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Does anybody have a good reference to classify all the subgroups of the additive group $\mathbb{R}^d$? In fact, I am looking for a probably simple, if it exists, classification result stating something like: "any subgroup of $\mathbb{R}^d$ is either discrete or is a product of discrete or dense subgroups in subspaces" i.e. mimicking the classical result in $\mathbb{R}$.

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    $\begingroup$ I am no expert, but I think this is hopeless. People are always on a hunt for nice lattices (for sphere packings etc.), and it would be much easier if they could be listed that easily. $\endgroup$
    – M. Winter
    Commented Jul 7, 2019 at 22:02
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    $\begingroup$ A cleaner statement to aim for is to ask what the closed subgroups of $\mathbf R^n$ look like up to isomorphism (as a topological group). $\endgroup$
    – KConrad
    Commented Jul 7, 2019 at 22:40
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    $\begingroup$ In fact, the closed subgroups of $\Bbb{R}^n$ are easily classified, see for instance Bourbaki's General Topology VII, §1, Theorem 2. $\endgroup$
    – abx
    Commented Jul 8, 2019 at 4:53
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    $\begingroup$ Abstractly (and ignoring topology), the groups $\mathbb{R}$ and $\mathbb{R}^n$ are isomorphic: both are $\mathbb{Q}$-vector spaces of dimension $2^{\aleph_0}$. A group is isomorphic to a subgroup of such iff it is torsionfree and of rank bounded by $2^{\aleph_0}$. But given what you are looking for, KConrad's suggestion makes a lot of sense. $\endgroup$
    – Todd Trimble
    Commented Jul 8, 2019 at 6:06
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    $\begingroup$ @KConrad we have a classification modulo conjugation by $\mathrm{GL}_n(\mathbf{R})$, which is finer than a classification as topological groups, yet coincides here. Since it has not been said explicitly every closed subgroup is in the $\mathrm{GL}_n(\mathbf{R})$-orbit of $\mathbf{R}^a\times\mathbf{Z}^b\times\{0\}^{n-a-b}$ for a unique pair $(a,b)$ with $0\le a+b\le n$. It's not hard to prove (after passing to the spanned subspace and modding out the zero component, it boils down to proving that a discrete subgroup of $\mathbf{R}^n$ spanning the whole space is in the orbit of $\mathbf{Z}^n$). $\endgroup$
    – YCor
    Commented Jul 8, 2019 at 6:54

1 Answer 1

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Let $G\subset \mathbf{R}^n$ be subgroup of $\mathbf{R}^n$, $rank(G)$ is defined to be maximum number of linearly independent vector in $G$.
Local rank of $G$ defined as follows
Let $r(\epsilon)$ be the maximum linearly independent vector in $G$ of length less then $\epsilon$, As $\epsilon \rightarrow 0$ say $r(\epsilon) \rightarrow s$ then $s$ is called $local\; rank\; of \; G$. Since $r(\epsilon)$ is integer valued, $s$ is attained for some $\epsilon>0$.
Let $s$ be local rank of group, if $s$ is zero then $G$ is discrete,hence $G$ is $lattice$. Suppose $s> 0$, then their exists $\epsilon > 0$ such that $r(\epsilon)=s$ and we can find vectors $ y^{(1)},y^{(2)},...,y^{(s)}$ in $G$ of length less then $\epsilon$. Let $E=Span \{ y^{(1)},y^{(2)},...,y^{(s)}\}$ and $D=E\cap G$. Then $E$ is independent of the choices of vector $y^{(1)},y^{(2)},...,y^{(s)}$ and $D$ is dense in $E$ (see siegel's geometry of numbers, lecture VI. In this lecture VI, proof of theorem 21 is not correct. Theorem 22 can be proved without assuming theorem 21. In fact in following proof we will prove theorem 21 using theorem 22 ).
Choose vector $x^{(1)},x^{(1)},...,x^{(q)}$ in $G$ such that they are linearly independent and linearly independent from $y^{(1)},y^{(2)},...,y^{(s)}$, $(s+q=rank(G))$.
Now given $x\in G$ we can write $$ x=\sum_{j=1}^{s} \nu_j y^{(j)} +\sum_{i=1}^{q} \mu_i x^{(i)}$$ then define $$L= \{ \sum_{i=1}^{q} \mu_i x^{(i)}| \exists x\in G \; s.t \; x=\sum_{j=1}^{s} \nu_j y^{(j)} +\sum_{i=1}^{q} \mu_i x^{(i)} \} $$ $L$ is projection of $G$ onto space spanned by $x^{(1)},x^{(1)},...,x^{(q)}$. Then $L$ is lattice(see siegel's geometry of numbers, lecture VI)

$\textbf{Theorem}$ Let $G\leq \mathbf{R}^n$ be closed subgroup, then $G=E\oplus L$ where $E$ and $L$ defined above
$\textbf{prrof}$ Since $G$ is closed $D=E$, hence $E\subset G$.
By definition of $L$ and $E \subset G$ it implies $L\subset G$, thus theorem will follow.

