Une formule du produit

Two positive definite hermitian lattices over $\mathcal O_E$, $E$ imaginary quadratic, lie in the same genus if they become isometric over $\mathcal O_E\otimes\Z_p$ for every prime $p$. The mass of the genus of $L$ is $\sum_i1/|\Aut(L_i)|$, summed over representatives $L_i$ of its isometry classes. The same definitions apply to positive definite $\Z$-lattices, with $\Aut=\mathrm O$.

Let $E=\Q(\sqrt{-d})$ have discriminant $d_E$ and quadratic character $\chi$, let $t$ be the number of primes dividing $d_E$, and let $L$ lie in the genus of $\mathcal O_E^m$ with the form $\sum_ix_i\bar y_i$. Then

where $\alpha_p=|\GL_m(\F_p)|/p^{m^2}$ if $p$ splits in $E$, $\alpha_p=|\mathrm U_m(\F_p)|/p^{m^2}$ if $p$ is inert, and $\alpha_p=|\mathrm O_m(\F_p)|/p^{m(m-1)/2}$ if $p$ is odd and ramified and $m$ is odd. For $m$ odd, $\operatorname{mass}(L)=2^{1-t}\prod_{j=1}^m|B_{j,\chi^j}|/2j$, with $\chi^j=\chi$ for odd $j$ and the trivial character for even $j$. (Hashimoto and Koseki; Gan and Yu prove the product form for every genus of unimodular lattices.)

The genus of $\mathcal O_E^3$ has exactly two classes, the standard lattice and Klein’s lattice $L_\infty$, and $\operatorname{mass}=\tfrac1{42}=\tfrac1{48}+\tfrac1{336}$, with $|\Aut(\mathcal O_E^3)|=48$ and $|\Aut(L_\infty)|=336$. The fourteen Kneser neighbours of $L_\infty$ at $(\alpha)$, seven for the points and seven for the lines of the Fano plane $\mathbb P(L_\infty/\alpha L_\infty)$, are standard; for each, $\bar\alpha$ times its six vectors of norm 1 is a frame of $L_\infty$, a bijection onto the fourteen frames; and its group is the stabilizer of the frame in $\Aut(L_\infty)$, of order 48, the stabilizer of a point or of a line of the Fano plane.

The genus of Mumford’s lattice $\mathcal O_L=\Z[\zeta_7]$, of type 2 at 7 and unimodular elsewhere, has exactly one class, and $\operatorname{mass}(\mathcal O_L)=\tfrac1{42}=1/|\Aut(\mathcal O_L)|$, $|\Aut(\mathcal O_L)|=42=|7{:}3|\cdot|\mu_E|$. The eight self-dual lattices between $\mathcal O_L$ and its dual are one Klein lattice and seven standard ones, permuted by $\Aut(\mathcal O_L)$ in orbits of sizes 1 and 7; $\mathcal O_E^3$ and $L_\infty$ each have exactly eight sublattices of type 2 and colength one, all isometric to $\mathcal O_L$; and the masses of the two genera are equal, $\tfrac8{42}=\tfrac8{48}+\tfrac8{336}$.

Let $W=E_8^\rho$ be the $z\in\Z[\omega]^4$ whose reduction modulo $\theta=\sqrt{-3}$ lies in the tetracode, with $h(x,y)=\sum_ix_i\bar y_i$. Then $W^\vee=\theta^{-1}W$, $\tfrac13\operatorname{Tr}h$ is $E_8$, and $|\Aut(W)|=155520$. The genus of $W$ has mass $\tfrac1{155520}$, so $W$ is alone in its genus and the product of local factors returns $1/|\mu_6\cdot\mathrm{Sp}(4,3)|$. Reduction maps $\Aut(W)$ onto $\mathrm{GU}_4(2)$ with kernel $\{\pm1\}$ and onto $\mathrm{Sp}_4(3)$ with kernel $\mu_3$. So the two lives of $\mathrm{PSp}(4,3)\cong\mathrm{PSU}(4,2)$ are two local factors of one mass, and the global group fills both.

$\lambda=(1+\theta)/2$ preserves $E_8^a$ and $E_8^b$; each is a free $\Z[\lambda]$-module of rank 4 with a basis of roots, and $\operatorname{Tr}_{K/\Q}h=2\langle\,,\rangle$. The roots in each nonzero residue of $E_8/\lambda E_8\cong\F_2^4$ form a cross, and likewise modulo $\bar\lambda$. The $\Z[\lambda]$-linear isometries of $E_8^a$ form a group $U\cong2{\cdot}A_7$ of order 5040 with centre $\{\pm1\}$, two-transitive on the fifteen crosses modulo $\lambda$ and transitive on the 240 roots, and the stabilizer of the cross of the sixteen vectors is $\SL(2,7)$.