## 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 is a map satisfying for and , and if we define by the formula , we also require that is a group homomorphism for all . The *dual group* of is , and we call *nondegenerate* if the homomorphism which sends is an isomorphism.

The *Gauss sum* associated with is the number

.

**Lemma** If is a nondegenerate quadratic form, then .

*Proof.* We begin with the calculation

.

Next make the substitutions and , so the above expression becomes

.

Now if then is a nontrivial character of , whence . So the sum simplifies to

.

Now let us prove the theorem. Recall our situation: is a non-Archimedean local field of characteristic not equal to , is a finite-dimensional vector space over , and is a nondegenerate quadratic form. Fix a nontrivial continuous character , and let be the Haar measure on which is self-dual with respect to . Consider the function , which is locally constant and therefore may be regarded as a distribution on . Then the statement of the theorem is

**Theorem (Weil)** There exists , called the *Weil index of* , such that

.

Furthermore, we have , and for any sufficiently large lattice ,

.

*Proof.* A straightforward calculation shows that for , and then the first part of the lemma from last time tells us that .

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

,

so is a translation-invariant distribution. If we consider the action of on by translations, then this means factors through the coinvariants for this representation, which are one-dimensional. Since integration against , i.e. the constant distribution , also has this property, we see that is a constant distribution and as desired.

Before we prove , let us pass to the more concrete formula for . It is easy to see that for any we have

,

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

.

We can simplify this even further: first find a sufficiently large lattice so that is trivial. This can be done by starting with any lattice and taking . Then let be the indicator function, so

.

Now if , then is a nontrivial smooth character of , whence its integral is zero. On the other hand, if then by definition for all . Thus

,

so by substituting into the last formula for we get

,

which then yields .

Now by our choice of , the last integral factors through and becomes

.

It is not hard to see that induces a nondegenerate quadratic form on in the sense of the discussion above, so it follows that

,

where we used the fact that because is self-dual.