omega-Commuting matrices vs Stone-von Neumann Theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:14:56Z http://mathoverflow.net/feeds/question/78584 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78584/omega-commuting-matrices-vs-stone-von-neumann-theorem omega-Commuting matrices vs Stone-von Neumann Theorem Denis Serre 2011-10-19T15:05:11Z 2011-10-20T09:03:35Z <p>Let me first recall the <a href="http://en.wikipedia.org/wiki/Stone%25E2%2580%2593von_Neumann_theorem" rel="nofollow">Stone-von Neumann theorem</a> 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 <a href="http://mathoverflow.net/questions/78573/classical-analogue-of-the-stone-von-neumann-theorem" rel="nofollow">related question</a>.</p> <p>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 $\omega$-<em>commute</em>, which means that $UV=\omega VU$. In this case, $\omega$ is some root of unity, say $\omega\ne1$, and we have the Weyl relation $U^jV^k=\omega^{jk}V^kU^j$.</p> <blockquote> <p>Is there any such pair with the property that $\|I_n-U\|&lt;1$ and $\|I_n-V\|&lt;1$, where $\|\cdot\|$ is the operator norm?</p> </blockquote> <p>My gess is <em>No</em> : 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 $\omega=1$.</p> <p><strong>Edit</strong>. 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 <em>H</em> is for Harry?).</p> <p>I replaced the spectral radius by the operator norm because it is equal for normal matrices, such as $I_n-U$.</p> <p><strong>re-Edit</strong>. 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 $\|I_n-U\|\ge\sqrt2$.</p> http://mathoverflow.net/questions/78584/omega-commuting-matrices-vs-stone-von-neumann-theorem/78593#78593 Answer by Mikael de la Salle for omega-Commuting matrices vs Stone-von Neumann Theorem Mikael de la Salle 2011-10-19T16:48:10Z 2011-10-20T08:37:52Z <p>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 $\omega \neq 1$.</p> <p>First note that for a couple of unitaries $(U,V)$ that $\omega$-commute, all the couples <code>$(V,U^*)$</code>, <code>$(V^*,U)$</code> and <code>$(U^*,V^*)$</code> also <code>$\omega$</code>-commute. In particular, by linearity the trace of <code>$(U+U^*)(V+V^*)$</code> is zero if <code>$\omega \neq 0$</code>.</p> <p>But the condition <code>$\rho(1-U)&lt;\sqrt 2$</code> exactly means that <code>$U+U^*&gt;0$</code> (as usual <code>$&gt;0$</code> means that the matrix is positive definite). Therefore if <code>$\rho(1-U)&lt;\sqrt 2$</code> and <code>$\rho(1-V)&lt;\sqrt 2$</code>, the trace of <code>$(U+U^*)(V+V^*)$</code> is positive.</p> <p>This proves the claim.</p>