--- title: CRAG subtitle: The Weil Conjectures author: D. Zack Garza date: April 2020 theme: Berkeley colortheme: default fontfamily: noto-sans aspectratio: 169 header-includes: - \usepackage{cmbright} - \usepackage{caption} fontsize: 9pt --- Tags: #projects/my-talks #homotopy #arithmetic-geometry/Weil-conjectures #projects/my-talks/slides Refs: [Weil Conjectures](Weil%20Conjectures.md) [Weil Conjectures Talks](Weil%20Conjectures%20Talks.md) ## A Quick Note - A big thanks to Daniel Litt for organizing this reading seminar, recommending papers, helping with questions!! - Goals for this talk: - Understand the Weil conjectures, - Understand *why* the relevant objects should be interesting, - See elementary but concrete examples, - Count all of the things! # Generating Functions ## Varieties Fix $q$ a prime and $\FF_q$ the (unique) finite field with $q$ elements, along with its (unique) degree $n$ extensions $$ \FF_{q^n} = \theset{x\in \bar \FF_q \suchthat x^{q^n} - x = 0} \quad \forall~ n\in \ZZ^{\geq 1} $$ Definition ([Projective Algebraic Varieties](Projective%20Algebraic%20Varieties)) Let $J = \gens{f_1, \cdots, f_M} \normal k[x_0, \cdots, x_n]$ be an ideal, then a *projective algebraic* variety $X\subset \PP^n_\FF$ can be described as $$ X = V(J) = \theset{\vector{x} \in \PP^n_{\FF_q} \suchthat f_1(\vector x) = \cdots = f_M(\vector x) = \vector 0} $$ where $J$ is generated by *homogeneous* polynomials in $n+1$ variables, i.e. there is a fixed $d = \deg f_i \in \ZZ^{\geq 1}$ such that \[ f(\vector x) = \sum_{\substack{\vector I = (i_1, \cdots, i_n) \\ \sum_j i_j = d}} \alpha_{\vector I} \cdot x_0^{i_1}\cdots x_n^{i_n} \qtext{ and } f(\lambda \cdot \vector x) = \lambda^d f(\vector x), \lambda \in \FF\units .\] ## Point Counts - For a fixed variety $X$, we can consider its $\FF_q\dash$points $X(\FF_q)$. - Note that $\size X(\FF_q) < \infty$ is an integer - For any $L/\FF_q$, we can also consider $X(L)$ - For $[L: \FF_q]$ finite, $\size X(L) < \infty$ and are integers for every such $n$. - In particular, we can consider the finite counts $\size X(\FF_{q^n})$ for any $n\geq 2$. - So we can consider the sequence \small \[ [N_1, N_2, \cdots , N_n, \cdots ] \definedas [\size X(\FF_q), ~ \sizeX (\FF_{q^2}), \cdots, ~\size X(\FF_{q^n}), \cdots] .\] \normalsize - Idea: associate some generating function (a formal power series) encoding the sequence, e.g. $$F(z) = \sum_{n=1}^\infty N_n z^n = N_1z + N_2 z^2 + \cdots .$$ ## Why Generating Functions? For an *ordinary* [generating function](generating%20function), the coefficients are related to the real-analytic properties of $F$: \[ [z^n] \cdot F(z) = [z^n]\cdot T_{F, z=0}(z) = \frac{1}{n!} \qty{\dd{}{z}}^{n} F(z) \Bigg\rvert_{z = 0} = N_n \] and also to the complex analytic properties: \[ [z^n] \cdot F(z) \definedas \frac{1}{2\pi i} \oint_{\SS^1} {F(z) \over z^{n+1}}~dz = \frac{1}{2\pi i} \oint_{\SS^1} \frac{N_n}{z} ~dz = N_n .\] > Using the Residue theorem. The latter form is very amenable to computer calculation. ## Why Generating Functions? - An OGF is an infinite series, which we can interpret as an analytic function $\CC\to \CC$ - In nice situations, we can hope for a closed-form representation. - A useful example: by integrating a geometric series we can derive \[ {1 \over 1-z} &= \sum_{n=0}^\infty z^n \hspace{5em} \qty{ = 1 + z + z^2 + \cdots} \\ \implies \int {1 \over 1-z} &= \int \sum_{n = 0}^\infty z^n \\ &= \sum_{n=0}^\infty \int z^n \quad{for } \abs{z} < 1 \qtext{by uniform convergence} \\ &= \sum_{n=0}^\infty \frac{1}{n+1}z^{n+1} \\ \implies -\log\qty{1-z} &= \sum_{n=1}^\infty {z^n \over n} \hspace{5em} \qty{= z + {z^2 \over 2} + {z^3 \over 3} + \cdots} .