# Finite subgroups of unitary groups

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 $[u^m,v^m] = 1_n$. (In fact, he showed that any finite subgroup of $GL(n)$ has a abelian normal subgroup of finite index, bounded independently of the subgroup.)

In particular, the set of pairs that generate a finite subgroup cannot be dense in $U(n) \times U(n)$ for $n \geq 2$. Indeed, a generic pair of unitaries in $U(2)$ generates a free group. I am asking whether a stable version could still be true.

Question Let $n$ be an integer, let $u,v$ be a pair of unitaries in $U(n)$ and let $\varepsilon>0$. Is there some integer $k \in {\mathbb N}$ and unitaries $u',v' \in U(nk)$, such that $\|u \otimes 1_k-u'\| \leq \varepsilon$ and $\|v \otimes 1_k - v'\| \leq \varepsilon$ (with respect to the standard embedding $U(n) \subset U(nk)$, $u \mapsto u \otimes 1_k$), and the unitaries $u',v'$ generate a finite group?

The question does only make sense once one has fixed a norm on $M_n {\mathbb C}$ for all $n \in {\mathbb N}$. I would either take the normalized Hilbert-Schmidt norm, i.e. $\|x\| = \frac1n Tr(x^*x)^{\frac12}$ for $x \in M_n \mathbb C$, or the operator norm.

(More globally seen, the questions are equivalent to the questions whether for the unitary group of certain UHF-algebras or the hyperfinite $II_1$-factor, pairs of unitaries generating a finite subgroup are dense in the natural topologies.)

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I don't believe this can be true (i.e., I think the answer to the question has to be "no") but I don't see how to approach it. –  Peter Shor Aug 17 '10 at 14:31

This might help: If $u$ and $v$ are both close enough to scalars in the operator norm, and $\varepsilon$ is small enough, then no matter how large $k$ is, if the group generated by $u^{\prime}$ and $v^{\prime}$ is finite, then it still has to be Abelian. The reasoning is something like this: I use the operator norm : given a unitary matrix $M$, there is a unique scalar $\lambda$ ( in the interior of the unit disk unless $M$ itself is a scalar matrix) with $\|M- \lambda I\|$ minimal, and this only depends on the spectrum of $M$. In particular, we can only have $\|M -\lambda I\| < \frac{1}{2}$ for a scalar $\lambda$ if the eigenvalues of $M$ all lie on an arc of length less than $\frac{\pi}{3}$ on the unit circle.It is essentially a Theorem of Frobenius that if $G$ is a finite group of unitary matrices, then the subgroup $H$ generated by the matrices at distance less than $\frac{1}{2}$ from a scalar is an Abelian normal subgroup. If $u$ and $v$ are close enough to scalars, and $\varepsilon$ is small enough then we can make both $\|u^{\prime} - \lambda I\|$ and $\| v^{\prime} - \mu I\|$ less than $\frac{1}{2}$ for suitable scalars $\lambda,\mu$. So if they generate a finite group, it must be Abelian. Note added: I will add a little more detail. One proof of Jordan's theorem uses a compactness argument and a kind of contraction Lemma. Let $[a,b]$ denote $a^{-1}b^{-1}ab$. If $a$ and $b$ are unitary, then $$\| I-[a,b] \| = \|I - a^{-1}b^{-1}ab \| = \|ab-ba \|$$ $$= \| (a-I)(b-I) - (b-I)(a-I)\| \leq 2 \|a-I\| \|b-I\|$$. Hence if $\| I-a\| < \frac{1}{2}$, then $\|I- [a,b] \| < \|I-b\|$ for all $b.$ Hence in a finite group $G$ of unitary matrices, let $H$ be the subgroup generated by those elements of $G$ with $\|I-h\|< \frac{1}{2}$. For any $h \in H$ with $\|I - h \| < \frac{1}{2}$, we have $[g,h,h, \ldots ,h] = 1$ for all $g \in G$. By a theorem of Baer, $H$ is nilpotent. With a little more work, you actually get $H$ Abelian. (This is a kind of hybrid of arguments of Frobenius and a little more finite group theory. More or less the same line of reasoning led to a similar theorem by Zassenhaus on discrete groups, and this led to the idea of a Zassenhaus neighbourhood in a Lie group). Note that if $x,y$ are in different cosets of $H$ in $G$, we must have $\|x-y\| \geq \frac{1}{2}$, so the compactness of the unit ball of $M_{n}(\mathbb{C})$ gives a fixed bound on $[G:H]$. But the same argument works when $\| a - \lambda I \| < \frac{1}{2}$ for some scalar $\lambda$, using $$\|ab - ba \| = \| (a-\lambda I)(b-\mu I) - (b-\mu I)(a-\lambda I)\|$$ for any scalars $\lambda,\mu$, and replacing $H$ by the group generated by elements at distance less than $\frac{1}{2}$ from any scalar. (Added in response to comment below): You do need a little argument to get from $H$ nilpotent to $H$ Abelian, but the statements above are all accurate. It is important to remember that $H$ is generated by elements within distance $\frac{1}{2}$ of a scalar. Any such element has all its eigenvalues on an arc of length less than $\frac{\pi}{3}$ on $S^{1}$. No subset of the "multiset" of eigenvalues of such an element has zero sum (it's clear eg, if you rotate so that all eigenvalues have positive real part). Let $\chi$ be the character of the given representation of $H$. Let $\mu$ be an irreducible constituent of $\chi$. By the remark above, if $\|h -\alpha I \| < \frac{1}{2}$ for some scalar $\alpha$, then $\mu(h) \neq 0.$ But if $\mu$ is not linear, it is induced from a character of a maximal subgroup $M$, which is normal as $H$ is nilpotent. But then $\mu$ vanishes on all elements of $H$ which are outside $M$. Hence the elements at distance less than $\frac{1}{2}$ from a scalar must all lie within $M$. But since $M$ is proper, and such elements generate $H$, this is a contradiction. Hence $\mu$ must be linear after all. Since $\mu$ was arbitrary, $H$ is Abelian.

