} + \frac{1}{2(b-1)}\right)}dx \newline &=& \frac{1}{\pi b \sqrt{b-1}}\end{eqnarray}$$ 
So, $h(b) = \int_1^b \frac{dy}{\pi y \sqrt{y-1}} = 1-\frac{2}{\pi} \arcsin \frac{1}{\sqrt{b}}$.
$\mathbb{P}(E_{j+1,n})$ is the probability that the Brownian motion returns to $0$ on $\left[\frac{j}{2^n},\frac{j+1}{2^n}\right]$ which is $h(\frac{j}{2^n},\frac{j+1}{2^n}) = h(1 + \frac{1}{j}) = 1-\frac{2}{\pi} \arcsin \frac{1}{\sqrt{1+1/j}}.$ That can be simplified to $1-\frac{2}{\pi}(\frac{\pi}{2} - \arctan \frac{1}{\sqrt{j}}) = \frac{2}{\pi}\arctan \frac{1}{\sqrt{j}}$. For large $j$, this is approximately $\frac{2}{\pi} \frac{1}{\sqrt{j}}$. 
$$\mathbb{E}(K_n) \sim \sum_{j=2^n}^{2^{2n}-1} \frac{2}{\pi} \frac{1}{\sqrt{j}} \approx \frac{2}{\pi} \int_{2^n}^{2^{2n}} \frac{1}{\sqrt{x}} dx =\frac{4}{\pi}(2^n -2^{n/2}) \sim \frac{4}{\pi} 2^n.$$<|endoftext|>
TITLE: Intersection of rotating regular polygons
QUESTION [5 upvotes]: This question has a recreational flavor, but may not be
entirely uninteresting.
Let $P_k$ be a unit-radius regular polygon of $k$ sides,
and $P_n$ a unit-radius regular polygon of $n \ge k$ sides.
Both are origin-centered.
Fix $P_k$, and let $P_n$ rotate about its center by angle $\theta$.
Define $s(k,n)$ to be the sequence of the number of sides
of $P_k \cap P_n(\theta)$ as $\theta$ varies from $0$ to $2\pi$.
For example, I believe that $s(4,8)=(\overline{4,8})$,
with the overbar indicating repetition, while
$s(5,8)=(\overline{10,9})$.
Both are illustrated below.
(Reload the page to repeat the [limited] animations.)

          


          


It is an elementary puzzle to 

Q. Determine $s(k,n)$ precisely for all $n \ge k$.

I was interested in a higher-dimensional version of this question,
but already, although elementary, it seems to need some care to 
nail down $s(k,n)$ explicitly.

REPLY [7 votes]: There are two situations that can happen around a vertex of $P_k$, either the two nearest vertices of $P_n$ are going to crop off a small corner (thus adding one edge to the intersection $P_k\cap P_n$), or a vertex of $P_n$ coincides with the vertex of $P_k$. In the second situation, once two vertices coincide, $\gcd(n,k)$ of them will coincide. Therefore
$$s(k,n)=(\overline{2k,2k-\gcd(n,k)}).$$<|endoftext|>
TITLE: $f$-vector of simple convex polytope via directions of facets
QUESTION [8 upvotes]: Let $P$ be a simple convex polytope in $\mathbb{R}^d$ (that is, any vertex belongs to exactly $d+1$ facets). Given the collection of outer normals to facets of $P$, combinatorics of $P$ may be different, of course. But is this information enough to reconstruct number of faces of $P$ of all dimensions? If yes, what is specific procedure to do this? I am looking at what happened when we move facets parallel and pass through `singularity', i.e. not-simple polytope, and on first glance it looks that $f$-vector preserves, but I am always not sure with such things. And this is not good proof anyway, I would prefer more direct argument. 
If the statement is however false, I wonder for which specific collections it is still true.

REPLY [8 votes]: Consider a simplex $S$ in five dimensions, and one more face normal $\pi$ in general position with respect to $S$ (no hyperplane perpendicular to $\pi$ passes through more than one vertex of $S$). Intersect $S$ with a halfspace $H$ perpendicular to $\pi$, and translate $H$ continuously towards $S$, starting from a position where it contains all of $S$.
Then when the intersection with $H$ first cuts off one vertex of $S$, the one cut-off vertex is replaced by five new vertices (on the edges of $S$ connecting the cut-off vertex to the others) so the total number of vertices becomes $6-1+5=10$. When the intersection with $H$ cuts off two vertices of $S$, they are replaced by eight new vertices (again, one for each edge of $S$ that connects one of the two cut-off vertices to the four remaining ones), so the total number of vertices becomes $6-2+8=12$. And when the intersection with $H$ cuts off three vertices of $S$, they are replaced by nine new vertices, so the total number of vertices becomes again $6-3+9=12$. All of these are simple $5$-polytopes.
So no, the $f$-vector is not preserved. Not even the number of vertices is preserved.<|endoftext|>
TITLE: Number of squares in a finite group
QUESTION [13 upvotes]: This was asked at MSE but never answered.
Let $G$ be a finite group and denote by $sq(G)$ the number of squares in $G$ i.e. the number of elements in $G$ which possess a square root.  For example, if
$G$ is a group of odd order, then $sq(G) =|G|$, since each element has a square root, while at the opposite extreme, $sq(G)= 1$ when $G$ is an elementary abelian 2-group.  For the symmetric groups, the values of $sq(\mathrm{Sym}(n))$ are listed at $\,$https://oeis.org/A003483 $\,$.  Finally, we note that both the dihedral and quaternion groups of order 8 share the value $sq(G)=2$.
Questions:
1) Can $sq(G)$ be determined from information in the character table of $G$?
2) Is it true that $sq(\mathrm{Sym}(n))$ is divisible by every prime which is less than or equal to $n$ for $n>3$?

REPLY [8 votes]: The answer to Question 1) is "No" (at least when "information from the character table" is understood as only the matrix containing the values in the character table).  The following is an example of two groups with "the same" character table, but different numbers of squares. Define
$$ G_1 := \langle a_1, b_1, c_1\mid 
           a_1^2=b_1^4=c_1^8=1,\; b_1^{a_1}=b_1^{-1},\; 
           c_1^{a_1} = c_1^3,\; c_1^{b_1} = c_1^{-1}\; \rangle $$
and 
$$ G_2 := \langle a_2, b_2, c_2\mid 
           a_2^2=b_2^4=c_2^8=1,\; b_2^{a_2}=b_2^{-1},\; 
           c_2^{a_2} = c_2^{-1},\; c_2^{b_2} = c_2^3 \; \rangle .$$
Both groups are a semidirect product of $D_8 \cong \langle a_i, b_i \rangle$ acting on $C_8\cong \langle c_i \rangle$, but the action is different. (These are $\texttt{SmallGroup(64,149)}$ and $\texttt{SmallGroup(64,150)}$ from GAP's Small Groups Library.)  
Then $G_1$ has $5$ squares, namely $\langle c_1^2 \rangle \cup \{b_1^2\}$, but $G_2$ has $6$ squares, namely $\langle c_2^2 \rangle \cup\{b_2^2, b_2^2c_2^4 =(b_2c_2)^2\}$, as one sees by computing a "general square" $(a_i^k b_i^l c_i^m)^2$ in both groups.  
Finally, one can check (even without GAP) that the bijection $G_1\to G_2$ sending $a_1^ib_1^jc_1^k$ to $a_2^i b_2^j c_2 ^k$ induces a bijection between $\DeclareMathOperator{\Irr}{Irr} \Irr(G_1)$ and $\Irr(G_2)$ (the sets of irreducible characters, viewed as functions on $G_i$): The bijection between the groups yields an isomorphism $G_1/\langle c_1^4 \rangle \cong G_2/ \langle c_2^4 \rangle$, which takes care of the characters with $c_i^4 $ in the kernel. In both groups, the remaining characters are two characters of degree $4$ which vanish outside the center $\mathbf{Z}(G_i)=\langle b_i^2, c_i^4 \rangle$, so that the bijection above maps these characters onto each other: Indeed, $G_i/\langle b_i \rangle $ is isomorphic to the holomorph of $C_8$ in both cases, which has a unique faithful irred. character of degree $4$. The groups $G_1/\langle b_1^2c_1^4\rangle$ and $G_2/ \langle b_2^2 c_2^4 \rangle$ are not isomorphic, but both have a unique character of degree $4$ and central type.
(The $\chi\in \Irr(G_1/ \langle b_1^2 c_1^4 \rangle)$ with $\chi(1)=4$ has Frobenius-Schur indicator $\nu(\chi)=-1$, but the $\chi\in \Irr(G_2/\langle b_2^2c_2^4\rangle)$ with $\chi(1)=4$ has $\nu(\chi)=+1$. Such a thing must happen somewhere, as can be seen from the answer of Geoff Robinson.)<|endoftext|>
TITLE: Computing quantum cohomology for total spaces of vector bundles over $\mathbb{P}^m$
QUESTION [5 upvotes]: Let $X$ be the total space of $\mathscr{O}(-1)^{\oplus{n}}\rightarrow\mathbb{P}^m$. I think there should be some general way to compute its quantum cohomology $QH^\ast(X)$.
However, since I'm not familiar with these things, I found it hard to carry out the computations directly by hand without assuming that $m\gg n$, or both of $m$ and $n$ are very small, and of course it's trivial when $n=1$. When $m\gg n$, by checking dimensions one sees easily that the degrees of the holomorphic curves must be 0 or 1 in order to obtain non-trivial Gromov-Witten invariants, but this simplification does not seem to happen in general.
Is there any coherent way to compute $QH^\ast(X)$ for all the possible $m,n\geq1$? I think the problem should be trivial to experts and it's quite possible that it can be found in the literature.

REPLY [4 votes]: I'm not sure whether what I'm going to describe is good enough for your purpose. But the following will be the first thing that I would think about if I'm asked to compute your quantum rings. I apologize if you have been already aware of this stuff.
The keyword is Quantum Lefschetz hyperplane theorem (for concave bundle). Although computing the quantum cohomology might be hard, there is a beautiful formula for the small $I$-function associated to these (non-equivariant limit of) local Gromov-Witten invariants. Once the $I$-function is given, there is a systematic algorithm to reconstruct all genus-$0$ Gromov-Witten invariants if the cohomology of the variety is generated by divisors. I don't think I know a reference for the whole algorithm written down in one particular paper. But I think it should be done in the following order: 
Birkhoff factorization -> Mirror transformation -> "Divisor relation" + Topological recursion relation
(by "divisor relation" I mean Theorem 1 of this paper)
Come back to your question. Here is how to write the (small) $I$-function in your case. Let $h$ be the hyperplane class in $H^2(\mathbb P^m)$. Let $t_0, t_1$ be the formal parameters parametrizing the fundamental class $1$ and the hyperplane class $h$ (just like what you would do in the small quantum cohomology). Your $I$-function is the following series:
$$I(z)=e^{\frac{1t_0+ht_1}{z}}\left( 1+ \sum\limits_{d\geq 1}\frac{\prod\limits_{i=1}^{d-1}(-h-iz)^n}{\prod\limits_{i=1}^d(h+iz)^{m+1}}q^de^{dt_1}\right) $$
where the denominators should be expanded into $z^{-1}$ series. (Depending on the author, $I$ functions might differ by a factor $z$ and/or by a sign on the variable $z$)
As far as I can recall, I don't think this $I$-function has any exception for $m,n\geq 1$. And I believe the rest of the procedures will be the same  for different $m,n$. Therefore I would say it's a coherent way, though extremely complicated.
As an example, when $n=m+1$, the $I$-function and a few other invariants are described in section 3.3 of this paper. 
But I still want to make a few further remarks about the computation. As you may know, there is the small $J$-function as a generating series of one point descendant invariants in the form of $\langle \tau_i h^j\rangle_{0,n,d}$. It has a variable $z$ as well (sometimes denoted by $\hbar$). The above $I$-function agrees with the $J$-function when the factor in the bracket (after $e^{\frac{1t_0+ht_1}{z}}$) doesn't have $z^{-1}$ term, where all the denominators in the fractions are already expanded into $z^{-1}$ series. In your case, I believe it happens when $n\leq m+1$ and $m$ is not too small. In this case, you can read off enough amount of one-point invariants (i.e., skip the Birkhoff factorization and Mirror transformation). One can also read off some of the relations in the small quantum ring by looking at the differential operators in $t_1$ that annihilates the $J$-function. I'm not sure how nice and explicit your quantum rings can be described. But these might be a good start.<|endoftext|>
TITLE: Sums of reciprocals of products of factorials
QUESTION [6 upvotes]: Let $d,m, r$ be positive integers, and define
$$
S = \left\{ (i_1, i_2, \dots, i_m) \in {\bf Z}_{+}^{m} \left | \sum_j i_j = d; \& \forall j, i_j \leq r \right. \right\};
$$
Here ${\bf Z}_+$ denotes the set of nonnegative integers (that is, including zero). So we are taking all (ordered) integer partitions of $d$ with $m$ constituents (permitting zero as a constituent), with the constituents bounded above by r.
I would like a good upper bound (in terms of $m$, $d$, and $r$) for
$$
\sum_{(i_j) \in S} \frac 1{i_1! \cdot i_2! \cdot \dots \cdot i_m!}.
$$
When there is no $r$ condition (e.g., if $r \geq d$), then it looks like the sum is exactly $m^d/d!$, arising from representing $m^d$ as a combination of descending products. And of course, if $mr < d$, then there are no terms.
However, I am most interested in the case when $d$ is less than $mr$ (special cases: if $d = mr-1, mr-2, mr-3, mr-4,  $, etc.; exact results are easily obtained). But I would like a general result or a reference for this type of problem (with $d$ much less than $mr$); this must have been considered before.

REPLY [4 votes]: What you have is the coefficient of $x^d$ in $f(x)^m$, where
$f(x) = \sum_{j=0}^r x^j/j!$. This is the sort of problem for which the saddle-point method works readily. For sure it has all been worked out before, but I don't remember where.
You can usually get the right answer by a probabilistic device. For any $\alpha>0$, $f(\alpha x)/f(\alpha)$ is the probability generating function for a random variable $X$, so $f(\alpha x)^m/f(\alpha)^m$ is the pgf for the sum of $m$ independent copies of $X$.  Choose $\alpha$ so that the mean of $X$ is about $d/m$; this involves solving some nonlinear equation but there is no escape since the solution really appears in the answer. Then apply the central limit theorem (say Berry-Esseen) to the sum of $m$ copies of $X$, together with the fact that the sequence is log-concave, to estimate the coefficient of $x^d$.  It should be possible to get precise asymptotics over the full range (perhaps with ad hoc methods at the extremes).<|endoftext|>
TITLE: First formulation of the Dedekind and Hasse-Weil conjectures
QUESTION [15 upvotes]: I'm looking for the original statement of two important conjectures in number theory concerning L-functions. I'm particularly interested in pinning down the year in which they were first formulated:

The Dedekind conjecture (that the quotient $\zeta_K(s)/\zeta(s)$ is entire for any number field $K$)

The only information I've been able to find is that Dedekind himself proved the conjecture in the case of pure cubic fields in 1873. This definitely gives an upper bound. I'm also curious to know for which extensions $K$ over $\mathbb{Q}$ he expected the conjecture to be true.
Edit (1) I've found the paper in question:
Richard Dedekind, Über die Anzahl von Idealklassen in reinen kubischen Zahlkörpern (1900)
Here is a link to his complete works. I can't read german, so the question is still the same, but I thought the link might help. Discussion about the zeta function of number fields seems to start in page 166 (section 6). 
Edit (2) Another clue, mentioned by Vesselin Dimitrov in the comments, and suggesting that the conjecture might be due to Landau:
"On page 34 of Bombieri's article The classical theory of zeta and L-functions in the Milan Journal of Mathematics, the conjecture is attributed to Landau. So it probably appears in Landau's Handbuch, much earlier than in Artin"

The Hasse-Weil conjecture (the zeta function of an algebraic variety has a meromorphic continuation to the complex plane and a functional equation)

(note: this has been nicely answered by Francois Ziegler in the comments)
The problem with this one seems to be that it applies to several different kinds of functions, and both definitions and result came in several stages. I think that the first and most basic definitions is Artin's zeta function of a curve over a finite field. This was generalized to arbitrary algebraic varieties, and also to schemes of finite type over $\mathbb{Z}$. Yet the most common definition is over number fields. There's also the local/global and L/zeta distinction.
In this case I'm just trying to get a feeling of when/where the conjecture was first formulated in some generality.

REPLY [5 votes]: Artin (1923) wrote that Dedekind proved the case of this conjecture concerning pure cubic number fields. Dedekind published his article in 1900, but writes that it is a reworking of a draft he had written in 1871/72. The conjecture was named after Dedekind by van der Waall in 1974. 
Despite Artin's claim, the integrality of the quotient $\zeta_K(s)/\zeta(s)$ is not addressed at all by Dedekind (Hecke proved that $\zeta_K(s)$ extends to a meromorphic function on the complex plane in 1917). Dedekind proved that  $\zeta_K(s)/\zeta(s)$ for pure cubic extensions may be written as a linear combination of Epstein zeta functions. Epstein proved in 1903 that these are meromorphic with simple poles in $s = 1$, and Artin realized in 1923 that this implies that the quotient is entire - I guess this explains his remark. 
In any case Dedekind had nothing to do with this conjecture.
Edit. Landau [Über die Wurzeln der Zetafunktion eines algebraischen Zahlkörpers, 
Math. Ann. 79 (1919), 388-401] writes on p. 390: "I have to be careful with its formulation since it is not known whether $\zeta_k(s)/\zeta(s)$ is always an entire function".<|endoftext|>
TITLE: What is the most efficient way to factor a matrix into a given set of generators?
QUESTION [8 upvotes]: I am studying finite index subgroups of certain finitely presented groups. The particular conditions on my groups make this problem easier than I phrase it here, but I am curious about a more general answer. After the main question I will indicate a simplified case for which there is probably a simplified answer.
Suppose you have a group of matrices $G\subset\mathrm{GL}_n(R)$
over a ring $R$,
and suppose it is finitely generated so that
$G=\langle g_1,\dots,g_n\rangle$.
Next suppose you are given a matrix $m\in G$
and you want to know how to write it as a word in the generators,
as short as possible.
What is an efficient way of computing this?
An inefficient solution that will give you a word
(not necessarily the shortest)
is to systematically go through all possible words and check by multiplying them together until $m$
is a result,
which is possible since the set of words on $n$ letters is countable.
Perhaps a smarter approach is to compute the Jordan canonical form of $m$ and of each of the $g_i$, then find a basis for $G$
in which you can write each of the $g_i$
as well as $m$,
upon which the solution can be found just be piecing words together
rather than having to multiply matrices each time.
I'm uncertain if this would lead one to discover the shortest word.
Even if it did, perhaps there is a more efficient process.
The easier sub-case:
Suppose $G\subset\mathrm{PSL}_2(\mathbb{C})$
is discrete and arithmetic, i.e. is Kleinian and has a representation into $\mathrm{GL}_n(\mathbb{Q})$ for some $n$.
Moreover, suppose $m=\overline{g}^{\top}g$ (it is Hermitian)
for some $g\in G$.
Is there an especially nice choice of generators for $\mathrm{PSL}_2(\mathbb{C})$
into which $m$
and the $g_i$
can be factored?
Or, perhaps a better approach than that?
I'm feeling like the eigendecomposition could be useful,
perhaps by using $m$'s pair of orthogonal eigenvectors
and the limited ways of splitting their eigenvalues (which are real)
over the coefficient ring.

REPLY [6 votes]: For the general case (i.e. no restrictions on the set $\{ g_1,\ldots,g_n \}$), one cannot get a computable upper bound on the runtime of any algorithm for the dimension $m\geq 4$. 
It follows from the undecidability of the membership problem in $SL_4(\mathbb{Z})$ due to Mikhailova. If one had an algorithm with a computable upper bound for the running time, then we could run it on any matrix $g\in SL_m(\mathbb{Z})$ and stop if it takes longer than our bound. After that we could check if the output indeed gives a word such that $g=g_{i_1}g_{i_2}\ldots g_{i_s}$ and hence we would solve the membership problem.<|endoftext|>
TITLE: In CGWH, is every cofibration an inclusion with closed image?
QUESTION [9 upvotes]: As the title suggests, in CGWH, is every cofibration an inclusion with closed image?

REPLY [6 votes]: The main obstruction in the CGWH case (as opposed to the Hausdorff case) is in proving that retracts have closed image. But if $r: X \to X$ is idempotent for a compactly generated space $X$, then$$r(X) = \{x : x = r(x)\} = (r \times \text{Id})^{-1} \Delta X,$$which is closed by continuity and the following lemma in $\S6.1$ of May's A Concise Course in Algebraic Topology. 

Lemma. If $i: A \to X$ is a cofibration and $g: A \to B$ is any map, then the induced map $B \to B \cup_g X$ is a cofibration.

Basically, it is all shuffled into the $k$-fication of products.

Update. In more detail, suppose $i: A \to X$ is a cofibration. By the homotopy extension property, the map $r: X \times I \to Mi$ described on page 44 of May's A Concise Course in Algebra Topology exists, along with $j: Mi \to X \times I$ such that $r \circ j = \text{Id}$. This implies $j$ is a monomorphism. Since $j|_{A \times I} = i \times \text{Id}$, it follows that$$i(a) = i(b) \implies (i \times \text{Id})(a, 0) = (i \times \text{Id})(b, 0) \implies a = b.$$Thus, $i$ is injective.
We claim $j(Mi) \subset X \times I$ is closed. To see this, let$$S = \{(x, t) \in X \times I : j \circ r(x, t) = (x, t)\}.$$Then since$$j \circ r \circ j = j \circ \text{Id} = j,$$we see that $j(Mi) \subset S$. Conversely, if $(x, t) \in S$, then$$(x, t) = j \circ r(x, t) \in j(Mi),$$so $S \subset j(Mi)$. Thus, it suffices to prove $S$ is closed. This follows from the following lemma.

Lemma. Suppose $Y$ is compactly generated. Let $f: Y \to Y$ be a map. Then $$\text{Fix}(f) = \{y \in Y : f(y) = y\}$$is closed in $Y$.

Proof. Let $\Delta Y = \{(y, z) \in Y \times Y : y = z\}$. Consider the following lemma in $\S6.1$ of May's A Concise Course in Algebra Topology.

Lemma. If $X$ is a $k$-space, then $X$ is weak Hausdorff if and only if $\Delta X$ is closed in $X \times X$.

By this lemma, we know $\Delta Y$ is closed in $Y \times Y$. Consider the map $\phi: Y \to Y \times Y$ given by $y \mapsto (y, f(y))$. This is continuous, so $\text{Fix}(f) = \phi^{-1}(\Delta Y)$ is closed in $Y$.$$\tag*{$\square$}$$Since $X \times \{1\} \subset X \times I$ is closed, we see$$j(Mi) \cap (X \times \{1\}) = (i \times \text{Id})(A \times \{1\}) = i(A) \times \{1\}$$is closed in $X \times I$. Let $\pi: X \times I \to X$ be the projection. We claim this is a closed map.

Lemma. If $Y$ is compact, the projection $\pi: X \times Y \to X$ is a closed map.

Proof. Let $S \subset X \times Y$ be closed. Choose $x \in X \setminus \pi(S)$. For any $y \in Y$, we know $(x, y) \notin S$, so there exists a neighborhood $U_{(x, y)} \times Y_{(x, y)} \subset S^c$ containing $(x, y)$. The $V_{(x, y)}$ cover $Y$, so there is some finite subcover $V_{(x, y_1)}, \dots, V_{(x, y_n)}$. Let $U = \bigcap_{k=1}^n U_{(x, y_k)}$. Then $U \subset \pi(S)^c$ is an open neighborhood of $x$. Thus, $\pi(S)^c$ is open, i.e. $\pi(S)$ is closed.$$\tag*{$\square$}$$Thus, we conclude $\pi(i(A) \times \{1\}) = i(A)$ is closed in $X$. 
Lastly, we claim $i: A \to X$ is a homeomorphism onto its image. To see this, let $h_t: A \to Mi$ be the map $h_t(A) = (a, t)$. Note that $h_0(a) = (a, 0)$ so $h_0$ restricts the inclusion $\nu: X \hookrightarrow Mi$. Thus, by the homotopy extension property, we get a homotopy $\widetilde{h}_t: X \to Mi$ with $\widetilde{h}_t \circ i = h_t$. Since $h_1$ is a homeomorphism $A \to A \times \{1\}$, we have $h_1^{-1} \circ \widetilde{h}_1 \circ i = \text{Id}$. Thus, $h_1^{-1} \circ \widetilde{h}_1 : i(A) \to A$ is a continuous inverse for $i$.<|endoftext|>
TITLE: Regularity up to the boundary for the Poisson problem
QUESTION [5 upvotes]: It seems that the following assertion is widely accepted:
For $k\in\mathbb N$, $p\geq 2$, $\Omega \subset \mathbb R^n$ bounded with $\partial\Omega\in C^{k+2}$ and $f\in W^{k,p}(\Omega)$, the weak solution $u\in H^1_0(\Omega)$ of the problem
$$
\begin{equation}
\left\{
\begin{aligned}
-\Delta u =f \text{ in } \Omega\\
u=0 \text{ on } \partial\Omega
\end{aligned}
\right.
\end{equation}
$$
satisfies $u\in W^{k+2,p}(\Omega)$ and $\|u\|_{W^{k+2,p}}\leq C_{\Omega,k,p}\|f\|_{W^{k,p}}$ for some $C_{\Omega,k,p}>0$.
The above is proved in Evans using difference quotients for $p=2$. For $k=0$ it appears to be true due to an interpolation argument (Theorem 7.1 of Giaquinta's and Martinazzi's book on regularity theory). For Hölder continuous domains one can use the classical Schauder theory. But is there a reference for the complete result?

REPLY [2 votes]: P. Grisvard, Elliptic Problems in Nonsmooth Domains, 1985:
Thm. 2.5.1.1 (you even need to impose less regularity on $\partial\Omega$).<|endoftext|>
TITLE: Logical complexity of algebraically closed fields
QUESTION [14 upvotes]: One can define fields using a finite list of axioms that quantify over the field itself.  However, the obvious way to define algebraically closed fields involves either an infinite list of axioms, or a more logically complex framework where one can quantify over natural numbers.  It would be interesting if one could avoid this, and give a finite list of axioms that quantify only over the field, but my guess is that that is impossible.  Are there known theorems in that direction?

REPLY [10 votes]: Certainly Todd answered the question, but the following "theorem" shows that this sort of problem is not as hard as it originally appears.
Suppose $\Sigma$ is an infinite set of statements such that no finite subset of $\Sigma$ implies all of $\Sigma$.  Then the theory $T$ which is the set of consequences of $\Sigma$ is not finitely axiomatizable.
The proof of this is immediate; if there were a finite set of axioms, they would be implied by some finite subset $\Sigma_0$ of $\Sigma$, so by transitivity, $\Sigma_0$ would imply all of $T$, and thus all of $\Sigma$.
Now to think about algebraically closed fields, see that the first set of axioms you come up with satisfies this.  We simply say that $F$ is a field, and for every $n\geq 2$, every monic polynomial of degree $n$ has a solution (this is a single sentence for each $n$; you quantify across the coefficients).  This works, and no finite subset works (to verify this one does need to show that you can have fields which have no extensions of degree $\leq k$, but has some irreducible polynomial of degree $>k$, which should be elementary).
A lot of other questions along these lines can be answered similarly.<|endoftext|>
TITLE: connected stable rank
QUESTION [5 upvotes]: There is a beautiful formula by Leonid Vaserstein relating the Bass and topological stable rank of a commutative unital Banach algebra A to 
that of the matrix algebra M_n(A). Is there something similar true for
the connected stable rank (defined to be the least integer n such that
for all m bigger than n the set U_m(A) of unimodular/invertible m-tuples
is connected).Thanks.

REPLY [4 votes]: In [1] Proposition 2.10, and also [2] Theorem 4.7, it is shown that the connected stable rank satisfies the following estimate for the matrix algebra over a (not necessarily commutative) C*-algebra $A$:
$$
\mathrm{csr}(M_n(A)) \leq \left\lceil \frac{\mathrm{csr}(A)-1}{n}\right\rceil+1.
$$
The reverse inequality seems to be unclear.
[1] Nistor. Stable range for tensor products of extensions of K by C(X), J. Operator Th. 16 (1986)
[2] Rieffel. The homotopy groups of the unitary groups of non-commutative tori, J. Oper. Theory 17 (1987)<|endoftext|>
TITLE: Convergence in the proof of Crofton's Formula
QUESTION [5 upvotes]: Let $\mathcal{L}$ be the set of oriented lines in $\mathbb{R}^2$ and let $\mu$ be the Kinematic Measure on $\mathcal{L}$; up to scaling, $\mu$ comes from the unique (up to scaling) volume form on $\mathcal{L}$ that is invariant under the action of the group of isometries of $\mathbb{R}^2$.  Given any function $f:[0,1] \rightarrow \mathbb{R}^2$, define a function $n(f):\mathcal{L} \rightarrow \mathbb{Z} \cup \{\infty\}$ by setting $n(f)(L)$ to be the number of elements of the set $\{\text{$x \in [0,1]$ $|$ $f(x) \in L$}\}$.
The key step in one of the standard proofs of Crofton's formula asserts the following.  Let $f:[0,1] \rightarrow \mathbb{R}^2$ be a rectifiable curve.  For any partition $P$ of $[0,1]$, let $f_P:[0,1] \rightarrow \mathbb{R}^2$ be the associated polygonal approximation to $f$, so $f_P(x) = f(x)$ for $x \in P$ and $f_P$ restricts to a straight line segment on each component of $[0,1] \setminus P$.  The set of partitions of $[0,1]$ is partially ordered by refinement, and by definition the length of $f$ is the limit (with respect to this partial order) of the lengths of the $f_P$.  
The "key step" referred to above is the following: the limit (over the set partitions $P$ of $[0,1]$) of $\int_{\mathcal{L}} n(f_P) d\mu$ equals $\int_{\mathcal{L}} n(f) d\mu$.
I don't see how to prove this and cannot find a source that has a complete proof.  Can anyone either give me a proof or a good reference for it?

REPLY [6 votes]: Rectifiability is a very weak condition, and the method of approximation of length you use is not the best one. Normally one uses  circumscribed polygons rather than inscribed polygons as you suggest. For higher dimensional objects the famous example of Schwartz shows why the approach based on inscribed polyhedra   fails spectacularly. In particular, the approach to Crofton formula you suggest   does not work in higher dimensions.
The typical proof of Crofton formula  relies on the double fibration trick aided by the coarea  formula.   Here is a proof of this formula for curves. The same proof extends to  higher dimensions as well; see  Chapter 9 of these notes.<|endoftext|>
TITLE: mapping class group relations
QUESTION [6 upvotes]: The question I want to ask is vague in a sense. We have examples of mapping class relations, e.g. lantern relation, chain relations, etc. For instance the latern relation on a disk with three boundary components $a$, $b$ and $c$ is: $t_a t_b t_c t_d=t_{ab}t_{bc}t_{ca}$, where $d$ is the curve enclosing all three boundary components and $t_{ij}$ encloses $i$ and $j$ in the most obvious way. While I know how this relation is proved, I have no idea how one comes up with this kind of relations.I want to know if there is general principle to obtain these relations, and whether it is possible to figure out different factorizations of a given element in a mapping class group (of a surface with or without boundary). Is there a more intuitive way to explain for instance the lantern relation?

REPLY [8 votes]: I can answer your last question: there exist natural ways of embedding the fundamental group of the unit tangent bundle of a surface into the mapping class group, and the lantern relation is the image of an "obvious" relation in the unit tangent bundle group (coming from a visually obvious relation in the fundamental group of a surface).  This way of coming up with the lantern relation was discovered independently by myself and by Margalit--McCammond; see our papers
A. Putman
An infinite presentation of the Torelli group
Geom. Funct. Anal. 19 (2009), no. 2, 591-643. 
and
Margalit, Dan; McCammond, Jon Geometric presentations for the pure braid group. J. Knot Theory Ramifications 18 (2009), no. 1, 1–20.
In my paper, this can be found in Section 3.1.2.  I also discuss this in my survey paper
A. Putman
The Torelli group and congruence subgroups of the mapping class group
in "Moduli spaces of Riemann surfaces (Park City, UT, 2011)", 167-194,
IAS/Park City Math. Ser., 20 Amer. Math. Soc., Providence, RI.
This occurs towards the beginning (on page 6 in the version on my webpage).  All of my papers can be found here.<|endoftext|>
TITLE: What are some important papers that use complex analytic techniques to get good bounds?
QUESTION [7 upvotes]: The motivation behind this question is somewhat similar to that of the tricky project launched by Gowers et al, but is certainly a specialization. My work tends to rely on both exact formulae and analytic techniques, most notably from complex analysis and Fourier analysis. It appears that search engines and fora like this is a great way to discover useful exact formulae that at least partially fit the bill, but analytic techniques, especially those magic ones like stationary phase approximation, are harder to come by, perhaps due to their subtle and amorphous nature. Thus there seems a genuine need to enumerate some of the most exemplary articles that exploit these techniques, from which I can draw inspirations. 
Note that I am specifically interested in getting bounds, as opposed to exact formulae. So for instance Riemann's use of complex analytic continuation to derive the functional equation would not count, even though one could argue exact formulae are a special case of bounds. An example that helped me move forward is Nazarov's extension of Remez's inequality. Another is Lucia's solution to this problem of mine, as well as the paper he cited. 
Not to give the wrong impression, I do think most ground-breaking results, including bounds, are proved by exploiting idiosyncratic properties pertaining to the problem itself. These could be some exact formulae, some symmetries, or some geometric interpretation. However it has become clear to me that analytic number theorists often have an edge in situations involving bounds of algebraic formulae.

REPLY [4 votes]: A simple but effective upper bound technique is to have a quantity expressed  by a Cauchy integral, choose an optimal contour $\gamma$, and proceed by inserting the absolute values along $\gamma$. For example, a clean  estimate $\log{\binom{a}{b}} \leq b \log(b/a) + (a-b)\log(1-b/a)$ on the binomial coefficients follows from expressing 
$$
\binom{a}{b} = \Big| \int_{|z| = R} \frac{(1-z)^a}{z^b} \, d\theta \Big| \leq R^{-b}
(1+R)^a
$$
 and noting that the bound is optimized for the choice $R = b/(a-b)$. This is the same idea as in Lucia's solution to your problem, with the added point that the choice of the contour is on  disposal. Many papers on diophantine approximations involve this idea in extrapolation arguments, e.g. Gelfond's solution to Hilbert's 7th problem.<|endoftext|>
TITLE: Recurrence of Poisson binomial distributed random walk
QUESTION [6 upvotes]: Let $X_n$ be the outcome of a Bernoulli trial where the probability of getting 1 is $p_n$ and the probability of getting 0 is $1-p_n$, and let $S_n = \sum_{i=1}^n \left(X_i - \textrm{E} X_i \right)$. In general, $p_i \neq p_j$, so $S_n$ is Poisson binomial distributed, but with the mean subtracted. Since $S_n$ has finite variance and expectation value 0, I would assume that $S_n$ is recurrent and that this should be a fairly well-known result. However, I have been browsing the literature and asking around, and despite the problem being apparently rather simple, I have not yet been able to find a proof. 


QUESTION: Does anyone have a proof that $S_n$ is recurrent, alternatively, references to relevant literature where I could find proof(s) and discussion(s) of this problem?

REPLY [7 votes]: $S_n$ is a martingale with bounded jumps, and there is a result that it should either converge to a finite limit, or fluctuate, in the sense that $\limsup S_n=+\infty$, $\liminf S_n=-\infty$ (this, I guess, should be understood as "recurrence"). See e.g. Theorem (3.1) of Chapter 4 of [Durrett, "Probability, Theory and Examples" (2004)]. If $\sum p_n = \sum (1-p_n) = \infty$, this should rule out the first possibility, so the walk will have to be recurrent.<|endoftext|>
TITLE: Reinterpreting Galois descent over finite fields
QUESTION [10 upvotes]: This question is indirectly related to my previous question Is an elliptic curve that is isomorphic to its Frobenius conjugate defined over $\mathbb{F}_p$?
Let $\mathbb{F}_{q^n}/\mathbb{F}_q$ be an extension of finite fields and $X$ a quasi-projective variety over $\mathbb{F}_{q^n}$. I suppose that I have an $\mathbb{F}_{q^n}$-isomorphism $i\colon X^{(q)} \simeq X$ between the $q$-Frobenius pullback of $X$ and $X$ itself. I want a criterion for when $i$ is a descent datum with respect $\mathbb{F}_{q^n}/\mathbb{F}_q$, a criterion specific to finite fields.
Of course, the cocycle condition is a criterion of this sort, but I have seen on p. 133 of a paper of Deligne and Rapoport "Les schemas de modules de courbes elliptiques" a different criterion, which I do not completely understand. Thus my question is:
Is it true that $i$ is such a descent datum if and only if the $n$-th power of the composition $i\circ \mathrm{Fr} \colon X \rightarrow X^{(q)} \rightarrow X$ is equal to the absolute $q^n$-power Frobenius of $X$ (here $\mathrm{Fr}$ is the relative $q$-power Frobenius)?
Note that $\mathrm{Fr}$ is not an isomorphism (same for the absolute Frobenius of $X$), so the Galois descent criterion I am asking for cannot be a tautological consequence of the cocycle condition (which involves only isomorphisms).

REPLY [11 votes]: Yes, assuming $X$ is reduced (so that its absolute Frobenius endomorphism is schematically dominant); this really is a formal consequence of the usual descent formalism (despite whatever red herring appears with Frobenius maps not being isomorphisms). 
As a warm-up, consider the case $n=1$ (so $X^{(q)} = X$ tautologically and ${\rm{Fr}}$ is thereby identified with the absolute $q$-Frobenius):  for this you're asking that $i \circ {\rm{Fr}} = {\rm{Fr}}$, which is to say that $i = {\rm{id}}$ since ${\rm{Fr}}$ is schematically dominant, which is indeed the only possible Galois descent datum when $n=1$.  So below we will assume $n>1$ to make things interesting.
By the universal functoriality of the relative $q$-Frobenius, the $n$th power of that composition is the same as $X \rightarrow X^{(q^n)} \simeq X$ where the first step is the relative $q^n$-Frobenius and the second step is the composition $i \circ i^{(q)} \circ \dots i^{(q^{n-1})}$ (notation meaningful as written since $n>1$).  But the relative $q^n$-Frobenius of $X$ is identified with its absolute Frobenius via the canonical isomorphism $X \simeq X^{(q^n)}$ defined via how $X$ is equipped with a structure of $\mathbf{F}_{q^n}$-scheme. Hence, again by the dominance condition (using reducedness of $X$) we can "cancel" as when $n=1$ to see that the criterion in your question is exactly the condition that this latter composite isomorphism coincides with the canonical isomorphism $X^{(q^n)} \simeq X$ already mentioned.
Now all of the usual confusion that permeates every discussion of absolute and relative Frobenius maps has been eliminated.  And to clarify matters let's get rid of all mention of finite fields by instead considering a cyclic extension of fields $K/k$ of degree $n>1$ with a chosen generator $\gamma$ of ${\rm{Gal}}(K/k)$ and an arbitrary $K$-scheme $X$ (no need for reducedness anymore) equipped with a $K$-isomorphism $i:\gamma^{\ast}(X) \simeq X$.  Any $K/k$-descent datum on $X$ gives rise to such an $i$ which moreover uniquely determines it (by the cocycle condition), and we can ask to characterize those $i$ which arise in that way.  Namely, a necessary condition is that the composite isomorphism 
$$i \circ \gamma^{\ast}(i) \circ \dots \circ (\gamma^{n-1})^{\ast}(i): (\gamma^n)^{\ast}(X) \simeq X$$
coincides with the canonical such isomorphism (using that $\gamma^n = {\rm{id}}_K$).
Let's make one final maneuver, to turn this into entirely a question about actual automorphisms of $X$ (though only over $k$, not $K$), exactly in the spirit of the elegant discussion of Galois descent as a special case of Grothendieck's descent formalism in Example B in 6.2 of the book "Neron Models".  Namely, by the universal property of fiber products a $K$-isomorphism $i: \gamma^{\ast}(X) \simeq X$ is the "same" as a scheme isomorphism $i':X \simeq X$ over the inverse of the automorphism ${\rm{Spec}}(\gamma)$ of ${\rm{Spec}}(K)$ (so $i'$ is over $k$), or in other words $i^{-1}$ corresponds to an automorphism of the scheme $X$ over ${\rm{Spec}}(\gamma)$.  In this way we see that the condition we are imposing on that $n$-fold composition is precisely the statement that $i'$ has $n$-fold composition equal to the identity (or same for ${i'}^{-1}$ if you prefer).  That in turn is precisely saying that we are giving a $k$-action on $X$ by a cyclic group of order $n$, or more specifically an action by ${\rm{Gal}}(K/k)$ covering its action on ${\rm{Spec}}(K)$ over ${\rm{Spec}}(k)$ (I leave to you the pleasure of getting left-actions and right-actions straightened out by thinking about the analogue of these matters with a possibly non-abelian Galois group).  Referring again to the exposition given in "Neron Models" as cited above, we are done.
QED<|endoftext|>
TITLE: How to compute second homology of a group given by presentation with two relators
QUESTION [9 upvotes]: I am interested in calculating of $H_2(G,\mathbb{Z})$, where $G$ is a group given by presentation with two relators $\langle a,b| r_1 = r_2 = 1\rangle$. 
Moreover, I am interested in such presentations, that $H_2(G,\mathbb{Z}) = \mathbb{Z}^2$, it would be great to learn how to construct a lot of such examples.

REPLY [4 votes]: If $G$ is a perfect group (i.e. $G=[G,G]$, i.e. $G_{ab}=0$) which admits a presentation with two generators and two relations, then $H_2G=0$.
More generally, if $G$ is a group which admits a presentation with $n$ generators and $m$ relations then $H_2G$ can be generated by $m-n+\text{rk}_\mathbb{Z}(G_{ab})$ elements. This appears as Exercise II.5.5(b) in Ken Brown's group cohomology book. Here is my old solution:
With the presentation $G=\langle s_1,\cdots,s_n\,|\,r_1,\cdots,r_m\rangle$ we associate the $2$-complex $\;Y=(\bigvee_s S^1)\cup_{r_1}e^2\cdots\cup_{r_m}e^2\;$ so that $\pi_1Y\cong G$.  By computing the Euler characteristic $\chi(Y)$ two different ways we obtain the equation $\sum(-1)^i\text{rk}_\mathbb{Z}(H_iY)=\sum(-1)^ic_i$, where $c_i$ is the number of $i$-cells.  Then $1-\text{rk}_\mathbb{Z}(G_{ab})+\text{rk}_\mathbb{Z}(H_2Y)=1-n+m$ and so $\text{rk}_\mathbb{Z}(H_2Y)=m-n+r$, where $r=\text{rk}_\mathbb{Z}(G_{ab})=\dim_\mathbb{Q}(\mathbb{Q}\otimes G_{ab})$.  Now $H_2Y=\text{ker}(\partial_2)$ is a free abelian group (subgroup of cellular 2-chain group), and by applying Theorem II.5.2[Brown] we get a surjection $H_2Y\rightarrow H_2G$ (from the exact sequence in that theorem -- this theorem is a play on the Hurewicz theorem).  Thus $H_2G$ $\textit{can}$ be generated by $m-n+r$ elements. Now in my special scenario, $m-n+r=2-2+0=0$.<|endoftext|>
TITLE: Is an entire function, with nowhere vanishing derivative, always a covering map?
QUESTION [10 upvotes]: Assume that $f:\mathbb C\to\mathbb C$ is entire, and also that $f'(z)\ne 0$, for all $z\in\mathbb C$. Does that imply that $f$ is a covering map of $f[\mathbb C]$?
Clearly, $f$ is a local homeomorphism. Hence, the question reduces to examining whether such an $f$ has the curve lifting property.

REPLY [6 votes]: To put the questions and answers in a wider context, the set of singular values $S(f)$ of an entire function $f$ is the set of points $w$ near which $f$ is not a covering map.
That is, $w$ is regular if there is a neighbourhood $U$ of $w$ such that $f\colon V\to U$ is a homeomorphism for every connected component $V$ of $f^{-1}(U)$, and singular if $w$ is not regular.
So $S(f)$ is the smallest closed set $S$ such that $\newcommand{\C}{\mathbb{C}}f\colon \C\setminus f^{-1}(S)\to \C\setminus S$ is a covering map.
As discussed in the answers above, an entire function with $S(f)=\emptyset$ must be affine, and an entire function with $\# S(f)=1$ is an exponential map (by basic covering theory). In particular, no other function can be a covering over its range, by Picard's theorem. 
On the other hand, the set of functions having two singular values is extremely rich. 
(The above notions generalise immediately to functions between arbitrary Riemann surfaces. Of course here it is possible to have covering maps with larger numbers of omitted values.)<|endoftext|>
TITLE: Is there a canonical split signature metric on $\mathbb{P}^n\times\mathbb{P}^{n\,\ast}$?
QUESTION [6 upvotes]: Let
$$
M:=\{(P,\pi)\mid P\not\in\pi\}\subset\mathbb{P}^n\times\mathbb{P}^{n\,\ast}
$$
be the open and dense (and as such $2n$-dimensional) subset of non-incident point-hyperplane pairs. If $P=\mathbb{P}(\ell)$ and $\pi=\mathbb{P}(W)$, with $\ell$ and $W$ linear subspaces of $V:=\mathbb{K}^{n+1}$ of dimension $1$ and $n$, respectively, then the condition $P\not\in\pi$ is equivalent to $\ell\cap W=0$, i.e., $M$ is the complement of the flag manifold $\mathrm{Fl(1,n;V)}$.
According, there is a chain of obvious and canonical identifications:
$$
T_{(P,\pi)}M=T_P \mathbb{P}^n \oplus T_\pi \mathbb{P}^{n\,\ast}=\left(\ell^\ast\otimes\frac{V}{\ell}\right)\oplus\left(\pi^*\otimes\frac{V}{\pi}  \right)=(\ell^*\otimes\pi)\oplus(\pi^*\otimes\ell)=(\ell^*\otimes\pi)\oplus(\ell^*\otimes\pi)^*\, ,
$$
showing that $M$ is equipped with a canonical metric of split signature (insofar as each linear space of the form $Z\oplus Z^*$ carries a scalar product of split signature).

QUESTION 1: what is the origin of such a metric? does it reflect some elementary properties of points and hyperplanes in projective spaces? what is the context where it plays the most relevant role? how it depends on the ground field $\mathbb{K}$? (e.g., when $\mathbb{K}=\mathbb{C}$ is there some relationship with the Fubini-Study metric?) does it generalizes to non-incident pairs in $\mathbb{G}(k,n)\times\mathbb{G}(n-k-1,n)$?

Motivations: I have found this metric in a beautiful preprint, where it goes under the name of "dancing metric" and it is defined in an unnecessarily (at least for me) complicated way, and only for $n=2$ (the above definition - I hope it is correct - I figured out by myself). I confessed to the author that such a structure is so elementary that it should have already appeared and studied long ago, but he said that he just know that it is "somehow related to the para-Fubini-Study metric". Also searching this forum, I could not find any clue, in spite of the many posts concerning metrics on Grassmannians.

QUESTION 2: can I extend this metric to the whole of $\mathbb{P}^n\times\mathbb{P}^{n\,\ast}$ and characterize $\mathrm{Fl(1,n;V)}$ as its degeneracy locus? if yes, would this idea work also for $k>1$?

REPLY [6 votes]: Regarding Question 1:  Of course, this generalizes to the space $\mathrm{S}_{pq}(V)$ of all splittings of a vector space $V$ into subspaces $V = P\oplus Q$ where $\dim P = p>0$ and $\dim Q = q>0$ and, of course, $\dim V = p+q$.  
The tangent space at $(P,Q)\in \mathrm{S}_{pq}(V)$ is canonically isomorphic to $(P{\otimes}Q^*)\oplus (Q{\otimes}P^*)$, for the same reasons that you list in the case where $p$ or $q$ is $1$, so the same reasoning applies.  
When the ground field is $\mathbb{R}$, this is an irreducible pseudo-Riemannian symmetric space (of split type), and, as such, appears in Berger's classification of pseudo-Riemannian symmetric spaces as $\mathrm{SL}(V)/\mathrm{S}(\mathrm{GL}(P){\times}\mathrm{GL}(Q))$.
When the ground field is $\mathbb{C}$, this space can be regarded naturally as a complexification of the space $\mathrm{Gr}_p(V)$ of $p$-planes in $V$.  This canonical metric then turns out to be, in an appropriate sense, the holomorphic extension of the Fubini-Study metric on $\mathrm{Gr}_p(V)$ (defined relative to any positive definite Hermitian inner product on $V$) to this complexification.
Regarding Question 2:  The metric (as a quadratic form) does not extend continuously to the product $\mathrm{Gr}_p(V)\times\mathrm{Gr}_q(V)$.  The reason is that the volume form of the canonical metric on $\mathrm{S}_{pq}(V)$ gives $\mathrm{S}_{pq}(V)$ infinite volume, which it could not do if the quadratic form extended continuously across the incidence hypersurface.  (Just look at the case $p=q=1$ to convince yourself of this.)<|endoftext|>
TITLE: Is this subset-sum-type problem discussed in the literature?
QUESTION [5 upvotes]: Let $y \in \mathbb{Z}_+^n$, with $y_1 < \dots < y_n$. I am interested in finding a 0-1 matrix $A$ and $x \in \mathbb{Z}_+^m$ s.t. $m$ is minimal and $Ax = y$, where I am guaranteed that at least one such pair $(A,x)$ exists by requiring that $\sum_{j = 1}^m x_j = y_n$. I am "really" only interested in $x$.
For example, say I have $y = (3,5,7,10,12,26)^*$. Then $x = (3,5,7,11)^*$.
This strikes me as similar to a subset sum or number partitioning problem, but I can't find a reference for it and (after trying a bit nevertheless) don't really know how to look for one. Is this problem studied--or better yet efficiently solved--in the literature?

REPLY [3 votes]: David Moulton looked at a related problem in Representing powers of numbers as subset sums of small sets, J Number Theory 89 (2001) 193-211, MR1845235 (2002j:11015), only he allowed negative entries in $x$. Mike Develin followed up in On optimal subset representations of integer sets, J. Number Theory 89 (2001)   212–221, MR1845235 (2002k:11032), and then Donald Mills wrote Some observations on subset sum representations, Integers 6 (2006) A25, MR2264840 (2007g:11027).<|endoftext|>
TITLE: Do all simple factors of jacobians of curves come from correspondences?
QUESTION [18 upvotes]: For this question I will let the overly ambiguous word curve mean: smooth projective and connected curve over $\mathbb C$ (or equivalently a smooth compact Riemann-Surface). 
Let $C$ be a  curve over $\mathbb C$. And suppose that $f: D \to C$ and $g:D \to E$ are dominant maps of curves. Then the pair $(f,g)$ forms a correspondence from $C$ to $E$, and one gets a map between Jacobians $f_* \circ g^* : J(E) \to J(C)$. Then the image of $J(E)$ under this map is going to be a sub-abelian variety of $J(C)$. 
My questions is, is it possible to get all simple sub-abelian varieties of $J(C)$ from correspondences  in this way?

REPLY [8 votes]: This argument is mostly contained in t3suji's comments, but with some of the proofs somewhat expanded.
As a convention (consistent with that of the theory of Chow motives), all actions of correspondences $\alpha \in \operatorname{CH}^*(X \times Y)$ on cohomology (or Jacobians) are covariant: $\alpha_* \colon H^*(X) \to H^*(Y)$, etc.
Lemma. If $X$ and $Y$ are smooth projective varieties over $\mathbb C$, then there is an isomorphism
$$\operatorname{NS}(X \times Y) \cong \operatorname{NS}(X) \times \operatorname{NS}(Y) \times \operatorname{Hom}(\operatorname{Alb}_X,\operatorname{Pic}^0_Y).\label{Eq NS}\tag{1}$$
Proof. The Künneth formula gives an isomorphism of $\mathbb Z$-Hodge structures
$$H^2(X \times Y) = \big( H^2(X) \otimes H^0(Y)\big) \oplus \big( H^0(X) \otimes H^2(Y) \big) \oplus \big( H^1(X) \otimes H^1(Y) \big).\tag{2}\label{Eq Künneth}$$
By Poincaré duality, the last term equals $\operatorname{Hom}(H^{2\dim X-1}(X),H^1(Y))$. Taking $(1,1)$-parts in (\ref{Eq Künneth}) gives the result by Lefschetz's (1,1) theorem and the theory of abelian varieties over $\mathbb C$. $\square$
There is also a version for $\operatorname{Pic}$ for smooth projective varieties over an arbitrary field (or even over a base, under some mild hypotheses to ensure existence of $\operatorname{\textbf{Pic}}_{X/S}$ and $\operatorname{\textbf{Alb}}_{X/S}$). The above analytic proof has the advantage that it shows us how classes act on Jacobians: if $X$ and $Y$ are curves, then $\operatorname{Alb}_X \cong \operatorname{Pic}_X^0 \cong \operatorname{Jac}_X$, and under these identifications, a class $\alpha \in \operatorname{CH}^1(X \times Y)$ induces the map $\alpha_* \colon \operatorname{Jac}_X \to \operatorname{Jac}_Y$ given by the corrresponding component of $\alpha$ under (\ref{Eq NS}).
Construction. Let $A \subseteq \operatorname{Jac}_C$ be any abelian subvariety. Let $E$ be any curve admitting a surjection $\operatorname{Jac}_E \to A$; for example, $E = C$ with the projection $\operatorname{Jac}_C \to A$ given by $n \pi$ for $\pi \in \operatorname{End}^\circ(\operatorname{Jac}_C)$ the projector corresponding to the isogeny factor $A$ of $\operatorname{Jac}_C$, and $n \in \mathbb Z_{> 0}$ such that $n\pi$ is integral. The composition gives a morphism $\phi \colon \operatorname{Jac}_E \to \operatorname{Jac}_C$, corresponding by (\ref{Eq NS}) to a class $\alpha \in \operatorname{NS}(X \times Y)$.
The class $(1,1,0)$ in (\ref{Eq NS}) is ample, hence for some $n \gg 0$ there exists a very ample line bundle $\mathscr L$ on $X \times Y$ mapping to $(n,n,\alpha) \in \operatorname{NS}(X \times Y)$. Let $D \in |\mathscr L|$ be a smooth member. By the discussion above, the cycle $[D] \in \operatorname{CH}^1(E \times C)$ induces the map $\phi$.
The graphs of the maps $g \colon D \to E$ and $f \colon D \to C$ give cycles $[\Gamma_g]^\top \in \operatorname{CH}^1(E \times D)$ and $[\Gamma_f] \in \operatorname{CH}^1(D \times C)$, whose actions on $\operatorname{Jac}$ are given by $g^*$ and $f_*$ respectively. Moreover, we have
$$[\Gamma_g]^\top \circ [\Gamma_f] = [D] \in \operatorname{CH}^1(E \times C),$$
which shows that indeed $\phi = f_* \circ g^*$. $\square$<|endoftext|>
TITLE: Parameterize unitary without transpose
QUESTION [7 upvotes]: For all unitary matrices, i.e. $A \overline{A}^T = I$, there is a skew-Hermitian matrix $X$ so that $A = exp(X)$. So the unitary group has $n^2$ dimensions.
Is there any similar parameterisation of all matrices with $A \overline{A} = I$? What can be said about the number of dimensions? (Btw. is there any name for this type of matrices in literature?)

REPLY [3 votes]: I totally lied about this not being a natural thing to ask! As loup blanc alludes to, in fact $n \times n$ matrices such that $M^{-1} = \overline{M}$ can be interpreted as Galois descent data for descending the complex vector space $\mathbb{C}^n$ to a real vector space, or equivalently as a $1$-cocycle in 
$$Z^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C})).$$
The statement that any such matrix must have the form $U \overline{U}^{-1}$ says that any such $1$-cocycle is cohomologous to zero, or equivalently that the Galois cohomology set
$$H^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C}))$$
is trivial. This just says that there is only one real form of $\mathbb{C}^n$ up to isomorphism, namely $\mathbb{R}^n$. The space $GL_n(\mathbb{C})/GL_n(\mathbb{R})$ appearing in Robert Bryant's answer can then be interpreted as the space of real forms of $\mathbb{C}^n$.<|endoftext|>
TITLE: Center of quantum affine algebras
QUESTION [5 upvotes]: Are there some references about the center of quantum affine algebras? I searched on google and only find the paper. In particular, what is the center of $U_q(\widehat{\mathfrak{sl}_2})$. Thank you very much.

REPLY [2 votes]: For affine (quantum or not) algebras the center appears only for special value of element "c" - the so-called "critical level".
The second order central element related to "Sugavara construction" is well-known for a long time. 
Somewhat nice explicit formulas for the  center of affine (not quantum) algebras for type A_n  first appeared in  our paper, sorry for self-advertisement.
For the generalization to quantum case see paper.
It is somewhat surprising that in quantum case there is an alternative approach to describe the center - it is presented in the paper you cited,
which appeared before cited above developments.
Let me briefly sketch the ideas behind these works.
Let us start with gl_n. Basic linear algebra tells that characteristic polynomial of a matrix det(L-t) gives invariant polynomials.
The algebra U(gl_n) is somehow a quantization/deformation of commutative Poisson algebra S(gl), so we may hope that "det(L-t)" can be somehow deformed to give generators of the center of U(gl_n), that indeed can be done - there exists Capelli determinat which does the job. We need to deal with the matrix with non-commuting elements, but nevertheless it works.
It appears that in affine case somewhat similar formula "det(d/dz - L(z))"
does the job. 
Going further to quantum case, it is well-known that quantum algebras can be described in the so-called RLL = LLR formalism, it is actually not one line to describe U_q(gl_n) (or affine) version since one has L^+, L^- operators,
however this is standard. Roughly speaking the formula $det(1-q^{d/dz} L(z) )$ will do the job for U_q(gl_n) both affine or not. For non-affine case one should take only the first term of L(z) in "z" expansion.
Sorry for being too sketchy.<|endoftext|>
TITLE: Cyclotomic polynomials: $\Phi_n(p)$ is like $p^{\phi(n)}$ for big enough $p$, right?
QUESTION [19 upvotes]: Apologies in advance if this turns out to be simple.  So far I haven't found a proof or a reference.
Although I like $p$ to be a prime, I can ask the following for positive integers $n$ and $p$, using what should be clear notations for the $n$th cyclotomic polynomial and Euler's totient function: Given $p \gt 1$, is
$$ \mid \Phi_n(p)/p^{\phi(n)} \mid \lt p/(p-1)$$
for every $n$?
Indeed, if $n$ is a power of a prime $q$, we have the left hand quantity bounded by $p^{n/q}/(p^{n/q} -1)$, and the lim sup over all primes $n$ achieves $p/(p-1)$. The case for composite $n$ is not clear to me, thus the question, but I would hope for a tighter bound (perhaps involving the smallest prime power factor of $n$) than $p/(p-1)$.  
An equivalent question asks to verify the bound
on $\Phi_n(1/p)$.  Of course the product of such quantities (to an appropriate power as $n$ runs over divisors of some $m$) will satisfy the bound, but this does not seem to help. If there is a reference offered that says (something like) the coefficients of cyclotomic polynomials grow slowly enough to exhibit the bound, I will read that.  I am hoping for a simpler proof than that.
I am looking at (the moral equivalent of)  prime factors of $\Phi_n(p)$ and wanted to make sure these values aren't much bigger than I think they are.
I would be satisfied with a coarse bound (replace $p/(p-1)$ by $2$, say), but
I think much more can be said.
UPDATE 2015.10.23 More now has been said, with my revised take on Jameson's presentation posted as a separate answer.  For me, the key parts will be that $p \geq 2 $ prime and $p/(p-1)$ can be replaced by real $x \gt (2 - \epsilon)^{r/n}$ and $x^{n/r}/(x^{n/r} - 1)$, where $r=$rad$(n)$.  Thanks again
all, and special thanks to Peter Mueller. END UPDATE 2015.10.23
UPDATE 2015.10.21: Thanks to Peter Mueller, I read from notes of G.J.O. Jameson at http://www.maths.lancs.ac.uk/~jameson/cyp.pdf on
cyclotomic polynomials of a sharper result, which indeed is simpler but also more challenging.  I remove some of the challenge by interpreting some highlights here (hopefully without errors), but I recommend following the development of the notes as it proceeds in small but useful steps, with a certain degree of economy that takes ones breath away.
First, Jameson notes in 1.3 an inversion relation involving $\Phi_n(1/x)$ 
and real,nonzero $x$ that appears below.
Jameson also prepares in 1.12 to work with squarefree indices through 
using $n_0=$rad$(n)$ and the identity $\Phi_n(x) = \Phi_{n_0}(x^{n/n_0})$.  I modify and sketch a strict inequality (Lemma 1.19) which is used:
For $0 \lt x \lt 1, m, a,\ldots,b$ positive integers (so also $0 \lt x^{powers} \leq x$),
\begin{eqnarray*}
(1 -x^m)(1-x^{m+a})\ldots(1-x^{m+b}) & \geq &
(1 - x^m - x^{m+a} - \ldots - x^{m+b}) \\
& \gt &  1 - ( x^m + x^{m+1} + \ldots ) = 1 - x^m/(1-x) \\
\end{eqnarray*}
Then Jameson has 1.20, which I rewrite and restrict to squarefree integers $n$, as one actually gets better bounds/ranges for when $n$ is not squarefree.  
1.20 (rewritten) Let $n>1$ be squarefree with $j=1$ if the number $k$ of distinct prime factors
of $n$ is an even number, and $j=-1$ if $k$ is odd. Let $0 \lt x  \leq 1/2$.  Then
$$1-x \lt \Phi_n(x)^j \lt 1.$$
Note when $n$ is 1, one has $\Phi_1(x)= x-1$ which is negative on the domain considered.
Using the inversion $\Phi_n(x)/x^{\phi(n)} = \Phi_n(1/x)$ when $n \gt 1$, this gives for $2 \leq x$
\begin{eqnarray*}
(x-1)/x && \lt \Phi_n(x)/x^{\phi(n)} \lt  1, && k=2m \\
1 && \lt \Phi_n(x)/x^{\phi(n)} \lt  x/(x-1), && k=2m+1 . \\
\end{eqnarray*}
Using the relation for general non-squarefree indices, one can improve the $x$ in $1-x$ to $x$ to a fractional power, as well as extend the range a little.  I am still working this part out.  Even working out the statement using the inversion requires care.  I think the results are both simple and challenging, and I am glad to share this on MathOverflow.
Jameson uses the tools carefully, working out the squarefree case in about half a page of elementary reasoning which I am still perusing.  I am joyed. I'm also willing to buy Jameson two hot beverages.  Peter Mueller can drop by 
and ask me for a toasted bagel.
END UPDATE 2015.10.21
Gerhard "Wants To Stop Spinning Head" Paseman, 2015.10.19

REPLY [2 votes]: I've decided to simplify the argument found in notes of Jameson, and at the same time improve the bounds and ranges of applicability.  I'm rewriting for the purpose of understanding and the specific goal of improving the answer; for other applications I still recommend the notes.
For $n=1$, and $x$ a positive real, we have the desired quantity $\Phi_n(x)/x^{\phi(n)} = (x-1)/x$ which we
take as understood, and will focus on $n \gt 1$.  By inclusion-exclusion or some other means of summing
over positive divisors of $n$, we have $\phi(n) = \sum_{d \mid n} d\mu(n/d)$, where I use the Moebius
function $\mu(m)$, and rewrite the quantity as done in the answer of Venkataramana: 
$$\frac{\Phi_n(x)}{x^{\phi(n)}}=
\frac{\prod_{d\mid n} (x^d -1)^{\mu(n/d)}}{\prod_{d \mid n} x^{d\mu(n/d)}}=
\prod_{d \mid n}(1 - x^{-d})^{\mu(n/d)}=
\left( \frac{P(x)}{Q(x)} \right)^j,$$
where I explain the last term below.
In the general case, $n$ is not always squarefree, so for some divisors $d$ of $n$ $\mu(n/d)$ is $0$ and
the associated base $1 - x^{-d}$ "drops out".  We collect the terms that don't drop out and arrange
for the term with the largest value of $-d$ (smallest $d$) to be on the top.  As a result, $j$ will be $1$ 
when the number
of distinct prime factors of $n$ is even, and $-1$ when this number is odd.  Letting $n_0=$rad$(n)$ and
$n_1= n/n_0$, $P(x)$ will contain those factors of the form $(1 - x^{-an_1})$, where $a$ runs over the
divisors of $n_0$ with $\mu(a)=1$ (so $1-x^{-n_1}$ is a factor of $P(x)$), and $Q(x)$ will contain the
rest (factors $1 - x^{-an_1}$ where $a\mid n_0$ and $\mu(a)=-1$).  However, even when $n$ is squarefree,
$n_1$ will be $1$ and the argument will apply in this case also.  By using $-n_1$, I avoid at some notational
cost the inversion and squarefree reductions used in the argument of Jameson.
As a quick check, let us take $n= q^k$, a prime power with $k \geq 1$.  Then $j=-1, n_1=n/q,$
$P(x)=(1-x^{-n_1}),$ and $Q(x)=(1-x^{-n})$, and
$$1 < \frac{\Phi_n(x)}{x^{\phi(n)}} = \frac{x^n - 1}{x^n - x^{n-n_1}} \lt 
\frac{x^{n_1}}{x^{n_1} - 1}= \frac{x^{n/q}}{x^{n/q} - 1},$$
and this last is bounded by $x/(x-1)$, and gets better when $k$ gets larger.  Note however for $k=1$ that as $n$ runs through larger primes,
$\Phi_n(x)/x^{\phi(n)}$ approaches $x/(x-1)$: we can't expect significant improvement for those $n$.
We continue now assuming $n$ is not a prime power.  I use a simple estimate to bound both
$P(x)$ and $Q(x)$.  Actually, a key feature of Jameson's argument which I emphasize here is that
we bound $P(x)/(1-x^{-n_1}) = (1- x^{-pqn_1})R(x)$ where $R(x)$ is the rest of the product (and could be 1), which is needed for the inequality in my question. $p$ and $q$ are the smallest and second smallest prime factors of $n$.
I introduce a familiar-looking inequality which I dub Lemma 91.1.  For $1 \lt x$ a real, $m \lt 0$ an
integer, and integers $0 \lt a \lt b$ and perhaps other distinct integer exponents coming from $[a,b]$
\begin{eqnarray*}
(1-x^{am}) & \gt & (1- x^{am})\ldots(1-x^{bm}) \gt  1 - x^{am} - \ldots - x^{bm} \\
& \geq & 1 - x^{am} - x^{(a+1)m} - \ldots - x^{bm} = 1 - \frac{x^{am} (1- x^{m(1+b-a)})}{1- x^m} \\
\end{eqnarray*}
Note that if $a=b$ in the above, then we have just $(1- x^{am})$ to bound, and we get equalities in
this case.
This Lemma is the algebraic replacement of fedja's idea which is the heart of the accepted answer.  It gives 
$1- x^{-n_1} \gt P(x) \geq (1 - x^{-n_1})(1 - x^{-pqn_1}(1- x^{-{\beta}n_1})/(1 - x^{-n_1}))$, for
some integer $1 \leq \beta \leq n_0 - pq$ sufficiently large, and picking $\gamma \leq n_0 - p$ similarly to $\beta$
$1 - x^{-pn_1} \gt Q(x) \gt 1 - x^{-pn_1}(1-x^{-{\gamma}n_1})/(1-x^{-n_1})$.  (If $1 \gt n_0 - pq$, pick
$\beta=1$ anyway.  One can always choose larger $\beta$ and $\gamma$ to weaken the inequality.)
To get $P(x) \lt Q(x)$, we look at when $1-Q(x) \lt 1 -P(x)$, or the sufficient condition
$1- Q(x) \lt x^{-pn_1}(1-x^{-{\gamma}n_1})/(1-x^{-n_1}) \leq x^{-n_1} \lt 1 -P(x)$.
For readability we substitute $y=x^{-n_1}$ and ask for this $y$ to satisfy
$y^p(1 - y^{\gamma}) \leq y- y^2$, or $ y + y^{p-1}(1- y^{\gamma}) \leq 1$. $p$ is at least $2$,
so this holds when $x^{-n_1} = y \leq 1/2$, however it can hold for slightly larger $y$.  So
for $x^{-n_1} \leq 1/2 + \epsilon(p,\gamma)$, we have $P(x) \lt Q(x)$.  We can't
expect a large $\epsilon(2,\gamma)$ (for we need $y \lt (2 - y^{\gamma})^{-1}$), 
but already $0.1 \lt \epsilon(p, \gamma)$ for primes $p \gt 2$.
We use the other inequalities to get immediately
$$\frac{P(x)}{Q(x)} \gt \frac{(1 - x^{-n_1})(1 - x^{-pqn_1}(1- x^{-{\beta}n_1})/(1 - x^{-n_1}))}{1 - x^{-pn_1}},$$
and we reuse $y$ to write this last as 
$$(1-y)\frac{1 - y^{pq}(1 - y^{\beta})/(1-y)}{1-y^p}=
(1-y)\frac{1- y - y^{pq}(1-y^{\beta})}{1 - y - y^p + y^{p+1}}.$$
As $q\geq 3$, $y^{p(q-1)} ( 1- y^{\beta}) \lt 1 - y$ whenever $y^4 + y \lt 1$, and
one can use $\beta$ to tweak the range further, so for such $y$ we have
$-y^{pq}(1- y^{\beta}) \gt - y^p(1-y)$
so the last displayed term is $(1-y)$ times something larger than $1$.  Thus
$P(x)/Q(x) \gt 1-y = 1- x^{-n_1}$ for these $x$, which include $x^{n_1} \geq 2$.
Thus we have $\Phi_n(x)/x^{\phi(n)}$ sandwiched between 
$1 - x^{-n/\textrm{rad}(n)}$
and $1$, or between $1$ and $(1 - x^{-n/\textrm{rad}(n)})^{-1}$ for $x^{n_1} \gt 2 - \epsilon$,
where we can tune epsilon based on $n$.
Gerhard "Apologies For The Simple Parts" Paseman, 2015.10.23<|endoftext|>
TITLE: Groups with trivial rational homology and their finite index subgroups
QUESTION [10 upvotes]: For a short exact sequence $0 \to G \to H \to K \to 0$ of (discrete) groups with $K$ finite we have, as a consequence of the Hochschild-Serre spectral sequence, that $H^{\ast}(H;\mathbb Q) = H^{\ast}(G;\mathbb Q)^K$. This can be used to see that free and free abelian groups embedd with finite index into groups with trivial rational cohomology: $$H^{\ast}(F_n \rtimes_{\text{sign}} \mathbb Z/2;\mathbb Q) = H^{\ast}(\bigvee_n S^1;\mathbb Q)^{\mathbb Z/2} = H^{\ast}(\text{pt};\mathbb Q).$$
$$H^{\ast}(\mathbb Z^n \rtimes_{\text{sign}} (\mathbb Z/2)^n;\mathbb Q) = H^{\ast}(\prod_n S^1;\mathbb Q)^{(\mathbb Z/2)^n} = H^{\ast}(\text{pt};\mathbb Q).$$
Does this work for every group? Or writing down the opposite:
Is there a group $G$ such that every group $H$ which contains $G$ with finite index has nontrivial rational cohomology?

REPLY [8 votes]: Thompson's group $T$ gives an example, i.e. if $T \le H$ has finite index, then $H^\ast(H;\mathbb{Q}) \neq 0$. 
More specifically, there is always a non-trivial class in $H^4(H;\mathbb{Q})$. 
Proof: Step 1: $T$ is normal in $H$ 
Since $T$ has finite index in $H$, $T_0 := \bigcap_{h \in H/T}hTh^{-1}\le T$ is a finite index, normal subgroup of $H$ (and $T$). But $T$ is infinite simple, so $T=T_0$ is normal in $H$. 
Step 2: Since $T$ is normal, $H^\ast(H;\mathbb{Q})=H^\ast(T;\mathbb{Q})^H$. 
By work of Ghys & Sergiescu, $H^2(T;\mathbb{Z})= \mathbb{Z}^2$ is generated by the Euler class $x'$ of $T$ and the Godbillion-Vey class $y'$. The corresponding rational classes $x=i(x')$ and $y=i(y')$ where $i: H^\ast(T;\mathbb{Z}) \to H^\ast(T;\mathbb{Q})$ generate the rational cohomology ring as $H^\ast(T;\mathbb{Q})=\mathbb{Q}[x,y]/(xy),\,\,|x|=|y|=2$. 
I'll show that $x^2 + y^2$ is invariant under the action of $H$. 
Let $c: T \to T$ be conjugation by an element from $H$. Hence $c^\ast$ induces an isomorphism on $H^2(T;\mathbb{Z})$. Write $c^\ast(x')=ax'+by',\, c^\ast(y')=cx' + dy'$ with integers $a,b,c,d$ s.t. $ad-bc= \pm 1$. Since $c^\ast$ commutes with $i$, we find $c^\ast(x)c^\ast(y)=acx^2 + bdy^2$. But also $c^\ast(x)c^\ast(y)=c^\ast(xy)=0$ because $xy=0$. Hence $ac=0$ and $bd=0$. Thus 
$$c^\ast(x)=\pm x,\, c^\ast(y)=\pm y\quad  \text{or}\quad c^\ast(x)=\pm y\,,c^\ast(y)=\pm x$$
In either case, $c^\ast(x^2+y^2)=x^2+ y^2$. qed<|endoftext|>
TITLE: Counting matrices over finite fields of a given order
QUESTION [8 upvotes]: How can I count/enumerate matrices in ${\rm GL}(2,{\rm GF}(2^5))$
of order $3$? In general, how can I obtain the number of matrices in
${\rm GL}(2,{\rm GF}(q))$, where $q$ is a power of a prime, of order,
say, $t$?

REPLY [5 votes]: In GAP, you can find the number of elements of order $t$
in ${\rm GL}(2,q)$ by the following function:
NumberOfElementsOfGivenOrderInGL2q := function ( q, t )
  return Sum(List(Filtered(ConjugacyClasses(GL(2,q)),
                           cl->Order(Representative(cl))=t),Size));
end;

For example in your case, i.e., $q = 2^5$ and $t = 3$, you obtain
gap> NumberOfElementsOfGivenOrderInGL2q(32,3);
992<|endoftext|>
TITLE: Automorphisms of curves in positive characteristic
QUESTION [8 upvotes]: It is well known that over an algebraically closed field of characteristic zero a general curve (for an open subset of $M_g$) of genus $g\geq 3$ is automorphism-free. 
Is this result still true over a non algebraically closed field and over a field of positive characteristic?

REPLY [6 votes]: Jason Starr already answered your question in the comments above.
Let me just point that you can also use a standard deformation argument to prove that the general curve of genus at least three has no non-trivial automorphisms, see e.g. Dan Petersen's answer to Examples where it's useful to know that a mathematical object belongs to some family of objects
This argument can also be used to prove that the general complete intersection of general type in some projective space has no non-trivial automorphisms.
Explicit examples of some varieties with no non-trivial automorphisms are given in Poonen's articles 
Varieties without extra automorphisms I: curves  
Varieties without extra automorphisms II: hyperelliptic curves  
Varieties without extra automorphisms III: hypersurfaces  

They are available on his website http://www-math.mit.edu/~poonen/
Addendum: As Jason Starr points out in the comment below, using singular specializations to deduce facts about smooth fibres requires a bit of care. Let me explain where one should be careful in applying the argument given by Dan Petersen.
Fix an integer $g\geq 3$ and an algebraically closed field $k$. Let $X$ be a stable curve of genus $g$ over $k$ with no non-trivial automorphisms. (You construct this by pinching $g-2$ well-chosen points on some genus $2$ curve, as explained by Dan Petersen.) Now, let $R = k[[t]]$ and let $\mathcal X\to \mathrm{Spec} R$ be a stable curve such that 

The special fibre $\mathcal X_k$ is isomorphic to $X$ over $k$, and
the generic fibre $\mathcal X_K$ is a smooth genus $g$ curve.

The existence of such a stable curve requires a bit of thought. A quick argument is to say that stable curves form a boundary of $\mathcal M_g$ (as mentioned by Dan Petersen in his answer). You can also try to give a more explicit construction (of a smooth curve $\mathcal X_K$ which specializes to $X$.) On the other hand, the easiest approach might be to just prove that any singular curve can be smoothened. (I can't remember whether that's a difficult fact to prove  at the moment.)
Now, let $G :=\mathrm{Aut}_R(\mathcal X)$ be the scheme of automorphisms of $\mathcal X$. Since $\omega_{\mathcal X/R}$ is relative ample, the group scheme $G$ is affine and finite type over $R$. Since stable curves are ``unique'' (by the theory of minimal regular models of curves, say) the group scheme $G$ is proper. (Note: this is the subtle point that Jason Starr is alluding to below. Uniqueness of limits of all stable limits (in the preceding sense) is enough, as it implies by the valuative criterion for properness, applied to the diagonal, that the (affine) diagonal of $\overline{\mathcal M_g}$ is finite.) So $G\to \mathrm{Spec} R$ is a finite group scheme, whose special fibre is the trivial group scheme. You can now conclude that $G$ is itself the trivial group scheme over $R$.<|endoftext|>
TITLE: Alternating elements in free graded-commutative algebras
QUESTION [5 upvotes]: It is classical that every alternating polynomial is (uniquely) the product of a symmetric polynomial with the Vandermonde polynomial, in particular the alternating polynomials are a free rank-one module over the symmetric polynomials. 
For some cohomology computations I am doing, I would like to know some analogous statement, not for the polynomial ring, but for the free graded-commutative algebra
$$\mathfrak{F}(n)=\mathbb{F}_\ell[X_1,\dots,X_n]\langle A_1,\dots,A_n\rangle,$$ 
i.e., polynomial variables $X_i$ and exterior variables $A_j$. For peace of mind, $\ell$ is a prime different from $2$, but I would be willing to assume $\ell> n$. Now consider the "diagonal permutation" action of the symmetric group $\Sigma_n$, i.e., an element $\sigma$ acts simultaneously on $\{X_1,\dots,X_n\}$ and $\{A_1,\dots,A_n\}$ via the natural permutation of indices. 
I am interested in the submodule of $\mathfrak{F}(n)$ where $\Sigma_n$ acts via the sign representation (may I call the elements alternating superpolynomials?). 

I would like to know that this submodule is free over the ring of symmetric polynomials in $\{X_1,\dots,X_n\}$, and I would like to have a general method of computing the rank and explicit generators. 

So far, the case $n= 2$ is clear and I also worked out the case $n=3$. This is based on first working out the $\Sigma_3$ decomposition of the polynomial ring and the exterior algebra separately and then tensoring the representations (using the assumption $\ell\neq 2,3$). The result is a free rank 8 module. This could of course be done in further cases, but will probably fail to produce a  general result because it involves tensoring representations of the symmetric group. So the parts of my question are: 

Has this question already been considered somewhere in the literature? (I would guess, it seems natural enough).
Is there a slick general argument dealing with all $n$ at once (and possibly with $\ell\leq n$ as well), like in the classical case of alternating polynomials? 

Any ideas or literature references would be much appreciated.

REPLY [3 votes]: If you assume $\ell>n$, then you may as well work over the field $F=\mathbb{Q}$, since the representation theory is exactly the same (the group algebra is semisimple, all Young symmetrisers in the group algebra over $\mathbb{Q}$ have denominators dividing $n!$, etc.). 
Under this (much) simplifying assumption, we have $F[x_1,\ldots,x_n]\cong FS_n\otimes F[x_1,\ldots,x_n]^{S_n}$, and the exterior algebra $\Lambda(a_1,\ldots,a_n)$ can be written as $\Lambda(V\oplus F)$, where $V$ is the standard $(n-1)$-dimensional representation of $S_n$, and $F$ is the trivial one, and that is the same as $\Lambda(V)\otimes\Lambda(F)$, where $\Lambda(F)\cong \Lambda(a_1,\ldots,a_n)^{S_n}$. Now, we see that your ring is, as a representation of $S_n$, 
 $$
FS_n\otimes\Lambda(V)\otimes F[x_1,\ldots,x_n]^{S_n}\otimes \Lambda(a_1,\ldots,a_n)^{S_n} .  
 $$
To extract the isotypic component of the sign representation, you should take that component in $FS_n\otimes\Lambda(V)$, and tensor with the invariants. This already gives freeness as a module over the symmetric polynomials, of course. But also, note that $FS_n$ is the sum of all irreducible modules with multiplicities equal to dimensions, and $\Lambda(V)$ is the sum of all irreps corresponding to hooks with all multiplicities equal to one. Also, the sign representation multiplicity in $U\otimes V$ is the trivial representation multiplicity in $U\otimes V\otimes\mathrm{sign}$ is the dimension of $\mathop{\mathrm{Hom}}(U,V\otimes\mathrm{sign})$ because of self-duality of representations of symmetric groups. Finally, tensoring with sign replaces each Young diagram by its transpose, which does not change either $FS_n$ or $\Lambda(V)$ as a representation. Thus, the corresponding multiplicity is equal to the sum of all dimensions of irreps corresponding to hooks; those dimensions are $\binom{n-1}{k}$ for various $k$, so the multiplicity is $2^{n-1}$. Multiplying that by $2=\dim\Lambda(a_1,\ldots,a_n)^{S_n}$, we finally see that the module is free of rank $2^n$. 
I am not sure what to expect in the case $\ell\le n$: most of the statements I made above would fail for rather trivial reasons.<|endoftext|>
TITLE: Hausdorff open image of a Polish space
QUESTION [9 upvotes]: Let $f\colon X\to Y$ a continuous open and surjective function, where   $X$ is Polish. 
It is known that $Y$ is Polish if: $f$ is closed or $Y$ is metric.
Suppose that we know that $Y$ is Hausdorff, does it implies that $Y$ is Polish? or Regular?
I think no, but i have not been able to find a counterexample.

REPLY [4 votes]: As a counterexample one can consider the projective space $P\mathbb R^\omega$ of the countable product of lines. This is a quotient space of the Polish space $\mathbb R^\omega_\circ=\mathbb R^\omega\setminus\{0\}^\omega$ by the equivalence relation $x\sim y$ iff $\mathbb Rx=\mathbb Ry$.
It is easy to see that the quotient map $q:\mathbb R^\omega_\circ\to P\mathbb R^\omega$ is open and the space $P\mathbb R^\omega$ is Hausdorff but not Urysohn and hence is not metrizable and not Polish.
Moreover, the space $P\mathbb R^\omega$ is superconnected in the sense that for any nonempty open sets $U_1,\dots,U_n$ in $P\mathbb R^\omega$ the intersection of thier closures $\bar U_1\cap\dots\cap\bar U_n$ is not empty.<|endoftext|>
TITLE: Fiber vs homotopy fiber in model categories: simple question
QUESTION [7 upvotes]: I have a concrete problem with the homotopy fiber and I am getting lost with the
literature. I state my question and, to avoid confusions, I state downwards the
definitions I am using.
Let $C$ be a pointed model category. To make things simpler, assume every object
is fibrant. Recall that the definition of the homotopy fiber of a morphism $f\colon
Y\to X$ is the homotopy pullback of the diagram
$$\begin{matrix}& & Y \\
&&\downarrow^f \\ *&\to & X\end{matrix}
$$
That is to say, we replace $*\to X$ and $f$ by fibrations (in the category of diagrams) and take the classic
pullback.
Now, if I have a map $Z\to X$ that fits in a commutative square
$$
\begin{matrix}
Z&\to &Y\\
\downarrow &&\downarrow^f\\
*&\to & X.
\end{matrix}\quad (1)
$$
then it factors through the fiber. More concretely, there is a map $Z\to
\mathrm{fib}$ such that the diagram
$$
\begin{matrix}
Z&\to&\mathrm{fib}&\to &Y\\
&&\downarrow &&\downarrow^f\\
&&*&\to & X.
\end{matrix}
$$
commutes in $C$ and the composed maps $Z\to *$ and $Z\to X$ are those of (1). 
My question: is there an analogue map to the homotopy
fiber?, i. e., is there a commutative square in $C$
$$
\begin{matrix}
Z&\to&\mathrm{hofib}&\to &Y'\\
&&\downarrow &&\downarrow^{Rf}\\
&&*'&\to & X'.
\end{matrix}\quad (2)
$$
where $*'\to X'$ and $Rf\colon Y'\to X'$ are the fibrant replacements?

Let me recall the general definition of the homotopy limit. Denote $D$ the above
diagram. Then $C^D$ has a model structure. Note $\mathrm{lim}\colon C^D\to C$
the functor taking pullback. The homotopy pullback is the right derived
functor of $\mathrm{lim}$ in the sense of Quillen in "Homotopical algebra".
Therefore, in order to have such a map I should have a functor $F$ such that for
every diagram $d\colon D\to C$ I have an object $F(d)$ with maps as in (1) and
that in my concrete diagram gives $Z$. Then I would have a diagram like (2). Right?
I have read in n-category lab that this definition is called global and there is another definition of the homotopy fiber called local and that, under certein hypothesis, they seem coincide. Do this terminology refer to this problem? I am unable to understand the notations in the literature.

REPLY [9 votes]: I work in a general pointed model category. The homotopy fibre of a morphism $f : Y \to X$ can be defined as follows: first, choose a fibrant replacement $w_X : X \to \hat{X}$ and then factor $w_X \circ f : Y \to \hat{X}$ as a weak equivalence $w_Y : Y \to \hat{Y}$ followed by a fibration $\hat{f} : \hat{Y} \to \hat{X}$; then the homotopy fibre (with respect to these choices) is the ordinary fibre of $\hat{f} : \hat{Y} \to \hat{X}$.
Thus, we have the following commutative diagram,
$$\require{AMScd}
\begin{CD}
\operatorname{fib} f @>>> Y @>{w_Y}>> \hat{Y} @<<< \operatorname{hfib} f\\
@VVV \mathrm{pb} @VV{f}V @V{\hat{f}}VV \mathrm{pb} @VVV \\
0 @>>> X @>>{w_X}> \hat{X} @<<< 0
\end{CD}$$
and in particular, we get a comparison $\operatorname{fib} f \to \operatorname{hfib} f$ making the following diagram commute:
$$\begin{CD}
\operatorname{fib} f @>>> \operatorname{hfib} f \\
@VVV @VVV \\
Y @>>{w_Y}> \hat{Y}
\end{CD}$$
If you can choose $(\hat{X}, w_X, \hat{Y}, w_Y, p)$ functorially, then you get a homotopy fibre functor, and it can be shown to be a right homotopical approximation of the fibre functor. This is the "global" universal property. The "local" universal property is more or less built into the definition I give above. One way of getting a feel for it is to think about what happens at the level of the homotopy category: in this case, it says that, for any morphism $g : Z \to Y$ such that $f \circ g = 0$ in the homotopy category, there is a (not necessarily unique) morphism $Z \to \operatorname{hfib} f$ making the evident diagram commute (in the homotopy category). But, of course, this is not a complete characterisation of $\operatorname{hfib} f$.<|endoftext|>
TITLE: Distribution of the Error term in GH Hardy's "curious result" $\sum_{\nu \leq n } \{ \nu \theta \}^2 = \tfrac{1}{12} n + O(1)$
QUESTION [22 upvotes]: In an early paper, GH Hardy talks about the distribution of "curious" sum:
$$ \sum_{\nu \leq n } \{ \nu \theta \}^2 = \tfrac{1}{12} n  + O(1)$$
where $\{x\}:=x-\left \lfloor x \right \rfloor -1/2$. With a computer it was not hard to verify the linear growth, the factor of $\frac{1}{12}$ or the constant error term.  Here are my experiments:


The line is rather easy to prove with Weyl equidistribution theorem - without the $O(1)$ term.
$$ \frac{1}{n}\sum_{\nu \leq n } \{ \nu \theta \}^2 \approx
\int_{-\frac{1}{2}}^{\frac{1}{2}} x^2 \, dx = \frac{1}{12}   $$
Are there any easy ways to understand the noise?  It clearly has no limit... in the figure $\theta = \sqrt{7}$ the $O(1)$ error term is distributed between 0.05 and 0.30 with clear bands at indeterminate values.  

Obviously $\theta \notin \mathbb{Q}$ and even then the uncertainty might be too large.  
I had computed the Fourier series in order to trace the proof of the Von Neumann ergodic theorem.  
We can plot the sum of the sawtooth functions.  The first 10 and the first 100 terms.  The limit of $\sum_{\nu \leq n} \{ \nu \theta \}^2$ is highly oscillatory but does not converge at some points.

REPLY [7 votes]: This is an approach on the error term. The idea is to use Koksma's inequality and Erdos-Turan inequality. I used this on a problem in MSE: $\sum \frac{(-1)^n|\sin n|}n$.
To recall, 

Theorem [Koksma]  
Let $f$ be a function on $I=[0,1]$ of bounded variation $V(f)$, and suppose we are given $N$ points $x_1, \ldots , x_N$ in $I$ with discrepancy
  $$
D_N:=\sup_{0\leq a\leq b\leq 1} \left|\frac1N \#\{1\leq n\leq N: x_n  \in (a,b) \} -(b-a)\right|.
$$
  Then
  $$
\left|\frac1N \sum_{n\leq N} f(x_n) - \int_I f(x)dx \right|\leq V(f)D_N.
$$
Theorem [Erdos-Turan] 
Let $x_1, \ldots, x_N$ be $N$ points in $I=[0,1]$. Then there is an absolute constant $C>0$ such that for any positive integer $m$,
  $$
D_N\leq C \left( \frac1m+ \sum_{h=1}^m \frac1h \left| \frac1N\sum_{n=1}^N e^{2\pi i h x_n}\right|\right).
$$

We are using these on the sequence $x_n = n \theta \ \mathrm{mod} \ 1$ and $f(x) = (x -\frac12)^2$.  By Erdos-Turan and Lemma 3.3 in Kuipers & Neiderrater 'Uniform Distribution of Sequences', we obtain the following bound of discrepancy:

Lemma 
Let $\theta$ be an irrational real number. If $\mu$ is the irrationality measure of $\theta$, then the discrepancy $D_N$ for the sequence $n\theta \ \mathrm{mod} \ 1$ satisfies
  $$
D_N \ll N^{-\frac1{\mu-1}+\epsilon}.
$$

Therefore, we obtain the following estimate:

Let $\theta$ be an irrational real number with irrationality measure $\mu$, then we have 
  $$
\sum_{\nu\leq n} \{ \nu\theta \}^2 = \frac n{12} + O\left( n^{1-\frac1{\mu-1} + \epsilon } \right). 
$$

Note that $\mu=2$ for almost all $\theta\in\mathbb{R}$ yielding the error term bound $O(n^{\epsilon})$. 
Kuipers & Neiderrater's book also contains the following result:

Theorem
  If $\theta\in\mathbb{R}\backslash \{0\}$ has a bounded partial quotients in the continued fraction, then 
  $$
D_N\ll \frac{\log N}N.
$$

For $\theta=\sqrt 7$, it has periodic partial quotients, so we have
$$
\sum_{\nu\leq n} \{ \nu\theta \}^2 = \frac n{12} + O\left( \log n \right)
$$
which is better than $O(n^{\epsilon})$.<|endoftext|>
TITLE: Reference Request: Conductors of Twists of Hyperelliptic Curves
QUESTION [17 upvotes]: It is my understanding that if I twist a hyperelliptic curve of genus 2 whose Jacobian has conductor $N$ by a prime $p$ with $p\nmid N$, that the conductor of the Jacobian of the twist is expected to be $Np^4$, but I am unable to find a reference for this.  Is anyone aware of a proof of this "fact"?

REPLY [4 votes]: (This answer is Community Wiki and is extracted from comments by user eric, who has declined to post them as an answer. The CW is to invite others to contribute, especially to provide suitable references for these facts.) 
If the curve has good reduction, then the Jacobian has good reduction. The standard reference for this is SGA 7. Now you need that the $l$-adic representation is unramified; this is one direction of Neron-Ogg-Shaferevich. (One reference for that off the top of my head is the Serre-Tate Annals paper, although note that it's only the easy direction of NOS we use here.) Meanwhile, a quadratic twist is tamely ramified (this is for $p$ odd -- when $p=2$ the result you want is probably false in fact). If you need a reference for this, you can try the Artin-Tate notes on class field theory, but much like the previous reference, I'd regard this as overkill for what you are asking for. 
In brief: if you have an unramified representation of dimension $d$ and you twist it by a tamely ramified character, the resulting representation has conductor $p^d$. I don't really know what would be a great reference for this specific observation, but it's not a difficult calculation.<|endoftext|>
TITLE: $H^{*}$ algebras as a generalization of $C^{*}$ algebras
QUESTION [7 upvotes]: Let $H$ be the quaternions algebra. An $H^{*}$ algebra is a normed ring $A$ which is simultaneously a unital  left $H$ module and has an involution $*$ with the following properties:
$\forall \lambda \in H, a,b \in A$
1.$\;\lambda(ab)=(\lambda a)b$

$\; \parallel ab\parallel \leq  \parallel a \parallel \parallel b \parallel,\;\;\; \parallel \lambda a\parallel=\parallel \lambda \parallel \parallel a\parallel$

$\;(ab)^{*}=b^{*}a^{*}$


4.$\;\;\parallel ab\parallel \leq \parallel a \parallel \parallel b \parallel$

$\;\; \parallel aa^{*} \parallel= \parallel a  \parallel^{2}$

(If A is unital) $\lambda 1 \lambda' 1= \lambda \lambda' 1$


There is  a natural definition of spectrum of an element $a\in A$ (as a subset of $H$). There is also a natural definition of morphism and isomorphisms between two $H^{*}$ algebras.
Example: For  a  compact Hausdorff space $X$ put $A=H(X)=$ The space of all continuous  $f:X \to H$ with obvious structures
Questions:


Is it true to say that the spectrum is always  non empty and compact?

Is it true to say that two compact space $X$ and $Y$ are homeomorphic if and only if $H(X) \simeq H(Y)$?



The motivation for this question is that we search for an alternative proof for the Borsuk Ulam  theorem for $f;S^{4} \to \mathbb{R}^{4} \simeq H$ via consideration of  a $\mathbb{Z}/2\mathbb{Z}$ graded structure for $H(S^{4})$. see the following related  post
Banach algebraic proof of the Borsuk Ulam theorem
Edit: According to the answer of Andre Henriques we ask:


Assume that $A$ is a real $C^{*}$ algebra which contains the Quaternions $H$. Under what algebraic conditions, $A$ is in the form $H(X)$=The space of continuous functions $f:X \to H$ for some compact Hausdorff $X$?

REPLY [4 votes]: Let assume that you consider unital algebra only (one can still study non unital algebra by unitarizing them, but notion of spectrum is always a little annoying when one want to consider non unital algebra) and that a $H^*$ algebra is a real $C^*$-algebra with morphism of real $C^*$-algebra $\mathbb{H} \rightarrow A$
First two remarks:

The name $H^*$ is IMHO not a good idea at all, first the $C$ in $C^*$-algebra is not for "complex", and second $H^*$-algebra are actually already a thing, see for example here. I would rather call them $\mathbb{H}$-$C^*$-algebra or maybe quaternionic $C^*$-algebras (as we say real $C^*$-algebras and not $R^*$-algebras)
This definition might not be completely satisfying in the sense that it does not generalizes complexe $C^*$-algebras: a complexe $C^*$-algebra is the same as a real $C^*$-algebra together with an injection of $\mathbb{C}$ into its center. but when you replace $\mathbb{C}$ by $\mathbb{H}$ you need to remove the assumption on the center (because $\mathbb{H}$ is non-commutative) and it is not clear that it is a good idea to just remove it.  For example, it might be a good idea to add all sort of commutativity conditions that holds in $\mathbb{H}$, like " if $x$ is self adjoint then $x$ commute to all elements of $\mathbb{H}$ which would simplify some answer below...

The only part I don't know how to answer is the non-emptyness of the spectrum... I'm still thinking about it, and it might be related to those possible additional condition I'm mentioning just above. I will edit if I find something.
So, let $A$ be a quaternionic $C^*$-algebra, and $a \in A$ I guess you want to define the spectrum as $\{ h \in \mathbb{H} | a - h \}$ is non-invertible.
Then the proof that is compact is exactly the same as in the usual case: you prove that it is bounded and closed using that if $X$ is invertible and $Y$ is such that $\Vert  Y \Vert < 1/\Vert X^{-1} \Vert $ then $X+Y$ is invertible using a series argument (that work for arbitrary complete normed algebra)
As I said, 'non-empty' is trickier because the spectrum in a real $C^*$-algebra don't have to be inhabited and we don't have holomorphic calculus at disposition for quaternionic $C^*$-algebra (althoug it might be an idea to develop it a little, the resolvant is still locally a formal series so maybe one can use a kind of quaternionic Liouville's theorem).
But it seems difficult to produce a counter example: For example, as soon as you have an element which commute to some purely imagnary unit $u$, then $x,x^*,u$ generates a complexe $C^*$-algebra (with $u$ as $i$) hence $x$ has a non-empty spectrum inside of it, and it implies that $x$ has a non-empty sepctrum (as in real $C^*$-algebra it also true that $x$ is invertible is a subalgebra if and only if it is invertible in the larger algebra: it is proved by taking the complex form of the algebras).
For your question 2, the usual proof caries over easily and one can see that $X$ is the space of $\mathbb{H}$-algebra morphism from $H(X)$ to $\mathbb{H}$ hence if $H( X) \simeq H(Y)$ then $X \simeq Y$ and the isomorphism between $H(X)$ and $H(Y)$ is induced by the homeomorphism between $X$ and $Y$..
For your question 3), you always have a morphism from $A$ to $H(Spec A)$ where $Spec A$ is the set of character of $A$ (i.e. morphisms of $\mathbb{H}$-algebra from $A$ to $\mathbb{H}$). So "the algebras for which this morphism is an isomorphism" is an answer to your question. of course one can hope to obtain something better... If I'm correct the quaternionic Stone-Weierstrass theorem work, and the only unital $\mathbb{H}$-*-sub-algebra of $H(X)$ which separates points is $H(X)$ itself, hence the comparison map $A \rightarrow H(Spec A)$ is always surjective. 
But one can give a more explicit condition: A quaternionic $C^*$-algebra is of the form $H(X)$ if and only if its self adjoint element ($x^*=x$) are central (or just commute between themselves and to elements of $\mathbb{H}$).
Indeed, it is then obvious that the set of self-adjoint element is a real $C^*$-algebra with trivial involution hence that it is of the form $C(X,\mathbb{R})$ for some $X$.
Moreover, using a kind of polarization formula, one can write any element in a unique way in the form $x+iy+jz+kt$ where $x,y,z,t$ are self-adjoint, and because self-adjoint are central, those multiply exactly as function on $X$ with values in $H$.<|endoftext|>
TITLE: Tiling the plane with incongruent isosceles triangles
QUESTION [34 upvotes]: It is not difficult to tile the plane with incongruent triangles.
One could tile with equilateral triangles, and then partition
each equilateral into three triangles, displacing their common
centerpoint so that no two triangles are congruent (left below).

          



Q1.
  Is it possible to tile the plane with isosceles triangles,
  no two of which are congruent?

It is easy to tile the plane with congruent isosceles triangles,
as illustrated right above.
But I don't see how to achieve a tiling with incongruent isosceles triangles.
Perhaps it is easier to answer this question:

Q2.
  Is it possible to tile the plane with equilateral triangles,
  no two of which are congruent?


Added 14Feb2020:
Q2 has been answered (negatively) in two papers
(independently). These results were presaged 
(by @Wojowu)
below.

(1) Pach, János, and Gábor Tardos. "Tiling the plane with equilateral triangles." arXiv:1805.08840 abstract (2018).

Corollary 4. There is no tiling of the plane with pairwise noncongruent equilateral triangles
whose side lengths are bounded from below by a positive constant.

(2) Richter, Christian, and Melchior Wirth. "Tilings of convex sets by mutually incongruent equilateral triangles contain arbitrarily small tiles." Discrete & Computational Geometry 63, no. 1 (2020): 169-181.
  Springer link.



Question Q1 was inspired by (but not addressed in) this paper:

Malkevitch, J. "Convex isosceles triangle polyhedra." Geombinatorics 10 (2001): 122-132.

REPLY [17 votes]: Simpler construction: a non-isosceles right triangle $T_0$
can be divided into two isosceles triangles not congruent to each other,
or into two right triangles similar to $T_0$.  Reversing the latter
construction, let $T'_n$, $T_n$ ($n = 1, 2, 3, \ldots$) be right triangles
similar to $T_0$ such that $T_n = T_{n-1} + T'_n$, and divide $T_0$ and
each $T'_n$ into a pair of isosceles triangles.  There are two choices
at each stage, most of which yield a tiling of the plane; alternatively,
choose the $T'_n$ and $T_n$ to all share a vertex with $T_0$, getting
a geometric progression [sic] of triangles that tiles a wedge,
and then arrange several wedges around a common vertex to fill the plane.
Here's a picture of such a wedge using $(3:4:5)$ triangles:

    (source)<|endoftext|>
TITLE: Two fold optimization: is there an established approach for this kind of problem?
QUESTION [5 upvotes]: I have a list of thousands of linear expressions, with less than 100 total variables. Each expression is associated with a positive point value. I want to make as many of the expressions greater than zero as possible, maximizing the total points of the expressions I can make positive.
For example:
x1 + x2 > 0 (5 points)
x1 - 3*x2 > 0 (1 point)
-x1 + x2 > 0 (4 points) 
I can choose x1=1, x2=2 to maximize the score (9 points).
Is there an established algorithmic approach to solve this kind of problem?

REPLY [3 votes]: If the coefficients of your linear expressions are integers and you want the variables $x_i$ to be integers as well then the problem can be written as an integer program:
\begin{align*}
\text{Maximize }\sum_{i=1}^m&c_iy_i\\
\text{subject to }My_i &\leqslant M-1+\sum_{j=1}^na_{ij}x_j &&i\in\{1,\ldots,m\},\\
x_j &\in\mathbb{Z}&&j\in\{1,\ldots,n\},\\
y_i &\in\{0,1\} && i\in\{1,\ldots,m\}. 
\end{align*} 
Here $\displaystyle\sum_{j=1}^na_{ij}x_j$ is the $i$-th linear expression, $c_i$ is its point value, and $M$ is a sufficiently large number.
Throwing this formulation at a MIP solver might be worth a try.<|endoftext|>
TITLE: How are holomorphic and real-analytic Eisenstein series related?
QUESTION [5 upvotes]: This is certainly not a research level question, but I didn't get an answer to my question on MSE, so here goes:
The holomorphic Eisenstein series can be given as
$$G_{2k}(z)=\sum_{(c,d)\in{\bf Z}^2\backslash(0,0)}\frac{1}{(cz+d)^{2k}}$$
while the real-analytic Eisenstein series is
$$E(z,s)=\frac{1}{2}\sum_{(c,d)=1}\frac{y^s}{|cz+d|^{2s}}$$
satisfying the relation
$$2\zeta(2s)E(s,z)=\sum_{(c,d)\in{\bf Z}^2\backslash(0,0)}\frac{\text{Im}(z)^s}{|cz+d|^{2s}}.$$
They look to be very similar, but have different analytic behaviours. I haven't found a discussions on how they are related, so I'm hoping to find some answers.

REPLY [3 votes]: [Comment reposted as an answer]
Up to the scaling by $2 \zeta(2s)$, which is just a matter of conventions, both are special cases of a single more general object: the series
$$ E_k(z, s) = \sum_{(c, d) \in \mathbf{Z}^2 \setminus (0,0)} \frac{y^s}{(cz + d)^k |cz + d|^{2s}}, $$
which converges for $Re(s) \gg 0$ (actually for $k + 2 Re(s) > 2$ if I remember correctly) and has meromorphic continuation to all $s \in \mathbf{C}$ (analytic if $k \ne 0$). 
For a fixed value of $s$, the function $E_k(-, s)$ transforms like a modular form of weight $k$ under the modular group (although it is not usually holomorphic in the $z$ variable).<|endoftext|>
TITLE: Does every smooth manifold carry a gaussian random field?
QUESTION [9 upvotes]: Let $M$ be an arbitrary finite-dimensional smooth manifold. For simplicity, let's assume that $M$ has no boundary. Does there always exist a gaussian random field with constant variance on $M$? If not, does there exist a theorem which states sufficient (or necessary and sufficient) conditions on $M$ under which a gaussian random field will always exist? 
My definition of a gaussian random field is a random function $f:M\to \mathbb{R}^{n}$ such that for arbitrary points $t_1, \dots, t_k$ in $M$, the images $f(t_1),\dots, f(t_k)$ form a multivariate guassian random vector. Adler, Taylor, and Worsley in Applications of Random Fields and Geoemtry assume that $f(t)$ has constant variance for all $t$.

REPLY [11 votes]: You can construct   examples a dime a dozen. $\newcommand{\bR}{\mathbb{R}}$
Here  is a first  simple way. Fix $N$ smooth functions $f_1,f_2,\dotsc, f_N:M\to\bR$ and $N$ independent Gaussian random variables $X_1,\dotsc, X_N$. Then
$$f(x)=\sum_{k=1}^N X_k f_k(x) $$
is a Gaussian random field and the sample functions are a.s. smooth.
You can   allow  infinitely many functions  in the above example, i.e., $N=\infty$, but then  you need to make some assumptions on the functions and on the variances of  the variables $X_k$ to guarantee  the convergence of the resulting series. 
We know from Kolmogorov that the convergence is a $0-1$ phenomenon. The  three-series theorem  tells us what these conditions should be. (In the Gaussian case we can  say a bit more.)
Here is an example of this kind when $M$ is compact. Fix a  Riemann metric $g$ on $M$, denote by $\Delta$ the resulting Laplacian. Its eigenvalues (multiplicities included) are
$$  0=\lambda_0<\lambda_1\leq \lambda_2\leq \cdots. $$
Fix an orthonormal eigenbasis $(\psi_k)_{k\geq 0}$ of $L^2(M, dV_g)$ 
$$\Delta \psi_k=\lambda_k,\;\; \Vert\psi_k\Vert_{L^2}=1,\;\;\forall k\geq 0. $$
Next choose  independent  Gaussian random variables $(X_k)_{k\geq 0}$. We denote by $v_k$ the variance of $X_k$.   If $v_k$ goes to $0$ sufficiently fast, then the random series 
$$f(x)=\sum_{k\geq 0} X_k \psi_k(x) $$
defines a  Gaussian random field on $M$.  The regularity of the sample functions  of this random field   depends on the decay rate of  $v_k$. The faster $v_k$ decays as $k\to\infty$, the more regular is the random function. For example, if 
$$\lim_{k\to \infty}k^\alpha v_k=0,\;\;\forall \alpha>0, $$
then the random function $f(x)$ is a.s. smooth.
Ultimately, the most general construction  of a Gaussian random function on a manifold  is via Gaussian measures on the space of   distributions (i.e. generalized functions) on $M$.  I refer to   this paper  for  more details and additional references. I would start with reference [6] in this paper.<|endoftext|>
TITLE: What is the value of $p$-adic $\zeta$-function at positive integer point?
QUESTION [12 upvotes]: $p$-adic zeta function is a $p$-adic interpolation of the Riemann $\zeta$-function for the values $\zeta(1−k)$, $k\ge 1$ (see $p$-adic Numbers, $p$-adic Analysis, and Zeta-Functions by Neal Koblitz) or, more precisely, it's corrected values
$$\zeta_p(1−k):=-(1-p^{k-1})\frac{B_k}{k}.$$
What is known about the values $\zeta_p(k)$ for $k\ge 1$?

REPLY [6 votes]: Let me mention some irrationality results about p-adic zeta values:
In 2005, Frank Calegari arxiv link proved that for $p=2,3$, the p-adic zeta values $\zeta_p(3)$ is irrational (hence nonzero). It can be viewed as the p-adic analogue of Apéry's theorem. Later, Frits Beukers arxiv link gave an alternating proof of Calegari's results (and the irrationality for some other p-adic L-values).
In 2010, Pierre Bel arxiv link obtained some partial results on the irrationality of p-adic Hurwitz zeta values. See also his 2018 paper on $\zeta_p(4,x)$.
Very recently (in 2020), Johannes Sprang arxiv link showed that
$$ \dim_{K}(K+\zeta_p(2)K+\zeta_p(3)K+\cdots+\zeta_p(s)K) \geqslant \frac{(1-o(1))\log s}{2[K:\mathbb{Q}](1+\log 2)}, $$
which is the analogue of Ball-Rivoal theorem for p-adic zeta values.<|endoftext|>
TITLE: Change of groups for naive G-spectra
QUESTION [6 upvotes]: Let $H$ be a subgroup of $G$ a compact Lie group and
let $\text{spectra}[G]$ be the category of naive $G$-spectra (ie G-objects in the category of spectra). Then there is a forgetful functor $i^*$ from
$\text{spectra}[G]$ to $\text{spectra}[H]$ with a left adjoint $G_+ \wedge_H -$.
What can be said about the adjunction $(G_+ \wedge_H - , i^*)$ on homotopy categories?
For example:

I believe the forgetful functor $i^*$ to be faithful, does anyone have a reference?
Is the left adjoint also faithful?
I've been told this adjunction (probably) satisfies some form of homotopy descent. What does this tell me? (Presumably that some spectral sequence converges.)
Is every $O(n)$-spectrum in the image of the left adjoint (on homotopy categories)?

(The specific case I'm interested in is $H$ is the symmetric group on $n$ letters and $G$ is the orthogonal group $O(n)$.)

REPLY [6 votes]: The following will assume that you are using the underlying weak equivalence structure on $G$-spectra and $H$-spectra, so that an equivariant map $X \to Y$ is an equivalence if and only if it is so after forgetting the action. Some of what I will say below can be altered if you instead use maps which are weak equivalences on fixed points, but it makes things harder.

The functor $i^*$ is not faithful. Take the inclusion $\{e\} \to C_2$ from the symmetric group on one letter to the 1-by-1 orthogonal group, giving both the Eilenberg-Mac Lane spectrum $H\Bbb F_2$ and the sphere spectrum $\Bbb S$ the trivial $C_2$-action. We can calculate and find that
$$
[\Bbb S, \Sigma H\Bbb F_2]_{C_2} \cong H^1(C_2, \Bbb F_2) \cong \Bbb F_2
$$
but 
$$
[\Bbb S, \Sigma H\Bbb F_2]_{\{e\}} \cong H^1(\{e\}, \Bbb F_2) = 0.
$$
Therefore, the restriction cannot be faithful.
The left adjoint is also not faithful. For this we can take the inclusion $\Sigma_3 \to O(3)$ from the symmetric group on 3 letters to the 3-by-3 orthogonal group. Take the two $\Sigma_3$-spectra given by $\Bbb S[\Sigma_3] = \Sigma^\infty_+(\Sigma_3)$ and $M = H\Bbb Z^3$, where $\Bbb Z^3$ is given the permutation action of $\Sigma_3$. We find that the left adjoint sends these to $\Sigma^\infty_+(O(3))$ and $\Sigma^\infty_+(O(3)) \wedge_{\Sigma_3} M$ respectively. We can calculate and find that
$$
[\Bbb S[\Sigma_3], M]_{\Sigma_3} \cong [\Bbb S, H\Bbb Z^3] \cong \Bbb Z^3
$$
by the adjunction, but
$$
[\Bbb S[O(3)], \Sigma^\infty_+(O(3)) \wedge_{\Sigma_3} M]_{O(3)} \cong [\Bbb S, \Sigma^\infty_+(O(3)) \wedge_{\Sigma_3} H\Bbb Z]
$$
is a proper quotient of $\Bbb Z^3$, roughly because smashing with $O(3)$ over $\Sigma_3$ explicitly adds a path between $(x,y,z)$ and $\sigma x = (y,z,x)$ for any $(x,y,z) \in \pi_0 M$ and $\sigma$ generating the alternating subgroup.
The adjunction does satisfy a form of descent. If $H$ is normal in $G$, then this takes the form of a homotopy fixed-point spectral sequence (analogous to the Lyndon-Hochschild-Serre spectral sequence) with $E_2$-term
$$
H^s(G/H; [\Sigma^t i^* M,i^* N]_H) \Rightarrow [\Sigma^{t-s} M,N]_G.
$$
However, you've listed groups that certainly don't satisfy normality. We do know that there always exists some descent-type spectral sequence for taking a group $G$ with a subgroup $H$ and calculating the group cohomology of $G$; however, the terms in this spectral sequence are built out of some kind of tangled web involving the group cohomologies of finite intersections of conjugates of $H$. In less obtuse terms, there's a classifying space $E{\cal F}$ for the family of subgroups of $G$ which are conjugate to a subgroup of $H$, and $E{\cal F}$ is weakly equivalent to a point. Therefore, we get an isomorphism
$$
[X,Y]_G \xrightarrow{\sim} [E{\cal F}_+ \wedge X, Y]_G,
$$
and the cellular filtration on $E{\cal F}$ gives us a spectral sequence of mixed practicality for calculating the left-hand side.
No, this is not true. For example, again when $n=2$ every $O(2)$-spectrum in the image of this map is obtained by extending a $C_2$-action freely to a little bit more than a circle action, and the image of such a spectrum always has Euler characteristic $0$ when defined. The spectrum $\Bbb S$ with the trivial $O(2)$-action cannot be in the image.<|endoftext|>
TITLE: How big can a set of integers be if all pairs have small gcd?
QUESTION [22 upvotes]: Suppose $A\subset[1,N]$ is a set of integers. If for any distinct $a,b\in A$ we have $(a,b)\leq M$ then how big can $|A|$ be? 
If $M=1$ then $|A|$ is at most $\pi(N)$ since the map $a\mapsto P_+(a)$ (which sends $a$ to its largest prime factor) is an injection to the primes, and so the primes themselves are the most efficient. If $M=N$ then we can take $|A|=N$. What about $M$ in between? Can you beat the primes? It would be safe to assume that $A$ consists solely of $M$-smooth numbers since we can decompose $A=A_1\cup A_2$ where each $a\in A_1$ is $M$-smooth and each $a\in A_2$ has a prime factor which at least $M+1$. The large prime factors appearing as divisors in $A_2$ cannot be repeated, so $|A_2|\leq \pi(N)$.

REPLY [13 votes]: We'll prove that the maximal cardinality of such a set for $M^2\leq N$ has size equal to $$\pi(N)+\sum_{1<n\leq M} \pi(p(n))$$ where $p(n)$ is the smallest prime factor of $n$. Since $$\sum_{1<n\leq M} \pi(p(n))\sim \frac{M^2}{2\log^2 M},$$ this proves that Ilya Bogdanov's example of including all pairwise products of primes not exceeding $M$ is nearly optimal in terms of asymptotics.

