Cancellation theorem for lattices - MathOverflow

Cancellation theorem for lattices

2012-08-27T17:32:45Z

By a lattice, we mean a finitely generated, free $\mathbb{Z}$-module together with a symmetric bilinear form. Typical examples are the hyperbolic lattices $U$ and the root lattices $A_{n}, D_{n}, E_{n}$ associated to Dynkin matrices. In general we cannot say that for lattices $L,M$ and $N$ $$ L\oplus M \cong L\oplus N \Longrightarrow M\cong N. $$ In other words, cancellation does not hold over $\mathbb{Z}$.

I wonder when this cancellation holds. Are there any criteria? I am particularly interested in the case $L=U$.

Answer by Will Jagy for Cancellation theorem for lattices

2012-08-27T18:27:03Z

Right. If $L = U$ is the lattice of the quadratic form $u(x,y) = 2 xy,$ and $M,N$ are positive definite, the conclusion is that $M,N$ are in the same genus. That is, they are rationally equivalent "without essential denominator." There is no complete proof printed in one place. I first saw this on page 378 of SPLAG by Conway and Sloane, first edition. The observation may be due to Conway. This is a small part of finding certain automorphism groups, and is first apparent in the articles on the automorphism group of the Leech Lattice. Anyway, click on my name and just go through my question with promising titles. In a minute I will find the one with a sketch of a proof, put a link here.

Found it, http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus

I also checked with Wai Kiu Chan about the case of "odd" lattices such as the sum of squares, it turns out it does not matter, same outcome.

Meanwhile, it is exactly this observation that allows one to conclude, given a positive "even" lattice with covering radius strictly below $\sqrt 2,$ such as $\mathbb E_8,$ that there is only one class in the genus, i.e. that your integral cancellation holds. See http://mathoverflow.net/questions/69444/a-priori-proof-that-covering-radius-strictly-less-than-sqrt-2-implies-class-nu

Answer by Atsushi Kanazawa for Cancellation theorem for lattices

2012-08-28T01:28:40Z

The genus of an even lattice is characterized by its signature and discriminant form. So if $M$ and $N$ are even, they belong to the same genus. Moreover, if an even lattice $K$ is non-degenerate and indefinite with $rank(K)>h(A_{K})+1$ (where $h(A_{K})$ is the number of minimal generators of the discriminant group $A_{K}$ of $K$), then the genus of $K$ consists of only one class. So if your $M$ (equivalently $N$) satisfies the condition above, the cancelation holds. The results above are proved in "Integral symmetric bilinear form and some of their applications" by V.V. Nikulin. I don't think much is known about odd lattices.