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I was reviewing some C*-algebra theory in Bruce Blackadar's Operator Algebras - Theory of C*-Algebras and von Neumann Algebras, when I came upon what seems to be a typo. On page 61, the following corollary is stated:

Let $A$ be a C*-algebra, and $x$ a normal element of $A$. Then $C^*(x)$ is isometrically isomorphic to $C_0(\sigma_A(x))$ under an isomorphism which sends $x$ to the function $f(t) = t$.

Here, $C^*(x)$ is the C*-subalgebra generated by $x$. My confusion is about the notation $C_0(\sigma_A(x))$. The index of the book refers to page 52 for this notation, where it says $C_0(X)$ is the space of continuous (complex-valued) functions vanishing at infinity. However, in the context of the above result, it seems to me that this should instead be the space of continuous functions with $f(0)=0$.

Question 1: Is this use of notation a typo in Blackadar's book?

I realize my question is similar to this question about the same bad notation, but I thought it was worth drawing attention to this specific typo in this book for others who might be confused.

I would also like to ask a follow-up question to ensure that I have understood this material correctly. As far as I understand, there are two closely related isomorphism theorems for the functional calculus of C*-algebras:

  1. If $A$ is a C*-algebra and $x \in A$ is normal, then $C^*(x) \cong \{f : \sigma_A(x) \rightarrow \mathbb{C}$ continuous, with $f(0)=0$ if $0 \in \sigma_A(x)$ $\}$ via an isomorphism sending $x$ to the function $f(t)=t$.

  2. If $A$ is a unitary C*-algebra (with unit $1$) and $x \in A$ is normal, then $C^*(1,x) \cong \{f : \sigma_A(x) \rightarrow \mathbb{C}$ continuous $\}$ via an isomorphism sending $x$ to the function $f(t)=t$ and sending $1$ to the function $f(t)=1$.

Question 2: Are the above statements correct? If so, is there a reference that states these results explicitly?

Sambo
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1 Answers1

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I'm not sure if it is a typo, as the notation \ kind of makes sense.

In the unital case, the natural isomorphism that the Gelfand transform gives is not onto $C(\sigma(x))$ but onto $C(\Sigma)$, where $\Sigma$ is the locally compact space of the characters with the weak$^*$-topology. Afterwards, one identifies the characters with the point evaluations.

When $A$ is not unital, the Gelfand transform gives you $A\simeq C_0^{\phantom{0}}(\Sigma)$, where here one indeed means the continuous functions vanishing at infinity. When one identifies the characters with the point evaluations at elements of $\sigma(x)$, $C_0(\Sigma)$ is mapped to $C_0(\sigma(x)\setminus\{0\})$; this in turn can be identified with the functions such that $f(0)=0$. Hence the notation, while not great, kind of makes sense.

You can find this stuff for instance in pages 7 and 8 in Davidson's C$^*$-Algebra by Example.

Martin Argerami
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