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For questions about or involving complex manifolds.
8
votes
2
answers
373
views
Real analytic subvariety in complex manifold which is complex outside of its singular set
Let $M$ be a complex manifold, and $Z \subset M$ a closed real analytic subvariety. Suppose that the set of smooth points in $Z$ is complex analytic in $M$. Will it follow that $Z$ is complex analytic …
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 …
5
votes
“Logarithmic” form of Kodaira Embedding
Sure, you should assume that there are sufficiently many functions which grow polynomially. Here is an example of such a result.
THEOREM: Let $M$ be a Stein variety equipped with a Kahler metric
outsi …
2
votes
Holomorphic vector fields acting on Dolbeault cohomology
Klemyatin proved that this action is trivial if the corresponding
${\Bbb C}$-flow is compatible with some metric (hence can be extended
to a compact torus action),
https://arxiv.org/abs/1909.04075,
(N …
7
votes
0
answers
126
views
holomorphic functions on Stein manifolds reaching maxima in a given point on the boundary (S...
Let $M$ be a Stein manifold with smooth, strictly
pseudoconvex boundary, and $x$ a point on its
boundary. Is there a holomorphic function $f$ on
$M$, smooth on the boundary, with strict
maximum of $|f …
1
vote
HKT manifolds with non trivial canonical bundle
For $SU(3)$ and Hopf it is the anticanonical
bundle which has many sections. In other
examples of Joyce, it is also the anticanonical
bundle or its powers. For nilmanifolds, the
canonical bundle is ho …
6
votes
Cohomology of neighborhood of $\mathbb{C}\mathbb{P}^1$ in $\mathbb{C}\mathbb{P}^n$
It is not hard to see that the
first cohomology is infinite-dimensional.
Take a complement $M:=CP^3 \ CP^1$
and consider a projection
$\pi:\; M \mapsto CP^1$. It is not
hard to see that $M$ is
isomor …
5
votes
2
answers
383
views
fibers of birational contraction for complex manifolds - are they Moishezon?
Let $X$ be a smooth complex manifold and
$\phi:\; X \mapsto Y$ a proper holomorphic
map which is birational ("birational contraction"),
and $Z= \phi^{-1}(y)$ its fiber in a point $y$.
The variety $Y$ …
5
votes
Different notions of convergence of complex subvarieties
For Kahler manifold this is true.
First, notice that limits in the sense of Barlett or Douady
clearly (up to multiplicity) coincide with the limits
in Hausdorff topology on the set of compact subsets. …
3
votes
Foliations by holomorphic curves on complex surfaces
This is a question which is related to Lagrangian foliations on hyperkaehler manifolds, but much of the results are conjectures, or unpublished. Let $M$ be a K3, and $\omega$ a cohomology class on a b …
5
votes
Accepted
Is it always possible to extend a closed (1, 1)-form on a divisor to a closed (1, 1)-form on...
Yes. For the proof, see e.g. http://arxiv.org/abs/math/0609617
Theorem 4.1. Here a stronger result is actually proven:
Theorem: Let $(M, \omega)$ be a compact Kahler manifold,
and $Z\subset M$ a clos …