User angelo - MathOverflow

Answer by Angelo for When is the intersection of an isolated normal singularity with a generic linear subspace through that singularity normal?

2013-05-20T20:20:05Z

I don't think so. There are examples of isolated normal threefold singularities that are not Cohen-Macaulay. A hyperplane section is not Cohen-Macaulay, hence it can not be normal, because a normal surface is Cohen-Macaulay.

Answer by Angelo for Homological characterization of smooth maps

2013-05-19T15:54:20Z

No: if $B$ is a quotient of $A$, then $B \otimes_A B = B$, but $B$ is very rarely smooth over $A$. The correct homological characterization of smooth maps involves André-Quillen cohomology.

Answer by Angelo for Rigidification and good moduli space (morphism) in the sense of Alper

2013-05-15T18:18:13Z

It is certainly not true that $\mathcal X \to \mathcal X^H$ is a good moduli morphism, unless $H$ is linearly reductive, because when you push forward the cohomology of $H$ will come into play.

On the other hand $\mathcal X^H \to X$ is a good moduli space, because the pushforward $QCoh(\mathcal X^H) \to QCoh(X)$ can be factored as the pullback $QCoh(\mathcal X^H) \to QCoh(\mathcal X)$ followed by the pushforward $QCoh(\mathcal X) \to QCoh(X)$, and both are exact.

Answer by Angelo for Can Inequivalent Topologies Have Same Sheaves/Cohomology?

2013-04-24T06:52:27Z

I would think so.

Two pretopologies are equivalent when they generate the same topology, that is, when the have the same sieves. If $A$ is an object of $C$, a sieve on $A$ is a subfunctor of the functor $h_A$ represented by $A$; and a subfunctor $S$ of $h_A$ is a sieve with respect to a topology $T$ if and only if the induced morphism $S^{\rm sh} \to h_A^{\rm sh}$ of the sheafifications is an isomorphism. Of course the category of sheaves determines the sheafifications; so if two topologies define the same sheaves, they admit the same sieves, so they are equivalent.

Answer by Angelo for What is "Data" involved in a mathematical construction?

2013-04-15T04:54:11Z

"Data" is the plural form of the Latin word "datum", which means, among other things, "thing that is given". Viewed this way, it makes perfect sense, doesn't it?

Answer by Angelo for Weil restriction of abelian schemes along finite étale (resp. finite locally free) morphisms

2013-04-10T19:21:48Z

Let $X \to S'$ be a smooth projective morphism, and $S' \to S$ a finite étale morphism.

The Weil restriction $Y \to S$ of $X \to S'$ is a closed subscheme of the scheme of morphisms $S' \to X$ on $S$, which is projective; hence it is projective (this is Grothendieck's original construction).

Let us show that $Y \to S$ is again smooth. Construction of the Weil restriction is étale local on $S$; hence we may assume that $S' = S_1 \sqcup \dots \sqcup S_m$ is a disjoint union of $m$ copies $S_1, \dots, S_m$ of $S$. Call $X_i$ the restriction of $X$ to $S_i$. It follows readily from the definition that $Y \to S$ is the product $X_1 \times_S \dots \times_S X_m$, and this proves smoothness.

If $X \to S'$ is an abelian scheme, the group scheme structure of $X \to S'$ induces a group scheme structure on $Y \to S$, and this completes the proof.

Answer by Angelo for Can one pick generators for the ring of invariants of binary n-ic forms which have rational coefficients?

2013-04-09T10:18:53Z

Sure. If you have an algebraic group $G$ defined over $\mathbb Q$ acting on a $\mathbb Q$-algebra $A$, the $\mathbb Q$-algebra of invariants $A^G$ is the equalizer of two usual homomorphisms of algebras $A \to A \otimes_{\mathbb Q}\mathbb Q[G]$. Tensoring with $\mathbb C$ is exact, so $(A \otimes \mathbb C)^{G_{\mathbb C}} = A^G\otimes \mathbb C$.

