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

Theorem - Let $\mathcal{A}$ and $\mathcal{B}$ be ${C}^{*}$-algebras (http://planetmath.org/CAlgebra) and $\mathrm{\Phi}:\mathcal{A}\u27f6\mathcal{B}$ an injective^{} *-homomorphism^{}. Then $\parallel \mathrm{\Phi}(x)\parallel =\parallel x\parallel $ and $\sigma (\mathrm{\Phi}(x))=\sigma (x)$ 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 $\sigma (\mathrm{\Phi}(x))\subseteq \sigma (x)$ since if $x-\lambda e$ invertible^{} for some $\lambda \in \mathcal{C}$, then so is $\mathrm{\Phi}(x)-\lambda e=\mathrm{\Phi}(x-\lambda e)$. Suppose the inclusion is strict, then there is a non-zero function $f\in C(\sigma (x))$ whose restriction^{} to $\sigma (\mathrm{\Phi}(x))$ is zero (here $C(\sigma (x))$ denotes the ${C}^{*}$-algebra of continuous functions^{} $\sigma (x)\u27f6\u2102$). Thus we have, by the continuous functional calculus, that $f(x)\ne 0$ and also that

$\mathrm{\Phi}(f(x))=f(\mathrm{\Phi}(x))=0$ |

by the continuous functional calculus and the result on this entry (http://planetmath.org/CAlgebraHomomorphismsPreserveContinuousFunctionalCalculus). Thus, we conclude that $\mathrm{\Phi}$ is not injective and which is a contradiction^{}. Hence we must have $\sigma (\mathrm{\Phi}(x))=\sigma (x)$.

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

${\parallel x\parallel}^{2}={R}_{\sigma}({x}^{*}x)$ |

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

${\parallel x\parallel}^{2}={R}_{\sigma}({x}^{*}x)={R}_{\sigma}\left(\mathrm{\Phi}{(x)}^{*}\mathrm{\Phi}(x)\right)={\parallel \mathrm{\Phi}(x)\parallel}^{2}$ |

i.e. $\parallel \mathrm{\Phi}(x)\parallel =\parallel x\parallel $.

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

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

Canonical name | InjectiveCalgebraHomomorphismIsIsometric |

Date of creation | 2013-03-22 18:00:35 |

Last modified on | 2013-03-22 18:00:35 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 6 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 46L05 |