As you suggest in the question, this problem is equivalent to constructing the largest subset $A\subset S_M(N)$ such that $\gcd(a,b)\leq M$ for every $a,b\in A$ where $S_M(N)$ is the set of $n\leq N$ whose largest prime factor is at most $M$. 
To see why, suppose that $A$ satisfies $\gcd(a,b)\leq M$ for every $a,b\in A$. Then every prime that is greater than $M$ can divide at most one element of $A$. If $p>M$ divides $a\in A$, then making $a=p$ only helps create a larger set $A$. Since these primes do not interact with the $M$-smooth numbers, the proof is complete.
Let $T(N,M)$ denote the maximum size of such a set $A\subset S_M(N)$. Then the size of the largest subset of $[1,N]$ with pairwise $\gcd$'s bounded by $M$ is $$\pi(N)-\pi(M)+T(N,M).$$ 
As mentioned by Fedja in the comments s, the reasoning above for the primes extends to all integers. By considering those primes $p,q\leq M$ such that $pq>M$, we see that there can only be exactly one such number divisible by $pq$. Similarly for any $p,q,r$ with $pq,qr,rp\leq M$ and $pqr>M$ there can only be one number in our set divisible by $pqr$. Thus we find that the maximal size is the sum over those integers whose largest proper divisor is less than $M$. Since the largest proper divisor of $n$ equals $n/p(n)$ where $p(n)$ is the least prime factor, we can group things based on this, and we have that $$T(N,M)=\sum_{1<p\leq M}\sum_{n\leq M:\ p(n)\geq p}1.$$ Rearranging this equals 
 $$\sum_{1<n\leq M}\sum_{p\leq p(n)}1=\sum_{n\leq M}\pi\left(p(n)\right),$$ and for composite $n$ $p(n)\leq\sqrt{n}$, so the primes dominate this sum. Thus asymptotically we have $$T(N,M)\sim\sum_{p\leq M}\pi(p)\sim\frac{M^{2}}{2\log^{2}M}.$$<|endoftext|>
TITLE: Do Hausdorff locally convex inductive limits always exist?
QUESTION [5 upvotes]: The following is from Schaefer, "Topological Vector Spaces", 1999, p. 56/57:
Let $(E_\alpha)_{\alpha \in A}$ be a family of locally convex spaces with $\alpha$ in a directed poset $A$ and $h_{\beta \alpha} : E_\alpha \to E_\beta$ continuous linear maps for $\alpha \leq \beta$. Set $F := \oplus_\alpha E_\alpha$ and let $g_\alpha : E_\alpha \to F$ be the canonical imbedding. Denote by $H$ the subspace of $F$ generated by the ranges of all the maps $g_\alpha - g_\beta h_{\beta \alpha} : E_\alpha \to F$ for $\alpha \leq \beta$.
If $H \subseteq F$ is closed then one can represent the inductive limit of the family $E_\alpha$ in the category of Hausdorff locally convex spaces by $F / H$ which is then a Hausdorff locally convex space.
Now Schaefer writes:
"It appears to be unknown whether $H$ is necessarily closed in $F$."
It seems that this statement is equivalent to say that the inductive limit does not necessarily exist.
Does one know here whether this is still an open issue?
EDIT: Summarizing two answers: The inductive limit $F / H$ in the cat. of l.c.s. does exist and is not necessarily Hausdorff. Also, the inductive limit in the cat. of Hausdorff l.c.s. does also exist and is equal to $F / \overline{H}$. For closed $H$ they coincide.

REPLY [7 votes]: It is quite well-known that locally convex inductive limits need not be Hausdorff. Here is my favorite example. It is the (countable and injective) limit of nuclear Frechet spaces: 
Let $U_n$ be a decreasing sequence of non-empty, bounded, connected, open sets in $\mathbb C$ with empty intersection and set $X_n=H(U_n)$, the space of holomorphic functions on $U_n$ 
 endowed with the topology of uniform convergence on compact subsets ($X_n$ is thus a nuclear Frechet space). 
 Since the restriction $X_{n}\to X_{n+1}$ $f\mapsto f|_{U_{n+1}}$ is injective we may consider
 $X_n$ as a subspace of $X_{n+1}$ with continuous inclusion. Then $X_\infty=$ind $X_n$ has the trivial topology.
By the Hahn-Banach theorem we have to show that every continuous linear functional $\varphi$ on $X_\infty$ is zero. To see this we define, for each $z\in\mathbb C$,
 the function $g_z(\omega)=  1/({z-\omega})$ which belongs to $X_n=H(U_n)$ if $z\notin U_n$. Then $h(z)=\varphi(g_z)$ defines a function on $\mathbb C$ which is
 holomorphic (since $(g_{z_k} - g_z)/(z_k-z)$ converges in $X_n$ if $z_k\to z$ in $U_n^c$) with $h(z_k)\to 0$ for $|z_k|\to\infty$ (since $g_{z_k}\to 0$ even in $X_1$).
 By Liouville's theorem, $h$ vanishes identically. Moreover, by Runge's theorem, $\{g_z: z\notin U_n\}$ has dense linear span in $X_n$ which implies that
 $\varphi$ vanishes on $X_n$.

REPLY [2 votes]: The limit always exists wether or not $H$ is closed: The inductive limit in the category of hausdorff lc vector space will be the quotient of $F$ by the closure of of $H$: a cone for the inductive limit is the same as a continuous map on $F$ which vanish on $H$, so if the target is hausdorff as well then it vanish on the closure.
The question of whether $H$ is closed or not is only relevant to know if the forgetful functor to the category of vector spaces or of general locally convexe vector space commute to this specific inductive limit.<|endoftext|>
TITLE: Summation of an infinite q-series
QUESTION [5 upvotes]: When calculating a Partition function, I encounter the following summation
$$\sum_{n=0}^{\infty} x^n q^{n^2}.$$
I know that the sum $\sum_{n=-\infty}^{\infty} x^n q^{n^2}$ is a Theta function, but I do not know how to perform the sum from 0.
Can anyone help?
By the way I would like to have the result in an infinite product form, which I can easily do in the Theta case using Jacobi's triple product identity.
I want to make a small addition to this question which might or not be related but is also relevant for the physical problem  I want to solve. The usual formulas for the infinite product generating functions involve products over 1 integer. To be more concrete one has for example
$$\sum_{n=0}^{\infty} \frac{x^n q^{n(n-1)/2}}{(1-q) ... (1-q^n)} = \prod_{m=0}^{\infty} (1+ x q^m).$$
which is the Partition for fermions.
Do you know any reference where people study double products like
$$\prod_{m,n=0}^{\infty} (1+ x q^{(n \pm m)}).$$
or maybe
$$\prod_{m,n=0}^{\infty} (1+ x q^{(n m)}).$$
Can then one write them as infinite sums with one label as
$$\sum_{n=0}^{\infty} x^n A^{n}?$$

REPLY [11 votes]: Those are usually called partial theta functions. They were probably first studied in some depth by Ramanujan, and seem to be closely related to general q-series, mock modular forms. They also show up in combinatorics and statistical physics.
They are not necessarily as well behaved as the complete ones, and usually not modular, but some basic properties such as convergence for $|q|<1$ still holds.
In particular, you still have an infinite product. Using q-Pochhammer notation, I think you can recover what you are looking for from the following Jacobi-type identity:
$$1+\sum_{n=1}^\infty (-1)^nq^{n(n-1)/2}(a^n+b^n)=(q)_\infty(a)_\infty(b)_\infty \sum_{n=1}^\infty \frac{(ab/q)_{2n}q^n}{(q)_n(a)_n(b)_n(ab)_n}$$
This is proved in section 2 of the paper:

S. Ole Warnaar, Partial Theta Function I. Beyond the Lost Notebook (2003)

See that you can get Jacobi's triple product identity by taking $b=q/a$.
For more general information about partial thetas here are some references:

Vladimir Petrov Kostov, A property of a partial theta function (2015)
Krishnaswami Alladi, A partial theta identity of Ramanujan and its number-theoretic interpretation (2009)
Jeremy Lovejoy, Ramanujan-type partial theta identities and conjugate Bailey pairs (2012)
Kathrin Bringmann, Amanda Folsom, Robert C. Rhoades, Partial theta functions and mock modular forms as q-hypergeometric series (2012)<|endoftext|>
TITLE: Plot of Ramanujan tau function
QUESTION [10 upvotes]: There is a picture on wikipedia of Ramanujan tau function. At first I noticed that there are exceptional red point (where the red points are sparse in the lower part), this should be due to Sato-Tate conjecture. 
But it seems strange since where red points are dense blue points  are sparse!
Is there any explanation about these? 
(Wiki: The blue line picks only the values of n that are multiples of 121.)

By the way, is there any conjecture about the growth （a lower bound） of this function? (Surely stronger than the Lehmer's conjecture.)

REPLY [3 votes]: The best lower bound known seems to be due to Ram Murty:
$$\tau(n)=\Omega (n^{11/2}e^{c \log n / \log \log n})$$
for some $c>0$ absolute and effective (note: thanks to the Sato-Tate conjecture might be able to take $c<\log 2$).
This result is proved in:

Ram Murty, Some Omega results for Ramanujan's tau function (1982)

In their book on Ramanujan, Ram and Kummar mention:

This result is essentially best possible since we know that
$$d(n)<e^{c' \log n / \log \log n}$$

It might be worth mention that this holds for arbitrary cusp forms, for some $c>0$,
$$a(n)=\Omega (n^{(k-1)/2}e^{c \log n / \log \log n})$$
The exceptional values you observe seem to be related to congruences modulo $11$ and $121$, but I have no idea about that. Ramanujan himself certainly had much to say on the matter:

Srinivasa Ramanujan, Properties of $p(n)$ and $\tau (n)$... (unpublished)

Hopefully someone can answer that part of your question better.<|endoftext|>
TITLE: Rank versus free-rank of a module
QUESTION [6 upvotes]: Suppose $M$ is a finitely generated left module over a ring $R.$ 
We define the rank of $M$ as the minimal number of generators of $M.$ 
If in addition $M$ is free, then we define the free-rank of $M$ as minimal cardinality of a basis of $M.$ 
It is clear that $\text{rank}(M)\leq\text{free-rank}(M)$. 

I want to know an example when $\text{rank}(M)<\text{free-rank}(M)$. 

If $R$ is commutative, then they are equal; so $R$ must be non-commutative.

REPLY [7 votes]: There are abelian groups $A$ such that $A\cong A\oplus A \oplus A$ but $A\not\cong A\oplus A$.
Let $E=\operatorname{End}(A)$. The functor $F=\operatorname{Hom}(A,-)$ is an equivalence from the category of finite direct sums of copies of $A$ to the category of finitely generated free $E$-modules.
So $E\oplus E$ is a module with free rank $2$ but is a direct summand of $E\cong E\oplus E\oplus E$, and so has rank $1$.
For an example where the ring has the IBN property:
For any $n>2$ there is an abelian group $A_n$ such that $A_n^n\cong A_n$ but $A_n^k\not\cong A_n$ for $1<k<n$. Let $E_n=\operatorname{End}(A_n)$ and $E=\prod_{n>2}E_n$.
Then $E^2$ has free rank $2$ but rank $1$.<|endoftext|>
TITLE: Is the Quot-scheme over non-singular curve reduced
QUESTION [6 upvotes]: Let $k$ be an algebraically closed field, $C$ a non-singular projective curve over $k$ of genus at least $2$ and $\mathcal{F}$ a locally free sheaf on $C$. Let $r,d$ be two integers satisfying $\mathrm{gcd}(r,d)=1$ and $r>0$. Denote by $Q$ the Quot-scheme parametrizing quotients of $\mathcal{F}$ of rank $r$ and degree $d$. Is every irreducible component of $Q$ (generically) reduced i.e., for a general point $x$ of the irreducible component of $Q$, the local ring $\mathcal{O}_{Q,x}$ is reduced?

REPLY [3 votes]: This is false, at least for special $C$ (perhaps it is true for $C$ that are sufficiently general in moduli).  The simplest counterexample I know of is a genus $4$ curve $C$ that is non-hyperelliptic and whose canonical image is contained in a singular quadric hypersurface in $\mathbb{P}^3$.  In this case, the scheme $G^1_3$ equals $W^1_3$ inside $\text{Pic}^3_C$, and this is one nonreduced point.  For the invertible sheaf $\mathcal{L}$ on $C$ parameterized by this point, $\mathcal{L}$ is globally generated.  Thus for $\mathcal{F}$ defined to be $H^0(C,\mathcal{L})\otimes_k \mathcal{O}_C$, the natural homomorphism, $$\phi:\mathcal{F} \to \mathcal{L},$$ is surjective.  
Since being invertible is an open condition on flat families of coherent $\mathcal{O}_C$-modules, every small deformation of the quotient $\mathcal{L}$ is an invertible $\mathcal{O}_C$-module of degree $3$ that has a $2$-dimensional vector space of global sections, and conversely.  Thus, there is a smooth morphism from a Zariski open neighborhood $U$ of $[\phi]$ in the Quot scheme to $W^1_3$.  Since $W^1_3$ is nonreduced with nilradical a rank $1$ vector space over the residue field, also $U$ is nonreduced with nilradical an invertible sheaf on the reduced scheme of $U$ (which is smooth).<|endoftext|>
TITLE: $\frac{d}{dt} (A+t B)^p\,\text{ for } p\geq 1$
QUESTION [13 upvotes]: Given two positive self-adjoint operators $A,B$ on a Hilbert space. Let $p\geq 1$.
I would like to calculate $$\frac{d}{dt}|_{t=0} (A+tB)^p,$$
where the power is defined through the spectral theorem. The difficulty is that $A$ and $B$ do not commute in general. Is there a nice formula for that? 
I thought about using some integral representation of $x^p$, which hopefully would make the differentiation easier, but I was not successful. 
Update:
Thanks for your answers. In my case, I actually only need it for $p\in [1,2)$. Using $y^p = \frac{\sqrt 3}{2\pi}\int_{0}^\infty dx \,\frac{y^2 x^{p-2}}{x+y}$, I thought about writing $C:=A+tB$ as $$C^p = \frac{\sqrt 3}{2\pi}\int_0^\infty C^2 x^{p-2} (x+C)^{-1} dx.$$
That might also work. I am still happy about other suggestions.

REPLY [4 votes]: I give you the answer in the finite dimensional case, and let you adapt it to the Hilbert space case. Let $f$ be an analytic function; in your case $f(t)=t^p$.
First of all, choose an orthonormal basis in which $A=:D$ is diagonal. Then form the matrix
 $f^{[1]}(D)\in{\bf M}_n({\mathbb C})$ by
$$f^{[1]}(D)_{jk}=\left\{\begin{array}{lr}
f'(d_j), & k=j, \\ \\ \frac{f(d_j)-f(d_k)}{d_j-d_k}, & k\ne j,\end{array}\right.$$
where we identify
$$\frac{f(b)-f(a)}{b-a}:=f'(a),$$
if $b=a$. Then the differential of $H\mapsto f(H)$, taken at $D$, is
$${\rm D}f(D)B=f^{[1]}(D)\circ B,
$$
where $A\circ B$ denotes the Hadamard product.
Edit. As noticed by Suvrit, this is Daleckii-Krein Formula.<|endoftext|>
TITLE: When is the flatness locus non-empty
QUESTION [5 upvotes]: Let $k$ be an algebraically closed field, $f:X \to Y$ be a surjective proper $k$-morphism locally of finite presentation between irreducible noetherian schemes. Assume that $Y$ is reduced. Under what additional condition on $f$ (other than flatness/ generic flatness) does there exist a non-empty open set $U$ of $X$ such that $f|_U$ is flat?
If I further assmume that $X$ is reduced, then does there exists a non-empty open subset $U$ of $Y$ such that $f|_{f^{-1}(U)}$ is flat?

REPLY [5 votes]: Let $f:X\to S$ be a morphism of schemes.
Assume that $S$ is integral, and let $K$ be its function field. As "everything is flat over a field", the generic fibre $f_K:X_K\to \mathrm{Spec} \ K$ is a flat morphism.
In particular, the locus of flatness is non-empty.
As Laurent Moret-Bailly points out below, if $f$ is of finite presentation (and $S$ still integral), then $f$ is flat over a dense open of $S$.<|endoftext|>
TITLE: Reference for restriction formula in terms of double Schubert polynomials
QUESTION [5 upvotes]: Everyone (that is, everyone who cares) knows that double Schubert polynomials represent Schubert classes in equivariant cohomology in type $A$. We also know that we can restrict Schubert classes to fixed points of the torus action to obtain a realization of the equivariant cohomology ring as a subring of a direct product of polynomial rings. Positive formulas for the restrictions in terms of the negative roots are well known.
There's another formula for the restrictions, however, that is not positive, that I figured out but don't remember seeing anywhere. Namely, the restriction of the Schubert class $[X_u]^T$ to the fixed point $t_w$ is given by
$$S_u(x_{w(1)},x_{w(2)},\ldots,x_{w(n)};x_1,x_2,\ldots,x_n)$$
where $S_u$ is the corresponding double Schubert polynomial. This can be proved with Macdonald's skew divided difference operators. This formula is pretty logical considering the equivariant cohomology ring as the dual of the nil-Hecke ring, but it's not immediately obvious that the relationship between the rings is so literal that you can just plug things into the polynomials to get the answer. 
Any references/insights/comments are welcome.

REPLY [3 votes]: Instead of thinking of the double Schubert polynomial $S_u$ as representing $[\overline{B_- uB}/B] \in H^*_T(GL_n/B)$, equivalently think of it as representing $[\overline{B_- uB}] \in H^*_{T\times B}(GL_n) \cong H^*_{T\times T}(GL_n)$. Then setting $y_i \mapsto x_{w(i)}$ corresponds to restricting the $T\times T$ action to the $w$-twisted diagonal $T$, i.e. the map $H^*_{T\times T}(GL_n) \twoheadrightarrow H^*_{w\cdot T_\Delta}(GL_n)$.
The reason to restrict is that $T\times T$ has no fixed points on $GL_n$, i.e. the $T$-fixed points on $GL_n/B$ are not images of $T$-fixed points upstairs. Rather, $wB/B$ is the image of many $w\cdot T_\Delta$-fixed points $p$. Now we can restrict from $H^*_{w\cdot T_\Delta}(GL_n) \to H^*_{w\cdot T_\Delta}(p) \cong H^*_{w\cdot T_\Delta}$ and get the polynomial you wanted.<|endoftext|>
TITLE: Realizing symmetric groups by diffeomorphisms
QUESTION [13 upvotes]: Let $M$ be a (closed, smooth) manifold of dimension $d$. For $n$ a positive integer, fix $n$ points $x_1, \dots, x_n \in M$. The group of diffeomorphisms of $M$ that permutes the points $x_i$ surjects to the symmetric group $\Sigma_n$, at least for $d \geq 2$:
$$\Phi\colon \text{Diff}(M,\{x_1, \dots, x_n\}) \to \Sigma_n$$
Is it always true that $\Phi$ does not have a section for $n$ big enough? 
How big do we have to choose $n$ for $M = S^d$? (The naive guess would be $d+3$.)

REPLY [28 votes]: I'm going to reverse the roles of $n$ and $d$ (since otherwise I will screw things up in this answer).  If $M^n$ is a connected $n$-dimensional smooth manifold, then there is no section of the map $Diff(M^n,\{x_1,\ldots,x_d\}) \rightarrow \Sigma_d$ if $d \geq n+3$ (nb: you forgot in your question to assume that $M^n$ is connected).  Indeed, assume that $\Sigma_d$ acts by diffeomorphisms on $M^n$ such that the action restricts to the permutation action on $\{x_1,\ldots,x_d\}$.  The group $\Sigma_{d-1} \subset \Sigma_d$ then fixes $x_d$, so we get a linear action of $\Sigma_{d-1}$ on the tangent space $T_{x_d} M^n = \mathbb{R}^n$.  Every element of $\Sigma_{d-1}$ must act nontrivially on this tangent space; indeed, by averaging we can find a Riemannian metric on $M^n$ that is preserved by the action of $\Sigma_d$, and an isometry of a connected Riemannian manifold that fixes a point and all tangent vectors at that point must be the identity.  But the smallest dimensional vector space on which $S_{d-1}$ acts faithfully is $(d-2)$-dimensional.  We thus deduce that $n \geq d-2$, so $d \leq n+2$, as desired.<|endoftext|>
TITLE: Brownian motion, "increase interval", exists constants, bound,
QUESTION [5 upvotes]: Let $B_t$ be a standard Brownian motion. Let $J(j, n) = [j/n, (j+1)/n]$. We will call $J(j, n)$ an increase interval if$$B_s \le B_t,\text{ }0 \le s \le {j\over{n}},\text{ }{{j+1}\over{n}} \le t \le 3.$$I am wondering, do there exist constants $0 < c_1$, $c_2 < \infty$ such that for $j = 1, 2, \dots, 2n$,$${{c_1}\over{\sqrt{jn}}} \le \textbf{P}\{J(j, n) \text{ is an increase interval}\} \le {{c_2}\over{\sqrt{jn}}}?$$Thanks!

REPLY [4 votes]: Rescaling, splitting, reversing time, etc., we arrive at the following reformulation: 
Let $B_1(t)$ and $B_2(t)$ be two independent Brownian motions and $A,B\ge 2$. Estimate the probability that $\min_{1\le t\le A}B_1(t)\ge \max_{1\le t\le B}B_2(t)$.
The lower bound is trivial: $B_1(1)\ge 1$ with fixed positive probability, after which the probability that we never go down to $0$ is about $A^{-1/2}$. Similarly, $B_2$ stays non-positive with probability about $B^{-1/2}$.
The upper bound is similar. Condition upon $B_1(1)=x$, $B_2(1)=y$. Obviously, we must have that $x>y$ and that $B_2\le x$, $B_1\ge y$. This immediately gives the bound like 
$$
\iint_{x>y}e^{-x^2/2}e^{-y^2/2} (x-y)^2A^{-1/2}B^{-1/2} \,dx\,dy
$$
(the probability to never go above $r$ before time $T$ is at most $r/\sqrt T$), which is enough to yield the desired estimate.<|endoftext|>
TITLE: Have "sturdy squares" been studied before?
QUESTION [11 upvotes]: Over on PPCG I've just made a question that involves arranging the numbers 1 to 9 on a 3 by 3 grid such that every 2 by 2 subgrid has the same sum. I'm calling such 3 by 3 grids and their N by N counterparts "sturdy squares".
For example
1 5 3
9 8 7
4 2 6

is a sturdy square because the four 2 by 2 subgrids
1 5
9 8
5 3
8 7
9 8
4 2
8 7
2 6
all sum to 23.
When I first broached the topic in our chatroom the other day, a lengthy discussion ensued (and continues to do so) about the properties of these squares and how to generate larger ones and so on. An OEIS sequence has even been drafted.
My question is, are there any mathematical publications that have considered this problem (the general N by N case and variants, not just 3 by 3)? Or else are there related topics in mathematics that could aid in our discussion of these "sturdy squares"?
(I'd also be interested to know what gleanings any of you have about sturdy squares.)
(My apologies that I'm very uncertain how to tag this.)

REPLY [2 votes]: I don't know what exists in literature (probably recreational math, if anywhere; this construction does not seem natural elsewhere). However, here are some observations that seem very relevant to what I perceive to be the most natural relaxation of the problem. (I'm assuming the generalization means $2 \times 2$ subsquares are checked, as opposed to $k \times k$ subsquares, although some of my observations can be extended to these generalizations as well)
(as I'm typing this, I realized Tim Chow made a comment that may have been spurred by similar instincts) Conditions like "have $1$ through $N^2$ each appear exactly once" are not very mathematically natural. They can be forced (which is why people do sometimes study magic squares) but are likely to be fairly ad-hoc in the attacks. A natural extension is to make sure the entries are arbitrary; i.e. the more general class of matrices with same $2 \times 2$ sums, without the uniqueness constraint. Let's call these "semi-sturdy squares", or "SSS." I believe I can describe these much more easily and naturally, and will do so below. I understand (and apologize!) that this is not exactly the problem you are solving, but it does provide structure that may be useful for you, since you are studying a special case).
One good strategy to start is to get some canonical form. Consider the following $4$ matrices, labeled by $M(a,b)$, where $a, b \in \{0,1\}$. Define $M(a,b)$ to have a $1$ in a coordinate $(x,y)$ if and only if $(x,y) = (a,b) \pmod{2}$ coordinatewise. Note that adding a linear combination of the $M(a,b)$ to your matrix does not change the sum condition (although it does change things like uniqueness of numbers). Thus, one thing we can do for free is to change the upper left $2 \times 2$ matrix to all $0$'s by shifting by whatever $M(a,b)$ you desire. 
Now, I claim the following (somewhat nice!) statement: 

Suppose you fixed the entire first row and first column (i.e. coordinates $(x,0)$ and $(0, y)$) to any
  integers of your choice. Then, selecting any integer as the value of $(1,1)$ gives a unique SSS.

Proving this isn't that hard, I'll give a short sketch: consider the case where you selected entries for coordinates $(0, y)$ and $(x, 0)$ for all $x$ and $y$, and you also fill in $(1,1)$. Then you now have the $2 \times 2$ sum, and from it you can determine $(1,2)$ (from $(1,1), (0,1),(0,2)$), $(1,3)$, etc. until you get that entire column. Similarly, you can get all the $(x,1)$ in the second row as well. Repeating this, you can fill the entire square. Note that if you chose the value of $s$ for $(1,1)$, what you are really doing is assigning a linear function of $s$ for every other coordinate, where $s$ appears with coefficient $1$ or $0$. (the $0$'s correspond to entries which do not depend on $s$, which in our case corresponds to those entries with either coordinate being even; for example, the moment you know $(0,0)$ and $(0,1)$ and $(2,0)$, you also know $(2, 1)$ without having to know $(1,1)$ or even $(1,0)$ explcitly).
Combining these two observations, the classification of SSS is quite easy: any SSS comes from a unique "zeroed SSS" which corresponds to making the top $2 \times 2$ matrix all $0$. The remaining choices are the first row and column, giving $2n-4$ remaining squares. Alternatively, 

SSS's are in bijection with ordered lists of $2n$ integers; $2n-1$ integers for the choices of the first row and column, and the last integer for selecting the coordinate $(1,1)$. Any such set of values can be completed into a unique SSS.

I hope this is helpful as a first step. If I have more thoughts I'll add them. The next natural extension (although from my experiences working with things like anti-magic graphs, I would bet on this being very difficult) is to have the entries be all distinct. I suspect using the fact that they are $1$ through $N^2$ may be hard to encode in a useful way outside of simple modular arithmetic, so the real "meat" is the distinctness. But I'm happy to be proven wrong.
(for those with theoretical interest, what I'm doing in somewhat colloquial language above is linear algebra; I'm in effect considering a vector space where the vectors are $2 \times 2$ submatrices, and looking at the dimension of this vector space)<|endoftext|>
TITLE: Parametrized Dold-Kan correspondence?
QUESTION [8 upvotes]: The stable Dold-Kan correspondence says that for every commutative ring $R$, there is an equivalence of $\infty$-categories between the category $Ch(R)$ of (unbounded) chain complexes of $R$-modules and the category $Mod(HR)$ of module spectra over the Eilenberg-Maclane ring spectrum $HR$. This can be realized as a Quillen equivalence of suitable model categories (see here).
My question is whether there is a parameterized version of this equivalence. Namely, given a based space $B$, there is a way to construct the category of parametrized spectra over $B$ with a suitable model structure (see here). In this setting there is a monoidal structure given by smash product "over $B$". We therefore can construct the parametrized ring spectrum $HR\times B$ and I guess also the category of parametrized modules over it (with suitable model structure). I wonder if this is equivalent (as an $\infty$-category) to something we can construct on the level of chains. 
My guess is that it should be equivalent to something like co-modules over the co-algebra $C_* (B)$ in $Ch(R)$. Is there indeed some algebraic model like this?

REPLY [12 votes]: We Discussed this privately but for future generations let me write this here.
This is false. As an example take the $B$ to be the classifying space of an acyclic group. Then $C_*(B) = \mathbb{Z}$. So if what you write was true, every local system on such a space would have been constant. but this is false. For example consider the regular representation.<|endoftext|>
TITLE: Exchange determinant and integral of a matrix-valued function
QUESTION [11 upvotes]: Assume $A(x)=(a_{ij}(x))_{k\times k}$ is a Hermitian matrix function on some manifold $M$, is there any inequality relates the integral of its determinant $\int_M \det(A)$ and the determinant of its integral $\det(\int_M A)$? Here $\int_M A$ is the matrix obtained by integrating each entries, i.e. $\int_M A=(\int_M a_{ij}(x)dx)_{k\times k}$. Thanks.

REPLY [13 votes]: Of course, there are. Suppose $A(x)$ is semi-positive definite. Then, because of the concavity of $A\mapsto(\det A)^{1/n}$, Jenssen's Inequality gives
$$\frac1{{\rm vol}M}\int_M(\det A)^{1/n}\le\det\left(\frac1{{\rm vol}M}\int A(x)\right)^{1/n}.$$
On the other hand, if $A$ is a constant $B$ plus the Jacobian matrix of a compactly supported field $\phi$ (for instance if $M$ has no boundary), then the integral and the determinant do commutte:
$$\frac1{{\rm vol}M}\int_M\det(B+ \nabla\phi)=\det B.$$
There are variations on this theme (quasi-convexity in the sense of Morrey, polyconvexity in the sense of Ball).<|endoftext|>
TITLE: Dirichlet's approximation only using prime power as denominator
QUESTION [5 upvotes]: I am not sure whether this is a suitable question for MO. We know the classical version of Dirichlet's approximation theorem that if $x$ is a real number and $Q>0$ there exist $p,q\in \mathbb{Z}$ with $0<q\le Q$ such that $|x-p/q|<1/qQ$.
I am looking for a version of this theorem where $q$ is only a prime power, say $2^n$. Then by binary approximation of real number one can immediately say that $|x-p/2^n|<1/2^n$. But can we do better along the line of Dirichlet?
Thanks for any reference.

REPLY [5 votes]: Minor observations:
If $x$ is rational with non-prime power denominator, you certainly can't do much better. If $x = 1/6$, then $|x-p/q| \geq 1/(6q)$ for any prime power $q$. 
If $x$ is irrational, then Vinogradov showed that $\{ x q \}$ is equidistributed as $q$ runs over the primes. So, for any $\epsilon>0$, we can find a prime $q$ and an integer $p$ such that $|xq-p| < \epsilon$ or, in other words, $|x-p/q|<\epsilon/q$.
I can't decide whether or not I think, for all irrational $x$, that there should be a prime $q$ such that $|x-p/q|<c q^{-2}$. (For some constant $c$ independent of $x$.) I definitely think it should be true except for $x$ of measure zero -- heuristically, the odds that a particular $q$ works are $2c/q$, and the sum $\sum 1/q$ diverges.<|endoftext|>
TITLE: Kernel of skew-symmetric matrix of rank $n-1$ with $n$ odd: is this a known result?
QUESTION [14 upvotes]: When $n$ is odd, the kernel of a skew-symmetric matrix $M$ of size $n\times n$ and rank $n-1$ is the span of $v$, where $v$ is a vector whose $i$-th component is the Pfaffian of the matrix obtained by removing the $i$-th line and $i$-th column of $M$, multiplied by $(-1)^i$. 
Example with $n=3$:
$$\begin{bmatrix} 0 & a&b \\ -a&0&c \\ -b&-c&0\end{bmatrix} \begin{bmatrix} c\\-b\\a\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$
Example with $n=5$:
$$\begin{bmatrix}
 0 & a & b & c & d \\
 -a & 0 & e & f & g \\
 -b & -e & 0 & h & i \\
 -c & -f & -h & 0 & j \\
 -d & -g & -i & -j & 0 \\
\end{bmatrix}\begin{bmatrix} -e j+f i-g h \\ b j-c i+d h \\ -a j+c g-d f \\a i-b g+d e \\ -a h+b f-c e\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
I suppose this is a known result: would you know a reference which mentions it?

REPLY [2 votes]: There is a proceeding by D. Knuth about that. You can see the relevant formula in a) of Exercise 93 in my list of exercises on matrices. 
Let $n$ be odd and $1\le i,j\le n$ be given. Then the minor $M[\hat i|\hat j]$ equals the product of the Pfaffians ${\rm Pf}M[\hat i]$ and  ${\rm Pf}M[\hat j]$ ; I have used the notation that $\hat i$ is the complement of $i$ in the set of indices. Application: because the rank of $M$ is $n-1$, we can find a index $j$ such that  ${\rm Pf}M[\hat j]\ne0$. Then the kernel is spanned that the vector of coordinates $(-1)^i M[\hat i|\hat j]$. Simplifying by ${\rm Pf}M[\hat j]$, we obtain your claim.
I apologize that I could not find the reference of Knuth's paper. If I remember well, he claimed in it there is no determinant, there are only Pfaffians.<|endoftext|>
TITLE: Compact hyperbolic 3-manifolds with prescribed quaternion algebra, quaternion parameters as ramification condition
QUESTION [6 upvotes]: What is an interesting class of examples of hyperbolic 3-manifolds,
each of which satisfies the following conditions?
1. It is compact
2. Its trace field contains a unique imaginary quadratic extension.
3. Its quaternion algebra is isomorphic to one of the form $\Big(\frac{a,b}{K}\Big)$, where $a,b\in K\cap\mathbb{R}$.
Since the word 'interesting' is not well-defined, I'll settle for any examples, but it would be cool if they have some nice combinatorial or geometric characterization.
Then there is a follow-up question (more for algebraic number theorists).
How could I replace conditions 2 and 3 with something in terms of the algebra's ramification set? That is, there is a number field $F=K\cap\mathbb{R}$ so that the algebra is isomorphic to $\Big(\frac{a, b}{F(\sqrt{-d})}\Big)$
where $a,b\in F$, and $d\in F^+$.
But yet it is still a division algebra (else the manifold is most likely not compact).
Can this be (at least for some $F, d$ choices) phrased in terms of the divisors of $a$ and $b$?
I expect one would need to already know the definitions involved to answer this, but I will supply the arithmetic ones below for the sake of readers.
Afterall the more people who know this, the more people I have to talk to!

To a hyperbolic 3-manifold $M$
is associated a Kleinian group $\Gamma\cong\Pi_1(M)$
represented in $\mathrm{PSL}_2(\mathbb{C})$.
The trace field of $M$
is the field $\mathbb{Q}(\{\mathrm{tr}(\gamma)\mid\gamma\in\Gamma\})$,
which I'll denote by $k_0 M$.
Using the character variety and algebraic geometry, it follows that this is a number field (a finite extension of $\mathbb{Q}$),
and by Mostow rigidity it is a manifold invariant.
Using the same setup,
the quaternion algebra
of $M$
is defined as
$$\big\{\sum_{i=t}^nt_i\gamma_i\mid t_i\in k_0M,\gamma_i\in\Gamma,n\in\mathbb{N}\big\}$$
and is commonly denoted by $A_0M$.
A quaternion algebra is a 4-dimensional central-simple algebra,
and it can be proven that $A_0M$
is such a thing using the Skolem-Noether theorem.
This is a stronger manifold invariant.
These algebras (provided the field is not characteristic 2, which doesn't matter here since we're using number fields) necessarily take the following form. If $K$ is the field it is over, then the algebra looks like
$$K\oplus Ki\oplus Kj\oplus Kij$$
where $i^2=a, j^2=b, ij=-ji$, with $a,b\in K\setminus\{0\}$.
And we denote this by the Hilbert symbol $\Big(\frac{a,b}{K}\Big)$.
A property of these algebras is that they are identified up to isomorphism by ramification of their places (field embeddings and prime ideals) over $K$.
Quaternion algebras of non-compact manifolds never have ramification over their primes. Quaternion algebras of compact manifolds typically do have ramification of their primes, but there are some strange examples where they don't.

REPLY [5 votes]: Suppose that $M$ is a compact arithmetic hyperbolic $3$-manifold which is derived from a quaternion algebra. Let $k$ denote the trace field of $M$ and $B=\left(\frac{a,b}{k}\right)$ be the associated quaternion algebra. 
It is known (see Maclachlan-Reid, Theorem 8.3.2) that $k$ has a unique complex place. Noting that every proper subfield of a number field with a unique complex place is totally real, we see that the only way for $k$ to contain an imaginary quadratic field is for $k$ to actually be an imaginary quadratic field.
Furthermore, Theorem 8.2.3 of Maclachlan-Reid implies that $M$ will be compact so long as $B\neq \mathrm{M}_2(\textbf{Q}(\sqrt{-d}))$. Combining this with the observation from the previous paragraph, we see that your conditions 1 and 2 will be satisfied if and only if $B$ is a quaternion division algebra which is defined over an imaginary quadratic field. 
Now we need to incorporate your third condition, which seems to me to be the most interesting. Suppose that $M$ contains an immersed totally geodesic surface. Then the results of Section 9.5 of Maclachlan-Reid imply that there is an indefinite quaternion algebra $B'$, defined over $\textbf{Q}$, such that $B\cong B'\otimes_\textbf{Q} k$. Because $\left(\frac{a,b}{\textbf{Q}}\right)\otimes_\textbf{Q} k\cong \left(\frac{a,b}{k}\right)$, the Hilbert symbol of $B$ satisfies your condition 3.
Putting all of this together, we see that if $M$ is an arithmetic hyperbolic $3$-manifold which is derived from a quaternion division algebra $B$ defined over an imaginary quadratic field and $M$ contains an immersed totally geodesic surface then $M$ satisfies your three conditions. An arithmetic hyperbolic $3$-manifold contains one totally geodesic surface if and only if it contains infinitely many commensurability classes of totally geodesic surfaces (this also follows from the results of Maclachlan-Reid, Section 9.5), so the aforementioned class of manifolds all contain infinitely many commensurability classes of totally geodesic surfaces. Whether this makes these manifolds interesting...I can't say.
I do not know of a geometric characterization of your third condition, though would like to point out Proposition 5 of Chinburg and Reid's paper Closed hyperbolic $3$-manifolds whose closed geodesics all are simple:
Proposition: Let $M$ be an arithmetic hyperbolic $3$-manifold derived from a quaternion algebra $B=\left(\frac{a,b}{k}\right)$. If $M$ has a non-simple closed geodesic then $a,b$ can be chosen so that $a\in k$ and $b\in k\cap \textbf{R}$.
This proposition shows that if the Hilbert symbol of $B$ cannot be written in a certain form then all closed geodesics of $M$ are simple. (Providing examples of such manifolds was the point of Chinburg and Reid's paper.) It is not known whether this Hilbert symbol obstruction is the only thing which prevents $M$ from having non-simple closed geodesics. If you believe that it is the only obstruction, then you would expect any arithmetic hyperbolic $3$-manifold $M$ satisfying your conditions to have lots of non-simple closed geodesics.
Regarding your request to reinterpret condition 3 in terms of the primes which ramify in $B$, the results of Section 4 of the Chinburg-Reid paper should be very helpful and are meant to provide exactly such an interpretation.
Added: There are infinitely many hyperbolic 3-manifolds which satisfy the OP's three conditions. One such family may be obtained as follows. Let $B$ be a rational quaternion division algebra which is split at the real place of $\bf{Q}$. Let $k$ be an imaginary quadratic field which does not embed into $B$. Then $A:=B\otimes_{\textbf{Q}} k$ is a quaternion division algebra over $k$. (Here I have used the fact that a quadratic field $L$ embeds into $B$ if and only if $B\otimes_\textbf{Q} L \cong \mathrm{M}_2(L)$.) Let $\mathcal O$ be a maximal order of $A$ and $\mathcal{O}^1$ the multiplicative subgroup of $\mathcal{O}^*$ generated by elements of reduced norm $1$. Let $\psi: A\hookrightarrow \mathrm{M}_2(\textbf{C})$ be the map induced by the inclusion $A\hookrightarrow A\otimes_k \textbf{C}\cong \mathrm{M}_2(\textbf{C})$. Finally, let $\Gamma_\mathcal{O}$ denote the image in $\mathrm{PSL}_2(\textbf{C})$ of $\mathcal{O}^1$ under the map $\psi$ composed with the projection $P: \mathrm{SL}_2(\textbf{C})\rightarrow \mathrm{PSL}_2(\textbf{C})$. Then $\Gamma_\mathcal{O}$ is a discrete subgroup of $\mathrm{PSL}_2(\textbf{C})$ of finite covolume which is cocompact and has trace field $k$. Let $\Gamma$ be a finite index subgroup of $\Gamma_\mathcal{O}$ which is torsion-free and $M=\textbf{H}^3/\Gamma$ be the corresponding hyperbolic $3$-manifold. The arguments that I gave in my original response show that $M$ satisfies the three conditions of the OP's question. There are infinitely many commensurability classes of such $M$ because there are infinitely many imaginary quadratic fields $k$ which do not embed into $B$.<|endoftext|>
TITLE: what's the formula of the inradius of a general simplex?
QUESTION [5 upvotes]: As the title, I just want to know whether there is a general formula for calculating the inradius of a n-simplex. Thank you!

REPLY [3 votes]: Apparently not in complete generality; here is a a formula for a hyperbolic truncated $n$-simplex, and here for a regular simplex. There is also a formula for a triangle.<|endoftext|>
TITLE: Are compact topological $n$-manifolds recursively enumerable?
QUESTION [21 upvotes]: Earlier this year it was asked on MO, "Are there only countably many compact topological manifolds?"  Thanks to Cheeger and Kister, the answer is yes.  On the other hand, Manolescu recently debunked the triangulation conjecture.  A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.
Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$.  It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.
If the answer is no, is it known how hard the problem of enumerating manifolds is?  Is it harder than the halting problem?
Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finitely computable description of topological manifolds.  As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations.   If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known.  (I also removed the madness about reference to Turing degrees above.)

REPLY [4 votes]: In a note of Freedman and Zuddas, they show that this is true for dimensions $\geq 4$.
In the "Background" section of the paper, they describe the solution in the higher dimensional case using surgery theory, but without any references.
Then they proceed to describe the 4-dimensional case. Here they use the fact that the complement of a point in a 4-manifold is smoothable. Hence one can describe a triangulation of a finite part of the complement of a point, together with a certificate of a 3-sphere tamely embedded.
It's frustrating that they don't give any references for the higher-dimensional case, but since you're in Santa Barbara, you could probably saunter over to Station Q to get the details from Mike once campus opens again.<|endoftext|>
TITLE: What is the criterion for a matrix containing vectors and their permutations being invertible?
QUESTION [5 upvotes]: Consider the matrix $A\in\mathbb{R}^{m\times 2m}$. Let any arbitrary choice of $m$ columns of $A$ be linearly independent. Together with a permutation $P\in\mathcal{P_{2m}}$, one can build the matrix $$B:=\begin{bmatrix}A\\AP\end{bmatrix}\in\mathbb{R}^{2m\times 2m}\text{ .}$$
Which conditions does $P$ need to satisfy, so that $B$ is invertible? 
I would like to have just a criterion for the matrix P.
The rest of this post shows an example and conjectures I already have from experiments. I had a look at $4\times 4$, $6\times 6$, $8\times 8$ and $10\times 10$ matrices $A$, all permutations $P$ and the determinants of $B$.
Example
$$B_1=\begin{bmatrix}
a_1&b_1&c_1&d_1\\
a_2&b_2&c_2&d_2\\\hline
\mathbf{b_1}&\mathbf{a_1}&c_1&d_1\\
\mathbf{b_2}&\mathbf{a_2}&c_2&d_2\\
\end{bmatrix}\qquad ,\qquad
B_2=\begin{bmatrix}
a_1&b_1&c_1&d_1\\
a_2&b_2&c_2&d_2\\\hline
\mathbf{c_1}&\mathbf{a_1}&\mathbf{b_1}&d_1\\
\mathbf{c_2}&\mathbf{a_2}&\mathbf{b_2}&d_2\\
\end{bmatrix}$$
$B_1$ is always singular ($\det B_1=0$) and $B_2$ always invertible ($\det B_2\neq 0$).
Conjectures
What helps is looking at derangements. A derangement is a permutation which does not assign an element its original position. It is said that the elements are deranged.
Conjecture 1:
Matrix $B$ is invertible if at least $2m-1$ columns of $A$ are deranged by $P$.
Conjecture 2:
The matrix $B$ is regular if at least $m+1$ columns of $A$ are deranged by $P$ and the permutation consists of a cycle of at least size $m$.
Both conjectures are no "if and only if" relations. There are definitely permutations which do not fulfill the criterion but lead to invertible $B$'s. What I also saw when I was investigating small $m$'s was, that all permutations with the same number and sizes of cycles always lead to the same result. In the example $B_1$ has one cycle of size 2 and every permutation with a single cycle and size 2 leads to a singular matrix. $B_2$'s permutation has one cycle of size 3 and every such permutation lead to an invertible matrix $B_2$.
I would be happy if I could show at least conjecture 1 for all $m$. Conjecture 2 would be even better. But I don't have an idea how. Any other criterion of $P$ leading to an invertible matrix $B$ are fine as well.

REPLY [3 votes]: Sorry, these are perhaps two remarks, but they are a bit large for a comment.
Generally, you may multiply $B$ by $\begin{bmatrix}X^{-1}&0\\0&X^{-1}\end{bmatrix}$ where $X$ is the left half of $A$, thus assuming that $A=\begin{bmatrix}E&Y\end{bmatrix}$. In this case, the formula
$$
  \det\begin{bmatrix}E&Y\\Z&T\end{bmatrix}
  =\det\begin{bmatrix}E&Y\\0&T-ZY\end{bmatrix}
  =\det(T-ZY)
$$
may be helpful.
Specifically, I do not see that $B_2$ is always invertible. Putting $Y=\begin{bmatrix}-1&1\\-1&2\end{bmatrix}$ (all pairs of columns of $A$ are independent!), then you get
$$
  \det B_2=\det\begin{bmatrix}1&0&-1&1\\0&1&-1&2\\-1&1&0&1\\-1&0&1&2\end{bmatrix}=0
$$
(the formula above may help). This seems to disprove both conjectures...<|endoftext|>
TITLE: Finite sets of vectors closed under an orthogonality property
QUESTION [12 upvotes]: Say that a set $X \subset \mathbf{S}^{d-1}$ is ortho-closed if for any set $\{x_1,\dots,x_{d-1}\} \subset X$ there exists $x \in X$ such that $\langle x,x_i \rangle = 0$ for $i=1,\dots,d-1$. (Apologies for the terrible name - other suggestions are welcome.) Thus, as a silly example, the set of standard basis vectors is ortho-closed.
My rather vague question is, what (non-trivial) finite ortho-closed sets are there? Much more concretely (and actually what I'm most interested in), with $d=3$, is there a finite ortho-closed set containing the normalizations of the seven non-zero 0-1 vectors in $\mathbb{Z}^3$? (Or indeed is there a finite ortho-closed set containing an arbitrary fixed set?)
It seems to me that the answer ought to be `no', and that one ought to be able to construct an infinite sequence of vectors that would have to be in any such set. However I can't yet see how to do this, and it occurred to me that this is a fairly natural property which might already be known. Thus, any references would also be greatly appreciated.

REPLY [8 votes]: You can get a nice combinatorial proof of Ilya's result if you use the Friendship Theorem.
As Ilya did, we can identify opposite vectors. Your condition says that for any two distinct vectors $v_1$ and $v_2$ of $X$, there exists a vector $v \in X$ which is orthogonal to both $v_1$ and $v_2$. However, since you are in $\mathbb{R}^3$, the vector $v$ must be unique.
Now let us consider the vectors in our set $X$ to be vertices of a graph $G$, such that two vectors are adjacent if they are orthogonal. Then, by the above, this graph has the property that every pair of vertices has a unique common neighbor. The Friendship Theorem says that any such graph is a windmill graph, meaning that $G$ consists of a vertex $v$ adjacent to all other vertices of $G$, and the remaining vertices form a perfect matching.
Translating the orthogonalities encoded in the edges of $G$ back to the vector setting gives you exactly Ilya's result.<|endoftext|>
TITLE: Limiting shape for Brillouin zones
QUESTION [13 upvotes]: Is it true that the limiting shape for Brillouin zones (for any lattice) is a circle?
You can find the definition and the step by step construction of Brillouin zones here. This picture is taken from the Graphics Gallery for The Mathematica GuideBook for Graphics.
It is known that all Brillouin zones have the same area, so if we have a picture with $n$ zones and we want to see it in a unit square then it is necessary to multiply it by $1/\sqrt{n}$. But they could be close to the circle of radius $c\sqrt{n}$ even without rescaling.
edit by j.c.: 
This question is stated in 2D but makes sense in arbitrary dimensions.

Suppose we have a lattice $\Gamma$ in $d$-dimensions (i.e. a discrete subgroup of $\mathbb{R}^d$ that spans $\mathbb{R}^d$) (in crystallography terms, $\Gamma$ is the reciprocal lattice of some Bravais lattice).  A Bragg plane of $\Gamma$ is a hyperplane that perpendicularly bisects a line segment in $\mathbb{R}^d$ between the origin and any other element of $\Gamma$. Then the $n$th Brillouin zone of $\Gamma$ is defined to be the closure of the set of points $P$ in $\mathbb{R}^d$ such that the line segment between $P$ and the origin intersects exactly $n-1$ Bragg planes.

REPLY [11 votes]: Yes, it is. Take a point P, and let us check if it belongs to Nth Brillouin zone. The Bragg planes (rather, lines, as we are in $R^2$) that we have to cross while going from the origin $O$ to $P$, correspond to the points $L$ of the lattice $\Gamma$ such that $P$ is closer to $L$ than to $O$.
In other words, we are looking for the number of the points $L$ of the lattice $\Gamma$ that belong to the $P$-centered ball of radius $|OP|$. 
This number is approximately equal to $\pi \cdot |OP|^2/\mathop{\mathrm{covol}} \Gamma$. Hence, $N$th Brillouin zone is very close to a circle of radius $\sqrt{N/\pi \cdot \mathop{\mathrm{covol}} \Gamma}$.<|endoftext|>
TITLE: A kind of anti-Ramsey result
QUESTION [6 upvotes]: In contrast to classic results for arithmetic progressions of arbitrary length in one set at least of any finite partition of $\mathbb N$, it is easy to construct a partition in two sets of integers $A$ and $B$ such that neither $A$ nor $B$ contains an infinite arithmetic sequence. Actually (with choice, of course), one has a much stronger result : if $(A_i)_{i\in I}$ is a family of subsets of $E$ with $\#A_i=\#I=\#E$ (infinite), there exists a partition of $E$, $E=F\cup G$, such that for all $i\in I$, $A_i$ is not included in $F$ nor in $G$ (the not very hard proof using a diagonal argument must be well-known). Is this kind of anti-Ramsey result part of some more general theory?

REPLY [5 votes]: According to P. Erdõs, A. Hajnal: On a property of families of sets, Acta Math. Acad. Sci. Hungar. 12 (1961), 87--123 (see page 90)  the stronger result you formulated is a theorem of Bernstein from F. Bernstein, "Zur Theorie der trigonometrischen Reihen", Leipz. Bet. (Berichte über die Verhandlungen der Königl. Sächsischen Gesellschaft der Wissenschaften zu Leipzig. Math.phys. Klasse) 60 (1908), 325--338" MR, eudml.   If you are looking for further results, "property B" is the keyword you need. 
I presented a minicourse on some related properties of families of sets on the 7th Young Set Theory Workshop in 2014, so  my slides may contain  some results you are interested in.<|endoftext|>
TITLE: Looking for a modern source about Ulm Invariants
QUESTION [7 upvotes]: I'm looking for a modern, approachable text (preferably a website, textbook, or expository article, and preferably one easily available online or at a library) which can explain the concept of Ulm invariants and how they factor into the classification of (countable) abelian groups.
The theorem I'm trying to understand is the following, apparently due to Mackey and Kaplansky:

A countable periodic reduced abelian group is determined uniquely up to isomorphism by its Ulm invariants for all prime numbers p and countable ordinals α.

Google covered the definitions of periodic (all elements of finite order; with abelian this is equivalent to locally finite) and reduced (no divisible elements but zero), but the definition of Ulm invariant is still a bit dodgy.  The proof itself is not summarized, so it's unclear to what extent the following questions are answered or unanswered:

What if the group is not reduced?  Do we have a characterization of (countable) divisible periodic abelian groups?

(apparently the divisible part is always a direct summand, so it's enough to do each part separately)

What if the group is not periodic?  Can we use (e.g.) the finitely generated abelian groups theorem to sort of separate out the $\mathbb Z$-pieces?
What if the group is not countable?  Can we see it canonically as made up of countable pieces, and assemble them?

I should say that I'm a logician, not an algebraist, so I don't know the "standard sources" in this field.  But I'm looking to learn.

REPLY [7 votes]: A standard reference for abelian groups is Laszlo Fuch's books Infinite Abelian Groups. The Ulm-Kaplansky Theorem, characterising countable $p$-groups, is Theorem 77.3 on page 63 of Volume II. The Ulm sequence and Ulm type is defined on page 57; I'm only browsing it on-line, but it looks like it contains all relevant proofs.
That every abelian group can be written as $A=D\oplus R$ with $D$ divisible and $R$ reduced is fairly standard; it can be found in Rotman's Introduction to the Theory of Groups 4th Edition, Theorem 10.26, page 322. The fact that every divisible group is a direct sum of copies of $\mathbb{Q}$ and copies of Prüfer $p$-groups is Theorem 10.28, page 323. (If $D$ is divisible, then $D\cong \mathrm{tor}(D)\oplus V$, with $V$ torsion free and divisible; this means that $V$ is a vector space over $\mathbb{Q}$, hence a direct sum of copies of $\mathbb{Q}$. The $p$-primary component of  $\mathrm{tor}(G)$ is divisible, and if you look at the subgroup of elements of exponent $p$ you get a vector space over $\mathbf{F}_p$,  and that gives you the copies of the Prüfer group). The uniqueness of the decomposition is only stated and not proven, but it follows from Linear Algebra by considering suitable subgroups that determine the cardinality of the number of direct summands of the appropriate groups.
Rotman also has some results on torsion groups, and states (but does not prove) Ulm's Theorem. A couple of exercises show that the characterization via Ulm invariants does not extend to uncountable groups. The example suggested in the exercises is to consider $\mathrm{tor}(\prod_{n=1}^{\infty}\mathbf{Z}_{p^n})$ and $\oplus_{n=1}^{\infty}\mathbf{Z}_{p^n}$; the former is not a direct sum of cyclic groups. 
In general you cannot separate a non-finitely generated mixed abelian group into "copies of $\mathbb{Z}$" and the rest in a satisfactory manner. Chapter XIV in volume II of Fuch's book discusses some results on mixed abelian groups.<|endoftext|>
TITLE: Noninvariance for a specific nonlinear oscillator
QUESTION [5 upvotes]: Consider the nonlinear system
\begin{align*}
  \frac{d}{dt} \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix}  = \begin{pmatrix} x_2(t) \\ -4x_1(t) + x_1^2(t) \end{pmatrix},
\end{align*}
which admits periodic orbits near the origin. Consider two initial states $x(0) = x_0$ and $z(0)=z_0$ where $x_0$ and $z_0$ are nonzero and satisfy $x_2(0) = - z_2(0)$. We can think of the situation as putting two particles in the vector field and watch them evolve.
I am trying to show that for any combination of initial states satisfying the above requirements, at some point in time it must happend that $x_2(t) + z_2(t) \ne 0$. This is due to the specific nonlinearity of the periodic orbits. But it seems to be very difficult to show by just looking at the equations
\begin{align*}
  x_2(t) + z_2(t) \equiv 0, \frac{d}{dt} (x_2(t) + z_2(t)) \equiv 0, \dots
\end{align*}
So I was hoping that there might be a more succinct argument involving the geometry of the periodic orbits. Any hints in the direction of a simpler argument are very welcome.

REPLY [2 votes]: As we're interested in two particles, it seems natural to introduce the coordinates $y = x + z$ and $w = x - z$. In these coordinates, we obtain the system
\begin{align}
\dot{y}_1 &= y_2,\\
\dot{y}_2 &= -4 y_1+\frac{1}{2}y_1^2+\frac{1}{2}w_1^2,\\
\dot{w}_1 &= w_2,\\
\dot{w}_2 &= -4 w_1 + w_1 y_1.
\end{align}
If I understand you correctly, the question is if, given an initial condition of the form $(y_1,y_2,w_1,w_2) = (a,0,c,d)$, there will be a time $t$ when $y_2(t) \neq 0$. The negation of this statement is equivalent with stating that $y_2(t) = 0$ for all time $t$, i.e. that the set $(a,0,c,d)$ is invariant under the above flow. However, for such an initial condition, we see that $\dot{y}_2(0) = -4a+\frac{1}{2}a^2 + \frac{1}{2}c^2 \neq 0$ for general $a,c$. Hence, the vector field at $(a,0,c,d)$ is not parallel to $(a,0,c,d)$, so the set $(a,0,c,d)$ is not invariant under the flow. Hence, there exists a time $t>0$ such that $y_2(t) \neq 0$, even when $y_2(0) = 0$.<|endoftext|>
TITLE: Weights on cyclic orderings
QUESTION [5 upvotes]: Are there standard or known weights/metrics on cyclic orders?
Cyclic orderings are different ways of listing elements from a finite set, where you call two lists the same if they differ only by a rotation.  For example, the ways that 4 people can sit at a circular table, where you ignore rotation of the table.
So the following 4 rankings of a 4-item set should be considered the same:
[2,1,4,3]
[1,4,3,2]
[4,3,2,1]
[3,2,1,4]

You can think of this as arranging the 4 items on a circular list, with no particular top, but a definite direction.
I would like to compare different cyclic orderings in a meaningful way.  I have learned that there are many well-known ways of assigning distance or weight to non-cyclic orderings (a.k.a. permutations!).  For example:

Kendall $\tau$
Transposition distance (Cayley)
Ulam metric
Hamming weight
$\ell_1$ norm

But these give different weights to permutations which correspond to the same cyclic ordering.

A first simple idea is the following: given a permutation $a$, count the number of $i$ such that $a(i+1) \neq a(i) +1$, where $0 \leq i \leq n$ and index addition is performed modulo $n$.
A natural variation on this idea would sum $|a(i+1) - (a(i)+1)|$.
I haven't thought about either of these too carefully yet.  Do they appear in the literature somewhere?  Some nontrivial internet searching hasn't turned anything up.

REPLY [2 votes]: cyclic permutations, and a need to define distances on them, pops up in literature on graph crossing numbers, e.g. we used it in http://arxiv.org/abs/math/0404142 
It probably goes back all the way to at least 
D.J. Kleitman, The crossing number of $K_5$. J. Combinatorial Theory 9 (1970), 315–323.<|endoftext|>
TITLE: Is the game Hanabi NEXPTIME-complete?
QUESTION [16 upvotes]: The game Hanabi is a cooperative, hidden-information game. You can read the rules elsewhere, but broadly speaking the players are attempting to cooperatively build a fireworks display by playing cards from their hand. The goal is to get as high a score as possible, by playing as many valid fireworks card as possible. The catch is that you hold your cards facing away from you, so without information from other players you are just blindly guessing which card to play. On your turn, you can either

Play a card (without looking at it). If it is not valid, there is a penalty.
Give a clue to another player. This uses up a clue token, and the clues are of a restricted form.
Discard a card. This gets a clue token back.

My question is: is the game NEXPTIME-complete? In order to put it into the framework of a formal game, we should think of it as a game with two sides: all the players (collectively), and Nature, who shuffled the deck and deals out the cards when players draw new cards. This then fits it into the context of a team game with hidden information, in the language of Hearn and Demaine's book Games, Puzzles, and Computation. More precisely, in their language, it is a bounded team game with hidden information. According to the handy chart in their book, a "generic" such game is expected to be NEXPTIME-complete:
$$
\begin{array}{rcccc}
 & \mathrm{0\ player} & \mathrm{1\ player} & \mathrm{2\ player} & \mathrm{Team} \\
\mathrm{Bounded} & \mathrm{P} & \mathrm{NP} & \mathrm{PSPACE} & \mathrm{NEXPTIME} \\
\mathrm{Unbounded} & \mathrm{PSPACE} & \mathrm{PSPACE} & \mathrm{EXPTIME} & \mathrm{RE\ (undecidable)}
\end{array}
$$
For everything that isn't in the "Team" column, many examples of real games of the given complexity class are known. But as far as I know, no games of the "generic" complexity class are know in the team case. Hanabi strikes me as a good candidate, as it has (a) a fairly simple rule set but (b) in practice, requires quite complicated deductions about other player's states of knowledge.
Of course, to make this a precise question, the game needs to be suitably generalized, increasing the number of colors and/or cards of a given color. A precise problem also has to be given: probably the problem of "given this initial play state, common knowledge to all players, do the players have a strategy that guarantees a win?"
One complication is going to be that the optimal strategy will almost certainly be quite artificial, of the form of giving a clue that is a hash function of all the cards that you can see. Maybe there's some way to vary the rules to forbid such conventions. (Perhaps just restricting to the two player case.)
Has anyone thought about this problem?

REPLY [4 votes]: There's an article at https://arxiv.org/pdf/1603.01911.pdf showing that a cheating solitaire variant is NP-complete.  It's not very explicit about the membership of NP, but here's an argument that works: a witness is a sequence of moves; whether a sequence wins can be computed in linear time in the deck size (it's O(1) to make each update to the game state; it's linear time in the number of colors to check that the maxnum in each color has beed player, and there are no more colors than cards).
At https://sites.google.com/site/rmgpgrwc/research-papers/Hanabi_final.pdf?attredirects=0 there's an article with an idea for how to clue efficiently: map each possible clue you can give onto $\{1, \ldots, n\}$.  Then, for each player, determine based on common knowledge in the current game state (including past clues, plays and discards) which card is most likely to be playable, and which suit/rank combination that card can possibly be; divide the set of possibilities into $n$ subsets (in some publicly known fashion), and assign to each player $p$ the number $k_p$ of the subset that contains the actual suit/rank combination.  Give the hint which encodes as the sum of all the $k_p$ values, modulo $n$.  Each player $p$ can then look at all other hands and compute $k_p = clue - \sum_{i \ne p} k_i\ (mod\ n)$.
It's not the most interesting hash function ever, but I think it qualifies as one.  An expansion on this idea: once enough cards have been played, the number of non-dead rank/suit combinations is at most $\sqrt n - 1$, so you can clue each player two cards independently. Or, before this point, if the set of possibilities for a particular card is very narrow, you can divide the possibility set into $m$ singleton subsets and say one out of $n/m$ things about the next likely card .
Note that with at least 4 players and 0-card clues allowed, $n \ge 30$ and there are at most 30 different rank/suit combinations across the standard variants, so this allows the perfect identification of one card (perhaps with a tiny bit of side information in the 5-player case or 6 clue colors).
Since the information channel is very narrow and the set of possibilities is large, I'm not convinced that using a more complicated hash function (say, poly-31 or some very weak cryptographic hash) is a big gain, but every little win is still a win.<|endoftext|>
TITLE: Self-avoiding/reflecting geodesics on a convex surface
QUESTION [6 upvotes]: Let $S$ be the surface of a convex body embedded in $\mathbb{R}^3$.
For me $S$ is a convex polyhedron,
but I am happy to view $S$ as a smooth body with positive Gaussian curvature
at each point, or with non-negative Gaussian curvature
at each point. Likely these various versions do not affect the question I am posing. Maybe even convexity is not required.
Let $x \in S$ be an arbitrary point on $S$, and $u$ an arbitrary direction
vector tangent to $S$ at $x$.
Shoot off a geodesic $\gamma$ from $x$ in direction $u$, and let it proceed
until it intersects its own path at some point $y$.
Rather than let the geodesic cross itself at $y$, have it instead
reflect from $\gamma$ like a mirror: angle of incidence $=$ angle of reflection.
And continue: every time $\gamma$ would cross itself, instead it reflects.
Call this $\gamma$ a reflecting geodesic.
("Self-avoiding" is eye-catching but inaccurate.)

Q. For generic $x$ and $u$, is it true that for "most" surfaces $S$,
  a reflecting geodesic $\gamma$, emanating from $x$ in direction $u$,
  converges to a point?

I would prefer not to attempt to define precisely what "most surfaces" means,
but in the polyhedral world, it would suffice for $S$ to be the convex hull
of random points in space. (Vertices would be hit with probability zero.)
I know that, without the "generic" qualifier, a reflecting geodesic
might get caught in an infinite loop.
For example, on a cube, $\gamma$ intersects itself at $90^\circ$
(essentially: because of the Gauss-Bonnet theorem):

          


   

A geodesic starting at the center
of the left-front face
eventually
intersects itself at $90^\circ$ on the bottom face.


These reflecting geodesics may seem contrived,
but something close to these came up in my research,
and I am hoping that an answer to Q might help.

REPLY [4 votes]: It is true.
If the set of limit points of your trajectory is not a single point 
it has to be a (limit set of) simple closed geodesics.
On the other hand, from Gauss--Bonnet formula, it follows that most of convex closed polyhedral surfaces do not admit simple closed geodesic.<|endoftext|>
TITLE: Covering of a surface of a cube $n\times n \times n$ by pieces of paper $1\times 6$
QUESTION [23 upvotes]: When I was too young one of my problems was in the list of problems of All-Russian Olympiad. The problem is the following:
Problem. We have a surface of a cube $n\times n \times n$ such that each face is divided in $n^2$ unit squares. Then this surface was covered (once) by $3n^2$ pieces of paper $1\times 2$ (domino) without overlapping.
Prove that if $n$ is odd than the number of bent "dominos" is also odd.
Comment. This problem is not very difficult therefore it was solved by 90% participants in 10th grade (about 63 of 70). The solution you can find here.
You really need it?)

"Hm..." - I thought - "What will be with $1\times 6$?".
It was obvious for myself that it is impossible to cover the surface for $n=1$, but possible for $n=2, 3, 4$... $n=6$...
Aga, if $n$ is even than we can divide faces in two groups (in each group $3$ faces) in such way that faces in one group forms on surface of the cube rectangle $3n\times n$ with $2$ "bends". Each rectangle can be easily divided in pieces of paper $1\times 6$.
The case when $3$ is a divisor of $n$ I am leaving for a reader.
After that I conjectured the following.

Conjecture. For $n$ such that $(n, 6)=1$ there is no such covering of the surface of the cube $n\times n \times n$ by $n^2$ pieces of paper $1\times 6$ without overlapping.

Then I prove that it is true for $n=5$ and some my friends found a more or less suitable proof but it does not work in general case (they understood that vertices of this cube are weak points in this problem).

Finally, this conjecture was disproved (for $n=11$) by Peter Mueller(see his answer below)! 
Also I like to add one problem (Russian olympiad folklore):
Problem. Let $n$ be such natural number that $(n,3)=1$, then there is no covering of three faces of a cube $n\times n \times n$ with a common vertex (of this cube) by strips $1\times 3$.

REPLY [33 votes]: Indeed, there is no solution for $n=5$, and also none for $n=7$. However, for $n=11$ there is a tiling of the requested form. I found it using a straightforward exact cover formulation and Knuth's original exact cover solver. I'm not sure how to best visualize a solution, here is an attempt using random colors for the tiles. It is a little hard to check that all tiles continue correctly across the cut edges of the cube:

Added later: Fedja suggested in his comment to more generally look at $1\times 6$ tilings of a $m\times n\times k$ cuboid, for it is easy to see that such a tiling yields a tiling of a $(m+6)\times n\times k$ cuboid.
Indeed, the covering of the $11\times11\times11$ cuboid can be obtained from the way faster to compute covering of the $5\times11\times11$ cuboid:

As to the $13\times13\times13$ case, the programs (Knuth's cover and alternatively integer linear program solvers) are still running. There are no solutions for $13\times7\times7$ and $13\times13\times1$. I don't know yet about $13\times13\times7$ and $13\times13\times13$.<|endoftext|>
TITLE: Is there a Galois theory for $\mathbb R_{\geq 0}$?
QUESTION [15 upvotes]: The broadest version of my question is the following:

Where can I find algebrogeometric abstract nonsense that handles "rings" and "fields" like $\mathbb R_{\geq 0}$ in which there is no subtraction?

The reason I think that such a theory might have been developed is that I know that such "rings" appear in algebraic geometry: in tropical geometry, in the study of semialgebraic sets, and so on.
Some specific questions that I hope such literature might answer (but probably this is too optimistic):

Is there an "etale site" over $\mathbb R_{\geq 0}$?
What is the "etale homotopy type" or "Galois group" of $\mathbb R_{\geq 0}$?

I'm under the vague impression that $\mathbb R_{\geq 0}$ is an approximation of the absolute field $\mathbb F_1$.  I also speculate that $\mathbb R_{\geq 0} \hookrightarrow \mathbb R$ is "etale" but not a cover, so that $\mathbb R_{\geq 0}$ has multiple components in etale topology.  This is in spite of the fact that $\mathbb R_{\geq 0}$ seems like a "field".

REPLY [10 votes]: The question seems to be about algebraic geometry of commutative semirings (these are rings without subtraction).
The theory by Toen-Vaquié (and others) in "Au-dessous de $Spec \mathbb{Z}$" develops (functorial) algebraic geometry relative to any bicomplete closed symmetric monoidal category. The usual case is $\mathcal{V}=(\mathsf{Ab},\otimes)$, since then $\mathsf{CommMon}(\mathcal{V}) \cong \mathsf{CommRing}$. Now take $\mathcal{V}=(\mathsf{CommMon},\otimes)$ (the tensor product is defined and constructed in the same fashion as the one of abelian groups), then $\mathsf{CommMon}(\mathcal{V})\cong \mathsf{CommSemiRing}$. The category of schemes relative to $(\mathsf{CommMon},\otimes)$ is called the category of $\mathbb{N}$-schemes by Toen-Vaquié (since $(\mathbb{N},+,\cdot)$ is the initial commutative semiring). They also compare this category to the category of $\mathbb{F}_1$-schemes (which corresponds to $\mathcal{V}=(\mathsf{Set},\times)$, or rather $\mathcal{V}=(\mathsf{Set}_*,\wedge)$).
Alternatively, we may view $\mathbb{N}$ as a generalized commutative ring (commutative algebraic monad on $\mathsf{Set}$) and hence use Durov's notion of a generalized scheme ("New Approach to Arakelov Geometry"), which has a more geometric flavor; these are locally ringed spaces which are locally isomorphic to affine schemes, which in turn consist of prime ideals and a structure sheaf defined by localizations. Thus, the construction of the category of $\mathbb{N}$-schemes is very similar to the construction of the category of $\mathbb{Z}$-schemes, except that we ignore the non-existing subtraction.
A commutative semiring is simple (i.e. has exactly two quotients) if and only if it is non-trivial and every non-zero element is invertible with respect to multiplication; such objects are sometimes called semifields. Thus, $\mathbb{R}_{\geq 0}$ is an example of a semifield.
I don't know how much has been done on étale morphisms and Galois theory in relative algebraic geometry, specifically for $\mathbb{N}$-schemes (hopefully, others can write something about this), but the notion of smooth morphisms is the main topic in Florian Marty's thesis "Des Ouverts Zariski et des morphisms formalement lisse en géométrie relative", and in the introduction he hints at a definition of étale morphisms as smooth morphisms with a smooth diagonal.
Here is another idea (which should work at least for $\mathbb{N}$-schemes): The étale morphisms are precisely the flat unramified morphisms. The notion of flatness is defined as usual for affine relative schemes and then by glueing to arbitrary relative schemes. Unramifiedness may be characterized as being locally of finite presentation, which has a common functorial characterization, and the vanishing of the quasi-coherent sheaf of differentials, which may be constructed as usual via glueing. 
I don't know anything specifically about the étale $\mathbb{R}_{\geq 0}$-schemes, though. My hope is that my "answer" bumps your question and invites others go give proper answers, which I would enjoy to read, too.
Added: $\Omega^1_{\mathbb{R}/\mathbb{R}_{\geq 0}}$ is indeed trivial. Since we have the commutative semiring presentation$$\mathbb{R} \cong \mathbb{R}_{\geq 0}[X]/\langle X^2=1,X+1=0\rangle,$$we obtain the $\mathbb{R}$-semimodule presentation
$$\Omega^1_{\mathbb{R}/\mathbb{R}_{\geq 0}} \cong \mathbb{R} \cdot d(X) / \langle X \cdot d(X)=0,d(X)+0=0 \rangle = 0.$$
However, $\mathbb{R}_{\geq 0} \to \mathbb{R}$ is not flat: Notice that $\mathbb{R}$ is the $\mathbb{R}_{\geq 0}$-semimodule freely generated by $1$ and $-1$ modulo the relation $1+(-1)=0$. Thus, if $M$ is some $\mathbb{R}_{\geq 0}$-semimodule, then $M \otimes_{\mathbb{R}_{\geq 0}} \mathbb{R}$ is the quotient of $M \oplus M$ by the congruence generated by $\{(m,m) : m \in M\}$. This is exactly the Grothendieck group $G(M)$ with the induced $\mathbb{R}$-action. Now if $M \to N$ is an injective homomorphism of $\mathbb{R}_{\geq 0}$-semimodules, it does not follow that $G(M) \to G(N)$ is an injective homomorphism of $\mathbb{R}$-modules. For example, let $M=\mathbb{R}_{\geq 0}$ and $N$ be the $\mathbb{R}_{\geq 0}$-semimodule freely generated by two elements $x,y$ with $x+y=y$. Then $M \to N$, $1 \mapsto x$ is injective, but $G(M) \to G(N)$ vanishes.<|endoftext|>
TITLE: Computing millions of coefficients of non self-dual modular forms
QUESTION [10 upvotes]: To test some conjectures made by some colleagues, I need to compute millions of coefficients of non self-dual modular forms, preferably in low weight (say 2 or 3).  A form such as this.
For elliptic curves over $\mathbb{Q}$ I can compute millions of coefficients and for newforms given as eta products I can, as well.  One example of a non self-dual form we can do is to compute a self-dual form and twist by a character, but we are looking for forms that don't arise in the way.
In Sage and MAGMA I can only compute hundreds of thousands of coefficients of the form linked to above.

REPLY [10 votes]: One shortcut you could use for computing the level 17 form you link to would be the following. There are exactly 8 Eisenstein series of weight 1 for $\Gamma_1(17)$ and they are all given by completely explicit $q$-expansion formulae (involving sums of divisors, etc). The pairwise products of these series span the space of forms of weight 2. So if $F$ is the form from your link, you can find some finite collection of weight 1 Eisenstein series $f_i, g_i$ and constants $\lambda_i$ such that 
$$ F = \sum_i \lambda_i f_i g_i .$$
So if you can compute the $f_i$ and $g_i$ up to the $q^N$ term, which just amounts to computing the divisors of the integers up to $N$ (about 30 seconds in Sage for $N = 10^6$ on my machine), and then evaluate the pairwise products to the same precision (again, this should take a couple of minutes at most in that sort of range), then this will allow you to compute $F$ up to precision $N$.
EDIT. I implemented this algorithm (in Sage), mostly out of curiosity to see how it would perform. See here for the code. It turns out that it doesn't perform too badly: it took 38 minutes to compute the first million coefficients. As an illustration, if $p = 1,000,003$ is the next prime after 1 million, then 
$$ a_p(F) = 277 \zeta_8^3 + 277 \zeta_8^2 - 109 \zeta_8 + 109. $$<|endoftext|>
TITLE: Optimal exponent in the Lojasiewicz-Simon gradient inequality
QUESTION [10 upvotes]: Lojasiewicz's theorem asserts that if $F: \mathbb{R}^n\to \mathbb{R}$ is a real-analytic function in a neighborhood of its critical point $0$, then there exist constants $\theta\in (0,1/2]$, $\gamma\ge2$ and $\delta>0$ such that
$$(1)~~~\operatorname{dist}(x,V)^{\gamma}\leq \|\nabla F(x)\|$$
$$(2)~~~|F(x)-F(0)|^{1-\theta}\leq \|\nabla F(x)\|$$
for all $x$ with $\|x\|\le \delta.$ Here $V=\{y\in\mathbb{R}^n:\|\nabla F(y)\|=0 \}.$ The constants $\theta,\gamma$ and $\delta$ depend on the functional $F$ at $0.$
An important generalization by L. Simon to a class of analytic functionals on certain Holder spaces is given here in Theorem 3, which has become known as the Lojasiewicz-Simon gradient inequality.
Simon's work:
Consider the energy functional $\mathcal{E}(u)=\int_{\Sigma}E(x,u,\nabla u)$ on a $C^{\infty}$ Riemannian manifold $\Sigma$, where $E$ is assumed to have analytics dependence on $u,\nabla u$ for $|u|_{C^1(\Sigma)}$ sufficeintly small, and where $\mathcal{M}(0)=0$ ($\mathcal{E}$ has some more properties by I will ignore them here). Here $\mathcal{M}(u)=-(\operatorname{grad}\mathcal{E}(u))$ is the Euler-Lagrange operator for $\mathcal{E};$ that is, the unique function with the following property
$$-(\mathcal{M}(u),\xi)_{L^{2}(\Sigma)}=\frac{d}{ds}\mathcal{E}(u+s\xi)\Big|_{s=0}$$
for all $u,\xi\in C^{2}(\Sigma).$
Theorem 3 There are constants $\theta\in(0,\frac{1}{2}]$, $\gamma\ge 2$ and $\eta,\sigma,\beta$ such that if $u$ is an arbitrary function in $C^{2,\eta}(\Sigma)$ with $\|u\|_{C^{2,\mu}(\Sigma)}<\sigma$ then
$$(1)~~~\inf_{\{\xi\in C^{\Sigma}: |\xi\|_{C^2(\Sigma)}<\beta,~ \mathcal{M}(\xi)=0\}}\|u-\xi\|_{L^2(\Sigma)}^{\gamma}\leq \|\mathcal{M}(u)\|_{L^2(\Sigma)}.$$
$$(2)~~~|\mathcal{E}(u)-\mathcal{E}(0)|^{1-\theta}\leq \|\mathcal{M}(u)\|_{L^{2}(\Sigma)}.$$
My Question:
Under what conditions $\gamma=2$ and $\theta=1/2?$
I would like to know a criterion that also works for the
gradient when it is a fully nonlinear elliptic operator (rather than a quasilinear one).

REPLY [6 votes]: For Inequality (2) in Simon's Theorem 3, a discussion of when the optimal exponent, $\theta=1/2$, is attained can be found (along with many references) in my paper Lojasieicz-Simon gradient inequalities for analytic and Morse-Bott functionals on Banach spaces and applications to harmonic maps with Manos Maridakis. Briefly, there are essentially three (not necessarily distinct) situations where one has the optimal exponent. The first (and simplest) situation is illustrated by Proposition 1.2 in our paper. The second situation occurs when the potential function $\mathcal{E}$ obeys a type of Morse-Bott condition at the critical point and is illustrated by Theorem 2 in our paper. The third situation occurs when the critical point is `integrable'. The recent paper Slowly converging Yamabe flows by Carlotto, Chodosh, and Rubinstein contains a very thorough analysis and discussion of the integrable case in the setting of Yamabe scalar curvature gradient flow.<|endoftext|>
TITLE: Almost idempotent approximate units in C*-algebras
QUESTION [13 upvotes]: As in Blackadar's "Operator Algebras" Definition II.4.1.1., call an approximate unit $(a_\lambda)$ in the positive unit ball of a C*-algebra almost idempotent if $a_\lambda a_\gamma=a_\lambda$ whenever $\lambda<\gamma$.

Does every C*-algebra have an almost idempotent approximate unit?

REPLY [6 votes]: Akemann has constructed a C*-algebra that does not contain an approximate unit of commuting elements. See Example 2.1 in
Akemann. Approximate units and maximal abelian C*-subalgebras. Pacific J. Math. 33 (1970)
As pointed out by Tristan, a C*-algebra might have no approximate unit of commuting elements but nevertheless have an almost idempotent approximate unit. (Thanks for pointing out the mistake in my original answer.) So the original question remains open.<|endoftext|>
TITLE: Stratification of complex algebraic varieties
QUESTION [6 upvotes]: Let $V$ be a complex  quasi-projective variety, we know from H. Whitney's and B Teissier works on stratifications of algebraic varieties that $V$ has an intrinsic stratification
$$X_0\subset X_2\subset\ldots\subset X_{2n}$$
made out by complex quasi-projective smooth varieties that satisfy Whitney conditions $a$ and $b$.
This stratification induces a structure of topological stratified pseudomanifold on $V$ meaning in particular that any point $x\in X_i$ has a conical chart, a stratified homeomorphism:
$$\phi:U_x\rightarrow V_x\times cL$$
where $L$ is a topological stratified pseudomanifold of dimension $2n-i-1$, $cL$ is the cone on $L$, $V_x$ is an open neighbourhood of $x\in X_i$ and $U_x$ is an open neighbourhood of $x\in V$.  
A priori $\phi$ is just a stratified homeomorphism (for a proof one can look at N. A'Campo survey in Armand Borel et al. lecture notes "Intersection cohomology" chapter IV Birkhauser), but we can ask wether if this intrinsic stratification gives us a Piecewise Linear stratification.
In 1984 A'Campo explains that this question is open for analytic spaces, I was wondering if an answer is known in the case of complex algebraic varieties.

REPLY [4 votes]: So i turned my comment into an answer after reading [1] again.
A Whitney stratification, i.e. a stratification satisfying Whitney's condition b (and so automaticly a), induces a triangulation compatible with the stratification. Especially this gives a PL-stratification compatible with the Whitney stratification:
See the comment: " By [Gor78], a Whitney stratified pseudomanifold X can be triangulated so that the Whitney stratification defines a PL stratification of X." in [2]
[1] Goresky, Mark; Triangulation of stratified objects. (Proc. Amer. Math. Soc. 72 (1978), 193-200)
[2] Banagl, Markus; Topology of stratified spaces.<|endoftext|>
TITLE: Graded Hopf algebras and H-spaces
QUESTION [6 upvotes]: Let $k$ denote an algebraically closed field of characteristic $0$. Suppose  $K=\bigoplus_{i\geq 0}K(i)$ is a Hopf $k$-algebra which admits a connected Hopf-grading (that is, a grading which is both an algebra and coalgebra grading, with $K(0)=k$). Call such a Hopf algebra a connected Hopf-graded Hopf algebra.
Connected Hopf-graded Hopf $k$-algebras arise naturally in algebraic toplogy when studying the cohomology rings (with coeffecients in $k$) of $H$-spaces. I assume (although I'm not 100% sure), that not all such Hopf algebras arise as cohomology algebras in this way. My question is therefore the following:
Let $K$ be a connected Hopf-graded Hopf algebra. Which additional properties on $K$ guarantee that it can be viewed as the cohomology ring $K^{*}(X;k)$ of some $H$-space $X$?

REPLY [6 votes]: Any Hopf algebra of the form $H^{\bullet}(X, k)$ is necessarily graded commutative, and in addition to the conditions you've given so far, the only remaining condition is a mild cardinality condition. (You do not need to assume that $k$ is algebraically closed, only that it has characteristic zero.) 
Any such Hopf algebra $K^{\bullet}$ is isomorphic, as an algebra, to the symmetric algebra (in the graded sense) of some graded vector space $V = \bigoplus_{n \ge 1} V_n$ over $k$. If this graded vector space has a graded predual $W = \bigoplus_{n \ge 1} W_n$ (so that $V_n \cong W_n^{\ast}$), which in particular is the case if each $V_n$ is finite-dimensional, then this is the cohomology of the product of Eilenberg-MacLane spaces 
$$X = \prod_{n \ge 1} K(W_n, n)$$ 
and now the comultiplication $\psi : K^{\bullet} \to K^{\bullet} \otimes K^{\bullet}$ induces an H-space structure $X \times X \to X$ in the obvious way. Otherwise, I think some fiddling with universal coefficients shows that no candidate $X$ exists. 
This correspondence can be upgraded to an equivalence of categories. In the case $k = \mathbb{Q}$ see, for example, May and Ponto's More Concise Algebraic Topology, Theorem 9.1.4.<|endoftext|>
TITLE: Distribution of $\max_{n \ge 0} S_n$, random walk
QUESTION [5 upvotes]: Say we have a random walk that is a nearest neighbor random walk on the integers where at each step the probability of moving one step to the right is $p$ and the probability of moving one step to the left is $q = 1-p$. Let $S_n$ be such a random walk started at $0$ for some $p \in (0, {1\over2})$. Let $M = \max_{n \ge 0} S_n$. What is the distribution of $M$?

REPLY [3 votes]: More generally, for any upward skip–free random walk on the integers (such that $P(X>1)=0$), the distribution of $M$ is geometric (when finite): $P(M=n)= (1-\theta) \cdot \theta^n$, where $\theta$ is the unique solution in $(0,1)$ of $1=E[\theta^{-X}]$.
The distribution of $M$ can also be calculated inductively for downward skip free random walk (such that $P(X<-1)=0$).
See e.g. Corollary 5.5 and Corollary 5.6 in Asmussen, S. (2003). Applied probability and queues, 2nd ed. New York: Springer, and http://people.bu.edu/pekoz/skipfree.pdf for other interesting information concerning skip free random walks.<|endoftext|>
TITLE: irreducibility of discriminant
QUESTION [34 upvotes]: This must be well-known to everyone but me, but here goes: take a general (monic) polynomial $p(x) = x^d + a_{d-1} x^{d-1} + \dotsc + a_0.$ The discriminant is a polynomial $D(a_0, \dotsc, a_{d-1}).$ Is this irreducible (over $\mathbb{C},$ or in general over algebraic closure of whatever the polynomial is defined over)?

REPLY [17 votes]: Some remarks:

Exception: In some cases it is actually reducible: Over a finite field of characteristic 2, the discriminant of $x^2+a_1x+a_0$ is $D(a_0,a_1)=a_1^2$.
(Almost) Irreducibility over finite fields: Standard $\zeta$-function manipulations show that over $\mathbb{F}_q[T]$ there are exactly $q^{d-1}$ monic non-squarefree polynomials (hint: the Dirichlet series of $\mu^2(f)=1_{f\text{ squarefree}}$ is $\frac{\zeta(u)}{\zeta(u^2)}$ where $\zeta(u) = \sum_{f \text{ monic}} u^{\deg f}$). 
In other words, the zero locus of $D(a_0,\cdots,a_{d-1})=0$ is of size $q^{d-1}$. On the other hand, Weil-Lang bounds express the size of the zero locus of $D=0$ over $\mathbb{F}_{q_0}$ as $q_0^{d-1} \cdot (1+O(\frac{1}{\sqrt{q_0}}))$ times the number of geometrically irreducible components of $D=0$ of dimension $d-1$. So using the information from the zeta functions for $q=q_0^k, k \to \infty$, we find that $D$ is a power of an irreducible polynomial. 
(Almost) Irreducibility over $\overline{\mathbb{Q}}$: This reduces to irreducibility over number fields, which reduces to irreducibility over rings of integers (Gauss lemma) which follows from irreducibility over finite fields by working modulo some prime ideal.<|endoftext|>
TITLE: what are the finite etale covers of $\mathbb{Z}_p((x))$?
QUESTION [8 upvotes]: Let $R$ be the ring of integers of some $p$-adic field $K$ (finite over $\mathbb{Q}_p$) with uniformizer $\pi$ and residue field $k$. I'd like to understand the finite etale extensions of $R((x)) := R[[x]][x^{-1}]$.
Note that $R((x))$ is a non-local PID with uncountably many prime ideals, all of which are generated by elements of the form $\pi^n + r_1x + r_2x^2 + \cdots$
One special prime is just the prime $\pi$, at which the residue field is $k((x))$.
Question: Is every connected finite etale extension of $R((x))$ given by $R'((x^{1/e}))$ where $R'$ is the ring of integers in some finite unramified extension $K'$ of $K$ and $e$ is coprime to $p$? If not, is it at least dominated by such an extension?
I'd also appreciate any relevant references.

REPLY [7 votes]: The answer is affirmative in the "dominated by" aspect, with $R$ any complete discretely valued field whose fraction field $K$ has characteristic 0 and residue field $k$ has characteristic $p \ge 0$ (using $e$ not divisible by $p$; i.e., $e \in k^{\times}$). One cannot do better since the absence of various roots of unity from $K$ can mess up the "easy" Galois theory description of intermediate extensions. This is a standard application of Abhyankar's Lemma, as follows.
In the ring $A = R[\![x]\!]$, the height-1 prime $P = (x)$ has residue field $K$ of characteristic 0 (and uniformizer $x$), so all ramification at $P$ in any finite extension $L'$ of $L := {\rm{Frac}}(A) = {\rm{Frac}}(R(\!(x)\!))$ is tame. Hence, if $A'$ is the (module-finite) normalization of $A$ in $L'/L$ and $A'[1/x]$ is etale over $A[1/x] = R(\!(x)\!)$ then there is only tame ramification over ${\rm{Spec}}(A)$. It follows by Abhyankar's Lemma (as proved in an appendix of the book of Freitag-Kiehl on etale cohomology, for example) that ${A'}^{\rm{sh}} = A^{\rm{sh}}[x^{1/e}]$ for an integer $e \ge 1$ with $e \in k^{\times}$. Thus, $A' \subset A^{\rm{sh}}[x^{1/e}]$. 
Since $A$ is complete local noetherian, so $A^{\rm{sh}}$ is exhausted by local finite etale extensions of $A$, which in turn correspond to finite separable extensions of $k$, as we vary through the local finite etale extensions $R'$ of $R$ the resulting finite etale $A$-algebras $R'[\![x]\!]$ exhaust $A^{\rm{sh}}$.  Hence, for big enough such $R'$ we see that $A' \subset R'[\![x]\!][x^{1/e}]$.  
The extension of Dedekind domains 
$$R(\!(x)\!) = A[1/x] \rightarrow R'(\!(x^{1/e})\!)$$
is clearly finite etale, so any intermediate module-finite Dedekind domain is also finite etale over $A[1/x]$.  Thus, the finite etale $R(\!(x)\!)$-algebra domains are precisely the module-finite subalgebras of such extensions $R'(\!(x^{1/e})\!)$ for $R'$ local finite etale over $R$ and $e \ge 1$ not divisible by ${\rm{char}}(k)$.  That is the "dominated by" assertion we wanted to prove.<|endoftext|>
TITLE: power sums and formal divisibility by the Euler totient function
QUESTION [8 upvotes]: For integers $s \geq 0$ and $n \geq 1$ let $\sigma_s(n) := 1^s + \dots + n^s$ and let 
\begin{equation} \sigma_s^*(n) \ : = \ \sum_{\stackrel{\scriptstyle 1 \, \leq \, m \, \leq \, n}{\text{gcd}(m,n) = 1}} m^s \end{equation} 
It is not to difficult to check that $\sigma^*$ can be expressed
as the Dirichlet convolution $\big( \mu \cdot \Bbb{i}^s \big) * \sigma_s$
where $\mu$ is the Möbius funtion, $\Bbb{i}$ is the identity function, namely $\Bbb{i}(n) = n$ for all integers $n \geq 1$, and
$\mu \cdot \Bbb{i}^s$ is the point-wise product, namely $\big( \mu \cdot
\Bbb{i}^s \big)(n) = \mu(n) \, n^s$ for all integers $n \geq 1$.
A cute little exercise in Vinogradov's text asserts that when $s=0, 1, 2$ 
formula for $\sigma_s^*(n)$ with $n > 1$ are given by 
\begin{equation} \begin{array}  $\phi(n) &\text{when} \quad s = 0 \\  \\ \Big( {\displaystyle 1 \over \displaystyle 2} \,n \Big) \, \phi(n) &\text{when} \quad s=1 \\ \\ 
\Big( {\displaystyle 1 \over \displaystyle 3 } \, n^2  + {\displaystyle 1 \over \displaystyle 6} \, \mu\big(\text{rad}(n) \big) \, \text{rad}(n)\Big) \, \phi(n) &\text{when} \quad s=2 
\end{array} \end{equation}
Where $\phi$ is the Euler totient function and $\text{rad}(n)$ denotes the 
radical of the integer $n$.
${ \bf \text{Question}}$: Is $\sigma_s^*$ always "formally" divisible by $\phi$ for any integer $s \geq 0$ ?

REPLY [7 votes]: Nice question. The answer is positive. I will give a very constructive answer.
First, write $\sigma_s(n)$ as a polynomial in $n$ of degree $s+1$: 
$$\sigma_s(n) = \sum a_i n^i$$
The numbers $a_i$ are closely related to Bernoulli numbers, see this Wikipedia page. In particular, $a_0=0,a_{s}=\frac{1}{2}, a_{s+1}=\frac{1}{s+1}$.
By the convolution identity relating $\sigma_s^{*}(n)$ to $\sigma_s(n)$ via the Möbius function, we find:
$$\sigma_s^{*}(n) = \sum_{i} a_i \sum_{d \mid n} (\frac{n}{d})^i\mu(d)d^s$$
It is enough to demonstrate that $\frac{\sum_{d \mid n}(\frac{n}{d})^i\mu(d)d^s }{\phi(n)}$ has some "nice" expression for any $i\le s+1$. Since $n \mapsto n^i, n\mapsto \mu(n) n^s$ are multiplicative functions, this numerator is a multiplicative function of $n$ and so is the ratio itself. We will evaluate it on the prime power $p^k$ ($k>0$):
If $i=s+1$ we get $(p^k)^s$, and so by multiplicativity, $\frac{\sum_{d \mid n}(\frac{n}{d})^i\mu(d)d^s }{\phi(n)}=n^s$.
If $i=s$ we get $0$ ($\sum_{d \mid p^k} \mu(d)=0$), and so by multiplicativity, $\frac{\sum_{d \mid n}(\frac{n}{d})^i\mu(d)d^s }{\phi(n)}=\delta_{n,1}$.
If $i<s$ we get
$$\frac{\sum_{d \mid p^k}(\frac{p^k}{d})^i\mu(d)d^s }{\phi(p^k)} = (-p) \cdot (p^k)^{i-1} (1+p+p^2+\cdots+p^{s-i-1}).$$
By multiplicativity, $\frac{\sum_{d \mid n}(\frac{n}{d})^i\mu(d)d^s }{\phi(n)}=\text{rad}(n) \mu(\text{rad}(n))n^{i-1} \cdot \sigma(\text{rad}^{s-i-1}(n))$, where $\sigma$ is the sum-of-divisors function.
This settles your problem.<|endoftext|>
TITLE: $p$-th Fourier coefficients of newforms of level $\Gamma_1(N)$ with $p|N$
QUESTION [8 upvotes]: Let $f$ be a newform of level $\Gamma_1(N)$  and character $\chi$ which is not induced by a character mod $N/p$. I learned from these notes by Ribet and Stein that $|a_p|=p^{(k-1)/2}$ where $k$ is the weight of $f$. So I wonder
1, the proof of this statement,
2, is $a_p$ for $p|N$ in the number field $K_f$ of $f$, i.e the number field generated by all $a_q$ for $(q,N)=1$?
Now I see that 2 is true. Why I thought it's wrong is because  $p^{(k-1)/2}$ is of degree 2, which will implies in this case $2|[K_f:\mathbb{Q}]$. Seems strange...

REPLY [2 votes]: This is really a comment, but the system won't let me post one. I think there's a confusion and that the absolute value is $p^((k-2)/2)$ where $k$ is the weight. For example there's a level $\Gamma_0 (2)$ newform of weight 8, $f=(\eta(z)\eta(2z))^8$, with $f^3 =\delta(z)\delta(2z)=q^3-24q^4+...$ Then $f=q-8q^2+...$, the $U_2$ eigenvalue is -8, and the exponent is $(8-2)/2$.
EDIT: The confusion was mine--I missed the condition on the character. My example illustrated case (iii) of Li's Theorem 3, while the question related to case (ii). I'd be grateful for an explicit example from the databases of case (ii) to further clear up my confusion.<|endoftext|>
TITLE: In the plane, does complement of Brownian path have infinitely many connected components?
QUESTION [6 upvotes]: Let $d = 2$. Do we have that with $P_x$—probability $1$, for every $T> 0$ the complement $W[0, T]^c$ of the Brownian path up to time $T$ has infinitely many connected components?
I had seen this post, and I was wondering if this result is also covered in the 1954 Dvoretsky/Erdos/Kakutani paper. If not, can somebody provide me a reference/supply an answer?

REPLY [4 votes]: Yes. Just observe that
(1) on any fixed time interval the Brownian path intersects itself with positive probability (easy to see);
(2) but the above implies that on any time interval the Brownian path intersects itself with probability 1 (divide that interval into many sub-intervals and use the Markov property);
(3) so, with probability 1 in any neighbourhood of a fixed point on the Brownian path there will be at least one connected component of the complement, and this implies the claim.<|endoftext|>
TITLE: Constructing a function over a metric space through given points
QUESTION [8 upvotes]: Suppose there is a compact metric space $(X,\rho)$ and a Euclidean space $\mathbb{R}^n$. 
There is a sequence of unequal points $\{x_1,...x_N\}$ in $X$ such that all metrics $\rho(x_i,x_j)$ are known and $f(x_i)=a_i$ for some $a_i$ in $\mathbb{R}^n$ whereas:
$$ \forall x_i,x_j \in \{x_1,...x_N\} . ||a_i-a_j|| \leq L \cdot \rho(x_i, x_j) $$
for a fixed $L$.
Suppose also that $\{x_1,...x_N\}$ form a finite cover of $X$ by balls of some suitable (known) radius.
If $X$ were just a compact interval, $\{x_1,...x_N\}$ its partition, and $n$ were $1$, we could certainly construct a piecewise-linear function $f:X \rightarrow \mathbb{R}^n$ such that:
$$ f(x_i)=a_i \\ \forall x,y \in X . ||f(x)-f(y)|| \leq L \cdot \rho(x, y) $$ 
with a slope controlled by $L$.
It should be also possible in certain cases if $X \subset \mathbb{R}^m$.
Is it possible to construct such a function for a general compact metric space? And if $X$ is a subset of a Euclidean space?

REPLY [6 votes]: Applying rescaling you can assume that $L=1$.
We look for a 1-Lipschitz piecewise linear maps $f\colon\mathbb{R}^m\to\mathbb{R}^n$. 
If $m=n$ we get $f$ from Brehm's theorem;
it says that there is a piecewise distance preserving map of that type (in particular 1-Lipschitz and piecewise linear). 
See Brehm, U., Extensions of distance reducing mappings to piecewise
congruent mappings on $\mathbb{R}^m$ and also our paper written for kids.
If $m<n$ we can think that $\mathbb{R}^m$ is a subspace of $\mathbb{R}^n$; in this case apply Brehm's theorem and restrict the obtained map to $\mathbb{R}^m$. 
If $m>n$ we can think that $\mathbb{R}^n$ is a subspace of $\mathbb{R}^m$; in this case apply Brehm's theorem and compose the obtained map with the projection to $\mathbb{R}^n$.
Since projection linear and 1-Lipschitz, so it the composition.

REPLY [6 votes]: Anton's answer is for the case where $X$ is a subset of a Hilbert space.  For $X$ a general metric space (compactness does not help, BTW), $L$ must be increased by a factor of order $\log^{1/2}N$; see
$$$$
 Johnson, William B.; Lindenstrauss, Joram Extensions of Lipschitz mappings into a Hilbert space. Conference in modern analysis and probability (New Haven, Conn., 1982), 189–206, Contemp. Math., 26, Amer. Math. Soc., Providence, RI, 1984.<|endoftext|>
TITLE: How much can an Eisenstein series be truncated?
QUESTION [11 upvotes]: For ease of exposition, I will stick to the simplest case: consider the Eisenstein series for $SL_2(\bf R)$
$$E(z,s)=\sum_{\gamma\in P_{\bf Z}\backslash SL_2(\bf Z)}\text{Im}(\gamma z)^s=\sum_{(c,d)\in{\bf Z}\backslash (0,0)}\frac{y^s}{|cz+d|^{2s}}$$
and its truncation used to compute the inner product of Eisenstein series (called the Maaß-Selberg relation) over $SL_2(\bf Z)\backslash H$,
$$\Lambda^TE(z,s)=\begin{cases} E(z,s) & y<T\\E(z,s)-y^s-\varphi(s)y^{1-s}&y>T\end{cases}$$
(I am being sloppy with the difference between the 'naive' truncation and Arthur's truncation, but I think in this case it does not make a difference; please correct me otherwise.)
My question concerns the parameter $T$. The first condition is that $T>0$, large enough that the truncated fundamental domain is such that the curve $y=T$ cuts the fundamental domain only on the two sides that pass through the cusp, and such that the truncated function has rapid decay at cusps to be automorphic.

What is the smallest value that $T$ can take so that the inner product
  formula remains valid?

Why I am asking: from what I understand, Selberg in Harmonic Analysis (Collected works, p.633) says that $T=1$ will suffice, while Garrett in his note Simplest Example of Truncation and Maaß-Selberg Relations claims $T=\sqrt3/2$ (p.2). Unfortunately, a computation I am making seems to fail for such small values, but holds for some other given $T\gg0$. I would like to know if there is a computational error, or I should expect the Maass-Selberg relation to fail or alter at some 
 limit.
EDIT: I am starting a bounty, as after making certain computations with Hejhal's calculation and my own formulas I am sure something is fishy. From Hejhal  in The Selberg Trace Formula and the Riemann Zeta Function one sees that
$$\frac{1}{4\pi}\int^{\infty}_{-\infty}h(r)\int_{\Gamma\backslash\bf H}|\Lambda^T E(z,\frac{1}{2}+ir)|^2dz\ dr$$
is equal to the sum of
$$\frac{1}{4}m(\frac{1}{2})h(0)-g(0)\log\pi+\frac{1}{2\pi}\int^\infty_{-\infty}h(r)\frac{\Gamma'}{\Gamma}(1+ir)dr-2\sum_{n=1}^\infty\frac{\Lambda(n)}{n}g(2\log n)$$
where $\Lambda(n)$ is the von Mangoldt function, and
$$+\frac{1}{4\pi i}\int^\infty_{-\infty}m(\frac{1}{2}-ir)h(r)e^{2ir \log T}\frac{dr}{r}+g(0)\log T$$
where $m(s)=\xi(2-2s)/\xi(2s),$ the quotient of completed Riemann zeta functions. Then taking as a test function $h(r)=e^{-r^2}$ (which is allowable), and $T$ very close to 1, one finds that the second line is negligible, the digamma function is negative, and the resulting expression is negative, whereas the original integral is clearly positive. More generally, if one chooses $g(0)$ or $h(0)$ to be zero, the expression is even more likely negative.
Certainly for $T$ large enough, the expression is positive for positive test functions. I have been unable to find accurate information on what $T$ is allowable for the inner product formula to hold.

REPLY [10 votes]: I think the answer truly does depend on whether you use the naive truncation, as in Iwaniec's book, or the Arthur truncation, as in Paul Garrett's note. In particular, the method of proof for the Maaß-Selberg relation with the naive truncation is proved in Iwaniec's book using Green's identity, and as GH from MO mentioned in his comment, the geometric picture here suggests that the proof only requires that $T \geq \sqrt{3}/2$.
On the other hand, I believe that with the Arthur truncation, one really does require that $T \geq 1$, with the proof of the Maaß-Selberg relation now instead via an unfolding argument as in Paul Garrett's notes. Here the Arthur truncation is
\[\Lambda^T E(z,s) = E(z,s) - \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} c_0 E(\gamma z,s),
\]
where for $z = x + iy \in \mathbb{H}$,
\[c_0 E(\gamma z,s) = \int_{0}^{1} E(\gamma z,s) \, dx,\]
so that $c_0 E(z,s) = y^s + \varphi(s) y^{1-s}$.
One can easily show that if $\gamma = \begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})$ with $\gamma \notin \Gamma_{\infty}$, so that $c \neq 0$, then
\[\Im(z) \Im(\gamma z) = \frac{y^2}{(cx + d)^2 + c^2 y^2} \leq 1,\]
while $\Im(\gamma z) = \Im(z)$ for $\gamma \in \Gamma_{\infty}$.
So for $T \geq 1$, the Arthur truncation is equal to
\[\Lambda^T E(z,s) = \begin{cases}
E(z,s) & \text{if $1/T \leq y \leq T$,} \\\
E(z,s) - y^s - \varphi(s) y^{1-s} & \text{if $y > T$.}
\end{cases}\]
If $y < 1/T$, then there may be more terms (and which terms are also present will now depend on $x$ as well), as there may be more coset representatives $\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z})$ other than just the identity for which $\Im(\gamma z) > T$.
If $T < 1$, then $\Lambda^T E(z,s) = E(z,s) - y^s - \varphi(s) y^{1-s}$ if $y \geq 1/T$, but there may be more terms if $y < 1/T$.
In particular, the Arthur truncation coincides with the naive truncation on the standard fundamental domain if $T \geq 2/\sqrt{3}$, but if $T < 2/\sqrt{3}$ then this is no longer the case.
Now a key step in proving the Maaß-Selberg relation is showing that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0.\]
One shows this by unfolding to find that this is
\[-\int_{T}^{\infty} \overline{c_0 E(z,r)} \int_{0}^{1} \Lambda^T E(z,s) \, dx \, dy,\]
using the fact that $c_0 E(z,r) = y^r + \varphi(r) y^{1-r}$ does not depend on $x$.
If $T \geq 1$, then $\Lambda^T E(z,s) = E(z,s) - c_0 E(z,s)$ for $T < y < \infty$ and $0 < x < 1$, and so the inner integral vanishes. But if $T < 1$, then there may be other terms, and so this is no longer the case. So the condition $T \geq 1$ truly is necessary here.
EDIT: With regards to your comment, note that the Maaß-Selberg relation states that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{\Lambda^T E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}.
\]
Here
\[\varphi(s) = \frac{\Lambda(2 - 2s)}{\Lambda(2s)}, \qquad \Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s),\]
so that $|\varphi(1/2 + it)|^2 = 1$ for all $t \neq 0$, and taking logarithmic derivatives shows that
\[\frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) = -2\Re\left(\frac{\Gamma'}{\Gamma}\left(\frac{1}{2} + it\right)\right) - 4\Re\left(\frac{\zeta'}{\zeta}(1 + 2it)\right) + 2 \log \pi.\]
Setting $s = r = 1/2 + it + \varepsilon$ for $t \neq 0$ and $\varepsilon > 0$, taking the limit as $\varepsilon \to 0$, and using the Laurent expansions of each term, we find that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + it\right)\right|^2 \, d\mu(z) = 2 \log T - \frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) - \Im \left( \varphi\left(\frac{1}{2} + it\right) \frac{T^{-2it}}{t}\right).
\]
Now it's not obvious to me either that this is nonnegative, and of course if $T$ is very close to zero then you wouldn't expect it to be. But for $T$ near $1$ (because as mentioned earlier, even using the naive truncation we can't have $T < \sqrt{3}/2$) it's not obvious to me either whether this is negative, so I'm not sure if there's truly a problem here; certainly for large $t$, the digamma function should dominate. Of course, I could be missing something here.
SECOND EDIT: I still don't understand your objection; as I have already stated, one requires that $T \geq 1$ for the Arthur truncation, and this is also sufficient (i.e. with the Arthur truncation, the Maaß-Selberg relation holds for all $T \geq 1$). Indeed, I already showed that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0\]
when $T \geq 1$, so it remains to show that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}.
\]
By the definition of the Arthur truncation, the left-hand side is
\[ \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_ {\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) \leq T}} \Im(\gamma z)^s \overline{E(z,r)} \, d\mu(z) - \varphi(s) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} \Im(\gamma z)^{1-s} \overline{E(z,r)} \, d\mu(z),
\]
which, by the $\mathrm{SL}_2(\mathbb{Z})$-invariance of $E(z,r)$, unfolds to
\[ \int_{0}^{T} y^s \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2} - \varphi(s) \int_{T}^{\infty} y^{1-s} \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2}.
\]
By the definition of the constant term of $E(z,r)$ (namely that it is $y^r + \varphi(r) y^{1-r}$), one can evaluate these integrals, with the result being the Maaß-Selberg relation.
Of course, to ensure convergence of all the integrals involved, this is initially only valid for $\Re(s), \Re(r) > 1$ with $\Re(s) - \Re(r) > 1$, but then by analytic continuation this extends to all $s,r \in \mathbb{C}$ with $s \neq \overline{r}$ and $s + \overline{r} \neq 1$ provided $\varphi(s), \varphi(r)$ are well-defined. As before, one can extend this to $s = r = 1/2 + it$ with $t \neq 0$ by setting $s = r = 1/2 + it + \varepsilon$ and taking the limit as $\varepsilon \to 0$.
All this is valid if $T \geq 1$, so one certainly should be able to take $T = 1$ in the Maaß-Selberg relation, unless I'm missing something.
With regards to the integral
\[\int_{-\infty}^{\infty} h(r) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + ir\right)\right|^2 \, d\mu(z) \, dr,\]
to me it looks like there may be an issue at $r = 0$, and one must be careful when breaking up the integral and blindly using Weil's explicit formula. (In particular, I don't see why you say the second line is negligible - doesn't the integrand blow up at $r = 0$?) But I haven't looked too closely at this.<|endoftext|>
TITLE: Why going to number fields in number field sieve help beat quadratic sieve?
QUESTION [11 upvotes]: To factor an $n$ bit integer number field sieve roughly takes $$e^{c{(\ln\ln n)^{\frac23}}({\ln n})^{\frac13}}$$ time while quadratic sieve takes  $$e^{c{(\ln\ln n)^{\frac12}}({\ln n})^{\frac12}}$$ time.
What does number field sieve improve on quadrative sieve? I know we work with number fields and other details. However what exact complexity reason are we able to improve and why cannot we improve any further?
I know wiki has a few sentence explanation in comments. However it is difficult to pinpoint both why quadratic sieve was beaten and why no other alternative has been known for two decades.

In other words why does looking at abelian extensions of number fields help? 
Which extensions help and why?

REPLY [4 votes]: Each relation in the quadratic sieve has to factor integers of size near the square root of N. In contrast, each relation in the number field sieve has to factor two things, a number and an element of a number field. That number field can be chosen so that the two factorizations are much easier than the factorization in the quadratic sieve. The number field is chosen in terms of a polynomial of degree d, and as N increases in size the value of d is increased. The asymptotic complexity of NFS assumes the optimal value of d<|endoftext|>
TITLE: Flat dimension of injectives over a Gorenstein ring
QUESTION [6 upvotes]: Let $A$ be a Gorenstein noetherian local ring, and let $M$ be an $A$-module of finite injective dimension.
If $M$ is a finite $A$-module, it is easy to show these assumptions imply that $M$ has finite flat dimension (just use the fact that $A$ is both perfect and a dualizing complex). 
What if $M$ is not finitely generated? is it still true that finite injective dimension implies finite flat dimension?

REPLY [2 votes]: Suppose that $A$ is a not necessarily commutative ring, which is both left and right Noetherian. Given a left $A$-module $N$, and a right $A$-module $M$, there is a general (fourth quadrant, cohomological) spectral sequence due to Ischebeck, with $E_2$ term
$$E_2^{p,-q} = \operatorname{Ext}^p_A(\operatorname{Ext}^q_A(N,A),M)$$
which converges to
$$\operatorname{Tor}^A_{q-p}(M,N)$$
whenever $M$ has finite injective dimension. If, in addition, $A$ has finite injective dimension $d = \operatorname{id}(A) < \infty$, then whenever $n = q-p$ with $0 \leq q \leq d$ and $p \geq 0$, we must have $n \leq d$. Thus $\operatorname{Tor}^A_n(M,N)$ vanishes for any left $A$-module $N$ and any $n > d$. So, if $M$ has finite injective dimension, then
$$\operatorname{fd}(M) \leq \operatorname{id}(A)$$
which gives a positive answer to your question because your ring $A$ is assumed to be Gorenstein, hence has finite self-injective dimension. 
You can find an English translation of Ischebeck's paper here; see Theorem (1.8).<|endoftext|>
TITLE: A generalization of residual finiteness to topological groups
QUESTION [12 upvotes]: Consider the following generalization of residual finiteness to
topological groups.
A locally compact Hausdorff group $G$ is called residually compact if
for every compact $K \subseteq G$ there is a normal cocompact lattice
$\Lambda \subseteq G$ such that
the projection $G \to G/\Lambda$ is injective on $K$. 
Notice that this clashes with the normal definition of a residual property, but I don't know a better name.
Q1: Has this property been studied under another name?
Up to now we know the following.
If $G$ contains a normal cocompact lattice which is residually finite, then the
group is residually compact.
On the other hand a residually compact group embeds (algebraically)
into a compact group, hence every finitely generated lattice in $G$ is
residually finite by Malcev's theorem and the fact that the
irreducible linear presentations of a compact group are finite
dimensional. If $G$ is compactly generated, then every lattice is finitely
generated. In the compactly generated case, residual finiteness is
thus the same as containing a residually finite normal cocompact
lattice.
Q2: Is there an example of a residually compact topological group that
does not contain a residually finite normal lattice?

REPLY [7 votes]: Here's an answer to the mathematical part of the question (namely Q2).
An example is the group of adeles. Start from $\mathbf{A'}$ defined as the inverse image of $\bigoplus_p\mathbf{Q}_p/\mathbf{Z}_p$ in $\prod\mathbf{Q}_p$, with $\prod\mathbf{Z}_p$ prescribed to be a compact open subgroup (the sum being over all primes); then $\mathbf{A}=\mathbf{R}\oplus\mathbf{A}'$. This is a locally compact ring with $\mathbf{Q}$ standing as a discrete cocompact subring through the diagonal embedding (let $Q$ denote its image).
Then $\mathbf{A}$ is "residually compact". Indeed, if we pick a compact subset, then there exists an integer $n\ge 0$, a finite set of primes $I$ with complement $J$ such that this compact subset is contained in the compact subset $$K_{I,n}=[-n,n]\times\bigoplus_{p\in I}p^{-n}\mathbf{Z}_p\oplus\prod_{p\in J}\mathbf{Z}_p.$$ Define $m=\prod_{p\in I}p^n$. The intersection $Q\cap K_{I,n}$ consists of those rationals of the form $m^{-1}k$ with $k\in\mathbf{Z}$ with real absolute value $|m^{-1}k|\le n$. So if we define $\phi$ as the topological group automorphism which the identity on $\mathbf{A}'$ and multiplies the real component by, say, $2mn$, then $\phi(Q)\cap K_{I,n}=\{0\}$. So $\mathbf{A}$ is "residually compact".
On the other hand, $\mathbf{A}$ has no residually finite lattice. Indeed, let $R$ be a lattice. The image of $R$ in $\bigoplus\mathbf{Q}_p/\mathbf{Z}_p$ has finite index, and since the latter group has no proper finite index subgroup, it is all of $\bigoplus\mathbf{Q}_p/\mathbf{Z}_p\simeq\mathbf{Q}/\mathbf{Z}$. The kernel of this projection is $R_1=R\cap (\mathbf{R}\oplus\prod\mathbf{Z}_p)$ and is a cocompact lattice in the open subgroup $\mathbf{R}\oplus\prod\mathbf{Z}_p$, which is torsion-free and compactly generated. In particular $R_1\cap\prod\mathbf{Z}_p$ is trivial and the projection of $R_1$ in $\mathbf{R}$ is injective and its image is a cocompact lattice in $\mathbf{R}$ (discreteness follows from compactness of $\prod\mathbf{Z}_p$). So $R_1$ is an infinite cyclic group. Now let us show that $R$ is divisible by any $p$ (this will imply that $R$ is not residually finite). Let $x\in R$. Then since $R/R_1$ is divisible, we can write $x=py+z$ with $z\in R_1$ and $y\in R$. Now let $M$ be the inverse image in $R$ of the subgroup of order $p$ of $R/R_1$. Then $M$ is torsion-free and contains $R_1$ with index $p$. Hence every element of $R_1$ has a $p$-root in $M$, hence in $R$. So $z=pz'$ with $z'\in R$, thus $x\in pR$. Thus $pR=R$ and $R$ is divisible, hence not residually finite (and actually, necessarily is isomorphic to $\mathbf{Q}$).<|endoftext|>
TITLE: Existence of non-homeomorphic pair of bijectively related closed subsets in $\mathbb{R}$
QUESTION [5 upvotes]: I want to find two closed, non-homeomorphic subsets $A$ and $B$ of $\mathbb{R}$ (with subset topology), with the property that there exist two continuous bijections
$$f:A\to B,~~~~g:B\to A.$$
Clearly $A$ or $B$ cannot be bounded. But I didn't find more restrictions. Do we have some results on this question?

REPLY [6 votes]: Using a relatively recent theorem, we can get a whole family of examples.
Suppose $A$ and $B$ are two closed subsets of the real line satisfying the following properties:

zero-dimensional
$\sigma$-compact, but not compact
no isolated points

Then they are "bijectively related" in the sense of your question. A proof can be found in Section 3 of this paper: https://wrbrian.files.wordpress.com/2012/01/cumet2.pdf.
It can be shown that there are $2^{\aleph_0}$ homeomorphism types of spaces satisfying these conditions (though, sadly, I don't have a reference). So now you have uncountably many examples!

