diff --git "a/stack-exchange/math_overflow/shard_12.txt" "b/stack-exchange/math_overflow/shard_12.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_12.txt" +++ /dev/null @@ -1,30739 +0,0 @@ -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 -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 $10$. -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^2+1)^2>m^4+1.$$ So we found a new pair $(m,\ell)$ with $\max(m,\ell)1$. -(Note: without the coprime assumption the conclusion fails, as $(m,n)=(m,m^3)$ for $m\ge 2$, e.g. $(m,n)=(2,8)$, satisfies $mn+1|n^4+1$.) -Proposition If $p$ is prime then $p^4+1$ has no divisor of the form $kp+1$ except $1$ and $p^4+1$. -Proof: write $p^4+1=(kp+1)(k'p+\ell)$ with $0\le\ell\le p-1$; then $\ell=1$. So exchanging $k$ and $k'$ if necessary we can suppose $k\le k'$. -If by contradiction $k$ is divisible by $p$ then $k'\ge k\ge p$, so $p^4+1\ge (p^2+1)^2>p^4+1$, contradiction. So $k$ and $p$ are coprime, and the lemma yields a contradiction. - -REPLY [2 votes]: Let $p$ be prime. Assume that $q \mid p^4+1$, where $q = ap+1$ for some -$a \in \mathbb{N}$. -Then $(p^4+1)/q \equiv 1 \!\! \mod p$, hence -$p^4+1 = (ap+1)(bp+1)$ for some $b \in \mathbb{N}$. -Now we have $p^4 = abp^2+(a+b)p$, respectively, $p^3-abp-a-b = 0$. Therefore, effectively -we are looking for solutions to the diophantine equation $x^3-xyz-y-z=0$ in positive -$x$, $y$, $z$, and prime $x$. -When allowing negative (or zero) $x, y, z$, we would get a lot of solutions, -and for positive $x, y, z \leq 1000$, we still find (up to switching of $y$ and $z$) -$$ - (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 -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 -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 -$$ \sigma_a(n) \triangleq \sum_{\substack{ d|n \\ 1 -TITLE: What is the analytic conductor of this Hecke L-function? -QUESTION [5 upvotes]: Following Iwaniek and Kowalski, S5.10, page 130 we consider an angle character $\xi_k$ on the Gaussian integers $\mathbb Z[i]$ defined by -$ \xi_k(\mathfrak a) = \left(\frac{\alpha}{|\alpha|}\right)^k $ -where $k \equiv 0 \pmod 4$. -This gives an $L$-function $L(s,\xi_k)$ and a functional equation described by the following data: -\begin{align*} - L(s, \xi_k) &= \sum_{\mathfrak a} \xi_k(\mathfrak a) (N\mathfrak a)^{-s} \\ - \gamma(s, \xi_k) &= \pi^{-s} \Gamma\left( s + \frac{\left\lvert k \right\rvert}{2} \right) \\ - \Lambda(s, \xi_k) &= \pi^{-s} \Gamma\left( s + \frac{\left\lvert k \right\rvert}{2} \right) L(s, \xi_k) = \Lambda(1-s, \xi_k) -\end{align*} -Here $L(s,\xi_k)$ has degree $d=2$ and conductor $q=4$ (the discriminant of the Gaussian integers). -I&K had earlier written a prime number theorem $\psi(f,x) = rx - \frac{x^{\beta_f}}{\beta_f} + O\left(\sqrt{\mathfrak q(f)}x e^{ -\frac{c}{2d^4}\sqrt{\log x}} \right)$ which held for all $L$-functions, and apply this to deduce that -$$ \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}.$$ -For $k\geq0$ the $B_k$ are even polynomials of degree $2k$ that take integer values on integers. One has $B_k(k)=1$ for $k\geq1$, but $B_0(0)=2$. Further $B_k(i)=0$ for $|i|2|k|-2\geq0$ because all terms in the sum vanish. It can also be shown that $P_m(0)=0$ for $m\geq2$, but that is more tricky. - Indeed we will show that $d(m,0)=0$ for $m\geq2$. -Note that $P_m(x)$ visibly lies in the local ring $\mathbb{Z}_{(2)}$ for integer $x$. -So it suffices to show that $d(m,k)$ lies in $\mathbb{Z}_{(p)}$ for any odd prime $p$. -In fact we will find that the $d(m,k)$ are integers for $m\geq1$. And $d(0,0)=3/2$ lies in $\mathbb{Z}_{(p)}$ for our odd prime $p$. - For $m$ not too large one may simply compute all $d(m,k)$. -The matrix $$(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)}.$$ -In particular $d(m,k)=0$ for $m<0$ and for $k>m$. We will see that $m(m-1)d(m,k)$ also vanishes for -$2k-20$ then $\binom{x}{j}=\frac{\Gamma(1+x)}{\Gamma(1+j)\Gamma(1+x-j)}$ and the bimeromorphic function $$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-22k-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 $j k-1\geq i\geq m+1-k\geq 0$. -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 $0 k-1\geq i\geq m+1-k\geq 0$. - 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( -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 -$$|\mu_n(A)-\mu_n(B)| \le \frac{\frac1n+\sum_{k -TITLE: How slowly can a power of an ideal grow? -QUESTION [23 upvotes]: For a polynomial ideal $I\subset \mathbb{C}[x_1,x_2]$, let $D(I)$ be the smallest degree of any polynomial in $I$. - -How slowly can $D(I^n)$ grow as a function of $n$? For example, if $D(I^n)\leq 1.01n$ for some $n$, does it imply that $I$ contains a linear polynomial? - -Note that the single-variable case is trivial: $D(I^n)=n\cdot D(I)$. -I am interested in a more general situation than in the question, but the version is the simplest case where I am stuck. - -REPLY [13 votes]: Consider $I = (y- x^k, x^{k+1})$. -For $k>1$ this does not contain any linear functions. It contains $xy$ so $D(I)=2$. But I claim $y^{k+1} \in I^k$, so $D(I^k) = k+1$. -By the binomial theorem -$$ 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 -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{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 $n_{k+2} -TITLE: Gauss-Milgram formula for fermionic topological order? -QUESTION [7 upvotes]: For Bosonic topological order, a very useful formula was proved to be true: -$\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 -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 -$$V_n:=\prod_{0\le i0$. - -REPLY [5 votes]: I am amazed that nobody noticed that Iosif asks for the calculation of the transfinite diameter of the interval $[0,1]$. This notion applies to arbitrary compact domains $K$ in ${\mathbb R}^n$ : -$$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$$ -for $0 -TITLE: When is a sequence the sum of two Beatty sequences? -QUESTION [7 upvotes]: In other words, given a sequence $(s_n)$, how can we tell if there exist irrationals $u>1$ and $v>1$ such that -$$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)$$ -This is the sum of $\prod_{i=1}^k g(y_i)$ over the lattice in $\mathbb Z^k$ satisfying the condition $\sum_{i=1}^k a_i y_i \equiv \xi $. -Now we know that $\prod_{i=1}^k g(y_i)$ has $L_1$ norm $ (4/\pi)^k$, and at most $(2/3) (4/\pi)^k$ of its mass is in the box where $|y_i| < Ck$, because $1-O(1/k)$ of the mass of $g(y)$ is in the box where $|Y| < k $. -So as long as no equation of the form $\sum_{i=1}^k y_i a_i =0$ for $y_i$ not all $0$ but $|y_i| < 2 C k$ holds, then no two points in that box get sent to the same $\chi$, so the $L_2$ norm goes down by at most $1-2/3$ (in the worst case where everything outside the box cancels something inside the box.) -When can we find $a_1, \dots, a_k$ such that no such equation holds? If $n=p$ is a prime then by selecting $a_1, \dots, a_k$ randomly, the probability of any given equation holds is $1/p$, so as long as $p > (2 C k)^k$ then sometimes none of the equations hold. -If we're not a prime then we can just do the Chinese Remainder Theorem thing, or do the probabilistic argument more carefully.<|endoftext|> -TITLE: Equivalent Norms on Sobolev Spaces -QUESTION [6 upvotes]: When $k$ is a positive integer and $10$ such that for all $u\in W^{k,p}\left(\mathbb{R}^{N}\right) :$ -$$ -\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}.$$ -If you're only after characterizing when $A$ is singular, then you need only determine when any of the $\lambda_{ij}$ can be zero, which looks fairly straightforward.<|endoftext|> -TITLE: When is alternating sum $\sum_{i}f(a_i)-\sum_{i0$ for all $a_1,\ldots,a_n$. If we perturb $f$ a tiny bit, say $f(x)=\frac{1}{x}-\frac{1}{100x^{100}}$, I would imagine that $S(f)>0$ still always holds. But the proof method for $f(x)=\frac1x$ is hard to generalize to other functions. Can we prove it in some other way? -More generally, is there a theorem out there stating sufficient conditions under which $S(f)>0$ always holds? - -REPLY [3 votes]: If $f$ is a polynomial with $\deg f< n$, then $S(f)=f(0)$. - -ADDED. More generally, for an analytic function $f(x)$, -$$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.$$ -It can be seen that $[x^k]\ g(x)=0$ for $k -TITLE: Kullback Leibler "variance": does that divergence have a name? -QUESTION [6 upvotes]: If you consider two probability distributions $p$ and $q$, one way to measure the distance between the two is the Kullback-Leibler divergence: -$$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 <...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 $ -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 $0$, 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 -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 -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 -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)