Tags: #arithmetic-geometry/zeta-functions #arithmetic-geometry/L-functions #arithmetic-geometry #projects/my-talks

Recall that the Unsorted/Riemann Zeta function has a product expansion

\begin{align*} \zeta(s) = \sum n^{-s} = \prod_{p\in P} (1-p^{-s})^{-1} \end{align*}

where the product is taken over all primes \(P\).

Let \(X = V(\left\{{f_i}\right\})\coloneqq V(f)\) be the vanishing locus of a family of polynomials in \(F = { \mathbf{F} }_q[x_1, \cdots, x_n]\) for some prime power \(q\).

Let \(N_m = {\left\lvert {\left\{{ \mathbf{x} \in X({ \mathbf{F} }_q) {~\mathrel{\Big\vert}~}f_i(\mathbf{x}) = 0 }\right\}} \right\rvert} = {\left\lvert {V(f)} \right\rvert} \subset F\), the number of \({ \mathbf{F} }_q\) points, or equivalently just the size of this variety.

Then the Hasse-Weil Zeta function is defined as

\begin{align*} \zeta_X(t) = \exp{\displaystyle\sum_{m\geq 1} \frac {N_m} m t^m} \end{align*}

We immediately make a change of variables and send \(t\to q^{-s}\) to obtain

\begin{align*} \zeta_X(s) = \exp{\sum_{m\geq 1} \frac {N_m} m \qty{q^{-s}}^m } .\end{align*}

Why? Turns the zeta function into a Dirichlet series in \(s\). Yields \({\left\lvert {t} \right\rvert} = q^{-\Re(s)}\). Defined for \({\left\lvert {t} \right\rvert} < \frac 1 q\) in \({\mathbf{C}}\), extended to all of \({\mathbf{C}}\) as a rational function in \(x\). Converts “All zeros of \(\zeta_X\) have absolute value \(\frac{1}{\sqrt q}\)” to “All zeros of \(\zeta_X\) have real part \(\frac 1 2\)”.

Explanation of why exponential appears

Rough explanation: Take a bad first approximation and then correct. Let \(X\) be a fixed variety, for \(p\in X\) define \({\left\lVert {p} \right\rVert}_X = q^n\) where \(n\) is the \(n\) occurring in the minimal field of definition of \(p\), which is \({ \mathbf{F} }_{q^n}\).

Attempt to define \begin{align*} \zeta_{X, q}(s) = \prod_{p\in X} \frac{1}{1-{\left\lVert {p} \right\rVert}_X^{-s}} .\end{align*}

Note that \(-\log(x+1) = \sum_{n\geq 1} \frac{x^n}{n}\).

Now fix one \(p\in X\) and consider the factor it contributes, and take its logarithm:

\begin{align*} \log\qty{\frac{1}{1-{\left\lVert {p} \right\rVert}_X^{-s}}} &= - \log(1-{\left\lVert {p} \right\rVert}_X^{-s}) \\ &= - \log(-{\left\lVert {p} \right\rVert}_X^{-s} + 1) \\ &= \sum_{j\geq 1} \frac{{\left\lVert {p} \right\rVert}_X^{-js}}{k} \\ &= \sum_{j\geq 1} \frac{q^{-nks}}{k} \\ &= \sum_{j\geq 1} \frac{n}{nk}(q^{-s})^{nk} \\ (m=nk)\quad &= \sum_{j\geq 1} \frac{n}{m}(q^{-s})^{m} ,\end{align*}

so we see this single point contributes \(n\) to \(N_m\), when instead we’d like it to contribute exactly 1.

Fix: If \(p\) is minimally defined over \({ \mathbf{F} }_{q^n}\), consider its Galois orbit (taking automorphisms of \({ \mathbf{F} }_{q^n}\)). There are exactly \(n\) points in the orbit of \(p\) – namely, the conjugates of \(p\) – so if we redefine

\begin{align*} \zeta_{X, q}(s) = \prod_{\text{One } p \text{ in each Galois orbit}} \frac{1}{1-{\left\lVert {p} \right\rVert}_{X}^{-s} } .\end{align*}

Then the above argument shows that each orbit now contributes \(n\), and each orbit is of size \(n\), so the contribution now accurately reflects the number of points.

Examples

1: \(f(x) = x\) over \({ \mathbf{F} }_q\).

Define \(X_q = V(f)\), then this has exactly \(q\) points over \({ \mathbf{F} }_q^{n}\) point for every \(n\), so \(N_n = 1\) and

\begin{align*} \zeta_{X_q}(s) = \exp{\sum_{n\geq 1} \frac 1 n (p^{-sn})} &= e^{-\log(1 - p^{-s})}= (1-p^{-s})^{-1} .\end{align*}

Note that the usual \(\zeta_s = \prod_{p\text{ prime}} \zeta_{X_p}(s)\), i.e. Riemann Zeta is a product of Hasse-Weil zetas over all primes.

• \(V = {\mathbf{CP}}^1\) the projective line.

Here \begin{align*} \zeta_V(s) = \frac{1}{(1-q^{-s})(1-q^{1-s})} .\end{align*}

Corresponds to Riemann sphere, can check Betti numbers.

• \(V = {\mathbf{CP}}^n\):

\begin{align*} \zeta_V(s) = \prod_{j=0}^n \frac{1}{1-q^{j-s}} .\end{align*}

• An elliptic curve :

\(N_m\) is given by \(1 - \alpha^m - \beta^m + q^m\) where \(\alpha = \mkern 1.5mu\overline{\mkern-1.5mu\beta\mkern-1.5mu}\mkern 1.5mu\) are complex conjugates with absolute value \(\sqrt{q}\).

\begin{align*} \zeta(E, s)=\frac{\left(1-\alpha q^{-s}\right)\left(1-\beta q^{-s}\right)}{\left(1-q^{-s}\right)\left(1-q^{1-s}\right)} .\end{align*}