Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G := SO_n(R)$ be equipped with the Euclidean metric on vectors of length $n^2$. Is it true that for any $\epsilon >0$, there is a finite subgroup of $G$ which intersects every metric ball of radius $< \epsilon$ in $G$?

share|improve this question
1  
Just out of curiosity, what does the "Lie" tag mean, that the tags lie-groups, lie-algebras etc. don't? –  Joel Fine Jul 24 '10 at 7:06
    
@Joel: That tag is essentially superfluous; but would keep popping up regardless. This time I have fixed it. –  Anweshi Jul 24 '10 at 7:14
1  
The title says "discrete subgroup" and the question says "finite subgroup". Which one do you mean? –  Bruce Westbury Jul 24 '10 at 7:50
1  
@Bruce: SO_n(R) is compact, so it doesn't matter, does it? –  Qiaochu Yuan Jul 24 '10 at 7:52
    
@Qiaochu, Good point. –  Bruce Westbury Jul 24 '10 at 9:24
add comment

2 Answers

up vote 10 down vote accepted

Generalizing Robin's answer to arbitrary $n$:

Jordan's theorem implies that for any $n$ there is an integer $J(n)$ such that the index of a normal abelian subgroup of a finite subgroup of $GL(n,\mathbf{C})$ and hence $SO(n)$ is $\leq J(n)$. A theorem by Boris Weisfeiler (based on the classification of finite simple groups) implies that there are real $a,b$ such that $J(n)<(n+1)!n^{b+a\log n}$. See http://www.pnas.org/content/81/16/5278.short?related-urls=yes&legid=pnas;81/16/5278

So any finite subgroup of $SO(n)$ is included in $\leq J(n)$ copies of the maximal torus and so if one takes $n\geq 3$ and small enough $\varepsilon>0$, then for any finite subgroup $G$ of $SO(n)$ there will there elements $\varepsilon$ or further away from $G$ (with respect to any say left-invariant metric)

share|improve this answer
    
Thank you very much. I guess I should have checked the wiki for the case SO_3 before asking :). Are there are any Lie groups apart from tori where this is true? –  Hari Jul 25 '10 at 2:34
1  
Hari -- welcome. I think the same argument shows that there aren't any connected examples apart form tori (in characteristic 0). –  algori Jul 25 '10 at 2:59
add comment

The only finite subgroups of $G=SO_3(\mathbf{R})$ are cyclic, dihedral or of order 12, 24 or 60. The latter three can't work for small enough $\epsilon$ but neither can the cyclic or dihedral groups as each one of the these lie inside a $1$-dimensional Lie subgroup of $G$ which is not dense in $G$.

share|improve this answer
    
Thank you very much. –  Hari Jul 25 '10 at 5:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.