PlanetMath: injective $C^*$-algebra homomorphism is isometric

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injective

-algebra homomorphism is isometric

(Theorem)

Theorem - Let $\mathcal{A}$ and $\mathcal{B}$ be -algebras and $\Phi : \mathcal{A} \longrightarrow \mathcal{B}$ an injective *-homomorphism. Then $\Vert\Phi(x)\Vert=\Vert x\Vert$ and $\sigma(\Phi(x)) = \sigma(x)$ for every $x \in \mathcal{A}$ , where $\sigma(y)$ denotes the spectrum of the element .

$\,$

Proof: It suffices to prove the result for unital -algebras, since the general case follows directly by considering the minimal unitizations of $\mathcal{A}$ and $\mathcal{B}$ . So we assume that $\mathcal{A}$ and $\mathcal{B}$ are unital and we will denote their identity elements by , being clear from context which one is being used.

Let us first prove the second part of the theorem for normal elements $x \in \mathcal{A}$ . It is clear that $\sigma(\Phi(x)) \subseteq \sigma(x)$ since if $x - \lambda e$ invertible for some $\lambda \in \mathcal{C}$ , then so is $\Phi(x) - \lambda e = \Phi(x - \lambda e)$ . Suppose the inclusion is strict, then there is a non-zero function $f \in C(\sigma(x))$ whose restriction to $\sigma(\Phi(x))$ is zero (here $C(\sigma(x))$ denotes the -algebra of continuous functions $\sigma(x) \longrightarrow \mathbb{C}$ ). Thus we have, by the continuous functional calculus, that $f(x) \neq 0$ and also that

$\displaystyle \Phi(f(x))=f(\Phi(x))=0$

by the continuous functional calculus and the result on this entry. Thus, we conclude that $\Phi$ is not injective and which is a contradiction. Hence we must have $\sigma(\Phi(x)) = \sigma(x)$ .

Let $R_{\sigma}(z)$ denote the spectral radius of the element . From the norm and spectral radius relation in -algebras we know that, for an arbitrary element $x \in \mathcal{A}$ , we have that

$\displaystyle \Vert x\Vert^2=R_{\sigma}(x^*x)$

Since the element

is normal, from the preceding paragraph it follows that $R_{\sigma}(x^*x)=R_{\sigma}(\Phi(x^*x))$ , and hence we conclude that

$\displaystyle \Vert x\Vert^2=R_{\sigma}(x^*x)=R_{\sigma}\big(\Phi(x)^*\Phi(x)\big)= \Vert\Phi(x)\Vert^2$

i.e. $\Vert\Phi(x)\Vert=\Vert x\Vert$ .

Since $\Phi$ is isometric, $\Phi(\mathcal{A})$ is closed *-subalgebra of $\mathcal{B}$ , i.e. $\Phi(\mathcal{A})$ is a -subalgebra of $\mathcal{B}$ , and it is isomorphic to $\mathcal{A}$ . Using the spectral invariance theorem we conclude that $\sigma(x)=\sigma(\Phi(x))$ for every $x \in \mathcal{A}$ . $\square$

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Cross-references: spectral invariance theorem, isomorphic, closed, isometric, normal, spectral radius, contradiction, continuous functional calculus, continuous functions, restriction, function, strict, inclusion, invertible, normal elements, clear, identity elements, minimal unitizations, unital, proof, spectrum, *-homomorphism, injective
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This is version 3 of injective $C^*$

-algebra homomorphism is isometric, born on 2008-04-20, modified 2008-04-21.
Object id is 10524, canonical name is InjectiveCAlgebraHomomorphismIsIsometric.
Accessed 209 times total.

Classification:

AMS MSC:

46L05 (Functional analysis :: Selfadjoint operator algebras :: General theory of $C^*$-algebras)

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