Let $C \in \mathfrak{gl}(\mathbb{Z},n)$ be a symmetric full rank integer valued matrix (in my case it is the symmetric part of a Cartan matrix). Let $\Lambda \subseteq \mathbb{Z}^n$ be a full rank sublattice, s.t. $C\mathbb{Z}^n \subseteq\Lambda \subseteq \mathbb{Z}^n$. One can think of $\Lambda$ as a lattice in between of a root lattice $\Lambda_R=C\mathbb{Z}^n$ and the co-weight lattice $\Lambda_W^\vee=\mathbb{Z}^n$. Question: Is there a symmetric matrix $C'$, s.t. $\Lambda=C'\mathbb{Z}^n$?
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$\begingroup$ So, $C^{-1}$ should not have integral entries? $\endgroup$– Ilya BogdanovCommented Sep 30, 2016 at 13:48
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3$\begingroup$ Isn't the answer "yes" through the magic of the smith normal form? $\endgroup$– Igor RivinCommented Sep 30, 2016 at 13:50
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1 Answer
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First of all, every full rank sublattice $\Lambda\subseteq \mathbb Z^n$ contains $\Lambda'=c\mathbb Z^n$ for some positive integer $c$. This $\Lambda'$ may serve as $C\mathbb Z^n$; so the question can in fact be reformulated as follows: Does every full rank sublattice $\Lambda\subseteq \mathbb Z^n$ have a symmetric base, i.e. a matrix $B=B^T$ with $\Lambda=B\mathbb Z^n$?
The answer seems to be yes, and, as Igor Rivin has suggested, the Smith normal form indeed helps. Let $\Lambda=A\mathbb Z^n$, and let $A=PDQ$ be the Smith decomposition of $A$, so $D$ is diagonal, and $P,Q\in SL(\mathbb Z,n)$. Then also $Q^{-1}P^T\in SL(\mathbb Z,n)$, and $B=AQ^{-1}P^T=PDP^T$ is a symmetric matrix satisfying $B\mathbb Z^n=AQ^{-1}P^T\mathbb Z^n=A\mathbb Z^n=\Lambda$.
(We did ot use the full rank condition; however, if $A$ is degenerate, then $B$ is degenerate as well, so surely it does not provide a base.)
