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Let $(M,g,\omega)$ be a $d$-dimensional manifold equipped with a metric $g$ of signature $(t,s)$, $d = t+s$, and a symplectic form $\omega$. Let us assume that a Lie group $G\subset Isometries(M,g)$ acts on $(M,g,\omega)$ with corresponding moment map $\mu\colon M\to\mathbb{R}^{\ast}$. The symplectic reduction manfiold (Marsden-Weinstein reduction)

$M_{r} = \mu^{-1}(0)/G$

is a symplectic manifold with symplectic form induced by $\omega$.

My question is, what is the signature of the induced metric on $M_{r}$ by $g$? Is it possible, starting with a metric of indefinite signature, to obtain a Riemannian metric on $M_{r}$?

Thanks.

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Just take $R^4=C^2$ with complex coordinates $z_1=x_1+ i y_1, z_2= x_2 + i y_2$ and the flat
metric $-dz_1 d\bar z_1 + dz_2 d \bar z_2$. It has signature (2,2). As the group of isometries take the group of $x_1 $-translations, it preserves the canonic symplectic form $dx_1\wedge dy_1 + dx_2 \wedge dy_2$, and actually is generated by the function $y_1$. The symplectic reduction gives you a positively definite metric $dz_2d\bar z_2$.

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  • $\begingroup$ thanks for the example. Do you know of any pseudo-Kahler example? If I am not mistaken, in your example the symplectic form that you use is compatible with the Riemannian metric, not the (2,2) metric. $\endgroup$
    – Bilateral
    Commented Mar 19, 2015 at 22:38
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    $\begingroup$ It depends of course what you understand by compatible. It is parallel, and the corresponding endomorphism is a complex structure, though not the standard one. Alternatively, you make take another symplectic form $-dx_1\wedge dy_1+ dx_2\wedge dx_2$; it corresponds to the standard complex structure $\endgroup$
    – Vladimir S Matveev
    Commented Mar 20, 2015 at 10:06

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