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 :)

  • 1,426
  • 2
  • 6
  • 18
  • A rather famous example from physics is the concept of a quasicrystal (ordered but not periodic structure in some $n$-dimensional space). [Read this Wikipedia section:](https://en.wikipedia.org/wiki/Quasicrystal#Mathematics) "Bohr showed that quasiperiodic functions arise as restrictions of high-dimensional periodic functions to an irrational slice (an intersection with one or more hyperplanes), and discussed their Fourier point spectrum. These functions are not exactly periodic, but they are arbitrarily close in some sense, as well as being a projection of an exactly periodic function. ..." – Andreas Nov 30 '21 at 21:18
  • @Andreas. Nice, is there an example with an explicit basis? – miraunpajaro Nov 30 '21 at 21:21
  • @miraunpajaro Check these sources: [Quasicrystals: Projections of 5-d Lattice into 2 and 3 Dimensions](https://www.researchgate.net/publication/2093388_Quasicrystals_Projections_of_5-d_Lattice_into_2_and_3_Dimensions) and [Constructing quasicrystalline lattices](https://arxiv.org/ftp/arxiv/papers/1104/1104.1143.pdf). You can also google "cut and project method" to get many more examples. – Andreas Nov 30 '21 at 21:40
  • Aren't there much simpler examples? E.g., take $L$ to be the integer lattice in $\Bbb{R}$ and $E$ to be the span of a point like $(1, \sqrt{2})$ that is not in $\Bbb{Q}^2$? – Rob Arthan Nov 30 '21 at 22:55
  • @Rob Artha Yes, a classmate pointed this out to me, you can write this as a solution if you want. Relevant to this is: https://math.stackexchange.com/questions/73262/proving-that-mn-sqrt2-is-dense-in-mathbb-r#73268 – miraunpajaro Nov 30 '21 at 23:03

2 Answers2


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

Will Jagy
  • 130,445
  • 7
  • 132
  • 248

This answer expands on my comment and on this question Proving that $m+n\sqrt{2}$ is dense in $\mathbb R$ first.

If $\xi$ is an irrational real, then the set $G$ of all $m + n\xi$, where $m, n \in \Bbb{Z}$ cannot be a lattice in $\Bbb{R}$. (To see this, assume that $G$ is a lattice, so that it has least positive element, say $L = m_1 + n_1\xi$ for some $m_1,n_1 \in \Bbb{Z}$. But then as $\xi$ is irrational, we cam approximate $\xi$ by a rational $a/b$ (with $a, b \in \Bbb{Z}$), such that $0 < a/b - \xi < L$. But that gives us $0 < a - b \xi < L$, contradicting our assumption that $L$ is the least element of $G$.)

Now, returning to the present question. If we take $L = \Bbb{Z}^2$ and $E$ to be the span of $(1, \xi)$ where $\xi$ is any irrational number, then the orthogonal projection of $L$ onto $E$ is the set of all points of the form:

$$\frac{((m, n) \cdot (1, \xi))}{\|(1, \xi)\|^2} (1, \xi) = \frac{(m + n\xi)}{1 + \xi^2}(1, \xi)$$ where $m, n \in \Bbb{Z}$. By the previous paragraph (identifying $\Bbb{R}$ and the span of $(1, \xi)$ in $\Bbb{R}^2$, by the linear mapping that maps $1$ to $ \frac{(1, \xi)}{\sqrt{1 + \xi^2}}$, the orthogonal projection of $L$ onto $E$ is not a lattice.

Rob Arthan
  • 42,571
  • 4
  • 38
  • 84