## omega-Commuting matrices vs Stone-von Neumann Theorem

Let me first recall the Stone-von Neumann theorem that if two one-parameter groups of unitary operators $U_t$ and $V_s$ over a Hilbert space satisfy $U_tV_s=e^{ist}V_sU_t$ for every $s,t\in{\mathbb R}$ (Weyl relations), then their generators $P$ and $Q$ satisfy the canonical commutation relation $[Q,P]\psi=i\psi$ for all $\psi$ in the common dense domain. See also this related question.

I am interested in discrete one-parameter groups $U^p$ and $V^q$ where $p,q\in{\mathbb Z}$, and $U$ and $V$ are unitary matrices. There are simple examples of pairs $U,V\in{\mathbb U}_n$ that $\omega$-commute, which means that $UV=\omega VU$. In this case, $\omega$ is some root of unity, say $\omega\ne1$, and we have the Weyl relation $U^jV^k=\omega^{jk}V^kU^j$.

Is there any such pair with the property that $\|I_n-U\|<1$ and $\|I_n-V\|<1$, where $\|\cdot\|$ is the operator norm?

My gess is No : if such a pair existed then we could define the logarithms of $U$ and $V$ by the converging series $\log(1-x)=-x-\frac{x}2-\cdots$. It seems to me that we should obtain two matrices $X$ and $Y$ satisfying $[X,Y]=\alpha I_n$ where $e^\alpha=\omega$. But a finite dimensional commutator has zero trace, thus $\alpha=0$ and $\omega=1$.

Edit. Just to let you know that $\omega$-commuting matrices are not abstract non-sense, here is a nice relation when $\omega^p=1$ : if $(A,B)$ $\omega$-commute, then $(A+B)^p=A^p+B^p$. I could be due to H. S. A. Potter (does anyone knows if H is for Harry?).

I replaced the spectral radius by the operator norm because it is equal for normal matrices, such as $I_n-U$.

re-Edit. I realize that there is a trivial answer to my question: $U$ is unitary equivalent to $\omega U$, thus its spectrum is a union of regular $m$-agons over the unit circle, where $m$ is the order or the root of unity $\omega$. Then there must be an eigenvalue with non-positive real part, which implies $\|I_n-U\|\ge\sqrt2$.

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I'm a bit confused about finite dimensions. If you want $U$ and $V$ in finite dim, then taking the determinant (which is the elder brother of the trace) of the relation $UV = \omega VU$ gives $\omega = 1$??? What have I got wrong here? – Stefan Waldmann Oct 19 2011 at 15:48
Yes you got wrong. What you obtain is $\omega^n=1$, because $\det(aM)=a^n\det M$. – Denis Serre Oct 19 2011 at 16:03
Oh bulls* I'm sorry. It seems to be already late... Just ignore what I wrote :) – Stefan Waldmann Oct 19 2011 at 16:55
Since the relation you mention is quite magical it could be well due to H. Potter... – François Brunault Oct 20 2011 at 8:47
It seems $H$ stands for Harold : calms.abdn.ac.uk/DServe/… – François Brunault Oct 20 2011 at 8:52

I did not try to follow your argument, but here is another proof of your claim. In fact it proves a slightly stronger result, i.e. that if $\rho(1-U)$ and $\rho(1-V)$ are strictly smaller that $\sqrt 2$, $U,V$ cannot $\omega$-commute if $\omega \neq 1$.

First note that for a couple of unitaries $(U,V)$ that $\omega$-commute, all the couples $(V,U^*)$, $(V^*,U)$ and $(U^*,V^*)$ also $\omega$-commute. In particular, by linearity the trace of $(U+U^*)(V+V^*)$ is zero if $\omega \neq 0$.

But the condition $\rho(1-U)<\sqrt 2$ exactly means that $U+U^*>0$ (as usual $>0$ means that the matrix is positive definite). Therefore if $\rho(1-U)<\sqrt 2$ and $\rho(1-V)<\sqrt 2$, the trace of $(U+U^*)(V+V^*)$ is positive.

This proves the claim.

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 I have a doubt. I find that two of these pairs are $\omega$-commuting and the two others $\bar\omega$-commute. We cannot say anything about the pair $(U+U^*,V+V^*)$. – Denis Serre Oct 19 2011 at 17:40 Maybe you are right (but I do not think so, I tried to be precise in the way I order the couples), but anyway my point is that $Tr(U^{\varepsilon_1} V^{\varepsilon_2})= 0$ whenever $\varepsilon_1,\varepsilon_2 \in \{1,*\}$, and this holds because the couple $(U^{\varepsilon_1},V^{\varepsilon_2})$ something-commute, with something different from $1$. By linearity, this implies that $Tr( (U+U^*)(V+V^*) )=0$. – Mikael de la Salle Oct 19 2011 at 21:18 To expand my comment: $(U,V)$ $\omega$-commute if and only if $(V,U)$ $\bar \omega$-commute. – Mikael de la Salle Oct 19 2011 at 22:08