User xandi tuni - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:21:46Z http://mathoverflow.net/feeds/user/5952 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120289/second-difference/120294#120294 Answer by Xandi Tuni for Second difference Xandi Tuni 2013-01-30T10:05:19Z 2013-01-30T10:05:19Z <p>Yes, i think. </p> <p>Pick an integer $n>1$ and let $f$ be the function given by $f(x) = x^n\sin(x^{-n})$ for $x>0$ and $f(x)=0$ for $x\leq 0$. It is continuous, and smooth except at $x=0$. It is not $C^1$ at zero, because the derivative for $x>0$ is $nx^{n-1}\sin(x^{-n}) - nx^{-1}\cos(x^{-n})$, which enters in a state of flailing tantrum close to $0$.</p> <p>On the other hand, we have $|f(0+t)+f(0-t)-2f(0)| = |f(t)|$, so every $\alpha &lt; n$ is good.</p> http://mathoverflow.net/questions/115647/torsion-of-elliptic-curves-is-finite/115652#115652 Answer by Xandi Tuni for Torsion of elliptic curves is finite Xandi Tuni 2012-12-06T23:27:50Z 2012-12-06T23:27:50Z <p>I have a vague idea that this might be wrong. </p> <p>First, it seems to me that you can even assume that $S$ is the spectrum of a local ring. Then, if $p$ is the resicual characteristic, the reduction map will in general not be injective on $p$--torsion. So, if I had to produce couterexamples to this, I would start with an $E$ over $\mathbb Z_p$ with supersingular reduction, and then base change $E$ to some big, totally ramified extension of $\mathbb Z_p$ where lots of $p$--power torsion points of $E$ are defined.</p> http://mathoverflow.net/questions/114066/singular-points-on-the-hilbert-scheme-of-a-product Singular points on the Hilbert scheme of a product Xandi Tuni 2012-11-21T15:48:06Z 2012-11-21T16:43:34Z <p>Let $X$ and $Y$ be smooth projective varieties, say over $\mathbb C$. Fixing a point $y\in Y$, we obtain a smooth, closed subvariety $X\times\{y\}$ of $X\times Y$, which in turn corresponds to a point $P_y$ on the Hilbert scheme $\mathcal{Hilb}(X\times Y)$.</p> <blockquote> <p>What technology can I use to decide whether $P_y\in \mathcal{Hilb}(X\times Y)$ is smooth or not?</p> </blockquote> http://mathoverflow.net/questions/108282/closed-subgroups-of-a-p-adic-algebraic-group/108288#108288 Answer by Xandi Tuni for Closed subgroups of a $p$-adic algebraic group Xandi Tuni 2012-09-27T21:28:19Z 2012-09-27T21:28:19Z <p>Let $G$ be a $p$--adic Lie group. There is a 1:1 correspondence between $p$-adically closed subgroups up to finite index of $G$ and Lie subalgebras of its Lie algebra (which is a Lie algebra over $\mathbb Q_p$). The correspondence works just as in the classical setting. This is shown in a paper by Mattuck from the fifties, and in much greater generality in Lazard's thesis (beware, beware).</p> <p>Attention: The image of a closed subgroup $H \subseteq G$ under the logarithm map is not always a $\mathbb Z_p$ submodule of the Lie algebra. Example: take $p=2$ and $H \subseteq \mathrm{GL}_3$ the unipotent radical of the Borel. Then $\mathrm{log}(H(\mathbb Z_p))$ is not stable under $+$. </p> <p>You cannot hope for such a correspondence for strictly all closed subgroups. Although the exponential map is a homeomorphism locally around zero, it can in general not be extended to a surjective map. There is already a problem with $\mathbb Z_p^\ast$. </p> <p>Also, this does not make much sense over extensions of $\mathbb Q_p$. If $F$ is some nontrivial finite extension of $\mathbb Q_p$, then $F$ viewed as a Lie group under addition has many closed subgroups (all the $\mathbb Q_p$-linear subspaces), but the Lie algebra (which is also $F$) has no proper $F$--linear subalgebra.</p> http://mathoverflow.net/questions/93743/are-algebraic-groups-defined-by-their-invariants-in-tensor-spaces Are algebraic groups defined by their invariants in tensor spaces? Xandi Tuni 2012-04-11T08:38:05Z 2012-05-29T22:23:25Z <p>Let $K$ be a field of characteristic zero, and let $G \subseteq \mathrm{GL}_V$ be an algebraic group over $K$, acting faithfully on a finite dimensional vector space $V$. Let $H \subseteq \mathrm{GL}_V$ be the largest algebraic subgroup with the following propertes:</p> <p>(1) If a subspace $V_1 \subseteq V$ is invariant under $G$, then it is also invariant under $H$.</p> <p>(2) Given $G$-invariant subspaces $V_1$ and $V_2$ of $V$, and integers $a,b\geq 0$, the equality $$\mathrm{Hom}_G(V_1^{\otimes a}, (V/V_2)^{\otimes b}) = \mathrm{Hom}_H(V_1^{\otimes a}, (V/V_2)^{\otimes b})$$ holds.</p> <p>The second condition means that $G$ and $H$ have the same fixed points in any tensor space that can be formed out of subquotients of $V$. The inclusion $G\subseteq H$ is tautological, and my question is:</p> <blockquote> <p>do we have $G=H$?</p> </blockquote> <p>If $G$ is reductive, then the answer iy yes, because in that case $V$ and all its tensor powers are semisimple, but the equality $G=H$ also holds for example if $G$ is the group of upper triangular matrices.</p> http://mathoverflow.net/questions/92290/about-tates-computation-of-k-2-rm-m-mathbb-q About Tate's computation of $K_2^{\rm M}(\mathbb Q)$ Xandi Tuni 2012-03-26T16:41:58Z 2012-03-26T17:36:13Z <p>For any field $F$, there is a natural group homomorphism $K_n^{\rm M}(F) \to K_n(F)$ from Milnor's $K$-theory to Quillen's $K$-theory. If $n=2$, it is an isomorphism, by Matsumoto's theorem. It is a well known theorem of Quillen that if $F$ is a number field, then the groups $K_n(F)$ are finitely generated for $n\geq 2$. It is known in particular, that the group $K_2(\mathbb Q)$ is finite, isomorphic to $\mathbb Z/2\mathbb Z$.