Sufficient condition for a *-homomorphism between C*-algebras being isometric

Let $\mathcal{A},\mathcal{B}$ be two unital C*-algebras and consider a *-homomorphism $\pi: \mathcal{A} \rightarrow \mathcal{B}$. I know that in general $\pi$ is contractive, i.e. $\vert\vert \pi(A) \vert\vert \leq \vert\vert A \vert\vert, \forall A\in \mathcal{A}$.

I want to show that under the additional assumption that $\pi(A)>0, \forall A>0$ one has equality, i.e. $\pi$ is an isometry: $\vert\vert \pi(A) \vert\vert = \vert\vert A \vert\vert, \forall A\in \mathcal{A}$

The crucial step in establishing the previous inequality lies in the fact that $\forall A\in\mathcal{A}$

$$r(\pi(AA^*)) \leq r(AA^*),$$

where $r$ denotes the spectral radius, respectively. Since $AA^*$ is positive, I sense that the additional condition enters at this point, but I can't finish the proof.

Am I on the wrong foot? Help is highly appreciated!

Solution 1:

Your assumption is that $\pi$ is faithful. Being a $*$-homomorphism, this implies that $\pi$ is injective, because $$ \pi(X)=0\ \implies\ \pi(X^*X)=\pi(X)^*\pi(X)=0\ \implies\ X^*X=0\ \implies\ X=0. $$ Now we need to use that the image of $\pi$ is closed and so it is a C$^*$-algebra (I added a proof at the end).

We showed that $\pi$ is injective, so bijective onto its image. Then $\pi^{-1}$ is a $*$-homomorphism, thus contractive. Then, for any $X\in\mathcal A$, $$ \|X\|=\|\pi^{-1}\pi(X)\|\leq\|\pi(X)\|\leq\|X\|. $$

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Proof that a homomorphic image of a C$^*$-algebra is closed (this is how Kadison-Ringrose prove it in Theorem 4.1.9 of their first volume).

Let $Y\in\overline{\pi(\mathcal A)}$. We want to show that $Y\in\pi(\mathcal A)$.

We have that $Y=\lim\pi(A_n)$ for some sequence $\{A_n\}$ in $\mathcal A$. As $Y^*=\lim\pi(A_n^*)$, the real and imaginary parts of $Y$ are limits of the corresponding real and imaginary parts of the $A_n$. So we may assume, without loss of generality, that $Y$ is selfadjoint.

Also, by choosing a subsequence if necessary, we may assume that $\|\pi(A_{n+1})-\pi(A_n)\|<2^{-n}$ for all $n$. In particular, $\sigma(\pi(A_{n+1}-A_n))\subset[-2^{-n},2^{-n}]$.

Choose continuous functions $f_n$ with the property that $f_n(t)=t$ when $t\in[-2^{-n},2^{-n}]$ and $|f_n(t)|\leq2^{-n}$ for all $n$. Let $$ A=A_1+\sum_nf_n(A_{n+1}-A_n)\in\mathcal A $$ (the series converges because $\|f_n(A_{n+1}-A_n)\|<2^{-n}$ for all $n$).

The key point is that, since $f_n$ is the identity on the spectrum of $\pi(A_{n+1}-A_n)$, it is the identity when evaluated on the operator. Then, as $\pi$ is a continuous $*$-homomorphism, $$ \pi(A)=\lim_m\pi(A_1)+\sum_{n=1}^m\pi(f_n(A_{n+1}-A_n))\\ =\lim_m\pi(A_1)+\sum_{n=1}^mf_n(\pi(A_{n+1}-A_n))\\ =\lim_m\pi(A_1)+\sum_{n=1}^m\pi(A_{n+1}-A_n)\\ =\lim_m\pi(A_1)+\sum_{n=1}^m\pi(A_{n+1})-\pi(A_n)\\ =\lim_m\pi(A_m)=Y. $$

Solution 2:

I would like to give an alternative approach since I think that the other answer is logically incomplete. Assuming that the algebraic * -isomorphism between $A/\ker(\pi)$ and the image of $\pi$ carries over the norm already includes what we would like to show. In fact, in every standard C*-lecture I have seen, the assertion is needed first in order to show that the image of a C*-algebra under a * -homomorphism is again a C*-algebra.

