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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
1
vote
K3 surfaces and density of rational curves
Rational curves are dense on K3 for all K3 outside of a Baire set
(countable union of closed, nowhere dense). Here is the reference:
https://arxiv.org/abs/1004.5167, Density of Rational Curves on K3 S …
4
votes
1
answer
241
views
Equivariant projective embeddings with optimal dimension
Let $X$ be a complex projective manifold, and $f\in Aut(X)$ an automorphism, which is linearizable, that is, can be extended to an ambient projective space ${\mathbb P}^m$. I am interested to find wha …
4
votes
0
answers
98
views
Existence of a rational curve in the center of a birational contraction for symplectic singu...
Let $M$ be a holomorphically symplectic
complex manifold, and $f: M \to X$
a holomorphic, birational contraction to a Stein
variety $X$, contracting a subvariety $E$
to a point, and bijective outside …
5
votes
What are meromorphic line bundles?
You can define "meromorphic vector bundle" as locally free sheaf of modules
over a sheaf of meromorphic functions. This is a highly non-trivial object, because (in contrast with rational functions) me …
3
votes
0
answers
231
views
Kawamata BPF applied to a semi-positive line bundle using Demailly's holomorphic Morse inequ...
Let $M$ be a compact complex manifold equipped with a line bundle $L$ which has curvature which is non-negative and strictly positive outside of a measure zero set $Z$. In his paper "Holomorphic Mors …
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 …
5
votes
Structure of Kähler cone
Explicit description of a Kahler cone for all hyperkahler manifolds is here:
https://arxiv.org/abs/1401.0479 (Rational curves on hyperkahler manifolds,
Ekaterina Amerik, Misha Verbitsky)
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 …
5
votes
When Atiyah class and Chern class coincide?
I guess this is always true, if you adjust the statement appropriately.
Consider the Bott–Chern cohomology $H^*_{BC}(M):=\dfrac{\ker d\cap \ker d^c}{\operatorname{im} dd^c}$. Since the curvature of a …
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 …
3
votes
Blowing up of a singular subvariety
Just blow up the singular
point in a variety $X_1$ which is obtained from
a smooth, irreducible manifold $X$ by identifying
points $x$ and $y$. The blow-up divisor
is ${\Bbb P} T_xX\coprod {\Bbb P}T_y …
1
vote
Different algebraic structures on complements to divisors
Do you know other examples of non-isomorphic algebraic
structures on complements to square-zero curves
The easiest example is
the twisted cotangent bundle to an elliptic curve.
This space can be rea …
5
votes
1
answer
228
views
A group in a neighbourhood of a Zariski dense subgroup
By Borel's theorem, lattices in simple Lie groups are Zariski dense. I expect that a small (in metric sense) deformation of a lattice in a Lie group is also Zariski dense.
Suppose we have a Zariski de …
1
vote
A group in a neighbourhood of a Zariski dense subgroup
This is an update, sorry for the trouble.
Let $M$ be a metric space. We say that $X \subset M$ is coarse equivalent to $Y \subset M$ if the the Hausdorff distance $d_H(X, Y)$ is finite, that is, there …
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))= …