A pair of groups $(G,H)$ is called a symmetric pair if $H$ is the group of fixed points of an involutive automorphism of $G$, for example $(GL(2n,\mathbb{F}_q),Sp(2n,\mathbb{F_q}))$ is a symmetric pair with respect to the involution $\theta(A)=J(A^{-1})^tJ^{-1}$ where:

$$ J=\left(\begin{array}{cc} 0&id_n\\-id_n&0 \end{array}\right) $$

We also say that a pair $(G,K)$ where $K$ is a subgroup of $G$ is a Gelfand pair if for any irreducible representation $\pi$ of $G$, we have that $\pi^K$ (the vectors fixed by $K$) is at most one dimensional. Alternatively, $Hom_K(\pi, 1)\leq 1$.

It is known that many symmetric pairs of finite groups of lie type are Gelfand pairs, as in many we have that an associated anti-involution $\sigma(x)=\theta(x^{-1})$ preserves $H-H$ double cosets. I was wondering if it was known that any such symmetric pair of finite groups of Lie type is a Gelfand pair, or if there is a known counterexample to that assertion.

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    $\begingroup$ If the symmetric space if Galois, i.e. of the form $H({\mathbb F}_{q^2})/H({\mathbb F}_q)$, then it is known that there is multiplicity $1$ for the "stable" representations of $H({\mathbb F}_{q^2})$. For instance for ${\rm GL}_n$ or ${\rm U}_n$ all representations are stable. Cf. D. Prasad Compositio 1999. $\endgroup$ Jun 4 '16 at 22:14

The symmetric space ${\rm GL}(2,{\mathbb F}_q)/T$, where $T$ is the diagonal torus, is not a Gelfand pair : the Steinberg representation contains the trivial character of $T$ with multiplicity $2$.

Here is a proof that the multiplicity of this character is $\geqslant 2$. The Steinberg representation may be viewed as the space of functions $f$ on the projective line ${\mathbb P}^1 ({\mathbb F}_q)$ satisfying $$ \sum_{x\in {\mathbb P}^1 ({\mathbb F}_q)}f(x)=0\ . $$ The torus $T$ fixes exactly two points $x$, $y$ of the projective line and acts transitively on the $z\not= x,y$. Let $f_0$ the functions defined by $f_0 (x)=1=-f_0 (y)$ and $f_0 (z)=0$, if $z\not= x,y$. Similarly let $f_1$ the function defined by $f_1 (x)=f_1 (y)=1$ and $f(z)=-2/(q-1)$, if $z\not= x,y$. Then ${\rm Vect} (f_0 ,f_1 )$ is a two dimensional subspace of the Steinberg representation fixed by $T$.


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