\] ## Exponential - For completeness, also recall that $$ \exp(z) \definedas \sum_{n=0}^\infty {z^n \over n!} $$ - We can regard $\exp, \log$ as elements in the ring of formal power series $\QQ[z](z)$. - In particular, for any $p(z) \in z\cdot \QQ[[z]]$ we can consider $\exp(p(z)), \log(1 + p(z))$ - Since $\QQ \injects \CC$, we can consider these as a complex-analytic functions, ask where they converge, etc. # Zeta Functions ## Definition: Local Zeta Function Problem: count points of a (smooth?) projective variety $X/\FF$ in all (finite) degree $n$ extensions of $\FF$. Definition (Local Zeta Function) The [local zeta function](local%20zeta%20function) of an algebraic variety $X$ is the following formal power series: \[ Z_X(z) &= \exp\qty{ \sum_{n=1}^\infty N_n {z^n \over n} } \in \QQ[[z]] \qtext{where} N_n \definedas \size X(\FF_n) .\] Note that \[ z \qty{\dd{}{z}} \log Z_X(z) &= z \dd{}{z} \qty{N_1z + N_2 {z^2 \over 2} + N_3 {z^3 \over 3} + \cdots} \\ &= z \qty{N_1 + N_2 z + N_3 z^2 + \cdots } \hspace{5em} \text{(unif. conv.)}\\ &= N_1 z + N_2 z^2 + \cdots = \sum_{n=1}^\infty N_n z^n ,\] which is an *ordinary* generating function for the sequence $(N_n)$. # Examples ## Example: A Point Take $X = \pt = V(\theset{f(x) = x-c})/\FF$ for $c$ a fixed element of $\FF$. This yields a single point over $\FF$, then \[ \sizeX(\FF_q) &\definedas N_1 = 1 \\ \sizeX(\FF_{q^2}) &\definedas N_2 = 1 \\ &\vdots \\ \sizeX(\FF_{q^n}) &\definedas N_n = 1 \] and so \[ Z_\pt(z) &= \exp\qty{ 1\cdot z + 1 \cdot {z^2 \over 2} + 1 \cdot {z^3 \over 3} + \cdots } = \exp\qty{\sum_{n=1}^\infty {{z}^n \over n} } \\ &= \exp\qty{ -\log\qty{1-z} } \\ &= {1 \over 1 - z} .\] > Notice: $Z$ admits a closed form **and** is a rational function. ## Example: The Affine Line Take $X = \AA^1/\FF$ the affine line over $\FF$, then We can write $$ \AA^1(\FF_{q^n}) = \theset{\vector x = [x_1] \suchthat x_1 \in \FF_{q^n}} $$ as the set of one-component vectors with entries in $\FF_n$, so \[ X(\FF_q) &= q \\ X(\FF_{q^2}) &= q^2 \\ &\vdots \\ X(\FF_{q^n}) &= q^n .\] Then \[ Z_X(z) &= \exp\qty{\sum_{n=1}^\infty q^n {z^n \over n} } \\ &= \exp\qty{\sum_{n=1}^\infty {\qty{qz}^n \over n} } \\ &= \exp(-\log(1 - qz)) \\ &= \frac 1 {1 - qz} .\] ## Example: Affine m-space Take $X = \AA^m/\FF$ the affine line over $\FF$, then We can write \[ \AA^m(\FF_{q^n}) = \theset{\vector x = [x_1, \cdots, x_m] \suchthat x_i \in \FF_{q^n}} \] as the set of one-component vectors with entries in $\FF_n$, so \noindent \begin{minipage}{.45\textwidth} \small \[ X(\FF_q) &= q^m \\ X(\FF_{q^2}) &= (q^2)^m \\ &\vdots \\ X(\FF_{q^n}) &= q^{nm} .\] \normalsize \end{minipage} \begin{minipage}{.45\textwidth} \centering \includegraphics[width = .5\linewidth]{figures/image_2020-04-23-00-21-46.png} \captionof{figure}{$\AA^2 / \FF_3 ~(q=3, m = 2, n = 1)$} \end{minipage} Then \small \[ Z_X(z) &= \exp\qty{\sum_{n=1}^\infty q^{nm} {z^n \over n} } = \exp\qty{\sum_{n=1}^\infty {\qty{q^m z}^n \over n} } \\ &= \exp(-\log(1 - q^mz)) \\ &= \frac 1 {1 - q^mz} .\] \normalsize ## Example: Projective Line Take $X = \PP^1/\FF$, we can still count by enumerating coordinates: \small \[ \PP^1(\FF_{q^n}) &= \theset{[x_1: x_2] \suchthat x_1, x_2 \neq 0 \in \FF_{q^n}}/\sim ~= \theset{[x_1: 1] \suchthat x_1 \in \FF_{q^n}} \disjoint \theset{[1: 0]} .\] \normalsize Thus \noindent \begin{minipage}{.45\textwidth} \small \[ X(\FF_q) &= q + 1\\ X(\FF_{q^2}) &= q^2 + 1 \\ &\vdots \\ X(\FF_{q^n}) &= q^n + 1 .\] \normalsize \end{minipage} \begin{minipage}{.45\textwidth} \centering \includegraphics[width = .5\linewidth]{figures/image_2020-04-23-00-07-42.png} \captionof{figure}{$\PP^1 / \FF_3 ~(q=3, n = 1)$} \end{minipage} Thus \small \[ Z_X(z) &= \exp\qty{\sum_{n=1}^\infty \qty{q^n + 1} {z^n \over n} } \\ &= \exp\qty{\sum_{n=1}^\infty {q^n } {z^n \over n} + \sum_{n=1}^\infty 1 \cdot {z^n \over n}}\\ &= {1 \over (1-qz)(1-z)} .