</p> <p>I am now confused about the following result of Tate (I got it from Milnors "Introduction to Algebraic K-theory", Theorem 11.6): There is a canonical, split exact sequence of commutative groups $$0 \to \{\pm1\} \to K_2^{\rm M}(\mathbb Q) \to \bigoplus_p (\mathbb Z/p\mathbb Z)^\ast \to 0$$ and in particular, $K_2^{\rm M}(\mathbb Q)$ is an infinite torsion group. The first term in the sequence should be interpreted as $K_2^{\rm M}(\mathbb Z)$, and each $(\mathbb Z/p\mathbb Z)^\ast$ should be seen as $K_1$ of the finite field with $p$ elements. </p> <p>What did I misunderstand? Is it not true that $K_2^{\rm M}(\mathbb Q) \cong K_2(\mathbb Q)$ -- or is it not true that $K_2(\mathbb Q)$ is finitely generated --- or did I misquote Tate's theorem? </p> http://mathoverflow.net/questions/86537/an-elementary-linear-algebra-problem An elementary linear algebra problem Xandi Tuni 2012-01-24T14:46:02Z 2012-01-24T14:46:02Z <p>Let $K$ be a field, and let $E$ be the algebra of $n\times n$ matrices over $K$. Let $V_0$ and $V_1$ be the (left) $E$-modules of matrices of size $n\times n_0$ and $n\times n_1$. Let $W \subseteq V_0$ be a $K$-linear subspace, and define $$\overline W := \{ v\in V_0 \:|\: f(v)\in f(W) \mbox{ for all } f\in \mathrm{Hom}_E(V_0,V_1) \}$$ Is there a method to compute $\overline W$? More precisely: </p> <blockquote> <p>Given a $K$-basis $w_1,\ldots,w_s$ of $W$, is there an algorithm that produces a $K$-basis of $\overline W$?</p> </blockquote> <p>This problem arises in the computation of certain cohomology groups attached to abelian varieties over number fields. In practice, the relevant fields are $K = \mathbb Q$ and $K = \mathbb Q_\ell$. Here are a few loose remarks: </p> <p>(0) The problem is only interesting if the inequalities $0 &lt; n_1 &lt; n_0$ hold.</p> <p>(1) One can always assume that $W$ generates $V_0$ as an $E$-module. Instead of a basis of $\overline W$, one can also ask for a finite list of $E$-module homomorphisms $f_i:V_0\to V_1$ which cut $\overline W$ out. Such homomorphisms are given by $n_0\times n_1$-matrices. </p> <p>(2) The problem depends on the field $K$, in the sense that the formation of $\overline W$ does not in general commute with base change to a field extension of $K$. Even in the case where $K$ is algebraically closed, I don't know of a solution to the problem.</p> <p>(3) One can replace the matrix-algebra $E$ by any finite dimensional semisimple $K$-algebra, and the question makes still sense. However, this more general question boils down to the matrix case, at least in characteristic zero.</p> <p>I have tagged the question "algebraic geometry" for the following reason: We can look at the set of $E$-linear homomorphisms $V_0\to V_1$ as the set of $K$-rational points of the $n_0n_1$-dimensional affine space over $K$. The homomorphisms that impose strong conditions on $\overline W$ are those for which $f(V_0)/f(W)$ is large. This leads to interesting subvarieties of the affine (and in fact projective) space over $K$, called <em>determinantal varieties</em>. These have usually many singularities and irreducible components. In view of remark (1), we are interested in rational points on these varieties.</p> http://mathoverflow.net/questions/86089/two-definitions-of-character-of-topological-groups/86093#86093 Answer by Xandi Tuni for Two Definitions of "Character" of topological groups Xandi Tuni 2012-01-19T11:47:01Z 2012-01-19T11:47:01Z <p>I am assuming all groups we are talking about are locally compact and commutative.</p> <p>The two definitions indeed do ageree on profinite groups. To prove it, you have to check that the functors $Hom(-,\mathbb Q/ \mathbb Z)$ and $Hom(-,\mathbb R/ \mathbb Z)$ both transform limits of compact groups into colimits of discrete groups. That's routine.</p> <p>The two definitions also agree on discrete torsion groups, hence on all groups which contain a closed and open profinite subgroup such that the quotient is torsion. I guess that this is exactly the class of groups on which the two definitions agree.</p> http://mathoverflow.net/questions/85805/a-question-about-abelian-scheme/85822#85822 Answer by Xandi Tuni for a question about abelian scheme Xandi Tuni 2012-01-16T16:42:05Z 2012-01-16T16:42:05Z <p>If I understand the question correctly, the answer is Yes. It suffices in fact to have a group structure on the generic fibre $X_K$ of $X$. Since $X$ is smooth and proper, it is then the Néron model of $X_K$, which carries a group structure by its universal property.</p> <p>By the way, there is only one group structure an an abelian variety with a given element as unit, so not only has $X$ a global group structure, but it induces the given group structure on each closed fibre.</p> http://mathoverflow.net/questions/81063/splitting-of-vector-bundles-on-a-complex-torus Splitting of vector bundles on a complex torus Xandi Tuni 2011-11-16T12:51:46Z 2011-11-16T12:51:46Z <p><em>Let $X$ be a complex torus (a finite dimensional complex vector space modulo a lattice) and let $E$ be a smooth (not necessarily holomorphic) complex vector bundle over $X$. Is it true $E$ is isomorphic to a sum of smooth line bundles? If not, what would be a minimal counterexample?