Petersson-Rigid Lattices in a Census of 100 Rank-Three Root Bases

Abstract

The first paper worked out one family of four lattices in full -- $|\det| = 12, 24, 36$ and $72$. This paper generalizes it. We enumerate the 44 symmetrizable rank-three hyperbolic Cartan matrices and their 56 depth-one edits, 100 root bases in all, and compute the obstruction space $S_{5/2}(\rho_L)$ of the 98 within our weight-$5/2$ budget: 10 vacuous, 34 unobstructed, 54 obstructed. The vacuous ten are the lattices Bruinier, Ehlen and Freitag call simple, read in signature $(2,3)$. Of their fifteen, our ten realize five; of the rest, five need more than three generators and cannot be the discriminant form of a rank-three lattice at all, three fail $|\det L| = 2k^2$, and two are simply not root bases. Within the rank-three hyperbolic world the simple lattices are exactly the Feingold-Frenkel neighbours of index $k \le 4$. The discriminant group $L'/L$ carries a finite quadratic form, and the finite group of its isometries acts on the weight-$3/2$ cusp forms for $\rho_L$, the bottom antisymmetric rung. We survey the lattices on which that action is absolutely irreducible of dimension at least two, so that the Petersson pairing there is pinned down up to a single scalar. The condition alone cuts the census to three: $L_4$ of the first paper, and two new ones at $|\det| = 40$ and $88$. The quaternion discriminants that occur are 6, 10 and 22, the three for which the Shimura curve $X^D$ has genus zero. Rigidity uses no quaternion input, so we record it as an observation, not a characterisation. Both invariants of the first paper -- the shadow norm $\|\Xi\|^2$ and the Petersson scalar $t$ -- were single points there. Three rigid lattices instead of one are where their special faces come off: $\|\Xi\|^2$ generalizes against $L(f,1)$ where the first paper read $L(f,2)$, and $t$ against an elliptic curve's imaginary period where it read $\Gamma(1/3)$. Both numerically, to 26-58 digits.