\] \normalsize # The Weil Conjectures ## Weil 1 (Weil 1949) Let $X$ be a smooth projective variety of dimension $N$ over $\FF_{q}$ for $q$ a prime, let $Z_X(z)$ be its zeta function, and define $\zeta_X(s) = Z_X(q^{-s})$. 1. (Rationality) $Z_X(z)$ is a rational function: \[ Z_X(z) &= {p_1(z) \cdot p_3(z) \cdots p_{2N-1}(z) \over p_0(z) \cdot p_2(z) \cdots p_{2N}(z)} \in \QQ(z),\quad\text{i.e. }\quad p_i(z) \in \ZZ[z] \\ \\ P_0(z) &= 1-z \\ P_{2N}(z) &= 1 - q^N z \\ P_j(z) &= \prod_{j=1}^{\beta_i} \qty{1 - a_{j, k} z} \qtext{for some reciprocal roots} a_{j, k} \in \CC \] where we've factored each $P_i$ using its reciprocal roots $a_{ij}$. \small In particular, this implies the existence of a meromorphic continuation of the associated function $\zeta_X(s)$, which a priori only converges for $\Re(s)\gg 0$. This also implies that for $n$ large enough, $N_n$ satisfies a linear recurrence relation. \normalsize ## Weil 2 2. (Functional Equation and Poincare Duality) Let $\chi(X)$ be the Euler characteristic of $X$, i.e. the self-intersection number of the diagonal embedding $\Delta \injects X\cross X$; then $Z_X(z)$ satisfies the following *functional equation*: \[ Z_X\qty{1 \over q^N z} = \pm \qty{q^{N \over 2} z}^{\chi(X)} ~~Z_X(z) .\] Equivalently, \[ \zeta_X(N-s) = \pm \qty{q^{\frac N 2 - s}}^{\chi(X)} ~\zeta_X(s) .\] > \small Note that when $N=1$, e.g. for a curve, this relates $\zeta_X(s)$ to $\zeta_X(1-s)$.\normalsize Equivalently, there is an involutive map on the (reciprocal) roots \[ z &\iff {q^N \over z} \\ \alpha_{j, k} &\iff \alpha_{2N-j, k} \] which sends interchanges the coefficients in $p_j$ and $p_{2N-j}$. ## Weil 3 3. (Riemann Hypothesis) The reciprocal roots $\alpha_{j,k}$ are *algebraic* integers (roots of some monic $p\in \ZZ[x]$) which satisfy \[ \abs{\alpha_{j,k}}_\CC = q^{j \over 2}, \quad\quad 1 \leq j \leq 2N-1,~ \forall k .\] 4. (Betti Numbers) If $X$ is a "good reduction mod $q$" of a nonsingular projective variety $\tilde X$ in characteristic zero, then the $\beta_i = \deg p_i(z)$ are the Betti numbers of the topological space $\tilde X(\CC)$. Moral: - The Diophantine properties of a variety's zeta function are governed by its (algebraic) topology. - Conversely, the analytic properties of encode a lot of geometric/topological/algebraic information. ## Why is (3) called the "Riemann Hypothesis"? Recall the Riemann zeta function is given by \[ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p\text{ prime}} {1 \over 1 - p^{-s}} .\] After modifying $\zeta$ to make it symmetric about $\Re(s) = \frac 1 2$ and eliminate the trivial zeros to obtain $\hat \zeta(s)$, there are three relevant properties 1. "Rationality": $\hat \zeta(s)$ has a meromorphic continuation to $\CC$ with simple poles at $s=0, 1$. 2. "Functional equation": $\hat \zeta(1-s) = \hat \zeta(s)$ 3. "Riemann Hypothesis": The only zeros of $\hat \zeta$ have $\Re(s) = \frac 1 2$. ## Why is (3) called the "Riemann Hypothesis"? Suppose it holds for some $X$. Use the facts: a. $\abs{\exp\qty{z}} = \exp\qty{\Re(z)}$ and b. $a^z \definedas \exp\qty{z ~\Log(a)}$, and to replace the polynomials $P_j$ with \[ L_j(s) \definedas P_j(q^{-s}) = \prod_{k=1}^{\beta_j} \qty{1 - \alpha_{j, k} q^{-s}} .\] ## Analogy to Riemann Hypothesis Now consider the roots of $L_j(s)$: we have \small \[ L_j(s_0) = 0 \iff (1 - \alpha_{j, k} q^{-s}) &= 0 \qtext{for some} k\\ \iff q^{-s_0} &= {1 \over \alpha_{j, k}} \\ \iff \abs{q^{-s_0}} &= \abs{1 \over \alpha_{j, k}} \quad\quad \stackrel{\text{\tiny by assumption}}{=} q^{ -{j \over 2}} \\ \\ \iff q^{-\frac j 2} \stackrel{(a)}= \exp\qty{- \frac j 2 \cdot \Log(q)} &= \abs{ \exp\qty{-s_0 \cdot \Log(q)} } \\ &\stackrel{(b)}= \abs{ \exp\qty{-\qty{\Re(s_0) + i\cdot \Im(s_0)} \cdot \Log(q)} } \\ &\stackrel{(a)}= \exp\qty{-\qty{\Re(s_0)} \cdot \Log(q)} \\ \\ \iff - \frac j 2 \cdot \Log(q) &= -\Re(s_0) \cdot \Log(q) \qtext{by injectivity} \\ \iff \Re(s_0) = \frac j 2 .