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I am struggling to understand the following passage on Weil's text "Sur la formule de Siegel dans la théorie des groupes classiques" on page 16.

enter image description here

If I understood correctly, in the second sentence Weil writes that a "convenable" (suitable) relatively invariant measures on $g(\omega)_\mathbb{A}/g(\omega)_k$ and $G_\mathbb{A}/g(\omega)_\mathbb{A}$ will allow us to unfold the integral at (10). For convenience, I reproduce the integral of Equation (10) here:

enter image description here

I am a bit confused here because there must be justification of why such relatively invariant measures must exist. In Bourbaki cited there, and also in the standard literature here, a relatively invariant measure exists on a homogeneous space $G/H$ only if the modular character on $H$ can be extended to be made a character of $G$. I don't see where Weil justifies that this will be true for $G_\mathbb{A}/g(\omega)_\mathbb{A}$.

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    $\begingroup$ Maybe they're both reductive, so unimodular? $\endgroup$
    – paul garrett
    Mar 18, 2023 at 19:42
  • $\begingroup$ The assumption he works with is that $G$ is a connected algebraic group acting on a variety $X$ (both over a field $k$) and $g(\omega)$ is the stabilizer of a point $\omega \in X$. $\endgroup$
    – Breakfastisready
    Mar 18, 2023 at 19:52
  • $\begingroup$ @paulgarrett, I have seen in the surrounding literature that the working assumption is that stabilizers in an algebraic group $G$ acting on a vector space are unimodular. Does this assumption fail for some pathological examples? Weil seems to somehow prove that when $G$ is semisimple, stabilizers of rational vectors are also unimodular. $\endgroup$
    – Breakfastisready
    Mar 19, 2023 at 23:54
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    $\begingroup$ Well, no, stabilizers of rational vectors can fail to be unimodular in interesting cases, for example, already for $SL(2,\mathbb Q)$. Many traditional overviews of such things pointedly ignore this possibility, indeed. And, yes, such subgroups do cause trouble in making precise "trace formula" assertions, etc. $\endgroup$
    – paul garrett
    Mar 20, 2023 at 0:07
  • $\begingroup$ @paulgarrett I'm still trying to understand your comment. If you mean the upper-triangular matrices in $SL_2$ then this is a stabilizer of a projective representation, but not a stabilizer of a usual representation as pointed out here: mathoverflow.net/questions/409499/… $\endgroup$
    – Breakfastisready
    Mar 20, 2023 at 3:53

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