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Dec 17, 2019 at 13:51 comment added David Loeffler Splendid, thanks!
Dec 16, 2019 at 12:54 vote accept David Loeffler
Dec 16, 2019 at 12:47 comment added Friedrich Knop 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.
Dec 16, 2019 at 8:45 comment added David Loeffler 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?
Dec 15, 2019 at 20:28 history answered Friedrich Knop CC BY-SA 4.0