User vivek shende - MathOverflow

Semicontinuity for complexes

2013-02-20T22:13:34Z

In algebraic geometry, the very useful semicontinuity theorem tells you the following:

Let $X \to Y$ be a projective morphism of schemes, and $F$ a coherent sheaf on $X$ which is flat over $Y$. Then the dimensions of the cohomology on fibers, $h^i(X_y)$ are upper semicontinuous functions of $y$, i.e., the locus where in $Y$ where $h^i(X_y) \ge n$ is Zariski closed for any $n$.

What condition on a complex $F$ with coherent cohomology ensures that $\dim \mathbb{H}^i(X_y, F \otimes^L \mathcal{O}_{X_y})$ is an upper semicontinuous function of $y$?

How do I find abelian subcategories of periodic triangulated categories?

2013-05-19T21:12:13Z

If $T$ is a triangulated category, then the formalism of $t$-structures gives a way to find abelian subcategories inside. You're supposed to find two strictly full subcategories, $T^{\le 0}, T^{\ge 0}$ so that

1) $T^{\ge 0}[-1] \subset T^{\ge 0}$ and $T^{\le 0}[1] \subset T^{\le 0}$

2) There are no nonzero maps from an object in $T^{\le 0}[1]$ to an object in $T^{\ge 0}$

3) For any object $K$ there's a triangle $A \to K \to B$ with $A \in T^{\le 0}[1]$ and $B \in T^{\ge 0}$.

But if $T$ is periodic, say $[n]$ is the identity, this does not seem like a recipe for success.

How do you find abelian subcategories of a periodic triangulated category?

When do two positive braids represent the same link?

2013-05-14T05:31:08Z

Let $B_n$ be the braid group on $n$ strands, with the usual generators: $s_1, \ldots, s_{n-1}$ and their inverses, where $s_i$ is a positive half-twist interchanging the strands labelled $i$ and $i+1$. Markov's theorem says that two elements of the braid group have closures giving the same knot in $S^3$ when they are related by a sequence of moves, each of which is either a conjugation $\beta \to s_i^{\pm} \beta s_i^{\mp}$ or the inclusion $B_n \to B_{n+1}$ taking $\beta \to \beta s_n^{\pm}$.

Recall a braid is positive if it can be written as $s_{i_1} \cdots s_{i_k}$. Conjugation does not generally preserve positive braids, but it can: $s_1^{-1} (s_1 \beta) s_1$ is of course positive when $\beta$ is.

If two positive braids represent the same knot, can they be related by a sequence of Markov moves in which only positive braids appear?

Answer by Vivek Shende for Riemann-Roch theorem for arbitrary 1-dim schemes

2013-03-15T17:53:54Z

You may find this paper of Hartshorne to be helpful.

Answer by Vivek Shende for Injective morphism from curves to $\mathbb CP^2$

2013-02-26T08:04:25Z

Update.

Estimates coming from the Bogomolov-Miyaoka-Yau inequality yield:

Corollary. (to Theorem A in this paper of Sakai) There is no $M$ such that every curve admits an injective morphism to a curve in $\mathbb{P}^2$ whose points all have multiplicity at most $M$.

Proof. The geometric genus $g$ curves of degree $d$ move in a family of dimension $3d + g - 1$, so if these are to fill the moduli space we must have $2g \le 3d + 2$. On the other hand, Theorem A of this paper of Sakai gives after rearranging that for any genus $g$, degree $d$, unibranch plane curve, $$ 2g \ge \frac{d^2}{2m+1} + O(md) $$ where $m$ is the maximal multiplicity of a point on the curve. For any bounded $m$, these inequalities cannot both be satisfied for large $g$. $\square$

Remark: the argument also shows that the same statement holds restricted to hyperelliptic curves. This is not incompatible with Jeremy's answer below, which requires a cusp of multiplicity $2g$ to inject a hyperelliptic curve of genus $g$.

I leave the older argument below because it is a different and possibly still useful approach.

Theorem. There is no finite set $\Sigma$ of topological types of unibranch singularities such that every smooth curve admits an injective map to a curve in $\mathbb{P}^2$ with only singularities from $\Sigma$.

Proof. I will argue that curves with unibranch singularities of total cogenus $\delta$ occur in the linear system $|\mathcal{O}(d)|$ with codimension at least $\mathrm{min}(\kappa, 3d - 1 + \delta)$, where $\kappa$ is the codimension of the equiclassical locus in the versal deformation of the singularities; it is $2\delta$ in the case of curves with only cusps, and for any finite set of singularities is bounded below by $(1 + \epsilon)\delta$ for $0 < \epsilon \le 1$ some constant depending on the set of singularities. Let us first see why this implies the result.

