Neighbors become isometric after adding a hyperbolic plane #
Let L be an even lattice, p a positive integer and v ∈ L a vector whose norm is divisible
by 2p², such that some lattice vector pairs with v coprimely to p. Then
L ⊕ U ≅ L.neighbor p v ⊕ U, where U is the hyperbolic plane.
Write M = L.neighbor p v, b = v / p ∈ M and e, f for the standard isotropic basis of U.
The vector z = p (b + e - (b · b / 2) f) of L ⊕ U is primitive and isotropic, and its orthogonal
complement modulo z is identified with M by x + α e + β f ↦ x - α b. Choosing y ∈ L ⊕ U
with z · y = 1 and correcting it to an isotropic vector, the plane spanned by z and y is a
copy of U whose orthogonal complement is a copy of M. Assembled, this is an explicit rational
isometry of M ⊕ U onto L ⊕ U carrying the integral carriers onto each other.
In particular two even unimodular lattices that are neighbors become isometric after adding U,
although they need not be isometric themselves; this is how cancellation fails over ℤ.
Main results #
TauCeti.IntegralLattice.nonempty_isometry_orthogonalSum_hyperbolicPlane_neighbor:L ⊕ U ≅ L.neighbor p v ⊕ U.
References #
- M. Kneser, Quadratische Formen, revised with R. Scharlau, Springer (2002), §28, for neighbors.
- J.-P. Serre, A Course in Arithmetic, Chapter V, for splitting off hyperbolic planes from indefinite unimodular lattices and the resulting failure of cancellation.
Neighbors become isometric after adding a hyperbolic plane. Let L be even and let
v ∈ L have norm divisible by 2p², with some w ∈ L pairing with v coprimely to p. Then
L ⊕ U and L.neighbor p v ⊕ U are isometric, where U is the hyperbolic plane.