# Showing isomorphism of two $C^*$ algebras

It seems that quite a standard trick of showing two $$C^*$$ algebras are as follows:

Let $$A$$ be a $$C^*$$ algebra $$B$$ another $$C^*$$ algebra. $$A' \subseteq A$$ be a subalgebra that is closed under $$*$$. (Not necessarily closed) and is dense in $$A$$. Let $$L:A' \rightarrow B$$ be a $$*$$-isometric homomoprhism which has dense image. Then $$L$$ extends to a $$*$$-isometric isomorphism .

So my aim is showing that $$L$$ actually extends to an injective map - since it is not necessarily case that $$L$$ is injective in its closure.

## 2 Answers

Given $$a\in A$$ take $$\{a_j\}\subset A'$$ with $$a_j\to a$$. Since $$L$$ is isometric, we have $$\|L(a_j)-L(a_k)\|=\|L(a_j-a_k)\|=\|a_j-a_k\|,$$ so $$\{L(a_j)\}$$ is Cauchy. As $$B$$ is complete, there exists $$b_a=\lim L(a_j)$$. We want to extend $$L$$ by defining $$L(a)=b_a$$. For this we need to show that $$b_a$$ is unique for each $$a$$; and this follows easily: if $$c_j\to a$$, then $$\|L(c_j)-L(a_j)\|=\|c_j-a_j\|\leq\|c_j-a\|+\|a-a_j\|\to0,$$ so $$L(c_j)\to b_a$$. So the extension $$L$$ is well-defined. Also $$\|L(a)\|=\|\lim_j L(a_j)\|=\lim_j\|L(a_j)\|=\lim_j\|a_j\|=\|a\|,$$ so $$L$$ is isometric.

If you don't require $$L$$ to be isometric, the result is not true. Let $$A=C[0,1]$$, $$B=\mathbb C\oplus C[0,1]$$, and take $$A'\subset A$$ to be the $$*$$-algebra of complex polynomials. Define $$L:A'\to B$$ by $$L(p)=p(2)\oplus p.$$ This is clearly an injective $$*$$-homomorphism. And it has dense image: Given $$(\lambda,f)\in B$$, we can choose a sequence of polynomials $$\{p_j\}$$ such that $$p_j\to f$$ uniformly on $$[0,1]$$ and $$p_j(2)=\lambda$$ (simply use Stone-Weierstrass on a continuous function $$g$$ that agrees with $$f$$ on $$[0,1]$$ and with $$g(2)=\lambda$$).

So $$L$$ satisfies all the hypotheses; but it is not bounded. If you consider $$p_n(x)=x^n$$, then $$\|p_n\|=1$$, while $$p_n(2)=2^n$$, so $$\|L(p_n)\|=2^n$$.

• I suppose you mean $\pi(A')=L(A')$? The statement I know requires $\rho$ to be bounded is that its domain is a Banach $*$ algebra i.e. closed. It seems that the claim is: $L:A \rightarrow B$ is bounded, if $A$ is a $*$-algebra (with a norm such that $A$ is not necessarily complete), then $B$ is a $C^*$ algebra. Please correct me if I mistinterpret you. – CL. Apr 16 at 19:24
• Martin, I have edited my problem, I noticed I have quite a lot of typos. – CL. Apr 16 at 19:34
• Yes, I was glossing over the part that $*$-homomorphisms are bounded, and that doesn't work when the domain is not closed. I have rewritten the answer with a counterexample. – Martin Argerami Apr 16 at 20:09
• Hi Martin, thanks a lot, I have added the condition that $L$ is an isometric isomorphism, I wonder if the argument holds from here. In fact, I am just trying to generalize the trick you did here: math.stackexchange.com/questions/1918086/… – CL. Apr 16 at 20:34
• And please do keep the counter example in your edits : ) – CL. Apr 16 at 20:35

The answer by Martin is of course correct, I just want to emphasise that this statement and the way to see it is essentially elementary functional analysis. In my opinion is very helpful to be able to differentiate when a statement is about the linear structure and when a statement needs the C*.

Central is the following statement: Let $$A:V\to W$$ be a continuous linear map between normed vector spaces, then there is a unique continuous linear extension $$\overline A :\overline V\to\overline W$$ between the completions. This can be done as an elementary exercise and is basically the first half of Martin's answer.

(As a remark: The above statement is itself an expression of a more general statement, namely that Lipschitz maps between metric spaces $$A,B$$ so that $$B$$ is complete and $$A$$ is dense in some other metric space $$A'$$ always have a unique continuous extension $$A'\to B$$.)

The rest of what you need follows from:

1. If a continous linear map is isometric on a dense sub-set, then it is isometric on its entire domain.

2. If a continuous linear map is multiplicative wrt continuous multiplications on a dense sub-set, then it is multiplicative on its entire domain.

3. If a continuous linear map $$A$$ satisfies $$A(v^*)= A(v)^*$$ on a dense sub-set and the $$*$$ operation is continuous on domain and codomain then the above equality holds on the entire domain.

Note that this is basically the same statement $$3$$ times: Continuous linear maps that preserve a continuous structure on a dense sub-set automatically do so on the entire domain.