Saying a geometric genus $g$ curve appears in degree $d$ in codimension $3d - 1 + \delta = 3d + (d-1)(d-2)/2 - g - 1$ means it comes in a $g + 1 + (d+1)(d+2)/2 - (d-1)(d-2)/2 -3d = g + 3$ dimensional family. On the other hand $\kappa$ is given by $\delta + r$, where $r$ is the total ramification divisor on the normalization of the curve (for the normalization map), i.e. the sum over the branches of the singularities of (multiplicity - 1). So in this case the curves come in a family of dimension $g + 4 + 3d - r$. I don't know what to do with this, but under the further assumption that the set of singularities from which we choose must be finite and so we get $\kappa = (1 + \epsilon) \delta$ as above (e.g., for only cusps, $\epsilon = 1$), then the argument in Will's answer above shows this can fill the moduli space for only finitely many $g$.

To get the advertised codimension estimate, let $C$ be such a curve of degree $d$. Let $\Lambda$ be the tangent space to the versal deformation of the singularities of $C$. Then there exists a linear subspace $L \subset \Lambda$ such that

equigeneric ideal $\ge$ L $\ge$ equiclassical ideal
aside from the above restriction, L may be taken so that the dimension of the equigeneric ideal mod L is up to $3d - 1$.
the image of $T_C|\mathcal{O}(d)| \to \Lambda$ is transverse to $L$.

This is essentially classical, for the sort of argument to prove it you can see e.g. the statement of Lemma 8 and the proof of Lemma 10 in my paper with Steve. $\square$

It's not yet clear to me how to improve this to answer the OP, maybe the better sort of bounds of Greuel, Lossen, and Shustin will help? (Or maybe there's just some easy argument not involving deformation theory...)

Which monomial subschemes are limits of smooth subschemes?

2013-03-13T04:12:05Z

Say $X \subset \mathbb{P}^n$ is monomial, by which I mean that it's cut out by a homogenous ideal generated by monomials, or in other words it's a fixed point on the Hilbert scheme for the action of the big torus. What is known about the following question:

When is $X$ in the closure of the locus in the Hilbert scheme parameterizing smooth subvarieties?

I'm particularly curious about the case when $X$ is zero dimensional, about which I would also hope more is known than in general...

Chern numbers via Euler characteristics?

2013-03-07T21:18:55Z

Let $X$ be a space good enough to have a fundamental class, and $E$ a complex vector bundle on $X$. Let $P$ be some polynomial expression, and say I want to evaluate $P(c_i(E)) \cap [X]$.

Is there some naturally associated space, $\mathbf{P}(X, E)$ such that (e.g.) the topological Euler characteristic $\chi(\mathbf{P}(X, E)) = P(c_i(E)) \cap [X]$?

Answer by Vivek Shende for Is it true that the only singularities upto codimension seven are the ADE singularities?

2013-03-02T05:20:36Z

Nonreduced curves in $|\mathcal{O}(d)|$ occur in codimension $2d-1$, so you will want $d > 4$ to get rid of them.

Fix a topological singularity type, and let $s$ be the codimension of its equisingular family inside its versal deformation. Then in $\mathcal{O}(d)$, singularities of this kind will occur in codimension at least $\mathrm{min}(d+1, s)$ because any length $d+1$ subscheme imposes indepedent conditions on $\mathcal{O}(d)$, and so the former can be shown to be transverse to a general $d+1$ codimensional subspace of the versal deformation of a singularity. Thus once $d \ge 7$, any singularity with $s > 8$ will not appear. This certainly limits the ADE singularities which may appear to the ones you listed. And if you have a handy table saying the only singularities with $s \le 7$ are ADE, then you could conclude.

Because I don't have such a table (and because I'm trying to stay awake long enough to get over my jetlag), let's show it. Certainly $s \ge \delta$, and equality holds only for a node. So you only have to worry about singularities which have $\delta < 7$. In fact there's a better estimate coming from considering the 'equiclassical locus': $s \ge \kappa = \delta + \sum(m_i -1)$ where $m_i$ are the multiplicities of the components.

Recall that $\delta$ is the sum over all infinitely near points of $m(m-1)/2$, where $m$ is the multiplicity of the curve at that point. In particular, if $\delta \le 6$, then certainly $m \le 4$, and moreover if the multiplicity is $4$ then all branches must be smoothed and made disjoint after one blowup. Now I list topological types of singularities with $\delta < 7$, by number of branches:

$y^2 = x^{2n+1}$ for $n \le 6$, or $y^3 = x^4, x^5, x^7$, or $x^4 = y^5$.
$y^2 = x^{2n}$ for $n \le 6$, or a union of $y^2 = x^{2n+1}$ with a transverse line for $n \le 4$, the union of $y^2 = x^3$ with its tangent cone, or two ordinary cusps $(y^2 = x^3)$ with different tangent directions.
If any branches are singular, it must be the union of an ordinary cusp with two transverse lines. One can have $x(y^2 - x^{2n})$ for $n = 1, 2, 3$. Also there is the possibility $y(y+ax^2)(y+bx^2)$.
Four mutually transverse lines.

