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$L\subset \mathbb{R}^{n}$ lattice of max. rank

$\Leftrightarrow \mathbb{R}^{n}/L$ compact in quotient topology

$\Leftrightarrow \exists$ bounded subset $B\subset\mathbb{R}^{n}$ s.t. $L+B = \mathbb{R}^{n}$

I just managed to do the first implication:

$L = \mathbb{Z} \omega_{1} + ... + \mathbb{Z} \omega_{n}$.
$P:=\{\lambda_{1}\omega_{1} + ...+ \lambda_{n}\omega_{n} | 0 \leq \lambda_{i} \leq 1\}.$
Then there exists an equivalence relation on $\mathbb{R}^{n}$, s.t. two numbers $x,x'$ are equivalent modulo $L$ $\Leftrightarrow$ $x-x'\in L$. $\pi:\mathbb{R}^{n}\rightarrow\mathbb{R}^n/L$ with $x\mapsto x+L$. We can find for all $x\in\mathbb{R}^{n}$ a $x'\in P$ s.t. $x-x'\in L$.
$\Rightarrow \pi(x)=\pi(x')$ $\Rightarrow \pi\big|_{P} \rightarrow \mathbb{R}^n$/L bijective.
Hence $\pi(P)=\mathbb{R}^n/L$ and since $P$ is compact, so is $\mathbb{R}^n/L$

For the next implication I have no idea yet.
I guess I have to use that $\mathbb{R}^n = \bigcup\limits_{i=1}^n U_{i}$, but I don't know how to get to $L+B = \mathbb{R}^{n}$. Maybe I must also use that $L$ is discrete? But do I only know that from the lattice, or does my proof for the first implication show that as well?

Hope someone feels like helping :)

best, Kathrin

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2 Answers 2

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(ii) $\to$ (iii): Let $B_\epsilon(x)$ denote the open $\epsilon$-ball around $x$. Denote by $\pi \colon \mathbb R^n\to \mathbb R^n/L$ the projection. As $\mathbb R^n/L$ is compact, and $\pi$ is open, there is a finite set $I \subseteq\mathbb R^n$ such that $\mathbb R^n \subseteq \bigcup_{i\in I} B_\epsilon(x_i)$, taking preimages under $\pi$, we get $$ \mathbb R^n \subseteq \bigcup_{i\in I} B_\epsilon(x_i) + L $$ and $B := \bigcup_{i\in I} B_\epsilon(x_i)$ is bounded.

(iii) $\to$ (i): If $L$ had not full rank, there were an $x \in \mathbb R^n\setminus \{0\}$ such that $x \perp L$ (we may assume that $0 \in L$). So foreach $\lambda \in \mathbb R$ and each $l \in L$: $$ \|\lambda x - l\| = \bigl(\lambda^2 \|x\|^2 + \|l\|^2\bigr)^{1/2} \ge |\lambda| \|x\| $$ If $B$ is bounded by $M$ say, choose $\lambda$ so large that $\lambda\|x\| > M$, then $$ \|\lambda x - (l+b)\| \ge \|\lambda|\|x\| - M > 0 $$ for each $l \in L$, $b \in B$. So $\lambda x \not\in L+B$.

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Thanks a lot!! :) –  Kathrin Nov 22 '12 at 1:21

HINT: Although it’s efficient to prove a ring of implications, in this case that may not be the approach that offers the most insight into what’s really going on.

After proving $(i)\to(ii)$, you might consider proving $(ii)\to(i)$. If $L$ is generated by $\omega_1,\dots,\omega_m$ with $m<n$, choose $\omega\in\Bbb R^n\setminus\operatorname{span}\{\omega_1,\dots,\omega_m\}$, and show that $\pi\upharpoonright \Bbb R\omega$ is injective and hence that $R/L$ is not compact. For $(i)\to(iii)$ just observe that $\Bbb R^n=L+P$. Finally, the same idea that I suggested for $(ii)\to(i)$ can be used to show $(iii)\to(i)$ (or rather, to be precise, the contrapositive $\lnot(i)\to\lnot(iii)$).

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ah, nice. Thank you as well!! I'll have a look at this! –  Kathrin Nov 22 '12 at 1:23
    
@Kathrin: You’re welcome. –  Brian M. Scott Nov 22 '12 at 1:26

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