--- date: 2020-01-27 --- Tags: #arithmetic-geometry/zeta-functions #arithmetic-geometry/L-functions #arithmetic-geometry #projects/my-talks Recall that the [Unsorted/Riemann Zeta](Unsorted/Riemann%20Zeta.md) function has a product expansion \[ \zeta(s) = \sum n^{-s} = \prod_{p\in P} (1-p^{-s})\inv \] where the product is taken over all primes $P$. Let $X = V(\theset{f_i})\definedas V(f)$ be the vanishing locus of a family of polynomials in $F = \FF_q[x_1, \cdots, x_n]$ for some prime power $q$. Let $N_m = \abs{\theset{ \vector x \in X(\FF_q) \suchthat f_i(\vector x) = 0 }} = \abs{V(f)} \subset F$, the number of $\FF_q$ points, or equivalently just the size of this variety. Then the [Hasse-Weil Zeta](Hasse-Weil%20Zeta.md) function is defined as \[ \zeta_X(t) = \exp{\displaystyle\sum_{m\geq 1} \frac {N_m} m t^m} \] We immediately make a change of variables and send $t\to q^{-s}$ to obtain \[ \zeta_X(s) = \exp{\sum_{m\geq 1} \frac {N_m} m \qty{q^{-s}}^m } .\] > Why? Turns the zeta function into a Dirichlet series in $s$. > Yields $\abs{t} = q^{-\Re(s)}$. > Defined for $\abs{t} < \frac 1 q$ in $\CC$, extended to all of $\CC$ 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](https://mathoverflow.net/questions/325186/motivation-for-zeta-function-of-an-algebraic-variety) Rough explanation: Take a bad first approximation and then correct. Let $X$ be a fixed variety, for $p\in X$ define $\norm{p}_X = q^n$ where $n$ is the $n$ occurring in the minimal field of definition of $p$, which is $\FF_{q^n}$. Attempt to define \[ \zeta_{X, q}(s) = \prod_{p\in X} \frac{1}{1-\norm{p}_X^{-s}} .\] 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: \[ \log\qty{\frac{1}{1-\norm{p}_X^{-s}}} &= - \log(1-\norm{p}_X^{-s}) \\ &= - \log(-\norm{p}_X^{-s} + 1) \\ &= \sum_{j\geq 1} \frac{\norm{p}_X^{-js}}{k} \\ &= \sum_{j\geq 1} \frac{q^{-nks}}{k} \\ &= \sum_{j\geq 1} \frac{n}{nk}(q^{-s})^{nk} \\ (m=nk)\quad &= \sum_{j\geq 1} \frac{n}{m}(q^{-s})^{m} ,\] 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 $\FF_{q^n}$, consider its Galois orbit (taking automorphisms of $\FF_{q^n}$). There are exactly $n$ points in the orbit of $p$ -- namely, the conjugates of $p$ -- so if we redefine \[ \zeta_{X, q}(s) = \prod_{\text{One } p \text{ in each Galois orbit}} \frac{1}{1-\norm{p}_{X}^{-s} } .\] 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 $\FF_q$. Define $X_q = V(f)$, then this has exactly $q$ points over $\FF_q^{n}$ point for every $n$, so $N_n = 1$ and \[ \zeta_{X_q}(s) = \exp{\sum_{n\geq 1} \frac 1 n (p^{-sn})} &= e^{-\log(1 - p^{-s})}= (1-p^{-s})\inv .\] > 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. 2. $V = \CP^1$ the projective line. Here \[ \zeta_V(s) = \frac{1}{(1-q^{-s})(1-q^{1-s})} .\] Corresponds to [Riemann sphere](Riemann%20sphere), can check Betti numbers. 3. $V = \CP^n$: \[ \zeta_V(s) = \prod_{j=0}^n \frac{1}{1-q^{j-s}} .\] 4. An [elliptic curve](elliptic%20curve.md) : $N_m$ is given by $1 - \alpha^m - \beta^m + q^m$ where $\alpha = \bar\beta$ are complex conjugates with absolute value $\sqrt{q}$. \[ \zeta(E, s)=\frac{\left(1-\alpha q^{-s}\right)\left(1-\beta q^{-s}\right)}{\left(1-q^{-s}\right)\left(1-q^{1-s}\right)} .\]