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See Weil Conjectures Talks

Reference: Andre Weil, Numbers of Solutions of Equations in Finite Fields

External Background

Here we fix a prime \(p\) and a system of polynomials \(S = \left\{{f_i}\right\}\) of degree \(n\), take the variety \(V(S)\) and let

• \(a_1\) be its number of points of \({ \mathbf{F} }_p\)
• \(a_2\) be its number of points of \({ \mathbf{F} }_{p^2}\)
• \(\cdots a_n\) be its number of points of \({ \mathbf{F} }_{p^n}\)

Idea: assemble them into a generating function.

For unknown reasons, we put them in a zeta function instead: \(\zeta(x) = \exp\qty{\sum \frac{a_n x^n}{n} }\).

Conjectures:

• \(\zeta(x) = \frac{P(x)}{Q(x)}\) is a rational function.
• There is an explicit formula \(P(x) = \prod_{i\text{ odd}}^{2n-1} P_i(x)\) and \(Q(x) = \prod{i\text{ even}}P_i(x)\) with each \(P_i \in {\mathbf{Z}}[x]\).

• For every root \(r\) of every \(P_i\), \(\frac 1 r\) is algebraic
• (Riemann Hypothesis) Every root has modulus equal to \(p^{-i/2}\) (???)

• (Functional Equation) The function \(z \mapsto \frac{1}{p^n z}\) interchanges roots of \(P_i\) with roots of \(P_{2n-i}\).
• (Under some conditions) \(\deg P_i = \beta_i(V)\), the \(i\)th Betti number of \(i\).

Relation to fixed points: In \({ \mathbf{F} }_{p^m}\), every point is a fixed point of the Frobenius \(\Phi_{p^m}\). So for any field \(F \supset { \mathbf{F} }_{p^m}\), the points in \({ \mathbf{F} }_{p^m}\) are precisely the fixed points of \(\Phi_{p^m}: F\to F\) (because enlarging the field can not add more solutions).

Claim: If \(S\subset F^d\) is any subset defined by polynomial equations and \(x = (x_1, x_2, \cdots, x_n) \in S\) is a point, then \(\Phi_{p^m}(x) = (\Phi_{p^m}(x_1), \cdots) \in S\). Moreover, the fixed points of \(\Phi_{p^m}\) restricted to \(S\) are precisely \(S \cap{ \mathbf{F} }_{p^m}^d\).

Compare 2b above: Riemann says roots are along critical strip \(\Re(z) = \frac 1 2\); this says roots of \(P_i\) are on a circle of radius \(p^{i/2}\) about the origin. (Note: there is a (conformal?) map that takes the circle to the line, so we can send the roots of \(P_i\) to the line \(\Re(z) = \frac 1 2\)….but not for all \(i\) at once.)

Consequences: Riemann zeta function : error estimates in the prime number theorem agree with probabilistic models Weil: error estimates in Ramanujan’s \(\tau\) is as small as hoped.

Proofs: Grothendieck: 1,3, and 4 with Unsorted/etale cohomology. Notably not Weil 2. Deligne: Weil 2, The Riemann Hypothesis

Cohomology of Complex Grassmannian: Schubert cells exhibit structure as a CW complex with only even-dimensional cells, and \(H^{2d}({\operatorname{Gr}}(k, {\mathbf{C}}^{n+k})) \cong {\mathbf{Z}}^\ell\) where \(\ell\) is the number partitions of \([d]\), i.e. solutions to \(\sum_{j=1}^k x_j = d\) with \(x_j\) weakly increasing, i.e. \(x_1 \leq x_2 \leq \cdots x_k\). The ring structure is isomorphic to the ring of symmetric polynomials and is generated by Chern classes. I.e. \(H^*({\operatorname{Gr}}(k, {\mathbf{C}}^\infty)) \cong {\mathbf{C}}[a_1, \cdots, a_k]\) (with \(a_k\) Chern classes) which is invariant under the obvious action of the symmetric group \(S_k\).

