TITLE: Can we separate the almost-disjointness sunflower numbers? QUESTION [12 upvotes]: This question concerns a new cardinal characteristic of the continuum that arose out of issues in my answer to the question, Sunflowers in maximal almost disjoint families. A family $\cal A$ of infinite subsets of $\omega$ is almost disjoint, if any two members of the family have finite intersection. Such a family is a maximal almost disjoint family if it cannot be extended to a larger almost disjoint family. A $\Delta$-system, also known as a sunflower, is a family of sets with all pairs having the same pairwise intersection. In his earlier question, Dominic had asked whether every maximal almost disjoint family must contain an infinite sunflower. In the general case, this seems still to be open, but my answer there shows that under the continuum hypothesis, there is a maximal almost disjoint family containing no sunflowers even of size $3$. Indeed, the construction there shows that there is a maximal almost disjoint family $\langle A_\alpha\mid\alpha<\omega_1\rangle$ such that every $A_\alpha$ has different intersections with every earlier $A_\beta$, for $\beta<\alpha$. This property implies that there can be no sunflower of size $3$ in this family. (But notice by a simple pigeon-hole argument that it will be impossible to extend this stronger property to enumerations beyond $\omega_1$.) My questions concern the property of almost disjoint families that are maximal with respect to the property of not containing any sunflower of a certain size. Question 1. If an almost disjoint family of infinite subsets of $\omega$ is maximal amongst almost disjoint families with respect to the property of not containing a sunflower of size $3$, is it a maximal almost disjoint family? And more generally, I ask the same for sunflowers of any particular size. The question leads naturally to new cardinal characteristics of the continuum. Namely, let us define almost-disjointness sunflower number, officially denoted $\frak{a}_{\kappa}^\Delta$, but let me immediately drop the superscript and write just $\frak{a}_\kappa$, to be the size of the smallest almost-disjoint family that is maximal among almost-disjoint families with respect to the property of not containing a sunflower of size $\kappa$. (We consider only $\kappa\geq 3$.) The construction on my other answer shows that $\omega_1\leq\frak{a}_{\kappa}$. Question 2. Can we separate these various cardinal characteristics $\frak{a}_\kappa$ from each other, and from the almost-disjointness number $\frak{a}$? For example, is it consistent with ZFC that $\frak{a}_{3}<\frak{a}$? This would be a strong refutation of question 1. Is it consistent that $\frak{a}_{3}\neq\frak{a}_{4}$? At first I had though it was clear that $\frak{a}_{\kappa}\leq\frak{a}$, the almost-disjointness number, which is the smallest size of any maximal almost disjoint family. But upon reflection, this no longer seems clear to me, since perhaps there could be a small maximal almost disjoint family, but it contains a lot of sunflowers, and the smallest maximal sunflower-free family might be larger. Or strictly smaller, since a maximal sunflower-free family might not be a maximal almost disjoint family. I had similarly expected that if $\kappa<\lambda$, then there should be some trivial provable relation between $\frak{a}_{\kappa}$ and $\frak{a}_{\lambda}$. But unless I am mistaken, this now also doesn't seem to be immediate. Question 3. What are the provable relations between $\frak{a}_\kappa$, $\frak{a}_\lambda$, and $\frak{a}$ when $\kappa<\lambda$? Can we say something even about the relation of $\frak{a}_{3}$ and $\frak{a}_{4}$, or their relation to $\frak{a}$? REPLY [5 votes]: The answer to Question 1 is NO: Split $\omega$ into two disjoint infinite sets $X_0$ and $X_1$, fix $\mathcal A_0$ a countable AD family of subsets of $X_0$ such that $\mathcal A_0$ does not contain a $\Delta$-system of size $3$, yet for every finite $F\subseteq X_0$ there are $A,B\in \mathcal A_0$ such that $F=A\cap B$. Added after comments below (To construct $\mathcal A_0$ start with a partition $\{B_n^{i}: n\in\omega, i\in 2\}$ of $\omega$ into infinite pieces and enumerate $[\omega]^{<\omega}$ as $\{F_n: n\in\omega\}$ . We shall recursively construct the family $\mathcal A_0=\{A_n^{i}:n\in\omega\}$ so that (1) $A_n^{i}\subseteq^* B_n^{i}$, (2) $F_n=A_n^0\cap A_n^1$ and (3) $|A_n^{i}\cap A_m^{j}|\not = |A_n^{i'}\cap A_{m´}^{j'}|$ for every $m,m' TITLE: Is there a unique "natural" action of $\mathsf{SL}_{n+1}$ on $\mathbb{R}^n$? QUESTION [10 upvotes]: Context By acting naturally via $\mathsf{SL}_3$ on $\mathbb{RP}^2=\{[x:y:z]\}$ and by taking the induced action on the affine hyperplane $z=1$ (which we identify with $\mathbb{R}^2$), one can realize the corresponding Lie algebra $\mathfrak{sl}_3$ as a sub-algebra of the algebra $\mathfrak{X}(\mathbb{R}^2)$ of vector fields of $\mathbb{R}^2$: by carrying out this simple exercise, one ends up with 8 vector fields, 6 of which are linear, and two contain quadratic coefficients, namely $$ X_A=-xy\partial_x-y^2\partial_y\, ,\quad X_B=-x^2\partial_x-xy\partial_y\, , $$ that correspond to the matrices $$ A=\left(\begin{array}{ccc}0&0&0\\ 0&0&0\\0&1&0\end{array}\right)\,,\quad B=\left(\begin{array}{ccc}0&0&0\\ 0&0&0\\1&0&0\end{array}\right)\, , $$ respectively. Image below shows how to use the flow of $X\in\mathfrak{sl}_3$ to obtain a vector field on the plane $z=1$: The 6 linear vector fields are of course complete, whereas the flows of $X_A$ and $X_B$ display a singularity at $t=-\tfrac{1}{y}$ and $t=-\tfrac{1}{x}$, respectively. This remark motivated the first question: For dimensions reasons (as observed in the comments), for every $n\ge 1$ there is no embedding $$ \mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, , $$ whose image is made of linear (allowing a constant) vector fields. However for any $n\geq 1$ it is easy to find an embedding $$ i_{\textrm{nat}}:\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, , $$ which we may call natural, because it is induced by the natural action of $\mathsf{SL}_{n+1}$ on $\mathbb{RP}^n$ by the same construction as above, one may wonder the following (initially "Question 2"): Question Given an arbitrary embedding $$ i :\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, , $$ it is true that $i$ must necessarily be equivalent to $i_{\textrm{nat}}$? If not, which conditions must $i$ satisfy, to be such? In other words, one has a collection $X_1,X_2,\ldots, X_{d(n)}$, where $d(n)=(n+1)^2-1=\dim \mathsf{SL}_{n+1}$ of vector fields on $\mathbb{R}^n$, that commute according to the commutation relations of $\mathsf{SL}_{n+1}$ and the question can be recast as follows: is there a diffeomorphism of $\mathbb{R}^n$ that transforms the aforementioned vector fields into the images via $i_{\textrm{nat}}$ of a suitable set of generators of $\mathfrak{sl}_{n+1}$? What kind of obstruction (if any) one should expect? REPLY [10 votes]: There are two aspects to this question, the global question and the local question. Also, the case $n=1$ is different from $n>1$. Basically, the answer is 'essentially yes, but with some caveats'. Here's a sample of the kind of results one might consider as an answer in the case $n=1$: First, an old theorem of Sophus Lie: Theorem: (Lie) Let $L\subset\mathfrak{X}(\mathbb{R}^1)$ be a Lie algebra of vector fields that is locally transitive, i.e., for every $p\in\mathbb{R}$, there is a vector field $X\in L$ such that $X(p)\not=0$. Then either $\dim L=1$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x}$, $\dim L=2$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x},\ x\,\frac{\partial}{\partial x}$, $\dim L=3$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x},\ x\,\frac{\partial}{\partial x},\ x^2\,\frac{\partial}{\partial x}$, or $\dim L=\infty$, and for every $k$ and every $p\in\mathbb{R}$, there is a vector field $X_k\in L$ such that $X_k$ vanishes to order exactly $k$ at $p$. Then there is a global result that is not too hard to prove: Theorem: Suppose that $L\subset\mathfrak{X}(\mathbb{R})$ is a locally transitive Lie algebra of dimension $3$. Then there is a diffeomorphism $u:\mathbb{R}\to u(\mathbb{R})\subseteq\mathbb{R}$ that carries $L$ into the span of the restrictions to $u(\mathbb{R})$ of the vector fields $\frac{\partial}{\partial x}$, $\cos x\,\frac{\partial}{\partial x}$, and $\sin x\,\frac{\partial}{\partial x}$. Note that all the vector fields in $L$ are complete if and only if $u(\mathbb{R})=\mathbb{R}$. In particular, if $L\subset\mathfrak{X}(\mathbb{R})$ is the span of $\frac{\partial}{\partial x}$, $\cos x\,\frac{\partial}{\partial x}$, and $\sin x\,\frac{\partial}{\partial x}$, then there is an $L$-preserving diffeomorphism $\phi:(a,b)\to(c,d)$ between two bounded intervals $(a,b),(c,d)\subset\mathbb{R}$ if and only if either $b-a=d-c=2\pi k$ or else $2\pi(k{-}1) < b-a, d-c <2\pi k$ for some integer $k\ge1$. Thus, this gives a complete answer in the case $n=1$ and the locally transitive case. The need for the hypothesis of local transitivity is shown by this example: Let $\phi:\mathbb{R}\to [0,1)$ be a smooth function such that $\phi(x)>0$ when $|x|<1$ and $\phi(x)=0$ when $|x|\ge 1$. Let $u:(-1,1)\to \mathbb{R}$ satisfy $u'(x) = 1/\phi(x)$ when $|x|<1$. Then $u:(-1,1)\to\mathbb{R}$ is a diffeomorphism. Let $L\subset\mathfrak{X}\bigl((-1,1)\bigr)$ be the Lie algebra spanned by the vector fields $$ \phi(x)\,\frac{\partial}{\partial x},\quad \cos\bigl(u(x)\bigr)\,\phi(x)\,\frac{\partial}{\partial x},\quad \sin\bigl(u(x)\bigr)\,\phi(x)\,\frac{\partial}{\partial x}. $$ These vector fields extend smoothly to the entire real line as zero vector fields where $|x|\ge1$, but one could also extend them smoothly to be periodic of period $2$. In either case, they vanish to infinite order at the odd integers. Clearly, there are infinitely many globally inequivalent ways to embed $\mathfrak{sl}_2$ as a subalgebra of $\mathfrak{X}(\mathbb{R})$ if one does not assume local transitivity. Classification is pretty hopeless. Now assume that $n>1$. Here, the situation is somewhat more straightforward. The first thing to notice is the classical result that $L=\mathfrak{sl}_{n+1}=\mathfrak{sl}_{n+1}(\mathbb{R})$ has no proper subalgebras of codimension less than $n$. (This was known to Lie, at least in the complex case, i.e., for $\mathfrak{sl}_{n+1}(\mathbb{C})$, and the result for the real case follows immediately from this by complexification.) In fact, there is a complete descriptions of the set $S_n$ of subalgebras of $L$ that have codimension $n$. This is an $n$-manifold with two components $$ S_n = \mathrm{SL}(n{+}1,\mathbb{R})/P_1 \cup \mathrm{SL}(n{+}1,\mathbb{R})/P_2\simeq \mathbb{RP}^n\cup \mathbb{RP}^n\, $$ where $P_1$ is the maximal parabolic consisting of the elements of $\mathrm{SL}(n{+}1,\mathbb{R})$ that preserve a given $1$-dimensional subspace of $\mathbb{R}^{n+1}$ and $P_2$ is the maximal parabolic consisting of the elements of $\mathrm{SL}(n{+}1,\mathbb{R})$ that preserve a given $n$-dimensional subspace of $\mathbb{R}^{n+1}$. (These two subgroups are not conjugate in $\mathrm{SL}(n{+}1,\mathbb{R})$, but there is an outer automorphism ($a\mapsto (a^{-1})^\mathsf{T}$) that carries one to the other. (Of course, when $n=1$, the space $S_1=\mathbb{RP}^1\simeq S^1$ has only one component.) Now, suppose that $L$ is embedded as a subalgebra of $\mathfrak{X}(M^n)$, where $M^n$ is a smooth manifold of dimension $n$. Then, for any $p\in M$, the subalgebra $L^0_p\subset L$ consisting of the vector fields that vanish at $p$ has codimension at most $n$, so either $L^0_p = L$, in which case, all the vector fields in $L$ vanish at $p$, or else the codimension of $L^0_p$ in $L$ is exactly $n$. Let's remove the closed set of points where all the vector fields in $L$ vanish (which will not affect the completeness of any of the vector fields in $L$) and assume that $L^0_p$ always has codimension $n$. Thus, we have a canonical map $\phi:M\to S_n$ given by $\phi(p) = L^0_p$. Theorem: If $L\subset\mathfrak{X}(M^n)$ is a locally transitive subalgebra isomorphic to $\mathfrak{sl}_{n+1}$, then $\phi:M\to S$ is a smooth immersion (in particular, $\phi$ is a local diffeomorphism). Moreover, $\phi_*$ carries $L$ to the 'natural' copy of $L$ in $\mathfrak{X}(S_n)$ induced by the action of $\mathrm{SL}(n{+}1,\mathbb{R})$ on $S_n$. The proof of this theorem is not hard, but it consists of a number of steps, whose details would take up a lot of space. The main points are these: First, let $L^i_p\subset L$ denote the subset of vector fields that vanish to order $i{+}1$ at $p$, then one first shows that $L^i_p = 0$ for $i$ suffciently large (a priori depending on $p$). Then, using the fact that $L^0_p$ contains a simple subalgebra isomorphic to $\mathfrak{sl}_n$, one uses its representation theory to show that this simple subalgebra cannot be contained in $L^1_p$ (which is a solvable ideal of $L^0_p$), and hence $L^0_p/L^1_p$ is an algebra containing $\mathfrak{sl}_n$. Using the nondegeneracy of the Killing form, one concludes that $L^1_p$ is dual, as a $\mathfrak{sl}_n$-representation, to $L/L^0_p\simeq\mathbb{R}^n$ and hence has dimension $n$. Thus, $L^0_p/L^1_p$ has dimension $n^2$ and must be isomorphic to $\mathfrak{gl}_n$. From this, it follows easily that $L^2_p = (0)$. Now, using these facts, it is easy to explicitly compute the differential of $\phi$ at $p$ and show that $\phi'(p):T_pM\to T_{\phi(p)}S_n$ is an isomorphism. Once one has the fact that $\phi$ is a local diffeomorphism, the final statement follows relatively easily. Note that because $S_n$ has two components there are essentially two inequivalent global realizations of $L=\mathfrak{sl}_{n+1}$ as vector fields on $\mathbb{RP}^n$. By this, I mean that there are two Lie algebra homomorphisms $\psi_i:L\to\mathfrak{X}(\mathbb{RP}^n)$ such that there is no diffeomorphism $u:\mathbb{RP}^n\to\mathbb{RP}^n$ such that $\psi_2(X) = u_*\bigl(\psi_1(X)\bigr)$ for all $X\in L$. What is true instead is that there is an outer automorphism $\tau:L\to L$ such that $\psi_2\bigl(\tau(X)\bigr) = u_*\bigl(\psi_1(X)\bigr)$. This is, perhaps, a subtle point, but it shows that there really are two essentially different ways that $L$ can be realized as vector fields in dimension $n$. It is not clear which one should be called 'natural'. Also, note that, if $L\subset\mathfrak{X}(M)$ consists entirely of complete vector fields, then $\phi:M\to S_n$ must be a covering space, in particular, if $M$ is connected, then it must be either $S^n$ or $\mathbb{RP}^n$. (Note that this is another place where $n>1$ differs from the case $n=1$.) Remark: This non-uniqueness can be even more dramatic. A famous example is the split form of type $\mathfrak{c}_2$ (usually denoted as $\mathfrak{sp}_4(\mathbb{R})$), a Lie algebra of dimension $10$ that has two non-conjugate families of subalgebras of minimal codimension $3$, leading to two inequivalent ways that it can appear as a Lie algebra of vector fields on a $3$-manifold. One is as contact vector fields on $\mathbb{RP}^3$, and the other is as the conformal vector fields on $\mathcal{N}^{2,1}$, the manifold of null subspaces of dimension $1$ in Minkowski $5$-space, $\mathbb{M}^{4,1}$. This is the basis of the classical Klein correspondence. For more exotic examples, the split form of the exceptional Lie algebra $\mathfrak{g}_2$ (of dimension $14$) has two non-conjugate subalgebras of the minimal codimension $5$. Since $\mathfrak{g}_2$ has no outer automorphisms, they are not even equivalent up to isomorphisms of the algebra. Thus, as both Cartan and Engel realized in 1893, $\mathfrak{g}_2$ can appears as two essentially different subalgebras of the vector fields in $\mathbb{R}^5$. This also happens for the split form of the exceptional Lie algebra $\mathfrak{f}_4$, and, correspondingly, it can be realized as two essentially different subalgebras of the vector fields in $\mathbb{R}^{15}$.<|endoftext|> TITLE: Dedekind-"finiteness" for arbitrary limit cardinals QUESTION [7 upvotes]: In $\mathbf{ZF}$, it is possible for a set $A$ to be infinite but not to admit a countable set. In other words, for any $\alpha\in\omega$, there is an injection from $\alpha$ into $A$, but there is no injection from $\omega$ into $A$. If we replace $\omega$ by a successor cardinal $\kappa^+$ in the above statement, any $A$ of cardinality $\kappa$ works, so there exists an example in $\mathbf{ZFC}$. However, for limit cardinals, the question seems unclear to me. Therefore: If $\kappa>\omega$ is a limit cardinal and $A$ is a set such that for any $\alpha\in\kappa$, there is an injection from $\alpha$ into $A$, does it follow in $\mathbf{ZF}$ that there is an injection from $\kappa$ into $A$? Additionally, does the existence of such an $A$ imply the existence of a Dedekind-finite set? REPLY [5 votes]: Start with your favourite model of $\sf ZFC$, your favourite regular cardinal $\mu$, and your favourite limit cardinal $\lambda>2^\mu$. Now consider the ${<}\mu$-support product $\prod_{\alpha<\lambda}\operatorname{Add}(\mu,\alpha)$. With automorphism groups that act on each individual component in the product, and with a filter of subgroups generated by fixing a bounded1 number of component pointwise, we get a symmetric system. It is easy to see that in the symmetric extension $2^\mu$ has an injection from any $\alpha<\mu$, since that is given by the $\alpha$th component. But if there is an injection from $\lambda$ into $2^\mu$, then this injection codes a set of ordinals, and by a standard homogeneity argument we can show that any set of ordinals added to the model must have been added by a bounded part of the product, and so that is impossible. Now, since this forcing is $\mu$-closed, has $\mu^+$-cc and the filter of groups has $\operatorname{cf}(\lambda)$-completeness we instantly get $\sf DC_{<\operatorname{cf}(\lambda)}$ to hold in the extension as well. Picking $\mu=\omega$ and $\lambda=\beth_{\omega_1}$, for example, provides us with $\sf DC$, and with $\aleph(\Bbb R)=\beth_{\omega_1}$. We can actually replace bounded by any other support, as long as it does not contain any cofinal subsets. This will affect how much $\sf DC$ holds in your model, though.<|endoftext|> TITLE: Convex polyhedra with non-congruent faces QUESTION [5 upvotes]: Question: Are there convex polyhedra wherein all faces are convex polygons with same area and perimeter and no two faces are mutually congruent? Remarks: If the answer to above is "no", then, one can ask if there are other weaker conditions which can say, ensure that at least a pair of faces are congruent and also about cases with quantities other than area and perimeter shared among the faces. And if the answer is "yes", one can ask if equalizing more but finitely many quantities among faces will guarantee at least one pair of congruent faces or all faces congruent. And one could ask about polyhedrons with convex faces with every pair of faces non-congruent and faces constrained only to have same perimeter OR same area(*). Another direction in which the question can be 'relaxed': let the polyhedron be non-convex with its faces remaining convex and non-congruent. (*) Guess: 'perturbed' pyramids with quadrilateral base (all four sides and all angles of base slightly different from each other) and 4 triangular faces can be built with faces pairwise non-congruent and all 5 faces having equal area or equal perimeter - but not both equal in any obvious way. REPLY [4 votes]: Most likely. Start with something like this polytope and denote by $u_1,\ldots,u_{12}$ the unit normals to faces. As given this polytope has all faces equilateral triangles with the same area $a$ and the same perimeter. Perturb all $u_i$ but keep all areas equal to $a$. This is always possible by the Minkowski theorem. The equal perimeter condition gives $12$ equations. The space of perturbations of normals is $24$-dimensional, giving you $12$-dimensional freedom of choice (probably). To ensure that all faces are non-congruent you will need to kill the relatively small symmetry group of order $8$, which should be possible. An actual computation is needed in this case to give a convincing answer.<|endoftext|> TITLE: Groups which maintain all their subgroups’ automorphisms as inner automorphisms QUESTION [15 upvotes]: Are there any groups, finite or infinite, other than the first three symmetric groups which maintain all their subgroups’ automorphisms as inner automorphisms (every automorphism of every subgroup extends to an inner automorphism of the whole group)? REPLY [4 votes]: Yes. By this paper of Minasyan one can construct a finitely generated group $G$ with all proper subgroups cyclic of prime order $p\gg 1$, two conjugacy classes and trivial $Out(G)$. This group obviously satisfies the conditions of OP.<|endoftext|> TITLE: Tensor product of perverse sheaves on flag varieties QUESTION [5 upvotes]: I am interested in computing tensor products of perverse sheaves on (partial) flag varieties. For a specific example - consider the product of the big projective on $\mathbb{P}^1$ with itself (This is the projective cover of the skyscraper sheaf on the 0-dimensional stratum). Does this have a simple description? How can I compute its cohomology? Any general tips or computational tricks in this context are very welcome. I am especially interested in tilting perverse sheaves such as the one above. REPLY [5 votes]: First a general comment: as Sasha alludes to, there are two tensor products of complexes of sheaves. Let $i : X \hookrightarrow X \times X$ denote the diagonal. We have, and given complexes of sheaves $\mathcal{F}$ and $\mathcal{G}$ on $X$ we can form their external tensor product $\mathcal{F} \boxtimes \mathcal{G}$ on $X \times X$. The two tensor products are \begin{gather*} \mathcal{F} \otimes \mathcal{G} := i^*(\mathcal{F} \boxtimes \mathcal{G}) \quad \text{and} \\ \mathcal{F} \otimes^{!} \mathcal{G} := i^!(\mathcal{F} \boxtimes \mathcal{G}). \end{gather*} (I think I learnt this notation from some very nice old notes of Ginzburg.) This description makes it clear that $\mathbb{D}(\mathcal{F} \otimes \mathcal{G}) = \mathbb{D}\mathcal{F} \otimes^{!} \mathbb{D}\mathcal{G}$ etc. In particular, we have control of the stalks of $\mathcal{F} \otimes \mathcal{G}$ in terms of the stalks of $\mathcal{F}$ and $\mathcal{G}$, and costalks of $\mathcal{F} \otimes^{!} \mathcal{G}$ in terms of costalks of $\mathcal{F}$ and $\mathcal{G}$, but for example the costalks of $\mathcal{F} \otimes \mathcal{G}$ are potentially very complicated. Now to your specific question: consider the big tilting sheaf $T_s$ on $\mathbb{P}^1$. (This is isomorphic to the projective cover you ask about.) Its stalks are $\mathbb{Q}[1]$ on the open locus, and $\mathbb{Q}$ at the point stratum. In particular, the stalks of $T_s \otimes T_s$ are $\mathbb{Q}[1] \otimes \mathbb{Q}[1] = \mathbb{Q}[2]$ on the open locus, and $\mathbb{Q}$ on the point stratum. In particular we have a distinguished triangle \begin{equation} j_!\mathbb{Q}[2] \to T_s^{\otimes 2} \to \mathbb{Q}_{0} \stackrel{[1]}{\to} \end{equation} where $j$ denotes the inclusion of the open stratum, and $0$ denotes the point stratum. Because the appropriate Ext group vanishes, we deduce that \begin{equation} T_s^{\otimes 2} \cong j_!\mathbb{Q}[2] \oplus \mathbb{Q}_{0}. \end{equation} (Already in this example we see the failure of the tensor product to be self-dual.) What about in general? We know the stalks of tilting sheaves on $G/B$ (see Yun's "weights of mixed tilting sheaves" for a lovely account). They are concentrated in a single degree, with dimension given by Kazhdan-Lusztig polynomials at 1. Hence the stalks of $T_x \otimes T_y$ along the stratum $BzB/B$ are concentrated in degree $-2\ell(z)$ and are computable via a product of KL polynomials at $1$. I am not sure if the tensor product splits as above. But maybe that is your job :) An aside: if one instead considers tensor products of $IC$ sheaves several small example suggest that the answer is always pure. (This has gotten me excited at least twice in my life, and I know I am not the only one!) However this is not the case in general, as the example of $IC_{st} \otimes IC_{ts}$ in $SL_3/B$ shows. (In this example the tensor product is the constant sheaf on the union of the two closed Schubert curves corresponding to $s$ and $t$.)<|endoftext|> TITLE: Modules over the integral dual Steenrod algebra as linear functors QUESTION [12 upvotes]: Let $\text{Latt}$ denote the category of lattices, i.e., finitely generated free abelian groups. In the appendix to Lecture 4 of Condensed.pdf, Scholze considers functors $F \colon \text{Latt} \to \mathcal D(\mathbb Z)$ that are additive: $F(A \oplus B) \cong F(A) \oplus F(B)$. It seems to be a folklore result that the category of such functors is equivalent to the category of modules over the integral dual Steenrod algebra $\mathbb Z \otimes_{\mathbb S} \mathbb Z$. (Here $\mathbb S$ denotes the sphere spectrum.) As I'm not very well-versed in homotopy theory/higher algebra, I have no idea how this equivalence works. Q: How does this equivalence work? Is there a reference that explains it? REPLY [11 votes]: First I should say that Clausen-Scholze are not considering functors which preserve direct sums, but rather all functors. (This is likely why, in the end, one needs to know something about the homology of Eilenberg-MacLane spaces rather than the homology of the Eilenberg-MacLane spectrum. That guess/observation I learned from Piotr Pstragowski.) But if you still want to know about functors that preserve direct sums, then it is indeed true that there's an equivalence $\mathsf{Fun}^{\oplus}(\mathsf{Latt}, \mathsf{D}(\mathbb{Z})) \simeq \mathsf{Mod}_{\mathbb{Z}\otimes_{\mathbb{S}}\mathbb{Z}}$. The 'intuitive' explanation is the following: on the one hand, if we have a functor $F$ on the left hand side, then its value on $\mathbb{Z}$ is some $\mathbb{Z}$-module $M$ and it's equipped with a bunch of extra structure, as a $\mathbb{Z}$-module, coming from the functoriality. This extra structure turns out to be a second $\mathbb{Z}$-module structure which `commutes' with the first, hence $M$ gets the structure of a bimodule, or a $\mathbb{Z}\otimes_{\mathbb{S}} \mathbb{Z}$-module. In the other direction, given a bimodule $M$, we can define a functor by $\Lambda \mapsto M \otimes_{\mathbb{Z}} \Lambda$, using the leftover $\mathbb{Z}$-module structure on $M$ to make this a functor from $\mathbb{Z}$-modules to $\mathbb{Z}$-modules. To be more precise, the equivalence proceeds in three steps. There is an inclusion $\mathsf{Latt} \to \mathsf{Mod}_{\mathbb{Z}}^{\mathrm{cn}}$ into connective $\mathbb{Z}$-modules (i.e. chain complexes concentrated in nonnegative degrees). Left Kan extending along this gives an equivalence $\mathsf{Fun}^{\oplus}(\mathsf{Latt}, \mathsf{Mod}_{\mathbb{Z}})\stackrel{\simeq}{\to} \mathsf{Fun}^{\mathrm{colim}}(\mathsf{Mod}^{\mathrm{cn}}_{\mathbb{Z}}, \mathsf{Mod}_{\mathbb{Z}})$ There is a canonical equivalence $\mathsf{Fun}^{\mathrm{colim}}(\mathsf{Mod}^{\mathrm{cn}}_{\mathbb{Z}}, \mathsf{Mod}_{\mathbb{Z}}) \simeq \mathsf{Fun}^{\mathrm{colim}}(\mathsf{Mod}_{\mathbb{Z}}, \mathsf{Mod}_{\mathbb{Z}})$. Every $\mathbb{Z}$-bimodule, i.e. $\mathbb{Z} \otimes_{\mathbb{S}} \mathbb{Z}$-module, $M$, defines a colimit preserving functor $N \mapsto M\otimes_{\mathbb{Z}} N$. This procedure gives a functor $\mathsf{Mod}_{\mathbb{Z}\otimes_{\mathbb{S}}\mathbb{Z}} \to \mathsf{Fun}^{\mathrm{colim}}(\mathsf{Mod}_{\mathbb{Z}}, \mathsf{Mod}_{\mathbb{Z}})$ which turns out to be an equivalence. Here (1) is a formal consequence fo the fact that Latt gives compact projective generators of $\mathsf{Mod}^{\mathrm{cn}}_{\mathbb{Z}}$ (see, e.g., Higher Topos Theory 5.5.8.22 for the equivalence once you know these are compact projective generators), (2) is a formal consequence fo the fact that $\mathsf{Mod}_{\mathbb{Z}}$ is stable (so there is a unique way to extend a colimit-preserving functor on connective modules to all modules), and (3) can be proven using the Barr-Beck theorem, but a reference (for a stronger statement) would be Higher Algebra 4.8.5.16 (though you'll have to do some translation of the notation...). REPLY [7 votes]: Theorem A from Jibladze and Pirashvili's 1991 paper "Cohomology of Algebraic Theories," http://www.rmi.ge/~jib/pubs/jipicat.pdf , identifies Mac Lane cohomology with the cohomology of additive functors from finitely generated projective $R$-modules to $R$-modules. In the case $R=\mathbb{Z}$, the finitely generated projective $R$-modules are simply the lattices that you mentioned. As far as I know, that 1991 paper is the correct one to cite for the result you mentioned. But the result also needs the connection between Mac Lane cohomology and the Steenrod algebra, and that came earlier. Sometimes people define Mac Lane cohomology in a way which does not make any obvious reference to Eilenberg-Mac Lane spaces or the Steenrod algebra at all, so it is nontrivial to see the relationship. Mac Lane cohomology is isomorphic to the stable cohomology of Eilenberg-Mac Lane spaces with appropriate coefficients; when those coefficients are suitably trivial, then of course the stable cohomology of Eilenberg-Mac Lane spaces is simply the Steenrod algebra. The right citation here is probably Eilenberg and Mac Lane "Homology theories for multiplicative systems" from 1951, or Mac Lane "Homologie des anneaux et des modules" from 1956, but the result is described a nice concise way on pg. 257 of the Jibladze-Pirashvili paper. Another nice modern exposition of the relationship between Mac Lane cohomology and Eilenberg-Mac Lane spaces is (the dual of) Definition 1.1 in Pirashvili-Waldhausen's 1992 "Mac Lane homology and topological Hochschild homology", here: https://core.ac.uk/download/pdf/82212669.pdf .<|endoftext|> TITLE: A question on regularity of the Legendre transform QUESTION [7 upvotes]: Let $f(x)$ be a strictly convex real-valued $C^{\infty}$ function on an open neighborhood of the origin in $\mathbb R^n$ with $f (0, \ldots , 0)= \partial_j f(0, \ldots , 0)=0$ for all $j$. If the second derivative $f''(0, \ldots , 0)$ is non-degenerate as a quadratic form, then it is easy to see that the Legendre transform $\tilde f(\xi)$ of $f$ is $C^{\infty}$ in some neighborhood of the origin. On the other hand, if $f''(0, \ldots , 0)$ is degenerate, then $\tilde f(\xi)$ is never in $C^{1,1}$ in a neighborhood of the origin. For instance, if $f(x_1, x_2 ) = \frac 12 x_1^2 + \frac 14 x_2^4$, then $\tilde f(\xi_1, \xi_2 ) = \frac 12 \xi_1^2 + \frac 34 |\xi_2|^{4/3}$. Does anyone know a reference for this fact? REPLY [2 votes]: Jean-Baptiste Hiriart-Urruty has pointed out that Theorem 4.2.2 in his book (together with C. Lemaréchal), Convex analysis and minimization algorithms, Volume 2, implies the assertion in the question. That theorem says (in particular) the following. If $f(x)$ is convex and the gradient $f'(x)$ is Lipschitz continuous, then $\tilde f(\xi)$ is strongly convex. To see that this implies the assertion just note that $\tilde{\tilde f} = f$ if $f$ is convex and that if $f$ is convex and $f''(x)$ is degenerate, then $f$ is not strongly convex.<|endoftext|> TITLE: Escaping from infinitely many pursuers QUESTION [18 upvotes]: The fugitive is at the origin. They move at a speed of $1$. There's a guard at $(i,j)$ for all $i,j\in \mathbb{Z}$ except the origin. A guard's speed is $\frac{1}{100}$. The fugitive and the guards move simultaneously and continuously. At any moment, all guards move towards the current position of the fugitive, i.e. a guard's trajectory is a pursuit curve. If they're within $\frac{1}{100}$ distance from a guard, the fugitive is caught. The game is played on $\mathbb{R}^2$. An animation with guards' speed $\frac{1}{4}$ looks something like this (source): Question: can the fugitive avoid capture forever? What I know: The fugitive will be caught if they remain in a bounded area. The distance between two guards is strictly decreasing unless the fugitive and the guards remain collinear. But the further away the guards are, the slower that distance decreases. Even if there're only 2 guards, a straight-line dash into the gap between the pair will lead to capture, as long as they're sufficiently far away (see radiodrome). The fugitive can escape from arbitrarily large encirclement, provided the wall of guards is not too "thick" (4 or 5 layers are fine), such as this (3 layers): The shape of the wall doesn't matter (doesn't have to be rectangular). I asked the question sometime ago on math.stackexchange, where I received the cool animation above, but got no definite answer. I was inspired by a very similar problem here on MO, with additional complication of randomness. REPLY [3 votes]: In the case that the pursuers have to actually catch the fugitive, this was answered in the article Escaping an infinitude of lions by Mikkel Abrahamsen, Jacob Holm, Eva Rotenberg, and Christian Wulff-Nilse. They prove the following much more general theorem showing that the fugitive can always escape. Theorem. For any $\epsilon >0$, a fugitive running at speed at most $1+\epsilon$, can always escape countably infinitely many pursuers running at speed at most $1$. This holds for any starting position (provided no pursuer starts at the same point as the fugitive). The pursuers are even allowed to follow any pursuit strategy they like. Note that in their version the fugitive and pursuers are allowed to slow down, but as pointed out by Alessandro Della Corte, this can be simulated by constant speed strategies.<|endoftext|> TITLE: What are some consequences of zero free strip of the Riemann zeta function? QUESTION [27 upvotes]: A weaker version of the Riemann hypothesis is the claim that if $\zeta(s) = 0$ then $Re(s) \leq 1 - h$ for some constant $h> 0$. What would the consequences be of a result of this type? REPLY [14 votes]: Many number-theoretic functions have error bounds contingent on the supremum of the real parts of zeroes of the Riemann zeta function. Call this $\Theta$. If we write $f(x)=\Omega_{\pm}(g(x))$ for the notion that \begin{align*} \limsup_{x\to\infty}\frac{f(x)}{g(x)}&>0,\text{ and}\\ \liminf_{x\to\infty}\frac{f(x)}{g(x)}&<0, \end{align*} then (e.g. in Montgomery and Vaughan, Section 15.1), for any $\epsilon>0$, \begin{align*} \psi(x)-x&=\Omega_{\pm}(x^{\Theta-\epsilon}),\\ \pi(x)-\operatorname{li}(x)&=\Omega_{\pm}(x^{\Theta-\epsilon}),\\ M(x)&=\Omega_{\pm}(x^{\Theta-\epsilon}). \end{align*} Here, $\psi(x)=\log\operatorname{lcm}(1,2,\dots,n)$ is the Chebyshev Psi function, $\pi(x)$ is the prime counting function, $\operatorname{li}(x)$ is the logarithmic integral, and $M(x)$ is the Mertens function. In other words, the maximal error of these summatory functions from the standard "simple" approximations is (somewhat) precisely determined by how far the zeroes of $\zeta$ may stray from the line $\Re s=1/2$. (In fact, sharper results involving $\Omega_{\pm}(x^\Theta)$ directly can be shown if there is a zero $\rho$ with $\Re\rho=\Theta$). So, knowing that $\Theta<1$ gives quite strong information about the distribution of prime numbers, and as $\Theta\to1/2$ such information approximates the best possible bounds, at least as far as functions of the form $x^\alpha$ are concerned.<|endoftext|> TITLE: Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra? QUESTION [9 upvotes]: In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}[2]{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. In the quantized enveloping algebra $\mathrm{U}_q(\mathfrak{g})$ where we define $\bracket{x}{y}_q := xy-qyx$ is there a "nice" most general $q$-analogue to the Jacobi? I've found that $$ \bracket{x}{\bracket{y}{z}_q} \;\;=\;\; \bracket{\bracket{x}{y}}{z} _q + \bracket{y}{\bracket{x}{z}}_q \quad\text{ and }\quad \bracket{x}{\bracket{y}{z}_q} + \bracket{z}{\bracket{x}{y}_q} + \bracket{y}{\bracket{z}{x}_q} =0 $$ (which work of any power of $q$ really) but in general, allowing for twistings by different powers of $q$, the cleanest equation I've been able to write is that for any $m,n,k \in \mathbf{Z}$, $$ \bracket{x}{\bracket{y}{z}_{q^n}}_{q^m} + q^{m+k}\bracket{z}{\bracket{x}{y}_{q^{n-k}}}_{q^{-m-k}} + q^{m}\bracket{y}{\bracket{z}{x}_{q^{n-m-k}}}_{q^k} =0\,. $$ But that's not very pretty or symmetric! Is there a better way to write this thing? Or a better way to think about what this $q$-analogue should be? (I'll keep playing with this off-and-on, and edit if I figure anything out) REPLY [4 votes]: This question might be better understood in the context of the well-known problem of whether there exists a ``quantum Lie algebra'' inside $U_q(\frak{g})$. See this old M.O. question for a discussion. While there is no consensus about what a ``quantum Lie algebra'' should be, one interesting proposed solution is Majid's notion of a braided Lie algebra, as discussed for example in his Quantum Groups Primer book. Braided Lie algebras are also discussed in the old answer M.O. linked above.<|endoftext|> TITLE: Algorithms to factorize words into product of powers QUESTION [5 upvotes]: I came across this problem, which I guess is well known to combinatorialists of words, so I write here to see if someone can help me with some references. Let $A$ be a finite set of symbols, are there efficient algorithms that take as input a word $w$ over $A$ and return as output a representation of $w$ in terms of product of powers? If so, what are their computational complexities? Of course, this representation is not unique, for example for $A = \{0,1\}$ we have $$0101010010010 = (01)^3 (001)^2 0 = (01)^2 (010)^3 ,$$ so they might be algorithms that impose further restrictions on the exponents or on the lengths of the bases of the powers. Also, to avoid trivialities, if $w$ can be written as a product of powers of words with at least a power with exponent $>1$, then such algorithms should be able to find one of such representations (otherwise, obviously, we always have $w = w^1$). Thanks for help. REPLY [3 votes]: Unfortunately, I couldn't easily find a paper that solves exactly this problem. But I believe the algorithm from Factorizing Strings into Repetitions can be adapted to solve the problem. They find a factorization into repetitions, meaning that $s = s_1 s_2 \dots s_k$, where each factor $s_i$ can be represented as $s_i = x^k x'$ for some string $x$, where $k \geq 2$ and $x'$ is a prefix of $x$. Their algorithm consists of an efficient dynamic programming simulation and it likely can be modified to get rid of $x'$ and $k \geq 2$ constraint, while asking for the minimum number of factors.<|endoftext|> TITLE: Is it true that $\det\big[\sin 2\pi\frac{(j-k)^2}p\big]_{1\le j,k\le p-1}=-\frac{p^{(p-1)/2}}{2^{p-1}}$ for each prime $p\equiv3\pmod4$? QUESTION [9 upvotes]: Question. Does the equality $$\det\left[\sin 2\pi\frac{(j-k)^2}p\right]_{1\le j,k\le p-1}=-\frac{p^{(p-1)/2}}{2^{p-1}} $$ hold for every prime $p\equiv3\pmod4$? I have checked the equality numerically for $p=3,7,11$. I conjecture that the equality holds for each prime $p\equiv3\pmod4$, but I don't know how to prove this. Your comments are welcome! REPLY [6 votes]: We will use the notation $e_p(t)=\exp\left(\frac{2\pi it}{p}\right)$. First, let us show that for any $1\leq m\leq \frac{p-1}{2}$ there is an eigenvector of your matrix $A_p$ with eigenvalue $$ \lambda_m=\sqrt{p}\cos\frac{2\pi m^2}{p}. $$ To do so, for any $m\in \mathbb Z$ denote by $v_m$ the vector from $\mathbb C^{p-1}$ with $k$-th coordinate $e_p(2mk)$ for all $k\leq p-1$. Computation of $j$-th coordinate of $A_pv_m$ gives $$ (A_pv_m)_j=\sum_{k=1}^{p-1}e_p(2mk)\sin\frac{2\pi(j-k)^2}{p}= $$ $$ =\frac{1}{2i}\sum_{k=1}^{p-1}e_p(2mk)(e_p((j-k)^2)-e_p(-(j-k)^2))=\frac{1}{2i}(S_{+}(m,j)-S_{-}(m,j)). $$ Next, completing the sum, we get $$ S_{+}(m,j)=\sum_{k=0}^{p-1}e_p(2mk+(j-k)^2)-e_p(j^2). $$ Noticing that $2mk+(j-k)^2=(k+m-j)^2-m^2+2mj$, shifting the variables and using periodicity, we obtain $$ S_+(m,j)=e_p(-m^2+2mj)\sum_{k=0}^{p-1}e_p(k^2)-e_p(j^2). $$ It is known that the Gauss sum here equals $i\sqrt p$, thus $S_+(m,j)=i\sqrt pe_p(-m^2+2mj)-e_p(j^2)$. To compute $S_{-}(m,j)$, notice that $S_{-}(m,j)=\overline{S_+(-m,j)}=-i\sqrt{p}e_p(m^2+2mj)-e_p(-j^2)$. Combining these, we see that $$ (A_pv_m)_j=\sqrt{p}\cos\frac{2\pi m^2}{p}e_p(2mj)-\sin\frac{2\pi j^2}{p}. $$ Therefore, $A_pv_m=\lambda_m v_m-s$, where $s$ is a fixed vector whose $j$-th coordinate is $\sin\frac{2\pi j^2}{p}$. Noticing that $v_m\neq v_{-m}$ for $1\leq m\leq \frac{p-1}{2}$ and $\lambda_m=\lambda_{-m}$, we see that $$ A_p(v_m-v_{-m})=\lambda_m (v_m-v_{-m}). $$ Additional bonus here is the fact that coordinates of $v_m$ lie in $\mathbb Q(e_p(1))$. Let us take advantage of this. Denote $v_m-v_{-m}=w_m$. The cyclotomic field $\mathbb Q(e_p(1/4))=\mathbb Q(e_p(1),i)$ has an automorphism $\sigma$ that sends $e_p(1)$ to $e_p(-1)$ and $i$ to itself. One can define $\sigma$ via formula $\sigma e_p(1/4)=e_p((2p-1)/4)$ (which works, because $(2p-1,4p)=1$). This automorphism sends $\sin\frac{2\pi l}{p}$ to $\sin\frac{2\pi(p-1)l}{p}=-\sin\frac{2\pi l}{p}$ for integer values of $l$, so, for instance $A_p^\sigma=-A_p$. Therefore, $$ A_p(\sigma^{-1}w_m)=\sigma^{-1}(A_p^\sigma w_m)=-\sigma^{-1}(A_pw_m)=-\lambda_m \sigma^{-1}w_m. $$ Thus, all eigenvalues of $A_p$ are $\pm \lambda_m$ for $1\leq m\leq \frac{p-1}{2}$. This means that your identity is equivalent to $$ (-1)^{(p-1)/2}\left(\prod_{m\leq \frac{p-1}{2}}\sqrt{p}\cos\frac{2\pi m^2}{p}\right)^2=-\frac{p^{(p-1)/2}}{2^{p-1}}. $$ Finally, this identity is true if and only if $$ \left(\prod_{m\leq \frac{p-1}{2}}2\cos\frac{2\pi m^2}{p}\right)^2=1. $$ To establish this formula, notice that $p-m^2$ is not a square and it gives us the same cosine, so the identity is equivalent to $$ \prod_{m\leq p-1}2\cos\frac{2\pi m}{p}=1. $$ This last identity follows from the fact that $2\cos\frac{2\pi m}{p}=\frac{\sin\frac{4\pi m}{p}}{\sin\frac{2\pi m}{p}}$ and $m\mapsto 2m$ is a bijection on the set of invertible residues mod $p$. EDIT: The proof is wrong, because I forgot that $\sigma$ also acts on $\lambda_m$. However, it seems that $-\lambda_m$ is still always an eigenvalue and then the rest of the proof works just fine. EDIT 2: So, it turns out that the actual proof that $-\lambda_m$ is always an eigenvalue is quite difficult. I'll outline the proof here. First of all, we notice that for any $k$ we have $$ \sum_{j=0}^{p-1}\frac{1}{\cos\frac{2\pi j}{p}+\cos\frac{2\pi k}{p}}=0. $$ To prove that, one can notice that $$ \sum_{j=0}^{p-1}\frac{1}{z-\cos\frac{2\pi j}{p}}=\frac{T_p'(z)}{T_p(z)}=\frac{pU_{p-1}(z)}{T_p(z)}, $$ where $T_p$ and $U_{p-1}$ are Chebyshev polynomials. Also, $U_{p-1}(-\cos\frac{2\pi k}{p})=U_{p-1}(\cos\frac{\pi(p-2k)}{p})=0$, which concludes the proof of this observation. Now, the multisets $\{\cos\frac{2\pi m^2}{p}\}_{0\leq m\leq p-1}$ and $\{\cos\frac{2\pi m}{p}\}_{0\leq m\leq p-1}$ are the same, so we get, for instance, for all $m\not\equiv 0\pmod p$ $$ \sum_{j=0}^{p-1}\frac{1}{\lambda_j+\lambda_m}=0. $$ From this we see, for example, that $$ \sum_{j=1}^{p-1}\frac{\lambda_j-\lambda_0}{\lambda_j+\lambda_m}=p. $$ This is a very useful identity for our proof. Next, we already showed that $$ A_pv_m=\lambda_mv_m-s. $$ One should also notice that $v_0=-v_1-\ldots-v_{p-1}$ and $$ s=\frac{1}{p}\sum_{m=1}^{p-1}(\lambda_m-\lambda_0)v_m. $$ Next, take $$ w_m=\sum_{j=1}^{p-1}\frac{\lambda_j-\lambda_0}{\lambda_j+\lambda_m}v_j. $$ Applying formulas for $A_pv_j$, we get $$ A_pw_m=\sum_{j=1}^{p-1}\frac{\lambda_j-\lambda_0}{\lambda_j+\lambda_m}\lambda_jv_j-s\sum_{j=1}^{p-1}\frac{\lambda_j-\lambda_0}{\lambda_j+\lambda_m}. $$ Due to our formula, the last coefficient is $p$, so $$ A_pw_m=\sum_{j=1}^{p-1}\frac{\lambda_j-\lambda_0}{\lambda_j+\lambda_m}\lambda_jv_j-\sum_{j=1}^{p-1}(\lambda_j-\lambda_0)v_j. $$ Therefore, $A_pw_m=-\lambda_mw_m$, which finally concludes the proof.<|endoftext|> TITLE: Thurston universe gates in knots: which invariant is it? QUESTION [5 upvotes]: Today I discovered this nice video of a lecture by Thurston: https://youtu.be/daplYX6Oshc in which he explains how a knot can be turned into a "fabric for universes". For example, the unknot can be thought as a portal to Narnia, and when you pass again you switch back to the Earth. This forms in a sense a $\mathbb{Z}/2\mathbb{Z}$. Then he proceeds to explore what fabric one gets with the treefoil and you get an $S_3$. I am sure there is some real geometry behind but I can't grasp how to translate a portal into something homotopic. A way to formalize this would be the following. Take a knot $K$ in $\mathbb{R}^3$. Fix a tubular neighborhood $N$ of $K$. For each point $x$ in the knot take the loop $L_x$ obtained as the sphere bundle of $N$ at $x$ (a small circle around $x$ that jumps into the portal). Then there exist a connected 3-manifold $M$ with a (finite?) cover $M \to \mathbb{R}^3 \setminus K$ such that the "monodromy in small circles" around $L_x$ has order two for all $x$. Then we set the "group of universes" as the group of cover automorphisms. I think this captures the previous idea in the following sense: to each locus of $\mathbb{R}^3 \setminus K$ we have $n$ counterimages that represent the different worlds. Some branch should be chosen to make distinguishing between worlds possible. The constraint on monodromy ensures that if you jump twice through the same portal (at least for the portals very close to the boundary) you get back. Does such a manifold exist for all knots? Is this construction just some simplification of the fundamental group of the complement? REPLY [6 votes]: Here is a higher-quality video of the same material. My answer is a more algebraic version of Thurston's presentation, but I will tie this back to Thurston's "intention" at the end. Suppose that $L$ is an oriented knot diagram of a knot $K$. Let $A_i$ enumerate the "over-strands" of the diagram $L$. There is a presentation of $\pi_1(S^3 - K)$ with the $A_i$ as generators; here $A_i$ represents a meridional loop that goes (using the righthand rule) about the corresponding strand. The crossings give relations; if $A$, $B$, and $C$ are the strands at a crossing of $L$ then we deduce the relation $BA = AC$ or $CA = AB$ (depending on the sign of the crossing). This is called the Wirtinger presentation Thurston is adding the relation $A_i^2 = 1$ for all $i$. This gives a quotient of $\pi_1(S^3 - K)$. The quotient group has a presentation with, again, the $A_i$ as generators and with relations $A_i^2 = 1$ and $BA = AC$ (with $A$, $B$, and $C$ as above). In the second half of the video he builds a Cayley graph of the quotient group in the case of the trefoil. It is an exercise (using the Reidemeister moves) to show that the quotient group is independent of the given diagram $L$. Thus the quotient is a knot invariant. As mme points out, in their comment above, we can also obtain this group by quotienting by the smallest normal subgroup containing the square of any meridional element. This gives the orbifold fundamental group of the “orbifold” Dehn filling of the knot complement $S^3 - K$ along the slope $(2, 0)$. This orbifold has a universal cover which is a manifold. The covering degree is the number of lands in Thurston's discussion. For "most" hyperbolic knots this Dehn filling is hyperbolic and the covering degree is infinite. It is a nice exercise to show that the covering degree is finite for all two-bridge knots.<|endoftext|> TITLE: Local optimum for Sendov's conjecture QUESTION [7 upvotes]: For Sendov's conjecture, the distance 1 appears in the conjecture is tight, if one consider the polynomials $f_{n}(z) = z^{n} - 1$ for all $n\geq 2$. I wonder if this polynomial is the local optima for the conjecture. More precisely, I want to know if the following statement is true: For all $n\geq 2$, there exists $\epsilon = \epsilon_{n} > 0$ such that for all degree $n$ polynomials $f(z) \in \mathbb{C}[z]$ with $\| f - f_{n}\| < \epsilon$, $f$ satisfies the Sendov's conjecture. Here the norm is given as the $L^\infty$-norm of the coefficient vector ($\|a_{n}z^{n} + \cdots + a_{1}z + a_{0}\| = \max_{0\leq i\leq n} \{|a_{i}|\}$). I couldn't find any works on Sendov's conjecture in this direction. REPLY [6 votes]: This follows from the work of Miller, Michael J., On Sendov’s conjecture for roots near the unit circle, J. Math. Anal. Appl. 175, No. 2, 632-639 (1993). ZBL0782.30007. and independently Vâjâitu, Viorel; Zaharescu, A., Ilyeff’s conjecture on a corona, Bull. Lond. Math. Soc. 25, No. 1, 49-54 (1993). ZBL0796.30004. who established Sendov's conjecture when the distinguished zero is sufficiently close to the unit circle. By Rouche's theorem, any sufficiently small perturbation of $f_n$ will have its zeroes close enough to the unit circle for one of these two results to apply. If one uses the more recent result of Kasmalkar, Indraneel G., On the Sendov conjecture for a root close to the unit circle, Aust. J. Math. Anal. Appl. 11, No. 1, Article No. 4, 34 p. (2014). ZBL1293.30018. then one can obtain an explicit value of $\epsilon_n$ for your question, probably of polynomial type in $n$ (although the asymptotic behavior in $n$ is not so relevant now, due to my recent result establishing the conjecture for all sufficiently large $n$).<|endoftext|> TITLE: Continuously varying norms QUESTION [9 upvotes]: Let $V$ be a finite-dimensional real vector space with its Euclidean topology. Then all norms on $V$ are equivalent and consequently given two norms $\lVert-\rVert$, $\lVert-\rVert'$, the number $$ d = d(\lVert-\rVert, \lVert-\rVert') := \sup_{0 \neq v \in V}\big| \log\lVert v\rVert - \log\lVert v\rVert'\big| $$ is finite. This equips the set of all norms on $V$ with the structure of a complete metric space. You can think of $C := \exp(d)$ as the smallest real number $\ge 1$ such that $1/C \cdot \lVert-\rVert' \le \lVert-\rVert \le C \cdot \lVert-\rVert'$. Now let $\{\lVert-\rVert_t\}_{t \in \mathbf{R}}$ be a family of norms on $V$, parametrized by the real numbers (I would also be interested in more general parameter spaces). Let us say that this is a continuously varying family of norms on $V$ if the map $$ \mathbf{R} \times V \to \mathbf{R}_{\ge 0}, \qquad (t, v) \mapsto \lVert v\rVert_t $$ is continuous. Question: Let $\{\lVert-\rVert_t\}_{t \in \mathbf{R}}$, $\{\lVert-\rVert'_t\}_{t \in \mathbf{R}}$ be two continuously varying families of norms on $V$. Will the function $$ \mathbf{R} \to \mathbf{R}_{\ge 0}, \qquad t \mapsto d(\lVert-\rVert_t, \lVert-\rVert'_t) $$ be continuous? REPLY [4 votes]: I expand my comment where I claim that, on the space (call it $N(V)$) of all norms on $V$, the smallest topology making continuous the evaluations $\|\cdot\| \mapsto \|v\|$ (for $v \in V$) coincides with the topology defined by the distance $d$. This clearly implies that, under your hypothesis, the maps $t \mapsto \|\cdot\|_t$ and $t \mapsto \|\cdot\|'_t$ are continuous for the distance $d$, and in particular your question has a positive answer. This holds more generally if $T$ is an arbitrary topological space, $\{\|\cdot\|_t\}_{t \in T}$ is a family indexed by $T$ of norms on $V$, and if for every $v \in V$, $t\mapsto \|v\|_t$ is continuous. Let me justify the initial claim. A first observation is that, by homogeneity, if $B$ denotes the closed unit ball for a fixed norm on $V$, the topology given by $d$ coincides with the topology of uniform convergence on $B$. Moreover, $V$ being finite dimensional, $B$ is compact and therefore the topology given by $d$ coincides with the topology of uniform convergence on compact subsets of $V$. On the other hand let $v_1,\dots,v_n$ be a basis for $V$. Observe that, for an arbitrary $v=\sum_i t_i v_i$, $v'=\sum_i t'_i v_i$ and a norm $\|\cdot\|$ on $V$, we have $$ \left|\|v\| - \|v'\| \right| \leq \|v-v'\| \leq \sum_i |t_i-t'_i| \|v_i\|.$$ This implies that, for every constant $C$, $\{ \|\cdot\| \in N(V) | \forall i\leq n, \|v_i\| < C\}$ is made of equicontinuous functions on $V$. By the the Arzelà-Ascoli Theorem, in restriction to this set, the uniform convergence on compact subsets of $V$ coincides with pointwise convergence. This proves the claim because the sets $\{ \|\cdot\| \in N(V) | \forall i\leq n, \|v_i\| < C\}$ form a exhaustion of $N(V)$ by open sets.<|endoftext|> TITLE: Moduli space of germs of riemannian metrics QUESTION [11 upvotes]: Let $S$ be the set of germs of riemannian metrics near $0$ on $\mathbb R^n$. It is acted on by the group $\textrm{Diff}$ of germs of diffeomorphisms of $\mathbb R^n$ preserving $0$. Let's denote by $S^{(k)}$ the set of $k$-jets of riemannian metrics at $0$ (first $k$ terms of the Taylor expansion). The $k$-th jet group $\textrm{Diff}_k$ is acting on it. It is natural to consider the quotients $S^{(k)}/\textrm{Diff}_k$ $S^{(k)}$ is a convex cone in an affine space of all jets so there should not be problems with the notion of dimension. I don't know if $S^{(k)}$ is semi-algebraic or no. My questions are: Is the Poincaré series $\sum \textrm{dim}(S^{(k)}/\textrm{Diff}_k)\frac{t^k}{k!}$ rational? Is there any reasonable description of "the moduli space of germs of riemannian metric" $S/\textrm{Diff}$ or at least $S^{(k)}/\textrm{Diff}_k$? I know that in normal coordinates the Taylor expansion of a metric has identity first term and the rest of the terms expressed as complicated contractions of curvature and its covariant derivatives, but it didn't help me to answer question 1. Have the above questions been considered for other geometric structures (Ricci-flat metrics, hyper-Kahler structures etc.)? Any buzzword would be appreciated. REPLY [21 votes]: The answers to these questions are known, but, perhaps, not well-known. The typical approach is to first divide only by the local diffeomorphisms $\phi:\mathbb{R}^n\to\mathbb{R}^n$ that fix the origin and for which $\phi'(0):\mathbb{R}^n\to\mathbb{R}^n$ is the identity. This quotient is essentially sectioned by geodesic normal coordinates $(x^i)$, in which a given metric $g=g_{ij}(x)\,\mathrm{d}x^i\mathrm{d}x^j$ satisfies $$ g_{ij}(x)\,x^j = g_{ij}(0)\,x^j. $$ Such a $g$ can be expanded in Taylor series $$ g_{ij}(x) = g_{ij}(0) + g^1_{ij}(x) + \cdots + g^k_{ij}(x) + \cdots $$ where $g^k_{ij}=g^k_{ji}$ is homogeneous in $(x^i)$ of degree $k$. For $k>0$, the dimension of the vector space of such $\bigl(g^k_{ij}(x)\bigr)$ that satisfy the relations $g^k_{ij}(x)\,x^j=0$ is easily computed to be $$ D_{k}(n) = {n\choose2}\,{{n+k-1}\choose{k}}-n\,{{n+k-1}\choose{k+1}}=\frac{n\,(k{-}1)}2\,{{n+k-1}\choose{k+1}}, $$ while $D_0(n) = {{n+1}\choose2}$. The group $\mathrm{GL}(n,\mathbb{R})$, as linear changes of the coordinates $(x^i)$, acts on the sum of these vector spaces. Since $g_{ij}(0)$ is positive definite, the action is transitive on the $0$-degree piece, with stabilizer $\mathrm{O}(n)\subset\mathrm{GL}(n,\mathbb{R})$, so this piece can be discarded and one can assume that $g_{ij}(0) = \delta_{ij}$. Thus, one can regard the quotient space of diffeomorphism-equivalence classes of $k$-jets for $k\ge 2$ as the quotient of the sum of the $j$-th homogeneous pieces for $2\le j\le k$ divided by a natural action of $\mathrm{O}(n)$. For $n>2$, the generic element has a finite stabilizer, so the dimension of the named quotient space is $$ \dim (S^{(k)}/\mathrm{Diff}_{k+1}) = D_2(n) + \cdots + D_k(n) - \dim\mathrm{O}(n), $$ for $k\ge 2$. In particular, the Poincaré series of the graded vector space of the positive degree terms, i.e., $D_2(n)\,t^2 +D_3(n)\,t^3 + \cdots$ is a rational function $p_n(t)/(1{-}t)^n$, where $p_n(t)$ is a polynomial of degree $n$ with leading term $(-1)^n{n\choose2}\,t^n$, corresponding to the fact that, in Cartan's sense, Riemannian metrics in dimension $n$ depend, modulo diffeomorphism, on $n\choose2$ functions of $n$ variables. (For $n=2$, there is a slight difference in the numbers because $O(2)$ does not act with finite stabilizer on the generic $2$-jets, but it does act with finite stabilizer on the generic $k$-jets for $k>2$, so the answer to Question (1) is 'yes' in this case as well.) As far as the answer to Question (2), for $S^{(k)}/\mathrm{Diff}_{k+1}$, it reduces to finding a description of the moduli space of orbits of the compact group $\mathrm{O}(n)$ on a finite dimensional vector space, which is a question in the invariant theory of representations of compact groups. (Generally, such a moduli space is embedded into some $\mathbb{R}^m$ as a closed semi-algebraic set by a set of $m$ generators of the ring of $\mathrm{O}(n)$-invariant polynomials on the vector space, but, for all but the simplest cases, finding such a set of generators and describing their relations is very difficult.) Finally, answers for Question (3) are generally available. They depend on the theory of involutive systems of PDE, which I'm most familiar with in the form developed by Élie Cartan. For example, for hyper-Kähler metrics in dimension $4$ (which, interestingly, Cartan treated in 1926, long before the term 'hyper-Kähler' was invented), one finds that the $k$-jets of such metrics can be written in 'normal coordinates' in a manner analogous to that above for general metrics in such a way that they are the orbit space of $\mathrm{SU}(2)$ acting on a finite dimensional vector space $\mathcal{H}^k$. These vector spaces have the property that $$ \dim \mathcal{H}^k - \dim \mathcal{H}^{k-1} = (k+3)(k-1) $$ for $k\ge 2$, and $\mathcal{H}^{0}= \mathcal{H}^{1}= (0)$. The generic $\mathrm{SU}(2)$-stabilizer of a $2$-jet is finite, so the dimension of the $k$-jet orbit space is $\dim\mathcal{H}^k - 3$ for $k\ge2$. In particular, the analogous Poincaré series is indeed rational (which turns out to be a quite general phenomenon). Remark: For more details on how this works, one might consult these notes of mine on exterior differential systems and their applications to problems of this kind. As a particular example (just to keep the discussion simple), consider the case of describing the space of germs of the $G$-structures for $G\subset\mathrm{O(n)}$ that have vanishing intrinsic torsion and whose curvature tensor $R$ is required to satisfy some identity of the form $f(R) = 0$ (which may be trivial). Some examples are (i) Riemannian metrics ($G=\mathrm{O}(n)$; no relation on the curvature), (ii) Ricci-flat metrics ($G=\mathrm{O}(n)$; $\mathrm{Ric}(R)=0$), (iii) Conformally flat metrics in dimensions $n\ge 4$ ($G=\mathrm{O}(n)$; $\mathrm{Weyl}(R)=0$), (iii) Self-dual Einstein metrics in dimension $n=4$ ($G=\mathrm{O}(4)$; $\mathrm{Weyl}_-(R)=\mathrm{Ric}^0(R)= 0$), (v) Kähler metrics ($G=\mathrm{U}(\tfrac12n)\subset\mathrm{O}(n)$; no relation on the curvature), (vi) torsion-free $\mathrm{G}_2$-structures $n=7$ ($G=\mathrm{G}_2\subset\mathrm{O}(7)$; no relation on the curvature), etc. The possible curvature tensors for such a $G$-structure take values in a subset $A\subset \mathrm{Sym}^2\bigl(\Lambda^2(\mathbb{R}^n)\bigr)$ that is frequently a subspace invariant under the natural action of $G$ on $\mathrm{Sym}^2\bigl(\Lambda^2(\mathbb{R}^n)\bigr)$. In such cases, $A$ will often have the property that it is an involutive tableau of second order when regarded as a subspace of $\Lambda^2(\mathbb{R}^n)\otimes \Lambda^2(\mathbb{R}^n)$, with Cartan characters $s_2,\ldots,s_n$. Assuming that the PDE represented by $f(R)=0$ is, itself, involutive (again, true in all the cases listed above), one finds that the space of diffeomorphism classes of $k$-jets of $G$-structures satisfying $f(R)=0$ can be identified with the quotient (by the natural action of $G$) of the vector space $$ A\oplus A^{(1)}\oplus \cdots\oplus A^{(k-2)} $$ where $A^{(i)}$ is the $i^{th}$-prolongation of $A$ (as a second order tableau). Consequently, the Poincaré series of the graded vector space $$ \mathcal{A} = A^{(0)}\oplus A^{(1)}\oplus \cdots\oplus A^{(k)}\cdots $$ (where $A = A^{(0)}$ and where $A^{(i)}$ is assigned degree $i{+}2$) is $$ P(t) = t^2\left(\frac{s_2}{(1{-}t)^2} + \cdots +\frac{s_n}{(1{-}t)^n}\right). $$ So, for example, for the general metric in dimension $n$, we find $$ s_k = \tfrac12\,n(k{-}1)(n{-}k+1)\qquad 2\le k\le n $$ while, for Ricci-flat metrics in dimension $n$, we find $$ s_k = \tfrac12\,n(k{-}1)(n{-}k+1)\ \ (2\le k\le n{-}2); \quad s_{n-1} = n(n{-}3);\quad s_n = 0. $$ For the case of hyperKähler metrics in dimension $n=4m$, one finds that $$ s_k = \tfrac12\,(k{-}1)(2m{+}2{-}k)(2m{+}3{-}k)\ \ (2\le k\le 2m{+}1)\ \text{and} \ s_k = 0\ (2m{+}2\le k). $$ Finally, for the case (vi) above (torsion-free $\mathrm{G}_2$-structures in dimension $7$), the system is involutive with $$ (s_2,\ldots,s_7) = (14,21,21,15,6,0). $$<|endoftext|> TITLE: Borel equivariant homology of a suspension QUESTION [8 upvotes]: Let $G$ be a discrete group. For a $G$-CW complex $X$, let $H^G_{\bullet}(X)$ denote the Borel equivariant homology of $X$. There are also relative versions of this. Here's my question. Let $X$ be a $G$-CW complex. The suspension $\Sigma X$ is then a $G$-CW complex in a natural way, and has two $G$-invariant base points coming from the two suspension points. Let $p_0$ be one of the suspension points. How can we compute $H^G_{\bullet}(\Sigma X,p_0)$ from $H^G_{\bullet}(X)$? More generally, is there a good reference on Borel equivariant (co)homology where these sorts of foundational questions are worked out? All the references on equivariant homotopy theory I've looked at seem focused on different things, but Borel equivariant homology is what comes up in the things I'm working on right now. REPLY [12 votes]: I assume that by Borel equivariant homology of $X$ you mean the ordinary homology of the "Borel construction" $X\times_G EG$. There is a homotopy cofibration sequence $$ X\times_G EG \to *\times_G EG \to \Sigma X\wedge_G EG_+. $$ It induces a long exact sequence in homology, which you can interpret as a long exact sequence of Borel homology groups $$\cdots \to H_n^G(X)\to H_n^G(*)\to H_n^G(\Sigma X, p_0)\to H_{n-1}^G(X) \to \cdots $$ Of course, $H_\bullet^G(*)$ is the same as the homology of $BG$, or the group homology of $G$. If $X$ is a pointed $G$ space (in other words, if $X$ has a point $p$ fixed by $G$), then there is an isomorphism $$H_\bullet^G(X)\cong H_\bullet^G(p) \oplus H_\bullet^G(X, p).$$ In this case the long exact sequence simplifies to an isomorphism $$H_\bullet^G(\Sigma X, p_0)\cong H_{\bullet-1}^G(X, p).$$ This is just an equivariant version of the suspension isomorphism for homology. But if $X$ does not have a $G$-fixed point then I think the long exact sequence above is the best general result you can have.<|endoftext|> TITLE: Heuristic lower bounds on small sums of roots of unity QUESTION [11 upvotes]: Let $f(k,n)$ be the smallest non-zero absolute value of a sum of $k$ complex $n$th roots of unity. Asking for bounds in either direction, Tao suggested that a polynomial lower bound seemed plausible based on probabilistic intuition. I can imagine a few things this might mean: for example, when $n$ and $k$ are even, a pigeonhole argument proves an $n^{-(1+o(1))k/4}$ upper bound, and maybe we should expect this to be roughly tight if the pigeons are distributed randomly. The best lower bound we have is $k^{-n}$. What heuristic lower bounds are there on $f(k,n)$? REPLY [19 votes]: One heuristic is to replace the $n^{th}$ roots of unity by $n$ iid elements $\zeta_1,\dots,\zeta_n$ of the unit circle, drawn uniformly at random. For any sum $\zeta_{i_1} + \dots + \zeta_{i_k}$ of $k$ of these with $i_1 < \dots < i_k$, a standard Fourier analytic calculation (Esseen concentration inequality) shows that if $k \geq 5$ (to make the $k$-fold convolution power of the Fourier transform of the unit circle have absolutely integrable Fourier transform), then $$ {\bf P}( |\zeta_{i_1} + \dots + \zeta_{i_k}| < r ) \ll_k r^2 $$ for $0 < r < 1$. Taking the union bound over all $\binom{n}{k}$ different sums, we see that the probability that $k$ distinct elements of the $\zeta_1,\dots,\zeta_n$ sum to something of magnitude less than $r$ is $O_k( \binom{n}{k} r^2 )$. This shows that with probability $\gg 1$, we should have the lower bound $$ |\zeta_{i_1} + \dots + \zeta_{i_k}| \gg_k \binom{n}{k}^{-1/2} \asymp_k n^{-k/2}$$ for all $i_1 < \dots < i_k$. This is for a fixed $n$ (which after averaging corresponds to predicting a lower bound $\gg_k n^{-k/2}$ for a positive density fraction of $n$, rather than for all $n$). To get a bound that is true almost surely for all but finitely many $n$ (in the spirit of the "strong" law of large numbers, as opposed to the "weak" law), the Borel-Cantelli lemma suggests that we need to get the failure probability for a fixed $n$ down to something like $1/n^{1+\varepsilon}$, and so the lower bound now worsens from $\gg_k n^{-k/2}$ to $\gg_{k} n^{-(k+1+\varepsilon)/2}$. Some remarks: The gap between the $k/2$ type exponents for the lower bound and the $k/4$ type exponent for the upper bound is related to the square root present in the birthday paradox. These bounds are fairly reversible in this model due to all the independence, and with a little more effort one should be able to show that the bounds of $\gg_k n^{-k/2}$ (for a positive density set of $n$) and $\gg_k n^{-(k+1+\varepsilon)/2}$ (for all but finitely many $n$) cannot be significantly improved except possibly for the $\varepsilon$ factor, but I have not attempted to work this out rigorously. To complete the heuristic analysis one should also treat the cases where there are collisions amongst the $i_1,\dots,i_k$, which can be handled by the same methods as long as there are at least $5$ distinct values of the $i_1,\dots,i_k$; the cases of only four or fewer values need to be treated by more direct computations but I think they should give lower order contributions in the $k \geq 5$ regime. EDIT: while the analysis of the above random model remains accurate within the scope of that model, I have just realised that the rotation invariance symmetry of the $n^{th}$ roots of unity leads to a larger lower bound prediction than the above naive model, as it indicates that the roots of unity have some relevant structure to them that prevents them from behaving like completely random elements of the unit circle. The key point is that for the original $n^{th}$ roots of unity problem, one can always rotate one of the roots to equal $1$. So one should really be looking at sums of the form $1 + \zeta_{i_2} + \dots + \zeta_{i_k}$ with $1 < i_2 < \dots < i_k$. This effectively lowers $\binom{n}{k}$ to $\binom{n-1}{k-1}$ and raises all the exponents in the previous analysis by $1/2$, thus the revised prediction would be a lower bound of $\gg_k n^{-(k-1)/2}$ for a positive fraction of $n$ and $\gg_k n^{-(k+\varepsilon)/2}$ for all but finitely many $n$. There are some additional symmetries coming from cyclic permutation of the roots and conjugation symmetry but these only affect the $k$-dependent implied constants and not the exponents. In principle, the further Galois group symmetries of the problem could lead to additional refined corrections to the model, but I tentatively am of the opinion that this will not change the predicted exponents further. REPLY [2 votes]: Not an answer but Philipp Habegger has done some interesting work that could be characterized as evidence in support of the (folklore) polynomial conjecture which you can find at: https://arxiv.org/abs/1611.07287 The work is exposited in his talk at: https://www.youtube.com/watch?v=dqIVzojueww<|endoftext|> TITLE: Does sine interact equationally with addition alone? QUESTION [20 upvotes]: $\DeclareMathOperator\Eq{Eq}\DeclareMathOperator\Th{Th}$Originally asked at MSE without success: For a structure $\mathcal{A}$ whose signature only contains function and constant symbols, let $\Eq(\mathcal{A})$ be the equational theory of $\mathcal{A}$ — that is, the set of all universal closures of atomic formulas which are satisfied by $\mathcal{A}$. Via trigonometric identities like the double angle identity, we have that $\Eq(\mathbb{R};+,\times,\sin)$ does not "reduce" to $\Eq(\mathbb{R};+,\times)$; formally, we have $\Eq(\mathbb{R};+,\times)\not\models \Eq(\mathbb{R};+,\times,\sin)$. Notably, all such examples I know crucially involve all three operations available. I'm curious whether this is necessary. Since this question seems harder than I originally suspected, I'll focus on addition specifically: Does $\Eq(\mathbb{R};+)\models \Eq(\mathbb{R};+,\sin)$? I suspect that the answer is yes; in a sense this would amount to saying that trigonometric identities have to use multiple arithmetic operations, which matches at least my experience. However, I don't see how to prove this. Note that $\Eq(\mathcal{A})$ is generally far less informative than $\Th(\mathcal{A})$. For example, suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is "sufficiently wild." Then $\Th(\mathbb{R};f)$ will be nontrivial (e.g. it will say "$f$ is not injective") but no nontrivial equation will hold in $(\mathbb{R};f)$ so $\emptyset\models \Eq(\mathbb{R};f)$. So the relative tameness of $(\mathbb{R};+)$ from the "coarse" perspective of first-order logic can't help us here. REPLY [11 votes]: Here is the outline of a proof that $Eq(\mathbb{R},+)$ proves all the equalities in both $Eq(\mathbb{R},+,\sin)$ and $Eq(\mathbb{R},+,\exp)$. Step 1, defining $<$: For any term $t$ in $L(+,\sin)$, define a real function $t_R$ by $$t_R(r_1,\ldots r_n)=\sup(\{|t(z_1,\ldots,z_n)|: z_1,\ldots,z_n \in \mathbb{C}, |z_1|=r_1, \cdots, |z_n|=r_n\})$$ Then for terms $t$ and $u$ in $L(+,\sin)$, define $t1$, then $\sum_{i=1}^m t_i<\sin(u)$ iff $t_i < \sin(u)$ for all $i\le m$. if $m>1$, then $\sum_{i=1}^m t_i<\exp(u)$ iff $t_i < \exp(u)$ for all $i\le m$. if $m\ge 1$, $n>1$ with $t_1 \ge \cdots \ge t_m$ and $u_1 \ge \cdots \ge u_n$, and none of the $t$'s and $u$'s are sums, then $\sum_{i=1}^m t_i<\sum_{j=1}^n u_j$ iff either $t_k < u_k$ at the first index $k$ where the $t$'s and $u$'s differ, or $mN(u)$. In the first and last cases, $t$ and $u$ are not equal over $\mathbb{R}$; for $\sin$ the definition of $<$ tells us that they are different over $\mathbb{C}$, and then the identity theorem tells us that they are also different over $\mathbb{R}$. In the middle case, $Eq(\mathbb{R},+)$ is enough to establish $t=N(t)=N(u)=u$.<|endoftext|> TITLE: Representations of $U_q(\mathfrak{sl}(2))$ as differential / difference operators QUESTION [12 upvotes]: $\mathfrak{sl}(2)$ (over $\mathbb{C}$) with basis $E_\pm, H$ with commutation relations $$ [H,E_{\pm}]=\pm 2 E_\pm,\quad [E_+,E_-]=H $$ admits the well-known representation on $\mathbb{C}[x]$ with $$ E_+ = \partial_x,\quad E_- = -x^2 \partial_x + s\,x,\quad H = -x \partial_x - s $$ where $\partial_x = \frac{d}{dx}$. This representation is highest-weight with highest-weight vector given by $1\in \mathbb{C}[x]$. The parameter $s$ is free to take any value in $\mathbb{C}$. Furthermore this differential operator realisation can also be used if one changes the space of functions on which we act. For example, the principle series representations can be realised in this way. Question: Can highest-weight representations of $U_q(\mathfrak{sl}(2))$ be realized in this way? I am aware that $U_q(\mathfrak{sl}(2))$ admits a representation on the quantum plane $\mathbb{C}_q[x,y]$ - this is not the representation I am looking for since that has a direct counterpart for usual $\mathfrak{sl}(2)$, which is not the representation I described at the beginning. I am familiar with the textbooks by Chari and Pressley and by Kassel but have not come across such a representation. I have tried to construct it directly, granted using a rather naive approach and substituting the differential operator $\partial_x$ with the $q$-differential operator $D_q$. This did not work due to the fact that the Liebniz rule for q-differentiation involves a multiplication by $q$ in the functions which are not differentiated. q-derivative Does such a representation exist? If so please provide a reference - I am also interested in the generalisation to higher-rank cases. Edit: The representation on $\mathbb{C}_q[x,y]$ is not the representation I am looking for as in the $q\rightarrow 1$ limit this reduces to a representation on $\mathbb{C}[x,y]$, not $\mathbb{C}[x]$ as I described in the initial paragraph of the question. The representation I am looking for should reduce to the one described initially for $q\rightarrow 1$. REPLY [4 votes]: Although i have some doubts as to what the OP is exactly looking for (see my comments above), i hope that the following will be of some interest for its purposes. In: $U_q(sl(n))$ Difference Operator Realization, A. Shafiekhani, the author introduces a unified scheme for constructing differential operator realizations for any irreducible representation of $U_q(sl(n))$. These come as $q$-deformations of the corresponding realizations for $sl(n)$ irreducible representations. What is actually done in this paper, is the generalization of results shown in: q-difference realizations of quantum algebras, R. Floreanini, L. Vinet, Phys. Let. B, v. 315, 3–4, p. 299-303, 1993 where formally similar realizations are obtained as $q$-deformations of the differential/difference realizations of totally symmetric $sl(2)$ and $sl(3)$ representations. (in my understanding, though i admit i have given just a quick look at this last reference, the realizations given there are some quotient of the $U_q(sl(2))$ representations on the quantum plane $\mathbb{C}_q[x,y]$ mentioned in the OP). Edit: One can find similar deformed differential/difference realizations for $q$-deformations of other Lie algebras as well. For some more cases in a similar spirit one can see for example: Realization of $U_q(so(N))$ within the differential algebra on $R^N_q$, G. Fiore, Comm. in Math. Phys. v. 169, p. 475–500, (1995), Realization of $U_q(sp(2n))$ within the Differential Algebra on Quantum Symplectic Space, J. ZHANG, N. HU, SIGMA 13 (2017), 084 Edit 2: For the completeness of this answer, i think it might be interesting to mention that there are more general, systematic ways of building $q$-deformed differential realizations of the various deformations $U_q$ of the UEA of (super)Lie algebras. These are using some $q$-deformed CCR/CAR (super)algebra $W_q$ (that is $q$-deformed Weyl/Clifford (super)algebras) in order to realize the quantum group with an algebra $D_q$ of deformed differential operators, acting on spaces spanned by polynomials mixing usual (commutative) and Grassman (non-commutative) variables, according to the general scheme: $$ U_q\stackrel{(I)}{\longrightarrow} W_q\stackrel{(II)}{\longrightarrow} D_q $$ In this sense, the $D_q$ representations are pulled back to representations of the quantum group $U_q$. There is a significant amount of literature on similar techniques (usually found in mathematical physics journals). For example, one can find realizations (of type $(I)$) of quantum groups in terms of $q$-deformed bosons (or fermions) at: A superalgebra $U_q[(osp(3/2)]$ generated by deformed paraoperators and its morphism onto a $W_q(1/1)$ Clifford–Weyl algebra, A SUPERALGEBRA MORPHISM OF $U_q[OSP(1/2N)]$ ONTO THE DEFORMED OSCILLATOR SUPERALGEBRA $W_q(N)$, see also: arXiv:hep-th/9303142, The quantum group $SU_q(2)$ and a $q$-analogue of the boson operators The $q$-deformed boson realisation of the quantum group $SU(n)_q$ and its representations Quantum lie superalgebras and q-oscillators And deformed differential realizations (of type $(II)$) of $q$-bosons at: The $q$‐analog of the boson algebra, its representation on the Fock space, and applications to the quantum group<|endoftext|> TITLE: Available frameworks for homotopy type theory QUESTION [16 upvotes]: I am thinking about trying to formalise some parts of classical unstable homotopy theory in homotopy type theory, especially the EHP and Toda fibrations, and some related work of Gray, Anick and Cohen-Moore-Neisendorfer. I am encouraged by the successful formalisation of the Blakers-Massey and Freudenthal theorems; I would expect to make extensive use of similar techniques. I would also expect to use the James construction, which I believe has also been formalised. Some version of localisation with respect to a prime will also be needed. My question here is as follows: what is the current status of the various different libraries for working with HoTT? If possible, I would prefer Lean over Coq, and Coq over Agda. I am aware of https://github.com/HoTT/HoTT, which seems moderately active. I am not clear whether that should be regarded as superseding all other attempts to do HoTT in Coq such as https://github.com/UniMath. I am also unclear about how the state of the art in Lean or Agda compares with Coq. REPLY [11 votes]: The HoTT libraries in Lean can be considered dead. Since Lean has moved away from HoTT, I don't think it's a more convenient system to do HoTT in than - for example - Coq. There is still some formalization material in the Lean 2 library that hasn't been formalized in another proof assistant, but probably the best thing to do with that is to port it to other libraries instead of working on top of it. About the Coq-HoTT vs Unimath libraries: I've used both a bit, and I found the Coq-HoTT library way more pleasant to work with. Also, it focuses more on synthetic homotopy theory than Unimath, which focuses more on doing "ordinary mathematics" in a univalent style. I haven't used Agda much, but it has a good library for synthetic homotopy theory. Last time I checked it had quite a bit more in that direction that Coq-HoTT, but it has way less activity.<|endoftext|> TITLE: Gaussian Integer a+2bi as sum of 3 squares QUESTION [9 upvotes]: Let $a, b \in \mathbb{Z}$. Then is it true that the Gaussian integer $a+2bi$ can be expressed as a sum of three squares? REPLY [4 votes]: The question is not new. See my old question and its answer available from Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?<|endoftext|> TITLE: Matrix-valued ordinary differential equation with symmetry QUESTION [6 upvotes]: I am considering the following equation $$\begin{pmatrix} -\frac{d}{dx} + \lambda \sin(2\pi x) & \lambda - \lambda \cos(2\pi x) \\ -\lambda-\lambda \cos(2\pi x) & -\frac{d}{dx} - \lambda \sin(2\pi x) \end{pmatrix}\varphi(x)=0$$ and I am wondering if there is an explicit characterization of $\lambda \neq 0$ for which there exist $1$-periodic solutions to this ODE. This ODE, though matrix-valued has a nice symmetry in the sense that the off-diagonal entries have only even Taylor polynomials whereas the diagonal ones have odd ones. However, it is of course difficult to say based on a Taylor series, if this will yield a periodic function. Observation: Close to zero, i.e. $x=0$, the solution should heuristically look like $\varphi(x)=(0,e^{-\lambda \pi x^2}),$ as it solves the equation "to leading-order" REPLY [7 votes]: Your equation in fact admits a simple explicit solution. As noted in the previous answer, the general solution is encoded in a matrix ODE, $$ M'(x) = \lambda A(x)M(x), \qquad M(0)=I\,. $$ Now, the simplification becomes apparent if change to a new basis, using the $x$-dependent rotation $$ R(x) = \begin{pmatrix} \cos(\pi x) & -\sin(\pi x) \\ \sin(\pi x) & \cos(\pi x)\end{pmatrix}. $$ Define $N(x)$ by $M(x)=R(x)N(x)$, and the ODE becomes $$ N' = \left(\lambda R^{-1}AR-R^{-1}R' \right)N = \begin{pmatrix} 0 & \pi \\ -\pi-2\lambda & 0\end{pmatrix} N. $$ Writing $\lambda = \frac{\pi}{2}(\mu^2-1)$, the solution is $$ N(x) = \begin{pmatrix} \cos (\pi\mu x) & \frac{1}{\mu} \sin (\pi\mu x) \\ -\mu \sin (\pi\mu x) & \cos (\pi\mu x) \end{pmatrix} \implies M(1) = \begin{pmatrix} -\cos (\pi\mu) & -\frac{1}{\mu} \sin(\pi\mu) \\ \mu \sin(\pi\mu) & -\cos(\pi\mu) \end{pmatrix}. $$ We get periodic solutions when $\mu=2n+1$, $n=1,2,3,\ldots$, so $\lambda=2\pi n(n+1)$.<|endoftext|> TITLE: Non-enriched Bousfield localizations QUESTION [5 upvotes]: We know that whenever we have a Bousfield localization between two simplicial model categories, this gives rise to a reflective subcategory in $\infty$-categories (or coreflective, depending on the direction of the Bousfield localization). I'm interested in what happens if the model categories at issues are not simplicial, or even in the intermediate case when the categories themselves are simplicial, but we don't know that the functors that comprise the Bousfield localization are simplicial. Does this still give rise to a reflection of $\infty$-categories? At least, when the model categories are combinatorial? Related question: we know by a result by Dugger that every combinatorial model category is Quillen equivalent to a combinatorial simplicial model category. Is this assignation functorial? Do functors and adjunctions between combinatorial model categories get promoted to simplicial functors and adjunctions? REPLY [2 votes]: This is not a full answer, but too long for a comment. Here's a relevant paper. Mazel-Gee proves the folklore claim that a Quillen adjunction induces an adjunction on underlying $\infty$-categories with very little assumption on the model categories involved. I think it's proved (or mentioned) in the appendix (at least) that upon passing to homotopy categories, the co/unit is the same as the one in the derived adjunction introduced by Quillen, in particular, in the case of a Bousfield localization, the (co)unit is an equivalence, and so it is an equivalence also at the level of $\infty$-categories ($C\to ho(C)$ is conservative), and so it also induces a reflective subcategory. In other words, I think you don't need properness or anything like that. For your question 2, I'm not sure about functoriality per se, but Quillen adjunctions definitely induce Quillen adjunctions: that's because Dugger's universal homotopy theory satisfies some almost universal property. You can have a look e.g. at propositions 2.3, 5.10 and theorem 6.3 in Dugger's paper : suppose $M\rightleftarrows N$ is a Quillen adjunction, then by 6.3 you can make it into a Quillen adjunction $UC/S \rightleftarrows N$ for some $C$ and $S$ and then with 5.10 you can lift it to $UC/S \rightleftarrows UD/T$ To ask whether it can be made simplicial is to ask whether 2.3 can be made simplicial, i.e. suppose given $\gamma : C\to M$ where $M$ is a simplicial model category, can the functor $Re: UC\to M$ from 2.3 be made simplicial ? Now I'm not entirely sure the answer is yes, but I would guess that it is if $M$ is nice enough. At least, Lurie seems to indicate something similar in the last sentence of the proof of A.3.7.6. from Higher topos theory : "The proof given in [19] [ Dugger's paper ] shows that when $\mathbf A$ is a simplicial model category , then $F$ and $G$ can be chosen to be simplicial functors" - the context is not quite the same, so that is not claimed in HTT, but something similar is. Hopefully someone can comment on whether this is the case (that's why this answer is not a complete answer)<|endoftext|> TITLE: What are prime number values of the trinomial $q(n) = n^2 + n + 41$? Assuming $n$ is a positive integer QUESTION [6 upvotes]: Are there infinitely many integer values $n$ such that $q(n)$ is a prime number? Numerical evidence points to a yes answer. This is similar to Landau's 4th problem from 1912. (The conjecture that there are infinitely many primes $p$ of the form $p=n^2+1 $?) Of course, Landau did not have a computer. Given n a positive number, for what values of $n$ is $q(n)=n^2 + n + 41 $ a prime number? This is known as Prime-Generating Polynomial. see link https://mathworld.wolfram.com/Prime-GeneratingPolynomial.html also Wikipedia https://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions also my document on this quadratic polynomial https://mersenneforum.org/showthread.php?p=581027#post581027 There are 3 .pdf files hosted at mersenne.org. I characterize all the cases when n^2 + n + 41 can be a composite number. Assuming positive integer n. I show a data table, graph, and curve fit to characterize all the cases when this trinomial is a composite number, up to a certain numerical limit. Also, I have found some algebraic factorizations for q(n) https://sites.google.com/site/mattc1anderson/prime-producing-polynomial A leading question is, "If we can know whenever n^2+n+41 is composite, what does that tell up about when that trinomial is a prime number?" Let me know if there are any questions. Matt REPLY [11 votes]: Since the ring of integers of $\mathbb{Q}[\sqrt{-163}]$ is a PID, it follows that a rational prime $p \neq 163$ may be expressed in the form $x^{2} + xy + 41y^{2}$ for rational integers $x$ and $y$ if and only if $p$ is a quadratic residue (mod $163$).(This is well-known). But, as you point out yourself, your question is comparable to asking how many primes $p$ have the form $n^{2} +1$ for integer $n$, which is well known to be open and Wojowu confirms in comments that your question is open too. Later edit: I find it mildly interesting that the prime $p$ is expressible in this way (ie $p = n^{2}+n+41$) if and only if $p$ is expressible as the sum of four integer squares in one of the following ways: If $n$ is odd, we find that $p = \left( \frac{n-9}{2} \right)^{2} + \left( \frac{n+1}{2} \right)^{2} +\left( \frac{n+1}{2} \right)^{2} + \left( \frac{n+9}{2} \right)^{2}$ and if $n$ is even we find that $p = \left( \frac{n-8}{2} \right)^{2} + \left( \frac{n}{2} \right)^{2} +\left( \frac{n}{2} \right)^{2} + \left( \frac{n+10}{2} \right)^{2}.$ Even later edit: For any prime $p \neq 41$ which is a quadratic residue (mod $163$), there is a unique integer $h$ with $1 \leq h \leq \frac{p-1}{2}$ such that $p$ divides $h^{2}+h+41$, and then $p$ is necessarily the largest prime divisor of $h^{2}+h+41.$ An inductive argument of a type which dates back to Euler and/or Fermat then shows that $p$ is necessarily of the form $x^{2}+xy+41y^{2}$ for integers $x$ and $y$, and allows you to explicitly determine $x$ and $y$, given such an expression for the other (smaller) prime divisors $q$ of $h^{2}+h+41$ (all of which are also necessarily quadratic residues (mod $163$)).<|endoftext|> TITLE: Relation between Ind-completion and "additive"-ind-completion QUESTION [11 upvotes]: Suppose that $\mathcal{C}$ is a skeletally small additive category. To enlarge $\mathcal{C}$ and produce a bigger category whose "small" objects can be identified with those in $\mathcal{C}$, one may consider the ind-completion $\operatorname{Ind}\mathcal{C}$ of $\mathcal{C}$ in the sense of Grothendieck and Verdier. Denote by $\operatorname{Fun}(\mathcal{C}^{op},\operatorname{Set})$ the category of all functors from $\mathcal{C}^{op}$ to $\operatorname{Set}$. Then $\operatorname{Ind}\mathcal{C}$ is defined as the full subcategory of $\operatorname{Fun}(\mathcal{C}^{op},\operatorname{Set})$ whose objects are the filtered colimits (in $\operatorname{Fun}(\mathcal{C}^{op},\operatorname{Set})$) of representable functors. We may also consider the category $\operatorname{AddFun}(\mathcal{C}^{op},\operatorname{Ab})$ of additive functors from $\mathcal{C}^{op}$ to the category of abelian groups and define $\operatorname{AddInd}\mathcal{C}$ as the full subcategory of $\operatorname{AddFun}(\mathcal{C}^{op},\operatorname{Ab})$ whose objects are the filtered colimits (in $\operatorname{AddFun}(\mathcal{C}^{op},\operatorname{Ab})$) of representable functors. This construction is thoroughly studied, for instance, in the paper "Locally finitely presented additive categories", by W. Crawley-Boevey. I am trying to understand the relation between $\operatorname{Ind}\mathcal{C}$ and $\operatorname{AddInd}\mathcal{C}$. I have seen in two papers statements claiming that $\operatorname{Ind}\mathcal{C}$ and $\operatorname{AddInd}\mathcal{C}$ are equivalent, but do not yet understand why this should be true. Any references or hints would be very welcome. REPLY [6 votes]: The equivalence you mention holds more generally whenever your base of enrichment has a finitely presentable unit. This certainly includes $Ab$ but also many other examples: $Cat$, $sSet$, $GAb$, $DGAb$, etc. Assume that $\mathcal V=(\mathcal V_0,\otimes,I)$ is a symmetric monoidal closed complete and cocomplete category. Then you can define the Ind-completion of a $\mathcal V$-category $\mathcal C$ as the closure of the representables in $[\mathcal C^{op},\mathcal V]$ under (conical) filtered colimits; let's call it simply $Ind(\mathcal C)$, and denote by $J:\mathcal C\to Ind(\mathcal C)$ the inclusion. This free cocompletion can be recognised as follows (see e.g. 4.2 from "Notes on enriched categories with colimits of some class" by Kelly and Schmitt): A $\mathcal V$-functor $F:\mathcal C\to\mathcal K$ exhibits $\mathcal K$ as the Ind-completion of $\mathcal C$ if and only if: $F$ is fully faithful; $\mathcal K$ has filtered colimits; $Fc$ is finitely presentable in $\mathcal K$ for any $c\in\mathcal C$ (i.e. $\mathcal K(Fc,-):\mathcal K\to \mathcal V$ preserves filtered colimits); The closure of $\mathcal C$ in $\mathcal K$ under filtered colimits is $\mathcal K$ itself. Now assume that the unit $I$ of $\mathcal V$ is finitely presentable, that is $\mathcal V_0(I,-):\mathcal V_0\to Set$ preserves filtered colimits. Let $\mathcal C$ be a $\mathcal V$-category and $J:\mathcal C\to Ind(\mathcal C)$ be the inclusion in its enriched Ind-completion (then $J$ satisfies (1)-(4) above). We can then consider the underlying ordinary functor $J_0:\mathcal C_0\to Ind(\mathcal C)_0$ of $J$; then the properties (1), (2), and (4) still hold for $J_0$ (independently of the unit) and (3) holds since $$ Ind(\mathcal C)_0(J_0c,-)\cong \mathcal V_0(I,Ind(\mathcal C)(Jc,-)_0) $$ is a composite of filtered-colimit-preserving functors. It follows by the characterization above of Ind-completions (for $\mathcal V=Set$) that $Ind(\mathcal C)_0$ is the Ind-completion of $\mathcal C_0$; in other words $$ Ind(\mathcal C)_0\simeq Ind(\mathcal C_0).$$<|endoftext|> TITLE: Structural properties of $\mathcal{V}$-$\mathsf{Cat}$ QUESTION [10 upvotes]: In this question $(\mathcal{V}, \otimes, e)$ is a (bi)complete symmetric monoidal category. We have an adjunction $$\mathscr{l}: \mathsf{Cat} \leftrightarrows \mathcal{V}\text{-}\mathsf{Cat} :(-)_0,$$ induced by the change of enrichment as discussed for example between Ex. 3.2 and 3.3 here. Let me call $\mathsf{T}_{\mathcal{V}}$ the induced monad on $\mathsf{Cat}$. Q1. Do we know, or can we characterize, when this adjunction is monadic? Q2. When it is not monadic, how should I think about an algebra for $\mathsf{T}_{\mathcal{V}}$? Sometimes one reads that being an additive category (i.e. being $\mathsf{Ab}$-enriched) is more a property than a structure, see for example this question which actually motivated the present question, but I am sure you met such statements in your life. Q3. Does this mean that the monad $\mathsf{T}_{\mathsf{Ab}}$ is lax-idempotent, or it is some different behavior? Q4. Do you have other examples, or a characterization, or a sufficient condition such that being a $\mathcal{V}$-category is a property, in the same way this is true for additive categories? (I am aware this is not a well-posed question). REPLY [10 votes]: I'm going to assume that $\mathcal{V}$ is closed as well, or at least that its tensor product preserves colimits in each variable; I'm not sure that you get the left adjoint from $\mathsf{Cat}$ to $\mathcal{V}\text{-}\mathsf{Cat}$ otherwise. In this situation the underlying adjunction $\mathsf{Set} \rightleftarrows \mathcal{V}$ is monoidal, i.e. its left adjoint is strong monoidal and its right adjoint lax monoidal. Therefore, the induced monad $S$ on $\mathsf{Set}$ is a monoidal monad and hence the category $S\text{-}\mathsf{Alg}$ is a multicategory. The adjunction on enriched categories is just obtained by applying this adjunction on hom-objects, and so a $T_{\mathcal{V}}$-algebra is just a category enriched over $S\text{-}\mathsf{Alg}$. Thus: Q1: If $\mathcal{V}$ is monadic over $\mathsf{Set}$ via this adjunction, then so is $\mathcal{V}\text{-}\mathsf{Cat}$ over $\mathsf{Cat}$. For instance, this is the case for $\mathcal{V} = \mathsf{Ab}$. Q2: It depends a lot on $\mathcal{V}$. For instance, if $\mathcal{V} = \mathsf{Top}$, then $S$ is the identity monad, hence so is $T_{\mathcal{V}}$. Q3: Because $T_{\mathcal{V}}$ acts as the identity on objects, any lax $T_{\mathcal{V}}$-morphism is automatically a strict $T_{\mathcal{V}}$-morphism, which means in turn that it is homwise an $S$-morphism. Thus, $T_{\mathcal{V}}$ cannot be lax-idempotent unless $S$ is idempotent. The statement that being additive is more a property than a structure depends on the fact that an additive category is not just an $\mathsf{Ab}$-enriched category, but also has finite products (which are therefore biproducts). This is what enables the enrichment to be characterized in terms of the biproducts, and it makes the forgetful functor from additive categories to ordinary categories pseudomonic, both as a 1-functor and as a 2-functor; thus being additive is a "property-like structure". But I'm not sure if this can be characterized in 2-monadic terms. Q4: To start with, the fact about additive categories has nothing to do with subtraction, so it is equally true of semiadditive categories, which are enriched only over commutative monoids. Moreover, conversely any category with biproducts is automatically enriched over commutative monoids. In fact, $\mathsf{CMon}$ is the "closed monoidal locally presentable category freely generated by the fact that finite coproducts are absolute in $\mathsf{CMon}$-enriched categories". Other monoidal categories with analogous properties, for which being a $\mathcal{V}$-category with the appropriate kind of colimits is also a property-like structure, include: Pointed sets, for initial objects (which are then zero objects) Suplattices, for arbitrary coproducts (which are then biproducts) In the $\infty$-categorical world, spectra, for arbitrary finite colimits (which then satisfy the stability property that pushout squares coincide with pullback squares). Moritz Rahn (nee Groth) and I wrote something about this phenomenon in Generalized stability for abstract homotopy theories, and intended to write a sequel, but haven't gotten around to it yet.<|endoftext|> TITLE: Rational homotopy groups of $S^2\vee S^2$ QUESTION [18 upvotes]: From what I understand $\pi_n(S^2\vee S^2)\otimes\mathbb{Q}\neq 0$ for $n\geq 2$. My question is: Is there a "hands-on" proof of this fact using differential forms? I am sure I will receive answers like: that is Hilton's theorem or use Sullivan's minimal model or check the section in Bott and Tu about the rational homotopy theory. However, all these answers are useless for me because I am an analyst and not topologist and in order to use this fact in my research I need a straightforward construction that I could use to get integral estimates of forms. Rational homotopy theory of Sullivan is build on differential forms and you can represent elements in rational homotopy using the Hopf-Novokov integral as explained in: Hardt, Robert; Rivière, Tristan Connecting rational homotopy type singularities. Acta Math. 200 (2008), no. 1, 15–83. The points is that I do not really understand what they do and I do not even know if their machinery can be used to answer my question. Thus I am asking if there is a simple way to prove the result I am quoting using integration of differential forms. To be more precise: One can use differential forms to prove that $\pi_{4n-1}(S^{2n})\neq 0$ and that can be explained on a couple of pages with all details. Is there a similar proof of the fact I mentioned? REPLY [7 votes]: Ryan's answer generalizes. I prefer $\iota_1, \iota_2$ for the inclusions of wedge summands, and then $\omega_1, \omega_2$ for forms which generate cohomology supported on each of the wedge summands. Then I claim that a numerical homotopy invariant which evaluates non-trivially on $[[\cdots[\iota_1,\iota_2],\iota_2],\cdots],\iota_2]$ is $f \mapsto \int_{S^n} d^{-1} \left( \cdots d^{-1} \left( d^{-1} \left(d^{-1} f^*(\omega_1) \wedge f^*(\omega_2)) \wedge f^*(\omega_2) \right) \wedge f^*(\omega_2) \right) \cdots \right) \wedge f^*(\omega_2),$ where each $d^{-1}$ denotes a choice of some form to cobound what is in each case a closed and thus exact form of degree less than $n$ on $S^n$. (As Sullivan and others point out, such choices can be made universally through a constructive proof of the Poincare Lemma.) Since there is such a Whitehead product in every degree greater than or equal to two, this claim answers the Question. The proof that this integral is homotopy invariant is elementary, along the lines which Ryan gives in the example of his answer. There is also probably a more direct "hands-on" proof that this integral evaluates to one on the Whitehead product given, but I am going to cite my work with Ben Walter, namely https://arxiv.org/abs/0809.5084 (not the paper Ryan linked to, though I appreciate his seeing the relevance of our ideas). This integral is a case of a Hopf invariant, which is the main object of study of the paper. This integral is the Hopf invariant associated to the element of the bar construction $|\omega_1|\omega_2| \omega_2| \cdots |\omega_2|$. To see it evaluates as claimed I prefer to use the bracket-cobracket compatibility theorem, namely Theorem 1.12 of the paper, from which this is quick inductive combinatorics. While the present MO answer is thus not self-contained, note that Theorem 1.12 occurs just on page 7, and I would call its proof "hands-on". The cited paper was written to solve exactly these types of questions. We give an algorithm called weight reduction which can be used to associate integral(s - non-unique) to any cycle in the (Lie coalgebraic) bar construction on the cochains (forms), which are homotopy periods. This association defines an isomorphism between the homology of this bar construction and the linear dual of homotopy. Though Sullivan in his seminal paper pointed to how one can produce complete rational homotopy periods (I wish we could call this "rational cohomotopy") through minimal models, what Ben and I find is that the Quillen functor approach is, perhaps surprisingly, more readily tied to geometry. The general principle which follows from our work is that one can detect all rational homotopy through numerical invariants which are "higher linking with correction," governed by the Lie coalgebraic bar construction. Thus, geometry, combinatorics, algebra and functorial formalism all work together as well as one could hope for in giving a sharp resolution to the homotopy periods question in the simply connected setting. (The non-simply connected case is ongoing research that I am having PhD students work on.) And as Ryan says, I have recorded lectures on this subject. The best place to see a summary of and links to the lectures is here https://pages.uoregon.edu/dps/GeometricAlgebraicTopology/ The first three lectures cover these Hopf invariants.<|endoftext|> TITLE: Viewing exceptional Lie algebras via the classical ones QUESTION [7 upvotes]: I've been trying to understand the exceptional Lie algebras through the classical ones that I am more familiar with. In particular I wanted to get a handle on the root spaces and most discussions that I've read focus on the compact case (e.g. Baez's approach via the octonions and the magic square constructions) while I am more interested in the complex/split story. To make it more clear what I mean, a choice of Cartan subalgebra in $\mathfrak{sl}(V)$ is equivalent to a choice of decomposition of $V$ into lines $L_1 \oplus \cdots \oplus L_n$ (The Cartan subalgebra being the diagonal guys with respect to this decomposition). Then the root spaces are all the $L_i^* \otimes L_j = \hom(L_i,L_j) \leq \mathfrak{sl}(V)$. We can also do this for $\mathfrak{so}(V)$ and $\mathfrak{sp}(V)$ by choosing decompositions into isotropic lines and identifying them with $\bigwedge^2V$ and $S^2V$. For $\mathfrak{e}_7$ we can make use of the above by taking a $\mathfrak{sl}_8$ subalgebra. In particular we can choose the Cartan subalgebra to be a Cartan subalgebra of this subalgebra. Let $V$ be 8-dimensional and $\mathfrak{sl}_8 = \mathfrak{sl}(V)$. Then there is a complement to $\mathfrak{sl}_8$ in $\mathfrak{e}_7$ which is $\mathrm{ad}(\mathfrak{sl}_8)$ invariant and it is isomorphic to $\bigwedge^4V$ as a $\mathfrak{sl}_8$ representation. So $\mathfrak{e}_7 = \mathfrak{sl}(V) \oplus \bigwedge^4V$ (not direct as a sum of Lie algebras) and we can see its root system this way. Again the Cartan subalgebra defines a decomposition $V = L_1 \oplus \cdots\oplus L_8$. The root spaces in the $\mathfrak{sl}(V)$ part are just as before and the root spaces in $\bigwedge^4V$ are all the $L_i \wedge L_j \wedge L_k \wedge L_l$. I know it is possible to do this for $\mathfrak{e}_8$ as well viewing it as a copy of $\mathfrak{so}_{16}$ together with one of the spin representations of $\mathfrak{so}_{16}$. Is there a way to see the other exceptional Lie algebras in this manner? I don't think $\mathfrak{e}_6$ contains a simple subalgebra with the same rank as it but it does have subalgebras like $\mathfrak{sl}_3 \oplus\mathfrak{sl}_3 \oplus\mathfrak{sl}_3 $ and $\mathfrak{sl}_6 \oplus\mathfrak{sl}_2 $. Are there any good references on this style of approach to the exceptional Lie algebras or a good way to see what the decompositions and root systems would be? REPLY [4 votes]: Élie Cartan himself, recognized and used the following description of $\mathfrak{e}_6$: Let $V$ be a vector space of dimension $6$ and let $W$ be a vector space of dimension $2$. Then there is a vector space splitting $$ \mathfrak{e}_6 = \mathfrak{sl}(V)\oplus\mathfrak{sl}(W)\oplus \bigl(\Lambda^3(V)\otimes W\bigr) $$ and, moreover, $\mathfrak{e}_6$ is the Lie algebra of linear transformations of $A = \Lambda^2(V^*)\oplus (V\otimes W)$ that preserve a certain cubic form on $A$ that is invariant under the obvious representation of $\mathrm{SL}(V)\times\mathrm{SL}(W)$ on $A$. (This $A$ is one of the two lowest dimensional linear representations of $\mathfrak{e}_6$, the other is its dual.) I don't know that Élie Cartan ever noticed this, or wrote about it, but I think there is a description along the following lines: (maybe in Freudenthal or Dynkin): Let $V_1$, $V_2$, and $V_3$ be three vector spaces of dimension $3$. Then there is a decomposition $$ \mathfrak{e}_6 = \mathfrak{sl}(V_1)\oplus\mathfrak{sl}(V_2)\oplus \mathfrak{sl}(V_3)\oplus (V_1\otimes V_2\otimes V_3)\oplus (V^*_1\otimes V^*_2\otimes V^*_3) $$ as a module over the group $\mathrm{SL}(V_1)\times\mathrm{SL}(V_2)\times \mathrm{SL}(V_3)$. I'll try to find the reference.<|endoftext|> TITLE: Regularity of the Radon transform with respect to the original function QUESTION [5 upvotes]: Consider a function $f: \mathbb{R}^{d} \rightarrow \mathbb{R}$ (whose properties are to be specified). I note $\mathbb{S}^{d-1}$ the hypersphere and the Radon transform of $f$ defined for $(t,\theta) \in \mathbb{R} \times \mathbb{S}^{d-1}$: \begin{equation} Rf(t,\theta):=\int_{x:\langle x,\theta\rangle=t} f(x)dx=\int_{\mathbb{R}^{d}} f(x)\delta_{t-\langle x,\theta\rangle}(x)dx \end{equation} and I also consider the Sobolev norm $\|.\|_{W^{s}(\mathbb{R}^{d})}$ defined by: \begin{equation} \|f\|_{W^{s}(\mathbb{R}^{d})}:=\sum_{|\alpha|\leq s} \int_{\mathbb{R}^{d}} |\partial^{\alpha}f(x)|dx \end{equation} where $\alpha$ is a multi-index and $\partial^{\alpha}$ the weak-derivative. In the same way on $\mathbb{R}$ I define for $f:\mathbb{R} \rightarrow \mathbb{R}$ the norm $\|f\|_{W^{s}(\mathbb{R})}:=\sum_{k=0}^{s} \int_{\mathbb{R}} |f^{(k)}(t)|dt$ where $f^{(k)}$ stands for $\frac{d^{k}}{dt^{k}}f$. My question is the following: can we relate the Sobolev norm of $\|Rf(\cdot,\theta)\|_{W^{s}(\mathbb{R})}$ for any $\theta \in \mathbb{S}^{d-1}$ with the norm $\|f\|_{W^{s}(\mathbb{R}^{d})}$ of the original function ? More precisely under which reasonable conditions on $f$ we have something like for $\theta \in \mathbb{S}^{d-1}$ $\|Rf(\cdot,\theta)\|_{W^{s}(\mathbb{R})}\leq C_\theta \|f\|_{W^{s}(\mathbb{R}^{d})}$ for some constant $C_\theta$ ? I known that there are connections between the regularity of $f$ and $Rf$ for the Sobolev $2$-norms but I am looking for reference in this case. REPLY [7 votes]: $\newcommand{\IR}{\mathbb{R}}$It is sufficient to consider $\theta=e_n$, the $n$-th standard basis vector, because the Sobolev norms are all rotationally invariant. For a compactly supported, smooth functions $f$: $$\begin{align*} \| \partial^k Rf(\cdot,\theta)\|_{L^1(\IR)} &= \int_\IR \left| \partial^k \int_{\IR^{n-1}} f(t\theta+y) dy \right|dt \\&= \int_\IR \left| \int_{\IR^{n-1}} \partial^{(0,...,0,k)} f(t\theta+y) dy\right|dt \\&\leq \int_\IR \int_{\IR^{n-1}} \left| \partial^{(0,...,0,k)} f(t\theta+y) \right| dy dt \\\implies \|Rf(\cdot,\theta)\|_{W^{k,1}(\IR)} &\leq \|f\|_{W^{k,1}(\IR^n)}\end{align*}$$ The compactly supported, smooth functions are dense in $W^{k,1}(\IR^n)$ so that $f\mapsto Rf(\cdot,\theta)$ is a continuous operator $W^{k,1}(\IR^n)\to W^{k,1}(\IR)$ with norm $\leq 1$.<|endoftext|> TITLE: Geometric meaning of coherent sheaves $\mathcal{F} \otimes \mathcal{O}_{\mathbb{P}^n}(d)$ over $\mathbb{P}^n$ QUESTION [6 upvotes]: Maybe it sounds like a silly question but I'm not able to figure out in my head the geometric meaning of "twisting" vector bundles (or more generaly coherent sheaf) over $\mathbb{P}^n$. I explain myself better: if I take $X=\mathbb{P}^n$ and then I consider a subscheme $Y \subset X$, then to $Y$ corresponds an ideal (saturated if you want) $$I_Y=(F_1,\dots,F_r) \subset \mathbb{C}[x_0,\dots,x_n]$$ Sheafifying I can view $\mathcal{I}_Y$ as a coherent $\mathcal{O}_X-$module. I can also consider its twist by $\mathcal{O}_X(d)$, i.e. $$\mathcal{I} \otimes \mathcal{O}_X(d)=\mathcal{I}(d)$$ I have a very clear geometric meaning of sections of $\mathcal{I}(d)$, i.e. $H^0(\mathcal{I}(d))$ corresponds to hypersurfaces $F \in \mathbb{P}^n$ of degree $d$ containing $Y$. Now, if for example I take the Tangent sheaf $\mathcal{T}_X$, it is also a coherent $\mathcal{O}_X-$module, and geometrically $H^0(\mathcal{T}_X)$ is the space of vector fields over $\mathbb{P}^n$. Now, if I consider the twist $$\mathcal{T}_X \otimes \mathcal{O}_X(d)=\mathcal{T}_X(d)$$ what is the geometric meaning of the sections in $H^0(\mathcal{T}_X(d))$? Are they a sort of vector fields with special properties? And more generally if I have a nice geometric description of sections of a coherent sheaf $H^0(\mathcal{F})$, how can I find a geometric interpretation of the sections of $H^0(\mathcal{F}(d))$? Thanks in advance for the help REPLY [8 votes]: Let $x_0,\ldots, x_n$ be homogeneous coordinates, then a section of $H^0(\mathbb{P}^n,\mathcal{T}_{\mathbb{P}^n}(d))$ can be expanded as a sum $$\sum_i f_i(x_0,\ldots, x_n) \frac{\partial}{\partial x_i}$$ where $f_i$ are homogenous polynomials of degree $d$. This follows more or less immediately from the Euler sequence $$0\to \mathcal{O}\to \bigoplus_0^n \mathcal{O}(1)\to \mathcal{T}\to 0$$ after twisting. Does this help?<|endoftext|> TITLE: How many Hamming spheres of radius 1 does it take to cover the cube? QUESTION [6 upvotes]: I am looking for the sharpest known upper bound on $K(n, 1)$ as $n \rightarrow \infty$. This is the minimal cardinality of a (not-necessarily linear) covering code of $\{0, 1\}^n$ of radius 1. In elementary terms: Using how few (possibly non-disjoint) Hamming balls of radius 1 can we cover $\{0, 1\}^n$? I am interested in upper-bounds to this quantity, especially asymptotically as $n \rightarrow \infty$. For example, even the statement $K(n, 1) \in o(2^n)$ I have not been able to find proven. (It's impossible to do strictly better than $\lceil{\frac{2^n}{n + 1}} \rceil$, by the sphere-covering bound.) As of 1998, exact values of $K(n, 1)$ were only known for specific values of $n$. For example: $K(2^k - 1, 1) = 2^{2^k - k - 1}$ (Hamming code) $K(2^k, 1) = 2^{2^k - k}$ (Johnson 1972) $K(2n + 1, 1) \leq 2^n \cdot K(n, 1)$ (Cohen–Lobstein–Sloane 1986) $K(K(n, 1) - 1, 1) \leq (n + 1) \cdot 2^{K(n, 1) - n - 1}$ (Cor. 1 of Östergård and Kaikkonen) REPLY [4 votes]: Clearly $K(n, 1) \le 2K(n-1, 1)$ by the construction of taking each code word of length $n-1$ and adding one copy with a suffix of $0$ and one with a suffix of $1$. (This is comment r to table 1 in the Cohen-Lobstein-Sloane paper referenced in the question). Then by taking the largest $k$ such that $2^k - 1 \le n$ and iterating this construction on the Hamming code we get $K(n, 1) \le 2^{n - \lfloor \lg (n+1)\rfloor}$. The sphere-covering lower bound can be written as $2^{n - \lg(n+1)} \le K(n,1)$, so the upper bound is less than twice the lower bound and we have the asymptotic $$K(n, 1) \in \Theta(2^{n - \lg(n+1)})$$ When $n = 2^k - 1$ the lower and upper bounds coincide; the gap is greatest when $n = 2^k - 2$, when the upper bound coincides with the size of the Hamming code for $n+1$.<|endoftext|> TITLE: Does the morphism of composition have some universal property? QUESTION [8 upvotes]: Let $A$, $B$ and $C$ be three objects in the category Set. For simplicity, assume that their underlying sets contain a finite number of elements, a, b and c respectively. Using the usual Haskell notation for types, there exists exactly $(c^a)^{c^b×b^a}$ morphisms from the object/type ($b\rightarrow c$, $a\rightarrow b$) to the object/type $a\rightarrow c$. Among this potentially huge set of morphisms, exactly one is the (uncurried) composition function, defined as ${\rm compose}\, (g, f) = g\circ f$. My question is, does this particular morphism have any universal property that make it possible to identify using only category theory? This is the computer-scientist-toying-with-categories-version of the question. If you prefer a version written by a proper mathematician, please refer to The universal property of composition of morphisms REPLY [5 votes]: First of all, as David Roberts pointed out in a comment above, the composition map c :: hom a b -> hom b c -> hom a c c f g = g . f and the evaluation map ev :: (a -> b) -> a -> b ev f x = f x share a common property, that of forming the components of a cowedge. I will now switch to category-theory notation, I'm confident you'll be able to translate back in pseudo-Haskell-ish/generic FP syntax. Being a cowedge means that fixed objects $a,c$ of your category, the correspondence $(b,b')\mapsto \hom_{\cal A}(a,b)\times\hom_{\cal A}(b',c)$ is a functor ${\cal A}^\text{op}\times {\cal A} \to {\sf Set}$; let's call it $T$ the family of maps $c_b : T(b,b)=\hom(a,b)\times\hom(b,c)\to \hom(a,c)$ behaves as follows, once a morphism $f : b\to b'$ is given: the square $$\require{AMScd}\begin{CD} T(b',b) @>T(f,b)>> T(b,b) \\ @VT(b',f)VV @VVc_bV \\ T(b',b') @>>c_{b'}> \hom(a,c) \end{CD}$$ is commutative. (I'll leave this verification to you). Not only this; there is a category of cowedges for $T$, i.e. families $\underline g = \{g_b : T(b,b) \to E\}$ pointing to generic sets $E$ and fitting into the same commutative squares. A morphism between a cowedge $\underline g$ and a cowedge $\underline h$ is a map between their codomains, with the property that the obvious diagram commutes (I'll leave it to you to find what is the obvious diagram). Now, Claim. $\underline c = \{c_b : T(b,b) \to \hom(a,c)\}$ is the initial object in the category of cowedges for $T$ (a functor that, let me remind you, depends on $a,b$!). I'll leave to you this verification. This is the universal property of the composition maps of a category. PS: what about the evaluation map $\text{ev} : B^A\times A \to B$? Well, in a cartesian closed category its components also form not only a cowedge (this is true in every monoidal closed category, and in fact it is true for every parametric adjunction, but nevermind); they form an initial one! In a sense, feeding a function with an element in its domain is "the simplest operation one can perform", given a function and an element (of course, this intuition deeply relies on the fact we can resort to elements; in category theory, this is often not possible). Similarly, given two arrows in a category that can be composed, composing them is the most natural thing you could do.<|endoftext|> TITLE: Bousfield's distributive lattice DL and non-ring spectra QUESTION [8 upvotes]: Bousfield, in his paper "The Boolean algebra of spectra" (Comm Math Helv 54, 368–377 (1979), https://doi.org/10.1007/BF02566281), defined $\mathbf{DL}$, a sublattice of the Bousfield lattice, to consist of all Bousfield classes $\langle X \rangle$ such that $\langle X \rangle \wedge \langle X \rangle = \langle X \rangle$. He pointed that if $X$ is a wedge of ring spectra, then $\langle X \rangle \in \mathbf{DL}$. Are there classes in $\mathbf{DL}$ which are known not to be Bousfield equivalent to a wedge of ring spectra? If the telescope conjecture fails, then that would yield examples. Are there others? REPLY [7 votes]: My paper A combinatorial model for the known Bousfield classes defines an complete ordered semiring $\mathcal{A}$ and a homomorphism from $\mathcal{A}$ to the Bousfield lattice mod the telescope conjecture, whose image contains most of the Bousfield classes that have been named and studied. In $\mathcal{A}$, all elements satisfying $x\wedge x=x$ correspond to wedges of unital ring spectra. (In fact, they all correspond to unital ring spectra, except for the case of $\bigvee_{i\in U}K(i)$, which is not obviously a unital ring if $U$ is infinite.) Although the main results are stated in a quotient of the Bousfield lattice where the telescope conjecture is forced to be true, many intermediate results are stated in the Bousfield lattice itself. I am therefore fairly confident that the literature does not contain any unconditional counterexamples for your question.<|endoftext|> TITLE: When is the augmentation ideal projective as RG-module? QUESTION [9 upvotes]: Let $G$ be a finite group and let $R$ be a commutative ring. I'd like to ask, if there is a theorem of the following kind: The augmentation ideal $I_G$ is projective as RG-module, if and only if ... ? This should happen only in rare cases, but I was wondering, if there exists an if-and-only-if criterion. Thank you very much for the help. REPLY [16 votes]: Okay, this happens precisely in the obvious case, namely if all primes dividing the order $|G|$ are invertible in $R$. To see this, note that $\operatorname{Ext}^*_{R[G]}(R,R)$ is group cohomology of $G$ with coefficients in $R$. If $I_G$ were projective, $I_G\to R[G]$ would be a projective resolution of $R$ and thus the cohomology would be $1$-dimensional. However, the cohomology of $G$ with coefficients in $\mathbb{Z}$ has $p$-torsion in arbitrarily high degrees for all primes dividing $|G|$ (this is a well-known fact about cohomology of finite groups). So by the universal coefficient theorem, the cohomology with $R$-coefficients is unbounded, except if $p$ is invertible in $R$ for each $p$ dividing $|G|$.<|endoftext|> TITLE: Cohomology of Grothendieck topology QUESTION [8 upvotes]: My naïve cartoon picture of the construction of étale cohomology is this: start with a scheme, associate to it a Grothendieck topology (making a site). A functor from the Grothendieck topology to abelian groups (say) has all the relevant properties of a presheaf (by the definition of a Grothendieck topology) and so one gets cohomology by sheafifying and taking (as it were) sheaf cohomology. My question is: is there a “minimal” reference describing the second step above without caring about schemes or étale cohomology (the first step)? Of course, I don't mind if the reference covers étale cohomology, as long as steps 1 and 2 are separated. Having formulated this question and anticipating the answer let me ask a second question: are there, among the first six exposés of SGA4, parts that I can skip while trying to learn about step 2? REPLY [11 votes]: Artin, M. Grothendieck topologies. (English) Zbl 0208.48701 Cambridge, Mass.: Harvard University. 133 p. (1962). (pdf copy) These notes seem to fit your description precisely. They are concise, start from first principles, assuming basically only knowledge of Grothendieck's Tôhoku paper. The focus is specifically on how to define cohomology in a topos, as opposed to many references on topos theory aimed more in the direction of algebraic stacks, logic, motivic homotopy... (This text was recommended in a now deleted answer by another user. In my opinion Artin's notes are quite nice and I thought the recommendation was worth preserving.)<|endoftext|> TITLE: A conjecture on binomial coefficients and roots of unity QUESTION [10 upvotes]: Is the following true? Let $p$ be a prime and let $w$ be a $(p-1)$st root of unity (not necessarily primitive). Then $$\binom{w}{n}=\frac{w(w-1)\cdots(w-n+1)}{n!}$$ is $p$-integral; i.e., it can be expressed as a polynomial in $w$ with $p$-integral coefficients. In other words, if $\Phi_k(x)$ is the $k$th cyclotomic polynomial and $k$ divides $p-1$ then the remainder of $\binom{x}{n}$ modulo $\Phi_k(x)$ is $p$-integral. REPLY [9 votes]: Here is an elementary and explicit way to see this: Suppose we have a set of $p$ integers $A=\{a_1, a_2,\dots, a_p\}$ which forms a complete set of residues modulo p. Then we have $$\prod_{a\in A}(x-a)=x^p-x \pmod{p}$$ which means that we can write $\prod_{a\in A}(x-a)=x^p-x+pF(x)$ for some polynomial $F(x)\in \mathbb Z[x]$. Since we have $w^p-w=0$, we can conclude that $$p^{-1}\prod_{a\in A}(w-a)=F(w)$$ so it is equal to a polynomial in $w$ with integer coefficients. Next, let's look at what happens when we have $A=\{a_1,a_2,\dots,a_{p^k}\}$ which forms a complete set of residues modulo $p^k$. The polynomial $x^p-x$ has $p$ distinct roots modulo $p^k$, let's call them $B_0=\{b_1,b_2,\dots,b_p\}\subset A$. Let's define $B_r$ as the subset of $A$ which coincides with $\{b_1+r, b_2+r, \dots, b_p+r\}\pmod{p^k}$. Then we can write $A$ as a disjoint union: $$A=B_0\cup B_p \cup B_{2p}\cup\cdots \cup B_{p^k-p}.$$ Suppose that $\ell_r$ is defined as the exponent of $p$ in the prime factorization of $r$, i.e. $p^{\ell_r}|| r$. Then we have $$\prod_{a\in B_r} (x-a)=(x-r)^p-(x-r)+p^k G_r(x)=x^p-x+p^{\ell_r} H_r(x)$$ for some polynomials $G_r,H_r$ with integer coefficients. Therefore we can calculate by grouping $$(p^k!)^{-1}\prod_{a\in A}(w-a)=\frac{1}{p^k!}\prod_{i=0}^{p^{k-1}-1}\prod_{a\in B_{pi}}(w-a)=\frac{1}{p^k!}(p^kG_0(w))\prod_{i=1}^{p^{k-1}-1}p^{\ell_{pi}}H_{pi}(w)$$ and this is $p$-integral because the exponent of $p$ in $p^k\cdot p^{\ell_p+\ell_{2p}+\cdots+\ell_{p^k-p}}$ matches that of $p$ in $p^k!$. In order to finish the general case in the problem, let $n$ be written in base $p$ as $n_kp^k+n_{k-1}p^{k-1}+\cdots+n_0$ and then split the set $\{0,1,\dots,n-1\}$ into $n_k$ complete sets of residues modulo $p^k$, $n_{k-1}$ complete sets of residues modulo $p^{k-1}$ and so on, and then apply the argument from the previous paragraph to each of them.<|endoftext|> TITLE: Question about functions $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $x$ is prime whenever $f(x)$ is prime QUESTION [8 upvotes]: Let $f: \mathbb{\mathbb{Z}^+} \to \mathbb{Z^+}$ be a function and suppose $(\star)$ For all integers $x \geq 3$, if $f(x)$ is prime, then $x$ is prime. A trivial example of such a function is the identity $f(x) = x$. However, a possible non-trivial example which I have come across is \begin{align*} f(x) = \left\lfloor \frac{\cosh(x\ln(2 + \sqrt{3}))}{2}\right\rfloor. \end{align*} This function seems to satisfies $(\star)$ (see OEIS A198196). I have two questions: How might one go about proving $f$ satisfies $(\star)$? I'm not sure where to begin with this, and it seems like a difficult task. Have functions with property $(\star)$ been studied before? Thanks for any information you can give me. Edits: I've composed the floor function with $f$, and added the condition to $(\star)$ that $x$ must be an integer. REPLY [15 votes]: As observed in comments, we have $f(n) = \lfloor g(n) \rfloor$ where $g(n) = \frac{\alpha^n + \alpha^{-n}}{4}$ and $\alpha = 2 + \sqrt{3}$. From the recurrence $g(n+1) = 4 g(n) - g(n-1)$ we see that $g(n)$ is a half-integer when $n$ is even and an integer when $n$ is odd. In fact we see from induction that for even $n$ we have $g(n) = \frac{1}{2} \hbox{ mod } 3$ and for odd $n$ we have $g(n) = 1 \hbox{ mod } 3$. Hence for even $n$, $f(n)$ is divisible by $3$ and thus not prime except when $n=2$. For odd $n$, we have $f(n)=g(n)$, and for odd $n,m$ we then have $$ f(nm) = f(n) (\alpha^{n(m-1)} + \alpha^{n(m-3)} + \dots + \alpha^{n(1-m)})$$ thanks to the formula $a^m+b^m = (a+b)(a^{m-1} + a^{m-2} b + \dots + b^{m-1})$. From the Galois group action interchanging $\alpha$ and $\alpha^{-1}$ we see that $\alpha^{n(m-1)} + \alpha^{n(m-3)} + \dots + \alpha^{n(1-m)}$ is an integer, and for $n,m \geq 3$ this integer is larger than $1$. Thus $f(nm)$ is composite when $n,m \geq 3$ are odd, so the only remaining possible values of $n$ for which $f(n)$ can be prime are the primes. In the language of divisibility sequences, $f(n)$ is a divisibility sequence on the odd natural numbers, though not on the even ones. In retrospect this is not so surprising given that $f$ is so similar to the Fibonacci sequence $F_n = \frac{\phi^n - \phi^{-n}}{\sqrt{5}}$, which is well known to be a divisibility sequence.<|endoftext|> TITLE: Cotangent spaces of finite flat group schemes in short exact sequences QUESTION [9 upvotes]: Fix $(R,m)$ a complete DVR of mixed characteristic $(0,p)$ with perfect residue field, and consider finite flat commutative group schemes $G = Spec(A)$ over $R$. One can associate a differential invariant to $G$, an integer $d \geq 0$, in two ways: the "absolute different", such that the dual $A^*$ of $A$ under the $R$-trace pairing is equal to $m^{-d}A$, or the "cotangent length", equal to the $R$-length of the cotangent space $e^*\Omega_{G/R}$ (where $e$ is the unit section). My question is: why is the cotangent length $d$ additive in short exact sequences of $G$? A direct argument for the cotangent length would be great, though I'm open to an argument that compares the two definitions and proves additivity for the absolute different. For what it's worth, the comparison is clear in some cases by the "differential characterization of the different", and the additivity for the absolute different is clear in some cases by the "transitivity of the different", but I don't know where either of these classical facts is stated in the full generality needed here. (Context: In Raynaud's famous "... type $(p,\ldots,p)$" paper, the theorem in Section 4.1 computes the action of inertia on the determinant of the generic fiber of $G$ in terms of the absolute different. The argument given proceeds in two steps. First, in the case where $G$ is simple, the result is explicitly verified using the classification worked out earlier in the paper. I have no issues with this argument, and "differential characterization of the different" even applies to compare the two definitions of $d$ in this case. Second, there is a reduction to the simple case by dévissage. But the required additivity of $d$ is never explicitly addressed, and moreover my needs would prefer the cotangent definition. Hence the question.) REPLY [4 votes]: By Proposition 5.1(i) in Mazur and Roberts, Local Euler characteristics, Invent. Math. 9 (1970), 201-234, $G$ fits in an exact sequence $0 \to G \to A \to B \to 0$, where $A$ and $B$ are smooth commutative affine group schemes over $R$ of the same relative dimension, say $n$. Then the cotangent space $e^* \Omega_{G/R}$ is isomorphic to the cokernel of the induced homomorphism $\psi \colon e^* \Omega_{B/R} \to e^* \Omega_{A/R}$ of free $R$-modules of rank $n$. Its length $d(G)$ equals the valuation of $\det \psi$. Given a short exact sequence $0 \to G' \to G \to G'' \to 0$, one can choose compatible resolutions of $G'$, $G$, $G''$ by using the same pushout constructions one would use for modules, to get a $3 \times 3$ commutative diagram with short exact rows and columns. The homomorphisms $\psi'$, $\psi$, $\psi''$ define a morphism of short exact sequences of cotangent modules, so $(\det \psi) = (\det \psi')(\det \psi'')$ as ideals of $R$. Hence $d(G)=d(G')+d(G'')$.<|endoftext|> TITLE: Group cohomology of $\mathbf{R}^\ast$ acting on $\mathbf{R}$ QUESTION [5 upvotes]: I am interested in computing the first group cohomology $H^1(\mathbf{R}^\ast, \mathbf{R})$, where $\mathbf{R}^\ast$ is acting on $\mathbf{R}$ by multiplication (here $\mathbf{R}$ denotes the real numbers). One should probably take into account continuous cochains or something like that, but I would already be very happy to know if it is $0$ or not. Of course this is equivalent to find a nontrivial crossed homomorphism $\mathbf{R}^\ast \to \mathbf{R}$, and I think that $\log$ or $\exp$ can help here, but I am stuck. Any help is appreciated! REPLY [6 votes]: There is no cohomology. No continuity condition is needed. The cocycle condition is $$a \phi(b) + \phi(a) = \phi(ab)$$ Since $\mathbb R^\times$ is commutative, $$ a\phi(b) + \phi(a) = \phi(ab)=\phi(ba) = b\phi(a) + \phi(b)$$ so $$ (a-1) \phi(b) = (b-1)\phi(a)$$ and thus $$ \frac{ \phi(b)}{b-1} =\frac{\phi(a)}{a-1}.$$ So the only cocycles are scalar multiples of $\phi(x)=x-1$, i.e. the one JLA wrote down. These scalar multiples are indeed coboundaries, so there is no cohomology. An alternate proof uses the fact that cocycles classify extensions of the trivial representation of $\mathbb R^\times$ by the standard representation, and all such extensions split, as we can see by diagonalizing the action of any nontrivial element of $\mathbb R^\times$.<|endoftext|> TITLE: Simple example of nontrivial simplicial localization QUESTION [8 upvotes]: Does anyone has a simple example of a 1-category $\mathcal{C}$ and a collection of morphisms W such that the infinity-categorical / simplicial localization $\mathcal{C}\left[W^{-1}\right]$ is not a 1-category? Of course there are obvious “big” examples like CW complexes / derived categories, I’m looking for a small example that I’ll be able to understand combinatorially. Thanks! REPLY [12 votes]: For any $1$-category $C$ the localization $C[C^{-1}]$ at all arrows is an $\infty$-groupoid homotopy equivalent to the nerve of $C$, so it can be any $\infty$-groupoid. For example take $C$ to be the poset with 6 elements ordered as a,b < c,d < e,f and when you localize at all arrows you get the $2$-sphere $\mathbb{S}^2$. So the simplicial localization will have all objects isomorphic and having a simplicial set of endomorphisms for each object equivalent to $\Omega(\mathbb{S}^2)$.<|endoftext|> TITLE: The square root of natural number expressed by an infinite series QUESTION [6 upvotes]: Can you prove or disprove the following claim: Let $U(n,P,Q)$ be the nth generalized Lucas number of the first kind and let $m$ be a natural number. Then, $$\sqrt{m}=1+\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \cdot (m-1)^n}{U(n,2,1-m) \cdot U(n+1,2,1-m)}$$ The SageMath cell that demonstrates this claim can be found here. REPLY [10 votes]: So, here we have $P=2$ and $Q=1-m$. Notice that $$\frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = \frac{U_{n+1}(P,Q)}{U_n(P,Q)}-\frac{U_{n+2}(P,Q)}{U_{n+1}(P,Q)}.$$ By telescoping, it follows that $$\sum_{n=1}^k \frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = P - \frac{U_{k+2}(P,Q)}{U_{k+1}(P,Q)}.$$ Taking the limit over $k\to\infty$, we get $$\sum_{n=1}^\infty \frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = \frac{P-\mathrm{sgn}(P)\sqrt{P^2-4Q}}2 = 1 - \sqrt{m}.$$ QED<|endoftext|> TITLE: Relationship between Lambert $W$ function and Hypergeometric function QUESTION [5 upvotes]: The Lambert $W$ Function is defined in this Wikipedia entry, while the Hypergeometric Function is defined in this other Wikipedia entry. There exists also a multivariate generalization which solves the following equation $$ e^{-c x}=d \frac{\left(x-a_{0}\right)\left(x-a_{1}\right) \cdots\left(x-a_{n}\right)}{\left(x-b_{0}\right)\left(x-b_{1}\right) \cdots\left(x-b_{m}\right)} $$ as I read from Quora post. This equation has some analogies in Hypergeometric functions as well. I also would like to know if the Lambert $W$ Function can be written as an inverse of Hypergeometric functions: is it so? Or are there any other kind of relationship about them? Thanks for your answers and references. REPLY [3 votes]: Q: Can the Lambert $W$ function be written as an inverse of a hypergeometric function? A: $x=W(y)$ is the solution of $_1F_1(2;1;x-1)=y/e$.<|endoftext|> TITLE: Are large powers of polynomials linearly independent? QUESTION [27 upvotes]: Let $P_1,\dots,P_k$ be polynomials over $\mathbf{C}$, no two of them being proportional. Does there exist an integer $N$ such that $P_1^N,\dots,P_k^N$ are linearly independent? REPLY [24 votes]: The answer is yes. In fact, an even stronger claim is true: there exists some $N$ such that for all $n \geq N, \ P_{1}^{n}, \dots, P_{k}^n$ are linearly independent over $\mathbb{C}$. For this we will use a generalization of the Mason-Stother's theorem which appears on the Wikipedia page (though I have taken the special case of the curve $C = \mathbb{P}^{1} (\mathbb{C})$ and written it in slightly different language.): Let $q_1, \dots, q_{k}$ be polynomials such that $q_1 + \cdots + q_{k} = 0$ and every proper subset of $q_1, \dots, q_{k}$ is linearly independent. Then, $$\max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right)$$ Now, we can prove the claim by induction on $k$. For $k = 2$ it is obvious. Now, by induction for all $n$ large enough every proper subset of $P_{1}^{n}, \dots, P_{k}^{n}$ is linearly independent. Suppose for contradiction that there exist constants $\lambda_{1}, \dots, \lambda_{k}$ such that $$\lambda_1 P_{1}^n + \cdots + \lambda_{k} P_{k}^n = 0$$ Letting $q_i = \lambda_i P_{i}^{n}$, notice that $q_1, \dots, q_k$ satisfy the requirements of the lemma (we have assumed that $\lambda_i \neq 0$), and therefore $$n \leq \max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right) = \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( P_1 \cdots P_k \right) \right) - 1 \right)$$ but the right hand side is constant, and so for $n$ large we get a contradiction.<|endoftext|> TITLE: Expected absolute value of the average of two points from the disc QUESTION [6 upvotes]: Looking at Average distance of the mean of n random complex numbers in a unit disc, I tried to figure out  what is the expected absolute value $|\frac{z_1 + z_2}{2}|$ of two numbers $z_1, z_2\in\mathbb{C}$ drawn uniformly from the unit disc.  We can look at the quadruple integral $$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2})| d\phi_2 dr_2 d\phi_1dr_1 $$ This can be reduced to two integrals, so that we can actually calculate the values with high precision (see the following code) from mpmath import mp, sqrt, ellipe, quad, pimp.dps = 15 def ellipe_integrand(r, s):     return  sqrt(r*r + 2*r*s + s*s)*(ellipe(+pi, 4*r*s/(r*r + 2*r*s + s*s))) def integrand_2(s):     return quad(lambda r: 2*r*ellipe_integrand(r, s), [0, s, 1])mp.dps=150quad(lambda s: s*integrand_2(s), [0,1]) # after ~40 minutes on my machine we get: mpc(real='1.42222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222182628758433670366371075596783355537463385287071932198285251', imag='4.76485167346043265351378514322413241842941120327748264250820934786695559445396622361703586368896605436500687259315655829464952759337107787438513621789267e-246') The result of that integration looks suspiciously like $\frac{64}{45}$. To get the expected absolute volume we still have to divide by the area of the disc:$$ \frac{64}{45\pi} \approx 0.45270739368361339952038048248181862978690743677285389866003155$$ (I haven't tried running numerical simulation of simply drawing two points repeatedly, but I would hope that one could recover the first few digits here..) Let's denote the expected absolute value of the mean of n points by $\operatorname{exp\_abs}(n)$. It is easy to see that $\operatorname{exp\_abs}(1) = \frac{2}{3}$. Question 1 Is $ \operatorname{exp\_abs}(2) = \frac{64}{45\pi}$, like the numerical evidence would suggest? How to prove that? Question 2 Is there some reason to expect that $\operatorname{exp\_abs}(n)\in\mathbb{Q}(\pi)$? Question 3 What is the value of  $\operatorname{exp\_abs}(3)$? Here I would be interested in numerical evidence if the exact value is not easily derived. REPLY [4 votes]: (Too long for a comment) For the record: $$\mathrm{exp\_abs}(3)=\frac{4}{3 \pi^2}\,I_3=0.3671989447$$ where $$I_3=\int_{-1}^1\int_{-1}^1 \int_{-1}^1 |x+y+z|\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}\,dx\,dy\,dz=2.718081241$$ (I couldn't express by known constants) UPDATE: The exact value of $\mathrm{exp\_abs}(3)$ can be deduced from the following (amazing!) result: if $X,Y,Z$ are independent and identically uniformly distributed on $S^3$ (the unit sphere in $\mathbb{R}^4$), then \begin{align*} \mathbb{E}|X+Y+Z|=W_3(1,1) &=\frac{476}{525}A+\frac{52}{7\,\pi^2}\frac{1}{A}\\ \end{align*} with $A:=\frac{3}{16}\frac{2^{1/3}}{\pi^4}\Gamma(\frac{1}{3})^6$. ( See https://scholarship.claremont.edu/jhm/vol6/iss1/7/ (on page 100). Timothy Budd pointed to this paper in the related post.) If $X=(X_1,X_2,X_3,X_4)$ is uniform on $S^3$, and $U=(U_1,U_2)$ is uniform on $D^2,$ (the unit disk in $\mathbb{R}^2$), the distributions of $X_1$ resp. $U_1$ coincide, and the same will then hold for the first coordinate of i.i.d. sums. Now, for any rotationally symmetric random vector $V=(V_1,V_2,...,V_k)$ with finite expectation of $|V|:=\sqrt{V_1^2+V_2^2+\ldots +V_k^2}$ it holds that $$\mathbb{E}(|V|)=\frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{k+1}{2})}{\Gamma(\tfrac{k}{2})}\mathbb{E}(|V_1|)$$ Thus if $U,V,W$ are uniform on $D^2$, and $X,Y,Z$ are uniform on $S^3$ we have \begin{align*}\mathbb{E}(|U+V+W|)=\frac{\pi}{2} \mathbb{E}(|(U+V+W)_1|\\ \mathbb{E}(|X+Y+Z|)=\frac{3\pi}{4} \mathbb{E}(|(X+Y+Z)_1|\end{align*} and therefore \begin{align*} \mathbb{E}(|U+V+W|)=\frac{2}{3}\,\mathbb{E}(|X+Y+Z|)=\frac{2}{3}W_3(1,1) \end{align*} and $$\mathrm{exp\_abs}(3)=\frac{2}{9}\,W_3(1,1)\;\;.$$ (And $I_3=\frac{\pi^2}{6}\, W_3(1,1)$.)<|endoftext|> TITLE: Is there any structural characterization of the rings in which every element other than the identity is a (two-sided) zero divisor? QUESTION [6 upvotes]: [I fear that I'm missing something obvious here, but I'll dare to ask anyway.] As we all know, a division ring is a (unital, associative, non-zero) ring where every non-zero element is a unit. So, let an anti-division ring be a ring where any element other than the identity is a (two-sided) zero divisor. There are many simple things one can actually say about these objects: Their class is closed under direct products; they all have characteristic $2$; every boolean ring is a member of the family; any anti-division ring is, in fact, a subdirect product of domains (see Jose Brox's answer to a related question on MSE); the only nilpotent element in such a ring is zero and, more generally, the Jacobson radical is trivial (as noted by Benjamin Steinberg in the comments below); if a ring in the family is semilocal, then it's a direct product of copies of the field with two elements (see znc's answer from the same thread on MSE); etc. Question. Is it perhaps the case that a more precise characterization (say, than the one provided by Jose Brox for rings whose non-units are all zero divisors) is indeed possible? If so, it's clear to me that this must be very well known (at least in some large circles) and I would much appreciate it if you could offer a reference. REPLY [3 votes]: Sorry for answering my own question, but it turned out that what I'm calling "anti-division rings" in the OP were already studied by P.M. Cohn under the name of "$0$-rings" (though Cohn's work on this stuff is seemingly restricted to the commutative setting), see P.M. Cohn, Rings of zero divisors, Proc. Amer. Math. Soc. 9 (1958), 914-919; T. Porter, Cohn's rings of zero divisors, Arch. Math. 43 (1984), 340-343. In particular, Cohn's paper (Sect. 3) contains a couple of structural results for commutative $0$-rings, and it seems unlikely that one can do any better (without throwing in further conditions on the ring).<|endoftext|> TITLE: Quote by Thurston on the Ricci flow QUESTION [11 upvotes]: I recall seeing a quote by William Thurston where he stated that the Geometrization conjecture was almost certain to be true and predicted that it would be proven by curvature flow methods. I don't remember the exact date, but it was from after Hamilton introduced the Ricci flow but well before Perelman's work. Unfortunately, most of the results for Geometrization and Ricci flow are from 2003 or after. Does anyone know if the quote I'm referring to actually exists, and if so, where to find it? There is a quote from Thurston lauding Perelman's work, which suggests that he thought the Ricci flow was a promising approach, but I thought there was one from before as well. That the geometrization conjecture is true is not a surprise. That a proof like Perelman's could be valid is not a surprise: it has a certain rightness and inevitability, long dreamed of by many people (including me). What is surprising, wonderful and amazing is that someone – Perelman – succeeded in rigorously analyzing and controlling this process, despite the many hurdles, challenges and potential pitfalls. Thanks in advance. REPLY [7 votes]: There is a video of Thurston's talk "A discussion on geometrization" from May 7, 2001. In the last part of this talk he speaks about possible approaches to proving geometrization. Starting from 46:55 he spends 20 seconds mentioning heat type flows (probably Ricci flow) and says that it might or not work. But then he switches to describing a different approach which he thinks is more robust. The part of the talk where he starts to speak about future predictions in the sense of proving geometrization starts at 39:32.<|endoftext|> TITLE: Explanation of several unpublished remarks of Gauss on representations of a given number as sums of two, three and four squares QUESTION [10 upvotes]: Remark 1: On p.384 of volume 3 of Gauss's Werke, which is a part of an unpublished treatise on the arithmetic geometric mean, Gauss makes the following remark: On the theory of the division of numbers into four squares: The theorem that the product of two sums of four squares is itself a sum of four squares, is most simply represented as follows: let $l,m,\lambda,\mu,\lambda',\mu'$ be six complex numbers such that $\lambda,\lambda'$ and $\mu,\mu'$ are conjugate. Let $N$ denote the norm, than $$(Nl+Nm)(N\lambda + N\mu)=N(l\lambda+m\mu)+N(l\mu'-m\lambda')$$ and also $$(N(n+in')+N(n''+in'''))(N(1-i)+N(1+i))=N((n+n'+n''-n''')+i(-n+n'+n''+n'''))+N((n+n'-n''+n''')-i(n-n'+n''+n'''))$$ From this it is easy to derive the following two propositions, in which different representation of a number by a sum of four squares refer to the different value systems of the four roots, taking into account both the signs and the sequence of the roots. 1. If the fourfold of a number of the form $4k+1$ can be represented by four odd squares, then it can be represented half as often by one odd and three even squares, and vice versa, if a number can be represented in this way, that it can be represented twice as often in the first way. 2. If the fourfold of a number of the form $4k+3$ can be represented by four odd squares, then it can be represented half as often by one even and three odd squares, and vice versa, if a number can be represented in this way, than its quadruple can be represented twice as often in the first way. Gauss than says that a certain identity of theta functions can be derived by these two theorems, namely the assertion (written here in modern notation): $$\vartheta_{00}(0;\tau)^4 = \vartheta_{01}(0;\tau)^4 + \vartheta_{10}(0;\tau)^4$$ (Gauss denotes the three theta functions by $p(y),q(y),r(y)$). Notes on remark 1: The first identity in this passage was verifed algebrically by me, and I also suspect it might be intimately connected with quaternions. The second identity follows directly from the first by substituting $\lambda = 1-i$ and $\mu = 1+i$. Since $N(1-i)+N(1+i)=4$, this identity enables one to generate new representations of an integer $4s$ as sum of four squares by simply changing the signs and order of the different numbers in the representation of $s$ as sum of four squares. For example, if $s = 13 = 2^2+2^2+2^2+1^2$ than this identity implies $52=4s = 5^2+3^2+3^2+3^2$. Remark 2: On p. 1-2 of volume 8 of Gauss's Werke there is an additional note on the representation of numbers as sums of squares. According to Dickson's "history of the theory of numbers": Gauss noted that every decomposition of a multiple of a prime $p$ into $a^2+b^2+c^2+d^2$ corresponds to a solution of $x^2+y^2+z^2\equiv 0 \pmod{p}$ proportional to $a^2+b^2,ac+bd,ad-bc$ or to the sets derived by interchanging $b$ and $c$ or $b$ and $d$. For $p\equiv 3 \pmod{4}$, the solutions of $1+x^2+y^2\equiv 0 \pmod{p}$ coincide with those of $1+(x+iy)^{p+1}\equiv 0 \pmod{p}$. From one value of $x+iy$ we get all by using: $$(x+iy)\frac{(u+i)}{(u - i)}$$ (where $u = 0,1,\cdots, p-1$). For $p\equiv 1 \pmod{4}, p = a^2+b^2$; then $b\frac{(u+i)}{a(u-i)}$ give all values of $x+iy$ if we exclude the values $a/b$ and $b/a$ of $u$. Notes on remark 2: The result of Gauss on the correspondence between the representation of a multiple of a prime number $p$ as sum of four squares and the solution to the congruence $x^2+y^2+z^2\equiv 0 \pmod{p}$ is straitforward to prove: $$x^2+y^2+z^2 = (a^2+b^2)^2+(ac+bd)^2+(ad-bc)^2=(a^2+b^2)^2+(a^2+b^2)(c^2+d^2)= (a^2+b^2)(a^2+b^2+c^2+d^2)\equiv 0 \pmod{p}$$. Therefore, what remains to be settled is the other results mentioned in remark 2. The result on the correspondence between the solutions $(x,y)$ of the congruence $1+x^2+y^2\equiv 0 \pmod{p}$ and the solution of a certain imaginary congruence of degree $p+1$ was checked by me by taking specific examples: for example, if $p=7$, than $x = 3, y = 2$ is a solution, and $1+(3+2i)^8 = -238-28560i = 7\cdot(-34-4080i)$ is a Gaussian integer multiple of $7$. The papers "On the Computation of Representations of Primes as Sums of Four Squares" and "The circle equation over finite fields" mention results equivalent to Gauss's results in this remark. In particular, Gauss's method of generating new solutions to the congruence $1+x^2+y^2\equiv 0 \pmod{p}$ by $x+iy = (x_0+iy_0)(\frac{u+i}{u-i})$ is refered to as "the method of diophantus" in section 2.2 of the first paper i mentioned. Since both papers appear not to be very advanced, i believe that explaning the results in remark 2 is an easy task in the standards of mathoverflow. Questions The two propositions which Gauss mentions in remark 1 are unclear to me, and I mean that the propositions themself are unclear, not their derivation. It is simply not formulated clearly. Therefore, I'd like to get an explanation of the two propositions which Gauss mentions, as well as an explanation of its proof. I'd like to understand how the theta function identity in remark 1 follows from the two propositions. What is the expalanation of the results in remark 2? they are complicated and i don't have a clue of understanding its proof. REPLY [2 votes]: Let me add a few remarks concerning 2. If $p \equiv 3 \bmod 4$, then ${\mathbb F}_p(i) = {\mathbb F}_{p^2}$. The relative norm of $x+iy$ is the product of $x+iy$ and its conjugate $x-iy$, but the latter is the image of the Frobenius automorphism, i.e., $x-iy = (x+iy)^p$. This shows that $x^2 + y^2 = (x+iy)(x-iy) = (x+iy)^{p+1}$. If $x+iy$ is an element with norm $-1$, i.e., with $x^2 + y^2 = -1$ in ${\mathbb F}_p$, then you get all other elements with norm $-1$ by multiplying $x+iy$ by an element with norm $1$. By Hilbert's Theorem 90, such elements have the form $\frac{c+di}{c-di}$; if $d = 0$, this quotient is $1$, so you may assume $d \ne 0$ and cancel $d$; writing $u = c/d$ then proves the second claim.<|endoftext|> TITLE: Embedding $S^1\sqcup S^2$ in $S^4$ and its variants QUESTION [5 upvotes]: We consider smooth embeddings of $S^1\sqcup S^2$ inside $S^4$. Let $f$ be such an embedding with the property that $f|_{S^2}$ is isotopic to the standard embedding, say in the first 3 coordinates. Let $\textrm{Emb}(S^1\sqcup S^2,S^4)_\textrm{std}$ denote the isotopy class of such embeddings as above. This is also an abelian group via embedded connect sum operation. [The notion of isotopy and ambient isotopy may not coincide as the codimension is $2$. Hence, the questions I am asking below are applicable for both notions, if they differ.] (1) Is it true that $\textrm{Emb}(S^1\sqcup S^2,S^4)_\textrm{std}$ is infinite cyclic? Remark: While reading Habegger's 1986 paper Knots & links in codimension greater than 2 (available here), it seems that one can deduce (1) via Theorem 1.2 although this is contradicted when one uses Corollary 1.3. This may be due to some typo's in the formula or lack of my understanding. (2) If we remove the assumption of the embedding being standard on $S^2$, then we are led to consider $\textrm{Emb}(S^1\sqcup S^2,S^4)$. How much of the group structure of $\textrm{Emb}(S^1\sqcup S^2,S^4)$ is known in terms of $\textrm{Emb}(S^2,S^4)$? (3) More generally, is it known that $\textrm{Emb}(S^1\sqcup S^{n-2},S^n)_\textrm{std}$ is infinite cyclic? [It seems to be the case based on Habegger's paper but I am stuck at the same confusion/typo as pointed above.] How much of $\textrm{Emb}(S^1\sqcup S^{n-2},S^n)$ can be captured by $\textrm{Emb}(S^{n-2},S^n)$? Edit: As pointed out by D. Ruberman, $\textrm{Emb}(S^1\sqcup S^{n-2},S^n)$ need not be a group. However, even for $n=2$, I'm interested in seeking a characterization of the embeddings in terms of $\textrm{Emb}(S^2,S^4)$ and the linking number of $S^1$ and $S^2$ in $S^4$. REPLY [4 votes]: In dimensions at least 4, one has that homotopy of circles implies isotopy of circles. So if you fix an embedding of $S^2$ (eg the standard embedding), then you are really asking about embeddings of the circle in the complement of that fixed $2$-sphere, i.e. about the fundamental group of the complement. If you fix the standard embedding, then the fundamental group is $\mathbb{Z}$. (By the way, these groups are typically not abelian, so you have to fix a base point in order to get a group.) The same comments would apply if $4$ is replaced by $n$ and $n-2$ by $2$. I don't really understand what you mean more generally about the group structure on $\operatorname{Emb}(S^1 \sqcup S^2,S^4)$. Beyond the need for a base point to use in multiplying circles by ‘connected sum’, there doesn't seem to be an inverse with respect to connected sum of $2$-spheres. It's different in Habegger's situation, because the codimension is bigger than two.<|endoftext|> TITLE: Homotopically equivalent compact Lie groups are diffeomorphic QUESTION [9 upvotes]: I have the following conjecture: Two homotopically equivalent compact Lie groups will be diffeomorphic. It may be necessary to restrict ourselves to only semisimple Lie groups. For simply connected compact Lie groups, such an assertion was known to Toda (1976), he even proved that they will be isomorphic. Any commentary (for or against)? REPLY [6 votes]: The OP seems to be interested in evidence "for or against", so I will give a partial result. Definition: Let $G$ be a compact connected semisimple Lie group, $\varphi_G:\tilde{G}\to G$ its universal cover, and $\Delta\cong \pi_1(G)$ the Deck group of $\varphi_G$. There exists simple simply-connected compact groups $G_1,...,G_n$ so $\tilde{G}\cong \prod_{i=1}^nG_i$. If $\Delta=\prod_{i=1}^n\Delta_i\leq \prod_{i=1}^nZ(G_i)\cong Z(\tilde{G}),$ we will say that $G$ is adjoint-like. Remark: If $G$ is adjoint-like then it is necessarily a cartesian product of simple groups. At one extreme, when $\Delta$ is trivial, it includes the simply-connected case, and at the other extreme when $\Delta$ is the full center, then $G$ is of adjoint-type (centerless); which motivates the name. Notation: For a Lie group $G$ we will use the notation $G_0$ for its identity component, $DG:=[G,G]$ for its derived subgroup, and $Z(G)$ for its center. Theorem: Let $G$ and $H$ be homotopic compact Lie groups. If both $DG_0$ and $DH_0$ are adjoint-like, then $G$ and $H$ are diffeomorphic. Proof: First note that both $G$ and $H$ have a finite number of connected components. Since $G/G_0\cong \pi_0(G)\cong\pi_0(H)\cong H/H_0$ they have the same number of components. And since each component of $G$ (resp. $H$) is diffeomorphic to $G_0$ (resp. $H_0$), it follows that $G$ and $H$ are diffeomorphic iff $G_0$ and $H_0$ are diffeomorphic. Second, by a result of Borel (Proposition 3.1 in Sous-Groupes Commutatifs et Torsion des Groupes de Lie Compacts Connexes, 1960), $G_0$ is diffeomorphic to $DG_0\times Z(G_0)_0$ and likewise $H_0$ is diffeomorphic to $DH_0\times Z(H_0)_0$. In each case the identity component of the center is a torus (a product of circles). Since $G$ and $H$ are homotopic their identity components are as well, and in particular, the fundamental groups are isomorphic. Consequently, since the fundamental group of a semisimple Lie group is finite, the ranks of the central tori $Z(G_0)_0$ and $Z(H_0)_0$ can be determined from the corresponding fundamental groups. Thus, $G_0$ and $H_0$ are diffeomorphic iff $DG_0$ is diffeomorphic to $DH_0$. Lastly,since $G_0$ and $H_0$ are homotopic, and the central tori $Z(G_0)_0$ and $Z(H_0)_0$ are diffeomorphic (from the previous step), we conclude that $DH_0$ and $DG_0$ are homotopic as well. From Theorem 2 in Compact Lie Groups with isomorphic Homotopy Groups (1998) by Boekholt, we have that $DH_0$ and $DG_0$ are locally isomorphic and hence their universal covering spaces $\tilde{DH_0}$ and $\tilde{DG_0}$ are diffeomorphic and their deck groups are abstractly isomorphic. However, since $DH_0$ and $DG_0$ are both adjoint-like, they are cartesian products of simple groups, arising from the quotient of corresponding simple factors in $\tilde{DH_0}$ and $\tilde{DG_0}$ by central subgroups of those simple factors. We conclude that each pair of corresponding simple factors are homotopic and hence by Theorem 9.3 in The Cohomology of Quotients of Classical Groups by Baum & Browder (1963) each pair of corresponding simple factors are diffeomorphic. The result follows. $\Box$ Corollary: If $G$ and $H$ are homotopic semisimple compact Lie groups that are simply-connected or adjoint-type, then they are diffeomorphic. Remark: There is some tautological reasoning with this corollary since the simply-connected case is Toda's 1976 theorem in A note on compact semi-simple Lie groups, and I think it was used in one of the references I quote above (but there is nothing inconsistent). But as far as I know the adjoint-type case seems new (although not by much). Also, I am being slightly sloppy with the difference between "homeomorphism" and "diffeomorphism" since they are equivalent here, and similarly I am being sloppy with the difference between "weak homotopy equivalence" and "homotopic" since compact Lie groups are homotopic to CW complexes and Whitehead's theorm. Remark: The analysis shows exactly how a counter-example to the general question could arise, if it exists. One needs to look at simply-connected semisimple groups with at least two factors and quotient by isomorphic central subgroups that do not (both) arise as a product of the centers of the simple factors. Any two groups constructed this way will be homotopic, and conversely the above proof shows that any two such groups that are homotopic must arise this way if they have any chance of not being diffeomorphic.<|endoftext|> TITLE: a filter of subsets intersecting the cartesian power of each infinite subset QUESTION [6 upvotes]: Is the following filter known in set theory, and does it have a name ? For $k=1$ it is the filter of cofinite subsets. Fix a natural number $k$ and a linear order $I$. Define a filter on the set of strictly increasing k-tuples $I^{(k)}:=\{(i_1,...,i_k): i_1<...1$ the proof that it is a filter uses the Ramsey theorem. The notion is not quite trivial: for $k=2$ and $I=\Bbb Z$ the subset of pairs with even sum is big, even though the subset of pairs with odd sum is not (take $J$ to be the subset of even numbers). Perhaps an equivalent description in terms of subsets is clearer. A subset $U$ of the set $P^k(I)$ of subsets of size $k$ is big iff each infinite subset has a subset of size $k$ in $U$, i.e. in notation, $P^k(J)\cap U \neq \emptyset$ for each infinite $J\subset U$. The motivation for the question is that this filter is implicit in the definition of dividing in model theory: a formula $\phi(-,b)$ does not k-divide in a model $M$ iff the subset $\{(b_1,...,b_k)\in M^k: tp(b_i)=tp(b), M\models\exists x \wedge_i \phi(x,b_i)\}$ is big in $p(M)^k=\{(b_1,...,b_k)\in M^k: tp(b_i)=tp(b)\}$, for any (eqv.,each) linear order on $p(M)$, where $p(M)$ denotes the set of realisations of $tp(b)$ in $M$. Indeed, if it is not big, by definition there is an infinite subset $(b_J)_J$ of realisations of $tp(b)$ such that each $k$-tuple in $\phi$-inconsistent, i.e. $\phi(-,b)$ $k$-divides. REPLY [3 votes]: Unfortunately I don't have a good answer to your question. In particular, I don't know a place where this filter is directly discussed or given a name. But I still think there are some interesting comments to make. Maybe they will be of use to you, or perhaps jog something in the knowledge of other people who read this answer. For $k>1$ it seems natural to view this filter as consisting of $k$-uniform hypergraphs. Specifically, if $P^k(I)$ is the set of $k$-element subsets of $I$, then a subset $E\subseteq P^k(I)$ is a $k$-uniform hypergraph with vertex set $I$. Let $\mathcal{F}^k(I)$ be the filter defined in the original post: $E\in\mathcal{F}^k(I)$ if and only for any infinite $J\subseteq I$, $E$ contains a $k$-element subset of $J$. If we reword this under the hypergraph viewpoint, then we get alternate description of the filter, which I think is illuminating. $\mathcal{F}^k(I)$ is the collection of $k$-uniform hypergraphs on the vertex set $I$ which contain no infinite independent sets. It follows from Ramsey's Theorem that a $k$-uniform hypergraph $H=(I,E)$ is in $\mathcal{F}_k(I)$ if and only if every infinite induced subgraph of $H$ contains an infinite complete set. As noted by the OP, this provides an easy way to see that $\mathcal{F}^k(I)$ is actually a filter. Assuming $I$ is infinite, it is a proper filter. A remark on the even sum example. As noted by the OP, $\mathcal{F}_2(\mathbb{Z})$ contains the graph on $\mathbb{Z}$ in which an edge is drawn between two integers whose sum is even. Indeed, this graph has no independent set of size $3$. In light of the previous remark, I think one can't help but define $\mathcal{F}^k_n(I)$ to be the set of $k$-uniform hypergraphs on $I$ with no independent set of size $n$. So $\mathcal{F}^k_n(I)\subseteq \mathcal{F}^k(I)$. Let $\mathcal{F}^k_{\text{fin}}(I)=\bigcup_{n>0}\mathcal{F}^k_n(I)$ be the set of $k$-uniform hypergraphs with a uniform finite bound on the size of an independent set. This is also a filter, since if $E_1,E_2\in \mathcal{F}^k_n(I)$ then $E_1\cap E_2\in \mathcal{F}^k_{R(n,n;2,k)}(I)$ where $R(n,n;2,k)$ is the relevant hypergraph Ramsey number (as defined here). Of course $\mathcal{F}^k_{\text{fin}}(I)$ is properly contained in $\mathcal{F}^k(I)$. A remark on cofinite filters. The cofinite filter $\mathcal{F}(P^k(I))$ of cofinite subsets of $P^k(I)$ corresponds to the $k$-uniform hypergraphs with cofinite edge set. This filter is contained in $\mathcal{F}_{\text{fin}}^k(I)$, but of course not equal to it (e.g., the even-sum graph on $\mathbb{Z}$ above does not have a cofinite edge set). In general, an ultrafilter on an infinite ground set is nonprincipal if and only if it contains the cofinite filter. Is there something interesting to be said about ultrafilters on $P^k(I)$ that contain $\mathcal{F}^k(I)$ (or $\mathcal{F}^k_{\text{fin}}(I)$)? Remarks on Dividing The underlying hypergraphs involved in this viewpoint of dividing appear in The Characteristic Sequence of a First-Order Formula by M. Malliaris. In this paper, Malliaris fixes a theory $T$ and a formula $\varphi(x,y)$ and, for each $k\geq 1$ defines the hypergraph $P_k(y_1,\ldots,y_k)$ given by $\exists x\bigwedge_{i=1}^k\varphi(x,y_i)$. So given some (saturated enough) model $M\models T$ and parameter $b\in M^y$, if $p=\text{tp}(b)$, then $\varphi(x,b)$ does not $k$-divide (over $\emptyset$) if and only if the hypergraph $(p(M),P_k)$ is in $\mathcal{F}^k(p(M))$. Something similar to this is stated in Observation 2.4(4) of Malliaris's paper. A theme of the paper is that many model-theoretic notions can be reformulated and studied by means of these hypergraphs.<|endoftext|> TITLE: Unknotted $S^{n-2}$ in $S^n$ QUESTION [12 upvotes]: I wonder is it still an open question that a smooth sphere $\Sigma^{2}\subset S^4$ is unknotted in $S^4$ iff its complement is homotopy equivalent to $S^1$? If it is an open question, how is it related to other known conjectures in 4D? I know for all the other $n$ this has been settled by Levine 1965 "Unknotting spheres in codimension 2" and Wall 1965 "Unknotting tori in codimension one and spheres in codimension two", see relatedly Status of a conjecture of C.T.C. Wall?. REPLY [12 votes]: My understanding is this remains an open problem in the smooth category. I believe there have been a few claims of proofs of this statement in the literature over the years, but as far as I know none of these arguments have been robust. As I believe you are aware, in the topological category this was done by Mike Freedman. If you jazz up the conjecture a little you could turn it into a recognition principle for $S^1 \times D^3$, and that could in turn be turned into a proof of the smooth 4-dimensional Poincare conjecture. Specifically, say you have a homotopy 4-sphere. Remove a small unknotted $S^2$ (i.e. in some embedded $D^4$ in your homotopy 4-sphere), then its exterior presumably would satisfy the recognition principle for $S^1 \times D^3$. From this you could argue the homotopy $4$-sphere is the standard smooth $S^4$ by filling in $D^2 \times S^2$.<|endoftext|> TITLE: Fractional derivative notation in wave turbulence QUESTION [6 upvotes]: This is my first question in MathOverflow and I will do my best to format it correctly and make it clear. I am reading a paper on dispersive wave turbulence which introduces the following family of equations: $$i\psi_t=|\partial_x|^{\alpha}\psi+|\partial_x|^{-\beta/4}\left(\left||\partial_x|^{-\beta/4}\psi\right|^2|\partial_x|^{-\beta/4}\psi\right)$$ Where $|\cdot|$ denotes the $L^2$ norm in relevant places. This equation is stated to have the dispersion relation $\omega=|k|^{\alpha}$ and becomes an NLS equation when $\alpha=2$ and a water-like dispersion law when $\alpha=1/2$. I am not terribly comfortable with fractional derivatives, and in my Googling have been unable to find the use of the particular absolute value notation in $|\partial_x|^{\alpha}$. Could anyone help me interpret the equation above or point me towards a solid source? Many thanks in advance! (reference: https://www.semanticscholar.org/paper/A-one-dimensional-model-for-dispersive-wave-Majda-McLaughlin/75056874558c915a68f9cb53fc0dc989148e6db5) REPLY [4 votes]: The fractional derivative $|\partial_x|^\alpha$ is discussed in One-dimensional wave turbulence by Zakharov, Dias, and Pushkarev. (Zakharov introduced the notation.) As they explain below Eq. 2.1, it is indeed defined via the Fourier transform, such that the Fourier transform of $|\partial_x|^\alpha\psi(x)$ is $|k|^\alpha\psi(k)$. Their appendix A contains a few more details.<|endoftext|> TITLE: Largeness of the set of zeroes of a Brownian motion QUESTION [5 upvotes]: Definitions: A measurable subset $S$ of $\mathbb R$ is said to be mesoscopic if there exists a continuous function $f: \mathbb R \to \mathbb R$ such that $f(S)$ is Lebesgue measurable and has nonzero Lebesgue measure. Question: Is the set of zeroes of a Brownian motion almost surely a mesoscopic set? Remark: Note that there exist mesoscopic sets of Lebesgue measure zero - for example the Cantor set with $f$ being the Cantor staircase function. REPLY [9 votes]: Yes, the local time (at zero) maps the zero set of Brownian motion to an interval. See e.g. Lemma 6.9 page 159 in [1] for continuity. [1] Brownian motion, by Peter Mörters and Yuval Peres. Cambridge University Press, 2010 https://people.bath.ac.uk/maspm/book.pdf<|endoftext|> TITLE: Who are the owners of the compactness theorem in $L^p(\Bbb R^d)$? QUESTION [5 upvotes]: As the title says, I am interested to know Who are the owners of the compactness theorem in $L^p(\Bbb R^d)$. There is some confusion in the literature. Let recall that the compactness theorem in $L^p(\Bbb R^d)$ is somewhat a generalization of the Ascoli-Arzelà compactness theorem for space $C(X)$ of continuous functions on a compact metric space $(X,d)$. The theorem can be phrased as follows for $1\leq p<\infty$ Theorem A set $\mathcal{F}\subset L^p(\Bbb R^d)$ is compact if and only if (Boundedness) $\sup_{u\in \mathcal{F}}\|u\|_{L^p(\Bbb R^d}<\infty,$ ($p$-equicontinuity) $$\lim_{|h|\to 0} \sup_{u\in \mathcal{F}}\|u(\cdot+h)-u(\cdot)\|_{L^p(\Bbb R^d}=0,$$ $p$-tighness $$\lim_{R\to \infty} \sup_{u\in \mathcal{F}}\int_{|x|>R}|u(x)|^pd x=0,$$ The book by Haim Brezis (Functional Analysis, Sobolev Spaces, and Partial Differential Equations) names this Theorem as "the Riesz-Fréchret-Kolmogorov Theorem" However, further research led me to this article with more historical details and the name of Fréchet does not appear therein. The theorem is thus merely named as "Kolmogorov-Riesz Theorem". Is there any reason why Brezis added the name of Fréchet? Does any experience professor here have some additional information about this? I am actually writing a manuscript and I would like to make sure the right owners are clearly mentioned. REPLY [18 votes]: I consulted the compendum on such topics: Dunford & Schwartz. Dunford, Nelson; Schwartz, Jacob T., Linear operators. I. General theory. (With the assistence of William G. Bade and Robert G. Bartle), Pure and Applied Mathematics. Vol. 7. New York and London: Interscience Publishers. xiv, 858 p. (1958). ZBL0084.10402. The compactness results for $L_p$ are IV.8.18, IV.8.20, IV.8.21. From the notes at the end of chapter IV, it seems: Fréchet [1908] characterized compact sets in $L_2[0,1]$ Kolmogorov [1931] characterized compact sets in $L_p(S)$, $1 < p < \infty$, $S$ a bounded set in finite-dimensional Euclidean space Tamarkin [1932] extended to unbounded sets (There were also papers with cases $p=\infty, p=1, 0 TITLE: Is there any non-commutative ring such that every element other than the identity is a zero divisor? QUESTION [14 upvotes]: A (unital) ring $R$ with the property that every element other than the identity $1_R$ is a (two-sided) zero divisor, seems to be commonly called a "$0$-ring" or "$\mathcal O$-ring". These rings were first studied by P.M. Cohn (though only in the commutative setting) in Rings of zero divisors, Proc. Amer. Math. Soc. 9 (1958), 914-919. Moreover, every right (or left) artinian $\mathcal O$-ring is, in fact, a boolean ring (and hence commutative), see H.G. Moore, S.J. Pierce, and A. Yaqub, Commutativity in rings of zero divisors, Amer. Math. Monthly 75 (1968), 392 Thence, the question is: Does there exist any non-commutative $\mathcal O$-ring? If so, can you provide a reference where this is discussed? I've tried to track the citations of Cohn's paper, but couldn't find an answer to my question. (See also here and there.) REPLY [11 votes]: [Sorry for answering my own question, and the more so because this is happening for the second time in 24 hours.] The question might be open. In fact, a positive answer would imply an equally positive answer to a question stated in the introduction of Melvin Henriksen's paper "Rings with a unique regular element", pp. 78-87 in B.J. Gardner (ed.), Rings, modules and radicals (Proc. Conf., Hobart/Aust. 1987), Pitman Res. Notes Math. Ser. 204, Longman Sci. Tech., Harlow, 1989, where Henriksen writes: We do not know if there is a ring with a unique regular ring [sic] that fails to be commutative. In Henriksen's paper, a ring need not be unital; and a regular element is nothing else than a cancellative element of the multiplicative semigroup of the ring (loc. cit., Definition 2.1). The question is marked as open by David Feldman in a 2012 post from the "Not especially famous, long-open problems which anyone can understand" big list (see also the comments under the same post), where Feldman writes: Must a non-commutative ring (with identity) contain a non-zero-divisor aside from the identity?<|endoftext|> TITLE: Jacobi fields on non-symmetric spaces QUESTION [5 upvotes]: I am searching for examples of manifolds which are not symmetric spaces but where Jacobi fields can be computed in closed form. For now, I am aware of Gaussian distribution with the Wasserstein metric: https://arxiv.org/pdf/2012.07106.pdf Kendall shape space: https://arxiv.org/pdf/1906.11950.pdf Are there many others? Thank you for your help. REPLY [3 votes]: Damek-Ricci spaces are obtained by equipping certain solvable Lie groups with appropriate invariant Riemannian metrics. This class is larger than the class of symmetric spaces (this was precisely the point of their construction as counterexamples to the Lichnerowicz conjecture ), and the Jacobi fields on them are explicitly described in Section 4.2 of Isoparametric hypersurfaces in Damek–Ricci spaces by Díaz-Ramos and Domínguez-Vázquez.<|endoftext|> TITLE: Is there always a simple module whose Green correspondent is a simple module under some conditions? QUESTION [8 upvotes]: Let $G$ be a finite group and $KG$ its group algebra over some field $K$ with $\mathrm{char}\ K$ dividing the order of $G$. It's well-known that the Green correspondence is compatible with the Brauer correspondence. Suppose we are dealing with Green correpondence between indecomposable modules of a block $B$ and its Brauer correpondent $b$. My question is the following: Is there always a simple $B$-module whose Green correspondent is a simple $b$-module? Here a module always means a left module. And the question is trivial for the principal block. For blocks with Klein four defect group, Craven and his coauthor have proved it in a published paper. I guess the question is true for blocks with abelian defect groups since all examples I have found in the literature satisfy this property. Or there might be counterexamples due to my lack of knowledge. I will be very grateful if anyone could provide me with one. REPLY [7 votes]: The answer is "no" in general. I presume you mean that $B$ is a block of $KG$, and $b$ is its local Brauer correspondent. Consider the case $G = {\rm SL}(2,3)$ with $p = 3.$ Then $G$ has three $3$-blocks. One is the principal $3$-block, one is $3$-block of defect zero. The third block is a non-principal $3$-block $B$ of full defect. Because $G$ is a $3$-nilpotent group, each $3$-block of $G$ contains just one simple module. In the case of the block $B$ the unique simple $B$-module is the $2$-dimensional ``natural" module, which we label $V$. Let $D$ be a Sylow $3$-subgroup of $G$, which is a defect group for $B$, and a vertex for $V$. Let $H = N_{G}(D)$. Then Green correspondence tells us that ${\rm Res}^{G}_{H}(V) = U \oplus P$ where $U$ is indecomposable with vertex $D$ and $P$ is projective (or zero). By dimension, $P = 0.$ Hence $U$ is indecomposable, but $U$ is not simple ( since $O_{3}(H) = D$ acts non-trivially on $U$). Thus no simple $B$-module has a simple Green correspondent.<|endoftext|> TITLE: Pushout of group schemes (question on a lemma in SGA3) QUESTION [15 upvotes]: $\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\diag{diag}$In SGA3, Expose XXIV, Lemme 7.2.2 it says (let's say our base scheme $S$ is an algebraically closed field $k$): if $G$ is reductive algebraic group, $T$ a maximal torus, $G'$ the derived subgroup of $G$ and $T' = T \cap G$ a maximal torus of $G'$, then $G$ is a pushout of $G'$ and $T$ over $T'$ (even in the category of group sheaves). Doesn't the following very basic example contradict this? Or what have I misunderstood? Suppose $G = \GL_2$, $G' = \SL_2$, $T$ the diagonal torus and $T'$ the diagonal torus with determinant 1. Define homomorphisms $\alpha : T = \mathbb G_m^2 \to \GL_2$ sending $\diag(x,y)$ to $\diag(x,x^{-1})$ and $\beta : G' = \SL_2 \to \GL_2$ the inclusion. Then $\alpha$ and $\beta$ agree on the intersection $T'$, but there is no homomorphism $G = \GL_2 \to \GL_2$ that restricts to $\alpha$ and $\beta$ because otherwise $\alpha(\diag(x,x))$ and $\beta(y)$ would commute for any $x$, $y$. (See pages 39-40 of the pdf of Expose XXIV at the SGA3 Réédition project page, or the official published version) REPLY [9 votes]: The Lemma is clearly wrong. There is no way to recover $G$ from $G'$, $T$ and $T'$ alone (not even up to isomorphism) since that data do not determine the radical $R:={\rm rad}(G)\subseteq T$. Maybe the authors had in mind the amalgamated product of $T$ and $R\times G'$ over $R\times T'$.<|endoftext|> TITLE: Diagonalizing against $\omega_1$-sequences of functions mod finite QUESTION [16 upvotes]: The following statement is a direct consequence of the Continuum Hypothesis: There exists a sequence $\langle f_\alpha:\omega_1\rightarrow\omega_1 ~ \vert ~ \alpha<\omega_1\rangle$ of functions such that there is no function $f:\omega_1\rightarrow\omega_1$ with the property that the sets $\{\xi<\omega_1 ~ \vert ~ f(\xi)=f_\alpha(\xi)\}$ are finite for all $\alpha<\omega_1$. Moreover, since a failure of this statement can be used to obtain an $\omega_2$-sequence of subsets of $\omega_1$ with pairwise finite intersection, results of Baumgartner in Baumgartner, James E., Almost-disjoint sets, the dense set problem and the partition calculus, Ann. Math. Logic 9, 401-439 (1976). ZBL0339.04003. show that the statement is not equivalent to CH. Question: Can the above statement consistently fail? REPLY [6 votes]: The arguments in Section 6 of my paper "The nonstationary ideal in the $\mathbb{P}_{\mathrm{max}}$ extension" show that there is a proper forcing adding a function from $\omega_{1}$ to $\omega_{1}$ which agrees with each such ground model function in only finitely many places. I say there that Todorcevic had done something similar in his "A note on the Proper Forcing Axiom". It follows that under PFA, and in the $\mathbb{P}_{\mathrm{max}}$ extension, the statement above fails. You don't need any large cardinals, however.<|endoftext|> TITLE: Are there more paths exiting a box in $\mathbb{Z}^2$ to the right if I remove some edges to the left QUESTION [9 upvotes]: Suppose that I am given the graph $G = (V,E)$ where $V = \{ 1, 2, \dots 2N+1 \} \times \{ 1, 2, \dots 2N+1 \} $ and there is an edge between two vertice $(n,m)$ and $(n',m')$ if and only if $\vert n-n'\vert + \vert m-m'\vert = 1$. Suppose that we remove some arbitrary edges between vertices $(n,m)$ and $(n',m')$ with $n, n' \leq N$. Prove or disprove that there are more edge-selfavoiding paths from $(N+1,N+1)$ to a vertex of the form $(2N+1, m)$ for $m \in \{ 1, 2, \dots 2N+1 \}$ than paths to a vertex of the form $(1, m)$ for $m \in \{ 1, 2, \dots 2N+1 \}$. REPLY [5 votes]: Expanding on Anthony Quas' comments above, it is indeed possible to show that there are cases when there are more paths to the left than to the right. Let $H$ be the graph obtained from $G$ by removing all edges from $(N-1,i)$ to $(N,i)$ except one (which then clearly is a bridge in $H$). Any walk in $H$ that ends on the right cannot cross the bridge and thus uses at most $2N^2 + O(N)$ vertices. One way to encode such a walk is to store its length and remember the behavior (turn left/turn right/continue straight) every time we encounter a previously unvisited vertex; the behavior at previously visited vertices is determined by this encoding since we are not allowed to use any edge twice. This shows that there are at most $3^{2N^2 + O(N)}$ walks that end on the right. For a lower bound on the number of walks in $H$ ending on the left, we observe that there is a connected subgraph of $H$ with $4N^2-O(N)$ vertices of degree $4$ in which $(N+1,N+1)$ and $(1,1)$ are the only vertices of odd degree. Starting with a Eulerian walk from $(N+1,N+1)$ to $(1,1)$ we can obtain $2^{4N^2-O(N)}$ different such Eulerian walks by iteratively performing local modifications at each vertex. Indeed, observe that if we fix how the walk traverses every vertex apart from one vertex of degree $4$, then there are always two options at this last vertex completing it to a Eulerian walk.<|endoftext|> TITLE: Given some recursive function, can we effectively associate it a polynomial as in the DPRM theorem? QUESTION [6 upvotes]: I'm interested in the following assertion about the Davis-Putnam-Robinson-Matijasevich theorem Given a recursive function $f:\mathbb{N}\rightarrow\mathbb{N}$, i.e. its index, we can effectively get a polynomial $p\in\mathbb{Z}[x,y,z_1,\dots,z_n]$ that satisfies $$f(x)=y\iff\exists z_1\dots z_n \in \mathbb{N} \\ p(x,y,z_1\dots z_n )=0.$$ In other words After indexing recursive functions and integer polynomials, there is a recursive function whose input is the index of a recursive function $f:\mathbb{N}\rightarrow\mathbb{N}$ and whose output is the index of a polynomial $p\in\mathbb{Z}[x,y,z_1,\dots,z_n]$ that satisfies $$f(x)=y\iff\exists z_1\dots z_n \in \mathbb{N} \\ p(x,y,z_1\dots z_n )=0.$$ I believe it's true, and that it's implicit in the proof of the DPRM theorem (both in Matijasevic proof and Martin Davis proof). If this is stated somewhere and you know it, please let me know :). On the other hand, it may be too obvious that this is implicit in the proofs, but in that case I would like to hear that from someone else. Why I believe this is true: both proofs are constructive, i.e., they indicate how to get the desired polynomial from an arbitrary recursive functions. I develop a bit more: Matijasevic proof As I understand the proof, given a register machine, i.e. its instructions, it specifies how to coded each possible instruction as a polynomial. If we index each possible register machine by its instructions, the polynomial is effectively obtained from the index of the function, and we could put this procedure in a recursive function. Martin Davis proof As I understand the proof (theorem 6.1), it proves that diophantine functions are closed under composition, primitive recursion, and minimalization (i.e. the $\mu$ operator). In each case, it gives an explicit and concrete diophantine formula (well, formulas with abbreviations, but all of those can be expanded). This proves that recursively enumerable functions are diophantine. Now, if we index each possible recursively enumerable function by its "factorization" using composition, primitive recursion, and minimalization (this is possible and effective), then the polynomial we get is effectively obtained from the index of the function. Thanks for reading :) REPLY [7 votes]: Davis (also available on the MAA site) begins section 7 by saying "an explicit enumeration of all the Diophantine sets of positive integers will now be described". He then gives theorem 7.1, from which one can get a polynomial form for a universal Diophantine set. In the proof of theorem 7.3, he says "write $$x\in D_n \iff (\exists z_1,\cdots, z_n) [P(n,x,z_1,\cdots, z_n)=0]$$ where P is some definite (though complicated) polynomial." I find all that plenty explicit about the explicitness.<|endoftext|> TITLE: Invertible 2-knots in $S^4$ QUESTION [6 upvotes]: Q1: Is it true that a knot $S^2\hookrightarrow S^4$ has an inverse iff it is trivial? Or it is also an open question? See relatedly Unknotted $S^{n-2}$ in $S^n$. Q2: It is easy to see that if a knot $f\colon S^2\hookrightarrow S^4$ has an inverse than its complement $C_f\simeq S^1$. Has the converse been proved? Both questions are answered below by Daniel Ruberman. REPLY [7 votes]: Q1: This is true in the topological category and unknown in the smooth setting. In the topological setting, the fundamental group of $S^4 - K_1 \# K_2$ is $G_1 *_\mathbb{Z} G_2$ where $G_i$ are the fundamental groups of $S^4 -K_i$. If this is $\mathbb{Z}$ then I think the $G_i$ are both $\mathbb{Z}$. By the arguments in your earlier question this means that both $K_i$ are unknotted. Q2: It is also true in the topological category and unknown in the smooth category that the complement being a homotopy circle implies that the knot in question has an inverse. The argument in the topological case is that the knot is trivial (since the group is $\mathbb{Z}$ as noted previously.)<|endoftext|> TITLE: Modulo $3$ calculations for a binomial-sum sequence QUESTION [5 upvotes]: Introduce the sequence (this is A047781 on OEIS) $$t_n=\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k}k$$ and denote the set $T(ij)=\{n\in\mathbb{N}: \text{the ternary digits of $n$ contain $i$ or $j$ only}\}$. QUESTION. Is this true modulo $3$? $$t_n\equiv_3\begin{cases} 1 \qquad \text{if $\lfloor\frac{n}2\rfloor\in T(01)$} \\ 0 \qquad \text{otherwise}. \end{cases}$$ REPLY [2 votes]: Here is a comment following Max Alekseyev's resolution. It has to do with his generating function for $t_n$ and working out directly on the Taylor's expansion (Binomial Theorem). Namely, $$\frac14\left(\frac{1+x}{\sqrt{1-6x+x^2}}-1\right) =\sum_{n=1}^{\infty}\frac{h(n-1)+h(n)}4\,x^n$$ where $$h(n)=\sum_{k=\lfloor\frac{n}2\rfloor}^n(-1)^{n+k}\binom{2k}k\binom{k}{n-k}3^{2k-n}2^{-n}.$$ To get to the conclusion we seek, let's just take $n\rightarrow 2n$ for instance. This leads to \begin{align*}t_{2n}=\frac{h(2n-1)+h(2n)}4&\equiv_3h(2n-1)+h(2n) \\ &\equiv_3\sum_{k=n}^{2n-1}(-1)^{k-1}\binom{2k}k\binom{k}{2n-1-k}3^{2k-2n+1}2^{-2n+1} \\ &+\sum_{k=n}^{2n}(-1)^k\binom{2k}k\binom{k}{2n-k}3^{2k-2n}2^{-2n} \\ &\equiv_3\sum_{k=n}^{2n}(-1)^k\binom{2k}k\binom{k}{2n-k}3^{2k-2n}2^{-2n} \\ &\equiv_3(-1)^n\binom{2n}n2^{-2n} \\ &\equiv_3(-1)^n\binom{2n}n. \end{align*}<|endoftext|> TITLE: Is there a mapping from Euclidean space to real numbers which relatively preserves distance? QUESTION [5 upvotes]: Motivation: I need to find a mapping from $n$-dimensional Euclidean space to real numbers such that the distance between each pair of points in the quoted space is relatively-preserved after the application of the mapping. Question: Given $a, b, c \in \mathbb{R}^{n}$ and assuming that $||a-b|| \le ||a-c|| \le ||b-c||,$ what can the mapping $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be such that the following property holds? $$|f(a)-f(b)| \le |f(a)-f(c)| \le |f(b)-f(c)|$$ REPLY [10 votes]: If I understood correctly, such a mapping must be constant if $n\geq 2$. Permuting the names of the variables, the condition implies that $f$ must send every equilateral triangle in $\mathbb{R}^n$ to an "equilateral triangle" in $\mathbb{R}$, which can only be a single point. Since every pair of points in $\mathbb{R}^n$ forms one side of an equilateral triangle, the mapping must collapse every pair of points, and hence be constant.<|endoftext|> TITLE: Egyptian number theory QUESTION [9 upvotes]: Might there be a good historical reference on Egyptian number theory ($ \sim 2000$ B.C.)? The following online reference by a professor at the UCLA indicates that they were aware of the Pythagorean theorem [1]. This makes me wonder whether Egyptian scientists and engineers might have done fundamental work in number theory as well. In particular, I'd like to know whether they were aware of Euclid's theorem of the infinitude of primes. References: Allen Klinger. Right Triangles - Pythagorean Theorem . http://web.cs.ucla.edu/~klinger/dorene/math1.htm Thomas Eric Peet. Mathematics in Ancient Egypt. 1931. REPLY [4 votes]: Besides the Pythagorean theorem, it appears that the Egyptians had general algorithms for computing the volume of a pyramid [1]. As for number theory, it appears that the theory of Egyptian numbers(i.e. integers which are the sum of Egyptian fractions(circa 3500 BC)) has motivated non-trivial developments in combinatorial number theory: There is a theorem due to Erdős which shows that it is not possible for a Harmonic progression to form an Egyptian number [3]. In an analysis of integer partitions, Ronald Graham found that if $n$ is an integer exceeding 77 there are positive integers $\{a_i\}_{i=1}^k$ such that $n = \sum_{i=1}^k a_i$ and $\sum_{i=1}^k \frac{1}{a_i} = 1$ [2]. More recently, Steve Butler, Erdős, and Graham found that any integer may be expressed as a sum of Egyptian fractions with denominators $a_i < a_{i+1}$ and where $a_i$ is the product of three distinct primes [4]. While it is not clear why the Egyptians developed Egyptian fractions [3], the methods used to compute Egyptian numbers in the Rhind papyrus suggest that they had a partial understanding of the difference between prime and composite numbers [5]. Moreover, from my analysis of the available papyri it appears that Egyptian mathematicians focused strictly on algorithmic/constructive methods which did not include a notion of infinity. This precludes methods such as Archimedes' method of exhaustion or Euclid's existence proof that there are infinitely many primes. References: Struve, Vasilij Vasil'evič, and Boris Turaev. 1930. Mathematischer Papyrus des Staatlichen Museums der Schönen Künste in Moskau. Quellen und Studien zur Geschichte der Mathematik; Abteilung A: Quellen 1. Berlin: J. Springer R. L. Graham. A theorem on partitions. 1963. Graham, Ronald L. (2013), "Paul Erdős and Egyptian fractions" (PDF), Erdös centennial, Bolyai Soc. Math. Stud., 25, János Bolyai Math. Soc., Budapest, pp. 289–309 Steve Butler, Paul Erdős, and R.L. Graham. Egyptian fractions with each denominator having three distinct prime divisors. 2015. Abdulrahman A. Abdulaziz. On the Egyptian method of decomposing 2/n into unit fractions. Elsevier. 2007.<|endoftext|> TITLE: Images of complemented subobjects in toposes QUESTION [6 upvotes]: Let ${f : E \rightarrow S}$ be a geometric morphism (between toposes). For $s$ in $S$ and $x$ in $E$ let ${\pi : f^* s \times x \rightarrow x}$ be the obvious projection in $E$. Let ${u \rightarrow f^* s \times x}$ be a complemented subobject of ${f^* s \times x}$. Is the image of $u$ along $\pi$ complemented as a subobject of $x$? (See also Images of complemented subobjects in hyperconnected toposes over Boolean bases) REPLY [4 votes]: A different flavour of counterexample from Andreas Blass’s answer, showing this can fail when $S$ is Boolean and satisfies choice: Take $S$ to be sets, and $E = Sh(2^{\mathbb{N}})$, where $2^{\mathbb{N}}$ is the Cantor space (and $f$ is the unique geometric morphism $(\Gamma,\Delta) : Sh(2^{\mathbb{N}}) \to S$). Now for $n \in \mathbb{N}$, we have $U_n \subseteq 1$ in $E$ corresponding to the clopen $\{ x \in 2^\mathbb{N}\ |\ x_n = 1 \}$. Now each $U_n \subseteq 1$ is complemented, so $\coprod_n U_n \subseteq \Delta(n) \times 1$ is complemented; but $\pi_!(\coprod_n U_n) = \bigcup_n U_n \subseteq 1$ is not complemented, since it’s dense but not equal to $1$. Generally, sheaf toposes give many counterexamples to the principle “set-indexed unions of complemented subobjects are complemented”.<|endoftext|> TITLE: Conjectures or Results? QUESTION [17 upvotes]: There is a paper (not accepted for publication yet) that contains several conjectures. Some of these conjectures were proven recently. The referee of the original paper requires to substitute the proven "Conjectures" with the "Results". However, there are several papers that cite these conjectures, so I feel it would be wrong to rename them.  What is the best (or standard) way to indicate in the original paper that the conjectures were proven in the subsequent papers? Are there any good examples of doing that? REPLY [41 votes]: The standard way is to leave the conjectures as they are, and add a remark, or a footnote, saying that "after this paper was written (or after it was submitted for publication) this conjecture was proved" and give a reference.<|endoftext|> TITLE: Does any projective bundle on a compact complex manifold have an associated holomorphic vector bundle? QUESTION [6 upvotes]: Let $X$ be a compact complex manifold, and $f: Y\to X$ a proper surjective holomorphic map with fiber $\mathbb{CP}^n$. Is there always a holomorphic vector bundle $E$ of rank $n+1$ such that $Y$ is biholomorphic to $\mathbb{P}(E)$ over $X$? REPLY [9 votes]: It is classically known that this is true when $\dim X=1$ ("ruled surfaces" = "geometrically ruled surfaces"). It is also true when $\dim X=2$, provided that $H^2(X, \, \mathcal{O}_X)=H^3(X, \, \mathbb{Z})=0$. However, it fails for a general smooth basis. You can see the discussion at p. 190 of Barth, Wolf P.; Hulek, Klaus; Peters, Chris A. M.; Van de Ven, Antonius, Compact complex surfaces, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge 4. Berlin: Springer (ISBN 3-540-00832-2/hbk). xii, 436 p. (2004). ZBL1036.14016.<|endoftext|> TITLE: What is the discrete quantum group associated to a compact group? QUESTION [10 upvotes]: Let $G$ be a compact topological group. Then $G$ is a CQG with function algebra $C(G)$ and the usual comultiplication on $C(G)$. Is there an easy description of the dual discrete quantum group $\widehat{G}$? In the case of commutative compact groups, I would hope that there is a connection with the usual dual of a compact group (in the sense of Pontryagin duality), but since the function algebra of the discrete quantum group $\widehat{G}$ is a von Neumann algebra, it won't be something nice like $C(\widehat{G})$. REPLY [4 votes]: It's not my area, but I think the answer is very simple: when $G$ is compact abelian the W${}^*$ dual quantum group is just $L(G) \cong l^\infty(\hat{G})$.<|endoftext|> TITLE: Fixed subspaces of a family of representations $\rho_t: F_2\to GL(n,\mathbb C)$ QUESTION [5 upvotes]: Suppose we have a real analytic family of matrices $A_t, B_t\in GL_n(\mathbb C)$, with $t\in \mathbb R$. Suppose that for $t\in (0,1)$ there is a common non-zero eigenvector $v_t\in \mathbb C^n\setminus 0$ such that $A_t(v_t)=B_t(v_t)=v_t$. Is it true that such a common (non-zero) vector exists for all $t\in \mathbb R$? REPLY [4 votes]: Define the sets $$M = \{ (A,B,l)\in GL_n(\mathbb{C})\times GL_n(\mathbb{C})\times\mathbb{P}^{n-1}(\mathbb{C}) : Aw=Bw=w \text{ for all } w\in l \},$$ $$N = \{ (A,B)\in GL_n(\mathbb{C})\times GL_n(\mathbb{C}):\exists v\in\mathbb{C}^n\setminus\{0\}, Av=Bv=v\}, $$ where a point of $\mathbb{P}^{n-1}(\mathbb{C})$ is thought of as a line in $\mathbb{C}^n$. The set $M$ is closed and analytic (actually algebraic). Let $$\pi: GL_n(\mathbb{C})\times GL_n(\mathbb{C})\times\mathbb{P}^{n-1}(\mathbb{C})\to GL_n(\mathbb{C})\times GL_n(\mathbb{C})$$ be the projection. Then $\pi(M)=N$, and since $\pi$ is proper, the set $N$ is closed and analytic.<|endoftext|> TITLE: A 3rd formula for the central Delannoy numbers? QUESTION [5 upvotes]: There are several in the literature proving the two alternative formulas for the (diagonal) Delannoy numbers; namely that $$d_n=\sum_{k=0}^n\binom{n}k\binom{n+k}k=\sum_{k=0}^n\binom{n}k^22^k.$$ Each formulation might have an advantage over the other, depending on the context. I have not seen (still willing to be referred to) the below bizarre-looking expression . So, I ask: QUESTION. Is there a combinatorial justification for this? $$d_n=\sum_{k=\lfloor\frac{n}2\rfloor}^n(-1)^{n+k}\binom{2k}k\binom{k}{n-k}3^{2k-n}2^{-n}.$$ The motivation comes from studying divisibility questions and a direct look at the generating function $$\frac1{\sqrt{1-6x+x^2}}.$$ REPLY [5 votes]: By convolution, $$ D_n=\sum_{k=0}^{n}\binom{n}{k}^2 2^k = \sum_{k=0}^{n}\binom{n}{k}2^k\binom{n}{n-k}= [z^n](1+2z)^n(1+z)^n = [z^n](1+3z+2z^2)^n $$ hence by Cauchy's integral formula $$ D_n = \frac{1}{2\pi i}\oint_{|z|=\varepsilon}(1+3z+2z^2)^n \frac{dz}{z^{n+1}} $$ where the only singularity of the integrand function is at the origin, so the value of the integral is unaffected if we consider $|z|=\frac{1}{\sqrt{2}}$ as a new integration path. A change of variable leads to $$ D_n = \frac{1}{2\pi}\int_{0}^{2\pi}(3+2\sqrt{2}\cos\theta)^n\,d\theta $$ from which it is simple to derive the asymptotic behaviour of $D_n$ (through Laplace's method). Additionally, by the binomial theorem $$ D_n = \sum_{k=0}^{n}\binom{n}{k}8^{k/2}3^{n-k}\cdot\frac{1}{2\pi}\int_{0}^{2\pi}\left(\cos\theta\right)^k\,d\theta $$ which simplifies into $$ D_n = \sum_{j=0}^{\lfloor n/2\rfloor} \binom{n}{2j}8^j 3^{n-2j}\cdot\frac{1}{4^j}\binom{2j}{j}= \sum_{j=0}^{\lfloor n/2\rfloor}\binom{n}{2j}\binom{2j}{j}2^j 3^{n-2j}=\sum_{j=0}^{\lfloor n/2\rfloor}\binom{n}{j}\binom{n-j}{j} 2^j 3^{n-2j}$$ that is equivalent to the claim.<|endoftext|> TITLE: When is a triangulation of sphere two-colorable? QUESTION [8 upvotes]: Let $T$ be a triangulation of sphere. We say that $T$ is $k$-colorable if the triangles of $T$ can be assigned with $k$ colors such that any two triangles with a common edge have different colors. I am interested in $2$-colorable triangulations. Easy examples are boundaries of bipyramids over even polygons such as the simplicial comlex $\{123,134,145,152, 623,634,645,652\}$ Question: Are there efficient criteria to tell if a triangulation is $2$-colorable? How to construct $2$-colorable triangulations? Do they arise in some contexts? I am also interested in triangulations of other manifolds, so any such references are welcome. REPLY [11 votes]: As Fedor Petrov mentions in the comments, a necessary and sufficient condition is that each vertex has even degree. Here is a proof. Let $T^*$ be the dual graph of the triangulation. That is, the vertices of $T$ are the faces of the triangulation, and two faces are adjacent if they share an edge. Rephrased, your question is asking when $T^*$ is bipartite. So, it suffices to prove that a planar graph $G$ is bipartite if and only if the dual graph $G^*$ is Eulerian (all vertices have even degree). For one direction, if $G$ is bipartite, then all cycles of $G$ are even. In particular, all facial cycles are even, and hence $G^*$ is Eulerian. For the other direction, suppose that the dual graph $G^*$ is Eulerian. Thus, each face of $G$ is even. Since the cycle space of $G$ is generated by the facial cycles of $G$ (take the symmetric difference of all faces inside the cycle), this implies that every cycle of $G$ is even. Hence $G$ is bipartite.<|endoftext|> TITLE: State of rigorous effective quantum field theories QUESTION [11 upvotes]: It's well-known that there are no rigorously constructed and physically relevant QFTs. There is, however, a lot of mathematical work on effective field theories and renormalization, such as the books by Costello and by Salmhofer. My question is: does this mathematical work allow one to give mathematically rigorous (albeit effective, possibly depending upon empirical parameters) derivations of the physical computations one does with QFT (such as the electron magnetic dipole moment, say, or Compton scattering, this sort of thing)? If not, how far are we from being able to do so? I ask because as somewhat of an outsider it seems hard for me to tell. Books that intend to give an account of these computations, like Folland's Quantum Field Theory, are very far from rigorous. The mathematical books on renormalization and effective field theories such as the ones I mentioned seem to be rigorous (although I have not read them in any amount of detail), but they also don't seem to discuss fundamental physics (or maybe I just missed it). REPLY [8 votes]: I will leave aside what is meant by "effective field theory" in a purely mathematical context and just presume that the question asks whether renormalized interactive perturbative QFT (using formal power series in $\hbar$ and the coupling constants) can be mathematically well-defined. The answer is Yes (in multiple different ways), which has been repeated on this website several times (it is sufficient to search with the corresponding keywords). But it seems that there was not much interest in references before. So here are some references for what I consider to be a fairly clean approach of perturbative algebraic QFT. The list mixes older and newer references, as well as more and less readable references. Scharf, G., Finite quantum electrodynamics. The causal approach., Texts and Monographs in Physics. Berlin: Springer-Verlag. x, 409 p. (1995). ZBL0844.53052. Steinmann, Othmar, Perturbative quantum electrodynamics and axiomatic field theory, Texts and Monographs in Physics. Berlin: Springer. ix, 355 p. (2000). ZBL0946.81079. Hollands, Stefan, Renormalized quantum Yang-Mills fields in curved spacetime, Rev. Math. Phys. 20, No. 9, 1033-1172 (2008). ZBL1161.81022. Brunetti, Romeo (ed.); Dappiaggi, Claudio (ed.); Fredenhagen, Klaus (ed.); Yngvason, Jakob (ed.), Advances in algebraic quantum field theory, Mathematical Physics Studies. Cham: Springer (ISBN 978-3-319-21352-1/hbk; 978-3-319-21353-8/ebook). xii, 453 p. (2015). ZBL1329.81022. Rejzner, Kasia, Perturbative algebraic quantum field theory. An introduction for mathematicians, Mathematical Physics Studies. Cham: Springer (ISBN 978-3-319-25899-7/hbk; 978-3-319-25901-7/ebook). xi, 180 p. (2016). ZBL1347.81011. The work of Costello and Gwilliam is a different formalism, but could also be cited as an example. Perhaps others can add answers with references to their own favorite approach.<|endoftext|> TITLE: What is the intuition behind the Kantorovich potential in optimal transport? QUESTION [6 upvotes]: From what I currently understand, under certain conditions one may turn the usual Kantorovich problem - a minimisation problem in terms of measures into a maximisation problem in terms of functions. By “turn into” I mean that the optimal values for both problems agree. The Kantorovich potential associated to the problem is the function $\phi$ that achieves the maximum in the latter problem, and can be chosen to be $c$-concave where $c$ is the cost function. To be more precise, the latter problem is a minimisation over pairs of functions, and the solution can be taken to be $(\phi, \phi^c)$ for a $c$-concave function $\phi$, where the superscript $c$ denotes the $c$-transform. Many results about the original problem can be proven by examining the Kantorovich potential. While I understand the proofs formally, I cannot visualise what exactly a Kantorovich potential is doing. Is there a geometric/analytic interpretation of the potential? For example, how does it relate directly to the optimal transport plan? Can one deduce the geometry of the plan from the potential and vice versa? REPLY [5 votes]: I recommend the interpretation with bakeries and cafes! In Villani's "Optimal Transport Old and New" in Chapter 5 "Cyclic monotonicity and Kantorovich duality you'll find this: I shall start by explaining the concepts of cyclical monotonicity and Kantorovich duality in an informal way, sticking to the bakery analogy of Chapter 3. Assume you have been hired by a large consortium of bakeries and cafés, to be in charge of the distribution of bread from production units (bakeries) to consumption units (cafés). The locations of the bakeries and cafés, their respective production and consumption rates, are all determined in advance. You have written a transference plan, which says, for each bakery (located at) $x_i$ and each café $y_j$ , how much bread should go each morning from $x_i$ to $y_j$. [...] The next key concept is the dual Kantorovich problem. While the central notion in the original Monge–Kantorovich problem is cost, in the dual problem it is price. Imagine that a company offers to take care of all your transportation problem, buying bread at the bakeries and selling them to the cafés; what happens in between is not your problem (and maybe they have tricks to do the transport at a lower price than you). Let $ψ(x)$ be the price at which a basket of bread is bought at bakery $x$, and $φ(y)$ the price at which it is sold at café $y$. On the whole, the price which the consortium bakery+café pays for the transport is $φ(y) − ψ(x)$, instead of the original cost $c(x, y)$. This of course is for each unit of bread: if there is a mass $μ(dx)$ at $x$, then the total price of the bread shipment from there will be $ψ(x) μ(dx)$. So as to be competitive, the company needs to set up prices in such a way that $$∀(x, y),\ φ(y) − ψ(x) ≤ c(x, y).$$ When you were handling the transportation yourself, your problem was to minimize the cost. Now that the company takes up the transportation charge, their problem is to maximize the profits. This naturally leads to the dual Kantorovich problem: $$\sup\{\int_Y \varphi(y) d\nu(y) - \int_X \psi(x) d\mu(x):\ φ(y) − ψ(x) ≤ c(x, y)\}$$.<|endoftext|> TITLE: Are groups with every proper, non-trivial subgroup infinite cyclic simple? QUESTION [15 upvotes]: In the 1970s Ol'shanskii constructed a non-cyclic finitely generated group $G$ with the following properties: Every proper, non-trivial subgroup of $G$ is infinite cyclic. If $X^m=Y^n$ for $X, Y\in G$ with $m,n\neq0$, then $\langle X, Y\rangle$ is cyclic i.e., any two maximal subgroups of $G$ have trivial intersection. Ol'shanskii gave an easy proof that such a group is simple, which roughly goes: Suppose $N$ is a proper, non-trivial normal subgroup of $G$. If $N$ is maximal then $G/N$ is cyclic of prime order, so $G$ is virtually-$N$, so $G$ is a torsion-free virtually-$\mathbb{Z}$ group, so must itself be cyclic. If $N$ is not maximal then $N$ is contained in a maximal subgroup $M$ such that $M^g\cap M\neq1$ for all $g\in G$, so as $M^g$ is also maximal and as maximal subgroups intersect trivially (by (2)) we have that $M^g=M$ for all $g\in G$, i.e. $M$ is normal in $G$, which is impossible by the previous case. Property (2) was used here. I was wondering if this can be dropped. So: Question. Suppose $G$ is a non-cyclic finitely generated group with every proper, non-trivial subgroup of $G$ infinite cyclic. Is $G$ simple? If $G$ instead satisfies that it is infinite and every proper, non-trivial subgroup has order $p$ for a fixed prime $p$ then $G$ is a "Tarski monster" group and is indeed simple: If $N$ is a proper, non-trivial normal subgroup of $G$ and $g\not\in N$ then $N\cap\langle g\rangle=1$, as both subgroups have prime order, so $N\langle g\rangle=N\rtimes\langle g\rangle$ has order $p^2$, a contradiction. However, this proof uses primality so does not extend to the setting here. REPLY [7 votes]: The answer is no, there exist non-simple torsion-free Tarski monsters. Theorem 31.4 of Ol’shanskii's book "Geometry of defining relations in groups" ZBL0676.20014 MR1191619 is: Theorem. There is a non-abelian group all of whose proper subgroups are infinite cyclic and the intersection of any two of them is non-trivial. Such a group necessarily has infinite cyclic centre: take any non-commuting $x$ and $y$, then the centralizer of $\langle x \rangle \cap \langle y \rangle \cong \mathbb{Z}$ contains $\langle x, y \rangle$, which is the whole group. Thanks to Ashot Minasyan for pointing out Ol’shanskii's theorem.