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Results tagged with complex-geometry
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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.
28
votes
Accepted
Why is $\mathbb{Z}$ not a Kähler group?
If $X$ is a compact Kähler manifold, then $h^{p,q}(X) = h^{q,p}(X)$ and $b_k(X) = \sum_{p+q=k}h^{p,q}(X)$, so in particular, $b_1(X) = h^{1,0}(X) + h^{0,1}(X) = 2h^{1,0}(X)$ is even. Now,
$$b_1(X) = …
- 19.4k
24
votes
Accepted
Does every open orientable even-dimensional smooth manifold admit an almost complex structure?
If $M$ admits an almost complex structure, then the odd Stiefel-Whitney classes vanish and the even Stiefel-Whitney classes admit integral lifts, namely $c_i(M) \equiv w_{2i}(M) \bmod 2$. These two co …
- 19.4k
20
votes
Accepted
An almost complex structure on $S^2\times ...\times S^2 / \mathbb{Z_2}$
Let $Y_n = X_n/\mathbb{Z}_2$.
If $G$ is a finite group acting freely on a manifold $M$, and $\pi : M \to M/G$ denotes the quotient map, then $\pi^* : H^*(M/G; \mathbb{Q}) \to H^*(M; \mathbb{Q})$ is in …
- 19.4k
16
votes
Accepted
Exact formula for $\chi(X, \, S^n \Omega^1_X)$
As you say, formulae for $c_1(\Omega_X^1)$ and $c_2(\Omega_X^1)$ can be obtained from the splitting principle. The following is a more general version of the calculation in this answer.
Lemma: Let $V …
- 19.4k
13
votes
Accepted
Does the Kähler form $\omega$ satisfy $d^*\omega=0$?
Recall that $d^* = -\ast d\ast$ and $\ast\omega = \frac{1}{(n-1)!}\omega^{n-1}$, see Example 1.2.32 of Huybrechts' Complex Geometry: An Introduction for example. Therefore
$$d^*\omega = -\ast d\ast\om …
- 19.4k
13
votes
Complex manifold with subvarieties but no submanifolds
The theorem that inkspot refers to in their answer is originally from Inoue's paper New Surfaces with No Meromorphic Functions, II which seems like a more complete reference for this question. In part …
- 19.4k
13
votes
Vector bundles on Stein manifolds
As has been established in the comments, the answer to your question is yes. It is a special case of a general result known as the Oka principle which has been strengthened over time. The key is that …
- 19.4k
12
votes
Complex structure on $S^4$
The argument via K-theory proceeds as follows.
There is a map $K(X) \to H^*(X; \mathbb{Q})$ given by the Chern character. If $X = S^{2n}$, then it follows from Bott periodicity that the image of the C …
- 19.4k
12
votes
Complex vector bundles on compact complex manifolds
This is an explanation of my comment above, namely: "Complex vector bundles over a CW complex of dimension $\leq 4$ are classified by their Chern classes and rank. Moreover, every possible choice of C …
- 19.4k
11
votes
Accepted
Inverse Hodge and inverse Betti problems for Kähler manifolds
The Hodge diamond \begin{array}{ccccc}&&1&&\\&0&&0&\\a&&1&&a\\&0&&0&\\&&1&&\end{array} is naively realisable.
Suppose $M$ is a compact Kähler surface with the given Hodge diamond with $a \geq 2$. As $ …
- 19.4k
11
votes
Examples of manifolds that do not admit scalar flat metrics
Bourguignon showed that if a compact manifold does not admit positive scalar curvature metrics, then any scalar flat metric (actually, any non-negative scalar curvature metric) is Ricci-flat; I suppos …
- 19.4k
10
votes
Accepted
Is the Bott-Chern/Aeppli cohomology determined by the de Rham and Dolbeault cohomologies?
I learnt all of the following from section $3.2$ of Angella's Cohomological Aspects in Complex Non-Kähler Geometry.
Let $R$ be a commutative ring with identity. The three-dimensional Heisenberg gro …
- 19.4k
10
votes
Accepted
The logarithm of Kähler metric is not globally defined
Question 1: Note, $\omega + \partial\bar{\partial}\phi$ cannot be zero. Recall that $\phi$ is chosen so that $\omega + \partial\bar{\partial}\phi$ is another metric (in particular, a Kähler-Einstein o …
- 19.4k
10
votes
Accepted
Status of a conjecture of Hirzebruch
Suppose $X$ is diffeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. Then $X$ is biholomorphic to a Hirzebruch surface.
Note that $b_1(X) = 0$, so $X$ admits a Kähler metric. …
- 19.4k
10
votes
Hodge diamonds of complex threefolds
Let $R$ be a commutative ring with identity. The three-dimensional Heisenberg group over $R$ is
$$\mathbb{H}(3, R) = \left\{\begin{bmatrix} 1 & z^1 & z^3\\ 0 & 1 & z^2\\ 0 & 0 & 1\end{bmatrix} : z^1, …
- 19.4k