Example from end of paper: The number of rational points on \({\operatorname{Gr}}(m, r, {\mathbf{P}}_{{ \mathbf{F} }_q})\) \begin{align*} F(x)=\frac{\left(x^{m+1}-1\right)\left(x^{m+1}-x\right) \cdots\left(x^{m+1}-x^{r}\right)}{\left(x^{r+1}-1\right)\left(x^{r+1}-x\right) \cdots\left(x^{r+1}-x^{r}\right)} \end{align*}

and so the Poincare polynomial for \({\operatorname{Gr}}(m, r, {\mathbf{P}}_{\mathbf{C}})\) is \(F(X^2)\).

Actual Paper

Considers equations of the form \(\sum_{i=1}^r a_i x_i^{n_i} = b\).

Examples:

• \(ax^3-by^3 = 1\) in \({ \mathbf{F} }_p\). (\(p = 3n+1\), Gauss, when studying “Gaussian sums”/ “cyclotomic periods”)
• \(ax^4 - by^4\) in \({ \mathbf{F} }_p\) (\(p = 4n+1\), Gauss)

Can consider corresponding variety \(V\) over \({\mathbf{C}}\), want to relate numbers of solutions to topological properties of \(V\).

Fix a finite field \(k\) with \(q\) elements, \(a_i \in k\setminus 0\), \(n_i\in {\mathbf{Z}}_{>0}\), and first discuss \(b=0\).

Definitions:

\begin{align*} f: k[x_0, \cdots, x_r] \to k \\ f(x_0, \cdots, x_r) &= a_0 x_0 ^{n_0} + \cdots + a_r x_r^{n_r} .\end{align*}

Only monomials appearing?

Example: Take \(k={\mathbf{Z}}_2\) and \(g: k[x, y] \to k\) where \(g(x, y) = x^2 + y^2\).

Non-example: \(h(x,y) = x^2 + y^2 + xy\).

Let \(N \coloneqq{\left\lvert {x\in k {~\mathrel{\Big\vert}~}f(x) =0} \right\rvert}\) the number of solutions over \(k\).

Note: shouldn’t this be the number of solutions in \(k^{r+1}\), since a “solution” is an \((r+1){\hbox{-}}\)tuple?

Example: For \(g\) above, \((x, y) = (0,0),~(1,1)\) are the only two solutions, so here \(N = 2\)

Define \(d_i \coloneqq\gcd(n_i, q-1)\)

Example: For \({\mathbf{Z}}_2\), \(q=1\) so \(d_1 = \gcd(2, 1) = 1\) and \(d_2 = \gcd(2, 1)\).

For an arbitrary \(u\in k\), define

\begin{align*} N_i(u) = {\left\lvert {\left\{{x\in k {~\mathrel{\Big\vert}~}x^{n_i} = u}\right\}} \right\rvert} ,\end{align*}

i.e. the number of solutions to \(x^{n_i} = u\) in \(k\), i.e. the number of \(d_i\)th roots of \(u\).

This is equal to:

• \(1\) if \(u = 0\),
• \(d_i\) if \(u\neq 0\) is a \(d_{i}\)th power in \(k\)
• \(0\) otherwise

Not entirely clear why case 2 holds. Try for an example in the case \(n_i = 2\) to compare to quadratic residues?

Define

\begin{align*} L: k^{r+1} &\to k \\ L(u) = L(u_0, \cdots, u_r) &= \sum_{i=0}^r a_i u_i .\end{align*}

We’ll consider the variety \(V(L)\) defined by \(L\).

This yields a decomposition

\begin{align*} N &= \sum_{u\in k^{r+1} {~\mathrel{\Big\vert}~}L(u) = 0} N_0(u_0) \cdots N_r(u_r) \\ &= \sum_{u \in V(L)} \prod_{i=0}^r N_i(u_i) ,\end{align*}

i.e. any solutions to \(f = 0\) over \(k^{r+1}\) can be found by first choosing a point \(u = (u_0, \cdots, u_r)\) in the variety cut out by \(L\), so \(L(u) = \sum a_i u_i = 0\), then picking an \(n_i\)the root \(s_i \in k\) of each \(u_i\) to obtain some \(s = (s_0, \cdots, s_r) \in k^{r+1}\). Then \(u_i = s_i^{n_i}\) implies that \(0 = \sum a_i u_i = \sum a_i s_i^{n_i}\), so \(s\) is a solution to \(f\).

