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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asked Jul 22, 2011 at 16:57
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$\begingroup$ 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? $\endgroup$– Donu ArapuraCommented Jul 22, 2011 at 20:37
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$\begingroup$ Yes, it should be invariant. $\endgroup$– Jean DelinezCommented Jul 24, 2011 at 14:15
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2 Answers
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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.
answered Jul 22, 2011 at 22:46
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1$\begingroup$ 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) $\endgroup$ Commented Jul 22, 2011 at 22:54
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$\begingroup$ The link to the article by Lichnerowicz and Medina at
springerlink.commentioned in the comment above is broken, but it can be found at doi:10.1007/BF00398959 (Zbl 0665.53046). $\endgroup$ Commented Jul 21, 2022 at 5:25
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See http://eom.springer.de/h/h047640.htm and references therein.
answered Jul 22, 2011 at 19:36
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$\begingroup$ How does this answer the question? $\endgroup$ Commented Jul 22, 2011 at 22:32
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$\begingroup$ The link to
eom.springer.deis broken, but the article can now be found at encyclopediaofmath.org/wiki/Homogeneous_complex_manifold. $\endgroup$ Commented Jul 21, 2022 at 5:22