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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.
9
votes
Accepted
Does $H^3\times I$ admit a Kähler metric?
Note that $S^3$ embeds in $S^4$ and $S^3$ is the total space of the Euler class one circle bundle over $S^2$. It follows that the Euler class one circle bundle over $\Sigma_g$ embeds in $S^4$ for all …
- 19.4k
5
votes
Accepted
The kernel of $H^{\bullet,\bullet}_{\bar\partial}(X)\to H_A^{\bullet,\bullet}(X)$
Let $\alpha$ be a $\bar{\partial}$-closed form. Denote its Dolbeault cohomology class by $[\alpha]_{\bar{\partial}}$ and its Aeppli cohomology class by $[\alpha]_A$; note that the map $g$ is given by …
- 19.4k
8
votes
Accepted
Compact complex non-Kähler manifolds with nef canonical bundle
Let $X$ and $Y$ be compact complex manifolds. Note that $K_{X\times Y} \cong \pi_1^*K_X\otimes \pi_2^*K_Y$. If $Y$ has trivial canonical bundle, then $K_{X\times Y} \cong \pi_1^*K_X$. Now the pullback …
- 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
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
$$d^*\omega = -\ast d\ast\om …
- 19.4k
12
votes
Complex structure on $S^4$
The argument via K-theory proceeds as follows.
There is a map $K(X) \to H^*(X; \mathbb{Q})$ given by the Chern character. If $X = S^{2n}$, then it follows from Bott periodicity that the image of the C …
- 19.4k
16
votes
Accepted
Exact formula for $\chi(X, \, S^n \Omega^1_X)$
As you say, formulae for $c_1(\Omega_X^1)$ and $c_2(\Omega_X^1)$ can be obtained from the splitting principle. The following is a more general version of the calculation in this answer.
Lemma: Let $V …
- 19.4k
10
votes
Accepted
Status of a conjecture of Hirzebruch
Suppose $X$ is diffeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. Then $X$ is biholomorphic to a Hirzebruch surface.
Note that $b_1(X) = 0$, so $X$ admits a Kähler metric. …
- 19.4k
24
votes
Accepted
Does every open orientable even-dimensional smooth manifold admit an almost complex structure?
If $M$ admits an almost complex structure, then the odd Stiefel-Whitney classes vanish and the even Stiefel-Whitney classes admit integral lifts, namely $c_i(M) \equiv w_{2i}(M) \bmod 2$. These two co …
- 19.4k
11
votes
Accepted
Inverse Hodge and inverse Betti problems for Kähler manifolds
The Hodge diamond \begin{array}{ccccc}&&1&&\\&0&&0&\\a&&1&&a\\&0&&0&\\&&1&&\end{array} is naively realisable.
Suppose $M$ is a compact Kähler surface with the given Hodge diamond with $a \geq 2$. As $ …
- 19.4k
10
votes
Hodge diamonds of complex threefolds
Let $R$ be a commutative ring with identity. The three-dimensional Heisenberg group over $R$ is
$$\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
6
votes
Accepted
Complex structure on product of two $n$-dimensional real manifolds
I'm going to take $M$ and $N$ to be connected. Note that they cannot have boundary. Furthermore, as complex manifolds are orientable, $M$ and $N$ must also be orientable.
If $n = 1$, then $M, N \in \ …
- 19.4k
3
votes
Status of global spherical shell conjecture for minimal complex surfaces?
Teleman has shown that the global spherical shell conjecture holds for $b_2 = 1$, $b_2 = 2$, and $b_2 = 3$ in the following papers respectively:
Teleman, Andrei, Donaldson theory on non-Kählerian su …
- 19.4k
3
votes
Accepted
Projection of an invariant almost complex structure to a non-integrable one
First of all, every almost complex structure on a two-dimensional manifold is integrable, see here.
Let $\pi : P \to X$ be a smooth principal $G$-bundle equipped with a connection. The connection det …
- 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