Showing that an injective $^*$-homomorphism between two $C^*$-algebras is isometric.

Question

Problem: "Let $(A,\|\cdot\|)$ and $(B,\|\cdot\|)$ be unital $C^*$-algebras and let $\phi:A\to B$ be an injective $^*$-homomorphism. Show that $\phi$ is isometric. Hint: Treat the case of self-adjoint elements firstly and use the fact that it suffices to consider the case when $A$ and $B$ are commutative."

I'll collect together the key facts that, I think, I need to use:

1. For $C^*$-algebras $(A,\|\cdot\|),(B,\|\cdot\|)$ we call $\phi: A\to B$ a $^*$-homomorphism if (i) $\phi$ is linear, (ii) $\phi(a_1a_2)=\phi(a_1)\phi(a_2)\,\forall a_1,a_2\in A$, and (iii) $\phi(a^*)=\phi(a)^*\,\forall a\in A$.

2. For $C^*$-algebras $(A,\|\cdot\|),(B,\|\cdot\|)$ and $\phi: A\to B$ we have that $r(a)=\|a\|\,\forall a\in A: a=a^*$, where $r(a)$ is the spectral radius of $a\in A$.

Attempt: I've made multiple attempts at this problem, and I'm not sure which, if any, are going to yield any fruit. I know that in order to show that this injective $^*$-homomorphism is isometric I need to show that $\|a\|=\|\phi(a)\|,\,\forall a\in A$.

Consider firstly those $a\in A:a=a^*$. Then we know that by the $C^*$-property $\|a\|^2=\|a^*a\|$ for self-adjoint $a\in A$. Then:

$$\|a\|^2=\|a^*a\|=r(a^*a)=r(a^*)r(a)=r(a)^2$$

But I don't see that this gets me anywhere, other than reiterating what I already know. Is there a connection between $r(a)$ and $\phi(a)$ that I can make use of? It seems to me, that in some sense, if the properties above for $\phi$ held for $r$ I might be able to get somewhere.

Alternatively, since we have an injective homomorphism we know that $\phi$ maps the identity element in $A$ to the identity element in $B$. Then, consider:

$$1=\|e_B\|=\|\phi(e_A)\|=\|\phi(a^*a)\|=\|\phi(a^*)\phi(a)\|=\|\phi(a)^*\phi(a)\|$$

And then $\phi(a)^*=b^*$ for some $b\in B:b=b^*$. Then we have that,

$$\|\phi(a)^*\phi(a)\|=\|b^*b\|=\|b\|^2$$

This, again, doesn't tell me anything that I already know.

Can anybody direct me on how best to proceed? In particular, in accordance with the hint, what exactly am I being told when hinted that "it suffices to consider the case when $A$ and $B$ are commutative?

Answer 1

The points you are missing are:

• Since $a^*a$ is positive, $C^*(a^*a)$ is abelian

• Since $\phi$ is a $*$-homomorphism, its image is closed (proof here), so we may replace $B$ with $\phi(B)$.

• Since $\phi$ is injective, $\sigma(\phi(a))\subset \sigma(a)$. Simply because $A-\lambda I$ invertible implies $\phi(A)-\lambda I$ invertible.

So, on $C^*(a^*a)$, $$ \|\phi(a^*a)\|=r(\phi(a^*a))\leq r(a^*a)=\|a^*a\|. $$ Thus $$ \|\phi(a)\|^2=\|\phi(a^*a)\|\leq\|a^*a\|=\|a\|^2, $$ and $\phi$ is contractive. Now we can apply the above to $\phi^{-1}:\phi(A)\to A$

Answer 2

The easiest way to show that $\phi$ is isometric goes as follows: Using that $\lVert a \rVert$ equals its spectral radius for self-adjoint $a$, one sees that the norm on $A$ is uniquely determined. Now, define a norm $$ \rho(a) := \lVert \phi(a)\rVert \qquad (a \in A). $$ Then the norm $\rho$ makes $A$ into a C*-algebra. Therefore, $\rho = \lVert \cdot \lVert$. It follows that $\phi$ is isometric.

Answer 3

Theorem. If $\rho:A\to B$ is an injective -homomorphism between unital abelian C-algebra, then \begin{align*} \rho^*: & K_B \mapsto K_A\\ & \tau\mapsto \tau\circ \rho. \end{align*} is a continuous surjection between character spaces $K_A$ and $K_B$(also called the maximal ideal space) of $A$ and $B$.

If $A$ is not abelian, consider the unital C*-algebra $C^*(a^*a)$ generated by $a^*a$.

\begin{align*} \|\phi(a)\| & =\sup_{\tau_B\in K_B}|\tau_B(\phi(a))|=\sup_{\tau_B\in K_B}|\phi^*(\tau_B)(a)|\\ & =\sup_{\tau_A\in K_A}|\tau_A(a)|=\|a\|. \end{align*}

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Proof of the theorem. If $\rho^* $ is not a surjection, then there is a $\tau_A\in K_A\backslash \rho^*(K_B)$. By Urysohn's lemma, there is a continuous function $f$ on $K_A$ such that $$f(\rho^*(K_B))=0,f(\tau_A)=1.$$ And by Gelfand representation, there is some $a\in A$ such that $f=\widehat{a}$. Hence $$\tau_B\circ\rho(a)=0\forall \tau\in K_B\tag{1}$$ and $$\tau_A(a)=1. \tag{2}$$ By (1) $\rho(a)=0$, and thus $a=0$ since $\rho$ is an injection. But this is contradict to (2).