</em></p> <p>A theorem of Atyiah and Hirzebruch tells me that the Chern-character is an isomorphism of $\mathbb Q$-algebras $K(X)\otimes\mathbb Q \to H^{2\ast}(X,\mathbb Q)$. Since $X$ is a torus, its even cohomology ring is generated by Chern-classes of complex line bundles, hence so is $K(X)\otimes\mathbb Q$. It follows that some power of $E$ is stably isomorphic to a sum of line bundles. If one can get rid of the denominators, it would follow that $E$ itself is stably isomorphic to a sum of line bundles, but that's probably as far as one gets with $K$-theory. </p> http://mathoverflow.net/questions/77210/logarithm-of-complex-matrices-in-holomorphic-families Logarithm of complex matrices in holomorphic families Xandi Tuni 2011-10-05T07:53:17Z 2011-10-10T17:22:05Z <blockquote> <p>Let $n,k\geq 1$ be integers, let $U \subseteq \mathbb C^n$ be a contractible open subset, and let $f:U\to \mathrm{GL}_k(\mathbb C)$ be a holomorphic function. Does there exist a holomorhpic function $F:U\to \mathrm{M}_k(\mathbb C)$ such that $\exp(F(u))= f(u)$ holds for all $u\in U$?</p> </blockquote> <p>Here, $\mathrm{M}_k(\mathbb C)$ means complex $k$ by $k$ matrices. The answer is of course "yes" if $k=1$.</p> <p>As soon as $k\geq 2$, the problem is that for some invertible matrices $A \in \mathrm{GL}_k(\mathbb C)$ the set of matrices $B\in \mathrm{M}_k(\mathbb C)$ with $\exp(B)=A$ is not discrete. This happens for example if $A$ is diagonalisable and has a double eigenvalue. If in the question we require that for all $u\in U$ the eigenvalues of $f(u)$ are pairwise distinct, the answer would again be yes. </p> http://mathoverflow.net/questions/77210/logarithm-of-complex-matrices-in-holomorphic-families/77671#77671 Answer by Xandi Tuni for Logarithm of complex matrices in holomorphic families Xandi Tuni 2011-10-10T07:46:06Z 2011-10-10T09:00:53Z <p>The answer is "no" in general. As Denis suspects, the problem is a global one, and it involves matrices with nontrivial Jordan blocks. These have, in a sense, "fewer" logarithms than the commoners. Concretely, I clain that the holomorphic function $$f(z) = \begin{pmatrix} e^{2\pi i z} &amp; 1 \\ 0 &amp; 1 \end{pmatrix}$$ has no holomorphic logarithm on $\mathbb C$. If it had one, there would also be a holomorphic square root of $f$ on $\mathbb C$, and not even that exists. Indeed, suppose by contradiction that there was a function $g:\mathbb C \to \mathrm{GL}_2(\mathbb C)$ such that $f(z) = g(z)^2$. The matrix $$f(0) = g(0)^2 = \begin{pmatrix} 1 &amp; 1\\ 0 &amp; 1 \end{pmatrix}$$ has only two square roots (a 2-by-2 matrix with distinct eigenvalues has four square roots!) differing by a sign, so we may suppose $$g(0) = \begin{pmatrix} 1 &amp; 1/2 \\ 0 &amp; 1 \end{pmatrix}$$ by changing $g$ to $-g$ if necessary. If we move $z$ on the real line from $0$ to $1$, we find by continuity of $g$ $$g(z) = \begin{pmatrix} e^{\pi i z} &amp; (e^{\pi i z}+1)^{-1} \\ 0 &amp; 1 \end{pmatrix}$$ and run into a pole as $z$ approaches $1$, end of story.</p> http://mathoverflow.net/questions/72180/what-dirichlet-doesnt-tell What Dirichlet doesn't tell... Xandi Tuni 2011-08-05T14:03:01Z 2011-08-25T02:38:29Z <p>Let $n>1$ be an integer, and let us consider the set $P(n)$ of all prime numbers $p$ such that $p$ is not congruent to $1$ modulo $n$. Dirichlet's Density Theorem tells us that $P(n)$ has a natural density, equal to $$1-\varphi(n)^{-1}$$ where $\varphi(n) = |(\mathbb Z /n)^\ast|$ is Euler's totient.</p> <p>From the Frobenian point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the ideal $(p)$ splits completely in the cyclotomic field $\mathbb Q(\zeta_n)$. </p> <p>From Chebotarev's point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the Frobenius element over $p$ in $\operatorname{Gal}(\mathbb Q(\zeta_n)|\mathbb Q) \simeq (\mathbb Z /n)^\ast$ is the identity.</p> <p>So far so good, now let us consider the set $P$ of all prime numbers $p$ which are not congruent to $1$ modulo $n^2$ for any $n>1$, that is $$P := \bigcap_{n>1}P(n^2) = \bigcap_{\ell\mathrm{ prime}}P(\ell^2)$$ Supposing that "the events $P(\ell^2)$ are uncorrelated" for different $\ell$'s, we can phantasise about the density of $P$, hoping it might be (at least up to a rational factor, I don't vouch for it) $$\operatorname{dens}(P) = \prod_\ell 1-\frac{1}{\ell(\ell-1)} \quad = 0.37395581361920228805...$$ a number called <em>Artin's constant</em> (it appears in Artins primitive root conjecture, which is similar in nature). The question whether $P$, or similarly constructed sets of primes, have a density and whether it is the expected one goes far beyond the density theorems of Dirichlet, Frobenius and Chebotarev. The corresponding Galois extension would be the maximal cyclotomic extension of $\mathbb Q$, which is ramified everywhere. </p> <blockquote> <p>Can you name this problem? Have you seen it before? Where? </p> </blockquote> <p>Hooley (1967) has shown that Artins primitive root conjecture follows from GRH. In principle, the problem of determining the density of $P$ should be simpler. </p> <blockquote> <p>Under GRH, is it true that the density of $P$ exists and is equal to Artin's constant?