Home > representation theory > Existence of the Weil index

## Existence of the Weil index

From now on, $F$ will be a non-Archimedean local field, with characteristic not equal to 2 as always, and $\psi : F \to \mathbb{T}$ a fixed nontrivial continuous character. For a finite-dimensional vector space $V$ over $F$, we will denote by $C^{\infty}_c(V)$ the space of locally constant, compactly supported $\mathbb{C}$-valued functions on $V$. If we are given a nondegenerate symmetric bilinear form $B : V \times V \to F$, there is a unique self-dual Haar measure $\mu$ with respect to the pairing $\psi \circ B : V \times V \to \mathbb{T}$. Concretely, given a subset $X \subset V$, we can define

$X^{\perp} = \{ v \in V \ | \ \psi(B(v,w)) = 1 \ \text{for all } w \in X \},$

and then $\mu$ is the unique Haar measure on $V$ such that $\mu(\Lambda) \cdot \mu(\Lambda^{\perp}) = 1$ for some (equivalently, for any) lattice $\Lambda \subset V$. So we obtain a canonical Fourier transform $\mathcal{F} : C^{\infty}_c(V) \to C^{\infty}_c(V)$ defined by the formula

$\mathcal{F}(f)(y) = \int_V f(x) \psi(B(x,y)) \ d\mu(x).$

Define the space of distributions $\mathcal{D}(V) = C^{\infty}_c(V)^*$ on $V$ to be the dual vector space to $C^{\infty}_c(V)$. There is a natural injection $C^{\infty}(V) \to \mathcal{D}(V)$, where $C^{\infty}(V)$ denotes the space of all locally constant $\mathbb{C}$-valued functions on $V$, which sends $f \in C^{\infty}(V)$ to the linear functional

$g \mapsto \int_V fg \ d\mu.$

Any invertible operator $T : C^{\infty}_c(V) \to C^{\infty}_c(V)$ induces an adjoint operator $(T^*)^{-1} : \mathcal{D}(V) \to \mathcal{D}(V)$ defined by the formula $T^*(\varphi)(f) = \varphi(T^{-1}(f))$. We will usually abuse notation and denote $(T^*)^{-1}$ by $T$ also.

Let $q(v) = B(v,v)$ be the associated quadratic form and consider the function $f_q : V \to \mathbb{T}$ given by $f_q(v) = \psi(\tfrac12 q(v))$. Obviously $f_q$ does not lie in $C^{\infty}_c(V)$, since $\psi$ never vanishes. However, since $q$ is continuous and $\psi$ is locally constant it follows that $f_q$ is locally constant, so we can think of $f_q$ as a distribution and take its Fourier transform.

Theorem There exists a scalar $\gamma(q) \in \mathbb{C}^{\times}$, called the Weil index of $q$, such that

$\mathcal{F}(f_q) = \gamma(q) \cdot f_{-q},$

and furthermore $|\gamma(q)| = 1$. In fact, for a sufficiently large lattice $\Lambda \subset V$, we have

$\gamma(q) = \int_{\Lambda} f_q \ d\mu.$

The proof of the theorem we will give is, unlike Weil’s original proof, completely independent from other results on the Weil representation and uses only the theory of Fourier transforms. The fact that $\gamma$ takes values in complex numbers of modulus one will follow from an elementary result on Gauss sums.

Unfortunately, finishing the proof this time would probably make this post unreasonably long, so I’ll content myself with a lemma on Fourier transforms. Let us define translation and multiplication operators on $C^{\infty}(V)$ by the formulas

$(T_af)(x) = f(x + a) \ \text{and} \ (M_af)(x) = \psi(B(a,x))f(x)$

for all $a \in V$. In particular, these operate on $C^{\infty}_c(V)$, so as discussed above we obtain adjoint operators $T_a$ and $M_a$ on $\mathcal{D}(V)$.

Observe also that $C^{\infty}_c(V)$ is a $C^{\infty}(V)$-module, so formally we see that $C^{\infty}(V)$ also operates on $\mathcal{D}(V)$ by the formula $(f \cdot \varphi)(g) = \varphi(f \cdot g)$.

Lemma We have the following equations of operators on $\mathcal{D}(V)$:

$\mathcal{F} \circ T_a = M_{-a} \circ \mathcal{F} \ \text{and} \ \mathcal{F} \circ M_a = T_a \circ \mathcal{F}.$

Also, the translation and multiplication operators interact with the $C^{\infty}(V)$-module structure on $\mathcal{D}(V)$ via

$T_a(f \cdot \varphi) = T_{-a}(f) \cdot T_a(\varphi) \ \text{and} \ M_a(f \cdot \varphi) = M_a(f) \cdot \varphi = f \cdot M_a(\varphi).$

Proof. These are trivial calculations.

$\Box$