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.



Is it known whether every symmetric pair of finite groups of Lie type is a Gelfand pair?
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 $T$. 


