Proof of Spectral Mapping theorem

Question

I'm following the lectures as seen here: https://www.youtube.com/watch?v=QptSpn7UPN4&list=PLFJf6jNakNII2h3sBMrIuLhN-vSQrvy2C&index=4

But I'm having trouble understanding the proof of Spectral Mapping Theorem as it is presented.

Here's the set up.

Let $A$ be an unital $C*$ algebra. And let $a \in A$ be normal. Let $C*(1,a)$ be the smallest $C*$ algebra generated by $1$ and $a$. And let $SP(D)$ for $C*$ algebra the set of bounded linear functional $\omega \colon D \to \mathbb{C}$, and for element $a \in C$, $SP(a)$ denote the spectrum of $a$. Let $C(X)$ denote the $C*$ algebra of $\mathbb{C}$-valued continuous function on $X$.

Then for $a$ normal, the function $\hat{a} \colon SP(C*(a,1)) \to SP(a)$ given by sending $\omega$ to $\omega(a)$ is a homeomorphism.

So we know $C(SP(C*(a,1))$ is isomorphic to $C(SP(a))$, via the map sending $f \in C(SP(a))$ to $f \circ \hat{a}$.

So $C*(a,1)$ is isomorphic to $C(SP(a))$, and we denote by $f(a)$ the element $b \in C*(a,1)$ such that $\hat{b} = f \circ \hat{a}$.

Now, the spectral theorem is to prove $SP(f(a)) = f(SP(a))$.

The proof goes that $SP(f(a)) = SP(\hat{f(a)})$, which I understand.

Then it says $SP(\hat{f(a)}) = f(SP(a))$, which I don't understand how it follows.

I know that $SP(\hat{f(a)}) = SP(f \circ \hat{a})$, so $\lambda \in SP(f \circ \hat{a})$ if and only if there exists $\omega$ such that $f(\omega(a)) - \lambda$ is not invertible, but I'm not sure how that helps.

Thanks!

Answer 1

Before embarking on a proof of the Spectral Mapping Theorem, it is helpful to have quite a clear picture of the following:

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1. Spectral Permanence: If $A$ is a unital $C^*$-algebra with a $C^*$-subalgebra $B$ containing the identity $1_A$ of $A$, then $\sigma_B(b)=\sigma_A(b)$ for all $b\in B$.

2. The Functional Calculus: If $A$ is a unital $C^*$-algebra and $a\in A$ is normal, then there exists a unique isometric $*$-isomorphism $\Phi:C(\sigma_A(a))\to C^*(\{1_A,a\})$ satisfying $\Phi(\text{id})=a$, where $\text{id}:\sigma_A(a)\to\mathbb{C}$ is given by $\text{id}(\lambda):=\lambda$. We write $\Phi(f)$ as $f(a)$.

3. If $X$ is a compact Hausdorff space, then $\sigma_{C(X)}(f)=f(X)$ for all $f\in C(X)$.

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Now we turn to the Spectral Mapping Theorem:

If $A$ is a unital $C^*$-algebra and $a\in A$ is normal, then $\sigma_A(f(a))=f(\sigma_A(a))$ for all $f\in C(\sigma_A(a))$.

Proof: Since $B:=C^*(\{1_A,a\})$ contains the identity $1_A$ of $A$ and $f(a)\in B$, spectral permanence gives that $\sigma_A(f(a))=\sigma_B(f(a))$.

Then the Functional Calculus identifies $B$ with $C(\sigma_A(a))$ as a $C^*$-algebra (via the isomorphism $\Phi$), so the spectrum of corresponding elements are identified: this gives that $\sigma_B(f(a))=\sigma_{C(\sigma_A(a))}(f)$, since $f$ is identified with $f(a)$ via $\Phi$.

Finally, since $\sigma_A(a)$ is a compact Hausdorff space, we use 3. above to establish that $$C_{C(\sigma_A(a))}(f)=f(\sigma_A(a)),$$ and we are done.

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I found the following notes by Dana P Williams helpful when learning about this.