REPLY [4 votes]: An example, I hope.
Write $X \sqcup Y $ for disjoint union, say a set made up of two disjoint closed parts, one homeomorphic to $X$ and one homeomorphic to $Y$.  And write $\bigsqcup_{i} X_i$ for a disjoint union of countably many closed sets homeomorphic to the $X_i$, "going to infinity" in the sense that any bouned interval meets only finitely many of them.  Write $X \prec Y$ to mean there is a continuous bijection from $X$ to $Y$.
Let $P$ (or with subscripts) be a single point.
Let $Q$ (or with subscripts) be homeomorphic to $\{0\}\cup\{3^{-n}:n=1,2,3,\cdots\}$; a convergent sequence together with its limit.
Let $C$ (or with subscripts) be homeomorphic to the middle-thirds Cantor set in $[0,1]$.
My two sets are
$$
A = \bigsqcup_i P_i \sqcup \bigsqcup_i C_i
\\
B = \bigsqcup_i P_i \sqcup \bigsqcup_i C_i \sqcup Q
$$
Of course $A$ and $B$ are homeomprphic to closed subsets of $\mathbb R$.
For example, $A$ could consist of the negative integers, together with
Cantor sets in each of the intervals $[2i,2i+1]$.
The derived set $A'$ of $A$ is $\bigsqcup_i C_i$, and has no isolated point.  But the derived set $B'$ of $B$ is $\bigsqcup_i C_i$ plus a single isolated point $Q'$.  So $A$ and $B$ are not homeomprphic.
Now we have to observe
$$
\bigsqcup_i X_i = \bigsqcup_i X_{2i} \sqcup \bigsqcup_i X_{2i+1}
\tag{0}
$$
$$
\bigsqcup_i P_i \prec Q
\tag{1}
$$
$$
\bigsqcup_i C_i \sqcup P \prec C
\tag{2}
$$
$$
\bigsqcup_{i} C_{i} \sqcup Q \prec C
\tag{3}
$$
From these we get $A \prec B$ and $B \prec A$:
$$
A = \bigsqcup_i P_i \sqcup \bigsqcup C_i
= \bigsqcup_i P_i \sqcup\bigsqcup_i P_i \sqcup \bigsqcup C_i \prec
Q \sqcup\bigsqcup_i P_i \sqcup \bigsqcup C_i = B 
$$
$$
B = \bigsqcup_i P_i \sqcup \bigsqcup C_i \sqcup Q =
\bigsqcup_i P_i \sqcup \bigsqcup C_i\sqcup \bigsqcup C_i \sqcup Q
\\ \prec\bigsqcup_i P_i \sqcup \bigsqcup C_i\sqcup C
=\bigsqcup_i P_i \sqcup \bigsqcup C_i = A
$$<|endoftext|>
TITLE: Hahn-Banach theorem with convex majorant
QUESTION [32 upvotes]: At least 99% of books on functional analysis state and prove the Hahn-Banach theorem in the following form: Let $p:X\to \mathbb R$ be sublinear on a real vector space, $L$ a subspace of $X$, and $f:L\to \mathbb R$ linear with $f\le p|_L$. Then there is a linear $F:X\to\mathbb R$ with $F\le p$ and $F|_L=f$.
However the theorem is true if the majorant $p$ is merely convex. This version has a very similar proof as the classical statement and several advantages. For instance, there is no need to introduce the new notion of sublinearity and the result is even interesting for $X=\mathbb R$.
The only reference I know is the book of Barbu und
Precupanu Convexity and Optimization in Banach Spaces.
Two questions:

Who first observed that sublinearity can be replaced by convexity?
Is there any (e.g. pedagocial) reason to prefer the sublinear version?

REPLY [2 votes]: Yet another reference is Theorem A on page 105 in:
A. Wayne Roberts, Dale E. Varberg,
Convex functions.
Pure and Applied Mathematics, Vol. 57. Academic Press , New York-London, 1973. 
This reference provides also an answer to a question in one of the comments:
What are some applications which can not be proved with the "sublinear" version? 
If $p:\mathbb{R}^n\to\mathbb{R}$ is convex, then the subdifferential $\partial p(x)$ is the set of all $v\in\mathbb{R}^n$ such that 
$$
(*)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p(y)\geq p(x)+\langle v,y-x\rangle \
\quad
\text{for all $y\in\mathbb{R}^n$.} 
$$
It is easy to prove that $\partial p(x)\neq\emptyset$ for all $x$. The result mentioned by OP has the following application:

Theorem. $p$ is differentiable at $x$ if and only if $\partial p(x)$ consists of one vector.

Sketch of the proof.
If $p$ is differentiable, then it easily follows that $\partial p(x)=\{\nabla p(x)\}$. Suppose now that $\partial p(x)=\{ v\}$. We want to show that $p$ is differentiable at $x$. It easily follows from the convexity that if partial derivatives exist at $x$, then $p$ is differebtiable. Indeed, if $A=[\frac{\partial p}{\partial x_1}(x),\ldots,\frac{\partial p}{\partial x_n}(x)]$, then $y\mapsto p(x+h)-p(x)-Ah$ is convex and one can use Jensen's inequality to estimate this remiander in terms of partial derivatives (very nice exercise, cf. p. 101 in the above book). Thus it remains to show that $p$ has partial derivatives at $x$. 
One-sided partial derivatives exist (due to monotonicity of difference quotients) and
it suffices to show that they are equal. For if not, they would give two different one dimensional linear functions satisfying $(*)$ in the direction on the partial derivative. Then the "convex" Hahn-Banach theorem mentioned by OP would give two different vectors $v$ satisfying $(*)$ for all $y\in\mathbb{R}^{n}$ which would contradict the assumption that $\partial p(x)=\{ v\}$.<|endoftext|>
TITLE: What are the ramifications of the Hodge conjecture to mathematical physics?
QUESTION [6 upvotes]: I read a little bit the survey:
http://www.claymath.org/sites/default/files/hodge.pdf
Well I understand its a conjecture in Algebraic Geometry, so it seems there should be some consequences of proving this conjecture in mathematical physics.
So are there any? and what are they?
Thanks.

REPLY [7 votes]: There are connections to string theory as explained in Open Strings and Extended Mirror Symmetry, by Johannes Walcher (2007). In one sentence: The Gromov-Witten theory of a Calabi-Yau manifold is solved by the Hodge theory of its mirror symmetric manifold.
A more elaborate quote from this 2011 conference:

The mathematical theory that describes how integrals and differential
  equations control the shape of algebraic spaces in various dimensions
  is known as Hodge theory. The most important conjecture in algebraic
  geometry -- the Hodge Conjecture -- can be thought of as "a metaphor
  for transforming transcendental computations into algebraic ones." The
  physical theory able to describe the universe at both micro- (quantum
  mechanics) and macro- (general relativity) scales, and at the same
  time thought to be a suitable candidate for unifying all known forces
  of nature, is string theory. There are several variants of this
  "theory of everything," linked by dualities which can radically alter
  mathematical formulations while preserving physical predictions.
  String dualities thus imply conjectures: seemingly unrelated pieces of
  mathematics must be related since they offer different descriptions of
  the same physical world. Although a role for Hodge theory in string
  theory has been hinted at for some time, only very recently has the
  depth and precision of this relationship begun to emerge. Recent
  results suggest that a mathematical "grand unification" relating
  arithmetic geometry and symplectic geometry is taking shape.<|endoftext|>
TITLE: Banach manifold of paths with endpoints on submanifolds
QUESTION [8 upvotes]: Fix a Riemannian metric on a manifold $M$.  Suppose that we fix two points $x,y \in M$.  We start with the space 
$C^{\infty}_{\searrow}(x,y) = \left\{\gamma: \mathbb{R}\to M:\,\lim_{t\to-\infty}\gamma(t) = x, \lim_{t\to\infty}\gamma(t)=y, \exists\,C,\delta > 0 \text{ such that} \vert\gamma'(t)\vert \leq Ce^{-\delta\vert t\vert}\right\}$
We can complete this space to a Banach manifold, $\mathcal{P}^{1,p}(x,y)$.  The charts are given by pairs $(U_{\gamma},\gamma^{*}\exp)$, where $\gamma \in C^{\infty}_{\searrow}(x,y)$, $\exp: TM \to M$ is the exponential map coming from our metric, and $U_{\gamma} \subseteq W^{1,p}(\gamma^{*}TM)$ is chosen small enough so that for any $X \in U_{\gamma}$, $\exp_{\gamma(t)}{X(t)}$ stays within an injectivity radius of $\gamma(t)$.
I'm interested in a generalization of this construction when the endpoints of my curves lie in submanifolds, $S_{1},S_{2}\subseteq M$.  I could try and do the same thing as above, varying over all pairs $(x,y) \in S_{1}\times S_{2}$:
$\mathcal{P}^{1,p}(S_1,S_2) := \coprod_{(x,y) \in S_1\times S_2} \mathcal{P}^{1,p}(x,y)$
But this gives me a space with the "wrong" topology.  For example, any two paths with different endpoints are not in the same path component of $\mathcal{P}^{1,p}(S_1,S_2)$ as I've defined it here.  I want paths with nearby endpoints to be considered close together.
To fix the previous problem, I could instead complete to a space where my local model is:
$C^{k}_{S_1,S_2}(\gamma^{*}TM) = \left\{X \in C^{k}(\gamma^{*}TM):\, \lim_{t\to-\infty} X(t) \in TS_1, \lim_{t\to\infty} X(t) \in TS_2\right\}$
This local model allows for variations of the end points; however, the drawback is that I can't use the same charts as I did before.  The problem is that $S_1$ and $S_2$ might not be totally geodesic manifolds, so my endpoints will detach from $S_1$ and $S_2$ when I try and deform through some $X \in C^{k}_{S_1,S_2}(\gamma^{*}TM)$ using the exponential map.  For the applications I have in mind, assuming that $S_1$ and $S_2$ are totally geodesic submanifolds is too restrictive.
Here are my questions:

Is there already in the literature a construction of a Banach manifold of paths whose endpoints are on fixed submanifolds? If it helps, I'm interested in this space in the context of Morse-Bott theory.
Is there any other construction which allows me to produce a 1-parameter family of curves $\gamma_s(t)$ such that $\partial_s\gamma = X$, and so that the endpoints of $\gamma_s$ are always on $S_1$ and $S_2$?

REPLY [2 votes]: For each point $s$ in a submanifold $S\subset M$, there exists
some Riemannian metric on $M$, which makes a neighborhood $U\subset S$ of $s\in S$ totally geodesic in $M$. The definition of (open sets in) these mapping spaces is local in $S_1,S_2$ (see below), so the existence of metrics which make $S_\pm$ near given points totally geodesic is sufficient for the construction of these mapping spaces. 
The auxiliary Riemannian metrics used here have nothing to do with other Riemannian metrics on M, which you may choose later for other purposes. 
(However these auxiliary metrics are all equivalent to any fixed Riemannian metric, so in the end, the exponential decay can still be specified by any fixed Riemannian metric on $M$.)
Now to the local model: Denote $S_-:=S_1,S_+:=S_2$.
For a smooth curve $\gamma:\mathbb{R}\rightarrow M$ with $lim_{t\rightarrow \pm \infty} \gamma(t)=s_\pm$ and (small) open subsets $U_\pm\subset S_\pm$ with
$s_\pm\in U_\pm$, one can consider the local model
$C^k_{U_-,U_+}(\gamma^*TM)=\{X\in C^k(\gamma^*TM) |\lim_{t\rightarrow \pm\infty}X(t)\in T_{s_\pm}S_\pm, exp^{U_\pm\subset M}_{s_\pm}(X(\pm \infty))\in U_\pm\}$
Here the exponential map is associated to some Riemannian metrics on tubular neighborhoods of $U_\pm\subset M$, which make $U_\pm\subset M$ totally geodesic.
In analogy to your space $C^\infty_{\searrow}(x,y)$, one may then consider variations of this space with some exponential decay condition on the component of $X$ normal to $S_\pm\subset M$ (in a tubular neighborhood of $S_\pm$, any vectorfield splits into two parts which are 'tangential resp. normal' to $S_\pm$).<|endoftext|>
TITLE: Publication rates in Mathematics
QUESTION [43 upvotes]: Have there been any studies of publication rates in Mathematics? 
We are trying to construct a workload model for the Faculty of Science and Engineering at my institution. Part of this involves assigning a fixed number of "points" for each published paper. It seems that our colleagues in some of the sciences publish many more papers than we do in Mathematics, which leaves us asking for the number of points per paper to be far higher in Mathematics than elsewhere. But we need to be able to back up our impressions with facts. 
What I would like to do is to get some idea of how many papers one might expect a research mathematician to publish over, say, a five-year period. I recognize that there are a lot of problems here with the words "expect" and "research mathematician", not to mention problems with counting a 100-page paper on the same footing as a 5-page paper, or a paper in a "top" journal on the same footing as a paper in a not-so-top journal; I want to stay away from all those subjective and opinion-based issues. 
I would like to know whether there are any publically-available figures along the following lines: pick a university where faculty are expected to be engaged in research; find out how many publications each member of the Math Department has had over (say) a five-year period; publish the median, or some other measure of the distribution of the publication numbers (not the mean, which could be skewed by a small number of members publishing a large number of papers). 
I'm aware of the paper by Jerrold Grossman, Patterns of collaboration in mathematical research, SIAM News 35 (2002), but that's a study of all papers listed in Math Reviews, which includes people who published a paper or two and then left research mathematics for other fields. I'm really interested only in people who are employed by departments where publication in refereed journals is expected.

REPLY [5 votes]: My impression is that some (scientific) fields tend to have many papers with very large number of authors, while mathematics papers tend to have fewer authors. (Although the number of co-authored math papers has been increasing, which has been documented by MathSciNet data.) Anyway, if your university is still discussing this, you might point out that less credit should be given for an N-person paper than for a 1-person paper. So maybe the way to assign "points" to a paper is to give a person 1/N points for an N-author paper. That might help math. Or, since some level of collaboration is to be encouraged, maybe use a weighting system that decreases in some other way, e.g., an N-person paper gets 1+2/N points or 1+3/N points.
A similar weighting system could be assigned to citations, you get 1/N of a citation for each time your N-author paper is cited!
Note that this wouldn't require you to get data about publication rates in different fields. It's a fairness argument about the amount of effort each individual is putting into their papers.<|endoftext|>
TITLE: Geometrically irreducible variety
QUESTION [7 upvotes]: Let $X\subseteq \mathbb{C}^n$ be an irreducible variety defined over $\mathbb{Q}$. I would like to show that for all but finitely many prime $p$ the variety $X(\mathbb{F}_p)$, defined over $\mathbb{F}_p$, is geometrically irreducible, i.e., $X(\overline{\mathbb{F}}_p)$ is irreducible. Can someone give me a hint or a reference?
Thanks

REPLY [10 votes]: Note that there is some small ambiguity here, as to talk about the reduction of $X/\mathbb{Q}$ modulo a prime $p$, one needs to choose a model $X$. i.e., a scheme $\mathcal{X} \to \mathbb{Z}$ whose generic fibre is isomorphic to $X$.
Anyway, if $X/\mathbb{Q}$ is geometrically integral, then the same holds over $\mathbb{F}_p$ for all but finitely many primes $p$. This follows from fact that being geometrically integral is a constructible property (see Section 9 of EGAIV). One applies this to the morphism $\mathcal{X} \to \mathbb{Z}$.
Precise references:
Definition of a constructible property: EGAIV Définition (9.2.1)
Proof that being geometrically integral is a constructible property: EGAIV Théorème (9.7.7).
You can also find a nice treatment of this in Chapter 10 of the book
Ulrich Görtz, Torsten Wedhorn - Algebraic Geometry: Part I: Schemes. With Examples and Exercises
 By<|endoftext|>
TITLE: Maximal $L_1$ norm of Fourier Transform of a Subset
QUESTION [8 upvotes]: Let $A_n$ be the following $n\times n$ matrix: $(A_n)_{i,j}= \frac{1}{\sqrt{n}}\omega_n^{i\cdot j}$ for all $0 \le i,j <n$, where $\omega_n=e^{\frac{2\pi i}{n}}$.
I want to understand how $A_n$ acts on the set $\{0,1\}^n \subseteq \mathbb{C}^n$ in the following specific sense:

Let $f(n)= \max_{v\in\{0,1\}^n} |A_n( v)|_1$. What upper bounds on $f$ exist? What is its asymptotic behavior? Is there a general form for some $v\in \{0,1\}^n$ at which $f$ attains its maximum, or is relatively close to its maximum?

Here are more delicate questions:

What can be said about $f(n,w):=\max_{v\in\{0,1\}^n, |v|_1=w} |A_n( v)|_1$?
What can be said about $g(n,C):=2^{-n} \{v \in \{0,1\}^n \mid |A_n (v)|_1 \le C\}$?

For the $L_2$ norm these questions are much easier as $A_n$ is unitary and preserves the norm.

REPLY [6 votes]: Regarding Q1, if one selects $v$ randomly and uses Khintchines's inequality one obtains a lower bound $f(n) \gg n$, which complements the trivial upper bound of $f(n) \leq n$ coming from Plancherel and Cauchy-Schwarz.  It looks like the same argument should also show that $f(n,w)$ is comparable to $w^{1/2} n^{1/2}$ for Q2.
For Q3, Theorem 1.3 of this paper of Green and Sanders should (in principle at least) answer the question in the regime where $C$ is a bounded multiple of $n^{1/2}$ (which is the smallest possible nontrivial value of $|A_n(v)|_1$, if I understand your normalisations correctly) and $n$ is large.
While not directly related to your questions (as they are more concerned with the min value of the $L^1$ norm rather than the max, and involve Fourier analysis on the unit circle rather than a cyclic group), you may find the work on Littlewood's conjecture on exponential sums (solved independently by McGehee-Pigno-Smith and by Konyagin) to be of interest.<|endoftext|>
TITLE: Is there a complex surface into which every Riemann surface embeds?
QUESTION [83 upvotes]: This question was previously asked on Math SE.

Every Riemann surface can be embedded in some complex projective space. In fact, every Riemann surface $\Sigma$ admits an embedding $\varphi : \Sigma \to \mathbb{CP}^3$. It follows from the degree-genus formula that the same is not true if we replace $\mathbb{CP}^3$ with $\mathbb{CP}^2$; for example, no Riemann surface of genus two can be embedded in $\mathbb{CP}^2$.
Said another way, there exists a compact complex manifold of dimension three into which every compact Riemann surface embeds (namely $\mathbb{CP}^3$), but the natural two-dimensional candidate (namely $\mathbb{CP}^2$) does not have this property. So my question is

Is there a compact complex surface into which every compact Riemann surface embeds?

The only other surface I have checked is $\mathbb{CP}^1\times\mathbb{CP}^1$. My hope was that every Riemann surface $\Sigma$ admitted an embedding $\varphi : \Sigma \to \mathbb{CP}^3$ with $\varphi(\Sigma)$ contained in the image of the Segre embedding $\mathbb{CP}^1\times\mathbb{CP}^1 \to \mathbb{CP}^3$. Many more Riemann surfaces embed in $\mathbb{CP}^1\times\mathbb{CP}^1$ than in $\mathbb{CP}^2$ - in particular, every genus can be realised. However, not all Riemann surfaces can be embedded in $\mathbb{CP}^1\times\mathbb{CP}^1$: any genus three Riemann surface which embeds in $\mathbb{CP}^1\times\mathbb{CP}^1$ must be hyperelliptic, but one can show using a dimension counting argument that there exist non-hyperelliptic genus three Riemann surfaces.

REPLY [84 votes]: The answer is negative. Suppose for contradiction that $S$ is such a surface, and let me first assume that it is smooth and projective.
Fix $g\geq 24$. Then the coarse moduli space of genus $g$ curves $M_g$ is of general type (this is due to Harris, Mumford and Eisenbud, see for instance [The Kodaira dimension of the moduli space of curves of genus $\geq 23$]), hence a fortiori not uniruled.
 Let $H$ be the Hilbert scheme of smooth curves of genus $g$ in $S$ and let $(H_i)_{i\in I}$ be its irreducible components: the index set $I$ is countable. By hypothesis, the classifying morphism $H\to M_g$ is surjective at the level of $\mathbb{C}$-points. By a Baire category argument, there exists $i\in I$ such that $H_i\to M_g$ is dominant. 
Suppose that the natural morphism $H_i\to \operatorname{Pic}(S)$ is constant. Then $H_i$ is an open subset of a linear system on $S$, hence is covered by (open subsets of) rational curves. By our choice of $g$, $H_i\to M_g$ cannot be dominant, which is a contradiction. Thus, $H_i\to \operatorname{Pic}(S)$ cannot be constant, which proves that $\operatorname{Pic}^0(S)$ cannot be trivial. Equivalently, the Albanese variety $A$ of $S$ is not trivial.
Consider the Albanese morphism $a:S\to A$. The curves embedded in $S$ are either contracted by $a$ or have a non-trivial morphism to $A$. Those that are contracted by $a$ form a bounded family, hence have bounded genus. Curves $C$ that have a non-trivial morphism to $A$ are such that there is a non-trivial morphism $\operatorname{Jac}(C)\to A$, but this is impossible if $\operatorname{Jac}(C)$ is simple of dimension $>\dim(A)$. Consequently, a smooth curve with simple jacobian that has high enough genus cannot be embedded in $S$. This concludes because a very general curve of genus $g$ has simple jacobian (there even exist hyperelliptic such curves by [Zarhin, Hyperelliptic jacobians without complex multiplication]). 
As pointed out in the comments, the argument needs to be modified if $S$ is non-algebraic or singular. I explain now these additional arguments.
If $S$ is smooth but non-algebraic, we can use the following (probably overkill) variant. We can consider the open space $H$ of the Douady space of $S$ parametrizing smooth connected curves in $S$. It is a countable union of quasiprojective varieties by [Fujiki, Countability of the Douady space of a complex space] and [Fujiki, Projectivity of the space of divisors on a normal compact complex space]. It has an analytic morphism to the analytic space $\operatorname{Pic}(S)$ parametrizing line bundles on $S$ [Grothendieck, Techniques de construction en géométrie analytique IX §3]. The dimension of $\operatorname{Pic}(S)$ is finite equal to $h^1(S,\mathcal{O}_S)$. The dimension of the fibers of $H\to\operatorname{Pic}(S)$ is at most $1$. Indeed, otherwise, we would have a linear system of dimension $>1$ on $S$ consisting generically of smooth connected curves, hence a dominant rational map $S\dashrightarrow\mathbb{P}^2$, showing that $S$ is of algebraic dimension $2$, hence that $S$ is algebraic. It follows that every connected component of $H$ has dimension $\leq h^1(S,\mathcal{O}_S)+1$. Now choose $g$ such that $\dim(M_g)>h^1(S,\mathcal{O}_S)+1$, and let $H_g$ be the union of connected components of $H$ parametrizing genus $g$ curves. A Baire category argument applied to the image of $H_g\to M_g$ shows that there are genus $g$ curves that do not embed in $S$, as wanted.
Finally, if $S$ is singular, consider a desingularization $\tilde{S}\to S$. The hypothesis on S implies that every smooth projective curve may be embedded in $\tilde{S}$, with the exception of at most a finite number of isomorphism classes of curves (namely, the curves that are connected components of the locus over which $\tilde{S}\to S$ is not an isomorphism). The arguments above apply as well in this situation.<|endoftext|>
TITLE: Monoidal cats and string diagrams for a semantics of object oriented programming languages
QUESTION [5 upvotes]: It seems to me that in a statically typed, object oriented language, there is a striking similarity to wiring diagrams.  Wires (objects) of type $X$ go into functions (boxes) of input type $X$.  Is the corresponding string calculus that of a symmetric monoidal category?
Edit:  it has been suggested that I formulate a more specific question.  Here goes:  let the programming language have private member variables.  At this point you cannot copy arbitrary objects because the data is hidden. I guess you also have to assume that there is no generic copy method for all types.  Becasue copying isn't allowed arbitrarily, we might need a monoidal category to handle the semantics.  Is this the case?

REPLY [4 votes]: No full answer but just some random thoughts on the issue:
Concerning the string diagram calculus: It should be that of a mutlicategory (see David Spivaks paper on the Wiring Operad: http://arxiv.org/abs/1305.0297).
Plus I think there's a hidden issue here: Private member variables are only of real concern if the objects are stateful; Otherwise - if they are not observable after object instantiation a.k.a. function call - they can just be considered implementation details. I don't think modelling the wiring between the objects using a multicategory / monoidal category is an approach powerful enough to reason about mutable state. But I guess by adding labels to the diagrams (similar to Petri nets) and let them vary over time this could be fixed. (Again, consult David Spivak for an idea on how to model such a labeling: http://math.mit.edu/~dspivak/informatics/olog.pdf)
EDIT: Concerning the labeling - a rough sketch: A typing of the diagrams at hand shoult propably be used as input to some sort of grothendieck construction and a labeling should then be a lift of the base diagram into the respective fibration. Including Maybe/Option/Exception Types should allow for partial diagrams while the program is still 'booting'.<|endoftext|>
TITLE: $k$-planar graphs and genus
QUESTION [5 upvotes]: Is there a simple function that connects $k$ in $k$-planar graphs and genus of such graphs?
If there is no simple function is there any non-trivial upper and lower bound?

REPLY [6 votes]: For fixed $k$ (even $k=1$) the genus can be arbitrarily large (linear in the number of vertices). Graphs with genus $g$ have $3n+O(g)$ edges by Euler's formula, while 1-planar graphs can have as many as $4n-O(1)$ edges. So dense 1-planar graphs (e.g. take a planar graph with all faces quadrilaterals and add two crossing edges in each face) have high genus.
In the other direction, even the graphs of bounded genus can have unbounded local crossing number (again linear in the number of vertices). For instance, take the nonplanar (but toroidal) graph $K_{3,3}$ and replace each edge by a collection of $n$ two-edge paths ($K_{2,n}$). Then the resulting toroidal graph has quadratic crossing number (it has $\Omega(n^9)$ subdivided copies of $K_{3,3}$, each with at least one crossing, and each crossing belongs to only $O(n^7)$ subdivided copies of $K_{3,3}$) so some edge must be crossed linearly many times.
So there is no functional relationship in either direction.<|endoftext|>
TITLE: are the coarse moduli schemes of finite etale covers of $\mathcal{M}_{1,1}$ smooth?
QUESTION [5 upvotes]: Let$\newcommand{\mM}{\mathcal{M}}$ $\mM_{1,1}$ be the moduli stack of elliptic curves. Let $R$ be a Dedekind domain, say $\mathbb{Z}[1/N]$ for simplicity, and suppose we have a finite etale cover:
$$\mM\rightarrow\mM_{1,1}[1/N]$$
Must the coarse moduli scheme $M$ of $\mM$ be smooth over $\mathbb{Z}[1/N]$? This is certainly true if $\mM$ is representable, since $\mM_{1,1}$ is smooth over $\mathbb{Z}$. Furthermore, for a geometric point Spec $k\rightarrow$ Spec $R$ of characteristic not 2 or 3, $M_k$ is the coarse moduli scheme of $\mM_k$, which is normal since it's the quotient of the scheme $\mM\times_{\mM_{1,1}[1/N]} \mM(N^2)$ by $GL_2(\mathbb{Z}/N)$, where $\mM(N^2)$ is the representable stack over $\mathbb{Z}[1/N]$ classifying elliptic curves with full level $N$ structure. Since $M_k$ is a normal curve, it's smooth, and thus since $R$ is a Dedekind domain, we find that $M[1/6]$ is smooth over $\mathbb{Z}[1/6N]$.
The same argument doesn't work at $p = $ 2 or 3, since for a residue field $k$ of $R$ of characteristic 2 or 3, $M_k$ is not necessarily the coarse moduli scheme of $\mM_k$ (the stack is not tame over 2 and 3). Certainly the only potential singularities are the preimages of $j = 0\equiv 1728\mod p$ in $M_k$.
Is this a real obstruction? Of course if $\mM$ is representable then this is a non-issue, and maybe for special problems like $\Gamma_0(N)$ one can use division polynomials, but that's kind of ad hoc. I'd like to understand better what can go wrong generally speaking. If this is a real problem, then can we at least say it's flat?

REPLY [5 votes]: Yes, $M$ is smooth. 
In proving this we may focus on a fixed residue characteristic, so we loose no generality by assuming that $N$ is divisible by a prime $\ell \ge 3$. The morphism $\mathcal{Y}(\ell)[1/N] \rightarrow \mathcal{M}_{1, 1}[1/N]$ is a $\mathrm{GL}_2(\mathbb{Z}/\ell\mathbb{Z})$-torsor, and so is its base change $\mathcal{Y}(\ell)[1/N]_{\mathcal{M}} \rightarrow \mathcal{M}$. Moreover, $\mathcal{Y}(\ell)[1/N]_{\mathcal{M}}$ is a smooth $\mathbb{Z}[1/N]$-curve (say, by IV.2.5 in Deligne--Rapoport) and $M$ is the categorical (ring of invariants) quotient $\mathcal{Y}(\ell)[1/N]_{\mathcal{M}}/\mathrm{GL}_2(\mathbb{Z}/\ell\mathbb{Z})$. In order to conclude it remains to recall the following useful theorem from the appendix of Katz--Mazur (see "Notes on Chapters 8 and 10").
Theorem. If $S$ is a Noetherian regular scheme and $X \rightarrow S$ is a relative curve that is affine over $S$ and that is equipped with an $S$-linear action of a finite group $G$, then the categorical quotient $X/G$ is a smooth $S$-curve.
The argument is the same one that also proves smoothness of coarse moduli spaces in the case of congruence level modular curves away from the level (see, for instance, the footnote on 10.10.3 (5) in Katz--Mazur) and works over any Dedekind domain $R$ as in your question.<|endoftext|>
TITLE: Commutative algebraic version of algebraic geometric object
QUESTION [13 upvotes]: In my work, I have to understand certain objects in commutative algebra (for example Gorenstein rings, Cohen–Macaulay rings e.t.c). I have a reasonable background in commutative algebra (I suppose!) and very basic knowledge of algebraic geometry. The problem is that most of the time I have tried to understand some properties about these objects from a paper, I found that it is written in the language of advanced algebraic geometry which is almost impossible for me to understand. Therefore my question is:
Are there notes, books, papers or survey articles which has parallel discussions about objects in commutative algebra and its counterpart in algebraic geometry?
Studying algebraic geometry is of course the answer but as both of these topics are not directly related to my work and I have already spent reasonable amount of time to understand one of them, this kind of literature would be extremely helpful for me. 
PS1: Any suggestion is welcome. Thanks in advance. 
PS2: I apologise if this question is not relevant in mathoverflow but I have asked people and searched but I could not find any advance literature like these.

REPLY [9 votes]: I think this question is quite important. For a commutative algebraist, some knowledge of algebraic geometry is extremely useful (and vice versa). Not only that, and more important to me personally, it makes life more fun, like being able to talk in another language. You can converse meaningfully with more people, appreciate more talks and papers, and get more insights/motivations for your own research. 
Unfortunately, with the explosion of knowledge and Balkanization of the areas, it is now virtually impossible for young people to learn deeply both subjects at the same time, unless if you are very motivated and is at a school with leading experts. So a lot of algebraists will go through their career without having to really learn what a "projective variety" or a "line bundle" is, and so are many geometers with "Gorenstein ring" or "integral closure of an ideal".
Below I will describe my own experience, hopefully it will help you. Unfortunately, I do not know any convenient text.  
First, you don't have to become an expert. Even being able to translate simple phrases will serve you well. For example, in my thesis I had a problem about primes ideals over a local hypersurface. I translated it in to a question about subvarieties of a projective hypersurface, and realized that my question can be solved there with the "moving lemma". While that is not available in my local situation, I was able to find a weak version to make things work.
Second, you still need to invest a certain amount of time on learning some basic geometry, probably by selecting certain topics from Hartshorne's first 3 chapters and compare them with bits from Eisenbud. I would focus my attention to the following to start:

The connections between the geometry of a projective scheme $X$ and it's various embeddings. Even when $X$ is the projective space. 
The connections between the graded  coordinate ring with respect to an embedding and the local ring at the origin. 
The connections between sheaf cohomology of a coherent sheaf and the local cohomology of the corresponding module.

Third, you can pick up more knowledge by occasionally looking at abstracts/papers, go to seminars/conferences, and try to match up what people are doing over there with what you know. For instance, you may learn that a smooth Calabi-Yau hypersurface is really a graded hypersurface of degree $n$ in $n$ variables with isolated singularity. So next time, when people talk about CY hypersurfaces, try to translate what they are interested in to your world.  
Let me give a concrete example of the simplest Calabi-Yau hypersurface, namely an elliptic curve $E$ embedded in $\mathbb P^2$. Say $X= Proj(R)$ with $R=k[x,y,z]/(x^3+y^3+z^3)$. Then $H^1(E,\mathcal O_E)=k$, a fact you can relate to algebraically by realizing that this is the $0$-graded piece of local cohomology $H^2_m(R)_0$. You may also learn that the Frobenius action on this cohomology is $0$ or bijective depending on what reduction you make ("ordinary" and "supersingular" curves). It is not hard to compute this piece of local cohomology, it is generated by the class of $\frac{z^2}{xy}$, and the Frobenius takes this to $\frac{z^{2p}}{x^py^p}$. See for yourself why this class is $0$ or not depending on residue of $p$. Such an example would help you to see the connections more clearly and build up your confidence.   
Last comment, make sure you read/watch people who are strong in both worlds and are good communicators. For instance, the papers by David Eisenbud and Karen Smith often have useful paragraphs that explain connections between local algebra and global geometry.<|endoftext|>
TITLE: Gromov-Hausdroff convergence for Alexandrov spaces
QUESTION [5 upvotes]: Let $\{X_n\}_{n=1}^\infty$ be a sequence of compact Alexandrov spaces (with curvature $\geq k$) converging to (in the sense of Gromov-Hausdroff convergence) an Alexandrov spaces $X$, and $f_n:X_n\mapsto X$ is the corresponding $\nu-$ approximation. The set of regular points in $X_n$ is denoted by $R_n$.
Question: (1) When $n$ sufficiently large, can we prove that $f_n$ is Lipschitz continuous on $R_n$ ? Or a weaker conclusion: $f_n$ is Lipschitz continuous on some $Y_n\subset R_n$ with $\mathrm{vol}(R_n-Y_n)=o(1)$ ,as $n\to\infty$ ?
(2) If question (1) is not true, can we construct a $g_n:X_n\mapsto X$ (maybe related to f_n), such that $g_n$ is Lipschitz continuous on some $Y_n\subset R_n$ with $\mathrm{vol}(R_n-Y_n)=o(1)$ ,as $n\to\infty$ ?

REPLY [4 votes]: The approximations $f_n$ (by def.) need not to be continuous, but you can approximate Lipschitz ones.
I will give a hint how to prove such thing, as far as I know the statement is not written anywhere.
If $X$ is Riemannian the statement follows from Yamaguchi theorem
and nearly the same proof works if all the tangent spaces of $X$ are close to Euclidean.
Now assume $X$ has an isolated strongly singular point $p$. Then the above argument gives an approximation which is Lipschitz approximations outside a fixed neighborhood of $p$. Contract all this neighborhood into $p$ and compose the resulting map with the approximation $X_n\to X$. This way you get a Lipschitz approximation.
The general case is none by backward induction on the dimension of strongly singular locus. Section 8 in my paper  "Semiconcave functions..." should help. Good luck.<|endoftext|>
TITLE: Geometric morphism of $\infty$ topos
QUESTION [9 upvotes]: I have a very simple question regarding geometric morphisms of $\infty$ topoi, but have been unable to find the answer in Lurie's HTT (although it seems likely that its there somewhere and I just can't find it). 
Classically, given two sites $C$ and $D$, a morphism of sites $f:C\to D$ induces a geometric morphism of topos $f_{*}:Shv(C)\to Shv(D)$. Does the statement still hold in the $\infty$ context? 
What I'm really after is the following: Let $C$ be an $\infty$ site and let $c$ be an object of $C$. Then $C_{/c}$ inherits a topology from $C$ by restricting the covering sieves to objects in the overcategory. The forgetful functor
$f:C_{/c}\to C$ is trivially a morphism of $\infty$ sites. Does $f$ induce a functor
$$
f^*:Shv(C)\to Shv(C_{/c})$$
preserving finite $\infty$ limits?

REPLY [12 votes]: I never found any discussion about this in HTT either, but it turns out to work exactly the same way as in SGA 4.
That is, let $f^* : P(D) \to P(C)$ denote the restriction functor on presheaves, and $f_!$ its left adjoint (given by left Kan extension).
Since $f$ preserves covering families, $f_!$ preserves local equivalences, so by adjunction $f^*$ preserves sheaves and induces a functor
  $$ f^*_s : Sh(D) \to Sh(C). $$
Define
  $$ f_!^s := a_D f_! i_C : Sh(C) \to Sh(D), $$
where $i_C$ is the inclusion $Sh(C) \hookrightarrow P(C)$ and $a_D : P(D) \to Sh(D)$ is the associated sheaf functor.
It is straightforward to check that this is left adjoint to $f^*_s$.
The fact that $f_!^s$ is left-exact follows from the fact that $i_C$ commutes with (small) limits, $f_!$ commutes with finite limits, and $a_D$ commutes with finite limits (I am assuming of course that $D$ is small).
The fact that $f_! : P(C) \to P(D)$ commutes with finite limits follows from the pointwise formula for the Kan extension
  $$ f_! F(d) = \varinjlim_{(c \in C, d \to f(c)) \in (d/f)} F(c) $$
where, since $f$ is left-exact, the comma category $(d/f)$ is filtered. (The commutativity of filtered colimits and finite limits in an $\infty$-topos is Example 7.3.4.7 in HTT.)<|endoftext|>
TITLE: Etale fundamental of a parahoric group scheme
QUESTION [7 upvotes]: Let $p:X\rightarrow Y$ be a double cover of curves, denote by $$SU_n:=(p_*SL_n(\mathcal O_X))^{\tilde{\sigma}}$$
i.e. the $\tilde{\sigma}-$invariant part, the action of $\tilde{\sigma}$ is given by $$\tilde{\sigma}(g)=\,^t(g\circ\sigma)^{-1}$$ where $\sigma$ is the involution induced by the double cover. $SU_n$ is well knowing to be a parahoric group shceme in the sens of Bruhat-Tits. 
My question: What is $\pi_1(({SU_n})_\eta)$ ? (the algebraic fundamental group). where $\eta$ is the generic point of $Y$.
Thanks

REPLY [2 votes]: There is an exact sequence
$$ \pi_1 ((SU_n)_{\overline{\eta}}) \to \pi_1((SU_n)_{\eta}) \to \operatorname{Gal}(\overline{\eta}|\eta)$$
Over $\overline{\eta}$, $p_* SL_n$ is just $SL_n \times SL_n$. The involution $\tilde{\sigma}$ acts by switching the two $SL_n$s and then doing an performing some automorphism, so the involution invariants $SU_n$ are just the diagonal $SL_n$ embedded by the graph of the automorphism.
In characteristic $0$, $\pi_1(SL_n)$ is trivial, so the fundamental group is just the Galois group of the base field. In characteristic $p$ things are more complicated as the fundamental group is infinite, as witnessed by the family of etale covers $g \to g^{-1} \operatorname{Frob}_q(g)$ which is Galois of group $SL_n(\mathbb F_q)$ for each $q$.<|endoftext|>
TITLE: Self intersection and deformations
QUESTION [5 upvotes]: Suppose I have a smooth projective surface $S$ and a curve $C\subset S$. The self-intersection of $C$ is by definition the degree of the restriction to $C$ of the normal bundle of $C$ inside $S$.
By the very definition, if I deform $C$ inside $S$ then the self intersection does not change. Now, I wonder if this stays constant if I deform the surface as well, but keeping it smooth. More precisely: let us fix $S$ smooth first and deform $C$ to $C'\subset S$ and then deform $S$ to $S''$ via a smooth family. Let $C''\subset S''$ the deformation of $C'$ obtained through the deformation $S \to S''$. Does $(C\cdot C)=(C'' \cdot C'')$ always hold?

REPLY [5 votes]: I think that, if the curve does not disappear, then the answer is positive.
In fact, let $\pi \colon \mathcal{X} \to \Delta$ your deformation of surfaces over a disk, with $S_1= \pi^{-1}(t_1)$ and $S_2 = \pi^{-1}(t_2)$. The assumption that $C_1 \subset S_1$ deforms to $C_2 \subset S_2$ along $\pi$ means that there exists a line bundle $\mathcal{L}$ on the total space $\mathcal{X}$ such that $$\mathcal{L}|_{S_1}=\mathcal{O}_{S_1} (C_1), \quad \mathcal{L}|_{S_2}=\mathcal{O}_{S_2} (C_2).$$
Therefore we have $$(C_1 \cdot C_1) = \mathcal{L} \cdot \mathcal{L} \cdot S_1 = \mathcal{L} \cdot \mathcal{L} \cdot S_2 = (C_2 \cdot C_2),$$
since any two fibers of $\pi \colon \mathcal{X} \to \Delta $ are clearly numerically equivalent.<|endoftext|>
TITLE: List of generic properties of Riemannian metrics
QUESTION [9 upvotes]: I am highly interested in compiling a list of generic properties of Riemannian metrics on a (may be compact) manifold in general, or under "relatively broad" assumptions, like generic properties of all non-positively curved metrics on a given manifold (broad curvature assumption), or maybe generic properties of metrics on a surface (broad dimension assumption). It is primarily for my own general knowledge, but may be it could help others as well to have a handy list lying around. As examples of what could be possible answers, let me begin the list with two generic properties that come to mind immediately:

Generic simplicity of spectrum and generic properties of eigenfunctions: heuristically, the eigenvalues of the Laplacian for the generic metric on a compact manifold are non-repeated. Also, the eigenfunctions are generically Morse and have no nodal critical points (Uhlenbeck and Albert).
The generic Riemannian metric on a smooth manifold is real analytic. This can either be realized by local approximation in charts, or by applying the Ricci flow and using a result of Bando (mr=888130).

Please add your own favorite generic property. Thanks!
Edit: As commented below, the word "generic" needs clarification. I am using it to refer to properties that are too strong for just any given metric to have, but if you perturb the metric slightly, you can produce a metric that now has the said property.

REPLY [3 votes]: Generic metric does not admit a lot of properties some special metrics admit.
A good demonstration of this is the examples listed in the question (no multiple eigenvalues) or in the answers of
Matheus, 
Ben and 
Holonomia  (and the property in the answer of Anton is of very different nature). 
My item of the same kind is as follows: 
(1) geodesic flow of the generic metric does not admit an integral which is polynomial or even analytic in momenta 
(see http://xxx.lanl.gov/abs/1510.01493 or 
Generic absence of non-trivial first integrals of geodesic flows).  This is a local property. 
(2) Morover, a geodesic flow of a generic metric on a closed manifold  has a hyperbolic basic set which in particular implies that it has positive topological entropy, see 
http://annals.math.princeton.edu/2010/172-2/p01
Since there was a discussion in the comments to the question whether the author means ''generic'' or ''dense'', my examples work in both meanings.<|endoftext|>
TITLE: "Squeezing" a finite group between symmetric groups
QUESTION [18 upvotes]: For a finite group $G$, take $m$ as the largest integer such that $G$ has a subgroup $H\cong S_m$ and $n$ as the smallest integer such that $G$ is itself isomorphic to a subgroup of $S_n$. We then define the "squeezing number" of $G$ as $s(G):=\dfrac nm$. This is probably not a new idea, so pointers are welcome. My question: 

Can each rational $q\ge1$ occur as squeezing number for an appropriate group? If not, what can be said about the set of squeezing numbers?

REPLY [32 votes]: Here is an attempt assuming Goldbach's Conjecture. Express the required ratio as $n/m$ with $n-m \ge 8$ even, and take $G = S_m \times C_{pq}$, where $p$ and $q$ are distinct primes with $p+q=n-m$.
In fact it appears to have been proved that every sufficiently large even integer is the sum of four distinct primes, so we could use that to complete the proof: choose $n/m$ such that $n-m$ is large and even, and take $G = S_m \times C_{pqrs}$ where $p,q,r,s$ are distinct primes with $p+q+r+s=n-m$.
But perhaps this is over-complicated. An alternative solution, that works whenever $n \ge m+2 \ge 4$ is to write $n=qm+r$ with $0 \le r < m$. If $r >1$, take $G = S_m^q \times S_r$, if $r=0$, $G=S_m^q$, and if $r=1$, $G = S_m^{q-1} \times A_{m+1}$.<|endoftext|>
TITLE: Between arithmetic and geometric Brownian motions: when are negative values possible?
QUESTION [5 upvotes]: Please note edits after original post changing the specific form of the setup
Let's say we have a stochastic differential equation:
$$
\mathrm{d}S_t = |S^\beta| {(\mu \mathrm{d}t + \sigma\mathrm{d}W_t)}
$$
where $W_t$ is a standard brownian motion.  
When $\beta=0$, the whole thing continues to be a brownian motion (albeit with drift and with non-unit variance).  In particular, there is a non-zero probability that $S_t$ will take on negative values.  Conversely, when $\beta=1$ this describes a particular instance of geometric brownian motion and as a result, there process almost certainly avoids taking on negative values.
My question: for $0 < \beta < 1$ (note: strict inequalties), what statements can be made about the probability of the process taking on negative values?  In particular, are negative values almost never achieved for whenever $\beta < 1$, or is there some critical $0 < \beta_c < 1$ such that processes with $\beta < \beta_c$ are able to take on negative values with non-zero probability while those with $\beta > \beta_c$ are not?  Does the answer depend on the value of $\mu$?  I'm most interested in the case where $\mu=0$ but left the drift term there to see if someone could offer insight about the more general setup.

REPLY [3 votes]: Assuming that $\mu$ and $S_0$ are positive, the process stays almost surely non-negative. This is easily seen as when $S$ hits zero, it has a deterministic drift upwards. However, the process does not necessarily stay strictly positive, the hitting probability of zero can be positive.
This model is well studied in the financial mathematics literature and goes there under the name CEV model. The generic reference on it and the close relationship to Bessel processes is the following paper by Delbaen and Shirakawa: A Note of Option Pricing for Constant Elasticity of Variance Model, https://people.math.ethz.ch/~delbaen/ftp/preprints/CEV.pdf<|endoftext|>
TITLE: Connectedness of units in finite-dimensional commutative complex algebras
QUESTION [6 upvotes]: In the following, an algebra will always mean a finite-dimensional associative commutative unital algebra (over some field $k$).
Let $A$ be a $\mathbb{C}$-algebra. I am trying to understand how its group of units $A^{\times}$ sits topologically in $A$ relative to the locus of non-invertible elements $A\setminus A^{\times}$.
In the answer to this question at MSE, Jeremy Rickard gives a nice argument showing that every $\mathbb{R}$-algebra $A$ with connected group of units $A^{\times}$ is necessarily complex.

(Q1) Is the converse true, i.e. does every $\mathbb{C}$-algebra have connected group of units?

If not,

(Q2) Can we characterize / classify those $\mathbb{C}$-algebras with connected group of units? Or is that too much to hope for? (Note that in general the classification of finite-dimensional commutative $\mathbb{C}$-algebras is a hard problem, see for example B.Poonen's work for more details on this.)

Since $A$ as above is necessarily complex, it follows that $A^{\times}$ is connected abelian complex Lie group, hence $A^{\times}\simeq\mathbb{C}^n/\Gamma$ for some lattice $\Gamma\subset\mathbb{C}^n$. (In fact, by Remmert-Morimoto $A^{\times}$ is isomorphic to a product of finite number of copies of $\mathbb{C}$, of $\mathbb{C}^{\times}$, and a Cousin group.)

(Q3) Conversely, when can a connected abelian complex Lie group be realized as the group of units of some $\mathbb{C}$-algebra $A$?

REPLY [10 votes]: Even without assuming commutativity, the set of non-units is the vanishing set of a not-identically-zero complex polynomial function, the function sending $x$ to the determinant of $y\mapsto xy$. This has complex codimension $\ge 1$, so its complement is connected.

REPLY [9 votes]: For (Q1): A finite dimensional $\mathbb{C}$-algebra $A$ is Artinian, so $A$ is a product of Artin local algebras. The units of a product of algebras is the product of the units, so we may assume $A$ is local, with maximal ideal $\mathfrak{m}$. The complement of a subspace in a complex vector space is connected, so $A^\times=A\backslash\mathfrak{m}$ is connected.
For (Q3): Tom's answer (and I guess mine too) shows that the group of units has the structure of an connected abelian linear algebraic group, so looks like a product of copies of $\mathbb{G}_m$ ($=\mathbb{C}^\times$) and $\mathbb{G}_a$ ($=\mathbb{C}$). Assumining $A\neq 0$, then $x\mapsto$ determinant of multiplication by $x$ is a surjective algebraic group homomorphism to $\mathbb{G}_m$, so there must be at least one factor of $\mathbb{G}_m$ (I confess I don't know what a Cousin group is, but we won't get them). If $G=\mathbb{G}_m^a\times \mathbb{G}_a^b$ with $a\geq 1$, then $G$ is isomorphic to the units of
$$
A:=\mathbb{C}[x_1,\ldots,x_b]/(x_ix_j)\times\mathbb{C}^{a-1}.
$$<|endoftext|>
TITLE: Twists, balances, and ribbons in pivotal braided tensor categories
QUESTION [6 upvotes]: Let $\mathcal{C}$ be a pivotal tensor category. Feel free to assume finiteness, semisimplicity, fusion, sphericality, unitarity or whatever makes things interesting. Which of the following structures come for free?

Does the pivotal structure canonically define a twist: a natural transformation $\theta$ from the identity functor to itself? Or must I to add this structure by hand?
Endow $\mathcal{C}$ with a braided structure $c$ (assume this is possible). Is $\mathcal{C}$ balanced: do $\theta$ and $c$ satisfy $\theta_{x\otimes y}=c_{yx}\cdot c_{xy}\cdot(\theta_y\otimes\theta_x)$? (ie does the ribbon diagram make sense?)
If so, is $\mathcal{C}$ ribbon: does $\theta$ satisfy $(\theta_x)^*=\theta_{x^*}$? (compatible rigidity, twist, and braiding)

I'm asking because premodular tensor categories appear in the literature with (at least) two definitions:

ribbon fusion categories 
spherical braided fusion categories

Ignoring the absence of sphericality from the first definition (is it implied by ribbon?), these definitions disagree unless the answers to the above are affirmative (perhaps assuming fusion and sphericality).

REPLY [11 votes]: Question 2: Given a pivotal braided category $\mathcal{C}$, there are 2 ways to endow $\mathcal{C}$ with twists under which $\mathcal{C}$ is a rigid balanced category. Conversely, given a rigid balanced category $\mathcal{C}$, there are 2 ways to endow $\mathcal{C}$ with a pivotal structure under which $\mathcal{C}$ is braided pivotal. These constructions are mutually inverse. 
This is all explained in detail in Appendix A.2 of my recent article with Henriques and Tener http://arxiv.org/abs/1509.02937. (These results are not original, but we give a comprehensive treatment there.) Of interest to you may also be Section 2.3, which gives a synoptic chart of the types of tensor category.
Question 3: No, not necessarily. For example, the center $\mathcal{Z}(\mathcal{C})$ of any pivotal fusion category $\mathcal{C}$ is pivotal braided, but it is not ribbon unless $\mathcal{Z}(\mathcal{C})$ is spherical. In general, there is no reason for this to be the case. In fact for a semisimple category $\mathcal{C}$, pivotal braided is ribbon if and only if it is spherical. This is Proposition A.4 in our preprint.
Question 1: I suspect the answer is that there is not necessarily any interesting twist in only the presence of a pivotal structure. I don't have an example off the top of my head though. But when you also have a braiding, you're in business as explained in the answer to Question 2 above.
So to sum up the bit at the end, yes, ribbon fusion is the same as spherical braided fusion, because the category is semisimple.<|endoftext|>
TITLE: Subsets of [1..N] with no three-term arithmetic progressions and no large gaps
QUESTION [8 upvotes]: Let S be a subset of [1..N] containing no three-term arithmetic progression, and let h(S) be the size of the largest gap between two consecutive elements of S.  By Roth's theorem, h(S) has to grow with N.  But how fast?  For instance, are there arbitrarily large N admitting a subset of [1..N] with no three-term AP and no gap longer than $N^\epsilon$? Does the subset constructed by Behrend have this property, or might it have unexpectedly long gaps?

REPLY [8 votes]: For the sake of having a reference, Ron Graham shows in "On the growth of a van der Waerden-like function" that for a fixed $k$, there exists a 3AP-free subset of $\{1,2,\dots,N\}$ with gaps bounded by $k$ and $N\geq k^{c\log k}$, where $c$ is some absolute constant that doesn't depend on $k$. Therefore you can get gaps to be even bounded by $\exp(c\sqrt{\log N})$.<|endoftext|>
TITLE: Beyond Brauer's theorem
QUESTION [8 upvotes]: Brauer's classic theorem states that any character of a finite group can be expressed as a linear combination with integer coefficients of characters induced by linear characters of p-elementary subgroups.
This is an improvement over Artin's theorem, in which the linear combination is with rational coefficients of characters induced by characters of cyclic subgroups.
So we get integer coefficients at the cost of using more complicated subgroups.
It is reasonable to try to improve Brauer's theorem, taking the coeffients to be positive integers. But this doesn't work: there are characters that can't be expressed as linear combination with positive coefficients of characters induced from elementary subgroups, even if we let the coefficients be real numbers (this is exercise 10.5 in Serre's book on representation theory).
In fact, Brauer's result is the best possible in a very concrete sense:

The only groups that have the property that all of its characters are expressible with non-negative coefficients in Brauer's theorem are solvable groups.
p-elementary subgroups is the minimal family of subgroups for which Brauer's theorem holds (Green).


What I was wondering is, if we replace p-elementary subgroups by a
  bigger (and more complicated) family of subgroups, much like Brauer
  had to let go of cyclic groups, can we get a similar theorem with positive
  coefficients that works for all finite groups?

$\require{AMScd}$
\begin{CD}
    \mathrm{cyclic} @>>> \mathrm{elementary} @>>> \mathrm{?} \\
    @V  V V\ @VV V\ @VV V\\
    \mathbb{Q} @. \mathbb{Z} @. \mathbb{N}
    \end{CD}

REPLY [10 votes]: Here is another way to think about this, where we consider nonnegative $real$ coefficients, and we do not restrict ourselves to integer coefficients as in Geoff Robinson's post.
If $\chi \in {\rm Irr}(G)$ satisfies $\chi = \sum_\mu a_\mu \mu^G$, where $\mu$ runs over some collection of characters of proper subgroups of $G$ and the coefficients $a_\mu$ are positive (real) numbers, then there exists a positive integer $m$ such that $m\chi$ is induced from a proper subgroup. (In the language of the paper of Le $et\,al$ (J. of Algebra 374 (2013), $\chi$ is "multiply imprimitive".) To see this, note that if $a_\mu > 0$ then $\mu^G$ can have no irreducible constituent other than $\chi$ because there can be no cancellation on account of the assumption that none of the coefficients is negative. It follows that $\mu^G = m\chi$ for some integer $m$, as claimed.
The paper referred to above shows that if every nonlinear irreducible character of a group is multiply imprimitive, then $G$ is solvable. The proof uses the classification of simple groups. 
A stronger property is that $\chi$ is "multiply monomial", which means that some multiple of $\chi$ is induced from a linear character. There is an elementary argument that shows that if every irreducible of $G$ is multiply monomial, then $G$ cannot be perfect. In fact, we argue that in a perfect group $G$, a nonlinear irreducible character of minimum degree cannot be multiply monomial.
Proof: Say $m\chi = \mu^G$, where $\mu \in {\rm Irr}(H)$ is linear, and note that $H < G$. Now write $(1_H)^G = 1_G + \sum a_\psi \psi$, where the $\psi$ are distinct nonprincipal and hence nonlinear irreducible characters and the coefficients $a_\psi$ are positive integers. Also, the set of $\psi$ in this sum is not empty. Then
$$
m\chi(1) = |G:H| = 1 + \sum a_\psi \psi(1) \ge 1 + \chi(1) \sum a_\psi\,,
$$
where the inequality holds because $\psi(1) \ge \chi(1)$ for all $\psi$. Then $m > \sum a_i > 0$, and we have $m^2 > (\sum a_i)^2$, and so
$m^2 \ge 1 + (\sum a_i)^2$. Now 
$$
m^2 = [\mu^G,\mu^G] \le [1_H^G,1_H^G] =
1 + \sum a_i^2 \le 1 + (\sum a_i)^2 \le m^2\,,
$$
Equality thus holds throughout, and thus $m^2$ exceeds the positive square integer $(\sum a_i)^2$ by $1$. This is a contradiction.<|endoftext|>
TITLE: Does it follow that any element of $J(A)$ is nilpotent?
QUESTION [5 upvotes]: Let $A[x]$ be the algebra of polynomials with coefficients in a $k$-algebra $A$. Assume that, for any simple $A[x]$-module $M$, we have $\text{End}_{A[x]} M = k$. Does it follow that any element of $J(A)$ is nilpotent?

REPLY [4 votes]: Let $a \in J(A)$, and suppose for a contradiction that the left ideal $A[x](1 - xa)$ is proper.
We can then choose a maximal left ideal $J$ of $A[x]$ containing $A[x](1-xa)$, so that $M := A[x] / J$ is a simple $A[x]$-module. Let $v := 1 + J$ be the canonical $A[x]$-module generator of $M$; then $(1 - xa)\cdot v = 0$ by construction.
Now $x$ acts on $M$ by $A[x]$-module endomorphisms, so by assumption, there exists $\lambda \in k$ such that $x - \lambda$ kills $M$. Hence
$$(1 - \lambda a) \cdot v = (1 - xa) \cdot v = 0.$$
However, since $a \in J(A)$, $1 - \lambda a$ is a unit in $A$, and it follows that $v = 0$. Hence $M = 0$, contradicting the simplicity of $M$.
So in fact $A[x](1 - xa) = A[x]$, and therefore we can elements $b_0, b_1,\ldots, b_n \in A$ such that
$$(b_0 + b_1x + \cdots + b_nx^n)(1 - xa) = 1.$$
Equating coefficients, we see that $b_i = a^i$ for all $i = 0, \ldots, n$, and also that $b_na = 0$. Hence $a^{n+1} = 0$ and $a$ is nilpotent.<|endoftext|>
TITLE: Ricci flow and isometry group
QUESTION [6 upvotes]: It is known (via Kotschwar's uniqueness of backwards Ricci flows) that the isometry group of a Riemannian metric remains unchanged under the Ricci flow. But, one can easily observe that it can change at the limit. For example, one can perturb a sphere $S^2$ slightly so that it has no symmetry. As long as the Ricci flow is defined, the resulting surfaces still have trivial isometry groups. But, in the limit, if one is using normalized Ricci flows, the isometry group suddenly jumps to $SO(3)$. My question is, can one say that if the isometry group under normalized Ricci flows changes at the limit, it actually has to get "bigger"? Heuristically, this sounds very plausible, but is it actually true?

REPLY [11 votes]: I think the phenomenon is much more general: If a sequence of metrics $d_i$ on a compact metric space $X$ converges (pointwise on $X\times X$) to a metric $d$, and if $h$ lies in the intersection of the isometry groups of $(X, d_i)$, then clearly $h$ is an isometry of $(X,d)$.<|endoftext|>
TITLE: Is every vector bundle over a noncompact finite-dimensional manifold a summand of a trivial bundle?
QUESTION [12 upvotes]: In the notes of Vector Bundles and K-theory by Prof Allen Hatcher, on page 12 he proved a Proposition that for each vector bundle $E\to B$ with $B$ compact hausdorff there exists a vector bundle $E'\to B$ such that $E\oplus E'$ is the trivial bundle.
Later in example 3.6 he showed that the compactness of $B$ is an important condition, otherwise canonical line bundle over $\mathbb{RP^\infty}$ would be a counter example.
My doubt is...  Is this Proposition still valid for finite dimensional manifold??? Or can someone provide me a counter example of a finite manifold where it is not true.

REPLY [15 votes]: The proposition you refer to holds for any space homotopy equivalent to a finite dimensional CW complex. Here are the main points.

The property that any vector bundle $E$ over a space has a complementary bundle $E^\prime$ is preserved by homotopy equivalences.
Any finite dimensional CW complex is homotopy equivalent to a smooth manifold. (A classical result of Whitehead shows that any finite dimensional CW complex is homotopy equivalent to a locally finite finite dimensional CW complex. Then simplicial approximation gives a homotopy equivalent finite dimensional simplicial complex. Embed it into a Euclidean space, take a regular neighborhood, and smooth it.)
The total space of any vector bundle $E$ over a smooth manifold $M$ smoothly embeds into a Euclidean space. The normal bundle  bundle of $E$ restricted to $M$ is the desired complementary bunlde $E^\prime$.  
Note that any topological manifold is homotopy equivalent to a finite dimensional CW complex.

REPLY [4 votes]: I think that the proposition still holds for smooth finite-dimensional manifolds. Here is a sketch of an argument.
First of all, for a bundle of rank $k$ the following statements are equivalent.

there exists a complementary bundle of rank $\ell$,
the classifying map $X\to BGL_k(\Bbbk)=G_k(\Bbbk^\infty)$ factors through $G_k(\Bbbk^{k+\ell})$.

Next, consider an $n$-dimensional smooth manifold, then $M$ admits a triangulation. The stars of its barycentric subdivision provide an open cover $\{U_0,\dots,U_n\}$ of $M$. Here each $U_i$ is the disjoint union of the stars of those vertices that come from the $i$-simplices of the original triangulation. In particular, each $U_i$ is a disjoint union of contractible open subsets of $M$. Fix trivialisations of $E|_{U_i}$ for each $i$.
Then the transition maps $g_{ij}\colon U_i\cap U_j\to GL_k(\Bbbk)$ can be used to construct an explicit classifying map from $M$ to the $n$-fold join $(GL_k(\Bbbk)*\cdots*GL_k(\Bbbk))/GL_k(\Bbbk)$ in Milnor's construction of a classifying space.
Now, there is a homotopy equivalence
$$\operatorname{colim}_n(\underbrace{GL_k(\Bbbk)*\cdots*GL_k(\Bbbk)}_{n+1\text {factors}})/GL_k(\Bbbk)
\to\operatorname{colim}_\ell G_k(\Bbbk^{k+\ell})$$
between these two models for $BGL_k(\Bbbk)$.
Because each of the joins on the left is a finite CW complex, the restriction ends up in a finite-dimensional Grassmannian. In particular, the classifying map $f$ above factors through one of them. Together with the observation above, that proves the claim.<|endoftext|>
TITLE: Faithful projective representations of symmetric groups
QUESTION [8 upvotes]: This is a reference request.
Do you know where I can find the dimensions of the faithful projective representations of $S_n$ and $A_n$ for $n\ge 5$?
Thank you in advance.

REPLY [11 votes]: Maybe it's worth adding a reference to a modern treatment of Schur's work in book form, which might be easier to read than the original paper of Schur (note that I'm unrelated to the authors):
P.N. Hoffman and J.F. Humphreys, Projective representations of the symmetric groups.
Q-functions and shifted tableaux.
Oxford Mathematical Monographs.
Oxford Science Publications.
The Clarendon Press, Oxford University Press, New York, 1992.

REPLY [8 votes]: See Schur's original paper
Schur, I. (1911), Über die Darstellung der symmetrischen und der 
alternierenden Gruppe durch gebrochene lineare Substitutionen,
Crelle's Journal 139: 155–250. (EuDML)<|endoftext|>
TITLE: internalization of the concept of large and small category
QUESTION [5 upvotes]: I have been poking around the internet and nlab looking at the concept of large and small categories.  My original focus was locally presentable categories of categories and I was thinking of finding categories like Hilb and Group and Set as colimits over diagrams of compact objects.  This lead to trying to define "largeness" and "smallness" in terms of colimits, ie, large categories as colimits of diagrams of small categories.
There is plenty of work done on internalizing the concept of "category".  We even know that a "small category" is one that is internal to set.  Is it possible to internalize how the concept of smallness is related to largeness?  I had one thought on the matter, which was to have some ambient category that could host SET, and so all the small categories are internal there.  We want to define largeness as maps from internal categories in other categories into this.  That is a stupid attempt, but its all I have.  Is it even possible to have a category which contains large categories and small categories?
I just don't see anywhere an attempt to internalize the concepts of how largeness relates to smallness.

REPLY [7 votes]: I am not sure exactly what sort of thing you are looking for, but you would probably find the work on algebraic set theory relevant and possibly interesting.
Algebraic set theory studies elementary set theory through category-theoretic methods. A central idea is an abstract notion of smallness which is axiomatized in a category-theoretic fashion. I recommend Steve Awodey's overview article A Brief Introduction to Algebraic Set Theory. Briefly, we set things up by postulating (I am quoting from Section 1.2 of Steve's paper):

a Heyting category $\mathcal{C}$ of "classes"
a subcategory $\mathcal{S} \hookrightarrow \mathcal{C}$ of "sets"
a "powerclass" functor $P : \mathcal{C} \to \mathcal{C}$ of subsets
a "universe" $U$ which is a free algebra for $P$.

The classes in $\mathcal{C}$ admit the interpretation of first-order logic; the sets $\mathcal{S}$ capture an abstract notion of “smallness” of some classes; the powerclass $P(C)$ of a class $C$ is the class of all subsets $A \rightarrowtail C$; and this restriction on $P$ to subsets (as opposed to subclasses) permits the assumption of a universe $U$ which, as a free algebra, has an isomorphism $i : P(U) \cong U$. 
There are many models of this setup apart from the canonical Zermelo-Fraenkel set theory.<|endoftext|>
TITLE: Isometry group of a compact hyperbolic surface
QUESTION [13 upvotes]: Consider a compact surface $M$ of genus $g \geq 2$ with a metric of constant negative curvature. My question is, is it known under what sorts of sufficient conditions such a metric will have non-trivial isometry group?

REPLY [13 votes]: I'll just point out an answer based on somewhat different criteria than explicitly knowing features of the hyperbolic metric:  As is well-known, every oriented, compact hyperbolic surface $C$ is canonically a Riemann surface of genus $g\ge2$.  
As Ian points out, if $C$ is hyperelliptic, it always has a symmetry, the hyperelliptic involution $\iota:C\to C$, and it can be written, almost canonically, as an algebraic plane curve
$$
y^2 = (x-\lambda_1)\cdots(x-\lambda_{2g+2}),
$$ 
where the $\lambda_i$ are distinct complex numbers.  Let $\Gamma\subset\mathrm{PSL}(2,\mathbb{C})$ be the (finite) group of linear fractional transformations that preserve the set $\Lambda =\{\lambda_1,\ldots,\lambda_{2g+2}\}$. Then the group of orientation preserving isometries of $C$ will be an extension of $\Gamma$ by a $\mathbb{Z}_2$ (because of the hyperelliptic involution).  If lets $\Gamma_+$ be the (possibly slightly larger) group of $\pm$-holomorphic linear fractional transformations of $\mathbb{P}^1$ that preserve $\Lambda$, then the full group of isometries of $C$ will be a $\mathbb{Z}_2$ extension of $\Gamma_+$.
If $C$ is not hyperelliptic, then its canonical mapping $\kappa:C\to\mathbb{CP}^{g-1}$ is an embedding, and the isometries of $C$ are exactly the $\pm$-holomorphic automorphisms of $\mathbb{CP}^{g-1}$ that preserve the image $\kappa(C)$.  For example, when $g=3$ and $C$ is not hyperelliptic, the image $\kappa(C)\subset\mathbb{CP}^2$ is a nonsingular plane quartic $Q(X,Y,Z)=0$, and every nonsingular plane quartic is a canonical curve of genus $3$.  The orientation-preserving isometries of $C$ are exactly the elements of $\mathrm{PSL}(3,\mathbb{C})$ that preserve the quartic $Q$.  Since one can write down explicit homogeneous nonsingular quartics in $3$ variables that have no nontrivial automorphisms, it follows that the generic curve of genus $3$ has no nontrivial automorphisms, and hence the conformal hyperbolic metric on $C$ has no nontrivial isometries.  
This gives one a way to write down 'explicit' examples, modulo, of course, the fact that there is no known way explicitly to 'write down' the conformal hyperbolic metric on a nonhyperelliptic curve of genus $3$.<|endoftext|>
TITLE: Why does genus control the number of points
QUESTION [12 upvotes]: Often number theorists can bound the number of solutions to a diophantine equation based on the size of the points and the size of the coefficients. But this, as I understand it, can be a bit of a red herring - the genus of a curve is what is really controlling the number of points. 
What I would like is some intuition as to why. I must admit that I don't understand genus very well, but perhaps someone can illustrate (in a somewhat elementary fashion) why genus is the right tool for bounding solutions.

REPLY [12 votes]: Even for curves, the situation becomes clearer if you also allow affine curves and look at integral points, or more generally, $S$-integral points in number field. If the curve is projective, that's the same as looking at rational points. Also, it's better to look at Euler characteristic $\chi(C)$ (so as to get integer values), so
$\chi(\text{projective curve of genus $g$ with $n$ points removed}) = 2-2g-n.$
Then we have
$\chi(C)=2$ for $\mathbb P^1$, which is projective and has lots of points.
$\chi(C)=1$ for $\mathbb A^1=\mathbb P^1\setminus\{pt\}$ is the additive group, so also has lots of points.
$\chi(C)=0$ means that $C$ has the structure of a group, either an elliptic curve (which is projective) or the multiplicative group $\mathbb G_m=\mathbb P^1\setminus\{pt_1,pt_2\}$. In both cases, the set of integral points forms a finitely generated group, so is often infinite, but much sparser than the $\chi(C)>0$ cases.
$\chi(C)<0$ means $C$ is either a projective curve of genus at least 2, or an elliptic curve minus one or more points, or $\mathbb P^1$ minus at least 3 points. In all three cases, there are finitely many integral points. These three major theorems are due, respectively, to Faltings, Siegel, and Siegel, although the last one (which is solutions to the unit equation) follows for units in number fields from Thue's theorem. 
So for quasi-projective curves, there fairly coarse Euler characteristic determines the qualitative behavior of the integral points. 
In higher dimensions, one has similar conjectures, where the Euler characteristic is replaced, roughly, by the extent to which the canonical bundle is ample, or anti-ample, or 0, or something "in between". This is all made quite precise in Vojta's conjectures. However, in higher dimensions one only gets statements that apply to all points outside of a proper Zariski closed subset. 
Summary: the intuition is that "as the geometry of the variety gets more complicated, the integral/rational points get more sparse."<|endoftext|>
TITLE: Algebraic surface of constant width?
QUESTION [15 upvotes]: Does there exist an irreducible polynomial $f \in \mathbb{R}[x, y, z]$ such that:
$$ V := \{ (x, y, z) \in \mathbb{R}^3 : f(x, y, z) \leq 0 \} $$
is a solid of constant width with a finite symmetry group?

The analogous result is true in two dimensions, with an explicit degree-$8$ example given in this paper. If we revolve this curve about its axis of symmetry (or equivalently replace every instance of $y^2$ with $y^2 + z^2$), we would obtain a degree-$8$ surface of constant width depicted below:

This has an infinite symmetry group isomorphic to $O(1)$, so does not answer the question. Similarly, the sphere:
$$ f(x, y, z) = x^2 + y^2 + z^2 - 1 \leq 0 $$
has symmetry group $O(3)$, which is again infinite. Does there exist an algebraic solid of constant width and finite automorphism group?

REPLY [14 votes]: This is a comment on Robert Bryant's answer, addressing the point in his last paragraph. It is substantially rewritten from a previous answer, because I realized that the best way to address the Bryant's question below is to rewrite the answer.
Bryant constructs constant width surfaces as the image of the sphere $S^2 = \{ (x,y,z) : x^2+y^2+z^2=1 \}$ under a polynomial map $b: S^2 \to \mathbb{R}^3$. He points out that the image will be a connected component of $F=0$ for some polynomial $F$, but says that he does not know whether it is the only component. 
I show that it is the only two dimensional component.
Let $\mathbb{C} S^2$ be the complex solutions to $x^2+y^2+z^2=1$. Let $\mathbb{R} Z$ be the real points of $F=0$, and $\mathbb{C} Z$ the complex points. We note that $b(\mathbb{C} S^2)$ is Zariski dense in $\mathbb{C} Z$, so the points of $\mathbb{C} Z \setminus b(\mathbb{C} S^2)$ are a complex variety of complex dimension $\leq 1$, and thus the real points of $\mathbb{R} Z \setminus b(\mathbb{C} S^2)$ have real dimension at most $2$.
It remains to bound the dimension of $(\mathbb{R} Z \cap b(\mathbb{C} S^2)) \setminus b(S^2)$.
Everywhere on $S^2$, we have the polynomial identity that the vectors $(x,y,z)$ and $(\nabla F)|_{b(x,y,z)}$ are parallel. Therefore, this identity remains true on $\mathbb{C} S^2$. (The zero vector is parallel to every vector.)
Let $(u,v,w) \in \mathbb{C} Z$ and suppose $(\nabla F)^2|_{(u,v,w)}$ is nonzero, then there at most two possible preimages for $(u,v,w)$ in $\mathbb{C} S^2$: the points $\pm \tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$. Moreover, the identity $(x,y,z) \cdot (b(x,y,z) - b(-x,-y,-z)) = 2 (\mathrm{width})$ holds everywhere on $\mathbb{C} S^2$, so we never have $b(x,y,z) = b(-x,-y,-z)$. So there is actually only one preimage of $(u,v,w)$ in $\mathbb{C} S^2$.
Furthermore, suppose that $(u,v,w)$ is real. Then the formula $\pm \tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$ is manifestly real. We have shown that, at any point of $\mathbb{R} Z$ where $\nabla F$ is nonzero, if that point is in $b(\mathbb{C} S^2)$, then it is in $b(S^2)$. Since $\nabla F$ vanishes along (at most) a curve, this concludes the proof.

A previous version of this answer explained more generally that, if a map $b: \mathbb{C} S^2 \to \mathbb{C}^3$ is generically injective, then the only two dimension component of $b(\mathbb{C} S^2) \cap \mathbb{R}^3$ is $b(S^2)$. (And this is true for other complex varieties defined over $\mathbb{R}$; it isn't special to the sphere.) But I did a crummy job explaining why $b$ is generically injective and, once I fixed that, it was easier to directly point out that $\tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$ is real when defined.<|endoftext|>
TITLE: Strong equivalence between intrinsic and extrinsic metrics on $GL_n^+$?
QUESTION [11 upvotes]: $\newcommand{\til}{\tilde}$
Lately, I have become interested in comparing intrinsic and extrinsic metrics on Riemannian manifolds.
Consider $GL_n^+$ (invertible matrices , $\det >0$) as an open Riemannian submanifold of $\mathbb{R}^{n^2}$.
We have two metrics on $GL_n^+$ (in the sense of metric spaces); intrinsic $d^{int}$ (when we only allow paths which stay inside $GL_n^+$) and extrinsic $d^{ext}$ (the standard Euclidean metric). 
Clarification: $d^{int}$ is the Riemann distance function on $GL_n^+$ induced by the standard Riemannian metric on $\mathbb{R}^{n^2}$ (i.e I consider $GL_n^+$ as a Riemannian submanifold of $\mathbb{R}^{n^2})$
Of course $d^{ext} \le d^{int}$. For some matrices $d^{ext}(A,B) < d^{int}(A,B)$;
If $\gamma(t)=A+t(B-A)$, then $\det(\gamma(t))$ is a polynomial which can be negative on a subinterval of $[0,1]$. (i.e the minimizing geodesic in the Euclidean space, is contained in $GL^{-}$ for some time)
(see this answer to this question for details on this specific case, and here for a cleaner proof that "nearly minimizing paths are nearly geodesic")

Note that $d^{ext},d^{int}$ both generate the same topology on $GL_n^+$.
Moreover, both are incomplete.
So, the following natural question arises:
Are $d^{ext},d^{int}$ strongly equivalent? i.e; Is it true that
$d^{int} \le C \cdot d^{ext} \text{ for some } C \in \mathbb{R}$?

REPLY [3 votes]: In two dimensions the condition $ad-bc=0$ translates into $(a+d)^2-(a-d)^2-(b+c)^2+(b-c)^2=0$ or to simplify notation $x^2+y^2=z^2+w^2$. Intersecting this with the unit sphere in $\mathbb{R}^4$ gives a flat 2-torus which decomposes the $3$-sphere into two solid tori. One solid torus consists of matrices of positive determinant and the other of matrices of negative determinant. Thus the determinant variety $ad-bc=0$ in this case is a cone on a 2-torus. 
Since the torus is smooth, the intrinsic distances in the torus are bilipschitz with ambient distance in the solid torus filling it.  In fact, one can probably get away with a Lipschitz constant not much bigger than $\frac{\pi}{2}$.  Therefore whenever one has a straight path joining two points in $GL^+(2,R)$ one can always replace its portions "dipping" into the (cone over the) "negative" solid torus by arcs following the (cone over the) flat 2-torus while retaining Lipschitz control over length of the path.  This proves that the intrinsic and the ambient distances are bilipschitz in the 2-dimensional case.
A complete answer appears at http://arxiv.org/abs/1602.01227<|endoftext|>
TITLE: How to calculate the principal graphs of a fusion ring with a given simple object?
QUESTION [6 upvotes]: My understanding is that the principal graphs are a pair of undirected bipartite graphs, $\Gamma_+,\Gamma_-$. They can be calculated from a fusion ring with a given simple object $X$. How to calculate such pair of graphs?
There is related dicussion  An embedding theorem for a fusion ring planar algebra? , but not enough detail.

REPLY [5 votes]: Probably what you really want is the fusion graph $\Gamma_X$ with respect to your simple object $X$. It is a directed graph with vertices labelled by the simple objects of your category. Between the vertices labelled by simples $Y$ and $Z$, there are $N^{Y,X}_Z=\operatorname{dim}(\operatorname{Hom}(Y\otimes X, Z))$ directed edges from $Y$ to $Z$. The fusion graph with respect to $X^*$ is obtained by reversing the orientation of all the edges, since by semisimplicity and Frobenius reciprocity,
$$
N^{Y,X}_Z
=
\operatorname{dim}(\operatorname{Hom}(Y\otimes X, Z))
=
\operatorname{dim}(\operatorname{Hom}(Y, Z\otimes X^*))
=
\operatorname{dim}(\operatorname{Hom}(Z\otimes X^*, Y))
=
N^{Y,X^*}_Z.
$$
This means that if $X$ is self-dual, then you can view this graph as unoriented.
In your question you've referred to a pair of unoriented graphs. These would arise if you were looking at the principal graphs of a subfactor. Given a finite index subfactor $A\subset B$, we have the semisimple tensor category of $A-A$ bimodules generated by $B$. The simples are obtained by decomposing the tensor powers $\bigotimes_A^n B$ into irreducibles. We can also view $\bigotimes_A^n B$ as an $A-B$ bimodule and decompose into irreducibles. This gives the semisimple module category of $A-B$  bimodules. Similarly, we have a tensor category of $B-B$ bimodules and a module category of $B-A$ bimodules.
We get the principal graphs $(\Gamma_+,\Gamma_-)$ as follows. The graph $\Gamma_+$ has even vertices the simple $A-A$ bimodules and odd vertices the simple $A-B$ bimodules. There are then $\operatorname{dim}(\operatorname{Hom}_{A-B}(Y\otimes_A B, Z))$ edges between the simple $A-A$ bimodule $Y$ and the simple $A-B$ bimodule $Z$. The dual principal graph $\Gamma_-$ is defined analogously with $B-B$ and $B-A$ bimodules.
Edit after the comments:
Given a unitary fusion category $\mathcal{C}$ and an object $X\in\mathcal{C}$, you can construct a subfactor standard invariant $\mathcal{P}_\bullet$ from which you can get a pair of principal graphs $(\Gamma_+,\Gamma_-)$. First, you let
\begin{align*}
\mathcal{P}_{n,+} &= \operatorname{End}(\underbrace{X\otimes X^*\otimes \cdots \otimes X^\pm}_{n \text{ copies}})
\\ 
\mathcal{P}_{n,-} &= \operatorname{End}(\underbrace{X^*\otimes X\otimes \cdots \otimes X^\mp}_{n \text{ copies}})
\end{align*}
where $X^\pm=X$ if $n$ is odd and $X^*$ if $n$ is even. Now the graph $\Gamma_+$ has even vertices the simple summands of objects of the form $(X\otimes X^*)^{\otimes n}$, and odd vertices the simple summands of objects of the form $(X\otimes X^*)^{\otimes n}\otimes X$. You get the edges similar to before, taking an even $Y$, tensoring with $X$, and decomposing into simples $Z$. The graph $\Gamma_-$ is defined similarly, but swapping the roles of $X$ and $X^*$.
So that's the formal explanation of how you get the principal graphs, but there is a simpler way to compute $(\Gamma_+,\Gamma_-)$ based on the fusion coefficients $N^{Y,X}_Z$.
As @Xiao-Gang points out in the comments, we can compute $\Gamma_+$ as the connected component containing $X$ in the bipartite graph with adjacency matrix
$$
\begin{pmatrix}
0 & N^X
\\
N^{X^*} & 0
\end{pmatrix}
$$
where $N^X$ is the matrix of fusion coefficients for tensoring on the right by $X$, and similarly for $X^*$.
The graph $\Gamma_-$ is then obtained by swapping the roles of $X$ and $X^*$.
Interestingly, if $X\in \mathcal{C}$ is self-dual and the fusion graph $\Gamma_X$ in $\mathcal{C}$ is already bipartite (here, $\mathcal{C}$ is $\mathbb{Z}/2\mathbb{Z}$-graded and $X$ is odd), then $(\Gamma_+,\Gamma_-) = (\Gamma_X, \Gamma_X)$. This occurs for many interesting examples.<|endoftext|>
TITLE: Removing singularities in generating functions
QUESTION [8 upvotes]: This is a problem about the practicalities of removing singularities in multivariable complex functions.
In trying to derive the generating function (in two variables) for a certain problem in combinatorics, we have obtained an expression of the form $S(x,u) = N(x,u)/D(x,u)$ where
$$
D(x,u) = 
(1 + u - u^2) x^2 + (3 u^3 - 2 u) x + u^2 (1 - u - u^2)
$$
and
$$
N(x,u) = P(x,u) + S(x) Q(x,u) + T(x) R(x,u)
$$
Here $P$, $Q$ and $R$ are low degree polynomials in $x$ and $u$, and $S$ and $T$ are known algebraic functions (having convergent power series expansions in a neighbourhood of 0). 
The theorem on singularity removal that we would hope to apply is that if $N/D$ is bounded on a neighbourhood of $(0,0)$ excluding the set where $D(x,u) = 0$, then the singularity which that set represents is actually removable, and there is an analytic $F(x,u) = N(x,u)/D(x,u)$ on a neighbourhood of $(0,0)$ wherever the right hand side is defined.
If it matters (probably not), note that setting $D(x,u) = 0$ and solving for $x$ actually gives $x$ as (one of two) rational functions of $u$ (since the discriminant is $5 u^6$). We can also verify (of course or this would be a very silly question indeed) that $N(x,u) = 0$ whenever $D(x,u) = 0$.
What's tripping us up is the practical issues of establishing the boundedness criteria. Can anyone make suggestions about how to do that?
For more specific details, the problem concerns the example discussed in section 4.1 of Generating permutations with restricted containers (pages 9 to 11) which, as it stands, is missing the final piece that an answer to this question would provide.

REPLY [6 votes]: I have a solution inspired by Fan Zheng's comment. There is a generalization of Study's lemma [1, $\S$6.13] (which is itself a special case of the Nullstellensatz), that says if $D(x,u)$ is irreducible and if $D(x,u)=0$ implies $N(x,u)=0$ then $D(x,u)$ divides $N(x,u)$. 
The zero condition ended up being not easy to check. Let $x_1(u)$ and $x_2(u)$ be two components of the curve on which the denominator is zero. We have $S(x)$ and $T(x)$ in terms of their minimal polynomials (of degree 4 -- explicit expressions are attainable, but quite messy). When substituting $x = x_1(u)$ or $x = x_2(u)$ into the numerator, Maple refuses to simplify to zero.
To get around this, I had to use resultant tricks to find a minimal polynomial for the entire numerator. (In theory, Groëbner bases can be used to do this, but I gave up on the computation after a few hours.) Once this expression was in hand, Maple was happy to verify that the numerator is zero after the substitutions were made.
[1] G. Fischer. Plane algebraic curves, volume 15 of Student Mathematical Library. American Mathematical Society, Providence, RI, 2001. Translated from the 1994 German original by Leslie Kay.<|endoftext|>
TITLE: T-equivariant cohomology of flag variety
QUESTION [10 upvotes]: Let $X=G/B$ , where $G=GL_n(\mathbb{C}^n)$ and $B$ be the upper triangular matrices. I am curoius about the structure of $H^*_T(G/B)$ which I consider  as a $H_T^*(pt)$-module. If we just consider ordinary cohmology, we know that the Schubert classes $[X_w]$ form a $\mathbb{Z}$-linear basis , their product is a positive combination of the others and it is described by Schubert polynomial. In Knutson and Tao's paper, they provided a nice puzzle to calculate the equivariant cohomology of $G/P$, the grassmanian variety.
Do we have something similar happening in $H_T^*(G/B)$? To be more precise, do we have the following properites?

$H_T^*(G/B)$ is graded module over $H_T^*(pt)$.
We have a natural basis $[\tilde{X_w}]$ for this $H_T^*$-module.
The forgetful functor from $H_T^*(-)$ to $H^*(-)$ maps $[\tilde{X_w}]$ to $[X_w]$.

If the above listed properties are true, is it possible to create a simlar puzzle as in Knutson and Tao's paper? 
What are the progress in this area?

REPLY [9 votes]: Points 1-3 are available for equivariant cohomology. As an economical link, this paper on the arXiv discusses three well-known available presentations for $T$-equivariant cohomology of the flag manifold $G/T$ and how to translate between them. It also contains the relevant literature references for all three presentations. (Although the reference is fairly new, the presentations relevant for the question are very classical, going back to work of Chevalley, Borel, Goresky-Kottwitz-MacPherson...)

Edit: as requested, I include some more details on points 1-3 (most of this can be found in the linked paper or the references therein). The presentation most relevant for the question is called Chevalley representation in this paper. 
Equivariant cohomology is defined via the Borel construction $ET\times_T G/B$. There is a fibre bundle $G/B\to ET\times_TG/B\to BT$, where $BT$ is the classifying space of the torus $T$; this fibre bundle is associated to the universal $T$-bundle $T\to ET\to BT$ and the standard left action of $T$ on $G/B$. The projection $\pi:ET\times_TG/B\to BT$ induces a graded ring homomorphism $H^\ast_T(\ast)\to H^\ast_T(G/B)$. This shows (1). 
For each Schubert variety $X_w$, there is an associated class $[\tilde{X}_w]$ in $H^\ast_T(G/B)$. Then the Chevalley presentation states that $H^\ast_T(G/B)$ is a free $H^\ast_T(\ast)$-module generated by the classes $[\tilde{X}_w]$. This is (2). 
For (3), it is possible to recover ordinary cohomology from equivariant cohomology via 
$$
H^\ast(G/B)\cong H^\ast_T(G/B)\otimes_{H^\ast_T(\ast)}\mathbb{Q}
$$
and under this identification, the Schubert classes in equivariant cohomology are mapped to the Schubert classes in ordinary cohomology as required. (Check the discussion of equivariant cohomology of equivariantly formal spaces in the paper of Goresky-Kottwitz-MacPherson.)<|endoftext|>
TITLE: When is the diagonal of a rational bivariate power series again rational
QUESTION [9 upvotes]: Given a rational bivariate power series $F(x,y)=\sum{a_{n,m}x^ny^m}$, the diagonal function $G(t):=\sum{a_{n,n}t^n}$ is known to be algebraic, although not rational in general. I was wondering if there were some known conditions for it to be rational. The setting I'm working with has $a_{n,m}$ non-negative integers; I'm interested in both characteristic 0 and positive characteristic.

REPLY [5 votes]: Assume for convenience that each $a_{n,m}\in\mathbb{C}$.  Consider the
proof in Enumerative Combinatorics, vol. 2, Theorem 6.3.3, that
$G(t)$ is algebraic. One computes the constant term (with respect to
$s$) of $F(s,t/s)$ (being careful of the meaning of $F(s,t/s)$). The
zeros of the denominator of $F(s,t/s)$ (with respect to $s$) are
algebraic functions of $t$ that enter into the formula for the
diagonal. Factor the denominator of $F(x,y)$ as $D_1(x,y)\cdots
D_k(x,y)$, where each $D_i(x,y)$ is irreducible (over
$\mathbb{C}$). Let $d_i$ be the least integer for which
$s^{d_i}D_i(s,t/s)$ has no negative exponents. The zeros of
$s^{d_i}D_i(s,t/s)$ (as a polynomial in
$s$ whose coefficients are polynomials in $t$) will all be rational
functions if and only if $s^{d_i}D_i(s,t/s)$ has degree $0$ or $1$ (in
$s$). This is equivalent to the following:  either (1)
every monomial of $D_i(x,y)$ has $x$-degree minus $y$-degree equal to
$0$ or $1$, or
(2) every monomial of $D_i(x,y)$ has $x$-degree minus $y$-degree equal
to $0$ or $-1$. An example is
 $$ F(x,y)=\frac{1}{(1-xy-x^2y+2x^3y^2)(1+xy^2-x^3y^3-5x^3y^4)}. $$
It seems plausible to me (and perhaps not so difficult to prove) that
the above sufficient condition is also "essentially necessary." This
means that we need to exclude such examples as (1) $F(x,y)=H(x^k,y^k)$,
where $k\geq 2$ and $H$ satisfies the condition above, (2)
"degenerate" examples like $F(x,y)=1/(1-xy^4+2x^2y^3-5x^5y^7)$, whose
diagonal is just $1$.<|endoftext|>
TITLE: Extending an infinite simple group
QUESTION [10 upvotes]: Maybe the question does not fit here.
Yesterday in my logic course, I presented a nice example about an application of model theory to group theory. The example is due to Hodges and as following: For any infinite simple group $G$, there is a countable simple group $H\subseteq G$.
Then a logician asked whether the upward version is true. i.e. $\mathbf{ \mbox{is it true that for every infinite simple group $H$ and cardinal }\kappa\geq |H|, \mbox{there is a simple group }G\supseteq H \mbox{ with } |G|=\kappa?}$
To give a positive answer to the question, it is sufficient to prove that every  infinite simple group $H$ is a proper subgroup of a simple group $G$. 
I was told by an algebraist that the question for finite simple groups has a negative answer.
$\mathbf{Remark}$: The question was already answered by Derek Holt. He proved that one may find such a $G$ with $|G|=2^{\kappa}$ for some cardinal $\kappa$. Then by Skolemization including $H$, one may obtain a full answer.

REPLY [11 votes]: I will make my comment into an answer. It is a standard result (see for example Chapter 8 of Dixon & Mortimer's book on Permutation Groups) that, for an infinite cardinal $\kappa$, the only normal subgroups of the symmetric group ${\rm Sym}(\Omega)$ on a set $\Omega$ of cardinality $\kappa$ are the subgroups
${\rm Sym}(\Omega,\lambda)$, consisting of those permutations of $\Omega$ with support strictly less than $\lambda$, for all cardinals $\lambda$ with $\aleph_0 \le \lambda \le \kappa$, and also the subgroup ${\rm Alt}(\Omega)$ of even permutations with finite support.
So, if $H$ is any infinite group with cardinality $\kappa$ then the standard regular embedding of $G$ into ${\rm Sym}(H)$, in which the image of each non-identity element has support equal to $H$, induces an embedding into the simple group ${\rm Sym(H)}/{\rm Sym}(H,\kappa)$, which has cardinality $2^\kappa$.<|endoftext|>
TITLE: Semistability of principal bundle vs vector bundle
QUESTION [5 upvotes]: Ramanathan has defined the semistability of a principal $G-$bundle $E$ over a curve $X$ as follows: 

$E$ is semistable iff for any parabolic subgroup $P\subset G$, for any reduction of the structure group of $E$ to $P$: $\sigma:X\rightarrow E(G/P)$, and for any dominant caracter $\chi:P\rightarrow \mathbb C^*$, one has $$\sigma^* E(\chi)\le 0$$
  where $E(\chi)$ is the line bundle associated to the $P$ bundle $E\rightarrow E(G/P)$ by $\chi$

Question: Taking $G=GL_r$, How could one proof that this definition is the same as the usual one (due to Mumford)? 
Thanks

REPLY [3 votes]: The following paper contains a proof (Corollary 1) of the identification of Ramanathan-semistability for principal $GL_n$-bundles and Mumford-semistability for the associated vector bundles: 

D. Hyeon and D. Murphy. Note on the stability for principal bundles. Proc. Amer. Math. Soc. 132 (2004), 2205-2213.

The basic point is that there is a direct relation between flags on the vector bundles and reductions to parabolic subgroups. This allows to relate slope of the vector bundles to the degree conditions for parabolic bundles. (I will not rewrite the proof in the above paper, it is really nice and short.) 
This does not completely answer the question, because the definition used in the paper I mentioned is different from the formulation of the question. The remaining identification of semistability conditions for the principal bundle is done in Ramanathan's original paper.<|endoftext|>
TITLE: Multiplicity of Laplace eigenvalues
QUESTION [9 upvotes]: Disclaimer: This is a very heuristic question and I will be satisfied with heuristic insights, if rigorous and precise answers are not possible.
All the examples of closed surfaces (or higher dimensional manifolds) whose spectrum I have seen evaluated explicitly have highly repeated eigenvalues. They also happen to be quite symmetric (non-trivial isometry group), which was instrumental in the calculation of the spectrum in the first place. I have heard many people quote this heuristic: the more symmetric a manifold is, the higher the chances of spectral multiplicity. I was wondering 

Is there a way to make this heuristic precise?
What about the converse? If a manifold has high spectral multiplicity, does it need to have a nontrivial isometry group?
If the answer to 2. is negative, and there are large classes of examples of surfaces/manifolds that are not symmetric at all, but have high spectral multiplicity, is there some other property shared by these examples which is causing the multiplicity? In other words, are there other criteria apart from symmetry that would make one suspect spectral multiplicity?  

Edit: Okay, I just found High multiplicity eigenvalue implies symmetry? So, I know that the answer to 2. is negative. Also, Liviu Nicolaescu answers 1. below. I am still confused about 3. In other words, is there any criterion other than symmetry, that, if you know that a metric satisfies, you would suspect high spectral multiplicity?

REPLY [9 votes]: The group of  isometries of a compact Riemann manifold is a compact Lie group.  This group acts on  the eigenspaces of the Laplacian. If the group  of isometries   is non-abelian, then it is natural to expect that that some eigenspaces will have large dimensions since  most of the nontrivial irreducible representations of  a compact  non-abelian Lie group have  dimensions $\geq 2$.
Looking for Laplacians with multiple eigenvalues is a bit like looking for a needle in a haystack since  a result of K. Uhlenbeck states  that for a generic Riemann metric on a given compact manifold the spectrum of the Laplacian  will not have multiple eigenvalues.<|endoftext|>
TITLE: General Linear Group as a Direct Product?
QUESTION [8 upvotes]: Let $K$ be a field and consider the surjective determinant homomorphism $\mathrm{GL}_n(K)\to K^\times$. Since the kernel is the special linear group $\mathrm{SL}_n(K)$ we obtain a short exact sequence
$$\require{AMScd}\begin{CD}
 1 @>>> \mathrm{SL}_n(K) @>i>> \mathrm{GL}_n(K) @>\det>> K^\times @>>> 1\\
\end{CD}$$
Since the determinant map has a section $s:K^\times\to\mathrm{GL}_n(K)$ defined by $$s(\alpha):=\begin{pmatrix} \alpha&&&\\ &1&& \\ &&\ddots& \\ &&&1\end{pmatrix},$$
we conclude from the splitting lemma that $\mathrm{GL}_n(K)\approx\mathrm{SL}_n(K)\rtimes K^\times$. Here's a question:
Under what conditions on $K$ and $n$ does the inclusion homomorphism $i:\mathrm{SL}_n(K)\to\mathrm{GL}_n(K)$ have a retraction?
Example: If $K=\mathbb{R}$ and $n$ is odd, then every $\alpha\in\mathbb{R}^\times$ has a unique real $n$-th root, so we obtain a group homomorphism $\sqrt[n]{\cdot}:\mathbb{R}^\times\to\mathbb{R}^\times$, and we can use this to define a retraction $r:\mathrm{GL}_n(\mathbb{R})\to\mathrm{SL}_n(\mathbb{R})$ by $$r(A):=\frac{1}{\sqrt[n]{\det(A)}}\cdot A.$$
Now it follows from the splitting lemma that $\mathrm{GL}_n(\mathbb{R})\approx \mathrm{SL}_n(\mathbb{R})\times\mathbb{R}^\times$. Here's another question:
Is there a topological/geometric explanation for this example? Is there a topological/geometric obstruction when $n$ is even?

REPLY [10 votes]: $\mathrm{GL}_n(K)\to\mathrm{SL}_n(K)$ has a retraction iff the following two conditions hold

The subgroup of $n$-root of unity in $K^*$ has a direct summand in $K^*$.
$x\mapsto x^n$ is surjective on $K$

Let us first check that (1.) is equivalent to the existence of a retraction in restriction to $K^*\mathrm{SL}_n(K)$ (the subgroup generated by homotheties and unimodular matrices). Clearly it implies it (consider the set of scalar matrices whose diagonal entry belongs to this direct summand). Conversely, if there is a retraction, its kernel has trivial intersection with $\mathrm{SL}_n(K)$, hence contained in its centralizer, which is reduced to scalar matrices, so it should form the set of scalar matrices with diagonal entry in some direct summand of the set $n$-roots of unity in $K^*$.
Now it is clear that $K^*\mathrm{SL}_n(K)$ equals $\mathrm{GL}_n(K)$ if and only if (2.) holds; so if both (1.) and (2.) hold it follows that we have a retraction; conversely if we have a retraction, its kernel is a normal subgroup with trivial intersection with $\mathrm{SL}_n(K)$, hence contained in its centralizer, which is reduced to scalar matrices, so $\mathrm{GL}_n(K)$ should be generated by unimodular and scalar matrices, i.e. (2.) holds, and then the first verification shows that (1.) holds.
Edit: as noticed by Julian Rosen, (2.) together with (1.) implies something much stronger than (1.), namely that the subgroup of $n$-roots of unity is actually trivial, which means that $x\mapsto x^n$ is injective on $K$. To conclude:

$\mathrm{GL}_n(K)\to\mathrm{SL}_n(K)$ has a retraction (in the category of groups) iff $x\mapsto x^n$ is a permutation of $K$<|endoftext|>
TITLE: Matsushima-Murakami Isomorphism for $L^2$-cohomology
QUESTION [9 upvotes]: Let $\mathbf{G}$ be a reductive connected linear algebraic group over a totally real global number field, say $\mathbb{Q}$. Let $\mathbb{A}=\mathbb{R}\times\mathbb{A}_f$ be the ring of rational adele.
 Fix a maximal compact subgroup $K_\infty$ of $\mathbf{G}(\mathbb{R})$ and an open compact subgroup $K_f$ of $\mathbf{G}(\mathbb{A}_f)$. Fix a finite dimensional complex representation $V$ of $\mathbf{G}(\mathbb{Q})$ with corresponding local system $\mathcal{V}$ on the Shimura variety $S_{K_f}=\mathbf{G}(\mathbb{Q})\backslash \mathbf{G}(\mathbb{A})/K_\infty K_f$.
Then there should be a Hecke-equivariant isomorphism
$$\boxed{H^\bullet_{(2)}(S_{K_f},\mathcal{V})\cong H^\bullet(\mathfrak{g},K_\infty;L^2(\mathbf{G}(\mathbb{Q})\backslash \mathbf{G}(\mathbb{A}))^{K_f}\otimes V)}$$
between the $L^2$-cohomology and the relative Lie algebra cohomology, where $\mathfrak{g}=Lie(\mathbf{G})$.
I have found several references for this and this seems to be well-known. However, everybody refers the article of Borel and Casselman: "$L^2$-cohomology of locally symmetric manifolds of finite volume" in Duke vol 50, p. 625-647. Indeed, Proposition 5.6 in this article makes almost exactly this statement, but only for semisimple $\mathbf{G}$. It seems to me that the proof does not use this extra assumption.

My question: Does this isomorphism hold for arbitrary connected reductive $\mathbf{G}$?

REPLY [4 votes]: Morally yes, by
Franke, Jens Harmonic analysis in weighted $L_{2}$-spaces. Ann. Sci. École Norm. Sup. (4) 31 (1998), no. 2, 181--279
See especially theorem 4 (and note that if you want an equivariant isomorphism, you need to restrict to the twisted action by some character).<|endoftext|>
TITLE: Structure of the group generated by two specific symplectic matrices
QUESTION [6 upvotes]: Consider the following two symplectic matrices
$$
A \ = \
\left(\begin{array}{rrrr}%
1&0&0&0\\%
0&1&0&0\\%
0&0&-1&1\\%
0&0&-1&0\\%
\end{array}\right), \ \ \
B \ = \
\left(\begin{array}{rrrr}%
-1&0&0&-1\\%
0&0&-1&0\\%
0&1&-1&0\\%
1&0&0&0\\%
\end{array}\right).
$$
Is it true that the (Zariski-dense) group $\langle A,B \rangle$ generated
by $A$ and $B$ has infinite index in ${\rm Sp}(4,\mathbb{Z})$
(i.e., $\langle A, B \rangle$ is thin in ${\rm Sp}(4,\mathbb{Z}))$?
This question emerged from some calculations performed by Vincent Delecroix 
and myself with the monodromy of certain square-tiled surfaces (and, in 
their turn, these calculations were motivated by a question posed by Peter Sarnak to Alex Eskin and Alex Wright).
More precisely, our considerations led to a representation
$p: G \to {\rm Sp}(4,\mathbb{Z})$ of the level $4$ congruence group
$G = \langle a,b \rangle$ generated by the order three matrices
$$
a \ = \
\left(\begin{array}{rr}%
0&-1\\%
1&-1\\%
\end{array}\right), \ \ \
b \ = \
\left(\begin{array}{rr}%
1&-3\\%
1&-2\\%
\end{array}\right)
$$
in ${\rm SL}(2,\mathbb{Z})$ such that $p(a) = A$ and $p(b) = B$.
As it turns out, $G = \langle a \rangle * \langle b \rangle$ is the free product of two copies of $\mathbb{Z}/3\mathbb{Z}$
(since $\{a,a^2\}$ and $\{b,b^2\}$ play ping-pong with some cones in $\mathbb{R}^2$). Moreover, Vincent and I believe that the group
$\langle A, B \rangle$ is thin because some numerical experiments with non-trivial words on $A, A^2$ and $B, B^2$ of length $< 25$ seem to indicate that the representation $p$ might be faithful (and, thus, $<A,B>$ would be thin as ${\rm Sp}(4,\mathbb{Z})$ doesn't contain finite-index subgroups isomorphic to free groups).
Nevertheless, after trying a couple of standard tricks (e.g., testing the injectivity of $p$ on finite-index free subgroups of $G$ or playing ping-pong in $\mathbb{R}^4$, its exterior powers [and $p$-adic variants], etc.), Vincent and I are still unable to establish the thinness of
$\langle A, B \rangle$ and/or the faithfulness of $p$, so that we would be thankful to any help with these problems!

REPLY [2 votes]: After talking to Gabriela Weitze-Schmithuesen, I think that we can show the arithmeticity of $\langle A, B\rangle$ using the argument in Section 2 of this paper of Singh and Venkataramana here (http://www.ams.org/mathscinet-getitem?mr=3165424). 
Indeed, let us consider the permutation matrix $P=\left(\begin{array}{cccc} 
 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0
\end{array}\right)$ exchanging the second and fourth basis vectors and let us show that the conjugate $P\cdot \langle A, B\rangle\cdot P$ of $\langle A, B \rangle$ is arithmetic, i.e., it has finite-index in $Sp(4,\mathbb{Z})$. 
For this sake, we asked Sage to look words on $A$, $B$, $A^2$ and $B^2$ of size $\leq 10$ fixing the first basis vector, and we found that the matrices $x=P(A^2 B)^2(AB^2)^2P$, $y=PABA^2BA(AB^2)^2P$ and $z=PA^2BA^2(B^2A)^2BP$ are interesting because 
$$[y,x]=yxy^{-1}x^{-1} = \left(\begin{array}{cccc} 
 1 & 0 & 0 & 18 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1
\end{array}\right), \quad x^6[y,x] = \left(\begin{array}{cccc} 
 1 & 0 & 18 & 0 \\ 0 & 1 & 0 & 18 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1
\end{array}\right)$$
$$y^6[y,x]^{-1} = \left(\begin{array}{cccc} 
 1 & 18 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -18 \\ 0 & 0 & 0 & 1
\end{array}\right), \quad z^6 \beta^{-1} = z^6 (x^6 [y,x])^{-1} = \left(\begin{array}{cccc} 
 1 & 0 & 0 & 0 \\ 0 & 1 & -18 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1
\end{array}\right)$$ 
generate the positive root groups of $Sp(4,\mathbb{R})$ and, thus, $P\cdot\langle A, B \rangle\cdot P$ intersects the subgroup $U(\mathbb{Z})$ of unipotent upper triangular matrices of $Sp(4,\mathbb{Z})$ in a finite-index subgroup. 
Since we know that $\langle A, B\rangle$ is Zariski-dense (see my comment above to a question of Venkataramana), we can apply a result of Tits (http://www.ams.org/mathscinet-getitem?mr=424966) saying that Zariski dense subgroups of $Sp(4,\mathbb{Z})$ containing a finite-index subgroup of $U(\mathbb{Z})$ are arithmetic to get the desired conclusion.<|endoftext|>
TITLE: Two Vinogradovs? Is one the son of the other?
QUESTION [13 upvotes]: Forgive me for my ignorance, but I'm very surprised to learn that there are two Vinogradovs, both famous in the field of analytic number theory. Guessing from their names and the Russian naming convention, is one the son of the other?

REPLY [14 votes]: The answer seems to be no. It is hard to find direct confirmation, but every time their names are mentioned together it is made clear that there's no relationship.

The Russian mathematician Askold Ivanovich Vinogradov is not to be
  confused with the other Russian mathematician (the mathematical
  great-grandchild of Pafnuty Lvovich Chebyshev) whose work is at the
  heart of this paper: Ivan Matveyevich Vinogradov.

(Newsletter of the European Mathematical Society, issue 90)

There is a theorem due to Askold Ivanovich Vinogradov (1929-2005, not
  to be confused with another and more famous Russian mathematician,
  Ivan Matveevich Vinogradov, 1891-1988) that states...

Ian Hacking, Why Is There Philosophy of Mathematics At All?
In this obituary in the Russian Mathematical Surveys there's also no mention of him having children, let alone another somewhat famous soviet mathematician.
Ivan Matveyevich Vinogradov's Wikipedia article also mentions that he never married.<|endoftext|>
TITLE: On a Sum of Gamma Functions
QUESTION [5 upvotes]: I am working on a problem where the following sum appears:
$$F(s, t)=\frac{1}{\Gamma(1+2\alpha)}\sum_{n=0}^{\infty}{\frac{s^{n} t^{n}}{\left[(s+1)(t+1)\right]^{n+1+\alpha}}\frac{\Gamma(n+1+2\alpha)}{\Gamma(n+1)}}$$
where $s, t$ are positive real numbers and $\alpha>-1/2$. If I let Wolfram Mathematica calculate this seemingly awful expression, I get the following:
$$F(s, t)=\frac{(s+1)^{\alpha}(t+1)^{\alpha}}{(s+t+1)^{1+2\alpha}}$$
and it seems it doesn't take that long to compute it too! So I assume it is a "sort of" well-known formula.
Does anybody know if this formula is well-known, and if it is, could they tell me where I could find it in the literature? Otherwise, could anybody point an approach for proving this? Thanks in advance for any help! 
This identity crops up in the study of some linear operators I have, whose kernel is a product of Gamma functions.
Sorry for the edits, but I realized too late I had adjust a tiny mistake.

REPLY [10 votes]: Expanding my comment.
Notice that $\frac{\Gamma(n+2\alpha+1)}{\Gamma(n+1)\Gamma(2\alpha+1)}=\binom{n+2\alpha}{n}=(-1)^n\binom{-2\alpha-1}{n}$ and thus the sum can be rewritten as
$$\frac{1}{[(s+1)(t+1)]^{\alpha+1}}\sum_{n=0}^{\infty} \binom{-2\alpha-1}{n} \left(-\frac{st}{(s+1)(t+1)}\right)^n$$
$$ = \frac{1}{[(s+1)(t+1)]^{\alpha+1}}\left(1-\frac{st}{(s+1)(t+1)}\right)^{-2\alpha-1}
 = \frac{[(s+1)(t+1)]^{\alpha}}{(s+t+1)^{2\alpha+1}}$$
as expected.<|endoftext|>
TITLE: Integral points on elliptic curves of the form $y^2=x^3+px$
QUESTION [10 upvotes]: As the title says. Can we determine all the integral points on elliptic curves of the form 
$$y^2=x^3+px$$
for a prime $p$? If yes, can someone explain me how? A good reference would also be sufficient.

REPLY [13 votes]: This is completely worked out in Walsh's paper:

P. G. Walsh, Integer Solutions to the Equation $y^2=x(x^2\pm p^k)$ (2008)

In particular, for your elliptic curve $y^2=x^3+px$, there are at most 2 primitive integer solutions if $p>3$, and 4 if $p=3$.
As for imprimitive solutions in this case ($k=1$, $d=p$), the proof of theorem 4 indicates that such a solution would be a counterexample to the Ankeny-Artin-Chowla conjecture.<|endoftext|>
TITLE: Can you simplify (or approximate) $\sum_{n=0}^{N-1} \binom{N-1}n \frac{(-1)^n}{n+1} e^{-\frac{n}{2(n+1)}\lambda}$?
QUESTION [5 upvotes]: Let $\binom x y$ be the binomial coefficient. I am trying to get a better understanding of the sum
$$
f(N,\lambda)=\sum_{n=0}^{N-1}\binom{N-1}n\frac{(-1)^n}{n+1} e^{-\frac{n}{2(n+1)}\lambda}
$$
as a function of $\lambda$ and $N$ without having to plot it for all possible values.
Do you know of a way to simplify it or approximate it? Although I'd rather obtain an approximation for positive $\lambda$ and $N$, an asymptotic approximation for $\lambda\rightarrow\infty$ or $N\rightarrow\infty$, or even an upper-bound would be helpful too.
Here is an equivalent expression:
$$
f(N,\lambda)=\frac{1}{N}\sum_{n=1}^{N}\binom N n(-1)^{n-1} e^{-\frac{n-1}{2n}\lambda}.
$$
In my opinion, this sum could be easily simplified if the exponential term would have an argument that was not a rational function $\frac{n}{n+1}$ or $\frac{n-1}{n}$. I have not found a way in the literature to deal with it.

REPLY [9 votes]: The factor $\exp(-n\lambda/(2n+2))$ probably precludes significant
simplification of $f(N,\lambda)$, but one can still construct
an integral representation that shows $f(N,\lambda) > 0$ for all
positive $N$ and $\lambda$, and makes it possible to estimate the sum.
The idea is to find, given $\lambda$, a representation of
$\exp(-n\lambda/(2n+2)) \, / \, (n+1)$ as a linear combination of
exponentials $\exp(-cn)$ for which the sum can be evaluated
in closed form as a binomial expansion.  In other words,
we want to write $\exp(-n\lambda/(2n+2)) \, / \, (n+1)$ as a Laplace transform.
We use the definite integral formula
$$
\int_0^\infty I_0(\sqrt x) e^{-ax} \, dx = \frac1a \exp \frac1{4a},
$$
where $I_0$ is the modified Bessel function
$I_0(z) = \sum_{m=0}^\infty (z/2)^{2m} / m!^2$.  Writing
$$
e^{-\frac{n}{2(n+1)}\lambda} = e^{-\lambda/2} e^{\frac\lambda{2(n+1)}},
$$
we soon find
$$
f(N,\lambda) = \frac{e^{-\lambda/2}}{2\lambda}
  \int_0^\infty I_0(\sqrt x) \left[
    \sum_{n=0}^{N-1} (-1)^n {N-1 \choose n} e^{-\frac{n+1}{2\lambda}x}
  \right] \, dx
$$
in which the bracketed sum is $\exp(-x/2\lambda)$ times a binomial expansion:
$$
f(N,\lambda) = \frac{e^{-\lambda/2}}{2\lambda}
  \int_0^\infty I_0(\sqrt x) \, e^{-\frac{x}{2\lambda}}
    \left(1 - e^{-\frac{x}{2\lambda}} \right)^{N-1}
  \, dx \, .
$$
The integrand is nonnegative, so $f(N,\lambda) > 0$ as claimed.
Moreover, its size can be estimated for any given $N,\lambda$
by determining at what $x$ the integrand is maximized and how rapidly
it decays from this maximum.<|endoftext|>
TITLE: The hyperbolic metric on a flat surface
QUESTION [6 upvotes]: Let $S$ be a closed oriented surface of genus $\geq 2$ and $\mathcal{F}_n(S)$ be the space of flat metrics with conic singularities on $S$ whose cone angles are of the form $2k\pi/n$ ($k\in\mathbb{N}$). Let 
$\mathcal{F}_n^1(S)$ denote metics $g\in\mathcal{F}_n(S)$ of total area $1$.
Each $g\in\mathcal{F}_n(S)$ yields a conformal structure on $S$ and hence a unique hyperbolic metric $g_\mathrm{hyp}$ in the conformal class, which can be written as
$$
g_\mathrm{hyp}=\lambda_g\cdot g
$$
outside the singularities of $g$.
I need some information on the relationship between $g$ and $g_\mathrm{hyp}$, for example,


Question: Fix $n\in\mathbb{N}$. When $g$ runs over $\mathcal{F}_n^1(S)$, does $\int_S\lambda_g^{-1}\mathsf{dvol}_g$ have a upper bound or a positive lower bound?


It seems that when $n=2$, a lower bound exists but upper bounds do not. see the remark below. 
It is also natural to ask whether $\max_S(\lambda_g^{-1})$ has upper or lower bounds. 
Remark: 
A metric $g$ belongs to $\mathcal{F}_n(S)$ if and only if it has local expression
$$
g=|a(z)|^\frac{2}{n}|dz|^2
$$
for some conformal structure on $S$ (of which $z$ is a conformal local coordinate) and some holomorphic $n$-differential $\alpha=a(z)dz^n$. 
Now assume $n=2$, then the integral in the above question is the Weil-Petersson co-norm of the quadratic differential $\alpha$ (viewed as a cotangent vector to the Teichmüller space). On the other hand, the total area of $g$ is the Finsler co-norm of $\alpha$ with respect to the Teichmüller metric, therefore, for $n=2$, the existence of a lower bound in the above question is equivalent to the majorization of the Weil-Petersson metric by a constant multiple of the Teichmüller metric. The latter majorization is established in a 1974 paper by Linch, but her method is based on Teichmüller geodesics and does not seem to be generalizable to other $n$'s. 
Also, I have read that the problem of understanding the relationship between $g$ and $g_\mathrm{hyp}$ is treated by many authors such as Masur, Minsky, Rafi, etc., however, in their papers I only find miscellaneous technical statements on this and can't get a guiding idea. I would appreciate any systematic explanation of that relationship.

