Let $G$ be a reductive algebraic group (over some alg. closed field $k$ of char 0), and $H$ a subgroup such that $(G, H)$ is spherical (i.e., the Borel $B$ of $G$ has an open orbit on $G/H$). Then $k[G/H]$ is multiplicity-free as a $G$-module. I'm looking for a nice description of the irreducible $G$-representations it contains.

More precisely, let $\Lambda(G)$ be the lattice of integral weights for $G$ and $\Lambda_+(G)$ be the cone of dominant integral weights. For each $\lambda \in \Lambda_+(G)$ we have an irreducible representation $V_\lambda$. Let me write $\Lambda_+(G, H)$ for the set of $\lambda \in \Lambda_+(G)$ such that $V_\lambda$ appears in $k[G/H]$ (equivalently, such that $(V_\lambda)^H \ne 0$). The set $\Lambda_+(G, H)$ is a sub-monoid of $\Lambda_+(G)$ (see this question). Playing around with some examples suggests the following:

Claim: $\Lambda_+(G, H)$ is a lattice cone in $\Lambda(G)$, and its $\mathbb{Z}$-span is exactly the weights trivial on the $B$-stabiliser of a point in the open $B$-orbit on $G/H$.

Is this true? If so, is there a good reference that is easily digestible by non-experts like myself?

(Brion's paper "Spherical varieties" in the 1994 ICM proceedings, page 755, has a very general assertion about the sections of any $G$-equivariant line bundle on any spherical $G$-variety, which seems to be closely related; but I couldn't make head or tail of the proof, and surely there must be an easier approach for homogenous spherical varieties $G/H$ and trivial line bundles.)

  • $\begingroup$ To the title question, I’m guessing there should be an orbit method prediction as in Kobayashi (2005, §9). In principle this is for unitary representations, but he somewhat cryptically notes that others can be discussed “by applying Weyl’s unitary trick” (p. 505). $\endgroup$ Dec 15 '19 at 19:18

Let $X_0=B/B_{x_0}=Bx_0$ be the open $B$-orbit in $X=G/H$. Then every character of $B$ which is trivial on $B_{x_0}$ yields a $B$-semiinvariant $f_\lambda$ on $X_0$. In general, it does not extend to a regular function on $X$. If it does, this means that $V_\lambda$ occurs in $k[X]$. So your second problems asks whether every $B$-semiinvariant on $X_0$ is a quotient of two regular $B$-semiinvariants on $X$. This is true if $X$ is quasiaffine (e.g., if $H$ is reductive) but false in general. Take, e.g., $G=SL(3,k)$ and $H=ker(\chi)$ where $\chi:B\to G_m$ is a character which is neither dominant nor antidominant (e.g. $\chi=a_{22}$). Then $V_\chi^H=0$ for all $\lambda$.

I don't know what you mean by lattice cone but if it means a monoid which is the intersection of a convex cone by a lattice then your first claim is true. This is a consequence of the normality of $X$. More precisely, let $\overline\Lambda_+$ be the intersection of the convex cone and the lattice generated by $\Lambda_+(G,H)$. It is called a saturation. An element $\lambda$ is in $\overline \Lambda_+$ if there is $m\ge1$ such that $m\lambda\in\Lambda_+(G,H)$. Now, since $\lambda$ is in the lattice generated by $\Lambda_+$ it corresponds to a rational $B$-semiinvariant $f_\lambda$ on $X$. The power $f_\lambda^m$ is regular. So $f_\lambda$ is regular and $\lambda\in\Lambda_+$.

  • $\begingroup$ This is great, thanks! Apologies for my vagueness about "lattice cone": the definition I had in mind was the monoid spanned by a linearly independent set of lattice vectors, i.e. a set of the form $\{ n_1 v_1 + \dots + n_r v_r : n_i \in \mathbb{Z}_{\ge 0}\}$, where $v_i$ are vectors in $\Lambda(G)$ that are linearly independent in $\Lambda(G) \otimes \mathbb{Q}$. Does $\Lambda_+(G, H)$ necessarily have this form, at least if $H$ is reductive? $\endgroup$ Dec 16 '19 at 8:45
  • $\begingroup$ The answer to this question is related to the factoriality of $G/H$ which is related to the character group of $H$. In particular, if $H$ is semisimple then the monoid is free. Otherwise, you can find counterexamples on p. 149 of Krämer's paper "Sphärische Untergruppen...", the easiest being $G=SL(5)$ and $H=\mathbf G_m\cdot Sp(4)$. There the cone is $3$-dimensional with $4$ extremal reays. $\endgroup$ Dec 16 '19 at 12:47
  • $\begingroup$ Splendid, thanks! $\endgroup$ Dec 17 '19 at 13:51

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