Let $X$ be a complex manifold with a holomorphic action of a complex reductive group $G$. Let $Y \subset X$ be a $G$-invariant reduced complex analytic subspace. Is the restriction $$ \mathcal{O}_X^G \to \mathcal{O}_Y^G $$ surjective (on stalks)? Here $\mathcal{O}_X^G$ is the sheaf of $G$-invariant holomorphic functions on $X$ and similarly for $Y$.
1 Answer
After writing an answer in terms of global sections, then realising the original question was in terms of stalks, I realised that I don't understand what "the sheaf of $G$-invariant holomorphic functions on $X$" means: what is $\mathscr O_X^G(U)$, if $U$ is a non-$G$-stable open subset of $X$? I wonder if you might mean the sheaf of holomorphic functions on the quotient $X/G$, which is defined by putting $\mathscr O_{X/G}(\overline U_X) = \mathscr O_X(U_X)^G$ for any open subset $\overline U_X$ of $X/G$ with pullback $U_X \subseteq X$. I'll answer the question that way, but, since the technique I describe is very robust, hopefully I can adapt it to whatever the correct interpretation is.
Fix a point $\overline y \in Y/G$, and a germ $f$ at $\overline y$ (that is, an equivalence class of $G$-invariant holomorphic functions on $Y$ defined on the pullback of an open neighbourhood in $Y/G$ of $\overline y$).
Fix an open subset $\overline U_X$ of $X/G$ containing $\overline y$ that is small enough that $f$ is defined on $\overline U_X \cap Y/G$ and that, if we write $U_X$ for the pullback to $X$ of $\overline U_X$, then $f$ extends to a holomorphic function on $U_X$. Then consider the space $\mathscr V$ of functions on $X$ whose restriction to $Y$ is proportional to $f$. This is a $G$-representation, hence also a $\mathfrak g$-representation.
Let $K$ be the compact form of $G$. Since $K$ is compact, the representation $\mathscr V$ has a non-$0$, $K$-fixed vector $F$. The span of $F$ is annihilated by $\mathfrak k$, hence by $\mathfrak g = \mathfrak k \otimes_{\mathbb R} \mathbb C$; so $F$ itself is $G$-fixed.
-
1$\begingroup$ Thanks a lot for your answer. Yes, that's the correct interpretation; sorry for not being very precise. On the last point: is it true that every complex representation (possibly infinite-dimensional) of a compact Lie group has a non-zero fixed-point? I wasn't aware of that. Any reference or easy argument? $\endgroup$ Aug 3, 2020 at 15:01
-
1$\begingroup$ hmm... that can't be right since that's equivalent to a 1-dimensional subrepresentation so the irreducible representations have no non-zero fixed points. I don't understand your argument then. What am I missing? $\endgroup$ Aug 3, 2020 at 15:47
-
$\begingroup$ No, not every representation has a non-$0$ fixed point, but it does have a canonical projection to its fixed space (i.e., its isotrivial component). In this concrete case, since we are restricting to a $G$-fixed function on $Y$, the canonical form of the restriction means we don't change the restriction. Since the restriction is non-$0$ (else extendibility is easy!), so too is the $K$-fixed function that extends it. $\endgroup$– LSpiceAug 3, 2020 at 19:19
-