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I would like to understand whether the following is true. Given a complex reductive group $G$ (for me $G = \operatorname{PGL}_n(\mathbb{C})$) acting on a vector space $V$ (for me $V = M_n(\mathbb{C})^{\oplus\, d}$, i.e, $d$-tuples of $n\times n$ complex matrices and the action is by simultaneous conjugation) we construct the quotient $\pi \colon V \to Q = V//G$. Let $X \in V$ be a stable point (in the case at hand it means that the $d$-tuple of matrices doesn't have a common non-trivial invariant subspace), does there exists an open neighborhood $\pi(X) \in U \subset Q$ and an analytic section $\sigma \colon U \to V$ of $\pi$, such that $\sigma(p)$ is in the Kempf-Ness set for $p \in U$. In the case at hand, we can take the $\operatorname{PU}_n$-invariant inner product to be $\operatorname{tr}(X Y^*) = \operatorname{tr}\left( \sum_{j=1}^d X_j Y_j^* \right)$. The Kempf-Ness set is the hypernormal points, i.e, $X$, such that $\sum_{j=1}^d [X_j, X_j^*] = 0$.

The motivation for this question is that for $d=1$, though the setting is a bit different, it works.

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This is a standard consequence of the local normal form for submersions (which is basically just a special case of the implicit function theorem; note this means you can work analytically); the map of the stable points in the Kempf-Ness locus to the quotient is a $PU_n$-principal bundle, and principal bundles are always locally trivial.

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  • $\begingroup$ Thanks for the answer, but I could use a clarification. I know that it is locally trivial, but why can you do it analytically?The Kempf-Ness locus is a real analytic subset and I want a section of the $\operatorname{PGL}_n$-principal bundle. What am I missing? $\endgroup$
    – shamovic
    Commented Apr 10, 2018 at 19:32
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    $\begingroup$ Because the local normal form for submersions is an analytic result. Pick a preimage $y$ of $x$ and pick $n^2-1$ functions that vanish at $y$ on a whose differentials at $y$ are dual to a basis of $\mathfrak{pgl}_n$; the common zeros of these functions are an analytic submanifold of a neighborhood of $y$ mapping isomorphically to a neighborhood of $x$. The inverse of this exists and is analytic by the inverse function theorem. $\endgroup$
    – Ben Webster ♦
    Commented Apr 11, 2018 at 0:20
  • $\begingroup$ I agree, you choose $y$ in the Kempf-Ness set and choose these functions, but you can't choose any $n^2-1$-tuple of functions that satisfies your condition. You need their common zeroes to lie in the Kempf-Ness set, which in itself is not cut out by analytic functions. $\endgroup$
    – shamovic
    Commented Apr 11, 2018 at 18:44

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