How did I make this list? Unibranch singularities are classified by sub semigroups of $\mathbb{N}$, so I listed the ones corresponding to semigroups with at most 6 gaps; then I tried gluing them together. Most of the above list were ADE singularities. The ones that weren't:

(a) $x^3 = y^7$ (b) $x^4 = y^5$, (c) two ordinary cusps with different tangents, (d) an ordinary cusp with two transverse lines, (e) $y(y+ax^2)(y+bx^2)$, and (f) four mutually transverse lines.

The items (a), (b), (c), are ruled out by the constraint $\kappa < 8$. In the other cases, there is a obvious $\delta$-constant deformation (move a smooth branch) which on the one hand changes the topological type of the singularity, and on the other does not reduce it to all nodes, so the equisingular locus must be of codimension at least 2 in the $\delta$-constant locus.

One reference for this kind of stuff is Diaz and Harris's ``Ideals associated to deformations of singular plane curves''.

Answer by Vivek Shende for Linearization instability and singular points of algebraic varieties

2013-02-24T04:09:54Z

Maybe the best answer I can give is ''learn about deformation theory''. But I will take a shot at writing the dictionary you request, in the case where $E(u)$ is a polynomial function

linearization stability <--> smoothness / instability <--> singularity
linearization stable/unstable point <--> singular / smooth point
linearized solutions <--> Zariski tangent space
functions like $Q[u]$ <--> obstructions
the zero set $Q[u]=0$ <--> unobstructed deformations I guess? or the formal completion of $X$ near the given point
functions like $S_i[u]$ <--> the functions $\partial E/\partial u$, or maybe it is better to say, a certain Fitting ideal of the sheaf of Kahler differentials on $X$
the zero set $S_i[u] = 0$ <--> on $X$, you would I guess call it the singular locus; in the whole ambient space I don't know.

Regarding ''learn about deformation theory'', the point is that you started with some very explicit algebraic variety $X$ and then decided to think about its structure near a point. But deformation theory lets you think about the following thing: say you want to think about the space of all algebraic ''solutions'' of some sort near a given ''solution'', but to a less explicit problem like ''describe all holomorphic maps from a genus 5 curve into projective space'', which of course can also be written as a PDE. It may be difficult or worse to construct such a space globally, but still you can start to think about the local infinitesimal structure near a given solution using infinitesimal methods; and then there are very powerful algebraic tools (the Artin approximation theorem) which let you under good conditions construct an algebraic neighborhood of your given solution.

When is a local Artin C-algebra a subring of C[t]/t^n

2010-11-25T17:40:29Z

Let $A$ be a local ring over $\mathbb{C}$, which moreover is a finite dimensional $\mathbb{C}$-vector space.

When is $A$ a subring of $\mathbb{C}[t]/t^n$?

What does the minimal such $n$ have to do with $A$?

Example: the ring $\mathbb{C}[x,y]/(x^3,y^2,xy)$ is a subring of $\mathbb{C}[t]/t^5$ by $x = t^2$ and $y = t^3$.

Answer by Vivek Shende for Math Annotate Platform?

2013-02-19T19:04:30Z

I think such a thing would provide immense value. In particular I can think of instances when the following sorts of comments would have saved me a great deal of time:

(1) No need to read pages XX-XXX, here is a one paragraph argument.

(2) This result has since been strengthened, see ...

(3) The following claims are not quite right, here is a counterexample, and here is how to fix it.

(4) The following claims actually are right, even though the following might at first seem like a counterexample.

(5) What the author really means by [SGA] is [SGA N, page XXX]

(6) This result has the following interesting applications ... (6a) What would be even better is an automated system where, not just can you see what papers cite a given paper as you can today, but you can even see where a given lemma or proposition is cited.

(7) The author has only cited the relevant papers of his friends, the following other work in the subject is closely related.

(8) This paper is actually much less / much more interesting than it sounds...

(9) The following seems to be a gap in the argument:

(10) This 200 page paper assumes along the way in places which are explicit but maybe you didn't notice the following conjectures...

I think it would be essential however to ensure that people post under their own names and other measures are taken to ensure responsibility and measure the credibility of authors, but I think at the present stage of development of the internet we know how to do that.

