Timeline for Do all symmetries of a Kähler quotient come from the original space?
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| Nov 5, 2017 at 17:37 | comment | added | user116804 | Great! What you wrote sounds right, I'll take the time to read it carefully. By the way, there is always a holomorphic structure on $\mu^{-1}(0)/K$, even when the action of $K$ doesn't extend to an action of $G=K_{\mathbb{C}}$. You define the complex structure by lifting vectors horizontally, apply the complex structure of $M$, and then project back down. This almost complex structure is always integrable and defines a Kahler structure on $\mu^{-1}(0)/K$ whose Kahler form is the reduced symplectic form. The details are in Hitchin-Karlhede-Lindström-Roček (1987). | |
| Nov 5, 2017 at 9:53 | comment | added | HYL | Set-theoretically the $M^{ss}/\sim$ you wrote is exactly $M//G$ and the latter is homoeomorphic to $\mu^{-1}(0)/K$. The reason I consider $M//G$ instead of $\mu^{-1}(0)/K$ is that $\mu^{-1}(0)/K$ doesn't have a natural holomorphic structure (because the moment map is not holomorphic), so we can't discuss $\operatorname{Iso}_{\mathbb{C}}(M_0)$ if $M_0 = \mu^{-1}(0)/K$. | |
| Nov 5, 2017 at 9:10 | comment | added | HYL | The moment map $\mu : L \times \mathfrak{u}(1) \simeq \mathbf{C} \times i \mathbf{R} \to \mathbf{R}$ for the $U(1)$-action on $L$ is $\mu(z,i\xi) = -\frac{1}{2}|z|^2\xi$. Since $K$ acts fiberwisely on $L \times \tilde{E}$, the moment map on $L \times \tilde{E}$ is just the composition of the projection $L \times \tilde{E} \to L$ with $\mu$. As the $\pi_1(E)$-action on $L$ is unitary, this moment map descends to the quotient, which is the moment map of the $U(1)$-action on $M$. | |
| Nov 5, 2017 at 7:03 | comment | added | user116804 | Thanks for your answer. How do you know that there is a moment map for the $U(1)$-action? And even so, the quotient $\mu^{-1}(0)/K$ is not always equal to the GIT quotient $M//G$ ($G=K_{\mathbb{C}}$) unless some additional conditions are proved. In general, we have $\mu^{-1}(0)/K=M^{ss}/{\sim}$, where $M^{ss}$ is the set of analytically semistable points (those points $p\in M$ for which the closure of $G\cdot p$ intersects $\mu^{-1}(0)$) and $\sim$ is the relation of $S$-equivalence ($p\sim q$ iff $\overline{G\cdot p}\cap \overline{G\cdot q}\cap M^{ss}\ne\emptyset$). | |
| Nov 5, 2017 at 0:34 | history | answered | HYL | CC BY-SA 3.0 |