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Suppose $G$ is a $\mathbb{Q}$-algebraic group (I am interested in the semisimple case) acting rationally on a vector space $V_\mathbb{Q}$. Let $x \in V_\mathbb{Q}$ be a non-zero rational vector. Consider the stabilizer group $G_{x}$ of $x$. This is a closed subgroup in $G$.

Consider a compactly supported continuous $f:V_{\mathbb{R}}\rightarrow \mathbb{R}$ and then consider the following integral \begin{equation} \int_{G(\mathbb{R}) / G_{x}(\mathbb{R})} f(gx )dg . \end{equation}

  1. When is this integral well defined for all rational points? That is, when can I perform an integration on the homogeneous space $G(\mathbb{R})/G_x(\mathbb{R})$ for all rational points. In the semisimple case, this is the same as asking if there any general conditions to guarantee that $G_x$ will be unimodular for any rational point $x \in V_\mathbb{Q}$.

  2. Once it is well-defined, when is it finite for any $f$?

For example, some very generous conditions are when $G(\mathbb{R})$ acts transitively on non-zero points, or when $G(\mathbb{R}) x$ forms the non-zero points of a subspace in $V_\mathbb{R}$.

My guess is that such questions must have been considered in representation theory of algebraic groups but I don't really know where to start looking.

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    $\begingroup$ What are $g$ and $dg$ in the displayed formula? According to the formula you are integrating over a homogeneous space of the group rather than over the group. $\endgroup$
    – R W
    Mar 20 at 10:27
  • $\begingroup$ That's what I want to define. Maybe "exists" is the wrong choice of word here. $\endgroup$
    – Breakfastisready
    Mar 20 at 10:35
  • $\begingroup$ I've edited to reflect this. $\endgroup$
    – Breakfastisready
    Mar 20 at 10:36

2 Answers 2

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One condition that I came across is that it is sufficient to have $G(\mathbb{C}) \cdot x $ a closed subvariety of $V_\mathbb{C}$. Then $G(\mathbb{C}) \cdot x$ is a closed affine variety. This guarantees in particular that $G_x(\mathbb{C})$ must be reductive from Matsushima's criterion which says if $G$ is a connected reductive $\mathbb{Q}$-group and $H \subseteq G$ is a $\mathbb{Q}$-subgroup then $G/H$ is affine if and only if $H$ is reductive.

This also applies if $G(\mathbb{C}) \cdot x$ is just affine but not closed but I don't know if there is a nice way to check this for some $x \in V_\mathbb{Q}$.

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    $\begingroup$ For reductive groups, this approach always guarantees that we have the existence of orbital integrals of semisimple elements, and then some miracle due to Ranga Rao (Orbital integrals in reductive groups) says that we can even handle non-semisimple orbital integrals (under some conditions that I can never remember, but that are certainly implied by, and perhaps equal, being characteristic-$0$). $\endgroup$
    – LSpice
    Mar 20 at 14:41
  • $\begingroup$ This is a nice paper. We can probably use the Ranga Rao result to answer my question for the adjoint representation of a reductive $\mathbb{Q}$-group $G$. A function $f:\mathfrak{g} \rightarrow \mathbb{R}$ can be lifted to become a function $f: G(\mathbb{R}) \rightarrow \mathbb{R}$ via the exponential map. $\endgroup$
    – Breakfastisready
    Mar 20 at 16:38
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    $\begingroup$ Re, if the adjoint representation is what's of interest to you, then that is definitely better understood than the general picture. I'm not sure why you'd want to lift $f$ to $G(\mathbb R)$, though; your whole set-up seems to happen on $V(\mathbb R)$, which in this case would be $\mathfrak g(\mathbb R)$, so no lifting seems necessary. $\endgroup$
    – LSpice
    Mar 20 at 18:07
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In the setup of general topological groups there is a necessary and sufficient condition for the existence of an invariant measure on the homogeneous space $G/H$, where $H$ is a closed subgroup of $G$, which amounts to the coincidence of the modular function of the group $H$ with the restriction of the modular function of $G$ to $H$ (this is Weil's theorem, see Section III.4 of Nachbin's book The Haar integral). If $G$ is semisimple, then it is unimodular, and therefore in this case the necessary and sufficient condition is that $H$ be unimodular as well.

PS Now I am actually puzzled by the interrogation mark at the end of the 3rd paragraph of your question. Are you aware of Weil's theorem and asking for the conditions that would provide unimodularity of a subgroup of $G$?

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    $\begingroup$ Yes, I'm aware of this general homogeneous space setting. This is exactly the sort of technical difficulty that obstructs you from defining the integral. Once you have the integral, there is also the question of if it is not infinite. Let me edit this to make it more precise. $\endgroup$
    – Breakfastisready
    Mar 21 at 20:10

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