Let $u, v$ be unitaries in a unital $C^*$-algebra satisfying $uv=e^{2\pi i \theta}vu$ where $\theta$ is irrational (so $\{e^{2 \pi i n \theta} : n \in \mathbb{Z} \}$ is dense in $\mathbb{T}$). For fixed $z \in \mathbb{T}$, finitely supported $a: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{C}$, and $\varepsilon >0$, show that

$||\sum_{m,n} a_{m,n} z^m v^m u^n|| \leq ||\sum_{m,n} a_{m,n} v^m u^n|| + \varepsilon$

There was a hint about being able to find $k$ such that $|e^{2 \pi i k \theta} - z|$ can be made small as desired.

So I thought to do something like

$||\sum_{m,n} a_{m,n} ((z^m - e^{2 \pi i k_m \theta}) + e^{2 \pi i k_m \theta}) v^m u^n|| $

$\leq \sum_{m,n} |a_{m,n}| |z^m - e^{2 \pi i k_m \theta}| + ||\sum_{m,n} a_{m,n} e^{2 \pi i k_m \theta} v^m u^n||$

choosing the $k_m$ such that the left sum is less than $\varepsilon$. Then if the right sum can be shown to be less than or equal to $||\sum_{m,n} a_{m,n} v^m u^n||$, that would be it. To do that I was thinking of using the commuting relation for $u, v$ to cancel out the $e^{2 \pi i k_m \theta}$, but then the $v, u$ are permuted from the desired $v^m u^n$ form...

  • $\begingroup$ The way the question is phrased, you can do it for any $\varepsilon>0$, which in the end means you can remove the $\varepsilon$ entirely. Are you sure the question is exactly like that? $\endgroup$ – Martin Argerami Aug 10 '18 at 4:40
  • $\begingroup$ Direct copy of problem statement: Prove that for every $z \in \mathbb{T}$, every finitely supported function $a : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{C}$ and every $\varepsilon >0$, we have $||\sum_{m,n} a_{m,n} z^n v^m u^n|| \leq ||\sum_{m,n} a_{m,n} v^m u^n|| + \varepsilon$ $\endgroup$ – Lucas Aug 10 '18 at 12:41
  • $\begingroup$ Context: I want a *-homomorphism $B \rightarrow B$, where $B=C^*(u,v)$, satisfying $v \mapsto zv$ and $u \mapsto u$. It can be shown that $B_0=\mathrm{span}(v^m u^n : m,n \in \mathbb{Z})$ is dense in $B$, so you can define a *-hom. $B_0 \rightarrow B$ satisfying the desired properties, and the norm inequality above would allow you to extend via continuity to one $B \rightarrow B$ $\endgroup$ – Lucas Aug 10 '18 at 12:44
  • $\begingroup$ I fail to see how that can be a homomorphism. It requires $(zv)^2=zv^2$, which only happens if $z=0$ or $z=1$. $\endgroup$ – Martin Argerami Aug 10 '18 at 12:50
  • $\begingroup$ Maybe I made a mistake. Going back a bit, I want a *-hom. $B \rightarrow B$ that satisfies $v^m u^n \mapsto z^m v^m u^n$. This would have to satisfy $v \mapsto zv$ and $u \mapsto u$, so I thought to consider a hom. determined on $B_0$ by this. Could you elaborate on why it must be that $(zv)^2 = zv^2$? I do not see why that would be required $\endgroup$ – Lucas Aug 10 '18 at 13:08

With $u, v$ as above, consider $B = C^*(u,v)$. Note that $u, zv$ are also unitaries, and still $u(zv) = e^{2 \pi i \theta} (zv) u$. By universality of the irrational rotation algebra, $B$ is *-isomorphic to $C^*(u,zv)$. In particular there is a *-isomorphism $\pi$ satisfying $u \mapsto u$ and $v \mapsto zv$. Hence, as an isometry,

$||\sum_{m,n}a_{m,n} z^m v^m u^n||= ||\pi (\sum_{m,n}a_{m,n} v^m u^n)|| = ||\sum_{m,n}a_{m,n} v^m u^n|| $


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