\] \normalsize ## Analogy with Riemann Hypothesis Roughly speaking, we can apply $\Log$ (a conformal map) to send the $\alpha_{j, k}$ to zeros of the $L_j$, this says that the zeros all must lie on the "critical lines" $\frac{j}{2}$. \begin{figure}[h] \centering \includegraphics[width = 1.0\textwidth]{figures/image_2020-04-23-02-26-27.png} \end{figure} In particular, the zeros of $L_1$ have real part $\frac 1 2$ (analogy: classical [../RH](../RH.md)). ## Precise Relation - Difficult to find in the literature! - Idea: make a similar definition for schemes, then take $X = \spec \ZZ$. - Define the "reductions mod $p$" $X_p$ for closed points $p$. - Define the *local* zeta functions $\zeta_{X_p}(s) = Z_{X_p}(p^{-s})$. - (Potentially incorrect) Evaluate to find $Z_{X_p}(z) = {1 \over 1 - z}$. - Take a product over all closed points to define \[ L_X(s) &= \prod_{p\text{ prime}} \zeta_{X_p}(p^{-s}) \\ &= \prod_{p\text{ prime}} \qty{ 1 \over 1 - p^{-s}} \\ &= \zeta(s) ,\] which is the Euler product expansion of the classical Riemann Zeta function. > If anyone knows a reference for this, let me know! # Weil for Curves ## Weil for Curves The Weil conjectures take on a particularly nice form for curves. Let $X/\FF_q$ be a smooth projective curve of genus $g$, then 1. (Rationality) $$Z_X(z) = {p_1(z) \over p_0(z) p_2(z)} = {p_1(z) \over (1-z)(1-qz)}$$ 2. (Functional Equation) $$Z_X\qty{1 \over qz} = \qty{z\sqrt q}^{2-2g} Z_X(z)$$ 3. (Riemann Hypothesis) $$p_1(z) = \prod_{i=1}^{\beta_1} (1 - a_iz ) \qtext{where} \abs{a_i} = \sqrt{q}$$ 4. (Betti Numbers) $\mcp_{\Sigma_g}(x) = 1 + 2g\cdot x + x^2 \implies \deg p_1 = \beta_1 = 2g$. \tiny > $\mcp$ here is the Poincaré polynomial, the generating function for the Betti numbers. > $\Sigma_g$ is the surface (real 2-dimensional smooth manifold) of genus $g$. \normalsize ## The Projective Line Recall $Z_{\PP^1/\FF_q}(z) = {1 \over (1-z)(1-qz)}$. 1. Rationality: Clear! 2. Functional Equation: $g=0 \implies 2g-2 = 2$ \[ Z_{\PP^1}\qty{{1 \over qz}} = {1 \over (1-{1 \over qz})(1 - {q \over qz}) } = {qz^2 \over (1-z)(1-qz)} = {\qty{\sqrt q z}^2 \over (1-z)(1-qz) } .\] 3. Riemann Hypothesis: Nothing to check (no $p_1(z)$) 4. Betti numbers: Use the fact that $\mcp_{\CP^1} = 1 + 0\cdot x + x^2$, and indeed $\deg p_0 = \deg p_2 = 1,~ \deg p_1 = 0$. > Note that even Betti numbers show up as degrees in the denominator, odd in the numerator. > Allows us to immediately guess the zeta function for $\PP^n/\FF_q$ by knowing $H^* \CP^\infty$! ## Elliptic Curves \begin{figure}[h] \caption{Some Elliptic Curves} \centering \includegraphics[width = 0.9\textwidth]{figures/image_2020-04-23-02-40-31.png} \end{figure} Consider $E/\FF_q$. - (Nontrivial!) The number of points is given by $$ N_n \definedas E(\FF_{q^n}) = (q^n + 1) - ( \alpha^n + {\bar \alpha}^n ) \qtext{where} \abs{\alpha} = \abs{\bar \alpha} = \sqrt{q} $$ - **Proof**: Involves trace (or eigenvalues?) of Frobenius, (could use references) - $\dim_\CC E/\CC = N = 1$ and $g=1$. The Weil Conjectures say we should be able to write \[ Z_E(z) = {p_1(z) \over p_0(z) p_2(z)} = {p_1(z) \over (1-z) (1 - qz)} = { (1 - a_{1}z)(1 - a_{2}z) \over (1-z)(1- qz)} .