Homotopy between unitary element and identity elements, Operator Theory

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 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 $$u$$.
• 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?) – Andreas Blass Apr 14 at 23:31