<|endoftext|> TITLE: Tensor of finite-dimensional algebra over perfect field is semisimple QUESTION [5 upvotes]: Let $K$ be a field and let $\Lambda_{1}$ and $\Lambda_{2}$ be two finite-dimensional $K$-algebras with Jacobson radicals $J_{1}$ and $J_{2}$ respectively. How to show or where can I find the proof of the following statement? $\Lambda_{1} / J_{1} \otimes_{K} \Lambda_{2} / J_{2}$ is always semisimple if $K$ is perfect or if $\Lambda_{1}$ and $\Lambda_{2}$ are path algebras of quivers factored by admissible ideals. Thank you. REPLY [3 votes]: Here is an alternate version of @Mare's answer. First recall that a $K$-algebra $A$ is separable if it is semisimple under all base extensions; its enough to check over an algebraic closure of $K$. Let us write $L\otimes_K A$ as $A^L$ for a $K$-algebra $A$ and field extension $L/K$. Any semisimple algebra over a perfect field is separable. By Wedderburn-Artin, it suffices to show that if $D$ is a finite dimensional division algebra over $K$, then $D^{\overline K}=\overline{K}\otimes_K D$ is semisimple, where $\overline{K}$ is an algebraic closure of $K$. Let $F$ be the center of $D$; then $F/K$ is a finite field extension and hence separable because $K$ is perfect. Then $D^{\overline K}\cong (\overline K\otimes_K F)\otimes_F D$. But since $F/K$ is separable, basic field theory says $\overline K\otimes_K F\cong \overline K^{[F:K]}$. Therefore, $D^{\overline K}\cong (\overline K\otimes_F D)^{[F:K]}$. But $D$ is central simple over $F$ and so by a basic result in the theory of central simple algebras, $\overline K\otimes_F D\cong M_n(\overline K)$ where $n^2=[D:F]$. Thus $D^{\overline K}$ is semisimple. Also, if $A$ is a split semisimple $K$-algebra (so isomorphic to a direct product of matrix algebras over a field), then $A$ is separable. In particular, if $\Lambda = KQ/I$ where $Q$ is a quiver and $I$ is an admissible ideal, then $\Lambda/J(\Lambda)\cong K^{|Q_0|}$ and hence is a separable $K$-algebra. Thus your question really boils down to proving that if $A$ and $B$ are separable $K$-algebras, then so is $A\otimes_K B$. It is enough to show that $(A\otimes_K B)^{\overline K}$ is semisimple where $\overline K$ is an algebraic closure. But $(A\otimes_K B)^{\overline K}\cong A^{\overline K}\otimes_{\overline K}B^{\overline K}$ as they both have the same universal property. The right hand side is a tensor product of direct sums of matrix algebras over $\overline{K}$ and hence is a direct sum of matrix algebras over $\overline{K}$ (as $M_n(\overline K)\otimes_{\overline K} M_m(\overline {K})\cong M_{mn}(\overline K)$) and thus semisimple. Therefore, $A\otimes_K B$ is separable and hence semisimple.<|endoftext|> TITLE: Defining rational numbers without using quotients or 0-truncations QUESTION [6 upvotes]: Most definitions of the rational numbers as a higher inductive type in univalent homotopy type theory (such as those in the cubical Agda library for example) require either the use of a quotient set or a 0-truncation constructor. Is there a way to define the rational numbers as a higher inductive type without using either quotient sets or 0-truncation? REPLY [12 votes]: One version of the theory of continued fractions is as follows. We can define operations $S,T,J\colon\mathbb{Q}^+\to\mathbb{Q}^+$ by $S(x)=x+1$ and $J(x)=1/x$ and $T(x)=JSJ(x)=x/(x+1)$, then we can define $M$ to be the free monoid generated by $S$ and $T$. We then have an evaluation map $M\to\mathbb{Q}^+$ given by $m\mapsto m(1)$, and it is not hard to check that this is a bijection. If we wanted, we could turn this around and essentially define $\mathbb{Q}^+$ to be the same as $M$ (and $\mathbb{Q}$ to be $\{0\}\amalg\mathbb{Q}^+\amalg -\mathbb{Q}^+$). This makes $\mathbb{Q}$ into an inductive type. The ordering on $\mathbb{Q}$ and the inclusion $\mathbb{Z}\to\mathbb{Q}$ work nicely in this picture, but the algebraic operations are awkward. You can also introduce the type $R$ of $2\times 2$ matrices $\begin{pmatrix} a&b\\c&d\end{pmatrix}$ with $a,c,d\in\mathbb{Z}^+$ and $b\in\mathbb{N}$ and $ad=bc+1$. Given a positive rational $q\in\mathbb{Q}^+$ we can write it as $q=a/c$ with $\text{gcd}(a,c)=1$. This means that there exist $b,d\in\mathbb{Z}$ with $ad=bc+1$. We can change the pair $(b,d)$ by adding multiples of $(a,c)$, so there is choice with $0 TITLE: Computing Massey products via intersection theory QUESTION [6 upvotes]: Let $K$ be an $n$-manifold with boundary and let $x,y,z \in H^*(K)$ be cohomology classes with $x\cup y=y\cup z=0$. The Massey product $\langle x,y,z \rangle$ is defined as the set of cohomology classes $[a\cup \tilde z - (-1)^{\deg (x)}\tilde x \cup b]$, where $a,b$ are cochains with $\partial a=\tilde x\cup\tilde y$, $ \partial b=\tilde y \cup\tilde z$ and $\tilde x,\tilde y, \tilde z$ are representatives of the cohomology classes $x,y,z$. In his paper "higher order linking numbers" Massey uses another approach to compute elements of Massey products using the Poincaré-duality of cohomology+cup-products and homology+intersection of manifolds: Let the fundamental classes $[M],[N],[P] \in H_*(K,\partial K)$ be the dual classes of the cohomology classes $x,y,z$, such that $X\cap P$ and $M \cap Y$ intersect transversally, for $X,Y$ manifolds with $M⫛ N= \partial X$ and $N ⫛ P= \partial Y$. Then the sum $X \cap P - (-1)^{n-\deg(x)}M\cap Y$ obviously represents the Poincaré-dual of a triple product. How exactly does this obvious duality work? In my attempt I didnt get very far... Using Bredon's intersection theory I get $[M \cap N]= D(x \cup y)$, for $D:H^*(K,\partial K) \to H_{n-*}(K)$ the duality isomorphism. This implies, that the cap product of a representative of $[K]$ and $\tilde x \cup \tilde y$ will be send to a representative of $[M \cap N]$, which we call $r_{mn}$. Since $x\cup y = 0$ it follows that $[M\cap N]=0$ and therefore $r_{m,n}$ is the boundary of a non-cyclic chain $r$. Since this chain is non-cyclic, it is not a representative of any homology class. Thus I dont see any way to continue, using the duality $[M \cap N]= D(x \cup y)$. REPLY [7 votes]: While not quite an answer from the perspective you're thinking of, one approach to this (at least if you're in a smooth manifold), would be to use a model of cohomology in which cocycles are represented by maps from other smooth manifolds. One such model is "geometric cohomology," which uses manifolds with corners. In this case, for a closed oriented manifold, homology and cohomology are tautologically the same and the cup product is the intersection product (at least for transverse things). Geometric cohomology was introduced by Lipyanskiy in https://arxiv.org/abs/1409.1121 and is being further developed by me, Medina, and Sinha. We have one preprint recently posted that included an outline of the main results at https://arxiv.org/abs/2106.05986. More details will be forthcoming in a more foundational paper.<|endoftext|> TITLE: Bounds on the expectation of $|X-Y|$ for $X,Y$ Poisson QUESTION [9 upvotes]: I would have a proof of the following fact; but it's a bit clunky, and am wondering if one can get a more elegant one (and/or improve the constants). I couldn't find this anywhere, and searching properties of the Skellam distribution didn't help much either. Let $X\sim\operatorname{Poi}(\lambda)$ and $Y\sim\operatorname{Poi}(\mu)$ be two independent random variables. Then $$ \max(|\lambda-\mu|, \frac{1}{40}\min(\sqrt{\lambda+\mu}, \lambda+\mu)) \leq \mathbb E[|X-Y|] \leq\min(|\lambda-\mu|+\sqrt{\lambda+\mu}, \lambda+\mu) $$ Here is the current proof I came up with: The upper bound follows from Cauchy–Schwarz, as $\mathbb E[(X-Y)^2] = (\lambda-\mu)^2 + \lambda+\mu$; and by the triangle inequality, as $\mathbb{E}[|X-Y|]\leq \mathbb{E}[X+Y]$. The lower bound $|\lambda-\mu|$ is a direct consequence of Jensen's inequality; we now proceed to establish the second term. Consider first the case $\lambda+\mu\geq 1$. We will use Paley–Zygmund, noting that for any $c\in(0,1)$ we have $$ \mathbb{E}[|X-Y|] \geq c\sqrt{\mu+\lambda}\cdot\mathbb{P}\{|X-Y|\geq c\sqrt{\mu+\lambda} \} $$ and $$ \begin{align*} \mathbb{P}\{|X-Y|&\geq c\sqrt{\mu+\lambda}\} \\ &= \mathbb{P}\{(X-Y)^2\geq c^2(\mu+\lambda)\} \\ &\geq \mathbb{P}\{(X-Y)^2\geq c^2\mathbb{E}[(X-Y)^2]\}\\ &\geq (1-c^2)^2\frac{\mathbb{E}[(X-Y)^2]^2}{\mathbb{E}[(X-Y)^4]}\\ &= \frac{ (1-c^2)^2(\lambda+\mu+(\lambda-\mu)^2)^2}{\lambda+\mu+6(\lambda^2+\mu^2)+6(\lambda-\mu)^2(\lambda+\mu)+(\lambda-\mu)^2+(\lambda-\mu)^4}\\ &\overset{(\dagger)}{\geq} (1-c^2)^2 \frac{(\lambda+\mu)^2+(\lambda-\mu)^4}{(\lambda+\mu)+10(\lambda+\mu)^2+3(\lambda-\mu)^4}\\ &\overset{(\ddagger)}{\geq} \frac{(1-c^2)^2}{11} \end{align*} $$ where $(\dagger)$ is using the AM-GM inequality and $(a-b)^2\leq (a+b)^2$; and $(\ddagger)$ relies on $\lambda+\mu \leq (\lambda+\mu)^2$, which holds since $\lambda+\mu\geq 1$. Taking $c=1/\sqrt{5}$ concludes the proof of this case, as then $\frac{c(1-c^2)^2}{11} > \frac{1}{40}$. If $\lambda+\mu \leq 1$, we will use properties of the modified Bessel function of the first kind $I_0$ (namely, that it is nondecreasing on $[0,\infty)$, with $e^{-x}I_0(x) \geq 1-\frac{x}{2}$ for $x\in[0,1]$) to conclude, as $$ \begin{align} \mathbb{E}[|X-Y|] &\geq 1 - \mathbb{P}\{|X-Y|=0\} \\ &= 1- e^{-(\lambda+\mu)} I_0(2\sqrt{\lambda\mu}) \\ &\geq 1- e^{-(\lambda+\mu)} I_0(\lambda+\mu) \geq \frac{\lambda+\mu}{2} \end{align}$$ where the first inequality is the AM-GM inequality. REPLY [9 votes]: $\newcommand{\la}{\lambda}$Let us show that \begin{equation*} E|Z|\ge J(1)\min[c,\sqrt c\,], \tag{1} \end{equation*} where \begin{equation*} Z:=X-Y,\quad c:=\la+\mu, \end{equation*} \begin{equation*} J(x):=\frac2\pi\,\int_0^\infty\frac{1-\exp\{-x\, (1-\cos t)\}}{t^2}\,dt. \end{equation*} Mathematica's command NIntegrate[] produces \begin{equation*} J(1)=0.67390\dots. \end{equation*} So, the constant factor $J(1)$ is an about 27-times improvement of the corresponding coefficient $\frac1{40}$ in the OP. Moreover, the constant factor $J(1)$ is the best possible. Inequality (1) is based on the Zolotarev identity \begin{equation*} E|Z|=\frac2\pi\,\int_0^\infty\frac{1-\Re Ee^{itZ}}{t^2}\,dt; \end{equation*} see e.g. formula (3.26) in this paper or formula (48) in the corresponding arXiv preprint. In our case, \begin{equation*} \Re Ee^{itZ}=\exp\{-c(1-\cos t)\}\cos((\la-\mu)\sin t)\le\exp\{-c(1-\cos t)\}; \end{equation*} Moreover, the latter inequality will turn into the equality when $\la=\mu$. So, \begin{equation*} E|Z|\ge J(c), \tag{2} \end{equation*} and the latter inequality will turn into the equality when $\la=\mu$. Note that, for each real $z>0$, \begin{equation*} \frac{1-e^{-zx}}x=\int_0^z e^{-xu}\,du \end{equation*} is decreasing in $x$. So, $J(x)/x$ is decreasing in real $x>0$. Moreover, as shown in this answer, $J(x)/\sqrt x$ is increasing in real $x>0$. So, $J(c)\ge cJ(1)$ if $c\in(0,1]$ and $J(c)\ge\sqrt c\,J(1)$ if $c\in[1,\infty)$. Thus, (1) follows from (2). Moreover, (1) turns into the equality when $\la=\mu=1/2$ (so that $c=1$).<|endoftext|> TITLE: Must a continuous $\varphi:\mathbb R^n\to\mathbb R^n$ with $\mathbb Q^n \subseteq \varphi[\mathbb Q^n]$ be surjective? QUESTION [8 upvotes]: Let $\varphi:\mathbb R^n \to \mathbb R^n$ be just some continuous function. If the image of $\varphi$ happens to contain $\mathbb Q^n$, does it follow that in fact all of $\mathbb R^n$ is contained in the image as well? No, it does not. For instance, the map given by $$(x,y)\overset{\varphi}{\longmapsto} (xy-1+\pi, x^2(xy-1)+y)$$ has as its image every point in $\mathbb R^2$ except for $(\pi,0)$. But now suppose instead that $\mathbb Q^n$ is actually contained in the image of just $\mathbb Q^n$. That is, instead of requiring only that $$\mathbb Q^n \subseteq \varphi[\mathbb R^n]\phantom{.}$$ we require that in fact $$\mathbb Q^n \subseteq \varphi[\mathbb Q^n].$$ Question. Does it then follow that $\varphi[\mathbb R^n]=\mathbb R^n$, or is there a counterexample? REPLY [13 votes]: A counterexample for $n=2$ is the map $\varphi(x,y) = (x,(x^2-2)y)$. Each point $(r,s)\in\mathbb{Q}^2$ is the image of $\left(r,\frac{s}{r^2-2}\right)\in\mathbb{Q}^2$, but e.g. $(\sqrt{2},1)\not\in\varphi(\mathbb{R}^2)$.<|endoftext|> TITLE: Formula for the anomalies of spin Chern-Simons theories? QUESTION [8 upvotes]: $\newcommand{\SH}{\mathit{SH}}\newcommand{\Z}{\mathbb Z}$Let $G$ be a compact Lie group and $\lambda\in H^4(BG;\Z)$. The data $(G, \lambda)$ determine a 3d topological field theory called Chern-Simons theory, except not quite: there is an obstruction to defining it on general closed, oriented $3$-manifolds called the anomaly. Freed-Teleman characterize anomalies of $n$-dimensional field theories as $(n+1)$-dimensional invertible field theories, which have been classified. I think in this case the anomaly field theory should be unitary, so is classified up to isomorphism by a torsion element of $\mathrm{Hom}(\Omega_4^{\mathrm{SO}}, \mathbb C^\times)$, by a theorem of Freed-Hopkins. If I choose $G$ and $\lambda$, is the isomorphism type of the anomaly field theory, as a bordism invariant, known? I'm actually interested in a slightly more general story, where $\lambda\in\SH^4(BG)$. (Here $\SH$ is a generalized cohomology theory called supercohomology: $\pi_0\SH = \Z$, $\pi_1\SH = \Z/2$, and the $k$-invariant is nonzero. When $G$ is simple and simply connected, using $\SH$ instead of $H$ amounts to choosing a half-integer rather than an integer.) Then there is a 3d spin TFT called spin Chern-Simons theory, which is again anomalous. Now the anomaly is a torsion element of $\mathrm{Hom}(\Omega_4^{\mathrm{Spin}}, \mathbb C^\times)$. For spin Chern-Simons theories, is the isomorphism type of the anomaly known? If not, is there an explicit conjectured description? I'm primarily interested in the spin case when $G$ is a torus, but any information (e.g. $G$ simple and simply connected, only for the oriented case, etc.) is helpful. REPLY [6 votes]: This is not a direct answer to your question, but I think it's relevant. One way of thinking about the anomaly for ordinary (oriented) Chern-Simons theories is that it's the evaluation of the associated 3+1-dimension Crane-Yetter theory (what Freed would call the anomaly theory) on a generator of 4d oriented bordism, i.e. $CP^2$. Choosing the standard handle decomposition of $CP^2$, one calculates that $$ Z_{3+1}(CP^2) = D^{-1/2} \sum_a \theta_a d_a^2 =: C $$ (The sum is over simple objects $a$ of the modular category associated to the Chern-Simons theory, $D$ is the global dimension of that category, $\theta_a$ is the ribbon twist of $a$, and $d_a$ is the quantum dimension (loop value).) The bordism invariant in this case is $C^{\sigma(W)}$, where $W$ is a closed 4-manifold and $\sigma$ is its signature. Imitating this approach for 2+1-dimensional Spin Chern-Simons TQFTs we can again evaluate the associated 3+1-dimensional Spin Crane-Yetter theory on a generator of 4-dimensional Spin bordism, such as the K3 surface. So $$ C := Z_{3+1}(K3) = \;\; ??? $$ The reason I wrote ??? above rather than an explicit formula is that the simplest handle decomposition of the K3 surface (which is in some sense the simplest generator of 4d Spin bordism) has 22 2-handles. Translating that complicated handle structure into a formula involving quantum dimensions, ribbon twists, $6j$-symbols, etc. would not be very enlightening. One way around this difficulty would be to enlarge the class of manifolds (on which the TQFT is defined) from Spin manifolds to "characteristic pairs" -- pairs $W^k \supset V^{k-2}$ such that $V$ is a Poincare dual to the second Stiefel-Whitney class of $W$ and $W\setminus V$ is equipped with a spin structure which does not extend over $V$. (See this paper by Kirby and Taylor for details.) The generators of the 4-dimensional characteristic pair bordism group are much simpler: $(CP^2,CP^1)$ and $(S^4,RP^2)$. The central charge associated to $(S^4,RP^2)$ can be normalized away. The central charge associated to $(CP^2,CP^1)$ looks very similar to the oriented case: $$ C:= Z_{3+1}(CP^2, CP^1) = D^{-1/2} \sum_v \frac{\theta_v d_v^2}{|\mbox{End}(v)|} $$ Here the sum is over only "vortex" simple objects (associated to the pair $(D^2,pt)$). $|\mbox{End}(v)|$ denotes the dimension of the endomorphism algebra of $v$ -- 2 for Majorana simple objects and 1 for ordinary simple objects. Can we always enlarge the domain of definition of a Spin TQFT associated to a fermionic modular category to characteristic/vortex pairs? I think that a recent paper of Johnson-Freyd and Reutter implies the answer is "yes", and Corey Jones, David Aasen and I think we have a more constructive proof (which we currently are in the process of writing up). More details on the above approach can be found in these two talks, and slides are here.<|endoftext|> TITLE: Combinatorial inequality involving alternating signs QUESTION [15 upvotes]: I would like to prove the following inequality. It arises from my study of random matrices. I have verified the inequality for $q\in \{0.01,0.02, \ldots, 0.99\}$ and $1\le n\le 100$. Let $n$ be any positive integer and $0\le q\le 1$. Then the following inequality is true. $$\sum_{k=0}^n(-1)^k\binom{n}{k}(q^k-q^n)^n\le (1-q^n)^n.$$ I suspect that the Wilf-Zeilberger method applies here. But I can't get it to work. REPLY [21 votes]: Mark each box of an $n\times n$ table with probability $q$. By inclusion-exclusion the difference RHS-LHS equals to the probability that there exists a full row (with all boxes marked) but there does not exist a full column: that's because for given $k$ rows the probability that (they are full but no column is full) equals $(q^k-q^n)^n$.<|endoftext|> TITLE: The justifiable universe QUESTION [7 upvotes]: Over the years of my study of set theory, I have encountered several sentences of the form V = X: V = L, V = HOD, V = WF (the exclusive assertion of the cumulative hierarchy), and (if I understand this paper) V = HW ("hereditarily winning"). Now, in general epistemology, there is a problem of the "regress of reasons," with three formal solutions: foundationalism, coherentism, and infinitism. (There is also an "empty solution," skepticism, which corresponds to J0 in justification logic.) It occurred to me that foundationalism seems to structurally correspond to well-founded sets; coherentism seems to correspond to looping set structures like Quine atoms or, "x ∈ y, y ∈ z, z ∈ x"; and infinitism to infinite descending ∈-chains. For the purposes of this post, then, assume that ∃-sentences regarding well-founded sets are justifiable in a foundationalist way, that ∃-sentences regarding looping sets are justifiable in a coherentist way, and that ∃-sentences regarding infinite descending ∈-chains are justifiable in an infinitist way. Implication: axioms of antifoundation are not justified; axioms are expressions of well-founded, not antifounded, justification. This is not to say that nonwell-founded set theories are unjustified altogether, only that their epistemological origins are of a different form than that of the cumulative hierarchy. V = X assertions can be understood as assertions about which ∃-sentences are true for V. For example, V = L can be interpreted as, "For all sets x, if x exists, then x is constructible." Let J be the class of all sets whose ∃-sentences are sufficiently justifiable. Is it intelligible, then, to propose something like V = J? In other words, "For all sets x, if x exists, then the sentence asserting the existence of x is sufficiently justifiable." If we allow all three of the non-empty solutions to the justificatory regress problem, then, we can have V include well-founded and nonwell-founded sets, while preserving the focus (in the context of axiomatic set theory) on WF-sets. Next, suppose a theory of justification values, along the lines of the theory of truth values. That is, let us have justification not be a predicate of sentences, but their reference. Let S stand for some or another sentence and have (S) be a function that takes sentences as inputs and outputs a number corresponding to the degree to which S is justified. Let 0 be the justification value for completely unjustified S and 1 be the justification value for completely justified S. Now, relax the symmetry with 0 and 1 as truth values and allow that sometimes (S) < 0 (antijustification) and sometimes (S) > 1 (hyperjustification). Peter Koellner says: ... when one restricts to [axiomatic] theories that “arise in nature” (that is, the kind of formal systems that one encounters in a mathematical text as opposed to examples of the type above, which are metamathematically manufactured by logicians to have the deviant properties in question) the interpretability ordering is quite simple: There are no descending chains and there are no incomparable elements. In other words, it is a well-ordering. He also says: We shall organize the statements in each interpretability degree by the relation of being more evident than and we shall speak of the collection of all statements so ordered as the evidentness order... In seeking axioms for theories within a given interpretability degree we seek axioms that are as low as possible in the evidentness order. Ultimately one would like to find axioms that are minimal in the evidentness order. With all these ideas in place, then... Two lines of reasoning in the V = J context A defense of axioms of infinity There isn't really an argument, here, so much as a novel formulation of the initial axiom of infinity, with an eye towards an account that justifies ∃-sentences concerning arbitarily larger cardinals. Viz., let us have the initial axiom of infinity be stated as ∃S((S) = ω), with the S in question being the very assertion of the existence of ω itself. This is rather impredicative, granted; but anyway, besides seeming to me to be a rather "superstitious" point of view, finitism lacks the resources to speak of infinite antijustification, after all. (The surreal number -ω (more on why we are using surreal numbers, here, below) cannot be assigned by the finitist, to any S.) In a slogan, "Having infinite numbers makes it possible to infinitely justify some S," which is far as I can go in otherwise justifying the initial axiom of infinity. But note that this scheme of things stands on its head the question of justifying the larger cardinal axioms: since these make ever-higher degrees of infinite hyperjustification possible, they are more and more infinitely justifiable. So far, the only filter I've come up with on these axioms is an exclusion of the choiceless sets: this paper by Hamkins, et. al. has it that the Reinhardt, etc. sets can be formed by varying the axioms of foundation and antifoundation, so since in the occurrent system there are no justifiable axioms of antifoundation, we relegate the choiceless sets to the antifounded section of J. Assigning a justification value to CH Let forcing be an S-justifier, i.e. if it is possible to force some S to hold, then S is justified to some extent. I won't go into the details right now, but in the occurrent system, I came up with a modal argument for the antecedent in Shelah's result that (2ℵn < ℵω) → (2ℵω < ℵω4). So there is an S-sequence 2ℵ0 = ℵ1, 2ℵ0 = ℵ2, etc. and since each can be forced to hold, each is justified to some extent. But modulo the modal argument and for reasons of cofinality, it is also true that 2ℵ0 ≠ ℵω. Let (2ℵ0 = ℵω) = 0, then. Using surreal numbers, we can rephrase this as (2ℵ0 = ℵω) = 0/ω. Now since any number minus itself equals zero, we can further refine this as (2ℵ0 = ℵω) = (ω - ω)/ω. Let this be the limit sentence in the relevant S-sequence. Abstracting over the form of the limit, then, seems to give us that (2ℵ0 = ℵn) = (ω - n)/ω, at least if n does not equal zero. Presto, we have that (2ℵ0 = ℵ1) = (ω - 1)/ω, or in other words, CH is only infinitesimally unjustified. This is about as close to a derivation of CH as we can get from this system as such, though. We also have that other Continuum hypotheses are only infinitesimally unjustified; at best, CH is minimal in the relevant order of justification. Note that there couldn't be a maximally justified maximum forcing. Now more or less by definition, the smallest strongly inaccessible cardinal, which we will here denote by I (with α as its initial ordinal), is not equal to the cardinality of the Continuum. So we might say that (2ℵ0 = I) = 0 = (α - α)/α. Then, waiving the modal argument for the antecedent in Shelah's result, we can recapitulate the relevant order of justification such that (2ℵ0 = ℵ1) = (α - 1)/α, etc. Granted, we have S in the α-sequence that interrupt the increase towards α; depending on your aesthetic sense of such "interruptions," you might be motivated to reject the α-picture, then, here. Again, I do have an argument in the system for restricting the powersets of the ℵn overall, so for what it's worth... Erotetic logic and forcing The logical background of the above is supposed to be some form of erotetic logic. Hamkins and Loewe have given us an understanding of forcing in terms of modal logic; I wonder if it is possible to interpret forcing in erotetic terms? The only progress I've made on this front is by representing forcing as a sort of disjunction elimination: with respect to the Continuum's cardinality, for example, we have an infinite disjunctive question, where each disjunct is a possible answer to the question, and we can fiat-eliminate the disjunction by forcing an answer. Then (here's a gap in the deduction), we say something like, "The larger the forcing, the less the abstract justification value," but how to frame this in terms of erotetic justification? (The template of an erotetic justifier is describable in the system, but I won't go into the details for now.) Is V = J a viable option when it comes to the V = X question? I feel like we could apply it to the issue of extrinsic vs. intrinsic justifiers. It also seems relevant to the multiverse issue. REPLY [10 votes]: I find the question interesting. But I believe that your division of set theories into justificatory types is undermined by the fact that we have instances of bi-interpretable theories that cross the type boundaries. For example, ZFC set theory is bi-interpretable with the antifoundational theory ZFC-foundation+AFA. Basically, inside ZFC, one can model anti-foundational set theory by considering the class of pointed extensional digraphs (up to isomorphism), and then proving that AFA holds for this interpretation. Conversely, in ZFC-foundation+AFA, one can consider the well-founded part of the model, which will satisfy ZFC. This is a bi-interpretation, because the AFA model can observe how it itself arises as the interpretation by means of extensional digraphs built using only vertices and edge relations in the well-founded part of the model. The usual attitude we have toward bi-interpretable theories is that they have essentially the same content. And therefore I would find them also to be justified to the same extent by any form of justification. The bi-interpretation extends to all extensions of these theories, such as ZFC + large cardinals, which will be bi-interpretable with a corresponding AFA theory. Meanwhile, your taxonomy would seem to place the ZFC set theories in the foundationalist justificatory realm and the anti-foundational theories in the coherentist realm. So it seems that we should be equally able to support either theory with either form of justification, and I take this to undermine the question. After all, if we can provide a foundationalist justification of theory T, which is bi-interpretable with theory $T'$, then this same justification also seems to provide justification for $T'$, even when you would seek only coherentist justifications for $T'$.<|endoftext|> TITLE: What's the number of solutions of the quadratic equation $x_1^2+\dots+x_m^2=0$ over finite ring $\mathbb{Z}/p^n$? QUESTION [10 upvotes]: I want to calculate the number of solutions to the quadratic equation $$x_1^2+\dots+x_m^2=0$$ where $m$ is odd (a given number) and $x_i\in\mathbb{Z}/p^n$ for a given prime number $p$ and a given positive integer $n$. I guess one can consider the projective variety over the $p$-adic field $\mathbb{Q}_p$ and count the point of some kind of projection but I didn't make it. REPLY [5 votes]: For what it's worth, if you want to investigate this more generally, let $f(x_1,\ldots,x_m)\in\mathbb Z[x_1,\ldots,x_m]$ be a homogeneous form of degree $d$, and let $$ N_n(f) = \Bigl(\text{# of solutions of } f(x_1,\ldots,x_m)\equiv0\pmod{p^n}\Bigr). $$ Then the Igusa zeta function of $f$ is (more-or-less) the generating function $$ Z(f,t) = \sum_{n=1}^\infty N_n(f) (t/p^m)^n . $$ A deep theorem of Igusa, with a subsequent alternative proof by Denef, say that $Z(f,t)$ is a rational function, but exact formulas are known in only a small number of cases. However, one case where a lot is known is "Fermat-type equations," i.e., diagonal equations, of which yours is the first interesting example.<|endoftext|> TITLE: Product-one sets in non-commutative groups QUESTION [7 upvotes]: A nonempty subset $D$ of a group $G$ is called $\bullet$ decomposable if $D\subseteq DD$, that is every element $x\in D$ is can be written as the product $x=yz$ of some elements $y,z\in D$; $\bullet$ product-one if there exists $n\in\mathbb N$ and pairwise distinct elements $x_1,\dots,x_n\in D$ such that $x_1\cdots x_n=1$. Problem 1. Let $D$ be a finite decomposable subset of a group. Is $D$ product-one? Remark 1. For commutative groups this problem was posed by Gjergji Zaimi and solved affirmatively by Lev, Nagy, and Pach. Remarks 2. For some non-commutative groups like generalized dihedral groups the answer to Problem is also affirmative, see my partial answer below. This partial answer suggests the following Problem 2. Let $G$ be a group containing an Abelian subgroup of index 2. Is every finite decomposable set in $G$ product-one? REPLY [3 votes]: GAP shows that the groups SmallGroup(27,3), SmallGroup(27,4), SmallGroup(36,11), SmallGroup(39,1) SmallGroup(48,3) do contain many 5-element decomposable sets, which are not product-one. So, the lower bound 5 for the smallest cardinality of a counterexample, obtained by @YCor in his answer, is the best possible. Below I write down 5-element decomposable non-product-one sets found by GAP in the groups SmallGroup(27,3): [ f1, f2, f1 * f2, f1^2 * f2, f1 * f2^2 ] SmallGroup(27,4): [ f1, f2, f1 * f2 * f3, f1^2 * f2 * f3, f2^2 * f3^2 ] SmallGroup(36,11): [ f1, f2 * f3, f1^2 * f3, f1 * f2^2 * f3, f1^2 * f2^2 * f4 ] SmallGroup(39,1): [ f1, f2, f1 * f2, f1^2 * f2, f2^4 ] SmallGroup(48,3): [ f1, f2, f1 * f2, f2 * f3, f1^2 * f2 ] These 5 groups are the only groups of order $\le 50$ that contain decomposable non-product-one sets.<|endoftext|> TITLE: Which projective varieties cannot appear as an exceptional divisor of a blow-up of the projective space QUESTION [5 upvotes]: I am looking for examples of projective varieties (over $\mathbb{C}$) of dimension, say $n$ which cannot appear as an exceptional divisor of a blow-up of $\mathbb{P}^{n+1}$ along some closed subscheme. Any idea/reference will be most welcome. REPLY [6 votes]: The exceptional divisor is the $\operatorname{Proj}$ of the normal sheaf, hence it is covered by rational curves. This excludes all the non-uniruled varieties.<|endoftext|> TITLE: Improving the error term in a classic sieving problem QUESTION [11 upvotes]: I'm new to asking questions on MathOverflow, so forgive me if this question is not the kind of thing to be asked here. Let $q$ be a positive integer and let $N$ be an integer with $1 \leq N \leq q$. The estimate $$ \sum_{\substack{n= 1\\ (n,q)=1}}^N 1 = N \frac{\phi(q)}{q} + O(2^{\omega(q)}) $$ is a classical and straightforward application of Mobius inversion. The error term can be given explicitly by $$ \sum_{d\mid q} \mu(d) \left\{ \frac{N}{d}\right\}. $$ If $q$ has few distinct prime factors, then the bound $$ \Big|\sum_{d\mid q} \mu(d) \left\{ \frac{N}{d}\right\}\Big| \leq 2^{\omega(q)} $$ is sharp. For instance, if $q=p^k$ for some prime $p$, then $2^{\omega(q)} = O(1)$. Writing $N=Mp+r$, where $0\leq r < p$, gives $$ \sum_{d\mid q} \mu(d) \left\{ \frac{N}{d}\right\} = - \frac{r}{p}, $$ and this can genuinely be size $O(1)$. On the other hand, I suspect that the bound $2^{\omega(q)}$ is quite wasteful when $q$ has many distinct prime factors. From some numerical calculations ($q \leq 10^6$), I suspect that $$ \tag{1} \Big|\sum_{d\mid q} \mu(d) \left\{ \frac{N}{d}\right\}\Big| \ll \log\log q. $$ If, if for instance, $q$ is the product of the first $k$ primes, then $q \ll (k\log k)^k$, and so $\log\log q \ll \log k$, whereas $2^{\omega(q)} = 2^k$. Thus my actual question: Does the estimate (1) hold in general? It might be possible to establish this estimate using sieve methods (via upper and lower bounds on the original sum), but I am not familiar enough with sieve theory to pursue this avenue myself. I have not been able to find any results of this kind in the literature, so I would gladly welcome any ideas and/or references on this topic. REPLY [14 votes]: In fact there are moduli $q$ with arbitrarily many prime factors where the error term can be shown to be as large as $2^{\omega(q)-2}$. The following construction is due to D.H. Lehmer, The distribution of totatives. Let $q$ be the product of $k$ distinct prime numbers all of which are $ 3 \pmod 4$, and take $N=q/4$. Then $\frac 12 -\{ N/d\} = \pm 1/4$ depending on whether $q/d$ has an even or odd number of prime factors. Therefore $$ \Big| \sum_{d|q} \mu(d) \{N/d\} \Big| = \Big| \sum_{d|q} \mu(d) (1/2- \{N/d\})\Big| = \frac 14 \sum_{d|q} |\mu(d)| = \frac 14 2^{\omega(q)}. $$ One interesting application of this idea (which is how I know it) is Montgomery's work on the error term in the counting function of $\phi(n)$: see Fluctuations in the mean .... REPLY [3 votes]: Let $p_1,\dots,p_k$ be the first $k$ primes. Let $q = p_1\dots p_k$. By CRT there's some $m \ge 1$ so that $m+j \equiv 0 \pmod{p_j}$ for $1 \le j \le k$. Then $$\sum_{\substack{n=1 \\ (n,q)=1}}^m 1 = \sum_{\substack{n=1 \\ (n,q) = 1}}^{m+k} 1,$$ while $$\left|(m+k)\frac{\phi(q)}{q}-m\frac{\phi(q)}{q}\right| = k\frac{\phi(q)}{q}.$$ So the error term at $m$ or at $m+k$ is at least $\frac{1}{2}k\frac{\phi(q)}{q}$, which is $\frac{1}{2}k\prod_{j=1}^k (1-\frac{1}{p_j})$, which is at least $ck/\log k$.<|endoftext|> TITLE: Is there a countable discrete infinite group $G$ over which the group algebra $\mathbb{C} G$ is semisimple? QUESTION [6 upvotes]: I am seeking for an Artin $k$-algebra (especially for group algebra) which is infinite-dimensional over some field $k$. It's known that any complex group algebra has trivial Jacobson radical. So I have the following question: Is there a countable discrete infinite group $G$ over which the group algebra $\mathbb{C} G$ is semisimple? REPLY [8 votes]: The answer is negative as any semisimple Hopf algebra is finite-dimensional. More generally, the same conclusion is true for all Artinian Hopf algebras. See e.g. Liu and Zhang - Artinian Hopf algebras are finite dimensional.<|endoftext|> TITLE: Stable Adams operations QUESTION [6 upvotes]: I have come across a paper by Adams, Harris and Switzer on the Hopf algebra of cooperations of real and complex K-theory. The Adams operations are stable in the $p$-local setting, however I have not come across a nicely written introduction on the stable Adams operations in complex topological K-theory. Could someone kindly explain the spectrum level maps used in defining the stable Adams operations or perhaps redirect me to some literature on the same? REPLY [5 votes]: We can define complex K-theory spectrum $K$ looking at even suspension maps $\Sigma^2:S^2\wedge BU\to BU$. The map $\Sigma^2$ has an expression as a classifying map for vector bundle $E\otimes \tau$, where $\tau=(\mathcal{O}(1)-1)\in \tilde{K}^0(S^2)$ is a generator and $E$ is the universal bundle over $BU$. Adams operations $\Psi^k$ are easily defined by universality and the splitting principle, thus giving $\Psi^k:BU\to BU$. To define them on the level of spectrum $K\to K$ we would like to have a commuting square: $\begin{align*} &S^2\wedge BU &&\overset{\Sigma^2}{\to}&& BU\\ &\downarrow{id\wedge \Psi^k} && &&\downarrow{\Psi^k}&\\ &S^2\wedge BU&&\overset{\Sigma^2}{\to}&&BU& \end{align*}$ Which fails, since $\Psi^k\circ \Sigma^2$ classifiyes $\Psi^k(E\otimes\tau)=\Psi^k(E)\cdot k\tau$, meanwhile $\Sigma^2\circ (id\wedge \Psi^k)$ classifies $\Psi^k(E)\cdot \tau$. Hence, to obtain a commutative square we have to adjust maps between corresponding terms of spectrum, i.e. putting $\Psi^k/k^n:BU\to BU$, where $K_{2n}=BU$ is 2n-th term of the spectrum $K$. Of course this make sense only after localisation of spectra inverting $k$ and does not provides a stable operation otherwise. This construction is sketched in Adams book on stable homotopy and generalized homology, see $K_*(K)$ computation there.<|endoftext|> TITLE: General formulas for derivative of $f_n(x)=\dfrac{ax^n+bx^{n-1}+cx^{n-2}+\cdots}{a'x^n+b'x^{n-1}+c'x^{n-2}+\cdots},\quad a'\neq0$ QUESTION [5 upvotes]: For the function $f_1(x)=\dfrac{ax+b}{a'x+b'},\quad a'\neq0$ , we have $$f_1'(x)=\dfrac{\begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix}}{(a'x+b')^2}$$ For $f_2(x)=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'},\quad a'\neq0$, we have $$f_2'(x)=\dfrac{{ \begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix} }x^2+2{ \begin{vmatrix}{a} && {c} \\ {a'} && {c'}\end{vmatrix} }x+{ \begin{vmatrix}{b} && {c} \\ {b'} && {c'}\end{vmatrix} }}{(a'x^2+b'x+c')^2}$$ Can we generalize the formula containing determinants to find $f_n'(x)$? REPLY [6 votes]: You can easily extend this, but for $n\geq 3$ you will end up with more than one term per monomial: For two functions $f$, $g$ rewrite the quotient rule using a determinant $$\frac{d}{dx} \frac{f}{g} = \frac{\frac{df}{dx}g-f \frac{dg}{dx}}{g^2} = \frac{\begin{vmatrix} \frac{df}{dx} & f \\ \frac{dg}{dx} & g \end{vmatrix}}{g^2}$$ Now assume that $f(x) := a_nx^n+\dots + a_0$, $g(x) :=b_n x^n+ \dots + b_0$ are polyonomials. Then $\frac{df}{dx}$ and $\frac{dg}{dx}$ can be calculated explicitly and you can use the multilinearity of the determinant to split it by monomials: $$\frac{d}{dx} \frac{f}{g} = \frac{\begin{vmatrix} \sum_{k=0}^n a_k k x^{k-1} & \sum_{j=0}^n a_j x^j \\ \sum_{k=0}^n b_k k x^{k-1} & \sum_{j=0}^n b_j x^j \end{vmatrix}}{g^2} = \frac{\sum_{k=0}^n \sum_{j=0}^n k\begin{vmatrix} a_k & a_j \\ b_k & b_j \end{vmatrix} x^{k+j-1} }{g^2}$$ Now in the last sum for $k=j$ the determinant vanishes and if $k\neq j$ the same determinant occurs again with flipped sign if their roles are reversed. So you can only count the cases $j0$. Also, we can further simplify $c_j$ as $$c_j=\sum_{k=j+1-n}^{\lfloor{\frac{j+1}{2}}\rfloor} k\begin{vmatrix}{a_k} && {b_k} \\ {a_{j+1-k}} && {a_{j+1-k}}\end{vmatrix}+ (j+1-k)\begin{vmatrix}{a_{j+1-k}} && {b_{j+1-k}} \\ {a_k} && {b_k}\end{vmatrix}$$ Hence, $$c_j=\sum_{k=j+1-n}^{\lfloor{\frac{j+1}{2}}\rfloor} (j+1-2k)\begin{vmatrix}{a_{j+1-k}} && {b_{j+1-k}} \\ {a_{k}} && {b_{k}}\end{vmatrix}$$ This gives $c_{2n-1}=0$<|endoftext|> TITLE: Factoring a projection of surface groups through a free group QUESTION [7 upvotes]: Consider the surface group $S_g=\langle a_1,b_1,a_2,b_2,\dots,a_g,b_g \mid [a_1,b_1][a_2,b_2]\cdots[a_{g},b_{g}]=1\rangle$, which is the fundamental group of the closed orientable genus-$g$ surface. Suppose $2\leq mn$, then no epimorphism $S_n\to F_r$ can exist. Apparently, when $r\leq n$, then there is only one epimorphism $S_n\to F_r$ "up to automorphism." The representative is this: choose $F:S_n\to F_r$ to send $a_i$, $1\leq i\leq r$ to the free generators and trivialize all other generators. This is unique in the sense that for any other epimorphism $f:S_n\to F_r$, we must have $F=f\circ \sigma$ for some $\sigma\in Aut(S_n)$. Considering this classification and the structure of $p$, my feeling is that the question has a negative answer but that I am perhaps overlooking something simple. REPLY [11 votes]: The homomorphism $p$ is non-trivial on second (co)homology, since it is induced by a degree one map of surfaces ("collapse $n-m$ holes to a point"). This can also be seen using the structure of the cohomology rings of orientable surfaces. Hence no such factorisation exists, as free groups have zero (co)homology above dimension one.<|endoftext|> TITLE: Is there a generalization of Du Bois singularities? QUESTION [6 upvotes]: Let $\underline{\Omega}_X^{\bullet}$ denote the Deligne -- Du Bois complex of a normal variety $X$. What kind sigularities satisfy $gr^k\underline{\Omega}_X^{\bullet}[k] \simeq \Omega_X^{[k]}$ where $\Omega_X^{[k]} := j_*\Omega^k_{X^{\operatorname{reg}}}$ and $j\colon X^{\operatorname{reg}} \hookrightarrow X$ is the inclusion of the regular locus of $X$? ADDED LATER: Is there even a map like $\Omega_X^{[k]} \to gr^k\underline{\Omega}_X^{\bullet}[k]$ to begin with? Recall that the case $k=0$ is by definition the Du Bois singularities and in order to define it, one does not need to assume that $X$ is normal. Log canonical singularities (normal) are known to be Du Bois (https://arxiv.org/abs/0902.0648). EDIT: As Donu pointed out, when $X$ is smooth $h^{-k}\omega_X^{\bullet} = 0$ for $k\neq -\dim X$ but $gr^k\underline{\Omega}_X^{\bullet}[k] = \Omega_X^k$. So the following question does not make sense. One could perhaps ask more generally when is $gr^k\underline{\Omega}_X^{\bullet}[k] \simeq h^{-k}\omega_X^{\bullet}$ where $\omega_X^{\bullet}$ is the dualizing complex of $X$? REPLY [8 votes]: Good question. Your first question has a positive answer when $X$ has finite quotient singularities, i.e. locally analytically $\mathbb{C}^n/G$, with $G\subset GL_n(\mathbb{C})$ finite. This was essentially proved in Du Bois' original paper "Complexe de de Rham filtré...", see also Steenbrink "Mixed Hodge structure on the vanishing cohomology". I have a hunch (see added comment) this true when $G$ is reductive. There are some other classes as well, like toric singularities. I can say more later, or perhaps Sándor will. Added A better reference for examples would be the book by Navarro Aznar et. al. "Hyperrésolutions cubiques et descente cohomologique". Also I overlooked your second question: Think about the case when $X$ is smooth. It's almost never true. Perhaps you meant something else? 2nd edit I suppose I may as well make my old notes public. See conjecture 2.8 of that for a more precise statement of what I'm expecting.<|endoftext|> TITLE: Do relatively prime polynomials $f$ and $g$ in $k[x,y]$ generate an ideal of finite codimension? QUESTION [9 upvotes]: Let $k$ be a field and $f,g \in k[x,y]$ be two non-constant polynomials in two variables. Is it true that the following conditions are equivalent: $f$ and $g$ are relatively prime in $k[x,y]$, in the sense that they have no non-constant common divisors, so if we have the identity $\alpha f + \beta g = 0$ in $k[x,y]$ for some polynomials $\alpha,\beta\in k[x,y]$, then $\alpha = -g \gamma$ and $\beta = f \gamma$ for some $\gamma \in k[x,y]$. The ideal $(f,g)$ generated by $f,g$ is of finite codimension in $k[x,y]$ over $k$, i.e. $\dim_{k} k[x,y] / (f,g) < \infty$. The following simple arguments prove that 2) implies 1). Suppose $f = a q$ and $g = b q$ for some $a,b,q\in k[x,y]$ and $q$ is non-constant, so 1) fail. Exchanging if necessary $x$ and $y$, one can assume that $q(x,y) \not= x^k$ for all $k$. Then neither of the following infinite and linearly independent over $k$ family of polynomials $\{x^i\}_{i\geq0}$ belongs to $(q) \supset (f,g)$. Whence $\dim_{k} k[x,y] / (f,g) \geq \dim_{k} k[x,y] / (q) = \infty$, and thus 2) fails as well. Thus the question is whether 1) implies 2)? Perhaps one should assume that $k$ has characteristic $0$. REPLY [9 votes]: A standard undergraduate maths approach is via resultants. I am not going to survey resultants here (but see below), I'll just say that an immediate consequence of 1) is that there exists a nonzero monic $u:=Res_y(f,g)\in k[x]\cap (f,g)$ and a nonzero monic $v:=Res_x(f,g)\in k[y]\cap (f,g)$. Therefore, any monomial $x^s y^t$ in $h\in R:=k[x,y]/(f,g)$ satisfies $s<\deg u$, $t<\deg v$, as you can reduce $x$-degree below $\deg u$ by replacing $x^{\deg u}$ with $u(x)-x^{\deg u}$, and analogously for $y$-degree (using $v$). Hence $R$ is finite-dimensional. EDIT: Remarks, definitions and explanations. Degree $d$ polynomials form $d+1$-dimensional vectorspace $k_d[t]\subset k[t]$ over $k$. The resultant $Res_t(p,q)$ of $p\in k_d[t]$, $q\in k_e[t]$ is the determinant $\det M$ of the linear map $M:k_d[t]\times k_e[t]\to k_{d+e}[t]$ defined by $M(w,z)=wq+zp$ (here $w\in k_d[t]$, $z\in k_e[t]$) - I am cheating here a bit with dimensions, as $\dim (k_d[t]\times k_e[t])=\dim(k_{d+e}[t])+1$, but we don't want this extra 1, so we assume w.l.o.g. that $z$ is monic, i.e. $z(t)=t^e+z_{e-1}t^{e-1}+\dots +z_0$. Assuming that $p,q$ have a common root $t^*$ in the algebraic closure of $k$, we see that $M$ cannot be 1-1 in this case, as $M$ would be divisible by $t-t^*$ for any $w,z$, i.e. $\det M=Res_t(p,q)=0$. Usually $M$ is written using monomial bases of $k_d[t]$ and $k_e[t]$, as Sylvester_matrix. We see that it readily generalises to rings, e.g. if we replace $k$ with $k[x]$ then $\det M\in k[x]$, and we can think of $Res_y(f,g)\in k[y]$, vanishing at common roots of $f$ and $g$ in the algebraic closure of $k$.<|endoftext|> TITLE: Number of 1's in binary expansion of $a_n = \frac{2^{\varphi(3^n)}-1}{3^n}$ QUESTION [11 upvotes]: My question is about the Hamming Weight (or number of 1's in binary expansion) of $a_n = \frac{2^{\varphi(3^n)}-1}{3^n}$ A152007 For example, $a_3 = 9709 = (10110111101001)_2 $ has nine 1's in binary expansion I guess the answer is $3^{(n-1)}$ but I can't prove it Is that correct? REPLY [21 votes]: It is, indeed, correct. Notice first that $2-(-1)=3$ is divisible by $3$, so by lifting-the-exponent lemma the number $$ A=\frac{2^{3^{n-1}}-(-1)^{3^{n-1}}}{3^n}=\frac{2^{3^{n-1}}+1}{3^n} $$ is an integer. Notice also that for $n>0$ it has less than $3^{n-1}$ binary digits. Assume that it has $m$ binary digits. We have $$ \frac{2^{\varphi(3^{n})}-1}{3^n}=\frac{2^{2\cdot 3^{n-1}}-1}{3^n}=A(2^{3^{n-1}}-1). $$ Next, let $l=3^{n-1}-m$, $B=2^l-1$ and $C=2^m-A$. We claim that your number's binary expansion looks like a concatenation of $A-1$, $B$ and $C$, where we also add some zeros in expansion of $C$ until we get exactly $m$ digits. So, we should have $$ a_n=C+2^mB+2^{m+l}(A-1). $$ This equality is true, because $$ C+2^mB+2^{m+l}(A-1)=2^m-A+2^m(2^l-1)+2^{m+l}(A-1)= $$ $$ =(A-1)(2^{m+l}-1)+2^{m+l}-1=A(2^{m+l}-1)=A(2^{3^{n-1}}-1). $$ Now, if the sum of digits of $A-1$ is equal to $s$, then the sum of digits of $C$ is $m-s$ and the sum of digits of $B$ is, of course, $l$, so the sum of digits of $a_n$ is $$ s+m-s+l=m+l=3^{n-1}, $$ as needed. For the particular case $n=3$, described in the post, $A=10011_2=19=\frac{2^9+1}{27}$ and $C=2^5-A=13=01101_2$ (notice the added zero)<|endoftext|> TITLE: Can all three numbers $\ n\ \ n^2-1\ \ n^2+1\ $ be fine (as opposed to coarse)? QUESTION [5 upvotes]: Let $\ n\ $ be an arbitrary natural number ($\ 1\ 2\ \ldots).\ $ Then $\ n\ $ is coarse $\ \Leftarrow:\Rightarrow\ $ there exists a prime divisor $p$ of $\ n\ $ such that $\ p^3>n.$; $\ n\ $ is a p-cube $\ \Leftarrow:\Rightarrow\ $ the positive cubical root of $\ n\ $ is a prime number; $\ n\ $ is fine $\ \Leftarrow:\Rightarrow\ p^3 TITLE: Proof that block matrix has determinant $1$ QUESTION [15 upvotes]: The following real $2 \times 2$ matrix has determinant $1$: $$\begin{pmatrix} \sqrt{1+a^2} & a \\ a & \sqrt{1+a^2} \end{pmatrix}$$ The natural generalisation of this to a real $2 \times 2$ block matrix would appear to be the following, where $A$ is an $n \times m$ matrix: $$\begin{pmatrix} \sqrt{I_n+AA^T} & A \\ A^T & \sqrt{I_m+A^TA} \end{pmatrix}$$ Both $I_n+AA^T$ and $I_m+A^TA$ are positive-definite so the positive-definite square roots are well-defined and unique. Numerically, the determinant of the above matrix appears to be $1$, for any $A$, but I am struggling to find a proof. Using the Schur complement, it would suffice to prove the following (which almost looks like a commutativity relation): $$A\sqrt{I_m + A^TA} = \sqrt{I_n + AA^T}A$$ Clearly, $A(I_m + A^TA) = (I_n + AA^T)A$. But I'm not sure how to generalise this to the square root. How can we prove the above? REPLY [24 votes]: Write the SVD of $A$, say $A=PDQ^T$ with $D$ diagonal with non-negative entries and $P\in O(n),Q\in O(m)$. Then $\sqrt{I_n + AA^T} = P\sqrt{1+D^2}P^T$ and $\sqrt{I_m+ A^TA} = Q\sqrt{1+D^2}Q^T$. This gives $$ \begin{pmatrix} \sqrt{I_n + AA^T} & A \\ A^T& \sqrt{I_m+A^TA} \end{pmatrix} = \begin{pmatrix} P & 0 \\ 0 & Q \end{pmatrix} \begin{pmatrix} \sqrt{I_n + D^2} & D \\ D & \sqrt{I_m+D^2} \end{pmatrix} \begin{pmatrix} P^T & 0 \\ 0 & Q^T \end{pmatrix}. $$ Up to permutation, the matrix in the middle is diagonal by block with $n$ blocks given by 2x2 matrices of the same form as in the question. REPLY [21 votes]: We have $Af(A^TA)=f(AA^T)A$ for any reasonable function $f$, including $f(x)=\sqrt{1+x}$. This suffices to check for $f(x)=x^k$ when it is obvious, then approximate your function by a polynomial.<|endoftext|> TITLE: What are the points (and generalized points) of the topos of condensed sets? QUESTION [15 upvotes]: What are the topos points of $\mathrm{CondSet}$, i.e. the geometric morphisms $\mathrm{Set} \to \mathrm{CondSet}$? More generally, is there a concise description of the geometric morphisms $\mathcal{E} \to \mathrm{CondSet}$ for any Grothendieck topos $\mathcal{E}$, for example in the form of a geometric theory classified by $\mathrm{CondSet}$? When asking these questions, we have to keep in mind that the category $\mathrm{CondSet}$ is not actually an honest topos, but is "approximated" by the toposes of $\kappa$-condensed sets (by my understanding of Peter Scholzes Lectures on Condensed Mathematics). So the above questions each have two variants: (1) What is the answer for $\mathrm{CondSet}_\kappa$? (2) How should we even interpret the notion of points and generalized points in the case of the full $\mathrm{CondSet}$? Partial answers are very much welcome, of course! REPLY [15 votes]: The category $\mathbf{Cond}$ of condensed sets is equivalent to the category of small sheaves over any of the following three large sites. (For small sheaves, see Mike Shulman's paper Exact completions and small sheaves in TAC.) The category $\mathbf{Comp}$ of compact Hausdorff spaces, equipped with the coherent topology (i.e. finite jointly epimorphic families are covering); The category $\mathbf{ProFin}$ of profinite sets, equipped with the coherent topology; The category $\mathbf{Proj}$ of projective compact Hausdorff spaces, equipped with the disjunctive topology (i.e. families of finite coproduct inclusions are covering). It follows that for any Grothendieck topos (and more generally any infinitary-pretopos) $\mathscr{E}$, the category of cocontinuous left exact functors from $\mathbf{Cond}$ to $\mathscr{E}$ is equivalent to the following three categories: The category of coherent functors from $\mathbf{Comp}$ to $\mathscr{E}$ (i.e. functors that preserve finite limits, finite coproducts, and epimorphisms); The category of coherent functors from $\mathbf{ProFin}$ to $\mathscr{E}$; The category of functors from $\mathbf{Proj}$ to $\mathscr{E}$ that preserve finite coproducts and weak finite limits. These are all instances of the universal property of the category of small sheaves on a (possibly large) site (for which, see Shulman's paper cited above). Namely, the category of condensed sets is the infinitary-pretopos completion of (i) the pretopos $\mathbf{Comp}$, (ii) the coherent category $\mathbf{ProFin}$, (iii) the weakly lextensive category $\mathbf{Proj}$. Now, essentially because the category $\mathbf{Cond}$ of condensed sets is "too big" (e.g. it is not locally presentable), cocontinuous functors $\mathbf{Cond} \to \mathscr{E}$ need not have right adjoints. For that reason, in place of geometric morphisms (defined as adjoint pairs with left exact left adjoint) from a Grothendieck topos $\mathscr{E}$ to $\mathbf{Cond}$, one should instead consider cocontinuous left exact functors from $\mathbf{Cond}$ to $\mathscr{E}$. In particular, the "correct" notion of point of $\mathbf{Cond}$ is a cocontinuous left exact functor $\mathbf{Cond} \to \mathbf{Set}$. Thus, by the above, points of $\mathbf{Cond}$ are equivalent to (i) coherent functors $\mathbf{Comp} \to \mathbf{Set}$, (ii) coherent functors $\mathbf{ProFin} \to \mathbf{Set}$, and (iii) functors $\mathbf{Proj} \to \mathbf{Set}$ that preserve finite coproducts and weak finite limits. Finally, for $\kappa$ a strong limit cardinal, the category of $\kappa$-condensed sets is equivalent to be the category of sheaves over the sites $\mathbf{Comp}_{\kappa}$, $\mathbf{ProFin}_{\kappa}$, and $\mathbf{Proj}_{\kappa}$, defined to be the full subcategories of the above sites spanned by the spaces of cardinality $< \kappa$, with the induced topologies. As above, it follows that points of the topos of $\kappa$-condensed sets are equivalent to certain kinds of functors (the same kinds as above) from these full subcategories to $\mathbf{Set}$.<|endoftext|> TITLE: How complex is the orbit equivalence relation of $\mathrm{Iso}_0(X)\curvearrowright S_X$ for $X=L^p([0,1])$? QUESTION [5 upvotes]: For a Banach space $X$ let $S_X$ denote its unit sphere and let $\mathrm{Iso}_0(X)$ denote the group of rotations of $X$, that is isometries fixing the origin. There is a natural continuous action $\mathrm{Iso}_0(X)\curvearrowright S_X$. When $X=L^p([0,1])$ the group $\mathrm{Iso}_0(X)$ is Polish, so we can ask how complex the orbit equivalence relation induced on $S_X$ is in the descriptive set theoretic sense. By a result of Pełczyński and Rolewicz every orbit is dense, but I don´t know much apart from this. Are orbits meagre? Does this equivalence relation admit classification by countable structures? What are good references for this kind of questions and for more general questions along those lines (for example when $X$ is the Gurarij space)? REPLY [5 votes]: The orbit structure is extremely simple. If $p=2$, there is one orbit (the isometry group of a Hilbert space acts transitively), whereas for $p\neq 2$ there are exactly $2$ orbits: the (classes of) functions that do not vanish on a set of positive measure and its complement, the functions that do vanish on a set of positive measure. This is certainly well-known, but I do not remember where this is written (I thought that it might in a paper by Ferenczi and Rosendal, but I could not locate it). Let me provide the proof in the interesting case $p\neq 2$. Let me first prove that if two functions $f$ and $g$ in $S_X$ do not vanish, then they are in the same orbit. The two probability spaces $([0,1], |f|^p d\lambda)$ and $([0,1], |g|^p d\lambda)$ (where $\lambda$ is the Lebesgue measure) are standard complete atomless probability spaces. They are therefore isomorphic. This implies that there is a bimeasurable bijection $\phi \colon [0,1] \to [0,1]$ which sends $|f|^p d\lambda$ to $|g|^p d\lambda$. The three maps $h \in L^p(d\lambda) \mapsto \frac{h}{f} \in L^p(|f|^p d\lambda)$, $h \in L^p(|f|^p d\lambda) \mapsto h \circ \phi^{-1} \in L^p(|g|^p d\lambda)$, $h \in L^p(|g|^p d\lambda) \mapsto gh \in L^p(d\lambda)$. are all surjective isometries (here we use that $|f|>0$ and $|g|>0$ a.s.). Their composition $h \mapsto g \frac{h \circ \phi^{-1}}{f \circ \phi^{-1}}$ is therefore a linear isometry which maps $f$ to $g$. If $f$ and $g$ both vanish (say on $A$ and $B$ respectively), we can find a linear isometry from $L^p(A,d\lambda)$ onto $L^p(B,d\lambda)$ and combine it with a linear isometry $L^p([0,1]\setminus A) \to L^p([0,1]\setminus B)$ given by the previous case to find a isometry of $L^p([0,1])$ that maps $f$ to $g$. The converse (that a function that almost surely does not vanish cannot be in the orbit of a function that vanishes) follows from the Banach-Lamperti theorem, which characterizes the linear isometries of $L^p([0,1])$: all such isometries are of the form $$ Uf(x) =\omega(x) f(T^{-1}(x))$$ for a measurable bijection $T \colon [0,1] \to [0,1]$ that preserves the class of the Lebesgue measure $\lambda$, and a function $\omega\colon[0,1] \to \mathbf{C}^*$ satisfying $|\omega(x)|^p = \frac{d T_*\lambda}{d\lambda}(x)$.<|endoftext|> TITLE: Do colimits of manifolds coincide with underlying colimits as topological spaces? QUESTION [10 upvotes]: Categories of manifolds (possibly with extra structure) tend not to have all colimits. Other questions have addressed when colimits of manifolds exist. I'd like to know what we can say in general about those colimits which do happen to exist. In particular: given a colimit which does exist in some category $C$ of manifolds, I'd like to know whether it necessarily coincides with the underlying colimit in $\mathsf{Top}$ (and hence also $\mathsf{Set}$). Hopefully, there is a general principle that I'm missing which answers this for many sensible categories of manifolds. But for the sake of a sharp question: does the forgetful functor $C \to \mathsf{Top}$ preserve colimits, for $C = \mathsf{Man}$, the category of topological manifolds and continuous functions [*]? $C = \mathsf{Diff}$, the category of smooth manifolds and smooth functions? $C = \mathsf{Rm}$, the category of smooth Riemannian manifolds and smooth local isometries? Or maybe the functor $\mathsf{Man} \to \mathsf{Top}$ fails to preserve colimits, while forgetful $\mathsf{Diff} \to \mathsf{Man}$ or $\mathsf{Rm} \to \mathsf{Diff}$ does preserve colimits? [*] Since being a topological manifold is a property of, rather than a structure on, a topological space, (1) isn't so much a forgetful functor as an inclusion. I don't see any obvious right adjoints to these forgetful functors, although I'd be happy to learn otherwise. A comment under this question suggests that there are examples of colimits which exist in $\mathsf{Man}$ (or maybe $\mathsf{Diff}$ in my notation?) but which don't agree with the underlying colimits in $\mathsf{Top}$. So a negative answer to (1) or (2) might spell out such an example. I know also that there are various categories of 'generalized smooth spaces' (Frölicher spaces, diffeological spaces, Chen spaces, ...) with $\mathsf{Diff}$ as a subcategory. If the inclusion functor is known to preserve colimits, and colimits of the generalized smooth spaces are understood, then this could be useful to address (2). But I currently know nothing about such generalized smooth spaces, so a very helpful answer along these lines might indicate e.g. which versions of generalized smooth space are best able to answer (2), and whether any of these generalizations have been extended to the Riemannian case for (3). Edit: It seems that the category $\mathsf{Haus}$ of Hausdorff topological spaces and continuous maps is important to this story. With this in mind, a better formulation of my question is as follows: we have a sequence of forgetful/inclusion functors $$ \mathsf{Rm} \rightarrow \mathsf{Diff} \rightarrow \mathsf{Man} \hookrightarrow \mathsf{Haus} \hookrightarrow \mathsf{Top}. $$ Which, if any, of these functors preserve all colimits? The comment below by Martin Brandenburg already shows that the functors with target $\mathsf{Top}$ do not preserve all colimits. REPLY [6 votes]: Here is an observation in the positive direction: Observation: Let $\mathcal C$ be a cocomplete category, and assume that $I \in \mathcal C$ is a strong cogenerator -- that is, $Hom_{\mathcal C}(-,I) : \mathcal C^{op} \to Set$ is faithful and conservative. Let $\mathcal D \subseteq \mathcal C$ be a full subcategory which contains $I$. Then the inclusion $\iota: \mathcal D \to \mathcal C$ preserves colimits. Proof: Suppose that $D = \varinjlim_j^{\mathcal D} D_j$ is a colimit in $\mathcal D$, and let $\underline D = \varinjlim_j^{\mathcal C} D_j$ be the colimit taken in $\mathcal C$. There is a canonical map $\varepsilon: \underline D \to D$. We have $Hom_{\mathcal C}(\underline D, I) = \varprojlim_j Hom_{\mathcal C}(D_j, I) = \varprojlim_j Hom_{\mathcal D}(D_j, I) = Hom_{\mathcal D}(D,I)$, so that $Hom(\varepsilon, I)$ is a bijection. Thus $\varepsilon$ is an isomorphism because $I$ is a strong cogenerator. Example: Euclidean space $\mathbb R^d$ (for $d \geq 1$) is a strong [1] cogenerator in the category $Tych$ of Tychonoff spaces, which is contained in the full subcategory $Man \subset Tych$ of topological manifolds. Therefore any colimit of topological manifolds is a colimit of Tychonoff spaces. [1] That it's a cogenerator follows almost by definition. So to see that $\mathbb R$ is a strong cogenerator in $Tych$, let $f : X \to Y$ be a map in $Tych$ which induces a bijection $f^\ast : C^0(Y) \to C^0(X)$; we wish to show that $f$ is a homeomorphism. First, $f$ is injective. For if $x \neq x' \in X$, then we may find a function $\varphi \in C^0(X)$ with $\varphi(x) \neq \varphi(x')$; since $\varphi$ extends along $f$ it follows that $f(x) \neq f(x')$. Next, $f$ is surjective. For if $y \in Y$ is not isolated and not in the image of $f$, then let $\psi \in C^0(Y)$ vanish uniquely at $y$; then $1/(f^\ast \psi) \in C^0(X)$ does not extend along $f$. If $f$ misses an isolated point $y \in Y$, then $f^\ast$ is not injective. Thus $f$ is a continuous bijection. If $f$ is not a homeomorphism, then pick a closed set $C \subseteq X$ whose image $f(C) \subseteq Y$ is not closed. Let $\varphi \in C^0(X)$ vanish uniquely on $C$. Then $\varphi$ does not extend continuously to $Y$.<|endoftext|> TITLE: Making the ($\infty$-categorical) Bar construction valued in (bi)-modules QUESTION [11 upvotes]: In Lurie's Higher Algebra, construction 4.4.2.7 presents a Bar construction in the setting of $\infty$-categories. The construction in 4.4.2.7 takes as input an $\mathcal{O}$-monoidal $\infty$-category $\mathcal{C}^\otimes \to \mathcal{O}^\otimes$ and a suitable pair of bi-modules in $\mathcal{C}^\otimes$ and gives a simplicial object in $\mathcal{C}$. In the proof of the Barr-Beck theorem, this construction is used in lemma 4.7.3.13, where Lurie uses a simplicial object $\mathrm{Bar}_T(T, M)_\bullet$ with values in $\mathrm{LMod}_T(\mathcal{C})$. All of the Bar constructions previously used in Higher-algebra are only defined as taking value in the underlying $\infty$-category. Looking at the construction, it feels obvious that the $\mathcal{C}$-valued simplicial object $\mathrm{Bar}_T(T, M)_\bullet$ should lift to a $\mathrm{LMod}_T(\mathcal{C})$-valued simplicial object, since the bar construction essentially comes from some functor $\mathbf{\Delta}^{op} \to \mathrm{Tens}_{\succ}$ in which the objects should have structure of modules and the maps should be linear, but I am unable to formally write down a lift along the forgetful functor $\mathrm{LMod}_T(\mathcal{C}) \to \mathcal{C}$ from this definition. Is there a way to see how to to produce such a lift? More generally, given an $(A,B)$-bimodule $M$ in $\mathcal{C}$ and a $(B,C)$-bimodule $N$ in $\mathcal{C}$, the simplicial object $\mathrm{Bar}(M,N)_\bullet$ should take value in the $\infty$-category of $(A,C)$-bimodules, rather than in $\mathcal{C}$. This extension to the bimodule case is also used without proof in HA 5.2.2.6. This question asks about the definition of $\mathrm{Bar}_T(T, M)_\bullet$ but seems to be only about it as a $\mathcal{C}$-valued object (that's what the answer in the commens gives), and do not address the issue of making it $\mathrm{LMod}_T(\mathcal{C})$-valued. REPLY [5 votes]: In Lurie's construction of the relative tensor product, there's an $\infty$-operad $\mathrm{Tens}_{[2]}^{\otimes}$ whose algebras in a monoidal $\infty$-category $\mathcal{C}$ are given by 3 associative algebras (say A,B,C) and two bimodules (say M for (A,B) and N for (B,C)). The simplicial bar construction is obtained by first restricting an algebra for this $\infty$-operad along a certain functor $\Delta^{\mathrm{op}} \to \mathrm{Tens}_{[2]}^{\otimes}$, so that the composite to $\mathcal{C}^\otimes$ takes $[n]$ to $(M, B,\ldots,B,N)$ (with $n$ copies of $B$), and then taking the cocartesian pushforward to the fibre $\mathcal{C}$ over $\langle 1 \rangle$ (to get the simplicial diagram in $\mathcal{C}$ that takes $[n]$ to $M \otimes B^{\otimes n} \otimes N$). To extend this to a simplicial diagram of bimodules, you want to first define a functor $\mathrm{Tens}_{[1]}^{\otimes} \times \Delta^{\mathrm{op}} \to \mathrm{Tens}_{[2]}^{\otimes}$ (where $\mathrm{Tens}_{[1]}^{\otimes}$ is just the operad for bimodules). If we call the 3 objects of the bimodule operad $a,b$ (the two algebras) and $m$, then the composite with our algebra in $\mathcal{C}^\otimes$ should take $((a,\ldots,a,m,b,\ldots,b), [n])$ (with $i$ $a$'s and $j$ $b$'s) to $(A,\ldots,A,M,B,\ldots,B,N,C,\ldots,C)$ with $i$ $A$'s, $n$ $B$'s and $j$ $C$'s. Next you again take a cocartesian pushforward, so that you get a diagram in $\mathcal{C}^\otimes$ of the same shape, but which now takes this object in the source to $(A,\ldots,A,M \otimes B^{\otimes n} \otimes N,C,\ldots,C)$. This new diagram is then adjoint to a simplicial object in algebras for the bimodule operad, and indeed factors through the fibre of $(A,C)$-bimodules.<|endoftext|> TITLE: Does an Ab-enriched category have a unique Ab-enrichment? QUESTION [5 upvotes]: I know that the group structure on Hom sets can be recovered from biproducts if they exit. Indeed, if $f, g : A \to B$ are two maps then there is a uniquely defined map $f \oplus g : A \oplus A \to B \oplus B$ and then there are diagonal and codiagonal maps giving $$ A \to A \oplus A \to B \oplus B \to B $$ so we get a composition law on Hom sets. However, if products/coproduct/biproducts don't exist in my category, then I don't see why the Ab-enrichment should be a "property" and not a "structure." Therefore, I am wondering if it possible for some wacky pre-additive category without biproducts to have multiple Ab-enriched structures? REPLY [11 votes]: As Martin Brandenburg and Maxime Ramzi suggest, it is easy to construct examples on small categories. For example, a one object Ab-enriched category is exactly a ring. The category corresponds to the monoid (multiplication) of the ring and the Ab-enrichment to the addition law. There are monoids which have multiple abelian group structures that make them a ring. Even more simply, consider the category with two objects $x,y$ and $n$ parallel morphisms $x \to y$. We need two morphisms $0, \mathrm{id} \in \mathrm{Hom}(x,x)$ so we may set $\mathrm{End}(x,x) \cong \mathbb{Z}$ as a ring generated by $\mathrm{id}$. Then we set $\mathrm{Hom}(y,x) = 0$ to be the trivial group. The only nontrivial compositions are $\mathrm{Hom}(y, y) \times \mathrm{Hom}(x, y) \to \mathrm{Hom}(x,y)$ and $\mathrm{Hom}(x,y) \times \mathrm{Hom}(x,x) \to \mathrm{Hom}(x,y)$ which must be the unique $\mathrm{Z}$-module structure on any abelian group because the generator $\mathrm{id}$ acts as a unit under composition. Therefore, an Ab-enrichment is exactly an abelian group structure on $\mathrm{Hom}(x,y)$ of which there are many.<|endoftext|> TITLE: Continuity of real functions QUESTION [5 upvotes]: The following question concerns that without $ZF+DC$, can every function be "a little bit" continuous? Question Is it consistent with $ZF+DC$ that for any function $f:[0,1]\to [0,1]$ and any positive measure set $A$, there is a positive measure closed set $B\subseteq A$ so that $f^{-1}(B)$ is closed? How about in Solovay's model? REPLY [7 votes]: Any Borel $f:[0, 1] \to [0, 1]$ satisfying $f^{-1}[\{y\}]$ is dense in $[0, 1]$ for every $y \in [0, 1]$ is a counterexample. For example, $f(x) = \limsup_n \frac{x_1 + x_2 + \dots + x_n}{n}$ where $x = 0.x_1 x_2 \dots$ is the binary expansion of $x$.<|endoftext|> TITLE: A method to bound distances between sets QUESTION [5 upvotes]: I have two sets of points, $X$ and $Y$, which are finite and disjoint. I want to compute the distance between them defined by: $$d(X,Y)= \frac{\sum_{x \in X} \|x-Y\| + \sum_{y \in Y} \|y-X\|}{|x|+|y|} $$ where $\|x-Y\|$ represents the distance between the point $x$ in $X$ to its closest point in $Y$, and $|x|$, $|y|$ are the cardinalities of $X$ and $Y$ respectively. Since both sets are large, the above formula is computationally expensive. How can I calculate it or bound it efficiently? REPLY [4 votes]: Let $d(x,Y):=\min\{\|x-y\|\colon y\in Y\}$ and $d(y,X):=\min\{\|y-x\|\colon x\in X\}$. Then $$d(X,Y)=(1-w)d_X(Y)+wd_Y(X),$$ where $$w:=\frac{|Y|}{|X|+|Y|},$$ $|X|$ and $|Y|$ are the cardinalities of $X$ and $Y$, respectively, and $$d_X(Y):=\frac1{|X|}\sum_{x\in X}d(x,Y)\quad\text{and}\quad d_Y(X):=\frac1{|Y|}\sum_{y\in Y}d(y,X)$$ are the average distances from $X$ to $Y$ and from $Y$ to $X$, respectively. Let $X_1$ and $Y_1$ be (say random) subsets of $X$ and $Y$, respectively, of large but not overly large sizes. Then, obviously, $d(x,Y)\le d(x,Y_1)$ for all $x$ and hence $d_X(Y)\le d_X(Y_1)$. Similarly, $d_Y(X)\le d_Y(X_1)$. So, $$d(X,Y)\le(1-w)d_X(Y_1)+wd_Y(X_1).$$ One could also take some large enough (say random) subsets $X_2$ and $Y_2$ of $X$ and $Y$ such that $w_2:=\frac{|Y_2|}{|X_2|+|Y_2|}\approx w$ and thus get a presumably upper estimate $$(1-w_2)d_{X_2}(Y_1)+w_2d_{Y_2}(X_1)$$ of $d(X,Y)$. These methods should work well if $X$ and $Y$ are well enough separated from each other on an average and if $w$ is known or can be estimated well enough.<|endoftext|> TITLE: Deligne's theorem on exponential sums QUESTION [12 upvotes]: I'm an analyst who needs to use Deligne's Theorem 8.4 in 1, but I feel lost in the maze of definitions, and I don't trust my geometric intuition here. Theorem 8.4: Let $Q$ be a polynomial in $n$ variables and of degree $d$ over $\mathbb{F}_q$, $Q_d$ the homogeneous part of degree $d$ of $Q$, and $\psi:\mathbb{F}_q\to\mathbb{C}^*$ a non-trivial additive character over $\mathbb{F}_q$. Suppose that: $d$ is comprime to $p$; The hypersurface $H_0$ in $\mathbb{P}^{n-1}_{\mathbb{F}_q}$ defined by $Q_d$ is smooth. Then $$\Big|\sum_{x_1,\ldots,x_n\in\mathbb{F}_q}\psi(Q(x_1,\ldots,x_n))\Big|\le (d-1)^nq^{n/2}$$ I'm only interested in the case $q = p$ a prime. I need clarification about the statement. How are smooth hypersurfaces defined? I guess smoothness here is equivalent to $\nabla Q_d(x) \neq 0$ for every $x\in \overline{\mathbb{F}}_q\setminus\{0\}$ (I'm using $x\cdot\nabla Q_d = dQ_d$). Am I right? Some books I've glanced (p. ej. Shafarevich or Hartshorne) define smoothness over closed fields, and in these notes (p. 91) it says that we should extend the field, I think. All sources define "smoothness" in two ways, one using regular rings and other using the gradient. These definitions, which should be equivalent in my case, seem to work over any field, so do I really need to extend to $\overline{\mathbb{F}_q}$? It would be nice to find references about all this which I could understand. I had the idea that to define smoothness we had to take $H_0 = \{Q_d = 0\}\subset \mathbb{P}^{n-1}(\overline{\mathbb{F}}_q)$, define the ideal $I_H = \langle P\rangle$ of polynomials vanishing there, where $P$ is a polynomial, and then to verify that $\nabla P$ doesn't vanish. It seems wrong. If I take $Q = X_1^d$, $d\ge 2$, it seems that Deligne's theorem doesn't hold, even though $\{X_1 = 0\}$ looks smooth. If $Q_d\in\mathbb{Z}[X_1,\ldots,X_n]$ and $\nabla Q_d(x)\neq 0$ for every $x\in\mathbb{C}^n\setminus\{0\}$, does Theorem 8.4 hold for all but finitely many primes? From what I've gathered until now, arithmetic-geometers find it trivial (should I too?), probably they'd point to proposition A.9.1.6 in 2, but again, I don't know if all these definitions of smoothness coincide. I'd be nice (for me at least) to have a proof without using schemes, but it is probably painful. I stated my question over the complex numbers because it is closed, but can I weaken the assumptions? On the other hand, is the converse true? I'd appreciate any reference or help. References 1 Deligne, Pierre. La conjecture de Weil. I. (French) Inst. Hautes Études Sci. Publ. Math. No. 43 (1974), 273--307. 2 Hindry, Marc; Silverman, Joseph H. Diophantine geometry. An introduction. Graduate Texts in Mathematics, 201. Springer-Verlag, New York, 2000. REPLY [17 votes]: Yes, smoothness is equivalent to the gradient being nonzero for every $x \in \overline{\mathbb F}_q^n \setminus \{0\}$. I would define smoothness of the hypersurface defined by $Q_d$ as the condition that for each such $x$, either the gradient or the value of $Q_d$ at $x$ is nonzero, but as you note, by homogeneity and the fact that $d$ is prime to $p$, if the value is nonzero then the gradient is nonzero. Checking over $\overline{\mathbb F}_q$ and not $\mathbb F_q$ is really necessary. For example, fix a basis for the field extension $\mathbb F_{q^n}$, giving a bijection between $\mathbb F_{q^n}$ and $\mathbb F_q$. The norm map that sends each element of $\mathbb F_{q^n}$ to the determinant over $\mathbb F_q$ of its multiplication action on $\mathbb F_{q^n}$ can be expressed as a polynomial of degree $n$ in the $n$ coordinates. Set $d=n$ and $Q_= Q_d=$ this polynomial. Then because the norm is a surjective homomorphism of multiplicative groups, there are $\frac{q^n-1}{q-1}$ elements of norm $a$ for each $a \in \mathbb F_q^\times $, so $$\sum_{x \in \mathbb F_q^n } \psi(Q(x))= \sum_{x\in \mathbb F_{q^n}} \psi (\operatorname{Norm}(x))= 1 + \frac{q^n-1}{q-1} \sum_{x \in \mathbb F_q^\times} \psi(x) = 1 - \frac{q^n-1}{q-1} $$ which does not satisfy Deligne's bound for $n>2$ and $q$ large. However, $Q_d$ never vanishes at any $x \in \mathbb F_q^n$, and thus neither does its gradient. So checking over the algebraic closure is really necessary! You can avoid working over the algebraic closure only by using definitions that involve checking point-by-point. For example, smoothness is equivalent to the claim that the ideal generated by $Q_d$, and $\frac{ \partial Q_d}{\partial x_i}$ for all $i$ contains $x_1^N, \dots, x_n^N$ for some $N$. You are right that we do not want to pass to the vanishing set of $Q_d$ and then to the ideal - we just want to look at $Q_d$ itself and whether it satisfies the gradient condition. The vanishing scheme of $Q_d$ is sensitive to taking powers in this sense, which is why this is referred to as a smoothness condition on the hypersurface rather than a smoothness condition on $Q_d$. If you like, this is equivalent to smoothness of the vanishing set plus the fact that $Q_d$ isn't divisible by the square of a nontrivial polynomial. It is indeed true that if the gradient of $Q_d$ is nowhere vanishing over $\mathbb C^n \setminus \{0\}$, then it is nowhere vanishing on $\overline{\mathbb F}_p^n \setminus \{0\}$ for all but finitely many $p$. The same is true for any algebraically closed field of characteristic zero. The proof actually doesn't require scheme there. Here are two approach: (1) By the Nullstellensatz, this implies that the ideal generated $\frac{ \partial Q_d}{\partial x_1}, \dots, \frac{ \partial Q_d}{ \partial x_n}$ contains a power of the ideal $(x_1,\dots, x_n)$ and in particular contains $x_1^N, \dots, x_n^N$ for some $N$. So some $\mathbb C[x_1,\dots,x_n]$-linear combination of those derivatives is equal to $x_1^N$, some linear combination is $x_2^N$, etc. We can choose a linear map $\mathbb C \to \mathbb Q$ whose composition with $\mathbb Q \to \mathbb C$ is the identity. (In fact we only need to do this for the finite-dimensional subset of $\mathbb C$ generated by the coefficients of the linear combination.) Plugging the coefficients of all the polynomials in $\mathbb C[x_1,\dots,x_n]$ into this map, we see that some $\mathbb Q[x_1,\dots,x_n]$-linear combination of those derivatives is equal to $x_1^N$, and so on. Clearing denominators, we see that there is a natural number $M$ such that some $\mathbb Z[x_1,\dots,x_n]$-linear combination of those derivatives is equal to $M x_1^N$, $\mathbb Z[x_1,\dots,x_n]$-linear combination of those derivatives is equal to $M x_2^N$, and so on. It follows for all $p$ not dividing $M$ that, if $x_1,\dots, x_n \in \overline{\mathbb F}_p$ are not all zero, then one of these derivatives is nonvanishing, as desired. (2) There exists a universal polynomial, the discriminant, in the coefficients of a homogeneous polynomial $f$ of degree $d$, which vanishes if and only if the derivatives of $f$ all vanish at some nonzero point over an algebraically closed field of characteristic not dividing $d$. This can be obtained as the Macauley resultant of the partial derivatives of $f$. If the nonvanishing condition holds over $\mathbb C$, then the discriminant is nonzero, so it is nonzero mod $p$ for all but finitely many $p$, so the nonvanishing condition holds for all but finitely many $p$.<|endoftext|> TITLE: Continuity of the period for a periodic dynamical system QUESTION [5 upvotes]: Let $v\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$ $(n\geq 1)$ a velocity field such that every solution $(x_t)_{t\geq 0}$ of $(d/dt)x_t=v(x_t)$ is periodic. Denote, for a non-stationary point $x\in\mathbb{R}^n$ (meaning $v(x)\neq0$), by $T(x)$ the period of such a solution $(x_t)_{t\geq0}$ such that $x(0)=x$. Quetion: Is $T$ continuous over the set of non-stationnary points? REPLY [4 votes]: The answer is trivially negative already for $n=3$: Start with the flow following along the "long" lines of a thickened (say with disc-shaped cut) Möbius strip. You can imagine this like a thick “rope” bent to a loop, making half a rotation on its way, and the flow follows the fibres of the rope (with a fixed speed). Obviously, the fibre in the "center" of the rope has half of the period than the nearby fibres. To extend the flow to $\mathbb R^3$, decrease the speed smoothly near the "boundary" of the rope to $0$, and then extend the field by letting it $0$ outside of the rope. However, this extension works only, because the question explicitly admits stationary points. If one wants to exclude stationary points, there is a topological obstacle in the extension of this particular flow to $\mathbb R^3$. After an embedding into $\mathbb R^4$ this obstacle trivially vanishes, as the Möbius strip can be "unentangled" in $\mathbb R^4$, so in $\mathbb R^4$ there is obiously a non-singular counterexample. I conjecture that for $n=3$ there is no non-singular counterexample at all, but I do not know how to prove this.<|endoftext|> TITLE: What are projective locales / injective frames? QUESTION [9 upvotes]: Judging by the compact regular case, and more generally the spatial case, regular projectivity of locales, resp. regular injectivity of frames, must have something to do with $\neg p\lor\neg\neg p$ and $\neg(x\land y)\to(\neg x\lor\neg y)$. On the other hand, existence of locales without points shows that the terminal locale is not regular projective. I am convinced somebody should already have found out which locales are regular projective, but who? The same question about projectivity with respect to arbitrary epimorphisms of locales is probably easier but less interesting. Still, I don't know anything about that either. PS As Simon Henry pointed out, I should rather pick some pullback stable class of locale epimorphisms. I guess I don't know an answer about any of them (except maybe the proper ones), so please just choose an as large as possible class of nicely behaved epimorphisms of your choice - say, quotients, or of effective descent, or triquotients, etc. REPLY [2 votes]: I am convinced by the answer of Simon Henry completely. This is just an addendum to it, mainly for myself: I want to look at these $I_\kappa$, $T_\kappa$ and $B_\kappa$ in as much detail as possible. In fact I will be just more or less repeating parts of what Simon wrote in the language more familiar to me. So, let first $F_\kappa$ be the free frame on generators $u_{n\alpha}$, $n\in\omega$, $\alpha\in\kappa$. Note that each element of $F_\kappa$ is a join of finite meets of generators $u_{n_1\alpha_1}\land\cdots\land u_{n_m\alpha_m}$. Then the frame of opens of $I_\kappa$ is the quotient of $F_\kappa$ by the obvious relations which ensure that locale maps $X\to I_\kappa$ are in one-to-one correspondence with families $U_{n\alpha}$ of opens of $X$ satisfying $$ \begin{aligned} \bullet\ &\bigvee_\alpha U_{n\alpha}=\top\text{ for all $n$};\\ \bullet\ &U_{n\alpha}\wedge U_{n\beta}=\bot\text{ for all $n$ and all $\alpha\ne\beta$};\\ \bullet\ &U_{m\alpha}\wedge U_{n\alpha}=\bot\text{ for all $m\ne n$ and all $\alpha$}. \end{aligned} $$ In particular, each embedding $i:\omega\hookrightarrow\kappa$ determines a map $X\to I_\kappa$ for any $X$, by declaring $U_{n\,i(n)}=\top$ for all $n$ and $U_{n\alpha}=\bot$ for all $n$, $\alpha$ with $\alpha\ne i(n)$. This implies that $u_{n_1\alpha_1}\land\cdots\land u_{n_m\alpha_m}=\bot$ if and only if there are either some $n_i=n_j$ with $\alpha_i\ne\alpha_j$ or some $\alpha_i=\alpha_j$ with $n_i\ne n_j$. For each $\alpha\in\kappa$ let $v_\alpha=\bigvee_nu_{n\alpha}$. Then by what we just said, $v_\alpha\land u_{n_1\alpha_1}\land\cdots\land u_{n_m\alpha_m}\ne\bot$ for any $u_{n_1\alpha_1}\land\cdots\land u_{n_m\alpha_m}\ne\bot$, hence $\neg v_\alpha=\bot$. It follows that imposing further relations $v_\alpha=\top$ for all $\alpha$ gives a nontrivial dense sublocale $T_\kappa$ of $I_\kappa$; in particular $T_\kappa$ contains $B_\kappa:=I_\kappa^{\neg\neg}$. Finally, given any map $X\to B_\kappa$ (or to $T_\kappa$ as well), all these relations imply that in the topos $\operatorname{Sh}(X)$ of sheaves on $X$, projections of the subobject $$ \coprod_{(n,\alpha)\in\omega\times\kappa}U_{n\alpha}\rightarrowtail\coprod_{\omega\times\kappa}1 $$ both to $\coprod_\omega1$ and to $\coprod_\kappa1$ are isomorphisms. And clearly also in fact such maps are in one-to-one correspondence with isomorphisms $\coprod_\omega1\cong\coprod_\kappa1$ in $\operatorname{Sh}(X)$. It follows that for each such map, if $X$ is nontrivial then there is an embedding $\kappa\hookrightarrow\hom_{\operatorname{Sh}(X)}(1,\coprod_\omega1)$. Hence for each nontrivial $X$ there is no such map for $\kappa$ exceeding the cardinality of $\hom_{\operatorname{Sh}(X)}(1,\coprod_\omega1)$, i. e. of the set of all countable partitions of $X$ into clopens.<|endoftext|> TITLE: Connection between determinant and quotient rule QUESTION [19 upvotes]: For the function $\dfrac{f(x)}{g(x)}$, we have, $\left(\dfrac{f}{g}\right)' = \dfrac{gf'-fg'} {g^2}$. We can write the numerator as $W(g,f) = \left|\begin{matrix} g & f \\ g' & f'\end{matrix}\right|$ which is called Wronskian. I wonder why determinant appears in the numerator ? Is there any mathematical relationship between derivative of $\dfrac fg$ and determinant that gives us $W(g,f)$ right away ? REPLY [7 votes]: With the risk of making a simple thing confusing, here's another point of view. Two functions $f,g$ on an interval $I$ give a map to the projective line $I \dashrightarrow \mathbb P^1$ by $x\mapsto [f(x):g(x)]$ (some indeterminacies can occur if both functions vanish at some point). You can ask: when does this map have a critical point? If you choose the chart $[y:1]$ on $\mathbb P^1$ you have to look at the vanishing of the derivative of $f/g$. If you lift the map to $I\to \mathbb R^2$ via $x\mapsto (f(x),g(x))$ then you get a curve in the plane, with tangent vector $(f'(x),g'(x))$. The map to $\mathbb P^1$ has a critical point precisely when the vectors $(f(x),g(x))$ and $(f'(x),g'(x))$ are proportional, so you get the Wronskian. The book of Ovsienko and Tabachnikov "Projective Differential Geometry, Old and New" contains a wealth of additional information.<|endoftext|> TITLE: The use of computers leading to major mathematical advances II QUESTION [67 upvotes]: I would like to ask about recent examples, mainly after 2015, where experimentation by computers or other use of computers has led to major mathematical advances. This is a continuation of a question that I asked 11 years ago. There are several categories: A) Mathematical conjectures or large body of work arrived at by examining experimental data B) Computer-assisted proofs of mathematical theorems C) Computer programs that interactively or automatically lead to mathematical conjectures. D) Various computer programs which allow proving automatically theorems or generating automatically proofs in a specialized field. E) Computer programs (both general purpose and special purpose) for verification of mathematical proofs. F) Large databases and other tools Of course more resources (like this Wikipedia page on experimental mathematics) are also useful. REPLY [5 votes]: The paper Constructions in combinatorics via neural networks by Adam Zsolt Wagner uses reinforcement learning algorithm and the deep cross-entropy method, to find explicit constructions and counterexamples to several open conjectures in extremal combinatorics and graph theory! Let me also mention the paper Refuting conjectures in extremal combinatorics via linear programming also by Adam Zsolt Wagner. (I thank Geordie Williamson for telling me about Adam's work.)<|endoftext|> TITLE: Papers on history and philosophy of mathematics suitable for master's students QUESTION [6 upvotes]: In the fall, I will give a course called "Perspectives in Mathematics". This is a mandatory course at our master's program in mathematics (including applied mathematics and statistics). The course contents are rather vague, but we are supposed to cover historical and philosophical perspectives on mathematics, as well as the role of mathematics in society. One thing I would like to do is to ask the students to read some short texts on topics such as history, philosophy and perhaps sociology of mathematics and discuss them in class. I would like these to be more than just popular science but still reasonably accessible. Also they should contain some interesting perspectives and perhaps challenge the beliefs of the students, so there is something to discuss. Finally, I want to cover not just the classical history of pure mathematics, but also applied mathematics and its role in society. I would be really grateful for tips on research papers or other short texts that you think may be suitable. REPLY [3 votes]: For the philosophy of mathematics, a standard reference is Benacerraf and Putnam's anthology Philosophy of Mathematics: Selected Readings. This will provide a good introduction to some of the "traditional" topics in the subject such as logicism, intuitionism, formalism, and platonism. There are of course many important topics not covered. I'd recommend that you also consider something written by Imre Lakatos, such as Proofs and Refutations, as well as Humanizing Mathematics and its Philosophy: Essays Celebrating the 90th Birthday of Reuben Hersh. For the history of mathematics, there is the encyclopedic book A History of Mathematics by Carl B. Boyer and Uta C. Merzbach. In particular, this book has some good chapters on the early history of mathematics in non-Western civilizations, a topic that was largely neglected in the West until relatively recently. Now, there's another take on the history of mathematics, which is to treat it not just as a subject in its own right but as something that can illuminate and inform the work of a research mathematician. As mentioned by someone in a comment, the work of Harold Edwards is exemplary in this regard. If your students have a strong enough mathematical background for it, I'd recommend the book Galois Theory. A rather neglected topic is the relationship between mathematics and religion. A good anthology is Mathematicians and their Gods: Interactions between mathematics and religious beliefs, edited by Snezana Lawrence and Mark McCartney. There are also the fascinating books Naming Infinity by Loren Graham and Jean-Michel Kantor and Equations from God by Daniel Cohen. Many of your students may be interested in the topic of women in mathematics. There are several books on this topic, such as Complexities: Women in Mathematics, edited by Bettye Anne Case and Anne M. Leggett. Finally, depending on how much you're willing to venture into controversial territory, you could consider devoting some time to the question of how to promote diversity, equity, and inclusion in mathematics. Especially in the realm of mathematics education, DE&I is a very hot topic—possibly hotter than you want to touch. I confess that I am not au courant with the literature, so I don't feel qualified to give recommendations, but one example I heard about recently (and which has generated heated debate) was A Pathway to Equitable Math Instruction. But perhaps others who are more knowledgeable than I am can provide better guidance in this area.<|endoftext|> TITLE: Why are $G$-Extensions of fusion categories rigid, when constructed via monoidal 2-functors $G \to \underline{\underline{BrPic}}(\mathcal{D})$? QUESTION [7 upvotes]: In arxiv:0909.3140 the $G$-extensions of a fusion category $\mathcal{D}$ are classified via monoidal 2-functors $G \to \underline{\underline{BrPic}}(\mathcal{D})$. A crucial part of the classification is Theorem 7.7 that shows that equivalence classes of such functors are in 1-1 correspondence with equivalence classes of $G$-extensions of $\mathcal{D}$. However in the proof they never show that the tensor categories they obtain from such functors are actually rigid. I was also unable to find this statement anywhere else in the paper, it is not even mentioned. Is there some obvious proof of this that I am missing or are they in fact classifying not necessarily rigid $G$-extensions? (The latter would however contradict Definition 2.1, which defines a G-extension to be in particular a fusion category, hence rigid.) REPLY [7 votes]: Great question! Yes, this is missing from the original paper and is certainly needed for the main result. It is true though. There's two different proofs currently in the literature, Deshpande and Mukhopadhyay Corollary 2.11 and Davydov and Nikshych proof of Theorem 8.5. You can also prove it using an argument I learned from Theo Johnson-Freyd, namely given an object $X$ in $C_g$ you can realize $C_g$ as $A$-mod in $C_e$ with $X$ corresponding to $A$ itself, then $C_{g^{-1}}$ can be identified with mod-$A$ and the dual object will be $A$ again and the evaluation and coevaluation maps are given by multiplication and the unit of the algebra structure on $A$. As to your last two sentences, rigidity is pretty essentially to the whole setup, otherwise the graded parts don't need to be invertible and then you leave the world of homotopy theory. For example, take the semisimple category with two objects 1 and X with tensor product defined by $X \otimes X = 0$ (aka the universal counterexample to all questions involving rigidity), this is $\mathbb{Z}/2\mathbb{Z}$ graded but the odd part is not invertible as a bimodule over the even part.<|endoftext|> TITLE: What properties should $C(M,\mathbb{R})$ have when $M$ is a $n$-dimensional manifold? QUESTION [6 upvotes]: Let $M$ be a n-dimensional manifold, $C(M,\mathbb{R})$ be the function space of continuous function from $M$ to $\mathbb{R}$. What kind of properties should $C(M,\mathbb{R})$ has, to reflect the manifold structure? Denote a L-shape line be $X$, we know it is not a manifold, but what difference can we observe from the algebras of continuous function compared to circle? i.e. What properties does $C(M,\mathbb{R})$ has and $C(X,\mathbb{R})$ does not? In this case their topological dimension is the same, so they should have same stable rank in terms of $C^*$-algebras. REPLY [8 votes]: For compact orientable manifolds, this is accomplished by the notion of a spectral triple. See the question Commutative spectral triples for additional information, including a precise statement of the theorem. See also the answer to Noncommutative smooth manifolds for additional remarks about orientability. As a side remark, if one is willing to use smooth functions instead of continuous functions, then there is a very satisfactory answer: the resulting contravariant functor from second countable Hausdorff smooth manifolds to commutative real algebras is a fully faithful functor, i.e., manifolds are (contravariantly) identified with a full subcategory of real algebras. This is known as Milnor's exercise. See also the answer How much of differential geometry can be developed entirely without atlases? for a discussion on how to characterize the essential image of the embedding functor. REPLY [4 votes]: This is not really an answer but rather a long-ish comment. First of all, if $M$ is not compact, $\mathfrak{A}=C(M,\mathbb{R})$ is not really a C${}^*\!$-algebra but actually only a locally C${}^*\!$-algebra, that is, a locally convex ${}^*\!$-algebra which admits a separating family of submultiplicative C${}^*\!$-seminorms, unless you require e.g. its elements to vanish at infinity. If $M$ is Hausdorff, paracompact and second countable, this family of seminorms can be in addition chosen to be countable (i.e. the underlying locally convex vector space is Fréchet). So, one could see non-compactness represented algebraically as the non-(C${}^*\!$-)normability of $\mathfrak{A}$. The second axiom of countability, on its turn, is characterized by the metrizability of $\mathfrak{A}$. If the elements of $\mathfrak{A}$ are instead required to vanish at infinity, you get the well-known characterization of non-compactness of $M$ by the fact that $\mathfrak{A}$ is then not unital. About the Hausdorff property, it is related to the capability of $\mathfrak{A}$ to separate the points of $M$. More precisely, $C(X,\mathbb{R})$ only actually sees the Tychonoffication of $X$ for any topological space $X$, which is defined by taking the quotient of $X$ modulo the equivalence relation $\sim$ which states that $x\sim y$ if $f(x)=f(y)$ for all $f\in C(X,\mathbb{R})$. Particularly, since a Tychonoff topological space is always Hausdorff, $\mathfrak{A}$ cannot see at all whether $M$ is Hausdorff or not. As for paracompactness, there is a nice characterization of it in terms of $C(X,\mathbb{R})$ if $X$ is locally compact (which is always the case if $X=M$ is finite dimensional): $X$ is paracompact if and only if the ${}^*\!$-ideal $C_c(X,\mathbb{R})\subset C(X,\mathbb{R})$ of continuous real-valued functions with compact support is a projective $C(X,\mathbb{R})$-module. This, of course, leaves open the question about how to characterize this compactly supported ideal in algebraic terms. The above discussion shows how the "size" and the "separation" of $M$ are encoded in the structure of $\mathfrak{A}$. Other aspects one may look for an characterization of in terms of properties of $\mathfrak{A}$ are: The (topological) dimension of $M$ (particularly, whether it is finite or not, which on its turn will determine whether $M$ is locally compact or not); The local structure of $M$, i.e. the fact that $M$ is locally Euclidean; The smooth structure of $M$ (if any). The local structure of $M$ requires one to look at not only $\mathfrak{A}$ alone, but at a whole (pre)sheaf of ${}^*\!$-algebras over $M$. Local algebras over (say) contractible domains should look like $\mathfrak{A}_0=C(\mathbb{R}^n,\mathbb{R})$. Therefore, at the local level, one should ask what is special about $\mathfrak{A}_0$ as a (Fréchet) locally C${}^*\!$-algebra. As for an occasional smooth structure on $M$, recall that such a structure is not necessarily fixed by the topology of $M$ alone. Typically this is reflected in the structure of (unbounded) derivations on $\mathfrak{A}$ in suitable dense domains, which on their turn should play the role of "smooth" functions on $M$. This structure is, of course, absent for topological manifolds which do not admit a smooth structure or only do so in (open) subsets thereof, which once again displays the need for looking at the local (i.e. presheaf) structure. They can also be used together to encode the dimension of $M$, as in e.g. the top de Rham cohomology of contractible domains. Edit: As pointed in Dmitri Pavlov's answer, a more precise formulation of this idea is provided by the notion of a spectral triple, which is pretty well developed in the case of a compact $M$.<|endoftext|> TITLE: Monotonicity of a parametric integral QUESTION [5 upvotes]: For real $x>0$, let $$f(x):=\frac1{\sqrt x}\,\int_0^\infty\frac{1-\exp\{-x\, (1-\cos t)\}}{t^2}\,dt.$$ How to prove that $f$ is increasing on $(0,\infty)$? Here is the graph $\{(x,f(x))\colon00$, the concavity would of course imply that $f$ is increasing. This question arises in a Fourier analysis of a certain probabilistic problem. REPLY [2 votes]: Here is another solution: Noting the identity $$\sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} = \frac{1}{2(1-\cos t)}$$ and taking advantage of the fact that the integrand is non-negative, we may apply Tonelli's theorem to get \begin{align*} f(x) &= \frac{1}{2\sqrt{x}} \int_{-\pi}^{\pi} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} \right) \bigl( 1 - e^{-x\left(1-\cos t\right)} \bigr) \, \mathrm{d}t \\ &= \frac{1}{4\sqrt{x}} \int_{-\pi}^{\pi} \frac{1 - e^{-x\left(1-\cos t\right)}}{1-\cos t} \, \mathrm{d}t \\ &= \frac{1}{8\sqrt{x}} \int_{-\pi}^{\pi} \frac{1 - e^{-2x\sin^2(t/2)}}{\sin^2(t/2)} \, \mathrm{d}t. \tag{1} \end{align*} Then by applying the integration by parts, we also get $$ f(x) = \frac{\sqrt{x}}{2} \int_{-\pi}^{\pi} \cos^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t. \tag{2} $$ Now using $\text{(1)}$, $$ \bigl( \sqrt{x}f(x) \bigr)' = \frac{1}{4} \int_{-\pi}^{\pi} e^{-2x\sin^2(t/2)} \, \mathrm{d}t. \tag{3} $$ Therefore by $\text{(2)}$ and $\text{(3)}$ altogether, \begin{align*} \sqrt{x}f'(x) &= \bigl( \sqrt{x}f(x) \bigr)' - \frac{1}{2\sqrt{x}} f(x) \\ &= \frac{1}{4} \int_{-\pi}^{\pi} e^{-2x\sin^2(t/2)} \, \mathrm{d}t -\frac{1}{4} \int_{-\pi}^{\pi} \cos^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t \\ &= \frac{1}{4} \int_{-\pi}^{\pi} \sin^2(t/2) e^{-2x\sin^2(t/2)} \, \mathrm{d}t \\ &> 0. \end{align*} This identity can also be used to show that $f''(x) < 0$, and hence $f$ is concave as expected.<|endoftext|> TITLE: A definition in poset theory QUESTION [10 upvotes]: I am working on a article in poset theory. In that article, I am defining a subposet of a poset. The definition is following: Let $P$ be a finite poset. A subposet $P'$ of $P$ is called closed under covering if for every $x,y \in P'$ with $x\lessdot y$ in $P'$, we have $x\lessdot y$ in $P$. Here, $x \lessdot y$ means $x$ is covered by $y$. I want to know weather the above definition is already in the literature? If yes, then what these subposets are called? If not, then the name I have given is correct? or what should I call such subposets? REPLY [10 votes]: I recall seeing in various sources the terminology "cover preserving embedding" and "cover preserving subposet". Googling it now (https://www.google.com/search?q=poset+%22cover+preserving%22) brings some 4000 results, many of which are research articles (with some repetitions - I am not implying there are 4000 distinct articles on this topic).<|endoftext|> TITLE: Group ring with infinite stable rank QUESTION [12 upvotes]: In searching for a counterexample in homological stability, I came across the following question: Is there a known example of a finitely presented group $G$, so that the group ring $\mathbb{Z}[G]$ has infinite Bass stable rank? REPLY [7 votes]: Yes, the integral group ring $\mathbb{Z}[F_2]$ of the free group $F_2$ on two generators has infinite stable rank. This can be deduced from [1, Corollary 3.6]: There exists a cyclic $\mathbb{Z}[F_2]$-module $M$ with the following property. For every $N \ge 1$, there exists an epimorphism $\theta_N: (\mathbb{Z}[F_2])^N \twoheadrightarrow M$ such that $(\mathbb{Z}[F_2])^N$ cannot be generated by $N$ elements one of which is contained in $\ker \theta_N$. It should be easy to see that the cyclic $\mathbb{Z}[F_2]$-module $M$ above has infinite stable rank. (Actually, explicit non-stable unimodular rows of length $N$ for every $N \ge 2$ can be extracted from the proof). Then combine the previous result with [2, Lemma 11.4.6] If $n$ is in the stable range of a finitely generated right $R$-module $M$ then $n$ is in the stable range of any factor module $M/K$. to conclude that the stable rank of $\mathbb{Z}[F_2]$ is infinite. [1] M. Evans, "Presentations of groups involving more generators than necessary", 1992. [2] J. McConnell and J. Robson, "Noncommutative Noethering rings", 1987.<|endoftext|> TITLE: reference to a theorem about a product of harmonic and parallel forms QUESTION [8 upvotes]: Let $\alpha$ be an exterior product of a harmonic and a parallel form on a Riemannian manifold. Then $\alpha$ is known to be harmonic. I have heard that this is an old result due to R. Bott, but I could never find a reference. I would be very grateful for any pointers to the early literature. REPLY [8 votes]: One place where (a generalization of) the desired result is stated explicitly is in a 1973 paper by J. H. Sampson, On a theorem of Chern. Sampson gives a simplified proof of Chern's main result in his 1957 paper On a generalization of Kähler geometry (Algebraic geometry and topology. A symposium in honor of S. Lefschetz, pp. 103–121. Princeton University Press, Princeton, N. J., 1957). There is also a very nice 1962 exposition (in French) of Chern's theorem by André Weil, Un théorème fondamental de Chern en géométrie riemannienne, in Séminaire N. Bourbaki, 1962, exp. no 239, p. 273-284. The form in which Sampson and Weil state Chern's result is as follows: Theorem: If a Riemannian manifold $(M^n,g)$ has holonomy $H\subseteq\mathrm{O}(n)$, then the $g$-Laplacian commutes with all of the linear operators on $\Omega^*(M)$ constructed from the ring of $H$-equivariant linear maps $L:\Lambda^*(\mathbb{R}^n)\to\Lambda^*(\mathbb{R}^n)$. The desired result is a special case of this theorem, since, if $\pi$ is a $g$-parallel $p$-form, then the holonomy $H$ of $g$ is contained in the stabilizer of $\pi$, and hence, by the above result, the operator $L(\alpha) = \alpha\wedge\pi$ commutes with the Laplacian of $g$. Of course, this implies the desired result since $\Delta(\alpha\wedge\pi) = \Delta\bigl(L(\alpha)\bigr) = L\bigl(\Delta(\alpha)\bigr) = 0$ when $\Delta(\alpha) = 0$, but is stronger. Comments: Chern proved the above theorem in his 1957 paper for the ring of operators $L$ as above that are degree-preserving, but, in fact, this implies the more general result. Both Chern and Sampson phrase the theorem in terms of $H$-structures without torsion, but, of course, this is the same as Riemannian manifolds with holonomy contained in $H$, a point explicitly made by Weil. Sampson does not refer to Weil's article; perhaps he was unaware of it. Meanwhile, Sampson goes on to apply Chern's idea to other Laplacians and proves new results. There are earlier results in special cases by André Lichnerowicz, e.g. Généralisations de la géométrie kählérienne globale. (French) Colloque de Géométrie Différentielle, Louvain, 1951, pp. 99–122. Georges Thone, Liège; Masson & Cie., Paris, 1951, but neither his articles nor those of Chern, Sampson, or Weil mention Bott.<|endoftext|> TITLE: Original reference for categories of presheaves as free cocompletions of small categories QUESTION [15 upvotes]: It is well known that, for a small category $\mathbf A$, the category $\widehat{\mathbf A} = [\mathbf A^\circ, \mathbf{Set}]$ of presheaves on $\mathbf A$ together with the Yoneda embedding $\mathbf A \to \widehat{\mathbf A}$ exhibits $\widehat{\mathbf A}$ as the cocompletion of $\mathbf A$ under small colimits. Where was this first observed? If an observation first appears independently of a proof, I would be interested in knowing where a proof first appears too. Bunge's 1966 thesis Categories of Set-Valued Functors seems a likely candidate, as it is concerned with properties of presheaf categories. However, I have been unable to obtain a copy of the thesis, so I do not know whether it appears here, and it is not mentioned in Bunge's summary of thesis, Regular categories. REPLY [12 votes]: The earliest reference I can find to the universal property of the presheaf construction is Remark 2.29 of Ulmer's Properties of Dense and Relative Adjoint Functors (1968). However, the proof is only lightly sketched, and in the introduction Ulmer states: As an application of relative adjoints we will show in a subsequent paper that every category $\mathbf M'$ admits a free right complete category. As far as I can tell, this paper never appeared. Note that Ulmer actually considers the universal property for arbitrary (possibly large) categories, by taking small presheaves rather than arbitrary presheaves. There is an earlier reference for the universal property of the Ind-completion (i.e. cocompletion under filtered colimits) in Proposition 8.7.3 of SGA4 (dated 1963–1964, but published in 1972), which suggests the universal property of free cocompletion was also known as this time, though an explicit statement does not appear. In the enriched context, the universal property first appears as Theorem 2.11 of Lindner's Morita equivalences of enriched categories (1974), where the Lindner attributes the unenriched result to Ulmer.<|endoftext|> TITLE: Filtered homotopy colimits and singular homology QUESTION [6 upvotes]: Suppose I have a functor $$ X_\bullet: I \to \text{Spaces} $$ where $I$ is a small filtered category. It seems to be a "folk theorem" that the homomorphism $$ \underset{\alpha\in I}{\text{colim }} H_\ast(X_\alpha) \to H_\ast(\underset{\alpha\in I}{\text{hocolim }} X_\alpha) $$ is an isomorphism, where $H_\ast$ denotes singular homology. Is there a reference for this? (Preferably standard?) REPLY [6 votes]: I think the usual reference here is chapter XII section 5 of the Bousfield-Kan book "Homotopy limits, completions, and localizations," where they build a spectral sequence $E_2^{*,*} \cong L_*colim_i H_*(X_i) \Rightarrow H_*(hocolim_i X_i)$. The rest of the argument is not explicit in the book (but it's very straightforward so I think it's still correct to cite their book for your result): since your diagram is filtered, the derived functors $L_*colim_i H_*(X_i)$ vanish in positive degrees, so the spectral sequence collapses to the $L_0colim_i H_*(X_i)$ line, yielding your isomorphism.<|endoftext|> TITLE: The history and original paper of the Rosser–Iwaniec sieve QUESTION [6 upvotes]: I'm trying to find Rosser's original paper where he introduces his eponymous sieve. I've already found https://arxiv.org/pdf/math/0505521 (where the reference isn't given, but where it is indicated that Rosser has priority) and http://matwbn.icm.edu.pl/ksiazki/aa/aa36/aa36210.pdf (which is called "Rosser's sieve", but where there is no reference to his paper). Any help would be very much appreciated. REPLY [6 votes]: This is an addition to Lucia's remark. Check out in Selberg's collected papers entry #36: Sieve methods (Proc. Symp. Pure Math., 1971). There he writes: Very important is also the unpublished work of Barkley Rosser who first settled the problem of the 1-residue sieve, and thus anticipated the work of Jurkat-Richert by about a decade. The part of §5 which deals with the Buchstab-Rosser sieve, refers to work done in the late 1950's after I had had the opportunity to see an unpublished manuscript by Rosser. [...] In the early 1950's Barkley Rosser devised a sieve procedure that essentially represents the limit of the Buchstab procedure. [...] Rosser's work was never published, but a paper about 10 years later by Jurkat and Richert [4] obtained essentially the same results for $k = 1$, by methods which combine Buchstab's ideas with estimations obtained by Brun's method and by a method mentioned in [6] and [7] for the upper bound. If one carefully analyzed their method, it would probably turn out that this is really the Buchstab-Rosser sieve again.<|endoftext|> TITLE: Origin of Laguerre geometry? QUESTION [5 upvotes]: Laguerre geometry is described as either the geometry of oriented lines and circles in the Euclidean plane, equipped with a certain unusual symmetry group (see https://en.wikipedia.org/wiki/Laguerre_transformations) or as the geometry of vertical parabolas and non-vertical lines (see https://en.wikipedia.org/wiki/Laguerre_plane). The group of symmetries of the Laguerre plane is the same as the group $PGL(2,\mathbb D)$ where $\mathbb D$ is the dual numbers. It makes me wonder: Did Laguerre start with the symmetry group $PGL(2,\mathbb D)$ and then work out the geometry, or did he formulate a geometry somehow and people later discovered that its symmetry group was $PGL(2,\mathbb D)$? (Don't know what tags to use.) REPLY [4 votes]: The formulation of Laguerre geometry in terms of dual numbers is a decidedly 'synthetic' one, meant to exhibit how this set of transformations can be regarded as a different 'real form' of the well-known linear fractional transformations of the complex plane. If you want to know more about the history of how Laguerre geometry (itself a special case of Lie sphere geometry) was developed (long before the introduction of 'dual numbers'), you might want to look at a few other articles, including the Wikipedia article on spherical wave transformations.<|endoftext|> TITLE: Balanced presentation of the fundamental group of a Seifert fiber space QUESTION [7 upvotes]: Is there any readily available reference for a balanced presentation of the fundamental group in terms of the classifying invariants of an arbitrary Seifert fibered space? I can't find it, and the usual presentation coming from the classification resists GAP (checked on four fibrations with three exceptional fibers over the sphere, got two generators and three relations each time). REPLY [9 votes]: You can get a balanced presentation of the group from a Heegaard diagram. The paper of Boileau-Zieschang, Heegaard genus of closed orientable Seifert 3-manifolds. Invent. Math. 76 (1984), no. 3, 455–468 shows how to find such a diagram. Following their discussion (too involved to reproduce here) you get a presentation that is determined by the Seifert invariants. If you only care about the group, then their lemma 1.5 shows how to go from the `usual' presentation (which is not balanced) to a balanced one. Except for some small examples, I think that these are in fact minimal (with respect to the number of generators) presentations of the fundamental group.<|endoftext|> TITLE: Relations between $3j$-symbols and intertwiners QUESTION [5 upvotes]: I am trying to understand the relation between Wigner's $3j$-symbols (or Clebsch-Gordan coefficients) and matrix coefficients of intertwiners. I am new to this topic and need some help to understand it properly. So, let us start with the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$, which is the complexification of $\mathfrak{su}(2)$. Now, as usual, all the irreducible representations of $\mathfrak{sl}(2,\mathbb{C})$ can be described by spins $j\in\mathbb{N}_{0}/2$ and have dimension $2j+1$. Now it is a general fact that for a simple and complex Lie algebra, ever finite-dimensional irreducible representation can be labeled with a heighest weight $\Lambda$, which in the case of $\mathfrak{sl}(2,\mathbb{C})$ are given by $2j$. In general, the tensor product of two such heighest weight modules is fully reducible and hence we can write $$V_{\Lambda}\otimes V_{\Lambda^{\prime}}=\bigoplus_{i}C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}V_{\Lambda_{i}}$$ with some coefficients, usually called multiplicities or Littlewood-Richardson coefficients. Now in order to relate this to intertwiners, we first of all know, according to the Lemma of Schur, that the space of intertwiners between two irreducible finite-dimensional representations of a complex Lie algebra $\mathfrak{g}$ is either $0$-dimensional (if they are not isomorphic) or $1$-dimensional (if they are isomorphic). As a consequence, we get that the Littlewood-Richardson coefficients are given by the dimension of the space of intertwiners from $V_{\Lambda_{i}}$ to $V_{\Lambda}\otimes V_{\Lambda^{\prime}}$, i.e. $$C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}=\mathrm{dim}_{\mathbb{C}}(\mathrm{Int}(V_{\Lambda_{i}},V_{\Lambda}\otimes V_{\Lambda^{\prime}}))$$ So in the case of $\mathfrak{sl}(2,\mathbb{C})$ (or equivalently $\mathfrak{su}(2)$), the coefficients are hence given by $$C_{\Lambda\Lambda^{\prime}}^{\Lambda_{i}}=\mathrm{dim}_{\mathbb{C}}(\mathrm{Int}(V_{2j_{i}},V_{2j}\otimes V_{2j^{\prime}}))$$ So far so good. In a textbook (Fuch, Schweigert — Symmetries, Lie algebras and representations), they then say that using this correspondence, it is clear that the Clebsch-Gordan coefficients for fixed $J$ are the matrix coefficients of intertwiners in $\mathrm{Int}(V_{2j_{i}},V_{2j}\otimes V_{2j^{\prime}})$ for "a definite choice of basis". Can anyone explain this step to me in more detail? Furthermore, what does "a definite choice of basis" mean in this context? Also, what is then the correspondence with the $3j$-symbols? Thank you very much! EDIT: In order to clarify the conventions I use, let me add some more details about the Clebsch-Gordan coefficients: Let $J_{\pm},J_{0}$ be the three generators of $\mathfrak{sl}(2,\mathbb{C})$, i.e. $$[J_{0},J_{\pm}]=\pm J_{\pm}\hspace{1cm}\text{and}\hspace{1cm}[J_{+},J_{-}]=2J_{0}$$ Then there is a basis $\{\mid j,m\rangle\}_{-j\leq m\leq j}$ of $V_{2j}$ satisfying $$J_{0}\mid j,m\rangle=m\mid j,m\rangle$$ $$J_{\pm}\mid j,m\rangle=\sqrt{j(j+1)-m(m\pm 1)}\mid j,m\pm 1\rangle$$ In order to define the Clebsch-Gordan coefficients, one usually defines to different bases of the tensor product $V_{2j}\otimes V_{2j^{\prime}}$: (1) The tensor product of the bases described above, i.e. $\{\mid j_{1},j_{2},m_{1},m_{2}\rangle\}_{-j_{1}\leq m_{1}\leq j_{1},-j_{2}\leq m_{2}\leq j_{2}}$ where $\mid j_{1},j_{2},m_{1},m_{2}\rangle:=\mid j_{1},m_{1}\rangle\otimes\mid j_{2},m_{2}\rangle$. (2) Define the total spin $\vec{J}:=\vec{J}_{1}+\vec{J}_{2}$. Then there is a basis $\{\mid J,M\rangle:=\mid j_{1},j_{2},J,M\rangle\}$ satisfying $$\vec{J}^{2}\mid J,M\rangle=J(J+1)\mid J,M\rangle$$ Then the Clebsch-Gordan coefficients are the coefficients of the change of basis matrix, i.e. $\langle j_{1},j_{2},m_{1},m_{2}\mid J,M\rangle$ REPLY [4 votes]: You can find a quick review of the relevant definitions of $3j$, $6j$, $9j$, symbold in Section 7 of my article with Chipalkatti "The higher transvectants are redundant", Annales de l'Institut Fourier, Tome 59 (2009) no. 5, pp. 1671-1713. I think the best best way to see the vector space corresponding to the spin $j$ representation (instead of spherical harmonics etc.) is the space of homogeneous polynomials with complex coefficients in two variables and of degree $2j$. The group $SU(2)$ obviously acts on $\mathbb{C}^2$ and thus has a natural action on functions on $\mathbb{C}^2$, in particular polynomial functions, and even more in particular, homogeneous polynomial functions. These are called binary forms. Finally, please correct the use of "Clebsch-Gordon" to "Clebsch-Gordan" with an "a". Half a century before physicists got interested in these coefficients etc. Paul Gordan and Alfred Clebsch figured out the decomposition into irreducibles of a tensor product of irreducibles for $SU(2)$, with explicit intertwiners. The analogous result in higher dimension is pretty much an open problem. For $SU(3)$, this was done recently by Böhning and Graf von Bothmer in this article. For higher dimension, and for tensor products of symmetric powers only, there are such explicit results, most notably by Jerzy Weyman and Peter Olver. See this article and references therein.<|endoftext|> TITLE: Connection on a Hilbert bundle QUESTION [5 upvotes]: Is there a well-defined notion of connection on a measurable bundle of Hilbert spaces? REPLY [3 votes]: This is an answer to the refined question formulated in the comments: the base space is the unitary dual of a Lie group $G$. The definition can be carried out in the setting of stacks in groupoids (or simplicial sets) on the site of cartesian spaces ($\def\R{{\bf R}} \R^n$ with smooth maps, for all $n≥0$). Specifically, we define a stack $R_G$ that gives the unitary dual of a Lie group $G$ and a stack $B_∇$ that gives bundles of Hilbert spaces with connectiom. Then a morphism $R_G→B_∇$ is precisely a bundle of Hilbert spaces with connection over $R_G$. To define the stack $R_G$ that gives the unitary dual of a Lie group, assign to a cartesian space $T$ the following groupoid $R_G(T)$. Objects of $R_G(T)$ are given by a Hilbert space $H$ together with a smooth $T$-indexed family of irreducible unitary representations of $G$ on $H$. This is simply a smooth map $\def\Hom{\mathop{\rm Hom}} f\colon T→\Hom(G,U(H))$ (landing in irreps), where $U(H)$ is equipped with the ultraweak topology and $\Hom(G,U(H))$ is the space of continuous group homomorphisms equipped with the compact-open topology. Smooth means that the adjoint map $T⨯G→U(H)$ composed with the inclusion $U(H)→B(H)$ is smooth as a map from a finite-dimensional smooth manifold to a topological vector space $B(H)$. Morphisms of $R_G(T)$ from $(H,f)$ to $(H',f')$ are given by a smooth map $h\colon T→U(H,H')$ that intertwines the action of $G$. Here $U(H,H')$ is the topological space of unitary isomorphisms $H→H'$ equipped with the ultraweak topology. Next, we define stacks $B$ and $B_∇$ of bundles of Hilbert spaces, (equipped with connection in the case of $B_∇$) as follows. Given a cartesian space $T$, we define a groupoid $B_∇(T)$ as follows. Objects of $B_∇(T)$ are pairs $(H,∇)$, where $H$ is a Hilbert space and $∇\colon T→Hom(T,I(H))$ is a smooth map, where $I(H)$ denotes the space of unbounded skew-adjoint operators on $H$ (equivalently, one-parameter unitary groups on $H$) and $\Hom(T,I(H))$ is the space of linear maps $T→I(H)$. Morphisms of $B_∇(T)$ from $(H,∇)$ to $(H',∇')$ are smooth maps $p\colon T→U(H,H')$ such that $∇'=p^*\theta+\mathop{\rm Ad}_{p^{-1}}∇$, where $\theta$ is the Maurer–Cartan form and Ad denotes the adjoint action. The stack $B$ is defined analogously, but dropping $∇$ and the condition on $p$. We have a canonical forgetful map $B_∇→B$. The obvious forgetful map $W\colon R_G→B$ defines the canonical bundle $W$ over $R_G$. In particular, the fiber of $W$ of a point of $R_G$ given by an irreducible representation $ρ$ of $G$ is simply $ρ$ itself. Now, a connection on $W$ is a lift of the map $W\colon R_G→B$ through the forgetful map $B_∇→B$.<|endoftext|> TITLE: Invariant theory over $\mathbb R$ QUESTION [5 upvotes]: $\DeclareMathOperator\SO{SO}$Suppose we have a (continuous) linear action of $\SO(n,\mathbb R)$ on a vector space $\mathbb R^N$. Consider the ring of invariants $A\subset \mathbb R[x_1,\ldots, x_N]$, which is an $\mathbb R$-algebra. Is it true that the orbits of the $\SO(n,\mathbb R)$ action are in one-to-one correspondence with the $\mathbb R$-algebra homomorphisms $A\to\mathbb R$? It is clear that for each orbit of the action we get such an evaluation homomorphism, by the value of an invariant polynomial on the orbit. But is the inverse true? Is there some nice book that considers that type of setting, over the real numbers? REPLY [9 votes]: The answer depends on what you mean by "one-to-one correspondence". Is it bijective or just injective? Robert Bryant's (standard) argument shows that $\mathbb R^N/\mathrm{SO}(n)\to \mathrm{AlgHom}_{\mathbb R}(A,\mathbb R)$ is injective. In general, this map is very far from being surjective, though. In other words, not every Homomorphism $A\to\mathbb R$ is an evaluation map. Already the standard action of $\mathrm{SO}(n)$ on $\mathbb R^n$ is a counterexample. In this case, $A=\mathbb R[q]$ where $q(v)=\|v\|^2$ is the norm square function. Thus a homomorphisms $f:A\to\mathbb R$ is given by the value $f(q)$ which can be an arbitrary real number. But the image of $\mathbb R^n/\mathrm{SO}(n)$ is obviously only $\mathbb R_{\ge0}$. So what is the "correct quotient"? The answer is "It depends". A topologist would say it is the orbit space. The embedding into $\mathrm{AlgHom}_{\mathbb R}(A,\mathbb R)$ gives the orbit space the structure of a semialgebraic set on which you have notions of continuous or smooth function. A theorem of G. Schwarz states that all continuous/smooth invariants are pull-backs of continuous/smooth function on the orbit space. An algebraist would prefer to call $\mathrm{AlgHom}_{\mathbb R}(A,\mathbb R)$ to be the quotient since it classified all closed orbits defined over $\mathbb R$ regardless whether the contain a real point or not. A very nice paper calculating the image of $\mathbb R^N/\mathrm{SO}(n)\to \mathrm{AlgHom}_{\mathbb R}(A,\mathbb R)$ (in particular) in terms of inequalities is Procesi, Claudio; Schwarz, Gerald: Inequalities defining orbit spaces. Invent. Math. 81 (1985), 539–554<|endoftext|> TITLE: Three squares in a rectangle QUESTION [15 upvotes]: One of my colleagues gave me the following problem about 15 years ago: Given three squares inside a 1 by 2 rectangle, with no two squares overlapping, prove that the sum of side lengths is at most 2. (The sides of the squares and the rectangle need not be parallel to each other.) I couldn't find a solution or even a source for this problem. Does anyone know about it? I was told that this was like an exercise in combinatorial geometry for high school contests such as IMO, but I'm not sure if this problem was listed in such competitions. Has anyone heard of this problem, or does anyone know how to solve it? Any kind of help will be appreciated. REPLY [5 votes]: Since the squares are convex, we can draw lines which separate them. In particular, if two separating lines go from $(b-a,0)$ to $(b,1)$ and from $(c,1)$ to $(c+d,0)$, then we can prove the result in terms of those lines and those variables. So: let the rectangle go from $(0,0)$ to $(2,1)$. Let $A$ be the leftmost square (or one such square). Let $C$ be the rightmost square (or one such square). Let $B$ be the other square. Draw a line separating $A$ and $B$, and let $(b,1)$ and $(b-a,0)$ be its intersections with the lines $y=1$ and $y=0$. Draw a line separating $B$ and $C$, and let $(c,1)$ and $(c+d,0)$ be its intersections with the lines $y=1$ and $y=0$. Reasoning as in user21820's answer, we assume wlog that: $00$, so $B$ is below both lines. (Since the lines may leave the rectangle, we do not assume $b-a>0$ or $b<2$ or $c>0$ or $c+d<2$.) Lemma (proved at the end): \begin{align} \text{sidelength of }A &\le \min\!\left(\frac{b}{a+1},\,1\right)\\ \text{sidelength of }B &\le \min\!\left((a-b+c+d)u,\,1\right)\\ \text{sidelength of }C &\le \min\!\left(\frac{2-c}{d+1},\,1\right) \end{align} where $$u=\max\left( \frac{1}{a+d+1}, \frac{\sqrt{a^2+1}}{a^2+a+d+1}, \frac{\sqrt{d^2+1}}{d^2+d+a+1} \right)$$ The factor $u$ satisfies $1/(a+1)>u$ and $1/(d+1)>u$ so long as $a<3.66$ and $d<3.66$ respectively. I will assume those inequalities for now to show that some functions are increasing or decreasing; I don't have a clean proof for those inequalities or without them yet. In the corner case of $b=a+1$ and $c=1-d$, the side lengths are $1$, $0$ and $1$, and they sum to exactly $2$. We now use this in analyzing four cases. Case I, $b\le a+1$ and $c\le 1-d$: The sum of the sidelengths is at most $$\frac{b}{a+1}+(a-b+c+d)u+1$$ This is increasing in both $b$ and $c$, so its value is at most the corner value of $2$. Case II, $b\le a+1$ and $c\ge 1-d$: The sum of the sidelengths is at most $$\frac{b}{a+1}+(a-b+c+d)u+\frac{2-c}{d+1}$$ This is increasing in $b$ and decreasing in $c$, so its value is at most the corner value of $2$. Case III, $b\ge a+1$ and $c\le 1-d$: The sum of the sidelengths is at most $$1+(a-b+c+d)u+1$$ This is decreasing in $b$ and increasing in $c$, so its value is at most the corner value of $2$. Case IV, $b\ge a+1$ and $c\ge 1-d$: The sum of the sidelengths is at most $$1+(a-b+c+d)u+\frac{2-c}{d+1}$$ This is decreasing in both $b$ and $c$, so its value is at most the corner value of $2$. So the sum of the sidelengths is at most $2$ in each case. Proof of Lemma: The sidelength of $A$ is clearly less than 1, and also clearly less than the maximum sidelength inscribed in the right triangle bounded by $x=0$, $y=1$, and the separator of $A$ and $B$. We use Polya's formula here to calculate the sidelength in the triangle as the maximum of $b/(a+1)$ and $b\sqrt{a^2+1}/(a^2+a+1)$; since $a>0$, the maximum is just $b/(a+1)$. A similar use of Polya's result bounds the sidelength of $C$. Yet another use of that result, now for a triangle which may be acute or obtuse, bounds the sidelength of $B$ by $|a-b+c+d|u$. Since we assumed $a-b+c+d>0$, we write this bound as $(a-b+c+d)u$.<|endoftext|> TITLE: Does there exist a "citation distance" calculator for papers or authors? QUESTION [7 upvotes]: This question is not directly a mathematical question, but I am interested in whether there exists a calculator akin to an Erdős number calculator. The main difference is that I am not interested in coauthorship, but in citation distance; that is, I would like to link papers or authors by a chain of papers $P_1 , \dots , P_n$ such that $P_i$ either cites or is cited by $P_{i+1}$. Considering how useful such a tool could be for detecting connections between different areas of math, I assume that someone has done this before. Any help is appreciated! REPLY [6 votes]: If you have access to the Web of Science you could use the CitNetExplorer tool to create a citation network (documentation). This tool is used quite extensively, but I have not used it myself. The arXiv has recently implented the LitMaps tool to create a citation network. I don't think Google Scholar offers a similar functionality. Incidentally, the average citation distance between papers published in the same year decreased from approximately 5.33 to 3.18 steps between 1950 and 2018 (source).<|endoftext|> TITLE: Open complement of hypersurfaces QUESTION [6 upvotes]: Let $k$ be an algebraically closed field. Let $H_1, H_2$ be two smooth hypersurfaces of the same degree $d$ in $P^n_k$. Let $U_1,U_2$ be their complements respectively. Are $U_1,U_2$ isomorphic as algebraic varieties? In $n=1,d=1$ case this is true, because the complement of any point is isomorphic to $A^1$. But $n=2$ case I guess this might be false. I want to prove that if $U_1,U_2$ are isomorphic then they must be induced by an automorphism of $P^n$, but this seems hard. I read something about the 'complement problem' on enter link description here, but this seems to be a more complicated question, and it focuses on $A^n$ instead. Maybe the $n=2,d=3$ case is easier? In this case, elliptic curves are isomorphic if and only if they have the same $j$-invariant. Can we read this from its complement? Are there any solutions/counterexamples? Any comments are welcome! REPLY [10 votes]: If $U_1$ and $U_2$ are isomorphic then $H_1$ and $H_2$ are equal in the Grothendieck ring of varieties and thus, by the Larsen-Lunts theorem, stably birational, which if $d>n$ implies that they are isomorphic. This is probably extreme overkill, but it seems to handle some different cases than the other answers. REPLY [9 votes]: The answer is no. Perhaps the simplest case is $n=2$, $d=4$. There is a unique double covering $\pi _i:S_i\rightarrow \mathbb{P}^2$ branched along $H_i$. If $U_1$ and $U_2$ are isomorphic, $S_1$ and $S_2$ are isomorphic; then $H_1$ and $H_2$ are isomorphic, because $H_i$ is the branch locus of the morphism $\pi _i$, which is given by the anticanonical system. REPLY [7 votes]: The easiest case is $n = 1$, $d = 4$. Indeed, the embeddings $U_i \to \mathbb{P}^1$ are canonical, hence an isomorphism $U_1 \cong U_2$ extends to an isomorphism of the ambient projective lines and induces an isomorphism $$ \mathbb{P}^1 \setminus U_1 \cong \mathbb{P}^1 \setminus U_2. $$ So, if the cross-ratio of the four points $\mathbb{P}^1 \setminus U_1$ differs from the cross-ration of $\mathbb{P}^1 \setminus U_2$, there can't be such an isomorphism.<|endoftext|> TITLE: Resource request on "$\in$-homomorphisms" in Set Theory QUESTION [6 upvotes]: Very loosely put, this is the intuitive idea behind an $\in$-homomorphism: Let $\mathcal{U}$ and $\mathcal{W}$ be universes of sets. A function $f \colon \mathcal{U} \to \mathcal{W}$ is said to be an $\in$-homomorphism if $f(X) = \{f(T) \in \mathcal{W} | T \in X \}$ for all $X \in \mathcal{U} $ Thomas Jech briefly mentions this concept in his book Set Theory on page 250 and 251. We note that he makes use of ZFA, an alternative axiomatisation of Set Theory which allows non-set objects called atoms. Could anyone perhaps point me towards any other resource that also mentions, or perhaps goes deeper into this concept? Ideally, I would like a resource, text or paper that also makes use of the ZFA axiomatisation, but I would also appreciate any resource based on any other axiomatisation, as I really can't find anything at all besides this brief mention by Thomas Jech. REPLY [4 votes]: I believe that you may have misstated the definition of what it means to be an $\in$-homomorphism. (I couldn't find your notion in Jech at your links — have I missed it?) For example, with your definition, $f(X)$ is always a set, and never an atom, and so if there are atoms, then the identity function will not be a $\in$-homomorphism. Your definition seems to require us to map all atoms to $\emptyset$. Further, your notion would require $f$ to fix all well-founded sets, since there could be no $\in$-minimal set that is moved. In ZFC, therefore, your notion trivializes. Your definition also seems to presume that the models are transitive, using the standard $\in$-relation, whereas the notion of homomorphism should be sensible with arbitrary models of set theory. I believe that the intended notion of homomorphism should be the one obtained by viewing models of set theory as relational structures $\langle M,\in^M\rangle$, a domain $M$ with a binary relation $\in^M$. In this case, the standard model-theoretic notion of (strong) $\in$-homomorphism or $\in$-embedding between two such structures would be a map $j:M\to N$ such that $$x\in^M y\iff j(x)\in^N j(y)$$ for all $x,y\in M$. The difference between this and your notion is that you require that $j(y)$ has no other $\in^N$ elements except the $j(x)$ objects. I analyzed this homomorphism notion in my paper Hamkins, Joel David, Every countable model of set theory embeds into its own constructible universe, J. Math. Log. 13, No. 2, Article ID 1350006, 27 p. (2013). ZBL1326.03046. Blog post One of the main results was that the countable models of set theory are linearly pre-ordered by embedding: Theorem. For any two countable models of set theory $\langle M,\in^M\rangle$, $\langle N,\in^N\rangle$, one of them is isomorphic to a submodel of the other. This notion of submodel is the model-theoretic notion of $\in$-homomorphism, which are necessarily injective and thus isomorphisms of the domain with the range. The theorem is often found surprising, but this is mainly because this notion of submodel is much weaker than we usually consider in set theory. In particular, submodels in this sense need not preserve much set-theoretic truth; they need not even be $\Delta_0$-elementary. This is a reason to view $\in$-homomorphisms as much weaker than we probably want. Set theorists typically want to consider emeddings that are at least $\Delta_0$-elementary, preserving $\Delta_0$ truth. To illustrate the difference, the map $j:V\to V$ defined by $$j(y)=\{j(x)\mid x\in y\}\cup\{\{\emptyset,y\}\}$$ has $x\in y\iff j(x)\in j(y)$, and it is therefore an $\in$-embedding of $V$ into $V$. And yet, it has no fixed points — it does not even fix the natural numbers to themselves, and it does not carry $\emptyset$ to $\emptyset$. Another part of my answer to your question is to mention my paper (updated from my initial post, which had mentioned the wrong paper): Daghighi, Ali Sadegh; Golshani, Mohammad; Hamkins, Joel David; Jeřábek, Emil, The foundation axiom and elementary self-embeddings of the universe, Geschke, Stefan (ed.) et al., Infinity, computability and metamathematics. Festschrift celebrating the 60th birthdays of Peter Koepke and Philip Welch. London: College Publications (ISBN 978-1-84890-130-8/hbk). 89-112 (2014). ZBL1358.03070. Blog post In that paper, we considered for various anti-foundational theories whether there can be nontrivial elementary embeddings $j:V\to V$. Some of the arguments involve Quine atoms in the anti-foundational theories, but these function essentially similarly to urelements in ZFA, and the relevant notion of embedding is your notion of $\in$-homomorphism. I'd encourage you to take a look there. REPLY [3 votes]: This answer has nothing to do with atoms, but on page 69 of Jech's book you'll find Mostowski's Collapsing Theorem. The map used there is exactly like the ones you mention. These collapsing functions are important tools for dealing with constructible sets, large cardinals, etc.<|endoftext|> TITLE: Cartan decomposition of loop group QUESTION [6 upvotes]: Let $G$ be a complex reductive group. Let $LG$ and $L^+ G$ denote the formal loop spaces given by maps from the punctured formal disk and the formal disk, respectively, to $G$. The quotient $LG/L^+ G$ is known as the affine grassmannian of $G$. I am looking for a reference or a proof of the well known fact that the $L^+ G$ orbits of the affine grassmannian are in bijection with dominant coweights of $G$. In other words, the double coset space $$ L^+ G \backslash LG / L^+ G = X_*(T), $$ for some maximal torus $T$. In some places it is stated that this is a consequence of the Cartan decomposition of the loop group $LG$, but I do not see exactly why that is true. REPLY [2 votes]: Apparently, this is based on a result by Iwahori and Matsumoto (Corollary 2.17 of [IM]). A modern proof of this result can be found on [DHLH]. This result is one of the main elements of the proof of the Hilbert-Mumford criterion (see page 52 of [GIT]). What the theorem of Iwahori and Matsumoto asserts is that any element $g\in LG$ can be written as $g=h_1 t^\mu h_2$, for $h_1, h_2 \in L^+G$, $\mu: \mathbb{G}_m\rightarrow G$ a 1-parameter subgroup and $t^\mu$ the image of $t\in \mathbb{C}((t))$ by the associated map $\mu(\mathbb{C}((t))): \mathbb{C}((t))^\times \rightarrow LG$. Since maximal tori are conjugate, the orbit $L^+G t^\mu L^+G$ associated to any 1-PS can be labeled by a coweight $X_*(T)$. Moreover, we can further identify two orbits if the coweights are conjugate, so we can quotient by the Weyl group and get $X_*(T)/W=X_*(T)^+$. A good reference for this topic, where I originally found about the paper of Iwahori-Matsumoto is section 4.5 of the book [BD]. I still do not know if there's an easier way of proving this (i.e. without using the theorem of Iwahori-Matsumoto) and do not get in what way this is a Cartan decomposition. Anyway, this solves my question. References: [AHLH] Alper, J., Halpern-Leistner, D., & Heinloth, J. (2019). Cartan-Iwahori-Matsumoto decompositions for reductive groups. arXiv preprint arXiv:1903.00128. [BD] Beilinson, A.; Drinfeld, V. Quantization of Hitchin's integrable system and Hecke eigensheaves. https://math.uchicago.edu/~drinfeld/langlands/QuantizationHitchin.pdf [IM] Iwahori, Nagayoshi; Matsumoto, Hideya. On some Bruhat decomposition and the structure of the Hecke rings of $p$-adic Chevalley groups. Publications Mathématiques de l'IHÉS, Tome 25 (1965) , pp. 5-48. http://www.numdam.org/item/PMIHES_1965__25__5_0/ [GIT] Mumford, D. ; Fogarty, J. ; Kirwan, F. Geometric invariant theory. Third edition.<|endoftext|> TITLE: Einstein metrics on connected sums QUESTION [6 upvotes]: Do there exist pairs of $n$-dimensional closed Einstein manifolds $(M_1,g_1)$ and $(M_2,g_2)$, $n\ge 3$, such that the connected sum $M_1\#M_2$ carries an Einstein metric which is conformal to $g_1$ and $g_2$ on the summands? REPLY [9 votes]: There are no nontrivial examples with $n\ge3$ beyond what I mentioned in my comment above, namely, either a conformal connected sum of a compact space form $(M_1,g_1)$ with the standard round $n$-sphere with the connected sum $M_1\# M_2$ being homothetic to $(M_1,g_1)$ or the case where both $M_1$ and $M_2$ are diffeomorphic to a (possibly exotic) $n$-sphere with homothetic metrics $g_1$ and $g_2$ and the resulting connected sum being again homothetic to $(M_1,g_1)$. (Note that there are many inequivalent Einstein metrics on manifolds homeomorphic to odd-dimensional spheres.) To begin, consider the following question: Given an Einstein $n$-manifold $(M^n,g)$ (that is connected but not necessarily complete), with Einstein constant $\lambda$ (i.e., $\mathrm{Ric}(g)= (n{-}1)\lambda g$), what are the possibilities for Einstein metrics on $M$ that are conformally equivalent to $g$? By the well-known formula for the Ricci tensor of $\tilde g = u^{-2}\,g$, where $u>0$ is a function on $M$, $$ \mathrm{Ric}(\tilde g) = \mathrm{Ric}(g) + (n{-}2)\,u^{-1}\,\nabla(\mathrm{d}u) - \bigl(u^{-1}\,\Delta u + (n{-}1)\, u^{-2}\,|\mathrm{d}u|^2\bigr)\, g\,, $$ it follows that $\tilde g$ is also Einstein if and only if $u$ is a non-vanishing function on $M$ that satisfies the equation $\nabla(\mathrm{d}u) = v\,g$ for some function $v$. Differentiating this equation, one finds that $v+\lambda u$ must be locally constant. Thus, by the connectedness of $M$, there exists a constant $c$ such that $\nabla(\mathrm{d}u) = (c{-}\lambda u)\,g$. Consequently, $\Delta u = n(\lambda u{-}c)$, which will be useful below. Moreover, one sees that the Einstein constant of $\tilde g$ is $\tilde\lambda = 2c\,u - \lambda\,u^2 - |\mathrm{d}u|^2$. (Also, one sees that the vector field $\nabla u$ must annihilate the Weyl tensor of $g$, i.e., $W(\nabla u, X,Y,Z) = 0$ for all vector fields $X,Y,Z$ on $M$. However, that will play no role in this argument.) For example, if $(M^n,g)=\bigl(\mathbb{R}^n,|\mathrm{d}x|^2\bigr)$ is the standard flat metric, then $\tilde g = u^{-2}\,g$ is Einstein (on the open set where $u$ is nonzero) if and only if $u = p\,|x|^2 + 2\,q{\cdot}x + r$ for some constants $p,r\in\mathbb{R}$ and vector $q\in\mathbb{R}^n$, not all zero. This shows how to essentially 'linearize' the problem of describing the conformal multiples of an Einstein metric that are, themselves, Einstein: On the bundle $E = \mathbb{R}\oplus\mathbb{R}\oplus T^*\!M$ over $M$, consider the connection $D$ that, for any vector field $X$ and any section $(c,u,\alpha)$ of $E$, satisfies $$ D_X\begin{pmatrix}c, & u, & \alpha\end{pmatrix} = \begin{pmatrix}\mathrm{d}c(X),& \mathrm{d}u(X)-\alpha(X),& \nabla_X\alpha - (c-\lambda\,u) X^\flat\end{pmatrix} $$ where $X^\flat$ is the $1$-form that satisfies $X^\flat(Y) = g(X,Y)$ for all vector fields $Y$ on $M$. Then we can rephrase the above equation on $u$ (and $c$) by saying that $u^{-2}\,g$ is Einstein if and only if there is a $c$ such that $(c,u,\mathrm{d}u)$ is a $D$-parallel section of $E$. Now, it turns out that, except for the metrics of constant sectional curvature on the $n$-sphere, for a compact, connected Einstein manifold $(M,g)$, the space of global $D$-parallel sections of $E$ consists only of the sections of the form $(\lambda u,\,u,\,0)$ where $u$ is constant. (Such sections are obviously $D$-parallel.) To see this, first note that if $(M^n,g)$ is compact, connected and Einstein with a non-positive Einstein constant, then the only global $D$-parallel sections of $E$ are those with $u$ constant. This is because the defining equation implies $\Delta u = n(\lambda u {-} c)$. When $\lambda=0$, this clearly implies $c=0$ and that $u$ be constant. When $\lambda<0$, this implies that $\lambda u {-} c$ is an eigenvalue of $\Delta$ with negative eigenvalue $n\lambda$ and hence must be $0$. Meanwhile, if one supposes that $\lambda>0$, the equation $\nabla(\mathrm{d}u) = (c{-}\lambda u)\,g$ implies that, if $p$ is a critical point of $u$, then the Hessian of $u$ at $p$ is $\bigl(c{-}\lambda u(p)\bigr)$ times the metric at $p$. If $c{-}\lambda u(p)=0$, then the $D$-parallel section $(c,u,\mathrm{d}u)$ agrees with the $D$-parallel section $(\lambda\,u(p), u(p),0)$ at $p$ and hence must equal it, i.e., $u$ must be constant. On the other hand, if $c{-}\lambda u(p)\not=0$, then $p$ is a non-degenerate critical point of $u$ that is either a maximum or a minimum. Thus, on a connected, compact Einstein manifold $(M^n,g)$ the only critical points of a nonconstant $u$ such that $(c,u,\mathrm{d}u)$ is $D$-parallel for some $c$ are either strict maxima or minima. By the Mountain Pass Lemma, it follows that $u$ can only have one maximum and one minimum, which implies that $M$ is homeomorphic to a $n$-sphere. Suppose that such a non-constant $u$ exists on $(M^n,g)$. By adding a constant to $u$, we can suppose that $u$ is strictly positive on $M$ and by scaling, we can assume that $\lambda=1$. Now by using the structure equations, one can prove that, on a punctured neighborhood of a maximum $p\in M$, $g$ can be written in the 'sine-cone' form $g = \mathrm{d}r^2 + (\sin r)^2\, h$, where $r$ is the distance from $p$, $h$ is an Einstein metric on $S^{n-1}$ with Einstein constant $1$, and $u = a + b \cos r$ for some constants $a$ and $b$. However, unless $h$ is conformally flat, the above sine-cone metric cannot be smooth at $r=0$. Thus, $h$ is conformally flat, which implies that $g$ is also conformally flat. Hence, the only possibility for $(M,g)$ is the round $n$-sphere, as was to be shown. Now, for any connected Einstein $(M^n,g)$, the sheaf of $D$-parallel sections of $E$ is well-behaved: There is an integer $k$ satisfying $1\le k\le n{+}2$ such that, for any 1-connected open set $U\subset M$, the dimension of the vector space $\mathcal{P}(U)\subset\Gamma(U,E)$ consisting of $D$-parallel sections of $E$ over $U$ is $k$. This follows from the well-known result of DeTurck and Kazdan that $g$, being Einstein, is real-analytic in harmonic coordinates: Since the linear connection $D$ is perforce real-analytic, any germ of a $D$-parallel section defined on an open neighborhood of $p\in M$ can be real-analytically continued as a $D$-parallel section along any real-analytic curve that starts at $p$. With all this understood, let's look at what might reasonably described as a 'conformal connected sum' of compact Einstein manifolds in dimension $n\ge 3$. I'll take it to be this: Three compact Einstein $n$-manifolds $(M_i,g_i)$ for $i=0, 1, 2$, with respective Einstein constants $\lambda_i$, together with smoothly embbeded compact submanifolds $N_i\subset M_i$ where $N_0$ is diffeomorphic to $[0,1]\times S^{n-1}$, and $N_i$ is diffeomorphic to the $n$-ball $B^n$ for $i=1,2$, and, finally, where $M_0{\setminus}N_0$ is conformally diffeomorphic (as Riemannian manfolds) to the disjoint union of $M_1{\setminus}N_1$ and $M_2{\setminus}N_2$. (Without lost of generality, one can take this diffeomorphism to be the identity map, so assume this.) Given these hypotheses, there exist smooth functions $u_i: M_i{\setminus}N_i\to (0,\infty)$ for $i=1,2$, such that $g_0 = {u_i}^{-2} g_i$ on $M_i{\setminus}N_i\subset M_0{\setminus}N_0$. Thus, there exist constants $c_i$ such that $(c_i,u_i,\mathrm{d}u_i)$ is a $D_i$-parallel section of $E_i$ on $M_i{\setminus}N_i$. Since $u_i$ is defined on the complement of the smoothly embedded ball $N_i\subset M_i$ in $M_i$, the 1-connectedness of $N_i$ implies that this $D_i$-parallel section extends uniquely as a $D_i$-parallel section of $E_i$ on all of $M_i$. In particular, $u_i$ extends to all of $M_i$ satisfying $\nabla(\mathrm{d}u_i)=(c_i{-}\lambda_i\,u_i)g_i$ for $i=1,2$. Consequently, for $i=1$ or $2$, $u_i$ must be constant unless $(M_i,g_i)$ is the round $n$-sphere. If both $u_1$ and $u_2$ are constant, then by scaling $g_1$ and $g_2$ by constants, we can assume that $g_0 = g_i$ on $M_i{\setminus}N_i\subset M_0{\setminus}N_0$ for $i=1$ and $2$, and hence $\lambda_0=\lambda_1=\lambda_2$. Moreover, because $N_i$ is simply-connected and $g_0$ and $g_i$ are real-analytic, it is not hard to show that the isometric inclusion $M_i{\setminus}N_i\subset M_0{\setminus}N_0$ extends to an isometry $\iota_i:(M_i,g_i)\to (M_0,g_0)$, and the compactness of $M_i$ and $M_0$ then implies that $\iota_i$ is a diffeomorphism, mapping $N_i$ diffeomorphically onto $M_0{\setminus}M_i = N_0\cup (M_{3-i}{\setminus}N_{3-i})$. Thus, $M_0$, $M_1$, and $M_2$ are all isometric and homeomorphic to an $n$-sphere. While $(M_0,g_0)$ is indeed a 'conformal connected sum' of $(M_1,g_1)$ and $(M_2,g_2)$ in the above sense, these ought to be regarded as trivial cases. As remarked at the beginning, though, there are many, many such examples that are not conformally flat. If, say, $u_1$ is constant and $u_2$ is not, we can still reduce to the case that $g_0=g_1$ on $M_1{\setminus}N_1\subset M_0{\setminus}N_0$, and hence $\lambda_0=\lambda_1$. Again, because $N_1$ is simply-connected and $g_0$ and $g_1$ are real-analytic, it follows that the isometric inclusion $M_1{\setminus}N_1\subset M_0{\setminus}N_0$ extends to an isometry $\iota_1:(M_1,g_1)\to (M_0,g_0)$, and the compactness of $M_1$ and $M_0$ then implies that $\iota_1$ is a diffeomorphism, mapping $N_1$ diffeomorphically onto $M_0{\setminus}M_1 = N_0\cup (M_2{\setminus}N_2)$. Meanwhile, $(M_2,g_2)$ must be isometric to the round $n$-sphere, which is conformally flat, implying that $(M_0,g_0)$ is conformally flat. Of course, examples such as this do exist with all of the $(M_i,g_i)$ being conformally flat, and $u_2$ non-constant, in which one attaches a conformally flat bubble to a compact space form, but $(M_0,g_0)$ is isometric (up to a constant multiple) to $(M_1,g_1)$. Finally, if neither $u_1$ nor $u_2$ is constant, then, $(M_1,g_1)$ and $(M_2,g_2)$ are both round $n$-spheres, so $M_0$ is as well.<|endoftext|> TITLE: Bialynicki-Birula decomposition for real analytic varieties QUESTION [8 upvotes]: Let $X$ be a smooth complex algebraic variety endowed with a $\mathbb{C}^*$ action. We assume also to have an antiholomorphic involution $\sigma$ over $X$ such that it anticommutes with the action above i.e $$\sigma(t \cdot x)=\bar{t}\cdot \sigma(x) .$$ Let us also assume that the $\mathbb{C}^*$-action respects the properties in order to get a well behaved Byalinicki-Birula decomposition. For every $x\in X$ there exists $\lim_{t \to 0} t \cdot x$ and the fixed point set is a projective variety. (I think $X$ is called semiprojective in this case). In this case, we know that the Poincare polynomial of $X$ can be expressed as $$P(X,t)=\sum_{i \in I}P(F_i,t)t^{d_i} $$ where $$X^{\mathbb{G}_m}=\bigsqcup_{i \in I} F_i $$ and $F_i$ are connected and $d_i$ are some positive integers associated to the action. The fixed point set $X^{\sigma}$ of the antiholomorphic involution is a smooth real manifold. The hypothesis tells us that $\sigma(X^{\mathbb{G}_m})=X^{\mathbb{G}_m}$. Is it still true somehow that $$P(X^{\sigma},t)=\sum_{i \in I}P(F_i^{\sigma},t)t^{h_i} $$ or not ? Is this known in the literature? I'm totally new to real algebraic geometry. EDIT: The answer below indicates this is not true in general. However, I'd be interested in the following more specific situation. $X$ should be given the structure of an hyperkahler manifold with complex structures $I,J,K$ such that the complex algebraic variety we are looking at is the one induced by $I$. The involution $\sigma$ should then be antiholomorphic with respect to structure $I,J$ and holomorphic with respect to $K$. The $\mathbb{C}^*$ action should be algebraic with respect to the structure induced by $I$ while in general it is not clear what happens with respect to the other structures. The setting to think of is that of Non Abelian Hodge theory: it is known that we have an hyperkahler manifold $M$ such that with respect to $I$ it is the moduli space of stable higgs bundle $(\mathcal{E},\phi)$ of fixed rank $n$ and degree $d$. There we have the action $$t(\mathcal{E},\phi)=(\mathcal{E},t\phi) .$$ With respect to the structure $J$ is the de Rham moduli space of connections and with respect to $K$ is the associated (twisted)character variety. As suggested below, it is likely that hyperkahler structure should help because of parity of dimension of the cells involved somehow, but I wasn't able to prove this neither to find some references. REPLY [4 votes]: There's some simple topology behind why this will work in complex cases, and not consistently in real ones: complex affine spaces are even dimensional as real manifolds, and real affine spaces are often not. So the Poincare polynomial equation you have depends on the differentials vanishing in a spectral sequence, in a way that is just not true if you can have cells of different parities (which the $\mathbb{RP}^n$ example illustrates well). This is why it should work in the hyperkahler case you mention. In this case, the flow-in sets of the fixed point sets are holomorphic and thus even dimensional. EDIT: Sorry to have been vague up above; in part my memory of the technical details has gotten a little misty. The point I was trying to make is this: BB decomposition breaks our manifold up into a finite number of locally closed submanifolds. Let $X_{\leq k}$ be the union of these pieces of dimension $\leq k$. If you have isolated fixed points, this will be the union of the cells of dimension $\leq k$. There's a spectral sequence whose $E_2$ page is given by $H^*(X_{\leq k}/X_{< k})$ (where quotient in the usual topological sense of crushing down to a point). The space $X_{\leq k}/X_{< k}$ is the wedge sum of the one point compactifications of the BB strata of dimension $k$; by the Thom isomorphism, this is the same as the sum of the cohomologies $H^*(F_i)$ of the fixed points sets giving this dimension, shifted by the dimension of the directions flowing in (I think you called these $h_i$'s). Thus, the RHS of your expression for Poincare polynomials is thus the $E_2$ page of this spectral sequence, and you'll get the equality you want if and only if all further differentials are trivial. In the case where each $F_i$ is a point, you're just computing cellular homology for the corresponding cell decomposition. If you're working in the complex world, all of these shifts are even, so if your fixed point set has only even cohomology, then all the differentials have to vanish for parity reasons and life is great. For example, in the cellular homology case, every other term in your complex is trivial, and so all differentials are trivial. It looks to me like this should still be OK if $F_i$ has odd cohomology, but I think you have to use that $H^*(F_i)$ has a pure Hodge structure (so all the differentials have to vanish since they are morphisms of between Hodge structures of the wrong weight). In the real world, this is all shot to pieces; you don't have any parity or Hodge structures to stop differentials from being non-zero, and the case $\mathbb{RP}^n$ shows lots of them won't be. In the hyperkähler situation, it could be salvageable, but you need to think carefully about exactly what hypotheses you have. The thing I worry about in the situation you've mentioned is that the flow-in sets might not be holomorphic with respect to $J$ and $K$, so the fixed points of $\sigma$ could be odd dimensional, and then you're done for.<|endoftext|> TITLE: Connes's absolute geometry and Lurie's spectral algebraic geometry QUESTION [28 upvotes]: Alain Connes and Caterina Consani seem to be currently working on "absolute algebraic geometry", which is a kind of "algebraic geometry over the sphere spectrum" (https://arxiv.org/abs/1909.09796, https://arxiv.org/abs/1502.05585). They seem to be mainly motivated by the idea that this helps with the Riemann hypothesis (see chapter 5 of Connes' essay on the Riemann hypothesis: https://arxiv.org/abs/1509.05576). There is an approach to the Riemann hypothesis via $\mathbb{F}_1$-geometry (Riemann hypothesis via absolute geometry), and Connes and Consani identify $\mathbb{F}_1$ with the sphere spectrum. Completely independently of that Jacob Lurie is developing his "Spectral algebraic geometry", which is also a kind of "algebraic geometry over the sphere spectrum". But here there is absolutely no mention of $\mathbb{F}_1$-geometry, Arakelov theory or the Riemann hypothesis. On the surface level the two theories sound quite similar, because both are "geometry over the sphere spectrum", and both are about studying spaces (Connes: topological spaces; Lurie: $\infty$-topoi) equipped with sheaves of ring spectra (Connes: $\Gamma$-sets with $S$-algebra structure; Lurie: $E_{\infty}$-ring spectra). My question is: How are these two theories related? Can Lurie's spectral algebraic geometry be useful for the Riemann hypothesis? It seems like each spectral scheme of Connes can be made into a spectral scheme of Lurie. If I have an affine spectral scheme à la Connes, then it is $\operatorname{Spec}$ of a $\Gamma$-set with $S$-algebra structure. This then gives rise to a connective spectrum with $E_{\infty}$-structure, and taking $\operatorname{Spec}$ of that gives an affine spectral scheme à la Lurie. For non-affine schemes à la Connes, like $\overline{\operatorname{Spec}(\mathbb{Z})}$ we can take an affine cover, then make each of those affine opens into a spectral scheme à la Lurie, and then glue them back together in Lurie's formalism. So it looks like one can construct the Arakelov-theoretic objects like $\overline{\operatorname{Spec}(\mathbb{Z})}$ that Connes cares about also in Lurie's formalism. Since Lurie's formalism seems much further developed already it is maybe easier to prove the Riemann-Roch theorem that Connes strategy requires in the setting of Lurie. Can the two theories be usefully connected, or are there obvious problems with this? REPLY [12 votes]: I know very little about the absolute/algebraic geometry side, but I think I understand the gist of the category theory going on here. I guess this answer might require one to know a bit of both the stable homotopy story and the $\mathbb{F}_1$-geometry story. tl;dr is that, no, algebra over $\mathbb{S}$ requires that the underlying objects be spectra, which are a derived version of abelian groups, in the sense that "discrete $\mathbb{S}$-modules" are exactly abelian groups; but algebra over $\mathfrak{s}$ is (i) entirely discrete or $1$-categorical and not derived at all (i.e. there's no notion of weak equivalence at all, only of isomorphism) and (ii) even if we took $\mathfrak{s}$-modules to be the discrete version of some framework for derived algebra, it wouldn't be $\mathbb{S}$-modules because $\mathfrak{s}$-modules contain things like hypergroups, tropical rings, and commutative monoids, so their "derived theory" will need to be based in $(\infty,n)$-categories rather than $(\infty,1)$-categories, as $\mathbb{S}$-modules is. I'll try to follow Connes and Consani and say $\mathfrak{s}$-modules instead of $\Gamma$-sets, but I'll almost certainly just start using them interchangeably at some point. So anyway, we've got $\mathfrak{s}$-modules and we've got $\mathbb{S}$-modules. First let's clarify the difference between $\mathfrak{s}$ and $\mathbb{S}$, as, in a sense, they are both the sphere spectrum, but I think this point of view is misleading. What we're calling $\mathfrak{s}$ here is the inclusion $\Gamma\hookrightarrow Set_\ast$, whereas, for the time being at least, $\mathbb{S}$ is the unit of the $\infty$-category of spectra, if you like, or the suspension spectrum of $S^0$. Recall that there is a model structure on $\Gamma$-spaces, called the stable model structure, whose fibrant objects are the very special $\Gamma$-spaces. Namely, those that satisfy the Segal condition, roughly $X[n_+]\simeq X[1_+]^n$ where the right hand side has $n$ factors, and such that $X[1_+]$ is an abelian group (here I'm using $n_+$ to denote the set with $n+1$ elements, one of which is the basepoint). These model "commutative topological groups up to homotopy," and therefore grouplike $\mathbb{E}_\infty$-spaces, and therefore connective spectra. Now, notice that the inclusion $\Gamma\to FinSet_\ast\hookrightarrow Set_\ast\hookrightarrow sSet_\ast$ does not define a special $\Gamma$-space. To see this, consider $1_+\times 1_+=\{\ast,1\}\times\{\ast,1\}=\{(\ast,\ast),(\ast,1),(1,\ast),(1,1)\}\cong3_+\neq 2_+$. So $\mathfrak{s}$ itself is not fibrant in $\Gamma$-spaces with the stable model structure and therefore doesn't correspond to any connective spectrum. Of course the way that we get around this in model categories is that we fibrantly replace. It's a theorem that when we fibrantly replace $\mathfrak{s}$ (thought of as a discrete $\Gamma$-space) we get $\mathbb{S}$. There's a concrete way to describe fibrant replacement in this category, which is to formally invert the so-called Segal maps and group complete $X[1_+]$ (thanks to Chris Schommer-Pries for explaining this part to me). In a very real way this deserves to be called "group completion." The first part is, basically, "commutative monoid completion" since it forces the multiplication to be defined everywhere and to be singly defined (an arbitrary $\Gamma$-space can be thought of as a sort of $\mathbb{E}_\infty$-monoid object in which the multiplication is either multi-valued or only partially defined). The second part is formally adding inverses. The second part is what we usually think of as group completion of course, but if you're starting with something that's not even a commutative monoid, you've got to do a bit more. So, if we believe that "fibrant replacement in the stable model structure" can reasonably be called group completion, then that's exactly what $\mathbb{S}$ is, it's the group completion of $\mathfrak{s}$ (or, if you accept Connes and Consani's contention that $\mathfrak{s}=\mathbb{F}_1$, then $\mathbb{S}$ is the group completion of $\mathbb{F}_1$, which is consistent with interpreting the Barratt-Priddy-Quillen as saying $K(\mathbb{F}_1)\simeq \mathbb{S}$; notice that although Barratt-Priddy-Quillen says that $K(FinSet)\simeq \mathbb{S}$, it's still true that $K(\mathfrak{s}-Mod)\simeq \mathbb{S}$). The category of $\Gamma$-spaces, or $\Gamma$-sets, has a symmetric monoidal structure coming from Day convolution. In this monoidal structure, $\mathfrak{s}$ is the monoidal unit. So everything in $\Gamma$-set deserves to be called an $\mathfrak{s}$-module, or an $\mathbb{F}_1$-module. But now, given an $\mathfrak{s}$-module $M$, one can further say that $M$ itself is a monoid with respect to the Day convolution monoidal structure, and this is what Connes and Consani call an $\mathfrak{s}$-algebra. The key insight here for them is that most of the existing models of "$\mathbb{F}_1$-algebra" are subsumed by this construction. In particular, commutative monoids give $\mathfrak{s}$-modules and semirings give $\mathfrak{s}$-algebras; a large class of abelian hypergroups and hyperrings embed into $\mathfrak{s}$-modules and $\mathfrak{s}$-algebras (for the two preceding examples we're using this fact that $\Gamma$-sets can be thought of as having either a partially defined or multivalued abelian group structure); and perhaps most excitingly, Durov's algebraic monads embed into $\Gamma$-sets, so we get things like tropical rings, and these generalized rings that Durov uses to do things like Arakelov geometry (Connes and Consani have also written about a connection between their $\mathfrak{s}$-modules and Borger's $\Lambda$-ring model for $\mathbb{F}_1$-geometry, but I haven't tried to understand that; it's also not at all clear to me, or anyone else as far as I know, that Lorscheid's blueprint stuff, and all the things related to it, can be described in terms of Connes and Consani's work). I should mention that one can take Durov's model of $\mathbb{F}_1$ and map it (via the so-called assembly map of Lydakis that Connes and Consani use) to $\Gamma$-sets, and one does, as an object, get $\mathfrak{s}$ (I think). But, if I'm not mistaken, the modules over Durov's $\mathbb{F}_1$, in Durov's setup, are less general than $\mathfrak{s}$-modules. One can of course talk about "simplicial $\mathfrak{s}$-modules," but if we think about what a simplicial $\mathfrak{s}$-module is going to be, well it's the same as a $\Gamma$-space (where by space I mean simplicial set) so if one wishes to do homotopy theory in the usual way, I think one is still going to end up back at something like $\mathbb{E}_\infty$-spaces, which loses all the examples that Connes and Consani are interested in (for instance, they seem pretty interested in this adèle class space construction, which gives a hyperring, so gets destroyed if you force the addition to be single-valued). Moreover, recall that if we have an $\mathbb{E}_\infty$-space $X$ then we get the associated $\Omega$-spectrum, i.e. connective $\mathbb{S}$-module, by taking the infinite delooping $HX=\{X,BX,B^2X,B^3X,\ldots\}$. This doesn't even make sense if $X$ isn't a very special $\Gamma$-space. So to do "homotopy theory over $\mathbb{F}_1$," we have to be a bit more clever. I.e., prima facie, "derived $\mathbb{F}_1$-algebra" or "derived $\mathbb{F}_1$-geometry" is going to be very different from spectral algebra(ic geometry). I think the way to get to derived $\mathbb{F}_1$-algebra, and have it be more general than spectral algebra (i.e. avoiding group completion), is that we're going to have to work with something like complicial $\Gamma$-sets, rather than simplicial. The idea here is that simplicial sets are pretty good at modeling $(\infty,1)$-categories, but pretty bad at modeling $(\infty,n)$-categories, and if you want to take the infinite delooping of, say, a commutative monoid $M$, you're going to need $B^nM$ to be an $n$-category of some sort, because you're going to have one object with an $M$'s worth of non-invertible $n$-cells, and complicial sets seem like a decent way to model categories like this (although there are other options too of course, and I haven't worked out any of the details at all). But whatever your category of $\mathbb{F}$-spectra is, there should be a group completion functor landing in $\mathbb{S}$-modules. So if we're following Connes and Consani's proposed path to a proof of the Riemann Hypothesis, we can't work over $\mathbb{S}$, because all of the machinery that Connes and Consani want to use disappears (because we're group completing). In particular, with respect to your plan of taking an affine scheme over $\mathbb{F}_1$ and producing the associated spectral scheme, what you're basically doing is group completing an $\mathbb{F}_1$-algebra to some commutative connective ring spectrum $A$, then taking spec of that. But, again, this is destroying most of the tools Connes and Consani are using or are interested in. If you wanted to start from scratch and say "Let's figure out how to rewrite Deligne's proof of the function fields Riemann Hypothesis by thinking of $H\mathbb{Z}$ as a curve over $\mathbb{S}$," well, I think you're going to struggle. I certainly wouldn't go on the record as saying it's impossible, but the sorts of machinery you'd want to use either don't exist over $\mathbb{S}$ or, at the time being at least, don't make sense (cf. Tyler Lawson's answer here: Dedekind spectra). Although you should take this last assertion of mine with a large grain of salt. Might make sense to ask Lars Hesselholt or Bjørn Dundas or somebody about this.<|endoftext|> TITLE: Uncountable homogeneous rectangles for subsets of $\omega_2\times\omega_2$ QUESTION [11 upvotes]: The background theory here is at least ZFC and probably ZFC+MA+$\neg$CH, if it matters. Here's the question: Suppose that $C\subseteq\omega_2\times\omega_2$. Must there exist uncountable sets $A,B\subseteq\omega_2$ such that either $A\times B\subseteq C$ or $(A\times B)\cap C=\emptyset$? Note that the usual issue with coloring ordered pairs, namely coloring them based on whether they agree with underlying ordering, doesn't present itself here since I'm not asking for $A=B$ or $|A|=|B|=\aleph_2$. My other thought for showing this to be false would be to do a kind of "Bernstein-like" construction, enumerating all pairs $A,B\subseteq\omega_2$ of size $\aleph_1$ and diagonalizing against the products $A\times B$. Assuming MA+$2^{\aleph_0}=\aleph_2$, we'd have that $|\mathcal{P}_{\aleph_1}(\omega_2)|=\aleph_2^{\aleph_1}=2^{\aleph_1}=2^{\aleph_0}=\aleph_2$, but the construction breaks down once $\geq\aleph_1$ many points have been chosen. If this statement turns out to be true, I'd also be interested in higher dimensional forms (sets of n-tuples, etc). Edit: After looking around some more, I think what I am really asking is whether the polarized partition relation $\binom{\aleph_2}{\aleph_2}\to\binom{\aleph_1}{\aleph_1}^{1,1}$ (just a restatement of my question above) is consistent with MA+$\neg$CH. Notably, Hajnal proved this relation under GCH. REPLY [2 votes]: I think Harvey Friedman in "A consistent Fubini-Tonelly theorem for nonmeasurable functions", Illinois Journ. Math., 24(1980), 390--395. proves that if $\aleph_2$ random reals are added to a model of CH, then each $C\subseteq [0,1]\times [0,1]$ has $A,B\subseteq [0,1]$ of outer Lebesgue measure 1 such that either $A\times B\subseteq C$ or else $(A\times B)\cap C=\emptyset$. This implies what you asked.<|endoftext|> TITLE: When can a surface in a 3-manifold be isotoped off a knot? QUESTION [10 upvotes]: Given a closed (perhaps irreducible) 3-manifold $M$ with an embedded surface $S$ and a knot $K$, what conditions allow the two to be isotoped to be disjoint? Obviously a necessary condition is that $[S] \cdot [K] = 0$, but when is this sufficient? This answer gives a condition when $K$ is homotopically trivial, a case I am particularly interested in. Are there any other conditions in this case? How about other cases? (Also would love references on the more general problem of making the algebraic and geometric intersections equal in a 3-manifold via isotopy) REPLY [8 votes]: Let’s assume that the manifold $M$ is irreducible and orientable and the surface $S$ is orientable. This is to avoid 1-sided surfaces. First let’s assume that the surface $S$ is fully compressible. That means that there is a sequence of compressions taking the surface to a collection of 2-spheres, for example a Heegaard surface. Since the manifold is irreducible, these spheres lie inside of a ball (but they might be nested). Thus we can make $K$ disjoint from the balls by an isotopy. Reversing the process of compression, we can iteratively tube these spheres together to get back $S$ up to isotopy. In this process, we can assume that each tube misses $K$ by general position (it is a small regular neighborhood of an arc) and hence get an isotopic surface which misses $K$. So this takes care of the case of fully compressible surfaces, which will be the case in a non-Haken irreducible 3-manifold such as lens spaces and the Poincaré dodecahedral space. In the general case, one compresses down the surface $S$ to a surface $S’$. By a similar argument, $K$ will miss $S$ up to isotopy if it misses $S’$ (although the converse is likely not true). In an irreducible 3-manifold, it must be Haken if $S’$ is not a collection of spheres (so a component is an incompressible surface of genus $>0$). Thus assume $S$ is incompressible. In principle then one can determine if $S$ can be made disjoint from $K$. Let me sketch the case that $M$ and $M-K$ are hyperbolic. In this case, one can enumerate the finitely many disjoint incompressible surfaces of the same genus as $S$ in $M$ and $M-K$. There are algorithms using normal surface theory or geometric group theory. Then there is an algorithm to tell if these surfaces are isotopic (in the non hyperbolic case, there may be infinitely many surfaces of a fixed genus, and one has to work harder). If so, then $K$ can be isotoped disjoint from $M$, otherwise not. Of course, this only gives a sufficient condition in the compressible case by reducing to $S’$. I suspect that there is an algorithm in general like Sam Nead suggests, but I expect that it isn’t written down in general.<|endoftext|> TITLE: Long exact sequence of cohomology from 2-groups QUESTION [6 upvotes]: I am trying to understand the following from Principal Infinity Bundles - General by Nikolaus, Schreiber and Stevenson. So following the reference there to Nikolaus-Waldorf tells us that given any (smoothly separable) Lie $2$-group (let's work in the model of crossed modules for convenience) $[d: H \to G]$, the following is a fiber sequence of Lie $2$-groups $$[\ker d \to 1] \to [H \to G] \to [1 \to \text{coker }d]$$ What this means precisely is explained in more detail in Lemma 2.2 of this paper. In the notation of the problem, this corresponds to $$BA \to G \to H$$ being a fiber sequence of $2$-groups, where $H$ is a group discretely embedded as a $2$-group and $A$ is an abelian group that is being delooped to a $2$-group. Now since $B$ is a right adjoint, it preserves this fiber sequence, and that gives us that $$B^2 A \to BG \to BH$$ is a fiber sequence. So here are my questions: What is this map $c$? In Definition 4.30 of the same paper they give a definition of what a central extension of infinity groups should be involving this $c$, but I don't know how to construct it in this example. In what ways can I present this map? Namely how would I present $B^3 A$? Can I use simplicial groups? Why is this the internal and only nontrivial stage of the postnikov tower? REPLY [9 votes]: $c$ is your crossed module, or 2-group, in a sense. Anything more concrete will depend on a choice of a cocycle description of the pointed set $[BH,B^3A]$. For example, $[BH,B^3A]$ classifies central 2-group extensions $BA \to G \to H$ up to weak equivalence. Hence, one could regard such extensions as the cocycles, and then $c=G$. A probably more concrete cocycle model is given by the smooth group cohomology $H^3_{sm}(H,A)$. By this I mean a smooth version of "Segal-Mitchison cohomology", like considered by Brylinksi. Recent references are: Schommer-Pries, Christopher J., Central extensions of smooth 2-groups and a finite-dimensional string 2-group, Geom. Topol. 15, No. 2, 609-676 (2011). ZBL1216.22005. Wagemann, Friedrich; Wockel, Christoph, A cocycle model for topological and Lie group cohomology, Trans. Am. Math. Soc. 367, No. 3, 1871-1909 (2015). ZBL1308.22006. Blanco, Jaider; Uribe, Bernardo; Waldorf, Konrad, Pontrjagin duality on multiplicative Gerbes, preprint. In the first reference it is described how to obtain the cocycle from the 2-group extension. As usual, this will depend on choices of open covers, sections, etc., so that one cannot expect a canonical formula for $c$. In several specific cases, $H^3_{sm}(H,A)$ can be identified with more familiar sets. For example, if $H$ is compact and $A=U(1)$, we have $H^3_{sm}(H,A)\cong H^4(BH,\mathbb{Z})$. If $G$ is the string 2-group, then $c \in H^4(BSpin,\mathbb{Z})=\mathbb{Z}$ is a generator. If $G$ is the T-duality 2-group, then $c\in H^4(B\mathbb{T}^{2n},\mathbb{Z})$ is $c=\sum_{i=1}^n c_i \cup c_{i+n}$, where $c_i \in H^2(BS^1,\mathbb{Z})$ is the first Chern class of the $i$th factor. Some further examples have been computed in the above-mentioned preprint in Sections 5.4 - 5.6.<|endoftext|> TITLE: Innovations in number theory leading to breakthroughs in statistical mechanics QUESTION [8 upvotes]: Might there be a good reference on the interaction of number theory with statistical physics? I am particularly interested in innovations in number theory that have led to breakthroughs in statistical physics. One such result that I am aware of is the Lee-Yang theorem. Marc Kac apparently used insights from Pólya's analysis of the Riemann Zeta function [2] in order to come up with an early proof of the Lee-Yang theorem(aka Lee-Yang Circle theorem) which states that the zeros of certain partition functions lie on the unit circle [1]. At present, I am more familiar with parallel developments such as the correspondence between Gaussian Unitary Ensembles(GUEs) and Montgomery's pair correlation conjecture [3]. References: Knauf, Andreas (1999), "Number theory, dynamical systems and statistical mechanics", Reviews in Mathematical Physics, 11 (8): 1027–1060, CiteSeerX 10.1.1.184.8685 Pólya, G.: Collected Papers, Vol. II: Locations of Zeros (R.P. Boas, ed.). Cambridge: M.I.T. Press 1974 Montgomery, Hugh L. (1973), "The pair correlation of zeros of the zeta function", Analytic number theory, Proc. Sympos. Pure Math., XXIV, Providence, R.I.: American Mathematical Society REPLY [9 votes]: The paper “The reasonable and unreasonable effectiveness of number theory in statistical mechanics” by George Andrews comes to mind. It is a nice survey that mentions some of the more striking appearances of number-theoretic results in statistical mechanics, such as the Rogers-Ramanujan identities and their connection to the hard hexagon model. It appeared in the book The Unreasonable Effectiveness of Number Theory (Ed. Stefan A. Burr, Amer. Math. Soc. 1993), which has a few other nice essays by different authors. The journal Communications in Number Theory and Physics might have some relevant material.<|endoftext|> TITLE: Are there any convex pentagonal rep-tiles? QUESTION [7 upvotes]: A rep-tile is a shape that can tile larger copies of the same shape. Question 1: Are there any convex pentagons that are also rep-tiles? Remarks: 15 convex pentagonal tiles of the plane are known and none of them appears to be a rep-tile. Assuming this observation is right, one can invoke a proof given in 2017 by Michel Rao - that these 15 are the only convex pentagonal tiles possible - to answer our question in the negative. However, I don't know if Rao's proof has been validated and if there is a simpler (elementary) proof that there are say no convex pentagonal rep-tiles. So, here one is actually asking if there is a simpler proof for a weaker claim. Definition: Let us say a multi-way rep-tile is a polygon P with the property: if P1 and P2 are magnified copies of P and P1 can be tiled with m units of P and P2 can be tiled with n units of P with n> m, then, a layout of n units can form P2 without m of the units in the layout together forming a P1. As shown on this page: https://en.wikipedia.org/wiki/Rep-tile, there are isosceles trapeziums with multi-way property (with m= 4 and n = 9). On the other hand, the square is obviously a rep-tile but not multi-way. Question 2: Are there other convex polygons with this multi-way rep-tile property? REPLY [6 votes]: Question 2 can be answered in the affirmative, at least: there are many triangles with the multi-way rep-tile property. Every triangle has a simple tiling of itself with $k^2$ copies (just take the affine image of the standard equilateral tiling); these are called quadratic tilings by Michael Beeson in his 2010 paper on tiling triangles by congruent triangular tiles. So for a triangle to have this property, it suffices to have any non-quadratic tiling of size $m$, since we can choose a larger quadratic tiling with some $n>m$ tiles: the quadratic tiling has no subtriangles that are not themselves tiled quadratically, as should be obvious from looking at which straight lines can be formed. (This condition is of course also necessary.) So we can then consult the linked paper for a variety of triangles with non-quadratic tilings: this is true of the $(30^\circ,60^\circ,90^\circ)$ triangle and any right triangle whose legs are in a rational ratio, which I think is exhaustive though I haven't read the full paper in a while to confirm this (it focuses mostly on the realizable numbers of tiles, which is related but not quite identical to our question here). If you want to revise your definition to the stricter condition that there are two self-tilings, neither of which contains any other nontrivial self-tiling, there are still triangular examples: consider the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling. As an addendum, the question of convex hexagonal rep-tiles ought to be much easier to resolve, since there are only three classes of convex hexagon which can tile the plane, and I believe none of them can even tile a half-plane (which is a prerequisite to being a polygonal rep-tile, though the proof involves a somewhat messy compactness argument). You can at least use the fact that any rep-tile must have an angle of at most $90^\circ$ to narrow the scope of possible tiles a little. Combined with a negative answer to Question 1, this would effectively reduce the classification problem of Question 2 to the convex quadrilaterals, which seems like it might prove rather difficult - it might end up being easiest to just classify all convex quadrilateral rep-tiles and then work out which ones are multi-way.<|endoftext|> TITLE: Is the affine closure of the basic affine space of a reductive algebraic group Cohen–Macaulay? QUESTION [9 upvotes]: Let $G$ be a reductive algebraic group with choices of Borel subgroup and maximal torus $B \supseteq T$ and unipotent radical $U$ over an algebraically closed field $k$ of characteristic zero. Is the affine closure $\overline{G/U} = \operatorname{Spec}(A)$ Cohen–Macaulay? If so, is there a reference to this fact? If not, is there a reference to a specific counterexample? Note: We have that $A$ is Cohen—Macaulay when $G = \operatorname{SL}_2$ since $\overline{G/U} \cong \mathbb{A}^2$ is smooth. However, in the paper Popov - Contractions of Actions of Algebraic Groups, the author seems to suggest that $A$ is not Cohen–Macaulay in general. Specifically, in section 6, Popov introduces the notion of a stable property of local rings of points of schemes, and explicitly states that "Among the examples of properties (P) of open type given in section 5 [which includes the property of being Cohen–Macaulay], the following properties are stable:" and goes on to give a list of properties which does not include the property of being Cohen–Macaulay. Furthermore, following Popov, an open property is stable if and only if it is closed under the invariants of a reductive group and the property is stable under tensor product of regular functions on the basic affine space of reductive groups, and the invariants of a reductive group acting on a Cohen–Macaulay ring is Cohen–Macaulay by the Hochster–Roberts theorem. If $A$ is not Cohen–Macaulay in general, I am interested in a specific counterexample. Edit: For $\operatorname{SL}_3$, we have $A \cong k[a,b,c,x,y,z]/(ax + by + cz)$ which is Cohen–Macaulay. To see this, note that using this problem set, the algebra of functions $A$ is the quotient of the free polynomial algebra in six variables (three coming from the standard representation $V(\omega_1)$ and three coming from the dual $V(\omega_2)$, respectively) modulo the quadratic relation given by the multiplication rule $V(\omega_1) \otimes V(\omega_2) \cong V(\omega_1 + \omega_2) \oplus V(0) \to V(\omega_1 + \omega_2)$, where the rightmost arrow is the quotient arrow. One can check (for example, this computation is carried out on p178 in Fulton and Harris's book Representation Theory: A First Course) that the kernel of this map is one dimensional and contains $e_1 \otimes e_1^{\ast} + e_2 \otimes e_2^{\ast} + e_3 \otimes e_3^{\ast}$. (Note the relations from the multiplication $V(\omega_1) \otimes V(\omega_1) \to V(2\omega_1) \cong Sym(V(\omega_1))$ are precisely expressing that the variables $a,b,c$ commute, and similarly for $V(\omega_2) \cong V(\omega_1)^{\ast}$.) REPLY [3 votes]: Update: It seems that every such quotient $G/U$ is Cohen-Macaulay. In the same list of properties as above, the author states that, by a theorem of Brion, the ring of functions of $\overline{G/U}$ has rational singularities. Immediately before this statement, the author also cites Elkik - Singularites rationnelles et deformations, which in particular explicitly states (Definition 1, p141) that varieties with rational singularities are in particular Cohen-Macaulay. Unfortunately, I have not been able to track down the referenced theorem of Brion cited. PS: One can also explicitly compute that, when $G := \operatorname{Sp}_4$ and $A$ is the ring of global functions on $G/U$, we have an isomorphism $$A \cong k[a,b,c,d,e,w,x,y,z]/(ad - bc + 2e^2, ex - 2cy + 2aw, bx - ey + 2az, dx - ew + 2cw, dy - bw + ez).$$ Here, the variables $a,b,c,d,e$ correspond to the five one-dimensional weight spaces of the representation $\Gamma_{0,1}$ associated to the long fundamental weight, with $a$ is a highest weight vector, and the first relation corresponds to the projection $$\Gamma_{0,1} \otimes \Gamma_{0,1} \to \text{Sym}^2(\Gamma_{0,1}) \cong \Gamma_{0,2} \oplus \Gamma_{0,0} \to \Gamma_{0,2}.$$ The variables $x,y,z,w$ correspond to the four one-dimensional weight spaces of the representation $\Gamma_{1,0}$ associated to the short fundamental weight, with $x$ a highest weight vector. The last four relations follow from the isomorphism $\Gamma_{1,0} \otimes \Gamma_{0,1} \cong \Gamma_{1,1} \oplus \Gamma_{1,0} \to \Gamma_{1,1}$. (Since $\text{Sym}^2(\Gamma_{1,0})$ is irreducible, the only relation given by $\Gamma_{1,0} \otimes \Gamma_{1,0}$ is that the variables $x,y,z,w$ commute.) This follows just as above, using computations carried out in 16.2 of Fulton and Harris.<|endoftext|> TITLE: Quasisplit but not split semisimple groups QUESTION [6 upvotes]: In section 35.1 of the book "Linear algebraic groups" by Humphreys, it is stated that the quasi-split but not split semisimple groups can only arise when the root system admits a nontrivial graph automorphism. Moreover, it seems that the relative root system in this case is obtained by adjoining the vertices of Dynkin diagram which are sent to each other by the graph automorphism. Also in the wikipedia page on quasi-split groups, it is stated that a quasi-split groups over a field correspond to actions of the absolute Galois group on a Dynkin diagram. In both, there is no reference about this statement. In what paper can I find some theory about this? REPLY [5 votes]: The quasi-split forms of a split reductive group $G$ over a field $k$ are classified by the elements of the first Galois cohomology group of $k$ with values in $Out(G)$ (see Theorem 23.51 of Milne's book Algebraic Groups). So no outer automorphisms means no nonsplit quasi-split forms. When $G$ is semisimple, the outer automorphisms correspond to graph automorphisms of the dynkin diagram.<|endoftext|> TITLE: When is the fiberwise compactification (not) equal to the compactification of the family? QUESTION [5 upvotes]: Suppose $\pi:\mathcal{X}\rightarrow S$ is a smooth family of complex affine varieties (to make things simpler, we can actually assume it is locally trivial). Let $P$ be a smooth projective variety, and assume there is an embedding $\mathcal{X}\hookrightarrow P\times S$ over $S$. Let $\overline{\pi}: \overline{\mathcal{X}}\rightarrow S$ be the projection for the closure of $\mathcal{X}$ in $P\times S$. Is it always true that for every (closed) point $s\in S$, $\overline{\pi^{-1}(s)}=\overline{\pi}^{-1}(s)$? Any references will be greatly appreciated. REPLY [4 votes]: No. Let $S = \mathbb{A}^1_{\mathbb{C}}$ and $\mathcal{X} = \mathbb{A}^1_S \rightarrow S$ be the constant family. Let $\mathcal{Y}$ be the blow-up of the surface $\mathbb{P}^1_S$ at the closed point over $0\in S$ lying at infinity. In other words: $\mathbb{P}^1_S = \mathbb{P}^1\times \mathbb{A}^1$ and $\mathcal{Y}$ is the blow-up at the point $(\infty,0)$. Since blowing-up is an isomorphism away from the exceptional locus, there is an open immersion $\mathcal{X} \hookrightarrow \mathcal{Y}$ of $S$-schemes. Since $\mathcal{Y}$ is irreducible, $\mathcal{X}$ is dense in $\mathcal{Y}$. Since the morphisms $\mathcal{Y} \rightarrow \mathbb{P}^1_S$ and $\mathbb{P}^1_S \rightarrow S$ are projective, the composite $\mathcal{Y} \rightarrow S$ is projective as well. Choose $n \geq 0$ large enough such that there exists a closed embedding $\mathcal{Y} \hookrightarrow P_S$, where $P = \mathbb{P}^n_{\mathbb{C}}$. The closure $\bar{\mathcal{X}}$ of $\mathcal{X}$ in $P_S$ is exactly $\mathcal{Y}$. The fibre of $\mathcal{Y} \rightarrow S$ above $0$ is a union of two projective lines. However, the fibre of $\mathcal{X} \rightarrow S$ above $0$ is just $\mathbb{A}^1$. Therefore $\mathcal{X}_0$ is not dense in $(\bar{\mathcal{X}})_0$, as required.<|endoftext|> TITLE: Are there non-homeomorphic 3-manifolds with the same Turaev-Viro-Barrett-Westbury invariants? QUESTION [7 upvotes]: The Turaev-Viro-Barrett-Westbury invariant of a closed oriented topological $3$-manifold $M$ for a spherical fusion category $\mathcal{C}$ is a number denoted $|M|_{\mathcal{C}}$ computed from (but independent of the choice of) a triangulation of $M$ and $\mathcal{C}$-data labelings. See the book Turaev-Virelizier (2017) on Section 13.1, and the papers Turaev-Viro (1991) and Barrett-Westbury (1996). Question: Are there non-homeomorphic closed oriented $3$-manifolds $M$ and $N$ such that $|M|_{\mathcal{C}} = |N|_{\mathcal{C}}$, for all spherical fusion category $\mathcal{C}$? This answer by Meng Cheng provides a formula for $|M|_{\mathcal{C}}$ with $M=L(p,q)$ a Lens space and $\mathcal{C}=Vec_{G,\omega}$ a pointed fusion category, and shows that it distinguishes $L(7,1)$ from $L(7,2)$, for a good choice of $G$ and $\omega$. REPLY [9 votes]: I asked Alexis Virelizier (coauthor of the book mentioned above) by e-mail. Here is his answer (reproduced with his authorisation): The answer is yes. See Theorem 1.1 (page 2291) in the following paper by Funar: https://www-fourier.ujf-grenoble.fr/~funar/2013geomtopol.pdf<|endoftext|> TITLE: Uniformization under AD QUESTION [7 upvotes]: Can the following uniformization statement be proved by $ZF+AD+DC$? For any binary relation $R\subseteq \mathbb{R}^2$ with the property that $\forall x (\{y\mid R(x,y)\}\mbox{ is at most countable and nonempty})$, then there is a function $f:\mathbb{R}\to \mathbb{R}$ uniformizing $R$. REPLY [7 votes]: Here's an argument under the further assumption that $V=L(\mathbb{R})$. Something like this is presumably recorded somewhere. The main point comes from Steel's paper "Scales in $L(\mathbb{R})$", but I don't find it quite explicitly there. Since $V=L(\mathbb{R})$, we can fix a real $x_0$ such that our relation $R$ is $\mathrm{OD}_{x_0}$. Then for each $x\in\mathbb{R}$, $R_x=\{y\bigm|R(x,y)\}$ is $\mathrm{OD}_{(x_0,x)}$. It suffices to see that $R_x$ contains some element which is $\mathrm{OD}_{(x_0,x)}$, since then we can set $f(x)=$ the least such (in the standard ordering of $\mathrm{OD}_{(x_0,x)}$). In fact, for each real $x$, every countable $\mathrm{OD}_x$ set of reals is $\subseteq\mathrm{OD}_x$. For let's assume $x=\emptyset$; the relativization to other $x$ is routine. Fix a $\Sigma_1$ formula $\varphi$ such that for some ordinals $\alpha<\beta$, the set $$A_{\alpha\beta}=\{y\in\mathbb{R}\bigm|L_\beta(\mathbb{R})\models\varphi(\alpha,\mathbb{R},y)\}$$ is countable; each OD set is of this form. We want to see that each such $A_{\alpha\beta}\subseteq\mathrm{OD}$ (assuming it's countable). Let $A'_{\alpha\beta}=A_{\alpha\beta}$ if $A_{\alpha\beta}$ is countable, and $A'_{\alpha\beta}=\emptyset$ otherwise. Claim: $A=\bigcup_{\alpha\beta}A'_{\alpha\beta}$ is countable. Proof: Otherwise we can define a prewellorder $<^*$ on $A$ such that the set of $<^*$-predecessors of any given real is countable, and then argue like in the proof that there is no $\omega_1$-sequence of pairwise distinct reals. QED (Claim). Note now that $A$ is $\Sigma_1^{L(\mathbb{R})}$ (here the standard notation allows the parameter $\mathbb{R}$ as default, but this is otherwise lightface, i.e. no other parameters). But by Proposition 2.11 of Steel's "Scales in $L(\mathbb{R})$" (applied at some sufficiently large and reflective level $\alpha\in\mathrm{OR}$), and since $A$ is countable, it follows that $A\subseteq\mathrm{OD}$, as desired.<|endoftext|> TITLE: Existence of a somewhat-smooth number in the interval $[x, x+ \log(x)]$ QUESTION [6 upvotes]: Smooth numbers in short intervals have been studied deeply in recent years with results of the form that $\psi(x, x^a)-\psi(x-x^b, x^a) \gg x^{b-\epsilon}$ for all $b>1-a-a(1-a)^3$ when $a \in (\frac 1 2, 1)$ (A.Weingartner, Somewhat smooth numbers in short intervals, https://arxiv.org/pdf/2105.13568.pdf). My question is whether there has been much progress on the short intervals to which at least one (somewhat)-smooth number belongs to? Specifically, should we expect at least one $x^{1-\epsilon}$-smooth number in the interval $[x, x+\log(x) ]$ for some $\epsilon > 0$? REPLY [5 votes]: It seems that the best we can prove is mentioned in the paper you quoted. A heuristic argument suggests that for every fixed $0< \epsilon < 1-\exp(-1/e)=0.307799...$ and all $x>x_0(\epsilon)$, the interval $[x,x+\log x]$ contains at least one $x^{1-\epsilon}$-smooth integer. On the other hand, for every fixed $\epsilon > 1-\exp(-1/e)$, the interval $[x,x+\log x]$ contains no $x^{1-\epsilon}$-smooth integer for infinitely many $x \in \mathbb{N}$. The heuristic argument goes like this: Let $0<\epsilon<1/2$. Assume each integer in the interval $(n, 2n]$ is $n^{1-\epsilon}$-smooth with probability $q=\rho(1/(1-\epsilon))=1+\log(1-\epsilon)$, independent from one another, where $\rho$ is the Dickman function. Let $p=1-q=-\log(1-\epsilon)$. The largest gap between $n^{1-\epsilon}$-smooth numbers is one more than the longest run of numbers that are not $n^{1-\epsilon}$-smooth. The longest such run is almost surely $\sim \log(nq)/\log(1/p)$. The threshold value $\epsilon= 1-\exp(-1/e)$ comes from solving $\log(1/p)=1$ for $\epsilon$. See The Surprising Predictability of Long Runs for a discussion of the distribution of the longest run of successes in $n$ independent Bernoulli trials.<|endoftext|> TITLE: Textbooks or lecture notes about mean field games QUESTION [5 upvotes]: I am looking for a good introductory level textbook (or lecture notes) on mean field games that would be suitable for a graduate course. Ideally, it would include some brief words about optimal control and dynamic programming. Thanks! REPLY [2 votes]: Here's my go to links: PDE flavor notes by Ryzhik: https://math.stanford.edu/~ryzhik/STANFORD/MEAN-FIELD-GAMES/notes-mean-field.pdf Probability flavor notes by Lacker: http://www.columbia.edu/~dl3133/MFGSpring2018.pdf Background material on mean-field interacting processes by Golse: https://arxiv.org/abs/1301.5494 Books: 1). There is a two volume textbook "Probabilistic Theory of Mean Field Games with Applications " by Carmona and Delarue 2). A very comprehensive textbook on MFG is being written by Tamer Basar, but I don't know if it is ready yet.<|endoftext|> TITLE: Why is the thing dual to a "meridian" called a "longitude"? QUESTION [15 upvotes]: A pair of distinguished generators of the fundamental group $\pi_1(\partial(S^3 \setminus K))$ of the boundary torus of a knot complement are usually called the "meridian" and "longitude". However, this terminology has always seemed a bit odd to me: in geography a meridian is a line of longitude, so shouldn't the curve it intersects be a "latitude"? Does anyone know the origin of this language? REPLY [13 votes]: There is a fundamental asymmetry between latitude and longitude on a sphere, whereas on a torus, there is a symmetry between the two generators. This symmetry could motivate the use of nearly synonymous words. The terminology may originate with the paper On the homology invariants of knots by H. Seifert (Quart. J. Math. Oxford (2) 1 (1950), 23–32). The torus $T$ of interest to Seifert is the boundary of a closed tubular neighborhood $V$ of a knot in $\mathbb{R}^3$. Seifert says that a Jordan curve on $T$ that bounds on $V$ (respectively, on $\mathbb{R}^3 - V + T$) but not on $T$ is called a meridian (respectively, a longitudinal circuit). Seifert phrases these definitions in a way that highlights the symmetry, which makes the use of near synonyms seem natural.<|endoftext|> TITLE: Fáry-Milnor theorem for positively curved metrics on $S^3$? QUESTION [6 upvotes]: I'm interested in generalizations the following well-known theorem of Fáry and Milnor. Theorem. (Fáry-Milnor) If a simple closed curve $\gamma \subset \mathbb{R}^3$ is knotted, then the total curvature of $\gamma$ is greater than $4\pi$. I have found some sources that prove essentially the same result when $\mathbb{R}^3$ is equipped with a non-positively curved metric (e.g. the usual hyperbolic metric), maybe satisfying some extra hypotheses (for example, see [1,2]). I'm interested in the opposite situation. Question. Is there a version of the Fáry-Milnor theorem for positively curved metrics on $S^3$? It seems that one could derive some inequality from the usual one using contraction maps to and from Euclidean space and estimating the resulting effect on the total curvature along $\gamma$. However, this would depend strongly on the metrics/contraction maps involved. I would really prefer something metric independent if possible. [1] F. Brickell and C. C. Hsiung. The total absolute curvature of closed curves in Riemannian manifolds. [2] C. Schmitz. The Theorem of Fáry and Milnor for Hadamard Manifolds REPLY [3 votes]: I expand a little on a comment made below Ian Agol's answer, including an improvement proposed by him. Claim: every knot can be realized on the round sphere $\mathbb{S}^3$ with arbitrarily small (but positive) curvature. Indeed, a knot can be presented as the closure of a braid; you an realize the closed braid inside a small tubular neighborhood of a closed geodesic, with the braid $C^2$-close to this core geodesic. This ensures the total curvature is as close to zero as you wish. This is especially easy to represent oneself with torus knots, but actually works for every link, as mentionned by Ian Agol.<|endoftext|> TITLE: Random walk on infinite graph QUESTION [9 upvotes]: Let $G$ be an infinite countable non-oriented connected graph with bounded degrees. Let $X(n)$ be the lazy random walk on $G$ and let $u,v$ be two vertices. Does the ratio $P(X(n)=v)/P(X(n)=u)$ tend to the ratio between the degrees of $u$ and $v$? REPLY [9 votes]: You haven't defined what "the lazy random walk" is. Since you refer to the vertex degrees, I presume that you mean that the transition probabilities are $$ p(x,y) = \begin{cases} \frac12 \;, & \text{if}\; x=y \;,\\ \frac1{2\,\mathbf{deg}\,x } \;, & \text{if}\; x,y\;\text{are neighbours} \;. \end{cases} $$ What you are asking about is known as the strong ratio limit property (SRLP) for Markov chains (one also uses the abbreviation SRLT where T stands for "theorem"). This terminology was introduced by Kingman and Orey in 1964, and the subject was popular in the 60s-70s. The first observation is that if the SRLP holds then the return probabilities $p^n(x,x)$ should decay subexponentially ($\equiv$ the spectral radius of the random walk is 1), which automatically excludes random walks on non-amenable graphs (these are precisely the ones for which the spectral radius is strictly less than one). An example of a non-amenable graph is provided by homogeneous trees (mentioned in one of the comments). In what concerns amenable graphs, the SLRP has been proved for random walks on amenable groups (in particular, for simple random walks on the Cayley graphs of finitely generated amenable groups) by Avez in 1973. I am not aware of any results for random walks on arbitrary amenable graphs. However, there is a counterexample to the SLPR for a general (not reversible) Markov chain due to Freeman Dyson and described on pp. 55-56 of Kai Lai Chung's book Markov chains with stationary transition probabilities.<|endoftext|> TITLE: Can second-order logic identify "amorphous satisfiability"? QUESTION [7 upvotes]: Recall that a set is amorphous iff it is infinite but has no partition into two infinite subsets. I'm interested in the possible structure (in the sense of model theory) which an amorphous set can carry. For example, if $T$ is a complete first-order theory with an amorphous model then $T$ must be strongly minimal; however, the converse fails badly (consider $Th(\mathbb{C};+,\times)$). I'm curious whether second-order logic does a better job. For $T$ a complete second-order theory with no finite models, say that $T$ is second-order strongly minimal iff every second-order-definable subset of every model $\mathcal{A}\models T$ is either finite or cofinite in $\mathcal{A}$. For example, $Th(\mathbb{C};+,\times)$ is not second-order strongly minimal since $\mathbb{Z}\subset\mathbb{C}$ is second-order definable. EDIT: by "complete" I mean "maximally satisfiable:" a second-order theory $T$ is complete iff it is satisfiable and for each second-order sentence $\varphi$, either $\varphi\in T$ or $\neg\varphi\in T$. To be reasonably concrete, my main question is the following: Question 1: Is it consistent with $\mathsf{ZF}$ that every second-order strongly minimal theory has an amorphous model? (More broadly, I'm interested in whether there is a reasonably natural logic $\mathcal{L}$ such that consistently with $\mathsf{ZF}$ every "$\mathcal{L}$-strongly-minimal" theory has an amorphous model. Second-order logic just seems like a good candidate at the moment.) There's also a "virtual" version of the question which makes sense even in the presence of choice: Question 2: Is it consistent with $\mathsf{ZFC}$ that for every second-order strongly minimal theory $T$, there is some (set) symmetric extension in which $T$ has an amorphous model? The exact relationship between these two questions isn't clear to me. That said, an affirmative answer to question 2 seems much more plausible than an affirmative answer to question 1. REPLY [6 votes]: Here's a monadic example. The second-order theory of the vector space $\mathbb F_2^{\oplus\omega}$ as a vector space over $\mathbb F_2$ is second-order strongly minimal because given a finite subspace $X$ of any model of this theory, any two points outside $X$ are related by an automorphism of the whole space fixing $X.$ There is a subtlety here pointed out by Emil Jeřábek in the comments. For Question 2, where we check strong minimality in ZFC, it is of course true that any automorphism of a subspace extends to the whole space. In ZF, this might not be true, but as Emil suggested we can use the axiom that there are complemented subspaces: $(\forall Y)(\exists Z)$ such that if $Y$ is a subspace then: $Z$ is a subspace with $Y\cap Z=\{0\},$ and $(\forall v)(\exists y\in Y)(\exists z\in Z)(v=y+z).$ This theory contains the second order axiom "($\exists P$) $P$ is a subspace of codimension 2", i.e. $P$ is closed under addition and there exists $v\notin P$ such that every $w\notin P$ satisfies $w+v\in P.$ So it can't have an amorphous model.