</p> </blockquote> http://mathoverflow.net/questions/69523/when-does-zariski-closure-commute-with-base-change When does Zariski closure commute with base change? Xandi Tuni 2011-07-05T08:35:52Z 2011-07-05T09:19:33Z <p>This should be an elementary question for anyone who knows SGA by heart (alas, not for me). It smells a lot like a descent problem. All schemes are supposed to be noetherian, and all morphisms to be locally of finite presentation.</p> <p>Let $X$ be a scheme of finite type over a (base) scheme $S$, and let $R \subseteq X(S)$ be a subset of the set of $S$--rational points of $X$. Denote by $\overline R$ the smallest closed subscheme of $X$ whose $S$--rational points contain $R$.</p> <p>Let $f:S'\to S$ be a (base--change) morphism of schemes, write $X' := X\times_SS'$ and denote by $R'$ the image of $R$ in $X(S') = X'(S')$. Again, let $\overline{R'}$ be the smallest closed subscheme of $X'$ whose $S'$--rational points contain $R'$. Then, $\overline{R'}$ is contained in $\overline R\times_SS'$, and the question is:</p> <blockquote> <p>Suppose $f:S'\to S$ is flat. Does the equality $\overline{R'} = \overline R \times_SS'$ hold?</p> </blockquote> <p>Clearly some hypothesis on $f$ is needed, and I just guess it's flatness. </p> http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem Converse to Banach's fixed point theorem? Xandi Tuni 2010-05-27T08:35:40Z 2011-06-23T10:09:24Z <p>Let $(X,d)$ be a metric space. Banach's fixed point theorem states that if $X$ is complete, then every contraction map $f:X\to X$ has a unique fixed point. A contraction map is a continuous map for which there is an real number $0\leq r &lt; 1$ such that $d(f(x),f(y))\leq rd(x,y)$ holds for all $x,y\in X$.</p> <blockquote> <p>Suppose $X$ is a metric space such that every contraction map $f:X\to X$ has a unique fixed point. Is $X$ complete?</p> </blockquote> http://mathoverflow.net/questions/65962/which-h-groups-satisfy-the-rigidity-property-of-abelian-varieties Which $H$--groups satisfy the rigidity property of abelian varieties? Xandi Tuni 2011-05-25T13:00:16Z 2011-05-26T03:23:16Z <p>Let us call a group object $G$ in a category $\mathcal C$ <em>rigid</em>, if it has the following property: For every group object $X$ in $\mathcal C$, every morphism $G\to X$ in $\mathcal C$ respecting the unit sections is already a morphism of group objects.</p> <p>As a formal consequence, rigid group objects are commutative (because the inversion is a group object morphism).</p> <p>This definition is motivated by the following fact: If $\mathcal C$ is the category of algebraic varieties over $\mathbb C$ (say), then the rigid group objects in $\mathcal C$ are precisely the complex abelian varieties. Indeed, that abelian varieties enjoy the rigidity property is a "standard fact", and the converse follows quickly from Chevalley's structure theorem (every algebraic group is an extension of an abelian variety by an affine group).</p> <p>If $\mathcal C$ is the category of sets or of hausdorff topological spaces, there are no interesting rigid group objects. I am puzzled with the case where $\mathcal C$ is the category whose objects are the topological spaces which are homotopy equivalent to CW-complexes and morphisms are continuous maps up to homotopy. A group object in this category is commonly called $H$--group. </p> <blockquote> <p>Is it true that the rigid $H$--groups are exactly the products of circles?</p> </blockquote> http://mathoverflow.net/questions/61496/mumford-tate-groups-and-hodge-structures/61511#61511 Answer by Xandi Tuni for Mumford-Tate groups and Hodge structures Xandi Tuni 2011-04-13T07:14:25Z 2011-04-13T07:14:25Z <p>Indeed. Look at Hodge structures as finite dimensional vector spaces $V$ over $\mathbb Q$ together with an action of the Deligne torus $\mathbb S$ on $V \otimes \mathbb C$, i.e. a morphism of algebraic groups $\rho: \mathbb S \to \mathrm{GL}(V \otimes \mathbb C)$. The Mumford--Tate group of $V$ is then just the Zariski closure over $\mathbb Q$ (!) of the imagr of $\rho$.</p> <p>Now take $V$ of dimension $\geq 4$ and a representation $\rho: \mathbb S \to \mathrm{GL}(V\otimes \mathbb C)$ of pure weight $k$ say, such that the Zariski closure over $\mathbb Q$ of the image of $\rho$ is $\mathrm{GL}(V)$. So the resulting Mumford--Tate group is $\mathrm{GL}(V)$ which is reductive. But the Hodge--structure $V$ is not polarisable, indeed, any polarisation $\psi: V \otimes V \to \mathbb Q(k)$ would confine the Mumford--Tate group to the Goup of symplectic similitudes $\mathrm{GSp}(V,\psi)$ which is strictly smaller than $\mathrm{GL}(V)$.</p> http://mathoverflow.net/questions/60912/dense-sphere-packings-which-are-not-lattice-packings Dense sphere packings which are not lattice packings Xandi Tuni 2011-04-07T10:18:48Z 2011-04-07T13:04:36Z <p>This question is about dense sphere packings in euclidean space $\mathbb R^n$. By a sphere packing I understand any arrangement of mutually disjoint solid open spheres in $\mathbb R^n$, all of the same radius. The density of a packing is $$\mathrm{lim}_{R \to \infty}\frac{\mathrm{vol }(B(0,R) \cap \mathrm{spheres})}{\mathrm{vol } B(0,R)}$$ if it exists. Here, $B(0,R)$ is the open ball of radius $R$ centered at $0 \in \mathbb R^n$.