\] ## Elliptic Curves: Weil 1 1. Rationality: using the point count, we can compute \small \[ Z_E(z) &= \exp \sum_{n=1}^\infty \size E(\FF_{q^n}) {z^n \over n} \\ &= \exp \sum_{n=1}^\infty \qty{q^n + 1 - \qty{\alpha^n + \bar\alpha^n}} {z^n \over n} \\ &= \exp \qty{ \sum_{n=1}^\infty q^n\cdot {z^n \over n} } \exp \qty{ \sum_{n=1}^\infty 1\cdot {z^n \over n} } \\ & \hspace{5em} \exp \qty{ \sum_{n=1}^\infty -\alpha^n \cdot {z^n \over n} } \exp \qty{ \sum_{n=1}^\infty -\bar\alpha^n\cdot{z^n \over n} } \\ \\ &= \exp\qty{-\log\qty{1-qz} } \cdot \exp\qty{-\log\qty{1-z} } \\ &\hspace{5em} \exp\qty{\log\qty{1- \alpha z} } \cdot \exp\qty{\log\qty{1 - \bar \alpha z} } \\ \\ &= {(1-\alpha z)(1-\bar \alpha z) \over (1-z)(1-qz)} \in \QQ(z) ,\] \normalsize which is a rational function of the expected form (Weil 1). \tiny > Note that the "expected" point counts show up in the denominator, along with the even Betti numbers, while the "correction" factor appears in the denominator and odd Betti numbers. \normalsize ## Elliptic Curves: Weil 2 and 3 2. Functional Equation: we use the equivalent formulation of "Poincaré duality": \[ {(1-\alpha z)(1-\bar \alpha z) \over (1-z)(1-qz)} = {p_1(z) \over p_0(z) p_2(z)} \implies \begin{cases} z &\iff {q \over z} \\ \alpha_{j, k} &\iff \alpha_{2-j, k} \end{cases} \] This amounts to checking that the coefficients of $p_0, p_2$ are interchanged, and that the two coefficients of $p_1$ are interchanged: \[ \mathrm{Coefs}(p_0) = \theset{1} &\mapsvia{z\to {q\over z}} \theset{1 \over q} = \mathrm{Coefs}(p_2) \\ \mathrm{Coefs}(p_1) = \theset{\alpha, \bar \alpha} &\mapsvia{z\to {q\over z}} \theset{{q \over \alpha}, {q \over \bar\alpha} } = \theset{\bar \alpha, \alpha} \qtext{using} \alpha\bar\alpha = q .\] 3. RH: Assumed as part of the point count ($\abs \alpha = q^{1\over 2}$) 4. Betti Numbers: $\mcp_{\Sigma_1}(x) = 1 + 2x + x^2$, and indeed $\deg p_0 = \deg p_2 = 1,~ \deg p_1 = 2$. ## History - 1801, Gauss: Point count and RH showed for specific elliptic curves - 1924, Artin: Conjectured for algebraic curves , - 1934, Hasse: proved for elliptic curves. - 1949, Weil: Proved for smooth projective curves over finite fields, proposed generalization to projective varieties - 1960, Dwork: Rationality via $p\dash$adic analysis - 1965, Grothendieck et al.: Rationality, functional equation, Betti numbers using étale cohomology - Trace of Frobenius on $\ell\dash$adic cohomology - Expected proof via *the standard conjectures*. Wide open! - 1974, Deligne: Riemann Hypothesis using étale cohomology, circumvented the standard conjectures - Recent: [Hasse-Weil conjecture](Hasse-Weil%20conjecture) for arbitrary algebraic varieties over number fields - Similar requirements on $L\dash$functions: functional equation, meromorphic continuation - 2001: Full [modularity](../modular%20form.md) theorem proved, extending Wiles, implies Hasse-Weil for elliptic curves - Inroad to [The Langlands Program](The%20Langlands%20Program.md): show every $L$ function coming from an algebraic variety also comes from an automorphic representation. # Weil for Projective m-space ## Setup Take $X = \PP^m/\FF$ We can write $$ \PP^m(\FF_{q^n}) = \AA^{m+1}(\FF_{q^n}) \setminus \theset{\vector 0}/\sim ~= \theset{\vector x = [x_0, \cdots, x_m] \suchthat x_i \in \FF_{q^n}}/\sim $$ But how many points are actually in this space? \begin{figure}[h] \caption{Points and Lines in $\PP^2/{\FF_3}$} \centering \includegraphics[width = 0.4\textwidth]{figures/image_2020-04-19-22-32-07.png} \end{figure} > A nontrivial combinatorial problem! ## q-Analogs and Grassmannians To illustrate, this can be done combinatorially: identify $\PP^m_\FF = \Gr_{\FF}(1, m+1)$ as the space of lines in $\AA^{m+1}_\FF$. Theorem : The number of $k\dash$dimensional subspaces of $\AA^N_{\FF_q}$ is the $q\dash$analog of the binomial coefficient: \[ \genfrac{[}{]}{0pt}{}{N}{k}_q \definedas \frac{(q^N - 1)(q^{N-1}-1) \cdots (q^{N - (k-1)} - 1)}{(q^k-1)(q^{k-1} - 1) \cdots (q-1)} .