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It is well known that (additive) subgroups of $\mathbb{R}^n$ are products of discrete subgroups (lattices) by dense subgroups in subspaces. My question is the following: given a generator set of $p$ vectors (obviously $p>n$), can one tell (i) whether there will be dense factors, if so (ii) what their dimensions will be (i.e. the dimension of the subspace it is dense in) and (iii) give a basis of these subsets in terms of the generator set. This answer is probably more algorithmic than explicit, but I am looking at specific examples in $\mathbb{R}^3$ (where the generators are two orthonormal frames), so hopefully one might say something conclusive there.

To clarify what I mean by an algorithm (thanks @Ycor), let me quote the (trivial) example of $n=1$, $p=2$. My "algorithm" would go as follows: are the two generators commensurable? If not, the subgroup is dense. Similarly, if all vectors in the generator set have rationals coordinates, then I know that the subgroup is a lattice. I do not wish for a computer automated procedure (otherwise, one would go into the problems of representing numbers, etc. which is not my goal). I want either a direct criterion, or a deciding procedure, in other words an algorithm.

Any reference will be welcome.

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    $\begingroup$ Even for $n=1$ and $p=2$, you cannot answer in a simple way because otherwise you would be able to determine whether a given number is irrational. Of course it depends on what you mean by "given a set" and "one can tell". If you ask at an algorithmic level, you have to wonder how you make the input. $\endgroup$
    – YCor
    Jan 7, 2016 at 10:01
  • $\begingroup$ Point taken. I wasn’t clear enough on what I meant by algorithm. I have corrected it in the question. $\endgroup$
    – Pascal Romon
    Jan 7, 2016 at 13:32
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    $\begingroup$ So let's say we set $n=1$ and $p=2$, and I give you $a=\pi$ and $b=e$. What does your "algorithm" give then? $\endgroup$
    – eric
    Jan 7, 2016 at 14:05
  • $\begingroup$ Algorithm is for me the same as computer automated procedure, except maybe that we describe its mathematical contents rather than the implementation. But the coding problem is essential then. Or you want a criterion, not an algorithm. $\endgroup$
    – YCor
    Jan 7, 2016 at 15:20
  • $\begingroup$ It's unknown whether $\pi/e$ is irrational? (I remember that it's unknown whether $(1,e,\pi)$ is $\mathbf{Q}$-linearly free but this is a bit stronger.) $\endgroup$
    – YCor
    Jan 7, 2016 at 15:35

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