</p> <p>In low dimensions, the highest possible densities of sphere packings are known to be attained by lattice packings, that is, packings such that the centers of the spheres form a discrete subgroup of $\mathbb R^n$ of rank $n$. One could speculate that this is so in all dimensions, but I doubt it very much...</p> <blockquote> <p>Is it true that for some (possibly very lagre) integer $n$, there is a sphere packing in $\mathbb R^n$ which has a higher density than any lattice packing?</p> </blockquote> <p>Edit -- Note: I didn't mean to ask about an explicit $n$, let alone about explicit packings. So i'm completely satisfied if somebody tells me that there is asymptotically such and such upper bound for lattice packing densities and this and that lower bound for general densest sphere packing densities.</p> http://mathoverflow.net/questions/60894/a-pair-of-subset-of-natural-numbers-having-density-but-whose-intersection-has-no/60900#60900 Answer by Xandi Tuni for A pair of subset of natural numbers having density, but whose intersection has no density Xandi Tuni 2011-04-07T07:33:50Z 2011-04-07T07:33:50Z <p>Let $A$ be the set of odd integers $\geq 0$, and let $B$ be the set of those integers $n$ which are odd if $2^m \leq n &lt; 2^{m+1}$ for an odd $m$ and even if $2^m \leq n &lt; 2^{m+1}$ for an odd $m$. Both sets $A$ and $B$ have naive density $\frac12$ (you call it Beurling density).</p> <p>The intersection $A\cap B$ has no density: its upper density (limsup...) is $\frac 12$ while its lower density is zero.</p> http://mathoverflow.net/questions/60673/basic-question-about-branch-points-on-riemann-surfaces/60678#60678 Answer by Xandi Tuni for Basic question about branch points on Riemann surfaces Xandi Tuni 2011-04-05T12:03:38Z 2011-04-05T12:03:38Z <p>This is true if $f$ is proper (the preimage of a compact is compact). Indeed, any compact neighborhood of $y\in Y$ contains only finitely many branch points because its preimage in $X$ contains only finitely many ramification points.</p> <p>If $f$ is not proper, ramification points in $Y$ may not be discrete. Take for $X$ the union of copies of $\mathbb C$ indexed by $n=1,2,3,...$ and let $f:X\to \mathbb C$ be the holomorphic map which on the $n$--th copy of $\mathbb C$ in $X$ is given by $f(z) = (z-1/n)^2$.</p> http://mathoverflow.net/questions/60434/are-chow-groups-generated-by-local-complete-intersections Are Chow groups generated by local complete intersections? Xandi Tuni 2011-04-03T13:07:59Z 2011-04-03T19:57:28Z <p>Let $X$ be a smooth projective variety over an algebraically closed field. The Chow group $\mathbb Q\mathrm{CH}^d(X)$ is $\mathbb Q$--linearly generated by irreducible subvarieties $Z \subseteq X$ of codimension $d$, modulo rational equivalence.</p> <p>I am interested in the linear subspace of $\mathbb Q\mathrm{CH}^d(X)$ which is generated by the subvarieties $Z\subseteq X$ of codimension $d$ which are locally complete intersections, so those which are locally the zero set of exactly $d$ regular functions. Let us denote this subspace by $\mathbb Q\mathrm{CH}^d_{\mathrm{lci}}(X)$. Then the question is:</p> <blockquote> <p>Are Chow groups generated by local complete intersections? I.e. does equality $$\mathbb Q\mathrm{CH}^d_{\mathrm{lci}}(X) = \mathbb Q\mathrm{CH}^d(X)$$ hold?</p> </blockquote> <p>If for instance $d=1$, equality holds indeed, as $X$ is smooth. I suspect this not so in general for $d\geq 2$... but where to look for a counter example?</p> http://mathoverflow.net/questions/58397/the-galois-group-of-a-random-polynomial The Galois group of a random polynomial Xandi Tuni 2011-03-14T08:22:49Z 2011-03-14T10:16:39Z <p>Intuitively, the Galois group (of a splitting field over $\mathbb Q$ of) a polynomial $f\in\mathbb Q[X]$ taken at random is most probably the full permutation group on the roots of $f$. This intuition can be made precise as follows.</p> <p><em>Elementary</em>: For inegers $d \geq 1$ and $N \geq 1$, consider the set $P_{d,N}$ of polynomials in one variable of degree $\leq d$ whose coefficients have absolute value $\leq N$. The set $P_{d,N}$ is finite, and we can consider the subset $Q_{d,N}$ of those $f\in P_{d,N}$ whose Galois group is the full symmetric group $S_d$. Then, is it true that the ratio $|Q_{d,N}|/|P_{d,N}|$ goes to $1$ as $N$ goes to infinity? If so, are there good estimates for the speed of convergence?</p> <p><em>Less elementary, but still the same</em>: Let $k$ be a number field. For an integer $d\geq 1$, indentify $k$--points of the $d$--dimensional projective space $\mathbb P^d$ with nonzero polynomials of degree $\leq d$ over $k$ up to scalars. This way, we can speak about the Galois group of a point $f\in \mathbb P^d(k)$. For an integer $N$, let $p_{d,N}$ be the number of points of $P^d(k)$ of height $\leq N$, and let $q_{d,N}$ be the number of points of height $\leq N$ and with Galois group $S_d$. Then, is it true that $$\frac{q_{d,N}}{p_{d,N}}\to 1$$ as $N$ goes to infinity? And again, are there good estimates for the speed of convergence?