Let me first recall the [Stone-von Neumann theorem](http://en.wikipedia.org/wiki/Stone%25E2%2580%2593von_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](https://mathoverflow.net/questions/78573/classical-analogue-of-the-stone-von-neumann-theorem).


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 $U^jV^k=\omega^{jk}V^kU^j$.



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> Is there any such pair with the property that $\|I\_n-U\|<1$ and $\|I\_n-V\|<1$, where $\|\cdot\|$ is the operator norm?
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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 $\|I\_n-U\|\ge\sqrt2$.



omega-Commuting matrices vs Stone-von Neumann Theorem
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.


