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A) Let $M$ is a non-compact manifold and $G$ be a compact Lie group which acts on $M$ and preserves complex structure then If $M$ has Kaehler manifold, then the symplectic quotient of $M$, i.e, $M//G$ has kaehler structure?

What about hyper-kaehler?

B) If $M$ has hyper-kaehler structure then $M//G$ has hyper-kaehler structure?

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If $(X,\omega)=(M,I,\omega)$ is Kaehler, and $G$ is a compact Lie group, which acts freely on $M$ and preserves both $I$ and $\omega$, then the symplectic quotient is Kaehler. More precisely, $M//G\simeq X^{st}/G^c$, where $G^c$ is the complexification of $G$, and $X^{st}\subset X$ are the stable points for the action. See, e.g., HKLR, Theorem 3.1 and Hitchin's Sem. Bourbaki 748, section 3.4.

In the hyperkaehler case, there is again compatibility with the symplectic quotient, but it is more subtle. Your suggestion in B) cannot possibly work, because $\dim_R M-2\dim_R G$ need not be divisible by $4$. The actual relation is as follows. Suppose $M$ is hyperkaehler, and $G$ (acts freely in a hamiltonian way by isometries and) preserves "the" three Kaehler forms $\omega_I$, $\omega_J$, $\omega_K$. Then we have a moment map $$ \mu=(\mu_I,\mu_J,\mu_K): M\to \mathbb{R}^3\otimes\mathfrak{g}^\vee, $$ and for a regular value $\xi$, the quotient $\mu^{-1}(\xi)/G$ is hyperkaehler (see HKLR, Theorem 3.2 and SB 748, Theorem 3.) added: provided $\xi$ is $G$-invariant.

If you want to link this with the symplectic or Kaehler quotient, you fix one of the complex structures, say $I$, and break the three moment maps into a real moment map $\mu_I$ and a complex one, $\mu_c=\mu_J+i\mu_K$. Then, if $\xi=(\xi_1,\xi_2,\xi_3)$, we have $$ \mu^{-1}(\xi) = \mu_I^{-1}(\xi_1)\bigcap\mu_c^{-1}(z) $$ where $z=\xi_2+i\xi_3$. The hyperkaehler quotient is the symplectic reduction of $\mu_c^{-1}(z)$, or equivalently, the Kaehler quotient $\mu_c^{-1}(z)^{st}/G^c$.

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  • $\begingroup$ But for the action of $S^1$ on $\mathbb C^2$, the symplectic quotient is $\mathbb P^1$, which is not Kahler,so which condition here does not occurs? $\endgroup$
    – user21574
    Jun 2, 2014 at 15:47
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    $\begingroup$ @HassanJolany Projective spaces are all Kahler (with the Fubini-Study metric), no? I suppose that in this example it should be the same as that produced by Thm 3.1 of HKLR. $\endgroup$
    – Stephen
    Jun 2, 2014 at 15:52
  • $\begingroup$ Yes, exactly, I copy/pasted Peter's comment, so by these answer $M$ is not required to be compact in general case for A) $\endgroup$
    – user21574
    Jun 2, 2014 at 15:56
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    $\begingroup$ I have seen this Kahler quotient called $M //// G$, in keeping with the idea that in $M//G$ we subtract $G$ off twice. $\endgroup$
    – Theo Johnson-Freyd
    Jun 3, 2014 at 3:54
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    $\begingroup$ @Peter Dalakov My point is that unless the stabilizer of $\xi$ is all of $G$, there is not reason for $\mu^{-1} (\xi)$ to be $G$-invariant. In symplectic setting this can be taken care of by the shifting trick. But in general the coadjoint orbits of compact Lie groups are only Kaehler, not hyperKaheler. So the shifting trick in the hyperKaehler setting does not work. $\endgroup$
    – Eugene Lerman
    Jun 4, 2014 at 14:55
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No, I think this need not be the case. Consider the usual action of $S^1$ on $\mathbb{C}^2$. The symplectic quotient is $\mathbb{P}^1$, which is not hyper-Kahler for dimension reasons.

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  • $\begingroup$ thanks PETER, So, I understand why Hitchin introduced another definition for symplectic quotient of hyper-kahler manifolds $\endgroup$
    – user21574
    Jun 2, 2014 at 11:01
  • $\begingroup$ But what about A) is it correct?, I know that when $M$ is compact then the reduced space certainly has kahler structure $\endgroup$
    – user21574
    Jun 2, 2014 at 11:12
  • $\begingroup$ But here in Theorem 3.3, he didn't mention that $M$ must be compact, I doubted and asked this question math.berkeley.edu/~alanw/277papers00/zhu.pdf $\endgroup$
    – user21574
    Jun 2, 2014 at 11:34
  • $\begingroup$ Okay, I apparently misinterpreted your question in part A). It did not sound like you were asking a question about the Kahler structure. $\endgroup$
    – Peter Crooks
    Jun 2, 2014 at 16:52
  • $\begingroup$ Peter Crooks, I am sorry, My question had not been written very well. $\endgroup$
    – user21574
    Jun 2, 2014 at 16:59
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There is a notion called hyperkahler quotient, if it is what you ask. (cf. Hitchin, etc., Hyperkahler Metrics and Supersymmetry, Commun. Math. Phys. 108, 535-589 (1987)). Assume that G is a compact Lie group acting freely on a hyperkahler manifold M, which preserves the complex structures I, J, K, and corresponding Kahler forms $\omega_{I}$, $\omega_{J}$ and $\omega_{K}$. If we have three moment maps $h_{I}$ $h_{J}$ $h_{K}$, then we denote $$h=(h_{I},h_{J},h_{K}): M \rightarrow g^{*} \otimes R^{3},$$ and $h^{-1}(0)/G$ is a hyperkahlerian. See Theorem 3.2 in the above paper.

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  • $\begingroup$ yes en.wikipedia.org/wiki/Hyperk%C3%A4hler_quotient, but, I am looking for noramal definition of symplectic quotient introduced by Marsdern and not this definition you mentioned. Your moment map is different. If you see I have used of $M//G$ and not $M///G$ which is for notation of Hitchin's reduced space for Hyperkahler manifolds $\endgroup$
    – user21574
    Jun 2, 2014 at 10:58

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