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I see that if $H$ is finite, then it is nilpotent. In fact it has an abelian subgroup of finite index, only depending on the dimension of the ambient $U(n)$. But $n$ is not fixed here. I do not see how you conclude that $H$ is abelian. –  Andreas Thom Apr 21 '11 at 6:56
Thanks a lot for your answer. It is a coincidence that only two weeks ago I could find myself an answer to the question above. You can find at least the statements in my answer below. My impression was that using only Frobenius-Baer-Jordan type arguments could not yield a negative answer to my question; but I might be wrong. I am really curious if this actually works. –  Andreas Thom Apr 21 '11 at 7:06
Glad that you found your own solution. I have added some detail to my original answer, because it was too long for a comment, but what I said was accurate. –  Geoff Robinson Apr 21 '11 at 10:25
Very nice! I did not know about the application of character theory to this problem. –  Andreas Thom Apr 21 '11 at 13:22
I do not understand (or care really) anything about bounties. I just like to help find solutions to problems. –  Geoff Robinson Apr 21 '11 at 19:27

In the meantime I could answer the question myself; at least if one considers the metric induced by the norm. This will be contained in a forthcoming joint preprint with Ken Dykema. We can show two theorems:

Theorem 1: For every $\varepsilon>0$, there exists $n \in \mathbb N$ and unitaries $a,t \in U(n)$ such that $\|a-1\|\geq 1$ and $$\|1-tat^{-1}ata^{-1}t^{-1}a^{-2}\| < \varepsilon.$$

and

Theorem 2: There exists $\varepsilon_0>0$ such that for all $n \in \mathbb N$, and any two unitaries $a,t \in U(n)$ which generate a finite group: $$\|1-tat^{-1}ata^{-1}t^{-1}a^{-2}\| < \varepsilon \leq \varepsilon_0$$ implies $$\|1-a\|< 1/4-\sqrt{1/16-\varepsilon}.$$

This together implies that the answer to my question above is negative. The details are not complicated but a bit too long for putting them in this answer. I will post a link to the preprint as soon as it is ready.

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is this also true for Hilbert-Schmidt norm? –  Kate Juschenko Apr 21 '11 at 12:45
Theorem 1 holds. The group $G=\langle a,t \mid tat^{-1}ata^{-1}t^{-1}=a^2 \rangle$ is known to be sofic; so Theorem 2 fails. –  Andreas Thom Apr 21 '11 at 12:59
I meant to ask the original question, it seems that H.-S. is natural framework for it. –  Kate Juschenko Apr 21 '11 at 13:28
I do not know about the answer to my question above for the Hilbert-Schmidt norm. –  Andreas Thom Apr 21 '11 at 13:47
Geoff: the constants of equivalence depend on $n$, so that in trying to answer questions about "all $n$ simultaneously" your reasoning doesn't always work –  Yemon Choi Apr 22 '11 at 10:25