<|endoftext|> TITLE: How to solve the following ODE with a parameter? QUESTION [6 upvotes]: I am considering the following ODE \begin{equation} \begin{split} &\frac{d^2}{dy^2}u + \frac{\alpha}{(1+y^2)^{\frac{r}{2}}}u = \delta(y)\\ &\lim_{|y|\to \infty}u(y) = 0. \end{split} \end{equation} Here $\alpha$ is a constant and $r > 0$ and $\delta(y)$ is the Dirac delta function. Is it possible to solve this ODE? I expect the solution to be singular in $\alpha$ as $\alpha$ approaches zero. If we can't get the explicit expression of the solution, can we analyze its singularity in $\alpha$ asymptotically as $\alpha \to 0$? For example, when $r = 0$, the solution behaves like $\frac{e^{-|y|\sqrt{-\alpha}}}{\sqrt{-\alpha}}$ which is singular in $\alpha$. REPLY [7 votes]: This is essentially an elliptic second-order ODE in $(0,\infty)$: $$ (1 + y^2)^{r/2} u''(y) = (-\alpha) u(y) , $$ with boundary conditions $u'(0) = -1$ and $u(\infty) = 0$. This kind of problems are well-studied, the corresponding theory is known as Krein's spectral theory of strings. The value $h(\alpha)$ of $u$ at $0$ for the given parameter $\alpha$ is the corresponding spectral function, and $h(-\alpha)$ is a Stieltjes function of $\alpha$ (other names: Nevanlinna–Pick function, Herglotz function). The relation between the coefficient (here: $(1+y^2)^{-r/2}$, roughly known as the string) and the spectral function $h$ is not explicit, but some comparison results are known. I do not have time now to search for the relevant results, but I would start by looking at the following paper: S. Kotani and S. Watanabe, Krein's spectral theory of strings and generalized diffusion processes, in: Functional Analysis in Markov processes (Katata/Kyoto, 1981), Lecture Notes in Math. 923, Springer, Berlin, 1982, 235–259, DOI:10.1007/BFb0093046. Another excellent reference with a chapter on Krein's theory is R. Schilling, R. Song, Z. Vondraček, Bernstein Functions: Theory and Applications. De Gruyter, Studies in Math. 37, Berlin, 2012, DOI:10.1515/9783110269338. Edit: some additional details. Strictly speaking, our case corresponds to the string $m(dy) = (1 + y^2)^{-r/2} dy$, which is commonly identified with its distribution function $$m(y) = m([0, y)) = \int_0^y (1 + s^2)^{-r/2} ds.$$ The spectral function can be defined in a number of equivalent ways: one of them inolves the integral of $(\phi_N)^{-2}$, where $\phi_N$ is the solution with "Neumann" initial condition $\phi_N(0) = 1$, $\phi_N'(0) = 0$. The one I am most familiar with defines it to be the reciprocal of $-\phi'(0)$, where $\phi$ is the solution with boundary conditions $\phi(0) = 1$ and $\phi(\infty) = 0$. This is clearly equivalent to the definition as $-u'(0)$ with $u$ as in the statement of the problem. I failed once to find a simple reference for the above calculation in analytical terms, so together with Jacek Mucha we included a very brief discussion in the appendix to our paper (see Section A.3 therein): J. Mucha, M. Kwaśnicki, Extension technique for complete Bernstein functions of the Laplace operator, J. Evol. Equ. 18(3) (2018): 1341–1379 DOI:10.1007/s00028-018-0444-4.<|endoftext|> TITLE: Reading list recommendation for a hep-ph student to start studying QFT at a more mathematically rigorous level? QUESTION [24 upvotes]: Edition On July 15 2021, the description of the question has been considerably modified to meet the requirement of making this question more OP-independent and thus more useful for general readers. But the central point is the same. One sentence summarization For a student initially working on a more phenomenological side of the high energy physics study, what is the recommendation of introductory reading materials for them to dive into a more mathematically rigorous study of the quantum field theory. Elaboration "phenomenology side of the high energy physics study" basically means that when this student uploads their article to arXiv, they will, in principle, choose hep-ph as the primary archive. This background also indicates that this student has, in general, already learned at least a. Undergraduate level physics classes; b. QFT from the first half of Peskin and Schroeder's classical textbook, or, maybe, Greiner's Field Quantization; c. Complex analysis like those covered in the first five chapters of Conway's Functions of One Complex Variable I; d. Special function; e. Elementary set theory; f. General topology like those covered in Amstrong's Basic Topology; g. Elementary vector space theory with no details about the operator algebra; h. Some elementary conclusions from group theory, with a focus on the Lie group and Lorentz group; i. Enough particle physics so that he/she is able to understand what the summary tables of the PDG are talking about, or he/she at least knows how to find the definition of some of those unknown symbols. "more mathematically rigorous study of quantum field theory" may have different meanings for different people. Just to give an example, Prof. Tachikawa from IPMU had a talk titled "Mathematics of QFT, by QFT, for QFT" on his website (https://member.ipmu.jp/yuji.tachikawa/transp/qft-tsukuba.pdf), which the OP find really tasteful, of course, from a rather personal point of view. One may agree that all those studies covered there may be considered mathematically rigorous studies of quantum field theory. Or, one may agree on only part of them. Or, maybe the answerer thinks that some other researches also count. That is all OK. But, please specify your altitude before elaborating your answer. The answerer may also want to specify explicitly for which subfield they is recommending references. In the OP's understanding, "more mathematically rigorous study of quantum field theory" basically covers those topics a. Algebraic/Axiomatic QFT, Constructive QFT or the Yang-Mills millennium problem, which tries to provide a mathematically sounding foundation of QFT. b. Topological QFT or Cconformal QFT, which utilizes some fancy mathematical techniques to study QFT. c. Even some math fields stemming from QFT like those fields-awarding work done by Witten. It is best that those "introductory reading materials" could focus on the mathematical prerequisites of that subfield. It will be really appreciated if the answerer could further explain why one has to know such knowledge before doing research in that subfield. Answers may choose from a really formal vibe or a physics-directing vibe when it comes to the general taste of their recommendation and may want to point out that difference explicitly. It is also a good recommendation if the answerer lists some review paper, newest progress, or commentary-like article in that subfield so that people wanting to dive into this subfield could get a general impression of that subfield. Additional comments In general, this is not a career advice post. But, if the answerer wants to say something about some real-life issues of both hep-ph or hep-th/math-ph research, or the transition between those two fields, please feel free and don't be shy. I understand that there already exist some good similar recommendation list posts on this site, but, at least as I know, none of them focuses on the transition from hep-ph to hep-th/math-ph, which is the central point of this post. If there does exist such a post and someone has already offered a quite good answer, please commet. Another way of formulating this question is the following. Please give a list of the introductory courses that a graduate student majoring in mathematical physics should learn, or maybe a list of those papers an advisor would recommend to a first-year graduate student majoring in mathematical physics. But, keep in mind that this student has already had accomplished a Master's level of hep-th study. This question has also been crossposted on Math SE (https://math.stackexchange.com/questions/4194712/how-should-i-start-to-study-qft-at-a-somewhat-mathematically-rigorous-level-lik). REPLY [12 votes]: As Igor said, it's a bit late for that. If you just finished undergrad and you discovered a passion for QFT from a rigorous mathematical standpoint, what you should do, for example, is apply for the math PhD program at UVa and then do a thesis with me ;) Without expert help, trying to get into the subject by reading material in topology, differential geometry or algebraic geometry, as preparation for introductory courses that should hopefully equip you with the necessary mathematical background to finally be ready to start thinking about a research problem related to rigorous QFT, sounds like a recipe for not getting anywhere. That being said, if you are serious about your goal, here is what you can do. Pick one of the QFT models that have been constructed rigorously and study that proof of existence until you understand it completely. I would recommend a result where the method is sufficiently general so by learning an example you actually get a feel for the general situation. This is in line with Hilbert's quote about the example that contains the germ of generality. In the present situation, this narrows the pick to a proof of construction of a QFT model using renormalization group methods (If you wonder why, see edit below). As a rule, Fermionic models are considerably easier that Bosonic models, when it comes to rigorous nonperturbative constructions. I therefore think the best pick for you would be the article "Gentle introduction to rigorous Renormalization Group: a worked fermionic example" by Giuliani, Mastropietro and Rychkov. It is pretty much self-contained. If you know the Banach fixed point theorem, you're in business. In the article, they construct an RG fixed point, which in principle corresponds to a QFT in 3d which conjecturally is a conformal field theory. What they do not do is construct the correlations from the knowledge of that fixed point. As a consequence, they also do not prove conformal invariance of correlations. So here are two contemporary research problems for someone who did the "homework assignment" I just mentioned and would like to go further and prove something worthwhile. If you prefer Bosonic models, then the other pick I would recommend is the article "Rigorous quantum field theory functional integrals over the p-adics I: anomalous dimensions" by Chandra, Guadagni and myself. It is a toy model for the Bosonic analogue of the example considered by Giuliani, Mastropietro and Rychkov. The spacetime on which the fields are defined has a hierarchical structure which facilitates the multiscale analysis. There we rigorously constructed the RG fixed point and also the correlations for two primary fields, the elementary scalar field and its square. What we did not prove is conformal invariance, which is also conjectured to hold. The definition of conformal invariance in this setting is the same as in my previous answer: What is a simplified intuitive explanation of conformal invariance? Namely, just change Euclidean distance to the maximum of the $p$-adic absolute values of the components. Our article is also self-contained and also needs the Banach fixed point theorem together with some complex analysis and very minimal knowledge of $p$-adic analysis. The basics of $p$-adic analysis needed take a weekend to learn. Edit addressing the OP's three new questions in the comments: Why didn't I mention say TQFT or other approaches? Topological QFTs (the stress tensor vanishes completely) is a small subset of Conformal QFTs (the trace of the stress tensor vanishes) which themselves form a tiny subset of general QFTs. The study of these particular cases is certainly interesting but this relies on different tools that are specific to these particular cases and once you invest in learning these tools you will most likely be stuck with these particular cases for the rest of your research career. I proposed RG methods because I believe they cast a wider net and also should broaden your understanding of the subject. I think it would be easier to later specialize in say TQFT if that is where your taste leads you, rather than go the other way around: first develop expertise in say TQFT, and then learn some other method like the RG in order to escape from the narrow realm of TQFTs and study QFTs which are not topological. Next, a comment on "does this mean that all other method than rigorous renormalization group method have failed currently to construct a sensible QFT satisfying Wightman’s axioms (or its equivalent)?". The quick answer is no, this is not why I said you should choose an article which uses RG methods. I wrote my answer not just for you but also for other young people interested in rigorous QFT, from the hypothetical perspective of a PhD advisor talking to a beginning PhD student, starting the PhD thesis work now in 2021. As far as what method has been successful in proving the Wightman axioms for specific models, there are several. RG is one, as in the work of Glimm, Jaffe, Feldman, Osterwalder, Magnen and Sénéor on $\phi^4$ in 3d, and the work of Feldman, Magnen, Rivasseau and Sénéor for massive Gross-Neveu in 2d. The earlier work of Glimm, Jaffe, Spencer on $\phi^4$ in 2d used a single scale cluster expansion. Methods based on correlation inequalities and the more recent ones based on stochastic quantization typically allow you to prove most of the Osterwalder-Schrader axioms but not all, e.g., not Euclidean invariance. Suppose we were having this discussion about geometric invariant theory instead of QFT, would you be asking me: I understand that Hilbert is the pioneering figure in the field of GIT, shouldn’t I start with some classical work at first? Sure, you could go read his 1893 Ueber die vollen Invariantensysteme but this might be more appropriate for a PhD in the history of math, rather than for doing research in this mathematical area now in 2021. More or less yes, although I do not like your choice of words when you said that Axiomatic QFT "put QFT on the rigorous mathematical ground" in contrast to Constructive QFT which merely tries to "propose actual QFT models which satisfy those axioms". The ones who propose models are theoretical physicists. They come to you and say: here look at this Lagrangian it describes a model which is important for physics. Then you, say the constructive QFT mathematician, your job is 1) to prove that this model makes sense rigorously, by controlling the limit of removing cutoffs with epsilons and deltas but certainly no handwaving, and 2) to prove the limiting objects satisfy a number of properties like the Wightman axioms. Then the axiomatic QFT person can come and say: as a consequence of satisfying the axioms, here are these other wonderful properties that your model also satisfies, by virtue of this general theorem I proved the other day. Hope this clarifies the logical articulation of these different subareas of rigorous QFT.<|endoftext|> TITLE: Is it true that this ideal must be principal? (proof verification) QUESTION [9 upvotes]: Let $L/K$ be a (abelian, Galois) quadratic extension of number fields with $\text{Gal}(L/K)$ generated by $\sigma$ and $\mathfrak{p} = \alpha\mathcal{O}_K$ a principal prime ideal of $\mathcal{O}_K$. Assume $\mathfrak{p}$ splits as $\mathfrak{P} \sigma(\mathfrak{P})$ in $\mathcal{O}_L$ and that $\alpha = \beta \sigma(\beta)$ for $\beta \in L^\times$ (so $\beta$ may not be integral though $\alpha$ is). I would like to show that $\mathfrak{P}$ is principal (and possibly that $\mathfrak{P} = \beta \mathcal{O}_L$). $\textbf{My attempt}$: It seems clear that if $\beta$ is not integral it must generate a fractional ideal of the form $$\beta\mathcal{O}_L = \frac{\mathfrak{P}I}{\sigma(I)}$$ for some $I \subset \mathcal{O}_L$. We can assume $\mathfrak{P}I \cap \sigma(I) = \mathcal{O}_L$, i.e. that the numerator and denominator are simplified: we can cancel common factors of $I$ and $\sigma(I)$ so that $I \cap \sigma(I) = \mathcal{O}_L$, and if $\mathfrak{P} \cap \sigma(I) = \mathfrak{P}$ then $\sigma(\mathfrak{P})$ divides $I$ and we get that $$\beta\mathcal{O}_L = \frac{\sigma(\mathfrak{P}) I'}{\sigma(I')}$$ where $I' = I/\sigma(\mathfrak{P})$. So, WLOG we can take the first expression for $\beta\mathcal{O}_L$. If $\beta$ is not integral we can find an integer $d \ne 1$ such that $d\beta$ is integral ($\beta$ is just a linear combination over the basis for $L$ and $d$ is the lcm of the denominators of the coefficients). Then $$ d\beta\mathcal{O}_L = (d)\frac{\mathfrak{P}I}{\sigma(I)} \subset \mathcal{O}_L.$$ Since this is integral, $\sigma(I)$ divides $(d)\mathfrak{P}I$. As $\mathfrak{P}I \cap \sigma(I) = \mathcal{O}_L$, $\sigma(I)$ divides $(d)$. But so does $I$ (as $\sigma$ fixes $d$). If $I$ is nontrivial then this contradicts $I \cap \sigma(I) = \mathcal{O}_L$, so either $I$ is trivial or $d = 1$. In either case we have $\beta\mathcal{O}_L = \mathfrak{P}$ as desired. This proof feels awkward and I suspect either it is wrong or just overly complicated. I'd appreciate any feedback! REPLY [10 votes]: I claim that under the given circumstances $\mathfrak{P}$ is not necessarily principal, i.e., the statement claimed in the question is wrong. Here is a counterexample. Consider $K = \mathbb{Q}$ and $L = \mathbb{Q}[\sqrt{-47}]$. Let $\alpha = -3$ and $\beta = \frac{1}{4}(1 \pm \sqrt{-47})$. Then the minimal polynomial of $\beta$ is $X^2 - \frac{1}{2} X + 3$. In particular, the norm of $\beta$ is $\beta\sigma(\beta) = -3$. (Note that $\beta$ is not integral, as its minimal polynomial is not integral.) Then with the help of SAGE one computes the factorization in $L$: $3\mathcal{O}_L =(3, \frac{\sqrt{-47}}{2} - \frac{1}{2})(3, \frac{\sqrt{-47}}{2} + \frac{1}{2})$. Say $\mathfrak{P}=(3, \frac{\sqrt{-47}}{2} - \frac{1}{2})$. Also, SAGE tells that the class group of $L$ is of order $5$ with generator $\mathfrak{a} = (2, \frac{\sqrt{-47}}{2} + \frac{1}{2})$, and that $\mathfrak{P} = \mathfrak{a}^2$ in the class group. Thus $\mathfrak{P}$ is not principal. I believe that a mistake in the posted proof is contained in the sentense "If $I$ is non-trivial, then ...". Indeed, that $I$ and $\sigma(I)$ divide (d) only means that all $p$-adic valuations of $I$ and $\sigma(I)$ are smaller than those of $(d)$. Why this should imply that $I \cap \sigma(I) = \mathcal{O}_L$?<|endoftext|> TITLE: Finite coproducts commute with which limits in Set? QUESTION [7 upvotes]: It is well known that (small) coproducts commute with connected limits in $\mathbf{Set}$. With which class of limits do finite coproducts commute? Ideally, we should furthermore like to know whether the class of finite coproducts is closed [1] in the sense that the class of finite coproducts is precisely the class commuting with the given class of limits in $\mathbf{Set}$. [1] Notes on Commutation of Limits and Colimits, Bjerrum–Johnstone–Leinster–Sawin (2015) REPLY [11 votes]: Indeed, let $D$ be a category. The canonical functor $D \to \pi_0(D)$ is both cofinal and coinitial. Therefore, if finite coproducts commute with $D$-limits in a category $\mathcal C$, then finite coproducts commute with $\pi_0(D)$-limits. And it is easily seen that the only discrete limit shapes with which finite coproducts commute in $Set$ are the singleton ones. So as Tom Goodwillie supposed, the only limit shapes with with finite coproducts commute in $Set$ are the connected ones. Finite coproducts are not closed -- they don't include splitting of idempotents, which commutes with any limit whatsoever. But I believe that the finite disjoint unions of absolute colimit shapes do form a closed class.<|endoftext|> TITLE: Minimal vs characteristic polynomial of geometric Frobenius QUESTION [7 upvotes]: Assume $X$ is a smooth projective variety over $\overline{\mathbf{F}}_p$ and fix a prime $\ell\neq p$. Let $F_i$ be the geometric Frobenius on $\ell$-adic cohomology $$H^i_{\rm ét}(X,\overline{\mathbf{Q}}_{\ell})$$ for fixed $i\ge 0$. What relation is expected to hold between the minimal polynomial $m_{F_i}(T)$ of $F_i$ and the characteristic polynomial $P_i(T)$ of $F_i$? (apart from $m_{F_i}\mid P_i$) Are $m_{F_i}(T)$ and $P_i(T)$ conjectured to agree? If so, is this a known result? Does $m_{F_i}(T)$ depend on the Weil cohomology theory chosen to compute $P_i(T)$? We know from Deligne's work on the Weil conjectures, that we have $P_i(T)\in\mathbf{Z}[T]$, $P_i(T)$ does not depend on the chosen Weil cohomology, and its roots are of the form $q^{-i/2}\rho$ for $\rho$ an algebraic number whose complex absolute value, for any complex embedding $\overline{\mathbf{Q}}\subset\mathbf{C}$, is one. I'm mostly interested to understand to what extent $m_{F_i}(T)$ is, or expected to be, intrinsic. REPLY [6 votes]: It's conjectured -- see e.g. this question -- that the Frobenius is always semisimple, so its minimal polynomial is the radical of its characteristic polynomial (the product of its distinct linear factors, each with multiplicity 1). So $m_{F_i}$ should be independent of $\ell$. This also shows that $m_{F_i}$ is different from $P_i$ iff $P_i$ has a root of multiplicity $> 1$. This can certainly occur, e.g. consider a supersingular elliptic curve over $\mathbf{F}_{p^2}$.<|endoftext|> TITLE: How did Euler calculate $i^i$? QUESTION [6 upvotes]: There is a well-know quotation of Euler in a letter from 1746 to Goldbach: Letztens habe ich gefunden, dass diese expressio $\sqrt{-1}^{\sqrt{-1}}$ einen valorem realem habe, welcher in fractionibus decimalibus $=0,2078795763$, welches mir merkwürdig zu seyn scheinet. For the principal value of the logarithm the expresssion is $i^i=\exp(i \log(e^{i\pi/2}))= e^{-\pi/2}$. My question is, how did Euler calculate this with such an accuracy? REPLY [10 votes]: On the numerical value of $i^i$ and Historical notes on the relation $e^{-\pi/2}=i^i$ describe how these accurate computations can be performed with logarithmic tables. Euler described how he arrived at the identity in a paper read at the Berlin Academy in 1746, giving more decimals (13) than in the letter to Goldbach. A later calculation by Gauss computed 35 decimal places. Euler did not present his computation, but Gauss did [source].<|endoftext|> TITLE: Metric TSP with integer edge cost QUESTION [7 upvotes]: Given a metric TSP with integer edge cost upper-bounded by a constant $C_{\max}$, can we find an poly-time algorithm solving this TSP instance? REPLY [7 votes]: No polynomial-time algorithm exists, unless P=NP. Indeed, even for TSP instances where all distances are $1$ or $2$ (note that these automatically satisfy the triangle inequality), Engebretsen and Karpinski showed that it is NP-hard to approximate TSP within a factor of $\frac{741}{740} - \epsilon$, for any $\epsilon > 0$.<|endoftext|> TITLE: Decidability of word problem for group admitting certain action QUESTION [13 upvotes]: Let $G$ be a group acting highly transitively (and faithfully) on a set $S$. Suppose that $G$ is finitely presented, and that every stabilizer in $G$ of a finite subset of $S$ is finitely generated. I think I can prove that $G$ embeds in a finitely presented simple group, which in particular implies $G$ has decidable word problem, but I'd like a better understanding of why such a $G$ should have decidable word problem. Is there a pre-existing (and/or more direct) reason that a group admitting such an action should have decidable word problem? (Edit: Here an action of a group $G$ on a set $S$ is called highly transitive if for all $n\in\mathbb{N}$ the induced action of $G$ on the set of $n$-tuples of distinct elements of $S$ is transitive.) REPLY [7 votes]: Yes. There is such a reason. I will write a subset of $G$ is RE if the set of those words over the generators for $G$ which represent elements of the subset is recursively enumerable. As IJL argued, since $G$ is finitely presented the subset $\{1\}$ of $G$ containing only the identity is RE. It remains to show that $G \setminus \{1\}$ is RE. Fix $s$ and $t$ in $S$ and let $H$ be the stabiliser of $s$. Since $H$ is finitely generated $H$ is RE. Let $f$ be some element of $G$ which moves $s$ and fixes $t$. Let $M$ be the set of elements of $G$ which conjugate $f$ into $H$. Note that $M$ is RE and $1 \notin M$ but any element $g$ of $G$ with $(t)g = s$ is in $M$. Let $N$ be the set of elements of $G$ conjugate to some element of $M$. Note that $N$ is RE. $G$ acts $2$-transitively on $S$ so $N$ is in the set of elements of $G$ which move at least one point of $S$. Which is to say that $N = G \setminus \{1\}$. In short: $G \setminus \{1\} = \left\{ g \in G \mid \textrm{there exists } h \in G \textrm{ with } f^{\left(g^h\right)} \in H \right\}$.<|endoftext|> TITLE: "The boat is not longer than it is." QUESTION [22 upvotes]: Bertrand Russell, I believe, somewhere presents a joke (if I remember correctly). Someone is shown the boat of another, and the first says: "I thought that your boat is longer than it is." The owner replies: "No, my boat is not longer than it is." Does someone know the reference? REPLY [19 votes]: Bertrand Russel, On denoting Mind, October 1905, pages 479-493. When we say: "George IV wished to know whether so-and-so", or when we say "So-and-so is surprising" or "So-and-so is true", etc., the "so-and-so" must be a proposition. Suppose now that "so-and-so" contains a denoting phrase. We may either eliminate this denoting phrase from the subordinate proposition "so-and-so", or from the whole proposition in which "so-and-so" is a mere constituent. Different propositions result according to which we do. I have heard of a touchy owner of a yacht to whom a guest, on first seeing it, remarked, "I thought your yacht was larger than it is"; and the owner replied, "No, my yacht is not larger than it is". What the guest meant was, "The size that I thought your yacht was is greater than the size your yacht is"; the meaning attributed to him is, "I thought the size of your yacht was greater than the size of your yacht".<|endoftext|> TITLE: On universally closed morphisms of reduced schemes QUESTION [5 upvotes]: In this question I'd like to examine some properties of universally closed morphisms. The question is self-contained. It can also be seen as a follow-up to this question. Let $R$ be a discrete valuation ring, $X$ and $S$ two $R$-flat and universally closed $R$-schemes. Let $f : X\to S$ be a universally closed morphism. Assume the special fiber of $f$ is an isomorphism. If $X$ and $S$ are reduced, is $f$ an isomorphism? If $X$ and $S$ have reduced fibers, is $f$ an isomorphism? Without either of the reducedness assumptions, the answer is "no", as shown by the example provided here. My expectation is that the answer is still "no" to both questions, but an example seems subtle to produce. Remarks In these remarks we further assume $f$ is flat, to see what more can be said in this case. After all, should the questions have positive answer, $f$ would have to be flat. The remarks are listed increasing the strength of the additional assumptions, to see what can be said in each ever-more-special case. From flatness we deduce $f$ is surjective, since generizations lift along $f$ (by flatness), and every point in $S$ specializes to a point in the special fiber (by universal closedness of $S$, if $s\in S$ is contained in the generic fiber, it maps to the generic point in $\text{Spec}(R)$. This specializes to the closed point in $\text{Spec}(R)$, and specializations lift along $S\to \text{Spec}(R)$ by universal closedness, so there is some $s_0\in S$ contained in the special fiber, to which $s$ specializes). If $f$ is, in addition, separated and locally of finite type, then it is also proper, since $f$ is quasi-compact by universal closedness. It is therefore an isomorphism (it is quasi-finite, hence finite, hence finite flat of degree $1$). The question is interesting only if $f$ is not locally of finite type. To summarize, the case when $f$ is flat and separated is already unanswered but one has access to more info on $f$. A non-separated example, flat or not, to answer the question in the negative, would be already what I'm looking for. A flat separated example, if any, would be the best way to settle this puzzle, and I expect it is very subtle to find. REPLY [2 votes]: Let $R$ be $\mathbb Z_p$ (or any other dvr). Let $S$ be obtained by gluing two copies of $\mathbb P^1_{\mathbb Z_p}$ away from the $0$-point in the special fiber, i.e. away from the vanishing locus of the ideal $(p,x)$ in local coordinates away from $\infty$. Let $X$ be obtained by gluing two copies of $\mathbb P^1_{\mathbb Z_p}$ away from the $0$-section, i.e. away from the vanishing locus of the ideal $x$. These are obtained by gluing two flat and universally closed schemes along an open set, hence flat and universally closed. Then we can map $X \to S$ by mapping each copy of $\mathbb P^1_{\mathbb Z_p}$ to a different copy of $\mathbb P^1_{\mathbb Z_p}$, which is compatible with the gluing since we glue along a larger open set in $S$ than in $X$. This map is an isomorphism in the special fiber since the constructions of $X$ and $S$, restricted to the special fiber, are identical. But it is manifestly not an isomorphism, e.g. because it makes a non-separated generic fiber to a separated one.<|endoftext|> TITLE: Renormalization in physics vs. dynamical systems QUESTION [18 upvotes]: I am studying complex dynamics, so to me renormalization of a dynamical system means something like a rescaled first-return map on (a subset of) the underlying space. I understand that in quantum field theory there are also notions of renormalization and universality having to do with eliminating divergences in certain perturbative integrals, and in statistical physics, where they help explain phase transitions. What I don’t see is the connection between the physicists‘ usage and what people in dynamical systems call renormalization. How are they related? Any references would be appreciated REPLY [4 votes]: There’s a bit of a terminological collision going on here. Physicists often use the term “renormalization” to refer the process of removing infinities from QFT calculations and to renormalization or “renormalization group” where you understand how theories behave at different scales, which is also how it is used in statistical physics. I prefer to call the removal of infinites “regularization” to distinguish the two. The connection between the two uses of the term is that the (UV) divergences of the theory arise due to the short scale behavior of the theory. However, we don’t actually know the theory at short distances. We can assume that some unknown theory cuts off the divergences and ask about how the theory behaves at lower energy scales using renormalization. You can show that only a subset of terms in the Lagrangian govern the theory and that other terms are suppressed at low energies. The terms that survive are precisely the “renormalizable” terms for which the various infinity removal processes work. In other words, understanding renormalization (a la Wilson) explains (some of) the divergences in QFT and what we’re really doing when we remove them.<|endoftext|> TITLE: Cohomology of quotient by free action QUESTION [6 upvotes]: Let $G$ be a finite group. Let $G$ act freely on a CW-complex $X$. I heard that the following fact is true. Claim. The canonical map $H^*(X/G,F)\to H^*(X,F)^G$ is an isomorphism, where $F$ is a field so that $|G|$ is invertible. I am looking for a reference to this fact. REPLY [15 votes]: This is true even if the group does not act freely. See Proposition 1.1 of my notes here. I deal with simplicial complexes and work over the rationals, but the statement you give can be proved the same way.<|endoftext|> TITLE: All two-point correlations equal to $0$, three-point correlation not $0$? QUESTION [8 upvotes]: Let $a_1,a_2,a_3,\dotsc \in \{-1,1\}$ be a sequence. Suppose that, for all $j>0$ and all $\epsilon, \epsilon'\in \{-1,1\}$, the proportion of $n\geq 1$ such that $(a_n,a_{n+j}) = (\epsilon,\epsilon')$ is $1/4$ (in the limit). I would like to see an example of such a sequence for which the average $S$ of $a_n a_{n+1} a_{n+2}$ is non-zero. (... or doesn't exist; that would also be interesting.) Reasons why this does not seem utterly trivial: (a) We cannot have $S=1$ or $S=-1$; in fact, it's not too hard to show (using Cauchy-Schwarz) that $|S|\leq 1/\sqrt{2}$. (b) Trying to produce such a sequence by a Markov chain where $a_{n+2}$ takes the value $1$ with probability $p_{(a_n,a_{n+1})}$ fails; that is, the only way in which the conditions are fulfilled (that is, all configurations $(a_n,a_{n+j})$ are equally likely) is if $p(a_n,a_{n+1}) = 1/2$ for every choice of $(a_n,a_{n+1})$. In other words, we would need a Markov chain of greater "depth". Further, more open-ended question: what would be a "magic sauce" condition such that, if all configurations $(a_n,a_{n+j}) = (\epsilon,\epsilon')$ are equally likely for every $j$, and $\{a_n\}_n$ has "magic sauce", then the average of $a_n a_{n+1} a_{n+2}$ is zero? REPLY [6 votes]: Let $\{b_n\}_{n \ge 1}$ be a sequence of independent random variables taking the values $\pm 1$ with probability $1/2$ each. Define $\{a_n\}_{n \ge 1}$ by $a_n=b_{n-1} b_{n-2}$ if $\; n \equiv 0 \mod 3$ and $a_n=b_n$ otherwise. Then the variables $a_n$ are pairwise independent. For example, if $\epsilon, \sigma \in {\pm1 }$ then $P(a_2=\epsilon,a_3=\sigma)=P(a_1=\sigma\epsilon, a_2=\sigma)=1/4$. Thus, by the Law of large numbers, for all $j>0$ and all $\epsilon, \sigma\in \{-1,1\}$, the limiting proportion of $n\geq 1$ such that $(a_n,a_{n+j}) = (\epsilon,\sigma)$ is $1/4$ with probability 1. (Formally, to apply the LLN, consider for each integer $\ell \in [0,2j)$ the limiting proportion of $k\geq 1$ such that $(a_{2kj+\ell},a_{2kj+\ell+j}) = (\epsilon,\sigma)$, then average over $\ell$.) However, the asymptotic average of $a_n a_{n+1} a_{n+2}$ is $1/3$ almost surely.<|endoftext|> TITLE: How to analytically prove chaos QUESTION [8 upvotes]: Consider the following map \begin{align*} T \colon \mathbb{R}\times\mathbb{S}^1 \to & \mathbb{R}\times\mathbb{S}^1 \\ (x,\theta) \mapsto & \left(\frac{x}{4}+ \sin^2\left(\pi\left(\theta+\frac{x}{2}\right)\right)+0.01, \theta+\frac{3x}{4}+\sin^2\left(\pi\left(\theta+\frac{x}{2}\right)\right)+0.01\right) \end{align*} I have been doing some numerical simulations and it seems this system presents a chaotic attractor. Here there is an approximation of the attractor. The colored points are the fixed points of the map. According by the following result that can be found in the Katok and Hasselblat book the chaos in this system might be generated by a transverse intersection of stable and unstable manifold of the orange saddle point. My Question: How do I check analytically that there is this intersection or how do I prove this map has chaotic behaviour? Could anyone help me ? REPLY [2 votes]: For specific systems at specific parameters, it can often be infeasible to give a pen-and-paper proof of chaos. However there is a rich literature of using computer-assisted-proofs based on interval arithmetic to demonstrate chaos, see for example Mischaikow, K., & Mrozek, M. (1995). Chaos in the Lorenz equations: a computer-assisted proof. Bulletin of the American Mathematical Society, 32(1), 66-72. The dynamical system you have looks similar to the Chirikov standard map. The following paper seems quite relevant to the problem you are studying. To quote from their abstract, they "show how interval analysis can be used to calculate rigorously valid enclosures of transversal homoclinic points in discrete dynamical systems." Neumaier, A., & Rage, T. (1993). Rigorous chaos verification in discrete dynamical systems. Physica D: Nonlinear Phenomena, 67(4), 327-346. In the intervening three decades there has been much progress and interest in developing computer-assisted-proofs of chaos in dynamical systems. For a more recent resource, I would recommend taking a look at the CAPD library: Kapela, T., Mrozek, M., Wilczak, D., & Zgliczyński, P. (2021). CAPD:: DynSys: a flexible C++ toolbox for rigorous numerical analysis of dynamical systems. Communications in Nonlinear Science and Numerical Simulation, 101, 105578.<|endoftext|> TITLE: Why do we have two theorems when one implies the other? QUESTION [23 upvotes]: Why do we have two theorems one for the density of $C^{\infty}_c(\mathbb{R}^n)$ in $L^p(\mathbb{R}^n)$ and one for the density of $C^{\infty}_c(\Omega)$ in $L^p(\Omega)$? with $\Omega$ an open subset of $\mathbb{R}^n$. Why not just the second one? I was asked by the prof what is the difference between the density of $C(\Omega)$ and $C(\mathbb{R}^n)$ but all I found when checking the demonstrations is that we take $\Omega \neq \mathbb{R}^n$ when giving the proof for the second theorem because for $\Omega = \mathbb{R}^n$ we already have the first theorem. REPLY [7 votes]: It may be helpful to think of a theorem not just as an expression of mathematical truth, but also as a tool that one mathematician or group of mathematicians have developed for use by others. It is like an API in a software library, that other authors/programmers can invoke when they have a need for it, without having to understand the low-level details of how the API does what it advertises itself as doing. To continue this analogy, a well-designed API will often offer ways of doing things at several levels of generality, one that can be used in simple/common cases, and another (or several other ones) that’s used more rarely and supplies more arguments/parameters, more customizability options, handles difficult edge cases better, etc. The more advanced API call does everything that the simple one does and more. But learning how to use it involves more effort on the part of the programmer, so in practice most programmers will rely on the simpler method (often labeled a “convenience method/function”). The same goes for theorems. Mathematicians sometimes need to “invoke” Euclid’s theorem on the infinitude of primes as part of a proof. Should we force all of them to learn about a much stronger fact, the prime number theorem, or one of the even stronger results that have been proved about the distribution of prime numbers? No, it’s much more convenient to keep Euclid’s theorem around as a “convenience method”. Even in your functional analysis example, I’m pretty sure the first theorem about $\mathbb{R}^n$ is useful (in the sense of being invoked as part of an argument) often enough to justify “having” it, even though in a formal sense it is superseded by the second one.<|endoftext|> TITLE: Fourier Series with Mobius coefficients QUESTION [16 upvotes]: I assume this question has been considered before, but I can't find an literature on it. Let $\mu(n)$ denote the usual Mobius function and define: $F(x) : = \sum_{n=1}^{\infty} \frac{\mu(n)}{n}e(nx)$ where $e(x):= e^{2\pi i x}$. The Prime Number Theorem is equivalent to the statement that $F(0)=0$. More generally, one can show that $F(x)$ is uniformly bounded. This follows from partial summation and an old theorem of Davenport. Two further questions naturally follow: Is $F(x)$ continuous? Davenport's theorem is ineffective due to the possible existence of Siegel zeros. Can one obtain an unconditional effective uniform bound on $F(x)$? REPLY [14 votes]: I can answer the first question. Davenport proved that $$S(N,x):=\sum_{n\leq N}\mu(n)e(nx)\ll\frac{N}{(\log N)^2},$$ with an upper bound independent of $x\in\mathbb{R}$. Therefore, by partial summation we see that $F(x)$ converges uniformly, and it equals $$F(x)=\sum_{n=1}^\infty\frac{S(n,x)}{n(n+1)}.$$ Uniform convergence implies continuity, since the terms in $F(x)$ are continuous (i.e. the partial sums of $F(x)$ are continuous).<|endoftext|> TITLE: Manifolds with trivial mapping class group and large $H^1$? QUESTION [8 upvotes]: Are there smooth closed manifolds $M^n$ in every dimension $n \geq 3$ with trivial mapping class groups and with $H^1(M^n;\mathbb{Z}/2\mathbb{Z})$ arbitrarily large? I am under the impression that "generically" a manifold will have no mapping class group (but maybe I am totally mistaken). There are presumably lots of constructions of such manifolds with trivial mapping class groups. So I guess I'm hoping to hear of some such construction where $H^1(M^n;\mathbb{Z}/2\mathbb{Z})$ can get very large. REPLY [5 votes]: I think that the construction in Belolipetsky--Lubotzky, Finite Groups and Hyperbolic Manifolds (https://arxiv.org/abs/math/0406607) provides a more algebraic construction of such manifolds for any $n$. One of the results in this paper (Theorem 3.1) is the following : given groups $\Gamma \triangleright \Delta \triangleright M$ with $\Gamma/\Delta$ finite and $F = \Delta/M$ a nonabelian free group, there exists $\Gamma \ge D \ge \Delta$ and $B \le D$ of finite index with $N_\Gamma(B) = B$ (in fact they prove that for any finite group $G$ there are infinitely many $B$ with $N_\Gamma(B)/B$ isomorphic to $G$). They use this to construct closed hyperbolic manifolds with trivial isometry group ; by Mostow rigidity the mapping class group of $X$ is isomorphic to the isometry group of $X$ (every homeomorphism is isotopic to an isometry), so these manifolds will have trivial mapping class group. Without too much details their construction uses a specific lattice $\Gamma$ in the isometry group of hyperbolic $n$-space $\mathbb H^n$, and subgroups $D, \Delta, M, B$ as in the previous paragraph ; the manifold is the quotient of $\mathbb H^n$ by $B$ (their construction ensures that $B$ is torsion-free though $\Gamma$ will likely not be). Now to look at Betti numbers, for which we need to dive a bit more into the details of this paper. Since $X$ is aspherical we have $$b_1(X, \mathbb Q) = b_1(B, \mathbb Q) \ge b_1(D, \mathbb Q).$$ On the other hand $b_1(D, \mathbb Q)$ is the dimension of the fixed subspace of $D/\Delta$ in $H^1(\Delta, \mathbb Q)$ (conjugation action on the normal subgroup $\Delta$ and its morphisms to $\mathbb Z$). By the definition of $D$ in the paper of Belolipetsky--Lubotzky (p. 4 in the arxiv version) we have that it acts trivially on the part of the cohomology coming from the surjective morphism $\Delta \to F$. So we have that $b_1(X, \mathbb Q)$ is larger than the rank of $F$. It remains to remark that in the data for Theorem 3.1 the quotient $\Delta/M$ can be taken to have arbitrarily large rank (since a non-abelian free group contains free subgroups of finite index with arbitrarily large rank), so we get examples with $b_1(B, \mathbb Q)$ arbitrarily large.<|endoftext|> TITLE: A sum over partitions involving "subpartitions" QUESTION [16 upvotes]: Consider the following sum over partitions of $n$: $$ S(n)=\sum_{\substack {j_1,\dots,j_n\geq 0\\j_1+2j_2+\dots+nj_n=n}} \prod_{t=1}^n \frac{1}{j_t!t^{j_t}}f_t(j_1,\dots,j_t),$$ where $$ f_t(j_1,\dots,j_t)=\begin{cases}\frac{j_t}{j_t+1}\,&\textrm{if } j_1+2j_2+\dots+(t-1)j_{t-1}=t-1 \\1 \,&\textrm{otherwise}\end{cases}.$$ I have strong numerical evidence that $$ S(n)=\frac{1}{n+1},$$ but I cannot prove it, I was wondering if anyone could give me ideas. Some observations: I wasn't able to compute it using a simple generating function, since I would need $f_t(j_1,\dots,j_t)$ to be a function only of $j_t$. The function $$f(j_1,\dots, j_n)=\prod_{t=1}^n f_t(j_1,\dots,j_t)$$ can also be intepreted as a function of the cycle structure of a permutation depending on which "subpartitions" $j_1,\dots, j_n$ contains, or on the invariant subsets of $\{1,\dots, n\}$ under the permutations with the cycle structure given by the partition. $S(n)$ can be seen as the average of $f$ over the permutation group, since the $\prod_{t=1}^n \frac{1}{j_t!t^{j_t}}$ is the probability of drawing a permutation with $j_k$ $k-$cycles in its decomposition. Ideally, I would like to compute, or at least bound $$ S(n,x)=\sum_{\substack {j_1,\dots,j_n\geq 0\\j_1+2j_2+\dots+nj_n=n}} \prod_{t=1}^n \frac{x^{j_t}}{j_t!t^{j_t}}f_t(j_1,\dots,j_t),$$ for $x>0$, but I don't have a good conjecture on the form of this sum, except for $x=1$. REPLY [2 votes]: This is just an extended comment providing an alternative view at $S(n)$, which (I hope) may lead to a solution. Notice that $f_t(j_1,j_2,\dots,j_t) = 1-\frac{\delta_{j,t}}{1+j_t}$, where $$\delta_{j,t} := \big[j_1+2j_2+\dots+(t-1)j_{t-1} = t-1\big]$$ is an Iverson bracket. Let $\bar n:=\{1,2,\dots,n\}$ and for a given partition exponents $j=(j_1,\dots,j_n)$ with $j_1 + 2j_2+\dots+nj_n=n$, define $J(j) := \{t\in\bar n\,:\,\delta_{j,t}=1\}$. Then \begin{split} (n+1)! S(n)&=(n+1)!\sum_{j:\ j_1 + 2j_2+\dots+nj_n=n} \prod_{t=1}^n \frac{1}{j_t!t^{j_t}} \sum_{T\subseteq J(j)} (-1)^{|T|} \prod_{t\in T} \frac1{j_t+1} \\ &= \sum_{T\subseteq \bar n} (-1)^{|T|} \sum_{j:\ j_1 + 2j_2+\dots+nj_n=n\atop J(j)\supseteq T} \prod_{t=1}^n \frac{(n+1)!}{(j_t + [t\in T])!\,t^{j_t}}. \end{split} The last formula may be viewed as application of the inclusion-exclusion principle under a suitable combinatorial interpretation of its terms, which then would likely imply the needed $(n+1)!S(n)=n!$ out of the box. Unfortunately, I was not able to find such an interpretation so far, but have a gut feeling it's out there.<|endoftext|> TITLE: Why should an algebraic geometer care about singular / simplicial (co)homology? QUESTION [22 upvotes]: I am a PhD student in algebraic / arithmetic geometry and I never took a formal course in algebraic topology, even though I have some basic knowledge. In algebraic geometry we deal exclusively with sheaf cohomology since we care about non-constant sheaves. But I feel, maybe in my naivety, that a lot of the important results (for usual topological spaces) are only true for singular and simplicial cohomology when they coincide with sheaf cohomology (Alexander duality and "$H^i=0$ for $i>$ the covering dimension" come to mind). With that in mind, I wonder if it is worth for someone with a similar background to study the details of a first course in algebraic topology. (Perhaps on the level of Hatcher's book.) Do I lose something by just thinking in terms of sheaf cohomology? REPLY [8 votes]: I'm turning my comment above into an answer, as I realized there was more that I wanted. IMO, if you're an arithmetic geometer the main motivation for learning singular cohomology is that it is intimately related to etale cohomology, in the sense that both of these are the simplest examples of a Weil cohomology theory. I won't recall all the axioms but state the first two (at least as defined in the wikipedia article here): Let $k$ be a field, $X/k$ a smooth projective variety of dimension $d$, and a $K$ a coefficient field of characteristic zero (for singular you often take $K = \mathbf{C}$ while for etale you take $K = \mathbf{Q}_\ell$) $H^i(X)$ is finite dimensional. $H^i(X)$ is zero for $i < 0$ and $i > 2d$. In the topological setting, each of these is somewhat straightforward to prove because $X(\mathbf{C})$ has the structure of a real $2d$-dimensional CW-complex: One deduces (1) and (2) from the corresponding simplicial cochain complex that computes the cohomology of $X(\mathbf{C})$. On the other hand in the etale setting these are not trivial to prove at all. For instance, the vanishing beyond degree $2d$ is a result of Artin for varieties over a separably closed field. To prove this result you need to fiber $X$ in curves to reduce to the case of a curve, and even then (may I say) the result is not trivial: Grothendieck was confused for a long time on how to compute the etale cohomology of curves! As you keep going down the list of axioms for a Weil cohomology theory, you realize immediately that the corresponding statements in singular cohomology are much easier to prove. This alone should be enough to motivate you to care about singular cohomology. Why is it important to realize that etale cohomology is a Weil cohomology theory? Well if you're an arithmetic geometer (with high probability) you will care about the Weil conjectures. The first two statements - rationality of zeta and the functional equation - pop out immediately as a consequence of the axioms of a cohomology theory. For instance, rationality is a consequence of the finite dimensionality of cohomology and the Lefschetz trace formula, while rationality from Poincare duality.<|endoftext|> TITLE: Pointwise convergence imples uniform convergence in an infinite subset QUESTION [10 upvotes]: I came upon this statement in a stack answer. Statement : If $f_n$ is a sequence of real valued functions (not necessarily continuous or measurable) on $[0,1]$ such that $f_n$ converges point-wise to $0$, then there exists an infinite subset of $[0,1]$ where the convergence is uniform. I couldn't prove it. I believe the claim is true because $[0,1]$ is uncountable but the set of sequences is countable only. Any help would be appreciated in assistance to how to prove it. REPLY [6 votes]: For every sequence $(F_n)_{n \in \omega}$ with $F_n:[0,1] \rightarrow \mathbb{R}$ converging pointwise to $0$, we can associate to every $x \in [0,1]$ an $f_x \in \mathbb{\omega}^\omega$ in the following way: Set $f_x(m):= \min\{n \in \mathbb{\omega}\,\, \colon \,\, \forall n' \geq n \,\,\,\, \vert F_{n'}(x) \vert < \frac{1}{m}\}$. Now the proof works in two steps. In the first step we show that there exists $f^* \in \omega^\omega$ such that for every $k \in \omega$ the set $\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ is uncountable. We will construct such an $f^* \in \omega^\omega$ by induction on $k \in \omega$: Assume that $f^* \restriction k$ has already been constructed, and we have that the set $\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ is uncountable. Since $$\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}= \bigcup_{l \in \omega}\,\, \{x \in [0,1] \,\, \colon \,\, f_x \restriction (k+1) \leq (f^* \restriction k)^\frown l\}$$ we find $l \in \omega$ such that $\{x \in [0,1] \,\, \colon \,\, f_x \restriction (k+1) \leq (f^* \restriction k)^\frown l\}$ is uncountable. Now set $f^*(k)=l$ and we see that $f^*\restriction (k+1)$ has the required properties. In the second step we inductively construct $(x_k)_{k \in \omega} \subseteq [0,1]$ injective and $(f_k)_{k \in \omega} \subseteq \omega^\omega$ increasing such that for every $k \in \omega$ we have $f^* \leq f_k$ , $f_{x_k} \leq f_k$ and $f_k \restriction (k+1) = f_{k+1} \restriction (k+1)$. Once we have shown this, we can define $g \in \omega^\omega$ such that $g(k):=f_k(k)$ and see that $f_{x_k} \leq g$ for every $k \in \omega$. But this proves that $(F_n)_{n \in \omega}$ converges uniformly on $(x_k)_{k \in \omega}$. To this end assume that $x_0,...,x_{k-1}$ and $f_0,...f_{k-1}$ with the required properties have already been constructed. Since $\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ is uncountable, we can find $x_k \in \{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ different from $x_0,...,x_{k-1}$. Set $f_k(m):=\max\{f_{k-1}(m), f_{x_k}(m)\}$, and note that since $f_{x_k} \restriction k \leq f^* \restriction k \leq f_{k-1} \restriction k$, we have $f_{k-1}\restriction k =f_k \restriction k$. This finishes the proof.<|endoftext|> TITLE: Artin-Rees lemma for multiplicative subsets? QUESTION [5 upvotes]: The classical Artin-Rees lemma tells the following. Let $R$ be a Noetherian commutative ring and $I\subset R$ be an ideal. Let $M$ be a finitely generated $R$-module and $N\subset M$ be a submodule. Then there exists an integer $m\ge0$ such that for all $n\ge0$ the following equality of two submodules in $N$ holds: $$ I^{n+m}M\cap N=I^n(I^mM\cap N). $$ The usual proof is based on the Hilbert Basis Theorem: essentially, one uses the fact that the Rees ring $\bigoplus_{n=0}^\infty I^n$ is graded Noetherian (since it is a quotient ring of the ring of polynomials over $R$ spanned by some finite set of generators of the ideal $I$). I would like to specialize the Artin-Rees lemma to principal ideals $I=(s)\subset R$, and then extend it from multiplicative subsets of the form $\{1,s,s^2,s^3,\dotsc\}\subset R$ to other multiplicative subsets $S\subset R$. So let $R$ be a Noetherian commutative ring and $S$ be a multiplicative subset in $R$. Assume for simplicity that $S$ is (at most) countable and all the elements of $S$ are regular (nonzero-divisors) in $R$. Let $M$ be a finitely generated $R$-module and $N\subset M$ be a submodule. Does there exist an element $t\in S$ such that, for every $s\in S$, the equality of two submodules in $N$ $$ stM\cap N=s(tM\cap N) $$ holds? Or, at least, can one find an element $t\in S$ such that the above equality holds for a cofinal subset of elements $s\in S$ (in the divisibility preorder)? A straightforward attempt to argue similarly to the Hilbert Basis Theorem proof of the Artin-Rees lemma does not seem to work, as the ring $\bigoplus_{s\in S}sR$ does not have to be graded Noetherian when the multiplicative set $S$ is not finitely generated. REPLY [3 votes]: This is straightforward . Define the submodule $P$ as $N\subset P\subset M$, the set of all elements $m\in M$ such that $sm\in N$ for some $s\in S$. Then, choose $t\in S$ such that $tP\subset N$. Now for what you need, clearly the right hand side is contained in the left. So, let $x\in stM\cap N$ for $s\in S$. Then, $x=stm$ and then, $m\in P$. Then $tm\in N$ by our choice of $t$. So, $tm\in tM\cap N$ and thus $x\in s(tM\cap N)$.<|endoftext|> TITLE: Is every continuous action of a compact topological group closed? QUESTION [6 upvotes]: I am reading Bredon's Introduction to compact transformation groups, and came across the following result and proof on page 34: Even though he writes "Recall our standing assumption that $X$ is Hausdorff," I cannot see where any properties of Hausdorffness of $X$ are used in this proof. I suspect it would be used to ensure that convergent nets in $X$ converge uniquely, but I do not see why this fact would be needed for this proof. Is it true that the continuous action of a compact topological group on a topological space $X$ is always closed, regardless of whether $X$ is Hausdorff? If not, then where in the above proof is Hausdorffness used? REPLY [4 votes]: Here is an elementary proof: Let $C\subseteq G\times X$ be closed. Note that $x_0\notin \Theta(C)$ if and only if $\{(g,g^{-1}x_0)\mid g\in G\}\cap C = \emptyset$ which is if and only if $G\times \{(e,x_0)\}\subseteq \Phi^{-1}((G\times X)\setminus C)$ where $\Phi: (g,(h,x))\mapsto (hg^{-1}, gx)$. Now, let $x_0\notin \Theta(C)$. By our observation, $G\times \{(e,x_0)\}\subseteq \Phi^{-1}((G\times X)\setminus C)$ and since $C$ is closed, $\Phi^{-1}((G\times X)\setminus C)$ is a neighborhood of $G\times \{(e,x_0)\}$. Hence, (by the tube lemma), we can find neighborhoods $U$ of $e$ and $V$ of $x_0$ such that $G\times (U\times V) \subseteq \Phi^{-1}((G\times X)\setminus C)$. It is easy to see that $\Theta(U\times V)$ is a neighborhood of $x_0$ which does not intersect $\Theta(C)$.<|endoftext|> TITLE: Motivation of the construction of $p$-adic period rings QUESTION [8 upvotes]: Let $B$ be either $B_{\text{dR}}$ or $B_{\text{crys}}$. For a $\mathbb{Q}_p$-representation $V$ of the absolute Galois group $\mathrm{Gal}(\overline{K}/K)$ of a $p$-adic field $K$ (a finite extension of $\mathbb{Q}_p$), there are subspaces of $H^1(K,V)$ defined by $\ker(H^1(K,V)\to H^1(K,B_\bullet\otimes V))$ for $\bullet\in\{\text{dR},\text{crys}\}$. Each element of the subspaces of $H^1(K,V)$ correponds of an extension of $\mathbb{Q}_p$ by $V$ in the category of $\mathbb{Q}_p$-representations of $\mathrm{Gal}(\overline{K}/K)$ having some nice properties -- being de Rham/crystalline. The constructions of the rings $B$ seem quite complicated. My question is How did we end up with the such complicated constructions of $B$'s so that elements of subspaces of $H^1$ having such representation-theoretic properties? Or, why do elements in $\ker(H^1(K,V)\to H^1(K,B\otimes V))$ have such representation theoretic properties? REPLY [4 votes]: How did we end up with the such complicated constructions of $B$? To add to Laurent's answer remark that "these rings did not, however, come out of nowhere", I believe that in the early 80s, Fontaine noticed that two seemingly very different constructions naturally lead to very similar (classes of) rings: The Honda systems he had introduced in his classification of $p$-divisible groups. Certain surjective evaluation maps and functorial properties in the theory of the field of norms he had established with Wintenberger. Once you have both, it is quite natural to look at the completion of some Witt vector rings with respect to the kernel of the evaluation maps in question, and that's $B_{\operatorname{dR}}^{+}$. You can have a look at the references given in the following answer. https://mathoverflow.net/a/342838/2284 The fact that an extension of Galois representation by a de Rham (or crystalline, or...) representation corresponds to a class in $$\ker\left(H^1(G_K,V)\longrightarrow H^1(G_K,V\otimes B_{\operatorname{dR},\operatorname{crys},\dots})\right)$$ follows formally from the notion of $B$-admissibility: an extension $\xi$ of $\mathbb Q_p$ by $V$ is an extension by a $B$-admissible representation if and only if it becomes split after applying $D_{B}(\cdot)$ if and only if the class of $\xi$ is in the kernel of $$\ker\left(H^1(G_K,V)\longrightarrow H^1(G_K,V\otimes B)\right).$$<|endoftext|> TITLE: Cutting of a regular polygon into congruent pieces QUESTION [20 upvotes]: Question. For which $N$ it is possible to cut a regular $N$-gon into congruent pieces such that the center of the regular polygon lies strictly inside one of the pieces? For $N=3,4$ there are trivial examples. For $N=6$ there is a nontrivial example by Peter Mueller (based on idea of Matt F.) below. Remark. There is also similar open problem about a disc. Is it possible to dissect a disk into congruent pieces, so that a neighborhood of the origin is contained within a single piece? It seems that the answer is NO, but there is no proof. REPLY [20 votes]: Using the tiles suggested in Matt F.'s answer, there is a solution for the regular $6$-gon:<|endoftext|> TITLE: Differentiability of the map $x\mapsto \delta_x$ in the Arens-Eells/Lipschitz-free space QUESTION [5 upvotes]: $\DeclareMathOperator\AE{AE}\DeclareMathOperator\Lip{Lip}$Let $\AE(X)$ denote the Arens-Eells space on a Banach space $X$. Consider the map: $$ \begin{aligned} \delta: X & \rightarrow \AE(X) \\ x&\mapsto \delta_x \end{aligned} $$ Is the map $\delta$ ever Gâteaux (or Fréchet) differentiable? Recall that $\AE(X)$ is the/a pre-dual of the Banach space $\Lip_0(X)$ whose elements are Lipschitz functions sending $0\in X$ to $0\in \mathbb{R}$ (with norm sending any $f\in \Lip_0(X)$ to its (unique) Lipschitz constant $\Lip(f)$), $\delta_x$ denotes the evaluation map defined on Lipschitz functions $f\in \Lip_0(X)$ by $$ \delta_x(f):= f(x), $$ and $\AE(X)$ is normed using the dual-norm construction; i.e.: $$ \|F-G\|:=\inf_{f \in \Lip_0(X),\, \Lip(f)\leq 1} F(f)-G(f). $$ REPLY [7 votes]: This fails for $X = \mathbb{R}$, and hence for every nonzero Banach space, since they all contain copies of $\mathbb{R}$. If the map $t \mapsto \delta_t$ were differentiable in either sense then for every bounded linear functional $F$ on $AE(\mathbb{R})$ the map $t\mapsto F(\delta_t)$ would be differentiable. Recalling that the dual of $AE(\mathbb{R})$ is ${\rm Lip}_0(\mathbb{R})$, differentiability at $t$ would imply that every Lipschitz function on $\mathbb{R}$ is differentiable at $t$, which is obviously false.<|endoftext|> TITLE: Term for theory that can be defined within another theory QUESTION [8 upvotes]: Suppose we have two formal axiomatic theories of mathematics and the objects and operators in one theory can be defined in terms of the objects and operators in another theory such that the axioms of the former are theorems in the latter. Is there a term for this type of relationship? Where would I go to learn more about questions about this type of relationship? My intuition is that goal of something like ZFC is to be a theory that's as simple and general as possible such that any other mathematics we were interested in could be reframed in terms of ZFC, something one might call a universal theory although I don't know what the actual terminology is. Maybe there are widely different types of universal theories. Perhaps ZFC or even specific universes within ZFC can be constructed from ZFC itself. REPLY [12 votes]: You are searching for the concept known as interpretation. A model $N$ is interpretable in another model $M$ (perhaps a totally different language signature) if inside $M$ we can define a domain (perhaps a quotient by a definable equivalence relation on $k$-tuples) and define the fundamental relations and structure of $N$ on that domain, in such a way that the defined structure is isomorphic to $N$. For example, the complex number field $\langle \mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field, by representing $a+bi$ with pairs of real numbers $(a,b)$. However, it turns out that the real number field is not interpretable in the complex field. A theory $T$ is interpreted in another theory $S$, if we have a uniform manner of defining a model of $T$ inside any model of $S$. Two models (or theories) are mutually interpretable, if each is interpreted in the other. A strictly stronger notion is bi-interpretation. Two models are bi-interpretable, if they are mutually interpretable, in such a way that the models can define the isomorphism of themselves with their copy inside (the copy of) the other structure. By iterating the interpretations, one obtains a fractal like situation: My recent paper Alfredo Freire, Joel David Hamkins, "Bi-interpretation in weak set theories," https://arxiv.org/abs/2001.05262. contains an elementary introduction of interpretation of models of set theory, including the proof that no two extensions of ZF set theory are bi-interpretable. Here is a talk I gave for the Oxford set theory seminar on the topic: "Bi-interpretation of weak set theories" https://youtu.be/8KxlpMOXoP4<|endoftext|> TITLE: Can one show that the real field is not interpretable in the complex field without the axiom of choice? QUESTION [35 upvotes]: We all know that the complex field structure $\langle\mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field $\langle\mathbb{R},+,\cdot,0,1\rangle$, by encoding $a+bi$ with the real-number pair $(a,b)$. The corresponding complex field operations are expressible entirely within the real field. Meanwhile, many mathematicians are surprised to learn that the converse is not true — we cannot define a copy of the real field inside the complex field. (Of course, the reals $\mathbb{R}$ are a subfield of $\mathbb{C}$, but this subfield is not a definable subset of $\mathbb{C}$, and the surprising fact is that there is no definable copy of $\mathbb{R}$ in $\mathbb{C}$.) Model theorists often prove this using the core ideas of stability theory, but I made a blog post last year providing a comparatively accessible argument: The real numbers are not interpretable in the complex field. The argument there makes use in part of the abundance of automorphisms of the complex field. In a comment on that blog post, Ali Enayat pointed out that the argument therefore uses the axiom of choice, since one requires AC to get these automorphisms of the complex field. I pointed out in a reply comment that the conclusion can be made in ZF+DC, simply by going to a forcing extension, without adding reals, where the real numbers are well-orderable. My question is whether one can eliminate all choice principles, getting it all the way down to ZF. Question. Does ZF prove that the real field is not interpretable in the complex field? I would find it incredible if the answer were negative, for then there would be a model of ZF in which the real number field was interpretable in its complex numbers. REPLY [21 votes]: (1) $\mathbb{C}$ is stable, (2) $\mathbb{R}$ is unstable, and (3) stability is preserved under interpretations. Of course the usual development of stability uses a lot of choice, but (2) and (3) are elementary. (1) follows from QE for algebraically closed fields. One of the reasons why model theorists care about the whole alphabet soup of classification-theoretic properties is that they enable this kind of argument. Furthermore we have a complete understanding of fields interpretable in $\mathbb{C}$: If $K$ is an algebraically closed field then any infinite field interpretable in $K$ is definably isomorphic to $K$. This is in Poizat's book on stable groups, but I forget who it is due to. I don't think this proof requires choice either. Edit: I just read Joel's post more carefully and realized that it contains "Model theorists often prove this using the core ideas of stability theory". But I think it's still worth pointing out that this argument does not require choice. Another edit: I thought about this a bit more today and I think that the following is the simplest argument I can think of. It uses ideas from stability, but no actual stability theory, just elementary model theory and algebra. Recall that a theory $T$ is $\omega$-stable if for any $M \models T$ and countable $A \subseteq M$, there are only countably many types in $M$ over $A$. It's easy to see that $\omega$-stability is preserved under interpretations - if $N$ is interpretable in $M$ then a witness of non-$\omega$-stability in $N$ lifts to one in $M$. It's also easy to see that a real closed field is not $\omega$-stable - there are uncountably many types over $\mathbb{Q}$. So it comes down to showing that algebraically closed fields are $\omega$-stable. Quantifier elimination for algebraically closed fields gives a bijection between $n$ types over $A \subseteq \mathbb{C}$ and prime ideals in $K [x_1,\ldots,x_n]$ where $K$ is the subfield generated by $A$, and every prime ideal in $K[x_1,\ldots,x_n]$ is finitely generated, so there are only countably many prime ideals. Yet another edit: Emil pointed out that the argument above actually uses choice - in the proof of preservation of $\omega$-stability under interpretations. He suggests replacing $\omega$-stability with the following property: there are only countably many types over any finite set of parameters. When you do that you only need to make finitely many choices in the preservation argument. Yet another another edit It turns out that this whole question had already been asked and answered on stack exchange. In particular Alex gave a worked out elementary version of the stability argument.<|endoftext|> TITLE: Almost evenly distributed spherical random vectors QUESTION [6 upvotes]: Consider $n$ i.i.d spherically distributed random vectors $z_1 ,\cdots , z_n \sim \text{Unif}(\mathbb{S}^{d-1})$. What is the best lower bound on $n$ for which whp there exists a constant $c>0$ such that the following bound holds for all $v\in \mathbb{R}^d\backslash \{0\}$: \begin{equation} cn\leq \left\vert\left\{i:\langle z_i,v \rangle>0 \right\} \right\vert \end{equation} REPLY [6 votes]: $\newcommand\PP{\mathbb P}$ Surely $n_\min \lesssim d$, because it works for $c = 1/4$ and $n=160d$. We use that the number of "distinct" $v$ with respect to the classifiers $\textrm{sgn}\langle \cdot, z_i \rangle$ is $$ \sum_{i=0}^{d-1} \binom{n-1}{i} \le \left( \frac{ne}{d} \right)^d $$ The proof can be found in [Bürgisser and Cucker (2013), Lemma 13.7]. (Anyone has a better reference?) Let $X = \lvert \left\{ i ~:~ \langle z_i, e_1 \rangle \right\} \rvert$ for a basis vector $e_1$. Then, by the union bound $$ \PP \left[ ~\exists v ~~ \lvert \left\{ i ~:~ \langle z_i, v \rangle > 0 \right\}\rvert < cn \right] \\ \le \left( \frac{ne}{d} \right)^d \PP \left[ X < cn \right] \\ \le \exp(d\log(n) - \frac{n}{16} + d - d \log{d}) \\ = \exp(\log(160)d - 9d) \xrightarrow{d \to \infty} 0$$ where we used the Chernoff bound on $\PP[X < cn]$, in the form $$ \PP\left[X \le (1 - \delta) \mathbb E[X] \right] \le \exp(-\frac12 \delta^2 \mathbb E[X]). $$ As noted in the comment by Sandeep Silwal, $n_{\text{min}} \ge d$ due to the strict inequality sign in the question. So the answer to the original question is $n_{\text{min}} = \Theta(d)$.<|endoftext|> TITLE: Hypergeometric function evaluation 4F3 QUESTION [5 upvotes]: I need to show that for $m$ being non-negative integer, the hypergeometric function ${}_4F_3$ below evaluates to $-1/2$ independent of $m$. This is Mathematica notation, but we have 4 and 3 sets of parameters, and evaluate at $z=1$. HypergeometricPFQ[{-1/3, 1/3, -3/2 - m, -m - 1}, {1/2, -2/3 - m, -1/3 - m}, 1] I have looked in W. Koepf's book, but have not found any identity which can be specialized to this. Also, mathematica does not simplify this further. As a side note, is there a nice list of hypergeometric identities in some good book? I know Gasper and Rahman has a nice book, but this deals with the q-analogs mainly, so I would prefer some reference which deals with the non-q-case. REPLY [6 votes]: From https://dlmf.nist.gov/15.4 one has $$ \sum_{k=0}^n\frac{(1/3)_k(-1/3)_k}{k!(1/2)_k}(-z^2)^k=\frac{1}{2}\left(\left(\sqrt{1+z^{2}}+% z\right)^{2/3}+\left(\sqrt{1+z^{2}}-z\right)^{2/3}\right) $$ and $$ \sum_{k=0}^n\frac{(1/3)_k(2/3)_k}{k!(3/2)_k}(-z^2)^k=\frac{3}{2z}\left(\left(\sqrt{1+z^{2}}+% z\right)^{1/3}-\left(\sqrt{1+z^{2}}-z\right)^{1/3}\right). $$ Multiplying both sides and calculating coefficient of $(-z^2)^n$ one obtains, e.g. using $(x)_{n-k}=(-1)^k\frac{(x)_n}{(1-x-n)_k}$, that $$ {}_4F_3\left({1/3,-1/3,-1/2-n,-n\atop 1/2,1/3-n,2/3-n};1\right)\cdot \frac{(1/3)_n(2/3)_n}{n!(3/2)_n}\\=-\frac{1}{2}\cdot\left[(-z^2)^n\right]\left\{\frac{3}{2z}\left(\left(\sqrt{1+z^{2}}+% z\right)^{1/3}-\left(\sqrt{1+z^{2}}-z\right)^{1/3}\right)\right\},\quad n\in\mathbb{N}, $$ from which the claim follows (here $\left[(-z^2)^n\right]\left\{f(z)\right\}$ denotes the coefficient of $(-z^2)^n$ of the power series $f(z)$). Gasper and Rahman is the best reference, even one is not working with $q$-series. Some other references, more suitable for combinatorists, might be Riordan's book, or Graham, Knuth, Patashnik's book (haven't read these two). Identities of this kind are called convolution type identites, because when written in terms of binomial coefficients the sum is a convolution.<|endoftext|> TITLE: Algebraic K-theory and intersection theory (Bloch's formula) QUESTION [5 upvotes]: It seems to be a well known fact that algebraic K-theory can be used to understand intersection theory, at least for varieties (or stacks!) over a field. A first glimpse of this result seems to be Bloch's formula (Quillen, 1972): $$ \operatorname H^p(X, \mathcal{K}_p) \cong \operatorname{CH}^p (X) $$ where $\mathcal{K}_p$ is the Zariski-sheafification of $K$-theory. This even induces an isomorphism of the respective graded algebras! But as beautiful as this formula is, I've got a question: can we actually extract information about the chow groups and its intersection products using this formula? Of course, $K$-theory is notoriously complicated, but sheafifying it might make things more amenable (I think $\mathcal{K}_p = \underline{\mathbf{Z}}$ and $\mathcal{K}_1 = \mathcal{O}_X^\times$). One example on how this might help, it seems that (Guillet, 1984) used the LHS to define an intersection theory for algebraic stacks, but it also seems to me (and I might be very wrong) that (Vistoli, 1989) approach bypassing this is what has become more prominent nowadays, and I don't know if Guillet's approach allows us to deduce anything that Vistoli's cannot. On a similar note, Quillen says on his paper on higher algebraic K theory that the left hand side of Bloch's formula is "manifestly contraviariant in X"; which may be one advantage, although I don't quite see how to prove this (maybe it follows from earlier remarks - i didn't read the entire paper yet....). It seems that Quillen has convinced himself by this point that the LHS will be important for future work. REPLY [2 votes]: Not an answer, but as you wrote "Of course, K-theory is notoriously complicated, but sheafifying it might make things more amenable", let me point out that Bloch's formula is also true for the Milnor $K$-theory sheaf instead of the Quillen $K$-theory sheaf. This does not help at all to compute any Chow group, but it makes it possible (and in easy examples possible in a concrete fashion) to write down an algebraic cycles as a representative in Cech cohomology of the Milnor $K$-theory sheaf. So this makes things even more amenable. Nonetheless, I am not sure this has any profound use. It is however definitely easier to work with such cycle representatives than trying to make a Cech representative of a Quillen K-theory sheaf explicit.<|endoftext|> TITLE: Upper bound an integral with exponential function QUESTION [5 upvotes]: I am working on my research about approximation a function. I come up with the following integral. I run some simulations and saw that the integral would converge to zero as n goes to infinty. Here is my prediction. Hopefully someone can give me the idea to deal with this kind of upper bound. Thanks a lot. Problem. Given $a \in [0,1]$. Prove that $\int_{0}^{1}e^{-n(t-a)^2} - e^{\frac{-n(t-a)^2}{1-(t-a)^2}}dt \leq \frac{c}{n},$ where $c >0$ is a constant. REPLY [9 votes]: The integral in question can be rewritten as $$ \begin{aligned} I&:=\frac1{\sqrt n}\,\int_{-a\sqrt n}^{(1-a)\sqrt n} e^{-u^2}\Big(1-\exp\Big\{-\frac{u^4/n}{1-u^2/n}\Big\}\Big)\,du \\ &\le\frac1{\sqrt n}\,\int_{-a\sqrt n}^{(1-a)\sqrt n} e^{-u^2}\min\Big(1,\frac{u^4/n}{1-u^2/n}\Big)\,du \\ &\le \frac1{\sqrt n}\,(I_1+I_2), \end{aligned}$$ where \begin{equation} I_1:=\int_{|u|\le\sqrt n/2} e^{-u^2}2u^4/n\,du\le\int_{|u|<\infty} e^{-u^2}2u^4/n\,du=O(1/n) \end{equation} and \begin{equation} I_2:=\int_{|u|>\sqrt n/2} e^{-u^2}\,du \le\int_{|u|>\sqrt n/2} \frac{u^2}{(\sqrt n/2)^2}e^{-u^2}\,du \le\int_{|u|>0} \frac{u^2}{(\sqrt n/2)^2}e^{-u^2}\,du =2\sqrt\pi/n =O(1/n) \end{equation} (as $n\to\infty$). So, \begin{equation} I=O(1/(n\sqrt n)), \end{equation} which is better than desired.<|endoftext|> TITLE: Can I find a bump function $\psi$ such that $\nabla\log\psi$ vanishes too? QUESTION [5 upvotes]: Consider a bump function supported in the ball of radius $1$, that is $\psi:\mathbb R^n\to\mathbb R$ such that $\ \psi(x)>0$ for $|x|<1$ $\ \psi(x)=0$ for $|x|\geq 1$ $\ \psi\in C^\infty$. Is it possible to find such a function $\psi$ that satisfies also one of the following conditions? For all $i,j=1,\dots,n$ $\ \displaystyle \frac{\partial_{x_i}\psi(x)}{\psi(x)}\to 0\ $, $\ \displaystyle\frac{\partial_{x_i}\partial_{x_j}\psi(x)}{\psi(x)}\to0\ $ as $|x|\to1\ $ $\ \displaystyle\lim\frac{\partial_{x_i}\psi(x)}{\psi(x)}\in\mathbb R\ $, $\ \displaystyle\lim\frac{\partial_{x_i}\partial_{x_j}\psi(x)}{\psi(x)}\in\mathbb R\ $ as $|x|\to1\ $. Intuitively this would mean that both $\psi$ and its derivatives vanish approaching the boundary of the ball, but the derivatives vanish faster than $\psi$ (or at least not slower). A typical example of function satisfying conditions 1.-3. is given by $$ \psi_0(x) = \begin{cases} e^{-\frac{1}{1-|x|^2}} &\textrm{ if }|x|<1 \\[2pt] 0 &\textrm{ if }|x|\geq1\end{cases} $$ but 4.,5. are clearly not satisfied by $\psi_0$ since $$ \frac{\nabla\psi_0(x)}{\psi_0(x)} \,=\, \frac{2x}{(1-|x|^2)^2}\,\xrightarrow[|x|\to1]{}\,\infty \,.$$ REPLY [6 votes]: Elaborating the comment by Wojowu: If we take a look at $n=1$ and $\psi\in C^\infty_{\text c}(\mathbb R)$ is a function satisfying conditions 1., 2. and 3. of your question, then for every $x\in]-1,1[$, we have, by smoothness of $\ln\psi$ on $]-1,1[$ and the fact that $\frac{\psi'}{\psi}$ is continuous on every $[-K,K]$ for $K\in]0,1[$, $$\ln\psi(x)=\ln\psi(0)+\int_0^x\frac{\psi'(s)}{\psi(s)}\,\mathrm ds.$$ Now, if we had $$\lim_{s\to 1}\frac{\psi'(s)}{\psi(s)}=r$$ for any real number $r\in\mathbb R$, then the function $$s\mapsto\frac{\psi'(s)}{\psi(s)}$$ would be well-defined and continuous on $[0,1]$, and therefore we would have $$\lim_{x\to1}\ln\psi(x)=\ln\psi(0)+\int_0^1\frac{\psi'(s)}{\psi(s)}\,\mathrm ds\in\mathbb R,$$ which is impossible since $\lim_{x\to1}\psi(x)=0$. Similarly, for a natural number $n>1$, any $\phi\in C_{\text c}^\infty(\mathbb R^n)$ with your properties 1., 2. and 3. defines a function $$\psi:\mathbb R\to\mathbb R, t\mapsto\phi(t,0,0,\dots, 0)$$ and we can proceed our argumentation as above.<|endoftext|> TITLE: Analog of Cerf theory in PL QUESTION [8 upvotes]: Is there an analog of Cerf theory in PL? More specifically, given two handle decompositions of a PL (relative) cobordism $W$, is it always possible to go from one handle decomposition to the other via a sequence of handle slides and handle cancellations? I think I have an argument, but I wanted to know if it is already known (and also check my argument): Choose some smoothing of the cobordism $W$. Construct Morse functions $f_0$ and $f_1$ that give the two handle decompositions (but smoothed). Find a homotopy $f_t$ that interpolates them such that $f_t$ only has at worst birth-death singularities - then $f_t$ gives a corresponding set of moves between handle decompositions. Approximate $W \times I$ by a triangulation, and approximate $f_t$ by a PL map. This should give the sequence of handle moves in PL. REPLY [4 votes]: PL manifolds are triangulable. The theorem you want is Pachner's theorem, that any two triangulations are related by stellar moves. You can turn triangulations into special PL handle decompositions, and PL handle decompositions you can triangulate. From this perspective, you could reduce your problem to showing that PL-triangulations could be viewed as very specific PL-handle decompositions. Then you argue that Pachner moves are obtainable by elementary handle operations, to complete the correspondence between handle decompositions up to elementary operations and triangulations up to stellar moves.<|endoftext|> TITLE: Definition of entropy and entropy flux of conservation laws: component-wise reasoning QUESTION [6 upvotes]: Consider the conservation law $$\DeclareMathOperator{\dvg}{\operatorname{div}} \partial_t u(x,t) + \dvg G(u(x,t)) =0, \\ u \in U\subseteq \mathbb R^m, x\in X\subseteq \mathbb R^n, G \subseteq \mathbb M^{n\times m}(\mathbb R). $$ We usually equip the system with an entropy (which is a scalar) $\eta$ with associate flux $Q \in \mathbb R^n$ related by $$ DQ_{\alpha} =D\eta \cdot DG_\alpha, \quad\alpha = 1,2,\ldots, n. \label{1}\tag{$\ast$} $$ Here are my concerns about this definition. In thermodynamics, the entropy flow is simply heat flux divided by temperature. It is NOT defined by the equation stated above via some other quantity $u.$ Therefore, I am puzzled by the following questions: Is the thermodynamic entropy/entropy flux an entropy in the sense of the definition above? In other words, does the tranditional physical entropy arise from some quantity $u$ via the above equation? If so, what is the physical meaning of $u?$ Are there other nonequivalent definitions of "mathematical" entropy? After all \eqref{1} is not necessary for $\eta(u)$ to be conserved for smooth solutions. To make sure $\eta(u)$ is conserved for smooth solutions, we only need $\text{div }Q = D\eta \cdot \dvg G(U)$ which is of course a much weaker condition. That leaves the possibility of defining other mathematical objects that looks somewhat like entropy. REPLY [3 votes]: Carlo's answer brings partial information. Let me complete it. First of all, a general gas is not polytropic, thus the pressure is not proportional to $e-\frac12\rho u^2=:\rho\varepsilon$. In other words, there is not such beast as $\gamma$, or it is not constant. Thus the entropy $\eta$ is not given by a log-algebraic formula in terms of $p$ and $\rho$. Instead, it is a function $\eta=-\rho s(\rho,p)$, where $s$ is obtained from the (Gibbs) differential equation $$\vartheta ds=d\varepsilon+pd\left(\frac1\rho\right)\,.$$ Even the above is not satisfactory, because the temperature $\vartheta$ is an integrating factor, which is far from unique. Actually, any $\bar s=h\circ s$ is another solution of the same equation, with $\bar\vartheta=\frac\vartheta{h'\circ s}$. Thus $\bar\eta=-\rho h\circ s$ is another entropy of the Euler system. This shows that the mathematical notion of entropy is far too wide. Which one is the correct entropy ? To answer this question, one cannot stay at the level of the Euler equation, but we must take in account a diffusive effect, namely the Fourier law of heat diffusion. This leads us to working with either Euler-Fourier system, or even with Navier-Stokes Fourier. Anyway, the conservation of energy is written as a second-order PDE. The heat diffusion occurs through the right-hand side $${\rm div}(\kappa\nabla\vartheta).$$ Now, the equations tell us what is the physical temperature, and thus what is the physical entropy. It is a theorem that ${\rm div}(\kappa\nabla\vartheta)$ cannot be recast as ${\rm div}(\bar\kappa\nabla\bar\vartheta)$, unless $h$ is an affine function. Thus every other entropy, admissible from the perspective of Physics, is a linear combination of $\eta$ and $\rho$.<|endoftext|> TITLE: Knot groups with big number of generators QUESTION [10 upvotes]: I start by saying that I am not an expert in this field and I apologize if the question is too elementary. Let $K$ be a knot in $S^3$. I denote by $\pi_1(K)$ the knot group, which is the fundamental group of its exterior: $$ \pi_1(K) = \pi_1(S^3 \smallsetminus K) .$$ The minimal number of generators of a knot $K$ is the minimal number of generators of $\pi_1(K)$. I am searching for knots with an arbitrarily high minimal number of generators. In particular: I found in https://arxiv.org/abs/1007.3175 , Lemma 4.17, a reference: Goodrick, R. (1968). Non-simplicially collapsible triangulations of In. In this article, the author proves that the connected sum of $n$ copies of a two-bridge knot is a $m$-bridge knot with $m$>$n$. A sharper result should hold from Schultens, Jennifer (2003). Additivity of bridge numbers of knots. I cannot understand, though, how this should prove the statement. In particular, by Knot theory question: bridge number vs. min generators of fundamental group of complement , this should not imply the result. Maybe I am missing something in the article, I did not go through the details. However, the article is quite dated and there is, hopefully, a simple way to prove this fact nowadays. So the first question is: is there a simple way to prove that there are knots with an arbitrarily high minimal number of generators? The techniques used seem to rely on the connected sum. What if we search for prime knots with an arbitrarily high minimal number of generators? What if we search for hyperbolic knots with an arbitrarily high minimal number of generators? Thank you in advance for the attention. REPLY [12 votes]: If $\pi_1(S^3\setminus K)$ has a presentation with $n$ generators then its representation variety $\mathrm{Hom}(\pi_1(S^3\setminus K),SL_2(\mathbb{C}))$ is a subvariety of $(SL_2(\mathbb{C}))^n$, which has complex dimension $3n$, so any component will have complex dimension at most $3n$. So if you want the minimal number of generators to be arbitrarily large, you just have to find knots with high-dimensional representation varieties. (You can replace $SL_2(\mathbb{C})$ with other groups if you prefer.) Cooper and Long ("Remarks on the A-polynomial of a knot", section 8) show how to construct hyperbolic knots where there are components of arbitrarily large dimension. Start with some $n$-fold connected sum, and realize it as the closure of a braid $\beta$ such that the union of the closure $\hat\beta$ with its braid axis is hyperbolic. Lift $\hat\beta$ to the $p$-fold cyclic branched cover of $S^3$, branched over the braid axis -- this will again be $S^3$, since the axis is unknotted -- to get a new knot, which is hyperbolic for all large enough $p$. (This knot will be the closure of the braid $\beta^p$.) Then the representation variety of this new knot turns out to have a component of dimension at least $n$ as well. REPLY [7 votes]: I think the rank of the Alexander polynomial gives you a lower bound on the number of generators of your fundamental group. i.e. just compute the Alexander module by lifting a 2-complex for the knot exterior to the universal cover. It's a very coarse bound, but it seems to answer your question, as there are hyperbolic knots with arbitrarily large Alexander polynomials. edit: my estimation was too quick. See Kyle Miller's comment below. His argument gets a lower bound and carefully avoids my mistake. But the statement needs to be changed, i.e. he uses full Alexander modules rather than Alexander polynomials.<|endoftext|> TITLE: Arithmetic groups and integral points of integral structures QUESTION [11 upvotes]: If $\mathbf{G}$ is an algebraic group defined over $\mathbb{Q}$, a subgroup of $\mathbf{G}(\mathbb{Q})$ is arithmetic if it is commensurable to $\mathbf{G}(\mathbb{Q}) \cap \operatorname{GL}_n(\mathbb{Z})$ where some representation $\mathbf{G} < \operatorname{GL}_n$ has been chosen (and the definition is made so that the choice does not matter). Fur the purpose of this question let us call a subgroup $\Gamma$ of $\mathbf{G}(\mathbb{Q})$ strictly arithmetic if there exists a group $\mathbb{Z}$-scheme $\mathbf{G}_\mathbb{Z}$ with generic fiber $\mathbf{G}$ such that $\Gamma = \mathbf{G}_\mathbb{Z}(\mathbb{Z})$. I was recently asked the natural question whether strictly arithmetic is the same as arithmetic. I suspect that the answer is "no". More specifically arithmetic groups can be arbitrarily small (for instance have arbitrarily large covolume in $\mathbf{G}(\mathbb{R})$) while I suspect that this is not true of strictly arithmetic groups. But I don't know enough about group schemes to underpin that intuition. So I'm asking here: Original question: Are there (resp. what are) examples of arithmetic groups that are not strictly arithmetic? The original question was answered in the comments by David Loeffler using a different obstruction so let me (following YCor's suggestion in the comments) specifically ask: Additional question: Do there exist "arbitrarily small" strictly arithmetic subgroups, for instance in the sense that the covolume or injectivity radius in $\mathbf{G}(\mathbb{R})$ is arbitrarily large? REPLY [11 votes]: First question (do non-strictly-arithmetic subgroups exist?): Any "strictly arithmetic" subgroup in your sense will, in particular, be a congruence subgroup, i.e. the intersection of $G(\mathbb{Q})$ with an open compact subgroup in $G(\mathbb{A}_f)$. Since non-congruence subgroups exist in $SL_2 / \mathbb{Q}$, and in lots of other groups too, these are examples of arithmetic subgroups which are not strictly arithmetic. Second question (can strictly arithmetic subgroups be small?): Start with your favourite $GL_n$-embedding $\iota$, defining some strictly arithmetic $\Gamma$. Take some $g \in G(\mathbb{Q})$ which isn't in $\Gamma$, and consider the embedding into $GL_{2n}$ sending $h$ to the block-diagonal matrix $\begin{pmatrix} \iota(h) \\& \iota(g^{-1} h g))\end{pmatrix}$. This defines a new $\mathbb{Z}$-model of $G$ whose integral points are $\Gamma \cap g \Gamma g^{-1}$. If $G = SL_2$, and probably for just about any $G$ which isn't abelian, the index of $\Gamma \cap g \Gamma g^{-1}$ in $\Gamma$ can be made arbitrarily large by a suitable choice of $\Gamma$.<|endoftext|> TITLE: A question on the monotonicity formula for minimal submanifolds QUESTION [5 upvotes]: I'm reading the proof of monotonicity formula from A Course in Minimal Surfaces by Colding-Minicozzi. The theorem says Suppose $\Sigma^k \subset \mathbb{R}^n$ is a minimal submanifold and $x_0\in\mathbb{R}^n$; then for all $00 \quad \mathcal{H}^k\mathrm{-a.e.} \text{ on } \Sigma,\label{1}\tag{$*$} $$ but didn't give a proof. How do I prove \eqref{1}? My idea: Since $\Delta_\Sigma f^2 = 2k>0$, $f^2$ is a subharmonic function on $\Sigma$. It suffices to show that the set of critical points of a smooth subharmonic function has $\mathcal{H}^k$ measure $0$. But I can't find a reference for the latter. REPLY [4 votes]: A stronger property is true: the critical set of a strictly subharmonic function $f: \Sigma \to \mathbf{R}$ is locally contained inside a codimension one submanifold. Explicitly: for every critical point $x \in \Sigma$ there is $r > 0$ and a $(k-1)$-dimensional submanifold $\Gamma \subset \Sigma \cap D_r(x)$ so that $\{ \nabla f = 0 \} \cap D_r(x) \subset \Gamma$. (Here $D_r(x)$ is the ball of radius $r$ in the intrinsic metric on the surface.) Let $x \in \Sigma$ be such a critical point. In local normal coordinates one can identify $\nabla^2 f(x) = (\partial^2_{ij} f(x))$. This matrix has positive trace because $\Delta f(x) > 0$. Therefore at least one partial derivative $\nabla \partial_i f(x) \neq 0$, and by the implicit function theorem the zero set $\{ \partial_i f = 0 \}$ is a $C^1$ embedded surface in a small enough neighbourhood of $x$. We conclude by setting $\Gamma = \{ \partial_i f = 0 \} \cap D_r(x)$ and observing that $\{ \nabla f = 0 \} \cap D_r(x) \subset \Gamma$.<|endoftext|> TITLE: Uses of Volodin's construction of algebraic K-theory QUESTION [12 upvotes]: There is a construction of the algebraic K-theory groups $K_i(R)$ of a ring $R$ by Volodin. He gave an explicit construction of the plus-construction $BGL(R)^+$ as the quotient of the bar construction $BGL(R)$ by the union $\bigcup_{n,\sigma} BU_n(R)^\sigma$ of the classifying space of the group $U_n(R)$ upper-triangular matrices with ones on the diagonal, conjugated by a permutation matrix $\sigma$. What are the uses of this construction in modern approaches to algebraic K-theory? What are its advantages and disadvantages? REPLY [12 votes]: It has several uses: Volodin $K$-theory was used by Igusa in the late 1970s/Early 1980s to define $K$-theoretic invariants of families of pseudoisotopies. In the 1980s, the Volodin construction (actually a variant of it) was used by Goodwillie to relate rational relative K-theory to rational relative THH. This was further developed by Dundas and McCarthy in their famous result that stable K-theory is isomorphic to THH. This is the genesis of trace methods in algebraic K-theory. In the early 1990s, Suslin and Wodzicki used Volodin's space to obtain excision results for algebraic $K$-theory.<|endoftext|> TITLE: An alternate definition of Sobolev space $W^{1,p}(\Omega)$ when $1 TITLE: Realizability for constructive Zermelo-Fraenkel set theory QUESTION [6 upvotes]: $ \def \CZF {\mathbf {CZF}} \def \IZF {\mathbf {IZF}} \def \A {\mathcal A} \def \then {\mathrel \rightarrow} \def \r {\mathrel \Vdash} \DeclareMathOperator \V V $ In "Realizability for Constructive Zermelo-Fraenkel Set Theory", Michael Rathjen shows that a notion of realizability due to Charles McCarty works well for $ \CZF $ (Constructive Zermelo-Fraenkel set theory), a more restricted theory than $ \IZF $ (Intuitionistic Zermelo-Fraenkel set theory), which is the theory of concern in McCarty's work. The notion is defined in terms of the realizability (class) structure $ \V ( \A ) $ over an applicative structure $ \A $. For $ e \in | \A | $ and a sentence $ \phi $ in the language of set theory extended by adding constants denoting members of $ \V ( \A ) $, realizability of $ \phi $ by $ e $ is defined recursively on the subformula tree of $ \phi $, and the notation "$ e \r \phi $" is used, read as "$ e $ realizes $ \phi $". The definition for the cases where $ \phi $ is not atomic is similar to the definitions for corresponding cases in the other well-known notions of realizability, e.g. that of Kleene defined for the theories of arithmetic. It is the definition of realizability for atomic sentences which I can't quite understand: \begin{align*} e \r a \in b \iff& \exists c \big( \langle ( e ) _ 0 , c \rangle \in b \land ( e ) _ 1 \r a = c \big) \\ e \r a = b \iff& \forall f , d \Big( \big( \langle f , d \rangle \in a \then ( e ) _ 0 f \r d \in b \big) \land {} \\ &\qquad\qquad \big( \langle f , d \rangle \in b \then ( e ) _ 1 f \r d \in a \big) \Big) \end{align*} Here $ a , b \in \V ( \A ) $, juxtaposition is used to denote application in the structure $ \A $, $ \langle . , . \rangle $ is a fixed pairing function, and $ ( . ) _ 0 $ and $ ( . ) _ 1 $ are the corresponding projections. Struggling to understand how this definition works, I ended up asking myself the following questions, which I couldn't figure out, and thus I decided to ask here. Isn't this definition circular? Something of the form $ e \r a = b $ appears in the definition of $ e \r a \in b $, and vice versa. Is this definition sensitive to the minimal language of set theory? More specifically, if one extends the language by adding function symbols to the language, say a unary symbol denoting the union of members of a set, would the notion cease to work? Would one need to break the case of atomic sentences into cases where the form of the terms appearing in the sentence is taken into account? Or would it be similar to Kleene's realizability where the atomic sentences are treated regardless of the addition and multiplication symbols appearing in the terms? In case where adding function symbols does not affect the way the definition works, does constructivity of the intended function really matter? This question comes for example from the fact that $ \IZF $ contains power sets, which may not be considered constructive (as they are rejected in $ \CZF $). To make this more specific and go even beyond $ \IZF $, would the notion of realizability work if we add a binary function symbol $ \chi $ to the language, with the intended meaning of the characteristic (class) function of membership, and add the following axioms to the language (one can either add $ \varnothing $ and $ \{ . \} $ to the language, or modify the following sentences in the obvious way so that they don't contain these symbols)? $ \forall x , y ( x \in y \then \chi ( x , y ) = \{ \varnothing \} ) $ $ \forall x , y ( \neg x \in y \then \chi ( x , y ) = \varnothing ) $ Rathjen, Michael, Realizability for constructive Zermelo-Fraenkel set theory, Stoltenberg-Hansen, Viggo (ed.) et al., Logic colloquium ’03. Proceedings of the annual European summer meeting of the Association for Symbolic Logic (ASL), Helsinki, Finland, August 14–20, 2003. Wellesley, MA: A K Peters; Urbana, IL: Association for Symbolic Logic (ASL) (ISBN 1-56881-293-0/hbk; 1-56881-294-9/pbk). Lecture Notes in Logic 24, 282-314 (2006). ZBL1102.03053. McCarty, Charles, Realizability and recursive set theory, Ann. Pure Appl. Logic 32, 153-183 (1986). ZBL0631.03035. REPLY [8 votes]: For your first question, the definition of $e\Vdash x\in y$ and $e\Vdash x=y$ seems circular, but $\mathsf{CZF}$ provides a way to avoid the circularity, called inductive definition. Definition. An inductive definition is a class $\Phi\subseteq \mathcal{P}(V)\times V$. For each inductive definition $\Phi$, define $\Gamma_\Phi(X)=\{a \mid \exists Y\subset X : (Y,a)\in \Phi\}$. (Note that $X$ may be a class, but $Y$ must be a set.) Sometimes we write $X/_\Phi a$ or $X\vdash_\Phi a$ instead of $(X, a)\in \Phi$, to emphasize the analog between deduction and inductive definition. In that view, we may think $\Gamma_\Phi(X)$ the least class containing $X$ that is closed under $\Phi$-deduction. It is known that $\mathsf{CZF}$ admits arbitrary inductive definition (unlike $\mathsf{KP}$ only admit $\Sigma$-ones): Theorem. (Class Inductive Definition Theorem) Let $\Phi$ be an inductive definition. Then there is a unique $\Gamma_\Phi$-least fixed point $I(\Phi)$, i.e., $I(\Phi)$ is least among classes such that $\Gamma_\Phi(I(\Phi))\subseteq I(\Phi)$. But justifying the circular definition with an inductive definition is sometimes a bit tricky. In our case, it seems that we need a 'mutual' inductive definition to justify it. Fortunately, it turns out that this is not the case: we may remove every formula of the form $f\Vdash x\in y$ in the definition of $e\Vdash a=b$ by replacing $f\Vdash x\in y$ to its definition, so we have a recursive definition of $e\Vdash a=b$. Then we can define $e\Vdash a\in b$ from $f\Vdash x=y$. For your second question, I do not get the point of your question clearly, but if my understanding ― whether adding a new predicate or function into the language affects the definability of $\Vdash$ ― is correct, then I think the answer is yes in general. The reason is that Class Inductive Definition Theorem works for arbitrary $\Phi$, although adding functional symbols would require more justifications. For example, consider the case of adding the binary unpaired order function $\{,\}$. For each $a,b\in V^{\mathcal{A}}$ for a pca $\mathcal{A}$, define $$\mathsf{up}(a,b) = \{\langle\underline{0}, a\rangle,\langle\underline{1}, b\rangle\}.$$ Now interpret $\{,\}$ by making use of $\mathsf{up}$, that is, for each $a,b\in V^\mathcal{A}$, take $\{a,b\}^{V(\mathcal{A})} := \mathsf{up}(a,b)$. This means we will replace every occurrence of $\{a,b\}$ in the realizability relation to $\mathsf{up}(a,b)$. A more dramatic example would be given by Chen and Rathjen: they defined $\Vdash$ over the extension of $\mathsf{CZF}$, which contains natural numbers and relevant operations as primitive notions. (Also it looks like a mere combination of the realizability à la McCarty and that of Kleene.) For the last question, I think the answer is yes if the ambient theory ($\mathsf{CZF}$ in our case) proves the functionality of our desired function. The following theorem seems relevant (Proposition 9.6.2 of Aczel and Rathjen.) Theorem. Let $T$ be a theory which comprises $\mathsf{BCST}$ (e.g. $\mathsf{CZF}$). Suppose $T \vdash ∀x ∃!yΦ(x, y)$. Let $T_Φ$ be obtained by adjoining a function symbol $F_Φ$ to the language, extending the schemata to the enriched language, and adding the axiom $∀x Φ(x, F_Φ(x ))$. Then $T_Φ$ is conservative over $T$. Focusing on your example, in fact, we can define $\chi$ over $\mathsf{CZF}$: take $\chi(x,y)=\{0\mid x\in y\}$. By using this definition, we have a conservative extension of $\mathsf{CZF}$, and we may simply reduce $\chi$ in the extended language to a corresponding definition over $\mathsf{CZF}$.<|endoftext|> TITLE: Checking if polynomial can be iterated and only take prime values QUESTION [5 upvotes]: I have the polynomial $f(x) = x^2-x+1$ and I am wondering if there is a positive prime value $p$ such that $f(p),f^2(p),f^3(p)\dots$ are all prime. I have ran some computer simulations and I feel like the answer should be "no" ( because looking at the map $x^2-x+1 \bmod p$ I get that the expected number of prime divisors of the first $M$ values should be larger than $M$). I feel that my analysis is not very good however. Does anyone know an approach which could be more fruitful? Note: $f^2(x) = f(f(x))$ Edit: I would be happy with any sort of reference regarding a polynomial staying inside the primes under iteration, if there isn't anything particularly useful which can be said about this sort of thing I understand that as well). REPLY [5 votes]: This is not an answer to your question, but will point you toward work on the number theoretic properties of such sequences. Iteration of $x^2-x+1$ starting at $a=2$ is called the Sylvester sequence. A theorem about primes that divide the terms in such sequences was proved by Rafe Jones (The density of prime divisors in the arithmetic dynamics of quadratic polynomials. J. Lond. Math. Soc. (2) 78 (2008), no. 2, 523–544. MR2439638) One of his results is that if $k\in\mathbb Z$ with $k\notin\{0,2\}$, then the set of primes dividing the integers in the orbit of any $a\in\mathbb Z$ under iteration of $x^2+kx-1$ has density $0$. In a different direction, one can prove that the numbers in your sequence, for any starting prime $p$, grow quadratically exponentially, and in fact $$ \lim_{n\to\infty} \frac{1}{2^n} \log f^n(p) \quad\text{converges to a positive real number.} $$ Thus your sequence is growing very rapidly, which means that we have very few tools at our disposal to prove statements of the sort you ask.<|endoftext|> TITLE: Contractible 4-manifold with an exhaustion QUESTION [6 upvotes]: Let $M^4$ be a noncompact, contractible, 4-dimensional, topological manifold. Suppose there exists an exhaustion $M=\bigcup_{i\ge 1} U_i$ by open sets such that (1) $\bar U_i \subset U_{i+1}$ and (2) each $U_i$ is homeomorphic to $\mathbb R^2 \times \mathbb T^2$. Can we prove that $M$ is homeomorphic to $\mathbb R^4$? REPLY [4 votes]: It seems to me that there is an obvious Whitehead-manifold inspired construction to try here to produce a counterexample. Let $w : S^1 \times D^2 \to S^1 \times \operatorname{int} D^2$ be the familiar Whitehead embedding, which is null-homotopic but not “null-isotopic”, for which the two inclusions $S^1 \times \partial D^2 \hookrightarrow S^1 \times D^2 \setminus w(S^1 \times \operatorname{int} D^2)$ and $w(S^1 \times \partial D^2) \hookrightarrow S^1 \times D^2 \setminus w(S^1 \times \operatorname{int} D^2)$ are each $\pi_1$-injective (google Whitehead manifold.) For $i = 1$ and $2$ let $w_i : T^2 \times D^2 \to T^2 \times D^2$ be the two obvious product embeddings engendered by $w$, namely, let $w_2 : = \operatorname{iden}(S^1) \times w : S^1 \times S^1 \times D^2 \to S^1 \times S^1 \times \operatorname{int} D^2$, and similarly let $w_1 =$ “$w \times \operatorname{iden}(S^1)$” (the quotes meaning that you do the appropriate factor permutations here to make it work). It looks as though the contractible direct-limit 4-manifold $M^4$ gotten from the sequence $w^1$, $w^2$, $w^1$, $w^2$, … is not simply-connected at infinity (like the Whitehead 3-manifold). True?<|endoftext|> TITLE: Permutations $\pi\in S_{p-1}$ with $\frac1{\pi(1)\pi(2)}+\frac1{\pi(2)\pi(3)}+\cdots+\frac1{\pi(p-2)\pi(p-1)}+\frac1{\pi(p-1)\pi(1)}\equiv0\pmod{p^2}$ QUESTION [5 upvotes]: A well known congruence of Wolstenholme states that $$\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{(p-1)^2}\equiv0\pmod{p}$$ for any prime $p>3$. For each $n=3,4,\ldots$ we clearly have $$\frac1{1\times2}+\frac1{2\times3}+\cdots+\frac1{(n-1)n}+\frac1{n\times1} = 1.$$ Motivated by the above, here I ask a new question. Question. Is it true that for each prime $p>3$ there is a permutation $\pi\in S_{p-1}$ with $\pi(p-1)=p-1$ and $\pi(p-2)=p-2$ such that the congruence $$\frac1{\pi(1)\pi(2)}+\frac1{\pi(2)\pi(3)}+\cdots+\frac1{\pi(p-2)\pi(p-1)}+\frac1{\pi(p-1)\pi(1)}\equiv0\pmod{p^2}$$ holds? For $p=5$, there is a unique permutation $\pi\in S_4$ meeting the requirement, namely, $$\frac 1{2\times1}+\frac1{1\times3}+\frac1{3\times4}+\frac1{4\times2}=\frac{25}{24}\equiv0\pmod{5^2}.$$ For $p=7$, there is also a unique permutation $\pi\in S_6$ meeting the requirement, namely, $$\frac1{2\times3}+\frac1{3\times4}+\frac1{4\times1}+\frac1{1\times5}+\frac1{5\times6}+\frac1{6\times2}=\frac{49}{60}\equiv0\pmod{7^2}.$$ For $p=11$ there are totally $323$ permutations $\pi\in S_{10}$ meeting the requirement. For $p=13$, the permutation $$(\pi(1),\ldots,\pi(12))=(1,2,3,7,4,9,5,8,10,6,11,12)$$ meets our purpose. Based on these data, I conjecture that the question has a positive answer. Your comments are welcome! REPLY [3 votes]: Let $$ (\pi(1),\pi(2),\ldots,\pi(p-1))=(2,3,\ldots,p-3,1,p-2,p-1). $$ Then \begin{align*} \frac1{\pi(1)\pi(2)}+\frac1{\pi(2)\pi(3)}+\cdots+\frac1{\pi(p-2)\pi(p-1)}+\frac1{\pi(p-1)\pi(1)}=&\frac{p^2}{2(p-1)(p-2)}\\ \equiv&0\pmod{p^2}. \end{align*}<|endoftext|> TITLE: Siegel zeros and other "illusory worlds": building theories around hypotheses believed to be false QUESTION [50 upvotes]: What are some examples of serious mathematical theory-building around hypotheses that are believed or known to be false? One interesting example, and the impetus for this question, is work in number theory based on the assumption that Siegel zeros exist. If there were such things, then the Generalized Riemann Hypothesis would be false, which it presumably isn't. So it's unlikely that there are Siegel zeros. Still, lots of effort has gone into exploring the consequences of their existence, which have turned out to be numerous, interesting, surprising and so far self-consistent. The phenomena generated by the Siegel zero hypothesis are sometimes referred to as an "illusory world" or "parallel universe" sitting alongside that of ordinary number theory. (There's some further MO discussion e.g. here and here.) I'd like to hear about other examples like this. I'd be particularly grateful for references, especially those that discuss the motivations behind and benefits of undertaking such studies. I should clarify that I'm mainly interested in "illusory worlds" built on hypotheses that were believed to be false all along, rather than those which were originally believed true or plausible and only came to be disbelieved after the theory-building was done. Further context: I'm a philosopher interested in counterfactual reasoning in mathematics. I'd like to better understand how, when and why mathematicians engage with counterfactual scenarios, especially those that are taken seriously for research purposes and whose study is viewed as useful and interesting. But I'd like to think this question might be stimulating for the MO broader community. REPLY [5 votes]: This is more of a theoretical/mathematical physics example, but it can happen in mathematical physics that a lot of stuff is built around hypotheses which are essentially known to be false from the beginning or objects which are known not to exist. Some examples: 1.) I know that you asked for hypotheses which were always known to be false, but Tait originally started to develop knot theory because Kelvin had hypothesised that atoms can be obtained as knots in the ether. The concept of the ether was invalidated by experiments of Michelson and Morley which obviously also invalidated the hypothesis of Kelvin regarding the physical basis of knot theory. The mathematics of knots has survived and flourished up to the present day although this application did not work. 2.) A huge amount of theoretical and mathematical papers have been published on magnetic monopoles (including Dirac and 't Hooft-Polyakov monopoles) although the consensus looks to be that magnetic monopoles likely do not exist in this Universe which we live in. I am not sure if Dirac regarded the hypothesis of existence of monopoles to be false ''all along'', but this is certainly possible, as magnetic monopoles are forbidden by the mathematical equations of classical electromagnetism and Dirac was considering what happened theoretically if you increased the amount of symmetry which the equations have.<|endoftext|> TITLE: Irreducible representations of product of profinite groups QUESTION [6 upvotes]: It is a standard fact in the representation theory of finite groups that for $G,H$ finite groups, all of the irreducible representations of $G \times H$ are the external tensor product of irreps of $G$ and $H$. Today I was talking to a friend about profinite groups and it got me thinking: "Is (some version of) this result still true?" The fact that so many results from the finite case carry over makes me think that this could be true, but I have no idea how to go about proving it. The standard proof for finite groups uses a counting argument to show that they are all of this form, so certainly some higher-level techniques will be required. Since we're considering profinite groups, we will definitely want to restrict ourselves to continuous representations on topological vector spaces. If the statement is not true in this generality, are there adjectives we can add that make it true? What if our representations are unitary, or the profinite groups are (topologically) finitely-generated? Any results, no matter the number of hypotheses, would be of interest to me. REPLY [7 votes]: This is not even true for finite groups, in this generality, and not even in characteristic $0$. Consider, for example, the group $Q_8 \times C_3$, where $Q_8$ is the quaternion group and $C_3$ is cyclic of order $3$, and consider $\mathbb{Q}$-representations of this direct product. The standard representation $\rho$ of $Q_8$ is not realisable over $\mathbb{Q}$, only $\rho\oplus \rho$ is. $C_3$ has an irreducible $\mathbb{Q}$-representation $\chi$, given by the sum of the two non-trivial irreducible complex characters of $C_3$. Now, $\rho\otimes \chi$ is realisable over $\mathbb{Q}$ and defines a simple $\mathbb{Q}[G]$-module, but it is not of the form $V\otimes W$ for any $\mathbb{Q}[Q_8]$-module $V$ and $\mathbb{Q}[C_3]$-module $W$. If you wanted to restrict to finite dimensional representations over $\mathbb{C}$, then the statement will be true also for profinite groups, because any continuous complex finite dimensional representation of a profinite group will factor through a finite quotient.<|endoftext|> TITLE: Induction arising in proof of Berry Esseen theorem QUESTION [9 upvotes]: I've been studying the paper An estimate of the remainder in a combinatorial central limit theorem by Bolthausen, which proves the Berry Essen theorem using Stein's method: Let $\gamma$ be the absolute third moment of a random variable $X$, and let $X_{i}$ be iid with the same law as $X$. Let $S_{n}=\sum_{i}^{n}X_{i}$, and suppose $E(X)=0$, $E(X^{2})=1$. The goal is to find some universal constant $C$ such that $|P(S_{n} \leq x) - P(Y\leq x)| \leq C\frac{\gamma}{\sqrt{n}}$. Let $\delta(n,\gamma) = \sup_{x}|P(S_{n} \leq x) - P(Y\leq x)|$. We would like to bound $\sup_{n}\frac{\sqrt{n}}{\gamma}\delta(n, \gamma)$. In the proof the following bound is derived: $\delta(n,\gamma) \leq c\frac{\gamma}{\sqrt{n}}+\frac{1}{2}\delta(n-1,\gamma)$ where $c$ is a universal constant. Noting that $\delta(1,\gamma) \leq 1$, the author claims that the result is implied. However when I try to use induction to get the result the constant $C$ increases without bound $n$ grows. If anyone has studied this paper before, I would love to hear from you. REPLY [14 votes]: Let $a_n = \frac{\sqrt{n}}{\gamma} \delta \left( n, y \right)$. The bound you have stated implies that $$a_n \leq c + \frac{2}{3} a_{n - 1}$$ where I replaced $\frac{\sqrt{n}}{\sqrt{n - 1}}$ with $\frac{4}{3}$ which is certainly true for $n > 2$. Therefore, $$a_n \leq c + \frac{2}{3} a_{n - 1} \leq c \left( 1 + \frac{2}{3} \right) + \left( \frac{2}{3} \right)^2 a_{n - 2} \leq \cdots \leq c \frac{1}{1 - \frac{2}{3}} + a_2$$ which is bounded as required. REPLY [9 votes]: A slight modification of the answer by Random: Given that $a_n\le c+\frac23\,a_{n-1}$ for $n\ge3$, take any real $b\ge\max(3c,a_2)$. Then $a_n\le b$ for all $n\ge2$, by induction on $n$.<|endoftext|> TITLE: Cohomological dimension of torsion-free groups and its subgroups QUESTION [6 upvotes]: In this thesis by Martin Hamilton on Finiteness Conditions in Group Cohomology there is on page 11 a reference to following result: Theorem 1.2.14. If $G$ is a torsion-free group and $H$ is a subgroup of finite index, then $$ \operatorname{cd} H = \operatorname{cd} G $$ where $\operatorname{cd} G $ is the cohomological dimension of $G$ defined as the projective dimension of $\mathbb{Z}$ considered as $\mathbb{Z}G$-module with trivial $G$ action, i.e. $g.1=1$ for every $g \in G$. That is $\operatorname{cd} G = \operatorname{proj.dim}_{\mathbb{Z}G} \mathbb{Z}$ and the latter is defined as the minimal length of all projecive resolutions $$ 0 \to P_n \to P_{n-1} \to ... \to P_1 \to P_0 \to \mathbb{Z} \to 0 $$ of projective $\mathbb{Z}G$-modules $P_j$. In the thesis the author gave as reference Jean-Pierre Serre's publication "Cohomologie des groupes discrets", can be found in this Bourbaki collection band: https://www.springer.com/gp/book/9783540057208 Unfortunatelly, this result cannot be found in this publication. So my concern is where I can find a complete proof of the quoted Theorem above. REPLY [13 votes]: This is Theorem 3.1, p. 190, in Brown, "Cohomology of groups". He also attributes it to Serre. As a remark, this is the reason that the virtual cohomological dimension (vcd) is well-defined.<|endoftext|> TITLE: Chromatic orientability of manifolds QUESTION [13 upvotes]: If a compact manifold $M$ with empty boundary is oriented with respect to all the connective Morava $K$-theories $k(n)_*$, localized at a prime $p$, can one conclude that $M$ is orientable with respect to $p$-local Brown-Peterson theory $BP_*$? [As an example, Chris Lloyd and I know that the hypothesis holds for the real Grassmanians $Gr_2(\mathbb R^m)$ with $m$ even (and $p=2$).] Added later: Chris and I have been having fun studying the Morava $K$-theory of $Gr_d(\mathbb R^m)$ (at $p=2$), and among other things, it seems that when $m$ is even, all of the spaces $Gr_d(\mathbb R^m)$ are $k(n)$-orientable for all $n$ and $d$. So I was just idly pondering what this means. REPLY [7 votes]: If $p$ is odd then this is easy but dull. Using the truncation $k(n) \to HF_p$, a $k(n)$-orientable vector bundle is orientable in the usual sense. On the other hand, $p$-locally the Thom spectrum $MSO$ has a cell structure with only even cells, so is orientable with respect to any even ring spectrum.<|endoftext|> TITLE: Does any cubic polynomial become reducible through composition with some quadratic? QUESTION [34 upvotes]: What I mean to ask is this: given an irreducible cubic polynomial $P(X)\in \mathbb{Z}[X]$ is there always a quadratic $Q(X)\in \mathbb{Z}[X]$ such that $P(Q)$ is reducible (as a polynomial, and then necessarily the product of 2 irreducible cubic polynomials)? I did quite some testing and always found a $Q$ that does the job. For example: $P=aX^3+b,\quad Q=-abX^2,\quad P(Q)=-b(a^2bX^3-1)(a^2bX^3+1)$ $P=aX^3-x+1,\quad Q=-aX^2+X,\quad P(Q)=-(a^2X^3-2aX^2+X-1)(a^2X^3-aX^2+1)$ and a particular hard one to find: $P=2X^3+X^2-X+4,\quad Q=-8X^2+5X+1,\quad P(Q)=(16X^3-18X^2+X+3)(64X^3-48X^2-11x-2)$ Could there be a formula for $Q$ that works for all cases? It feels to me that this may have a really basic Galois theoretic proof or explanation, but I can't figure it out. Update. Maybe a general formula for $Q$ is close. For $P=aX^3+cX+d$ taking $Q=-adX^2+cX$ works. REPLY [33 votes]: You should refer to Lemma 10 (page-233) in this paper by Schinzel where he proves that for any polynomial $F(x)$ of degree $d$ we have a polynomial $G(x)$ of degree $d-1$ such that their composition is reducible.<|endoftext|> TITLE: Does there exist an almost surely differentiable martingale? QUESTION [5 upvotes]: Does there exist a continuous time martingale $X_t$ not a.s. constant in $t$ that is almost surely everywhere differentiable? REPLY [2 votes]: Others have answered that there does not exist an a.s. differentiable martingale process that's not constant. One related fact (not asked for but interesting) is that an a.s. differentiable Markov process satisfies a deterministic ODE with random initial conditions. See https://www.ams.org/journals/notices/196808/196808FullIssue.pdf page 748 Somehow martingale/Markov don't work well with differentiability.<|endoftext|> TITLE: Construction of the universal covering space of the etale homotopy type $Et(X)$ QUESTION [5 upvotes]: Let $X$ be a nice scheme (additional assumptions could be added), and let $Et(X)$ be its (Artin-Mazur) etale homotopy type. I am looking for a/the scheme $Y$ over $X$ whose etale homotopy type $Et(Y)$ will be the topological universal cover of $Et(X)$. By definition $Et(X)$ is the geometric realization of a simplicial set, and it was pointed out to me that if $R \rightarrow S$ is the universal cover of a simplicial set $S$, then the geometric realization $|R|$ is the topological universal cover of $|S|$. What is the meaning of "Universal cover of a simplicial set"; Is there a reference for that, and also for the second assertion?. How could we apply this to find $Y$? Edit: If we want to avoid the simplicial method. Suppose that there is a scheme $Y$ over $X$ such that The first etale homotopy group $\pi_1^{et}(Y)$ is trivial. For all $n \geq 2$ we have $\pi_n^{et}(Y) \simeq \pi_n^{et}(X)$. Are these conditions sufficient to state that $Et(Y)$ is the topological universal cover of $Et(X)$? REPLY [6 votes]: Such an "étale universal cover" exists at least if $X$ is Noetherian and geometrically unibranch (and for all qcqs $X$ if one considers profinite étale homotopy types). Background. I will regard the étale homotopy type of a scheme as an object in the $\infty$-category $\mathrm{Pro}(\mathcal S)$ of pro-spaces. In the $\infty$-category $\mathcal S$, the universal cover of a pointed space $(X,x)$ can be characterized as the initial object in the $\infty$-category of pointed $0$-truncated morphisms to $(X,x)$ (i.e., morphisms with discrete fibers). Similarly, one can define $0$-truncated morphisms in $\mathrm{Pro}(\mathcal S)$, and every pointed object admits a universal cover. References for the étale homotopy type and for $n$-truncated/$n$-connected morphisms in $\mathrm{Pro}(\mathcal S)$ are Sections E.2 and E.4.2 in Spectral Algebraic Geometry. Let $\mathcal S_{<\infty}$ be the $\infty$-category of truncated spaces and let $\mathrm{Et}\colon \mathrm{Sch} \to \mathrm{Pro}(\mathcal S_{<\infty})$ be the protruncated étale homotopy type, which is the homotopy-coherent incarnation of the construction of Artin and Mazur. Under the equivalence $$\mathrm{Pro}(\mathcal S_{<\infty}) \simeq \mathrm{Fun}^\mathrm{acc,lex}(\mathcal S_{<\infty},\mathcal S)^\mathrm{op},$$ $\mathrm{Et}(X)$ is the functor $\mathcal S_{<\infty}\to\mathcal S$ sending a truncated space $K$ to the global sections $\Gamma(X_\mathrm{et},K)$ of the constant étale sheaf on $X$ with value $K$. Construction. Let $X$ be a scheme with a geometric point $x\colon \operatorname{Spec} k\to X$, with $k$ separably closed. Consider the category $\mathrm{FEt}(X,x)$ of pointed finite étale covers of $X$, that is, factorizations of $x$ through a finite étale morphism $X'\to X$. This category has finite limits and in particular is cofiltered (it is also essentially small). Thus, the limit $\tilde X$ of the forgetful functor $$ \mathrm{FEt}(X,x) \to \mathrm{Sch}_{/X} $$ exists (note that $\tilde X$ depends on $x$, so the notation is abusive). This is a natural candidate for the "universal cover" of $(X,x)$. Now, the statement that $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is the universal cover of the pointed pro-space $(\mathrm{Et}(X),x)$ is equivalent to the following three conditions: 1) the map $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is $0$-truncated, 2) $\mathrm{Et}(\tilde X)$ is connected, and 3) $\pi_1^\mathrm{et}(\tilde X,x)$ is trivial. Results. 1) always holds if $X$ is quasi-compact and quasi-separated. First, I claim that the functor $\mathrm{Et}$ preserves the limit of the diagram defining $\tilde X$. Under the above equivalence of $\infty$-categories, this is the statement that for $K$ a truncated space, $\Gamma((-)_\mathrm{et},K)$ transforms this limit into a colimit, which is a standard property of étale cohomology with respect to inverse limits of qcqs schemes. [Here it is important that $K$ is truncated, otherwise this may not be true.] Then, if $p\colon X'\to X$ is finite étale, I claim that the morphism $\mathrm{Et}(p)\colon \mathrm{Et}(X')\to \mathrm{Et}(X)$ is $0$-truncated. In fact, it is the pullback of a morphism of groupoids $\pi\colon\Xi'\to\Xi$ with finite discrete fibers. To see this, note that the morphism of étale $\infty$-topoi induced by $p$ is itself the pullback of such a morphism $\pi$. The point is then that for any space $K$, $\Gamma(X'_\mathrm{et}, K)$ is $\Gamma(X_\mathrm{et},p_*K)$ and $p_*K$ is a locally constant sheaf on $X$ in a strong sense: it is the pullback of the sheaf $\pi_*K$ on $\Xi$ (the fact that the fibers of $\pi$ are finite spaces is used here, to commute the pushforward with the pullback). One can thus apply Proposition 2.15 in Higher Galois theory to compute $\Gamma(X_\mathrm{et},p_*K)$ in terms of the étale homotopy type of $X$. Unpacking this formula gives $\mathrm{Et}(X')=\mathrm{Et}(X)\times_\Xi\Xi'$. [Here, $K$ need not be truncated, so $p$ induces a $0$-truncated morphism on actual étale homotopy types, and $X$ need not even be qcqs.] ETA: If $X$ is locally noetherian, another proof of this claim is Lemma 2.1 in Schmidt-Stix, Anabelian geometry with etale homotopy types. Finally, since $0$-truncated morphisms are stable under limits, $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is also $0$-truncated. 2) also holds if $X$ is qcqs. In this case $\tilde X$ is connected, as any clopen subset of $\tilde X$ lifts to a clopen subset of some finite étale $X'\to X$. 3) holds if we assume moreover that the pro-group $\pi_1^\mathrm{et}(X,x)$ is profinite, e.g., $X$ is Noetherian and geometrically unibranch. Then $\pi_1^\mathrm{et}(\tilde X,x)$ is also profinite by 1), so 3) is equivalent to the statement that every finite étale cover of $\tilde X$ is trivial, which holds by construction. Note that the additional assumption for 3) is not needed if we pass to the profinite completions. That is, if $X$ is qcqs, then $\mathrm{Et}(\tilde X)^\wedge \to \mathrm{Et}(X)^\wedge$ is the universal cover of the profinite étale homotopy type $(\mathrm{Et}(X)^\wedge,x)$. Example. Let $X=\mathbb G_m$ over a field $k$ of characteristic zero, pointed at $1$ by an algebraic closure $\bar k$ of $k$. Then $\tilde X=\operatorname{Spec}\bar k[t^{\pm 1}][t^{1/n}, n\geq 2]$.<|endoftext|> TITLE: Birman-Series for variable negative curvature QUESTION [11 upvotes]: A famous theorem of Birman and Series says that if $S$ is a compact hyperbolic surface, then the set of points that lie on simple geodesics is nowhere dense and has Hausdorff dimension one; in particular, it has measure zero. This is proved in Birman, Joan S.; Series, Caroline, Geodesics with bounded intersection number on surfaces are sparsely distributed. Topology 24 (1985), no. 2, 217–225. Question: Is this true for compact oriented surfaces equipped with metrics of variable negative curvature? REPLY [7 votes]: Here is another argument which reduces the general result to the constant curvature case: For any negatively curved metric $g$ on $S$, the set of geodesics of the universal cover $\tilde{S}$ is canonically identified with $\partial_\infty \tilde{S}^{(2)}$, the set of pairs of distinct points in the Gromov boundary of $\tilde{S}$, in which $\overline{\Lambda}$ (the subset of geodesics that are embedded in $S$) is a closed subset. Note that the Gromov boundary and the set $\Lambda$ are quasi-isometric invariants, so they don't really depend on the choice of the metric $g$. What choosing the metric $g$ does is it puts a $\mathcal C^1$ structure on $\partial_\infty \tilde{S}^{(2)}$. Let $g_0$ be another metric of constant curvature of minus one. Then Birman and Series' theorem tells you that $\overline{\Lambda} \subset \partial_\infty \tilde{S}^{(2)}$ has Hausdorff dimension $0$ with respect to the $\mathcal C^1$ structure defined by $g_0$. Now, one can prove that the two $\mathcal C^1$ structures on $\partial_\infty \tilde{S}^{(2)}$ induced respectively by $g$ and $g_0$ are Hölder regular with respect to each other, which implies that $\overline{\Lambda}$ also has Hausdorff dimension zero for the $\mathcal C^1$ structure defined by $g$.<|endoftext|> TITLE: Polynomial approximation for square root function with fast convergence and bounded coefficients QUESTION [7 upvotes]: Let $\delta, \varepsilon \in (0,1)$. I am interested in a sequence $\{f_n\}$ of polynomial approximations of the square root function $x \to x^{1/2}$ on $[\delta,1]$, of the form $$ f_n(x) = \sum_{i=0}^n \alpha_i x^i $$ which satisfies \begin{align*} \forall x \in [\delta,1],\ \left|f_n(x) - x^{1/2}\right|\ & \leq \varepsilon \\ n & = O\left(\frac{1}{\delta} \log\frac{1}{\varepsilon}\right),\\ \sum_{i=0}^\infty |\alpha_i| & \leq B, \\ \end{align*} where $B$ is a universal constant. (Obviously, I also want the sequence to satisfy $f_n(x) = f_{n-1}(x) + \alpha_n x^n$.) Does such a sequence of approximations exist? We can also relax the requirement that $B$ is a constant, and require $\sum_{i=0}^n |\alpha_i|$ to have a bound which is polynomial in $n$. More generally, I am interested in approximations satisfying the same properties for the function $x \to x^{\alpha}$ where $\alpha \in (0,1)$. REPLY [5 votes]: I have a strong impression that something like that has been asked before (perhaps, by somebody else) but it is easier to answer again than to find that old thread. What you ask for is patently impossible. Indeed, assume that you have a polynomial $f_n(z)$ that approximates $\sqrt z$ on the interval $[\frac 13,1]$ with precision $e^{-cn}$. Consider the domain $\Omega=\{z: \frac 13\le |z|\le 1, 0\le \arg z\le \frac{3\pi}2\}$. The function $g(z)=f_n(z)-\sqrt{z}$ is then analytic in $\Omega$, continuous up to the boundary, and bounded by $B_n+1$. By the standard two constant lemma, we have $$ |g(-2/3)|\le [\max_{\partial\Omega\setminus[\frac 13,1]}|g|]^{1-\gamma}[\max_{[\frac 13,1]}|g|]^{\gamma}\le (B_n+1)^{1-\gamma}e^{-\gamma cn} $$ with some constant $\gamma\in(0,1)$ (the harmonic measure of $[\frac 13,1]$ with respect to the domain $\Omega$ and the point $-2/3$). If $B_n$ is subexponential in $n$, the RHS tends to $0$ as $n\to\infty$, so we get $f_n(-2/3)$ close to $i\sqrt{\frac 23}$. Considering the domain symmetric to $\Omega$ with respect to the real axis, we conclude that $f_n(-2/3)$ must be also very close to $-i\sqrt{\frac 23}$. But those two numbers are rather far apart.<|endoftext|> TITLE: Exact formula for $\chi(X, \, S^n \Omega^1_X)$ QUESTION [7 upvotes]: I have a smooth, compact complex surface $X$, and I need an explicit formula for the Euler characteristic $$\chi(X, \, S^n \Omega^1_X),$$ where $S^n$ denotes the symmetric product, in terms of $c_1(X), c_2(X)$. I know how to do the computation, by using the splitting principle in order to calculate the Chern classes $c_i(S^n \Omega^1_X)$ and then the Hirzebruch-Riemann-Roch formula, and I did it for small values of $n$. However, extracting a formula for general $n$ seems quite tedious. Question. Is there a reference where I can find the value of $\chi(X, \, S^n \Omega^1_X)$? Related to this: is there a reference for the Chern classes $c_1(S^n \Omega^1_X)$, $c_2(S^n \Omega^1_X)$? Notice that I need an exact formula, not an asymptotic one (which I already know). REPLY [4 votes]: Just for reference: using the Schubert2 package from Macaulay2 this can be quite effortlessly done. i1 : needsPackage "Schubert2"; i2 : X = abstractVariety(2, QQ[n,c1,c2,Degrees=>{0,1,2}]); i3 : T = abstractSheaf(X, Rank=>2, ChernClass=>1+c1+c2); i4 : X.TangentBundle = T; i5 : chi symmetricPower(n, cotangentBundle X) 1 3 2 1 3 1 2 1 2 1 2 1 1 o5 = integral(-n c1 - -n c2 - --n*c1 - -n c2 + --c1 - -n*c2 + --c2) 6 6 12 2 12 4 12 If one looks at the implementation, the $n$-th symmetric power of a bundle $\mathcal F$ is computed using the pushforward of the relative $\mathcal{O}(n)$ on the projectivization $\mathbf{P}(\mathcal F^\vee)$ (as already mentioned in the other answer).<|endoftext|> TITLE: Which unimodular lattices $L\subset \mathbb R^2$ minimize $f_t(L):=\sum_{ v\in L} e^{-t \|v\|_2}$? (for parameters $t>0$) QUESTION [9 upvotes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$Consider the lattices in $\SL(2,\mathbb R)(\mathbb Z^2)$ up to rotation. The space of such lattices can be identified with the modular surface $\mathcal M:=\SO(2)\backslash \SL(2,\mathbb R)/\SL(2,\mathbb Z)$. We can then define a family of functions $f_t(A):\mathcal M\rightarrow \mathbb R^+$ by $$f_t(A)=\sum_{ v\in A(\mathbb Z^2)} e^{-t \|v\|_2},$$ for $t>0$, where $\|v\|_2$ denotes the Euclidean norm. I conjecture that for all $t>0$, $f_t(A)$ has a unique minimum at the lattice which can be tiled by equilateral triangles. I have computational evidence which suggests that this is true; I have plotted approximations of $f_t(A)$, $$f_{t,N}(A):=\sum_{v\in A(\mathbb Z_N^2)} e^{-t\|v\|_2},$$ where $\mathbb Z_N:= \{m\in\mathbb Z:-N\leq m\leq N\}$, and the conjecture seems to hold. Note that for $t$ small and $N$ small, $f_{t,N}(A)$ is not necessarily minimized by the equilaterally tiled lattice. However, for a fixed $t>0$, it appears that there exists an $N_t>0$ such that for all $N>N_T$, $f_{t,N}(A)$ is minimized by the equilaterally tiled lattice. REPLY [5 votes]: The answer is that $f_t(A)$ is uniquely minimized at the hexagonal lattice (up to rotation). The comment by Marco Golla led me to the following paper by Laurent Bétermin which proves the result in a more general setting: https://arxiv.org/abs/1502.03839<|endoftext|> TITLE: Examples of rings in monoidal categories QUESTION [13 upvotes]: Ring objects are usually defined on Cartesian monoidal categories, but one can define them more generally on non-Cartesian symmetric monoidal categories as follows: Let $(\mathcal{C},\otimes,\mathbf{1})$ be a symmetric monoidal category. When equipped with the tensor product of $\mathcal{C}$, the category $\mathsf{CCoMon}(\mathcal{C})$ of cocommutative comonoids in $\mathcal{C}$ becomes Cartesian monoidal ― note that this requires $\mathcal{C}$ to be symmetric. A ring object in $\mathcal{C}$ is then a ring object in $\mathsf{CCoMon}(\mathcal{C})$. Alternatively, if $\mathcal{C}$ and the category $\mathsf{HopfMon}^{\mathrm{bicomm}}(\mathcal{C})$ of bicommutative Hopf monoids in $\mathcal{C}$ have all co/limits, and there is a "free bicommutative Hopf monoid functor" $\mathcal{C}\to\mathsf{HopfMon}^{\mathrm{bicomm}}(\mathcal{C})$, then we can mimic the construction of tensor products of abelian groups in $\mathcal{C}$, obtaining a symmetric monoidal category $(\mathsf{Ab}(\mathcal{C}),\boxtimes)$, the monoids in which are then defined to be ring objects in $\mathcal{C}$. Note that Replacing $\mathsf{HopfMon}^{\mathrm{bicomm}}(\mathcal{C})$ by $\mathsf{BiMon}^{\mathrm{bicomm}}(\mathcal{C})$ and carrying out the second approach, one obtains a notion of a rig object in $\mathcal{C}$. The latter approach is the one developed in Part II of Goerss's Hopf Rings, Dieudonné Modules, and $E_*\Omega^2S^3$. See there for more details and arXiv:1804.10153 for the example of Hopf algebras and affine and formal abelian group schemes. When both approaches can be carried out, they agree. Examples Examples of ring and rig objects in monoidal categories are the following. When $\mathcal{C}=\mathsf{Sets}$, one recovers rings and rigs, as $\mathsf{CCoMon}(\mathsf{Sets})\cong\mathsf{Sets}$, or alternatively since \begin{align*} \mathsf{HopfMon}^{\mathsf{bicomm}}(\mathsf{Sets}) &\cong \mathsf{Ab},\\ \mathsf{BiMon}^{\mathsf{bicomm}}(\mathsf{Sets}) &\cong \mathsf{CMon}, \end{align*} with $\boxtimes$ recovering the tensor product of abelian groups or commutative monoids. More generally, rings in Cartesian monoidal categories coincide with the usual notion of a ring object in a category with finite limits. Rings in $(\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z})$ and $(\mathsf{Rings},\otimes_{\mathbb{Z}},\mathbb{Z})$ are plethories. A categorification of this approach, where one replaces monoids by pseudomonoids, recovers $2$-rigs and $2$-rings. Background Let $(\mathcal{C},\otimes,\mathbf{1})$ be a monoidal category. A monoid in $\mathcal{C}$ consists of an object $A$ of $\mathcal{C}$ together with maps $\mu\colon A\otimes A\to A$ and $\eta\colon\mathbf{1}\to A$ making the diagrams commute. For example, monoids in the Cartesian monoidal category of sets recover ordinary monoids, while monoids in $(\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z})$ recover (non-commutative) rings. Moreover, if $\mathcal{C}$ has a braided monoidal structure, we say that a monoid $(A,\mu,\eta)$ in $\mathcal{C}$ is commutative if the diagram commutes. Again, this recovers commutative monoids and commutative rings when applied to $\mathsf{Sets}$ and $\mathsf{Ab}$. Dually, a comonoid in $\mathcal{C}$ is a monoid in $\mathcal{C}^{\mathsf{op}}$: it is a triple $(C,\Delta,\epsilon)$ consisting of an object $C$ of $\mathcal{C}$ equipped with maps $\Delta\colon C\to C\otimes C$ and $\epsilon\colon C\to\mathbf{1}$ making the diagrams commute. Cocommutative comonoids are defined dually to commutative monoids. Any object of a Cartesian monoidal category is canonically a comonoid when equipped with the diagonal and projection to the unit maps. They are quite more interesting if the category in question is non-Cartesian, however: in $\mathsf{Mod}_{R}$, for instance, they give rise to $R$-coalgebras. Now, we can also consider bimonoids in (a braided monoidal category) $\mathcal{C}$. These are objects of $\mathcal{C}$ equipped with both a monoid and a comonoid structure in a compatible way (image from the bialgebra page in Wikipedia): A bimonoid in $\mathcal{C}$ is bicommutative if it is commutative and cocommutative. A Hopf monoid in $\mathcal{C}$ is a bimonoid $H$ in $\mathcal{C}$ together with a morphism $\sigma\colon H\to H$, called the antipode of $H$, making the diagrams commute. In a sense, Hopf monoids are bimonoids with inverses: bimonoids in $\mathsf{Sets}$ are monoids, but Hopf monoids in $\mathsf{Sets}$ are groups. The classical examples of bimonoids and Hopf monoids are bialgebras and Hopf algebras. The question What are some other examples of rings and rigs in monoidal categories, in particular non-Cartesian ones? REPLY [10 votes]: Below is a natural example of a ring object in the monoidal category of affine schemes. If $p$ is a prime and $R$ is a commutative ring, then the ring $W(R)$ of $p$-typical Witt vectors is also a commutative ring. The functor $W: CommRings \rightarrow CommRings$ is in fact co-representable by some commutative ring $\mathbb{W}$ which has two coproducts inducing the addition and multiplication, respectively, on $W(R)$ for each $R$. That is, the two coproducts give $hom_{CommRings}(\mathbb{W},R)$ a natural ring structure for all commutative rings $R$, and with that ring structure it is isomorphic to $W(R)$. Consequently, Spec of that co-representing ring is a ring object $Spec\ \mathbb{W}$ in the category of affine schemes. There is a nice, but unpublished, note about this affine ring scheme by Paul Pearson from the mid-2000s, which has a few calculations of the structure maps. Another class of examples is provided by Hopf rings, which are ring objects in the category of cocommutative coalgebras. These things arise naturally in topology from studying unstable operations on generalized homology theories. The point is that, if you have a suitably nice spectrum $E$, then when you evaluate $E$ on the spaces in the $\Omega$-spectrum for $E$ and take the direct sum over all those spaces in the $\Omega$-spectrum, the resulting bi-graded gadget inherits a lot of structure from the structure of $E$: in fact, it picks up the structure of a Hopf ring. This is pretty well-documented in various resources on the Web, e.g. here: https://encyclopediaofmath.org/wiki/Hopf_ring<|endoftext|> TITLE: A dichotomy for the quadratic variation of differentiable functions? QUESTION [8 upvotes]: For a real-valued function $f$ on $[0,1]$, define its quadratic variation by the formula $$[f]:=\limsup\sum_{j=1}^n(f(t_j)-f(t_{j-1}))^2,$$ where the $\limsup$ is taken over all "partitions" $0=t_0<\cdots TITLE: Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$? QUESTION [6 upvotes]: Let $\mathbb H= \{z \in \mathbb C\,:\, \textrm{Re}\,(z)\geq 0\}$ and for $j=1,2$ suppose that $F^{(j)}:\mathbb H\to \mathbb H$ is defined via $$ F^{(j)}(z) = \sum_{k=1}^{\infty} \frac{a_k^{(j)}}{z+\lambda^{(j)}_k},$$ where $|a_{k}^{(j)}|\leq k^{-2}$ and $0<\lambda^{(j)}_1<\lambda^{(j)}_2<...$ with $\lim_{k\to \infty} \lambda^{(j)}_k=\infty$. Suppose that $$ F^{(1)}(n)=F^{(2)}(n)\quad \text{for all $n \in \mathbb N$}.$$ Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$? REPLY [10 votes]: The answer is positive. Indeed, $f=F^{(1)}-F^{(2)}$ is a bounded analytic function in the right half-plane (this follows from your conditions $|a_k|\leq k^{-2}$ and $\lambda_k\to\infty$). But a bounded analytic function cannot be zero at positive integers, unless it is identically equal to zero. This follows from the Blaschke condition on zeros of bounded analytic functions. Your assumption that $F$ maps the right half-plane into itself is redundant, and the assumption that $\lambda_k>0$ can be very much relaxed; all you need is that your functions are bounded in the right half-plane.<|endoftext|> TITLE: On equibounded sequences in $L^\infty$ QUESTION [10 upvotes]: Let $f_n: [0, 1] \to \mathbb R$ be a sequence of positive functions in $L^\infty$ (hence a fortiori in $L^1$) that are equibounded in $L^\infty$ norm - that is $\sup_{n \in \mathbb N} \|f_n\|_{L_\infty} \leq M$ for some $M > 0$. Is it true that there exists some absolute positive constant $c < 1$ such that $$\inf_{n_k} \sup_{i, j > N} \|f_{n_i} - f_{n_j}\|_{L^1} \leq cM$$ for all such sequences $f_n$? Where the first infimum is taken over all increasing sequences $n_k$ of naturals. REPLY [10 votes]: The sharp constant is $c=1/2$. As in Jochen's answer, we may and shall assume that $M=1$. Proposition: Let $f_n: [0, 1] \to \mathbb R$ be a sequence of positive functions in $L^\infty$ such that $\sup_{n \in \mathbb N} \|f_n\|_{L_\infty} \leq 1$. Then $$\inf_{\{n_k\}} \sup_{i, j \ge 1} \|f_{n_i} - f_{n_j}\|_{L^1} \leq 1/2 \,,$$ where the infimum is over all strictly increasing sequences $\{n_k\}$. To see that this is sharp, consider the functions $\{b_n\}$ where $b_n(x)$ is the $n$th bit in the binary expansion of $x$. Note that $\|b_n-b_m\|_1=1/2$ for all $n \ne m$. Lemma 1: For any $k$ numbers $y_1,\ldots ,y_k$ in $[0,1]$, we have $$ \sum_{i=1}^{k-1}\sum_{j=i+1}^k |y_j-y_i| \le \lfloor k/2 \rfloor \cdot \lceil k/2 \rceil \le k^2/4 \,.$$ Proof: We may reorder the $y_i$ so that $y_1 \le y_2\le\ldots \le y_k$. Every interval $[y_\ell,y_{\ell+1}]$ is included in $\ell(k-\ell)$ intervals of the form $[y_i,y_j]$ with $i \le \ell 1/2$ there is a strictly increasing sequence ${n_k}$ such that $$ \quad \sup_{i, j \ge 1} \|f_{n_i} - f_{n_j}\|_{L^1} \leq c \,.\label{2}\tag{$\ast$}$$ By changing the values on a set of measure zero, we may assume that each $f_n$ takes values in $[0,1]$. Fix $c>1/2$ and find $k$ such that $\frac{k}{2(k-1)} c$. By Lemma 2, this graph does not contain a clique of $k$ nodes, so by Ramsey's Theorem [1] there is an infinite independent set (i.e., an anti-clique) in this graph, which proves \eqref{2}. Remark: Related considerations in Hilbert space are in [2]. [1] https://en.wikipedia.org/wiki/Ramsey%27s_theorem#Infinite_graphs [2] http://www.cs.tau.ac.il/~nogaa/PDFS/Publications/Euclidean%20Ramsey%20Theory%20and%20a%20construction%20of%20Bourgain.pdf<|endoftext|> TITLE: Does solving polynomial equations commute with tropicalization? (particularly for the field of Puiseux series) QUESTION [6 upvotes]: The field of Puiseux series over an algebraically closed field of characteristic zero is also an algebraically closed field, and furthermore it has a valuation so that our Puiseux series can be tropicalized. Is the tropicalization of the solutions of a monic polynomial equation over them the same as the solution of the tropicalization of the polynomial equation? I.e. informally, can we get the degree of the largest pole of the series that solves the equation only from the information of the largest poles in the coefficients of the polynomial? REPLY [6 votes]: Yes. This is normally expressed in terms of the Newton polygon of the polynomial. Specifically, given an arbitrary field $K$ with a valuation, and a polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, the Newton polygon of $f$ is the lower side of the convex hull of the set of points $(i, v(a_i))$. If $K$ is algebraically closed, then the valuations of the $n$ roots of $f$ are exactly minus the slopes of the Newton polygon over the $n$ intervals $[0,1], [1,2], \dots, [n-1,n]$. The most straightforward way I know to check this is to first, completely factor $f$ into linear factors, i.e. $a_n \prod_{i=1}^n (x- \alpha_i)$, with the $\alpha_i$ in order of increasing valuation, write $a_i$ in terms of the $\alpha_j$, and then check that $$v(a_{n-k}) \geq v(a_0) + \sum_{i=1}^k v(\alpha_i)$$ with equality if $v(\alpha_i) < v(\alpha_j)$, so that the Newton polygon is contained in the polygon with slopes $- v(\alpha_1),\dots, -v(\alpha_n)$, and contains the corners of that polygon, hence is equal to that polygon. This inequality and equality can be checked using the definition of valuation. Similarly, one can check that the tropicalization of a polynomial equation is a piecewise linear function with corners at minus the slopes of the Newton polygon (in fact, it is dual to the Newton polygon in a certain sense, so the edges of the tropicalization correspond to vertices of the Newton polygon and vice versa), and so the solutions of the tropicalization are also minus the slopes of the Newton polygon.<|endoftext|> TITLE: Subalgebras of singular matrices QUESTION [5 upvotes]: Is it true that any subalgebra of singular matrices have a common null-vector? In other words, is it true that, for any subalgebra $\cal S$ of the algebra of linear operators in a finite-dimensional vector space over a field, $$ \bigcap_{A\in\cal S}\ker A=\{0\}\quad\hbox{implies that} \quad\ker A=\{0\}\hbox{ for some $A\in\cal S\;$? } $$ (I am interested in finite prime fields mostly.) Note that for subspaces (instead of subalgebras) nothing similar is true. But for algebras, Burnside's theorem gives me some hope... REPLY [5 votes]: It's false. Take the subalgebra of $M_3(K)$ generated by the matrices $\begin{bmatrix} 0 & 0&0\\ 1 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ and $\begin{bmatrix} 0 & 0 & 0\\ 0 &1&0\\ 1& 0&1\end{bmatrix}$. These two elements form a two element right zero semigroup and so the algebra they generate is just their span which is $2$-dimensional and obviously singular since all elements have first row zero. These two matrices have no common zero except zero. Conceptually, let $S$ be the two-element right zero semigroup $\{a,b\}$. Then $KS$ acts faithfully on the left of $KS^1$ where $S^1$ is obtained by adjoining an identity. But these two right zeroes have no common vector they annihilate. For $a(c_11+c_2a+c_3b) = c_1a+c_2a+c_3b$ and $b(c_11+c_2a+c_3b) =c_1b+c_2a+c_3b$ and for these both to be zero you need $c_3=0=c_2$ and you need $c_1=-c_3$ and $c_1=-c_2$ and so $c_1=c_2=c_3=0$. The representation is by singular matrices since $1$ is not in $KS\cdot KS^1$. Update I guess you can go one dimension lower and consider the algebra of matrices of the form $\begin{bmatrix}a&b\\0&0\end{bmatrix}$.<|endoftext|> TITLE: In search of an alternative proof of a series expansion for $\log 2$ QUESTION [7 upvotes]: We all know the series expansion $$\log 2=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n. \tag1$$ I also am able to use the method of Wilf-Zeilberger to the effect that $$\log 2=3\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n\binom{2n}n2^n}. \tag2$$ QUESTION. Can you provide yet another proof of the formula in (2)? Remark. My motivation for this question goes beyond this particular series, hoping it paves a way forward in my study. Postscript. After those generous replies (see below), it appears that the idea rests on $$\log\left(1+\frac1x\right)=2\sinh^{-1}\left(\frac1{2\sqrt{x+x^2}}\right)$$ so that we may put $x=1$ to obtain (1) and (2). To reveal the background: (2) is found from (1) by a "series acceleration" method which does not even stop there. In fact, stare at these two \begin{align*}\log 2&=3\sum_{n=1}^{\infty}\frac{14n-3}{\binom{2n}2\binom{4n}{2n}2^{2n+1}}, \tag3 \\ \log 2&=3\sum_{n=1}^{\infty} \frac{(171n^2 - 111n + 14)(-1)^{n-1}}{\binom{3n}3\binom{6n}{3n}2^{3n+1}} \tag4 \end{align*} One may now ask: can you furnish an alternative proof for the formulae (3) or (4)? REPLY [13 votes]: Since you wish to develop techniques, you might want to consider the more general form $$S_k=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^k\binom{2n}n2^n}.$$ The arcsine representation $$\arcsin^2z=\frac12\sum_{n=1}^\infty\frac{(2z)^{2n}}{n^2{2n \choose n}}$$ directly gives $$S_2=\tfrac{1}{2}\ln^2 2,$$ (substitute $z=2^{-3/2}i$), upon differentiation one finds $$S_1=\tfrac{1}{3}\ln 2,$$ $$S_0=\tfrac{1}{9}+\tfrac{4}{27}\ln 2,$$ and upon integration, $$S_3=\tfrac{1}{4}\zeta (3)-\tfrac{1}{6}\ln^3 2 ,$$ $$S_4=4\operatorname{Li}_4\left(\tfrac12\right)-\tfrac72\zeta(4)+\tfrac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\tfrac5{24}\ln^42.$$ This method apparently fails to give a closed form expression for $k>4$, see this MSE posting. REPLY [11 votes]: Write the $n$-th term, $(-1)^{n-1} \!\left/ \bigl(n {2n \choose n} 2^n\bigr) \right.$, as the definite integral $$ \frac14 \int_0^1 \left(-\,\frac{x-x^2}{2} \right)^{n-1} dx $$ using the formula for the Beta integral $B(n,n) = \int_0^1 (x-x^2)^{n-1} dx$. Thus the sum over $n$ is $$ \frac14 \int_0^1 \left( 1 + \frac{x-x^2}{2} \right)^{\!-1} dx = \frac12 \int_0^1 \frac{dx}{2+x-x^2} \, dx, $$ which is elementary: the denominator factors as $(1+x)(2-x)$, so expand the integrand in partial fractions and integrate termwise, obtaining $$ \frac16 \int_0^1 \left(\frac1{x+1} + \frac1{2-x}\right) dx = \frac16 (\log 2 + \log 2) = \frac13 \log 2, $$ QED REPLY [8 votes]: It is well known that $$ 2\left(\sin^{-1}\frac{\sqrt{x}}{2}\right)^2 = \sum_{n\geq 1} \frac{x^n}{n^2{2n\choose n}}. $$ See e.g. here or Enumerative Combinatorics, vol. 1, second ed., Exercise 1.173. Differentiate with respect to $x$, put $x=-1/2$, and use $$ \sin^{-1}z = -i\log(iz+\sqrt{1-z^2}) $$ to deduce (2).<|endoftext|> TITLE: Extending Apéry's proof to Catalan's constant? QUESTION [14 upvotes]: I've been looking into Apéry's irrationality proof of $\zeta (3)$, and one of the first questions I instantly had, was how did he derive the following continued fraction? $$\begin{equation*} \zeta (3)=\dfrac{6}{5+\overset{\infty }{\underset{n=1}{\mathbb{K}}}\dfrac{-n^{6}}{34n^{3}+51n^{2}+27n+5}}\end{equation*}$$ Furthermore, is it possible to get a similar continued fraction for $\zeta(5)$, $\zeta(7)$ or $G$? A rapidly converging central binomial series was recently found for Catalan's constant: $$G = \frac{1}{2} \sum_{n=0}^{\infty} (-1)^n \frac{(3n+2) 8^n}{(2n+1)^3 \binom{2n}{n}^3}$$ which is in similar spirit to $$\zeta(3)=\frac{5}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3 \binom{2n}{n}}$$ which Apéry utilised. If I understand correctly, this gives us the first stage of an analogue of Apéry's proof for $G$. The next stage in his proof is to use a fast recursion formula that approaches $\zeta(3)$: $$n^3 u_n + (n-1)^3 u_{n-2} = (34n^3 - 51n^2 + 27n - 5) u_{n-1},\, n \geq 2$$ A similar recursive formula that approaches $G$ was found: $$(2n+1)^2 (2n+2)^2 p(n) u_{n+1}-q(n)u_n = (2n-1)^2 (2n)^2 p(n+1) u_{n-1}$$ where $$p(n) = 20n^2 - 8n + 1$$ $$q(n) = 3520n^6 + 5632n^5 +2064n^4-384n^3-156n^2+16n+7$$ Now as Zudilin mentions in his paper, 'the analogy is far from proving the desired irrationality of $G$', but why exactly is this recursive formula not good enough to prove the irrationality of $G$? A major part of Apéry's proof is to define the double sequence: $$c_{n,k} := \sum_{m=1}^{n} \frac{1}{m^3} + \sum_{m=1}^{k} \frac{(-1)^{m-1}}{2m^3 \binom{n}{m} \binom{n+m}{m}}$$ Where does this come from, and how could one create a similar sequence instead that follows from the above binomial series for $G$? REPLY [15 votes]: Summary: The continued fraction, the recurrence and the explicit form of the sequence are interchangeable and for the Apéry numbers, we don't know what come first. This extend to other constructions for other constants. The approximation for the Catalan's constant $G$ fails because it doesn't converge too fast. i.e the growth ratio between the inclusion function and the sequence diverges (or when raising it to the power of $\frac{1}{n}$ and taking limit to $\infty$, is $> 1$) This is in the spirit of Fischler's expository article, Zudilin's paper and Zeilberger-Zudilin paper on automatic discovery of proofs of irrationality. I will try to be rigorous to the extent possible for my knowledge. First, we will modify a bit the criteria for irrationality. How the criteria for irrationality that appears in Poorten's paper implies the criteria that appear in these papers? Let's recall it. Let $\beta$ a real number. If $\frac{p_{n}}{q_{n}}$ is a rational sequence with $\frac{p_{n}}{q_{n}} \neq \beta$ and $$\left|\beta-\frac{p_{n}}{q_{n}}\right| < \frac{1}{q_{n}^{1+\delta}}$$ then $\beta$ is irrational. If you multiply by $q_{n}$ $$\left|q_{n}\beta-p_{n}\right| < \frac{1}{q_{n}^\delta}$$ with $0<\delta<1$ that exists. If we make the same procedure (power to $\frac{1}{n}$ and then taking limit), for these kind of sequences, the right side simply yields a constant (e.g $\mu$) to the power of $\delta$. $$\lim_{n \rightarrow \infty} \frac{1}{q_{n}^{\frac{1}{n}\delta}}=\frac{1}{\mu^{\delta}}<1$$ The last part is just because we need only a $\delta$ to exist, and it could a very small one, close to 0, but no 0. In summary, the criteria get translated into $$\lim_{n \rightarrow \infty}\left|q_{n}\beta-p_{n}\right|^{\frac{1}{n}}=L<1$$ Second, it is necessary to mention that the explicit form of a sequence, its recurrence relation, its continued fraction, integral representations, etc. are all complementary descriptions of the sequence and each of them help us to prove the ingredients of a Apéry-like proof, namely: The existence of a linear form over two sequences satisfying the previous irrationality criteria The existence of an "inclusion" for the sequences, i.e, multiplying the sequences by a function give integer sequences Our approximation is different to 0 for infinite $n$. Point 1) requires to know asymptotics estimates of the sequences and for this purpose the recurrence description is useful. Also we need to estimate the error between the number and our approximation. Again you can use the recurrence or an integral representation (like in Beukers's proof). Point 2) is generally proved using the explicit form of the sequences. And for Point 3) you have many options. As you can see, it's flexible and depends of the nature of the sequence. The technical considerations are a) The asymptotics of a sequence given by a holonomic recurrence can be obtained using the algorithm in Wimp and Zeilberger paper or the method of Flajolet and Sedgewick presented in the book "Analytic Combinatorics". (Some cautious here, and indeed I would like to consider this in another post. The Wimp-Zeilberger algorithm is an explanation of the Birkhoff-Trjitzinsky method, which hasn't been verified completely. Also, since calculations are complicated, we require of a algorithmic package to find the terms, like Kauer's Mathematica package. It doesn't mean that some previously given asymptotics are wrong, but we need to check further details). These are the reasons why we will avoid calculating an full asymptotics (even if we can) and we will deal with expressions of the kind $\lim_{n \rightarrow \infty}a_{n}^{\frac{1}{n}}$ , that can be tackle using the Poincaré-Perron theorem plus some assumption of minimality of the solutions plus this paper of Perron, also knows as Perron's second theorem. Modern approaches to the proof use this setup to avoid more technicalities. For a summary of the full asymptotic of the Apéry numbers see this, page 379. b) it's a bit more tricky and it's required to calculate the $p$-adic valuation of sub-terms in the explicit form, which generally are combination of powers and binomial coefficients. Here also we need the inclusion function to be smaller in comparison to the sequence's asymptotics, to reach point 1). Let's try some examples with $\zeta (3)$ and the Catalan's constant to visualize this. For $\zeta(3)$, the recurrence is given by $$n^3 u_n -(34n^3 - 51n^2 + 27n - 5) u_{n-1}+ (n-1)^3 u_{n-2} = 0\quad n \geq 2$$ To use the Poincaré-Perron (PP) theorem, we need to transform it into a form $u_{n}+a(n)u_{n-1}+b(n)u_{n-2}=0$ with $$a=\lim_{n \rightarrow \infty} a(n) \quad b=\lim_{n \rightarrow \infty} b(n)$$ Dividing by $n^3$, we obtain $$u_n -(34 - 51n^{-1} + 27n^{-2} - 5n^{-3}) u_{n-1}+ (1 +3n^{-1}+3n^{-2}+n^{-3}) u_{n-2} = 0\quad n \geq 2$$ Then, we calculate the roots of a characteristic polynomial $x^2+ax^2+b=0$, i.e, $x^2-34x^2+1=0$, and the conclusion of the theorem is that if $x_{n}$ and $y_{n}$ are individual independent solutions of the recurrences, then $$\lim_{n \rightarrow \infty}\frac{x_{n+1}}{x_{n}}=\lambda_{1} \quad \lim_{n \rightarrow \infty}\frac{y_{n+1}}{y_{n}}=\lambda_{2}$$ where $\lambda_{1}$ and $\lambda_{2}$ and the roots and $|\lambda_{1}|\leq|\lambda_{2}|$. This alone doesn't give asymptotic estimates of the general solutions, so we will use some improvements of the theorem. If $a(n)\sim n^{C_{1}}$ , $b(n)\sim n^{C_{2}}$ and $C_{1}=C_{2}$, then the solution given by $y_{n}$ is minimal (page 370), in the sense that for any other independent solution of the recurrence, we can call it $h_{n}$ (i.e $h_{n}$ is not a multiple of $y_{n}$ for some initial conditions), we have: $$\lim_{n \rightarrow \infty}\frac{y_{n}}{h_{n}}=0$$ In this case, we can use the following estimates, $$\lim_{n \rightarrow \infty} a_{n}^{\frac{1}{n}} = \lim_{n \rightarrow \infty} b_{n}^{\frac{1}{n}}=17+12\sqrt{2}=\alpha$$ In the case of $G$, the recurrence is $$(2n+1)^2 (2n+2)^2 p(n) u_{n+1}- q(n)u_n - (2n-1)^2 (2n)^2 p(n+1) u_{n-1}=0$$ We have a similar limit, if we divide by $(2n+1)^{2}(2n+2)^2 p(n)$ and then taking limits on $a(n)$ and $b(n)$, we arrive to the characteristic polynomial $x^2-11x-1=0$ and therefore we obtain $$\lim_{n \rightarrow \infty} a_{n}^{\frac{1}{n}} = \lim_{n \rightarrow \infty} b_{n}^{\frac{1}{n}}=\frac{11}{2} + \frac{5 \sqrt{5}}{2}=\delta$$ From the recurrence, you can also estimate the error between the number and our approximation. In the case of $\zeta(3)$, using the recurrence, we have: $$a_{n}b_{n-1}-a_{n-1}b_{n}=\frac{6}{n^{3}}$$ $$\left|\zeta (3)-\frac{a_{n}}{b_{n}}\right| = \sum_{k=n+1}^{\infty}\frac{6}{n^{3}b_{k}b_{k-1}}$$ For $G$ we will find a similar expression since: $$(2n+1)^2 (2n+2)^2 p(n)(a_{n+1}b_{n}-a_{n}b_{n+1})=-(2(n-1)+1)^2 (2(n-1)+2)^2 p(n+1)(a_{n}b_{n-1}-a_{n-1}b_{n})$$ $$=+(2(n-2)+1)^2 (2(n-2)+2)^2 p(n+2)(a_{n-1}b_{n-2}-a_{n-2}b_{n-1})$$ $$=\textrm{.....}$$ $$\therefore a_{n+1}b_{n}-a_{n}b_{n+1}=\frac{13}{32}(-1)^{n}\frac{p(2n)}{(2n+1)^2 (2n+2)^2 p(n)}$$ and a similar bound holds. Now you have to prove the inclusions, for $\zeta (3)$ the inclusion is given by $2D_{n}^{3}a_{n} \in \mathbb{Z}$, and for $G$ Zudilin proved the inclusions $$2^{4n+3} D_{n} b_{n} \in \mathbb{Z} \quad 2^{4n+3}D_{2n-1}^3 a_{n} \in \mathbb{Z}$$ where $D_{n}=\textrm{lcm}(1,2,....,n)=e^{\psi(n)}$ ($\psi$(n) is the Chebyshev $\psi$ function). So, for $G$, if we multiply $a_{n}$ and $b_{n}$ by $2^{4n+3}D_{2n-1}^3$, they become integers. Note: Trivially we have that $D_{n}|D_{2n-1}^3$. We will define the inclusion of $a(n)$ as $\textrm{Incl}_{a,n}$ (or by analogy, $\textrm{Incl}_{b,n}$ for $b_{n}$) to be the function that we need to multiply to become $a_{n}$ (or $b_{n}$) an integer. I won't give too much detail on this step, but the proof of both of them relies on the technical argument explained before (the explicit form sometimes yields this easily by means of hypergeometric manipulations) Finally, we have that $$\left|q_{n}\beta-p_{n}\right|<\frac{Cq_{n}}{b_{n}^{2}}$$ Since for these sequences, we have $q_{n}=b_{n}\textrm{Incl}_{a,n}$ $$\left|q_{n}\beta-p_{n}\right|<\frac{C\textrm{Incl}_{a,n}}{b_{n}}$$ Let's define $\phi$ $$\phi=\lim_{n \rightarrow \infty}\textrm{Incl}_{a,n}^{\frac{1}{n}}$$ For $\zeta(3)$ $$\phi=\lim_{n \rightarrow \infty}e^{3\frac{\psi(n)}{n}}=e^{3}$$ by the prime number theorem $\lim_{n \rightarrow \infty}\frac{\psi(n)}{n}=1$ Thus, if you raise the approximation to the power of $\frac{1}{n}$ and them apply the limit to $\infty$, the behaviour of the right side if governed by $\frac{\phi}{\alpha}$ (or $\delta$ in denominator, in the case of $G$). If this ratio is bigger that one, the right bound is not useful. If it less that 1, we can prove irrationality. For $\zeta(3)$, $\frac{e^3}{\alpha} \approx 0.59$ and for $G$, $\frac{e^6 2^4}{\gamma}\approx 582$. The huge difference in the ratios is what Zudilin indicated. Since for $G$ the value is bigger that one, therefore the sequence is not useful to prove irrationality. Even if we use the conjectured inclusions in the conclusion remarks, we obtain $\frac{e^4 2^4}{\gamma}\approx 79$. This illustrates very well the interplay between the approximation, the asymptotics of the inclusion function and the asymptotics of our sequence. In the literature, you can find many approaches to construct these sequences: first you can work with an explicit form of the sequences (many of them are obtained from hypergeometric identities), and then you guess a recurrence for them (using the Gosper-Zeilberger algorithm; see this). If you have the recurrence, the continued fraction description is immediate. Or viceversa, you try to find recurrences that gives origin to integer or "quasi-integer" sequence, by searching the space of coefficients in the recurrence and then prove a closed form for the sequences using combinatorial techniques. Therefore, we still don't have a general way to come up with sequences like the Apéry numbers, i.e, we're looking for "well-poised" sequences in the sense of they almost satisfy the three ingredients, but the seed or the intuition to know where to look comes from the researcher. So the answer to your questions are: where does the continued fraction comes? From the recurrence. Where does the recurrence comes? From the explicit form with the $c_{n,k}$ of Poorten's paper. Where does the explicit form with the $c_{n,k}$ comes? From combinatorial identities and a bit of luck. You can start wherever you want, for example see this, specially Remarque 2. The interesting question is of course, what is common in the sequences that allow us to prove the irrationality of $\zeta(2)$ and $\zeta(3)$? This question is deep, and goes back to Zagier's paper: the generating functions of the sequences that are used in the proofs admitted a modular parametrization, a fact that was proved rigorously by Beukers and is related also to the differential equation satisfied by the generating function (Picard-Fuchs equations and variants). Some sequences that are "well-poised" in the sense of they almost satisfy the three ingredients don't possess this modular behaviour. In contrast, some family of sequences that arise in the massive search of recurrences exhibit this phenomenon. Surprisingly, the latter give sequences that satisfy the point 2) almost immediately with small inclusions, which is not trivial, and there's still work in progress to understand why. Also some of them satisfy the condition 1) qualitatively, depending of which constant are approximating, of course. But they fail in the final calculation of the irrationality criteria. For more details, see this and this post. I'm in the combinatorial side, so an expert in the work of Zagier and Beukers can complement this part better. Comment: In many parts, I'm using the positivity, rationality and monoticity of the sequences to simplify the argument. Point 3) is in general easy to prove.<|endoftext|> TITLE: An equivalent condition for differentiability almost everywhere? QUESTION [7 upvotes]: Given a function $f \in L^1 (\mathbb R)$, define the roughness $R_f$ of $f$ at $x \in \mathbb R$ by $$\DeclareMathOperator{\esssup}{\operatorname{esssup}} R_f (x) := \limsup_{r \to 0+}\dfrac{r \esssup_{y \in B_r (x)} |f(y) - f(x)|}{\displaystyle\int\limits_{B_r (x)} |f(s) - f(x)| ds} $$ where $\esssup$ denotes the essential supremum, and by convention we take $\frac{0}{0} = 1$. Question: Let $f$ be continuous. Is it true that $f$ is differentiable almost everywhere if and only if $R_f = 1$ almost everywhere? Remark: The “only if” direction is relatively straightforward, the “if” direction is the issue. REPLY [2 votes]: The answer is negative: If $f'(x) = 0$ "too often", then $R_f$ may fail to be equal to one almost everywhere. Let $C$ be a fat Cantor set, let $I_n = (a_n, b_n)$ ($n \geqslant 2$) be the sequence of all finite components of the complement of $C$, and let $f$ be a differentiable function with the following properties: $f(x) = 0$ for $x \in C$; on $I_n$, $f$ is a smooth bump of a fixed shape, supported in $\tfrac1n I_n$ (the middle $n$th part of $I_n$), and with maximum equal to $|I_n|^2$ (and hence the integral of $f$ over $I$ is equal to $\tfrac1n |I_n|^3$). Then $f'(x) = 0$ for $x \in C$ by the first property, so $f$ is everywhere differentiable. On the other hand, it is rather straigthforward to see that there is a constant $C$ such that for each $n$ and $t \in I_n = (a_n, b_n)$ we have $$ \int_{a_n}^t f(y) dy \leqslant \frac{C}{n} (t - a_n) \sup_{y \in [a_n, t]} f(y) $$ and $$ \int_t^{b_n} f(y) dy \leqslant \frac{C}{n} (b_n - t) \sup_{y \in [t, b_n]} f(y) . $$ It follows that if $x \in C$ and $r$ is small enough, so that $B_r(x)$ is disjoint with $\tfrac12 I_2 \cup \tfrac13 I_3 \cup \ldots \cup \tfrac1{n-1} I_{n-1}$, then $$ \int_{B_r(x)} f(y) dy \leqslant \frac{C}{n} |B_r(x)| \sup_{y \in B_r(x)} f(y) . $$ This, in turn, implies that $R_f(x) = \infty$ for every $x \in C$. Thus, $R_f(x) \ne 1$ on a set of positive Lebesgue measure.<|endoftext|> TITLE: When $p(x)^2 \mid f(g(x))$? QUESTION [5 upvotes]: Let $f(x),g(x),p(x)$ be non-constant polynomials with rational coefficients. Is it true that for all $f$ exist $g,p$ such that $p(x)^2 \mid f(g(x))$? Partial results: $f(g(x))$ is divisible by square iff the discriminant of $f(g(x))$ is zero. For variables $z_i$, write $g_0(x)=\sum_{i=0}^n z_i x^i$. Then the discriminant of $f(g_0(x))$ is polynomial $D$ in variables $z_i$. Solution $D=0$ will find $p(x)$, but finding points on variety is hard. If we allow $g,p$ to be with coefficients algebraic integers, the solution is easy: in $D$ fix all but one variables say $z_0$ to be integers and work in the number field with defining polynomial $D(z_0)$. As corollary to positive solution, we have $f(g(x))$ reducible. REPLY [15 votes]: Yes, it is true. Let $f_0$ be an irreducible divisor of $f$. It suffices to find $g$ such that $f_0^2$ divides $f_0(g(x))$ (which, in turn, divides, $f(g(x))$). Try to choose $g(x)=x+h(x)f_0(x)$. Then $f_0(g(x))=f_0(x+h(x)f_0(x))\equiv f_0(x)+f_0'(x)h(x)f_0(x) \pmod {f_0^2(x)}$, and we need $1+f_0'(x)h(x)$ to be divisible by $f_0$. Since $f_0'$ and $f_0$ are coprime (as $f_0$ is irreducible and $0\leqslant \deg f_0'<\deg f_0$), such $h$ exists. Actually by the same reasoning once we found $g_k$ for which $f_0^k$ divides $f_0(g_k)$, where $k\geqslant 1$, we may find $h$ such that $f_0(g_k+f_0^kh)$ is divisible by $f_0^{k+1}$.<|endoftext|> TITLE: A property of $C^2$ functions QUESTION [7 upvotes]: Let $f\in C^2(\Bbb R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate? REPLY [9 votes]: $\newcommand\R{\mathbb R}$Let $\R:=R$. Suppose that $|f''(x)(h,h)|\le C|h|^2$ for all $x$ and $h$ in $\R^m$ -- this is how we interpret the condition "Hessian matrix of $f$ is upper bounded by some constant $C$". Of course, here $f''(x)$ is the bilinear form that is the second derivative of $f$ at $x$, so that $f''(x)(h,k)=h^\top H(x)k$ for all $x,h,k$ in $\R^m$, where $H(x)$ is the Hessian matrix of $f$ at $x$. Consider first the case $m=1$. Take any $x\in\R$ and any real $h>0$. Then $$0\le f(x+h)\le f(x)+f'(x)h+Ch^2/2,$$ whence $$f'(x)\ge-\frac{f(x)}h-\frac C2\,h=-K\sqrt{f(x)}$$ if $h=\sqrt{2f(x)/C}$, where $K:=\sqrt{2C}$. Similarly (or by the left-right symmetry), $f'(x)\le K\sqrt{f(x)}$, and hence $$|f'(x)|\le K\sqrt{f(x)} \label{1}\tag{1}$$ for all $x\in\R$. Now take any natural $m$. Considering the restrictions of $f$ to all straight lines in $\R^m$, we see that \eqref{1} holds for all $x\in\R^m$, where now $|f'(x)|$ denotes the norm of the linear form $f'(x)$ that is the derivative of $f$ at $x$, so that $f'(x)(h)=h^\top\,\nabla f(x)$ for all $h\in\R^n$ and hence $|f'(x)|=|\nabla f(x)|$. Thus, the desired conclusion holds, in general.<|endoftext|> TITLE: Linear transport equation with unbounded coefficients QUESTION [5 upvotes]: Consider the PDE $$\partial_t f(x,t) = \langle q(x), \nabla \rangle f(t,x) + p(x),$$ with Schwartz initial data $f(0,x) = f_0(x) \in \mathscr S(\mathbb R^n).$ I am wondering then if $q$ and all its derivatives are polynomially bounded and $p$ is Schwartz, too: Does there exist a solution to this equation that decays faster than any polynomial in space $x$ at any fixed time $t>0$? This sounds plausible to me, but I am not sure how one argues for such an equation. I assume it must be a classical question. As there was apparently some confusion about the meaning of this question, let me ask it again: Fix a time $t>0$, then as a function of $x$, does the solution decay faster than any polynomial? This seems to be true in your case for example, as it is just a translation of a Schwartz function. REPLY [5 votes]: No. Consider the case $p=0$. In that case, solutions are constant along characteristics. But polynomial growth does not preclude characteristics from diverging to infinity in finite time.<|endoftext|> TITLE: Global sections of multiples of a divisor QUESTION [7 upvotes]: Let $D$ be an integral divisor on a smooth projective variety $X$. Consider the multiples $mD$ of $D$ for $m\geq 0$. Clearly, $h^0(X,mD) = 1$ for $m = 0$. Is there any example where $h^0(X,mD) = 0$ for $1\leq m\leq \overline{m}$, and $h^0(X,mD) = c$ for all $m\geq \overline{m}+1$, where $c$ is a constant? REPLY [6 votes]: If $c = 2$, let $D',D'' \in H^0(X,kD)$ be different divisors. Then $2D',D'+D'',2D''$ are three different divisors in $|2kD|$. Similarly, the case $c > 2$ is impossible. So, assume $c = 1$. Let $D' \in H^0(X,kD)$ and $D'' \in H^0(X,(k+1)D)$ (with $k > 1$). Then $(k+1)D'$ and $kD''$ are different divisors in $H^0(X,k(k+1)D)$. So, the case $c = 1$ is also impossible. REPLY [4 votes]: Not except in the trivial case $c =1$, $\overline{m}=0$. Assume for contradiction that $D_1$ and $D_2$ are two effective divisors, with $D_1$ equivalent to $m_1 D$ and $D_2$ equivalent to $m_2D$, and $$m_2 D_1 \neq m_1 D_2 $$ Then there exists an irreducible closed subset of $X$ of codimension $1$, call it $Y$, which $m_2 D_1$ and $m_1 D_2$ contain to different multiplicities. So the divisors $$ c m_2 D_1 , (c-1) m_2 D_1 + m_1 D_2, \dots, c m_1 D_2$$ are all equivalent to $m_1m_2 D$, hence all are the divisors of sections $s_0,\dots, s_c \in H^0 ( X, m_1m_2 D)$, and all contain $Y$ to different multiplicities, so the sections $s_0,\dots,s_c$ vanish to different orders at $Y$. It follows that $s_0, \dots, s_c$ are linearly independent, as any linear relation of them would have to express a leading term which vanishes to some order at $Y$ as a linear combination of functions which vanish to higher order at $Y$, which is impossible. This contradicts the assumption that $\dim H^0(X, m_1m_2 D) \leq c$. Thus, for any divisors $D_1, D_2$ equivalent to multiples $m_1 D$, $m_2 D$, respectively, we have $m_2 D_1 = m_2 D_1$. In particular, there is at most one divisor equivalent to $m D$ for each $m$, so $c=1$. Taking $m_1$ and $m_2$ sufficiently large that there indeed exist such $D_1, D_2$, and choosing $m_1$ and $m_2$ relatively prime so that there exist $a,b$ with $am_1 + bm_2 = 1$, we see that $a D_1 + b D_2$ is equivalent to $D$, and is effective because $m_1 ( a D_1 + b D_2) = D_1$, which is effective, so $H^0 (X, D) \neq 0$, and thus $\overline{m}=0$.<|endoftext|> TITLE: on the group generated by transitive permutation groups and diagonal groups QUESTION [5 upvotes]: Let $T$ be a transitive permutation group in $S_n$, embedded in $GL_n(F)$ as permutation matrices. Let $D$ be the group of diagonal matrices in $GL_n(F)$. Let $G$ be the group generated by $T$ and $D$. That is, $G$ is a subgroup of the monomial group in $GL_n(F)$. Question: is $G$ irreducible as a matrix group? Probably this is a very basic question, but I cannot figure it out or find a reference... Thank you for your help! REPLY [6 votes]: This is just a summary of the answers in the comments. If $|F|=2$, then the vector $(1,1,\ldots,1)$ spans a subspace invariant under $G$, so the group is reducible (assuming that $n>1$). Otherwise, if $|F|>2$ then, under the action of $D$, the natural module $V$ is the sum $V_1 \oplus V_2 \cdots \oplus V_n$ of $1$-dimensional irreducible submodules which are mutually nonisomorphic. (The mutual nonisomorphism follows from the fact the actions of $D$ on these submodules have different kernels.) It follows that the only submodules under the action of $D$ are sums of subsets of $\{V_1,\ldots,V_n\}$. But, the action of $T$ is transitive, so the $V_i$ are permuted transitively by $T$, and none of these subspaces except for $V$ itself is invariant under $T$. So the action of $G$ is irreducible.<|endoftext|> TITLE: Why are distributions "tempered"? QUESTION [25 upvotes]: Google N-Gram shows that both "tempered distribution" and "temperate distribution" are used in English, but the first version significantly prevails, and usage of the second term declines. Schwartz himself seems to have used this term for the first time in his paper of 1951, Analyse et synthèse harmoniques dans les espaces de distributions, Canad. J. Math. 3 (1951), 503–512, doi:10.4153/CJM-1951-051-5, and the French term was "distributions tempérées". Google translates "tempérées" as "temperate". I am not a native English speaker, but I understand "temperate" as a synonym of "moderate", which makes sense to me as a name of those distributions. For example, we say "temperate climate", not "tempered climate". While "tempered" seems to be related to metallurgy, namely to a process making steel hard. What does tempering of steel have to do with distributions? Why does the name "tempered" win in English? REPLY [2 votes]: I've always understood tempering as a process to enable something to be more easily molded into something useful--to make it more malleable and robust--in this case the Fourier transform. As nLab puts it: The main property is that the Fourier transform of a TD is well-defined, and is itself a TD; and that it naturally extends the standard FT. This makes TDs the natural setting for solving (linear) PDEs. Consistently, in metallurgy, tempering increases the ductility of a material, i.e., "the degree to which a material can sustain plastic deformation under tensile stress before failure," making it more useful. Similarly, to temper one's emotions is to guide them, shape them, mold them, into productive channels, or at least less destructive / disruptive ones. 'To temper' has a long history. From Oxford Languages: In Latin, temparare--to restrain, moderate $\to$ Old English, temprian, and Old French, temprer--to temper, moderate. Old English temprian ‘bring something into the required condition by mixing it with something else’, from Latin temperare ‘mingle, restrain’. Sense development was probably influenced by Old French temprer ‘to temper, moderate’. The noun originally denoted a proportionate mixture of elements or qualities, also the combination of the four bodily humors, believed in medieval times to be the basis of temperament, hence temper (sense 1 of the noun) (late Middle English). (I'm pretty sure Schwarz understood French.) We use, in modern English, 'a temperate climate' to mean a moderate climate between tropical and harsh northern climates--a comfortable mixture of the two. We can say, "The climate of the coastal regions is moderated by the cool waters of the Pacific," but never, "The climate of ... is temperated by ...," rather, "The climate of ... is tempered by ... ." Temperate is purely an adjective like 'mild' whereas tempered is a verb (past participle) used as an adjective meaning to have been tempered--same grammatically as burned in 'a burned / burnt car'. Even the pronunciation of moderate changes according to whether it is being used as an adjective or as a verb or the past participle adjective moderated. (Maybe a native French speaker can comment on parallels in the grammar in French.) From all the considerations above, I see the delta 'function' as an construct of Heaviside and Dirac that has been molded into one amenable to the tastes of purist mathematicians (It's NOT a function!--the shrill mantra)--a morphing not really necessary for pragmatic physicists and engineers--by Schwarz and his theory of distributions. In that sense the delta function as a distribution that has been tempered by Schwarz and his theory seems fitting, just as the climate of coastal southern California is tempered or moderated by the Pacific waters. Of course one is free to say the climate along the coast is temperate, but that doesn't stress an agent and action that results in that quality. To end with the Bard: 1591, William Shakespeare, Henry VI, Part 1, Act II, Scene 4, [22]: Between two blades, which bears the better temper: […] I have perhaps some shallow spirit of judgement; But in these nice sharp quillets of the law, Good faith, I am no wiser than a daw.<|endoftext|> TITLE: Motivation for birational geometry QUESTION [11 upvotes]: I'm interested in how do people that work in birational geometry view their field — specifically, what are the kinds of geometric questions (as opposed to commutative-algebraic questions) that interest them? Often I hear that the main objective of birational geometry is to classify algebraic varieties up to birational isomorphism. This is an interesting mathematical question, but do the people who study it view it as a geometric question or as a commutative-algebraic one? As a concrete example, we can take the birational classification of surfaces: it gives you very nice answers to questions like "when can we find a function with certain properties between two surfaces? what are the properties of these functions?" However, I don't see what kinds of geometric questions (i.e. questions about shapes) this theory gives you answers for, given that surfaces birational to each other can nevertheless look so radically different from each other. Do people who study birational geometry view their field as geometry or as algebra? What are the kinds of geometric questions that interest them? I originally asked this question on math.se, but was advised to move it here. REPLY [9 votes]: Will has already said many of the things I would have said trying to answer your extended question, but let me add a few things without trying to avoid overlap. In fact, let me start with an overlap: Indeed, a fundamental question of birational geometry is what you asked: "What are some interesting properties of varieties that are preserved under birational transforms?" Studying this question is already interesting, but let me say a bit more. Algebra vs. Geometry I know this is the original question, but I would like to record my favorite comment about this issue. I would actually say that (at least with respect to algebraic geometry, and hence to birational geometry), Algebra=Geometry! Let me give two quick examples: a) Resolution of singularities of curves, a.k.a., starting with an arbitrary (=possibly singular, non-compact) curve and constructing a smooth projective curve which is birational to the original one. You might agree that this is a "purely" geometric question. However, one way to do it is to consider the set of DVRs of its function field (plus some technical condition), define first a topology, then regular and rational functions (functions, again!) on this set and then prove that this set is actually isomorphic to an "actual" curve. b) Lüroth's theorem, which asserts that if $k$ is an algebraically closed field and $k\subsetneq K \subseteq k(x)$ is a subfield of the purely transcendental extension of $k$ of transcendence degree $1$ is isomorphic to $k(x)$ (i.e. $K\simeq k(x)$, but not necessarily equal). You might agree that this is a "purely" algebraic question. Yet, one way to prove this is to observe that $K$ corresponds to a smooth projective curve, say $X$, over $k$ (for instance by a)!) and the inclusion $K \subseteq k(x)$ induces a morphism $\mathbb P^1\to X$. The latter implies that $X\simeq \mathbb P^1$ and hence the statement of the theorem. These two examples also provide examples to your extended question: A few things where birational geometry is relevant/interesting/worthwhile: Resolution of singularities: Given an arbitrary variety $X$, find a proper/projective morphism $\tilde X\to X$ such that $\tilde X$ is non-singular. This has lots of variants, and the relevant existence theorems are probably the most often used ones in algebraic geometry. Rationality questions: Which algebraic varieties are birational to projective space? If you want, this is a generalization of Lüroth's theorem. This is also one that one can consider either a purely geometric or a purely algebraic question. Again, lots of variants. Interesting keywords to look up: unirational, rationally connected varieties. The minimal model program is a quintessential part of birational geometry. It is sometimes considered part of classification, although it is actually really a precursor of that. The main guiding question of the mmp is: How can we choose a nice representative from every birational class? These are the first three that come to my mind, but others might argue that I am forgetting some. So, let me add some random examples without claiming that they are the most important ones. Birational geometry comes up in string theory. Physicists like to work with smooth objects, so they are in trouble when they encounter a singular object, which is essentially inevitable. So they want to make is smooth. You can do that either by resolution of singularities (as above) or (sometimes) by deforming it to something smooth. Doing both leads to something they call the conifold transition (see here) Another place where birational geometry is extremely important is moduli theory. This is another huge subject, so I'm not going to get into it. In fact, I kind of feel that I could just keep going, so instead I will stop here. Let me just say this: birational geometry is everywhere in algebraic geometry and even beyond that. To respond to the question in the comments: I would somewhat disagree with your assessment of the message of the answer on math.se. IMHO, isomorphism and birational equivalence are not competing. A different way to say what the math.se answer is saying is this: A priori the classification problem is to classify all varieties up to isomorphism. This is essentially impossible in the sense of usual classifications (such as classification of finite simple groups). The method employed is the following: first find a nice representative in each birational class, and a way to obtain it, then classify these nice representatives (up to isomorphism). This does give you a sort of classification of all varieties up to isomorphism. Start with your favorite one, obtain the nice representative, look it up in the classification. Now, you can obtain the original one by reversing the way you obtained the nice representative starting with the one you found on the list. In fact, this gives you a better classification than if you just had a list, because using this method you might be able to compare two varieties and decide whether they are isomorphic and if not then whether they are birational. But this is not the main point. The main point is what I said already, isomorphism and birational equivalence are not in competition. They play for the same team. (Also, just to link this to the above. That "nice representative" I mention here is the minimal model and the way to get it is provided by the mmp.)<|endoftext|> TITLE: Convergence criterion in the domain of an unbounded operator QUESTION [6 upvotes]: Cross-post from math.sx. My question is somewhat close to this one, but the counterexamples given there do not apply here. Setup. Given a Hilbert space $\mathcal H$, a closed operator $A$ and a convergent sequence $(x_n)_{n\in\mathbb N}\subset \mathcal D(A)$, I want to ensure that $\lim_{n\to\infty}x_n \in \mathcal D(A)$.It is not necessary that $\lim_{n\to\infty} Ax_n = Ax$. My intuition says that something like a uniform upper bound on $\|Ax_n\|$ should be a sufficient condition. Let me illustrate that in the following setting for normal operators. Lemma. Let $A$ be a normal operator on $\mathcal H$, $x_n$ a convergent sequence in $\mathcal D(A)$ and assume $\liminf_{n\to\infty}\|Ax_n\|<\infty$. Then $\lim_{n\to\infty} x_n \in \mathcal D(A)$. Proof. We use the spectral theorem and Fatou's lemma for weakly convergent measures. Let $E_A$ be the projection valued measure associated to $A$. Then $$ \int_{\mathbb C} |\lambda|^2 d\langle x,E_A(\lambda)x\rangle \le \liminf_{n\to\infty} \int_{\mathbb C} |\lambda|^2 d\langle x_n,E_A(\lambda)x_n\rangle = \liminf_{n\to\infty}\|Ax_n\|^2<\infty. $$ Hence, $x\in\mathcal D(A)$. $\square$ Question. What extensions of above lemma are known? Does anyone have a counterexample? References or own proofs both are warmly welcome. REPLY [6 votes]: The answer by Lars van der Laan gives a positive answer for the Hilbert space case (which was considered in the question), and it also works on reflexive Banach spaces. It might be worthwhile to add that the answer is no, in general, on non-reflexive Banach spaces. A simple counterexample is: Example. Endow the space $E = C_0((0,1])$ of continuous functions on $[0,1]$ that vanish in $0$ with the supremum norm, and consider the operator $A$ given by \begin{align*} D(A) & = \{u \in E: \, u \text{ is differentiable and } u' \in E \}, \\ Au & = -u'. \end{align*} This is a closed operator with compact resolvent (and even a semigroup generator). Now, consider the function $f \in E$ given by $f(x) = x$. Then $f$ can easily be approximated in sup norm by functions from $D(A)$ whose derivatives are bounded in sup norm. Yet, $f \not\in D(A)$ since $f' \not \in E$.<|endoftext|> TITLE: Are there ill-founded "maximally wide" models of $\mathsf{ZFC}$? QUESTION [9 upvotes]: Throughout assume $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals." I'm also happy to strengthen the large cardinal hypothesis if that would help. Say that a model $M\models\mathsf{ZFC}$ is powerful iff every end extension satisfying $\mathsf{ZFC}$ is a top extension. The transitive powerful models are exactly the $V_\alpha$s where $\alpha$ is a worldly cardinal: that every $V_{\mathsf{worldly}}$ is powerful is trivial, and the large cardinal hypothesis above gives the other half of the result. The ill-founded situation is on the other hand completely unclear to me: Is there an ill-founded powerful model of $\mathsf{ZFC}$? I don't really see where to start with this. Certainly there are no countable powerful models of $\mathsf{ZFC}$, ill-founded or otherwise, since we can (genuinely) force over such models; however, I don't see any useful tools for analyzing uncountable ill-founded models. The existence of rigid ill-founded models (e.g. the substructure of any ill-founded model of $\mathsf{ZFC+V=L}$ consisting of the parameter-freely-definable elements) does make the existence of powerful models feel not totally implausible, but that said I don't see how any of the techniques for building examples of the former type are useful for attempting to build an example of the latter. REPLY [7 votes]: Back in the mid 1980s I remember convincing myself (alas, in unpublished work) that there is an ill-founded model $M$ of ZFC that has no end extension to another model of ZFC. Such a model $M$ by default is powerful and technically answers the question. My unpublished work above used techniques employed in the following related results from the same time period: 1. There are ill-founded models $M$ of any given consistent extension of ZFC that have no top extension to another model of ZFC; see Theorem 1.5 of this paper of mine. Note that top extensions are referred to as rank extensions in the paper. 2. There are ill-founded models $M$ of any given consistent extension of ZFC such that there are no $N \supsetneq M$ such that $N$ is a forcing end extension of $M$. This was noted in this other paper of mine (see the remark after Theorem 1.5, which itself relates to Theorem 1.4). In (1) and (2) above, the cardinality of $M$ can be arranged to be $\aleph_1$. This is optimum since every countable model of ZF (including ill-founded ones) has lots of top extensions and forcing extensions<|endoftext|> TITLE: Is there an analogue for Ramanujan–Serre derivative for Hilbert modular forms? QUESTION [8 upvotes]: If $f$ is a modular form of weight $k$, it is well known that $$ D(f)=f' -\tfrac k{12}E_2f $$ is modular of weight $k+2$. Here $E_2$ is the Eisenstein series. I wanted to ask if there is an extension of this fact for Hilbert modular forms. When looking up Hilbert modular forms (Shimura 1975), I was only able to find the following differential operators: for a Hilbert modular form $f$ on $\mathbb H^n$, we define $$ D_{j,t}(f)=\left(\frac{t}{2i y_j}+\frac{\partial}{\partial z_j}\right)f $$ where $z_j=x_j+iy_j$ is the $j$-th coordinate. I don't think this is the same operator (is it?) I am really looking for a many variable version of the operator $D(f)=f' -\frac k{12}E_2f$. I don't really have a background in analysis or number theory so this question may be really easy for experts. REPLY [3 votes]: I think there's not a good analogue. The operator $D_{j, k_j}$, acting on Hilbert modular forms of weight $k = (k_1, \dots, k_n)$ where $n = [F: \mathbf{Q}]$, is the "Maass--Shimura differential operator". It doesn't preserve holomorphicity, but it sends holomorphic Hilbert modular forms to "nearly-holomorphic" forms in Shimura's sense. For $n = 1$ something special happens, which is that the nearly-holomorphic forms are freely generated by $E_2$ as a polynomial ring over the holomorphic ones. So you can "adjust" the differential operator by a multiple of $E_2$ to make it preserve holomorphicity (at a cost of breaking its nice compatibility with Hecke operators), and this gives the Ramanujan--Serre operator. However, for $n > 1$ there isn't a tidy set of generators for nearly-holomorphic forms as an algebra over the holomorphic ones so I don't think there is an analogue. (Perhaps if you were more precise about what it is that you want this operator to be useful for, then you might get more useful answers in response.)<|endoftext|> TITLE: The polytope algebras generated by polytopes with rational vs arbitrary vertices QUESTION [8 upvotes]: The polytope algebra was defined by P. McMullen in "The polytope algebra" Adv. Math. 78 (1989) as follows. Let us denote by $\Pi'_\mathbb{R}$ the quotient of the free abelian group generated by the symbols $[P]$, where $P\subset \mathbb{R}^n$ is an arbitrary convex compact polytope, by the subgroup generated by elements \begin{eqnarray*} (1)\, [P\cup Q]+[P\cap Q]-[P]-[Q] \mbox{ where } P,Q,P\cup Q \mbox{ are convex compact polytopes};\\ (2)\, [P+x]-[P] \mbox{ where } x\in \mathbb{R}^n. \end{eqnarray*} Similarly let us define the analogous group $\Pi_{\mathbb{Q}}'$ generated by convex compact polytopes with rational vertices and the same relations (1)-(2) (in the relation (2) one assumes $x\in\mathbb{Q}^n$). We have the obvious group homomorphism $\Pi_{\mathbb{Q}}'\to \Pi'_\mathbb{R}$. Is it injective? A reference would be helpful. REPLY [5 votes]: Looking more carefully at the above McMullen's paper, I realized that the question has the positive answer due to Theorem 3 in the paper. McMullen constructs homomorphisms $\Pi'_{\mathbb{F}}\to \mathbb{F}$ (where $\mathbb{F}= \mathbb{R},\mathbb{Q}$) which all together separate points, i.e. if all of them vanish on some element, then the element vanishes. Let $u$ be a non-zero linear functional. For a polytope $P$ denote $$P_u:=\{x\in P|\, u(x)=\max_{y\in P}u(y).\}$$ Let us define the homomorphism $f_u\colon \Pi'_{\mathbb{F}}\to \mathbb{F}$ on generators by $f_u([P]):=vol(P_u)$ where $vol$ is $(n-1)$-dimensional volume. McMullen shows that $f_u$ is well defined. This construction can be generalized recursively. Let $u_1,\dots,u_k$ be linearly independent linear functionals. Define $P_{u_1,\dots,u_k}:=(P_{u_{1},\dots,u_{k-1}})_{u_k}$. Define the homomorphism $f_{u_1,\dots,u_k}\colon \Pi'_{\mathbb{F}}\to \mathbb{F}$ by $$f_{u_1,\dots,u_k}([P]):=vol(P_{u_{1},\dots,u_{k}}),$$ where $vol$ is $(n-k)$-dimensional volume. Theorem 3.(McMullen) All the homomorphisms $f_{u_1,\dots u_k}$ separate points in $ \Pi'_{\mathbb{F}}$, i.e. if on some element $w\in \Pi'_{\mathbb{F}} $ all such homomorphisms vanish then $w=0$. Let us show how Theorem 3 answers (immediately) the question in the post. Let $w\in\Pi'_{\mathbb{Q}} $ belongs to the kernel of the homomorphism in the question. That implies that for all $k$-tuples of linearly $\mathbb{R}$-independent real linear functionals $u_1,\dots,u_k$ one has $f_{u_1,\dots,u_k}(w)=0$. Now if $u_1,\dots,u_k$ are rational linear functionals which are linearly independent over $\mathbb{R}$ then they are linearly independent over $\mathbb{Q}$. For all of such $k$-tuples all $f_{u_1,\dots ,u_k}$ separate points in $\Pi'_{\mathbb{Q}}$. Hence $w=0$. QED.<|endoftext|> TITLE: Linear hyperbolic PDE on compact two dimensional domain QUESTION [5 upvotes]: Consider the equation $$ \begin{equation} \frac{\partial^2f}{\partial x\partial y}=f \end{equation} $$ on a Jordan domain (i.e. the interior of a simple, closed curve on the plane). The equation is hyperbolic, but we cannot formulate a Cauchy problem in the usual sense since the domain is finite, so the question is What kind of initial value problem can we formulate on such a domain ? Which initial data along the boundary do when need to establish existence ? REPLY [4 votes]: Section 4 of the following paper considers in some detail the 2D wave equation ($\partial_x\partial_y f = 0$ in your coordinates; not exactly the same but closely related) on compact domains with smooth boundary (with unpleasant cases where the boundary is too closely tangent to null/characteristic directions is also excluded): Cattaneo, Alberto S.; Mnev, Pavel, Wave relations, Commun. Math. Phys. 332, No. 3, 1083-1111 (2014). arXiv:1308.5592 ZBL1300.53069. A quick summary that doesn't do justice to all the details: The wave equation has a variational formulation, which indirectly means that the boundary data (value and normal derivative of $f$ on the boundary) gets an induced symplectic structure. The authors show that the restriction of solutions $f$ to the boundary gives a Lagragian subspace $L$ of this space of boundary data. Thinking abstractly, any set of boundary conditions identifies another (affine) subspace $C$ of the boundary data. The boundary conditions specify a unique solution when the two subspaces are complementary, the intersection $L\cap C$ is zero dimensional (just one point, the unique solution). The authors then consider some examples of such subspaces $C$. Heuristically (ignoring the infinite dimensional context), a generic second Lagrangian subspace would give a good $C$.<|endoftext|> TITLE: How to interpret this quote of Lin? QUESTION [8 upvotes]: I recently stumbled across a quote of Fang-Hua Lin that I have trouble understanding [1, page 42]. It is a well-known fact that a weakly converging sequence of stationary integral currents may have a limit which is not a stationary current. Question. How should I interpret this quote? What does Lin mean by a 'stationary current', and which sequence demonstrates this 'well-known fact'? My initial guess would be that an integral current $T$ is 'stationary' if the varifold $\lvert T \rvert$ obtained by forgetting orientations is stationary. If I am not mistaken, this should mean that $\partial T = 0$? However my impression is that a flat limit $T$ would be 'stationary' in this sense of the word. [1] F.-H. Lin. Mapping problems, fundamental groups and defect measures. Acta Math. Sin. 15 (1999), 25-52. REPLY [2 votes]: I believe the following sequence demonstrates the failure of flat limits to be stationary. This would be consistent with the natural interpretation of the quote, meaning: a current $T$ is called stationary if the varifold $\lvert T \rvert$ is. (A quick side remark before the construction: on second thought whether $\partial T = 0$ or not seems seems unrelated to the stationarity of $\lvert T \rvert$. For example, a triple junction has boundary as a current, but is stationary as a varifold. A slightly simpler variant of the example below would see the current $S$ replaced with a triple junction.) That being said, the currents in the constructed sequence $(T_n \mid n \in \mathbf{N})$ are one-dimensional cycles in the unit disc $D \subset \mathbf{R}^2$: $\partial T_n = 0$ for all $n$. They converge weakly as currents to another cycle, say $T_n \to T$ as $n \to \infty$. Most important: $\lvert T_n \rvert$ is stationary for all $n$, but $\lvert T \rvert$ is not. To construct the sequence, let $\{ v_1,\dots,v_6 \} \subset \partial D$ be unit vectors with \begin{equation} v_1 + \cdots + v_6 = 0, \end{equation} but which do not match up into antipodal pairs. For example \begin{equation} -v_1 \not \in \{ v_1,\dots,v_6 \}. \end{equation} Let $S \in I_1(D)$ be the current supported in the union of the segments $\{ t v_i \mid 0 \leq t \leq 1 \}$, oriented so that $\partial S = 0.$ The associated varifold $\lvert S \rvert$ is stationary by construction. Let $L$ be the current supported in the segment $\{ tv_1 \mid -1 \leq t \leq 1 \}$, which we orient in the opposite direction. In other words \begin{equation} \{ t v_1 \mid 0 < t \leq 1 \} \cap \mathrm{spt} \, (S + L) = \emptyset. \end{equation} This too has $\partial L = 0$ and $\lvert L \rvert$ stationary. The orientations are chosen so as to ensure that the current $T := S + L$ is not stationary; this is because $-v_1 + \cdots + v_6 = -2v_1 \neq 0$. Next we consider a sequence of positive angles $\theta_n \to 0$. We use these angles to rotate $L$, forming a sequence of currents \begin{equation} R_{\theta_n \#} L \to L \text{ as $n \to \infty$.} \end{equation} As long as $\theta_n$ is small enough that $\mathrm{spt} \, R_{\theta_n \#} L \cap \mathrm{spt} \, S = \{ 0 \}$, the cycles $T_n := S + R_{\theta_n \#} L$ are stationary. However \begin{equation} T_n = S + R_{\theta_n \#}L \to S + L = T \text{ as $n \to \infty$} \end{equation} in the current topology, which was pointed out above is not stationary.<|endoftext|> TITLE: Why is $\mathsf{D}_{qc}(X)$ the right notion, instead of $\mathsf{D}(\mathsf{QCoh}(X))$? QUESTION [10 upvotes]: Let $\mathsf{A}$ be an abelian category and $\mathsf{B}$ be a full abelian subcategory. More often than not, instead of being interested in the derived category $\mathsf{D}(\mathsf{B})$, we are interested in the full subcategory $\mathsf{D}_{\mathsf{B}}(\mathsf{A})$ of $\mathsf{D}(\mathsf{A})$ composed of the complexes whose cohomology is in $\mathsf{B}$. (The cases of quasi-coherent / coherent sheaves and of holonomic $\mathcal{D}$-modules come to mind.) This seems counter-intuitive for me since we usually have more tools for dealing with objects in the category $\mathsf{B}$ which could be applied to $\mathsf{D}(\mathsf{B})$. While both categories are usually equivalent, I feel that people usually think that $\mathsf{D}_{\mathsf{B}}(\mathsf{A})$ is the right object. Why is it so? REPLY [17 votes]: The triangulated category $\mathsf{D}_{\mathsf{B}}(\mathsf{A})$ can be promoted to a stable $\infty$-category. One of the many interests of working with stable $\infty$-categories is that we have a reasonable theory of descent for them: we can define sheaves of stable $\infty$-categories (it is possible to formulate such concepts using the language of dg categories). However, the assignment $$X\mapsto \mathsf{D}(\mathsf{QCoh}(X))$$ is unfortunately not a sheaf of stable $\infty$-categories on the big site of all schemes (for the Zariski or fppf topology, say) and its sheafification leads to the assignment $$X\mapsto \mathsf{D}_{qc}(X)\, .$$ We have $\mathsf{D}(\mathsf{QCoh}(X))\cong \mathsf{D}_{qc}(X)$ locally (in fact for $X$ any separated and quasi-compact scheme), so that $\mathsf{D}_{qc}(X)$ gets many of the nice features of $\mathsf{D}(\mathsf{QCoh}(X))$ by local-to-global principles.<|endoftext|> TITLE: Cohomology ring of grassmannian and Pieri rule QUESTION [5 upvotes]: I am sorry if this question is not for mathoverflow. I asked the same question on stackexchange (https://math.stackexchange.com/questions/4203667/cohomology-ring-of-grassmannian-and-pieri-rule), but I didn't get an answer: Let $X=OG(n,2n+1)$, where $OG(n,2n+1)$ denotes the variety of $n$-dimensional isotropic subspaces of a vector space $\mathbb{C}^{2n+1}$ with a nondegenerate symmetric bilinear form. According to Theorem 2.2 a) (Page 17, Anders Skovsted Buch, Andrew Kresch, Harry Tamvakis, Quantum Pieri rules for isotropic grassmannians, https://arxiv.org/pdf/0809.4966.pdf), the cohomology ring of X is given by $$ H^{*}(X,\mathbb{Z})=\mathbb{Z}[\tau_{1},\ldots, \tau_{n}]/I,$$ where $I$ is the ideal generated by $$ \tau_{r}^{2}-2\tau_{r+1}\tau_{r-1}+2\tau_{r+2}\tau_{r-2}+\cdots +(-1)^{r}\tau_{2r}$$ for $1\leq r\leq n$. In particular, if $n=4$, then the ideal $I$ is generated by the following four elements $$ \tau_{1}^{2}-\tau_{2},\quad \tau_{2}^{2}-2\tau_{3}\tau_{1}+\tau_{4}, \quad \tau_{3}^{2}-2\tau_{4}\tau_{2},\quad \tau_{4}^{2}.\tag{*}\label{*}$$ But if I apply Pieri rule for X (Theorem 2.1, Page 16, Anders Skovsted Buch, Andrew Kresch, Harry Tamvakis, Quantum Pieri rules for isotropic grassmannians) to $\tau_{2}\cdot \tau_{2}$, I get the following relation $$\tau_{2}^{2}-2\tau_{3}\tau_{1}-\tau_{4} \tag{**}\label{**}$$ Therefore, combining (\ref{*}) and (\ref{**}), I get $2\tau_{4}=0$ in $H^{*}(X,\mathbb{Z})$. I seems that some computation is wrong, but I don't know where I made a mistake. Thank you. REPLY [3 votes]: The equation $\tau_2^{2}-2\tau_{3}\tau_{1}-\tau_{4}$ (obtained by Pieri rule) is incorrect. It should be $$\tau_2^{2}-2\tau_{3,1}-\tau_{4},$$ so there is no problem for such computations.<|endoftext|> TITLE: Supersingular curves over $\mathbb{F}_q$ and the splitting of $p$ QUESTION [7 upvotes]: I'm looking at chapter 4 of Waterhouse's "Abelian varieties over finite fields"; and Theorems 4.1 and 4.2 seem to use the following fact: Suppose that $E/\mathbb{F}_q$ is an elliptic curve over a finite field with $q=p^n$ elements and let $\pi_E$ denote the $q$th-power Frobenius map acting on $E$. Suppose that $\pi_E$ does not act like multiplication-by-$N$ for any integer $N$ so that $\pi_E = [\alpha]_E$ where $\alpha$ is a root of the polynomial $$x^2 - \mathrm{tr}(\pi_E)x + q$$ Then $E$ is ordinary if and only if $p$ is splits in $\mathbb{Q}(\alpha)$. Waterhouse uses some very general theory (namely a theorem of Honda/Tate and the general theory of abelian varieties) and I find his reasoning/terminology vague. Does anybody know of a more down to earth proof for this result? Lang has a proof utilitizing the fact that $End(E)\to End(T_p(E))$ is injective; but i'm trying to find proofs that don't use local methods EDIT: Update on attempt: Here is a proof attempt; perhaps someone can help me finish it off. We prove $E$ is supersingular if and only if $p$ is non-split in $\mathbb{Q}(\alpha)$. First suppose that $p$ is non-split. If $p$ is inert, then since $\alpha$ has norm $q = p^n$, it follows that $p$ divides $(\alpha)$ and so $E[p]\subseteq E[\pi_E] = \{O \}$; thus $E$ is supersingular. If $p$ ramifies then $p$ divides the discriminant of $\mathbb{Q}(\alpha)$ and so $p$ divides $\mathrm{tr}(\pi_E)^2 - 4q$, which implies that $p$ divides $\mathrm{tr}(\pi_E)$ and so $E$ is supersingular; Conversely, suppose that $E$ is supersingular. Suppose that $p$ splits as $(p) = \mathfrak{p}\overline{\mathfrak{p}}$. Then $(\alpha) = \mathfrak{p}^a\overline{\mathfrak{p}}^b$ where $a+b=n$. However, since $E$ is supersingular $\alpha^N \in \mathbb{Z}$ for some positive integer $N$. Which implies that $(\alpha)^N = \mathfrak{p}^{Na}\overline{\mathfrak{p}}^{Nb}$ and so $Na=Nb$ since $(\alpha^N) = \overline{(\alpha^N)}$. Therefore $a=b$ and so $(\alpha) = \mathfrak{p}^a \overline{\mathfrak{p}}^a = (p^a)$ where $a=n/2$. This implies that $\alpha = \zeta p^{a}$ for some unit $\zeta$. This is where I am stuck. Any help would be appreciated. REPLY [5 votes]: Here is a solution that avoids explicit use of $\mathbf{Q}_p$ and in particular does not require knowing that $(\operatorname{End} E) \otimes \mathbf{Q}_p$ is a division ring. The key is to use inseparable degree of endomorphisms. Identify $\alpha$ with $\pi_E$. Let $a = \operatorname{tr} \alpha \in \mathbf{Z}$. Let $\mathcal{O}$ be the ring of integers of $\mathbf{Q}(\alpha)$. Let $\mathcal{O}' := (\operatorname{End} E) \cap \mathbf{Q}(\alpha)$. For $\beta \in \mathcal{O}'$, the inseparable degree $\deg_{\text{i}} \beta$ is multiplicative in $\beta$. The $\beta \in \mathcal{O}'$ with $\deg_{\text{i}} \beta \ge p^m$ are the $\beta$ that factor through the $p^m$-power Frobenius morphism $F_m \colon E \to E^{(p^m)}$; they form an additive subgroup. Thus $\beta \mapsto \log_p \deg_{\text{i}} \beta$ defines a valuation $v \colon \mathcal{O}' \to \mathbf{Z} \cup \{\infty\}$. Extend $v$ to the fraction field $\mathbf{Q}(\alpha)$. Case 1: $E$ is ordinary. Then $p \nmid a$, so $x^2-ax+q$ mod $p$ has two distinct factors, so $p$ splits in $\mathbf{Q}(\alpha)$. Case 2: $E$ is supersingular. Then $p \mid a$, so the equality $\alpha^2 - a \alpha + q = 0$ yields $\alpha^2 \in p \mathcal{O}$. First consider the case $\mathcal{O}'=\mathcal{O}$. If $\beta \in \mathcal{O}$ satisfies $v(\beta) \ge 2n$, then $\beta$ factors through $F_{2n} = \alpha^2 \in p \mathcal{O}$, so $\beta \in p \mathcal{O}$. This implication for all $\beta$ shows that $v$ is the unique place above $p$. In general, write $(\mathcal{O}:\mathcal{O}')=p^e d$ with $p \nmid d$. If $\beta \in \mathcal{O}$ satisfies $v(\beta) \ge (e+1)2n$, then the endomorphism $(p^e d) \beta$ is divisible (in $\mathcal{O}'$ or in $\mathcal{O}$) by $\alpha^{2(e+1)}$ and hence by $p^{e+1}$, but $p \nmid d$, so $\beta \in p\mathcal{O}$. Again, this shows that $v$ is the unique place above $p$.