\] > Remark: Note $\lim_{q\to 1} \genfrac{[}{]}{0pt}{}{N}{k}_q = {N \choose k}$, the usual binomial coefficient. **Proof:** To choose a $k\dash$dimensional subspace, - Choose a nonzero vector $\vector v_1 \in \AA^n_\FF$ in $q^N - 1$ ways. - For next step, note that $\size\spanof\theset{\vector v_1} = \size\theset{\lambda \vector v_1 \suchthat \lambda \in \FF_q} = \size \FF_q = q$. - Choose a nonzero vector $\vector v_2$ *not* in the span of $\vector v_1$ in $q^N - q$ ways. - Now note $\size \spanof\theset{\vector v_1, \vector v_2} = \size \theset{\lambda_1 \vector v_1 + \lambda_2 \vector v_2 \suchthat \lambda_i \in \FF} = q\cdot q = q^2$. ## Proof continued - Choose a nonzero vector $\vector v_3$ *not* in the span of $\vector v_1, \vector v_2$ in $q^N -q^2$ ways. - $\cdots$ until $\vector v_k$ is chosen in $$(q^N-1)(q^N-q) \cdots (q^N - q^{k-1}) \qtext{ways}.$$ - This yields a $k\dash$tuple of linearly independent vectors spanning a $k\dash$dimensional subspace $V_k$ - This overcounts because many linearly independent sets span $V_k$, we need to divide out by the number of ways to choose a basis inside of $V_k$. - By the same argument, this is given by $$(q^k-1)(q^k-q) \cdots (q^k - q^{k-1})$$ Thus \small \[ \size \text{subspaces} &= \frac{ (q^N-1)(q^N-q)(q^N - q^2) \cdots (q^N - q^{k-1}) }{ (q^k-1)(q^k-q)(q^k-q^2) \cdots (q^k - q^{k-1})}\\ \\ &= {q^N - 1 \over q^k - 1} \cdot \qty{q \over q} {q^{N-1} - 1 \over q^{k-1} - 1} \cdot \qty{q^2 \over q^2}{q^{N-2} - 1 \over q^{k-2} - 1} \cdots \qty{q^{k-1} \over q^{k-1}}{q^{N - (k-1)} - 1 \over q^{k - (k-1) - 1}} \\ &= \frac{(q^N - 1)(q^{N-1}-1) \cdots (q^{N - (k-1)} - 1)}{(q^k-1)(q^{k-1} - 1) \cdots (q-1)} .\] \normalsize $\qed$ ## Counting Points > Note that we've actually computed the number of points in any Grassmannian. Identify $\PP^m_\FF = \Gr_{\FF}(1, m+1)$ as the space of lines in $\AA^{m+1}_\FF$. We obtain a simplification (importantly, a *sum formula*) when setting $k=1$: \[ \genfrac{[}{]}{0pt}{}{m+1 }{1}_q = {q^{m+1}-1 \over q - 1} = q^{m} + q^{m-1} + \cdots + q + 1 = \sum_{j=0}^{m} q^j .\] Thus \[ X(\FF_q) &= \sum_{j=0}^{m} q^j \\ X(\FF_{q^2}) &= \sum_{j=0}^{m} \qty{q^2}^j \\ &\vdots \\ X(\FF_{q^n}) &= \sum_{j=0}^{m} \qty{q^n}^j .\] ## Computing the Zeta Function So \[ Z_X(z) &= \exp\qty{ \sum_{n=1}^\infty \sum_{j=0}^{m} \qty{q^n}^j {z^n \over n} } \\ &= \exp\qty{ \sum_{n=1}^\infty \sum_{j=0}^{m} {\qty{q^jz}^n \over n} } \\ &= \exp\qty{ \sum_{j=0}^{m} \sum_{n=1}^\infty {\qty{q^jz}^n \over n} } \\ &= \exp\qty{ \sum_{j=0}^{m-1} -\log(1 - q^j z) } \\ &= \prod_{j=0}^{m} \qty{1-q^jz}\inv \\ &= \qty{1 \over 1 - z} \qty{1 \over 1 - qz} \qty{1 \over 1 - q^2 z} \cdots \qty{1 \over 1- q^m z} ,\] > Miraculously, still a rational function! Consequence of sum formula, works in general. ## Checking the Weil Conjectures \[ Z_X(z) = \prod_{j=0}^m \qty{1 \over 1 - q^j z} .\] 1. Rationality: Clear! 2. Functional Equation: Less clear, but true: \[ Z_X\left(\frac{1}{q^{m} z}\right) &=\frac{1}{\left(1-1 / q^{m} t\right)\left(1-q / q^{m} t\right) \cdots\left(1-q^{m} / q^{m} z\right)} \\ &=\frac{q^{m} z \cdot q^{m-1} z \ldots q z \cdot z}{(1-z)(1-q z) \ldots\left(1-q^{m} z\right)} \\ &=q^{m(m+1) \over 2} z^{m+1} \cdot Z_X(z)\\ &= \qty{q^{m\over 2}z}^{\chi(X)} \cdot Z_X(z) \\ .\] ## Checking \[ Z_X(z) = \prod_{j=0}^m \qty{1 \over 1 - q^j z} .\] 3. Riemann Hypothesis: Reduces to the statement $\theset{ \alpha_i } = \theset{{q^m \over \alpha_j }}$. - Clear since $\alpha_j = q^j$ and every $\alpha_i$ is a lower power of $q$. 4. Betti Numbers: Use the fact that $\mcp_{\CP^m}(x) = 1 + x^2 + x^4 + \cdots + x^{2m}$ - Only even dimensions, and correspondingly no numerator. ## An Easier Proof: "Paving by Affines" Quick recap: \[ Z_{\pt} = {1 \over 1 - z} && Z_{\PP^1}(z) = {1 \over 1 - qz} && Z_{\AA^1}(z) = {1\over (1-z)(1-qz)} .