$\textbf{Theorem}$ Let $G\leq \mathbf{R}^n$ then $G$ is direct sum of $D$ (defined as above, is the dense set in $E$) and for some lattice $M$.

$\textbf{proof}$ $\bar{G}=E\oplus L$ from theorem 1, Let $E=Span \{ y^{(1)},y^{(2)},...,y^{(s)}\}$, and lattice is generated by $x^{(1)},x^{(1)},...,x^{(q)}$.
Define $G_i=\{ a+x^{(i)} | a \in E \} \subseteq \bar{G}$
$\textbf{Claim 1}$ $G_i\cap G \neq \emptyset$
Suppose $G_i\cap G = \emptyset$. Since $G_i\cap \bar{G}= G_i$, Given $a+x^{(i)}\in G_i$ we get sequence $\{w_k \}_{k\in \mathbf{N}}\subset G$ such that $w_k$ converges to $a+x^{(i)}$.
Then $x^{(i)}$'s coordinate in $w_k$ converges to 1 and all other $x^{(j)}$'s coordinate in $w_k$ converges to 0, since $x^{(j)}$'s coordinates in $w_k$ are all integers they are eventually $0$ if $j\neq i$ and $1$ if $j=i$
Then their exists $N\in \mathbf{N}$ such that $w_k\in G_i\; \forall k>N$, which shows that for $k>N$ we get $w_k \in G_i\cap G$ (contradiction).

let $z^{(i)} \in G_i\cap G$ for $i=1,2,...,q$ write $z^{(i)}=a_i+x^{(i)}$ where $a_i \in E$ and $M=span\{z^{(1)},z^{(2)},...,z^{(q)}\}\cap G$($M$ is lattice since it can't have vectors of arbitrary small length apart from 0 )
Note $span\{y^{(1)},y^{(2)},...,y^{(s)},x^{(1)},x^{(1)},...,x^{(q)}\} =span\{y^{(1)},y^{(2)},...,y^{(s)},z^{(1)},z^{(2)},...,z^{(q)} \}$
as $z^{(i)}=a_i+x^{(i)}\in LHS $ and $x^{(i)}= z^{(i)}-a_i\in RHS $
So $\bar{G}\subseteq span\{y^{(1)},y^{(2)},...,y^{(s)},x^{(1)},x^{(1)},...,x^{(q)}\} =span\{y^{(1)},y^{(2)},...,y^{(s)},z^{(1)},z^{(2)},...,z^{(q)} \}$ Given $x\in G$ it also in $\bar{G}$, so we can write $x=\sum \mu_i y^{(i)}+\sum \nu_i z^{(i)} $
$\textbf{Claim 2}$ $\nu_i \in \mathbf{Z}$ for $i=1,2,...,q$
sine $z^{(i)}=a_i+x^{(i)}$ where $a_i \in E$ so replacing $z^{(i)}$ by $a_i+x^{(i)}$ we get $$x=\sum \mu_i y^{(i)}+\sum \nu_i z^{(i)}=\sum \mu_i y^{(i)}+\sum \nu_i (a_i+x^{(i)}) $$ as $a_i \in E=span \{y^{(1)},y^{(2)},...,y^{(s)} \}$ we can write above expression as
$x=\sum \lambda_i y^{(i)}+\sum \nu_i x^{(i)}\in \bar{G}=E\oplus L$ for some$\lambda_i \in \mathbf{R}$
which implies $\sum \nu_i x^{(i)} \in L$, which in turn implies the claim.
Claim2 implies $M\subset G$,thus $D\oplus M \subset G$. Claim 2 shows that given $x \in G$ we can write $x=\sum \mu_i y^{(i)}+\sum \nu_i z^{(i)} $, where $\nu_i \in \mathbf{Z}$ for $i=1,2,...,q$. Thus $\sum \mu_i y^{(i)}=x-\sum \nu_i z^{(i)} \in G$ and $\sum \mu_i y^{(i)} \in E$, It implies $\sum \mu_i y^{(i)} \in G\cap E= D$

Hence $G=D\oplus M$ where $D$ is dense in $E$ and $M$ is lattice

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