<|endoftext|> TITLE: Convergence properties in dense subsets of $\omega^*$ QUESTION [8 upvotes]: The space $\omega^*$, the remainder of the Cech-Stone compactification of the integers, fails to have all convergence-type properties known to me. Sequentiality. (As a matter of fact $\omega^*$ does not even have any convergence sequences, because every infinite closed subset of $\omega^*$ has cardinality $2^\mathfrak{c}$). Pseudoradiality. (Every infinite compact pseudoradial space must contain a convergent sequence). Countable tightness (As proved by Kunen, $\omega^*$ even contains points which are not accumulation points of every countable subset of $\omega^*$). The "weak Whyburn property" (that means, for every non-closed $A \subset X$ there is $x \in \overline{A} \setminus A$ and $B \subset A$ such that $\overline{B} \setminus B=\{x\}$. Note that every compact weakly Whyburn space must contain a convergent sequence). Discrete generability. That follows from Proposition 2.4 of: Tkachuk, V. V.; Wilson, R. G., Box products are often discretely generated, Topology Appl. 159, No. 1, 272-278 (2012). ZBL1236.54005. or from the fact that $\omega^*$ contains a point which is in an accumulation point of a countable set, but not an accumulation point of any countable discrete set (see Theorem 4.4.1 in Jan van Mill's article in the Handbook of set-theoretic topology). Obviously a dense subset of $\omega^*$ cannot be sequential, but: QUESTION: Can a dense subset of $\omega^*$ be pseudoradial, countably tight, weakly Whyburn or discretely generated? Parts of the above question have already been asked in the literature and in the comment section of a question on Mathoverflow (I'll give references below), but I thought it would be nice to have them all in one place. Is there a compact space with no countably generated dense subspace? (see the comment section of my answer) Alas, Ofelia T.; Madriz-Mendoza, Maira; Wilson, Richard G., Some results and examples concerning Whyburn spaces, Appl. Gen. Topol. 13, No. 1, 11-19 (2012); corrigendum ibid. No. 2, 225-226 (2012). ZBL1245.54024. (Question 4.4 even asks whether $\omega^*$ can have a dense Whyburn subspace). EDIT (22/07/2021): Let me remark that under CH $\omega^*$ has a dense subspace which is radial, Whyburn and discretely generated. It suffices to take the set $D$ of all $P$-points. REPLY [3 votes]: Comments from Alan Dow: (1) under CH the set of P-points is dense and radial even, using $\omega_1$-sequences. (2) if $D$ is dense and pseudoradial then for every cozero set $C$ of $\omega^*$ the intersection $C\cap D$ is closed in $D$. Say $C=f^{-1}[(0,1]]$; a convergent sequence $\langle x_\alpha:\alpha<\kappa\rangle$ in $C\cap D$ should have uncountable cofinality, but then there is an $n$ such that $\{\alpha:f(x_\alpha)\ge2^{-n}\}$ is unbounded and hence the limit wound be in $\{x:f(x)\ge2^{-n}\}$. (3) if there are no P-points then every point of the dense set $D$ is in the boundary of some cozero set $C$, for that $C$ the intersection $C\cap D$ is not closed; hence $D$ is not pseudoradial.<|endoftext|> TITLE: Who is Mrs. Gerber? QUESTION [22 upvotes]: This question on a theorem in information theory called Mrs. Gerber's lemma piqued my curiosity. Who is this individual, and why the "mrs." ? A quick Google search was not informative, although it did produce a Mr. Gerber's lemma (arXiv:1802.05861) -- can someone enlighten me? REPLY [28 votes]: Check out the original reference "A theorem on the entropy of certain binary sequences and applications - I" by Wyner and Ziv: http://doi.org/10.1109/TIT.1973.1055107. Footnote 2 on page one explains This result is known as “Mrs. Gerber’s Lemma” in honor of a certain lady whose presence was keenly felt by the authors at the time this research was done. I'm not sure you're going to get more of an explanation than that.<|endoftext|> TITLE: Integrality of a sequence formed by sums QUESTION [14 upvotes]: Consider the following sequence defined as a sum $$a_n=\sum_{k=0}^{n-1}\frac{3^{3n-3k-1}\,(7k+8)\,(3k+1)!}{2^{2n-2k}\,k!\,(2k+3)!}.$$ QUESTION. For $n\geq1$, is the sequence of rational numbers $a_n$ always integral? REPLY [3 votes]: Here is another proof, inspired by Tewodros Amdeberhan's. We represent the sum as a constant term in a power series. To represent $(7k+8) \frac{(3k+1)!}{k!\,(2k+3)!}$ as a constant term, we need to express it as a linear combination of binomial coefficients. To do this we express $7k+8$ as a linear combination of $(2k+2)(2k+3)$, $k(2k+3)$, and $k(k-1)$ since each of these polynomials yields a binomial coefficient when multiplied by $\frac{(3k+1)!}{k!\,(2k+3)!}$. We find that $$(7k+8) \frac{(3k+1)!}{k!\,(2k+3)!}=\frac13 \left[4\binom{3k+1}{k} -7\binom{3k+1}{k-2} -2\binom{3k+1}{k-2}\right].$$ Using $\binom{n}{j} =\text{CT}\, (1+x)^n/x^j$, where CT denotes the constant term in $x$, we have $$ \begin{aligned} (7k+8) \frac{(3k+1)!}{k!\,(2k+3)!} &= \text{CT}\, \frac13\left( 4\frac{(1+x)^{3k+1}}{x^k} -7\frac{(1+x)^{3k+1}}{x^{k-1}}-2\frac{(1+x)^{3k+1}}{x^{k-2}}\right)\\ &=\text{CT}\,\frac{(1-2x)(x+4)(1+x)^{3k+1}}{3x^k}. \end{aligned} $$ Multiplying by $3^{3n-3k-1}/2^{2n-2k}$, summing on $k$ from 0 to $n-1$, and simplifying gives $$ a_n = 3\,\text{CT}\, \frac{(1+x)^{3n+1}}{x^{n-1}(1-2x)} -3\left(\frac{27}{4}\right)^n\!\text{CT}\,\frac{x(1+x)}{1-2x}. $$ The second constant term is 0, so $a_n = 3\,[x^{n-1}]\, (1+x)^{3n+1}/(1-2x)$, which is clearly an integer. In fact this gives the formula $$a_n = 3 \sum_{k=0}^{n-1}2^{n-k-1}\binom{3n+1}{k}.$$ Tewodros's formula can also be derived in the same way; if we represent his sum as a constant term and simplify we also get $3\,[x^{n-1}]\, (1+x)^{3n+1}/(1-2x)$.<|endoftext|> TITLE: Is there a square with all corner points on the spiral $r=k\theta$, $0 \leq \theta \leq \infty$? QUESTION [14 upvotes]: I've posted this question on Math Stack Exchange, but I want to bring it here too, because 1) the proof seems missing in the literature, although they are some sporadic mentions and 2) maybe it requires more sophisticated topological tools to prove it or disprove it, like bordism arguments that have been used before for some cases of the square peg problem, and these are out of my reach. The question is: Is there a square that all of its corner points lie in the spiral $$r = k\theta \quad 0 \leq \theta \leq \infty $$ ? REPLY [12 votes]: The points on the spiral with $\theta=$ $$ 3.4535999354657\\ 15.1248305526170\\ 22.0370015553781\\ 16.3950081067565 $$ form a square. The numbers can be generated from this Mathematica code, if anyone wants to play with the algorithm and the initial conditions: start = {{r, 15}, {s, 16}}; curve[r_] := {r Cos[r], r Sin[r]}; rotate[{x_, y_}] := {-y, x}; left[p_, q_] := (p + q + rotate[q - p])/2; right[p_, q_] := (p + q - rotate[q - p])/2; curvedist[p_] := Norm[p - curve[Norm[p]]]^2; test[p_, q_] := curvedist[left[p, q]] + curvedist[right[p, q]]; m = FindMinimum[test[curve[r], curve[s]], start]; {r, s, left[curve[r], curve[s]] // Norm, right[curve[r], curve[s]] // Norm} /. m[[2]] Update: Initial conditions with a Pythagorean form start = {{r, Pi(m^2+2mn-n^2)p}, {s, Pi(-m^2+2mn+n^2)p}} for positive integer $m,n,p$ with $m TITLE: Small simplicial set models for BG QUESTION [12 upvotes]: Let $F$ be a finite group. Is there a model for $BF$ as a simplicial set such that the number of nondegenerate $n$-simplices grows at most polynomially? For example the Bar construction has the property that there are exactly $(|F|-1)^n$ nondegenerate $n$-simplices. This answers the question affirmatively for $\mathbb{Z}/2$, but for other groups it still grows exponentially. A lower bound for the number of such simplices is of course given by the rank of the group homology and in all examples that I know this only grows polynomially. Of course it would be nice to have a functorial model, but that might be a follow up. REPLY [4 votes]: This question may be somewhat relevant: Small simplicial complexes with torsion in their homology. David Speyer's answer there shows that one can build a simplicial complex $X$ with $H_1(X)=\mathbb{Z}/p$ where the number of simplices of $X$ is $O(\log(p))$. It seems unlikely that one can do much better than that in the world of simplicial sets. With CW complexes, you only need a single $0$-cell, a single $1$-cell and a single $2$-cell. This gives an initial picture of how the simplicial set version of the question might deviate from the CW complex version.<|endoftext|> TITLE: About an application of BBD decomposition theorem QUESTION [5 upvotes]: There is a following proposition in the famous paper on Koszul duality by Beilinson, Ginzburg and Soergel: let $G$ denote a semisimple complex Lie group, let $B$, $Q$ and $W^Q$ denote a pair of a Borel and parabolic subgroups and the corresponding parabolic Weyl group respectively, let $\pi$ denote an evident morphism $G/B \to G/Q$. For a variety $X$ let $\mathbb C_X$ denote a constant sheaf of 1-dimensional vector spaces on X. Then the derived pushforward $\pi_* \mathbb C_{G/B}$ is $$\bigoplus_{x \in W^Q} \mathbb C_{G/B}[-2l(x)].$$ Authors claim that it's a consequence of BBD decomposition theorem which is known to me in the following formulation: *for a proper morphism $f: X \to Y$ of algebraic varieties there is an isomorphism $$R f_{*}\left[\mathrm{IC}_{X} \cdot\right] \simeq \bigoplus_{k}^{\text {finite }} i_{k *} \mathrm{IC}_{Y_{k}}\left(L_{k}\right)^{\cdot}\left[l_{k}\right],$$ where $Y_k$, $L_k$ and $l_k$ are some locally closed subvarieties, local systems on them and integer numbers respectfully.* (See also here for the stratified version.) Can someone explain me how to derive BGS's claim about $G/B$ and $G/Q$ from this theorem? (Using the obvious stratification (by B-orbits) one can see that it is sufficient to prove the desired result for a big cell $F$ in G/Q (i.e. to prove the claim that $Rf_*(\mathbb C_{G/B})|_F = \bigoplus_{x \in W^Q} \mathbb C_F[-2l(x)]$), isn't it?) REPLY [10 votes]: One approach: Because $\pi$ is smooth, the $Y_i$s in the decomposition theorem must all be the entire space $G/Q$. Because $G/Q$ is simply-connected, the local systems are all constant. To calculate the multiplicities in each degree, it therefore suffices to evaluate the stalk at any one point, i.e. to calculate the calculate the cohomology of any fiber. The fiber of $G/B \to G/Q$ at the identity is $Q/B$, and this is known to be a sum over $W^Q$ by the standard cell decomposition.<|endoftext|> TITLE: Atiyah duality without reference to an embedding QUESTION [11 upvotes]: Atiyah duality is the equivalence $M/\partial M \simeq (M^{-T(M)})^\vee$, i.e. the Spanier-Whitehead dual of the space $M/\partial M$ is the Thom complex of the stable normal bundle of $M$. The theorem is proven by taking an appropriate embedding $i$ of $M$ into a sphere $S^d$, identifying the complement with the Thom complex of the normal bundle of the embedding, and constructing a duality map $M_+ \wedge M^{N(i)}\rightarrow S^d$ using the embedding. In the case $M$ is a compact, framed manifold of dimension $n$, it is possible to describe a duality map $M_+ \wedge M_+ \rightarrow S^n$ without reference to an embedding by appealing to the fact that if we collapse everything outside a small ball around $p \in M$, we can use the framing to continuously identify it with $S^n$. This is the adjoint of a duality map $M_+ \wedge M_+ \rightarrow S^n$. $\bf Question:$ Is it possible to create such a duality map for a framed manifold with nonempty boundary, namely one that does not invoke an embedding? REPLY [9 votes]: Here is another short construction which is much simpler and just takes a few lines. Let $M$ be a closed $n$-manifold. Consider the diagonal $M \to M \times M$. It is an embedding. Take its Pontryagin-Thom construction to get a map $$ M_+ \wedge M_+ \to M^\tau $$ (we have identified a tubular neighborhood of the diagonal with the total space of the tangent bundle). If $M$ is (stably) framed, then $M^\tau \simeq M_+ \wedge S^n$ (stably). Then we have the map $M_+ \wedge S^n \to S^n$ induced by smashing $M_+ \to \text{pt}_+$ with $S^n$. The composition $$ M_+ \wedge M_+ \to M^\tau \simeq M_+ \wedge S^n \to S^n $$ is what you want: it's a duality map. This can be seen on the level of homology (but it is enough to check a map is a duality map on the level of homology). If $M$ is compact with non-empty boundary $\partial M$, then there is a map of pairs $$ (M\times M, M\times \partial M) \to (M\times M,M\times M - U) $$ where $U$ is a tubular neighborhood of the diagonal. To get this map, one might choose a collar neighborhood $C$ of $\partial M$ and thereafter identify $M$ with $M-C$. Then $M-C $ and $\partial M$ are disjoint. The map of pairs determines a map of quotients $$ M_+ \wedge M/\partial M \to M^\tau \, , $$ and one may then proceed as in the empty boundary case, assuming that $M$ is framed, to obtain a map $$ M_+ \wedge M/\partial M \to S^n $$ which will be an $S$-duality.<|endoftext|> TITLE: Classification of finite subgroups of $\mathrm{SL}_5(\mathbb{C})$ QUESTION [9 upvotes]: $\DeclareMathOperator\SL{SL}$I would like to know if there is a full list of isomorphic class of finite subgroups of $\SL_5(\mathbb{C})$. I have already found the classification for $\SL_2$ and $\SL_3$. Thank you very much! REPLY [13 votes]: The irreducible primitive finite subgroups of $SL_5(C)$ were classified up to isomorphism by Richard Brauer: "Über endliche lineare Gruppen von Primzahlgrad", Math. Ann. 169 (1967), 73–96.<|endoftext|> TITLE: Intermediate notions of bilinearity in higher algebra QUESTION [11 upvotes]: It is well-known that when passing to $\infty$-categories the notion of commutativity gets replaced by an infinite array of notions of commutativity: $\mathbb{E}_{1}$, $\mathbb{E}_{2}$, ..., $\mathbb{E}_{\infty}$. This is already apparent when passing from sets to categories and $2$-categories: For sets, we have monoids ($\mathbb{E}_{1}$) and commutative monoids ($\mathbb{E}_2=\cdots=\mathbb{E}_\infty$); For categories, we have monoidal ($\mathbb{E}_{1}$), braided ($\mathbb{E}_{2}$), and symmetric monoidal categories ($\mathbb{E}_{3}=\mathbb{E}_{4}=\cdots=\mathbb{E}_{\infty}$); For $2$-categories, we have monoidal ($\mathbb{E}_{1}$), braided ($\mathbb{E}_{2}$), sylleptic ($\mathbb{E}_{3}$), and symmetric monoidal categories ($\mathbb{E}_{4}=\mathbb{E}_{5}=\cdots=\mathbb{E}_{\infty}$). A similar phenomenon happens to bilinearity: A morphism $f\colon A\times B\to C$ of commutative monoids is bilinear if, for each $a,a'\in A$ and each $b,b'\in B$, we have \begin{gather*} f(a,b+b') = f(a,b)+f(a,b'),\\ f(a+a',b) = f(a,b)+f(a',b),\\ f(1_A,b) = 1_C,\\ f(a,1_B) = 1_C. \end{gather*} For categories, these relations are replaced by morphisms: we say that a strong bilinear structure on a functor $F\colon\mathcal{C}\times\mathcal{D}\to\mathcal{E}$ of symmetric monoidal categories is a collection of isomorphisms \begin{align*} F^{\mathsf{bil}}_{A,B\otimes_{\mathcal{D}}B'} &\colon F(A,B\otimes_{\mathcal{D}}B') \longrightarrow F(A,B)\otimes_{\mathcal{E}}F(A,B'),\\ F^{\mathsf{bil}}_{A\otimes_{\mathcal{C}}A',B} &\colon F(A\otimes_{\mathcal{C}}A',B) \longrightarrow F(A,B)\otimes_{\mathcal{E}}F(A',B),\\ F^{\mathsf{bil}}_{\mathbf{1}_{\mathcal{C}},B} &\colon F(\mathbf{1}_{\mathcal{C}},B) \longrightarrow \mathbf{1}_{\mathcal{E}},\\ F^{\mathsf{bil}}_{A,\mathbf{1}_{\mathcal{D}}} &\colon F(A,\mathbf{1}_{\mathcal{D}}) \longrightarrow \mathbf{1}_{\mathcal{E}} \end{align*} satisfying coherence conditions. Questions: Is there a similar array of notions of bilinearity in higher algebra? In particular, can we speak of "$\mathbb{B}_{k}$-morphisms of spectra"? Tensor products can be characterised/defined as universal bilinear maps; do we also have intermediate tensor products corresponding to "$\mathbb{B}_{k}$"-bilinearity? REPLY [9 votes]: Let me clarify a bit what I meant in my comment on how the notion of bilinearity will depends on "how commutative" are $A$, $B$ and $C$, and this is one way to define a hierarchy of notion of bilinear maps in higher algebras. The idea is that up to equivalence of $\infty$-categories, (and let's say in a cartesian monoidal $(\infty,1)$-category for simplicity), an $\mathbb{E}_{n+k}$-algebra is the same as an $\mathbb{E}_n$-algebra in the (cartesian monoidal) category of $\mathbb{E}_{k}$-algebra. So if $A$ is an $\mathbb{E}_k$-algebra, $B$ is an $\mathbb{E}_n$-algebra and $C$ is an $\mathbb{E}_{n+k}$-algebras then I can define a $(n,k)$-bilinear map $A \times B \to C$ to be a morphism of $\mathbb{E}_k$-algebra from $A$ to the $\mathbb{E}_k$-algebra $\operatorname{Map}_{\mathbb{E}_n}(B,C)$. Where by $\operatorname{Map}_{\mathbb{E}_n}(B,C)$ I mean the space of morphisms of $\mathbb{E}_n$-algebra morphism from $B$ to $C$, which has an $\mathbb{E}_k$-algebra structure induced by the fact that $C$ is an $\mathbb{E}_k$-algebra when seen as an object of the category of $\mathbb{E}_n$-algebra. But this is very different from the notion you mention in your coment when instead one goes to "lax/colax" notion of bilinearity.<|endoftext|> TITLE: Geometric interpretations of the exponential of entropy QUESTION [14 upvotes]: Question: Might there be a natural geometric interpretation of the exponential of entropy in Classical and Quantum Information theory? This question occurred to me recently via a geometric inequality concerning the exponential of the Shannon entropy. Original motivation: The weighted AM-GM inequality states that if $\{a_i\}_{i=1}^n,\{\lambda_i\}_{i=1}^n \in \mathbb{R}_+^n$ and $\sum_{i=1}^n \lambda_i = 1$, then: \begin{equation} \prod_{i=1}^n a_i^{\lambda_i} \leq \sum_{i=1}^n \lambda_i \cdot a_i \tag{1} \end{equation} As an application, we find that if $H(\vec{p})$ denotes the Shannon entropy of a discrete probability distribution $\vec{p} = \{p_i\}_{i=1}^n$ and $r_p^2 = \lVert \vec{p} \rVert^2 $ is the $l_2$ norm of $\vec{p}$ then: \begin{equation} e^{H(\vec{p})} \geq \frac{1}{r_p^2} \tag{2} \end{equation} This result follows from the observation that if $a_i = p_i$ and $\lambda_i = p_i$, \begin{equation} e^{-H(\vec{p})} = e^{\sum_i p_i \ln p_i} = \prod_{i=1}^n p_i^{p_i} \tag{3} \end{equation} \begin{equation} \sum_{i=1}^n p_i^2 = \lVert \vec{p} \rVert^2 \tag{4} \end{equation} and using (1), we may deduce (2) where equality is obtained when the Shannon entropy is maximised by the uniform distribution i.e. $\forall i, p_i = \frac{1}{n}$. A remark on appropriate geometric embeddings: If we consider that the Shannon entropy measures the quantity of hidden information in a stochastic system at the state $\vec{p} \in [0,1]^n$, we may define the level sets $\mathcal{L}_q$ in terms of the typical probability $q \in (0,1)$: \begin{equation} \mathcal{L}_q = \{\vec{p} \in [0,1]^n: e^{H(\vec{p})} = e^{- \ln q} \} \tag{5} \end{equation} which allows us to define an equivalence relation over states $\vec{p} \in [0,1]^n$. Such a model is appropriate for events which may have $n$ distinct outcomes. Now, we'll note that $e^{H(\vec{p})}$ has a natural interpretation as a measure of hidden information while $e^{-H(\vec{p})}$ may be interpreted as the typical probability of the state $\vec{p}$. Given (5), a natural relation between these measures may be found using the Hyperbolic identities: \begin{equation} \cosh^2(-\ln q) - \sinh^2(-\ln q) = 1 \tag{6} \end{equation} \begin{equation} \cosh(-\ln q) - \sinh(-\ln q) = q \tag{7} \end{equation} where $2 \cdot \cosh(-\ln q)$ is the sum of these two measures and $2 \cdot \sinh(-\ln q)$ may be understood as their difference. This suggests that the level sets $\mathcal{L}_q$ have a natural Hyperbolic embedding in terms of Hyperbolic functions. References: Olivier Rioul. This is IT: A Primer on Shannon’s Entropy and Information. Séminaire Poincaré. 2018. David J.C. MacKay. Information Theory, Inference and Learning Algorithms. Cambridge University Press 2003. John C. Baez, Tobias Fritz, Tom Leinster. A Characterization of Entropy in Terms of Information Loss. Arxiv. 2011. REPLY [2 votes]: With apologies also for promoting my own work, a colleague (Tim Hosgood) and I have recently published an article in Entropy, where we give a geometric interpretation for the exponentiated entropy of empirical distributions, namely as the length $L$ of a certain rectangle. What makes this rectangle interesting is that its width $W$ and area $A$ (from which you can calculate its length as $L=\frac{A}{W}$ and hence entropy as $H=\log(L)$) are additively and multiplicatively homomorphic in the original distribution. It's well known that the entropy of a product of distributions is the sum of their entropies, so the exponentiated entropy (sometimes called the perplexity) of a product is the product of the exponentiated entropies. Thus one says that exponentiated entropy is multiplicatively homomorphic in the distribution. What we're saying in this short note is that if you use the whole rectangle (its area and width) instead of just its length, you can get not only a multiplicative but also an additive homomorphism; that is you get a rig homomorphism from empirical distributions to rectangles.<|endoftext|> TITLE: Weak Mordell-Weil for EC using Chevalley-Weil theorem QUESTION [5 upvotes]: I am reading the book Applications of Diophantine Approximation to Integral Points and Transcendence by Zannier and Corvaja and, after their proof of the Chevalley-Weil theorem, in Example 3.8 they suggest that it is possible to prove the weak Mordell-Weil theorem using Chevalley-Weil. Honestly, I can't see that, but I am very curious. Does anyone have an idea about it? Thank you in advance! REPLY [11 votes]: The multiplication-by-$m$ map $[m]:E\to E$ is unramified, so there exists a finite set of primes $S$, depending only on $E$ and $m$, so that for every $P\in E(K)$, the field generated by the coordinates of the points in $[m]^{-1}(P)$ is unramified outside of $S$. (This is where we use Chevelley-Weil.) The degree of that extension is also bounded as a function of $m$. It follows from standard results in algebraic number theory that there are only finitely many such fields. Hence $[m]^{-1}\bigl(E(K)\bigr)$ is contained in $E(L)$, where the extension $L/K$ is a finite extension. I'll let you read elsewhere the fact that it suffices to prove the weak Mordell-Weil theorem under the assumption that $E[m]\subset E(K)$. Under that assumption, and with notation as in the first paragraph with $L=K\bigl([m]^{-1}\bigl(E(K)\bigr)\bigr)$, there is a well-defined injective homomorphism $$ E(K)/mE(K) \longrightarrow \operatorname{Hom}\bigl(\operatorname{Gal}(L/K),E[m]\bigr) $$ defined by $$ P \longmapsto \bigl(\sigma\mapsto \sigma(Q)-Q\bigr) \quad\text{for any choice of $Q\in E(L)$ satisfying $[m](Q)=P$.} $$ Since $L/K$ is a finite extension from Chevalley-Weil, and since $E[m]$ is a finite group, this gives the finiteness of $E(K)/mE(K)$.<|endoftext|> TITLE: Splitting a group into two subsets closed under multiplication QUESTION [5 upvotes]: Given a group $G$, find subsets $A,B$ such that $G=A\sqcup B$ and $A$ and $B$ are closed under multiplication: $x,y\in A$ (corr. $x,y\in B$) implies $xy\in A$ (corr. $xy\in B$). For example, if $G$ is finite then all splitting are trivial: if $1\in A$ then $A=G$ and $B=\emptyset$. If $G=\mathbb{Z}^d$ then any splitting is determined by a halfspace $\Pi^+$ in $\mathbb R^d\supset\mathbb Z^d$: $A=\Pi^+\cap\mathbb Z^d$. (More precisely, by a flag of halfspaces $\Pi^+\supset \Pi_1^+\supset\dots$, because we can distribute the points of the hyperplane $\partial\Pi^+$ between $A$ and $B$.) What about the free group $G=F_d$? Are there splitting which are not induced by the projection $F_d\to\mathbb Z^d$? REPLY [10 votes]: This is a way to rediscover a quite well-studied class of groups: Proposition Let $G$ be a group. Equivalent statements: (a) $G$ admits a partition $G=A\sqcup B$ with $1\in A$, $B$ nonempty, and both $A,B$ subsemigroups. (b) $G$ admits a nontrivial order-preserving action on some totally ordered set [which can be chosen to be the real line, or $\mathbf{Q}$, if $G$ is countable] (c) $G$ has a nontrivial left-orderable quotient. The equivalence between (b) and (c) is classical. If $G$ has a non-trivial orderable quotient $p:G\to Q$, there is (by definition) a submonoid $K$ of $Q$ such that $K\cap K^{-1}=\{1_Q\}$ and $K\cup K^{-1}=Q$; pulling back to $G$ yields the desired decomposition. Conversely, suppose that $G=A\cup B$ as above (with $1\in A$). Define $H=A\cap A^{-1}$: this is a subgroup (call it "core" of the decomposition). Observe that $HB=BH=B$: indeed if $h\in H$ and $b\in B$, suppose by contradiction $a:=hb\in A$: so $b=ah^{-1}\in AA\subset A$, contradiction, same contradiction if $bh\in A$. On $G/H$ define a total order by $gH\le g'H$ if $g^{-1}g'\in A$. Note that this doesn't depend on the choices of $g,g'$ modulo $H$ on the right. This relation is $G$-invariant, and is easily seen to be a total order. It is not reduced to a singleton (since $H=A$ would force $B$ empty, since the complement of a subgroup can't be closed under multiplication, unless empty). Edit: this argument shows something more precise, not only describing the class of groups, but describing these decompositions: Proposition 2: Let $G$ be a group. (1) If $G$ acts on a totally ordered set $X$ in a order-preserving and $x\in X$, define $A=\{g\in G:gx\le x$ and $B=\{g\in G:gx>x\}$. Then $G=A\sqcup B$ is a decomposition with the given axioms (and $1\in A$). (2) Conversely, any such decomposition of $G$ occurs in this way (for some such action and $x$, where we can assume in addition the action to be transitive). Note that $B$ is empty iff $G$ fixes $x$. Note that this shows that if $G$ has such a decomposition, it has one for which the core is a normal subgroup (while in general the core need not be normal). Also, since non-abelian free groups are themselves left-orderable, they admit such decomposition that are not of the prescribed form (namely have such a decomposition with trivial core).<|endoftext|> TITLE: Sum of the lengths of all diagonals in a regular polygon QUESTION [6 upvotes]: I believe that the sum of the lengths of diagonals of a regular polygon ($n$-gon) is always greater than or equal to any other irregular polygon ($n$-gon) inscribed in a circle.. For example for a 4-gon, if we consider a regular $4$-gon, then the two diagonals pass through the centre of a circle, resulting in the sum of the diagonals' lengths twice the diameter. Considering any other $4$-gon (irregular), the sum of the lengths of the diagonals is not greater than $2\times$ the diameter. Case : 5-gon I proved that the sum of the lengths of diagonals in a regular $5$-gon is maximum compared to any $5$-gon (irregular polygon). I used Lagrange's multiplier to prove it. Case : $n$-gon I tried proving the same for an $n$-gon using Lagrange's multiplier, where we need to maximize the sum of the lengths of all diagonals with respect to $$\theta_1+ \theta_2 + \dots + \theta_n =2\pi,$$ but unable to do the calculation as the number of variables increases. Here $\theta_i$ is the angle subtended to the centre of a circle by the side $a_i$ of the polygon. I believe that the sum of the lengths of diagonals is maximum if $$\theta_1 = \theta_2 = \dots = \theta_n,$$ which is possible only in a regular polygon. REPLY [4 votes]: I am going to assume that all the points lie on the complex unit circle. The points are denoted by $e^{i x_k}$ for $k=1$ to $n$ in the counter-clockwise direction. All indices are mod $n$ for simplicity. The sum of distances that we are looking to minimize is a sum in the following form: $$\sum_{r,t}\sqrt{2-2\cos(x_r-x_t)}$$ This sum can be broken down to sums of the following form where $m$ ranges from $1$ to $\lfloor n/2 \rfloor$: $$s_m = \sum_{r}\sqrt{2-2\cos(x_{r+m}-x_{r})}$$ I claim that each $s_m$ gets maximized when the polygon is regular. This is a constrained optimization problem, constrained by $\sum x_{r+m}-x_r=2m\pi$. The proof is very simple, by Jensen's inequality: the function $f(x)=\sqrt{1-\cos x}$ is concave as its second derivative is $\frac{-1}4\sqrt{1-\cos x}$. Jensens's inequality gives $s_m \le n f(2m\pi/n)$, with equality when $x_{r+m}-x_r=2m\pi/n$ for all $r$. This happens when the polygon is regular (there might be other configurations that this might happen for example in the quadrilateral case the sum of diagonals of rectangle is equal to a square).<|endoftext|> TITLE: Cohomology of Shimura varieties and coherent sheaves on the stack of Langlands parameters QUESTION [19 upvotes]: In Zhu's Coherent sheaves on the stack of Langlands parameters theorem 4.7.1 relates the cohomology of the moduli stack of shtukas to global sections of a certain sheaf on the stack of global Langlands parameters: Let $H_{I,V}^i=H^iC_c\left(\mathrm{Sht}_{\Delta(\bar\eta),K},\mathrm{Sat}(V)\right)$. By [Xu20, Xu1, Xu2], the natural Galois action and the partial Frobenii action together induce a canonical $W_{F,S}^I$-action on $H_{I,V}^i$. The following statement can be regarded as a generalization of the main construction of [LZ]. Theorem 4.7.1. Assume that $k=\mathbb Q_\ell$ and regard $\mathrm{Loc}_{^cG,F,S}$ as an algebraic stack over $\mathbb Q_\ell$. Then for each $i$, there is a quasi-coherent sheaf $\mathfrak A_K^i$ on $^{cl}\mathrm{Loc}_{^cG,F,S}$, equipped with an action of $H_K$, such that for every finite dimensional representation $V$ of $(^cG)^I$, there is a natural $(H_K\times W_{F,S}^I)$-equivariant isomorphism $$ H_{I,V}^i\cong\Gamma(^{cl}\mathrm{Loc}_{^cG,F,S},(w_{F,S}V)\otimes\mathfrak A_K^i)\tag{4.26} $$ where $w_{F,S}V$ is the vector bundle on $\mathrm{Loc}_{^cG,F,S}$ equipped with an action by $W_{F,S}^I$ as in Remark 2.2.7. This is a generalization of a construction in section 6 of Lafforgue-Zhu's Décomposition au-dessus des paramètres de Langlands elliptiques (also recapitulated in remark 8.5 of V. Lafforgue's ICM survey). The idea is that a regular function of the Langlands parameter determines an endomorphism of $H_{\lbrace 0\rbrace,\mathrm{Reg}}$, a subspace of the cohomology of the moduli of shtukas with one leg, with relative position of the modification bounded by the regular representation of $\widehat{G}$ (see section 7 of 4). Let $V$ be a representation of $\widehat{G}$, let $x\in V$, $\xi\in V^{*}$. Any function $f$ of $\widehat{G}$ can be expressed as a matrix coefficient $\langle\xi,g.x\rangle$, and for $\gamma\in\mathrm{Gal}(\overline{F}/F)$ the functions $F_{f,\gamma}:\sigma\mapsto f(\sigma(\gamma))$ are supposed to topologically generate all such functions on the stack of Langlands parameters, as $f$ and $\gamma$ vary. Letting $\underline{V}$ be the underlying vector space of the representation $V$, the tensor product $H_{\lbrace 0\rbrace,\mathrm{Reg}}\otimes \underline{V}$ will have an action of $\mathrm{Gal}(\overline{F}/F)$, obtained from an isomorphism with $H_{\lbrace 0,1\rbrace,\mathrm{Reg}\boxtimes V}$ (see remark 8.5 of 4). Since the data of $F_{f,\gamma}$ is the same as the data of $x$, $\xi$, and $\gamma$, we may now construct the endomorphism of $H_{\lbrace 0\rbrace,\mathrm{Reg}}$ as follows: $$H_{\lbrace 0\rbrace,\mathrm{Reg}}\xrightarrow{\mathrm{Id}\otimes x} H_{\lbrace 0\rbrace,\mathrm{Reg}}\otimes \underline{V}\xrightarrow{\gamma} H_{\lbrace 0\rbrace,\mathrm{Reg}}\otimes \underline{V}\xrightarrow{\mathrm{Id}\otimes\xi} H_{\lbrace 0\rbrace,\mathrm{Reg}}$$ This allows us to construct a sheaf of $\mathcal{O}$-modules on the stack of Langlands parameters, whose global sections is $H_{\lbrace 0\rbrace,\mathrm{Reg}}^{i}$. In several other places in 1, Zhu also hints at the hope of finding an analogue of theorem 4.7.1 for the cohomology of Shimura varieties, following ongoing work in progress of Zhu and Emerton on an analogous stack of Langlands parameters for number fields. How might such an analogue proceed? In the setting of Shimura varieties, as opposed to shtukas, it appears we do not have the notion of legs, or of bounding relative positions of modifications with representations of $\widehat{G}$ (I do not know whether there are any existing analogies). How might such an endomorphism be constructed? REPLY [12 votes]: The anticipated analogue is as follows: There is a map $f: \mathcal X \to \prod_{v \in S} \mathcal X_v,$ where $\mathcal X$ is the stack of $p$-adic representations of $G_{E,S}$ into ${}^LG$ (the $L$-group over $E$ of some group $G$ that is part of a Shimura datum, with reflex field $E$) unramified outside $S$, and each $\mathcal X_v$ is the corresponding local version at the place $v$. (We choose $S$ to include all primes $v$ over $p$, and over $\infty$. And probably any primes where $G$ is ramified.) Choosing a level structure at each place $v$ of $S$ will give rise to a coherent sheaf $\mathfrak A_v$ on $\mathcal X_v$ as in Xinwen's paper. (Strictly speaking, his paper only treats the case where $v \not\mid p$ or $\infty$, but there should be sheaves for the other $v$ too. The case of $v \mid \infty$ is conceptually a bit mysterious, but there are some ideas. The case of $v \mid p$ is related to the conjectural $p$-adic Langlands program; in this case $\mathcal X_v$ will be an appropriate EG-type stack.) We can then form $\mathfrak A := \boxtimes_{v \in S} \mathfrak A_v$, the exterior tensor product, a coherent sheaf on $\prod_{v \in S} \mathcal X_v$. We can consider $f^! \mathfrak A$ on $\mathcal X$. The Hodge cocharacter $\mu$ from the Shimura datum gives rise to a representation $V_{\mu}$ of ${}^LG$, and composing this with the universal representation over $\mathcal X$, we obtain a locally free sheaf $\mathcal V$ of Galois representations over $\mathcal X$. Now we have the conjectural formula: $$R\Gamma_c(\text{ Shimura variety at given level}, \mathbb Z_p)[\text{shift by dimension of Shimura variety}] $$ $$ = R\Gamma(\mathcal X, f^!\mathfrak A \otimes \mathcal V).$$ The analogous formula in the function field case, for cohomology of stacks of shtuka, is stated in Xinwen's paper, and this is the Shimura varieties version. The isomorphism should be compatible with Galois and Hecke actions on the two sides. The main difference with the shtuka case is that we just have the single $V_{\mu}$ that we can consider, so there is less structure in the Shimura variety case than in that case. (We can get a bit more structure by allowing non-trivial coefficient systems in the cohomology, or even going to $p$-adically completed cohomology; this will be reflected in the precise choice of sheaf $\mathfrak A_v$ at the places $v|p$.) This will all be discussed in (hopefully) forthcoming papers of mine and Xinwen's, and of mine, Toby Gee, and Xinwen.<|endoftext|> TITLE: Length of commutators in the free group QUESTION [10 upvotes]: Let $G=F_2$ be the free group of rank $2$. Is there a constant $c>0$ such that the word length $|[u,v]|$ of every commutator $[u,v]=uvu^{-1}v^{-1}$ where $u,v\in G$, $|u|,|v|>0$ is at least $c(|u|+|v|)$ unless $[u,v]=[u_1,v_1]$ for some $u_1,v_1$ with $|u_1|+|v_1|<|u|+|v|$? REPLY [7 votes]: Victor Guba sent me a proof. The proof is based on the old result by Wicks, Wicks, N. J. Commutators in free products. J. London Math. Soc. 37 (1962), 433–444., which describes all words which are commutators in a free group (Lemma 5 in the paper). By that result, a word is a commutator iff it is a conjugate of a reduced word of the form $ABCA^{-1}B^{-1}C^{-1}$. This immediately implies that one can take $c=1/4$ (possibly bigger).<|endoftext|> TITLE: Bijection from "black-white balanced" partitions to pairs of partitions QUESTION [5 upvotes]: Definition Call a partition $\lambda$ of an even integer $2n$ "black-white balanced" if the following equivalent conditions are satisfied: In the usual (Ferrers-)Young diagram of $\lambda$, if you color alternating squares black and white in a checkerboard pattern, there are an equal number of black and white squares. The Young diagram of $\lambda$ can be covered by dominoes. The partition $\lambda$ has empty 2-core. If you sort the parts of $\lambda$ (as is usual), then the number of odd parts in even places is equal to the number of odd parts in odd places. (A part $\lambda_i$ is odd if the number $\lambda_i$ itself is odd; the place is odd or even based on the parity of the index $i$.) If you sort the parts of $\lambda$, then $\sum_i (-1)^i(1-(-1)^{\lambda_i}) = 0$. The number of black-white balanced partitions of $2n$ is OEIS sequence A000712, which begins $1$, $2$, $5$, $10$, $20$, $36$, $65$, $110$. The name of this sequence is "Number of partitions of n into parts of 2 kinds". Alternatively, it is the convolution of the partition numbers with themselves. The description above appears on the linked page as "Also equals number of partitions of 2n in which the odd parts appear as many times in even as in odd positions." Question In other words, there appears to be a bijection between black-white balanced partitions $\lambda$ of $2n$ and ordered pairs of partitions ($\lambda_1$, $\lambda_2$) of a combined $n$ (that is, if $\lambda_1$ is a partition of $n_1$ and $\lambda_2$ is a partition of $n_2$, then $n=n_1+n_2$). Can anyone describe this bijection, or provide a reference for it? Remarks I'm fairly sure I'm just a Google search away from finding a reference, but I just haven't succeeded yet. I found this question on Mathematics Stack Exchange and this identical question on MathOverflow asking about a bijection between strict (ie, distinct parts) black-white balanced partitions of $2n$ and unrestricted partitions of $n$ (not pairs thereof). The paper "Balanced partitions" by Sam Vandervelde also proves this result. REPLY [3 votes]: Richard Stanley and Sam Hopkins have answered this question in the comments. (Thanks!) Professor Stanley mentions that this is a special case of exercise 7.59(e) in Enumerative Combinatorics, volume 2: 7.59(e) Let $\mu$ be a $p$-core. Let $Y_{p,\mu}$ be the set of all partitions whose $p$-core is $\mu$. Define $\lambda \le \nu$ in $Y_{p,\mu}$ if $\lambda$ can be obtained from $\nu$ by removing border strips of size $p$. Show that $Y_{p,\mu} \cong Y^k$, where $Y$ denotes Young's lattice. Deduce that if $f_{\mu}(n)$ is the number of partitions of $n$ with $p$-core $\mu$, then $$ \sum_{n\ge 0} f_{\mu}(n)x^n = x^{\lvert \mu\rvert} \prod_{i\ge 1} (1-x^{pi})^{-p}. $$ (Presumably the $k$ in $Y^k$ is meant to be a $p$ as in $Y^p$.) The proof sketched by the exercise is very nice and uses an infinite binary sequence called the code of a partition $\lambda$. The bijection from this question can be obtained by considering every other bit from this binary sequence. Professor Hopkins mentions that my desired bijection is known under the name "$2$-quotient". Specifically, every partition $\lambda$ has a $2$-core $\alpha$, which is a single partition, and a $2$-quotient $(\beta_1, \beta_2)$, which is an ordered pair of partitions. Two key facts are the following: $\lvert \lambda\rvert = \lvert \alpha \rvert + 2\lvert \beta_0\rvert + 2\lvert \beta_1\rvert. $ The triple $(\alpha; \beta_0, \beta_1)$ uniquely determines $\lambda$. In the case of this question, the $2$-core $\alpha$ is empty (by assumption), so the $2$-quotient $(\beta_0, \beta_1)$ is precisely the desired bijection. One way of describing the $2$-quotient is by coloring the Young diagram of $\lambda$ in a checkerboard pattern. Then, $\beta_0$ (say) is obtained by considering the subdiagram of $\lambda$ restricted to those rows ending in a black square and those columns ending in a white square. Oppositely, $\beta_1$ is obtained by restricting to those rows ending in a white square and those columns ending in a black square. One reference for this is James and Kerber, "The Representation Theory of the Symmetric Group," page 83, Theorem 2.7.30.<|endoftext|> TITLE: Categorical models for truncations of the sphere spectrum QUESTION [5 upvotes]: Picard $n$-groupoids are expected to model stable homotopy $n$-types. So far this has been proved for $n=1$ in Niles Johnson, Angélica M. Osorno, Modeling stable one-types. Theory Appl. Categ. 26 (2012), No. 20, 520–537; arXiv:1201.2686. and for $n=2$ in Nick Gurski, Niles Johnson, Angélica M. Osorno, The $2$-dimensional stable homotopy hypothesis. Journal of Pure and Applied Algebra, Volume 223, Issue 10, 2019, Pages 4348-4383. arXiv:1712.07218. In particular, for $n=1$, Johnson and Osorno have also shown that the free Picard groupoid on one object $\mathbb{S}$ models the $1$-truncation of the sphere spectrum. In detail, this is the Picard groupoid where The objects of $\mathbb{S}$ are the integers; For $m,n\in\mathrm{Obj}(\mathbb{S})$, we have $$ \mathrm{Hom}_{\mathbb{S}}(m,n) \overset{\mathrm{def}}{=} \begin{cases} \emptyset &\text{if $m\neq n$,}\\ \mathbb{Z}_{2} &\text{if $m=n$;} \end{cases} $$ The monoidal structure $\oplus$ on $\mathbb{S}$ is given by addition of integers; The symmetry of $\mathbb{S}$ at $(A,B)$ is the morphism $\beta^{\mathbb{S},\oplus}_{m,n}\colon m\oplus n\to n\oplus m$ defined by $$ \beta^{\mathbb{S},\oplus}_{m,n} \overset{\mathrm{def}}{=} \begin{cases} 0 &\text{if $mn$ is even,}\\ \eta_{m+n} &\text{if $mn$ is odd,} \end{cases} $$ where $\eta_{m+n}$ is the unique non-zero element of $\mathrm{Hom}_{\mathbb{S}}(n,n)$. In section 3 of the same paper, they also show that there is a symmetric monoidal functor $$\xi\colon\mathbb{F}\longrightarrow\mathbb{S}$$ from the symmetric monoidal category of finite sets and bijections $\mathbb{F}$, the categorification of the monoid of natural numbers $\mathbb{N}$, defined on objects by the inclusion $\mathbb{N}\hookrightarrow\mathbb{Z}$ and on morphisms by the sign map $\mathrm{sgn}\colon\Sigma_{n}\to\mathbb{Z}_{2}$. Questions. Here are some questions around these topics: Free Picard $n$-groupoids on one-object are supposed to model the $n$-truncations of the sphere spectrum. Is there some sense in which free symmetric monoidal $n$-categories on one object model the $n$-truncations of the "directed sphere spectrum"? The zeroth and first truncations of the sphere spectrum are given respectively by the abelian group of integers $\mathbb{Z}$ and by the Picard groupoid $\mathbb{S}$ above. What are the second and third truncations of the sphere spectrum, or, equivalently, how can we explicit describe the free Picard $2$- and $3$-groupoids on one object? The free symmetric monoidal $n$-categories on one object for $n=0,1,2$ are given by the commutative monoid $\mathbb{N}$ of natural numbers, the symmetric monoidal category $\mathbb{F}$, and the discrete monoidal bicategory on $\mathbb{F}$ (see arXiv:1210.1174, Corollary 1.11). What is an explicit description of the free symmetric monoidal tricategory on one object? REPLY [6 votes]: I don't understand what you mean about the "directed sphere" so will focus on the other questions. The free Picard $n$-category on one object has a description as a bordism $n$-category. Specifically it has: Objects are stably framed 0-manifolds; 1-Morphisms are stably framed 1-dimensional bordisms; 2-morphisms are stably framed 2-dimensional bordism between bordisms; etc until level $n$ where we take equivalence classes of stably framed $n$-dimensional bordisms between bordisms between ... and the equivalence relation is up to one more layer of stable bordism. Using the Pontyragin-Thom construction, one can see that this is just the same as the fundamental $n$-groupoid of $\Omega^\infty S^\infty$. The free (non-Picard) symmetric monoidal $n$-category is always just $\mathbb{F}$ when $n\geq 1$. This is explained in my answer to you other recent question. https://mathoverflow.net/a/398384/184<|endoftext|> TITLE: What is the free symmetric monoidal $\infty$-category on one object? QUESTION [10 upvotes]: It is well-known that the free symmetric monoidal category on one object is the category $\mathbb{F}$ of finite sets and bijections. This is supposed to be the categorification of the monoid of natural numbers, and its algebraic $K$-theory is given by the sphere spectrum $\mathbb{S}$, the free symmetric monoidal $\infty$-groupoid with inverses. Is there a natural description of the free symmetric monoidal $\infty$-groupoid (resp. $\infty$-category, $(\infty,\infty)$-category) on one object? REPLY [15 votes]: Yes, it is the same as $\mathbb{F}$. As John Baez points out, it is the same as the free symmetric monoidal $\infty$-groupoid on one object. (This can also be seen by playing around with the adjoints between groupoids and categories). Symmetric monoidal $\infty$-groupoids are the same as $E_\infty$-spaces (careful: not assumed to be "grouplike"). So we want the free $E_\infty$-space on a point. As a space this is homotopy equivalent to $\sqcup_{n\geq0} B \Sigma_n$. Note all the hom spaces are actually 1-types, so it can be modeled as a 1-groupoid, namely $\mathbb{F}$. It might seem surprising at first that the free symmetric monoidal $\infty$-category is actually an ordinary category. However on reflection we see that it follows from the fact that the spaces in the $E_\infty$-operad are actually 1-types. Equivalently, as John says, it is the space of all finite subsets of $\mathbb{R}^\infty$. If you pick an identification of $\mathbb{R}^\infty$ with the "interior" of an infinite cube, then you get a natural $E_\infty$ structure on this model of the space, too. You can also see that it is $\mathbb{F}$ in the way that Denis suggests in his comment to the OP.<|endoftext|> TITLE: Invariant section of a linearized sheaf QUESTION [5 upvotes]: I am struggling to understand what an invariant section with respect to a linearization of a line sheaf is. In Geometric Invariant Theory, given a $k$-scheme $X$ (being $k$ an algebraically closed field of characteristic zero) where a reductive algebraic group $G$ acts on by $\sigma$, and a line bundle $L\rightarrow X$ over $X$, a linearization on $L$ is just a lift of the $G$-action on $X$. This is equivalent to give an isomorphism of sheaves of $\mathcal{O}_{X\times G}$-modules $$\sigma^{\ast}\mathcal{L}=p_{1}^{\ast}\mathcal{L}$$ satisfying the 1-cocycle condition, where $\mathcal{L}$ is the invertible sheaf associated to $L$ and $p_{1}:X\times G\rightarrow X$ is the projection on the first factor. Mumford also says that from this definition we obtain a dual action of $G$ on the group of global sections of the line bundle $$H^{0}(X,\mathcal{L})\rightarrow H^{0}(G,\mathcal{O}_{G})\otimes H^{0}(X,\mathcal{L})$$ So my questions are: What is a dual action of the group? Can we talk about invariant sections of the line sheaf? How? We have a correspondence between sections of an invertible sheaf, and sections of the associated line bundle, that is, map of schemes $X\rightarrow L$ which are sections of the projection map $L\rightarrow X$. In this sense, what is an invariant section of the line bundle? Thank you very much for your comments in advance. It is really hard for a new geometer to understand fully what is inside of Mumford's head. REPLY [6 votes]: Question: "What is a dual action of the group?" Answer: For simplicity, if $X:=Spec(A)$ and $G:=Spec(R)$ is a linear algebraic group over a field $k$ acting on $X$ via $\sigma^*: G\times_k X \rightarrow X$ you get a "dual action" $$\sigma: A \rightarrow R\otimes_k A.$$ If $L\in Pic(A)$ is an invertible module with $\mathcal{L}:=\mathbb{V}(L^*):=Spec(Sym_A^*(L^*)):=Spec(B)$, a $G$-linearization of $L$ gives an action $$\gamma^*: G\times \mathcal{L}\rightarrow \mathcal{L}$$ and a "dual action" $$\gamma: B \rightarrow R\otimes_k B.$$ There are canonical maps $A\rightarrow B$ and $R\otimes_k A \rightarrow R\otimes_k B$ and all diagrams are to commute. The map $\sigma$ is a local version of your map on global sections $$H^{0}(X,\mathcal{L})\rightarrow H^{0}(G,\mathcal{O}_{G})\otimes H^{0}(X,\mathcal{L}).$$ Question: "Can we talk about invariant sections of the line sheaf? How? We have a correspondence between sections of an invertible sheaf, and sections of the associated line bundle, that is, map of schemes X→L which are sections of the projection map L→X. In this sense, what is an invariant section of the line bundle?" Answer: You formulate the notions "invariant function" and "invariant section" using the dual actions $\sigma, \gamma$ defined above. Example: If $X \subseteq \mathbb{P}^n_k$ is a quasi projective scheme with an action $\sigma^*:G\times_k X \rightarrow X$, and where you can give $X$ an open affine cover $U_i:=Spec(A_i)$ of $G$-invariant subschemes, you get induced dual actions $$ \sigma_i:A_i \rightarrow R\otimes_k A_i.$$ You use these dual actions $\sigma_i$ to study the action $\sigma$ and the "quotient" $X/G$. When $G$ is "finite" such an affine open cover always exist. Moreover the quotient $X/G$ "exist" and you define it locally as $Spec(A_i^G)$. Example: if $C:=\mathbb{P}^1_k$ is the projective line, there is an action of the symmetric group $S_n$ on $C^{\times_k n}$, and you may prove that if $n\neq char(k)$ there is an isomorphism $$Sym^n(C):=(C^{\times_k n})/S_n \cong \mathbb{P}^n_k.$$ You embed the product $C^{\times_k n}$ into a projective space and use an open affine $S_n$-invariant cover to construct the quotient $Sym^n(C)$ and to prove it is isomorphic to projective $n$-space. So you cannot construct the symmetric product $Sym^n(C)$ as $Spec(A^G)$ for some group $G$ acting on a ring $A$, since projective space is not affine. Example: An affine example and a "set theoretic paradox". If $G:=S_n$ is the symmetric group on $n$ letters acting on the polynomial ring $A:=k[x_1,..,x_n]$ it follows the invariant ring $A^G \cong k[s_1,..,s_n] \subseteq A$ where $s_i$ are the elementary symmetric polynomials. The polynomials $s_i$ are algebraically independent over $k$ hence $A^G$ is a polynomial ring and there is an isomorphism $$ \pi: \mathbb{A}^n_k \rightarrow \mathbb{A}^n_k/G \cong \mathbb{A}^n_k.$$ The map of schemes $\pi$ gives at the level of topological spaces a surjective and non-injective endomorphism $$\pi: \mathbb{A}^n_k \rightarrow \mathbb{A}^n_k$$ of affine $n$-space. Since the ring extension $A^G \subseteq A$ is an integral extension, it follows $\pi$ is surjective. Given any prime ideal $\mathfrak{p} \subseteq A$, it follows the orbit of $\mathfrak{p}$ under the action of $G$ has $n!$ elements in general. Hence the map $\pi$ is not injective. To some people (not all) this is a "paradoxical situation". Note: When working with schemes over fields that are not algebraically closed you will need this construction. Some people are "sloppy" when considering such actions and do not use the dual action.<|endoftext|> TITLE: Why are free Boolean topological groups Hausdorff? QUESTION [7 upvotes]: Assume $X$ is a Tychonoff space. Then $A(X)$ is the free topological abelian group over $X$. I know that $A(X)$ is Hausdorff and the canonical embedding from $X$ to $A(X)$ is a topological embedding. Now consider the subgroup $N:=2A(X)=\left\{g+g\ \colon\ g\in A(X)\right\}$. The quotient $B(X):=A(X)/N$ is the free topological Boolean group. I want to show that $B(X)$ is Hausdorff which is equivalent to say that $N$ is closed in $A(X)$. Unfortunately, I do not know how show the closedness of $N$ since I can only describe the topology on $A(X)$ by the universal property and I do not know how that helps me here... Or is this maybe not true for a general Tychonoff space $X$ and one needs additional properties on $X$ like hereditarily disconnected? I have found a reference on Free Boolean Groups Genze, L.V. Free Boolean topological groups. Vestn. Tomsk. Gos. Univ. 2006, 290, 11–13 which maybe could answer my question but I am unable to find this text on math.sci.net (or anywhere else). Thank you! REPLY [8 votes]: You can use the universality property with the following Boolean group as codomain: $B$ is the measure algebra over the unit interval (the quotient of the $\sigma$-algebra of Lebesgue measurable sets by the ideal of sets of measure zero), with symmetric difference as operation this is a Boolean group and $d(A,B)=\lambda(A\Delta B)$ defines a metric that will turn it into a topological group. This group contains a copy of the unit interval, namely the set $\{[0,t]:0\le t\le 1\}$. Since $X$ is Tychonoff you now have enough continuous functions to separate all non-trivial words from the empty word. This survey by Sipacheva gives more information: Free Boolean Topological Groups, Axioms 4 (2015) 492-517<|endoftext|> TITLE: Effect of the curse of dimension on collision detection QUESTION [11 upvotes]: I require a crude intersection analyzer or collision detector in about 15 dimensions. I am wondering if such a function is rendered difficult or impossible by the "curse of dimension". Collision detectors are widely used in 2 and 3 dimensions to determine if two shapes are in contact. A function that determines the intersection of two regions is one way to make a collision detector. The "curse of dimension" refers to difficulties encountered as the number of dimensions involved increases beyond about 5. For example, all points tend to be clustered near hyper-surfaces, and, statistically, all points behave like outliers. (See the Encyclopedia of Mathematics.) I haven't been able to find any literature on such collision detectors beyond 4 dimensions, and I would appreciate being pointed in the right (best?) direction. REPLY [11 votes]: An overview of high-dimensional collision detection for the purpose of motion planning is in Petrović - Motion planning in high-dimensional spaces: Grid-based approaches are resolution complete and often offer optimal solutions. However, the number of grid points grows exponentially in the configuration space dimension, which makes even the state-of-the-art methods inappropriate for very high-dimensional problems. Sampling-based approaches are efficient in most practical problems but offer weaker guarantees. They are probabilistically complete, however, they often require post-processing and can still be inefficient in very complex configuration spaces. Trajectory optimization approaches can solve high-dimensional motion planning problems quickly, but solutions are only locally optimal. REPLY [7 votes]: If your objects fit into ellipsoids, then one option might be the Perram-Wertheim distance; in dimension $n$ you do need to invert an $n \times n$ matrix, which might be do-able ...<|endoftext|> TITLE: What do you call the generalisation of the direct image? QUESTION [6 upvotes]: This question was posted on Math Stack Exchange, but did not attract an answer. Here is the question: Informal Description Let me start with an example. Let $X$ be the set $\{a, b, c, d, e\}$ and $E$ be the set $\{a, b, c\}$. Let $f$ be a function with domain $X$. Then the mapping that sends $E$ to $\{f(a), f(b), f(c) \}$ is called the direct image of $E$ under $f$, denoted by $f_*(E)$. Now, let $W$ be the set $\{\{a, b\}, \{c\}\}$. What do you call the mapping that sends $W$ to $\{\{f(a), f(b)\}, \{f(c)\}\}$ ? Informally, this mapping replaces every element of $X$ "inside" $W$ with its image under $f$. Slightly More Formal Description Let us assume the ZFA-axiomatisation of Set Theory, a variant of ZF that allows non-set objects called atoms. Let $A$ be a set of atoms. Define $P^0(A)$ = A. And for all $n \in \mathbb{N} \colon$ define $P^n(A)= \mathscr{P}(P^{n-1}(A)) \cup A $ , where $\mathscr{P}(-)$ denotes the powerset of a set. And define $P^{\infty}(A)$ as $\bigcup_{n \in \mathbb{N}} P^{n}(A)$ Let $X$ be a set of atoms. Let $f$ be a function with domain $X$. We could then inductively define a function $f_{**}$ such that $dom(f_{**}) = P^{\infty}(X) \backslash X$ , and $f_{**}$ has the property that for all $Z \in dom(f_{**}) \colon f_{**}(Z) = \{f_{**}(T) | T \in Z \backslash X \} \cup \{f(x) | x \in Z \cap X \}$. Examples Let $X = \{a, b, c, d, e\}$ be a set of atoms. Then $\{a, \{b\}\} \in P^{\infty}(X) \backslash X$ and $f_{**}(\{a, \{b\}\}) = \{f(a), \{f(b)\}\}$ Let $X = \{a, b, c, d, e\}$ be a set of atoms. Then $\{\{ \{c \} \}\} \in P^{\infty}(X) \backslash X$ and $f_{**}(\{\{\{c \} \}\}) = \{\{\{f(c) \} \}\}$ So $f_{**}$ can, in a sense, be thought of as a generalisation of the direct image. My quesiton is, what do you call this concept? And can you point me towards any resources regarding it? I would prefer if the resource makes use of the ZFA-axiomatisation, but I would also accept any other resource. Sidenote I am aware that this concept comes up in Permutation Models for the case when $f$ is a bijective endofunction. REPLY [4 votes]: This idea is commonly used in set theory with atoms. I'm not sure whether it has a standard name, but I would be inclined to call it the natural extension of $f$ to sets. There is no need to stop the iteration at $\omega$ as you do, for one can continue the cumulative hierarchy through the ordinals. If $A$ is any class of atoms, you can define the well-founded cumulative hierarchy over $A$ by transfinite recursion, just like the ordinary cumulative hierarchy, like this: $\newcommand{\WF}{{\rm WF}}$ $$\WF_0(A)=A,$$ $$\WF_{\alpha+1}(A)=\mathcal P(\WF_\alpha(A)),$$ $$\WF_\lambda(A)=\bigcup_{\alpha<\lambda}\WF_\alpha(A)$$ Specifically, we start with just the atoms, and then iteratively add all subsets of what we've got so far. It is straightforward to see that every function $\sigma:A\to V$ defined on the atoms extends naturally to a function $\bar\sigma$ defined on the entire hierarchy over those atoms by the following recursion: $$\bar\sigma(x)=\begin{cases} \sigma(x)&\text{if $x\in A$,}\\ \bar\sigma[x]&\text{otherwise} \end{cases}$$ You can view this as a $\in$-recursion or alternatively as a recursion on ordinals, defining $\bar\sigma$ on each $\WF_\alpha(A)$. This definition agrees with yours (on finite levels), since we apply the original function to the atoms, and otherwise apply the pointwise image recursively. Another way to think about the natural extension is that every set is having the atoms in its hereditary membership replaced with their values as specified by the original function. So this is a kind of hereditary application of the original function. You asked for references, and so let me mention that we use exactly this idea in our paper: Daghighi, Ali Sadegh; Golshani, Mohammad; Hamkins, Joel David; Jeřábek, Emil, The foundation axiom and elementary self-embeddings of the universe, Geschke, Stefan (ed.) et al., Infinity, computability and metamathematics. Festschrift celebrating the 60th birthdays of Peter Koepke and Philip Welch. London: College Publications (ISBN 978-1-84890-130-8/hbk). 89-112 (2014). ZBL1358.03070. In the proof of theorem 2, for example, we have a function $\sigma$ defined on the atoms (for us it is Quine atoms, but ZFA urelements would work the same), and then extend this to all of $\WF(A)$ by recursion. In the case that the original map was a permutation of the atoms, then the natural extension of it is an automorphism of the cumulative hiearchy over those atoms. Another application of the idea arises in the Jech-Sochor embedding theorem, which can be found in Jech's book Set Theory.<|endoftext|> TITLE: Permutation matrices with large distance QUESTION [5 upvotes]: We say that an order $n$ permutation matrix $P$ has distance $d$ if the Hamming distance between any two $1$-elements of $P$ is at least $d$. For example, the following matrix has distance 2: $$ \begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix} $$ I am interested in the following question: For which values of $d$ and $n$ do there exist such matrices? More generally, what is the (approximate) number of order $n$ permutation matrices of distance $d$? I am absolutely sure that these objects have been studied before - there are, after all, a natural variation of error correcting codes. But under what name? Are the answers to my questions already known? I would appreciate any reference. Thank you! REPLY [7 votes]: I assume that the interpretation with $\ell_\infty$-distance is correct, I.e., the distance between entries $(i,j)$ and $(i’,j’)$ is $\max\{|i-i’|,|j-j’|\}$. If $n=k^2$, you may achieve $d=k$ but not more. To achieve $d=k$, enumerate rows and columns from $1$ through $n$ and put ones in all cells of the form $(i, ik\mod (n+1))$, where $i=1,2,\dots,n$. To see this is optimal, assume that $d=k+1$ is achieved. Consider ones in the first $d$ columns; they must be spread vertically at distances at least $d$ from each other, so we should have $n-1\geq d(d-1)$ which is wrong.<|endoftext|> TITLE: When does the eikonal equation $\lvert Du \rvert^2 = f$ admit a local solution? QUESTION [7 upvotes]: Let $f$ be a smooth function defined on the unit disc $D \subset \mathbf{R}^2$ with \begin{equation} f \geq 0 \text{ in $D$ and } f(0) = 0. \end{equation} This is allowed to have a degenerate minimum at the origin, namely it is allowed that $D^2 f(0) = 0.$ Question. When is there $\rho \in (0,1)$ and $u \in C^1(D_\rho)$ so that $\lvert D u \rvert^2 = f$? I would be more than happy with an answer specialised to the case where $f$ is the polynomial $(xy)^{2N}$—say with $N \geq 2$—if a general discussion is too onerous. As far as I understand, the equation was initially considered with a strictly positive right-hand side. One classical example is where $\lvert \nabla u \rvert^2_g = 1$, with respect to the some Riemannian metric $g$ on $D$. One may attempt to rescale the Euclidean metric to $g = f^2 g_e$, in order to get $\lvert \nabla u \rvert_g^2 = f^{-2} \lvert \nabla u \rvert_{g_e} = 1$. However $g$ is unfortunately degenerate where $f = 0$. When the zero of $f$ at the origin is non-degenerate, then one can construct a solution to the eikonal equation via a sort of dynamic argument, as is explained in this answer. REPLY [6 votes]: For $n=1$ or $2$, there is no $u\in C^1(D_\rho)$ for any $\rho>0$ that satisfies $|\nabla u|^2 = (xy)^{2n}$. (Note that $f=(xy)^2$ has a degenerate minimum at $(0,0)$, so $n=1$ should be allowed in this discussion.) Meanwhile, for $n\ge 3$, there do exist $u\in C^1(\mathbb{R}^2)$ that satisfy $|\nabla u|^2 = (xy)^{2n}$. The above non-existence result is special to the case $f = (xy)^{2n}$. For example, in the case $f=(xy)^{2n}(x^2{+}y^2)$, which has an even more degenerate minimum at $(0,0)$, there is a real-analytic, global solution $u(x,y) = (xy)^{n+1}/(n{+}1)$. A few preliminaries before sketching the main argument are in order. First, note that the disk radius $\rho>0$ actually plays no role in the problem. If $u\in C^1(D_\rho)$ were to satisfy $|\nabla u|^2 = (xy)^{2n}$, then for any $r>0$ the scaled function $\tilde u(x,y) = r^{-(2n+1)}\,u(rx,ry)$ would satisfy $\tilde u\in C^1(D_{\rho/r})$ and $|\nabla \tilde u|^2 = (xy)^{2n}$. Thus, one can assume that $\rho$ be arbitrarily large. Second, one can assume, by adding a constant to $u$, that $u(0,0)=0$, so I will assume this normalization made henceforth. Then, the obvious integral inequality arising from $|\nabla u| = |xy|^n$ and the Cauchy-Schwartz inequality would imply that $$ |u(x,y)| \le \frac{|xy|^n\sqrt{x^2+y^2}}{(2n{+}1)}. $$ In particular, $u$ would vanish to order $2n{+}1$ at $(0,0)$ and would satisfy $u(x,0) = u(0,y) = 0$. It follows from this that $u$ could not be of differentiability class $C^{2n+2}$, since, if it were, the limit function $$ p(x,y) = \lim_{r\to0} \frac{u(rx,ry)}{r^{2n+1}} $$ would exist and be a polynomial homogeneous of degree $2n{+}1$ that satisfied $|\nabla p|^2 = (xy)^{2n}$, and it is easy to show that there is no such polynomial. However, as will be seen, when $n\ge3$, there exists a $u\in C^{n-1}(\mathbb{R}^2)$ satisfying $|\nabla u|^2 = (xy)^{2n}$ and the homogeneity condition $u(rx,ry) = |r|^{2n+1}\,u(x,y)$ for all $r$. This $u$ is real-analytic away from the lines $x\pm y = 0$ but fails to be $C^n$ on these two lines. From now on, I will assume that $u\in C^1(D_\rho)$ (with $\rho>>0$ as large as necessary for the argument) satisfies $u(0,0)=0$ and $|\nabla u|^2 = (xy)^{2n}$. In particular, as the OP points out, $u$ satisfies the eikonal equation $|\nabla^g u|^2 = 1$ for the singular 'metric' $g = (xy)^{2n}(\mathrm{d}x^2+\mathrm{d}y^2)$, so that the gradient flow lines of $\nabla^g u$ are $g$-geodesics in the four quadrants of the $xy$-plane where $xy\not=0$. Now, the metric $g$ has some interesting properties: First, it is homogeneous of degree $2n{+}2$, so that its family of geodesics is preserved under the scaling homothety, and, moreover, it is invariant under the discrete symmetries $(x,y)\to(-x,y)$, $(x,y)\to(x,-y)$, and $(x,y)\to(y,x)$. Consequently, it suffices to study the behavior of its geodesics in the 'first' quadrant, where $x>0$ and $y>0$, and it is immediate that the lines $y\pm x = 0$ are geodesics in the quadrants. Let the ray $\{(x,x)\ |\ x>0\}$ in the first quadrant be known as the fundamental geodesic. Of course, $g$ is not complete in the first quadrant, as the two boundary rays can be reached from, say, $(1,1)$, by curves of finite $g$-length. Now, computation shows that the Gauss curvature of $g$ is $K = n(x^2{+}y^2)/(xy)^{2n+2}>0$, which suggests that nearly all of the geodesics of $g$ will avoid going to the singular boundary where $xy=0$, and, indeed, this turns out to be the case (see below). To parametrize the geodesics, it turns out to be convenient to use a parameter $t$ other than arc length. A curve $\bigl(x(t),y(t)\bigr)$ in the first quadrant parametrizes a $g$-geodesic when there is a function $\phi(t)$ satisfying the ODE system $$ \dot x = xy\cos\phi, \quad \dot y = xy\sin\phi, \quad \dot\phi = n\,(x\cos\phi-y\sin\phi),\tag1 $$ and every $g$-geodesic in the first quadrant has such a parametrization, unique up to replacing $t$ by $t+t_0$ for some constant $t_0$. In this case, arclength $s(t)$ along the geodesic satisfies $\dot s = (xy)^{n+1}$. [The advantage of writing the geodesic equations this way is that they extend smoothly across the singular locus $xy=0$.] Note that these equations are invariant under the homothetical scaling $(t,x,y,\phi)\to(t/r,rx,ry,\phi)$. Because of the scaling symmetry of the equations, one can extract a 2D phase portrait that makes clear the behavior of the geodesics as follows: Let $x+iy = \mathrm{e}^{u+iv}$. Then the above equations become (after a change of independent variable) $$ u' = \cos(\phi{-}v)\,\cos v\sin v,\qquad v' = \sin(\phi{-}v)\,\cos v\sin v,\qquad \phi' = n\,\cos(\phi{+}v).\tag2 $$ One can now draw the $v\phi$-phase portrait, concentrating on the strip $0\le v\le \pi/2$, which represents the first quadrant in the $xy$-plane, and bearing in mind that these equations are invariant under the involution $(v,\phi)\to(\tfrac12\pi{-}v,\tfrac12\pi{-}\phi)$ and reverse under $(v,\phi)\to (v,\phi{+}\pi)$. There are a sink at $S_- = (v,\phi)=(\pi/4,\pi/4)$, a source at $S_+ = (v,\phi)=(\pi/4,-3\pi/4)$, and saddles at $S_1 = (v,\phi)=(0,\pi/2)$, $S_2 = (v,\phi)=(\pi/2,0)$, $S_3=(v,\phi)=(0,-\pi/2)$, and $S_4=(\pi/2,-\pi)$. In addition to the 'trivial' separatrices that make up the boundary lines $v=0$ and $v=\tfrac12\pi$, there are four 'non-trivial' separatrices: $L_1$ leaving $S_1$ and going to $S_-$, $L_2$ leaving $S_2$ and going to $S_-$, $L_3$ leaving $S_+$ and going to $S_3$, and $L_4$ leaving $S_+$ and going to $S_4$. Here is where the difference between the cases $n=1,2$ and the cases $n\ge 3$ becomes evident. When $n\le 2$, the fixed points $S_\pm$ are spiral, i.e., the linearization of the flow at these two points has eigenvalues that are non-real (and complex conjuate), while, when $n\ge 3$, the linearizations of the flow at these two points have distinct real eigenvalues whose ratio is a real number strictly between $n{-}2$ and $n{-}1$. One then finds that, when $n\ge 3$, the union of the two separatrices $L_1$ and $L_2$, together with their endpoints $S_1$, $S_2$, and $S_-$ is the graph $\phi = f_n(v)$ of a function $f_n:[0,\tfrac12\pi]\to[0,\tfrac12\pi]$ that is real-analytic except at the midpoint $v=\tfrac14\pi$, where it is $C^{n-2}$. One then shows that the corresponding $g$-geodesics in the first quadrant define a $C^{n-2}$ foliation by geodesics that meet the boundary rays $x=0$ and $y=0$ orthogonally. Moreover, by reflecting this foliation across the lines $x=0$ and $y=0$, one can construct a foliation $\mathcal{F}$ of $\mathbb{R}^2$ minus the origin by $g$-geodesics that is real-analytic except along the lines $x\pm y=0$, where it is $C^{n-2}$. It then follows easily that there is a unique $C^{n-1}$ function $u$ that vanishes on the axes $x=0$ and $y=0$, satisfies $|\nabla u|^2=(xy)^{2n}$ and, away from the axes, the gradient lines of $u$ are the leaves of the foliation $\mathcal{F}$. Meanwhile, when $n\le 2$, the spiral nature of the two fixed points $S_\pm$ leads to an analysis that shows that there is no foliation of the 'first quadrant' quarter of a disk $D_\rho$ by $g$-geodesics, which leads to the conclusion that there is no $C^1$ solution $u$ on any $D_\rho$ to the equation $|\nabla u|^2=(xy)^{2n}$. If there is interest, I can supply details of these arguments when I get the time.<|endoftext|> TITLE: Space of solutions to a fourth order wave equation QUESTION [6 upvotes]: I'm interested in finding solutions a fourth order version of the standard wave equation in $d$ dimensional Minkowski spacetime $\mathcal{M}^d$. Defining $\Box := \partial_0^2 - \sum_{i = 1}^{d-1} \partial_{i}^2$, I want to find solutions to $$\Box^2\Phi(x) = 0,$$ which are not also solutions to $\Box \Phi(x) = 0$, and which transform as scalars under Lorentz transformations. I'm mostly interested in solutions for $d=4$, but I don't think that's likely to be relevant at the level of analysis I'm currently working at so I want to stick to general $d$ if possible. I know that in solving $\Box \Phi(x) = 0$ with boundary conditions that require some kind of fast enough fall off at infinity, then the solution space can be taken to be Lorentz invariant $L^2$ integrable functions. A basis of solutions for functions with these boundary conditions can be given by plane waves $\Phi_k(x) := e^{i k\cdot x}$ where $k^2 = 0$. My understanding is that as $\Box \Phi(x) = 0$ is seperable, Sturm-Liouville theory tells us that for some choices of boundary conditions we can find a function space of solutions to the equation spanned by some basis of solutions. (Disclaimer: I don't have a very in-depth understanding of analysis. I can see that $\Phi_k \not\in L^2(\mathcal{M}^d)$, so I understand that at some level calling this a `basis' for the space is not correct. I'm happy to call it this given that two such functions are orthogonal in the sense that $\int_{\mathcal{M}^d}dx \;\Phi_k(x)\Phi_{k'}(x) \propto \delta^d(k-k')$, and that any (I think?) function in this space can be decomposed in terms of an integral over all possible $\Phi_k$ by the Fourier transform. I appreciate that to answer my question it may be necessary to go to some more detailed level of analysis where it doesn't make sense to think of $\Phi_k$ as a basis.) I note that, given any $f$ such that $\Box f(x) = 0,$ we can construct a solution to $\Box^2 \Phi(x) = 0$ by taking $\Phi(x) = a \cdot x f(x)$, where $a$ is a vector. For some choices of $a$ this new solution may also satisfy $\Box\left( a\cdot x f(x)\right) = 0,$ but I think it's always possible to choose $a$ such that $\Box\left( a\cdot x f(x)\right) \neq 0.$ Then my question is, given a function space of solutions corresponding to a choice of boundary conditions to $\Box \Phi(x) = 0$, is there a canonical way to extend this to a larger function space of soluitons to $\Box^2 \Phi(x) = 0$, which includes functions which satisfy $\Box^2 \Phi(x) = 0$ and $\Box \Phi(x) \neq 0$, which can be expanded in terms of a basis of functions which are orthogonal under some inner product? If not a canonical way, is there some set of ways of doing this? So then in the case of $L^2$ functions, is it possible to do something like extend to functions which increase like $x$ as $x\rightarrow \infty$? Then I could imagine that this new space could be spanned by $\Phi_k$, and $\chi_{k,a}(x) := a \cdot x e^{i k \cdot x}$, with some inner product under which $\Phi_k, \Phi_{k'}$, $\chi_{k,a}$ and $\chi_{k',a'}$ are orthogonal to each other? I think what I've written here is probably a little naive, but I'm hoping there may be some kind of result similar to this which makes sense. After some searching online I've found that there exists a fourth order Sturm-Liouville theory, but I wasn't able to find anything accesible for me to read, and it appears to me that $\Box^2 \Phi(x) = 0$ is anyway not seperable, at least in Cartersian coordinates. Any references to a simple introduction or review article of fourth-order Sturm-Lioville theory would be appreciated, as well as any local coordinate transformation which makes $\Box^2\Phi(x) = 0$ seperable, if you think this would be relevant to my problem. I appreciate that you could read this question and say, 'why would you expect there to be a canonical way to extend solution spaces for $\Box \Phi(x) = 0$ to $\Box^2 \Phi(x) = 0$ with boarder boundary conditions?'. So here is some evidence that I have from solving the equation in 2D. In 2D in lightcone coordinates $u = x^0 + x^1, v = x^0-x^1$ the wave operator factorises so that $\Box = \partial_u\partial_v$. This makes it simple to write down d'Alembert's general solution to $\Box\Phi(x) = 0$ for all possible boudary conditions, $\Phi(x) = f^+(x^0 + x^1) + f^-(x^0-x^1).$ It's also simple to solve $\Box^2\Phi(x) = 0$ in this case. Writing a seperable ansatz $\Phi(u,v) = U(u)V(v)$ the equation becomes $U'' V''=0$, which is solved non-trivially by either $U' = 0$, $V' = 0$, $U''=0$ or $V''=0$. Then the general solution is given by $\Phi(u,v) = f^+_1(u) + v f^+_2(u) + f^-_1(v) + u f^-_2(v).$ Introducing null vectors $k_+ = k(1,1)$ and $k_- = q(1,-1)$, then we see that $u \propto k_+ \cdot x$, and $v \propto k_- \cdot x.$ Then the general solution can be written as a sum over solutions of the form $$\Phi(x) = f_1(k\cdot x) + a \cdot x f_2(k \cdot x),$$ where $k^2 = 0$ and $a$ is any vector. If $a \propto k$ then the second term solves $\Box\Phi(x) = 0$, otherwise it produces new solutions that satisfy $\Box^2\Phi(x) = 0$ and $\Box\Phi(x) \neq 0$. (I took a linear combination of the lightcone coordinate solutions to produce the new solution). I believe this to be the general solution to $\Box^2\Phi(x) = 0$ in 2D for all possible boundary conditions, and I've intentionally written it in a form that extends in a natural way to general dimension $d$. So this is some kind of motivation for why I think my question may have a solution; it appears to me that what I'm asking in general dimensions happens in 2D based on the form of the general solution I gave here. REPLY [3 votes]: You talk about the non-separability of the $\Box^2 \phi = 0$ equation, which I don't understand. Each plane wave $e^{ik\cdot x} = \prod_{j=0}^{d-1} e^{i k_j x^j}$ is already in separated form with the components of the null vector $k=(k_j)$ playing the role of the separation constants. But rather than get into technicalities about what does or does not constitute a separable equation or a basis of separated solutions, why not just solve the equation via the Fourier transform? Recall first the ordinary wave equation. If $\phi(x)$ is smooth and grows no faster than polynomially in any direction, its Fourier transform $\hat{\phi}(k)$ exists as a distribution (a subclass of tempered distributions in this case). The equation $\Box \phi(x) = 0$ translates to $k^2 \hat{\phi}(k) = 0$, where of course $k^2 = k_0^2 - \sum_{j=1}^{d-1} k_j^2$. No non-vanishing continuous, let alone smooth function of $k$ can satisfy that condition. Hence, $\hat{\phi}(k)$ must be a distribution. A natural candidate is $\hat{\phi}(k) = f(k) \delta(k^2)$, with $f(k)$ an at least continuous function, because of the simple identity $u \delta(u) = 0$. The function $f(k)$ is of course not unique, any functions $f_1,f_2$ such that the difference $f_1(k)-f_2(k)$ vanishes smoothly on the locus of $k^2$ define the same $\hat{\phi}(k)$. The level of regularity (including behavior at infinity) of $f(k)$ translates in a certain way to the level of regularity of $\phi(x)$ via the Fourier transform. The above parametrization of $\hat{\phi}(k)$ is actually exhaustive when these regularity classes are appropriately fixed. To see how the data in $f(k)$ translates to initial data for $\phi(x)$, try the $d=1$ example. Now, on to the squared wave equation. Under the Fourier transform, $\Box^2 \phi(x) = 0$ translates to $(k^2) \hat{\phi}(k) = 0$. Taking derivatives of the $\delta$-function identity, we can get $u \delta'(u) + \delta(u) = 0$ or more importantly $u^2 \delta'(u)=0$. Hence, a natural candidate for a solution is \begin{align*} \hat{\phi}(k) &= f(k) \delta(k^2) + g(k) \delta'(k^2) , \\ &= F(k) \delta(k^2) + ia\cdot\partial_k [G(k) \delta(k^2)] \end{align*} for functions $f(k), g(k)$ of appropriate regularity and $G(k) = g(k)/(2ia\cdot k)$, $F(k) = f(k) - (a\cdot\partial_k) G(k)$. To make it easier to go between $g(k)$ and $G(k)$, the vector $a$ should not be null, so that $a\cdot k$ only vanishes at $k=0$. The corresponding formula in real space is $$ \phi(x) = \phi_1(x) + (a\cdot x) \phi_2(x) , $$ where $\phi_1(x), \phi_2(x)$ are independent solutions of the wave equation, $\Box \phi_{1,2}(x)=0$. This is I think the parametrization of solutions that you were looking for. As before, it is a matter of figuring out the right regularity classes for $F(k), G(k)$ and the corresponding initial data for $\phi(x)$ to make sure that the above parametrization is exhaustive.<|endoftext|> TITLE: Weak descent and effective equivalence relations QUESTION [10 upvotes]: I want to prove that weak descent of a $1$-category implies the classical Giraud axioms. More precisely, let $\mathsf{C}$ be a cocomplete, finitely complete $1$-category. We say that $\mathsf{C}$ satisfies weak descent if the following conditions are satisfied: $(\mathbf{D1}a)$-(Universal coproducts): Given a collection of objects $\{ Y_i \}_{i \in I}$, let $Y = \coprod_i Y_i$. Let $f: X \to Y$ be a morphism, and let $X_i = Y_i \times_Y X$. Then the induced map $\coprod_i X_i \to X$ is an isomorphism, $(\mathbf{D1}b)$-(Universal pushouts): Given a span $Y_0 \leftarrow Y_1 \to Y_2$, let $Y = Y_0 \coprod_{Y_1} Y_2$. Let $f: X \to Y$ be a morphism and let $X_i = Y_i \times_{Y} X$. Then the induced map $X_0 \coprod_{X_1} X_2 \to X$ is an isomorphism. $(\mathbf{D2}a)$-(Effective coproducts): Given a collection of maps $\{ f_i: X_i \to Y_i \}$, let $X = \coprod_i X_i$, and $Y = \coprod_i Y_i$, and let $f: X \to Y$ be the coproduct $\coprod_i f_i$. Then the natural maps $X_i \to Y_i \times_Y X$ are isomorphisms for each $i$. $(\mathbf{D2}b)$-(Weak effective pushouts): Given a map of spans: $\require{AMScd}$ \begin{CD} X_0 @<<< X_1 @>>> X_2\\ @Vf_0VV @Vf_1VV @Vf_2VV\\ Y_0 @<<< Y_1 @>>> Y_2 \end{CD} such that each square is a pullback square, let $X = X_0 \coprod_{X_1} X_2$ and $Y = Y_0 \coprod_{Y_1} Y_2$, and let $f:X \to Y$ denote the induced map of pushouts. Then the natural maps $X_i \to Y_i \times_Y X$ are regular epimorphisms. Condition $(\mathbf{D2}b)$ is the real difference between $1$-topoi and $\infty$-topoi, and I am trying to better understand this comparison. Now recall the classical Giraud Axioms: $(\mathbf{G1})$ Coproducts are disjoint, namely $A \times_{A \coprod B} B \cong \varnothing$, $(\mathbf{G2})$ For any morphism $f: X \to Y$, the base change functor $f^*: \mathsf{C}_{/Y} \to \mathsf{C}_{/X}$ preserves colimits, $(\mathbf{G3})$ Equivalence relations are effective. Rezk sketches how to prove that $(\mathbf{D1}) = (\mathbf{D1}a) \wedge (\mathbf{D1}b)$ is equivalent to $(\mathbf{G2})$, and that $(\mathbf{D2}a) \implies (\mathbf{G1})$. My suspicion is that $(\mathbf{D2}b) \implies (\mathbf{G3})$ or is possibly equivalent to it, but I can't quite see how to prove it. I also suspect that the way one can prove it by is proving that $(\mathbf{D2}b)$ is equivalent to the coequalizer defining equivalence relations being effective. Namely if $R \xrightarrow{(s,t)} X \times X$ is an equivalence relation, then $R \rightrightarrows X \to X/R$ is a coequalizer iff $X/R \cong R \coprod_{R \coprod R} X$, and my idea is to show that having condition $(\mathbf{D2}b)$ hold, but this time up to isomorphism rather than regular epi but only for pushouts of equivalence relations as above, and this would be equivalent to $(\mathbf{D2}b)$ and from this somehow it would be easier to see that it implies $(\mathbf{G3})$, but I've had no luck with this either. Any ideas or insight would be appreciated. REPLY [5 votes]: I find it a bit surprising, but I think you are correct. The proof I have is maybe a little too long for MO, so I'm only sketching it, but I'll be happy to provide more details if needed. First one observe a form of "weak descent" for coequalizer diagram: Lemma: Assume that the category $C$ satistifes all four condition of the op, then for any pair of coequalizer diagram $Y_0 \rightrightarrows Y_1 \to \text{coeq }Y_i$ and $X_0 \rightrightarrows X_1 \to \text{coeq }X_i$ and a cartesian natural transformation $X_0 \to Y_0$ , $X_1 \to Y_1$; the natural map $X_1 \to Y_1 \times_{\text{coeq } Y_i} (\text{coeq } X_i)$ is a regular epimorphism. Proof: We use that a coequalizer $Y_0 \rightrightarrows Y_1$ can be written as a pushout $Y_1 \coprod_{Y_0 \coprod Y_1} Y_1$. Using that coproduct are disjoint and universal, one obtains that if $X_i \to Y_i$ is cartesian the map of spans from $X_1 \leftarrow X_0 \coprod X_1 \rightarrow X_1$ to $Y_1 \leftarrow Y_0 \coprod Y_1 \rightarrow Y_1$ is also cartesian, indeed, $(Y_0 \coprod Y_1) \times_{Y_1} X_1 = (Y_0 \times_{Y_1} X_1) \coprod (Y_1 \times_{Y_1} X_1) = X_0 \coprod X_1$. and one can apply condition $(D2b)$ to this pushout. With a bit more work, one can actually prove this also for $X_0 \to..$ and not just for coequalizer but for all colimits. Now, we apply this to coequalizer of equivalence relation in a way inspired from the usual proof that descent for all colimits (in $\infty$-topos) implies that equivalence realtion are effective. All the claim below are proved in the same way: they are clear for sets and we prove them for a general category with finite limits by interpreting everything in terms of "generalized" elements (or if you prefer by doing everything in terms of presheaves). Consider now the case of an equivalence relation $R \rightrightarrows X$. Let $R^{(2)}$ be the subobject of $X^3$ corresponding to $\{ x_1,x_2,x_3 \in X^3 | x_1 R x_2 \text{ and } x_2 R x_3 \}$ $R^{(2)}$ has two maps to $R$ that sends $(x_1,x_2,x_3)$ to $(x_1,x_2)$ and $(x_1,x_3)$ and this makes $R^{(2)}$ into an equivalence relation on $R$. The coequalizer $R/R^{(2)}$ is $X$ because there is a split coequalizer diagram $R^{(2)} \rightrightarrows R \rightarrow X$ (with $X \to R$ and $R \to R^{(2)}$ defined in the obvious way). Finally, we have a cartesian transformation of equalizer: $(R^{(2)} \rightrightarrows R) \to ( R \rightrightarrows X)$ where $R^{(2)} \to R$ is $(x_1,x_2,x_3) \mapsto (x_2,x_3)$ and $R \to X$ is $(x_1,x_2) \mapsto x_2$. It then follows from the version of $(D2b)$ for coequalizer proved above that $R \to X \times_{E} X$ is a regular epimorphism (where $E =X/R$ is the coeqalizer). But given that both $R$ and $X \times_E X$ are subobject of $X \times X$, this map is both a regular epimorphism and a monomorphism, hence it is an isomorphisms.<|endoftext|> TITLE: Stability properties of essential geometric morphisms QUESTION [7 upvotes]: Notation. $\mathsf{Topoi}$ is the bicategory of topoi, geometric morphisms and natural transformations between left adjoints. $\mathsf{Topoi}_{\text{ess}}$ is the bicategory of topoi, essential geometric morphisms and natural transformations between left adjoints. $\mathsf{Presh}$ is the full subcategory of $\mathsf{Topoi}$ spanned by presheaf topoi. $\mathsf{Presh}_{\text{ess}}$ Is the full subcategory of $\mathsf{Topoi}_{\text{ess}}$ spanned by presheaf topoi. Questions. Is there a reference for the bicategorical properties of $\mathsf{Topoi}_{\text{ess}}$, $\mathsf{Presh}$, $\mathsf{Presh}_{\text{ess}}$? Which (pseudo)(co)limits are preserved by the inclusion $\mathsf{Presh}_{\text{ess}} \subset \mathsf{Topoi}$? (This is my motivating question.) REPLY [5 votes]: This is only a partial answer. With '(co)limit' I will always mean pseudo(co)limit. If $\mathcal{C}$ and $\mathcal{D}$ are Cauchy-complete, then the category of essential geometric morphisms $\mathbf{PSh}(\mathcal{C}) \to \mathbf{PSh}(\mathcal{D})$ (and geometric transformations between them) is equivalent to the opposite of the category of functors $\mathcal{C} \to \mathcal{D}$ (and natural transformations between them). This is in “Sketches of an Elephant”, Part A, Example 4.1.4 and Lemma 4.1.5. So in this sense $\mathsf{Presh}_\mathrm{ess}$ is a full subcategory of $\mathsf{Cat}^\mathrm{co}$, the bicategory of small categories, functors, and natural transformations (with the natural transformations in the opposite direction). By using the above, I think it follows that $$\mathrm{colim}_i\, \mathbf{PSh}(\mathcal{C}_i) ~\simeq~ \mathbf{PSh}(\mathrm{colim}_i \,\mathcal{C}_i)$$ in $\mathsf{Presh}_\mathrm{ess}$, as long as $\mathrm{colim}_i\, \mathcal{C}_i$ is still Cauchy-complete. In particular, coproducts are computed as $\bigsqcup_i \mathbf{PSh}(\mathcal{C}_i) \simeq \mathbf{PSh}(\bigsqcup_i \mathcal{C}_i)$, and this agrees with the coproduct in $\mathsf{Topoi}$. So the inclusion $\mathsf{Presh}_\mathrm{ess} \subset \mathsf{Topoi}$ preserves coproducts, in particular the initial object.

 Similarly, I think that $$\mathrm{lim} \, \mathbf{PSh}(\mathcal{C}_i) \simeq \mathbf{PSh}(\mathrm{lim}_i \mathcal{C}_i)$$ in $\mathsf{Presh}_\mathrm{ess}$, as long as $\lim_i \mathcal{C}_i$ is still Cauchy-complete. In particular, the terminal object is $\mathbf{PSh}(1) \simeq \mathbf{Sets}$, just like in $\mathsf{Topoi}$. Also, the product of $\mathbf{PSh}(\mathcal{C})$ and $\mathbf{PSh}(\mathcal{D})$ is $\mathbf{PSh}(\mathcal{C}\times\mathcal{D})$. This is also the product in $\mathsf{Topoi}$ (see Pitts, "On product and change of base for toposes"). I don’t know whether $\mathsf{Presh}_\mathrm{ess} \subset \mathsf{Topoi}$ preserves pullbacks. Examples seem to suggest that pullbacks are preserved, I would be very interested in a proof (here by 'pullback' I mean pseudo-pullback). Update: here is an example showing that arbitrary products are not preserved. Consider the categories $(\mathcal{C}_n)_{n \in \mathbb{N}}$ with $\mathcal{C_n}$ given by the discrete category on two objects, for each $n \in \mathbb{N}$. Then in $\mathsf{Presh}_\mathrm{ess}$ the product $\prod_{n \in \mathbb{N}} \mathbf{PSh}(\mathcal{C_n})$ is given by $\mathbf{PSh}(\mathcal{D})$, where $\mathcal{D} \simeq \prod_{n \in \mathbb{N}} \mathcal{C}_n$ is the discrete category with $2^{|\mathbb{N}|}$ objects. However, in $\mathsf{Topoi}$ the product is $\prod_{n \in \mathbb{N}} \mathbf{PSh}(\mathcal{C_n}) \simeq \mathbf{Sh}(X)$, where $X$ is the Cantor space (the product in the category of toposes/locales agrees with the product in the category of topological spaces, because it is a countable product of completely metrizable spaces, see Isbell, "Atomless parts of spaces"). Update 2: I believe the inclusion $\mathsf{Presh}_\mathrm{ess} \subset \mathsf{Cat}^\mathrm{co}$ has a left adjoint given by $\mathcal{C} \mapsto \mathbf{PSh}(\mathcal{C})$. This can be used to show that in $\mathsf{Presh}_\mathrm{ess}$ the colimit of $\mathbf{PSh}(\mathcal{C}_i)$'s, with each $\mathcal{C}_i$ Cauchy-complete, is given by $\mathbf{PSh}(\mathcal{D})$ where $\mathcal{D} = \mathrm{colim}\, \mathcal{C}_i$ is the colimit in $\mathsf{Cat}^\mathrm{co}$ (the category $\mathcal{D}$ does not have to be Cauchy-complete in order for this to work). To show that $\mathbf{PSh}(\mathcal{D})$ is also the colimit in $\mathsf{Topoi}$ (so colimits are preserved), we can use that colimits in $\mathsf{Topoi}$ are computed as the limit in $\mathsf{Cat}$ of the corresponding diagram of inverse image functors (see here). Further, in $\mathsf{Cat}$ we have that $$\mathrm{lim}_i\, \mathbf{PSh}(\mathcal{C}_i) \simeq \mathrm{lim}_i \, \mathrm{Fun}(\mathcal{C}_i^\mathrm{op}, \mathbf{Sets}) \simeq \mathrm{Fun}(\mathrm{colim}_i\, \mathcal{C}_i^\mathrm{op}, \mathbf{Sets}) \\ \simeq \mathrm{Fun}(\mathcal{D}^\mathrm{op}, \mathbf{Sets}) \simeq \mathbf{PSh}(\mathcal{D}).$$ Here we use that $\mathbf{Fun}(-,-)$ sends colimits to limits in the first argument. The proof above is based on the answer by Yonatan Harpaz here.<|endoftext|> TITLE: Situations where “naturally occurring” mathematical objects behave very differently from “typical” ones QUESTION [63 upvotes]: I am looking for examples of the following situation in mathematics: every object of type $X$ encountered in the mathematical literature, except when specifically attempting to construct counterexamples to this, satisfies a certain property $P$ (and, furthermore, this is not a vacuous statement: examples of objects of type $X$ abound); it is known that not every object of type $X$ satisfies $P$, or even better, that “most” do not; no clear explanation for this phenomenon exists (such as “constructing a counterexample to $P$ requires the axiom of choice”). This is often presented in a succinct way by saying that “natural”, or “naturally occurring” objects of type $X$ appear to satisfy $P$, and there is disagreement as to whether “natural” has any meaning or whether there is any mystery to be explained. Here are some examples or example candidates which come to my mind (perhaps not matching exactly what I described, but close enough to be interesting and, I hope, illustrate what I mean), I am hoping that more can be provided: The Turing degree of any “natural” undecidable but semi-decidable (i.e., recursively enumerable but not recursive) decision problem appears to be $\mathbf{0}'$ (the degree of the Halting problem): it is known (by the Friedberg–Muchnik theorem) that there are many other possibilities, but somehow they never seem to appear “naturally”. The linearity phenomenon of consistency strength of “natural” logical theories, which J. D. Hamkins recently gave a talk about (Naturality in mathematics and the hierarchy of consistency strength), challenging whether this is correct or even whether “naturality” makes any sense. Are there "natural" sequences with "exotic" growth rates? What metatheorems are there guaranteeing "elementary" growth rates? concerning the growth rate of “natural” sequences, which inspired the present question. The fact that the digits of irrational numbers that we encounter when not trying to construct a counterexample to this (e.g., $e$, $\pi$, $\sqrt{2}$…) experimentally appear to be equidistributed, a property which is indeed true of “most” real numbers in the sense of Lebesgue measure (i.e., a random real is normal in every base: those which are are a set of full measure) but not of “most” real numbers in the category sense (i.e., a generic real is not normal in any base: those which are are a meager set). What other examples can you give of the “most $X$ do not satisfy $P$, but those that we actually encounter in real life always do (and the reason is unclear)” phenomenon? REPLY [2 votes]: Euclidean lattices of high density are generic and are very difficult to construct in large dimensions.<|endoftext|> TITLE: Detecting uncountable cardinals in $(\mathbb{R};+,\times,\mathbb{N})$ QUESTION [24 upvotes]: For a structure $\mathcal{X}=(X;...)$, say that a cardinal $\kappa$ is $\mathcal{X}$-detectable iff there is some sentence $\varphi$ in the language of $\mathcal{X}$ together with a fresh unary predicate symbol $U$ such that for all $A\subseteq X$, the expansion of $\mathcal{X}$ gotten by interpreting $U$ as $A\subseteq X$ satisfies $\varphi$ iff $\vert A\vert\ge\kappa$. For example, $\omega_1$ is $(\omega_1;<)$-detectable since the uncountable subsets of $\omega_1$ are exactly the unbounded ones. By contrast, Alex Kruckman observed that by a result of Robinson no uncountable cardinal is $\mathcal{R}=(\mathbb{R};+,\times)$-detectable. I'm interested in the expansion $\mathcal{R}_\mathbb{N}:=(\mathbb{R};+,\times,\mathbb{N})$ of $\mathcal{R}$ gotten by adding a predicate naming the natural numbers (equivalently, adding all projective functions and relations). Since we can talk about one real enumerating a list of other reals, $\omega_1$ is $\mathcal{R}_\mathbb{N}$-detectable ("there is no real enumerating all elements of $U$"). More pathologically, if $\mathfrak{c}=2^\omega$ is regular and there is a projective well-ordering of the continuum of length $\mathfrak{c}$ then $\mathfrak{c}$ is $\mathcal{R}_\mathbb{N}$-detectable. So for example it is consistent with $\mathsf{ZFC}$ that $\omega_2$ is $\mathcal{R}_\mathbb{N}$-detectable. I'm curious whether this type of situation is the only way to get $\mathcal{R}_\mathbb{N}$-detectability past $\omega_1$. There are multiple ways to make this precise, of course. At present the following two seem most natural to me: Is it consistent with $\mathsf{ZFC}$ that there are at least two distinct regular cardinals $>\omega_1$ which are $\mathcal{R}_\mathbb{N}$-detectable? Is it consistent with $\mathsf{ZFC}$ that there is a singular cardinal which is $\mathcal{R}_\mathbb{N}$-detectable? Note that an affirmative answer to either question requires a large continuum, namely $\ge\omega_3$ and $\ge\omega_{\omega+1}$ respectively. Although my primary interest is in first-order definability, I'd also be interested in answers for other logics which aren't too powerful (e.g. $\mathcal{L}_{\omega_1,\omega}$). REPLY [6 votes]: Farmer S asked in the comments about the consistency of $\aleph_2$ being undetectable. I hope to dispel any doubt: adding enough Cohen reals does work. Specifically, adding $\kappa$ Cohen reals will ensure that no cardinal $\aleph_1<\lambda\leq \kappa$ is $\mathcal R_{\mathbb N}$-detectable, at least. Let $\dot{G}$ be the canonical name for a function $\kappa\to\mathbb R$ enumerating the new generic reals. A name $\dot{r},$ thought of as a name for a real, is "supported by" a set $I\subseteq\kappa$ if the Boolean value of each statement "$\check{i}\in \dot{r}$" ($i\in\omega$) is invariant under permutations that fix every element of $I.$ Every name has a countable support. Countability comes from the fact that these Boolean values can be represented by a regular open set in $2^{\kappa\times\omega},$ which is the union of some countable maximal antichain of basic opens, and the indices used in the basic opens constitute a support. For the purposes of induction I'll use a relative kind of indetectability. Fix disjoint sets $F,U,V\subseteq \kappa$ with $U,V$ uncountable, and names $\dot{r_1},\dots,\dot{r_k}$ of reals supported by $\kappa\setminus(U\cup V).$ Fix also a formula $\phi$ in prenex normal form in the language of $\mathcal R_{\mathbb N}$ with a unary predicate, translated to a statement of set theory by restricting quantifiers to the reals. The unary predicate is written as the first parameter. I will argue that $$\Vdash S\iff T\tag{*}$$ with $$S=\phi(\dot{G}[F\cup U],\dot{r_1},\dots,\dot{r_k}),$$ $$T=\phi(\dot{G}[F\cup V],\dot{r_1},\dots,\dot{r_k}).$$ To get indetectability of $\lambda$ plug in $F=\emptyset$ and $k=0,$ with $U$ a subset of $\kappa$ of cardinality $\lambda,$ and $V$ a disjoint subset of $\kappa$ of cardinality $\aleph_1.$ I think it might not be too hard to extend this to any formula $\phi$ in $L(\mathbb R)^{M[G]},$ and analogous models using HOD or symmetric extensions. Suppose (*) does not hold. Negate $\phi$ if necessary so that the first quantifier, if any, is existential. Swapping $(S,U)$ and $(T,V)$ if necessary we can assume there is a condition $p’$ such that $p’\Vdash S\wedge \neg T.$ Let $p$ be the condition obtained by ignoring indices in $U\cup V,$ i.e. $p=p’\setminus((U\cup V)\times\omega\times 2).$ The following standard argument shows that $p\Vdash S\wedge \neg T.$ Permutations of $\kappa$ act on forcing conditions and on names by their action on the Boolean algebra of regular opens of $2^{\kappa\times \omega}.$ For any $q\leq p$ we can find a permutation $\pi$ of $\kappa$ fixing $\kappa\setminus(U\cup V)$ elementwise and fixing $U$ and $V$ setwise, so that no index $i\in U\cup V$ is used by both $q$ and by $\pi p’.$ This means that the conditions $q$ and $\pi p’$ are compatible. Applying symmetry we have $\pi p’\Vdash S\wedge \neg T$ and hence $q \cup \pi p’\Vdash S\wedge \neg T,$ and since $q$ was arbitrary we have $p\Vdash S\wedge \neg T.$ If $\phi$ is quantifier-free then (*) is easy to see. Otherwise we arranged that $\phi$ is of the form $(\exists x\in\mathbb R)\psi.$ Then there exists a name $\dot{x}$ for a real, and a condition $q\leq p,$ such that $$q\Vdash \psi(\dot{G}[F\cup U],\dot{r_1},\dots,\dot{r_k},\dot{x}).$$ Partition $U$ as $U_1\cup U’$ with $U_1$ countable, and similarly $V=V_1\cup V’,$ such that $q$ and $\dot{x}$ are supported by $\kappa\setminus(U’\cup V’)$ (extending the definition of "supported by" to conditions). By induction on quantifiers we can apply (*) to $\psi$ to get $$q\Vdash \psi(\dot{G}[F\cup U_1\cup V’],\dot{r_1},\dots,\dot{r_k}, \dot{x}).$$ Pick a permutation $\pi$ fixing each element of $\kappa\setminus(U\cup V)$ but with $\pi[U_1\cup V’]=V.$ Then $$\pi q\Vdash \psi(\dot{G}[F\cup V],\dot{r_1},\dots,\dot{r_k}, \pi\dot{x}).$$ The condition $\pi q$ and name $\pi \dot{x}$ therefore witnesses $p\not\Vdash \neg T.$<|endoftext|> TITLE: Functions over monoids which factor in two different ways QUESTION [8 upvotes]: This is a follow-up question to this MO question, which was asked by Richard Stanley in a comment to my answer there. Let $S$ be a commutative monoid and $f(x_1, \dots, x_n)$ be a function from $S^n$ to $S$. Given a partition $\alpha$ of $[n]$, we say that $f$ factors with respect to $\alpha$, if for each $A \in \alpha$ there exists a function $f_A$ (which only depends on the variables $x_i$ for $i \in A$) such that $f=\prod_{A \in \alpha} f_A$. Given two partitions $\alpha$ and $\beta$ of $[n]$, $a \wedge b$ is the partition of $[n]$ whose sets are the non-empty sets of the form $A \cap B$ for $A \in \alpha$ and $B \in \beta$. Question. Is it true that if $f$ factors with respect to both $\alpha$ and $\beta$, then $f$ also factors with respect to $\alpha \wedge \beta$? My answer to the linked question shows that the answer is yes if $S$ is a group, but the proof uses the fact that inverses exist. My proof also works for non-abelian groups (as long as you are careful what factoring means), but for this question I am happy to assume that $S$ is commutative. REPLY [4 votes]: Take $S$ to be the monoid on the set $\{0,2,3,4,5,6\}$ with operation $x\oplus y=\min(x+y,6).$ Let $g:\{0,1\}^3\to S$ be the function $(x,y,z)\mapsto \min(x+y+z+4,6).$ Using any surjective $h:S\to \{0,1\}$ this can be converted to $f(x,y,z)=g(h(x),h(y),h(z)).$ I'll just work with $g.$ $g$ factors with respect to $12,3$ and $1,23$: $g(x,y,z)=(x+y+2)\oplus(z+2)=(x+2)\oplus(y+z+2).$ But $g$ does not factor with respect to $1,2,3.$ Suppose for contradiction that $g(x,y,z)=f_1(x)\oplus f_2(y)\oplus f_3(z)$ for all $x,y,z.$ We must have $f_i(1)=f_i(0)+1$ for each $i,$ because $g(0,0,1)=g(0,1,0)=g(1,0,0)=5$ and $g(0,0,0)=4.$ This implies $f_i(0)\geq 2.$ But then $f_1(0)\oplus f_2(0)\oplus f_3(0)\geq 6> g(0,0,0).$<|endoftext|> TITLE: Projecting onto space of matrices with spectral radius less than one QUESTION [7 upvotes]: Consider the space $$ S = \left\{ A \in \mathbb{R}^{n \times n} : \mathrm{SpectralRadius}(|A|) \leq 1 \right\}$$ where $|A|$ is the entry-wise absolute value. Given a matrix $M \in \mathbb{R}^{n\times n}$ such that $M \not \in S$. Is there a way to solve $$\arg\min_{\hat M \in S} \Vert M - \hat M \Vert$$ for some matrix norm $\Vert \cdot \Vert$? That is, is there a (hopefully somewhat simple) operator that projects onto $S$ for some matrix norm? For context, I'm interested in optimising a function $f:\mathbb{R}^{n\times n} \rightarrow \mathbb{R}$ over this space using stochastic gradient descent. This means that I'd have to perform this projection after every optimisation step. I know that in general $S$ is non-convex and so I won't have any guarantees. Edit: I'm interested in the case with the absolute value. However, someone noted in a comment that the scenario may simplify if we consider the space $$S' = \left\{ A \in \mathbb{R}^{n \times n} : \mathrm{SpectralRadius}(A) \leq 1 \right\}$$ Maybe to start, is there a simpler answer in this case? REPLY [2 votes]: For the version without the absolute value, a colleague and I published a few months ago an algorithm to solve that optimization problem and other related ones (the problem you need is called "Schur stable" case in the paper). Some competitors are mentioned in the introduction and in the section with numerical experiments, but I believe ours is the fastest algorithm available as of today. It is not too simple as it relies on optimization on manifolds, however this is not a simple problem in general: the constraint on the eigenvalues is non-smooth and tricky to handle numerically. We have also published on https://github.com/fph/nearest-omega-stable some proof-of-concept Matlab code that you can use.<|endoftext|> TITLE: Limit on a certain double sum QUESTION [5 upvotes]: While working with multi-zeta functions, I encountered the below (experimental) value for a certain evaluation (in a limit sense). Notice first this well-known fact in context $$\sum_{n,m\geq1}\frac1{nm(n+m)}=2\zeta(3).$$ QUESTION. Does this hold true? If yes, how? $$\lim_{s\rightarrow\frac12^+}\sum_{n,m\geq1}\frac{2s-1}{n^sm^s(n+m)}=\pi.$$ Edited. $s\rightarrow\frac12$ has been replaced by $s\rightarrow\frac12^+$. REPLY [9 votes]: The limit (as $s\downarrow1/2$) will not change if the double sum is replaced by the corresponding double integral, for which we find this, as desired: Alternatively, the limit (as $s\downarrow1/2$) will not change if the double sum is replaced by the double integral $$I(s):=(2s-1) \iint\limits_{x,y>0,\,x+y>1}\frac{dx\,dy}{x^s y^s(x+y)} \\ =(2s-1)\int_1^\infty \frac{du}u\,\int_0^u \frac{dx}{x^s (u-x)^s} \\ =(2s-1)\int_1^\infty \frac{du}{u^{2s}}\,J(s) =J(s),$$ where $$J(s):=\int_0^1 \frac{dt}{t^s (1-t)^s}\to J(1/2)=\pi.$$ So, $I(s)\to\pi$, as desired. Details on why the limit (as $s\downarrow1/2$) will not change if the double sum is replaced by an appropriate double integral: The reasons for this are as follows: $f(x,y):=\dfrac1{x^s y^s(x+y)}$ is decreasing in $x>0$ for each real $y>0$ and in $y>0$ for each real $x>0$. So, for any natural $i$ and $j$, $f(i,j)\ge\int_{(i,i+1]\times(j,j+1]}f\ge f(i+1,j+1)$. For each natural $m$, $\sum_{n\ge1}f(m,n)<\infty$ and, similarly, for each natural $n$, $\sum_{m\ge1}f(m,n)<\infty$. Each of such ordinary sums will get multiplied by $2s-1$, and so, will not contribute to the limit as $s\downarrow1/2$. 2a. Similarly to point 2, for each real $x>0$, $\int_{\max(0,1-x)}^\infty f(x,y)\,dy\le\dfrac1{x^s}\int_0^\infty \min\Big(\dfrac1{y^s},\dfrac1{y^{s+1}}\Big)\,dy =\dfrac1{(1-s)sx^s}$, and $\int_{\max(0,1-y)}^\infty f(x,y)\,dx\le\dfrac1{(1-s)sy^s}$ for each real $y>0$.<|endoftext|> TITLE: Noncompact three-manifold with fundamental group isomorphic to a surface group QUESTION [9 upvotes]: Let $M$ be an open orientable three-manifold such that $\pi_1 (M)$ is isomorphic to the fundamental group of a closed orientable surface $S\ncong \mathbb{S}^2$. Furthermore, suppose that $\tilde{M} \cong \mathbb{R}^3$. Is it true that $M \cong S \times \mathbb{R}$? Without making any assumptions about $\tilde{M}$, there are counterexamples to the question, (see for instance... ), but in those cases the universal cover is not very nice. [As remarked by several commentators, you need $S$ to be orientable. I added this hypothesis and fixed some of the notation. - Sam] REPLY [8 votes]: I'll restate the question for the convenience of the reader.$\newcommand{\RR}{\mathbb{R}}$ Suppose that $M$ is a non-compact, connected, oriented three-manifold without boundary, with universal cover homeomorphic to $\RR^3$. Suppose that $F$ is a compact, connected, oriented surface with genus at least one. Suppose that the fundamental groups of $M$ and $F$ are isomorphic. Is $M$ homeomorphic to $F \times \RR$? The answer is "no". We can build a counterexample using the paper Some examples of exotic non-compact three-manifolds by Scott and Tucker. Here I closely follow their wording and notation, starting at the bottom of page 488. Consider their example manifold $M_5$. This is a three-manifold with boundary a closed surface $F$. The universal cover of $M_5$ is homeomorphic to $\RR^2 \times \RR_{\geq 0}$, and the inclusion of $F$ into $M_5$ is a homotopy equivalence. However, $M_5$ is not "almost compact" (that is, it is not tame), and $M_5$ is not homeomorphic to $F \times \RR_{\geq 0}$. Now double $M_5$ across its boundary to obtain $M = D(M_5)$. This does not change the fundamental group. Note that $M$ has two ends, separated by (the image of) $F$. By Scott-Tucker, neither end is tame. Thus $M$ is not homeomorphic to $F \times \RR$. However, the universal cover of the double is (in this case) the double of the universal cover; thus it is $\RR^2 \times \RR$, as desired.<|endoftext|> TITLE: Intuitive reason that the regular representation is a uniform function QUESTION [8 upvotes]: Corollary 12.14 of Digne-Michel's book Representations of finite groups of Lie type gives various decompositions of the regular representation $\operatorname{reg}_G$ in terms of the Deligne-Lusztig characters. The two I am primarily interested in are $$\operatorname{reg}_G = \frac{1}{|G^F|_p} \sum_{T \in \mathcal{T}} \epsilon_G \epsilon_T R^G_T(\operatorname{reg}_T) = \frac{1}{|G^F|_p} \sum_{\substack{T \in \mathcal{T}\\ \theta \in \operatorname{Irr}(T^F)}} \epsilon_G \epsilon_T R^G_T(\theta).$$ My question is: can someone give an intuitive reason for why these decompositions exist? I am not asking for a proof, but for a moral reason that this should be true. For context: the Deligne-Lusztig characters form a core part of my thesis on the representation theory of finite groups of Lie type, but I do not have the space to completely develop all of the results. I have therefore chosen a few proofs which I feel to be illustrative of the techniques in the field, and for the rest I am trying to offer an intuitive explanation. If I had to guess a decomposition of the regular representation then this would certainly be it (possibly modulo the signs), but I am having difficulty coming up with a way to justify this to readers. An example of the type of argument I am looking for: the Mackey decomposition for Deligne-Lusztig induction/restriction can be interpreted as `pushing forward' the Bruhat decomposition of a finite group of Lie type onto its representations. Indeed, you can make this argument rigorous (at least on the level of characters, I'm not focusing on the actual representations) by looking at the action of $G^F$ on a certain finite affine variety related to the double cosets, then using properties of the Lefschetz number to make combinatorial simplifications, so to me this gives a good intuition for the Mackey decomposition. REPLY [5 votes]: This is an interesting question, but the kind of geometric or structural intuition your are looking for may not exist. To put it another way, the reason behind the fact in the OP is a non-trivial combination of several other central facts, but can't conceptually be reduced to either of them. If we want to guess that some virtual character such as $$ \frac{1}{\lvert G^F\rvert_p} \sum_{T, \theta} \epsilon_G \epsilon_T R^G_T(\theta)$$ equals the regular character $\operatorname{reg}_G$, then, as a first step, we had better make sure that they have the same degree. That this is true is non-trivial and the proof actually uses some of the same steps as the proof that the two characters are equal. More precisely, to compute the degree of the above character, one has to, as far as we know, at some point use the fact that the Steinberg character $\mathrm{St}$ is zero on all non-semisimple elements. This is based on properties of Bruhat decomposition/BN-pairs and Curtis's alternating sum formula for $\mathrm{St}$. The above fact about the values of $\mathrm{St}$ together with the fact that values of $\sum_{\theta} R^G_T(\theta)$ are Lefschetz numbers and hence zero on non-trivial semisimple elements, gives an easy expression for the inner product of $\mathrm{St}$ and $\sum_{\theta} R^G_T(\theta)$. The alternating sum formula for $\mathrm{St}$ together with the fact that $\mathrm{St}(1)=|G^F|_p$, can then be used to show that $$\sum_{\theta} \epsilon_G \epsilon_T R^G_T(\theta)(1) = \frac{|G^F|}{\lvert G^F\rvert_p}.$$ Finally, a non-trivial result of Steinberg says that there are $|G^F|_p^2$ $F$-stable maximal tori, so $\sum_{T,\theta} \epsilon_G \epsilon_T R^G_T(\theta)$ indeed has degree $|G^F|$. One might ask whether lifting the question to isomorphism of representations or some other categorified objects would make the fact in the OP more 'geometric'. It is not clear to me that this can be done and the fact about the values of the character $\mathrm{St}$ highlighted above is used in all of the proofs I know of the statement in the OP. Note that the proof of 12.14 in Digne--Michel uses this fact (when it refers to 9.4), so in particular I am not aware of any character-free proof (as in, does not use anything about character values). In summary, I think the result in the OP can be made plausible by combining the observation in LSpice's answer with a summary of why the two characters have the same degree (e.g., as given above), but this doesn't reduce the result to any straightforward conceptual or geometric principle. Instead, like many non-trivial proofs, it is a combination of several main ingredients. In this case the main ingredients are the existence and properties of the Steinberg character (the alternating sum formula, character values), Bruhat decomposition, the character formula for Deligne--Lusztig characters, etc.<|endoftext|> TITLE: What is the correct definition of weak map between 2-term $L_\infty$-algebras? QUESTION [7 upvotes]: The definition of $L_\infty$-algebra is by now pretty standard. I gather that the sign conventions given in Lada–Markl's paper Strongly homotopy Lie algebras, Communications in Algebra 23 Issue 6 (1995) (arXiv:hep-th/9406095) are widely used, and I will keep to them here. I will not rehash the definition of $L_\infty$-algebra, because I'm sure the people who can answer this question will know it well-enough, or could look it up. I emphasise that in my case, I am not assuming anything is finite-dimensional, so that I am unwilling to dualise from dg coalgebras to dg algebras in case there is something funny going on. What I want to know is what is the correct set of conditions on a weak map of 2-term $L_\infty$-algebras. Such a thing is a dg-coalgebra map of the graded cocommutative cofree coalgebra equipped with the differential that is the sum of all the brackets of the $L_\infty$-algebra. I have found several published papers that all use the definition of Lada–Markl for $L_\infty$-algebras, specialise to the case of 2-term $L_\infty$-algebras or slight variations, then give conflicting definitions for a morphism. None of them have commented on potential errors in the others, or if there are conventions that differ leading to different signs. The coherence conditions have various signs flipped in various ways, and it's not clear to me, a complete uninitiate for dealing with cofree coalgebras of infinite-dimensional vector spaces, how to even arrive at even one of these sets. All of the sources I have seen so far do not actually calculate these coherence conditions, merely state them as an definition, rather than deriving them from the definition I just gave (if there is a published source working this stuff out, I'd love to see it). Note: The SE software is suggesting to me this question, except that uses an apparently different sign convention compared to Lada–Markl. However, that's exactly the sort of formula I'm after - just with the signs sorted (which is of course the whole difficulty in this game), and I don't trust myself to not miss something. Edit I will point out that Lada and Markl do provide an example of an explicit formula for a class of weak maps, namely from a $L_m$-algebra (including $m=\infty$) to a dg-Lie algebra (i.e. an $L_\infty$-algebra with no brackets of arity higher than 2). This is Definition 5.2, and in Remark 5.3 they give their definition of weak map, merely as a map of dg coalgebras between the free coalgebras generated from the $L_\infty$-algebras, with the differential coming from extending the collection of brackets as a derivation. The Remark then claims that the collection of formulas in Definition 5.2 defines a weak map. What I would like is to know if this is indeed consistent. No derivation is given, and maybe it's obvious. However, the signs are the tricky part, and I have no idea how the particular combination of signs in Definition 5.2 are arrived at. REPLY [2 votes]: In the article Classification of 2-term $L_\infty$-algebras (arXiv:2109.10202), Kevin van Helden gives the definition of a morphism of 2-term $L_\infty$-algebras (Definition 2.3), and he was kind enough to share with me some private calculations that go through and checks the definition agrees with the one given in Lada and Markl's Remark 5.3.<|endoftext|> TITLE: Divisors whose restriction is big QUESTION [5 upvotes]: Let $f:X\rightarrow Y$ be a flat morphism of smooth projective varieties, and $\mathcal{L}$ an effective and ample line bundle on $Y$. For a divisor $A\in H^0(Y,\mathcal{L})$ set $X_A := f^{-1}(A)$. Let $D\subset X$ be a divisor such that $D_{|X_A}$ (the restriction of $D$ to $X_A$) is big for $A\in H^0(Y,\mathcal{L})$ general. Under these conditions, might $D$ be not pseudo-effective? REPLY [5 votes]: Consider a product $X = \mathbb{P}^n\times\mathbb{P}^1$, with projections $g:X\rightarrow\mathbb{P}^n$ and $f:X\rightarrow\mathbb{P}^1$. Set $H_1:= g^{*}\mathcal{O}_{\mathbb{P}^n}(1)$ and $H_2:= f^{*}\mathcal{O}_{\mathbb{P}^1}(1)$. The effective cone of $X$ is closed and generated by $H_1,H_2$. Now, take a divisor $D = aH_1+bH_2$ with $a > 0$ and $b < 0$. Then $D_{|f^{-1}(p)} = \mathcal{O}_{\mathbb{P}^n}(a)$, which is ample since $a > 0$, for all $p\in\mathbb{P}^1$. However, since $b < 0$ the divisor $D$ is not pseudo-effective.<|endoftext|> TITLE: Nonstandard proofs of the fundamental theorem of arithmetic QUESTION [13 upvotes]: Thirty or so years ago, someone showed me a clever proof of the Fundamental Theorem of Arithmetic that did not make use of the lemma "If $p\mid ab$ then $p\mid a$ or $p\mid b$". I'm unable to reconstruct the argument; all I remember is that it used induction and that it didn't generalize to other number rings. Can anyone provide such a proof, or provide other offbeat elementary proofs of unique factorization of natural numbers into primes? REPLY [21 votes]: To summarize the comments, this is also known as Zermelo's proof. A version can be found on wikipedia. I will give the proof here to avoid link rot. The proof is by contradiction. If FTA did not hold, then use the well ordering principle to select the smallest number $s$ which can be factored in two distinct ways into products of primes, $s = p_1p_2 \dots p_m = q_1q_2\dots q_n$. No $p$ can be equal to a $q$, for otherwise $s/p = s/q$ would give a smaller example, violating minimality. Assume without loss of generality that $p_1 < q_1$. Let $P = p_2 \dots p_m$ and $Q = q_2 \dots q_n$. Then $s = p_1P = q_1Q$, so that $$t=(q_1-p_1)Q = p_1(P-Q) < s$$ Since $t$ is less than $s$, then by minimality of $s$ there is only one way to factor it into a product of primes, namely the prime factorization of $q_1-p_1$ times the (known) prime factorization of $Q$. Since $p_1$ is a prime factor of $t$, $p_1$ must either be one of the prime factors of $q_1 - p_1$ or be one of the prime factors $q_i$ of $Q$. We already said that $p_1$ is not equal to any of the $q_i$, so $p_1$ is one of the prime factors of $q_1-p_1$. In particular, $p_1$ divides $q_1$. But $q_1$ is a prime not equal to $p_1$, a contradiction.<|endoftext|> TITLE: Subtraction-free identities that hold for rings but not for semirings? QUESTION [26 upvotes]: Here is a concrete, if seemingly unmotivated, aspect of the question I am interested in: Question 1. Let $a$ and $b$ be two elements of a (noncommutative) semiring $R$ such that $1+a^3$ and $1+b^3$ and $\left(1+b\right)\left(1+a\right)$ are invertible. Does it follow that $1+a$ and $1+b$ are invertible as well? The answer to this question is "yes" if $ab=ba$ (because in this case, $1+a$ is a left and right divisor of the invertible element $\left(1+b\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$). "yes" if $R$ is a ring (because in this case, $1+a$ is a left and right divisor of the invertible element $1+a^3 = \left(1+a\right)\left(1-a+a^2\right) = \left(1-a+a^2\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$). "yes" if $1+a$ is right-cancellable (because in this case, we can cancel $1+a$ from $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(\left(1+b\right)\left(1+a\right)\right) = 1+a$ to obtain $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = 1$, which shows that $1+a$ is invertible), and likewise if $1+b$ is left-cancellable. I am struggling to find semirings that are sufficiently perverse to satisfy none of these cases and yet have $\left(1+b\right)\left(1+a\right)$ invertible. (It is easy to find cases where $1+a^3$ is invertible but $1+a$ is not; e.g., take $a = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$ in the matrix semiring $\mathbb{N}^{2\times 2}$.) The real question I'm trying to answer is the following (some hopefully reasonably clear handwaving included): Question 2. Assume we are given an identity that involves only positive integers, addition, multiplication and taking reciprocals. For example, the identity can be $\left(a^{-1} + b^{-1}\right)^{-1} = a \left(a+b\right)^{-1} b$ or the positive Woodbury identity $\left(a+ucv\right)^{-1} + a^{-1}u \left(c^{-1} + va^{-1}u\right)^{-1} va^{-1} = a^{-1}$. Assume that this identity always holds when the variables are specialized to arbitrary elements of an arbitrary ring, assuming that all reciprocals appearing in it are well-defined. Is it then true that this identity also holds when the variables are specialized to arbitrary elements of an arbitrary semiring, assuming that all reciprocals appearing in it are well-defined? There is a natural case for "yes": After all, the same claim holds for commutative semirings, because in this case, it is possible to get rid of all reciprocals in the identity by bringing all fractions to a common denominator and then cross-multiplying with these denominators. However, this strategy doesn't work for noncommutative semirings (and even simple-looking equalities of the form $ab^{-1} = cd^{-1}$ cannot be brought to a reciprocal-free form, if I am not mistaken). Question 1 is the instance of Question 2 for the identity \begin{align} \left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1} \left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = \left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1} \end{align} (where, of course, the only purpose of the $\left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1}$ factors is to require the invertibility of $1+a^3$ and $1+b^3$). Indeed, if $\alpha$ and $\beta$ are two elements of a monoid such that $\beta\alpha$ is invertible, then we have the chain of equivalences \begin{align} \left(\alpha\text{ is invertible} \right) \iff \left(\beta\text{ is invertible} \right) \iff \left( \alpha \left(\beta\alpha\right)^{-1} \beta = 1 \right) \end{align} (easy exercise). REPLY [11 votes]: Tim Campion's idea works, though his example needs a little fixing. As in Tim's answer, we will find a rig with two elements $X$ and $Y$ such that $X+Y=1$ but $XY \neq YX$. Let $(M,+,0)$ be any commutative monoid. Let $R$ be the set of endomorphisms of $M$ obeying $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(0)=0$. Then $R$ is a rig, with $(\alpha+\beta)(x) = \alpha(x) + \beta(x)$ and $(\alpha \beta)(x) = \alpha(\beta(x))$. Let $M$ be $\{ 0,1,2 \}$ with $x+y \mathrel{:=} \max(x,y)$. Define \begin{gather*} \alpha(0) = 0,\ \alpha(1) = 0,\ \alpha(2) = 2 \\ \beta(0) = 0,\ \beta(1) = 1,\ \beta(2) = 1. \end{gather*} Then $\alpha+\beta=\mathrm{Id}$ but $\alpha \beta \neq \beta \alpha$.<|endoftext|> TITLE: Scalar curvature and the degree of symmetry QUESTION [10 upvotes]: Let $M$ be a closed connected smooth manifold. We define the degree of symmetry of $M$ by $N(M):=\sup_\limits{g}\mathrm{dim}\,\mathrm{Isom}(M,g)$, where $g$ is a smooth Riemannian metric on $M$ and $\mathrm{Isom}$ is the isometry group of the Riemannian manifold $(M,g)$. The torus $T^n$ does not admit a Riemannian metric with positive scalar curvature and has $N(T^n)\neq 0$. Is there a closed connected manifold $M$ with $N(M)=0$ such that $M$ admits a metric with positive scalar curvature? That is, does admitting a metric with positive scalar curvature imply its degree of symmetry is nonzero? REPLY [17 votes]: It seems that there are examples. By a theorem of Gromov and Lawson every simply connected manifold of dimension $n \geq 5$ which is not spin admits a metric of positive scalar curvature. There are many examples of simply connected, non-spin, closed $6$-manifolds which cannot admit a smooth circle action, constructed by Puppe. Theorem 7 of https://arxiv.org/pdf/math/0606714.pdf. Then, since the isomotetry group of a closed manifold is a compact Lie group, if $N(M)>0$ then taking a maximal torus gives a non-trivial circle action, which contradicts the above. So every metric has isometry group of dimension $0$. Edit: A specific example would be a quartic $3$-fold $X \subset \mathbb{CP}^4$. It admits a metric with positive Ricci curvature (since it is Fano), or alternatively since it is not spin we can apply Gromov-Lawson. It does not admit any smooth circle action due to a Theorem of Dessai and Wiemeler https://arxiv.org/pdf/1108.5327.pdf.<|endoftext|> TITLE: Count of non-trivial ergodic measures of a topological dynamical system QUESTION [7 upvotes]: Given a compact Hausdorff space $X$ and a continuous mapping $\varphi: X \to X$. We denote by $C(X)$ the space of continuous functions $f: X \to \mathbb{C}$. A probability measure $\mu$ on the Borel-$\sigma$-algebra of $X$ is said to be ergodic for $\varphi$ if it is $\varphi$-invariant, i.e., $$\int_X f \, d\mu = \int_X f \circ \varphi \, d\mu$$ for all $f \in C(X)$, and if all Borel sets $A \subseteq X$ with $\phi(A) \subseteq A$ satisfy $\mu(A) \in \{0,1\}$. It would be of interest to know how large the set of all ergodic measures $\mu$ that have a support $\mathrm{supp}(\mu)$ that contains at least 2 elements, let's denote it by $M^\mathrm{erg}_{\varphi, \geq 2}(X)$, can become. More precisely, is $M^\mathrm{erg}_{\varphi, \geq 2}(X)$ always countable? And if not in general, are there conditions that assure that $M^\mathrm{erg}_{\varphi, \geq 2}(X)$ is countable, e.g., $\varphi$ is a homeomorphism, $X$ is metrizable or both. REPLY [4 votes]: I know that you were mostly asking about cardinality, but more generally, you can ask about the structure of this space $\mathcal{M}(X,T)$ of invariant measures as a topological space with the weak topology (as Vaughn mentioned). It's known that $\mathcal{M}(X,T)$ must be a nonempty convex compact metrizable Choquet simplex, meaning that every point can be written as an "average" (integral) over extreme points. (This is because extreme points are precisely the ergodic measures, and the ergodic decomposition theorem states that every invariant measure is an "average" of ergodic measures). In fact EVERY possible nonempty compact metrizable Choquet simplex is realizable as $\mathcal{M}(X,T)$ for some topological dynamical system $(X,T)$. There are all kinds of crazy Choquet simplices (in fact one where the extreme points are dense, as Vaughn mentioned), so you can get many different structures for the ergodic measures. There are some hypotheses that guarantee that the set of ergodic measures is small. For instance, linear growth of the word complexity function implies this for subshifts. So does a uniform bound on the topological sequence entropy for all sequences. But the examples with huge $\mathcal{M}(X)$ are quite ubiquitous; for instance, there are several results showing that all possible Choquet simplices are realizable as spaces of invariant measures for topological systems from restricted classes (e.g. minimal systems, Toeplitz subshifts, logistic maps, etc.)<|endoftext|> TITLE: Is this definability principle consistent? QUESTION [11 upvotes]: (Below I'm thinking only about computably axiomatizable set theories extending $\mathsf{ZFC}$ which are arithmetically, or at least $\Sigma^0_1$-, sound.) Say that a theory $T$ is omniscient iff $T$ proves that the following holds: For every formula $\varphi(x,y)$ there is some formula $\psi(z,y)$ such that $T$ proves: "For all $y$, if $\varphi(-,y)^V=\varphi(-,y)^{V[G]}$ for every set generic extension $V[G]$, then $\psi(-,y)$ is a truth predicate for $\varphi(-,y)^V$." (That's not a typo, I do want a "$\vdash\vdash$"-situation. Note that by the soundness assumption on $T$, there really does exist such a $\psi$ for every $\varphi$.) Two points about omniscience are worth noting: For each $\varphi,\psi$, the statement in quotes is indeed expressible in the language of set theory - in particular, although we can't talk about the full theory of a generic extension via a single sentence, for each $\varphi,\psi$ we only need to talk about a bounded amount of those theories. So it does in fact make sense. $\mathsf{ZFC}$ itself is not omniscient - since $\mathsf{ZFC}$ proves that $L$ is forcing-invariant, any omniscient theory must prove that $V$ is not a set forcing extension of $L$. Informally speaking, any omniscient theory must prove that $V$ is "much bigger" than any canonical inner model. Omniscience strikes me as an implausibly strong property. However, I don't see an immediate reason why no consistent omniscient theory can exist. So my question is: Is there a consistent omniscient theory at all? REPLY [8 votes]: There is a consistent omniscient theory, at least assuming the consistency of a Woodin limit of Woodin cardinals. The Maximality Principle (MP) asserts that if a sentence is forceable in $V$, it is forceable in every generic extension of $V$. In other words, if a sentence can be forced to be indestructible by set forcing, then the sentence was true all along. Variants of the principle were discovered independently by many people, including Stavi, Väänänen, Bagaria, Chalons, and Hamkins. The main reference is Hamkins's paper. The Boldface MP (due to Hamkins) asserts the same but allowing hereditarily countable parameters. The Necessary MP (NMP, also due to Hamkins) asserts that the Boldface MP is true in all forcing extensions. MP and BMP are fairly weak, but Woodin [unpublished] showed the consistency strength of NMP lies between $\text{AD}$ and $\text{AD}_\mathbb R + \Theta \text{ is regular}$. I claim ZFC + NMP is an omniscient theory. Since $\text{AD}_\mathbb R + \Theta\text{ is regular}$ is consistent (by a theorem of Sargsyan its consistency follows from a Woodin limit of Woodin cardinals), so is ZFC + NMP. The proof is essentially due to Hamkins, who showed that under NMP, if $W$ is a forcing invariant inner model, then for all cardinals $\lambda$, $H(\lambda)\cap W\preceq W$. His proof adapts to the case that $W$ is an arbitrary class with a forcing invariant definition using a parameter $x$, although of course one must assume $\lambda$ is above the hereditary cardinality of $x$. This implies that the truth predicate for $W$ is definable: $W\vDash \varphi(\overline p)$ if and only if $W\cap H(\lambda) \vDash \varphi(\overline p)$ for some/all $\lambda > |\text{tc}(p)|$. I recite Hamkins's proof below since it's nice and I wanted to check that it works, but there are really no new ideas. We want to show that the truth predicate for $(W,\in)$ is definable from $x$. By homogeneity, it suffices to show it is definable from $x$ over $V[G]$ where $G$ is generic for $\text{Col}(\omega,\text{tc}(x))$. We may therefore pass to $V[G]$ and assume without loss of generality that $x$ is hereditarily countable. (This is ok because $V[G]$ is also a model of NMP.) Now fix a cardinal $\lambda$, and I will show $W\cap H(\lambda) \preceq W$. By induction on the complexity of $\varphi$, I'll show that $\varphi$ is absolute between $W\cap H(\lambda)$ and $W$. The only nontrivial step is to show that if $\varphi(\overline u) \equiv \exists t\, \psi(t,\overline u)$ and $\psi$ is absolute between $W\cap H(\lambda)$ and $W$, then $\varphi$ is absolute as well. So fix $\overline p\in W\cap H(\lambda)$ and suppose $W\vDash \varphi(\overline p)$. Let $H$ be generic for $\text{Col}(\text{tc}(\overline p))$. In $V[H]$, one can force so that the minimum hereditary cardinality of a set $z\in W$ such that $W\vDash \psi(z,\overline p)$ is at most $\aleph_0$. Once this is true, it is of course true in any outer model, so applying the Boldface MP in $V[H]$, $V[H]$ satisfies that the minimum hereditary cardinality of a set $z\in W$ such that $W\vDash \psi(z,\overline p)$ is at most $\aleph_0$. Fix such a set $z\in W$. Then $z\in V$ (since $W\subseteq V$) and the hereditary cardinality of $z$ is at $|\text{tc}(\overline p)| < \lambda$. Thus there is some $z\in W\cap H(\lambda)$ such that $W\vDash \psi(z,\overline p)$, and so by our induction hypothesis, $W\cap H(\lambda)\vDash \psi(z,\overline p)$, and hence $W\cap H(\lambda)\vDash \varphi(\overline p)$ as desired.<|endoftext|> TITLE: Explicit constant in Green/Tao's version of Freiman's Theorem? QUESTION [15 upvotes]: Green and Tao's version of Freiman's theorem over finite fields (doi:10.1017/S0963548309009821) is as follows: If $A$ is a set in $\mathbb{F}_2^n$ for which $|A+A| \leqslant K|A|$, then $A$ is contained in a subspace of size $2^{2K+O(\sqrt{K}\log(K))}|A|$. Does anybody know of a version of this bound with an explicit constant for the error term, or just an explicit exponent with more or less the same order of magnitude? For example, replacing the $O(\cdot)$ term by another multiple of $K$ might be sufficient for my purposes. (I would like to apply the bound to some specific sets $A$ and values of $n$. Roughly speaking, my $n$ range from about $50$ to $100$, and my $K$ range from about $50$ to $1000$, in case that helps.) There is an earlier result of Ruzsa and Green, in which the upper bound is given explicitly as $K^22^{2K^2-2}$, but this is far too large for what I need it for. I suppose one might be able to infer an explicit bound by reading Green and Tao's proof with enough care, but of course it would be easier not to have to do this. REPLY [16 votes]: Even-Zohar (On sums of generating sets in $\mathbb{Z}_2^n$, Combin. Probab. Comput. 21 (2012), no. 6, 916–941, available at https://arxiv.org/abs/1108.4902) has proved a completely explicit and sharp version of Frieman's theorem in $\mathbb{F}_2^n$ - see Theorem 2 of that paper. As a corollary, one gets that if $\lvert A+A\rvert\leq K\lvert A\rvert$ then $A$ is contained a subspace of size at most $$ (1+o(1))\frac{4^K}{2K}\lvert A\rvert.$$ (Precise bounds with no hidden constants are in the Theorem 2, but are quite cumbersome to write out here - but they are trivial to calculate for any specific values of $K$.)<|endoftext|> TITLE: Induced subgraphs of any given smaller chromatic number QUESTION [8 upvotes]: Let $G = (V,E)$ be a simple, undirected graph with $\chi(G)$ infinite. Given a cardinal $\kappa$ with $0 < \kappa < \chi(G)$, is there an induced subgraph $S$ of $G$ with $\chi(S) = \kappa$? What I tried: Let ${\cal S}$ be the collection of all subgraphs of $G$ colorable with $\kappa$ colors. I hoped to find a maximal element $M$ in ${\cal S}$ using Zorn's Lemma, establishing $\chi(M) = \kappa$. But this approach does not work. REPLY [6 votes]: As noted in a comment by Robert Furber, the original question about induced subgraphs is answered by the following simple counterexample due to F. Galvin, Chromatic Numbers of Subgraphs, Periodica Mathematica Hungarica 4 (1973), 117–119. Paraphrasing: Suppose there are cardinals $\lambda\lt\kappa$ such that $2^\lambda=2^\kappa$. Let $(S,\lt)$ be a totally ordered set of cardinality $|S|\gt2^\kappa$. Given any set $X\subseteq[S]^2$ we define a graph $G_X$ with vertex set $V(G_X)=S$ and edge set $E(G_X)=X$, and another graph $H_X$ with vertex set $V(H_X)=X$ and edge set $E(H_X)=\{\{x,y\},\{y,z\}\}:\{x,y\}\in X,\ \{y,z\}\in X,\ x\lt y\lt z\}$. Lemma. For any set $X\subseteq[S]^2$ and any infinite cardinal $m$, the graph $H_X$ is $m$-colorable if and only if the graph $G_X$ is $2^m$-colorable. (In particular, if $H_X$ is $\kappa$-colorable, then $H_X$ is also $\lambda$-colorable.) Proof. If $h$ is a proper $m$-coloring of $H_X$, then $x\mapsto\{h(\{x,y\}):\{x,y\}\in X,\ x\lt y\}$ is a proper $2^m$-coloring of $G_X$. For the other direction, let $M$ be a set of colors of cardinality $|M|=m$. Since $m\ge\aleph_0$, there is a family $\mathcal M$ of subsets of $M$ such that $|\mathcal M|=2^m$ and no element of $\mathcal M$ is a subset of another. If $G_X$ is $2^m$-colorable, then there is a proper coloring $g:S\to\mathcal M$. Now, for any $\{x,y\}\in X$, since $g(x)\not\subseteq g(y)$, we can choose an element $h(\{x,y\})\in g(x)\setminus g(y)$ and assign it as a color to $\{x,y\}$. In this way we get a proper $m$-coloring of $H_X$ Since $|S|\gt2^\kappa$, the graph $H_{[S]^2}$ is not $\kappa$-colorable. The induced subgraphs of $H_{[S]^2}$ are the graphs $H_X$ where $X\subseteq[S]^2$. By the lemma, any $\kappa$-colorable induced subgraph of $H_{[S]^2}$ is also $\lambda$-colorable, so its chromatic number cannot be exactly $\kappa$.<|endoftext|> TITLE: High-dimensional pointless varieties over finite fields QUESTION [16 upvotes]: Let $\mathbb{F}_q$ be a finite field. Do there exist smooth projective varieties over $\mathbb{F}_q$ of arbitrarily high dimension that have no $\mathbb{F}_q$-points and no non-constant maps to lower-dimensional varieties? Two lines of attack that don't work: Brauer-Severi varieties (the Brauer group of $\mathbb{F}_q$ is trivial) and varieties that become simple abelian over $\mathbb{F}_q$ (they have a rational point to begin with as mentioned by Wojowu). REPLY [10 votes]: Just take the Weil restriction with respect to $\mathbb{F}_{q^r} / \mathbb{F}_q$ of a pointless curve. This will typically admit no $\mathbb{F}_q$-morphism to a variety of lower dimension. REPLY [8 votes]: Not an answer, this is just a (cw) out-of-curiosity example of the recipe from Will Sawin's answer. The following polynomial defines a smooth projective surface without any $\mathbb F_3$-points: $$\scriptstyle -t^{40}-zt^{39}-x^3t^{37}-x^4t^{36}+y^4t^{36}-xy^3t^{36}-yz^3t^{36}-x^3yt^{36}-x^9t^{31}+y^9t^{31}-z^9t^{31}+x^{10}t^{30}-y^{10}t^{30}-xz^9t^{30}-yz^9t^{30}-x^9yt^{30}-y^{12}t^{28}-z^{12}t^{28}-x^3y^9t^{28}+x^3z^9t^{28}+x^9y^3t^{28}+x^9z^3t^{28}+x^{13}t^{27}-z^{13}t^{27}-xy^{12}t^{27}+yz^{12}t^{27}-x^3y^{10}t^{27}+x^4y^9t^{27}+y^4z^9t^{27}+xy^3z^9t^{27}-x^3yz^9t^{27}+x^9y^4t^{27}-x^9z^4t^{27}+y^9z^4t^{27}+x^{10}z^3t^{27}+y^{10}z^3t^{27}-xy^9z^3t^{27}+x^9yz^3t^{27}+x^{12}yt^{27}-x^{12}zt^{27}-y^{12}zt^{27}-x^9y^3zt^{27}-x^{27}t^{13}-y^{27}t^{13}-z^{27}t^{13}-xy^{27}t^{12}+xz^{27}t^{12}-yz^{27}t^{12}+y^{30}t^{10}+z^{30}t^{10}-x^3y^{27}t^{10}+y^3z^{27}t^{10}-y^{27}z^3t^{10}-x^{31}t^9-y^{31}t^9-z^{31}t^9-xy^{30}t^9+xz^{30}t^9+yz^{30}t^9+x^3y^{28}t^9-x^3z^{28}t^9-y^3z^{28}t^9+x^4z^{27}t^9+y^4z^{27}t^9+x^3yz^{27}t^9-x^{27}y^4t^9-x^{27}z^4t^9-y^{27}z^4t^9+xy^{27}z^3t^9-x^{30}yt^9-x^{30}zt^9+x^3y^{27}zt^9+x^{27}y^3zt^9-y^{36}t^4-x^9y^{27}t^4-x^9z^{27}t^4+y^9z^{27}t^4-y^{27}z^9t^4-x^{37}t^3+y^{37}t^3-z^{37}t^3-xy^{36}t^3+xz^{36}t^3+yz^{36}t^3-x^{10}z^{27}t^3-xy^9z^{27}t^3-x^{27}z^{10}t^3+y^{27}z^{10}t^3+x^{28}y^9t^3-x^{28}z^9t^3+y^{28}z^9t^3-x^{27}yz^9t^3+x^{36}yt^3-x^{36}zt^3-y^{36}zt^3-x^{27}y^9zt^3-x^{39}t-y^{39}t+z^{39}t-x^3y^{36}t+x^3z^{36}t-x^9z^{30}t-y^9z^{30}t-x^{12}z^{27}t+y^{12}z^{27}t-x^3y^9z^{27}t+x^9y^3z^{27}t-x^{27}y^{12}t-x^{27}z^{12}t+y^{30}z^9t-x^3y^{27}z^9t-x^{36}y^3t-x^{36}z^3t-y^{36}z^3t+x^{27}y^9z^3t+x^{40}-y^{40}+z^{40}-xy^{39}+yz^{39}-x^4z^{36}+xy^3z^{36}-x^9y^{31}-x^9z^{31}+y^9z^{31}+x^{10}z^{30}+xy^9z^{30}-x^{12}y^{28}+x^{12}z^{28}+y^{12}z^{28}-x^{13}y^{27}+y^{13}z^{27}-xy^{12}z^{27}-x^4y^9z^{27}-x^9y^4z^{27}-x^{10}y^3z^{27}+x^{12}yz^{27}-x^{27}y^{13}+x^{27}z^{13}+y^{27}z^{13}+xy^{27}z^{12}+x^{30}y^{10}-x^{30}z^{10}-y^{30}z^{10}-x^3y^{27}z^{10}+x^{27}y^3z^{10}-x^{31}y^9+x^{31}z^9+y^{31}z^9-xy^{30}z^9+x^3y^{28}z^9-x^4y^{27}z^9+x^{27}y^4z^9+x^{36}z^4-x^9y^{27}z^4+x^{27}y^9z^4-x^{37}y^3-x^{37}z^3+y^{37}z^3-xy^{36}z^3-x^{27}y^{10}z^3-x^{28}y^9z^3+x^{36}yz^3+x^{39}y+x^{39}z-y^{39}z+x^9y^{30}z-x^{12}y^{27}z-x^{30}y^9z+x^{36}y^3z $$ (this is the trace of $(x+y\alpha+z\alpha^2+t\alpha^3)^{40}\alpha$ where $\alpha$ is the generator of the multiplicative group of $\mathbb F_{81}$ and $x,y,z,t\in\mathbb F_3$). In fact, using that $x=x^3=x^5=...$ and $x^2=x^4=x^6=...$ for any $x\in\mathbb F_3$, one can turn the above into a much shorter example: $$ -t^4-t^3 x+t^3 y+t^2 x z-t^2 y^2-t x^3-t x^2 y+t x^2 z+t x y^2+t y^3-t y z^2+t z^3+x^4+x^3 y-x^3 z+x^2 z^2+x y^2 z-x z^3-y^4-y^3 z+z^4 $$ However the latter might be non-smooth, as Will Sawin points out in the comment.<|endoftext|> TITLE: A generalization of Rubio de Francia's inequality QUESTION [11 upvotes]: Suppose that $\{I_m\}$ is a sequence of pairwise disjoint intervals in $\mathbb{Z}$. The well known Rubio de Francia's inequality says that for any function $f\in L^p(\mathbb{T})$, $2\le p<\infty$, we have \begin{equation} \Big\| \Big( \sum_{m} |(\hat{f}\chi_{I_m})^{\vee}|^2 \Big)^{1/2} \Big\|_{L^p}\lesssim \|f\|_{L^p}. \end{equation} The constant which hides under the sign ''$\lesssim$'' depends only on $p$. Using duality, it is not difficult to see that this inequality is equivalent to the following: \begin{equation} \Big\| \sum_j f_j \Big\|_p\lesssim \Big\|\Big(\sum_j |f_j|^2\Big)^{1/2}\Big\|_p, \qquad 1 TITLE: Best way to introduce B-splines? QUESTION [5 upvotes]: I have the option of mentoring pure math undergrads in a topic lying within Approximation Theory and I really want to do $B$-splines. Mostly because I have recently found applications of them in my own research and I think it's a good opportunity for me to further learn the material. (And to show them cool stuff as well of course.) Suppose we are given a sequence of knots t $= (t_i)_{i \in \mathbb{Z}} \subset \mathbb{R}$. I am aware of two ways in which $B$-splines can be defined. Method 1: First define the $B$-splines of order $1$ (or degree $0$) to be the characteristic functions $B_{i1} = \chi_{[t_i,t_{i+1})}$. Then we define the $B$-splines of higher order by the recurrence relation $$B_{ik} = \lambda_{ik}B_{i,k-1} + (1 - \lambda_{i+1,k})B_{i+1,k-1}$$ where \begin{equation*} \lambda_{ik}(t) = \left\{ \begin{array}{ll} \frac{t - t_i}{t_{i+k-1} - t_i} & \quad \text{if} \ \ \ t_i \neq t_{i+k-1} \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation*} I understand that this is a computationally practical way of defining $B$-splines and that many of the early theorems about $B$-splines have simple proof when given this definition. However, I am of the opinion that you wouldn't introduce $B$-splines this way unless you really want to bore your audience as this recurrence relation is highly unmotivated and, until you begin to actually prove theorems with it, it simply doesn't look interesting. The next way is longer but the idea is more natural (at first at least). I don't want to make this post too long so I will skip details. I include some details for completion, but I suspect someone with an answer to this post is likely familiar with everything I mention below. Method 2: Suppose we are investigating the problem of finding a basis for the space of piece-wise polynomials of order $k$ (or degree $k-1$) with breakpoints at $(t_i)$ with some specified smoothness condition at each $t_i$. To find this basis we first make the problem easier by finding a basis for the space of piece-wise polynomials on $\mathbb{R}$ with a finite set of breakpoints and some specified smoothness condition at these breakpoints. If this finite knot sequence is $\{x_1, \dots x_n\} $, we get led to the truncated power basis $\{(t - x_i)_+^{j} | 1 \leq i \leq n , 0 \leq j \leq k-1\}$. We then find some linear combination of these truncated powers to begin constructing compactly supported piece-wise polynomials supported on closed intervals with end points belonging to our knot sequence t. (I have skipped many details here). But in order to actually define the $B$-splines, we need to find out the coefficients of the truncated powers that yield them. This takes us to the divided difference operators and I am not a fan of these operators either. I also find considering them to be somewhat unmotivated (albeit not as unmotivated as the first idea). My question: Are there other ways to introduce $B$-splines aside from the $2$ methods I have given? I suspect the solution to my dilemna is to understand these divided difference operators more in depth, but I want to know if there are other ways. I've been reading through a book on Box Splines which are defined as distributions. It seems interesting, but I've only begun and don't yet fully see how they generalize $B$-splines. Even if this approach would work as well, I am unsure whether it would be accessible to undergrads. REPLY [2 votes]: The undergrad course I took which included B-splines spent a lot of time first on Bézier curves. You might not necessarily want to spend much time on them, but I think they can motivate a variant on method 1 by defining Bézier curves with de Casteljau's algorithm and B-spline curves with de Boor's algorithm. This assumes that you're at least as interested in the curves as in the basis functions, but since you talk about applications... Note that I haven't tested this on anyone (undergrad or no) and the latter half has somewhat the flavour of experimental mathematics, which may not suit your pedagogical style. Basics We start from the concept of a spline curve as a curve defined by some control points and weighting functions, and we parameterise the weighting functions over an interval of the real line. So generically we have $P(t) = \sum_k w_k(t) P_t$ where $w_k$ is the $k$th weighting function and $P_k$ is the $k$th control point. One highly desirable property of the weighting functions is that for all $t$ in the interval on which the line is defined, $\sum_k w_k(t) = 1$, because this is necessary and sufficient for an affine transformation of the control points to transform the entire line consistently. Recursive interpolation A straightforward way to achieve this unit weight property is to choose weighting functions corresponding to recursive interpolation. In other words, the evaluation process is to define a series of sequences of control points starting with the original control points ($P_{0,k} = P_k$) and repeatedly interpolating between adjacent control points, reducing the number of points by one each time, until we get down to a single point. In general, $$P_{i,k} = (1 - f_{i,k}(t)) P_{i-1,k} + f_{i,k}(t) P_{i-1,k+1}$$ (This may be a good point at which to show that if the $f_{i,k}$ all map into $[0,1]$ then we get the convex hull property). Example 1: Bézier The simplest such scheme is to take $[0,1]$ as our interval of the real line and $f_{i,k}(t) = t$. Then by induction we can show that the corresponding weighting functions are $w_k(t) = \binom{n}{k} (1-t)^{n-k} t^k$ when we have $n+1$ control points numbered $P_0$ to $P_n$. This is the Bézier spline, which is very popular but has some important drawbacks. In particular, we can see that each weighting function has support over the entirety of $(0,1)$. We can attempt to fix this and achieve local control by chaining a series of Bézier curves, which we now treat as segments in a larger curve. For simple continuity we require the last control point of one segment to be the first control point of the next segment; if we want higher level continuity then we impose conditions on more control points. (There's an elegant umbral calculus approach to Bézier curves as $P(t) = (1 - t + t \mathcal{P})^n$ where $\mathcal{P}^k$ corresponds to $P_k$ which allows us to derive the condition for $c$th order continuity from segment $P$ to subsequent segment $Q$ as $$\forall j \le c: Q_j = 2^j \sum_{k=0}^j (-2)^k \binom{j}{k} P_{n-k}$$ so that if we want a degree of continuity at the joins which is half of the degree of the segments then moving a control point in one segment can have knock-on effects beyond the adjacent segment; but I suspect that this level of detail is too much of a digression for you. It's more straightforward to show that the $c$th derivative of segment $P$ at $P_n$ depends on the last $c+1$ control points in the segment, and similarly the $c$th derivative of segment $Q$ at $Q_0$ depends on the first $c+1$ control points, so that equating them necessarily establishes constraints between those control points). Example 2: cardinal B-spline A natural solution is to say that if we're going to have constraints between the control points which directly affect adjacent segments of our curve, we should take the simplest possible constraints: identification. Therefore we'll have a single set of control points $P_k$ and we'll say that for $t \in [a, a+1)$ we'll perform a recursive interpolation involving points $P_a$ to $P_{a+n}$. (This may not be the standard indexing, for which I apologise). Still in the interests of simplicity, we'll take the $f_{i,k}$ as linear functions: $f_{i,k}(t) = \alpha_{i,k} t + \beta_{i,k}$. And in the interests of uniformity we'll refine the interpolation to take into account that we're handling multiple segments, and update it to $$P_{i,k} = (1 - f_{i,k}(t-a)) P_{i-1,k} + f_{i,k}(t-a) P_{i-1,k+1}$$ Then the question is what values of $\alpha_{i,k}$ and $\beta_{i,k}$ maximise the continuity between the segments without imposing constraints between the $P_k$. In $[a, a+1)$ we have (introducing some hopefully transparent notation for brevity) $P_{0,k} = P_{a+k}$ for $0 \le k \le n$. $P_{1,k} = \overline{f_{1,k}} P_{a+k} + f_{1,k} P_{a+k+1}$ for $0 \le k \le n-1$. $P_{2,k} = \overline{f_{1,k}} \overline{f_{2,k}} P_{a+k} + (f_{1,k} \overline{f_{2,k}} + \overline{f_{1,k+1}} f_{2,k}) P_{a+k+1} + f_{1,k+1} f_{2,k} P_{a+k+2}$ for $0 \le k \le n-2$ etc. In the interests of having a digestible example, consider the case $n=2$. $P(t) = P_{n,0} = \overline{f_{1,0}} \overline{f_{2,0}} P_{a} + (f_{1,0} \overline{f_{2,0}} + \overline{f_{1,1}} f_{2,0}) P_{a+1} + f_{1,1} f_{2,0} P_{a+2}$ In the segment $[a, a+1)$, $$\lim_{t \to a+1} P(t) = \overline{f_{1,0}} \overline{f_{2,0}} (1) P_{a} + (f_{1,0} \overline{f_{2,0}} + \overline{f_{1,1}} f_{2,0}) (1) P_{a+1} + f_{1,1} f_{2,0} (1) P_{a+2}$$ and in the segment $[a+1, a+2)$ we have $$P(a+1) = \overline{f_{1,0}} \overline{f_{2,0}} (0) P_{a+1} + (f_{1,0} \overline{f_{2,0}} + \overline{f_{1,1}} f_{2,0}) (0) P_{a+2} + f_{1,1} f_{2,0} (0) P_{a+3}$$ So for simple continuity without imposing constraints on the control points we require $$\begin{eqnarray*} (1 - \alpha_{1,0} - \beta_{1,0})(1 - \alpha_{2,0} - \beta_{2,0}) &=& 0 \\ (\alpha_{1,0} + \beta_{1,0}) (1 - \alpha_{2,0} - \beta_{2,0}) + (1 - \alpha_{1,1} - \beta_{1,1}) (\alpha_{2,0} + \beta_{2,0}) &=& (1 - \beta_{1,0})(1 - \beta_{2,0}) \\ \alpha_{1,1} \alpha_{2,0} + \alpha_{1,1} \beta_{2,0} + \alpha_{2,0} \beta_{1,1} &=& \beta_{1,0} - \beta_{1,0} \beta_{2,0} + \beta_{2,0} \\ 0 &=& \beta_{1,1} \beta_{2,0} \end{eqnarray*}$$ This still leaves some degrees of freedom, so we can consider continuity of the first derivative, which (some algebra-bashing omitted) yields \begin{eqnarray*} \alpha_{1,0}(1 - 2 \alpha_{2,0} - \beta_{2,0}) + \alpha_{2,0} (1 - \beta_{1,0}) &=& 0 \\ \alpha_{1,0} (2 - 2 \alpha_{2,0} - 2 \beta_{2,0}) - \alpha_{1,1} (2 \alpha_{2,0} + \beta_{2,0}) + \alpha_{2,0} (2 - 2 \beta_{1,0} - \beta_{1,1}) &=& 0 \\ \alpha_{1,0} (1 - \beta_{2,0}) - \alpha_{1,1} (2 \alpha_{2,0} + 2 \beta_{2,0}) + \alpha_{2,0} (1 - \beta_{1,0} - 2 \beta_{1,1}) &=& 0 \\ \alpha_{1,1} \beta_{2,0} + \alpha_{2,0} \beta_{1,1} &=& 0 \end{eqnarray*} and combining the two sets of continuity constraints we get $\alpha_{1,0} = \tfrac12$, $\alpha_{1,1} = \tfrac12$, $\alpha_{2,0} = 1$, $\beta_{1,0} = \tfrac12$, $\beta_{1,1} = 0$, $\beta_{2,0} = 0$. If we now suggest considering $\alpha_{i,k} = \frac1{n+1-i}$, $\beta_{i,k} = \frac{n-k-i}{n+1-i}$ then it's no longer entirely unmotivated... The full B-spline Thus far we haven't mentioned knots, but there's no fundamental reason why we should segment the parameter space at the integers... Once the full generality has been introduced, and since we came via Bézier curves, it's probably worth throwing out a comment about the Bézier basis functions corresponding to B-spline basis functions with knot vectors $0^u 1^v$.<|endoftext|> TITLE: Weakened version of the Artin's primitive root conjecture QUESTION [16 upvotes]: $\DeclareMathOperator{\ord}{ord}$Artin's conjecture stipulates that $\ord_p(2) = p -1$ for infinitely many primes $p$, where $\ord_p(2)$ denotes the multiplicative order of $2$ modulo $p$. More generally one expects that $\ord_p(2)$ is often quite large. I'm looking for a weakened version of this, namely:    Does the sum $\displaystyle\sum_{p \leq x} \frac{1}{\ord_p(2)^2}$ converge as $x\to\infty$? I would prefer unconditional results, but results conditional on e.g. GRH are still welcome. REPLY [17 votes]: Since Joe Silverman raised the possibility of variations of the question, I want to point out (as a long comment) that even a mild variation $$ \sum_{p } \frac{1}{ \operatorname{ord}_p(2)^2 \log ( \operatorname{ord}_p(2))^\epsilon}$$ provably converges, since (following an argument given by Julian Rosen in the comments) $\operatorname{ord}_p(2)=d$ only if $p$ divides $2^d-1$, and the number of such is $$\omega(2^d-1) = O \left( \frac{\log(2^d-1)}{ \log (\log(2^d-1))} \right) = O \left( \frac{d}{ \log d} \right)$$ so $$ \sum_{p } \frac{1}{ \operatorname{ord}_p(2)^2 \log ( \operatorname{ord}_p(2))^\epsilon} \leq \sum_d \frac{1}{ d^2 \log(d)^\epsilon} O \left( \frac{d}{ \log d} \right) = O \left( \sum_d \frac{1}{ d( \log d)^{1+\epsilon}}\right) <\infty.$$ One can try to improve this argument by taking advantage of the fact that one $p$ can't divide $2^d-1$ for too many different $d$s, but I don't think you will solve it with such reasoning. A bad scenario you would need to rule out is that for each $d$ there are $ \sim c_1 d / \log d$ primes, all of size $\sim d^{c_2}$, that divide $2^{d}-1$, for some arbitrary constants $c_1,c_2$ with $c_2 > 3$ and $c_1 c_2 < \log 2 $. We want $c_2>3$ so the number of primes between $d^{c_2}$ and $(d+1)^{c_2}$ that are congruent to $1$ modulo $d$ is still at least $\sim d/\log d$, and we want $c_1 c_2 < \log 2$ so that the product of all these primes is still less than $2^{d}-1$. REPLY [14 votes]: Not quite what you've asked for, but in case it helps in whatever application you have in mind: $$ \sum_{p~\text{prime}} \frac{\log p}{p \operatorname{ord}_p(a)^\epsilon} \le \log\log a + \frac{2}{\epsilon} + C $$ for all $\epsilon>0$ and an absolute constant $C$. Here $a\in\mathbb Z$ is any integer with $|a|\ge2$. This will at least tell you that $\operatorname{ord}_p(a)$ cannot be too small, too often. The proof, which is fairly elementary, is in Variations on a theme of Romanoff, Internat. J. Math. 7 (1996), 373-391 [MR1395936].<|endoftext|> TITLE: Is there a algorithm to compute the Schur multiplier of a finite group from a group presentation QUESTION [8 upvotes]: Suppose we have a finite group $G$ whose presentation or Cayley table is given. Is there an algorithm (at least theoretically - without considering computational complexity) to compute the Cayley table or a presentation of the Schur multiplier? If possible please refer me to a paper which talks about the algorithm. REPLY [10 votes]: A reference is: D.F. Holt, The calculation of the Schur multiplier of a permutation group. In: Michael D. Atkinson, Edotor, Computational Group Theory (Conference proceedings, Durham, 1982), Academic Press, 1984, pages 307-319. But you might have difficulty finding it! The idea of this algorithm is to find the $p$-parts of the multiplier separately for the primes $p$ dividing $|G|$. To do that, we first find the multiplier $M(P)$ of a Sylow $p$-subgroup $P$ of $G$ using part of the $p$-quotient algorithm, and then find the $M(G)_p$ as the subgroup of $G$-stable elements of $G$. (I programmed this myself first in ALGOL60 and then in C in the early 1980s, partly motivated by the fact that there had been so many errors in the calculation of the multipliers of the finite simple groups - it had taken three attempts to get $M(M_{22})$ right!) There is a much simpler algorithm available in Magma as $\mathtt{Darstellungsgruppe}$ that takes as input a finite presentation $\langle X \mid R \rangle$ of the finite group $G$, and finds a presentation of a Schur-covering group $C(G)$ of $G$ by factoring out a free abelian subgroup of $R/[F(X),R]$ in $F(X)/[F(X),R]$ using the Hopf formula. The multipler can then be calculated as the kernel of the natural map $C(G) \to G$. This works OK but only for moderately small groups $G$. Here is an example with $G=A_5$. > G:=Group; > C,phi:=Darstellungsgruppe(G); > K:=Kernel(phi); > #K; 2<|endoftext|> TITLE: Can you solve the listed smallest open Diophantine equations? QUESTION [48 upvotes]: In 2018, Zidane asked What is the smallest unsolved Diophantine equation? The suggested way to measure size is substitute 2 instead of all variables, absolute values instead of all coefficients, and evaluate. For example, the size $H$ of the equation $y^2-x^3+3=0$ is $H=2^2+2^3+3=15$. Below we will investigate only the solvability question: does a given equation has any integer solutions or not? Selected trivial equations. The smallest equation is $0=0$ with $H=0$. If we ignore equations with no variables, the smallest equation is $x=0$ with $H=2$, while the smallest equations with no integer solutions are $x^2+1=0$ and $2x+1=0$ with $H=5$. These equations have no real solutions and no solutions modulo $2$, respectively. The smallest equation which has real solutions and solutions modulo every integer but still no integer solutions is $y(x^2+2)=1$ with $H=13$. Well-known equations. The smallest not completely trivial equation is $y^2=x^3-3$ with $H=15$. But this is an example of Mordell equation $y^2=x^3+k$ which has been solved for all small $k$, and there is a general algorithm which solves it for any $k$. Below we will ignore all equations which belong to a well-known family of effectively solvable equations. Selected solved equations. The smallest equation neither completely trivial nor well-known is $ y(x^2-y)=z^2+1$ with $H=17$. As noted by Victor Ostrik, it has no solutions because all odd prime factors of $z^2+1$ are $1$ modulo $4$. The smallest equation not solvable by this method is $ x^2 + y^2 - z^2 = xyz - 2 $ with $H=22$. This has been solved by Will Sawin and Fedor Petrov On Markoff-type diophantine equation by Vieta jumping technique. The smallest equation that required a new idea was $y(x^3-y)=z^2+2$ with $H=26$. This one was solved by Will Sawin and Servaes by rewriting it as $(2y - x^3)^2 + (2z)^2 = (x^2-2)(x^4 + 2 x^2 + 4)$, see this comment for details. Equation $ y^2-xyz+z^2=x^3-5 $ with $H=29$ has been solved in the arxiv preprint Fruit Diophantine Equation (arXiv:2108.02640) after being popularized in this blog post. Equation $ x(x^2+y^2+1)=z^3-z+1 $ with $H=29$ has solution $x=4280795$, $y=4360815$, $z=5427173$, found by Andrew Booker. This is the smallest equation for which the smallest known solution has $\min(|x|,|y|,|z|)>10^6$. Equation $ x^3 + y^3 + z^3 + xyz = 5 $ with $H=37$ has been listed here as the smallest open symmetric equation, but then I found solution $x=-3028982$, $y=-3786648$, $z=3480565$, see the answer for details how it was found. Smallest open equations. The current smallest open equation is $$ y(x^3-y)=z^3+3. $$ This equation has $H=31$, and is the only remaining open equation with $H\leq 31$. Also, the only open equations with $H \leq 32$ are this one and the two-variable ones listed below. One may also study equations of special type. For example, the current smallest open equations in two variables are $$ y^3+xy+x^4+4=0, $$ $$ y^3+xy+x^4+x+2=0, $$ $$ y^3+y=x^4+x+4 $$ and $$ y^3-y=x^4-2x-2 $$ with $H=32$. The current smallest open cubic equations are listed in a separate question here. The current smallest open symmetric equation is $$ x^3+x+y^3+y+z^3+z = x y z + 1 $$ with $H=39$, while the current smallest open 3-monomial equation is $$ x^3y^2 = z^3 + 6 $$ with $H=46$. The shortest open equations. I was told that it would be interesting to order equations by a more "natural" measure of size than $H$. Define the length of a polynomial $P$ consisting of $k$ monomials of degrees $d_1,\dots,d_k$ and integer coefficients $a_1,...,a_k$ as $l(P)=\sum_{i=1}^k\log_2|a_i|+\sum_{i=1}^k d_i$. This is an approximation for the number of symbols used to write down $P$ if we write the coefficients in binary, do not use the power symbol, and do not count the operations symbols. Note that $2^{l(P)}=\prod_{i=1}^k\left(a_i2^{d_i}\right)$ while $H(P)=\sum_{i=1}^k\left(a_i2^{d_i}\right)$. If we order equations by $l$ instead of $H$, then the current "shortest" open equations are $$ y(x^3-y) = z^4+1, $$ $$ 2 y^3 + x y + x^4 + 1 = 0 $$ and $$ x^3 y^2 = z^4+2 $$ of length $l=10$. For each of the listed equations, the question is whether they have any integer solutions, or at least a finite algorithm that can decide this in principle. The paper Diophantine equations: a systematic approach devoted to this project is available online: (arXiv:2108.08705). Paper last updated 13.04.2022. The plan is to list new smallest open equations once these ones are solved. The solved equations will be moved to the "solved" section. REPLY [15 votes]: The equation $$ x^3 + y^3 + z^3 + xyz = 5 $$ is solvable in integers. For example, take $$ x=-3028982, \quad y=-3786648, \quad z=3480565. $$ Verification is straightforward, but I would like to add more details how this solution has been found. If we just try $x$ and $y$ in order and then solve for $z$, then there are $10^{12}$ pairs $(x,y)$ even up to a million, hence finding the solution above was out of reach, at least for my computer. Instead, I have noticed that linear change of variables $$ y \to y - z, \quad x \to x + 3 y $$ reduces the equation to $$ -5 + x^3 + 9 x^2 y + 27 x y^2 + 28 y^3 + x y z - x z^2 = 0 $$ from which we can see that $28y^3-5$ is divisible by $x$. Hence, we may choose any $y$, then choose $x$ among the divisors of $28y^3-5$ and solve the resulting quadratic equation in $z$. This allows to find the above solution in just a few hours on standard PC.<|endoftext|> TITLE: Is Vopenka's Principle + "ORD has the tree property" consistent? QUESTION [10 upvotes]: Vopenka's principle implies the existence of weakly compact cardinals (a proper class of them, I believe). My question is whether Vopenka's principle is consistent with the assertion that the universe itself is weakly compact. Alternatively, can a Vopenka cardinal be weakly compact? There are several versions of this question, depending on how classes are treated in the formulation of the two statements. I'm hopeful that the answer is not too different for the different formulations. I suspect that VP does not imply that ORD is weakly compact. After all, VP implies that ORD is Woodin, and I've read that the first Woodin cardinal is not weakly compact. So if consistent, the conjunction VP + ORD is weakly compact might have higher consistency strength than either of its conjuncts. REPLY [11 votes]: Yes, a Vopenka cardinal can be weakly compact, at least assuming the consistency of a huge cardinal (though this is certainly a bit of an overkill). A huge cardinal is a weakly compact (in fact, measurable) Vopenka cardinal. EDIT: Actually almost huge cardinals suffice to get measurable Vopenka cardinals. Theorem 24.18 of Kanamori's "Higher Infinite" says that if $\kappa$ is almost huge, then there is a normal ultrafilter $U$ on $\kappa$ such that there are $U$-many $\alpha < \kappa$ such that $\alpha$ is a Vopenka cardinal. The argument actually shows that there are $U$-many $\alpha$ such that $\alpha$ is Vopenka AND measurable. To see why, note that part (b) of that theorem follows from the fact that if $j: V \to M$ witnesses the almost-hugeness of $\kappa$, then $M \models$ "$\kappa$ is Vopenka". But $M$ also models "$\kappa$ is measurable" (in fact $U$ itself is in $M$), because $\kappa$ is measurable in $V$ $j(\kappa)$ is inaccessible in $V$; so in particular all $\kappa$-complete ultrafilters on $\kappa$ are in $V_{j(\kappa)}$; and $V_{j(\kappa)} \subset M$ because $M$ is closed under $ TITLE: Explicit isomorphism between $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$? QUESTION [7 upvotes]: As Hilbert spaces, $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$ are isomorphic. Of course the isomoprhism is vastly not unique. I wonder if there are any particularly nice explicit isomorphisms. E.g. I wonder if there is an integral transform $$ f(x,y) \mapsto (K f)(z)=\int dx\, dy K(x,y,z) f(x,y) $$ with a nice explicit kernel $K(x,y,z)$ which maps $L^2(\mathbb{R}^2)$ isometrically onto $L^2(\mathbb{R})$? Any example would be appreciated. REPLY [4 votes]: The following result was obtained by an "explicit" construction in [1]. It is related to the comment of Terry Tao. A modification of the argument allows one to replace the cube by the whole space. Theorem. If $k\geq n$ and $1\leq p\leq \infty$, then there is an isometric isomorphism $\Phi: L^p([0,1]^k)\to L^p([0,1]^n)$ such that $\Phi(u)$ is continuous on $(0,1)^n$ for each $u\in L^p([0,1]^k)$ that is continuous on $(0,1)^k$. I do not know if the result is true for $k TITLE: Non-commutative complex geometry QUESTION [7 upvotes]: I was reading a physics paper where it was mentioned that the basic framework of Connes' differential non-commutative geometry (or actually, a slight modification of Connes in that paper) would need some extensions in order to arrive at a theory of non-commutative complex geometry, where one now works with complex manifolds. Are there specific technical obstacles which would make such an extension of non-commutative geometry particularly difficult or involved? What would be the main issues with trying to do this? REPLY [6 votes]: I don't think there are any obvious ``technical obstacles'' to extending noncommutative geometry to the a theory of noncommutative complex geometry. Instead I would say that there are a number of differing points of view on what form noncommutative complex geometry should take, and it's not completely clear how the different approaches relate to each other. Let's start with Connes' theory of noncommutative geometry, in particular spectral triples, which are usually thought of as ``noncommutative Riemannian manifolds''. A basic spectral triple is an unbounded representative of a $K$-homology class of a unital $C^*$-algebra. Such unbounded representatives exist for all classes of any unital $C^*$-algebra, so in particular, they exist for the $K$-homology classes of a compact Hausdorff space. Now in general a compact Hausdorff space will may not admit a differential structure, so in the commutative case a spectral triple is a more general structure than a smooth structure. To address this Connes introduced a number of higher axioms, and then proved his reconstruction theorem, showing that spectral triples satisfying these extra axioms are equivalent to compact Riemannian manifolds. The question of how well-suited these higher axioms to the noncommutative setting is the subject of debate . . . but that is a discussion for another day. So following this point of view, a "complex spectral triple" would be spectral triple (satisfying the higher axioms) plus "something extra". Connes' proposal was to look at positive Hochschild cocycles since in the classical case, for surfaces, such cocycles are equivalent to complex structures. This is explained in Section VI.2 of Connes' book. The motivating noncommutative example is, as usual, the noncommutative torus $\mathbb{T}_{\theta}^2$. This is a $\theta$-deformation of a classical complex manifold (in fact a Calabi--Yau manifold) and since its structure is quite close to the classical situation, a lot of the classical complex and Kähler geometry carries over. In another direction, there is the work of Fröhlich, Grandjean, and Recknagel:. They start with a spectral triple, and then try to build a noncommutative version of complex geometry on the associated differential graded algebra. Their approach takes globally defined classical identities in complex geometry and makes them into noncommutative axioms. For example, the Lefschetz identities and the Kähler identities play a major role in their work. Their main example is again the noncommutative torus. In a very different approach one should mention noncommutative projective algebraic geoemtry. This is an entirely algebraic approach to noncommutative geometry based on Serre's characterisation of the quasi-coherent sheaves over a projective variety. This approach has been hugely successful in recent years, see here for a nice introduction. A connection with the differential geometric approach to noncommutative complex geometry (analogous to the classical GAGA correspondence) has been postulated, see here for example. However, we are still very far from any kind of a noncommutative GAGA. A large and very important family of motivativing examples comes from Drinfeld--Jimbo quantum groups: the quantum flag manifolds. As shown in the seminal papers of Heckenberger and Kolb, the irreducible quantum flag manifolds admit a direct $q$-deformation of their classical de Rham complex. These complexes have a remarkable structure, $q$-deforming the classical Kähler geometry of the flag manifolds. These examples are crucial for understanding both the noncommutative geoemtry of the quantum groups, and as a forum for reconciling the various approaches to noncommutative complex geometry. Finally, we note the approach of Pirkovskii based on topological algebras. The motivating examples here are the quantum $n$-polydisk and the quantum $n$-ball.<|endoftext|> TITLE: Fields in monoidal categories QUESTION [7 upvotes]: We can speak of rings in monoidal categories, including also the non-Cartesian case. What about fields? Question 1: Definitions What are some possible notions of a (skew or commutative) field in a symmetric monoidal category $\mathcal{C}$? So far, I've found the following: Approach #1: Taking the point of view of fields as rings where nonzero elements have inverses, one could start by considering groups of units, as done in the nLab page on topological fields. Approach #2: Viewing fields representation-theoretically, we could define a field in $\mathcal{C}$ to be a ring object $k$ such that every $k$-module in $\mathcal{C}$ is (isomorphic to a) free one. Approach #3: The notion of an ideal makes sense in any "nice" monoidal category; see Section 4.2 of Martin brandenburg's PhD thesis. One could define a field in $\mathcal{C}$ as a ring object $k$ in $\mathcal{C}$ having only $k$ and $(0)$ as ideals. How do these approaches compare to each other? What are other possible definitions? Question 2: Examples Finally, what are some examples of fields in monoidal categories? In particular: When $\mathcal{C}=\mathsf{Top}$, the first approach above recovers topological fields (i.e. topological rings $k$ which are fields but also whose inverse map $a\mapsto a^{-1}$ defines a continuous map $(-)^{-1}\colon k^{\times}\to k^{\times}$). Do the other approaches recover this continuity condition too? For $\mathcal{C}=\mathsf{Sch}$, we have a well-studied notion of a ring scheme, of which a very important example is the ring scheme $\mathbb{W}$ of Witt vectors (for an introduction, see Eric's translation of Grothendieck's Groupes de Barsotti–Tate et Cristaux de Dieudonné). What are some examples of field schemes? For $\mathcal{C}=\mathsf{CCoAlg}_{R}$, rings in $\mathcal{C}$ give the notion of a Hopf ring. What are examples of Hopf fields? REPLY [2 votes]: Paul Balmer, Henning Krause, and Greg Stevenson have investigated the concept of 'field' in tensor-triangulated geometry (see Definition 1.1).<|endoftext|> TITLE: Which abelian groups are $\aleph_1$-filtered colimits of finitely-generated abelian groups? QUESTION [10 upvotes]: Observation: Every $\aleph_1$-directed colimit $\varinjlim_i X_i$ of finite sets is finite. Proof: Because the $X_i$'s are finite, the Mittag-Leffler condition holds, so by passing to the diagram of essential images, we may assume that the transition maps are injective. Therefore the cardinalities of the $X_i$ must be bounded (here is where we use $\aleph_1$-directednes), and by passing to a cofinal sequence we may assume that the cardinalities are constant. By the pigeonhole principle, all the transition maps are bijections. So the colimit is given by evaluation at any of its terms, and is finite. Question: Is every $\aleph_1$-directed colimit of finitely-generated abelian groups finitely-generated? How about not-necessarily-abelian groups? More generally, let $\mathcal C$ be a locally finitely-presentable category. Is every $\aleph_1$-directed colimit of finitely-presentable objects finitely-presentable? REPLY [2 votes]: Ycor answered the question, so I shall add a couple of remarks providing some insight about why these results should be true at all. Ind-completions are nested $$\text{Ind}(C) \supset \text{Ind}_{\aleph_1}(C) \supset \text{Ind}_{\aleph_2}(C) \supset \dots \supset \bigcap_{\lambda \in \text{RegCard}} \text{Ind}_{\lambda}(C).$$ Via the classical representation theorem $\text{Ind}_{\lambda}(C) \simeq \text{Flat}_\lambda(C^\circ,\text{Set}) \simeq \text{Cocont}\lambda\text{cont}(\text{Psh}(C^\circ),\text{Set})$, we see that their intersection actually coincides with the category of tiny objects in the presheaf category, that is it coincides with the Cauchy completetion $\hat{C}$ of $C$. So, it is indeed true that, in a sense, $\text{Ind}_\lambda(C)$ converges to $\hat{C}$ when $\lambda$ grows. Moreover, for a small category $C$, there exists a $\lambda$ such that this matrioska stabilizes. I think one can show that such $\lambda$ is $\kappa^{++}$, where $\kappa$ is the cardinality of $C$. As a result, for every small category, there must be a regular cardinal $\lambda$ such that $\text{Ind}_\lambda(C) \simeq \hat{C}$. Of course, if $C$ is already cauchy complete, we get $\text{Ind}_\lambda(C) \simeq \hat{C} \simeq C$. I do not have a precise reference for this, but I think some shadow of these ideas sit at the very core of a recent work by Tendas: On continuity of accessible functors. Also 2.6 in LPAC actually relies on this kind of ideas.<|endoftext|> TITLE: Ordinary abelian varieties and Frobenius eigenvalues QUESTION [5 upvotes]: Say $A_0$ is an ordinary abelian variety over ${\mathbf{F}}_q$. Call $\mathcal{A}$ the canonical lift of $A_0$ over $R := W({\mathbf{F}}_q)$. It carries a lift of the $q$-th power map on $A_0$. We call $\phi : \mathcal{A}\to\mathcal{A}$ this lift. It exists by functoriality of the canonical lift. Call $K = \text{Frac}(R)$ and choose a field embedding $K\subset\mathbf{C}$. Call $A$ the complex torus $(\mathcal{A}\times_K\mathbf{C})(\mathbf{C})$ and $F$ the endomorphism of $A$ induced by the Frobenius lift on $\mathcal{A}$, i.e. $$F = (\phi \times_{\text{Spec}(R)}\text{id}_{\mathbf{\text{Spec}(C)}})^{\rm an}:A\to A.$$ $F$ acts on the Betti cohomology $H^*(A,\mathbf{C})$. By the Weil conjectures (or by an argument of Serre, in this case), its eigenvalues are of the form $q^{*/2}\zeta$ for algebraic numbers $\zeta$ of complex absolute value $1$. Suppose $v \in H^{2m}(A,\mathbf{C})$ is an eigenvector of $F$ whose eigenvalue is of the form $q^m\zeta$ for $\zeta$ a root of unity. Is $v$ a class of type $(m,m)$? REPLY [4 votes]: Let's examine how $\phi$ acts on the algebraic Dolbeaut cohomology $$H^1(\mathcal A_K , \mathcal O_{\mathcal A})+ H^0 ( \mathcal A_K, \Omega^1_{\mathcal A}).$$ I claim its eigenvalues on $H^1(\mathcal A_K , \mathcal O_{\mathcal A})$ are units and its eigenvalues on $H^0 ( \mathcal A_K, \Omega^1_{\mathcal A})$ are $q$ times units. For the first claim, we can use the Artin-Schreier exact sequence $$H^1 ( A_{0, \overline{\mathbb F_q}}, \mathbb Z/p) \to H^1 ( A_{0, \overline{\mathbb F_q}} , \mathcal O_{ A_0} )\to H^1 ( A_{0, \overline{\mathbb F_q}} , \mathcal O_{ A_0} ) ,$$ the image of whose first arrow is a basis for $H^1 ( A_{0, \overline{\mathbb F_q}} , \mathcal O_{ A_0} )$ on which Frobenius acts invertibly, showing that Frobenius acts invertibly on $H^1 ( A_0, \mathcal O_{A_0})$ and thus invertibly on $H^1 (\mathcal A, \mathcal O_{\mathcal A})$. For the second claim, we can use the fact that the pullback of a polarization along $\phi$ is $q$ times that polarization, so Frobenius is a symplectic similitude with similitude character $q$ for the form on $H^1(\mathcal A_K)$ induced by that polarization, hence for each eigenvalue $\lambda$ that is a $p$-adic unit, $q/\lambda$ must also be an eigenvalue. Using that claim and the isomorphism $$\wedge ^a H^1(\mathcal A_K , \mathcal O_{\mathcal A})\otimes \wedge^b H^0 ( \mathcal A_K, \Omega^1_{\mathcal A}) \to H^a ( \mathcal A_K, \Omega^b_{\mathcal A})$$ we conclude that all eigenvalues of $\Phi$ on $$H^a ( \mathcal A_K, \Omega^b_{\mathcal A})$$ are $q^b$ times a $p$-adic unit. So any eigenvalues in degree $2m$ of the form $q^m$ times a unit must occur in $H^{m,m}$, as desired.<|endoftext|> TITLE: A question on an equivalence of RH QUESTION [8 upvotes]: In page 6, RH Equivalence 5.3. An equivalence of the Riemann Hypothesis says that $$\sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}= 2 + \gamma - \log 4\pi$$ where $\rho$ is over nontrivial zeros of the Riemann zeta function. It's not hard to see that RH implies $$\sum_{\rho} \frac{1}{|\rho|^2}= 2 + \gamma - \log 4\pi.$$ But conversly I don't see how the equality above implies RH and there is no reference in page 6, RH Equivalence 5.3. REPLY [12 votes]: Note that if $1/2< \sigma <1, t \in \mathbb R$ one has $\frac{2\sigma-1}{\sigma^2+t^2} < \frac{2\sigma-1}{(1-\sigma)^2+t^2}$. By a little manipulation, one gets: $\frac{2\sigma}{\sigma^2+t^2} + \frac{2(1-\sigma)}{(1-\sigma)^2+t^2} < \frac{1}{\sigma^2+t^2} + \frac{1}{(1-\sigma)^2+t^2} $ But if RH is false and there is $\rho=\sigma+it, 1/2<\sigma<1$, the above gives that $2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}}=2\Re{\frac{1}{\rho (1-\rho)}} < \frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$, so if we group together the four terms $\frac{1}{\rho (1-\rho)}, {\frac{1}{\bar \rho (1-\bar \rho)}}$ corresponding to the four roots $\rho, 1-\rho, \bar \rho, 1-\bar \rho$ we get that their sum is srictly less than the sum of the corresponding reciprocal of the respective four roots square modulus, so RH false implies $\sum_{\rho} \frac{1}{|\rho|^2} >\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ and the equivalence is established Edit later - per comments - note that if $\Re \rho =1/2$ then $\bar \rho=1-\rho$ so roots group naturally in pairs only and $\frac{1}{\rho (1-\rho)}=\frac{1}{|\rho|^2}$ so the corresponding (two) terms on both sides of the equality $\sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ are equal When there is a root with $\Re \rho_0 >1/2, t>0$ the roots group into four as noted $\rho_0, \bar \rho_0, 1-\rho_0, 1-\bar \rho_0$ and now the corresponding terms in $\sum_{\rho} \frac{1}{|\rho|^2}$ are $2(\frac{1}{|\rho_0|^2}+\frac{1}{|1-\rho_0|^2})$, while the terms in the sum $\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ are also four and since they are conjugate in pairs add to $2(2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}})=4\Re{\frac{1}{\rho (1-\rho)}}$ and the inequality above applies<|endoftext|> TITLE: In search of a $q$-analogue of a Catalan identity QUESTION [10 upvotes]: Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How): \begin{equation} \label1 \sum_{k=0}^n\binom{2n-2k}{n-k}C_k=\binom{2n+1}n \qquad \iff \qquad \sum_{i+j=n}\binom{2i}iC_j=\binom{2n+1}n. \tag1 \end{equation} I like to ask QUESTION. Is there a $q$-analogue of \eqref{1}? Possibly, a combinatorial proof of \eqref{1} would shed some light into this. REPLY [9 votes]: This identity is known as Jonah's formula (special case with $n\rightarrow 2n$ and $r\rightarrow n$, see "Catalan Numbers with Applications" by Thomas Koshy, pg. 325-326 for a combinatorial proof) $$\sum_{k=0}^r\binom{n-2k}{r-k}C_k=\binom{n+1}r$$ and a $q$-analogue was obtained by Andrews in "$q$-Catalan identities" in the book "The legacy of Alladi Ramakrishnan in the Mathematical Sciences". It's Theorem 3, pg. 186. $$\frac{(1+q^{n-r+1})}{(1+q^{r+1})}\left[ {\begin{array}{c}n+1\\r\end{array} } \right]_{q^2}=-(-q\;;q)_{n+1}\sum_{k=0}^r\left[ {\begin{array}{c}n-2k\\r-k\end{array} } \right]_{q^2}\frac{\textrm{C}_{k+1}(-1;q)}{(-q\;;q)_{n-2k}}q^{-k-1}$$ where $\textrm{C}_n(\lambda,q)$ is a $q$-analogue of the Catalan numbers considered also by Andrews here. $$\textrm{C}_n(\lambda,q)=\frac{q^{2n}(-\lambda/q; q^{2})_{n}}{(q^2;q^2)_{n}}$$ In the paper, he says that the general strategy is to go from a binomial coefficient identity to a generalized hypergeometric identity, and then we can look for a $q$-analogue of the latter. In this case, he used the Pfaff-Saalschütz summation formula and then he searched for a $q$-analogue of this one with the help of Bailey's and Gasper and Rahman's books. I can't help much more, I'm not familiar with these kind of hypergeometric identities. If $n\rightarrow 2n$ and $r\rightarrow n$, the limit $q\rightarrow 1$ recovers the identity (1).<|endoftext|> TITLE: A family of Diophantine equations with no integer solutions but solutions modulo every integer QUESTION [7 upvotes]: Selmer's curve is the equation $3x^3 +4y^3 +5z^3=0$. This equation is famous for having non-trivial solutions in every completion of $\mathbb{Q}$ but only having the trivial solution in the rationals. This curve has been discussed on Mathoverflow before such as here and here. A nice proof that the curve always have solutions for all $p$-adics is in this writeup by Kevin Buzzard. I have two questions related to this curve. Question I: How much worse can we get for cubics if the number of variable is increased. That is: For every $n \geq 3$ is there a list of non-zero integers $a_1, a_2 \cdots a_n$ such that the equation $$a_1x_1^3 +a_2x_2^3 \cdots a_n x_n^3 =0$$ has solutions in every completion of $\mathbb{Q}$ but no non-trivial integer solutions? Question II: can we make a family of such equations which is nested? That is is there a sequence of non-zero integers $a_1, a_2, a_3 \cdots $ such that for any $n \geq 3$ the equation $$a_1x_1^3 +a_2x_2^3 \cdots a_n x_n^3 =0$$ has solutions in every completion of $\mathbb{Q}$ but no non-trivial integer solutions? And if so, can we take $a_1=3$, $a_2=4$ and $a_3=5$ (that is using Selmer's curve as the start of our family). REPLY [18 votes]: The answer is no. Heath-Brown has shown every cubic form over the integers in at least 14 variables represents zero nontrivially. The Wikipedia article on Hasse principle contains references as well as related results for forms in higher odd degrees.<|endoftext|> TITLE: Is "almost-solvability" of Diophantine equations decidable? QUESTION [20 upvotes]: Say that a Diophantine equation is almost-satisfiable iff for each $n\in\mathbb{N}$ it has a solution mod $n$. Trivially genuine satisfiability over $\mathbb{N}$ implies almost-satisfiability, but the converse fails - see the discussion here, or for a fun "nuke" note that almost-satisfiability is $\Pi^0_1$ and so cannot coincide with satisfiability as the latter is properly $\Sigma^0_1$ (and as far as I can tell that's actually non-circular! :P). My question is the following: is almost-satisfiability known to be decidable? It's plausible to me that one could whip up a Diophantine equation $\mathcal{D}_T$ such that the behavior of a given Turing machine $T$ over the first $s$ steps is connected to the behavior of $\mathcal{D}_T$ over something like $\mathbb{Z}/s\mathbb{Z}$ (sort of a "Diophantine Trakhtenbrot theorem"), but I don't actually see how to do that. Certainly I don't see how to lift any of the MRDP analysis to almost-satisfiability in a useful way. On the other hand, I also don't see how to get a $\Sigma^0_1$ definition of almost-satisfiability. Work of Berend/Bilu shows that almost-satisfiability of single-variable Diophantine equations is decidable, which is nontrivial (in contrast to genuine solvability for single-variable equations which is a trivial application of the rational roots theorem), but at a glance I don't see how to generalize their arguments to multiple variables. REPLY [27 votes]: A Diophantine equation is almost-satisfiable iff it is satisfiable over the ring $\widehat{\mathbb Z}$, the profinite completion of $\mathbb Z$ (also called by some the Prüfer ring), by a standard compactness argument (the solutions over each $\mathbb Z/n\mathbb Z$ form an inverse system of finite sets whose inverse limit is the set of solutions over $\widehat{\mathbb Z}$, and hence if the set of solutions is non-empty for each $n$, then the inverse limit is non-empty). From the direct product decomposition $\widehat{\mathbb Z}=\prod_p \mathbb Z_p$, this in turn reduces things to solvability over the $p$-adic integers $\mathbb Z_p$ for all $p$. This question is solved in the paper J. Ax, Solving diophantine problems modulo every prime. Ann. Math. 85, 161–183 (1967) on pages 170,171. So the problem is decidable.<|endoftext|> TITLE: An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$ QUESTION [8 upvotes]: Can you prove or disprove the following claim: Claim: $$\frac{\sqrt{3} \pi}{24}=\displaystyle\sum_{n=0}^{\infty}\frac{1}{(6n+1)(6n+5)}$$ The SageMath cell that demonstrates this claim can be found here. REPLY [12 votes]: Let the sum be $S$. First of all, it is easy to see that $$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$ This is true because $f(z)=f(-1-z)$. Then by the summation formula (see any undergraduate Complex Variables textbook) $$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a \left(f(z)\pi\cot\pi z\right),$$ where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$. Computing the residues (each of them equals $-\pi\sqrt{3}/24$) we obtain the result.<|endoftext|> TITLE: What do nearby/vanishing cycles have to do with Fourier transforms? QUESTION [6 upvotes]: Let $E$ be a vector bundle on some smooth algebraic variety and $E^*$ its dual. Suppose $A$ is a sheaf (constructible or a $D$-module) on $E$. Given a linear function $f$ on $E$, we may compute the stalk at $f$ of the Fourier transform of $A$. Now I’ve heard a slogan along the lines of “the stalk of the Fourier transform of $A$ is the vanishing cycles of $A$ along $f$.” Obviously this doesn’t really make sense; the stalk is a sheaf on the base and the vanishing cycles a sheaf on $f^{-1}(0)$. Nevertheless, my guess is that somehow this idea can be made precise. So, is there a precise statement relating nearby and vanishing cycles to Fourier transforms of sheaves? REPLY [4 votes]: You probably want to say $A$ is a constructible sheaf / $D$-module with regular singularities. $D$-modules with irregular singularities (for example, those created by the Fourier transform) will behave very differently, even if $E$ is a vector bundle of rank $1$ on a point. I'm going to pretend you asked the question for $\ell$-adic sheaves and describe the relevant phenomena there. I think the situation for $D$-modules will be analogous. Let's start with the Fourier transform on $\mathbb A^1$. For an $\ell$-adic sheaf $A$ with tame ramification (analogue of regular singularities), the correct statement is: The nearby cycles of the Fourier transform of $A$ at $\infty$ is isomorphic to a sum over points of $\mathbb A^1$ of the vanishing cycles of $A$ at that point. Furthermore, the Fourier transform is locally constant away from $0$, so its stalk at any point is non-canonically isomorphic to the nearby cycles of the Fourier transform at $\infty$ and thus non-canonically isomorphic to this sum. Now let's consider the Fourier transform on $\mathbb A^n$, keeping the base still a point. The stalk of the Fourier transform at $\lambda f$ for a scalar $\lambda$ is going to be the stalk of the Fourier transform of $R f_* A$ at $\lambda$. Thus it will be non-canonically isomorphic to the sum of the stalks of the vanishing cycles of $R f_* A$ at each point of $\mathbb A^1$. If $f$ were proper, it would then by isomorphic to the sum over $a$ of the cohomology of $f^{-1}(a)$ with coefficients in the vanishing cycles complex of $A$ along $f$. Maybe we can say that this is morally true in general. So we need to take $f^{-1}(a)$ for all $a$, not just $f^{-1}(0)$, and take the cohomology, to get the desired isomorphism. If the vanishing cycles are supported at finitely many points, this would just be the sum of the vanishing cycles at those points. Maybe that's what people who told you this were thinking of? However, while this may be morally true, it will not be literally true in every case. The simplest counterexample I can think of is on $\mathbb A^n$, take $A$ to be the constant sheaf on the curve $xy=1$ and $f$ to be $y$. Then $A$ has no vanishing cycles on any point of $\mathbb A^2$ (none on $f^{-1}(0)$ since $A$ vanishes in a neighborhood of $f^{-1}(0)$, and none elsewhere because $A$ in each other fiber looks identical to the generic fiber) but its Fourier transform has a stalk of rank one at $y$. Over a nontrivial base, I guess you want to take the vanishing cycles sheaf of $A$ along $f$ on $f^{-1}(a)$, push forward to the base, and then sum over $a$. This will be even more morally and less factually correct, but maybe there is some genericity condition under which it is literally true.<|endoftext|> TITLE: Is $\operatorname{PSL}(2,q)$ the most quasirandom group? QUESTION [9 upvotes]: Is the following statement true? Every finite group $G$ has a non-trivial irreducible representation of dimension $O(\lvert G\rvert^{1/3})$. Context: Groups with no small irreducible representations are called quasirandom in Gowers - Quasirandom groups as these groups always have good mixing properties. The standard example of a quasirandom group used everywhere is $\operatorname{PSL}(2,q)$, which has $\sim q^3/2$ elements and no irreducible representation of dimension less than $(q-1)/2$. Is $\operatorname{PSL}(2,q)$ known to be quantitatively the best example? REPLY [12 votes]: The answer is likely yes. Here is a partial proof: Every finite group has a maximal proper normal subgroup, and the quotient by this gives a surjection to a finite simple group. Any nontrivial irreducible representation of that finite simple quotient gives a nontrivial irreducible representation of the original group. So it suffices to prove such a bound for finite simple groups. Next it is natural to invoke the classification of finite simple groups. Cyclic and alternating groups are obviously fine, having irreps of much lower dimension. We can absorb the sporadics into the big $O$. So we really only have to consider finite simple groups of Lie type. Let's consider for simplicity the finite simple groups of Lie type arising from split algebraic groups $G$ over finite fields $\mathbb F_q$. These have dimension roughly $q^{ \dim G}$ and we can construct a nontrivial permutation representation of size roughly $q^n$ where $n$ is the dimension of any flag variety of $G$. So it suffices to find such a flag variety where $\frac{n}{ \dim G} \leq \frac{1}{3}$. For $G = SL_n$, of dimension $n^2-1$, projective space is a flag variety of dimension $n-1$, for a ratio of $\frac{1}{n+1} \leq \frac{1}{3}$. For $G = Sp_{2n}$, of dimension $n (2n+1)$, projective space is a flag variety of dimension $2n-1$, for a ratio of $\frac{2n-1}{ n (2n+1)} \leq \frac{1}{ n+1} \leq \frac{1}{3}$ since $\frac{2n-1}{2n+1} \leq \frac{2n-2}{2n}$. For $G= SO_n$, of dimension $n (n-1)/2$, the set of rank one maximal isotropic subspaces is a flag variety of dimension $n-2$, for a ration of $\frac{2n-4}{n (n-1)} = \frac{1}{3} - \frac{(n-3)(n-4)}{ 3 n(n-1)} \leq \frac{1}{3}$ For $G = G_2, F_4, E_6, E_7, E_8$, the dimension is $14, 52, 78, 133, 248$ and they have (lowest-dimensional) flag varieties of dimension $5, 15, 16, 33, 78 $. So each one satisfies the inequality except $G_2$. Unless I have misread the tables from Character Degrees and their Multiplicities for some Groups of Lie Type of Rank < 9, regardless of the congruence class of $q$ mod $6$, $G_2(q)$ has an irreducible representation of degree $O(q^3)$ or $O(q^4)$ and thus does satisfy this inequality. (In fact, there are always exactly one or two irreducible representations in that range.) So one is reduced to considering the various non-split groups...<|endoftext|> TITLE: Kernel of a matrix and the Catalan numbers QUESTION [11 upvotes]: Let $B_n$ denote the Boolean lattice of a set with $n \geq 2$ elements and $C_n$ the matrix with entries $c_{i,j}=1$ if $i \leq j$ and $c_{i,j}=0$ else, where $i,j\in B_n$. Let $M_n:=C_n+C_n^T$ (this is also the Cartan matrix of a certain Frobenius algebra associated to $B_n$), which is a symmetric matrix. Thus geometric and algebraic multiplicity should coincide. Let $I$ denote the identity matrix. Question: Is there a bijective proof that for $n$ even (odd) we have that the basis of the kernel of $M_n - 2 I$ ($M_n-3 I$) is enumerated by the Catalan numbers? This is true for $n \leq 15$. (see also Factorisation of a polynomial from the Boolean algebra ) Here a bijective proof asks for a bijection of a basis of the kernel to known combinatorial objects that are enumerated by the Catalan numbers. For example for $n=6$, the kernel has dimension 5 and basis vectors are given by (0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 1, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, -1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0) I was not able to see a pattern so far for a nice basis of the kernel as the vectors are so big for larger n. For example for $n=2$, $C_2$ is given by $\begin{bmatrix} 1 & 1 & 1 &1 \\ 0 & 1 & 0 &1\\ 0 & 0 & 1 &1\\ 0 & 0 & 0 &1\\ \end{bmatrix}$ and $M_2$ is given by $\begin{bmatrix} 2 & 1 & 1 &1 \\ 1 & 2 & 0 &1\\ 1 & 0 & 2 &1\\ 1 & 1 & 1 &2\\ \end{bmatrix}$. REPLY [6 votes]: Another suggestion: let $A_{2n}=M_{2n}-2I$ ($I$=identity matrix), so we are interested in $\ker A_{2n}$. Let $V$ be the vector space on which $A_{2n}$ acts, so we can regard $V$ as having a basis consisting of all subsets of $[2n]=\{1,2,\dots,2n\}$. Thus $V$ has a grading $V_0\oplus V_1\oplus\cdots \oplus V_{2n}$, where $V_i$ has a basis consisting of all $i$-element subsets of $[2n]$. The symmetric group $\mathfrak{S}_{2n}$ acts on $V$ and preserves the grading. Write $M_\lambda$ for the irreducible $\mathfrak{S}_{2n}$-module indexed by the partition $\lambda$ of $2n$. Then (as is well-known) $V_i$ decomposes as $M_{(2n)}\oplus M_{(2n-1,1)} \oplus \cdots\oplus M_{(2n-i,i)}$ for $0\leq i\leq n$, and $V_i\cong V_{2n-i}$ (as $\mathfrak{S}_{2n}$-modules). Assuming that $\dim\ker A_{2n}=C_n$, it is natural to conjecture that as an $\mathfrak{S}_{2n}$-module, $\ker A_{2n}$ is isomorphic to the irreducible module $M_{(n,n)}$. The operator $\frac{1}{(2n)!}\sum_{w\in\mathfrak{S}_{2n}}\chi^{(n,n)}(w)w$ will then project $V_n$ to $\ker A_{2n}$. Here $\chi^{(n,n)}$ is the irreducible character of $\mathfrak{S}_{2n}$ indexed by the partition $(n,n)$. There is something similar for $M_{2n+1}$.<|endoftext|> TITLE: Fibonacci embedded in Catalan? QUESTION [7 upvotes]: Given a partition $\lambda$ and its Young diagram $\pmb{Y}_{\lambda}$, we say $\lambda$ is a $(t,s)$-core partition provided that neither $t$ nor $s$ is a hook length in $\pmb{Y}_{\lambda}$. We now recall a conjecture (now a theorem) of D. Armstrong which states "the total number $(s,t)$-core partitions is $\frac1{s+t}\binom{s+t}s$". Focusing on the special case of $(s,s+1)$-core partitions, Stanley and Zanello proved the enumeration $C_s=\frac1{2s+1}\binom{2s+1}s$. Their argument utilizes a result of J. Anderson: $(s,s+1)$-cores correspond bijectively to the order ideals of the poset $P(s,s+1)$ (that is, positive integers that are not contained in the numerical semigroup generated by $s$ and $s+1$) by associating the lower ideal $\{a_1,\dots,a_j\}$, where $a_1>\cdots>a_j$, to the $(s,s+1)$-core partition $(a_1-(j-1), a_2-(j-2),\dots,a_{j-1}-1,a_j)$. A few years ago, I considered core-partitions with distinct parts (see page 13) to the effect that the number of such $(s,s+1)$-core partitions equals the Fibonacci number $F_{s+1}$. This conjecture is now a theorem due to different people (see this paper and references therein). QUESTION. Viewing $(s,s+1)$-cores of distinct partitions as embedded in the ordered ideals of the poset $P(s,s+1)$ (the above Catalan case), do they have any "interesting" structure? REPLY [5 votes]: The order ideal corresponding to a core partition with distinct parts cannot contain an element $x \geq s+2$, otherwise it must also contain $x-s, x-s-1$, resulting in two equal parts. For the same reason, it also can't contain two consecutive elements in $\{1,2,\dots, s-1\}$. Therefore the poset in questions is simply the poset of subsets of $\{1,2,\dots, s-1\}$ that do not contain consecutive elements, ordered by inclusion. The underlying Hasse diagram of this poset is called the Fibonacci cube.<|endoftext|> TITLE: Residue field of point on an algebraic stack QUESTION [10 upvotes]: $\DeclareMathOperator{\Spec}{Spec}$ Let $X$ be an algebraic stack. Is there is a well-defined notion of the residue field of a point $x \in |X|$? Attempts: Recall that a point on a stack is an equivalence class of morphisms $\Spec k \to X$ from fields $k$. The issue is that it is not clear that there is a minimal choice of $k$ to warrant being called the residue field. There is also the notion of a residual gerbe on a stack; but again it is not clear whether this comes with some kind of canonical field of definition which is compatible with 1. If $X$ has a coarse moduli space $X^c$, then one could define the residue field of $x$ to be the residue field of the image of $x$ in $X^c$. This is well-defined, but seems to lose some of the subtle properties of stacks and again it's not clear whether it is compatible with 1. and 2. I'm happy to assume my stack is sufficiently nice (e.g. smooth, DM,..) REPLY [5 votes]: By definition, a residue field is an equivalence class of morphisms $\operatorname{Spec} k \to X$, i.e. of pairs of a field $k$ and an object in $X(k)$ We can upgrade that equivalence class into a category: Given fields $k$, $L$ and objects $a \in X(k) , b\in X(L)$, a morphism is a map $s \colon k \to L $ together with an isomorphism $s^* a \to b$. The key property that the residue field $F$ should have is that for every $k$-point of $X$ we obtain a map $F \to k$. I claim we should define the residue field as the universal object with this property. In other words, an element of the residue field is an assignment to each pair $k, a \in X(k)$ an element $\alpha_a \in k$, compatible in the obvious way with morphisms: For $s \colon (k, a) \to (L,b)$ a morphisms, we have $s(\alpha_a) = \alpha_b$. Elements of the residue field form a ring as there is an obvious notion of addition and multiplication. To check that they form a field, we need to check that every element is either invertible on every $k$-point or zero on every $k$-point, but this follows from the fact that we are working with a single equivalence class. So we indeed have a universal notion of the residue field.<|endoftext|> TITLE: Minimal set of assumptions for set theory in order to do basic category theory QUESTION [15 upvotes]: Consider a normal first course on category theory (say up to and including the statement and proof) of the adjoint functor theorem (AFT). What are the minimal assumptions for the definition of a set one needs to make in order that everything works? As far as I understand, up to and including the AFT there is very little one needs besides that fact that sets should have elements and that we have avoided Russell's paradox. So what is a minimal set of axioms allowing this to work? REPLY [17 votes]: To complement Tom Leinster's answer, let me try to be specific: To form the product category $\mathcal{C} \times \mathcal{D}$, we need ordered pairs, which we can get from the axiom of unordered pairs. It's probably a good idea to have the empty set $\emptyset$, so that the initial category exists. My experience from type theory leads me to believe that we want function extensionality, or else we cannot reasonably work with functors and natural transformations (which are functions). Function extensionalty is equivalent to set-theoretic extensionalty. To form the hom-set $\mathrm{Hom}(A,B) = \{f \in \mathcal{C}_1 \mid \mathrm{dom}(f) = A \land \mathrm{cod}(f) = B\}$ we seemingly need bounded separation. It's a little more difficult to see whether we need unbounded separation (my guess would be that we can work pretty nicely without it). To form functor categories, we need powersets. Indeed, given any set $A$, its powerset may be generated as the set of objects of the functor category $2^A$, where $2$ is the discrete category on two objects. There are two functors form the terminal category $\mathbf{1}$ to the arrow category $\bullet \to \bullet$. If we think their coequalizer exists (in the category of small categories) then we believe in the axiom of infinity, because the coequalier is the monoid of natural numbers. I am pretty sure the axiom of choice and excluded middle are not needed for general category theory, and foundation also seems quite irrelevant. How about union and replacement?