I also think items like (3), (4), (9), (10) will become increasingly important; already it seems that people who consider themselves sufficiently famous don't necessarily bother publishing in journals (and so are not subjected to the review system), or even if they do are perhaps sufficiently famous to override or intimidate the reviewers, perhaps by sheer number of pages, etc...

What does "$H^*(X)$ is Hodge-Tate" mean?

2013-02-04T00:50:30Z

Let $X$ be a (let us say smooth to obscure any confusions I have between $H(X)$ and $H_c(X)$) algebraic variety defined over some subfield of $\mathbb{C}$. I have occasionally overheard the expression "$H^*(X)$ is Hodge-Tate" used to mean something which, as far as I could tell from context, resembled one of the following:

(1) $H^*(X)$ is generated by $(p,p)$ classes, i.e. those in some intersection $W_{2p} H^i(X,\mathbb{Q}) \cap F^p H^i(X,\mathbb{C})$, where $W$ and $F$ are the weight and Hodge filtrations from the mixed Hodge structure. In particular were $X$ smooth and proper, $H^*(X) = \bigoplus H^{p,p}(X)$.

(2) Spread $X$ out as appropriate and reduce mod a good prime, then it is `polynomial count', i.e. the number of points over $\mathbb{F}_{p^n}$ is a polynomial in $p^n$.

(3) Spread $X$ out as appropriate and reduce mod a good prime, then all the eigenvalues of Frobenius are powers of $p$.

(4) The class of $X$ in the Grothendieck group of varieties is in $\mathbb{Z}[\mathbb{A}^1]$

But when I searched for "Hodge-Tate" on google, I arrived at some description of "Hodge-Tate numbers" etc which seemed to have something to do with p-adic Hodge theory and apply to any variety. Anyway my question is as in the title,

What does it mean for $H^*(X)$ to be Hodge-Tate?

Also I guess (4) => (3) => (2) and I vaguely recall from some appendix of N. Katz that => (1) can be tacked on the end (?) I would also like to know

Which of the reverse implications is false, and what are some counterexamples?

Resolving the universal sheaf on the Quot scheme

2012-12-16T18:02:04Z

Let $X$ be a smooth projective variety, and $Quot$ some quot scheme, i.e. it parametrizes quotients of some fixed sheaf $F$ on $X$ of some fixed Hilbert polynomial. There is a universal quotient sheaf $\mathcal{Q}$ on $X \times Quot$ such that $\mathcal{Q}|_{X \times [E]} = E$. Because $X$ is smooth and $F$ is flat over $Quot$, there is a finite resolution of the universal quotient $\mathcal{Q}$ by finite dimensional locally free sheaves on $X \times Quot$.

Recall that by construction, $Quot$ enjoys a Plucker embedding into a Grassmannian of sections of $F(n)$ for some large $n$.

(How) can the resolution of $\mathcal{Q}$ be described explicitly in terms of restrictions of tautological bundles on the Grassmannian?

Answer by Vivek Shende for What are the fixed points of the jacobian acting on the compactified jacobian ?

2012-11-23T09:39:56Z

I think that's the only fixed point.

The compactified Jacobian is a homeomorphic to a product of the Jacobian of the normalization times some local factors from the singularities; likewise the Jacobian splits into the Jacobian of the normalization times some local factors. In particular this means that there are certainly no fixed points if the geometric genus is > 0, since the Jacobian of the normalization acts freely; also the splitting into the product of the contributions for the singularities means you might as well consider a curve with a unique singularity.
If $u:C' \to C$ is the minimal unibranch partial normalization (i.e. you separate all the branches but don't do anything else), then I believe the entire fixed locus lies in the pushforward of torsion free sheaves on $C'$; you should find a proof in Beauville's article on rational curves on K3 surfaces. In particular if the curve is immersed, there are no other fixed points than the one you describe.
Certainly if the normalization is $n:\mathbb{P}^1 \to C$, then $n_* \mathcal{O}$ is a fixed point. By the projection formula, $L \otimes n_*(\mathcal{O}) = n_{*} (n^*L \otimes \mathcal{O}) = n_{*}(\mathcal{O})$.
Passing to the complete local ring $R$ at the (now unibranch) singularity, the space of torsion free sheaves (of some fixed degree) is constructibly (with respect to the orbits of the Jacobian) in bijection with the space of $R$-modules $R'$ with $\mathbb{C}[[t]] > R' > \mathrm{Conductor}(R)$, with the same relative dimension as $R$. The action of the Jacobian is just the action of the invertible power series (modulo the action of $R^*$). From this it is clear the only fixed point is some $t^k \mathbb{C}[[t]]$ itself, i.e. the pushforward of the normalization.

p.s. I may have assumed characteristic zero at some point.

can an nonzero IC sheaf have zero hypercohomology?

