Presentation: The Functional Equation for Riemann Zeta Over Function Fields

This is a presentation I gave for my Algebraic Number Theory class during the Spring 2019 semester. In it I prove the functional equation for the Riemann zeta function for global function fields of transcendence degree one.

The Functional Equation for Riemann Zeta Over Function Fields

Using the Riemann-Roch theorem we’re going to prove the functional equation for the \zeta-function associated function fields of transcendence degree one over \mathbb{F}_{q}. As a consequence we will show that the \zeta-function satisfies a rational function in u = q^{-s}.

Theorem (The Functional Equation): Let K be a function field of transcendence degree one over \mathbb{F}_{q}, and let \zeta_{K} be the corresponding \zeta-function. Define \xi(s) = q^{(g-1)s}\zeta_{K}(s) where g is the genus of K. Then

\xi(s) = \xi(1-s).

proof: Let \deg(A) = n, so that if NA is the norm of A, NA^{-s} = q^{-ns}. Summing over n \ge 0 and making the substitution u = q^{-s} we may write

\zeta_{K}(s) = \sum_{n \ge 0}b_{n}u^{n} := Z_{K}(u) \qquad \zeta_{K}(1-s) = \sum_{n \ge 0}b_{n}\left(\frac{1}{qu}\right)^{n} := Z_{K}\left(\frac{1}{qu}\right),

where we define b_{n} to be the number of effective divisors A \in \mathcal{D}_{K} of degree n (a divisor A is effective if A \ge 0). The functional equation now takes the form

u^{1-g}Z_{K}(u) = \left(\frac{1}{qu}\right)^{1-g}Z_{K}\left(\frac{1}{qu}\right).

Now notice if we substitute \frac{1}{qu} for u on the left-hand side we get the right-hand side. Denoting the left-hand side as F'(u), the theorem is equivalent to showing F'(u) = F'(\frac{1}{qu}). To make the computations easier to work with we’re going to show the equivalent statement F(u) = F(\frac{1}{qu}) where F(u) = (q-1)F'(u). We will accomplish this by rewriting the sum on the left-hand side as a sum over divisor classes and using Riemann-Roch. Define Cl_{K}^{+} to be the set of all divisor classes \overline{A} such that \deg(\overline{A}) \ge 0 (we can do this because all the divisors in a class have the same degree) and let Cl_{K}^{i} be the set of all divisor classes \overline{A} with degree 0 \le \overline{A} \le i. Then we have the following:

\begin{aligned} F(u) &\overset{1}{=} (q-1)u^{1-g}\sum_{n \ge 0}b_{n}u^{n} \\ &\overset{2}{=} (q-1)u^{1-g}\left(\sum_{\overline{A} \in Cl_{K}^{+}}\frac{q^{l(\overline{A})-1}}{q-1}u^{\deg(\overline{A})}\right) \\ &\overset{3}{=} u^{1-g}\left(\sum_{\overline{A} \in Cl_{K}^{+}}q^{l(\overline{A})}u^{\deg(\overline{A})}-\sum_{\overline{A} \in Cl_{K}^{+}}u^{\deg(\overline{A})}\right) \\ &\overset{4}{=} u^{1-g}\left(\sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})}+\sum_{\{\overline{A} \mid \deg(\overline{A}) \ge 2g-1\}}q^{l(\overline{A})}u^{\deg(\overline{A})}-\sum_{\overline{A} \in Cl_{K}^{+}}u^{\deg(\overline{A})}\right) \\ &\overset{5}{=} \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})-g+1}+h_{k}\left(\sum_{n = 2g-1}^{\infty}q^{n-g+1}u^{n-g+1}-\sum_{n = 0}^{\infty}u^{n-g+1}\right) \\ &\overset{6}{=} \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})-g+1}+h_{k}\left(\frac{q^{g}u^{q}}{1-qu}-\frac{u^{1-g}}{1-u}\right) \end{aligned}

