Timeline for Is $(G\rtimes H,H)$ a Gelfand pair iff $G$ is abelian?

6 events

when toggle format	what		by	license	comment
Jun 15, 2016 at 18:45	comment	added	Sebastien Palcoux		"$X:=G\rtimes H/H$ equals $G$" means that they can be naturally identified as set: $X=\{ (g,H) \ \| \ g \in G \}$. Then $H\backslash X = \{ (\sigma_H(g),H) \ \| \ g \in G \}$, and $\{ \sum_{h \in H}\sigma_h(g) \ \| \ g \in G \}$ generates $\mathbb C[G]^H$.
Jun 15, 2016 at 18:29	comment	added	Friedrich Knop		The equality holds because $X:=G\rtimes H/H$ equals $G$ with $H$ acting by automorphisms. Thus $H\backslash X$ is the set of $H$-orbits on $G$. But those also give a basis of $\mathbb C[G]^H$ (take the sum over each orbit).
Jun 15, 2016 at 18:29	vote	accept	Sebastien Palcoux
Jun 15, 2016 at 18:28	comment	added	Sebastien Palcoux		The fact that $\mathbb{C}[H \backslash (G\rtimes H) / H] \simeq \mathbb C[G]^H$ explains also your first paragraph because (for the action by conjugation) $\mathbb C[G]^G = Z(\mathbb C[G])$ which is commutative by definition.
Jun 15, 2016 at 14:39	history	edited	Friedrich Knop	CC BY-SA 3.0	Extra remark added
Jun 15, 2016 at 14:23	history	answered	Friedrich Knop	CC BY-SA 3.0