Homotopy between Unitary with gap in spectrum

Question

In the comments of this question it is mentioned that when a unitary operator has a gap in the spectrum one can show it is homotopic by 'dragging the eigenvalues to $1$ along the unit circle, away from the gap'. How can this be done explicitly? I expect that there will be constructed some continuous function with a branch trough the gap, but how do we get such a transformation, and how does one prove that it's images are unitary operators?

Answer 1

Note that in infinite dimension the spectrum is more than just eigenvalues (and it may have no eigenvalues at all).

There are a couple things here. If you are talking about homotopy over the full unitary group in $B(H)$, then what you do is you define a branch of the logarithm, and then you use Borel functional calculus to show that $U=e^{iS}$ for some selfadjoint $S$. And then $U(t)=e^{i(1-t)S}$ is a homotopy.

In the question you mention the context is different, though, as the homotopy is considered inside a C$^*$-algebra. That means that Borel functional calculus is not available, and here is where the gap in the spectrum becomes relevant. The gap allows you to construct a continuous branch of the logarithm. Then you can find $S=-i\log(U)$, selfadjoint, and an element of the C$^*$-algebra. So again you can construct a homotopy $U(t)=e^{i(1-t)S}$.