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 |
Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
7
votes
Hom between Brody hyperbolic varieties
I assume that for $\operatorname{Hom}(X,Y)$ you mean $\operatorname{Hol}(X,Y)$, that is the family of all holomorphic maps from $X$ to $Y$, endowed with its universal complex structure (which exists s …
27
votes
Accepted
Is the complex structure of $\mathbb CP^n$ unique?
Let me write this too long comment as an answer.
As abx says, what we do know is
Theorem 1. If a Kähler manifold $X$ is homeomorphic to $\mathbb{CP}^n$, then $X$ is biholomorphic to it.
This is due to …
5
votes
Accepted
Existence of a connection $A$ on a holomorphic line bundle $L$, s.t $F(A)=(\deg L)\omega$
Let me expand a bit Henri's (hi Henri!) answer, even if this is completely standard. In general, given a compact Kähler manifold $X$ of any dimension, given a holomorphic line bundle $L\to X$, and giv …
2
votes
Examples of surfaces with negative Kahler curvature operator
Examples of compact Kähler manifolds with non positive holomorphic bisectional curvature are given by:
Closed submanifolds of complex tori.
Smooth compact quotients of bounded symmetric domains.
M …
22
votes
1
answer
1k
views
Relationship between the signs of different notions of curvature in complex geometry
Let $(X,\omega)$ be a complex hermitian manifold, and call $\Theta$ its Chern curvature tensor. Out of this we can consider different notions of curvature, namely the holomorphic bisectional curvature …
0
votes
A question about nef classes on compact Kähler manifolds
Just for fun, here is an answer in the purely algebraic setting.
So, suppose that $X$ is irreducible projective algebraic of dimension $n$, $\alpha=c_1(\mathcal O_X(D))$ is the class of a nef diviso …
5
votes
Is hyperbolicity a Zariski open condition?
Kobayashi hyperbolicity (or Brody hyperbolicity, the two notions coincide for compact complex spaces), is an open condition with respect to the analytic topology. Thus, for instance, once you find an …
2
votes
What is the holomorphic sectional curvature?
Maybe you already were aware of that, or maybe it really doesn't answer to your question, but I'll try anyhow...
Take a look at this, Subsection 7.5 on page 39. The construction you talk about in you …
2
votes
Accepted
Complex manifolds with trivial canonical bundle
I think the answers you are looking for are in this paper by V. Tosatti, see in particular Proposition 1.1, point (4) and Proposition 1.3.
Warning (in view of the comment below by S.S.): the holonom …
3
votes
Accepted
holomorphic sectional curvature and total scalar curvature
The answer to your question is exactly the same of the answer to this older question of mine. Enjoy!
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 …
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) …
4
votes
What is the DGLA controlling the deformation theory of a complex submanifold?
This paper by Donatella Iacono could be of some help for you.
5
votes
A question on Ricci curvature and Ricci form.
The whole point is that if the metric is Kähler, then the Chern connection coincides with the complexification of Levi-Civita connection of the riemanian metric on the underlying real manifold given b …
5
votes
Accepted
Holonomy of a Kähler manifold
The answer is yes.
The holonomy principle states that a given a riemannian manifold $(M,g)$ and a point $x\in M$, the datum of a parallel tensor field of a given type is equivalent to the datum of a …