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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 $L=U$.

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If $L \oplus M \cong L \oplus N$ then at least you can say that $M, N$ have the same theta function. –  Qiaochu Yuan Aug 27 '12 at 17:43
...assuming that $L,M,N$ are positive definite, which apparently was not the intention because the question indicates interest in the special case $L = U$. Indeed in that case all we can conclude is that $M,N$ are in the same genus, which in general does not imply $M \cong N$. For example: if $M = E_8^2$, and $N$ is the unimodular lattice of rank $16$ that contains $D_{16}$ with index $2$, then $U \oplus M \cong U \oplus N$ (both are the even unimodular lattice II$_{17,1}$). Likewise: $M = {\bf Z}^9$, $N = {\bf Z} \oplus E_8$, $U \oplus M \cong U \oplus N \cong {\rm I}_{10,1}$. –  Noam D. Elkies Aug 27 '12 at 18:32

2 Answers 2

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, 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 A Priori proof that Covering Radius strictly less than $\sqrt 2$ implies class number one

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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.

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Your second sentence needs "unimodular". –  S. Carnahan Aug 28 '12 at 4:41
I deleted "unimodular" from the third sentence (otherwise the discriminant form is trivial). I don't think we need unimodularity anywhere. –  Atsushi Kanazawa Aug 28 '12 at 5:40

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