Let $\mathcal{T}$ be the Toeplitz algebra. I.e. the $C^*$ algebra generated by the shift operator $S\in B(l^2(\Bbb N))$.

In page 6, line 8 of a proof we have a unitary element $u \in \mathcal{T} \otimes \mathcal{T}$, and it is claimed that $u$ is homotopic to the identity by a path of unitaries.

The claim seems to be quite general. Is there any reference/ similar result, which I could read to understand more about this?


The unitaries you consider there are self-adjoint. In that particular case, you can write down an easy formula for the path. Let $u$ be a self-adjoint unitary in a C*-algebra. Define $h := (1-u)/2$. Then $e^{\pi i h} = u$ and the path $t \mapsto e^{\pi i th}$ connects $1$ to $u$.

  • $\begingroup$ You commented on an earlier answer, now deleted, and I think at least part of that comment is worth recording for the benefit of readers who can't see deleted answers: A similar idea works whenever $u$ has a gap in its spectrum (on the unit circle). Namely, drag all the eigenvalues toward $1$ by moving them along the circle away from the gap.(What I don't immediately see: Is a one-point gap enough, or do you need a whole arc?) $\endgroup$ – Andreas Blass Apr 14 at 23:31
  • $\begingroup$ One point missing is enough since the spectrum is closed. Hence there exists a branch of logarithm which is continuous on the spectrum of the unitary. $\endgroup$ – Epsilon Apr 15 at 7:34
  • 1
    $\begingroup$ Thanks for pointing out that "one point missing" implies "a whole arc". (I hope I'm awake now.) $\endgroup$ – Andreas Blass Apr 15 at 11:27

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