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## 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 $F$ be any local field of characteristic not equal to $2$.

Given $a,b \in F^{\times}$ we can define an $F$-algebra $Q(a,b)$ to be the $F$-vector space with basis $1,i,j,k$ and multiplication given by

$i^2 = a, j^2 = b, k^2 = -ab, ij = -ji = k, jk = -kj = -bi, \text{ and } ik = -ki = aj$.

Proposition Either $Q(a,b) \cong \text{Mat}_2(F)$ is the $2 \times 2$ matrix algebra over $F$ or $Q(a,b)$ is a division algebra.

Proof. It is easy to see by direct calculation that $Q(a,b)$ is a central simple $F$-algebra, meaning it has center $F$ and the only two-sided ideals are $0$ and $Q(a,b)$. 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 $D$ over $F$ and $n \geq 0$ such that $Q(a,b) \cong \text{Mat}_n(D)$. Clearly the only possibilities are $D = Q(a,b)$ and $n = 1$ or $D = F$ and $n = 2$.

$\Box$

Definition Let $a,b \in F^{\times}$ and define the Hilbert symbol $(a,b)$ to be $1$ if $Q(a,b) \cong \text{Mat}_2(F)$ or $-1$ if $Q(a,b)$ is a division algebra.

It is not hard to see that $(a,b) = 1$ if and only if $a$ is in the image of the norm $F[\sqrt{b}] \to F$, or equivalently $z^2 = ax^2 + by^2$ has a solution with $x,y,z \in F$ not all zero, whence the Hilbert symbol is trivial when $F$ is the field of complex numbers or finite. If $F$ is non-Archimedean, recall that we defined the Weil index $\gamma : W(F) \to \mathbb{C}^{\times}$. For $a \in F^{\times}$ we will abuse notation and write $\gamma( \langle a \rangle ) = \gamma(a)$.

Theorem (Weil) Let $F$ be a non-Archimedean local field of characteristic not equal to $2$. For all $a,b \in F^{\times}$ we have

$(a,b) = \frac{\gamma(1) \gamma(ab)}{\gamma(a)\gamma(b)}$.

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 $\gamma : W(F) \to \mathbb{C}^{\times}$ takes values in $8$th roots of unity.

Proof. Taking $a = b = -1$ in the equation from the theorem, we get

$\gamma(1)^8 = \big( \frac{\gamma(1)^2}{\gamma(-1)^2} \big)^2 = (1,-1)^2 = 1$.

Then we can deduce that for any $a \in F^{\times}$, we have

$\gamma(a)^8 = (\gamma(a)^2)^4 = (\gamma(1)^2 (a,a))^4 = 1$.

The result follows for general quadratic spaces by diagonalization.

$\Box$

The algebra $Q(a,b)$ has an antiautomorphism called conjugation, which sends $\xi = x + yi + zj + wk$ to $\overline{\xi} = x - yi - zj - wk$. Then the reduced norm $\nu : Q(a,b) \to F$ is given by $\nu(\xi) = \xi \overline{\xi}$, and it is not hard to see that $\nu$ is a nondegenerate quadratic form on $Q(a,b)$. In fact, $Q(a,b) \cong \langle 1 \rangle \oplus \langle -a \rangle \oplus \langle -b \rangle \oplus \langle ab \rangle$ with respect to the form $nu$.

Proof of the theorem. Note that if $Q(a,b) \cong \text{Mat}_2(F)$ then $\nu$ coincides with the determinant, and from this one deduces that $Q(a,b) \cong \mathfrak{H} \oplus \mathfrak{H}$ as quadratic spaces and hence $\gamma(\nu) = 1$.

