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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.

Thanks.

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 $W$?2012-08-26
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    @JyrkiLahtonen Hmmm perhaps you're right my idea will not work.2012-08-26
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    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...2012-08-26
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    @JyrkiLahtonen Are you saying I should abandon my approach here?2012-08-26
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    But if $\mathrm{Im}(H)$ is a finitely generated free abelian group, then the existence of a splitting homomorphism is a gimme. Just select the images for a set of generators, and extend linearly. In other words, a free abelian group is a projective $\mathbb{Z}$-module, and if a module has a projective module as a quotient, the said module is also a direct summand.2012-08-26
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    (cont'd) So if you can show that $p(H)$ is discrete, then your argument will run merrily to a positive conclusion.2012-08-26
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    @JyrkiLahtonen What if I choose $W$ to be the orthogonal complement of $L$, will that help?2012-08-26
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    Then you must show that the lengths of projections of non-zero elements of $H$ in $w$ are bounded away from zero. Or in other words, elements of $H$ cannot be arbitrarily close to any point of the line $L$ (not just $L\cap H$).2012-08-26
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    @JyrkiLahtonen I'm wondering if this will help me: Choose some $x \in \textrm{Im}(p)$ so write $x = p(y)$ for some $y \in H$. Now I claim that there is a ball $B$ such that $B \cap \textrm{Im}(p) = \{x\}$. Now suppose that $||x-y|| = d$. Then we have that the ball of radius $d + 1$ about $y$ will contain $x$. Now I believe it is possible to prove that such a ball contains only finitely many points of $H$ and so we can choose a ball $B$ about $x$ ***in the image of $p$*** such that it contains no points of $p(H)$.2012-08-26
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    You have to somehow address the potential problem that if $U$ is a neighborhood of $p(y)$ in $W$, then $p^{-1}(U)$ contains arbitrarily long vectors, and infinitely many points from $H$. The trick is to show (somehow) that they can have only finitely many distinct projections. The common theme with my question is that distant points may have nearby projections. As a lithmus test to your arguments do the following: If the projection mapping does not have an infinite kernel in $H$, the image $p(H)$ may be dense in $W$ (see the image on my question for an example).2012-08-26
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    What I was trying to say is the following. When you consider using an approach to proving it this way (completing your proof by showing that the image under projection is discrete), do the following check in advance: The approach must fail, if the kernel of $p$ is trivial, because we then have counterexamples. So I am asking you: *How did you plan to use the kernel together with the ball of radius $d+1$?*2012-08-26
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    @JyrkiLahtonen I am not too sure I about your last question but I do see the problem that I need to prove. For example, suppose I am projecting onto the $x-y$ plane in $\Bbb{R}^3$. Then suppose $p(U)$ is some circle in the $x-y$ plane. For each point in $H$ above such a circle, draw a verticle line down onto the circle. I now need to check that there are only ***finitely*** many distinct such lines yes?2012-08-26
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    @JyrkiLahtonen I don't understand how because $||p(y)|| \geq ||x||/2$, that the image of $H$ under the projection is discrete. What I mean is, I understand there is a "cylinder" about the line $L$ such that no point of $H\setminus \Bbb{Z}\alpha$ is mapped under $p$ into that cylinder. However does it tell me the existence of an open set $U$ such that $U \cap \textrm{Im}(H) = \{x\}$ given any $x \in \textrm{Im}(H)$?2012-08-26
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    If a single point (here the origin) of **an additive group** is an isolated point, so is every other: $$d(p(y_1),p(y_2))=||p(y_1)-p(y_2)||=||p(y_1-y_2)||$$ must be either zero (when $y_1$ and $y_2$ are in the same coset of $\mathbb{Z}x$) or at least $||x||/2$ (by the lemma in my answer).2012-08-26
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    (cont'd) Or if $x\in p(H)$ is a fixed element, then the mapping $f:y\mapsto y+x$ is a homeomorphism from $p(H)$ to itself. But $f$ maps the isolated point $0$ to the point $x$ that must then also be isolated. I added a bit about this into my answer.2012-08-26
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    Or yet in other words. Parallel translate that tube around. Where it goes, it will contain only the points of a single coset of $\mathbb{Z}x$.2012-08-26
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    @JyrkiLahtonen Can I ask you something? I have added above that my problem is actually something specific, the discrete subgroup I am dealing with is actually the kernel of the exponential map that is also a local homeomorphism. Now can I use $\exp$ being a local homeomorphism to prove that the image of $H$ under $p$ is a discrete space? $H$ is now the kernel of the exponential map.2012-08-27
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    I don't see a way of doing that. That propety of exp was already used in proving that $H$ itself is discrete, no? Consider the following example. Let $H$ be the subgroup of the $xy$-plane generated by $(1,0)$ and $(0,\sqrt2)$. $H$ is the kernel of the Lie homomorphism $$(x,y)\mapsto(e^{2\pi ix},e^{\sqrt2\pi iy}).$$ If $p$ is the orthogonal projection to the line $L:x+y=0$, then $p(H)$ is dense in $L$. What I want to stress is that there is no general principle allowing you to get the discreteness of $p(H)$ free of charge.2012-08-27
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    (cont'd) The fact that kernel of $p$ contains a non-trivial subgroup of $H$ has to be used at some point. There is nothing about $H$ alone that would allow us to make this deduction. $p$ has to be tied to $H$ for the claim to hold. I don't think that it is necessary to select $x$ to be a shortest vector (see Alex's answer). But it is necessary for my argument to work. That's why I like Alex's approach better myself.2012-08-27

2 Answers 2

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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).2012-08-26
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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?2012-08-26
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    @BenjaLim I'm not sure. I just saw this approach and ran with it, since I really *really* like tori.2012-08-26
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    @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.2012-08-26
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    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.2012-08-26
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    @BenjaLim I think sentences 3 and 4 in my answer should help with showing that.2012-08-26
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    +1 Using a known compact space is a nice substitute for a counting argument :-)2012-08-26
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    @AlexBecker I could only accept one answer and I accepted Jyrki's because I felt it was directly related to what I was stuck in. However thanks a lot for your answer and I learned from it.2012-08-27
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    @BenjaLim Naturally you accepted his, as it better answered your question. Thanks for your kind words though.2012-08-27