Talk:Kähler manifold

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Kähler manifolds are usually thought of as finite dimensional manifolds. There is a generalized theory of infinite dimensional manifold, but it would be good to give at least one reference to literature dealing primarily with finite dimensional Kähler manifold. Hottiger 17:26, 6 March 2006 (UTC)

Why are they important?[edit]

I think this entry would benefit if the importance of Kahler manifolds is explained. I don't know it.

I second this comment. Kähler manifolds need to be motivated. LTX2mny (talk) 10:55, 22 April 2008 (UTC)

A special class of Kaehler manifolds are Calabi-Yau manifolds, as is mentioned in the article. Those are very important in stringtheory. Besides the n-dimensional complex projective space is a Kaehler manifold, as is every closed submanifold. Since by Chows theorem every closed submanifold of projective space is algebraic, all smooth complex algebraic varieties are Kaehler manifolds. So they are of tremendous importance in (classical) algebraic geometry and in fact many modern theorems were motivated by those classical considerations. Another very important result is hodge and lefshetz decomposition, which is not mentioned at all in the article. I feel however not qualified enough to change the article accordingly, although in its current state it is less then satisfactory. —Preceding unsigned comment added by (talk) 17:56, 2 August 2010 (UTC)


Many sources define Kahler manifolds as symplectic manifolds with integrable compatible almost complex structures. The relation between this definition and the definition via Kahler metrics should be made explicit. I would be tempted to give the definition mentioned here first in the article, and then later mention the relation with Kahler metrics. Of course, being a symplectic geometer I may be biased. What do other thinks? 23:23, 19 April 2007 (UTC)

Kahler identity[edit]

I'm really not a specialist, but is there a reason why the article makes no mention of the fundamental Kahler identity? Does it only hold for a certain class of Kahler manifold? Just wondering -- Taku (talk) 22:53, 15 June 2008 (UTC)

Partial Rewrite[edit]

I have rewritten a fair amount of the article and tried to take into consideration all of the points addressed on this talk page. It is thus far only a cursory rewrite.

I have removed the original author's definition of a Kähler manifold, since it seemed incredibly wishy-washy. For example, the article often referred to "an additional integrability condition." What does it mean for a Hermitian metric to be integrable? This is not addressed in the linked article. Did the author mean that the -subbundle of the tangent frame bundle was integrable? Is this equivalent to the (in my opinion) more common definitions which I have included? I could not find this definition in the references I was able to access.

I would love to re-incorporate the definition wherein we endow the tangent frame bundle with an integral subbundle. If the article on G-structures made the idea more explicit that would be sufficient. If anyone could provide a reference for this, I would be more than happy to do the write-up. Kreizhn (talk) 21:52, 3 August 2012 (UTC)

I think that the potential might not be global though someone wrote "the (global) potential." I will confirm it. And I will add Laplacians on Kahler manifolds.--Enyokoyama (talk) 13:22, 28 January 2013 (UTC)

I decide to delete the word "global" because a compact Kahler manifold has not global potential. addressed here refers to the same thing except for the difference except for the difference in real constants.

And it is also necessary in terms of the holonomy. So I add a reference to Kahler manifolds in Ricci tensor titled as "Ricci tensor and Kahler manifolds." Then it naturally leads to relate Kahler manifolds with Einstein manifolds in the next topic. --Enyokoyama (talk) 13:45, 3 February 2013 (UTC)

I will add some explanations about "The Laplacians on Kahler manifolds."--Enyokoyama (talk) 13:56, 3 February 2013 (UTC)

Dear AHusain314: Thank you!--Enyokoyama (talk) 14:39, 6 February 2013 (UTC)

The Kahler condition is originally independent on the Einstein condition, i.e. the Riemannian metric tensor is proportional to Ricci tensor for constant real number. The important point is that if X is Kahler then Christfel symbols vanish and Ricci curvature is much simplified. Therefore, the Kahler condition is closely related with Ricci curvature. In fact Aubin and Yau prove the Calabi conjecture using the fact that on a compact Kahler manifold with the first Chern class c_1=0 there is an unique Ricci-flat Kahler metric in each Kahler class, though in non-compact case the situation turns to be more complicated.

I think that in "application" we should refer not only "Einstein manifold" but also "Calabi conjecture". In "application" I would point out the above important facts and would make a relevant link in order to naturally fill some citations.--Enyokoyama (talk) 15:17, 7 February 2013 (UTC)

Of course, the above points are found in the Part 4 in Andrei Moroianu, Lectures on Kähler Geometry (2004)

URL: --Enyokoyama (talk) 15:33, 7 February 2013 (UTC)

Thanks everybody! I had translated this article into Japanese.--Enyokoyama (talk) 09:04, 9 February 2013 (UTC)

entanglement between 3 articles[edit]

I recognized some entanglements between 3 articles, Hermitian mfd., Kahler mfd. and Ricci tensor. Firstly, I found a unexpected translation, Kahler metric (redirect)-> Hermitian manifold. The Kahler condition is stronger than the Hermitian condition because a closed hermitian form is Kahler. I think that Kahler metric should translate to Kahler manifold, not to Hermitian manifold. But I don't know the way to release or delete the redirection Kahler metric to Hermitian manifold. Another entanglments might are between them.--Enyokoyama (talk) 17:32, 9 February 2013 (UTC)

I have corrected the destination of redirection "Kahler metric" from "Hermitian mfd." to "Kahler mfd.".--Enyokoyama (talk) 04:12, 10 February 2013 (UTC)

paragraph written by someone[edit]

15 Feb 2013 someone, whose address is, has written on this article and the article "Hodge conjecture."

This is incorrect. Kahler manifold is not absolute, as proven in the Hodge conjecture when one cannot assume X is a Kahler manifold due to decomposition not being constant. Hp, q(X) is not a subgroup of cohomology classes being that (X is not a Kahler manifold) and cannot be represented by harmonic forms of (p, q).

On this article, "Kahler manifold," the definition from the symplectic viewpoint is not dependent on harmonic forms as someone wrote.

On another article, "Hodge conjecture," the original author as saying at the head of this paragraph assumes Kahler manifold in order to be clear for the origin of Hodge conjecure, which is the properties of Kahler manifolds and fortunately on compact Kahler manifolds there exist such harmonic forms.

I'll undo them in a few days.--Enyokoyama (talk) 06:53, 16 February 2013 (UTC)