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Complex geometry is the study of complex manifolds, complex algebraic varieties, complex analytic spaces, and, by extension, of almost complex structures. It is a part of differential geometry, algebraic geometry and analytic geometry.
6
votes
Accepted
Parameterization of complex analytic subvarieties
The answer to your question is yes, even in a much broader setting and moreover the parametrizing open set can be chosen always to be the unit polydisc!
Theorem. (Fornaess-Stout '77). If $X$ is an $n …
3
votes
Siu decomposition
I am not sure if I understand correctly your question, but anyway I'll try to answer: the point is that Siu's decomposition is in fact unique, but let me explain better.
Let $X$ be a complex manifold …
11
votes
Bott Chern cohomology via currents
When $X$ is a compact complex manifold, its Bott-Chern cohomology groups can be computed either by smooth forms or by currents. The proof of this fact can be found for instance in Demailly's book (lin …
2
votes
Accepted
Decomposition of hermitian form used in the definition of Griffiths/Nakano positivity
With your notations, the hermitian form $\theta_E$ on $T_X\otimes E$ defined by $\Theta_E$ is given in a somewhat more extrinsic way by
$$
\theta_E(v\otimes\sigma,v\otimes\sigma):=h(\Theta_E(v,\bar v) …
2
votes
Accepted
Local expression involved in the definition of positivity of vector bundles
My answer will consist mainly of a collection of trivial facts but which nonetheless often generate some confusion.
I begin by fixing some notation. Let $V$ be a complex vector space of complex dimen …
9
votes
Accepted
Negative holomorphic sectional curvature
Here is the answer.
Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/ …
11
votes
Algebraic Geometry versus Complex Geometry
Concerning your example, there is definitely no analytic proof of the existence of rational curves on Fano manifolds. This is one of the dream of complex geometers...
You can also consider this weaker …
2
votes
Uniformity of ampleness
1. Here is an elementary and constructive proof from a hermitian point of view.
I will reproduce it only in the case of curves and blow-up equal to the identity, the general case being just more comp …
3
votes
Accepted
The asymptotic growth of global sections of powers of a complex line bundle
Hi,
An enlightening and very elementary proof of this fact can be found in the very complete book of X. Ma and G. Marinescu "Holomorphic Morse inequalities and Bergman kernels".
You will find this i …
15
votes
Specific line bundle over complex manifold implies Kähler?
As Dima said, it is much more: in fact it is projective. But let me give you some more insights on this kind of questions.
I shall give you the definition of four different classes of compact comple …
5
votes
Accepted
A big line bundle in complex compact manifold
At least in the projective setting the following holds true (this is taken from J. Kollár "Shafarevich maps and automorphic forms", Proposition 13.14.2).
Proposition. Let $X$ be a smooth projective va …
1
vote
Accepted
extending biholomorphic maps to bimeromorphic maps
The answer is no. Please check this question I asked some time ago.
5
votes
Why Calabi-Yau manifolds should be complex?
I think that one possible answer is that a Calabi-Yau manifold is a Riemannian manifold $M$ with $SU(n)$ Riemannian holonomy, where $2n=\dim_\mathbb R M$.
Such a manifold is then necessarily complex, …
9
votes
Kähler metric on projectivised bundle
Let me recall you briefly how to obtain such a Kähler metric on the total space of the projectivized bundle.
Start with any given hermitian metric $h$ on $E$ and consider on the projectivized bundle …
6
votes
Examples of Brody hyperbolic affine varieties which are not Kobayashi hyperbolic
Here is an answer to the second part of the first question. It has been communicated to me by Leandro Arosio.
The answer is: yes, there does exist a Stein manifold which is Brody hyperbolic but not K …