Answer by Angelo for The sum of same powers of all matrices modulo p

2013-04-07T20:11:10Z

My answer was nonsense, sorry.

Answer by Angelo for How to refer to a theorem that you have shown to be wrong

2013-04-05T17:00:01Z

I once saw a mathematician giving a talk about a theorem that he thought he had proved, for which a counterexample had later been found. He stated the "result" as follows:

Theorem (1983–1987): Let $A$ be $\dots$

(I made up the dates of birth and death)

Answer by Angelo for Finite-type Artin Stack over $\mathbb C$

2013-04-04T20:51:11Z

Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective, hence open. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $S \times_{\frak M} M$ surjecting onto $S$. The image of $U$ in $M$ is quasi-compact; let $V$ be a quasi-compact open subscheme of $M$ containing it. Since $S$ surjects onto $\frak M$, so does $U$. So $V \to \frak M$ is smooth and surjective, and $V$ is of finite type, so $\frak M$ is of finite type.

Answer by Angelo for A criterion for freeness over a local ring

2013-04-04T16:00:51Z

No, this is false as soon as $n ≥ 3$. A second syzygy $M$ of the residue field $K$ gives a counterexample: each $M[1/X_i]$ is projective, hence free, and it is reflexive, so the second condition is satisfied. On the other hand the projective dimension of $K$ as a module is $n$, so $M$ can not be free.

Two ways of getting a cohomology class from an extension of a discrete group by $\mathbb C^*$

2013-04-03T09:50:38Z

Suppose that $\overline G$ is a Lie group such that the connected component of $1$ is $\mathbb C^*$ . Assume that $\mathbb C^*$ is central in $\overline{G}$, and set $G := \overline G/\mathbb C^*$. There are two ways of constructing a class in $\mathrm H^3(G, \mathbb Z)$.

The first is to consider the central extension $$ 1 \longrightarrow \mathbb C^{*} \longrightarrow \overline G \longrightarrow G \longrightarrow 1 $$ as an extension of discrete groups; this gives a class in $\mathrm H^{2}(G, \mathbb C^{*})$ , and we can take its image in $\mathrm H^{3}(G, \mathbb Z)$ via the connecting homomorphism $\mathrm H^{2}(G, \mathbb C^{*}) \to \mathrm H^{3}(G, \mathbb Z)$.

The other involves the fibration $B \overline{G} \to B G$, with fiber $B \mathbb C^{*}$ . Since $B \mathbb C^{*}$ is a $K(\mathbb Z, 2)$ and the action of $G$ on $B \mathbb C^{*}$ is trivial, we obtain an obstruction to the existence of a section $BG \to B\overline{G}$, which lives in $\mathrm H^{3}(G, \mathbb Z)$. This is the second way.

An alternate construction of the second class (up to sign) is via the Leray-Serre spectral sequence $$ E^{ij}_{2} = \mathrm H^{i}(G,\mathbb Z)\otimes \mathrm H^{j}(B\mathbb C^{*}, \mathbb Z) \Longrightarrow \mathrm H^{i+j}(B \overline{G}, \mathbb Z) $$ in which the first possibly non-zero differential is $$ d_{3}\colon \mathrm H^{2}(B\mathbb C^{*}, \mathbb Z) \longrightarrow \mathrm H^{3}(G,\mathbb Z)\,. $$ Then we take the image of the canonical generator of $\mathrm H^{2}(B\mathbb C^{*}, \mathbb Z)$ under $d_{3}$.

Do these two classes coincide, up to sign? I am fairly sure that they do, and it's only my deep ignorance in algebraic topology that prevents me from seeing clearly why this is so. A reference would be better than an argument, but I'll be grateful for any hint.

Answer by Angelo for deRham cohomology of a manifold with covering space $S^{n}$

2013-03-22T06:43:51Z

A finite covering map induces an injection on de Rham cohomology. Try searching for "integration along the fibers"; yours is an easy case, as you integrate on a finite set of points, which means you just sum.