</p> <p><em>Remark</em>: The subset of $\mathbb Q^{d+1} = \mathbb A^{d+1}(\mathbb Q)$ of those elements $(a_0,\ldots, a_d)$ such that the Galois group of the polynomial $f = a_0+a_1X+\cdots+a_dX^d$ is $S_d$ is Zariski dense in $\mathbb A^{d+1}$. This is so because the Galois group of the generic polynomial $f = t_0+t_1X+\cdots+t_dX^d$ with indeterminates $t_0,\ldots, t_d$ over $\mathbb Q(t_0,\ldots, t_d)$ is $S_d$ and the fact that $\mathbb Q$ (or any number field) is a Hilbertian field. </p> http://mathoverflow.net/questions/24718/singular-homology-of-a-graph Singular homology of a graph. Xandi Tuni 2010-05-15T10:02:58Z 2010-12-28T18:11:39Z <p>By a graph I will understand an undirected graph without multiple edges or loops. By a morphism of graphs I will understand a map $f$ between the underlying sets of vertices, such that if $x$ and $y$ are adjacent, then $f(x)$ and $f(y)$ are either adjacent or equal.</p> <p>Let $G$ be a finite graph. One can realise $G$ as a CW-complex $|G|$ and look at topological invariants, such as singular homology. But this captures only very little information about $G$, because except from $H_0(|G|)$ and $H_1(|G|)$ all homology groups are zero.</p> <p>Consider the following alternative construction: Let us write $\Delta_n$ for the complete graph on $n$ vertices, and let us re-baptise this graph by the name "standard $n$-simplex". There are obvious codegeneracy and coface maps between standard simplices, so that we obtain a cosimplicial object $\Delta_\bullet$ in the category of graphs. Now, proceed as usual: Morphisms $\Delta_\bullet \to G$ form a simplicial set, applying the free group construction yields then simplicial group, and the associated chain complex is the one whose homology $H_i^{\mathrm{sing}}(G)$ I shall call "singular homology of $G$".</p> <p>Obvious properties of $H_i^{\mathrm{sing}}(G)$ are: It is a finitely generated commutative group ($G$ is finite), covariantly functorial in $G$. In particular, if we work with coefficients in a field, we obtain representations of the automorphism group of $G$. The homology of the point is $\mathbb Z$ in degree $0$ and trivial in higher degrees. We can define singular cohomology accordingly, and get then a natural pairing between homology and cohomology.</p> <p>The list of all natural questions one must ask after making such a definition is long, so I will not ask everything. </p> <p>(a) Is there a comparison map $H_i(|G|) \to H_i^{\mathrm{sing}}(G)$, maybe even on the level of chain complexes? Is there some more elaborate CW-complex $||G||$ one can naturally associate with $G$ such that $H_i(||G||)$ gives back singular homology of $G$? In that case, one would ask for a natural map $|G| \to ||G||$.</p> <p>(b) Given a graph $G$, is there a largest integer $i$ such that $H_i^{\mathrm{sing}}(G)$ is nonzero? Assuming yes, is this integer less or equal the size of the largest complete subgraph of $G$.</p> <p>(c) Is there a Künneth morphism in singular cohomology? --is there a natural ring strucure on cohomology?</p> <p>(d) What is a homotopy between morphisms of graphs? Given an answer to that, do homotopically equivalent morphisms induce the same maps in homology?</p> <p>(z) Can you give an example of a graph with nontrivial $H_2^{\mathrm{sing}}(G)$?</p> http://mathoverflow.net/questions/46948/automorphic-form-encoding-the-orders-of-n-modulo-p Automorphic form encoding the orders of $N$ modulo $p$. Xandi Tuni 2010-11-22T13:30:50Z 2010-11-24T22:05:12Z <p>Let $N$ be a nonzero rational number. For every prime number $p$ with $v_p(N)=0$, let $a_p$ denote the index in $\mathbb Z/p\mathbb Z$ of the subgroup generated by $N$ modulo $p$. So we have $a_p=1$ if and only if $N$ is a primitive root mod $p$.</p> <blockquote> <p>Is there an automorphic form which encodes the numbers $a_p$?</p> </blockquote> <p>Let me explain: To give $N$ is the same thing as to give a group homomorphism $\mathbb Z \to \mathbb G_m$ over $\mathrm{spec}\mathbb Q$. Such a morphism is also the same as an extension $$0\to \mathbb Q(1) \to M \to \mathbb Q \to 0$$ of motives over $\mathrm{spec}\mathbb Q$. The motive $M$ is an example of a "mixed Tate motive", and also an example of a 1--motive. Its weights are $0$ and $-2$. The condition $v_p(N)=0$ means then that $M$ has good reduction at the prime $p$, i.e. extends to a 1--motive over the local ring at $p$. The number $a_p$ is then given by $$a_p = H^0(\mathbb F_p,M)$$ where we interpret $M = [\mathbb Z \to \mathbb G_m]$ as a complex concentrated in degrees $-1$ and $0$. With $M$ are associated "realisations", in particular $M$ comes with integral $\ell$--adic representations. So to restate the question:</p> <blockquote> <p>Is there an automorphic form corresponding to $M$? </p> </blockquote> <p>We could look at the L-function associated with the $\ell$--adic representation of $M$ to get a hint. The problem with this is that the L-function does not see the extension structure, it only depends on the semisimplification of the representation. This is clear because it is constructed by taking <em>traces</em> of Frobenius elements -- the L-function is in fact $\zeta(s)\zeta(s-1)$, independently of $N$.