REPLY [2 votes]: The answer for the case $n=2$ of your question is described in my comments above. 
In summary, the lower bound can be derived from Linch's paper on the comparison of Teichmuller and Weil-Petersson distances, while an essentially sharp upper bound involving the inverse of the square root of the systole function can be derived from Wolpert's asymptotic formulas in Bers regions. 
Also, a non-sharp upper bound can be derived from Lemma 5.4 of a paper of Burns-Masur-Wilkinson (quoted above): the advantage of this estimate is that its proof uses only Cauchy formula and collar lemma, and, thus, it can be generalized to any other $n\in\mathbb{N}$.<|endoftext|>
TITLE: Primes in arithmetic progression with a moduli equal to a power of 2
QUESTION [8 upvotes]: I am currently looking for a result stronger than Siegel-Walfisz theorem, which gives an upper bound on the error term $|\pi(x,a,b)-\frac{\pi(x)}{\phi(a)}|$ for particular $a$.
The Siegel Walfisz is true for $a<(\ln x)^A$. Relying on this work :
https://eudml.org/doc/207492,
I conclude that I can assume that $|\pi(x,a,b)-\frac{\pi(x)}{\phi(a)}| < \frac{Cx}{\phi(a)(\ln x)^A}$ uniformly for $a = p^k$ ($p\geq 3$) and $a^{8/3+\epsilon} < x$ (correct me if I misunderstand something).
Regarding this recent work, http://link.springer.com/article/10.1007/s11139-006-0250-4, can I assume the same thing for $a=2^k$ and $a^3<x$ ?
What is so special about $p=2$ ? Is there a reference on this case ?

REPLY [7 votes]: There is nothing special about $p=2$; just that there are some technical differences, e.g. the group of reduced residues looks a little different than for odd prime powers.  In the case of powers of a fixed prime, one has a stronger Siegel-Walfisz theorem because the characters can essentially be thought of as polynomials and Vinogradov's method (and Vinogradov's zero free region) applies.  A generalized version of this was worked out in an early paper of Iwaniec On zeros of Dirichlet's L-series in Inventiones, 1974 (following the work of Postnikov, and Gallagher), and he does allow for $q$ to be a power of $2$.  He gives a good zero free region for integers $q$ for which the radical of $q$ (namely $\prod_{p|q} p$) is small, with the exception of a possible Siegel zeros for a real character, see Theorem 2.  Of course for powers of $2$ (or any fixed prime), there are only primitive real characters $\pmod 4$ or $\pmod 8$ and so the Siegel zero question is moot.  Once a Vinogradov type zero free region is at hand, one can combine this with the usual zero density results and then obtain the desired stronger Siegel-Walfisz results.<|endoftext|>
TITLE: A functional inequality about log-concave functions
QUESTION [20 upvotes]: Let $f,g$ be  smooth even log-concave functions on $\mathbb{R}^{n}$, i.e.,$f=e^{-F(x)}, g=e^{-G(x)}$ for some even convex functions $F(x),G(x)$. Is it true that:
$$
 \int_{\mathbb{R}^{n}}  \langle \nabla f(x), x\rangle\; g  dx \cdot \int_{\mathbb{R}^{n}}  \langle \nabla g(x), x\rangle\; f  dx \leq \int_{\mathbb{R}^{n}}\langle \nabla f(x), x\rangle\ \langle \nabla g(x), x\rangle\ dx \cdot \int_{\mathbb{R}^{n}}fg\,dx\; \quad (*)  
$$
The answer seems to be positive. 
Motivation: There is an interesting question which asks whether the function $\varphi(t)=\mu(e^{t} K)$ is log-concave on $\mathbb{R}$ where $K$ is a symmetric convex body in $\mathbb{R}^{n}$ and $d\mu$ is  an even log-concave probability measure. Now if you consider the expression $(\ln \varphi(t))''_{t=0}$ and ``do integration by parts'' several times and simplify few expressions you will arrive to an inequality which follows from (*) but definitely is weaker than (*). So I was wondering if there is a simple counterexample to (*).

REPLY [2 votes]: Not sure whether it is still of interest after three years, but anyway, I suggest the following:
ensure that your assumptions on $F$ and $G$ imply
1. $x\mapsto\langle \nabla F,x\rangle$ (same for $G$) is increasing with respect to some partial order $\le$ on $\mathbb{R}^{d}$
2. $(\mathbb{R}^{d}, \mathcal{B}(\mathbb{R}^{d}), Q, \le)$ with $Q(dx)=\frac{fg}{\langle f,g\rangle}dx$ is an FKG-space, c.f. https://projecteuclid.org/download/pdf_1/euclid.ecp/1465058067.
Hope that helps. Regards<|endoftext|>
TITLE: Slight variation on law of the iterated logarithm
QUESTION [6 upvotes]: Let$$M_t = \max\{B_s : 0 \le s \le t\},\text{ }m_t = \min\{B_s : 0 \le s \le t\},$$where $B_t$ is a standard Brownian motion. My question is, does there exist $r$ such that with probability one,$$\limsup_{t \to \infty} {{M_t - m_t}\over{\sqrt{t \log \log t}}} = r?$$If so, what is $r$?

REPLY [4 votes]: You can use the fact that $B_t-m_t$ is a reflected Brownian motion (see e.g. Revuz-Yor, Chapter VI, Theorem 2.3). I think it shouldn't be difficult to show that
$$
\limsup_{t\to\infty} \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup_{t\to\infty} \frac{B_t-m_t}{\sqrt{t\log\log t}}.
$$<|endoftext|>
TITLE: Reference for de Rham cohomology in positive characteristic
QUESTION [25 upvotes]: It is known in characteristic $0$ that (algebraic) de Rham cohomology is a Weil cohomology theory. However, in characteristic $p > 0$ it isn't, if only because it has mod $p$ coefficients, whereas Weil cohomology theories should take values in characteristic $0$. It turns out that de Rham cohomology picks up torsion from Crystalline cohomology, so the dimensions are actually not the correct ones.
However, even if it's not a Weil cohomology theory, one could still try to prove some of the properties. The one I am interested in is Poincaré duality, including the construction of the cup product pairing, say for smooth proper varieties.
I think I know how to do this, but it is a bit technical. I was wondering if there is a place in the literature where it is carried out. It seems that most sources on algebraic de Rham cohomology immediately assume characteristic $0$, and don't prove the results for characteristic $p$ (presumably because de Rham cohomology is 'not the correct thing to consider').
Edit. Maybe I should have stated more clearly what I already know, and what it is that I am looking for. I am certainly convinced of the truth of the statement. For $X$ projective, there is a nice Čech-theoretic approach, worked out below by David Speyer. In general, there is a dévissage approach where we truncate the de Rham complex and induct on the number of terms. One place where one can read about this is these notes (p. 3-4) by Johan de Jong. In the notes, it is only carried out for $X$ projective, but the same approach should work for $X$ proper. (The notes assume characteristic $0$, but this is not used in the construction.)
However, the reason I posted this reference request is that I want to use the result in a paper. I prefer using peer reviewed sources to informal notes, and this is the type of result that should have been known for more than 30 years. I was hoping someone happens to know where to find it.

REPLY [4 votes]: Prompted by user Yai0Phah's comment, let me answer my question: I did indeed find a reference. It turns out the correct place to look for these things is any book on crystalline cohomology, because
$$H^i_{\operatorname{dR}}(X) = H^i_{\operatorname{cris}}(X/k).$$
One canonical reference for Poincaré duality for smooth proper varieties in the more general setting of crystalline cohomology over $W_n(k)$ is [Berthelot, Ch. VII, Thm. 2.1.3]. The proof for $H^*_{\operatorname{cris}}(X/W_n(k))$ actually immediately reduces to the case of $H^*_{\operatorname{cris}}(X/k)$. This is then carried out by a careful dévissage on the successive truncations $\tau_{\geq i} \Omega_X^*$ and $\tau_{\leq n-i} \Omega_X^*$, which in the end deduces the statement from Serre duality. $\square$

References.
[Berthelot] Berthelot, Pierre, Cohomologie cristalline des schemas de caractéristique $p > 0$, Lecture Notes in Mathematics 407 (1974). ZBL0298.14012.<|endoftext|>
TITLE: For a linear recurrence sequence $(u_n)_{n\geq 0}$, can $\{i \mid u_i > 0\}$ be the set of Fibonacci numbers?
QUESTION [18 upvotes]: Question: Is there a linear recurrence sequence $(u_n)_{n\geq0}$ (on the rationals, but I would also be interested by reals) for which $\text{Pos}(u) = \{i \mid u_i > 0\}$ is precisely the set of Fibonacci numbers?
Additional infos:

It is well known that the set $\{i \mid u_i = 0\}$ is quite easy to describe (semilinear), but the sets with $>$ instead of $=$ can get quite weird.  For instance, there is a $(u_n)_{n\geq 0}$ with  $\{{k_1} < {k_2} < \cdots \} = \text{Pos}(u)$ such that $\lim_{i \to \infty} k_{i+1} - k_i = \infty$.
This question is somewhat linked to the question "For a trigonometric polynomial $P$, can $\lim \limits_{n \to \infty} P(n^2) = 0$ without $P(n^2) = 0$?"
The use of Fibonacci in the current question is not all that
relevant—take the squares or the powers of two if it's easier.

REPLY [27 votes]: Solutions to (complex) linear recurrences are of the form 
$$\sum_i c_i n^{e_i} \alpha_i^n.$$
To build such a function that is positive only on Fibonacci numbers, take $n^2(\alpha^n + \bar{\alpha}^n-2)+c$ where $\alpha = \exp(2\pi i / \phi), \phi = \frac{\sqrt{5} + 1}{2}$. For this to be positive, $\Re(\alpha^n)$ has to be close to $1$, which means $\alpha^n$ is close to $1$, which means $n/\phi$ is close to an integer. 
Numerically, $c=8$ works. The values of $n^2(\alpha^n+\bar{\alpha}^n-2)+8$ are positive and drop toward $0.1043$ at Fibonacci numbers and they are quite negative at non-Fibonacci indices, with the nearest misses at indices of the form $\operatorname{Fibonacci}(n) + \operatorname{Fibonacci}(n-3)$, where it seems to takes the value $-118$ followed by indices of the form $\operatorname{Fibonacci}(n) + \operatorname{Fibonacci}(n-2)$, $-189$. A proof using standard results on simple continued fractions shouldn't be hard, and I'll try to fill in the details later.

I'll prove that $c=9$ works.
If $n = F_k$, the $k$th Fibonacci number, then we have an explicit formula $n = \frac{1}{\sqrt{5}}\left( \phi^k - (-\phi)^{-k}\right)$ and the distance between $F_k/\phi$ and the nearest integer is $1/\phi^k$ as long as $k \ge 2$. The formula is incorrect for $k=1$, but $F_1=1=F_2$, so we can simply assume $k\ne 1$. 
$$\begin{eqnarray}n^2 (\alpha^n + \bar\alpha^n - 2) &=& 2n^2(\cos 2\pi n/\phi - 1) \newline &=& 2n^2(\cos 2\pi \phi^{-k} -1) \newline & \ge & 2n^2 \left(-\frac{1}{2}(2\pi \phi^{-k})^2\right) \newline &=& \left(1 - (-1)^k\phi^{-2k}\right)^2 \left(-\frac{4 \pi^2}{5}\right) \end{eqnarray}$$
The estimate was that $\cos x \ge 1-\frac{x^2}{2}$.
When $k$ is even, we get that $-4\pi^2/5 = -7.90$ is a lower bound, so if we add $9$, $n^2(\alpha^n+\bar\alpha^n-2)+9 \gt 0$. For $k$ odd, $k \ge 3$ and then 
$$-\frac{4 \pi^2}{5}(1-(-1)^k\phi^{-2k})^2 \ge -\frac{4 \pi^2}{5} (1+\phi^{-6})^2 = - 8.8.$$
So, if $n$ is a Fibonacci number, then $n^2 (\alpha^n + \bar\alpha^n - 2) + 9 \gt 0$.

Suppose $n$ is not a Fibonacci number. Let $m$ be the closest integer to $n/\phi$. Then $\alpha^n + \bar\alpha^n = 2\cos 2\pi n/\phi = 2 \cos(2 \pi m - 2 \pi n/\phi) = 2 \cos 2\pi n(m/n - \phi)$. 
If $m/n$ is reduced, then because the denominator is not a Fibonacci number, $m/n$ is not a convergent to the simple continued fraction for $1/\phi$. This implies that 
$$\left|\frac{m}{n} - \frac{1}{\phi}\right| \ge \frac{1}{2 n^2}$$ 
by Theorem 184 in Hardy and Wright, An Introduction to the Theory of Numbers, 4th edition. (In fact, this inequality is far from sharp for approximations to $\phi$ or $1/\phi$, so much better estimates could be used.) Multiply both sides by $2\pi n$: 
$$\left|2 \pi m - 2\pi n/\phi \right| \ge \frac{\pi}{n}.$$
If $\left|2 \pi m - 2\pi n/\phi \right| \gt 1$, then $-2n^2 (\cos \left|2 \pi m - 2\pi n/\phi \right|-1) $ can only be small if $n$ is small, and those cases are easy to check numerically. Otherwise, we can use that when $|\theta| < 1, \cos \theta \le 1-\frac{11}{24}\theta^2$ from the first three terms of the power series, which implies that 
$$\begin{eqnarray}2n^2 (\cos \left| 2 \pi m - 2\pi n/\phi \right| -1) &\le& 2n^2\left(-\frac{11}{24} \left(\frac{\pi}{n} \right)^2\right) \newline &\le & -\frac{11}{12}\pi^2 \newline &\le& -9.04.\end{eqnarray}$$
If $m/n$ is not reduced, then $m$ and $n$ share a common factor, and then even if $m/n$ is a convergent to the simple continued fraction for $1/\phi$, $\cos 2\pi n (m/n - 1/\phi)$ is too far from $1$. 
Thus, if $n$ is not a Fibonacci number, then $n^2(\alpha^n + \bar\alpha^n - 2)+9 \le 0$.<|endoftext|>
TITLE: A technical question about root systems
QUESTION [7 upvotes]: I'm studying root systems and coming up with an observation:
Let $\Phi$ be an irreducible root system and $\Phi^+$ be a positive root system. Denote by $\Delta=\{\alpha_1, \alpha_2,\ldots,\alpha_n\}$ be the corresponding base. For $\beta, \gamma \in \Phi^+$, $\alpha_1, \alpha_2 \in \Delta$ with the properties that $\beta=\gamma +\alpha_1+\alpha_2$, $\gamma +\alpha_1 \in \Phi^+, \gamma +\alpha_2\in \Phi^+, \alpha_1+ \alpha_2 \notin  \Phi^+$ then

Statement: There exists $\alpha_3 \in \Delta\setminus\{\alpha_1,\alpha_2\}$ such that either $\gamma +\alpha_1-\alpha_3 \in \Phi^+$ or $\gamma +\alpha_2-\alpha_3 \in \Phi^+$.

The motivation is to investigate the relationship between a positive positive root and other lower positive roots. That is to understand the so-called Hasse diagram root systems (for instance see definition of Hasse diagram in here)
Any help would be much appreciated.

REPLY [4 votes]: Consider the root system $\mathsf{B}_n$ with the standard numbering of the fundamental roots (that is, $\alpha_n$ is short). Take $\alpha_{n-2}$ and $\alpha_n$ as a pair of orthogonal roots (indeed, $\alpha_{n-2}+\alpha_n$ is not a root), and take $\gamma=\alpha_{n-1}+\alpha_n$. Then $\beta_1=\alpha_{n-2}+\alpha_{n-1}+\alpha_n$, $\beta_2=\alpha_{n-1}+2\alpha_n$ and $\alpha_{n-2}+\alpha_{n-1}+2\alpha_n$ are all roots. But among the differences of the form $\beta_i-\alpha_j$ the only roots are $\beta_1-\alpha_{n-2}$, $\beta_1-\alpha_n$ and $\beta_2-\alpha_n$, so this gives a counter-example.
The beautiful pictures of Hasse diagrams you refer to provide a good way to spot such examples, but for this one should draw them in a different way, which is easier to read. The keyword here is a weight diagram. For examples, see this collection of the diagrams along with a description of their various usages.
Namely, Figure 14 on page 35 is the weight diagram of the adjoint representation of a group (or of a Lie algebra) of type $\mathsf{B}_n$, which also describes the structure of the root system (its vertices are the roots, plus the "zero weights" corresponding to the fundamental roots). One looks for a square which has non-consecutive label on its sides (say $i$ and $j$), such that the bonds joining its top or bottom vertex to something on the right are also labeled by either $i$ or $j$. Such a square is immediately found on the very bottom of the picture slightly to the left from the middle.
By the way, the quick inspection of the weight diagrams shows that there is no counter-examples besides the one above (I'm not sure, but I haven't spotted any).<|endoftext|>
TITLE: Examples of differential cohomology in cohesive $\infty$ topos
QUESTION [14 upvotes]: I might direct this question to Urs Schreiber directly, but just in case someone else has some interesting examples, I'll make the question public.
The formulation of differential cohomology in certain cohesive $\infty$ topos seems, to me, to be an incredibly powerful generalization of the concepts surrounding classical Chern-Weil theory. I have, for a while, been curious as to what forms differential cohomology may take in topos other than the typical one modeled on cartesian spaces (or differentiable manifolds).
Question: Can someone provide some examples of what form differential cohomology may take in topos modeled on different local data? More precisely, what does the differential cohomology diamond look like in these other topos? 
To give the beginning of a possible example, suppose we take as our starting point the p-adics $\mathbb{Q}_p$ instead of the reals and form the cohesive site whose objects are $(\mathbb{Q}_p)^n$, $n\in \mathbb{N}$ and morphisms are $p$-adic smooth maps (for example locally constant when the support is compact). If I'm not mistaken, we should be able to form the differentially cohomology diamond in the resulting $\infty$ topos. Can we identify the components of this diamond with something more familiar from the world of number theory?

REPLY [9 votes]: If we consider just shape modality "$\Pi$" (or "ʃ") and flat modality $\flat$ (which are sufficient for the differential cohomology hexagon, then the traditional arithmetic fracture squares are the left half of the differential hexagon for a kind of pointed $E_\infty$-derived cohesive geometry modeled on Greenlees-May duality. For more on this see my talk  Differential cohesion and Idelic structure.
If we consider in addition also the sharp modality $\sharp$ (which is needed to construct moduli stacks of differential cocycles) then an "exotic" model for cohesive homotopy theory is global equivariant homotopy theory. This was found by Charkes Rezk, see his writeup "Global Homotopy Theory and Cohesion".
For these "exotic" models shape is not the localization at an interval object, and hence these exotic differential cohomology theories need not satisy the fundamental theorem of calculus/Stokes theorem (nLab:Stokes theorem -- Abstract formulation in cohesion)
By the way, even if the "standard model" with interval object the real line feels very standard, there are things still to be learned here. Readers with an interest in formal logic might enjoy Mike Shulman's recent "Brouwer's fixed-point theorem in real-cohesive homotopy type theory".
A mild but important variant of the standard model over smooth manifolds which still does satisfy the integration theorem is the model over complex analytic manifolds (see at complex analytic infinity-groupoid). In here ordinary differential cohomology comes out as the theory of holomorphic $p$-gerbes with connection (which of course was what Deligne considered first, back in 1971.) This is also the context of Hopkins-Quick 12.
Then, both the model over smooth and over complex-analytic manifolds may be refined to supergeometry. When doing so, then the adjoint triple of cohesion is accompanied by a progression of two more adjoint triples, each "resolving" the former one. The second one gives what I had called "differential cohesion", which serves to axiomatized manifolds and PDE theory. The third reflects the two possible ways of projecting out bosonic formal geometry from supergeometry. This gives a second cohesive structure on supergeometric homotopy theory. 
Details for this super-cohesion are in v2 of my book, some key points are in nLab:super Cartan geometry; exposition includes Modern physics formalized in Modal homotopy theory. Maybe see also the chapter Higher prequantum geometry which I am preparing for "New Spaces in Mathematics and Physics".<|endoftext|>
TITLE: $(n-2)$-blocking sets in $AG(n,2)$
QUESTION [5 upvotes]: Let's define $k$-blocking set in affine space $AG(n,q)$ a set that meets every coset (translate of subspace) of dimension $k$. 
I have seen a lot work related to minimal $(n-1)$-blockings set.
Covering finite fields with cosets of subspaces. 
The Blocking Number of an Affine Space 
In these articles it is proved that minimal $(n-1)$-blocking has $n(q-1)+1$ points. 
I can't find any result about minimal $(n-2)$-blocking sets (even for $AG(n,2)$). I have managed to prove following bounds about $(n-2)$-blocking set in $AG(n,2)$. It has at least $2n-1$ and no more than $3n^{\log_{2} 3}+1$ points.
Here is a link to my paper.
http://ysu.am/files/8.On%20The%20Minimal%20Coset%20Covering%20of%20Solutions%20of%20a%20Boolean%20Equation.pdf
I am very interested in the solution of the problem. Does anyone have any information about it?
I have also asked following questions related to problem:
https://math.stackexchange.com/questions/869308/blocking-set-for-cosets-of-codimension-2
https://math.stackexchange.com/questions/863592/find-minimal-set-of-cosets-c-so-that-each-2-vectors-in-a-n-are-in-one-cos

REPLY [5 votes]: Not much is known for the general case. 
Let $m(k, n, q)$ denote the minimum size of an $k$-blocking set in $AG(n, q)$. Trivially we have $m(0, n, q) = q^n$ and $m(n, n, q) = 1$. By Jamison/Brouwer-Schrijver we get $m(n-1, n, q) = 1 + n(q-1)$ as you have mentioned. To at least give bounds on other values we can prove the following inequality $$qm(k, n-1, q) \leq m(k, n, q) \leq (q-1)m(k, n-1, q) + m(k-1, n - 1, q).$$
The lower bound follows from taking $q$ parallel hyperplanes that cover the space and then observing that the intersection of the blocking set with each of these hyperplanes is $k$-blocking set in $AG(n-1, q)$. For the upper bound, again take $q$ parallel hyperplanes, $H_1, \dots, H_q$. For $1 \leq i \leq q - 1$, let $B_i$ be a $k$-blocking set in $H_i$ and let $B_q$ be a $(k-1)$-blocking set in $H_q$.  Then one can argue that $B = B_1 \cup \dots \cup B_q$ is a $k$-blocking set in $AG(n, q)$ as follows. If $S$ is a $k$-flat in $AG(n,q)$ which is contained in any $H_i$ for $i < q$, then it is blocked by $B_i$. If it is contained in $H_q$, then it is blocked by $B_q$ since $B_q$ blocks all $(k-1)$-flats, it must block all $k$-flats as well. If $S$ is not contained in any of the $H_i$'s then its intersection with $H_q$ is a $(k-1)$-flat, which is again blocked by $B_q$. 
For your particular case of $(n-2)$-blocking sets, this tells us that $$q(1 + (n-1)(q-1) \leq m(n-2, n, q) \leq (q-1)^2\binom{n}{2} + (q-1)n + 1.$$
So, for a constant $q$ there is a linear lower bound and a quadratic upper bound. I don't know if there has been any improvement ``in general'' besides the one that you mention for $q = 2$. (Do your methods extends to $q > 2$? If yes, then do they improve these bounds?) I have asked a couple of colleagues who work in finite geometry and they don't know either. 
For some particular values, note that in $AG(3,3)$ the complement of such a blocking set is a cap (see https://www.mathi.uni-heidelberg.de/~yves/Papers/CapSurvey.pdf). And we know that the maximum size of a cap in $AG(3,3)$ is $9$. Thus, $m(1, 3, 3) = 18$, which is one less than the upper bound we get above. I guess some other specific values can also be computed. Similarly, the problem of finding smallest 1-blocking sets in $AG(n,3)$ is equivalent to finding the largest caps in $AG(n,3)$, on which there is significant progress in general. See New bounds on cap sets. 
Note that the problem is completely solved for projective spaces. We know that a $k$-blocking set in $PG(n,q)$ has size at least $\big[ {n-k + 1 \atop 1} \big]_q = 1 + q + \dots + q^{n-k}$ with equality if and only if the blocking set is an $(n-k)$-dimensional subspace. But if you want to talk about ``minimal'' blocking sets instead of the smallest ones, then the problem again becomes much harder. 
For a recent survey on projective blocking sets see Chapter 3 in current research topics in galois geometry. There is also a small section on affine blocking sets where they mention the same recursive bound. They follow a different terminology though: what you have called an $(n-k)$-blocking set is called a $k$-blocking set. 
Edit When $n = 3$ there are some nice improvements to the bounds. Peter Sziklai, and later Simeon Ball proved  general results on $1$-blocking sets in affine spaces, which imply that the size of such a set in $AG(3,q)$ is at least $2q^2 - 1$. See Blocking sets in three dimensional finite affine spaces for the details. Also, the lower bound can be improved in general to $m(n-2, n, q) \geq (n-1)(q^2 - 1) + 1$. See Section 3 in http://www-ma4.upc.es/~simeon/polynomialmethod.pdf<|endoftext|>
TITLE: Arithmetical results to help study arithmetic geometry?
QUESTION [13 upvotes]: I'm very keen to deepen my understanding of arithmetic and diophantine problems. In the past I studied some algebraic, analytic and sieve based number theory. Recently I've been reading Weil - Basic Number Theory which covers some early results of Fermat and Euler in their original forms and then discusses modern interpretations of the work in terms of curves and genus.
Researching into this I think that the next thing I should do is learn arithmetic algebraic geometry: I have been looking for guides and this one was helpful Road map for learning arithmetic algebraic geometry. (Open to other things if they would be more appropriate too).
I am very keen on learning concrete results about the integers, so I have been finding it hard to do things like work through all of Hartshorne before I can learn more number theory. I'm trying to study An Invitation to Arithmetic Geometry - Dino Lorenzini instead, but it seems like it doesn't cover much concrete results either, instead being the background to start studying something like Falting's theorem.

So my question is please can you help recommend to me some results in number theory that I can set as goals to understand the proofs of, and motivate me to follow what looks like a very hard and long path of learning arithmetic geometry. If you could suggest any research papers or books.

I'm having a difficult getting back into mathematics a couple of years after finishing my masters. Since I'm not at an institution I don't have anyone to look to for guidance so I would greatly appreciate help.
Thank you kindly.

REPLY [13 votes]: That will depend  on your more specific interests. One point of entrance into arithmetic geometry could be to have a good understanding of Szpiro's discriminant-conductor inequality for non-isotrivial elliptic curves over a function field, particularly the proof embodied by the Kodaira-Spencer class, and to see how this result (but not its proof!) leads, by direct translation over arithmetic surfaces, to the statement of the $abc$ conjecture  (in the averaged form). You may start with Szpiro's papers, particularly  Discriminant et conducteur des courbes elliptiques from 1990. This requires a modest algebro-geometric background and will motivate  a deeper study of the subject, leading up to works on Arakelov theory and to the Vojta conjectures, which have informed Vojta's diophantine approximations reproof of Faltings's theorem.
Some landmark results in this spirit:

Faltings's Big theorem: For $A/\mathbb{C}$ an abelian variety and $\Gamma \subset A(\mathbb{C})$ a finitely generated subgroup, the Zariski closure of any subset of $\Gamma$ is a finite union of cosets of algebraic subgroups of $A$. 
The Manin-Mumford conjecture, proved by Raynaud: The same for $\Gamma = A_{\mathrm{tors}}$, the torsion subgroup of $A$. Szpiro, Ullmo, and Zhang have discovered an Arakelov theoretic proof of a stronger form (Bogomolov's conjecture) that has the torsion points replaced by a sequence of points with canonical height approaching zero.
Andre-Oort conjecture (now a theorem in this case, after papers by Tsimerman and Yuan-Zhang): The description as special subvarieties of Zariski closures of sets of CM points of $\mathcal{A}_g(\mathbb{C})$. This is in the spirit of the other two problems but has especially interesting ties with analytic number theory, which you will easily appreciate given your background.

A guide to current developments on this topic is Zannier's book Some Problems of Unlikely Intersections in Arithmetic and Geometry. And, as for setting a systematic and rigorous learning path, it seems to me that the book you might be looking for is Heights in Diophantine Geometry, by Enrico Bombieri and Walter Gubler. Following it, you could set the proof of Faltings's theorem as a more distant goal while reaping many small rewards continually along the way. The fourth and fifth chapters, for example, prove real results about the integers that do not require any algebro-geomeric background, and yet have extensions leading quickly to arithmetic geometry. Such are the uniform finiteness of solutions to the two-variable $S$-unit equation or the Bogomolov property of the field $\mathbb{Q}^{\mathrm{ab}}$, with a beautiful application to a reproof of Smyth's theorem: an integer non-reciprocal polynomial has Mahler measure bounded away from $1$.  All proofs are given in full detail, and all the involved algebro-geometric background is collected in an appendix, that you can learn as you go through the book. After the diophantine approximations proof of Faltings's theorem you may turn with fresh motivation to the specialized literature and engage, for example, with a study of either of the three topics I mentioned earlier.
I have only touched on one facet of arithmetic geometry. But I think it is one that will give you a good taste for the subject. It is then only natural to look into Faltings's Endlickeitsatze paper (that contains among other things the first, and very different, proof of the Mordell conjecture), leading up to other problems and results that, in turn, stimulated developments in arithmetic geometry. In this particular case we have

Fontaine's proof that there is no abelian variety over $\mathbb{Z}$,

another major result of arithmetic geometry that is especially well motivated from an analytic number theory background (cf. the Odlyzko bound and Theorem 5.51 in Iwaniec and Kowalski's book).<|endoftext|>
TITLE: Motivation for equivariant homotopy theory?
QUESTION [15 upvotes]: I'm in the process of learning equivariant homotopy theory, so I was wondering: what is the importance of equivariant homotopy theory, and what has it been applied to so far? I know of HHR's solution to the Kervaire invariant problem, but what are some other applications?

REPLY [11 votes]: One of the applications that motivated the development of the theory is the solution of the Segal conjecture which states that the zeroth cohomotopy of a classifying space of a finite group is isomorphic to the completion of the Burnside ring.  You can read about this in 

G. Carlsson. A survey of equivariant stable homotopy theory. Topology 31 (1992), 1-27.

There is also a book by J.P. May "Equivariant homotopy and cohomology theory" which discusses such applications.

Another application of equivariant cohomology is to the computation of ordinary cohomology of spaces with group actions. Equivariant cohomology has a nice property referred to as "localization", which allows to get cohomology information from just looking at fixed points. This can make computations of ring structures on equivariant cohomology easier than in ordinary cohomology (but one can then recover the ordinary cohomology ring from the equivariant one). You can read about this in the applications section of Tu's "What is ... equivariant cohomology?", the paper by Goresky-Kottwitz-MacPherson on equivariantly formal spaces, or the paper of Brion-Vergne on localization in equivariant cohomology

Finally, equivariant homotopy theory can also be used for classification of group actions on e.g. spheres. One can use equivariant versions of obstruction theory to classify maps between representation spheres of finite groups, cf. e.g.

S.J. Willson, Equivariant maps between representation spheres. Pacific J. Math.  56, (1975), 291-296.

See also the works of tom Dieck, Petrie,... This application of equivariant homotopy theory is sort of a non-abelian version of representation theory.<|endoftext|>
TITLE: Triangle centers from curve shortening
QUESTION [21 upvotes]: The curve-shortening flow transforms curves in the plane by moving each point perpendicularly to the curve at a speed proportional to the curvature at that point. It is usually defined for smooth curves (so you can determine the curvature) but according to a survey by White (Math. Int. 1989) it can be extended to rectifiable varifolds. In particular, according to this extension, it should be well-defined for triangles, which should instantly turn into smooth approximations of themselves and then evolve as normal, shrinking to a point. (This would also follow immediately from a positive answer to a previous question about continuity of the curve-shortening flow, assuming that the continuity is in a sufficiently general space of curves to include triangles and not just smooth curves.) Since the flow isn't affected by Euclidean transformations, this point must be a triangle center. Is it a known triangle center (one of the ones listed at the Encyclopedia of Triangle Centers)? If so, which one? If we could just calculate the flow accurately enough we could look it up in ETC but I don't know how to do the calculation.
If this turns out to be known, a reference would be helpful, so that it could be added to the Wikipedia article.

REPLY [2 votes]: I formerly had a handwavy argument here showing that the vanishing point is always near (but not equal) the centroid, but have now replaced it by a more rigorous argument. This doesn't entirely answer the question (which center is the vanishing point), but should at least help eliminate many centers as definitely not being it, because most centers do not have the same property of always being near the centroid.
First, to eliminate the centroid itself: intuitively, the vanishing point shouldn't be the centroid by the idea presented in the answer by foliations: the centroid has a local formula and the vanishing point shouldn't have such a formula. (This should also be true of the Spieker center, the centroid of the triangle's perimeter.) But a little more directly: in a highly-obtuse isosceles triangle, all triangle centers lie on the altitude of the triangle, but the centroid is always exactly 1/3 of the way along the altitude from the base (and the Spieker center is almost on the base), while the vanishing point should be closer to the incenter, near the midpoint of the altitude.
Next, let's show that the vanishing point is near the centroid. The previous handwavy argument used a vague concept of "aspect ratio" (see comments); to make this more concrete let's use the isoperimetric ratio $\rho=L^2/A$ which is known to decreases monotonically through the evolution of a convex curve (Gage 1984).
Consider a smooth convex curve $C$ with diameter $D$, and let $w$ be the width perpendicular to the diameter; then $D=\Theta(L)$ and $A=\Theta(wD)$ so
$w=\Theta(D/\rho)$.
Let $xy$ be a chord of $C$ perpendicular to $D$ through the centroid. Then $xy$ must cross $D$ somewhere within its middle third, and the tangents to $C$ at $x$ and $y$ must form an angle with each other (and with the diameter segment) that is $O(1/\rho)$, else $xy$ would be longer than $w$. We consider all of the symbols $D$, $w$, $x$, and $y$ to vary continuously as the curve evolves, and use $w_0$ (etc.) to refer to their initial values. Note that, because $C$ remains confined to a rectangular box with width $w_0$, it will always be the case that $w=O(w_0)$.
At any point in the evolution of the curve, the same calculation shows that the angle between the two tangents is small, $O(w/D)$. It follows that the rates of area loss on the two sides of $xy$ differ from each other by at most a factor of $1+O(w/D)$, and we know that the total rate of area loss is constant, so the rate at which the area difference between the two sides changes is $O(w/D)$. It follows that the speed at which segment $xy$ can move is $O(1/D)$, because a faster movement would create too much area difference from one side to the other.
Now consider any time period within which the diameter decreases from $D_t$ to $D_t/2$. Surrounding the curve by pair of width-$w$ grim reaper curves (one in each direction), and using the avoidance principle for curve shortening (if a curve is surrounded by another curve, the two curves cannot cross) shows that this happens in time $O(w_tD_t)$. Within this time period, the centroid can only move a distance of $O(w_t)$: the speed of $xy$ controls its motion parallel to the diameter and the length of $xy$ controls its perpendicular motion. After repeating this argument $\Theta(\log\rho)$ times we will have $D\le w_0$, after which the vanishing point is bounded within the $O(w_0)$ remaining diameter of the curve. Therefore, the vanishing point of any smooth curve is within distance $O(D\frac{\log\rho}{\rho})$ of the centroid. 
Triangles aren't smooth but immediately become smooth as soon as you start evolving them, so the same argument applies. I'm not sure whether the logarithmic factor in the bound above is necessary, or whether the vanishing point is always within the closer distance $O(D/\rho)$ of the centroid.<|endoftext|>
TITLE: Unramified extension of number fields
QUESTION [7 upvotes]: Any finite field extension (in particular Galois extension) of $\mathbb{Q}$ is ramified. Is there an intuitive geometric explanation of this fact?
Suppose we have an number field $K$, is any Galois extension of $K$ ramified? I think the answer is no, but I do not have a clear picture, examples will be appreciated. 
My main question is the following:
Can we reconstruct the absolute Galois group $\mathrm{Gal}_{\mathbb{Q}}(\overline{\mathbb{Q}})$ using only unramified Galois extensions?

REPLY [2 votes]: I am far from being an expert, but I can confirm that there exist number fields $K\neq\mathbb{Q}$ which have no nontrivial unramified extensions. For example, imaginary quadratic number fields of class number one have this property. See this paper by Yamamura for more examples and background information.
On the other hand, I don't understand your main question. What do you mean by "reconstructing" the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$?<|endoftext|>
TITLE: Are there varieties with non finitely generated Picard group and vanishing irregularity?
QUESTION [8 upvotes]: Let $X$ be a smooth projective variety over an algebraically closed field $k$.
Can it happen that $q(X) := \dim H^1(X,\mathcal O_X) =0$ and $\textrm{Pic} \,X$ is not finitely generated?
Certainly, if $q(X) =0$, then $\textrm{Pic}^0 \,X =0$. But is that enough to conclude that the abelian group $\textrm{Pic} \, X$ is finitely generated?

REPLY [10 votes]: As pointed out by Jason Starr, the answer to your last question is yes, so the answer to the question in your title is no. Let me give a quick (and straightforward) proof in the case $k= \mathbb{C}$. 
First of all, any projective variety over $\mathbb{C}$ has the homotopy type of a finite CW-complex, hence all (co)homology groups of $X$ with coefficients in $\mathbb{Z}$ are finitely generated abelian groups.
Next, let us consider the exponential sequence $$0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^* \to 0.$$
Passing to cohomology, and using the assumption $q(X)=0$, we deduce that $\textrm{Pic}\,X=H^1(X, \, \mathcal{O}_X^*)$ injects into $H^2(X, \, \mathbb{Z})$. 
Since the latter is a finitely generated abelian group, the same holds for $\textrm{Pic}\,X$.

REPLY [2 votes]: I will quote Jason Starr's comment:
"Yes, that is enough to conclude. Classically this is the theorem of Lang-Neron, sometimes also called the "Theorem of the Base". An exposition in modern terms is contained in SGA 6, one of the exposes by Steven Kleiman (I do not have my copy of SGA 6 handy) . . . I just looked it up: Expose XIII."<|endoftext|>
TITLE: Positivity of coefficients of the inverse of a certain power series
QUESTION [17 upvotes]: Consider the unique formal power series $g(z)$ with $g(0)=0$ and $g'(0)=1$ satisfying the equation 
 $$
g(z)-g(z)^8+g(z)^{15}=z,
 $$
that is the inverse of
 $$
z-z^8+z^{15} 
 $$
in the group of formal power series of the form $z+O(z^2)$. Does this series have any non-positive coefficients of powers $z^{7k+1}$? (Other coefficients are manifestly zero).
Background.  Existence of non-positive coefficients among those would imply that a certain operad fails the Ginzburg-Kapranov test for Koszulness. (See http://arxiv.org/abs/0907.1505). 
Comments.  The first ten thousand or so of those coefficients are positive. Some elementary complex analysis approaches didn't work, but maybe I am not aware of some useful tricks.

REPLY [19 votes]: Combining Omar and fedja's comments gives a quick solution. By the Lagrange Inversion formula, we want to show that the coefficient of $t^k$ in $(1-t+t^2)^{-(7k+1)}$ is positive. Writing this as a contour integral, we want to compute
$$\frac{1}{2 \pi i} \oint \frac{dt}{t (1-t+t^2)} \left( \frac{1}{t (1-t+t^2)^7} \right)^k.$$
Set $f(t) = \tfrac{1}{t(1-t+t^2)}$ and $g(t) = \frac{1}{t(1-t+t^2)^7}$, so we want $\tfrac{1}{2 \pi i} \oint f(t) g(t)^k dt$.
The critical points of $g$ are at $t=1/5$ and $t=1/3$, so take our contour to be a circle of radius $1/5$ around the origin. We want $\tfrac{1}{10 \pi} \int_{\theta=0}^{2 \pi} f(e^{i \theta}/5) e^{i \theta} g(e^{i \theta}/5)^k d \theta$.
We have $g(1/5) = \tfrac{30517578125}{1801088541} \approx 17$, and $g(e^{i \theta}/5)$ smaller than that for all other $\theta$. We set $a=g(1/5)$. Here is a plot of $|g(e^{i \theta}/5)|$; I leave rigorous proof for others. So values of $\theta$ very near $0$ dominate the integral. The Taylor series for $g(e^{i \theta}/5)$ starts $a-b \theta^2$ where $b=\tfrac{152587890625}{37822859361}>0$. (Note the absence of an $i \theta$ term, because we chose a path through the critical point.)

Also, $f(1/5)=\tfrac{121}{25}>0$, call this $c$. So the integral is roughly $\int c (a-b \theta^2)^k d \theta \approx \int c a^k e^{-(b/a)k \theta^2} d \theta = \frac{C}{\sqrt{k}} a^k$ for some positive $C$. In particular, it is positive for large enough $k$ and, as fedja says, $10000$ is going to be way more than enough. I leave rigorous bounds to you.<|endoftext|>
TITLE: A specific linear differential equation on $\mathbb{C}-\{0,1\}$ whose monodromy group represents the fundamental group of $\mathbb{C}-\{0,1\}$
QUESTION [8 upvotes]: Is there a linear differential equation on $\mathbb{C}$ with singularities at $0$ and $1$ whose monodromy group represents the fundamental group of $\mathbb{C}-\{0,1\}$? If so, can someone give a specific example?

REPLY [13 votes]: If the word "represents" in the question means "isomorphic to", one can take
the hypergeometric equation: $w''+Qw=0$ with
$$Q(z)=\frac{1}{4z^2}+\frac{1}{4(z-1)^2}-\frac{1}{4z(z-1)}.$$
See Hurwitz-Courant, Ch7, or (with much more detail) Caratheodory, vol. II.
The ratio $f=w_1/w_2$ of two linearly independent solutions of this equation
maps the upper half-plane onto a triangle with all angles $0$. One can normalize
(choose particular solutions $w_1,w_2$) so that this triangle is inscribed in the unit disk. Then the well-known picture (see any course of complex variable) shows that $f$ is the inverse to the universal covering of $C\backslash\{0,1\}$ by the unit
disk. Which implies that the monodromy group is the fundamental group of $C\backslash\{0,1\}$.
EDIT. Written in the standard hypergometric form, the equation is 
$$z(z-1)y''+(1-2z)w'-w/4=0.$$<|endoftext|>
TITLE: Is Max (R) a Hausdorff space?
QUESTION [10 upvotes]: I asked the following question at < https://math.stackexchange.com/questions/1508168/is-max-r-a-hausdorff-space  >, but I pose it here for any help.

Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Now let $R $ be a commutative ring with identity such that $\operatorname{Max}(R)$ is a totally disconnected space, in the sense of the $Zariski$ topology. I want to know if $\operatorname{Max}(R)$ Hausdorff in this case?

Thanks for any help.

REPLY [10 votes]: For a ring $R$, the space $\operatorname{MaxSpec} R$ -- the set of maximal ideals of $R$ endowed with the Zariski topology -- has two evident properties: its singleton sets are closed (i.e., the T1 separation property) and it is quasi-compact.  In his thesis, M. Hochster showed that conversely any quasi-compact T1 topological space $X$ is homeomorphic to $\operatorname{MaxSpec} R$ for some commutative ring $R$.  
A totally disconnected topological space is automatically T1: closures of points are connected sets!  So your question reduces to one of general topology: is there a totally disconnected, quasi-compact non-Hausdorff topological space $X$?  The answer is yes and indeed Eric Wofsey's answer to your earlier math.stackexchange.com question provides an example.
In summary: no, not necessarily.
Added: On the other hand, $\operatorname{Spec} R$ -- the set of prime ideals of $R$ endowed with the Zariski topology -- is T1 iff it is Hausdorff iff it is totally disconnected iff $\operatorname{Spec} R = \operatorname{MaxSpec} R$.  See e.g. $\S$13.3 of these notes.  Maybe this is helpful for you.<|endoftext|>
TITLE: Determining geodesics between two points in curved space
QUESTION [5 upvotes]: In order to determine the geodesics between two points, one must solve the geodesic differential equations, which are as following
\begin{align}
p &= u'(s)\\
q &= v'(s)\\
p' + \Gamma^0_{00}p^2 + 2\Gamma^0_{01}pq + \Gamma^0_{11}q^2 &= 0\\ 
q' + \Gamma^1_{00}p^2 + 2\Gamma^1_{01}pq + \Gamma^1_{11}q^2 &= 0
\end{align}
where $\Gamma^i_{jk}$ is the Christoffel symbol of second kind, and the initial conditions are $u(s_0) = u_0,\, \, v(s_0) = v_0,\, \,  u'(s_0) = p_0,\, \, v'(s_0) = q_0$.
In order to solve these equations, one can only provide one point $(u,v)$, and the direction $(u',v')$. How is it possible to determine the geodesics between two points ?
For a sphere, both points will lie on the great circle. So here it is sufficient to provide the initial point and direction. But is it possible to accomplish this by only providing both points, if information such as direction is not known a priori ?

REPLY [6 votes]: I assume that   imranal     asks  how to  find numerically a geodesic connecting two given points if the  connection is given. 
One way to do it  is to implement the solution of the ODE system he wrote in his question numerically (there are many effective ways for it) and then use this implimintation  to find the correct initial velocity vector $(u', v')$  such that the geodesic starting from $(u, v)$ in the direction $(u', v')$ hits the second point.  One can  slighly improve this search but the principle remains the same. 
Alternative method  which actually needs that the connection you have comes from a Riemannian metric 
would be to use the curve shortening flow: start with any curve connecting these two points (say, a straight line)  and then
 start to deform  the curve  such that the velocity vector of the deformation is orthogonal to the velocity vector of the curve and its length is    the  geodesic curvature times some fixed function which vanishes at the first and  last points of the curve. 
This deformation  should  normally converge to a geodesic. 
I expect that there should be  effective implimentation of this  idea as a numerical scheme but I am not an expert in these questions. 
Of course geodesic does not always exist  (as explained  by Igor) and   may be is not unique (as explained by Narasimham) or/and the second procedure may converge to a geodesic which is not a minimal one.<|endoftext|>
TITLE: How does Poincare duality interact with the Serre spectral sequence?
QUESTION [22 upvotes]: Suppose $F^m \to E^{m + n} \to B^n$ is a fiber bundle of closed oriented manifolds. I'm interested in understanding how the Serre spectral sequences for homology and cohomology of $E$ interact with each other via Poincaré duality. Specifically:

Does the Poincaré duality isomorphism for $E$ (as realized by the cap product with the fundamental class) respect the filtrations on $H_*(E)$ and $H^*(E)$ induced by the filtration on $E$ by the sub-bundles over the $p$-skeleta in $B$? Were this to be the case, I would expect that $F_*^p$ (in the homological filtration) would be dual to $F^{m+n-*}_{n-p}$ (in the cohomological filtration). Does this happen?
Amalgamating Poincaré dualities for $B$ and for $F$ will yield isomorphisms between the $E_2$-page and the $E^2$-page: $E_2^{p,q} \cong E^2_{n-p, m-q}$. Is this isomorphism induced from Poincaré duality on $E$ in some way?
Does the above isomorphism on $E_2$-pages commute with the differentials? Is there a Poincaré duality $E_\infty \cong E^\infty$ induced from the Poincaré duality on the $E_2$-page?

In a related question, Tyler Lawson explains how the homology and cohomology versions of the Serre spectral sequence (over a field) are dual to each other under the universal coefficient theorem. The key observation is that dualizing sends the exact couple giving rise to the homology SSS to the exact couple giving rise to the cohomology SSS. I'd like to be able to apply this principle in this setting, except that the pieces of the filtration on $E$ are not in general submanifolds of $E$ so I don't even know how to define a Poincaré duality map between these exact couples.

REPLY [6 votes]: After thinking about this for a little while, it seems like there is a satisfactory way of carrying this out in group cohomology using the Hochschild-Serre spectral sequence of a group extension 
$$1 \to H \to G \to Q \to 1$$ 
(henceforth we will call this the HSSS), and since my original motivating situation happened to be a fibration of aspherical manifolds, this will suffice for my purposes. I will refrain from selecting this as the answer, though, since I would still be curious to know how to carry this out in the topological setting. 
For simplicity, I'll restrict attention to the trivial coefficient module $\mathbb Z$. Suppose that $H$ is a $PD_{m}$-group, $Q$ is a $PD_{n}$-group, and $G$ is a $PD_{m+n}$-group. Chapter VII of Brown's Cohomology of Groups contains a construction of the HSSS. In the homological setting, the idea is to show that there is a particular chain complex $C$ for which the homology with coefficients in the chain complex $H_*(Q, C)$ is isomorphic to $H_*(G)$; the HSSS is then derived from one of the filtrations on the bigraded chain complex underlying the definition of $H_*(Q, C)$.
In fact, letting $F$ be a projective resolution of $\mathbb Z$ over $\mathbb Z G$ (which can be assumed to be of length $m+n$), Brown explains why $C = F_H$ suffices. If we let $F'$ be a projective resolution of $\mathbb Z$ over $\mathbb Z Q$, the upshot is that the tensor product of chain complexes $F' \otimes_Q F_H$ computes $H_*(G)$.
Turning to cohomology, similar arguments show that the homology of the cochain complex
$$
\operatorname{Hom}_Q(F', \operatorname{Hom}_H(F, \mathbb Z))
$$
is given by $H^*(G)$. 
The crucial observation is that for a $PD_k$-group $\Gamma$, if $F$ is a projective resolution of finite length $k$, then $\overline F : = \operatorname{Hom}_\Gamma(F, \mathbb Z \Gamma)$ is also a projective resolution of $\mathbb Z$ over $\mathbb Z \Gamma$. (A technical point which explains how the shift in degrees arises is that one must monkey with the grading on $\overline F$ in the proper way).
We can apply tensor-Hom adjunction to see that
$$
(*)\quad \operatorname{Hom}_Q(F', \operatorname{Hom}_H(F,\mathbb Z)) \cong \overline{F'} \otimes_Q \operatorname{Hom}_H(F,\mathbb Z) \cong \overline{F'} \otimes_Q (\operatorname{Hom}_H(F,\mathbb Z H) \otimes_H \mathbb Z).
$$
As $Q$ is a $PD$-group by assumption, we may use $\overline{F'}$ as our projective resolution in the computation of $H_*(Q,F_H)$. I next claim that after an adjustment to the grading, (a portion of) the chain complex $\operatorname{Hom}_H(F,\mathbb Z)$ is weakly equivalent to $F_H$. This follows from the fact that $H$ is also a $PD$-group. More specifically, the assumption that $H$ is a $PD_m$ group is equivalent to the condition that $H^i(H, \mathbb Z H)$ vanish unless $i = m$, in which case it should be infinite cyclic. This says that the portion 
$$
\operatorname{Hom}_H(F_0, \mathbb Z H) \to \dots \to \operatorname{Hom}_H(F_{m-1}, \mathbb Z H) \to \operatorname{Hom}_H(F_{m}, \mathbb Z H)
$$
of the chain complex $\operatorname{Hom}_H(F, \mathbb Z H)$ forms a projective resolution of $\mathbb Z$ over $\mathbb Z H$, from which the claim follows.
Paying attention to the bigradings, the $(p,q)$-component of the left-hand side in $(*)$ corresponds to the $(n - p, m-q)$-component on the right.
Now the properties I was looking for above will follow automatically from the spectral sequence machinery.<|endoftext|>
TITLE: A conjecture about $\lfloor n!\cdot q/e\rfloor-\,!n\cdot q$
QUESTION [14 upvotes]: I was thinking about this question asked at Math.SE, when I came up with the following conjecture.
For every $q\in\mathbb Q$ consider a sequence $s_n^{(q)}$ (terms within the sequence are indexed by $n\in\mathbb N$):
$$s_n^{(q)}=\left\lfloor\frac{n!\cdot q}e\right\rfloor-\,!n\cdot q$$
($!n$ denotes the subfactorial).
Conjecture: For every $q$ the sequence $s_n^{(q)}$ is periodic. Furthemore, if $q$ is a positive integer is a reciprocal of a positive integer $m$, then the period is:
$$p_{1/m}=\begin{cases}m,\quad m\text{ is even}\\2m,\quad m\text{ is odd}\end{cases}$$
Could you suggest any ideas how to prove this conjecture?

REPLY [7 votes]: Write
$$
\frac{n!\,e^{-1}}m=x_n+\frac{(-1)^{n+m+1}y_n}m+\frac{(-1)^{n+1}\delta_n}m
\qquad\text{and}\qquad
\frac{!n}m=x_n+\frac{(-1)^{n+m+1}y_n}m,
$$
where
$$
x_n=\frac{n!}{m\cdot(n-m)!}\cdot(n-m)!\sum_{k=0}^{n-m}\frac{(-1)^k}{k!}\in\mathbb Z,
\\
y_n=n!\sum_{k=n-m+1}^n\frac{(-1)^{k-(n-m+1)}}{k!}\in\mathbb Z,
$$
and
$$
\delta_n=\frac1{n+1}\biggl(1-\frac1{n+2}+\frac1{(n+2)(n+3)}-\dotsb\biggr)
$$
satisfies $0<\delta_n<1$ (even this rough estimate is sufficient). Note that
$$
y_n=n(n-1)\dotsb(n-m+1)-n(n-1)\dotsb(n-m+2)+\dotsb+(-1)^{m-1}
$$
is $m$-periodic modulo $m$, because each factor in each term of the latter sum is;
in other words, $\{y_n/m\}$ (the fractional part) is periodic with period $m$. It remains to observe that
$$
\biggl\lfloor\frac{n!\,e^{-1}}m\biggr\rfloor-\biggl\lfloor\frac{!n}m\biggr\rfloor
=\biggl\lfloor\frac{(-1)^{n+m+1}(y_n+(-1)^m\delta_n)}m\biggr\rfloor-\frac{(-1)^{n+m+1}y_n}m
$$
and that
$$
\biggl\lfloor\frac{\pm(y_n+(-1)^m\delta_n)}m\biggr\rfloor
=\biggl\lfloor\frac{\pm y_n}m\biggr\rfloor \quad\text{or}\quad \biggl\lfloor\frac{\pm(y_n-1)}m\biggr\rfloor,
$$
respectively, depending on whether $m$ is even or odd. The above argument works for $1/m$ replaced with $\ell/m$,
though to guarantee the estimate for $\delta_n$ one needs $|\ell|\ge n$, so that the periodicity will be eventual;
the period can double if $\ell<0$.
Proving that $m$ or $2m$ is the minimal period is equivalent to showing that the $m$ residues $(-1)^{m-1}y_0,(-1)^{m-1}y_1,\dots,(-1)^{m-1}y_{m-1}$
are not periodic modulo $m$:
$$
(-1)^{m-1}y_0=1, \quad (-1)^{m-1}y_1=1-1, \quad \ldots, \\
(-1)^{m-1}y_k=1-k+k(k-1)-k(k-1)(k-2)+\dots+(-1)^kk!\,, \quad \ldots
$$
This seems to be correct but the distribution of those residues modulo $m$ is quite chaotic.<|endoftext|>
TITLE: An Inequality of KL Divergence
QUESTION [8 upvotes]: Given two probability distributions $P$ and $Q$ defined over a finite set $\mathcal{X}$, one can define the KL divergence between $P$ and $Q$ as 
$$D(P||Q):=\sum_{x\in \mathcal{X}}P(x)\log\frac{P(x)}{Q(x)}.$$
It is known that KL divergence is not a distance (not symmetric and also does not satisfy the triangle inequality).  
Let $\lambda\in (0, \frac{1}{2})$ and let $R_{\lambda}:=\lambda P+(1-\lambda)Q$. 
Two distributions $P$ and $Q$ defined over $\mathcal{X}=\{\pm k,\pm(k-1), \dots, \pm 2, \pm 1\}$, is said to be symmetry of each other if $P(x)=Q(-x)$ for $x\in \mathcal{X}$. 
I want to show that the following inequality always hold for symmetric distributions $P$ and $Q$
$$\frac{D(P||R_{1-\lambda})}{D(P||R_{\lambda})}< \frac{\log(1-\lambda)}{\log(\lambda)}.$$
It is easy to see that $\lambda\mapsto D(P||R_{\lambda})$ is convex and tends to zero as $\lambda\to 1$ and hence it is strictly decreasing on $[0, 1]$ and similarly $\lambda\mapsto D(P||R_{1-\lambda})$ is convex and strictly increasing and therefore $\lambda\mapsto \frac{D(P||R_{1-\lambda})}{D(P||R_{\lambda})}$ is strictly increasing.
The following is the plot of two sides of above inequality when 
$$Q=[0.06, 0.03, 0.05,  0.04, 0.02,  0.01,   0.04,  0.03, 0.02 , 0.04, 0.04, 0.09, 0.03, 0.25, 0.15, 0.1]$$ and 
$$P=0.1,  0.15, 0.25, 0.03, 0.09, 0.04, 0.04, 0.02, 0.03, 0.04, 0.01, 0.02, 0.04, 0.05, 0.03, 0.06]$$ 
Red is the plot of the left-hand side and blue represents the right-hand side.
  
Thanks to Iosif Pinelis, now we know that the above inequality does not hold for arbitrary $P$ and $Q$.  
I will be very thankful if you can let me know of any reference related to this problem.

REPLY [8 votes]: The inequality 
$$\frac{D(P||R_{1-\lambda})}{D(P||R_{\lambda})} < \frac{\log(1-\lambda)}{\log\lambda} \quad\text{for all $\lambda\in(0,1/2)$}$$
is not true in general. E.g., take $P=\frac1{100}\,(57,42,1)$, $Q=\frac1{100}\,(41, 12, 47)$, and $\lambda=\frac1{10}$. Then the left-hand side of this inequality and its right-hand side are $0.0537\dots$ and $0.0457\dots$, respectively. 
However, it seems likely that this inequality holds if it is additionally assumed that $P(x)=Q(-x)$ for all $x$. 

Addendum 
The previous answer was given not assuming the "symmetry" condition.
Given now this condition, write
$$D(P||R_t)=\sum_{|x|=1}^k P(x)\ln\frac{P(x)}{t P(x)+(1-t)P(-x)}$$
$$=\sum_{x=1}^k \Big(P(x)\ln\frac{P(x)}{t P(x)+(1-t)P(-x)}
+P(-x)\ln\frac{P(-x)}{t P(-x)+(1-t)P(x)}\Big)
$$
$$=\sum_{x=1}^k P(x)F(t,r(x))\,\ln\frac1t  
$$
for $t\in(0,1)$, where $r(x):=P(-x)/P(x)$ and 
$$F(t,u):=\frac1{\ln(1/t)}\Big(\ln\frac1{t +(1-t)u}
+u\ln\frac1{t +(1-t)/u}\Big). 
$$
Here we are assuming that $P(x)>0$ for all $x$; otherwise, extend by the continuity. 
So, the problem is reduced to showing that 
$$(*)\qquad \Delta(t,u):=F(t,u)-F(1-t,u)>0$$ 
for $t\in(0,1/2)$ and $u\in(0,1)\cup(1,\infty)$ (clearly, $\Delta(t,1)=0$). By plotting, inequality $(*)$ appears quite likely, and it should be not very hard to prove it by calculus, as the problem now involves only two real variables, $t$ and $u$. 
Here the graphs of $\Delta(t,u)$ and $\text{sgn}\,\Delta(t,u)$ are shown for $t\in(0,1/2)$ and $0<u<5$: 

Addendum 2 
It is indeed not very hard to show that $(*)$ is true. 
Note that $\Delta(t,1/u)=\Delta(t,u)/u$ for $u>0$. So, without loss of generality $u>1$. That is, it suffices to show that 
$$h(u):=\Delta(t,u)\ln(1-t)\ln t
$$
is positive for $t\in(0,1/2)$ and $u>1$. 
Let 
$$h_2(u):=h''(u)\frac{(1 -t + t u)^2 (t + (1- t) u)^2 u}{1 + u}
$$
$$=
 t^2 [(1- t)^2 -(1 - 4 t + 2 t^2) u + (1- t)^2 u^2] \ln t
 -(1- t)^2 (t^2 (u-1)^2 + u) \ln(1 - t) . 
$$
So, $h_2$ is a quadratic polynomial in $u$, with the coefficient of $u^2$ equal to
$-(1- t)^2 t^2 \ln(-1+1/t)<0$ for $t\in(0,1/2)$. 
So, $h_2$ is concave and $h_2(\infty-)=-\infty$. 
Next, $h_2(1)=g(t):=t^2 \ln t-(1- t)^2 \ln(1-t)$, $g(0+)=g(1/2)$ and $g''(t)=-2\ln(-1+1/t)<0$, so that $g$ is concave and hence $h_2(1)>0$ for $t\in(0,1/2)$. Because $h_2$ is concave and $h_2(\infty-)=-\infty$, it follows that $h_2(u)$ and hence $h''(u)$ switch its sign exactly once, from $+$ to $-$, as $u$ increases from $1$ to $\infty$. 
So, for some real $c=c(t)>1$, the function $h$ is convex on $[1,c]$ and concave on $[c,\infty)$. 
At that, $h(1)=h'(1)=0$ and $h(\infty-)=\infty$. So, $h>0$ on $(1,\infty)$. 
That is, indeed $\Delta(t,u)>0$ for $t\in(0,1/2)$ and $u>1$. \qed<|endoftext|>
TITLE: Computing Homotopy Fixed Point Spectral Sequences related to Morava E thories
QUESTION [13 upvotes]: Given a finite subgroup of $G$ sitting inside the Morava stabilizer group $S_n$, we can form the homotopy fixed point spectrum $E_n^{hG}$. There is a spectral sequence with $E_2^{s,t} = H^s(G;\pi_t(E_n)) \Rightarrow \pi_{t-s}(E_n^{hG})$. 
For resolving further differentials in this spcetral sequence I have seen a way by claiming some elements in the $E_2$ page is actually the image of certain elements of $\pi_*(S)$ under Hurewicz map and then by the known multiplicative relation(for example $h_1^3 = h_0^2 h_2$) we shall know that some elements in this spectral sequence must die and try to figure out who kills it or who it kills. 
Question: what is the general way, or is there any general way of proving this kind of claim? i.e. how to prove that something in HFPSS $E_2$ is actually corresponding to the image of an element from $\pi_*(S)$?
I only know some classical results that might be related to this: If we start with $Ext_{BP_*BP}$ and keep track of the element we are interested in inside the chromatic spectral sequence, we will end up with continuous group cohomology of $S_n$(after tensoring with $\mathbb{F}_{p^n}$ at some point) thus cohomology of finite subgroup of it. Then we can look at how elements $t_i\in BP_*BP$ works as functions from $S_n$ to $\mathbb{F}_{p^n}$ to tell the image of $t_i$ in group cohomology and finally detecting elements of $\pi_*(S)$.
The second question comes from a concrete example of this. By classical result, we know that the non-exist elements $\beta_{p^j/p^j}$ and their monomials for $p \geq 5$ can be detected in the way above, in group cohomology of a finite group of order $p$ inside $S_{p-1}$ with coefficient $\mathbb{F}_{p^{p-1}}$. And my second question is the following:
Question: I believe that if I want to rephrase the detection of $\beta_{p^j/p^j}$ in modern language, it will becomes things like "If $\beta_{p^j/p^j}$ survives to $\pi_*(S)$, then it can be detected in the $C_p$ homotopy fixed point of $E_{p-1} $ for $p \geq 5$". So is this true? If so, if I can draw the $E_2$ of this spectral sequence, could I actually identify elements in $H^*(C_p;\mathbb{F}_{p^{p-1}})$ (where the classical detection happens) inside this $E_2$? Or is there a suitably defined map from $E_2$ to $H^*(C_p;\mathbb{F}_{p^{p-1}})$ which maps image of $\beta_{p^j/p^j}$ to non-zero elements?
Any help or hint is really appreciated.

REPLY [7 votes]: In Ravenel's paper on the Arf invariant, he shows (p. 439) that there is a composite of maps
$$
\mathrm{Ext}_{BP_*BP}(BP_*,BP_*) \to H^*_c(\mathbb{S}_n,E_*) \to H^*(C_p,E_*/\frak{m}),
$$
under which the images of $\alpha_1,\beta_1$ and $\beta_{p,p}$ are non-zero. You can use these to determine the Hurewicz image of classes under the composite
$$
\mathrm{Ext}_{BP_*BP}(BP_*,BP_*) \to H^*_c(\mathbb{S}_n,E_*) \to H^*(G,E_*).
$$
Since this is probably the argument you outline in the first paragraph, it is useful to point out another method, via the functoriality of the Greek letter construction; this is the argument used in Proposition 7 of this paper by Goerss, Henn, and Mahowald.  
I'm afraid I don't fully understand the second part of the question; if a class survives in the $BP$-Adams spectral sequence, and is detected in the $C_p$-HFPSS spectral sequence under the map
$$
\mathrm{Ext}_{BP_*BP}(BP_*,BP_*) \to H^*(C_p,E_*),
$$
then it must also survive in the latter spectral sequence.<|endoftext|>
TITLE: What was Hilbert's view of Gödel's Incompleteness Theorems?
QUESTION [65 upvotes]: According to Solomon Feferman, in his slide presentation "Three Problems for Mathematics", Hilbert wrote (in regards to Gödel's second incompleteness theorem):

...the end goal [is] to establish as consistent all our usual methods
  of mathematics.  With respect to this goal, I would like to emphasize
  the following:  the view which temporarily arose and which maintained
  that certain results of Gödel showed that my proof theory can't be
  carried out, has been shown to be erroneous.
In fact that result shows only that one must utilize the finitary standpoint in a sharper way for the further reaching consistency proofs..." 

(Hilbert, Einführung to [Hilbert and Bernays 1934]).
What did Hilbert mean by that statement?
Is there a more complete account from Hilbert regarding that statement and what is it?

REPLY [5 votes]: Here is a relevant reference not mentioned so far, which in a sense (taking the strong statement I cite below at face value) gives an answer to the OP,  and is so new that perhaps this is unknown to the OP. I am posting this since I just stumbled about this thread while reading the chapter that I quote from below. It definitely seems on-topic for this thread. 
In the very recent Sammelband on Gerhard Gentzen's legacy

Gentzen’s Centenary: The Quest for Consistency.
  edited by Reinhard Kahle and Michael Rathjen
  Springer, x+561 pages, Springer, 2015, 
  ISBN: 978-3-3191-0103-3

on pp. 5-6 of the contribution

[Reinhard Kahle: Gentzen’s Consistency Proof in Context. op. cit. p. 3-24]

one reads

"But Hilbert was, by no means, a philosophical hardliner. The only piece of written evidence which we have about Hilbert's reception of Gödel's result [emphasis added] is the cryptic short preface in the first volume of the Grundlagen der Mathematik [52], saying that Gödel's result "shows only that---for more advanced consistency proofs---the finitistic standpoint has to be exploited in a manner that is sharper [...],"11 i.e., the philosophical starting point was to change. Bernays and Ackermann provide us with two additional testimonies that Hilbert soon adapted his "meta-mathematical standpoint." [emphasis added] Based on Bernay's reports, Reid [whose writing about Hilbert seems to be thought of as substantiated by actual historical research; see e.g. the AMS obituary] writes about Hilbert's reaction to Gödel's result [  Constance Reid: Hilbert, p. 198] "At first he was only angry and frustrated, but then he began to try to deal constructively1 with the problem. Bernays found himself impressed that even now, at the very end of his career, Hilbert was able to make great changes in his program." Ackermann writes in a letter to Hilbert (August 23rd, 1933)12: "I was particularly interested in the new meta-mathematical standpoint which you now adopt and which was provoked by Gödel's work." Unfortunately, we have no sources which explicate in detail Hilbert's new standpoint, but it goes without saying that Gentzen's work was in line with it.

I here give the footnote '11' from the citation above: 

"[Hilbert and Bernays, p. VII]. German original: "Jenes Ergebnis zeigt in der Tat auch nur, daß man für die weitergehenden Widerspruchsfreiheitsbeweise den finiten Standpunkt in einer schärferen Weise ausnutzen muß, [...]."

I here give the footnote '12' from the citation above

German original [Kahle here gives a reference to a letter of Ackermann to Hilbert, apparently kept by the Universitätsbibliothek Göttingen]: "Besonders interessiert hat mich der neue meta-mathematische Standpunkt, den Sie jetzt einnehmen und der durch die Gödelsche Arbeit veranlaßt worden ist." The letter was written after Ackermann visited Göttingen, but didn't meet Hilbert and spoke only with Arnold Schmidt, who informed him about "everything" [Kahle does not indicate what sort of quotation marks these are, i.e., whether these are scare quotes meant to indicate that 'everything' is used ironically, or quotation marks indicating that someone is cited here] going on in Göttingen.

So, to summarize
(0) Kahle makes a strong statement 'the only written evidence we have' about what is, in a sense, only a semi-decidable problem (to think, e.g., of Hilbert's boxes Reid found in Göttingen), and this statement of Kahle's essentially says that what the OP quoted is the only historical evidence of how Hilbert reacted. 
(1) The answer to the question "What did Hilbert mean by that statement?" in the OP seems to be: "No one really knows, yet much can be extrapolated from what Hilbert did in the ten years after this statement."
1 Reid's use of 'to deal constructively with the problem' is a little awkward in this context, for obvious reasons. Here, scare-quotes would have been in order.<|endoftext|>
TITLE: How to extend Morley's omitting type theorem to uncountable languages?
QUESTION [7 upvotes]: In his 1965 paper Omitting Classes of Elements (found in The Theory of Models: Proceedings of the 1963 International Symposium at Berkeley, published by North-Holland Publ. Co., Amsterdam (1965)), Morley proved the following omitting types theorem:
For a complete countable theory $T$ and a 1-type $\Sigma$, if for every $\alpha<\omega_1$ there exists a model of $T$ with cardinality $>\beth_\alpha$ which omits $\Sigma$ then there is a model of $T$ omitting $\Sigma$ in every infinite cardinality.
Morley then claims without proof that this theorem still holds if $T$ is of cardinality $\lambda>\aleph_0$ by replacing $\omega_1$ in the statement by $(2^\lambda)^+$, but the proof given in the paper appears to fail for uncountable languages. Specifically, even after replacing $\omega_1$ with $(2^\lambda)^+$, the inductive step of the proof (Theorem 3.1 in the paper) fails at the limit cases.
So is there a way to prove that this statement indeed holds for uncountable languages? Or is the statement actually false when $T$ is uncountable?

REPLY [5 votes]: The most general theorems, covering also uncountable languages and uncountable sets of types can be found in Shelah's Classification Theory in section 5 of Chapter VII.  Theorems VII.5.3 together with Theorem VII.5.5(2) prove not only the result of Morley you are asking about but at the same time extend two-cardinal ("wide apart") theorems named after Vaught and Morley.<|endoftext|>
TITLE: Group acting on two different sets, can isomorphisms be computed efficiently?
QUESTION [5 upvotes]: For the permutation group isomorphism problem, a permutational isomorphism between two permutation groups is sought. A closely related but seemingly easier problem arises, if an isomorphism between the groups is already fixed, and only a compatible (bijective) mapping of the sets where the groups act must be computed. Instead of fixing an isomorphism between the groups, we can also say that we have just a single group acting on two different sets.
This problem seems to be easier (than the permutation group isomorphism problem), because it seems to allow to directly exploit the fact that any group of size $n$ has a generator set of size at most $\log_2 n$. Such a generator set allows a construction similar to a Cayley graph, such that all that remains is to compute an isomorphism of the corresponding graphs. But I haven't found a suitable data structure for representing the automorphisms of such graphs yet, so I'm stuck at that point.
Because the problem doesn't feel too difficult, I wonder whether it has been solved already. An algorithm with run time polynomial in the size of the set and the size of the group (i.e. exponential in the size of a minimal generator set) would qualify as a solution, in the context of this question.

REPLY [2 votes]: I'm still not sure in what form your input is given. I'll assume you have permutations $\lbrace g_1,\ldots,g_k\rbrace$ generating $G$, and permutations $\lbrace h_1,\ldots,h_k\rbrace$ generating $H$, and you know there is an isomorphism $\phi:G\to H$ defined by $\phi(g_j)=h_j$ for each $j$. Now you want to find a conjugacy in $S_n$ that implements that isomorphism, i.e. an element $\ell\in S_n$ such that $g_j^\ell=h_j$ for all $j$, or to prove that there is none.
If this is the question, then the question seems to be identical to this one. You will see that I gave an algorithm there with time requirements $O(n^2k)$.
If your input doesn't consist of permutations, something else is needed.<|endoftext|>
TITLE: Duality between compactness and Hausdorffness
QUESTION [53 upvotes]: Consider a non-empty set $X$ and its complete lattice of topologies
(see also this thread).
The discrete topology is Hausdorff. Every topology that is finer than a Hausdorff topology is also Hausdorff. A minimal Hausdorff topology is such that no strictly coarser topology is Hausdorff.
The trivial topology is compact. Every topology that is coarser than a compact topology is also compact. A maximal compact topology is such that no strictly finer topology is compact.
A compact Hausdorff topology is both minimal Hausdorff and maximal compact.
A minimal Hausdorff topology need not be compact and a maximal compact topology need not be Hausdorff (Steen & Seebach, Examples 99 and 100).
Question:
It seems that there is some duality between Hausdorffness and compactness? Can this kind of duality be stated more explicitly (e.g. in some category theoretic formulation)? Is a compact topology on a fixed set $X$ some "dual" of a Hausdorff topology on $X$?
Here it is stated that a Hausdorff topology need not contain a minimal Hausdorff topology. If there is some duality these notions then we could also automatically deduce that a compact topology must not be contained in a maximal compact topology.

REPLY [15 votes]: Hausdorff is dual to discrete. Compact is dual to overt.
A space $X$ is Hausdorff if and only if the diagonal $\Delta_X = \{(x,x) \mid x \in X\}$ is closed in $X \times X$. A space $X$ is discrete if and only if $\Delta_X$ is open in $X \times X$.
Given a space $X$ let $\mathcal{O}(X)$ be its topology, seen as a topological space equipped with the Scott topology. Let $\Sigma = \{\bot, \top\}$ be the Sierpinski space.
Now, a space $X$ is compact if and only if the map $\forall_X: \mathcal{O}(X) \to \mathcal{O}(1)$, defined by
$$\forall_X (U) = \begin{cases}
\top & \text{if $U = X$} \\
\bot & \text{else}
\end{cases}$$
is continuous. A space is overt if and only if the map $\exists_X : \mathcal{O}(X) \to \mathcal{O}(1)$, defined by
$$\exists_X (U) = \begin{cases}
\top & \text{if $U \neq \emptyset$} \\
\bot & \text{else}
\end{cases}$$
is continuous. See Part II of Martin Escardo's Synthetic topology
of data types and classical spaces for a good tutorial on these ideas, or for a deep dive into the subject Paul Taylor's work on Abstract Stone Duality, and in particular his Foundations for Computable Topology.<|endoftext|>
TITLE: Estimates of a sum involving both the Möbius function and Mertens function
QUESTION [5 upvotes]: I want to ask on the estimates of the sum
$$ \sum_{n=1}^{\infty} \mu(n)M\Big(\frac{x}{n}\Big)=\frac{1}{2\pi i }\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{x^s}{s\zeta(s)^2}ds.$$
It seems that the estimate of the sum be $x\exp(-c\sqrt{\log x})$ for some positive constant $c$, but it is little known to the sign-changes of the Mertens function. Also, the sum can be represented as
$$ \sum_{1 \leq n \leq x}\bigg( \sum_{n=ab}\mu(a)\mu(b) \bigg). $$
It is easy to find the condition that $(the \  inner \  sum)=0$ if every degree of the prime factors of $n$ is bigger than $1$, where $k(n)$ is the square-free kernel of an integer. But also the representation has some difficulties. 
So I thought that it is more good to deal with the integral, then there have to be some informations on the integrations with the integrand $\zeta(s)^{-2}$. I want to ask the estimates of the sum or the informations on the integrals which involve the integrand $\zeta(s)^{-2}$.
Additionally, the exact expression of the complex integral above with the zeros of the Riemann zeta function is given by
$$ 4+2\sum_{|\text{Re } \rho| <1}\frac{x^{\rho}}{\zeta'(\rho)\rho}\phi_{(0,\rho)}+\sum_{n=1}^{\infty}\text{Res}\bigg[\frac{x^s}{s\zeta(s)^2},-2n\bigg],$$
where
$$\phi_{(0,\rho)}=\sum_{k=1}^{\infty}\bigg\{ \frac{\mu(k)}{k^{\rho}}-\frac{1}{\zeta'(\rho)}\log\Big(\frac{k+1}{k}\Big) \bigg\} .$$
(Ref: N. A. Carella, "Simple Zeros Of The Zeta Function", page 3, http://arxiv.org/ftp/arxiv/papers/1306/1306.0458.pdf)

REPLY [9 votes]: One can treat this sum with the Dirichlet hyperbola method:
$$\begin{align} \sum_{ab\leq x}\mu(a)\mu(b) &= \sum_{\substack{a\leq\sqrt x\\b\leq x/a}}\mu(a)\mu(b)+\sum_{\substack{b\leq\sqrt x\\a\leq x/b}}\mu(a)\mu(b)-\sum_{a,b\leq\sqrt x}\mu(a)\mu(b)\\
&= 2\sum_{n\leq\sqrt x}\mu(n)M\left(\frac{x}{n}\right)-M(\sqrt x)^2\\
&\ll\sum_{n\leq\sqrt{x}}M\left(\frac{x}{n}\right)+M(\sqrt x)^2\\
&\ll\sum_{n\leq\sqrt{x}}\frac{x}{n}\exp\left(-c_1\sqrt{\log\frac{x}{n}}\right)+x\exp\left(-2c_1\sqrt{\log\sqrt{x}}\right)\\
&\ll\sum_{n\leq\sqrt{x}}\frac{x}{n}\exp\left(-c_2\sqrt{\log x}\right)+x\exp\left(-c_2\sqrt{\log x}\right)\\
&\ll x\left(\log x\right)\exp\left(-c_2\sqrt{\log x}\right)\\
&\ll x\exp\left(-c_3\sqrt{\log x}\right).
\end{align}$$
Alternately, one can start from your Mellin transform representation and apply the usual technique as in the proof of the prime number theorem (truncation, contour shift, zero free region and lower bounds for $\zeta(s)$) to get the same result. In fact one can go the other way, i.e. upper bounds for your sums can be turned into a zero free region and a lower bound for $\zeta(s)$, so the above bound cannot be improved significantly. Specifically, one can replace $\sqrt{\log x}$ by something close to $(\log x)^{3/5}$, but that is the limit of current knowledge.<|endoftext|>
TITLE: Multiplicative cohomology theories and smash products
QUESTION [15 upvotes]: In his student guide on page 154, Adams gives a construction of products for cohomology using "pairings" of spectra (now known as maps from $E\wedge E\to E$). But then he says

However, G. W. Whitehead did not say (because it is not
  true) that every product in generalised cohomology theory arises
  from a pairing in this sense.

On the other hand, if $E$ is a CW-spectrum (as in Adams' Chicago lecture notes), one can turn it into an $\omega$-spectrum in the category of CW-spectra using Milnor's theorem. Then each product in the corresponding generalized cohomology theory tautologically defines maps $E_k\wedge E_\ell\to E_{k+\ell}$.
Thus my questions: What examples have G. W. Whitehead and J. F. Adams been thinking of? In which situations would the naïve Argument above fail?

REPLY [11 votes]: I do not claim that I know what Adams meant; for all I know his definition of "product" may include products which are not bilinear. Take the following with a grain of salt.
What I suspect is the following. Adams says, just before your quote:
"In order to construct such products, G.W. Whitehead introduced the notion of a 'pairing of spectra'. A pairing of spectra from $E$ and $F$ to $G$, in the sense of G.W. Whitehead, consists of maps
$$
\mu_{n,m}: E_n \wedge F_m \to G_{n+m}
$$
which satisfy suitable axioms..." (Said suitable axioms appearing in detail in Whitehead's paper "Generalized homology theories".)
Adams then describes the pairing on complex cobordism as arising from maps $MU_n \wedge MU_m \to MU_{n+m}$.
I suspect that what Adams meant is that, given spectra $E$, $F$, and $G$, there may be pairings from $E$ and $F$ to $G$ which do not arise from maps of the above specific type: for example, if you skipped the step of replacing $G$ with an $\Omega$-spectrum, or don't adopt the "cells now, maps later" methodology, some needed targets for maps from $E_n \wedge F_m$ may be missing. For example, the stable Hopf map $\eta \in \pi_2^{stable}(S^1)$ cannot be represented by a pairing from $\Sigma^\infty S^1$ and $\Sigma^\infty S^1$ to $\Sigma^\infty S^1$.<|endoftext|>
TITLE: geodesic of $\rm SO(3)$ as a compact Lie group vs as a Riemannian symmetric space
QUESTION [6 upvotes]: I got a little bit confused about the definition of geodesic for $\rm SO(3)$ as

a compact Lie group
a Riemannian symmetric space

In the former case, it is given by the usual matrix exponential:
$$
\exp_{g}tX=ge^{tX}\quad g\in{\rm SO(3)}, X\in\mathfrak{so}(3)
$$
In the latter case, the geodesic is given by the transvections $\tau(\exp t(X',-X'))$:
$$
{\rm Exp}_{g}~t(\frac{1}{2}X'g+\frac{1}{2}gX')=\tau(\exp\frac{t}{2}(X',-X'),g)=e^{tX'/2}ge^{tX'/2}
$$
where $\tau$ is given by:
$$
\begin{aligned}
\tau:{(\rm SO(3)\times SO(3))}\times{\rm SO(3)}&\to{\rm SO(3)}\\
((g,h),x)&\mapsto gxh^{-1}
\end{aligned}
$$
The two only coincide at the identity. Now my question is, we have two types of geodesics, all defined by the same bi-invariant metric (though in difference sense). Is there anything wrong with my calculation?
If the two types of geodesics are indeed different, which one will be shorter in length?

REPLY [8 votes]: I think the second formula is wrong. The map $\tau$ should be given as $$\tau\colon (SO(3)\times SO(3))/SO(3)\to SO(3)\;;\quad [(g,h)]\mapsto gh^{-1}\;.$$
A geodesic in this picture is then given by
$$\tau(ge^{tX/2},he^{-tX/2})=ge^{tX}h^{-1}=gh^{-1}e^{t\mathrm{Ad}_hX}\;.$$
This also proves the equivalence of both constructions.<|endoftext|>
TITLE: Explicit $2$-descent on elliptic curves
QUESTION [8 upvotes]: Let $k$ be a field of characteristic $0$ and let 
$$E: y^2 = f(x)$$
be an elliptic curve over $k$, with $\mathrm{deg}(f) = 3$. Kummer theory yields a map
$$\varphi:\mathrm{H}^1(k, E[2]) \to \mathrm{H}^1(k, E)[2].$$ In particular, to each element $c \in \mathrm{H}^1(k, E[2])$ we may associate a principal homogeneous space for $E$.


Can one make this map explicit if $f$ is irreducible? Namely, write down equations for the associated principal homogeneous space determined by $\varphi(c)$?


Under the assumption that $f$ is completely split (so that $E$ has full $2$-torsion), this is done in many books. Here we have $\mathrm{H}^1(k, E[2]) \cong k^*/k^{*2} \times k^*/k^{*2}$, and given such a pair one can write down explicit equations for the associated homogeneous space.
If $f$ is irreducible, then I believe that $E[2] \cong \mathrm{R}_{K/k}(\mu_2)/ \mu_2$ as group schemes, where $K/k$ is the cubic extension associated to $f$. Hence we have $\mathrm{H}^1(k, E[2]) \cong K^*/k^*K^{*2}$. Given an element $c \in K^*$, I would like to be able to write down explicit equations of the associated principal homogeneous space. I'm guessing that one would need to write down these equations in the Weil restriction of an affine space.

REPLY [13 votes]: It is perhaps easier to use the isomorphism of $E[2]$ with the kernel of the 
`norm' map $R_{K/k} \mu_2 \to \mu_2$, which translates into
$H^1(k, E[2]) \cong \ker(N \colon K^\times/K^{\times 2} \to k^\times/k^{\times 2})$. In this case you start with $c \in K^\times$
whose norm is a square: $N(c) = a^2$. The descent map $E(k) \to H^1(k, E[2])$
is given in these terms by $(x,y) \mapsto x-\theta$ (mod squares), where
$\theta$ is the image of $x$ in $K = k[x]/\langle f\rangle$. Now, given $c$,
we want to sort of reverse this and parametrise the preimages of $c$.
This translates into
$$ y^2 = f(x), \quad x - \theta = c (z_0 + z_1 \theta + z_2 \theta^2)^2. $$
The right hand side (using $c = c_0 + c_1 \theta + c_2 \theta^2$)
can be expanded as
$$ Q_0(z_0,z_1,z_2) + Q_1(z_0,z_1,z_2) \theta + Q_2(z_0,z_1,z_2) \theta^2 $$
with quadratic forms $Q_j$ over $k$. Comparing coefficients, we obtain
$$ Q_2(z_0,z_1,z_2) = 0, \quad Q_1(z_0,z_1,z_2) = -1 $$
and $x = Q_0(z_0,z_1,z_2)$ can be eliminated; $y$ is given (up to a sign
that one has to fix) by $y^2 = N(c) N(z_0 + z_1 \theta + z_2 \theta^2)^2$,
so (say) $y = a N(z_0 + z_1 \theta + z_2 \theta^2)$.
Homogenizing, we obtain an intersection of two quadrics in $\mathbb P^3$:
$$ C \colon \quad Q_2(z_0,z_1,z_2) = 0, \quad Q_1(z_0,z_1,z_2) + z_3^2 = 0 ,$$
which is a model of the principal homogeneous space you are looking for.
(You also get the 2-covering map to $E$ via $x$ and $y$ as above.)
Note that when $k$ is a number field and $C$ has points over all completions
of $k$, then the conic given by the first equation has a $k$-point and
can be parametrised by binary quadratic forms. Plugging this into the
second equation gives you the $y^2 = \text{quartic}$ model.<|endoftext|>
TITLE: Status of Zeeman's collapsability Conjecture
QUESTION [10 upvotes]: Zeeman's conjecture  in  topological  combinatorics  states  that if $K$ is a contractible polyhedron of dimension 2, then $K\times I$ has a collapsible subdivision.
What  is  the  status  of  this  conjecture  as  of  2015?

REPLY [15 votes]: These two very recent surveys state the conjecture as open:

Robert Kropholler, Zeeman’s collapsibility conjecture (2013)
A. P. M. Kupers, Zeeman's conjecture (2014)

It doesn't seem like there's been any progress since. In particular, the most important results towards the conjecture are:
(Cohen, 1975) If $K$ is a contractible $2$-polyhedron, then $K\times I^6$ is collapsible.  Similarly, if $K$ is a contractible $n$-polyhedron for $n \geq 3$, then $K \times I^{2n}$ is collapsible.
(Perelman, 2003, corollary to the Poincaré conjecture) Zeeman's conjecture is true for standard $2$-polyhedra that are spines of $3$-manifolds<|endoftext|>
TITLE: Unipotent radical of minimal parabolic subgroup of a unitary group over an arbitrary field
QUESTION [5 upvotes]: I am looking for an explicit description of the unipotent radical of a minimal parabolic subgroup of a unitary group, i.e. the group of isometries of a hermitian form, over an arbitrary field.
In his notes "Linear algebraic groups" (Boulder, 1966), Section 6.6 "Examples", Borel gives such a description for an orthogonal group $SO(F)$ for a non-degenerate quadratic form $F$, and subsequently writes

When one starts with a hermitian form, the same considerations apply, except that one gets a root system of type $\mathbf{BC}_q$.

Working out the details seems rather tricky to me, in particular because a hermitian form is defined over a skew field with involution and not just over a commutative field.
Has this been worked out somewhere in the literature?

REPLY [2 votes]: I deal with the simpler  case: $G=U(h)$ where $E/F$ is a separable quadratic extension of fields and $h: E^n\times E^n\rightarrow E$ is Hermitian with respect to $E/F$. Suppose $W$ is a maximal isotropic subspace of $E^n$ with respect to the Hermitian form $h$. WE have the partial flag (where $W^{\perp}$ is the orthogonal complement of $W$ w.r.t. the Hermitian form $h$)
$$0\subset W \subset W^{\perp} \subset E^n.$$ The unipotent radical of the minimal parabolic is precisely the subgroup of $G=U(h)$ which preserves this flag and acts trivially on successive quotients. 
When we have a Hermitain form over a skew field, a similar description obtains, but sometimes, you may get that the unitary group becomes a symplectic or orthogonal group (these details are in Tits' article on classification of algebraic groups in the same Boulder conference volume)<|endoftext|>
TITLE: Elementary chains in forcing extensions of $M_1$
QUESTION [6 upvotes]: Let $M_1$ be the canonical inner model with one Woodin cardinal $\delta$. Now suppose that $\mathbb{P}$ is a forcing notion of size $< \delta$, which preserves $\omega_1$  and that $G$ is a generic filter for the forcing. Note that, assuming $M_1^{\sharp}$ exists, there exists a formula which defines $M_1$ in the generic extension properly. Suppose that in $M_1[G]$ the $(M_1)$-initial segment $(M_1)_{\alpha}$ has size $\aleph_1$. 
My question is whether it is possible, for unboundedly many $\alpha < \aleph_2^{M_1[G]}$ to find an elementary continuous chain $(N_i \, : \, i \in \omega_1)$ of elementary submodels of $(M_1)_{\alpha}$ such that $\bigcup_{i < \omega_1} N_i = (M_1)_{\alpha}$, and such that for every model $N_i$ in the chain, its transitive collapse is an initial segment of $M_1$ again?
Note that the statement is true if $\mathbb{P}$ is just the trivial forcing, however I would be interested in the case when we collapse cardinals down to $\omega_1$.

REPLY [3 votes]: i think the answer is yes.
let $M=M_1$. let $\lambda=\omega_2^M$ and say $G$ collapses $\omega_2$ to $\omega_1$. let $f\colon\omega_1\to\lambda$ be an increasing cofinal function. let construct a continuous chain $(M_i: i<\omega_1)$ of submodels of $M|\lambda$ such that $f(i)$ is in $M_i$. 
let $N_i$ be the transitive collapse of $M_i$. then $N_i$ is an initial segment of $M$, this follows from comparison. 
let then $M_{\omega_1}$ be the direct limit of $N_i$. there is a map $\pi\colon M_{\omega_1}\to M_i|\lambda$. this map has the range of $f$ in its range and its critical point is bigger than $\omega_1$, so it is identity. 
you can do this with any $\gamma<\omega_3^M$. fix $f\colon\omega_1\to\lambda$ and $g\colon\omega_1\to\gamma$ that are increasing and cofinal. then construct a chain that exhaust $f$ and $g$. similar argument would show that $M_{\omega_1}$ is $M|\gamma$.<|endoftext|>
TITLE: Cyclic groups acting on balls, and interior fixed points
QUESTION [7 upvotes]: Let a finite cyclic group $G = \mathbb Z/n$ act continuously on an open $d$-ball $B^d$. Suppose further that this action extends to the closed ball $\overline{B^d}$. Is there necessarily a fixed point in the interior?
Note that by Brouwer's fixed point theorem, there has to be some fixed point in the closed ball.
It is also true that for $n$ a prime power, every action of $\mathbb Z/n$ on an open ball has a fixed point, this can be seen from Smith theory. 
On the other hand, for $n$ not a prime power, there are actions 
of $\mathbb Z/n$ on open balls without fixed points, see e.g. Theorem 8.3 in Bredon: Introduction to Compact Transformation Groups. But I am not aware of examples of such actions that extend to the closed ball.

REPLY [5 votes]: There need not be a fixed point in the interior. As in Bredon's book, p. 61, there exist smooth actions of a cyclic group $C_{r}$, of order $r$, without fixed points on $\mathbb{R}^{n}$ for large enough $n$ (e.g. $n\ge 8$) whenever $r$ is not a prime power. By one-point compactification one obtains a continuous, one-fixed-point action on the $n$-sphere $S^{n}$. Then by coning from the fixed point one obtains a continuous action on the closed ball $D^{n+1}$, smooth on the interior, but with no interior fixed points. One can push $n$ down at least to $7$ (and down to $6$ if $r$ involves at least $3$ primes), but not down to $4$, in this argument. See 
Richard Haynes, Slawomir Kwasik, Jerrel Mast, and Reinhard Schultz, Periodic maps on $\mathbb{R}^7$ without fixed points, Math. Proc. Cambridge Philos. Soc. 132 (2002), no. 1, 131--136.<|endoftext|>
TITLE: Convergence of an oscillatory integral
QUESTION [8 upvotes]: Given $f\in H^1(\mathbb{R}^3)$ and $t>0$, consider the following integral:
$$I_f(t):=\int_{\mathbb{R}^3}\int_0^{+\infty}e^{-s+i\frac{(s+|x|)^2}{t}}f(x)dsdx$$
I need to show that $I_f(t)$ is finite, and to find how fast it grows as $t\rightarrow +\infty$.
If $f$ decays fast enough at infinity so that $f\in L^1(\mathbb{R}^3)$, we easily get:
$$|I(f)|\lesssim\int_{\mathbb{R}^3}|f(x)|dx<+\infty$$
so in this case $I_f(t)$ is finite, uniformily in $t$.
If $f$ doesn't dacays fast enough, we can't simply pass to absolute value. We must take into account the oscillatory term $e^{i(s+|x|)^2/t}$, which should produce some polynomial growth in $t$. How to proceed in this case?
Thank you for any suggestions.

REPLY [2 votes]: Here is a more detailed version of my sketch in the comments above.
First of all, let $\rho$ be a compactly supported smooth function (cutoff function) such that $\rho=1$ on $B_1$. Write $f=\rho f+(1-\rho)f$. Since $\rho f$ is compactly supported, it is $L^1$, and your old argument applies to it, so it remains to bound that for $(1-\rho)f$, whose support is disjoint from $B_1$, which we simply denote as $f$ henceforth.
Then we do the $s$ integral first. Integration by parts once gives
$$ \int_0^\infty e^{-s}e^{i(s+|x|)^2/t}ds=\frac{te^{i|x|^2/t}}{2i|x|}-\int_0^\infty t\frac{d}{ds}(\frac{e^{-s}}{2i(s+|x|)})e^{i(s+|x|)^2/t}ds. $$
Integrate by parts a second time gives
$$ \int_0^\infty e^{-s}e^{i(s+|x|)^2/t}ds
=\frac{te^{i|x|^2/t}}{2i|x|}+\frac{t}{2i|x|}(\frac{t}{2i|x|}+\frac{t}{2i|x|^2})e^{i|x|^2/t}
-\int_0^\infty t^2\frac{d}{ds}(\frac{1}{2i(s+|x|)}\frac{d}{ds}(\frac{e^{-s}}{2i(s+|x|)}))e^{i(s+|x|)^2/t}ds. $$
Write $I(t)=I_1(t)+I_2(t)+I_3(t)$, corresponding to the three terms above.
We need oscillation in $x$ to bound $I_1$; we don't need it for $I_2$ and $I_3$.
$$I_1(t)=\frac{t}{2i}\int \frac{e^{i|x|^2/t}}{|x|}f(x)dx=-\frac{t}{2i}\int e^{i|x|^2/t}\vec x\cdot\nabla(\frac{f(x)}{|x|(3+2i|x|^2/t)})$$
is bounded by a constant (depending on $t$) times
$$\int \frac{|\nabla f(x)|}{|x|^2}+\frac{|f(x)|}{|x|^3}\le(\int |\nabla f|^2)^{1/2}(\int_{|x|\ge1} |x|^{-4})^{1/2}+(\int |f|^2)^{1/2}(\int_{|x|\ge1} |x|^{-6})^{1/2}\le C.$$
For $I_2$ we have
$$ |I_2(t)|\le t^2\int |x|^{-2}f(x)\le t^2(\int |f|^2)^{1/2}(\int_{|x|\ge1} |x|^{-4})^{1/2}.$$
For $I_3$ we note that both $e^{-s}$ and $1/(s+|x|)$ are decreasing and convex in $s$, so is their product. Hence $\frac{d}{ds}\frac{e^{-s}}{s+|x|}\le0$ and is increasing. Since $1/(s+|x|)>0$ is decreasing, their product is negative increasing. Hence
$$ \int_0^\infty |\frac{d}{ds}(\frac{1}{2i(s+|x|)}\frac{d}{ds}(\frac{e^{-s}}{2i(s+|x|)}))|ds=\frac{t}{2|x|}(\frac{t}{2|x|}+\frac{t}{2|x|^2}) $$
and the bound for $I_3$ follows the same way as $I_2$.
P.S. Going through the above estimates with an explicit dependence on $t$ shows that the bound growth like $t^2$, though I think the actually growth may well be slower.
P.P.S. A place to study oscillatory integrals systematically is Chapter 6 of Stein, Harmonic Analysis.<|endoftext|>
TITLE: Whitney sum formula for Pontryagin classes I
QUESTION [10 upvotes]: I have read in several places that the total Pontryagin classes of real vector bundles satisfy a Whitney sum formula $p(E\oplus F) = p(E)\cdot p(F)$ modulo 2-torsion. I would like to understand this better, and have two precise questions.

Using the definition $p_i(E)=(-1)^{i}c_{2i}(E_\mathbb{C})$, the fact that complexification respects direct sums, and the Whitney sum formula for Chern classes, I find for example that
$$
p_1(E\oplus F) = -c_2(E_\mathbb{C}\oplus F_\mathbb{C}) = -c_2(E_\mathbb{C}) - c_1(E_\mathbb{C})c_1(F_\mathbb{C}) - c_2(F_\mathbb{C}),
$$
and
$$p_1(E) + p_1(F)=-c_2(E_\mathbb{C})-c_2(F_\mathbb{C}).$$
I do not see why the difference $c_1(E_\mathbb{C})c_1(F_\mathbb{C})$ between these two expressions is necessarily 2-torsion. What am I missing?
edit: I've moved the second question to a separate post Whitney sum formula for Pontryagin classes II

REPLY [15 votes]: I do not see why the difference $c_1(E_\mathbb{C})c_1(F_\mathbb{C})$ between these two expressions is necessarily 2-torsion. What am I missing?

Real line bundles are classified by $H^1(X, \mathbb{Z}_2)$, which is a 2-torsion group (geometrically, because real line bundles are self-dual, their tensor squares are trivializable). Complex line bundles are classified by $H^2(X, \mathbb{Z})$. Complexification therefore describes a cohomology operation $H^1(X, \mathbb{Z}_2) \to H^2(X, \mathbb{Z})$, so the image of this cohomology operation (the first Chern classes of complexifications of real line bundles) is necessarily also 2-torsion. In fact this cohomology operation is a Bockstein.<|endoftext|>
TITLE: A result of Sierpiński on non-atomic measures
QUESTION [10 upvotes]: There is a classical result commonly attributed to W. Sierpiński that reads as follows:

Theorem 1. If $f: \Sigma \to \bf R$ is a non-atomic (*) measure on a set $S$, then for every $X \in \Sigma$ and $a \in [0, f(X)]$ there exists $A \in \Sigma$ such that $A \subseteq X$ and $f(A) = a$. 

In an attempt of finding out where this was first proved, I bumped into the Wiki article on atomic measures (here), which sketches a proof of a stronger result and provides a reference to [3].
However, the reference doesn't seem to be correct, as the paper above proves, to the best of my understanding, that: 

Theorem 2. If $E_0$ is a bounded subset of ${\bf R}^m$ and $f$ a finitely additive function $\mathcal P(E_0) \to \bf R$ that is continue (**), then for all $E_1, E_2 \subseteq E_0$ and every $a \in [0,1]$ there exists a (Lebesgue-)measurable set $A \subseteq E_0$ such that $f(A) = a f(E_1) + (1-a) f(E_2)$. 

(Caveat lector: I'm essentially sticking to Sierpiński's notation and terminology, and hoping to interpret them correctly.) So my question is:

Q. Assuming that I'm not missing something, do you know of a correct reference to Sierpiński's work where Theorem 1 was first proved?

Update (Nov 04, 2015). It follows from [2, Theorem 1] that the range of a countably additive non-atomic vector measure $\mu:\Sigma\to{\bf R}^n$ (viz., an $n$-tuple of countably additive measures $\Sigma\to\bf R$ each of which is non-atomic) is convex. This, yet, doesn't seem to have a clear relation with Theorem 1 above, as it implies that for all $X\in\Sigma$ and $a\in [0,1]$ there exists $A\in\Sigma$ s.t. $\mu(A)=a\mu(X)$, while for Theorem 1 we should also have $A\subseteq X$. Lyapunoff's theorem appears as (part of) Theorem 13 in [1, Chapter 2].
In particular, it is noted on p. 39 of Fryszkowski's book that "There is a long story of results concerning the range of vector measures. The first result of this type belongs to Sierpiński [...]," and here a reference is given to the 1922 paper I cited before, "who showed that the range of a real non-atomic measure is a compact interval." This, however, is not the content of Theorem 2 above, is it?! So I feel all the more confused by the entire story, and continue missing the link with Theorem 1 above. 
Notes. 
(*) It means that if $X \in \Sigma$ and $|f(X)| > 0$ then there exists $Y \in \Sigma$ such that $Y \subseteq X$ and $0 < |f(Y)| < |f(X)|$. 
(**) It means that $f(E) \to 0$ as $E \subseteq E_0$ and the diameter of $E$ tends to $0$.
Bibliography.
[1] A. Fryszkowski, Fixed Point Theory for Decomposable Sets, Topological Fixed Point Theory and Its Applications 2, Dordrecht: Kluwer Academic Publishers, 2004.
[2] A. A. Liapounoff, Sur les fonctions-vecteurs complètement additives, Izv. Akad. Nauk SSSR Ser. Mat. 4 (1940), No. 6, 465-478.
[3] W. Sierpiński, Sur les fonctions d'ensemble additives et continues,  Fund. Math. 3 (1922), No. 1, 240–246 (in French).

REPLY [6 votes]: I don't yet have a reference, but it seems the result might have been first proved by Fichtenholz and Sierpiński, independently from each other. This should be mentioned in a remark to Problem 12 in:

R. Sikorski, Real Functions, Vol. 1, PWN: Warsaw, 1958 (in Polish),

at least according to the historical remark on p. 28 in:

P. Lorenc and R. Wituła, Darboux property of the nonatomic $\sigma$-additive positive and finite dimensional vector measures,  Matematyka Stosowana 3 (2013), 25-36.

Unfortunately, I couldn't retrieve either a hard copy of Sikorski's book or simply a scan of the relevant page(s).
Update (Nov 27, 2015). Thanks to Martin Sleziak and Jacek Jendrej, I found out that the "remark to Problem 12" referred to by Lorenc and Wituła is actually a footnote on p. 225 of Sikorski's book. 
Now, Problem 12 reads, "If $\mu$ is a non-atomic measure and $0 < s < \mu(A) < \infty$, then there exists $B \subseteq A$ such that $\mu(B) = s$", so we are really talking of Theorem 1 in the OP. And the footnote on p. 225  really makes reference to "Sierpiński [7] and Fichtenholz [${}$1]". But the copy of Sikorski's book I could retrieve is incomplete and doesn't include the bibliography... Could anyone having access to a complete copy fill in this answer?
Update (Dec 08, 2015). Here is some more evidence that the common (?) attribution of Theorem 1 in the OP to W. Sierpiński may be wrong. 
With the invaluable help of Nathalie Granottier and Fabienne Grosjean (Bibliothèque du CIRM, Institut de Mathématiques de Luminy) and Cyril Mauvillain (Bibliothèque de Recherche Mathématiques et Informatique, Université Bordeaux 1), I've finally got a scan copy of the bibliography of Sikorski's book. 
The final outcome is that "Sierpiński [7]" (from the footnote on p. 225 of Sikorski's book) is nothing but the paper of Sierpiński that was already cited in the OP, namely

W. Sierpiński, Sur les fonctions d'ensemble additives et continues, Fund. Math. 3 (1922), No. 1, 240–246 (in French),

while "Fichtenholz [${}$1]" is

G. Fichtenholz, Sur les fonctions d'ensemble additives et continues, Fund. Math. 7 (1925), No. 1, 296–301 (in French).

Moreover, neither of these two papers deals with Theorem 1 in the OP, and this is not in contradiction to Sirkoski's statement (though it is in contradiction to statements made in a bunch of places, where the result is attributed to Sierpiński and a reference is made to the above paper of his), as the text in the footnote on p. 225 of Sikorski's book says "por. także", which sounds more like a "see also", which I interpret as a "cf." (and makes perfect sense), than like a "see", which I would rather interpret as "the result appears in". 
So, my conclusion is that, if anyone, then it's Sikorski who should deserve credit for the result, which in fact applies to countably additive measures (as of the statement in the OP), whereas measures considered in Sierpiński's and Fichtenholz' papers are finitely additive. 
And to confirm that Sikorski is really talking of countably additive measures, let me cite from the incipit of Section 3, Chapter VI of his book:

§3. Miara. Zgodnie z uwagami wstępnymi z §1 miarą nazywamy każdą nieujemną, przeliczalnie addytywną funkcję zbioru $\mu$, określoną na pewnym przeliczalnie addytywnym ciele $\mathfrak M$ podzbiorów przestrzeni $X$, tzn. nieujemna funkcję rzeczywistą na $\mathfrak M$, nierówną tożsamościowy $\infty$, taką, że dla każdego ciągu nieskończonego zbiorów rozłącznych $A_n \in \mathfrak M$
  $$\mu(A_1 + A_2 + \ldots) = \mu(A_1) + \mu(A_2) + \ldots$$

More or less, this should translate into:

§3. Measures. Based on §1, we let a measure be any nonnegative, countably additive set function $\mu$, defined on a sigma-algebra $ \mathfrak M $ of subsets of $ X $, i.e. a non-negative real function on $\mathfrak M $, possibly taking also the value $ \infty $, such that for every countable sequence of disjoint sets $ A_n \in \mathfrak M $ we have
  $$ \mu (A_1 + A_2 + \ldots) = \mu (A_1) + \mu (A_2) + \ldots $$<|endoftext|>
TITLE: Factorization of conformal maps between annuli
QUESTION [6 upvotes]: Consider two doubly-connected open subsets $A$ and $A'$ of the Riemann sphere. We assume these two domains to be of same modulus (the moduli space being one real parameter), i.e. we assume that there exists a holomorphic bijection $\phi:A\rightarrow A'$. Note that the map $\phi$ is then unique up to precomposition by the automorphisms of the annulus $A$, loosely speaking the choice of whether we switch the two boundary components, plus a rotation.
Question
Is it possible to factor $\phi$ as a composition of holomorphic maps $\phi=f_1\circ g_1 \circ \cdots \circ f_n \circ g_n$, where the maps $f_i$ and $g_i$ are defined on simply-connected domains?
If yes, can we find such a factorization with $n=1$?
Motivation: case $\phi=f_1$.
To fix ideas, suppose the annuli $A$ and $A'$ separate $0$ and $\infty$, and that the map $\phi$ sends outer boundary to outer boundary. Assume that the map $\phi$ can be extended to the inside of $A$, i.e. $\phi$ is the restriction to $A$ of a holomorphic function defined on a simply-connected domain containing $0$ and bounded by the outer boundary of $A$. I claim that this gives no information on the outer boundary of the domain $A'$. However, once we fix the outer boundary of $A'$, its inner boundary has to live in a (small) family of three real parameters. Inded, fix the outer boundary of $A'$ to what you wish, and see by Riemann uniformization that the possible maps $\phi$ form a three real parameters family.
Thinking in terms of degrees of freedom, it seems it could be possible, for any couple of annular domains $A$ and $A'$, to write $\phi$ as a composition $f\circ g$, where $f$ extends to the inside, and $g$ extends to the outside: either one of these maps $f$ and $g$ allows complete freedom on either the exterior or the interior boundary of its image.
Note: uniqueness if $n=1$.
In the $n=1$ case, the factorization, if it exists is unique up to Moebius transformation.
Indeed, suppose $\phi = f\circ g = \tilde f \circ \tilde  g$.
Let us consider the holomorphic function $\tilde f^{-1} \circ f = \tilde  g \circ g^{-1}$, a priori a holomorphic bijection $g(A)\rightarrow \tilde g(A)$. However, $\tilde f^{-1} \circ f$ extends inside $g(A)$ i.e. it is actually defined on a simply-connected domain bounded by the outer boundary of $g(A)$. Whereas $\tilde  g \circ g^{-1}$ extends outside of $g(A)$. The function $\tilde f^{-1} \circ f$ can thus be extended to the whole sphere, and hence is a Moebius transformation.

REPLY [7 votes]: There is such a factorization, with $n=1$.
Let us suppose first that the annuli are non-degenerate, and the boundaries are sufficiently smooth, and $\phi$ sends the outer boundary of $A$ to the outer boundary of $A'$.
Then $\phi$ has a quasiconformal extenson, say $\Phi$ which is a homeomorphism between the spheres. Let $\mu=\Phi_{\overline{z}}/\Phi_z$ be the Beltrami coefficient of $\Phi$. It naturally splits into two parts $\mu^+$ and $\mu^-$ supported on the outside and on the inside. Now consider a homeomorphic solution $\Phi^+$ of the Beltrami equation $\Phi^+_{\overline{z}}=\mu^+\Phi_z$. It is a homeomorphism of the sphere, conformal inside the outer component of $A$. It maps $A$ onto some $A_1$
conformally. Using this map, we push forward $\mu^-$ to the inside component of 
$A_1$ and obtain a new Beltrami coefficient $\nu$ on the inside component of $A_1$. Then solve the Beltrami equation ${\Phi^-}_{\overline{z}}=\nu{\Phi^-}_z$.
The composition $\psi=\Phi^-\circ\Phi^+$ is a quasiconformal map with the same Beltrami coefficient as $\phi$. Therefore $\psi=L\circ\phi$ where $L$ is fractional-linear, and we obtain a required factorization: $\Phi^+$ is conformal
on the complement of the outside component of $A$ and $\Phi^-$ is conformal on
the complement of the inside component of $A_1$.
To do the general case, consider an exhaustion of $A$ by annuli with anaytic boundaries, say $A=\lim A_n$.Then $A_n'=\phi(A)_n$ also have analytic boundaries,
and the previous argument applies. Now, it is not difficult to prove, using a normal family argument that the limit of these factorizations will give the required factorization of $\phi$.
The degenerate cases (when one boundary component is a point) are easy, by the removable singularity theorem.
Reference for this techniques: Ahlfors, lectures on quasiconformal mappings.
(It also has a complete proof of the existence of a homeomorphic solution of Beltrami equation, though we did not use the full strength of this theorem: in our setting all Beltrami coefficients can be chosen smooth in which case the required existence theorem is just the uniformization theorem for the special case of the sphere.)<|endoftext|>
TITLE: Labeling edges of an icosahedron with sum constraints
QUESTION [9 upvotes]: The question is inspired by this previous MO question. There it was shown that it's possible to label the edges of a cube by the numbers $\{1,2,\ldots,6,8,9, \ldots, 13\}$ in such a way that:

Three edges meeting at a vertex sum to 21.
Four edges constituting a face sum to 28.

The solution is unique up to isometry (numbers in red correspond to invisible edges) :

Similary, is it possible the label the edges of a regular icosahedron (or dodecahedron) with distinct integers in such a way that the two conditions above are satisfied (with different sums of course). If yes, is the solution unique up to isometry?
Note that for the regular tetrahedron, the answer is no: a computation shows that two opposite edges must be equal.
Moreover, what would be natural generalizations of this problem for polytopes in $n$-dimensional space?

REPLY [9 votes]: Alas there's no such labeling by
$\{1, 2, 3, \ldots, 15, 17, 18, 19, \ldots, 31\}$,
assuming I made no computational error somewhere along the way.
The vertex and face conditions give $32$ linear equations on the $30$ edge labels,
but with enough redundancy that there is a $6$-dimensional space of solutions.
The obvious solution is to label all the edges $1$.  A less trivial solution
is obtained by fixing two opposite vertices $v,v'$ of the icosahedron, and
using the edges that connect a neighbor of $v$ to a neighbor of $v'$: 
there are $10$ such, forming a cycle, and we can label them alternately
$+1$ and $-1$, and label the remaining $20$ edges zero.  This gives $6$
generators (in addition to the all-$1$ vector), satisfying one relation 
(they sum to zero if the signs are chosen consistently), for a total
dimension of $1 + (6-1) = 6$.  These turn out to generate the space of
solutions even over the integers; that is, every integer solution of
the linear equations is an integer linear combination of the generators.
Subtracting off $16 \cdot (1,1,1,\ldots, 1)$, we are looking for a vector in
the $5$-dimensional span of the remaining generators whose coordinates
are a permutation of $(\pm 1, \pm 2, \pm3, \ldots, \pm15)$.
There is no obvious obstruction: no two of the $30$ coordinates
are identically equal (and coordinates at opposite edges are
automatically each other's negatives, which should only help).
It would now be feasible to choose $5$ independent coordinates,
try all $30 \cdot 29 \cdot 28 \cdot 27 \cdot 26$ choices of labeling,
and check whether any of them results in a solution of the desired kind.
But first I tried the shortcut of looking at all solutions modulo $2$ and $3$.
The former failed: there are $15$ linear combinations with the requisite 
$16$ odd and $14$ even entries.  But modulo $3$, (unwelcome) success:
none of the $121 = (3^5-1) / 2$ pairs of nonzero linear combinations has
coordinates equally split among $0,1$, and $2 \bmod 3$.  (The possible
counts of $0 \bmod 3$ coordinates are $6, 8, 12, 14, 20$, occurring with
multiplicities $15, 60, 25, 15, 6$ respectively.)  Hence there is no
edge labeling satisfying all the conditions, QED.  
[added later: I should have noticed that none of $6,8,12,14,20$
is $1 \bmod 3$.  This can be seen directly from the matrix of inner products
of our five nontrivial generators, which has rank $1 \bmod 3$.
So no exhaustive search is required at all, though there's still
the integer linear algebra to find our generators.]<|endoftext|>
TITLE: Almost commuting unitary matrices
QUESTION [20 upvotes]: Suppose that $A_1,\dots, A_k$ are unitary matrices such that any two of them can be approximated by commuting unitary matrices. i.e. for any $i$ and $j$, there are unitary matrices $A_i'$ and $A_j'$ such that $\|A_i-A_i'\|<\varepsilon$, $\|A_j-A_j'\|<\varepsilon$ and $A_i'A_j'=A_j'A_i'$. Can we find unitary matrices $X_1,\dots,X_k$ such that any two of them commute and $\|X_i-A_i\|<O(\varepsilon)$ for all $i$?
What if any three (or small number) of them can be simultaneously approximated by commuting unitary matrices?
Here the matrix norm could be any unitary invariant norm. I'm specially interested in the operator norm and Hilbert-Schmidt norm.

REPLY [13 votes]: Edit Now this answers the first question for the operator norm and the normalized Hilbert-Schmidt norm.
The answer depends on the norm you are considering. The answer is no for the operator norm, but is yes for the normalized Hilbert-Schmidt norm (at least if you replace $O(\varepsilon)$ by $o(1)$, see the answers to this question). 
Here are some details on the counterexample for the operator norm.

By a theorem of Lin (see here), for a pair of self-adjoint matrices of norm less than $1$, they approximately commute if and only if they can be approximated by commuting matrices.
Voiculescu proved that the preceding does not hold for triples of  self-adjoint matrices of norm less than $1$ (see the link I gave here, or the references in the paper by Exel and Loring given in the comments).

1+2 imply that there is a sequence of triples $A_1^n,A_2^n,A_3^n$ of matrices of norm less than $1$ which are pairwise close to (self-adjoint) commuting matrices, but whose distance to the triples of commuting matrices is bounded below.

