When does projecting a lattice onto a subspace give another lattice?

Question

For a lattice $L$ and vector $v$ in $\mathbb{R}^n$ take $L\backslash v := \{l-(\textstyle \frac{v^\dagger l}{\|v\|^2})v: l\in L\}.$ When is $L\backslash v$ a lattice?

Answer 1

Let $p\colon \Bbb R^n\to \Bbb R^{n-1}$ be a linear map of rank $n-1$. and $L\subset \Bbb R^n$ a lattice. We wonder when $p(L)$ is a lattice in $\Bbb R^{n-1}$.

Pick a basis $e_1,\ldots, e_n$ of $L$. Then the $p(e_i)$ span $\Bbb R^{n-1}$, but are linearly dependent in an essentially unique way, $$\tag1 c_1p(e_1)+\ldots +c_np(e_n)=0$$ with not all $c_i=0$. We may assume wlog. that $c_n=1$.

If all $c_i$ are rational, say $c_i=\frac {a_i}{b_i}$ with $a_i\in\Bbb Z$, $b_i\in\Bbb N$, then $p(L)$ is contained in $L':=\frac {p(e_1)}{b_1}\Bbb Z+\ldots + \frac {p(e_{n-1})}{b_{n-1}}\Bbb Z$. As the listed vectors span $\Bbb R^{n-1}$, $L'$ is a lattice and $L$ then is a sublattice of it.

On the other hand, if $p(L)$ is a lattice than there is a linear dependence among $p(e_1),\ldots,p(e_n)$ that has only rational coefficients. By the essential uniqueness of $(1)$, it follows that the $c_i$ are rational.

We conclude that $p(L)$ is a lattice if and only if the coefficients in the produced linear dependency $(1)$ are (after normalizing) rational. In the setting of the OP, we find the $c_i$ by expressing the projecting vector $v$ because $v=\sum c_ie_i$ turns intor $0= \sum c_ip(e_i)$. The rationality criterion is then equivalent to the following:

$L\setminus v =p(L)$ is a lattice if and only if $v$ is a multiple of a lattice vector.