<|endoftext|> TITLE: A question on motivic zeta-function QUESTION [5 upvotes]: It's well-known that over $\mathbb F_q$ every smooth projective conic $C$ is isomorphic to a projective line. So the formula for the motivic zeta-function $Z_{mot}(C)$ is evident since $S^n\mathbb P^1 \simeq \mathbb P^n$. But what can one say when the finite field in the formulation of this question is replaced by an arbitrary field? Are there any references which can help me to determine the answer? REPLY [6 votes]: $S^nC$ will equal $\mathbb P^n$ for $n$ even and a Severi-Brauer variety with the same Brauer class as $C$ for $n$ odd. In the $n$ odd case, its class in the Grothendieck group will be $[C] ( 1+ L^2 + \dots + L^{n-1})$. Combining the even and odd cases, you get a formula for the motivic zeta function. For details, and a much more general result, see Zeta Functions of Curves with No Rational Points by Daniel Litt.<|endoftext|> TITLE: Verdier duality under more general conditions QUESTION [9 upvotes]: In the section 3.2 of Sheaves in Topology by A. Dimca, the author explains that if $f:X\to Y$ is a continuous map (between locally compact, $\sigma$-compact topological spaces with finite homological dimension) such that $f_!$ has finite cohomological dimension, then the following holds: (Verdier duality, local form) There is an additive functor of triangulated categories $f^!:\mathsf{D}^+(Y)\to \mathsf{D}^+(X)$ such that there is a functorial isomorphism $$\mathsf{R}\underline{\operatorname{Hom}}^\bullet(\mathsf{R}f_! \mathscr{F}^\bullet,\mathscr{G}^\bullet)\cong \mathsf{R}f_*\mathsf{R}\underline{\operatorname{Hom}}^\bullet(\mathscr{F}^\bullet,f^!\mathscr{G}^\bullet)$$ in $\mathsf{D}^+(Y)$ for any $\mathscr{F}^\bullet\in\mathsf{D}^b(X)$ and $\mathscr{G}^\bullet\in\mathsf{D}^+(Y)$. Do we really need all those hypotheses? Perhaps we can use Brown's representability theorem to prove it under more general conditions for the unbounded derived category? (There's a post here on MO about using this theorem but there it is under less general conditions.) Edit: Let me be clear about my "proposed" proof. Let $f:X\to Y$ be a morphism of (locally compact) ringed spaces and $\mathsf{D}(X),\mathsf{D}(Y)$ be their derived categories of modules. The tag 0F5Y on the Stacks Project implies that any triangulated functor $\mathsf{D}(X)\to\mathsf{D}(Y)$ which preserves direct sums has a right adjoint. If we could prove that $\mathsf{R}f_!$ preserves infinite direct sums, then we would conclude the existence of a functor $f^!:\mathsf{D}(Y)\to\mathsf{D}(X)$ such that $$\hom_{\mathsf{D}(Y)}(\mathsf{R}f_!\mathscr{F}^\bullet,\mathscr{G}^\bullet)\cong \hom_{\mathsf{D}(X)}(\mathscr{F}^\bullet,f^!\mathscr{G}^\bullet)$$ naturally in $\mathscr{F}^\bullet$ and $\mathscr{G}^\bullet$. This yields the local form as usual (for example, prop. 3.1.10 in Sheaves in Manifolds). REPLY [6 votes]: The status of the following answer is a bit speculative, unfortunately. To be precise, I believe that all of what I say below is true; I also believe that there is no proof in the literature of some of the things I state below, and I am not enough of an expert to supply those proofs. Nevertheless I'm putting it out there in the hope that it can be helpful. First of all, when dealing with unbounded complexes of sheaves one runs into the issue of hypercompleteness, and I have to say something about this. Let's first do the purely psychological change that instead of complexes of sheaves, we think of sheaves of complexes. Chain complexes are most naturally thought of as an $\infty$-category, once we localize at quasi-isomorphisms, so one is then led to thinking about sheaves valued in an $\infty$-category. Now a sheaf on a space $X$ in a complete 1-category $C$ is is a functor $F \colon \mathrm{Op}(X)^{op} \to C$ such that if $\{U_i \to U\}$ is an open cover of a subset $U$, then $F(U)$ is the equalizer (limit) of the two arrows $\prod_i F(U_i) \to \prod_{i,j} F(U_i \cap U_j)$. If $C$ is an $\infty$-category then the limit must be a homotopy limit, taking higher coherences into account, and then the proper definition of a $C$-valued sheaf turns out to be that the natural map from $F(U)$ to the (homotopy) limit of the cosimplicial diagram which in level $n$ is given by $\prod_{i_1,\ldots,i_n} F(U_{i_1} \cap \ldots \cap U_{i_n})$, is an isomorphism. This specializes to the usual $1$-categorical sheaf axiom when $C$ is a 1-category. When we take $C$ to be the $\infty$-category of bounded below cochain complexes modulo quasi-isomorphism, then the $\infty$-category of $C$-valued sheaves on $X$ is an $\infty$-categorical enhancement of the derived category $D^+(X)$. But if we consider instead unbounded complexes, the analogous statement is false. Namely, the $\infty$-categorical sheaf axiom from the previous paragraph describes what's known as descent for Cech covers. To recover the classical unbounded derived category one must instead impose descent for hypercovers. In a sense this has been known since the 60's, and traditionally this has been interpreted as meaning that Cech descent produces the "wrong" answer, and hypercovers "correct" this deficiency. After Lurie, a more modern perspective is that Cech descent is for many purposes more natural. In any case, the upshot is that for a space $X$ there are two typically inequivalent notions one can consider: $\infty$-sheaves on $X$ valued in unbounded complexes of sheaves, and hypersheaves on $X$ valued in unbounded complexes. The latter category can be recovered from the former via the process of hypercompletion, and the latter produces an $\infty$-categorical enhancement of the classical unbounded derived category. Now you mention the condition $(\ast)$ in Spaltenstein's paper, which looks like it is just an annoying technicality, and whether it can be removed using more modern homotopical machinery. I believe instead that the result is just plain false without some condition like $(\ast)$. More specifically I think that if a space $X$ satisfies condition $(\ast)$ then this forces Cech descent and hyperdescent to coincide for abelian sheaves, and that this is fundamentally the reason that $(\ast)$ appears in Spaltenstein's paper. Namely, Spaltenstein's Theorem B concerns the classical unbounded derived category, and I believe that all of these results fail in general when one works with hypersheaves. But a very general version of Spaltenstein's Theorem B should hold if one works with $\infty$-sheaves throughout, with no condition like $(\ast)$ or finiteness. Here's one reason to believe this. Part of Spaltenstein's Theorem B is proper base change. In Higher Topos Theory, Lurie proves a very general nonabelian version of proper base change, which implies the classical one. Crucially, Lurie's version of proper base change is a theorem for $\infty$-sheaves, not hypersheaves (and he gives an example where proper base change fails for hypersheaves). In particular it implies a form of proper base change for $\infty$-sheaves of unbounded complexes on a space $X$, and not for hypersheaves. Now I should add that in Higher Topos Theory Lurie only considers proper morphisms, so he works in the setting where $f_! = f_\ast$. So he does not introduce the functors $f_!$ or $f^!$ to state his result. Another indication that $(\ast)$ is actually about hypercompleteness is that Spaltenstein remarks that locally finite dimensional spaces satisfy $(\ast)$. But locally finite dimensional things should also be hypercomplete. More precisely, Lurie proves this statement in HTT, if "dimension" is interpreted as "homotopy dimension", a notion that he introduces. For paracompact topological spaces, "homotopy dimension" coincides with "covering dimension". Spaltenstein doesn't elaborate on what notion of dimension he's thinking of but I assume cohomological dimension. In any case, it is true that higher category theory can be used to give constructions of $f_!$ and $f^!$, under milder hypotheses than what Spaltenstein uses. Namely, $f_!$ and $f^!$ should exist for any continuous map between locally compact Hausdorff spaces, for $\infty$-sheaves valued in any complete and cocomplete stable $\infty$-category. Unlike Lurie's proper base theorem which is fully nonabelian (ie works for sheaves of spaces), this part of the story uses stability in a crucial way. In Higher Topos Theory, Lurie proves that on a locally compact Hausdorff space $X$, $\infty$-sheaves on $X$ can be described equivalently in terms of functors taking values on open subsets of $X$, or taking values on compact subsets of $X$. If the target category is moreover stable then this can be used to construct an equivalence of $\infty$-categories between sheaves and cosheaves on $X$. There is a natural pushforward operation on cosheaves, much like the pushforward of sheaves. Translating the pushforward functor via the sheaf-cosheaf equivalence to an operation on sheaves, one recovers precisely the functor $f_!$. By the adjoint functor theorem one directly gets $f^!$, too. This is sketched in https://www.math.ias.edu/~lurie/282ynotes/LectureXXI-Verdier.pdf In your second question it sort of sounds like you are asking about whether versions of the functors $f_!$ and $f^!$ can be useful in coherent cohomology (and you are not just asking about Grothendieck duality). Something like this is true in the setting of condensed mathematics. One version of six-functors formalism is in the final chapter of Scholze's "Condensed mathematics", but even closer to what you're looking for is I think the material from Dustin Clausen's final lecture in the Copenhagen Masterclass (you can find it on Youtube), which I will not attempt to summarize.<|endoftext|> TITLE: SPOT as a conservative extension of Zermelo–Fraenkel QUESTION [8 upvotes]: In Infinitesimal analysis without the Axiom of Choice, Hrbacek and Katz have shown that it is possible to formulate an axiomatic theory which provides a formalisation of calculus procedures which make use of infinitesimals (known as SPOT, an acronym of its axioms). Elsewhere in another somewhat related article by Katz I have read that SPOT is conservative over traditional Zermelo–Fraenkel set theory and so does not depend on the axiom of choice or on the existence of ultra-filters. Can someone explain in terms more suitable for a non-expert what it means for SPOT to be “conservative” over ZF, and also why this implies no dependence on the axiom of choice? REPLY [11 votes]: Ali's answer contains all the required technical explanation. My answer is more "sociological background " in nature: For various reasons, there have been rather strong negative reactions to Nonstandard Analysis. Bishop and Connes come to mind as good examples, as discussed at length in papers by M. Katz and his co-authors. By the usual standard of mathematics, the negative responses to Nonstandard Analysis can only be called "fierce" and "highly emotional". To dress up this semi-religious war with intellectual arguments, one often hears that "Nonstandard Analysis more heavily makes use of the Axiom of Choice (than usual mathematics)." The point of SPOT is to show that this commonplace is actually false.<|endoftext|> TITLE: An infinite game possibly due to Ernst Specker QUESTION [14 upvotes]: I have a vague memory of an infinite game due to Ernst Specker with the following properties: (1) It is a two-person perfect information game, where the players move alternately. (2) The possible moves depend only on the current position. (3) There is no winning strategy where each move is based only on the current position (4) There is a winning strategy (for one of the players) if the strategy uses the current position and the history of previous moves. Can someone provide a reference or explanation for this game? REPLY [21 votes]: I don't know about the game attributed to Specker, but here is a simple game with your desired features. Let us call it the Chocolatier's game. There are two players, the Chocolatier and the Glutton. To begin play, the Chocolatier serves up finitely many unique and exquisite chocolate creations on a platter, and then the Glutton chooses one of them to eat. Play continues — at each stage the Chocolatier adds finitely many additional chocolates to the platter, and the Glutton consumes one of those available. The excess uneaten chocolates accumulate on the platter as play progresses. After infinite play, the Glutton wins if every single chocolate that was served was eventually consumed. Otherwise, the Chocolatier wins. The Glutton can easily win simply by paying attention to the order in which the chocolates were added, and consuming them in that order. If only finitely many chocolates are added at each stage, the Glutton should simply make sure to consume them before moving on to the chocolates that were added at later stages. Indeed, the Glutton can win even if the Chocolatier places countably many chocolates on each turn. At turn $n=2^k(2m+1)$, let the Glutton eat the $k$th chocolate added at stage $m$, if any, and otherwise eat arbitrarily. But what about strategies that depend only on the current position, that is, the assortment of chocolates on the platter? In one sense, it is easy to see that there can be no such winning strategy for the Glutton. If the Glutton will choose a particular chocolate from a given assortment, then let the Chocolatier simply replace it with an identical chocolate type on the next move, and again on all subsequent moves. If the strategy does not know the history, then the Glutton will choose it again every time, and the other chocolates will never be eaten. Perhaps it makes a more interesting game, however, to say that the Chocolatier loses if chocolate types are ever repeated. And for this version of the game, there are some interesting things to say. If there are only countably many chocolate types available in all, then again the Glutton has a winning strategy that depends only on the current assortment on offer. Namely, the Glutton should fix an enumeration of the possible chocolate types that might appear, and at each stage select the chocolate that appears earliest in this order — it is the tastiest-looking chocolate as defined by that priority. With this strategy he will succeed in eating all the chocolates, because at the limit, if any chocolate was left, there would have to be a tastiest-looking one (earliest in the enumeration), and this would have been eaten once the finitely many tastier chocolates had been consumed. If the Chocolatier were uncountably creative, however, and able to serve up uncountably many different chocolate types, then this argument breaks down. Indeed, in this case I claim there is no winning strategy for the Glutton that depends only on the chocolate assortment on offer. And indeed, there is no such strategy that works even if we should insist that the Chocolatier present an assortment only of size two at each stage. To see this, suppose the Glutton will follow a fixed strategy that selects a particular chocolate to eat from amongst any two chocolates. For any given chocolate $c$, if there are infinitely many others $d_n$ that would be preferred to it by the strategy, if presented as a pair $\{c,d_n\}$, then the Chocolatier can present these pairs $\{c,d_0\}$, $\{c,d_1\}$, $\{c,d_2\}$, in turn. At each stage, $d_n$ would be consumed and the Chocolatier can present $\{c,d_{n+1}\}$. In the end, the inferior chocolate $c$ would never have been selected and so the Chocolatier will win. So if this is a winning strategy for the Glutton, then we may assume that every chocolate is preferred to all but finitely many of the others. But this is simply impossible with an uncountable set. To see this, take any countably infinite set of chocolates and close under the finite witnesses of strictly preferred chocolates. One thereby constructs a countably infinite set $D$ of chocolates so that any chocolate in $D$ is preferred to any chocolate not in $D$. Since there were uncountably many chocolate types, there is some $c\notin D$. This contradicts our assumption that there are only finitely many chocolates preferred to a given chocolate. Let me remark that in the version of the game where we do not allow the Chocolatier to repeat chocolate types (as opposed to saying that this is allowed, but causes the Chocolatier to lose), then we should really include the list of already-consumed chocolates as part of the position, and this complicates the arguments above. (I asked a followup question about this at The Chocolatier's game.) I suspect but do not yet know that the Glutton can have no winning strategy that depends only on these more general kinds of positions in the uncountable case.<|endoftext|> TITLE: $\mathit{NP}$-hard statements which are $\mathit{NP}$-complete under the Riemann Hypothesis QUESTION [7 upvotes]: $\newcommand\NP{\mathit{NP}}\newcommand\SAT{\mathit{SAT}}\newcommand\CH{\mathit{CH}}\newcommand\PSPACE{\mathit{PSPACE}}$Are there $\NP$-hard problems which are $\NP$-complete under the Riemann Hypothesis corresponding to the following cases? Superpolynomial certificate length becomes polynomial sized. Randomized polynomial time reduction to $\SAT$ becomes deterministic polynomial time reduction. Reduction to $\SAT$ becomes polynomial deterministic time from superpolynomial deterministic time. Certificate becomes polynomial time verifiable. An infinitely often reduction always holds. The problem was originally in $\CH$ (counting hierarchy) or below and strictly in $\PSPACE$. I think 2. and 6. are impossible unless RH is false and so might be difficult to identify while 1., 3. and 4. might be similar. Bounty is for 2. or 6. I would be happy to award for 2. and 6. if there is a reduction for these under the condition of failure of RH and in my opinion such a twisted reduction is allowed but just 2. and 6. are disallowed under correctness of RH. It would be amazing if there are reductions to 2. and 6. under both truth and falseness of RH and this I think is unlikely since I think reduction under truth of RH is unlikely. REPLY [3 votes]: Here's an example, but you may consider it to be a bit artificial Consider the following language: s is in $L$ if $S=(x,y)$ where $x$ is a 3-SAT instance, $y$ is a satisfying instance for $y$, and also satisfying that if $n$ is the number of true variables in $y$, then the $n$th non-trivial zero of the zeta function has real part equal to $1/2$. But this is a bit artificial. My guess is that you specified your five examples in order to avoid that. Edit: The following does not work. Possibly it can be fixed, but see Emil's comment below. But we can modify this is a little bit to cheat and get an example of 1, provided one is willing to use GRH rather than RH; my guess is that a similar construction can be done with RH, but it doesn't immediately jump out at me. The first detail is that assuming GRH, the smallest positive primitive root $u$, of a prime $p$ is bounded above by a power of $\log p$. In particular, Shoup showed that $u = O( (\log p)^6)$. Given positive constants $c_1$, $c_2$, and a function $f(n)$ Let $A_{c_1,c_2, f(n)}$ be the language given by the following: $a$ is in $A$ if $a=(x,y,z,w)$ where $x$, $y$, and $z$ are defined as follows. We have $x$ is a 3-SAT instance and $y$ is a satisfying set of variables for $x$. $z$ is a prime $p$ such that $y< p < (c_1)p^{c_2}$ (where we are thinking of our list $y$ of variable assignments as just a binary number). We have $w$ is a string of 1s where the number of 1s is a primitive root of $p$, and $w$'s length is bounded by $f(p)$. Now, we can choose a $c_1$ and $c_2$ such that there is always such a prime $p$ when there is a $y$. (Baker, Harman and Pintz showed that one can take $c_2 = 0.525$). So choose such a $c_1$ and $c_2$. Then, note that if we believe GRH, we can always find such a $w$ which is polynomial in the length of $x$, and we can choose $f(n)$ so that $w$'s length is guaranteed to be less than $f(p)$ if GRH is true. So now our NP-hard question then is just is a given $x$ a valid $x$ for an $(x,y,z,w)$ in $A$? If GRH is true, we can always find a $y$, $z$ and $w$ which is guaranteed to be just polynomial in length when $x$ is true. But without GRH we can only get that this is NP-hard since our $w$ may be too large. This is essentially just a padding trick. My guess is that a similar padding trick can be used with just RH as an assumption but I don't immediately see how to do it. And this is still a pretty artificial problem. I'd be interested to see if someone has a more natural example.<|endoftext|> TITLE: Existence of curves of arbitrary genus on some K3 surface QUESTION [5 upvotes]: Voisin uses the fact "If $X$ is a K3 surface with an ample line bundle $\mathcal L$ such that $\mathcal L$ generates $\mathop{\mathrm{Pic}}(X)$ and $(\mathcal L^2) = 4t - 2$, then every smooth curve $C \in \lvert\mathcal L\rvert$ satisfies $K_{t, 1}(C, K_C) = 0$." to prove the Green conjecture holds for generic curves of even genus. My question is why given an even integer $g$, there always exists a K3 surface $X$ and a smooth curve $C \subseteq X$ of genus $g$ satisfying the conditions above. REPLY [4 votes]: This should be a consequence of the surjectivity of the period map for K3 surfaces. I believe with this in mind the reasoning is somewhat standard, but it's useful to try and make it explicit. The underlying strategy is as follows: 1) identify a non-empty locus $\mathcal{W}$ in the period domain to which a K3 surface $X$ with the desired property could be mapped into via the period map, then 2) use surjectivity of the period map to assert existence of such $X$. To this end Let's make the following definitions of, respectively, the K3 lattice, the moduli space of marked K3 surfaces[1], and the period domain: $$\Lambda := E_8(-1)^{\oplus 2}\oplus U^{\oplus 3}$$ $$N := \{(X,\varphi)\}/\sim$$ $$ D := \{[v] \in \mathbb{P}\Lambda_{\mathbb{C}} : (v)^2 = 0\text{ and } (v,\overline{v}) > 0\}. $$ Then the period map $\mathcal{P} : N \rightarrow D \subset \mathbb{P}\Lambda_{\mathbb{C}}$ sending $(X,\varphi)$ to $[\varphi_{\mathbb{C}}(H^{2,0}(X))]$ is surjective (see e.g. Theorem 4.1 in "Lectures on K3 surfaces" by D. Huybrechts). Now suppose $V \subset \Lambda$ is any sublattice. Then $V^{\perp}$ (in particular) determines a closed slice $$D_V := D\cap\mathbb{P}(V^{\perp})_{\mathbb{C}}$$ of the period domain. If $\mathcal{P}(X,\varphi) \in D_V$ this then means $\varphi_{\mathbb{C}}(H^{2,0}(X)) \subset (V^{\perp})_{\mathbb{C}}$ which[2] implies $\varphi_{\mathbb{C}}(H^{1,1}(X)) \supset V_{\mathbb{C}}$ and thus[3] $$ \varphi(NS(X)) \supset V. $$ Finally, let $$ D_V^{\circ} := D_V \setminus \bigcup_{V'\not\subset V}D_{V'}. $$ Then clearly if $\mathcal{P}(X,\varphi) \in D_V^{\circ}$ we have $$ \varphi(NS(X)) \cong V. $$ Since $D_V^{\circ}$ is the complement of a countable union of proper closed subsets, it is non-empty. So for the desired result, it now suffices to take $\mathcal{W} = D_V^{\circ}$ where $V = \langle \lambda \rangle$ for $\lambda \in \Lambda$ such that $(\lambda)^2 = 4t - 2$. The existence of such $\lambda$ can be verified for any $t \geq 1$ using our explicit knowledge of the lattice $\Lambda$ (as indicated above) - e.g. $\lambda = (2t-1,1) \in U$ has the desired property, thinking of $U \cong \mathbb{Z}^2$ with intersection form $\left[\begin{array}{cc}0 & 1\\1 & 0\end{array}\right]$. [1] if $X$ is a K3 surface with marking $\varphi : H^2(X;\mathbb{Z}) \xrightarrow \cong \Lambda$ (an isometry) then $(X,\varphi) \sim (X',\varphi')$ if and only if $\varphi' = \varphi\circ f^*$ for some isomorphism $f : X \rightarrow X'$. [2]using the fact that $\varphi$ is an isometry and is defined over $\mathbb{Z}$ (and thus commutes with conjugation on $\Lambda_{\mathbb{C}}$), one sees this as follows: $$ \begin{array}{rcl} \varphi_{\mathbb{C}}(H^{2,0}(X)) & \subset & (V^{\perp})_{\mathbb{C}}\\ \implies \varphi_{\mathbb{C}}(H^{0,2}(X)) = \varphi_{\mathbb{C}}(\overline{H^{2,0}(X)}) & \subset &(V^{\perp})_{\mathbb{C}} = \overline{(V^{\perp})_{\mathbb{C}}}\\ \implies V_{\mathbb{C}} = (V^{\perp})_{\mathbb{C}}^{\perp} & \subset & \varphi_{\mathbb{C}}(H^{2,0}(X)\oplus H^{0,2}(X))^{\perp}\\ & = & \varphi_{\mathbb{C}}((H^{2,0}(X)\oplus H^{0,2}(X))^{\perp})\\ & = & \varphi_{\mathbb{C}}(H^{1,1}(X)) \end{array} $$ [3]using the same facts as above, one sees this as follows: $$ \begin{array}{rcl} \varphi(NS(X)) & = & \varphi_{\mathbb{C}}(H^{1,1}(X)\cap H^2(X;\mathbb{Z}))\\ & = & \varphi_{\mathbb{C}}(H^{1,1}(X))\cap \varphi_{\mathbb{C}}(H^2(X;\mathbb{Z}))\\ & = & \varphi_{\mathbb{C}}(H^{1,1}(X))\cap \Lambda\\ & \supset & V_{\mathbb{C}}\cap \Lambda\\ & = & V. \end{array} $$<|endoftext|> TITLE: Do powers of the shift operator applied to a non-zero vector always yield a total set? QUESTION [5 upvotes]: Let $S$ be the (say, left) shift operator on $\ell^2(\mathbb{Z})$. For a non-zero vector $x \in \ell^2(\mathbb{Z})$, consider the set $$X = \{ S^n v \mid n \in \mathbb{Z} \}.$$ Is this always a total set, i.e., is its span dense in $\ell^2(\mathbb{Z})$? REPLY [9 votes]: Such sets are not always total. The shift operator $S$ is unitarily equivalent to multiplication by $z$ on $L^2(S^1)$. From this perspective you can see vectors for which the set you write is not total, for example the characteristic function of an interval.<|endoftext|> TITLE: Finite CW complex with finite non-abelian fundamental group and higher homologies zero QUESTION [6 upvotes]: I want to build a finite CW complex such that $\pi_1$ is non-abelian and $H_i$ are zero for $i\geq 2.$ From Hatcher for a given group G, one can create an example of a 2-complex $X_G$ with $\pi_1(X_G)=G.$ I also checked from Mayer-Vietoris that if $G$ is cyclic such complex won't have any higher homology for $i\geq 2.$ I tried to take $G=S_3,$ the symmetric group of order 6 and from Mayer-Vietoris I get $H_2$ is $Z.$ I believe this was a correct calculation. Or is there a way to get the groups $G$, with $G$ non-abelian, such that we can get $\pi_1 =G$ and $H_i=0$ for $i\geq 2.$ Any reference or idea to create such an example? Or is there a way to claim such a finite complex can't exist? REPLY [7 votes]: Theorem. Let $G$ be a group. There exists a finite 3-complex $X_G$ with $\pi_1 X_G = G$ and $H_i X_G = 0$ for $i > 1$ if, and only if, $G$ is finitely presentable and has second group homology $H_2(G) = 0$. The more interesting question to me is whether this is possible for a finite 2-complex, and Jens Reinhold's answer gives us the first step there. This is possible for the group $2I$ mentioned in my comment above. Lemma: Let $X$ be any CW complex. Then the cokernel of the Hurewicz map $\pi_2 X \to H_2 X$ is isomorphic to $H_2(\pi_1 X)$. (Proof: attach cells to make $X$ into a $K(\pi_1 X, 1)$; doing so kills off precisely $\pi_2 X$ inside of $H_2 X$.) This is exercise 23 in Hatcher section 4.2. Proof of theorem. Pick any finite presentation $P$ of $G$ and construct the presentation complex $X_P$. This is a finite complex with $\pi_1 X_P = G$, and with $H_2(X_P)$ a free abelian group $\Bbb Z^k$ on a finite number of generators. Because the cokernel of $\pi_2(X_P) \to H_2(X_P)$ is precisely $H_2(\pi_1 X_P) = H_2(G) = 0$, it follows that $\pi_2(X_P) \to H_2(X_P) = \Bbb Z^k$ is surjective. Now choose $k$ maps $\rho_i: S^2 \to X_P$ so that these give a basis of the second homology group, and set $$X_G = X_P \cup_{i=1}^k D^3,$$ attaching these 3-cells along the $\rho_i$. The resulting complex has $H_i X_G = 0$ for $i > 1$ by a Mayer-Vietoris argument, while $\pi_1 X_G = \pi_1 X_P \cong G$ by the van Kampen theorem.<|endoftext|> TITLE: About the sum of rectangular power sums QUESTION [7 upvotes]: Let $n \geq 1$ be an integer and consider the symmetric function $$D_n = \sum_{d|n} p_d^{n/d},$$ where $p_{d}$ are the power-sum symmetric functions. It can be checked up to $n=35$ that the symmetric function $D_n$ is Schur-positive. The multiplicity of the Schur function $s_n$ in $D_n$ is given by the number of divisors of $n$ (sequence A000005 in the OEIS). One can also easily compute the multiplicity of the Schur function $s_{1^n}$ in $D_n$: $$\sum_{d|n} (-1)^{n+d},$$ which is sequence A112329 in the OEIS. Q1: Is the Schur-positivity of $D_n$ for all $n$ known ? Q2: Has this been considered somewhere in the literature ? I have alas not much more context for this. REPLY [4 votes]: Just adding that the expression for the Frobenius characteristic of $\theta_d$ is classically attributed to H. O. Foulkes, see e.g. Richard Stanley's EC2, Problem 7.88.<|endoftext|> TITLE: The 4th Lemoine circle QUESTION [5 upvotes]: The first and second Lemoine circles are well-known to geometers. According to this article the third Lemoine circle has been first discovered by Jean-Pierre Ehrmann in 2002. It is worth noting that the centres of these circles belong to a line that is going through the Lemoine point and the circumcenter of the triangle. I would like to present the construction of another circle, which I believe might be called the 4th Lemoine circle. The idea behind its construction is slightly similar to that proposed by Ehrmann: Symmedians AA', BB', CC' intersect each other at the point M (the Lemoine point of the triangle ABC). The circumcircle of the triangle A'B'C' was drawn (also known as the 'symmedial circle'). A'', B'', C'' are the first intersection points of the symmedians with the symmedial circle. Finally, the circumcircles of the triangles A''B''M, B''C''M, A''C''M always intersect the sides of the original triangle ABC at six points, that are conclycic. Theorem illustration As far as I can tell the point O (the center of our six point circle) is not included into the C.Kimberling's Encyclopedia, so it must be unknown or uncatalogued. O belongs to the line that contains the centres of the first three Lemoine circles, so by my reckoning, the dotted red circle that is shown in the picture above perfectly qualifies for being called the 4th Lemoine circle. Could you please give a synthetic proof of this theorem ? Geogebra dynamic sketch. REPLY [3 votes]: A notational preamble: Writing $K$ for the circumcenter, and $L_1$, $L_2$ (OP's $M$), $L_3$ for the centers of the First, Second, and (Ehrmann's) Third Lemoine Circles, all of which are collinear, we have this happy terminological coincidence: $$\frac{KL_i}{KL_1}=i \tag{1}$$ (This invites dubbing the circumcenter $L_0$, and the circumcircle itself the "Zero-th Lemoine Circle"; but I digress.) To the problem at hand ... Some ugly coordinate bashing confirms that OP's center $O$ lies on the line-of-$L$s, a rather remarkable property that provides the corresponding circle some "Numbered Lemoine Circle" credibility. Further bashing shows that the counterpart of $(1)$ is $$\frac{KO}{KL_1}=\frac{3 (a^2 + b^2 + c^2)^3 - (-a^2 + b^2 + c^2) (a^2 - b^2 + c^2) (a^2 + b^2 - c^2)}{8\,(a^2 + b^2) (a^2 + c^2) (b^2 + c^2)} \tag2$$ The dependence of this value upon the shape of the triangle distinguishes it from the $L_i$, perhaps so much so that OP's circle establishes a new category, "Numbered Non-Tucker Lemoine Circles". I'll take this opportunity describe another point on the line-of-$L$s with a shape-independent ratio $(1)$: For each vertex $V=A, B, C$, construct the circle through $V$ and $L_2$ whose center lies on $\overleftrightarrow{VK}$. The six "other" points where the three circles meet the triangle's side-lines are concyclic, and their circumcenter —known to Kimberling as $X(585)$— wants to be denoted $L_{3/2}$, because $$\frac{KL_{3/2}}{KL_1} = \frac32 \tag3$$ This may-or-may-not earn the circle (which, incidentally, is a Tucker circle) the title of "$\frac32$-th Lemoine Circle".<|endoftext|> TITLE: Does the small object argument need replacement? QUESTION [7 upvotes]: Does one need the axiom of replacement in the small object argument and in the transfinite construction of free algebras? My motivation for the question is that I heard that the axiom of replacement is never needed in practice. If the answer is "yes", how can one avoid the axiom of replacement in practical applications of these two theorems? Note that one can't use ordinals if one doesn't allow the axiom of replacement (I think). REPLY [7 votes]: The way it is usually presented, certainly yes. As you point out it usually refers to possibly uncountable regular ordinals, which would usually mean von Neumann ordinals. Once you have the regular ordinal, say $\kappa$, then regardless of whether $\kappa$ is a von Neumann ordinal or just a well ordered set, you need to define a sequence of objects $K_\alpha$ for each $\alpha < \kappa$, and take colimits at each limit stage, and overall at the end. That kind of argument does seem to require replacement, as far as I can see. Certainly it is possible to define a sequence of sets, even of length $\omega$ whose colimit does not provably exist in the category of sets under $\mathbf{ZFC}$ minus replacement (since $V_{\omega + \omega}$ is a model of that theory, and we can define the sequence $K_n := V_{\omega + n}$). On the other hand, there are two alternative versions of the small object argument in Swan, W-types with reductions and the small object argument that don't use replacement (and also don't use definitions by recursion into universes of small types, which is, roughly speaking, the type theoretic version of replacement). The first can be carried out in a topos with natural number object and satisfying the choice principle $\mathbf{WISC}$, which I believe holds for any Grothendieck topos under the assumptions of $\mathbf{ZFC}$ minus replacement. The second is specific to the case of monic generating cofibrations in presheaf toposes, but does not require $\mathbf{WISC}$, so would work in a metatheory of $\mathbf{ZF}$ minus replacement. Both arguments are for a slightly non standard definition of cofibrantly generated, and they have a different format to the usual proof using ordinals. Instead of defining a sequence of objects along an ordinal, there are just two steps, first define a $W$-type, and then either quotient it (for the first argument) or carve out a subobject (for the second).<|endoftext|> TITLE: Computing the complex roots of a monic polynomial QUESTION [6 upvotes]: The map from monic complex polynomials to the unordered tuples of their roots (each appearing according to its multiplicity) is computable. This seems to have been known for a long time, and with modern technology is quick and easy to prove. What I am interested in is whether there is a canonical reference for this (and if so, whom to credit). The weaker observation that if the coefficients are computable numbers, then the roots are computable numbers, too, doesn't count here. The uniformity is important. There is a paper from 2002 proving the result I mentioned (Lester, Chambers, and Lu - A constructive algorithm for finding the exact roots of polynomials with computable real coefficients), but there are older references to the result being well-known. In fact, I vaguely remember having read a pre-LaTeX paper containing a proof — unfortunately, I don't remember where I may have come across it. Weihrauch's Computable Analysis has this as Exercise 6.3.11. REPLY [8 votes]: I'm worried that I'm misunderstanding your question, but I think one could argue there is no satisfactory answer here except to say that this result likely predates "computability" itself. From the wikipedia article on the Fundamental theorem of algebra: None of the proofs mentioned so far is constructive. It was Weierstrass who raised for the first time, in the middle of the 19th century, the problem of finding a constructive proof of the fundamental theorem of algebra. He presented his solution, which amounts in modern terms to a combination of the Durand–Kerner method with the homotopy continuation principle, in 1891. Moreover, it's been known since Brouwer and Weyl that the theorem was constructive. Brouwer and deLoor in 1924 showed that the Fundamental Theorem of Algebra was Intuitionistically provable: L. E. J. Brouwer and B. deLoor, Intuitionistischer Beweis des Fundamentalsatzes der Algebra, Amsterdam Kon. Akad. van Wetenschappen, Proc., vol. 27, 1924, pp. 186-188 In the same year, Weyl also gave a constructive proof of some sort: H. Weyl, Randbemerkungen zu Hauptproblemen der Mathematik, Math. Zeit., vol. 20, 1924, pp. 131-150 In that case, it is really just a matter of connecting constructive math with computable math. (Also, constructive math has the advantage that uniformity is built in.) I don't know if there is a clear first mention of a uniformly computable FTA, but here is what I've found. Rosenbloom gives an alternative elementary constructive proof avoiding the heavy machinery of Brouwer, deLoor and Weyl: P. C. Rosenbloom. An Elementary Constructive Proof of the Fundamental Theorem of Algebra. The American Mathematical Monthly, Vol. 52, No. 10 (Dec., 1945), pp. 562-570 In this paper we shall give a method of constructing from a given polynomial $P(z)$ by rational operations a sequence of complex numbers ${z_n}$ which satisfies the Cauchy convergence criterion and such that $P(z_n)$ converges to zero. All of this part of the proof is constructive and completely elementary. Rice cites Rosenbloom's construction to show that the recursively complex numbers are algebraically closed (but not explicitly mentioning uniformity) in H. G. Rice. Recursive real numbers. Proc. Amer. Math. Soc. 5 (1954), 784-791 And I believe, Bishop also has a constructive proof of the FTA in his book. (And I haven't looked at the Russian school of constructive math or Martin-Lof's book on constructive math, but I wouldn't be surprised if both discuss this.) At this point in history, the 1960s, I would say the theorem is basically folklore if there hasn't already been a canonical reference to it yet, since it is one short step away from a lot that is already known. I think all computable analysts understand, at least now if not then, that if something is in Bishop's book, it is (no more than an exercise to check that it is) uniformly computable. Update 1: Actually, maybe this is the paper you want. P. Henrici and I. Gargantini, Uniformly convergent algorithms for the simultaneous approximation of all zeros of a polynomial, in [2], pp. 77–113 where [2] is B. Dejon and P. Henrici, Editors, Constructive Aspects of the Fundamental Theorem of Algebra, Proceedings of a symposium at IBM Research Lab, Zürich-Rüschlikon, June 5-7, 1967, Wiley-Interscience, London I found this mentioned in "A Constructive Proof of the Fundamental Theorem of Algebra without using the Rationals" (Herman Geuvers, Freek Wiedijk, Jan Zwanenburg): The first constructive proof of FTA (for monic polynomials) is from Weyl, where the winding number is used to simultaneously find all zeros of a (monic) polynomial. A similar but more abstract proof, also using the winding number, occurs in [Bishop and Bridges], where FTA is proved for arbitrary non-constant polynomials. Based on Weyl’s approach, [Henrici and Gargantini] presents an implementation of an algorithm for the simultaneous determination of the zeros of a polynomial. So maybe the best answer is the first uniform construction is due to Weyl, but Henrici and Gargantini first put it in terms of uniform computability. Update 2: Actually, maybe this paper is more canonical looking. (Note, I don't actually have access to a lot of these papers, so you should check them yourself.) It is in the same book [2] mentioned above: E. Specker, The Fundamental Theorem of Algebra in Recursive Analysis, in [2], pp. 321–329. Again, more from the the same section of Geuvers-Wiedijk-Zwanenburg: In [Brouwer and De Loor], Brouwer and De Loor give a constructive proof of FTA for monic polynomials by first proving it for polynomials with rational complex coefficients (which have the advantage that equality is decidable) and then make the transition (viewing a complex number as the limit of a series of rational complex numbers) to general monic polynomials over C. This proof – and also Weyl’s and other FTA proofs – are discussed and compared in [deLoor 1925]. Brouwer [1924] was the first to generalize the constructive FTA proof to arbitrary non-constant polynomials (where we just know some coefficient to be apart from 0). In [Specker] it is shown that, for general non-constant polynomials, there is a continuous map from the coefficients to the set of zeros. Also, on the Google Books page for Ernst Specker Selecta p 375, I found more description about this paper: So this looks like exactly the sort of thing you want! (Nonetheless, I still think the pre-Turing works--namely Weyl, Brouwer-deLoor, and Brouwer--likely deserve the most credit, unless I'm mistaken on the uniformity of those constructions.)<|endoftext|> TITLE: Are there exotic polynomial bijections from $\mathbb N^d$ onto $\mathbb N$? QUESTION [15 upvotes]: The Cantor bijection given by $$(x,y)\longmapsto {x+y\choose 2}-{x\choose 1}+1$$ is a bijection from $\{1,2,3,\dotsc\}^2$ onto $\{1,2,3,\dotsc\}$. It can be generalized to bijections $\varphi_d:\{1,2,3,\dotsc\}^d \longrightarrow \{1,2,3,\dotsc\}$ given by $$(x_1,\dotsc,x_d)\longmapsto (d+1\bmod 2)+(-1)^d\sum_{k=1}^d(-1)^k{x_1+\dotsb+x_k\choose k}$$ where $(d+1\bmod 2)$ equals $1$ if $d$ is even and $0$ otherwise. (The proof is a sort of double induction on $d$ and on the sum $x_1+x_2+\dotsb+x_d$.) It is of course possible to consider compositions of the above formulæ in order to get additional, more complicated polynomial bijections. A straightforward counting argument shows that we obtain in this way $d! s_d$ different polynomial bijections between $\{1,2,\dotsc\}^d$ and $\{1,2,\dotsc\}$ where $s_1,s_2,\dotsc$ are the little Schroeder numbers with generating series $$\sum_{n=1}^\infty s_nq^n=\frac{1+q-\sqrt{1-6q+q^2}}{4}\ .$$ Are there other "exotic" polynomial bijections (between $\{1,2,\dotsc\}^d\longrightarrow \{1,2,\dotsc\}$)? (The answer is obviously "no" for $d=1$ and unknown for $d=2$. I do not know if an "exotic" bijection is known for $d=3$.) Added for completeness: Sketch of a proof that $\varphi_d$ is a bijection: We set $$A_d(n)=\{(x_1,\ldots,x_d)\in\{1,2,\ldots\}^d\ \vert\ x_1+x_2+\ldots+x_d=n\}$$ and $A_d(\leq n)=A_d(d)\cup A_d(d+1)\cup \ldots\cup A_d(n)$. It is enough to prove that $\varphi_d$ induces a bijection between $A_d(\leq n)$ and $\{1,\ldots,{n\choose d}\}$. This is clearly true for $d=1$ (and arbitrary $n$) and for $n=d$ with arbitrary $d$. Since $(x_1,\ldots,x_d)\longmapsto (x_1,\ldots,x_{d-1})$ is a bijection between $A_d(n)$ and $A_{d-1}(\leq n-1)$ we have (using slightly abusing notations for sets) \begin{align*}\varphi_d(A_d(n))&= {n\choose d}-\left(\varphi_{d-1}(A_{d-1}(\leq n-1))-1\right)\\ &=\left\{{n\choose d}-{n-1\choose d-1}+1,\ldots,{n\choose d}\right\}\\ &=\left\{{n-1\choose d}+1,\ldots,{n\choose d}\right\} \end{align*} which ends the proof of the induction step. REPLY [14 votes]: Wikipedia says "The generalization of the Cantor polynomial in higher dimensions" is $$(x_1,\ldots,x_n) \mapsto x_1+\binom{x_1+x_2+1}{2}+\cdots+\binom{x_1+\cdots +x_n+n-1}{n}$$ Note that this is not equivalent to your generalisation $$(x_1,\ldots,x_d)\mapsto (d+1\bmod 2)+ \sum_{k=1}^d(-1)^{k+d}{x_1+\dotsb+x_k\choose k}$$ so the word "The" is misleading. Lew, Morales, and Sánchez proved (Diagonal polynomials for small dimension, Math. Sys. Th. 29 (1996) 305–310) that these two families give the only "diagonal polynomials" up to permutation of variables for dimension 3, where a "diagonal polynomial" is a bijection $\mathbb{N}^d \to \mathbb{N}$ which orders all tuples satisfying $x_1 + \cdots + x_d = k$ before any tuple satisfying $x_1 + \cdots + x_d = k + 1$. In the same edition Morales and Lew gave $2^{d-2}$ inequivalent (i.e. not related by permutation of variables) diagonal polynomials for higher $d$ (An enlarged family of packing polynomials on multidimensional lattices, Math. Sys. Th. 29 (1996) 293-303). In follow-up papers, Morales (Diagonal polynomials and diagonal orders on multidimensional lattices, Th. Comput. Sys. 30 (1997) 367–382) showed that this was an undercount, and in particular listed 6 inequivalent diagonal polynomials for $d=4$; and Sánchez described A family of $(n-1)!$ diagonal polynomial orders of $\mathbb{N}^n$, Order 12 (1997). Fetter, Arredondo, and Morales proved in The diagonal polynomials of dimension four, Adv. Applied Math. 34 (2005) 316-334 that the six polynomials which Morales and Sánchez describe are all the diagonal polynomials of dimension 4 (up to permutation of variables). As far as I know it's an open question whether there are diagonal polynomials outside Sánchez's family for $d > 4$. (Disclosure: I have not been able to read all of these papers, as I don't currently have access to a library with journal subscriptions, so I'm relying on the reports of the two I've found online). I'm not sure to what extent these diagonal polynomials count as "exotic", but it is at least interesting to note that there's more than one per dimension, and this is far too long for a comment anyway.<|endoftext|> TITLE: How small can the support of a nontrivial $\mathbb F_p$-cocycle on $C_p$ be? QUESTION [9 upvotes]: Let $p$ be a prime, and let $\phi : C_p^n \to \mathbb F_p$ be an $\mathbb F_p$-valued $n$-cocycle on $C_p$ (the cyclic group of order $p$) which is not an $n$-coboundary, i.e. $\phi$ represents a nontrivial element of $H^n(C_p;\mathbb F_p)$. Define the support $\operatorname{supp}(\phi) \subseteq C_p^n$ to be the set of elements $\vec g$ such that $\phi(\vec g) \neq 0$. Question: What is a good lower bound on the cardinality of $\operatorname{supp}(\phi)$? In particular, is there a lower bound which grows exponentially in $n$? (of course, $p^n$ is an upper bound -- I believe $(p-1)^n$ is also an upper bound; so this part of the question assumes that $p \neq 2$) Remarks: When $n=1$, observe that we have a lower bound of $p-1$, which is optimal. When $n=2$, the "canonical" cocycle $\phi_\text{carry}$ which defines the extension $\mathbb F_p \to \mathbb Z / p^2 \to C_p$ via usual carry arithmetic has support of cardinality about $p^2 / 2$. I don't know how tight this is; one might try to play with the fact that, perhaps after multiplying by a scalar, we have $\phi = \phi_\text{carry} + d\psi$ for some function $\psi : C_p \to \mathbb F_p$. In order for $\operatorname{supp}(\phi)$ to be small, it would need to be the case that $d\psi$ vanishes on most of the complement of $\operatorname{supp}(\phi_\text{carry})$ (so that in some sense $\psi$ is close to being a homomorphism) and that $d\psi = -1$ on most of $\operatorname{supp}(\phi_\text{carry})$. This sounds like a promising tension, but I don't know how to leverage it. For $n \geq 3$ I don't even know where to start. Of course, it would be interesting to know something about how this works for other finite groups / other coefficients. I've chosen the above ones for simplicity. I'm not sure what the best formulation would be in the case where $H^n(G;k)$ is not a cyclic $k$-module. REPLY [8 votes]: Here is a bound in the case $n=2$. Suppose $\phi$ is a cocycle with nontrivial cohomology class and $\phi(x,y)=0$ for at least $(1-\epsilon) p^2$ pairs $x,y$. Then $\epsilon \geq 1/8$. Proof: Without loss of generality, the class of $\phi$ is the standard generator of $H^2(\mathbb Z/p, \mathbb Z/p)$. It follows that $\phi$ arises from a lift $l \colon \mathbb Z/p \to \mathbb Z/p^2$ where $l(x) \equiv x \mod p$, taking $\phi(x,y) = (l(x+y)-l(x) - l(y))/p$. So we have $l(x+y) = l(x)+l(y)$ for all but $\epsilon p^2$ pairs $x,y$. Now fix $a \in \mathbb Z/p$. What is the probability that $$ l(b) + l(a-b) = l(c) + l(a-c)$$ for $b,c$ random? Well, it is the probability that $$ l(b) - l(c) - l(b-c) = l (a-c) - l(a-b) - l(b-c)$$ and both pairs $(c,b-c)$ and $(a-b,b-c)$ are uniformly distributed, so both sides vanish with probability $1-\epsilon$ by assumption, so both sides are equal with probability $1 - 2 \epsilon$. By linearity, it follows that there exists a single $c$ such that $$ l(b) + l(a-b) = l(c) + l(a-c)$$ with probability at least $1-2 \epsilon$. So there exists $f(a)$ such that $$l(b) + l(a-b)=f(a)$$ with probability at least $1-2\epsilon$. Next let's check that $$f(a) + f(b) = f(a+b).$$ Well, we know for $c,d$ random that we have both $f(a) = l(c) + l(a-c)$ and $f(b) = l(d) + l(b-d)$ with probability at least $1-2 \epsilon$ each, so at least $1-4\epsilon$ together. We have $l(c) + l(d) = l(c+d)$ and $l(a-c) + l(b-d) = l(a+b-c-d)$ with probability $1-\epsilon$ each, so all four events have probability at least $1-6 \epsilon$ together. In this case, we have $$ f(a) + f(b) = l(c) + l(a-c) + l(d) + l(b-d) =l (c+d) + l(a+b-c-d) $$ which equals $f(a+b)$ with probability at least $1-2\epsilon$. So $$f(a) + f(b)= f(a+b)$$ with probability at least $1-8 \epsilon$. If $\epsilon< 1/8$, then both sides are equal with positive probability. But both sides are deterministic, so both sides are equal. Hence $f: \mathbb Z/p \to \mathbb Z/p^2$ is a group homomorphism with $f(x) \equiv x \mod p$. But this is impossible, so $\epsilon \geq 1/8$. Let me explain the connection to computer science. This proof relies on an algorithm which, given a black box that computes the function $l \colon \mathbb Z/p \to \mathbb Z/p^2$, computes the homomorphism $f \colon \mathbb Z/p \to \mathbb Z/p^2$ (i.e. sample $l (b) + l(a-b)$ a large but bounded number of times and take the plurality vote). The algorithm as stated is not very useful, because its output cannot exist and thus its input does not exist, but a similar algorithm works for general groups (constructed by Blum, Luby, and Rubinfield in their paper Self-testing/correcting with applications to numerical problems). The advantage of this algorithm is that the input needs only be approximately linear (which means it could fail to be linear on exactly the inputs you care about) whereas the output is exactly linear (and can be calculated, with high probability, on the inputs you care about). Because approximate linearity can be tested by checking enough examples, this allows an untrustworthy source to prove to you the existence of a linear function with certain properties in a short amount of time.<|endoftext|> TITLE: How many divisors of $n$ are below $n^{1/3}$? QUESTION [14 upvotes]: I am trying to bound a function that includes $\sum\limits_{\substack{d < n^{1/3} \\ d \mid n}} 1$. Is there an upper bound known for this sum, either in general or in terms of $\sum\limits_{\substack{d \mid n}} 1$? Or in general is there a bound for $\sum\limits_{\substack{d < n^{1/k} \\ d \mid n}} 1$? Any help is appreciated. Edit: I am realizing that a lower bound for this sum in terms of $\sum\limits_{\substack{d \mid n}} 1$ would also be useful if anyone can help with that. REPLY [11 votes]: One thing you asked for is a lower bound. Following FusRoDah, I will let $d_k(n)$ be the number of divisors of $n$ of size less than $n^{1/k}$, and $d(n)$ be the number of divisors of $n$. Then I claim $$ d_1(n) \leq d_3(n) (d_3(n)+5),$$ giving an explicit lower bound of size roughly $d_1(n)^{1/2}$. Proof: First note that $2 d_2(n) =d_1(n)$ since, for $d$ a divisor, $n/d$ is also a divisor, so at least half the divisors have size at most $\sqrt{n}$. Then $d_2(n)$ is at most the number of divisors of $n$ of size $\leq \sqrt{n}$ that can be written as a product of two divisors of size $< n^{1/3}$ plus the number that cannot be written as a product. We bound both separately. The number that can be written as a product of two is certainly at most $\frac{d_3(n) (d_3(n)+1)}{2}$. If $d$ cannot be written as a product, then writing it as a product of a sequence of primes and taking the products of initial segments of the sequence, we must skip straight from $\leq d / n^{1/3}$ to $\geq n^{1/3}$, so one of the prime divisors must be at least $n^{2/3}/d \geq n^{1/6}$. Then the product of the remaining prime divisors is $\leq d/ n^{1/6} \leq n^{1/3}$, so is $ TITLE: Vanishing rate of a harmonic function near a boundary point QUESTION [8 upvotes]: Let $u(x, y)$ be a harmonic function on the upper half-plane $\mathbb{R}\times \mathbb{R}^+$, that is, $$\partial_x^2 u(x, y) + \partial_y^2 u(x, y) = 0$$ for $x \in \mathbb{R}, y>0$. Assume $u(x, 0)$ is smooth for $x \in \mathbb{R}$. In addition, we assume that both $u(x, 0)$ and $\partial_yu(x, 0)$ vanish at $x = 0$ to infinite order, i.e., for every $k \in \mathbb{Z}^+$, $$\lim_{x \to 0}\frac{u(x, 0)}{x^k} = \lim_{x\to 0}\frac{\partial_yu(x,0)}{x^k} = 0.$$ More explanation: As pointed out by Alexandre Eremenko, here we assume that $u(x, y)$ is continuous up to the boundary $\mathbb{R}\times \{0\}$, and we treat $-\partial_yu(x, 0)$ as the outer normal derivative of $u$ at the boundary point $(x, 0)$. Question: can we conclude that $u(x, y)$ vanish to infinite order at the origin with respect to interior points? In other words, does the following limit hold for every $k \in \mathbb{Z}^+$? $$\lim_{\substack{(x, y) \to (0, 0)\\ x\in \mathbb{R}, y>0}}\frac{u(x, y)}{(|x|+|y|)^k} = 0.$$ If not, is there a counter-example? REPLY [5 votes]: There is a counterexample. Consider the harmonic function $$u(x,y) = Re\left(e^{-1/z^2}\right) = e^{-\frac{x^2-y^2}{r^4}}\cos\left(\frac{2xy}{r^4}\right),$$ where $r^2 = x^2 + y^2$. We have that $$u(x,\,0) = e^{-1/x^2}$$ vanishes to infinite order in $x$, and that $$u_y(x,\,0) \equiv 0$$ since $u$ is even in $y$. However, $$u(x,\,x) = \cos(2/x^2)$$ does not vanish to infinite order. REPLY [4 votes]: Assuming $u$ is smooth enough, the answer seems to be affirmative. The $k$-th term in the power series of $u$ near the origin must be a solid harmonic polynomial $P_k$ of degree $k$, satisfying two independent conditions: $\partial_x^k P_k = 0$ and $\partial_x^{k-1} \partial_y P_k = 0$. The space of harmonic polynomials of degree $k$ is two-dimensional, so this essentially tells us that $P_k = 0$. Consequently, the power series of $u$ near $(0, 0)$ is zero, and hence all partial derivatives of $u$ vanish at the origin. Edit: Here are some additional details. Suppose that $u$ is the Poisson integral of the boundary data $f$ (so, for example, it suffices to know that $u$ is bounded, or non-negative — this is a rather mild condition). Suppose, furthermore, that $f$ is infinitely smooth in a neighbourhood of $0$. Then it is an easy exercise to see that $u$ is infinitely smooth in a neighbourhood of $(0,0)$ (intersected with $\mathbb R \times [0, \infty)$, of course). In particular, we can develop $u$ into the power series at $(0, 0)$. Of course, this power series need not be convergent, it is just a convenient formal way to speak about the partial derivatives of $u$. The $k$-th term of this power series, call it $P_k$, is a homogeneous polynomial of degree $k$, and using smoothness of $u$ it is easy to check that $P_k$ is a harmonic polynomial. The remaining part of the argument is given in the original answer.<|endoftext|> TITLE: Unknot recognition - how tangled does it get? QUESTION [15 upvotes]: A recent algorithm unknots in quasipolynomial time. But I want to know what happens to the crossing number. Assuming your unknot has $n$ crossings, if I remember correctly it might be necessary to increase $n$ temporarily. But to what? $n+C$? $C*n$? Even worse? (I don't even know if the mentioned algorithm can be translated to actual Reidemeister moves - it might not be necessary if the goal is just recognizing the unknot.) REPLY [23 votes]: Joel Hass and Jeff Lagarias proved that one can transform any unknot diagram with $n$ crossings into the standard unknot diagram using not more than $2^{cn}$ Reidemster moves. They were able to obtain the explicit value for $c$ of $c=10^{11}$. See here. This was subsequently improved by Marc Lackenby to a polynomial bound in the paper in the Annals "A polynomial upper bound on Reidemeister moves" In that paper, an unknot with at most $n$ crossings is shown to have a transformation into the standard unknot diagram with at most $(236n)^{11}$ Reidemeister moves; at no stage does the diagram have more than $(7n)^2$ crossings.<|endoftext|> TITLE: $(q,t)$-Fibonacci polynomials: area & bounce statistics QUESTION [7 upvotes]: This is related to my earlier (unanswered) MO post. Preserve notations from there. We take advantage of the one-to-one correspondence between the $(s,s+1)$-core partitions and $(s,s+1)$-Dyck paths. Let $\mathcal{F}_s$ denote the set of $(s,s+1)$-Dyck paths whose associated core partitions have distinct parts. Also, we recall the two notions area and bounce statistics defined on Dyck paths. We write $[n]_t=\frac{1-t^n}{1-t}$ and the $t$-binomials $\binom{n}k_t=\frac{[n]_t}{[k]_t[n-k]_t}$. Experimental observations lead to the following QUESTION 1. Is this true? $$\sum_{\lambda\in\mathcal{F}_s}q^{\text{area}(\lambda)}\,t^{\text{bounce}(\lambda)} =\sum_{j\geq0}\binom{s-j}j_tq^j\,t^{\binom{s}2-j(s-j)}.$$ QUESTION 2. Alternatively, if $H_s(q,t):=\sum_{\lambda\in\mathcal{F}_s}q^{\text{area}(\lambda)}\,t^{\text{bounce}(\lambda)}$ then does this hold? $$H_{s+1}(q,t)=t^{s+1}\,H_s(q,t)+q\,t^s\,H_{s-1}(q,t)$$ with initial conditions $H_1(q,t)=1$ and $H_2(q,t)=q+t$. REPLY [3 votes]: I mentioned in my previous answer that the order ideals corresponding to $(s,s+1)$-cores with distinct parts are precisely the subsets of $\{1,2,\dots,s-1\}$ that contain no consecutive elements. This means that the Dyck paths under investigation are all of height 2, with peaks at each element of the order ideal. Since there are no consecutive elements, the paths will always be "bouncing", in the sense that the associated bounce path is the same as the Dyck path itself. Let's restrict our attention to the Dyck paths with exactly $j$ peaks. There are exactly $\binom{s-j}{j}$ of these. They will all have area $j$, so it remains to prove that the generating function for their bounce statistic is given by $$t^{\binom{s}{2}-j(s-j)}\binom{s-j}{j}_t=t^{\binom{j}{2}+\binom{s-j}{2}}\binom{s-j}{j}_t.$$ To each subset $\{a_1,a_2,\dots,a_j\}$ of $\{1,2,\dots,s-1\}$ we can associate the subset $\{b_1,b_2,\dots,b_j\}$ of $\{0,1,\dots,s-j-1\}$ given by $b_i=a_i-i$. We can check that the bounce statistic of this subset is equal to $\binom{s-j}{2}+b_1+b_2+\cdots+b_j$, therefore the generating function we want is the coefficient of $x^j$ in the expression $$t^{\binom{s-j}{2}}\prod_{i=0}^{s-j-1}(1+t^ix).$$ From here the desired result follows from the q-binomial theorem.<|endoftext|> TITLE: Diffeomorphism of an open set and almost all of $\mathbb{R}^n$ QUESTION [8 upvotes]: (Question reposted from Math Stackexchange) I am aware of the statement that a open set in $\mathbb{R}^n$, if it is star-like, is diffeomorphic to $\mathbb{R}^n$, although this is apparently not so easy to prove. I am wondering if a weaker statement exists. Namely, if $U$ is an open set of $\mathbb{R}^n$, does there always exist a diffeomorphism $\phi : U\to\mathbb{R}^n\setminus N$ where $N$ is a Lebesgue-negligible set? Edit: Here is a reference the claim that star-shaped open sets are diffeomorphic to $\mathbb{R}^n$. REPLY [5 votes]: Yes, there always exists such a diffeomorphism (provided $U$ is nonempty). It can be constructed using the following lemma. Lemma. Let $K\subset V\subset(0,1)^n$ be a pair of sets, with $K$ compact and $V$ nonempty open. Then there exists a diffeomorphism $\psi$ of $(0,1)^n$ such that $\psi$ is the identity over $K$ and the measure of $\psi((0,1)^n\setminus V)$ is at most half of that of $(0,1)^n\setminus V$. Using the lemma, we can construct your diffeomorphism using a compact exhaustion of $U$. For instance, up to a diffeomorphism $\mathbb R^n\to(0,1)^n$, we can assume that $U$ is an open subset of $(0,1)^n$, and we want a diffeomorphism $\phi:U\to\phi(U)\subset(0,1)^n$ such that $(0,1)^n\setminus\phi(U)$ has zero measure. Write $U=\bigcup_{i\geq0}K_i$ with $K_i$ compact and $K_i\subset\operatorname{int}K_{i+1}$ for all $i\geq0$.¹ Then, using the lemma, we can define $\psi_0$ that fixes $K_0$ and such that the measure of $(0,1)^n\setminus\operatorname{int}K_1$ is halved under the action of $\psi_0$. In the next step, we define $\psi_1$ that fixes $\psi_0(K_1)$ and such that the measure of $(0,1)^n\setminus\psi_0(\operatorname{int}K_2)$ is halved. Iteratively, we define a sequence of diffeomorphisms of $(0,1)^n$, with the property that $\psi_k\circ\cdots\circ\psi_0$ stabilises over each fixed $K_i$. The limit diffeomorphism is well-defined over $U$, and sends it to a set of full measure. It remains to prove the lemma. Let $K\subset V\subset(0,1)^n$ be as described. Set $m:=|(0,1)^n\setminus V|$ the Lebesgue measure of the complement of $V$. We want to get it down to at most $m/2$. If $m=0$, then we can just choose the identity; assume then that $m>0$. Step 0: cubes and blow-ups. One can cover $(0,1)^n\setminus V$ by a finite number of open cubes $C_1,\ldots,C_L\subset(0,1)^n$ that do not intersect $K$.² Cubes that intersect are called neighbours. The diffeomorphism $\psi$ will be constructed by (finitely many) successive blow-ups: if $V$ intersects $C_\ell$ and $C\subset C_\ell$ is a closed cube (in the sequel, $C$ will fill $C_\ell$ almost entirely), there is a diffeomorphism $\theta:(0,1)^n\to(0,1)^n$ (that I will call a blow-up) whose support (the closure of the set of $x$ with $\theta(x)\neq x$) is compact in $C_\ell$, and such that $\theta(V)$ contains $C$. The existence of such a $\theta$ is clear on a picture, and to construct it explicitly is not too difficult. I will write $V$ instead of $V_t=\theta_{t-1}(V_{t-1})$ for the successive blow-ups of $V$ for notational clarity; however, $m$ is fixed and is the volume of the complement of the initial $V_0=V$. Step 1: connectedness. To be free, in the next step, to blow up any cube, we want $V$ to intersect every cube $C_\ell$. We work by induction, proving that for all $k$, up to a finite number of blow ups, there exists at least $k$ cubes $C_\ell$ such that $V$ intersects $C_\ell$ together with all its neighbours. The base case $k=0$ is clear. Suppose the result is true for some $kk}C_\ell \end{align*} $$ are two open sets that cover the connected set $(0,1)^n$. Since none of them are empty, they must intersect, so one of the $C_\ell$ with $\ell>k$ must intersect $V$ (since intersecting a good cube means also intersecting $V$). Let $C\subset C_\ell$ be a closed cube large enough that $C$ intersects all neighbours of $C_\ell$. Blowing up the inside of $C_\ell$ so that $V$ fills $C$, we get another good cube and conclude the induction. In particular, at this point every cube $C_\ell$ intersects $V$ and we can blow up cubes as we wish. A caveat: we must always blow up cubes with enough power that the new $V$ continues to intersect all their neighbours; we do so in step two. Step 2: volume expansion. The point, in this section, is that there exists a constant $\delta>0$ such that if the volume of the complement of $V$, after the few blow-ups, is still at least $m/2$, then there exists a blow-up that decreases the volume further by $\delta$. Then it is clear that, choosing a good blow-up at each step, we will reach a volume at most $m/2$ after a finite number of steps. If the volume of the complement of $V$ is at least $m/2$, then by the pigeonhole principle there must be a cube $C_\ell$ such that the complement of $V$ in $C_\ell$ has measure at least $$ \frac m2\cdot\frac{|C_\ell|}{\sum_{\ell'}|C_{\ell'}|} =: \mu\cdot|C_\ell|. $$ This $\mu$ is fixed (independently of the blow-ups) and positive. Choose $0<\lambda<\mu\leq1$. For a closed cube $C\subset C_\ell$ of volume at least $(1-\lambda)\cdot|C_\ell|$ and intersecting all neighbours of $C_\ell$, a corresponding blow up will decrease the volume of the complement of $V$ in $C_\ell$ by at least $(\mu-\lambda)\cdot|C_\ell|$. Since we can choose $\lambda$independently of the blow-ups, this concludes the argument, showing the existence of $\delta$ for $$ \delta := (\mu-\lambda)\cdot\min_\ell|C_\ell|>0. $$ ¹ Choose a proper function $f:U\to\mathbb R$ and set $K_i=f^{-1}[-i,i]$. ² For $\varepsilon>0$ with $K\subset (\varepsilon,1-\varepsilon)^n$, such cubes cover the compact set $[\varepsilon,1-\varepsilon]^n\setminus V$. Choose a finite subcover, and add a few cubes of side $\varepsilon$ tangent to the boundary of the cube to cover $(0,1)^n\setminus[\varepsilon,1-\varepsilon]^n$.<|endoftext|> TITLE: Bipartite graph with exactly one perfect matching QUESTION [5 upvotes]: $\textbf{Problem:}$ Find all bipartite graphs $G[X,Y]$ satisfying the following properties: $1.$ $|X|=|Y|$, where $|X|\ge 2$ and $|Y|\ge 2$. $2.$ All vertices have degree three except for two vertices in $X$ and two vertices in $Y$, which have degree two. $3.$ $G$ has exactly one perfect matching. $\textbf{My approach:}$ By trying out some examples it seems fair enough to conjecture that there doesn't exist such a graph $G$. I haven't yet found a proof to the above conjecture, but for the sake of contradiction if I assume that there exists such a graph $G$, then it seems that the bipartiteness of $G$ along with property $1$ and $2$ would end up with $G$ having at least two distinct perfect matchings, a contradiction to property $3$. There is a very close result to the above problem which states the following: Let $G[X,Y]$ be a bipartite graph on $2n$ vertices in which all vertices have degree three except for one vertex in $X$ and one vertex in $Y$, which have degree two. Then $pm(G)\ge 2(4/3)^{n-1}$, where $pm(G)$ is the number of perfect matchings in $G$. Is there any way to use this result to solve the above problem? Actually, even any kind of hint to proceed along my lines of approach would be of great help. Thanks! REPLY [3 votes]: There are indeed no such bipartite graphs. In fact: Proposition. Let $G=(X,Y,E)$ be a bipartite graph with $n:=|X|=|Y|$ in which the degree of each vertex is at least $2$ and at most $3$. Then, $G$ has either no perfect matching or at least two perfect matchings. Observation. Under the hypotheses of the proposition, the number of degree-$2$ vertices is the same in $X$ and $Y$. (Count the number of edges in $G$ as $\sum_{x\in X}d(x)$ or as $\sum_{y\in Y}d(y)$, and equate the two expressions.) Proof of the proposition. We use induction on $n$. The base is when $n=2$, in which case the claim is trivial, because the only bipartite graph satisfying the condition is the cycle of length $4$. Let $n>2$, and suppose, for the sake of getting a contradiction, that $G$ has exactly one perfect matching. Then, $G$ satisfies Hall's condition, and moreover, there is a set $\varnothing\neq A\subsetneq X$ which is critical, in the sense that $|N(A)|=|A|$, where $N(A)$ is the set of neighbors of the vertices in $A$. Namely, if no critical set exists, then Hall's condition remains satisfied even if we remove a single edge from the graph (in particular, one of the edges of the perfect matching), and this contradicts the uniqueness of the perfect matching. Note that if $A$ is a critical set, then the vertices of $A$ are necessarily matched with $N(A)$, and the vertices in $X\setminus A$ are necessarily matched with $Y\setminus N(A)$. Moreover, every critical set in our graph has at least two vertices. Let $A$ be a minimal critical set. Let $k$ denote the number of degree-$2$ vertices in $A$, and $\ell$ the number of degree-$2$ vertices in $N(A)$. Again, a counting argument shows that $k\geq\ell$, and there are exactly $k-\ell$ vertices between $N(A)$ and $X\setminus A$. Let $Q$ denote the set of vertices in $N(A)$ that have at least one neighbor in $X\setminus A$, that is $Q:=N(A)\cap N(X\setminus A)$. Note that for every $y\in Q$, $d(y)=3$ and there is exactly one edge between $y$ and $X\setminus A$. Indeed, if this is not the case, then $y$ is connected to exactly one vertex $x\in A$. But then $A\setminus\{x\}$ will also be a critical set, contradicting the minimality of $A$. It follows that in the subgraph $G'$ which is induced on $A\cup N(A)$, the degree of each vertex is at least $2$ and at most $3$. Therefore, by the induction hypothesis, $G'$ has either no perfect matching or at least two perfect matchings. In either case, this leads to a contradiction because we already know that $A$ can be matched with $N(A)$ and $X\setminus A$ can be matched with $Y\setminus N(A)$. $\square$<|endoftext|> TITLE: Visualizing genus-two Riemann surfaces: from the three-fold branched cover to the sphere with two handles QUESTION [10 upvotes]: I am trying to visualize the genus-two Riemann surface given by the curve $$ y^3 = \frac{(x-x_1)(x-x_2)}{(x-x_3)(x-x_4)}. $$ We can regard this surface as a three-fold cover of the sphere with four branch points. An image of the surface is here. Represented like this, it is not straightforward to me that this surface is topologically equivalent to a sphere with two handles. I am wondering if there is a way to deform this surface into something that resembles a double torus. As an example, we can take the following deformation of the torus. In particular, I would like to map the non-contractible cycles of the genus-two surface from one picture to the other (for the torus, two of these cycles are depicted in red and blue). REPLY [13 votes]: The picture (produced by Nick Schmitt) of the Lawson surface of genus 2 might help: It shows the genus 2 Riemann surface given by the algebraic equation $$y^3=\frac{z^2-1}{z^2+1}.$$ The lines show the vertical and horizontal trajectories of a holomorphic quadratic differential (given by a constant multiple of $\frac{(dz)^2}{z^4-1}$). You can deform the surface by Moebius transformations in the [webgl applet][2] (also by Nick Schmitt; use 2 fingers to apply Moebius transformations) to the following biholomorphic Riemann surface: [![enter image description here][3]][3] [2]: https://www.discretization.de/gallery/model/45)/ [3]: https://i.stack.imgur.com/aud2m.png<|endoftext|> TITLE: The Chocolatier's game: can the Glutton win with a restricted form of strategy? QUESTION [21 upvotes]: I have a question about the Chocolatier's game, which I had introduced in my recent answer to a question of Richard Stanley. To recap the game quickly, the Chocolatier offers up at each stage a finite assortment of chocolates, and the Glutton chooses one to eat. At each stage of play, the Chocolatier can extend finitely the chocolate assortment on offer, and the Glutton chooses from those currently available. After infinite play, the Glutton wins if every single chocolate that was offered is eventually consumed. In this post, I am interested in the version of the game where the Chocolatier is not allowed to repeat chocolate types — each new chocolate on offer is a uniquely exquisite new creation. As I explain at the other the post, the Glutton clearly has a winning strategy, simply by keeping track of when new chocolates are added and making sure to organize the consumption so that every chocolate is eventually eaten. (And this idea works even when the Chocolatier is allowed to extend the offers countably infinitely, and not necessarily just finitely.) Furthermore, in the case where there are only countably many possible chocolate types, then the Glutton has a winning strategy that depends only on the chocolates currently on offer, not requiring any knowledge of the game history. The strategy is simply to fix an enumeration of all the possible chocolate types and then eat the chocolate available that appears earliest in that enumeration. None could be left at the limit, since it would have been chosen once the earlier ones had been consumed. Meanwhile, in the case where there are uncountably many chocolate types available, I had proved that the Glutton has no such strategy that depends only on the chocolates currently on offer. My question is whether we can extend this argument also to allow the strategy to depend on the set of chocolates already eaten. Question. Does the Glutton have a winning strategy in the Chocolatier's game which depends only on the set of chocolates currently on offer and the set of chocolates already eaten? I conjecture a negative answer on sufficiently large uncountable sets and perhaps on all uncountable sets. Here is an alternative equivalent formulation of the game, the catch-up covering game on a set $X$. The first player plays an increasing chain of finite subsets $$A_0\subset A_1\subset A_2\subset\cdots\subset X$$ and the second player chooses elements $a_n\in A_n$. After infinite play, the second player wins if $\bigcup_n A_n=\{a_0,a_1,a_2,\ldots\}$. Of course, the second player can win, by looking at the history of how elements were added to the sets, but the question is whether there is a winning strategy that at move $n$ depends only on the current set $A_n$ and the set of already-chosen elements $\{a_k\mid k TITLE: Does the functor $\mathrm{Sh}\colon\mathbf{Top}\to\mathbf{Topos}$ have an adjoint? QUESTION [7 upvotes]: Consider the category $\mathbf{Top}$ of topological spaces, the category $\mathbf{Topos}$ of toposes and geometric morphisms, and the category $\mathbf{Loc}$ of locales. Let $$\mathrm{Sh}\colon\mathbf{Top}\to\mathbf{Topos}$$ be the functor sending a space $X$ to the topos of sheaves on $X$. Does this functor have a left or a right adjoint? Of course, $\mathrm{Sh}$ factorizes as $$\mathbf{Top}\to\mathbf{Loc}\to\mathbf{Topos},$$ since the sheaves on $X$ are the sheaves on the locale of open subsets of $X$. It is well-known that $\mathbf{Top}\to\mathbf{Loc}$ has a right adjoint and $\mathbf{Loc}\to\mathbf{Topos}$ has a left adjoint ($\mathbf{Loc}$ is reflective in $\mathbf{Topos}$). So the question whether $\mathrm{Sh}$ has a left adjoint reduces to the question whether $\mathbf{Top}\to\mathbf{Loc}$ has a left adjoint and the question whether $\mathrm{Sh}$ has a right adjoint reduces to the question whether $\mathbf{Loc}\to\mathbf{Topos}$ has a right adjoint. Remark. I already asked this on math.SE but I didn't get an answer. REPLY [9 votes]: In this answer, Topos is interpreted as a 2-category. (As a side remark, the 1-category of toposes does not make sense until one picks a specific model for toposes and geometric morphisms, and different models need not be equivalent as 1-categories. For the 1-categorical framework to make sense, at the very least one needs to organize toposes into a relative category, so that different models can be shown to be Dwyer–Kan equivalent as relative categories.) whether Top→Loc has a left adjoint The functor Top→Loc does not have a left adjoint because it does not preserve finite products. For example, the product of rational numbers with themselves as locales and as topological spaces produces nonisomorphic locales. In particular, rational numbers form a topological group, but not a localic group. whether Loc→Topos has a right adjoint Loc→Topos does not have a right adjoint because it does not preserve homotopy colimits. For example, suppose G is a discrete group acting on a point. Then the colimit of this action in locales is again a point. But the homotopy colimit of this action in toposes is the delooping of G, which is not equivalent to a point. Thus, both Loc→Topos and Top→Topos do not have a right adjoint functor.<|endoftext|> TITLE: Tensor products of $\mathbb{E}_\infty$-spaces QUESTION [9 upvotes]: In the $\infty$-world, connective spectra play the role of abelian groups, while $\mathbb{E}_\infty$-spaces play that of commutative monoids. This may be rephrased by saying that we may identify the $\infty$-categories of spectra and $\mathbb{E}_\infty$-spaces with the $\infty$-categories $\mathsf{Mon}_{\mathbb{E}_\infty}(\mathcal{S})$ and $\mathsf{Grp}_{\mathbb{E}_{\infty}}(\mathcal{S})$ of $\mathbb{E}_\infty$-monoids/groups in the $\infty$-category of anima $\mathcal{S}$. Now, the $1$-categories $\mathsf{Ab}$ and $\mathsf{CMon}$ come equipped with tensor products $\otimes_{\mathbb{Z}}$ and $\otimes_{\mathbb{N}}$. These correspond in homotopy theory to the tensor products $\otimes_{\mathbb{S}}$ and $\otimes_{\mathbb{F}}$ of connective spectra and $\mathbb{E}_\infty$-spaces. While the tensor product of connective spectra is widely discussed in the literature, I'm finding it a bit difficult to find references for that of $\mathbb{E}_\infty$-spaces. So far I've found the following: Gepner–Groth–Nikolaus, Universality of multiplicative infinite loop space machines, arXiv:1305.4550, which establishes in Theorem 5.1 a universal property for the tensor product $\otimes_{\mathbb{F}}$ as the unique functor making the free $\mathbb{E}_\infty$-monoid functor $$ \mathcal{S}_*\to\mathsf{Mon}_{\mathbb{E}_{\infty}}(\mathcal{S}) $$ into a symmetric monoidal functor. Blumberg–Cohen–Schlichtkrull, Topological Hochschild homology of Thom spectra and the free loop space, arXiv:0811.0553, which establishes a point-set model for $\mathbb{E}_{\infty}$-spaces, called $*$-modules, rectifying $\mathbb{E}_\infty$-spaces to strict monoids in $*$-modules. See also MO 92866. Sagave–Schlichtkrull, Diagram spaces and symmetric spectra, arXiv:1103.2764, which establishes another point-set model for $\mathbb{E}_{\infty}$-spaces, called $\mathcal{I}$-spaces, similarly rectifying $\mathbb{E}_\infty$-spaces to strict monoids in $\mathcal{I}$-spaces. See also arXiv:1111.6413. Lind, Diagram spaces, diagram spectra, and spectra of units, arXiv:0908.1092, which proves that $\mathcal{I}$-spaces and $*$-modules define equivalent homotopy theories. Questions: What are some other references discussing the tensor product of $\mathbb{E}_\infty$-spaces? What is the unit of this tensor product? Finally, what are some concrete examples of it? REPLY [10 votes]: The article by Gepner-Groth-Nikolaus is the canonical reference for the tensor product of $E_\infty$-spaces. In the end it is quite a formal construction so there is not that much to say. A useful point of view that does not appear in loc. cit. is that this tensor product comes from the Lawvere theory of commutative monoids. To explain this, consider the $(2,1)$-category $\mathrm{Span}(\mathrm{Fin})$ whose objects are finite sets and whose morphisms are spans $I\leftarrow K\rightarrow J$. It has the following universal property: for any $\infty$-category $\mathcal C$ with finite products, there is an equivalence $$ \mathrm{CMon}(\mathcal C) = \mathrm{Fun}^\times(\mathrm{Span}(\mathrm{Fin}),\mathcal C), $$ where $\mathrm{Fun}^\times$ is the $\infty$-category of functors that preserve finite products. Since $\mathrm{Span}(\mathrm{Fin})$ is self-dual, this means that $E_\infty$-spaces are finite-product-preserving presheaves on $\mathrm{Span}(\mathrm{Fin})$: $$ \mathrm{CMon}(\mathcal S) = \mathcal P_\Sigma(\mathrm{Span}(\mathrm{Fin})). $$ This was first studied in the thesis of J. Cranch. From this perspective, the direct sum and tensor product are the Day convolutions of $\sqcup$ and $\times$ on $\mathrm{Span}(\mathrm{Fin})$ (here $\times$ means the usual product of finite sets, which is not the categorical product in $\mathrm{Span}(\mathrm{Fin})$; the latter is the same as the categorical coproduct, i.e., the disjoint union $\sqcup$). For example, $E_\infty$-semirings can be described as right-lax symmetric monoidal functors $(\mathrm{Span}(\mathrm{Fin}),\times)\to(\mathcal S,\times)$ that preserve finite products. The unit. As Rune already explained, the unit for the tensor product of $E_\infty$-spaces is the free $E_\infty$-space on a point, that is the groupoid $\mathrm{Fin}^\simeq$ of finite sets with the $E_\infty$-structure given by disjoint union. This is equivalently the presheaf on $\mathrm{Span}(\mathrm{Fin})$ represented by the point, which is the unit for $\times$ on $\mathrm{Span}(\mathrm{Fin})$. Here are a few examples I could think of. Let $E\in \mathrm{CMon}(\mathcal S)$. Tensoring with a free $E_\infty$-space. Let $X\in\mathcal S$. Then $$ \left(\coprod_{n\geq 0} (X^n)_{h\Sigma_n}\right) \otimes E = \operatorname{colim}_X E, $$ where the colimit is taken in $\mathrm{CMon}(\mathcal S)$. This follows from the case $X=*$ using that $\otimes$ preserves colimits in each variable. Tensoring with $\mathbb S$. Tensoring with the sphere spectrum $\mathbb S$ is the same as group-completing: $$ \mathbb S\otimes E = E^\mathrm{gp}. $$ For example, for a ring $R$, $$ \mathbb S\otimes \mathrm{Proj}(R) = K(R). $$ where $\mathrm{Proj}(R)$ is the groupoid of finitely generated projective $R$-modules, and $K(R)$ is the K-theory space. Tensoring with $\mathrm{Fin}^\simeq[n^{-1}]$. Another localization of $\mathrm{CMon}(\mathcal S)$ is obtained by inverting integers (or rather, finite sets). The inclusion of the full subcategory of $E_\infty$-spaces on which multiplication by $n$ is invertible has a left adjoint $E\mapsto E[n^{-1}]$, which is equivalent to tensoring with $\mathrm{Fin}^\simeq[n^{-1}]$. But unlike in the cases of either abelian monoids or spectra, $\mathrm{Fin}^\simeq[n^{-1}]$ is not just the sequential colimit of multiplication by $n$ maps; it is obtained from the latter by killing suitable perfect subgroups of its fundamental groups, in the sense of Quillen's plus construction, to ensure that $n$ acts invertibly. Tensoring with $\mathbb N$. Let $\mathrm{FFree}_{\mathbb N}$ be the 1-category of finite free $\mathbb N$-modules. There is a functor $$ \mathrm{Span}(\mathrm{Fin}) \to \mathrm{FFree}_{\mathbb N} $$ sending a finite set $I$ to $\mathbb N^I$, inducing an adjunction $$ \mathrm{CMon}(\mathcal S) = \mathcal P_\Sigma(\mathrm{Span}(\mathrm{Fin})) \stackrel{\mathrm{str}}\rightleftarrows \mathcal P_\Sigma(\mathrm{FFree}_{\mathbb N}). $$ Objects in the right-hand side are sometimes called strictly commutative monoids (the group-complete ones are connective $H\mathbb Z$-module spectra). Tensoring with $\mathbb N$ amounts to strictifying a commutative monoid in this sense: $$ \mathbb N\otimes E = E^\mathrm{str}. $$ Unlike $\mathbb S$, $\mathbb N$ is not an idempotent semiring, that is, strictifying is not a localization. Indeed, $\mathbb N\otimes\mathbb N$ is an $E_\infty$-space whose group completion is the "integral dual Steenrod algebra". Tensoring with $\mathrm{Vect}_\mathbb{C}^\simeq$. Let $\mathrm{Vect}_\mathbb{C}^\simeq=\coprod_{n\geq 0} BU(n)$, where $U(n)$ is regarded as an $\infty$-group (despite the notation, this is not really the core of an $\infty$-category of vector spaces). This is an $E_\infty$-space whose group completion is $\mathrm{ku}$. There is a related $\infty$-category $2\mathrm{Vect}_{\mathbb C}$ whose objects are finite sets and whose morphisms are matrices of complex vector spaces. As in the previous example we get an adjunction $$ \mathrm{CMon}(\mathcal S) = \mathcal P_\Sigma(\mathrm{Span}(\mathrm{Fin})) \rightleftarrows \mathcal P_\Sigma(2\mathrm{Vect}_{\mathbb C}). $$ An object in the right-hand side is roughly speaking a commutative monoid such that $U(n)$ acts on the multiplication by $n$ map in a coherent way. Tensoring with $\mathrm{Vect}_\mathbb{C}^\simeq$ gives the free commutative monoid with such structure.<|endoftext|> TITLE: Convergence of Fourier series QUESTION [5 upvotes]: Say $f \in L^p[a,b]$, with $p \in \mathbb{N}, p > 1 $. Does its Fourier Series converge in the metric space $L^p[a,b]$? Does the series converge pointwise? And at which conditions? Say now $p = 1$, Does its Fourier Series converge in the metric space $L^1[a,b]$? Does the series converge pointwise? And under which conditions? REPLY [24 votes]: Convergence in $L^p$, $p>1$. True, by M. Riesz's Theorem. This is a standard topic in every harmonic analysis course, with several readable proofs. Convergence pointwise almost everywhere, $p>1$. True, by the Carleson-Hunt Theorem. This result is over 50 years old, but famously difficult. Though the techniques used are now standard, there aren't really any easy proofs, even for continuous functions. Convergence in $L^1$. False, and not very difficult to prove. Suppose that Fourier series converged in $L^1$. Let $S_N$ be the operator that takes a (Schwartz, say) function and returns the $N$th partial Fourier sum. I claim that these operators are not bounded uniformly in the $L^1$ norm, which is enough to give a contradiction by the uniform boundedness principle. For indeed applying it to the Fejer kernel $K_M$ gives $$\|S_NK_M\|_{L^1}\to\|D_N\|_{L^1}$$ as $M\to\infty$. Where $D_N$ is the Dirichlet kernel. But $\|D_N\|_{L^1}$ is unbounded, so $S_N$ cannot be uniformly bounded. Convergence pointwise almost everywhere, $p=1$. False, due to Kolmogorov almost 100 years ago. In fact, it is possible to find an $L^1$ function whose Fourier series diverges everywhere. This is not an easy counterexample, but it is presented in some courses in harmonic analysis.<|endoftext|> TITLE: Group completion of $\mathbb{E}_{\infty}$-monoids via tensor products QUESTION [9 upvotes]: $\newcommand{\K}{\mathrm{K}}$The abelian group completion functor $\K_0\colon\mathsf{CMon}\to\mathsf{Ab}$ satisfies $$ \K_0(A) \cong \mathbb{Z}\otimes_{\mathbb{N}}A, $$ naturally in $A\in\mathrm{Obj}(\mathsf{CMon})$, where $\mathbb{Z}$ is the additive monoid of integers (i.e. $\K_0(\mathbb{N})$, the group completion of $\mathbb{N}$); $\otimes_\mathbb{N}$ is the tensor product of commutative monoids. Question. Does the $\mathbb{E}_{\infty}$-group completion functor $\K_0\colon\mathsf{Mon}_{\mathbb{E}_\infty}(\mathcal{S})\to\mathsf{Grp}_{\mathbb{E}_\infty}(\mathcal{S})$ similarly satisfies $$\K_0(X)\cong QS^{0}\otimes_\mathbb{F}X,$$ where now $QS^0$, the stabilization of $S^0$, is the $\mathbb{E}_{\infty}$-group completion of $\mathbb{F}=\coprod_{n\in\mathbb{N}}\mathbf{B}\Sigma_{n}$, the groupoid of finite sets and permutations; $\otimes_{\mathbb{F}}$ is the tensor product of $\mathbb{E}_{\infty}$-spaces? REPLY [9 votes]: Yes, for the same reason. Let me sketch a proof. 1- $QS^0\otimes X$ is group-complete. Indeed, its $\pi_0$ is $\mathbb Z\otimes \pi_0(X)$, and that's a group for the usual reasons. Another way to prove it is to prove that the shear map for $X\otimes Y$ is (the shear map of $X)\otimes Y$, which can be seen by noting that $\otimes$ commutes with coproducts and hence finite products in each variable. 2- There is a natural transformation $X\to QS^0\otimes X$ given by tensoring $\mathbb F\to QS^0$ by $X$, and this induces a natural transformation $X^{gp}\to QS^0\otimes X$. 3- Both sides commute with colimits (a colimit of grouplike $E_\infty$-spaces is grouplike so I don't have to worry about whether I'm talking about colimits in monoids or grouplike monoids), therefore to check that this map is an equivalence, it suffices to do so for $X= \mathbb F$, and for that one it is a tautology. Another way to phrase this is to use the following sequence of natural equivalences (and using point 1- for the last one): $X^{gp} = QS^0\otimes_{QS^0} X^{gp} = (QS^0\otimes_\mathbb F X)^{gp}= QS^0\otimes X$ The second natural equivalence comes from the fact that group completion is symmetric monoidal, and $(QS^0)^{gp}\simeq QS^0$.<|endoftext|> TITLE: Can the numerator in Weyl's character formula be written as a determinant? QUESTION [10 upvotes]: I paraphrase part of the wikipedia article on the Weyl character formula: Weyl character formula. If $\pi$ is an irreducible finite-dimensional representation of a complex semisimple Lie algebra $\mathfrak{g}$ and $\mathfrak{h}$ is a choice of Cartan subalgebra of $\mathfrak{g}$, then the Weyl character formula states that the character $\operatorname{ch}_\pi$ of $\pi$ is given by $$ \operatorname{ch}_\pi(H) = \frac{\sum_{w \in W} \epsilon(w) e^{w(\lambda+\rho)(H)}}{\prod_{\alpha \in \Delta^+}(e^{\alpha(H)/2} - e^{-\alpha(H)/2})},$$ where $W$ is the Weyl group, $H \in \mathfrak{h}$, $\epsilon(w)$ is the determinant of the action of $w \in W$ on the Cartan subalgebra $\mathfrak{h}$, $\Delta^+$ denotes the set of positive roots of $(\mathfrak{g}, \mathfrak{h})$, $\lambda$ denotes the highest weight of $\pi$ and $\rho$ is half the sum of all positive roots of $(\mathfrak{g}, \mathfrak{h})$ (i.e. half the sum of all the elements of $\Delta^+$). If $\mathfrak{g} = \mathfrak{sl}(n)$, it is known that the denominator of the RHS of the Weyl character formula can be written as a determinant (actually a Vandermonde determinant). While it does seem counterproductive, since the numerator looks simple enough (well, to some extent), yet I am interested whether or not the numerator can also be written as a determinant, at least for $\mathfrak{g} = \mathfrak{sl}(n)$, though I am also interested in the other cases too. After all, the Weyl group in this special case is $S_n$, the symmetric group on $n$ elements, and an $n \times n$ determinant can be expanded as an alternating sum over $S_n$. So this does seem promising. I apologize if it turns out to be a trivial question perhaps (I currently have limited access to online journals etc.). REPLY [14 votes]: The classical definition of the Schur polynomials, which considerably predates the Weyl character formula, is as a ratio of two determinants (a so-called "bialternant"): see, e.g., https://en.wikipedia.org/wiki/Schur_polynomial#Definition_(Jacobi's_bialternant_formula). Of course the Schur polynomials are $\mathfrak{sl}_n$ characters. There are similar things in other types too, if you are interested in that (for example, see Proposition 1.1 in Okada's "Applications of Minor Summation Formulas to Rectangular-Shaped Representations of Classical Groups", https://doi.org/10.1006/jabr.1997.7408 ).<|endoftext|> TITLE: A topological version of the Lowenheim-Skolem number QUESTION [18 upvotes]: This is a continuation of an MSE question which received a partial answer (see below). Given a topological space $\mathcal{X}$, let $C(\mathcal{X})$ be the ring of real-valued continuous functions on $\mathcal{X}$ (with the ring operations given pointwise as usual). For $\varphi$ a first-order sentence in the language of rings, write "$\mathcal{X}\models_C\varphi$" if $C(\mathcal{X})\models \varphi$ in the usual sense, and say that $\varphi$ is $C$-satisfiable iff this holds for some $\mathcal{X}$. My question is: What is the least cardinal $\kappa$ such that for every $C$-satisfiable sentence $\varphi$ there is some space $\mathcal{X}$ with $\mathcal{X}\models_C\varphi$ and $\vert\mathcal{X}\vert<\kappa$? Note that we can characterize disconnectedness by a single sentence; for example, either "$\exists x,y(x+y=1\wedge xy=0)$" or "$\exists x(x^2=x\wedge x\not=0\wedge x\not=1)$" will do the trick. This has two consequences. First, it means that the "subspace" analogue of $\kappa$ does not exist at all since there are connected spaces of arbitrarily large cardinality with no connected subspaces of strictly smaller cardinality but still having more than one point (take the Dedekind completion of an appropriately homogeneous dense linear order of large cardinality). Second, with a bit more thought it leads to a proof that $\kappa>2^{\aleph_0}$ (pointed out by Atticus Stonestrom at MSE). However, I don't see how to go any further than that at the moment. Ultimately I'm interested in what happens when we replace the ring of real numbers with the usual topology, $\mathbb{R}$, with an arbitrary topological structure (= all functions continuous, all relations closed): given a first-order signature $\Sigma$ and a topological $\Sigma$-structure $\mathfrak{A}=(A,\tau,\Sigma^\mathfrak{A})$ we get a natural way to assign to each topological space $\mathcal{X}$ a $\Sigma$-structure with underlying set $C(\mathcal{X}, (A,\tau))$ and so a corresponding satisfaction relation $\mathcal{X}\models_{C:\mathfrak{A}}\varphi$. I'd like to understand how the choice of "target" $\mathfrak{A}$ impacts the resulting abstract model theoretic properties of $\models_{C:\mathfrak{A}}$. But already the case $\mathfrak{A}=\mathbb{R}$ seems nontrivial. REPLY [3 votes]: In response to Noah's request after my earlier comments: this doesn't answer the question, but may shed some light. If $\kappa$ is supercompact, then for every $C$-satisfiable sentence $\phi$, there is a topological space $\mathcal{X}$ of size $<\kappa$ such that $\mathcal{X} \models_C \phi$. In fact, the entire powerset of (the underlying set of) $\mathcal{X}$---in particular, its topology---can be taken to be of size $<\kappa$. Proof sketch: suppose $\mathcal{Y}=(Y,\tau) \models_C \phi$. Fix a regular $\theta$ such that $\mathcal{Y} \in H_\theta$, where $H_\theta$ denotes the collection of sets of hereditary cardinality $<\theta$ (you could work with rank initial segments if you prefer). Since $\kappa$ is supercompact, there is some $\mathfrak{N} \prec (H_\theta,\in)$ of size $<\kappa$ such that $\mathcal{Y} \in \mathfrak{N}$ and the transitive collapse $\mathfrak{H}(\mathfrak{N})$ of $\mathfrak{N}$ is equal to $H_\mu$, where $\mu$ is the ordertype of $\mathfrak{N} \cap \theta$. (This is folklore; see Fact 7.7 of my paper for a proof. In fact existence of such models capturing all possible sets is equivalent to supercompactness, by a classic theorem of Magidor). In particular, this ensures that $\mathfrak{H}(\mathfrak{N})$ is correct about lots of things and closed under lots of operations. For each $b \in \mathfrak{N}$, let $b_{\mathfrak{N}}$ denote the image of $b$ under the transitive collapsing map. By elementarity of the collapsing map and the fact that the statement ``$\mathcal{Y} \models_C \phi$" is downward absolute from the universe to the structure $(H_\theta,\in)$, it follows that $\mathfrak{H}(\mathfrak{N}) \models$ ``$\Big( Y_{\mathfrak{N}}, \tau_{\mathfrak{N}} \Big)$ is a topological space, and the ring of continuous real-valued functions on it satisfies the formula $\phi_{\mathfrak{N}}$". Since $\phi$ is just a formula in a countable language, the transitive collapsing map fixes $\phi$, i.e., $\phi = \phi_{\mathfrak{N}}$. So $\mathfrak{H}(\mathfrak{N}) \models$ ``$\Big( Y_{\mathfrak{N}}, \tau_{\mathfrak{N}} \Big)$ is a topological space, and the ring of continuous real-valued functions on it satisfies the formula $\phi$". Now $\mathfrak{H}(\mathfrak{N})$, being of the form $H_\mu$, is correct about the quoted statement (this is standard stuff; in fact, in this particular case the quoted statement is even $\Pi_1$, and sets of the form $H_\mu$ are always $\Pi_1$ elementary in the universe). And the space $\Big( Y_{\mathfrak{N}}, \tau_{\mathfrak{N}} \Big)$ is of size $\le |\mathfrak{H}(\mathfrak{N})|= |\mathfrak{N}|<\kappa$.<|endoftext|> TITLE: Is there a measurable isomorphism ${\mathbb C}\to {\mathbb C}_p$? QUESTION [5 upvotes]: Let $p$ be a prime and ${\mathbb C}_p$ be the completion of the algebraic closure $\overline{{\mathbb Q}_p}$. This field is isomorphic to $\mathbb C$. Both fields come with natural absolute values but have very different topologies. There are no continuous field isomorphisms, but that does not exclude Borel-measurable ones (measurable in both directions, that is). Do those exist? A related question is the question for measurable automorphisms of $\mathbb C$. Are there more than the complex conjugation? REPLY [9 votes]: $\mathbb C$ and $\mathbb C_p$ are separable complete metric spaces (Polish spaces). They are Polish groups under addition. A Borel-measurable homomorphism between Polish groups is continuous. (Sometimes called "automatic continuity".) So any such Borel isomorphism must be a homeomorphism. But $\mathbb C$ is connected while $\mathbb C_p$ is totally disconnected, so they are not homeomorphic. We did not require field isomorphism, only group isomorphism for addition.<|endoftext|> TITLE: Open problem in analysis with just one quantifier? QUESTION [15 upvotes]: I'm looking for an open problem in analysis or number theory with just one "genuine" or "second order" quantifier. E.g. "Every continuous function $\mathbb{R} \rightarrow \mathbb{R}$ has the property $\theta$", where $\theta$ is expressible using only quantifiers over rationals. "Every set $S$ of natural numbers has the property $\theta$", where $\theta$ is expressible using only quantifiers over rationals. No cheat examples like "For every real number, Goldbach's conjecture holds"! That's an arithmetical problem. In technical terms, I'm looking for a $\Pi^1_1$ sentence that we don't know how to reduce to an arithmetical sentence. I'd also like it to be easy to state and obviously $\Pi^1_1$, so that it can be included in a logic paper without requiring much explanation. REPLY [4 votes]: The Littlewood conjecture is an example that meets all my requirements. It is easy to state and obviously $\Pi^1_1$. Furthermore this comment by Christian Reiher gives me confidence that it has no known reduction to an arithmetical sentence. (Hopefully it even lacks a known reduction to a $\Sigma^1_1$ sentence.)<|endoftext|> TITLE: Is this quiver with relations of finite representation type QUESTION [5 upvotes]: Let $Q=(Q_0,Q_1)$ be the following quiver, $Q_0$ consist of 2 vertices, denoted by 1,2. $Q_1$ consist a loop at 1 called $\gamma$, an arrow $\alpha$ from 1 to 2 and an arrow $\beta$ from 2 to 1. The relation $\rho$ is $\{\beta\alpha, \beta\gamma, \gamma\alpha, \gamma^m\}$ for some integer $m\geq 2$. Given a field $k$, is the quiver algebra $kQ$ always of finite representation type? REPLY [8 votes]: An easy way to see that the algebra is of infinite representation type (for any $m \geq 2$) is to observe that it is a string algebra, and that you have strings of arbitrary length, each corresponding to an indecomposable module. For instance the strings $(\alpha \beta \gamma^{-1})^t$, $t \geq 1$, correspond to indecomposable modules of vector space dimension $3t+1$.<|endoftext|> TITLE: Closed formula for a certain infinite series QUESTION [28 upvotes]: I came across this problem while doing some simplifications. So, I like to ask QUESTION. Is there a closed formula for the evaluation of this series? $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}$$ where the sum runs over all pairs of positive integers that are relatively prime. REPLY [44 votes]: Let me add the details for Henri Cohen's nice answer, without claiming any originality. We have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\zeta(4)\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}.$$ On the other hand, by the identity (cf. Ron Gordon's answer here) $$\sum_{m\geq 1}\frac{\cos(mx)}{m^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\zeta(2),\qquad 0\leq x\leq 2\pi,$$ we also have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\sum_{n\geq 1}\frac{1}{n^2}\left(\frac{n^{-2}}{4}-\frac{\pi n^{-1}}{2}+\zeta(2)\right)=\frac{\zeta(4)}{4}-\frac{\pi\zeta(3)}{2}+\zeta(2)^2.$$ Comparing the right hand sides of the first and third equation, we conclude that $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{1}{4}-\frac{\pi\zeta(3)}{2\zeta(4)}+\frac{\zeta(2)^2}{\zeta(4)}.$$ Here we have $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$, therefore $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{11}{4}-\frac{45\zeta(3)}{\pi^3}.$$<|endoftext|> TITLE: Fourier cosine transform from Erdélyi's Tables of Integral Transforms QUESTION [9 upvotes]: I’d like confirmation that $$ \frac{\cos⁡(b \sqrt{a^2+x^2})}{(a^2+x^2)^{3/2}} $$ has the Fourier cosine transform $$ \frac{\pi}{2 a} \, \exp(-ay) \qquad \text{if $y>a$,} $$ as found in Tables of Integral Transforms by Arthur Erdelyi et al. equation (35) in Sect. 1.7. I am puzzled that it is independent of $b$, making me wonder whether the inequality should be $y>b$. Since the numerator is a kind of frequency swing when $x$ is time, I would expect the spectrum to reflect the width $b$ of the swing, (including the case $b=0$) and not its suddenness $a$. Comparison with equation (29) shows some similarity, but many comparable functions have Fourier transforms with a frequency limit at $b$ rather than $a$. A hint about evaluating the integral might be what I need. REPLY [7 votes]: Suppose that $a>0$ and $b$ are real numbers. Then the value of this Fourier cosine transform at a real $y$ is $$\sqrt{\frac2\pi}\int_0^\infty dx\,\cos(xy)\frac{\cos⁡(b\sqrt{a^2+x^2})}{(a^2+x^2)^{3/2}} \\ =2\sqrt{\frac2\pi}\int_a^\infty dt\,\cos\big(y\sqrt{t^2-a^2}\,\big)\frac{\cos⁡(bt)}{t^{3/2}}\to0$$ as $b\to\infty$, by the Riemann--Lebesgue lemma. Also, the value of this Fourier cosine transform at $b=0$ is ${\sqrt{\frac{2}{\pi }}}\frac y{a}\, K_1(a y)>0$ if $y>0$. So, this Fourier cosine transform must depend on $b$. Mathematica cannot find this Fourier cosine transform, which therefore seems unlikely to exist in closed form: REPLY [6 votes]: According to the Table Errata reported in Mathematics of Computation, vol. 65, no. 215, 1996, pp. 1379–1386, this entry in Erdélyi's Tables of Integral Transforms is flawed. The exponent in the denominator should be $1$ instead of $3/2$, and the condition should be $y>b>0$ instead of $y>a$. Unfortunately, therefore, this table entry does not actually address the Fourier transform required by the OP. REPLY [2 votes]: For what it is worth, the following related integral $$ C(y;b,c) := \int_0^\infty \frac{\cos(c\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}} \cos(bx) \, dx = \begin{cases} K_0(y \sqrt{b^2-c^2}) & [b>c>0; y>0] \\ -\frac{\pi}{2} Y_0(y \sqrt{c^2-b^2}) & [c>b>0; y>0] \end{cases} . $$ appears as formula 2.5.25.15 in Prudnikov, A. P.; Brychkov, Yu. A.; Marichev, O. I., Integrals and series. Vol. 1. Elementary functions, Moscow: Fiziko-Matematicheskaya Literatura (ISBN 5-9221-0323-7). 632 p. (2003). ZBL1103.00301. A quick numerical check with Mathematica suggests that the formula checks out. Twice integrating $-C(y;b,c)$ should give your desired cosine transform, up to fixing integration constants. A once-integrated $C(y;b,c)$ appears in the cited evaluation of another related cosine transform in this answer, which was helpfully linked by MathOverflow under related questions.<|endoftext|> TITLE: Is $\mathbb{Q}$ "equivalent" to a structure with transitive automorphism group action? QUESTION [7 upvotes]: Say that structures $\mathfrak{A},\mathfrak{B}$ with the same underlying set are parametrically equivalent iff every primitive relation/function in one is definable (with parameters) in the other. For example, every group $\mathfrak{G}=(G;*,{}^{-1})$ is parametrically equivalent to its "torsor reduct" $\mathfrak{T}_\mathfrak{G}=(G; (x,y,z)\mapsto x*y^{-1}*z)$. Parametrically equivalent structures can still have very different combinatorial properties. In analogy with the notion of vertex transitive graphs, say that a structure $\mathfrak{A}$ is point-transitive iff the natural action of its automorphism group is $1$-transitive - that is, if for every $a,b\in\mathfrak{A}$ there is some $f\in Aut(\mathfrak{A})$ with $f(a)=b$. If $\mathfrak{G}$ is a nontrivial group then $\mathfrak{G}$ is not point-transitive (no automorphism can move the identity) but $\mathfrak{T_G}$ is point-transitive (for each $g\in G$ the map $a\mapsto g*a$ is in $Aut(\mathfrak{T_G}$)). I'm broadly interested in understanding the information captured by the family of automorphism groups of parametric equivalents of a given structure (see also here). Motivated by the group/torsor example, one question which seems like it should be easy to answer is: which structures have a parametric equivalent whose automorphism group acts $1$-transitively? But even for simple structures things aren't very clear to me. So I'd like to start with a simple example: Is the field of rationals $\mathfrak{Q}=(\mathbb{Q};+,\times)$ parametrically equivalent to a point-transitive structure? Per the torsor example above, the answer is affirmative for $(\mathbb{Q};+)$. It's also affirmative for $(\mathbb{Q};+,<)$ and various similar expansions, by the same construction, so the existence of such a parametric equivalent isn't connected to any obvious model-theoretic tameness property. However, multiplication seems to complicate things. REPLY [6 votes]: Consider $(\mathbb{Q},R)$, where $R$ is the 6-ary relation defined by $$R(a,b,c,d,e,f) \iff (a-b)(c-d)=(e-f)$$ The automorphism group of this structure includes at least the translations $x\to x+h$. Since those translations include $x \to x+(b-a)$, there is always a translation taking $a$ to $b$, so the group of automorphisms is point-transitive. We can define $x+y$ and $xy$ in terms of $R$ as the unique $s$ and $t$ for which $R(1,0,s,x,y,0)$ and $R(x,0,y,0,t,0)$ respectively -- i.e. using the equations \begin{align} (1-0)(s-x)&=(y-0)\\ (x-0)(y-0)&=(t-0) \end{align} Finally, we can define $R$ in terms of $+$ and $\times$ directly by $ac+bd+f=ad+bc+e$.<|endoftext|> TITLE: Conjecture about the density of primes QUESTION [8 upvotes]: Conjecture For any sufficiently large integer $kn$ , the sequence representing the number of primes in each block obtained by splitting $kn$ into $k$ equal blocks, is a strictly decreasing sequence, i.e: $$\pi\left(n\right)>\pi\left(2n\right)-\pi\left(n\right)>\pi\left(3n\right)-\pi\left(2n\right)\ldots>\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)$$ My approach For any given $k$, we need to prove that for all sufficiently large $n$, the last block contains less primes than the block before: $$\underbrace{\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)}_{\text{before last}}>\underbrace{\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)}_{\text{last}}$$ Let $f( k,n )$ denote the difference between the prime count of before last and last block of $kn$: $$f\left(k,n\right) = \left(\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)\right)-\left(\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)\right)$$ $$ = 2\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)-\pi\left(kn\right)$$ Therefore, an equivalent formulation of the conjecture is that $f(k,n)>0$ for any integers $k,n$ with sufficiently large $n$. Does anybody have ideas or references on how to prove this? $$$$ Computation of $f( k,n )$ I made some computation that strongly suggests the conjecture is true, at least for small $k$: Failing case for small $n$ Let's take the case $k=2$. We see that the conjecture fails for $n=1,2,4$ & $10$: I computed $f( k,n )$ for $2 ≤ k ≤ 30$ with $1 ≤ n ≤ 10^{8}$, in order to find what appears to be the last failing case of each $k$ , after which $f( k,n ) >0$ for all $n$: REPLY [9 votes]: The conjecture is true, and it follows from the following form of the Prime Number Theorem: $$\pi(x)=\frac{x}{\log x}+(1+o(1))\frac{x}{\log^2 x},\qquad x\to\infty.$$ Indeed, this implies for any fixed $m\geq 1$ and sufficiently large $n$ that $$2\pi(mn)>\pi(mn+n)+\pi(mn-n).$$ Why? We have \begin{align*}\pi(mn)&=\frac{mn}{\log(mn)}+(1+o(1))\frac{mn}{\log^2(mn)}\\ &=m\frac{n}{\log n}\frac{1}{1+\frac{\log m}{\log n}}+(m+o(1))\frac{n}{\log^2 n}\\ &=m\frac{n}{\log n}+(m-m\log m+o(1))\frac{n}{\log^2 n}. \end{align*} Similarly (replacing $m$ by $m-1$ and $m+1$, respectively), \begin{align*} \pi(mn+n)&=(m+1)\frac{n}{\log n}+(m+1-(m+1)\log(m+1)+o(1))\frac{n}{\log^2 n},\\ \pi(mn-n)&=(m-1)\frac{n}{\log n}+(m-1-(m-1)\log(m-1)+o(1))\frac{n}{\log^2 n}. \end{align*} Here, in case of $m=1$, we understand $(m-1)\log(m-1)$ as zero. It follows that $$2\pi(mn)-\pi(mn+n)-\pi(mn-n)=(f(m)+o(1))\frac{n}{\log^2 n},$$ where $$f(m):=(m+1)\log(m+1)+(m-1)\log(m-1)-2m\log m.$$ So we only need to verify that $f(m)$ is positive, that is, $$(m+1)\log(m+1)+(m-1)\log(m-1)>2m\log m.$$ However, this one is clear, because the function $x\mapsto x\log x$ is strictly convex (its derivative is strictly increasing).<|endoftext|> TITLE: Self-dual Orlicz sequence spaces QUESTION [7 upvotes]: Suppose that $\ell_\phi$ is a reflexive Orlicz sequence space such that its dual space $\ell_\phi^*$ is isomorphic to $\ell_\phi$. Is $\ell_\phi$ isomorphic to $\ell_2$? REPLY [6 votes]: For a given $1 TITLE: Uniform upper bounds for the return probability of random walks on $ \mathbb{R}$ QUESTION [6 upvotes]: Let $\mu$ be a probability measure with finite support on integers or the real line with the property that $\mu( 0) \le p$ for a fixed $0

0$ and all real $t>0$ $$P(S_n=0)\le C/\sqrt{n(1-Q(t))},$$ where $$Q(t):=\sup_{x\in\mathbb R}P(x\le X\le x+t).$$ Letting now $t\downarrow0$, we get $$P(S_n=0)\le f(n):=C/\sqrt{n(1-q)},$$ where $$q:=Q(0)=\sup_{x\in\mathbb R}P(X=x).$$ Assuming now that $q<1$, we get $f(n)\to0$ as $n\to\infty$.<|endoftext|> TITLE: Lewis's convenience argument for $\mathbb{E}_{\infty}$-spaces QUESTION [6 upvotes]: The 1991 paper of Lewis, “Is there a convenient category of spectra?” proved that it is impossible to have a point-set model for spectra satisfying the following criteria: There is a symmetric monoidal smash product $\wedge$; We have an adjunction $\Sigma^\infty\dashv\Omega^\infty$; The sphere spectrum $\mathbb{S}$ is the unit for $\wedge$; There is either a natural transformation $$(\Omega^\infty E)\wedge(\Omega^\infty F)\Rightarrow\Omega^\infty(E\wedge F)$$ or a natural transformation $$\Sigma^\infty(E\wedge F)\Rightarrow(\Sigma^\infty E)\wedge(\Sigma^\infty F),$$ either of which is then required to satisfies the usual coherence conditions for monoidal functors. There is a natural weak equivalence $\Omega^\infty\Sigma^\infty X\dashrightarrow\lim_{n\in\mathbb{N}}(\Omega^n\Sigma^nX)$. Since Lewis's paper, a number of model categories of spectra have appeared, each of which satisfies some, but not all, of the requirements (1)–(5). For instance, the category of $\mathbb{S}$-modules of Elmendorf–Kriz–Mandell–May satisfy (1)–(4), but not (5). A modern point of view regarding spectra is that they are the $\infty$-categorical analogue of abelian groups in the sense that $\mathsf{Grp}_{\mathbb{E}_{\infty}}(\mathcal{S})\cong\mathsf{Sp}_{\geq0}$. Similarly, the $\infty$-categorical analogue of commutative monoids are $\mathbb{E}_{\infty}$-spaces, the $\mathbb{E}_{\infty}$-monoids in the symmetric monoidal $\infty$-category of spaces $\mathcal{S}$. Is there an analogue of Lewis's argument for $\mathbb{E}_{\infty}$, giving a similar list of nice properties we may expect of a point-set model of $\mathbb{E}_{\infty}$-spaces, but such that there is no such point-set model satisfying all of them? Moreover, in this case, how do the current point-set models for $\mathbb{E}_{\infty}$-spaces (such as $*$-modules, $\Gamma$-spaces, $\mathcal{I}$-spaces) fare in such a list? REPLY [9 votes]: For $E_\infty$ spaces, homotopy-theoretically there is a functor $L: \mathcal{S} \to E_\infty \mathcal{S}$ with a right adjoint $R$. The only property on this list that really needs replacing on this list is property (5): the unit $$ X \to RL(X) $$ should be homotopy equivalent to the natural inclusion $$ X \to Free_{E_\infty}(X) \simeq \coprod_{k \geq 0} E \Sigma_k \times_{\Sigma_k} (X^k) $$ into the free $E_\infty$-space on $X$. (Yes, yes, possibly a version with basepoints, I know) I believe that all three of the models of $E_\infty$ spaces that you list (commutative monoids in $*$-modules, $\Gamma$-spaces, commutative $\mathcal{I}$-space monoids) satisfy properties (1)-(4) and fail the analogue of property (5), each due to an issue about whether an input to an adjunction is cofibrant/fibrant. For $\Gamma$-spaces, for example, the map $X \to RL(X)$ only adds a disjoint basepoint. Perhaps someone with more experience with the other models would be able to fill in those stories better.<|endoftext|> TITLE: Is there a recognition principle for $\mathbb{E}_{\infty}$-spaces with zero? QUESTION [6 upvotes]: A commutative monoid with zero is a commutative monoid $A$ together with an element $0_{A}$ such that $0_{A}a=a0_{A}=0_{A}$ for all $a\in A$. They are precisely the monoids (in the sense of monoidal categories) in the category of pointed sets, equipped with the smash product. Passing to the derived world, commutative monoids get replaced by $\mathbb{E}_{\infty}$-monoids in spaces, while pointed sets get replaced by pointed spaces. So the natural analogue of a commutative monoid with zero in homotopy theory should be an $\mathbb{E}_{\infty}$-monoid in the symmetric monoidal $\infty$-category $(\mathcal{S}_*,\wedge,S^0)$ of pointed spaces, called an $\mathbb{E}_{\infty}$-space with zero. Question. The May recognition principle states that a space is (weakly equivalent to) an infinite loop space iff it is a grouplike $\mathbb{E}_{\infty}$-monoid in spaces. Is there a recognition principle for $\mathbb{E}_{\infty}$-spaces with zero? (Or of a subclass of them, such as some appropriate version of “grouplike” that works for the non-Cartesian monoidal $\infty$-category $\mathcal{S}_*$) REPLY [2 votes]: I see no real point in considering $E_{\infty}$ spaces with $0$ except in the context of the multiplicative structure of an $E_{\infty}$ ring space or, essentially equivalently, the multiplicative structure of an $E_{\infty}$ ring spectrum. $E_{\infty}$ ring spaces or even just $E_{\infty}$ spaces with $0$, come with two basepoints, say $0$ and $1$, of which only $0$ should be thought of as truly the basepoint. $E_{\infty}$ spaces with zero then give the multiplicative structure of $E_{\infty}$ ring spaces, in which one has two interrelated $E_{\infty}$ space structures, one "additive'' and one "multiplicative'' but with $0$. The $0$ in the multiplicative structure allows one to form two monads in the same category, the category of based spaces, which are interrelated in the sense prescribed in a beautiful 1969 paper "Distributive laws" of Beck. Beck's theory is recalled in section 15 of "The construction of $E_{\infty}$ ring spaces from bipermutative categories". That paper and its companions "What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra" and "What are $E_{\infty}$ ring spaces good for?'' date to 2009. They modernize the theory first developed in the 1977 Springer volume "$E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra'' that has already been referred to. It was first proven there that grouplike $E_{\infty}$ ring spaces, those for which $\pi_0$ is a group under addition and therefore a commutative ring, are equivalent to connective $E_{\infty}$ ring spectra. This and the whole of infinite loop space theory, additive, multiplicative, and especially equivariant are being reworked in a paper provisionally titled ``Group completions and the homotopical monadicity theorem'', by Hana Jia Kong, Foling Zou, and myself.<|endoftext|> TITLE: Unbounded resolutions for Grothendieck abelian categories QUESTION [12 upvotes]: Consider the following result: Theorem 1: Let $\mathsf{A}$ be a Grothendieck abelian category. Then every complex in $\mathsf{C}(\mathsf{A})$ has a $K$-injective resolution. As far as I know, the first proof of this result was in Localization in categories of complexes and unbounded resolutions. In 2003 another proof was published in Resolution of unbounded complexes in Grothendieck categories. There also seems to be a proof in Categories and Sheaves by M. Kashiwara and P. Shapira, but it also relies on some set theory and is quite long. I found no other proofs. The second proof is possibly the shortest but, due to a lot of set theory involved, seems very strange to me. A more conceptual route seems to be using the Gabriel-Popescu theorem and Brown representability to prove corollary 5.2 in Localization in categories of complexes and unbounded resolutions and then conclude the proof. That's better, but it is still quite long and difficult. Given the importance of Theorem 1, I wonder if there exist other, perhaps simpler, proofs of this result. REPLY [11 votes]: It appears that this result may go back to a 1984 letter of Joyal to Grothendieck. The reference to this letter, as well as some other early references, can be found in Example 3.2 in the paper Cotorsion pairs, model category structures, and representation theory. Later references using this approach (based on the small object argument) include Corollary 7.1 in the 2007 paper Kaplansky classes and derived categories. There is also a very recent paper of mine Derived, coderived, and contraderived categories of locally presentable abelian categories, Corollary 8.5.<|endoftext|> TITLE: Why do we care about $(\infty,2)$-categories? QUESTION [13 upvotes]: crossposted from MSE as suggested by Igor Sikora Homotopy theory provides much motivation for studying $(\infty,1)$-categories in their relations to homotopical algebra, derived geometry, stable homotopy stuffs, cohomology, physics, and so on. As for $2$-categories, one doesn't even have to motivate them since they're all over the place. However, I'm having a hard time motivating myself to study $(\infty,2)$-categories. I've been learning a bunch of facts about them: how the Duskin nerve can be regarded as an embedding from bicategories to the complicial sets model, how the Lack-Paoli nerve can be regarded as an embedding to a "simplicially enriched model", but in the end I can't see why we would want to deal with $(\infty,2)$-categories in the first place. All I've seen so far is their use in low dimensional TQFT, and also as a way to encode the $(\infty,2)$-category of $(\infty,1)$-categories, both in the context of specific models as well as in $\infty$-cosmological contexts. So (do we care, and if so) why do we care about $(\infty,2)$-categories? REPLY [10 votes]: One place where $(\infty,2)$-categories shows up is the geometric Langlands program. (As in David Ben-Zvi's comment, this is again related to the TFT example.) Indeed, local geometric Langlands is often stated as an equivalence of $(\infty,2)$-categories: $$F:D(G((t)))\operatorname{-mod}\cong\operatorname{QCoh}(\operatorname{LocSys}_{\check{G}}(\overset{\circ}{D}))\operatorname{-mod}.$$ (Actually, this is false as stated - the RHS is modified in the actual conjecture, but it's a little hard to state this modification.) Here $G((t))$ is the loop group, and $\operatorname{LocSys}_{\check{G}}(\overset{\circ}{D})$ is the moduli space of $\check{G}$-local systems on the punctured disk. $D(X)$ and $\operatorname{QCoh}(X)$ denote the derived ($\infty-$) categories of D-modules and quasicoherent sheaves on $X$, respectively. The $(\infty,2)$-categories in question are categories of modules over the respective monoidal $(\infty,1)$-categories. By the way, an $(\infty,2)$-category is an enormous amount of data, and so an equivalence of $(\infty,2)$-categories is an extremely powerful statement. For any objects $X,Y\in D(G((t)))\operatorname{-mod},$ the above conjecture predicts an equivalence (of ($\infty,1)$-categories) $\operatorname{Hom}(X,Y)\cong\operatorname{Hom}(F(X),F(Y)).$ At this point we know what $F(X)$ should be for quite a few $X,$ which makes it possible to extract a lot of nontrivial equivalences from local geometric Langlands. The simplest example is to take $X=Y=D(\operatorname{Gr}),$ $\operatorname{Gr}$ the affine Grassmannian. In this case $\operatorname{End}_{G((t))}(D(\operatorname{Gr}))$ can be identified with the derived Satake category, i.e., the derived category of $G[[t]]$-equivariant D-modules on $\operatorname{Gr}$, and you recover the derived geometric Satake equivalence.<|endoftext|> TITLE: Coefficients of modular forms and the Sato-Tate distribution QUESTION [5 upvotes]: Let $a(n)$ be the $n$th Fourier coefficient of a normalized Hecke eigenform $f(z)=\sum_{n=1}^{\infty}a(n)q^n$ of weight $k$ with respect to the full modular group, where $q=e^{i2\pi z}$. A new paper [1] has the following result. "We prove for $100\%$ of primes $p$ that \begin{equation} 2p^{\frac{k-1}{2}}\frac{\log \log p}{ \sqrt{\log p}} < |a_f (p)| < 2p^{\frac{k-1}{2}}. \end{equation} The authors claim that this is true on a subset of primes of density one, the complete details are given in Theorem 1.1. Another recent paper [3] has the following result. \begin{equation} |a_f (n)| < 2p^{\frac{k-1}{2}}\left (\log n\right)^{-1/2+o(1)} \end{equation} on a subset of integers of density one, the complete details are given in Theorem 1. And an old paper [2] has the following result. "Corollary 2. For a positive density of primes $p$, we have \begin{equation} |a(p)|> \left(\sqrt{2}-\epsilon\right)p^{\frac{k-1}{2}}. \end{equation} In [3], page 441, the author shows that \begin{equation} |a(n)|> \left(\sqrt{2}-\epsilon\right)n^{\frac{k-1}{2}}e^{c\log n /\log \log n}, \end{equation} for some constant $c>0$, which is true on a subset of positive density. Question. Can you have a pair of disjointed subsets of primes of density one as in the Theorem 1.1, and positive density as in the Corollary 2? In other words, does Corollary 2 contradict Theorem 1.1 in [1] or Theorem 1 in [3]? Thanks for your valuable comments. [1] Ayla Gafni, Jesse Thorner, Peng-Jie Wong; Almost all primes satisfy the Atkin-Serre conjecture and are not extremal; arXiv:2003.09026; doi 10.1007/s40993-021-00258-w [2] M. Ram Murty; Oscillations of Fourier Coefficients of Modular Forms; Math. Ann. 262, 431--446 (1983. [3] Florian Luca, Maksym Radziwill, Igor E. Shparlinski; On the Typical Size and Cancelations Among the Coefficients of Some Modular Forms; arXiv:1308.6606. Note: The question was edited per comments. REPLY [10 votes]: Since $(\sqrt{2} - \varepsilon) p^{(k-1)/2} < 2 p^{(k-1)/2}$, there is no contradiction between the first and third statements. The second statement as written certainly contradicts the third. On the other hand the second statement is not what is proven in the paper; that paper talks about the values for arguments at all integers $n$ not prime arguments. Since the set of primes has density zero there is no contradiction. Because of the multiplicative property of these coefficients, it is not surprising that that the behavior for general $n$ should not be exactly the same as for prime $n=p$. The entire point of the second paper is about how assuming a Sato-Tate conjecture for the prime coefficients and assuming a multiplicative condition implies various results for general coefficients.<|endoftext|> TITLE: A natural construction of real numbers? QUESTION [30 upvotes]: Summary Someone claims $\mathbb{R}$ can be constructed as the following intriguing quotient, which is related to Gromov's bounded cohomology. I want to find out if it is true. $$\frac{\bigl\{f:\mathbb{Z} \to \mathbb{Z} \mathrel| \mbox{ the set } \{f(m+n)-f(m)-f(n) \mathrel| m, n \in \mathbb{Z}\} \mbox{ is bounded}\bigr\}}{\{f:\mathbb{Z} \to \mathbb{Z} \mathrel| f \mbox{ is bounded}\} }.$$ EDIT See KConrad's comment below. A similar construction, described in Hermans - An elementary construction of the real numbers, the $p$-adic numbers and the rational adele ring, yields $\mathbb{Q}_p$ and the rational adele ring $\mathbb{A}$. Main text In A'Campo - A natural construction for the real numbers, a natural construction of the real numbers is given as follows. (EDIT: In this post I only address the bijection and the ring structure. The correspondence is complete; for that please refer to the paper.) Definition (Bounded cochains) Define $C^{n} = C^{n}(\mathbb{Z})$ to be $\operatorname{Map}(\mathbb{Z}^{\times n}, \mathbb{Z})$ and $C^n_b = C^{n}_b(\mathbb{Z})$ to be the subset consisting of functions $f$ having bounded image, i.e. $\operatorname{Card}(\operatorname{Im}(f)) < \infty$. Definition (Differentials) Define $d: C^n \to C^{n+1}$ to be such that $$df(x_1,\dotsc,x_{n+1}) = f(x_2,\dotsc,x_{n+1}) + \sum_{k=1}^{n}(-1)^{k} f(x_1, \dotsc, x_{k-1}, x_k+x_{k+1}, \dotsc, x_{n+1}) + (-1)^{n+1}f(x_1,\dotsc,x_n).$$ Obviously, $d(C^n_b) \subseteq C^{n+1}_b$, so $C^n_b \subseteq d^{-1}(C^{n+1}_b)$. Algebraic Operations Clearly, $C^1$ has a ring structure, where addition is given by point-wise addition, and multiplication is given by function composition. Claim. $\mathbb{R} \simeq d^{-1}(C^2_b)/(C^1_b)$ This claim is made in page 1 (definition of $\mathbb{R}$) and page 6 (that $\mathbb{R}$ is the usual $\mathbb{R}$) of the paper. An explicit map $\Phi: d^{-1}(C^2_b) \to \mathbb{R}$ is given in page 4 as $$\lambda \mapsto \left[\left(\frac{\lambda (n+1)}{n+1}\right)_{n \in \mathbb{N}}\right]$$ using Cauchy sequences. Question Why is $\ker(\Phi) = C^1_b$? By the definition of the equivalence on the set of Cauchy sequences, $\Phi(\lambda)$ represents $0 \in \mathbb{R}$ if and only if For each $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\frac{|\lambda(n+1)|}{(n+1)} < \epsilon$ whenever $n > N$. However, $\lambda: \mathbb{Z} \to \mathbb{Z}$ that sends $n$ to $\lfloor\sqrt{|n|}\rfloor$ is one such element that is not in $C^1_b$ (namely, not bounded). EDIT As Anthony Quas points out below, such $\lambda$ isn't in the preimage of $d$. You can see this by taking $m = n \to \infty$. Still, I'm curious about a direct proof for the kernel being $C^1_b$. This is given in Anthony Quas's answer. Related Category-theoretic description of the real numbers (Mathematics Stack Exchange) Gromov's bounded cohomology, see Ivanov - Notes on the bounded cohomology theory and the 9th page of A'Campo's paper. REPLY [10 votes]: This is an extended comment. A natural (in my mind) way to capture this construction is by the following categorical interpretation. Consider the category $\text{ACG}$ of Abelian Coarse Groups, which are abelian group objects in the category of coarse spaces, as defined here*. This is an additive category. Note that every abelian locally compact topological group $G$ could be seen as a coarse group, by setting $E\subset G\times G$ to be controlled iff $\{x-y\mid (x,y)\in E\}$ is precompact in $G$. One sees easily that the inclusion $\mathbb{Z}\hookrightarrow \mathbb{R}$ is an isomorphism in $\text{ACG}$, thus we get a natural isomorphism of rings, $\text{End}_\text{ACG}(\mathbb{Z})\simeq \text{End}_\text{ACG}(\mathbb{R})$. One also sees easily that the natural map of rings $\mathbb{R} \simeq \text{End}_\text{AG}(\mathbb{R}) \to \text{End}_\text{ACG}(\mathbb{R})$ is an isomorphism (here $\text{AG}$ stands for Abelian Groups). One gets a natural ring isomorphism $\mathbb{R} \simeq \text{End}_\text{ACG}(\mathbb{Z})$ by composing the above maps. Further, Hom sets in the category of abelain coarse groups could be naturally topologized, and this is an isomorphism of topological rings. Similarly, one gets that $\mathbb{Q}\hookrightarrow \mathbb{A}$ is an isomorphism in $\text{ACG}$ (here $\mathbb{A}$ stands for the rational adels) and concludes an isomorphism of topological rings $\mathbb{A}\simeq \text{End}_\text{ACG}(\mathbb{Q})$. $*$ For our discussion here, we take for morphisms in the category of coarse spaces classes of controlled maps, that is maps $f:X\to Y$ such that for every controlled $E\subset X\times X$, $f\times f(E)\subset Y\times Y$ is controlled, and $f\sim g:X\to Y$ if $f\times g(X\times X)$ is bounded in $Y$. We note that the category thus obtained has products, so talking about group objects makes sense.<|endoftext|> TITLE: Cyclic amalgamation of solvable groups QUESTION [6 upvotes]: Given a nonempty set of groups $\{G_\lambda|\lambda \in \Lambda\}$ together with a group $H$ which is isomorphic with a subgroup $H_\lambda$ of $G_\lambda$ by means of monomorphism $\phi_\lambda: H \to G_\lambda$. There is a frequently used term known as the free product of the $G_\lambda$'s with the amalgamated subgroup $H$. Roughly speaking, this is the largest group generated by the $G_\lambda$'s in which the subgroups $H_\lambda$ are identified with $H$ by $\phi_\lambda$. Such groups are known as generalized free products. My question concerns a result proved by Kahorbaei and Majewicz claiming the following. Question. I'm uncertain about this proof. Why must the cosets $a\delta_{m+1}A$ and $b\delta_{n+1}B$ have the same order in its respective factor group? I'm worried about the infinite cyclic subgroup. Here is the definition for residually solvable group. Edit: I suspect there are counterexamples to the statement (or maybe I misunderstood their definition of the generalized free product). It seems that it requires extra hypothesis such as cosets of same order or something called compatible filtration in Proposition 1 of Baumslag (as suggested by Carl-Fredrik Nyberg Brodda). REPLY [3 votes]: The elements do not have to have the same order it would seem. According to the paper ON TORSION-FREE METABELIAN GROUPS WITH COMMUTATOR QUOTIENTS OF PRIME EXPONENT by Narain Gupta and Said Sidki there are torsion-free metabelian groups whose commutator quotients are nontrivial with prime exponent for any odd prime $p$. So choosing examples $A$ and $B$ with different primes, say $3$ and $5$, gives infinite order elements $a\in A$ and $b\in B$ (not in the commutator subgroup) that have orders $3$ and $5$, respectively, modulo the commutator subgroup.<|endoftext|> TITLE: The 4th vertex of a triangle? QUESTION [12 upvotes]: I was immensely surprised and amused by the idea of the fourth side of a triangle that was introduced by B.F.Sherman in 1993. 'Sherman's Fourth Side of a Triangle' by Paul Yiu is available here. Naturally a question arises whether it is possible to determine the '4th vertex' of a triangle in a somewhat similar way? Basically a vertex of the given triangle can be described as a point that is lying on its circumcircle. Then the middle point of the segment connecting this point to the Orthocenter must belong to the nine point circle of the triangle. (This holds true for any point that is lying on the circumcircle.) Finally it appears that we should just add another equation somehow linking our 'vertex' to the inscribed circle. In other words, there might exist some triangle center on the circumcircle that is also satisfying a certain condition related to the inscribed circle, so that eventually this point can be called the fourth vertex of the triangle. Luckily I found in my files a construction of a triangle center X that in a sense satisfies these conditions: It belongs to the circumcircle of ABC The midpoint of the segment XX(4) belongs to the nine point circle. The constriction of X primarily relies on the Incenter of the triangle (i.e. the inscribed circle) Last but not least, this point X is not included in Kimberling's encyclopaedia: A',B',C' is the circumcevian triangle with respect to the Incenter I. Lines AB and A'B' intersect at point C'', points B'', A'' are defined cyclically. Circumcircles for the triangles IRA, IRB, IRC were drawn. These 3 circles intersect the circumcircle at the points A''', B''', C'''. Finally A''A''',B''B''',C''C''' always cross each other at some point X that conveniently belongs to the circumcircle of the original triangle ABC. Geogebra dynamic sketch. It is highly speculative to call our point X the fourth vertex of a triangle, of course. Presumably this attribution will be outright dismissed or proven wrong. However I assume that perhaps a better fit for the 4th vertex of a triangle can be found? What would its construction be? REPLY [17 votes]: I propose an alternative "fourth vertex". To paraphrase Sherman's result: The "fourth side" ($w$) of a triangle is a chord of the circumcircle, is a tangent to the incircle ($\bigcirc I$), and is bisected by the nine-point circle ($\bigcirc N$). (The last aspect is equivalent to $w$ intersecting the nine-point circle at the foot of the perpendicular from circumcenter $K$.) Thus, this "fourth side" fits a description that applies to the (standard) three; importantly, there is exactly one such "fourth side" that can do so. My proposal: The "fourth vertex" ($W$) of a triangle determines a line through the orthocenter ($H$) that intersects the nine-point circle at the foot of a perpendicular tangent to the incircle, and is such that the midpoint of $\overline{WH}$ is the "other" intersection with the nine-point circle. (The last part distinguishes $W$ from the other endpoint of the chord determined by $\overleftrightarrow{WH}$.) Note. As observed in comments, point $W$ is Kimberling's triangle center X(1309). As before, the ostensible "fourth vertex" fits a description that applies to the (standard) three; also, as it turns out, there is exactly one "fourth vertex" that can do so. The simplicity of this proposed definition has some appeal, but what makes it particularly relevant is a fact the reader may have suspected from the images: The two "fourth" elements involve the same construction. The line of segment $w$ serves as the tangent to $\bigcirc I$ perpendicular to $\overline{WH}$ where they meet on $\bigcirc N$; that is, Sherman could have —and may have (I haven't checked)— noted: $w$ intersects the nine-point circle again at the foot of the perpendicular from orthocenter $H$. Proof isn't complicated (essentially all that is needed is to rotate some elements around the center of the nine-point circle), but getting too caught-up in the triangle context is unnecessarily limiting. It's better to see these "three elements and a spare" results as special cases of a broader "four elements" result. Lemma. Suppose tangents at $A$, $B$, $C$ of $\bigcirc P$ meet $\bigcirc Q$ at feet ($A_+$, $B_+$, $C_+$) of perpendiculars from a common point $R_+$. Then there is a unique point $D$ of $\bigcirc P$ such that the tangent at $D$ meets $\bigcirc Q$ at the foot ($D_+$) of the perpendicular from $R_+$. Moreover, reflecting $R_+$ in $Q$ gives a point $R_-$ such that the feet ($A_-$, $B_-$, $C_-$, $D_-$) of perpendiculars to the tangents lie on $\bigcirc Q$. (In the triangle context, $\bigcirc P$ and $\bigcirc Q$ are respectively the incircle and nine-point circle; lines $\overleftrightarrow{A_+A_-}$, etc, contain the "four sides"; and points $R_+$ and $R_-$ are the circumcenter and orthocenter.) The "moreover" follows from recognizing that a $180^\circ$ rotation about $Q$ effectively creates four inscribed rectangles. This rotational symmetry guarantees that sides of these rectangles concur iff the opposite sides concur. The main part of the Lemma can be proven with some light coordinate bashing. For instance, taking $P=(0,0)$ and $R_+=(r,0)$, defining angles $\alpha:=\angle R_+PA$, $\beta:=\angle R_+PB$, $\gamma:=\angle R_+PC$, $\delta:=\angle R_+PD$, and letting the radius of $\bigcirc P$ be $p$, then $$A_+ = R_+ + (p - r \cos\alpha) (\cos\alpha,\sin\alpha), \qquad B_+=\cdots, \qquad C_+=\cdots, \qquad D_+=\cdots$$ We find that $A_+$, $B_+$, $C_+$, $D_+$ are concyclic iff $$2 p \sin\sigma = r \left(\;\sin(\sigma-\alpha) +\sin(\sigma-\beta) + \sin(\sigma-\gamma) + \sin(\sigma-\delta)\;\right) \tag{1}$$ where $\sigma:=\frac12(\alpha+\beta+\gamma+\delta)$. Condition $(1)$ determines any one angle from the other three; eg, we can write $$\begin{align} &\;\phantom{-}\cos\tfrac12\delta\left(\; 2 p \sin\tau - r ( \sin(\tau-\alpha) + \sin(\tau-\beta) + \sin(\tau-\gamma) + \sin\tau) \;\right) \\[0.5em] = &-\sin\tfrac12\delta \left(\; 2 p \cos\tau - r (\cos(\tau-\alpha) + \cos(\tau-\beta) + \cos(\tau-\gamma) - \cos\tau)\;\right) \end{align} \tag{1'}$$ where $\tau:=\frac12(\alpha+\beta+\gamma)$. This identifies $\delta/2$ up to a half-turn, hence $\delta$ up to a full turn, guaranteeing the unique $D$ claimed in the Lemma. $\square$ It is perhaps worth noting that (barring degeneracies) $(1)$ allows for determining distance $r=|PR_+|$ to make any chosen collection of distinct angles (ie, any chosen collection of points $A$, $B$, $C$, $D$ on $\bigcirc P$) "work" to give the concurrencies shown in the figure.<|endoftext|> TITLE: Chain complexes split in the derived category over rings of global dimension 1 QUESTION [8 upvotes]: Let $R$ be a ring of global dimension $1$. Then I have seen the claim (in a paper, and in this MO post When do chain complexes decompose as a direct sum?) that any chain complex over $R$ is equivalent to its cohomology as an object of the derived category of $R$-module. How does one prove such a thing? (Either a sketch of the argument or a reference would be appreciated). REPLY [8 votes]: One reference is H. Krause, "Derived categories, resolutions, and Brown representability", Contemporary Math. vol.436, AMS, 2007, p.101-139 or https://arxiv.org/abs/math/0511047 , Section 1.6. Another possible reference is L. Positselski, O.M. Schnürer, "Unbounded derived categories of small and big modules: Is the natural functor fully faithful?", J. Pure Appl. Algebra 225 (2021) Paper No. 106722 or https://arxiv.org/abs/2003.11261 , Appendix A.<|endoftext|> TITLE: Smooth rank one foliations with closed leaves QUESTION [8 upvotes]: Let $F$ be a smooth rank one foliation on a manifold $M$. Suppose that all leaves of $F$ are compact (that is, circles). Then its leaf space (edit: when additional assumptions are taken) is an orbifold. When the foliation is (transversally) Riemannian, this is proven in the book "Riemannian Foliations" by P. Molino. I suppose that this result is classical without the Riemannian assumption, but I could not find its proof in the literature. Any help with references is highly appreciated. REPLY [5 votes]: The result is not true without additional assumptions. See A counterexample to the periodic orbit conjecture by Sullivan. Added later. The paper by Sullivan linked above exhibits a foliation by circles on a compact manifold of dimension $5$ with non-Hausdorff leaf-space. Later, Epstein and Vogt constructed a foliation by circles on a compact manifold of dimension $4$ with the same property, see A Counterexample to the Periodic Orbit Conjecture in Codimension 3. The problem makes sense for foliations of arbitrary dimensions. No need to be restricted to foliation by curves. There are also examples of holomorphic foliations having all its leaves compact but with non-Hausdorff leaf-space, on non-compact complex manifolds, see On the stability of holomorphic foliations by Holmann. As far as I know, there are no examples in the literature of holomorphic foliations with all its leaves compact and with non-Hausdorff leaf-space on a compact complex manifold. REPLY [3 votes]: The question was already answered by Jorge Vitório Pereira, but let me add here what I have already found. Recall that a foliation on a Riemannian manifold is called "Riemannian foliation" if the restriction of the Riemannian metric to the normal bundle is preserved by the holonomy of the foliation. The leaf space of a Riemannian foliation with compact leaves is an orbifold, a result which is probably due to Molino. My question was related to the following question, which Sullivan calles "the periodic orbit conjecture". Suppose that $F$ is a rank foliation on a compact manifold with compact fibers. Is it true that there exists a circle action tangent to $F$? The answer is positive in dimension 3 (D. B. A. Epstein, Periodic Flows on Three-Manifolds, Ann. M ath., 95 (1972), 66--82.) and false in dimension 5 (D. Sullivan, A counterexample to the periodic orbit conjecture Publications Mathématiques de l'IHÉS, Tome 46 (1976) , pp. 5-14.) When the foliation is Riemannian, the periodic flow conjecture is proven in the thesis of A. W. Wadsley, a student of Epstein (A. W. Wadsley, Geodesic foliations by circles, J. Differential Geometry {\bf 10} (1975), no. 4, 541--549.)<|endoftext|> TITLE: What is a Serre-smooth algebra? QUESTION [5 upvotes]: Let $A$ be an $R$-algebra. In the book "Noncommutative Geometry and Cayley-smooth Orders" by Le Bruyn one can find the notion of "Serre-smooth" in the introduction. But no formal definition seems to be given in this (very long) introduction. It is just mentioned that $A$ is Serre-smooth if $A$ has finite global dimension together with some extra features such as Auslander regularity or the Cohen-Macaulay property. Then it is refered to the article https://www.cambridge.org/core/journals/glasgow-mathematical-journal/article/some-properties-of-noncommutative-regular-graded-rings/244AE0A8E9C8AB6782515B504F1AC5C0 but there I can not find the word "Serre" (searching with Strg+F). Question: Is there a reference for the complete definition of being Serre-smooth? REPLY [5 votes]: No, there is no such reference. The introduction to that book is based on a couple of lectures I gave in Luminy and there I had to distinguish between several notions of 'smoothness', formal smoothness a la Kontsevich-Rosenberg, Cayley-smoothness, and Artin-Schelter- or Auslander-Gorenstein-regularity as used by people working in NAG. For the later category I then used the term 'Serre-smoothness' as it is the noncommutative equivalent of smoothness for commutative affine algebras (finite global dimension) to the noncommutative world. If one only considers (maximal) orders in central simple algebras which are finite modules over their centers then one can do with less than the full repertoire of these homological conditions, I think. Please read 'Serre-smoothness of A' as 'A is an Auslander-Gorenstein regular ring'. My apologies for the confusion this ad-hoc terminology may have caused.<|endoftext|> TITLE: Theorems with many distinct proofs QUESTION [36 upvotes]: I was told that whenever one learns a new technique, it is a good idea to see if one can prove a well-known theorem using the new technique as an exercise. I am hoping to build a list of such theorems to test my technique. My Question. What are some theorems with a.) many different proofs, or b.) proofs using strikingly different techniques from mathematics? Theorems with multiple proofs: Fundamental theorem of algebra Fundamental theorem of arithmetic Existence of Jordan normal form. (Standard minimal polynomial proof, PID proof, Terry Tao's proof, proof using complex analysis) (Related to prev.) Cayley-Hamilton theorem Spectral theorem Quadratic reciprocity law Pólya's recurrence theorem Basel problem Stirling's formula (Arguably:) Uniform boundedness principle. (Pf 1. Baire category, Pf 2. Gliding hump) Brouwer fixed point theorem (although all the proofs I know of boil down to showing that the ball is not homeomorphic to the sphere) I think the Pythagorean theorem also satisfies my description, but it is a bit too elementary. Clarification 1: In order to prevent this question from being "what are some theorems that have higher level generalizations," a trivial specialization of a harder theorem (e.g. Hilbert space Pythagorean theorem implying Euclidean Pythagorean theorem) will not be considered a new proof, for the purposes of this question. Clarification 2: This question differs from this similar stackexchange question in the following sense. The stackexchange post asked for a.) very elementary theorems (lower level undergrad), and b.) short proofs. My question asks for theorems from all levels of mathematics (up to say third year graduate level), which are (preferably) central to the theory. REPLY [2 votes]: The local monodromy theorem is a fundamental result on the topology of families of complex algebraic varieties. Let $X \rightarrow B$ be a smooth projective family of complex algebraic varieties over a smooth complex algebraic curve $B$, complement of a finite set $S$ of points in a compact curve $\overline{B}$. Then the local monodromy theorem states that for every point $s \in S$, the monodromy action around $s$ on the cohomology (with rational coefficients) of the fibers of $X \rightarrow B$ is quasi-unipotent (i.e. all its eigenvalues are roots of unity). There are many proofs of this theorem, using very different ideas (a summary is given in section 3 of https://www.ams.org/journals/bull/1970-76-02/S0002-9904-1970-12444-2/S0002-9904-1970-12444-2.pdf ): there is a "global geometric" proof due to Landman, using Lefschetz pencils and induction on the dimension. there is a "local geometric" proof due to Grothendieck in the algebraic setting and Clemens in the Kähler setting, based on a local study of vanishing cycles. there is a "complex hyperbolic geometric" proof due to Borel using the differential geometric fact that moduli spaces of polarized Hodge structures are negatively curved. there is a proof by Brieskorn using the regularity of the Gauss-Manin connection and the Gelfond-Schneider theorem on transcendental numbers. there is an "arithmetic" proof due to Grothendieck based on étale cohomology, reduction to positive characteristic, and general properties of l-adic Galois representations. there is another "arithmetic" proof due to Katz, using the study of the Gauss-Manin connection in positive characteristic.<|endoftext|> TITLE: Role of univalence in homotopy group calculations QUESTION [15 upvotes]: This book has a section with proofs of the fact $\pi_1(S^1)=\mathbb Z$ using the univalence axiom. They are a bit too technical for me at the moment to read, but I want to understand the following (vague but conceptual) question: What is the role of the univalence axiom in these proofs? Usually, univalence is motivated with a slogan like "isomorphic structures are equal". However, it seems the application of univalence in the proofs of $\pi_1(S^1)=\mathbb Z$ must be of a completely different kind than "treating isomorphic algebraic structures as the same". So I really would like to get a hint at how univalence can be relevant to such concrete questions, as a motivation for reading the proofs in more detail. REPLY [11 votes]: This is essentially the same content as earlier answers, but I’ll try to emphasise the aspect OP is asking about a bit more explicitly. $\newcommand{\Z}{\mathbb{Z}}$I’ll use the terminology of “paths” rather than “equalities”, to emphasise the space-like viewpoint. Classically, how do you investigate the structure of a free group, or a group presented by generators and relations? In particular, how do you know it’s non-trivial? You look at actions or representations of it — concretely presented groups, into which the presented group maps. The circle $S^1$ is defined as the type freely generated by a point and a loop at that point. $\pi_1(S^1)$ is the total type of loops at the basepoint. There’s a clear map $\Z \to \pi_1(S^1)$, by taking integer powers of the generating loop. But how do you show this is non-trivial? Since it’s a freely presented structure, you need to find maps into concretely presented types. Univalence says that paths between types correspond to equivalences between them. It follows fairly directly that paths between types-with-structure correspond to structure-preserving-equivalences — e.g. paths in the type of groups correspond to group isomorphisms. Take $\Z$, equipped with the successor endomorphism $s$. As a type-with-endomorphism $(\Z,s)$, it has $\Z$-many automorphisms, the integer powers of $s$ — you can calculate this in HoTT exactly the same way you would calculate it classically. By univalence, it follows that it has $\Z$-many paths to itself, all the integer powers of the basic loop corresponding to $s$. $\newcommand{\TyEnd}{\textrm{TyEnd}}$ In other words, $\pi_1(\TyEnd,(\Z,s)) \cong \Z$, where $\TyEnd$ is the type of types-with-endomorphism. So the universal property of $S^1$ gives a map into $\TyEnd$, taking its basepoint to $(\Z,s)$ and its generating loop to (the path corresponding to) $s$. This shows that the $\Z$-many powers of the loop in $\pi_1(S^1)$ must be distinct, since they get mapped to the automorphisms of $(\Z,s)$, which we know are distinct. (And with a little more work in similar tools, we can show the map $\Z \to \pi_1(S^1)$ is surjective.) So the take-home in all this is: $S^1$ is the free type-with-a-loop. So to investigate its structure (in particular, $\pi_1(S^1)$), we want to find types with non-trivial loop spaces, or better still, where we can calculate loop spaces explicitly. Univalence tells us that loops in types of types-with-structure correspond to automorphisms of such types-with-structure; so it gives us types where there are interesting loop spaces that we can calculate explicitly.<|endoftext|> TITLE: Classifying Hopf algebras that admit a single irreducible comodule QUESTION [5 upvotes]: Is it possible to classify Hopf algebras $H$, over a field $k$, which admit a unique (up to isomorphism) irreducible comodule, namely the trivial $1$-dim comodule $$ k \to k \otimes H, ~~ k \mapsto k \otimes 1_H. $$ REPLY [3 votes]: There have been several papers published on such Hopf algebras (which are referred to as "connected" Hopf algebras) over the past decade. In particular, over an algebraically closed field of characteristic 0, the discovery of non-commutative + non-cocommutative examples has helped spur on the interest in classifying these algebras. See, for example, the following: Properties of pointed and connected Hopf algebras of finite Gelfand-Kirillov dimension Quantum homogeneous spaces of connected Hopf algebras Connected (graded) Hopf algebras In general there is no known concrete classification of such connected Hopf algebras. In particular, the last of the papers listed above presents a non-cocommutative, non-commutative connected Hopf algebra which is not isomorphic as an algebra to the universal enveloping algebra of any lie algebra.<|endoftext|> TITLE: Examples of $\mathbb{E}_{k}$-semiring spaces QUESTION [5 upvotes]: Semirings, also called rigs, are rings without negatives: their underlying additive monoids are not groups (in other words, while rings are monoids in $(\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z})$, semirings are monoids in $(\mathsf{CMon},\otimes_{\mathbb{N}},\mathbb{N})$). Examples include all rings, but also objects like The natural numbers semiring $(\mathbb{N},+,\cdot,0,1)$, the monoidal unit of $(\mathsf{Semirings},\otimes_{\mathbb{N}},\mathbb{N})$; The Boolean semiring $\mathbb{B}=\{0,1\}$, with semiring structure characterised by $1+1=1$; The tropical semiring $\mathbb{T}=(\mathbb{R}\cup\{\infty\},\min,+)$, and its sibling $\mathbb{A}=(\mathbb{R}\cup\{-\infty\},\max,+)$, sometimes called the Arctic semiring; The semiring $(\mathrm{Idl}(R),+,\cdot,(0),R)$ of ideals of a ring $R$. When we pass to the $\infty$-world, we replace the symmetric monoidal category $(\mathsf{Sets},\times,\mathrm{pt})$ of sets and maps between them by the symmetric monoidal $\infty$-category $(\mathcal{S},\times,\mathrm{pt})$ of spaces, and also commutative monoids and groups by $\mathbb{E}_{\infty}$-monoids and $\mathbb{E}_\infty$-groups. As a result, the immediate analogues of $\mathsf{CMon}$ and $\mathsf{Ab}$ become the $\infty$-categories $\mathsf{Mon}_{\mathbb{E}_{\infty}}(\mathcal{S})$ and $\mathsf{Grp}_{\mathbb{E}_{\infty}}(\mathcal{S})$ of $\mathbb{E}_\infty$-spaces and grouplike $\mathbb{E}_{\infty}$-spaces. The latter of these is equivalent to the $\infty$-category of connective spectra $\mathsf{Sp}_{\geq0}$, which embeds into the $\infty$-category of all spectra $\mathsf{Sp}$, the “true” analogue of $\mathsf{Ab}$ in homotopy theory. The $\infty$-categories $\mathcal{S}_*$, $\mathsf{Mon}_{\mathbb{E}_{\infty}}(\mathcal{S})$, $\mathsf{Grp}_{\mathbb{E}_{\infty}}(\mathcal{S})$, and $\mathsf{Sp}$ all admit symmetric monoidal structures, determined uniquely by the requirement that the free functors from $\mathcal{S}$ to them can be equipped with a symmetric monoidal structure (see Theorem 5.1 of Gepner–Groth–Nikolaus's Universality of multiplicative infinite loop space machines, arXiv:1305.4550). We can thus consider $\mathbb{E}_{k}$-monoids in these categories: For $(\mathsf{Sp},\otimes_{\mathbb{S}},\mathbb{S})$, we get $\mathbb{E}_{k}$-ring spectra; For $(\mathsf{Grp}_{\mathbb{E}_{\infty}}(\mathcal{S}),\otimes_{QS^0},QS^0)$, we get $\mathbb{E}_{k}$-ring spaces, which are equivalent to connective $\mathbb{E}_{k}$-ring spectra; Finally, for $(\mathsf{Mon}_{\mathbb{E}_{\infty}}(\mathcal{S}),\otimes_{\mathbb{F}},\mathbb{F})$, we obtain $\mathbb{E}_{k}$-semiring spaces. The last of these provides a partial (i.e. connective) analogue of semirings in homotopy theory. However, examples of $\mathbb{E}_{k}$-semiring spaces (that are not $\mathbb{E}_{k}$-ring spectra) are harder to come by. A motivating example of them is given by the $\mathbb{E}_{\infty}$-space $\mathbb{F}\overset{\mathrm{def}}{=}\coprod_{n=0}^{\infty}\mathbf{B}\Sigma_{n}$―the classifying space of the groupoid of finite sets and permutations―which can be given the structure of an $\mathbb{E}_{\infty}$-semiring space, becoming the monoidal unit of the symmetric monoidal $\infty$-category $\mathsf{Semirings}_{\mathbb{E}_{\infty}}(\mathcal{S})$ of $\mathbb{E}_{\infty}$-semiring spaces. Additionally, $\mathbb{F}$ is the spectral analogue of the semiring $\mathbb{N}$ of natural numbers, and, by the multiplicative Barratt–Priddy–Quillen–Segal theorem, its $\mathbb{E}_{\infty}$-ring space completion (i.e. $QS^0\otimes_{\mathbb{F}}\mathbb{F}$) is $QS^0$, corresponding to the sphere spectrum $\mathbb{S}$ under the equivalence $\mathsf{Ring}_{\mathbb{E}_{\infty}}(\mathcal{S})\cong\mathsf{RingSp}_{\geq0}$. Questions: What are some other examples of $\mathbb{A}_{k}$-semiring or $\mathbb{E}_{k}$-semiring spaces? What are some examples of homotopy associative/commutative semiring spaces, i.e. monoids or commutative monoids in $(\mathsf{Ho}(\mathcal{S}),\otimes_{\mathbb{F}},\mathbb{F})$? Finally, do we have semiring space analogues of $\mathbb{B}$, $\mathbb{T}$, $\mathbb{A}$, and $\mathrm{Idl}(R)$? REPLY [4 votes]: Here are some interesting examples of symmetric bimonoidal groupoids $\mathcal{C}$. In each case, the resulting classifying space $B\mathcal{C}$ is an $E_\infty$ ring space, whose group completion is $\Omega^\infty K(\mathcal{C})$. Fix a finite group $G$, and let $\mathcal{F}_G$ be the groupoid of finite $G$-sets. Then $K(\mathcal{F}_G)=\bigvee_{(H)}\Sigma^\infty_+BW_G(H)$, where $H$ runs over the conjugacy classes of subgroups of $G$, and $W_G(H)=N_G(H)/H$ is the Weyl group. This can also be described as the Lewis-May fixed point spectrum $(S_G)^G$ of the $G$-equivariant sphere spectrum. More generally, if $X$ is any finite $G$-set and $\mathcal{F}_G/X$ is the groupoid of finite $G$-sets equipped with a map to $X$, then $K(\mathcal{F}_G/X)=\left(\Sigma^\infty_+X\right)^G$. Now let $\mathcal{F}_G^f$ be the subgroupoid of free $G$-sets, or equivalently, the groupoid of finite sets equipped with a $G$-torsor. Then $B(\mathcal{F}_G^f)\simeq\coprod_nB(G\wr \Sigma_n)$, which is the total extended power of $BG$. We also have $K(\mathcal{F}_G^f)=\Sigma^\infty_+BG$. Fix a field $F$, and let $\mathcal{V}_F$ be the groupoid of finite-dimensional vector spaces over $F$. Then $B(\mathcal{V}_F)\simeq\coprod_nBGL_n(F)$ and $K(\mathcal{V}_F)$ is just the algebraic $K$-theory spectrum $K(F)$. By a trellis in $V$ I mean an unordered set of one-dimensional subspaces of $V$ whose direct sum is $V$. Let $\mathcal{T}_F$ be the groupoid of finite-dimensional vector spaces over $F$ equipped with a trellis; this has a fairly evident symmetric bimonoidal structure. Then $B(\mathcal{T}_F)\simeq\coprod_nB(GL_1(F)\wr\Sigma_n)$ and $K(\mathcal{T}_F)=\Sigma^\infty_+BGL_1(F)$. The forgetful functor $\mathcal{T}_F\to\mathcal{V}_F$ gives rise to a well-known map $\Sigma^\infty_+BGL_1(F)\to K(F)$ of ring spectra. By a multiset I mean a finite set $X$ equipped with a multiplicity function $m_X\colon X\to\mathbb{N}$. A morphism of multisets is a function $f\colon X\to Y$ that is bijective up to multiplicity in the sense that $\sum_{f(x)=y}m_X(x)=m_Y(y)$ for all $y\in Y$. We write $\mathcal{M}$ for the category of multisets, which has a fairly evident symmetric bimonoidal structure. There is an adjunction linking $\mathcal{M}$ to the discrete category $\mathbb{N}$, so $B(\mathcal{M})\simeq\mathbb{N}$ and $K(\mathcal{M})\simeq H$. The Eilenberg-MacLane spectrum $H$ can also be interpreted as the infinite symmetric product $SP^\infty(S)$, so it is filtered by subspectra $SP^k(S)$, which have been important for a number of applications. We can define $\mathcal{M}_k=\{X\in\mathcal{M}\;:\;m_X(x)\leq k\text{ for all }x\}$, then it turns out that $K(\mathcal{M}_k)\simeq SP^k(S)$. This is a reinterpretation of work of Kathryn Lesh, from which one can also extract some information about $B(\mathcal{M}_k)$. The product functor $\mathcal{M}\times\mathcal{M}\to\mathcal{M}$ sends $\mathcal{M}_p\times\mathcal{M}_q$ to $\mathcal{M}_{pq}$. Using this, we can make $\coprod_n\mathcal{M}_n$ into a symmetric bimonoidal category with a symmetric bimonoidal functor to $\mathcal{M}$. This corresponds to a well-known ring map $\bigvee_nSP^n(S)\to SP^\infty(S)=H$.<|endoftext|> TITLE: Converse of mean value theorem almost everywhere? QUESTION [5 upvotes]: Let $f: \mathbb R \to \mathbb R$ be a $C^1$ function. We say a point $c \in \mathbb R$ is a mean value point of $f$ if there exists an open interval $(a,b)$ containing $c$ such that $f’(c) = \frac{f(b) - f(a)}{b-a}$. Question: Is it true that (Lebesgue) almost every point in $\mathbb R$ is a mean value point of $f$? REPLY [5 votes]: Let $U$ be an open and dense subset of $\mathbb{R}$ with finite measure. Let $g: \mathbb{R} \to \mathbb{R}^{\ge 0}$ be a continuous function with $\{g = 0\} = U^c$. Then define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = g(0)+\int_0^x g(t)dt$. Then $f \in C^1$, and $f' \equiv g$ means $f$ is strictly increasing (since any interval $(x,y)$ contains an interval lying in $U$ on which $g$ is strictly positive). So we have a strictly increasing $C^1$ function with zero derivative everywhere except for a finite measure set.<|endoftext|> TITLE: Roadmap for L-Theory QUESTION [12 upvotes]: Background: I spent sometime reading about algebraic K-theory and started reading research papers on the subject with relative facility at least I do understand constructions, statements of the Theorems and (some) proofs. As a student, I feel more comfortable with standard algebraic topology and topological manifolds. I start to read some classical papers on L-theory (essentially Ranicki papers ). The problem is that I can't make a connection with algebraic K-theory (Quillen, Waldhausen, Thomason,...) One example is the multiple variation of The L-Theory (Symmetric, Quadratic, Hermitian,...). Here is my question: Could someone indicate how can a student be initiated to the subject of L-theory. I need to understand some basic ideas and more importantly I wish to know how to learn L-Theory assuming that I can understand K-Theory. REPLY [10 votes]: I apologize for the self promotion -- I hope the content of this answer can be useful anyway... My favourite introduction to L-theory is Lurie's notes on Algebraic L-theory and surgery (warning: aggressively modern). Working from the ideas in this notes my coauthors and I have built a series of papers (I, II, III) developing L-theory and hermitian K-theory in a fashion that closely mirrors the development of algebraic K-theory. One of the main point of our work is to find a home for all (well, almost all) variants of L-theory in a single unified framework, that of Poincaré structures. These papers are probably not very good for a student though -- while we do all the details, this results in many pages of technical results and it's easy to lose the forest for the trees. If you are interested in this circle of ideas some of my coauthors have given a series of minicourses on the topic (Fabian Hebestreit's I, II, Markus Land I, II, III, IV; Yonatan Harpaz I, II, III) which are probably more accessible than the papers.<|endoftext|> TITLE: Different smooth structures on the infinite jet bundle (for the purposes of calculus of variations) QUESTION [10 upvotes]: Let $\pi:Y\rightarrow X$ be a (smooth, finite dimensional) fibred manifold. Since no other fibrations will be considered on $Y$, I will identify $(Y,\pi,X)$ with $Y$. The finite order jet bundles are denoted $J^rY$ ($r\in\mathbb N$, $J^0Y=Y$), and the infinite jet bundle $J^\infty Y$ is - as a topological space - the projective limit of the system $(J^rY,\pi^r_s)$ where $\pi^r_s:J^rY\rightarrow J^sY$, $r\ge s$ is the standard projection. In the literature, there exists several ways through which $J^\infty Y$ acquires a smooth manifold-like structure. What is considered a smooth function on $J^\infty Y$ depends heavily on the choice of this smooth structure. I am confused by the wide variety of these definitions and I often have trouble establishing equivalence. First I will list some examples I have encountered. In [Saunders], $J^\infty Y$ is defined as a Fréchet manifold built on the vector space $\mathbb R^\infty$ which arises as the projective limit of $\mathbb R^n$s with respect to the system $p^n_{n-1}:\mathbb R^n\rightarrow\mathbb R^{n-1}$, $p^n_{n-1}(x^1,\cdots,x^{n-1},x^n)=(x^1,\cdots,x^{n-1})$. Smooth functions are determined by what smooth functions are on $\mathbb R^\infty$, and their "finite-orderness" is characterized by the fact that such smooth functions have only a finite number of nonzero partial derivatives at each point. I don't know whether this structure admits partitions of unity (Saunders doesn't go into it) but I find it likely that it does. In [Takens], no systematic smooth structure is given on $J^\infty Y$, he takes the smooth functions to be those which are locally the pullbacks of smooth functions from a finite order jet bundle, i.e. $f:J^\infty Y\rightarrow\mathbb R$ is smooth if for each $q\in J^\infty Y$ there is an $r$) and an $U^r\in J^rY$ open set with $q\in (\pi^\infty_r)^{-1}(U^r)$ and a smooth function $f_r\in C^\infty(U^r)$ such that $f|_{(\pi^\infty_r)^{-1}(U^r)}=f_r\circ\pi^\infty_r$. Takens then proves that there is a partition of unity subordinate to every open cover of $J^\infty Y$. I also think that this definition allows one to define smooth functions over every $O\subseteq J^\infty Y$ open set as the open sets of $J^\infty Y$ are generated by open sets of $J^rY$ for all $r$. In [Anderson], no systematic smooth structure is given either. Instead, he defines smooth functions to be the (globally) finite order ones, i.e. $f:J^\infty Y\rightarrow\mathbb R$ is smooth if and only if there is an $r$ and a smooth function $f_r\in C^\infty(J^rY)$ such that $f=f_r\circ\pi^\infty_r$. Anderson states that if the notion of smooth function was extended to the locally finite order ones then there would be smooth functions that aren't continuous. This definition seems to be the most convenient for calculus of variations, but I have some concerns. For example it seems non-obvious me how to define the space $C^\infty(O)$ where $O$ is an arbitrary open set of $J^\infty Y$. I guess a Takens-like approach can be taken but then it seems the association $O\mapsto C^\infty(O)$ defines only a presheaf as the glueing of finite order functions only need to be locally finite order. Moreover it seems that the association $W\mapsto C^\infty(W^\infty)$ where $W\subseteq Y$ is open and $W^\infty=(\pi^\infty_0)^{-1}(W)$ also defines only a presheaf over $Y$ for the same reason. In [Güneysu] pro-finite dimensional manifolds are defined and the infinite jet bundle is given as a pro-finite dimensional manifold.The definition of a smooth function seems to be same as in [Takens], with the added difference that a smooth function is demanded to be also continuous (Takens does not postulate continuity and based on the aforementioned remark in [Anderson] it seems that this added assumption is not superfluous). And now come the problems: There are a couple of remarks on nlab which seems to contradict many of these. For example in https://ncatlab.org/nlab/show/jet+bundle#concrete they say but one has to decide in which category of infinite-dimensional manifolds to take this limit: 1) one may form the limit formally, i.e. in pro-manifolds. This is what is implicit for instance in Anderson, p.3-5; 2) one may form the limit in Fréchet manifolds, this is farily explicit in (Saunders 89, chapter 7). See at Fréchet manifold – Projective limits of finite-dimensional manifolds. Beware that this is not equivalent to the pro-manifold structure (see the remark here). It makes sense to speak of locally pro-manifolds. and in the link in this remark states Beware, that infinite jet bundles are also naturally thought of as pro-manifolds. This differs from the Frechet manifold structure of example 4.5: A morphism of pro-manifolds is equivalently a function that is “globally of finite order”, in that [...] But by prop. 4.4 a morphisms of Fréchet manifolds is only restricted to have finite order of partial derivatives at every point. This is a weaker condition. In fact it seems to be also weaker than the condition of being “locally of finite order” considered in Takens 79. Hence it makes sense to speak of locally pro-manifolds. So it seems that nlab's definition of a pro-manifold (I prefer saying pro-finite dimensional manifold but I meant what is supposed to be the same thing) is different from that of [Güneysu], on the other hand the pro-manifold page on nlab has no references, so I don't know what is precisely the framework they are working in. The second problem I have is the paper [GMS]. I will try to briefly summarize its contents. The authors seem to be comparing the cohomology of the variational bicomplex in the Takens-like approach (differential forms are locally finite order) and the Anderson-like approach (differential forms are globally finite order). They make the claim that so far only the locally finite order variational bicomplex (LFOVB) had its cohomology calculated and most approaches that work in this case do not work for the globally finite order variational bicomplex (GFOVB). They then proceed to prove that the GFOVB has the same cohomology as the LFOVB by relating the cohomology of the former to the latter. $$ \ $$ Their argument seems to be around the fact that most cohomology calculations (eg. in [Takens] but also in the paper by Anderson and Duchamp which predates [Anderson] and also in [Krupka] for finite order jet bundles) are taken by considering a fine sheaf of differential forms that are defined on a jet bundle but the sheaf is over $Y$, i.e. a sequence of the form $$ 0\longrightarrow\mathbb R_Y\longrightarrow \mathcal O^0\longrightarrow\cdots\longrightarrow\mathcal O^r\longrightarrow\cdots, $$ where each $\mathcal O^r$ is a sheaf over $Y$ whose sections are differential forms over $J^\infty Y$ or $J^sY$ (for some $s$) and using sheaf cohomology together with the abstract de Rham theorem. However the globally finite order forms form a separated presheaf whose sheafification is the sheaf of locally finite order forms (I think?) and they don't have partitions of unity which throws a wrench in this argument for the GFOVB. $$ \ $$ However [GMS] postdates [Anderson] by quite a few years and Anderson does calculate the cohomology of the GFOVB in [Anderson]. He uses techniques other than the aforementioned sheafy approach though. [GMS] never refers to [Anderson] though (they do refer to Anderson/Duchamp). $$ \ $$ In light of the above I question how correct the global calculations are in [Anderson]. Anderson only uses paritions of unity on $Y$ but upon thinking about it I get the feeling his various glueing arguments are faulty as they might result in locally finite order forms rather than globally finite order ones. At any rate it seems to me that [Anderson] and [GMS] contradict one another thus cannot be both correct (unless of course I am missing something). Questions: I'd like to make some sense into this mess so lets see. How many inequivalent definitions of $J^\infty Y$ as a manifold-like structure exists? Based on the nlab remarks at least the Saunders/Takens/Anderson approaches seem to be inequivalent to one another. Are there multiple inequivalent definitions of pro-finite dimensional manifolds in use? The nlab comments seem to imply that whatever definition they are working with contains the Anderson approach to $J^\infty$ as a subcase but the Güneysu approach seems to be different and closer (but not necessarily equivalent) to the Takens approach. Although this is subjective, what structure on $J^\infty Y$ is the best suited for calculus of variations/the variational bicomplex? Basically, I just want to properly formalize the variational bicomplex for myself in a way that it includes only finite-order differential forms (I don't care about Lagrangians with unbounded order) but I got really confused by all these stuff. It also seems to me the Anderson approach might not actually be the best even if it only contains finite order forms if convenient arguments cannot be used. I am basically looking for the most hassle-free and lazy approach possible. Assuming Anderson wasn't being sloppy with the definitions in [Anderson], what is the generalization of finite dimensional manifolds into which Anderson's definition of $J^\infty Y$ fits into? (eg. what nlab calls pro-manifolds. But since this terminology seems to be ambigous I cannot just rely on the name alone) I'd like to read more on that structure, even outside jet bundles. Regarding the [Anderson]-[GMS] conflict, are the global cohomology calculations in [Anderson] correct? What these to publications say seem to be in conflict and I cannot decide which one is correct. Are they even in conflict actually? References: [Saunders]: D. Saunders - The geometry of jet bundles [Takens]: F. Takens - A global version of the inverse problem to the calculus of variations [Anderson]: I. M. Anderson - The Variational Bicomplex [Güneysu]: B. Güneysu, M. J. Pflaum - The Profinite Dimensional Manifold Structure of Formal Solution Spaces of Formally Integrable PDEs [Krupka]: D. Krupka - Introduction to Global Variational Geometry [GMS]: Giachetta, Mangiarotti, Sardanashvily - Cohomology of the variational bicomplex on the infinite order jet space REPLY [2 votes]: The following remarks are based on having previously gone through the literature that you've mentioned also for the purposes of figuring out these differences. It has been a while since then, but the state of our understanding that we reached with Urs Schreiber at the time was recorded in the following work, to be more precise in Sec 2.2 and possibly other parts referenced there. The remarks that you find on the nLab are based on that work. [KhS]: Igor Khavkine, Urs Schreiber Synthetic geometry of differential equations: I. Jets and comonad structure [arXiv:1701.06238] If you happen to find any inconsistencies, they might have been fixed in a more current version (though I don't think there were many), which hasn't been uploaded to the arXiv. If you think you've found some, just let me know and I can share the updated version with you. I haven't rechecked all the references that you cite, so these answers are mostly going by memory and what is written in our paper. I think your zoo of manifold structures on $J^\infty Y$ is overly complicated. There are really only two versions, distinguished by the smooth functions they admit: those of locally finite order ([Saunders], [Takens], [KhS], [Güneysu] + private communication from Pflaum) and those of globally finite order ([Anderson], [DF]). The relation of the Fréchet manifold structure with the locally finite order version is captured by Prop 2.29 in [KhS] (which reproduces the proof by Michor from an earlier reference). [DF]: P. Deligne and D. S. Freed, “Classical field theory,” in Quantum fields and strings: a course for mathematicians, P. Deligne, D. Kazhdan, P. Etingof, J. W. Morgan, D. S. Freed, D. R. Morrison, L. C. Jeffrey, and E. Witten, eds., vol. 1, pp. 137–225. AMS, Providence, RI, 1999. The notion of a pro-finite dimensional manifold depends on the ambient category where you take the projective limit. In the standard category of finite dimensional manifolds this limit does not exist. If you formally throw in projective limits, you get the globally finite order version. If you take the limit in the category of Fréchet manifolds (or any larger category where Fréchet manifolds are faithfully included) then you get the locally finite order version. You asked for a subjective opinion. My preference is the locally finite order version. [Anderson] fits into the globally finite order camp, which is the same as treating pro-finite dimensional manifolds as pro-objects in a standard categorical way. I think that the literature on general pro-objects (you can use the nLab link as a starting point) is much larger than on pro-finite dimensional manifolds specifically. Your best bet here might be to try to understand the general theory. I can't personally vouch for all arguments in [Anderson], but his book is so well-known that I would expect any known errors to have already been mentioned in the literature. If you find such a claim, it should be evaluated on its own merits. But the cohomology calculations in Anderson should be correct, since they agree with what is reported in [GMS] and [DF] (see the Appendix to Ch 2 specifically). Even if you think Anderson's arguments are not completely precise, I think that the other two references should be detailed enough, so that you should be able to verify them with enough study.<|endoftext|> TITLE: Taking the category of sheaves is symmetric monoidal QUESTION [10 upvotes]: Let $M$ and $N$ be topological spaces. Let $\operatorname{Sh}(M)$ denote the presentable $\infty$-category of space-valued sheaves on $M$. It seems to me that the equivalence $$\operatorname{Sh}(M) \otimes \operatorname{Sh}(N) = \operatorname{Sh}(M \times N)$$ where $\otimes$ is the standard symmetric monoidal structure of $\operatorname{Pr^L}$, the $\infty$-category of presentable categories with left adjoints, is well-known. However, I cannot find a proof in the literature. My question is if there is a standard reference for this statement? Or if there is a "simple proof" of it by assuming some well-known results? By the way, I believe that I was once told that it's in Lurie but I've failed to find it in either Higher Topos Theory or Higher Algebra. REPLY [12 votes]: Provided at least one of $M$ and $N$ is locally compact, the $\infty$-topos $\mathrm{Sh}(M \times N)$ is the product of $\mathrm{Sh}(M)$ and $\mathrm{Sh}(N)$ in $\mathrm{RTop}$. This is HTT 7.3.1.11. Products in $\mathrm{RTop}$ can be computed as tensor products in $\mathrm{Pr^L}$. This is HA Example 4.8.1.19. HA doesn't include a complete proof of the latter, but the case where one of the factors is of the form $\mathrm{Sh}(M)$ is essentially HTT 7.3.3.9, up to the matter of identifying $\mathrm{Sh}(M; \mathrm{Sh}(N))$ with the tensor product $\mathrm{Sh}(M) \otimes \mathrm{Sh}(N)$, for which HA 4.8.1.17 should be useful.<|endoftext|> TITLE: What are the consequences of an ineffective proof of the Riemann Hypothesis? QUESTION [21 upvotes]: Suppose a proof came out (and was verified by credible peer review) of the following statement: There is a $T_0$ such that for all $t>T_0$, all zeros $\zeta(\beta+it)=0$ have $\beta=1/2.$ where $T_0$ is totally ineffective. What interesting consequences would this partial result have? Of course you could ask this sort of question for all kinds of weakenings/strengthenings/relatives of RH: Zero-density estimates (which already has its own question here) Density Hypothesis Lindelöf Hypothesis Generalized Riemann hypotheses for various L-functions Grand Lindelöf Hypothesis But so far all the uses I have seen of $\zeta$ zeros has been in the strip $0 1} + \frac{1}{2} 1_{x = 1}$$ for any $c > 0, \ x \in \mathbb{R}$. Taking $c > 1$, and interchanging summation and integration we get (up to an error of $\frac{d (n)}{2}$, which is negligible) $$\sum_{n = 1}^{N} d(n) = \intop_{c - i \infty}^{c + i \infty} \sum_{n = 1}^{\infty} d(n) \left( \frac{N}{n} \right)^{s} \frac{\mathrm{d} s}{s} = \intop_{c - i \infty}^{c + i \infty} \zeta \left( s \right)^2 \frac{N^s \mathrm{d} s}{s}$$ Now, shift the contour to $\mathrm{Re(s) = \frac{1}{2}}$. The residue picked up at the pole $s = 1$ is exactly $N \log N + (2 \gamma - 1) N$, so all we have left to proveis to show that the expression $$N^{\frac{1}{2}} \intop_{-\infty}^{\infty} \zeta \left( \frac{1}{2} + i t \right)^2 N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$ is $\mathcal{O}_{\varepsilon} \left( N^{1/2 + \varepsilon} \right)$, or equivalently that the integral is $\mathcal{O}_{\varepsilon} \left( N^{\varepsilon} \right)$. Here is the point where I tell you that I actually lied beforehand: it turns out that using the full inverse Mellin transform is, although very elegant, not necessarily the best choice to get a good analytic bound. What is usually done is approximate it by integrating not from $c - i \infty$ to $c + i \infty$, but from $c - i N$ to $c + i N$, where $c$ is say something like $1 + \frac{1}{\log N}$. I don't remember the details off the top of my head (they appear for example in Montgomery's book, and in a few expositions of proofs of the Prime Number Theorem), so just trust me here when I say that it is sufficient to bound the integral $$\intop_{- N}^{N} \zeta \left( \frac{1}{2} + i t \right)^{2} N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$ But now (and here we finally use the Lindelöf Hypothesis!) we can bound pointwise this integral, and get that it is $\mathcal{O}_{\varepsilon} \left( N^{\varepsilon} \right)$ as required. This example, although somewhat stupid, shows the power of the Lindelöf Hypothesis. Indeed, see Tao's answer The relationship between the Dirichlet Hyperbola Method, the prime counting function, and Mertens function, where he points out the fundamental difference between arithmetic functions with zeta in their denominator (whose behaviour is controlled very much by the zeroes of zeta) and arithmetic functions with zeta in the numerator. Despite that, we still managed to use information about the zeroes of zeta to get a nontrivial estimate.<|endoftext|> TITLE: Are there examples of Einstein manifolds with unbounded curvature? QUESTION [12 upvotes]: Are there any known examples of Einstein manifolds $(M, g)$ such that $$\sup_{x \in M} \|\text{Rm}(x) \| = \infty$$ I'm looking for these examples because they might provide a counter-example to a problem of mine, but I can't think of any. Obviously such a manifold can't be compact (and therefore it can't be complete with positive scalar curvature) but I haven't been able to think of any concrete example. REPLY [12 votes]: If you don't care about completeness, here's a fairly simple way to construct such examples: Start with a compact Einstein manifold $(M^n,g)$ with Einstein constant $1$ (i.e., $\mathrm{Ric}(g) = (n{-}1)\,g$) that is not conformally flat. Now take the sine-cone, i.e., $\bigl(M\times(0,\pi),h\bigr)$ where $h = \mathrm{d}r^2 + (\sin r)^2\,g$. Then one easily computes that this is also an Einstein manifold with Einstein constant $1$, i.e., $\mathrm{Ric}(h) = n\,h$, but the Weyl curvature of $h$ (which is nonzero since $g$ is not conformally flat) blows up as $r$ approaches either $0$ or $\pi$. (Also, if one just takes the ordinary cone, $\bigl(M\times(0,\infty),h\bigr)$ where $h = \mathrm{d}r^2 + r^2\,g$, then $h$ will be Ricci-flat, but the Weyl curvature of $h$ will blow up as $r\to 0$.) Finally, as Anton Petrunin pointed out in the comment below, if $(M,g)$ is complete and Ricci-flat (i.e., Einstein with Einstein constant 0), but not conformally flat (equivalently, not flat), the Riemannian manifold $\bigl(M\times\mathbb{R}, h = \mathrm{d}r^2 + \mathrm{e}^{2r}\,g\bigr)$ will be a complete Einstein manifold with Einstein constant $-1$ whose Weyl curvature has unbounded norm as $r\to-\infty$.<|endoftext|> TITLE: Failure of "directedness" for second-order logic? QUESTION [6 upvotes]: Say that a logic $\mathcal{L}$ is directed iff whenever $\mathfrak{A}\equiv_\mathcal{L}\mathfrak{B}$ there is some $\mathfrak{C}$ with $\mathcal{L}$-elementary substructures $\mathfrak{A}'\preccurlyeq_\mathcal{L}\mathfrak{C}$, $\mathfrak{B}'\preccurlyeq_\mathcal{L}\mathfrak{C}$ with $\mathfrak{A}\cong\mathfrak{A}',\mathfrak{B}\cong\mathfrak{B}'$. It's a standard exercise to show that $\mathsf{FOL}$ is directed - or more generally, that every compact logic is directed. On the other hand, it's easy to whip up artificial logics demonstrating that this joint embeddability isn't equivalent to compactness. I'm curious about the situation with second-order logic $\mathsf{SOL}$. It's consistent with $\mathsf{ZF}$ that there are $\mathsf{SOL}$-equivalent structures which do not $\mathsf{SOL}$-elementarily embed into the same structure (see below), but I don't see how to get this result outright in $\mathsf{ZFC}$ (much less $\mathsf{ZF}$). However, I recall seeing an easy argument (due to Mostowski?) that in fact this is a $\mathsf{ZF}$-theorem. Question: Does $\mathsf{ZF}$ prove that $\mathsf{SOL}$ is not directed? Here's a proof that the non-directedness of $\mathsf{SOL}$ is consistent with $\mathsf{ZF}$. Suppose there is a family $\mathbb{A}$ of amorphous sets of pairwise incomparable cardinality such that there is no injection from $\mathbb{A}$ into $2^{\aleph_0}$. Thinking of each element of $\mathbb{A}$ as a structure in the empty language, we must have $X,Y\in\mathbb{A}$ with $X\equiv_\mathsf{SOL}Y$ but $X\not\cong Y$. But any set into which both $X$ and $Y$ inject must be non-amorphous, hence cannot satisfy $Th_\mathsf{SOL}(X)=Th_\mathsf{SOL}(Y)$ since amorphousness is second-order-expressible. Of course this doesn't help at all without a background assumption of lots of amorphous sets, so it's not really relevant to the question I'm asking here, but it's still neat. Note that a positive answer will have to crucially involve uncountable structures, since it's consistent with $\mathsf{ZFC}$ that $\equiv_{\mathsf{SOL}}$ implies $\cong$ for countable structures. REPLY [8 votes]: The answer to the question is yes. Let $\alpha_0<\alpha_1$ be the least ordinals (in the reverse lex order, say) such that $V_{\alpha_0}$ and $V_{\alpha_1}$ have the same second order theory $T$. Assume towards a contradiction that $V_{\alpha_0}$ and $V_{\alpha_1}$ are elementarily embeddable into a common structure $M$, which we may assume is transitive. Note that the embeddings fix $T$. Since $V_{\alpha_1}$ satisfies that there is an ordinal $\beta$ such that $V_\beta$ satisfies $T$, so does $M$, and hence so does $V_{\alpha_0}$. But this means that some $\beta < \alpha_0$ has the same theory as $V_{\alpha_0}$, contrary to the minimality of the pair $\alpha_0 < \alpha_1$. (Edit: I now see this is very close to what Trevor was doing.)<|endoftext|> TITLE: Graphs resembling the math genealogy graph must have concentration in a small number of families? QUESTION [6 upvotes]: I was talking with a non-mathematician the other week at a workshop about the fact that many mathematicians, like myself, are indexed in the math genealogy database. We talked a little about how many people tend to have family trees linking back to a few influential/well-known mathematicians (Newton/Gauss/Euler/etc...). I looked online later and this casual observation seems to have been examined closer (in this paper) and about 65 percent of the +200k nodes of the network fits within 24 families. I have some hypotheses for the cause of this concentration and expect that sociological factors play in heavily, but I still wondered if a random graph model with similar properties to the genealogy network would explain this effect at least partially. For simplicity maybe it would be best to assume a forest (collection of trees) structure on the random model $G$ (ignoring the case where someone has more than one advisor). Some potentially useful properties are below: It seems reasonable that the maximum out-degree of $G$ increases as one descends down the generations of the tree (corresponding to time), since more Ph.D.'s tend to be awarded now than before. Also, there are more Ph.D. students now that do not go onto direct Ph.D. students themselves (since the number of math Ph.D.'s is about as high as ever, but there are only a select number of Ph.D. awarding institutions at which one can supervise Ph. D.'s). I think these first two conditions can be emulated by adjusting the distribution for $G$ at each generation. There also should be a much lower probability of someone getting a Ph.D. supervised by someone who is not a mathematician (a disconnected node being generated ) vs the standard case of a descendent being generated in the graph $G$. There should probably be a small number of nodes to start with, but this seems less important. Altogether, there are two questions that I have. Question 1: Is there a standard random graph model that emulates these properties of the math genealogy graph? Question 2: If so, does that model have concentration of the network in a small number of families, if the network is allowed to generate for a sufficient length of time? I don't have enough intuition about random graphs to answer the second question with some very simple random graph models, so would be interested if anyone can point out results for a different setting than the one outlined here. REPLY [13 votes]: Precisely this question was the starting point of the Galton - Watson theory of branching processes. To quote the opening paragraph of their 1875 paper On the Probability of the Extinction of Families: The decay of the families of men who occupied conspicuous positions in past times has been a subject of frequent remark, and has given rise to various conjectures. It is not only the families of men of genius or those of the aristocracy who tend to perish, but it is those of all with whom history deals, in any way, even of such men as the burgesses of towns, ...<|endoftext|> TITLE: Non-compact Dirichlet fundamental domains and free Fuchsian groups QUESTION [6 upvotes]: Let $G$ be a finitely generated Fuchsian group, and let $\mathcal{F}$ denote the Dirichlet fundamental domain of $G$ with respect to $0$ in the Poincaré disc model. Assume throughout that $\mathcal{F}$ is non-compact. I am interested in properties of $G$, and how these properties are connected with Poincaré's theorem. Here are my questions: Is $G$ the free product of elementary hyperbolic, parabolic and elliptic subgroups? It is true for the modular group PSL(2,Z). Moreover, it holds under the stronger assumptions in 2). Assume that $G$ has no elliptic elements. Then $G$ is free (because $G$ is fundamental group of a non-compact surface) and thus, $\mathcal{F}$ has vertices only at the boundary of hyperbolic space. Can we derive this property of the vertices from Poincaré's theorem? Assume that $G$ is of the second kind (that is, the limit set of $G$ is not equal to the boundary of hyperbolic space). Then in particular $\mathcal{F}$ is non-compact. Is it true that the sides of $\mathcal{F}$ are pairwise disjoint, except (possibly) the sides paired by elliptic elements? UPDATE: Sam Nead answered 3) in the negative. Is the answer in 3) also negative if we allow an arbitrary reference point for the Dirichlet fundamental domain? Or does there always exist a suitable reference point for a Dirichlet fundamental domain such that the sides are pairwise disjoint, except (possibly) the sides paired by elliptic elements? REPLY [2 votes]: For (1) the answer is "yes". Since the surface has finitely generated fundamental group, there is a finite collection of disjoint embedded bi-infinite geodesics that cut the surface into a collection of ideal (or hyperideal) polygons, each with at most one cone point in its interior. (There is a special case when the cone point has angle $\pi$, which I leave as an exercise.) The statement now follows from the Seifert-van Kampen theorem. For (2) I read your question as "can a Dirichlet domain for a surface (as above) have material vertices?" and the answer is "yes, it may". To see this, consider the surface made by doubling an ideal triangle across its boundary. If we take the origin to be the centre of one of the triangles, then the Dirichlet domain has three ideal vertices and three material vertices. For (3) the answer is "not in general". To see this, consider the surface of (2) where we "open" the cusps, replacing them by (identical) funnels. Again the surface has a three-fold symmetry about the centres of the (now hyperideal) triangles. The Dirichlet domain has three material vertices, six ideal vertices, six material edges, and three ideal edges. The property you want - disjoint material edges glued in pairs - feels very similar to "Schottky, with all circles perpendicular to a single circle". Even this only gets you some Dirichlet domain with the desired form, not all. I am not sure if this property has a name, but I would look at Marden's book, or Maskit's, as a first reference. For (3) "updated" the answer is "not in general". That is, there is a hyperbolic surface $S$ with infinite volume, with finitely generated (and free) fundamental group, so that all Dirichlet domains have material vertices. Consider the surface $T$ given by (3) (first version) above. Let $x$ in $T$ be one of the points with three-fold symmetry. Let $D \subset T$ be a very small round disk about $x$. Let $S$ be the surface obtained by doubling $T - D$ across $\partial D$. Uniformise $S$. Let $\gamma$ be the image of $\partial D$ in $S$. Since $\gamma$ is fixed by a reflection in $R$, it is also a hyperbolic geodesic. Since the radius of $D$ was very small, the geodesic $\gamma$ is very short. I claim that $S$ has the desired property - that is, all Dirichlet domains have material vertices. The proof is harder than that for (3).<|endoftext|> TITLE: Question on a constructive proof that space projective curves are the intersections of three hypersurfaces QUESTION [5 upvotes]: $\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$I am a bit confused by a proof I am reading on the fact that a projective algebraic space-curve (i.e. an algebraic curve in $\P^3(k)$, where $k$ is an algebraically closed field) is the intersection of 3 hypersurfaces. I am trying to constructively create the hypersurfaces given the vanishing ideal of the curve. The proof I am reading is from an old source (written in 1960) by Martin Kneser. Since then, many generalizations of this fact exists (e.g. Storch, Forster, Eisenbud–Evans etc.). However, I would like to understand this proof by Kneser because it is quite constructive and supposedly the simplest of all. The text is written in German. I just don't quite understand how the proof starts. This is how Kneser starts the proof (I will do my best by adding my interpretation and translation to the original German text): Let $\C$ be the curve and suppose $I=\langle f_1,\dotsc,f_n \rangle$ ($f_i$ homogenous and nontrivial) is the vanishing ideal of the curve. And suppose that our coordinates are $(t:x:y:z)$ in $\P^3$. Without loss of generality, one may assume that the point $P:=(1:0:0:0)$ is in $\C$. Then Kneser continues by letting $f\in k[x,y,z]$ be a generator of the vanishing (principal) ideal of the "cone" of $\C$ with $P$ as its vertex. First question: Is a generator of the vanishing ideal of the cone the elimination ideal $I\cap k[x,y,z]$? Kneser then continues by assuming that $g \in k[t,x,y,z]$ is (my interpretation) a homogenous polynomial among the generators $f_1,\dotsc,f_n$ with minimum degree (say $d$) in $t$. One may write $$g = \sum_{i=0}^d g_i t^i$$ for (homogenous) polynomials $g_i \in k[x,y,z]$ such that $g_d\not\equiv 0$. Then comes the most confusing part of the whole construction. Kneser states (without a proof) something like Lemma: For any $p\in I$ with degree $m$ over $t$ one has the polynomial division (by $g$) $$g_d^m p = qg + r$$ where the remainder $r$ is divisible by $f$. Second question: Why must $r$ be divisible by $f$? From the above "Lemma" Kneser concludes that $Z(f,g) = \C\cup V$ where $V$ is a subset of the vanishing of $(f,g_d)$. He denotes the vanishing set $D:=Z(f,g_d)$ and states that $D$ is the union of finitely many lines passing $P$. I will not continue the proof here (unless someone needs to see all of it in my broken translation). I am already confused on why $V\subset Z(f,g_d)$ and that $Z(f,g_d)$ is the union of lines. Can anyone who understands this better than me clarify my confusion? The title of the paper (which is only 2 pages) in the original German is: Über die Darstellung algebraischer Raumkurven als Durchnitte von Flächen, Arch. Math. 11, 157–158 (1960). REPLY [3 votes]: $\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$Using cohomology is interesting when dealing with higher dimensions and more general results but I don't see why it would result into a shorter proof. In fact the generalization of this result of Kneser (Eisenbud and Evans) avoids any sheaf cohomology as well and the result is just as short as this specific case. Here is Kneser's algorithm with some explanation: The initial assumption is that, without loss of generality, $P=(1:0:0:0)$ is on the curve. We use the same notation $(t:x:y:z)$ as Kneser for the coordinates. Let $\mathbb k$ be your (algebraically closed) ground field. One also assumes that $\C$ is not a line (otherwise the algorithm won't work, but a line is trivially a set-theoretic complete intersection). There exists a $g \in I(\C)$ with the property that: a.) $d > 0$ where $d=\deg_t(g)$ b.) $g_d$ is not in $I(\C)$ (where $g_d\in \mathbb k[x,y,z]$ is the coefficient of $t^d$). c.) $d$ is minimum with the property a.) and b.) As you noted $I(\C)\cap \mathbb k[x,y,z]$ is principal, and is generated by say $f$. The vanishing of $f$ is therefore a cone with "vertex" at $P$ and "base" $\C$. Note that $Z(f,g_d)$ is a finite union of lines because: If $Q\in Z(f,g_d)\backslash\{P\}$ then the line $PQ$ is on the surface $Z(g_d)$ and on the cone $Z(f)$ which means that this line meets $\C$. We also know that $g_d\notin I(\C)$ so the curve $\C$ cuts the surface $Z(g_d)$ finitely many times. Kneser then states that for any $p\in I(\C)$ we can find a $m\ge 1$ such that $g_d^m p \in \spn{f,g}$. As pointed out in the comments by Kapil, you can use degree (of $t$) argument to prove this. Suppose now $g(Q)=0$ and $f(Q)=0$, then either $Q\in \C$ or $Q\notin \C$ and in this case there is a $p\in I(\C)$ such that $p(Q)\ne 0$. By 3. $Q\in Z(g_d)$. So, $$Z(f,g) \subset \C \cup Z(f,g_d).$$ In words, $Z(f,g)$ is contained in the union of the curve $\C$ and finitely many lines whose common intersection is $P$. This already provides an immediate and nice proof that all irreducible projective space (algebraic) curves are intersections of an analytic surface and an algebraic surface, just perturb the cone (or the surface $Z(g)$) so that it intersect these lines only at $\C$. We want to now find a third algebraic surface $Z(h')$ (the two previous surfaces are $Z(f)$ and $Z(g)$) that avoids these finite lines $Z(f,g_d)$ except at $\C$. This can be done in several ways, Kneser himself provides an algorithm with a proof. I believe his algorithm is inefficient but didactically the easiest one to follow. I will not explain this part, but you can usually get away using a surface defined by products and sums of linear and quadratic forms avoiding the lines and having some higher multiplicity at $P$. By Hilbert's Nullstellensatz there is therefore an $N\in \mathbb N$ and $h\in I(\C)$ such that $$h'^N - h \in \spn{g_d}.$$ You can obtain $N$ by just iterating through the powers of $h'$ and reducing via Groebner basis (ideal-membership test) of $\spn{g_d}+I(\C)$ and you can obtain $h$ as a by-product. Therefore we choose $h$ and get $\sqrt{\spn{f,g,h}} = I(\C)$.<|endoftext|> TITLE: Tiling a Jordan polygon QUESTION [5 upvotes]: I saw this problem some years ago, don't remember the source: Let $P$ be a Jordan polygon (i.e. the only points of the plane belonging to two edges are the polygon vertices) that can be tiled with parallelograms. Does it follow that the sum of the areas of all the rectangles among those parallelograms is the same independent of the tiling? Does it follow that the sum of the areas of all parallelograms with angles $\alpha$ and $\beta$ is the same independent of the tiling? I couldn't find a solution, but I believe the answer to both questions is positive. Does anyone know how this problem can be solved? REPLY [4 votes]: When you tile a simple polygon by parallelograms, the tiling can be partitioned into "zones" of rectangles with parallel sides, meeting end-to-end, and forming a path from one edge of the polygon to a parallel edge of the opposite orientation. Two parallel zones cannot cross, so the pairing of opposite edges into zones is uniquely determined. Additionally, two non-parallel zones cross either zero times or one time, according to whether the edges at the ends of the zones appear in nested or alternating order around the polygon. That is, the number of crossings is again uniquely determined by the polygon. You get exactly one rectangle for each two zones that are determined by perpendicular edges of the polygons and that cross each other. The shape of this rectangle is just the Cartesian product of the two perpendicular edges, so it is also determined. So the shapes of all rectangles in the tiling, and not just the sum of their areas, is equal for all tilings. The same goes for parallelograms with other angles.<|endoftext|> TITLE: Comparing Kripke-Joyal semantics of toposes to model-theoretic satisfaction QUESTION [6 upvotes]: Let $\mathcal E$ be a topos and $\varphi$ a statement formulating a property of toposes. There are two ways of checking whether $\mathcal E$ satisfies $\varphi$: Consider the first-order language $L$ of a category. Each topos can be considered as an $L$-structure. So, using standard model-theoretic notions, one can consider the satisfaction relation, $\mathcal E\models \varphi$. Using the Kripke-Joyal semantics, one can look at whether $\varphi$ is true in the internal language of $\mathcal E$. Roughly, one difference between 1. and 2. is that in 1. only the notions "object", "arrow", and "composition" occuring in $\varphi$ are interpreted in $\mathcal E$, while the logic is interpreted on the meta level. However, in 2. also the logic (i.e., the quantifiers $\forall$ and $\exists$ and $\land, \lor, \neg, \dots$) are interpreted in $\mathcal E$. Strictly speaking, this is not an appropriate comparison because in 1. $\varphi$ is a first-order $L$-sentence and in 2. $\varphi$ is a sentence in higher-order logic. However, in spirit 1. and 2. feel similar, because the give a way of looking whether a statement is true in a topos. Is there some way of comparing 1. and 2., and can one say anything interesting about this comparison? In particular, I wonder whether the following works: given a sentence $\varphi$ in higher-order logic, can one assign to it a sentence $\varphi'$ in first-order logic such that $\varphi$ is true in the internal language of $\mathcal E$ if and only if $\mathcal E\models \varphi'$ in the model-theoretic sense? (And what about the other way round?) REPLY [7 votes]: As you observe yourself, the question does not quite make sense as $\phi$ in 1. is a formula in the first order language of a category and in 2. $\phi$ is a formula in higher order logic (something like the Mitchel-Benabou language). The only framework I can think of where this question makes perfect sense is if you interpret "internal logic" in the sense of Mike Shulman "Stack semantics" as in his paper https://arxiv.org/abs/1004.3802. Here you interpret "the first-order language of categories" as in def. 3.1 of the paper linked above, and the stack semantics is an extention of the Kripke-Joyal semantics that is also formulated in that language (where the higher order part come from the fact that you can quantify on objects). In this case, as pointed out by Andrej Bauer, the Kripke-Joyal semantics (or rather its extention called the stack semantics) turns a formula $\phi$ in this language to a formula $\phi'$, so that "$\phi$ hold internally in $\mathcal{E}$" if and only if $\mathcal{E}$ satisfies $\phi'$). Of course $\phi'$ is in general different from $\phi$, though I believe $\phi'' = \phi'$. In any case, not every formula arise as a $\phi'$ (i.e. corresponds to an "internal property"). For example, the validity of such formulas are always local properties: if $\mathcal{E}$ satisfies $\phi'$ then every slice $\mathcal{E}/X$ also satisfies $\phi'$; and conversely if $\mathcal{E}/X$ satisfies $\phi'$ for a family of objects covering the terminal, then $\mathcal{E}$ also satisfies $\phi'$.