1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.