By continuity of the functional calculus and the fact that $t \in [-2,2] \mapsto e^{it}$ is a homeomorphism on its image, this implies that the unitary matrices $(e^{i A_1^n}, e^{i A_1^n},e^{i A_1^n})$ are pairwise close
to pairs of commuting unitaries, but are at positive distance from triples of commuting unitaries. This is what you were looking for.<|endoftext|>
TITLE: Is it consistent with ZFC that no nontrivial forcing notion has automatic mutual genericity?
QUESTION [15 upvotes]: A nontrivial forcing notion $\newcommand\Q{\mathbb{Q}}\Q$ exhibits
automatic mutual genericity, if whenever $G,H\subseteq\Q$ are
distinct $V$-generic filters (existing, say, in some forcing
extension of $V$), then they are mutually generic.
In any forcing extension $V[G]$ by such forcing, $G$ must be the
only $V$-generic filter; so this kind of forcing leads to unique
generics. Thus, it is a rigidity concept. In particular, $\Q$
cannot be forcing equivalent below incompatible conditions, even
in a forcing extension by a filter containing one of them, since
then the isomorphic filter would be distinct but not mutually
generic with it.
Meanwhile, it is consistent with ZFC that there are forcing
notions $\Q$ with automatic mutual genericity.
For example, consider a Suslin tree $T$ that is
Suslin-off-the-generic-branch, which means that after forcing to
add a $V$-generic branch $g$ through $T$, then $T\upharpoonright
p$ remains Suslin in $V[g]$ for any node $p$ not on $g$. If $g$
and $h$ are two $V$-generic branches through $T$, then consider a
node $p$ on $h$ that is not on $g$. Since $T\upharpoonright p$ is
Suslin in $V[g]$, every maximal antichain in $V[g]$ is refined by
a level of the tree, and since $h$ goes through every level of the
tree, it follows that $h$ is $V[g]$-generic, and so they are
mutually generic. Gunter Fuchs and I discuss this property of Suslin trees in our
paper Degrees of rigidity for Suslin trees,
 J. Symbolic Logic, vol. 74, iss. 2, pp. 423-454, 2009. We prove
there that a $V$-generic Suslin tree is
Suslin-off-the-generic-branch in $V[T]$, and also one can
construct such trees from the $\Diamond$ principle.
Because all the examples of automatic mutual genericity of which I
am aware have the character of these Suslin tree examples, which
do not exist in every model of ZFC, I was wondering whether it might be possible that there are no forcing notions with automatic mutual genericity.
Question. Is it relatively consistent with ZFC that no
nontrivial forcing notion exhibits automatic mutual genericity?
Alternatively, I would be delighted if someone could exhibit in ZFC that there is a forcing notion with automatic mutual genericity.
This is question 10 of my recent paper, Upward closure and
amalgamation in the generic multiverse of a countable
model of set theory.
That paper is focused on various amalgamation and upward closure
results for countable models of set theory. For example, if $W$ is
any countable transitive model of set theory, then there are
$W$-generic Cohen reals $c$ and $d$ for which $W[c]$ and $W[d]$
have no common forcing extension. One can generalize the argument
(see the paper) to many other notions of forcing, including any
forcing notion $\Q$ that is wide, in the sense that $\Q$ is not
$|\Q|$-c.c. below any condition. I had wondered which other
forcing notions exhibit this non-amalgamation property, and by
means of the rigid Suslin tree examples was led to the concept of
automatic mutual genericity, which have amalgamation rather than
non-amalgamation.

REPLY [10 votes]: Omer Ben-Neria and I found out that Joel's example, and analogues at larger cardinals, are the only ones. Therefore, Joel's hunch that MA proves there are no c.c.c. forcings with automatic mutual genericity is correct, although it is not entirely clear to me how to give a positive answer to the question with no extra restrictions.
I'll sketch the proof here. First we showed that a poset $\mathbb{Q}$ satisfies automatic mutual genericity if and only if for every $p\perp q$ in $\mathbb{Q}$, forcing below $p$ does not add a new maximal antichain below $q$. This is the main part of the proof: one direction is trivial and the other involves collapsing $2^\mathbb{Q}$ to be countable and using a name for a new antichain $\dot{A}$ to construct $V$-generics $G,H$ below $p$ and $q$ which do not meet the dense-below-$(p,q)$ subset 
$$\{(p',q'):\exists q^* \textrm{ such that }q'\le q^*\textrm{ and }p'\Vdash q^*\in\dot{A}  \}.$$
Now suppose $\mathbb{Q}$ has automatic mutual genericity, and let $\kappa$ be least so that $\mathbb{Q}$ adds a new subset of $\kappa$. For simplicity, assume that $\mathbb{Q}$ is a complete Boolean algebra. If $\mathbb{Q}$ had an antichain $A$ of size $\ge\kappa$, then in the generic extension by $\mathbb{Q}$ we could take $p\in A$ and use the new subset of $\kappa$ to define a new antichain below $-p$. So $\mathbb{Q}$ is $\kappa$-c.c., and therefore it is a $\kappa$-Suslin algebra.
Now the characterization of posets with automatic mutual genericity implies that $\mathbb{Q}$ must remain "Suslin off the generic branch," since otherwise this is witnessed by a new antichain.<|endoftext|>
TITLE: how do you visualize characteristic class?
QUESTION [26 upvotes]: For cohomology, there are some equivalent definitions when the object we consider is sufficiently nice. Since I mainly work with algebraic variety, I will restrict the objects I am considering to be algebraic variety.
There are several ways I think of cohomology namely:

singular cohomology (valuation of simplices)
de Rham cohomology (differential form)
Chow ring (closed subvariety)

I personal prefer thinking in the third way since it seems to be more geometric and easier for me to work with.
However, for characteristic classes, I find it hard to visualize. I know the technical details about them and some of their useful properties but I just don't get a feeling of it. Since I usually work over complex number, I personally more interested in how people think of Chern class.
Thank you!

REPLY [8 votes]: This is a variation on standard answers (to this standard question).  I'll start with a bit of philosophy since you mentioned different ways to think of cohomology.  Singular cohomology is fairly "transcendental" since to define a cochain one must assign a value to each and every chain.  When we teach basic topology, homology seems easier since we can draw/ conjure up finite linear combinations of such chains, at least in first examples.  
The latter two ways to define cohomology you cited make the assignments of values to chains geometrically - de Rham theory through integration, and sub manifolds through intersection.
So indeed, one can use your preferred intersection theory approach (which is mine too) to define characteristic classes as vanishing/independence loci.  But one can be slightly more direct in this setting.  First, triangulate your manifold and restrict to the simplicies, or if you wish just pull-back your bundle over a smooth chain (with some transversality assumptions so what's below works).
For the Stiefel-Whitney-Chern classes, take $n - i + 1$ sections of the original bundle.  The value of the $i$th SW or Chern class on a simplex will be the number of times these are linearly dependent over that simplex.
Schubert classes can be defined by embedding the bundle in a trivial bundle, and then counting the number of times a fiber over a simplex respects a given flag.
This Schubert style of geometry for characteristic classes is something I've also seen in my work on cohomology of symmetric groups, where Giusti, Salvatore and I give similar interpretations of characteristic classes of covering spaces.<|endoftext|>
TITLE: Is $AA+A$ always at least as large as $A/A$?
QUESTION [8 upvotes]: Let $A$ be a finite set of real numbers. Is it always the case that $|AA+A| \geq |A/A|$?
In the line above, $AA+A:=\{ab+c:a,b,c \in A \}$, while $A/A:=\{a/b:a,b \in A, b\neq 0 \}$ is the ratio set.
This is closely related to a previous question about the relative sizes of the sets $AA+A$ and $A+A$. See Is the set $ AA+A $ always at least as large as $ A+A $?.

REPLY [6 votes]: Just in case anyone out there was wondering, it turns out that the answer to this question is "no".
Earlier this year, a paper of myself, Imre Rusza, Chun-Yen Shen and Ilya Shkredov was posted. One of the results in the paper was a construction of a set of reals $A$ with $|AA+A|=o(|A|^2)$. For this particular example, one has $|A/A| = \Omega(|A|^2)$.
I suspect that the following weaker conjecture holds: for all $\epsilon >0$ there is an absolute constant $c=c(\epsilon)$ with $|AA+A| \geq c|A/A|^{1-\epsilon}$.<|endoftext|>
TITLE: Whitney sum formula for Pontryagin classes II
QUESTION [6 upvotes]: I have read in several places that the total Pontryagin classes of real vector bundles satisfy a Whitney sum formula $p(E\oplus F) = p(E)\cdot p(F)$ modulo 2-torsion. I would like to understand the 2-torsion part better.

Is there a reference which describes the difference between $p(E\oplus F)$ and $p(E)\cdot p(F)$, perhaps in terms of Bocksteins of Stiefel-Whitney classes of $E$ and $F$?

This question was previously part of Whitney sum formula for Pontryagin classes I; Qiaochu Yuan's answer to that question might be helpful.

REPLY [9 votes]: Brown, Edgar H., Jr.
The cohomology of BSOn and BOn with integer coefficients. 
Proc. Amer. Math. Soc. 85 (1982), no. 2, 283–288. 
Theorem 1.6, last sentence:
Under Whitney sum,
$p_q\mapsto  \sum_j r_{2q-j}\otimes r_j$, where
$r_{2s} = p_s$ and $r_{2s+1} = (\delta w_{2s})^2+ p_s\delta w_1$.<|endoftext|>
TITLE: global section of affine $C^\infty$-scheme
QUESTION [5 upvotes]: I'm reading Algebraic Geometry over $C^\infty$-rings.
It is written that "If $\mathfrak{C}$ is not finitely generated then $\Phi_{\mathfrak{C}}:\mathfrak{C}\rightarrow \Gamma(\text{Spec}\mathfrak{C})$ need not surjective, so $\Gamma(\text{Spec} \mathfrak{C})$ can be larger than $\mathfrak{C}$"(p.26).
However, I think as follows.
Let $U_1=\{ x\in \text{Spec}\mathfrak{C}|x(1)\neq0\}$, then $U_1 = \text{Spec}\mathfrak{C} $ because for every element $x:\mathfrak{C}\rightarrow \mathbb{R}$ in $\text{Spec}\mathfrak{C}, x(1)=1\neq 0$. Hence $\Gamma(\text{Spec}\mathfrak{C})=\mathcal{O}_{\text{Spec}\mathfrak{C}}(\text{Spec}\mathfrak{C})=\mathcal{O}_{\text{Spec}\mathfrak{C}}(U_1)=\mathfrak{C}[1^{-1}]=\mathfrak{C}$.
Where am I wrong?

REPLY [4 votes]: My apologies, Omar is correct. There was an mistake in Definition 4.12 of version 5 of my paper Algebraic Geometry over $C^\infty$-rings (posted 24 September 2015). I have replaced this by version 6 (posted 11 November 2015), which I hope is correct.<|endoftext|>
TITLE: Travelling salesman: can the furthest-neighbour algorithm beat the nearest-neighbour?
QUESTION [12 upvotes]: This is a problem that has bugged me for quite some time, and I have not been able to find any documentation about it online. It is well known that the NN algorithm can yield the worst possible route - for small cases it would seem any (undirected) complete weighted graph with this property forces the FN algorithm to produce an equally bad route. 
Naturally this raises the question whether FN can ever produce a shorter path than NN - I am mostly interested in the symmetric TSP here.
Any input would be greatly appreciated.
Posted on MSE over a year ago, with no answers.

REPLY [9 votes]: Label the vertices of the NN path by $0,1,2,\dots,n=0$. Let $x_0=0,x_1,\dots,x_n=0$ be the FN path. Define $A=\{i\in \{1,\dots,n-1\}| \exists j>i \text{ with } x_j>x_i\}$. Then for any $i\in A$ we have by construction $d(x_i,x_i+1)\leq d(x_i,x_j)\leq d(x_i,x_{i+1})$. Let $b_0=0< b_1\dots < b_k=n-1$ be the elements of $\{0\}\cup A^c=\{0,1,\dots,n-1\} \backslash A$. Again by construction we have $d(x_{b_{i+1}},x_{b_{i+1}}+1)\leq d(x_{b_{i+1}},x_{b_{i}})\leq d(x_{b_i},x_{b_i+1})$ for all $i\in \{0,\dots,k-1\}$. Furthermore we have $d(x_{b_0},x_{b_0}+1)=d(0,1)\leq d(n-1,0)=d(x_{b_k},x_{b_k+1})$. Summing up all inequalities yields that the length of the NN path is at most as long as the path of the FN path.<|endoftext|>
TITLE: Expansion in strongly regular graphs
QUESTION [9 upvotes]: Have you seen the following statement proven anywhere?
Let $G$ be a strongly regular graph with parameters $(n,k,\lambda,\mu)$ with $\lambda,\mu>0$. Then there is no set $A$ of at least $n/4$ vertices in $G$ such that the neighborhood of $A$ is of size less than $3n/4$.
A vertex in $A$ in this case would be included in the neighborhood of $A$ iff it is adjacent to another vertex in $A$. It seems like this should be known, if it's true.

REPLY [5 votes]: Consider the complete tripartite graph $G = \langle X\cup Y\cup Z,E\rangle$ with $|X|=|Y|=|Z|=\frac{n}{3}$, and $E = X\times Y\cup X\times Z \cup Y\times Z$. This graph is strongly (n,2n/3,n/3,2n/3)-regular, and e.g. $X$ has more than $\frac{n}{4}$ vertices and less than $\frac{3}{4}n$ neighbors, so this is a counterexample to your conjecture.
On the other hand, since the spectrum of strongly regular graphs is known precisely (see https://en.wikipedia.org/wiki/Strongly_regular_graph#Eigenvalues), one can get bounds on the edge expansion of such a graph via Cheeger's inequality, which implies results like the one you conjecture.
Question: What is the minimal $\alpha>0$ such that in every strongly regular graph, every set $S \subseteq V$ whose size is at least $\alpha n$ has at least $(1-\alpha) n$ neighbors, and which strongly regular graph achieves this $\alpha$?
This seems like a natural question, but I wouldn't be surprised if no one had considered it. It looks like a good research question to me. My example implies $\alpha \geq 1/3$.<|endoftext|>
TITLE: Consistency Strength of "HC is elementary in V[G]"
QUESTION [8 upvotes]: Let $P$ be the Levy-collapse of the ordinals, so $P$ is a class forcing notion that makes every ordinal countable. 
Note that since $P$ is weakly homogeneous, for any formula $\phi(\overline{a})$ with parameters from $\mathbf{V}$, $P$ decides $\phi(\hat{\overline{a}})$. Hence it makes sense to ask whether $(HC, \in)$ is an elementary substructure of $(\mathbf{V}[G], \in)$ for some or any $P$-generic extension $\mathbf{V}[G]$.
Consider the axiom schema $\Phi$ that asserts that $(HC, \in)$ is elementary in $(\mathbf{V}[G], \in)$. 
Question: Is $\Phi$ consistent relative to large cardinals, and if so, what is its strength? (Or at least get an upper bound)
As an example,  $\Phi$ implies that $\omega_1$ is inaccessible to the reals. To see this, fix a real number $x$; then $\mathbf{L}[x]$ satisfies the powerset axiom, so $HC \models `` \mathbf{L}[x] \mbox{ satisfies the powerset axiom}"$, so $\mathbf{L}_{\omega_1}[x]$ satisfies the powerset axiom.

REPLY [8 votes]: $\newcommand\HC{\text{HC}}$Allow me to denote your theory $\Phi$ by "$\HC\prec V[G]$". As you mention, this is not a single statement, 
since we have no truth predicate able to express truth in the
extension $V[G]$, but rather it is expressed as a scheme, asserting of each formula $\varphi$ with parameters
$z\in\HC$ that $\HC\models\varphi[z]$ if and only if
$\Vdash\varphi(\check z)$, where the forcing relation is over the
Levy collapse of all ordinals.
Theorem. The following theories are equiconsistent:

ZFC plus your theory, $\HC\prec V[G]$, expressed as a scheme.
ZFC plus the axiom scheme Ord is
Mahlo, which asserts that every definable (from
parameters) closed unbounded class $C$ of ordinals contains a
regular cardinal.
ZFC plus the assertion scheme that $\kappa$ is an inaccessible reflecting cardinal. That is, $V_\kappa\prec V+\kappa$ is inaccessible, expressed as a
scheme.

Proof. The equiconsistent of 2 and 3 is explained on the
Cantor's attic page for Ord is Mahlo, to which I linked.
If your axiom holds, then we have $\HC\prec V[G]$. If
$\kappa=\omega_1^V$, it follows easily that $L_\kappa\prec L$,
since $L_\kappa=L^{\HC}$ and $L=L^{V[G]}$. And since also $\kappa$ is
inaccessible in $L$ as you observed, this gives us a model of
statement 3.
Conversely, if statement $3$ holds, then consider the Levy
collapse $V[G]$ of all ordinals. Let $g=G\upharpoonright\kappa$, so that $V[g]$ is the usual Levy collapse of $\kappa$, and $\HC^{V[g]}=V_\kappa[g]$.
Since $V_\kappa\prec V$, it is not difficult to see that
$V_\kappa[g]\prec V[G]$, as follows. If
$V_\kappa[g]\models\varphi[x]$, then by taking names, we get that
some $p\in g$ forces $\varphi(\dot x)$ in $V_\kappa$, and so $p$
forces $\varphi(\dot x)$ in $V$, and so $V[G]\models\varphi[x]$. So
$\HC^{V[g]}\prec V[G]$ holds, verifying your theory in $V[g]$.
QED
In particular, the axiom is stronger than just an inaccessible cardinal. For example, you could have also continued your observation about inaccessibility to reals to notice that $\omega_1$ must be a limit of inaccessible cardinals in $L$, since there can be no bound below $\omega_1$, as that bound would not work in $L^{V[G]}$. Similarly, it must be $\alpha$-inaccessible for every countable $\alpha$, and $\alpha$-hyperinaccessible and so on. Ultimately, of course, you get that $\kappa$ is an inaccessible fully reflecting cardinal in $L$, as I argued.
The axiom Ord is Mahlo is stronger than inaccessible and
$\alpha$-inaccessible and hyperinaccessible and various strong
hyper-degrees of inaccessibility, but strictly weaker than the
existence of a Mahlo cardinal.<|endoftext|>
TITLE: Transitive models and CH
QUESTION [9 upvotes]: The following was asked on stackexchange but I think it also belongs here:
https://math.stackexchange.com/questions/1513446/transitive-models-and-ch
Suppose $M, N$ are two countable transitive models of ZFC which have same ordinals, cofinalities and reals (but not necessarily same sets of reals!). Suppose $M$ models the continuum hypothesis. Can we conclude that $N$ also models continuum hypothesis? I can see that if this fails, then neither one of $M, N$ is included in the other. But what if $M, N$ are incomparable.

REPLY [12 votes]: This strange situation can indeed happen.
Start with a countable transitive model of set theory $W$,
satisfying CH, and let $G$ be $W$-generic for the forcing
$\newcommand\Add{\text{Add}}\Add(\omega,\omega_1)^W$. So the model
$M=W[G]$ continues to satisfy CH. Now, let $g$ be $W[G]$-generic
for the collapse of $\omega_2^W$ to $\omega_1^W$, using the
collapse forcing as it is defined in $W$, and temporarily consider
$W[G][g]$. Notice that the collapse forcing was countably closed
in $W$, and so it remains at least countably distributive in
$W[G]$. So we have added no new countable sequences of ordinals in
going from $W[G]$ to $W[G][g]$. Inside $W[G][g]$, the ordinals
$\omega_1^W$ and $\omega_2^W$ are now in bijection, and so there
is an isomorphism $\pi:\Add(\omega,\omega_1)^W\cong
\Add(\omega,\omega_2)^W$ between these two forcing notions as posets. Furthermore, since by design
the bijection of the ordinals has the property that every
countable piece of it is in $W$, the same property is true for
this isomorphism $\pi$. Let
$H=\pi[G]\subset\Add(\omega,\omega_2)^W$ be the isomorphic copy of
$G$ induced by the isomorphism $\pi$. I claim that $H$ is
$W$-generic for $\Add(\omega,\omega_2)^W$, because if $A$ is any
maximal antichain in this forcing in $W$, then by the c.c.c. it
follows that $A$ is countable, and so
$\pi^{-1}A\subset\Add(\omega,\omega_1)^W$ is a maximal antichain
in $W$ for the first forcing, since this much of $\pi$ is in $W$.
And so, since $G$ must meet $\pi^{-1}A$, it follows that $H$ meets
$A$; so $H$ is $W$-generic, and we may let $N=W[H]$. So $N$ is a
model of $\neg\text{CH}$, since we've added $\omega_2$ many Cohen
reals. Both $M$ and $N$ are c.c.c. extensions of $W$, and so they
have the same cardinals and cofinalities.
Let's now argue that they have the same reals. First off, every
real of $N=W[H]$ is certainly in $W[G][g]$, where $H$ is
constructed, and the reals of $W[G][g]$ are the same as the reals
of $W[G]=M$, since the $g$ forcing is countably distributive. So
every reals of $N$ is in $M$. Conversely, if $x$ is a real in
$M=W[G]$, then $x$ is in $W[G\upharpoonright\alpha]$ for some
$\alpha<\omega_1$. And since
$\pi\upharpoonright\Add(\omega,\alpha)$ is in $W$, it follows that
$x$ is in $W[H\upharpoonright\pi[\alpha]]$, which is contained in
$W[H]=N$. So they have the same reals.
In conclusion, $M$ and $N$ have the same reals, the same
cardinals, the same cofinalities, but one has CH and one does not,
as desired.
One may cast the argument in terms of forcing over $V$; there is no need to go to countable transitive models. Namely, start in $V$, with CH, and then force to add $\omega_1$ many Cohen reals to form $V[G]$. Now collapse $\omega_2$ to $\omega_1$ using the ground model collapse, and in $V[G][g]$ define $H$ as the copy of $G$ induced by that isomorphism. It now follows by the argument above that $V[G]$ and $V[H]$ have the same reals, the same cardinals and cofinalities, but one has CH and the other does not.
And of course, there is nothing special about $\omega_2$ in these arguments, we could have used $\omega_3$ or any other regular cardinal in $W$ just as easily.<|endoftext|>
TITLE: Convergence of functions on Alexandrov spaces
QUESTION [7 upvotes]: Consider a sequence of $n-$dim Alexandrov spaces with curvature $\geq$ -1 $\{(M_i,p_i)\}$ Gromov-Hausdroff converging to an $n-$dim Alexandrov space $(M,p)$. Let $f:M\mapsto \mathbb R$ be a Lipschitz function, and $R>0$ fixed. For $0<r<R$, can we construct Lipschitz functions $f_i:B(p_i,R)\subset M_i\mapsto \mathbb R$ ,such that $\int_{B(p_i,r)}|\nabla f_i|^2d\mathcal H^n\to \int_{B(p,r)}|\nabla f|^2d\mathcal H^n$ ?

REPLY [5 votes]: First note that if $B(p,R)$ lies in a distance chart then you can lift the function using the chart and it will satisfy the needed convergence.
Now assume you have few overlapping distance charts;
they cover subset $\Omega$ and it might have nonempty complement $C=M\backslash \Omega$.
You can lift the function in each chart and use partition of unity to glue these liftings together. 
Further you can extend the obtained function keeping it Lipschitz to whole $M_n$.
This produce a sequence of Lipschitz functions for which you have the needed convergence with an error depending on your Lipshcitz constant and volume of $C\cap  B(p,R)$.
The latter can be made as small as you want.
So applying diagonal procedure you get the result.<|endoftext|>
TITLE: Volume of the convex hull of the set of all graphic sequences of a given length
QUESTION [5 upvotes]: Consider the set of all graphic sequences with $n$ elements as a subset of $\mathbb{R}^{n}$, namely let
$$D(n)=\{(d_{1},\dots,d_{n})\in\mathbb{Z}_{+}^{n}:d_{1}\geq\dots\geq d_{n},\ \sum_{i=1}^{n}d_{i}\ \text{is even},\ \sum_{i=1}^{k}d_{i}\leq k(k-1)+\sum_{i=k+1}^{n}\min\{k,d_{i}\}\ \text{for all}\ 1\leq k\leq n \}.$$
One can observe that the diameter of $D(n)$ equals $(n-1)\sqrt[]{n}$. The question is:

How to compute the volume of the convex hull of $D(n)$?

REPLY [2 votes]: The polytope considered by Richard Stanley is the polytope of degree sequences.
Sergiy Kozerenko's question is about the polytope of degree partitions for which see paper R46 of Volume 13 (2006) of the Electronic Journal of combinatorics. It is shown there that the volume of the polytope of degree sequences is n! times the volume of the polytope of degree partitions.<|endoftext|>
TITLE: Let $X$ be a projective variety and let $Pic(X) \cong \mathbb Z$, then $(X, −K_X) $ is $K$-semistable
QUESTION [9 upvotes]: Is there any approach for the following conjecture?
Let $X$ be a projective Fano manifold and let  $Pic(X) \cong \mathbb Z$, then $(X, −K_X) $ is $K$-semistable.

REPLY [9 votes]: The answer correspond to weighted stability of Fedor Bogomolov and result of Gang Tian. We explain it here as an additional comment to previous answer
Tian introduced a stronger notion of strong stability around 90 as follows and proved the following theorem albeit strong stability does not satisfied for any Fano manifold. But it can has its own interest.
Definition(Tian 92): Let $E_1$ and $E_2$ be two coherent holomorphic sheaves on $X$. An extension of $E_1$ by $E_2$ is a coherent sheaf $E_3$ with the following short exact sequence $$0\to E_2\to E_3\to E_1\to 0$$
A pair $(E_1,E_2)$ of coherent sheaves is said to be stable (resp semi-stable) with respect to Kahler class $\omega$ if the generic extension $R$ of $E_1$ by $E_2$ is stable(resp semi-stable) with respect to same Kahler class.
Now let $E$ be a holomorphic vector bundle then we say $E$ is strongly stable (resp semi-stable)with respect to $\omega$ if both $E$ and the pair $(E,\mathcal O_X)$ are stable with respect to $\omega$. Here $\mathcal O_X$ is the structure sheaf of $X$. i.e sheaf of local holomorphic functions.
Now Bogomolov introduced the following $\mu$-stability as weighted stability.
Definition: Given any positive number $\mu<1$. A bundle $E$ is called $\mu$-stable (resp $\mu$ semi-stable)if for any subsheaf $L$ of $E$ with rank $rank L\leq rank E$,
$$\frac{deg L}{rank L}<\mu\frac{deg E}{rank E}$$ $$\text{resp}\leq$$
Theorem (Tian 92)If $X$ is a Kahler-Einstein manifold with positive scalar curvature and $Pic(X)\cong \mathbb Z$ then the tangent bundle is $\mu$-semi-stable where here $\mu=\frac{n}{n+1}$.
Moreover if $X$ has no holomorphic vector field then $TX$ is $\frac{n}{n+1}$-stable. 
Theorem: Tian proved that if $TX$  be $\frac{n}{n+1}$-stable, then $TX$ is strongly stable<|endoftext|>
TITLE: Brownian motion in $n$ dimensions
QUESTION [5 upvotes]: Consider a particle starting at the origin in $\mathbb{R}^n$ and undergoing Brownian motion. Is there an expression known for the probability of the particle hitting the sphere $S^{n - 1}_r = \{x \in \mathbb{R}^n : \Vert x\Vert = r\}$ within time $t$ (obviously such an expression, if known, will be in terms of $r$ and $t$)? In other words, I am looking for a formula for $\mathbb{P}(\sup _{s \leq t}\Vert B(s)\Vert \geq r)$. Thanks!

REPLY [2 votes]: The distributions and properties  of the maximum of a Bessel processes and  Bessel bridges are discussed in details in
Jim Pitman: The law of of the maximum of a Bessel bridge<|endoftext|>
TITLE: Relation between Weil Conjecture and Langlands Program
QUESTION [15 upvotes]: Recently I read Gelbart's An Elementary Introduction To The Langlands Program, which explained the origin of the program, and this question came to me. For an elliptic curve over finite field, the associated L-function will coincide with the L-function associated to a certain automorphic cusp form (in Silverman's "Advanced topics in the arithmetic of elliptic curves" there is an alternative statement using Grössencharakter).
Since both Weil conjecture and Langlands program consist rationality and functional equation and by the preceding result (BTW, some approaches are similar), are there some inner relations between the two conjectures?

REPLY [6 votes]: The Weil conjectures asks for rationality, functional equation, Riemann hypothesis, and degree relating to the Hasse-Weil zeta function of a nonsingular projective variety $X$ over a finite field $k$, which is a product of L-functions. Since you are reading Gelbart, as Artin you can think of these as L-functions associated to the Galois representation of Gal$(\bar k/k)$ acting on each degree of the (etale) cohomology of $X$ respectively. 
The Langlands program does not ask for rationality, but to each automorphic representation it does ask for a functional equation and meromorphic continuation for the associated automorphic $L$-function. (This also depends on the representation of the $L$-group.) It is generally believed that these $L$-functions also satisfy the Riemann hypothesis, but this I do not think is explicitly a part of the program.
What you do get is that the functoriality conjecture for $H=\{1\}$ and $G=GL_n$ (p.208 of Gelbart) is what some call Langlands reciprocity: to every Galois representation is associated an automorphic representation, so if you like, you may think of the Hasse-Weil zeta function as (conjecturally) a product of automorphic $L$-functions instead, though I am not sure what one gains from this.
EDIT: Also see the answer to this question. Particularly, Kevin Buzzard's comment that RH given by the Weil conjectures leads to bounds on Satake parameters, i.e., towards the Ramanujan for the corresponding automorphic form.<|endoftext|>
TITLE: Length of Weyl group element mapping highest root to a simple root
QUESTION [5 upvotes]: Let $\Phi$ be an irreducible root system and $\Delta$ a simple system (base). Let $W$ be the Weyl group of $\Phi$. Let $\theta$ be the highest root and $h^\vee$ be the dual Coxeter number. Choose the shortest $w \in W$ such that $w(\alpha_i)=\theta$ for some simple root $\alpha_i$. I got the result that the length of $w$ can be determined by $l(w) = h^\vee −2$ (for instance see Lemma 5.1 in here). However I'm not given yet any explicit proof for that result. Can anyone give me proof or references where this formula is introduced and proved?
Thanks in advance.

REPLY [3 votes]: Ok I'm late on this, but I've just seen the question! Another way to look at it: the W-orbit of the highest root consists of the long roots, so the long roots are in bijection with the minimal coset representatives for the stabilizer of the highest root (parabolic subgroup generated by the simple roots orthogonal to the highest roots, i.e. corresponding to nodes in the Dynkin diagram not connected to the affine root in the affine Dynkin diagram). One may ask: what is the relation between the length of a minimal coset representative and the height of the corresponding long root? Actually there is a simple relation between that length and the height of the corresponding coroot (hence the dual Coxeter number...). I needed this in my first article, "Cohomology of the minimal nilpotent orbit". Before submitting the paper, I found a paper of Carter ("Weyl groups and finite Chevalley groups") which dealt with precisely this problem (exchanging the role of roots and coroots, so he was talking about the highest short root); so I quoted it. This is the kind of problem which must have been worked out independently many times! :)<|endoftext|>
TITLE: Selecting subsets with size $\frac{n}{2}$ covering every pair of the elements
QUESTION [5 upvotes]: Given a set $S$ of $n$ elements. Let $T$ be the set of all subsets of $S$, with size $\frac{n}{2}$ ($n$ is even). We want to select a subset $T'$ of $T$, with the property that for any pair of the elements $x,y \in S$, there exists a subset $Q \in T'$, such that $x \in Q$ and $y \in Q$. What is the minimum size of $T'$? In particular, is $|T'| < n$?

REPLY [7 votes]: The answer is $6$ for $n \ge 4$. This is problem 1.3 on IMC 2013:
http://www.imc-math.org.uk/imc2013/IMC2013-day1-solutions.pdf

It takes $6$ subsets. 
Lower bound: For any element $a$, to cover all $n-1 \gt 2(n/2 -1)$ pairs of elements including $a$, there must be at least $3$ subsets containing $a$. This is true for all elements so there must be at least $3n$ pairs $(e,Q), e\in Q$. Each subset contributes $n/2$ pairs $(e,Q), e \in Q$ so there must be at least $6$.
Construction: Write $n/2$ as $2s+3t$ with $s,t$ nonnegative integers. Break the elements into sets $A,B,C,D$ of size $s$ and $X',X'',Y',Y'',Z',Z''$ of size $t$, with $X= X' \cup X'', Y=Y'\cup Y'',Z = Z' \cup Z''$. Then consider the unions of $\{A,B,X,Y'\},$ $\{C,D,X,Y''\},$ $\{A,C,Y,Z'\},$ $\{B,D,Y,Z''\},$ $\{A,D,Z,X'\},$ and $\{B,C,Z,X''\}.$<|endoftext|>
TITLE: Seeking an explanation for a peculiar factorization
QUESTION [6 upvotes]: Recently during my work, I encountered the following family of sextic polynomials
$$\displaystyle x^6 - 3x^5 + cx^4 + (5 - 2c)x^3 + cx^2 - 3x + 1.$$
By plugging in various values of $c$, I noticed that this sextic always has all real roots or all non-real roots. When the roots are all complex, I noticed the following: there is always a conjugate pair $\alpha, \overline{\alpha}$ such that the real part of $\alpha$ is $1/2$, the other two pairs $\beta, \overline{\beta}$ and $\gamma, \overline{\gamma}$ have the same imaginary parts, and the real part of $\beta$ and $\gamma$ always sum to $1$. Using this data, I formulated relations between the roots and found that in fact the sextic always factors as follows:
 $$\displaystyle x^6 - 3x^5 + cx^4 + (5 - 2c)x^3 + cx^2 - 3x + 1 =$$
$$\displaystyle (x^2 - x + a)(x^2 - x/a + 1/a)(x^2 - (2 - 1/a)x + 1),$$
where $a$ is a real root of the cubic equation
$$\displaystyle x^3 + (3 - c)x^2 + 3x - 1 = 0.$$
My proof is by brute force. Is there any more enlightened explanation of this fact?
Edit: in my latest experiments, I have discovered that polynomials of the shape
$$\displaystyle x^6 + bx^5 + cx^4 + (2c + 10 - 5b)x^3 + (c + 15 - 5b)x^2 + (6 - b)x + 1$$ 
also have the strange property that it either has all real roots or all complex roots, and if all of the roots are complex, then there is a pair with real part equal to $1/2$ and the other two pairs have real parts summing to one and identical imaginary parts. This should lead to a similar factorization once one works out the details, but there is no obvious folding that occurs!

REPLY [6 votes]: You have reciprocal polynomial and after standart 
change of variable $t=x+1/x$ you'll get the cubic equation
$$t^3-3t^2+(c-3)t+11-2c=0.$$<|endoftext|>
TITLE: Regularity of Hodge Laplacian on bounded domains
QUESTION [11 upvotes]: I need a reference for the $W^{s,p}$ regularity of the Hodge boundary value problem on bounded domains. I need estimates $\lVert \omega \rVert_{W^{s+2,p}} \leq c\lVert f\rVert_{W^{s,p}}$, $s\geq 0$ for the system
$$\begin{split}\Delta \omega &=f\text{ in }\Omega ,\\   
\nu\wedge\omega &=0\text{ on }\partial\Omega\ ,\\      
\nu\wedge\delta\omega &=0\text{ on }\partial \Omega\ .
\end{split}$$
I need to know:  

A reference which actually verifies the Agmon-Douglis-Nirenberg condition for this system for general boundary.... most references either do not verify or verifies the condition only when $\partial\Omega$ is flat.  
Whether regularity results extend to the scale of negative Sobolev spaces - e.g.
is  $\lVert \omega \rVert_{W^{1,p}} \leq c \lVert f \rVert_{W^{-1,p}}$ true? 
Whether there is such a result for the system
$$ \begin{split} \delta ( A d\omega) + d\delta\omega &=0\text{ in }\Omega\ ,\\  
\nu\wedge\omega &=0\text{ on }\partial\Omega ,\\      
\nu\wedge\delta\omega &=0\text{ on }\partial\Omega\ ,\end{split}$$
where $A$ is elliptic.

REPLY [2 votes]: Morrey's book [2] mentioned in another answer covers the $L^2$ theory only. It is surprisingly difficult to find good, detailed and reliable references for the Hodge decomposition in $L^p$ for $1<p<\infty$. 
The $L^p$ Hodge decomposition on compact manifolds is done carefully in the papers [1,3]. Although the authors do not discuss directly the higher order regularity,  it is shown in [3] (see Definition 5.23) that the Green operator maps $L^p$ to $W^{2,p}$. The paper [1] says very little about the boundary conditions (which for forms are far from obvious), but the boundary conditions are carefully studied in [1].
[1] T. Iwaniec, C. Scott, B. Stroffolini, 
Nonlinear Hodge theory on manifolds with boundary.
Ann. Mat. Pura Appl. (4) 177 (1999), 37–115. 
(MathSciNet review.)
[2] C. B. Morrey, Jr., Multiple integrals in the calculus of variations. Die Grundlehren der mathematischen Wissenschaften, Band 130 Springer-Verlag New York, Inc., New York 1966. 
(MathSciNet review.)
[3] C. Scott, $L^p$ theory of differential forms on manifolds.
Trans. Amer. Math. Soc. 347 (1995), no. 6, 2075–2096. 
(MathSciNet review.)<|endoftext|>
TITLE: Pair of square matrices related by traces formulas
QUESTION [8 upvotes]: Let $A$ and $B$ be two $n\times n$ matrices over $\mathbb{C}$. Assume that for every $k\geq 1$ it holds  $tr(A^k) = tr(B^{2k-1})$. What can we say about the possible eigenvalues of $A$ and of $B$? How limited is the set of possible eigenvalues? Must it be, for example, finite?

REPLY [4 votes]: Suppose that all of the nonzero eigenvalues of $A$ are the same, so $\text{tr}(A^k) = m \lambda^k$ for some $\lambda$ and some $m \le n$. If $\lambda = \frac{1}{m^2}$ then $\text{tr}(A^k) = \frac{1}{m^{2k-1}}$ and so $B$ can have eigenvalue $\frac{1}{m}$ and $\pm$ pairs as in Mike Jury's comment.
Similarly, suppose that all of the eigenvalues of $B$ that don't come in $\pm$ pairs are the same, so $\text{tr}(B^{2k-1}) = m \lambda^{2k-1}$ for some $\lambda$ and some $m \le n$. If $\lambda = m$ then $\text{tr}(B^{2k-1}) = m^{2k}$ and so $A$ can have eigenvalue $m^2$and all other eigenvalues $0$ as in Mike Jury's comment.
On the other hand, if you require that all of the nonzero eigenvalues of $A$ and $B$ have distinct absolute values then I think you can show that $A$ has no nonzero eigenvalues by a comparison of the growth rates of the LHS and the RHS as $k \to \infty$.<|endoftext|>
TITLE: The limit of the following product? What is the closed form of the value?
QUESTION [5 upvotes]: Assume that $P_n$ is the $n$'th prime: Please help me solve the following  $$\lim_{k\to\infty} {k}\prod_{n=1}^k \frac{P_{2n-1}}{P_{2n}}$$
I am not really sure quite where to start here as I am dealing with primes, one of my weaknesses I want to make the important point: I am not a professional mathematician, in fact only a high school student, but am desperate to learn all the info I can. So if it would be possible, for me, could you please namedrop theorems in your answer? I also want to apologize if it is rude to ask questions here not being remotely professional.
Back to the problem, however, I did note that the values do seem to converge. If they do converge, if possible, find a closed form. If there is no closed form, simply ignore my last sentence.
EDIT: if there is a factor that generates a convergent product, please let me know, as k alone doesn't do it, I would want to know what does.

REPLY [4 votes]: Clarification: This is not a full answer to the original question (which might be a very hard one), but rather an heuristic argument.
Let $F_k = \prod_{n=1}^{k}  \frac{p_{2n-1}}{p_{2n}}, G_k = \prod_{n=1}^{k}  \frac{p_{2n}}{p_{2n+1}}$. You are interested in $F_k$, but heuristically at least, they should be close to each other, although it might be hard to prove:
$$ F_k \sim G_k$$
Fortunately, $F_k \cdot G_k$ has a nice form - it is $\prod_{i=1}^{2k} \frac{p_i}{p_{i+1}}$, which telescopes to $\frac{p_1}{p_{2k+1}}$. By the prime number theorem,
$$p_{2k+1} \sim (2k) \log k$$
Hence we'd expect 
$$F_k \sim \sqrt{F_k \cdot G_k} \sim \sqrt{\frac{2}{2k \log k}} = \sqrt{\frac{1}{k \log k}}$$
This implies $kF_k \sim \sqrt{\frac{k}{\log k}}$, which goes to infinity as $k$ goes to infinity.<|endoftext|>
TITLE: curve over higher dimensional basis with 0-dimensional locus of bad reduction
QUESTION [5 upvotes]: Is there an example of a flat proper relative curve $X/S$ with geometrically connected fibres and with $\mathrm{dim} S > 1$ and $S$ regular and connected with $0$-dimensional locus of bad reduction $S_{\mathrm{bad}} = \{s \in S: X_s/s \text{ not smooth}\}$?

REPLY [8 votes]: The answer is no if $f:X\to S$ is locally projective and the genus $g$ of the general fiber is $\geq1$. (About these restrictions, see remarks at the end). 
Assuming this, put $U:=S\smallsetminus S_\mathrm{bad}$. By assumption, $S$ is regular and $U$ contains all points of codimension $\leq1$ of $S$. I do not assume $\dim S_\mathrm{bad}=0$, but of course we may reduce to this case if we prefer.
The main point is:

$X_U$ extends uniquely to a proper and smooth $f':X'\to S$. 

This follows from my paper:  Un théorème de pureté pour les familles de courbes lisses, CRAS Paris vol 300 issue 14 (1985), 489-492.
Now let us prove that the identification $X_U\cong X'_U$ extends to an isomorphism $X\cong X'$. Fix an invertible sheaf $L$ on $X$, very ample relative to $S$ and such that $\mathscr{E}:=f_*L$ is locally free on $S$, and commutes with base change in the usual sense. Moreover, assume that the degree of $L$ in the fibers is large enough ($\geq2g+2$ ?) to automatically ensure these properties on all smooth curves of genus $g$. So, we have a closed immersion $i:X\hookrightarrow \mathbb{P}(\mathscr{E})$.

Claim: The pair $(X,L)$ (subject to the conditions of flatness, smoothness over $U$, and strong ampleness) is determined up to unique isomorphism by $(X_U,L_U)$.

Indeed, $(X_U,L_U)$ determines  $\mathscr{E}_U$ and $i_U:X_U\hookrightarrow \mathbb{P}(\mathscr{E}_U)$. Then we recover $\mathscr{E}$ as $j_*\mathscr{E}_U$ (by the codimension assumption, $j$ being the inclusion of $U$), and then $X$ must be isomorphic to the schematic closure of $i_U(X_U)$ in $\mathbb{P}(\mathscr{E})$. Of course, $L$ is then the restriction of $\mathscr{O}_{\mathbb{P}(\mathscr{E})}(1)$ to $X$.
Now we can "do the same" with $X'$. More precisely, over $X'_U=X_U$ we have the sheaf $L_U$, and since $X'$ is a regular scheme this extends to an invertible sheaf $L'$ on $X'$. Since the degree in the fibers is locally constant, this $L'$ satisfies all our ampleness requirements. So we get a pair $(X',L')$ whose restriction to $U$ is isomorphic to $(X_U,L_U)$, and we conclude by the above claim.
Remark 1. One can probably get rid of the projective assumption: we may assume $S$ local henselian, in which case projectivity should be automatic.
Remark 2. My purity result might be true in genus $0$, but I never worked this out.<|endoftext|>
TITLE: Is the Veronese variety "enough" to describe all the $SL(V)$-orbits in $\mathbb{P}(\textrm{Sym}^dV)$?
QUESTION [5 upvotes]: I apologise in advance if the question will look ridicolous to experienced eyes: in this case a good reference will be enough to clarify my doubts.
Let $V$ be a complex vector space of dimension $n$, and $\textrm{Sym}^dV$ the space of homogeneous polynomial of degree $d$ (on $V^*$). Assume also that $SL_n$ acts naturally on $$\mathbb{P}^{M(n,d)}:=\mathbb{P}(\textrm{Sym}^dV)\, .$$
Motivating the prelim. question. If $n=2$ and $d=3$, then $M(n,d)=3$. In this case, there are two special orbits in $\mathbb{P}^{M(n,d)}=\mathbb{P}^3$: a one-dimensional one, given by the twisted cubic $\gamma$, and a two-dimensional one, given by $T\gamma\smallsetminus\gamma$, which is the set of the smooth points of the tangent variety to $\gamma$. I don't know if $\mathbb{P}^3\smallsetminus T\gamma$ is an orbit as well, but surely it is an invariant subset: let me call it the "open cell".
This example can be easily generalised, by replacing the twisted cubic $\gamma$ with the Veronese variety $v_d(\mathbb{P}(V))$ in $\mathbb{P}^{M(n,d)}$, and by noticing that the subsets
$$
X_{i,n,d}:=\textrm{Osc}_i(v_d(\mathbb{P}(V)))\smallsetminus \textrm{Osc}_{i-1}(v_d(\mathbb{P}(V)))\subset \mathbb{P}^{M(n,d)}\quad \quad (^*)
$$
are $SL_n$-invariant (by "$\textrm{Osc}_i$" I mean the $i^\textrm{th}$ osculating variety, i.e., the union of the $i^\textrm{th}$ order osculating spaces).

PRELIM. QUESTION: for which values of $n$ and $d$ all the subsets $X_{i,n,d}$ (possibly discarding the "open cell") are orbits?

The  example above (if I did it correctly) shows that $n=2$ and $d=3$ are ok, but what next?
Motivating the main question. Observe that $X_{0,n,d}$ is the Veronese itself, which is always an orbit (as long as $SL_n$ acts transitively on $\mathbb{P}V$). Observe also that the next $X_{i,n,d}$'s are constructed from $X_{0,n,d}$ in a "natural way" (e.g., by taking tangent lines).
Idea. Maybe there are more sophisticated "natural operations" allowing to "enlarge" the Veronese variety $X_{0,n,d}$ in such a way that the difference between subsequent "enlargements" is an orbit.
I call "natural enlargement" the construction of a larger invariant subset out of a smaller one, by means of natural objects associated to the latter (lines/subspaces with a certain tangency, (multi)secant subspaces, sections of natural bundles, etc.). 

MAIN QUESTION: Is there any other example of a decomposition of $\mathbb{P}^{M(n,d)}$ into invariant subsets, simliar to $(^*)$, but finer?

By "similar" I mean a formula like $(^*)$ where the osculating varieties are replaced by something else, and by "finer" I mean that the so-obtained  invariant subsets are smaller than the $X_{i,n,d}$'s.

REPLY [6 votes]: one can take the  secant varieties of the Veronse, where in general
$\sigma_r(X)=\overline{\langle x_1,...,x_r\rangle \mid x_j\in X}$, where
$\langle \rangle$ denotes linear span. These sets are $GL(V)$-invariant, as are the sets constructed by mixing your osculating construction and these (e.g. take $X$ the tangential variety of the Veronese instead of just the Veronese). The cases where they are all orbits are just $d=2$ any $n$ or
$d=3$ $n=2$. Another way to get invariant sets is to take the dual variety
of the Veronese, then its singular locus, then the singular locus of the singular locus etc.. and again one can mix and match constructions.<|endoftext|>
TITLE: Embedding abelian cancellative Hausdorff topological semigroups into abelian Hausdorff topological groups
QUESTION [8 upvotes]: An abelian cancellative semigroup embeds (via a semigroup monomorphism) into an abelian group. What about an abelian cancellative Hausdorff topological semigroup that does not embed (via a monomorphism of topological semigroups) into an abelian Hausdorff topological group? And what if the embedding is required to be a homeomorphism when corestricted to the image (endowed with the relative topology induced by the overlying topological group)? 
I'm aware of work of E. Schieferdecker (Math. Ann. 131, 1956), N. J. Rothman (Math. Ann. 139, 1960), and a few more authors on the subject, but couldn't find any counterexample in print.

REPLY [3 votes]: Clearly, there are examples for the second question. Each Hausdorff abelian paratopological group (that is, a group endowed with a topology making the multiplication continuous) which is not a topological group (for instance, the Sorgenfrey arrow, that is the real line endowed with the Sorgenfrey topology generated by the base consisting of half-intervals $[a,b)$, $a<b$) is a counterexample. Less trivial and locally compact counterexample (which necessarily is not a group, because each locally compact paratopological group is a topological group) should be the additive semigroup of  non-negative real numbers, endowed with the standard topology and then with isolated zero. (Maybe I even wrote a proof somewhere).
Update. A counterexample to your first question. For our purpose it is important that each Hausdorff space which admits a continuous injective map into a Hausdorff (para)topological group is functionally Hausdorff, because each Hausdorff topological group is Tychonoff (and each Hausdorff paratopological group is functionally Hausdorff). I recall that a space is  functionally Hausdorff if for each distinct points $x,y\in X$ there exists a continuous function $f:X\to[0;1]$ such that $f(x)=0$ and $f(y)=1$. Clearly, each functionally Hausdorff space is Hausdorff. Let $X$ be an arbitrary  Hausdorff space which is not functionally Hausdorff (see, for instance, quotations from “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989) added at the end of the answer).  Let $S(X)$ be a free abelian semigroup over a the space $X$, that is $S(X)$ is a set of formal sums $x=x_1+\cdots+x_n$ of elements of the set $X$. Clearly, $S(X)$ is a cancellative abelian semigroup. Determine a topology on the set $S(X)$ by its base consisting of all sets of the form $U_1+\cdots+U_n$, where $U_i$ are arbitrary open subsets of the space $X$ and $U_1+\cdots+U_n$ is a set of formal sums $x=x_1+\cdots+x_n$, where $x_i\in U_i$ for each $i$. Clearly, $S(X)$ is a topological semigroup. Since the space $X$ is naturally homeomorphic to a subspace of $S(X)$, the space $X$ is not functionally Hausdorff. It rests to show that the space $S(X)$ is Hausdorff. Let $x=x_1+\cdots+x_n$ and $y=y_1+\cdots+y_m$ be arbitrary distinct points of the set $S(X)$.  Let $\{z_1,\dots, z_k\}=\{x_1,\dots, z_n\}\cup\{y_1,\dots, y_m\}$. Since the space $X$ is Hausdorff, there exist mutually disjoint neighborhoods $U(z_i)\ni z_i$. Put $U(x)=U(x_1)+\cdots+U(x_n)$ and $U(y)=U(y_1)+\cdots+U(y_m)$. Define a (not necessarily continuous) homomorphism $h:S(X)\to S^0(X)$ (where $S^0(X)=S(X)\cup \{0\}$ is the semigroup $S(X)$ with an added zero element $0$ such that $0+x=x+0=x$ for each element $x\in S(X)$) on the set $X$ of generators of the semigroup $S(X)$ by putting $h(z)=z_i$ if $z\in U(z_i)$ for some $i$ and $h(z)=0$, otherwise. Since the map $h$ is homomorhism, $h(U(x))=\{x\}$ and $h(U(y))=\{y\}$, so the neighborhoods $U(x)$ and $U(y)$ are disjoint.<|endoftext|>
TITLE: Zero scheme of global sections of vector bundles on affine varieties
QUESTION [9 upvotes]: I want to understand better the notion of zero scheme of a section of a vector bundle. For simplicity I will consider the case of affine varieties.
Let $\mathbb{K}$ be an algebraically closed field, $V\subset\mathbb{K}^n$ be an affine algebraic variety (maybe reducible) and $A(V)=\mathbb{K}[x_1,...,x_n]/I(V)$ be its coordinate ring.
By Serre's theorem, taking global sections gives a bijective correspondence between isomorphism classes of algebraic vector bundles over $V$ and projective $A(V)$-modules of finite type.
So consider a projective $A(V)$-module $M$ of finite type, and let $s\in M$. Then we have:

the zero set of $s$, $$Z(s):=\{z\in V\ |\ s(z)=0\},$$ where $s(z)$ is the image of $s$ in the fiber $M(\mathfrak{m}_z)=M\otimes_{A(V)}k(\mathfrak{m}_z)$, $\mathfrak{m}_z\subset A(V)$ being the maximal ideal corresponding to $z\in V$ and $k(\mathfrak{m}_z)=A(V)/\mathfrak{m}_z$ being its residue field;
the ideal associated to $s$, $I_s:=\text{im}\, \iota_s$, where $\iota_s:M^\vee\rightarrow A(V)$ is the $A(V)$-linear map given by evaluation on $s$; this defines a (maybe nonreduced) closed subscheme of $V$.

I expect that $Z(s)$ coincides with the zero set of $I_s$, i.e. with $$Z(I_s):=\{z\in V\ |\ a(z)=0\ \forall a\in I_s\}.$$
How to see this (possibly without passing through the localizations of $A(V)$ and $M$)? Does everything make sense even when $M$ is not projective? Does the dual $M^\vee$ have some geometric interpretation in this context?

REPLY [4 votes]: Consider the following conditions:

$z\in Z(s)$
$s(z)=0$ 
$s\in\mathfrak{m}_zM$
the image of the map  $A(V)\stackrel{\varphi_s}{\longrightarrow}M$ defined by $1\mapsto s$, is in $\mathfrak{m}_zM$ 
$\forall f\in M^{\vee}$ we have $(f\circ\varphi_s)(1)\in\mathfrak{m}_z$ 
$\forall f\in M^{\vee}$ we have $ f(s)\in \mathfrak{m}_z$ 
$I_s\subseteq\mathfrak{m}_z$. 

Then it is clear that $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Rightarrow(5)\Leftrightarrow(6)\Leftrightarrow (7)$. Assuming $M$ is projective we will show $(6)\Rightarrow (3)$, which will achieve what you want to prove. Assume condition $(6)$ holds. If $s\not\in\mathfrak{m}_zM$, then $\overline{s}$, the image of $s$ in $M/\mathfrak{m}_zM$ is not zero. Since $M/\mathfrak{m}_zM$ is a finite-dimensional vector space over $A(V)/\mathfrak{m}_z$, one can define a linear map $\sigma\colon M/\mathfrak{m}_zM\rightarrow A(V)/\mathfrak{m}_z$ such that $\sigma(\overline{s})\neq0$. This $\sigma$ can also be considered as a map of $A(V)$-modules. Composing $\sigma$ with the map $M\rightarrow M/\mathfrak{m}_zM$ we get an $A(V)$-linear map $\overline{f}\colon M\rightarrow A(V)/\mathfrak{m}_z$ such that $\overline{f}(s)\neq0$. Since $M$ is projective, $\overline{f}$ can be lifted to an $A(V)$-linear map $f\colon M\rightarrow A(V)$ and we have $f(s)\not\in\mathfrak{m}_z$. This contradict with condition $(6)$, which we had assumed holding. 

P.S. If $M$ is not projective you will get $Z(s)\subseteq Z(I_s)$, but I don't know how one would get $Z(I_s)\subseteq Z(s)$ without using that $M$ is projective.<|endoftext|>
TITLE: Symmetric algebras of given dimension
QUESTION [6 upvotes]: Fix an algebraically closed field $F$. Are there only finitely many symmetric algebras with unit over $F$ of a given finite dimension (up to isomorphism)? By symmetric I mean a Frobenius algebra where the form is symmetric. For a general associative algebra I know it is false:
Are there only finitely many associative algebras of fixed dimension?
However, the example in the above thread does not give an infinite number of non-isomorphic symmetric algebras.

REPLY [8 votes]: The answer is certainly "no", but I can't immediately think of an example where it's easy to prove all the details.
In Karin Erdmann's (almost) determination of the structure of tame blocks, there are one-parameter families of algebras, where it's not known which parameters give algebras that actually occur as blocks of finite group algebras. I'm pretty sure those will give a counterexample.
Alternatively, "trivial extensions" give a construction of symmetric algebras. If $A$ is any finite-dimensional algebra, then $TA=A\oplus DA$, where $DA$ is the vector space dual of $A$ considered as an $A$-bimodule in the obvious way, and with multiplication
$$(a,\theta)(b,\phi)=(ab,a\phi+\theta b)$$
is symmetric, and although non-isomorphic algebras may have isomorphic trivial extensions, I'm sure that doesn't typically happen, and that you could could construct infinite families of non-isomorphic trivial extension algebras starting from infinite families of non-isomorphic non-symmetric algebras.
Edit: Actually, here's a fairly simple example in the same spirit as my answer to the question you linked to.
For $0\neq a\in F$, let 
$$S(a)=F\langle x,y,z\vert x^2=y^2=z^2=0,xy=ayx,yz=azy,zx=axz\rangle.$$
So $S(a)$ is an $8$-dimensional algebra with basis $\{1,x,y,z,xy,yz,zx,xyz\}$, and $\operatorname{rad}^iS(a)$ is spanned by the basis elements of degree at least $i$ in $\{x,y,z\}$. 
It is symmetric, with a symmetrizing form that kills all the basis elements except $xyz$.
Unless $a=-1$, the elements $w$ of $\operatorname{rad}S(a)/\operatorname{rad}^2S(a)$ such that $w^2=0$ in $\operatorname{rad}^2S(a)/\operatorname{rad}^3S(a)$ are precisely the scalar multiples of $x$, $y$ and $z$. Choosing two linearly independent such elements $w_1,w_2$, either $w_1w_2=aw_2w_1$ or $w_2w_1=aw_1w_2$ in $\operatorname{rad}^2S(a)/\operatorname{rad}^3S(a)$. So $S(a)\not\cong S(b)$ unless $a=b$ or $a=b^{-1}$.<|endoftext|>
TITLE: What are some very important papers published in non-top journals?
QUESTION [121 upvotes]: There has already been a question about important papers that were initially rejected. Many of the answers were very interesting. The question is here.
My concern in this question is slightly different. In the course of a discussion I am having, the question has come up of the extent to which the perceived quality of a journal is a good reflection of the quality of its papers. The suggestion has been made that because authors tend to submit their best work to the best journals, that makes it easy for those journals to select papers that are on average of a high standard, but it doesn't necessarily solve the reverse problem -- that they miss other papers that are also very important. (Note that the situation more generally in science is different, because there is a tendency for prestigious journals to value papers that make exciting claims, and not to check too hard that those claims are actually correct. So there one has errors of Type I and Type II, so to speak.) 
I am therefore interested to know of examples of papers that are very important, but are published in middle-ranking journals. I am more interested in recent papers than in historical examples, since it is the current journal system that we are discussing.
Just in case it doesn't go without saying, please do not nominate a paper that you yourself have written...

REPLY [2 votes]: There are a couple of examples already of papers only "published" on the arXiv; here is another one.
"Total positivity, Grassmannians, and networks" by Alexander Postnikov https://arxiv.org/abs/math/0609764, preprint from 2006.
In this paper Postnikov introduces the positroid cell decomposition of the totally nonnegative Grassmannian. The totally nonnegative Grassmannian and its positroid stratification have become an important topic in many diverse areas, from cluter algebras to scattering amplitudes in physics. In particular, the main combinatorial tool introduced by Postnikov here, namely, plabic graphs, has been applied to everything from soliton solutions of integrable systems (https://arxiv.org/abs/1105.4170) to knot theory and sympletic geometry (https://arxiv.org/abs/1512.08942). (Incidentally, for a textbook treatment of the theory of plabic graphs, see Chapter 7 - https://arxiv.org/abs/2106.02160 - of the book-in-progress on cluster algebras by Fomin, Williams, and Zelevinsky.)
According to Google scholar this paper has been cited over 500 times, which is pretty good for 15 years. I am not totally sure why this paper was never published (even though, full disclaimer, Postnikov was my PhD adviser). But it is my understanding that it was passed around as a manuscript many years before being put on the arXiv, and he was asked to put it on the arXiv and make it publicly available before his tenure review. Considering its impact, I certainly think it could have been published in a top journal.<|endoftext|>
TITLE: Is there a (first-order) sentence which admits $(\aleph_2,\aleph_0)$ iff a Kurepa tree exists?
QUESTION [10 upvotes]: In Chang and Keisler's Model Theory I came across the following theorem (Theorem 7.2.13):
Theorem There exists a (first-order) sentence $\sigma$ such that for all infinite cardinals $\alpha$, $\sigma$ admits $(\alpha^{++},\alpha)$ iff there exists an $\alpha$-Kurepa tree.
Definitions 
(1) Fix a predicate $P$ in the language of $\sigma$. Then $\sigma$ admits $(\kappa,\lambda)$ if there is a model $M$ of $\sigma$ with $|M|=\kappa$ and $|P^M|=\lambda$. Obviously, $\kappa\ge\lambda$.
(2) An $\alpha$-Kurepa tree has height $\alpha^+$, each level has at most $\alpha$ elements and there are at least $\alpha^{++}$ elements of height $\alpha^+$.
Although $\sigma$ is not explicitly defined in Chang and Keisler, I am assuming it says something like:
There is a predicate $P$, there is a surjection from $P$ to each level, the levels of the tree are linearly ordered, and there is a surjection from $P$ to every initial segment of this linear order.
I see this sentence works under GCH, but I have a hard time when GCH fails. 
Consider for instance the case when $\alpha=\aleph_0$ and, say $2^{\aleph_0}=\aleph_2$. Then we can cook up a model of the above sentence that resembles $(\omega^{<\omega},\subset)$ with $\omega^\omega$ many cofinal branches. This model will be have type $(\aleph_2,\aleph_0)$. 
By a theorem of 
Devlin, we can find a model of ZFC where there is no Kurepa tree and the continuum is $\aleph_2$. Thus, it is consistent that $\sigma$ will have models of type $(\aleph_2,\aleph_0)$, while there are no Kurepa trees.
My question: Do I miss something "obvious", or is the above theorem true only under GCH?

REPLY [5 votes]: The problem, as I see it, is to build into the sentence $\sigma$ something that makes a certain definable set $A$ (in this case a set indexing the levels of your tree) have cardinality $\aleph_1$.  You've already ensured that $A$ isn't any bigger than $\aleph_1$ by having surjections from $P$ to all of the initial segments of $A$.  I claim the same idea can be used to ensure that $A$ isn't any smaller than $\aleph_1$, as follows.  Since the universe of your model is guaranteed to have cardinality $\aleph_2$, you can ensure that $A$ is big enough by having a linear ordering of the universe and requiring surjections from $A$ onto all the initial segments of this ordering.  The point is that there is no linear ordering of cardinality $\aleph_2$ in which all proper initial segments are countable.  (Here, and also in the $\sigma$ that you described in the question, "initial segment" should mean a set of the form $\{x:x\leq a\}$ for some $a$, not just a downward-closed set.)<|endoftext|>
TITLE: Not sure how to fix an error in the Handbook of Categorical Algebra (vol 3)
QUESTION [5 upvotes]: In the proof of Theorem 1.5.7 (in which it is shown the the nuclei on a locale, when ordered by pointwise partial order, themselves form a locale) in the computation at the bottom of p.35, there is used an inequality
$$ j\Bigl( \bigl( j(b)\Rightarrow k(b)\bigr)\wedge b\Bigr) \Rightarrow k(b)  ~~\leq~~ j(b)\Rightarrow k(b)$$
however since $j(bla\wedge b) \leq j(b)$ and $- \Rightarrow k(b)$ is contravariant, I don't think that that's true.
I'm not sure what to do to avoid using this.

REPLY [11 votes]: You have a point that the last step is unclear. It might have been clearer if the last inequality were written as an equality instead, since that would essentially force the reader to verify that $b = (j(b) \Rightarrow k(b)) \wedge b$, or simply that $b \leq j(b) \Rightarrow k(b)$, which is true since it is equivalent to $b \wedge j(b) \leq k(b)$, which follows from $b \leq k(b)$.<|endoftext|>
TITLE: Gluing two 3 manifolds along their boundary
QUESTION [5 upvotes]: Let $X,Y$ be two compact, smooth, orientable 3 manifolds, each with an incompressible  boundary component diffeomorphic to some genus $g $ surface $S_g$. Under an orientation-reversig diffeomorphism $f:S_g \to S_g$, those two manifolds can be glued together to obtain a new smooth, orientable manifold $X \cup_f Y$. I wonder now in how far the diffeomorphism type of this result depends on the choice of $f $. Using a collar, one can show that if $f $ and $g $ are two isotopic diffeomorphisms of $S_g $, then the corresponding gluings are diffeomorphic . Is this also a necessary condition ? 
Some thoughts I have made so far: 
1) If both X and Y are irreducible, then so is $X \cup_f Y$, and if $X \cup_f Y $ still has some boundary component, it must be an aspherical manifold. Hence, all important Information is contained in the fundamental group. Is it true that the homeomorphism type of aspherical 3 - manifolds obtained this way is already determined by their fundamental group ?
2) Also, I wonder if its true that the isotopy type of two orientation-reversing diffeomorphisms on a surface $S_g$ is determined by their action on the fundamental group. 
Edit: This is probably false, since mapping class groups of surfaces are very distinct from the corresponding fundamental group.
Any help is appreciated. 
Edit: I apologize for missing and/or inaccurately placed capital letters. I wrote this question yesterday evening on my phone, and it was impossible to fix all the mistakes made by auto-correct (I am writing on a german phone).
Edit 2: I have also updated this question, according to what has already been solved by the answers and what is still open.

REPLY [4 votes]: The original question: 

Using a collar, one can show that if $f$ and $g$ are two isotopic diffeomorphisms of $S_g$, then the corresponding gluings are diffeomorphic. Is this also a necessary condition? 

No, it is not.  Suppose we have glued to obtain $M = X \cup_f Y$.  Suppose that $X$ admits a self-homeomorphism $\Phi$.  Define $\phi = \Phi|\partial X$.  Then the map $g = \phi \circ f$ gives the manifold $N = X \cup_g Y$ and this is homeomorphic to $M$.  
Now, it is simple to find such $\Phi$ if $X$ has compressible boundary - namely we can do a Dehn twist on a disk. You've ruled that out.  But we can still find examples by twisting along an essential properly embedded annulus in $X$.  For example, if $X$ is a twisted $I$-bundle over a non-orientable surface.  
If you further assume that $X$ is "acylindrical" then there are still examples, but they are harder to find.  We can build a hyperbolic manifold $X$ which has a self-homeomorphism $\Phi$ of finite order (eg Thurston's knotted Y).
If you further assume that $X$ (and $Y$) has no symmetries, then examples should still exist, but they will be very hard to find.  Basically, we need to find a manifold $M$ that contains homeomorphic surfaces $S$ and $S'$, but where there is no homeomorphism of $M$ taking $S$ to $S'$.  We then need to "get lucky" and find that $M - n(S)$ and $M - n(S')$ are homeomorphic and win.  One way to do this is by a search through one of the many censuses of closed three-manifolds (eg snappy or regina).  Another way that should work is to think deeply about hyperbolic three-manifolds with "corners".<|endoftext|>
TITLE: Homology 3-sphere with a unique Stein-fillable contact structure
QUESTION [7 upvotes]: Are there any known examples of oriented integer homology 3-spheres $Y$ (besides $S^3$) which have exactly one Stein-fillable contact structure up to isotopy? Failing that, what are the known examples with the smallest number of Stein-fillable contact structures where one exists? Note here I am only counting Stein-fillable contact structures not other contact structures on $Y$.
There are situations where I'd like to be able to compute bounds on the maximum of the adjunction number $ad(K)=tb(K)-1+|r(K)|$ for $K$ a knot in $Y$ allowing both the Legendrian representative of $K$ to vary as well as the contact structure. I was wondering if there were cases where this was at all tractable. I have not spent much thought on the classification of contact structures, so I do not know how hard this should be.

REPLY [4 votes]: One example you might want to start with is the Poincaré homology sphere. It is a Seifert fibred space over $S^2$ with three singular fibres $M(\frac{1}{2}, - \frac{1}{3}, -\frac{1}{5})$ supporting exactly one Stein fillable contact structure. Note that the the Poincaré homology sphere with reversed orientation $\overline{\Sigma}(2,3,5)$ was the first example of a three-manifold with no positive tight contact structure (Etynre and Honda, 1999). 
I'm not sure if a proof ever appeared in the literature, but the fact that the Poincaré homology sphere supports one Stein fillable contact structure is not hard to prove using the techniques from Etnyre and Honda. A proof follows exactly the lines further examples were classified, which gives a second example: 
The Brieskorn homology sphere $\overline{\Sigma}(2,3,11)$ supports exactly one Stein fillable contact structure. Again, this is a Seifert fibred space over $S^2$ with three singular fibres $M(-\frac{1}{2}, \frac{1}{3}, \frac{2}{11})$. The proof of this was written up by Paolo Ghiggini in  2003. 
As Marco Golla remarks, you might be able to come up with many more examples from the classification results on Seifert fibred spaces by Ghiggini, Lisca, Stipsicz, Wu. I hope the two concrete examples are of any help for you.<|endoftext|>
TITLE: Extending a left fibration along an inner horn
QUESTION [11 upvotes]: Let $\Lambda^n_i \subseteq \Delta^n$ be an inner horn, and let $X \rightarrow \Lambda^n_i$ be a left fibration. Does there exist a left fibration $Y \rightarrow \Delta^n$ such that $X = Y \times_{\Delta^n} \Lambda^n_i$?
The only results in that direction that I am aware of are the following two lemmas from Left fibrations and homotopy colimits by Heuts, Moerdijk:

Lemma 7.3. Consider a pullback square of simplicial sets 
  $\require{AMScd}$
  \begin{CD}
    X \times_Y Z @>g>> Z\\
    @VVV @VV p V\\
    X @>>f> Y
\end{CD}
  in which $f$ is inner anodyne and $p$ is a left fibration. Then $g$ is a trivial cofibration in the Joyal model structure.
Lemma 7.4. Let $0 < k < n$ and let $p : A \rightarrow \Lambda^n_k$ be a left fibration. Then there exists a left fibration $q : B \rightarrow \Delta^n$ and an equivalence
  \begin{CD}
    A @>g>> \Lambda^n_k \times_{\Delta^n} B\\
    @VpVV @VVV\\
    \Lambda^n_k @= \Lambda^n_k
\end{CD}
  in the covariant model structure over $\Lambda^n_k$.

REPLY [6 votes]: Yes. This is shown using minimal left fibrations in Cisinski's book Higher categories and homotopical algebra, see the proof of Theorem 5.2.10 therein. This theorem states that the simplicial set $\mathscr{S}$, whose $n$-simplices are (essentially) the left fibrations over $\Delta^n$ with small fibres, is a quasi-category.<|endoftext|>
TITLE: Two (new?) variants of convex functions
QUESTION [6 upvotes]: I find that the following two types of functions are useful to my research.
(i) We know that a function $f: \mathbb{R}_+^m\rightarrow \mathbb{R}$ is called convex  if  for all ${\bf x,y}\in \mathbb{R}_+^m$ and all $\lambda_1,\lambda_2\geq 0$ s.t. $\lambda_1+\lambda_2=1$, it holds that  \begin{equation}f(\lambda_1{\bf x}+\lambda_2 {\bf y})\leq \lambda_1f({\bf x})+\lambda_2f({\bf y}).\end{equation}
The first type of function is to drop the restriction of $\lambda_1+\lambda_2=1$, i.e. to replace convex combination with conic combination.
(ii) The second type of function: $f({\bf 0})=0$, and for all ${\bf x},{\bf y},{\bf z}\in \mathbb{R}_+^m$ such that ${\bf y}\geq {\bf x}$, it holds that \begin{equation}f({\bf y})-f({\bf x})\leq f({\bf y}+{\bf z})-f({\bf x}+{\bf z}).\end{equation}
Could anyone tell me whether they are new or not? Are there any useful references? Many thanks!

REPLY [5 votes]: If you throw in positive homogeneity, then the first class of functions is what is called sublinear, see for instance Proposition 1.1.4 ("Fundamentals of Convex Analysis"; Hiriart-Urruty, Claude Lemaréchal).
The second class of functions that you mention is essentially the same as the Wright convex functions, see for instance pg. 223 of A.W. Roberts, D.E. Varberg, "Convex functions", Academic Press, 1973.<|endoftext|>
TITLE: Examples of abstractions that did *not* turn out to be useful
QUESTION [11 upvotes]: I’ve read (but cannot find any reference now) that new abstract mathematical concepts like set theory and – not too long ago – category theory were in their time often considered too abstract to be useful.
But now, all of them have turned out to be useful. What are examples of abstractions that were once somewhat popular, but have in fact not turned out be useful (or even turned out to be not useful)?

REPLY [5 votes]: I don't have an opinion of my own about Fuzzy Set Theory, but someone whom I respect, Saunders Mac Lane, seemed to think it wasn't a very fruitful development. (I freely acknowledge he could be opinionated at times, and he may have had his blind spots.) He mentions this example in his book Mathematics: Form and Function. 
I invite others to explain to me what solutions to problems (or insightful points of view on problems) Fuzzy Set Theory has enabled, in the spirit of the proposed working definition of useful theory: something used to solve an existing problem that was posed before the theory was invented and that recordedly has withstood at least one previous attempt not using the new theory. Of course, fuzzy set theory may itself be a 'fuzzy' or at least broad term, in that it is cousin to other forms of set theory which do have their purpose, such as Heyting- or Boolean-valued set theory.<|endoftext|>
TITLE: Automorphisms of Eilenberg-Mac Lane spaces and semidirect products (and the odd line)
QUESTION [14 upvotes]: If $A$ is an abelian group, we have
$Aut\left(K\left(A,n\right)\right)=Aut(A) \ltimes K\left(A,n\right),$
where the left hand side is the space of self-homotopy equivalences. Is there an easy way to see this abstractly rather than by computation?
Also, it seems there is an analogy with the following: if $\mathbb{R}^{0|1}$ is the odd-line (regarded as a supermanifold),
$Aut\left(\mathbb{R}^{0|1}\right)=\mathbb{R}^\times \ltimes \mathbb{R}^{0|1}$
The vague analogy is that $\mathbb{R}^{0|1}=\mathbb{R}[1]$, and under Dold-Kan, $K\left(A,n\right)=A\left[-n\right].$
It seems to me there is unifying categorical principle at play. Can anyone shed some light on it? Thanks!
Edit: I have accepted the answer below as it does answer my first question. However I am still interested in seeing if there is any general principal at play.

REPLY [10 votes]: As per Qiaochu Yuan's comment we need to only understand the space of based maps between $K(A,n)$ with a chosen base point.
The loop-deloop pair of functors establish an equivalence between the categories of $A^\infty$-groups and connected spaces with a base point:
$$\Omega: \mathrm{Top}_{*,\ \pi_0=0} \simeq \mathrm{Grp}: \mathbb{B}$$
More generally, $\Omega^n$ and $\mathbb{B}^n$ give an equivalence between $(n-1)$-connected basepointed spaces and $E_n$-groups. This means that 
$$\mathrm{Top}_*(K(A, n), K(A, n)) = \mathrm{Grp}_{E_n} (A, A) $$
But $A$ is discrete, so all higher coherence data vanishes and a map of $E_n$-groups is the same as a map of set-theoretic groups, thus the answer.
I don't think your second statement is formally analogous since it mixes up the additive and multiplicative groups and I can't think of any inductive construction that links them, but who knows.
EDIT: Regarding the semidirect product. I don't think it is actually true.
A group $G$ is a semidirect product of groups $H \rtimes F$ iff there is a split exact sequence
$$0 \to H \to G \stackrel{\leftarrow}{\to} F \to 0$$
If X is a connected space with a chosen point $*\in X$ for the mapping space $\mathrm{Top}(X, X)$ we have in general a fiber sequence of spaces $$ \mathrm{Top}_* (X,X) \to \mathrm{Top}(X,X) \to X$$
The right arrow maps $f: X\to X$ to $f(*)$. There is the same sequence if $\mathrm{Top}(\cdot,\cdot)$ will denote invertible or homotopy invertible maps. Lets assume in the following that we consider invertible maps and all relevant spaces are topological spaces (we do not lose any generality since any homotopy fiber sequence is equivalent to a topological fibration and any homotopy group is equivalent to a topological group). Note that the left arrow in the sequence is always an inclusion of subgroup and the right arrow is the coset space of the action of based maps on all maps. If $X$ is a group $G$ (in particular, $G = K(A,n)$), then the right map has a section which maps an element $g: G$ to the shift operator $t_g : h: G \mapsto gh: G$ and this is a homomorphism of topological groups. Any $f: Aut(G)$ also can be uniquely represented as $f= t_g \circ \hat f$, where $g := f(*)$ and $\hat f : Aut_*(G)$.
On the level of spaces we thus certainly have that $Aut(G) = Aut_*(G) \times G$. However, while it looks like a semidirect product in fact it isn't, since neither $Aut_*(G)$ nor $G$ are normal subgroups in $Aut(G)$. While we have a factor map $Aut(G) \xrightarrow{/Aut_*(G)} G$ it is not a homomorphism of groups. If we represent some element of $f: Aut(G)$ as $t_g \cdot \hat f$, then the product of $f$ and $f^\prime$ looks like $$t_g \cdot \hat f \cdot t_{g^\prime} \cdot \hat f^\prime = t_g t_{g^\prime} \left( t_{g^\prime}^{-1} \hat f t_{g^\prime} \right) \hat f^\prime$$ which looks like a semidirect product but isn't, since conjugation by $t_{g^\prime}$ maps $Aut_*(G)$ to $Aut_{g^{\prime -1}}(G)$.
EDIT 2: As Achim Krause noted below, we actually have a homomorphism  $Aut (K (A,n)) \to Aut (A) $ given by the action on n-th homology, thus the stated semidirect product decomposition  $Aut (A) \ltimes K (A, n) $ is valid.<|endoftext|>
TITLE: Alternate proofs of Hilberts Basis Theorem
QUESTION [17 upvotes]: I'm interested in proofs using ideas from outside commutative algebra of Hilbert's Basis Theorem.

If $R$ is a noetherian ring, then so is $R[X]$.

or its sister version

If $R$ is a noetherian ring, then so is $R[[X]]$.

I am very much aware of the standard non-construtive proof by contradiction given by Hilbert as well as the direct version using Groebner basis.
Most important theorems in mathematics that are old enough have several very different proofs. Comparing different ideas can be very enlightening and also give a hint to possible generalizations in different areas. For the Basis Theorem however, I am not aware of such.

REPLY [3 votes]: There is a proof of the theorem for $R[[x]]$ that uses the well-known result of I.S. Cohen that a ring is noetherian if and only if its prime ideals are finitely generated.  Such a proof is given by Kaplansky in his 1970 book Commutative Rings, Theorem 70.  
This method of proof can be generalized to noncommutative rings to show that the power series ring $R[[x]]$ over a left noetherian ring is left noetherian, as long as one is careful when putting forth a definition of "prime left ideal."  Gerhard Michler gave such a proof in a paper that's not so easy to locate (but it seems to be described here), and I gave a similar proof in Theorem 3.9 in this paper.<|endoftext|>
TITLE: Why are integer points on elliptic curves interesting and useful?
QUESTION [7 upvotes]: I read some papers which dealed with integer points on elliptic curves. One of these papers are
http://projecteuclid.org/euclid.rmjm/1214947612.
My question is: Why are integer points on elliptic curves useful? One thing I could imagine: 
If there is a method which tells you that a give set of integer points:
$P_1,\dots,P_l$ is independent (in terms of generation). So this would give a lower bound for the rank.
Is this possible? Even for special cases?
If not, is there anything else useful?

REPLY [12 votes]: To answer the second part of your question. there is a conjecture of Serge Lang which says that the number of integer points (on a minimal Weierstrass equation) is bounded solely in terms of the rank. The conjecture is known to be true if one assumes the $ABC$-conjecture, in the stronger form
$$ \#E(\mathbb Z) \le C^{1+\operatorname{rank}E(\mathbb Q)} $$
for an absolute constant $C$. So a sequence of curves with $\#E_i(\mathbb Z)\to\infty$ would (conjecturally) give a sequence with unbounded rank. However, there are those who believe that there should be an absolute upper bound for $\#E(\mathbb Z)$, and recently some who believe that there should be an absolute upper bound for $\operatorname{rank}E(\mathbb Q)$. Anyway, lots of interesting open questions.
For the first part of your question, there are many natural problems that end up coming down to integer points on elliptic curves. Just as there are many natural problems that come down to rational points on curves of genus $g\ge2$. But I'm not sure that's the right answer to your question. For me, it comes down to (1) integers and their relationships to one another are interesting, (2) the basic operations on the integers are addition and multiplication, so the basic functions are polynomial functions, (3) hence Diophantine equations (polynomial equations to be solved in integers) are intrinsically interesting. If you'll grant that, then people eventually realized that the "right" way to classify equations of the form $f(x,y)=0$ was by genus, the first case ($g=0$) is now well understood, and the second case ($g=1$) still presents challenges.<|endoftext|>
TITLE: Smallest eigenvalue gap of a non-symmetric random matrix
QUESTION [6 upvotes]: The question: Let $A$ be the matrix whose each element is an independently generated random variable which is uniform on $[0,1]$. One can see that the eigenvalues of $A$ will be distinct almost surely. Let $\delta = \min_{i,j} |\lambda_i(A) - \lambda_j(A)|$ be the smallest distance between any two eigenvalues of $A$ in the complex plane. What sort of lower bounds does $\delta$ satisfy with high probability? 
Background: looking at the literature, it seems that for some symmetric random matrices $A$ the question has been settled, in the sense that the exact distribution of $\delta$ is known. This does not, however, seem to apply to the non-symmetric case above, and it seems reasonable to guess that the exact distribution of $\delta$ for the random matrix I am asking about is an open question. 
However, here I am asking for something much less than an exact distribution of $\delta$ -- for example, can one make a statement along the lines of $P(\delta \geq 1/n^{10}) \geq 1-1/n^{0.1}$, or something like that?

REPLY [4 votes]: This paper treats the case of iid complex Gaussian entries:
http://arxiv.org/abs/1207.4240
In particular, the smallest eigenvalue gap is shown to be of order $n^{-3/4}$ when the entries are normalized to have variance $1/n$. One might expect this to extend to more general entry distributions, but as far as I am aware this is unknown. 
Edit: As to whether one can show $P(\delta\ge n^{-C})\rightarrow 1$ for some $C>0$, for general (possibly discrete) entry distributions it seems difficult to even show $P(\delta> 0)\rightarrow 1$. However, as is the case for controlling the smallest singular value, it may be easier to lower bound $\delta$ when the entries have a bounded density (perhaps this is why you posed the question for the uniform distribution on $[0,1]$).<|endoftext|>
TITLE: Efficient SVD of a matrix without some of the columns
QUESTION [8 upvotes]: I have a matrix $A \in \mathbb{R}^{p \times q}$ of rank $r$ and its SVD decomposition, i.e,
$$
A = U S V^\top,
$$
where $U \in \mathbb{R}^{p \times r}$ and $V \in \mathbb{R}^{q \times r}$ are orthonormal, $S = \mathrm{diag}\left(s_1, \dots, s_r\right)$, and $s_i \in \mathbb{R}$. I would like to compute SVD for $B = [A_{i_1}, \dots, A_{i_k}] \in \mathbb{R}^{p \times k}$, and $k \leq r$, i.e., for a matrix that consists of a full rank selection of columns from $A$.
If SVD is straightforwardly applied to $B$, then the time complexity would be $\mathcal{O}(pk\min(p,k))$. Can I do that more efficiently?
Note:

If you consider the the right singular vectors of $A$, column selection effectively leads to projecting these to a subspace in the canonical basis. Projection does not preserve the angles between the vectors, and hence one cannot do better than just re-doing the SVD on $B$.
However, does there exist a rotation of the $V$ singular vectors that preserves the angles and it could be done more efficiently than just running SVD on $B$?

REPLY [4 votes]: After some research, I managed to find a reasonable answer. The operation is called updating (in case of adding new columns to the original matrix) or donwdating (in case of removing) of the SVD. Full update/downdate of SVD with a single column can be done in $\mathcal{O}\left(r^2(1 + p + q)\right)$ time [1].
Such modifications of SVD can be also parallelized.
Here are a few links to papers that deal with this problem:

Fast low-rank modifications of the thin singular value decomposition
A stable and fast algorithm for updating the singular value decomposition
A fast and stable algorithm for downdating the singular value decomposition<|endoftext|>
TITLE: Is there a modification of Martin's Axiom which admits non-measurable sets of size less than continuum?
QUESTION [9 upvotes]: Background/Motivation. We know that some of the usefulness of Martin's Axiom lies in giving certain "smallness" properties to sets of size less than continuum, e.g. we have
that for all infinite cardinals $ \lambda < 2^{\aleph_0}, 2^{\lambda} = 2^{\aleph_0}$; also all sets of reals of cardinality less than $2^{\aleph_0}$ are Lebesgue measure 
zero sets.
But I note at the end of the article "Between Martin's Axiom and Souslin's Hypothesis" (by K. Kunen and F. Tall) that the topological characterization of 
MA restricted to compact hereditarily separable spaces
is consistent with having $\aleph_1 < 2^{\aleph_0} < 2^{\aleph_1}$.   Which leads me to wonder...
Question: Is there any known restriction/modification of MA ( + not-CH, of course) which is consistent with existence of a non-Lebesgue measurable set of reals of 
cardinality less than $2^{\aleph_0}$?
(The proof of small sets being Lebesgue null (assuming MA) in Jech's set theory book seems so simple and compelling, it would make an affirmative answer rather intriguing.)

REPLY [5 votes]: Adding on to Andreas's comment above:
In Section 3 of Chapter XVIII of "Proper and Improper forcing", Shelah proves a very general preservation theorem.  "Application 3.8" on page 912 deals with preserving positive outer measure, and in Claim 3.8C(2) he pins down the (very technical) condition he needs to prove that a countable support iteration preserves the property of "being of positive outer measure".  
This is an extension of the Judah-Shelah result mentioned by Blass, as in the paper 
The Kunen-Miller chart (Lebesgue measure, the Baire property, Laver reals and preservation theorems for forcing) J. Symbolic Logic 55 (1990), no. 3, 909–927,
Judah and Shelah prove an iteration theorem strong enough to prove that the ground model reals have positive outer measure after an $\omega_2$-length iteration of Laver forcing.  That particular situation is covered by the more general theorem from Proper and Improper Forcing.<|endoftext|>
TITLE: 1st Chern class is invariant under choice of section?
QUESTION [5 upvotes]: How do I see that the 1st Chern class is invariant under choice of section? I know metric invariance follows from how two metrics on line bundle have to be conformally equivalent, but how do we show invariance under choice of section? Thanks in advance.

REPLY [6 votes]: There are many definitions of  the  $1$-Chern class of a complex line bundle $L\to M$, $M$ compact $CW$-complex.    The topological one goes as follows.  The line bundle $L$  is an oriented rank $2$ real vector bundle over $M$. As such it has a Thom class $\tau_L\in H^2(D(L), S(L))$, where $D(L)$ and $S(L)$ are  the disk and respectively the unit sphere bundle of $L$.  If $\zeta:M\to L$ is the zero section, then the  first Chern class of $L$ coincides with the Euler class of $L$  defined as the pullback 
$$  e(L):=\zeta^*\tau_L\in H^2(M). $$
If additionally $M$  is an oriented $m$-dimensional manifold, then the Poincare dual is a homology class $e(L)^*\in H_{m-2}(M)$.  This class is also represented  by the zero set of a generic section of $L$; see  Chapter 4  of these notes for details and proofs.<|endoftext|>
TITLE: How to calculate expected value of matrix norms of $A^TA$?
QUESTION [8 upvotes]: Let $A$ be a random $m$ by $n$ rectangular sign matrix, chosen uniformly at random, with $m < n$. Let $B = A^T A$.  We know, for example, that $B$ is a square and symmetric $n$ by $n$ matrix with all its  diagonal entries equal to $m$ exactly.   I am trying to work out how to calculate (or estimate) the expected Frobenius and spectral norm  of $B$. We can assume both $m$ and $n$ are large.

How can you calculate $\mathbb{E}(||B||_F)$ and
  $\mathbb{E}(||B||_2)$?

The expected Frobenius norm of $B$ is defined to be
$$
\mathbb{E}(||B||_F)=\mathbb{E}\left(\sqrt{\sum_{i=1}^n\sum_{j=1}^n |b_{ij}|^2}\right)
$$
where $b_{ij}$ are the elements of $B$.
The expected spectral norm of $B$ is defined to be
$$
\mathbb{E}(||B||_2)= \mathbb{E}\left(\max_{|x|_2 \ne 0}\frac{|Bx|_2}{|x|_2}\right).
$$
I apologize if this turns out to be simple to do. I previously asked at https://math.stackexchange.com/questions/1517103/how-to-calculate-expected-value-of-matrix-norms-of-ata with no answer.

REPLY [6 votes]: Expanding a bit on Yemon Choi's comment: concentration is indeed the key. First,
$\|B\|_F^2$ is simply the sum of the squares of singular values of $A$, and 
$E\|B\|_F^2=m(m-1)n+mn^2$. On the other hand by standard concentration inequalities (using that $\|B\|_F^2=\sum \sigma_i^4$ where $\sigma_i$ are the singular values of $A$), you obtain an exponential (in fractional power of m) concentration around the mean, which is good enough to also conclude that 
$E\|B\|_F/\sqrt{m^2n+mn^2}\to 1$ (as both $m$ and $n$ grow)..
For the spectral norm, one can argue similarly: the top singular value also concentrates around its mean, and converges to the edge of the spectrum of the pastur-marchenko law. This is explained in great detail in Bai and Silverstein's book on spectral analysis of random matrices.<|endoftext|>
TITLE: What's wrong with my understanding of the scheme $\text{Isom}(E_\lambda, E_{\lambda'})$?
QUESTION [12 upvotes]: Let $\mathcal{M}_{1,1}$ be the moduli stack of elliptic curves (over the complex numbers). Define
$$\begin{eqnarray*} X &:=& \Bbb{A}^1_{\lambda} - \{0,1\}\\
X' &=& \Bbb{A}^1_{\lambda'} - \{0,1\}.\end{eqnarray*} $$
There are morphisms $X \to \mathcal{M}_{1,1}$ and  $X' \to \mathcal{M}_{1,1}$ given by  the families of curves
$$\begin{eqnarray*} E_\lambda := V(y^2 - x(x-1)(x-\lambda)) \\
E_{\lambda' }:= V(y^2 - x(x-1)(x-\lambda')).
\end{eqnarray*}$$
By results of Grothendieck, we know that the fiber product $\text{Isom}(E_{\lambda}, E_{\lambda'}) :=  X \times_{\mathcal{M}_{1,1}} X'$
is a scheme. Its $T$-points are given by 
$$\text{Isom}(E_{\lambda}, E_{\lambda'})(T) = (T \to X ,T \to X',  {E_{\lambda}}_T \stackrel{\simeq}{\to} {E_{\lambda'}}_T   ) .$$
The isomorphism above between ${E_{\lambda}}_T$ and ${E_{\lambda'}}_T$ is a $T$-isomorphism.


My goal  is to try and understand why $X \to \mathcal{M}_{1,1}$ is not étale. To do this, it is enough to show that $\text{Isom}(E_{\lambda},E_{\lambda'}) \to X$ is not \'{e}tale.


Since the automorphisms of any elliptic curve in Legendre form are given by $y \mapsto cy$ and $x \mapsto ax +b$, I can see that the scheme $\text{Isom}(E_{\lambda},E_{\lambda'})$ is given by the following conditions in $\Bbb{A}^5:$ 
$$\text{Isom}(E_{\lambda},E_{\lambda'}) = \operatorname{Spec} \frac{\Bbb{C}[\lambda, \frac{1}{\lambda}, \frac{1}{1-\lambda},\lambda', \frac{1}{\lambda'}, \frac{1}{1-\lambda'},a,\frac{1}{a},b,c]}{(j(\lambda) - j(\lambda'), f_1,f_2,f_3,f_4)}. $$
The polynomials $f_1,f_2,f_3,f_4$ are obtained from equating the coefficients of the relation
$$ x(x-1)(x-\lambda') = \frac{(ax+b)(ax+b-1)(ax+b-\lambda)}{c^2}.$$
Explicitly, they are given by:
$$\begin{eqnarray*}
f_1 &=& a^3 - c^2 \\
f_2 &=& 3a^2b - a^2 \lambda - a^2 + a^3(\lambda' + 1) \\
f_3 &=& 3ab^2 - 2ab\lambda - 2ab + a\lambda -a^3\lambda'\\
f_4 &=& b^3 - b^2\lambda - b^2 + b\lambda.
\end{eqnarray*} $$
Now if I compute the fiber of the map $\text{Isom}(E_{\lambda}, E_{\lambda'}) \to X$ over the $\lambda = -1$ ($j = 1728$), I get the non-reduced scheme
$$\text{Isom}(E_{\lambda}, E_{\lambda'})_{-1} = \operatorname{Spec} \frac{\Bbb{C}[\lambda', \frac{1}{\lambda'}, \frac{1}{1-\lambda'},a,\frac{1}{a}, b,c]}{ \left((2 \lambda'-1)^2 (\lambda'+1)^2 (\lambda'-2)^2,f_1,f_2',f_3',f_4'\right)}$$
where
$$\begin{eqnarray*}
f_1 &=& a^3 - c^2 \\
f_2' &=& 3b +a (\lambda'+1) \\
f_3' &=& 3b^2 - a^2\lambda' - 1\\
f_4' &=& b^3 - b.
\end{eqnarray*} $$
Hence $X \to \mathcal{M}_{1,1}$ is ramified.


However: On the other hand, I have computed the cardinality of the fiber $\text{Isom}(E_{\lambda},E_{\lambda'}) \to X$ to always be 12.


Indeed consider the $\Bbb{C}$-point of $X$ corresponding to $\lambda = -1$. There are three possibilities for $\lambda'$, namely $-1,2,1/2$. An elliptic curve with $j$-invariant 1728 has automorphism group of order 4, and so the fiber over $-1$ has cardinality $3\times 4 = 12$. The story is the same for the other values of $\lambda$.


My question is: Why am I always getting 12? I am not taking into account some non-reduced issue here? I am also confused because in my head, the fiber cardinality should jump for a ramified morphism. 
Edit I was wrong previously. The fiber over $\lambda = -1$ is reduced, as Macaulay2 tells me (using the command isNormal) that the same ring with coefficients in $\Bbb{Q}$ is normal, hence reduced. Tensoring with $\Bbb{C}$ over $\Bbb{Q}$ still preserves reducedness (since $\Bbb{Q}$ is perfect). The key point is that the element $(2\lambda'-1)(\lambda'+1)(\lambda'-2)$ is already in the ideal $(f_1,f_2',f_3',f_4')$ (also confirmed by Macaulay2).

REPLY [8 votes]: Let $Leg: \mathbb P^1-\{0,1,\infty\}\to \mathcal M_{1,1}$ be the Legendre map. (This associates to $\lambda$ the elliptic curve given by $y^2 = x(x-1)(x-\lambda)$.) 

The coarse moduli space map $j:\mathcal M_{1,1}\to \mathbb A^1$ is of degree $1/2$.
The morphism $\mathbb P^1-\{0,1,\infty\}\to \mathbb A^1$ is of degree $6$. (Indeed, given a $j$-invariant, there are precisely 6 possibilities for the $\lambda$-invariant of that curve: $\lambda$, $1/\lambda$, $1-\lambda$, $1/(1-\lambda)$, $\lambda/(1-\lambda)$ and $(1-\lambda)/\lambda$.) In other words, the degree of $j\circ L$ is $6$.

It follows that the degree of $L$ is the degree of $j\circ L$ divided by the degree of $j$. This gives $6\times 2 = 12$.<|endoftext|>
TITLE: Conjugacy of matrices of order three in $PGL(2,k)$, where $k$ is any field
QUESTION [9 upvotes]: We know that every matrix of order three in $PGL(2,\mathbb Z)$ is conjugate to the following matrix 
$$
\left(
\begin{array}[cc]
 &1 & -1 \\
 1&0
\end{array}
\right)
$$
I want to know if the same result holds in $PGL(2,k)$, where $k$ is any field, under some/no constraints.

REPLY [9 votes]: The following argument does not need any assumption on the field $\Bbbk$. 
Let $[A]\in PGL(2,\Bbbk)$ be of order three. Then $A$ is invertible and not a multiple of $E=\binom{1\;0}{0\;1}$, so there exists a vector $v$ such that $v$ and $Av$ form a basis. With respect to this basis, $A$ is of cyclic normal form
$$\begin{pmatrix}0&b\\1&a\end{pmatrix}\quad\text{with}\quad
\begin{pmatrix}0&b\\1&a\end{pmatrix}^3=\begin{pmatrix}ab&b(a^2+b)\\a^2+b&ab+a(a^2+b)\end{pmatrix}\;.$$
So $[A]$ is of order three if and only if $b=-a^2$. Conjugating (in $PGL(2,\Bbbk)$) gives
$$\begin{pmatrix}0&-a\\1&0\end{pmatrix}\begin{pmatrix}0&-a^2\\1&a\end{pmatrix}\begin{pmatrix}0&-a\\1&0\end{pmatrix}=\begin{pmatrix}a^2&-a^2\\a^2&0\end{pmatrix}\;,$$
which represents the matrix in the question.<|endoftext|>
TITLE: A tricky tractrix question about vertical tangents
QUESTION [10 upvotes]: This is raised by a recent question occurring in combinatorial geometry. 
It is about a sort of tractrix, but instead of a line, the pulling end moves along a circle of radius $r>\frac12$ (typically around $0.8$), starting with a chord if length 1. Imagine this chord horizontally on the bottom and take its center as the origine of a cartesian coordinate system. If the circle has center $(0,h)$, we have $h^2+\frac14=r^2$, so only one of $h$ and $r$ is needed to define the tractrix.
 
The tractrix is approximately the green line in the picture, where all blue segments (the tangents) have unit length.
Here is what I have got so far :
For a point $(x,y)$ on the tractrix where it has slope $f’$, we have 3 equations for the point $(u,v)$ where the tangent intersects the circle on the right:
$(i)\ \ \ \ v-y=(u-x)f’\ $ (equation of the tangent)
$(ii) \ \ \ u^2+(h-v)^2=r^2\ $ (intersection with the circle), equivalently $u^2+v^2-2hv-\frac14=0$.
$(iii)\ \ (u-x)^2+(v-y)^2=1\ $ (constant length of the tangent between tractrix and circle).  
Eliminating $u$ and $v$  yields the following differential equation for the tractrix:
$$x+\sqrt{1-s^2}=\sqrt{2h(y+s)-(y+s)^2-\frac14}$$ where $$s :=\frac{f'}{\sqrt{1+f’^2}}.$$ (Thus $s$ is the sine of the slope angle.)

I don’t think there is a closed form of the tractrix equation. But is there a way to determine at which point $(x,y)$ the tangent is vertical? I'd expect it to be a not-too-involved function of $h$.

Note that, unless $r$ is too small for reaching the vertical position at all, the tractrix will carry on winding beyond the vertical for a finite time. That is, under the assumption that at each moment, the segment is tangent to the tractrix at its (the segment's) endpoint. (Imagine $r$ a bit smaller than in the picture, such that the vertical tangent of the tractrix coincides with the $y$-axis. From there on, the top of the blue segment cannot move further to the left without "breaking the smoothness" of the tractrix. I think that for each $r$, the movement will eventually arrive at such a point, which we will naturally consider as the endpoint of the tractrix.)  
But where is that point? I have no idea how far it can go if $r$ is big, but I don't think it will go further than becoming horizontal again. All this is easier to perceive if we keep $r$ constant and require the 'rotating' segment of length $\epsilon$ instead of unit length. So:

Where does the tractrix stop if $\epsilon\to0$?

REPLY [15 votes]: In fact, using the moving frame, it is easy explicitly to solve the equations and get the formula for the slope $\tan\bigl(\theta(s)\bigr)$ as a function of arc-length along the curve.  However, one sees immediately that it is a transcendental function, so explicitly solving for the value of $s$ for which $\theta(s) = \tfrac12\pi$ is not going to be easy.  
Here are the steps:  Let 
$$
X(s) = \pmatrix{x(s)\\y(s)},\qquad \text{with}\qquad
X(0) = \pmatrix{-\tfrac12\\-\sqrt{r^2-\tfrac14}}
$$
be the arclength parametrization of the curve, and set
$$
E_1 = X'(s) = \pmatrix{\cos\bigl(\theta(s)\bigr)\\\sin\bigl(\theta(s)\bigr)}
 \quad\text{and}\quad
E_2 = \pmatrix{-\sin\bigl(\theta(s)\bigr)\\\cos\bigl(\theta(s)\bigr)}.
$$ 
We then have structure equations $X'(s) = E_1(s)$,  $E_1'(s) = \kappa(s)E_2(s)$,
and $E_2'(s) = -\kappa(s)E_1(s)$, where $\kappa(s) = \theta'(s)$.    We also have $\theta(0) = 0$, by the above normalization.
We are requiring that $\bigl|X(s)+E_1(s)\bigr| = r$.  (This is the tractrix equation.)
Let us set
$$
X(s) = u(s)\ E_1(s) + v(s)\ E_2(s),
$$
where the initial conditions imply $u(0) = -\tfrac12$ and $v(0) = -\sqrt{r^2-\tfrac14}$.
Then the above structure equations imply, since $X'(s) = E_1(s)$, that
$$
u'(s) = 1 + \kappa(s) v(s)\qquad\text{and}\qquad v'(s) = -\kappa(s) u(s).
$$
Moreover, the tractrix equation implies that $(u(s)+1)^2 + v(s)^2 = r^2$.  Differentiating this equation and using the above differential equations then yields
$$
1 + u(s) + \kappa(s)v(s) = 0,
$$
which implies $u'(s) = -u(s)$, which, with the initial condition above implies $$
u(s) = -\tfrac12\,e^{-s}
$$ 
and then, consequently, that 
$$
v(s) = -\sqrt{r^2 - \bigl( 1-\tfrac12\,e^{-s} \bigr)^2}.
$$
Moreover, we have
$$
\theta'(s) = \kappa(s) = -\frac{(1+u(s))}{v(s)} 
= \frac{1-\tfrac12\,e^{-s}}{\sqrt{r^2 - \bigl( 1-\tfrac12\,e^{-s} \bigr)^2}},
$$
i.e.,
$$
\theta(s) = \int_0^s
\frac{2-e^{-\sigma}}{\sqrt{4r^2-\bigl( 2-e^{-\sigma}\bigr)^2}}\,\mathrm{d}\sigma.
$$
(Note, by the way, that this is an elementary integral, but the integral is a transcendental function.)
Armed with the knowledge of $u(s)$, $v(s)$, and $\theta(s)$, the above formulae give an explicit arc-length parametrization of the tractrix.  Note, that, of course, we must have $r>\tfrac12$ or the equations (and the condition) don't make sense.  
When $r\ge1$, these formulae exist for all $s\ge0$ and, of course, $u(s)$ goes to zero as $s\to\infty$, while $v(s)$ goes to $-\sqrt{r^2-1}$, i.e., the tractrix asymptotically approaches a circle of radius $\sqrt{r^2-1}$.  Moreover, not surprisingly, $\theta(s)$ increases without bound as $s\to\infty$, so there is a first value of $s$ for which $\theta(s) = \tfrac12\pi$.
When $\tfrac12<r<1$, the curvature of the curve goes to infinity in finite arclength, in fact, when $s = -\log\bigl(2(1{-}r)\bigr)>0$. At this point, the curve will have a cusp, and the tangent line will point directly towards the origin.  In fact, when $r < 0.724651057$, the (strictly increasing) function $\theta$ will not actually reach $\tfrac12\pi$ in the interval $0\le s\le -\log\bigl(2(1{-}r)\bigr)$, so the curve does not turn vertical in this range before its curvature goes to infinity.
Addendum:  Continuing the curve past the cusps
It turns out that there is a reasonable way to continue the curve past the cusps, and this gives a more reasonable picture.  The way to do this is to realize that one can reparametrize the curve smoothly (not by arclength) in a smooth way as follows:  Consider the equations
$$
\mathrm{d}u = -u\,\mathrm{d}s,\quad 
\mathrm{d}v = \frac{u(u+1)}{v}\,\mathrm{d}s,\quad
\mathrm{d}\theta = -\frac{(u+1)}{v}\,\mathrm{d}s\,.
$$
Setting $\mathrm{d}s = v\,\mathrm{d}t$, these can be rewritten as
$$
\mathrm{d}u = -uv\,\mathrm{d}t,\quad 
\mathrm{d}v = u(u+1)\,\mathrm{d}t,\quad
\mathrm{d}\theta = -(u+1)\,\mathrm{d}t\,.
$$
Now, regarding $(u,v,\theta)$ as functions of $t$, we get the formula for $X(t)$ in the form
$$
X(t) = u(t)\ \pmatrix{\cos\bigl(\theta(t)\bigr)\\\sin\bigl(\theta(t)\bigr)} 
     + v(t)\ \pmatrix{-\sin\bigl(\theta(t)\bigr)\\\cos\bigl(\theta(t)\bigr)}.
$$
Note that $(\dot u,\dot v) = \bigl(-uv,u(u{+}1)\bigr)$ defines a vector field on the circle $(u{+}1)^2+ v^2 = r^2$, and, when $\tfrac12<r<1$, this vector field has no zeros.  Hence $u$ and $v$ are smooth periodic functions of $t$ of some period $T(r)$, and $\theta$ satisfies $\theta(t+T(r)) = \theta(t) + P(r)$, where 
$$
P(r) = \int_0^{T(r)} (1+u(t))\,\mathrm{d}t.
$$
(The cusps occur where $v$ vanishes.)  In fact, it is easy to compute these period integrals when $r^2<1$.  One finds that
$$
T(r) 
= \int_{1-r}^{1+r}\frac{2\,\mathrm{d}\rho}{\rho\sqrt{r^2-(1{-}\rho)^2}}
= \frac{2\pi}{\sqrt{1-r^2}}
$$
and
$$
P(r) 
= \int_{1-r}^{1+r}\frac{2(1{-}\rho)\,\mathrm{d}\rho}{\rho\sqrt{r^2-(1{-}\rho)^2}}
= \frac{2\pi\bigl(1-\sqrt{1-r^2}\bigr)}{\sqrt{1-r^2}}\,.
$$
In particular, note that, when $\sqrt{1-r^2}$ is rational, the curve closes periodically, with a finite number of cusps.  The length of the curve between consecutive cusps is
$$
S = \log\left(\frac{1+r}{1-r}\right)\,.
$$
A further remark on integration of the tractrix
In fact, by a change of variable, one can integrate the equations completely when $r^2<1$.  The equation $(u{+}1)^2+v^2=r^2$ together with the differential equations 
$$
\mathrm{d}u = -uv\,\mathrm{d}t,\quad 
\mathrm{d}v = u(u+1)\,\mathrm{d}t,\quad
\mathrm{d}\theta = -(u+1)\,\mathrm{d}t\,.
$$
show that we can actually write 
$$
(u,v) = \bigl(r\,\cos 2\phi -1,\ r\,\sin 2\phi)
$$
for some function $\phi$ on the curve.  From the first two differential equations, this implies $\mathrm{d}t = -2\,\mathrm{d}\phi/(1-r\,\cos 2\phi)$, so 
$$
\mathrm{d}\theta = -(u+1)\,\mathrm{d}t 
= \frac{2r\,\cos 2\phi\,\mathrm{d}\phi}{(1-r\,\cos 2\phi)}
$$ 
and this integrates to give
$$
\theta(\phi) = 2\left( 
\frac{\arctan\bigl(\sqrt{\frac{1{+}r}{1{-}r}}\,\tan\phi\bigr)}{\sqrt{1-r^2}}-\phi\right).
$$
Hence,
$$
X(\phi) = \pmatrix{r\,\cos\bigl(2\phi{+}\theta(\phi)\bigr)-\cos(\theta(\phi))\\
                   r\,\sin\bigl(2\phi{+}\theta(\phi)\bigr)-\sin(\theta(\phi))},
$$
As a result, it follows that, when $\sqrt{1-r^2}$ is rational, the curve not only closes periodically, but it is algebraic.  I suspect that this was known classically.<|endoftext|>
TITLE: A different kind of divisor sums
QUESTION [6 upvotes]: Define, for a number $n$ the function $\mathcal{T}(n)$ which equals the number of different sums $d_1 + d_2,$ where $d_{1, 2}$ are divisors of $n.$ Now, one assumes that $\mathcal{T}(n)$ is approximately $\tau^2(n)/2$ for most $n$ but is there any analytic machinery to analyse the average/maximal behavior of something like this?

REPLY [3 votes]: One thing you can try to use to get estimates for $\mathcal T$ is bounds for the number of divisors $2^k\leq d\leq 2^{k+1}$. Since pairs of divisors from different dyadic ranges have distinct sums, you can get a rough idea for a lower bound of $\mathcal T$. 
The maximum $$\Delta(n)=\max_{k}|\{d|n:2^k\leq d\leq 2^{k+1}\}|$$ has been estimated by Hooley [On a new technique and its application to the theory of numbers, 1977] - something like $$\sum_{n\leq x}\Delta(n)\ll x(\log x)^{4/\pi-1}.$$ Thus, the typical number $n$ has $\Delta(n)=O((\log x)^{4/\pi-1})$ and so has about $(\log n)^{2/3}$ divisors with each from distinct dyadic ranges. Any pair of such divisors has a distinct sum so you get $\mathcal T(n)\gg (\log n)^{4/3}$. This is some way away from your conjecture however...<|endoftext|>
TITLE: Galois cohomologies of an elliptic curve
QUESTION [12 upvotes]: I asked this question on Mathematics Stack Exchange but did not get any answer and I was suggested to post the question here.
I am studying basic theory of elliptic curves. I encountered Galois cohomology. But two introductory textbooks I read used only $H^0$ and $H^1$. I am curious why higher cohomologies did not appear. I even do not know the definition of higher Galois cohomologies.

Is there any interesting application of higher cohomologies ($H^2, H^3, H^4$ etc) in the study of elliptic curves?
If so, what are these? Can you describe the idea?

Thank you.

REPLY [14 votes]: $H^2$ appears quite prominently in the duality theory of elliptic curves. For a local field such as a finite extension of $\mathbb Q_p$, one has $H^2(\text{Gal}(\bar K/K),\bar K^*)=\mathbb Q/\mathbb Z$, and for a global field such as a number field, one similarly has $H^2(\text{Gal}(\bar K/K),\mathsf{A}_{\bar K}^*)=\mathbb Q/\mathbb Z$, where $\mathsf{A}_{\bar K}^*$ is the group of ideles. This makes these groups ideal targets for pairings via cup product from, e.g., two $H^1$ groups. Tate's local pairing on $E(K)\times H^1(\text{Gal}(\bar K/K),E(\bar K))$ and the Cassels-Tate pairing on the Shafarevich-Tate group can be formulated in these terms; see Tate's address to the International Congress. An interesting side note is that Tate's definition of the global pairing relies at one point on the fact that $H^3(\text{Gal}(\bar K/K),\bar K^*)=0$ in order to modify a 2-cochain and turn it into a cocycle. So there's a situation where an $H^3$ appears.
You ask "But two introductory textbooks I read used only H0 and H1. I am curious why higher cohomologies did not appear." I'm guessing one of those textbooks is mine. The reason is that the material in the text only needed $H^0$ and $H^1$, and I wanted to keep the exposition as accessible as possible, so I did not see the need to develop the general theory. In principle, the whole book could have been written without any cohomology. But cohomology is very important in modern mathematics, so my idea was to introduce the $H^0$ and $H^1$ very concretely and show the reader how useful they are, which I hoped would help motivate the more general (but more complicated) theory of higher $H^i$'s when the reader eventually ran across them. I will note that in the first paragraph of Appendix B of The Arithmetic of Elliptic Curves, I did provide 4 references for people who wanted to see more of the theory of group cohomology, including the definition and construction of the higher $H^i$'s.<|endoftext|>
TITLE: Is every complex rational algebraic variety simply connected for the Euclidean topology?
QUESTION [5 upvotes]: Is it true that every quasi-projective rational irreducible algebraic  complex variety is simply connected for the Euclidean topology?
Of course, this is false if we replace "complex" with "real" or if we forget "rational".

REPLY [12 votes]: I want to mention the positive direction.  Let $X$ be a smooth, projective variety over $\mathbb{C}$, resp. over an algebraically closed field of arbitrary characteristic.  Let $Z\subset X$ be a proper closed subset.  If $X$ is (separably) rationally connected and if $Z$ has codimension $\geq 2$ in $X$, then $X\setminus Z$ is simply connected for the Euclidean topology, resp. the algebraic fundamental group is trivial.  This was first proved by Campana over $\mathbb{C}$: there is an excellent explanation in Debarre's book, "Higher dimensional algebraic geometry".  There is another proof by Kollár that extends to positive characteristic: there is an excellent explanation in one of Debarre's Bourbaki seminars.  
One might hope to drop the hypothesis that $Z$ has codimension $2$ if we assume that every general pair of points of $X\setminus Z$ is connected by a rational curve completely contained in $X\setminus Z$.  Unfortunately there are many counterexamples, such as the smooth locus of the singular cubic surface with equation $xyz-w^3=0$.  What is true is that the fundamental group is finite.  However, it is quite open to understand the fundamental group of the smooth locus of $\mathbb{Q}$-log-Fano varieties, cf. work of Chenyang Xu, Zhiyu Tian, etc.<|endoftext|>
TITLE: Blumenthal and Kolmogorov 0-1 law
QUESTION [5 upvotes]: Blumenthal's 0-1 law see theorem 5.8/5.9 tells us that an event in the germ $\sigma-$ algebra has either probability zero or one with respect to a measure induced by a Brownian motion starting in some point $x.$ (Theorem 5.8)
Conversely, one can show that this is also the case for events in the tail $\sigma-$ algebra. This is not the standard Kolmogorov 0-1 law, but I want to call it that way, as it is also name in the used in the link. (Theorem 5.9)
Now, as you can see in the proof of theorem 5.9, that the two sigma algebras (tail and germ) can be mapped onto each other. Thus, the Kolmogorov 0-1 law for Brownian motion is in some sense the same as Blumenthal's 0-1 law. 
My question is now the following: I am almost sure, that the probability $P^x(A)$ does not depend on $x$ for events in the tail sigma algebra, but afaik it is in general dependend on $x$ in Blumenthal's 0-1 law. Now, since the two sigma algebras agree (according to the proof given in the link), this does not make sense. Either both should depend on $x$ or they are both independent.
Does anybody know where my reasoning is wrong? I have been thinking about this for quite a time now but since I am not a probability theorist, I thought that somebody who is familiar with these things can easily help me out here.

