Cancellation theorem for lattices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:57:18Z http://mathoverflow.net/feeds/question/105644 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105644/cancellation-theorem-for-lattices Cancellation theorem for lattices M Koerner 2012-08-27T17:32:45Z 2012-08-28T05:38:31Z <p>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 <strong>cannot</strong> 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 $\mathbb{Z}$. </p> <p>I wonder when this cancellation holds. Are there any criteria? I am particularly interested in the case $L=U$. </p> http://mathoverflow.net/questions/105644/cancellation-theorem-for-lattices/105649#105649 Answer by Will Jagy for Cancellation theorem for lattices Will Jagy 2012-08-27T18:27:03Z 2012-08-27T18:44:56Z <p>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. </p> <p>Found it, <a href="http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus" rel="nofollow">http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus</a> </p> <p>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. </p> <p>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 href="http://mathoverflow.net/questions/69444/a-priori-proof-that-covering-radius-strictly-less-than-sqrt-2-implies-class-nu" rel="nofollow">http://mathoverflow.net/questions/69444/a-priori-proof-that-covering-radius-strictly-less-than-sqrt-2-implies-class-nu</a> </p> http://mathoverflow.net/questions/105644/cancellation-theorem-for-lattices/105671#105671 Answer by Atsushi Kanazawa for Cancellation theorem for lattices Atsushi Kanazawa 2012-08-28T01:28:40Z 2012-08-28T05:38:31Z <p>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. </p>