There is a theorem in Murphy's book on operator theory and $C^\ast$-algebras:

Let $u$ be a unitary element in a unital $C^\ast$-algebra $A$. Then if $\sigma(u) \subsetneq S^1$ then there exists a self-adjoint element $a\in A$ such that $u = e^{ia}$.

This theorem comes after a discussion of some properties of $C^\ast$-algebras and the Gelfand representaiton theorem. The theorem is followed by a proof of the existence of a functional calculus at a normal element $a$. The theorem does not appear to be used in any of the two proofs of the theorems that follow it.

I don't understand where this theorem fits into the theory: what is it used for? Why does it appear in a seemingly random place with no relation to adjacent theorems and proofs in the book?

I understand that it gives a sufficient condition for a unitary element to have a logarithm. I also understand its proof. I don't know anything about functional calculus so perhaps this is a very important theorem in functinoal calculus. But if it is this is not mentioned in the book and I'd be very grateful for context!

• It seems to be a variation on Stone's theorem on one-parameter families of unitaries, which is important in the mathematical study of quantum mechanics. (Here $u$ is the time evolution of a system after a short length of time, say, and $a$ is a multiple of the Hamiltonian of the system.) – Qiaochu Yuan Nov 1 '14 at 5:33
• Thank you for your comment. I am still hoping for a connection to functional calculus because it seems to be used in star algebra theory. – user167889 Nov 1 '14 at 7:26

• He uses it for example in 2.3.3, 2.5.6, 3.5.1, 7.2.3. Functional calculus is a way to evaluate functions (continuous, in this context; although later Murphy considers the Borel functional calculus) on a normal operator. Theorem 2.1.13 is the key because it assigns an operator in $C^*(a)$, in a natural way, to each $f\in C(\sigma(a))$. We denote this operator by $f(a)$. – Martin Argerami Nov 2 '14 at 15:17
• Of course. But $\varphi$ is indeed continuous, as any $*$-homomorphism is. – Martin Argerami Nov 3 '14 at 5:16