Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
18
votes
1
answer
810
views
Cohomology of real analytic coherent sheaves
Let $M$ be a real analytic variety
(if someone is concerned about distinction between
"real analytic spaces" and "real analytic varieties"
in real analytic geometry, let's assume that $M$
is both "va …
13
votes
Accepted
Action of automorphisms of a $K3$ surface on its $(-2)$-curves
The group of symplectomorphisms $Aut(X)$ of a K3 is the group $O(\Lambda)$ of
automorphisms of its period lattice $\Lambda=H^{1,1}(M,{\Bbb Z})$. For each
(-2)-cohomology class $\eta\in H^{1,1}(M,{\B …
13
votes
1
answer
642
views
Does a resolution of a rational singularity have rationally connected fibers?
A rational singularity is a singularity of a
complex variety $X$ such that for any
resolution $\pi:\; \tilde X\rightarrow X$ the
higher direct images $R^i\pi_*(O_{\tilde X})$
vanish for all $i>0$. Sup …
12
votes
Accepted
Deformations of Kähler manifolds where Hodge decomposition fails?
This is known, for projective (even Moishezon)
manifolds as shown by Dan Popovici in his
paper http://arxiv.org/abs/1003.3605
For general Kaehler manifold, this is conjectured.
Popovici has proved t …
9
votes
Accepted
Different occurences of the word 'period' in algebraic geometry
The second and the third are pretty much equivalent.
Indeed, "the period" in XIX century sense is essentially
the same as the discrepancy between the branches of a
multi-valued function, obtained as a …
8
votes
moduli spaces are kahler?
The corresponding Kahler metric is called "Weil-Petersson metric"; it is often constructed using infinite-dimensional determinants of the corresponding Laplace operators. The standard reference is a s …
8
votes
Accepted
What is the moduli of an algebraic torus
There is just no definition of the moduli for complex structures on non-compact manifolds, but by any reasonable definition, it would be (generally) very bad space, certainly infinite-dimensional. For …
8
votes
Coincide between Chern-connection and Levi-Civita connection
It is easier to prove this result for 1-forms, instead of vector fields. On (1,0)-forms, $\nabla^{0,1}=\bar\partial$ because the Levi-Civita connection is torsion-free, hence $\bigwedge(\nabla(\eta))= …
8
votes
Accepted
Proper family deformation retracts onto special fiber
Here is the reference:
Persson, Ulf,
On degenerations of algebraic surfaces,
Mem. Amer. Math. Soc. 11 (1977), no. 189.
Clemens, C. H. Degeneration of Kähler manifolds. Duke Math. J. 44 (1977), no. …
8
votes
2
answers
901
views
Gorenstein varieties: why the two definitions are equivalent?
There are two definitions of Gorenstein singularities
in the literature. Using Grothendieck's (or Serre's) duality, one
defines the "dualizing sheaf" an object $\hat K_M$ of derived category
of cohere …
7
votes
Accepted
Finiteness of De Rham cohomology of smooth quasi-projective varieties
I think there are proofs which are much easier.
For example, you can try to compute the cohomology using the Morse theory. For that you need existence of Morse functions having finitely many critical …
7
votes
Accepted
Is the generic deformation of a symplectic variety affine?
Being "affine" in this case does not make much sense,
because the hyperkaehler deformation is a complex manifold, without
a fixed algebraic structure. Simpson produced an example of a
hyperkaehler …
7
votes
Accepted
"Simple" Kahler manifolds
A generic deformation of a Hilbert scheme of K3 and a generic torus have no
subvarieties, hence they are "simple" in the above sense. For a torus it's
well known, for a Hilbert scheme of K3 it's in my …
7
votes
Complex manifolds whose tangent and cotangent bundles are isomorphic as complex vector bundles
There are many such examples, for instance, all complex nilmanifolds (and most complex solvmanifolds) have tangent bundle which is topologically trivial. The Hopf manifolds also have topologically tri …
7
votes
0
answers
250
views
K3 surfaces with no −2 curves
I seem to remember that a K3 surface with big Picard rank always
has smooth rational curves.
This question is equivalent to the following question about integral quadratic lattices. Let us call a vect …