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Let $G$ be a Lie group acting on a complex manifold $M$. Let $p$ be an isolated fixed point. Let us look at the representation of $G$ on $T_pM$. Suppose $T_pM = \bigoplus V_i^{\oplus n_i}$ where $V_i$ is an irreducible complex representation and $V_i\ne V_j$. Let us denote $V_i^{\oplus n_i}$ by $W_i$ and call $W_i$ an isotypic component of the representation.

Question 1: Under what conditions on $G$ and/or on the isotypic component $W_i$ there locally exists a $G$-invariant complex submanifold $M_i$ such that $T_pM_i = W_i$? If it exists, how one can construct it? It is clear that we should look exactly on the isotypic components and not on arbitrary subrepresentations as one can easily imagine an example when the representation is trivial but for every complex subspace of $T_pM$ there doesn't exist an invariant submanifold tangent to it. In fact, when $W_i$ is trivial, I suppose, many bad things can happen.

Question 2: If such a submanifold exists, is it unique?

I'd be grateful if you can answer these questions just for $\mathbb G_m^n$ but of course it is much more interesting for arbitrary Lie groups. In the case of $\mathbb G_m^n$ one can look on the attractor and the repellent of the fixed point but that gives as only invariant submanifolds tangent the direct sum of subrepresentations with negative and positive weights accordingly.

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    $\begingroup$ I think that the following answer settles the question in the case when $G$ is a compact group or a complexification of a compact group: mathoverflow.net/questions/272120/… by reducing this question to the case of linear action on $\mathbb C^n$ $\endgroup$ – Dmitri Panov Dec 18 '18 at 14:42
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By Bochner's theorem, in the complex analytic setting, if $G$ is a reductive complex Lie group acting holomorphically, or is a compact Lie group, then there such submanifolds, because the action is linearisable.

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    $\begingroup$ Dmitri Panov's link gives the proof of Bochner's theorem. $\endgroup$ – Ben McKay Dec 18 '18 at 14:43

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