Tags: #projects/notes/reading #projects/mytalks #arithmeticgeometry/Weilconjectures
Refs:
Zeta function Weil Conjectures GRH
About variety : defined by polynomial equations, Hypersurfaces are defined by one equation Let \(X/K: f(x_0, \cdots, x_{n+1}) = 0, \mathbf{x} \in {\mathbf{P}}_k^{n+2}\), so the variety will have dimension.
For any \(L/K\) we can consider the \(L\) points \(X(L) = \mathbf{x}\in L^{n+2} / \sim\) where identify \(\mathbf{x} = \lambda \mathbf{x}\) for any \(\lambda \in L^{\times}\)?
Example: \(n=1\), \(f(x,y,z) = y^2 z  x^3  ax^2 z  bz^3 \in {\mathbf{P}}^2_K\) for \(K = {\mathbf{Q}}\), where \(a, b \in {\mathbf{Z}}\), is the homogeneous Weierstrass equation for an elliptic curve.
Example: Taking \(K = {\mathbf{C}}\) for an arbitrary \(X\) yields a complex analytic space (manifold with singularities), which is in fact a manifold when \(\operatorname{Jac}(f_1, \cdots, f_n) \neq 0\) is nonvanishing on \(X({\mathbf{C}})\) (so the tangent space is full dimension everywhere).
Example: For \(K = { \mathbf{F} }_p\), \(X({ \mathbf{F} }_p)\) is a finite set, which we can count.
We can consider extensions \(F_{p^r} = { \mathbf{F} }_p[\left\{{\zeta_{p^r}}\right\}]\) given by adjoining all of the roots of \(x^{p^r}  1\). We now obtain a sequence of numbers: \begin{align*} {\sharp}X({ \mathbf{F} }_p), {\sharp}{ \mathbf{F} }_{p^2}, \cdots .\end{align*}
Can we determine them completely (without a bruteforce count)? Can we say anything about the asymptotics?
The simplest case: \(X = {\mathbf{A}}^n\), we have \({\mathbf{A}}^n({ \mathbf{F} }_{p^r}) = p^{rn}\), so in general \(X({ \mathbf{F} }_{p^r}) = p^{rn}  \text{ error terms }\).
 Theorem (Weil)
 If \(n=1\), so there is 1 defining equation, with \(X\) smooth and connected, then \({\sharp}X({ \mathbf{F} }_{p^r}) = p^r + O(p^{r/2})\).
We in fact know \({\sharp}X({ \mathbf{F} }_{p^r})\) is \(p^r + 1  \sum_{i=1}^g \alpha+i^r\) where \(g\) is the genus of ths curve and \(\alpha_i \in \mkern 1.5mu\overline{\mkern1.5mu{\mathbf{Q}}\mkern1.5mu}\mkern 1.5mu\) with \({\left\lvert {\iota(\alpha)} \right\rvert}= \sqrt{2}\) for any embedding \(\mkern 1.5mu\overline{\mkern1.5mu{\mathbf{Q}}\mkern1.5mu}\mkern 1.5mu\hookrightarrow{\mathbf{C}}\) where the \(\alpha_i\) are algebraic.
Note that this condition is much stronger than \({\left\lvert {\alpha_i} \right\rvert} = \sqrt{p}\), since e.g. \(1 \pm \sqrt 2\) have two different norms.
How to generalize to higher dimensions: do combinatorics! Take a certain generating function, the zeta function: [[ Z(X, t) =
\exp(
\sum^
\infty {t^r \over r}
{\sharp}X(
{ \mathbf{F} }{p^t}))
\in
{\mathbf{Q}}
{\left[\left[ t \right]\right] }]]
Note that we’re not using an OGF.
Example: For \(X= {\mathbf{P}}^n\), take the hyperplane \(f(\mathbf{x})  x_{n+2} = 0\). Then \(\sizeX({ \mathbf{F} }_{p^r}) = p^rn + p^{r(n1)} + p^{r(n2)} + \cdots + 1\) where we count the embedded \({\mathbf{A}}^n\) in the first term, an \({\mathbf{A}}^{n1}\) in the hyperplane at infinity, the embedded \({\mathbf{A}}^{n2}\) in its hyperplane at infinity, etc.
Note that the sum will turn this into a product, and we get \begin{align*} Z({\mathbf{P}}^n, t) = \prod_{j=0}^n \exp(\sum_t {t^r \over r}p^{rj}) = \prod_{j=0}^n {1 \over 1  p^jt} .\end{align*} which is in fact a rational function.
In general, for \(X\) a curve, we obtain \begin{align*} Z(X, t) = {(1  \alpha_1 t) \cdots(1  \alpha_j t \over (1t) (1pt) )} .\end{align*}
Weil Conjectures for \(X\) smooth and connected of dimension \(n\):
 \(Z(X, t) \in {\mathbf{Q}}(t)\)

(RH) Explicitly, \(Z(X, t) = {P_1(t) \cdots P_{2n1}(t) \over P_0(t) \cdots P_{2n}(t)}\)
 \(P_i(t) \in {\mathbf{Z}}[t]\), so the roots are algebraic integers. Normalize to \(P_i(0) = 1\) so the constant term is 1.
 \(P_i(t) = \prod_j (1  \alpha_{ij} t)\) (i.e. the roots), so the absolute value of every embedding is \(p^{i/2}\)
 (Functional equation) \(Z(X, 1/p^nt) = \pm t^\chi p^{c/2} Z(X, t)\) where \(c = \sum_{i=0}^{2n} (1)^\chi \deg P_i\). Note that \(\chi\) will be the EulerPoincare characteristic.
 If \(X\) comes by reduction mod \(p\) from some \(X'/{\mathbf{Q}}\), then \(\deg P_i = \beta_i = \dim_{\mathbf{Q}}H^i(X({\mathbf{C}}); {\mathbf{Q}})\).
Note that the genus is half of the first Betti number. See also Poincare polynomial
Why are they called the RH for varieties over finite fields? Identify \({\mathbf{Z}}\) as the ring of functions over a curve \(\operatorname{Spec}{\mathbf{Z}}\). We think of \(x\in \operatorname{Spec}{\mathbf{Z}}\) as a map \(\operatorname{ev}_x: {\mathbf{Z}}\to { \mathbf{F} }_p\) of rings, so points correspond to evaluating at the point.
Recall that we can write the Riemann zeta function as the Dirichlet series \(\zeta(s) = \sum_n {1 \over n^s}\) and expand as an Euler product \(\prod_{p\text{ prime}} {1 \over 1  p^{s}\).