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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
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 …
6
votes
How do you define the Euler Characteristic of a scheme?
Here are some comments on the questions of the posting.
Every complex algebraic variety has the homotopy type of a finite CW-complex, so the Betti numbers are finite and the Euler characteristic is w …
3
votes
Endomorphisms of the cohomology of a projective variety
Here is an extended version of a comment above.
First, one can consider $R\Gamma\Omega^\bullet_X$ as a complex of vector spaces. Any complex of vector spaces is quasi-isomorphic to its cohomology. Ta …
1
vote
Hodge-to-de Rham and algebraic analogues of the ddbar lemma
I think the answer to the question is yes. More precisely, let $X$ be a smooth compact complex algebraic variety and suppose $(F^{\ast,\ast},d',d'')$ is a bounded below double complex of $\Gamma$-acyc …
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. …
11
votes
1
answer
541
views
Vector fields on complete intersections
Here is a question the answer to which I've been trying to locate for some time.
Let $X$ be a smooth projective complete intersection over an algebraically closed field $k$; assume that $X$ is not co …
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 …
3
votes
Accepted
Divisor Intersections and Chern Class Products
I'm not sure what you mean by a line bundle on a real algebraic variety and its Chern classes, but for smooth complex analytic manifolds the Chern class of a line bundle corresponding to a divisor is …
2
votes
A specific branched cover of S^2 as a subgroup of Pi_1
The sphere minus 4 points is doubly covered by the torus minus 4 points. This double cover gives a representation of $\pi_1(S^2\setminus\mbox{4 points})$ in the symmetric group on two letters. Namely, …
3
votes
Accepted
Determining if two algebraic sets are homeomorphic
I think the answer to the "real" version of the question is no. Here are some remarks.
One can realize each smooth manifold as a real algebraic variety in a Euclidean space. So one can realize each …
16
votes
Birational invariants and fundamental groups
True since topologically blowing up a point is taking a connected sum with $\overline{ \mathbf{P}^2(\mathbf{C})}$.
True for smooth surfaces (generally, for smooth projective hypersurfaces, but it suf …
4
votes
2
answers
967
views
Local freeness of direct images
This question arose from an unsuccessful attempt to settle another question of mine: Vector fields on complete intersections
Let $X\to Y$ be a smooth projective morphism of noetherian schemes and let …
1
vote
Accepted
Second cohomology group with finite coefficients of the product of two varieties
The identity map of $H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)$ can be decomposed as $$H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)\to H^2(X\times Y,\mathbb{Z}/n)\to H^2(X,\mathbb{Z}/n)\oplus H^2 …
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 …
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.