most recent 30 from http://mathoverflow.net 2013-05-19T11:51:54Z http://mathoverflow.net/feeds/question/100764 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100764/globally-irreducible-lattices DavidLHarden 2012-06-27T10:42:06Z 2012-06-27T10:42:06Z

<p>Here I am only interested in globally irreducible lattices over $\mathbb{Z}$. </p> <p>The basic theorem concerning these says that a globally irreducible lattice is similar to a lattice which is integral and unimodular. With that scaling, the lattice will also be even, except in the degenerate case where the dimension is 1 and the lattice is just $\mathbb{Z}$ itself.<br> I see how the proof goes, except for one gap: </p> <p>Let $\Lambda$ be globally irreducible. This means that, for any prime $p \in \mathbb{Z}$, $Aut(\Lambda)$ acts irreducibly on $\Lambda / p\Lambda$. It is easy to show (so I will skip the proof) that this condition implies every invariant sublattice of $\Lambda$ is of the form $k\Lambda$ for some nonnegative integer $k$.<br> Global irreducibility is not affected by scaling $\Lambda$. The gap: I am assuming any two nonzero inner products of vectors in $\Lambda$ have a rational quotient.<br> Assuming that, let $v_{1}, \ldots , v_{n}$ be an integral basis for $\Lambda$ (so $n = \mathrm{dim}(\Lambda)$). Then it is possible to scale $\Lambda$ so that all $v_{i} \cdot v_{j}$, where $1 \leq i \leq j \leq n$, are integers whose greatest common factor is 1. Then $\Lambda$ is integral.<br> Evenness is immediate: Since $\Lambda$ is integral, the identity $|u+v|^{2} = |u|^{2} + |v|^{2} + 2u \cdot v$ means that the set of vectors $v$ such that $|v|^{2}$ is even is an invariant sublattice of $\Lambda$ (whose index, as an additive subgroup, is 1 or 2). But also this sublattice is of the form $k\Lambda$ for some $k \geq 1$ and the index of this sublattice is $k^{\mathrm{dim}(\Lambda)}$. If $\mathrm{dim}(\Lambda) > 1$, $k^{\mathrm{dim}(\Lambda)} = 2$ is impossible so $k^{\mathrm{dim}(\Lambda)} = 1$ and $k = 1$.<br> Unimodularity sounds not so hard to prove: If there is a prime $p$ dividing the determinant of the Gram matrix of $\Lambda$, the choice of scaling means $p$ does not divide all the entries in the Gram matrix. Then the row space of the Gram matrix should lead to (though I am unclear on the details of this, I am not asking about this at the moment) an invariant subspace of $\Lambda / p\Lambda$ (which should be proper since the Gram matrix is nonzero modulo $p$ but the determinant modulo $p$ is 0). </p> <p>What is the easiest way to fill in the gap?</p>