\normalsize

\(\hfill\blacksquare\)

Counting Points

Note that we’ve actually computed the number of points in any Grassmannian.

Identify \({\mathbf{P}}^m_{ \mathbf{F} }= {\operatorname{Gr}}_{{ \mathbf{F} }}(1, m+1)\) as the space of lines in \({\mathbf{A}}^{m+1}_{ \mathbf{F} }\).

We obtain a simplification (importantly, a sum formula) when setting \(k=1\): \begin{align*} \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 .\end{align*}

Thus \begin{align*} X({ \mathbf{F} }_q) &= \sum_{j=0}^{m} q^j \\ X({ \mathbf{F} }_{q^2}) &= \sum_{j=0}^{m} \qty{q^2}^j \\ &\vdots \\ X({ \mathbf{F} }_{q^n}) &= \sum_{j=0}^{m} \qty{q^n}^j .\end{align*}

Computing the Zeta Function

So \begin{align*} 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}^{-1}\\ &= \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} ,\end{align*}

Miraculously, still a rational function! Consequence of sum formula, works in general.

Checking the Weil Conjectures

\begin{align*} Z_X(z) = \prod_{j=0}^m \qty{1 \over 1 - q^j z} .\end{align*}

• Rationality: Clear!

• Functional Equation: Less clear, but true:

\begin{align*} 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) \\ .\end{align*}

Checking

\begin{align*} Z_X(z) = \prod_{j=0}^m \qty{1 \over 1 - q^j z} .\end{align*}

• Riemann Hypothesis: Reduces to the statement \(\left\{{ \alpha_i }\right\} = \left\{{{q^m \over \alpha_j }}\right\}\).

• Clear since \(\alpha_j = q^j\) and every \(\alpha_i\) is a lower power of \(q\).

• Betti Numbers: Use the fact that \({\mathcal{P}}_{{\mathbf{CP}}^m}(x) = 1 + x^2 + x^4 + \cdots + x^{2m}\)

• Only even dimensions, and correspondingly no numerator.

An Easier Proof: “Paving by Affines”

Quick recap: \begin{align*} Z_{{\operatorname{pt}}} = {1 \over 1 - z} && Z_{{\mathbf{P}}^1}(z) = {1 \over 1 - qz} && Z_{{\mathbf{A}}^1}(z) = {1\over (1-z)(1-qz)} .\end{align*}

Note that \({\mathbf{P}}^1 = {\mathbf{A}}^1 {\textstyle\coprod}\left\{{\infty}\right\}\) and correspondingly \(Z_{{\mathbf{P}}^1}(z) = Z_{{\mathbf{A}}^1}(z) \cdot Z_{{\operatorname{pt}}}(z)\). This works in general:

Lemma (Excision)

If \(Y/{ \mathbf{F} }_q \subset X/{ \mathbf{F} }_q\) is a closed subvariety, for \(U = X\setminus Y\), \(Z_X(z) = Z_Y(z) \cdot Z_{U}(z)\).

Proof: Let \(N_n = {\sharp}Y({ \mathbf{F} }_{q^n})\) and \(M_n = {\sharp}U({ \mathbf{F} }_{q^n})\), then \begin{align*} \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) .\end{align*}

An Easier Proof: “Paving”

Note that geometry can help us here: we have a decomposition \({\mathbf{P}}^n = {\mathbf{P}}^{n-1} {\textstyle\coprod}{\mathbf{A}}^n\), and thus inductively a stratification \begin{align*} {\mathbf{P}}^m = {\textstyle\coprod}_{j=0}^m {\mathbf{A}}^j = {\mathbf{A}}^0 {\textstyle\coprod}{\mathbf{A}}^1 {\textstyle\coprod}\cdots {\textstyle\coprod}{\mathbf{A}}^m .\end{align*}

Recalling that \begin{align*} Z_{X{\textstyle\coprod}Y}(z) = Z_X(z) \cdot Z_Y(z) \end{align*}

and \(Z_{{\mathbf{A}}^j}(z) = {1 \over 1 - q^j z}\), we have \begin{align*} Z_{{\mathbf{P}}^m}(z) = \prod_{j=0}^m Z_{{\mathbf{A}}^j}(z) = \prod_{j=0}^m {1 \over 1 - q^j z} .\end{align*}

Grassmannians

Motivation

Consider now \(X = {\operatorname{Gr}}(k, m) / { \mathbf{F} }\) – by the previous computation, we know \begin{align*} X({ \mathbf{F} }_{q^n}) = \genfrac{[}{]}{0pt}{}{m}{k}_{q^n} \coloneqq\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)} \end{align*}

but the corresponding Zeta function is much more complicated than the previous examples: \begin{align*} Z_X(z) &= \exp\qty{ \sum_{n=1}^\infty \genfrac{[}{]}{0pt}{}{m}{k}_{q^n} {z^n \over n} } = \cdots ? .\end{align*}

Since \(\dim_{\mathbf{C}}{\operatorname{Gr}}_{\mathbf{C}}(k, m) = 2k(m-k)\), by Weil we should expect \begin{align*} Z_X(z) = \prod_{j=0}^{2k(m-k)} \frac{p_{2(j+1)}(z)}{p_{2j}(z)} \end{align*} with \(\deg p_j = \beta_j\).

Grassmannian

It turns out that (proof omitted) one can show \begin{align*} \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)} \end{align*}

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 \(q{\hbox{-}}\)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 \begin{align*} {\mathcal{P}}_{{\operatorname{Gr}}_{\mathbf{C}}(m, k)}(x) = \sum_{n=1}^{k(m-k)} \lambda_{m, k}(n) x^{2n}, \end{align*}

In particular, the \(H^* {\operatorname{Gr}}_{\mathbf{C}}(m, k)\) 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 \(X/{ \mathbf{F} }_q\) defined by \begin{align*}X = V(a_0x_0^{n} + \cdots + a_r x_r^{n}) \subset { \mathbf{F} }_q^{r+1}.\end{align*}
• We want to compute \(N = {\sharp}X\).
• Set \(d_i = \gcd(n_i, q-1)\).
• Define the character \begin{align*} \psi_q: { \mathbf{F} }_q & \to {\mathbf{C}}^{\times}\\ a &\mapsto \exp\qty{2\pi i ~\operatorname{Tr}_{{ \mathbf{F} }_q/{ \mathbf{F} }_p}(a) \over p} .\end{align*}
  • By Artin’s theorem for linear independence of characters, \(\psi_q \not \equiv 1\) and every additive character of \({ \mathbf{F} }_q\) is of the form \(a \mapsto \psi_q(ca)\) for some \(c\in { \mathbf{F} }_q\).
• Shorthand notation: say \(a\sim 0 \iff a \equiv 0 \operatorname{mod}1\).

A Diagonal Hypersurface

• Fix an injective multiplicative map \begin{align*} \phi: \mkern 1.5mu\overline{\mkern-1.5mu{ \mathbf{F} }\mkern-1.5mu}\mkern 1.5mu_q^{\times}\to {\mathbf{C}}^{\times} .\end{align*}

• Define \begin{align*} \chi_{\alpha, n}: { \mathbf{F} }_{q^n}^{\times}&\to {\mathbf{C}}^{\times}\\ x & \mapsto \phi(x)^{\alpha\qty{q^n-1}} \\ \\ \quad {\quad \operatorname{for} \quad} \alpha \in {\mathbf{Q}}/{\mathbf{Z}}, n\in {\mathbf{Z}}, & \quad \alpha\qty{q^n-1} \equiv 0 \operatorname{mod}1 .\end{align*}

  • Extend this to \({ \mathbf{F} }_{q^n}\) by \begin{align*} \begin{cases} 1 & \alpha \equiv0 \operatorname{mod}1 \\ 0 & \text{else} \end{cases} .\end{align*}
  • Set \(\chi_\alpha = \chi_{\alpha, 1}\).

• Proposition: \begin{align*}\alpha(q-1) \equiv 0 \operatorname{mod}1 \implies \chi_{\alpha, n}(x) = \chi_\alpha(\mathrm{Nm}_{{ \mathbf{F} }_{q^n} / { \mathbf{F} }_q }(x) )\end{align*}

• Proposition: \begin{align*}d \coloneqq\gcd(n, q-1), u \in { \mathbf{F} }_q \implies {\sharp}\left\{{x\in { \mathbf{F} }_q {~\mathrel{\Big\vert}~}x^n = u}\right\} = \sum_{d\alpha \sim 0} \chi_\alpha(u)\end{align*}

