I have the following problem: Let $L$ be a $\mathbb{R}^n$ lattice (that is a discrete closed $\mathbb{R}^n$ subgroup). Let $E$ be a vector subspace of $\mathbb{R}^n$ and consider $\pi$ to be orthogonal projection on $E$.
Apparently, it is well known that $\pi(L)$ need not be a lattice. I can think of examples of discrete subsets whose projection is not discrete (e.g $\{(n,1/n):n\in\mathbb{N}\}$). But not an example of a lattice.
Does someone know (or can find) an example?
P.S: For completeness, in the case where $L=span_\mathbb{Z}(b_1,\ldots,b_n)$ and $E=(b_i,\ldots,b_n)^\bot$ it is well known that $\pi(L)$ is a lattice since one can "lift" elements from $\pi(L)$ to $L$ and guarantee that the norm doesn't increase much).
Thanks to everyone for your help 🙂
Best Answer
For subspace $y = x \sqrt 2$ and lattice point $(m,n)$ the projected point is $$ \left( \; \; \frac {m+n \sqrt 2}{3} \; , \; \; \; \frac {m \sqrt 2+2n }{3} \; \; \right) $$
By carefully choosing $m,n$ we can get a point arbitrarily close to the origin