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Let $G=GL_n(K)$ where $K$ is an algebraically closed field of characteristic zero. Let $V$ be a finite dimensional rational representation of $V$. Assume that $v\in V$ has a reductive stabilizer $H\subseteq G$. I would like to ask for a reference for the following fact:

There is a rational finite dimensional representation $W$ of $ G$ and a point $w\in W$ such that the stabilizer of $(v,w)$ in $V\oplus W$ is also $H$, and such that the orbit $G\cdot (v,w)$ is closed in $V\oplus W$.

A simple instance of the above statement is the case where $n=1$, $V=K$ is the one dimensional representation upon which $x\in G$ acts by $x$, and the element $v$ is $1\in K$. The representation $W$ will then be $V^*$, and $w\in W$ can be any nonzero vector. The orbit of $(v,w)$ is then the hyperbola in $K^2$ which is closed (even though the orbit of $v$ is not closed).

Is it known if this is also true in case we replace $GL_n$ with some other reductive group?

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    $\begingroup$ Hi, Udi. Observe that it is enough to find $(W,w)$ for which the stab of $w$ is $H$ and its orbit is closed. This is done by thm 1.12 in Borel's book. $\endgroup$
    – Uri Bader
    Commented Feb 24, 2017 at 14:35
  • $\begingroup$ ... Using Matsushima's theorem. $\endgroup$
    – Uri Bader
    Commented Feb 24, 2017 at 14:39
  • $\begingroup$ Two small comments. 1) "Borel's book" refers to the first section (after AG) of his 1969 Benjamin lecture notes (with Bass), or equally to the expanded second edition published by Springer as GTM 126 in 1991. 2) In Ehud's second sentence, the last symbol should be $G$ rather than $V$. $\endgroup$
    – Jim Humphreys
    Commented Feb 24, 2017 at 15:06
  • $\begingroup$ @JimHumphreys, thanks a lot. Let me also take this opportunity and refer to the enlightening remarks in mathoverflow.net/q/225681/89334 regarding Matsushima's theorem. $\endgroup$
    – Uri Bader
    Commented Feb 24, 2017 at 15:15

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I don't know a reference to the question as asked, but below I give a brief argument.

Let $G$ be an algebraic group and $H$ a reductive subgroup. By Matsushima's theorem, $G/H$ is affine. Theorem 1.12 in Borel's "Linear algebraic groups" tells that there any $G$ affine action could be $G$-equivariantly closedly embedded in a rational $G$ representation. Applying to $G/H$ we obtain a $G$-representation $W$ and $w\in W$ having stabilizer $H$ and a closed orbit isomorphic to $G/H$.

Given any $G$-representation $V$ and $v\in V$ with stabilizer $H$, the stabilizer of $(v,w)\in V\oplus W$ will be $H$. We are left to argue that the orbit will be closed. Since $Gw\subset W$ is closed, it is enough to show that $G(v,w)\subset V\times Gw \simeq V\times G/H$ is closed. For this you need that the $H$-orbit of $v$ is closed, which is clear.

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    $\begingroup$ Many thanks. I am not sure I understand the last argument: why do you need to know that the $H$-orbit of $v$ is closed? Couldn't we just say that the orbit is the closed subset defined by the equation $\{(v',gH)|gv=v'\}$ ? $\endgroup$
    – Ehud Meir
    Commented Feb 24, 2017 at 15:05
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    $\begingroup$ Sure, you are right. I was referring to a more general observation: There is a natural homeomorphism (when you regard the spaces of orbits with the quotient topologies) $(V\times G/H)/G \simeq V/H$. In particular closed orbits in one correspond to closed orbit in the other. $\endgroup$
    – Uri Bader
    Commented Feb 24, 2017 at 15:10

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