To keep things simple, let $G$ be a finite group and $K$ a subgroup of it. The simplest definition of the Hecke algebra associated to this pair $(G, K)$ is that it is the algebra of $G$-endomorphisms $\text{End}_G(\mathbb{C}[G/K])$ of the permutation representation of $G$ on $G/K$.

The significance of this algebra is that $\mathbb{C}[G/K]$ represents the functor sending a $G$-representation $V$ to its $K$-fixed points $V^K$, and hence the Hecke algebra is the endomorphism ring of this functor: it naturally acts on $V^K$ for each $V$ and is the largest thing to do so.

Say that the pair $(G, K)$ is a **Gelfand pair** if any of the following equivalent conditions hold:

- whenever $V$ is irreducible, $\dim V^K \le 1$;
- the representation $\mathbb{C}[G/K]$ is multiplicity-free;
- the Hecke algebra is commutative.

Then we can diagonalize the action of the Hecke algebra on $V^K$ and hence associate to any $K$-fixed vector its Hecke eigenvalues, which together give a character of the Hecke algebra. If $V$ is irreducible and $\dim V^K = 1$ (which, by Frobenius reciprocity, is equivalent to $V$ being a component of $\mathbb{C}[G/K]$) then we only have one $K$-fixed vector to choose from up to scale and hence only one collection of Hecke eigenvalues.

One thing that is really nice about the Gelfand pair condition is that it can sometimes be verified "geometrically," by thinking explicitly about multiplication in the Hecke algebra in terms of double cosets and "relative positions" as described here and here. For example, $(S_n, S_{n-1})$ is a Gelfand pair because double cosets in this case correspond to relative positions of two elements of $\{ 1, 2, ... \dots n \}$ under the action of $S_n$, and there are two such relative positions: "equal" and "not equal." "Equal" is the identity so the Hecke algebra is generated by "not equal," and in particular must be commutative. This simple argument already implies that branching for the symmetric groups is multiplicity-free, allowing us to construct Gelfand-Tsetlin bases etc.

More generally, relative positions have a natural involution which switches the order of the two cosets involved; this gives an involution on the Hecke algebra, and the Hecke algebra is commutative iff every element is fixed by this involution. To check this condition we just need to check whether each relative position is the same when the two cosets involved are switched.

A more explicitly geometric example involving Lie groups is $(SO(3), SO(2))$: here the Hecke algebra involves relative positions of two points on the sphere $S^2$, which are labeled by the length of the shortest path between them. All of these relative positions are invariant under switching the two points involved, and so the Hecke algebra, whatever exactly that means, is also commutative in this case.

of representationsare! $\endgroup$ – Wolker Oct 17 '17 at 19:53