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For a Kähler manifold $M$, let $\operatorname{Iso}_{\mathbb{C}}(M)$ denote the group of holomorphic isometries.

Suppose that $K$ is a compact subgroup of $\operatorname{Iso}_{\mathbb{C}}(M)$ and there is a moment map $\mu:M\to\mathfrak{k}^*$ for the $K$-action such that the Kähler quotient $M_0:=\mu^{-1}(0)/K$ is smooth. Then, if $N_{\operatorname{Iso}_{\mathbb{C}}(M)}(K)$ denotes the normalizer of $K$ in $\operatorname{Iso}_{\mathbb{C}}(M)$, there is a natural group homomorphism $$N_{\operatorname{Iso}_{\mathbb{C}}(M)}(K)/K\to\operatorname{Iso}_{\mathbb{C}}(M_0).$$ Are there instances where this map is not surjective? I.e. $M_0$ posseses more symmetries than those comming from $M$?

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  • $\begingroup$ Is there any condition on $K$? If $K$ is trivial then the map is not surjective whenever $\mathrm{Iso}_{\mathbb{C}}(M)$ is nontrivial. $\endgroup$
    – HYL
    Commented Nov 4, 2017 at 10:18
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    $\begingroup$ @HYL If $K$ is trivial, then $M_0=M$ and the map is the identity map on $\operatorname{Iso}_{\mathbb{C}}(M)$. $\endgroup$
    – user116804
    Commented Nov 4, 2017 at 10:26

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Let $f : M \to E$ be a line bundle over an elliptic curve such that $\deg(M)<0$ and let $G = \mathbf{C}^*$ act on $M$ (on the left) by scalar multiplications. The maximal compact subgroup $K$ of $G$ is $U(1)$ and let $M$ be endowed with the $K$-invariant Kähler metric $h$ constructed as follows.

There exists a unitary representation of $\pi_1(E)$ of a hermitian vector space $L$ of dimension 1 such that $M = (L \times \tilde{E})/\pi_1(E)$ where $\tilde{E} = \mathbf{C}$ is the universal cover of $E$. This identification is compatible with the $G$-actions where on the right hand side, $G$ acts on $L$ on the left by scalar multiplications. Let $L \times \tilde{E}$ be endowed with the product metric $\tilde{h}= (h_L,h_{\mathrm{std}})$ where $h_L$ is the constant hermitian metric coming from the hermitian structure of $L$ and $h_{\mathrm{std}}$ the standard hermitian metric on $\mathbf{C}$. The metric $\tilde{h}$ is Kähler and is at the same time $K$-invariant and $\pi_1(E)$-invariant. So $\tilde{h}$ descends to a $K$-invariant Kähler metric $h$ on $M$.

The Kähler quotient $M_0 = M // G$ is isomorphic to the elliptic curve $E$ and the induced Kähler metric on $M_0$ is invariant under translations. So the group of translations of $E$ (still denoted by $E$) is contained in $\mathrm{Iso}_\mathbf{C}(M_0)$. We shall show that not every element of $E$ can be lifted to $N_{\operatorname{Iso}_{\mathbf{C}}(M)}$.

The $N_{\operatorname{Iso}_{\mathbf{C}}(M)}$-action on $M$ preserves fibers of $f$. As $H^0(E,M)= 0$, the action also preserves the $0$-section. Thus for each $g \in N_{\operatorname{Iso}_{\mathbf{C}}(M)}$, there exist a biholomorphic map $\phi : E \to E$ and an isomorphism $i:\phi^*M \simeq M$ such that $g = i\circ \phi^*$, and the image of $g$ under the map $N_{\operatorname{Iso}_{\mathbf{C}}(M)} \to \operatorname{Iso}_{\mathbf{C}}(M_0)$ is $\phi$. Finally since $\deg(M) \ne 0$, the line bundle $M$ is not invariant under translations, which shows that $N_{\operatorname{Iso}_{\mathbf{C}}(M)} \to \operatorname{Iso}_{\mathbf{C}}(M_0)$ is not surjective.

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  • $\begingroup$ 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 $\overline{G\cdot p}\cap \overline{G\cdot q}\cap M^{ss}\ne\emptyset$). $\endgroup$
    – user116804
    Commented Nov 5, 2017 at 7:03
  • $\begingroup$ 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 $M$. $\endgroup$
    – HYL
    Commented Nov 5, 2017 at 9:10
  • $\begingroup$ 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 $M_0 = \mu^{-1}(0)/K$. $\endgroup$
    – HYL
    Commented Nov 5, 2017 at 9:53
  • $\begingroup$ 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). $\endgroup$
    – user116804
    Commented Nov 5, 2017 at 17:37

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