Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
1  
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
add comment

2 Answers

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.

share|improve this answer
    
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
1  
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
show 2 more comments

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.

share|improve this answer
    
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
1  
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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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