# Why is this a rational lattice?

I've usually seen a rational lattice $\Gamma \subset \mathbb{R}^n$ defined like this:

$$\displaystyle \Gamma = \{Ax : x \in \mathbb{Z}^n \},$$

where $A \in \mathbb{Q}^{n \times n}$ is a matrix of full rank. However, some texts call $(2 \pi \mathbb{Z})^n$ a rational lattice, which does not seem to fit this definition. Is there an alternative definition of a rational lattice, or have I misunderstood the definition I've posted above? And if $(2\pi\mathbb{Z})^n$ is a rational lattice, then what distinguishes it from an irrational lattice?

This is likely an "alternate" definition, but really it's effectively just a notational change. Instead of writing $1$, $2$, $3$, you write $2\pi$, $4\pi$, $6\pi$. In other words, $(2\pi\mathbb{Z})^n \cong \mathbb{Z}^n$. Essentially all you're doing is allowing $A$ to be scaled by an arbitrary non-zero real factor. Presumably this is done to make some other equations look nicer, e.g. complex exponentials. An "irrational lattice" would be one where there are points that aren't a linear combination with rational weights of the basis vectors. Concretely, for the $\mathbb{Z}^2$ case, we have some vector $v_1$ that we identify with $(1,0)$ and another vector $v_2$ that we identify with $(0,1)$ and we want to know that every other point in the lattice is $pv_1 + qv_2$ for some pair of rationals $p$ and $q$. That $v_1$ and $v_2$ have irrational coordinates in some other basis isn't really important. If we choose a coordinate system where $v_1$ and $v_2$ are written $(2\pi,0)$ and $(0,2\pi)$ respectively, that's just changed how we talk about it but not the actual situation/interrelationships.