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 answers only
not deleted
user 297
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.
7
votes
Accepted
Dolbeault Cohomology of $\mathbb{P}^1$
I wrote a blog post about almost exactly this question. I'll give a summary here:
Since $H^{1,1}(X)$ is one dimensional, I could answer your question by giving anythng with the correct integral. Howe …
- 156k
22
votes
Accepted
Are all holomorphic vector bundles on a contractible complex manifold trivial?
No, even for line bundles. We have the short exact sequence of sheaves
$$0 \to \underline{\mathbb{Z}} \overset{2 \pi i}{\longrightarrow} \mathcal{O} \overset{\exp}{\longrightarrow} \mathcal{O}^{\ast} …
- 156k
5
votes
Do some kind of maximum principle exist on complex manifold?
If $f_1$, $f_2$, ..., $f_n$ are holomorhic functions on an open set $U$ in $\mathbb{C}^k$, then $\sum_{i=1}^n |f_i|^2$ has no local maximum. Since complex manifolds are locally $\mathbb{C}^k$, this is …
- 156k
0
votes
Alternative construction of the first Chern class map
$\def\cO{\mathcal{O}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}$Yes, this is right. This answer will prove the equality of three maps $H^1(\cO^{\ast}) \to H^2(\CC)$.
(a) The map $$H^1(\cO^{\ast}) \overs …
- 156k
3
votes
Accepted
Bounding the value of a polynomial map by the distance from the variety
Here is a counter-example. Take
$$f(x,y) = (x(1-x),xy-1)$$
So $Z = \{ (1,1) \}$. Consider now consider the family $(1/R,R)$ as $R \to \infty$. The distance from $(1/R,R)$ to $(1,1)$ goes to $\infty$, …
- 156k
7
votes
Divisors, extensions of functions
There is another very nice theorem: if $K \subset \mathbb{C}^n$ has complex codimension $ \geq 2$ (real codimension $\geq 4$), and $f$ is an analytic function on $\mathbb{C}^n \setminus K$, then $f$ e …
- 156k
3
votes
Logarithm of complex matrices in holomorphic families
It might be worth mentioning that there is an analogous problem with $C^{\infty}$ functions, even when all the matrices are diagonalizable.
Let
$$F(x,y) = \begin{pmatrix} e^{ix} & 0 \\ 0 & e^{-ix} \ …
- 156k
4
votes
Accepted
Is $\mathbb{C}[x^{\pm 1},y]/\langle y^3-(x^2+ax+b) \rangle$ a $n$-point ring?
No (assuming that $a^2 \neq 4b$, so the quadratic is not a square). In fact, the fields $\mathrm{Frac}(R)$ and $\mathbb{C}(t)$ are not isomorphic. From a sophisticated perspective, the point is that t …
- 156k
6
votes
Are holomorphic vector bundles over Kähler manifolds Kähler
Proposition 3.18 of Voisin's Hodge Theory and Complex Algebraic Geometry I says that, if $X$ is compact Kahler and $E$ is a holomorphic vector bundle over $X$, then $\mathbb{P}(E)$ is Kahler. Since $E …
- 156k
4
votes
Rational functions on reduced complex varieties that extend to global holomorphic functions
The answer is yes. Ariyan Javanpeykar has contributed the hard part; here are the easy parts.
Let $\tilde{A}$ be the integral closure of $A$ in $\mathrm{Frac}(A)$ and let $\tilde{X} = \mathrm{Spec}(\ …
3
votes
Example of a variety with explicit cohomology ring and Kahler cone
You could look at surfaces with maximal Picard rank. A surface is said to have maximal Picard rank if $H^{1,1}(X) \cap H^1(X, \mathbb{R})$ is spanned by curve classes. So the Kahler cone is the same a …
- 156k
18
votes
Accepted
$\partial \bar{\partial}$ lemma for contractible domains
Okay, here is a counter-example. Let $X$ be the following open subset of $\mathbb{C}^2$:
$$X:= \{ (z_1, z_2) : |z_1| < 2,\ |z_2| < 1 \} \cup \{(z_1, z_2): |z_1| < 1,\ |z_2| < 2 \}$$
This is the stand …
- 156k
8
votes
Accepted
Second betti number of compact analytic spaces
$\def\ZZ{\mathbb{Z}}$ Let $X$ be a complete toric variety. I will show that $H^2(X, \mathbb{Z}) \cong A^1(X, \mathbb{Z}) \cong \mathrm{Pic}(X)$. Since there are examples of complete toric $3$-folds wi …
- 156k
5
votes
Relation between $h^1(\mathcal{O}_X)$, $h^0(\Omega_X)$, and first betti number for general c...
Let $h^{10} = \dim H^0(X, \Omega^1)$, $z^{10} = \dim H^0(X, Z^1)$ where $Z^1$ is the closed holomorphic $1$-forms, $h^{01} = \dim H^1(X, \mathcal{O})$ and $b_1 = \dim H^1(X, \mathbb{C})$. We claim tha …
- 156k
1
vote
Complexifications of Real Flags are dense in complex Flag Varieties
$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$This yields to a standard lemma: Let $X$ (in your situation, the flag manifold) be a smooth connected complex variety equipped with an anti-holomorphic involuti …
- 156k