So we need to treat the case where $Q(a,b) = D$ is a division algebra. First observe that the absolute value on $F$ extends to a non-Archimedean absolute value on $D$ as follows: for any $\alpha \in D \setminus F$, we can consider the degree two extension $F[\alpha] / F$ of local fields, which is automatically separable because $F$ has odd (or zero) characteristic. It is well-known that in this situation the absolute value on $F$ extends to $F[\alpha]$ via the formula $| \alpha | = |N(\alpha)|^{1/2} = |\nu(\alpha)|^{1/2}$, where $N : F[\alpha] \to F$ is the field norm. Thus by letting $\alpha$ vary we get the desired non-Archimedean absolute value on $D$. Moreover, $\nu^{-1}(\mathcal{O}_F) = \{ x \in D \ | \ |x| \leq 1 \}$ is a maximal $\mathcal{O}_F$-order in $D$, and in particular is an $\mathcal{O}_F$-lattice.

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

$\gamma(\nu) = \int_{\Lambda} \psi(\frac12 \nu(x)) \ d\mu(x)$

for any sufficiently large lattice $\Lambda \subset D$ (we fixed a continuous additive character $F : \psi \to \mathbb{T}$ a long time ago). Choose a uniformizer $\pi \in \mathcal{O}_F$: if we make $n \geq 0$ large enough, then the lattice $\Lambda = \nu^{-1}(\pi^{-n}\mathcal{O}_F)$ will work.

We would like to rewrite the integral above as an integral in $F$ by pushing forward along $\nu$, but first we must pass to the multiplicative Haar measure $d\mu^{\times}(x) = | \nu(x) |^{-2} d\mu(x)$, so the integral becomes

$\int_{\nu^{-1}(\pi^{-n}\mathcal{O}_F)} \psi(\frac12 \nu(x) ) |\nu(x)|^2 \ d\mu^{\times}(x)$.

Since the reduced norm $\nu$ is a proper (meaning preimages of compact sets are compact) homomorphism, it follows that $\nu_*\mu^{\times}$ is a multiplicative Haar measure on $F^{\times}$. Now if we write $\alpha$ for the additive Haar measure on $F$ satisfying $\alpha(\mathcal{O}_F) = 1$, then $d\alpha^{\times}(z) = |z|^{-1} d\alpha(z)$ defines a multiplicative Haar measure on $F^{\times}$, so it follows that $\nu_*\mu^{\times}$ is a scalar multiple of $\alpha^{\times}$, which we will write as $\nu_*\mu^{\times} \sim \alpha^{\times}$. Thus

$\gamma(\nu) \sim\int_{\pi^{-n}\mathcal{O}_F} \psi(\frac12 z) | z |^2 \ d\alpha^{\times}(z) \sim \int_{\pi^{-n}\mathcal{O}_F} \psi(z) | z | \ d\alpha(z)$.

Decompose $\pi^{-n}\mathcal{O}_F = \coprod_{k = -n}^{\infty} \pi^k\mathcal{O}_F^{\times}$, so the integral can be rewritten as

$\sum_{k = -n}^{\infty} q^{-k} \int_{|z| = q^{-k}} \psi(z) \ d\alpha(z)$.

Suppose $\pi^r\mathcal{O}_F$ is the conductor of $\psi$, meaning the largest $\mathcal{O}_F$-submodule of $F$ on which $\psi$ is trivial. Then an easy computation shows that $\int_{|x| = q^{-k}} \psi(x) \ dx = q^{-k}(1-q^{-1})$ if $k \geq r$, $-q^{-r}$ if $k = r-1$, and $0$ if $k < r-1$, so for sufficiently large $n$ we see that

$\sum_{k = -n}^{\infty} q^{-k} \int_{|z| = q^{-k}} \psi(z) \ d\alpha(z) = -q^{1-2r} + (1-q^{-1})\sum_{k = r}^{\infty} q^{-2k} = \frac{-q^{1-2r}}{1+q^{-1}}$

is negative. But we know that $|\gamma(\nu)| = 1$, so $\gamma(\nu) < 0$ implies $\gamma(\nu) = -1$.

$\Box$

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