2012-11-19T17:27:54Z

Can someone tell me which of the following are true? Let $X$ be a reasonable space.

Suppose $F$ is a complex whose cohomology groups are constructible sheaves, at least one of which is nontrivial.

Can $\mathbb{H}(X, F) = 0$?

If so, can it still happen assuming $F$ is really just

(1) a constructible sheaf (2) a local system (3) a perverse sheaf (4) an intersection cohomology complex ?

Are Donaldson-Thomas invariants "A-model" or "B-model" ?

2012-07-04T15:29:09Z

Donaldson-Thomas invariants are the (virtual) Euler characteristics of moduli spaces of elements of the derived category of coherent sheaves (with some fixed Chern class, satisfying some stability condition, etc.) which bear some relation to "holomorphic Chern Simons theory", whatever that is.

Should I think of these as "A-model" or "B-model" invariants?

On the one hand, DT invariants come from the bounded derived category of coherent sheaves, which is what features in the B-model. On the other, there is the MNOP conjecture which tells me that the DT invariants of a CY 3-fold are ``the same'' as the Gromov-Witten invariants of the SAME 3-fold, which are A-model things.

As I understand it, according to Costello, if I take the cyclic A-infinity category built out of d-b-coh and run it through his machinery, I should get the topological string corresponding to the B-model. But (modulo my confusion on the matter) according to Kontsevich and Soibelman, if I take the cyclic A-infinity category built out of d-b-coh and run it through their machinery, I should get DT invariants. So what is going on?

Bounding the size of stalks of IC sheaves

2012-02-10T21:01:45Z

Say $X$ is a smooth algebraic variety, $U$ is a Zariski open set in $X$, $L$ is a local system on $U$, and $IC(L)$ the intersection cohomology sheaf on $X$ which restricts to $L$ on $U$. Then is:

$$\dim L_u \ge \sum_i \dim \mathrm{H}^i(\mathrm{IC}(L)_x) $$

for $u \in U$ and all $x \in X$?

If not, is it true after e.g. putting in some scalar depending only on the dimension of $X$?

Which blowups of a smooth surface are smooth?

2012-01-30T06:42:21Z

Say $X$ is a smooth algebraic surface and $Z$ is some subscheme. When is the blowup of $X$ along $Z$ smooth?

Decomposition theorem and virtual Poincare polynomial

2011-07-08T18:30:12Z

Let $f:X \to Y$ be a proper morphism between smooth algebraic varieties (say over $\mathbb{C}$), let me write $A_X$ for the constant sheaf on $X$ with coefficients of the appropriate type. Then the decomposition theorem of [Beilinson-Bernstein-Deligne] tells me that $f_* A_X$ splits as a direct sum of shifted semisimple perverse sheaves.

So suppose I have collected some of the summands, say into $\mathcal{F}$, and want to know if I have found everything. Invoking proper base change, it is enough to ask, for every point $y \in Y$, whether

$\dim H^*(y,\mathcal{F}_y) = \dim H^*(X_y)$,

since any missing summands would have to contribute something somewhere. But computing the RHS requires actually thinking about the topology of $X_y$. Something which is easier to compute are the virtual Betti numbers, defined for an arbitrary variety in terms of the weight filtration as

$b^j_{vir}(Z) = \sum_i (-1)^{i+j} \dim Gr^j_W H^i_c(Z)$

At first glance these may not look easier, but the point is that they are motivic, i.e., additive under cutting into constructible pieces (and multiplicative under Zariski locally trivial fibrations); evidently they agree with the ordinary Betti numbers for smooth proper varieties. Thus for instance since a rational curve with a single node can be reassembled into $\mathbb{A}^1$, it has a single nonvanishing virtual Betti number, $b^2_{vir} = 1$.

To determine whether $f_* A_X = \mathcal{F}$, the latter known to be a summand of the former, is it enough to compare virtual Betti numbers? To be precise, does it suffice to know, for every $y \in Y$ and all $j$, that $b^j_{vir}(X_y) = \sum_i (-1)^{i+j} \dim Gr^j_W H^i(y,\mathcal{F}_y)$?

I sketch an argument in the affirmative which unfortunately due to my lack of knowledge about what weights are, I cannot seem to make precise. It is enough to show that for any nonzero summand $G$ of $f_* A_X$, there is some $y \in Y$ where not all the virtual Betti numbers of $G_y$ vanish. Let me take $y$ which has a neighborhood where $G$ is a sum of shifted local systems supported on some smooth subvariety $Y'$, and here it seems to me that restricting to $y$ must act as some global shift on the weights. Since $G$ was pure to begin with, being a summand of $f_* A_X$, so is $G_y$ (up to a shift) and hence its virtual Betti numbers are its actual Betti numbers, which see $G$.

I have put the "reference request" tag in the hopes that if the answer to the question is yes, it is some well known thing, maybe somewhere in [BBD], to which someone can point me.

Cohomology of projective space bundles

2011-07-05T09:17:08Z

Suppose $Y$ is an algebraic variety and $\mathcal{E}$ a coherent sheaf on $Y$. Suppose $f:X=\mathrm{Proj}(\mathrm{Sym}(\mathcal{E})) \to Y$ is a morphism of algebraic varieties with all fibres scheme theoretically projective spaces.

If the fibres all had the same dimension, I would have $\mathrm{R}f_* \mathbb{C}_X = \mathrm{H}^*(\mathbb{P}^n) \otimes \mathbb{C}_Y$.

In the case that the fibre dimension varies, let $Y_k$ be the locus where the fibre dimension is at least $k$. Then is it true that $\mathrm{R}f_* \mathbb{C}_X = \bigoplus \mathbb{C} _{Y_k}[-2k](k)$?

(and if not in general, are there any reasonable assumptions which make it true?)

What is the replacement for a "sufficiently small disc" in characteristic p?

2011-06-23T08:59:24Z

How do I make the following sort of argument work in characteristic p?

Let $f:X \to Y$ be a proper equidimensional map of smooth algebraic varieties, assume all fibres are reduced. Say at some point $y \in Y$, I have computed the differential and know that $df(T_x X) \supset V$ for all $x \in X_y$ and some vector subspace $V \subset T_y Y$. Then over a sufficiently small polydisc $D$ such that $T_y D + V = T_y Y$, the total space of $X_D$ is smooth. Thus the same is true of a (say) one-dimensional thickening $\tilde{D}$ of $D$, and so moving $D$ a little bit in $\tilde{D}$ produces many deformation equivalent smooth manifolds (with boundary) $X_{D'}$.

In particular if I want to know something about $H^*(X_y)$, I can first thicken to the smooth $X_D$ which is homotopy equivalent to it, then move $X_D$ to the diffeomorphic $X_{D'}$, which I prefer because say $D'$ avoids some bad points in $Y$.

Answer by Vivek Shende for Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof

2011-06-24T02:29:18Z

This is worked out in excruciating detail in the article Jacobians and Symmetric products by Schwarzenberger. I think the arguments there are perfectly good in the families setting as well.

Tangent cones to Severi strata

2011-06-24T01:58:43Z

Let $\mathbb{C}[[x,y]]/f(x,y)$ be a reduced plane curve singularity. The base of a versal family can be taken to be (an open subset in) $\Lambda = \mathbb{C}[x,y]/(f,\partial_x f, \partial_y f)$; the curve over $g \in \Lambda$ being the locus cut out by $f+g$. One can consider the closed loci $\Lambda_h$ where the sum of the $\delta$-invariants ("virtual number of nodes near the origin") of the fibre is at least $h$.

The smallest of these -- where the $\delta$-invariant is the same as that of the central fibre -- is sometimes called the equigeneric stratum, and it was shown by Diaz and Harris that the reduced subvariety of $T_0\Lambda$ underlying the tangent cone to $\Lambda_\delta$ is the image of the conductor ideal $I \subset \mathbb{C}[x,y]/f$ inside $ \mathbb{C}[x,y]/(f,\partial_x f, \partial_y f)$.

Is there a description of the tangent cones of the other $\Lambda_h$ ?

Intersection theory over non algebraically closed fields

2011-06-17T17:13:31Z

Let $k$ be a field, consider intersecting schemes over $k$.

Is there a version of intersection theory which keeps track of the extension of $k$ over which the intersections are happening?

Ideally I would imagine that $A_0(pt)$ was the representation ring of $Gal(\overline{k}/k)$, so that I could take various traces to extract information about who is living where.

Transversals to singular subvarieties

2011-06-12T15:11:35Z

Say $\mathbb{C}^d \subset Y^{N-k} \subset \mathbb{C}^N$ are closed imbeddings of complex analytic subvarieties of the indicated dimension, $Y$ is not smooth. At a point $y \in Y$, a generic, sufficiently small polydisc $\mathbb{D}^k \ni y$ will satisfy $\mathbb{D}^k \cap Y = y$. I would like to do this continuously along the $\mathbb{C}^d$.

For $p \in \mathbb{C}^d$, does there exist an analytic neighborhood $p \in U \subset \mathbb{C}^d$, and a subvariety $\widetilde{U} \cong U \times \mathbb{D}^k \subset \mathbb{C}^N$ such that $\widetilde{U} \cap Y = U$?

(I am not really sure what the right tag for this question is. Actually if someone would clue me in as to where to find some basic treatment of whatever subject this question belongs to, that would be great.)

Euler characteristics and characteristic classes for real manifolds?

2011-05-21T02:13:56Z

Consider an oriented manifold $X$. To calculate its Euler characteristic, one might integrate the Euler class. Now if $X$ were a complex manifold, and given as a section of some complex bundle $E$ over $Y$, we would have

$\chi(X) = \int\limits_X \mbox{ } c_* (TY) / c_* (E)$

where $c_*(\cdot)$ is the total Chern class.

Can anything of the sort be said if $X$ is a real manifold?

Presumably, if one wants only the Euler characteristic modulo 2, one can use the Stiefel-Whitney classes instead of the Chern classes. On the other hand, it seems to me that the topology of $TY$ and $E$ as bundles over $Y$ cannot suffice to carry the information of the Euler characteristic of the zero locus of a section of $E$. So I guess what I'm really asking is:

What should I know about a section $\sigma:Y\to E$ in order to know the Euler number of its intersection with the zero section, assuming this is transverse?

Positivity of braid monodromy of curve singularities

2011-05-19T19:37:01Z

I recall the notion of braid monodromy. Let $C \subset \mathbb{C}^2$ be an algebraic curve, and choose a projection $\pi: \mathbb{C}^2 \to \mathbb{C}$ such that the restriction $\pi: C \to \mathbb{C}$ is finite of degree $n$. Let $B \subset \mathbb{C}$ be the branch divisor, i.e., the locus where the preimage consists of fewer than $n$ points. Let $\mathrm{Conf}_n(\mathbb{C})$ denote the space parameterizing subsets of $\mathbb{C}$ of cardinality $n$. Then there is a map

$\mathbb{C} \setminus B \to \mathrm{Conf}_n(\mathbb{C})$ given by $z \mapsto [\pi|_C^{-1}(z) \subset \pi^{-1}(z)]$.

Picking a basepoint $p \in \mathbb{C}$ and lifting loops gives a map:

$BM: \pi_1(\mathbb{C} \setminus B, p) \to \pi_1(\mathrm{Conf}_n(\mathbb{C}), [\pi|_C^{-1}(p) \subset \pi^{-1}(p)] )$

which is called the braid monodromy since $\pi_1(\mathrm{Conf}_n(\mathbb{C}))$ is the braid group on $n$ strands.

Suppose now that $C$ is smooth and moreover $\pi|_C$ has only simple ramification; i.e., the preimage of every point $B$ has cardinality $n-1$. Consider a disc $\mathbb{D} \subset \mathbb{C}$ with no ramification points on the boundary, and the braid $BM(\partial \mathbb{D})$. This is ``quasipositive'' -- i.e., it can be decomposed into a product of braids in which two of the points are exchanged by a counterclockwise half-twist. To see this, just factor $\partial \mathbb{D}$ into a sequence of loops each containing exactly one ramification point. Since a small deformation of any $C$ will be of this form, in fact the braid monodromy is always quasipositive.

However, in the knot theory literature, there is a rather stronger notion of positivity. The braid group on $n$ strands is generated by the $n-1$ counterclockwise half-twists $\tau_1,\ldots,\tau_{n-1}$ which exchange adjacent strands, and their inverses. (In the description of the braid group as $\pi_1(\mathrm{Conf}_n(\mathbb{C}))$, numbering the strands is done by fixing an ordering on the set of points corresponding to the basepoint.) A braid is said to be ``positive'' if it can be written as a product of the $\tau_i$.

Let $C_0$ be a (reduced) plane curve with singularity at the origin, and fix a projection $\pi: \mathbb{C}^2 \to \mathbb{C}$ as above. Consider a sufficiently small disc $\mathbb{D} \subset \mathbb{C}$ encircling the image of the singularity at the origin; in particular it should not contain the images of any other singularities or ramification points. It is known that $BM(\partial \mathbb{D})$ is not just quasipositive, but in fact positive. I would like a factorization into positive half-twists to be given in the following manner: first deform $C_0$ to some smooth $C$ such that $\pi|_C$ has simple ramification. Then factor $\partial \mathbb{D}$ into loops which each encircle one of these ramification points. Can this be done?

Symmetric group irreps in tensor products of exterior products of the standard representation

2011-04-21T13:14:49Z

Let $\mathbb{C}^n = V + \mathbb{C}$ be the defining representation of the symmetric group. Is there a nice formula for how $\Lambda^i V \otimes \Lambda^j V$ splits into irreps?

What is known about Higgs bundles with sections?

2011-03-10T18:22:50Z

Let $C$ be a complex curve. Recall that a Higgs bundle on $C$ is a vector bundle $E$ on $C$ equipped with a morphism $E \to E \otimes K_C$. The space of (stable) Higgs bundles is much studied, and is in particular known to be smooth. Moreover there is a "nonabelian Hodge theorem" giving a diffeomorphism between the moduli of Higgs bundles and a certain character variety of $\pi_1(C)$.

What is known about the moduli space of Higgs bundles with a section, i.e., the space parameterizing triples $(E, s \in \mathrm{H}^{0}(E), \phi: E\to E \otimes K_C)$ ? Is it smooth (after imposing some appropriate stability condition)? Is there an analogue of the "nonabelian Hodge theorem"?

Cancellation of contractibility of fibres

2011-03-09T01:05:19Z

Suppose given maps $f:X \to Y$ and $g:Y \to Z$ such that $f$ and $g \circ f$ both have contractible fibres. Then does $g$ have contractible fibres?

And, the same question, but with the maps assumed to be morphisms of algebraic varieties and contractible replaced with isomorphic to $\mathbb{C}^n$ (for varying $n$).

Comment by Vivek Shende

2013-05-20T16:43:02Z

for $M$ some module category, I can recover $M$ from $D^b(M)$ by a t-structure. How would I recover $M$ from just the periodic complexes?

Comment by Vivek Shende

2013-05-18T17:33:25Z

en.wikipedia.org/wiki/…

Comment by Vivek Shende

2013-05-17T06:31:51Z

thanks for all the comments and suggestions!

Comment by Vivek Shende

2013-05-11T04:17:32Z

One does not need 'mixed Hodge theory' for the alternating sum formula, just the long exact sequence in compactly supported cohomology and the fact that the compactly supported cohomology has the same Euler characteristic as the usual cohomology for algebraic varieties. Over $\mathbb{C}$ this can be seen from the fact that links of points on algebraic varieties have Euler number zero, which in turn one can see by using resolution of singularities to reduce to the same statement for odd dimensional spheres.

Comment by Vivek Shende

2013-05-10T22:25:05Z

why should the desired birational map factor through an imbedding into $\mathbb{P}^4$?

Comment by Vivek Shende

2013-04-22T01:03:51Z

I propose we move this discussion to room101.mathoverflow.net.

Comment by Vivek Shende

2013-03-14T12:21:40Z

Having asked around a bit, I also think it's open. The fact that hyperelliptic curves inject, but not with bounded multiplicity, suggests that giving an answer of 'yes' to your original question requires a subtler analysis that sees the difference between 2g and 3g which sounds discouraging. I haven't given up yet though!

Comment by Vivek Shende

2013-03-13T21:41:53Z

For anyone coming across this ancient post, the reason I wanted to know what these diagrams meant was I was hoping for a relation to what is going on in pages 11, 12 of arxiv.org/pdf/1208.1973.pdf . Alas, I at least never found one...

Comment by Vivek Shende

2013-03-12T06:26:48Z

$\chi(\mathcal{O}_X) = \int_X \mathrm{ch}(\mathcal{O}_X) \mathrm{Td}(T_X) = \int_X \mathrm{ch}(f^* \mathcal{O}_Y) \mathrm{Td}(f^* T_Y) = d \int_Y \mathrm{ch}(\mathcal{O}_Y) \mathrm{Td}(T_Y) = d \chi(\mathcal{O}_Y)$. Here, the first and last equality are HRR, the third is push-pull, and the second is Allen's remark (which is better decompressed by pulling back than by pushing forward).

Comment by Vivek Shende

2013-03-12T05:50:55Z

In algebraic geometry, 'deformation' usually means 'flat deformation', which in a projective variety will preserve the Hilbert polynomial and hence, by Riemann-Roch, the arithmetic genus. On the other hand it is certainly possible to find two smooth curves with different genera in the same homology class, for instance the twisted cubic in $\mathbb{P}^3$ (parameterized by $(t^3, t^2 s, t s^2, s^3)$) has genus zero, whereas any cubic in a plane $\mathbb{P}^2 \subset \mathbb{P}^3$ has genus one; nonetheless they are both in the same homology class as thrice a line.

Comment by Vivek Shende

2013-03-12T05:18:59Z

I think, if I had to give three words of advice to such a high schooler, they would be "Go to Ross."

Comment by Vivek Shende

2013-03-03T13:18:59Z

I agree with the established mathematicians.

Comment by Vivek Shende

2013-03-03T12:55:46Z

saying mathy words does not a mathematical explanation make

Comment by Vivek Shende

2013-02-27T23:53:29Z

what's a reference (or argument) for the above statement that cabling preserves $k$?

Comment by Vivek Shende

2013-02-26T23:10:42Z

I'll try and write about the deformation theory when I'm less lazy (& assuming no-one puts a proof of the whole thing...)