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Results tagged with complex-geometry
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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.
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
1
vote
When is a six-dimensional manifold the twistor space of a four-dimensional manifold?
Let $Z$ be the twistor space of a four-manifold $M$. The six-manifold $Z$ can be equipped with an almost complex structure. Furthermore, Hitchin showed that $Z$ is in fact spin.
This may look like t …
- 19.4k
7
votes
Accepted
Are there compact complex manifolds with non-constant pluriclosed functions?
A function $f : M \to \mathbb{C}$ satisfying $\partial\bar{\partial}f = 0$ is called pluriharmonic.
In any holomorphic coordinates $(z^1, \dots, z^n)$, a pluriharmonic function satisfies $\partial_{ …
- 19.4k
4
votes
References on almost complex structures on spheres
Konstantis & Parton - Almost complex structures on spheres is a self-contained paper which explains the proof very clearly.
- 19.4k
7
votes
Accepted
Deform a compact Kähler manifold to a non Kähler one
(Just so this question has an answer. All manifolds are compact.)
In dimension one every deformation of Kähler manifolds is Kähler because every Riemann surface is Kähler.
In dimension two the same …
- 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
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
5
votes
Accepted
Examples of compact complex manifolds for which the $dd^c$ lemma does not hold
Gauduchon proved that a compact complex manifold satisfies the $dd^c$ lemma for $(1, 1)$-forms if and only if $b_1 = 2h^{0,1}$. As a compact complex surface is Kähler if and only if $b_1$ is even, a c …
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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
4
votes
Why does the Lefschetz Operator not Square to Zero?
(Just so this question has an answer.)
If $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^l(M)$, $\alpha\wedge\beta = (-1)^{kl}\beta\wedge\alpha$.
So if one of $\alpha$ or $\beta$ has even degree (i …
- 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 …
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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 …
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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 …
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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) = …
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