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can any subgroup of the unitary group of full matrix alg $M_d(\mathbb{C})$ be approximated on finite sets by a finite subgroup?

i.e. is the following True or false? Let $n, d$ be positive integers and let $u_1,..., u_n$ be in the unitary group $U_d=U (M_d(\mathbb{C}))$ of $M_d(\mathbb{C})$. Then for every $\epsilon > 0$ there are $v_1, ..., v_n$ in $U_d$ such that $\| u_k - v_k \| < \epsilon$ for $k = 1, .., n$ and such that the subgroup of $U_d$ that $v_1, ..., v_n$ generate is finite.

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can you write more details, what kind of approximation do you mean? – Kate Juschenko Mar 14 '11 at 17:23
regard M_n(C) as B(C^n) the bounde operators on the f.d. Hilbert space C^n – Paulo Mar 14 '11 at 18:15
Any such group is residually finite... But it sounds more like you're asking whether any subgroup is a Gromov--Hausdorff limit of finite subgroups. I doubt this is true - for instance, it's not true in SO(3), where the only infinite families of subgroups are cyclic and dihedral. – HJRW Mar 14 '11 at 18:17
I think you should edit your question to: (a) format it correctly, with words spelled in full and sentences capitalized; and (b) fully explain every definition you are using. By asking a question here, you are asking other people to do work for you. Phrasing your question clearly and correctly shows that you respect this fact. – HJRW Mar 14 '11 at 18:20
the following question is related:… – Kate Juschenko Mar 14 '11 at 18:34
up vote 12 down vote accepted

no, it is not true. the following is contained in Andreas Thom question.

from the first paragraph of his question:

Let $n$ be an integer. Camille Jordan showed that there exists some $m \in > {\mathbb N}$ (depending on $n$), such that for any pair of $n \times > n$-unitaries $u,v \in U(n)$ which generate a finite group, one has $[u^m,v^m] = 1_n$.

Take $u_1,u_2\in{U}(n)$ that generate a free group (easy to construct for $n\geq{2}$), and let $m$ be as above. Then, since $v_1^m,v_2^m$ commute, $$ \|u_1^mu_2^m-u_2^mu_1^m\|\leq{}2\|u_1^m-v_1^m\|+2\|u_2^m-v_2^m\|\leq{}4m\varepsilon$$

Since $\epsilon$ is arbitrary we have a contradiction.

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