Let $(U,x)$ be an open complex $n$-manifold (say an $n$-ball) with an action of $S^1$ by holomorphic transformations that fix $x$. How to prove that there is a neighbourhood $U_1\subset U$ of $x$ where the action is linearisable? I.e. it is conjugated to some linear (diagonal) action of $S^1$ on $\mathbb C^n$.
1 Answer
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The tangent space $T_xU$ has a linear action of $G=S^1$. Choose a holomorphic map $\Phi$ from a neighborhood of $x$ to the tangent space $T_xU$ whose derivative is the usual isomorphism of vector spaces $T_xU\cong T_0(T_xU)$. Then the average of $g\circ\Phi\circ g^{-1}$ over all $g\in G$ (defined on some smaller neighborhood) has the same properties as $\Phi$ but now commutes with the action of $G$.
This works for any compact group $G$, and of course also works in the real case.