REPLY [2 votes]: The issue, informally, is this.  Suppose $W_t$ is a Brownian motion started at some $x \in \mathbb{R}^d$.  Let $X_{t} = t W_{1/t}$ as in the linked notes.  Then $X_t$ is a Brownian motion, but started at 0, not $x$.  (By the strong law of large numbers, we have $\lim_{t \to 0} X_t = \lim_{s \to \infty} \frac{1}{s} W_s = 0$ almost surely.)
To see the issue more formally, it's helpful to work with an explicit sample space $\Omega$.  Let's take $\Omega = C((0, \infty), \mathbb{R}^d)$, to be the set of all continuous $\omega : (0, \infty) \to \mathbb{R}^d$, equipped with the Borel $\sigma$-algebra coming from the topology of uniform convergence on compact sets.  For each $t > 0$, let $W_t : \Omega \to \mathbb{R}^d$ be simply $W_t(\omega) = \omega(t)$.  And for each $x$, let $P^x$ be the unique probability measure on $\Omega$ under which $\{W_t\}$ is a Brownian motion started at $x$.
Now consider the map $\Phi : \Omega \to \Omega$ given by $(\Phi\omega)(t) = t \omega(1/t)$.  I agree with your claim that $\Phi$ maps the tail $\sigma$-algebra to the germ $\sigma$-algebra and vice versa.  But it sounds like you are thinking that $\Phi$ also preserves all the measures $P^x$, and that is not true.  What is true is that $P^x(\Phi^{-1}(A)) = P^0(A)$ for all events $A$.
The Kolmogorov 0-1 law says that for any $x$, the tail $\sigma$-algebra is $P^x$-almost trivial.  So a consequence is that the germ $\sigma$-algebra is $P^0$-almost trivial.  In other words, Kolmogorov for $P^x$ directly implies Blumenthal for $P^0$ only.  If you want to prove Blumenthal for $P^x$, you need one more (easy) step; say, that $P^x$ is the pushforward of $P^0$ under translation by $x$, and this translation preserves the germ $\sigma$-algebra.<|endoftext|>
TITLE: Mathematics of Chiral Rings
QUESTION [10 upvotes]: Let $A$ be a graded vector space, and suppose that two commuting differentials $d_1$ and $d_2$ of degree +1 act on $A$, such that $A$ equipped with either is a chain complex.
We now construct $C(A)$, the "chiral ring" of $A$, defined as follows: Let $A^\prime$ be the intersection of $\textrm{Ker}(d_1)$ and $\textrm{Ker}(d_2)$ inside $A$. That is, it is the space of vectors that are annihilated by both differentials. $C(A)$ is then a particular quotient of $A^\prime$. Then, we take its quotient by $\textrm{Im}(d_1)$ and $\textrm{Im}(d_2)$. Because space may not be a subspace of $A^\prime$, the quotient is taken by the intersection of ($\textrm{Im}(d_1)\cup \textrm{Im}(d_1)$) with $A^\prime$. In fact, the chiral ring $C(A)$ is NOT the cohomology of $A$, with respect to any differential.
If $A$ is endowed with a ring structure (commutative differential graded algebra), $A^\prime$ is obviously a subalgebra. So the product of two chiral operators is chiral. It remains to check whether or not the quotient is by an ideal in $A^\prime$. Since the kernel of the quotient $A^\prime \to C(A)$ is an ideal, $C(A)$ inherits a multiplication from $A$. The chiral ring is a ring.

Question:  Is this construction well-known in the mathematical literature, and does it have a name? 

In physics, the chiral ring is defined in this way (see the paper for a nice review). In supersymmetric gauge theories, chiral operators are particularly interesting because a correlation function of chiral operators is independent of positions of the operators. I wonder if mathematicians study properties of a chiral ring.

REPLY [4 votes]: [Edit: I assumed you were talking about the case of commuting differentials. Are the examples in physics you're interested in not of this kind? If not, the commutator of the supercharges is a translation - are you not restricting to modes invariant by that translation (effectively making them commute in the reduced theory)? if not I'll delete this answer. Of course if they do commute there's the full theory of spectral sequences eg addressing the situation, of which the mixed complexes I mention are a special case, so maybe should delete anyway.]
I don't have an answer but a few comments:
It's clear that there's more structure in local operators of SUSY QFT than has been fully exploited in the math literature. In making twisted field theories one usually fixes a supercharge $Q^2=0$ and considers its cohomology. One then often asks how this depends on the choice of $Q$, e.g. in the work of Kapustin-Witten on Geometric Langlands, resulting in an interesting family of [in their case topological] field theories. But as you say these chiral rings are different objects, they are ways of extracting more structure out of the collection of available supercharges. 
I'm not sure about the comment about independence of positions of operators - I thought that had to do with specific properties of the Q's rather than the abstract construction of chiral rings: if a given translation operator is Q-exact then correlation functions in Q-cohomology are independent of that translation, and I assume you're saying something similar for this bi-Q-cohomology, but that feels like an independent feature that one finds.
The closest structure to working with many Q's at once I've seen studied in math I learned about in a talk by Vera Serganova at MSRI last year. Look in the attached notes on p.4, where one considers (following Serganova and Duflo) the cone C of all square nilpotents in a Lie superalgebra, and for any module M we localize it over C by considering the various Q-cohomologies. The support of this is a version of the theory of support varieties, a key tool in many contexts in representation theory. In any case I assume one can describe the multi-Q-cohomology similarly over a space of commuting pairs. This also has a flavor common in representation theory, where "higher characters" are functions of k-tuples of commuting elements in a group.
Finally I should mention the theory of mixed complexes, namely complexes with two commuting differentials, which is the standard algebraic setting for the theory of cyclic homology. In this case we have a preserved charge (ie a graded vector space) and the two differentials have opposite charges (cohomological and homological differentials), which won't always be the case in physics. These model complexes with an action of the circle: the cohomological remnant of a circle action on a space is an action of the homology of $S^1$, which (over the reals) is the enveloping algebra of the Lie superalgebra ${\mathbb R}^{0|1}$, i.e. a square zero differential, which is indeed how cyclic homology arises in physics. e.g. we can start with a SUSY field theory (with fixed supercharge $Q$) in two dimensions and compactify on the circle (leading to a circle-rotation action) -- this is how de Rham cohomology arises out of the B-model say. So your setting with more commuting $Q$'s is an algebraic version of a torus action, and I think the cohomological invariants like chiral rings should have a natural interpretation in this setting.<|endoftext|>
TITLE: Polynomial $g:\mathbb R^n \rightarrow\mathbb R^n$ with no critical point may have no root
QUESTION [29 upvotes]: Version 1 (solved): If $g$ : $\mathbb R^n \rightarrow \mathbb R^n$ is a polynomial, $Dg(x)$ is non-degenerate for every $x$, then there exists $x$, such that $g(x)=0$.
Version 2: If $f$ : $\mathbb R^n \rightarrow \mathbb R$ is a polynomial, $D^2f(x)$ is non-degenerate for every $x$, then $f(x)$ has at least one critical point.
The problem is how to prove or disprove Version 1. The result of Version 2 is what I need to prove another problem.

REPLY [3 votes]: Building on David Speyer's answer, and as the problem (version 2) seems hard, I venture a (nonconclusive) argument for the case where $D^2f(x)$ has signature $+-$, on the plane. (I was meaning this as a comment, but it was too long...)
We probably can select smooth vector fields $e^+(x)$ and $e^-(x)$ in the positive and negative (orthogonal) directions (eigenvectors of $D^2f(x)$). This defines $X(s,t)\in\mathbb R^2$, $s,t\in\mathbb R$, with $\partial_s X=e^+(X)$ and $\partial_t X=e^-(X)$ (plus $X(0,0)=0$ for definiteness). Now $f(X(s,t))$ is strictly convex in $s$ and strictly concave in $t$, isn't it?
This doesn't prove $f$ has a saddle point (example $\mathrm{Re} (e^z)$), but maybe polynomials have something more that allows a "mountain pass lemma" to apply ?
EDIT That seemed plausible, but no: maybe there exists $X(s,t)$ such that $f(X(s,t))$ is convex in $s$ and concave in $t$, but it cannot satisfy $\partial_s X=e^+(X)$ and $\partial_t X=e^-(X)$, because the necessary integrability condition $(e^-\nabla) e^+=(e^+\nabla) e^-$ is (generically) incompatible with the orthogonality $e^+\cdot e^-=0$ (if I'm not mistaken again...)<|endoftext|>
TITLE: Are quasi-Möbius maps always quasi-conformal?
QUESTION [5 upvotes]: The article "Quasimöbius maps" by Jussi Väisälä states that one always has the implication QM $\implies$ QC. But a proof is only given in for maps of the form
$f:\dot{A} \to \dot{Y}$ where $A \subset \dot{X}$ and where $\dot{X} = X \cup \{\infty\}$ denotes the one-point extension of the metric space $X$.
Does the statement hold for all QM maps between arbitrary metric spaces? Does anyone know a reference to a paper where the relationship between QM and QC for arbitrary metric spaces is shown more clear?

REPLY [2 votes]: It holds in general. I can't see exactly what Vaisala writes, since it's behind a paywall, but since quasiconformality is an infinitesimal property and quasi-Mobius is a global one, it is easy to prove the former from the latter. (The general idea to prove quasiconformality at $x$ is to use the cross-ratio condition on four points: $x$, 2 points on (or near) a small sphere around $x$, and a fourth point which is extremely far away from all of this action.)
Igor Rivin's answer above is a bit backwards; the Loewner condition is used for the harder task of improving the infinitesimal property of quasiconformality to the global quasisymmetry or quasi-Mobius condition.
(Since my answer is not very precise, this should be a comment, but I hope it was mildly helpful.)<|endoftext|>
TITLE: Does the notion of provably total function depend on the chosen representation?
QUESTION [5 upvotes]: A typical definition of "provably total function in a theory $T$" goes like this (paraphrased from Odifreddi, Classical Recursion Theory II):

A function $f : \mathbb{N}^n \to \mathbb{N}$ is provably total in a theory $T$ (over a language that includes $0$ and successor) iff
  $$T\models (\forall x_1)...(\forall x_n)(\exists y) \varphi(x_1,...,x_n,y)$$
  where for any $\vec{x}\in\mathbb{N}^n$, $y \in \mathbb{N}$,
  $$f(\vec{x})=y \Leftrightarrow T \models \varphi(\underline{x_1},...,\underline{x_n},\underline{y})$$
  where for any $n \in \mathbb{N}$ we have denoted by $\underline{n}$ a canonical term representing $n$ (like $s^{(n)}(0)$).

I haven't been able to find any reference where this definition is fully formalized, in the sense that the first relation must happen for any $\varphi$ that "weakly represents" $f$ according to the second relation, or only for some $\varphi$, or even a (short?) proof that "$\exists$ implies $\forall$".
Can anyone enlighten me?

REPLY [8 votes]: It's probably useful to note that every property that somehow depends on provability depends quite critically on the representation. From On Provable Recursive Functions (H. B. Enderton, 1968)

Theorem: (a) We can choose $\mathrm{M}$ as above such that no function is provable.

where $\mathrm{M(e,x,y,z)}$ is a formula that is true iff $e$ is a code for a machine that on input $x$ returns $y$ in fewer than $z$ steps.
It's useful to note that even the second incompleteness theorem is subject to encoding trouble: surely $\mathrm{Con}(PA)$ is equivalent to $0=0$ if $PA$ is consistent! But $PA\vdash 0=0$, so all formulae expressing $\mathrm{Con}(PA)$ must not be created equal.
In practice one fixes an encoding of the computational model in the hopes that the encoding is straightforward enough to have the "right" meaning. The Enderton paper discusses this to some extent.<|endoftext|>
TITLE: Triviality of a fiber bundle
QUESTION [5 upvotes]: Is the principal fiber bundle $GL^+(6,\mathbb R)$ over $GL^+(6,\mathbb R)/SL(3,\mathbb C)$ trivial ?

REPLY [6 votes]: No. One can reduce to the maximal compact subgroups. So the question is, if $U(3)\rightarrow SO(6)\rightarrow SO(6)/U(3)$ is trivial. But $SO(6)/U(3)\cong \mathbb{C}P^3$ (this can be seen by using $Spin(6)\cong SU(4)$, for instance). But $\pi_2(U(3)\times \mathbb{C}P^3) = \pi_2(\mathbb{C}P^3) \cong \mathbb{Z}$ and $\pi_2(SO(6)) \cong \pi_2(SO(3))\cong \pi_2(S^3) = 0$.<|endoftext|>
TITLE: How to divide a square into three similar rectangles
QUESTION [9 upvotes]: Preparing some exercises for my High School pupils I came across this question: How can you tile a square into three similar (ie., same shape, different size) rectangles?
With a bit of algebra it can be easily shown that there is one non-trivial solution (I mean, apart from three equal stripes) involving the Plastic number (aka Padovan constant).
It has to be a very old problem but I hadn't been able to find on the web any references or any real example (eg, in architecture) of this "Plastic proportion"... Any hint?

REPLY [3 votes]: I discovered and obtained the plastic-constant-related solution of the division of a square into three similar, mutually non-congruent rectangles independently, and possibly earlier than anyone else, circa 1987 in connection with my researches into the problem of rectangular tripartite similarity divisions of generalized rectangles.  In May of 1996 I encountered the article entitled "Tales of a Neglected Number" on pages 102 and 103 of Ian Stewart's Mathematical Recreations column in the June 1996 issue of Scientific American Magazine.  My reading of that article disabused me of the mistaken notion that the number (1.324717957) was something that I alone was familiar with and served to spur me to share my discovery and solution of the division of a square into three similar, mutually non-congruent rectangles as well as a number of other plastic-constant-related discoveries with Ian Stewart in a letter dated May 22, 1996 and sent to him in care of Scientific American Magazine. In the November 1996 issue of this same Scientific American Magazine, in the feedback section of Ian' Stewart's Mathematical Recreations column entitled a Guide to Computer Dating, on page 118, a published diagram of my result with the following acknowledgement can be found: John H. Bonnett, Jr. of Livingston, N.J., sent me a wealth of information, and I offer one example.  If a square is divided into three similar (same shape, different size) rectangles, as in the figure, then the ratio of the two pieces along the vertical edge is the plastic number.
Unbeknownst to me at the time, the question of this tripartite division evidently had at least two, earlier 'in the print record' occurrences.  It appeared in the Canadian Journal of Mathematics: Crux Mathematicorum, Volume 15, #7, September 1989, Problem No. 1350, pages 215 thru 218, posed by Peter Watson-Hurthig, Columbia College, Burnaby, British Columbia, in the following form: (a) Dissect an equilateral triangle into three polygons that are similar to each other but all of different sizes. (b) Do the same for a Square. (c) Can you do the same for any other regular polygon? (Allow yourself more than three pieces if necessary.) It was solved for (b) 'a square' in the case of similar rectangles by L. F. Myers, The Ohio State University and by Richard K. Guy, University of Calgary.  Also, it evidently was posed (along with five other dissection tasks) by Karl Scherer some time prior to 1994 and disclosed by Martin Gardner (who evidently also discovered this dissection on his own) in his Mathematical Surprises column in the May/June 1994 issue of Quantum magazine in an article entitled "Six challenging dissection tasks". There it was discussed as correlated with the p^2 = 1.754877666 value (which Gardner proposed calling "high-phi") with no apparent appreciation of this number's relation to Dom Hans van der Laan's plastic constant (p = 1.324717957), now commonly referred to as "psi."
As far as I've been able to ascertain, the relationship of this tripartite division (in particular the ratio of the edge division of the square) to psi, the plastic constant (p = 1.324717957), as fundamental and not to high-phi (p^2 = 1.754877666) had not hither been noted, disclosed, or published in any form prior to Stewart's November 1996 Scientific American column publication of my disclosure to him of this fact and of my disclosure to him of the dissection itself and of my understanding of the logic of the associated p edge ratio plastic rectangle and its p^2 edge ratio gnomon (the very ratio of the tripartite dissected square's similar rectangles).   
Gardner's July 18th 2001 publication of this dissection on page 124 of his Workout book was perhaps only the fourth time that this dissection had thus far appeared in print.  I have only been able to find a handful of other, subsequent early instances of its appearance in print, including: 1) a paper by Paul Yiu, Department of Mathematics, Florida Atlantic University, Summer 2003, Chapters 1–44, Version 031209 entitled Recreational Mathematics 2003, Project: Cutting a Square into three similar parts, p. 317.  2) a paper by Federico Ardila and Richard P. Stanley (circa 2004) entitled Tilings, page 10.  3) The paper by de Spinadel, Vera W., and Antonia Redondo Buitrago. "Towards van der Laan’s Plastic Number in the Plane." Journal for Geometry and Graphics, 13.2 (2009).<|endoftext|>
TITLE: Coxeter exchanges in non-reduced words
QUESTION [14 upvotes]: Let $Q$ be a word in the generators of some Coxeter group, and consider a subword $R$ (not necessarily reduced, though I might want $Q$ to be).
Define the greedy or Demazure product of $R$ as follows: while multiplying generators left-to-right, insist on only going upward in Bruhat order; any letter in $R$ that would take you downward should be skipped over.
For abstruse geometric reasons,
I want to divide the set of all $2^{|Q|}$ subwords of $Q$ into equivalence classes $\{R\}$, according to (the greedy product of $R$, the locations in $Q$ of the skipped letters). That is, for each pair $(w \in W, S\subseteq Q)$, I consider the set of $R$ with greedy product $w$ and skip locations at exactly $S$.
For example, each $(w,\emptyset)$ corresponds to the class of reduced subwords with (ordinary) product $w$, a well-studied object (e.g. in my paper with Ezra Miller on subword complexes). If $Q=121$ in $S_3$, the $2^3$ subwords break into a singleton for each $w$, except for $\{1--,--1\}$ and $\{1-\underline{1}\}$ for $w=(12)$, where the underlined $\underline{1}$ is the only skipped letter.

Has anyone studied these equivalence classes of subwords before?
If so, are standard results like the Coxeter exchange condition known for them?

EDIT: here's a larger example, hopefully to make clearer the equivalence relation.
Let $Q = 1231$, a word in the generators of $S_4$.
Then the classes are
$$ ---- $$
$$ 1---, \quad ---1 $$
$$ -2-- $$
$$ --3- $$
$$ 12-- $$
$$ 1-3-, \quad --31 $$
$$ 1--\underline{1} $$
$$ -23- $$
$$ -2-1 $$
$$ 123- $$
$$ 12-1 $$
$$ 1-3\underline{1} $$
$$ -231 $$
$$ 1231 $$
where I've underlined the skip positions.

REPLY [4 votes]: (in contrary to what I thought first,) here is a proof that the "exchange condition" holds in the following sense.
It is based on the root configuration in http://arxiv.org/abs/1111.3349 [1].
Let $(W,S)$ be a Coxeter system, $Q \in S^*$ a word in $S$, $w \in W$, $P$ a subword of $Q$. We everywhere consider subwords as being indexed by positions in $Q$, so if $Q = ss$, then the two subwords $s{-}$ und ${-}s$ are considered to be different.
Let $(Q,w,P)$ be the set of subwords $X$ of $Q \setminus P$ such that the complement $R = Q \setminus X$ has greedy product $Dem(R) = w$ and the skips in the greedy product are in exactly the positions in $P$, we call these positions the greedy skips. (So this is the situation Allen introduced in the question, except that he did not mention the part about the complement.)
Let $D(Q,w,P)$ be the simplicial complex with facets given by $(Q,w,P)$. This complex is clearly pure since every facet contains exactly $len(Q)-\ell_S(w)-len(P)$ many letters.
Observation 1: For every facet $X$ of $D(Q,w,P)$, the disjoined union $X \cup (P\setminus P')$ is also a facet of $D(Q,w,P')$ for $P' \subseteq P$.
Construction 2: The root configuration in Definition 3.1 in [1] is given for a facet $X$ of $D(Q,w,\{\})$ as follows. Associate to each letter $q_i$ in $Q$ a root $R(X,i)$ by applying the prefix up to position $q_{i-1}$ of the word $Q \setminus X$ of $w$ to the simple root $\alpha_{q_i}$.
In my example $Q = tsstst$, the element $w = sts = tst$ and the facet $t{-}s{--}t$ with complement ${-}s{-}ts{-}$ which is a reduced word for $w$, the root configuration is
$$
  R(X,\cdot) =\beta,\alpha,s(\alpha),s(\beta),st(\alpha),sts(\beta)
$$
where $\alpha = \alpha_s, \beta = \alpha_t$ which is equal to
$$
  23,12,-12,13,23,-12
$$
where I write $12$ for $\alpha$, $23$ for $\beta$, and $13$ for $\alpha+\beta$.
Observation 3: The collections of roots of the root configuration of the complement of $X$ is exactly the inversion set of $w$ (in particular, that's all positive roots).
Observation 4: The negative roots in the root configuration could be used as greedy skips. This also means that picking a a facet $X$ of $D(Q,w,\{\})$, and a subset $P$ of the letters for which the root configuration is negative results in a facet $X \setminus P$ of $D(Q,w,P)$.
Observation 5: If a root in the root configuration is negative, than all appearances of the same root "to the right" are also negative. Analogously, if a root there is positive then all appearances "to the left" are positive.
Statement 6: The simplicial complex $D(Q,w,P)$ is vertex-decomposable has the exchange property in the following sense.
Let $q_1$ be the first letter in $Q$
If $\ell_S(q_1w) < \ell_S(w)$, then for any facet $X$ of $D(Q,w,P)$, there exists a facet $Y$ of $D(Q,w,P)$ containing $X \setminus q_1$.
Proof:
Consider $X' = X \cup P$ as a facet of $D(Q,w,\{\})$ as in Observation 1.
We now know from Observation 4 that $P$ is a subset of the negative roots in the root configuration of $X'$.
We know from $\ell_S(q_1w) < \ell_S(w)$ that $\alpha_{q_1}$ is in the inversion set of $w$ which is by Observation 3 equal to the root configuration of the complement of $X$.
This provides to obtain $Y'$ by "flipping" $q_1$ in $X'$ to the unique position not in $X'$ for which the root configuration is given by the same simple root $\alpha_{q_1}$.
Call this position $q_i$ and we have $Y' = (X' \setminus q_1 ) \cup q_i$.
It remains to show that $Y'$ contains $P$ and that all roots in the root configuration of $Y'$ at positions in $P$ are negative, since Observation 1 then tells us that $Y = Y' \setminus P$ is the desired facet of $D(Q,w,P)$.
Lemma 3.3(3) in [1] tells us how the root configuration changes when doing the flip $X'$ to $Y'$.
This is indeed easy to see: the roots in the root configuration after $q_i$ are not changed, while we apply $s_{q_1}$ to all the roots before that,
$R(Y',j) = R(X',j)$ for $j>i$, and $R(Y',j) = s_{q_1}\big(R(X',j)\big)$ for $j \leq i$.
Since we only care about the signs, observe that $s_{q_1}$ does only affect the sign of $\alpha_{q_1}$, while all other signs are unchanged.
But since $R(X',i)$ is positive (and ind.eed in equal to $\alpha_{q_1}$), Observations 4 and 5 finally tell us that $P$ is a subset of the letters for which the root configuration of $Y'$ is negative. As desired, we conclude that $Y=Y' \setminus P$ is a facet of $D(Q,w,P)$.<|endoftext|>
TITLE: Pursuit-Evasion type game on graph ("Flyswatter game")
QUESTION [11 upvotes]: An instance of the "flyswatter game" is defined by a graph $G$ and positive integer $k$. There are two players, A (the 'fly') and B (the 'swatter'). Essentially, the fly moves around $G$ and the swatter tries to guess where the fly will be $k$ steps in the future.
Formally, the game is played in two stages:
(1) Player A chooses an infinite walk $w_1, w_2, \dots$ on $G$ (but player B doesn't see it).
(2) Player B goes through the walk one step at a time. At step $t$ (having already observed $w_1, w_2, \dots, w_t$) he gets to either go to step $t+1$ (revealing $w_{t+1}$) or he can 'swat' by picking a vertex $v$ (which ends the game -- so he can only do it once). Player B is allowed to observe as many steps of the walk as he wants before swatting.
Player B wins if and only if $w_{t+k} = v$, where $t$ is the step where he decided to swat $v$ (i.e. the fly gets hit).
Even simple examples (e.g. $G$ is a 3-cycle, $k = 2$) give interesting behavior. I'd be surprised if nobody has ever studied this game, but I can't find it anywhere. Has anyone seen this before?

EDIT: As per Kaveh's suggestion (on the CSTheory copy -- now deleted), here's a bit of background about this.
I'm interested in this question as a simplification of a continuous problem about finding a maximally unpredictable random process for generating a path on the real line (for a particular measure of unpredictability). Thus, one thing I'd like to know in particular is an optimal strategy for the fly on the integer line (which can be simulated by using a cycle of size $2k + 1$ instead).
What I've proven:
Let $p_A(G, k)$ be the probability of winning for player A. Then (for any $G$) $k_1 < k_2$ implies that $p_A(G, k_1) \leq p_A(G, k_2)$ (player A prefers larger $k$).
If $G$ has $n$ vertices, $p_A(G, k) \leq \frac{n-1}{n}$ (this is trivial, as player B can simply guess a random vertex).
What I strongly suspect to be true (not yet proven):
There is always an optimal strategy for player A which generates the walk using an order-$k$ Markov chain.
If $k = 1$, the optimal strategy for player A is to find the subgraph $G^*$ of $G$ which has largest minimum degree (there is a poly-time algorithm to find this), and then do a random walk on $G^*$.
An interesting case: $C_3$ with $k = 2$. 
The best strategy I've found so far for A to generate his walk is:

begin at any vertex; at first move, move CW or CCW with probability $1/2$ each.
for all subsequent moves, with probability $\frac{\sqrt{5} - 1}{2}$ repeat the previous move (CW or CCW); otherwise switch.

This gives player B a winning probability of $1 - \frac{\sqrt{5} - 1}{2} \approx 0.382$ (the probability for A to be back where he started after 2 moves, or to have moved twice in the same direction as his previous move).

REPLY [4 votes]: When the graph is just the integer lattice ${\bf Z}$, this problem was proposed by Isaacs, and solved independently by Dubins [1] and Karlin [2]. 
To paraphrase Math reviews, [1] and [2] deal with the following game: At each unit of time, the Evader may move either one unit of distance to the left or one to the right. The Pursuer, knowing the past moves of the Evader, has one move—he predicts two units of time in advance the position of the Evader. The payoff to the Pursuer is 1 if he predicts correctly; otherwise it is zero. Then [1], [2] show that this game has the value   $(3-\sqrt{5})/2$ and that the Evader has a unique optimal strategy which depends on the previous move only. However, the Pursuer does not have an optimal strategy. 
Of course the value on ${\bf Z}$ is an upper bound for the value the game on any cycle. Extensions to some other graphs are in Ferguson [3].
An asymptotic version of the problem is considered in [4]. In that paper, given $\alpha\in (1/2,1)$, a nearest-neighbor process $\{S_n\}$ on the integers  that is less predictable than simple random walk is constructed, in the sense that given the process until time $n$, the conditional probability that  $S_{n+k}=x$ is uniformly bounded above by $Ck^{-\alpha}$ for some constant $C<\infty$. Such a process does not exist for $\alpha=1$; the exact predictability profiles possible were determined in [5] and [6].
[1] Dubins, L. E.  A discrete evasion game.
Contributions to the theory of games, Vol. 3, pp. 231–255; Ann. of Math. Studies, No. 39, Princeton Univ. Press, Princeton, N.J., 1957.  
[2] Karlin, Samuel An infinite move game with a lag. Contributions to the theory of games, vol. 3, pp. 257–272.  Annals of Mathematics Studies, no. 39. Princeton University Press, Princeton, N. J., 1957 \newline
[3]  Ferguson, Thomas S. On discrete evasion games with a two-move information lag. 1967  Proc. Fifth Berkeley Sympos. Math. Statist. and Probability (Berkeley, Calif., 1965/66), Vol. I: Statistics  pp. 453–462 Univ. California Press, Berkeley, Calif. \newline
[4] Benjamini, Itai ; Pemantle, Robin ; Peres, Yuval.
Unpredictable paths and percolation.
Ann. Probab. 26 (1998), no. 3, 1198–1211.  
[5] Häggström, Olle; Mossel, Elchanan Nearest-neighbor walks with low predictability profile and percolation in 2+ϵ  dimensions. Ann. Probab. 26 (1998), no. 3, 1212–1231.  
[6] Hoffman, Christopher Unpredictable nearest neighbor processes. Ann. Probab. 26 (1998), no. 4, 1781–1787.<|endoftext|>
TITLE: Replacing Axiom of Choice with Axiom of Countable Choice
QUESTION [5 upvotes]: Many people find ACC more intuitive than AC ("Pick something from the first set, then something from the second set, then...) and it also doesn't lead to "controversial consequences" (See for eg: Peculiar examples with Axiom of Countable Choice ?)
My question is:
What are the consequences for Set Theory if we replace AC by ACC -as in ZF and ACC assumed true but cannot assume AC ?  
Specifically:
1) Are all ZFC ordinals - $\aleph_1$, $\aleph_\omega$, aleph and beth fixed points etc - still well-defined ?
2) Is the Continuum Hypothesis still undecidable ?
3) Are there any striking changes to Large Cardinal properties ? (For eg: "The smallest measurable cardinal can be equal to the smallest strongly inaccessible cardinal")  
PS: If the question is too broad, I'd be very happy to be referred to a book/paper.
Edited: Realized from a comment below that ZF + ACC + Not AC was the system I had in mind, otherwise question 2 becomes trivial.

REPLY [17 votes]: First of all, the axiom of countable choice says that given a countable family of non-empty sets, you can choose from each set simultaneously. If you want to choose from one, then from another, then from another, and so on you need a strictly stronger form of choice called Dependent Choice, abbreviated as $\sf DC$.
To your questions, the definition of the $\aleph$ numbers uses absolutely no choice, although the proof that any of them is regular does use the axiom of choice (well, except $\aleph_0$). So it is consistent, for example, that $\aleph_1$ is singular. But assuming $\sf DC$ will prevent that. Whether or not you can have every cardinal $\geq\aleph_2$ singular with $\sf ZF+DC$ is still open.
The definition of $\beth$ numbers, on the other hand, goes out the window. The axiom of choice is equivalent to saying that the power set of a well-ordered set is well-ordered. So if the axiom of choice fails, there will be some ordinal $\alpha$ whose power set cannot be well-ordered. This means that at some point $\beth$ cardinals will not be $\aleph$ numbers. You can still talk about $\beth$ numbers, of course, as iterated powers of $\omega$, but that gives you significantly less information in that sense.
As for the continuum hypothesis, as noted by others it is still unprovable. But now you even get different forms of the continuum hypothesis. $2^{\aleph_0}=\aleph_1$ is no longer equivalent to "Every uncountable set of reals is equipollent with the reals themselves". Indeed, even without $\sf DC$, it is consistent that $2^{\aleph_0}$ and $\aleph_1$ are incomparable, and every uncountable set of reals has cardinality $2^{\aleph_0}$.
Finally, for large cardinals, most properties which are reserved for very large cardinals can be made compatible with $\aleph_1$ and $\sf ZF+DC$. For a more complete survey, look at What sort of large cardinal can $\aleph_1$ be without the axiom of choice?.<|endoftext|>
TITLE: Relationship between the Lie functor applied to a Lie group action, and the fundamental vector field mapping?
QUESTION [6 upvotes]: Let $M$ be a smooth manifold, and $G$ a Lie group with Lie algebra $\mathfrak{g}$. The Lie algebra of the diffeomorphism group of $M$ is the Lie algebra of vector fields on $M$; that is $\text{Lie}(\text{Diff}(M)) = \Gamma(TM)$. 
Given a smooth left-action of $G$ on $M$ expressed as a Lie group homomorphism $\lambda : G \to \text{Diff}(M)$, we obtain a Lie algebra homomorphism $\text{Lie}(\lambda): \mathfrak{g} \to \Gamma(TM)$. 
What is the relation between this Lie algebra homomorphism and the fundamental vector field mapping $\zeta: \mathfrak{g} \to \Gamma(TM)$, which is an anti-homomorphism of Lie algebras? Recall that the fundamental vector field mapping is given by
$$\zeta(X)_x = T_e \lambda(-, x).X$$
for any $x \in M$ and $X \in \mathfrak{g}$. Here I have written the group action as a smooth map $\bar{\lambda}: G \times M \to M$.
I don't know much about infinite-dimensional Lie groups and manifolds, personally, but it seems that there should be some simple relationship between these two mappings.
(X-post to stackexchange)

REPLY [5 votes]: So I have discovered that indeed that indeed $\zeta = T\lambda: \frak{g} \to \frak{X}$$(M)$. 
There is a simple reason why this mapping is an antihomomorphism, even though the Lie functor should take Lie group homomorphism to Lie algebra homomorphisms.
The Lie bracket of most Lie algebras is defined via left-invariant vector fields on the Lie group, but the bracket on $\frak{X}$$(M)$ is usually defined in such a way that it corresponds to right-invariant vector fields on $\text{Diff}(M)$. Thus, if we chose to define Lie brackets for all of our Lie algebras consistently using the right-invariant convention (as Olver does, for example), then the fundamental vector field mapping will be a homomorphism.<|endoftext|>
TITLE: Are the primary parallelotopes classified? (equivalently, Voronoi cells of lattices)
QUESTION [7 upvotes]: A primary parallelohedron is a polyhedron that can fill space with infinite translated copies.
It is known (e.g., Coxeter, H. S. M. Regular Polytopes, 3rd ed. New York: Dover, pp. 29-30, 1973; or, Tutton, A. E. H. Crystallography and Practical Crystal Measurement, 2nd ed. London: Lubrecht & Cramer, 1964.) that the primary parallelohedra are the cube, hexagonal prism, elongated dodecahedron, rhombic dodecahedron, and truncated octahedron.
Is there is a classification for any higher dimensions? What are the primary d-parallelotopes?
The following is a conjecture of mine regarding the case of $d=4$.

Conjecture: There are exactly 7 primary 4-parallelotopes:
(1) Hypercube
(2) 16-cell
(3) 24-cell
(4) Hexagonal Square Duoprism
(5) Prismatic Elongated Dodecahedron
(6) Prismatic Rhombic Dodecahedron
(7) Prismatic Truncated Octahedron

EDIT: Based upon the recent answer, an improved question would concern the classification of combinatorial equivalence classes of Voronoi cells of lattices. Have these been classified or are there known classifications which could prove or disprove my conjecture?

REPLY [11 votes]: Voronoi conjectured that every parallelotope is combinatorially equivalent to a Voronoi cell of a lattice. The conjecture was proved for $d\le 4$ by Delone.
Reference: Handbook of Convex Geometry, P. M. Gruber and J. M. Wills (eds.) page 1005
To answer the added question: there are 52 inequivalent Voronoi cells in 4 dimensions. I do not know whether this classification proves or refutes your stated conjecture, but it should not be hard to check.
Reference: Computational Geometry of Positive Definite Quadratic Forms, A. Schuermann, page 60<|endoftext|>
TITLE: Geometry of convex subsets in Alexandrov space/ Riemannian manifold
QUESTION [6 upvotes]: Let $X^n$ be an $n$-dimensional complete Alexandrov space with curvature bounded below (or a smooth Riemannian manifold, possibly with boundary). Let $U\subset X$ be an open dense subset with the following convexity property: Any shortest path connecting any two points of $U$ is contained in $U$.
Question 1. Is the Hausdorff dimension of $X\backslash U$ at most $n-1$?
Question 2. If the answer to question 1 is yes, whether the corresponding Hausdorff measure of $X\backslash U$ is locally finite?
The answer is not known to me even in the case of Riemannian manifolds. 
A reference would be helpful.

REPLY [4 votes]: In the Riemannian case $U$ has to be whole manifold.
In Alexandrov space it is expected that $U$ has full measure, 
if true it would be a partial answer to your question.<|endoftext|>
TITLE: What happened to the fourth paper in the series "On the classification of primitive ideals for complex classical Lie algebras" by Garfinkle?
QUESTION [18 upvotes]: In a series of papers in Compositio Math. entitled On the classification of primitive ideals for complex classical Lie algebras I, II and III, Garfinkle describes an algorithm that allows one to determine the fibers of the Duflo map, and hence the left (or right) Kazhdan-Lusztig cells in the Weyl gorup for a Lie algebra of type $B$ or $C$.
Several places, both in reviews and in the papers themselves, mention is made of a fourth part which will deal with the additional things needed to treat type $D$, but I have not been able to find this paper anywhere (it is certainly not listed on MathSciNet).
What happened to this paper? Was it ever written? And if not, is there some other paper that describes what happens in type $D$?

REPLY [24 votes]: I am the author of this series of papers.  Thank you for your interest in my work. I have continued to be active in math when I have had time for it, but most of my time in the past years has been taken with parenting responsibilities.  
After completing the third paper, I began work on the fourth paper and made partial progress before setting it aside for an number of years, though I had hoped to return to it.
Because of interest in this work by researchers in this area, I had shared an incomplete draft of the fourth paper with some colleagues at their request.  Last year, to my surprise, I discovered that a link to a modified version of that draft had been posted to this website.  At my request, it has been taken down.  To be totally clear, this version did not and does not have my authorization to have my name on it, nor to use my copyrighted content.  
Meanwhile, I have been working on completing the fourth paper.  While much of the proof parallels the argument in the third paper, there are new difficulties that need to be resolved.  Many of them have to do with the fact that the Weyl group for type D_n is of index 2 in the Weyl group of type C_n. Thus, you have fewer pairs of tableaux to work with, and the proofs are a lot more difficult.  However, as a result of recent progress, I am now optimistic that in the near future I will have a manuscript that can be posted to the arXiv.<|endoftext|>
TITLE: Integer matrix that does not belong to a free group of rank 2
QUESTION [9 upvotes]: I'm given two matrices in $SL_2(\mathbb{Z})$
$$
A = \left(\begin{array}{cc}
       2 & 3\\
       3 & 5
     \end{array}\right), \ \
B = \left(\begin{array}{cc}
       5 & 3\\
       3 & 2
     \end{array}\right).
$$
Then the group $\langle A, B \rangle$ is free of rank 2. Now my problem is to prove that $\langle A, B \rangle$ does not contain any matrix of the form
$$
  \left(\begin{array}{cc}
       1 & n\\
       0 & 1
  \end{array}\right)
$$
with $n\in\mathbb Z$ nonzero. That is, no such matrix can be obtained from products of $A$ and $B$ and their inverses.
I started by writing
$$
  \left(\begin{array}{cc}
       0 & 1\\
       - 1 & 0
     \end{array}\right) \longrightarrow x, \left(\begin{array}{cc}
       0 & 1\\
       - 1 & 1
     \end{array}\right) \longrightarrow y,
A \longrightarrow (y x y^{- 1} x^{-1})^2,
B \longrightarrow (y^2 x y^2 x^{- 1})^2,
$$
but then did not get any further. 
Any help is highly appreciated.

REPLY [13 votes]: This can be proved using the modular diagram in the upper half plane, and the dual tree $T$ of the modular diagram (see below for a discussion of the picture in the link). The modular diagram has vertices at the rational numbers together with $\infty = \frac{1}{0}$. The hyperbolic line $\overline{\frac{p}{r}, \frac{q}{s}}$ is an edge if and only if $ps-qr=\pm 1$ (all rational numbers are in lowest form).
The dual tree $T$ is bipartite, its $0$-cells of valence 2 vertices being where it crosses the edges of the modular diagram, and its $0$-cells of valence 3 vertices being the barycenters of the ideal triangles of the modular diagram. The $0$-cell of valence 2 on the side $\overline{\frac{1}{0},\frac{0}{1}}$ is 
$$v_0 = 0 + 1i
$$
and $v_0$ is connected by a 1-cell, denoted $E_+$, to a 0-cell of valence 3 vertex at
$$v_1 = \frac{1}{2} + \frac{\sqrt{3}}{2} i
$$
Let $E_-$ denote the reflection of $E_+$ across the $y$-axis, joining $v_0$ to the point $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$.
Consider paths in $T$ with initial edge $E_+$ that terminate on a $0$-cell of valence 2 (that $0$-cell having positive real part). These paths can be identified with words in the letters $L$ and $R$. For example, the path $RLRL$ starts at $v_0$, goes along $E_+$, then turns Right at the valence 3 vertex $v_1$, then continues past the next valence 2 vertex located at $\frac{1}{2} + \frac{1}{2} i$ to turn Left at the next valence 3 vertex, then Right, then Left, and then stops at the next valence 2 vertex.
What to show is the following list of properties of the fractional linear action of $GL(2,\mathbb{Z})$ restricted to $T$:

$\pmatrix{2 & 3 \\ 3 & 5}$ has an axis with fundamental domain $RLRL$
$\pmatrix{5 & 3 \\ 3 & 2}$ has an axis with fundamental domain $LRLR$
$\pmatrix{1 & n \\ 0 & 1}$ has an axis with fundmantal domain $\underbrace{LLLL…L}_{\text{$n$ times}}$. That axis is the unique line in $T$ passing through all the points $k + i$, $k \in \mathbb{Z}$.

By constructing the entire axes of these elements (just continue forward and backward in $T$ with further copies of the fundamental domain), one sees that the intersection of the axis of $\pmatrix{2 & 3 \\ 3 & 5}$ and the axis of $\pmatrix{5 & 3 \\ 3 & 2}$ is therefore equal to $E_- \cup E_+$. 
From this, it follows easily that the entire minimal subtree of the free group $\langle A,B\rangle$ intersects the axis of $\pmatrix{1 & n \\ 0 & 1}$ solely in a finite segment, namely the union of the segment $E_- \cup E_+$ connecting $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$ to $v_1 = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ together with the translate of $E_- \cup E_+$ one unit to the right, connecting $v_1$ to $\frac{3}{2} + \frac{\sqrt{3}}{2} i$, and the translate of $E_- \cup E_+$ one unit to the left, connecting $-\frac{3}{2} + \frac{\sqrt{3}}{2}$ to $-\frac{1}{2} + \frac{\sqrt{3}}{2}$.
From this it follows that $\pmatrix{1 & n \\ 0 & 1} \not\in \langle A,B\rangle$ for all $n \ne 0$, because those elements all have the same axis in the minimal tree of $\langle A,B \rangle$ as described above, and that axis would have to be contained in the minimal tree of $\langle A,B \rangle$.
What remains is to show 1 and 2 above (3 being obvious). The general statement is as follows. Denote 
$$L = \pmatrix{1 & 1 \\ 0 & 1} \quad R = \pmatrix{1 & 0 \\ 1 & 1}
$$
Every matrix $\pmatrix{p & q \\ r & s} \in SL(2,\mathbb{Z})$ with $p,q,r,s > 0$ can be factored uniquely as a product of $L$ and $R$; simply use column reduction. The word you get from that factorization, when interpreted as a path in $T$ starting with $E_0$, is a fundamental domain for the axis of $\pmatrix{p & q \\ r & s}$.
So what's left is a column reduction calculation, to verify that
$$\pmatrix{2 & 3 \\ 3 & 5} = RLRL, \qquad \pmatrix{5 & 3 \\ 3 & 2} = LRLR
$$ 
Added bonus: Apropos of some of the comments, one can show, by examining the minimal subtree of $\langle A,B \rangle$ constructed above, that every ray in the minimal subtree oscillates infinitely often between $L$ and $R$. Therefore, this group has no cusps, because each ray in $T$ that limits on a cusp is eventually $RRRRRRR…$ or $LLLLLLL….$.
Added: I finally found a good picture on the web, linked to in the first line of the answer. The hyperbolic lines of the modular diagram are in blue, the dual tree $T$ is in red, and one can label the cusps along the bottom of the picture as $-1,0,1,2,3$.<|endoftext|>
TITLE: Is the embedding dimension minus the dimension upper semicontinuous?
QUESTION [6 upvotes]: For a Noetherian local ring $R$ with maximal ideal $\mathfrak{m}$ and residue class field $K$, consider the invariant $$\operatorname{def}(R) := \operatorname{dim}_K(\mathfrak{m}/\mathfrak{m}^2) - \operatorname{dim}(R).$$ Here are some questions:

If $P \subseteq R$ is prime ideal, is it always true that $\operatorname{def}(R_P) \le \operatorname{def}(R)$?
For which Noetherian rings is the function $\operatorname{Spec(}R) \to \mathbb{N}$, $P \mapsto \operatorname{def}(R_P)$ upper semicontinuous?
Is there a name (and/or notation) that is commonly used for this invariant?

By a variant of the Jacobian criterion one can see that question 2 has an affirmaitve answer if $R$ is a finitely generated algebra over a perfect field. But are there more general results?

REPLY [2 votes]: Edit. Finally the proof was not so long, so I include it complete:
Question 3. Embedding codimension (sometimes simply codimension).
Question 1. I don't have access here to "Lech, Inequalities related to certain couples of local rings, Acta math. 112 (1964), 69-89", but maybe that paper will answer it better, I can't remember. However, here is a proof when $A$ is quasi-excellent:
Notation: dim = Krull dimension, edim = embedding dimension, codim = edim - dim = embedding codimension.
Lemma 1. Let $(A,m) \to (B,n)$ be a flat homomorphism of noetherian local rings. Then dim $A$ + dim $B/mB$ = dim $B$.
Lemma 2. Let $(A,m) \to (B,n)$ be a flat homomorphism of noetherian local rings with regular closed fiber $B/mB$. Then edim $A$ + edim $B/mB$ = edim $B$.
Deduction of Lemma 2 from Andre-Quillen homology (references (x.y) are to Result y from Chapter x in Andre, Homologie des algebres commutatives, Springer, 1974): let $f:(A,m,k) \to (B,n,l)$ be a flat homomorphism of noetherian local rings with regular closed fiber. Let $p=n/mB$ be the maximal ideal of $B/mB$. It is sufficient to prove that we have an exact sequence of $l$-vector spaces
$$0 \to m/m^2 \otimes_k l \to n/n^2 \to p/p^2 \to 0.$$
But this exact sequence is a part of the Jacobi -Zariski exact sequence in Andre-Quillen homology (5.1) associated to $B \to B/mB \to l$:
$$H_2(B/mB,l,l) \to H_1(B,B/mB,l) \to H_1(B,l,l) \to H_1(B/mB,l,l) \to H_0(B,B/mB,l).$$
$H_2(B/mB,l,l)=0$ since $B/mB$ is regular by (6.26).
$H_1(B,B/mB,l)=H_1(A,k,l)= m/m^2 \otimes_k l$ by (4.54) (since $f$ is flat) and (6.1) respectively.
$H_1(B,l,l)= n/n^2$, and $H_1(B/mB,l,l)= p/p^2$ by (6.1).
$H_0(B,B/mB,l)=0$ by (4.60).
Another proof without Andre-Quillen homology can be found in arXiv:1205.2119v3, Lemma 3.1.
Corollary 3. Let $(A,m) \to (B,n)$ be a flat homomorphism of noetherian local rings with regular closed fiber. Then codim $A$ + codim $B/mB$ = codim $B$. In particular, codim $A \leq $ codim $B$. Another (trivial) particular case is codim $A$ = codim $\hat{A}$.
Definition. We say that a noetherian local ring is a G-ring if the completion homomorphism $A \to \hat{A}$ is regular. By Matsumura, Commutative Algebra, (33.C) Theorem 75 page 251 this is equivalent to the usual definition of G-ring, and by Theorems, 73, 76 and 77 this is also equivalent to be quasi-excellent (since it is local).
Corollary 4. Let $A$ be a local G-ring. Then for any prime ideal $p$ of $A$ we have codim $A_p \leq $ codim $A$.
Proof. The result is valid for a complete ring $A$ (choose a regular local ring $S$ with the same embedding dimension of $A$ such that $A=S/I$ and localize), so codim $\hat{A}_q \leq $ codim $\hat{A}$ = codim $A$. Since $A \to \hat{A}$ is faithfully flat, there exists a prime ideal $q$ of $\hat{A}$ contracting to $p$. The local homomorphism $A_p \to \hat{A}_q$ is regular and so by Corollary 3 codim $A_p \leq $ codim $\hat{A}_q$.
Question 2.
For excellent rings. An indirect proof, but short enough to be written here, is as follows. Embedding dimension is the diference between the invariant $\delta_2$ in "Ragusa, On Opennes ..." which is upper semicontinuous by Proposition 3.6 in that paper, and the complete intersection defect which is upper semicontinuous by the main theorem in "Alonso-Rodicio, On the upper ...". So for each $n$ the set of primes $p$ such that codim $A_p > n$ is constructibe. By Corollary 4 it is also stable under specialization. So by Matsumura, Commutative Algebra, (6.G) Lemma, page 46, it is closed.<|endoftext|>
TITLE: Why do we need ampleness in the definition of stability/semistability
QUESTION [8 upvotes]: This is a general question that I have. Let $X$ be a projective variety over an algebraically closed field $k$. Let $L$ be an ample line bundle over $X$. Let $F$ be a vector bundle on $X$. We say that $F$ is $L$-semistable if for any coherent subsheaf $0\neq E\subset F$ of strictly lesser rank, $\mu_L(E)\leq\mu_L(F)$. Here $\mu_L(F)=\frac{c_1(F).L^{n-1}}{r}$, where $r=$ rank$(F)$ and $n=dim\ X$. 
My question is, why do we need $L$ to be ample? The definitions and basic properties seem to go through without ampleness. Even the fact that semistability is an open condition does not seem to depend on the fact that $L$ is ample, and existence of Harder-Narasimhan filtration/Jordan Holder filtration too seems to hold without ampleness. Why can't we just take $L$ to be a globally generated or nef and big line bundle for example.
$\textbf{Edit:}$
If $L$ is a big and nef bundle, it seems like the following will go through still,
a) Definition and basic properties of like there is no nontrivial homomorphism of semi stable sheaves from one with higher slope to one with lesser slope etc.
b) Existence and uniqueness of Harder-Narasimhan and Jordan-Holder filtrations
c) stability/semistability is an open condition.
Am I mistaken. Is there some subtle/obvious point where we need ampleness for this?

REPLY [7 votes]: You are right, the condition that $L$ be ample can be weakened. In fact, on an $n$-dimensional normal projective variety $X$ one can measure (semi-)stability with respect to an arbitrary movable curve class $\alpha \in N_1(X)_{\mathbb R}$. A numerical curve class $\alpha$ is said to be movable if $D \cdot \alpha \ge 0$ for all effective divisors $D$ on $X$. Then most of the basic properties you mentioned continue to hold.
The classical case ($L$ ample) is recovered by taking $\alpha$ to be the class of a complete intersection curve $C = H_1 \cap \cdots \cap H_{n-1}$, where the $H_i$ are general members of some very ample linear system $|mL|$, $m$ sufficiently large.
This (along with several applications) is worked out in the paper "Movable curves and semistable sheaves" by Greb, Kebekus, and Peternell.
You might also be interested in the survey "Moduli of vector bundles on higher-dimensional base manifolds - Construction and Variation" by Greb, Ross, and Toma, where this machinery is applied.<|endoftext|>
TITLE: Intermediate value for a vector-valued function
QUESTION [7 upvotes]: Consider a vector-valued function $f: [0,1]^n\rightarrow[0,1]^n$.  Write $f(x)=\{f_1(x), ..., f_n(x)\}$ with $x\in[0,1]^n$, where the $f_i: [0,1]^n\rightarrow[0,1]$ are  continuous functions with the following properties:
1) $f_i(\{x_1, x_2, ..., x_n\})=0$ if $x_i=0$.
2) $f_i(\{x_1, x_2, ..., x_n\})=1$ if $x_i=1$.
Fix $p\in[0,1]^n$. I want to prove that there exists a solution $x\in[0,1]^n$,  such that $f(x)=p$.
If feel that this should be true, since by the intermediate value theorem, we know that for fixed $x_2$, ..., $x_n$ we can find $x_1$ such that $f_1(\{x_1, x_2, ..., x_n\})=p_1$ (and similarly for $f_2$ we can find an $x_2$ if we fix the other $x_i$'s).
Any help is appreciated.

REPLY [7 votes]: This has its own Wikipedia entry Poincaré–Miranda theorem and is similar to the Brouwer fixed-point theorem.

REPLY [5 votes]: This is a straightforward application of degree theory.
First of all, by induction you can reduce to the case when $p\in(0,1)^n$.
Then, note that the a linear homotopy between $f$ and the identity map such that the image of the boundary of the unit box remains in the boundary. Thus
$$\deg(f,(0,1)^n,p)=\deg(I,(0,1)^n,p)=1$$.
It follows that $f(x)=p$ has at least one solution.
Reference: Ambrosetti and Machiodi Nonlinear Analysis and Semilinear Elliptic Equations<|endoftext|>
TITLE: Discrete Wavelets
QUESTION [5 upvotes]: I am looking for research that has been done in Discrete wavelets. Let me be specific as Google doesn't give me what I want when I say "discrete wavelets". I don't want countable basis for $ L^2(\mathbb{R}) $, Daubechies book, "Ten Lectures on Wavelets", already has this.
I am looking for research on $ \ell^2(\mathbb{R})$ vector spaces and wavelets that form either orthogonal or preferably non-orthogonal basis for them with a compact support in the frequency and space domains.
In the standard wavelet theory, the major benefit of the vectors is their compact space and frequency resolution. In addition to compact space support, I would like "over-completeness" (ie when $ A>1 $ for unit vectors, see below for $ A $) for the noise reduction property. Having a chain of vector spaces like in wavelet multi-resolution analysis would be nice as well, ie $ V_0 \subset V_1 \subset ... \subset \ell^2 $.
Or in math terms, I am looking for research on sets $W = \{ \psi_\alpha) \}_\alpha \subset \ell^2(\mathbb{C})$ with the following properties:

$\operatorname{span}(W) = \ell^2(\mathbb{C})$;
$\sum_\alpha \langle f, \psi_\alpha \rangle \langle \psi_\alpha, g \rangle = A \langle f, g \rangle$ for all $f, g \in \ell^2(\mathbb{C})$ and $A \geq 0$.

Recall $\ell^2(\mathbb{C}) = \left\{f\colon \mathbb{Z} \rightarrow \mathbb{C} \ \middle\vert \  \sum_{n\in \mathbb{Z}} |f(n)|^2 < \infty \right\}$ with inner product $\langle f, g \rangle = \sum_{n \in \mathbb{Z}} \overline{f(n)} g(n)$.
If it doesn't make sense, please let me know.

REPLY [2 votes]: I followed the comment by mkreisel and discovered that I should have been Googling "Discrete-time Wavelet transform"
Here are links to a paper and a book,

A Discrete-time Multiresolution Theory
Wavelets and Sub-band Coding (see Chapter 3)

Thank you commenters<|endoftext|>
TITLE: Reference to a Don Zagier Result and the Congruent Number Problem
QUESTION [14 upvotes]: I was looking for a reference/explanation as to how Don Zagier managed to find the side lengths of a rational right triangle with area 157. There have been many literature references to the fact that Zagier was the first to find the triangle but no hard reference as to where one can look up the ideas used.
Does anyone here know where I could find this result? Thanks!

REPLY [16 votes]: Don Zagier's own explanation is here, page 4+5 (in German, my translation).

Consider the elliptic curve given by the equation $y^2=x(x+n)(x-n)$.
  If $P=(x,y)$ is an arbitrary nontrivial solution (meaning $y\neq 0$)
  of this equation, the point $P'=(x',y')$ constructed using the
  Diophantine tangent method has the property, that not only the product
  $x'(x'+n)(x'-n)$ ($=y'^2$), but also all three factors $x'$, $x'+n$,
  $x'-n$ are squares, hence $n$ is congruent. 
For the original solution
  $P$ the numbers $x$ and $x\pm n$ need not be squares, but they are
  strongly restricted up to quadratic factors. If for example $n$ is
  prime and $\equiv 5$ (mod 8), then one can easily show that each of
  these three numbers is of the form $\pm\square$, $\pm 2\cdot\square$,
  $\pm n\cdot\square$ or $\pm 2n\cdot\square$ (with $\square$ a
  rational square). This leads to the consideration of a finite number
  of cases, that have to be examined one by one. 
If for example
  $x=-A^2$, $x+n=B^2$, $x-n=-C^2$ (and in our case of $n$ prime,
  $n\equiv 5$ (mod 8) one can easily show that if there is any solution
  if must be of this form), then we need to solve the set of equations
  $C^2-B^2=2A^2$, $C^2-A^2=n$. The first equation can be immediately
  solved using the Diophantine method: it must hold that
  $A=2RS/M$, $B=(R^2-2S^2)/M$, $C=(R^2+2S^2)/M$ for suitable integers
  $R,S,M$. In this way the problem is reduced to the solvability of $M^2
> n=R^4+4S^4$. For $n=5$ the solution is evidently $M=R=S=1$
  ($\Rightarrow x=-4$, $y=6$, $x'=6\frac{97}{144}$). For other prime
  numbers $n$ one must occasionally repeat the descent one or more
  times, when in the first step one applies the Diophantine method to
  the quadratic equation $n=U^2+4V^2$ and tries to find a solution with
  $UV=\square$ ($\Rightarrow U=R^2/M$, $V=S^2/M$). For $n=157$ this
  method leads to a solution in a few steps. 
This is quite remarkable,
  as Fermat himself might have said, but the page is unfortunately too
  small to record the solution: The three rational squares have in
  numerator and denominator each almost 100 digits.<|endoftext|>
TITLE: Jacquet's approach to Rankin--Selberg L-functions
QUESTION [14 upvotes]: In his book "Automorphic Forms on GL(2), II", Springer Lecture Notes vol. 278, Jacquet defines the Rankin--Selberg L-function of $\pi_1 \times \pi_2$, where $\pi_i$ are automorphic representations of $\operatorname{GL}_2$ over a global field, and proves its analytic continuation and functional equation.
Is there any reference which explains the ideas of Jacquet's theory in a somewhat more concrete and explicit way? In particular, I'm interested in the case where the field is $\mathbf{Q}$ and the $\pi_i$ correspond to holomorphic modular eigenforms -- is there any account of the theory which makes explicit the dictionary between Jacquet's representation-theoretic approach and more "classical" objects?

REPLY [3 votes]: I'm not an expert on integral representations, but I don't think such a reference should exist.  The method of automorphic representations are crucial to Jacquet's approach which, analogous to Tate's thesis, is to define the local Rankin-Selberg L-factor as a gcd of zeta integrals $Z(s,\phi_v \times \phi_v')$ where $\phi_v, \phi_v'$ run over vectors in the associated local representations.  (There are also parameters for additive characters and Eisenstein series, which I supress.)  It seems to me there should be no classical translation of this approach, as there is no classical version of varying $\phi_v, \phi_v'$ locally in a $GL_2(\mathbb Q_v)$-representation.
So now you may ask: how can you relate the $L$-function to a classical Rankin-Selberg integral?
In the unramified case, Jacquet proves that $Z(s,\phi_v \times \phi_v') = L(s, \phi_v \times \phi_v')$ where $\phi_v, \phi_v'$ are new vectors.  This means the classical approach to Rankin-Selberg should give the right factors at the unramified primes.  However, Jacquet does not determine "test vectors" $\phi_{v}, \phi_{v}'$ where the zeta integral spits out the $L$-factor on the nose in general.  Indeed, determining test vectors is a nontrivial problem and they are known not to always exist in more general settings (i.e., the gcd may not be attained for any choice of $\phi_{v}, \phi_v'$).  In this setting however, test vectors probably do exist, and as I recall under some conditions they were determined in the (unpublished but accessible) thesis of Kim Mi-Kyung, a fairly recent student of Cogdell.)  
The problem is that without knowing how $Z(s,\phi_v \times \phi_v')$ relates to $L(s, \phi_v' \times \phi_v')$ when $\phi_v, \phi_v'$ are newvectors for ramified representations, you can't relate the complete $L$-function to a classical Rankin-Selberg integral exactly.  (One also needs the analogous archimedean theory; in this case Jacquet does determine test vectors.)  Possibly one can do more with the classical Rankin-Selberg integrals now in light of Kim's thesis (I don't remember exactly what she did), but I'm not aware of any such work so far. 
If you still want to understand the idea of Jacquet's approach, besides Bump's book and his surveys (discussing more general Rankin-Selberg cases), Cogdell has several nice notes on the Rankin-Selberg method which include the more general setting of $GL(n) \times GL(m)$ (e.g., his Fields lecture notes).<|endoftext|>
TITLE: Lemma 2 from Beilinson's "Coherent Sheaves on $\mathbb{P}^n$ and Problems of Linear Algebra", intuition?
QUESTION [7 upvotes]: This is a followup to here.
Consider Lemma 2 from Beilinson's paper "Coherent Sheaves on $\mathbb{P}^n$ and Problems of Linear Algebra", as follows.

Lemma 2. For any pair $i$, $j$ such that $0 \le i$, $j \le n$, and $l \ge 1$:$$\text{Hom}(\Omega^i(i), \Omega^j(j)) = \Lambda^{i-j}(V^*),\text{ Ext}^l(\Omega^i(i), \Omega^j(j)) = 0,$$$$\text{Hom}(\mathcal{O}(-i),\mathcal{O}(-j)) = S^{i-j}(V),\text{ Ext}^l(\mathcal{O}(-i),\mathcal{O}(-j)) = 0,$$where composition of morphisms coincides with multiplication in $\Lambda^\bullet(V^*)$ and $S^\bullet(V)$, respectively.
The lemma is proved by induction with the aid of the exact sequence$$0 \to \Omega^i(i) \to \Lambda^i(V) \otimes \mathcal{O} \to \Omega^{i-1}(i) \to 0.$$

My question is, how do others think about it/the (hopefully underlying geometric) intuition for this? What role does the "dual exceptional collection" intuition play here?

REPLY [4 votes]: $\newcommand\Ocal{\mathcal{O}}\newcommand\Hom{\mathop{\mathrm{Hom}}}\newcommand\Ext{\mathop{\mathrm{Ext}}\nolimits}$I think the following is a rather clumsy way of proving it. It is an induction on two variables, but probably there are better ways to set up the induction. The idea is to prove it for

$i$ between 0 and $n$, $j=0$
$i=0$, $j$ between 0 and $n$
show that the cases $(i,j)$ and $(i-1,j-1)$ are isomorphic

I will denote the sequence that Beilinson suggests to do the induction with by $(\#_i)$.
First part The case $i=0$ is clear. To see that $\Hom(\Omega^i(i),\Ocal)=\bigwedge^iV^*$ for $i\geq 1$ we apply $\Hom(-,\Ocal)$ to $(\#_i)$. One gets a four-term exact sequence
$$0\to\Hom(\Omega^{i-1}(i),\Ocal)\to\Hom(\bigwedge\nolimits^iV\otimes\Ocal,\Ocal)\to\Hom(\Omega^i(i),\Ocal)\to\Ext^1(\Omega^{i-1}(i),\Ocal)\to 0$$
and isomorphisms $\Ext^l(\Omega^i(i),\Ocal)\cong\Ext^{l+1}(\Omega^{i-1}(i),\Ocal)$ for $l\geq 1$. The second term in the sequence is isomorphic to $\bigwedge^i V^*$, which is what we are after. So we would like to have that $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$.
Lemma $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$, and $i=1,\dotsc,n$.
Proof The case $i=1$ is clear. So assume that $i\geq 2$, and now we apply $\Hom(-,\Ocal(-1))$ to $(\#_{i-1})$. In the long exact sequence we get we see that $\Ext^l(\bigwedge\nolimits^{i-1}V^*\otimes\Ocal,\Ocal(-1))=0$ for all $l$. But then we apply the induction hypothesis, and conclude that everything is zero. $\square$.
So we get the desired isomorphism in this case.
Second part The case $j=0$ is clear. The case $j=n$ is also clear, because $\Omega^n(n)=\Ocal(-1)$. Then we rewrite the short exact sequence $(\#_{j+1})$ as
$$0\to\Omega^{j+1}(j)\to\bigwedge\nolimits^{j+1}\otimes\Ocal(-1)\to\Omega^j(j)\to 0$$
and then applying $\Hom(\Ocal,-)$ to this modified sequence gives a long exact sequence where $\Ext^l(\Ocal,\bigwedge\nolimits^{j+1}V\otimes\Ocal(-1))=0$, and similar to the previous lemma we can sandwich the desired term between zeroes.
Third part By now the ideas should be clear. Assume that $i\geq 1$ and $j\geq 1$. Apply $\Hom(-,\Omega^j(j))$ to $(\#_i)$. By the second part we see that the second (and fifth, etc.) term vanishes. We get another $\Ext^l=\Ext^{l+1}$ isomorphism.
Now apply $\Hom(\Omega^{i-1}(i),-)$ to $(\#_j)$. We get yet another long exact sequence, where by induction the third term $\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j))$ is isomorphic to $\bigwedge\nolimits^{i-j}V^*$ (and higher $\Ext^l$ vanish).
Now we do yet another induction on as in the lemma, to see that the second (and fifth, etc.) term vanishes. Hence tracing back all the isomorphisms we see that
\begin{align}
\bigwedge\nolimits^{i-j}V^*
&\cong\Hom(\Omega^{i-1}(i-1),\Omega^{j-1}(j-1)) \\
&\cong\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j)) \\
&\cong\Ext^1(\Omega^{i-1}(i),\Omega^j(j)) \\
&\cong\Hom(\Omega^i(i),\Omega^j(j))
\end{align}
as desired. Hurray!
The geometric intuition should correspond to its interpretation as a dual exceptional collection, but I don't see a way to make this into an actual explanation. The above is just a proof by tedious and boring computation. I hope I didn't make mistakes.<|endoftext|>
TITLE: Inverse limit of $\Bbb Q/q\Bbb Z$ isomorphic to finite adeles?
QUESTION [6 upvotes]: Let $\Bbb Q/q\Bbb Z$, for some positive rational $q$, denote the quotient group of the discrete rationals by the subgroup of integers times $q$.
For any $q_1, q_2 \in \Bbb Q^+$ and $n \in \Bbb N^+$ such that $q_2 = n \cdot q_1$, we can form a surjective homomorphism $\Bbb Q/q_2\Bbb Z \to \Bbb Q/q_1\Bbb Z$.
The set of all such $\Bbb Q/q\Bbb Z$ forms a poset with these surjections, and we can take the inverse limit to get something like a "rational solenoid."

Is this group isomorphic to the finite adele ring $\Bbb A_\Bbb Q^f \cong \Bbb Q \otimes \hat{\Bbb Z}$?

Is this group equal to the same thing you'd get if you took the inverse limit of $\Bbb Q/n\Bbb Z$ for a natural number $n$ instead of a positive rational $q$?


For #1, I think it is, because the dual group should be a direct limit of localizations of $\hat{\Bbb Z}$, the group of profinite integers, which I believe is isomorphic to $\Bbb Q \otimes \hat{\Bbb Z}$, which is self-dual. The same reasoning applies for #2.
This question was originally asked on MSE here. It got upvoted but no answers, so I'm giving it a shot to repost here.

REPLY [7 votes]: For 1 you can indeed use duality, but you can also give an explicit isomorphism:
First for each rational number $q$ it is easy to see that $\mathbb{A}^{f}/(q \hat{\mathbb{Z}}) \simeq \mathbb{Q}/(q\mathbb{Z})$: indeed, $(q \hat{\mathbb{Z}})$ is open and $\mathbb{Q}$ is dense in the finite adeles, so the image from $\mathbb{Q}$ to this quotient is surjective and its kernel is $q \mathbb{Z}$.
So any finite adele gives you an element of the projective limit,  this element is zero if and only if the adele $x$ you started from is in $q \hat{\mathbb{Z}}$ for each $q$ which implies it is zero hence the adele inject into the projective limit.
Conversely, you need to work just a little more but an element $y$ of the projective limit is the data of a compatible familly of $y_q \in \mathbb{Q}/(q \mathbb{Z})$. pick for each $y$ a lifting of $y_q$ in $\mathbb{Q}$ the compatiblity of the familly implies that $q_y$ is a cauchy net (for the divisibility order on rational numbers of course, if you don't like net then $y_{n!}$ is going to be a cauchy sequence that works equally well) in $\mathbb{A}^f$ which hence converge to an adele $x$ whose images in $\mathbb{Q}/(q \mathbb{Z})$ are the $y_q$.
For your second question, the answer is also yes: you can either do the exact same proof as above and check that the integer are enough for the arguments to work. Or observe that in your projective limit $\mathbb{N}^* \subset \mathbb{Q}^*$ is cofinal for the ordering and hence the projective limit indexed by $\mathbb{N}^*$ and by $\mathbb{Q}^*$ are the same.<|endoftext|>
TITLE: Probability distribution derived from gamma function - does it have a name?
QUESTION [11 upvotes]: Consider the complex gamma function, denoted by $\Gamma(\sigma+it)$.
Now, let's fix $\sigma$ and let t vary. Then consider the following expression:
$$|\Gamma(\sigma+it)|^2$$
For any choice of $\sigma$ such that $\Gamma(\sigma)$ isn't a pole, this will appear to be (almost) a two-sided probability density function, save for that it isn't normalized. It decays around as quickly as $e^{-|t|}$, somewhat resembles the function $e^{-\sqrt{1+t^2}}$, and appears related to the hyperbolic distribution.

Is there a precise closed-form expression for this function in terms of elementary or Louvillian functions?
Is there a name for this probability distribution (assuming it's normalized)?

This question was originally asked on MSE here. It got some attention but no answers; someone commented I should repost here.

REPLY [4 votes]: This is called Generalized Hyperbolic Secant Distributions. See  Generalized Hyperbolic Secant Distributions of W. L. Harkness and M. L. Harkness (http://www.jstor.org/stable/2283852). 
In particular, see equation (3) in that paper and use the fact that $\overline{\Gamma(z)}=\Gamma(\bar z)$<|endoftext|>
TITLE: The role of the index set in the product of uncountably many topological spaces
QUESTION [6 upvotes]: Let $‎\langle‎ ‎‎X_i,\mathcal{T}_i \rangle_{i\in I}$ be a family of topological spaces. Consider $X=\prod_{i\in I} X_i$ with product topology.
Question. Is there a topological property that holds in $X$ when $|I|=\aleph_{\beta}$, but does not hold whenever $|I|=\aleph_{\alpha}$, for some $0<\alpha<\beta$ ?
Any reference or helpful comment will be appreciated.

REPLY [2 votes]: The splitting number $\mathfrak{s}$ is the minimal cardinal $\kappa$ such that if $\mathcal{A} \subseteq \mathcal{P}(\omega)$ has size $< \kappa$, there exists some $X \subseteq \omega$ such that for all $A \in \mathcal{A}$, $X \setminus A$ is finite or $X \cap A$ is finite.
Then if all $X_i$ are sequentially compact Tychonoff spaces, and $|I| <\mathfrak{s}$ then $\prod_{i \in I} X_i$ is also sequentially compact. And $\{0,1\}^{\mathfrak{s}}$ is not sequentially compact. What $\aleph_\alpha$ equals $\mathfrak{s}$ cannot be said in ZFC. It could be $\alpha = 1$ (under CH, e.g.) or much bigger, depending on the size of the continuum. See van Douwen's paper in the Handbook of Set Theoretic Topology on cardinal invariants of the continuum.<|endoftext|>
TITLE: Removing large cardinals from an uncountable transitive model
QUESTION [19 upvotes]: The usual way of removing large cardinals from a given model of set theory is to cut off the model below the least large cardinal of interest. But this method may have dramatic effects on the external properties of the model, the simplest of which is cardinality.
Question: Suppose there is an uncountable transitive model of ZFC. Is there also an uncountable transitive model of ZFC + "There are no inaccessibles" ?
I don't have a strong intuition about which way this should go, but I would expect the answer to be positive. Having spent some time thinking about the problem, here are some observations.
Fix an uncountable transitive model $M$ and write $M_\alpha$ for $V_\alpha^M$. We may assume that for any $\eta$ inaccessible in $M$, the initial segment $M_\eta$ is countable, for if not we can replace $M$ by the first uncountable such initial segment. Since we have excluded the possibility of cutting off the model $M$ and passing to an inner model clearly will not help with getting rid of the inaccessibles, the only alternative that seems to be left is some sort of forcing construction.
The problem now splits into a number of cases:

$M$ has boundedly many inaccessibles. If we let $\eta$ be the supremum of the inaccessibles of $M$ then $\eta$ (and $M_\eta$) is countable. To destroy the inaccessibles we could now try to force with a poset of size $\eta$ in $M$, for example $\mathrm{Add}(\omega,\eta)$. The issue, of course, is that, for this to have any bearing on the original question, the $M$-generic for this forcing needs to exist in $V$. This is fine if $(2^\eta)^M$ is countable, since then $M$ only has countably many dense subsets of the forcing (as seen in $V$), but if this is not the case we seem to be stuck. Alternatively, if we could at least arrange that $(2^\eta)^M<\mathfrak{c}^V$ then an appropriate fragment of MA in $V$ would still give an $M$-generic.
$M$ has class many inaccessibles and $M\models\mathrm{Ord\ is\ not\ Mahlo}$. In this case we can fix a definable over $M$ class club $C$ in $\mathrm{Ord}^M$ not containing any of $M$'s inaccessibles. We now force over $M$ with a class-length Easton product $\mathbb{P}$ of the posets $\mathbb{Q}_\delta=\mathrm{Add}(\delta^+,\delta^*)^M$ where $\delta\in C$ and $\delta^*$ is the next point of $C$ above $\delta$. The key point now is that $V$ has $M$-generics for $\mathbb{P}$. This is because $M$ is $\omega_1$-like, meaning that all of its elements are countable, and remains such through all of the initial stages of the forcing $\mathbb{P}$. This means that, given any $\delta\in C$, the extension $M^{\mathbb{P}_\delta}$ constructed thus far only has countably many
dense subsets of $\mathbb{Q}_\delta$, which allows $V$ to build a $M^{\mathbb{P}_\delta}$-generic $G_\delta\subseteq\mathbb{Q}_\delta$. Finally, it follows that $G=\prod_\delta G_\delta\subseteq \mathbb{P}$ is $M$-generic. As it is well known that forcing with $\mathbb{P}$ preserves ZFC, the model $M[G]$ is an uncountable transitive model without inaccessibles.
$M\models\mathrm{Ord\ is\ Mahlo}$. In this case we would like to undertake the same construction as in the previous bullet point after first forcing over $M$ to add a class club avoiding the inaccessibles. The problem again is with the generic club existing in $V$. The model $M$ is $\omega_1$-like again, so the club shooting forcing has $\omega_1$ many definable dense classes. The forcing is $\alpha$-strategically closed in $M$ for any $\alpha$, but I don't see that this would allow $V$ to build a generic.

REPLY [4 votes]: i added an answer and then realized it was a complete nonsense, i was getting at the following question.
Suppose kappa is an inaccessible cardinal. Is there a transitive model M of ZFC such that

M has height kappa

for some stationary in kappa set S, for every alpha in S, M computes alpha^+ correctly,

for some stationary set D, for every alpha in D, alpha is inaccessible in M

there are no inacessibles below kappa.
One can show that if the answer is yes then there is an inner model with a proper class of measurables.


Assume M is as in the statement. Assume there is no inner model with a proper class of measurables. Then clause 2 implies that K^M and K must coiterate (K is the core model). This means that K has stationary set of inaccessibles, so by Trevor's trick V must also have inaccessibles. So we must have covering fails in V which then implies that K doesn't exist.<|endoftext|>
TITLE: 3-manifolds homotopy equivalent to a surface
QUESTION [6 upvotes]: I have heard that an open, orientable 3-manifold $X$ (non-compact, without-boundary) that is homotopy equivalent to an orientable surface $S_g$ must itself already be homemorphic to $S_g \times \mathbb R$. There seems to be a very deep theorem behind it, however, I couldn't find any reference which would bring this up. Does anybody know a good reference for this ?
Edit: The original question has been answered well. I wonder in how far the assumptions have to be strengthened in order for this to be true. The Tameness Theorem shows that $X = S_g \times \mathbb R$ whenever $X$ additionally admits a complete hyperbolic metric. However, for my purposes, it would be nice to have purely topological assumptions. I have the following suggestions:
$X$ is as above, but additionally we require $X$ to have two ends, such that for each end, there exists a sequence of end-neighborhoods $U_1 \supset U_2 \supset U_3 \supset ....$ with $\bigcap cl(U_i) = \emptyset$ and the following properties:

$U_1$ is homotopy equivalent to $S_g$
the natural map $i_*: \pi_1(U_{n+1}) \to \pi_1(U_n)$ is an isomorphism for all $n$.

Intuitively, this will prevent $X$ from "going crazy" at one end. Is this enough now ?

REPLY [2 votes]: This paper gives some other very nice counter-examples. As I recall, for open manifolds you really need to think about proper homotopies, not just homotopies.
MR1033220 (91b:57021) Reviewed 
Scott, Peter(1-MI); Tucker, Thomas(1-COLG)
Some examples of exotic noncompact 3-manifolds. 
Quart. J. Math. Oxford Ser. (2) 40 (1989), no. 160, 481–499. 
57N10 (57M10)<|endoftext|>
TITLE: Intuition for Zagier's theorem for $\zeta_K(2)$
QUESTION [29 upvotes]: In 1986, Don Zagier generalized Euler's theorem ($\zeta_\mathbb{Q}(2)=\pi ^2 /6$) to an arbitrary number field $K$:
$$\zeta_K(2)=\frac{\pi^{2r+2s}}{\sqrt{|D|}}\times \sum_v c_v A(x_{v,1})...A(x_{v,s})$$
where
$$A(x)=\int^x_0 \frac{1}{1+t^2}\log \frac{4}{1+t^2}dt$$
In this sense, the result is conjectured to hold for $2<s\in \mathbb{N}$, with the $A(x)$ replaced by more complicated functions $A_m(x)$
This might seem rather unenlightening, but we can also state Zagier's result like this:

$$\zeta_K(2)=\text{the volume of a hyperbolic manifold}$$

This amazing fact doesn't seem to have a direct analogue for $\zeta_K(2m)$ with $m \neq 1$

What I'd like to know is if there is any big picture explanation for the appearance of hyperbolic manifolds in this context.

Zagier's calculation is quite geometrical, but as far as I understand gives no clear explanation of "what the manifold is doing here".
Don Zagier, Hyperbolic manifolds and special values of Dedekind zeta-functions (1986)

REPLY [12 votes]: There is a K-theory reformulation of Zagier's theorem which explains both the appearance of the integrals $A(x)$ as well as the appearance of hyperbolic geometry. I will try to give a sketch what is behind the relation of the zeta-value with volumes of hyperbolic simplices. 
First, due to the work of Borel, there is a relation between zeta-values and regulators for K-theory. For a number field $K$, the Borel regulator embeds $K_{2n-1}(\mathcal{O}_K)$  as a lattice in some $\mathbb{R}$-vector space (later reinterpreted by Beilinson as Deligne-cohomology group $H^1_{\mathcal{D}}(\operatorname{Spec} K\otimes\mathbb{C},\mathbb{R}(n))$). The covolume of the lattice is, up to multiplication by a power of $\pi$ and an algebraic number, identified with the zeta-value $\zeta_K(-n+1)$. So, Borel's theorem states: The zeta-value is the covolume of the regulator embedding $K_3(\mathcal{O}_K)\to\mathbb{R}^{r_2}$.
In a second step, there is a relation between K-theoretic regulators and dilogarithms (following the work of Bloch, Suslin, Dupont-Sah...). Recall that the pre-Bloch group $\mathcal{P}(K)$ is given by generators $x\in K^\times\setminus\{1\}$ subject to the five-term relation $$[x]-[y]+[y/x]-[(1-x^{-1})/(1-y^{-1})]+[(1-x)/(1-y)]$$ for $x, y\in K^\times\setminus\{1\}$. The Bloch group $\mathcal{B}(K)$ is the kernel of the map
$$
\mathcal{P}(K)\to\bigwedge^2 K^\times:[x]\mapsto x\wedge(1-x).
$$
The five-term relation is a version of the functional equation for the dilogarithm. Hence, by definition, the single-valued real-analytic version of the dilogarithm yields a function mapping the Bloch group into $\mathbb{R}^{r_2}$ by taking a class $[x]$ to the values of the dilogarithm under the various complex embeddings $K\to\mathbb{C}$. The Bloch group can (up to 2-torsion) be identified with the indecomposable quotient of $K_3(\mathcal{O}_K)$. The important thing is that under this identification, the regulator on $K_3$ is mapped to the dilogarithm function on the Bloch group. (There are some factors, algebraic numbers and powers of $\pi$, but this does not cause problems for the identification.)
It is mentioned in Zagier's paper that $A(x)$ is essentially the dilogarithm. The combination of the above two steps yields the reformulation of Zagier's theorem: Expressing the regulator embedding by dilogarithms, the zeta-value is a determinant of dilogarithms of elements of $K$.
In a third step, the relation to hyperbolic geometry can now be obtained from the Bloch group. Via the cross-ratio, the pre-Bloch group is isomorphic to the third homology of the complex of points on $\mathbb{P}^1(K)$ (with the omit-one-point-differential) modulo the diagonal action of $GL_2(K)$. This reinterprets elements of the Bloch group as ordered $4$-tuples of points on $\mathbb{P}^1(K)$. Now four points on $\mathbb{CP}^1$ span a simplex in $\mathbb{H}^3\cup\partial\mathbb{H}^3$. One can talk about hyperbolic $3$-polytopes up to scissors congruence. It is possible to prove that the group of scissors congruence classes in $\mathbb{H}^3$ is isomorphic to the group of scissors congruence classes of polytopes in $\mathbb{H}^3\cup \partial\mathbb{H}^3$ whose vertices lie on $\partial\mathbb{H}^3$ (this is done in the book "Scissors congruences, group homology and characteristic classes" by J.-L. Dupont). 
We can now explain the link between the Bloch group and hyperbolic geometry by: $\partial\mathbb{H}^3\cong \mathbb{CP}^1$. More precisely, the Bloch group is a group of scissors congruence classes of hyperbolic $3$-polytopes with vertices on $\partial\mathbb{H}^3$.
Combining the above three steps means that $\zeta_K(2)$ can be expressed as a combination of volumes of hyperbolic simplices. This statement actually has a conjectural generalization for all $\zeta_K(n)$. Zagier formulated the corresponding conjecture in "Polylogarithms, Dedekind zeta functions and the algebraic K-theory of fields". There are analogues of the Bloch group which conjecturally can be identified with higher K-groups $K_{2n-1}(\mathcal{O}_K)_{\mathbb{Q}}$, such that the regulator on the K-group can be expressed in terms of $n$-logarithms. If true, this conjecture would allow to express $\zeta_K(n)$ as a determinant of $n$-logarithms, or equivalently as a linear combination of volumes of hyperbolic $(2n-1)$-simplices with vertices on the boundary. The trilogarithm case of Zagier's conjecture was proved by Goncharov, see

A. Goncharov. Volumes of hyperbolic manifolds and mixed Tate motives. J. Amer. Math. Soc. 12 (1999), 569-618. 

As a final note, the relation between polylogarithms and volumes of hyperbolic simplices goes back to Lobachevskiy, Schläfli,...<|endoftext|>
TITLE: why don't (can't?) we sheafify the structure presheaf of an adic space
QUESTION [8 upvotes]: In the definition of an adic space, usually there is a presheaf defined by first saying what it is on a particular basis of the topology of the underlying space, the so called rational subsets. One then extends this to arbitrary opens by taking the limit over all the rational subsets inside the given open. However, in all the references I looked into so far, it is then quickly pointed out that this is in general not a sheaf, followed by a list of particular cases in which it is one. 
In Algebraic Geometry, there are many occasions where some construction does not give a sheaf and one simply forces it to be one by saying "Sheafify!" So the Question is:
-Why doesn't one sheafify the structure presheaf of an adic space? Is there no sheafification functor in this case? If not, what goes wrong?

REPLY [2 votes]: There is actually a sheafification process used in classical construction of rigid geometry, which is used to pass from the weak G-topology to the strong one (see section 9.2.2 of Bosch, Güntzer, Remmert's Non-Archimedean Analysis, for instance). 
As to why people don't use it in the case of non-sheafy spaces, I would say that if depends on what you want to do. If it's enough for you to know the stalks of the structure sheaf, then fine. But, as Dylan Wilson pointed it out, you are likely not to end up with the global sections you expect. For an explicit example, see Theorem 3.15 in Mihara's paper "On Tate Acyclicity and Uniformity of Berkovich Spectra and Adic Spectra" (http://arxiv.org/abs/1403.7856).<|endoftext|>
TITLE: Are the integer matrices in SO(3,2) "boundedly generated"?
QUESTION [6 upvotes]: Let $G$ be the subgroup of integer matrices in $\mathrm{SO}(3,2)$.

(The invertible linear maps from a $5$ dimensional real vector space to itself which leave invariant a nondegenerate symmetric billinear form of signature $(3,2)$, and have determinant $1$.)

$G$ is a group of real rank $2$, so conjecturally, it should be
  boundedly generated.

(There exists an $m \in \mathbb{N}$, and cyclic subgroups $C_1, \dots, C_m \leq G$ such that $G = C_1 \cdots C_m$.)
I would like to know whether the following much weaker statement holds:

There exists a positive integer $m$, and finitely generated subgroups
  $H_1, \dots, H_m \leq G$ of infinite index, such that $G = H_1 H_2
 \cdots H_m$.

REPLY [6 votes]: $\mathrm{SO}(3,2)_{\mathbb{Z}}$ is an orthogonal group of $\mathbb{Q}$-rank 2, so it is boundedly generated. This was proved by Tavgen [Math. USSR. Izvestiya 36 (1991) 101-127; MR1044049] for the quasi-split case, which includes $\mathrm{SO}(3,2)_{\mathbb{Z}}$, but the general case is due to Erovenko and Rapinchuk [J. Number Theory 119 (2006) 28-48; MR2228948]. For most arithmetic groups of real rank $\ge 2$, it is not known whether they are boundedly generated.
The Erovenko-Rapinchuk proof is by induction, and the induction step writes the group as a product of finitely many subgroups $H_i$, as in the question's weaker statement. (Each $H_i$ is the stabilizer of a vector.)  Although it does not seem to be spelled out in their paper, I have heard from the authors (in talks and elsewhere) that their methods to write the group as a product also work for unitary groups (and maybe others?), not just orthogonal groups. So the methods may suffice to establish the weaker statement for isotropic arithmetic groups of classical type. Or perhaps the weaker statement is fairly easy for isotropic groups?<|endoftext|>
TITLE: Motivation for cyclotomic units
QUESTION [5 upvotes]: I am wondering the original motivation for considering cyclotomic units. Maybe one can rephrase the question as:

Why did people initially consider such units in $\mathbb{Q}(\zeta_p)$ specially?

There are many interesting results we can get with the notion of cyclotomic units, for example, comparison between the class number and the index of the subgroup of cyclotomic units. However, I think that they are not the philosophical origin of cyclotomic units. One another but stupid rephrase of the question is:

Why was the name cyclotomic taken only for such units, not for every unit in $\mathbb{Q}(\zeta_p)$?

REPLY [7 votes]: I'm quite sure cyclotomic units were first introduced by Kummer in his 1847 paper, where he proved his very famous partial solution to Fermat's Last Theorem:

Theorem (Kummer) If $p$ is an odd prime that does not divide the class number of the field $\mathbb{Q}(\zeta_p)$ and $(xyz,p)=1$, then
$$x^p+y^p=z^p$$
has no rational solutions.

In order to get the proof started, he needed to study $\mathbb{Z}[\zeta_p]$, and in particular, certain units, which we know call cyclotomic.
As for the name, I guess they wanted to single out the particularly relevant ones (and only those).
The first chapter in Lawrence Washington's "Introduction to Cyclotomic Fields" is a great exposition of the proof.<|endoftext|>
TITLE: When a compact topological manifold with boundary is a ball?
QUESTION [15 upvotes]: Let $X$ be a compact topological manifold with boundary. Suppose its interior is homeomorphic to  $\mathbb{R}^n$. Is $X$ homeomorphic to a (closed) ball?
Context: I want to show that a certain compactification of moduli of convex $n$-polygons (up to scaling and rotations) is a $(2n-4)$-cell. I can degenerate all sides (but $2$, of course), keeping records of the slopes, and all angles to $\pi$ (again except $2$). If more structure is needed, like smoothness on the interior, PL at the boundary - you have it.

REPLY [7 votes]: Igor's answer is spot-on, avoiding the h-cobordism theorem, which is fragile in low dimensions. Here is a proposal to tweak his answer (which works in all dimensions), so as to use the Poincare Conjecture only once, just where it is required.
Step 1. Boundary X is a homotopy sphere (noted already), and hence a sphere, by the PC. (Big machinery here, which varies with the dimension, but there's no way to avoid it. If you know in advance that boundary X is a sphere, so much the better.)
The next two steps use two classic theorems of Morton Brown (both mentioned already), and they work uniformly in all dimensions.
Step 2. Boundary X is collared in X (Brown, beautifully redone by Connelly). And so, by pulling X inward into its interior, we can assume that X lies in R^n, hence in S^n.
Step 3. Now Brown's Schoenflies Theorem (pure magic) implies that boundary X bounds a closed ball. Done.<|endoftext|>
TITLE: Intuitive reasons for the existence of modular parametrizations
QUESTION [10 upvotes]: Whenever I encounter anything about modular parametrizations, I have a feeling it is something very unnatural: you have some kind of moduli space and all of a sudden it parametrizes an object represented by a point of some similar moduli space (sometimes even of itself?). Why on earth should this happen? How such a strange possibility could possibly occur to anybody?
And yet I know this is one of the most deep and important achievements in current mathematics. Can it be given some sort of heuristic/intuitive justification an outsider could understand?
Proper reaction to such questions probably is "why don't you read this or that paper"? However I would only accept such answer if the corresponding paper contains a good noob-friendly introduction since I certainly am a noob in this case.

REPLY [4 votes]: Take an elliptic curve. At each prime p you have some local information. It either has bad reduction, and there is a reduction type, or good reduction, and then you count the points and the number is $1+p-a_p$ with $a_p$ an integer between $-2 \sqrt{p}$ and $2 \sqrt{p}$. Also if the reduction is multiplicative, one should remember split or nonsplit reduction. One should also ignore the difference between reduction types that can be isogenous to each other. This information turns out to be the natural information about the elliptic curve to use in a whole lot of arithmetic situations.
Suppose I give you, for each prime, something that looks like this bit of information about the reduction of the elliptic curve - either a fiber type or a number $a_p$. How can you ever hope to tell if there is an elliptic curve that stitches all these disparate numbers together? Well, you can write down some necessary conditions, like that there are only finitely many primes of bad reduction. But there will still be only uncountably many "plausible" sequences, of which only countably many come from an elliptic curve. What makes them special?
Now take a modular form. This is an object that has primes of good reduction and bad reduction. The good reduction primes will just be the primes not dividing the level. At these primes, we have a Hecke eigenvalue, which is a real number between $-2 \sqrt{p}$ and $2\sqrt{p}$ - in fact it lies in the ring of integers of some number field, depending on the modular form. What's more, we can look at the action of the automorphism group of the modular curve on our modular form. If we look at the different representations that can appear, they turn out to be in perfect correspondence with the different reduction types of an elliptic curve (at least when the coefficient field is $\mathbb Q$).
So you have two different ways of stitching together this local information to get global objects - elliptic curves and modular forms whose coefficient field happens to be $\mathbb Q$. In both, you choose only a countable set of special sequences out of an uncountable set of possibilities. Could they be the same?
That would be quite impressive, because the ways they arise are so different! One comes from looking arithmetically at a natural class of algebraic geometry objects, and the other from doing analysis on some special symmetric spaces.
It would be particularly impressive because some facts (like the bound on the $a_p$) are much easier to prove on the elliptic curve side, while others (like a bound on the average of $a_p/\sqrt{p}$) are much easier to prove on the modular curve side.
Of course this sounds totally ridiculous. Those two things should be totally unrelated! Except that you can get, by a not-too-difficult construction, elliptic curves from modular forms...
So you have two ways to solve this incredibly mysterious problem of patching together local data at each prime to get a single global picture. They behave the same, and one is a subset of the other. Might the subset in fact be the whole - could these two pictures be the same? Once you came up with the question, and understood its importance, you would start searching for evidence, counterexamples, etc., and like the mathematical community in the mid-late 20th century, be convinced that they are.
Now this correspondence between elliptic curves and modular forms does not mention the actual map from the modular curve to the elliptic curve. Constructing it, giving the equality of $a_p$s, is nontrivial (the Tate conjecture for morphisms of abelian varieties) and doesn't really generalize. That is one reason you shouldn't think about this too geometrically. This is an arithmetic statement, not a geometric one. Note that it is only true for elliptic curves over $\mathbb Q$.
Also note that the fact that the modular curves are moduli spaces of elliptic curves is a coincidence, akin to the small number coincidences from Lie groups, like the fact that $PGL_2=SO_3$.<|endoftext|>
TITLE: Is every open convex subset of a Riemannian manifold necessarily contractible?
QUESTION [9 upvotes]: Question: Is every open convex subset $C$ of a Riemannian manifold $M$, necessarily contractible?
Here by a "convex subset" I mean a set $C$ having the property that between each  pair of points in $C$ there is a unique geodesic contained in $C$. (For example an open hemisphere is convex inside the sphere)
I have tried to construct a deformation retract along the unique geodesic connecting every point in $C$ to a fixed point in $C$. But is this map always continuous? If the answer is not positive in general, I'm also interested in the special case of complete manifolds with non-positive curvature.
Added. For the case of complete manifolds with non-positive curvature one can argue as follows: Let's $q\in C$ be the fixed point and $p\in C$. Now if $\gamma(t) = \mathrm{exp}(tv) (t\in[0,1])$ is the unique geodesic in $C$ connecting $q$ to $p$, there is a neighborhood $U$ of $v$ in $T_q M$ s.t. $\exp([0,1]*U)$ is contained in $C$. Now since $\exp$ has no critical points in non-positive curvature, $\exp(U)$ contains an open neighborhood of $q$ and we can use inverse mapping theorem.
Thanks!

REPLY [5 votes]: Fix $q \in C$. The set $D\subseteq T_q M$ of all tangent vectors $v$ such that $[0,1]\ni t \mapsto \exp_q(tv)$ is the unique geodesic in $C$ connecting $q$ with $p$ is open and star shaped, hence contractible.
Then $\exp_q: D \to C$ is a continuous map between manifods with the same dimension. It is also one-to-one, by your convexity hypothesis. By Browuer invariance of domain it is necessarily an homeomorphism.<|endoftext|>
TITLE: Compact Quantum Groups and FRT-Algebras
QUESTION [6 upvotes]: As is well known, every compact quantum group in the sense of Woronowicz has a dense Hopf $*$-sub-algebra. For the case of $q-SU(n)$ (among others) this Hopf $*$-sub-algebra is an FRT-algebra, which is to say, roughly, that they can be constructed from an R-matrix in the Yang--Baxter sense. (This goes back to the roots of quantum groups as objects in mathematical physics - the quantum inverse scattering problem to be exact.) Can we tell from the compact quantum group itself when the Hopf $*$-sub-algebra has an R-matrix structure?

REPLY [2 votes]: Already for ${\rm SU}_q(n)$ you have to remember that there is an extra relation about the quantum determinant being $1$. Anyhow, the correspondence you mention comes the fact that intertwiners of finite dimensional representations of ${\rm SU}_q(n)$ can be constructed from eigenspace decomposition of the braiding acting on $(\mathbb{C}^n)^{\otimes k}$ (and the extra embedding $\mathbb{C} \to (\mathbb{C}^n)^{\otimes n}$ corresponding to the q-determinant). Because of the Tannaka-Krein duality, the relations between matrix coefficients correspond to intertwiners in the representation category exactly like for usual compact groups. So, if the representation category is (almost) generated by braiding, you will have an "R-matrix" structure on the regular functions. The free orthogonal groups are such examples, since they have the same representation category as ${\rm SU}_q(2)$.<|endoftext|>
TITLE: A property of the derivatives of a function
QUESTION [6 upvotes]: Suppose that $f,g_1,g_2,\dots$ are functions from $\mathbb{R}$ to $\mathbb{R}$ such that $f'=f\,g_1$ and $g'_j=g_j^2-g_j g_{j+1}$. Here and in what follows, $j$ is any natural number. Then, by induction, $f^{(j)}=f\, P_j(g_1,\dots,g_j)$, where 
$$P_j(u_1,\dots,u_j)=\sum_{k_1=0}^j\cdots\sum_{k_j=0}^j c_{k_1,\dots,k_j}
u_1^{k_1}\cdots u_j^{k_j}
$$
is a polynomial in $\mathbb{R}[u_1,\dots,u_j]$ of degree $j$. 
The problem is to show that 
$$\sum_{k_1=0}^j\cdots\sum_{k_j=0}^j |c_{k_1,\dots,k_j}|\, 
1^{k_1}2^{k_2}\cdots j^{k_j}=2^{j - 1} j! 
$$
for all natural $j$. 
This equality has been verified for $j=1,\dots,10$.

REPLY [9 votes]: We have, writing for short $P_n=P_n(g_1,\dots,g_{n })$,
$$\big(f^{(n)}\big)' =\big(fP_{n }\big)'=f' P_{n }+\sum_{i=1}^n f\partial_iP_ng_i'=fg_1 P_{n }+\sum_{i=1}^n f\partial_iP_n(g_i^2-g_ig_{i+1})$$
Comparing with $f^{(n+1)}=fP_{n+1}$ we obtain a linear recursion for the sequence of polynomials $P_n=P_n(x_1,\dots,x_{n })$
\begin{cases} P_1=x_1  \\ P_{n+1}=x_1P_n  +\sum_{i=1}^n (x_i - x_{i+1}) x_i\partial_i P_n, \end{cases} 
whence it is clear that $P_n$ is a homogeneous polynomial of degree $n$, $P_ n:=\sum_{|\alpha|=n} c(\alpha)x^\alpha$, the sum being extended over all multi-indices $\alpha:=(\alpha_1,\alpha_2,\dots )$ of weight 
$|\alpha|:=\alpha_1+\alpha_2+\dots=n.$
Equating the coefficients of the monomial $x^\alpha:=x_1^{\alpha_1}\dots x_{n+1}^{\alpha_{n+1}} $ of degree $n+1:=|\alpha|$ in the above recursive formulas  we obtain:
$$c(\alpha)= \alpha_1 c(\alpha-\delta_1)- \sum_{i=2}^{n} (\alpha_{i-1}-\alpha_{i }+1)c(\alpha-\delta_i) \; - \alpha_{n}c(\alpha-\delta_{n+1}),$$
where $\delta_i:=(\delta_{i1},\delta_{i2},\delta_{i3}\dots)$ and $\delta_{ij}$ is the usual Kronecker symbol.
From this formula for $c(\alpha)$ it follows easily by induction that
(i) If $c(\alpha)\neq0$ then $\alpha_1\ge\alpha_2\ge \dots  $;
(ii) $\operatorname{sgn}c(\alpha)=(-1)^{|\alpha|-\alpha_1}$.
As a consequence, the  sum we are interested in is  $$ \sum_{|\alpha|=n} \big|c(\alpha)\big|1^{\alpha_1}2^{\alpha_2}\dots n^{\alpha_n}=P_n( 1, - 2,- 3,\dots,- n).$$
To evaluate it, consider $\sigma_n(t):=P_n( t, - 2,- 3,\dots,- n).$ Then, by the recurrence relation of the polynomials $P_n$ 
and by the Euler formula for homogeneous polynomials we find
\begin{cases} \sigma_1=t \\ \sigma_{n+1} =(t+n)\sigma_n +  (t+t^2)\sigma_n' , \end{cases} 
Now it is easy to check that
$$\sigma_n(t)=n!t(t+1)^{n-1},$$
whence $\sigma_n(1)=n!2^{n-1}.$ 
$$*$$
rmk. Also note that the polynomials $Q_n:=  \sum_{|\alpha|=n} \big|c(\alpha)\big|x^\alpha$ are simply $P_n(x_1, -x_2,-x_3,\dots,-x_n),$ whence one finds a linear recurrence relation for them 
\begin{cases} Q_1=x_1  \\ Q_{n+1}=x_1Q_n+2x_1^2\partial_1Q_n  +\sum_{i=1}^n (x_{i+1} - x_{i}) x_i\partial_i Q_n. \end{cases}<|endoftext|>
TITLE: Is there a proof of the uniformization theorem using circle packing?
QUESTION [12 upvotes]: In this paper: http://www.dm.unipi.it/~benedett/rodin-sullivan.pdf 
Rodin and Sullivan show that circle packings converge to the Riemann map. Later, Scharmm and He found another proof of the same result that did not rely on the Riemann mapping theorem. 
Start with a simply connected domain, fill it with coins making an hexagonal pattern (or any other packing). Add an extra vertex to their intersection graph, that will correspond to the boundary. Now apply the Koebe-Andreev-Thurston theorem to this planar graph. This defines a map from the centres of the coins to the centres of some circle packing, filling in each triangle by affine maps gives a map that converges to a conformal one. 
See https://en.wikipedia.org/wiki/Circle_packing_theorem for details. 
Can this approach be generalised to to obtain the full uniformization theorem?
https://en.wikipedia.org/wiki/Uniformization_theorem
Is there some discouraging reason why this might be false or hard?

REPLY [6 votes]: Firstly, the Rodin-Sullivan argument can not, in principle, be used to give a proof of uniformization, since it uses uniformization of domains of infinite type (due to Marden, if I recall) as an ingredient.
Secondly, there are many circle-packing approaches to uniformization, the first of them given by G. Leibon, using variational ideas of Y. Colin de Verdiere and of I. Rivin, see
MR1903777 (2003d:57038) Reviewed 
Leibon, Gregory(1-DTM)
Random Delaunay triangulations and metric uniformization. 
Int. Math. Res. Not. 2002, no. 25, 1331–1345

Leibon's argument is philosophically (and a bit more than that) very similar to the argument of Osgood-Phillips-Sarnak (who do things by maximizing log det of the laplacian).
Since then, there has been a mini-industry in approximating conformal maps by circle packing maps (based on the same variational ideas), where the actors are G. Brock Williams, A. Bobenko, B. Springborn, F. Luo, B. Chow, and others. A mathscinet search for these names will reveal much.<|endoftext|>
TITLE: Even Galois representations "mod p"
QUESTION [20 upvotes]: Consider an irreducible $\mathrm{mod}$ $p$ representation:
$$\rho: \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\to\mathrm{GL}_2(\bar{\mathbb{F}}_p)$$
If $\rho$ is odd, it was conjectured by Serre in the 70s and proved by Khare-Wintenberger in the 00s that $\rho$ comes from a modular form.
But much like in the characteristic 0 case, for $\rho$ even the situation is even more complicated. In particular, it can be easily proved that they never arise form modular forms.

The question is, where are (irreducible) even mod p Galois
  representations conjectured to come from? What is the analogue of Serre's conjecture in the even case?

This issue has been mentioned several times on MO. For example, this 2012 question:
Is there a theory of Maaß forms over finite fields?, by Chandan Singh Dalawat.
asks something a bit weaker, whether those representations come from Maass forms, for which I'm quite sure the answer is no, but I wouldn't know how to justify it.
Kevin Buzzard briefly addressed this issue in his reply to this 2010 question:
Is there a canonical notion of “mod-l automorphic representation”?, by David Hansen.
And Jim Stankewicz in his answer to this other question:
The significance of modularity for all Galois representations, by Jonah Sinick
although it seems to be some confusion with mod l representations, see the comments.
The closest to an answer I've come across is this blog post, which seems to suggest that even representations are either dihedral or automorphic from $U(3)$, but I don't quite follow the argument.
Any information or reference would be appreciated.

REPLY [2 votes]: To lift $\bar{\rho}$ to a geometric representation in the sense of Fontaine-Mazur, the standard technique requires that $\bar{\rho}$ is balanced, i.e. the dimension of a certain Selmer group must equal the dimension of its dual Selmer group (associated to a certain deformation problem). This is not the case for even representations. However, if one relaxes the $p$-adic Hodge theoretic condition at $p$ (crystalline or de Rham for example) then it becomes possible to sometimes lift an even $\bar{\rho}$ to a characteristic-zero representation $\rho:G_{\mathbb{Q}}\rightarrow \text{GL}_2(\mathbb{Z}_p)$ which is ramified at only finitely many primes as was demonstrated by Ramakrishna in "Deforming an Even Galois reprentation". In this paper, some even representations were lifted to characteristic zero for $p=3$. These lifts should not satisfy the de Rham condition at $p$. On the other hand, you're not expected to get geometric lifts even for small primes ($p>7$ was deduced in Calegari's paper). 
If you're looking for lifts which satisfy a local condition at $p$ you will get lifts which are ramified at an infinite density zero set of primes. These may be constructed through an application of the lifting strategy of Khare, Larsen and Ramakrishna. You may be interested in knowing where such representations in general come from. 
Also, it is worth noting that the standard geometric lifting technique does not apply for residual representations $\bar{\rho}:G_{K}\rightarrow \text{GL}_2(\mathbb{F}_{p^m})$ if $K$ is not totally real since in this case the associated Selmer and dual Selmer groups do not match up in dimension. This in no means implies that there are no geometric lifts when $K$ is imaginary quadratic (for instance).<|endoftext|>
TITLE: Which surfaces admit unbounded-length simple geodesics?
QUESTION [11 upvotes]: Let $S$ be a surface embedded in $\mathbb{R}^3$.
A simple geodesic on $S$ is one that does not self-intersect.
Some surfaces have simple geodesics whose length exceeds any
given bound $L$. For example, a cylinder or a torus allows tight
winding geodesics that are arbitrarily long before they cross themselves.
But a sphere, or a Zoll surface,
does not admit arbitrarily long simple geodesics, because every geodesic
forms a simple closed loop.

Q. Which surfaces $S$ admit arbitrarily long simple geodesics?

To be specific: Do ellipsoids possess such geodesics?

Update (11 May 2017).
This paper settles a version of my 2-yr-old question by
proving that "if the surface of a convex body $K$ contains arbitrary long closed simple geodesics, then $K$ is an isosceles tetrahedron":

Akopyan, Arseniy, and Anton Petrunin. "Long geodesics on convex surfaces." arXiv preprint arXiv:1702.05172 (2017).

REPLY [2 votes]: Anosov constructed (Anosov, Section 8) a smooth Riemannian metric $g$ on $S^2$ such that the surface $(S^2,g)$ admits arbitrarily long non-intersecting closed geodesics. His construction starts with a standard sphere $S^2$. Split it along its equator, and insert a surface of revolution with a thinner waist. Then he deformed the metric on the upper hemisphere along with a geodesic such that the directions around the picked path get twisted faster. Then he did the same deformation on the lower hemisphere. Those long geodesics spend most of their time around the waist, while the twisting on the two hemispheres guarantees the nonintersection property. The resulting surface is very nice, and I enjoyed picturing it in my mind very much.<|endoftext|>
TITLE: What are the implications of the new quasi-polynomial time solution for the Graph Isomorphism problem?
QUESTION [65 upvotes]: This week, news came out that Laszlo Babai has found a quasi-polynomial time algorithm to solve the Graph Isomorphism problem (that is: $O(\exp(polylog(n)))$). He is giving a series of talks this week, and the abstracts are here. I'm not an expert in this field (for instance, I had to look up what polylog means), but clearly this is a very significant result. Babai is a giant in this field, so let's assume for now that the claimed result is correct.

(1) What are some implications of this result in complexity theory? 

For example: This problem is NP, but is not known to be NP-complete. If it were NP complete, and if the algorithm were polynomial time, then we'd know P = NP and that would be huge. Does this new result give us any insight into the class of problems that are NP but not known to be NP complete? Does it prove that Graph Isomorphism is not in some other class of problems (e.g. does anyone study the class of exponential but not quasi-polynomial time problems)?

(2) Are there other important problems that reduce to the graph isomorphism problem that are now known to be solvable in quasi-polynomial?

Babai's abstract mentions the Coset Intersection problem and the String Intersection problem. I can guess what those are from the name. Are there other well-studied problems that reduce to graph isomorphism?

(3) How far is quasi-polynomial from polynomial?

Based on the linked thread above, polylog generally means $O(\log(n)^k)$ in complexity theory.  If so, then exp[\log(n)^k]) = exp[k]*n, which doesn't look all that much like polynomial time, unless you can bound $k$ by a constant that doesn't depend on $n$. But perhaps I'm misunderstanding the whole story, in which case I'd love to be set straight. What would be the next step to getting the bound lower, from the perspective of complexity theory?
Obviously, the proof hasn't appeared in print yet, so I'm not asking if the proof can somehow be made polynomial instead of quasi-polynomial, and I'm definitely not asking if it's correct. I'm asking what it implies if it's correct, and how much it changes the current playing field for complexity theory.
EDIT (based on the comments):

(4) Is there any known implication of this new result in Quantum Computing?

REPLY [15 votes]: An update from Babai:
http://people.cs.uchicago.edu/~laci/update.html
The main result, which appears to have been vetted at this point with extreme care, has weakened somewhat. His bound is now $\exp(n^{o(1)})$, a big improvement over the previous $\exp(n^{1/2 + o(1)})$, but no longer quasi-polynomial.
UPDATE: Babai is now claiming a new Split-or-Johnson subroutine restores the quasipolynomial claim!  http://people.cs.uchicago.edu/~laci/update.html<|endoftext|>
TITLE: Does the category of categories-mod-natural-isomorphism have any nonobvious autoequivalences?
QUESTION [15 upvotes]: The simpler question is to study the 2-groupoid $\mathrm{Aut}(\mathsf{Cat})$ of all autoequivalences of the category of (small) categories (i.e. the groupoid of equivalences of categories $\mathsf{Cat} \to \mathsf{Cat}$). This is equivalent to $\mathbb{Z}/2$ considered as a locally discrete 2-groupoid: the terminal object $\mathbb{1}$ is fixed up to unique isomorphism, and the arrow category $\mathbb{2}$ is the unique minimal generator (i.e. it is a generator and no proper subobject is a generator) so also fixed up to isomorphism. Since every category is functorially a colimit of copies of $\mathbb{2}$, once we decide whether our autoequivalence fixes or swaps the two maps $\mathbb{1} \to \mathbb{2}$ we've determined the autoequivalence up to a canonical isomorphism. And of course the nontrivial autoequivalence is the "opposite category" functor $\mathcal{C} \mapsto \mathcal{C}^\mathrm{op}$.
This is a somewhat artificial question, since an autoequivalence of the category $\mathsf{Cat}$ must respect the notion of isomorphism of categories, whereas all we should really want to respect is equivalence of categories. One way to fix this would be to consider $\mathsf{Cat}$ as a 2-category, and ask about the 3-groupoid of autobiequivalences of $\mathsf{Cat}$. But this starts to involve heavy machinery, and the problem will only get worse if we want to go up the categorical ladder. Hence a more "homotopical" approach:
Question: What is the biequivalence type of the 2-groupoid $\mathrm{Aut}(\mathrm{ho}\mathsf{Cat})$ of autoequivalences of $\mathrm{ho}\mathsf{Cat}$, the category of small categories localized at the equivalences of categories? Equivalently, $\mathrm{ho}\mathsf{Cat}$ is the category of small categories modulo the congruence identifying naturally isomorphic functors. In particular, is $\mathcal{C} \mapsto \mathcal{C}^\mathrm{op}$ isomorphic to the identity (presumably the answer is no)? Are there any other autoequivalences, essentially (probably not, but you never know)?
A variant would be to ask about the 2-groupoid of functors $\mathsf{Cat} \to \mathsf{Cat}$ which preserve equivalences of categories and induce equivalences on $\mathrm{ho}\mathsf{Cat}$.
I'm aware that Toën showed in some sense that the $(\infty,1)$-category of $(\infty,1)$ categories has an automorphism group of $\mathbb{Z}/2$, and Barwick and Schommer-Pries generalized this to $(\infty,n)$-categories for larger $n$. This question is meant to be an easier / more elementary warmup to those questions.
The approach that worked for $\mathsf{Cat}$ certainly isn't going to work for $\mathrm{Ho}\mathsf{Cat}$: Freyd showed ("On the concreteness of certain categories." Symposia Mathematica. Vol. 4. 1968-9, pp. 431-56) that $\mathrm{Ho}\mathsf{Cat}$ is not concrete, so it certainly can't have a generating object, never mind a colimit-dense object.
By playing around, I've convinced myself at least that the indiscrete categories (the equivalence class of the terminal category) can be characterized in $\mathrm{Ho}\mathsf{Cat}$ by the fact that all functors into it are isomorphic (consider functors from the walking equalizer diagram to see that such a category is a preorder and functors from the terminal category to see that all objects are isomorphic). So $\mathrm{Ho}\mathsf{Cat}(\mathbb{1},-)$ tells us how many isomorphism classes of objects a category has, and what isomorphism classes of functors are doing to them. But that's about all I've got.

REPLY [6 votes]: $\newcommand{\hoCat}{\mathrm{ho}\mathsf{Cat}}
\newcommand{\Cat}{\mathsf{Cat}}
\newcommand{\Gaunt}{\mathsf{Gaunt}}
\newcommand{\hoGaunt}{\mathrm{ho}\mathsf{Gaunt}}
\newcommand{\Free}{\mathsf{Free}}
\newcommand{\hoFree}{\mathrm{ho}\mathsf{Free}}
\newcommand{\comp}{\mathrm{comp}}
\newcommand{\inv}{^{-1}}
\require{AMScd}$
Chris Schommer-Pries' argument in the comments, can be beefed up to show that indeed, essentially the only autoequivalences of $\hoCat$ are the identity and taking the opposite category.
Summary.
Use Chris's argument to show that an autoequivalence $F: \hoCat \to \hoCat$ restricts to the identity on the free categories $\Free = \hoFree$ (possibly after composing with the known autoequivalence $(-)^\mathrm{op}: \hoCat \to \hoCat$). As Chris then argues, $\mathsf{Free}$ forms a (large) regular generator in $\hoCat$, and it follows that $F$ is essentially the identity on objects. We just need the slightly stronger statement that $\Free$ is a (large) dense generator to show that $F$ is also the identity on morphisms. It turns out that the $\Cat$-functoriality of the presentations of categories is enough to guarantee this, even though these presentations are not actually functorial on $\hoCat$. In fact, this argument generalizes to show that if the category of algebras for a monad is modded out by a congruence relation, then the free algebras still form a dense generator in the resulting category $\widetilde{\mathcal{C}^T}$ if the canonical coequalizer diagrams descend.
Details.

As I said in the question statement, the terminal category $\mathrm{pt}$ remains terminal in $\hoCat$, so it is preserved by any autoequivalence. My argument in the question statement is overly complicated: it's clear that when modding out by a congruence, the terminal object remains terminal.
As Chris argues, the subcategory $\mathrm{ho}\mathsf{Pre}$ of preorders is preserved by any autoequivalence because it is the class of categories $P$ with $\hoCat(D,P) \to \mathsf{Set}(\hoCat(\mathrm{pt},D),\hoCat(\mathrm{pt},P))$ injective for every category $D$. One direction is clear, the other comes from taking $D$ to be the walking equalizer diagram. Now, since $\mathrm{ho}\mathsf{Pre} = \mathrm{ho}\mathsf{Pos} = \mathsf{Pos}$, our autoequivalence of $\hoCat$ restricts to either the identity or opposite on $\mathsf{Pos}$ by the same argument I gave in the question statement characterizing the autoequivalences of $\Cat$. After composing with the "opposite" autoequivalence of $\hoCat$, we can assume our autoequivalence is the identity on $\mathsf{Pos}$.
As Chris argues, in particular the walking arrow $\mathbb{2}$ is fixed. So the groupoids are also fixed: they are characterized as the categories $G$ such that every functor $\mathbb{2} \to G$ extends up to natural isomorphism along the functor $\mathbb{2} \to \mathrm{pt}$. The connected groupoids are also fixed because $\hoCat(\mathrm{pt},-)$ tells us how many isomorphism classes a category has. Note that we don't know the autoequivalence fixes each individual groupoid, but only that the class of groupoids is preserved, but this will be good enough for the next step.
As Chris argues, the gaunt categories $\hoGaunt$ are fixed because they are the categories such that every functor from a connected groupoid extends along the functor to $\mathrm{pt}$. We have $\Gaunt = \hoGaunt$, and again the argument given in the question statement for $\mathsf{Cat}$ shows that the autoequivalence restricts to the identity on $\Gaunt$. In particular, it's the identity on the subcategory $\Free$ of free categories.
As Chris argues, the $\Cat$ - coequalizer $P^2X \stackrel{\to}{\to} PX \overset{\comp_X}{\to} X$ descends to $\hoCat$ (in particular, $\Free$ is a regular generator in $\hoCat$). Here $P$ is the "path" comonad on $\Cat$, induced by the forgetful functor to graphs, and the coequalizer is the canonical one: one leg $\mu_X$ de-parenthesizes a path of paths, while the other $P\comp_X$ composes each individual path in a path of paths; the coequalizer $\comp_X$ composes a path in $X$. The reason the coequalizer descends is that if $F: PX \to Y$ coequalizes $\mu_X$ and $P\comp_X$ up to a natural isomorphism $\alpha$, then for every path $\gamma_1, \dots, \gamma_k$ in $X$ we have $F(\gamma_k \circ \cdots \circ \gamma_1) = \alpha \circ F(\gamma_k) \circ \cdots \circ F(\gamma_1) \circ \alpha\inv$. But in the unary case, this tells us that $F(\gamma_i) = \alpha \circ  F(\gamma_i) \circ \alpha\inv$. So we can commute the $\alpha$ down to the end to get $F(\gamma_k \circ \cdots \circ \gamma_1) = F(\gamma_k) \circ \cdots \circ F(\gamma_1) \circ \alpha \circ \alpha\inv = F(\gamma_k) \circ \cdots \circ F(\gamma_1)$. That is $F$ already coequalizes the fork strictly, so it descends to a functor $X \to Y$. Uniqueness of the factorization in $\hoCat$ is easy to see.
Let $i_\Free: \Free \to \hoCat$ be the inclusion functor. Let $\alpha: \hoCat(i_\Free - , X) \implies \hoCat(i_\Free - , Y)$ be a natural transformation. To show that $\Free$ is dense in $\hoCat$, we just need to show that $\alpha$ is induced by some $\bar\alpha: X \to Y$ (we know that such a $\bar\alpha$ is unique because we've shown that $\Free$ is a regular generator, and hence a generator, in $\hoCat$). There is an obvious candidate: by naturality $\alpha_{PX}(\comp): PX \to Y$ coequalizes the two functors $P^2X \to PX$ up to natural isomorphism, so by the previous part it descends to a functor $\bar\alpha: X \to Y$. So much is true of any regular generator; the trick is to see that $\bar\alpha$ really induces the natural transformation $\alpha$, i.e. that $\bar\alpha \circ f \cong \alpha_A(f)$ for any $f: A \to X$ with $A \in \Free$, leading to the following diagram:
$$
\begin{CD}
PA @>\comp_A>> A;\\
@VPfVV @VfVV \searrow^{\alpha_A(f)} \\
PX @>\comp_X>> X @>\bar\alpha>> Y;
\end{CD}
$$
Since the square commutes and the outside commutes up to natural isomorphism and $\comp_A$ is epi in $\hoCat$, the triangle also commutes up to natural isomorphism as desired.

