I apologize if this notion is well-known, but I couldn't find anything useful and I am not sure what key words to look for.


Suppose we have a lattice $\Lambda \subset \mathbb{Z}^n$, given by in the form 


$$\displaystyle \Lambda = \left\{M \mathbf{u} : \mathbf{u} \in \mathbb{Z}^n \right \}$$


for some matrix $M$ with integer entries and non-zero determinant. By 'transpose' lattice I mean the corresponding lattice given by


$$\displaystyle \Lambda^T = \left\{M^T \mathbf{u} : \mathbf{u} \in \mathbb{Z}^n \right \}$$.


Is there a name for $\Lambda^T$? What properties can be deduced about $\Lambda^T$ given $\Lambda$?


For example, it is clear that $\det \Lambda = \det M = \det M^T = \det \Lambda^T$. 



Given a lattice in $\mathbb{Z}^n$, what can be said about its 'transpose' lattice?
Let $V = \mathbb{Z}^n/(M \mathbb{Z}^n), V' = \mathbb{Z}^n/(M^T \mathbb{Z}^n)$. I claim that $V \simeq V'$ as abelian groups.


By the [classification of finite abelian groups](https://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Classification), $V \simeq \oplus \mathbb{Z}/d\_i \mathbb{Z}$ for some $d\_1 | d\_2 | \dots | d\_n$, and with this condition, the set $\{d\_i\}$ is unique. Let $D$ be the diagonal matrix with $\{d\_i\}$ on the diagonals; then it's not hard (but it is slightly tedious) to show that $M = ADB$ for some $A, B \in SL\_n(\mathbb{Z})$ (in fact, this is one proof of the above classification). Then $M^T = B^TDA^T$, so by reversing the above reasoning, $V' \simeq \oplus \mathbb{Z}/d\_i \mathbb{Z} \simeq V$.


In fact, we can use the above reasoning for the converse: if $\Lambda, \Lambda'$ are sublattices of $\mathbb{Z}^n$ such that $\mathbb{Z}^n/\Lambda \simeq \mathbb{Z}^n/\Lambda'$, then there are some $M\_0, M'\_0$ such that $\Lambda = M\_0 \mathbb{Z}^n, \Lambda' = M'\_0 \mathbb{Z}^n$. If $D$ is the common Smith Normal Form, then $M\_0 = ADB, M'\_0 = A'DB'$; let $M = ADA'^T$. Then $\Lambda = M\_0 \mathbb{Z}^n = ADB\mathbb{Z}^n = AD\mathbb{Z}^n = ADA'^T\mathbb{Z}^n = M\mathbb{Z}^n$, and similarly $\Lambda' = M^T\mathbb{Z}^n$. So two lattices come from transpose matrices iff their quotients are isomorphic.


As Sam Hopkins said in a comment, the key word here is Smith Normal Form.


