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 | |
Results tagged with complex-geometry
Search options questions only
not deleted
user 21564
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.
36
votes
2
answers
2k
views
Complex manifold with subvarieties but no submanifolds
I previously asked this question on MSE and offered a bounty but received no responses.
There are examples of compact complex manifolds with no positive-dimensional compact complex submanifolds. Fo …
- 19.4k
24
votes
5
answers
4k
views
Weitzenböck Identities
I asked this question at Maths Stack Exchange, but I haven't received any replies yet (I'm not sure how long I should wait before it is acceptable to ask here, assuming there is such a period of time) …
- 2,821
21
votes
1
answer
1k
views
Does every group arise as the fundamental group of a complete Kähler manifold?
The fundamental group of a manifold is countable, and every countable group $G$ arises as the fundamental group of a (smooth) manifold; see this comment or this answer for a construction of an open su …
- 63.9k
7
votes
4
answers
2k
views
$E$ is a holomorphic vector bundle if and only if there is a Dolbeault operator $\bar{\parti...
I am looking for a reference which shows that the following statements are equivalent for a complex vector bundle $E$:
$E$ is a holomorphic vector bundle.
There is a Dolbeault operator $\bar{\partia …
12
votes
0
answers
1k
views
How much algebraic geometry do I need to study complex geometry?
As one can deduce from the questions I have asked on MO, I'm interested in complex geometry. I am aware that there are many facets to the field, some of which I am more comfortable with than others. T …
- 1,422
14
votes
0
answers
704
views
Is Hironaka's example the only known deformation of Kähler manifolds with non-Kähler central...
A well-known example in the deformation theory of compact complex manifolds is the one given by Hironaka in his 1962 paper An Example of a Non-Kählerian Complex-Analytic Deformation of Kählerian Compl …
CommunityBot
- 1
8
votes
4
answers
1k
views
Hermitian Christoffel Symbols
Does anyone know of some good references for computing Christoffel symbols for Hermitian metrics?
A quick Google search turns up this. The following formula appears on page 4:
$$\Gamma_{AB}^C = …
- 26.3k
24
votes
4
answers
2k
views
Examples of compact complex non-Kähler manifolds which satisfy $h^{p,q} = h^{q,p}$
The existence of a Kähler metric on a compact complex manifold $X$ imposes restrictions on it's Dolbeault cohomology; namely, $h^{p,q}(X) = h^{q,p}(X)$ for every $p$ and $q$. I am looking for some exp …
9
votes
2
answers
1k
views
Alternative Almost Complex Structures
Originally posted on Maths Stack Exchange.
Let $V$ be a real vector space. An almost complex structure on $V$ is a map $J : V \to V$ such that $J^2 = -\mathrm{id}_V$. An almost complex structure give …
CommunityBot
- 1
11
votes
2
answers
3k
views
Riemannian metrics as sections of a vector bundle
Let $\pi \colon E \to M$ be a smooth vector bundle. A Riemannian metric on $E$ can be regarded as a global section of the vector bundle $(E\otimes E)^{\ast}$, or more specifically, the subbundle $(S^2 …
CommunityBot
- 1
84
votes
1
answer
5k
views
Is there a complex surface into which every Riemann surface embeds?
This question was previously asked on Math SE.
Every Riemann surface can be embedded in some complex projective space. In fact, every Riemann surface $\Sigma$ admits an embedding $\varphi : \Sigma …
CommunityBot
- 1
18
votes
2
answers
1k
views
Does equality of Laplacians imply Kähler?
This question follows on from this one.
Let $(X, \omega)$ be a Hermitian manifold and define the Laplacians $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$ and $\Delta_{\bar{\partial}} …
CommunityBot
- 1
29
votes
1
answer
4k
views
Almost Complex Structure approach to Deformation of Compact Complex Manifolds
I don't know much about the deformation of compact complex manifolds, I've only read chapter 6 of Huybrechts' book Complex Geometry: An Introduction. There are two parts to this chapter. The second go …
- 19.4k
2
votes
1
answer
216
views
For compact complex surfaces $h^{1,0}$ is either $h^{0,1}$ or $h^{0,1} - 1$. Do we need to u...
In the Wikipedia article on the Enriques-Kodaira classification, before the classification itself, the following sentence appears:
For compact complex surfaces $h^{1,0}$ is either $h^{0,1}$ or $h^ …
- 6,871
9
votes
1
answer
1k
views
Non-compact Kähler manifolds which admit a positive line bundle
A complex manifold which admits a positive line bundle is automatically Kähler. Furthermore, if the manifold is compact, then it is projective by the Kodaira Embedding Theorem. In particular, not ever …
- 9,237