A 'generalized Four Squares Theorem'? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:23:16Z http://mathoverflow.net/feeds/question/73314 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73314/a-generalized-four-squares-theorem A 'generalized Four Squares Theorem'? DavidLHarden 2011-08-21T07:28:09Z 2011-08-21T13:38:11Z <p>The $4$-dimensional lattice $\mathbb{Z}^{4}$ has vectors of length $\sqrt{n}$ for any positive integer $n$ by the Four Squares Theorem, but this need not be true for higher-dimensional integral, unimodular lattices for two reasons: </p> <p>(i) Some small positive integers could be skipped as squared lengths of lattice vectors. For example, the odd Leech lattice has no $v$ with $v \cdot v = 1$ or $2$.<br> (ii) The lattice may be even. </p> <p>Therefore, the way to word the question to recognize these possibilities is: </p> <p>Let $\Lambda$ be an integral unimodular lattice of dimension $d$, where $d \geq 4$.<br> (i) If $\Lambda$ is odd, then is it true that every sufficiently large positive integer arises as the squared length of a vector in $\Lambda$?<br> (ii) If $\Lambda$ is even, then is it true that every sufficiently large even positive integer arises as the squared length of a vector in $\Lambda$?</p> http://mathoverflow.net/questions/73314/a-generalized-four-squares-theorem/73328#73328 Answer by Abhinav Kumar for A 'generalized Four Squares Theorem'? Abhinav Kumar 2011-08-21T13:26:25Z 2011-08-21T13:38:11Z <p>Yes, both are true. For example, see Theorem 1.6 of Chapter 11 of Cassels's "Rational Quadratic Forms", which says that if $Q$ is a positive definite integral quadratic form, then there is an integer $N$ depending on $Q$ such that if $a > N$ and $a$ is represented primitively by $f$ over all $\mathbb{Z}_p$ then $a$ is represented by $Q$. The local primitive representability is easy to show using the fact that $f$ is unimodular, and the classification of forms over $\mathbb{Z}_p$ by invariants.</p> <p>For instance, if $p$ is odd and $p \nmid a$, then $f$ is equivalent over $\mathbb{Z}_p$ to $(a, a \det(f), 1, 1, \dots, 1)$, which obviously represents $a$ primitively. If $p | a$ you could use $((a-1), (a-1) \det(f), 1, 1, \dots, 1)$ which represents $a$ primitively. I won't do the analysis for $p = 2$, but see section 4 of chapter $8$ of Cassels.</p>