Answer by Angelo for Generalizing the square theorem

2013-03-21T06:58:42Z

There are many examples. Take $X := \mathbb A^1$ and $Y := \mathbb A^2 \smallsetminus \{(0,0)\}$ ; since all vector bundles on $X$ and $Y$ are trivial, it is sufficient to give an example of a vector bundle on $U := X \times Y$ that is not trivial.

Consider the embedding $j \colon U \subseteq \mathbb A^3$. The are reflexive sheaves $F$ on $\mathbb A^3$ that are locally free on $\mathbb A^3 \smallsetminus \{(0,0, 0)\}$ , but not at the origin. For example, you can take the sheaf associated with a second syzygy of the $\mathbb C[x,y,z]$-module $\mathbb C[x,y,z]/(x,y,z)$; this is not projective, because the projective dimension of $\mathbb C[x,y,z]/(x,y,z)$ over $\mathbb C[x,y,z]$ is 3, but it is reflexive, because every second syzygy is. The restriction $G$ of $F$ to $U$ is a non-trivial locally free sheaf on $U$. In fact, since $F$ is reflexive we have $F = j_*G$, so if $G$ were trivial $F$ would also be a free sheaf, which is not the case.

Answer by Angelo for Factoriality: local or global?

2013-03-14T13:05:51Z

It's a standard result in commutative algebra that every noetherian integral domain is a UFD if and only if every prime ideal of height 1 is principal. When applied to the local rings of X this gives exactly the equivalence above.

Answer by Angelo for Constants sheaves on an open subset

2013-03-08T13:29:23Z

This is not true; for example, take $X = \mathbb R^2$, $U = \mathbb R^2 \smallsetminus \{(0,0)\}$ . Then your $\mathbb Z_U$ coincides with $\mathbb Z_X$, and $Hom(\mathbb Z_U, F)$ is $F(X)$, not $F(U)$.

For the formula to hold you have to take as $\mathbb Z_U$ the extension of the constant sheaf by $0$, which is a very different animal.

Answer by Angelo for Rigid monoidal abelian category without an exact tensor functor to Vect

2013-02-22T04:59:58Z

Rigid symmetric monoidal abelian categories over a field $k$ with a fiber functor (i.e., a strong monoidal exact functor) to $K$-vector spaces for some extension $K$ of $k$ (the functor is assumed to exist, but not fixed) are are often called tannakian categories; a tannakian category with a fixed fiber functor to $k$-vector spaces is referred to as a neutral tannakian category. I will adhere to this terminology.

It was proved by Grothendieck, Saavedra-Rivano and Deligne that the 2-category of tannakian categories over a fixed field $k$ is equivalent to the 2-category of affine gerbes over $k$. With each affine gerbe $\Gamma$ you associate the category of vector bundles on $\Gamma$. Conversely, given a tannakian category $\mathcal A$, the objects of the corresponding gerbe over a $k$-algebra $A$ are fiber functors from $\mathcal A$ to finitely generated projective $A$-modules.

So, if $\Gamma$ is an affine gerbe, fiber functors from the category of vector spaces on $\Gamma$ to vector spaces on $k$ correspond to objects of $\Gamma(k)$; hence, to give an example of a tannakian category without fiber functors it is enough to give an example of an affine gerbe without a rational point. There are many such examples; for example, you can take a surjective homomorphism of algebraic groups $E \to G$, and a $G$-torsor $P$ over $k$ that does not lift to an $E$-torsor. The gerbe of liftings of $P$ to an $E$ torsor is an affine gerbe, and does not have any rational points.

As to examples of symmetric rigid monoidal abelian categories without fiber functors to any extension of $k$, the standard one is the category of $\mathbb Z/2\mathbb Z$-graded vector spaces.