\] Note that $\PP^1 = \AA^1 \disjoint \theset{\infty}$ and correspondingly $Z_{\PP^1}(z) = Z_{\AA^1}(z) \cdot Z_{\pt}(z)$. This works in general: Lemma (Excision) : If $Y/\FF_q \subset X/\FF_q$ is a closed subvariety, for $U = X\setminus Y$, $Z_X(z) = Z_Y(z) \cdot Z_{U}(z)$. **Proof**: Let $N_n = \size Y(\FF_{q^n})$ and $M_n = \size U(\FF_{q^n})$, then \[ \zeta_X(z) &= \exp\qty{\sum_{n=1}^\infty \qty{N_n + M_n} {z^n \over n} } \\ &= \exp\qty{\sum_{n=1}^\infty N_n \cdot {z^n \over n} + \sum_{n=1}^\infty M_n \cdot {z^n \over n}}\\ &= \exp\qty{\sum_{n=1}^\infty N_n \cdot {z^n \over n}} \cdot \exp\qty{\sum_{n=1}^\infty M_n \cdot {z^n \over n}} = \zeta_Y(z) \cdot \zeta_U(z) .\] ## An Easier Proof: "Paving" Note that geometry can help us here: we have a decomposition $\PP^n = \PP^{n-1} \disjoint \AA^n$, and thus inductively a stratification \[ \PP^m = \disjoint_{j=0}^m \AA^j = \AA^0 \disjoint \AA^1 \disjoint \cdots \disjoint \AA^m .\] Recalling that $$ Z_{X\disjoint Y}(z) = Z_X(z) \cdot Z_Y(z) $$ and $Z_{\AA^j}(z) = {1 \over 1 - q^j z}$, we have \[ Z_{\PP^m}(z) = \prod_{j=0}^m Z_{\AA^j}(z) = \prod_{j=0}^m {1 \over 1 - q^j z} .\] # Grassmannians ## Motivation Consider now $X = \Gr(k, m) / \FF$ -- by the previous computation, we know \[ X(\FF_{q^n}) = \genfrac{[}{]}{0pt}{}{m}{k}_{q^n} \definedas \frac{(q^{nm} - 1)(q^{{nm}-1}-1) \cdots (q^{{nm} - n(k-1)} - 1)}{(q^{nk}-1)(q^{n(k-1)} - 1) \cdots (q^n-1)} \] but the corresponding Zeta function is much more complicated than the previous examples: \[ Z_X(z) &= \exp\qty{ \sum_{n=1}^\infty \genfrac{[}{]}{0pt}{}{m}{k}_{q^n} {z^n \over n} } = \cdots ? .\] Since $\dim_\CC \Gr_\CC (k, m) = 2k(m-k)$, by Weil we should expect $$ Z_X(z) = \prod_{j=0}^{2k(m-k)} \frac{p_{2(j+1)}(z)}{p_{2j}(z)} $$ with $\deg p_j = \beta_j$. ## Grassmannian It turns out that (proof omitted) one can show \[ \genfrac{[}{]}{0pt}{}{m}{k}_{q} = \sum_{j=0}^{k(m-k)} \lambda_{m, k}(j) q^j \implies Z_X(z) = \prod_{j=0}^{k(m-k)} \qty{ \frac{1}{ 1 - q^j x} }^{\lambda_{m, k}(j)} \] where $\lambda_{m, k}$ is the number of integer partitions of of $[i]$ into at most $m-k$ parts, each of size at most $k$. - One proof idea: use combinatorial identities to write $q\dash$analog $\genfrac{[}{]}{0pt}{}{m}{k}_{q}$ as a *sum* - Second proof idea: "pave by affines" (need a reference!) This lets us conclude that the Poincare polynomial of the complex Grassmannian is given by $$ \mcp_{\Gr_\CC(m, k)}(x) = \sum_{n=1}^{k(m-k)} \lambda_{m, k}(n) x^{2n}, $$ > In particular, the $H^* \Gr_\CC(m, k)$ vanishes in odd degree. # Weil's Proof ## Diagonal Hypersurfaces Proof of rationality of $Z_X(T)$ for $X$ a diagonal hypersurface. - Set $q$ to be a prime power and consider $X/\FF_q$ defined by $$X = V(a_0x_0^{n} + \cdots + a_r x_r^{n}) \subset \FF_q^{r+1}.$$ - We want to compute $N = \size X$. - Set $d_i = \gcd(n_i, q-1)$. - Define the character \[ \psi_q: \FF_q & \to \CC\units \\ a &\mapsto \exp\qty{2\pi i ~\Tr_{\FF_q/\FF_p}(a) \over p} .\] - By Artin's theorem for linear independence of characters, $\psi_q \not \equiv 1$ and every additive character of $\FF_q$ is of the form $a \mapsto \psi_q(ca)$ for some $c\in \FF_q$. - Shorthand notation: say $a\sim 0 \iff a \equiv 0 \mod 1$. ## A Diagonal Hypersurface - Fix an injective multiplicative map \[ \phi: \bar{\FF}_q\units \to \CC\units .\] - Define \[ \chi_{\alpha, n}: \FF_{q^n}\units &\to \CC\units \\ x & \mapsto \phi(x)^{\alpha\qty{q^n-1}} \\ \\ \quad \qtext{for} \alpha \in \QQ/\ZZ, n\in \ZZ, & \quad \alpha\qty{q^n-1} \equiv 0 \mod 1 .\] - Extend this to $\FF_{q^n}$ by \[ \begin{cases} 1 & \alpha \equiv0 \mod 1 \\ 0 & \text{else} \end{cases} .