Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hallo,

Let $G$ be a semi-simple, compact Lie Group. Consider its complexification $G_{\mathbb{C}}$. Does there exist a Kähler structure on $G_{\mathbb{C}}$ which is $G$-invariant (maybe in a neighbourhood of $G$ in $G_{\mathbb{C}}$)?

hapchiu

share|improve this question
add comment

1 Answer

up vote 13 down vote accepted

Yes, such a Kähler form always exists: Embed $G$ as a matrix group in $\mathrm{SU}(n)$ for some $n$ and then let $G_\mathbb{C}\subset \mathrm{SL}(n,\mathbb{C})\subset M_n(\mathbb{C})$ be the complexification. Choose a Kähler form on this latter vector space, pull it back to $G_\mathbb{C}$ and then, using the compactness of $G$, average its pullbacks under left and right multiplications. This will yield a Kähler form on $G_\mathbb{C}$ that is $G$-invariant.

share|improve this answer
    
Is there any reference where I can see how the proof is done? –  hapchiu Dec 2 '12 at 8:27
    
what do you mean by average its pullbacks under left and right multiplications if $G$ is compact??? –  hapchiu Dec 2 '12 at 8:59
    
@hapchiu: It's a standard technique. If you have a compact Lie group $H$ acting on the left as biholomorphic transformations on a Kähler manifold $M$, then, letting $dh$ be Haar measure on $H$ and $\Omega$ be a Kähler form on $M$, one can construct an $H$-invariant Kähler form on $M$ by averaging: $$\bar\Omega = \int_H L_h^*(\Omega)\ dh\ ,$$ where $L_h:M\to M$ is left action by $h\in H$. In your case, $M=G_\mathbb{C}$ and $H = G\times G$, with $L_{(g_1,g_2)}(g) = g_1g{g_2}^{-1}$. –  Robert Bryant Dec 2 '12 at 13:21
    
You can see arxiv.org/abs/1307.0454 –  Hassan Jolany Nov 5 '13 at 23:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.