</p> <p>I don't know if morally the association Motives $\to$ Automorphic forms which is classically conjectured for <em>pure</em> motives should extend to mixed motives... also, I don't know what a mixed automorphic form is. Maybe these are simply automorphic forms for nonreductive groups? There is a theory of "mixed modular forms" by Min Ho Lee, but I don't think this is what I am looking for.</p> http://mathoverflow.net/questions/47208/is-projectivity-local-on-the-base/47211#47211 Answer by Xandi Tuni for Is projectivity local on the base? Xandi Tuni 2010-11-24T08:27:21Z 2010-11-24T08:27:21Z <p>No, there should be at least some noetherian hypothesis on $Y$. Take for example $Y$ to be an infinite disjoint union, say indexed by natural numbers, of points $x_i = \mathrm{spec}k$, and take for $X\to Y$ over each point $x_i$ the $i$--dimensional projective space. Then $f$ is not projective although "locally projective" in your sense.</p> http://mathoverflow.net/questions/25337/lifting-varieties-to-characteristic-zero Lifting varieties to characteristic zero. Xandi Tuni 2010-05-20T09:54:28Z 2010-10-25T20:51:59Z <p>If you want to compute crystalline cohomology of a smooth proper variety $X$ over a perfect field $k$ of characteristic $p$, the first thing you might want to try is to lift $X$ to the Witt ring $W_k$ of $k$. If that succeeds, compute de Rham cohomology of the lift over $W_k$ instead, which in general will be much easier to do. Neglecting torsion, this de Rham cohomology is the same as the crystalline cohomology of $X$.</p> <p>I would like to have an example at hand where this approach fails: Can you give an example for</p> <blockquote> <p>A smooth proper variety $X$ over the finite field with $p$ elements, such that there is no smooth proper scheme of finite type over $\mathbb Z_p$ whose special fibre is $X$.</p> </blockquote> <p>The reason why such examples <em>have</em> to exist is metamathematical: If there werent any, the pain one undergoes constructing crystalline cohomology would be unnecessary.</p> http://mathoverflow.net/questions/43280/integral-decomposition-of-the-diagonal-chow-motives Integral decomposition of the diagonal (Chow motives) Xandi Tuni 2010-10-23T10:01:13Z 2010-10-23T10:01:13Z <p>Let $k$ be a field of characteristic zero and let $X$ be a smooth proper varity over $k$ of dimension $d$. The Künneth standard conjecture conjectures that there exist projectors $e_0, e_1, \ldots, e_n$ in the Chow group $\mathrm{CH}^d(X\times X)\otimes \mathbb Q$ which sum up to the diagonal $\Delta \subseteq X\times X$, and induce the projections $H^\ast(X,F) \to H^i(X,F)$ in any Weil cohomology theory with coefficients in a field of characteristic zero $F$ (or any $\mathbb Q$-algebra $F$ for that matter).</p> <p>Although still a conjecture in general it is known to be true for curves (trivially) for surfaces (Murre) and for abelian varieties (Shermenev, and Deninger-Murre). The constructions of these decompositions make essential use of the fact that we are working with $\mathrm{CH}^d(X\times X)\otimes \mathbb Q$ and not just $\mathrm{CH}^d(X\times X)$. For instance, an essential ingredient for the construction of the decomposition of the diagonal of an abelian variety is Fourier-Mukai transform, so there is somewhere an exponential with $\frac{1}{m!}$ coefficients.</p> <blockquote> <p>Question: Is there any reason why the Künneth standard conjecture should fail with <em>integral</em> coefficients? That is, why should there not exist a decomposition of the diagonal in a sum of idempotents $e_i \in \mathrm{CH}^d(X\times X)$ inducing the projections $H^\ast(X,F) \to H^i(X,F)$ in any reasonable cohomology, say singular cohomology with $F=\mathbb Z$ and $\ell$-adic cohomology with $F = \mathbb Z_\ell$?</p> </blockquote> http://mathoverflow.net/questions/40499/extensions-of-an-infinite-product-of-copies-of-z-by-z Extensions of an infinite product of copies of Z by Z Xandi Tuni 2010-09-29T17:13:02Z 2010-10-05T13:08:58Z <p>The question is simple:</p> <p><em>Let $P$ be an infinite direct product of copies of $\mathbb Z$. Do there exist any nontrivial extensions $$0 \to \mathbb Z \to E \to P \to 0$$ in the category of commutative groups?</em></p> <p>In other words, I am asking whether the group $\mathrm{Ext}^1(P,\mathbb Z)$ is trivial. The problem here is of course that the group $P$ is not a free group.</p> <p>Already a funny thing happens with $\mathrm{Hom}(P,\mathbb Z)$. For any finite or infinite index set $I$, the canonical evaluation map $$\bigoplus_{i\in I}\mathbb Z \to \mathrm{Hom}\Big(\mathrm{Hom}\Big(\bigoplus_{i\in I}\mathbb Z,\:\mathbb Z \Big),\:\mathbb Z \Big) \cong \mathrm{Hom}\Big(\prod_{i\in I}\mathbb Z,\:\mathbb Z \Big)$$ is an isomorphism! That is a nontrivial statement (due to??), whose proof is not a formality at all. Replacing $\mathbb Z$ by, say, $\mathbb Z/p\mathbb Z$, the corresponding statement is wrong for infinite $I$.</p> http://mathoverflow.net/questions/37239/dense-cyclic-subgroup/37242#37242 Answer by Xandi Tuni for Dense cyclic subgroup Xandi Tuni 2010-08-31T09:58:07Z 2010-08-31T09:58:07Z <p>How about the infinite cyclic group itself with the discrete topology? Or p-adic integers?