Given a lattice in $\mathbb{Z}^n$, what can be said about its 'transpose' lattice?

Question

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 $\det \Lambda = \det M = \det M^T = \det \Lambda^T$.

Answer 1

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, $V \simeq \oplus \mathbb{Z}/d_i \mathbb{Z}$ 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.

Answer 2

To follow up on user44191's comment, here is a concrete example: consider the matrices $$ M = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \qquad M' = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}.$$ The columns of $M$ and $M'$ span the same lattice $\Lambda$. However, for the transpose matrices $$ M^T = M,\qquad (M')^T = \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix},$$ the columns do not span the same lattice.

This shows that the definition of $\Lambda^T$ is not well-defined, as a subset of $\mathbb Z^n$. It depends on the choice of basis $M$ for $\Lambda$.