4
$\begingroup$

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.

$\endgroup$
8
  • 1
    $\begingroup$ can you write more details, what kind of approximation do you mean? $\endgroup$ Mar 14, 2011 at 17:23
  • $\begingroup$ regard M_n(C) as B(C^n) the bounde operators on the f.d. Hilbert space C^n $\endgroup$
    – Paulo
    Mar 14, 2011 at 18:15
  • 3
    $\begingroup$ 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. $\endgroup$
    – HJRW
    Mar 14, 2011 at 18:17
  • 2
    $\begingroup$ 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. $\endgroup$
    – HJRW
    Mar 14, 2011 at 18:20
  • 2
    $\begingroup$ the following question is related: mathoverflow.net/questions/34625/… $\endgroup$ Mar 14, 2011 at 18:34

1 Answer 1

12
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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