\] - Set $\chi_\alpha = \chi_{\alpha, 1}$. - Proposition: $$\alpha(q-1) \equiv 0 \mod 1 \implies \chi_{\alpha, n}(x) = \chi_\alpha(\mathrm{Nm}_{\FF_{q^n} / \FF_q }(x) )$$ - Proposition: $$d \definedas \gcd(n, q-1), u \in \FF_q \implies \size\theset {x\in \FF_q \suchthat x^n = u} = \sum_{d\alpha \sim 0} \chi_\alpha(u)$$ ## A Diagonal Hypersurface - This implies \[ N &= \sum_{\substack{\alpha = [\alpha_0, \cdots, \alpha_r] \\ d_i \alpha_i \sim 0}} \quad \sum_{\substack{\vector u = [u_0, \cdots ,u_r] \\ \Sigma~ a_i u_i = 0}} \quad \prod_{j=0}^r \chi_{\alpha_j}(u_j) \\ \\ &= q^r + \sum_{\substack{\alpha,~ \alpha_i \in (0, 1) \\ d_i \alpha_i \sim 0}} \qty{ \prod_{j=0}^r \chi_{\alpha_j}(a_j \inv ) \sum_{\Sigma~ u_i=0} \quad \prod_{j=0}^r \chi_{\alpha_j}(u_j) } .\] since the inner sum is zero if some *but not all* of the $\alpha_i \sim 0$. - Evaluate the innermost sum by restricting to $u_0 \neq 0$ and setting $u_i = u_0 v_i$ and $v_0 \definedas 1$: \[ \sum_{\Sigma~ u_i=0} \quad \prod_{j=0}^r \chi_{\alpha_j}(u_j) &= \sum_{u_0 \neq 0} \chi_{_{\Sigma ~ \alpha_i}}(u_0) \sum_{\Sigma ~v_i = 0} ~\prod_{j=0}^r \chi_{\alpha_j} (v_j) \\ &= \begin{cases} \qty{q-1} \sum_{\Sigma~ v_i = 0} ~\prod_{j=0}^r \chi_{\alpha_j}(v_j) & \qtext{if} \sum \alpha_i \sim 0 \\ 0 & \qtext{else} \end{cases} .\] ## A Diagonal Hypersurface - Define the *Jacobi sum* for $\alpha$ where $\sum \alpha_i \sim 0$: \[ J(\alpha) \definedas \qty {1 \over q-1} \sum_{\Sigma~ u_i = 0} ~\prod_{j=0}^r \chi_{\alpha_j}(u_j) = \sum_{\Sigma~ v_i = 0} ~\prod_{j=1}^r \chi_{\alpha_j}(v_j) \] - Express $N$ in terms of Jacobi sums as \[ N = q^r + \qty{q-1} \sum_{\substack{\Sigma \alpha_i \sim 0 \\ d_i \alpha_i \sim 0 \\ \alpha\in (0, 1)}} \prod_{j=0}^r \chi_{\alpha_j}(a_j\inv ) J(\alpha) .\] - Evaluate $J(\alpha)$ using Gauss sums: for $\chi: \FF_q \to \CC$ a multiplicative character, define \[ G(\chi) &\definedas \sum_{x\in \FF_q} \chi(x) \psi_q(x) .\] - Proposition: for any $\chi \neq \chi_0$, - $\abs{G(\chi)} = q^{1 \over 2}$ - $G(\chi) G(\bar \chi) = q \chi(-1)$ - $G(\chi_0) = 0$ \[ \chi(t) = {G(\chi) \over q} \sum_{x\in \FF_q} \bar \chi(x) \psi_q(tx) .\] ## A Diagonal Hypersurface - Proposition: if \[ \sum \alpha_i \sim 0 \implies J(\alpha) = {1 \over q} \prod_{k=1}^r G(\chi_{\alpha_k}) \qtext{and} \abs{J(\alpha)} = q^{r - 1\over 2} .\] - We thus obtain \[ N = q^r + \qty{q-1 \over q} \sum_{\substack{\Sigma \alpha_i \sim 0 \\ d_i \alpha_i \sim 0 \\ \alpha\in (0, 1)}} ~\prod_{j=0}^r \chi_{\alpha_j}(a_j\inv ) G(\chi_{\alpha_j}) .\] - We now ask for number of points in $\FF_{q^\nu}$ and consider a point count \[ \bar{N}_\nu = \size \theset{[x_0: \cdots : x_r] \in \PP^r_{\FF_q^\nu} \suchthat \sum_{i=0}^r a_i x_i^n = 0} .\] - Theorem (Davenport, Hasse) $$\qty{q-1}\alpha \sim 0 \implies -G(\chi_{\alpha, \nu}) = \qty{-G(\chi_\alpha)}^\nu.$$ ## A Diagonal Hypersurface - We have a relation $\qty{q^\nu - 1} \bar N_\nu = N_\nu$. - This lets us write \[ \bar N_\nu = \sum_{j=0}^{r-1} q^{j\nu} + \sum_{\substack{\sum \alpha_i ~\sim 0 \\ \gcd(n, q^\nu - 1)\alpha_i \sim 0 \\ \alpha_i \in (0, 1) }} \prod_{j=0}^r \bar \chi_{\alpha_{j, \nu}}(a_i) J_\nu(\alpha) .\] - Set \[ \mu(\alpha) &= \min\theset{\mu \suchthat \qty{q^\mu - 1} \alpha \sim 0} \\ C(\alpha) &= (-1)^{r+1} \prod_{j=1}^r \bar \chi_{\alpha_0, \mu(\alpha)}(a_j) \cdot J_{\mu(\alpha)}(\alpha) .\] - Plugging into the zeta function $Z$ yields \[ \exp\qty{\sum_{\nu = 1}^\infty \bar N_\nu {T^\nu \over \nu} } = \prod_{j=0}^{r-1} \qty{1 \over 1 - q^j T} \prod_{\substack{\sum \alpha_i \sim 0 \\ \gcd(n, q^\nu - 1)\alpha_i \sim 0 \\ \alpha_i \in (0, 1) }} \qty{1 - C(\alpha) T^{\mu(\alpha)}}^{(-1)^r \over \mu(\alpha)} ,\] which is evidently a rational function! $\qed$