Multiplicity one theorem for $(\mathrm {GL}_{n+1},\mathrm {GL}_n)$ over a local field of positive characteristic

Abstract

Let ${\mathbb {F}}$ be a non-archimedean local field of characteristic different from 2. We consider distributions on $\mathrm {GL}(n+1,{\mathbb {F}})$ which are invariant under the adjoint action of $\mathrm {GL}(n,{\mathbb {F}})$. We prove that any such distribution is invariant under transposition. This implies that the restriction to $\mathrm {GL}(n,{{\mathbb {F}}})$ of any irreducible smooth representation of $\mathrm {GL}(n+1,{{\mathbb {F}}})$ is multiplicity free. The characteristic zero case was proven in [5], and we use many ideas of that proof.

Appendix A. Lemmas in linear algebra

Our aim in this appendix is to prove Theorem 4.7. In all of the following discussion, V may be a finite dimensional vector space over any field ${{\mathbb {F}}}$, and we assume we have $A\in \mathrm {gl}(V)$, $v\in V$, and $\phi \in V^*$.

Notation A.1

Let $B\in \mathrm {gl}(V)$. We denote the characteristic polynomial of B by

$$\begin{aligned} P_B(x)=\sum _{k=0}^n {c_k(B) x^{n-k}}. \end{aligned}$$

Theorem A.2

For any $1\le k\le n$,

$$\begin{aligned} c_k(A+v\otimes \phi )=c_k(A)-\sum _{j=0}^{k-1}{c_{k-1-j}(A)\phi A^j v}. \end{aligned}$$

With this theorem, we can prove the desired Theorem 4.7 easily.

Proof of Theorem 4.7

By replacing v with $\lambda v$ we get

$$\begin{aligned} c_k(A+\lambda v\otimes \phi )=c_k(A)-\lambda \sum _{j=0}^{k-1}{c_{k-1-j}(A)\phi A^j v}. \end{aligned}$$

Now $(1)\Rightarrow (2)$ is a direct consequence. As this formula is linear in $\lambda$, one gets $(2)\iff (3)\iff (4)$. For $(3)\Rightarrow (1)$, use induction on k to show $\phi A^k v=0$ for all $k\ge 0$. If the claim is true for all non-negative integers smaller than k (it might be that $k=0$ and so this condition is trivial), then

$$\begin{aligned} \begin{aligned} c_{k+1}(A)&=\; c_{k+1}(A+\lambda v\otimes \phi )=c_{k+1}(A)-\lambda \sum _{j=0}^k{c_{k-j}(A)\phi A^j v}\\&= \; c_{k+1}(A)-\lambda \phi A^k v, \end{aligned} \end{aligned}$$

hence the claim. $\square$

1.1 A.1 Proof of Theorem A.2

Recall that for any element $A\in \mathrm {gl}(V)$, one defines $\mathrm {adj}(A):=(-1)^{n+1}\sum _{k=0}^{n-1} {c_k(A) A^{n-k-1}}$. It is characterized by $\mathrm {adj}(A)A=A\mathrm {adj}(A)=\det (A)I$.

Lemma A.3

(Matrix Determinant Lemma, see [14, Theorem 2 for k=1]).

$$\begin{aligned} \det (A+v\otimes \phi )=\det A+\phi \,\mathrm {adj}(A)v. \end{aligned}$$

Notation A.4

Let p(x) be a polynomial. We denote

$$\begin{aligned} \Delta _p(x,y)=\frac{p(x)-p(y)}{x-y}. \end{aligned}$$

Direct computation shows that if $p(x)=\sum _{i=0}^n a_i x^i$ then

$$\begin{aligned} \Delta _p(x,y)=\sum _{0\le i+j<n}a_{i+j+1} x^i y^j. \end{aligned}$$

In particular, $\Delta _p(x,y)$ is a polynomial in x, y.

Proposition A.5

( [9, p. 85]). We have $\mathrm {adj}(\lambda I-A)=\Delta _{P_A}(\lambda I, A)$.

Proof

Since both sides are polynomials in $\lambda$ and A, it is enough to check for a Zariski open subset. If $\lambda I-A$ is invertible then

$$\begin{aligned} \mathrm {adj}(\lambda I-A)=\frac{\det (\lambda I-A)}{\lambda I-A}=\frac{P_A(\lambda I)-P_A(A)}{\lambda I-A}=\Delta _{P_A}(\lambda I, A), \end{aligned}$$

thus equation follows for all $\lambda ,A$. $\square$

Proof of Theorem A.2

Applying Lemma A.3 to $\lambda I-A$ (and to $-v$ instead of v), we get

$$\begin{aligned} P_{A+v\otimes \phi }(\lambda )=\det (\lambda I - A-\phi \otimes v)=\det (\lambda I-A) -\phi \,\mathrm {adj}(\lambda I-A)v. \end{aligned}$$

Substituting Proposition A.5, we get

$$\begin{aligned} P_{A+v\otimes \phi }(\lambda )=P_A(\lambda )-\sum _{0\le i+j<n}c_{n-i-j-1}(A)\lambda ^i\phi A^j v. \end{aligned}$$

Isolating the coefficients of $\lambda ^i$ we get the equation in Theorem A.2. $\square$