Answer by Angelo for Group cohomology of orthogonal groups with integer coefficient

2013-02-17T22:26:19Z

When $n$ is a prime, the additive structure of $H^*(BPU_n, \mathbb Z)$ has been computed independently in Kameko, Masaki; Yagita, Nobuaki, The Brown-Peterson cohomology of the classifying spaces of the projective unitary groups ${\rm PU}(p)$ and exceptional Lie groups. Trans. Amer. Math. Soc. 360 (2008), no. 5, 2265–2284 and in Vistoli, Angelo, On the cohomology and the Chow ring of the classifying space of ${\rm PGL}_p$. J. Reine Angew. Math. 610 (2007), 181–227. For $n = 3$, the second paper contains a computation of the multiplicative structure.

Answer by Angelo for When are the Smooth Sections of a Bundle Generated as a Module (over Smooth Functions) by the Holomorohic Sections

2013-02-10T06:30:37Z

Swan has proved that taking global section gives an anti-equivalence between finitely generate projective $\Gamma^{\infty}(M)$-modules and $C^{\infty}$ vector bundles on $M$; this correspondence is functorial in $M$. Hence a set of section of a $C^{\infty}$ vector bundle $E$ on $M$ generated $\Gamma^{\infty}(E)$ if and only if it generates each fiber of $E$. So if $E$ is holomorphic, the holomorphic sections generate if and only if $E$ is globally generated. In particular, this is always true if $M$ is Stein. In the case of $L_k$ on $\mathbb{CP}^N$, this is true if and only if $k ≥ 0$.

Answer by Angelo for purity for finite flat group schemes

2013-02-04T16:57:07Z

Here is a counterexample for $G =\alpha_p$. Suppose that $k$ is a field of characteristic $p > 0$, and set $X = \mathbb A^2_k$, $Z= \{0\}$ . Set $U := X \smallsetminus Z$. Then $\mathrm H^1(U, \mathbb{G}_{\rm a}) = \mathrm{H}^1(U, \mathcal{O})$ is an infinite dimensional vector space over $k$; hence it is $p$-torsion, and from the long exact sequence associated with the exact sequence $$ 0 \longrightarrow \alpha_p\longrightarrow \mathbb{G}_{\rm a}\longrightarrow \mathbb{G}_{\rm a}\longrightarrow 0 $$ we see that $\mathrm H^1(U, \alpha_p)$ surjects onto $\mathrm H^1(U, \mathbb{G}_{\rm a})$ . If we take any class in $\mathrm H^1(U, \alpha_p)$ such that its image in $\mathrm H^1(U, \mathbb{G}_{\rm a})$ is $\neq 0$, this represents a torsor that does not extend to $X$, since $\mathrm H^1(X, \mathbb{G}_{\rm a}) = 0$.

[Edit:] Anon is right, my construction is nonsense. I apologize.

Answer by Angelo for flatness of power series rings

2013-01-31T13:11:47Z

As a module, $A[[X]]$ is the product of a countable family of copies of $A$. It is known that the product of flat $A$-modules is flat if and only if the ring $A$ is coherent, that is, every finitely generated ideal is finitely presented ( http://www.ams.org/journals/tran/1960-097-03/S0002-9947-1960-0120260-3/S0002-9947-1960-0120260-3.pdf ).

If you look at the proof of Theorem 2.1 in that paper, you can show that if $k$ is a field and $A$ is the quotient a polynomial ring $k[t_1, t_2, \dots]$ in countably many variables by the ideal generated by the products $t_it_j$, the product of countably many copies of $A$ is not flat over $A$.

Answer by Angelo for Galois action on special fiber of a stable model

2013-01-27T14:31:37Z

This is correct in characteristic $0$. Let $G$ be the kernel of the action; then one has to show that $X_{R_L}/G$ is a stable curve over $R_L^G$. This is Lemma 3.2 in my paper with Dan Abramovich Complete moduli for fibered surfaces, http://arxiv.org/abs/math/9804097.

I am pretty sure that it is false in positive characteristic, though.

[Edit:] I guess I was wrong about positive characteristic.

