TITLE: Is the representation of $GL_n(\mathcal{O})$ in functions on Grassmannian multiplicity free?
QUESTION [9 upvotes]: Let $\mathbb{F}$ be a local non-Archimedean field. Let $\mathcal{O}\subset \mathbb{F}$ be its ring of integers. Let $GL_n(\mathcal{O})$ be the (compact) group of $n\times n$ invertible matrices with entries in $\mathcal{O}$ such the inverse matrix also has entries in $\mathcal{O}$. Let $Gr_{i,n}$ be the Grassmannnian of linear $i$-dimensional subspaces in $\mathbb{F}^n$.

Is it true that the natural representation of $GL_n(\mathcal{O})$ in the space of locally constant functions on $Gr_{i,n}$ is multiplicity free? In other words is it true that any irreducible representation of $GL_n(\mathcal{O})$ appears with multiplicity at most 1? 

A reference would be helpful.
Remark. Archimedean analogue of the above statement, i.e. $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, is true; here $GL_n(\mathcal{O})$ should be replaced by $O(n)$ or $U(n)$ respectively. (That follows from the fact that the Grassmannian is a symmetric space.)

REPLY [5 votes]: This may be redundant as a complete answer with references has already been posted.
Just in case this is still useful: I think multiplicity one can be proved by the usual Gefland trick: we need to check that endomorphisms of the induced representation is a commutative ring, this follows once we check that the identity automorphism of that algebra is an anti-involution. The endomorphism algebra coincides with the Hecke algebra ${\mathbb{C}}[P(O)\backslash G(O)/P(O)]$ where $P$ is the group of block upper triangular matrices with blocks of sizes $i, \, n-i$. This can also be realized as the space of $G(O)$-invariant functions on pairs of rank $i$ summands in $O^n$. Switching the two elements in the pair induces an anti-involution. 
Looking at relative positions of two direct summands in $O^n$ one sees that the relative position of a pair $(N,M)$ is the same as the relative position of $(M,N)$, which yields the statement.