The principle used to analyze the auotequivalences of $\Cat$ in the question statement, and then $\mathsf{Pos}$, and then $\Gaunt$, and finally $\hoCat$, is that an equivalence (or any colimit-preserving functor) is determined up to natural isomorphism by its action on a dense generator. One way to see this is that a dense generator $\mathcal{A} \subseteq \mathcal{C}$ embeds $\mathcal{C}$ as a full subcategory of $\mathrm{Fun}(\mathcal{A}^\mathrm{op},\mathsf{Set})$ containing $\mathcal{A}$ as the representables. Each object is (canonically) a colimit of representables, and each morphism is (canonically) a colimit of maps between representables, so the value of the functor on them is determined by colimit preservation and where it sends the representables.<|endoftext|>
TITLE: How bad is the modular space?
QUESTION [6 upvotes]: I'm wondering if there is some results about the quotient space $\mathbb{H}^{3}/PSL(2,\mathcal{O}_{K})$?
Do we know something about its homology or homotopy groups ?
$\mathbb{H}^{3}$ is the hyperbolic space of dimension 3 and $\mathcal{O}_{K}$ the ring of integers associated to a the number field $K$.

REPLY [5 votes]: I assume you mean for $K$ to be an imaginary quadratic field (see Misha's comment). There have been computations of the homology for a substantial array of small values of the discriminant of $K$, by Alexander Rahm (see here), in fact there are programs to compute cell decompositions of the quotient orbifolds (see loc.  cit. and also Aurel Pages's work here). 
There are also more theoretical works about the asymptotics of the homology when the discriminant goes to infinity, see here, or for local systems of rank going to infinity and finite-index subgroups.<|endoftext|>
TITLE: Probability of two vectors lying in the same orthant
QUESTION [19 upvotes]: Let $S^{d-1} = \{x \in \mathbb R^d: \|x\| = 1\}$ denote the unit sphere in $\mathbb R^d$. Let $v$, $w$ be drawn uniformly at random from $S^{d-1}$, conditioned on their inner product being equal to $\langle v, w \rangle = \cos \theta$. In other words, $v$ and $w$ have a fixed common angle $\theta$. I am interested in the probability that these two vectors lie in the same orthant, say the all-positive orthant, averaged over all possible $v$ and $w$:
$$f_d(\theta) = \Pr(v > 0 \mid w > 0, \ \langle v, w \rangle = \cos \theta).$$
Here $v > 0$ means that all $d$ coordinates of $v$ are positive. 
In particular, I am interested in the asymptotics of large $d \to \infty$ of $f_d(\theta)$ for $\theta \in (0, \frac{1}{2} \pi)$. Equivalently, as $f_d$ scales exponentially in $d$, I'm interested in the function
$$g(\theta) = \lim_{d \to \infty} f_d(\theta)^{1/d}.$$
Note that an approximation to the above probabilities can be obtained by replacing the uniform distribution over the sphere by a multivariate Gaussian distribution, where each coordinate is independently drawn from a Gaussian $\mathcal{N}(0, \frac{1}{d})$. For large $d$, with overwhelming probability such a random vector will have norm $1 \pm o(1)$, and with overwhelming probability two such random vectors will have inner product $o(1)$. If we ignore the fact that the norms of such vectors may not exactly be equal to $1$ and that two random vectors may not be exactly orthogonal (which is why this is only an approximation), then two vectors $v$ and $w$ from the sphere with angle $\theta$ can be generated by taking $v = n_1$ and $w = (\cos \theta) n_1 + (\sin \theta) n_2$ for two independent random Gaussian vectors $n_1, n_2 \sim \mathcal{N}(0, \frac{1}{d})^d$. These lie on the sphere exactly if $n_1, n_2$ have norm $1$, and their angle is then equal to $\theta$ if and only if $n_1$ and $n_2$ are orthogonal. 
With this approximation, probabilities can be computed quite easily, as different coordinates are independent and probabilities multiply. However, I'm looking for more precise estimates than using this Gaussian approximation of the uniform distribution on the sphere.
So far I've tried sharing this problem with a few others in the department, and rewriting the probability to computing the expected volume of the intersection of the sphere with $d$ orthogonal hyperplanes, but so far nothing led anywhere. Any pointers on how to solve this would be greatly appreciated!

Update: To verify/compare different approaches, simulations for $\theta = \arccos 0.9$ in dimensions $d \in \{10, \dots, 40\}$ show the following trends for $\ln f$:

The points are the simulation results (using 100.000-500.000 experiments each, so that the number of successes was at least a few hundred) and the line is the linear fit $-0.0133857 - 0.170074 d$, or equivalently $f(\arccos 0.9) \approx C \cdot 0.8436^d$ for a constant $C \approx 1$. This suggests that $g(\arccos 0.9) \approx 0.8436$. The answers given so far say:

Carlo's answer: $0.66$
Other answer (main): $1.08$
Other answer (alternative): $0.71$

So perhaps all answers so far are still far off when $\theta$ is small and the inner product between $v$ and $w$ is large.

REPLY [3 votes]: Thanks to Ofer Zeitouni, unless I made some serious mistakes somewhere in the calculations, I've been able to find the exact answer using large deviations theory. The derivation of the final result is a rather tedious affair, so I will just state the result here. Below $\mathcal{S}^{d-1}$ denotes (the uniform distribution over) the unit ball in $\mathbb{R}^d$. The proof can be found in the appendix here and is about 5 pages long.

Theorem:
Let $\mathbf{X}, \mathbf{Y} \sim \mathcal{S}^{d-1}$, let $\phi(\mathbf{X}, \mathbf{Y})$ denote the angle between $\mathbf{X}$ and $\mathbf{Y}$, and let
\begin{align}
f_d(\theta) = \mathbb{P}(\mathbf{X} > 0 \ | \ \mathbf{Y} > 0, \ \phi(\mathbf{X}, \mathbf{Y}) = \theta).
\end{align}
For $\theta \in (0, \arccos \frac{2}{\pi})$ (respectively $\theta \in (\arccos \frac{2}{\pi}, \frac{\pi}{3})$), let $\beta_0 \in (1, \infty)$ (respectively $\beta_1 \in (1, \infty)$) be the unique solution to:
\begin{align}
\arccos\left(\frac{-1}{\beta_0}\right) = \frac{(\beta_0 - \cos \theta) \sqrt{\beta_0^2 - 1}}{\beta_0 (\beta_0 \cos \theta - 1)} \, , \qquad \arccos\left(\frac{1}{\beta_1}\right) = \frac{(\beta_1 + \cos \theta) \sqrt{\beta_1^2-1}}{\beta_1 (\beta_1 \cos \theta + 1)} \, . \label{eq:beta}
\end{align}
Then, for large $d$,
\begin{align}
f_d(\theta) = \begin{cases}
\left(\displaystyle\frac{(\beta_0 - \cos \theta)^2 }{\pi \beta_0 (\beta_0 \cos \theta - 1) \sin \theta}\right)^{d + o(d)}, & \qquad \text{if } \theta \in [0, \arccos \tfrac{2}{\pi}]; \\[3ex]
\left(\displaystyle\frac{(\beta_1 + \cos \theta)^2}{\pi \beta_1 (\beta_1 \cos \theta + 1) \sin \theta}\right)^{d + o(d)}, & \qquad \text{if } \theta \in [\arccos \tfrac{2}{\pi}, \tfrac{\pi}{3}]; \\[3ex]
\left(\displaystyle\frac{1 + \cos \theta}{\pi \sin \theta}\right)^{d + o(d)}, & \qquad \text{if } \theta \in [\tfrac{\pi}{3}, \tfrac{\pi}{2}); \\[3ex]
0, & \qquad \text{if } \theta \in [\tfrac{\pi}{2}, \pi].
\end{cases}
\end{align}

To illustrate, the following figure shows $g(\theta) = \lim_{d\to \infty} f_d(\theta)^{1/d}$ against $\theta$. The constant $\alpha$ on the y-axis is $\alpha = \pi / (2 \sqrt{\pi^2 - 4}) \approx 0.648$. The dashed lines indicate the boundaries of the piece-wise parts in the theorem.

As stated in the OP, experiments suggested that $g(\arccos 0.9) \approx 0.8436$, and indeed from this theorem we find $g(\arccos 0.9)$ to be approximately $0.843$. This is quite different from Carlo's approximate solution based on Gaussians - if $\theta$ is close to $0$, the Gaussian approximation seems to be further off.
For the case $\theta = \frac{\pi}{3}$, the exact solution is $g(\frac{\pi}{3}) = \frac{\sqrt{3}}{\pi} \approx 0.5513$. Other answers based on different approximations all suggested $g(\frac{\pi}{3}) \in [0.55, 0.57]$ so here the Gaussian approximations are pretty accurate.
And as $\theta$ approaches $\frac{\pi}{2}$ from below, we curiously get $g(\theta) \to \frac{1}{\pi}$. In other words, two almost-orthogonal vectors in high dimensions are in the same orthant with probability proportional to $(\frac{1}{\pi})^{d + o(d)}$. I'm not sure if there is a nice intuitive explanation for this.<|endoftext|>
TITLE: $\pi e$ and an unfamiliar polynomial
QUESTION [13 upvotes]: Ever since my exposure to this integral involving $\pi e$, I've conjectured and set about evaluating the possible nature of the following integral
$$\int_0^1 x^m \sin(\pi x) x^x (1-x)^{1-x} \ dx, \quad m = 0, \, 1, \, 2, \, 3, \dots$$
Observations
I've inputted a few values in WolframAlpha and NumberEmpire engine, and I discovered this interesting beauty:
\begin{align}
 f_q{(x)} = \begin{cases} 
      -\left(\frac{1}{e}(1-x)^{1-\frac{1}{x}} \right)^q, &  x\neq 0 \\
      -1, & x = 0 
   \end{cases}
 = \sum_{n = 0}^{\infty} b_n{(q)} x^n
\end{align}
By using clever manipulation (Putnam style), a recurrence relation was derived for the coefficients of $f_q(x)$, namely
\begin{align}
 b_n{(q)} = -\frac{q}{n} \sum_{k=1}^{n}  \frac{b_{n-k}{(q)}}{k+1}, \; \; b_0{(q)} = -1.
\end{align}
$b_n{(q)}$ generates the following polynomials
\begin{align*}
b_0{(q)} &= -1 \\
b_1{(q)} &= \frac{q}{2} \\
b_2{(q)} &= \frac{-3q^2+4q}{24} \\
b_3{(q)} &= \frac{q^3-4q^2+4q}{48} \\
b_4{(q)} &= \frac{-15q^4+120q^3-320q^2+288q}{5760} \\
b_5{(q)} &= \frac{3q^5-40q^4+200q^3-448q^2+384q}{11520} \\
b_6{(q)} &= \frac{-63q^6+1260q^5-10080q^4+40544q^3-82656q^2+69120q}{2903040} \\
b_7{(q)} &= \frac{9q^7-252q^6+2940q^5-18368q^4+65184q^3-125568q^2+103680q}{5806080} \\
\vdots
\end{align*}
Interestingly, the denominators seem to obey a more-or-less known sequence, brought about by Manjul Bhargava's work "The Factorial Function and Generalizations", although proving the connection between the two isn't easy, or at least isn't obvious.
Thus, based on the nature of the $b_n(q)$, I was then prompted to consider this generalization:
$$I_m(q) = \int_0^1 x^m \sin(\pi x) (x^x (1-x)^{1-x})^q \ dx$$
for non-negative integer $m$ and real $q$.
Evaluating the Integral
Looking back in the forum where I found the integral, I set about to apply the same "dumbbell" contour used for a simple case, first considering the following function
$$f(z) = \exp{(-i\pi q z+(m+qz)\log(z) + q(1-z)\log(1-z))} $$
and defining branch cuts for the logarithms, specifically
$$z^{m+qz} = \exp{((m+qz)\log(z))}, \quad -\pi \leq \arg(z) < \pi $$
$$(1-z)^{q(1-z)} = \exp{(q(1-z)\log(1-z))}, \quad 0 \leq \arg(1-z) < 2\pi $$
Now, since $f(z)$ is continuous on $(-\infty, 0)$ and $(1, \infty)$, our given cut is $[0,1]$, with singularities at $z = 0, \, 1$; thus, the "dumbbell" contour, which we define as $\Gamma$, would consist of 

$\Gamma_1$: Line segment slightly above $[0,1]$
$\Gamma_2$: Line segment slightly below $[0,1]$
$\Gamma_3$: Circle of small radius enclosing $z = 0$
$\Gamma_4$: Circle of small radius enclosing $ z= 1$

If done properly, for $\Gamma_1$, with $0 \leq x \leq 1$, we have
$$\int_{\Gamma_1} f(z) \ dz  = \int_0^1 x^m e^{i\pi(-qx+2q(1-x))}  \left(x^x (1-x)^{1-x}\right)^q \ dx $$
For $\Gamma_2$, with $0 \leq x \leq 1$, we have
$$\int_{\Gamma_2} f(z) \ dz  = \int_0^1 x^m e^{-i\pi qx}  \left(x^x (1-x)^{1-x}\right)^q \ dx $$
Since $f$ is bounded, the radius will tend to zero, so thus
$$\int_{\Gamma_3} f(z) \ dz  = \int_{\Gamma_4} f(z) \ dz = 0$$
Thus, we have
$$\int_{\Gamma} f(z) \ dz = \int_{\Gamma_2} f(z) \ dz - \int_{\Gamma_1} f(z) \ dz $$
$$\int_0^1 x^m \left(e^{-i\pi qx} - e^{i\pi(-qx+2q(1-x))}\right)  \left(x^x (1-x)^{1-x}\right)^q\ dx $$
$$= -2ie^{i\pi q}\int_0^1 x^m e^{-2\pi i qx} \sin\left(\pi q  (1-x) \right)  \left(x^x (1-x)^{1-x}\right)^q\ dx $$
On the other hand, with residue at infinity calculations, we have
$$\int_{\Gamma} f(z) \ dz = 2 \pi i \mathrm{Res}_{z = 0} \frac{1}{z^2} f\left( \frac{1}{z} \right)$$
Now $f(z)$ has the following Laurent series expansion
$$\frac{1}{z^2} f\left( \frac{1}{z} \right) = -e^{q(1-\pi i)} \sum_{n=0}^{\infty} b_n(q) z^{n-(q+m+2)}$$
Thus, by the Cauchy residue theorem,
$$\int_{\Gamma} f(z) \ dz = 2 \pi i \mathrm{Res}_{z = 0} \frac{1}{z^2} f\left( \frac{1}{z} \right) = -2 \pi i e^{q(1-\pi i)} b_{m+q+1}(q)$$
where we are restricting $q$ to be an integer (unfortunately).
Thus, with some manipulations, our final result is
$$ \int_0^1 x^m e^{2\pi i q(1-x)} \sin\left(\pi q  (1-x) \right)  \left(x^x (1-x)^{1-x}\right)^q\ dx =  \pi  e^{q} b_{m+q+1}(q).$$
I can't help but look at this and think I've done goofed somewhere---something isn't right. The $e^{2\pi i q(1-x)} $ term especially bugs me; I've attempted comparing real and imaginary components, to no avail. Maybe I didn't pick the best contour? $f(z)$? 
Your input is much appreciated!

REPLY [5 votes]: Ahhh, I see where something went "off":
Consider 
$$f(z) = \exp{\left( i \pi q z+ (m+qz)\log(z) + q(1-z)\log(1-z)\right)} $$
where $m$ is a non-negative integer and $q$ is a integer. Using the exact same branch cuts and contour (just about every single piece of information stays true, with the exception of some algebraic manipulation), we have
$$\int_{\Gamma_1} f(z) \ dz  = \int_0^1 x^m e^{-i\pi qx}  \left(x^x (1-x)^{1-x}\right)^q \ dx $$
and 
$$\int_{\Gamma_2} f(z) \ dz  = \int_0^1 x^m e^{i\pi qx}  \left(x^x (1-x)^{1-x}\right)^q \ dx $$
This $f$ is bounded, so it remains that the circles around the singularities will vanish. We have then
$$\int_{\Gamma} f(z) \ dz = \int_{\Gamma_2} f(z) \ dz - \int_{\Gamma_1} f(z) \ dz $$
$$\int_0^1 x^m \left(e^{i\pi qx} - e^{-i\pi qx}\right)  \left(x^x (1-x)^{1-x}\right)^q\ dx $$
$$= 2i\int_0^1 x^m \sin\left(\pi q x\right)  \left(x^x (1-x)^{1-x}\right)^q\ dx $$
We have the following Laurent series for this $f(z)$:
$$f(z) = (-1)^{q+1} e^q \sum_{n=0}^\infty \frac{b_n(q)}{z^{n-q-m}}$$
Applying the same residue at infinity calculations, and applying Cauchy's residue theorem, we finally have
$$ \int_0^1 x^m  \sin\left(\pi q x \right)  \left(x^x (1-x)^{1-x}\right)^q\ dx =  (-1)^{q+1} \pi  e^{q} b_{m+q+1}(q).$$
and we are done.<|endoftext|>
TITLE: Classification of 2-dimensional Alexandrov spaces
QUESTION [5 upvotes]: Is it possible to classify explicitly compact 2-dimensional Alexandrov spaces with curvature bounded below (either with or without boundary)?
If yes, a reference would be helpful.
EDIT: If the question is too general, may be one can classify somehow non-negatively curved compact 2-dimensional Alexandrov spaces, e.g. as pieces of convex surfaces (see my comment below). In general I would be interested to know as precise result as possible.

REPLY [5 votes]: Applying doubling theorem, we can get rid of boundary.
Then pass to the universal cover.
The obtained space is isometric to convex surface in the model space with curvature $\kappa$ and any motion of your original space can be extended as a motion of this surface.
So your space is a quotient of convex surface by a discrete group of motions of the model space.
In particular, if your space had a boundary it will be lifted to plane curves in the model space.
In general, the surface is not unique (up to congruence), 
so it is not exactly a classification, but close.<|endoftext|>
TITLE: How to calculate log or exp of a value in GF(2^n) using log/exp table of GF((2^k)^m) where n=k*m?
QUESTION [5 upvotes]: Consider Galois fields $\mathbb{F}_{2^n}$ and $\mathbb{F}_{2^k}$, where $n=km$, and $\mathbb{F}_{2^k}$ is a ground field of $\mathbb{F}_{2^n}$. 
I’d appreciate pointers to papers or suggestions on:

How to find $\log(a)$ and $\exp(a)$ for $a\in \mathbb{F}_{2^n}$, given log/exp look-up tables of $\mathbb{F}_{2^k}$?
How to convert the values from $\mathbb{F}_{2^n}$ to $\mathbb{F}_{(2^k)^m}$ .

Specifically I need solution for n=16 (with any combination of integer m and k, e.g. k=8, m=2) such that amount of calculations used for conversion is minimal. Generator polynomials for all three fields can be assumed to be known, for example for case of n=16, k=8, m=2:
 - GF(2^16):     x^16 + x^5 + x^3 + x^2 + 1
 - GF(2^8):      x^8  + x^4 + x^3 + x^2 + 1
 - GF((2^8)^2):  x^2  + 3x  + 1

Additional background info: Generally I have log and exp look-up tables for GF(2^n) and can avoid the conversion problem all together, but 2^n tables don't fit into memory-constrained CPU I’m using. Thus i'm interested to calculate log and exp of GF(2^16) using log/exp tables of GF(2^8) or GF(2^4).
I came across this paper (http://web.eecs.utk.edu/~plank/plank/papers/UT-CS-13-717.pdf), but it explicitly says that GF(2^km) is not identical to GF((2^k)^m), but doesn’t offer a way to convert between the two: the result of multiplication using ground and extension fields doesn’t match the multiplication result using any other method (presumably because the composite field is not identical to the original field).
Thanks in advance for any help

REPLY [2 votes]: Efficient algorithms to compute in finite fields are based on
Conway polynomials. Such algorithms are implemented in several
computer algebra systems, for example in GAP. For details, see e.g.
the chapter on finite fields in the GAP Reference Manual.
In particular, GAP can compute very efficiently in the field
${\rm GF}(2^{16})$ you ask about -- to cite the GAP Reference Manual:

GAP can represent elements of all finite fields ${\rm GF}(p^d)$
  such that either (1) $p^d \leq 65536$ (in which case an extremely
  efficient internal representation is used); (2) $d = 1$,
  (in which case, for large $p$, the field is represented using the
  machinery of residue class rings (see Section 14.5) or (3) if the
  Conway polynomial of degree $d$ over the field with $p$ elements
  is known, or can be computed (see ConwayPolynomial (59.5-1)).<|endoftext|>
TITLE: Club-guessing at $\omega_2$
QUESTION [9 upvotes]: Let $S=\{\delta<\omega_2:\text{cf}(\delta)=\omega\}$.  A well-known theorem of Shelah tells us that we can find $\langle C_\delta:\delta\in S\rangle$ such that
for every club $C\subseteq\omega_2$ there are stationarily many $\delta\in S$ with $C_\delta\subseteq C$, that is, $\langle C_\delta:\delta\in S\rangle$ is a type of club-guessing sequence.
Given $\sigma<\omega_1$, can we in ZFC find a club-guessing sequence as above so that the order-type of each $C_\delta$ is at least $\sigma$?  Could we even demand that $C_\delta$ is of order-type $\delta$?
Note that we get such sequences $\langle C_\delta:\delta\in S\rangle$ if $\diamondsuit_S$ holds.

REPLY [11 votes]: YES for any given $\sigma<\omega_1$. And you can even have the clubs (somewhat) cohere. I worked out the details in here: http://blog.assafrinot.com/?p=2133<|endoftext|>
TITLE: Equation $x^2=y^p + 1$
QUESTION [11 upvotes]: can you help me please for solving this diophantine equation : $x^2=y^p+1$
and if you can give me a general method to studying such equation : $x^2=y^p+t$
Thanks

REPLY [2 votes]: For $t = 1$, your question is about a special case
of Catalan's conjecture, which has been proved in 2002
by Preda Mihăilescu.
In particular, for $t = 1$ the only solution is $3^2 = 2^3 + 1$.<|endoftext|>
TITLE: classifying maps of Whitney sums of vector bundles
QUESTION [7 upvotes]: For an $n$-dimensional vector bundle $\xi$  with structure group $G\leq O(n)$ over a $CW$-complex $B$, we have a classifying map up to homotopy
$$
f(\xi): B\longrightarrow BG, 
$$
$f(\xi)\in [B;BG]$, and the composition up to homotopy
$$
g(\xi): B\overset{f}{\longrightarrow} BG\overset{i}{\longrightarrow}BO(n)\overset{j}{\longrightarrow}BO$$
where $BO=\lim _{n\to\infty} BO(n)$ and $g(\xi)\in [B;BO]$.
Suppose we have two such $n$-dimensional vector bundles $\xi_1$, $\xi_2$.
Question: Are there any formulas
$$
g(\xi_1\oplus\xi_2)=? \text{  in terms of } g(\xi_1), g(\xi_2)?
$$
And
$$
f(\xi_1\oplus\xi_2)\in [B,BG\times BG]=?\text{  in terms of } f(\xi_1), f(\xi_2)?
$$

REPLY [9 votes]: $BO(n)$ is the infinite-dimensional Grassmannian $Gr(n,\infty)$ of $n$-planes in ${\mathbf R}^\infty$. There is a natural direct sum operation
$$\oplus\colon Gr(n,\infty)\times Gr(m,\infty)\to Gr(n+m,\infty)$$
(just taking the direct sum of linear subspaces) and it gives you the desired map 
$$BO(n)\times BO(m)\to BO(n+m).$$<|endoftext|>
TITLE: how to prove the $n$-times self-product of a map is null-homotopic
QUESTION [5 upvotes]: Let $k$ be a fixed positive integer and $\Sigma_k$ the $k$-th symmetric group. By letting $\Sigma_k$ permuting an orthonormal basis of a $k$-dimensional Euclidean space, there is a "regular representation" 
$$
r_k: \Sigma_k\longrightarrow O(k)
$$
which induces a map between classifying spaces
$$
\rho_k: B\Sigma_k\longrightarrow BO(k).
$$
Let $X$ be a finite CW-complex and a map
$$
f: X\longrightarrow B\Sigma_k.
$$
For any positive integer $n$, we produce a map by taking self-product of $f$ for $n$-times
$$
\prod_n f: X\longrightarrow \prod_n B\Sigma_k
$$
and have a composition
$$
g_n: X\overset{\prod_n f}{\longrightarrow}\prod_n B\Sigma_k\overset{\prod_n \rho_k}{\longrightarrow}\prod_n BO(k)\longrightarrow BO(nk)=G_{nk}(\mathbb{R}^\infty).
$$
where the last map is induced by the inclusion
$$
\prod_n O(k)\longrightarrow O(nk).
$$
Question: I want to prove that for any $k$, any finite CW-complex $X$ and any map $f$,  there exists a positive integer $n$ such that $g_n$ is null-homotopic. Is it true? How to prove?

REPLY [5 votes]: Yes, this is true.
Your map $B\Sigma_k\rightarrow BO(k)$ gives rise to a map $B\Sigma_k\rightarrow BO$, i.e. an element in the reduced K-theory group $\tilde{ko}^0(B\Sigma_k)$.
Now that whole group is probably not torsion, but the Atiyah-Hirzebruch spectral sequence tells you that, if $K$ is a finite-dimensional $CW$ complex with torsion homology groups, $\tilde{ko}^0(K)$ is torsion.
So if you take $K$ to be a suitably modified skeleton of $B\Sigma_k$ (see below), it follows that there is $n$ such that the map 
$$K\rightarrow B\Sigma_k\rightarrow BO(k)\rightarrow BO(nk)$$
is null homotopic, where the last map is the times $n$ map.
But if $X$ is a finite-dimensional $CW$-complex, the map $X\rightarrow B\Sigma_k$ will factor over $K$ if we choose $K$ to be a sufficiently high-dimensional "skeleton".
I was imprecise above about $K$ being a "suitably modified skeleton", let me clarify this:
If $Y$ is a CW-complex with torsion homology groups, then, for each $n$, there is an $n+1$-dimensional CW complex $K$ with torsion homology groups and a map $K\rightarrow Y$ that induces isomorphisms on homology up to degree $n-1$.
To build that $K$, simply take $\tilde{K}$ to be the $n$-skeleton of $Y$. Now the $n$-th homology of that is not necessarily torsion, but since rationalized homotopy and rational homology agree, we can choose maps from $S^n$ to $\tilde{K}$ such that they are nullhomotopic in $Y$, and their images in $H_n(\tilde{K})$ generate it rationally. But then form $K$ by attaching cells along those maps. $K$ comes with a map to $Y$, and has torsion homology.<|endoftext|>
TITLE: Which philosophy for reductive groups?
QUESTION [46 upvotes]: I am just beginning to look further into trace formulas and automorphic forms in a quite general setting. For long I have noticed that the natural assumption on the group $G$ we work on is to be reductive.
Hence my question : why is reductive interesting?
Of course I have already seen some formal definitions, like to have a trivial unipotent radical or some discussions about root systems. But it does not seem very satisfactory to me: why is this assumption the one that seems to be the Grail of all generalizations done in this field?  why are those reductive groups so deeply interesting? what are the implied or equivalent properties making them so desirable? (I have heard of the complete reducibility of representations -- is it true? without other assumption on the representations? is it an equivalent or a weaker statement?)
Any clue helping me to cloth off reductive groups will be of great use :)

REPLY [15 votes]: Here is a natural interpretation of complex reductive groups:
A complex algebraic group is reductive if and only if it is the complexification of a compact Lie group.
More precisely, every compact Lie group $G $ admits a (unique?) real algebraic structure and thus we can speak of its complexification $ G \otimes_\mathbb R \mathbb C $.  The groups obtained this way are precisely the complex reductive groups.  In fact, there is an equivalence of categories between compact Lie groups and complex reductive groups.<|endoftext|>
TITLE: States and left ideals
QUESTION [8 upvotes]: Given a nontrivial left ideal $J$ of a unital $C^*$ algebra $A$, is there a state on $A$ which vanishes on all elements of $J$? (Left or right doesn't matter, just not 2-sided.) 
The problem came from the idea of a state as evaluation at a 'point' of a noncommutative space. If an ideal corresponds to a vanishing 'set', then can we look at 'points' of that 'set'? I must admit that this problem as I stated it sounds unlikely to be true, but is there another version which might work?
I would also be interested in any related references which people could suggest about the theory of 1-sided ideals in operator algebras.

REPLY [6 votes]: Just a couple of minor comments on Tomek's answer.
1) I don't think you need to restrict to positive elements.  Indeed, if $I$ is a closed left ideal then $I=AI_+$ where $I_+=I\cap A_+=$ the positive elements of $I$.  And for any $a\in A$, $b\in I_+$ and state $\phi$ vanishing on $I_+$, Cauchy-Schwarz yields
$$|\phi(ab)|^2\leq\phi(aba^*)\phi(b)=0,$$
so $\phi$ also vanishes on $I$.
2) For separable C*-algebras, each proper closed left ideal is the left kernel of a single state, but not in general.  For example, there is no faithful state on the Calkin algebra $\mathcal{B}(H)/\mathcal{K}(H)$ or even its commutative C*-subalgebra $C(\beta\mathbb{N}\setminus\mathbb{N})$, i.e. $\{0\}$ is a proper closed (left) ideal but not the left kernel of a single state.  Rather, in general the closed left ideals correspond to (weak*-)closed faces of states ($F$ is a face of a convex subset $C$ of a vector space if, for all $x,y\in C$ and $t\in(0,1)$, $x,y\in F\Leftrightarrow tx+(1-t)y\in F$).  Specifically, for any $F\subseteq A^*$ and $I\subseteq A$ define
$$\begin{align*}
{}^\perp F &=\{a\in A:\forall \phi\in F(a\phi=0)\}\text{ and}\\
I^\perp &=\{\phi\in A^*:\forall a\in I(a\phi=0)\text{ and }\phi(1)=1=||\phi||\},
\end{align*}$$
where $a\phi\in A^*$ is defined by $a\phi(b)=\phi(ba)$.  Then the maps $F\mapsto{}^\perp F$ and $I\mapsto I^\perp$ are mutually inverse bijections between closed faces of states and closed left ideals.
Personally, I think the best reference for this kind of thing is the following.

Pedersen, Gert K.  $C^∗$-algebras and their automorphism groups. 
  London Mathematical Society Monographs, 14. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], London-New York, 1979. ix+416 pp. ISBN: 0-12-549450-5.

It sounds like you might be interested in the non-commutative topological theory of C*-algebras, in which case you could also take a look at various papers by Akemann et al.
But if you don't want to go through the references, here's an outline of a proof answering your original question:
Take a proper left ideal $I$ of $A$.  Note no $a\in I$ can be (left) invertible as then we would have $1=a^{-1}a\in I$ and hence $I=A$.  This means $0\in\sigma(a)$ and hence $\lambda\in\sigma(a+\lambda1)$ so $|\lambda|\leq||a+\lambda1||$, for all $a\in I$ and $\lambda\in\mathbb{C}$.  Thus the linear functional $\phi$ defined on $I+\mathbb{C}1$ by $\phi(a+\lambda1)=\lambda$ has norm $1$ and, by Hahn-Banach, has a norm $1$ extension to $A$.  Thus this extension is a state on $A$ vanishing on $I$.
The same is true even for non-unital $A$ so long as you are talking about proper closed left ideals.<|endoftext|>
TITLE: The cooperations algebras Johnson-Wilson theory and truncated BP-theory
QUESTION [5 upvotes]: Given a prime $p$, there is a well known homology theory $BP$, known as Brown-Peterson homology. Has several related theories, namely the Johnson-Wilson theories $E(n)$ and the truncated Brown-Peterson theories $BP\langle n\rangle$. Their homotopy groups are given by 
$$
E(n)_* = \mathbb{Z}_{(p)}[v_1, ..., v_{n-1}, v_n^{\pm 1}]
$$
and $BP\langle n \rangle$ is a connective version of $E(n)$. In the case $n=1$, $p$-local $KU$ splits as a wedge of suspensions of $E(1)$, and $ku$ splits as a wedge of $BP\langle 1 \rangle$, and in this case the algebras $E(1)_*BP\langle 1\rangle$ and $E(1)_*E(1)$ are understood and the elements have interpretations as numerical polynomials. 
Does anyone know of methods to compute these algebras for $n>1$, or even $n=2$? I think I can compute $E(1)_*E(2)$ and $E(1)_*BP\langle 2\rangle$, so I would also be interested to know if there is a method of producing $E(2)_*E(2)$ from $E(1)_*E(2)$ and $K(2)_*E(2)$.

REPLY [5 votes]: Regarding $E(n)_*E(n)$, see "On the Structure of the Hopf Algebroid $E(n)_*E(n)$" by Keith Johnson. Johnson shows that $$E(n)_*E(n) \otimes \mathbb{Q} \simeq \mathbb{Q}[v_1,\cdots,v_{n-1},v_n^{\pm 1},\overline{v}_1,\cdots,\overline{v}_{n−1},\overline{v}_n^{\pm 1}],$$
where $v_i = \eta_L(v_i)$ and $\overline{v}_j = \eta_R(v_j)$.
Moreover, let $A$ be the ring of integers in an unramified degree $n$ extension of p-adic numbers, and let $\mathbb{S}_n$ be the $n$-th Morava stabilizer group. Johnson also shows that there is an embedding of $E(n)_*E(n)$ in $C(\mathbb{S}_n \times \mathbb{S}_n ,A)$.<|endoftext|>
TITLE: The intuition behind the Hilbert projective metric and the Perron Frobenius Theorem
QUESTION [9 upvotes]: Recently I have read a proof of the Perron Frobenius Theorem for positive aperiodic matrices. In this proof, the trick is to put a metric in the "positive quadrant" of $\mathbb{R}^n$, $\mathbb{R}^{n}_+$, in order to make the map $\mathbb{R}^n_+\ni p\mapsto pA $ a $\lambda$-contraction, $0<\lambda<1,$ and in this way applies the Contraction Fixed Point Theorem to obtain the only one fixed point for $A.$
To obtain this kind of result, $\mathbb{R}^n_+$ is "projectivezed" and the
 $\textbf{Projective Hilbert Metric}$ is considerate.  https://en.wikipedia.org/wiki/Hilbert_metric
This proof was given by G. Birkhoff, http://www.jstor.org/stable/1992971 
My Question: What is  the intuition behind the Hilbert Projective metric, there exists an easy way to see that the projective metric $d$ is the right metric  to make $(\mathbb{R}^n_+\ni p\mapsto pA, d) $ a  $\lambda$-contraction ?

REPLY [2 votes]: There is actually a paper written on this topic by E. Kohlberg and W. Pratt, named The  contraction  mapping  approach  tothe  Perron-Frobenius  theory  :  why  Hilbert’s  metric ? published in Math. Oper. Res.7.2 (1982). They explain there why the Hilbert metric is contracting.
They also prove something which is worth noting here. Let $K$ be a convex cone in $\mathbb{R}^d$. Say that $K$ is pointed if $K\cap -K=\{0\}$. A projective metric on a pointed cone $K$ is an application $D:K\times K \to \mathbb{R}_{\geq 0}$ which is symmetric, satisfies the triangle inequality and the following separation condition : if $D(x,y)=0$, then $x=\lambda y$ for some $\lambda\geq 0$. Typically, the Hilbert metric is a projective metric on the pointed cone of non-negative vectors.
Letting $K$ be a cone, say that a linear transformation $A$ is positive (with respect to $K$) if $Ax\in K$ whenever $x\in K$. When given a projective metric $D$ on $K$, define then
$$k_D(A)=\mathrm{sup}  \left ( \frac{D(Ax,Ay)}{D(x,y)},D(x,y)>0 \right ).$$
Then, $A$ induces a contraction on $K$ for the metric $D$ if and only if $k_D(A)<1$.
E. Kohlberg and W. Pratt define in general a Hilbert metric on any pointed cone $K$, which is the same as the metric described by @VaughnClimenhaga. They prove that any positive linear transformation $A$ induces a contraction for this Hilbert metric. They also prove the following.
Theorem (Kohlberg, Pratt) Let $K$ be a pointed cone in $\mathbb{R}^d$ and let $d$ be the Hilbert metric on $K$ and let $D$ be another projective metric on $K$.
Let $A$ be a positive linear transformation.
Then, there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that for any $x,y$, $D(x,y)=f(d(x,y))$.
Moreover, $k_d(A)\leq k_D(A)$.
Consequently, not only the Hilbert metric yields a contraction, but this is the optimal metric doing so (since it has the best possible contraction coefficient).<|endoftext|>
TITLE: Calculation of integral using Gamma function when the imaginary part is zero
QUESTION [7 upvotes]: Consider the following expression of Gamma function
$$\frac{\Gamma(z)}{p^z}=\int_{0}^{\infty}e^{-pt}t^{z-1}dt  \ \ \ \ \ \ \ \ \ (1)$$
where $Re(z)>0$ and $Re(p)>0$.
In Lebedevs book "special functions and their applications" he uses that integral to calculate
$$\int_{0}^{\infty}e^{2izs}s^{\alpha-1}ds=\frac{\Gamma(\alpha)}{(-2iz)^{\alpha}} \ \ \ \ \ \ \ \ (2)$$
but what about case when $z=x$ is real? This means $Re(p)=0$ in $(1)$ expression.
P.S.
The Formula $(2)$ must be valid for Real $z$ because of final result of calculation (asymptotic expression of Hankels' function $H_{\nu}^{(1)}(z)$).
I just want to know how to integrate (2) integral if z=x is real variable.

REPLY [4 votes]: The identity can be deduced by the Euler integral by homotopy invariance of path integrals. 
The function $e^{-z}z^{\alpha}$ is holomorphic on $\mathbb{C}\setminus \{\operatorname{Re}z\le0,\operatorname{Im}z=0 \}$, so the value of the path integral along the boundary of the domain $\{z: \operatorname{Re}z>0,\operatorname{Im}z>0, \epsilon< |z|< \rho \}$ is zero. Hence we get
$$\int_\epsilon^{\rho}i^\alpha e^{-it}t^{\alpha-1}dt=\int_\epsilon^{\rho}e^{-t}t^{\alpha-1}dt+  \int_0^{\pi/2}ie^{-\rho e^{it} }(\rho e^{it})^{\alpha}dt- \int_0^{\pi/2}ie^{-\epsilon e^{it}}(\epsilon e^{it} )^{\alpha}dt $$
We consider separately the three integrals on the RHS.

The first integral, of course, converges to the Euler integral for $\Gamma(\alpha)$ as $\epsilon\to0$ and $\rho\to+ \infty$, provided $\operatorname{Re}\alpha>0$. 
The second integrand has absolute value $  e^{-\rho\cos(t) -\operatorname{Im}\alpha t}\;  \rho^{ \operatorname{Re}\alpha} $, and it is a bit singular at $t=\pi/2$;  to evaluate the corresponding integral, it is convenient to spit it further in the integrals over $[0,x]$ and $[x,\pi/2]$ with a free $x$, optimizing then the bound over the choice of $x$. This way one gets a bound for this integral  of order 
$O\Big(\rho ^{\operatorname{Re}\alpha -1} \log(\rho) \Big)$, which is $o(1)$ as $\rho\to+ \infty$, provided $\operatorname{Re}\alpha <1$.
The third integrand  has absolute value $  e^{-\cos(t)  \epsilon-\operatorname{Im}\alpha t}\;  \epsilon^{ \operatorname{Re}\alpha}=O(\epsilon^{ \operatorname{Re}\alpha}) $.

We conclude that for   $0< \operatorname{Re}\alpha<1$ the integral on the LHS converges, and 
$$\int_0^{+\infty}  e^{-it}t^{\alpha-1}dt=\frac{\Gamma(\alpha)}{i^\alpha},$$
which is the wanted identity for $z=-1/2$. For any real $z<0$, with a linear change of variable, 
$t=(-2z)s$ we plainly get  
$$\int_0^{+\infty}  e^{2izs}s^{\alpha-1}dt=\frac{\Gamma(\alpha)}{(-2iz)^\alpha}.$$
Finally, for a real positive $z>0$, we take  complex conjugate to both sides to the latter identity, written for $-z$ and $\overline \alpha$, which yields to the identity for $z$ and $\alpha$.<|endoftext|>
TITLE: Hilbert schemes and moduli of ideal sheaves
QUESTION [11 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{C}$. The Hilbert scheme on $X$ parametrizes quotients $\mathcal{O}_X \to E$ with fixed Hilbert polynomial. Let us fix the Hilbert polynomial to lead to subschemes of codimension at least $2$. Then we can also look at the moduli space parametrizing semistable rank $1$ sheaves with corresponding Hilbert polynomial and trivial determinant, i.e. ideal sheaves.
Whenever we have a family of quotients, we can take the kernel to get a family of ideal sheaves. That should induce a bijective morphism. However, it seems unclear to me how to get an inverse morphisms.
Under what kind of hypotheses are these two moduli spaces the same? I am in particular interested in the case of Hilbert schemes of curves in $\mathbb{P}^3$.
Edit: Since I never wrote it down and it came up in the comments, let me write down what the functors are I am exactly talking about.
The Hilbert scheme represents the functor that maps $S$ to quotients $\mathcal{O}_{X \times S} \to E$ that are flat over the base $S$.
The moduli of semistable sheaves represents the functor that maps $S$ to flat families $E \in \operatorname{Coh}(X \times S)$ of semistable sheaves modulo tensoring with line bundles pulled back from $S$.
We then look at the corresponding connected components coming from fixing the Hilbert polynomial and ask whether they are the same.

REPLY [7 votes]: It seems to me that $$\mathscr{I}_Z^{\vee\vee}\simeq \mathscr{O}_X,$$ because the left-hand side is reflexive and isomorphic to $\mathscr{O}_X$ away from $Z$, which has codimension at least $2$ by assumption.  In other words, the abstract sheaf $\mathscr{I}_Z$ is canonically embedded in $\mathscr{O}_X$ via the double-dual map $$\mathscr{I}_Z\to \mathscr{I}_Z^{\vee\vee}\simeq \mathscr{O}_X.$$
So the inverse map you are looking for is induced by $$\mathscr{I}_Z\mapsto (\mathscr{O}_X\simeq\mathscr{I}_Z^{\vee\vee}\to \text{coker}(\mathscr{I}_Z\to \mathscr{I}_Z^{\vee\vee})\simeq \mathscr{O}_Z).$$
I suppose I used that $X$ is smooth in the identification $$\mathscr{I}_Z^{\vee\vee}\simeq \mathscr{O}_X,$$ but you can probably relax this a bit. (I'm a little worried about making this work in families, which is, strictly speaking, necessary; so this answer is not quite complete.)<|endoftext|>
TITLE: List of finitely presented groups with undecidable word problem
QUESTION [8 upvotes]: Is there any reasonably updated list of (representative) examples of finitely presented groups with undecidable word problem?
By "representative" I mean "avoiding obvious redundancy", i.e. examples different from each other in providing (from some point of view) new insight into the problem.
A brief comment on the ideas and methods used, and even the concrete presentations when possible, would also be of interest.
In case there is not such list, I wonder if it is a suitable big-list for MO.

REPLY [5 votes]: In addition to Miller's book cited above you can look at "Algorithmic problems in varieties" and "Asymptotic invariants, complexity of groups and related problems" for more recent examples. These include  results about  complexity of the word problem. Even more recent results are in papers by Olshanskii and myself (see the arXiv).<|endoftext|>
TITLE: Is there an integer for every Euler rank?
QUESTION [10 upvotes]: Let $φ$ be the Euler function, then $φ(n)+1 = n$ iff $n$ is prime.
Let the map $\psi: n \mapsto φ(n)+1$  
An integer $m$ is of Euler rank $0$ if it is prime, and of Euler rank $r$ if $\psi^{(r)}(m)$ is prime whereas $\psi^{(r-1)}(m)$ is not. The following list shows the smallest integer $n \ge 2$ of rank $r \le 15$:  
$\scriptsize{ \begin{array}{c|c}
 r  &0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline 
                             \hline
n &2&4&15&35&69&255&535&949&1957&2513&2923&4531&17701&22957&54589&79421  
\end{array} }  $  
Question: Let $r \ge 0$,  is there an integer $n \ge 2$ of  Euler rank $r$?

REPLY [5 votes]: Not an answer, but an extension of the sequence with PARI/GP
? for(j=0,40,m=2;s=0;while(s<>j,m=m+1;n=m;s=0;while(isprime(n)==0,s=s+1;n=eulerp
hi(n)+1));print(m," ",s))
2 0
4 1
15 2
35 3
69 4
255 5
535 6
949 7
1957 8
2513 9
2923 10
4531 11
17701 12
22957 13
54589 14
79421 15
80029 16
84493 17
98581 18
102827 19
115243 20
239111 21
291149 22
310813 23
362621 24
398893 25
598341 26
801923 27
838307 28
1063493 29
1079833 30
1123813 31
1311121 32
1329403 33
1582439 34
1677931 35
1751831 36
2382469 37
7754613 38
70817623 39
94342651 40

The sequence of the smallest numbers with a given rank must strictly increase because for
every composite number $n$, we have $\phi(n)+1<n$. If the smallest number with
rank $r$ is $a$ and the smallest number with rank $r+1$ is $b$, then the assumption $a\ge b$ together with rank $(\phi(b)+1)=r$ and $\phi(b)+1<b\le a$ contradicts the choice of $a$.<|endoftext|>
TITLE: Is the Feichtinger's algebra $(S_0(\mathbb{R^d}),||\cdot||_{S_0})$ reflexive?
QUESTION [5 upvotes]: The Feichtinger's algebra
$S_0(\mathbb{R^d})=M^{1,1}(\mathbb{R^d}):=\{f\in L^2(\mathbb{R^d}):V_g(f)\in L^1(\mathbb{R^{2d}})\}$, where
$V_g(f)(x,\omega)$ is the short-time Fourier transform of $f$ with respect to $g \in S(\mathbb{R^d})$, which is the Schwartz space, defined by
$$
V_g(f)(x,\omega)=\int_\mathbb{R^d} f(t)\, \overline{g(t-x)}\, e^{-2\pi i\omega \cdot t} \, dt.
$$ 
Then we define the norm on $S_0(\mathbb{R^d})$:
$$
||f||_{S_0}=\int_{\mathbb{R^d}}  \left(\int_{\mathbb{R^d}}|V_gf(x,\omega)| \, dx\right)\, d\omega
$$
It's known that $(S_0(\mathbb{R^d}),||\cdot||_{S_0})$ is a Banach space, and $S(\mathbb{R^d})$ is a dense subspace of $S_0(\mathbb{R^d}).$ My question: is $(S_0(\mathbb{R^d}),||\cdot||_{S_0})$ reflexive?

REPLY [7 votes]: No. Wilson bases (or more generally local Fourier bases) allow to identify $S_0$ with the sequence space $\ell^1$ (i.e. the elements of $S_0$ are just the functions in $L^2$ having an absolutely convergent Wilson series expansion). HENCE the space is a dual space (the pre-dual is so to say $c_0$ at the sequence space level), resp. the closure of the test functions in the dual space $S_0'$, resp. the space of all tempered distributions  which have STFTs (short time Fourier transforms) which vanish at infinity (in phase space). So, NO, it is not a reflexive space.   HGFei<|endoftext|>
TITLE: How many random sieve operations to decimate the set {2,...,n}?
QUESTION [11 upvotes]: Let $S$ be the set of integers $\{2,3,4,\ldots,n\}$.
Consider the following process:

Select a random element $k \in S$. 
Remove from $S$ every number divisible by $k$. 
Repeat with this reduced $S$.

I am interested in the number of repetitions $\sigma(n)$ needed to reduce $S$ to the empty set.
Example, $n=10$:
\begin{eqnarray}
\# &:& 2,3,4,5,6,7,8,9,10\\
10 &:& 2,3,4,5,6,7,8,9\\
9 &:& 2,3,4,5,6,7,8\\
3 &:& 2,4,5,7,8\\
4 &:& 2,5,7\\
5 &:& 2,7\\
7 &:& 2\\
2 &:& \varnothing
\end{eqnarray}
Limited data suggests that the number of sieve operations $\sigma(n)$ needed
to decimate the set $S$ might grow approximately linearly with $n$:

          


          

Number of random sieve iterations to decimate the set $\{2,3,\ldots,n\}$.


          

Each point plotted is the average of $10$ simulations.


The slope is about $0.28$ (without any careful statistics).
Perhaps it is possible to estimate $\sigma(n)$ for large $n$?

REPLY [12 votes]: An equivalent way of describing the process: We start with a randomly chosen permutation $\tau$ of $\{2, \dots, n\}$.  At each step we choose the first number in $\tau$ which is still in our set $S$, and apply the deletion process with that as our $k$.  A key property here:
An integer $x$ is chosen at some point if and only if x appears before all of its divisors in $\tau$
This is simply because if some divisor $y$ of $x$ appears before $x$, then either $y$ itself is chosen, or some divisor of $y$ (which is also a divisor of $x$) is chosen.  
In particular, the probability that $x$ is chosen at some point is $\frac{1}{d(x)-1}$, where $d(x)$ is the number of divisors of $x$.  This implies that the expected number of steps to decimation is 
$$\sum_{x=2}^n \frac{1}{d(x)-1}$$
I feel like this sum (or at the very least $\frac{1}{d(x)}$) has to be well-studied, but I don't have any references.<|endoftext|>
TITLE: Proof without distributions
QUESTION [5 upvotes]: I was wondering whether there is a way to show this identity 
$$\pi \int_{\mathbb{R}^3} \frac{f(x)}{|x|} dx = \int_{\mathbb{R}^3} \frac{\widehat{f(x)}}{|x|^2} dx $$ without using distributions for $f \in \mathcal{S}(\mathbb{R}^n)$.
The reason I ask is the following: Obviously this identity makes perfect sense in the classical way, but the most obvious proof would use the fourier transform of $\frac{1}{|x|^2}$ which is clearly not defined in $L^1$ or $L^2$.
Thus, I would be interested, whether one can proof this identity with classical methods, i.e. no distribution theory,  too?

REPLY [2 votes]: @Phil Isett's device also does apply to arbitrary $\int_{\mathbb R^n} f(x)/|x|^\alpha\;dx$ with $0< \Re(\alpha)<n$, as follows, disregarding constants and normalization of Fourier transforms:
$$
\int_{\mathbb R^n} {f(x)\over |x|^\alpha}\;dx
\;=\;
{1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} f(x)\,e^{-t|x|^2}\,t^{\alpha/2}\;dx\;{dt\over t}
$$
$$
= 
{1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} \widehat{f}(x)\,e^{-{1\over t}|x|^2}\,t^{{\alpha-n\over 2}}\;dx\;{dt\over t}
$$
by Plancherel and by knowing how to take Fourier transform of Gaussians. Replace $t$ by $1/t$ to obtain (up to constants)
$$
{1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} \widehat{f}(x)\,e^{-t|x|^2}\,t^{{n-\alpha\over 2}}\;dx\;{dt\over t}
= {\Gamma({n-\alpha\over 2})\over \Gamma(\alpha/2)} \int_{\mathbb R^n} {\widehat{f}(x)\over |x|^{n-\alpha}}\;dx
$$<|endoftext|>
TITLE: Is any function taking compact sets to compact sets, and connected sets to connected sets, necessarily continuous?
QUESTION [17 upvotes]: It is well-known that continuous image of any compact set is compact, and that continuous image of any connected set is connected.
How far is the converse of the above statements true?
More precisely: Let $X$ be a topological space. Suppose there is a map $f$ from $X$ into itself which takes compact sets to compact sets and connected sets to connected sets. Do these conditions imply that $f$ is continuous?
If not in general, at least when $X$ is a metric space? or, when $f$ is a bijection? or both? or under any other additional conditions?
My apologies if my question in boringly elementary.
Thanks in advance!

REPLY [22 votes]: There are highly non-trivial positive results. 
Let us call a function $f$ from a space $X$ into a space $Y$ preserving if the image of every compact subspace of $X$ is compact in $Y$ and the image of every connected subspace of $X$ is connected in $Y$. 
McMillan  [On continuity conditions for functions, Pacific J. Math. 32 (1970) 479-494] proved the following result:
Theorem: If $X$ is Hausdorff, locally connected and Fréchet, and $Y$ is Hausdorff (e.g. if $X=Y=\mathbb R$), then any preserving function $f:X\to Y$ is continuous.
For further  results see Gerlits, Juhasz, Soukup, Szentmiklossy: Characterizing continuity by preserving compactness and connectedness, Top. Appl, 138 (2004), 21-44
Our main result is the following:
Theorem:  If $X$ is any product of connected linearly ordered spaces (e.g., if $X = \mathbb R^\kappa$ ) and $f:X \to Y$ is a preserving function into a regular space $Y$, then $f$ is continuous.<|endoftext|>
TITLE: The number of integral solutions to $x^2+y^2-az^2=0$
QUESTION [6 upvotes]: I think this must be well-known (and probably not hard to prove either), but I cannot find a reference: for a (positive) rational number $a$, the number of integral solutions to the equation
$$ x^2+y^2-az^2=0, $$
with $|x|,|y|,|z|<T$ is $C(a) T \log T$, where $C(a)$ is a constant depending only on $a$. I would very much appreciate a reference which also includes a proof.

REPLY [8 votes]: You need some further assumptions on $a$ to guarantee the existence of a solution to this equation, as there might not be solutions modulo some prime, say. The theorem of Hasse and Minkowski tells you exactly when solutions exist.
But anyway, the result you want is Theorem 8 in the paper:
Heath-Brown - A New Form of the Circle Method, and its
Application to Quadratic Form.
You can find this here:
http://eprints.maths.ox.ac.uk/144/1/circle.pdf (Wayback Machine)<|endoftext|>
TITLE: Reference request for simple $q-$ identities
QUESTION [5 upvotes]: I stumbled upon the following  simple $q-$identities:
$$\frac{1}{(-q;q)_\infty}\sum \limits_{j =0}^{\infty}\frac{q^{(2r+1)j}}{(q^2;q^2)_j}=(q;q^2)_r$$
and
$$\frac{1}{(q;q^2)_\infty}\sum \limits_{j =0}^{\infty}(-1)^{j}\frac{q^{(r+1)j}}{(q;q)_j}=(-q;q)_r,$$
where $r\in\mathbb{N} $ and $\binom{r}{j}_q$ denotes a $q-$binomial coefficient. 
Probably these are well known. But where can I find such identities?  Are there tables of $q-$identities in the literature?

REPLY [5 votes]: You can use eqn. (1), (2) and (5) in this file to show the identities.<|endoftext|>
TITLE: Model independent proof of colimit formula for left Kan extensions
QUESTION [6 upvotes]: I am interested in finding a proof of the colimit formula for left Kan extensions $(\infty,1)$-categories which does not rely on a chosen model. (By a model inpendent proof I mean a proof which uses only categorical reasoning, which should be able to be performed in any given model +  straightening/unstraightening).
Moreover, I would like to prove the reciproque: If the considered colimits exist in the target category then the left Kan extension exists and is given by the colimit formula. I've tried different approaches, namely:
1. Establishing the coend formula: However, to give a proof of this result, one needs to establish first the end formula for natural transformations, and it seems that a model independent of this fact cannot be given, since at some point one needs to do some calculations, which cannot be done in a model-free way. (For a reference of the end formula for natural transformations, using the theory of quasicategories, one has http://arxiv.org/pdf/1501.02161.pdf)
2. The colimit formula would follow from the Beck-Chevalley condition for coCartesian fibrations. However, even though that Lurie's HTT has a treatment about this, a model independent proof of this seems far from reachable.
3. To explicitly construct a functor, whose on objects would give the desired colimit formula, satisfying the universal property for LKE, using straightening/unstraightening in a clever way. But still, to prove such result, one needs to explicit do some calculations which do not seem at all feasible, even if one adopts a given model, (e.g., quasicategories).
Even if such approaches seem to be in vain, I am convinced that it should be possible to give a model independent proof of this result.

REPLY [18 votes]: Dominic Verity and I have been working to develop model independent foundations of $(\infty,1)$-category theory. Our aim at present isn't to cover all models of $(\infty,1)$-categories but only the "best-behaved ones" (which include quasi-categories, complete Segal spaces, Segal categories, and naturally marked simplicial sets). The benefit to restricting to these models is that it allows us to work more strictly so that the development of the basic category theory is more similar to that for ordinary 1-categories, working in the strict 2-category CAT. A survey of the present state of this project can be found here:
$\infty$-category theory from scratch
Let me now sketch how our approach deals with Kan extensions. The details can be found in section 5 of our most recent paper:
Kan extensions and the calculus of modules for $\infty$-categories
For each of the well-behaved models of $(\infty,1)$-categories there is a strict 2-category whose objects are the $(\infty,1)$-categories of this variety, whose morphisms are the $(\infty,1)$-functors, and whose 2-cells I'll call natural transformations. For concision let me now call the objects and morphisms $\infty$-categories and $\infty$-functors. We call this the homotopy 2-category because it can be thought of as a categorification of the usual homotopy category associated to a model category. The homotopy 2-categories defined for each of the four models mentioned above are connected by 2-functors that are biequivalences. Each homotopy 2-category has certain additional structures that are derived from a common axiomatization which we use to develop the theory of $\infty$-categories: namely each of the well-behaved models defines what we call an $\infty$-cosmos. Everything is proven relative to the $\infty$-cosmos axiomatization, which is why this approach is "model agnostic."
Some pleasant features of the homotopy 2-category is that the equivalences and adjunctions in there are exactly the correct notions, meaning they recover the model-theoretic notions of weak equivalence between $\infty$-categories and also Lurie's notion of adjunctions in the case of quasi-categories (though this isn't obvious). Inside the homotopy 2-category you can also define (co)cartesian fibrations (which again specialize to the Lurie ones in the case of quasi-categories) and also define limits and colimits of diagrams valued in an $\infty$-category indexed by either a simplicial set or by another small $\infty$-category (because the models just mentioned happen to all be cartesian closed).
Now Kan extensions of one $\infty$-functor along another can also be defined in any 2-category but some care is needed here. The universal property of being a mere Kan extension in the homotopy 2-category is not strong enough: instead the correct notion is of a pointwise Kan extension. We give two equivalent definitions of this notion but the simplest for the present purpose is that a Kan extension of an $\infty$-functor $f \colon A \to C$ along an $\infty$-functor $k \colon A \to B$ is pointwise if when you paste an exact square whose bottom edge is $k$ on top, the result is still a Kan extension. Exact squares are squares with a natural transformation in them that are characterized by a property that makes use of the modules of the title.
Examples of exact squares include pullback squares one of whose edges is (co)cartesian; perhaps this is the Beck-Chevalley result you were referring to? Another example of an exact square is a comma square. This proves the easy half of the colimit formula for left Kan extensions that you were referring to: the value of the pointwise left Kan extension of $f$ along $k$ at some $b : 1 \to B$ is given by the limit of the restriction of $f$ along the projection $k \downarrow b \to A$.
The converse direction is more subtle because we need to assemble these colimit values, one for each "object" $b$ in $B$, into an $\infty$-functor $lan_k f : B \to C$. What's needed to do this is something like the principle that "universal properties for $\infty$-categories are determined objectwise." For quasi-categories, we prove a precise theorem expressing this idea. The proof involves some horn-filling inductive argument over simplices, so it's very particular to the quasi-categorical model. In the very last part of our paper, we use this to sketch a proof that any cocomplete quasi-category has all Kan extensions of small functors given by the colimit formula (full details will appear in our forthcoming sixth paper, which specializes to the quasi-categorical case).
Because our argument uses the quasi-categorical model there is some model independence lost at this step, but happily it's only lost in the proof, not in the conclusion. The reason for this is a model independence theorem (which is sketched in the "scratch" lecture notes and will be presented in more detail in our forthcoming seventh paper, on model independence). Not only are the homotopy 2-categories for the models mentioned above biequivalent, but there is a more sophisticated categorical structure called a virtual equipment that we construct for each model, and these virtual equipments are also biequivalent. The virtual equipment encodes the "calculus of modules" for $\infty$-categories; in particular, this captures the universal properties of cartesian fibrations and Kan extensions and the like. So the "universal properties for $\infty$-categories are determined objectwise" result also holds for the other models and the rest of our argument (which takes place in the virtual equipment) now applies.
Sorry for the length of this post!<|endoftext|>
TITLE: Characterization of finite groups using sum of the orders of their elements
QUESTION [11 upvotes]: Notation: If $G$ is a finite group, $o(g)$ denotes the order of the element $g\in G$. 
Motivation: Some finite groups could be uniquely determined by the size of the group. For example given a prime number $p$ there is just one group $\mathbb{Z}_p$ up to isomorphism. As there are infinitely many prime numbers, the number of finite groups that are completely determined with their size is infinite.  
On the other hand there are finite groups that could be uniquely determined by the sum of the orders of their elements. For example $A_5$ is the unique finite group $G$ with $\Sigma_{g\in G}o(g)=211$ (see this paper by Habib Amiri, S. M. Jafarian Amiri & I. M. Isaacs). 
Inspired by this fact the following question arises:  

Question: For which natural numbers like $n$, there is a unique finite group up to isomorphism with $n=\Sigma_{g\in G}o(g)$? Is there infinitely many such natural numbers?

REPLY [5 votes]: As a counterbalance to the question,  note  that for any odd prime $p$, there are two non-isomorphic groups of ordér $p^{3}$ with sum of element orders $p^{4} -p+1$.
Any $p$-group with that element order sum has order at most $p^{3}$. Groups of exponent $p$ are dealt with, and a $p$-group with more than one cyclic subgroup of order $p^{2}$ has element order sum too large- likewise for a $p$-group of order $p^{3}$ with a unique subgroup of order $p^{2}$. Cyclic $p$-groups are easily dealt with. So there are no more $p$-groups with that element order sum.<|endoftext|>
TITLE: Natural operators in differential geometry - why are they natural?
QUESTION [7 upvotes]: I'm reading bits and pieces of Kolar, Michor, & Slovak's Natural Operations in differential Geometry, and I'm having "doubt" about some of the definitions. All I'm trying to do is sheafify some of the concepts and see how natural they seem to me. I don't know any differential geometry so please don't kill me.
Let $l\mathsf{Diff}_m$ denote the category of smooth manifolds and local diffeomorphisms. A fibered manifold is a surjective submersion.

Definition 14.1 A natural bundle is a functor $l\mathsf{Diff}_m\rightarrow \mathsf{FM}$ satisfying

Prolongation $BF=1$ where $B$ is the base functor sending a fibered manifold into its base space.
Locality - If $i:U\hookrightarrow M$ is an inclusion of an open submanifold, then $FU=p^{-1}_M (U)$ and $Fi:FU\hookrightarrow FM$ is
  the inclusion of $FU$ into $FM$.


Denote by $\Gamma(Y)$ the (global) sections of a fibered manifold $p:Y\rightarrow M$.

Definition 14.13 Let $p:Y\rightarrow M,\bar p:\bar Y\rightarrow M$ be fibered manifolds. A local operator $A:\Gamma(Y)\rightarrow
 \Gamma(\bar Y)$ is a map such that for every global section $s$ and
  every point $x\in M$, the value of $As(x)$ depends only on the germ of
  $s$ at $x$. If moreover, for some $k\in \mathbb N$ we have
  $j_x^ks=j^k_sq\implies As(x)=Aq(x)$, $A$ is said to be of order $k$.
  A regular operator is a local operator which sends smoothly
  parametrized section into smoothly parametrized sections into smoothly
  parametrized sections.

First batch: So it seems a local operator is just a map $A:\Gamma(Y)\rightarrow \Gamma(\bar Y)$ which lifts to stalks $A_x:\Gamma(Y)_x\rightarrow \Gamma(\bar Y)$. Do we get a lift $\Gamma(Y)_x\rightarrow \Gamma(\bar Y)_x$?
For the following definition, it seems the value of a natural bundle $F$ at $N$ is a bundle $FN\rightarrow N$ and that we no longer fix a base space $M$.

Definition 14.15 A natural operator $A:F\rightsquigarrow G$ between two natural bundles $F$ and $G$ is a system of regular
  operators $A_M:\Gamma(FM)\rightarrow \Gamma(GM),\;M\in
 l\mathsf{Diff}_m$ satisfying

For each global section and each diffeomorphism we have $A_N(Ff\circ s\circ f^{-1})=Gf\circ A_Ms\circ f^{-1}$
$A_U(s|_U)=(A_Ms)|_U$ for each global section $s$ and every open submanifold $U\subset M$


Second batch: First of all, why is $f$ taken to be a diffeo? Shouldn't we ask for commutation for all local diffeos? They're the arrows in our category after all... If so, then it seems a natural operator is natural in two different ways: Given a natural bundle $F$, define the functor $\Gamma(-,F-):l\mathsf{Diff}_m\longrightarrow \mathsf{Set}$ on objects by global sections, and on arrows by $\Gamma(f,Ff):s\mapsto Ff\circ s\circ f^{-1}$. In fact, each $\Gamma(M,FM)$ is really the sheaf of sections of the bundle $FM\rightarrow M$, so $\Gamma(-,F-)$ seems to yield a functor taking values in sheaves (I don't know into what category one should stick all sheaves). Now it seems a natural operator is a natural transformation $A:\Gamma(-,F-)\Rightarrow \Gamma(-,G-)$ which also respect the sheaf structure in that the components $(A_U)$ for open submanifolds $U$ of a fixed manifold $M$ also give a sheaf morphism $\Gamma(M,FM)\Rightarrow \Gamma(M,GM)$. Assuming I am not too far off, and that this really is equivalent data to a natural operator, I have to admit that this notion doesn't seem all that natural to me at all! Could someone geometrically motivate this notion? Why shouldn't we be satisfied with mere natural transformations $F\Rightarrow G$?

REPLY [3 votes]: This is a long comment on some of the questions from the second batch. Before I start, here is a typical example of a natural operator. Take $F=\Lambda^kT^*$, $G=\Lambda^{k+1}T^*$ to be the natural bundles of differential forms of degrees $k$ and $k+1$. They are clearly local, and they behave naturally under pullback (which is not asked for here). But local diffeos have local inverses, by which we pull back, so we can regard them as covariant functors as in Definition 14.1 (this part looks unnatural to me in a nontechnical sense, but the OP did not complain about this point).
The exterior differential $d$ is a natural operator sending $\alpha\in\Gamma(\Lambda^kT^*N)$ to $d\alpha\in\Gamma(\Lambda^{k+1}T^*N)$. Again, it is local, and compatible with pullback by smooth maps. In particular, it is compatible with pullback by inverses of global diffeomorphisms, which is the first point in Definition 14.15.
If you are used to think in terms of sheafs, you want to rewrite 14.15 in terms of sheafs, and you will get an equivalent notion. However, some differential geometers don't linke to think in sheafs, but rather in terms like "local" and "natural under diffeomorphisms". So up to the very last few sentences, I think you understood everything perfectly.
Now to the last remark. In the example above, you are not allowed to write $d\colon\Lambda^kT^*-\Rightarrow\Lambda^{k+1}T^*-$ because elements of $\Lambda^kT^*M$ are differential $k$-forms at some single point $x\in M$ (not germs), so you cannot differentiate. Indeed, a natural operator in the sense above can be written as a natural transformation $F-\Rightarrow G-$ if and only if it is of degree $0$.<|endoftext|>
TITLE: Automorphisms and infinitesimal deformations of a smooth complete intersection
QUESTION [5 upvotes]: Let $X\subset\mathbb{P}^{n+c}$ be a smooth complete intersection of dimension $n$. 
Is it known when $Aut(X)$ is finite ?
Does there exist a formula for the dimension of the tangent space to the space of first order infinitesimal deformations $H^1(X,T_X)$ ?
I am especially interested in the case of a codimension two complete intersection of two quadrics.

REPLY [4 votes]: I can suggest you a shorthand for your second question, in case you need a practical answer.
First, if $X$ is a smooth complete intersection, let me denote by $A_X$ the affine cone over it, by $U_X= A_X \smallsetminus v$ the punctured affine cone and $\pi: U_X \to X$ the natural projection. We have then a standard short exact sequence (see for example Schlessinger's paper "On Rigid Singularities") $$ 0 \to \mathcal{O}_{U_X}\to T_{U_X} \to \pi^* T_X \to 0 $$
since we have a trivialization of the relative tangent by the Euler vector field.
Now, for a smooth complete intersection of dimension $\geq 2$ we have $$H^1(U_X, T_X) \cong T^1_{A_X}$$ where the object on the right hand side is the (graded) module of first order deformations of the affine cone $A_X$. Now, passing in cohomology and since  in general $$H^q(U_X, \pi^* \mathcal F)= \bigoplus_{k \in \mathbb Z} H^q(X, \mathcal F(k))$$ we have (when we restrict to the degree 0) $$ \cdots \to H^1(X, \mathcal{O}_X) \to (T^1_{A_X})_0 \to H^1(X, T_X) \to H^2(X, \mathcal O_X) \to \cdots $$
For a complete intersection of dimension $>2$ $H^i(X, \mathcal O_X)=0$, $i=1,2$, and thus we have $$(T^1_{A_X})_0 \cong H^1(X, T_X).$$
Now, the point is that the dimension of the graded component of the $T^1_{A_X}$ are super-easily computable using a bit of computer algebra: for example SINGULAR is extremely efficient in doing this (see for example https://www.singular.uni-kl.de/Manual/latest/sing_874.htm#SEC925), and basically you can get your answer with a couple of 'one-line command'.<|endoftext|>
TITLE: Compact surface with arbitrarily large eigenvalue
QUESTION [5 upvotes]: Consider a compact surface $M$ with genus $\gamma \geq 2$ and fix a positive real number $V$. Is it known whether it is possible to produce a metric $g$ on the surface $M$ such that $(M. g)$ has volume $V$ and the first eigenvalue of the Laplacian $-\Delta_g$ is as high as one wants? If the answer is yes, what could one make the same conclusions if the metric were additionally required to be negatively curved (but of course not constant negative curvature)? This is mainly a reference request. Somehow, from general heuristics, I am very much willing to believe that the answer to the first question is yes, but I am not at all sure about the second one.
Update: Okay, I just found this paper. See in particular Corollary 1 on page 4, which gives very general bounds on the eigenvalues depending only on the conformal class on a compact manifold of dimension $n$. More explicitly, given a compact Riemannian manifold $M$ and a fixed conformal class $[g]$, it says that the $k$-th eigenvalue is bounded above by a constant $c^{(k)}_{[g]}$ depending on the conformal class, and the aforementioned corollary gives lower bounds for this constant $c^{(k)}_{[g]}$. However, it is not at all clear that this constant $c^{(k)}_{[g]}$ remains bounded as we vary the conformal class.
Furthermore, being not familiar with literature on this kind of investigation, there are a few more questions: (1) If we consider the above Corollary on a closed surface and in the particular case $k =1$, does there exist a smooth metric $g$ in every conformal class such that $\lambda_1(g) = c^{(1)}_{[g]}$? (2) What can we say about the curvature properties of such maximizing metrics?
I am hoping that someone sufficiently familiar with the literature can answer this. Thanks, and sorry if I the question is too long.

REPLY [4 votes]: For the first eigenvalue, there is a maximal  metric for infinitely many genus: http://arxiv.org/pdf/1310.4697.pdf
The author also proved a similar result  for other eigenvalues but the paper is not yet available: http://math.univ-lyon1.fr/homes-www/petrides/research.html<|endoftext|>
TITLE: Center of $k$-algebra with two generators and sole defining relation $yx - xy = 1$ when $\text{char}\,k > 0$
QUESTION [5 upvotes]: Let $A(k)$ be a $k$-algebra with two generators, $x$, $y$, and one defining relation: $yx - xy = 1$. What is the center of the algebra $A(k)$ in the case $\text{char}\,k > 0$?

REPLY [14 votes]: This is the Weyl algebra. The center is $k[x^p,y^p]$, where $p$ is the characteristic. See, for example, here for a proof.<|endoftext|>
TITLE: Is being rational decidable?
QUESTION [21 upvotes]: Given a real number uniquely defined by a finite system of equations and inequalities with rational coefficients involving the standard elementary functions only. Is it decidable whether this number is rational?
Edit: The system of equations and inequalities may have $n$ real-valued variables, and the real number to be defined is $\xi$. By decomposition using intermediate variables, we may suppose that each inequality is of the form $x_i\,\omega\, c$ where $\omega\in\{<,>,\le,\ge\}$ and $c$ is an integer, and each equation is of the form $x_i=\xi$, or $x_i=c$ where $c$ is an integer, or $x_i=x_j\circ x_k$ with $\circ\in\{+,-,*,/\}$, or $x_i=\phi(x_k)$ where $\phi\in E$ with a specified set $E$ of elementary functions, such as $E=\{\exp,\log\}$. The answer to the question might depend on the choice of $E$.
I assume that $\xi$ is uniquely determined (hence defined) by this system of equations and inequalities. One may even assume that one has a proof for this, should this help.
The algorithm may be defined according to any of various traditional models, specifying a sensible set of primitive operations (including the rational operations [e.g., on bits or on reals] and branching on an expression). I am more interested in the applicable techniques than in an answer for a particular model.
Note that the Wikipedia article on the constant problem 
claims (by reference to 
D.H. Bailey, Mathematics of Computation 50 (1988), 275–281)
a result from which decidability would follow, but this claim is spurious.

REPLY [4 votes]: I found a paper by Babai, Just, and Meyer auf der Heide that answers a closely related question. 
They prove among others (see the discussion on p.101) that there are no anlytic computation trees (ACTs) that accept the complement of the rationals in the reals, and hence would prove the irrationality of a number.<|endoftext|>
TITLE: A systematic canonical construction of the Hodge star operator
QUESTION [6 upvotes]: I'm struggling to make sense of the Hodge star as a global canonical object. Here are my struggles so far and some questions:
Let $M$ be a finitely generated projective $R$-module (hence locally free and finitely presented). Suppose the $M$ is of rank $n$ and is equipped with a nondegenerate symmetric bilinear form $g(-,-): M \times M \to R$. We first extend the bilinear form to the exterior powers of $M$.
Let $\bigwedge^k m_{j}$,$\bigwedge^k l_{j} \in  \bigwedge^k M$ be given. There's a corresponding "orthogonal projection" map $T_{m,l}: span\{m_j\} \to span\{l_j\}$ defined by: 
$$T_{m,l} : m \mapsto \sum_{j} g(m,l_j)l_j$$
So we obtain a map $T_{m,l} \in Hom(span^k_{j=1}\{m_j\},span^k_{i=1}\{l_i\})$. Denote the induced map on the top exterior powers by $det(T_{m,l})\in$ $Hom(\bigwedge^k span\{m_j\},\bigwedge^k span\{l_i\})$. By local freenes $det(T_{m,l})$ is locally just multiplication by the determinant of the local transformation. 
Define the extended inner product: $<\bigwedge^k m_{j}$,$\bigwedge^k l_{j}> := det(T_{m,l})$. Symmetry and nondegeneracy are inherited from $g(-,-)$. Extend to non simple elements by linearity. 
This was the construction I arrived at when trying to find a cannonincal geometric definition for the extension of an inner product to exterior powers of a vector bundle. My problem now is that by this definition the extended bilinear form is only locally a bilinear form. differential-geometrically there's no real problem because things only need to make sense locally but what's really happening here?

1) Can the above construction be tweaked to get an honest bilinear form on $\bigwedge^k M$?

Assuming such a tweak is possible we'll now proceed to construct an unoriented analogue of the Hodge star out of the following maps:
$$\phi:\bigwedge^k M \to ( \bigwedge^{n-k}M \to\bigwedge^n M), u \mapsto u \wedge (-)$$
$$\psi:\bigwedge^{n-k} M \otimes\bigwedge^n M\to ( \bigwedge^{n-k}M \to\bigwedge^n M), u\otimes\omega \to <-,u>\otimes \omega$$
Define the "pre-Hodge" operator $\rho := \psi^{-1} \circ \phi : \bigwedge^k M \to \bigwedge^{n-k} M \otimes \bigwedge^n M$
Evaluating at an element $\bigwedge_{J} m_j \in \bigwedge^k M$ we get:
$$\rho(\bigwedge_{J} m_j)= \psi^{-1} \circ \phi (\bigwedge_{J} m_j) = \psi^{-1}(\sum_{|I|=n-k} ((\bigwedge_{J} m_j)\wedge (\bigwedge_{I} l_i)) \otimes (\bigwedge_{I} l_i)^*) = \sum_{|I|=n-k} ((\bigwedge_{J} m_j)\wedge (\bigwedge_{I} l_i)) \otimes (\bigwedge_{I} l_i)$$
The summing part is the problem here. It has to be some sum over all $(n-k)-$ collections with signs attached to them but i'm not sure how it should be done...


How does the map $\rho$ act on simple elements?


To apply the above construction to $\bigwedge^k M^*$ there has to be some duality pairing between them. Assume the pariring is as in the following question by Qiaochu Yuan:
$$(m_1 \wedge ... \wedge m_k) \otimes (\theta_1 \wedge ... \wedge \theta_k) \to \frac{1}{k!} \sum_{\sigma \in S_k} \text{sgn}(\sigma) \prod_{i=1}^k    \theta_i(m_{\sigma(i)}).$$
Now we have two choices for extension to $\bigwedge^k M^*$. We can extend to $M^*$ by the canonincal evaluation pairing and then extend to $\bigwedge^k M^*$ by the above. Or we could first extend to $\bigwedge^k M$ and then extend to $\bigwedge^k M^*$ by the above pairing. 

3. Do these extensions agree? If not can it be fixed?

Finally Supposing $\bigwedge^n M \cong R$. By choosing an orientation we get a trivilization of $\bigwedge^n M$ and composing with $\rho$ gives a map $\bigwedge^k M \to \bigwedge^{n-k} M$ and by the previous paragraph we get a certain "Hodge star" $\star: \bigwedge^k M^* \to \bigwedge^{n-k} M^*$

4. Does the above map $\star$ coincide with the Hodge star?

And finally:

5. Does it have to be so complicated? Or is there a simple construction i'm unaware of?

REPLY [2 votes]: The bilinear form $g$ need not be symmetric. There is a canonical Hodge operator  associated to symplectic manifolds.  Here is a general construction of a Hodge type operator. 
Suppose that  $V$ is a   finite  dimensional real vector space, $\dim V=n$. To construct a Hodge type operator  you need two things.  

A linear isomorphism $D: V\to V^*$.
An orientation, i.e.,  a linear isomorphism $\newcommand{\bR}{\mathbb{R}}$ $\omega: \det V\to \bR$, where $\det V:=\Lambda^{\dim V} V=\Lambda^n V$.

The  existence of a linear isomorphism $D: V\to V^*$  is equivalent with the existence of a  nondegenerate bilinear map
$$ B: V\times V\to\bR. \tag{1} $$
(The bilinear map $B$ need not be symmetric.)
The duality $D$ induces canonical isomorphisms $ D^{\wedge k}: \Lambda^k V\to\Lambda^k V^*$, $k=1,\dotsc, n$.
We obtain   nondegenerate bilinear maps
$$\Xi_k :\Lambda^k V^*\times \Lambda^{n-k} V\to \bR,\;\;\Xi_k(\alpha_k,\beta^{n-k})=
\omega\Bigl(\, \bigl(D^{\wedge k}\bigr)^{-1}\alpha_k\wedge \beta^{n-k}\,\Bigr). $$
This  nondegenerate bilinear map produces  a linear isomorphism
$$H_k:  \Lambda^k V^*\to \bigl(\,\Lambda^{n-k} V\,\bigr)^* $$
which is a generalization of the  Hodge operator.  
More concretely,  if we generically denote by $\langle-,-\rangle$ the canonical pairing between a vector space and its dual, then
$$ \langle H_k (\alpha_k), \beta^{n-k}\rangle =  \Xi_k(\alpha_k,\beta^{n-k}). $$
If $\underline{e}:=(e_j)$ is a basis of $V$ and $e^j)$ is  the $D$-dual basis of $V^*$, $e^j= D(e_j)$. We set
$$ \omega_e:=\omega( e_1\wedge \cdots \wedge e_n)\in\bR\setminus \{0\}. $$
Then 
$$ H_k (e^1\wedge \cdots \wedge e^k)=  Ce^{k+1}\wedge \cdots \wedge e^n, $$
where the constant $C$ is found from the equality
$$ C \det(\, \langle e_i,e^j\rangle )_{k<i,j\leq n}=\omega_e. $$
Now observe that  $$\langle e_i, e^j\rangle = B(e_i,e_j),$$ where $B$ is the bilinear form (1) associated to $D$.<|endoftext|>
TITLE: Pushforward of line bundle under "toric isogeny"
QUESTION [8 upvotes]: Let $(X,T)$ be a smooth complex toric variety of dimension $d$ with torus $T$ and toric boundary $D=X\setminus T$. Let $\phi : X\to X$ be a finite endomorphism of $X$ such that the restriction
$$\phi|_{T}:T\to T$$
is the power map $t\mapsto t^{\ell}$.
Let $\mathscr{L}$ be a line bundle on $X$. Then its pushforward $\phi_*\mathscr{L}$ will still be locally free, and will have rank $\delta=\ell^d$.
Is it possible to describe such pushforward? For example, is it reasonable to expect something like $\phi_*\mathscr{L}\simeq \mathscr{L}^{\oplus\delta}\otimes\mathcal{O}(D)$?
What about the case $X=\mathbb{P}^d$, $\mathscr{L}=\mathcal{O}(k)$  and $\phi:[x_0:\ldots:x_d]\mapsto[x_0^\ell:\ldots:x_d^\ell]$ ?

REPLY [11 votes]: It has been proved by Thomsen [Thomsen J. F.. “Frobenius direct images of line bundles on toric varieties” Journal of Algebra 226, no. 2 (2000)] that such a push-forward is a direct sum of line bundles. To be precise, his theorem applies to the Frobenius map on a smooth toric variety in characteristic $p$, but the same result holds for the "toric Frobenius" map you describe. For an explicit description of $\phi_*\mathscr{L}$, see Section 2 of [Achinger P. "A Characterization of Toric Varieties in Characteristic p" Int Math Res Notices (2015)]. See also Bondal's article, p. 284 in Oberwolfach Reports from 2006: https://www.mfo.de/document/0605/OWR_2006_05.pdf .
In case $X=\mathbb{P}^d$, the description of $\phi_* (\mathscr{O}(n))$ is easy to obtain without consulting the above sources. First, note that $\phi^* \mathscr{L} = \mathscr{L}^q$ for any line bundle $\mathscr{L}$. Second, recall the Horrocks' splitting criterion: a coherent sheaf $\mathscr{F}$ on $\mathbb{P}^d$ is a direct sum of line bundles if and only if $H^i(\mathbb{P}^d, \mathscr{F}(k)) = 0$ for all $0<i<d$ and all $k\in\mathbb{Z}$. Apply this to $\mathscr{F}= \phi_* (\mathscr{O}(n))$ and use the projection formula $H^i(\mathbb{P}^d, \phi_* (\mathscr{O}(n))(k)) = H^i(\mathbb{P}^d, \phi_*(\mathscr{O}(n+qk)) = H^i(\mathbb{P}^d, \mathscr{O}(n+qk)) = 0$ for $0<i<d$ and all $k$. So we know that $\phi_* (\mathscr{O}(n)) = \bigoplus_k \mathscr{O}(k)^{a_{n,k}}$ for some integers $a_{n, k}$ which we will now compute. Apply $h^0(-) = \dim H^0(\mathbb{P}^d, -)$ to the above isomorphism to obtain:
$$ \binom{n+d}{d} = \sum_{k \in \mathbb{Z}} a_{n,k} \binom{k+d}{d}. $$
Now note $a_{n, k} = a_{n-qk, 0}$ (again by the projection formula). Let $a_m = a_{m, 0}$, then the above equality says $\binom{n+d}{d} = \sum_{k \in \mathbb{Z}} a_{n-qk} \binom{k+d}{d}$. Now if $S(x) = \sum_{m \in \mathbb{Z}} \binom{m+d}{d}x^m = (1-x)^{-d-1}$, $M(x) = \sum_{m \in \mathbb{Z}} a_m x^m$, then the previous equality yields $S(x) = S(x^q)M(x)$, so $M(x) = (1+x+\ldots+x^{q-1})^{d+1}$, thus $a_{n,k}$ is the coefficient of $x^{n-qk}$ in the polynomial $(1+x+\ldots+x^{q-1})^{d+1}$.
The case of a general toric variety is similar: instead of using the Horrocks criterion, we could have noticed that $\phi$ is the quotient morphism by the action of the $q$-torsion $T[q]$ of the torus $T$, which is a diagonalizable group, and that every line bundle on a toric variety has a $T$-equivariant structure, hence also a $T[q]$-equivariant structure. Thus $\phi_* \mathcal{L}$ is the direct sum of its $T[q]$-eigensheaves, which are easily seen to be line bundles.
For example, for $d=1$ and $n=0$ one gets $\phi_* \mathscr{O} = \mathscr{O}\oplus\mathscr{O}(-1)^{q-1}$.<|endoftext|>
TITLE: Hecke characters and Conductors
QUESTION [7 upvotes]: Motivation: Let $\ell$ be an odd prime. There is a conductor-preserving correspondence between primitive Dirichlet characters of order $\ell$
and cyclic, degree $\ell$ number fields $K/\mathbb{Q}$.
The proof of this correspondence can be found in Chapter 3 of Washington's ``Introduction to Cyclotomic Fields".
Question: Let $k$ be a CM field. Is there a conductor-preserving correspondence between primitive Hecke characters of order $\ell$ and cyclic, degree $\ell$ extensions of $k$?
I have tried to construct a proof of this correspondence for Hecke characters using global class field theory and the theory of complex multiplication, however, I am not totally confident with my answer.
Any answers/references on the topic would be greatly appreciated.
This question is a duplicate from the one I posted on the Stackexchange, which can be found here.

REPLY [6 votes]: Let $k$ be a number field. By class field theory, any Hecke character $\chi$ of $k$ of order $\ell$ determines a cyclic extension $k'/k$ of degree $\ell$. Moreover, the set of Hecke characters determining this cyclic extension $k'/k$ equals $\{\chi,\chi^2,\dots,\chi^{\ell-1}\}$. These $\ell-1$ Hecke characters have the same conductor $\mathfrak{q}$, and the discriminant of $k'$ over $k$ equals their product $\mathfrak{q}^{\ell-1}$. In fact the product of the corresponding $L$-functions $L(s,\chi),\dots,L(s,\chi^{\ell-1})$ equals $\zeta_{k'}(s)/\zeta_k(s)$.<|endoftext|>
TITLE: Difference stencils approximating Laplacian
QUESTION [6 upvotes]: Let $\Delta$ be the Laplace operator on the interval $[0,1]\subset \mathbb{R}$. 
Divide $[0,1]$ into small intervals of size $h$ to get an equidistant grid. One can approminate $-\Delta$ on this grid by 
$$
\frac{u_{i-1} - 2u_i + u_{i-1}}{h^2}
$$
or 
$$
\frac{- u_{i+2} + 16 u_{i+1} - 30u_i + 16 u_{i-1} - u_{i-2}}{12h^2}
$$
where the weights -30, 16, -1 are calculated from Taylor approximations with radius $h$ and $2h$ around a point, and 
optimized in order to get a higher accuracy, in the 1d case $O(h^4)$.
But we could also take wider stencils, and they might not be symmetric. 
My question:Which are the (minimal) conditions on a finite difference stencil  $\Delta^h$  such that in the limit as $h \to 0$, for sufficiently regular initial condition, a solution $u^h$ to the ODE system (discrete in space, continuous in time heat equation)
$$
\partial_t u^h(t) = \Delta^h u^h(t) \qquad t \in [0,T]
$$
converges to a solution $u$ to the heat equation on $[0,1] \times [0,T]$
My thoughts:

$\Delta^h$ must be diagonally dominant and negative semidefinite
$\Delta^h$ must satisfy a discrete coercivity condition or a strong monotonicity condition

I am grateful for any hints/comments/literature advice.

REPLY [5 votes]: You have assumed an equidistant grid in one dimension, but the answer below can be formulated (and is true) for general grids in any number of dimensions.  You also haven't specified the boundary conditions, but as long as the boundary conditions lead to a well-posed problem, the result below applies (assuming a stable discretization of the boundary conditions).
The necessary and sufficient condition is that the semi-discretization $\Delta^h$ be consistent.  In other words, given a twice-differentiable function $u(x)$, let $u^h$ denote the vector of values of $u(x)$ evaluated at the points $x_i$.  Then $\Delta^h$ is said to be consistent if
$$\lim_{h\to 0} \left((\Delta^h u^h)_i - \Delta u(x_i)\right) = 0.$$
Any consistent $\Delta^h$ will give an ODE system whose solution converges to that of the heat equation.
In fact, we can make an even stronger statement.  If $\Delta^h$ is consistent and if the resulting ODE system is discretized in time in a consistent, stable way, then the fully-discrete solution will converge to that of the heat equation provided the CFL condition is satisfied.  For details, see the original paper of Courant, Friedrichs, and Lewy (which doesn't actually cover parabolic equations, but the tools are all there) and the paper of Lax and Richtmyer.  For a more modern treatment, I recommend chapter 9 of LeVeque's book.
I'm not aware of a reference dealing directly with convergence of the semi-discrete system.  But the result on convergence of the fully-discrete system implies the convergence of the semi-discrete system.
Note that the consistency condition can be written as an algebraic condition on the finite difference coefficients by using Taylor series.<|endoftext|>
TITLE: Is the moduli space of graphs simply connected?
QUESTION [16 upvotes]: The moduli space of graphs $MG_n$ is the quotient of Culler-Vogtmann's outer space $X_n$ by the action of $\mathrm{Out}(F_n)$. It can be thought of as the space of metric graphs homotopy equivalent to a wedge of $n$ circles with all vertices of valence $\geq 3$. A metric graph can be thought of as simply a graph with an assignment of positive edge lengths to all edges, and when an edge length goes to $0$, the edge itself also contracts, yielding a new graph type in $MG_n$. One also often assumes that the sum of edge lengths is $1$, since this condition yields a homotopy equivalent space.
One of the useful properties of $MG_n$ is that it is a rational $K(\pi,1)$ for $\mathrm{Out}(F_n)$: $H_*(MG_n;\mathbb Q)\cong H_*(\mathrm{Out}(F_n);\mathbb{Q})$. If $MG_n$ were an actual $K(\pi,1)$, the fundamental group would be $\mathrm{Out}(F_n)$ and that would be that, but it seems in practice that $MG_n$ is fairly highly connected, and in particular simply connected. $MG_2$ is certainly contractible. I'm not sure about $MG_3$; I would guess it's contractible. I suspect that $MG_4\simeq S^4$.
My question is whether $MG_n$ is indeed simply connected. I am unaware of any particular consequences this would have, but it also seems we ought to know such a basic property about such a fundamental space.

REPLY [18 votes]: Yes, it is.  If $G$ is a discrete group acting on a simply-connected simplicial complex $X$, then a theorem of M. A. Armstrong says that there is a short exact sequence
$$1 \longrightarrow H \longrightarrow G \longrightarrow \pi_1(X/G) \longrightarrow 1,$$
where $H$ is the subgroup of $G$ generated by elements that fix a point in $X$.  The original reference is
M. A. Armstrong, On the fundamental group of an orbit space, Proc. Cambridge Philos. Soc. 61 (1965),
639–646.
See also my paper
A. Putman,
Obtaining presentations from group actions without making choices,
Algebr. Geom. Topol. 11 (2011), 1737-1766.
which proves a more specific result and contains a streamlined version of Armstrong's proof in Section 3.1 (which is really just a meditation on the usual facts about covering spaces, which the above restricts to if the action is free and thus $H = 1$).
In any case, if we apply this with $G = \text{Out}(F_n)$ and $X = X_n$ (Outer Space), then we see that $H = G$ since $G$ is generated by torsion elements, which fix points of Outer space.  The quotient is thus simply-connected, as desired.
A similar argument shows that the (ordinary, non-orbifold) fundamental group of the moduli space of curves is trivial.  This was originally proved (by this argument) in
Maclachlan, Colin, Modulus space is simply-connected, Proc. Amer. Math. Soc. 29 1971 85–86.<|endoftext|>
TITLE: Automorphism group of a free product
QUESTION [6 upvotes]: Suppose that $G$ and $H$ are groups (not isomorphic) and $G\ast H$ the free  product. Let $Aut(G)$, $Aut(H)$ be the automorphism groups of $G$ and $H$. What is $Aut(G\ast H)$ ?

REPLY [10 votes]: This is not an answer, but it's too long for a comment and makes an important point that I hope is useful.
It's not enough to think of your group just as some free product. To understand its automorphism group you need to think of it as a free product in a canonical way, and for this you need the Grushko decomposition.  Recall:

Grushko's decomposition theorem: If $G$ is finitely generated then
$G=G_1*\ldots* G_n*F_k$
where each $G_i$ is freely indecomposable (ie isn't $\mathbb{Z}$ and doesn't split as a free product) and $F_k$ is free of rank $k$. Furthermore, the $G_i$ are determined up to conjugacy (and reordering), and $k$ is determined.

The key thing to notice here is that the $F_k$ factor is not determined up to conjugacy, and therefore if $k>0$ then there are complicated automorphisms that arise essentially from the different possible ways to choose $F_k$. However, the picture is relatively simple if $k=0$.
A modern reference on the (outer) automorphism groups of free products is this nice paper of Guirardel and Levitt.<|endoftext|>
TITLE: Under what conditions is the induced map of etale fundamental groups surjective?
QUESTION [9 upvotes]: Let $f:X \to Y$ be a morphism of schemes. I am interested in sufficient conditions on $f$ which would ensure that the induced map $\pi_1^{et}(X) \to \pi_1^{et}(Y)$ of etale fundamental groups is surjective. According to this question 
https://math.stackexchange.com/questions/491305/etale-fundamental-group 
SGA 1 X Corollary 1.4 says that it is true when $f$ is proper, separable, $Y$ is connected, and $f_*(\mathcal O_X)=\mathcal O_Y$ (in particular, $f$ has geometrically connected fibers). Specifically I am interested if the condition that $f$ be a (not necessarily proper) Zariski locally trivial fibration with connected fibers (where $X$ and $Y$ are smooth algebraic varieties) is sufficient for the map $\pi_1^{et}(X) \to \pi_1^{et}(Y)$ induced by $f$ to be surjective?

REPLY [11 votes]: There is a very general criterion for a map on $\pi_1$ to be surjective. Recall that for $X$ connected, the category of finite étale covers of $X$ is equivalent to the category $\pi_1(X)\text{ -}\operatorname{Set}_f$ of finite sets with a continuous $\pi_1(X)$-action. Under this correspondence, the $Y \to X$ finite étale with $Y$ connected correspond to the connected $\pi_1(X)$-sets $S$ (i.e. $\pi_1(X)$ acts transitively on $S$).
Lemma. Assume $X$, $Y$ connected, and $f \colon X \to Y$ a morphism. Then the induced morphism $\pi_1(f) \colon \pi_1(X) \to \pi_1(Y)$ is surjective if and only if for every $Z \to Y$ finite étale with $Z$ connected, the pullback $Z_X \to X$ is connected.
Proof. If $\pi_1(f)$ is surjective, then clearly any connected $\pi_1(Y)$-set is connected as $\pi_1(X)$-set. Conversely, if $\pi_1(f)$ is not surjective, then some $\gamma \in \pi_1(Y)$ is not in the image. Since fundamental groups are profinite, the image of $\pi_1(f)$ is closed, so the image of $\pi_1(f)$ misses some open neighbourhood of $\gamma$. Thus, there exists an open subgroup $U \subseteq \pi_1(Y)$ such that
$$\gamma U \cap \operatorname{im} \pi_1(f) = \varnothing.$$
Then the finite $\pi_1(Y)$-set $S = \pi_1(Y)/U$ is not connected as $\pi_1(X)$-set. But it is clearly connected as $\pi_1(Y)$-set. $\square$
To apply this to the specific geometric setting you are interested in, just note that if $f \colon X \to Y$ has connected geometric fibres, then the same holds for the base change to any finite étale covering $Z \to Y$. It is then clear that if $Z$ is connected, so is $Z \times_Y X$.
Remark. There are more equivalent criteria for surjectivity; see for example Tag 0B6N. The one I gave above is amongst the ones listed, but this was not the case at the time of writing; hence my writing out the proof. My proof above was originally part of the proof of Tag 0BTX.<|endoftext|>
TITLE: classifying space of orthogonal groups
QUESTION [6 upvotes]: Let $O(n)$ be the $n$-th orthogonal group and $O$ be the direct limit of $O(n)$ with respect to $n$. Let $BO(n)$ and $BO$ be the classifying spaces.
Question: 
Why $BO$ is an $H$-space? My supervisor said "$BO=\Omega^\infty\mathbb{E}$ where $\mathbb{E}$ is a spectrum." What does this mean?

REPLY [10 votes]: $BO$ can be defined as the colimit over $(k,n)$ of Grassmanians $G_k(\Bbb R^n)$ of $k$-dimensional linear subspaces of $\Bbb R^n$ (the limit over $n$ is defined by standard inclusions $\Bbb R^n \subset \Bbb R^{n+1}$, and the limit over $k$ is defined by the operation $X \mapsto X\oplus \Bbb R$ which takes a $k$-plane in $\Bbb R^n$ to the $(k+1)$-plane $X\oplus \Bbb R$ inside $\Bbb R^n \oplus \Bbb R = \Bbb R^{n+1}$. 
The direct sum operation defines pairings
$$
G_k(\Bbb R^n) \times G_{\ell}(\Bbb R^m) \to G_{k+\ell}(\Bbb R^{n+m})
$$
that are compatible with respect to taking colimits.
Taking colimits induces a map $BO \times BO \to BO$ defining the $H$-space structure.<|endoftext|>
TITLE: Conjugate transpose and discreteness, for Kleinian groups
QUESTION [5 upvotes]: Let $G=\langle g_1,\dots g_n\rangle<\mathrm{PSL}_2(\mathbb{C})$ be discrete,
i.e. a finitely generated Kleinian group.
Let $H=\langle g^{\dagger}g\mid g\in G\rangle$,
(the group generated by the $g^{\dagger}g$)
where $g^{\dagger}$ is the conjugate transpose.
Under what conditions is $H$ discrete?
Edit: In light of comments from @Misha, below I am removing my claims about what I can prove and adding some motivation for the question.
In the spinor representation of the restricted Lorentz group, $\mathrm{PSL}_2(\mathbb{C})$
is faithfully represented as the group of orientation-preserving isometries of the hyperboloid model $\mathbb{I}$ for hyperbolic 3-space.
In this construction, we identify the points on $\mathbb{I}$ with the Hermitian matrices of norm 1, up to identification of $\pm1$.
Then for $(g,p)\in\mathrm{PSL}_2(\mathbb{C})\times\mathbb{I}$
the action can be defined by $g(p)=g^{\dagger}pg$.
Skipping some details, the point $p=(1,0,0,0)\in\mathbb{I}\subset\mathbb{R}^4$
is identified with the identity matrix.
So in this construction, the set $\{g^{\dagger}g\mid g\in G\}$
is identified with the orbit under $G$
of $p$.
It is for this reason that I said this set is certainly discrete.
I am working on a more elaborate duality between points in hyperbolic space and isometries of that space using quaternion algebras. I am holding back some info because this is part of my thesis which is not published, and my school would not be too happy if I posted certain ideas online at this point. So suffice it to say I have interesting applications using the group $H$ generated by that set.

REPLY [3 votes]: Here is a thought about your question. Suppose that $G$ contains a unipotent element $u$. Then we can conjugate $G$ to $G'=hGh^{-1}$ such that $u$ conjugates to $u'=huh^{-1}$ which is close to $1\in SL(2,C)$. Note that if $u_1, u_2$ are two such commuting unipotent elements then $v_i=(u_i')^{\dagger} u_i'$, $i=1, 2$, will not (generically) commute. Therefore, by Jorgensen's inequality (or Zassenhaus Lemma if you prefer), the subgroup generated by $v_1, v_2$ will be nondiscrete. I suspect, one can play a similar game in greater generality and prove that for any infinite discrete subgroup, its conjugate $hG h^{-1}$ leads to  nondiscrete subgroup. This definitely applies to "generic" geometrically infinite finitely generated Kleinian groups. But for convex-cocompact groups cannot use Jorgensen inequality argument and one needs a different approach.<|endoftext|>
TITLE: Minimal Hausdorff topologies compatible with a bunch of functions
QUESTION [6 upvotes]: Let $X$ be an infinite set, let ${\cal F}$ be a set of functions $f: X\to X$. We say that a topology $\tau$ is compatible with ${\cal F}$ if every $f\in {\cal F}$ is a continuous function $f:(X, \tau)\to (X,\tau)$. We denote the collection of $T_2$-topologies compatible with ${\cal F}$ by $T_2({\cal F})$.
Note that $\big(T_2({\cal F}), \subseteq\big)$ is a poset; it is always non-empty and the discrete topology ${\cal P}(X)$ is its largest element.
Question: Is there an infinite set $X$ and a set ${\cal F}$  of functions $f:X\to X$ such that $\big(T_2({\cal F}\big), \subseteq)$ does not contain minimal elements?

REPLY [3 votes]: The answer to this question is affirmative:
Theorem. There exists a countable set $X$ and an uncountable family $\mathcal F$ of self-functions of $X$ such that the poset  $T_2(\mathcal F)$ has no minimal elements. 
Here $T_2(\mathcal F)$ is the poset of all Hausdorff topologies on $X$ making all functions $f\in\mathcal F$ continuous.  
Proof. To construct the space $X$ and the family $\mathcal F$, take any Hausdorff $(\omega_1,\omega_1)$-gap on $\omega$, which is a pair $\big((A_\alpha)_{\alpha\in\omega_1},(B_\alpha)_{\alpha\in\omega_1}\big)$ of families of infinite subsets of $\omega$ satisfying the following two conditions:
(H1) for any $\alpha<\beta<\omega_1$ we have $A_\alpha\subset^* A_\beta\subset^* B_\beta\subset^* B_\alpha$;
(H2) for any set $C\subset\omega$ one of the sets $\{\alpha\in \omega_1:A_\alpha\subset^* C\}$ or $\{\alpha\in\omega_1:C\subset^* B_\alpha\}$ is at most countable.
Here the notation $A\subset^* B$ means that the complement $A\setminus B$ is finite.
It is well-known that Hausdorff $(\omega_1,\omega_1)$-gaps do exist in ZFC.
Let $X=\{-\infty,+\infty\}\cup\omega$ and for every $\alpha\in\omega_1$ consider the functions $f_\alpha,g_\alpha:X\to X$ defined by 
$$f_\alpha(x)=\begin{cases}
x&\mbox{if $x\in \{-\infty\}\cup B_\alpha$};\\
+\infty&\mbox{otherwise};
\end{cases}
$$ 
and
$$g_\alpha(x)=\begin{cases}
x&\mbox{if $x\in \{+\infty\}\cup (\omega\setminus A_\alpha)$};\\
-\infty&\mbox{otherwise}.
\end{cases}
$$
For every $a,b\in\omega$ let $p_{a,b}:X\to X$ be the function defined by
$$p_{a,b}(x)=\begin{cases}
b&\mbox{ if $x=a$},\\
a&\mbox{ if $x=b$},\\
x&\mbox{otherwise}.
\end{cases}
$$
We claim that the family of functions $\mathcal F=\{f_\alpha\}\cup\{g_\alpha\}_{\alpha\in\omega_1}\cup\{p_{a,b}\}_{a,b\in\omega}$ has the required property: the poset $T_2(\mathcal F)$ has no minimal elements.
Indeed, take any Hausdorff topology $\sigma$ on $X$ in which every map $f\in\mathcal F$ is continuous. The continuity of the maps $p_{a,b}$, $a,b\in\omega$, implies that each point of the set $\omega$ is isolated in the topological space $(X,\sigma)$.
We claim that for every $\alpha\in\omega_1$ the set $W_\alpha^-:=B_\alpha\cup\{-\infty\}$ is a neighborhood of $-\infty$ and $W_\alpha^+:=(X\setminus A_\alpha)\cup\{+\infty\}$ is a neighborhood of $+\infty$ in the topology $\sigma$.
For this choose any two disjoint neighborhoods $U_-,U_+\in\sigma$ of $-\infty$ and $+\infty$, respectively. By the continuity of the maps $f_\alpha$ and $g_\alpha$, there are neighborhoods $V_-,V_+\in\sigma$ of $-\infty$ and $+\infty$ such that $f_\alpha(V_-)\subset U_-$ and $g_\alpha(V_+)\subset U_+$. Then $$V_-\subset f_\alpha^{-1}(U_-)\subset f_\alpha^{-1}(X\setminus\{+\infty\})\subset \{-\infty\}\cup B_\alpha$$
and $$V_+\subset g_\alpha^{-1}(U_+)\subset g_\alpha^{-1}(X\setminus\{-\infty\})\subset\{+\infty\}\cup(\omega\setminus A_\alpha).$$
Therefore, the topology $\sigma$ contains the topology $\tau$ defined in the answer to this MO problem. Repeating the argument from this answer it can be shown that $\sigma$ is not a minimal element of the poset $T_2(\mathcal F)$.
However, for the interested reader let us write more details.
By the condition (H2), one of the sets $A=\{\alpha\in\omega_1:A_\alpha\subset^* U_-\}$ or $B=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$ is countable.
First we assume that the set $A$ is countable. In this case we can find a countable ordinal $\alpha$ such that $A_\alpha\not\subset^*U_-$.
Consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:
$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\cup A_\alpha\subset^* W$;
$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\subset^* W$.
It is clear that $\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ (because $U_-\in\sigma\setminus\sigma'$).
The topology $\sigma'$ is Hausdorff since $U_-\cup A_\alpha$ and $U_+\setminus A_\alpha=U_+\cap W^+_\alpha$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively.
To see that $\sigma'\in T_2(\mathcal F)$, observe that for any $a,b\in \omega$ the permutation $p_{a,b}$ is continuous as the points $a,b$ are isolated in the topology $\sigma'$. Next, for every ordinal $\beta\in\omega_1$ the inclusion $A_\alpha\subset^* B_\beta$ implies that the map $f_\beta$ is continuous at $-\infty$. The continuity of the map $f_\beta$ at $+\infty$ in the topology $\sigma$ implies the continuity of this map at $+\infty$ in the topology $\sigma'$.
The continuity of the map $g_\beta$ at $-\infty$ in the topology $\sigma'$ follows from the inclusion $g_\beta(x)\in\{-\infty,x\}$ holding for all $x\in X$.
The continuity of $g_\beta$ at $+\infty$ in the topology $\sigma'$ follows from the continuity of $g_\beta$ at $+\infty$ in the topology $\sigma$ and the definition of the topology $\sigma'$.
Now consider the case of countable set $B:=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$. In this case we can find a countable ordinal $\alpha$ such that $U_-\setminus\{-\infty\}\not\subset^* B_\alpha$ and hence $\omega\setminus B_\alpha\not\subset^* U_+$.
In this case we can consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:
$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\subset^* W$;
$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\cup(\omega\setminus B_\alpha)\subset^* W$.
It is clear that $\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ (because $U_+\in\sigma\setminus\sigma'$).
The topology $\sigma'$ is Hausdorff since $U_-\cap B_\alpha$ and $U_+\cup(\omega\setminus B_\alpha)$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively. By analogy with the case of countable set $A$, we can show that each map $f\in\mathcal F$ remains continuous in the topology $\sigma'$.
In both cases we have constructed a strictly weaker topology $\sigma'\subset \sigma$  in $T_2(\mathcal F)$ witnessing that the topology $\sigma$ is not minimal in the poset $T_2(\mathcal F)$.<|endoftext|>
TITLE: Coefficients of Ehrhart polynomials, in the binomial-coefficient basis
QUESTION [5 upvotes]: Let $P$ be the convex hull of a finite set of points in $\mathbb Z^d$, and $p(n) = \#\{nP \cap \mathbb Z^d\}$ be its Ehrhart polynomial, which is also the Hilbert polynomial of the corresponding projective toric variety.
If we write it as $\sum_{k=0}^d c_k {n \choose k}$, then since $p$ is $\mathbb Z$-valued, the $c_k$ are also in $\mathbb Z$; one can compute $c_k$ as $(\Delta^k p)(0)$ where $(\Delta q)(m) = q(m)-q(m-1)$.

Is there a succinct description of the coefficients $c_k$ in terms of the original $P$?

REPLY [8 votes]: One possible answer: if you write the Ehrhart polynomial as $\sum_{k=0}^{d} c_k\binom{n-1}{k}$, then the coefficients $c_k$ are called the $f^{\ast}$-vector by Felix Breuer in the paper Ehrhart f*-coefficients of polytopal complexes are non-negative integers. He gives a combinatorial interpretation in terms of counting 'atomic lattice points' by 'height'. The relevant material is in Section 3. Of course, if you want to use $\binom{n}{k}$, then I do not know if there is any obvious combinatorial interpretation, aside from the fact that the resulting coefficients can be easily computed from the $f^{\ast}$-vector using the Pascal relation for binomial coefficients.<|endoftext|>
TITLE: What is the exact statement about uniform boundedness of rational points on curves of genus greater than one? Singular points can be unbounded
QUESTION [7 upvotes]: According to several sources, it is conjectured (or at least believed)
that the rational points of curves over the rationals of genus $g > 1$
are uniformly bounded by $g$. E.g. here p. 1.
Assuming the curve is irreducible, the singular points (which are also
rational) can easily be made unbounded for fixed $g > 1$.
For natural $n$, define $j(n)=\prod_{i=1}^n(x-i)$.
Consider the curve $C_n : j(n)^2(x^5+13)=y^2$.
It birationally equivalent to $x'^5+13=y'^2$ which is genus $2$.
$C_n$ has the rational (singular) points $(1,0),(2,0),\ldots(n,0)$
which are unbounded, since $n$ is unbounded.
The same applies for $2$ replaced by integer $d>1$.
For $d > 2$ the curve need not be hyperelliptic.

Q. What is the exact statement about uniform boundedness?

According to both sage and magma, $C_n$ is irreducible for small
values of $n$.

REPLY [6 votes]: The precise statement of the conjecture is:

Uniformity Conjecture. Let $K$ be a number field and $g\geq2$ an integer. There exists a number $B(K,g)$ such that for any smooth curve
  $X$ of genus $g$ defined over $K$
$$|X(K)|\leq B(K,g)$$

and the original reference for it is:

Lucia Caporaso, Joe Harris, Barry Mazur, Uniformity of rational points (1997)

The conjecture refers indeed to geometric genus, following Faltings early results on finiteness of rational points.
Also note (here of example) that this can be proved conditionally on Lang's conjecture.<|endoftext|>
TITLE: Ergodic theory: from Dynamics to Gibbs measure
QUESTION [6 upvotes]: I'm trying to understand the ergodic theory approach to statistical mechanics, namely how ergodic measure preserving dynamics lead to the Gibbs measure.
I have a compact space $X$, a probability measure $\mu$ on $X$ and a one parameter family of transformations $T_t:X\rightarrow X$ ergodic on $\mu$. As I understand, ergodic theory states that $\mu$ is unique. Which result in ergodic theory shows that $\mu$ is the Gibbs measure? 
More specifically, if my dynamics conserves average energy $\langle E\rangle$, how, in the framework of ergodic theory, do I show that $\mu$ looks something like this:
$$\mu(x) \propto \exp[-\beta E(x)],$$
where $\beta$ is the inverse temperature?
From what I've seen so far in physics and maths textbooks, some sort of entropy maximisation hypothesis is invoked to construct $\mu$. If this is the only way to go about it, is there a justification for the principle of maximum entropy within ergodic theory?

