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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, …
Glorfindel's user avatar
  • 2,821
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 …
Gunnar Þór Magnússon's user avatar
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 …
Gunnar Þór Magnússon's user avatar
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 $ …
Gunnar Þór Magnússon's user avatar
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 …
Gunnar Þór Magnússon's user avatar
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 …
Gunnar Þór Magnússon's user avatar
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) …
Gunnar Þór Magnússon's user avatar

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