Definition: Let \(G\) be a group and \(V\) a vector space over a field \(F\), then a representation is morphism of groups \(\rho: G \to \operatorname{GL}(V)\). For \(V\) finite-dimensional, a character of \(\rho\) is the function \(\chi_\rho: G\to F\) where \(g\mapsto {\mathrm{tr}}(\rho(g))\). (Recall that the trace can be defined by choosing a basis for \(V\) and taking the trace of the image of \(g\), and is basis-independent.) A character is irreducible iff ?

Lemma: Let \(G = {\mathbf{Z}}/n{\mathbf{Z}}\) and define \(\lambda: G \to {\mathbf{C}}^{\times}\) where \(1 \mapsto \zeta_n\) a primitive \(n\)th root of unity, then \(\left\{{\lambda^i {~\mathrel{\Big\vert}~}0\leq i \leq n-1}\right\}\) is a complete set of irreducible characters.

Aside, maybe not useful: The irreducible characters span the space of class functions \(\mathcal{C}(G)\), so we can define a surjective map

\begin{align*} \Phi: {\mathbf{C}}[x] \to \mathcal{C}(G) \\ f \mapsto f(\lambda) \end{align*}

and since \(\lambda^n = \operatorname{id}_{\mathbf{C}}\), we have \(\ker \Phi = (x^n - 1)\), so \(\mathcal{C}(G) \cong {\mathbf{C}}[x]/(x^n-1)\) is a polynomial algebra.

Let \(G = k^{\times}\cong {\mathbf{Z}}/(q-1){\mathbf{Z}}\), and let \(\chi: G \to {\mathbf{C}}\) be any character.

Note: are the representations actually taking values in \({\mathbf{C}}\) here?

Since \(G\) is cyclic, let \(\omega\) by any generator; then \(\chi\) is fully determined by \(\chi(\omega)\).

For \(\alpha \in {\mathbf{Q}}\) any rational such that \((q-1)\alpha \in {\mathbf{Z}}\), define a character

\begin{align*} \chi_\alpha: k^{\times}\to {\mathbf{C}}\\ \omega \mapsto e^{2\pi i \alpha} .\end{align*}

We extend this to a character on \(k\) by setting \(\chi_\alpha(0) = 1 \iff \alpha\in {\mathbf{Z}}\) and \(0\) otherwise.

Let \(S_i = \left\{{\alpha \in {\mathbf{Q}}\cap[0, 1) {~\mathrel{\Big\vert}~}d_i \alpha \in {\mathbf{Z}}}\right\}\). We can then write

\begin{align*} N_i(u) &= \sum_{\alpha \in S_i}\chi_\alpha(u) .\end{align*}

Note: no clue why!

A priori, this is a countable infinite sum. The claim is that it can in fact be reduced to a finite sum. (?)

We can then let \(\zeta = \chi_{\frac{1}{d_i}}(u)\), which is \(d_i\)th root of unity. Then \(\zeta = 1 \iff u\) is a \(d_i\)th power in \(k^{\times}\).

Since both sides equal 1 if \(u=0\), we can rewrite this as

\begin{align*} N_i(u) = \sum_{j=1}^{d_i - 1} \zeta^j ,\end{align*}

and thus

\begin{align*} N = \sum_{u\in V(L)} \chi_{\alpha_0}(u_0) \cdots \chi_{\alpha_r}(u_r) \quad \text{ where } \alpha_i \in [0, 1), ~ d_i\alpha_i \in ZZ .\end{align*}

Definitely countable due to the previous equation, hence the \(i\) index. But where did the \(\zeta\)s go?