# $C^{*}$-algebra homomorphisms preserve continuous functional calculus

Let us setup some notation first: Let $\mathcal{A}$ be a unital $C^{*}$-algebra  (http://planetmath.org/CAlgebra) and $z$ a normal element of $\mathcal{A}$. Then

$\,$

Let $\mathcal{A}$, $\mathcal{B}$ be unital $C^{*}$-algebras (http://planetmath.org/CAlgebra) and $\Phi:\mathcal{A}\longrightarrow\mathcal{B}$ a *-homomorphism       . Let $x$ be a normal element in $\mathcal{A}$. If $f\in C(\sigma(x))$ then

 $\displaystyle\Phi(f(x))=f(\Phi(x))$

$\,$

Proof: The identity elements  of $\mathcal{A}$ and $\mathcal{B}$ will be both denoted by $e$ and it will be clear from the context which one we are referring to.

First, we need to check that $f(\Phi(x))$ is a well-defined element of $\mathcal{B}$, i.e. that $\sigma(\Phi(x))\subseteq\sigma(x)$. This is clear since, if $x-\lambda e$ is invertible   for some $\lambda\in\mathbb{C}$, then $\Phi(x)-\lambda e=\Phi(x-\lambda e)$ is also invertible.

Let $\{p_{n}\}$ be sequence of polynomials in $C(\sigma(x))$ converging uniformly to $f$. Then we have that

• $\Phi(p_{n}(x))\longrightarrow\Phi(f(x))$, by the continuity of $\Phi$ (see this entry (http://planetmath.org/HomomorphismsOfCAlgebrasAreContinuous)) and the continuity of the continuous functional calculus mapping.

• $p_{n}(\Phi(x))\longrightarrow f(\Phi(x))$, by the continuity of the continuous functional calculus mapping.

It is easily checked that $\Phi(p_{n}(x))=p_{n}(\Phi(x))$ (since $\Phi$ is an homomorphism). Hence we conclude that $\Phi(f(x))=f(\Phi(x))$ as intended. $\square$

Title $C^{*}$-algebra homomorphisms preserve continuous functional calculus CalgebraHomomorphismsPreserveContinuousFunctionalCalculus 2013-03-22 18:00:50 2013-03-22 18:00:50 asteroid (17536) asteroid (17536) 5 asteroid (17536) Theorem msc 47A60 msc 46L05