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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.

1 vote
0 answers
119 views

Quotients of open subsets of the semi-stable locus

This is a rewrite of a deleted question. I've decided to focus on one particular example mentioned in that question. Below a point means a closed point. Let $U$ be the set of irreducible non-cuspidal …
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2 votes
0 answers
163 views

Can components vanish without a trace?

Let $H_{P,n}$ be the Hilbert scheme of subschemes of $\mathbb{P}^n(\mathbb{C})$ with Hilbert polynomial $P\in\mathbb{Q}[t]$, and let $U_{P,n}\to H_{P,n}$ be the flat universal family. Are there $n,P$ …
algori's user avatar
  • 23.5k
10 votes
1 answer
2k views

Extending holomorphic functions

Suppose $K\subset \mathbb{C}^n$ is a compact subset and $f:\mathbb{C}^n\setminus K\to \mathbb{C}$ is a holomorphic function. Then, provided $n>1$, $f$ extends to a holomorphic function defined on the …
algori's user avatar
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6 votes

A_infinity structure on cohomology and the weight filtration

This was meant to be a comment before it got too long. One of the ways to define a minimal $A_\infty$ model of a cdga (more generally, an $A_\infty$-algebra) $A$ is Merkulov's recipe (see Merkulov ht …
algori's user avatar
  • 23.5k
12 votes

Pullback map in homology

The wrong-way maps in homology are in fact less mysterious than they look. Suppose that $ f: X \to Y $ is a continuous map of oriented closed topological manifolds. Then there is a composition of maps …
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15 votes

Is it true that every projective curve is a union of two affine curves?

One way to see this would be as follows: if $X\subset\mathbb{P}^n$ is a curve one can find hyperplanes $H_1,H_2$ such that $X\cap H_1\cap H_2=\varnothing$. Then $X$ will be the union of two affine cur …
algori's user avatar
  • 23.5k
5 votes
Accepted

Second lowest weight piece of the cohomology of an algebraic variety

Here is a geometric description of $W_{k+1}H^k(U)/W_k H^k(U)$. Set $d=\dim U$. Suppose $X$ is a compactification of $U$ such that the complement $X\setminus U$ is the union of normal crossing divisors …
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4 votes

cohomology and $j_!$

If $X$ is smooth and we consider sheaves of $k$-modules, $k$ a commutative ring, then $H^*(X, j_!\mathcal{S})\cong H^*_c(U,\mathcal{S})$; the latter is equipped with a non-degenerate pairing $$H^*_c( …
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4 votes

Strong Kodaira vanishing

Re question 1: yes, see Deligne, Cohomologie des intersections compl`etes, SGA 7 II, th\'eor`eme 1.1. The proof is by "force brutale" as Deligne himself puts it, so I'm not sure this generalizes.
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5 votes

Are there any known formulas about the Hodge-Deligne structure of quotients by actions of gr...

The rational cohomology of the quotient of a variety $X$ by an action of a finite group $G$ is the $G$-invariant part of $H^*(X,\mathbb{Q})$; this is a Hodge substructure if $G$ acts by automorphisms. …
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3 votes
0 answers
217 views

Extending intersection bundles

Let $X$ be the product $Gr_i(V)\times Gr_j(V)$ of two Grassmannians where $V$ is a complex vector space of dimension $d$. There is an open $U\subset X$ formed by all those $(V',V'')\in X$ such that $V …
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8 votes

Given a polynomial f, can there be more than one constant c such that every root of f(x)-c i...

The question can be restated as follows: can there be a branched $n$-fold cover $S^2\to S^2$ with at least 3 critical values, one of index $n$ and all preimages of critical values being critical point …
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  • 23.5k
7 votes

Twisted cohomology of the mapping class group

I think the answer is 0. Indeed, $G=SL_2(\mathbb{Z})$ contains a normal $H=\mathbb{Z}/2\cong\{\pm I\}$. So if $V$ is set to be $\mathbb{Z}^2$ with the standard $G$-action, then the restriction of $H$ …
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3 votes

Reference for the derived category of $X$, $[X/G]$ and $X/G$

A very readable reference for constructible derived categories on quotient stacks is Bernstein-Lunts, Equivariant sheaves and functors. In particular, Bernstein and Lunts construct the bounded derived …
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11 votes
Accepted

How Gr(2,7) and Gr(3,6) are related?

I don't know whether the ideals have the same kind of free resolutions, but $Gr(3,6)$ is definitely not a hyperplane section of $Gr(2,7)$. Otherwise, their $H^{\leq 8}$ would be the same by the Lefsch …
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