Undergrad Proof regarding congruence relation

[Peter_Newman]

Let $\Lambda$ be a lattice and $a, b \in \mathbb{R}^n$, then

$$a \equiv b \text{ mod } \Lambda \Leftrightarrow a- b \in \Lambda$$

I want to prove the statement.

For the left to right direction I would say, $a \equiv b \text{ mod } \Lambda \Leftarrow a = b +k\Lambda$, where $k \in \mathbb{Z}^n$ and therefore $a - b = k\Lambda$, but how does one follow from this, that $a - b \in \Lambda$?

For the other direction $a - b \in \Lambda$, say $u \in \Lambda$, then $a - b = u$ that is $a = b + u$ and this is relative close to the other expression above, but how can we introduce the mod operation here?

 

[fresh_42]

Let $\Lambda$ be a lattice and $a, b \in \mathbb{R}^n$, then

$$a \equiv b \text{ mod } \Lambda \Leftrightarrow a- b \in \Lambda$$

I want to prove the statement.

For the left to right direction I would say, $a \equiv b \text{ mod } \Lambda \Leftarrow a = b +k\Lambda$, where $k \in \mathbb{Z}^n$ and therefore $a - b = k\Lambda$, but how does one follow from this, that $a - b \in \Lambda$?

For the other direction $a - b \in \Lambda$, say $u \in \Lambda$, then $a - b = u$ that is $a = b + u$ and this is relative close to the other expression above, but how can we introduce the mod operation here?

This is a definition to me and cannot be proven. But if it can be proven, then what does $a\equiv b\pmod{\Lambda}$ mean?

 

[Peter_Newman]

Oh yes, that is indeed a definition. Therefore, I take back the "proof".

But how does one come to define it that way?

Let's take a lattice e.g. $\mathbb{Z}^2$. And $a = (1,1)$ and $b = (0,1)$, then $a - b = (1,0)$ would be $\in \Lambda$. That is the case but why are these vectors also congruent and vice versa? What does it mean to use modulo in the congruence anyway?

 

[fresh_42]

But how does one come to define it that way?

The answer lies in the concept of the fundamental domain, the first basic area, uni matrix one. In your example, it is $\left\{\mathrm{x}\in \mathbb{R}^2\,|\,0\leq x_1<1\text{ and }0\leq x_2 <1\right\}$ and in general $\mathcal{F}_{\Lambda} =\left\{\left.\textstyle \sum \limits _{i=1}^{m}r_{i}b_{i} \, \right|\,0\leq r_{i}<1\right\}.$ It is the cell that you can copy to get the entire lattice. And this copying process is the answer.

Any arbitrary point in $p\in\mathbb{R}^2$ lies in one specific cell. If we undo the copy process, i.e. translate the point, say $x$ times in the vertical direction (up or down) and $y$ times in the horizontal direction (left or right) until we end up at a point $q\in \mathcal{F}_\Lambda$ in the fundamental domain, then we have represented $p$ by $q$ together with the translations $(x,y).$ In fact we have
$$
p=q+\begin{pmatrix}x\\y\end{pmatrix} \;\Longleftrightarrow \; p-q=\begin{pmatrix}x\\y\end{pmatrix} \in \Lambda \;\Longleftrightarrow \; p\equiv q \pmod{\Lambda}
$$
This works in any dimension with accordingly more directions, i.e. coordinates. It says literally: every point $p\in \mathbb{R}^n$ can be represented by a point $q\in \mathcal{F}_\Lambda$ of the fundamental domain of the lattice modulo some movements along the lattice points (the undo of the copy process). It is also the reason why this domain, region, or cell is called fundamental.

 

[Peter_Newman]

Hello @fresh_42 , I really like your answer! As you explain it with the fundamental region, the definition also makes absolutely sense! Thank you!