<|endoftext|> TITLE: Jon Beck's untitled manuscript containing the "tripleability theorem" (i.e. the monadicity theorem) QUESTION [17 upvotes]: Many papers refer to an untitled manuscript of Jon Beck (Cornell, 1966) for the origin of the monadicity theorem (originally called a "tripleability theorem"). An early proof is in Manes's 1967 thesis A Triple Miscellany: Some Aspects of the Theory of Algebras over a Triple (Theorem 1.2.9). Manes cites Beck's 1967 thesis Triples, Algebras, and Cohomology as a reference, but the monadicity theorem does not actually appear there. Where can one find a copy (preferably digitised) of the untitled manuscript of Beck containing the monadicity theorem? (Considering that the manuscript is cited, presumably a copy exists and was circulated, rather than passed on by word of mouth.) Evidence for the existence of the manuscript is given by an email of Marta Bunge on the categories mailing list (dated 4th November 2007): There is an unpublished (untitled and undated) four-pages manuscript which John Beck gave to me (and I supposed also to many ohers) when he was at McGill. In it, he states and proves two theorems, the CTT (crude tripleableness theorem), and the PTT (precise tripleableness theorem). There is a connection between triples and descent implicit in the PTT. But this is not the same connection with descent as the Benabou-Roubaud theorem. REPLY [21 votes]: After reaching out to every researcher who cited the manuscript, John Kennison was kind enough to find and scan his copy of the untitled manuscript containing the crude and precise monadicity theorems. I have uploaded it to the nLab for posterity: Jon Beck's untitled manuscript. This copy was distributed at the Conference Held at the Seattle Research Center of the Battelle Memorial Institute in June – July 1968, though evidence from citations suggests it was first distributed as early as 1966.<|endoftext|> TITLE: Spectrum of matrix involving quantum harmonic oscillator QUESTION [14 upvotes]: The quantum harmonic oscillator relies on two classical objects, the so-called creation and annihilation operator $$a ^* = x- \partial_x \text{ and }a = x+\partial_x.$$ Fix two numbers $\alpha,\beta \in \mathbb R.$ Can we explicitly compute the spectrum of $$H = \begin{pmatrix} aa^* + \alpha^2 & \alpha a+ \beta a^* \\ \alpha a^* + \beta a & aa^* +\beta^2 \end{pmatrix}?$$ This looks like an innocent problem, but somehow I find it hard to determine what exactly the spectrum of this operator is. The operator is acting on a suitable domain of $L^2(\mathbb R) \oplus L^2(\mathbb R).$ REPLY [14 votes]: The Hamiltonian $$H=\begin{pmatrix} \alpha^2+a^\ast a&\alpha a+\beta a^\ast\\ \alpha a^\ast+\beta a&\beta^2+a^\ast a \end{pmatrix} $$ is known in the physics literature as the anisotropic Rabi Hamiltonian. (In the most general case there is an additional term $\Delta\sigma_z$.) I give some pointers to the literature in this Physics SE posting. The eigenvalues can be computed from a recursive scheme, but closed-form expressions for the spectrum only exist for either $\alpha=\beta$ or $\alpha\beta=0$. $\bullet$ Consider first the case $\alpha=\beta$. A unitary transformation $H'=UHU^\ast$ with $U=e^{i\pi\sigma_y/4}$ brings the Hamiltonian to the diagonal form $$H'=\begin{pmatrix} b_+^\ast b_+&0\\ 0&b_-^\ast b_- \end{pmatrix},\;\; b_\pm=a\pm\alpha. $$ The eigenvalues are the integers $N=0,1,2,\ldots$, each twofold degenerate. The corresponding eigenstates $|N,\pm\rangle$ are obtained from the eigenstates $|N\rangle$ of the harmonic oscillator by acting on these with the displacement operator, $$| N,\pm\rangle=e^{\pm\alpha(a-a^\ast)}|N\rangle.$$ $\bullet$ At the other extreme, we can take one of the two parameters $\alpha, \beta$ much smaller than the other. Let me set $\beta=0$, $\alpha\neq 0$. (The spectrum is the same for $\alpha=0$, $\beta\neq 0$.) The Hamiltonian $$H=\begin{pmatrix} \alpha^2+a^\ast a&\alpha a\\ \alpha a^\ast&a^\ast a \end{pmatrix}=(\tfrac{1}{2}\alpha^2+a^\ast a)I+\alpha(\sigma_+a+\sigma_-a^\ast)+\tfrac{1}{2}\alpha^2\sigma_z$$ is known in physics as the Jaynes-Cummings Hamiltonian. The eigenvalues can be computed exactly, because the Hamiltonian decomposes into an infinite direct product $H_n$ of $2\times 2$-matrix Hamiltonians in the basis $|g,n+1\rangle$ (two-level system in the lower state, $n+1$ quanta excited in the oscillator) and $|e,n\rangle$ (two-level system in upper state, $n$ quanta excited): $$H_n\begin{pmatrix} |g,n+1\rangle\\ |e,n\rangle \end{pmatrix}= \begin{pmatrix}n+1&\alpha\sqrt{n+1}\\ \alpha\sqrt{n+1}&n+\alpha^2 \end{pmatrix} \begin{pmatrix} |g,n+1\rangle\\ |e,n\rangle \end{pmatrix}.$$ The eigenvalues then follow directly, $$\Omega_{n,\pm}=\tfrac{1}{2}(\alpha^2+1)+n\pm\tfrac{1}{2}\sqrt{(\alpha^2+1)^2+4\alpha^2 n}.$$ This sequence of eigenvalue pairs $\Omega_{n,+},\Omega_{n,-}$ exists for $n\in\{0,1,2,\ldots\}$. In addition, there is an unpaired ground state $|g,0\rangle$ without any excitations, which is annihilated by $H$ so it has eigenvalue $\Omega_{0}=0$ independent of $\alpha$. $\bullet$ The general case of arbitrary $\alpha,\beta$ does not have a simple closed form expression for the eigenvalues. For $\alpha,\beta\ll 1$ a perturbative solution is given in Appendix C of arXiv:1008.1317. The ground state does have a simple exact result: the lowest eigenvalue equals zero for any $\alpha,\beta\in\mathbb{R}$. The zero-mode – the eigenstate which is annihilated by $H$ – is a coherent state $|\xi\rangle=e^{\xi(a^\ast-a)}|0\rangle$ for the harmonic oscillator (such that $a|\xi\rangle=\xi|\xi\rangle$). Substitution of $H{p\choose q}\otimes|\xi\rangle=0$ gives the two zero-modes $$\Psi_\pm={{\sqrt\beta}\choose{\mp\sqrt\alpha}}\otimes|\!\pm\!\!\sqrt{\alpha\beta}\rangle.$$ (Thanks to Michal Pacholski for helping me with this.) Notation: $I$ is the $2\times 2$ unit matrix, $\sigma_x$, $\sigma_y$, $\sigma_z$ are the three Pauli matrices, and $\sigma_\pm=\tfrac{1}{2}(\sigma_x\pm i\sigma_y)$. Also please note that the eigenvalues in the OP are shifted by one unit relative to those given here, because I reordered $aa^\ast\mapsto a^\ast a$. Since there was an issue with the appearance of spurious "parabolas" in the comparison with the numerics by yarchik (in a now deleted answer), I show here a numerical check of the spectrum as a function of $\alpha$ for $\beta=0$, using a slightly modified version of yarchik's code. Solid curves are the numerical result, dashed and dotted curve are $\Omega_{n,\pm}$. Note that the ground state is at zero energy for any $\alpha$ when $\beta=0$. The plot below, for $\beta=2\alpha$ shows this also holds for nonzero $\beta$.<|endoftext|> TITLE: Which Lie groups have finitely many conjugacy classes of subgroups of fixed isomorphism type? QUESTION [11 upvotes]: Let $G$ be a real Lie group. What conditions must $G$ satisfy so that the following is true: For any finite group $\Gamma$ there exist finitely many conjugacy classes of subgroups of $G$ that are isomorphic to $\Gamma$. I believe that for $G=GL(n,\mathbb{R})$ this is true because: subgroups of $GL(n,\mathbb{R})$ are conjugate if and only if the restriction of the standard representation of $GL(n,\mathbb{R})$ to the subgroups are isomorphic representations. Finite groups have finitely many irreducible representations, and that proves the claim. I also believe that for $G=SO(n)$ this is true because of this answer: https://mathoverflow.net/a/17074/164084. It would be enough for my application to know that every real, compact, connected Lie group that has a faithful representation has the property stated above ("For any finite group $\Gamma$..."). This is a crosspost from Stackexchange Mathematics, see here: https://math.stackexchange.com/questions/4219091/which-lie-groups-have-finitely-many-conjugacy-classes-of-subgroups-of-fixed-isom. REPLY [12 votes]: A natural condition is that $G$ has finitely many connected components. One can easily reduce this case to the connected group case, and then to the compact group case, as all maximal compact subgroups in a connected Lie group are conjugated. Then the representation variety $\text{Hom}(\Gamma,G)$ is compact and local rigidity, that is vanishing of $H^1(\Gamma,\mathfrak{g})$, guarantees the finiteness of the number of $G$-orbits. Here $\mathfrak{g}$ denotes the Lie algebra of $G$. The fact that $H^1$ vanishes could be deduced from the fact that every isometric action of $\Gamma$ on $\mathfrak{g}$ has a fixed point, by averaging an orbit.<|endoftext|> TITLE: Berge-Fulkerson conjecture --- the planar case QUESTION [7 upvotes]: A well-known conjecture of Berge and Fulkerson says that every bridgeless cubic graph has a collection of six perfect matchings that together cover every edge exactly twice. Is this still open for bridgeless cubic planar graphs? REPLY [12 votes]: The Berge-Fulkerson conjecture holds for planar graphs. Here is a proof. Let $G$ be a bridgeless cubic planar graph. The dual graph $G^*$ is a triangulation. By the Four Colour Theorem, $G^*$ has a 4-colouring $c$. We will use $\mathbb{Z}_2 \times \mathbb{Z}_2$ as the set of colours for $c$. Now for each edge $e \in E(G)$, colour $e$ with colour $c'(e):=c(f_1)+c(f_2)$ where $f_1$ and $f_2$ are the two faces of $G$ incident to $e$. Since $c$ is a proper colouring of $G^*$, $c'(e)$ is a non-zero element of $\mathbb{Z}_2 \times \mathbb{Z}_2$ for all $e \in E(G)$. Moreover, if $v \in V(G)$ and $e_1, e_2$, and $e_3$ are the edges of $G$ incident to $v$, then the dual edges $e_1^*, e_2^*$, and $e_3^*$ are a triangular face $\Delta$ in $G^*$. Since the vertices of $\Delta$ receive different colours in $c$, $c'(e_1), c'(e_2)$, and $c'(e_3)$ are all distinct. That is, $c'$ is a proper 3-edge colouring of $G$. In other words, $E(G)$ can be partitioned into three perfect matchings. So you can just use each of these perfect matchings twice.<|endoftext|> TITLE: How to treat Puiseux series as functions? QUESTION [6 upvotes]: I have been reading about Puiseux series in the context of the Newton–Puiseux algorithm for resolution of singularities of algebraic curves in $\mathbb{C}^2$. Given a curve $f(x,y)=0$ with $f$ a convergent power series and such that the curve has a singularity at the origin, the algorithm produces convergent Puiseux series $\varphi_1, \dotsc, \varphi_m$ verifying $f(x,y) = (y-\varphi_1(x))\dotsm(y-\varphi_m(x))$. All of the books I have consulted start by constructing the Puiseux series as a formal power series and then invoking the Implicit Function Theorem to show that there exist analytic solutions to the problem, and therefore the constructed series must be convergent (in a small enough neighbourhood of the origin). Then they continue by studying the properties of the singularity using $\varphi_1, \dotsc, \varphi_m$ as parametrizations of the curve. Here is where I have a problem. Say for instance that one of the resulting Puiseux series is $\varphi_1(x) = x^{3/2} + x^{7/4}$. This is not a well-defined complex function, is it? For $x=1$ we could have $x^{3/2}$ equal to $1$ or $-1$, and $x^{7/4}$ equal to $\pm1$ or $\pm i$, so there are eight different determinations of this particular Puiseux series. In general, if the Puiseux series has infinitely many terms, there would be infinitely many determinations. As far as I understand, only one of these determinations of $\varphi_1$ is the actual function you want to consider, which is the function that the Implicit Function Theorem gives you — but because of the nature of the proof, you can't really know which one it is! It is not clear to me how do these books account for this problem, but their treatment of the series is as if they were well-defined functions (see for instance Brieskorn–Knörrer). REPLY [10 votes]: One standard way to bring actual functions in the picture is the following formulation of the existence + convergence results of Puiseux roots: "Given an irreducible power series $f \in \mathbb{C}[[x, y]]$ which is a Weirstrass polynomial in $y$ of degree $d$ (i.e. $f = y^d + \sum_{i=1}^d f_i(x)y^{d-i}$ with $f_i(x) \in x\mathbb{C}[[x]]$), there is $\phi(t) \in \mathbb{C}[[t]]$ such that $$f(t^d, y) = \prod_{i = 1}^d (y - \phi(\zeta^it))$$ where $\zeta$ is a $d$-th primitive root of $1$. In addition, and here is where you have actual functions, if $f$ is convergent, then so is $\phi$." Added later (prompted by OP's comment about choices for $\phi$): in this formulation the uniqueness of Puiseux roots is the statement that if $\psi(t)$ is any other power series such that $y - \psi(t)$ divides $f(t^d, y)$, then $\psi(t) = \phi(\zeta^i t)$ for some $i$. In other words, there are precisely $d$, not infinitely many, Puiseux roots of $f$, namely $\phi(\zeta^it)$, $i = 1, \ldots, d$. This is covered, e.g. in chapter one of Singularities of Plane Curves by Casas-Alvero. REPLY [10 votes]: Responding to the question somewhat tangentially: there is no obligation to think of a Puiseux series as any sort of "global" pointwise-valued function. Luckily so, because, as you observe, it's impossible. :) But/and the degree of impossibility is probably not as severe as I suspect you are worried about, namely, a power series in $x^{1/N}$ does not have ambiguities of which $N$th root in every term, but only one choice. So, for example, $x^{1/2}+x^{3/2}=x^{1/2}+(x^{1/2})^3$, in its role as Puiseux series, has just two possible values, depending on the choice of sign for $x^{1/2}$. Not four choices. Of course, we could declare that we interpret such a series so that it has all possible ambiguities in terms independently of each other. But the way that Puiseux series arise, as elements of finite algebraic extensions of convergent (or formal) power series, dictates that we only need a single choice. This is a consequence of a theorem, that needs some proof. Namely, ... maybe up to changes of coordinates..., all the algebraic extensions of $\mathbb C[[z]]$ are of the form $\mathbb C[[z^{1/N}]]$... with a single "function" $z^{1/N}$, ... not different ones in different terms.<|endoftext|> TITLE: How can I find all integer solutions of $3^n - x^2 = 11$ QUESTION [6 upvotes]: I know that $n$ can't be even because of the following argument: Let $n = 2p$. Then we can use the difference of two squares and it becomes like this : $(3^p + x)(3^p - x) = 11; 3^p + x = 11 , 3^p - x = 1$. $3^p = 6$ which is not possible if $p$ is an integer. I also found out that $x$ has to be an even number. I think that the only solution is $3^3 - 4^2 = 11$ but how can I find the true answer? REPLY [6 votes]: $$3^{n} - x^2 = 11$$ According to Silverman's answer, we take the three cases $n=3a, n=3a+1,$ and $n=3a+2.$ The problem can be reduced to finding the integer points on elliptic curves as follows. $\bullet n=3a$ Let $X = 3^a, Y=x$, then we get $Y^2 = X^3 - 11.$ According to LMFDB, this elliptic curve has integer points $(X,Y)=(3,\pm 4), (15,\pm 58)$ with rank $2.$ Hence $(X,Y)= (3,\pm 4) \implies (n,x)=(3,\pm4).$ $\bullet n=3a+1$ Let $X = 3^{a+1}, Y=3x$, then we get $Y^2 = X^3 - 99.$ This elliptic curve has rank $0$ and has no integer point, so there is no integer solution $(n,x).$ $\bullet n=3a+2$ Let $X = 3^{a+2}, Y=9x$, then we get $Y^2 = X^3 - 891.$ This elliptic curve has integer points $(X,Y)=(31,\pm 170)$ with rank $1.$ Hence there is no integer solution $(n,x).$ Thus, there are only integer solutions $(n,x)=(3,\pm4).$<|endoftext|> TITLE: Existence of a winning strategy in the $\boldsymbol{\Delta}_2^0$ game QUESTION [6 upvotes]: Consider the following infinite perfect information game with two players (the name I gave in the title of the post it totally made up): at each round $i \in \omega$, player $\mathrm{I}$ picks a natural number $x_i$ and a $\boldsymbol{\Delta}_2^0$ subset of the Baire space $\omega^\omega$ denoted by $A_i$ such that $A_{i-1}\subseteq A_i$ for all $i$; player $\mathrm{II}$ then picks a boolean value $y_i \in \{0,1\}$. So at the end of the game player $\mathrm{I}$ will have built an infinite sequence $x = (x_n)_{n \in \omega}$ and an increasing chain (wrt set inclusion) of $\boldsymbol{\Delta}_2^0(\omega^\omega)$ sets $(A_n)_{n \in \omega}$ whilst player $\mathrm{II}$ will have built a boolean sequence $y = (y_n)_{n \in \omega}$ (an element of the Cantor space). Player $\mathrm{II}$ wins the play if at least one of these conditions is satisfied: $\bigcup_{n \in \omega} A_n \not\in \boldsymbol{\Delta}_2^0(\omega^\omega)$ The sequence $y$ is eventually constant, i.e. $\exists k \ \forall n \ge k \ y_n = y_k$. Moreover The sequence $y$ is eventually equal to $1$ if and only if $x \in \bigcup_{n \in \omega} A_n$. Intuitively we require player $\mathrm{II}$ to "guess" whether the real played by $\mathrm{I}$ will or won't belong to the set $\mathrm{I}$ is building. Now I'm wondering whether player $\mathrm{II}$ has a winning strategy in this game, or if the game is determined, and, eventually, under which hypotheses. In a simpler game, in which player $\mathrm{I}$ does not keep changing the sets $A_i$, player $\mathrm{II}$ has a winning strategy (see, for this result and a wider discussion, this paper by Raphael Carroy Playing in the first Baire class, specifically Proposition 3.10). Any idea? Thanks REPLY [9 votes]: Player I has a winning strategy: First play a singleton $A_0=A_1=\ldots=\{z_0\}$, for some real $z_0$, and the $x_n$'s consistent with $z_0$, until player II plays their first 1, if they ever do. After they play a 1 at stage $n$, continue with $A_{n+1}=A_{n+2}=\{z_0\}$, but make $x_{n+1}$ inconsistent with $z_0$, and proceed in this way until player II plays their first 0 after this stage, say at stage $m$. Then play $A_{m+1}=\{z_0,z_1\}$ with some $z_1$ where $(x_0,x_1,\ldots,x_m)\subseteq z_1$, and keep playing $A_{m+2}=A_{m+3}=\ldots=\{z_0,z_1\}$ and the $x_i$'s consistent with $z_1$ until player I again plays a 1. Then continue playing $\{z_0,z_1\}$ but play the $x_i$'s inconsistent with $z_1$, until player II next plays a $0$ at stage $m_1$, etc. The $A_n$'s and $\bigcup_{n<\omega}A_n$ are easily boldface-$\Delta^0_2$, and it's easy to see that player I wins. (In the first edit I said something suggesting that every countable set is boldface-$\Delta^0_2$, but that's false, as e.g. the rationals are not boldface-$\Pi^0_2$, as they are dense but not comeager. If $\bigcup_{n<\omega}A_n$ is infinite above, note that its complement is the union of an open set with a singleton (the singleton contains the limit of the $z_n$'s), which is therefore boldface-$\Sigma^0_2$.)<|endoftext|> TITLE: L-theory of additive category QUESTION [5 upvotes]: Reading some articles in the field, I found the following statement: Proposition: Let $\mathcal{B}$ be an additive category and $\mathcal{A}$ a full additive subcategory of $\mathcal{B}$. If $\mathcal{A}\subset \mathcal{B}$ induces an isomorphism in $K$-theory $$K_{n}(\mathcal{A})\rightarrow_{\cong}K_{n}(\mathcal{B}) $$ for all $n\in \mathbf{Z}$, then $$L_{n}(\mathcal{A})\rightarrow_{\cong}L_{n}(\mathcal{B}) $$ for all $n\in \mathbf{Z}$. Is there a proof of such statement? The proposition in question is lemma 4.17 of the article: in their proof the authors say that under the hypothesis of lemma 4.17, any chain complex in $\mathcal{B}$ is homotopy equivalent to a chain complex in $\mathcal{A}$! My question is why it is true? Here is the exact statement of the lemma: Lemma 4.17. Let $\mathcal{B}$ be an additive category. If $\mathcal{A}$ is a full subcategory, inducing isomorphism on $K_{0}$, then $L(\mathcal{A}) \rightarrow L(\mathcal{B})$ is a weak homotopy equivalence (any decoration on the $L$-theory). Proof. We need to show the map induces isomorphism on homotopy groups, but any chain complex in $\mathcal{B}$ is homotopy equivalent to a chain complex in $\mathcal{A}$, so from this it follows that $L^{h}(\mathcal{A})\rightarrow L^{h}(\mathcal{B})$ is an isomorphism, and since the inclusion induces isomorphism on $K$-theory Lemma 4.15 finishes off the proof. I do not understand the meaning of the following argument in their proof : "but any chain complex in $\mathcal{B}$ is homotopy equivalent to a chain complex in $\mathcal{A}$". How do they make this statement in such general stituation. It seems to me that Achim Krause has given a convincing counterexample in his answer. So I'm wondering if I'm missing something in the statement of Lemma 4.17 ? REPLY [2 votes]: I did find a reference. Sort of. I think the way you stated it, the statement (I'm referring to the statement about complexes in $\mathcal{B}$ and $\mathcal{A}$, not about the $L$-theory statement) is incorrect. For example, there are rings such that $R\cong R\oplus R$ as modules. For such a ring, $K(R)\simeq 0$, so if you take $\mathcal{B}=\operatorname{Mod}(R)$ and $\mathcal{A}=0$, the map on $K$-theory is an isomorphism, but certainly not every complex in $\mathcal{B}$ is exact. I believe you additionally need that $\mathcal{A}$ has enough objects, in the sense that every object of $\mathcal{B}$ admits an epimorphism from an object of $\mathcal{A}$ (and possibly more, I'm rarely thinking about bare additive categories). Under the required technical assumptions, you should be able to apply Lemma 2.2 from [Thomason, "The classification of triangulated subcategories", Compositio Mathematica 105: 1–27, 1997] (to the derived categories of $\mathcal{A}$ and $\mathcal{B}$).<|endoftext|> TITLE: Any hints on how to prove that the function $\lvert\alpha\;\sin(A)+\sin(A+B)\rvert - \lvert\sin(B)\rvert$ is negative over the half of the total area? QUESTION [11 upvotes]: I have this inequality with $0 TITLE: Does GCH for alephs imply the axiom of choice? QUESTION [10 upvotes]: GCH for alephs means the statement that, for any aleph $\kappa$, there are no cardinals $\mathfrak{r}$ such that $\kappa<\mathfrak{r}<2^\kappa$. Does GCH for alephs imply the axiom of choice? Remark. Lindenbaum and Tarski assert in ``Communication sur les recherches de la théorie des ensembles'' without proof that GCH for alephs is equivalent to Cantor's aleph hypothesis that $2^{\aleph_\alpha}=\aleph_{\alpha+1}$ for all ordinals $\alpha$ (see page 188, No. 96). But as we know (although Lindenbaum and Tarski possibly do not kown) Cantor's aleph hypothesis implies AC. This means that Lindenbaum and Tarski in fact also assert that GCH for alephs implies the axiom of choice. But I do not see how to prove it. REPLY [9 votes]: The answer is positive, yes. Note that $2^\kappa\leq 2^{\kappa^+}$, and therefore $\kappa^+\leq\kappa^++2^\kappa\leq 2^{\kappa^+}$. So either $2^\kappa=2^{\kappa^+}$ or $2^\kappa=\kappa^+$. In the first case $\kappa^+$, $\kappa<\kappa^+<2^\kappa$ is impossible. So the latter case holds. Therefore the power set of an ordinal can be well-ordered, and the Axiom of Choice follows in $\sf ZF$.<|endoftext|> TITLE: The "canonical fibration" for the Lie group $G_2$ QUESTION [8 upvotes]: $\DeclareMathOperator\SO{SO}$At the very begining of Akbulut and Kalafat - Algebraic topology of $G_2$ manifolds, the authors stated that there is a "canonical fibration" for $G_2$ of the form $$G_2\to \SO(7)\to \mathbb{R}P^7,$$ where the map $G_2\to \SO(7)$ is obtained by regarding $G_2$ as the automorphisms of the imaginary octonions, which, as a real vector space, is of dimension $7$. So, how does one construct a homotopy equivalence $\SO(7)/G_2\to\mathbb{R}P^7$? REPLY [5 votes]: In Spinors and Calibrations by F. Reese Harvey, you can find proof (p. 283) of $S^7 \simeq Spin(7)/G_2.$ It takes the same approach as Bryant's notes mentioned in the comments but it is much more detailed in this case. The main idea is to write down the spinor representation of $Spin(7)$ using octonionic multiplication. It takes some work to see that the stabilizer of a point is then isomorphic to the automorphism group of the octonions, which is the compact Lie group $G_2.$<|endoftext|> TITLE: The finite groups with a zero entry in each column of its character table (except the first one) QUESTION [15 upvotes]: $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Aut{Aut}$Consider the class of finite groups $G$ having a zero entry in each column of its character table (except the first one), i.e. for all $g \neq e$ there is an irreducible character $\chi$ such that $\chi(g) = 0$. I have been led to consider such character table, here are the examples found (see the GAP codes in Appendix): at order less than $384$, this class reduces to $A_5$, $S_5$, $\PSL(2,7)$, $\Aut(\PSL(2,7))$ and $A_6$, every simple group of order less than $3000000$ is in this class except $A_7$, $M_{22}$, every perfect group of order less than $3600$ in this class is simple, if $5 \le n \le 19$, then $A_n$ is not in this class iff $n \in \{ 7, 11, 13, 15, 16, 18, 19\}$. Idem for $S_n$. Question: What are the finite groups in this class? Which simple groups are not in? The group $\PSL(2,q)$ is in this class iff it is simple, iff $q \ge 4$ (the generic character table is known). Observe that all the examples found above are almost simple, but: Proposition: This class is stable by direct product. proof: Immediate by Theorem 4.21 in Isaacs' Character Theory of Finite Groups stating that the irreducible characters of a direct product are exactly the product of the irreducible characters of the components. $\square$ Note that $A_5 \times A_5$ is a non almost-simple finite group in this class, maybe the smallest one. Appendix Small groups: gap> for o in [2..383] do n:=NrSmallGroups(o);; for d in [1..n] do G:=SmallGroup(o,d);; L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=0 then Print([G,[o,d]]); fi; od; od; [ AlternatingGroup( [ 1 .. 5 ] ), [ 60, 5 ] ][ SymmetricGroup( [ 1 .. 5 ] ), [ 120, 34 ] ][ Group( [ (3,4)(5,6), (1,2,3)(4,5,7) ] ), [ 168, 42 ] ][ Group( [ (1,4,6,8,5,2,7,3), (1,3,8,6,5,4,7) ] ), [ 336, 208 ] ][ AlternatingGroup( [ 1 .. 6 ] ), [ 360, 118 ] ] Simple Groups: gap> it:=SimpleGroupsIterator(10,3000000);; for G in it do L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([G]); fi; od; [ A7 ][ M22 ] Perfect Groups: gap> for o in [60..3599] do n:=NumberPerfectGroups(o);; for d in [1..n] do G:=PerfectGroup(o,d);; L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=0 then Print([G,[o,d]]); fi; od; od; [ A5, [ 60, 1 ] ][ L3(2), [ 168, 1 ] ][ A6, [ 360, 1 ] ][ L2(8), [ 504, 1 ] ][ L2(11), [ 660, 1 ] ][ L2(13), [ 1092, 1 ] ][ L2(17), [ 2448, 1 ] ][ L2(19), [ 3420, 1 ] ] Alternating groups: gap> for n in [5..19] do s:=Concatenation("A",String(n));; L:=Irr(CharacterTable(s));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([s]); fi; od; [ "A7" ][ "A11" ][ "A13" ][ "A15" ][ "A16" ][ "A18" ][ "A19" ] REPLY [3 votes]: By using the answer of Geoff Robinson, we can prove that every non-abelian and non-alternating finite simple group is in this class except $M_{22}$, $M_{24}$ and $C_3$. The result follows from the following computation together with Corollary 2 in Granville and Ono - Defect zero $p$-blocks for finite simple groups. gap> S:=["M12","M22","M24","J2","HS","Suz","Ru","C1","C3","BM"];; gap> for s in S do L:=Irr(CharacterTable(s));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([s]); fi; od; [ "M22" ][ "M24" ][ "C3" ] Corollary 2. Every finite simple group $G$ has a $p$-block of defect $0$, for every prime $p$, except in the following cases: $G$ has no $2$-block of defect $0$ if it is isomorphic to $M_{12}$, $M_{22}$, $M_{24}$, $J_2$, $\mathit{HS}$, $\mathit{Suz}$, $\mathit{Ru}$, $C1$, $C3$, $\mathit{BM}$, or $A_n$ where $n \ne 2m^2 + m$ nor $2m^2 + m + 2$ for any integer $m$. $G$ has no $3$-block of defect $0$ if it is isomorphic to $\mathit{Suz}$, $C3$, or $A_n$ with $3n + 1 = m^2 r$ where $r$ is squarefree and divisible by some prime $q \equiv 2 \bmod 3$.<|endoftext|> TITLE: Abelian variety with CM defined over real numbers QUESTION [5 upvotes]: Is there an abelian variety $A/\mathbb R$ of dimension $n$ such that $End_{\mathbb R}(A)\otimes \mathbb Q$ contains a field $K$ of degree $[K:\mathbb Q]=2n$? ($End_{\mathbb R}(A)$ is the ring of $\mathbb R$-endomorphisms of $A$) REPLY [5 votes]: (Essentially the same argument as the one given by Will Sawin, but perhaps a bit simpler. Further clarification included thanks to comment by Wojowu.) If $A$ is an abelian variety over a field $k\supset\mathbb{Q}$, then the tangent space $T_0(A)$ at identity is a module over $\mathrm{End}_{k}(A)\otimes\mathbb{Q}$. Now, if the latter contains a field $K$, then $T_0(A)$ has to have dimension at least 1 over $K$. On the other hand $T_0(A)$ has dimension $\dim(A)$ over $k$. Thus $[K:\mathbb{Q}]\leq\dim(A)$. Added: Alternate explanation of above. Consider $A(\mathbb{R})$ as a Lie group with connected component $A(\mathbb{R})_0$. The exponential map $T_0(A)\to A(\mathbb{R})_0$ is the universal covering of a compact torus of real dimension $n=\dim(A)$. It is clear that elements of $\mathrm{End}_{\mathbb{R}}(A)$ lift to this cover; Let $K$ be a subfield of $\mathrm{End}_{\mathbb{R}}(A)$. Note that $\mathcal{O}_K$ is a domain and the covering group (which is $\mathbb{Z}^{n}$) is a module over $\mathcal{O}_K$. Thus the rank of $\mathcal{O}_k$ as a $\mathbb{Z}$ module is at most $n$.<|endoftext|> TITLE: For what LCH groups is the Haar measure $\mu(U x U)$ bounded? QUESTION [14 upvotes]: Let $G$ be a locally compact Hausdorff (LCH) topological group with left Haar measure $\mu$. Given a compact unit neighborhood $U$, consider the function $$ \Phi: \quad G \to (0,\infty), \quad x \mapsto \mu(U x U). $$ My question is: Can one give a natural characterization of the groups for which this function is bounded? One conjecture (see below): This happens exactly for IN groups, which are groups for which there exists a compact unit neighborhood $U$ satisfying $x U x^{-1} = U$ for all $x \in G$. Thoughts/observations: The question is independent of the choice of $U$. Indeed, if $U,V$ are both compact unit neighborhoods, then $U \subset \bigcup_{i=1}^n x_i V$ and $U \subset \bigcup_{j=1}^m V y_j$ for suitable $x_i,y_j \in G$, and this easily allows to bound $\mu(U x U)$ in terms of $\mu(V x V)$. If $G$ is not unimodular, then $\Phi$ is not bounded, since $\Phi(x) \geq \mu(U x) = \Delta(x) \cdot \mu(U)$, so that $\Phi$ is bounded from below (up to a constant) by the modular function, which is unbounded for non-unimodular groups. If $G$ is an IN group, then $\Phi$ is bounded. Indeed, by the first observation from above we can choose $U$ to satisfy $x U x^{-1} = U$ for all $x$, and then $\Phi(x) = \mu(U x U) = \mu(x U U) = \mu(U U)$ for all $x \in G$. What I have not been able to show is that if $\Phi$ is bounded, then $G$ needs to be IN. Of course, it could be that this simply does not hold. REPLY [13 votes]: Your conjecture is correct. Suppose that we have a compact unit neighbourhood $U$ such that $\mu(UxU) \ll 1$ for all $x$. As you have already noted, we can take $\mu$ to be unimodular, and the choice of neighbourhood is not relevant, so we may assume without loss of generality that $U$ is symmetric: $U^{-1} = U$. (This makes $U$ what is called an approximate group in arithmetic combinatorics, and the intuition that $U$ should behave like a subgroup of $G$ is underlying the arguments below.) We allow implied constants in asymptotic notation to depend on $U$, thus for instance $\mu(U) \asymp 1$. Note that the conjugate $x U x^{-1}$ of $U$ is commensurate with $U$ in the sense that $\mu( U \cdot x U x^{-1} ) = \mu( U x U ) \ll 1$. (In the language of arithmetic combinatorics, $U$ stays close to its conjugates $xUx^{-1}$ in Ruzsa distance.) In the spirit of the isomorphism theorems (or the Ruzsa covering lemma in arithmetic combinatorics), one now expects $U$ and $xUx^{-1}$ to have large intersection, and this can be accomplished (at the cost of enlarging $U$ to $U^2$) by the following convolution argument (cf. the double counting argument that shows that $|H \cdot K| = |H| |K| / |H \cap K|$ for finite subgroups $H,K$ of $G$). Observe that the convolution $1_U * 1_{xUx^{-1}}$ has an $L^1(G,\mu)$ norm of $\mu(U) \mu(x U x^{-1}) \asymp 1$ and is supported on $U x U x^{-1}$, which has measure $O(1)$. Thus there must exist a point $y \in G$ where $1_U * 1_{xUx^{-1}}(y) \gg 1$, thus $$ \mu( yU \cap x U x^{-1} ) \gg 1$$ which implies $$ \mu( (yU \cap x U x^{-1})^{-1} \cdot (yU \cap x U x^{-1}) ) \gg 1$$ and hence $$ \mu( U^2 \cap x U^2 x^{-1} ) \gg 1.$$ To put it another way, the inner products of the functions $1_{x U^2 x^{-1}}$ with $1_{U^2}$ are uniformly bounded from below. We can now use an "ergodic" argument to extract an invariant object (in the spirit of the Alaoglu--Birkhoff ergodic theorem). Let $S$ be the closed convex hull in $L^2(G,\mu)$ of the functions $1_{x U^2 x^{-1}}$. By the Hilbert projection theorem, this set has a unique element $f$ of minimal norm. All elements of $S$ have inner product with $1_{U^2}$ uniformly bounded from below, so $f$ does also; in particular, $f$ is non-trivial. Since $S$ is conjugation-invariant, symmetric, bounded and consists of non-negative functions, $f$ must be non-negative, symmetric, and conjugation-invariant. At this point one could already extract a conjugation-invariant set of positive finite measure by taking level sets of $f$, but this is not quite regular enough for the conjecture, so we take a convolution to achieve an additional smoothing (in the spirit of the Steinhaus theorem). The convolution $f*f$ is then in $C_0(G)$ (this follows from a standard limiting argument, approximating $f$ in $L^2(G,\mu)$ by $C_c(G)$ functions and using Young's inequality), conjugation-invariant, and strictly positive at the origin; taking level sets, we obtain a non-trivial conjugation-invariant compact unit neighbourhood, as required.<|endoftext|> TITLE: Looking for an electronic copy of Holmgren's old paper QUESTION [7 upvotes]: I would like to know if anyone has an electronic copy of the following paper: "Holmgren, E.: Über Systeme von linearen partiellen Differentialgleichungen. Översigt Vetensk. Akad. Handlingar 58, 91–105 (1901)" In my search, the best result I found was the (possible) statement of the main result of this article which can be found in the following article: https://people.kth.se/~haakanh/publications/Hed-MZ2.pdf. More precisely, Theorem (Holmgren) Suppose $I$ is a real-analytic nontrivial arc of $\partial \Omega$. Then if $u$ is smooth on a planar neighbohood $\mathcal{O}$ of $I$ and $\Delta^N u=0$ holds on $\mathcal{O} \cap \Omega$ with $\partial_{n}^{j-1}|_I=0$ for $j=1, \dots, 2N$, then $u(z)=0$ on $\mathcal{O} \cap \Omega$, provided that the open set $\mathcal{O} \cap \Omega$ is connected. Any information is welcome, for example, if this article is published in a book. REPLY [12 votes]: The full text of the article can be found scanned here.<|endoftext|> TITLE: Example of a $\Pi^2_2$ sentence? QUESTION [12 upvotes]: This question is about logical complexity of sentences in third order arithmetic. See Wikipedia for the basic concepts. Recall that the Continuum Hypothesis is a $\Sigma^2_1$ sentence. Furthermore (loosely speaking) it can't be reduced to a $\Pi^2_1$ sentence, as stated in Emil Jeřábek's answer to Can we find CH in the analytical hierarchy?. Is there an example of a $\Sigma^2_2$ sentence with no known reduction to a $\Pi^2_2$ sentence? (Equivalently, a $\Pi^2_2$ sentence with no known reduction to a $\Sigma^2_2$ sentence.) I mean that there should be no known reduction even under large cardinal assumptions. I'd prefer an example that's either famous or easy to state. But to begin, any example will do. Update: Sentences such as "$\mathfrak{c} \leqslant \aleph_2$" and "$\mathfrak{c}$ is a successor cardinal" are $\Delta^2_2$, meaning that they're simultaneously $\Sigma^2_2$ and $\Pi^2_2$. The reason is that each such sentence (and also its negation) can be expressed in the form "$\mathbb{R}$ has a well-ordering $W$ such that $\phi(W)$" where $\phi$ is $\Sigma^2_2$. REPLY [9 votes]: The Suslin hypothesis is $\Pi^2_2,$ and $T = ZFC + GCH + LC$ (LC an arbitrary large cardinal axiom) does not prove it to be equivalent to any $\Sigma^2_2$ sentence. Suppose toward contradiction $T$ proves SH to be equivalent to $\exists A \subset \mathbb{R} \varphi(A),$ where $\varphi$ is $\Pi^2_1.$ Assume $V \models T.$ We'll use several results from Chapters VIII and X of Devlin and Johnsbraten's The Souslin Problem. There are generic extensions $V[G] \models T+\diamondsuit^*$ and $V[G][H] \models T+SH$ which do not add reals to or collapse cardinals of $V.$ In $V[G][H],$ there is $A \subset \mathbb{R}$ such that $\varphi(A)$ holds and $A' \subset \omega_1$ which codes a bijection between $\mathbb{R}$ and $\omega_1$ as well as $A.$ By downwards absoluteness of $\varphi,$ $A$ witnesses that $V[G][A'] \models SH.$ But we also have $V[G][A'] \models \diamondsuit^*$ by Lemma 4 (pg. 79), which is a contradiction since $\diamondsuit$ negates SH.<|endoftext|> TITLE: CE(g) for g infinite dimensional QUESTION [6 upvotes]: On the nlab page for Chevalley–Eilenberg algebras, it defines $\operatorname{CE}(\mathfrak g)$ for $\mathfrak g$ finite dimensional, and then says "This has a more or less evident generalization to infinite-dimensional Lie algebras", and provides no more details. Please could someone outline this generalisation, and if the theory follows through (e.g. is there still a contravariant equivalence of categories?). REPLY [6 votes]: A definition that always works and does agree with that one in the finite-dimensional case is the following: put $$ C^k(\mathfrak{g})=({\Lambda}^k\mathfrak{g})^*=\operatorname{Hom}({\Lambda}^k\mathfrak{g}, \mathbb{F}). $$ (Here $\mathbb{F}$ is the ground field, of course.) The differential is given by the formula $$ (d\phi)(g_1\wedge\dotsb\wedge g_{k+1})=\sum_{1\le i TITLE: Can we embed a closed manifold into a homotopy equivalent CW complex? QUESTION [11 upvotes]: Suppose $X$ is a CW complex and $M$ is a closed manifold and suppose further that there exists a homotopy equivalence $X \simeq M$. Does there exists an embedding of $M$ into $X$ (i.e. an injective (potentially cellular) map)? If this setting is to broad, I'm specifically interested in the case, where $M$ is a surface and $X$ is also $2$-dimensional (maybe even restrict it to aspherical surfaces). Edit: mme provided a counterexample in dimension 3 (homotopy equivalent but not homeomorphic lens spaces), which can probably be generalized to higher dimensions. So only the two-dimensional case remains. REPLY [15 votes]: Pick a torus, and add two discs along a meridian and a longitude. You get a 2-complex homotopic to a sphere that does not contain a sphere. This generalises easily to any genus by picking a genus-$g$ surface. More generally, a finite 2-complex contains finitely many surfaces, and there are some moves (like the Matveev - Piergallini move) that preserve the (simple) homotopy type of the 2-complex, but can modify the surfaces it contains. REPLY [6 votes]: While thinking about it with a friend, we came up with the following two dimensional counter example: Take the standard knot diagram of the trefoil knot (as a self-intersecting curve in $\mathbb{R}^2$) and let $\bar{X}$ denote the "inner" of this curve i.e. the curve together with the 4 areas bounded by it. Let $X$ denote $\bar{X} \cup_\phi D^2$, where $\phi \colon S^1\to \bar{X}$ follows the trefoil knot. Since $\bar{X}$ is contractible, $X$ is homotopy equivalent to $S^2$. Using cellular homology, one can see that the "fundamental class" of $X$ hits the middle cell of $\bar{X}$ twice, hence there are no injective homotopy equivalences. By taking "connected sums" of this counterexample with surfaces, one obtains counter examples for all closed surfaces.<|endoftext|> TITLE: Infinitely many rigid and non-rigid reductions $\mathrm{mod}\:p$ QUESTION [8 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{Q}$. Choose a model $\mathcal{X}$ over $\mathbb{Z}$. Can it be that for infinitely many primes $p$ there are non-trivial automorphisms of $\mathcal{X}\otimes \overline{\mathbb{F}_p}$ and for infinitely many primes there aren't? Maybe it happens for K3 surfaces but I can't think of an example. REPLY [2 votes]: $\newcommand{\bQ}{\mathbb{Q}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\fp}{\mathfrak{p}}\newcommand{\bF}{\mathbb{F}}\newcommand{\bP}{\mathbb{P}}$Here is a variation on the theme of Will Sawin's answer which allows for a non-conditional example. The construction is based on the following theorem of François Charles: Theorem (Theorem 1.1 in Exceptional isogenies between reductions of pairs of elliptic curves,) If $E_1,E_2$ are elliptic curves over a number field $K$ then there exist infinitely many primes $\mathfrak{p}$ of $K$ such that the reductions $E_{1,\mathfrak{p}}$ and $E_{2,\mathfrak{p}}$ are isogenous over $\overline{\mathbb{F}}_p$. So let $E_1,E_2$ be elliptic curves over $\bQ$ that are not geometrically isogenous. There must be infinitely many primes at which the reductions of $E_1, E_2$ are not geometrically isogenous, as otherwise Chebotarev density would imply that there is a finite extension $K/\mathbb{Q}$ over which Galois representations $V_lE_1,V_lE_2$ are isomorphic, contradicting the assumption. Charles's theorem says, on the other hand, that there are also infinitely many primes $p$, such that $E_{1,p}$ and $E_{2,p}$ are geometrically isogenous. For the variety $Y=E_1\times E_2$ the reduction $Y_p$ has more automorphism when $E_{1,p}$ and $E_{2,p}$ are isogenous than for other primes $p$. We are going to rigidify $Y$ in a way that there will be no non-trivial automorphisms modulo those primes for which $E_{1,p}$ and $E_{2,p}$ are non-isogenous while keepong an infinite automorphism group for the rest of the primes of good reduction. We'll start by replacing $\bQ$ by a finite extension and will construct the desired variety there, so that the ultimate example will be obtained by considering that variety as a (non-geometrically connected) scheme over $\bQ$. Let $K/\bQ$ be a finite extension such that $E_1[3](K)=(\bZ/3)^2,E_2[3](K)=(\bZ/3)^2$. Pick an injective function $a:(E_1\times E_2)[3](K)\to \bZ_{\geq 1}$ and define the variety $X$ as the following birational modification of the surface $Y_K$: for each $3$-torsion point $x\in (E_1\times E_2)[3](K)$ perform a blow-up of $E_1\times E_2$ at an infinitely near point of order $a(x)$ supported at $x$. The fiber of the map $X\to Y_K$ at a point $x$ is thus a chain of $a(x)$ copies of the projective line. There is also a natural integral model $\mathcal{X}$ of $X$ over $\mathcal{O}_K$, constructed by blowing up the product of models of $E_1$ and $E_2$. Lemma If $\fp\not\mid 3$ is a prime of $K$ such that $E_1,E_2$ have good reduction at $\fp$, then $X_{\fp}\times\overline{\bF}_p$ has non-trivial automorphisms if and only if $E_{1,\fp}$ and $E_{2,\fp}$ are geometrically isogenous. Proof. An automorphism $g$ of $X_{\overline{\bF}_p}$ induces a permutation of closed subvarieties of this surface that are isomorphic to $\bP^1$, preserving the incidence relation between them. In particular, for each $x\in Y[3](\overline{\bF}_p)$ the fiber of $X\to Y$ above $x$ is preserved by $g$. An automorphism of a chain of $\bP^1$s must have a fixed point, so $g$ admits a fixed point lying in the fiber of $X\to Y$ above $0\in Y_{\overline{\bF}_p}=E_{1,\overline{\bF}_p}\times E_{2,\overline{\bF}_p}$ and therefore $g$ descends to an automorphism of $Y_{\overline{\bF}_p}$ because $Y_{\overline{\bF}_p}$ can be established as the Albanese variety of $X_{\overline{\bF}_p}$ wrt that fixed point. If $E_{1,\overline{\bF}_p}$ and $E_{2,\overline{\bF}_p}$ are not isogenous then any automorphism $h$ of $Y_{\overline{\bF}_p}$ must be a product of automorphisms of $E_{1,\overline{\bF}_p}$ and $E_{2,\overline{\bF}_p}$ and, in particular, has finite order. For it to lift to an automorphism of $X_{\fp}$ the action of $h$ on $Y_{\fp}[3]$ must be trivial, but this forces $h$ to be trivial because any matrix in $GL_n(\bZ_3)$ congruent to $1$ mod $3$ has infinite order if it is not the identity. Conversely, suppose that $f:E_{1,\overline{\bF}_p}\to E_{2,\overline{\bF}_p}$ is an isogeny. Let $N$ be an integer larger than any of the values of the function $a$. Then $g:(t,s)\mapsto (t,s+3p^N\cdot f(t))$ is an automorphism of $Y_{\overline{\bF}_p}$ that preserves every $3$-torsion point and, moreover, for every $3$-torsion point $x\in Y_{\fp}[3]$ the action of $g$ on the quotient $\mathcal{O}_{Y,x}/\mathfrak{m}_x^N$ is trivial. Therefore $g$ lifts to an automorphism of the blow-up $X_{\overline{\bF}_p}\to Y_{\overline{\bF}_p}$, as desired.<|endoftext|> TITLE: Upper bound for maximal gap between consecutive numbers consisting only $4k+1$ primes QUESTION [5 upvotes]: Denote $A$ $-$ set of positive numbers with only prime factors of the form $4k+1$ and $B$ $-$ set of positive numbers that can be represented as sum of two squares. $A$ is a subset of $B$ and there is upper bound $2 \sqrt{2} n^{\frac{1}{4}}$ for maximal gap between consecutive elements of $B$. Question: Is any upper bound for maximal gap between consecutive elements of $A$ known? REPLY [6 votes]: One can prove a bound comparable to the $O(n^{1/4})$ bound for $B$. To derive it, notice that any odd number of the form $a^2+b^2$ with $(a,b)=1$ lies in $A$. Now, for a given large number $N$, choose the largest even $a$ such that $a^2 TITLE: Examples of vector spaces with bases of different cardinalities QUESTION [15 upvotes]: In this question Sizes of bases of vector spaces without the axiom of choice it is said that "It is consistent [with ZF] that there are vector spaces that have two bases with completely different cardinalities". Now I've had a hard time finding a reference for this fact, so my first question would be whether someone knows a proper sources detailing the construction of a vector space with bases of different cardinalities (in some suitable model of $ZF+\neg AC$). Secondly, is there a general procedure for generating such spaces? (E.g. by starting with a model $M$ of $ZF+\neg AC$ and a vector space $V$ which has no basis in $M$ and then adding several bases of $V$ to $M$ via iterated forcing.) And lastly, are there only "artifical/exotic" examples of such vector spaces or are there also models where some "standard" vector spaces such as $\mathbb{F}_2^\omega$ over $\mathbb{F}_2$ or $\mathbb{R}$ over $\mathbb{Q}$ have bases of different cardinalities. (Pretty sure that $\mathbb{R}$ over $\mathbb{Q}$ is still an open problem but maybe there are some other "nice" examples.) REPLY [14 votes]: This is not a very thoroughly studied problem. So to start from the end, there is no standard procedure for this sort of construction. We know of one, it can maybe be adapted slightly to get a mildly more general result, but it's not something like "let's add a new vector space without a basis" or "let's add an amorphous set" that has a very well-understood and common constructions. The original construction is due to Läuchli and you can find it in German in his paper Läuchli, H., Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, 1-18 (1962). ZBL0108.01002. It also appears as an exercise in Jech "The Axiom of Choice" as Problem 10.5 with an elaborate explanation of the proof. What you suggest is, in principle, a valid approach. Start with no bases, add one, then add another one with a different cardinality. Unfortunately, we don't really understand the mechanism of adding subsets to models of $\sf ZF$ as well, so it's not clear as to how to do that without: Making the space well-orderable to begin with; or collapsing the two bases to have the same cardinality after all; or giving the space a well-orderable basis; or one of many other unforeseen problems that can crop up when you add sets to your universe. Frankly, I do not understand the construction, mainly because Jech's explanation of it is a bit wishywashy to my taste (it is marked with an asterisk, which at least says that it's not very obvious).<|endoftext|> TITLE: What are the main open problems in the theory of amenability of groups? QUESTION [16 upvotes]: I have read the Paterson and Runde books about amenability of groups, but I do not know what are the most intriguing questions in this area today. A survey or a list of questions would be welcome. REPLY [3 votes]: The Part C of Appendices of the book "Amenability of discrete groups by examples" by Kate Juschenko, contains open problems in amenability.<|endoftext|> TITLE: Question about estimating random symmetric sums modulo p QUESTION [8 upvotes]: Let $n > 0$ be a positive integer (large) and $p > 2$ a fixed prime number. What is the probability that $$\sum_{ 1 \leq i < j \leq n} a_ia_j = 0 \mod p$$ where $a_1, a_2, \dots a_n$ are chosen uniformly from the set $S = \{-1, 1\}$. Does this sum equidistribute mod $p$ as $n$ goes to infinity? What would be the speed of equidistribution in terms of $n$? Is there any literature in this type of random sums? I would be surprised if not but I am unable to find anything related or similar to this. One can also ask what is the probability of this sum being actually zero, but I also have no idea how to deal with it and thought that modulo a prime would be simpler. REPLY [14 votes]: The condition $$\sum_{ 1 \leq i < j \leq n} a_ia_j \equiv 0 \pmod p$$ is equivalent to $$\left(\sum_{ 1 \leq i\leq n} a_i\right)^2 \equiv n \pmod p.$$ So a necessary condition is that $n$ is a quadratic residue modulo $p$ (including the zero residue). If $n$ is divisible by $p$, then the above condition says that the sum of the $a_i$'s is divisible by $p$. Otherwise, the condition says that the sum of the $a_i$'s is congruent to one of the two square-roots of $n$ modulo $p$. Now it is easy to see that the sum of the $a_i$'s is equidistributed modulo $p$ (think about what happens when an $a_i=1$ is switched to $a_i=-1$), hence in the first case the probability is $1/p+o(1)$, in the second case it is $2/p+o(1)$, as $n$ tends to infinity. In fact the probabilities can be calculated explicitly as a linear combination of $n$-th powers of $p$ complex numbers (which only depend on $p$), since the sum of the $a_i$'s modulo $p$ is determined by $\#\{i:a_i=1\}$ modulo $p$, and vice versa. Compare with this post, where the role of $p$ is played by $4$. It follows, in particular, that the $o(1)$ terms above decay exponentially fast. For a more complete reference, see Theorems 8.7.2 & 8.7.3 in Wagner: A first course in enumerative combinatorics (AMS, 2020).<|endoftext|> TITLE: Partitioning a set of lattice points in the plane into rectangles QUESTION [9 upvotes]: The "long comment" by Pietro Majer on Reference for puzzle on dividing piles and scoring products suggests the following problem. Let $S$ be a finite subset of $\mathbb{Z}\times \mathbb{Z}$. By a rectangle, I mean an $a\times b$ array of contiguous elements of $\mathbb{Z}\times \mathbb{Z}$ ($a,b\geq 1$). E.g., $\{(0,0),(2,0)\}$ is not a rectangle since $(1,0)$ is missing. Let $\nu(S)$ be the greatest number of elements of $S$ such that no two lie in a rectangle contained in $S$. Let $\rho(S)$ be the least number of disjoint rectangles whose union is $S$. For what $S$ do we have $\nu(S)=\rho(S)$? Clearly $\nu(S)\leq \rho(S)$. Suppose we remove a maximal rectangle $R$ from $S$, then a maximal rectangle $R'$ from $S-R$, etc., until after $k$ steps we have removed all the elements of $S$. For what $S$ do we have that $k=\rho(S)$ (for any choice of $R,R',\dotsc$)? These questions can easily be extended to higher dimensions. REPLY [4 votes]: Here is a generalization of the original puzzle that also satisfies conditions (1) and (2). Let $k, \ell,n \in \mathbb{N}$, and $S$ be the set of lattice points contained in the convex hull of $\{(-\ell, 0) , (-\ell, n), (0,n), (nk, 0)\}$. Note that the original puzzle corresponds to the case that $\ell=0$ and $k=1$. I claim that $S$ also satisfies (1) and (2). For (1), note that $S$ can be partitioned into $n+1$ disjoint rectangles (take maximal horizontal line segments, for example). On the other hand, let $S':=\{(0,n), (k,n-1), (2k, n-2), \dots, (nk, 0)\} \subseteq S$. Since no two points in $S'$ can be covered by the same rectangle, we have $\nu(S)=\rho(S)=n+1$. For (2), note that removing any maximal rectangle from $S$ yields two smaller instances $S_1$ and $S_2$ of the same problem, such that the sum of the 'heights' of $S_1$ and $S_2$ is one less than the height of $S$. Thus, any sequence of rectangles chosen by the procedure given in (2) will terminate after exactly $n+1$ steps. Note that not all subsets of $\mathbb{Z} \times \mathbb{Z}$ satisfy (1) and (2). In fact, we claim that the set $S=([3] \times [6]) \setminus \{(1,3), (1,6), (3,1), (3,4)\}$ fails both (1) and (2). For (1), it is easy to check that $\rho(S)=5$, but $\nu(S)=4$. For (2), if we choose $R=\{(1,4), (1,5), (2,4),(2,5)\}$ as the first maximal rectangle to remove, then procedure (2) will produce a partition of $S$ into six rectangles. However, as already noted, $\rho(S)=5$.<|endoftext|> TITLE: Distribution of some sums modulo p QUESTION [5 upvotes]: Fix a finite set of integers $S$ and a prime number $p$. Let $(a_1, a_2, \dotsc, a_n)$, $(b_1, b_2, b_3, \dotsc, b_n)$ be two sequences of integers where the numbers $a_i$ and $b_i$ are chosen uniformly from the set $S$ (In the case I care about S = {-2,-1,1,2 }). I would like to understand what is the probability that $$\sum_{i_1 < i_2 < \dotsb < i_k } a_{i_1}a_{i_2}\dots a_{i_k} = 0 \mod p$$ an the probability that $$\sum_{i_1 \leq j_1 < i_2 \leq j_2 < \dotsb < i_k \leq j_k} a_{i_1}b_{j_1}\dots a_{j_k}b_{j_k} = 0 \mod p.$$ I am interested in the asymptotic probabilities as $n$ increases and $k$ is fixed (or $k$ grows much slower than $n$) and expect some type of equidistribution as $n$. One particular case is Question about estimating random symmetric sums modulo p, where one particular sum is reduced to the linear case. Can this trick be for the general case? Any suggestion, trick or vaguely related information related to this is highly appreciated. P.S. Can the second type of sum (the one involving $a_i$, $b_i$) be somehow related to a sum of the first type?, the ordering makes things much more tricky and I am wondering whether there is a procedure to reduce this sum to the first type. (Also the second sum even though not so natural is the one that I really need to understand). Thanks! REPLY [7 votes]: Using the Newton identities, one can (in the high characteristic regime $p>k$) express the elementary symmetric polynomial $\sum_{i_1 < \dots < i_k} a_{i_1} \dots a_{i_k}$ in terms of the moments $\sum_{i=1}^k a_i^j$ for $j=1,\dots,k$. The question then boils down to the equidistribution of $\sum_{i=1}^k (a_i^j)_{j=1}^k$ in ${\mathbf F}_p^k$. There is however an obstruction to equidistribution because the points $(a^j)_{j=1}^k$ for $a=-2,-1,1,2$ do not span the entirety of ${\mathbf F}_p^k$ once $k \geq 4$, due to the existence of non-trivial polynomials $P$ of degree $4$ or higher that vanish at $-2,-1,1,2$. For instance because $P(x) = (x+2)(x+1)(x-1)(x-2) = x^4 - 5 x^2 + 4$ vanishes at these points, the vector $(m_j)_{j=1}^k := \sum_{i=1}^k (a_i^j)_{j=1}^k$ is surely constrained to the hyperplane $m_4 - 5 m_2 + 4k = 0$. This is going to create some distortions to the probability that $\sum_{i_1 < \dots < i_k} a_{i_1} \dots a_{i_k}$ vanishes mod $p$ that can be explicitly calculated for each $p,k$, but the formula is going to be messy (these errors will be of lower order in the limit $p \to \infty$ holding $k$ fixed though, by the Lang-Weil estimates). (To put it another way, one can use the Newton identities to express the elementary symmetric polynomial as some polynoimal combination of the frequencies $d_{-2}, d_{-1}, d_1, d_2$ that count how often the random sequence $a_1,\dots,a_k$ attains each of its permitted values $-2, -1, 1, 2$. To count the probability that this polynomial vanishes mod $p$, one has to count the points in some variety over ${\bf F}_p$ which in general is a task for the Lang-Weil estimate.) Your second sum seems to be expressible in terms of a symmetric polynomial in a suitable matrix algebra. If one had $i_k < j_k$ in place of the constraint $i_k \leq j_k$ then one just needs to take the order $2k$ elementary symmetric polynomial of the matrices $\begin{pmatrix} 0 & b_i \\ a_i & 0 \end{pmatrix}$ and extract the top left coefficient. With $i_k \leq j_k$ the situation is more complicated but I expect there is still some sort of matrix representation. However being non-abelian I doubt there is a reduction to the abelian elementary symmetric polynomial considered earlier, and given how complicated that formula already was I'm afraid it is not going to be fun to try to control this sum (except possibly in the asymptotic regime where $p$ is somewhat large and $k$ goes to infinity; actually the nonabelian case might be substantially more "mixing" than the abelian one and one could conceivably get better asymptotics by using some of the theory of expansion of Cayley graphs, but this looks like a lot of work...).<|endoftext|> TITLE: Radio-playing sequence QUESTION [5 upvotes]: Motivation. (Please skip if you are not in the mood for "chitchat".) Last night I listening to a classical radio station, and for the umpteenth time, they played Mendelssohn's Psalm 42, a composition that I like very much. Luckily, a week ago, when they played it, it was followed by a different piece (Rodeo by Copland) than yesterday (Bach d-minor piano concerto). I wondered how long they can proceed so that piece $X$ is never immediately followed by piece $Y$ two separate times. Which led to the following little problem. Formalization. We regard any positive integer $n$ as the set of its predecessors, so $n = \{0,\ldots,n-1\}$. For positive integers $m, n\in \mathbb{N}$ we say that a map $f : m\to n$ is a radio-playing function if whenever $a,b \in m-1$ with $a\ne b$ and $f(a) = f(b)$, then $f(a+1) \neq f(b+1)$. Using the pigeonhole principle, it is easy to see that if $m > n^2$ there cannot be a radio-playing function $f : m\to n$. So, given $n\in \mathbb{N}\setminus \{0\}$, let $A_n$ be the largest integer such that there is a radio-playing function $f: A_n \to n$. What is the value of $A_n$ in terms of $n$? REPLY [7 votes]: Although we have by now a precise answer, I'd like to keep the summer mood of the question and play a little more with it by an elementary arithmetic approach. The solution I wish to sell is good for any $n$, and (just in case your favorite classic radio station entrusts you with the organization of the schedule of the year, which I think would be a finest choice) includes the issues: what piece $f(t)$ will be played at time $t$, and conversely, at what time $h(x,y)$ the piece $x$ is to be played and followed by piece $y$. Consider the functions $(f,g):\mathbb{N}\to\mathbb N\times\mathbb N$ and $h:\mathbb N\times\mathbb{N}\to\mathbb N$ defind by $$ f(t):=\cases{ \frac{t-\lfloor\sqrt t\rfloor^2}2 &if $\; t-\lfloor\sqrt t\rfloor^2$ is even\\ \\ \lfloor\sqrt t\rfloor &if $ \;t-\lfloor\sqrt t\rfloor^2$ is odd}\qquad\qquad g(t):=f(t+1),$$ and $$h(x,y):=\cases{x^2+2x&if $\;y=0 ,$\\\\y^2+2x&if $\;0\le x n^2$, as you observed).<|endoftext|> TITLE: Squares in a triquadratic field QUESTION [12 upvotes]: I would like to know (as part of an attempt to streamline some calculations in the cohomology of a Morava stabiliser group) whether $1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$ is a square in $\mathbb{Q}(\sqrt{-3},\sqrt{5},\sqrt{-7})$. What is an efficient method for this kind of question? REPLY [10 votes]: Set $K = \mathbf Q(\sqrt{-3},\sqrt{5},\sqrt{-7})$ and $\alpha = 1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$. I think the "best" way to show $\alpha$ is not a square in $K$ is not to show something else is not a square in a (smaller) number field, but to show something else is not a square in a finite field, and more precisely in $\mathbf Z/p\mathbf Z$ for some prime $p$. We'll wind up doing this with $p = 79$. To prove $\alpha$ is not a square in $K$, or equivalently in $\mathcal O_K$, it suffices to find a (nonzero) prime ideal $\mathfrak p$ in $\mathcal O_K$ such that $\alpha \bmod \mathfrak p$ is not a square. There are many such $\mathfrak p$ if $\alpha$ is not a square: the Chebotarev density theorem implies that a nonsquare in $\mathcal O_K$ is not a square mod $\mathfrak p$ for a set of $\mathfrak p$ with density $1/2$ (natural density or Dirichlet density). That means if such $\mathfrak p$ exist, it should not take very long to find one, just like for a nonsquare integer $b$ it should not take long to find a prime $p$ such that $b \bmod p$ is not a square (the set of such $p$ has density $1/2$). How can we go about searching for a prime ideal $\mathfrak p$ where $\alpha \bmod \mathfrak p$ is not a square? In practice it's a good idea to look for $\mathfrak p$ lying over a prime number $p$ that splits completely in $\mathcal O_K$ since then $\mathcal O_K/\mathfrak p \cong \mathbf Z/p\mathbf Z$ and it's computationally easy to work in a field of prime order. The set of such $p$ has (natural) density $1/[K:\mathbf Q] = 1/8$ and $p$ splits completely in $K$ if and only if it splits completely in the quadratic fields $\mathbf Q(\sqrt{-3})$, $\mathbf Q(\sqrt{5})$, and $\mathbf Q(\sqrt{-7})$. These splitting conditions are the congruences $$ p \equiv 1 \bmod 3, \ \ p \equiv 1, 4 \bmod 5, \ \ p \equiv 1, 2, 4 \bmod 7. $$ By the Chinese remainder theorem, these conditions correspond to six congruence conditions mod $105$, which turn out to be $$ p \equiv 1, 4, 16, 46, 64, 79 \bmod 105. $$ The first prime fitting each of these six conditions is, respectively, $$ 211, \ \ 109, \ \ 331, \ \ 151, \ \ 379, \ \ 79. $$ The smallest of these primes is $79$, so $79$ is the smallest prime splitting completely in $K$. Let's work with $p = 79$. Since it splits completely in $K$, there are $8$ primes $\mathfrak p_1, \ldots, \mathfrak p_8$ lying over $79$ in $\mathcal O_K$. For each $\mathfrak p_i$, $\mathcal O_K/\mathfrak p_i \cong \mathbf Z/79\mathbf Z$. We will reduce $\alpha$ modulo some of these $\mathfrak p_i$ until we find a nonsquare in $\mathbf Z/79\mathbf Z$. Since $1170 \equiv 64 \bmod 79$ and $19110 \equiv 71 \bmod 79$, for each $\mathfrak p_i$ lying over $79$ we have $$ \alpha \equiv 64\sqrt{-3}\sqrt{5}\sqrt{-7}-71 \bmod \mathfrak p_i. $$ How can we describe the primes $\mathfrak p_i$? The congruences $a^2 \equiv -3 \bmod 79$, $b^2 \equiv 5 \bmod 79$, and $c^2 \equiv -7 \bmod 79$ each have two solutions: $a \equiv 32, 47 \bmod 79$, $b \equiv 20, 59 \bmod 79$, and $c \equiv 25, 54 \bmod 79$. Picking one solution for each of these three congruences mod $79$ distinguishes the $8$ different prime ideals $\mathfrak p_i$ from each other. Example 1. There is a prime $\mathfrak p_1$ over $79$ where $$ \sqrt{-3} \equiv 32 \bmod \mathfrak p_1, \ \ \sqrt{5} \equiv 20 \bmod \mathfrak p_1, \ \ \sqrt{-7} \equiv 25 \bmod \mathfrak p_1. $$ That makes $$ \alpha \equiv 64(32)(20)(25)-71 \equiv 10 \bmod 79, $$ but $(\frac{10}{79}) = 1$ by quadratic reciprocity. Explicitly, $10 \equiv 22^2 \bmod 79$, so $\alpha \equiv 22^2 \bmod \mathfrak p_1$. That didn't work out as we had hoped. Example 2. There is a prime $\mathfrak p_2$ over $79$ where $$ \sqrt{-3} \equiv 47 \bmod \mathfrak p_2, \ \ \sqrt{5} \equiv 20 \bmod \mathfrak p_2, \ \ \sqrt{-7} \equiv 25 \bmod \mathfrak p_2. $$ That makes $$ \alpha \equiv 64(47)(20)(25)-71 \equiv 6 \bmod 79, $$ and $(\frac{6}{79}) = -1$ by quadratic reciprocity. Thus $\alpha \bmod \mathfrak p_2$ is not a square, so $\alpha$ is not a square in $\mathcal O_K$ and we are done.<|endoftext|> TITLE: Is there a categorical version of the splitting principle? QUESTION [18 upvotes]: One of many places we see a "splitting principle" at work is in the category $\mathsf{Vect}(X)$ of complex vector bundles over a compact connected Hausdorff space $X$. For any object $E$ in this category, we can find a bundle of compact Hausdorff spaces $p: F \to X$ such that $$ p^* : \mathsf{Vect}(X) \to \mathsf{Vect}(F) $$ maps $E$ to a vector bundle $p^*(E)$ that splits as a direct sum of line bundles. Given that this same idea shows up in many other contexts, I'm hoping there's a general version that subsumes a lot of examples. Maybe something like the following. Just for brevity, let's say a 2-rig is a symmetric monoidal $k$-linear category with absolute colimits. (Such 2-rigs were studied in a paper I wrote with Joe Moeller and Todd Trimble, but there we were assuming the field $k$ had characteristic zero, and here I'd rather not—unless it turns out to be helpful.) Having absolute colimits is the same as having biproducts and having splittings of all idempotents. A map of 2-rigs is a symmetric monoidal $k$-linear functor; such functors automatically preserve absolute colimits. I'm using 2-rigs to generalize categories of vector bundles: $\mathsf{Vect}(X)$ is an example of a 2-rig when $k = \mathbb{R}$ or $\mathbb{C}$, and the above map $p^\ast : \mathsf{Vect}(X) \to \mathsf{Vect}(F)$ is a map of 2-rigs. (Note that $\mathsf{Vect}(X)$ is not an abelian category.) Given an object $E$ in a 2-rig $\mathsf{R}$, I would like to find a map of 2-rigs $$ f: \mathsf{R} \to \mathsf{R}' $$ such that $f^*$ is a direct sum of line objects, meaning objects $L$ with duals $L^*$ such that $L \otimes L^* \cong I$, $I$ being the unit object. Are any general theorems along these lines already known? Of course such theorems become less interesting if $f$ is "far from one-to-one". It may be too much to demand that $f$ is fully faithful, though I'd like to if I could. But we can define the Grothendieck ring $K(\mathsf{R})$ for any 2-rig $\mathsf{R}$, and in the vector bundle example $$ K(p^\ast) : K(\mathsf{Vect}(X)) \to K(\mathsf{Vect}(F)) $$ is monic. So, in the general case, a fallback position would be to ask for a map $f: \mathsf{R} \to \mathsf{R}'$ such that $K(f)$ is monic. Are there general conditions under which we can achieve this? I expect we'll need some finiteness condition to get $f(E)$ to split as a finite direct sum of line objects. Furthermore, as pointed out by Simon Henry below, the splitting principle for $\mathrm{Vect}(X)$ as stated above fails when $X$ is not connected: this is a clue as to the further conditions we will need. For example, it may help to assume that $\mathrm{End}(I)$ has no idempotents other than $0$ and $1$. This is true for $\mathrm{Vect}(X)$ when $X$ is connected, but not otherwise. For each connected component $C \subseteq X$, multiplication by the characteristic function of that component gives an idempotent in $\mathrm{End}(I)$, and splitting this idempotent we obtain a vector bundle $E$ that restricts to a trivial line bundle on $C$ and a 0-dimensional vector bundle on $X-C$. Thus, when $X$ has more than one component, $E$ is not a direct sum of line bundles. REPLY [6 votes]: I don't think there can be a reasonable such splitting principle, even in the weakest sense you asked about (i.e. $K_0(\mathsf R) \to K_0(\mathsf R')$ is injective) and in the case when the ground ring $k$ is a field of characteristic zero. Let us make the standing assumption that $\mathrm{End}(\mathbf 1)$ has no nontrivial idempotent, as suggested in the question. Then if $L$ is a line object, we must have either $\wedge^n(L)=0$ for all $n>1$, or $\mathrm{Sym}^n(L)=0$ for all $n>1$. Indeed consider $$ L^{\otimes 2} \cong \wedge^2(L) \oplus \mathrm{Sym}^2(L). $$ If both summands here are nontrivial then we get nontrivial orthogonal idempotents in $\mathrm{End}(L)\cong \mathrm{End}(\mathbf 1)$. The argument generalizes to higher $n$. From this it follows that any finite sum of line objects is annihilated by some Schur functor. More precisely the sum of $n$ even and $m$ odd line objects is killed by the Schur functor given by an $(n +1) \times ( m +1 )$ Young diagram. This follows from Pieri's formula. So let for example $\mathsf{R}$ be the 2-rig $\mathsf{Poly}$ of polynomial functors $\mathsf{Vect}\to \mathsf{Vect}$, and let $\mathrm{id} \in \mathsf R$ be the identity functor. If $f:\mathsf{R} \to \mathsf R'$ is such that $f(\mathrm{id})$ is a sum of line bundles then $K_0(\mathsf R) \to K_0(\mathsf R')$ can not be injective, since all Schur functors applied to $\mathrm{id}$ are nonzero in $K_0(\mathsf R) \cong \Lambda$.<|endoftext|> TITLE: On intersections of several convex regions QUESTION [5 upvotes]: Question: Given n convex planar regions. Required to place them (in suitable position and orientation) so that that part of the plane lying under all the regions (their common intersection) is of maximum area. If we first place any two of the planar regions such that their intersection is maximized, then, place a third region so that its max intersection with the max intersection of the first two is maximized and so on, are we guaranteed to find the maximum common intersection of all n regions - irrespective of the order in which we consider them? I have no counterexample to this simple method. Note: I don't know if the algorithmic problem of placing two convex polygonal regions so that the area under both is maximized has been optimally solved. The above question can also be asked replacing area with (say) perimeter. Further Question: One can ask a similar question with the union of planar convex regions - how to 'stack' n regions one above the other so that the convex hull of their union has least area / perimeter. Guess: One could begin by placing the region with maximum width above the region with maximum diameter such that the width or the former is perpendicular to diameter of the latter. REPLY [5 votes]: The simple method will not always work. Consider the trapezoids $$A: (1,0), (2,1), (-2,1), (0,0)$$ $$B: (1,0), (2,1), (0,1), (-2,0)$$ and their pentagonal intersection $$C: (1,0), (2,1), (0,1), (-1,\frac12), (0,0)$$ Clearly the maximal overlap of shapes $A$, $B$ and $C$ is in these positions and orientations; no rotation, translation, or reflection can get an overlap with area greater than $C$. But if we had looked first for the maximal overlap of $A$ and $B$, we would have found $|A\cap B|<|A\cap B’|$, where $B’$ is a version of $B$ reflected about the line $y=\frac12$: $$B’: (2,0), (1,1), (-2,1), (0,0)$$ As a result, the simple method will not get the maximal area for these $A$, $B$, $C$. The same argument shows that the simple method will not get the maximal perimeter either. Update, without needing reflections Suppose that in polar coordinates: $A$ is the convex hull of the unit circle together with $(\sec\frac\pi8,0),(\sec\frac\pi9,\frac\pi2),(\sec\frac\pi9,\pi)$. $B$ is the convex hull of the unit circle together with $(\sec\frac\pi8,0),(\sec\frac\pi9,\frac{3\pi}{4}),(\sec\frac\pi9,\frac{5\pi}{4})$. $C$ is the convex hull of the unit circle together with $(\sec\frac\pi8,0)$. Each point at radius $\sec(\theta)$ increases the area by $f(\theta)=\tan(\theta)-\theta$, and covers $2\theta$ of the circumference. We calculate that $2f(\pi/9)>f(\pi/8)$. So again, the maximal overlap of shapes $A$, $B$ and $C$ is in these positions and orientations; no rotation, translation, or reflection can get an overlap with area greater than $C$. But if we had looked first for the maximal overlap of $A$ and $B$, we would have found $|A\cap B|<|A\cap B'|$, where $B'$ is a rotation of $B$ by $\pi/4$. As a result, the simple method will not get the maximal area for these $A$, $B$, $C$. The same argument shows that the simple method will not get the maximal perimeter either.<|endoftext|> TITLE: About Lie group $G$ has this escape property？ QUESTION [6 upvotes]: Every Lie group $G$ has the following escape property: For every $x \ne e$ in a sufficiently small neighborhood $U$ of the identity $e$ in $G$, there is a integer $n$ such that $x^n$ is not in $U$. The $\textbf{Question one }:$ is if we can find a sufficiently small neighborhood $V$ of $e$ in $G$, for any two different points $a,b \in V$ , there is a integer $m$ such that $a^m(b^{-1})^m$ is not in $V$. Thanks for Scholar's answer about the above question If $G$ is a commutative group the question actually has a positive answer, since $G$ has the escape property. I guess it is true for some larger class of Lie groups. The problem is to state that the power mapping can enlarge the distance between two different points. But I didn't find a good property of power mapping by looking up data. meanwhile，Thanks Michael Albanese for editing of the question. REPLY [3 votes]: This is false in the affine group of matrices $\begin{pmatrix} \theta & \eta\\0 & 1\end{pmatrix}$, $\theta,\eta\in\mathbf{R}$, $\theta>0$. Indeed $V$ being fixed, choose $a=\begin{pmatrix} \theta & \eta\\0 & 1\end{pmatrix}$, $b=\begin{pmatrix} \theta & 0\\0 & 1\end{pmatrix}$, so $$a^mb^{-m}=\begin{pmatrix} 1 & (\theta^{m-1}+\theta^{m-2}+\dots +\theta+1)\eta\\0 & 1\end{pmatrix}.$$ Indeed choose $\theta<1$, close enough to $1$, and then $\eta$ small enough, so that $a,b\in V$ and so that $S=\begin{pmatrix} 1 & [0,\eta/(1-\theta)]\\0 & 1\end{pmatrix}$ is contained in $V$. Then $a^mb^{-m}\in S$, hence remains in $V$. On the other hand the property is true when $G$ is connected nilpotent. Indeed if $G$ is simply connected nilpotent, $a^mb^{-m}$ tends to infinity for all $a\neq b$, by a simple application of the BCH formula (details upon request). In the case of $G$ connected nilpotent Lie group, the same property follows if $ab^{-1}$ is not central, and if $ab^{-1}$ is central, one argues as in the abelian case.<|endoftext|> TITLE: Quantifier elimination for abelian groups QUESTION [5 upvotes]: In the Wikipedia article (https://en.wikipedia.org/wiki/Quantifier_elimination#cite_note-4) it is said that every abelian group has quantifier elimination property and a long old paper of W. Szmielew (1955) is given as a reference. But as far as I know, this is true for special classes of abelian groups (like divisible, ordered, ... groups). I am not an expert of model theory but I need to apply quantifier elimination for reduced abelian groups (abelian groups the only divisible subgroup in which is the trivial one), so I need to know if really we have quantifier elimination for any abelain group and if not, what is the most complex quantifier combination of formulas in such groups. REPLY [10 votes]: Abelian groups are the same thing as $\mathbb Z$-modules. In general, for any ring $R$, the theory of left $R$-modules has quantifier elimination down to Boolean combinations of primitive positive formulas and certain sentences (expressing so-called Baur–Monk invariants). This is the Baur–Monk quantifier elimination theorem; see e.g. §A.1 in Hodges, Model Theory. In particular, since the theory of any particular abelian group or $R$-module is complete, it has quantifier elimination down to Boolean combinations of p.p. formulas. Note that a p.p. formula expresses solvability of a linear system; in the case where $R$ is a PID (including abelian groups with $R=\mathbb Z$), the Smith normal form shows that you can reduce to the case of p.p. formulas with one quantifier. That is, every formula $\phi(\vec x)$ is equivalent to a Boolean combination of invariant sentences and the divisibility formulas $a\mid\sum_ia_ix_i$ for some $a,a_i\in R$. (For $a=0$, this amounts to the original atomic formulas $\sum_ia_ix_i=0$.)<|endoftext|> TITLE: Embedding linklessly embeddable graphs without Borromean rings QUESTION [5 upvotes]: A linklessly embeddable graph is a graph which can be embedded into $\Bbb R^3$ so that no two of its cycles are linked. For example, the Petersen graph is not such a graph. Now, I can think of another type of "linkedness" that is not already addressed by this notion, namely, whether an embedding contain a Borromean rings configuration. By that I mean three cycles of the graph, no two of which are linked, but all three can still not be "entangled into three separate cycles" (see the image from Wikipedia). $\qquad\qquad$ Question: if a graph is linklessly embeddable, does it also have a linkless embedding without a Borromean rings configuration? Or the other way around, are there linklessly embeddable graphs that cannot be linklessly embedded without Borromean rings? Some thought Maybe one can show that whenever a graph cannot be embedded without Borromean rings, then the following two (black) paths must be present. This would then already imply the presence of a link: This does not address the possibility that a graph might be linklessly embeddable, and embeddable without Borromean rings, but not both at the same time. REPLY [2 votes]: Isn't it that linklessly and flatly embeddable are the same family, and that a flat embedding can not contain a Borromean ring? upd - clarification: From the wiki article on linkless embeddings: "A flat embedding is an embedding with the property that every cycle is the boundary of a topological disk whose interior is disjoint from the graph." As far as I remember (and the wikipedia seems to confirm), every linklessly embeddable graph has a flat embedding (every flat embedding is trivially linkless). Finally, that a flat embedding can't contain a Borromean ring seems trivial to me from the above definition of a flat embedding (still can be wrong)<|endoftext|> TITLE: On the homological dimension of a Borel construction QUESTION [14 upvotes]: Let $M$ b a closed connected smooth manifold with fundamental group $\Gamma$. Suppose $G$ is a simply-connected Lie group that acts smoothly on $M$. Then the Borel construction $$M//G = M \times_G EG$$ has fundamental group $\Gamma$ as well. Can the 1-truncation (i.e., the map classifying the universal cover) $$M//G \to B\Gamma$$ be non-trivial in rational cohomology in degrees bigger than the dimension of $M$? At first I thought perhaps one could produce a counterexample along the following lines: Take $\Gamma < G$ a discrete group ($G$ some big simply-connected Lie group), and let $M = G/\Gamma$. Then $M//G = B\Gamma$, and the map in question is the identity. But then I realized that the rational cohomological dimension of any such $\Gamma$ is at most $\text{dim}(G) - \text{dim}(K)$, where $K < G$ is a maximal compact subgroup, so this cannot possibly give a counterexample... REPLY [3 votes]: I think $f: M /\!\!/ G \to B\Gamma$ cannot be nontrivial on $\mathbb{Q}$-(co)homology in degrees beyond the dimension of $M$, because I think one can find a factorisation $$f_* : H_*(M /\!\!/ G ; \mathbb{Q}) \longrightarrow H_*(M / G ; \mathbb{Q}) \overset{\theta}\longrightarrow H_*(B\Gamma ; \mathbb{Q})$$ and the middle term vanishes for $* > dim(M)$ (and in fact often lower, depending on the dimension of the principal orbit). Theorem 1.1 of W. Browder, W.-C. Hsiang, G-actions and the Fundamental Group, Invent. Math. 65 (1981/82), no. 3, 411–424. gives such a factorisation after precomposing with $H_*(M ; \mathbb{Q}) \to H_*(M /\!\!/ G ; \mathbb{Q})$, but looking at their proof it seems to prove the strengthening I have claimed above. Their strategy is to consider $M$ as a stratified space, stratified by orbit type, and take the induced stratifications of $M \times EG$, $M/G$, and $M /\!\!/ G = (M \times EG)/G$. They then consider the functor $k$ sending a stratified space to its stratified 1-type. This allows us to form the commutative diagram $$\require{AMScd} \begin{CD} (M \times EG)/G @>>> k((M \times EG)/G) @>>> B\Gamma\\ @VVV@VV{\bar{t}}V \\ M/G @>>> k(M/G) \end{CD}$$ where the map to $B\Gamma$ is the canonical comparison between the stratified 1-type of $(M \times EG)/G$ and its ordinary 1-type. Now Lemma 3.4 (a) of the paper shows that $\bar{t}$ induces an isomorphism on rational homology, which gives the required factorisation.<|endoftext|> TITLE: Methods to bound the number of solutions to $x^x \equiv 1 \mod p$ with $1 \le x \le p$ QUESTION [10 upvotes]: For a prime $p$, let $N(p)$ be the number of solutions $1 \le x \le p$ to $x^x \equiv 1 \mod p$. I am interested in methods to bound $N(p)$. Background: This quantity appears in Problem 1 of the Miklós Schweitzer Competition 2010 where it was asked to prove $N(p) \ll p^{c}$ for some $c<\frac{1}{2}$. Let me quickly explain how one can solve this problem and why the exponent $\frac{1}{2}$ is critical. For each divisor $d$ of $p-1$, let $A_d$ be the set of numbers $1 \le x \le p$ for which $(x;p-1)=d$ and $x^x \equiv 1 \mod p$ so $N(p)=\sum_{d \mid p-1} \vert A_d\vert$. The condition $x^x \equiv 1 \mod p$ now just says that $x$ is a $e$-th power modulo $p$ where $e=\frac{p-1}{d}$ is the complementary divisor. So trivially $\vert A_d\vert \ll \min(d,e) \ll p^{1/2}$ and hence $N(p) \ll p^{1/2+\varepsilon}$. To improve this exponent, it clearly suffices to improve the bound for $\vert A_d\vert$ when $d \approx e \approx p^{1/2}$. Here the sum-product theorem over finite fields comes in handy: Since $A_d+A_d$ still contains essentially only multiples of $d$, it is easy to see that $\vert A_d+A_d\vert \le 2e$ and similarly $A_d \cdot A_d$ still contains only $e$-th powers so that $\vert A_d \cdot A_d\vert \le d$. Hence $\max(\vert A_d +A_d\vert, \vert A_d \cdot A_d\vert) \ll \max(d,e)$. But by the (currently best-known version of the) sum-product theorem the LHS is at least $\vert A_d\vert^{5/4-\varepsilon}$ so that we get the bound $\vert A_d\vert \ll \max(d,e)^{4/5+\varepsilon}$ and we win. Indeed, working out the exponents, we can prove $N(p) \ll p^{4/9+\varepsilon}$ this way. Now I would be curious to learn about Question 1: What are some other techniques that can be applied to get a non-trivial bound for $N(p)$? To be clear, I would be equally interested in (possibly more difficult) techniques that lead to an exponent $c<\frac{4}{9}$ as well as more elementary techniques that lead to a (possibly worse, but) non-trivial result. Now even if one assumes a best possible sum-product conjecture to be true, it seems that by the method described above we could only prove $N(p) \ll p^{1/3+\varepsilon}$. On the other hand, it seems natural to conjecture that even $N(p) \ll p^{\varepsilon}$ is true, albeit very hard to prove. Given this gap, I am wondering about Question 2: Are there some "natural/standard" conjectures that would imply an exponent less than $\frac{1}{3}$, possibly even as small as an $\varepsilon$? Or is there a good heuristic why the exponent $\frac{1}{3}$ is a natural barrier here? EDIT: As pointed out in the answers, Cilleruelo and Garaev (2016) proved $N(p) \ll p^{27/82}$. This leaves us with the question of whether there is a natural/standard conjecture that would imply that $N(p) \ll p^{\varepsilon}$. PS: To be clear, I don't claim that this is a very important problem on its own right. It just seems like a good toy problem to test our understanding of the interference between multiplicative and additive structures. REPLY [3 votes]: Another answer already gives a reference to a smaller exponent, but since the OP states that they are interested in an elementary proof, this is a proof I came up with which shows $c < \frac{3}{7} + \varepsilon$ (which is smaller than $\frac{4}{9}$). For $A \subset \left( \mathbb{Z} / p \mathbb{Z} \right)^{*}$ we define $A A = \left\{ xy : x, y \in A \right\} \subset \mathbb{Z} / p \mathbb{Z}$ and $r (a) = \# \left\{ xy \equiv a : x, y \in A \right\}$, where throughout we will let $\equiv$ denote equality $\bmod p$ (just to simplify writing). We define the multiplicative energy of $A$ to be $E(A) = \# \left\{ x, y, z, w \in A : xy \equiv zw \right\}$. Notice that $$\sum_{x \in A A} 1 = \lvert A A \rvert$$ $$\sum_{x \in A A} r(x) = \lvert A \rvert^2$$ $$\sum_{x \in A A} r(x)^2 = E(A)$$ and so by Cauchy-Schwarz, $\lvert A \rvert^4 \leq \lvert A A \rvert E(A)$. As stated in the post, it is sufficient to show $\lvert A_d \rvert \leq p^{3/7 + \varepsilon}$. Our strategy will be to bound $\lvert A_d A_d \rvert$ and $E(A)$. Notice that since every element of $A_d$ is a $d$-th root of unity, so is every element of $A_d A_d$, and therefore $\lvert A_d A_d \rvert \leq d$. Also, every element of $A_d$ is of the form $d x$, where $x < \frac{p}{d}$. This means that the multiplicative energy of $A_d$ is at most $$\# \left\{ x, y, z, w < \frac{p}{d} : xy \equiv zw \right\}$$ We split into cases according to the larger among $d$ and $\sqrt{p}$. Case 1: $d > \sqrt{p}$. In this case, $xy \equiv zw$ is equivalent to $xy = zw$. Fixing $x, y$, the number of solutions is at most the number of divisors of $xy$, which is $\ll p^{\varepsilon}$, and so $E(A_d) \ll \frac{p^{2 + \varepsilon}}{d^2}$. This means that $$\lvert A_d \rvert \ll \left( d \cdot \frac{p^{2 + \varepsilon}}{d^2} \right)^{1/4} \ll p^{3/8 + \varepsilon}$$ which is even stronger than what we require. Case 2: $d < \sqrt{p}$. Now, $xy \equiv zw$ means that $xy + pk = zw$ for some $k \leq \frac{p}{d^2}$. Fixing $x, y, k$ we have as before $\ll p^{\varepsilon}$ solutions, and so the multiplicative energy is at most $$\frac{p^{3 + \varepsilon}}{d^4}$$ which as before gives that $$\lvert A_d \rvert \ll \left( \frac{p}{d} \right)^{3/4 + \varepsilon}$$ This bound is relatively good for $d \approx \sqrt{p}$, however when $d$ is small this is quite bad. Luckily, we always have the trivial bound $\lvert A_d \rvert \leq d$. Optimizing, we get that $\lvert A_d \rvert \ll p^{3/7 + \varepsilon}$, as required.<|endoftext|> TITLE: Is there any version of the Banach-Tarski paradox in ZF? QUESTION [38 upvotes]: The Banach-Tarski paradox states that for a solid ball in 3‑dimensional space, there exists a decomposition into a finite number of disjoint subsets, which can then be put back together in a different way to yield two identical copies of the original one. Obviously it is based on AC. I was wondering if anyone here knew if analysis under the axioms of ZF has been developed to invent a version of Banach-Tarski which is independent of AC. What does the Banach-Tarski paradox look like without AC? Are there any versions of it? (For an example, one of the theorems that has been proven without AC is the Heine-Borel theorem.) REPLY [57 votes]: According to Dougherty and Foreman's 1992 PNAS paper, Banach-Tarski paradox using pieces with the property of Baire (doi:10.1073/pnas.89.22.10726), the following result can be shown without AC: the unit ball has a finite collection of disjoint open subsets that transforms by rigid motions to another collection of disjoint open sets such that the closure of the union is two unit balls.<|endoftext|> TITLE: End formulas for sets of monoidal natural transformations QUESTION [5 upvotes]: Perhaps the characteristic feature of the theory of ends is that they are extremely useful for computing sets of transformations between two functors. For example, one has the formulas \begin{align*} \mathrm{Nat}(F,G) &\cong \int_{A\in\mathcal{C}}\mathrm{Hom}_{\mathcal{C}}\left(F_{A},G_{A}\right),\\ \mathrm{DiNat}(F,G) &\cong \int_{A\in\mathcal{C}}\mathrm{Hom}_{\mathcal{C}}\left(F^{A}_{A},G^{A}_{A}\right), \end{align*} see Coend Calculus, Theorem 1.4.1 and Example 1.4.4. Is there a similar end formula for the set $\mathrm{Nat}^\otimes(F,G)$ of monoidal natural transformations between two strong monoidal functors $F,G\colon\mathcal{C}\rightrightarrows\mathcal{D}$? REPLY [2 votes]: Monoidal ends Let $(\mathcal{C},\otimes)$ be a monoidal category and let $(\mathcal{D},\times)$ be a cartesian monoidal category. Let $(X,\eta,\mu) : (\mathcal{C}^{\mathrm{op}},\otimes) \times (\mathcal{C},\otimes) \to (\mathcal{D},\times)$ be a lax monoidal functor. A wedge to $(X,\eta,\mu)$ is a wedge to the underlying functor $X : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathcal{D}$, thus consisting of an object $T \in \mathcal{D}$ and a family of morphisms $(w_A : T \to X(A,A))_{A \in \mathcal{C}}$, such that the following properties hold: $w_1 : T \to X(1,1)$ is equal to $$T \xrightarrow{\exists!} 1 \xrightarrow{\eta} X(1,1).$$ $w_{A \otimes B} : T \to X(A \otimes B, A \otimes B)$ is equal to $$T \xrightarrow{~(w_A,w_B)~} X(A,A) \times X(B,B) \xrightarrow{~~\mu~~} X(A \otimes B, A \otimes B).$$ A universal wedge, i.e. end, is defined as usual. It is easy to see that if $\mathcal{C}$ is small and $\mathcal{D}$ is complete, then any lax monoidal functor has an end. We can denote it by $\int (X,\eta,\mu)$. Since it is a common practice (sigh) to ignore forgetful functors and just write $X$ both for the functor and the monoidal functor, some people will prefer to call this a "monoidal wedge" and a "monoidal end", the latter then being denoted by something like $\int^{\otimes} X$. I do not know if this concept has appeared elsewhere, I just made it up to answer the question below. Monoidal natural transformations Now let $(\mathcal{C},\otimes)$, $(\mathcal{C}',\otimes)$ be two monoidal categories and $(F,\eta_F,\mu_F),(G,\eta_G,\mu_G) : (\mathcal{C},\otimes) \to (\mathcal{C}',\otimes)$ be two strong monoidal functors. Consider the functor $X : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set}$ defined on objects by $$X(A,B) := \mathrm{Hom}(F(A),G(B)).$$ We equip it with the following lax monoidal structure: $$\eta_X : 1 \to X(1,1)$$ corresponds to the isomorphism $\eta_G \circ \eta_F^{-1} : F(1) \to 1 \to G(1)$, and $$\mu_X : X(A,A') \times X(B,B') \to X(A \otimes B,A' \otimes B')$$ maps a pair of morphisms $f : F(A) \to G(A')$, $g : F(B) \to G(B')$ to the morphism $$F(A \otimes B) \xrightarrow{\mu_F^{-1}} F(A) \otimes F(B) \xrightarrow{f \otimes g} G(A') \otimes G(B') \xrightarrow{\mu_G} G(A' \otimes B').$$ One needs to check the coherence conditions in the definition of a lax monoidal functor, I will not do this here. It is straight forward to check that $\int (X,\eta,\mu)$ is the set of morphisms $(F,\eta_F,\mu_F) \to (G,\eta_G,\mu_G)$ (aka monoidal natural transformations).<|endoftext|> TITLE: Existence of a polynomial $Q$ of degree $\geq (p-1)/4$ in $\mathbb F_p[x]$ such that $QQ'$ factorizes into distinct linear factors QUESTION [21 upvotes]: For all primes up to $p=89$ there exists a product $Q=\prod_{j=1}^d(x-a_j)$ involving $d\geq (p-1)/4$ distinct linear factors $x-a_j$ in $\mathbb F_p[x]$ such that $Q'$ has all its roots in $\mathbb F_p$. Since $Q$ has only simple roots, the product $QQ'$ has also only simple roots. (It is obvious that a factorization into distinct linear factors of $QQ'$ implies that $Q$ has degree at most $(p+1)/2$. My computations suggest however that it seems to be hard to go substantially beyond $p/4$.) An example for $p=53$ is given by $$Q=x(x^2-5^2)(x^2-8^2)(x^2-10^2)(x^2-12^2)(x^2-14^2)(x^2-16^2)$$ with derivative $$Q'=13(x^2-4^2)(x^2-13^2)(x^2-19^2)(x^2-20^2)(x^2-21^2)(x^2-22^2).$$ Similar examples of the form $x\prod_{j=1}^7(x^2-a_j^2)$ for $59$ and $61$ are given by $\{a_1,\ldots,a_7\}=\{1,3,12,15,17,18,22\}$ for $59$ and by $\{1,3,6,8,11,21,22\}$ for $61$. For $p=67$ we have $x\prod_{j=1}^8(x^2-a_j^2)$ with $\{a_1,\ldots,a_8\}=\{1,2,5,9,15,23,31,32\}$. For $p=71$ and $73$, we have $\prod_{j=1}^9(x^2-a_j^2)$ with $\{a_1,\ldots,a_9\}$ given by $\{1,2,3,5,18,20,29,30,33\}$ for $p=71$ and by $\{1,2,5,17,19,22,27,32,33\}$ for $p=73$. For $p=79$, we have $\prod_{j=1}^{10}(x^2-a_j^2)$ with $\{a_1,\ldots,a_{10}\}=\{1,2,6,10,15,17,18,20,33,37\}$. For $p=83$, we have $x\prod_{j=1}^{10}(x^2-a_j^2)$ with $\{a_1,\ldots,a_{10}\}=\{1,2,3,5,12,16,21,33,36,41\}$. For $p=89$, we have $\prod_{j=1}^{11}(x^2-a_j^2)$ with $\{a_1,\ldots,a_{11}\}=\{1,10,14,15,18,33,34,36,37,43,44\}$. (I did a search over all polynomials up to $p=23$ and restricted then my attention to polynomials which are either even or odd for all primes up to $89$.) Adopting a very naive viewpoint, the existence of such a polynomial of degree $\geq \lambda p$ for some $\lambda>0$ and for almost all primes would be somewhat surprising: There are ${p\choose \lambda p}$ such polynomials of degree $\lambda p$ given as a product of $\lambda p$ distinct linear factors in $\mathbb F_p[x]$. Adopting the very naive viewpoint that derivatives of these polynomials behave randomly with respect to decomposition into irreducible factors, the probability for such a polynomial $Q\in\mathbb F_p[x]$ (with $QQ'$ factorizing completely into distinct linear factors) to exist should be roughly given by $$\left(\lambda^{2\lambda}(1-\lambda)^{2(1-\lambda)}p^\lambda\right)^{-p}$$ (up to polynomial contributions) which decays exponentially fast. Question: Can this naive argument be made rigourous or do such polynomials of degree at least $(p-1)/4$ (or perhaps of degre at least $\lambda p$ for some $\lambda >0$) exist for all primes? Remark (added after adding additional examples up to $83$): The outlook for the existence of such polynomials (of degre at least $(p-1)/4$) for all primes starts to look better: Indeed, for $p=83$, the naive argument given above (slightly improved) yields a fairly small probability: $${41\choose 10}{31\choose 10}83^{-10}\sim 0.0032\ .$$ This suggests that we should have ended up empty-handed at this point. (End of added remark.) Motivation: The analogous question is open for degree $7$ over $\mathbb Z[x]$ (existence of a polynomial $Q\in\mathbb Z[x]$ of degree $7$ such that $QQ'$ has $13$ distinct roots in $\mathbb Z$), see proposition 27 of Proposals for polymath projects. A stronger question (over finite fields) is to ask for polynomials of maximal degree in $\mathbb F_p[x]$ such that $Q$ and all its derivatives $Q^{(1)},Q^{(2)},\ldots$ factorize into distinct linear factors. One can either admit or forbid common roots among different derivatives. Forbidding them gives of course the trivial upper bound $O(\sqrt{p})$ on the maximal degree. I guess that admitting them does not allow substantially higher degrees. Given an integer $d$, I guess that such polynomials of degree $\geq d$ exist in $\mathbb F_p[x]$ except for a finite number of small primes. (A bug in my code made me think that there are no such polynomials of degree $\geq 6$ which prompted me to post a now deleted question on MO.) It would of course be interesting to have information on the maximal degree $d_p$ for (the two versions of) this stronger question. (Interestingly, the stronger question is computationally slightly easier since the set of all such polynomials (not necessarily of maximal degree) is closed under derivation.) REPLY [14 votes]: Such polynomials always exist, examples are Dickson polynomials of the first kind (with parameter $1$). For a positive integer $n$ these degree $n$ polynomials $D_n$ are most conveniently defined implicitly by $D_n(z+1/z)=z^n+1/z^n$. For $p=4m\pm1$ the polynomial $D_m(x)$ has the required property, which is to say that $D_m(x)\cdot D_m'(x)$ divides $x^p-x$. In fact, using $D_n'(z+1/z)=n\frac{z^n-1/z^n}{z-1/z}$ we verify the identity (which holds in any characteristic) \begin{equation} D_m(x)\cdot D_m'(x)\cdot D_{2m\pm1}'(x)\cdot(x^2-4)=m\cdot(2m\pm1)\cdot(D_{4m\pm1}(x)-x), \end{equation} and the claim follows since $D_p(x)=x^p$ in characteristic $p$. Remark: I have no idea about the strong version of the other question for the higher derivatives. The polynomials which achieve maximal degrees for given $p$ do not seem to be of a particular form (even in those cases when they are unique up to trivial transformations).<|endoftext|> TITLE: $\operatorname{PSL}(2,\mathbb{F}_p) $ does not embed in $\mathfrak{S}_p$ for $p>11$ QUESTION [25 upvotes]: A famous result of Galois, in his letter to Auguste Chevalier, is that for $p$ prime $>11$ the group $\operatorname{PSL}(2,\mathbb{F}_p) $ does not embed in the symmetric group $\mathfrak{S}_p$. The standard proof nowadays goes through the classification of subgroups of $\operatorname{PSL}(2,\mathbb{F}_p) $ (Dickson's theorem), which is far from trivial. Does anyone know a simpler argument? REPLY [8 votes]: Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time). If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$ Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$ Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$. Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above. Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and $H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$ (It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ for $p = 5,7$ and $11$ respectively. Later edit: To digress on a remark above, I point out that a theorem of Blichfeldt (actually left as an exercise in his book), can be used to show that if a finite primitive subgroup $H$ of ${\rm SL}(2,\mathbb{C})$ contains an element $x$ of order $p+1$ for some prime $p$, then $p \leq 7.$ For certainly $x \not \in Z(H)$, so Blichfeldt's result shows that the eigenvalues of $x$ can't all lie on an arc of length less than $\frac{2 \pi}{5}$ on the unit circle $S^{1}.$ Replacing $x$ by a suitable generator of $\langle x \rangle$ if necessary, we may suppose that the eigenvalues of $x$ are both on an arc of length $\frac{4\pi}{p+1}.$ Hence $\frac{p+1}{4} \leq \frac{5}{2}$ and $p \leq 9$ (so $p \leq 7$ as $p$ is prime). We remark that the binary octahedral group of order $48$ contains an element of order $8$, and is isomorphic to a subgroup of both ${\rm SL}(2,7)$ and ${\rm SL}(2,\mathbb{C})$ (primitive in the latter case). Similarly for $H = {\rm SL}(2,3)$, which contains an element of order $6$, and is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and to a subgroup of ${\rm SL}(2,5).$ However, ${\rm SL}(2,5),$ (which is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and a subgroup of ${\rm SL}(2,11)$) indeed contains no element of order $12$. That same result of Blichfeldt also can be used to prove (in the manner alluded to above) that a finite primitive subgroup of ${\rm SL}(2,\mathbb{C})$ has order $24,48$ or $120.$<|endoftext|> TITLE: Killing form: clean proof that two spaces are orthogonal? QUESTION [10 upvotes]: Let $G$ be a linear algebraic group whose Lie algebra $\mathfrak{g}$ is semisimple. Let $x$ be a regular semisimple element of $G$. Write $\mathfrak{t}$ for the Lie algebra of the maximal torus $T = C(x)$ centralizing $x$, and let $\mathfrak{h}$ be the image of $\mathfrak{g}$ under the map $g\mapsto \textrm{Ad}_x(g)-g$. Since $g$ is semisimple, we may diagonalize it; then all the elements of $\mathfrak{h}$ have only zeroes in the diagonal. It is not hard to see that, for $G$ a classical group ($\textrm{SL}_n$, $\textrm{SO}_n$, $\textrm{Sp}_{2 n}$, over an arbitrary field), the spaces $\mathfrak{t}$ and $\mathfrak{h}$ are orthogonal under the Killing form: the Killing form is then a multiple of $(X,Y)\mapsto \textrm{tr}(X Y)$, and, for $X$ diagonal and $Y$ having only zeroes in the diagonal, $X Y$ clearly has only zeroes in the diagonal, and thus has trace $0$. Is there a clean proof that works for all linear algebraic groups $G$ with $\mathfrak{g}$ semisimple -- preferably one with as little computation or casework as possible? REPLY [12 votes]: $$(X, \mathrm{Ad}_x(Y) - Y) = (X,\mathrm{Ad}_x(Y)) - (X,Y) = (\mathrm{Ad}_{x^{-1}}(X),Y) - (X,Y) = (X,Y) - (X,Y) = 0$$ Using the fact that the Killing form is invariant under the adjoint action and that $\mathrm{Ad}_x(X) = X$ for $X \in \mathfrak{t}$. I can't think of a reason why this wouldn't work in any characteristic although I'm much more familiar with characteristic 0.<|endoftext|> TITLE: Compactness of symmetric power of a compact space QUESTION [5 upvotes]: Suppose I have a compact metric space $(X,d)$ and let $\mathcal{X}=X^K$ be the product space. Consider the equivalence relation $\sim$ on $\mathcal{X}$ given as: for $\alpha,\beta\in \mathcal{X}$, $\alpha\sim\beta$ iff there exists a permutation $\tau$ of $\{1,\dots,K\}$ such that $\alpha_i = \beta_{\tau(i)}$ for all $i$. Consider the quotient space $\mathcal{X}/\sim$ and define a metric $\rho$ on this quotient space as $\rho([\alpha],[\beta]) = \min_{\tau} \sum_i d(\alpha_i, \beta_{\tau(i)})$. I could show this defines a well-defined valid metric. I also know that product of compact spaces under the product metric is compact (Tychonoff's theorem) and for a compact topological space $Q$ with an equivalence relation $R$, the quotient space $Q/R$ is also compact. But are there any results showing whether $(\mathcal{X}/\sim, \rho)$ is compact under the topology induced by this metric (necessary and/or sufficient conditions if any)? REPLY [2 votes]: Here's a direct proof (although showing your metric yields the topology of $\mathcal X/\sim$ is more useful). In a metric space, compactness is equivalent to sequential compactness: every infinite sequence of points has a convergent subsequence. Let $x_n\in X^K$ correspond to an arbitrary sequence $[x_n]\in\mathcal X/\sim$. Since $X^K$ is compact, let $y_n$ be a subsequence converging to $x\in X^K$. I claim $[y_n]$ converges to $[x]$. To see this, let $\epsilon>0$, and choose $N$ with $d(y_n(i),x(i))<\epsilon/K$ for each $1\leq i\leq K$ and $n\geq N$. Then $\rho([y_n],[x])\leq\sum_i d(y_n(i),x(i))<\epsilon$ for all $n\geq N$. This also basically shows that the map $x\mapsto[x]$ is continuous, which is another angle, since continuous images of compact spaces are compact.<|endoftext|> TITLE: Is $\oplus$ the only monoidal structure on the simplex category? QUESTION [7 upvotes]: Simplicial sets are presheaves on the simplex category $\Delta$, while augmented simplicial sets are presheaves on $\Delta_+$, the augmented simplex category. Because Day convolution allows us to lift monoidal structures on a category $\mathcal{C}$ to its category of presheaves $\mathrm{Sets}^{\Delta^\circ}$, it is therefore of interest to find monoidal structures on $\Delta$ and $\Delta_+$, as these then provide "natural" monoidal structures on simplicial sets. The only monoidal structure I know of is the ordinal sum of $\Delta_+$ (which is not braided), whose Day convolution gives the join of simplicial sets, and whose internal hom is given by $$[X,Y]_n=\mathrm{hom}_{\mathrm{Sets}^{\Delta^\circ_+}}(X,\mathrm{Dec}^{n+1}Y)$$ Is this the only monoidal structure on $\Delta_+$? If not, what other monoidal structures are there on $\Delta_+$, and what are there on $\Delta$? REPLY [6 votes]: Here is half of a classification. Let $\otimes$ be a monoidal structure on $\Delta_+$. As I mentioned in a comment, the monoidal unit must be $[-1]$ or $[0]$ because these are the only objects with commutative endomorphism monoids. Suppose that the monoidal unit is $[0]$. Let us consider $[1] \otimes [1]$. We have that $[0]$ is a retract of $[1]$ in 2 ways, and as a result we obtain 4 retracts of $[1] \otimes [1]$ with support $[0] \otimes [0] = [0]$. Consider the induced linear ordering on these 4 points. We also have 4 ways that $[1] = [1] \otimes [0] = [0] \otimes [1]$ is a retract of $[1] \otimes [1]$, and from this we can deduce most of the ordering. It must have the following relations $\require{AMScd} \begin{CD} 0 \otimes 0 @>>> 0 \otimes 1\\ @VVV @VVV\\ 1 \otimes 0 @>>> 1 \otimes 1 \end{CD}$ To complete this to a linear order, without loss of generality we must have $0 \otimes 1 \leq 1 \otimes 0$. But now, one of our 4 projections onto $[1]$ is the coordinate projection onto the right column of the above square. The fact that this projection is order-preserving implies that $1 \otimes 0 = 0 \otimes 1 = 1 \otimes 1$. This contradicts the fact that the right column exhibits $[1]$ as a retract of this subset of $[1] \otimes [1]$. Therefore the monoidal unit is not $[0]$; it must be $[-1]$. I also think I'm ready to conjecture that $\oplus$, $\oplus^{rev}$ (as mentioned by Peter) and the degenerate monoidal structure are probably the only ones. You could imagine a classification starting as follows. Consider the maps $[0] = [0] \otimes [-1] \to [0] \otimes [0]$ and $[0] = [-1] \otimes [0] \to [0] \otimes [0]$. If these are the same, then we should have the degenerate monoidal structure. If they are different, then one is less than the other, and those two cases should correspond to $\oplus$ and $\oplus^{rev}$.<|endoftext|> TITLE: Are representations in computable analysis the equivalent to countably-generated condensed sets? QUESTION [6 upvotes]: This is the first in a pair of questions. For the other see here. Dustin Clausen and Peter Scholze have a theory of condensed sets, which is a slightly different take on topology. For most cases, the behavior of the usual topology and the “condensed” one, align. However, for some quotient spaces, like $\mathbb{R} / \mathbb{Q}$, the usual quotient topology is indiscrete, so every map $f : \mathbb{R} / \mathbb{Q} \to \mathbb{R} / \mathbb{Q}$ is continuous. However, as a condensed set, more structure is preserved. Indeed, the “continuous” maps are exactly those coming from continuous functions $\mathbb{R} \to \mathbb{R}$ which commute with the equivalence relation. *[Edit: I was a bit careless here. The maps actually come from continuous multifunctions. See Arno's answer below.) (See here for more motivation.) I’m trying to understand condensed sets better and how they relate to topology as it comes up in computable analysis, which I’m quite familiar with. I think the theory of “represented spaces" in computable analysis, is essentially the study of "countably generated condensed sets", and I want to know if anyone has worked this out, or thought about this? In computable analysis, we want to encode every point $x$ in a space $X$ with a value in Cantor space $2^\omega$. This makes it possible to do computations on the points using computations on $2^\omega$ which are well understood (say via Turing machines). But not only is it a way to do computability theory, it is also a way to talk about continuous functions. (Aside: This latter viewpoint is also a big part of descriptive set theory.) In computable analysis, a representation of a set $X$ is a partial, but surjective map $\rho : {\subseteq}2^\omega \to X$. (See sections 2 and 3 here.) There are four types of morphisms for the category of representations: (partial/total) (computable/continuous) maps. Let’s just consider total continuous maps even though usually in computable analysis one considers computable maps. Given two sets $X$ and $Y$ with representations $\rho : {\subseteq}2^\omega \to X$ and $\sigma : {\subseteq}2^\omega \to Y$, then a total function $f : X \to Y$ is $(\rho, \sigma)$-continuous if there is a corresponding continuous function $f' : \text{dom}(\rho) \to \text{dom}(\sigma)$ such that $$\rho \circ f = f' \circ \sigma$$ Conversely, it is also not hard to see that any continuous $f' : \text{dom}(\rho) \to \text{dom}(\sigma)$ generates a $(\rho, \sigma)$-continuous function $f$ iff $$\rho(x) = \rho(y) \implies f'(\sigma(x)) = f'(\sigma(x))$$ What we would like, is that if $X$ is a topological space, then there is a representation $\rho : {\subseteq}2^\omega \to X$ such that all partial continuous maps, $f : {\subseteq}2^\omega \to X$ are also $(\text{id}_{2^\omega}, \rho)$-continuous. This is often true, but there are indiscrete quotient spaces like $\mathbb{R} / \mathbb{Q}$ or $2^\omega / \text{fin}$ where this does not hold. For this reason, it is common to only consider admissible representations which have this nice property. However, it seems to me that actually the category of representations is very close to, if not in some sense equal to, the category of condensed sets. Let me explain. Representations only use a partial map ${\subseteq}2^\omega \to X$ to represent a space, where as condensed sets use a family of (total) maps $S \to X$ for all profinite sets $S$. A profinite set is just a space homeomorphic to a closed subspace of $2^\kappa$ for some $\kappa$. In particular, the seperable profinite sets are homeomorphic to closed subspaces of $2^\omega$. If one only uses separable profinite sets in the definition of condensed set in place of profinite sets, let’s call those countably generated condensed sets. I conjecture the category of representations is the same as the category of countably generated condensed sets. Has anyone worked something like this out? [Edit: This definition of countably generated condensed sets is definitely not what I'm looking for. Restricting the size of the profinite sets doesn't restrict the size of the set $X$.] I think I've verified the following: Every seperable profinite set $S$ has an admissible total representation $\sigma_S : 2^\omega \to S$. Using these representations, the generated category of $(\sigma_S, \sigma_{S’})$-continuous functions is equivalent to the category of separable profinite sets with continuous maps. For every set $X$ with representation $\rho : {\subseteq}2^\omega \to X$, and every separable profinite set $S$ with admissible representation $\sigma_S : 2^\omega \to S$, one can talk about the set of $(\sigma_S, \rho)$-continuous functions. I believe this set obeys the axioms of condensed sets (which I've included at the end), but restricted to separable profinite sets. Conversely, I still need to check that every countably generated condensed set has a representation. Also, I need to check that every countably generated condensed set is indeed also a condensed set. (Clausen and Scholze have a notion of $\kappa$-condensed sets, but it is only for uncountable limit cardinals.) It should be possible to replace $2^\omega$ with $2^\kappa$ for any cardinal $\kappa$, giving a way to connect $\kappa$-condensed sets and with a generalization of representations, and more formally give a representation-centric definition of all condensed sets. I think everything should carry over nicely. Anyway, there is more to work out, but I want to see if anyone has thought about this connection before. It is very possible I'm mixing up something subtle or this connection is already well known. On the other hand, if this does work out, it hints that computable analysis can very easily adapt to condensed sets (and condensed groups/rings/vector spaces/etc.). Appendix: Axioms of condensed sets For better or worse, here are the axioms for condensed sets stated in a way that it might be easier for a computable analyst or (non-categorical) logician to understand. (Warning: I may have stripped out too much category theory and forgot a necessary condition.) The "topology" on a condensed set $X$ is given by describing the class of all $T(S,X)$ of all the "continuous maps” (homomorphisms) $S \to X$ for each profinite set $S$. (Recall, again the profinite sets are exactly the closed subspaces of $2^\kappa$ for some cardinal $\kappa$. To avoid proper class issues, Scholze considered $\kappa$-condensed sets for uncountable limit cardinals $\kappa$ where we only use the profinite sets of cardinality less than $\kappa$.) A condensed set is a set $X$ with a class of sets $T(S,X)$ for every profinite set $S$. Each $T(S,X)$ must satisfy the following axioms: For the empty profinite set $\varnothing$, there is exactly one map $\varnothing \to X$ in $T(\varnothing,X)$. For the singleton profinite set $*$, the set $T(*,X)$ are exactly the set of maps $f_x : * \to X$ such that $f_x(*) = x$ for all $x \in X$. For profinite sets $S_1, S_2$, and their disjoint union $S_1 \sqcup S_2$: For every $f_1 : S_1 \to X$ in $T(S_1,X)$ and $f_2 : S_2 \to X$ in $T(S_2,X)$, the map $f : S_1 \sqcup S_2 \to X$ where $f(x) = f_1(x)$ if $x \in S_1$, else $f(z)=f_2(x)$ if $x \in S_2$ is in $T(S_1 \sqcup S_2,X)$. Conversely, every map $f : S_1 \sqcup S_2 \to X$ in $T(S_1 \sqcup S_2,X)$ is of this form for maps $f_1$ in $T(S_1,X)$ and $f_2$ in $T(S_2,X)$. Let $g : S' \to S$ be continuous and surjective map between profinite sets. For every map $f' : S' \to X$ in $T(S',X)$ such that $g(x) = g(y)$ implies $f'(x) = f'(y)$, there is a map $f : S \to X$ in $T(S,X)$ such that $f' = f \circ g$. If $f : S \to X$ in $T(S,X)$ and $g : S' \to S$ is continuous, then $f \circ g$ is in $T(S',X)$. For condensed sets $X$ and $Y$, a homomorphism from $X$ to $Y$ is a function $f : X \to Y$ such that for all profinite sets $S$ and all $g : S \to X$ in $T(S,X)$ the composition $f \circ g : S \to Y$ is in $T(S, Y)$. REPLY [5 votes]: This seems like a very interesting comparison, and to my knowledge there hasn't been a systematic investigation yet. The two settings seem similar, but I would be surprised if there is a wide-ranging coincidence. A few comments: Similar to how the condensed sets are defined, we can seek to understand represented spaces by characterising a represented space $\mathbf{X}$ via the collection $C(S,\mathbf{X})$ where $S$ ranges of subspaces of $2^\omega$. These satisfy the axioms for condensed sets sans the penultimate one (regarding quotients), but in addition they satisfy that the quasiorder defined by $f \leq g$ iff $\exists \text{ continuous } H : dom(f) \rightarrow dom(g) \ f = g \circ H$ has a maximum (the representatives of this degree are the representations). The question should probably be whether there is a "big" natural full subcategory of both the category of represented spaces and the category of condensed sets. This would be akin how to the admissible represented spaces/the $\mathrm{QCB}_0$ spaces are simultaneously subcategories of the represented spaces and the topological spaces. Quotients of represented spaces carry non-trivial structure, but can be tricky to understand. For example, it is not clear to me that every represented space morphism $f: \mathbb{R}/\mathbb{Q} \to \mathbb{R}/\mathbb{Q}$ comes from a continuous map on $\mathbb{R}$ compatible with the group-action. On the other hand, there cannot be many extra morphisms - the extra morphisms would come from continuous multivalued functions on $\mathbb{R}$ without a continuous choice function which get turned into functions by taking the quotient. For other quotients, this phenomenon definitely happens. Restricting to closed subsets of $2^\omega$ doesn't "feel" to be near enough. This might already happen when we only look into $\mathrm{QCB}_0$-spaces. For example, the space $\mathbb{R}[X]$ of polynomials is not even countably-based, but any of its compact subsets is Polish. It is also known that representations of the Kleene-Kreisel spaces ($K_0 = \mathbb{N}$, $K_{n+1} = \mathcal{C}(K_n,\mathbb{N})$) require domains of their representations going up the projective hierarchy. For context, see "Base-Complexity Classifications of $\mathrm{QCB}_0$-Spaces." by de Brecht, Schröder and Selivanov (Computability). The result itself was recently obtained by Hoyrup (Preprint).<|endoftext|> TITLE: Are the “topologies” arising from constructive type theories with quotients actually condensed sets? QUESTION [11 upvotes]: This is the second in a pair of questions. For the other see Are representations in computable analysis the equivalent to countably-generated condensed sets?. Dustin Clausen and Peter Scholze have a theory of condensed sets (see Lectures in condensed mathematics), which is a slightly different take on topology. For most cases, the behavior of the usual topology and the “condensed” one, align. However, for some quotient spaces, like $\mathbb{R} / \mathbb{Q}$, the usual quotient topology is indiscrete, so every map $f : \mathbb{R} / \mathbb{Q} \to \mathbb{R} / \mathbb{Q}$ is continuous. However, as a condensed set, more structure is preserved. Indeed, the “continuous” maps are exactly those coming from continuous functions $\mathbb{R} \to \mathbb{R}$ which commute with the equivalence relation. (See this answer to Examples of the difference between Topological Spaces and Condensed Sets for more motivation.) I’m trying to understand condensed sets better and how they relate to topology as it comes up in constructive mathematics. In constructive theories, for example dependent type theory, there is an interesting phenomenon that I’ve always found fascinating. Just from the axioms of type theory, we automatically get topological structure on types. For example, every constructively definable function $f : 2^\mathbb{N} \to 2^\mathbb{N}$ is continuous for the usual product topology applied to $2^\mathbb{N}$. Now, many constructive theories also allow quotients. For example, in Coq you can add quotients, and similarly in Homotopy Type Theory quotients are implied by univalence and higher inductive types. While quotients may break computability in some theories, if I understand correctly, they don’t break continuity. (For example, you still can’t construct a term for a non-continuous function $f : 2^\mathbb{N} \to 2^\mathbb{N}$ in a constructive type theory with quotients, right?) Now my question is as follows: Does it seem that the induced topological structure on quotient spaces in constructive type theory is actually the condensed set structure of Clausen and Scholze? Here are some ways to make my question more formal: Consider the quotient space $2^\mathbb{N} / \text{fin}$ of all binary sequences, mod two sequences being equivalent if they agree on all but finitely many sequences. Can we (in say HoTT with univalence, or MLTT with quotients) construct a term for a function $f : 2^\mathbb{N} / \text{fin} \to 2^\mathbb{N} / \text{fin}$ which is not a condensed set endomorphism? I assume not. Is it possible to make this “induced topology” theory formal? For example, I’ve heard folks say that it is consistent with constructive mathematics that every function is continuous, but I don’t know what that means formally. For example, can we add an axiom to MLTT or some other type theory that says that every $f : A \to B$ is continuous for the induced topologies on $A$ and $B$? Or better, is there a topos model of MLTT where the types are topological spaces and the functions are continuous? (Sorry, if I’m not using the right terms here. I’m still getting used to categorical logic.) If there is a way to make this formal, can this formalism be extended to type theories with quotients, and then can one replace topological spaces with condensed sets? (Bonus) In HoTT, if indeed, one can endow all sets (homotopy 0-types) with a condensed set structure, what about higher homotopy types? If the sets can be though of as condensed sets, can say the n-groupoids be interpreted as condensed n-groupoids and so forth? REPLY [4 votes]: Not quite a complete answer, but: There are models of constructive mathematics where sets are $T_0$ second-countable (optionally zero-dimensional) topological spaces equipped with equivalence relations, morphisms are continuous maps which respect the equivalence relation, and pointwise-equivalent morphisms are equal. In these two models, quotients are taken simply by inflating the equivalence relation, and the result satisfies all of the usual rules of MLTT + Quotient Types. (We need to add an equivalence relation structure so that the quotients are well-behaved enough for exponents to exist.) In particular, each functions $f : 2^{\mathbb{N}}/\operatorname{fin} \to 2^{\mathbb{N}}/\operatorname{fin}$ arises from some continuous function $g:2^{\mathbb{N}} \to 2^{\mathbb{N}}$ These models don't have a set of truth values (hence no power-set operation), but they can be upgraded to contain one. (Roughly speaking - the above models arise as categories of PERs over two different PCAs, and there's a general technique for upgrading these, called realisability topoi.) This is the smallest change I know to the category of topological spaces that results in a model of MLTT. As for what "every function is continuous" means - there are three ways that I know of. The first is, to every set, assign a topology (we could say that the opens are the unions of preimages of arbitrary functions into $\mathbb{R}$, for instance), and then every function is continuous with respect to this topology. This could also be done in classical mathematics, but there all sets would be given the discrete topology. Constructively, it may be the case that we get nontrivial topologies, and may even happen to get the 'correct' topologies for spaces such as $2^{\mathbb{N}}$ or $\mathbb R$. The second way gives a number of examples where this holds. It is consistent that all functions $2^{\mathbb{N}} \to \mathbb{R}$ are continuous. In the presence of countable choice (which is consistent with this; iirc both (and MP, mentioned below) hold in Johnstone's topological topos, which has FirstCountableTop as a full subcategory and which I suspect is a full subcategory of that of condensed sets), this implies the stronger theorem that If $M$, $N$ are metric spaces with $M$ complete, then all functions $f:M \to N$ are continuous. (If MP ($\not \geq_{\mathbb{R}}$ implies $<_{\mathbb{R}}$) holds, we can weaken "complete" to "complement of a subset of a complete metric space".) The third way is to generalise to grothendieck topologies, which is the main way to make (grothendieck) topoi, which model MLTT + quotients. Here, we specify a small category of "test spaces" and specify some families of morphisms to be "coverings" (for example, we could take all compact second-countable spaces for test spaces, and finite joint surjections as coverings) satisfying some technical conditions. The resulting grothendieck topos includes the spaces from the test category, and doesn't add any new morphisms between them - so if all of your morphisms between test spaces were continuous, they will continue to all be in the larger topos. The condensed sets are formed very similarly to this, except that the category of test spaces is no longer small, so we need to deal with size issues to get a grothendieck topos, like Simon mentioned.<|endoftext|> TITLE: Can we determine the game-theoretically best first move by White in chess without solving chess? QUESTION [5 upvotes]: In turn-based board games with high branching factor (such as chess) are there any arguments that could ascertain the ideal first move but not solve the entire game? I am asking because solving chess entirely seems infeasible but conceivably one could hope to determine the ideal first moves. REPLY [8 votes]: Setting aside chess in particular, we can certainly come up with games for which we can determine the game-theoretically best move without solving the game in general. For example, it is easy to prove, via a strategy-stealing argument, that a game such as Hex is a first-player win. So you can invent a game called "meta-Hex" in which the first player's first move is to choose whether to be the first player or the second player of a game of Hex, and then to play a game of Hex accordingly. Obviously, the game-theoretically optimal best first move in meta-Hex is to choose to be the first player in a game of Hex. We have figured this out without figuring out how to play Hex (or meta-Hex) optimally in general. For chess in particular, any kind of rigorous proof about the best first move appears to be hopelessly out of reach. As Carlo Beenakker noted in a comment, the conventional wisdom is that chess is a draw; if true, this might seem to be a favorable situation for mathematical analysis since it would mean that any reasonable first move (and perhaps any first move at all) would probably be game-theoretically optimal. But in practical terms, we cannot expect a mathematical proof. (It is also morally certain that if White gives Black queen odds, then Black has a win, but a mathematical proof of even this seemingly obvious fact seems out of reach.) At the same time, of course, it is even more hopeless to prove that we cannot find such a proof. It's perhaps worth mentioning Fraenkel and Lichtenstein's paper Computing a perfect strategy for $n\times n$ chess requires time exponential in $n$. This result provides strong circumstantial evidence that there is not going to be a short proof that $8\times 8$ chess is a draw, but of course their result does not actually say anything about $8\times 8$ chess.<|endoftext|> TITLE: An interesting Hankel determinant QUESTION [5 upvotes]: Let $h(n,t) = \sum\limits_{j = 0}^n {\binom {\lfloor {\frac{n}{2}} \rfloor }{j}\binom {\lfloor {\frac{n+1}{2}}\rfloor }{j}t^j \\ }.$ I am interested in the Hankel determinants $${D_k}(n,t) = \det \left( {h(k + i + j,t)} \right)_{i,j = 0}^{n - 1}.$$ These can easily be computed for $0 \leq k \leq 3.$ It seems that $${D_4}(n,t) = {t^{\lfloor {\frac{{{n^2}}}{4}} \rfloor }}b(n,t)$$ with $b(n,t) = \sum\limits_{j = 0}^{2n} \min({\binom{3+j}{3},\binom{2n+3-j}{3}})t^j.$ In order to prove this, I need the identity $$a{(n,t)^2} = b(n - 1,t)\sum\limits_{j = 0}^{n - 1} {{t^j}} - tb(n - 2,t)\sum\limits_{j = 0}^n {{t^j}} $$ with $a(n,t) = \sum\limits_{j = 0}^{n - 1} {(j + 1){t^j}} .$ Any idea how to prove this identity? REPLY [5 votes]: Denote $a_n=a(n,t)$ and $b_n=b(n,t)$. To help avoiding the min function, write $$b_n=\binom{n+3}3t^n+\sum_{j=0}^{n-1}\binom{3+j}3\left[t^{2n-j}+t^j\right].$$ Notice that $a_n=\frac{nt^{n+1}-(n+1)t^n+1}{(1-t)^2}$ and $\sum_{j=0}^nt^j=\frac{1-t^{n+1}}{1-t}$. Your identity takes the form $$(nt^{n+1}-(n+1)t^n+1)^2=(1-t)^3[(1-t^n)b_{n-1}-t(1-t^{n+1})b_{n-2}].$$ Now, as Mark Widon mentioned, try to read-off the coefficients of $t^k$. UPDATE. Resorting back the original formulation of the claim $$a_n^2=b_{n-1}\sum_{j=0}^{n-1}t^j-t\,b_{n-2}\sum_{j=0}^nt^j,$$ I was able (after lots of routine algebraic simplification and reorganization) to rewrite the right-hand side as \begin{align*} \left(\sum_{j=0}^{n-1}(j+1)t^j\right)^2 &=\sum_{j=0}^{n-1}\binom{3+j}3t^j+\sum_{j=0}^{n-2}\left[\beta_n-\beta_{j+1}-\binom{n-j}3\right]\,t^{n+j} \\ &=\sum_{j=0}^{n-1}\binom{3+j}3t^j+\sum_{j=0}^{n-2} \frac{(n - j - 1)(j^2 + 4jn + n^2 + 5j + 7n + 6)}6\, t^{n+j} \end{align*} where $\beta_k=\frac{k(k+1)(2k+1)}6$ (the sum of squares function). Once we got this far, the next step is to compare the coefficients of $t^k$.<|endoftext|> TITLE: Does simple theory of types + ambiguity prove axiom of infinity? QUESTION [6 upvotes]: Does simple theory of types + ambiguity prove axiom of infinity? The simple theory of types known as $\sf TST$ is a multi-sorted first order theory, syntactical restrictions include $\in$ being a dyadic symbol where the symbol on the right of it is one sort (type) higher than the one on the left, while the two symbols linked by $=$ must be of the same type. Axioms (on top of multi-sorted axioms of identity): Extensionality: $\forall x^{i+1} \, \forall y^{i+1}:\\ \forall z^i (z^i \in x^{i+1} \leftrightarrow z^i \in y^{i+1}) \to x^{i+1}=y^{i+1}$ Comprehension: if $\phi$ is a formula in which $x^{i+1}$ doesn't occur, then: $$\exists x^{i+1} \, \forall y^i (y^i \in x^{i+1} \leftrightarrow \phi)$$ Now the schema of ambiguity is: Ambiguity: If $\phi^+$ is the formula obtained from formula $\phi$ by raising all type indices in $\phi$ by one, then: $$\phi \iff \phi^+$$ It's well known that $\sf TST +Ambg$ is equi-interpretable with Quine's $\sf NF$ [Specker]. The latter is known to prove Infinity [Specker]. Would that entail that the fomer must also prove Infinity? That was the first question. The second question is if the answer is to the positive, then clearly the above theory is purely motivated by a logic of types, and types are only needed in a relative manner, and clearly this doesn't need the particular values of types to matter other than their relative positions. There is a remote resemblance between ambiguity and axiom of reducibility of Russell's, although of course they are not the same principle. If Infinity is provable from the above purely logically motivated axioms for set theory, then in some sense this could be seen as a motivation for logicism, since infinity, a clearly mathematical axiom, is not axiomatized here! So my second question is: Had $\sf TST + Ambg$ been considered as motivating the program of logicism? REPLY [3 votes]: The development of TST + Amb is arguably motivated by the logicist program, ultimately. But Amb is not a purely logical principle, it is a conjecture past the logically provable facts, with unexpected consequences. I dont think that TST + Amb can be taken to motivate logicism. It is a by-product of this program. TST + Ambiguity proves every stratified theorem of NF, so it proves Infinity, disproves Choice, and so forth.<|endoftext|> TITLE: Superconnected spaces QUESTION [6 upvotes]: Question 1. Let $\epsilon > 0$ and $V > 0$. Is there always a complete connected Riemannian manifold $M$ with $$ \operatorname{diam} M < \epsilon\quad\text{ (small diameter)} \quad \text{and} \quad \operatorname{vol}M > V\quad\text{(large volume)}? $$ In other words, can we construct worlds with room for arbitrarily many planets, but where any two planets are arbitrarily close to each other? Note that $M$ has to be compact by Hopf-Rinow. I'm almost certain that the answer is 'yes', but I'm having a hard time finding an explicit sequence $M_1, M_2, M_3,\ldots$ such that $\operatorname{diam}(M_n)$ goes to zero while $\operatorname{vol}(M_n)$ goes to infinity. Question 2. (Fixed dimension) Is there a sequence of complete connected Riemannian manifolds $M_1, M_2, M_3, \ldots$ with $$ \dim M_n = m, \quad \lim_{n \to \infty}\operatorname{diam} M_n = 0, \quad \text{and} \quad \lim_{n \to \infty}\operatorname{vol}M_n = \infty $$ for some fixed $m \in \mathbb{N}$. For this one, I have less intuition, but I'm more inclined towards 'no'. Note that for $m = 0, 1$, this is clearly impossible. For $m = 2$ my intuition is very strongly in favour of 'no'. REPLY [6 votes]: The answer to both questions is positive, even in dimension 2. Take a round sphere of diameter $\epsilon$, and make many, say $N$ little holes in it. Then take $N$ spheres of diameter $\epsilon$ and make one little hole in each. Then glue these $N$ spheres to the first sphere along the boundaries of the holes. The diameter of the resulting surface is about $3\epsilon$ while the volume (area) is approximately $(N+1)\epsilon$. Now make $\epsilon$ as small as you wish, while $N>V/\epsilon$. Here is another construction. Take a round sphere $S$ of diameter $\epsilon$, think of it as the Riemann sphere, and consider the $N$-fold ramified covering $S_1\to S,\;z\mapsto z^n$. Here $S_2$ is also a sphere, and equip it with the pullback of the metric from $S$. Then the solume of $S_1$ will be $N\epsilon$ while diameter is at most $2\epsilon$. If a smooth metric is desirable, approximate it with a smooth one.<|endoftext|> TITLE: Category with few endofunctors? QUESTION [7 upvotes]: Let $\mathcal C$ be a category whose skeleton has $\lambda$-many objects and $\kappa$-many morphisms. Then the skeleton of the endofunctor category $\mathcal C^{\mathcal C}$ has at most $\kappa^{3 \times \kappa}$ many morphisms. My guess is that in most cases, this upper bound is achieved. Question: What is an example of a category $\mathcal C$ (other than the terminal category or equivalents) whose skeleton has $\lambda$ many objects and $\kappa$ many morphisms, but the skeleton of whose endofunctor category $\mathcal C^{\mathcal C}$ has Strictly fewer than $\kappa^\kappa$ many morphisms? Strictly fewer than $\kappa^\kappa$ many objects? As few as $\kappa$-many morphisms? As few as $\lambda$-many objects? EDIT: Neil Strickland's example of $B\mathbb N$ in the comments below affirmatively answers (1), (2), and (3). So it remains to see about (4). Note that $\mathcal C^{\mathcal C}$ always has (skeletally) at least $\lambda+1$ many objects and $\kappa+1$-many morphisms, given by constant morphisms between constant endofunctors, along with the identity functor. Note also that if $\mathcal C$ is accessible, and has as many objects and morphisms as the size of the universe, then the number of accessible endofunctors and morphisms between them is also the size of the universe. So if my guess is correct, then most endofunctors are usually non accessible. See also here for an argument which shows that if $\mathcal C$ has small products and coproducts and is not a preorder, then $\mathcal C^{\mathcal C}$ has (skeletally) at least $2^\kappa$-many objects, where $\kappa$ is the size of the universe. REPLY [5 votes]: An example of (1),(2),(3),(4) with $\kappa=\lambda=|\mathbb R|$ is to take $\mathcal C$ to be the posetal category $\mathbb R.$ This is already skeletal. Its category of endofunctors is the poset of non-decreasing functions $f:\mathbb R\to\mathbb R,$ with the pointwise order. There are at most $|\mathbb R^{\mathbb Q}|=|\mathbb R|$ such functions: $f$ is determined by the preimages $f^{-1}((-\infty,q))$ for $q\in\mathbb Q,$ each of these intervals is convex, and there are at most $|\mathbb R|$ convex subsets of $\mathbb R.$ (I don't know if there are, unconditionally, examples with $\lambda$ and $\kappa$ regular. I think the $\mathbb R$ example generalizes to $2^\mu$ with the lex order, where $\mu$ is an ordinal satisfying $|2^{2^{<\mu}}|=|2^\mu|,$ for example $\mu=\beth_\alpha$ with $\alpha$ a limit ordinal. Here $2^{<\mu}$ means the functions $\mu\to 2$ with bounded support.)<|endoftext|> TITLE: Scott's trick without regularity QUESTION [11 upvotes]: In ZF(C), one can easily get a class partition of $V$, we can even get an $\mathrm{Ord}$-partition using the Cumulative hierarchy: $P=\{V_{α+1}\setminus V_α\mid α∈\mathrm{Ord}\}$, such a partition let us do stuff like Scott's trick: given a class $A$, we can look at $A∩V_β$ where $β$ is the minimal $β$ so that intersection is not empty. But the fact that $\bigcup P=V$ is equivalent to the axiom of regularity. I remember somewhere reading that in $ZF\text{-regularity}$ we can't have Scott's trick like trick. We can formulate Scott's trick as: There exists an $\mathrm{Ord}$-partition of $V$ (or equivalently - there exists cumulative hierarchy that sums up to the universe). In a sense this version of Scott's trick is a "global trick". While the intuition tells me that $\text{regularity}$ does not follow from this version of Scott's trick, is this true? REPLY [4 votes]: Yes! Extend $\sf ZF - Reg.$ with the existence of a unique Quine atom $\sf Q=\{Q\}$, take $V$ to be the hierarchy over $\sf Q$, that is: $$\begin{align} & V_\emptyset = \sf Q \\ & V_{\alpha+1}= \mathcal P (V_{\alpha}) \\ & V_\lambda= \bigcup_{\alpha < \lambda} V_\alpha, \text {for limit } \lambda \\ & V= \bigcup _{\alpha \in \mathrm{Ord}} V_\alpha \end {align}$$ clearly this $V$ has an $\mathrm{Ord}$-partition.<|endoftext|> TITLE: Lemoine-Lozada circles QUESTION [5 upvotes]: I made some rookie attempt to define the 4th Lemoine circle recently. The alternative name for this circle was suggested yesterday. Further investigation revealed a family of circles associated with the Brocard axis. Supposedly the name of Ehrmann-Lozada circles (or Lemoine-Lozada circles) can be safely attached to them. Specifically, it appears that for each Lozada circle it is possible to construct at least one circle of the Lemoine type, with its center on the line that is going through the Symmedian point and the circumcenter of the triangle. step 1: Let A'B'C' be the midheight triangle of ABC and let BA and CA be the orthogonal projections of B' and C' on BC, respectively. Build AB, CB, AC and BC similarly. These last six points lie on a circle with center X(9729). The circle is called the 1st Lozada circle. (César Lozada, March 17, 2016) step 2: Circumcircles of the triangles BcAcX(6), AbCbX(6), BaCaX(6) cut the sides of the original triangle at six points, that define the 1st Ehrmann-Lozada circle. Its center O_1 is not in the ETC. Ehrmann-Lozada circles can be constructed in the similar manner for each(?) of the ten remaining Lozada circles: https://www.geogebra.org/geometry/pg9mrmzx 2 new circles of the Lemoine type https://geogebra.org/geometry/uqnpqjeh 2 new circles of the Lemoine type https://www.geogebra.org/geometry/t4jpn9dp only 1 new circle of the Lemoine type ... 11. https://www.geogebra.org/m/kppjdcny gives us only 1 new circle of the Lemoine type... More links will be added soon. None of these points O_i is included in Clark Kimberling's encyclopaedia. Finding a strict proof of existence of these new circles is the main question here. Bit I am also interested in what makes Lozada points/circles so special? (Taylor point belongs to the Brocards Axis as well, but for some reason it is impossible to construct an Ehrmann-Taylor circle in the same way!) Are these Ehrmann-Lozada circles also Tucker circles? REPLY [5 votes]: Once you have worked through even a handful of examples of a construction, it can be worthwhile to abstract things a bit in hopes of glimpsing a general principle. A cyclic sextuple of points —two on each side-line of a triangle— is determined by a triple of of those points (one on each side-line). We can ask when such a triple gives rise to a secondary cyclic sextuple by the OP's construction, recapped below. (In what follows, "$X_Y$" and "$X_Y'$" indicate points on the side-line opposite triangle vertex $X$.) Given $A_B$, $B_C$, $C_A$, define $A_C$, $B_A$, $C_B$ as the "other" points where $\bigcirc A_BB_CC_A$ meets the (side-lines of) the triangle. Further, given a specified point $S$, let $B_C'$, $C_B'$ be the "other" points where $\bigcirc SB_AC_A$ meets the triangle; likewise, we have $C_A'$ and $A_C'$ via $\bigcirc SC_BA_B$, and $A_B'$ and $B_A'$ via $\bigcirc SA_CB_C$. Definition. Triple $A_B$, $B_C$, $C_A$ is magical with respect to $S$ if the six derived points $A_B'$, $A_C'$, $B_C'$, $B_A'$, $C_A'$, $C_B'$ lie on a circle (which I'll call the triple's (First) Magic Circle). (Note that the definition says nothing about the centers of the circumcircles involved. I'll get to that.) For general $S$, the condition of magicality is complicated to express, involving a quartic relation. The condition happens to be replete with side-squares $a^2$, $b^2$, $c^2$, and the relation factors nicely when $S$ is the symmedian point, with barycentric coordinates $(a^2:b^2:c^2)$. For this case, we can state Theorem. Triple $A_B$, $B_C$, $C_A$ is magical with respect to symmedian point $S$ if and only if $$\frac{\alpha}{c^2}=\frac{\beta}{a^2}=\frac{\gamma}{b^2} \qquad\text{or}\qquad \frac{1-\alpha}{b^2}=\frac{1-\beta}{c^2}=\frac{1-\gamma}{a^2} \tag{1}$$ for the signed ratios $$\alpha := \frac{|BA_B|}{|BC|}\qquad \beta :=\frac{|CB_C|}{|CA|} \qquad \gamma := \frac{|AC_A|}{|AB|}$$ Proof is by Mathematica-assisted symbol crunching (that I won't reproduce here), fueled mostly by power-of-a-point relations among all the points on the side-lines; the relations encoding the role of $S$ are what bog the calculations down. For the symmedian case, I suspect there's a not-unreasonable synthetic proof of $(1)$. (I should acknowledge: I don't doubt $(1)$, and much (most? all?) of what follows, already appears in the literature.) If the first proportion in $(1)$ holds for $A_B$, $B_C$, $C_A$, then the second holds for $A_C$, $B_A$, $C_B$ (redefining $\alpha$, $\beta$, $\gamma$ appropriately). Consequently, the latter triple is also magical, which allows us to say If triple $A_B$, $B_C$, $C_A$ is magical with respect to symmedian point $S$, then the six "other" points where $\bigcirc SA_BB_A$, $\bigcirc SB_CC_A$, $\bigcirc SC_AA_C$ meet the triangle are also concyclic, giving a Second Magic Circle. Moreover, the construction is symmetric in the given points $A_B$, $B_C$, $C_A$ and derived points $A_B'$, $B_C'$, $C_A'$; thus, the derived triple is necessarily also magical (as is derived triple $A_C'$, $B_A'$, $C_B'$). We don't get any new circles, however; rather, we get the circumcircle of our original points (their "Zeroth Magic Circle"), and Second Magic Circle noted above. We effectively have a magical triple of circles: any one gives rise to the other two. For a bit of metric specificity ... Define $\kappa_0$ as the (normalized) constant of proportionality in $(1)$. $$\frac{\kappa_0}{a^2+b^2+c^2} \;:=\; \frac{\alpha}{c^2}=\frac{\beta}{a^2}=\frac{\gamma}{b^2}$$ Likewise defining $\kappa_1$ and $\kappa_2$ relative to the First and Second Magic Circles, we find $$\kappa_{i+1} = \frac{3 - \kappa_i}{2 - \kappa_i}$$ Also, defining $k_i$ as the radius of $i$-th Magic Circle, we have $$k_i^2 = \frac{r^2}{(a^2 + b^2 + c^2)^2} \left(\;(a^2 + b^2 + c^2)^2 (1-\kappa_i) \;+\; (a^2 b^2+b^2c^2+c^2a^2)\kappa_i^2\;\right)$$ where $r$ is the circumradius of $\triangle ABC$. Recall that the definition of a magical triple of points made no assumptions about the centers of the original or derived circles; in particular, it seems to have ignored what seems to be the key property motivating OP's question. It happens that the symmedian case gives us that property for free: If triple $A_B$, $B_C$, $C_A$ is magical with respect to symmedian point $S$, then its circumcenter lies on the Brocard axis (joining $S$ to the circumcenter $O$ of $\triangle ABC$). Since the construction transfers magic to the derived points, we can say the Zeroth, First, and Second Magic Circles have centers $K_i$ on the Brocard axis. And, specifically, $$\frac{|OK_i|}{|OS|}=\frac{\kappa_i}{2}$$ This, and the radius calculation, tell us how to construct a Magic Circle about any point on the Brocard axis. (Sanity check: $\kappa=0$ corresponds to $k=r$. The circle is the triangle's circumcircle. Points $A_B$, $B_C$, $C_A$ coincide with the vertices, so $\alpha=\beta=\gamma=0$.) Also, one can confirm that each of the three Magic Circles is a Tucker circle. Converting the above results into Tucker/Brocard parameters is left as an exercise to the reader. Okay, that's enough of that. To address OP's particular interest in Lozada circles ... I don't know if there's a unifying principle in that family of circles, or if "the $n$-th Lozada Circle" is simply "the $n$-th circle that Lozada happened to notice had a neat construction and a center on the Brocard axis". So, I can't say whether it's surprising —or maybe it's obvious— that members of the family are magical. Nevertheless, I'll explicitly confirm the magical nature of three cases illustrated by OP. (I'm afraid I don't have the patience to work through all eleven.) To avoid a little clutter, define $\lambda := \kappa/(a^2+b^2+c^2)$ as the un-normalized proportionality factor in $(1)$. Revealing magic is simply a matter of showing that the value of $\lambda$ is symmetric in triangle metrics. First Lozada Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are orthogonal projections of mid-height points. Straightforward right-triangle trig yields $$|BA_B|=r\sin^2C\sin A, \; |CB_C|=r\sin^2A\sin B, \; |AC_A|=r\sin^2B \sin C\;\to\; \lambda=\frac{1}{8r^2}$$ Second Lozada Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are orthogonal projections of points where altitudes meet the circle having the orthocenter and centroid as a diameter. A coordinate bash gives $$|BA_B| = \frac{8|\triangle ABC|^2}{3ab^2}, \quad |CB_C|=\cdots, \quad |AC_A|=\cdots \quad\to\quad \lambda = \frac{8|\triangle ABC|^2}{3a^2b^2c^2}$$ Fourth Lozada Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are points where the side-lines meet lines through "opposite" sides of Pythagoras-esque squares erected on the triangle's sides. Here again, simple trig allows us to write $$|BA_B|=-\frac{c}{\sin B}, \quad |CB_C|=\cdots, \quad |AC_A|=\cdots \quad\to\quad \lambda=-\frac{1}{2|\triangle ABC|}$$ (I don't know why OP asserts the existence of "only one" circle derived from the Fourth Lozada Circle. There are two.) OP seems to be suggesting that the Taylor circle doesn't quite fit with this investigation. In fact, the Taylor Circle (rather, each of two triples of points it determines on the triangle) is magical: Taylor Circle: $A_B$, $B_C$, $C_A$ (and $A_C$, $B_A$, $C_B$) are orthogonal projections of the feet of triangle's altitudes onto the side-lines. Specifically, with $X_Y$ the projection of the foot from vertex $Y$ projected onto the side-line opposite vertex $X$, we find $$|BA_B|= a\sin^2C \qquad |CB_C|=b\sin^2A \qquad |AC_A|=c\sin^2B \quad\to\quad \lambda=\frac{1}{4r^2}$$ Here's a figure showing a Taylor Circle (blue) and associated First and Second Magic Circles: I'll stop typing now.<|endoftext|> TITLE: Can perfect numbers be seen $p$-adically? QUESTION [8 upvotes]: It is well known that all even perfect numbers are of the form $N=(2^{q}-1).2^{q-1}$ with $M_{q}:=2^{q}-1$ a Mersenne prime. As the very defining property of such a perfect number is to fulfill the equality $\sigma(N)=2N$, one can see that this value is almost the sum of a geometric series. But another conceptual framework can emerge from this: namely that $\sigma(N)$ is (close to) a $2$-adic series. As Euler showed unrigorously, one has $\sum_{k=0}^{\infty}2^{k}=-1$, the latter being interpreted as convergence in $\mathbb{Z}_{2}$, the ring of $2$-adic integers. So a proof of the existence of infinitely many even perfect numbers should be morally equivalent to the convergence of the sequence $(2N)_{N}$ where $N$ runs over the even perfect numbers towards $0$ (edited after Wojowu's comment, I wrote $-1$ at first) in $\mathbb{Z}_{2}$. My question is: can this be generalized to hypothetical odd perfect numbers? Namely, should $\sigma(N)$ for $N$ an OPN be "close" to some $p$-adic series for some $p>2$? Is there some hint that this should indeed be the case? Edited after JoshuaZ' supposedly incomplete but insightful answer: The number $2$ in the perfect number defining equality $\sigma(N)=2N$ may be viewed as the Euler factor at $p=2$ of $\zeta(s)$ as $s$ tends to $1$, that is, $N=\sigma(N)\lim_{s\to 1}\frac{\zeta_{2}(s)}{\zeta(s)}$ where $\zeta_{2}$ is the $2$-adic zeta function. Could we be able to prove that any perfect number $N$ whose smallest prime factor is $p$ fulfills $N=\sigma(N)\lim_{s\to 1}\frac{\zeta_{p}(s)}{\zeta(s)}$, this would imply $p=2$ and that no OPN exists. Second edit: one can define the "numerical $p$-adic transform" of an L-function $F$ whose sequence of Dirichlet coefficients is $(a_{n}(F))_{n>0}$, so that $F(s)=\sum_{n>0}a_{n}(F).n^{-s}$ whenever $\Re(s)>1$, by $\mathcal{L}_{p}(F):=\sum_{n>0}a_{n}(F)p^{n}$ in $\mathbb{Z}_{p}$. That way perhaps a link between $\mathcal{L}_{2}(\zeta)$ and RH could be made. REPLY [8 votes]: (Not a complete answer but a bit too long for a comment.) There's a fundamental difficulty here in proving the sort of result you envision. If there were some prime $p$ which had to divide every odd perfect number and larger odd perfect numbers had to be divisible by higher powers, that would work. But we can't right now rule out now even that for any given odd prime $p$, there are infinitely many odd perfect numbers with smallest prime factor $p$. As far as I'm aware, the closest thing we have to a restriction on that are the results in this recent paper of mine, especially Theorem 8 on page 47. The closest thing in the literature might be some of what Tomohiro Yamada has done. One goal that Tomohiro Yamada has been working on has been to rule out certain families of exponents. One of his strongest results of this sort is the following: Let $N$ be an odd perfect number and write $$N= q^ap_1^{2e_1}p_2^{2e_2}\cdots p_{k-1}^{2e_{k-1}}$$ where $q \equiv a \equiv 1$ (mod 4) and $p_1$, $p_2 \cdots $ $p_{k-1}$ are distinct primes none equal to $q$. (Note that by Euler's theorem for odd perfect numbers we can always put one in this form.) Then if all the $e_i$ are the same. That is, $e_1=e_2...=e_{k-1}=e$, then we have $k \leq 2e^2 + 8e+3$. Note that since there are bounds on the size of an odd perfect number in terms of its number of distinct prime factors, this automatically says that there are only finitely many odd perfect numbers which all share the same exponent for all but one prime. The obvious stronger goal to aim for is a theorem of the same flavor as above, but where instead of assuming that all the exponents are the same, construct a function $f(x)$ such that if all the $e_i$ are bounded above by some constant $E$, then $k \leq f(E)$. The results from Tomohiro use high powered sieve theory, but it seems like not many people have thought that hard about extending them, so I can't speak about how plausible this goal is given the current power of machinery. It might be the sort of thing that a group of very dedicated people might prove or it might be completely out of reach with current technology. That said, if one did have a theorem of the sort envisioned in the last paragraph, one would then have as a corollary that for any $m$ there are only finitely many odd perfect numbers not divisible by a perfect $m$th power of a prime. So this would move sort of in the direction you want. But even this would be a very weak move in that direction because there's no guarantee that the primes are the same in each case. Finally, let me add one piece of almost completely unfounded speculation, of how one might be able to get a result of the sort you want. It isn't hard to write down a Dirichlet series in terms of $\zeta(s)$ where the series has a non-zero $n$th term if and only if $n$ is an odd perfect number. Sketch of argument: write down the Dirichlet series for $(\sigma(n)-2n)^2$ using that $$\sum_{n=1}^{\infty} \frac{\sigma(n)^2}{n^s} = \frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)}. $$ (The above is a special case of a general theorem for a Dirichlet series for $\sigma_a(n)\sigma_b(n)$. See e.g. Theorem 305 in Hardy and Wright.) Then remove all the even terms from the Dirichlet series, and you now have a series which has a term which is zero exactly where the odd perfect numbers are. One would hope that one could then maybe turn analytic methods to estimate the coefficients of this Dirichlet series. I've never been successful in getting anything non-trivial out of this, but it is possible that someone who is better at complex analysis than I am could make this work. But your question brings up a related idea had not occurred to me until now and which might be worth someone thinking about: how does the resulting series behave $p$-adically? Vague hand-wave to Kubota-Leopoldt goes here. If that could work it might be possible to go in the other direction, do some sort of $p$-adic approach to the Dirichlet series in question and see if that gives us new information about odd perfect numbers.<|endoftext|> TITLE: About the maximum number of leaves adjacent to a vertex in a tree QUESTION [5 upvotes]: Let $T$ be a finite tree graph with the set of vertices $V(T)$. For an arbitrary vertex $ v \in V(T)$, I define $l(v)$ to be the number of leaves connected to $v$. In my study, I need to define the following concept: $$D(T) = \max_{v \in V(T)}l(v). $$ Obviously, $1 \leq D(T) \leq \Delta(T)$, which are achieved by (for example,) the path graphs and the star graphs, respectively. Here, I have two questions: Question 1. Is there a common notation for $D(T)$? Question 2. For a random tree $T$, is there some bounds (especially lower bound) for $$ \dfrac{D(T)}{\vert V(T) \vert}, $$ or, what can one say about the average of the above ratio for the trees of a given order $n$? Thanks in advance. REPLY [2 votes]: Expanding on a comment upon OP's request: To calculate the average over all unlabelled trees on $n$ vertices we can exploit the property that every finite tree has either a centroid or a bicentroid. This allows us to do most of the work with rooted trees. If $R(n, \ell, n')$ denotes the number of rooted trees on $n \ge 1$ vertices with leaf degree no greater than $\ell$ and largest subtree of the root having no more than $n'$ vertices, it's straightforward to observe that $$R(n, \ell, n') = \sum_{\substack{a_1 + 2a_2 + \cdots + n'a_{n'} = n-1 \\ a_i \ge 0, \, a_1 \le \ell}} \prod_i \binom{R(i, \ell, i-1) + a_i - 1}{a_i}$$ and then the number of unrooted trees on $n \ge 1$ vertices with leaf degree no greater than $\ell$ is $$U(n, \ell) = R(n, \ell, \lfloor \tfrac{n-1}{2} \rfloor) + [n \bmod 2 = 0] \binom{R(\tfrac n2, \ell, \tfrac n2-1) + 1}{2}$$ where the first term counts centroidal trees and the second term counts bicentroidal trees. The average value of $D$ can then be calculated from the sum of $\ell \times (U(n, \ell) - U(n, \ell - 1))$. $$\begin{array}{lll}n & \textrm{Average value of }D \\ 10 & \tfrac{ 273 }{ 106 } & 2.5754716981132075 \\ 20 & \tfrac{ 2442856 }{ 823065 } & 2.9679988822267984 \\ 30 & \tfrac{ 8019420586 }{ 2471811967 } & 3.2443489606262594 \\ 40 & \tfrac{ 139811830066568 }{ 40443361975927 } & 3.4569784319559744 \\ 50 & \tfrac{ 19140270453206756713 }{ 5272616851455754767 } & 3.6301273148497755 \\ 60 & \tfrac{ 1280181154745986534991397 }{ 339028211512423891688777 } & 3.7760313486450774 \\ 70 & \tfrac{ 45682358753537258004625609117 }{ 11707975085109065447981840723 } & 3.901815507930055 \\ 80 & \tfrac{ 854579683870799320987268538948428 }{ 212996951362776435118713593401621 } & 4.012168617452553 \\ 90 & \tfrac{ 1052333348443216545883200707797525831 }{ 256017690896079594101576682057492020 } & 4.1103930933834185 \\ 100 & \tfrac{ 2645869542867297259202984820861933788535463 }{ 630134658347465720563607281977639527019590 } & 4.198895438962389 \\ 110 & \tfrac{ 108056089072218514932463235689898358206661706828 }{ 25249910851886927651512825423644528049790467849 } & 4.279464181321791 \\ 120 & \tfrac{ 4497776622994307502713890704059983902896242709764300 }{ 1033152520048854877807430091756234770759903219425807 } & 4.353448823588622 \\ 130 & \tfrac{ 190236537379552120425249444657539623940185621178617944373 }{ 43021618115267465696869237333175709309993559077318319227 } & 4.421882432916701 \\ 140 & \tfrac{ 4078309455388167468312120474298736087566287820459068991699340 }{ 909206958384622679578660502467356659913467956631833840699407 } & 4.485567799254476 \\ 150 & \tfrac{ 353859940291181481656671781458954968017223608594806127078280803439 }{ 77854597863245593803382458410206749993828687629640764332291246132 } & 4.545138630254686 \\ 160 & \tfrac{ 3101913783575326643405047412687839902474001481340929721745853929687909 }{ 674167392856385067467973832042845361202496256060353873446102172056969 } & 4.601103251868655 \\ 170 & \tfrac{ 228641621023362551935592604670190188836671021946402528467487288932247650053 }{ 49129289916495568808202244198612600438761664756803878655268859041109764997 } & 4.65387595489334 \\ 180 & \tfrac{ 30578444138206708760484247282898045586908979222565303716833585814401220038148136 }{ 6500796542785541223708685574350747551917347770343511085479226714248864725046787 } & 4.703799593934697 \\ 190 & \tfrac{ 1372910069939378604123094002938718076965411651529591781815624762766787704284684359397 }{ 288963010503327141303436810960435570375562139390295662777413482341210574122229958615 } & 4.751161982801847 \\ 200 & \tfrac{ 31017734225457597933395862767739806385786293974719248494783415780971891125944778400164787 }{ 6467137160511131588828576867929731880706314086113664851345986440188472113397739601846462 } & 4.796207882346214 \\ 210 & \tfrac{ 352410710751392741057157079578720601593230993520260043369781528118511482826645957294214586621 }{ 72824952200684480254538422820789504157985441774873950082120417000877246368175883433716800531 } & 4.839147848394749 \\ 220 & \tfrac{ 128797282048478184965954075556518131019553531235842408067945212679480184453626360695959830028141707 }{ 26391994208692151383402347175682684406164602489756559480467014002726855337575585275144893746100626 } & 4.880164834457983 \\ 230 & \tfrac{ 5912096543789000433038220579892213157945368680400433791633085366777082281086933708715575870840324133777 }{ 1201787513172706269909923669062213294532536027184453127982144782354836198036307712699780197770445921657 } & 4.919419181000748 \\ 240 & \tfrac{ 272563652060306897395747185358655803738754221966330221551036734394657565391873216737292786227592630120780361 }{ 54985024966026897870269075763444229658301263057496157167354274092275126613889793234171034121437216543398430 } & 4.957052437981311 \\ \end{array} $$<|endoftext|> TITLE: Existence and uniqueness of an Euler-type ODE with varying parameters QUESTION [6 upvotes]: Consider this ODE on $[1, \infty)$ $(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - ({4a} + m(m+1))f(r) = -4af(1) $ with initial conditions $\frac{a}{1-2a} f(1) + f'(1) = C, \qquad \lim_{r\to \infty} f(r) = 0$ where $0\leq a < \frac{1}{2}$, $m$ is a positive integer, and $C \in \mathbb{R}$. I want to ask if there exists a unique solution (at least for $a$ small enough). If $a=0$, then this becomes the Euler equation: $r^2f''(r) + 2r f'(r) - (m(m+1))f(r) = 0 $ $ f'(1) = C, \qquad \lim_{r\to \infty} f(r) = 0$ which we know the unique solution is: $f(r) = \frac{-C}{\alpha r^{\alpha}}$ where $\alpha = \frac{1}{2} + \frac{\sqrt{1+4m(m+1)}}{2}$ Can I prove existence and/or uniqueness for the $a>0$ case using some kind of continuity method? I know for instance that injectivity is a continuous property for elliptic operators, and one has the method of continuity to prove surjectivity of a 1-parameter family of elliptic operators. Is there something similar in this context? Any help or references is appreciated. $\textbf{EDIT} $: I wrote the equations incorrectly above. I apologize for that. I allowed the right side to decay to $0$ and so I believe it's possible to prove existence now. Here are the correct equations: $(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{4a^2}{r(r-2a)}f(1)+ \frac{4a(1-2a)}{(1-a)r(r-2a)} C $ with initial conditions $f'(1) = C, \qquad \lim_{r\to \infty} f(r) = 0$ There is an ODE that is somehow related to the above non-local differential equation. $(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{2a}{r(r-2a)} D $ with initial conditions $\frac{2a}{1-2a}f(1) + \frac{2}{1-a}f'(1) = D, \qquad \lim_{r\to \infty} f(r) = 0$ where $D$ is any real number. REPLY [6 votes]: As Iosif said, in general the system you specified does not admit a solution. Here we will give a more pedestrian argument using only comparisons. Monotonicity Claim: if a solution exists, and $f(1) > 0$, then the function is monotonically decreasing; if $f(1) < 0$, then the function is monotonically increasing. Proof: we will focus on the positive case. The negative case is similar. Let $\zeta = \frac{4a}{4a + m(m+1)} \in (0,1)$ (if $a\in (0,\frac12)$). The second derivative test shows that $f$ cannot have a local maximum with $f(r) > \zeta f(1)$ or a local minimum with $f(r) < \zeta f(1)$. This immediately implies monotonicity in light of $f(1) > 0 = \lim f(r)$. More details on monotonicity proof: Given $f(1) > f(\infty) = 0$. Suppose $f$ were not monotonic. Then there exists $r_m, r_M$ with $1 < r_m < r_M < \infty$ such that $f(r_m) < f(r_M)$. I claim that $f(r_M) > 0$ and $f(r_m) < f(1)$. Suppose not: if $f(r_M) \leq 0$ then $f(r_m) < 0$ and $f$ would have a negative local minimum, in contradiction to the second derivative test. If $f(r_m) \geq f(1)$ then $f$ would have a local maximum with value $> f(1)$, against in contradiction to the second derivative test. Therefore there must exist a local minimum $s_m\in (1,r_M)$ with $f(s_m) \leq f(r_m)$ and a local maximum $s_M\in (r_m,\infty)$ with $f(s_M) \geq f(r_M)$. We have therefore established $$ \zeta f(1) \leq f(s_m) < f(s_M) \leq \zeta f(1) $$ which is a contradiction. In particular, we must have $f' \leq 0$ on $[1,\infty)$. Comparison We have then $$ f'(1) - f'(r) = -\int_1^r f''(s) ~ds = \int_1^r \frac{4a}{s^2 - 2as} f(1) - \frac{4a + m(m+1)}{s^2 - 2a s} f(s) + \frac{2(s-a)}{s^2 - 2as} f'(s) ~ds $$ Since $f'$ is signed, we know that it is absolutely integrable on $[1,\infty)$. Since $f$ is monotonic (and hence bounded) the second integrand is also absolutely integrable. We conclude then that $\lim_{r\to\infty} f'(r)$ exists. Since $\lim_{r\to\infty} f(r) = 0$, we must have also $\lim_{r\to\infty} f'(r) = 0$. But now writing $f'(r) = - \int_r^\infty f''(s) ~ds$ using the above formula, we see that asymptotically $|f'(r)| \sim \frac1r$ (coming from the $4a f(1)$ term if it is non-zero; the other two terms can both be bounded by $O(1/r) f(r) = o(1/r)$). But this contradicts the integrability of $f'(r)$. And hence we have proved: Claim: no solution can exist with $f(1) \neq 0$. Uniqueness When $f(1) = 0$, the same maximum principle argument shows that $f$ must be identically zero. This shows that Theorem The only solution to your system is $f \equiv 0$, with $f(1) = f'(1) = 0$ and $C = 0$. Final remark Heuristically, if you want to look for asymptotically constant solutions to your equation, you probably want it to converge to $\zeta f(1)$ in the limit. REPLY [5 votes]: First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE \begin{equation*} (r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1} \end{equation*} where $p$ is a real number; your equation corresponds to (1) with \begin{equation} p=-4af(1). \tag{2} \end{equation} The general real solution of (1) is given by \begin{equation*} f(r)=f_p(r):= c_1 P_s\left(\frac{r}{a}-1\right)+c_2 Q_s\left(\frac{r}{a}-1\right)-\frac{p}{4 a+m^2+m}, \end{equation*} where $c_1$ and $c_2$ are arbitrary real constants; $P_s$ and $Q_s$ are, respectively, the Legendre functions of the first and second kinds whose values are real on $(1,\infty)$; and \begin{equation*} s:={\frac{1}{2} \left(\sqrt{4 m^2+4 m+16 a+1}-1\right)}>1. \end{equation*} Obtaining now the root (say $p_*$) of the equation $p=-4af_p(1)$ (cf. (2)) for $p$ and substituting $p_*$ for $p$, we get the general solution $F$ of your main equation: \begin{equation*} F(r):=f_{p_*}(r)= c_1 P_s\left(\frac{r}{a}-1\right) +c_2 Q_s\left(\frac{r}{a}-1\right)+c_1 A+c_2 B, \end{equation*} where \begin{equation*} A:= \frac{4 a P_s\left(\frac{1}{a}-1\right)}{m (m+1)},\quad B:=\frac{4 a Q_s\left(\frac{1}{a}-1\right)}{m(m+1)}. \end{equation*} According to Sections 15.23 and 15.33 of Whittaker and Watson, 4th ed., \begin{equation*} P_s(\infty-)=\infty,\quad Q_s(\infty-)=0,\quad Q_s>0\text{ on }(1,\infty). \tag{3} \end{equation*} So, the condition $F(\infty-)=0$ implies that $c_1=0$ and hence \begin{equation*} F(\infty-)= c_2 B. \end{equation*} Also, if $a\in(0,1/2)$, then the inequality in (3) implies $B>0$. So, $F(\infty-)\ne0$ -- unless $c_2=0$ and hence $F=0$. Thus, there is no nonzero solution to your differential problem.<|endoftext|> TITLE: An infinite series involving harmonic numbers QUESTION [9 upvotes]: I am looking for a proof of the following claim: Let $H_n$ be the nth harmonic number. Then, $$\frac{\pi^2}{12}=\ln^22+\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n(n+1) \cdot 2^n}$$ The SageMath cell that demonstrates this claim can be found here. REPLY [16 votes]: Denoting $H_0=0$, we have $$\sum_{n=1}^\infty \frac{H_n}{n(n+1)2^n}=\sum_{n=1}^\infty \left(\frac1n-\frac1{n+1}\right)\frac{H_n}{2^n}=\sum_{n=1}^\infty \frac{1}n\left(\frac{H_n}{2^n}-\frac{H_{n-1}}{2^{n-1}}\right)\\=\sum_{n=1}^\infty\frac1{n^22^n}-\sum_{n=1}^\infty\frac{H_n}{(n+1)2^{n+1}}.$$ It is well known that $\sum_{n=1}^\infty\frac1{n^22^n}=\frac{\pi^2}{12}-\frac{\log^2 2}2$ (see the value of ${\rm Li}_2(1/2)$ here). Thus, it remains to show that $$\sum_{n=1}^\infty\frac{H_n}{(n+1)2^n}=\log^2 2.$$ For this, we take the square of the series $$ \log 2=-\log\left(1-\frac12\right)=\frac12+\frac1{2\cdot 2^2}+\frac1{3\cdot 2^3}+\ldots$$ to get $$\log^2 2=\sum_{a,b=1}^\infty \frac1{ab2^{a+b}}=\sum _{a,b=1}^\infty \frac1{(a+b)2^{a+b}}\left(\frac1a+\frac1b\right)=\sum_{n=2}^\infty\frac1{n2^n}2H_{n-1}=\sum_{n=1}^\infty\frac{H_n}{(n+1)2^{n}}.$$<|endoftext|> TITLE: Why are W-types called "W"? QUESTION [10 upvotes]: Why are W-types called "W"? Probably "W" means either "wellordered" or "wellfounded". (Martin-Löf uses the term "wellorder".) But these are notions associated to order theory, whereas W-types don't directly have to do with order relations (if at all). REPLY [17 votes]: You write: Probably "W" means either "wellordered" or "wellfounded". […] But these are notions associated to order theory, whereas W-types don't directly have to do with order relations (if at all). I don’t know an official source for this, but I’ve always assumed W stands for “well-founded” as you suggest. (Edit: I asked Per Martin-Löf, and he confirmed that this was what he meant by it.) The justification for this is several-fold. On the one hand, W-types do naturally carry well-founded partial orders (the structurally smaller relation), and the recursion/induction principles of W-types can be seen as well-founded induction over these orders. On the other hand, well-foundedness was originally defined and developed to analyse and explain induction principles — it ended up becoming part of “order theory”, but it was motivated by Cantor’s analysis of induction. W-types give an alternative analysis of induction principles, not starting with an order relation — so as such, they can be seen as an alternative exploration of the original intention of well-foundedness. And on the third hand, an important way of viewing W-types is as types of well-founded labelled trees, in exactly the traditional order-theoretic sense of well-founded trees. This can be made precise in two directions. If you are modelling type theory with W-types inside set theory (or indeed any other foundation without native inductive types), one way to model W-types is to define them as sets of isomorphism classes of suitably-labelled well-founded trees. On the other hand, inside type theory, you can associate to each element of a W-type a well-founded labelled tree, and show that elements of the W-type correspond precisely to isomorphism classes of such trees.<|endoftext|> TITLE: Drinfeld center of a Deligne tensor product QUESTION [5 upvotes]: Let $\mathcal{C}$ and $\mathcal{D}$ be two tensor categories (if necessary, assume they are fusion categories). Is the canonical braided monoidal functor $$\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})\rightarrow\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$$ an equivalence? NB: The two monoidal categories $\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})$ and $\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$ have the same Frobenius-Perron dimension so it would be enough to show that the above functor is either injective or surjective in the sense of EGNO. REPLY [4 votes]: Here is another way to see this. As noticed by Theo in the comments to the OP, the center of $\mathcal{E}$ is the endomorphism category of $\mathcal{E}$ as and $\mathcal{E}$-$\mathcal{E}$-bimodule category. That it is, its objects consist of the functors $\mathcal{E} \to \mathcal{E}$ equipped with coherence data with respect to the left and right $\mathcal{E}$-actions. When you take the tensor product of tensor categories $\mathcal{C} \boxtimes \mathcal{D}$, then the left and right actions break apart in the usual way. For example the $\mathcal{C}$ actions "only act on the $\mathcal{C}$ components". What this means is that we have a formula $$\mathcal{Z}(\mathcal{C} \boxtimes \mathcal D) = End_{C \boxtimes D, C \boxtimes D} (\mathcal{C} \boxtimes \mathcal{D}) $$ $$ \cong Fun_{C,C}(\mathcal{C}, Fun_{D,D}(\mathcal{D}, \mathcal{C} \boxtimes \mathcal{D}))$$ $$\cong Fun_{C,C}(\mathcal{C}, \mathcal{C} \boxtimes Fun_{D,D}(\mathcal{D}, \mathcal{D}))$$ $$\cong Fun_{C,C}(\mathcal{C}, \mathcal{C}) \boxtimes Fun_{D,D}(\mathcal{D}, \mathcal{D})) $$ $$ = \mathcal{Z}(\mathcal{C}) \boxtimes \mathcal{Z}(D)$$ I think the only one of these isomorphisms which, perhaps, isn't straigtforward is the second one, where we identify $Fun_{D,D}(\mathcal{D}, \mathcal{C} \boxtimes \mathcal{D}) \cong \mathcal{C} \boxtimes Fun_{D,D}(\mathcal{D}, \mathcal{D}))$. Note that this doesn't use anything about the tensor category structure of $\mathcal{C}$. So for fusion categories this is totally obvious since as underlying categories then are simply finite sums of $Vect$. It also holds for finite linear categories, though this takes more work. I think it is likely to hold for some larger classes of categories as well, but I am not sure - you might have to be careful about what tensor product you are taking.<|endoftext|> TITLE: Fundamental group of the space of smooth embeddings of $S^1$ into $\mathbb R^3$ QUESTION [7 upvotes]: Has the fundamental group of the space of smooth embeddings of $S^1$ into $\mathbb R^3$ been completely computed? Say the basepoint is an unknot. Maybe something is known for other components? If yes, I would really appreciate any reference for the computation of it. To be absolutely precise, I am interested whether for every smooth knot there is a non-contractible loop of smooth knots based at it. REPLY [4 votes]: Your question is a bit ambiguous, but if you really are interested in the fundamental group with the base point the unknot, then you only care about the homotopy type of the component of the space of embeddings containing the unknot. This was computed entirely in Hatcher's classical paper Hatcher, Allen E., A proof of the Smale conjecture, Diff(S3)≃O(4), Ann. of Math. (2) 117 (1983), no. 3, 553–607. See the the 15th equivalent form of the Smale conjecture on the last page of text before the bibliography. Budney's work generalizes this (and I assume probably depends on it?). A different kind of generalization that might interest you is in Brendle, Tara E., Hatcher, Allen, Configuration spaces of rings and wickets, Comment. Math. Helv. 88 (2013), no. 1, 131–162. which computes the homotopy type of the component of the space of embeddings of k disjoint circles containing the unlink.<|endoftext|> TITLE: Are there only finitely many $m$ such that $m$ is the number of elliptic curves with a given conductor? QUESTION [15 upvotes]: Let $f:\mathbb{N}\to\mathbb{N}$ be the map sending $n$ to the number of isomorphism classes of elliptic curves over $\mathbb{Q}$ with conductor $n$. Is $f(\mathbb{N})$ finite? REPLY [16 votes]: The answer is no. Indeed, counterexamples are given by Mordell curves $E_d:y^2=x^3+d$, where $d$ is a sixth-power-free integer. These curves are pairwise non-isomorphic over $\mathbb Q$. Furthermore, this Weierstrass equation is minimal, which implies $E_d$ has additive reduction at all primes dividing $d$. Therefore the conductor of $E_d$ has the form $2^a3^b\prod_{p\mid d,p>3}p^2$, where $a,b$ range over finitely many possibilities (at most $24$ in this case if memory serves me right). But if we take $d$ to range over integers of the form $p_1^{e_1}\dots p_k^{e_k}$ where $1\leq e_i\leq 5$, we get $5^k$ curves with the same conductor give or take powers of $2,3$, and hence in total at least $5^k/24$ curves with the same conductor. Of course, these curves all arise as twists of a single curve, and it could be interesting to ask for nontrivial examples where this doesn't hold. Alternatively we could restrict to squarefree conductors, that is semistable elliptic curves.<|endoftext|> TITLE: Statement of local geometric Langlands QUESTION [5 upvotes]: A precise statement of the global geometric Langlands conjecture is well-known. However, I am unable to find a statement of the local Langlands conjecture. Does anyone have a modern statement or a reference to such? In particular, the statement should account for temperedness issues. REPLY [7 votes]: There is no precise formulation of local geometric Langlands in the literature, but the rough outline is known and goes back to the papers of Frenkel-Gaitsgory starting with https://arxiv.org/abs/math/0508382 and further refined by Gaitsgory and collaborators, see https://arxiv.org/abs/1601.05279. The state of the art at the time (mostly about the quantum version, which is actually better developed) is laid out in the lectures at the Paris winter school https://sites.google.com/site/winterlanglands2018/home. [BTW you write "geometric Langlands is well known" and I assume you mean the de Rham version, but there are also the Betti version and the restricted version, which is the intersection of the other two but has the advantage of making sense over any field -- in particular is the only one currently relevant to the Langlands program over function fields. Really one would like to know what the restricted version of local GL is, I believe Sam Raskin and others have thought about it, but nothing is written AFAIK.] In any case the rough idea is you categorify the local Langlands correspondence, which (these days) very roughly relates smooth representations of a group over a local field with (ind-)coherent sheaves on the stack of local Langlands parameters (well the spectral side as I state it is way bigger but there's a way to enlarge the automorphic side to correct that, which is unnecessary once we categorify). So a categorical analog is the following: $\bullet$ There's an equivalence of ($\infty$,2)-categories between [de Rham] categorical representations of the loop group $LG$ of a reductive group $G$ and sheaves of categories over the stack $Conn_{G^\vee}(D^\star)$ of local Langlands parameters. Automorphic side: a [de Rham] categorical representation of a group means an algebraic action of the group for which the Lie algebra action has been trivialized (this is a good categorified analog of a smooth representation, where matrix coefficients are locally constant), or equivalently module category for the monoidal category of D-modules on the group under convolution. Examples include categories of D-modules on G-spaces and the category of reps of the Lie algebra. Here the group in question is the infinite-dimensional LG so one has to deal with a lot of infinite-dimensional complications, but these are I believe largely understood now (see e.g. papers of Beraldo and Raskin about the theory of Whittaker models in this setting). Spectral side: the stack $Conn_{G^\vee}$ of local Langlands parameters means flat $G^\vee$ connections on the punctured disc. By a "sheaf of categories" one might initially mean module categories for $QC(Conn_{G^\vee})$, or quasicoherent sheaves of categories on the stack (the two are the same thanks to a surprising result of Raskin https://arxiv.org/abs/1511.01378) [Edit: according to the "History" section of that paper, the lack of such an identification is a reason the local geometric Langlands conjecture wasn't formulated earlier.] But just as in global geometric Langlands quasicoherent sheaves are not the final answer, and that's where we run into the cutting edge (the reason why the conjecture hasn't appeared in print yet). The notion of quasicoherent sheaf of categories is "too smooth" - for example when calculating its categorical trace you end up with QC not ind-coherent sheaves on the singular space given by its inertia stack. So we're running into the categorified analog of the relation distinction between perfect complex and coherent sheaf (or cohomology vs homology or function vs distribution or...) This problem is also familiar in topological field theory, in trying to define boundary conditions for Rozansky-Witten theory of a cotangent bundle. So to properly formulate the spectral side of the conjecture you need a notion of "ind-coherent sheaf of categories" and a microlocal understanding thereof (so you can define "nilpotent singular support" in this setting). There are lectures of Arinkin and Gaitsgory presenting such a notion, but more recently (2021) the Berkeley PhD thesis work of German Stefanich gives a thorough development of this notion of "2IndCoh". Anyway to conclude I believe the correct statement of local geometric Langlands is an equivalence of 2-categories $$D(LG)-mod \simeq 2IndCoh_{\mathcal N}(Conn_{G^\vee}(D^\star))$$ (plus of course a load of compatibilities, with Hecke operators, with Whittaker normalization, and more.)<|endoftext|> TITLE: English name and references for a combinatorial puzzle from Japan QUESTION [15 upvotes]: I am looking for the name and references of the following puzzle. There are n intersecting circles in a row. At the center of the circle and at the intersection of the two circles, fill the numbers 1, 2, 3, ..., 2n-1. The number at the intersection should be the sum of the numbers at adjacent two centers. An example for n=2: An example for n=3: The case for n=4: (A,B,C,...,G) is a permutation of (1,2,3,...,7) satisfying A+C=B, C+E=D, E+G=F. This puzzle is called "連環の数" in Japanese language. (Pronunciation is "ren-kan-no-kazu". "ren" means "linked". "kan" means "circle" or "ring". "no" means "of" etc. "kazu" means "number".) I asked the site owner who uses this Japanese name, but he said he did not know any references and English name. I would like to discuss or to publish a certain combinatorial property of this puzzle in English, but I am having trouble finding the name and references. Thank you for reading. PostScript: Unfortunately, it was determined to be off-topic. Can I repost the same question on puzzling.SE? REPLY [8 votes]: Tsuyoshi Uema refers to it as a "renkan" puzzle, and has written some code to solve small instances at http://prolog.web.fc2.com/src_017_renkan.html<|endoftext|> TITLE: How do we give mathematical meaning to 'physical dimensions'? QUESTION [26 upvotes]: In so-called 'natural unit', it is said that physical quantities are measured in the dimension of 'mass'. For example, $\text{[length]=[mass]}^{-1}$ and so on. In quantum field theory, the dimension of coupling constant is very important because it determines renormalizability of the theory. However, I do not see what exactly the mathematical meaning of 'physical dimension' is. For example, suppose we have self-interaction terms $g_1\cdot \phi\partial^\mu \phi \partial_\mu \phi$ and $g_2 \cdot \phi^4$, where $\phi$ is a real scalar field, $g_i$ are coupling constants and we assume $4$ dimensional spacetime. Then, it is stated in standard physics books that the scalar field is of mass dimension $1$ and so $g_1$ must be of mass dimension $-1$ and $g_2$ is dimensionless. But, these numbers do not seem to play any 'mathematical' role. To clarify my questions, What forbids me from proclaiming that $\phi$ is dimensionless instead of mass dimension $1$? What is the exact difference between a dimensionless coupling constant and a coupling constant of mass dimension $-1$? These issues seem very fundamental but always confuse me. Could anyone please provide a precise answer? REPLY [54 votes]: Mathematically, the concept of a physical dimension is expressed using one-dimensional vector spaces and their tensor products. For example, consider mass. You can add masses together and you know how to multiply a mass by a real number. Thus, masses should form a one-dimensional real vector space $M$. The same reasoning applies to other physical quantities, like length, time, temperature, etc. Denote the corresponding one-dimensional vector spaces by $L$, $T$, etc. When you multiply (say) some mass $m∈M$ and some length $l∈L$, the result is $m⊗l∈M⊗L$. Here $M⊗L$ is another one-dimensional real vector space, which is capable of “storing” physical quantities of dimension mass times length. Multiplicative inverses live in the dual space: if $m∈M$, then $m^{-1}∈M^*$, where $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} M^*=\Hom(M,\R)$. The element $m^{-1}$ is defined as the unique element in $M^*$ such that $m^{-1}(m)=1$, where $-(-)$ denotes the evaluation of a linear functional on $M$ on an element of $M$. Observe that $m ⊗ m^{-1} ∈ M⊗M^* ≅ \R$, where the latter canonical isomorphism sends $(f,m)$ to $f(m)$, so $m^{-1}$ is indeed the inverse of $m$. Next, you can also define powers of physical quantities, i.e., $m^t$, where $m∈M$ is a mass and $t∈\R$ is a real number. This is done using the notion of a density from differential geometry. (The case $\def\C{{\bf C}} t\in\C$ works similarly, but with complex one-dimensional vector spaces.) In order to do this, we must make $M$ into an oriented vector space. For a one-dimensional vector space, this simply means that we declare one out of the two half-rays in $M∖\{0\}$ to be positive, and denote it by $M_{>0}$. This makes perfect sense for physical quantities like mass, length, temperature. Once you have an orientation on $M$, you can define $\def\Dens{\mathop{\rm Dens}} \Dens_d(M)$ for $d∈\R$ as the one-dimensional (oriented) real vector space whose elements are equivalence classes of pairs $(a,m)$, where $a∈\R$, $m∈M_{>0}$. The equivalence relation is defined as follows: $(a,b⋅m)∼(a b^d,m)$ for any $b∈\R_{>0}$. The vector space operations are defined as follows: $0=(0,m)$ for some $m∈M_{>0}$, $-(a,m)=(-a,m)$, $(a,m)+(a',m)=(a+a',m)$, and $s(a,m)=(sa,m)$. It suffices to add pairs with the same second component $m$ because the equivalence relation allows you to change the second component arbitrarily. Once we have defined $\Dens_d(M)$, given $m∈M_{>0}$ and $d∈\R$, we define $m^d∈\Dens_d(M)$ as the equivalence class of the pair $(1,m)$. It is easy to verify that all the usual laws of arithmetic, like $m^d m^e = m^{d+e}$, $m^d n^d = (mn)^d$, etc., are satisfied, provided that multiplication and reciprocals are interpreted as explained above. Using the power operation operations we just defined, we can now see that the equivalence class of $(a,m)$ is equal to $a⋅m^d$, where $m∈M_{>0}$, $m^d∈\Dens_d(M)_{>0}$, and $a⋅m^d∈\Dens_d(M)$. This makes the meaning of the equivalence relation clear. In particular, for $d=-1$ we have a canonical isomorphism $\Dens_{-1}(M)→M^*$ that sends the equivalence class of $(1,m)$ to the element $m^{-1}∈M^*$ defined above, so the two notions of a reciprocal element coincide. If you are dealing with temperature without knowing about the absolute zero, it can be modeled as a one-dimensional real affine space. That is, you can make sense of a linear combination $$a_1 t_1 + a_2 t_2 + a_3 t_3$$ of temperatures $t_1$, $t_2$, $t_3$ as long as $a_1+a_2+a_3=1$, and you don't need to know about the absolute zero to do this. The calculus of physical quantities can be extended to one-dimensional real affine spaces without much difficulty. None of the above constructions make any noncanonical choices of physical units (such as a unit of mass, for example). Of course, if you do fix such a unit $μ∈M_{>0}$, you can construct an isomorphism $\R→\Dens_d(M)$ that sends $a∈\R$ to $aμ^d$, and the above calculus (including the power operations) is identified with the usual operations on real numbers. In general relativity, we no longer have a single one-dimensional vector space for length. Instead, we have the tangent bundle, whose elements model (infinitesimal) displacements. Thus, physical quantities no longer live in a fixed one-dimensional vector space, but rather are sections of a one-dimensional vector bundle constructed from the tangent bundle. For example, the volume is an element of the total space of the line bundle of 1-densities $\Dens_1(T M)$, and the length is now given by the line-bundle of $λ$-densities $\Dens_λ(T M)$, where $λ=1/\dim M$.<|endoftext|> TITLE: Amenable action intuition QUESTION [7 upvotes]: Let $\Gamma$ be a discrete group and $A$ be a $C^*$-algebra. Consider an action $\alpha: \Gamma \to \operatorname{Aut}(A)$. There is a notion of amenability for such an action (see e.g. Brown and Ozawa's book "C*-algebras and finite-dimensional approximations", section 4.3), but how should one think intuitively about an amenable action? How was the definition originally motivated? REPLY [14 votes]: It is impossible to understand the motivation behind the definition of an amenable action without first understanding the definition of amenable groups, so let me first talk about groups (for simplicity, just countable ones). Finite groups are precisely the ones for which there is an invariant probability measure (the uniform one). Infinite groups also have invariant measures (the counting ones), but they can not be normalized and made finite (for topological groups one should talk about the Haar measures instead of the counting ones). There are two natural ways of relaxing the above condition of the existence of a finite invariant measure, and therefore of extending the class of finite groups. One consists in extending the space of measures and looking for invariant objects in this extended space. In our context it amounts to passing from the "usual" sigma-additive probability measures to finitely additive ones (called means). This is precisely the original 1929 definition of von Neumann: a group is amenable if it has an invariant mean. The other way consists in keeping measures, but replacing, as a trade off, exact invariance with an approximate one. In spite of being more constructive, this approach was only developed in the 50s, and turned out to be equivalent to the original one (I skip the historical details and just mention that the most important contributor was Day who also coined the modern term "amenability"). More precisely, a group $G$ is amenable if and only if it carries an approximately invariant sequence of probability measures $m_n$, i.e., such that $$ \| g m_n - m_n \| \to 0 \qquad\forall\,g\in G \;. $$ This is the Day condition (sometimes one also adds the name of Reiter, but this is not really correct from the historical point of view). In order to pass to group actions, it is actually more convenient to go a bit further and first talk about groupoids instead. They are like groups with the only difference that the multiplication is defined only subject to a certain condition. Very succinctly, a groupoid is a "small category with invertible morphisms". In plain language it means that there is a set of points ("objects") and a set of arrows ("morphisms") between objects endowed with a composition operation with the same properties as for groups and with the only additional requirement that for composing two arrows the endpoint of the first must coincide with the starting point of the second. In particular, a part of the formal definition of a groupoid is the presence of a target map from morphisms to objects which assigns to any arrow its endpoint. Therefore, over each object $x$ of a groupoid $\mathbf G$ there is the corresponding fiber $\mathbf G^x$ of the target map which consists of all arrows ending at $x$. A group $G$ can be considered as the groupoid $\mathbf G$ for which there is only one object $\bullet$, so that all arrows are composable, and the fiber $\mathbf G^\bullet$ is just the whole group $G$. One can easily see that the groupoid moves around the fibers of the target map, namely, given any arrow $\mathbf g$ with the starting point $x$ and the endpoint $y$, one can compose $\mathbf g$ with any arrow whose endpoint is $x$, and the result will be an arrow whose endpoint is $y$. Formally, $$ \mathbf g \mathbf G^x = \mathbf G^y \qquad \forall\,\mathbf g:x\to y \;. $$ One can now talk about the systems $(m^x)$ of measures on the target fibers of a groupoid invariant in the sense that $\mathbf g m^x=m^y$. They are called Haar systems, being a straighforward generalization of the Haar measures in the group case. In precisely the same way as for groups, there are two ways of relaxing the condition of the existence of a finite Haar system: either to talk about systems of means instead of measures, or to replace exact invariance with an approximate one and to require the existence of a sequence of probability measures $m_n^x$ on the fibers $\mathbf G^x$ such that $$ \| \mathbf g m_n^x - m_n^y \| \to 0 \qquad\,\forall\mathbf g:x\to y \;. $$ In the same way as for groups, these two definitions happen to be equivalent (Renault 1980), and produce what is called amenable groupoids (I skip some technical details here). Now back to actions. An action of a group $G$ on a space $X$ gives rise to the action groupoid whose objects are the points from $X$, and the arrows are the triples $(x, g ,y)$ with $y=gx$, so that for any $y\in X$ the corresponding fiber $\mathbf G^y$ can be identified just with the group $G$, and the aforementioned action of the groupoid on the fibers of the target map amounts just to the left action of the group on itself. Thus, the amenability of the action groupoid amounts precisely to the existence of a system of means on the group indexed by the action space and invariant with respect to the group action, or, in the approximate terms, to the condition that there exists a sequence of systems $(m_n^x)$ of probability measures on $G$ indexed by the action space, and such that $$ \| g m_n^x - m^{gx}_n \| \to 0 \;. $$ This is precisely the definition from Brown - Ozawa you are asking about (I skip the details concerning the difference between the definitions in the topological and in the measure "worlds"). One more comment. The way the notion of an amenable action was first introduced by Zimmer in 1977 is actually different. His motivation was the fixed point characterization of amenable groups (the existence of a fixed point for any continuous affine action on a compact) - this is the main application of amenability, but is not so convenient for establishing amenability - and given in pretty convoluted terms of existence of invariant sections for certain Banach bundles over the action space (this is the reason some of the papers on amenable action from that period are so excessively long). For more historical and motivational details see the book by Anantharaman-Delaroche and Renault or a recent preprint of Bühler and Kaimanovich.<|endoftext|> TITLE: Complexity of rectangular matrix multiplication QUESTION [5 upvotes]: I am interested in the complexity of multiplying two matrices $A$ and $B$, i.e. to compute $AB$. From [Le Gall and Urrotia], I know that: if $A$ and $B$ are square-matrices of size $n$, then this can be done in $O(n^{\omega})$ where $\omega\approx 2.372$. if $A$ has size $n\times n^{k}$ and $B$ has size $n^k \times n$ this can be done in $O(n^{\omega(k)})$ with $\omega(k)<\omega$ for $k<1$ (typically, $\omega(k)=2$ for $k\le0.3$). I am wondering if there an efficient algorithm when $A$ has size $n^k\times n$ and $B$ has size $n\times n$ (or when $B$ has size $n\times n^{k}$), for $k\in(0,1)$. Remarks: when I say "efficient" algorithm, I mean an algorithm that a complexity smaller than the naïve $O(n^{2+k})$ algorithm. I am not refering to an actual implementation, just the existence of an algorithm. I suppose that additing or multiplying two entries of a matrix is $O(1)$. [Le Gall and Urrotia] Improved Rectangular Matrix Multiplication using Powers of the Coppersmith-Winograd Tensor by François Le Gall† Florent Urrutia REPLY [3 votes]: Thanks to Jukka's answer, I found the paper https://www.sciencedirect.com/science/article/pii/S0885064X98904769, which cites a result of http://www.mathnet.ru/links/70088989aead859688ca36af8d30e329/rm5125.pdf (unfortunately in Russian). If I understand this paper correctly (and in particular around Equation (2.7)), this paper claims that for any given exponent $r,s,t>0$, the complexity of the multiplication $$ (i.e. multiply a $n^r\times n^s$ matrix by a $n^s\times n^t$) is independent by permutation of $r,s,t$. This implies for my original question that the complexity of the multiplication $$ is upper bounded by $\omega(k)$.<|endoftext|> TITLE: Asymptotic estimate for an integral involving the squared modulus of the Riemann zeta function QUESTION [6 upvotes]: For any fixed $\frac{1}{2} < \sigma < 1$, let $$\int_0^T \frac{|\zeta(\sigma+it)|^2}{\sqrt{1+t^2}} \ dt = O(T^\theta), \qquad T \uparrow \infty. $$ It is clear that $\theta > 0$, since we have the classical asymtotic $$\int_0^T \frac{|\zeta(\sigma+it)|^2}{T} \ dt \sim \zeta(2\sigma), \qquad T \uparrow \infty. $$ Is there more precise information about the value of $\theta$? REPLY [9 votes]: Let us introduce the notation $$M(T):=\int_0^T|\zeta(\sigma+it)|^2\,dt.$$ Then $$\int_0^T \frac{|\zeta(\sigma+it)|^2}{\sqrt{1+t^2}} \,dt=\int_0^T\frac{dM(t)}{\sqrt{1+t^2}}=\frac{M(T)}{\sqrt{1+T^2}}+\int_0^T\frac{tM(t)}{(1+t^2)^{3/2}}\,dt$$ by writing this as a Riemann-Stieltjes integral and then integrating by parts. Using that $M(t)$ is asymptotically $\zeta(2\sigma)t$, we conclude that the left-hand side is asymptotically $\zeta(2\sigma)\log T$.<|endoftext|> TITLE: Does derived hom commute with homotopy limits? QUESTION [8 upvotes]: Suppose that $\mathcal{V}$ is a symmetric monoidal model category, and that $\mathcal{C}$ is a $\mathcal{V}$-enriched model category. Write $\Bbb{R}\!\operatorname{Hom}(-,-)$ for the derived Hom functor $$ \Bbb{R}\!\operatorname{Hom}(-,-) : \operatorname{Ho}(\mathcal{C})^{\textrm{op}}\times\operatorname{Ho}(\mathcal{C})\to\mathcal{V}. $$ Suppose also that $\mathcal{C}$ (respectively $\mathcal{V}$) has functorial cofibrant and fibrant replacement functors, denoted $Q_\mathcal{C}$ and $R_\mathcal{C}$ (respectively $Q_\mathcal{V}$ and $R_\mathcal{V}$), so that $\Bbb{R}\!\operatorname{Hom}(X,Y)$ may be computed as $\mathcal{C}(Q_\mathcal{C}(X),R_\mathcal{C}(Y)).$ Let $X\in\mathcal{C},$ let $F : D\to\mathcal{C}$ be a diagram in $\mathcal{C},$ and suppose that $\operatorname{holim}F$ and $\operatorname{holim}\Bbb{R}\!\operatorname{Hom}(X,-)\circ F$ exist. Question: Is it true that $$ \operatorname{holim}\left(\Bbb{R}\!\operatorname{Hom}(X,-)\circ F\right)\simeq\Bbb{R}\!\operatorname{Hom}(X,\operatorname{holim}F)? $$ If this is true, what is the proof (preferably a proof using the language of model categories, as opposed to a proof using $\infty$-categories). If this is not true in general, I would be interested in knowing what general hypotheses could be placed on the objects/functors/categories involved which would guarantee that there is such an equivalence. I've heard that in the language of $\infty$-categories, we do have commutativity of limits and homs, which makes me suspect that this statement at the level of model categories should hold. However, I'm not sure if the translation is so direct, and it makes me a bit suspicious that I have not been able to find a statement like the above anywhere in my searches of the literature, and my attempts to prove the statement have not been fruitful. Thanks! REPLY [10 votes]: Yes, this is always true. Replacing $X$ by its cofibrant replacement if necessary, we can assume $X$ to be cofibrant. In this case, $\def\Hom{\mathop{\rm Hom}} \Hom(X,-)\colon C→V$ is a right Quillen functor. The right derived functor of this right Quillen functor computes the derived hom $\def\RHom{\mathop{\rm RHom}} \RHom(X,-)$. We can now conclude by invoking the fact that right derived functors of right Quillen functors preserve homotopy limits. The latter fact can be proved as follows. For simplicity of exposition, suppose that the injective model structure on D-indexed diagrams valued in $C$ and $V$ exists. (This is true in almost all practical cases, and the assumption can be eliminated anyway by working with more complicated models for homotopy limits.) Replacing $F$ by its fibrant replacement if necessary, we can assume $F$ to be injectively fibrant. Then $\RHom(X,-)∘F$ is also injectively fibrant. Homotopy limits of injectively fibrant diagrams can be computed as ordinary limits. Thus, the statement becomes $$\lim(\Hom(X,-)∘F) ≅ \Hom(X,\lim F),$$ which is a true 1-categorical fact.<|endoftext|> TITLE: Maximizing sum of vector norms QUESTION [7 upvotes]: Given matrices $A, B \in \mathbb{R}^{n\times n}$, I would like to solve the following optimization problem, $$\begin{array}{ll} \underset{v \in \mathbb{R}^n}{\text{maximize}} & \|Av\|_2+\|Bv\|_2\\ \text{subject to} & \|v\|_2 = 1\end{array}$$ I'm hoping to solve this with some sort of convex optimization approach, such as SDP or SOCP. When $B=0$, the problem reduces to simply asking for the largest singular value of $A$, which can be solved by an SDP (see here for instance although it's very well known). It can also be computed with SVD of course. I've tried lots of different approaches but can't seem to write it in a convex way. I would be okay with an SVD-like solution, that is, one that is iterative not a convex program, but I would much prefer a convex program because ultimately I would like to use this as an inner program to something else, ideally. As a note on one attempt that got a decent ways, I did manage to establish that it was equivalent to the problem, $$\begin{array}{ll} \underset{u, v \in \mathbb{R}^n}{\text{maximize}} & \|A^T u + B^T v\|_2\\ \text{subject to} & \|u\|_2 = \|v\|_2 = 1\end{array}$$ REPLY [5 votes]: I have no doubt that someone will come with some brighter idea but here are my 2 cents anyway. If you don't aim at something very fast, I would just use the inequality $(a+b)^2\le (ta^2+(1-t)b^2)(t^{-1}+(1-t)^{-1})=F_t(a,b)$ and try to find $\min_{t\in[0,1]}\max_v F_t(\|Av\|,\|Bv\|)$. The maximum inside is just $(t^{-1}+(1-t)^{-1})$ times the maximal eigenvalue of $tA^*A+(1-t)B^*B$ and the function $F_t$ is convex in $t$, so the maximum in $v$ is also convex, which makes simple one-dimensional tools for finding the minimum like bisection quite efficient. The reason it will work is simple: the maximal unit eigenvector $v$ will move continuously (when you have a multiple eigenvalue, you can slowly move it from one limiting position to the other keeping the parameter $t$ constant), so there will be a moment in that process where $\|Av\|/\|Bv\|=(1-t)/t$. At that moment the inequality in the Cauchy-Schwarz will become equality, so the sum of the norms will achieve the upper bound coming from the quadratic relaxation (I assume that $A,B$ are both non-degenerate in this argument). Thus, the minimax we created really equals the maximin. Note that it will all work nicely if you want just the value of the maximum, not the maximizing vector itself. The difficulty with the latter is that you may achieve the minimum in $t$ where the maximal eigenvalue is multiple, in which case not every eigenvector will be good but only the one for which the ratio of the norms is just right. Fortunately, generically it should not happen. However, you should be ready for this nuisance.<|endoftext|> TITLE: Equivariant implicit function theorem QUESTION [5 upvotes]: Let $f:\mathbb{R}\times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be a smooth function and $G\subset \operatorname{SO}(n)$ be a $1$-dimensional compact Lie group (diffeomorphic to the circle). Moreover let $G$ act on $\mathbb{R}^{n}$ by standard left multiplication. We assume that $f$ is equivariant with respect to $G$, i.e. for all $g\in G$ and all $(t,x)\in \mathbb{R}\times \mathbb{R}^{n}$ we have $f(t,g\cdot x) = g\cdot f(t,x)$. Let now $(t_{0},x_{0}) \in \mathbb{R}\times \mathbb{R}^{n}$ such that $x_{0} \not= 0$, $f(t_{0},x_{0}) = 0$ and $$\ker \left ( \frac{\partial f}{\partial x}(t_{0},x_{0}) \right ) = T_{x_{0}}(G\cdot x_{0}),$$ i.e. the kernel of the Jacobian of $f$ is only in the direction of the action. QUESTION: Is there any version of the implicit function theorem in this setting? Does one need more additional conditions to be able to construct near-by solutions? If yes, which conditions are these? Thanks in advance. REPLY [6 votes]: The equivariant version of the implicit function theorem is the following. Let $f: \mathbb{R}^p \times \mathbb{R}^n \to \mathbb{R}^m$ be a smooth function (possibly only defined on open neighborhoods) which is equivariant with respect to the action of a compact Lie group $G$ on $\mathbb{R}^n$ and $\mathbb{R}^m$. Let $(t_0, x_0)$ be such that $f(t_0, x_0) = 0$. Assume that the stabilizer of $x_0$ is trivial and that $0$ is a fix point for the action on $\mathbb{R}^m$ (these assumptions are not essential but simplify the argument, see below). Consider the so-called deformation complex $$\mathfrak{g} \to \mathbb{R}^n \to \mathbb{R}^m,$$ where the first arrow is the action of the Lie algebra at the point $x_0$, i.e. $\xi \mapsto \xi \,. x_0$ and the second arrow is the differential of $f$ at $(t_0, x_0)$ with respect to the second slot (i.e. the Jacobian). If this complex is exact, then there exist a smooth function $x: \mathbb{R}^p \to \mathbb R^n$ such that $$ \{ (t, g \cdot x(t)) | t \in \mathbb R^p, g \in G \} = f^{-1}(0). $$ (I'm a bit sloppy here and in the proof below: everything needs to be restricted to open neighborhoods of $t_0$, $e \in G$ and $0 \in \mathbb{R}^m$ etc). Remark: Your assumption about the derivative is equivalent to the exactness of the complex in the first arrow. However, since you assume that $n = m$, the complex is never exact in the second arrow. What you could do is apply this result to the function $pr \circ f$, where $pr$ is the projection onto the image of the Jacobian. Proof: Since $G$ is compact, and the action is free at $x_0$, there exist slice coordinates around $x_0$, i.e. there is a map $\iota: \mathbb{R}^{n-d} \to \mathbb{R}^n$ such that $\iota(0) = x_0$ and such that $\chi: G \times \mathbb{R}^{n-d} \to \mathbb{R}^n, (g, y) \mapsto g \cdot \iota(y)$ is a local diffeo (here $d$ is the dimension of $G$). Define the map $F: \mathbb{R}^p \times \mathbb{R}^{n-d} \to \mathbb{R}^{n}$ by $F(t, y) = f(t, \iota(y))$. The assumption about the exactness of the deformation complex is equivalent to the invertibility of the Jacobian of $F$. Thus, using the ordinary implicit function theorem, there exist $y(t)$ such that $F(t, y(t)) = 0$ and every such point in the zero level set is of this form. Set $x(t) = \iota(y(t))$ and the claim follows as $\chi$ is a local diffeo. Final remark: The statement generalizes directly to actions on manifolds, and properness of the action is enough (instead of compactness of $G$). In fact, they generalize even to the infinite-dimensional setting. Moreover, the assumptions about the stabilizers of $x_0$ and $f(t_0, x_0)$ can be relaxed. You can find the details in my PhD thesis https://arxiv.org/abs/1909.00744 and in the paper https://arxiv.org/abs/2010.10165.