Answer by Angelo for Commutativity of the Chow ring in positive characteristic

2013-01-25T17:35:05Z

De Jong's theorem proves that the ring is commutative when tensored with $\mathbb Q$. For the torsion part, I am pretty sure that the question is still open.

[Edit:] Mikhail is absolutely right, with Gabber's theorem (http://www.math.u-psud.fr/~illusie/refined_uniformization3.pdf) one can show that the bivariant Chow ring tensored with $\mathbb Z[1/p]$ is commutative in characteristic $p > 0$.

Answer by Angelo for Extending line bundles

2013-01-25T17:11:52Z

This is not even true when all fibers are smooth. For example, see Example 5.22 in Kleiman's paper on the Picard scheme in "FGA explained".

Answer by Angelo for Automorphism groups of general type varieties

2013-01-21T16:55:18Z

William Lang produced examples of surfaces of general type in positive characteristic with non-zero vector fields. Since surfaces of general type must have finite automorphism groups, this gives examples. See William Lang, "Examples of surfaces of general type with vector fields", Arithmetic and geometry, Vol. II, Progr. Math., 36, Birkhäuser, pp. 167–173

Answer by Angelo for Top chern class under finite, unramified, dominant morphism

2013-01-20T12:56:41Z

Yes, it does hold in positive characteristic. You can show that the degree of $c_n(X)$ equals its Euler characteristic with respect to étale cohomology with coefficients in $\mathbb{Q}_{\ell}$, where $\ell$ is a prime different from the characteristic ( http://mathoverflow.net/questions/80697/top-chern-class-in-positive-characteristic ), and this is has the required behavior under étale covers ( http://mathoverflow.net/questions/41576/behaviour-of-euler-characteristics-in-characteristic-p-for-finite-etale-covers ).

[Edit:] the equality can be proved directly; in fact the proof is much easier. Since $\pi\colon Y\to X$ is étale, the tangent bundle $T_Y$ is the pullback $\pi^*T_X$ . By functoriality of Chern classes, $c_n(Y) = \pi^*c_n(X)$ (at the level of Chow groups). But it easy to see that the composite $\pi_*\pi^*$ from the Chow group of $X$ to itself is just multiplication by $d$.

Answer by Angelo for A more general form of Grauert's Theorem on Higher Direct Image Sheaves?

2013-01-09T21:35:06Z

There is a very clear exposition of the base change theory in Mumford's book on abelian varieties.

Answer by Angelo for Origin of notion of "split Grothendieck group"?

2013-01-07T13:33:21Z

The split Grothendieck group for vector bundles on a complete variety appears in Nori's PhD thesis on the fundamental group scheme; this was published in the Proceedings of the Indian Academy of Science in 1981. It is used to define and study finite vector bundles. Nori does not give any references, so as far I know the construction might be due to him.

Answer by Angelo for Projection formula for immersions

2012-12-11T15:38:00Z

Set $X = \mathbb A^2_k$, $Y = X \smallsetminus \{(0,0)\}$ , and suppose that $M$ is non-zero and supported at the origin.

Answer by Angelo for Profinite groups as étale fundamental groups

2012-12-01T11:04:06Z

[Edit:] The answer should be positive, that is, every profinite group appears as the fundamental group of a scheme. Here is a sketch of proof.

First of all, I claim that for any finite group $G$ there exists a complex affine simply connected variety $X$ with a free action of $G$. Start from a faithful finite-dimensional representation $G \to \mathrm{GL}(V)$ of $G$, such that there is an open subset $U \subseteq V$ where the action of $G$ is free, and such that $V \smallsetminus U$ has codimension at least 2 in $V$. Then $X$ is obtained with an easy equivariant extension of Jouanolou's trick. This yields a $G$-covering $X \to X/G$; then $X/G$ is an affine variety, and its fundamental group is $G$.

Now take a profinite group $G = \projlim_{i\in I}G_i$; identify $G$ with the corresponding affine group scheme over $\mathbb C$, in the usual fashion. For each finite subset $J\subseteq I$ denote by $G_J$ the image of $G$ in $\prod_{j\in J}G_j$; clearly we have $G = \projlim_{J \subseteq I}G_J$.

