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I am trying to prove that every closed discrete subgroup of $\Bbb{R}^n$ under addition is a free abelian group of finite rank. I have tried to do this by induction on the dimension $n$.

Base case: We claim that a subgroup $H \subset \Bbb{R}$ has a least positive element $\alpha$. The completeness axiom tells us that there is a least positive real number $\alpha$ among all positive reals in $H$, and being closed implies that $\alpha \in H$.

Now assume the inductive hypothesis and consider a closed, discrete subgroup $H$ of $\Bbb{R}^n$. Choose some element $x \in H$ of least positive distance to the origin. This can be done because otherwise we get a contradiction like the above. Now let $L$ be the subspace spanned by $x$ and write $\Bbb{R}^n = L \oplus W$ for some complement $W$. Consider the projection (as a linear map) $p : \Bbb{R}^n \to W$. Now this is also a group homomorphism and so we have

$$\textrm{Im} (H) \subset W$$

being a subgroup and so by the induction hypothesis is isomorphic to $\Bbb{Z}^l$ for some $1 \leq l \leq n-1$. However from here I am having some trouble concluding that $H$ itself must be isomorphic to $\Bbb{Z}^l \oplus \Bbb{Z}x$. If I know the existence of a map $l : \textrm{Im}(H) \to H$ such that

$$p \circ l = \textrm{id}_{\textrm{Im}(H)}$$

then by the splitting lemma I can conclude my problem. However, I don't have such a map so how do I conclude the problem? Please do not post complete solutions.


Edit: Perhaps I should add some context. I am trying to conclude that the kernel of the exponential map $\exp : \mathfrak{g} \to G$ where $G$ is a connected abelian matrix Lie group is a free abelian group of finite rank. We know that the assumptions on our matrix Lie group mean that $\exp$ is a group homomorphism, so that $\ker \exp$ is a subgroup of the Lie algebra $\mathfrak{g}$.

Edit: The problem is reduced to showing that the image of $H$ under the projection is indeed discrete, because otherwise we cannot apply the inductive hypothesis.

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How did you conclude that $\mathrm{Im}(H)$ is a discrete subset of $W$? – Jyrki Lahtonen Aug 26 '12 at 6:20
@JyrkiLahtonen Hmmm perhaps you're right my idea will not work. – user38268 Aug 26 '12 at 6:21
It is easy to see that any $n+1$ elements of $H$ must be linearly dependent over the integers, for otherwise a large $n$-dimensional box of size $k\cdot N^n$ ($k$ constant, $N$ an integer parameter) would contain $N^{n+1}$ points of $H$, and letting $N\to \infty$ would then yield accumulation points. That may be another start... – Jyrki Lahtonen Aug 26 '12 at 6:31
@JyrkiLahtonen Are you saying I should abandon my approach here? – user38268 Aug 26 '12 at 6:39
Or yet in other words. Parallel translate that tube around. Where it goes, it will contain only the points of a single coset of $\mathbb{Z}x$. – Jyrki Lahtonen Aug 26 '12 at 12:45

2 Answers 2

up vote 4 down vote accepted

I fully endorse Alex Becker's answer. I just want to address the question of discreteness of $p(H)$ in the original post.

Let $W$ be the orthogonal complement to $L$, so $p$ is the orthogonal projection onto $W$.

Lemma. For all elements $y\in \left(H\setminus\mathbb{Z}x\right)$ we have $||p(y)||\ge ||x||/2.$

Proof. Let $y\in \left(H\setminus\mathbb{Z}x\right)$ be arbitrary. Let us write $$ y=rx+w, $$ where $r\in\mathbb{R}$ and $w\in W$. Because $x$ generates $L\cap H$, we must have $y\notin L$, so $w\neq0$. There exists an integer $m$ such that $|m-r|\le1/2$. Consider the vector $y'=y-mx=(r-m)x+w\in H$. Because $y\notin L$, $y'\neq0$. Therefore $||y'||\ge ||x||$ as $x$ was selected to be (one of) the shortest non-zero vectors in $H$. By construction $||(r-m)x||\le ||x||/2$. By triangle inequality $$ ||p(y)||=||w||=||y'-(r-m)x||\ge ||y'||-||(r-m)x||\ge ||x||-\frac{||x||}2=\frac{||x||}2. $$ Q.E.D.

The discreteness of the image follows immediately from the Lemma. If $p(y)$ and $p(y')$ are two distinct points in the group $p(H)$, then $y-y'\notin\mathbb{Z}x$, and therefore $$ d(p(y),p(y'))=||p(y)-p(y')||=||p(y-y')||\ge\frac{||x||}2. $$

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The general principle in the last step is that a topological group is homogeneous. The translation by a fixed element is a homeomorphism, and therefore we have a group of homeomorphisms acting transitively on the group. Consequently the topology about each point "is similar" (a suitable translation gives a bijective correspondence between neighborhoods of any point and the neighborhoods of the neutral element). – Jyrki Lahtonen Aug 26 '12 at 12:20

Let $H\subset \mathbb R^n$ be such a subgroup. Note that no element of $H$ has finite order. Suppose that there exist elements $x_1,\ldots,x_{n+1}\in H$ which are linearly independent over $\mathbb Z$. They are also linearly independent over $\mathbb Q$ as clearing denominators shows, and over $\mathbb R$ since any real dependence relation can be approximated by rational ones, making $0$ an accumulation point. Then $H/\mathbb Zx_1\oplus\cdots \oplus \mathbb Zx_n$ is a subgroup of the $n$-torus $\mathbb R/\mathbb Zx_1\oplus\cdots \oplus \mathbb Zx_n$, and the image $\overline{x_{n+1}}$ of $x_{n+1}$ under the quotient map has infinite order. By compactness of the $n$-torus, the set $\mathbb Z\overline{x_{n+1}}$ has an accumulation point. Thus the set $\mathbb Zx_{n+1}$ comes arbitrarily close to the lattice $\mathbb Zx_1\oplus\cdots \oplus \mathbb Zx_n$, so $0$ is an accumulation point of $H$, a contradiction. Thus $H$ is generated by at most $n$ elements. The conclusion follows by applying the fundamental theorem of finitely generated abelian groups.

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Thanks for your answer. Do you think there is a way to salvage my approach above? – user38268 Aug 26 '12 at 7:25
@BenjaLim I'm not sure. I just saw this approach and ran with it, since I really really like tori. – Alex Becker Aug 26 '12 at 7:26
@BenjaLim I re-read your post and noticed you asked us not to post complete solutions. Sorry! I missed that the first time around. I suppose the damage is done now. – Alex Becker Aug 26 '12 at 7:46
It's ok as I don't think I can use your approach above anyway. I am in the process of trying to show that the image of $H$ under the projection is a discrete space, because from there I think my problem will follow. – user38268 Aug 26 '12 at 7:48
@BenjaLim I think sentences 3 and 4 in my answer should help with showing that. – Alex Becker Aug 26 '12 at 8:02

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