Do all symmetries of a Kähler quotient come from the original space? 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 $M$?
 A: 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.
