Let $C \in \mathfrak{gl}(\mathbb{Z},n)$ be a symmetric full rank integer valued matrix (in my case its 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. $\Lambda=C'\mathbb{Z}^n$?



Intermediate lattices $C\mathbb{Z}^n \subseteq \Lambda \subseteq \mathbb{Z}^n$
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.)