So let us show that an injective * -homomorphism $\pi: A\to B$ is isometric. Because of the C*-condition on the norm, it suffices to show that $\|a\|=\|\pi(a)\|$ for all $a\geq 0$ in $A$. Let us assume that this is false. Find $a\geq 0$ with $\|a\|=1$, but $\|\pi(a)\|=r<1$. Define the continuous function $$f: [0,1]\to [0,1]\quad\text{by}\quad f(t)=\begin{cases} 0 &,\quad t\leq r \\\\ \frac{t-r}{1-r} &,\quad t>r. \end{cases} $$ Since $1\in\operatorname{Sp}(a)$ and $f(1)=1$, we get $\|f(a)\|=1$. On the other hand, $\operatorname{Sp}(\pi(a))\subset [0,r]$, so $f(\pi(a))=0$. But since functional calculus is compatible with *-homomorphisms, we get $\pi(f(a))=f(\pi(a))=0$, which is a contradiction to injectivity.

Solution 3:

Simpler. Assume $\pi:\mathcal{A}\to\mathcal{B}$ is an injective $\ast$-Algebra Homomorphism. (No assumption of continuity.) It is shown, that $\pi$ is automatically an isometry.

Lemma 1. In a $C^{\ast}$-Algebra $a\leq M\Longleftrightarrow \|a\|\leq M$, if $a\geq 0$.

A simple proof involves observing the spectrum $\sigma(M-a)$ in relation to $\sigma(a)$.

Lemma 2. $\pi(a)\geq 0\Longleftrightarrow a\geq 0$.

Proof. $(\Longleftarrow)$: If $a\geq 0$, then $\pi(a)=\pi(\sqrt{a}^{\ast}\sqrt{a})=\pi(\sqrt{a})^{\ast}\pi(\sqrt{a})\geq0$.

$(\Longrightarrow)$: If $\pi(a)\geq 0$, then $\pi(a)=\sqrt{\pi(a)^{\ast}\pi(a)}=\sqrt{\pi(a^{\ast}a)}$.

Now for $c\in\mathcal{A}$ positive, $\pi(\sqrt{c})^{2}=\pi(\sqrt{c}^{2})=f(c)$. Now $\pi(c), \pi(\sqrt{c})\geq 0$, since $c,\sqrt{c}\geq 0$ (by the first part). Hence $\pi(\sqrt{c})=\sqrt{\pi(c)}$.

Thus $\pi(a)=\sqrt{\pi(a^{\ast}a)}=\pi(\sqrt{a^{\ast}a})$. Since $\pi$ is injective, it follows that $a=\sqrt{a^{\ast}a}\geq 0$. $\blacksquare$

Corollary. $\pi$ is an isometry.

Proof. Consider $a\in\mathcal{A}$ positive. Set $M:=\|a\|$ and $N:=\|\pi(a)\|$.

(1) It holds that $M-a\geq 0$, thus $M-\pi(a)=\pi(M-a)\geq 0$, thus $\pi(a)\leq M$, thus $\|\pi(a)\|\leq M=\|a\|$.

(2) Similarly $\pi(N-a)=N-\pi(a)\geq 0$, thus $N-a\geq 0$, thus $a\leq N$, thus $\|a\|\leq N=\|\pi(a)\|$.

Hence $\|\pi(a)\|=\|a\|$ for $a$ positive.

For $a\in\mathcal{A}$ a general element: $\|a\|^{2}=\|a^{\ast}a\|=\|\pi(a^{\ast}a)\|=\|\pi(a)^{\ast}\pi(a)\|=\|\pi(a)\|^{2}$. Hence $\|a\|=\|\pi(a)\|$. $\blacksquare$