Reduction modulo p and zeta functions

Let \(L\) be an integral lattice of rank \(r\) that is nondegenerate, with Gram tensor \(G\) in the chosen basis. Let \(Q(x) = b(x, x) = x^{t} G x\), and for an integer \(n\) let \(X_n\) be the closed subscheme \(Q(x) = n\) of the affine space \(\mathbb{A}^r_{\mathbb{Z}}\). \(X_0\) is the cone of isotropic vectors.

Primes of bad reduction

The primes of bad reduction of \(L\) are the primes \(\Sigma_L = \{p : p \mid 2 \det L\}\).

For an odd prime \(p\), \(Q\) modulo \(p\) is a nondegenerate quadratic form over \(\mathbb{F}_p\) exactly when \(p \nmid \det L\). Modulo 2, \(Q(x) = \sum_i b(e_i, e_i) x_i^2\) is the square of a linear form, so \(2 \in \Sigma_L\) for every \(L\). For \(n \neq 0\) and \(p \notin \Sigma_L\) with \(p \nmid n\), the gradient \(2 G x\) of \(Q\) is zero modulo \(p\) only at \(x = 0\), which is not on \(X_n\); so \(X_n\) is smooth over \(\mathbb{Z}[1/(2 n \det L)]\). The record states \(\Sigma_L\) as integral.bad_reduction_primes.

The character of the discriminant

Let \(r = 2m\) be even and \(D = (-1)^m \det L\). The record states as integral.quadratic_character the discriminant \(d\) of the field \(\mathbb{Q}(\sqrt{D})\), and \(1\) when \(D\) is a square. For \(p \notin \Sigma_L\), \(\chi_D(p)\) is the Kronecker symbol \((d / p)\), which is the Legendre symbol \((D / p)\).

Every nondegenerate quadratic space over a finite field \(\mathbb{F}_q\) of odd characteristic is isomorphic to exactly one of \(mH \oplus c x^2\) in dimension \(2m + 1\), and \(mH\) or \((m - 1)H \oplus N\) in dimension \(2m\), where \(H\) is the hyperbolic plane and \(N\) is the norm form of \(\mathbb{F}_{q^2}\) (Casselman, Quadratic forms over finite fields, 2018, Theorem 1.6, PDF). The determinant of \(H\) is \(-1\) and that of \(N\) is \(-1\) times a nonsquare, up to squares. So \(Q\) modulo \(p\) is \(mH\) exactly when \(\chi_D(p) = 1\). Over \(\mathbb{F}_{p^k}\) the same holds with \(\chi_D(p)^k\), because an element of \(\mathbb{F}_p^{\times}\) is a square in \(\mathbb{F}_{p^k}\) exactly when it is a square in \(\mathbb{F}_p\) or \(k\) is even.

Points over finite fields

Let \(p \notin \Sigma_L\), \(q = p^k\) and \(\chi = \chi_D(p)^k\). Casselman (§3, cases 4 and 5) counts the points of \(mH\) and \((m - 1)H \oplus N\), which gives for \(r = 2m\):

	\(n = 0\)	\(n \neq 0\), \(p \nmid n\)
\(\lvert X_n(\mathbb{F}_q) \rvert\)	\(q^{2m - 1} + \chi\, q^{m} - \chi\, q^{m - 1}\)	\(q^{2m - 1} - \chi\, q^{m - 1}\)

For \(r = 2m + 1\), \(Q \cong mH \oplus c x^2\) with \(c = (-1)^m \det L\) up to squares. A point of \(mH \oplus c z^2\) with value \(n\) is a \(z\) and a point of \(mH\) with value \(n - c z^2\). The value \(0\) is taken \(1 + \left(\frac{nc}{q}\right)\) times by \(c z^2 - n\), and \(mH\) takes the value \(0\) at \(q^{2m - 1} + q^m - q^{m - 1}\) points and each other value at \(q^{2m - 1} - q^{m - 1}\) points, so for \(p \nmid n\):

\[\lvert X_n(\mathbb{F}_q) \rvert = q \left(q^{2m - 1} - q^{m - 1}\right) + \left(1 + \left(\tfrac{nc}{q}\right)\right) q^m = q^{2m} + \left(\tfrac{D_n}{p}\right)^k q^m, \qquad D_n = (-1)^m n \det L.\]

Also \(\lvert X_0(\mathbb{F}_q) \rvert = q^{2m}\). Casselman’s case 6 prints \(\operatorname{sgn}(-x/c)\) in place of \(\left(\frac{xc}{q}\right)\); the count above gives the sign.

For \(G\) the identity matrix, \(\det L = 1\) and the counts for \(n = 1\) are those of Ireland and Rosen, A Classical Introduction to Modern Number Theory, 2nd edition, Springer, Proposition 8.6.1.

Zeta functions

For a finite set \(\Sigma\) of primes that contains \(\Sigma_L\), the partial zeta function of \(X_n\) is \(\zeta^\Sigma(X_n, s) = \prod_{p \notin \Sigma} Z_p(X_n, p^{-s})\), with \(Z_p(X_n, T) = \exp\left(\sum_{k \geq 1} \lvert X_n(\mathbb{F}_{p^k}) \rvert\, T^k / k\right)\). Also \(\zeta^\Sigma(s) = \prod_{p \notin \Sigma} (1 - p^{-s})^{-1}\) and \(L^\Sigma(s, \chi) = \prod_{p \notin \Sigma} (1 - \chi(p) p^{-s})^{-1}\).

A count \(\sum_j \pm a_j^k\) gives \(Z_p = \prod_j (1 - a_j T)^{\mp 1}\), because \(\sum_k a^k T^k / k = -\log(1 - aT)\). So the counts of the points give, with \(\Sigma = \Sigma_L\) for \(n = 0\) and \(\Sigma = \Sigma_L \cup \{p : p \mid n\}\) for \(n \neq 0\):

	\(\zeta^\Sigma(X_0, s)\)	\(\zeta^\Sigma(X_n, s)\), \(n \neq 0\)
\(r = 2m\)	\(\dfrac{\zeta^\Sigma(s - 2m + 1)\, L^\Sigma(s - m, \chi_D)}{L^\Sigma(s - m + 1, \chi_D)}\)	\(\dfrac{\zeta^\Sigma(s - 2m + 1)}{L^\Sigma(s - m + 1, \chi_D)}\)
\(r = 2m + 1\)	\(\zeta^\Sigma(s - 2m)\)	\(\zeta^\Sigma(s - 2m)\, L^\Sigma(s - m, \chi_{D_n})\)

When \(D\) is a square, \(\chi_D\) is trivial and \(L^\Sigma(s, \chi_D) = \zeta^\Sigma(s)\). For example, \(E_8\) has \(r = 8\) and \(\det = 1\), so \(\zeta^\Sigma(X_n, s) = \zeta^\Sigma(s - 7) / \zeta^\Sigma(s - 3)\) with \(\Sigma = \{2\} \cup \{p : p \mid n\}\). \(A_2\) has \(r = 2\) and \(\det = 3\), so \(D = -3\) and \(\zeta^\Sigma(X_0, s) = \zeta^\Sigma(s - 1)\, L^\Sigma(s - 1, \chi_{-3}) / L^\Sigma(s, \chi_{-3})\) with \(\Sigma = \{2, 3\}\).

The page of a lattice states these functions for its \(r\), \(\det L\) and \(\chi_D\).