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If $G$ is a reductive group, $T$ a maximal torus and $W$ its Weyl group the Chevalley restriction theorem (in its "multiplicative" version) gives an isomorphism between the GIT quotient of $G$ by the conjugation action on itself and the quotient $T/W$.

This result has several generalisations. In particular, in Orbits, Invariants, and Representations Associated to Involutions of Reductive Groups, Richardson proved a similar theorem: $X//G^{\theta} \cong A/W_{\theta}$ for $X=G/G^{\theta}$ a symmetric variety, $A$ a maximal $\theta$-anisotropic torus and $W_{\theta}$ the so called "little Weyl group".

A well-known generalisation of symmetric varieties are spherical varieties. I was wondering if a similar result exists in this situation, namely, if $G/H$ is a spherical homogeneous space,

  1. does there exist an isomorphism between the GIT quotient of $X$ by $H$ and the quotient of the torus $A$ associated to the spherical variety and the corresponding Weyl group?

Related to the theory of spherical varieties is the theory of spherical embeddings, and in particular, the wonderful compactification: a projective variety compactifying a spherical variety $X$.

In a comment in the blog post The Toric Variety Associated to the Weyl Chambers, Jason Starr mentions some "extension" of the Chevalley map, from the wonderful compactification of $G$ to the toric $T$-variety defined by the fundamental Weyl chamber. Regarding this I have two questions:

  1. Is there a reference for this last fact? That is, a reference for the fact that there is an isomorphism between the GIT quotient of the wonderful compactification by $G$ and the quotient of that toric variety by the Weyl group.

  2. Can it be generalized to any spherical variety? The result I have in mind is the existence of an isomorphism between the GIT quotient of the wonderful compactification of the spherical homogeneous space $G/H$ by the spherical subgroup $H$ and the toric variety defined by the associated torus and the corresponding dominant cocharacters.

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    $\begingroup$ Jason Starr had a few comments on that nCat blog post that you link, so I guessed which one you meant and edited accordingly. I hope that it was correct. $\endgroup$
    – LSpice
    May 5, 2022 at 14:12
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    $\begingroup$ @LSpice Exactly, thank you very much :-) $\endgroup$
    – G. Gallego
    May 5, 2022 at 14:16
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    $\begingroup$ In your Question 1. and Question 3., could you please clarify how $H$ is acting on $G/H$? Is it just the left regular action? If I understand correctly, the subgroup $H$ contains a unipotent radical of a Borel subgroup. In that case, the action of $H\times H$ on the open Bruhat cell is "almost transitive" in the sense that the multiplication map from $H\times T \times H$ to the Bruhat cell is surjective. So the quotient variety for the left $H$-action on the open orbit in $G/H$ is already a quotient torus of $T$. $\endgroup$ May 5, 2022 at 14:31
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    $\begingroup$ I see now that $H$ need not contain a unipotent radical, e.g., a maximal torus in $\textbf{SL}_2$ has a dense orbit on $\textbf{SL}_2/B \cong \mathbb{P}^1$. $\endgroup$ May 5, 2022 at 14:39
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    $\begingroup$ @JasonStarr Yes! It's the natural left action on cosets! Thanks for your comment, I'll think about it. $\endgroup$
    – G. Gallego
    May 5, 2022 at 14:47

1 Answer 1

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Edit: The answer to question 1 is yes if $G/H$ is a symmetric variety as the OP pointed out.

For arbitrary spherical varieties the answer is no in general. If my memory serves me right, the spherical variety $Sp(4,\mathbb C)/(\mathbb C^*\times SL(2,\mathbb C))$ is a counterexample. As far as I know, the $H$-orbit structure of $G/H$ is still unknown in full generality.

Added: The actual theorem of Chevalley is not a statement about conjugacy classes on the group $G$ but rather on its Lie algebra $\mathfrak g$: there is an isomorphism $\mathbb C[\mathfrak g]^G\overset\sim\to\mathbb C[\mathfrak t]^W$ where $\mathfrak t\subseteq\mathfrak g$ is a Cartan subalgebra. This theorem has been extended by Kostant-Rallis to symmetric spaces in the form $\mathbb C[\mathfrak p]^H\overset\sim\to\mathbb C[\mathfrak a]^{W_X}$ where $X=G/H$ is a symmetric variety. Here $\mathfrak p$ is the tankent space of $X$ in $eH$.

The point is that Kostant-Rallis does generalize to arbitrary spherical varieties $X=G/H$ if one replaces the tangent space by the cotangent space $\mathfrak h^\perp=(\mathfrak g/\mathfrak h)^*\subseteq\mathfrak g^*$ of $X$ in $eH$. Of course this makes only a difference if $H$ is not reductive but then it is essential.

Theorem: There is a subspace $\mathfrak a^*\subseteq\mathfrak h^\perp$ and an action of the little Weyl group $W_X$ of $X$ on $\mathfrak a^*$ such that $\mathbb C[\mathfrak h^\perp]^H\overset\sim\to\mathbb C[\mathfrak a]^{W_X}$.

The proof of this theorem is much more involved than Kostant-Rallis. First of all, the little Weyl group $W_X$ is not easy to define since it is in general not a subquotient of $H$. It was discovered by Brion while studying compactifications of $X$. Second, the subspace $\mathfrak a^*$ is not at all canonical (even up to conjugation by $H$). Finally, the action of $W_X$ on $\mathfrak a^*$ is not induced by elements of $H$. It is rather a monodromy action.

References: My Inventiones papers in vols. 99 and 116. In the first paper the theorem is proved where $W_X$ is some monodromy group. The second paper shows that $W_X$ coincides with the little Weyl group defined by Brion in J. Algebra vol. 134.

Coming back to the original question: The change from the tangent space to the cotangent space prevented so far all attempts to get a global Chevalley theorem. This change makes only a difference if $H$ is not reductive but that affects even the reductive case since most methods involve non-reductive subgroups in an essential manner.

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  • $\begingroup$ Isn't your first paragraph (the case where $H$ is a symmetric variety) the special case already mentioned in the question? $\endgroup$
    – LSpice
    May 6, 2022 at 14:48
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    $\begingroup$ Yes, don‘t know how I missed that. Thanks for pointing that out. I‘ll fix that later. $\endgroup$ May 6, 2022 at 16:26

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