# Lattice definition as finite density infinite set of vectors closed under addition?

This question came up in the definition of lattices (in the crystallography/group theory sense, not the ordered set sense) in our condensed matter lectures, but I believe it's more appropriate here than on the Physics StackExchange; please move it if I'm wrong.

Suppose we have an infinite set of vectors $V$ in real space $\mathbb{R}^n$ which we interpret as the position vectors of a set of points $S$. If we may write $V$ in the form $\{ \sum_{i=1}^n c_i \mathbf{e}_i : c_i \in\Bbb{Z}\}$ where the $\{\mathbf{e_i}\}$ are a set of $b \le n$ linearly independent vectors then we call $S$ a lattice. Is there a set $V$ which is closed under addition such that $S$ is not a lattice but has finite density of points (i.e. a finite number of points in any given finite volume)?

An example of what I mean in case I am still not formulating the problem correctly: The set of integers $\mathbb{Z}$ in $\mathbb{R}^1$ is a lattice in that we may find a basis vector $(1)$ for which all the vectors in $V$ may be written as integer multiples and further all integer sums of the basis $\{(1)\}$ appear in $V$. On the other hand the set $B=\{m+n\sqrt2 : m,n \in\mathbb{Z}\}$ is not a lattice because one needs two linearly dependent vectors, e.g $(1)$ and $(\sqrt 2)$ in order to write all the vectors in $B$ as an integer sum of basis vectors. $B$ is also closed under addition. However there are an infinite number of points in $B$ in the interval $(0,1)$ say, so it has infinite density. I would like to find a set which is like $B$ in that it satisfies the first two properties listed but has finite density.

• The relation between $S$ and $V$ seems a bit unclear. Are $V$ all pairwise differences between points in $S$? If so, then $v\in V$ and $s\in S$ does not imply $s+v\in S$.
– MvG
Feb 12, 2014 at 7:12
• Yes, the integer multiples of a single nonzero vector in $\mathbb R^2$ is closed under addition and has "finite density", but is not an $2$-dimensional lattice. Feb 13, 2014 at 15:38
• Oh, yes, you're quite right @MvG ! It probably makes more sense to start from $V$ to construct $S$; I've tried to edit the question appropriately. Hopefully it now makes more sense; I'm sorry if it's still not right. Feb 13, 2014 at 15:38
• @Samuel, that's certainly true! However in some not-very-mathematical sense it's basically a 1-dimensional lattice embedded in $\mathbb{R}^2$, which isn't quite what I'm looking for. I'll add the caveat that the set can also have less than $n$ basis vectors into the question, sorry for not being precise and keep having to change the question. Also sorry about "finite density"; I know it's not right but don't know the correct terminology! Feb 13, 2014 at 15:48

## 1 Answer

Another example is the set $\{n: n\geq 0\}$, which is closed under addition but is not a lattice. I believe that is not a satisfactory answer either, so I think what you are looking for is Theorem 17 from Lectures on the Geometry of Numbers by Carl Ludwig Siegel:

Theorem: A non-empty set which is closed under subtraction and which does not contain arbitrarily short vectors, can be written as $V=\{\sum_1^k c_i e_i, c_i\in\mathbb Z\}$ for some finite set of vectors $e_1,\ldots,e_k$. (That we can choose $k\leq n$ is guaranteed by Theorem 18 in the same book.)

Such a set will have "finite density", for if there were an infinite number of points inside a small ball, you could for any $\epsilon>0$ find two vectors, say $x$ and $y$, at a distance at most $\epsilon>0$ from each other, and then $x-y$ would be a vector in your set with length $\epsilon$.

• You might want to dd an argument as to why $k\le n$, since by the edited definition, $k>n$ would not be a lattice in that definition, but still satisfy the above theorem.
– MvG
Feb 13, 2014 at 16:31
• @MvG: Fair enough, added a reference to a theorem which guarantees that. Feb 13, 2014 at 16:44
• @Samuel Sorry again about the terminology. Would good practice be for me to edit the q again to read 'closure under subtraction' or should I just leave it? Anyway, I'll track down those Siegel Lectures, thanks. Feb 13, 2014 at 17:51
• @maxwelldecoherence: If you change your question, the first sentence in my answer will not make sense. It's fine as it is now. The book is on Google Books. Search for "Theorem 17" and you'll get the proof. Feb 14, 2014 at 22:46