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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
16
votes
4
answers
1k
views
Wanted: an example of a natural non-Kähler metric on a Kähler manifold
Let $X$ be a Kähler manifold. Associated to any hermitian metric $h$ on $X$ is a smooth real $(1,1)$-form $\omega = -\text{Im } h$, called the Kähler form of $h$. One of several equivalent conditions …
6
votes
Kähler manifold with Ricci-flat Kähler form
You need some more hypotheses for the existence of the $(n,0)$-form, in general it will exist only up to finite torsion. For example, an Enriques surface does not admit a nowhere zero holomorphic $(2, …
5
votes
1
answer
570
views
Translation of Kähler's "Über eine bemerkenswerte Hermitesche Metrik"
Has anyone translated Erich Kähler's "Über eine bemerkenswerte Hermitesche Metrik" into English or French? (Preferably, but I'll take anything.)
1
vote
Sign of $\int_X\operatorname{Tr}(F_h^2)$
The short answer: No.
The longer answer: The $(2,2)$-form $\operatorname{Tr}(F_h^2)$ represents the second Chern class of $E$, or $c_2(E)$. In general, the integral of that class over the manifold doe …
18
votes
2
answers
4k
views
What is the holomorphic sectional curvature?
Let $X$ be an $n$-dimensional complex manifold, let $\omega$ be a Kahler metric on $X$ and let $R$ be the $(4,0)$ curvature tensor of $\omega$. We can simplify the tensor $R$ in different ways, two of …
16
votes
3
answers
3k
views
References for holomorphic foliations
I'm looking for an introduction to holomorphic foliations and foliations of complex manifolds.
Any little helps, but I'm particularily interested in problems of the type where we have a hermitian man …
2
votes
Accepted
Yau-Uhlenbeck inequality works for higher Chern class?
This is not true, as we can check by calculating the intersection number $\int_X c_4(\operatorname{End}(T_X)) \cup \omega^{n-4}$ for some easy examples of Kahler-Einstein spaces:
This intersection nu …
9
votes
2
answers
1k
views
Is the deformation limit of Ricci-flat Kahler manifolds Kahler?
Let $X$ be a compact complex Kahler manifold with first real Chern class $c_1 = 0$. Consider a family $\pi : \mathcal X \to \Delta$ over the unit disc in $\mathbb C$, where the fibers $X_s$ are compac …
33
votes
2
answers
6k
views
Which almost complex manifolds admit a complex structure?
I was reading Yau's list of problems in geometry, and one of them is to prove that any almost complex manifold of complex dimension $n \geq 3$ admits a complex structure. It's been some time since Yau …
9
votes
Accepted
Questions on the Hodge Dual of the Kähler Class
(1) Yes. If $M$ is $n$-dimensional and $K$ is the Kahler form of a Kahler metric $g$ (or just the Kahler form of a hermitian metric) we have
$$
*_g \frac{K^p}{p!} = \frac{K^{n-p}}{(n-p)!}
$$
for any $ …
9
votes
1
answer
1k
views
Calculating a second fundamental form in the space of hermitian metrics
Let $X$ be a compact Kahler manifold and let $\mathcal M$ denote the space of hermitian metrics on $X$. We'll identify a hermitian metric with a smooth, real and positive $(1,1)$-form $\omega$. Let $\ …
8
votes
When is a Form a Kähler Form?
It's possible to approach the question in a slightly different way if $X$ is compact. Donu and Spiro are of course right in that the condition for a smooth closed $(1,1)$-form $\omega$ to be a Kahler …
11
votes
3
answers
1k
views
Can a metric conformal to a Kahler metric be Kahler?
Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X …
11
votes
Accepted
recognizing Kahler manifolds of complex dimension n
I thought this had already been answered here on MO but my searches didn't turn anything up. It might be good to have a general purpose answer here somewhere.
I'll restrict myself to compact manifold …
0
votes
Adjoint of a Connection Using the Hodge Map?
Unfortunately this does not work, because the Hodge $*$ operator commutes with the Levi-Civita connection. Indeed, we have
$$ \nabla (\langle u,v \rangle dV) =
\langle \nabla u, v \rangle dV + (-1) …