# Semi-Simple Kahler Groups?

We say that a Kahler manifold is a Kahler group if it is also a Lie group. I would like to know which semi-simple Lie groups are also Kahler groups?

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I was about to reply, but then I saw that you meant something quite different than what I thought. Kahler group usually means fundamental group of a compact Kahler manifold. But anyway, for the question you asked, do you want the metric to be invariant under the group action? – Donu Arapura Jul 22 '11 at 20:37
Yes, it should be invariant. – Jean Delinez Jul 24 '11 at 14:15

Semisimple Lie groups admit bi-invariant metrics (although not necessarily positive-definite) and it is not hard to show that if a Lie group admits a bi-invariant metric and also a left-invariant Kähler structure, then the group is abelian, contradicting the assumption that it was semisimple. Hence no semisimple Lie group admits a left-invariant Kähler structure.

In the case where the Kähler structure is not left-invariant, the two structures do not talk to each other and hence you are asking whether a manifold which admits the structure of a semisimple Lie group could also admit a Kähler structure. The identity component of such a manifold is (rationally) homotopy equivalent to a product of odd spheres (of dimension at least 3), so $H^2$ vanishes and thus, if compact, they again cannot admit a Kähler structure.

I'm not sure about the noncompact case, though; but it looks unlikely to me at this time.

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In fact, it is enough for the group to be unimodular in order to deduce that if it has a left-invariant Kähler structure it is abelian. This is proved in a paper by Lichnerowicz and Medina (springerlink.com/content/p7p8gl6h5163j465) – José Figueroa-O'Farrill Jul 22 '11 at 22:54

See http://eom.springer.de/h/h047640.htm and references therein.

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How does this answer the question? – José Figueroa-O'Farrill Jul 22 '11 at 22:32