<|endoftext|> TITLE: Does there exist a definition of equivalence of functors? QUESTION [5 upvotes]: I have two functors $F_1,F_2$ from a category $C$ into two distinct categories $D_1,D_2$. I would like to say that $F_1$ and $F_2$ are equivalent if there exists a commutative square $\require{AMScd}$ \begin{CD} C @>F_1>> D_1\\ @V F_2 V V @VV E_1 V\\ D_2 @>>E_2> D \end{CD} of functors such that the $E_i$ are equivalences. Can one say that this notion is well-known?:) Do there exist any similar notions in the literature? Upd. Possibly, it would be better to reverse the arrows here, that is, to demand that there exists a functor $F$ (that is bijective on objects) and two equivalences of categories $E_1$ and $E_2$ such that $F_i=E_i\circ F$. REPLY [8 votes]: It's worth first understanding the 1-categorical analogue: what would it mean for two arrows $f \colon X \to Y$ and $f' \colon X \to Y'$ in a category $C$, which share the same domain, to be isomorphic? One way of making sense of this is by saying that they are isomorphic in the coslice category $X/C$, whose objects are pairs $(Y,f)$ where $Y$ is an object of $C$ and $f \colon X \to Y$ is an arrow; morphisms are commutative triangles. Your question is really about the analogue of the above when $C$ is not a 1-category but the 2-category $Cat$. Now there are multiple inequivalent ways of defining a "coslice 2-category" under an object of a 2-category, depending on whether you want the 2-morphisms to be invertible, or identities, or if they are not invertible, in which case you need to choose a direction for them. You want invertible 2-cells, which is also what people will generally expect by default if you only say "coslice 2-category". More precisely, the following are equivalent: there exists a square as in your question which commutes strictly there exists a square as in your question which commutes up to a natural isomorphism there exists a commuting triangle which commutes up to a natural isomorphism However, asking for a triangle which commutes strictly is in general strictly stronger (and seems like a rather unnatural notion). https://ncatlab.org/nlab/show/slice+2-category<|endoftext|> TITLE: Euler numbers and permanent of matrices QUESTION [21 upvotes]: Motivated by Question 402249 of Zhi-Wei Sun, I consider the permanent of matrices $$e(n)=\mathrm{per}\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]_{1\le j,k\le n-1},$$ where $n$ is an odd integer greater than 1 and $ \operatorname{sgn} $ is the sign-function. When $n=3,5,7,$ the matrices are $$\left[ \begin {array}{cc} -1&0\\ 0&1\end {array} \right] , $$ $$ \left[ \begin {array}{cccc} 1&-1&-1&0\\ -1&-1&0&1 \\ -1&0&1&1\\ 0&1&1&-1\end {array} \right], $$ $$ \left[ \begin {array}{cccccc} 1&1&-1&-1&-1&0\\ 1&-1 &-1&-1&0&1\\ -1&-1&-1&0&1&1\\ -1&- 1&0&1&1&1\\ -1&0&1&1&1&-1\\ 0&1&1& 1&-1&-1\end {array} \right]. $$ Numerical computation indicates that $$e(n)=E(n-1)$$ for $3 \leq n \leq 23 $, where $E(n)$ are the Euler numbers. [e(3) = -1, E(2) = -1] [e(5) = 5, E(4) = 5] [e(7) = -61, E(6) = -61] [e(9) = 1385, E(8) = 1385] [e(11) = -50521, E(10) = -50521] [e(13) = 2702765, E(12) = 2702765] [e(15) = -199360981, E(14) = -199360981] [e(17) = 19391512145, E(16) = 19391512145] [e(19) = -2404879675441, E(18) = -2404879675441] [e(21) = 370371188237525, E(20) = 370371188237525] [e(23) = -69348874393137901, E(22) = -69348874393137901] Conjecture. For any odd integer $n>1$, $e(n)=E(n-1)$. Question 1. How to prove this conjecture? Edit. By T. Amdeberhan's recent MO post, $$\mathrm{det}\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]_{1\le j,k\le n-1}=(-1)^{\frac{n-1}{2}}=\sin(\frac{n\pi}{2})\neq 0,$$ we can consider the permanent of inverse matrices $$e'(n)=\mathrm{per}(A^{-1}),$$ where $A=\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]_{1\le j,k\le n-1}$ and $n$ is an odd integer greater than 1. When $n=3,5,7,$ $A^{-1}$ equals $$\left[ \begin {array}{cc} -1&0\\ 0&1\end {array} \right], $$ $$ \left[ \begin {array}{cccc} 1&-1&1&0\\ -1&-1&0&-1 \\ 1&0&1&1\\ 0&-1&1&-1\end {array} \right], $$ $$ \left[ \begin {array}{cccccc} 1&1&-1&1&1&0\\ 1&-1&- 1&1&0&-1\\ -1&-1&-1&0&-1&-1\\ 1&1&0 &1&1&1\\ 1&0&-1&1&1&-1\\ 0&-1&-1&1 &-1&-1\end {array} \right] $$ respectively. Numerical computation indicates that all elements of $A^{-1}$ are $-1, 0, 1 $ and $$e'(n)=E(n-1)$$ for $3 \leq n \leq 23 $. Added Conjecture 1. For any odd integer $n>1$, $e'(n)=E(n-1)$. The following are some properties of the $A$ and $A^{-1}$ that may be easy to prove: Added Conjecture 2. For any odd integer $n>1$, all elements of $A^{-1}$ are $-1, 0, 1 $. Added Conjecture 3. For any odd integer $n>1$, the Characteristic Polynomials of $A$ and $A^{-1}$ are $$ \mathrm{Char}(A)=\mathrm{Char}(A^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}}, $$ where $i^2=-1$ and $\mathrm{Char}(A)=\mathrm{det}(\lambda I-A)$. Question 2. How to prove these added conjectures? Edit. The connection between characteristic polynomials and recurrences of Euler numbers: Let $$\mathrm{Char}(A)|_{\lambda=iE}=\mathrm{det}(\lambda I-A)|_{\lambda=iE}={\frac { \left( iE+i \right) ^{n-1} + \left( iE-i \right) ^{n-1}}{2}}=0,$$ we deduce $$(E+1)^{n-1}+(E-1)^{n-1}=0,$$ which is the recurrence formula for the Euler numbers ($E^n≡E_n≡E(n)$ in symbolic notation). Question 3. Is this a coincidence? Edit. Let $B=\left[\operatorname{sgn} \left(\sin\pi\frac{j+k}n \right)\right]_{1\le j,k\le n-1}$ where $n>1$ is an integer. Zhi-Wei Sun and Nemo guessed in the comments that $$\mathrm{per}(B)=E(n-1)$$ for any integer $n>1$. Numerical calculations show that this is correct for $11$. When $n=3,5,7,$ $B$ and $B^{-1}$ are equal to $$\left[ \begin {array}{cc} 1&0\\ 0&-1\end {array} \right] , \left[ \begin {array}{cc} 1&0\\ 0&-1 \end {array} \right] ,$$ $$ \left[ \begin {array}{cccc} 1&1&1&0\\ 1&1&0&-1 \\ 1&0&-1&-1\\ 0&-1&-1&-1 \end {array} \right] , \left[ \begin {array}{cccc} 1&-1&1&0 \\ -1&1&0&-1\\ 1&0&-1&1 \\ 0&-1&1&-1\end {array} \right] ,$$ $$ \left[ \begin {array}{cccccc} 1&1&1&1&1&0\\ 1&1&1& 1&0&-1\\ 1&1&1&0&-1&-1\\ 1&1&0&-1& -1&-1\\ 1&0&-1&-1&-1&-1\\ 0&-1&-1& -1&-1&-1\end {array} \right] , \left[ \begin {array}{cccccc} 1&-1&1&-1 &1&0\\ -1&1&-1&1&0&-1\\ 1&-1&1&0&-1&1\\ -1&1&0&-1&1&-1\\ 1&0&-1&1&-1 &1\\ 0&-1&1&-1&1&-1\end {array} \right] $$ respectively. A natural question is: Does $B$ and $B^{-1}$ have the same properties as $A$ and $A^{-1}$ for any odd integer $n>1$. Added Conjecture 4. For any odd integer $n>1$, $$\mathrm{per}(B)=\mathrm{per}(B^{-1})=E(n-1)$$ and $$ \mathrm{Char}(B)=\mathrm{Char}(B^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}}. $$ Question 4. If $$\mathrm{per}(C)=\mathrm{per}(C^{-1})=E(n-1)$$ and $$\mathrm{Char}(C)=\mathrm{Char}(C^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}},$$ where $n$ is an odd number greater than $1,$ $C$ is an $(n-1)\times (n-1)$ matrix and all elements of $C$ are $-1, 0, 1 ,$ what can we say about $C$? REPLY [5 votes]: The conjecture and the added Conjectures 1 and 2 have been proved. See the preprint Proof of five conjectures relating permanents to combinatorial sequences by Fu, Lin and me available from http://arXiv.org/abs/2109.11506.<|endoftext|> TITLE: Can $[0,1]^4$ be partitioned into copies of $(0,1)^3$? QUESTION [32 upvotes]: Is there a partition of $[0,1]^4$ such that every member of the partition is homeomorphic to $(0,1)^3$? More generally, I would like to know for which values of $m$ and $n$ there is a partition of $[0,1]^n$ into topological copies of $(0,1)^m$. For such a partition to exist, it is necessary that $m < n$. Another simple observation is that if $[0,1]^n$ can be partitioned into copies of $(0,1)^m$, then $[0,1]^{n+k}$ can be partitioned into copies of $(0,1)^m$ for all $k$. (This is because $[0,1]^{n+k}$ can be partitioned into copies of $[0,1]^n$, each of which can then be further partitioned into copies of $(0,1)^m$.) I have solved the two simplest instances of this general problem already: $\bullet$ $[0,1]^2$ can be partitioned into copies of $(0,1)$ (shown below), and this implies $[0,1]^n$ can be partitioned into copies of $(0,1)$ for all $n > 1$, and $\bullet$ $[0,1]^3$ can be partitioned into copies of $(0,1)^2$, and this implies $[0,1]^n$ can be partitioned into copies of $(0,1)^2$ for all $n > 2$. The title question is the simplest instance of this more general problem for which I don't already know the answer. A partition of $[0,1]^2$ into copies of $(0,1)$: The first three pictures in the bottom row show three partition pieces. The last picture is what's left over after removing these three, and because the leftover bit is simply four copies of $(0,1)^2$ (up to homeomorphism), it's easy to partition the leftovers into copies of $(0,1)$. To get a partition of $[0,1]^3$ into copies of $(0,1)^2$, take the partition of $[0,1]^2$ into copies of $(0,1)$ illustrated above, and rotate everything about the vertical line through the center of the square, in the same plane as the square (the axis of symmetry for the M-shaped bit of the partition). Upon rotation, our square transforms into a cylinder, which is (homeomorphic to) $[0,1]^3$. Each of the first three partition pieces transforms into a copy of $(0,1)^2$. The leftovers in the rightmost picture transform into two copies of $(0,1)^3$ plus one copy of $S^1 \times (0,1)^2$, and it is easy to further partition these into copies of $(0,1)^2$. The same trick does not seem to work for stepping up another dimension to $(0,1)^3$ and $[0,1]^4$. Lastly, let me mention that if we reverse the roles of $(0,1)$ and $[0,1]$ in my general question, then the answer is known (and not too difficult to prove): there is a partition of $(0,1)^n$ into topological copies of $[0,1]^m$ if and only if $m < n$. (This is proved as Theorem 2.2(i) in this 1979 paper of Bankston and McGovern.) REPLY [8 votes]: Almost exactly the same picture can be drawn for the 2-dimensional and 3-dimensional cases (simpler than the OP's) and then it's easy to generalize to higher dimensions by induction. Here $I$ indicates the open interval and $\bar{I}$ the closed one. The general case $n\ge 4$ is now easy: as above one can decompose ${\bar I}^n$ as follows: $I^{n-1}\sqcup I^{n-1}\sqcup I^n \sqcup S^{n-2}\times I^2$ and $S^{n-2}$ itself can be decomposed as union of copies of $I^{n-3}$ by inductively using the decomposition of $\bar{I}^{n-2}$ joined to its complement in $S^{n-2}$ which is another copy of $I^{n-2}$ and therefore also a union of copies of $I^{n-3}$. NOTE: I didn't fully understand Ben Johnsrude's earlier answer, but I think that mine may be essentially the same as his.<|endoftext|> TITLE: Is there a concrete description of the nontorsion elements in the homotopy groups of spheres? QUESTION [11 upvotes]: By Serre's theorem, we know the only nontorsion parts of the homotopy groups of spheres occur as $\pi_n(S^n)$ and $\pi_{4n-1}(S^{2n})$. The first of these are trivial to describe, but the second have very interesting, symmetric incarnations, they are the generalised hopf fibrations, at least for $n=1,2,4$, associated to the real normed division algebras. Are there a similar explicit descriptions for representatives of these higher nontorsion elements too? Even if we don't have explicit descriptions, do we know anything about the values of the hopf invariants associated to these maps? REPLY [21 votes]: For $n \neq 1,2,4$, the minimal positive Hopf invariant of an element of $\pi_{4n-1}(S^{2n})$ is $2$. An explicit element of Hopf invariant $2$ can be constructed as follows: consider the attaching map $S^{4n-1} \to S^{2n} \vee S^{2n}$ of the $4n$-cell of the CW-complex $S^{2n} \times S^{2n}$ and compose it with the codiagonal $S^{2n} \vee S^{2n} \to S^{2n}$. That no elements of Hopf invariant $1$ exist outside the "Adams dimensions" $n = 1,2,4$ is known as the "Hopf invariant 1 problem". It was studied and solved by Adams. Later Adams and Atiyah found a shorter proof. For a very accessible write-up, see this REU paper.<|endoftext|> TITLE: Is the one-point compactification of $\mathbb{N}$ computably countable? QUESTION [17 upvotes]: The one-point compactification $\mathbb{N}_\infty$ of $\mathbb{N}$ is obtained from the discrete space $\mathbb{N}$ by adjoining a limit point $\infty$. It may be identified with the subspace of Cantor space $$ \mathbb{N}_\infty = \{ \alpha \in \{0,1\}^\mathbb{N} \mid \forall n \,.\, \alpha_n \geq \alpha_{n+1} \}. $$ Indeed, we may embed $\mathbb{N} \to \mathbb{N}_\infty$ by mapping $n$ to the sequence $$\overline{n} = \underbrace{1 \cdots 1}_n 0 0 \cdots,$$ and taking $\infty = 1 1 1 \cdots$. Classically of course adjoining a single point to a countable set has no effect on countability. How about the computable version? If we adjoin the new point as an isolated one then of course we again obtain a countable set. This question is about adjoining $\infty$ as a limit point in the sense of metric spaces. Let $\varphi$ be a standard enumeration of partial computable maps. Question: Do there exist a total computable map $q$ and a partial computable map $s$ such that: $\varphi_{q(n)} \in \mathbb{N}_\infty$ for all $n \in \mathbb{N}$ For all $k \in \mathbb{N}$, if $\varphi_k \in \mathbb{N}_\infty$ then $s(k)$ is defined and $\varphi_{q(s(k))} = \varphi_k$. The map $q$ realizes an enumeration $\mathbb{N} \to \mathbb{N}_\infty$, and $s$ the fact that $q$ is surjective. Clarification: The following map $q : \mathbb{N} \to \mathbb{N}_\infty$ comes to mind: $$q(n)(k) = \begin{cases} 1 & \text{if $T_n$ has not terminated within $k$ steps of execution}\\ 0 & \text{if $T_n$ has terminated within $k$ steps of execution} \end{cases} $$ However, it seems hard to get the corresponding map $s$ witnessing surjectivity of $q$. (I should say that $q$ works as a computable enumeration for yet a third way of adjoining a point to $\mathbb{N}$, namely $$\mathbb{N}_\bot = \{ S \subseteq \mathbb{N} \mid \forall i, j \in S \,.\, i = j \}.$$ We embed $n \in \mathbb{N}$ into $\mathbb{N}_\bot$ as a singleton $\{n\}$, while the extra point is $\emptyset$. Think of $\mathbb{N}_\bot$ as the set of enumerable subsets of $\mathbb{N}$ with at most one element. REPLY [15 votes]: The answer is no. Suppose that there are computable functions $q$ and $s$ as you describe. Let $k$ be a program that performs the following task. It starts enumerating $1$s at the start of the sequence until it discovers that $s(k)$ is defined. (We use the Kleene recursion theorem to know that there is such a self-referential program $k$.) Note that this must eventually happen, since otherwise $\varphi_k$ would be $\infty$, in which case $s(k)$ should be defined. When it finds that $s(k)$ is defined, then the program pauses the enumeration of its output and starts computing $\varphi_{q(s(k))}$. This will definitely produce an element of $\mathbb{N}_\infty$. And so the program waits until either it produces more $1$s than we have put on $\varphi_k$, in which case program $k$ switches to $0$s immediately, causing $\varphi_{q(s(k))}\neq\varphi_k$; or else $\varphi_{q(s(k))}$ produces a $0$, in which case we can let $\varphi_k$ produce all $1$s, again causing $\varphi_{q(s(k))}\neq\varphi_k$.<|endoftext|> TITLE: Is computer algebra or symbolic computation an active area of research? QUESTION [13 upvotes]: I'm interested in doing PhD in computer algebra or symbolic computation, and was wondering if this is an active area of research? Would this area of research also help me in the transition to industry after my PhD? I'm also based in the USA. I apologize if this question is kind of "bare", I'm a PhD student at a university that does computer algebra, and I just want to know what I'm getting into. Thanks! REPLY [7 votes]: Just answering to your question in the title: computer algebra is definitely an active area of research! For example, there is a huge research program across several German universities funded by the DFG (German version of the NSF): https://www.computeralgebra.de/sfb/ and https://oscar.computeralgebra.de/. Among all those people involved I’m sure you’ll find someone you know (by name) or who’s doing what you’re interested in.<|endoftext|> TITLE: Is there a universal way to construct a bimonoidal category structure on $\mathsf{PSh}(\mathcal{C})$ from such a structure on $\mathcal{C}$? QUESTION [5 upvotes]: The Day convolution monoidal category structure $(\mathsf{PSh}(\mathcal{C}),\circledast,\mathsf{h}_{\mathbf{1}_{\mathcal{C}}})$ on the category of presheaves of a monoidal category $(\mathcal{C},\otimes,\mathbf{1}_{\mathcal{C}})$ satisfies the following universal properties: In A universal property of the convolution monoidal structure, Im–Kelly prove that it is the free monoidal cocompletion of $\mathcal{C}$. Here, a monoidal category is monoidally cocomplete if it is cocomplete and the functors $A\otimes-$ and $-\otimes B$ preserve colimits for all $A,B\in\mathrm{Obj}(\mathcal{C})$. Im–Kelly then prove: The monoidal category $(\mathsf{PSh}(\mathcal{C}),\circledast,\mathsf{h}_{\mathbf{1}_{\mathcal{C}}})$ is the universal monoidally cocomplete category on $\mathcal{C}$ in that, given any monoidally cocomplete monoidal category $(\mathcal{D},\otimes,\mathbf{1}_{\mathcal{D}})$, precomposition with $よ\colon\mathcal{C}\hookrightarrow\mathsf{PSh}(\mathcal{C})$ defines an equivalence of categories $$よ^*\colon\mathsf{Fun}^{\otimes,\mathsf{strong}}_{\mathsf{cocont.}}(\mathsf{PSh}(\mathcal{C}),\mathcal{D})\longrightarrow\mathsf{Fun}^{\otimes,\mathsf{strong}}(\mathcal{C},\mathcal{D}).$$ That is, $(\mathsf{PSh}(\mathcal{C}),\circledast,\mathsf{h}_{\mathbf{1}_{\mathcal{C}}})$ is uniquely determined by the following requirements: The Yoneda embedding $よ\colon\mathcal{C}\hookrightarrow\mathsf{PSh}(\mathcal{C})$ is strong monoidal. $\circledast$ is cocontinuous in each variable. Given another monoidal category $(\mathcal{D},\otimes_{\mathcal{D}},\mathbf{1}_{\mathbf{D}})$, we have an equivalence of categories $$ \left\{ \begin{gathered} \text{symmetric strong monoidal}\\ \text{functors $\mathcal{C}\times\mathcal{D}\to\mathsf{Sets}$} \end{gathered} \right\} \cong \left\{ \begin{gathered} \text{symmetric strong monoidal}\\ \text{functors $\mathcal{D}\to\mathsf{PSh}(\mathcal{C})$} \end{gathered} \right\}, $$ natural in $\mathcal{D}$. The analogue of the first of these for bimonoidal categories, however, doesn't work. Question. Given a bimonoidal category $(\mathcal{C},\otimes,\oplus,\mathbf{1}_{\mathcal{C}},\mathbf{0}_{\mathcal{C}})$, is there a universal way to put a bimonoidal category structure on $\mathsf{PSh}(\mathcal{C})$? In particular, is there a bimonoidal category structure on $\mathsf{PSh}(\mathcal{C})$ such that, given another bimonoidal category $(\mathcal{D},\otimes_{\mathcal{D}},\oplus_{\mathcal{D}},\mathbf{1}_{\mathcal{D}},\mathbf{0}_{\mathcal{D}})$, we have an equivalence of categories $$ \left\{ \begin{gathered} \text{symmetric strong }\color{red}{\text{bi}}\text{monoidal}\\ \text{functors $\mathcal{C}\times\mathcal{D}\to\mathsf{Sets}$} \end{gathered} \right\} \cong \left\{ \begin{gathered} \text{symmetric strong }\color{red}{\text{bi}}\text{monoidal}\\ \text{functors $\mathcal{D}\to\mathsf{PSh}(\mathcal{C})$} \end{gathered} \right\}, $$ natural in $\mathcal{D}$? REPLY [3 votes]: $\def\C{\mathcal{C}}\def\Set{\mathsf{Set}}$ (What follows comes from a private chat with Todd Trimble) Notation. If $(\C,\otimes,\oplus)$ is a bimonoidal category, I will call $\otimes$ the multiplicative structure and $\oplus$ the additive structure; if $\oplus$ is the cocartesian monoidal structure, I will call $\C$ a 2-rig, following https://arxiv.org/abs/2103.00938. I am convinced that in general (=for a general bimonoidal category) little can be said, because one needs some compatibility between the multiplicative structure and co/products. Even when $\C$ is a 2-rig the most I can formulate until now is a Conjecture. When $\C$ is a 2-rig, the category $[\C^o,\Set]_\times$ of functors $\C^o \to \Set$ that are product-preserving (=sending coproducts in $\C$ to products in $\Set$) is the free 2-rig on $\C$. In order for this freeness property to be legitimate, the least we can ask is that $F\C=[\C^o,\Set]_\times$ is a 2-rig if the multiplicative structure is Day convolution; The Yoneda embedding $y : \C \to F\C$ is a morphism of 2-rigs. Unfortunately, I am still unable to prove that the Day convolution restricts to "models" of the "theory" $\C$ (it is a fruitful intuition to think of $\C$ like it was a Lawvere theory even if it's not, were it only because it's easier to query google with questions ;-) ) Update: I couldn't because it's not true, but it falls very short from being true, in the sense that the conjecture is "true up to reflecting the monoidal structure": observe that $[\C^o,\Set]_\times$ has many desirable properties for the free cocomplete 2-rig on $\C$: $[\C^o,\Set]_\times$ is a cocomplete[¹], reflective subcategory of the entire $[\C^o,\Set]$. the yoneda embedding $y : \C \to [\C^o,\Set]$ clearly factors through $[\C^o,\Set]_\times$. $F\C=[\C^o,\Set]_\times$ has the following universal property, if $\mathcal D$ is cocomplete: $$\{\text{cocontinuous } g : F\C \to\mathcal D\}\cong \{\text{coproduct preserving } h : \C \to \mathcal D\}$$ Another useful universal property for $[\C^o,\Set]_\times$ is that it is the cocompletion of $\C$ under sifted colimit. All these facts turn out to be useful to establish that $[\C^o,\Set]_\times$ is the free 2-rig over $\C$: the coend that expresses the usual Day convolution formula is to be interpreted in a way that we first take the coend in the usual category of presheaves, but then to that apply the reflection functor $r$ that is left adjoint to the full inclusion. So, it's not true that the ordinary Day convolution takes a pair of product-preserving functors to a product-preserving functor; you have to sheafify. After you do that, everything falls into place. Look how neat everything becomes! Suppose we have a functor $F$ in $[\C^o,\Set]_\times$; by 4 above, it's a sifted colimit of representables $\C(-, c)$. Since we assume $\C$ is a 2-rig, we have a composite of (finite) coproduct-preserving functors $$Y : \C \stackrel{C \otimes -}{\to} \C \stackrel{y}{\to} [\C^o,\Set]_\times$$ under the equivalence of point 3, this becomes a colimit preserving functor $[\C^o,\Set]_\times \to [\C^o,\Set]_\times$, exactly the reflected convolution $\C(-,C)\ast \_$. But since the convolution we want to build must be cocontinuous, it is uniquely determined by this construction! The missing detail is some result ensuring that $\ast$ is a monoidal structure. All in all I expect this to be a consequence of a theorem about transport of monoidal structures $\otimes$ (Day convolution of presheaves) into $\ast$ ("reflected" Day convolution) given some lax monoidality assumptions on $r$, so I won't enter the details. Let me just add a final neat detail: the coend formula expressing the Day convolution product is a reflexive coequalizer; a reflexive coequalizer is a sifted colimit, so it can be interpreted as the usual, pointwise colimit in Set. Moreover, the sums = coproducts involved in the coend formula are filtered colimits of finite coproducts; again, filtered colimits are sifted colimits. So the only "re-interpretation" involved with the reflector resides in the way you compute finite coproducts in models, i.e.: you don't compute finite coproducts set-wise, but at the level of models, and every other colimit as usual. [¹] But colimits in $[\C^o,\Set]_\times$ are not computed as colimits in the presheaf category (thikn again to the case of Lawvere theories and coproducts of monoids...), and this will be crucial, because for example, in $[\C^o,\Set]_\times$ the object $\C(-,A+B)$ has the universal property of the coproduct $\C(-,A)+\C(-,B)$.<|endoftext|> TITLE: Is there a degenerate simplex in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths? QUESTION [16 upvotes]: The Cayley-Menger determinant gives the squared volume of a simplex in $\mathbb{R}^n$ as a function of its $n(n+1)/2$ edge lengths: $$v_n^2 = \frac{(-1)^{n+1}}{(n!)^2 2^n} \begin{vmatrix} 0&d_{01}^2&d_{02}^2&\dots&d_{0n}^2&1\\ d_{01}^2&0&d_{12}^2&\dots&d_{1n}^2&1\\ d_{02}^2&d_{12}^2&0&\dots&d_{2n}^2&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ d_{0n}^2&d_{1n}^2&d_{2n}^2&\dots&0&1\\ 1&1&1&\dots&1&0 \end{vmatrix}$$ where $d_{i j}$ is the distance from vertex $i$ to vertex $j$ of the simplex. For the regular simplex with all edge lengths 1, i.e. $d_{i j}=1$ for all $i,j$, we have: $$(n!)^2 2^n v_n^2 = (-1)^{n+1} \begin{vmatrix} 0&1&1&\dots&1&1\\ 1&0&1&\dots&1&1\\ 1&1&0&\dots&1&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 1&1&1&\dots&0&1\\ 1&1&1&\dots&1&0 \end{vmatrix} = n+1$$ This follows from the fact that the $(n+2)\times(n+2)$ matrix here has eigenvector $(1,1,1,\dots,1)$ with eigenvalue $n+1$, along with $n+1$ eigenvectors of the form $(1,0,0,\dots,-1,0,0,\dots)$ with eigenvalue $-1$. Any odd integer $2k+1$ has a square that is equal to 1 modulo 8, since: $$(2k+1)^2 = 4k(k+1)+1$$ and either $k$ or $k+1$ must be even. It follows that, for any simplex whose edge lengths are all odd integers, the quantity $(n!)^2 2^n v_n^2$ must equal $n+1$ modulo 8. When $n+1$ is not a multiple of 8, this means a simplex in $\mathbb{R}^n$ whose edge lengths are all odd integers cannot be degenerate, i.e. it cannot have zero volume. However, this does not settle the same question when $n=8k-1$. Are there known examples of degenerate simplices in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths? Or is there a proof for their nonexistence that completes the proof that applies in other dimensions? Acknowledgement: This question arose from a discussion on Twitter between Thien An and Ian Agol. Edited to add: For all cases where $n = 7$ mod 16, it is possible to rule out a degenerate simplex by working modulo 16, where any squared odd integer must equal either 1 or 9. Computing the determinant when $x$ is added to any single squared edge length gives a quadratic in $x$ that has even coefficients for $x$ and $x^2$ (given that all the original entries are integers), from which it follows that adding 8 to any squared edge length preserves the determinant modulo 16. Since the determinant when all squared edge lengths are equal to 1 is $n+1$, changing any number of the squared edge lengths from 1 to 9 can never yield a determinant divisible by 16. REPLY [12 votes]: This is answered in a paper by R. L. Graham, B. L. Rothschild & E. G. Straus "Are there $n+2$ Points in $E_n$ with Odd Integral Distances?". Such simplexes exist iff $n+2 \equiv 0 \pmod {16}$. They also consider the related problem of integral distances relatively prime to 3 and 6.<|endoftext|> TITLE: An infinite series involving the mod-parity of Euler's totient function QUESTION [6 upvotes]: Can you prove or disprove the following claim: First, define the function $\xi(n)$ as follows: $$\xi(n)=\begin{cases}-1, & \text{if }\varphi(n) \equiv 0 \pmod{4} \\ 1, & \text{if }\varphi(n) \equiv 2 \pmod{4} \\ 0, & \text{if otherwise } \end{cases}$$ where $\varphi(n)$ denotes Euler's totient function. Then, $$\frac{\pi^2}{72}=\displaystyle\sum_{n=1}^{\infty}\frac{\xi(n)}{n^2}$$ The SageMath cell that demonstrates this claim can be found here. REPLY [9 votes]: Reinforcing Nemo's negative answer. $$ \sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} - \sum_{n=10^4+1}^\infty \frac{1}{n^2} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq \sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} + \sum_{n=10^4+1}^\infty \frac{1}{n^2} $$ The first and third of these are directly computable: $$ 0.13712{\dots} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq 0.13732\dots \text{.} $$ However, $$ \frac{\pi^2}{72} = 0.137077{\dots} \text{,} $$ which does not fit between those bounds. (Aside: There may be smaller cutoffs than $10^4$ that exhibit this negative result. $10^3$ does not.)<|endoftext|> TITLE: Is this set dense in [0,+∞)? QUESTION [11 upvotes]: We define $A=\{ \frac{c}{rad(abc)}: a, b > 0, c=a+b, gcd(a, b)=1 \}$. Is the set $A$ dense in $[0, +\infty)$? Does $\overline{A}$ have interior? Here $\overline{A}$ is the closure of $A$. A well-known fact is that $\inf A=0$ and $\sup A=+\infty$. REPLY [3 votes]: This is not a full answer, but a pair of soft arguments suggesting that $A$ is dense in $[0, +\infty)$. First Argument Given any triple $(a,b,c)$, let $\displaystyle r(a,b,c)=\frac{c}{\text{rad}(abc)}$. One can generate two new triples $$t_1=(a(c+b),b^2,c^2)$$ $$t_2=(a^2,b(c+a),c^2)$$ with ratios $$r(t_1)=r(a,b,c)\cdot\frac{c}{c+b}\cdot\frac{c+b}{\text{rad}(c+b)}$$ $$r(t_2)=r(a,b,c)\cdot\frac{c}{c+a}\cdot\frac{c+a}{\text{rad}(c+a)}$$ Clearly $\displaystyle\frac{1}{2}<\frac{c}{c+a}, \frac{c}{c+b}<1$. On the other hand $\displaystyle \frac{n}{\text{rad}(n)}$ is on multiplicative average equal to $\prod_{p} p^{1/(p^2-p)}\approx2.128$. So the tree of triples generated by repeated application of the transformations above, will form paths of ratios that drift (mostly slowly) to increasingly higher values. $0$ being an obvious accumulation point of $A$, one has infinitely many starting points for the process above, making it plausible that any presumed gap in $A$ will be cut by infinitely many paths. Second Argument This is via an example specifically targeting the gap around $0.87$ that was mentioned in a comment to the question. Start by approximating $0.87$ with a rational with squarefree denominator (and preferably multiple small prime factors). $61/70$ is a good candidate. Take the triple $(1, 2^{100}\cdot 5^7\cdot 7^4, 2^{100}\cdot 5^7\cdot 7^4+1)$ where the exponents $100,7,4$ are picked so that $61^2$ divides $c$. There is then a good chance that $c/\text{rad}(c)=61$, which is indeed the case here, and the ratio of the triple is therefore $61/70$.<|endoftext|> TITLE: Why are VOA characters modular forms (geometrically)? QUESTION [8 upvotes]: In Zhu's seminal paper, he proves (5.3.2) that if $V$ is a vertex algebra the character of all of its modules are modular forms! (This is not literally true- there are conditions). I have always found this statement very mysterious. How on earth do modular forms/elliptic curves come into the picture? Is there a proof using the moduli space of elliptic curves $\mathcal{M}_{1,1}$ more directly? e.g. if $M$ is a vertex algebra module, it can be localised onto any elliptic curve $E$. Is Zhu's proof saying we can vary $E$ and thereby get some structure living over $\mathcal{M}_{1,1}$, from which we can take a character (which will just be a modular form because it's a section of a line bundle on $\mathcal{M}_{1,1}$)? I am also very confused about the conditions ($C_2$-finiteness, rationality). Are these conditions necessary for the theorem? If they are, is there any "geometric" meaning to them, say in terms of the associated chiral algebras, or the structure in 1. living over $\mathcal{M}_{1,1}$? Hopefully, with all the progress on vertex/chiral algebras in the 25 years since the proof was published, there is a clean modern answer to the above. REPLY [10 votes]: I'm sure someone here can handwave the intuition behind these statements much better than me. This handwaving version goes like this, one is interested in computing vacuum 1-point functions on a torus, but you can cut a torus along an $S^1$ making it into a cylinder, put a module M on the boundaries, now you start with a vector $m \in M$, you evolve it with $q^{L_0}$ and get a vector $m' \in M$. Gluing back the torus corresponds to taking the trace. Now to actually answer the different parts of the question. if MM is a vertex algebra module, it can be localised onto any elliptic curve EE. Is Zhu's proof saying we can vary EE and thereby get some structure living over M1,1M1,1, from which we can take a character There are a number of misinterpretations here. It is true that given a vertex algebra module (with very mild conditions) you can get a sheaf on an elliptic curve corresponding to it (a module for the corresponding chiral algebra). However, this has nothing to do with coinvariants of $V$. Coinvariants of $V$, or conformal blocks, have to do with the bundle/sheaf that you get on an elliptic curve by localizing $V$ itself, and not a module. The statement from Zhu can be translated in a way very similar to what you say though: For each genus g curve $X$ with $n$ marked points $\{x_i\} \subset X$, one can construct a vector space $C(X, V, \{x_i\})$ of coinvariants. In the language of Frenkel-Ben-Zvi cited in the comments these are coinvariants of $V$ itself: you put the vacuum module $V$ at each of the marked points $x_i$. So far no module $M$ is involved. Now as you move $X$ in the moduli space $\mathcal{M}_{g,n}$, these spaces arrange into a twisted $\mathcal{D}$-module on this moduli space. In the particular case of 1-marked elliptic curves, ie. $g=1$ and $n=1$, under some conditions (below we'll deal with that part of your question) these vector spaces are finite dimensional and you get an actual vector bundle with a flat connection. Notice that this is simply the statement that the twisted $D$-modules are coherent sheaves, since the space is one dimensional, this has to be a vector bundle with a flat connection. Zhu's statement is that characters of irreducible representations of $V$ are flat sections of these bundles (rather their duals), and moreover they form a basis of the space of flat sections. So the role of the module $M$ is to produce a section of a geometric object constructed from $V$, not from $M$ directly. Is there a proof using the moduli space of elliptic curves M1,1M1,1 more directly? Yes indeed, and this was already included in Emile's comments, but I'll repeat them essentially here: We have a vector bundle with a flat connection on the moduli space of elliptic curves given by the dual bundle to coinvariants of $V$. This flat connection is rather explicit (I'll describe it below), so to have a flat section is to solve an explicit differential equation. This in turn leads to prove that flat sections give rise to a solution of an explicit ODE. This ODE has a singularity in the boundary of the moduli space at $q = 0$, but it is a mild singularity in the sense that it is a regular singular point. What you do is prove that the formal character of a module $M$ gives a formal power series that solves this differential equation (there is a subtlety here between taking $\tau \in \mathbb{H}$ as the parameter or $q = e^{2 \pi i \tau}$). Once you have this, the general theory of Frobenius expansions of solutions of these ODEs tell you that these characters converge (the hard part) and therefore they form vector valued modular forms (the easy part: we already knew that they were sections of a bundle on the moduli space of elliptic curves). I am also very confused about the conditions (C2C2-finiteness, rationality). Are these conditions necessary for the theorem? Not exactly. What $C_2$ gives you is the finite dimensionality of the space of coinvariants (in fact you need less than this, as we'll see below), what rationality gives you is the fact that you can construct all flat sections starting from irreducible modules. The idea here is geometric and goes back to studying the behavior of flat sections near the nodal curve limit $q=0$. This idea roughly goes as follows: If you give me a flat section, it gives a solution to an ODE as we talked above. As such you can expand it in power series of $q$ (and in theory possibly a polynomial in $\tau = \log q$). The lowest coefficient of this series defines a symmetric function of the Zhu algebra $Z(V)$ of $V$. This is a remarkable associative algebra that has a few properties: The list of its irreducible modules up to isomorphism is in bijection with the irreducible modules V. And in addition under the finitenes conditions listed above $Z(V)$ happens to be finite dimensional and semisimple. So the conformal block being a symmetric function, is a linear combination of traces of irreducible $Z(V)$-modules. This is the direct connection: one can start from an irreducible finite dimensional module $M_0$ for the Zhu algebra $Z(V)$ of $V$. One induces a module $M$ for $V$ having $M_0$ as it lowest graded piece. Then by what we have talked above we know that the character of $M$ is a flat section, and its restriction to its lowest graded piece $M_0$ is the symmetric function we started from. There are some holes here and there in the description I just gave, but it is essentially correct. Now, if $V$ is not rational, but it is $C_2$ cofinite, you still have finite dimensional space of coinvariants and finitely many irreducibles. Just that now you may have extensions between these modules. Miyamoto carried out Zhu's program in this scenario: instead of looking at irreducible $V$ modules induced from $Z(V)$ we need to look at modules induced from projective modules of higher Zhu algebras. Finite dimensionality of coinvariants on elliptic curves. The $C_2$ condition and its role in having finite dimensionality of coinvariants is a bit confusing still (at least to me). It appears in different ways that are somewhat independent. I'll describe in detail the situation for elliptic curves which is what your question is about. The space of coinvariants of the elliptic curve $X_q = \mathbb{C}^* / \mathbb{Z} \simeq C / \mathbb{Z} + \mathbb{Z}\tau$ (where the action of $Z$ is given by multiplication by $q$), with coefficients in the chiral algebra associated to a vertex algebra $V$ with supports in its vacuum module ($V$ itself) in the marked point $0 \in X_q$ is explicitly given by the cokernel of the map $$ d: V \otimes V \otimes \Gamma(\mathcal{O}_{X_q}, X_q \setminus 0) \rightarrow V, \qquad a \otimes b \otimes f(z) \mapsto \mathrm{res}_z f(z) Y(a,z) b.$$ So these functions $f(z)$ are biperiodic functions of $z$ with possible poles at $m + n \tau$ with $m,n \in \mathbb{Z}$. It turns out that $d(V \otimes V \otimes f') \subset d(V\otimes V \otimes f)$ so it is enough to kill a basis of functions modulo total derivatives. This is the top de Rham cohomology of $X_q \setminus 0$ and this has a basis given by the constant function $1$ and the Weierstrass $\wp(z,q)$ function. So the space of coinvariants is simply $$ V / \langle \mathrm{res}_z \wp(z) Y(a,z)b, \, \mathrm{res}_z Y(a,z)b \rangle $$ I am lying slightly here in that the vertex operation that is needed is denoted by $Y[a,z]$ in Zhu's paper and has to do with the fact that we are using an isomorphism $\mathbb{C}^*/\mathbb{Z} \simeq \mathbb{C} / \mathbb{Z}^2$, but that doesn't really change the exposition. The key point is that every vertex algebra is filtered, the relevant filtration is known as the Li filtration. And the above two-term complex is compatible with this filtration. So by looking at the associated graded we get an immediate bound for the homology of that two term complex in degree 0. It turns out that the operation $$ a \otimes b \mapsto \mathrm{res}_z \wp(z;q) Y(a,z)b$$ in this associated graded is simply the operation $a \otimes b \mapsto a_{(-2)}b = \mathrm{res}_z z^{-2} Y(a,z)b$, that is, in this filtration, we only care about the singular part of $f(z)$, not $f(z)$ itself. There you go, just killing $d(V \otimes V \otimes \wp(z))$ we get, in the associated graded, the $C_2$ quotient of $V$. So if $V$ is $C_2$ cofinite, we know that the space of coinvariants is finite dimensional. But there is more. We didn't kill the constant functions $f(z)= 1$. Zhu didn't really care since he needed rationality and finite dimensionality of $Z(V)$, but in his paper is clear that what actually is enough to prove finite dimensionality of coinvariants is not $R_V := V/C_2(V)$ being finite dimensional, but rather the quotient of $R_V$ by the image of $a \otimes b \mapsto \mathrm{res}_z Y(a,z)b = a_{(0)}b$. Now $R_V$ is a Poisson algebra, with Poisson bracket given by $\{a,b\} = a_{(0)}b$ so the actual sufficient condition to have finite dimensionality of coinvariants is that the zeroth Poisson homology of $R_V$, $R_V / \{ R_V, R_V\}$ is finite dimensional. This condition fits nicely in a much wider context: it was proved that the finite dimensionality of the first Poisson homology of $R_V$ also plays a role in the finite dimensionality of the first chiral homology of the elliptic curve with coefficients in $V$ (there is a complex whose degree $0$ homology is coinvariants, the general homology is known as chiral homology). We expect this to be the case for arbitrary chiral homology in fact, not only degree $0$ and $1$. Now that I've set up the above notation, I can quickly tell you about the flat connection that I promised above it was explicit. What is a class in the dual to the coinvariants: it's a functional $\varphi: V \rightarrow \mathbb{C}$ that satisfies $$ \varphi ( \mathrm{res}_z f(z) Y(a,z)b ) = 0$$ For every biperiodic $f$ with possible poles at $z=0$ and every $a,b \in V$. Now there is the Weierstrass $\zeta$ function, which is not really bi-periodic, but it almost is. It's derivative is the Weierstrass $\wp$ function. The flat sections of coinvariants are $\varphi$ that satisfy the equation above and in addition satisfy: $$ \frac{d}{d\tau } \varphi(a) = \varphi ( \mathrm{res}_z \zeta(z) Y(\omega, z)a ) $$ where $\omega \in V$ is the conformal vector. Notice that I never talked about $V$ being conformal up to this point and indeed the conformal vector only appears in the geometric picture just to give us the flat connection. $C_2$ appears in a different way as a sufficient condition to prove finite dimensionality of coinvariants in arbitrary genus. I'll point you to recent work of Damiolini, Gibney and Tarasca. They use this condition to check that you indeed have a vector bundle in the moduli space. But I have a strong suspicion that this can be relaxed as well to finite dimensionality of the Poisson homology. A final remark in an already long answer. When you ask: say in terms of the associated chiral algebras, Not as far as I know actually. The key point was the Li filtration. And the Li filtration is a filtration of Vertex algebras, but it is not compatible with the chiral algebra structure. There is no literature that I am aware about this filtration in relation to chiral algebras. And up to not long ago at least one of the leading experts in the field of chiral algebras was not aware of the existence of this filtration. It is picking up pace and there have been many exciting results in the last couple of years, so I expect to see some new idea popping up in the arxiv soon regarding this.<|endoftext|> TITLE: Disjoint perfect matchings in complete bipartite graph QUESTION [7 upvotes]: Let $K_{n,n}$ be a complete bipartite graph with two parts $\{u_1,u_2,\ldots,u_n\}$ and $\{v_1,v_2,\ldots,v_n\}$, and let $K^-_{n,n}$ be the graph derived from $K_{n,n}$ by delete a perfect matching $\{u_1v_1,u_2v_2,\ldots,u_nv_n\}$. Since $K^-_{n,n}$ is now $(n-1)$-regular, it has $n-1$ disjoint perfect matchings. My question is whether the edges of $K^-_{n,n}$ with $n\geq 4$ can be decomposed into $n-1$ disjoint perfect matchings in such a way that in each matching $M$, if $u_iv_j\in E(M)$ then $v_iu_j\not\in E(M)$. REPLY [8 votes]: The answer is that this is possible for all $n>4$. Your question is equivalent to asking whether there exists a unipotent Latin square $L$ of order $n$ with $L_{ij}\ne L_{ji}$ for $i\ne j$. The equivalence is obtained by using $L_{ij}$ to record the index of the matching that contains the edge $u_i v_j$ (and putting $L_{ii}=n$ for each $i$). The existence of such a Latin square follows from a stronger property. It is a theorem (collectively due to the work of Kotzig, McLeish, Turgeon and others) that for all $n\notin\{2,4\}$ there exists a Latin square of order $n$ that has no intercalates (an intercalate is a $2\times2$ submatrix that is itself a Latin square). This property of Latin squares is called $N_2$ in the literature. If you permute the rows of any $N_2$ Latin square you can make every symbol on the main diagonal equal to $n$. You then have what you need.<|endoftext|> TITLE: Large sieve inequality for sparse trigonometric polynomials QUESTION [13 upvotes]: Let $S(\alpha) = \sum_{n\leq N}f(n) e^{2\pi i \alpha n}$ for some arithmetic function $f$. Suppose $\alpha_1, \ldots, \alpha_R$ are real numbers that are $\delta$-spaced modulo $1$, for some $0 < \delta < 1/2$. The large sieve inequality then gives $$ \sum_{r=1}^R \left| S\left(\alpha_r \right)\right|^2 \ll (N + \delta^{-1}) \sum_{n\leq N}|f(n)|^2. $$ The $N$ term is satisfactory when the support of $f$ is dense enough in $[1,N]$. However it becomes worse the sparser the support of $f$ is. Are there any methods or inequalities or examples known in literature to tackle the case when the support of $f$ is sparse, say when $\#(\text{supp}(f) \cap [1,N]) \asymp \sqrt{N}$. Even in this extreme of $\sqrt{N}$ the large sieve can give non-trivial cancellation, but can one do better? REPLY [16 votes]: If $f$ is $M$-sparse, then from Cauchy-Schwarz one has $|S(\alpha)|^2 \leq M \sum_{n \leq N} |f(n)|^2$ which gives the bound $$ \sum_{r=1}^R |S(\alpha_r)|^2 \leq R M \sum_{n \leq N} |f(n)|^2$$ which is superior in the regime $RM \leq N$ (i.e., below the range of the Heisenberg uncertainty principle). For $RM \geq N$ one is now consistent with the uncertainty principle and one cannot hope to do better than the large sieve in general. Indeed, if $M$ divides $N$ and $f$ is the indicator function of the multiples $\{ mN/M: m=1,\dots,M\}$ of $N/M$, then $S(\alpha)=M$ for $\alpha$ a multiple of $M/N$, so if one chooses $\alpha_1,\dots,\alpha_R$ to contain the $N/M$ different multiples of $M/N$ mod $1$ (and sets $\delta$ to be anything less than or equal to $M/N$) then equality in the large sieve is essentially attained. Note that in this example $f$ had a lot of additive structure (in particular the support was basically closed under addition). If $f$ is additively unstructured (e.g., if it is supported on a dissociated set) then one can do better by taking higher moments. For instance by applying the large sieve to $f*f$ rather than to $f$ one has $$ \sum_{r=1}^R |S(\alpha_r)|^4 \ll (N + \delta^{-1}) \sum_{n \leq 2N} |f*f(n)|^2$$ and hence by Cauchy-Schwarz $$ \sum_{r=1}^R |S(\alpha_r)|^2 \ll R^{1/2} (N + \delta^{-1})^{1/2} (\sum_{n \leq 2N} |f*f(n)|^2)^{1/2}.$$ If $f$ is not very additively structured, then $f*f$ will be quite spread out and this can be a superior bound to the previously mentioned bounds. Similarly if one plays with higher convolutions such as $f*f*f$. Basically one is convolving one's way out of sparsity in order to make more efficient use of the large sieve. (A similar technique is frequently employed to study large values of Dirichlet polynomials, taking advantage of the fact that the logarithms $\log n$ of the natural numbers only have a very limited amount of additive structure, thanks to the divisor bound.) As a variant of the above strategy, if $f$ is supported in some set $E$ on which good bounds are known on exponential sums $\sum_{n \in E} e(\alpha n)$ (or weighted versions of such sums) - which is a measure of lack of additive structure in $E$ - then one can sometimes get good bounds from a duality argument (sometimes known as a "large values" argument, particularly in the context of controlling the large values of Dirichlet polynomials). See for instance Green, Ben; Tao, Terence, Restriction theory of the Selberg sieve, with applications, J. Théor. Nombres Bordx. 18, No. 1, 147-182 (2006). ZBL1135.11049. for an example of this (where the function $f$ is supported on something like the set of primes).<|endoftext|> TITLE: Sidon sets with k >1 QUESTION [7 upvotes]: Let $S$ be a subset of ${1,2,...,n}$ such that for every $a,b$ in $S$ the numbers of form $a^k+b^k$ are distinct ($k$ is positive integer) What is the maximum cardinality of $S$ REPLY [10 votes]: The $k=2$ case was considered by Alon and Erdős (European J. Combin. 6 (1985) 201-203, MR0818591) and improved by Lefmann and Thiele (Combinatorica 15 (1995) 379-408, MR1357284). They expressed the problem as looking for the largest Sidon set of integers squared. The 1995 result: There exists a Sidon set $S \subset \{1^2, 2^2, \ldots, n^2\}$ with $$|S| \ge c \cdot n^{2/3}$$ where $c>0$ is constant. Alon and Erdős explain that a result of Landau on the density of the sums of two squares implies that, for any Sidon set $S$, $$|S| \le \frac{c'n}{(\log n)^{1/4}}.$$ Scanning the MathSciNet citations from references for the two articles, I don't see subsequent improvements of these bounds or treatments of $\{1^k, 2^k, \ldots, n^k\}$ for $k \ge 3$. (In the related case of starting from an infinite set of integers to a fixed power, there are more results; see a recent arXiv article by Kiss and Sándor and the references there.)<|endoftext|> TITLE: Topology on cohomology of a sheaf of topological groups QUESTION [6 upvotes]: Let $X$ be a topological space and $\mathcal{F}$ be a sheaf of commutative topological groups on $X$. I am interested in the following question: Is there a natural way to introduce topology on $H^i(X, \mathcal{F})$? My guess is that for each open covering $\mathcal{U}$ the space of Čech cochains $\check{C}(\mathcal{U}, \mathcal{F})$ can be endowed with the compact-open topology. This topology can be restricted to the space of closed cochains, descends to $H^i(\mathcal{U}, \mathcal{F})$ and induce direct limit topology on $H^i(X, \mathcal{F})$. But does this construction make sense? Say, is it functorial? Is it true that $H^i(X, \mathcal{F})$ are commutative topological groups (with respect to the natural group operation)? Are the other reasonable choices of topology on Čech complex? Can they lead to other topologies? And what is the best reference on this topic? I don't think that this is important, but in the situation I am interested in, the space $X$ is a (finite-dimensional) manifold and $\mathcal{F}$ is a sheaf of Lie groups. REPLY [5 votes]: Both cases ($F$ is a sheaf of abelian topological groups or abelian Lie groups) can be treated using the same machinery. The Yoneda embedding embeds abelian Lie groups as a fully faithful subcategory of the category of sheaves of abelian groups on the site of smooth manifolds, and the embedding functor preserves small limits. I refer to the latter category as the category of abelian smooth groups. This category is complete and cocomplete. It is, in fact, better behaved than the category of abelian topological groups, since the latter category is not an abelian category: a morphism of abelian topological groups can have a trivial kernel and cokernel, without being an isomorphism. On the other hand, the category of smooth groups is abelian. From now on, we work either with presheaves of chain complexes of abelian topological groups or presheaves of chain complexes of smooth groups. The category of such chain complexes can be equipped with the projective model structure transferred from simplicial topological spaces respectively simplicial sheaves on the site of smooth manifolds. The category of presheaves of such chain complexes can itself be equipped with the projective model structure, which can then be further localized with respect to Čech nerves of open covers. Fibrant objects in the resulting model category are presheaves of chain complexes that satisfy the homotopy descent condition. The fibrant replacement functor computes the (hyper)cohomology of sheaves. More precisely, if $F→RF$ is a fibrant replacement of the sheaf $F$, then $H^i(X,F)=H^i(Γ(X,RF))$. The resulting cohomology theory is essentially (a reformulation of) the Segal–Mitchison cohomology in the topological case, or its smooth version by Brylinski. In particular, the cohomology group $H^i(X,F)$ is by definition a smooth (respectively topological) group, since $H^i$ is computed in the category of smooth (respectively topological) groups.<|endoftext|> TITLE: Lower bound $\langle (I + A)^{-1}x, x \rangle$ given that $\sigma_\text{max}(A) < 1$ QUESTION [6 upvotes]: Let $A$ a square real matrix such that the largest singular value $\sigma_\text{max}(A) = \sigma < 1$. I want to find a lower bound on $\langle (I + A)^{-1}x, x\rangle$ where $x$ is a vector of euclidean norm $1$: $\langle x, x\rangle=1$. I empirically find that a seemingly tight lower bound is $$ \langle (I + A)^{-1}x, x\rangle \geq \frac{1}{1+\sigma} $$ which is reached for $A= \sigma I$. I cannot prove the above result. Note that it is pretty straightforward to prove that $\sigma_\text{min}((I + A)^{-1}) \geq \frac1{1+\sigma}$ but that does not suffice to conclude, since I do not assume that $A$ is symmetric. PS: $\sigma_\text{max}$ and $\sigma_\text{min}$ are the largest and smallest singular values: $\sigma_\text{max} = \sqrt{\lambda_\text{max}(AA^T)}$ is the operator norm of $A$. REPLY [6 votes]: The map $f(z)=(1+z)^{-1} - (1+\sigma)^{-1}$ maps the disk of radius $\sigma$ into the right half plane as a function of one complex variable. Therefore, essentially by von Neumann's inequality, we get that $$\frac{f(A)+f(A)^*}{2}=\mathrm{Re }f(A)\geq 0$$ since $\|A\|\leq \sigma.$ Assuming $A$ has real entries, this implies the claim as $$\langle (1+A)^{-1}x,x\rangle = \langle \mathrm{Re} (1+A)^{-1}x,x\rangle.$$ To see the calculation with von Neumann's inequality more explicitly, let $\psi(z) = \frac{z-1}{z+1}.$ Note $\psi$ takes the right half plane to the disk. So, $\psi \circ f$ takes the disk of radius $\sigma$ into the unit disk. Therefore, von Neumann's inequality states that $\|\psi \circ f(A)\|\leq \sup_{z\in \sigma\mathbb{D}} |\psi\circ f(z)| \leq 1.$ Note that $$1-(\psi \circ f(A))^*(\psi \circ f(A)) \geq 0.$$ (Here by $T\geq 0$ we mean that $T$ is positive semi-definite.) Writing out what that means $$1-(f(A)^*+1)^{-1}(f(A)-1)^*(f(A)-1)(f(A)+1)^{-1} \geq 0.$$ So, $$(f(A)^*+1)(f(A)+1)-(f(A)^*-1)(f(A)-1)=2(f(A)+f(A)^*)\geq 0.$$ Results of the above form (positivity of noncommutative rational functions) always have to have "algebraic proofs," many of which can be done algorithmically. See, e. g., Helton, Klep, and McCullough - The convex Positivstellensatz in a free algebra and Pascoe - Positivstellensätze for noncommutative rational expressions.<|endoftext|> TITLE: Variety of commuting matrices QUESTION [18 upvotes]: Let $G=\operatorname{GL}(n,\mathbb{C})$ and $\mathfrak{g}=\operatorname{Mat}(n,\mathbb{C})$ and let us consider the two varieties $X,Y$ defined as $$X=\{(x,y) \in G \times G \ | \ xy=yx\} $$ and $$Y=\{(x,y) \in \mathfrak{g} \times \mathfrak{g} \ | \ xy=yx\} .$$ The group $G$ acts on both of them by conjugation: I'd like to find out what is known in the literature for the $G$-equivariant cohomology of $X,Y$ (an the mixed Hodge structure on it).Moreover, is the cohomology of their GIT quotients $X//G$, $Y//G$ known too? Is there a relation between them? REPLY [5 votes]: My new paper with Carlos Florentino and Jaime Silva answers this question: Mixed Hodge structures on character varieties of nilpotent groups. In particular, see Section 4.4. For an implementation of the MHS in some special cases, please see the Mathematica NB here: https://github.com/seanlawton/Mixed-Hodge-structures-on-character-varieties-of-nilpotent-groups<|endoftext|> TITLE: Can a dodecahedron be deformed into a great stellated dodecahedron? QUESTION [6 upvotes]: Can a convex regular dodecahedron be deformed into a great stellated dodecahedron while keeping all pentagons planar and all edges of nonzero length the whole time? REPLY [4 votes]: Talking with Saul Schleimer, we came up with the following: Orthogonally project the great stellated dodecahedron into the $z=0$ plane, choosing a direction that does not result in any zero length edges. Do the same for the dodecahedron. We can realise each of these projections as the endpoint of a homotopy through affine maps $(x,y,z) \to (x,y,(1-t)z)$, so planarity is preserved. Now that everything is in the plane, the planarity of faces is easy to maintain. Note that the condition that an edge has length zero is codimension two. Therefore we can move the vertices of the flat dodecahedron to the vertices of the flat great stellated dodecahedron in a generic way.<|endoftext|> TITLE: Are differential rings monoids in a monoidal category? QUESTION [9 upvotes]: $\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$A number of algebraic structures can be defined as monoids in some appropriate monoidal category: A monoid is a monoid in $(\mathsf{Sets},\times,\pt)$; A semiring is a monoid in $(\mathsf{CMon},\otimes_{\N},\N)$; A ring is a monoid in $(\mathsf{Ab},\otimes_\Z,\Z)$; An $R$-algebra is a monoid in $(\mathsf{Mod}_R,\otimes_R,R)$; A graded $R$-algebra is a monoid in $(\mathsf{Gr}_\Z\mathsf{Mod}_R,\otimes_R,R)$; A differential graded $R$-algebra is a monoid in $(\mathsf{Ch}_\bullet(\mathsf{Mod}_R),\otimes_R,R)$. Is this also the case for differential rings? REPLY [7 votes]: $\newcommand{\defeq}{\overset{\mathrm{def}}{=}}\newcommand{\id}{\mathrm{id}}\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\d}{\mathrm{d}}\newcommand{\dAb}{\mathsf{End}(\mathsf{Ab})}$DGAs are monoids in chain complexes. To get differential rings as monoids in some monoidal category, it suffices to remove the grading and the $\d^{2}=0$ condition. In detail, consider the category $\mathsf{End}(\mathsf{Ab})\defeq\mathsf{Fun}(\mathbf{B}\N,\mathsf{Ab})$ whose Objects are pairs $(A,\d)$ with $A$ an abelian group and $d\colon A\to A$ a morphism of abelian groups. Morphisms $(A,\d_A)\to(B,\d_B)$ are morphisms of abelian groups preserving the derivation, i.e. such that the diagram $$ \require{AMScd} \begin{CD} A @>\d_A>> A\\ @V f V V @VV f V\\ B @>>\d_B> B \end{CD} $$ commutes. We can then put a monoidal structure $\otimes_\Z$ on $\dAb$ by defining $$(A,\d_A)\otimes(B,\d_B)=(A\otimes_\Z B,\d_A\otimes_\Z1_B+1_A\otimes_\Z\d_B),$$ where the unit is given by the pair $(\Z,\d_\Z)$ with $\d_\Z\overset{\mathrm{def}}{=} 0$. Note that a morphism in $\dAb$ from $(\Z,\d_\Z)$ to $(A,\d_A)$ is just a "constant" element of $A$, i.e. an element with $\d_A a = 0$. A monoid in $(\dAb,\otimes_\Z,(\Z,\d_\Z))$ will then be a triple $((A,\d),\mu,\eta)$ with $(A,\d)$ an object of $\dAb$; this accounts for the underlying additive abelian group of a differential ring and the derivation $\d$, which is $\Z$-linear; $\mu\colon(A,\d_A)\otimes_\Z(A,\d_A)\to(A,\d_A)$ a morphism of $\dAb$; this accounts for the multipication and the Leibniz rule: asking for the diagram $$ \require{AMScd} \begin{CD} A\otimes_\Z A @>\d_A\otimes_\Z1_A+1_A\otimes_\Z\d_A>> A\otimes_\Z A\\ @V \mu V V @VV \mu V\\ A @>>\d_A> A \end{CD} $$ to commute is equivalent to asking $$\d(ab)=\d(a)b+a\d(b)$$ to hold for all $a,b\in A$; $\eta\colon(\Z,\d_\Z)\to(A,\d_A)$ a morphism of $\dAb$, determining an element $1_A$ of $A$; such that the usual associativity and unitality diagrams commute, which makes $(A,\mu,\eta)$ into a ring, and together with $\d$, this makes the quadruple $((A,\d),\mu,\eta)$ into a differential ring.<|endoftext|> TITLE: Submanifolds of Lie groups with abelian normal bundle QUESTION [7 upvotes]: Let $M$ be a submanifold of a symmetric space $Q$. The normal bundle $NM$ is called abelian if $\exp(N_{p}M)$ is contained in some totally geodesic and flat submanifold of $Q$ for all $p \in M$; see Terng & Thorbergsson, "Submanifold geometry in symmetric spaces", J. Differential Geom. 42 (1995), 665–718. It is clear that every codimension-one submanifold (i.e., hypersurface) of $Q$ has abelian normal bundle. I am interested in the case where $Q$ is a Lie group $G$ equipped with a bi-invariant metric. Questions: Are submanifolds with abelian normal bundle (and codimension greater than one) plentiful or rare in $G$? In particular, do two-dimensional examples exist? REPLY [2 votes]: Let $M$ be a submanifold of a symmetric space $Q$, $G = I^0(Q)$, $\mathfrak{g} = \mathrm{Lie}(G)$. Take $p \in M$ and write $K$ for the isotropy subgroup of $G$ at $p$ and $s_p \in I(Q)$ for the geodesic symmetry at $p$. We have $\Theta = C_{s_p} \colon g \mapsto s_p g s_p$ an involutive automorphism of $G$ and $\theta = \Theta_*$ the corresponding automorphism of $\mathfrak{g}$. As usual, we decompose $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ into the $(+1)$- and $(-1)$-eigenspaces of $\theta$, where $\mathfrak{k} = \mathrm{Lie}(K)$. Note that $(G,K)$ is a Riemannian symmetric pair representing $Q$. We identify $\mathfrak{p}$ with $T_pQ$ in a standard way. Since $p \in M$, the normal space $N_pM$ becomes a subspace of $\mathfrak{p}$ and thus $\mathfrak{g}$. First of all, a nice algebraic way to rephrase the condition that $\exp(N_pM)$ is contained in a flat immersed totally geodesic submanifold of $Q$ is to say that $N_pM$ is an abelian subspace of $\mathfrak{g} = \mathrm{Lie}(G)$. This follows from the standard correspondence between complete totally geodesic immersed submanifolds of $Q$ passing through $p$ and Lie triple systems in $\mathfrak{p}$ and the fact that flat such submanifolds correspond precisely to abelian Lie triple systems, i.e. just abelian subspaces of $\mathfrak{p}$. As noted in the comments, it is necessary for $M$ to be of codimension $\leqslant \mathrm{rk}(Q)$ if you want it to have an abelian normal bundle, simply by definition of the rank of a symmetric space. As Robert Bryant noticed in his comment, one can easily find submanifolds with abelian normal bundles when $Q$ is reducible by taking products of submanifolds of the factors. In other words, if $Q = Q_1 \times \ldots \times Q_k$ is a Riemannian product of smaller symmetric spaces and $M_i \subseteq Q_i$ are submanifolds with abelian normal bundles, then $M_1 \times \ldots \times M_k$ has an abelian normal bundle as a submanifold of $Q$. As a special case, you can take $M_i$'s to be any hypersurfaces. Now, the question of abundance and classification of such submanifolds is difficult in general, but it becomes easier in case $M$ is homogeneous. Recall that $M$ is called homogeneous if the subgroup of $I(Q)$ of isometries preserving $M$ acts transitively on $M$. An equivalent definition as that $M$ is homogeneous if it is an orbit of a closed connected subgroup of $G$ (here we assume that $M$ is properly embedded and connected). If $M$ is homogeneous and has an abelian normal bundle, then it is an orbit (maybe singular) of a hyperpolar action on $Q$. Somewhat conversely, if $H$ is a closed connected subgroup of $G$ acting on $Q$ hyperpolarly, then any nonsingular orbit of $H$ ($=$ of the maximal dimension) has an abelian normal bundle. Recall that the action of $H$ on $Q$ is called polar if it admits a section, i.e. a complete immersed submanifold $S \subseteq Q$ that intersects all $H$-orbits and does so orthogonally. It is called hyperpolar if it is polar and admits a flat section. Sections are always totally geodesic and their dimension is equal to the cohomogeneity of the action. So if you have a hyperpolar action of cohomogeneity greater than one, its nonsingular orbits are examples of submanifolds of codimension $> 1$ with an abelian normal bundle. Classification of polar and hyperpolar actions on symmetric spaces has a long history and is quite hard in general. When $Q$ is irreducible of compact type (in particular, when it is a simple compact Lie group), a classification was obtained by A. Kollross. He explicitly classified hyperpolar actions and their special case of actions of cohomogeneity one here and then, in a series of papers, he showed (together with A. Lytchak) that polar actions are automatically hyperpolar on such spaces. So what you need is hyperpolar actions in Kollross's classification whose cohomogeneity is greater than one in case $Q$ is a simple compact Lie group. This is discussed in Section 3 of his paper. If I'm not mistaken and got the paper correctly, in his notation (since $Q$ is a Lie group, he denotes it by $G$), every such action is orbit equivalent to the action of a subgroup $H \times K \subseteq G \times G = I^0(G)$, where $H$ and $K$ are as in the following tables , , , or the action of the isotropy subgroup $\Delta \subseteq G \times G$ at $e$ (i.e. the adjoint action of $G$ on itself; the cohomogeneity of this action equals the rank of $G$), or a so-called $\sigma$-action (whose construction is explained in the paper): . Let me finally remark that if you're interested in such submanifolds in symmetric spaces other than compact Lie groups, there are some classification results as well. In this paper Berndt, Tamaru, and Diaz-Ramos classified hyperpolar actions without singular orbits on symmetric spaces of noncompact type (possibly reducible). Since it is very easy to compute the cohomogeneity of the actions they construct, their result gives a wealth of examples of submanifolds of noncompact symmetric spaces with abelian normal bundles.<|endoftext|> TITLE: Plane partitions as irreducible representations QUESTION [10 upvotes]: The irreducible representations of the symmetric group algebras $A_n=KS_n$ over a the complex numbers (or any field of characteristic 0) $K$ satisfy the following properties: The irreducible representations of $A_n$ are in natural bijection to paritions of $n$. We have natural subalgebra inclusions $A_k \subseteq A_{k+1}$ for all $k$ and an irreducible $A_n$-representation $V$ restricts to a direct sum of distinct irreducible $A_{n-1}$-representations giving a poset structure (Hasse diagram has arrows from those restricted irreducible representations to $V$) that is isomorphic to the Young lattice for partitions (paritions ordered by "inclusion" via their Young diagrams). Question: Is there a sequence of algebras $B_n$ that satisfy the same properties when we replace "partitions" (which are 2-dimensional) by "plane paritions" (which are 3-dimensional)? So the following properties should be satisfied for the irreducible representations of those algebras $B_n$: The irreducible representations of $B_n$ are in natural bijection to plane partitions with $n$ blocks. We have natural subalgebra inclusions $B_k \subseteq B_{k+1}$ for all $k$ and an irreducible $B_n$-representation $V$ restricts to a direct sum of distinct irreducible $B_{n-1}$-representations giving a poset structure that is isomorphic to the lattice of plane partitions (plane paritions ordered by "inclusion"). It would be especially interesting whether this is possible when choosing $B_n$ to be a semigroup algebra over a finite semigroup or at least finite dimensional algebras. If it is not possible with irreducible representations alone for algebras, maybe it is possible with indecomposable representations instead for algebras $B_n$ that might be not semisimple. REPLY [6 votes]: The "path model" described in Section 2.3.11 of Goodman, de la Harpe, and Jones' book "Coxeter graphs and towers of algebras" gives a construction for a sequence of finite-dimensional semisimple algebras whose branching rules give any graded poset with finite ranks that you like (or more generally, weighted versions too). Furthermore, this sequence is essentially uniquely determined. It is an interesting question whether the algebras one gets from this construction have other natural descriptions, for example as semigroup algebras as you suggest, but they definitely exist.<|endoftext|> TITLE: Math overdose? Professional advise how to cope with it QUESTION [13 upvotes]: I'm a PhD student, currently working on my Thesis. Over the years I have many time encountered a problem. Maybe professional mathematicians know what I'm talking about? When I study a math topic, frequently I reach a stage when I stop understanding anything in this topic anymore. Like when one repeats a word many times over and over it becomes unintelligible and meaningless. I read a book or a series of papers on a topic and everything which I understood before I started actively studying the topic, are now not understandable anymore, even though I understood (or at least I thought I understood) them when I started reading. So I give it up for a while, because continuing this topic is not possible anymore. And then exactly the same thing happens with the next topic. The study is not lost, because when I return to the topic it is much easier and I'm able to progress deeper into it, until I reach a new stage where I have to give it up again. Has this happened to other professional mathematicians? Or maybe I'm not meant to be a professional mathematician? Any professional advice how to cope with this situation? REPLY [16 votes]: I can give three bits of advice. One that has been given by many other people is to take a break. People think that endlessly hitting the books without a break is the mark of a hard working mathematician, and that to do anything else is lazy. This is far from true. Doing so only depletes ones creativity. The second bit of advice is to form hobbies and interests that may be mathematically related but that give one an opportunity to make connections outside of ones department. Doing a PhD. in a ruthlessly competitive environment can be draining. Getting an opportunity to play pool, MTG, chess, to exercise, play video games, go camping, go hiking or something of the sort may restore a PhD. student's confidence, and also restore one's creativity. The third bit of advice is not to judge one's worth solely based on one's mathematical achievements. You may say "What else is there to judge oneself by, and why should I stop this judgement if I want to increase my productivity?" The answer is that this judgement does not increase your productivity, in fact it decreases it. When you are highly productive this judgement does not truly increase your productivity. When your productivity is lower, which happens to every mathematician, then this can tank productivity as all elements that further creativity are reduced. Realize that you have innate worth beyond your work, and this will also help to shorten those times when creativity is lacking. I hope that this is helpful as your question is certainly worthwhile and I suspect that your experience is universal among mathematicians.<|endoftext|> TITLE: Polytope where each vertex belongs to all but two facets QUESTION [24 upvotes]: Let $P$ be a (convex, bounded) polytope with the following property: for every vertex $v$, there are exactly two facets which do not contain $v$. Does it follow that $P$ is (combinatorially) a Cartesian product of two simplices? Some remarks by "facet" I mean "face of codimension $1$" a Cartesian product of two simplices has the desired property this is trivially true for $\dim P = 2$ this is true for $\dim P = 3$: by playing with the Euler formula and the list of polyhedra with small $f$-vectors, one checks that only the triangular prism satisfies the condition it is a classical fact that a $n$-dimensional simple polytope with $n+2$ facets is a Cartesian product of two simplices. So the answer is positive if $P$ is assumed to be simple. REPLY [18 votes]: There are other polytopes with this property that can be obtained via the free join construction. Given two polytopes $P_1\subset\Bbb R^{d_1}$ and $P_2\subset\Bbb R^{d_2}$, the free join $P_1\bowtie P_2$ is obtained by embedding $P_1$ and $P_2$ into skew affine subspaces of $\Bbb R^{d_1+d_2+1}$ and taking the convex hull. The claim is that if $P_1$ and $P_2$ have your property, then so does $P_1\bowtie P_2$. To see this, note that the facets of $P_1\bowtie P_2$ are exactly of the form $P_1\bowtie f_2$ and $f_1\bowtie P_2$, where $f_i$ is a facet of $P_i$. Now, if $v$ is a vertex in, say, $P_1\subset P_1\bowtie P_2$ that is in all facets of $P_1$ except for $f,f'\subset P_1$, then $v$ is in all facets of $P_1\bowtie P_2$ except for $f\bowtie P_2$ and $f'\bowtie P_2$. Equivalently for vertices in $P_2$. Example. Take the free join of two squares, which is a 5-dimensional polytope with 8 vertices, 8 facets and vertex degree 6. This polytope is not simple and can thus not be the cartesian product of two simplices. This construction yields counterexamples in dimension $d\ge 5$. Concerning $d=4$, there should not be any counterexamples in this dimension. Suppose that $P\subset\Bbb R^4$ is a counterexample. According to Moritz Firsching's comment we can assume that it has $n\ge 10$ facets. Each vertex is then contained in exactly $n-2\ge 8$ of them. More generally, each set of $k$ vertices is contained in at least $n-2k$ facets. But let $f\subset P$ be a 2-face of $P$ and $v_1,v_2,v_3\in f$ three affinely independent vertices, then this set of vertices, and thus also $f$, is contained in $n-6\ge 4$ facets. This cannot be, since each 2-face of $P$ is contained in exactly two facets. I believe this generalizes to show that any polytope with your property must have $\le 2d$ facets.<|endoftext|> TITLE: Minimum number of generators for quotients of congruence subgroups of SL(2, Z) QUESTION [5 upvotes]: For a given positive integer $N$ let $L(N)$ denote the principal congruence subgroup of $\operatorname{SL}(2, \mathbb{Z})$ of level $N$. It is known that $L(N)$ is a finitely generated free group. Let $T(N)$ denote the minimum number of generators. Let $r$ be a number strictly less than $T(N)$. Question: Is there a congruence subgroup $L$ of $L(N)$ such that the minimum number of generators of $L(N)/L$ is at least $r$? REPLY [9 votes]: $\DeclareMathOperator\SL{SL}$Not always. For simplicity, take $N$ a large prime $p$. Following user44191's suggestion in the comments, take $r$ to be $T(N)-1$. Since any congruence subgroup contains a principal congruence subgroup, we may as well assume $L$ contains a principle congruence subgroup. We can write this as $L(p^e M)$ where $M$ is relatively prime to $M$. Then $$L(p)/ L(p^eM) = \SL_2( \mathbb Z/M) \times \ker ( \SL_2(\mathbb Z/p^e) \to \SL_2( \mathbb Z/p))$$ Now $\SL_2(\mathbb Z/M)$ is a quotient of $\SL_2(\mathbb Z)$ and so can be generated by $2$ elements. The kernel $\ker ( \SL_2(\mathbb Z/p^e) \to \SL_2( \mathbb Z/p))$ is a $p$-group, so it can be generated by a number of elements equal to its $p$-rank, which is $3$. Thus the product can be generated by $5$ elements. (In fact, if we're careful, I'm pretty sure we can reduce this down to $3$, but this is unnecessary.) Taking $p$ large enough that the principle congruence subgroup $L(p)$ has more than $5$ generators (i.e. taking $p>3$), we get a counterexample. In the comments you ask whether, for each congruence subgroup $L$, there are congruence subgroups $L'' \subseteq L' \subseteq L$ such that the number of generators of $L'/ L''$ is at least half the number of generators of $L$. The answer is no. To see this, choose our congruence subgroup $L$ to be $L(p)$ for $p$ a large prime as above, and $T(p)$ the number of generators of $L$. Let $[L:L']$ be the index of $L'$ in $L$. Because $L(p)$ is a free group on $T(p)$ generators, its subgroup $L'$ is a free group on $(T(p)-1) [L: L']+ 1$ generators and thus has generator rank at least $(T(p)-1) [L: L']+ 1$. On the other hand, because $L/ L''$ can be generated by $5$ elements, it admits a surjection from a free group on $5$ generators, so its subgroup $L'/L''$, of index $[L:L']$, admits a surjection from a free group on $(5-1)[L:L']+1$ generators and thus has generator rank at most $4 [L:L']+1$. To produce a counterexample, it suffices to have $$ 4 [L:L']+1 < \frac{ (T(p)-1) [L: L']+ 1}{2} $$ for which, given $[L:L']\geq 1$, it suffices to have $T(p)>10$, something that is easy to achieve for $p$ large enough.<|endoftext|> TITLE: Absolute values of two functions and absolute values of their Fourier transform coincides QUESTION [6 upvotes]: Let $f, g \in L^2(\mathbb{R})$. Is it true that if both $|f|=|g|$ and $|\hat f|=|\hat g|$ hold, then there exists $\theta \in \mathbb{R}$ such that $f=ge^{i\theta}$? I am not able to prove it or disprove it. I suspect that this is true. Do you have a reference for this? REPLY [9 votes]: This is true. This was proved by Hardy and Littlewood and the proof is reproduced in Zygmund's Trigonometric Series (which I don't have access to at the moment). (Contradicting my prior "answer") The answer to this question is negative. Such counterexamples are known as "Pauli partners" and are studied in, among other places, the quantum mechanics literature. See, for example: (J. V. Corbett and C. A. Hurst) Are Wave Functions Uniquely Determined by their position and momentum distributions? [https://www.cambridge.org/core/services/aop-cambridge-core/content/view/228E4A34D0B3C63C54B1A01006278C42/S0334270000001569a.pdf/are-wave-functions-uniquely-determined-by-their-position-and-momentum-distributions.pdf] (P. Jaming) Phase Retrieval Techniques for Radar Ambiguity Problems [https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.521.4906&rep=rep1&type=pdf]<|endoftext|> TITLE: Contraction of some surfaces over a ring of algebraic integers QUESTION [6 upvotes]: The situation: Let $X$ be a 2 dimensional normal quasi-projective $\mathcal{O}_K$-scheme, where $K$ is an algebraic number field. Assume the following conditions on $X$: $X$ is integral. $X_K$ is geometrically integral. $X \to \textrm{spec}(\mathcal{O}_K)$ is surjective. Let $X\to \bar{X}$ an open immersion into a projective scheme. (this exist since $X$ is quasi-projective). In particular, 1-dimensional irreducible closed subschemes of $\bar{X}$ are either Horizontal: spectra of finite flat extensions of $\mathcal{O}_K$ Vertical: curves over $\mathcal{O}_K/p\mathcal{O}_K$, which is a finite field, for nonzero ideals $p\in \textrm{spec}(\mathcal{O}_K)$. The claim I want to prove: Let $V$ be the vertical part of $\bar{X}\backslash X$. Theorem 2 of (Points entiers des variétés arithmétiques, Moret-Bailly) states (withouth proof) that there exists a "contraction" of $V$. i.e. a map $\bar{X}\to Y$ which is surjective, $\bar{X}\backslash V\to Y$ is an open immersion and the image in $V$ is a set of isolated points. I am looking for a proof of such a claim. If that helps, you may assume that $K=\mathbb{Q}$. You may assume that $\bar{X}$ is regular. What I have already found: The paper states states that this follows similarly from a paper of Artin, but I couldn't understand how. I also found a paper of Moret-Bailly which uses the existance of integral points on $X$ to prove what I am surching for. But I am looking for a proof which does not rely on this fact, since I am trying to write the proof of theorem 1 of (Points entiers des variétés arithmétiques, Moret-Bailly) using theorem 2 of the same paper. Conditions of the existance of such a contraction can also be found in theorem 27.1 of this work , so you may just help me to find why these conditions apply to my case. REPLY [6 votes]: The paper [2] is a seminar talk announcing the results of [3]. In this talk, I explain how to deduce the existence of integral points (theorem 1) from the contraction theorem (theorem 2). In turn, the contraction theorem is due to Artin [1] in the geometric case (surfaces over finite fields), but Raynaud explained me the arithmetic analogue (see remark 4.5/2 in [2]). Thus, the arithmetic case is not in Artin's paper. Then Szpiro and I found it more convenient to prove the existence of integral points directly (and then deduce the contraction theorem): this is what I did in [3], redirecting Raynaud's arguments for the contraction theorem, especially the crucial proposition 3.8 of [3], for which I must confess that Raynaud does not get proper credit. [1] M. Artin, Some numerical criteria for contractability of curves on algebraic surfaces, Amer. J. Math. 84 (1962), 485-496. doi:10.2307/2372985 [2] L. Moret-Bailly, Points entiers des variétés arithmétiques, Séminaire de Théorie des Nombres, Paris 1985–86, p. 147-154, Progress in Mathematics vol. 71 (Birkhäuser) [3] L. Moret-Bailly, Groupes de Picard et problèmes de Skolem I, Ann. scient. Ec. Norm. Sup. 22 (1989), 161–179. doi: 10.24033/asens.1581<|endoftext|> TITLE: Are complex-oriented ring spectra determined by their formal group law? QUESTION [24 upvotes]: To every complex-oriented ring spectrum $E$ there is associated a formal group law, which is a power series $F_E(x,y)\in E_*[[x,y]]$. Suppose $E$ and $F$ are two complex-oriented ring spectra and suppose I have an isomorphism of coefficient rings $\phi:E_*\rightarrow F_*$ that carries $F_E(x,y)$ to $F_F(x,y)$. Does this imply that $E$ and $F$ are homotopy equivalent spectra? Note that if $F_E(x,y)$ and $F_F(x,y)$ are "Landweber exact" formal group laws then the answer is yes. REPLY [20 votes]: The following is a communal answer from the algebraic topology Discord [1], primarily put forward by Irakli Patchkoria (correcting previous half-answers by Tyler Lawson and me). Kiran suggested it be recorded here to ease future reference. The idea is to produce two topological realizations $M$, $N$ of a single $MU_*$–module by finding two distinct resolutions whose effect on homotopy is the same. The two associated square-zero extensions then give a counterexample. We'll reduce complexity first by considering $ku$–modules rather than $MU$–modules, and second by aiming for a $ku$–module whose homotopy cleaves into small even and odd parts, forcing its $ku_*$–module structure to trivialize. $\DeclareMathOperator{\Sq}{Sq} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\HFtwo}{H\F_2} \newcommand{\Susp}{\Sigma} \newcommand{\co}{\colon\thinspace}$ The nice homotopy groups of complex $K$–theory, $ku_* = \Z[u]$, can be used to show that its bottom $k$–invariant $\kappa_{ku}$ is $ku$–linear: in the diagram $$ \begin{array}{ccccc} & & \Susp^4 ku \\ & & u \downarrow \\ \Susp^{-1} H\Z & \to & \Susp^2 ku & \to & \Susp^2 H\Z \\ & & u \downarrow \\ & & ku & \to & H\Z, \end{array} $$ the vertical maps are multiplication by homotopy elements, hence are $ku$–linear; in turn the horizontal co/fibers are also $ku$–linear; and, finally, the $k$–invariant appears as the middle composite, hence is also $ku$–linear. Similarly, we can show the $ku$–linearity of the bottom $k$–invariant of $ku/2$ and of the Bockstein map $\beta\co \HFtwo \to \Susp H\Z$ (relying on the $ku$–linearity of $2\co H\Z \to H\Z$). Stringing some of these together gives a $ku$–linear composite $$\HFtwo \xrightarrow{\kappa_{ku/2}} \Susp^3 \HFtwo \xrightarrow{\beta} \Susp^4 H\Z \to \Susp^4 \HFtwo.$$ The bottom $k$–invariant $\kappa_{ku/2}$ of $ku/2$ is given as the Milnor primitive $Q_2 = \Sq^3 + \Sq^2 \Sq^1$, the composite of the latter two maps is given as $\Sq^1$, and hence the whole composite is the nontrivial Steenrod operation $$\Sq^1 Q_2 = \Sq^1(\Sq^3 + \Sq^2 \Sq^1) = \Sq^3 \Sq^1.$$ Meanwhile, the homotopy groups of the cofiber $M$ of this composite are $\Susp \F_2 \oplus \Susp^4 \F_2$, which splits as a $ku_*$–module — hence this $ku_*$–module could alternatively be modeled by $N = \Susp H\F_2 \oplus \Susp^4 \HFtwo$ (i.e., the cofiber of the zero map). To finish, set $E = ku \oplus M$ and $F = ku \oplus N$.<|endoftext|> TITLE: Frequency of large values of the Mertens function QUESTION [5 upvotes]: It is known that with $M(x) = \sum_{n\le x}\mu(n)$, there are infinitely many $x$ s.t. $|M(x)|\ge x^{\frac{1}{2} - \varepsilon}$ (see Chapter 15 of Montgomery-Vaughan, for example). Is there any way to make this result effective in the following sense: show that for large $X$, there exists some $x\in [X, f(X)]$ s.t. $|M(x)|\ge x^{\frac{1}{2} - \varepsilon}$ (or $x^{\sigma - \varepsilon}$ if $\zeta(\sigma + it) = 0$ for some $\sigma > 1/2$), where $f(X)$ is some explicit function in $X$? (Potentially naive) heuristics suggest one can take $f(X) = X + X^{2\sigma}$. REPLY [3 votes]: It was proved by Kaczorowski and Pintz (Acta Math. Hungar. 48 (1986), 173-185, doi: 10.1007/BF01949062) that there exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\geq x^{\sigma-\varepsilon}$, and there also exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\leq -x^{\sigma-\varepsilon}$. See Corollary 1 in their paper.<|endoftext|> TITLE: Rozendorn's Article QUESTION [5 upvotes]: I'm researching the isometric dips of the hyperbolic plane and in particular I'm interested in reading the results of Rozendorn who proved that the hyperbolic plane is isometrically immersed in $\mathbb R^5$. The demonstrations that I have been able to find in this regard, for example in Gromov's book, do not go with Rozendorn's initial idea, I say it because he demonstrates it in: but I have not been able to locate that article anywhere. Does anyone know where I can find it? or maybe someone knows other results in the same line of the original demonstration of him? From what I have been able to read, the proof that Rozendorn gives is based entirely on the functions that Blanusa gives in this Article in which he proved that the hyperbolic plane is isometrically embedded in $\mathbb R^6$. REPLY [5 votes]: If you don't read Russian: Rozendorn's construction is presented by Aminov in Extrinsic geometric properties of the Rozendorn surface, which is an isometric immersion of the Lobachevskiĭ plane into $E^5$. Now Aminov's paper was also in Russian, but the article does have an English translation available, which you can find at https://iopscience.iop.org/article/10.1070/SM2009v200n11ABEH004051. On pages 1576-1578 there's a pretty complete reproduction of Rozendorn's arguments.<|endoftext|> TITLE: Unsplitting sequence of vector bundles QUESTION [5 upvotes]: Let $V$ be a $n$-dimensional complex vector space. Using Grothendieck's notation, we define the Grassmannian $G(k,V)$ as the space of $k$-quotients of $V$ or, equivalently, as $$ G(k,V)=\{ \mathbb P W \subset \mathbb P V: \dim W=k\}. $$ On $G(k,V)$ we have the tautological bundle sequence: $$ 0 \to \mathcal S^\vee \to V \otimes \mathcal O_{G(k,V)} \to \mathcal Q \to 0. $$ We suppose that the following short exact sequence holds: $$ 0 \to \mathcal Q \otimes \mathcal S^\vee \to N \to \wedge^2 \mathcal Q \to 0, $$ where $N$ is another vector bundle on $G(k,V)$. I know that in the case $k=2$ $$ Ext^1(\wedge^2 \mathcal Q, \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V), \mathcal Q \otimes \mathcal S^\vee \otimes (\wedge^2 \mathcal Q)^\vee) $$ become, using $\wedge^2 \mathcal Q=\mathcal O(1)$ and $\mathcal Q(-1)=\mathcal Q^\vee$, $$ Ext^1(\mathcal O(1), \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V),\mathcal Q^\vee \otimes \mathcal S^\vee)=H^1(G(k,V), \Omega^1)=\mathbb C, $$ hence the short exact sequence do not split. I expect that this happens also in the general case, but I cannot prove it. How can I proceed? REPLY [5 votes]: $\def\CC{\mathbb{C}}$A splitting would be a global map $f : G(k,n) \times \CC^n \to \CC^n$ such that $f(L,v) \in L$ for all $L \in G(k,n)$ and $v \in \CC^n$. But, since $G(k,n)$ is projective and connected and the target $\CC^n$ is affine, any map $f : G(k,n) \times \CC^n \to \CC^n$ is constant on each $G(k,n) \times \{ v \}$. So we would have a map $f : \CC^n \to \CC^n$ such that $f(v) \in L$ for each $v \in \CC^n$ and each $L$ in $G(k,n)$. The only such map is the zero map.<|endoftext|> TITLE: A multiple integral that seems related to the $\zeta$ function at even integers QUESTION [20 upvotes]: I came across this integral that seems related to the Riemann zeta function $\zeta(2n)$ evaluated at even integers $2n \in 2\mathbb{Z}$. Letting $n$ be an even integer, define the multiple integral over $(2n+1)$ variables $u_1 \cdots u_{2n+1}$ \begin{equation} \mathcal{I}_{2n} = \int_0^1 du_1 \cdots \int_0^1 du_{2n+1} \frac{1}{1+u_1} \frac{1}{u_1+u_2} \frac{1}{u_2+u_3} \cdots \frac{1}{u_{2n} + u_{2n+1}} \frac{1}{u_{2n+1}+1}. \end{equation} For example, the case $2n=4$ has integrand $\frac{1}{1+u_1} \frac{1}{u_1+u_2} \frac{1}{u_2+u_3} \frac{1}{u_{3} + u_{4}} \frac{1}{u_{4}+u_5} \frac{1}{u_{5}+1}$. Below are listed some exact and numerical results for the integrals for the first few values of $2n$ that could be numerically evaluated on my laptop. The $2n=0$ case is easy, and the $2n=2$ case could evaluated exactly by Mathematica in terms of a complicated expression of polylogarithms (although below, I show an alternate way to explicitly derive this). $\mathcal{I}_0 = \frac{1}{2} = - \zeta(0)$ $\mathcal{I}_2 = \frac{\pi^2}{6} = \zeta(2)$ $\mathcal{I}_4 \approx 8 \frac{\pi^4}{90} = 8 \cdot \zeta(4)$ $\mathcal{I}_6 \approx 54 \cdot \zeta(6)$ $\mathcal{I}_8 \approx 384 \cdot \zeta(8)$ $\mathcal{I}_{10} \approx 2880 \cdot \zeta(10)$ These integrals are almost exactly integer multiples of the zeta function at even integers, up to the small errors given by Mathematica. The series of integers $1,8,54,384,2880$ don't appear in the OEIS, although the first terms $1,8,54,384$ go as $k^2 \cdot k!$, and $2880$ isn't far off from $5^2 \cdot 5! = 3000$. This integral seems closely related to the one considered in this question. However, their method for their integral seemed quite magical and I was not able to generalize it. While I'm specifically interested in this integral, it would be nice to know if there's a general method to deal with these integrals. Below are alternate formulations of the integral that may help, as well as showing the result for $2n=2$. First, one can change variables to $u_i = \frac{1-v_i}{1+v_i}$ to rewrite it as \begin{equation} \mathcal{I}_{2n} = \frac{1}{2} \int_0^1 dv_1 \cdots \int_0^1 dv_{2n+1} \frac{1}{1 - v_1 v_2} \frac{1}{1 - v_2 v_3} \cdots \frac{1}{1 - v_{2n} v_{2n+1}}, \end{equation} which is similar to a form found in the other MathOverflow question linked above. As such, the same substitutions used there done backwards yield other expressions \begin{equation} \begin{split} \mathcal{I}_{2n} &= \frac{1}{2} \int_0^1 dy_{2n+2} \int_0^{y_{2n+2}} dy_{2n+1} \cdots \int_{0}^{y_3} dy_{2} \frac{1}{y_3} \frac{1}{y_4-y_2} \cdots \frac{1}{y_{2n+2}-y_{2n+1}} \frac{1}{1-y_{2n+1}} \\ &= \frac{1}{2} \int_0^1 d \tilde{u}_{1} \cdots \int_0^1 d \tilde{u}_{2n+2} \frac{\delta(1-\tilde{u}_1 - \cdots - \tilde{u}_{2n+2})}{(\tilde{u}_1 + \tilde{u}_2)(\tilde{u}_2 + \tilde{u}_3)\cdots(\tilde{u}_{2n+1} + \tilde{u}_{2n+2})} \end{split} \end{equation} which almost match the integral considered in linked question with $2n+2$ variables, except missing the last factor $\frac{1}{\tilde{u}_{2n+2} + \tilde{u}_{1}}$ For $2n=2$, one can use Feynman Parameterization to write \begin{equation} \begin{split} \mathcal{I}_{2} &= \frac{1}{2} \int_0^1 dv_1 \int_0^1 dv_2 \int_0^1 dv_3 \frac{1}{1 - v_1 v_2} \frac{1}{1 - v_2 v_3} \\ &= \frac{1}{2} \int_0^1 du \int_0^1 dv_1 \int_0^1 dv_2 \int_0^1 dv_3 \frac{1}{\left[(1 - v_1 v_2)u + (1 - v_2 v_3)(1-u)\right]^2} \\ &= \frac{1}{2} \int_0^1 du \left( \frac{\log(1-u)}{u} + \frac{\log(u)}{1-u} \right) = \int_0^1 du \frac{\log(1-u)}{u} = \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}, \end{split} \end{equation} where the second-to-last equality follows from term-by-term integration of the series expansion. I haven't found a way to generalize the method for any of the other cases and still don't have analytical proof for the next equalities. EDIT: See below for a beautiful answer! It turns out that it's not related to the $\zeta$ function except for the first few values of $n$. REPLY [12 votes]: We have $$I_{2n}=\frac{(2n)!!}{(2n+1)!!}\cdot \frac1{2n+2}\cdot \pi^{2n}.$$ To see this, we follow the suggestion by Terry Tao in the comments and apply the diagonalization of the integral operator with the kernel $1/(x+y)$ on $[0,1]$. Change the variable to $1/x\in [1,\infty)$ and use (1.18) here (this is Mehler integral operator, as I understand) to diagonalize. Substituting the value of Legendre functions at 1, we get $$I_{2n}=\pi^{2n}\int_0^\infty x\tanh x/\cosh^{2n+2} x dx.$$ (The above part is suggested by Vladimir Petrov, I understand nothing about all this special functions and integral transforms stuff. But the answer for small $n$ coincides, so I guess everything is ok:) Well, you are free to ask for more details if necessary.) For evaluating these integrals, integrate by parts noting that $$\tanh x/\cosh^{2n+2} x dx=\tanh x\cosh^{-2n} xd\tanh x=\frac12(1-\tanh^2 x)^nd\tanh^2 x\\=\frac{-1}{2(n+1)}(1-\tanh^2x)^{n+1},$$ thus $$I_{2n}=\pi^{2n}\cdot \frac1{2(n+1)}\int_0^\infty (1-\tanh^2x)^{n+1}dx=\pi^{2n}\cdot \frac1{2(n+1)}\int_0^1 (1-t^2)^{n}dt\\=\frac{\pi^{2n}}{2(n+1)}\cdot \frac {(2n)!!}{(2n+1)!!}$$ (here $t=\tanh^2 x$, the last integral is well known and may be calculated by induction or using Beta function or how do you prefer.)<|endoftext|> TITLE: Permanent identities QUESTION [9 upvotes]: The permanent $\mathrm{per}(A)$ of a matrix $A$ of size $n\times n$ is defined to be: $$\mathrm{per}(A)=\sum_{\tau\in S_n}\prod_{j=1}^na_{j,\tau(j)}.$$ Let $$A=\left[\tan\pi\frac{j+k}n\right]_{1\le j,k\le n-1},$$ $$B=\left[\sin\pi\frac{j+k}n\right]_{1\le j,k\le n-1},$$ $$C=\left[\cos\pi\frac{j+k}n\right]_{1\le j,k\le n-1},$$ $$D=\left[\sec\pi\frac{j+k}n\right]_{1\le j,k\le n-1}.$$ Motivated by Question 402249, I found the following Conjecture 1. For any odd integer $n>1,$ $$(-1)^{(n-1)/2}\mathrm{per}(A)=\frac{2(n!!)^2}{n+1}\sum_{k=0}^{\frac{n-1}{2}}\frac{(-1)^k}{2k+1}. \tag{1}$$ Numerical calculations show that this is correct for $3 \leq n \leq 33$. See Question 402249 for details. Inspired by Question 402572, I also found the following identities Conjecture 2. For any odd integer $n>1,$ \begin{align} (-1)^{(n-1)/2}\mathrm{per}(B)&=\frac{n!}{2^{n-2}(n+1)}，\tag{2} \\ \mathrm{per}(C)&={\frac{(n-1)!}{2^{n-1}}}\sum_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}},\tag{3} \\ \mathrm{per}(D)&= (n-2)!!^2\left( (-1)^{\frac{n+1}{2}}+2n\sum_{k=0}^{\frac{n-1}{2}} {\frac {\left( -1 \right) ^{k}}{2k+1} } \right) .\tag{4} \end{align} Numerical calculations show that it is correct for $3 \leq n \leq 21$. Question. Are these identities correct? How to prove them? REPLY [10 votes]: Let $\zeta$ be a primitive $n$-th root of unity. Then $$\prod_{j=1}^{n-1}(x-\zeta^j)=\frac{x^n-1}{x-1}=1+x+\cdots+x^{n-1}$$ and hence $$\sigma_k=\sum_{1\le i_1<\cdots TITLE: How to determine if you've discovered a new identity for a special function QUESTION [11 upvotes]: Often times, we consult resources, like Abramowitz and Stegun's Handbook of Mathematical Functions https://www.math.ubc.ca/~cbm/aands/, NIST's database on special functions https://www.nist.gov/programs-projects/special-functions, or Mathematica to find identities which aid us with some kind of computation. However, what if we want to know if we have found a new identity, want to systematically check against the above resources, and want to add to the library in the case the identity is new? Also, are there journals which, even today, still consider mathematical effort toward discovering identities of classical functions? REPLY [11 votes]: Q: Are there journals which would publish identities of classical functions? A: Elsevier's Applied Mathematics and Computation has published quite a number of papers in that category, see this search listing. It is ranked as a Q1 journal, open access is optional.<|endoftext|> TITLE: Algebraic microlocal analysis and nonlinear PDE QUESTION [18 upvotes]: Though originating in the study of linear partial differential equations, microlocal analysis has become an invaluable tool in the study of nonlinear pde. Of particular importance has been the application of paraproducts/paradifferential operators and related tools developed by Bony, Coifman, Meyer and others. The theory of paradifferential operators fits nicely within the framework of microlocal analysis pioneered by Calderón, Zygmund, Hörmander, Kohn, Nirenberg, etc. However, there is another rather distinct school of (or approach to) microlocal analysis. This second school, developed by Sato, Kashiwara, Kawai and others, makes liberal use of tools from algebra as well as the theory of sheaves (hence algebraic microlocal analysis). Additionally, analytic functions (as opposed to $C^\infty$ functions) play a much more prominent role in algebraic microlocal analysis. As far as I'm aware (and this isn't terribly far as concerns algebraic microlocal analysis), one can obtain very similar theories of linear pde using either microlocal analysis or algebraic microlocal analysis (though, of course, some differences surely exist). If I'm wrong about this, I'd certainly be interested to hear more. However, I'm not familiar with any work applying algebraic microlocal analysis to the study of nonlinear pde. This leads to my question(s). I have heard that there is at least some work applying algebraic microlocal analysis to nonlinear pde. Could anyone point me towards some references? Additionally, I'd love to hear any thoughts or intuition anyone has about the ability to apply (or the limitations of the applicability of) algebraic microlocal analysis to the study of nonlinear pde. Edit: So, I was able to locate a couple references that are somewhat relevant to this question. In a set of lecture notes [1] by Pierre Schapira, he indicates that chiral algebras can be used to construct a theory of nonlinear pde in the spirit of $\mathcal{D}$-module theory. He gives [2] and [3] as references, which he says sketch the aforementioned theory. [1] P. Schapira. An Introduction to $\mathcal{D}$-Modules. Lecture Notes, https://perso.imj-prg.fr/pierre-schapira/wp-content/uploads/schapira-pub/lectnotes/Dmod.pdf, 2017. [2] A. Beilinson and V. Drinfeld. Chiral Algebras. Amer. Math. Soc. Coll. Publ., 51, Providence, RI, 2004. [3] M. Kapranov and Y. Manin. Modules and Morita Theorem for Operads. Amer. J. Math., 123:811-838, 2001. REPLY [11 votes]: Edit. I'have added a few other infos taken from the historical notes by Goro Kato and Daniele C. Struppa in [A1], to point out that the application of algebraic microlocal analytic (and thus cohomological) methods to nonlinear (system of) PDEs is nevertheless an old problem. This is not an answer but perhaps, as an extended comment, it can clarify a bit the status of recent relations between algebraic microlocal analysis and nonlinear PDEs. At the end of his BAMS review [2] (p. 118) of the monograph [1] by François Trèves and Paulo Domingos Cordaro, Pierre Schapira remarks: However, even if cohomological methods are in many situations a really efficient tool, one should keep in mind that they have their own limits. They rapidly become very hard to manipulate when dealing with precise growth conditions, and so far, don’t apply at all to non-linear problems, contrary to traditional methods which often are suitable with slight modifications. This is a good reason not to reject approaches based on explicit representations of cohomology, and this book is an attempt in this direction. Despite not being an expert in (algebraic) microlocal analysis, it seems to me that the status of the theory is still the one described by Schapira (who incidentally is one of the leading mathematicians and contributors in that area): algebraic microlocal analysis (in the sense of Kawai, Kashiwara and Schapira) is not (perhaps not yet) able to deal with nonlinear PDEs. However, it is possibly a good idea to ask directly prof. Schapira, since he and his colleague and coauthor Masaki Kashiwara are currently very active in developing this field of research. Addendum While searching for applications of algebraic microlocal analysis, I remembered this remark by Kato and Struppa ([A1], chapter 3, §3.5 p. 180): Around 1957, Sato began the cohomological study of systems of partial differential equations. The first occasion in which he publicly announced his work on this topic was the series of talks which he delivered in the Spring of 1960 at the Kawada Friday-Seminar just before his departure for the Institute for Advanced Study. During his talks, Sato put emphasis on the importance of cohomological treatments of systems of linear and non-linear partial differential equations. We can say that Sato’s algebraic analysis began in that moment. Thus, despite applications of the methods from microlocal algebraic analysis to nonlinear PDE currently seem to lack, this possibility was in the mind of Mikio Sato from the very first years of his introduction of the concept of hyperfunction. References [1] Cordaro, Paulo D.; Treves, François, Hyperfunctions on hypo-analytic manifolds. (English) Annals of Mathematics Studies, vol. 136, Princeton, NJ: Princeton University Press, pp. xx+377 (1994), ISBN: 0-691-02993-8, MR1311923, Zbl 0817.32001. [2] P. Schapira, "Book review: Hyperfunctions on hypo-analytic manifolds, by Paulo D. Cordaro and Francois Trèves, Annals of Mathematics Studies, vol. 136, Princeton University Press, Princeton, NJ, 1994, pp. xx+377, ISBN 0-691-92992-X", Bulletin of the American Maththematical Society, vol. 33 (1996) 115-118. Addendum reference [A1] Goro Kato, Daniele C. Struppa, Fundamentals of algebraic microlocal analysis. (English) Pure and Applied Mathematics, Marcel Dekker, 217. New York: Marcel Dekker, Inc. pp. x+296 (1999), ISBN: 0-8247-9327-7, MR1703357, Zbl 0924.35001.<|endoftext|> TITLE: Is there an increasing function on $[a, b]$ which is differentiable, but not absolutely continuous？ QUESTION [7 upvotes]: Is there an increasing function on $[a, b]$ which is differentiable, but not absolutely continuous？ REPLY [5 votes]: Such a function does not exist. Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f$ is Henstock–Kurzweil integrable on $[a,b]$. Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and hence $$f(x)-f(a)=(L)\int_a^x f'(t)\,dt$$ for all $x$, where $(L)\int$ is the Lebesgue integral. So, $f$ must be absolutely continuous.<|endoftext|> TITLE: A conjectural infinite series for $\frac{\pi^2}{5\sqrt{5}}$ QUESTION [12 upvotes]: I am looking for a proof of the following claim: First define the function $\chi(n)$ as follows: $$\chi(n)=\begin{cases}1, & \text{if }n \equiv \pm 1 \pmod{10} \\ -1, & \text{if }n \equiv \pm 3 \pmod{10} \\ 0, & \text{if otherwise } \end{cases}$$ Then, $$\frac{\pi^2}{5\sqrt{5}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n^2}$$ The SageMath cell that demonstrates this claim can be found here. REPLY [11 votes]: This is asking for the value of an $L$-function of an even Dirichlet character $\chi$ at a positive even integer, and these have known values. It is analogous to the explicit expressions for the Riemann zeta-function at positive even integers (definitely not at positive odd integers!), but instead of the values being rational multiples of powers of $\pi$, they are algebraic multiples of powers of $\pi$. The particular Dirichlet character $\chi$ in this question is defined modulo 10, but it is a lifting of a character mod 5. Let $\psi(n) = (\frac{n}{5})$, which is a nontrivial even Dirichlet character mod 5. For odd $n$, $\chi(n) = \psi(n)$, so for ${\rm Re}(s) > 1$, $$ L(s,\chi) = \sum_{n \geq 1} \frac{\chi(n)}{n^s} = \prod_{p > 2} \frac{1}{1 - \psi(p)/p^s} = \left(1 - \frac{\psi(2)}{2^s}\right)L(s,\psi) = \left(1 + \frac{1}{2^s}\right)L(s,\psi). $$ Taking $s = 2$, $$ \sum_{n \geq 1} \frac{\chi(n)}{n^2} = \frac{5}{4}L(2,\psi). $$ I will show $L(2,\psi) = \frac{4}{25\sqrt{5}}\pi^2$, and multiplying this by $5/4$ gives the desired value for $\sum \chi(n)/n^2$. The character $\psi \bmod 5$ is primitive since every nontrivial Dirichlet character modulo a prime is primitive. It is a quadratic character: its nonzero values are $\pm 1$. For each even primitive quadratic Dirichlet character $\eta \bmod m$, let's work out a formula for $L(k,\eta)$ when $k$ is a positive even integer and then apply it to $\eta = \psi$ and $k = 2$. (There are analogous formulas for $L(k,\eta)$ when $\eta \bmod m$ is an odd primitive quadratic Dirichlet character and $k$ is a positive odd integer, and also formulas when $\eta$ is not quadratic, but I omit all of this for simplicity.) For even primitive quadratic $\eta \bmod m$ and ${\rm Re}(s) > 1$, the completed $L$-function of $\eta$ is $$ \Lambda(s,\eta) := \left(\frac{\pi}{m}\right)^{-s/2}\Gamma\left(\frac{s}{2}\right)L(s,\eta). $$ This turns out to be an entire function, so $L(s,\eta)$ is also entire, and we have a functional equation $\Lambda(s,\eta) = \Lambda(1-s,\eta)$. (If $\eta$ were not quadratic, the functional equation would be more complicated.) When $s = k$ is a positive even integer, unraveling the formula $\Lambda(k,\eta) = \Lambda(1-k,\eta)$ gives us $$ \left(\frac{\pi}{m}\right)^{-k/2}\Gamma\left(\frac{k}{2}\right)L(k,\eta) = \left(\frac{\pi}{m}\right)^{-(1-k)/2}\Gamma\left(\frac{1-k}{2}\right)L(1-k,\eta), $$ so $$ L(k,\eta) = \left(\frac{\pi}{m}\right)^{k - 1/2}\frac{\Gamma((1-k)/2)}{\Gamma(k/2)}L(1-k,\eta). $$ The first factor on the right is a known expression. Using the reflection and duplication formulas for the Gamma-function, $$ \frac{\Gamma((1-s)/2))}{\Gamma(s/2)} = \frac{2^{s-1}\sqrt{\pi}}{\cos\left(\frac{\pi}{2}s\right)\Gamma(s)}. $$ Letting $s = k$ be an even integer, $\cos((\pi/2)k) = \cos((k/2)\pi) = (-1)^{k/2}$, so $$ L(k,\eta) = \left(\frac{\pi}{m}\right)^{k - 1/2}\frac{(-1)^{k/2}2^{k-1}\sqrt{\pi}}{(k-1)!}L(1-k,\eta) = \frac{(-1)^{k/2}2^{k-1}}{(k-1)!m^{k-1/2}}L(1-k,\eta)\pi^k $$ and the number $L(1-k,\eta)$ for positive integers $k$ and nontrivial primitive $\eta \bmod m$ turns out to be algebraic. (That generalizes the Riemann zeta-function at negative integers being rational). Explicitly, $$ L(1-k,\eta) = -\frac{B_{k,\eta}}{k} $$ where the "twisted" Bernoulli number $B_{k,\eta}$ has exponential generating function $$ \sum_{k \geq 0} B_{k,\eta}\frac{x^k}{k!} = \sum_{j=1}^m \eta(j)\frac{xe^{jx}}{e^{mx}-1}, $$ so $B_{k,\eta}$ is a $\mathbf Q$-linear combination of values of $\eta$. (See Larry Washington's "Introduction to Cyclotomic Fields", esp. Theorem 4.2.) Let's specialize all this to the case when $\eta$ is the (even primitive) quadratic character $\psi \bmod 5$ above and $k = 2$. We get $$ L(2,\psi) = \frac{-2}{5^{3/2}}L(-1,\eta)\pi^2 = \frac{B_{2,\psi}}{5\sqrt{5}}\pi^2. $$ We had said the coefficient of $\pi^2$ turns out to be $4/(25\sqrt{5})$, and we'll get this by showing $B_{2,\psi} = 4/5$. For the character $\psi \bmod 5$, $$ \sum_{k \geq 0} \frac{B_{k,\psi}}{k!}x^k = \frac{x(e^x - e^{2x}-e^{3x}+e^{4x})}{e^{5x}-1} = \frac{xe^x(e^x-1)^2(e^x+1)}{e^{5x}-1} = \frac{2}{5}x^2 - \frac{1}{3}x^4 + \cdots, $$ so looking at the coefficient of $x^2$ on both sides tells us $B_{2,\psi} = 4/5$.<|endoftext|> TITLE: Numbers without prime factors in a set of positive relative density QUESTION [14 upvotes]: Let $\mathcal{P}$ be a set of prime numbers of relative density $\kappa \in (0, 1)$, which means that $$\#\left(\mathcal{P} \cap [1, x]\right) = \kappa \,\pi(x) + E(x) \quad (x \to \infty)$$ for a "suitable" error term $E(x)$ (of course, $E(x) = o(\pi(x))$). Let $\mathcal{S}$ be the set of natural numbers having no prime factor in $\mathcal{P}$. Question: Do we have an asymptotic formula for $\#\left(\mathcal{S} \cap [1, x]\right)$ ? For $\kappa \in (0, 1/2)$, Corollary 1.2 of [1] gives that $$\#\left(\mathcal{S} \cap [1, x]\right) \sim C(\mathcal{P}) \frac{x}{(\log x)^\kappa} \quad (x \to +\infty)$$ where $C(\mathcal{P})$ is a positive constant depending on $\mathcal{P}$. I guess that something similar holds in the whole range $\kappa \in (0, 1)$, but I could not find such a result. Thank you for any help. P.S. Although I am interested in general set of primes $\mathcal{P}$ of relative density $\kappa$, results concerning sufficiently general $\mathcal{P}$ of relative density $\kappa$ are welcome. For example, in light of Daniel Loughran's comment, sets like those of Chebotarev density theorem $$\mathcal{P} = \{p \text{ unramified in $K$}, \mathrm{Frob}_p \subseteq C\} ,$$ where $K$ is a finite Galois extension of $\mathbb{Q}$ and $C$ is a conjugacy class of its Galois group, are certainly interesting. P.S. This is very related to question: Natural density of set of numbers not divisible by any prime in an infinite subset , as pointed out by user Hhhhhhhhhhh. [1] Iwaniec and Kowalski, Analytic Number Theory (2004) REPLY [20 votes]: You basically ask about the sum $$ \sum_{n \le x} \alpha(n)$$ where $\alpha$ is a completely multiplicative function with $\alpha(p) = \mathbf{1}_{p \notin \mathcal{P}}$. This is addressed by Wirsing in his famous paper ``Das asymptotische Verhalten von Summen über multiplikative Funktionen'' (Math. Ann. 143 (1961), 75–102). The only requirement on $E$ is $E(x)=o(\pi(x))$, and it gives the asymptotic result $$\tag{$\star$} \sum_{n \le x} \alpha(n) \sim\frac{ e^{-\gamma (1-\kappa)}}{\Gamma(1-\kappa)} \frac{x}{\log x} \prod_{p \le x,\, p \notin \mathcal{P}}(1-1/p)^{-1},$$ where $\gamma$ is the Euler-Mascheroni constant (appearing also in Mertens' theorem). Remark 1: Suppose $\sum_{p \le x, p \in \mathcal{P}} 1/p = \kappa\sum_{p \le x} 1/p +C + o(1)$, which holds if $E(x)$ is small enough, say $O(x/\log^{1+\varepsilon} x)$ (by partial summation). Then $$C(\mathcal{P}) :=\frac{ 1}{\Gamma(1-\kappa)} \prod_{p \notin \mathcal{P}}(1-1/p)^{-1} \prod_{p}(1-1/p)^{1-\kappa}$$ converges and the last result may be simplified as $$C(\mathcal{P})\frac{x}{(\log x)^{\kappa}},$$ by Mertens. This should recover the result from Iwaniec and Kowalski. Remark 2: In Wirsing's sequel to his own paper, ``Das asymptotische Verhalten von Summen über multiplikative Funktionen. II'' (Acta Math. Acad. Sci. Hungar. 18 (1967), 411–467) he relaxes the condition on $\mathcal{P}$ even further, requiring less than positive relative density, while still retaining $(\star)$.<|endoftext|> TITLE: Proving a lemma for a decomposition of orthogonal matrices QUESTION [7 upvotes]: Setting Consider two independent orthogonal matrices, which are decomposed into 4 blocks: \begin{align} Q_{1} = \left[\begin{array}{cc} A_{1} & B_{1}\\ C_{1} & D_{1} \end{array}\right] , \,Q_{2}=\left[\begin{array}{cc} A_{2} & B_{2}\\ C_{2} & D_{2} \end{array}\right]\in \mathbb{R}^{d\times d} \end{align} where $A_i \in \mathbb{R}^{r\times r}$ and $D_i \in \mathbb{C}^{(d-r)\times (d-r)}$ and $d\ge 3r$. Also notice that \begin{align} \forall i=1,2:I_{d}=\left[\begin{array}{cc} I_{r} & 0\\ 0 & I_{d-r} \end{array}\right] &= Q_{i}^{\top}Q_{i}=\left[\begin{array}{cc} A_{i}^{\top}A_{i}+C_{i}^{\top}C_{i} & A_{i}^{\top}B_{i}+C_{i}^{\top}D_{i}\\ B_{i}^{\top}A_{i}+D_{i}^{\top}C_{i} & B_{i}^{\top}B_{i}+D_{i}^{\top}D_{i} \end{array}\right] \\ &=Q_{i}Q_{i}^{\top}=\left[\begin{array}{cc} A_{i}A_{i}^{\top}+B_{i}B_{i}^{\top} & A_{i}C_{i}^{\top}+B_{i}D_{i}^{\top}\\ C_{i}A_{i}^{\top}+D_{i}B_{i}^{\top} & C_{i}C_{i}^{\top}+D_{i}D_{i}^{\top} \end{array}\right] \end{align} Goal I am trying for a few days to prove (or refute) the following lemma. In my numerical simulations, it always holds for random $Q$s. Prove (or refute) that if $v\in\mathbb{C}^{d-r}$ is an eigenvector of $D_1^\top M$ (with $\lambda=1$), i.e., \begin{align} D_{1}^{\top} \underbrace{D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)}_{\triangleq M}v&=v \end{align} then it is in the nullspace of $C_2^\top M$, i.e., \begin{align} C_{2}^{\top} \underbrace{D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)}_{=M}v&=0_{r} \end{align} Update: Alternative goal If one could show that any eigenvector such that $D_{1}^{\top} D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v=v$ is also in the null space of $B_1$, i.e., $B_1 v = 0_r$, then I know how to prove the rest of the lemma above using it. And again, empirically, this also seems to always hold. I feel like I may be missing some simple trick here. Any help would be greatly appreciated. REPLY [3 votes]: Using the direction Federico Poloni gave, I was finally able to prove it. Notice that the matrices are real, so I use $A^\top$ and $A^H$ interchangeably. Let $v\in\mathbb{\mathbb{C}}^{d-r}$ be a normalized eigenvector as required. Lemma: The norm of $v$ is preserved under the following transformations $\left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}=\left\Vert D_{2}^{\top}D_{1}v\right\Vert _{2}=\left\Vert D_{1}v\right\Vert _{2}$. Proof. We use a simple trick using the Euclidean (operator) norms, \begin{align} v&=D_{1}^{\top}D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v\\v^{H}v&=v^{H}D_{1}^{H}D_{2}\left(B_{2}^{H}B_{1}+D_{2}^{H}D_{1}\right)v\\1&=\left\Vert v^{H}D_{1}^{H}D_{2}\left(B_{2}^{H}B_{1}+D_{2}^{H}D_{1}\right)v\right\Vert _{2}\\&\le\underbrace{\left\Vert v^{H}D_{1}^{H}D_{2}\right\Vert _{2}}_{\le1}\underbrace{\left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}}_{\le1}\\&\le\underbrace{\left\Vert v^{H}D_{1}^{H}\right\Vert _{2}}_{\le1}\underbrace{\left\Vert D_{2}\right\Vert _{2}}_{\le1}\underbrace{\left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}}_{\le1}\le1, \end{align} where we used the fact that $\left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)$ is the bottom-right block of the orthogonal matrix $Q_{2}^{\top}Q_{1}$. As a conclusion, we get the following equalities: \begin{align} \left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}&=1\\\left\Vert v^{H}D_{1}^{H}D_{2}\right\Vert _{2}=\left\Vert D_{2}^{H}D_{1}v\right\Vert _{2}=\left\Vert D_{2}^{\top}D_{1}v\right\Vert _{2}&=1\\\left\Vert v^{H}D_{1}^{H}\right\Vert _{2}=\left\Vert D_{1}v\right\Vert _{2}&=1. \end{align} Lemma: The vector $v$ is in the null space of $B_{2}^{\top}B_{1}$, i.e., $B_{2}^{\top}B_{1}v=0_{r}$. Proof. We get back to the equality above and show \begin{align}1&=v^{H}D_{1}^{H}D_{2}\left(B_{2}^{H}B_{1}+D_{2}^{H}D_{1}\right)v\\&=v^{H}D_{1}^{H}D_{2}B_{2}^{H}B_{1}v+v^{H}D_{1}^{H}D_{2}D_{2}^{H}D_{1}v\\&=v^{H}D_{1}^{H}D_{2}B_{2}^{H}B_{1}v+\underbrace{\left\Vert D_{2}^{H}D_{1}v\right\Vert _{2}}_{=1}\\0&=v^{H}D_{1}^{H}D_{2}B_{2}^{H}B_{1}v. \end{align} We use the fact that $\left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v$ is a normalized vector to obtain \begin{align} 1&=\left\Vert \left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v\right\Vert _{2}^{2}=\left\Vert B_{2}^{\top}B_{1}v\right\Vert _{2}^{2}+\underbrace{2v^{\top}D_{1}^{\top}D_{2}B_{2}^{\top}B_{1}v}_{=0}+\underbrace{\left\Vert D_{2}^{\top}D_{1}v\right\Vert _{2}^{2}}_{=1}\\0&=\left\Vert B_{2}^{\top}B_{1}v\right\Vert _{2}^{2}\\0_{r}&=B_{2}^{\top}B_{1}v. \end{align} Lemma: The vector $v$ is in the null space of $C_{2}^{\top}D_{1}$. Proof. We use the above results, \begin{align} 1&=v^{H}D_{1}^{H}D_{2}D_{2}^{H}D_{1}v=v^{H}D_{1}^{H}\left(I_{d-r}-C_{2}C_{2}^{H}\right)D_{1}v \\&=\underbrace{v^{H}D_{1}^{H}D_{1}v}_{=1}-v^{\top}D_{1}^{H}C_{2}C_{2}^{H}D_{1}v\\0&=\left\Vert C_{2}^{\top}D_{1}v\right\Vert _{2}^{2}\\0_{r}&=C_{2}^{\top}D_{1}v. \end{align} Finally, we are ready to conclude the theorem we wanted to prove. \begin{align} C_{2}^{\top}D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v &=C_{2}^{\top}D_{2}D_{2}^{\top}D_{1}v \\&=C_{2}^{\top}\left(I_{d-r}-C_{2}C_{2}^{T}\right)D_{1}v \\ &=\underbrace{C_{2}^{\top}D_{1}v}_{=0_{r}}-C_{2}^{\top}C_{2}\underbrace{C_{2}^{\top}D_{1}v}_{=0_{r}}=0_{r} \end{align}<|endoftext|> TITLE: Artin vanishing for Stein manifolds and restriction maps QUESTION [13 upvotes]: In the setting of complex Stein manifolds $X$ of complex dimension $d$, the theorem of Andreotti--Frankel implies the vanishing of the singular cohomology group $H^i(X,\mathbb Z)=0$ for $i>d$. With complex coefficients, a simple argument for this is to compute the cohomology in terms of the cohomology of the de Rham complex. Their theorem gives a more precise Morse-theoretic statement. Now let $U\subset X$ be another Stein manifold open in $X$, and assume that the map $\mathcal O(X)\to \mathcal O(U)$ has dense image. (This condition is not automatic, and necessary for the following.) Conjecture. In top degree, the map $H^d(X,\mathbb Z)\to H^d(U,\mathbb Z)$ is surjective. Is this known? The same result should also be true with constructible coefficients. With $\mathbb C$-coefficients, it follows from the comparison with de Rham cohomology (at least when cohomology groups are finite-dimensional, which I'm happy to assume). Is there some argument using Morse theory? The motivation for the question is that the analogue in rigid-analytic geometry is true (but I found it quite surprising); it is essentially equivalent to a version of Artin vanishing for affine schemes over absolutely integrally closed valuation rings stated by Gabber in Oberwolfach last year. REPLY [12 votes]: The pairs (U,X) are called Runge pairs. The homology version of your statement is proved in the paper of Andreotti and Narasimhan Annals of Math vol 76 no 3 (1962) 499-509 using Morse Theory.The title of the paper is "A Topological property of Runge pairs" The paper by Coltoiu Mihalache titled On the Homology Groups of Stein spaces and Runge pairs, Journal fur reine und angewandte Mathematik volume 371 no 5 pages 215-220 proves the homology statement for Runge pairs of Stein spaces.<|endoftext|> TITLE: John von Neumann's remark on entropy QUESTION [20 upvotes]: According to Claude Shannon, von Neumann gave him very useful advice on what to call his measure of information content [1]: My greatest concern was what to call it. I thought of calling it 'information,' but the word was overly used, so I decided to call it 'uncertainty.' When I discussed it with John von Neumann, he had a better idea. Von Neumann told me, 'You should call it entropy, for two reasons. In the first place your uncertainty function has been used in statistical mechanics under that name, so it already has a name. In the second place, and more important, no one really knows what entropy really is, so in a debate you will always have the advantage.' What I am curious about is what von Neumann meant with his last point. I find it particularly surprising given that he axiomatised what we now call the von Neumann entropy in Quantum Mechanics around two decades prior to Shannon's development of Classical Information Theory. Might modern information theorists know of the specific difficulties he had in mind and whether these have been suitably addressed? References: McIrvine, Edward C. and Tribus, Myron (1971). Energy and Information Scientific American 225(3): 179-190. von Neumann, John (1932). Mathematische Grundlagen der Quantenmechanik (Mathematical Foundations of Quantum Mechanics) Princeton University Press., . ISBN 978-0-691-02893-4. Shannon, Claude E. (1948). A Mathematical Theory of Communication Bell System Technical Journal 27: 379-423. doi:10.1002/j.1538-7305.1948.tb01338.x. Olivier Rioul. This is IT: A Primer on Shannon’s Entropy and Information. Séminaire Poincaré. 2018. E.T. Jaynes. Information Theory and Statistical Mechanics. The Physical Review. 1957. John A. Wheeler, 1990, “Information, physics, quantum: The search for links” in W. Zurek (ed.) Complexity, Entropy, and the Physics of Information. Redwood City, CA: Addison-Wesley. REPLY [2 votes]: It seems to me that it is actually the first reason put forward by von Neumann (your uncertainty function has been used in statistical mechanics under that name) that fully justifies Timothy Chow's remark concerning the second one: Surely von Neumann's comment was partially a joke, so it may be misguided to try to interpret it too literally. Let me outline the context in which von Neumann gave his advice, and to what extent Shannon's entropy $$ \tag{H} H(p) = - \sum p_i \log p_i $$ was by that time (the late 40s) known and used in statistical mechanics. Boltzmann gave a probabilistic interpretation of the Clausius' thermodynamic entropy as the negative logarithm of the "probability" for the system to be in a given state. I write "probability", because actually this quantity (called Permutabilitätmass, "permutability measure" by Boltzmann, "thermodynamic probability" by Planck, and now usually referred to as Boltzmann's entropy) was obtained by a certain limit procedure involving combinatorial calculations with infinitesimal regions in the phase space of a single molecule, and therefore was defined up to an unspecified additive constant. It results in the differential entropy (in modern terms) of the empirical distribution on the phase space. Gibbs further considered the integral of Boltzmann's entropy over the state space (nowadays called Gibbs' entropy). Neither Boltzmann nor Gibbs were interested in the sums like (H) per se, and they appear in their work only as a technical tool (with $p_i$ not necessarily being probability weights). For instance, the proof of Theorem VIII in Gibbs' Elementary Principles in Statistical Mechanics amounts to establishing what is now known as Gibbs' inequality ($\equiv$ positivity of the Kullback - Leibler divergence for discrete distributions), but Gibbs feels no need to look at it as a property of the discrete entropy. It is Planck who explicitly states and addresses the problem of the physical meaning of the limit procedure used by Boltzmann which leads him to the idea that quantization is a physical reality. I can't resist the temptation to quote his 1925 lecture The Origin and Development of the Quantum Theory where he recalls how, after having announced his "radiation formula" at the October 19, 1900 meeting of the Berlin Physical Society, he then faced the necessity to give it a conceptual interpretation: If, however, the radiation formula should be shown to be absolutely exact, it would possess only a limited value, in the sense that it is a fortunate guess at an interpolation formula. Therefore, since it was first enunciated, I have been trying to give it a real physical meaning, and this problem led me to consider the relation between entropy and probability, along the lines of Boltzmann's ideas. After a few weeks of the most strenuous work of my life, the darkness lifted and an unexpected vista began to appear. This vista was the first glimpse of the quantum theory. To enjoy very lucid Planck's style again (from the introduction to the second edition of The Theory of Heat Radiation, German original 1913, English translation 1914): ... the hypothesis of quanta as well as the heat theorem of Nernst may be reduced to the simple proposition that the thermodynamic probability of a physical state is a definite integral number, or, what amounts to the same thing, that the entropy of a state has a quite definite, positive value, which, as a minimum, becomes zero, while in contrast therewith the entropy may, according to the classical thermodynamics, decrease without limit to minus infinity. For the present, I would consider this proposition as the very quintessence of the hypothesis of quanta. To put it in modern mathematical language, termodynamic entropy is a function, not a cocycle, so that not only its increments, but also its "absolute value" has a physical meaning. The "absolute entropy" $S$ of Planck is equal to the product of "Shannon's entropy" (H) of the arising physically meaningful discrete probability distribution $(w_i)$ by the Boltzmann constant $k$ (also introduced by Planck) and the number of molecules $N$. This is very explicit formula (173) in the aforementioned Planck's book: $$ S = - kN \sum w_i \log w_i \;. $$ It is this very formula that is directly and equally explicitly quoted by von Neumann in Mathematische Grundlagen der Quantenmechanik on p. 210 when claiming that his quantum entropy is analogous to the thermodynamic one. Needless to say, Planck's work was in the very mainstream of the 20th century physics. By the late 30s the definition (H) of the entropy of a probability distribution appears as standard in books on statistical mechanics (for instance, not only in the quoted by Shannon Tolman's book The principles of statistical mechanics, 1938, but also in Slater's Introduction to chemical physics, 1939). For a physical discussion of entropy in classical statistical vs quantum mechanics see very instructive Chapter 14 (written by Goldstein, Lebowitz, Tumulka, and Zanghì) of the recent book Statistical Mechanics and Scientific Explanation. Its preprint is available at Gibbs and Boltzmann Entropy in Classical and Quantum Mechanics.<|endoftext|> TITLE: Homeomorphism with finitely many fixed points acting by -1 on $H^2(CP^2)$ QUESTION [7 upvotes]: Is there a homeomorphism $f:CP^2\to CP^2$ that has finitely many fixed points and acts by -1 on $H^2$? REPLY [12 votes]: Let $a_1$, $a_2$ and $a_3$ be three distinct positive real numbers. Take the map $$[z_1 : z_2 : z_3] \mapsto [a_1 \overline{z_1}: a_2 \overline{z_2} : a_3 \overline{z_3}]$$ where the bar is complex conjugation. I claim that the only fixed points are $[1:0:0]$, $[0:1:0]$ and $[0:0:1]$. Suppose that to the contrary that $[z_1:z_2:z_3]$ is fixed and that $z_i z_j \neq 0$ for some indices $i \neq j$. Then there is a scalar $\lambda$ such that $$\lambda = \frac{a_i \overline{z_i}}{z_i} = \frac{a_j \overline{z_j}}{z_j}.$$ But then we have $$|\lambda| = a_i = a_j,$$ contradicting that the $a$'s are distinct.<|endoftext|> TITLE: Class-theoretic sentences that are $\Pi^1_1$ or $\Pi^1_2$ QUESTION [5 upvotes]: I'm looking for the following: (1) an example of a $\Pi^1_1$ class-theoretic sentence that has no known equivalence to a $\Sigma^1_1$ sentence, even if large cardinal hypotheses or reflection principles are assumed. (2) an example of a $\Pi^1_2$ class-theoretic sentence that has no known equivalence to a $\Sigma^1_2$ sentence, even if large cardinal hypotheses or reflection principles are assumed. Admttedly, "large cardinal hypothesis" and "reflection principle" are not well-defined, so these questions are rather vague. I am assuming impredicative Comprehension (Kelley-Morse class theory) and all forms of choice. The best answer I currently have for (1) is Vopenka's principle ("every proper class of directed graphs has either a member with a nontrivial endomorphism or two distinct members with a homomorphism between them"). But this is often considered a large cardinal hypothesis, so it's not a good answer. I don't have any answer for (2). REPLY [4 votes]: Let's work in GBC for definiteness. In analogy with combinatorics on $\omega_1$, define an $\mathsf{ORD}$-tree to be a (proper class sized) tree of height $\mathsf{ORD}$ so that each level is set sized. Such a tree is Souslin if all of its chains and antichains are set sized as well. By essentially the same arguments as in the case of any regular $\kappa$, the class-sized forcing consisting of normal binary trees ordered by end-extension adds an $\mathsf{ORD}$-Souslin tree generically while preserving GBC (and KM if it held in the ground model). Moreover the forcing does not add sets so it cannot change any first order property of the model. Conversely, in unpublished work, Joel David Hamkins and I discovered that any countable model of GBC has an expansion with the same first order part satisfying GBC plus ``there are no $\mathsf{ORD}$-Souslin trees". This model won't satisfy KM even if it held in the ground model, for what it's worth. The point being, similar to the case of any regular $\kappa$, the existence of an $\mathsf{ORD}$-Souslin tree is independent of GBC and neither the existence of such a tree nor the non-existence of such a tree has any first order consequences over GBC as a base theory. Now, for a fixed $\mathsf{ORD}$-tree $T$, to say that "$T$ is Souslin" is a $\Pi^1_1$ statement (with $T$ as a parameter of course) since we're really saying ``for all $A \subseteq T$ if $A$ is a chain or an antichain then it is set sized". It's also properly $\Pi^1_1$. If it were equivalent to a $\Sigma^1_1$ sentence then it would be absolute to class-forcing extensions preserving GBC (or KM if that's your ground model theory). But, since we can add a class-sized branch (and antichain) by forcing with $T$, just like in the case on $\omega_1$ this is not true. Some work is needed to show that forcing with the tree preserves GBC/KM but this is in fact true and is essentially a consequence of the fact that the tree is $\mathsf{ORD}$-c.c. and adds no sets. It follows that to say, for a fixed $\mathsf{ORD}$-tree $T$ that "$T$ is not Souslin" is properly $\Sigma^1_1$. Clearly we have that to say "there is an $\mathsf{ORD}$-Souslin tree" is $\Sigma^1_2$. I would assume, given the above that it is properly so, but I haven't been able to dot the i's and cross the t's on such a proof.<|endoftext|> TITLE: Every homeomorphism isotopic to one with finitely many fixed points QUESTION [10 upvotes]: Is every homeomorphism from a compact manifold to itself isotopic to a homeomorphism with finitely many fixed points? REPLY [4 votes]: This is always true for surfaces. For dimension 3 and higher it is known that the homeomorphism may always be homotoped to a homeomorphism with finitely many fixed points. Since this question has been unanswered for a while I thought these references might be of interest. For surfaces it follows from the main theorem of "Fixed points of surface diffeomorphisms" Boju Jiang and Jianhan Guo, Pacific J. Math. Vol. 160, No. 1, 1993. ProjectEuclid link Quoting from the paper: Theorem: (Jiang-Guo) Let $f: M \rightarrow M$ be a homeomorphism of a compact surface. When $M$ is closed, then $f$ is isotopic to a diffeomorphism with $N(f)$ fixed points, where $N(f)$ is its Nielsen number. When $M$ has boundary, $N(f)$ should be replaced by the relative Nielsen number $N(f;M,dM)$ defined by Schirmer. Note that they prove something stronger that the number of fixed points can always taken to be the Nielsen number, for finitely fixed points it may have been known earlier (I am not an expert in this field so I don't know). In dimension $\geq 3$ the statement up to homotopy follows from a theorem of Wecken (this is mentioned in the introduction of Jiang and Guo's article): W] F. Wecken, Fixpunktklassen, I, Math. Ann., 117 (1941), 659-671; II, Math. Ann., 118 (1942), 216-234; III, Math. Ann., 118 (1942), 544-577: Theorem (Wecken): Any self-homeomorphism $F$ of a manifold of dimension at least 3 may be homotoped to a map with exactly $N(F)$ fixed points. I don't know whether it is known up to isotopy, since Jiang and Guo don't mention such a result I think it is reasonable to guess that it was open as of 1993 (although again I am not an expert so I cannot say for sure).<|endoftext|> TITLE: Possible small mistake in Bilu-Hanrot-Voutier paper on primitive divisors of Lehmer sequences (?) QUESTION [8 upvotes]: I think that I might have spotted I small mistake (a missing $5$-defective Lehmer pair) in the classification of terms of Lehmer sequences without primitive divisors given in: 1 Bilu, Hanrot, and Voutier, Existence of primitive divisors of Lucas and Lehmer numbers, Journal für die reine und angewandte Mathematik (Crelles Journal), 2001. (available on the web) Since I might very much be wrong myself, I would like to know if my reasoning is correct or not. I recall below the main definitions of 1. A Lehmer pair is a pair of complex numbers $(\alpha, \beta)$ such that $(\alpha + \beta)^2$ and $\alpha\beta$ are non-zero coprime integers and $\alpha / \beta$ is not a root of unity. Two Lehmer pairs $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$ are said to be equivalent if $\alpha_1 / \alpha_2 = \beta_1 / \beta_2 \in \{-1,+1,\sqrt{-1},-\sqrt{-1}\}$. Given a Lehmer pair $(\alpha, \beta)$, the associated Lehmer sequence is $$\widetilde{u}_n(\alpha, \beta) := \begin{cases} (\alpha^n - \beta^n) / (\alpha - \beta) & \text{ if $n$ is odd} \\ (\alpha^n - \beta^n) / (\alpha^2 - \beta^2) & \text{ if $n$ is even}\end{cases}$$ for every positive integer $n$ (it is an integer sequence). A prime number $p$ is a primitive divisor of $\widetilde{u}_n(\alpha, \beta)$ if $p$ divides $\widetilde{u}_n(\alpha, \beta)$ but does not divide $(\alpha^2 - \beta^2)^2 \widetilde{u}_1(\alpha, \beta) \cdots \widetilde{u}_{n-1}(\alpha, \beta)$. If $\widetilde{u}_n(\alpha, \beta)$ has no primitive divisor then the Lehmer pair $(\alpha, \beta)$ is $n$-defective. One of the claims of Theorem 1.3 of 1 is that, up to equivalence, all $5$-defective Lehmer pairs are of the form $((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with $$(1) \qquad (a, b) = (\phi_{k-2\varepsilon}, \phi_{k-2\varepsilon} - 4\phi_k) \quad (k \geq 3)$$ or $$(2) \qquad (a, b) = (\psi_{k-2\varepsilon}, \psi_{k-2\varepsilon} - 4\psi_k) \quad (k \neq 1) ,$$ where $k$ is a nonnegative integer, $\varepsilon \in \{-1, +1\}$, $(\phi_n)$ is the sequence of Fibonacci numbers, and $(\psi_n)$ is the sequence of Lucas numbers. Claim 1: The Lehmer pair $(\alpha_0, \beta_0) := ((1 - \sqrt{5}) / 2, (1 + \sqrt{5}) / 2)$ is $5$-defective. First, note that $(\alpha_0 + \beta_0)^2 = 1$ and $\alpha_0\beta_0 = -1$ are non-zero coprime integers and $\alpha_0 / \beta_0 = (\sqrt{5}-3) / 2$ is not a root of unity, so that $(\alpha_0, \beta_0)$ is indeed a Lehmer pair. Second, for the associated Lehmer sequence we have $\widetilde{u}_5 = 5$ and $(\alpha_0^2 - \beta_0^2)^2 = 5$, thus $\widetilde{u}_5$ has no primitive divisor and $(\alpha_0, \beta_0)$ is $5$-defective. Claim 2: The Lehmer pair $(\alpha_0, \beta_0)$ is not equivalent to a pair of the form $(\alpha, \beta) = ((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with $(a, b)$ as in (1) or (2). For the sake of contradiction suppose $(\alpha_0, \beta_0)$ is equivalent to a pair of the form $(\alpha, \beta) = ((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with $(a, b)$ as in (1) or (2). Then $(a - b) / 4 = \alpha\beta = \pm \alpha_0 \beta_0 = \pm 1$ so that $a - b = \pm 4$. In case (1), we have $a - b = 4\phi_k \geq 8$, because $k \geq 3$. In case (2), we have $a - b = 4\phi_k \geq 8$, because $k \neq 1$. Absurd. Possible source of the error: I think that the missing $5$-defective pair is lost in the last paragraph of case $n = 5$ in section "Small $n$" of 1. It is said that: "By (28), we have $k \geq 3$ in the case (34), and $k \neq 1$ in the case (35)." But, in case (35), $k = 1$ (and $\varepsilon = 1$) are not in contradiction with (28). In other words, (2) should allow $k = 1$ (and $\varepsilon = 1$). This would lead to the $5$-defective pair $((\sqrt{-1} + \sqrt{-5}) / 2, (\sqrt{-1} - \sqrt{-5}) / 2)$, which is equivalent to $(\alpha_0, \beta_0)$. Thank in advance to anyone who takes the time to check. REPLY [9 votes]: Yes, there are some omissions in the lists in the original BHV article. I think all of them were fixed by Mourad Abouzaid Mourad Abouzaid, Les nombres de Lucas et Lehmer sans diviseur primitif, J. Théor. Nombres Bordeaux 18 (2006), no. 2, 299–313.<|endoftext|> TITLE: Surjective linear isometries on $\ell_\infty(\mathbb{N})$ QUESTION [10 upvotes]: In Volume 1 of "Classical Banach Spaces" Lindenstrauss and Tzafriri note that all surjective linear isometries on $\ell_\infty$ are of the from $(a_i) \mapsto (\varepsilon_i a_{\pi(i)})$ where $\pi$ is a permutation of $\mathbb{N}$ and $\varepsilon_i \in\{\pm 1\}$. They mention that the proof is a consequence of the more general result on isometries on $C(K)$ spaces that it will appear in Volume 3 (which does not exist). Does anyone know a reference for the proof of this fact? REPLY [14 votes]: I don't know a reference (other users will probably provide one). Here's a (quite immediate) proof anyway. I'll replace $\mathbf{N}$ with an arbitrary set $X$ since the integers play no role. Fact The extremal points of the closed 1-ball of $\ell^\infty(X)$ precisely consists of the maps $X\to\{-1,1\}$. Proof: Given $f,g,u:X\to [-1,1]$ if $u\in [f,g]$ and for some $x\in X$, $u(x)\in\{-1,1\}$, then we see that $f(x)=g(x)=u(x)$. Hence, all $u:X\to\{-1,1\}$ are extremal. Conversely, let $u$ not be of this form. So $-10$ small enough, both $u_\pm$ belong to the closed 1-ball. So $u$ is not extremal. Corollary: every bijective self-isometry of $\ell^\infty$ has the required form (call this "standard", and "diagonal-standard" when the permutation is trivial). Proof: Write $W_A=1_A-1_{A^c}$. Then, by the fact, the isometry group acts on the power set $2^X$ by $g\cdot W_A=W_{gA}$. Given an isometry, after composing with a diagonal-standard isometry, we can suppose that $f(1_X)=1_X$, i.e., $f\in G$, where $G$ is the stabilizer of $1_X$. We have to show that $G$ is reduced to the permutation group. Now for $A,B,C\subseteq X$, write $F(A,B,C)$ if $W_A+W_B-W_C=1_X$; since $G$ acts linearly and fixes $1_X$, this ternary relation is preserved by $G$. Also it is easy to check that $F(A,B,C)$ holds if and only if $C=A\cap B$. Hence, we see that $G$ (acting on $2^X$ by $f(W_A)=W_{gA})$ preserves the intersection operation. In particular, it preserves the ordering of $2^X$, and hence is an automorphism of the Boolean algebra $2^X$. Such an automorphism is necessarily induced by a permutation. Now for $f\in G$, composing with a permutation (which is standard), we can suppose that $f$ pointwise fixes all maps $X\to\{-1,1\}$. By a $\mathbf{Q}$-linear combination, it therefore preserves all characteristic functions (=maps $X\to\{0,1\}$), and in turn, it therefore preserves all simple functions (= maps $X\to\mathbf{R}$ with finite image). Since these form a dense subspace of $\ell^\infty(X)$, the resulting isometry is the identity, which means that the original isometry is standard. Note: an analogous result also holds for all $p\in [1,\infty]\smallsetminus\{2\}$. The case $1$ is similarly easy, but in the other cases this easy argument doesn't work since then the all sphere consists of extremal points. As pointed out by @GiorgioMetafune the initial argument was incorrect and some combinatorial argument was needed. REPLY [14 votes]: This is a simple consequence of the Banach-Stone theorem. Identify $l^\infty$ with $C(\beta \mathbb{N})$ and note that any self-homeomorphism of $\beta \mathbb{N}$ must take $\mathbb{N}$ to itself, since this is the set of isolated points of $\beta\mathbb{N}$.<|endoftext|> TITLE: Is every category a localization of a poset? QUESTION [25 upvotes]: Question 1: Let $C$ be a small category. Does there exist a poset $P$, a set of $W$ of morphisms in $P$, and an equivalence $P[W^{-1}] \simeq C$? Here $P[W^{-1}]$ is the universal category receiving a functor from $P$ which carries each morphism of $W$ to an isomorphism. I'm hoping for an affirmative answer. I'm also interested in the following variation: Question 2: Let $C$ be a small category with finite colimits. Does there exist a join-semilattice $P$, a set $W$ of morphisms in $P$, and an equivalence $P[W^{-1}] \simeq C$? Version control: There are actually two versions of question 2 -- in one version we require that $P \to P[W^{-1}]$ preserves finite colimits, and in the other we don't. As I'm hoping for an affirmative answer, it should be easier to do this without requiring the preservation of the finite colimits, and I'd be happy with an answer to that version. On top of that, I am interested in two versions of these questions: the 1-categorical version and the $\infty$-categorical version (the term "poset" means the same thing in both versions). Other Notes: I'm thinking a good way to try to construct such a $P$ in general may be via some sort of of subdivision of $C$. But I'm a bit unclear as to when the barycentric subdivision, say, of a category is a poset. REPLY [22 votes]: Yes, this is true. It follows form the work of Barwick and Kan on relative categories as a model for $\infty$-categories. The idea is similar to how Thomason's work shows that every homotopy type can be modeled by a poset (by taking subdivisions of the category of simplicies of the simplicial set). Specifically in ArXiv:1011.1691 and ArXiv:1101.0772 they construct a Quillen equivalence between the Joyal model structure and the model structure of "Relative Categories". This latter is a model structure on the category of small ordinary categories equipped with a collection of "weak equivalences". Importantly they show that the cofibrant objects in the latter are relative posets - relative categories whose underlying category is a poset. This means that every $\infty$-category can be modeled by a relative poset. If you are just interested in ordinary categories, then you don't even need the Quillen equivalence. View the ordinary category $C$ as a relative category with trivial weak equivalences. Then its cofibrant replacement in the Barwick-Kan model structure will be a relative poset $(P,W)$, and it will satisfy $C \simeq P[W^{-1}]$. Added for clarification From the comments to the OP is seems that people want to see a bit more about how this works. In particular how can we get a cofibrant replacement? Is it functorial? In the first paper Barwick and Kan construct an adjunction which they then show is a Quillen equivalence: $$K_\xi: ssSet \leftrightarrows RelCat: N_\xi$$ Here $N_\xi$ is a sort of nerve functor. Claim: for any relative category $C$, $K_\xi N_\xi(C) \to C$ is a cofibrant replacement of $C$ (hence a relative poset modeling $C$). This is clearly functorial in $C$ by construction. Proof: $N_\xi(C)$ will automatically be a cofibrant object in bisimplicial sets in the Reedy(=injective) model structure, and so $K_\xi$, being a left Quillen functor, will send it to a cofibrant relative category. We just need to know that (1) the counit $\epsilon_\xi :K_\xi N_\xi C \to C$ is a weak equivalence in Relative categories. Weak equivalences in the model category of relative categories are detected by $N_\xi$ (by construction - it is a transferred model structure). Thus (1) will be true if the map $N_\xi\epsilon_\xi: N_\xi K_\xi N_\xi C \to N_\xi C$ is a weak equivalence. Prop 10.3 in 1011.1691 states that the unit map $\eta_\xi: id \to N_\xi K_\xi$ is always a weak equivalence. Thus $\eta_\xi N_\xi$ is a weak equivalence. This is a left inverse of $N_\xi \epsilon_\xi$, so by two-out-of-three $N_\xi \epsilon_\xi$, and hence $\epsilon_\xi$ are weak equivalences. $\square$ You can also get an easier description using the functor "$N$" rather than the more cumbersome $N_\xi$. See section 7-8 of that same paper.<|endoftext|> TITLE: Distinct integer roots for a degree 7+ polynomial and its derivative QUESTION [17 upvotes]: Question: Is there a polynomial $f \in \mathbb{Z}[x]$ with $\deg(f) \geq 7$ such that all roots of $f$ are distinct integers; and all roots of $f'$ are distinct integers? Background: I asked a related question about four years ago on MSE (2312516) that led to the largest known examples having degree six. Shortly after, I suggested an investigation into this problem in the Polymath Proposal big-list on MO (219638). Since I haven't seen any subsequent traction post-2017, I would be interested as to whether anyone knows of progress on this problem or ways of dealing with it in this particular direction. (A different direction could be e.g. requiring that the distinct integer roots are a feature of all derivatives – not just the first derivative.) As a clarifying example, here is one of the degree six polynomials: $$f(x) = (x^2−3130^2)(x^2−3590^2)(x^2−10322^2)$$ has distinct integer roots at $x = \pm 3130, \pm 3590, \pm 10322$, and differentiating yields: $$f'(x) = 6x(x^2 - 3366^2)(x^2 - 8650^2)$$ which has distinct integer roots at $x = 0, \pm 3366, \pm 8650$. (So, it is straightforward to verify examples but seemingly difficult to generate them!) REPLY [18 votes]: Suppose one wishes to solve a system $P_1(n_1,\dots,n_k) = \dots = P_m(n_1,\dots,n_k)=0$ of diophantine equations involving polynomials $P_1,\dots,P_m$ of various degrees. We have the following probabilistic heuristic (discussed for instance in this blog post of mine): if we pick $n_1,\dots,n_k$ to be random integers of size $N$ (here we make the assumption that the dominant scenario is one in which the $n_1,\dots,n_k$ all have comparable magnitude), then each equation $P_j(n_1,\dots,n_k)=0$ heuristically has a probability $\sim N^{-\mathrm{deg} P_j}$ to hold, so (assuming no unusual correlations) the whole system should be satisfied with probability $\sim N^{-\mathrm{deg} P_1 - \dots - \mathrm{deg} P_m}$. On the other hand, we have $\sim N^k$ choices of candidates $(n_1,\dots,n_k)$ here. Thus, if we define the degree surplus (or degree-of-freedom deficit) of this system to be $\mathrm{deg} P_1 + \dots + \mathrm{deg} P_m - k$, we arrive at the following naive predictions: For a negative surplus, solutions should be quite abundant (unless there are some obvious local obstructions, or maybe some less obvious obstructions such as Brauer-Manin obstructions), and locatable by brute force search. For a positive surplus, one would typically only expect a small number of sporadic or degenerate solutions, or no solutions at all, unless there is some algebraic miracle that allows one to lower the surplus, for instance through some birational embedding. If the surplus is close to zero, e.g., if it is equal to $1$, then it might not be too much to ask for such a miracle, but when the surplus is large then this begins to look unreasonable. For zero surplus, the situation is delicate and could go either way. Very roughly speaking, this surplus is a crude measure of how many "algebraic miracles" would be needed to be present in this problem in order to produce a large number of non-trivial integer solutions. This heuristic has to be taken with several grains of salt - for instance, one needs the polynomials $P_1,\dots,P_m$ to be "independent" in some sense for this prediction to be accurate, one is implicitly assuming that the $n_1,\dots,n_k$ are probably of comparable magnitude, and this notion of surplus is not a birational invariant, in contrast to more intrinsic notions such as the genus of an algebraic curve - but it can still serve as a starting point for making initial guesses. Some examples: The Fermat equation $a^n + b^n = c^n$ has surplus $n-3$ and so one expects a lot of integer solutions for $n=2$, almost no non-trivial solutions for $n>3$, and no definitive conclusion for $n=3$. (Of course, as it turns out, there are no non-trivial solutions for $n=3$, but the story for say $a^3 + b^3 = c^3 + h$ for other constants $h$, e.g., $h=33$, can be different; see https://en.wikipedia.org/wiki/Sums_of_three_cubes .) Rational points on elliptic curves $y^2 = x^3 + ax + b$ in Weierstrass form, after clearing out denominators, becomes a cubic equation in three integer variables with a surplus of $3-3=0$, which is consistent with the fact that some elliptic curves have positive rank and thus infinitely many rational points, and others have rank zero and only finitely many rational points (or none at all). An elliptic equation in quartic form $y^2 = a_4 x^4 + \dots + a_0$ has a surplus of $4-3=1$ so one would naively expect very few rational points here; but (assuming the existence of a rational point) one can use a (rational) Mobius transformation (moving the rational point to infinity in a suitable fashion) to convert the equation into Weierstrass form, thus lowering the degree surplus down to $0$. The question here is that of finding integer solutions $a_1,\dots,a_d,b_1,\dots,b_{d-1}$ to the equation $$ \frac{d}{dx} (x-a_1) \dots (x-a_d) = d (x-b_1) \dots (x-b_{d-1}).$$ Comparing coefficients, the naive surplus here is $$ 1 + 2 + \dots + d-1 - (d + d-1) = \frac{d^2-5d+2}{2}.$$ As noted in comments, this is negative for $d=1,2,3,4$, suggesting plentiful solutions in those cases. For $d=5$ the surplus is $1$, suggesting a minor miracle is needed to find solutions. For $d>5$ the surplus is quite large and one would not expect many solutions. However, with symmetry reductions one can do a bit better. For $d$ even one can look at the symmetrised equation $$ \frac{d}{dx} (x^2-a^2_1) \dots (x^2-a^2_{d/2}) = d x(x^2-b^2_1) \dots (x^2-b^2_{d/2-1})$$ and now the surplus is $$ 2 + 4 + \dots + (d-2) - (\frac{d}{2} + \frac{d}{2}-1) = \frac{d^2-6d+4}{4}.$$ In particular for $d=6$ the surplus is reduced from $4$ to $1$, so one is only "one miracle away" from finding many solutions in some sense. But it gets rapidly worse, for instance for $d=8$ this symmetrised problem has surplus $5$ (though this still improves substantially from the non-symmetric surplus of $13$). In the paper of Choudry linked by the OP, some ad hoc substitutions and birational transformations are used in the non-symmetric $d=5$ case and the symmetric $d=6$ case (both of which have a surplus of $1$, as noted before) to extract a subfamily of solutions parameterized by a specific elliptic curve in quartic form; the surplus is still $1$, but as discussed previously one can then use a Mobius transform to send the surplus to $0$, giving hope that solutions exist. (One still has to be lucky in that the rank of the elliptic curve is positive, but this happens to be the case for both of the elliptic curves Choudry ends up at, so that paper in fact succeeds in obtaining an infinite family of solutions in both cases.) Thus I would be doubtful of any non-degenerate solutions (beyond perhaps some isolated sporadic ones) for $d \geq 7$, even if one imposes symmetry; one would need quite a long sequence of miraculous surplus-lowering transformations to be available, and absent some remarkable algebraic structure in this problem (beyond the affine and permutation symmetries) I don't see why one should expect such a sequence to be present. Establishing this rigorously (or even conditionally on conjectures such as the Bombieri-Lang conjecture) looks quite hard though. Being homogeneous, the problem is one of finding rational points to a certain projective variety (avoiding certain subvarieties corresponding to degenerate solutions); if one could somehow show this variety to be of general type, Bombieri-Lang would then say that one should only expect solutions that are trapped inside lower dimensional subvarieties, but a complete classification of such subvarieties looks extremely tedious at best given the relatively high dimension and degree. (And in view of the undecidability of Hilbert's tenth problem, which already shows up at relatively low choices of degree and dimension, there is little chance of a systematic way to resolve these questions even with the aid of conjectures such as Bombieri-Lang.) Still I would very much doubt that there are any non-trivial infinite families of solutions for $d>6$.<|endoftext|> TITLE: Finite Galois module whose Ш¹ is nonzero? QUESTION [12 upvotes]: In algebraic number theory, we constantly make use of the nine-term Poitou-Tate sequence: Let $K$ be a number field and $M$ a finite $K$-Galois module. Then we have the nine-term exact sequence $$ H^0(K, M) \to \prod' H^0(K_v,M) \to H^2(K, M^\vee)^\vee \mathop{\to}\limits^\delta H^1(K, M) \mathop{\to}^{\operatorname{loc}} \prod' H^1(K_v,M) \to \cdots $$ The kernel of the localization map $\operatorname{loc}$, which is also the image of the connecting map $\delta$, is traditionally called $Ш^1(K, M)$. It is frequently remarked (e.g. in Neukirch, Schmidt, and Wingberg's Cohomology of Number Fields, Definition 8.6.2), that $Ш^1$ is finite but not necessarily $0$. However, in all the examples I've been able to compute explicitly (e.g. $M$ cyclic of prime order), $Ш^1 = 0$. Is there a ready example of a number field $K$ and a finite module $M$ such that $Ш^1(K, M) \neq 0$? REPLY [12 votes]: Wang's conterexample to Grunwald's theorem: $K=\mathbb{Q}(\sqrt{7})$ and $M=\mu_8$. Then $H^1(K,M) \cong K^\times/(K^\times)^8$. Now $16$ is not an $8$-th power in this field but locally an $8$-th power everywhere. Your group is cyclic of order $2$. See wikipedia.<|endoftext|> TITLE: Semisimple super Lie algebras QUESTION [6 upvotes]: There is a classification of simple Lie algebras in $\text{Vec}_{\mathbb{C}}$ given by Dynkin diagrams. We then have 4 families of simple lie algebras, plus some exceptional ones. Question: How about simple Lie algebras in the bigger category $\text{sVec}_{\mathbb{C}}$ of super vector spaces? Is there a known classification for simple Lie algebras there? REPLY [11 votes]: Yes there is a complete classification of finite dimensional, simple Lie superalgebras (over $\mathbb{C}$), which -up to a certain extent- goes very much in parallel with the corresponding case of Lie algebras and incorporates the later as a special case. There are significant conceptual differences though (as to the role and uniqueness of Dynkin diagrams, Cartan matrices etc). Historically it has been fully developed by Kac. The original references are: V.G. Kac, Lie Superalgebras, Adv. Math. 26 (1977) 8. V.G. Kac, A sketch of Lie superalgebra theory, Commun. Math. Phys. 53 (1977) 31. V.G. Kac, Representations of classical Lie superalgebras, Lectures Notes in Mathematics 676 (1978) 597; Springer-Verlag, Berlin. The above provide a self-contained full account of the work. You can also find a succint description of the classification with lots of detailed references at: the Dictionary of Lie superalgebras, arXiv:hep-th/9607161v1.<|endoftext|> TITLE: $2$-adic valuation of Schur $P$-functions in the power-sum basis QUESTION [10 upvotes]: For a partition $\lambda$, let $P_\lambda$ be the Schur $P$-functions (case $t=-1$ of Hall-Littlewood symmetric functions) and let $p_\lambda=p_{\lambda_1}p_{\lambda_1}\cdots p_{\lambda_k}$ be the power-sum symmetric functions. It is known that the space $\Gamma$ spanned by the $P_\lambda$ for $\lambda$ with distinct parts is the same as the one spanned by the $p_\rho$ for $\rho$ with odd parts (see e.g. Macdonald's Symmetric Functions and Hall Polynomials). I noticed that the coefficients of the transition matrix between the two bases for $\Gamma$ are all in $\mathbb{Z}_{(2)}$; that is, if $\lambda$ is a partition with distinct parts and $$P_\lambda = \sum_\rho a^\lambda_\rho p_\rho$$ then $v_2(a^\lambda_\rho) \ge 0$. Is this fact known? Note that we have $a^\lambda_\rho=2^{\ell(\rho)-\ell(\lambda)}z_\rho^{-1}X^\lambda_\rho(-1)$, where $\ell(\cdot)$ is the number of parts of a partition and $X^\lambda_\rho \in \mathbb{Z}[t]$ are the Green polynomials (see Macdonald chapter III.7). In particular, the question comes down to showing that $$v_2(X^\lambda_\rho(-1))\ge v_2(z_\rho)+\ell(\lambda)-\ell(\rho).$$ REPLY [3 votes]: Here is a proof of the generalization suggested by Richard Stanley in the comments and even of a more general result (with "odd" replaced by "not divisible by a given prime $q$"). It is completely different from the argument I sketched in the comments, and is completely elementary (using no Macdonald polynomials). Unfortunately, it is also somewhat awkward and way too long (much of it devoted to a fight with notations). For any commutative ring $R$, we let $\Lambda_{R}$ be the ring of symmetric functions over $R$; this is a commutative $R$-algebra. Let $\operatorname{Par}$ be the set of all partitions. We shall use the standard notation $p_{\lambda}$ for the power-sum symmetric function indexed by a partition $\lambda$. Fix a prime $q$. Let $\mathbb{Z}_{\left( q\right) }$ denote the ring of all rational numbers that can be written in the form $\dfrac{a}{b}$ for two integers $a$ and $b$ such that $b$ is coprime to $q$. These numbers are known as $q$-integers. Obviously, $\mathbb{Z}_{\left( q\right) }$ is a subring of $\mathbb{Q}$, so that $\Lambda_{\mathbb{Z}_{\left( q\right) }}$ is a subring of $\Lambda_{\mathbb{Q}}$. Now, Richard Stanley's generalization (generalized a bit further) claims: Theorem 1. We have \begin{align*} \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right] =\mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] . \end{align*} The proof requires some preparations, which include showing some results of independent interest. First, we introduce a few more classical notations from symmetric function theory: For any partition $\lambda$ and any $i\geq1$, we let $m_{i}\left( \lambda\right) $ denote the multiplicity of $i$ in $\lambda$ (that is, the number of parts of $\lambda$ that equal $i$). For any partition $\lambda$, we define the positive integer \begin{align*} z_{\lambda}:=\prod_{i=1}^{\infty}\left( \left( m_{i}\left( \lambda\right) \right) !\cdot i^{m_{i}\left( \lambda\right) }\right) . \end{align*} Let $V$ be the $\Lambda_{\mathbb{Z}}$-subalgebra of $\Lambda_{\mathbb{Q}}$ generated by the fractions $\dfrac{p_{i}}{i^{k}}$ for all positive integers $i$ and all nonnegative integers $k$. In other words, let \begin{align*} V=\Lambda_{\mathbb{Z}}\left[ \dfrac{p_{i}}{i^{k}}\ \mid\ i>0\text{ and } k\geq0\right] \subseteq\Lambda_{\mathbb{Q}}. \end{align*} Now, we claim the following: Theorem 2. We have \begin{align*} z_{\lambda}^{-1}p_{\lambda}\in V \end{align*} for each partition $\lambda$. To prove this, we need a simple arithmetic lemma: Lemma 3. Let $c$ and $d$ be two integers with $d\neq0$. Then, there exist some integers $a$ and $b$ and some nonnegative integer $i$ such that $c^i =ad+bc^{i+1}$. Proof of Lemma 3. The ring $\mathbb{Z}/d\mathbb{Z}$ is finite (since $d\neq0$). For any integer $m$, we let $\overline{m}\in\mathbb{Z}/d\mathbb{Z}$ denote the residue class of $m$ in this ring. The infinitely many residue classes $\overline{c^{0}},\overline{c^{1}},\overline{c^{2}},\ldots$ all belong to the finite ring $\mathbb{Z}/d\mathbb{Z}$, and thus cannot all be distinct (by the pigeonhole principle). In other words, there exist two nonnegative integers $i$ and $j$ satisfying $i1$ (since $n$ is a positive integer). Thus, $nv > v$ (since $v > 0$). In other words, $v < nv$. From $p_{\lambda}=p_n^v $ and $z_{\lambda}=v!\cdot n^v $, we obtain \begin{equation} z_{\lambda}^{-1}p_{\lambda}=\left( v!\cdot n^v \right) ^{-1}p_n ^v =\dfrac{p_n^v }{v!\cdot n^v }. \label{darij1.pf.t2.c2.1} \tag{2} \end{equation} We must prove that $z_{\lambda}^{-1}p_{\lambda}\in V$. In other words, we must prove that $\dfrac{p_n^v }{v!\cdot n^v }\in V$ (since $z_{\lambda} ^{-1}p_{\lambda}=\dfrac{p_n^v }{v!\cdot n^v }$). We shall now use Lemma 3 to decompose $\dfrac{p_n^v }{v!\cdot n^v }$ into a sum of partial fractions -- one with a denominator of $v!$ and another with a power of $n$ in the denominator. We will then prove that both of these fractions belong to $V$. Indeed, Lemma 3 (applied to $c=n^v $ and $d=v!$) yields that there exist some integers $a$ and $b$ and some nonnegative integer $i$ such that $\left( n^v \right) ^i =a\cdot v!+b\left( n^v \right) ^{i+1}$. Consider these $a$, $b$ and $i$. Multiplying both sides of the equality $\left( n^v \right) ^i =a\cdot v!+b\left( n^v \right) ^{i+1}$ by $\dfrac {p_n^v }{\left( n^v \right) ^{i+1}\cdot v!}$, we obtain \begin{align} \dfrac{p_n^v }{v!\cdot n^v } &=\left(a\cdot v!+b\left( n^v \right) ^{i+1}\right) \cdot \dfrac{p_n^v }{\left( n^v \right) ^{i+1}\cdot v!} \\ &=a\cdot\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}+b\cdot\dfrac{p_n^v }{v!} . \label{darij1.pf.t2.c2.parfrac} \tag{3} \end{align} Thus, in order to prove that $\dfrac{p_n^v }{v!\cdot n^v }\in V$, it suffices to show that $\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}\in V$ and $\dfrac{p_n^v }{v!}\in V$ (because $a$ and $b$ are integers, and $V$ is a ring). The first of these two claims is easy: We have $\dfrac{p_n}{n^{i+1}}\in V$ (since $\dfrac{p_n}{n^{i+1}}$ is one of the designated generators of the $\Lambda_{\mathbb{Z}}$-algebra $V$). Hence, $\left( \dfrac{p_n}{n^{i+1} }\right) ^v \in V$ (since $V$ is a ring). In other words, $\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}\in V$ (since $\left( \dfrac{p_n}{n^{i+1} }\right) ^v =\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}$). It remains to prove that $\dfrac{p_n^v }{v!}\in V$. To do this, we will need the Frobenius endomorphism $\mathbf{f}_n$. It is defined as follows: For any commutative ring $R$, we let \begin{align*} \mathbf{f}_n:\Lambda_{R}\rightarrow\Lambda_{R} \end{align*} be the $R$-algebra homomorphism that sends each symmetric function $f$ to $f\left( x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right) $ (where we regard $f$ as a symmetric formal power series in countably many indeterminates $x_{1},x_{2} ,x_{3},\ldots$). This homomorphism $\mathbf{f}_n$ is called the $n$-th Frobenius endomorphism and is functorial in $R$ (that is, it commutes with the morphisms $\Lambda_{R}\rightarrow\Lambda_{S}$ induced by ring homomorphisms $R\rightarrow S$). If you like to think in terms of plethysm, $\mathbf{f}_n$ can be described as sending each $f\in\Lambda_{R}$ to the plethysm $f\left[ p_n\right] $. The functoriality of $\mathbf{f}_n$ in $R$ entails that the $\mathbf{f}_n$ defined for $R=\mathbb{Z}$ is a restriction of the $\mathbf{f}_n$ defined for $R=\mathbb{Q}$. Thus, we can safely denote both of these maps by $\mathbf{f}_n$ without risking confusion. They both are ring homomorphisms (since they are $R$-algebra homomorphisms for appropriate $R$). Of course, $\mathbf{f}_n\left( \Lambda_{\mathbb{Z}}\right) \subseteq\Lambda _{\mathbb{Z}}$ (since the $\mathbf{f}_n$ defined for $R=\mathbb{Z}$ is a restriction of the $\mathbf{f}_n$ defined for $R=\mathbb{Q}$). It is easy to see that $\mathbf{f}_n\left( p_{i}\right) =p_{in}$ for each $i>0$. Hence, for each positive integer $i$ and each nonnegative integer $k$, we have \begin{align*} \mathbf{f}_n\left( \dfrac{p_{i}}{i^{k}}\right) =\dfrac{p_{in}}{i^{k} }=n^{k}\cdot\dfrac{p_{in}}{\left( in\right) ^{k}}\in V \end{align*} (since $\dfrac{p_{in}}{\left( in\right) ^{k}}$ is one of the designated generators of the $\Lambda_{\mathbb{Z}}$-algebra $V$). Therefore, \begin{align*} \Lambda_{\mathbb{Z}}\left[ \mathbf{f}_n\left( \dfrac{p_{i}}{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \subseteq V \end{align*} (since $V$ is a $\Lambda_{\mathbb{Z}}$-algebra). Now, from $V=\Lambda_{\mathbb{Z}}\left[ \dfrac{p_{i}}{i^{k}}\ \mid\ i>0\text{ and }k\geq0\right] $, we obtain \begin{align*} \mathbf{f}_n\left( V\right) & =\mathbf{f}_n\left( \Lambda _{\mathbb{Z}}\left[ \dfrac{p_{i}}{i^{k}}\ \mid\ i>0\text{ and }k\geq0\right] \right) \\ & =\underbrace{\left( \mathbf{f}_n\left( \Lambda_{\mathbb{Z}}\right) \right) }_{\subseteq\Lambda_{\mathbb{Z}}}\left[ \mathbf{f}_n\left( \dfrac{p_{i}}{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \\ & \qquad\left( \text{since }\mathbf{f}_n\text{ is a ring homomorphism} \right) \\ & \subseteq\Lambda_{\mathbb{Z}}\left[ \mathbf{f}_n\left( \dfrac{p_{i} }{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \subseteq V. \end{align*} Let $\mu$ be the partition $\underbrace{\left( 1,1,\ldots,1\right) }_{v\text{ entries}}$. Then, $p_{\mu}=p_{1}^v $ and $z_{\mu}=v!\cdot \underbrace{1^v }_{=1}=v!$ and $\left\vert \mu\right\vert =v$ and $\ell\left( \mu\right) =v$. Hence, \begin{align*} \underbrace{\left\vert \mu\right\vert }_{=v}+\underbrace{\ell\left( \mu\right) }_{=v} & =\underbrace{v}_{< nv} + v\\ & <\underbrace{nv}_{=\left\vert \lambda\right\vert }+\underbrace{v} _{=\ell\left( \lambda\right) }=\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N. \end{align*} Hence, $z_{\mu}^{-1}p_{\mu}\in V$ (by our induction hypothesis, applied to $\mu$ instead of $\lambda$). In view of $z_{\mu}=v!$ and $p_{\mu}=p_{1}^v $, this rewrites as $v!^{-1}\cdot p_{1}^v \in V$. Hence, \begin{equation} \mathbf{f}_n\left( v!^{-1}\cdot p_{1}^v \right) \in \mathbf{f} _n\left( V\right) \subseteq V. \label{darij1.pf.t2.c2.5} \tag{4} \end{equation} However, since $\mathbf{f}_n$ is a $\mathbb{Q}$-algebra homomorphism, we have \begin{align*} \mathbf{f}_n\left( v!^{-1}\cdot p_{1}^v \right) =v!^{-1}\cdot\left( \mathbf{f}_n\left( p_{1}\right) \right) ^v =\dfrac{\left( \mathbf{f}_n\left( p_{1}\right) \right) ^v }{v!}=\dfrac{p_n^v }{v!} \end{align*} (since $\mathbf{f}_n\left( p_{1}\right) =p_n$). Thus, \eqref{darij1.pf.t2.c2.5} rewrites as $\dfrac{p_n^v }{v!}\in V$. Hence, \eqref{darij1.pf.t2.c2.parfrac} becomes \begin{align*} \dfrac{p_n^v }{v!\cdot n^v }=a\cdot\underbrace{\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}}_{\in V}+b\cdot\underbrace{\dfrac{p_n^v }{v!}}_{\in V}\in V \end{align*} (since $V$ is a ring and since $a$ and $b$ are integers). In view of \eqref{darij1.pf.t2.c2.1}, this rewrites as $z_{\lambda}^{-1}p_{\lambda}\in V$. Hence, $z_{\lambda}^{-1}p_{\lambda}\in V$ has been proved in Case 2. Let us finally consider Case 3. In this case, not all parts of $\lambda$ are equal. We need another notation: If $\alpha=\left( \alpha_{1},\alpha_{2} ,\ldots,\alpha_{m}\right) $ and $\beta=\left( \beta_{1},\beta_{2} ,\ldots,\beta_n\right) $ are two partitions, then $\alpha\sqcup\beta$ shall denote the partition obtained by sorting the tuple $\left( \alpha_{1} ,\alpha_{2},\ldots,\alpha_{m},\beta_{1},\beta_{2},\ldots,\beta_n\right) $ in weakly decreasing order. For instance, $\left( 3,2,2\right) \sqcup\left( 5,3,2\right) =\left( 5,3,3,2,2,2\right) $. It is easy to see that if $\alpha$ and $\beta$ are two partitions that have no part in common, then \begin{equation} z_{\alpha\sqcup\beta}=z_{\alpha}z_{\beta}. \label{darij1.pf.t2.c3.zaub} \tag{5} \end{equation} Moreover, if $\alpha$ and $\beta$ are any two partitions, then \begin{equation} p_{\alpha\sqcup\beta}=p_{\alpha}p_{\beta}. \label{darij1.pf.t2.c3.paub} \tag{6} \end{equation} Now, recall that not all parts of $\lambda$ are equal. Hence, we can write $\lambda$ in the form $\lambda=\alpha\sqcup\beta$ where $\alpha$ and $\beta$ are two nonempty partitions that have no part in common. (Indeed, we can define $\alpha$ and $\beta$ by choosing an arbitrary part $i$ of $\lambda$, then letting $\alpha$ be the partition consisting of all parts of $\lambda$ equal to $i$, while $\beta$ is the partition consisting of all remaining parts of $\lambda$.) Consider these $\alpha$ and $\beta$. It is easy to see that $\left\vert \alpha\right\vert <\left\vert \alpha\sqcup\beta\right\vert $ (since $\beta$ is nonempty) and $\ell\left( \alpha\right) <\ell\left( \alpha\sqcup\beta\right) $ (for the same reason). Since $\alpha\sqcup \beta=\lambda$, these two inequalities rewrite as $\left\vert \alpha \right\vert <\left\vert \lambda\right\vert $ and $\ell\left( \alpha\right) <\ell\left( \lambda\right) $. Adding these two inequalities together, we obtain \begin{align*} \left\vert \alpha\right\vert + \ell\left( \alpha\right) <\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N. \end{align*} Hence, $z_{\alpha}^{-1}p_{\alpha}\in V$ (by our induction hypothesis, applied to $\alpha$ instead of $\lambda$). Similarly, $z_{\beta}^{-1}p_{\beta}\in V$. However, from \eqref{darij1.pf.t2.c3.zaub} and \eqref{darij1.pf.t2.c3.paub}, we obtain \begin{align*} z_{\alpha\sqcup\beta}^{-1}p_{\alpha\sqcup\beta}=\left( z_{\alpha}z_{\beta }\right) ^{-1}p_{\alpha}p_{\beta}=\underbrace{z_{\alpha}^{-1}p_{\alpha}}_{\in V}\cdot\underbrace{z_{\beta}^{-1}p_{\beta}}_{\in V}\in V \end{align*} (since $V$ is a ring). In view of $\alpha\sqcup\beta=\lambda$, this rewrites as $z_{\lambda}^{-1}p_{\lambda}\in V$. Hence, $z_{\lambda}^{-1}p_{\lambda}\in V$ has been proved in Case 3. We have now proved $z_{\lambda}^{-1}p_{\lambda}\in V$ in all three cases 1, 2 and 3. Thus, the induction step is complete. Thus, Theorem 2 is proved by induction. $\blacksquare$ We are not quite ready to prove Theorem 1 yet. We first need some more notations. We let $\operatorname{QPar}$ be the set of all partitions that have no part divisible by $q$. (If $q=2$, these are precisely the partitions into odd parts.) We let $J$ be the ideal of the ring $\Lambda_{\mathbb{Q}}$ generated by the $p_{i}$ with $i\equiv0\mod q$. In other words, $J=\sum\limits_{i=1}^{\infty}p_{iq}\Lambda_{\mathbb{Q}}$. Recall that the family $\left( p_{\lambda}\right) _{\lambda\in\operatorname{Par}}$ is a basis of the $\mathbb{Q}$-vector space $\Lambda_{\mathbb{Q}}$. Thus, the $\mathbb{Q} $-vector subspace $J$ of $\Lambda_{\mathbb{Q}}$ has basis $\left( p_{\lambda }\right) _{\lambda\in\operatorname{Par}\setminus\operatorname{QPar}}$ (because multiplying any $p_{\mu}$ by a $p_{iq}$ yields a $p_{\lambda}$ with $\lambda\in\operatorname{Par}\setminus\operatorname{QPar}$, and conversely, any $p_{\lambda}$ with $\lambda \in \operatorname{Par}\setminus\operatorname{QPar}$ can be obtained in such a way). A well-known fact (or easy exercise) in abstract algebra says the following: Lemma 4. Let $B$ be a subring of a ring $A$. Let $I$ be a (two-sided) ideal of $A$. Then, $B+I$ is a subring of $A$. Applying Lemma 4 to $A=\Lambda_{\mathbb{Q}}$, $B=\Lambda_{\mathbb{Z}_{\left( q\right) }}$ and $I=J$, we conclude that $\Lambda_{\mathbb{Z}_{\left( q\right) }}+J$ is a subring of $\Lambda_{\mathbb{Q}}$. We denote this subring $\Lambda_{\mathbb{Z}_{\left( q\right) }}+J$ by $W$. We note that $W$ is furthermore a $\mathbb{Z}_{\left( q\right) }$-subalgebra of $\Lambda _{\mathbb{Q}}$ (since $W$ is a subring of $\Lambda_{\mathbb{Q}}$ and is preserved under scaling by $\mathbb{Z}_{\left( q\right) }$). Next, we observe: Proposition 5. We have $V\subseteq W$. Proof of Proposition 5. We have \begin{align} \Lambda _{\mathbb{Z}}\subseteq \Lambda _{\mathbb{Z}_{\left( q\right) }}\subseteq \Lambda_{\mathbb{Z}_{\left( q\right) }}+J=W . \end{align} Now, $W$ is a commutative ring (since it is a subring of $\Lambda_{\mathbb{Q}}$) and contains $\Lambda_{\mathbb{Z}}$ as a subring (since $\Lambda _{\mathbb{Z}}\subseteq W$). Hence, $W$ is a $\Lambda_{\mathbb{Z}}$-algebra. Thus, in order to prove that $V\subseteq W$, it suffices to show that $\dfrac{p_{i}}{i^{k}}\in W$ for each positive integer $i$ and each nonnegative integer $k$ (by the definition of $V$). So let us show this. Fix a positive integer $i$ and a nonnegative integer $k$. We must prove that $\dfrac{p_{i}}{i^{k}}\in W$. If $i\equiv0\mod q$, then this follows from the obvious fact that $\dfrac{p_{i}}{i^{k}}\in J\subseteq\Lambda_{\mathbb{Z}_{\left( q\right) }}+J=W$. Thus, we WLOG assume that $i\not \equiv 0\mod q$. Hence, $i$ is coprime to $q$. Hence, $\dfrac{1}{i}\in \mathbb{Z}_{\left( q\right) }$, so that $\dfrac{1}{i^{k}}\in\mathbb{Z}_{\left( q\right) }\subseteq W$. Now, $\dfrac{p_{i}}{i^{k}} =\underbrace{\left( \dfrac{1}{i}\right) ^{k}}_{\in W} \underbrace{p_{i}}_{\in \Lambda_{\mathbb{Z}} \subseteq W}\in W$ (since $W$ is a ring). This completes our proof of Proposition 5. $\blacksquare$ At last, we can prove Theorem 1. Proof of Theorem 1. It is clear that $\mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] \subseteq \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right] $. Hence, it suffices to prove the reverse inclusion, i.e., to prove that \begin{align*} \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] . \end{align*} Thus, we fix an arbitrary $f\in\Lambda_{\mathbb{Z}_{\left( q\right) }} \cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] $. We must prove that $f\in \mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right]$. We have $f\in\Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] $. Hence, $f$ is a $\mathbb{Q}$-linear combination of the family $\left( p_{\lambda}\right) _{\lambda\in\operatorname{QPar}}$ (since this family $\left( p_{\lambda}\right) _{\lambda\in\operatorname{QPar}}$ is a basis of the $\mathbb{Q}$-vector space $\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] $). In other words, we can write $f$ in the form \begin{equation} f=\sum_{\lambda\in\operatorname{QPar}}c_{\lambda}p_{\lambda} \label{darij1.pf.t1.f=} \tag{7} \end{equation} for some $c_{\lambda}\in\mathbb{Q}$. Consider these $c_{\lambda}$. We shall prove that they all belong to $\mathbb{Z}_{\left(q\right)}$. We let $\left\langle \cdot,\cdot\right\rangle $ denote the Hall inner product on $\Lambda_{\mathbb{Q}}$. This is a $\mathbb{Q}$-bilinear form sending $\Lambda_{\mathbb{Q}}\times\Lambda_{\mathbb{Q}}$ to $\mathbb{Q}$ and sending $\Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z}_{\left( q\right) }}$ to $\mathbb{Z}_{\left( q\right) }$ (since its restriction to $\Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z}_{\left( q\right) }}$ is the Hall inner product on $\Lambda_{\mathbb{Z}_{\left( q\right) }}$). It is well-known (see, e.g., Corollary 2.5.17(b) in Grinberg/Reiner, arXiv:1409.8356v7) that the families $\left( p_{\lambda}\right) _{\lambda\in\operatorname{Par}}$ and $\left( z_{\lambda}^{-1}p_{\lambda }\right) _{\lambda\in\operatorname{Par}}$ are dual bases of $\Lambda _{\mathbb{Q}}$ with respect to the Hall inner product. Hence, \begin{equation} \left\langle p_{\lambda},z_{\mu}^{-1}p_{\mu}\right\rangle =\delta_{\lambda ,\mu} \label{darij1.pf.t1.dualbases} \tag{8} \end{equation} for any $\lambda\in\operatorname{Par}$ and $\mu\in\operatorname{Par}$ (where the $\delta_{\lambda,\mu}$ is a Kronecker delta). In other words, \begin{equation} \left\langle p_{\lambda},p_{\mu}\right\rangle =z_{\mu}\delta_{\lambda,\mu } \label{darij1.pf.t1.dualbases2} \tag{9} \end{equation} for any $\lambda\in\operatorname{Par}$ and $\mu\in\operatorname{Par}$. Now, fix a partition $\mu\in\operatorname{QPar}$. Then, Theorem 2 (applied to $\lambda=\mu$) yields $z_{\mu}^{-1}p_{\mu}\in V\subseteq W$ (by Proposition 5). Hence, $z_{\mu}^{-1}p_{\mu}\in W=\Lambda_{\mathbb{Z}_{\left( q\right) } }+J$. In other words, we can write $z_{\mu}^{-1}p_{\mu}$ in the form $z_{\mu }^{-1}p_{\mu}=w_{1}+w_{2}$ for some $w_{1}\in\Lambda_{\mathbb{Z}_{\left( q\right) }}$ and some $w_{2}\in J$. Consider these $w_{1}$ and $w_{2}$. Now, it is easy to see that $\left\langle f,w_{1}\right\rangle \in \mathbb{Z}_{\left( q\right) }$. [Proof: We have $f\in\Lambda _{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i}\ \mid \ i\not \equiv 0\mod q\right] \subseteq\Lambda_{\mathbb{Z} _{\left( q\right) }}$ and $w_{1}\in\Lambda_{\mathbb{Z}_{\left( q\right) }}$. Thus, $\left(f, w_1\right) \in \Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z}_{\left( q\right) }}$. Hence, $\left\langle f,w_{1}\right\rangle \in\mathbb{Z}_{\left( q\right) }$, because the Hall inner product $\left\langle \cdot,\cdot\right\rangle $ sends $\Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z} _{\left( q\right) }}$ to $\mathbb{Z}_{\left( q\right) }$.] Furthermore, it is easy to see that $\left\langle f,w_{2}\right\rangle =0$. [Proof: We have $w_{2}\in J$; thus, we can write $w_{2}$ as a $\mathbb{Q} $-linear combination of the family $\left( p_{\lambda}\right) _{\lambda \in\operatorname{Par}\setminus\operatorname{QPar}}$ (since this family is a basis of the $\mathbb{Q}$-vector space $J$). In other words, we can write $w_{2}$ in the form \begin{equation} w_{2}=\sum_{\beta\in\operatorname{Par}\setminus\operatorname{QPar}}d_{\beta }p_{\beta} \label{darij1.pf.t1.w2=} \tag{10} \end{equation} for some coefficients $d_{\beta}\in\mathbb{Q}$. Consider these $d_{\beta}$. From \eqref{darij1.pf.t1.f=} and \eqref{darij1.pf.t1.w2=}, we obtain \begin{align*} \left\langle f,w_{2}\right\rangle & =\left\langle \sum_{\lambda \in\operatorname{QPar}}c_{\lambda}p_{\lambda},\sum_{\beta\in \operatorname{Par}\setminus\operatorname{QPar}}d_{\beta}p_{\beta }\right\rangle \\ & =\sum_{\lambda\in\operatorname{QPar}}\ \ \sum_{\beta\in\operatorname{Par} \setminus\operatorname{QPar}}c_{\lambda}d_{\beta}\underbrace{\left\langle p_{\lambda},p_{\beta}\right\rangle }_{\substack{=z_{\beta}\delta _{\lambda,\beta}\\\text{(by \eqref{darij1.pf.t1.dualbases2})}}}\\ & =\sum_{\lambda\in\operatorname{QPar}}\ \ \sum_{\beta\in\operatorname{Par} \setminus\operatorname{QPar}}c_{\lambda}d_{\beta}z_{\beta}\underbrace{\delta _{\lambda,\beta}}_{\substack{=0\\\text{(since }\lambda\neq\beta \\\text{(because }\lambda\in\operatorname{QPar}\\\text{whereas }\beta \in\operatorname{Par}\setminus\operatorname{QPar}\text{))}}}\\ & =\sum_{\lambda\in\operatorname{QPar}}\ \ \sum_{\beta\in\operatorname{Par} \setminus\operatorname{QPar}}c_{\lambda}d_{\beta}z_{\beta}0=0, \end{align*} qed.] From $z_{\mu}^{-1}p_{\mu}=w_{1}+w_{2}$, we obtain \begin{align*} \left\langle f,z_{\mu}^{-1}p_{\mu}\right\rangle =\left\langle f,w_{1} +w_{2}\right\rangle =\underbrace{\left\langle f,w_{1}\right\rangle } _{\in\mathbb{Z}_{\left( q\right) }}+\underbrace{\left\langle f,w_{2} \right\rangle }_{=0}\in\mathbb{Z}_{\left( q\right) }. \end{align*} On the other hand, from \eqref{darij1.pf.t1.f=}, we obtain \begin{align*} \left\langle f,z_{\mu}^{-1}p_{\mu}\right\rangle & =\left\langle \sum _{\lambda\in\operatorname{QPar}}c_{\lambda}p_{\lambda},z_{\mu}^{-1}p_{\mu }\right\rangle =\sum_{\lambda\in\operatorname{QPar}}c_{\lambda} \underbrace{\left\langle p_{\lambda},z_{\mu}^{-1}p_{\mu}\right\rangle }_{\substack{=\delta_{\lambda,\mu}\\\text{(by \eqref{darij1.pf.t1.dualbases})} }}=\sum_{\lambda\in\operatorname{QPar}}c_{\lambda}\delta_{\lambda,\mu}\\ & =c_{\mu}. \end{align*} Hence, \begin{align*} c_{\mu}=\left\langle f,z_{\mu}^{-1}p_{\mu}\right\rangle \in\mathbb{Z}_{\left( q\right) }. \end{align*} Forget that we fixed $\mu$. We thus have shown that $c_{\mu}\in\mathbb{Z} _{\left( q\right) }$ for each $\mu\in\operatorname{QPar}$. In other words, $c_{\lambda}\in\mathbb{Z}_{\left( q\right) }$ for each $\lambda \in\operatorname{QPar}$. Hence, \eqref{darij1.pf.t1.f=} becomes \begin{align*} f=\sum_{\lambda\in\operatorname{QPar}}\underbrace{c_{\lambda}}_{\in \mathbb{Z}_{\left( q\right) }}p_{\lambda}\in\sum_{\lambda\in \operatorname{QPar}}\mathbb{Z}_{\left( q\right) }p_{\lambda}=\mathbb{Z} _{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] \end{align*} (by the definition of $\operatorname{QPar}$). Forget that we fixed $f$. We thus have shown that $f\in\mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] $ for each $f\in\Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] $. In other words, \begin{align*} \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] . \end{align*} As explained, this completes the proof of Theorem 1. $\blacksquare$<|endoftext|> TITLE: How to prove the second Korn inequality? QUESTION [5 upvotes]: $\textbf{Theorem}.1$ (The first Korn inequality) Suppose that $ \Omega $ is a bounded domain in $ \mathbb{R}^d $ with Lipschitz boundary. Then\ \begin{eqnarray} \sqrt{2}\left\|\triangledown u\right\|_{L^2(\Omega)}\leq \left\|\triangledown u+(\triangledown u)^T\right\|_{L^2(\Omega)} \end{eqnarray} for any $ u\in H_{0}^{1}(\Omega;\mathbb{R}^d) $, where $ (\triangledown u)^T $ denotes the transpose of $ \triangledown u $. $\textbf{Theorem}.2$ (The second Korn inequality) Suppose that $ \Omega $ is a bounded domain in $ \mathbb{R}^d $ with Lipschitz boundary. If $ u\in H^{1}(\Omega,\mathbb{R}^d) $ is a function with the property that $ u\perp R $ in $ H^{1}(\Omega;\mathbb{R}^d) $, then \begin{eqnarray} \int_{\Omega}|\triangledown u|^2dx\leq C\int_{\Omega}|\triangledown u+(\triangledown u)^T|^2dx \end{eqnarray} where $ R=\left\{\phi=Bx+b:B\in\mathbb{R}^{d\times d} \text{ is skew-symmetric and }b\in\mathbb{R}^d\right\} $ and $ C $ is a constant. I recently see the two theorems in a book about elliptic equations. I tried to get the estimate for the second inequality by direct computation which works in the proof of the first Korn inequality, but for this inequality, I cannot combine the condition $ u\perp R $ with the final results. Can you give me some hints or references? REPLY [4 votes]: You can find a full proof (to my knowledge the simpler one currently known) in the paper [1] and in the book [2], chapter I, §2.1 pp. 14-21. The original proof of Arthur Korn is so long and involved that K.O. Friedrichs, who gave a much simpler yet sophisticated proof, had doubts on his validity: starting from the work of Friedrichs, several authors gave their (in general quite complex) proofs, until Olga Oleĭnik gave a much shorter and simpler one (despite being still not elementary). References [1] Vladimir Alexandrovitch Kondratiev, Olga Arsenievna Oleĭnik, "On Korn’s inequalities" (English), Comptes Rendus de l’Académie des Sciences, Série I, 308, No. 16, pp. 483-487 (1989), MR0995908, Zbl 0698.35067. [2] Olga Arsenievna Oleĭnik, Alexei Stanislavovich Shamaev, Grigorii Andronikovich Yosifian, Mathematical problems in elasticity and homogenization. (English) Studies in Mathematics and its Applications. 26. Amsterdam-London-New York-Tokyo: North- Holland, pp. xiii+398 (1992), ISBN: 0-444-88441-6, MR1195131, Zbl 0768.73003.<|endoftext|> TITLE: A property of rapid sequences of natural numbers QUESTION [6 upvotes]: $\newcommand{\IR}{\mathbb R}$ $\newcommand{\IT}{\mathbb T}$ $\newcommand{\w}{\omega}$ $\newcommand{\e}{\varepsilon}$ Taras Banakh and me proceed a long quest answering a question of ougao at Mathematics.SE. Recently we encountered a notion of a remote sequence. We are interested whether it was studied before and want to know a solution of a problem below. Recall that a circle $\mathbb T=\{z\in\mathbb C:|z|=1\}$, endowed with the operation of multiplication of complex numbers and the topology inherited from $\mathbb C$ is a topological group. Let $(r_n)_{n\in\w}$ be an increasing sequence of natural numbers. The sequence $(r_n)_{n\in\w}$ is called remote if there exists $z\in\IT$ such that $\inf_{n\in\w}|z^{r_n}-1|>0$. A minimum growth rate of $(r_n)_{n\in\w}$ is a number $\inf_{n\in\w} r_{n+1}/r_{n}$. Problem. Find the maximal set $M\subseteq [1,\infty)$ such that if the minimum growth rate of $(r_n)_{n\in\w}$ belongs to $M$ then $(r_n)_{n\in\w}$ is remote. Our try. It is easy to see that $1\not\in M$. On the other hand, we can show that any real number $m>2$ belongs to $M$ as follows. Let $m$ be the minimum growth rate of $(r_n)_{n\in\w}$. Find $\e>0$ such that $2+\frac{\e}{1-\e}\le m$. Then $\frac{r_{n+1}}{r_n}\ge 2+\frac\e{1-\e}=\frac{2-\e}{1-\e}$ and hence $\frac{r_n}{r_{n+1}}\cdot\frac{2-\e}{1-\e}\le 1$ for every $n\in\w$. Let $W=\{e^{it}:|t|<\pi \e\}$ and for every $n\in\w$ consider the set $U_n=\{z\in\IT:z^{r_n}\in W\}$. Consider the exponential map $\exp:\IR\to \IT$, $\exp:t\mapsto e^{2\pi t i}$, and observe that for every $n\in\w$ any connected component of the set $\exp^{-1}[U_n]$ is an open interval of length $\frac{\e}{r_n}$ and every connected component of the set $\IR\setminus \exp^{-1}[U_n]$ is a closed interval of length $\frac{1-\e}{r_n}$. Since $\frac{r_n}{r_{n+1}}\cdot\frac{2-\e}{1-\e}\le 1$ and $$\frac{1-\e}{r_{n+1}}+\frac\e{r_{n+1}}+\frac{1-\e}{r_{n+1}}=\frac{2-\e}{r_{n+1}}=\frac{1-\e}{r_n}\cdot\frac{r_n}{r_{n+1}}\cdot\frac{2-\e}{1-\e}\le \frac{1-\e}{r_n},$$every connected component of the set $\IR\setminus \exp^{-1}[U_n]$ contains a connected component of the set $\IR\setminus\exp^{-1}[U_{n+1}]$. Then for every $n\in\w$ we can choose a connected component $I_n$ of the set $\IR\setminus\exp^{-1}[U_n]$ such that $I_{n+1}\subseteq I_n$. By the compactness of the set $I_0$, the intersection $\bigcap_{n\in\w}I_n$ contains some real number $t$. Then the point $z=\exp(t)$ does not belong to $\bigcup_{n\in\w}U_n$, which implies that $z^{r_n}\notin W$ for every $n\in\w$. The definition of the neighborhood $W$ ensures that $\inf_{n\in\w}|z^{r_n}-1|>0$, that is the sequence $(r_n)_{n\in\w}$ is remote. Thanks. REPLY [4 votes]: A sequence is called lacunary if, in your terminology, its minimum growth rate is strictly greater than $1$. The following articles prove that every lacunary sequence is remote. If I understand your question correctly, this means that the (only) maximal $M$ you seek is the interval $(1,\infty)$. Pollington, Andrew D., On the density of sequences $\{n_k\xi\}$, Ill. J. Math. 23, 511-515 (1979). ZBL0401.10059. de Mathan, B., Numbers contravening a condition in density modulo 1, Acta Math. Acad. Sci. Hung. 36, 237-241 (1980). ZBL0465.10040.<|endoftext|> TITLE: Golden ratio in contemporary mathematics QUESTION [55 upvotes]: A (non-mathematical) friend recently asked me the following question: Does the golden ratio play any role in contemporary mathematics? I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today. I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question. My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it. REPLY [3 votes]: In numerical analysis, a basic problem consists in approaching a solution $x^*$ of $f(x)=0$, where $f:{\mathbb R}\rightarrow{\mathbb R}$ is at least continuous and more often $C^2$-smooth. The basic methods are those of dichotomy, secant and Newton. Dichotomy is more than elementary (based upon the intermediate value theorem) and its convergence is of order $1$, meaning that the error $|x-x^*|\sim2^{-n}$ decays exponentially. Newton's method is much faster - order $2$ if $x^*$ is non-degenerate, that is $f'(x^*)\ne0$, which means $|x-x^*|=O(\rho^{2^n})$ for some $\rho<1$. But it requires the knowledge of $f'$, which can be questionable in real world applications, where values of $f$ are given by measurements. A good compromise is the secant method, which ressembles a lot Newton, but the tangent to the graph at $x_n$ is replaced by the chord of the graph determined by the abscissæ $x_{n-1}$ and $x_n$. Then its intersection with the horizontal axis determines $x_{n+1}$. The secant method turns out to be of order $\phi$ whenever $x^*$ is non-degenerate: $|x-x^*|=O(\rho^{\phi^n})$ for some $\rho<1$. The main drawback of the secant method is that its extension to vector fields $f:{\mathbb R}^n\rightarrow{\mathbb R}^n$, which can be defined easily, behaves badly when $n\ge2$. Notice that the extension of the dichotomy, which involves triangulations and the Sperner Lemma, becomes very slow. Only the Newton method admits an efficient generalization (then called Newton-Raphson).<|endoftext|> TITLE: Why is the generalized flag variety a “variety”? QUESTION [6 upvotes]: In several places (for example, Chriss & Ginzburg’s book “Representation Theory and Complex Geometry”), the author says that the set $X$ of Borel subalgebras of a semi-simple Lie algebra $\mathfrak g$ forms a Zariski-closed subset of a Grassmannian on this semi-simple Lie algebra. A Borel subalgebra is a maximal solvable subalgebra, by general theorems, all of them have the same dimension, say $d$. Then the set $X$ can be identified with a subset of the Grassmannian $\mathrm{Gr}(d,\mathfrak g)$. My question is: why is $X$ a Zariski-closed subset? How to translate the condition “solvable” to an algebraic condition? I thought of Lie’s criterion on solvability, but it is not an equivalence condition, so it did not work. REPLY [6 votes]: I think, the statement "the set of Borel subalgebras of a semi-simple Lie algebra $\mathfrak g$ forms a Zariski-closed subset of a Grassmannian" can be understood in more than one way and then the question becomes less trivial than one thinks. Everything is over $\mathbb C$, by the way, otherwise things become even more complicated. First interpretation: This is the one presumably intentend by Chriss-Ginzburg and others. It consists of two statements: The set of Borel subgroups is in bijection with $G/B$. The natural morphism $G/B\to Gr_d(\mathfrak g)$ is a closed embedding. The first assertion uses the conjugacy of Borel subgroups and that Borel subgroups are selfnormalizing. The second uses that $G/B$ is projective and that $B$ is the normalizer of $\mathfrak b$ in $G$. Second interpretation: The statement could also mean that the moduli space of Borel subalgebras is precisely the schematic image of $G/B$. This is more subtle and is not addressed in any of the standard text, as far as I know. One can see this as follows: Step: A subalgebra $\mathfrak b\subseteq\mathfrak g$ is the Lie algebra of a Borel subgroup if and only if $\mathfrak b$ is solvable of maximal possible dimension $d=\dim B$. Here one has to prove mainly that a maximal dimensional solvable subalgebra is algebraic in the sense that it is the Lie algebra of some algebraic subgroup. Step: There is a closed subscheme $X\subseteq Gr_d(\mathfrak g)$ which is a moduli scheme of $d$-dimensional solvable subalgebras. For this one has to translate "solvable" into an algebraic condition. This condition comes from that fact that the degree of solvability of any $\mathfrak b$ is bounded by say $\ell$. Then $X$ is defined by the condition that all "multicommutators" $[\ldots[[\xi_1,\xi_2],[\xi_3,\xi_4]],\ldots]$ vanish for all $\xi_1,\ldots\xi_{2^\ell}\in\mathfrak b$. This is easily seen to be a closed condition. Step: Step 1 implies that $X$ is set theoretically the image of $G/B$ in $Gr_d(\mathfrak g)$. Step: It remains to prove that $X$ is smooth, hence reduced and normal. For this we compute the tangent space $Z$ of $X$ in $\mathfrak b$. First the tangent space of $Gr_d(\mathfrak g)$ in $\mathfrak b$ is $H:=\mathrm{Hom}(\mathfrak b,\mathfrak g/\mathfrak b)$. It turns out that $Z$ is the space of $1$-cocycles (=derivations) in $H$. Moreover, the tangent space of $G/B$ in $\mathfrak b$ is the space of all $1$-coboundaries (=inner derivations) $C\cong\mathfrak g/\mathfrak b$. So $Z/C=H^1(\mathfrak b,\mathfrak g/\mathfrak b)$. The latter is easily seen to be $0$ (take a filtration with $1$-dimensional quotients and use that $\mathfrak b$ and $\mathfrak g/\mathfrak b$ have no weights in common). Thus $Z=C$ and therefore $\dim Z=\dim X$ and we are done.<|endoftext|> TITLE: What happens to a closed manifold to ensure it is homeomorphic to a torus $T^{n}$? QUESTION [15 upvotes]: If $M$ is a smooth connected closed $n$-dimensional manifold, its universal covering space is homeomorphic to Euclidean space $R^{n}$, and its fundamental group is $Z^{n}$, then is it homeomorphic to a torus $T^{n}$? If not, why? REPLY [16 votes]: This answer is intended to give references of the cases for the case $n \leq 4$. In dimensions $n \leq 2$ this is covered in a first topology course so there are two interesting cases. In dimension 3, one has the prime decomposition theorem: every 3-manifold is uniquely the connected sum of prime 3-manifolds. Because $\Bbb Z^3$ has no non-trivial decomposition as a free product, it follows that $M \cong M_P \# M_S $, where $\pi_1 M_S = 0$, so $M_S$ is a homotopy sphere. This is the one place where one must use the geometrization theorem. The geometrization theorem implies the Poincare conjecture implies $M_S = S^3$ implies $M = M_P$. In Hempel's book you will see this called the "Poincare associate" $\mathcal P(M)$ of $M$; it is prime. The argument below shows $M_P \cong T^3$. The canonical reference for such pre-geometric techniques is Hempel's book "3-Manifolds". Hempel's Theorem 11.10 says that if $\pi_1 M$ fits into a sequence $1 \to N \to \pi_1 M \to \pi_1 B$, where $B$ is a closed surface of nonpositive Euler characteristic, then $N = \Bbb Z$ and $M_P$ is a circle bundle over $B$. Here we have $B = T^2$; if $Y_k$ is the circle bundle over $B$ with Euler class $k$, then $H_1(Y_k) = \Bbb Z^2 \oplus \Bbb Z/k$. Because $H_1(M_P) = \Bbb Z^3$, it follows that $M_P \cong Y_0 = T^3$, as desired. Hempel's proof of this statement is quite explicit and combinatorial. It is not analagous to the high-dimensional proofs. You may also be interested in Hempel Chapter 13, which describes how to show that 3-manifolds and their automorphisms are essentially determined by their fundamental group in certain cases ($P^2$-irreducible and "Haken"). In dimension 4, you may invoke the Theorem of section 11.5 of Freedman and Quinn's book "Topology of 4-Manifolds". This states: "Suppose $f: M \to N$ is a homotopy equivalence of compact aspherical 4-manifolds with polyfinite or polycyclic fundamental groups, which restricts to a homeomorphism of boundaries. Then $f$ is homotopic relative to the boundary to a homeomorphism." Finitely generated abelian groups are polycyclic and your manifold $M$ is aspherical. Thus the classifying map $f: M \to K(\Bbb Z^4, 1) = T^4$ for its fundamental group is a homotopy equivalence, and Freedman-Quinn's result implies that this map is homotopic to a homeomorphism. This result is very nice but much more inexplicit than the 3-dimensional case. It requires a complicated infinite procedure, salvaging as much of Whitney's ideas as possible (simplifying handle decompositions by finding non-intersecting discs). Because this is no longer guaranteed by transversality, the procedure is much more complicated. Oxford University Press has just published a new book on the disc embedding theorem, which will hopefully make its proof accessible to a wider audience.<|endoftext|> TITLE: Grading ring spectra over the sphere spectrum QUESTION [6 upvotes]: $\mathbb{Z}$-graded rings play an important role in algebra and algebraic geometry, so when moving to derived algebra and spectral algebraic geometry, it's natural to ask about ring spectra graded in the sphere spectrum $\mathbb{S}$. As discussed in Section 2 of Bunke–Nikolaus's Twisted differential cohomology, such an object may be defined as a symmetric lax monoidal functor $\mathbb{S}\to\mathsf{Sp}$, where $\mathbb{S}$ is considered as a monoidal $\infty$-groupoid. In particular, given an $\mathbb{S}$-graded ring spectrum $R$, it follows that the ring $\pi_0(R)$ is $\mathsf{Ho}(\mathbb{S})$-graded. By the description here, such a $\tau_{\leq1}\mathbb{S}$-graded ring consists of a pair $(R_\bullet,\{\sigma_k\}_{k\in\mathbb{Z}})$ with $R_\bullet$ a $\mathbb{Z}$-graded ring; $\{\sigma_k\colon R_k\to R_k\}_{k\in\mathbb{Z}}$ a family of order $2$ automorphisms, one for each $R_k$. Moreover, if $R$ is an $\mathbb{E}_{\infty}$-ring, then $\pi_0(R)$ is "$\mathsf{Ho}(\mathbb{S})$-graded commutative", in that we additionally have $$ ab = \begin{cases} ba &\text{if $\deg(a)\deg(b)$ is even,}\\ \sigma_{\deg(a)+\deg(b)}(ab) &\text{if $\deg(a)\deg(b)$ is odd} \end{cases} $$ for each $a,b\in R_\bullet$. This includes in particular $\mathbb{Z}$-graded commutative algebras by picking $\sigma_k(a)\overset{\mathrm{def}}{=}-a$ for each $k\in\mathbb{Z}$, as in that case the above condition becomes $$ab=(-1)^{\deg(a)\deg(b)}ba.$$ Question. So, are there any interesting/non-trivial "in nature" examples of $\mathbb{S}$-graded ring spectra? REPLY [9 votes]: One of the default examples of ordinary graded commutative rings is the polynomial ring $\mathbf Z[t]$. Let us first examine the analogue of that, and then see where else that leads! 1. $S$-grading on $S\{t\}$ For the sake of clarity, allow me to denote the underlying infinite loop space (equivalently: grouplike $\mathbb E_\infty$-space) of the sphere spectrum $S$ by $\Omega^\infty(S)$. Recall that, by the Barrat-Priddy-Quillen Theorem, the ininite loop space $\Omega^\infty(S)$ can be described as the group completion of the groupoid of finite sets and isomorphisms $\mathcal F\mathrm{in}^\simeq$. Consider the constant functor $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$ with value $S$ - since the latter is the monoidal unit, this is a symmetric monoidal functor. Left Kan extension along the canonical map $\mathcal F\mathrm{in}^\simeq \to (\mathcal F\mathrm{in}^\simeq)^\mathrm{gp}\simeq \Omega^\infty(S)$ produces a symmetric monoidal functor $\Omega^\infty(S)\to \mathrm{Sp}$, and as such exhibits an $S$-graded $\mathbb E_\infty$-ring. But which one? To figure out which one, note that the passage to the "underlying $\mathbb E_\infty$-ring" of an $S$-graded $\mathbb E_\infty$-ring is given by passage to the colimit. Thus the underlying $\mathbb E_\infty$-ring, which we have just adorned with an $S$-grading, is $$ \varinjlim_{\Omega^\infty(S)}\mathrm{LKan}^{\Omega^\infty(S)}_{\mathcal{F}\mathrm{in}^\simeq}(S)\simeq \varinjlim_{\mathcal F\mathrm{in}^\simeq}S, $$ where we used that the left Kan extension of a functor does not change its colimit. Now we can use the explicit description of the groupoid of finite sets $\mathcal F\mathrm{in}^\simeq \simeq \coprod_{n\ge 0}\mathrm B\Sigma_n$ to obtain $$ \varinjlim_{\mathcal F\mathrm{in}^\simeq} S\simeq \bigoplus_{n\ge 0}S_{h\Sigma_n}. $$ We may recognize this as the free $\mathbb E_\infty$-ring on a single generator $S\{t\}$, corresponding in terms of spectral algebraic geometry to the smooth affine line $\mathbf A^1$ (as opposed to the flat affine line $\mathbf A^1_\flat = \mathrm{Spec}(S[t])$ for the polynomial $\mathbb E_\infty$-ring $S[t]\simeq \bigoplus_{n\ge 0}S$). 2. $S$-grading on symmetric algebras The previous example can be easily generalized by observing that $\mathcal F\mathrm{in}^\simeq$ is the free $\mathbb E_\infty$-space ( = symmetric monoidal $\infty$-groupoid) on a single generator. That means that a symmetric monoidal functor $\mathcal F\mathrm{in}^\simeq \to \mathrm{Sp}$ (which always factors through the maximal subgroupoid $\mathrm{Sp}^\simeq\subseteq\mathrm{Sp}$) is equivalent to the data of a spectrum $M\in \mathrm{Sp}$ (the image of the singleton set). The functor is given by sending a finite set $I$ to the smash product $M^{\otimes I}$, and is evidently symmetric monoidal. The same Kan extension game as before now gives rise to an $S$-graded $\mathbb E_\infty$-ring spectrum, this time with underlying $\mathbb E_\infty$-ring $$ \mathrm{Sym}^*(M)\simeq \bigoplus_{n\ge 0}M^{\otimes n}_{h\Sigma_n}, $$ the free $\mathbb E_\infty$-ring generated by $M$. We recover the prior situation by setting $M=S$. For $M = S^{\oplus n}$, we obtain an $S$-grading on $S\{t_1, \ldots, t_n\}$, corresponding to the spectral-algebro-geometric smooth affine $n$-space $\mathbf A^n$. 3. $S$-grading on $S\{t^{\pm 1}\}$ Another way to generalize the example of the $S$-grading on $S\{t\}$ is to take directly the constant functor $\Omega^\infty(S)\to \mathrm{Sp}$ with value $S$, instead of starting with a constant functor on $\mathcal F\mathrm{in}^\simeq$ and Kan-extending it. This produces a perfectly good $S$-graded $\mathbb E_\infty$-ring, which let us denote $S\{t^{\pm 1}\}$. From the group-completion relationship between $\Omega^\infty(S)$ and $\mathcal F\mathrm{in}^\simeq$, it may be deduced that $S\{t^{\pm 1}\}$ and $S\{t\}$ are related in terms of $\mathbb E_\infty$-ring localization as $S\{t^{\pm 1}\}\simeq S\{t\}[t^{-1}]$, justifying our notation. Here we are localizing $S\{t\}$ at the element $t\in \mathbf Z[t] = \pi_0(S\{t\})$. In terms of spectral algebraic geometry, this is encoding the spectral scheme $\mathrm{GL}_1$, the smooth punctured line. 4. Remark on non-negative grading In algebraic geometry, we often prefer to think about non-negatively graded commutative rings than graded commutative rings. Just as the latter are equivalent to lax symmetric monoidal functor $\mathbf Z\to\mathrm{Ab}$, so are the former equivalent to lax symmetric monoidal functors $\mathbf Z_{\ge 0}\to \mathrm{Ab}$. By analogy, the "non-negatively $S$-graded $\mathbb E_\infty$-rings" are lax symmetric monoidal functors $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$. Indeed, just as $\mathbf Z_{\ge 0}$ is the free commutative monoid on one generator, so is $\mathcal F\mathrm{in}^\simeq$ the free $\mathbb E_\infty$-space on one generator. That is the reason why we were encountering such functors above (and Kan extending them along group completion, as we would to view as $\mathbf Z_{\ge 0}$-grading as a special case of a $\mathbf Z$-grading). 