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
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$\begingroup$ If $G$ be compact then $T^*G\cong G\times \mathfrak g^*\cong G^{\mathbb C}$, since $T^*G$ is K\"ahler, then it induces a Kähler structure on $G^{\mathbb C}$ which is $G$-invariant $\endgroup$– user21574Commented Jan 19, 2017 at 5:18
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1 Answer
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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.
answered Dec 1, 2012 at 19:48
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$\begingroup$ Is there any reference where I can see how the proof is done? $\endgroup$– hapchiuCommented Dec 2, 2012 at 8:27
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$\begingroup$ what do you mean by average its pullbacks under left and right multiplications if $G$ is compact??? $\endgroup$– hapchiuCommented Dec 2, 2012 at 8:59
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$\begingroup$ @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 $L_{(g_1,g_2)}(g) = g_1g{g_2}^{-1}$. $\endgroup$ Commented Dec 2, 2012 at 13:21
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$\begingroup$ You can see arxiv.org/abs/1307.0454 $\endgroup$– user21574Commented Nov 5, 2013 at 23:40