What does it mean to mod a vector by a set of points?

Question

If $\vec{y}$ is in $\mathbb{R}^n$, what does "$\vec{y}$ mod $P(L)$" mean? Where $P(L)$ is the fundamental parrallelepiped of a lattice given by the set:

$P(L)= \{x_1v_1 + \ldots + x_mv_m \ \vert \ x_i \in \mathbb{R}, 0 \leq x_i < 1\}$

Answer 1

In rough terms, modding out by a set means considering that any element in that set is $0$, hence that you can add or subtract them and stay in the same equivalence class.

Formally: $$\vec{y}\bmod P(L)=\{\vec{y} + p\ |\ p\in P(L)\}$$

and two vectors $\vec{y}$ and $\vec{y}'$ are said to be congruent $\bmod P(L)$ when $\vec{y}\bmod P(L)=\vec{y}'\bmod P(L)$, which is equivalent to $\vec{y}-\vec{y}'\in P(L)$ if $P(L)$ is a subgroup of the ambient additive structure, here $\Bbb R^n$.

This last sentence is the reason why it makes much more sense in general to mod out by the lattice itself rather than a fundamental parallelepiped.

Note also that if $P(L)$ was not a subgroup, then congruence $\bmod P(L)$ defined by $$\vec{y}\equiv \vec{y}'\pmod {P(L)}\Longleftrightarrow \vec{y}'\in \vec{y}\bmod P(L)$$ would not be an equivalence relation.