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.



Subgroups of U(M_n)
no, it is not true. the following is contained in Andreas Thom [question](https://mathoverflow.net/questions/34625/finite-subgroups-of-unitary-groups). 


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.


