Can SO_n(R) be approximated arbitrarily well using a discrete subgroup? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:21:36Z http://mathoverflow.net/feeds/question/33176 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33176/can-so-nr-be-approximated-arbitrarily-well-using-a-discrete-subgroup Can SO_n(R) be approximated arbitrarily well using a discrete subgroup? Hari 2010-07-24T06:55:16Z 2010-07-24T08:06:23Z <p>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 $&lt; \epsilon$ in $G$?</p> http://mathoverflow.net/questions/33176/can-so-nr-be-approximated-arbitrarily-well-using-a-discrete-subgroup/33179#33179 Answer by Robin Chapman for Can SO_n(R) be approximated arbitrarily well using a discrete subgroup? Robin Chapman 2010-07-24T07:04:11Z 2010-07-24T07:06:41Z <p>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 $G$.</p> http://mathoverflow.net/questions/33176/can-so-nr-be-approximated-arbitrarily-well-using-a-discrete-subgroup/33182#33182 Answer by algori for Can SO_n(R) be approximated arbitrarily well using a discrete subgroup? algori 2010-07-24T07:45:18Z 2010-07-24T08:06:23Z <p>Generalizing Robin's answer to arbitrary $n$:</p> <p>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 $J(n)&lt;(n+1)!n^{b+a\log n}$. See <a href="http://www.pnas.org/content/81/16/5278.short?related-urls=yes&amp;legid=pnas;81/16/5278" rel="nofollow">http://www.pnas.org/content/81/16/5278.short?related-urls=yes&amp;legid=pnas;81/16/5278</a></p> <p>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)</p>