## A connection with the Hilbert symbol

This time we will talk about an interesting arithmetical feature of the Weil index, namely its relationship with the Hilbert symbol from quadratic local class field theory. The proof of the theorem is based on the proof in Daniel Bump’s *Automorphic Forms and Representations*, and the ideas are due to Weil. For the moment, let be any local field of characteristic not equal to .

Given we can define an -algebra to be the -vector space with basis and multiplication given by

.

**Proposition** Either is the matrix algebra over or is a division algebra.

*Proof.* It is easy to see by direct calculation that is a central simple -algebra, meaning it has center and the only two-sided ideals are and . Thus by the structure theory for central simple algebras (see, for instance, *Noncommutative Algebra* by R. Keith Dennis and Benton Farb) there is a central division algebra over and such that . Clearly the only possibilities are and or and .

**Definition** Let and define the *Hilbert symbol* to be if or if is a division algebra.

It is not hard to see that if and only if is in the image of the norm , or equivalently has a solution with not all zero, whence the Hilbert symbol is trivial when is the field of complex numbers or finite. If is non-Archimedean, recall that we defined the Weil index . For we will abuse notation and write .

**Theorem** (Weil) Let be a non-Archimedean local field of characteristic not equal to . For all we have

.

Before we go on to the proof, which is somewhat difficult, let us give an easy but surprising application of this result.

**Corollary** The Weil index takes values in th roots of unity.

*Proof.* Taking in the equation from the theorem, we get

.

Then we can deduce that for any , we have

.

The result follows for general quadratic spaces by diagonalization.

The algebra has an antiautomorphism called conjugation, which sends to . Then the *reduced norm* is given by , and it is not hard to see that is a nondegenerate quadratic form on . In fact, with respect to the form .

*Proof of the theorem.* Note that if then coincides with the determinant, and from this one deduces that as quadratic spaces and hence .

So we need to treat the case where is a division algebra. First observe that the absolute value on extends to a non-Archimedean absolute value on as follows: for any , we can consider the degree two extension of local fields, which is automatically separable because has odd (or zero) characteristic. It is well-known that in this situation the absolute value on extends to via the formula , where is the field norm. Thus by letting vary we get the desired non-Archimedean absolute value on . Moreover, is a maximal -order in , and in particular is an -lattice.

Let be the canonical additive Haar measure on which is self-dual with respect to . Recall that we proved last time that

for any sufficiently large lattice (we fixed a continuous additive character a long time ago). Choose a uniformizer : if we make large enough, then the lattice will work.

We would like to rewrite the integral above as an integral in by pushing forward along , but first we must pass to the multiplicative Haar measure , so the integral becomes

.

Since the reduced norm is a proper (meaning preimages of compact sets are compact) homomorphism, it follows that is a multiplicative Haar measure on . Now if we write for the additive Haar measure on satisfying , then defines a multiplicative Haar measure on , so it follows that is a scalar multiple of , which we will write as . Thus

.

Decompose , so the integral can be rewritten as

.

Suppose is the conductor of , meaning the largest -submodule of on which is trivial. Then an easy computation shows that if , if , and if , so for sufficiently large we see that

is negative. But we know that , so implies .