Home > representation theory > Proving the existence of the Weil index

## Proving the existence of the Weil index

Before going on to the proof of the theorem from last time, we need to prove the basic result about Gauss sums on finite abelian groups. A quadratic form on a finite abelian group $A$ is a map $q : A \to \mathbb{C}^{\times}$ satisfying $q(n \cdot a) = q(a)^{n^2}$ for $a \in A$ and $n \in \mathbb{Z}$, and if we define $B : A \times A \to \mathbb{C}^{\times}$ by the formula $B(x,y) = q(x+y)q(x)^{-1}q(y)^{-1}$, we also require that $y \mapsto B(x,y)$ is a group homomorphism $A \to \mathbb{C}^{\times}$ for all $x \in A$. The dual group of $A$ is $A^* = \text{Hom}(A,\mathbb{C}^{\times})$, and we call $q$ nondegenerate if the homomorphism $A \to A^*$ which sends $x \mapsto B(x,-)$ is an isomorphism.

The Gauss sum associated with $q$ is the number

$S(q) = \sum_{a \in A} q(a)$.

Lemma If $q : A \to \mathbb{C}^{\times}$ is a nondegenerate quadratic form, then $|S(q)| = \sqrt{\# A}$.

Proof. We begin with the calculation

$|S(q)|^2 = S(q) \cdot \overline{S(q)} = \sum_{a,b \in A} q(a)\overline{q(b)} = \sum_{a,b \in A} q(a-b)B(a-b,b)$.

Next make the substitutions $x = a-b$ and $y = b$, so the above expression becomes

$\sum_{x \in A} q(x) \sum_{y \in A} B(x,y)$.

Now if $x \neq 0$ then $y \mapsto B(x,y)$ is a nontrivial character of $A$, whence $\sum_{y \in A} B(x,y) = 0$. So the sum simplifies to

$q(0) \sum_{y \in A} B(0,y) = \# A$.

$\Box$

Now let us prove the theorem. Recall our situation: $F$ is a non-Archimedean local field of characteristic not equal to $2$, $V$ is a finite-dimensional vector space over $F$, and $q : V \to F$ is a nondegenerate quadratic form. Fix a nontrivial continuous character $\psi : F \to \mathbb{T}$, and let $\mu$ be the Haar measure on $V$ which is self-dual with respect to $\psi \circ q$. Consider the function $f_q : V \to \mathbb{C}$, which is locally constant and therefore may be regarded as a distribution on $V$. Then the statement of the theorem is

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

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

Furthermore, we have $|\gamma(q)| = 1$, and for any sufficiently large lattice $\Lambda \subset V$,

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

Proof. A straightforward calculation shows that $T_af_q = f_q(a)M_{a}f_q$ for $a \in V$, and then the first part of the lemma from last time tells us that $M_{-a}\mathcal{F}(f_q) = f_q(a)T_a\mathcal{F}(f_q)$.

Now we apply the second part of that lemma to see that

$T_a(f_q \cdot \mathcal{F}(f_q)) = T_{-a}f_q \cdot T_a\mathcal{F}(f_q) = f_q(a)f_q(a)^{-1}M_{a}f_q \cdot M_{-a}\mathcal{F}(f_q) = f_q \cdot \mathcal{F}(f_q)$,

so $f_q \cdot \mathcal{F}(f_q)$ is a translation-invariant distribution. If we consider the action of $V$ on $C^{\infty}_c(V)$ by translations, then this means $f_q \cdot \mathcal{F}(f_q)$ factors through the coinvariants for this representation, which are one-dimensional. Since integration against $d\mu$, i.e. the constant distribution $1$, also has this property, we see that $f_q \cdot \mathcal{F}(f_q)$ is a constant distribution $\gamma(q)$ and $\mathcal{F}(f_q) = \gamma(q) \cdot f_{-q}$ as desired.

Before we prove $|\gamma(q)| = 1$, let us pass to the more concrete formula for $\gamma(q)$. It is easy to see that for any $g \in C^{\infty}_c(V)$ we have

$(f_q * g)(x) = f_q(x)\mathcal{F}(f_q \cdot g)(-x) = f_q(x)\mathcal{F}^{-1}(f_q \cdot g)(x)$,

and since the latter is compactly supported we see that $f_q * g$ is also. In fact, we can even conclude that

$\int_V f_q * g \ d\mu = \int_V f_q \mathcal{F}^{-1}(f_q \cdot g) \ d\mu = (f_q \cdot \mathcal{F}(f_q))(g) = \gamma(q)\int_V g \ d\mu$.

We can simplify this even further: first find a sufficiently large lattice $\Lambda$ so that $f_q|_{\Lambda^{\perp}}$ is trivial. This can be done by starting with any lattice $X \subset V$ and taking $\Lambda = (2 \cdot (X \cap X^{\perp}))^{\perp}$. Then let $g = e_{\Lambda^{\perp}}$ be the indicator function, so

$(f_q * e_{\Lambda^{\perp}})(x) = \int_{\Lambda^{\perp}} f_q(x+y) \ d\mu(y) = f_q(x) \int_{\Lambda^{\perp}} \psi(B(x,y)) \ d\mu(y)$.

Now if $x \notin \Lambda$, then $y \mapsto \psi(B(x,y))$ is a nontrivial smooth character of $\Lambda^{\perp}$, whence its integral is zero. On the other hand, if $x \in \Lambda$ then by definition $\psi(B(x,y)) = 1$ for all $y \in \Lambda^{\perp}$. Thus

$(f_q * e_{\Lambda^{\perp}})(x) = \mu(\Lambda^{\perp})f_q(x)e_{\Lambda}(x)$,

so by substituting into the last formula for $\gamma(q)$ we get

$\gamma(q)\mu(\Lambda^{\perp}) = \gamma(q) \int_V e_{\Lambda^{\perp}} \ d\mu = \int_V f_q * e_{\Lambda^{\perp}} \ d\mu = \mu(\Lambda^{\perp})\int_{\Lambda} f_q \ d\mu$,

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

Now by our choice of $\Lambda$, the last integral factors through $\Lambda/\Lambda^{\perp}$ and becomes

$\gamma(q) = \mu(\Lambda^{\perp}) \sum_{x \in \Lambda/\Lambda^{\perp}} f_q(x)$.

It is not hard to see that $f_q$ induces a nondegenerate quadratic form on $\Lambda/\Lambda^{\perp}$ in the sense of the discussion above, so it follows that

$|\gamma(q)| = \mu(\Lambda^{\perp}) \sqrt{\# (\Lambda/\Lambda^{\perp})} = \sqrt{\mu(\Lambda^{\perp})^2\mu(\Lambda)/\mu(\Lambda^{\perp})} = 1$,

where we used the fact that $\mu(\Lambda^{\perp})\mu(\Lambda) = 1$ because $\mu$ is self-dual.

$\Box$