A Diagonal Hypersurface

• This implies \begin{align*} N &= \sum_{\substack{\alpha = [\alpha_0, \cdots, \alpha_r] \\ d_i \alpha_i \sim 0}} \quad \sum_{\substack{\mathbf{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 ^{-1}) \sum_{\Sigma~ u_i=0} \quad \prod_{j=0}^r \chi_{\alpha_j}(u_j) } .\end{align*} 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 \(v_0 \coloneqq 1\): \begin{align*} \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) & {\quad \operatorname{if} \quad} \sum \alpha_i \sim 0 \\ 0 & {\quad \operatorname{else} \quad} \end{cases} .\end{align*}

A Diagonal Hypersurface

• Define the Jacobi sum for \(\alpha\) where \(\sum \alpha_i \sim 0\): \begin{align*} J(\alpha) \coloneqq\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) \end{align*}

• Express \(N\) in terms of Jacobi sums as \begin{align*} 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^{-1}) J(\alpha) .\end{align*}

• Evaluate \(J(\alpha)\) using Gauss sums: for \(\chi: { \mathbf{F} }_q \to {\mathbf{C}}\) a multiplicative character, define \begin{align*} G(\chi) &\coloneqq\sum_{x\in { \mathbf{F} }_q} \chi(x) \psi_q(x) .\end{align*}

• Proposition: for any \(\chi \neq \chi_0\),

  • \({\left\lvert {G(\chi)} \right\rvert} = q^{1 \over 2}\)
  • \(G(\chi) G(\mkern 1.5mu\overline{\mkern-1.5mu\chi\mkern-1.5mu}\mkern 1.5mu) = q \chi(-1)\)
  • \(G(\chi_0) = 0\) \begin{align*} \chi(t) = {G(\chi) \over q} \sum_{x\in { \mathbf{F} }_q} \mkern 1.5mu\overline{\mkern-1.5mu\chi\mkern-1.5mu}\mkern 1.5mu(x) \psi_q(tx) .\end{align*}

A Diagonal Hypersurface

• Proposition: if \begin{align*} \sum \alpha_i \sim 0 \implies J(\alpha) = {1 \over q} \prod_{k=1}^r G(\chi_{\alpha_k}) {\quad \operatorname{and} \quad} {\left\lvert {J(\alpha)} \right\rvert} = q^{r - 1\over 2} .\end{align*}

• We thus obtain \begin{align*} 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^{-1}) G(\chi_{\alpha_j}) .\end{align*}

• We now ask for number of points in \({ \mathbf{F} }_{q^\nu}\) and consider a point count \begin{align*} \mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu_\nu = {\sharp}\left\{{[x_0: \cdots : x_r] \in {\mathbf{P}}^r_{{ \mathbf{F} }_q^\nu} {~\mathrel{\Big\vert}~}\sum_{i=0}^r a_i x_i^n = 0}\right\} .\end{align*}

• Theorem (Davenport, Hasse) \begin{align*}\qty{q-1}\alpha \sim 0 \implies -G(\chi_{\alpha, \nu}) = \qty{-G(\chi_\alpha)}^\nu.\end{align*}

A Diagonal Hypersurface

• We have a relation \(\qty{q^\nu - 1} \mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu_\nu = N_\nu\).
• This lets us write \begin{align*} \mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu_\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 \mkern 1.5mu\overline{\mkern-1.5mu\chi\mkern-1.5mu}\mkern 1.5mu_{\alpha_{j, \nu}}(a_i) J_\nu(\alpha) .\end{align*}
• Set \begin{align*} \mu(\alpha) &= \min\left\{{\mu {~\mathrel{\Big\vert}~}\qty{q^\mu - 1} \alpha \sim 0}\right\} \\ C(\alpha) &= (-1)^{r+1} \prod_{j=1}^r \mkern 1.5mu\overline{\mkern-1.5mu\chi\mkern-1.5mu}\mkern 1.5mu_{\alpha_0, \mu(\alpha)}(a_j) \cdot J_{\mu(\alpha)}(\alpha) .\end{align*}
• Plugging into the zeta function \(Z\) yields \begin{align*} \exp\qty{\sum_{\nu = 1}^\infty \mkern 1.5mu\overline{\mkern-1.5muN\mkern-1.5mu}\mkern 1.5mu_\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)} ,\end{align*} which is evidently a rational function! \(\hfill\blacksquare\)