The first equality is by definition F(u). To get the second, sum over effective divisors instead of degree and note that the number of effective divisors in \overline{A} is exactly \frac{q^{l(\overline{A})-1}}{q-1} where l(\overline{A}) is the dimension of \overline{A}, really any representative, as a vector space over \mathbb{F}_{q}. Use the q-1 term to clear denominators to get the third. Breaking up the first sum by degree 2g-2 gives the fourth. The fifth follows since there are always h_{K} = |Cl_{K}^{0}| classes of degree n (h_{K} is independent of n), and a corollary of Riemann-Roch which says that if \deg(\overline{A}) \ge 2g-1 then l(\overline{A}) = n-g+1. Reindexing the middle sum to start from n = 0 and noticing the two latter sums are infinite geometric series which converge because u = q^{-s} with \Re(s) > 1 yields the sixth equality.

A direct substitution u \mapsto \frac{1}{qu} will show that the second term remains unchanged. In fact, the two terms in the difference switch under this substitution. So, all that remains to show is the summation remains unchanged. Now let \overline{C} \in Cl_{K} be the canonical class. By a corollary of Riemann-Roch, \deg(\overline{C}) = 2g-2 so that the map \varphi:Cl_{K}^{2g-2} \to Cl_{K}^{2g-2} given by \overline{A} \mapsto \overline{C}-\overline{A} is a well-defined involution. So in particular, is a bijection. Then

\begin{aligned} \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})-g+1} &= \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{C}-\overline{A})}u^{\deg(\overline{C} \overline{A})-g+1} \\ &= \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})-\deg(\overline{A})+g-1}u^{-\deg(\overline{A})+g-1} \end{aligned}

Since \varphi is a bijection, the first equality holds. We get the second as follows. Use Riemann-Roch thrice. The first two times use it for \overline{A}-\overline{C} and \overline{A}, and combine results. This gives the change in power for u. The change in power for q is just Riemann-Roch again applied to \overline{A}.

On the other hand, if we make the substitution u \mapsto \frac{1}{qu}:

\sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}\left(\frac{1}{qu}\right)^{\deg(\overline{A})-g+1} = \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})-\deg(\overline{A})+g-1}u^{-\deg\overline{A}+g-1},

and we get the same result as above. This proves F(u) = F(\frac{1}{qu}) as desired.

Having proven the functional equation we now want to show the existence of a polynomial L_{K}(u) of degree 2g such that

\zeta_{K}(s) = \frac{L_{K}(u)}{(1-u)(1-qu)}.

As before, take \zeta_{K}(s) = Z_{K}(u). It’s a consequence of Riemann-Roch that if n > 2g-2, then b_{n} = h_{K}\frac{q^{n-g+1}-1}{q-1}. If we substitute this into Z_{K}(u) and sum the geometric series for terms with n > 2g-2 we find

Z_{K}(u) = \sum_{n = 0}^{2g-2}b_{n}u^{n}+\frac{h_{K}}{q-1}\left(\frac{q^{g}}{1-qu}-\frac{1}{1-u}\right)u^{2g-1}.

If we combine the terms inside the parenthesis then their numerator divides q-1 since q^{g}(1-u)-(1-qu) = (q^{g}-1)-qu(q^{g-1}-1) (we can manually check the case g = 0). Collecting all terms under the common denominator gives the result. We now wish to find the degree of L_{K}(u). Start by noticing L_{K}(0) = 1 because b_{0} = 1. Since u^{1-g}Z_{K}(u) is invariant under u \mapsto \frac{1}{qu}, computing u^{1-g}Z_{K}(u) and using the substitution u \mapsto \frac{1}{qu} easily shows q^{-g}u^{-2g}L_{K}(u) = L_{K}(\frac{1}{qu}). If we take the limit as u \to \infty, then L_{K}(\frac{1}{qu}) \to 1 and hence q^{-g}u^{-2g}L_{K}(u) \to 1. This implies L_{K}(u) is of degree 2g and in fact its leading term is q^{g}u^{2g}. This rational expression for the \zeta-function has some other interesting properties such as giving an analytic continuation to all of \mathbb{C} with simple poles at s = 0 and s = 1. Also, L'_{K}(0) = a_{1}-1-q and L_{K}(1) = h_{K} where a_{1} is the number of primes of K of degree one. These facts are easy to check and we omit their proof.

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