5. Some actual "non-tautological" examples though So far, a not-completely-unreasonable complaint might be that all the examples of $S$-graded $\mathbb E_\infty$-rings were sort of tautological. For a very non-tautological example, see the main result of this paper of Hadrian Heine. It shows that there exists an $S$-graded $\mathbb E_\infty$-ring spectrum such that its $S$-graded modules are equivalent to the $\infty$-category of cellular motivic spectra. In fact, much more is proved: this situation is very common, and under some not-too-harsh compact-dualizable-generation assumptions, a symmetric monoidal stable $\infty$-category will be equivalent to $S$-graded modules over some $S$-graded $\mathbb E_\infty$-ring. So you may just as well view this result as a wellspring of potentially interesting examples of $S$-gradings "occurring in nature"! :)<|endoftext|> TITLE: What is the homotopy category of the sphere spectrum? QUESTION [7 upvotes]: Is there a known explicit description of the abelian $2$-group $\mathsf{Ho}(\mathbb{S})\overset{\mathrm{def}}{=}\mathsf{Ho}(QS^0)\cong\Pi_{\leq1}(QS^0)$? REPLY [12 votes]: This is the groupoid given by the 1-truncation $\tau_{\leq 1}(QS^0)$. This groupoid has $\mathbb Z$-many objects (since $\pi_0^s = \mathbb Z$), and each one has automorphism group $C_2$ (since $\pi_1^s = C_2$). The tensor product on objects is given by addition in $\mathbb Z$, and on morphisms by addition in $C_2$. One way to see this is to consider the universal functor $\Sigma \to QS^0$ given by the Barratt-Priddy-Quillen theorem (i.e. the fact that $K(\Sigma) = QS^0$; here $\Sigma$ is the groupoid of finite sets with the disjoint union monoidal structure), and to postcompose with the truncation functor $QS^0 \to \tau_{\leq 1} (QS^0)$; the fact that this functor is symmetric monoidal yields this description of the category. This perspective is discussed a bit more here. From the description I've given, I suppose it follows that $\tau_{\leq 1} (QS^0)$ splits symmetric monoidally as $\tau_{\leq 1} (QS^0) = \mathbb Z \times BC_2$, (where $\mathbb Z$ is a discrete symmetric monoidal groupoid and $BC_2$ is a 1-object symmetric monoidal groupoid), which is maybe a little surprising. This is not to say that $\tau_{\leq 1} \mathbb S$ splits...<|endoftext|> TITLE: Is there a simple formula to compute the Casson invariant of an homology $3$-sphere from its Heegaard diagram? QUESTION [8 upvotes]: Let $(S_g,\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2)$ be a Heegaard diagram of a Heegaard splitting $\Sigma=H_g \cup_{\phi_1\phi_2^{-1}}H_g$ of an integral homology sphere $\Sigma$, i.e. $S_g$ is a closed, orientable surface of genus $g$, $H_g$ is the $3$-dimensional $1$-handlebody of genus $g$, $\phi_1,\phi_2 \colon \partial H_g \rightarrow S_g$ are homeomorphisms and $\boldsymbol{\alpha}_i=\phi_i(\boldsymbol{\alpha})$ for $i=1,2$, where $\boldsymbol{\alpha}$ is the set of boundaries of a maximal system of meridians for $H_g$. I know that the Casson invariant of $\Sigma$ can be defined in terms of its Heegaard splitting, but I do not see clearly how to compute it from a Heegaard diagram. Is there a simple way to do that? REPLY [5 votes]: I don’t know a nice recipe from a Heegaard diagram, but every Heegaard splitting of a homology 3-sphere is equivalent to one whose gluing map lies in the Torelli subgroup of the mapping class group (ie the subgroup acting trivially on homology), and Morita has a beautiful formula for the Casson invariant in terms of this gluing map in his paper S. Morita, On the structure of the Torelli group and the Casson invariant, Topology 30 (1991), no. 4, 603–621. REPLY [4 votes]: I don't know if this qualifies as a simple formula, but here at least is a procedure. You can convert the Heegaard diagram into a framed link description of your homology sphere, and then use Hoste's formula (A formula for Casson's invariant. Trans. Amer. Math. Soc. 297 (1986), no. 2, 547–562). There is an intermediate step, because Hoste requires that your surgery description have all linking numbers 0, so you have to arrange that first.<|endoftext|> TITLE: Is there a "spectral exterior algebra" construction in higher algebra? QUESTION [5 upvotes]: Given a ring spectrum $R$ and an $R$-module $E$, we have the spectral symmetric algebra $\mathrm{Sym}_R(E)$ of $E$ over $R$, defined by $$ \begin{align*} \mathrm{Sym}_R(E) &\overset{\mathrm{def}}{=} \mathrm{colim}_{\mathbb{F}}(\Delta_{E})\\ &\cong \bigoplus_{n\in\mathbb{N}}E^{\otimes_\mathbb{S}n}_{\mathsf{h}\Sigma_{n}}, \end{align*} $$ where $\mathbb{F}\overset{\mathrm{def}}{=}\mathsf{FinSets}^\simeq$ is the groupoid of finite sets and permutations. As A Rock and a Hard Place showed here, the $\mathbb{E}_\infty$-ring $\mathrm{Sym}_R(E)$ comes with a natural grading by the sphere spectrum, inducing a $\mathbb{Z}$-grading on $\pi_0(\mathrm{Sym}_R(E))\cong\mathrm{Sym}_{\pi_0(R)}(\pi_0(E))$. So e.g. picking $R=E=\mathbb{S}$, gives $$ \begin{align*} \pi_0(\mathrm{Sym}_{\mathbb{S}}(\mathbb{S})) &\overset{\mathrm{def}}{=} \pi_0(\mathbb{S}\{t\})\\ &\cong \mathbb{Z}[t], \end{align*} $$ which carries the natural $\mathbb{Z}$-grading. However, the $\pi_0$ of an $\mathbb{S}$-graded ring can be more complicated than just a commutative $\mathbb{Z}$-grading, and for instance allows for the multiplication on the $\pi_0$ to be supercommutative, satisfying $ab=(-1)^{\deg(a)\deg(b)}ba$. This led me to the following pair of questions: Is there an "spectral exterior algebra" construction $\bigwedge_RE$, whose $\pi_0$ is the $\mathbb{Z}$-graded supercommutative exterior algebra $\bigwedge_{\pi_0(R)}\pi_0(E)$? If so, does it come with an $\mathbb{S}$-grading? One of the more homotopy-theoretic points of view on symmetric and exterior algebras is that the passage from the former to the latter corresponds to considering a larger portion of the sphere spectrum. More generally, do we have an $\mathbb{N}$-indexed sequence of "higher exterior algebra" constructions $\mathrm{Sym}_R(E)$, $\bigwedge_R(E)$, $\bigwedge^{\mathbf{2}}_R(E)$, $\ldots$? REPLY [2 votes]: This is not an answer, but rather a comment to A Rock and a Hard Place's answer. Here's another idle musing: what about replacing the relation $ab=(-1)^{\deg(a)\deg(b)}ba$ with coherent homotopies, having an analogue of $\mathbb{E}_{\infty}$ for graded-commutativity? I'm not sure what's the best way to do this, but I see a possible (probably not totally satisfying) one. The basic idea is to generalise from the characterisation of exterior algebras as free graded commutative algebras. 1. The case of classical exterior algebras. Recall that: A $\mathbb{Z}$-graded $R$-algebra $A_\bullet$ is $\mathbb{Z}$-graded commutative if we have $$ab=(-1)^{\deg(a)\deg(b)}ba$$ for each $a,b\in A_\bullet$. The exterior algebra $\bigwedge_RM$ on an $R$-module $M$ is the free $\mathbb{Z}$-graded commutative algebra on $M$: the assignment $M\mapsto\bigwedge_RM$ defines a functor $$\textstyle\bigwedge_R\colon\mathsf{Alg}_R\to\mathsf{CommGr}_{\mathbb{Z}}\mathsf{Alg}_R$$ that is left adjoint to the forgetful functor $\mathsf{CommGr}_{\mathbb{Z}}\mathsf{Alg}_R\hookrightarrow\mathsf{Alg}_R$. The category of $\mathbb{Z}$-graded commutative algebras embeds fully faithfully into that of $\tau_{\leq1}\mathbb{S}$-graded commutative algebras, which are lax symmetric monoidal functors $(\tau_{\leq1}\mathbb{S},\otimes,0)\to(\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z})$. Question: is $\bigwedge_RM$ not only the free $\mathbb{Z}$-graded commutative algebra on $M$, but also the free $\tau_{\leq1}\mathbb{S}$-graded commutative algebra on $M$? 2. The graded commutativity condition. As I mentioned in my other question, a $\tau_{\leq1}\mathbb{S}$-graded $R$-algebra is a lax monoidal functor from the $1$-truncation of the sphere spectrum to $(\mathsf{Mod}_R,\otimes_{R},R)$, and it is $\tau_{\leq1}\mathbb{S}$-graded commutative when this functor is symmetric lax monoidal. Now, for a functor $(\tau_{\leq1}\mathbb{S},\otimes,0)\to(\mathsf{Mod}_R,\otimes_{R},R)$ to be (symmetric) lax monoidal is the same as for it to be a (commutative) monoid under the Day convolution monoidal structure on $\mathbf{Fun}(\tau_{\leq1}\mathbb{S},\mathsf{Mod}_R)$. So, given an $R$-module $M$, we can consider the constant functor $$\Delta_M\colon\tau_{\leq1}\mathbb{S}\to\mathsf{Mod}_R$$ on $M$ and consider the free commutative $\otimes_{\mathsf{Day}}$-monoid on $\Delta_M$. If the answer to the question above is yes, then this is the exterior algebra of $M$. 3. Spectral exterior algebras. Now we can repeat the same strategy for module spectra: given a ring spectrum $R$ and an $R$-module $M$, pick the constant functor $$\Delta_M\colon\tau_{\leq k}\mathbb{S}\to\mathsf{ModSp}_R$$ and apply the free $\mathbb{E}_{\infty}$-monoid functor to $\Delta_M$ with respect to the Day convolution monoidal structure on $\mathsf{Fun}(\tau_{\leq k}\mathbb{S},\mathsf{ModSp}_R)$. The result is then a possible candidate for the spectral "$k$th higher" exterior algebra of $M$ over $R$. 4. The catch. I'm not sure if this really gives a desirable result at all. For instance, does it recover spectral symmetric algebras when $k=0$?<|endoftext|> TITLE: Does $\mathrm{SO}(3)$ act faithfully on a countable set? QUESTION [29 upvotes]: Let $\mathrm{SO}(3)$ be the group of rotations of $\mathbb{R}^3$ and let $S_\infty$ be the group of all permutations of $\mathbb{N}$. Is $\mathrm{SO}(3)$ isomorphic to a subgroup of $S_\infty$? This question is due to Ulam. It is discussed in V.2 of Ulam's "A collection of mathematical problems". It is also discussed in de le Harpe's "Topics in Geometric Group Theory", Appendix III.B. According to de la Harpe it was open as of 2003. Is it still open? Ulam also asks the more general question of whether every Lie group is isomorphic to a subgroup of $S_\infty$. De la Harpe constructs homomorphic embeddings $\mathbb{R} \to S_\infty$ and $\mathbb{R}/\mathbb{Z} \to S_\infty$. There is an obvious embedding $S_\infty \times S_\infty \to S_\infty$, it follows that any connected abelian Lie group is a subgroup of $S_\infty$. So it is natural to look at $\mathrm{SO}(3)$ or other low dimensional non-abelian Lie groups. It is worth pointing out that we cannot hope to realize $\mathrm{SO}(3)$ as a subgroup of $S_\infty$ without using a lot of choice. Suppose that $G$ is a compact lie group and $f : G \to S_\infty$ is a homomorphism constructed without choice. Then $f$ is Baire measurable, so by the Pettis lemma $f$ is continuous, hence $f(G)$ is compact and connected. Finally, any compact connected subgroup of $S_\infty$ is trivial. (In general closed subgroups of $S_\infty$ are totally disconnected.) So a faithful action of $\mathrm{SO}(3)$ would be rather different than the usual kinds of actions. REPLY [16 votes]: From MathSciNet: Thomas, Simon Infinite products of finite simple groups. II. J. Group Theory 2 (1999), no. 4, 401–434. Summary: "(...) In the course of our classification proof, we also show that if $K$ is a field of cardinality $2^\omega$ and $G$ is a non-trivial linear group over $K$, then there exists a subgroup $H$ of $G$ such that $1<[G:H] \le \omega$." [This is almost the same since enlarging $G$, we can always assume $G$ is simple, say $G=\mathrm{PSL}_m(K)$, and then the induced homomorphism $G\to \mathrm{Sym}(G/H)$ is injective. The proof uses countable valuations as well.] Ershov, Yu. L. ; Churkin, V. A. On a problem of Ulam. (Russian) Dokl. Akad. Nauk 399 (2004), no. 3, 307-309. Theorem 1. Every linear group over a field with continuum cardinality (in particular over the field $\mathbf{R}$ of all real numbers or over the field $\mathbf{C}$ of all complex numbers) can be embedded (as an abstract group) in the permutation group of a countable set. (Unlike S. Thomas, they were aware of Ulam's question. On the other hand they were visibly not aware of S. Thomas paper.)<|endoftext|> TITLE: Software for detecting Brauer-Manin obstructions? QUESTION [9 upvotes]: In the context of another MO question, the following question arose: Does there exist any software for detecting Brauer–Manin obstructions to the existence of integer solutions to a single polynomial equation? Failing that, has anyone written down a detailed description of such an algorithm? It may be that one (ahem) obstruction to the existence of such software is that the standard accounts of the Brauer–Manin obstruction are written in the language of modern algebraic geometry, which is unfamiliar to many people who might otherwise have the right skills to write the software. To some extent, such language is unavoidable, but it would be nice to have an account that is as elementary as possible. Perhaps some of the building blocks have already been implemented in (say) Sage, and it is not too hard to explain what is needed to put them together. REPLY [11 votes]: I strongly disagree with the assertion that "the language of modern algebraic geometry [...] is unfamiliar to many people who might otherwise have the right skills to write the software". You are denying the existence of a flourishing research field, computational arithmetic geometry! See e.g. this paper: Bright, M. J.; Bruin, N.; Flynn, E. V.; Logan, A., The Brauer-Manin obstruction and $\text{Ш}[2]$, LMS J. Comput. Math. 10, 354-377 (2007). ZBL1222.11084. From the abstract: "We discuss the Brauer-Manin obstruction on del Pezzo surfaces of degree 4. We outline a detailed algorithm for computing the obstruction and provide associated programs in MAGMA." This isn't exactly the same question as you asked, since the relevant surfaces are intersections of two quadrics in $\mathbf{P}^4$, so not defined by a single equation; but it should serve to demonstrate that there is a substantial literature focussing on explicit algorithmic computations of the Brauer--Manin obstruction -- maybe you can find something matching your question more precisely among papers citing (or cited by) this one.<|endoftext|> TITLE: Hereditarily primary Banach spaces QUESTION [6 upvotes]: A Banach space $X$ is said to be prime if every infinite dimensional complemented subspace is isomorphic to the space $X$. The space $X$ is primary if it has an infinite dimensional subspace $Y$ such that every complemented subspace is either isomorphic to $X$ or to the subspace $Y$. The space $X$ is quasiprime if it is primary and the only decomposition of $X$, as a direct sum, into two infinite dimensional subspaces is $X +Y$. Prime spaces are quasiprime which are primary. Also there are examples that separate the above classes. A Banach space $X$ is hereditarily prime (primary, quasiprime) if every infinite dimensional subspace is prime (primary, quasiprime). Hilbert spaces are hereditarily prime. Also there exists a non-Hilbertian space $X$ which is hereditarily quasiprime. Question I Is every hereditarily prime space isomorphic to a Hilbert space? Question II Do there exist subspaces of $\ell_p$, which are not primary? REPLY [2 votes]: Actually the above three conditions yielding a counterexample are satisfied by any counterexample and this is easy to be checked. However such a space has extremely peculiar structure. In particular it does not contain any subspace of the form Y + Z with Y Hilbertian and Z indecomposable. This follows from the previous positive partial answer. That means that for every indecomposable subspace Z and every Hilbertian subspace Y dist( $S_Z$ , $S_Y$ )= 0. Further the same holds for all Y subspaces of X. The last follows from the fact that X is $l_2$ saturated. It is hard to think how such a space is defined. However if the following has a positive answer then the original question I has also a positive one. Question III Let X be a separable non Hilbertian and $l_2$ saturated Banach space.Does the space X contain Y +Z with Y Hilbertian and Z non Hilbertian?<|endoftext|> TITLE: Value of divergent sum $\sum_{n=0}^\infty (-1)^n n^n$ QUESTION [8 upvotes]: I'm hoping to find a reasonable value to assign to the divergent series $\sum_{n=0}^\infty (-1)^n n^n$ and $\sum_{n=0}^\infty (-1)^n (xn)^n$. For the first one, I have obtained something around 0.71, but I'm very unsure if this is correct. For the second series, I get a graph that looks about like this: . What sort of methods are powerful enough to sum this series? I think Borel summation is too weak, since this grows faster than any $\left(\alpha n\right)!$. Any help would be appreciated! EDIT: I obtained my values from a series of questionably valid approximations, but here is another method that seems to give a similar value. The idea is that illegally swapping summation signs, and then continuing the inner summation past its regular convergent range allows one to assign a value to a divergent series. Starting with $$\sum_{n=0}^\infty (-1)^n n^n$$ We want to get two summations in order to be able to swap, so we expand into $$\sum_{n=0}^\infty (-1)^n e^{n\ln(n)} = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty \frac{\left(n \ln(n)\right)^k}{k!}$$ Simplifying a bit more, we get $$\sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty \frac{n^k \ln(n)^k}{k!} =\sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty \frac{e^{\ln(n)k} \ln(n)^k}{k!}$$ Swapping the summations, we get $$\sum_{k=0}^\infty \frac{1}{k!} \sum_{n=0}^\infty (-1)^n e^{\ln(n)k} \ln(n)^k = \sum_{k=0}^\infty \frac{1}{k!} \sum_{n=0}^\infty \frac{d^k}{dk^k} (-1)^n n^k $$ (Note: I'm using $\frac{d^k}{dk^k}$ to represent taking the derivative with respect to k, k times). Now, we can continue the inner summation by doing $$\sum_{k=0}^\infty \frac{1}{k!} \frac{d^k}{dk^k}\sum_{n=0}^\infty (-1)^n n^k = \sum_{k=0}^\infty \frac{1}{k!} \frac{d^k}{dk^k}(1-\eta(-k))$$ To finally get $$1/2 - (\eta'(-1) - \frac{\eta''(-2)}{2!} + \frac{\eta'''(-3)}{3!} - \dots)$$ The sum of the first few terms seems to be around .71, which agrees fairly closely with the other method. EDIT2: It looks like most methods converge onto about the same shape eventually. The green line is the asymptotic expansion, meaning I only took the first 5 terms in the power series, and left the rest out. Usually, this asymptotic expansion converges to the right function within a very small radius. The black function is using something like FusRoDah's method. The orange graph is using the non-rigourous series of approximations. This next pictures shows the functions over larger intervals--the purple function is using that eta method I outlined above. To elaborate more about how I used FusRoDah's method, I started with $BA(t)=\sum_{k=0}^\infty \frac {(-kt)^k} {k!} = 1+\sum_{k=1}^\infty \frac {(-kt)^k} {k!}$. Then, I added in the approximation to get $$\operatorname{BA}(t) = 1+\sum_{k=1}^\infty \frac {(-kt)^k} {k!} + \frac {(-et)^k} {\sqrt{2\pi k}}-\frac {(-et)^k} {\sqrt{2\pi k}} = 1+\sum_{k=1}^\infty\frac {(-et)^k} {\sqrt{2\pi k}}+ \sum_{k=1}^\infty \frac {(-kt)^k} {k!} -\frac {(-et)^k} {\sqrt{2\pi k}}.$$ The first part of the sum can be written as an integral of the polylogirthmn with some other terms, but we are still left with the $\int_{0}^{\infty}e^{-t}\sum_{k=1}^{\infty}\left(\frac{\left(-ktz\right)^{k}}{k!}-\frac{\left(-etz\right)^{k}}{\sqrt{2\pi k}}\right)dt$ term. Since this doesn't converge on its own, we can approximate it with $$\int_{0}^{A}e^{-t}\sum_{k=1}^{B}\left(\frac{\left(-ktz\right)^{k}}{k!}-\frac{\left(-etz\right)^{k}}{\sqrt{2\pi k}}\right)dt.$$ Increasing $A$ and $B$ increases the accuracy, but decreases the range of convergence. I used $A = 2.5$, $B = 30$ for the first graph. What remains to be done then is either find a more accurate approxation, such that $\sum_k \frac{(-kt)^k}{k!} - f_t(k)$ converges for all values of t, or to find a way to continue the difference between the approximation and the original series. REPLY [13 votes]: The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function. Reference Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66. Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ as it can be seen in the following Pari GP v2.14.0 lines although, as it is pointed out, a rigorous proof is not yet known. And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$<|endoftext|> TITLE: Examples of solid abelian groups QUESTION [16 upvotes]: I am reading through Clausen's and Scholze's Lectures on condensed mathematics. I am struggling to understand the concept of solid abelian groups so I am looking for some examples. Is the underlying condensed abelian group of a finitely generated module (with the unique induced topology) over a Banach ring solid? Do you know of other examples? REPLY [28 votes]: Here's a rule of thumb: As long as the construction is nonarchimedean and does not involve noncompleted tensor products, it's solid. More precisely, anything you can build from discrete abelian groups by repeatedly forming limits (in particular, kernels), colimits (in particular, cokernels), extensions, and internal Hom's, is solid. This includes any nonarchimedean Banach ring: These are (usually) of the form $A=A_0[\tfrac 1t]$ where $A_0\subset A$ is the unit ball and $t\in A$ is some pseudouniformizer, and $A_0$ is $t$-adically complete, $A_0=\varprojlim_n A_0/t^n$. Here, all $A_0/t^n$ are discrete abelian groups, thus solid; thus the limit $A_0$ is solid; thus the colimit $A=A_0[\tfrac 1t]$ is solid. Now if $M$ is a finitely generated $A$-module, you can write it as a cokernel of a map $f: A^n\to A^m$, and this makes $M$ naturally into a solid $A$-module, again by the above principles. (Beware that if you first do this topologically, endowing $M$ with the quotient topology, you may run into the issue that the quotient topology may not be separated. In that case, the corresponding condensed $A$-module may not agree with the above cokernel of $f$ taken in condensed $A$-modules. But I'd simply argue that here the condensed perspective gives you a more sensible thing to do, namely just take the cokernel of $f$ in condensed $A$-modules.) Also beware that if $A$ is a Banach ring over the real numbers, then it's not solid (unless $A=0$).<|endoftext|> TITLE: Invariants of matrices (by simultaneous $\mathrm{GL}_n$ conjugation) over arbitrary rings QUESTION [14 upvotes]: $\DeclareMathOperator\GL{GL}$Let $R$ be a commutative ring, let $R[n] := R[M_d^{\oplus n}]$ be the polynomial ring on $nd^2$ variables corresponding to the coordinates of $n$-many $d\times d$ matrices. Let these matrix variables be $X_1,\ldots,X_n$. For an $d\times d$ matrix $A$, let $c_k(A)$ be the coefficient of $T^k$ in the characteristic polynomial $\det(A-TI)$. The group scheme $\GL_{d,R}$ acts on $R[n]$ by simultaneous conjugation on the $d$ matrices. Clearly for any product $X_{i_1}X_{i_2}\cdots X_{i_r}$ (where $i_j\in[1\ldots n], r\ge 1$), the function $c_k(X_{i_1}\cdots X_{i_r})\in R[n]$ is invariant under $\GL_{d,R}$. Is it true that for any commutative ring $R$, $R[n]^{\GL_{d,R}}$ is generated as an $R$-algebra by the functions $c_k(X_{i_1}\cdots X_{i_r})$? Does anyone have a reference for this? Remark - The statement over $\mathbb{C}$ is a classical result of Sibirski and Procesi. This was later extended to the case $R = \mathbb{Z}$ and $R$ any algebraically closed field by Donkin in Invariants of several matrices. In Concini-Procesi's The invariant theory of matrices, they also seem to obtain the result when $R$ is an infinite field. REPLY [4 votes]: For the benefit of myself and other novices in the area, I wanted to add some details to Wilberd's excellent answer, citing Donkin's Invariants of several matrices and Jantzen's book Representations of Algebraic Groups as appropriate. We work over a base ring $R$. First, one checks that Donkin's definition of a "good filtration" on a $G$-module $V$ coincides with Jantzens (defined in II, 4.16). The latter uses the isomorphism $H^i(M) := R^i\text{Ind}^G_B(V)\cong H^i(G/B,\mathcal{L}(M))$ (Jantzen I, 5.12), where $B\subset G$ is a Borel, $M$ a $B$-module, and $\mathcal{L}(M)$ denotes the quasi-coherent sheaf on $G/B$ associated to $M$ (I, 5.8). If $M = R$ with trivial $G$-action, then $\mathcal{L}(R)$ is the structure sheaf $\mathcal{O}_{G/B}$. Now write $R[n] = \bigoplus_{d\ge 0}R[n]_d$, where $R[n]_d$ consists of the polynomials of total degree $d$. One easily checks that each $R[n]_d$ is a $\text{GL}_{n,R}$-module. We claim that the $\mathbb{Z}$-finite $G = \text{GL}_{n,\mathbb{Z}}$-modules $\mathbb{Z}[n]_{\le D} := \bigoplus_{0\le d\le D} \mathbb{Z}[n]_d$ have good filtrations for any $D\ge 0$. We use the following Lemma Lemma (Appendix B.9 of Jantzen) Let $G$ be a split reductive algebraic group, and let $T$ be a maximal torus. Suppose that $R$ is a principal ideal domain. Let $M$ be a $G$-module which is free of finite rank over $R$. As explained in B.4, let $V(\lambda) := H^0(-w_0\lambda)^*)$ be the Weyl module, where $w_0$ is the longest element of the Weyl group (II, 1.5). The following are equivalent: (i) $M$ has a good filtration. (ii) $Ext^i_G(V(\lambda),M) = 0$ for all $\lambda\in X(T)_+$ and all $i > 0$ (iii) $Ext^1_G(V(\lambda),M) = 0$ for all $\lambda\in X(T)_+$. (iv) For each maximal ideal $\mathfrak{m}$ in $k$, the $G_{R/\mathfrak{m}}$-module $M\otimes R/\mathfrak{m}$ has a good filtration. By (iv), it suffices to check that the $G_{\mathbb{F}_p}$-modules $\mathbb{F}_p[n]_{\le D}$ all have good filtrations. Using flat base change for $G$-module cohomology (see Jantzen I, 4.13), (ii) (or (iii)) implies that it suffices to check the existence of good filtrations over algebraically closed fields $k$. By a result of Donkin (Donkin The normality of closures of conjugacy classes of matrices Proposition 1.2a (i) and (ii)), good filtrations behave well with respect to direct summands, so it suffices to check that $\overline{\mathbb{F}_p}[n]$ have good filtrations. This is done in $\S3$ of Donkin's Invariants of several matrices. Thus we conclude that $\mathbb{Z}[n]_{\le D}$ has a good filtration for any $D\ge 0$. Let $T$ be a maximal torus of $G = GL_{n,\mathbb{Z}}$, and let $\lambda = 0\in X(T)$, then $\lambda$ is dominant and we have $V(0) = H^0(G/B,\mathcal{L}(0)) = H^0(G/B,\mathcal{O}_{G/B})$. Since $G/B$ is smooth proper over $\mathbb{Z}$ (use Jantzen I, 5.6(9) and I,5.7 to reduce to the classical case over algebraically closed fields), Stein factorization implies that $V(0) = \mathbb{Z}$ (with trivial $G$-action). By (ii) or (iii), we conclude that $Ext^1_G(\mathbb{Z},M) = H^1(G,M) = 0$. By a "universal coefficient theorem" (see Jantzen I, 4.18), vanishing of $H^1(G,M)$ implies that for any ring $R$, $\mathbb{Z}[n]^{GL_{2,\mathbb{Z}}}\otimes R = R[n]^{GL_{2,R}}$, as desired.<|endoftext|> TITLE: Duhamel's formula QUESTION [6 upvotes]: Formula (12) in the paper Bauer, M., Chetrite, R., Ebrahimi-Fard, K., & Patras, F. (2013). Time-ordering and a generalized Magnus expansion. Letters in Mathematical Physics, 103(3), 331-350. expresses the derivative of an exponential of a parameter-dependent Lie algebra element. The authors call it 'Duhamel's formula', without giving a reference. Where and in which context did Duhamel prove this formula? (I know the formula as Kubo's formula, from his work in statistical mechanics.) REPLY [8 votes]: Duhamel's formula for the derivative of the exponent of a matrix refers to Jean-Marie-Constant Duhamel, who described it in Eléments de calcul infinitésimal (volume 2, 1856; page 36) as a method to obtain solutions to inhomogeneous linear differential equations in terms of the solution to the homogenous equation. In the context of Lie groups the formula for the derivative of the exponential map is attributed to Schur (1891). A modern proof is in these notes. or alternatively here (theorem 3). A formal proof is given by \begin{align} \frac{d}{dt}e^{A(t)} &= \lim_{N \to \infty}\frac{d}{dt}\left(1 + \frac{A(t)}{N}\right)^N\\ &= \frac{1}{N}\lim_{N \to \infty}\sum_{k=1}^N\left(1 + \frac{A(t)}{N}\right)^{k-1}A'(t)\left(1 + \frac{A(t)}{N}\right)^{N-k}~\\ &=\int_0^1 e^{sA(t)}A'(t)e^{(1-s)A(t)}\,ds. \end{align}<|endoftext|> TITLE: The devil's playground QUESTION [9 upvotes]: On the $\mathbb{R}^2$ plane, the devil has trapped the angel in an equilateral triangle of firewalls. The devil starts at the apex of the triangle. can move at speed $1$ to leave a trajectory of firewall behind, as this can teleport from one point to another along the firewall. The angel can teleport to any point that is not completely separated by firewalls from her current position. The devil catches the angel if their distance is $0$. Question 1: how should the devil move to catch the angel in the shortest amount of time? Question 2: if the devil is given a fixed length of firewalls to enclose the angel in the beginning, what shape maximizes the survival time of the angel? (The devil always starts on the firewall) REPLY [3 votes]: It seems there is value to think about about splitting the triangle into 3 or 4 pieces rather than just 2 in each stage. I give a natural approach that achieves 3, and another which gives $\frac{1}{3} +\sqrt{3}\le 2.06539$. For any two points $x,y$, let $xy$ denote the line segment from $x$ to $y$. For example, one natural approach is to try and split the triangle into 4 equilateral triangles. Let $a,b,c$ be the points of our triangle. We start by teleporting to the midpoint of $ab$. Then, we move from the midpoint of $ab$ to the midpoint of $bc$, and then to the midpoint of $ca$. This takes length 3/2. In doing so, we have scaled the triangle down by a factor of 1/2. We may now repeat, this time able to teleport to a midpoint, now only requiring length 3/4, and so on. Hence, this gives an upper bound of $3\sum_{i=1}^\infty 2^{-i} = 3$, as desired. Now, what if we split the triangle into three parts? Let $d,e,f$ denote the respective midpoints of $ab,bc,ca$. Let $o$ be the interesection of $dc$ and $ea$. We walk from $d$ to $o$, $e$ to $o$ and $f$ to $o$. Each segment has length $\frac{1}{2\sqrt{3}}$, thus this in total costs $\sqrt{3}/2\le 0.86603$. The angel will be trapped in a quadrilateral $Q$, since our triangle is split into 3 identical copies. WLOG, we assume the vertices of $Q$ are $a,d,f,o$. We have that $Q$ is contained in a equilateral triangle $T$ of height $1/\sqrt{3}$ (indeed, consider the line $\ell$ parallel to $df$ through $o$, we have that $Q$ is contained in the triangle whose vertices are $a$ and the intersection of $\ell$ with our initial triangle). We have that said triangle has side lengths 2/3. Now, suppose we try to iterate, and split our triangle $T$ into 3 parts. We first draw the line parallel to $do$, then the line parallel to $fo$. The angel has two choices. Either, it goes in the new quadrilateral $Q'$ that is formed, or it is on the other side. In the first case, it is easy to see that $Q'$ is similar to $Q$, and 2/3 as large (since $T$ is 2/3 as large as our initial triangle). In the other case, we add the third line (which will be will be in the same direction as $oe$); here the angel is stuck in one of two parts which are each covered by a copy of the $1/6\times \frac{1}{2\sqrt{3}}$ rectangle. We now can analyze the total length required to catch the angel once it is in $Q$. Let $L$ denote the total length needed if we follow the above strategy, (namely adding two lines, then scaling down if the angel stays in $Q'$, or adding the third line and doing a rectangle strategy otherwise), assuming the angel behaves optimally. If the angel chooses to stay in $Q$, then we only added two edges each of length $(2/3) \frac{1}{2\sqrt{3}}$, for a total length of $\frac{2}{3\sqrt{3}}$, and since $Q'$ is a 2/3 scaled down copy of $Q$, we will catch the angel having used length $\frac{2}{3}L+\frac{2}{3\sqrt{3}}$ here. In the other case, we add 3 lines, whose net length is $\frac{1}{\sqrt{3}}$. By the argument given by Sawin's comment below his answer, we can do the rectangle using length $2(1/6)+(1/2\sqrt{3})$. This totals to a length of $\frac{1}{3}+\frac{\sqrt{3}}{2}\le 1.19936$. Hence, we just solve for $L = \max\{\frac{1}{3}+\frac{\sqrt{3}}{2},\frac{2}{3}L+\frac{2}{3\sqrt{3}}\}$. Observing the solution to $x= (2/3)x +\frac{2}{3\sqrt{3}}$ is $2/\sqrt{3} \approx 1.15470$, we see that $L =\frac{1}{3}+\frac{\sqrt{3}}{2}\le 1.19936$. It follows that we can catch the angel in our triangle by adding our initial cost $\frac{\sqrt{3}}{2}$ to $L$, which gives $1/3 + \sqrt{3} \le 2.06539$. Perhaps we get get a slightly better solution by optimizing the second case, since they are not truly rectangles.<|endoftext|> TITLE: Sum of squares and divisibility QUESTION [25 upvotes]: Consider an integer of the form $$N = 1 + \sum_{i=1}^r d_i^2$$ where $d_i \in \mathbb{N}_{\ge 3}$ and $d_i^2$ divides $N$. Question: Must $r$ be greater than or equal to $9$? Checking (with SageMath): It is true for $N \le 500000$. Remark: If it is true in general then it is optimal because $$144 = 1 + 3^2 + 3^2 + 3^2+ 4^2 + 4^2 + 4^2 + 4^2 + 4^2 + 6^2$$ Motivation (from tensor category theory): If above question has a positive answer then by [1, Proposition 8.14.6] and [2, Theorem 3.4] a simple integral modular fusion category (over $\mathbb{C}$) would be of rank $ \ge 10 $. And for so, if required, we can also assume $d_i$ not a prime-power by [3, Corollary 6.16]. Then the (SageMath) checking is $N \le 10^6$; and above optimality would be unchanged because $$ 116964 = 1 + 6^2 + 18^2 + 38^2 + 38^2 + 114^2 + 114^2 + 171^2 + 171^2 + 171^2$$ We did not add this additional assumption at the beginning because it seems to be true without it (experimentally), and without it the question is (number theoretically) more interesting. With this additional assumption, the next example with $r=9$ is $$ 396900 = 1 + 18^2 + 30^2 + 70^2 + 70^2 + 210^2 + 210^2 + 315^2 + 315^2 + 315^2 $$ Above two examples with $r=9$ can be excluded from coming from a fusion ring because: First one: consider the simple object of FPdim $6$, multiply it with its dual (which must be itself here) and apply FPdim (ring homomorphism). Then you get $6^2=1+\cdots$, but there is no FPdim $\le 35$ (of a non-trivial simple object) which is not a multiple of $6$. Second one: as above, $18^2 - 1 = 323$ is odd and the only odd FPdim for a non-trivial simple object is $315$ , but $323-315 = 8$ and there is no non-trivial simple object of FPdim $\le 8$. Conclusion: if above two examples are the only ones for $r=9$ (I checked that there is no other one for $N \le 10^6$), and if above question has a positive answer (we can put the additional assumption if required), then a simple integral modular fusion category (over $\mathbb{C}$) would be of rank $ \ge 11 $. References [1]: P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205. [2]: J. Dong, S. Natale, L. Vendramin; Frobenius property for fusion categories of small integral dimension. J. Algebra Appl. 14 (2015), no. 2, 1550011. [3]: D. Nikshych, Morita equivalence methods in classification of fusion categories, Hopf algebras and tensor categories, 289-325, Contemp. Math., 585, Amer. Math. Soc. (2013). REPLY [8 votes]: I hope so. But please double check (or, better, simplify) the argument below. Denote $N=qs^2$ for $q$ squarefree. Then each $d_i$ divides $s$, say $d_i=s/m_i$ and we get $$q=1/s^2+\sum_{i=1}^r 1/m_i^2, \quad\quad\quad (\heartsuit)$$ and the sum of $r+1$ square reciprocals is integer. Since $d_i\geqslant 3$, we get $m_i\leqslant s/3$. Subtracting the summands with $m_i=1$ we get $(\heartsuit)$ with smaller $r$ and the same condition $m_i\leqslant s/3$ (possibly $q$ is no longer squarefree, but we do not care anymore.) So, now each $m_i$ is at least 2. So, we assume that $(\heartsuit)$ holds with $r\leqslant 8$, some integer $q$ and $s/3\geqslant m_i\geqslant 2$. If $r=8$ and all $m_i$ are equal to 2, then $1/s^2$ must be an integer which is absurd. Otherwise $\sum 1/m_i^2\leqslant 7\cdot 1/4+1/9$ and $1/s^2+\sum 1/m_i^2\leqslant 1/36+7/4+1/9<2$, thus $q=1$. Further we denote $s=m_{r+1}$, and $(\heartsuit)$ reads as $$1=\sum_{i=1}^{r+1} 1/m_i^2. \quad\quad\quad(\clubsuit)$$ And we have a Special Condition. $m_{r+1}\geqslant 3\max(m_1,\ldots,m_r)$ and $m_i$ divides $m_{r+1}$ for all $i$. What we do below is finding all solutions of $(\clubsuit)$ with $r\leqslant 8$. They all fail Special Condition. Let me start with listing them: (a) $r=3$, $(2,2,2,2)$; (b) $r=8$, $(2,2,2,3,6,6,6,6,6)$; (c) $r=8$, $(2,2,3,3,4,4,4,4,6)$; (d) $r=8$, $(2,2,2,4,4,4,6,6,12)$; (e) $r=6$, $(2,2,2,4,4,4,4)$; (f) $r=8$, $(2,2,2,3,3,12,12,12,12)$; (g) $r=7$, $(2,2,2,3,4,4,12,12)$; (h) $r=8$, $(2,2,2,3,9,4,4,36,36)$; (i) $r=8$, $(2,2,2,3,4,4,12,15,20)$; (j) $r=7$, $(2,2,3,3,3,3,6,6)$; (k) $r=5$, $(2,2,2,3,3,6)$; (l) $r=7$, $(2,2,2,3,3,7,14,21)$; (m) $r=7$, $(2,2,2,3,3,9,9,18)$; (n) $r=8$, $(3,3,3,3,3,3,3,3,3)$ (o) $r=0$, $(1)$. Now goes a rather lengthy and boring proof that there are no other solutions of $(\clubsuit)$. Denote by $2^\beta$ the maximal power of 2 which divides one of the numbers $m_i$, $i=1,\ldots,r+1$. We have $\beta\geqslant 1$: otherwise all $1/m_i^2$ are at most $1/9$, and not all equal to $1/9$, and RHS of $(\clubsuit)$ is less than 1. If we have exactly $h$ indices $i$ for which $2^\beta$ divides $m_i$, then multiplying $(\clubsuit)$ by $2^{2\beta}$ and considering the expression modulo 4 we get $4|h$. Thus $h=8$ or $h=4$ or $h=0$. Consider several cases. $h=0$, all $m_i$'s are odd. Then either $r=0$ and we get solution (0), or they all are not less than 3, and $\sum 1/m_i^2\leqslant 9\cdot 1/9=1$, with equality corresponding to solution (n). $h=8$ and $\beta \geqslant 2$. Then RHS of $(\clubsuit)$ does not exceed $8\cdot \frac1{16}+1\cdot \frac14<1$. $h=8$ and $\beta=1$. Then eight even $m_i$'s may be denoted by $2s_i$, $s_i$ are odd, $i=1,\ldots,8$, and $(\clubsuit)$ reads as $$ \frac14\sum_{i=1}^8\frac1{s_i^2}+\frac{\varepsilon}{\ell^2}=1,\quad\quad\quad(\smile) $$ where $\varepsilon=0$ if $r=7$ and $\varepsilon=1$ if $r=8$ (and $\ell$ is odd). At most three $s_i$'s may be equal to 1, others are at least 3, also $\ell\geqslant 3$, thus LHS of $(\smile)$ is at most $\frac14(3\cdot1+5\cdot \frac19)+\frac19=1$, the equality case is unique, it is the solution (b). $h=4$ and $\beta\geqslant 3$. Analogously, $(\clubsuit)$ reads as $$ \frac1{4^\beta}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-3}\frac1{\ell_i^2}=1,\quad\quad\quad(\ast) $$ where $s_i$ are odd. At most 3 $\ell_i$'s are equal to 2. Assume that at least one of $\ell_i$'s is at least 4. Then LHS of $(\ast)$ is at most $\frac1{64}\cdot 4+3\cdot\frac14+1\cdot\frac19+1\cdot\frac1{16}<1$, a contradiction. So, all $\ell_i$'s are equal to 2 or 3, and we get contradiction modulo 8 after multiplying by $4^\beta$. $h=4$, $\beta=2$. Now we get $$ \frac1{16}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-3}\frac1{\ell_i^2}=1,\quad\quad\quad(\diamond) $$ where $s_i$ are odd and $\ell_i$ not divisible by 4. Assume that at most one $\ell_i$'s is equal to 2, then LHS of $(\diamond)$ does not exceed $\frac1{16}\cdot 4+1\cdot \frac14+4\cdot \frac19<1$, a contradiction. Thus we get $r\geqslant 5$ and we may suppose $\ell_{r-3}=\ell_{r-4}=2$, and $(\diamond)$ reads as $$ \frac1{16}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-5}\frac1{\ell_i^2}=\frac12.\quad\quad\quad(\diamond\diamond) $$ If all $\ell_i$'s ($i=1,\ldots,r-5$) are odd, we get a contradiction modulo 8 after multiplying by $16$. If none of $\ell_i$'s ($i=1,\ldots,r-5$) equals to 2, then one of them is at least 6, and LHS of $(\diamond\diamond)$ does not exceed $\frac1{16}\cdot 4+2\cdot \frac19+\frac1{36}=1/2$. There exists unique equality case, it corresponds to solution (c). Thus, we may suppose that $r\geqslant 6$ and $\ell_{r-5}=2$, and $(\diamond\diamond)$ reads as $$ \frac1{16}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-6}\frac1{\ell_i^2}=\frac14.\quad\quad\quad(\diamond\diamond\diamond) $$ If exactly one of $\ell_i$'s ($i=1,\ldots,r-6$) is even (recall that it is not however divisible by 4), we get a contradiction modulo 8 after multiplying by 16. If two of $\ell_i$'s ($i=1,\ldots,r-6$) are even, then they are at least 6, and LHS of $(\diamond\diamond\diamond)$ is at most $\frac1{16}(3+\frac19)+2\cdot \frac1{36}=\frac14$. Again, the equality case is unique, it provides a solution (d). Thus, all $\ell_i$'s ($i=1,\ldots,r-6$) are odd. If $r=6$, we immediately get $s_1=\ldots=s_4=1$, this gives solution (e). So, $r\geqslant 7$, at most 3 of $s_i$'s are equal to 1 and $$\sum_{i=1}^{r-6}\frac1{\ell_i^2}\geqslant \frac14-\frac1{16}\left(3+\frac19\right)=\frac1{18}.$$ Thus there is 3 or 5 between $\ell_i$'s, $1\leqslant i\leqslant r-6$. Well, proceed with cases. Assume that $\ell_{r-6}=5$, but $\ell_{r-7}\ne 3$ (or $r=7$ and $\ell_1=5$). Then $\sum \frac1{s_i^2}=4-16\sum_{i=1}^{r-6}\frac1{\ell_i^2}\geqslant 4-\frac{32}{25}$. Therefore at least three $s_i$'s are equal to 1, say $s_2=s_3=s_4=1$, and $(\diamond\diamond\diamond)$ reads as $\frac1{16s_1^2}+\frac{\varepsilon}{\ell_{r-7}^2}=\frac1{16}-\frac1{25}=\frac9{16\cdot 25}$ (here $\varepsilon=0$ if $r=7$ and $\varepsilon=1$ if $r=8$). We see that $r=8$, and we get an equation $\frac1{\ell_1^2}=\frac{9}{16\cdot 25}-\frac1{16s_1^2}=\frac{(3s_1-5)(3s_1+5)}{16\cdot 25\cdot s_1^2}$. If 5 does not divide $s_1$, then $3s_1\pm 5$ is coprime to $25s_1^2$, and the numerator of $\frac{(3s_1-5)(3s_1+5)}{16\cdot 25\cdot s_1^2}$ may be equal to 1 only if both $3s_1-5$ and $3s_1+5$ are powers of 2, which is impossible (powers of 2 never differ by 10). So, $s_1$ is divisible by 5 and $\frac9{16\cdot 25}\geqslant \frac1{\ell_1^2}\geqslant \frac8{16\cdot 25}$, that is, $\ell_1=7$, but this does not provide a solution. Assume then that $\ell_{r-6}=3$. Analogously, we get $\sum \frac1{s_i^2}+\frac{16\varepsilon}{\ell_{r-7}^2}=16(\frac14-\frac19)=\frac{20}{9}$. If all $s_i$'s are at least 3, then LHS is at most $4/9+16/9=20/9$, and equality holds in the unique case, corresponding to solution $(f)$. So let us suppose that $s_4=1$, we get $\sum_{i=1}^3 \frac1{s_i^2}+\frac{16\varepsilon}{\ell_{r-7}^2}=\frac{11}9$. If all $s_1,s_2,s_3$ are at least 3, we get $r=8$ and $\frac{11}9\geqslant \frac{16}{\ell_1^2}\geqslant \frac{11}9-3\cdot \frac19=\frac{8}9$, that is impossible (recall that $\ell_1$ is odd). So, without loss of generality $s_3=1$ and $\frac1{s_1^2}+\frac1{s_2^2}+\frac{16\varepsilon}{\ell_1^2}=\frac29$. Clearly $s_1,s_2\geqslant 3$. If $r=7$, i.e., $\varepsilon=0$, this yields a solution with $s_1=s_2=3$, that is (g). So assume that $r=8$. Since $\frac{16}{\ell_1^2}<\frac2{9}$, we get $\ell_1\geqslant 9$. Thus $\frac1{s_1^2}+\frac1{s_2^2}\geqslant \frac29-\frac{16}{81}=\frac2{81}$. It yields $\min(s_1,s_2)\leqslant 9$. If now $s_1,s_2$ are at least 5, then $\frac{16}{\ell_1^2}\geqslant \frac29-\frac2{25 }=\frac{32}{225}$, or $\frac{225}2\geqslant \ell_1^2$, this yields $\ell_1=9$. Thus $\frac1{s_1^2}+\frac1{s_2^2}=\frac2{81}$. If $s_1=s_2=9$, we get solution (h). Otherwise, say, $s_2=7$, this does not lead to a solution. Finally, if $s_2=3$, we get an equation $\frac1{s_1^2}+\frac{16}{\ell_1^2}=\frac19$, or $\frac{16}{\ell_1^2}=\frac19-\frac1{s_1^2}=\frac{(s_1-3)(s_1+3)}{9s_1^2}$. If 3 does not divide $s_1$, then $s_1\pm 3$ are coprime to $9s_1^2$, so both $s_1-3$ and $s_1+3$ must be powers of 2. This happens only for $s_1=5$, and we get solution (i). If $s_1=3k$, we get $\frac{(k-1)(k+1)}{9k^2}=\frac{16}{s_1^2}$, thus $k^2-1=(k-1)(k+1)$ is a positive perfect square, which is impossible. Uff, now let $h=4$ and $\beta=1$. We get $$ \frac14\sum_{i=1}^4\frac1{s_i^2}+\sum_{i=1}^{r-3}\frac1{\ell_i^2}=1 $$ for odd $s_i$ and $\ell_i$. Multiplying by 4 and considering modulo 8 we see that $r-3$ must be even, thus $r$ is odd, so $r\leqslant 7$. Therefore $\sum\frac1{\ell_i^2}\leqslant \frac49$ and $\sum\frac1{s_i^2}\geqslant 4(1-\frac49)=\frac{20}9$. This implies that at least two of $s_i$'s are equal to 1, say $s_3=s_4=1$. We get from above $\frac1{s_1^2}+\frac1{s_2^2}\geqslant \frac29$, so either $r=7$, $\ell_1=\ldots=\ell_4=3$, $s_1=s_2=3$, that's solution (j), or, say, $s_2=1$, and we get $\frac1{4s_1^2}+\sum \frac1{\ell_i^2}=\frac14$. There is a solution (a) with $s_1=1$ and $r=3$. Otherwise $s_1\geqslant 3$ and we get $\sum \frac1{\ell_i^2}\in (\frac14-\frac1{36},\frac14)$. Thus, one of $\ell_i$'s is equal to 3, say $\ell_{r-3}=1$, and $\frac1{4s_1^2}+\sum_{i=1}^{r-4}\frac1{\ell_i^2}=\frac5{36}$. Assume that all remaining in LHS $\ell_i$'s are greater than 3, and that $s_1$ is greater than 3. Then $\frac1{4s_1^2}+\sum_{i=1}^{r-4}\frac1{\ell_i^2}\leqslant \frac1{100}+3\cdot \frac1{25}=\frac{13}{100}<\frac5{36}$. Next case. $s_1=3$. Then $\sum_{i=1}^{r-4}\frac1{\ell_i^2}=\frac19$. If $r=5$, we get solution (k). Otherwise $r\geqslant 6$, then remaining $\ell_i$'s are at least 5, not all equal to 5, thus $\sum_{i=1}^{r-4}\frac1{\ell_i^2}\leqslant \frac1{25}+\frac1{25}+\frac1{49}<\frac19$. Finally, let $s_1>3$, but $\ell_{r-4}=3$. Then $\frac1{4s_1^2}+\sum_{i=1}^{r-5}\frac1{\ell_i^2}=\frac1{36}$. If all $\ell_i$'s, $i\leqslant r-5$, are at least 11, then LHS is at most $1/100+2/121<1/36$. So, one of them equals to 7 or 9. This leads to equations $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{13}{36\cdot 49}$ and $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{5}{4\cdot 81}$, respectively. For $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{13}{36\cdot 49}$ note that 7 must divide both $s_1$ and $\ell_1$ (otherwise we get a contradiction modulo 7 multiplying by 49, due to 13 being quadratic non-residue modulo 7). If, say, $s_1=7z$, $\ell_1=7w$, we get an equation $\frac1{4z^2}+\frac1{w^2}=\frac{13}{36}=\frac14+\frac19$. This is only possible for $z=1$, $w=3$ (for $w=1$ LHS is too large, otherwise too small). Surprisingly, this solution does not enjoy SC again. For $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{5}{4\cdot 81}$, we analogously get that $s_1$, $\ell_1$ must be divisible by 9 (since 5 is not a quadratic residue modulo 3). So, we get an equation $\frac1{4z^2}+\frac1{w^2}=\frac{5}4$ which has the unique solution $z=w=1$, which leads to solution (m).<|endoftext|> TITLE: Are these theories of real and complex number biinterpretable? QUESTION [8 upvotes]: Let $T_R$ be the first-order theory of real closed fields. This is precisely the theory over the language $\{0,1,+,\times\}$ such that the theorems are the formulas that hold in $\Bbb R$. It can be effectively axiomatized by saying that it's a field in which every odd-degree polynomial has a root, and that for all $r$ either $r$ or $-r$ has a square root, but that $-1$ is not a square. Define $T_C$ to be the analogous theory for $\Bbb C$, but where we include complex conjugation in the language. So it's precisely the first-order theory over the language $\{0,1,+,\times,\bar{}\}$ such that a formula is a theorem if and only if it holds in $\Bbb C$. This can be axiomatized by saying that it's an algebraically closed field in which $z\mapsto\bar z$ is a self-inverse automorphism and $z\bar z=-1$ has no solutions. Then these two theories can interpret each other. The theory $T_R$ is interpretable in $T_C$ by considering the fixed points of $z\mapsto\bar{z}$. In the other direction, and $T_C$ is interpretable in $T_R$ by equipping pairs $(r_0,r_1)$ with operations to make them act like $r_0 + r_1i$. I'm interested in whether this pair of interpretations rises to the level of biinterpretation, as defined by Joel David Hamkins here. This definition requires that when we compose the two interpretations above (to get interpretations of $T_R$ and $T_C$ in themselves) these composite interpretations are isomorphic to the identity representations, in a way that is definable and provable within the theories themselves. At first it seemed to me that this would be true, since applying the two interpretations to a model does indeed yield the original model. And Hamkins himself seemingly calls this a biinterpretation here. What has me doubting is that the nLab says here that biinterpretations of theories induce equivalences between the categories of models. That can't be true here, since the above interpretations send $\Bbb R$ and $\Bbb C$ to each other, and yet $\Bbb R$ has only the trivial automorphism, whereas $\Bbb C$ also has complex conjugation. An equivalence of categories should preserve automorphism groups. So is this a biinterpretation or not? I also asked this question over on the category theory Zulip. The answers weren't conclusive, but they lead me to believe the subtlety lies in the interpretation of $T_C$ in itself. If we adjoin a square root of $-1$ to the fixed points of $z\mapsto\bar{z}$ then we would want to define an isomorphism back to the original field by sending this fixed point to $i$. But how do we define $i$ separately from $-i$? Is this indeed the problem, and is it unavoidable? Does it also prevent $\Bbb R$ and $\Bbb C$ from being biinterpretable as models? REPLY [10 votes]: $\DeclareMathOperator\Th{Th}\def\R{\mathbb R}\def\C{\mathbb C}\DeclareMathOperator\Aut{Aut}$Basically, all the statements you mention are correct, we just have to be careful about the details to avoid the apparent contradiction. So let us review how these interpretations work: There is a 1-dimensional interpretation $\rho$ of $\Th(\R,0,1,+,\cdot)$ in $\Th(\C,0,1,+,\cdot,\bar{})$ with domain $\{z:z=\bar z\}$ and all operations absolute. There is a 2-dimensional interpretation $\kappa$ of $\Th(\C,0,1,+,\cdot,\bar{})$ in $\Th(\R,0,1,+,\cdot)$ defined in the usual way so that a pair $(x,y)$ represents $x+iy$. The interpretation $\rho\circ\kappa$ of $\Th(\R,0,1,+,\cdot)$ in itself is definably isomorphic to the identity interpretation, the isomorphism being $(x,0)\mapsto x$. The interpretation $\kappa\circ\rho$ of $\Th(\C,0,1,+,\cdot,\bar{})$ in itself is isomorphic to the identity interpretation, with the isomorphism being $(x,y)\mapsto x+iy$. Now, here comes the trouble: this isomorphism is only definable with a parameter $i$ (which is not definable without parameters, as it is indistinguishable from $-i$). You can read this in two ways: $\Th(\R,0,1,+,\cdot)$ is bi-interpretable with $\Th(\C,0,1,i,+,\cdot,\bar{})$. $\Th(\R,0,1,+,\cdot)$ is bi-interpretable with $\Th(\C,0,1,+,\cdot,\bar{})$ via a bi-interpretation with parameters. (Actually, the bi-interpretation as such is still parameter-free, only the connecting isomorphism needs a parameter, as explained above.) Note that different people have different conventions about what exactly is the default definition of an interpretation, and in particular, whether it allows parameters. I’d guess that for syntactically minded people, it is more natural to consider paarameter-free interpretations, whereas for model-theorists, it is more natural to allow parameters. $\Th(\R,0,1,+,\cdot)$ is not bi-interpretable with $\Th(\C,0,1,+,\cdot,\bar{})$ without parameters. An interpretation $\iota$ of $T_1$ in $T_0$ induces a construction of a model $i(M)\models T_1$ from any model $M\models T_0$, and it is easy to check that if $\iota$ is a parameter-free bi-interpretation, then $\Aut(\iota(M))\simeq\Aut(M)$. Now, $\Th(\R,0,1,+,\cdot)$ has rigid models (i.e., with $\Aut(M)=1$) such as $\R$ itself, whereas every model of $\Th(\C,0,1,+,\cdot,\bar{})$ has at least one nontrivial automorphism: $\bar{}$.<|endoftext|> TITLE: Putnam 2020 inequality for complex numbers in the unit circle QUESTION [43 upvotes]: The following simple-looking inequality for complex numbers in the unit disk generalizes Problem B5 on the Putnam contest 2020: Theorem 1. Let $z_1, z_2, \ldots, z_n$ be $n$ complex numbers such that $\left|z_i\right| \leq 1$ for each $i \in \left\{1,2,\ldots,n\right\}$. Prove that \begin{align} \left| z_{1} + z_{2} + \cdots + z_n - n \right| \geq \left| z_{1} z_{2} \cdots z_n - 1 \right| , \end{align} and equality holds only if at least $n-1$ of the $n$ numbers $z_1, z_2, \ldots, z_n$ equal $1$. In the particular case when $n = 4$, the theorem can be proved using stereographic projection onto a line, followed by a longish computation. This is how both proposed solutions go. On the other hand, in the general case, the only elementary solution I know was given by @mela_20-15 on AoPS (spread over several posts). It has some beautiful parts (Cauchy induction), but also some messy ones (tweaking the points to lie on the unit circle in the induction step). There might also be a heavily analysis-based proof in Kiran Kedlaya's solutions (not sure if Theorem 1 is proved in full there). Question. What is the "proof from the book" for Theorem 1? Someone suggested to me to try to interpolate expressions of the form $\dbinom{n-1}{k-1}^{-1} \left|\sum\limits_{i_1 < i_2 < \cdots < i_k} z_{i_1} z_{i_2} \cdots z_{i_k} - \dbinom{n}{k}\right|$ between the left and the right hand sides in Theorem 1; but this does not work. For example, the inequality $\dfrac{1}{2} \left| z_1 z_2 + z_2 z_3 + z_1 z_3 - 3\right| \geq \left|z_1 z_2 z_3 - 1\right|$ fails quite often even on the unit circle. A warning: Inequalities like Theorem 1 are rather hard to check numerically. Choosing the $z_i$ uniformly will rarely hit close to the equality case; usually the left hand side will be much larger than the right. Near the equality case, on the other hand, it is hard to tell whether the answer comes out right legitimately or whether accumulated errors have flipped the sign. REPLY [22 votes]: Here is a detailed and self-contained proof for general $n$, which also covers the "equality" case. It is based on Fedja's post, but it only uses (a variant of) the Gauss-Lucas theorem once. Let $a_0,\dotsc,a_{n-1}\in\mathbb{C}$ be coefficients such that $$p(z):=(z-z_1)\dotsb(z-z_n)=(z-1)^n+\sum_{k=0}^{n-1}a_k z^k,$$ and assume that $(z-1)^{n-1}$ does not divide the left-hand side. Then, the $k$-sum on the right-hand side is a polynomial of degree $n-1$, because $$|a_{n-1}| = |n - (z_1 + \dotsb + z_n)|\geq n-(|z_1|+\dotsb+|z_n|)>0,$$ and we need to prove that $|a_{n-1}|>|a_0|$. Introducing the differential operator $$Df(z):=(1-z)f'(z)+nf(z),$$ we claim that every root of the polynomial $Dp(z)$ lies in the closed unit disk. Indeed, assume for a contradiction that $Dp(z)=0$ and $|z|>1$. Then $p'(z)/p(z)=n/(z-1)$, that is, $$\frac{1}{n}\sum_{j=1}^n \frac{1}{z-z_j}=\frac{1}{z-1}.$$ Let $K$ be the image of the closed unit disk under the Möbius transformation $s\mapsto 1/(z-s)$. Using $|z|>1$, we see that $K$ is a closed disk containing $1/(z-1)$ on its boundary. By the previous display, this boundary point is a convex linear combination of the points $1/(z-z_j)$, which also lie in $K$. This forces that all the $z_j$'s are equal to $1$, contrary to our assumption. So the claim is proved. Now observe that the polynomial $Dp(z)$ is the same as $$D\left(\sum_{k=0}^{n-1}a_k z^k\right)=a_{n-1}z^{n-1}+\sum_{k=0}^{n-2}((n-k)a_k+(k+1)a_{k+1})z^k.$$ We proved that this polynomial has all its roots in the closed unit disk, therefore $$|(n-k)a_k+(k+1)a_{k+1}|\leq\binom{n-1}{k}|a_{n-1}|,\qquad k\in\{0,\dotsc,n-2\}.$$ Applying the triangle inequality and re-arranging, we get that $$\frac{n-k}{k+1}|a_k|\leq|a_{k+1}|+\frac{1}{k+1}\binom{n-1}{k}|a_{n-1}|.\tag{$\ast$}$$ From here we derive by induction that $$\binom{n}{k}|a_0|\leq |a_k|+\binom{n-1}{k-1}|a_{n-1}|,\qquad k\in\{1,\dotsc,n-1\}.$$ In particular, the special case $k=n-1$ tells us that $n|a_0|\leq n|a_{n-1}|$. To finish the proof, we only need to show that some of our inequalities are strict, so that the final inequality is strict as well. Let us assume that equality holds in each of our inequalities. Then the roots of $Dp(z)$ are equal to a single $w$ on the unit circle, so that $$(n-k)a_k+(k+1)a_{k+1}=\binom{n-1}{k}a_{n-1}(-w)^{n-1-k},\qquad k\in\{0,\dotsc,n-2\}.$$ In particular, the case $k=n-2$ yields $$\frac{2}{1-n}a_{n-2}=(1+w)a_{n-1}.$$ However, we have equality in $(\ast)$ for $k=n-2$, whence $|1+w|=2$, which forces $w=1$. But then the recursion yields that $$a_k=\binom{n-1}{k}(-1)^{n-1-k}a_{n-1},$$ i.e., $$p(z)=(z-1)^n+a_{n-1}(z-1)^{n-1}.$$ This contradicts our initial assumption, and we are done. Added. As Malkoun pointed out in the comments, $(\ast)$ also implies the inequalities $$|a_k|\leq\binom{n-1}{k}|a_{n-1}|,\qquad k\in\{0,\dotsc,n-1\}.$$ Hence, renaming $k$ to $n-k$, we get the following generalization of the original inequality: $$\left|\sum_{1\leq j_1<\dotsb< j_k\leq n}z_{j_1}\dotsb z_{j_k}-\binom{n}{k}\right|\leq\binom{n-1}{k-1}|z_1+\dotsb+z_n-n|,\qquad k\in\{1,\dotsc,n\}.$$<|endoftext|> TITLE: Properties a triangulation must have in order to describe a manifold QUESTION [5 upvotes]: I am mainly interested in the $3$-dimensional case. It is a well-known fact, following from the work of E. E. Moise and R. H. Bing in the 1950s, that every $3$-dimensional topological manifold (with or without boundary) admits a triangulation, i.e. its homeomorphic to (the geometric realization of) an abstract simplicial complex. Furthermore, it is a well known fact that a manifold is piecewise-linear if and only if it admits a combinatorial triangulation, i.e. a triangulation in which the link of each simplex is Pl-homeomorphic to a sphere, and that in $d\leq 4$, every triangulation of a manifold is combinatorial. In other words, every $3$-manifold admits a PL-structure. I am interested in the other way round: Is there a bunch of properties an abstract simplicial complex has to have in order to define a topological manifold? Clearly, not all $3$-dimensional simplicial complexes which one can draw give rise to a manifold. The complex should be at least pure and non-branching, I guess. Is it maybe enough to assume that a complex is combinatorial? In the literature, I also have found the notions of ''pseudo-manifolds'', which are abstract simplicial complexes, which are pure, non-branching and strongly-connected. How is this related to my question? Any help is appreciated. If someone could provide some reference, I would be happy too. REPLY [5 votes]: From the comments: Suppose that $T$ is a triangulation. If all vertex links are PL $(n-1)$-dimensional spheres then the realisation space of $T$ is a PL manifold and thus a topological manifold. (In the compact case this is equivalent to the usual definition.) There are triangulations of topological manifolds (in fact, of $S^5$) that do not have this property. Examples come from the double suspension theorem of Cannon and also Edwards. Furthermore: There are triangulations of topological manifolds that have no PL structure. This is (essentially) the failure of the Hauptvermutung. There are closed topological manifolds that do not admit any triangulation. This is the failure of the Triangulation Conjecture, and is obtained by Casson in dimension four and to Manolescu in all higher dimensions.<|endoftext|> TITLE: Why is the ring of Grothendieck differential operators bad when $X$ is singular? QUESTION [7 upvotes]: $\DeclareMathOperator\Diff{Diff}$Suppose for simplicity that $X$ is affine, it is then possible to define $\Diff(X)$ — the ring of Grothendieck differential operators. When $X$ is smooth, then Definition. the category of $D$-modules on $X$ is defined to be modules over $\Diff(X)$. (Category 1) However, when $X$ is singular, this is not the right category to consider. One usually follows Kashiwara's approach: Definition. choose a closed embedding $X\hookrightarrow V$ and define $D$-modules be to modules over $\Diff(V)$ such that are (set-theoretically) supported $X$. (Category 2) The usual reason I heard for why to consider the second category is that $\Diff(X)$ behaves badly when $X$ is singular, and specifically people will point out that $\Diff(X)$ is not Noetherian. (Noetherian = left + right.) For example, this is the case when $X$ is the 'cubic cone' [BGG72]. However, I am no longer satisfied with this answer because of the following: (1), when $X$ is a curve then $\Diff(X)$ is Noetherian. [SS88] (2), when $X=V/G$ a quotient singularity then $\Diff(X)$ is Noetherian. But in these cases, one still considers Category 2 for these $X$. So it has to be the case that, in general and in these cases, $\Diff(X)$ is bad not just because it is not Noetherian, it is also bad for other reasons. So my question is: Question: Why do we work in category 2 in the situations above. Or a better questions, what is bad about $\Diff(X)$ besides not being Noetherian. Note my question is not how to work in category 2, but why it fails badly if we work in category 1 in situations (1) and (2). It is worth to remark that: in (1), if the curve is cuspidal then category 1 $\cong$ category 2. [SS88] generalised in [BZN04] in (2), if the $X=\mathbb{C}^2/(\mathbb{Z}/2\mathbb{Z})$ then category 1 $\cong$ category 2 (I think this is true, but do please correct me if I am wrong.) [BGG72] I. N. Bernˇste ̆ın, I. M. Gel’fand, and S. I. Gel’fand. Differential operators on a cubic cone. Uspehi Mat. Nauk, 27(1(163)):185–190, 1972. [BZN04] David Ben-Zvi and Thomas Nevins. Cusps and D-modules. Journal of the American Mathematical Society, 17.1:155–179, 2004 [SS88] S. P. Smith and J. T. Stafford. Differential operators on an affine curve. Proc. London Math. Soc. (3), 56(2):229–259, 1988. Noted later: actually it is not true that on $X=\mathbb{C}^2/(\mathbb{Z}/2\mathbb{Z})$ then category 1 $\cong$ category 2, sorry for the confusion. REPLY [9 votes]: There probably can be many answers to this question, but here is one: We want the category of $D$-modules to behave like categories of sheaves in other sheaf theories. In the complex setting, we want the Riemann-Hilbert correspondence, an equivalence of categories between constructible sheaves in the analytic topology and regular holonomic $D$-modules, and over an arbitrary base field, we want them to have similar behavior. Why do we want this? All these sheaf theories are expected to be various shadows of the category of motives, and probably motives are the really interesting thing we want to study, so we want our theories to be similar enough that they capture motives. Anyways, for the categories of constructible sheaves (algebraic/analytic), there is an equivalence of categories between sheaves on $X$ and sheaves on $V$ supported on $X$. So if we want the category of $D$-modules to have similar behavior, we must use category 2! It's possible we could use some other definition and prove the equivalence with a full subcategory of $D$-modules on $V$ supported on $X$. But since we know this is what we want, we might as well take it to be true by definition. So, the thing that is bad about the category 1 is simply that it is not 2. We absolutely can use definition 1 in the special cases where it agrees with 2, but doing so might leave us less prepared for the general case. REPLY [9 votes]: One thing that's wrong with the Grothendieck definition on singular varieties is the same thing that's wrong with defining the tangent space at a singular point rather than the tangent complex - it's not sufficiently derived (ie it's a strange truncated notion of the "true" derived object), which is why several different attempts to define D-modules give different answers there (the one quoted by OP, the notion of modules over vector fields - ie the algebra of differential operators generated by first order ones, which needn't be the full algebra in the singular case, and crystals - which amounts to the "option 2" one). If you define D-modules in a derived fashion, as is done e.g. in the book of Gaitsgory-Rozenblyum, then order is restored: all the different natural notions you might come up with agree [here it's important to be in characteristic zero, a whole lot more interesting stuff happens in positive characteristic]. If you replace the naive notion of the sheaf vector fields by the tangent complex and consider modules for it, or define the Grothendieck differential operators derivedly (as a groupoid algebra for the de Rham groupoid of X), or define crystals (option 2), you get the same notion.<|endoftext|> TITLE: Stallings' binding tie QUESTION [7 upvotes]: I came to know that the statement below could be proved using Stallings' binding tie argument, though I have no reference article proving the statement by the binding tie argument. Can anyone help me telling something related to this? Another question, except this, does there exist any other application of binding tie argument, maybe in other dimensions? Given a $\pi_1$-surjective map $f\colon S\to S'$ between two closed orientable surfaces and a smoothly embedded circle $C$ in $S$, we can homotope $f$ to make it transverse to $C$ so that $f^{-1}(C)$ is either empty or exactly one circle. REPLY [6 votes]: Several of the papers citing Stallings paper A topological proof of Grushko's theorem on free products are using the binding tie argument. Here are the ones I found using a Google search. More can probably be found using MathSciNet. Jaco [1968] Constructing 3-manifolds from group homomorphisms Heil [1972] On Kneser’s conjecture for bounded 3-manifolds Feustel [1972] A splitting theorem for closed orientable 3-manifolds Bowditch [1999] Connectedness properties of limit sets Bellettini, Paolini, Wang [2021] A complete invariant for closed surfaces in the three-sphere Several searches failed to find the precise statement you mention. However, Lemma 3.2 in Jaco's paper comes close (he maps a surface to a graph, pulls back midpoints, and so on).<|endoftext|> TITLE: Non-enumerative proof that, in average, less than 50% of tiles in domino tiling of 2-by-n rectangle are vertical? QUESTION [20 upvotes]: Is there a non-enumerative proof that, in average, less than 50% of tiles in domino tiling of 2-by-n rectangle are vertical? It is a nice exercise with rational generating functions (or equivalently, linear recurrence relations) to show that for a random domino tiling of a $2\times n$ rectangle, with $n$ large, we can expect about $\frac{1}{\sqrt{5}}\approx 44.7\%$ of the tiles to be vertical. In the spirit of Non-enumerative proof that there are many derangements?, I wonder if there is an easy way to see, without computing this fraction, that this average must be some constant < 50%. EDIT: Maybe to sharpen the request, is there any heuristic which works here and also carries over to the case of a $k\times n$ rectangle (with say $k$ fixed and $n$ large). REPLY [5 votes]: Essentially also a reformulation: Tilings of $n\times 2$ with vertical and horizontal dominos are in bijection with Zeckendorf expansions of integers in $[0,F_{n+1}-1]$ (perhaps up to a small shift of $n$), expanded to the correct length by adding leading zeroes: From left to right write $01$ for two parallel horizontal tiles and write $0$ for a vertical tile. The Zeckendorf expansion of a random integer in $[0,F_{n+1}-1]$ involves $F_n$ with asymptotic probability $1/\omega^2$ (for $\omega=(1+\sqrt{5})/2$ the golden number). The inequality $2/\omega^2>1/\omega$ implies now the result (and, more precisely, the probability for a vertical tile is thus indeed $\omega/(2+\omega)=1/\sqrt{5}$).<|endoftext|> TITLE: Sobolev spaces of differential forms and regular atlases QUESTION [10 upvotes]: In [1] (section 3), C. Scott introduces the following concept of regular atlas for closed $C^\infty$-smooth Riemannian manifolds. He says: When referring to a coordinate system $(U,\phi)$ as regular, we shall mean that there is another system $(V,\psi)$ with $\overline{U}$ compact, $\overline{U} \subset V$ and $\psi\vert_{U} = \phi$. His motivation to introduce such concept comes from the problem of defining Sobolev spaces of differential forms. In self-explanatory notation, he defines ($1 \leq l \leq n = \dim M$) $$\mathscr{W}^{1,p}_{\mathscr{A}}\left( \bigwedge^l M \right) := \left\{ \omega \in \left( \bigwedge^l M \right) : \omega, |\nabla{\omega}| \in L^p \right\},$$ endowed with the norm $\|\omega\|_p + \|\nabla \omega \|_p$. Here, $$|\nabla \omega(x)|^2 := \sum_{U \in \mathscr{A}} |\nabla_U \omega(x) |^2 = \sum_{U \in \mathscr{A}} \sum_{I,k} \left| \frac{\partial \omega_I}{\partial x^k}(x) \right|^2,$$ where $I = { 1 \leq i_1 < \dots < i_l \leq n}$. Then, he observes Simple examples demonstrate that it is possible to choose, in perfectly reasonable ways, two atlases which yield Sobolev spaces that are not equivalent as normed linear spaces. [...] From here on, classical Sobolev space refers to one constructed as above using a regular atlas. This is all fairly familiar and once again, Morrey [Multiple integrals in the calculus of variations] is a fine reference. Such observation is not proved nor references to the literature are provided for the counterexamples. Morever, I have not found a similar definition in Morrey's book (Morrey deals with admissible cooordinate systems, see pp. 288 and 300, which are defined differently and it appears that the only requirement about them is the standard compatibility between new and old coordinates). In addition, it seems to me (but I'm quite new to this topic) that this problem is not considered by other authors in this area. Exceptions are [2] and [3]. In [2] (section 2.2), Iwanienc, Scott and Stroffolini, dealing now with smooth Riemannian manifolds with boundary, give the following definition of regular atlas. If $\mathscr{R}$ is a $C^\infty$-smooth, closed Riemannian manifold, a regular open region $M \subset \mathscr{R}$ is one for which there exists a finite atlas $\mathscr{A}$ on the reference manifold $\mathscr{R}$ consisting entirely of coordinate charts $(U,\kappa) \in \mathscr{A}$ so that $\kappa$ is a $C^\infty$-diffeomorphism onto $\mathbb{R}^n$ and $\kappa(U \cap M) = \mathbb{R}^n_+$ whenever $U$ meets $\partial M$. In [3], section 2.1.1, the authors are concerned only with closed manifolds. The requirement is the same but obviously there is no need for the second part (as $\partial M = \varnothing$). It seems to me that no one of these requirements is common in standard differential geometry. For instance, I guess that the sphere $\mathbb{S}^n$ with the most common atlas consisting of the two coordinate patches associated with the stereographic projections from the north and south poles is not regular in the sense of Scott, and this is perhaps quite weird. Summarizing, these are my questions (they are multiple but strictly tied): Is it always possible to introduce such atlases? Are the two definitions in [1] and [3] equivalent? Are they necessary to give a meaningful definition of Sobolev spaces of differential forms? What are the easy counterexamples mentioned above? Are they necessary to prove the $L^p$-version of Gaffney's inequality and Hodge-de Rham-Kodaira decomposition, which are essentially the main results of [1] and [2]? [1] C. Scott, $L^p$ theory of differential forms on manifolds, Trans. Amer. Math. Soc. 347 (6), 1995. [2] T. Iwaniec, C. Scott, B. Stroffolini, Nonlinear Hodge theory on manifolds, Ann. Mat. Pura Appl. (IV), CLXXVII (1999), 37-115. [3] P. Hajlasz, T. Iwaniec, J. Maly, J. Onninen, Weakly differentiable mappings between manifolds, AMS. REPLY [10 votes]: There is a coordinate free way of defining Sobolev spaces of sections of a vector bundle $E$ over a manifold $M$. You need to make a few choices: a metric $g$ on $M$, a metric $h$ on $E$ and a connection $\nabla=\nabla^E$ on $E$ compatible with the metric $h$. The metric $g$ defines Levi-Civita connections $\nabla^g$ on the tensor bundles $T^*M^{\otimes k}$. In particular, we obtain connections on all the bundles $T^*M^{\otimes k}\otimes E$ and in particular a $k$-th order derivative $$ \nabla^k : C^\infty(E)\to C^\infty(T^*M^{\otimes k}\otimes E). $$ The bundle $T^*M^{\otimes k}\otimes E$ is also equipped with a natural metric induced from $g$ and $h$ For a smooth compactly supported section $u$ of $E$ we define the $(m,p)$ Sobolev norm $$ \Vert u\Vert_{m,p}=\left(\sum_{k=0}^m \int_M |\nabla^k u|^p dV_g\right)^{1/p}. $$ Define the Sobolev space as the completion of $C_0^\infty$ under this norm. The norm depends on $g,h,\nabla^E$, but if $M$ is compact all these norms are equivalent. If $M$ is noncompact, all bets are off. For details, see Chapter 10 of these notes.<|endoftext|> TITLE: Reductive groups over number rings QUESTION [5 upvotes]: Let $F$ be a number field. If $G$ is a reductive group over $\mathcal{O}_F$ then we can look where $G\otimes \mathbb{C}$ fits in the classification of complex reductive groups and get a "standard model" $G_{\mathrm{st}}$. The algebraic group $G$ will differ from $G_{\mathrm{st}}$ at only finitely many places. Is it true that if we fix $G\otimes \mathbb{C}$ and fix the places of difference then there is only finitely many reductive groups over $\mathcal{O}_F$ we could have started with? REPLY [5 votes]: Yes. See Javanpeykar and Loughran - Good reduction of algebraic groups and flag varieties.<|endoftext|> TITLE: Bound on $L^2$ norm of $1/\zeta(1+i t)$? QUESTION [20 upvotes]: What sort of bounds (explicit of preference) can one give for $$\int_T^{2 T} \frac{dt}{|\zeta(1+i t)|^2} \;\;\;\;\;?$$ Some obvious points: One can give a pointwise bound $\frac{1}{|\zeta(1+ it)|} \leq C \log t$ (with $C\leq 42.9$ for $t\geq 2$) and deduce a bound of the form $\leq K T (\log T)^2$ on the integral above. If possible, we would like to do better (asymptotically and also as far as the constants involved are concerned). To know that the integral is always finite, one needs to know that $\zeta(1+it)\ne 0$ for all $t$. Thus, it is not enought to just apply a mean-value theorem. REPLY [7 votes]: Balasubramanian, Ivić and Ramachandra ("An application of the Hooley-Huxley contour", Acta Arith. 65 (1993), no. 1, 45–51) prove the asymptotic result found by Lucia and Terry Tao. Their error term is $O(\log T)$, so slightly better than Lucia's error, but they make use of zero-density results. More generally, their equation (1) gives an analogous result for $\int_{1}^{T} |\zeta(1+it)^k|^2 dt$ for any complex $k$. In fact, this is itself a special case of their equation (4), $$\int_{1}^{T} |F(1+it)|^2 dt = T\sum_{n \ge 1}|a_n|^2 n^{-2} + O\left( \log \log T + \sum_{n \le T^C} |a_n|^2 n^{-1}\right)$$ where $F(s) = \sum_{n \ge 1} a_n/n^s$ is a Dirichlet series 'arising' from L-functions (concretely, a product of powers of L-functions, their logarithms and their derivatives) and $C$ is a constant depending only on $F$. Let me also mention that a quick conditional (on RH) proof of your required result is also given in Titchmarsh (2nd edition, p. 338).<|endoftext|> TITLE: Thom spectrum of $(\mathrm{Spin}\times_{Z_2} \mathrm{SO}(d))$ QUESTION [5 upvotes]: $\DeclareMathOperator\MSO{MSO}\DeclareMathOperator\MSpin{MSpin}\DeclareMathOperator\SO{SO}\DeclareMathOperator\BSO{BSO}\DeclareMathOperator\BG{BG}\DeclareMathOperator\BO{BO}\DeclareMathOperator\MG{MG}\DeclareMathOperator\Spin{Spin}\MG(d)$ is the Thom space of the pullback of the vector bundle $V_d$ over $\BO(d)$ along the map $\BG(d) \to \BO(d)$. The colimit of $\Sigma^{-d}\MG(d)$ is $\MG$. The $\MG$ is the Thom spectrum of $G$. It is known that $M(\Spin\times \SO(d))= \MSpin \wedge \BSO(d)_+$, where $\BSO(d)_+$ is a disjoint union of $\BSO(d)$ with a point. My question is: What is the Thom spectrum of $(\Spin\times_{Z_2} \SO(d))$? The hope is to break down the spectrum to familiar one of $\MSpin$. the smash product $\wedge$, the suspension $\Sigma^{-n}$, and perhaps the Thom spectrum $\MSO(d)$. For $d=4$, $M(\Spin\times_{Z_2}\SO(4))=\MSpin\wedge\Sigma^{-3}\MSO(3)\wedge\Sigma^{-3}\MSO(3)$, which breaks down to the familiar spectra. For $d=6$, $M(\Spin\times_{Z_2}\SO(6))$, could we do the similar break down? For $d=8$, $M(\Spin\times_{Z_2}\SO(8))$, could we do the similar break down? Note that $\SO(6)=\Spin(6)/Z_2=\operatorname{PSU}(4)$. $\SO(8)=\Spin(8)/Z_2$. EDIT: The $d$ is even for $\SO(d)$ here. Also the modded out $Z_2$ is shared between the identified normal of Spin and SO groups. Spin is the tangent structure (tangent spin bundle) of any dimensions of any base Spin-manifold. $\SO(d)$ is the principal SO structure and principal SO-bundle. REPLY [3 votes]: $\newcommand{\SO}{\mathrm{SO}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\PSO}{\mathrm{PSO}}\newcommand{\Z}{\mathbb Z}$This is a partial answer: I am not very familiar with the representation theory of $\PSO_{2d}$, and I have to make an assumption to get the splitting. I think the assumption is right, but I don't have a proof of it. In any case, once that assumption is in place, the argument is similar to the splitting argument made by Freed-Hopkins (see section 10). The central extension $$1\to \Z/2\to \SO_{2d}\to\PSO_{2d}\to 1$$ defines a class $c\in H^2(B\PSO_{2d};\Z/2)$. The assumption I have to make is: there is a representation $\rho\colon \PSO_{2d}\to \SO_m$ such that if $V_\rho\to B\PSO_{2d}$ denotes the associated vector bundle, $w_1(V_\rho) = 0$ and $w_2(V_\rho) = c$. Given that data, we can make the diagram where $\Spin_n\times_{\Z/2}\SO_{2d}\to\SO_n\times\PSO_{2d}$ is the quotient by the common $\Z/2$ and $\SO_n\times\SO_m\to\SO_{n+m}$ is the block diagonal embedding. The composition along the lower left $\Spin_n\times_{\Z/2}\SO_{2d}\to \SO_{n+m}$ kills the central $\Z/2$, and therefore we can choose a lift $\phi\colon \Spin_n\times_{\Z/2}\SO_{2d}\to \Spin_{n+m}$. Stabilizing in $n$, let $S$ denote a tautological (stable) vector bundle on a classifying space, thought of as a map to $B\mathrm O$. Then there is a commutative diagram It is possible to show that the horizontal arrow is a homotopy equivalence (the proof is essentially the same as the one given by Freed-Hopkins). The data defining a Thom spectrum is a space with a map to $B\mathrm O$, and a homotopy equivalence of that data induces a homotopy equivalence of Thom spectra: $$\mathit{MT}(\Spin\times_{\Z/2}\SO_{2d}) \overset\simeq\longrightarrow \mathit{MTSpin}\wedge (B\PSO_{2d})^{V_\rho - m}.$$ Here $(B\PSO_{2d})^{V-m}$ is the Thom spectrum of the rank-zero stable vector bundle $V_\rho-m\to B\PSO_{2d}$.<|endoftext|> TITLE: Bounds for a small cardinal QUESTION [5 upvotes]: $\newcommand{\w}{\omega}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x}{\mathfrak x}\newcommand{\cov}{\mathrm{cov}}\newcommand{\lac}{\mathrm{lac}}$Taras Banakh and I proceed a long quest answering a question of ougao at Mathematics.SE. Recently we encountered a small cardinal $\x_{\lac}$, which is the smallest cardinality of a family $\F$ of infinite subsets of $\w$ such that for any lacunary set $R$ there exists $F\in\F$ such that $F\cap R$ is finite. Recall that a set $R\subset\w\setminus\{0\}$ is called lacunary, if $\inf\{b/a:a,b\in R,\;a1$. It would be ideally for us to find a known small cardinal equal to $\x_{\lac}$. While $\x_{\lac}$ remains unknown, we are interested in bounds (especially lower) for it by known small cardinals. Our try. For any family $\I$ of sets let $\cov(\I)=\min\{|\J|:\J\subseteq\I\;\wedge\;\bigcup\J=\bigcup\I\}$. Let $\M$ and $\N$ be the ideals of meager and Lebesgue null subsets of the real line, respectively. We can prove that $\cov(\M)\le \x_{\lac}$ and are interested whether this bound can be improved and whether $\cov(\N)\le \x_{\lac}$. Lyubomyr Zdomskyy suggested that it is consistent that $\mathfrak d<\x_{\lac}$, where $\mathfrak d$ is the cofinality of $\w^\w$ endowed with the natural partial order: $(x_n)_{n\in\w}\le (y_n)_{n\in\w}$ iff $x_n\le y_n$ for all $i$. We are interested whether $\x_{\lac}\le \mathfrak a$, where $\mathfrak a$ is the minimum size of a maximal (with respect to inclusion) pairwise almost disjoint family of infinite subsets of $\omega$. Thanks. REPLY [4 votes]: EDIT: In my original post, I showed that $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model. Upon further reflection, I think we can prove a stronger result, with an arguably easier (but completely different) proof: Theorem: $\mathfrak{x}_{lac} \leq \mathrm{non}(\mathcal N)$. Note that this implies $\mathfrak{x}_{lac} < \mathrm{cov}(\mathcal N)$ in the random model, and it also implies the consistency of $\mathfrak{x}_{lac} < \mathfrak{d}$. In addition to this theorem, let me also point out that it is consistent to have $\mathfrak{a} < \mathrm{cov}(\mathcal M)$. (See Corollary 2.6 in this paper of Brendle.) Therefore the lower bound $\mathrm{cov}(\mathcal M) \leq \mathfrak{x}_{lac}$ mentioned in the post already implies $\mathfrak{a}$ is not an upper bound for $\mathfrak{x}_{lac}$. Proof of the theorem: Suppose we form an infinite set $B$ by choosing from each interval of the form $[2^k,2^{k+1})$ exactly one integer $b_k$ at random, and then taking $B = \{b_k :\, k \in \omega \}$. (By "at random" I mean that we choose with the uniform distribution, so each integer in $[2^k,2^{k+1})$ has probability $1/2^k$ of being selected.) I claim that if $A$ is lacunary, then it is almost surely true that $A \cap B$ is finite. To see this, fix some $c > 1$ and some $n_0 \in \omega$ such that if $a$ and $b$ are consecutive members of $A$ above $n_0$, then $b/a > c$. If $k$ is large enough that $n_0 < 2^k$, then this implies there are at most $log_c(2)$ members of $[2^k,2^{k+1})$ in $A$. This implies that the probability of choosing $b_k \in A$ is $\log_c(2)/2^k$ when $k$ is large enough. It follows that the probability of there being $>K$ members of $A$ in $B$ (when $K$ is large) is $\sum_{k > K} \log_c(2)/2^k = \log_c(2)/2^K$. Since this goes to $0$ as $K$ goes to infinity, the probability of $A \cap B$ being infinite is $0$. The idea of choosing $B$ randomly, one point at a time, like this can be formalized by defining a probability measure on a Polish space, where points of the space correspond to possible choices of the sequence of $b_k$'s. What the previous paragraph shows is that in this probability space, the set of all $B$'s with $A \cap B$ infinite form a null set, for any given lacunary set $A$. Hence any non-null subset of this probability space will contain a $B$ with $A \cap B$ finite. Since this holds for any $A$, we see that any non-null subset $X$ of this probability space contains, for any lacunary set $A$, some infinite $B$ with $A \cap B$ finite. The smallest possible size of such a set $X$ is $\mathrm{non}(\mathcal N)$. QED One more observation: The strict inequality $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ is also consistent, so this upper bound cannot be improved to an equality. To see this, begin with a model of Martin's Axiom $+ \, \neg \mathsf{CH}$, and then do a legnth-$\omega_1$, finite support iteration of the eventually different reals forcing. It is not difficult to see that this forcing will make $\mathfrak{x}_{lac} = \aleph_1$ in the extension. But the iteration is $\sigma$-centered, and forcing with a $\sigma$-centered poset over a model of $\mathsf{MA}$ does not change the value of $\mathrm{non}(\mathcal N)$. Thus we get $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ in the extension. Original post: It is not provable that $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}_{lac}$, because $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model. To see this, let me sketch an argument that after forcing to add any number of random reals (in the usual way, via a measure algebra), the collection $[\omega]^\omega \cap V$ of infinite subsets of $\omega$ in the ground model has the property described in the definition of $\mathfrak{x}_{lac}$. That is: for every lacunary set $A \subseteq \omega$ in the extension, there is an infinite $B \subseteq \omega$ in the ground model such that $A \cap B$ is finite. We work in the ground model. Suppose $\dot A$ is a name for an infinite lacunary set in the extension. There is some fixed $c > 1$ and $n_0 \in \omega$, and some forcing condition $p$, such that $p \Vdash$ if $a$ and $b$ are consecutive elements of $\dot A$ above $n_0$, then $b/a > c$. I claim that for every $\varepsilon > 0$, there are arbitrarily large $n \in \omega$ such that $m(p \wedge [n \in \dot A]) < \varepsilon$. (Note: Here I'm using the standard notation for forcing with measure algebras. If $\varphi$ is any well-formed formula in the forcing language, then we write $[\varphi]$ to mean the supremum of all the conditions forcing $\varphi$, and $m([\varphi])$ denotes the measure of this supremum. Roughly, you may think of $m([\varphi])$ as the probability that $\varphi$ ends up being true in the forcing extension.) To prove my claim, suppose, aiming for a contradiction, that it is false. Then there is some $\varepsilon > 0$ and some $N \in \omega$ such that $m([n \in \dot A]) \geq \varepsilon$ for all $n \geq N$. But this is just another way of saying that the "expected size" of $A \cap \{n\}$ is $\geq\varepsilon$ for all $n \geq N$. By the linearity of expectation, this means the expected size of $A \cap \{k,k+1,\dots,\ell-2,\ell-1\}$ is $\geq (\ell-k)\varepsilon$ whenever $\ell > k > N$. But by our choice of $p$, if $k \geq N$ and $\ell < ck$, then $p \Vdash |A \cap \{k,k+1,\dots,\ell-2,\ell-1\}| \leq 1$. Since $c > 1$, this yields a contradiction for sufficiently large $k$, namely $k > N,2/c\varepsilon$. Using this claim, we can now find an infinite ground model set almost disjoint from the set named by $\dot A$ in the extension. Using the claim, we may find for each $k \in \omega$ some $n_k > n_{k-1}$ such that $m([n_k \in \dot A]) < m(p)/2^{k+2}$. Now let $p' = p - \bigvee_{k \in \omega}[n_k \in \dot A]$. Then $m(p') \geq m(p) - \sum_{k \in \omega}m([n_k \in \dot A]) > m(p)/2 > 0$, so $p'$ is a condition in our measure algebra, and $p'$ forces $\dot A$ to be disjoint from $\{n_k :\, k \in \omega \}$ (because it extends the complement of each $[n_k \in \dot A]$). This shows that it is impossible to have a name $\dot A$ for a lacunary set such that $\dot A$ is forced to have infinite intersection with every infinite subset of $\omega$ from the ground model. Therefore there is no such set.<|endoftext|> TITLE: Vector bundles on the various sites of a preperfectoid QUESTION [6 upvotes]: Let $X$ be a preperfectoid space over $\mathrm{Spa}(\mathbb{Q}_p,\mathbb{Z}_p)$. It has several associated sites, with successively finer topologies: $$X_{an} \subset X_{et} \subset X_{proet} \subset X_v.$$ I was wondering: what is the relationship between vector bundles on these different sites? Here, by a vector bundle I mean a locally finite free $\mathcal{O}_X$-sheaf for $X_{an}, X_{et}$ and $X_v$, and a locally finite free $\widehat{\mathcal{O}}_X$-sheaf for $X_{proet}$. If $X$ is perfectoid, it is known that the categories of vector bundles on all of these sites are equivalent, by Theorem 3.5.8 of Kedlaya-Liu's paper "Relative p-adic Hodge Theory II". Is this also true for preperfectoid spaces? What about sousperfectoid spaces? If this statement is not true, is it still true that some of these categories are equivalent? I expect that at the very least the pullback functor from $X_{an}$ to $X_v$ is fully faithful. My apologies if there is some silly counterexample. Thanks! REPLY [7 votes]: There is some relevant work of Ben Heuer on this. In short, analytic and etale vector bundles agree on all sousperfectoid adic spaces (and much more generally, for all "etale sheafy" adic spaces), while proetale and v-vector bundles also agree (because proetale locally, the space is perfectoid, and so this reduces to the assertion for perfectoids). The pullback functor from etale vector bundles to proetale vector bundles is fully faithful. So in summary, there are two distinct categories (etale vector bundles and proetale vector bundles), one being a full subcategory of the other. However, etale and proetale vector bundles are very different. This is already a well-known phenomenon for $p$-adic fields $K$ (complete discretely valued with perfectly residue field), where letting $C=\widehat{\overline{K}}$ it is the difference between $K$-vector spaces and $G_K$-equivariant $C$-vector spaces, which forms the subject of Sen--Tate theory. By the way, there is some recent line of work (by Heuer, Mann, Werner, ...) that takes a fresh look at the $p$-adic Simpson correspondence from the perspective of proetale vector bundles. Roughly speaking, proetale vector bundles are what Faltings called "generalized representations" in his first paper on the $p$-adic Simpson correspondence, and should be more-or-less equivalent to Higgs bundles in case $X$ is smooth over an algebraically closed nonarchimedean $C/\mathbb Q_p$. Noth that Higgs bundles also contain usual vector bundles on $X$ fully faithfully, as those with vanishing Higgs field. (On the other hand, if $X$ is proper, then proetale vector bundles contain $C$-local systems fully faithfully, by the primitive comparison isomorphisms.)<|endoftext|> TITLE: Existence of a continuous ergodic dynamical system for a given distribution? QUESTION [6 upvotes]: It seems to me that given a distribution (which is well-behaved), there should be at least an ergodic dynamical system that its time average would create this distribution. Is this question already answered in the literature? or is it just too simple? Most of my search results talk about the measure that's the result of an ergodic system, and not the other way around. Could somebody shed light on this please? EDIT: Based on the comment of @WillSawin I am adding that, let's assume the distribution is defined over $\mathbb{R}^n$, or maybe more generally over a Riemannean manifold, with a compact support. Additionally the dynamical system has a flow which is a smooth function of time. REPLY [3 votes]: Let $\mu$ be Lebesgue measure on $S^1$, and $\delta_P$ be a point-mass at a point $P \in S^1$. Then there is no flow on $S^1$ whose time averages lead to $\frac{1}{2}(\mu + \delta_P)$. (Consider the orbit of $P$.) This distribution seems like it should count as "well-behaved." Its support is connected, and both $\mu$ and $\delta_P$ themselves arise as time-averages (by a rotational flow, and a flow to an attracting fixed point, respectively). You can generalize to $\mathbb{R}^n$ by looking at an embedded $S^1$, and generalize to continuous maps by replacing the rotational flow with an irrational rotation.<|endoftext|> TITLE: When did Grothendieck join Bourbaki? QUESTION [15 upvotes]: Bourbaki listed Grothendieck as a third-generation member. Nevertheless, it does not provide details on when he joined and when he left. Concerning his departure, there is a Letter from October 9, 1960 and a translation¹ I thank You for your letter², marked by both wisdom and clemency. Indeed it seems pointless that a personal disagreement could be the occasion for the departure of a disciple. I recognize that it was pointless for me to wait for the Master to arbitrate a quarrel that did not concern him and that such arbitration would resolve nothing. I have asked myself many times over the years of my collaboration with the Master whether my lack of social skill, my impassioned character, and my repugnance for overcoming the repugnance of others, did not render me unsuitable for a productive collaboration during the meetings. No longer wanting to search for the cause anywhere except in myself, I now think that it is better this way and that I reached earlier than the traditional age the moment when I would better serve the Master by my departure, rather than remaining as a result of His kind insistence. I will endeavor to remain worthy of the teachings that You for so long lavished upon me and not to betray the spirit of the Master who, I hope, will remain visible in my work as it has been in the past. With this, we can infer the date he left Bourbaki. Regarding his entrance, It is very likely that he entered with P Cartier, but I cannot prove that. According to Bourbaki that was in 1955. Is there a reference about when Grothendieck entered Bourbaki? By W Messing It would be very interesting to know this letter too. REPLY [6 votes]: In Notices of the AMS Volume 63, Number 3 we find in the section of P Cartier: It was in March 1955 at the Bourbaki seminar after a special lecture that Grothendieck gave about convexity inequalities¹. He told me: “Very soon, both of us will join Bourbaki.” I began regularly attending Bourbaki meetings in June 1955. Grothendieck joined soon and participated actively from 1956 to 1960. In June 1955 one of the most interesting pieces to read during our meeting was a first draft of his famous Tôhoku paper, where he gives a new birth to homological algebra. And from Grothendiec-Serre correspondence Grothendieck to Serre June 4, You will find enclosed a neat draft of the outcome of my initial reflections on the foundations of homological algebra. Serre to Grothendieck on July 13, Your paper on homological Algebra² was read carefully³ and converted everyone (even Dieudonné, who seems to be completely functorised) to your point of view. Réarrangements de fonctions et inégalités de convexité dans les algèbres de von Neumann munies d'une trace. Séminaire Bourbaki : années 1954/55 - 1955/56, exposés 101-136, Séminaire Bourbaki, no. 3 (1956), Exposé no. 113, 13 p. Some aspects of homological algebra At the Bourbaki meeting<|endoftext|> TITLE: Bézout matrices and interlacing roots QUESTION [6 upvotes]: Suppose that $f(t)$ is a polynomial of degree d with real roots $a_1 \leq a_2 \leq \dotsb \leq a_d$, and $g(t)$ is a polynomial of degree $d-1$ with real roots $b_1 \leq b_2 \leq \dotsb \leq b_{d-1}$. Let’s say that $g(t)$ "weakly interlaces" $f(t)$ if we have $$a_1 \leq b_1 \leq a_2 \leq b_2 \leq \dotsb \leq b_{d-1} \leq a_d,$$ and that $g(t)$ "strongly interlaces" $f(t)$ if the inequalities are all strict. For arbitrary $f(t)$ and $g(t)$ (not necessarily real rooted), let $B(f,g)$ be the Bézout matrix, whose $(i,j)$ entry is equal to the coefficient of $x^{i-1} y^{j-1}$ in the polynomial $$\frac{f(x)g(y) - f(y)g(x)}{x-y}.$$ It seems to be a well known fact that both polynomials are real rooted and $g(t)$ strongly interlaces $f(t)$ if and only if $B(f,g)$ is positive definite. For example, see Corollary 9.145 here: Fisk - Polynomials, roots, and interlacing. It follows that, if both polynomials are real rooted and $g(t)$ weakly interlaces $f(t)$, $B(f,g)$ must be positive semidefinite. (You can just perturb $f(t)$ and $g(t)$ so that they strongly interlace, and the limit of positive definite matrices is positive semidefinite.) I would like to know if the converse is true. That is, if $B(f,g)$ is positive semidefinite, does it follow that both polynomials are real rooted and $g(t)$ weakly interlaces $f(t)$? This assertion is made in Theorem 2.13 of this paper: Kummer, Naldi, and Plaumann - Spectrahedral representations of plane hyperbolic curves. However, the source that they cite (Krein and Naimark) only appears to treat the positive definite case. It may indeed be true that the positive semidefinite case follows from the positive definite case, but I would like to understand why. REPLY [3 votes]: It seems that the answer to my question is no. For example, if we take $g(t) = t^2 + 1$ and $f(t) = tg(t) = t^3 + t$, then neither $f(t)$ nor $g(t)$ is real rooted, but the matrix $$B(f,g) = \left(\begin{matrix}1 & 0 & 1\\ 0 & 0 & 0 \\ 1 & 0 & 1\end{matrix}\right)$$ is positive semidefinite. More generally, we will obtain a counterexample whenever $f(t)$ is a multiple of $g(t)$ and $g(t)$ is not real rooted. What is true is that, if the Bezout matrix is positive semidefinite, then the polynomials strongly interlace after dividing by their GCD. (Thanks to Mario Kummer for explaining this to me.)<|endoftext|> TITLE: Lattice structure (wrt dominance order) on the set of Young diagrams appearing in the decompositions given by the Littlewood-Richardson rule QUESTION [11 upvotes]: The irreducible decomposition of the tensor product of two irreducible representations of GL(n) is described by the Littlewood-Richardson rule. This same rule also governs the decomposition of the product of two Schur polynomials into a linear combination of Schur polynomials. In both cases, we label the components of the product and the decomposition with Young diagrams, or integer partitions. We noticed that the Young diagrams in the decomposition always seem to form a lattice w.r.t. dominance order. That is, for each pair of diagrams in the decomposition, the least upper bound ("join") and the greatest lower bound ("meet") of the pair is also part of the decomposition. For example, looking at the following decomposition (where the Young diagrams are denoted by the corresponding integer partitions): $$(2,1) \otimes (2,1)=(4,2) \oplus (4,1,1) \oplus (3,3) \oplus 2 (3,2,1) \oplus (3,1,1,1) \oplus (2,2,2) \oplus (2,2,1,1)$$ The two unordered pairs in this decomposition are: (2,2,2), (3,1,1,1); and (3,3), (4,1,1). The join and meet of the first pair are (3,1,1,1) and (2,2,1,1) respectively. For the second pair, these are (4,2) and (3,2,1). In both cases, the join and meet diagrams are parts of the decomposition. We checked it for many larger examples with extensive symbolic calculations, and the product had this lattice property in all cases. Thus my question is: Applying the Littlewood-Richardson rule to a pair of Young diagrams, does the set of Young diagrams appearing in the result (with nonzero multiplicity) always form a lattice wrt dominance order? REPLY [3 votes]: Please excuse that I answer with a link, I only have a phone right now. http://www.findstat.org/MapsDatabase/Mp00192/<|endoftext|> TITLE: On the density of the orders excluded by the Sylow theorems for simple groups QUESTION [7 upvotes]: If $G$ is a finite group whose order is divisible by a prime $p$ and $p^r$ is the maximal power of $p$ that divides it, the Sylow theorems tell us that the number $n_p$ of Sylow $p$-subgroups of $G$ is congruent to $1$ modulo $p$ and a divisor of $\lvert G\rvert/p^r$, and that if $n_p=1$ then $G$ is not simple. With this information alone we can decide that some numbers are not the order of a finite simple group. How big is the set of numbers that these two facts exclude as candidates to be the order of a simple group? By big here one can mean, say, the density in $\mathbb{N}$. One can also weigh numbers by the number of actual groups of each order, so as to turn the question into What proportion of finite groups are known to be non-simple by using only these two Sylow theorems on the number of they Sylow subgroups? Since most groups are apparently $p$-groups, the answer to this is probably one. REPLY [12 votes]: Such numbers have density $1$. In fact, for a density $1$ set of numbers it suffices to apply Sylow's theorem to the largest prime factor. In other words, for a density $1$ set of numbers $n$, with $p$ the largest prime factor of $n$, there is no number congruent to $1$ mod $p$ dividing $n$ other than $1$. Proof: We start from a classical fact of analytic number theory, that for each $\epsilon>0$ there is $\delta>0$ such that a density $1-\epsilon$ fraction of numbers have a prime factor of size $>n^{\delta}$. So it suffices to restrict attention to numbers with a prime factor of size $n^{\delta}$. The number of numbers between $X$ and $2X$ which have a large prime factor do not satisfy the condition is at most the number of triples $p,a,b$ where $p> X^{\delta}$ is prime, $a>0$, and $$X< p (1+pa) b< 2X$$ Indeed, take $p$ to be the largest prime factor, $1+pa$ a nontrivial divisor congruent to $1$ mod $p$, and $b$ to be whatever factor remains. Now note that for each $p,a$, the number of possible values of $b$ is at most $\frac{2 X}{ p (1+pa) }\leq \frac{2X}{p^2 a}$. Summing over $a$ from $1$ to $X$, the number of such triples with fixed $p$ is $O ( \frac{X }{p^2} \log X)$ by the harmonic series. Summing over $p$ from $X^\delta$ to $\infty$, the number of such triples is $O( \frac{X \log X}{ X^{\delta}} ) = o(X)$ so the set of numbers not satisfying the condition has density zero, as desired.<|endoftext|> TITLE: Polynomials which are functionally equivalent over finite fields QUESTION [7 upvotes]: Recall that two polynomials over a finite field are not necessarily considered equal, even if they evaluate to the same value at every point. For example, suppose $f(x) = x^2 + x + 1$ and $g(x) = 1$. Then $f$ and $g$ agree at every point in the finite field $\mathbb{F}_2$, but $f$ has degree 2 and $g$ has degree 0, hence $f$ and $g$ are distinct when viewed as polynomials, even though they are equivalent as functions $\mathbb{F}_2 \rightarrow \mathbb{F}_2$. Let $f(x) = \sum_{i=0}^d a_i x^i$ be a univariate polynomial of degree $d$ over the finite field $\mathbb{F}_p$ and let $S_f$ be the set of all polynomials of degree $\leq d$ which evaluate to the same value as $f$ at every point $x \in \mathbb{F}_p$; clearly $S_f$ is non-empty, since $f \in S_f$. My question is, can you characterize $S$? How big is it, as a function of $d$ and $p$? It is clear that $S_f$ includes the set $T_f$, where $$T_F = \left\{ \sum_{i=0}^d a_i x^{ip^{r_i} + k_i(p-1)} \mid k_i, r_i \in \mathbb{Z}, 0 \leq ip^{r_i} + k_i(p-1) \leq d \right\}, $$ by Fermat's Little Theorem. How much larger can $S_f$ be, relative to $T_f$? REPLY [17 votes]: The cardinality of $S_f$ is $p^{ \max(0, d+1-p)}$ because $S_f$ consists of polynomials of the form $f + (x^p-x) g$ with $g$ of degree $\leq d-p$. The fact that such polynomials lie in $S_f$ follows from Fermat's little theorem. The converse is because any polynomial in $S_f$, after subtracting $f$, must vanish at $0,1,\dots,p-1$, hence be divisible by $\prod_{i=0}^{p-1}(x-i) = x^p-x$, and the quotient by $x^p-x$ must have degree $\leq d-p$ (and in particular must vanish if $d TITLE: $\infty$-local systems QUESTION [7 upvotes]: Let $X$ be a "nice" topological space, $R$ a ring. I believe that there is an equivalence of $\infty$-categories betweeen: the full subcategory of $D(X,R)$ (derived category of sheaves of $R$-modules) spanned by objects with locally constant cohomology dg modules over $C_\ast(\Omega X,R)$ (or "$\infty$-local systems", or "parametrized $HR$-module spectra") Is something like this true, under some conditions? What's a reference? REPLY [2 votes]: This is to record that Dylan's suggestion in the comments was on the money: Sections A.1 and A.4 of Higher algebra are precisely what is needed. Specifically there is an equivalence like this whenever $X$ is a topological space locally of singular shape, and $\mathrm{Shv}(X)$ is hypercomplete, e.g. if $X$ is a metric ANR locally of finite dimension. Section A.1 defines in general what it means to have a locally constant sheaf valued in an $\infty$-category $C$ on a topological space $X$ (or more generally an $\infty$-topos). Now $D(X,R)$ is the $\infty$-category of hypercomplete $D(R)$-valued sheaves on $X$. The full subcategory of locally constant objects in Lurie's sense coincides precisely with the full subcategory of objects with locally constant cohomology. So if $\mathrm{Shv}(X)$ is hypercomplete then the $\infty$-category defined in my first bullet point is simply the category of locally constant $D(R)$-valued sheaves on $X$, as defined in Lurie. What Lurie then explains/defines in A.4 is that for $X$ locally of singular shape, there is an equivalence of $\infty$-categories between locally constant $C$-valued sheaves on $X$, and functors $\mathrm{Sing}(X) \to C$. When $C=D(R)$ these are precisely parametrized $HR$-module spectra as in the second bullet point.<|endoftext|> TITLE: What is the significance of the Jiang Su algebra in classification of C$^*$ -algebras? QUESTION [9 upvotes]: Something I've been thinking about for a while that I'm not sure I understand is why $\mathcal{Z}$ stability, as opposed to say $\mathcal{O}_\infty$-stability or even $\mathcal{K}$-stability is so important to representation theory. I know that the Jiang-Su algebra has a lot of interesting properties such as being strongly self-absorbing, or projectionless, simple, KK-equivalent to $\mathbb{C}$. I can certainly see that it is an interesting object, but I think I struggle to understand the relevance for classification. I was wondering if someone a bit closer to classification might be able to explain a bit is why $\mathcal{Z}$-stability is the type of stability we are interested in for Elliot classification. One thing I remember hearing in the YMC*A minicourse Chris Schafhauser gave this year is that it's the analogue to the hyperfinite $II_1$ factor in von Neumann algebras. If anyone could expand on this I would be really interested. Bit of a soft question I suppose, is more out of interest than anything. Probably there is an abstract in a paper/introduction which explains this and pointing me in the right direction would be good. REPLY [15 votes]: Diego's answer in the comments above is related to why we would expect any classifiable $C^\ast$-algebra to satisfy $A\cong A\otimes \mathcal Z$: Since $\mathcal Z$ is separable, nuclear, unital, simple and UCT (the properties of $C^\ast$-algebras we wish to classify by $K$-theory and traces), tensoring with any other $C^\ast$-algebras with these properties will have all the same properties. And since $\mathcal Z$ has the same $K$-theory and traces as $\mathbb C$, one doesn't change $K$-theory and traces by tensoring with $\mathcal Z$. Hence if we expect to have classification by $K$-theory and traces, we would expect $A \cong A \otimes \mathcal Z$ for all $C^\ast$-algebras classified by $K$-theory and traces. Now, I really want to address a misconception which I think is quite common amongst people not working in the classification or structure programme of nuclear C*-algebras: $\mathcal Z$-stability (essentially) has nothing to do with the Jiang-Su algebra $\mathcal Z$! For instance, the UHF algebras, the irrational rotation algebras, and the Cuntz algebras are all $\mathcal Z$-stable, but they have nothing to do with $\mathcal Z$. This is similar to how McDuff factors* don't really have anything to do with $\mathcal R$, or how $\mathcal O_\infty$-stable $C^\ast$-algebras have nothing to do with $\mathcal O_\infty$** One should consider the very natural and frequently occuring regularity property "$\mathcal Z$-stability" as being equivalent (by very deep, surprising, and non-constructive theorems!) to "$A \cong A\otimes \mathcal Z$". Similar to the McDuff property, $\mathcal Z$-stability can be characterised by the (norm-)central sequence algebra $\frac{\prod_{\mathbb N} A}{\bigoplus_{\mathbb N} A} \cap A'$ (for separable unital $A$) being suitably non-trivial, e.g. by requiring that it contains a unital copy of $Z(2,3)$; the $C^\ast$-algebra of continuous functions $f\colon [0,1] \to M_2(\mathbb C) \otimes M_3(\mathbb C)$ with $f(0) \in M_2(\mathbb C) \otimes 1$ and $f(1) \in 1 \otimes M_3(\mathbb C)$. Unfortunately, it would be much too technical to explain exactly how this property is used in the classification thereom. A key technique is that these "almost central homotopies" coming from $Z(2,3)$ can for instance be used to show that $\mathcal Z$-stable $C^\ast$-algebras have cancellation of full projections by $K_0$, that $K_1(A) = U(A)/U_0(A)$ (this was proved in an unpublished paper by Jiang in the late 90's; the paper can be found on arXiv), and to prove that the Cuntz semigroup is almost unperforated (Rørdam). $\mathcal Z$-stability turns out to be a very mild property that can be verified in an abundance of examples, such as through the Toms-Winter conjecture in the cases where this is known to hold, such as separable, simple, non-type I $C^\ast$-algebras with finite nuclear dimension. In fact, it is very hard to construct separable simple nuclear non-type I $C^\ast$-algebras which are not $\mathcal Z$-stable (cf. Villadsen, Rørdam, Toms). So if you have any natural construction that gives you a separable, nuclear, simple $C^\ast$-algebra, you are almost certain to obtain a $\mathcal Z$-stable $C^\ast$-algebra unless you tried very very hard to construct one without this property. (*) McDuff factor: separable $II_1$-factors such that $M \cong M\overline{\otimes} \mathcal R$. Equivalently, their ($W^\ast$-)central cequence algebras are non-abelian. (**) For separable, nuclear, simple $C^\ast$-algebras being $\mathcal O_\infty$-stable is equivalent to being purely infinite by a theorem of Kirchberg. In particular, the Cuntz algebras $\mathcal O_n$ all satisfy $\mathcal O_n \cong \mathcal O_n \otimes \mathcal O_\infty$ (this is quite spectacular; you can't express this isomorphism explicitly!). Being $\mathcal O_\infty$-stable is equivalent to the central sequence algebra $\frac{\prod_{\mathbb N} A}{\bigoplus_{\mathbb N} A} \cap A'$ (for separable unital $A$) containing two isometries $s_1,s_2$ such that $s_1^\ast s_2 = 0$. And this is the property which is used for classifying purely infinite $C^\ast$-algebras by "moving things around" in a paradoxical way.<|endoftext|> TITLE: Finite group with squarefree order has periodic cohomology? QUESTION [8 upvotes]: Is it true that a finite group with squarefree order has periodic group cohomology (with trivial coefficients)? I cannot see why this would be the case, but I'm looking at a paper which seems to implicitly say it's true [Vogel, ON STEENROD'S PROBLEM FOR NON ABELIAN FINITE GROUPS, p.1]. REPLY [10 votes]: All subgroups are then squarefree, and by FTAG all such abelian subgroups are cyclic. Now, a group has periodic cohomology iff all its abelian subgroups are cyclic (Theorem VI.9.5 of Brown's book) -- this is proved by first reducing to Sylow p-subgroups.<|endoftext|> TITLE: Counting $\pm 1$ and $0$'s in the character tables of $\frak{S}_n$ QUESTION [6 upvotes]: Let $\chi_{\mu}^{\lambda}$ denote a value of an irreducible character of the symmetric group $\frak{S}_n$, where $\mu, \lambda\vdash n$. When $\mu=(n)$, then it's known that $$\sum_{\lambda\vdash n}\chi_{\mu}^{\lambda}=\delta_\text{odd}(n).$$ I like to ask: QUESTION 1. Is there a formula for the enumeration (cardinality) of $a_n=\#\{\lambda\vdash n: \chi_{(n)}^{\lambda}=0\}$? How about $b_n=\#\{\lambda\vdash n: \chi_{(n)}^{\lambda}=1\}$ or $c_n=\#\{\lambda\vdash n: \chi_{(n)}^{\lambda}=-1\}$? Encouraged by Mark Wildom's prompt/neat reply, I like to update the question to counting all $0$'s. QUESTION 2. How about the cardinality (or generating function) of the below? $$d_n=\#\{(\lambda,\mu): \lambda\vdash n, \mu\vdash n, \chi_{\mu}^{\lambda}=0\}.$$ REPLY [9 votes]: It is an open problem to determine how many of the entries of the character table are zero — or, indeed whether the proportion of zeros tends to a positive constant, or to zero. One should be careful about the exact problem considered: if a character is picked at random, and a group element is picked at random (so $p(n)$ choices for the character and $n!$ choices for the group element) then almost all entries are zero. This is the partial result mentioned in Stanley's comment to the question. However this is a very different statistic from picking the character at random and a conjugacy class at random (the problem in the question). Miller (On parity and characters of symmetric groups) has done some numerics on this (see Table 3), and conjectured that almost all values were multiples of any given number $d$. This last conjecture has been established for all prime numbers $d$, in the work of Peluse (On even entries in the character table of the symmetric group) and Peluse and Soundararajan (Almost all entries in the character table of the symmetric group are multiples of any given prime) (consequently, $\pm 1$ — or indeed any fixed non-zero integer — appears with density zero in the character table). As for zero entries, the best lower bound that I know (which follows from these ideas of using Murnaghan—Nakayama and the structure of random partitions) is that at least a proportion $c/\log n$ of the entries are zero. REPLY [2 votes]: Since this boils down to the famous Murnaghan–Nakayama rule, I like to advertise an alternative formula due to Adin and Roichman, Ron M. Adin and Yuval Roichman. Matrices, characters and descents. Linear Algebra and its Applications, 469:381–418, March 2015. It was stated in a more explicit manner by Athanasiadis (Power sum expansion of chromatic quasisymmetric functions); see the exact formula and reference here: http://symmetricfunctions.com/gessel.htm#gesselPowerSum. (The statement is much more general, but for your application, just plug in a Schur function for $X(x)$. Its expansion in Gessel quasisymmetric functions is well-known.)<|endoftext|> TITLE: A variation of the Ryll-Nardzewski fixed point theorem QUESTION [12 upvotes]: Is there a fixed-point theorem that implies the following result? Let $F$ be a nonempty convex set of functions on a discrete group with values in $[0,1]$. Suppose $F$ is invariant with respect to left shifts and closed with respect to the pointwise convergence. Then $F$ contains a constant function. This statement looks like Ryll-Nardzewski fixed point theorem, but it does not seem to follow from the theorem. REPLY [9 votes]: The claim does not hold, as the nice example by Naratuka Ozawa shows. The purpose of this answer (or rather, extended comment) is to share a related fixed point theorem. Theorem: Let $X$ be a non-empty set endowed with two structures: The structure of a compact convex set, that is a convex structure and a compatible compact topology $T$ such that continuous affine functionals separate the points. The structure of a separable metric space $(X,d)$. Assume the following compatibility of the two structures: $d$-balls in $X$ are measurable with respect to the Borel $\sigma$-algebra associated with $T$. If $G$ is a group that acts on $X$ preserving both structures - the convex structure, the topology $T$ and the metric $d$ - then $G$ has a fixed point in $X$. Let me illustrate how the theorem above implies the fixed point theorem of Ryll-Nardzewski: for a non-empty weakly compact subset $C$ in a Banach space, there is a point in $C$ which is fixed by all affine isometries of $C$. It is enough to show that $C^G$ is non-empty for every countable group of affine isometries $G$. Given such a group $G$, fix a point $c\in C$ and set $X$ to be the norm closure of the convex hull of the orbit $Gc$. Take $T$ to be the weak topology and $d$ to be the norm metric on $X$. We are now in a position to apply the above theorem. Sketch op the proof: As above, we assume as we may that $G$ is countable. We fix a fully supported probability measure $\mu$ on $G$ and denote by $B$ the corresponding Furstenberg-Poisson boundary. Then the $G$ action on $B$ is amenable and metrically ergodic. By amenability and by 1 there exists a measurable, defined a.e, $G$-equivariant map $B\to X$. By the compatibility of 1 and 2, this map is $d$-measurable, thus by metric ergodicity and by 2 this map must be essentially constant. Its essential image is a fixed point.<|endoftext|> TITLE: A strong Borel selection theorem for equivalence relations QUESTION [6 upvotes]: In Kechris' book "Classical Descriptive Set Theory" there is the following theorem (12.16): Let $X$ be a Polish space and $E$ an equivalence relation such that every equivalence class is closed and the saturation of any open set is Borel. Then $E$ admits a Borel selector. So under the hypotheses of the theorem, one gets a set of representatives that is Borel. There are also stronger forms of this theorem that weaken the assumptions for instance to the classes being $G_\delta$ instead of closed (see Miller (1980)). I have the feeling that a theorem with an even weaker assumption as follows should be true. Question: Let $X$ be a Polish space and $E$ an equivalence relation such that every equivalence class is Borel and the saturation of any open set is Borel. Does $E$ always admit a Borel selector? I am not an expert in this field and therefore also happy if you can just give a reference. Thank you in advance! REPLY [11 votes]: Let $X$ be the Cantor space $2^\omega$, and let $E$ be the relation of "equivalence mod $\mathrm{Fin}$" -- i.e., $xEy$ if and only if $\{n \in \omega :\, x(n) \neq y(n) \}$ is finite. The equivalence classes for this relation are countable (hence Borel, and even $F_\sigma$). If $U \subseteq 2^\omega$ is open, then $U$ contains a basic clopen subset of $2^\omega$, which (by how these are defined) contains a representative of every equivalence class of $E$. So the saturation of any nonempty open set with respect to $E$ is $2^\omega$ itself. However, $E$ does not admit a Borel selector: any selector for $E$ is non-measurable with respect to the usual Haar measure on $2^\omega$, by Vitali's classical argument.<|endoftext|> TITLE: Should every modern day mathematician care about category theory? QUESTION [8 upvotes]: As far as I know, category theory is used mainly in topology. I have a dislike towards category theory, similar to my dislike of Bourbakism, and want to avoid it as much as I can. However, the head of our math department (where I have just started my PhD recently) made a speech where he sang praises for category theory and said that in the future every area of mathematics will be affected by it, and every mathematician who ignores category theory will be left in the gutter (his actual words). I was pretty depressed after this meeting. I want to get as objective an answer to this question: Is it possible to survive in the current mathematics (during the next several decades) as a successful mathematician without caring about category theory? If it is not possible, then what is the minimum required amount of knowledge that every mathematician should get about category theory? REPLY [37 votes]: When I was young I didn’t like sheaves or cohomology, so wanted to find something that was algebraic but didn’t involve too much sheaves or cohomology. I didn’t really need to know much about either to get a tenure track job. But now I’m a more mature person and a more mature mathematician, and I’ve learned to stop worrying and accept cohomology. All of this is to say that everyone in comments is right, you can certainly be a mathematician without caring about category theory, but strictly avoiding a subject entirely is going to make you an immature mathematician and hold back your development. You don’t have to love category theory, but it’s a good idea to stop hating it. REPLY [28 votes]: I say many (most?) mathematicians with thriving research careers completely ignore large parts of mathematics in their work. Probably, they don't even remember what they learned in some of their introductory graduate courses, unless they teach them, and would be unable to pass some comprehensive PhD exams without preparation. What you don't use you forget. Disliking some parts of mathematics is a way of finding what you really enjoy, a completely natural process. Being broadly educated helps, as long as it does not interfere with research. Learning and doing math are somewhat different activities. One cannot do math without learning some. On the other hand, it is possible to enjoy learning so much that you never actually do anything. There has to be a balance. In particular, most math research can surely be done without category theory. If you ever need to learn what is, say, a colimit, just read Wikipedia, and follow the references there. Short term, grad students should focus on finding the kind of math they enjoy doing, and also on passing their exams. Personally, I revere broadly educated mathematicians, and I strive to become one. Is it a must for a successful career? Not really. REPLY [26 votes]: You can look at the edit history of this post to see previous versions, which took a different tack whose thread I have honestly lost. I want to take a different tack, though. What makes this question peculiar is the fact that if you substitute any other area of math for "category theory" in the question, the resultant discussion would look quite different. That is consider the following dialog for various values of $X$: Professor : Any mathematician who ignores $X$ will be left in the gutter. Student : I have a distaste for $X$. What's the minimum I should know about $X$ to get by? I invite the reader to perform the thought experiment of considering the different reactions this exchange would elicit for various values of $X$, such as set theory, group theory, ring theory, combinatorics, functional analysis, topology, category theory. When I run this thought experiment, I find that in most cases, the professor's pronouncement admits basically two interpretations: a strong interpretation, where they mean you must be actively be keeping up with current research in $X$. a weak interpretation, where they mean that you must have an idea of what $X$ is good for, and that you should be prepared to reach for tools from $X$ when the situation calls for it in your own research. For most values of $X$, the strong interpretation is a clear stretch, and the onlooker will charitably assume that the weak interpretation is intended. For most values of $X$, that's all there is to it. But when $X$ is category theory, unlike other values of $X$, there's additionally a flame war among the onlookers. After surviving the latest flame war, I have a theory as to why this is so. My theory is that for most values of $X$, there's a general understanding of how to formulate a weak interpretation of the professor's statement. But when it comes to category theory, people may not be so clear on what kind of weak interpretation should be understood. I propose to remedy this situation with the following pronouncement: Category theory is good for understanding the naturality vs. choice-dependence of constructions. This is intended to be parallel to the following pronouncement, which I believe is widely-understood among mathematicans: Group theory is good for understanding symmetries. or Set theory is good for quotienting by equivalence relations. In each case, the pronouncement doesn't give a complete picture of what $X$ is good for, but gives some kind of launching-off point. Just as it's reasonable for the professor to say "questions of symmetry are everywhere in math -- be ready to reach for group-theoretic tools to help understand them" it's similarly reasonable to say "questions of naturality are everywhere in math -- be ready to reach for category-theoretic tools to help understand them". I hope we can all think of examples illustrating (1). Perhaps the situation is different in the case of (2), and perhaps this points to a shortcoming in general mathematical education. Here's a small example pulled from differential geometry: Let $f : X \to Y$ be a smooth map of manifolds, and let $\omega$ be a differential form on $Y$. Then there is a pullback form $f^\ast(\omega)$ on $X$. You might define $f^\ast(\omega)$ in terms of coordinates, and then wonder whether your definition depends on the choice of coordinates. You can prove that it doesn't, and you can prove things like $g^\ast \circ f^\ast = (f \circ g)^\ast$. The statements of each of these facts are very naturally stated category-theoretically (though the proofs are mostly geometry). There are various routine coordinate-based manipulations you can do on differential forms which are justified by these facts, which again can be nicely summarized in category-theoretic language. A couple of takeaways from this last example: The use of category-theoretic language here is not supposed to be earth-shattering or anything. It's pretty banal, really. We could continue the flame war by arguing about whether it's necessary to use category-theoretic language here (of course, strictly speaking it isn't). But we don't devolve into such arguments when it comes to examples of using group theoretic-language to understand symmetry. I have a dream that one day we will stop treating category theory differently from group theory in this respect!<|endoftext|> TITLE: Number of distinct normalized vectors from the center of a hexagon in a hexagonal grid QUESTION [5 upvotes]: Consider an infinite hexagonal grid composed of regular hexagons. Choose any hex to be the origin hex. Let n be a natural number. Find an expression, in terms of n, for the number of distinct normalized vectors from the center of the origin hex to the center of every hexagon whose distance from the origin (measured in hexagons) is less than or equal to n. I manually calculated up to n=15: n distinct normalized vectors increment 1 6 - 2 12 6 3 24 12 4 36 12 5 60 24 6 72 12 7 108 36 8 132 24 9 168 36 10 192 24 11 252 60 12 276 24 13 348 72 14 384 36 15 432 48 What I do know: The increment from n to n+1 has to be a multiple of 6, because of the simmetry of the hexagon. The expression has to be less than 3n^2+3n, since this is the formula for the total number of hexagons at a distance n, at most, from the origin hex (in hexagons); When n is prime, the increment from n-1 is 6*(n-1). REPLY [2 votes]: Let's move the centers of the hexagons to the grid $\mathbb{Z}\times\mathbb{Z}$ by an affine transformation, so that the centers of the hexagons at hexagonal distance less than or equal to $n$ are now represented by the more Cartesian-looking (but uglier) set $$H_n:=\{(x,y)\in\mathbb{Z}\times\mathbb{Z}: |x|\le n, |y|\le n, |x-y|\le n, \}.$$ We can partition $H_n$ into the area in the axis, a square on the first quadrant and a square on the third quadrant; a triangle in the second and a triangle in the fourth quadrant. Counting separately the corresponding normalized vectors from each set, we get, for your $H(n)$ for $n\ge1$ $H(n)=4+2A(n)+2B(n)$, where $$A(n):=\{(x,y)\in\mathbb N_+ \times \mathbb N_+ : x\le n, y\le n, (x,y)=1\}$$ is easily seen to be $A(n)= -1+\sum_{k=1}^n \phi(k)=$ A18805 , and so is $$B(n):=\{(x,y)\in\mathbb N_+ \times \mathbb N_+ : 1\le x TITLE: VC dimension of standard topology on the reals QUESTION [8 upvotes]: Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is an open set $U$ with $D=S\cap U$? I'm asking merely out of curiosity, but I'll mention that this would imply $2^{\aleph_1}=2^{\aleph_0}$. REPLY [12 votes]: An uncountable $S\subseteq \mathbb{R}$ has an accumulation point $x\in S$. Then for $D=\{x\}$ there is no such open set $U$.<|endoftext|> TITLE: Asymptotics of degree of $\textrm{SO}_n$? QUESTION [5 upvotes]: (This is an offshoot of Degree of parametrization of $\textrm{SO}_n$?) Consider $G=\textrm{SO}_n$ as an affine subvariety of the affine space of $N$-by-$N$ matrices. There is an explicit expression for $\deg(G)$ in Theorem 4.2. of Brandt, Bruce, Brysiewicz, Krone, and Robeva - The degree of $\operatorname{SO}(n)$: $\deg(G) = 2^{n-1} N(n)$, where $N(n)$ is the number of non-intersecting lattice paths from the points $(n-2 i,0)$, $1\leq i\leq \lfloor n/2\rfloor$, to the points $(0,n-2 j)$, $1\leq j\leq \lfloor n/2\rfloor$. What are the asymptotics of $N(n)$? REPLY [9 votes]: I can inject these lattice paths into cyclically symmetric plane partitions in an $(n-1) \times (n-1) \times (n-1)$ box. CSPP's grow at $(27/16)^{n^2/2 + O(n)}$. In the first draft this answer, I suggested that this would also be right for $N(n)$, but now I don't believe this; I think the right value should be $A^{n^2+O(n)}$ for some smaller $A$. However, numerically, they still seem close. I will follow the usual route to biject non-intersecting paths with rhombus tilings, which I wrote up in an earlier answer. Figure 1 in Brandt-Bruce-Brysiewicz-Krone-Robeva shows how to compute $N(5) = 24$ as noncrossing paths from $\{ (1,0), (3,0) \}$ to $\{ (0,1), (0,3) \}$. The image below shows the grid they use, and one such pair of paths on it, tilted and skewed for future convenience: Slide those paths down one unit and cover the spread out paths with rhombi with vertical sides. The result is shown shaded in the figure below. For each vertex that was not on the paths, add a horizontal rhomus. We get a rhombus tiling: The number of non-crossing paths $N(n)$ is the number of rhombus tilings of the resulting shape. Another way to think of this shape is that it is an $(n-1) \times (n-1)$ rhombus, with $\lceil \tfrac{n-1}{2} \rceil$ triangles deleted from each of its top two sides. I've shaded the deleted triangles below: Now, given a rhombus tiling of this shape, we can take three copies of it and arrange them centrally around a central vertex. The deleted triangles can be paired off with each other. Thus, we get a bijection with $(n-1) \times (n-1) \times (n-1)$ cyclically symmetric plane partitions which use the rhombi shaded below and don't cross the solid lines. (I missed the second condition the first time.) Thus, $N(n)$ is less than the number of CSPP's, and the number of CSPP's grows at rate $(27/16)^{n^2/2+O(n)}$. I used to think the growth rates would match, but now I don't. First, numerical data. Here is a table (taken from Table 1 in BBBKR and from OEIS) comparing the two: $$ \begin{array}{c|cccccccc} n & 1&2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline N(n) &1& 1 & 2 & 5 & 24 & 149 & 1744 & 26825 & 769408 \\ CSPP(n-1) &1 & 2 & 5 & 20 & 132 & 1452 & 26741 & 826540 & 42939620 \\ \end{array}.$$ Interestingly, $N(n)$ seems very close to $CSPP(n-2)$ for low values of $n$, but I think this is a coincidence. Now, theoretical arguments. The shape of a typical plane partition in a large $n \times n \times n$ box was worked out by Cohn, Larsen and Propp. There is a continuous function on the hexagon which gives the limiting probability that a rhombus in a given position will be used, and each of the fixed rhombi in our problem is in the most likely orientation, so I expect those not to impose a major restriction. However, the solid lines which may not be crossed are major problems. In the corners of the hexagon, these lie in the "frozen" region where, with probability $1$, all rhombi have the same orientation. Thus, it is highly unlikely that the solid lines will not be crossed. This rhombus tiling problem (with the deleted rhombi) fits into the general framework of papers like Kenyon and Okounkov "Planar dimers and Harnack curves" and Cohn, Kenyon and Propp "A variational principle for domino tilings", where we have a region of growing size with height function on the boundary approaching a limit. This would take a while to write up carefully, and I don't see how to get a closed answer from it so, at least for today, I'm going to stop. Will Sawin asks whether I can prove a lower bound of the form $A^{n^2}$ in this way. The answer is yes. I found lattice paths easier to draw than rhombi this time. Take the half of the paths closer to the corner at $(0,0)$ and make them hug the axes as closely as possible. Then connect the other half of the points to the points on the line segments $((n/2, n/4), (3/n4, n/2))$ and $((n/4, n/2), (n/2, 3n/4))$. (I may have off by one errors; see the image.) Then fill the shaded hexagon with paths however you please. The number of ways to fill the hexagon is the number of plane partitions in an $(n/4) \times (n/4) \times (n/4)$ box, which grows like $a^{n^2}$.<|endoftext|> TITLE: Divergent series summation beyond natural boundaries QUESTION [10 upvotes]: I'm hoping to investigate the effects of divergent summation methods on series which cannot be analytically continued due to a dense set of singularities. At least a priori, it doesn't seem that a natural boundary should necessarily mean there aren't analytic functions that continue the original function. For example, we can easily generate somewhat contrived examples such as $\sum_{n=0}^\infty \frac{1}{2^{n-1}(n+1)^2}\sum_{k=0}^{2^{n-1}} (z-e^{\frac{2k+1}{2^n} \tau i})^{-1}$ This series absolutely converges whenever z is not exactly equal to a dyadic rational. On the other hand, it has a natural boundary at the unit circle because the dyadic rationals are dense on the unit circle. So because the terms in this series eventually go to zero, the partial sums provide arbitrarily good approximations of the function when not around the unit circle, and so we have a natural way to "continue" this function to an analytic function using the partial sums. We can easily craft many of these examples by enumerating the rationals--or some other dense set-- in steps and causing the later steps to eventually go to 0. Each of these functions will similarly have a natural boundary, yet can be approximated by analytic functions. I'm interested in finding the validity of using divergent series methods to find continuations of functions with natural boundaries. For instance, one function I am looking at is $$\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{\frac{n(n+1)}{2}}}{\prod_{m=1}^{n}(1+x^{m})}$$ which is from one of Ramanujan's notebooks. It is special in that its terms grow quite slowly, so it's easier to handle with divergent series methods. In particular, applying different types of smooth cutoff functions to this function leads to the same thing. Furthermore, the smooth cutoff method agrees with my approximation method. Graphed together, these methods look like this: The green line is the smooth cutoff function (using $\frac{1}{1.005^{n^2}}$), the orange line is the approximate method and blue is approximating Borel method (which ends up essentially like applying a smooth cutoff function). Each of these functions agrees with the starting function when the starting function converges. Are there other ways to regularize this series that lead to this same result? Are there other methods that disagree? Also, because we don't have the identity theorem when we allow natural boundaries, is there a natural replacement of this theorem, possibly using properties of divergent series? My guess is that something like the Tauberian theorems could provide a good start for proving some sort of uniqueness, though it's not ideal because it requires somewhat strong restrictions on the properties of the terms in the summation. I'd appreciate any insight into this problem, or insight into similar problems with faster divergences (for instance, I'm also interested in $\sum_{n=0}^\infty (-1)^n x^{n^2}$). Finally, are there are any other proposed methods of continuing functions with natural boundaries that I can check the divergent series methods against? It's quite useful for me to be able to check my own results against the results of others to find where the methods agree and disagree, and hopefully why. REPLY [10 votes]: From those three examples, Rogers-Ramanujan's series belong to the class of basic hypergeometric series ($q$-series). It is a marginally logarithmic divergent series for $q > 1$ and it should be computed easily using traditional methods for alternating series. Just the third series, also belonging to this class, is really challenging and it deserves special attention. The following "simple" formal series $$s(\zeta,q)= \sum_{n=0}^\infty (-\zeta)^n q^{n^2}$$ is closely related to Jacobi's $ϑ$ functions Fourier expansion (See either DLMF-NIST Handbook or Wikipedia). A proper choice of $ζ$ allows to get every $ϑ$ function (See Addenda 2 below). These $ϑ$ series (having strong convergence) are defined if nome $q = e^{iπτ}$ fulfills $0 < |q| < 1$ with $\mathrm{Im}(τ) > 0$. As far as I know almost nothing is known off this range. For $q > 1$ the above series diverges far away factorial (Gevrey-$α$ series) divergence. This makes this kind of series really challenging and should be subject for a deep research. If you put $q^{n^{2}}$ as an exponential Taylor series in powers of $n^{2}\log(q)$ and swap the resulting sums you get a power series in $\log(q)$ whose coefficients are polylogarithms (See Addenda 1). Except for some particular values of $\zeta$, the final series is still divergent (although much slower) with an additional issue, summands have an irregular sign pattern making difficult they can be fitted into some known summation method. Taking few terms of this asymptotic sum you can get about 2-3 decimal digits at most. In my humble opinion no currently known summation method can handle the above series (nor Jacobi's $ϑ$ Fourier series) in the general case when nome $q > 1$, however they can be computed using numerical quadratures as follows. By means of Nachbin's Theorem (see Wikipedia for a brief introduction), these series can be analytically continued (beyond Borel's Summation. See Ovidiu Costin - Asymptotics and Borel Summability) using this Mellin Transform $$\int_0^\infty t^{s-1}\exp\left(-\frac{1}{a}\log^2(t)\right)\mathrm{d}t = \sqrt{\pi a} e^{\frac{a}{4}s^2}$$ valid for $\mathrm{Re}(a) > 0$. Lebesgue's dominance can be applied for $ζ ∈ ℂ$ \ $(-∞,0)$ to get $$s(\zeta,e^{a/4}) = \sum_{n=0}^\infty (-\zeta)^n e^{\frac{an^2}{4}} = \frac{1}{\sqrt{\pi a}} \int_0^\infty \frac{\exp(-\frac{1}{a}\log^2(t))}{1+\zeta t} \cdot \frac{dt}{t}$$ with $a = 4\log(q)$ and $|q|>1$, there are several interesting values for $ζ$ and $q$ to test, focused on Jacobi's thetas. For example $[ζ, q] → [ζ_{ } q^{±1/2},q^{±1/2}]$ with $q = 2$ or $q = e$. The multiprecision computation of this integral can provide as many digits as demanded. I hope this helps. Addenda 1 case $s(1,q) = \frac {1}{2}$ is proved using simple algebra $$s(\zeta,q) = 1 + \sum_{n=1}^\infty (-\zeta)^n e^{n^2\log(q)} = 1 + \sum_{n=1}^\infty (-\zeta)^n \sum_{k=0}^\infty \frac {n^{2k}\log^{k}(q)}{k!}$$ Swapping these sums we get $$s(\zeta,q) = 1 + \sum_{k=0}^\infty \frac {\log^{k}(q)}{k!} Li_{-2k}(-\zeta)$$ where $Li_{-2k}(-\zeta)$ are polylogarithms of nonpositive even integer indexes. Using $Li_0(-\zeta) = (1 + \zeta)^{-1} - 1$ $$s(\zeta,q) = \frac {1}{1 + \zeta} + \sum_{k=1}^\infty \frac {\log^{k}(q)}{k!} Li_{-2k}(-\zeta).$$ But all summands vanish at $\zeta = 1$ since $$Li_{-k}(-\zeta)+(-1)^k Li_{-k}(-\zeta^{-1}) = 0$$ holds for all $ζ ∈ ℂ$ and $k = 1, 2, 3, ...$ (Erdélyi et al. 1981, § 1.11-17 pg 31). Addenda 2 Jacobi's $\vartheta$ Functions Generalized Analytic Continuation (GAC) The above series $s(\zeta,q)$ was studied in Fredholm's first paper (1890) where it is proved that $i)$ $s$ is not analytically continuable across any point of $|q|=1$ and $ii)$ $s$ and all its derivatives extend continuosly for $|q|>1$. (Proof was amended over 100 years later. Details can be seen here). We will see now how this applies to Elliptic Theta functions $\vartheta_{\ell}(z,q)$, $\ell=1,2,3,4,$ defined as Fourier $q$-series $$\vartheta_{1}(z,q)=2\sum_{n=0}^\infty (-1)^{n}q^{(n+1/2)^{2}}\sin((2n+1)z)$$ $$\vartheta_{2}(z,q)=2\sum_{n=0}^\infty q^{(n+1/2)^{2}}\cos((2n+1)z)$$ $$\vartheta_{3}(z,q)=\vartheta_{4}(z,-q)=-1+2\sum_{n=0}^\infty q^{n^{2}}\cos(2nz)$$ $$\vartheta_{4}(z,q)=-1+2\sum_{n=0}^\infty (-1)^{n}q^{n^{2}}\cos(2nz)$$ which converge super-fast because of quadratic exponents if nome $q=e^{i\pi\tau}$ is located inside the unit disk $|q|<1$ (lattice parameter $\tau$ holds $\mathrm{Im}(τ)>0$). By Fabry Gap Theorem, $\vartheta$ functions have $|q|=1$ ($\mathrm{Im}(τ)=0$) as their natural boundary, this means that monodromy theorem does not apply (due to not connected regions), which makes that $\vartheta$ functions cannot be analytically continued beyond the unit $q$-disk. Such functions are lacunary in the sense of Fabry Gap (but they are not in the sense of Ostrowsky-Hadamard theorem). Fredholm's clause $ii)$ leaves a door open apparently, for the existence of an analytical extension or exo-analyticity of these functions beyond their natural boundaries. In fact, the above series $s(\zeta,q)$ satisfies conditions of Grecchi-Maioli Theorem -section II, p. 39-. This theorem uses Nachbin's Theorem implicitly to extend Borel's Summability, which allows us to derive this (I guess new) integral for $q\in\mathbb{C}\wedge|q|>1, n\in\mathbb{N}_{0}$ and $\zeta\in\mathbb{C}\backslash(-\infty,0)$ $$s^{(n)}(\zeta,q)=\frac{d^{n}}{d\zeta^{n}}s(\zeta,q)=\frac{n!}{2\sqrt{\pi\log q}}\int_{0}^{\infty}\frac{(-t)^{n}e^{-\frac{1}{4}\frac{\log^{2}t}{\log q}}}{(1+\zeta\,t)^{n+1}}\frac{dt}{t}$$ providing an off-boundary definition of elliptic $\vartheta_{\ell}(z,q)$ and derivatives $\vartheta_{\ell}^{(n)}(z,q)$. For $\vartheta_{\ell}(z,q)$ and $\vartheta'_{\ell}(z,q)=d\vartheta_{\ell}(z,q)/dz$ we have these expressions for $|q|>1$ (higher derivatives are found in the same way) $$\vartheta_{\ell}(z,q)=A\cdot s(\zeta_{a},q)+B\cdot s(\zeta_{b},q)+C$$ $$\vartheta'_{\ell}(z,q)=A'\cdot s(\zeta_{a},q)+B'\cdot s(\zeta_{b},q)+A\cdot s'(\zeta_{a},q)\cdot\zeta'_{a}+B\cdot s'(\zeta_{b},q)\cdot\zeta'_{b}$$ $$\begin{align*} \begin{array}{|c|cccccrcccccr|} \hline & & A & & B & & C & & A' & & B' & & D\\ \hline \vartheta_{1}|\vartheta'_{1}(z,q) & & -iq^{1/4}e^{iz} & & iq^{1/4}e^{-iz} & & 0 & & q^{1/4}e^{iz} & & q^{1/4}e^{-iz} & & q\\ \vartheta_{2}|\vartheta'_{2}(z,q) & & q^{1/4}e^{iz} & & q^{1/4}e^{-iz} & & 0 & & iq^{1/4}e^{iz} & & -iq^{1/4}e^{-iz} & & -q\\ \vartheta_{3}|\vartheta'_{3}(z,q) & & 1 & & 1 & & -1 & & 0 & & 0 & & -1\\ \vartheta_{4}|\vartheta'_{4}(z,q) & & 1 & & 1 & & -1 & & 0 & & 0 & & 1 \\\hline \end{array} \end{align*}$$ where $\zeta_{a,b}=D\cdot e^{\pm2iz}$ and $\zeta'_{a,b}=\pm2i\cdot\zeta_{a,b}$. This new domain of analyticity must exclude $\arg(\zeta_{a,b})=(2k+1)\cdot\pi$, $k \in \mathbb{\mathbb{Z}}$ where $\arg(\zeta_{a,b})$ is $$\begin{array}{|c|c|} \hline & \arg(\zeta_{a,b})\\ \hline \vartheta_{1}|\vartheta'_{1}(z,q) & \arg(q)\pm2\mathrm{Re}(z)\\ \vartheta_{2}|\vartheta'_{2}(z,q) & \pi+\arg(q)\pm2\mathrm{Re}(z)\\ \vartheta_{3}|\vartheta'_{3}(z,q) & \pi\pm2\mathrm{Re}(z)\\ \vartheta_{4}|\vartheta'_{4}(z,q) & \pm2\mathrm{Re}(z) \\\hline \end{array}$$ A big world is opened from this place. For instance, to determine which of this huge number of $\vartheta$ relationships are still fulfilled and, hopefully, find new ones by deeper analysis. Since $\vartheta$ functions are closely related to some number theory topics, this extended formulation might produce interesting consequences on that context as well.<|endoftext|> TITLE: VC dimension of Borel sets QUESTION [8 upvotes]: Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is a Borel set $U$ with $D=S\cap U$? I'm asking merely out of curiosity, but I'll mention that this would imply $2^{\aleph_1}=2^{\aleph_0}$. This is a hopefully more interesting adaption of a recent too easy question. REPLY [9 votes]: Yes! Martin's Axiom implies that if $S \subseteq \mathbb R$ and $|S| < \mathfrak{c}$, then every subset $D$ of $S$ is a relative $G_\delta$ in $S$: i.e., there is a $G_\delta$ set $X \subseteq \mathbb R$ with $X \cap S = D$. (And let me note that $2^{\aleph_0} = 2^{\aleph_1}$ is another consequence of Martin's Axiom.)<|endoftext|> TITLE: Does every open set contain a dense $F_{\sigma}$ subset? QUESTION [5 upvotes]: Let $U$ be a regular open set in a Tychonoff space $X$ (regular means that it is an interior of a closed set). [ In my specific situation $U$ is of the form $\operatorname{int} f^{-1}(0)$, where $f$ is a continuous real-valued function on $X$, and $X$ is a Baire space (a sequence of dense open sets has a dense intersection), but I am not sure if it helps. ] Is there a sequence $\{A_n\}_{n\in\mathbb{N}}$ of closed (in $X$) subsets of $U$ such that $\bigcup_{n\in\mathbb{N}} A_n$ is dense in $U$? Of course, this is the case if $X$ is perfectly normal (which is equivalent to every open set being $F_{\sigma}$), or separable, but I hope a less restrictive assumption will suffice, e.g. normality. REPLY [6 votes]: Not in $\beta\mathbb{N}\setminus\mathbb{N}$: if $A$ is an $F_\sigma$-subset of $\operatorname{int}f^{-1}(0)$ then there is even a clopen set $C$ such that $A\subseteq C\subseteq \operatorname{int}f^{-1}(0)$. Of course this is only an example if the zero-set is not clopen but you can get an example by working on the countable set $N=\mathbb{N}^2$ the clopen sets determined by the vertical lines $V_n=\{n\}\times\mathbb{N}$ union up to a cozero set: let $f$ have value $2^{-n}$ on $V_n$. That cozero set is not clopen and for every closed-set $F$ contained in $\operatorname{int}f^{-1}(0)$ there is a function $h:\mathbb{N}\to\mathbb{N}$ such that $F$ is in the clopen set determined by $L_h=\{(m,n):n\le h(m)\}$. If $A$ is an $F_\sigma$ then we get a sequence $\langle h_n:n\in\omega\rangle$ of such functions. Define $h(m)=1+\max\{h_n(m):n\le m\}$. Then the clopen set determined by $L_h$ contains $A$ and is a subset of $\operatorname{int}f^{-1}(0)$. We consider the behaviour of the extension of $f$ to $\beta N$ and its restriction to $N^*=\beta N\setminus N$ (all called $f$). The important thing to note is that if $X$ determines a clopen set, denoted $X^*$, in $N^*$ then $X^*\subseteq f^{-1}(0)$ iff $X\cap V_n$ is finite for all $n$. In general $X^*\cap Y^*=\emptyset$ iff $X\cap Y$ is finite, and in this case if $X\cap V_n$ is finite for all $n$ then $f[X^*]=\{0\}$. Furthermore: $X\cap V_n$ is finite for all $n$ iff $X\subseteq L_h$ for some $h$ as above. Finally: $N^*$ is compact and zerodimensional, so if $F$ is closed and contained in $\operatorname{int}f^{-1}(0)$ then there is a clopen set $C$ such that $F\subseteq C\subseteq \operatorname{int}f^{-1}(0)$. See this note for a short introduction.<|endoftext|> TITLE: Is a spin structure on a knot complement the same thing as an orientation of the knot? QUESTION [5 upvotes]: There are a variety of characterizations of spin structures on the tangent bundle of a manifold. Two facts about them: Spin structures on $TM$ are an affine space over $H^1(M; \mathbb{Z}/2\mathbb{Z})$, but in general there's no canonical way to identify them with $H^1$. Spin structures on $TM$ are the same as trivializations of $TM$ along the $1$-skeleton that extend to trivializations on the $2$-skeleton. I'm interested in the case where $M = S^3 \setminus L$ is a knot or link complement. (I'm not completely clear whether I want to think about $M$ as compact with torus boundary or non-compact, but I guess my proof below uses ideal triangulations.) I am suspicious that the following is true: Theorem: Spin structures on $S^3 \setminus L$ are in one-to-one correspondence with orientations of $L$, hence with $H^1(S^3 \setminus L; \mathbb{Z}/2\mathbb{Z})$. Proof idea: From an oriented diagram of $L$ we can use the octahedral decomposition to get an ideal triangulation of $S^3 \setminus L$. For such a triangulation it's possible to orient everything so that the obvious trivialization of the $1$-skeleton (coming from its orientation) extends over the $2$-skeleton. This is pretty vague, but I think the "compatible orientations" are a branching, in the sense of [1]. I believe [1] works out the details, since they have a "spin calculus" for triangulations of spin $3$-manifolds. My question: Is the theorem true? If so, is there a more direct proof than what I've given? It seems like there should be a more direct path than choosing some special triangulation. [1] R. Benedetti, C. Petronio, Branched Standard Spines of 3-manifolds, Lect. Notes Math. 1653, Springer (1997) EDIT: Some motivation for the question is the "well known" fact that spin structures on a hyperbolic knot complement are in natural 1-1 correspondence with lifts of the holonomy representation $\rho : \pi_1(S^3 \setminus K) \to \operatorname{PSL}_2(\mathbb C)$ to $\operatorname{SL}_2(\mathbb C)$. (The argument has to do with identifying $\mathbb H^3$ as $\operatorname{PSL}_2(\mathbb C)/\operatorname{SO}_3$.) Then for a hyperbolic knot it's straightforward to pick a canonical lift/spin structure, because in one lift the trace of the meridian is $2$ and in the other it's $-2$. It doesn't matter which meridian you pick, because $\mathfrak m$ and $\mathfrak m^{-1}$ have the same trace. However, upon further thought we can't identify this choice of lift with a choice of meridian ($\mathfrak m$ versus $\mathfrak m^{-1}$), which is the same as an orientation of the knot. This agrees with the answers that say that orientations are not naturally the same as spin structures. REPLY [7 votes]: Is the theorem true? There is an non-natural bijection. There is no natural bijection. A link exterior is homotopy-equivalent to a $2$-complex, so a trivialization of the tangent bundle over the $2$-skeleton is simply a map: $$S^3 \setminus L \to SO_3.$$ This is because there is a canonical trivialization of $TS^3$ coming from, say, a left-invariant framing. Similarly, if you had a trivialization over the $1$-skeleton that admits an extension to the $2$-skeleton, any two extensions $S^3 \setminus L \to SO_3$ have to be homotopic, due to $\pi_2 SO_3$ being trivial. So spin structures on link exteriors are canonically in bijection with $$\pi_0 Maps(S^3 \setminus L, SO_3).$$ $SO_3$ is diffeomorphic to $\mathbb RP^3$, which is the $3$-skeleton of $\mathbb RP^\infty$, which is a $K(\mathbb Z_2,1)$-space. So you have a natural map between the spin structures on the link exteriors and $$\pi_0 Maps(S^3 \setminus L, K(\mathbb Z_2,1)) \equiv H^1(S^3 \setminus L, \mathbb Z_2).$$ You can argue this map is injective. The argument that comes to mind is that $\pi_1 \mathbb RP^3 \to \pi_1 \mathbb RP^\infty$ is an isomorphism. I think the relation to Kevin Walker's observation is that diffeomorphisms of the link exterior act trivially on some of the spin structures, but not on all. A diffeomorphism that reverses a link component does not affect the spin structure. The only way a diffeomorphism can act non-trivially on spin structures is if they permute components of the link. Similarly the diffeomorphism group of a knot exterior acts trivially on the spin structures, even for invertible knots (where diffeomorphisms can reverse the knot). The action of the diffeomorphism group on the spin structures factors through the permutation representation, i.e. the map $\Sigma_n \to GL_n \mathbb Z_2$ given by the $n \times n$ matrices that permute the rows and columns, and that permutation representation is the permutation of the link components. Technically to get the above factorization you need to know that the link has no split Hopf link components. When you have a split Hopf link you get a $GL_2(\mathbb Z)$ in the mapping class group and this action would not belong to the above representation.<|endoftext|> TITLE: Reference request: Moore graphs QUESTION [15 upvotes]: It is clear that the term Moore graph was coined by Hoffman and Singleton in their paper On Moore graphs with diameters $2$ and $3$, where they write E. F. Moore has posed the problem of describing graphs for which equality holds in (2). We call such graphs "Moore graphs of type $(d,k)$". My question is: does anyone know where Moore posed this problem? REPLY [18 votes]: Moore posed this problem to Hoffman at a conference, so it is not in print. Hoffman makes the following remark (from "Selected Papers of Alan Hoffman with Commentary", pp. 367): After I discussed the preceding paper at an IBM summer workshop, E.F. Moore raised the graph theory problem described in the paper, and my GE colleague Bob Singleton and I pondered it. Moore told me the problem because he thought the eigenvalue methods I was using might find another "Moore graph" of diameter 2 besides the pentagon and the Petersen graph. Indeed, we found the Hoffman-Singleton graph with 50 nodes (and showed it was unique) and that any other Moore graph of diameter 2 had to have 3,250 nodes (and to this day, no one knows if such a graph exists). Moore declined joint authorship, so we thanked him by giving his name to the class of graphs. When it was later proved by Damerell, and also by Bannai and Ito, that there were no other Moore graphs other than the trivial odd cycles, I felt a twinge of guilt in giving Moore's name to such a small set. But I was wrong: Moore graphs, Moore geometries, etc. continue to be discussed in the profession.<|endoftext|> TITLE: How can I seek help in preparing a very long research article for publication? QUESTION [29 upvotes]: Some background first. I recently graduated with a master's degree in applied mathematics. During graduate school I began working on a paper, which I continued to work on post-graduation. A complete working copy of the paper is done and I have posted it on the arXiv here. The work contained in the paper is completely original and solves an open problem. It was of my opinion that the paper contained publishable material. To verify, I emailed a professor at my alma mater with a copy of the current draft (current as of approximately six months ago). The professor did reply stating that the work was publishable and even suggested some Q1/Q2 journals that might accept this type of work. While this was useful feedback, I received the reply in just a few days, so I doubt the professor in question had the ability to read my paper in depth. The problem: I have published a couple of papers before and thus have some experience in the world of academic publishing. That said, the scale and complexity of this paper is something I have never dealt with before so I am not comfortable with proceeding to publish it without help/guidance, i.e. on my own. In particular, I suspect I am going to have to divide the work into small portions and publish a few separate papers, but I don't know how best to do this and do not want to sink a lot of additional time into this without any direction. Also, the paper is very dense and I am concerned that its readability is not optimal. Given that I do not have a ton of experience with large papers like this and am essentially working in a vacuum, I also really desire to get feedback on the quality of my proofs, which I suspect are not as concise as they could be. My situation seems a little unusual and I suspect that the feedback I am looking for would typically be provided by an adviser in a Ph.D. program. Given that I do not have an adviser that can provide detailed feedback, what should I do? I have tried reaching out to researchers/experts with relevant backgrounds and offering authorship in exchange for the help I need. Since what I'm seeking entails a significant amount of work, this proposition seemed reasonable; however, my efforts have not lead to much fruit. As mathematicians in academia, how are these types of requests viewed and how might I go about reaching out for help? I do not have funding and so I considered the offering of authorship as a a reasonable incentive to get the help I need. Is this practice frowned upon and is there anything else I can do to increase my chances of getting a researchers attention? REPLY [10 votes]: I usually solve this problem by publishing research announcements. I highlight the main results, write them down (without proofs, but with the reference to the original article in arxiv.org) in some small article and send it to a journal where they publish this (there are some Russian and French journals where mathematicians can publish texts like this). But this is fatally important only if your employer requires you to publish regularly. If this is not necessary, you can simply send your big article to a journal which publish big texts, and if all is well, they will publish it (there are several journals for this, in particular, Journal of Mathematical Sciences). Of course, you will have to wait, most likely several years, but in any case, long articles are published with long delays. And of cource, no one bothers you to simultaneously publish a research announcement, as I wrote above. Offering co-authorship to an outsider is an absurd idea.<|endoftext|> TITLE: Revisiting Gordon-Luecke theorem QUESTION [11 upvotes]: $\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\GL{GL}$Here is an proof-sketch of a strengthened Gordon-Luecke theorem. This is presumably known, but is it written down somewhere? I am also curious if there is a pre-geometrization proof of this statement, in roughly this generality. Theorem: Let $L$ be an $n$-component link in the 3-sphere. Let $M$ be the exterior of the link, i.e. $S^3 \setminus \nu L$ where $\nu L$ is an open tubular neighbourhood of $L$. Then there is a natural representation given by restricting diffeomorphisms to the boundary $$\pi_0 \Diff(M) \to \pi_0 \Diff(\partial M).$$ This representation has infinite image if and only if the link $L$ contains a split unknot, or a split Hopf link. Notation: $\Diff(M)$ means to the full group of diffeomorphisms of the knot exterior, so $\pi_0 \Diff(M)$ means isotopy-classes, i.e. the mapping class group. Similarly $\pi_0 \Diff(\partial M)$ is the mapping class group of the boundary, i.e. a disjoint union of tori. So this group is the wreath product $\GL_2 \mathbb Z\wr\Sigma_n$, i.e. one can permute tori and act by automorphisms of tori. Proof: The unknot exterior is $S^1 \times D^2$ and the Hopf link exterior is $S^1 \times S^1 \times I$, and both admit mapping-classes of infinite order — generalized Dehn twists. So what remains to be shown is that if a link does not have one of these as split sublinks, then the representation is of finite order. The idea is to prove it using geometrization. For links whose exteriors are Seifert-fibered manifolds this is a direct computation using the Seifert data and their mapping class groups. For hyperbolic links the mapping class group is finite, so the image is finite. For satellite links, we can take the subgroup of finite index that preserves all the 3-manifolds in the JSJ-decomposition, and these are Seifert-fibered or hyperbolic link exteriors, so the result follows from the previous two steps. Lastly, unknot and Hopf link exteriors do not arise via torus splittings i.e. the JSJ decomposition. Off the top of my head the only extensions to Gordon-Luecke that I know of concern the question of if two distinct links can have diffeomorphic complements. While this is related to my question, it only would appear to be "in the neighbourhood" and not really on exactly the same topic. REPLY [6 votes]: I think that the correct theorem should be the following: If $L$ does not contain any split unknot or any two coaxial components, the image of the map is finite Following Cameron Gordon, two coaxial components are two components $K_1$ and $K_2$ such that $K_1$ is an essential curve in the boundary $\partial N(K_2)$ of a solid torus regular neighbourhood of $K_2$. We also suppose that no other component of $L$ intersects $N(K_2)$. Here is one example: If you have two coaxial components, there is an incompressible annulus in $S^3\setminus L$ which is contained in $N(K_2)$, which winds once along $K_1$ and some $n$ times along $K_2$. The Hopf link corresponds to the case where $K_2$ is the unknot and $n=0$. Every time you have an incompressible annulus that connects two different components of the link you can do a Dehn twist along it and act non-trivially on the boundary. It has infinite order. I think that the proof should go as follows: the only infinite-order automorphisms in a 3-manifold with boundary that arise as a link complement should be twists along discs or annuli. So $S^3\setminus L$ contains an essential disc or annulus. In the second case, since we are in $S^3$, the annulus arises necessarily from two coaxial components.<|endoftext|> TITLE: When does $C(X)$, $X$ a continuum, admit a continuous choice function? QUESTION [8 upvotes]: Given a continuum $X$ (compact metrizable connected $X$) let $K(X)$ denote the hyperspace of nonempty compact subspaces of $X$ with the Vietoris topology and let $C(X)$ denote the (closed) subspace of $K(X)$ consisting of connected sets. It is well known (and has been discussed on MO before) that if there is a continuous choice function $K(X)\to X$ then $X\simeq[0,1]$, indeed any other continua doesn’t even have a continuous choice function for the subspace of $K(X)$ of unordered pairs. The situation is different when looking at $C(X)$: for example the “star” $X$ obtained by joining $n$ arcs at their endpoint has a continuous selection for $C(X)$ by looking at the tree partial order induced by choosing the common point as root and then mapping $C\in C(X)$ to its $\min$ with respect to this partial order. In a similar way we get a continuous selection for $C(X)$ whenever $X$ is a dendrite. It is also easy to see that $S^1$ doesn’t have a continuous selection for $C(S^1)$ by using a combination of Borsuk-Ulam, $C(S^1)\simeq D^2$ and $\partial C(S^1)\cong F_1(S^1)$, where $D^2$ is the unit disk in $\Bbb R^2$ and $F_1(X)$ is the space of singletons of $X$. This suggests the following: suppose $X$ is a continuum with a continuous selection $C(X)\to X$. Must $X$ be a subcontinuum of a dendrite? If not is there a characterization of the continua $X$ with a continuous selection $C(X)\to X$? REPLY [2 votes]: I will post a CW answer to my own question to remove it from the unanswered list since Benjamin doesn't seem interested in turning his comment into an answer, but this is all informations taken from the section of Hyperspaces that he suggested. To begin with let's fix the terminology selectible to refer to a continuum $X$ for which there exist a continuous selection $C(X)\to X$. The simple answer to the question at the end, namely "is there a characterization of the selectible continua?" is no. What is known (according to the book by Illanes and Nadler, I'm not aware of any recent developement however) is more or less the following: If $X$ is a selectible continuum, then $X$ is a dendroid. Say that a selection $s\colon C(X)\to X$ is rigid if whenever $A,B\in C(X)$ are such that $s(B)\in A\subseteq B$, then $s(A)=s(B)$ (note that the definition of rigid selection in Hyperspaces has a typo, this is the correct definition as given in the original reference Rigid Selections and Smooth Dendroids by L.E.Ward). Then there is a characterization of the rigidly selectable continua: those are exactly the smooth dendroids by the paper of Ward mentioned above. The idea here is that if $X$ is a smooth dendroid then the "tree" partial order (with root a point witnessing smoothness) is closed in $X\times X$, so the map sending each subcontinuum to its minimum is a rigid selection. Conversely if $s\colon C(X)\to X$ is a rigid selection, then $\{\{s(A)\}\times A\mid A\in C(X)\}$ is a closed partial order on $X$ which agrees with the "tree" order with root $s(X)$, so that $s(X)$ is a point witnessing the smoothness of $X$. Not all dendroids are selectible. Given two points $x,y$ in a dendroid let $[x,y]$ denote the unique arc joining them and let $(x,y)$ denote $[x,y]\setminus\{x,y\}$. Following definition 75.10 in Hyperspaces say that a dendroid $X$ is of type $N$ (a drawing justifies this name and clarifies the definition) if there are two points $p,q\in X$ and two sequences of arcs $[p_np_n'],[q_n,q_n']$ with points $p_n''\in(q_nq_n')$ and $q_n''\in(p_n,p_n')$ such that: $$\begin{align} [p,q] &=\lim [p_n,p_n']=\lim [q_nq_n'] \\ p &= \lim p_n = \lim p_n' = \lim p_n'' \\ q &= \lim q_n = \lim q_n' = \lim q_n''\end{align}.$$ Then any dendroid of type $N$ is nonselectible. A simple example of a dendroid of type $N$ is the doubly infinite broom: start with the segment between $(0,0)$ and $(1,0)$ in $\Bbb R^2$ and add segments between $(0,0)$ and $(1,1/n)$ and between $(1,0)$ and $(0,-1/n)$ for all $n\in\Bbb N^+$. More surprisingly there are even contractible dendroids that are not selectible, this is example 75.9 in Hyperspaces and the construction is originally due to Maćkowiak. selectibility is not preserved by neither monotone nor open maps. Nonselectibility isn't either, this is exercise 75.28 in Hyperspaces. There are various open questions about selectibility apart from a classification of selectible continua, some of them are in chapter 76 of Hyperspaces.<|endoftext|> TITLE: Subobjects as an object in a topos QUESTION [5 upvotes]: Forgive me if this question turns out to be too elementary-then feel free to move it to stack exchange. I believe that this should be very basic fact from topos theory nevertheless being not familiar with topos theory let me ask it. One of the highlights of topos theory is the possibility to consider an analogue of the power set in the form of $Sub(A)=\{all \ subobjects \ of \ A \}$ (which are defined as equiavlence classes of monic arrows). In any topos there is a distinguished object $\Omega$ called the subobjects classifier. Moreover in any topos there is an analogue of the operation of taking all functions from $A \to B$-this is called exponentation. Given those facts we have a one to one correspondence $Sub(A) \cong \Omega^{A}$. Am I right in saying that: The left hand side of this correspondence is a priori not an object of our topos $\mathcal{T}$ but the right hand side is: therefore this natural correspondence is the way of making $Sub(A)$ again an object in our category $\mathcal{T}$ Please corect me if I'm wrong REPLY [4 votes]: To 'turn an object $A$ of a category $\mathcal{C}$ into a set' you can consider the set of objects mapping into/out of that object from/to some other object; if you consider maps out of a terminal object ${\bf 1}\to A$ these are called global elements of the object. If $\mathcal{C}$ is a well-pointed topos these global elements are sufficient to delineate between parallel arrows with domain $X$ and thusly contain 'all relevant information' about $X$ in the topos. As Paul pointed out in his answer and Andrej elaborated on in his comment, if you consider $\Omega^A$ as a set in this way by considering its set of global elements $\{f:{\bf 1}\to \Omega^A|f\in\mathcal{C}\}$, this set is isomorphic to $Sub(A)$ as a lattice. You may also want to read about the subobject fibration, which deals with the notion of categorical subobjects in a more satisfying way imo; you can turn the subobjects of an object into a category in an obvious way, and it turns out these categories are also the fibers of the fibration obtained by turning the codomain fibration into an indexed category via the Grothendieck construction, postcomposing with the skeleton endo $2$-functor on $\mathfrak{Cat}$, then turning the resulting indexed category back into a fibration via Grothendieck in the other direction.<|endoftext|> TITLE: Total sum of squares of characters of the symmetric group $\mathfrak{S}_n$ QUESTION [7 upvotes]: In my earlier MO post, I proposed the double sum $\sum_{\mu\vdash n}\sum_{\lambda\vdash n}\chi_{\mu}^{\lambda}$ regarding characters of the symmetric group $\mathfrak{S}_n$. Soon after, I started considering the sum of squares $\sum_{\mu\vdash n}\sum_{\lambda\vdash n}(\chi_{\mu}^{\lambda})^2$ hoping to gain a better formula. A further look into older MO posts here and also here shows a Burnside-type Lemma $$\frac{1}{n!} \sum_{\alpha \in \frak{S}_n} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$ After comparing the last two sums, I got prompted to ask: QUESTION. The numerics suggest the below equality. Why is this true? $$\sum_{\mu\vdash n}\sum_{\lambda\vdash n}(\chi_{\mu}^{\lambda})^2 =\frac{1}{n!} \sum_{\alpha \in \frak{S}_n} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$ REPLY [3 votes]: I think this is true. Let $X$ be the character table of $S_{n}$, viewed just as a $p(n) \times p(n)$ matrix. Then the left side is ${\rm trace}(XX^{T}) = {\rm trace}(X^{T}X).$ This is $\sum_{\alpha} |C_{S_{n}}(\alpha)|$, where $\alpha$ runs over a set conjugacy class representatives for $S_{n}.$ As mentioned in my comment, the right side is clearly $\frac{1}{n!} \left( \sum_{\alpha} |S_{n}| |C_{S_{n}}(\alpha)| \right),$ where $\alpha$ again runs through class representatives for $S_{n},$ so both sides are equal. Later edit: For this result, the only special property of $S_{n}$ used is that its complex irreducible characters are all real valued. The following similar equality is true for a general finite group $G$ by a slight modification of the argument above (a straightforward consequence of the orthogonality relations). $$\sum_{(\chi,x)}|\chi(x)|^{2} = \frac{1}{|G|}\sum_{y \in G} \left( \sum_{\chi} |\chi(y)|^{2} \right)^{2},$$ where $\chi$ runs through the distinct irreducible complex characters of $G$ and $x$ runs over a set of representatives for the conjugacy classes of $G$.<|endoftext|> TITLE: Does the language of fibred categories gives the commutativity of the diagrams in Residues and Duality? QUESTION [11 upvotes]: In Residues and Duality, R. Hartshorne and A. Grothendieck say that there are a plethora of compatibilities that need to be shown in order to have a six functor formalism. For example, if $f:X\to Y$, $g:Y\to Z$, and $h:Z\to W$ are morphisms of ringed spaces, we have isomorphisms and we would like to be sure that this diagram commutes. While it's simple to see that this diagram commutes, B. Conrad says in Grothendieck Duality and Base Change that some of these compatibilities are quite difficult to verify. Finally, R. Hartshorne and A. Grothendieck says that "perhaps the language of fibred categories or the techniques of Giraud's thesis will supply what is needed". I wonder what is the state of the art in this subject and, particularly, if fibred categories are any helpful. P.S.: the book Triangulated Categories of Mixed Motives by D. C. Cisinski and F. Déglise uses this language to study six functor formalisms but, even though I tried, I admit that it is not clear to me if it solves this problem or not. REPLY [5 votes]: If you have an abelian category $A$ with enough injective, for any functor defined on bounded below cochain complexes $\Phi\colon C^+(A)\to D$ sending cohain homotopy equivalences between degree-wise injective cochain complexes to isomorphism, there is a total right derived functor $R\Phi\colon D^+(A)\to D$. This means that there is a natural transformation $\eta_K\colon\Phi(K)\to R\Phi(K)$ for any bounded below cochain complex $K$ and that the pair $(R\Phi,\eta)$ is universal (initial) among all possible choices (i.e. it is the left Kan extension of the functor $\Phi$ along the localization functor from cohain complexes to the derived category). Furthermore, $R\Phi$ is not only a derived functor, but a universal one: this means that, for any functor $\Psi:D\to E$, the pair $(\Psi\circ R\Phi,\Psi(\eta))$ is the right derived functor of the composition $\Psi\circ\Phi$. All this works the same for unbounded complexes is $A$ is an abelian category and if we replace degree-wise injective cochain complexes by a suitable notion (e.g. fibrant of objects in a suitable Quillen model structure). In particular, in practice, an identification of the form $R(g_*\circ f_*)\cong Rg_*\circ Rf_*$ never comes out of the blue, but rather always as derived functors: the isomorphism above is the unique map which is compatible with the canonical map $(g_*\circ f_*)(K)\to R(g_*\circ f_*)(K)$ and the image by $Rg_*$ of the canonical map $f_*(K)\to Rf_*(K)$ (functorially in $K$). Furthermore, even better, for any functor $\Psi$ whose domain is the codomain of $Rg_*$, the isomorphism $$\Psi(R(g_*\circ f_*)(K))\cong \Psi(Rg_*\circ Rf_*(K))$$ is the unique map which is compatible with the image by $\Psi$ of the canonical map $(g_*\circ f_*)(K)\to R(g_*\circ f_*)(K)$ and of the image by $Rg_*$ of the canonical map $f_*(K)\to Rf_*(K)$ (functorially in $K$). For the compatibility problem you mention, we have to compare two maps from $R(h_* g_* f_*)$ to $Rh_* Rg_* Rf_*$. Using the universal property of $R(h_* g_* f_*)$, we just have to check that, for any cohain complex $K$, the two induced maps $$(h_* g_* f_*)(K)\to (Rh_* Rg_* Rf_*)(K)$$ agree, which comes right away from the compatibilities mentioned before. As final remark: all this reasoning can be made in much more general context (e.g. with derived functors defined on homotopy categories of model categories). This extra level of generality means that all the reasoning takes place in a context where there are no possiblities to make sign mistakes because signs do not exist. All the ingredients of the proof are available since the late 1960's.<|endoftext|> TITLE: A question related to fiber bundle QUESTION [5 upvotes]: Let $f:\mathbb{C}^3 \to \mathbb{C}$ be a morphism of varieties such that it is a smooth fiber bundle. Can I say that the fiber is $\mathbb{C}^2$? REPLY [2 votes]: Note that $f$ is smooth as a map of varieties. If $f$ comes from a direct product $\mathbb{C}^3\approx S\times \mathbb{C}$ then $S\approx \mathbb{C}^2$. See Section 5.1 of $\mathbb{A}^1$-homotopy theory and contractible varieties: a survey I do not see why $f$ must come from a direct product so this is not a complete answer.<|endoftext|> TITLE: Convex hull of a variety in real space QUESTION [5 upvotes]: I am a physicist currently working on a question posed as part of an algebraic geometric description of a physical set: I did not find a question that is closely related to what I am searching for yet, but please feel free to just post the link below. My question concerns semialgebraic sets that are the convex hulls of affine varieties. Imagine I am given an ideal $I_n$ of $k$ polynomials in $\mathbb{R}^n$ whose zero locus defines my variety. The variety is the image of all extreme points of a convex set in $\mathbb{R}^{m+n}$ given by the $n$-th elimination ideal of some ideal $I$. As I experienced, it is not possible to recover the full set of Boolean combinations needed to describe the projection of my original set this way.(If you have any experience in that and know something that could help me, please let me know. I already tried Cylindrical Algebraic Decomposition but it turned out to be too complex.) Weakening my expectations, is it correct that if I take a point $P\in\mathbb{R}^n$ that from my preliminaries cannot be part of the projection and demand the ideal $I_n$ to be prime, I can find a sign condition on each of the $k$ polynomials $f_i$ by evaluating $f_i(P)$? Then $Q\in\text{conv}(Z(I_n))$ if $\lnot f_i(Q)*_i0$ for all $i=1,...,k$ where $*_i$ is the inequality coming from $f_i(P)*_i0$. I thought it is possible to use Separation Theorem from Convex Geometry. The only subtlety is how to deal with singular points? I appreciate any kind of help! REPLY [5 votes]: These references may help? Or at least lead you to related literature. João Gouveia and Rekha Thomas. "Convex hulls of algebraic sets." In Handbook on Semidefinite, Conic and Polynomial Optimization, pp. 113-138. Springer, Boston, MA, 2012. Pre-pub arXiv version. "The main feature of the technique is that all computations are done modulo the ideal generated by the polynomials defining the set to the convexified."       Ranestad, Kristian, and Bernd Sturmfels. "The convex hull of a variety." In Notions of Positivity and the Geometry of Polynomials, pp. 331-344. Springer, Basel, 2011. Pre-pub arXiv version.<|endoftext|> TITLE: Are the fibers of a surjective holomorphic submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic? QUESTION [12 upvotes]: Are the fibers of a surjective holomorphic submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic? For $n=1$ this means that a surjective entire function $\mathbb{C}\to\mathbb{C}$ without critical points assumes each value infinitely often. Is this obvious? REPLY [14 votes]: The answer is no for $n=1$. Lemma: Suppose $f$ is an entire function with $f^{-1}(z_0)$ finite non-empty for some $z_0 \in \mathbb{C}$. Then $f$ is surjective. Proof: By Picard, $f$ misses at most one value. Up to translating $f$ by a scalar (which obviously preserves the hypothesis), we may assume $f$ misses $0$. Then $f = e^g$ for some entire $g$. By assumption, $z_0 \neq 0$, so $\exp^{-1}(z_0)$ is infinite, so by Picard, $g^{-1}(\exp^{-1}(z_0)) = f^{-1}(z_0)$ is infinite, which is a contradiction. Now take $q(z) := \sum_{n \geq 1} \frac{z^n}{n\cdot n!}$ and $f(z) := ze^{q(z)}$. I claim this $f$ yields a contradiction. Observe that $f(z) = 0$ if and only if $z = 0$. Therefore, the lemma applies, and we see $f$ is surjective. Moreover, we see that not all fibers of $f$ are in bijection: $f^{-1}(0)$ is a singleton, but by (great) Picard (applied to $f(\frac{1}{z})$), $f^{-1}(z)$ is infinite for any other value of $z$. Finally, let's check that $f$ is a submersion. We have: $$df = (1+zq')e^q dz.$$ We have $1+zq' = e^z$ by choice of $q$, so this is $e^{z+q}dz$, which is clearly a nowhere vanishing differential.<|endoftext|> TITLE: Can the theory of elliptic functions developed from purely geometric considerations? QUESTION [6 upvotes]: I always had this question, but was unable to get a definitive answer to it. There is the theorem of division of the arc length of the lemniscate with ruler and compass. So I always wondered, is it possible to reconstruct the theory of elliptic functions from purely geometric considerations? If this is possible, then to what extent? Geometric means geometric constructions to prove main formulas but without use of functions of complex variables. Note that I don't imply that this geometric theory would be practical. From modern standpoint, this construction probably will be considered as a waste of time. I'm interested in the theoretical possibility of such a development. Also I'm interested in understanding the intuition behind such a development, why it is possible? REPLY [3 votes]: Some elementary parts of the theory of elliptic functions can indeed be developed in this way. To those books listed by Alexey Ustinov I can add a large treatise by G. Halphen, Traité des fonctions elliptiques et de leurs applications (JFM 22.0447.01), (BnF) in 3 volumes, published in 1886-1890. He never misses a opportunity to give a geometric proof of some theorem, when available.<|endoftext|> TITLE: When are those polynomials irreducible? QUESTION [7 upvotes]: Let $f_n (x) := x^n - x^{n-1} - x^{n-2} - ... - x^2 - x - 1$, which is an irreducible polynomial by corollary 2.2 of https://www.sciencedirect.com/science/article/pii/S0022404903002457. Question: For which $m,n \geq 2$ is the polynomial $f_n(x^m)$ irreducible over the integers? Especially interesting is the case when $n$ (or $m$) is a prime. Question 2: Is $f_n(x^m)$ irreducible whenever $n$ is a prime? Computer experiments found no counterexample so far. REPLY [8 votes]: We show that for all $m, n \geq 2$ the polynomial $f_n \left( x^m \right)$ is irreducible. This answer is based on a wonderful technique I learnt a few years ago from an answer on MSE by Keith Conrad, and I encourage the reader to look at that answer first (it is not very long), as it shows the key ideas in a much nicer setup. I do here a lot of casework which can probably be shortened, but I didn't really attempt to streamline it. If $$p(x) = \sum_{i = 0}^{n} a_i x^i$$ is a polynomial of degree $n$, then we define $$\tilde{p} (x) = \sum_{i = 0}^{n} a_{n - i} x^i$$ that is the polynomial with coefficients reversed. Algebraically, $\tilde{p} (x) = x^{\mathrm{deg} (P)} p \left( \frac{1}{x} \right)$. Notice that $$(x - 1) f_n (x) = x^{n + 1} - 2 x^n + 1$$ Therefore, to understand the factorization of $f_n \left( x^m \right)$, it suffices to understand the factorization of $$x^{n m + m} - 2 x^{n m} + 1$$ Fix $n, m$ and from now on we will call this polynomial $f(x)$. Throughout, we will assume that $n > 2$: if $n = 2$ the method that we use shows that for all $m$ the polynomial $f_{n} \left( x^m \right)$ is irreducible. Suppose that there was a factorization into nonconstant monic polynomials $g, h$ $$f = g h$$ Then, taking $k = g \tilde{h}$ or $k = - g \tilde{h}$ (according to whether $g(0) = 1$ or $- 1$ respectively), we see that $k$ is a monic polynomial such that $$f \tilde{f} = k \tilde{k}$$ We will show that this implies that $k = f$ or $k = \tilde{f}$. Write $$k(x) = \sum_{i = 0}^{n m + m} a_i x^i$$ and then $$\tilde{k} (x) = \sum_{i = 0}^{n m + m} a_{n m + m - i} x^i$$ We know that $a_{n m + m} = 1$, and by looking at the constant coefficient of $k \tilde{k} = f \tilde{f}$ we see that $a_0 = 1$. Now, the key point: compare the coefficient of $x^{n m + m}$ in $f \tilde{f} = k \tilde{k}$: it is the sum of the squares of the coefficients of $f$ and $k$ respectively, therefore $$\sum_{i = 0}^{n m + m} a_{i}^2 = 6$$ It will be useful in the cases below to see explicitly what $f \tilde{f}$ is: $$f \tilde{f} = x^{2 m n + 2 m} - 2 x^{2 m n + m} - 2 x^{m n + 2 m} + 6 x^{m n + m} - 2 x^{m n} - 2 x^{m} + 1$$ From the coefficients of $k$ we already now, there are two possibilities: Case 1: There exist $m n + m > a > b > c > d > 0$ such that $$k(x) = x^{n m + m} \pm x^a \pm x^b \pm x^c \pm x^d + 1$$ Looking at the coefficient of $x^{2 n m + 2 m - 1}$ in $k \tilde{k}$ we see that we must have $a = n m + m - 1, \ d = 1$ and $x^a, x^d$ must both come with a minus sign. We must have $k(1) = 0$, and so without loss of generality (otherwise we look at $\tilde{k}$) $x^b$ comes with a plus sign and $x^c$ with a minus. Now, $$k \tilde{k} = \left( x^{m n + m} - x^{m n} + x^b - x^c - x^m + 1 \right) \times \\ \left( x^{m n + m} - x^{m n} - x^{m n + m - c} + x^{m n + m - b} - x^m + 1 \right)$$ Looking at the coefficient of $x^{2 m n}$, since $n > 2$ the positive contribution of $\left( - x^{m n} \right) \left( x^{- m n} \right)$ must cancel out, and the only options are $c = 2 m, m n - m$. In particular $c$ is divisible by $m$. Substituting $x$ to be a primitive $m$-th root of unity, we get that $x^b = 1$, and therefore $b$ is divisible by $m$ as well. Since $m n > b > c$, this rules out the option of $c$ being $m n - m$, and therefore we have $c = 2 m$. Now looking at the coefficient of $x^{m n}$ we see that we must have $b = m n - m$, but this means that the coefficient of $x^{2 m n}$ is wrong, and so we get a contradiction. Therefore, what happens is Case 2: for some $m n + m > a > 0$, we have $k(x) = x^{m n + m} \pm 2 x^a + 1$. It is immediate to see in this case that $a = mn, m$ and $x^a$ comes with a minus sign, that is $k = f, \tilde{f}$ and without loss of generality, $k = f$. Therefore $g h = \pm g \tilde{h}$, that is $h = \pm \tilde{h}$. However, notice that $$\gcd \left( f, \tilde{f} \right) = \gcd \left( x^{m n + m} - 2 x^{m n} + 1, x^{m n - m} - 1 \right) = \\ = \gcd \left( x^{2 m} - 2 x^{m} + 1, x^{m n - m} - 1 \right)$$ It is easy to see that the only roots of $x^{2 m} - 2 x^m + 1$ which are on the unit circle are the $m$-th roots of unity, and they each appear with multiplicity $1$, therefore $\gcd \left( f, \tilde{f} \right) = x^m - 1$. Since $h | f, \ \tilde{h} | \tilde{f}$ we have $h | \gcd \left( f, \tilde{f} \right) = x^m - 1$. Over all, we have shown that in every nontrivial factorization of $\left( x^m - 1 \right) f_n \left( x^m \right) = g h$, one of the factors on the right hand side divides $x^m - 1$, which immediately implies that $f_n \left( x^m \right)$ is irreducible: if there was a factorization $f_n \left( x^m \right) = u(x) v(x)$ then $g(x) = u(x), \ h(x) = \left( x^m - 1 \right) v(x)$ would be a counterexample. QED<|endoftext|> TITLE: What is a function field analog of Giuga's conjecture? QUESTION [7 upvotes]: Giuga's conjecture (1950), which is still open and has strong numerical support, reads : Let $n$ be a positive integer. If $1+\sum_{k=1}^{n-1}k^{n-1} \equiv 0\pmod{n}$ then $n$ is prime. What would an analog for function fields be? REPLY [11 votes]: Here is a proof of the experimental fact Joe Silverman discovered. Let $f$ be a polynomial of degree $d$ in $R:=\mathbb{F}_q[x]$. For $I=(f)$, the sum $$1+\sum_{a \in R/I} a^{\# R/I - 1}$$ is equal, modulo $f$, to $$1+\sum_{g \in R,\, \deg g < d} g^{q^d-1}.$$ Let $$F_d(x) := 1+\sum_{g \in R,\, \deg g < d} g^{q^d-1} \in R.$$ We shall show that if $f$ is irreducible than $f \mid F_d$ and if $f$ is reducible than $f \nmid F_d$. First, suppose that $f$ is reducible. Let $\alpha$ be a root of $f$, with minimal polynomial $m_{\alpha} \mid f$ and $\deg m_{\alpha} = e < d$. We can write any $g \in R$ of degree $ TITLE: Ideals of $F_2[x_1, x_2, \cdots, x_n]/(x_1^2, x_2^2, \cdots x_n^2)$ QUESTION [6 upvotes]: I am interested in the poset of all ideals of the local ring $$R_n = \mathbb{F}_2[x_1, x_2, \cdots, x_n]/(x_1^2, x_2^2, \cdots x_n^2).$$ $n=1$ is trivial. $n=2$ takes little work and it is shown below. But beyond that, it is getting tedious. Is there a description of this lattice of ideals in $R_n$? Or can this be computed using some software? I am actually interested in computing all quotients of $R_n$. Any help would be appreciated. Thank you! REPLY [3 votes]: The ideals correspond to submodules of the regular module and as this is a finite dimensional commutative quiver algebra, you can obtain all submodules using the GAP package QPA (at least for finite fields). Example for $n=2$ over a finite field with $2^t$ elements for $t \geq 1$: t:=1;;K:=GF(2^t);Q:=Quiver(1,[[1,1,"x1"],[1,1,"x2"]]);KQ:=PathAlgebra(K,Q);AssignGeneratorVariables(KQ);rel:=[x1^2,x2^2,x1*x2-x2*x1];A:=KQ/rel;RegA:=IndecProjectiveModules(A)[1];W:=AllSubmodulesOfModule(RegA);WW:=Flat(W);Size(WW); The output are the ideals with given as inclusions into the regular module, here: [ [ <<[ 0 ]> ---> <[ 4 ]>> ], [ <<[ 1 ]> ---> <[ 4 ]>> ], [ <<[ 2 ]> ---> <[ 4 ]>>, <<[ 2 ]> ---> <[ 4 ]>>, <<[ 2 ]> ---> <[ 4 ]>> ], [ <<[ 3 ]> ---> <[ 4 ]>> ], [ <<[ 4 ]> ---> <[ 4 ]>> ] ],[ <<[ 0 ]> ---> <[ 4 ]>>, <<[ 1 ]> ---> <[ 4 ]>>, <<[ 2 ]> ---> <[ 4 ]>>, <<[ 2 ]> ---> <[ 4 ]>>, <<[ 2 ]> ---> <[ 4 ]>>, <<[ 3 ]> ---> <[ 4 ]>>, <<[ 4 ]> ---> <[ 4 ]>> ] The number of ideals for a finite field with $2^t$ elements and $n=2$ starts for $t \geq 1$ with 7, 9, 13, 21, 37, 69, 133 and thus seems to be given by https://oeis.org/A168614 and depends strongly on the field (in general it seems the number of ideals for a finite field with $p^t$ elements is given by $p^t+5$). The number of ideals increases with the number of elements of the finite field. Thus there surely will be infinitely many ideals (even up to isomorphism) for an infinite field, for example over a field of characteristic zero the ideals $(x_1+x_2)$ and $(q x_1 + x_2)$ are not isomorphic for $q \neq 1$. You can also compute all quotients with QPA, having the inclusion maps.<|endoftext|> TITLE: Are there some interesting propositions independent with ZF+V=L that do not increase consistency strength? QUESTION [8 upvotes]: In some MO questions such as this and this, Hamkins gave some examples that is independent with ZF+V=L, however, all of them increase the consistency strength. Are there some propositions P, which is interesting in some field of mathematics, and is independent with ZF+V=L, and con(ZFC) proves con(ZF+V=L+P) and con(ZF+V=L+¬P)? REPLY [10 votes]: As per @TimothyChow 's suggestion, I'm posting my comment as an answer. This is essentially unknown. The only real way we know to produce independence results without increasing consistency is by forcing, which emphatically doesn't allow one to preserve V=L. This is discussed in some detail in Shelah's Logical dreams, see https://arxiv.org/pdf/math/0211398.pdf, 4.8 Dream.<|endoftext|> TITLE: Different Bialgebra/Hopf algebra structures on coalgebras QUESTION [7 upvotes]: Given a coalgebra $C$, can there exist more than one algebra structure on $C$ giving it the structure of a bialgebra? I will also ask the same question for Hopf algebras. REPLY [7 votes]: By dualizing, you are looking for an algebra $A$ such that there are two different coproducts $\Delta_1, \Delta_2: A \to A\otimes A$ that make $A$ into a bialgebra resp. Hopf algebra. As an example let $p$ be prime, $k$ a field of char. $p$ and let $A=k[x_1,...,x_n]/(x_1^p,...,x_n^p)$ be the truncated polynomial algebra. $A$ is the group algebra of the elementary abelian group $(\mathbb{Z}/p)^n=\langle g_1,...,g_n\rangle$ via $x_i=g_i-1$. Thus $A$ is a Hopf with coproduct $\Delta_1(x_i)=x_i\otimes 1 + 1\otimes x_i + x_i\otimes x_i$ and antipode $S_1(x_i)=g_i^{-1}-1=(-x_i)+\cdots +(-x_i)^{p-1}$. $A$ is also the restricted enveloping algebra of $k^n$ considered as trivial restricted $p$-Lie algebra. As such $A$ is a Hopf algebra with coproduct $\Delta_2(x_i)=x_i \otimes 1 + 1\otimes x_i$ and antipode $S_2(x_i)=-x_i$. These Hopf algebras where used by Avrunin and Scott in their proof of Carlson's conjecture on the rank variety in group cohomology. For a reference see section 4 of Carlson, Iyengar: Hopf algebra structures and tensor products for group algebras<|endoftext|> TITLE: Is there an even number $a$ such that $a^{2^{n}}+1$ is prime for infinitely many $n$? QUESTION [5 upvotes]: Is there an even number $a$ such that $\{n: a^{2^{n}}+1 \text{ is prime} \}$ is an infinite set? Let $a$ be even. Is there infinitely many $n$ such that $a^{2^{n}}+1$ is composite? REPLY [12 votes]: A remark that might be noteworthy... Either the sequence $\{2^{2^{n}}+1\}_{n \in \mathbb{N}}$ contains infinitely many composite numbers or the sequence $\{6^{2^{n}}+1\}_{n \in \mathbb{N}}$ contains infinitely many composite numbers. Proof. If there are only finitely many composite numbers in the first sequence (the sequence of Fermat numbers), we might assure the existence of an $n_{0} \in \mathbb{N}$ such that $2^{2^{n}}+1$ is a prime number for every $n$ that belongs to $\mathcal{I}:=[n_{0},\infty) \cap \mathbb{N}$. We claim that, in such a case, $6^{2^{n}}+1$ is a composite number for every $n \in \mathcal{I}$. Indeed, for any given $n \in \mathcal{I}$, the Fermat number $2^{2^{n}}+1$ is a prime and Pepin's test gives us that $$3^{2^{n-1}} \equiv -1 \pmod{2^{2^{n}}+1}.$$ Taking squares on both sides of the previous congruence we get $3^{2^{n}} \equiv 1 \pmod{2^{2^{n}}+1}$; from this information and the fact that $2^{2^{n}} \equiv -1 \pmod{2^{2^{n}}+1}$, we obtain that $$6^{2^{n}} \equiv -1 \pmod{2^{2^{n}}+1}$$ and the validity of our claim follows.<|endoftext|> TITLE: Does there exist some $p(x) \in \mathbb{Q}[x]$, deg$(p) > 1$, which maps $\mathbb{Q}$ onto itself surjectively? QUESTION [11 upvotes]: Clearly this is impossible for $p$ of even degree, and I imagine that Cardano’s formula quickly reveals it to be impossible in the cubic case, although I have not checked in detail. My guess is that no such $p$ exists. Does one exist? If so, is there an explicit example? Failing a general yes or no answer, are there sufficient conditions to identify a non-surjective polynomial function? REPLY [8 votes]: I claim that no polynomial $q$ of degree greater than $1$ and rational coefficients can be a surjective mapping from $\mathbb{Q}$ to $\mathbb{Q}$. Suppose that a polynomial $q$ of degree greater than $1$ is surjective from $\mathbb{Q}$ to $\mathbb{Q}$. For simplicity, by replacing $q(x)$ with $p(x)=\alpha(q(\beta x)-\gamma)$ where $\alpha,\beta,\gamma$ are rational with $\alpha,\beta\neq 0$, we can assume that $p(x)$ is a surjective monic polynomial with constant term $0$ and integer coefficients. Suppose now that $p(x)=x^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x$ where the coefficients $a_{1},\dots,a_{n-1}$ are integers. If $\alpha,\beta$ are integers with $p(x)=\frac{\alpha}{\beta}$, then $\beta x^{n}+\dots+\beta a_{1}x=\alpha$, so by the rational root theorem, $x$ must be of the form $\frac{r}{s}$ where $r$ is a factor of $\alpha$ and $s$ is a factor of $\beta$. In particular, in the case where $\beta=1$, if $p(x)=\alpha$, then $x$ must be a factor of $\alpha$. Therefore, $p$ must restrict to a surjective function from $\mathbb{Z}$ to $\mathbb{Z}$. This is impossible.<|endoftext|> TITLE: Banach spaces whose second conjugates are separable QUESTION [8 upvotes]: It was known that the James space $J$ has separable second conjugate, is non-reflexive and isometric to its second conjugate. I want to know whether there are Banach spaces $X$ with separable second conjugates $X^{**}$, but $X$ is not a dual space (the James space $J$ is a dual space). Furthermore, are there any references about Banach spaces with separable second conjuates ? Thank you ! REPLY [3 votes]: This post is intended not as an answer, but rather to list several Banach space properties of $X$ given above (that is $X^{**}$ is separable, $X$ is not a dual space). Clearly $X$ is not reflexive. Clearly $X$ and $X^{*}$ are also separable. Being separable dual spaces, $X^{*}$ and $X^{**}$ have RNP (Radon-Nikodym property). $X$ also has RNP, being a closed subspace of a space with RNP. $X$, $X^{*}$ do not contain copies of $\ell^1$. $X$, $X^{*}$, $X^{**}$ do not contain copies of $c_0$. $X$, $X^{*}$ do not possess an unconditional basis. $X$, $X^{*}$ are not w.s.c. (weakly sequentially complete) since any Banach space which is w.s.c. and contains no copy of $\ell^1$ is reflexive. $X$, $X^{*}$ do not have Schur property, for any Banach space that has Schur property and contains no copy of $\ell^1$ is finite dimensional. $X$ does not have DPP (Dunford-Pettis property), since otherwise $X^{*}$ would have Schur property. Consequently, $X^{*}$ and $X^{**}$ do not have DPP (and thus not have Schur property) $X$, $X^{*}$, $X^{**}$ do not have Pełczyński property (V), for any Banach space that has property (V) and contains no copy of $c_0$ is reflexive. $X$, $X^{*}$ have Dieudonne property, for any Banach space that contains no copy of $\ell^1$ has Dieudonne property. $X$, $X^{*}$, $X^{**}$ are not Grothendieck spaces, since every separable Grothendieck space is reflexive. The properties above are also shared by the James space, since we merely used that "$X$ is not a dual space" to deduce that $X$ is not reflexive.<|endoftext|> TITLE: Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgroups? QUESTION [5 upvotes]: Let $\mathrm{cov}_H(C_2^\omega)$ be the smallest cardinality of a cover of the Boolean group $C_2^\omega=(\mathbb Z/2\mathbb Z)^\omega$ by closed subgroups of infinite index. It can be shown that $$\max\{\mathrm{cov}(\mathcal M),\mathrm{cov}(\mathcal N)\}\le \mathrm{cov}(\mathcal E)\le\mathrm{cov}_H(C_2^\omega)\le \mathfrak r,$$ where $\mathcal E$ is the $\sigma$-ideal generated by closed subsets of Haar measure zero in $C_2^\omega$ (the upper bound $\mathrm{cov}_H(C_2^\omega)\le\mathfrak r$ is proved, for example, here). It is known that $\max\{\mathrm{cov}(\mathcal N),\mathrm{cov}(\mathcal M)\}$ can be strictly smaller than $\mathfrak b$ (this happens, for example, in the Laver and Mathias models). Problem. Is $\mathrm{cov}_H(C_2^\omega)<\mathfrak b$ consistent? What is the value of $\mathrm{cov}_H(C_2^\omega)$ in the Laver (or Mathias) model? REPLY [3 votes]: Lyubomyr Zdomskyy proved that in the Laver model $\mathrm{cov}_H(2^\omega)=\omega_1<\mathfrak b=\mathfrak c$. His argument used the following known Laver property of the Laver model $V'$: for every function $f:\omega\to\omega$ in $V'$ upper bounded by some function $h:\omega\to\omega$ in the ground model $V$, there exists $H:\omega\to [\omega]^{<\omega}$ in $V$ such that $|H(n)|\leq n+1$ and $f(n)\in H(n)$ for all $n\in\omega$. Choose any increasing sequence $(k_n)_{n\in\omega}\in \omega^\omega\cap V$ such that $k_0=0$ and $k_{n+1}>k_n+n+1$ for every $n\in\omega$. Consider the family $\mathcal H$ of closed nowhere dense subgroups of $2^\omega$ of the form $ \prod_{n\in\omega}H_n, $ where $H_n$ is a proper subgroup of $2^{k_{n+1}\setminus k_n}$ and $( H_n)_{n\in\omega}\in V$. We claim that for every $x\in 2^\omega$ there exists $H\in\mathcal H$ with $x\in H$. Indeed, since $x{\restriction} (k_{n+1}\setminus k_n)\in 2^{k_{n+1}\setminus k_n}$, the latter set has size $2^{k_{n+1}-k_n}$, and the sequence $\langle 2^{k_{n+1}-k_n}\rangle_{n\in\omega}$ lies in $V$, there exists a sequence $\langle X_n\rangle_{n\in\omega}\in V$ such that $$x{\restriction}(k_{n+1}\setminus k_n)\in X_n\in \big[ 2^{k_{n+1}\setminus k_n}\big]^{n+1}$$ for all $n\in\omega$. It suffices to denote by $H_n$ the subgroup of $2^{k_{n+1}\setminus k_n}$ generated by $X_n$ and note that $H_n\neq 2^{k_{n+1}\setminus k_n}$ because $|X_n|=n+1$ and $k_{n+1}-k_n>n+1$. Therefore in the model $V'$ we have $\mathrm{cov}_H(2^\omega)\le|\mathcal H|\le|V\cap 2^\omega|=\omega_1$.<|endoftext|> TITLE: Categories disguised as other structures QUESTION [14 upvotes]: It is common to hear that category theory unifies many apparently disparate areas of mathematics. One way it does so is by allowing us to take other mathematical structures and organize them into categories, then explore connections with other structures organized into categories and their opposites -- a famous example would be the equivalence between the category of Stone spaces and the opposite of the category of Boolean algebras. A second way it allows us to unify other areas of mathematics is by permitting the definition of other mathematical objects inside a category with sufficient structure, like Frobenius algebras or groups inside a category, in a way that reproduces their standard set-theoretical definitions when realized in the category of sets but canonically adds structure to them when realized in other categories. My question is not about either of the above methods of unification. A third way in which categories unify many apparently disparate areas of mathematics is by allowing us to 'study them all at once' by realizing the objects of study in these fields as particular types of categories; a famous example here is that a group 'is' a category with one object, or that a preordered/partially ordered set 'is' a thin/skeletal thin category. What are some other structures that can be realized as categories wearing a disguise? To be a bit more specific, for another structure to 'be a category in disguise' I mean that we can define a correspondence between the (first or higher order) language of the other structure and the first order language of category theory such that the axioms of the other structure are immediately satisfied as a consequence of the axioms of a category, potentially with additional constraints like being thin or having products etc. As an aside, I don't mean to suggest that thinking about these structures as categories will allow a category theorist to instantly gain the insights of mathematicians in these other fields; those treasures are hard won and likely unavailable through other routes. What I hope is that some of the insight of categorical reasoning could be turned towards these structures, to complement the existing formidable mental architecture around them. Any contributions are appreciated. REPLY [8 votes]: Algebraic theories can be identified with certain categories with finite products, a la Lawvere.<|endoftext|> TITLE: A question about mod $p$ local Langlands for $\mathrm{GL}_{2}(\mathbb{Q}_{p})$ QUESTION [8 upvotes]: In the mod $p$ local Langlands correspondence for $\mathrm{GL}_{2}(\mathbb{Q}_{p})$, the irreducible supercuspidal representation $\left(\mathrm{ind}^{\mathrm{GL}_{2}(\mathbb{Q}_{p})}_{\mathrm{GL}_{2}(\mathbb{Z}_{p})\mathbb{Q}_{p}^{\times}}\mathrm{Sym}^{r}\overline{\mathbb{F}}_{p}^{2}\right)/T$ of $\mathrm{GL}_{2}(\mathbb{Q}_{p})$ is mapped to the irreducible Galois representation $\mathrm{ind}(\omega_{2}^{r+1})$ for $r\in\lbrace 0,\ldots,p-1\rbrace$. See Definition 1.1 in https://www.imo.universite-paris-saclay.fr/~breuil/PUBLICATIONS/GL2%28Qp%29II.pdf (with $\eta=1$). However, I notice that this is not compatible with the local class field theory: on $\mathbb{Z}_{p}^{\times}$, the central character of the supercuspidal representation is the $r$-th power map after going modulo $p$, while the determinant character of the Galois representation is the $(r+1)$-th power map. So is it defined this way to satisfy some other compatibilities (for example, reduction modulo $p$ of the $p$-adic local Langlands for $\mathrm{GL}_{2}(\mathbb{Q}_{p})$)? REPLY [6 votes]: You seem to be expecting that mod $p$ local Langlands should satisfy the same compatibilities as "conventional" local Langlands (for smooth representations of $GL_2(\mathbf{Q}_p)$ and $WD(\mathbf{Q}_p)$ with coefficients in $\mathbf{C}$). However, before you can even talk about reduction mod $p$, you need to check that the coefficients can be descended from $\mathbf{C}$ to a number field. I.e., if $\pi$ is a smooth irred rep of $GL_n(\mathbf{Q}_p)$ on an $L$-vector space, where $L$ is some subfield of $\mathbf{C}$, then is its Langlands parameter $\phi_{\pi}$ an $L$-valued Weil-Deligne rep? The answer, annoyingly, is "no": if $n$ is even, you have to add $\sqrt{p}$ to $L$ in order to get this to work. So it's common to re-normalise by twisting the correspondence for $GL_n$ by $|\det|^{(n-1)/2}$; this makes it compatible with coefficient fields, and also works better for local-global compatibility. It's this same shift which you are seeing in the mod $p$ theory, and here it's completely impossible to get rid of, even if you extend the coefficient fields as much as you like: the character $|\cdot|^{1/2}$ of $\mathbf{Q}_p^\times$ is not $p$-adically unitary, so there is no way you can reduce it mod $p$.<|endoftext|> TITLE: Could a nice principle be extracted from this lemma of Gauss QUESTION [7 upvotes]: I asked the following question in the math SE, with a bounty of 200 pts, without result. question: To prove the quadratic reciprocity law, Gauss needed the following lemma: If $p$ is a prime number congruent to 1 modulo 8, then there exists a prime $q