1
$\begingroup$

Let $A$ be a unital $C^{*}$-algebra and $a \in A$ be normal, with spectrum $\sigma(a)$. Let $B = C^{*}(a)$ be the $C^{*}$-algebra generated by $1$ and $a$, which is abelian. Let $\hat{B}$ be the space of all homomorphisms $\chi: B \to \mathbb{C}$ with the weak* topology, which makes it a compact Hausdorff topological space.

Gelfand Isomorphism Theorem states that $\varphi: B \to C(\hat{B})$ is an isometric $*$-isomorphism. Moreover, $\hat{a}: \hat{B} \to \sigma(a) \subset \mathbb{C}$ is a homeomorphism, so that $\hat{B} \cong \sigma(a)$.

On one hand, given a positive functional $\omega: C(\hat{B}) \to \mathbb{C}$, Riesz Representation Theorem states that there exists a unique regular Borel measure $\mu$ on $\sigma(a)$ such that: $$\omega(f) = \int_{\hat{B}}f(\chi)d\mu_{\omega}(\chi).$$

On the other hand, given a function $g \in C(\sigma(a))$, there is a natural identification $g \mapsto g\circ \hat{a}$ from $C(\sigma(a))$ to $C(\hat{B})$. Hence, each $\mu_{\omega}$ induces a new measure $\nu_{\omega}$ on $\sigma(a)$ by using $\hat{a}^{-1}$ as a push-forward.

By the abstract change of variables formula, it must hold: $$\int_{\hat{B}} (g\circ \hat{a})(\chi)d\mu_{\omega}(\chi) = \int_{\sigma(a)}g(z)d\nu_{\omega}(z) \tag{1}\label{1}$$

My question is: can we extend this analysis to $A$ instead of $B$ (and $\hat{A}$ instead of $\hat{B}$ and so on) when $A$ is abelian? I don't think so because, in this case, $\hat{a}$ is not a homeomorphism between $\hat{A}$ and $\sigma(a)$, but I couldn't tell if a general version of formula (\ref{1}) holds from the literature I know on the topic.

Improve this question
$\endgroup$
4
  • $\begingroup$ The question is a bit vague as stated. Suppose $a=1$ so that $\sigma(a)$ and $\hat B$ are points. What kind of formula do you have in mind instead of (1) in this case? $\endgroup$
    – Dmitri Pavlov
    Commented Jan 25, 2023 at 4:18
  • $\begingroup$ @DmitriPavlov Although I started the analysis with an arbitrary $C^{*}$-algebra and studied the subalgebra $B= C^{*}(a)$, which is abelian. My question is: if $A$ was abelian, does the analysis follow for all $A$ instead of for an abelian subalgebra of it? $\endgroup$
    – MathMath
    Commented Jan 25, 2023 at 12:35
  • $\begingroup$ So what do you want to play the role of σ(a) in the absence of a? The most natural analogue of σ(a) is just  itself, so your formula becomes a tautology. Unless you have a different formula in mind, which is why I asked the above question. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 25, 2023 at 17:45
  • $\begingroup$ @DmitriPavlov what I meant is the following. Suppose $A$ is an abelian $C^{*}$-star and $a \in A$ be normal, with $\sigma(a)$ its spectrum. Can we prove formula (\ref{1}) in this case, when $B$ is replaced by $A$? The only difference is that, in my post, $A$ was not abelian, so I have to restrict to $B=C^{*}(a)$ to use Gelfand's Isomorphism Theorem. But when $A$ is abelian, do the same formulas hold? $\endgroup$
    – MathMath
    Commented Jan 25, 2023 at 20:36

1 Answer 1

Reset to default
1
$\begingroup$

To see that (1) continues to hold when B is replaced by A, it suffices (by linearity and density of step functions) to consider only the case when g is the characteristic function of some measurable subset S of σ(a).

Indeed, in this case the right side equals the ν-measure of S, whereas the left side equals the μ-measure of the â-preimage of S. Since ν was defined as the â-pushforward of μ, this equality holds by definition.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Dmitri, thank you for your answer. I was puzzled because, in the analysis of my post, I used the fact that $\hat{B} \cong \sigma(a)$. But if $A$ is abelian and $a \in A$ is normal, I believe the same analysis provide an isomorphism $\hat{A}\cong \sigma(a)$, right? And then formula (\ref{1}) follows from my analysis, which justifies your answer. $\endgroup$
    – MathMath
    Commented Jan 26, 2023 at 14:47
  • $\begingroup$ @MathMath: No, Â is isomorphic to σ(a) if and only if a generates A as a C*-algebra. Otherwise, you just have a continuous map Â→σ(a) provided by the Gelfand transformation, but it is no longer a homeomorphism. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 26, 2023 at 18:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .