--- title: Notes on Weil Conjectures date: 2021-04-28 18:11:25 --- #projects/notes/reading #projects/my-talks See [Weil Conjectures Talks](Weil%20Conjectures%20Talks.md) > Reference: > Andre Weil, [Numbers of Solutions of Equations in Finite Fields](https://projecteuclid.org/download/pdf_1/euclid.bams/1183513798) # External Background Here we fix a prime $p$ and a system of polynomials $S = \theset{f_i}$ of degree $n$, take the variety $V(S)$ and let - $a_1$ be its number of points of $\FF_p$ - $a_2$ be its number of points of $\FF_{p^2}$ - $\cdots a_n$ be its number of points of $\FF_{p^n}$ 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**: 1. $\zeta(x) = \frac{P(x)}{Q(x)}$ is a rational function. 2. 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 $P_i \in \ZZ[x]$. - For every root $r$ of every $P_i$, $\frac 1 r$ is algebraic - (Riemann Hypothesis) Every root has modulus equal to $p^{-i/2}$ (???) 3. (Functional Equation) The function $z \mapsto \frac{1}{p^n z}$ interchanges roots of $P_i$ with roots of $P_{2n-i}$. 4. (Under some conditions) $\deg P_i = \beta_i(V)$, the $i$th Betti number of $i$. Relation to fixed points: In $\FF_{p^m}$, every point is a fixed point of the Frobenius $\Phi_{p^m}$. So for any field $F \supset \FF_{p^m}$, the points in $\FF_{p^m}$ 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 $S \intersect \FF_{p^m}^d$. 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](Riemann%20zeta%20function.md) : 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](Unsorted/etale%20cohomology). 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 $H^{2d}(\Gr(k, \CC^{n+k})) \cong \ZZ^\ell$ 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. $H^*(\Gr(k, \CC^\infty)) \cong \CC[a_1, \cdots, a_k]$ (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 $\Gr(m, r, \PP_{\FF_q})$ \[ F(x)=\frac{\left(x^{m+1}-1\right)\left(x^{m+1}-x\right) \cdots\left(x^{m+1}-x^{r}\right)}{\left(x^{r+1}-1\right)\left(x^{r+1}-x\right) \cdots\left(x^{r+1}-x^{r}\right)} \] and so the Poincare polynomial for $\Gr(m, r, \PP_\CC)$ 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 $\FF_p$. ($p = 3n+1$, Gauss, when studying "Gaussian sums"/ "cyclotomic periods") - $ax^4 - by^4$ in $\FF_p$ ($p = 4n+1$, Gauss) Can consider corresponding variety $V$ over $\CC$, want to relate numbers of solutions to topological properties of $V$. Fix a finite field $k$ with $q$ elements, $a_i \in k\setminus 0$, $n_i\in \ZZ_{>0}$, and first discuss $b=0$. Definitions: \[ f: k[x_0, \cdots, x_r] \to k \\ f(x_0, \cdots, x_r) &= a_0 x_0 ^{n_0} + \cdots + a_r x_r^{n_r} .\] > Only monomials appearing? Example: Take $k=\ZZ_2$ 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 $N \definedas \abs{x\in k \suchthat f(x) =0}$ the number of solutions over $k$. > Note: shouldn't this be the number of solutions in $k^{r+1}$, since a "solution" is an $(r+1)\dash$tuple? Example: For $g$ above, $(x, y) = (0,0),~(1,1)$ are the only two solutions, so here $N = 2$ Define $d_i \definedas \gcd(n_i, q-1)$ Example: For $\ZZ_2$, $q=1$ so $d_1 = \gcd(2, 1) = 1$ and $d_2 = \gcd(2, 1)$. For an arbitrary $u\in k$, define \[ N_i(u) = \abs{\theset{x\in k \suchthat x^{n_i} = u}} ,\] 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 \[ L: k^{r+1} &\to k \\ L(u) = L(u_0, \cdots, u_r) &= \sum_{i=0}^r a_i u_i .\] We'll consider the variety $V(L)$ defined by $L$. This yields a decomposition \[ N &= \sum_{u\in k^{r+1} \suchthat L(u) = 0} N_0(u_0) \cdots N_r(u_r) \\ &= \sum_{u \in V(L)} \prod_{i=0}^r N_i(u_i) ,\] 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](representation) is morphism of groups $\rho: G \to \GL(V)$. For $V$ finite-dimensional, a [character](character) of $\rho$ is the function $\chi_\rho: G\to F$ where $g\mapsto \tr(\rho(g))$. (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 $G = \ZZ/n\ZZ$ and define $\lambda: G \to \CC\units$ where $1 \mapsto \zeta_n$ a primitive $n$th root of unity, then $\theset{\lambda^i \suchthat 0\leq i \leq n-1}$ 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 \[ \Phi: \CC[x] \to \mathcal{C}(G) \\ f \mapsto f(\lambda) \] and since $\lambda^n = \id_\CC$, we have $\ker \Phi = (x^n - 1)$, so $\mathcal{C}(G) \cong \CC[x]/(x^n-1)$ is a polynomial algebra. Let $G = k\units \cong \ZZ/(q-1)\ZZ$, and let $\chi: G \to \CC$ be any character. > Note: are the representations actually taking values in $\CC$ here? Since $G$ is cyclic, let $\omega$ by any generator; then $\chi$ is fully determined by $\chi(\omega)$. For $\alpha \in \QQ$ any rational such that $(q-1)\alpha \in \ZZ$, define a character \[ \chi_\alpha: k\units \to \CC \\ \omega \mapsto e^{2\pi i \alpha} .\] We extend this to a character on $k$ by setting $\chi_\alpha(0) = 1 \iff \alpha\in \ZZ$ and $0$ otherwise. Let $S_i = \theset{\alpha \in \QQ\intersect [0, 1) \suchthat d_i \alpha \in \ZZ}$. We can then write \[ N_i(u) &= \sum_{\alpha \in S_i}\chi_\alpha(u) .\] > 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 $k\units$. Since both sides equal 1 if $u=0$, we can rewrite this as \[ N_i(u) = \sum_{j=1}^{d_i - 1} \zeta^j ,\] and thus \[ N = \sum_{u\in V(L)} \chi_{\alpha_0}(u_0) \cdots \chi_{\alpha_r}(u_r) \quad \text{ where } \alpha_i \in [0, 1), ~ d_i\alpha_i \in ZZ .\] > Definitely countable due to the previous equation, hence the $i$ index. But where did the $\zeta$s go?