</p> http://mathoverflow.net/questions/30511/ebook-readers-for-mathematics/30517#30517 Answer by Xandi Tuni for eBook readers for mathematics Xandi Tuni 2010-07-04T13:34:41Z 2010-07-04T13:34:41Z <p>I tested the smaller version of Kindle a year ago or so - it really did not work too well. Kindle now supports pdf (that was not always so), and memory can probably be upgraded ad infinitum. The real problem is quite stupid, namely that there is no zoom-function. Rotating the page by 90 degrees and thereby displaying half of an A4 page is the best Kindle can do, and that leaves the text still awfully small. Reading text is already uncomfortable, reading a formula with a double superscript is impossible. This is especially frustrating if one third of the page witdh is occupied by margins.</p> http://mathoverflow.net/questions/124311/pullbacks-as-manifolds-versus-ones-as-topological-spaces/124317#124317 Comment by Xandi Tuni Xandi Tuni 2013-03-12T13:27:42Z 2013-03-12T13:27:42Z Reading Chris's answer, the first guess is nonsense. The pb doesn't exist. http://mathoverflow.net/questions/123994/tannaka-duality Comment by Xandi Tuni Xandi Tuni 2013-03-08T21:00:08Z 2013-03-08T21:00:08Z Nothing fails. In your example, the functor sending a representation to its conjugate by an non-unit element does not commute with the fibre functor. http://mathoverflow.net/questions/111059/relative-de-rham-cohomologies Comment by Xandi Tuni Xandi Tuni 2012-10-30T08:52:59Z 2012-10-30T08:52:59Z the otimes should be an oplus... http://mathoverflow.net/questions/93743/are-algebraic-groups-defined-by-their-invariants-in-tensor-spaces/93752#93752 Comment by Xandi Tuni Xandi Tuni 2012-04-11T11:47:59Z 2012-04-11T11:47:59Z I do not think that this answers the question. In order to use loc.cit., one should be able to deduce from the conditions (1) and (2) that that the inclusion $G\subseteq H$ induces an equivalence of categories $\mathrm{Rep}(H)\to \mathrm{Rep}(G)$. http://mathoverflow.net/questions/92290/about-tates-computation-of-k-2-rm-m-mathbb-q/92293#92293 Comment by Xandi Tuni Xandi Tuni 2012-03-26T17:37:08Z 2012-03-26T17:37:08Z thanks- that settles it. http://mathoverflow.net/questions/86089/two-definitions-of-character-of-topological-groups/86110#86110 Comment by Xandi Tuni Xandi Tuni 2012-01-19T18:49:18Z 2012-01-19T18:49:18Z N.b. The group $G$ in the 2nd paragraph is not locally compact. @georges: If $G$ is compact, its image under a character isn't necessarily discrete in the circle... http://mathoverflow.net/questions/78972/group-cohomology-of-symmetric-powers Comment by Xandi Tuni Xandi Tuni 2011-10-24T16:21:25Z 2011-10-24T16:21:25Z But isn't Sym(V) a direct factor of the tensor product as a G-representation? So you may just compute the cohomology of the tensor product, and then apply the projector. http://mathoverflow.net/questions/78460/learning-arakelov-geometry Comment by Xandi Tuni Xandi Tuni 2011-10-18T15:27:54Z 2011-10-18T15:27:54Z You should know about schemes in general, and a good deal about K-theory and intersection theory in particular (Fulton's book alone will not suffice). I suggest you have a look at &quot;Lectures on Arakelov Geometry&quot; by Soul&#233;, Abramovich etc..., and a check the references given there. http://mathoverflow.net/questions/60912/dense-sphere-packings-which-are-not-lattice-packings/60930#60930 Comment by Xandi Tuni Xandi Tuni 2011-04-07T13:08:52Z 2011-04-07T13:08:52Z thanks, that's already interesting news to me. http://mathoverflow.net/questions/60894/a-pair-of-subset-of-natural-numbers-having-density-but-whose-intersection-has-no/60900#60900 Comment by Xandi Tuni Xandi Tuni 2011-04-07T07:35:42Z 2011-04-07T07:35:42Z nah, now you have it twice. http://mathoverflow.net/questions/60816/want-to-get-back-into-maths-after-not-doing-it-since-age-18-whats-a-good-place-t Comment by Xandi Tuni Xandi Tuni 2011-04-06T14:19:12Z 2011-04-06T14:19:12Z that very much depends on how long is &quot;since the age of 18&quot;. http://mathoverflow.net/questions/60527/rank-of-subgroup-of-elliptic-curve Comment by Xandi Tuni Xandi Tuni 2011-04-04T16:08:10Z 2011-04-04T16:08:10Z Aside: Would it not be much easier to check linear independency of $p$ and $q$ by reducing modulo some prime? From theory you know there are plenty (a positive proportion) of primes $\ell$ such that $p$ and $q$ generate a noncyclic subgroup of $E$ mod $\ell$, provided $p$ and $q$ are independent. http://mathoverflow.net/questions/60434/are-chow-groups-generated-by-local-complete-intersections Comment by Xandi Tuni Xandi Tuni 2011-04-03T17:50:01Z 2011-04-03T17:50:01Z @Sasha: Just linear combinations. One could of course also look at the subring of the Chowring generated by lci's... why would the answer be yes then? http://mathoverflow.net/questions/60172/almost-clean-module Comment by Xandi Tuni Xandi Tuni 2011-03-31T12:25:57Z 2011-03-31T12:25:57Z At least one MO-user would appreciate if you'd recall briefly what &quot;almost clean&quot; means (it seems not to be a very difficult definition) and explain a bit your motivation. http://mathoverflow.net/questions/60101/density-of-irreducible-polynomials-in-mathbbzx Comment by Xandi Tuni Xandi Tuni 2011-03-30T21:03:24Z 2011-03-30T21:03:24Z See also this question: <a href="http://mathoverflow.net/questions/58397/the-galois-group-of-a-random-polynomial" rel="nofollow" title="the galois group of a random polynomial">mathoverflow.net/questions/58397/&hellip;</a>