For each $i$ take a complex affine simply connected variety $X_i$ with a free action of $G_i$. Consider the affine scheme $X := \prod_{i \in I}X_i$, with the resulting action of $G$, and the quotient $X/G$, which is the spectrum of the ring of invariants $\mathbb C[X]^G$. For each finite subset $J\subseteq I$ set $X^J := \prod_{j \in J}X_j$. The action of $G$ on $X^J$ factors through a free action of $G_J$. Furthermore, we have $\mathbb C[X]^G = \injlim_{J \subseteq I}\mathbb C[X^J]^{G_J}$, hence $X/G = \projlim X^J/G_J$.

Now, $X^J$ is simply connected, hence the fundamental group of $X^J/G_J$ is $G_J$. On the other hand, it is easy to see that the Galois category of $X$ is the inductive limit of the Galois categories of the $X^J$, so that its Galois group is precisely $\projlim G_J = G$.

[Edit2:] Let me clarify what I mean by the "equivariant Jouanolou trick". The theorem of Jouanolou says that if $U$ is a quasi-projective variety, there exists an locally trivial fibration in affine spaces $X\to U$, where $X$ is affine. What we need here is the statement that if $G$ is a finite group acting on $U$, then we can construct such a map $X \to U$ that is also $G$-equivariant. This is easy: start from a fibration in affine space $Y \to U$ with $Y$ affine, and take $X$ to be the fiber product over $U$ of all the $g^*Y$ for $g \in G$, with the obvious action of $G$.

Comment by Angelo

2013-05-21T07:30:14Z

If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours.

Comment by Angelo

2013-05-21T06:29:27Z

If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours. I voted to close.

Comment by Angelo

2013-05-19T17:24:04Z

Obviously it does not vanish anywhere on $\mathbb R^3$, but you are restricting it to $S^2$, and the restriction map is not pointwise injective. I voted to close as "too localized".

Comment by Angelo

2013-05-18T02:17:07Z

Please read "how to ask".

Comment by Angelo

2013-05-14T21:51:17Z

No. Take $A = k[x,y]$, and as $M$ the quotient field of $k[x] = k[x,y]/(y)$.

Comment by Angelo

2013-05-13T16:52:09Z

If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours. I voted to close as off topic.

Comment by Angelo

2013-05-12T11:43:46Z

I suppose you want $k$ to be also algebraically closed, otherwise the subgroup of elements fixing $\overline k \subseteq K$ would be a proper normal subgroup. I don't have access to Lascar's paper, but from the title I would guess he takes $k = \overline{\mathbb Q}$.

Comment by Angelo

2013-05-12T09:42:32Z

I meant this: it is very hard to imagine this might be true, but if it were, it would probably be extremely difficult to prove. In any case, it seems that you are being given counterexamples.

Comment by Angelo

2013-05-12T07:57:58Z

You want to do this for every possible family? That is very hard to imagine.

Comment by Angelo

2013-05-10T04:51:51Z

If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours.

Comment by Angelo

2013-05-09T13:50:18Z

If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours.

Comment by Angelo

2013-05-09T05:35:56Z

The algebraic cohomology of $\mathbb G_{\rm m}$ is very different from the analytic cohomology of $\mathcal O^*$. For a regular scheme the former is always torsion in degree at least 2 (this is a well-known result of Grothendieck), whereas the analytic cohomology of a complex manifold tends to contain positive-dimensional $\mathbb Q$-vector spaces.

Comment by Angelo

2013-05-09T05:23:20Z

If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours.

Comment by Angelo

2013-05-08T16:33:06Z

Please read the FAQ.

Comment by Angelo

2013-05-08T03:05:43Z

If you can't do your homework problems, I would suggest, first of all, studying, and, if this fails, going to your teacher's office hours.