# injective $C^{*}$-algebra homomorphism is isometric

Let $\mathcal{A}$ and $\mathcal{B}$ be $C^{*}$-algebras (http://planetmath.org/CAlgebra) and $\Phi:\mathcal{A}\longrightarrow\mathcal{B}$ an injective  *-homomorphism        . Then $\|\Phi(x)\|=\|x\|$ and $\sigma(\Phi(x))=\sigma(x)$ for every $x\in\mathcal{A}$, where $\sigma(y)$ denotes the spectrum of the element $y$.

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Proof: It suffices to prove the result for unital $C^{*}$-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 $e$, 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 $C^{*}$-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 (http://planetmath.org/CAlgebraHomomorphismsPreserveContinuousFunctionalCalculus). 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 $z$. From the norm and spectral radius relation   in $C^{*}$-algebras (http://planetmath.org/NormAndSpectralRadiusInCAlgebras) we know that, for an arbitrary element $x\in\mathcal{A}$, we have that

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

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

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

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

Since $\Phi$ is isometric, $\Phi(\mathcal{A})$ is closed *-subalgebra  of $\mathcal{B}$, i.e. $\Phi(\mathcal{A})$ is a $C^{*}$-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 $x\in\mathcal{A}$. $\square$

Title injective $C^{*}$-algebra homomorphism is isometric InjectiveCalgebraHomomorphismIsIsometric 2013-03-22 18:00:35 2013-03-22 18:00:35 asteroid (17536) asteroid (17536) 6 asteroid (17536) Theorem msc 46L05