REPLY [6 votes]: That's an excellent but highly unresolved question.  The problem is that the physicists tend to be not so interested in mathematical foundations once a theory (statistical mechanics in this case) is successful and there is some plausible heuristic justification for it, whereas ergodic theorists quite often have little clue about the physical relevance and have an altogether different reason for being interested in Gibbs measures (starting with a "non-physical" analogy with statistical mechanics made by Sinai, Ruelle, ...).  Meanwhile mathematical physicists have been busy with already highly challenging mathematical problems that arise if we take the physical meaning of Gibbs measures for granted.
I will tell you my take (à la Boltzmann) on how to approach the problem of the dynamical justification of Gibbs measures as "equilibrium states".  Others will correct me if they don't agree.
Statistical mechanics is all about the relationship between the microscopic and macroscopic properties of systems.  Let us say we have a set $X$ containing all the possible instantaneous microscopic states of a physical system, and a transformation $T:X\to X$ describing the time dynamics of the system, which we take to be discrete for simplicity.  We also have a collection $\mathscr{O}$ of macroscopic observables, which are simply functions $f:X\to\mathbb{R}$.  Knowing the macroscopic state of the system amounts to knowing the value of all the macroscopic observables, that is, a map $\pi:\mathscr{O}\to\mathbb{R}$.  Each macroscopic state $\pi$ corresponds to a set $X_\pi$ of microscopic states that realize $\pi$, that is, $X_\pi=\lbrace x\in X: \text{$f(x)=\pi(f)$ for each $f\in\mathscr{O}$}\rbrace$.
A typical setting in statistical mechanics is a lattice model, in which the microscopic states of a physical system are represented by symbolic configurations on the lattice $\mathbb{Z}^d$, that is, by functions $x:\mathbb{Z}^d\to\Sigma$ for some finite set of symbols $\Sigma$.  A natural candidate for the time dynamics $T$ is a cellular automaton, that is, a continuous map on $X=\Sigma^{\mathbb{Z}^d}$ that has all the translations as its symmetries (i.e., it commutes with the shifts).  The choice of the macroscopic observables in this setting is somewhat debatable, but let's say the frequency of a finite word in the configuration is a macroscopic observable, and everything that can be written as a linear combination of such frequencies is also a macroscopic observable.  Then, the macroscopic states can be represented by shift-invariant probability measures on $X$ (or better, shift-ergodic probability measures).
Suppose now that the system has a collection of conserved quantities like energy, that is, macroscopic observables $e\in\mathscr{O}$ such that $e(Tx)=e(x)$ "for all" $x\in X$.  What can we say about the macroscopic state of the system?  The second law of thermodynamics suggests that if the system is "in equilibrium", then the macroscopic state of the system has to maximize the "entropy" subject to the constraints imposed by the conservation laws, and if the system is not "in equilibrium", it gradually approaches the "equilibrium", provided the system is sufficiently "chaotic".
The physicists' explanation of the second law of thermodynamics in a finite-state system is that if the dynamics goes through all the configuration space in one cycle (it is "ergodic"), then in equilibrium (long long time after it has started its evolution) it is "most likely" to be found in a configuration whose macroscopic state encompasses the largest portion of the state space (i.e., $|X_\pi|$ is the largest), and if it is not in equilibrium, it is "most likely" to evolve in time towards macroscopic states with larger volume.  In this case, the entropy of a macroscopic state $\pi$ is $\log|X_\pi|$.
Translating this heuristic to infinite systems (say in our lattice setting with cellular automaton dynamics) is not obvious.  One is inclined to replace $|X_\pi|$ with the configuration space volume of $X_\pi$ according to some notion of volume, say the uniform Bernoulli measure in $\Sigma^{\mathbb{Z}^d}$ in case of a lattice model.  Although this is how the mathematical notion of ergodicity came to being, it doesn't help with the problem of identifying macroscopic equilibrium states, because for all but a single macroscopic state $\pi$, the measure of $X_\pi$ with respect to the uniform Bernoulli measure is zero.
The correct way is to measure the the size of $X_\pi$ is by the amount of information per site required to describe a microscopic configuration in $X_\pi$, that is, by the entropy per site of $\pi$.  Note that in the language of dynamical systems, this is the Kolmogorov-Sinai entropy of $\pi$ under the dynamics of the shift action (i.e., the space dynamics), and a priori has nothing to do with the time dynamics.  Equilibrium statistical mechanics (e.g., the book of Robert Israel) tells us that the macroscopic states maximizing entropy under the constraints of the conservation laws are Gibbs measures.
A reasonable ergodic-theoretic justification of considering Gibbs measures as equilibrium states involves (1) showing that they are invariant under the time dynamics (this is the easy part) and (2) showing that starting from other macroscopic states the system evolves towards states with larger entropy at least under reasonable assumptions on the starting state, and provided the dynamics has a sufficiently strong chaotic behaviour.
Like I said at the beginning, this is largely open, but there is a beautiful example (and its extensions) for which some mathematical results have been found.  The XOR cellular automaton is defined as $T:\lbrace 0,1\rbrace^{\mathbb{Z}}\to\lbrace 0,1\rbrace^{\mathbb{Z}}$ with $(TX)_i:=x_i+x_{i+1}\pmod{2}$.  This system has no conserved quantity, so the second law suggests maximum entropy measure (the uniform Bernnoulli measure) as the only macroscopic equilibrium state.  It is easy to see that the uniform Bernoulli measure is indeed invariant under $T$.  Miyamoto and Lind (and later others) have shown that starting from any macroscopic state that has sufficient mixing property under the space dynamics (e.g., any Bernoulli or Markov measure), the system gradually approaches under $T$ towards the uniform Bernoulli measure (in a suitable sense).<|endoftext|>
TITLE: Questions about the regularity of the "norm" associated to a convex set
QUESTION [5 upvotes]: Suppose $K\subset \mathbb{R} ^n$ is a closed convex set whose interior contains the origin. We can assign a gauge function to $K$ as $g_{K}(x):=\inf\{\lambda>0 \mid x\in\lambda K\}$. $g_K$ has all the properties of a norm on $\mathbb{R}^n$ except for $g_K(-x)=g_K(x)$, and it is a norm when $K=-K$. We can think of $K$ as the unit ball of the "norm" $g_K$. 
It is well known that when $\partial K$ is $C^2$ and its principal curvatures are all positive everywhere, then $g_K$ is $C^2$. It is easy to see that this condition is not necessary and we can allow the curvatures to be zero at least on a finite set, for example by looking at the $p$-norms for $p>2$.
Another notion here is that of the polar of $K$ which is defined as $K^\circ :=\{x\mid x\cdot y\le 1 \forall y\in K\}$. The "norm" $g_{K^\circ}$ is a "dual" to the "norm" $g_K$, much like $p,q$-norms when $1/p+1/q=1$. It is also well known that when $\partial K$ is $C^2$ and its principal curvatures are all positive everywhere, then $\partial K^\circ$ is $C^2$ and its principal curvatures are all positive everywhere too.
A good reference to all of these is the book by Rolf Schneider.
My questions are:
1) Are there any necessary and sufficient conditions on $\partial K$ that imply $g_K$ is $C^2$?
2) Is there any relation between the curvature of $\partial K$ and the curvature of $\partial K^\circ$ supposing both of them are $C^2$ except possibly at finitely many points, at least when $n=2$? 
3) Any reference that has more on the regularity of these "norms" than the above book, is also appreciated.
Edit: These "norms" appear naturally in studying gradient constraints in PDE and calculus of variations, and their regularity has implications about the regularity of the solutions of the aforementioned problems.

REPLY [3 votes]: Although the answers to these regularity questions are not addressed explicitly in the paper cited in the comments to the question, they all follow from the formulas derived in that paper.
1) The answer is yes. The boundary $\partial K$ is the radial graph of its gauge function over the unit sphere. A hypersurface is defined to be $C^2$ if it is locally the graph of a $C^2$ function, which in turn holds if and only if the gauge function is $C^2$.
2) If $\partial K$ has strictly positive Gauss curvature, then at each point $x \in \partial K$ there is a formula for the second fundamental form of $\partial K^*$ at the corresponding "dual" point (given by the Gauss map) in terms of the second fundamental form and gauge function of $\partial K$ at $x$. Taking the determinant gives an explicit formula for the corresponding Gauss curvatures.
An affine invariant version of this formula is given in the paper cited in the comments to the question. The Euclidean version of the formula can be derived with some effort, from the affine one.
Additional questions asked in comments:
(Aside: It's easier to work with the square of the gauge function.)
a) There is an explicit formula for the Hessian of the squared gauge function in terms of the Hessian of the squared polar gauge function. From this, it follows that $\partial K$ is $C^2$ and the Gauss curvature positive at a point $\iff$ the Hessian of the squared gauge function is positive definite and continuous $\iff$ the Hessian of the squared polar gauge function (i.e., the squared support function of $K$) is positive definite and continuous $\iff$ $\partial K^*$ is continuous and the Gauss curvature positive at the dual point on $\partial K^*$.
b) Since the support function of $K$ is the gauge function of $K^*$, the support function of $K$ at a dual point is $C^2$ if and only if $\partial K^*$ is $C^2$ at that dual point.
ADDED: One more comment regarding the use of gauge functions (i.e., convex functions homogeneous of degree 1) in PDE's and the calculus of variations. Often, the Euclidean norm plays no role in the setup of the problem (i.e., the definition of the constraints and objective functional), even though they are needed for defining the functional norms when proving estimates. In that situation it is easiest (at least for me) to work out affine invariant formulas, where the Euclidean norm and inner product are never used. When doing this, it is important to keep track of what lies in the vector space that contains the convex body and what lies in the dual vector space. All of this is discussed pretty carefully in the paper cited. I would add that when I was learning all of this, I found Rockafellar's book Convex Analysis to be extremely helpful.<|endoftext|>
TITLE: When a Riemannian manifold with boundary is an Alexandrov space?
QUESTION [7 upvotes]: Let $M$ be a smooth Riemannian manifold (without boundary). Let $X\subset M$ be a smooth compact submanifold with boundary, $\dim X=\dim M$. 
Under what conditions $X$, equipped with the induced intrinsic metric, is an Alexandrov space with curvature bounded below?
A simple sufficient condition is some convexity of $X$: any point of $X$ has a neighborhood such that any two points of $X$ in it are connected (in $M$) by a unique shortest geodesic, and this geodesic is contained in $X$. Is this condition also necessary?

REPLY [5 votes]: Yes, this condition also necessary.
Assume $X$ is an Alexandrov space.
At any point of $\partial_MX$ (the relative boundary of $X$ in $M$)
the space of directions is a half-sphere;
therefore $\partial_MX$ is also boundary of $X$ as it is defined for Alexandrov spaces.
If  curvature $\ge 0$ then the distance function $f$ to the boundary is concave. 
It follows that the hypersurface $\partial_MX$ can be approximated by level sets of convex functions and therefore locally convex.
If curvature $\ge -1$, we get that $h=\mathrm{sh}\circ f$ satisfies $h''\le h$ in the barier sence and the same conclusion holds.<|endoftext|>
TITLE: Minimal maximal subgroup of the symmetric group
QUESTION [15 upvotes]: The question is pretty much in the title: What is the maximal subgroup of $S_d$ of maximal index (so minimal size)? A slight variant (I am not sure if it leads to a different answer) is: what if we restrict to transitive subgroups? (Of course, in both cases, there might be massive ties, but at least the index is well-defined).
Clarification: I want both asymptotic results, and, if possible, a pointer to a table for small $d.$

REPLY [5 votes]: A table of the largest indices of maximal subgroups of ${\rm S}_d$ for small $d$
can be computed with GAP:
    d | largest index of a maximal subgroup of S_d 
------+----------------------------------------------------------------
    2 : 2
    3 : 3
    4 : 2^2
    5 : 2*5
    6 : 3*5
    7 : 2^3*3*5
    8 : 2^3*3*5
    9 : 2^3*3*5*7
   10 : 2^3*3^2*5*7
   11 : 2^7*3^4*5*7
   12 : 2^7*3^4*5*7
   13 : 2^8*3^4*5^2*7*11
   14 : 2^8*3^4*5^2*7*11
   15 : 2^3*5^2*7^2*11*13
   16 : 3*5^3*7^2*11*13
   17 : 2^11*3^6*5^3*7^2*11*13
   18 : 2^11*3^6*5^3*7^2*11*13
   19 : 2^15*3^6*5^3*7^2*11*13*17
   20 : 2^15*3^6*5^3*7^2*11*13*17
   21 : 2^14*3^8*5^4*7^2*11*13*17*19
   22 : 2^11*3^7*5^3*7^2*11*13*17*19
   23 : 2^18*3^9*5^4*7^3*11*13*17*19
   24 : 2^18*3^9*5^4*7^3*11*13*17*19
   25 : 2^17*3^9*5^3*7^3*11^2*13*17*19*23
   26 : 2^18*3^9*5^4*7^3*11^2*13*17*19*23
   27 : 2^18*3^7*5^6*7^3*11^2*13*17*19*23
   28 : 2^22*3^9*5^6*7^3*11^2*13*17*19*23
   29 : 2^23*3^13*5^6*7^3*11^2*13^2*17*19*23
   30 : 2^23*3^13*5^6*7^3*11^2*13^2*17*19*23
   31 : 2^25*3^13*5^6*7^4*11^2*13^2*17*19*23*29
   32 : 2^25*3^13*5^6*7^4*11^2*13^2*17*19*23*29
   33 : 2^12*5^5*7^3*11^2*13^2*17*19*23*29*31
   34 : 3^9*5^4*7^2*11^2*13*17*19*23*29*31
   35 : 2^7*3^6*7^4*11^3*13^2*17^2*19*23*29*31
   36 : 2^27*3^13*5^7*7^4*11^3*13^2*17^2*19*23*29*31
   37 : 2^32*3^15*5^8*7^5*11^3*13^2*17^2*19*23*29*31
   38 : 2^32*3^15*5^8*7^5*11^3*13^2*17^2*19*23*29*31
   39 : 2^12*5^6*7^4*11^2*13^2*17^2*19^2*23*29*31*37
   40 : 2^31*3^14*5^8*7^5*11^3*13^3*17^2*19^2*23*29*31*37
   41 : 2^35*3^18*5^8*7^5*11^3*13^3*17^2*19^2*23*29*31*37
   42 : 2^35*3^18*5^8*7^5*11^3*13^3*17^2*19^2*23*29*31*37
   43 : 2^38*3^18*5^9*7^5*11^3*13^3*17^2*19^2*23*29*31*37*41
   44 : 2^38*3^18*5^9*7^5*11^3*13^3*17^2*19^2*23*29*31*37*41
   45 : 2^34*3^17*5^9*7^6*11^4*13^3*17^2*19^2*23*29*31*37*41*43
   46 : 3^12*5^6*7^3*11^2*13^2*17*19*23*29*31*37*41*43
   47 : 2^41*3^21*5^10*7^6*11^4*13^3*17^2*19^2*23*29*31*37*41*43
   48 : 2^41*3^21*5^10*7^6*11^4*13^3*17^2*19^2*23*29*31*37*41*43
   49 : 2^41*3^20*5^10*7^5*11^4*13^3*17^2*19^2*23^2*29*31*37*41*43*47
   50 : 2^41*3^21*5^10*7^6*11^4*13^3*17^2*19^2*23^2*29*31*37*41*43*47
   51 : 2^15*5^9*7^6*11^3*13^2*17^2*19^2*23^2*29*31*37*41*43*47
   52 : 3^5*5^10*7^7*11^3*13^3*17^3*19^2*23^2*29*31*37*41*43*47
   53 : 2^47*3^23*5^12*7^8*11^4*13^3*17^3*19^2*23^2*29*31*37*41*43*47
   54 : 2^47*3^23*5^12*7^8*11^4*13^3*17^3*19^2*23^2*29*31*37*41*43*47
------+----------------------------------------------------------------

The GAP function used to compute these values is
LargestIndexOfMaximalSubgroupOfSn := function ( n )

  local  Sn, maxes, indices;

  Sn := SymmetricGroup(n);
  maxes := MaximalSubgroupClassReps(Sn);
  indices := List(maxes,G->Factorial(n)/Size(G));
  return Maximum(indices);
end;

and the body of the table has been produced with
gap> for n in [2..54] do
>      Print("       ",String(n,2)," : ");
>      PrintFactorsInt(LargestIndexOfMaximalSubgroupOfSn(n));
>      Print("\n");
>    od;

This loop takes about $4$ seconds on my laptop; for $n > 54$ there appears
to be a sharp increase in runtime.<|endoftext|>
TITLE: Where did the term "additive energy" originate?
QUESTION [16 upvotes]: A fundamental object in modern additive combinatorics and harmonic analysis is additive energy. Given a subset $A$ of (say) an abelian group $G$ the additive energy of $A$ is defined to be the quantity $E(A):=|\{(a,b,c,d) \in A^4 : a+b=c+d \}|$. 


Where in the literature did the term "additive energy" originate?

REPLY [33 votes]: Van Vu and I coined the term in our book because there did not seem to be a widely adopted name for it previously.  (Gowers, for instance, refers to "number of additive quadruples" rather than "additive energy", but this seemed to be too unwieldy to use for our purposes.)  I think we settled on "energy" due to the vaguely quadratic nature of the expression, and also because it was monotone (larger sets have larger energy), in contrast to other statistics measuring additive structure such as the doubling constant.  (Coincidentally, in the theory of nonlinear dispersive equations, there are sometimes quartic potential energy terms such as $\frac{1}{4} \int |u|^4$ which can be viewed as essentially being the additive energy of the Fourier transform of the field $u$, so additive energy can in fact be physically interpreted as an energy in some cases, though this was not our intent at the time.  Certainly I sometimes visualise an additive quadruple as a kind of "chemical bond" between four "atoms" in a set $A$, so that the additive energy measures something like a "latent energy" in that set.)
I have sometimes thought of referring to the energy $E(A,B) := | \{ (a,b,a',b') \in A \times B \times A \times B: a+b = a'+b' \}$ between two different sets $A$, $B$ as the "synergy" between $A$ and $B$ rather than the "energy", but perhaps it was better that we didn't propose this name as it was probably too clever by half.<|endoftext|>
TITLE: Progress on isospectral plane domains
QUESTION [7 upvotes]: Has there been any progress on the smooth isospectral plane  domains for Laplacian problem with Dirichlet data? In particular, are there known examples of domains which are isospectral to the unit disk? 
Related to this and of course this.
Edit 1: What if a non-local boundary condition is enforced? To be precise, suppose $D$ is an open disk, and $\Omega$ is a symmetric bounded domain with smooth boundary so that $D$ and $\Omega$ are isospectral w.r.t to Laplacian with non-local boundary condition. Does it follow that $D$ and $\Omega$ are congruent?
Update. Z.Lu and J.Rowlett [paper] recently proved the following:
Theorem. Let Ω be a simply connected planar domain with piecewise smooth Lipschitz
boundary. If Ω has at least one corner, then Ω is not isospectral to any bounded planar domain
with smooth boundary that has no corners.
Corollary. Amongst all planar domains of fixed genus with piecewise smooth
Lipschitz boundary, those that have at least one corner are spectrally distinguished.

REPLY [4 votes]: The Dirichlet/Neumann spectrum determines the domain among the class of analytic domains with some discrete symmetries by work of Zelditch.
The unit disk is determined by its Dirichlet spectrum (among any region, with sufficiently regular boundary; I'm not sure what the minimal assumptions are): First, by Weyl's law, "you can hear the area of a drum" (i.e., the area is a spectral invariant). Then, the claim follows from the rigidity statement in the Faber--Krahn inequality.<|endoftext|>
TITLE: covering map from spheres to projective spaces and the associated vector bundle
QUESTION [5 upvotes]: Let $S^n$ be the $n$-sphere and consider a $2$-sheeted covering
$$
S^n\longrightarrow\mathbb{R}P^n.
$$
We have an associated vector bundle
$$
\xi: \mathbb{R}^2\longrightarrow S^n\times_{\mathbb{Z}/2}\mathbb{R}^2\longrightarrow \mathbb{R}P^n
$$
where the nontrivial element of $\mathbb{Z}/2$ acts on $\mathbb{R}^2$ by reversing the order of coordinates.
Question: What is the smallest positive integer $k$ such that the Whitney sum
$
\xi^{\oplus k}
$
is stable equivalent to a trivial bundle?

REPLY [7 votes]: Theorem 7.4 of J. F. Adams, Vector fields on spheres, Ann. of Math. 75 (1962), 603–632
says that $$\tilde{KO}({\mathbb R} P^n)=\mathbb Z\,/\,2^{\phi(n)},$$ generated by $\xi-1$, where $\phi(n)$ is the number of integers $s$ such that $0 < s\le n$ and $s$ is congruent to 0,1, 2 or 4 modulo 8, and $\xi$ is the canonical line bundle over $\mathbb RP^n$ given as $S^n\times_{\mathbb Z/2} \mathbb R\to \mathbb RP^n$, where $\mathbb Z/2$ acts on $ \mathbb R$ by multiplication with $-1$.
Since your vector bundle is the direct sum of $\xi$ and the trivial 1-dimensional bundle, the answer to your question is $k=2^{\phi(n)}$.<|endoftext|>
TITLE: Representations of orthogonal groups over the field of two elements
QUESTION [6 upvotes]: I am looking for some references on modular representation theory of the orthogonal groups $O_{2n+1}(2)$, $O_{2n}^{+}(2)$, or $O_{2n}^{-}(2)$ over $\mathbb{F}_2$.

REPLY [6 votes]: The question is brief, but there are a great many unknown features of the modular representation theory of these families of groups (for the prime $p=2$).   The approach via algebraic groups pioneered by Steinberg is conceptually attractive but far from able to deal effectively with small primes at this point.   There might also be approaches via finite group theory and combinatorics.  It's hard to predict what will ultimately give the most explicit results.
Since the chosen prime  2 divides the order of each finite group involved, 
complete reducibility of representations breaks down badly.  So there is an open-ended problem of determining the indecomposable representations which are not irreducible; in your case there are infinitely many of these up to isomorphism.  Thus far the best systematic results have been obtained just for irreducible representations (= simple modules for the group algebra).   So I'll make a few comments about these.
1) As Nick Gill indicates, there is some supporting literature.  Note that the useful 1967-68 lectures on Chevalley groups given at Yale by the late Robert Steinberg are still downloadable from his webpage at 
UCLA.  Also, the reference of mine which is most useful is neither of those listed; instead try my survey with many references in the more recent LMS Lecture Note Series No. 326 (2006) Modular Representatins of Finite Groups of Lie Type.   For example, 5.3-5.4 deal in your case with the abstract group isomorphism between symplectic groups $Sp_{2n}$ and odd spin groups (the 2-fold covers of the groups $SO_{2n+1}$)  In characteristic 2 there are further identifications of related finite groups.
2) One of Steinberg's main contributions was his proof that irreducible (rational) representations of the ambient algebraic group restrict naturally to irreducible representations of the finite subgroups of Lie type.   Along with this is his tensor product theorem, which reduces the study of the latter representations in your case to those whose "highest weight" in Lie theory has $n$ coordinates equal to 0 or 1. 
3) Unfortunately, most of the successful further theory involving algebraic groups (especially building on Lusztig's ideas) only applies so far when the given prime is "sufficiently large".   In particular, for $p=2$ there is little guidance from the general theory.  Even so, there is some hope that affine Weyl groups of Langlands dual type (here relative to root lattices multiplied by suitable powers of 2) will turn out to predict the key data required for the modular representations of these finite groups.
4) Keep in mind that, from the Lie theory viewpoint, the groups you are interested in are of two distinct types: $B_n$ (which in characteristic 2 have group isomorphisms with type $C_n$ finite groups) and $D_n$.   Moreover, the Lie theoretic approach tends to start with simply connected groups (here Spin groups) and then yield results in a fairly mechanical way for the related groups you consider.   How to express the results is not immediately clear, since for the algebraic group a "formal character" involves just a knowledge of weight multiplicities, while for the finite groups one has to re-interpret the data as a Brauer character.   
5) The modular version of the Atlas of Finite Groups has a lot of detail about relatively small groups of all types, assembled by explicit computations.<|endoftext|>
TITLE: Unusual isoperimetry and maximizing the measure of unions of translates of a set
QUESTION [7 upvotes]: Let me state a standard result first. Let a $A\subset \mathbb{R}^d$ be a set of fixed volume. Define $A_t$ to be the set of all points at distance at most $t$ from $A$. Then the volume of $A_t$ is minimal if $A$ is a ball of the prescribed volume.
Another way to define $A_t$ is by $A_t=A+B(0,t)$, where $B(0,t)$ is the centered ball of radius $t$. We shall think of it as the union of translates of $A$ by all vectors in $B(0,t)$.
I am interested in extending such a result to the discrete setting. Say, we translate $A$ only in the $d$ orthogonal directions. That is, we look at the union $U(A)=\cup_v (A+v)$, where $v$ is either the zero vector or $\pm e_i$, where $e_i$ is an element of the standard orthonormal basis.
Given that the volume of $A$ is fixed, which $A$ minimize the volume of $U(A)$?

REPLY [8 votes]: Fix large $N$.
Take $A$ to be union of $\varepsilon$-balls which centers have integer  coordinates between $\pm N$. 
(You have to ajust $\varepsilon$ to get the needed volume.)
In this case $U(A)$ is a union $\varepsilon$-balls which centers have integer  coordinates between $\pm (N+1)$.
Therefore $\mathrm{vol}[U(A)]$ can be made arbitrary close to $\mathrm{vol}A$ and there is no minimizer.<|endoftext|>
TITLE: metric condition forcing convex position
QUESTION [7 upvotes]: Let $A_1,A_2,\ldots, A_n$ be distinct in the plane. For every $1\le i \le n$, let
$S_i=\sum\limits_{j=1}^n d(A_i,A_j)$ be the sum of distances from $A_i$ to all the other points.
Assume that $S_i=S_j$ for all $1\le i<j\le n$. Is it true that the points $A_1,A_2,\ldots, A_n$ must be the vertices of a convex $n$-gon?

REPLY [7 votes]: Suppose that the points are not in convex position, then there is a point (WLOG let it be $A_n$) that is a convex combination of other points:
$$A_n = \sum_{i=1}^{n-1} \lambda_i A_i\text,$$
where $\lambda_i\ge0$ and $\sum_i\lambda_i=1$. Now consider the function $f_i(X) = d(A_i,X)$. This function is convex: its value at a convex combination of points is no more than the same convex combination of the values at the points. It is strictly convex as long as the points are not on a line through $A_i$. Therefore, you have that
$$S_n = \sum_{i=1}^{n-1} d(A_i,A_n) < \sum_{i=1}^{n-1}\sum_{j=1}^{n-1} \lambda_j d(A_i,A_j) < \sum_{j=1}^{n-1} \lambda_j S_j\text.$$
This is impossible if $S_i = S_j$ for all $i,j$, so, by contradiction, the points must be in convex position.<|endoftext|>
TITLE: Show that the positive existential theory is undecidable
QUESTION [6 upvotes]: To show that the positive existential theory of $\mathbb{C}[t, e^{\lambda t} \mid \lambda \in \mathbb{C}]$ in the language $\{+, \cdot , ' , 0 , 1, t\}$ is undecidable we have to prove the following: $$n \in \mathbb{N} \leftrightarrow \exists x \left (n \in \mathbb{C} \land tx'=nx \land x(1)=1\right )$$  
So do we have to do the following? 
We suppose that the positive existential theory of $\mathbb{C}[t, e^{\lambda t} \mid \lambda \in \mathbb{C}]$ in the language $\{+, \cdot , ' , 0, 1, t\}$ is decidable, i.e., there is an algorithm that answers to positive existential questions over  $(\mathbb{C}[t, e^{\lambda t} \mid \lambda \in \mathbb{C}];+, \cdot , ' , 0, 1, t)$. 
We want to reduce the positive existential theory $\mathbb{N}$ in the language $\{+, \cdot , 0, 1\}$ into the positive existential theory of $\mathbb{C}[t, e^{\lambda t} \mid \lambda \in \mathbb{C}]$ in the language $\{+, \cdot , ' , 0, 1, t\}$. 
Is this correct? 
But which is the mapping of the reduction?

REPLY [11 votes]: I like this question very much. Before answering, let me try to
explain the question in my words.
You are considering the structure $\mathbb{C}[t,e^{\lambda t}]_{\lambda\in\mathbb{C}}$,
which is the ring of all polynomial expressions over $\mathbb{C}$ in the
indeterminate variable $t$ and $e^{\lambda t}$, for any $\lambda\in\mathbb{C}$.
So this ring has objects like this: $$t\qquad\qquad (t^2+5)e^{3t}+te^{\pi
t}\qquad\qquad t^2e^{\pi i t}+5t^{10}.$$ We consider this structure in the
language of rings, using addition, multiplication, 0, 1, augmented
with the differentiation operation $f'$ and a constant symbol for
the polynomial $t$. So we may freely make terms like $t^2x''+tx$
and so on, where $x$ is a variable ranging over the ring. The
exponential polynomials are simply points in the structure. The question is whether the positive existential theory of this structure is undecidable.
In order to show that the existential theory of this structure is
undecidable, it suffices to show that we may define the natural
numbers in it by an existential property, for then we would be
able to reduce any existential question in the language of
arithmetic, that is, in the structure
$\langle\mathbb{N},+,\cdot,0,1\rangle$, to an existential question
in your structure. Since the existential theory of arithmetic is
undecidable, it will follow that the existential theory of your
structure will be undecidable.
Notice that we can define the constant polynomials, that is, the
elements of $\mathbb{C}$ in your ring, since these are precisely
the $x$ that satisfy $x'=0$.
Let us view $\mathbb{N}$ as a subset of $\mathbb{C}$, which is a
subset of your ring $\mathbb{C}[t,e^{\lambda t}]_{\lambda\in\mathbb{C}}$. Using the idea
you suggested, we can define $\mathbb{N}$ in your ring by an
existential formula as follows:
Lemma. $n\in\mathbb{N}$ if and only if
$\mathbb{C}[t,e^{\lambda t}]_{\lambda\in\mathbb{C}}\models n'=0\wedge \exists x\
tx'=nx\wedge x\neq 0$.
Proof. If $n$ is a natural number, then it is constant and so
the derivative $n'$ is $0$. But furthermore, if $x=t^n$, then
$x'=nt^{n-1}$, and so $tx'=nx$. So $n$ satisfies the definition.
Conversely, suppose that $n\in\mathbb{C}$ and there is some nonzero $x$ in
your ring with $tx'=nx$. I claim that the only solutions of this
differential equation are of the form $\lambda t^n$, for
$\lambda\in\mathbb{C}$. Certainly any expression of this form is a
solution. And if $x$ and $y$ are both solutions of the
differential equation, then $(\frac
xy)'=\frac{x'y-y'x}{y^2}=\frac{\frac{nx}ty-\frac{ny}tx}{y^2}=0$.
So every solution is a constant multiple of $t^n$. Since this
occurs only for $n\in\mathbb{N}$ in your ring, it follows that $n$
is a natural number. QED
So we have defined $\mathbb{N}$ by an existential formula in your
structure, and therefore the existential theory of your structure
is undecidable. Any existential assertion in arithmetic, such as
the halting problem, can be transformed to an existential theory
in your structure.
The existential assertion in the lemma is not positive, however, because of the $x\neq 0$ part. But we can replace $x\neq 0$ there with the assertion that $(t-1)$ divides $x-1$, since once we know that $x$ is a solution of $tx'=nx$, we know that $x$ must have the form $x=\lambda t^n$, and the only way that $t-1$ can divide $\lambda t^n-1$ is if $\lambda=1$. Thus, we have:
$$n\in\mathbb{N}\iff \mathbb{C}[t,e^{\lambda t}]_{\lambda\in\mathbb{C}}\models n'=0\wedge \exists x\
tx'=nx\wedge \exists y\ (t-1)y=x-1.$$
So the natural numbers are defined by a positive existential assertion in your exponential polynomial ring.
Therefore, for any positive existential assertion $\exists n_0,n_1,\ldots,n_k\ \varphi(\vec n)$ in the language of arithmetic, where $\varphi$ is quantifier free, we may translate this to $\exists n_0,\ldots,n_k( n_0\in\mathbb{N}\wedge\cdots\wedge n_k\in\mathbb{N}\wedge\varphi(\vec n)$ in the language of your ring, where $n\in\mathbb{N}$ is an abbreviation for the formula displayed above.  The original assertion will be true in arithmetic just in case the translation is true in your exponential polynomial ring.<|endoftext|>
TITLE: abelian and nonabelian parts of Aut($\widehat{F_2}$)
QUESTION [5 upvotes]: Let $F$ be the free profinite group on two generators. Let $\text{IA}(F) := \ker\left(\text{Aut}(F)\rightarrow GL_2(\widehat{\mathbb{Z}})\right)$, the group of "IA automorphisms" of $F$. (I'm also happy to consider $\text{Out}(F)$ rather than $\text{Aut}(F)$, though there shouldn't be much difference.)
Here, the image of an automorphism of $F$ in $GL_2(\widehat{\mathbb{Z}})$ tells you how it acts on abelian quotients. It seems to me there should morally be another "part" of $\text{Aut}(F)$ which describes how automorphisms of $F$ act on nonabelian quotients of $F$, and moreover that for some notion of an "extremely nonabelian" quotient, the abelian and nonabelian "parts" of $\text{Aut}(F)$ should be "orthogonal".
To what extent is this correct/reasonable?
I also don't know what "extremely nonabelian" should mean, but we'll let ENA denote such a property. A reasonable candidate for ENA is "perfect", or "all simple composition factors are finite simple nonabelian groups". Let $\Delta$ be this second property, then $\Delta$ has the benefit that the set of $\Delta$-groups are a formation of finite groups (ie, closed under quotients and subdirect products), and so one may speak of the pro-$\Delta$ completion of a group, and so on. It's unclear to me if 2-generated perfect groups also form a formation.
My first specific question is this: Let $F\twoheadrightarrow G$ be an ENA quotient. Let $K_G$ be the intersection of all $G$-defining subgroups of $F$, then $K_G$ is characteristic in $F$ and we have a surjection
$$p_G : \text{Aut}(F)\longrightarrow\text{Aut}(F/K_G)$$
For which values of ENA is the restriction $p_G|_{\text{IA}(F)}$ also surjective?
My second question is this: What is known about the maximal pro-$\Delta$ quotient of $F$? (or similar objects. I don't even know what to google!) If $F^\Delta$ denotes the maximal pro-$\Delta$ quotient of $F$, then we get a surjection $F\twoheadrightarrow F^\Delta$ with characteristic kernel, and again there is a surjective map
$$p_\Delta : \text{Aut}(F)\longrightarrow\text{Aut}(F^\Delta)$$
Is $p_\Delta|_{\text{IA}(F)}$ surjective? What if we replace $\Delta$ with other possible values of ENA?
My last question is: Is it reasonable to expect that the map $\text{Aut}(F)\longrightarrow \text{Aut}(F^\Delta)\times GL_2(\widehat{\mathbb{Z}})$ is injective? Surjective? What if we replace $\Delta$ with other possible values of ENA?
I'd very much appreciate any references or results related to the questions.

REPLY [2 votes]: $\newcommand{\Zhat}{\widehat{\mathbb{Z}}}$
Here are some partial answers, though I would welcome any additional input.
There is a natural map
$$p : F\rightarrow F^\Delta\times\Zhat^2$$
The kernel of $F\rightarrow\Zhat^2$ is just $[F,F]$, and since all finite quotients of $F^\Delta$ are perfect, $[F,F]$ surjects onto $F^\Delta$, and thus the map $p$ above is surjective. Since the all finite simple quotients of $F$ must factor through $F^\Delta\times\Zhat^2$, the kernel of $p$ is contained in the intersection of all maximal normal subgroups of $F$, otherwise known as the "Jacobson radical" of $F$ (also known as the Baer radical, or small radical). This is similar to the Frattini subgroup $\Phi(F)$, but in this case it cannot equal to $\Phi(F)$, since by Corollary 8.7.5 in Ribes/Zalesskii, $\Phi(F) = 1$ (thanks to Benjamin Steinberg for the reference), and by Proposition 8.7.7, $F$ is not a direct product.
Nonetheless, we get maps
$\newcommand{\Aut}{\text{Aut}}$
$$\Aut(F)\twoheadrightarrow\Aut(F^\Delta\times\Zhat^2)\cong \Aut(F^\Delta)\times\Aut(\Zhat^2)$$
where the isomorphism is due to the fact that the projections of $F^\Delta\times\Zhat^2$ onto each direct factor is a characteristic quotient.
Thus, if $G$ is a 2-generated pro-$\Delta$ group, then any surjection $F\rightarrow G$ must factor through $F^\Delta$. This implies a positive answer to the first question if ENA = $\Delta$ and $G$ is pro-$\Delta$. Actually, by similar arguments, as long as $F/K_G$ is perfect, the answer to the first question is also positive.
This also gives a positive answer to the second question.
For the third question, the above shows that the map is definitely surjective, though not injective - for example, inner automorphisms by elements in the Jacobson radical of $F$ lie in the kernel.<|endoftext|>
TITLE: Suspension of the third Hopf map
QUESTION [9 upvotes]: There are Hopf maps in three dimensions, I denote their homotopy classes by $h_1 \in \pi_3(S^2)$, $h_2 \in \pi_7(S^4)$ and $h_3 \in \pi_{15}(S^8)$ respectively. For $h_1$ and $h_2$ it is true that their supensions generate the corresponding stable homotopy groups of spheres: $Sh_1$ generates $\pi^s(1) \approx Z_2$, and $Sh_2$ generates $\pi^s(3) \approx Z_{24}$.
Question: Is this true for $h_3$ too?. I.e. does $Sh_3$ generate $\pi^s(7) \approx Z_{240}$?

REPLY [6 votes]: I think this follows from elementary considerations, as in Chris Schommer-Pries's question How to see the quaternionic hopf map generates the stable 3-stem?
The long exact sequence of the Hopf fibration
$$
S^7\to S^{15}\stackrel{\sigma}{\to} S^8$$ leads to a short exact sequence $$0\to \pi_{15}(S^{15})\stackrel{\sigma_\ast}{\to} \pi_{15}(S^8)\to \pi_{14}(S^7)\to 0$$ in which the kernel is isorphic to $\mathbb{Z}$ and the quotient to $\mathbb{Z}_{120}$. In fact this is split, as can be seen by looking up $\pi_{15}(S^8)\cong \mathbb{Z}\oplus\mathbb{Z}_{120}$ in tables (there should be a better way to see this). It follows that $\sigma$ generates the $\mathbb{Z}$ summand. 
Now the Freudenthal Suspension Theorem implies that $$E: \mathbb{Z}\oplus\mathbb{Z}_{120}\cong\pi_{15}(S^8)\to \pi_{16}(S^9)\cong \pi_7^{st}\cong\mathbb{Z}_{240}$$ is surjective, and it is now an exercise in elementary algebra to check that $E\sigma$ generates.  
Update: As András points out, the above answer is incomplete: it does not follow from what's written above that $E\sigma$ generates $\pi_7^{st}=\mathbb{Z}_{240}$. Here is an attempt to rectify this. It seems to work, modulo a claim made below which hopefully is true and maybe even well known.
There is a short exact sequence
$$
0\to \pi_{15}(S^{15})\stackrel{P}{\to} \pi_{15}(S^8)\stackrel{E}{\to} \pi_{16}(S^9)\to 0
$$ which is the portion of the EHP sequence mentioned in András' comment. The map $P$ is composition with the Whitehead product $[\iota_8,\iota_8]:S^{15}\to S^8$. We can cross this with the short exact sequence coming from the Hopf fibration above to get a cross diagram with exact diagonals
$$
\begin{array}{ccccc}
\pi_{15}(S^{15}) & & & & \pi_{15}(S^{15}) \newline
 & \searrow & & \swarrow & \newline
\downarrow & & \pi_{15}(S^8) & & \downarrow \newline
 & \swarrow & & \searrow & \newline
\pi_{16}(S^9) & & & & \pi_{14}(S^7) 
\end{array}
$$
and by the Butterfly Lemma (Mosher and Tangora, Exercise 7.1) we are reduced to showing that $\partial [\iota_8,\iota_8]$ generates $\pi_{14}(S^7)$, where $\partial:\pi_{15}(S^8)\to \pi_{14}(S^7)$ is the connecting homomophism of the Hopf fibration.
Claim: The map $\partial: \Omega S^8\to S^7$ which induces this connecting homomorphism is an H-map.
Accepting this claim for a moment, we see that $\partial$ maps the Whitehead product $[\iota_8,\iota_8]$ to the Samelson product $\langle \iota_7,\iota_7\rangle\in \pi_{14}(S^7)$ which is defined using the multiplication on $S^7$. But this Samelson product generates $\pi_{14}(S^7)\cong \mathbb{Z}_{120}$, by a result of James (page 176 of
I. M. James, On $H$-spaces and their homotopy groups, Quart. J. Math. Oxford Ser. (2) 11 (1960), 161--179.)<|endoftext|>
TITLE: Pullback along Frobenius morphism
QUESTION [10 upvotes]: Let $X$ be a scheme over a finite field $\mathbb{F}_q$ and let $F : X \to X$ be the absolute Frobenius morphism. If $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module, then there is a natural isomorphism $F^*(\mathcal{L}) \cong \mathcal{L}^{\otimes q}$.

Is there also a general formula for $F^*(\mathcal{M})$ if $\mathcal{M}$ is a locally free $\mathcal{O}_X$-module of given rank $d$?

Of course, we have $F^*(\mathcal{M}) = \mathcal{M} \otimes_{\mathcal{O}_X} F_*(\mathcal{O}_X)$ (since $F$ is affine), but I would like to have a formula which is independent of the quasi-coherent algebra $F_*(\mathcal{O}_X)$ (unless you can describe this algebra via generators and relations), but only depends on $\mathcal{M}$ and uses the usual operations on quasi-coherent modules ($\otimes$, $\ker$, $\mathrm{coker}$, $\oplus$, $\mathrm{Sym}^n$, $\Lambda^n$, $\dotsc$).
For $q=2$ I have found that $F^*(\mathcal{M})$ is the kernel of the homomorphism $\mathrm{Sym}^2(\mathcal{M}) \twoheadrightarrow \Lambda^2(\mathcal{M})$, $m \cdot n \mapsto m \wedge n$. In general, there is an embedding$$F^*(\mathcal{M}) \to \mathrm{Sym}^q(\mathcal{M})$$ which maps $m \otimes 1 \mapsto m^q$.
My question may be also phrased as follows: On the $K$-theory of $X$, $F^*$ induces the Adams operation $\psi^q$. Hence, I ask for a bundle representative  of the Adams operation $\psi^q$. (Not just a formal difference of bundles.)

REPLY [8 votes]: Consider the $p$-th tensor power $\mathcal{M}^{\otimes p}$. The group $\mathbb{Z}/p\mathbb{Z}$ acts by cyclic permutations on it. Denote its generator by $\sigma$. There is a map from coinvariants to invariants $$(\mathcal{M}^{\otimes p})_{\sigma}\xrightarrow{1+\sigma+\dots+ \sigma^{p-1}}(\mathcal{M}^{\otimes p})^{\sigma}$$ Its kernel is canonically isomorphic to $F^*\mathcal{M}$ via the map $s\mapsto s^{\otimes p}$. This is proven in Lemma 6.9 here:http://arxiv.org/pdf/1509.08784v1.pdf for the affine case and the general case follows by functoriality.
Edit: By the way, its cokernel is also isomorphic to $F^*\mathcal{M}$.<|endoftext|>
TITLE: A point set of power series with coefficients in {-1, 1}. Connected or not?
QUESTION [27 upvotes]: Let $z$ be a fixed complex number with $|z|<1$ and consider the set
$$X_z := \Big\{\sum\limits_{i=1}^{\infty} a_i z^i \ \Big|\ a_i\in \{-1,1\} \forall i\Big\}.$$
What can be said about the set $M$ of those $|z|\lt 1$ such that $X_z \subset \mathbb{C}$ connected?

REPLY [7 votes]: What you are referring to is called the connectedness locus. The most recent important paper on the subject is probably this paper by Calegari, Koch and Walker. 
In particular, they show that the connectedness locus contains infinitely many holes. The paper also contains an extensive list of references.<|endoftext|>
TITLE: equivariant stable class of quaternionic Hopf fibration in RO(G)-degrees of ADE-type
QUESTION [11 upvotes]: Does the quaternionic Hopf fibration possibly represent a non-torsion element in the $G$-equivariant stable homotopy groups of spheres, for $G$ a finite subgroup of $SO(3)$ and in RO(G)-degree being its canonical 3d representation?
For the complex Hopf fibration the analog is true, as far as I see: In

Shôrô Araki, Kouyemon Iriye, Equivariant stable homotopy groups of spheres with involutions. I, Osaka J. Math. Volume 19, Number 1 (1982), 1-55. (Euclid)

it is shown (theorem 8.7 i) that the complex Hopf fibration -- canonically regarded as a representative for a $\mathbb{Z}/2$-graded stable homotopy group of spheres in $RO(G)$-degree being the canonical 1-dimensional representation of $\mathbb{Z}/2$ -- is a non-torsion generator.
In this case the construction of the $\mathbb{Z}/2$-equivariant structure is induced, via the Hopf construction, from (on top of their p. 24) the $\mathbb{Z}/2$-equivariance of the product operation on complex numbers with respect to complex conjugation.
Analogously, the product operation on quaternions is of course $SO(3)$-equivariant, again with respect to the canonical action on their imaginary part. So it would seem that restricting this to any finite subgroup $G$ of $SO(3)$ and then applying the Hopf construction to the quaternions will yield a $G$-equivariant quaternionic Hopf fibration that represents an element in the $G$-equivariant stable homotopy groups of spheres in $RO(G)$-degree the corresponding 3d rep. (Right?)
For which choice of $G \hookrightarrow SO(3)$, if any, is this element non-torsion?
Or more generally, what is known about equivariant stable homotopy groups of spheres in what one might call  RO(G)-degrees of ADE type?

REPLY [6 votes]: I think the Hopf construction gives a non-torsion class when G is dihedral or exceptional, but probably not when G is cyclic.
Non equivalently, we can perform Hopf constructions on the 0, 1, 3, and 7 spheres.  Only the 0 sphere gives a non torsion class in stable homotopy.
Suppose $G$ acts on our sphere, preserving the multiplication, with fixed points $(S^n)^G$ also a sphere $S^k$.  If $H: S^{2n+1} \to S^{n+1}$ is the Hopf map, then the restriction of $H$ to $G$ fixed points is also a Hopf map $H': S^{2k+1} \to S^{k+1}$.  If $k=0$, then $H'$ is stably non-torsion (non-equivariantly), and therefore so is $H$ in the  $G$ equivariant stable category.
Conversely, if $k>0$, I think $H$ must be stably torsion $G$ equivariantly: the rational stable equivariant homotopy type of a space $X$ (for finite groups) is (I think) completely determined by the $H^*(X^K, Q)$ as $WK=NK/K$ modules, where $K$ ranges over the conjugacy classes of subgroups of $G$, and rational stable maps are all detected by these groups.<|endoftext|>
TITLE: When is an erratum necessary?
QUESTION [33 upvotes]: A typo, a spelling error etc., in a published article, is definitely not enough for issuing an erratum.
If a mistake destroys a main result, then an erratum is definitely necessary, and the proof should be rewritten, if it's possible to fix.
What about cases in between?  What if a small lemma, remark is not correctly stated?  What about a mistake in the proof that is actually very easy to fix? What about a wrong number in the calculation? What if ... (you name the case) ...?

REPLY [9 votes]: Checking MathSciNet in the five-year period from 2010 to 2014, there were 1319 entries with "erratum" or "errata" in the title.  Sampling a few dozen of them, most cases involve a misstatement severe enough to affect the statement of a theorem.  In a few cases, something was inadvertently left out, such as MATLAB code or an author's name. For perspective, during this period, there were 525795 items added to MathSciNet.<|endoftext|>
TITLE: Thin large subspaces of $\ell^N_1$
QUESTION [8 upvotes]: Consider a sequence $V_N$ of subspaces of $\ell^N_1$ so that $\dim V_N = N- n$ and $n$ is $\mathsf{o}(N)$. Is it true that these spaces are "thick" (unofficial terminology), i.e. are there constants $K_1,K_2>0$ so that for $N$ large enough,
$$
\sup_{x \in V, \|x\|_1 \leq K_1} \|x\|_\infty \geq K_2
$$
where $\|x\|_\infty$ is (as usual) the supremum of the coordinates of $x$.
A relaxation (or equivalent formulation?) of the above: is there a sequence $x_i \in V_i \subset \ell_1^i \subset \ell_1\mathbb{N}$ which is convergent in the weak$^*$ topology of $\ell_1 \mathbb{N}$ to something else than $0$?
In the $\ell_2$ case, the answer is "overwhelmingly positive". Let $\lbrace e_i\rbrace_{i=1}^N$ be the "standard" basis of $\ell^N_1$. Consider $P_{V_N}$ to be the orthogonal projection on $V_N$, then 
$$
N-n = \dim V_N = \sum_{i=1}^N \langle P_{V_N} e_i \mid e_i \rangle
$$
So that, in fact, there are at least $N-2n$ coordinates $i$ so that  $\langle P_V e_i \mid e_i \rangle \geq \frac{1}{2}$ and $\langle P_{V_N} e_i \mid e_i \rangle = \| P_{V_N} e_i\|_2^2 \leq 1$.

REPLY [6 votes]: I discussed this question with Pisier over lunch.  He later called my attention to the paper
[GG] Garnaev, A. Yu.; Gluskin, E. D.;
The widths of a Euclidean ball. (Russian)
Dokl. Akad. Nauk 277 (1984), 1048–1052.
Pisier emailed me,
“They get an equivalent of the relevant  Kolmogorov numbers
of inclusion of $n$-dim Euclidean space into $\ell_\infty^n$.
It says (from my book on the volume p. 81 but there is a misprint- I checked it must be $+1/2$ in the exponent of the log not $-1/2$):
$ L_1^N$ contains a subspace of codimension $N-n$
with Banach-Mazur distance to Hilbert at most
$$
\min \{ \sqrt{N},  (N/n)^{1/2} [\log (1+  N/n)] ^{1/2}\ ”
$$
Thus the OP’s question has a negative answer if $n^{-1}\log (1+  N/n)$ tends to zero; i.e., if  $n^{-1}\log N \to 0$. What if $n\to \infty$ slower than $\log N$? 
Take an $n$ codimensional subspace $V_N$ of $\ell_1^N$.  Take an Auerbach basis $y_1,\dots, y_n$ for $V_N^\perp \subset \ell_\infty^N$.
Let $1/\epsilon $ be a positive integer; something like $\epsilon = 1/10$ is OK. Tile $[-1, 1]^n$ with boxes of side length $\epsilon/n$; the number of boxes is around $(2n\epsilon/n)^n$. Consequently, if $N> (2n\epsilon)^n$, there exist $1\le i < j \le N$ s.t. $(\langle y_k, e_i \rangle)_{k=1}^n$ and $(\langle y_k, e_j \rangle)_{k=1}^n$ lie in the same box.  Since $y_1,\dots, y_n$ is Auerbach, this means that 
$|\langle y, e_i -e_j\rangle| < \epsilon $ for all norm one $y \in V_N^\perp$.  That is, the distance of $e_i -e_j$ to $V_N$ is less than $\epsilon$. Since $\|e_i -e_j\|_1 = 2$ and $\|e_i -e_j\|_\infty =1$, this gives a positive answer to the OP’s question if $N > (2n\epsilon)^n$ for infinitely many $N$; i.e., if $n\log n < \delta \log N$ for a computable $\delta$.  
I don’t know where the break point occurs.<|endoftext|>
TITLE: Definition of étale (etc) for non-representable morphisms of algebraic stacks?
QUESTION [9 upvotes]: I've stumbled upon the statement that the morphism $\pi$ from a root stack of the form $\sqrt[r]{\mathscr{L}/\mathscr{Y}}$ (i.e. the "generic" version, not the one concentrated along a divisor) to its underlying stack $\mathscr{Y}$ is "étale".
Now, I know what "étale" means for representable morphisms (which the above $\pi$ is not). If $\boldsymbol{\mathrm{p}}$ is a property of morphisms of schemes (usually required to be stable under base change), then $f:\mathscr{X}\to\mathscr{Y}$ is $\boldsymbol{\mathrm{p}}$ if every base change $\mathscr{X}\times_\mathscr{Y} U\to U$ is $\boldsymbol{\mathrm{p}}$.
What about the case of a non representable $f:\mathscr{X}\to\mathscr{Y}$ when we have atlases $\alpha:X\to\mathscr{X}$, $\beta:Y\to \mathscr{Y}$  (with $X$ and $Y$,say, schemes)? I had a look to the stacks project but wasn't able to find it.
At least when $\alpha$ and $\beta$ are étale atlases (i.e. when $\mathscr{X}$ and $\mathscr{Y}$ are Deligne-Mumford), I think it would make sense to say that $f$ is $\boldsymbol{\mathrm{p}}$ if $f'':=f'\circ \alpha'$ is $\boldsymbol{\mathrm{p}}$, where $f':\mathscr{X}':=\mathscr{X}\times_{\mathscr{Y}}Y\to Y$ and $\alpha':\mathscr{X}'\times_{\mathscr{X}}X\to\mathscr{X}'$. Edit: This is equivalent to requiring that $X_Y:=X\times_{f\circ\alpha, \mathscr{Y},\beta}Y\to Y$ is $\boldsymbol{\mathrm{p}}$.
Is this one the right definition people use? Is it correct also for Artin stacks?
Edit: Also, how does "my" definition compare with others in the literature?
Edit: So far, it seems that for DM stacks all the definitions agree and work for $\boldsymbol{\mathrm{p}}=$"étale". On page 10 here Alper says "finite morphisms of stacks are necessarily representable": why?

REPLY [6 votes]: For non-representable morphisms of Artin (i.e. algebraic) stacks, different properties are defined in different ways, depending on their particular nature.
E.g. for properties which are smooth local on source and target, see here in the Stacks Project.
The property of being etale is not smooth local on source and target, so this approach doesn't work in that case.  (If we restrict to DM stacks, we can use the etale analogue of this definition, and so in that case we get the correct definition of etale; see David Carchedi's answer.)
This doesn't mean that we can't define the notion of etale morphisms, though; it just means that this particular framework doesn't apply.
One approach is via infinitesimal lifting properties: using these one can define what it means for a morphism of stacks (representable or not) to be formally smooth, formally etale, or formally unramified.  By imposing locally fin. pres. in the first two cases, or locally fin. type in the third case, one then gets notions of smooth, etale, and unramified. 
Alternatively, one can define a morphism to be unramified iff it is loc. finite type and has etale diagonal.  (Since diagonal morphisms are always representable, we know what etale means for the diagonal.) 
We can then define a a morphism to be etale if it is flat, loc. fin. pres., and unramified.
Note that one can also define a smooth morphism using the framework of smooth-local-on-source-and-target discussed above, because being smooth is smooth local.  
See Appendix B of this paper by David Rydh for a discussion of these various definitions and their equivalences.  (Note though that the affineness condition on the schemes involved in the infinitesimal lifting properties seems to have been accidentally omitted.)
Note also that in the non-representable context, etale is stronger than smooth and quasi-finite, or smooth and relative dimension zero. (Since smooth implies flat, etale is equivalent to smooth and unramified.)
E.g. if G is a positive dimensional smooth alg. group over Spec $k$, then $$BG := [ \mathrm{Spec} \, k / G] \to \mathrm{Spec} \, k$$ is smooth and quasi-finite, but not etale.  It is of negative relative dimension (equal to $- \dim G$.
The morphism 
$$ [\mathbb A^1/ \mathbb G_m] \to \mathrm{Spec }\,  k$$ is smooth and of relative dimension zero, but it is not etale.
(An etale morphism is unramified, thus has etale diagonal, thus has unramified diagonal, and thus is a DM morphism.  In particular, if the target is a scheme, then the source is a DM stack, which $BG$ (for positive dimensional $G$) and $[\mathbb A^1/\mathbb G_m]$ are not.)

Added later: here is a correct characterization of etale morphisms of Artin stacks as certain kinds of smooth morphisms (akin to the fact that etale morphisms of schemes are smooth morphisms that are locally quasi-finite).
Recall that a morphism of Artin stacks is called Deligne--Mumford if it has unramified diagonal, or equivalently (but non-obviously) if it's base-change over any scheme yields a Deligne--Mumford stack (see here in the Stacks Project, especially footnote 1).  (So this is a weakening of representability ---
which is equivalent to the diagonal being a monomorphism --- and is automatic for morphisms between DM stacks.)
Then a morphism $X \to Y$ is etale iff it is smooth, Deligne--Mumford, and locally quasi-finite.
(For the proof: note that an unramified morphism has (by definition) an etale, and so unramified, diagonal, hence is Deligne--Mumford.  Using this, 
one can also check that an unramified morphism is locally quasi-finite, because this reduces to checking that a Deligne--Mumford stack which is unramified over the Spec of a field is locally quasi-finite --- which is
easy.  Since etale morphisms are in particular smooth, we get the only if direction.
For the if direction, by pulling back over a chart of $Y$ we may assume that $Y$ is a scheme, and hence that $X$ is a DM stack.  Now we have to 
check that a smooth and loc. quasi-fin. morphism from a DM stack to a scheme is etale, which is again easy.)

Yet another formulation of the same notion, adopting the view-point expressed by user t3suji in a comment on another answer:
A morphism $X \to Y$ of Artin stacks is etale if it is DM, and if for any morphism $V \to Y$ whose source is a scheme, the base-changed morphism
$$X \times_Y V \to V$$ 
(which is now a morphism from a DM stack to a scheme) is etale (in the sense of morphisms of DM stacks, where we already know what etale means: choose an etale chart $U \to X\times_Y V$, and require that the composite $U \to V$ be etale).
(The equivalence with the various preceding definitions is easily checked.)<|endoftext|>
TITLE: Sections of the conormal bundle
QUESTION [6 upvotes]: Let $X\subset\mathbb{P}^N$ be a quadratic manifold. That is $I(X)$ is generated by quadratic polynomials $Q_1,...,Q_m$.
Let $\mathcal{I}_X$ be the ideal sheaf of $X$ and $\mathcal{I}_X/\mathcal{I}_X^2$ be the conormal bundle of $X$. Is it true that $H^0(X,\mathcal{I}_X/\mathcal{I}_X^2(2))$ has rank $m$ and that it is generated by $dQ_1,...,dQ_m$ ?

REPLY [5 votes]: No. The simplest example is the twisted cubic $C\subset \mathbb{P}^3$. The conormal bundle on $C\cong \mathbb{P}^1$ is $\mathcal{O}_{\mathbb{P}^1}(-5)^2$. Hence $H^0(C,\mathcal{I}_C/\mathcal{I}_C^2(2))$ has dimension 4, while $I(C)$ is generated by 3 quadrics.<|endoftext|>
TITLE: Assuming $\operatorname{depth}M\ge \operatorname{depth}N$, what can one say about $\operatorname{depth}M_p$ and $\operatorname{depth}N_p$?
QUESTION [6 upvotes]: Let $(R,m)$ be a Noetherian local ring, $M$ and $N$ finite $R$-modules, $p$ a prime ideal, and $I$ an ideal such that $IM\neq M$.
Definition: The common length of the maximal $M$-sequences in $I$ is called the grade of $I$ on $M$; denoted by $\operatorname{grade}(I,M)$.
$\operatorname{grade}(m,M)$ is denoted by $\operatorname{depth}M$. So by $\operatorname{depth}M_p$, we mean $\operatorname{grade}(pR_p,M_p)$.

Assuming  $\operatorname{depth}M\ge \operatorname{depth}N$, what can one say about $\operatorname{depth}M_p$ and $\operatorname{depth}N_p$? Is there any inequality between them?
What if we impose additional assumptions on $M$ and $N$? For example, if $M=R/I$ and $N=R/J$?

Thank you.

REPLY [8 votes]: Let $R$ be the localization of $k[x, y, z]$ at $(x, y, z)$ where $k$ is a field. Let $M = R/(x)$ and $N = R/(y)$. Then the depth of $M$ and $N$ is $2$.

If $\mathfrak p = (x, y)$, then the depth of $M_{\mathfrak p}$ is 1 and the depth of $N_{\mathfrak p}$ is $1$.
If $\mathfrak p = (x, z)$, then the depth of $M_{\mathfrak p}$ is $1$ and the depth of $N_{\mathfrak p}$ is $\infty$.
If $\mathfrak p = (y, z)$, then the depth of $M_{\mathfrak p}$ is $\infty$ and the depth of $N_{\mathfrak p}$ is $1$.
Take $M = R \oplus R/(x)$ and $N = R \oplus R/(y)$ and you can replace $\infty$ in the above by $2$.

Actually, you can get any integers subject to some constraints. Here is what I mean. We know that (a) $\dim(R) \geq \dim(M) \geq \text{depth}(M)$, (b) $\dim(R) \geq \dim(R_\mathfrak p) \geq \dim(M_\mathfrak p) \geq \text{depth}(M_\mathfrak p)$, and (c) $\dim(M) \geq \dim(M_\mathfrak p)$. I think you can make examples of $R, M, \mathfrak p$ such that the integers $\dim(R)$, $\dim(M)$, $\text{depth}(M)$, $\dim(R_\mathfrak p)$, $\dim(M_\mathfrak p)$, $\text{depth}(M_\mathfrak p)$ are an arbitrary $6$-tuple satisfying (a), (b), (c).
For example, take $6, 6, 1, 5, 4, 3$. Then you can take
$R = k[x_1,\ldots, x_5, y, z]/(yz)$ localized at $(x_1, \ldots, x_5, y, z)$, take $M = R/(y) \oplus R/(z) \oplus R/(z, x_2) \oplus R/(y, z, x_2, \ldots, x_5)$, and take $\mathfrak p = (z, x_2, \ldots, x_5)$. Observe that in this case the depth of $M_\mathfrak p$ is bigger than the depth of $M$.
You can do a similar game with a pair of modules and there will be no relationship whatsoever between the depth of the modules and the depth of their localizations at $\mathfrak p$.
If you want the modules to be cyclic I am pretty sure you can do this as well. In any case, if you are only interested in depth and not dimension, then you can convert any example as above into an example with cyclic modules as follows. Step 1: we can always assume our ring $R$ is regular by writing the example ring as a quotient of a regular ring. In the example above we can write $k[x_1, \ldots, x_5, y, z]/(yz)$ as a quotient of $k[x_1, \ldots, x_5, y, z]$. Step 2. Say $M$ is our module and is generated by $n$ elements $e_1, \ldots, e_n$ subject to some relations $\sum r_{ji} e_i = 0$ for $j = 1, \ldots, m$. Then we introduce a new ring $P = R[t_1, \ldots, t_n]_{(\mathfrak m, t_1, \ldots, t_n)}$ and we consider the cyclic $P$-module
$$
M_{new} = P/(t_it_{i'}, 1\leq i, i' \leq n; \sum r_{ji}t_i, 1 \leq j \leq m)
$$
Then there is a short exact sequence $0 \to M \to M_{new} \to R \to 0$. A computation shows that $\text{depth}_P(M_{new}) = \text{depth}_R(M)$ provided $M$ is not zero (this is where we need that $R$ is regular so that it has depth not less than $M$). Similarly for the depth at $\mathfrak p$ and the corresponding prime of $P$.<|endoftext|>
TITLE: Frucht's type theorem for Riemann surface
QUESTION [7 upvotes]: Frucht's theorem is a theorem in algebraic graph theory conjectured by Dénes Kőnig in 1936 and proved by Robert Frucht in 1939. It states that every finite group is the group of symmetries of a finite undirected graph. More strongly, for any finite group $G$ there exist infinitely many non-isomorphic simple connected graphs such that the automorphism group of each of them is isomorphic to $G$,You can see here.
I am not familiar with Riemann geometry, but I think my question has some meaning.
Do we have such Frucht's type theorem for Riemann surface? Precisely, for a finite group $G$, is there Riemann surface $M$, which its fundamental group is isomorphic to $G$?
I am interested in the special case, where $G=D_{2n}$ is the dihedral group. Also, when $G=D_{\infty}$ is interesting for me.
Thanks in adavanced.

REPLY [10 votes]: You seem to be asking about the group of isometries, not the fundamental group. If so, for every $n$ and every finite group $G$ there is a compact hyperbolic manifold of dimension $n$ whose isometry group is $G.$ See Belolipetsky and Lubotzky. They actually do $n\geq 4,$ but the paper has references to work in lower dimensions (by Greenberg and Kojima).<|endoftext|>
TITLE: Realization Functor From $SH$ to Derived Category of $Gal$-Modules
QUESTION [9 upvotes]: Let $k$ be a field. I would like a reference for realization functors from Morel-Voevodksy's stable category $SH(k)$ to the derived categories of $Gal(\bar{k}/k)$-modules. Has something like this been written down?
Thanks!

REPLY [13 votes]: The most general functor of this form was constructed by Ayoub in La réalisation étale et les opérations de Grothendieck.
Ayoub considers the ∞-category $DA^{et}(S,\Lambda)$ which is defined exactly as $SH(S)$ except that (1) spectra are replaced by chain complexes of $\Lambda$-modules and (2) the Nisnevich topology is replaced by the étale topology. So there's an obvious functor $SH(S) \to DA^{et}(S,\Lambda)$. There's an adic version of this when $\Lambda$ is the completion of a ring at an ideal.
On the other hand, if $D^{et}(S,\Lambda)$ denotes the derived category of sheaves of $\Lambda$-modules on the small étale site of $S$, there is also an obvious functor $D^{et}(S,\Lambda) \to DA^{et}(S,\Lambda)$. Ayoub's "rigidity theorem" (Theorem 4.1 in loc. cit.) combined with results of Gabber shows that this functor is an equivalence of categories if $S$ is excellent and if $\Lambda$ (or the quotient of $\Lambda$ by its ideal of definition) is a $\mathbb Z/n$-algebra for some $n$ invertible on $S$.
Under these assumptions, we therefore get a functor
$$ SH(S) \to D^{et}(S,\Lambda). $$
It is symmetric monoidal and colimit-preserving, since the two functors considered above are.
This étale realization functor sends a smooth $S$-scheme $p\colon X\to S$ to $p_!p^!\Lambda$. An alternative approach to this functor would be to directly use the universal property of $SH(S)$ established by Robalo in K-theory and the bridge from motives to noncommutative motives (Corollary 2.39). With no assumptions on $S$, it is clear that the assignment $(p\colon X\to S)\mapsto p_!p^!\Lambda$ satisfies Nisnevich descent, homotopy invariance, and $\mathbb P^1$-stability, but promoting it to a symmetric monoidal functor is where the difficulty lies, and where excellence is needed.<|endoftext|>
TITLE: Can a stochastic Turing machine output a consistent extension of PA with positive probability?
QUESTION [16 upvotes]: Suppose that we interpret the output tape of a Turing machine as an assignment of true or false to all sentences of PA, taking the $n$th output bit as the truth value of the sentence with Goedel number $n$. (We can ignore bits when no sentence has number $n$, or we can choose an encoding which is an onto function; it doesn't matter.) By the first incompleteness theorem, it is not possible for any Turing machine given finite input to output a complete and consistent assignment in this way. On the other hand, it's quite possible for a Turing machine given infinite input to do so, since this allows an oracle to be constructed.
Supposing that we give a Turing machine random input, is it possible for the machine's output to be complete and consistent with nonzero probability? While this seems implausible, I haven't been able to disprove it yet.
To be precise, suppose that we take some machine $T$ and feed it fair coin-flips whenever it asks for additional input. The outcomes of the process are the states of the output tape, which will be either finite or infinite binary strings depending on whether the machine ever stops producing output. To define the probabilities, consider the $\sigma$-algebra generated by finite input-sets: for each finite binary string $b$, there is a set of possible outputs which could result from the first random bits being $b$. Each of these sets is measurable. The remaining measurable sets are generated from these by closure under countable union and complement.
Using this $\sigma$-algebra, the set of complete and consistent extensions to $PA$ is measurable, as follows. Each possible contradiction with $PA$ is a finite truth-value combination, which is measurable as the countable union of all inputs to $T$ which lead to that particular combination of truth-values. (This may be the empty set.) The set of all such contradictions is also countable, and therefore measurable. We can also measure the finite bit-strings, since these possibilities are countable. The union of these two measurable sets are the possibilities which are either inconsistent with $PA$ or finite; by closure under compliment, we can measure those which are both infinite and consistent with $PA$.
Can that measure be nonzero?

REPLY [11 votes]: The answer is no, but it's almost yes. A stochastic Turing machine can find a diagonally non-computable ($\textsf{DNC}$) function ($f$ with $f(x)\ne\varphi_x(x)$ for all $x$) and finding a complete extension of PA is equivalent to finding a $\textsf{DNC}$ function with $f(x)\in\{0,1\}$ for all $x$.
Antonin Kucera, in Measure, $\Pi^0_1$ classes, and complete extensions of PA, in Recursion Theory Week, Lecture Notes in Mathematics 1141, 245-259, showed that a stochastic TM cannot find any completion of PA. (The fact that it cannot find any given one, as in Carl Mummert's answer, was known earlier.)

REPLY [8 votes]: Supposing that we give a Turing machine random input, is it possible for the machine's output to be complete and consistent with nonzero probability? 

The answer is no. By a theorem due to de Leeuw, Moore, Shannon, and Shapiro (1956), which was later stated in this terminology by Sacks, if a real $x \in 2^\omega$ is computable from every real $y\in 2^\omega$ in a set of positive measure, then $x$ is already computable.  (For references and background, see section 8.12 of Algorithmic Randomness and Complexity by Downey and Hirschfeldt).
Here the measure on $2^\omega$ is the standard fair-coin measure I will call $m$.  No completion of PA is computable, so no completion of PA is computable from a set of oracles of positive measure. 
The idea of the proof is that if $x$ is computable from a set of positive measure then there is an open subset $Z$ such that $x$ is uniformly computable from a subset $W$ of $Z$ such that $m(W) / m(Z) > 1/2$. But then we can compute $x$ up to any given precision by just enumerating computations from various elements of $Z$ until enough of them all give the same result (enough meaning more than $m(Z)/2$ of them). 
The set of completions of PA is closed in $2^\omega$, so it has to be measurable. It is easy to see that the set of completions of PA has measure zero in $2^\omega$, because we can list a sequence of sentences $(\phi_n)$ such that $\text{PA} \vdash \phi_n$ for each $n$. Each condition of the form $\text{PA} \vdash \phi$ divides the measure of the set of completions by $2$, so the overall measure must be zero.<|endoftext|>
TITLE: Koszul complex for non-Koszul algebras
QUESTION [17 upvotes]: Let $A$ be a graded, connected, locally finite, quadratic algebra over a field $k$; that is, $A$ may be presented as $T(V)/I$, where $V = A_1$ is a finite dimensional $k$ vector space, and the ideal $I$ of relations is generated in $T(V)_2 = V \otimes V$.  Let $A^!$ be the quadratic dual of $A$; $A^!$ is defined $T(V^*)/I^{\perp}$, where $I$ is the ideal generated by the complementary relations to those in $I$.
We may form the Koszul complex $K(A) = A \otimes (A^!)^*$, equipped with differential 
$$d(x \otimes \phi)= \sum_i v_i x \otimes v_i^* \phi$$
where $v_1$, ..., $v_n$ form a basis for $V$.  
When $A$ is Koszul, $K(A)$ is exact.  If $A$ is not Koszul, is there any reasonable or natural description of the homology of $K(A)$?

REPLY [7 votes]: OK, I think I worked this out on my own; I'm leaving the answer here in case others need it.
Let me write $B_*(A)$ for the bar complex of $A$, and $B^*(A) = Hom(B_*(A), k)$ for the cobar complex.  The latter is a differential (bi)graded algebra; its cohomology is $Ext_A^{*, *}(k, k)$.  The quadratic dual $A^!$ is isomorphic to the diagonal part $\sum Ext_A^{n, n}(k, k)$.  
In fact, there is a map $B^*(A) \to A^!$ of dgas (equipping $A^!$ with the trivial differential) which in cohomology induces the projection of the Ext algebra onto its diagonal part.  This makes $A^!$ a module for $B^*(A)$.  I claim that if we dualize the Koszul complex $K(A)$, we have
$$H^*(K(A)^*) \cong Tor_{B^*(A)}(k, A^!)$$
When $A$ is Koszul, the projection $B^*(A) \to A^!$ is a quasi-isomorphism (this is essentially one definition of Koszulity), so the right hand side is $k$, concentrated in degree $0$, recovering the fact that the Koszul complex is acyclic in this case.
We can see the claim as follows: Consider the bar resolution of $k$ over $A$; I may write this as the complex $(A \otimes B_*(A), d)$, where $d$ is the standard differential.  Notice that we can split $d$ up as $d = d_1 + d_2$, where $d_2$ is the internal differential in $B_*(A)$, and $d_1$ mixes together the two tensor factors.  We normally regard this as an $A$-resolution of $k$, free over $B_*(A)$.  However, we can turn this on its head and regard it as a cofree resolution of $k$ as a $B_*(A)$-comodule.  
Specifically, we map $k \to A \otimes B_*(A)$ by tensoring together the unit of $A$ with the unit of $B_*(A)$ (perhaps co-augmentation is a better term).  Since the unit is primitive, this is a map of $B_*(A)$-comodules.  We already know that this is an equivalence.  Thus this is a cofree resolution of $k$ over the dg-coalgebra $B_*(A)$, coinduced over $A$.
Dualizing, we get a free resolution $(A^* \otimes B^*(A), d^*)$ of $k$ over $B^*(A)$.  Then $Tor_{B^*(A)}(k, A^!)$ is the homology of the complex
$$[A^* \otimes B^*(A)] \otimes_{B^*(A)} A^! = A^* \otimes A^! = K(A)^*$$
The fact that the differential is the same as in the Koszul complex follows, for instance from Priddy's construction of the Koszul complex as a portion of the cobar resolution.<|endoftext|>
TITLE: A remarkable sum over partitions
QUESTION [8 upvotes]: While studying some seemingly unrelated topological questions, I have experimentally discovered what appears (to me) to be a remarkable sum over partitions.  I was wondering if anyone knows how to prove it.
Fixing $n \geq 1$, it can be stated as follows:
$$1=\sum_{(a_1^{k_1},\ldots,a_p^{k_p}) \vdash n} \left(\frac{1}{(a_1^{k_1}) (k_1 !)}\right) \left(\frac{1}{(a_2^{k_2}) (k_2 !)}\right)\cdots\left(\frac{1}{(a_p^{k_p}) (k_p !)}\right).$$
Here the sum is over all unordered partitions of $n$, and the symbol $(a_1^{k_1},\ldots,a_p^{k_p})$ denotes the partition where $a_1$ appears $k_1 \geq 1$ times, where $a_2$ appears $k_2 \geq 1$ times, etc, and where $a_1 > a_2 > \cdots > a_p > 0$.
I have numerically verified this up to $n=6$.

REPLY [15 votes]: Here is a more informative version of this identity. Let $Z_n$ denote the cycle index polynomial of the symmetric group $S_n$, namely
$$Z_n = \frac{1}{n!} \sum_{\sigma \in S_n} z_1^{c_1(\sigma)} z_2^{c_2(\sigma)} \dots $$
where $c_i(\sigma)$ denotes the number of $i$-cycles of $\sigma$. The observations in the comments boil down to the observation that
$$Z_n = \sum_{\sum ic_i = n} \prod_{i=1}^n \frac{z_i^{c_i}}{i^{c_i} c_i!}$$
and setting all $z_i = 1$ gives your identity, but without doing this, it is possible to arrange the $Z_n$ into a beautiful generating function, namely
$$\sum_{n \ge 0} Z_n t^n = \exp \left( \sum_{n \ge 1} \frac{z_i t^i}{i} \right).$$
This is the "permutation form" of the exponential formula, which has many applications in combinatorics. See, for example, Stanley's Enumerative Combinatorics: Volume 2 for an exposition, or this blog post.<|endoftext|>
TITLE: Categorical or simplicial introduction to modern homotopy theory
QUESTION [11 upvotes]: I've heard some people saying that it would be easier to study homotopy theory first through simplicial sets. First, i thought that these people were not being serious. But it caught me wondering...
It's not a secret that nowadays homotopy theory is not just a branch of algebraic topology, it's a actually an enourmous mathematical field of research subsuming category theory, homological algebra and having deep connections with algebraic geometry and commutative algebra.
I wonder if it's worth it to think about introducing homotopy theory as an independent subject, for example, build everything axiomatically relying on a category theory machinery rather than introduct it as a subject of algebraic topology.
I'm not sure such references do exist, maybe, Emily Riehl's book "Categorical homotopy theory" can be thought as a introduction to homotopy theory for category theorists(thought she advises to read on simplicial homotopy theory and homotopy limits and colimits before approaching the book, but says it's not a formal prerequisiste ).
P.G. Goerss, J.F. Jardine "Simplicial homotopy theory" seems to assume algebraic topology, though I haven't seriously researched whether it truly does so.
So, to sum it up, it is a question about both whether such approach can be viable and whether it's already used in some literature. Of course, anything else close to this topic will be appreciated as well.

REPLY [2 votes]: I think I should give more information on my comment. 
One of the starting points for homotopy was Poincaré's notion of the fundamental group, in which one took homotopy classes of loops at a base point. Later homology was proved homotopy invariant, and then Cech, and later more forcefully, Hurewicz, introduced the homotopy groups. One of the advantages of taking homotopy classes is that could reduce uncountable models to countable, or even finite ones. Another was modelling the intuitive idea of deformation. 
These both reside on the idea of composition: you compose paths, you compose homotopies. And in category theory, you compose morphisms. 
Thus it is reasonable to look for higher compositions. And this was the reason why Dan Kan's thesis and first paper were cubical. However cubical sets were found to have to have two disadvantages: 

Cubical groups were not necessarily Kan, unlike simplicial groups. 
The realisation of the cartesian product of cubical sets had the wrong homotopy type, though that of a tensor product was OK.

So cubical sets were kind of abandoned for the development of algebraic topology. 
However cubical methods still have  advantages some of which are discussed in this mathoverflow discussion on $n$-fold categories.  One is the rule that $I^m \times I^n \cong I^{m+n}$; another is the easy formulation of the idea of multiple higher composition, and so of "algebraic inverses to subdivision", and hence applications to "local-to-global" questions.  These advantages are exploited in the book I refer to; the main results on Higher Seifert-van Kampen Theorems would hardly have been conjectured, let alone proved,   in traditional terms. The results do  require the notion of cubical set with connections. This innovation has led to  the above specific difficulties being  resolved to some extent. 
There is a large literature on simplicial sets and their  applications, and it seems that all talk of $(\infty,n)$-categories is in simplicial terms. Also $n$-fold categories need a geometric aspect which is broader than the usual cubical theory, in which the faces in different directions are all of the same type. 
It seems to me that young people should try to be eclectic, and assess and evaluate different approaches. It all may depend on the problem at hand, or the next problems to come. Or simply for expressing intuitions. 
Nov 23: For a discussion of "Anomalies in algebraic topology" see this presentation in Galway, Dec  2014.<|endoftext|>
TITLE: Stable manifolds of a sequence of Morse functions
QUESTION [6 upvotes]: Let $\ f_n \ $ be a sequence of Morse functions on $\mathbb{R}^d$, adequately converging (in the $C^2$-topology, say) to a limit Morse function $\ f$:
$$ f_n \to f  \ .$$
At any critical point $\ p\ $ of the limit function $\ f$, it can be proved that there exists an open neighborhood $\ U\ $ such that, for $n \geq n_0$, 
$$ f_{n|U} \mbox{ has a unique critical point, $p_n$, of the same Morse index than } p. $$
Moreover, the sequence of these critical points $p_n$ converges to $p$.
On the other hand, let $\ W^s(p)\ $ stand for the stable manifold associated to $p$:
$$ W^s(p) :=  \left[ \ x \in \mathbb{R}^d \ \colon \  \lim_{t \to \infty} \tau_t (x) = p \ \right] , $$
where $\tau_t(x)$ denotes the integral curve of the gradient of $f$ passing through $x$ at $t=0$. Analogously, let $\ W^s(p_n)\ $ stand for the stable manifold associated to $p_n$ (and the gradient of $f_n$)
My question is:
Do the stable manifolds $\ W^s(p_n)\ $ converge (in an adequate sense) to $W^s(p)$?
For example, does the Hausdorff distance $d_H (W^s(p_n) , W^s(p))$ converge to zero?
In dimension 1, the biggest stable manifolds are intervals, which are delimited by critical points, and it is not difficult to check that this distance $\ d_H ( W^s(p_n) , W^s(p))\ $ converges to zero.
In dimension 2, analyzing the integral curves that separate stable manifolds, I also think we can prove $d_H(W^s(p_n), W^s(p))$ goes to zero.
Nevertheless, in the general case ($d \geq 3$), this method seems to fail.
I am working with a colleague on a quite far topic, but we have come to this problem. We are both totally new in this area, although our intuition is that the question above should have an affirmative answer, at least for a wide class of nice functions.

REPLY [5 votes]: The answer is no. Of course, in a neighbourhood of a critical point,
one has convergence.
Assume that the stable manifolds $W^s(p_n)$ converge to some set $W$ in the Hausdorff distance.
It can happen that there is a sequence $q_n\in W^s(p_n)$ that converges
to a critical point $q\in W$ that lies in the interior of $W$ if $W$ can be viewed as a submanifold of $\mathbb R^d$ with singularities. This can "break" the stable manifold, so $W\setminus W^s(p)$ can be a large set.
Here is an example with $d=2$ and two critical points, assuming that we take the positive gradient flow. Instead of $n$, let $t\in\mathbb R$ be the parameter. In a neighbourhood of $p_t=(t,1)$, $f_t$ is given by
$$f_t(x,y)=\frac{(x-t)^2-(y-1)^2}2\;,$$
so the stable manifold will contain a vertical
interval $\{t\}\times(1-\epsilon,1+\epsilon)$.
In a neighbourhood of the $x$-axis, the function is given by
$$f_t(x,y)=\frac{y^2-x^2}2-c$$
for some suitable $c>0$. Then there is a critical point $q_t=(0,0)$ with $W^s(q)=\mathbb R\times\{0\}$.
Finally, assume that $f_{-t}(x,y)=f_t(-x,y)$, and that
$$\frac{\partial f_t(x,y)}{\partial x}\le 0\text{ for $t>0$, $x=0$ and $y\in(0,1)$.}$$
For $t> 0$, the stable manifold of $p_t$ will descend and then "turn right", so that it becomes asymptotic to the positive $x$-axis.
For $t=0$, the stable manifold will run into the origin and stop there. But $\lim_{t\searrow 0}W^s(p_t)$ will be the union of $W^s(p_0)$ and the positive $x$-axis.
For $t<0$, the behaviour of $W_s(p,t)$ will be asymptotic to the negative $x$-axis. This tells you that one cannot even define an "extended stable manifold" of $p_0$ to which $W_s(p,t)$ will converge both for $t\searrow 0$ and for $t\nearrow 0$.<|endoftext|>
TITLE: No nonconstant coprime polynomials $a(t)$, $b(t)$, $c(t) \in \mathbb{C}[t]$ where $a(t)^3 + b(t)^3 = c(t)^3$
QUESTION [7 upvotes]: See David Speyer's answer here.

I saw Brian Conrad give an excellent one hour talk to undergraduates where he proved that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$ and $c(t) \in \mathbb{C}[t]$ such that
  $$a(t)^3 + b(t)^3 = c(t)^3.$$
  He gave an elementary proof, then outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve.

Could anyone expand a bit on the "outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve" part? That's not so clear to me. Thanks in advance!

REPLY [4 votes]: Let $a(t)$, $b(t)$, $c(t)$ have degree $n_1$, $n_2$, $n_3$, respectively, and $n = \text{max}(n_1, n_2, n_3)$. Then we can define $A(u, v)$, $B(u, v)$, $C(u, v)$ as homogeneous polynomials of degree $n$ such that$$A(u, 1) = a(u),\text{ }B(u, 1) = b(u),\text{ }C(u, 1) = c(u).$$Then, by construction,$$A(u, v)^3 + B(u, v)^3 = C(u, v)^3.$$Let$$E = \{(x, y, z) \in \mathbb{P}^2 : x^3 + y^3 = z^3\}.$$$E$ is a smooth curve of genus $1$, i.e. an elliptic curve. Now, define a map$$\varphi: \mathbb{P}^1 \to E,\text{ }(u, v) \mapsto (A(u, v), B(u, v), C(u, v)).$$This map is well-defined since $A(u, v)$, $B(u, v)$, $C(u, v)$ are homogeneous polynomials of the same degree which do not vanish simultaneously. Moreover, this map is nonconstant and proper since its source is projective. As the image of a proper map is closed and it is not a point, and $E$ is an irreducible $1$-dimensional variety follows that $\varphi$ is a surjective morphism. $E$ is a topologically a torus, i..e it has genus $1$ and there is a one up to scaling differential form $\tau$ on it. This will imply that its pullback $\varphi^*\tau$ is a differential form on $\mathbb{P}^1$, which is impossible since it has genus $0$. In more formal terms, a surjective map $\mathbb{P}^1 \to E$ gives rise to the injection $H^0(E, \Omega^1) \to H^0(\mathbb{P}^1, \Omega^1)$. But this is absurd, since the former is a vector space of dimension $1$ and the latter is a vector space of dimension $0$.

Perhaps you might want to say something about why the pullback of a non-zero holomorphic differential is non-zero.

$X$ and $Y$ are Riemann surfaces and $f: X \to Y$ holomorphic differential form looks like $g(z)\,dz$, where $g$ is a holomorphic function. Then its pullback is locally given by $g(f(z))\,df$. It is obviously nonzero as long as $f$ is nonconstant.<|endoftext|>
TITLE: Computer Algebra Systems that support variable sized matrices
QUESTION [8 upvotes]: I'm familiar with sympy, the matlab symbolic package, reduce, and have tried out a few other computer algebra systems. However, as far as I can tell, none of them seem to be able to do algebra on variable sized matrices - they can only work with fixed sized matrices.
Are there any that can do algebra for variable sized matrices? I understand there would be quite a few gross cases but I feel like there is a significant amount that is doable simply because of the ease of many simplifications/algebra one can do by hand in with matrices in R^nxn.
It is possible to just work with non-communiative algebraic elements in many of these, and so that covers addition and Hadamard product with matrices, which is useful and a start. However it covers a very small portion of what one actually does with matrices (say, transpose, inverses, eigenvalue decomposition, using matrices in R^nxm, etc.). Does any more general software exist?
(this question is reposted from here because mathoverflow seemed like a more appropriate place to ask this question)

REPLY [9 votes]: SymPy has a matrix expressions module that does this. Example:
>>> from sympy import MatrixSymbol, Matrix, symbols
>>> n, m = symbols('n m', integer=True)
>>> X = MatrixSymbol('X', n, m)
>>> Y = MatrixSymbol('Y', m, n)
>>> (X*Y).T
Y'*X'

Matrix expressions can have symbolic sizes (like n and m) or explicit integer sizes, in which case they can be combined with explicit matrices. 
It's also worth noting that there are a lot of things that aren't documented in the doc page I linked to, so take a look at https://github.com/sympy/sympy/tree/master/sympy/matrices/expressions for the full functionality. 
You can also look at this talk which uses this module, along with some modules that are separate from SymPy.<|endoftext|>
TITLE: BSD and congruent numbers
QUESTION [11 upvotes]: Let $n$ be a positive integer, and let $E_n$ denote the elliptic curve $y^2=x^3-n^2x$. By work of Tunnell, it's known that if $E_n$ satisfies the BSD conjecture, then there is an algorithm to decide whether $n$ is a congruent number or not. By work of Bhargava and Shankar, we also know that a positive proportion of elliptic curves satisfy BSD. Is it known whether a positive proportion of the curves $E_n$ satisfy BSD?

REPLY [12 votes]: Bhargava and Shankar's work is only part of the story.
In order to say anything about BSD, we certainly have to be able to say something about the analytic rank.  Sometimes we can get lucky and access the analytic rank directly: in quadratic twist families, Waldspurger's formula relates critical $L$-values to coefficients of half-integral weight modular forms.  Tunnell's theorem is proved using Waldspurger's result, and for some elliptic curves, this can go further and be used to say that a positive proportion of twists have non-vanishing $L$-value.  For example, if $E$ is the elliptic curve $X_0(14)$, then Kevin James (http://www.math.clemson.edu/~kevja/PAPERS/JAMS.pdf) showed that at least 7/64 of the (negative) quadratic twists have analytic rank 0.  By Kolyvagin's theorem, analytic rank 0 implies arithmetic rank 0, so BSD is satisfied for these twists.  James's work requires the associated half-integral weight modular forms to satisfy certain properties modulo 3, which I believe are not satisfied for the congruent number elliptic curve.  (I didn't check this carefully.)
One other way in which we know how to control the analytic rank is via results on $p$-adic $L$-functions, using work of Skinner, Urban, and W. Zhang.   Roughly speaking, these results say that if $E$ is moderately well-behaved at an odd prime $p$, and its $p$-Selmer group has rank either 0 or 1, then analytic rank = $p$-Selmer rank = arithmetic rank.  Bhargava and Shankar study the average size of the $p$-Selmer group in families defined by congruence conditions, $p\leq 5$, and they're able to find a large, well-behaved family of curves for which the average size of the $p$-Selmer group is $p+1 < p^2$.  Thus, a positive proportion of these curves have $p$-Selmer rank either 0 or 1, and we're able to apply the results of Skinner, Urban, and Zhang.  (More care is needed, but this is the gist.  A souped up version of this argument is in Bhargava's paper with Skinner and Zhang.)
You should now ask whether we can understand the distribution of $p$-Selmer groups in families of quadratic twists.  This set is too small for Bhargava and Shankar's approach to reach it, but if $p=2$, there's another way!  The key here is that, for any quadratic twist, $E_n[2] \simeq E[2]$, as Galois modules.  Now, the $2$-Selmer group of $E_n$ sits inside $H^1(\mathbf{Q},E_n[2])$, but this $H^1$ is isomorphic to $H^1(\mathbf{Q},E[2])$, so we can view the varying $2$-Selmer groups as landing in the same target space.  This gives us some traction, and indeed, if $E$ has full $2$-torsion, we can use this to find the full probability distribution on $\dim_{\mathbf{F}_2}\mathrm{Sel}_2(E_n/\mathbf{Q})$.  This was carried out for the congruent number elliptic curve by Heath-Brown, and for general curves with full two-torsion by Dan Kane.  In particular, for $5/16$ of the twists, the rank of $E_n(\mathbf{Q})$ is provably 0.  This says nothing about the analytic rank, though, other than that it's even (this follows from the proof).  To understand the analytic rank via Skinner, Urban, and Zhang, we'd need to be able to access $p$-Selmer for odd primes $p$, and here, we lose the fact that the $p$-Selmer groups of the twists all sit inside the same space.  Thus, we've lost our traction, and we have no idea how to start.
Thus, it seems very likely that this is still not known.  That said, I'm not familiar with the results of Shou-Wu Zhang that Jeremy mentioned, so I'd happily be corrected!<|endoftext|>
TITLE: Existence of a continuous section
QUESTION [7 upvotes]: Let $f\colon X\to Y$ be a surjective continuous map between two topological spaces such that $X,Y$ are path-connected and such that every fibre $f^{-1}(y)$ is connected, for each $y\in Y$. Is there always a continuous section $s\colon Y\to X$ (i.e. a continuous map $s$ such that $f\circ s$ is the identity on $X$)?
If no, what kind of sufficient conditions could we impose on $f$ ? For example, is it better if $Y$ is supposed to be simply connected? Or if we choose $Y$ to be the interval $[0,1]$ ?

REPLY [6 votes]: Asking when a continuous map $f : X \to Y$ has a continuous section is analogous to asking when a Diophantine equation over $\mathbb{Z}$ has a solution over $\mathbb{Z}$; see, for example, this blog post. This suggests that in general the problem will be hard and that there will be many obstructions, both local and global, and both of these are true.
For example, asking that $f$ be surjective, or equivalently that the fibers $f^{-1}(y)$ be nonempty, is the analogue of asking that the Diophantine equation have a solution $\bmod p$ for all primes. The next most local thing to ask for is that around each $y \in Y$ there is a neighborhood $U$ on which $f$ has sections; this is loosely the analogue of asking that the Diophantine equation have $p$-adic solutions for all primes. This is true, for example, for all covering maps, but of course there is a global obstruction to having a section in this case, namely that the induced map on $\pi_1$ must also have a section. 
Sufficient conditions are hard to come by, but I can list a bunch of necessary conditions. The problem becomes much more amenable to homotopy theory if $f$ is a fibration; then all of the obstructions are global and homotopy-theoretic in nature and it is possible in principle to completely describe them.<|endoftext|>
TITLE: What is an example of a colimit-dense generator which is not dense?
QUESTION [5 upvotes]: An object $G$ of a category $\mathcal{C}$ is a dense generator if every object $X$ is the colimit of the canonical diagram of copies of $G$ mapping to $X$. (This canonical diagram is indexed by the full subcategory of the slice $\mathcal{C}_{/X}$ on the objects of the form $G \to X$.)
An object $G$ is called a colimit-dense generator if every object $X$ is a colimit of some diagram of copies of $G$. (That means $X$ is the colimit of some functor $F : \mathcal{I} \to \mathcal{C}$ with $F(i) = G$ for all objects $i$.)
Is there an example of a colimit-dense generator which is not a dense generator?

REPLY [9 votes]: $\Bbb R$ is colimit-dense in the category of real vector spaces but not dense
(see 6.F, 6.34 in my book with J. Adámek "Locally presentable and accessible categories").<|endoftext|>
TITLE: Upper bound on level of a congruence subgroup of the modular group
QUESTION [5 upvotes]: Let $\Gamma = PSL(2,\mathbb{Z}) = \langle S,T \ | \ S^2=(ST)^3=1 \rangle$. Let $G$ be some mystery normal subgroup of $\Gamma$ that we happen to think may be congruence. Recall that a subgroup of $\Gamma$ is a congruence subgroup of level $N$ if it contains the principal congruence subgroup of level $N$.
Are there methods for establishing upper bounds on the possible levels of $G$, say given information about $\Gamma/G$? Establishing lower bounds is trivial: the minimal $N$ such that $T^N \in G$ (if such an $N$ exists) is the minimal possible level of $G$. 
For example, suppose $\Gamma/G \cong (\mathbb{Z}_3 \times \mathbb{Z}_9) \rtimes S_3$ or more generally of the form (finite abelian group)-semidirect product-(symmetric group). I'm generally thinking about the case where $G$ is the kernel of some unitary representation with finite image.

REPLY [4 votes]: If you only look at upper bounds, you can use the result by Lubotzky: For every algebraic group G, there is some $c$, such that the level of a congruence subgroup $U$ is bounded by the $c(G:U)$.
If you want to determine the level algorithmically, you can look at the image of the subgroup in $PSL(\mathbb{Z}/p^n\mathbb{Z})$. Determine the cusp split of the subgroup, if there are two cusp width $c_1, c_2$, and $p$ a prime with $p\nmid c_1c_2$, then $U$ contains different parabolic elements of order coprime to $p$, hence $U\Gamma(p)/\Gamma(p)$ contains two different parabolic subgroups, hence $U\Gamma(p)/\Gamma(p)=PSL(\mathbb{F}_p)$, thus $p$ does not divide the level.<|endoftext|>
TITLE: When is the closed unit ball in a smaller Banach space closed in a larger Banach space?
QUESTION [12 upvotes]: Recently I saw an interesting lemma:
For any $s>0$, the closed unit ball in $H^s$ is also closed in the $L^2$ norm. That is, suppose $u_j\in H^s$ and $\|u_j\|_{H^s}\le 1$. Suppose $u_j\to u$ in $L^2$. Then $u\in H^s$ and $\|u\|_{H^s}\le 1$.
A possible proof of the above lemma is to take the Fourier transform and use Fatou's lemma. Another possible proof is to use Riesz representation theorem to find out successively higher derivatives of $u$ (seems to work only when $s$ is a positive integer, though).
Contemplating the above proofs, it seems to me that there is some sort of compatibility between $H^s$ and $L^2$ for the above lemma to hold. For a nonexample, it is easy to see that the closed unit ball in $C^0([0,1])$ is not closed in $L^2$ norm simply by taking a sequence of continuous functions converging in $L^2$ to $1_{[0,1/2]}$, for example.
That brings me to the question: Let $Y$ be a dense subspace of a Banach space $X$. What condition on the pair $(X,Y)$ makes the closed unit ball in $Y$ also closed in $X$? If there is no simple iff condition, is there any nontrivial sufficient/necessary condition? Among the common spaces in Analysis (e.g. Holder spaces or Sobolev spaces), is there a list of all the pairs for the above property to hold?
Thanks for your time.

REPLY [6 votes]: Gonzalez pointed to some deeper results on semi-embeddings.  When $X$ is reflexive, there is an elementary characterization: $B_Y$ is closed in $X$ iff there is a $Z$ s.t. $Y = Z^\ast$  isometrically and the inclusion map $J$ from $Y$ into $X$ is weak$^\ast$ to weak continuous. This is automatic if $Y$ is reflexive and cannot happen if $Y$ is not isometrically isomorphic to a dual space; in particular, if $Y$ is $C[0,1]$, $c_0$, or $L_1$.<|endoftext|>
TITLE: Rational points on the "quintic circle" $x^5 + y^5 = 7$
QUESTION [20 upvotes]: I suspect that the curve $x^5 + y^5=7$ has no $\mathbb Q$ points, and a brief computer search verifies this hypothesis for denominators up to $10^4$. What techniques can be used to show that there are no solutions?

REPLY [40 votes]: There is an action of $\mu_5$, the group of fifth roots of unity, on your curve,
given by $\zeta \cdot (x,y) = (\zeta x, \zeta^{-1} y)$. The quotient by this
group action is the hyperelliptic curve
$$C \colon Y^2 = X^5 + \frac{49}{4},$$
the map being given by $(X, Y) = (-xy, x^5 - \frac{7}{2})$. So it is enough
to find all the rational points on $C$. $C$ is isomorphic to
$C' \colon Y^2 = 4 X^5 + 49$. By a 2-descent, one can show that the Jacobian
variety of $C'$, $J'$, has Mordell-Weil rank (at most) 1, and since one
finds a point of infinite order
($(x^2 - \frac{10}{9} x - \frac{10}{9}, \frac{200}{27} x - \frac{61}{27})$
in Mumford representation), Chabauty's method is applicable and shows
that $\infty$, $(0, \pm \frac{7}{2})$ are the only rational points on $C$.
This implies that there are no rational points on your affine curve.
Here is Magma code to check this:

P<x> := PolynomialRing(Rationals());
C := HyperellipticCurve(4*x^5 + 49);
J := Jacobian(C);
RankBound(J); // --> 1
ptsJ := Points(J : Bound := 500); // the first 5 are torsion
Chabauty(ptsJ[6]); // --> { (0 : -7 : 1), (1 : 0 : 0), (0 : 7 : 1) }

(If you do not have direct access to Magma, you can try this out
with the online Magma calculator at http://magma.maths.usyd.edu.au/calc/ .)

REPLY [14 votes]: In his 1825 paper, Lejeune Dirichlet proved that the equation $x^5 + y^5 = cz^5$ has no nontrivial solution with x and y coprime integers and z integer for a rather large class of integers c. His paper can be read on the site Gallica :
http://gallica.bnf.fr/ark:/12148/bpt6k5842885h/f2.item.zoom
but the visualization is very bad.<|endoftext|>
TITLE: Examples of Noetherian overkill
QUESTION [23 upvotes]: I have read in many places that the noetherian hypothesis is often overkill - both in commutative algebra and in ($\overset?=$) algebraic geometry. In particular, I've read that coherence and finite presentation are the really important properties. This is nicely stated in the foreword to Quitté and Lombardi's Commutative Algebra - Constructive Methods. An excerpt from it appears in Darij Grinberg's answer to this question. I have included it below. Martin Brandenburg's comment on the same question mentions some results true "mainly" for Noetherian rings, e.g $\dim(R[T])=\dim(R)+1$. In this question, an elementary characterization of Krull dimension by Lombardi and Coquand is mentioned in hopes it would provide a simpler proof. Since noetherianity is non-constructive, perhaps it is actually overkill here as well...
My problems are I don't know any examples, don't have enough time to dive into the proof of every result involving Noetherianity, and don't have any intuition to feel when it's overkill.
So I'm looking for nice examples of facts which are often stated for Noetherian rings/schemes but really only require, say, (quasi)coherence. If that's too optimistic, maybe some "right theorems" in the sense of the excerpt.

Finally, let us mention two striking traits of this work compared to
  classical works on commutative algebra.
The first is to have left
  Noetherianity on the backburner. Experience shows that indeed
  Noetherianity is often too strong an assumption, which hides the true
  algorithmic nature of things. For example, such a theorem usually
  stated for Noetherian rings and finitely generated modules, when its
  proof is examined to extract an algorithm, turns out to be a theorem
  on coherent rings and finitely presented modules. The usual theorem is
  but a corollary of the right theorem, but with two nonconstructive
  arguments allowing to deduce coherence and finite presentation from
  Noetherianity and finite generation in classical mathematics. A proof
  in the more satisfying framework of coherence and finitely presented
  modules is often already published in research articles, although
  rarely in an entirely constructive form, but “the right statement” is
  generally missing in the reference works.

REPLY [13 votes]: 1) It is sometimes stated that a finitely generated module $M$ over a Noetherian commutative ring $R$ is projective if for all maximal ideals $\mathfrak m\subset R$ the localized module $M_\mathfrak m$ is free over $R_\mathfrak m$.
However the noetherianity of $R$ is unnecessary if you add the hypothesis that $M$ is finitely presented over $R$.  
2) Similarly, is  a finitely generated  flat  module $N$ over  $R$ projective?
The answer is yes if $R$ is noetherian, but noetherianity is not necessary if you know that $R$ is an integral domain or if you know that $N$ is finitely presented.    
3) Finally let me mention Kaplansky's extraordinary theorem:   

Every projective module over a local ring is free 

The ring doesn't need to be noetherian (nor  commutative!) and the module needn't even be finitely generated!<|endoftext|>
TITLE: Subgroups of hyperbolic groups
QUESTION [5 upvotes]: Let $G$ be a finitely generated hyperbolic group, and let $H \leq G$ be a subgroup whose profinite completion is finitely generated. Must $H$ be finitely generated?

In view of Ian Agol's answer, I am ready to assume that $G$ is
  residually finite.

REPLY [5 votes]: Let $A$ be a finitely generated group, and let $\beta \colon A \to A$ be an injective homomorphism which is not surjective. Freely construct a group $G$ generated by $A$ and some formal element $t$ such that the equality $tat^{-1} = \beta(a)$ holds in $G$ for each $a \in A$. $G$ is called the strict ascending HNN extension of $(A, \beta)$. Set $$H := \bigcup_{n=0}^{\infty} t^{-n}At^n$$ where the union is taken in $G$. $H$ is a strictly ascending union of finitely generated subgroups of $G$ which are all isomorphic to $A$. It follows that $H$ is a subgroup of $G$ which is not finitely generated (if it were, the union could not be strictly ascending). On the other hand, every finite image of $H$ is clearly a finite image of one of the groups in the union (which is isomorphic to $A$) and can thus be generated by $d(A)$ elements. It follows that the profinite completion of $H$ is finitely generated.
To answer my question it suffices to choose $A$ and $\alpha$ in a way that will make $G$ hyperbolic. Take $A$ to be the free group on $x,y$ and define $\beta$ by $\beta(x) = xy$ and $\beta(y) = yx$. It is easy to see that $\beta$ is injective but not surjective. In this case, $G$ is the Sapir group, and its hyperbolicity is established in Theorem 4.1 of http://arxiv.org/pdf/1302.5370.pdf
If we want an answer to the extended question, i.e. with $G$ residually finite, then we can take $A$ to be the free group on $x,y$ and define $\beta$ by $\beta(x) = xy^{-1}x^2y$ and $\beta(y) = yx^{-1}y^2x$. Again, it is easy to see that $\beta$ is injective but not surjective. By Theorem 4.2 of http://arxiv.org/pdf/1302.5370.pdf the resulting $G$ is hyperbolic and linear over $\mathbb{Z}$, and thus, residually finite.<|endoftext|>
TITLE: How do I prove that compact-open topology is metrizable?
QUESTION [5 upvotes]: Let $X$ be a $\sigma$-compact topological space and $(Y,d)$ be a metric space.
Let $\{K_n\}$ be a sequence of compact subsets of $X$ whose union is $X$.
Define $\rho_n(f,g):=\sup \{d(f(z),g(z)): z\in K_n\}$ and $\rho(f,g)=\sum_{n=0}^\infty (\frac{1}{2})^n \frac{\rho_n(f,g)}{1+\rho_n(f,g)}$ for all $f,g\in C(X,Y)$.
Then, it can be directly checked that $\rho$ is a metric on $C(X,Y)$.
Assuming $K_n\subset int(K_{n+1})$ for all $n$, it can be easily shown that $\rho$ induces the compact-open topology on $C(X,Y)$. However, this conditions seems too strong.
In Conway's functions of one complex variable text, it's written there that if $X$ is Baire in addition, then it can be still shown that $\rho$ induces the compact-open topology. However, I don't underatand why. How is it so? Is there any reference for it?

REPLY [5 votes]: Some more remarks: also note that you can embed $(Y,d)$ at zero price into a Banach space by the Fréchet-Kuratowski isometric embedding, so w.l.o.g. you can assume $Y$ itself is  a Banach space. The compact–open topology makes $C(X,Y)$ a LCTVS, with the family of seminorms $ \{  \|\cdot\|_{\infty,K}  \}   $ for all $K\Subset X$. It is metrizable iff it is first-countable iff $X$ is hemicompact, that is the family of its compact subsets has countable cofinality.<|endoftext|>
TITLE: Analytic continuation for $L$-functions of elliptic curves
QUESTION [7 upvotes]: Let $E$ be an elliptic curve over a number field.
When $E$ has no CM and is a $\mathbb Q$-curve (i.e. it is $\overline{\mathbb Q}$-isogenous to all of its conjugates), it is nowadays known that $E$ is a factor up to isogeny of $J_1(N)$. In a nice paper X. Guitart and J. Quer define a subclass of $\mathbb Q$-curves, called strongly modular curves, which enjoy the property that their $L$-function is a product of $L$-functions of newforms. Thus, the $L$-function of these curves can be analitically continued to $\mathbb C$.
If $E$ has CM, the $L$-function of $E$ is known to admit an analytic continuation to $\mathbb C$ because it coincides with a product of $L$-functions of Hecke characters.
My first question is: are $L$-functions of elliptic curves with CM also products of $L$-functions of (classical) newforms? It is not clear at all to me what is the connection between $L$-functions of cuspforms and $L$-functions of Hecke characters.
My second question is: what is the state of art of the problem? Namely, is there any other class of elliptic curves whose $L$-function is known to have an analytic continuation?

REPLY [7 votes]: As said in the comment above, the answer to your first question is no. Hecke L-functions are expected to factor as a product of irreducible cuspidal automorphic representations of $\mathrm{GL}_n(\mathbb{A})$ (they are known to be a product of Artin L-functions, but we don't know those are entire in general). In the case of an elliptic curve with CM (using Deuring's theorem) this means
$$L_E(s)=L(s,\chi)L(s,\bar\chi)=\prod_{i=1}^r L(s,\pi_i)^{e_i}$$
But in general those representations $\pi$ are $n$-dimensional.
To see this in the simplest case, consider a Hecke L-function with trivial character, that is, $L(s,1)=\zeta_K(s)$, the Dedekind zeta function of the number field. We know from Aramata-Brauer (in black) and conditionally on the strong Artin conjecture (in red) that
$$\zeta_K(s)=\zeta(s)\prod_{\rho \neq 1} L(s,\rho)=\color{red}{\zeta(s)\prod_{\pi \neq 1} L(s,\pi)}$$
where the product goes over all the nontrivial irreducible representations of $\mathrm{Gal}(K/\mathbb{Q})$.
Now, in order for $\zeta_K$ to be expressible in terms of classical newforms, all (or some, depending on how one interprets your question) those representations of that Galois groups have to be of degree $2$. But that's definitely not true. Number fields with $\mathrm{Gal}(K/\mathbb{Q}) \cong S_n$, $n>4$ are alredy a counterexample.
Of course classical newforms also show up, but I'm not sure that there's a nice characterization of when that happens, depending on $E$, $K$ and $\mathrm{End}_K(E)\otimes\mathbb{Q}$.
Regarding the second question, analytic continuation (a special case of the Hasse-Weil conjecture) is only known in general when $E$ has CM (Deuring), when $E$ is defined over $\mathbb{Q}$ (Wiles-Breuil-Conrad-Diamond-Taylor) and when $K$ is real quadratic (Freitas-Hung-Siksek).<|endoftext|>
TITLE: $p$-torsion of an abelian variety of $p$-rank $0$
QUESTION [7 upvotes]: Let $k$ be an algebraically closed field of characteristic $p > 0$ and let $A$ be an abelian variety over $k$ such that $A[p](k) = 0$, i.e., such that $A$ has $p$-rank $0$. If I am not mistaken, this implies that $\mathrm{Ker}(F) \subset A[p]$, where $\mathrm{Ker}(F)$ is the Frobenius kernel of $A$. Is it true that $A[p] = \mathrm{Ker}(F^2)$, i.e., that $A[p]$ is the $p^2$-Frobenius kernel of $A$? Equivalently, is $I^{p^2} = 0$, where $I$ denotes the augmentation ideal of $A[p]$?
I know that the claim is true for elliptic curves but am confused about the higher dimensional case...

REPLY [2 votes]: An explicit example is the Jacobian of the hyperelliptic genus-$3$ curve in characteristic two, $Y^2+Y=X^7$. The vertices of the polygon are $(0,0)$, $(3,1)$, and $(6,3)$, giving slopes of $1/3$ and $2/3$.<|endoftext|>
TITLE: Is this almost-cosimplicial object familiar?
QUESTION [6 upvotes]: I have a sequence $X_0,X_1,\ldots$ of Abelian groups, along with face maps $d_0,\ldots, d_n: X_{n-1}\to X_n$ which satisfy the standard cosimplicial identities EXCEPT at the very bottom, where in $\mathrm{Hom}(X_0, X_2)$, I have
$$d_2d_0 \ne d_0d_1$$
However, I have one extra map $d_\infty:X_1\to X_2$ which satisfies the following relations in $\mathrm{Hom}(X_0,X_2)$
$$
d_\infty d_0 = d_2d_0
$$
$$
d_\infty d_1 = d_0d_1
$$
and the following relations in $\mathrm{Hom}(X_1,X_3)$
$$
d_0d_\infty = d_1d_\infty
$$
$$
d_2d_\infty = d_3d_\infty
$$
By hand, I can build a chain complex out of $X_\cdot$, adding a $d_\infty$ to the usual cosimplicial differential. The relations above imply that $d^2=0$.

Is this familiar to anyone?

At first I had hoped that I just had the "wrong" face maps and if I changed around my face maps $X_1\to X_2$ in a clever enough way that this would become a regular (semi)-cosimplicial Abelian group. But I don't see it. 
Another thought I had is that I have many other $d_\infty$ maps that I just haven't seen yet because I don't know the relations that they should satisfy and so am not looking in the right place.
I also noticed that $d_\infty$ satisfies the relations I'd expect from the map $[1]=\lbrace 0,1\rbrace$ to $[2]=\lbrace 0,1,2\rbrace$ which is the constant map to $1$. But this seems not to lead anywhere because of the failure of the face identities at the bottom; i.e., the maps $X_0\to X_1$ and $X_1\to X_2$ don't act like regular injections of ordered sets.

REPLY [6 votes]: I will also talk about simplicial things rather than cosimplicial things.  Furthermore, I will say "semi-trapezoidal" instead of "almost-simplicial". "Semi-" because there is no codegeneracy maps, and "trapezoidal" because an element $x$ of $X_2$ can be visualized as a trapezoid with four edges $d_2(x), d_\infty(x), d_0(x)$ and $d_1(x)$.
There is a way to make a semi-trapezoidal set into a semi-simplicial set. Given a semi-trapezoidal set $X$ define a semi-simplicial set $A$ setting
$A_0 = X_0, A_1 = X_1 + 2X_2, A_2 = 4X_2, A_3 = X_3 + X_2$ and $A_n = X_n$ for $n > 3.$
Roughly speaking, what we are doing is: We leave the $0$-simplices unchanged. For $1$-simplices we take those of $X$, and add to them the diagonals of the trapezoids (directed in a certain way). For the set of 2-simplices we take the set of triangles in which trapezoids are divided by their diagonals (there are four triangle for each trapezoid). Each trapezoid becomes a 3-simplex of $A$, as well as each original 3-simplex of $X$ becomes a 3-simplex of $A$. More concretely:  
Denote the two embeddings of $X_2 \rightarrow A_1$ by $i_a$ and $i_b$, and their respective images by $A_{a}$ and $A_{b}$. Denote the embeddings $X_2 \rightarrow A_2$ by $i_{aa}, i_{ab}, i_{ba}, i_{bb}$ and their respective images by $A_{aa}, A_{ab}, A_{ba}, A_{bb}$. Define the face maps $e$ of $A$ by:
On $X_1 \subset A_1$: $e_0 = d_0, e_1 = d_1$. 
On $A_{a}$: $e_0 = d_0d_0 = d_0d_1$, $e_1 = d_0d_\infty = d_0d_2$
On $A_{b}$: $e_0 = d_1d_\infty = d_1d_0$, $e_1 = d_1d_2 = d_1d_1$
On $A_{aa}$: $e_0 = i_a, e_1 = d_2, e_2 = d_2$, and similarly on $A_{ab}$, $A_{ba}$, $A_{bb}$.
On $X_2 \subset A_3$: $e_0 = i_{ab}, e_1 = i_{bb}, e_2 = i_{aa}, e_3 = i_{ba}$.
On $X_3 \subset A_3$: $e_0 = i_{ab}d_0, e_1 = i_{bb}d_1, e_2 = i_{aa}d_2, e_3 = i_{ba}d_3$.
On $A_n$: $e_i = d_i$.
This construction extends to a functor and works for simplicial objects in a category with coproduct.
If we are in an abelian category, we can define chain complexes $CX$ and $CA$. There is a morphism $CX \rightarrow CA$ whose component $CX_2 \rightarrow CA_2$ is defined by $i_{aa} + i_{ab}$, which should be a quasi-isomorphism if the intuition is not wrong.<|endoftext|>
TITLE: Algorithm to count the number of perfect matchings in non planar graph
QUESTION [7 upvotes]: I need to count the number of perfect matchings of a certain family of graphs. This family of graph is non planar and a type of snark. For the initial cases, it seems that this number is growing exponentially. My request is different from the one here because right now, I have no interest in knowing what these perfect matchings are.
The FKT algorithm gives a polynomial time algorithm for a planar graph and I am wondering if there is anything similar for non planar graphs. Does anyone have any ideas?

REPLY [8 votes]: Counting the number of perfect matchings in arbitrary graphs (i.e. non planar, non bipartite...) seems to be quite more difficult than the restricted cases that FKT-type algorithms can handle.
In particular, Valiant proved that the problem is in $\mathrm{\text{#}P}$.
With that in mind, you can find information on approximation algorithms, see for example:

Steve Chien, A determinant-based algorithm for counting matchings in a general graph
Martin Fürer, Shiva Prasad Kasiviswanathan, Approximately Counting Perfect Matchings in General Graphs<|endoftext|>
TITLE: Practical advantages of univalent foundations
QUESTION [13 upvotes]: I'm interested in the machine translation of mathematics from informal to formal (a la Ganesalingam in The Language of Mathematics). As a first step, I am designing a computer language for expressing mathematical definitions, theorems, and arguments that would at some later point be amenable to a small formalization kernel. (At this point, the language looks like a pared-down Isabelle/ZFC with support for functions not defined on full domains, realm-like objects, and a weak sense of parametric types and polymorphic functions).
This translation project, which is already sizable, would be the first step towards making a general mathematical knowledge management system/formalization system. The goal of this system would be to be sufficiently useful and easy to use that it sees adoption by those who don't intrinsically care for formalization. Cf. this old MO post for previous community thoughts about such a system.
Recently, the system of univalent foundations by Voevodsky has brought general interest back into formalization. The system of univalent foundations, like the type theoretic conceit that "a proof is an instance of a proposition type" that univalent foundations is ultimately based on, is illuminating and quite beautiful, and I found my brain exploding repeatedly while reading the IAS book on the subject (available here).
Questions:

Do univalent foundations provide a better structure for the introduction of definitions and the statement of theorems than the system I describe in the first paragraph? Can the intensional/extensional questions resolved by the univalence axiom in practice be handled by realms?
If univalent foundations provides this better foundation, will users who try to understand definitions and theorems in this system necessarily need to understand univalent foundations?
Which kinds of formal arguments are easier to make using univalent foundations, and are some kinds of sentence-by-sentence translations of informal proofs more feasible?
Is full-scale reflection in the sense of John Harrison's Metatheory and Reflection in Theorem Proving able to introduce the advantages of univalent foundations into a less well developed underlying type theory?

I am working on this problem because it seems important and not because I am particularly suited for it, so any thoughts would be very helpful.

REPLY [4 votes]: Regarding 2, structural set theory may be seen as a subtheory of the univalent foundations (Ch10 of the book, or our paper).
For 3, the nlab has a list of advantages.