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

As far as I can tell, the theorem is there because it allows Murphy to show an application of the Gelfand transform (2.1.10), and because it has to be somewhere in the book.

He later uses the theorem a couple times (in the proof of 7.3.2 and in the proof of 7.5.6).

• Thank you! So this theorem has nothing to do with functional calculus? I looked at the two theorems in chapter 7 and they seem to be unrelated to functional calculus. – user167889 Nov 1 '14 at 23:00
• I guess it depends on what you call "functional calculus"; the Gelfand transform is functional calculus. – Martin Argerami Nov 2 '14 at 0:44
• You seem to know the book really well. Do you think you could point me to where the theorem 2.1.13. is used later? It would help me understand what functional calculus is. – user167889 Nov 2 '14 at 4:16
• 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