Maximal norm of tensor product

Question

Definition 3.3.3. (Maximal Norm) Given $A$ and B (two C*-algebra), we define the maximal C*-norm on $A \odot B$ to be $$||x||_{max}=sup\{||\pi(x)||: \pi: A\odot B\rightarrow B(H) ~a~*-homomorphism\}$$ for $x\in A\odot B$. We let $A\otimes_{max} B$ denote the completion of $A\odot B$ with respect to $||.||_{max}$. (Here, the "$\odot$" means algebraic tensor product.)

Proposition 3.3.7 (Universality). If $\pi: A \odot B \rightarrow C$ is a $*$-homomorphism, then there exists a unique $*$-homomorphism $A \otimes_{max} B \rightarrow C$ which extends $\pi$.

Proof. Faithfully representing $C$ on some Hilbert space, this fact follows from the definition of $||.||_{max}$.

However, I can not understand the proof of the Proposition. Yes, we have a $*$-homomorphism $A\odot B\rightarrow B(H)$ for some $H$. But how can we extend $\pi$ from the definition?

Answer 1

$C$ is (by faithful representation) a closed $\ast$-subalgebra of $B(H)$. By the definition of $\lVert\cdot\rVert_{max}$, $\pi$ is continuous, $\lVert \pi(x)\rVert_{B(H)} \leqslant \lVert x\rVert_{max}$.

So there is a unique linear continuous extension $\widehat{\pi}\colon A \otimes_{max} B \to B(H)$ of the map $\pi$ to $A\otimes_{max} B$, and since $C$ is closed, $\mathcal{R}(\widehat{\pi}) \subset C$. One then only needs to verify that this extension is in fact a $\ast$-homomorphism.

The maps $x \mapsto \widehat{\pi}(x^\ast)$ and $x\mapsto \widehat{\pi}(x)^\ast$ are continuous and coincide on the dense subspace $A \odot B$, so they are the same map.

For every fixed $x\in A\odot B$, the maps $y \mapsto \widehat{\pi}(x\cdot y)$ and $y\mapsto \widehat{\pi}(x)\cdot \widehat{\pi}(y)$ are continuous and coincide on $A\odot B$, so they are the same map. Thus we know that $\widehat{\pi}(xy) = \widehat{\pi}(x)\widehat{\pi}(y)$ whenever $x \in A \odot B$. Now, for fixed $y \in A \otimes_{max} B$, consider $x \mapsto \widehat{\pi}(xy)$ and $x \mapsto\widehat{\pi}(x)\widehat{\pi}(y)$. Once again, the maps coincide on the dense subspace $A\odot B$ and are continuous, hence they coincide everywhere.

Thus it is established that $\widehat{\pi}$ is a $\ast$-homomorphism.

Answer 2

The proof looks too complicated.

Let $\sigma : C \hookrightarrow B(H)$ be a faithful $*$-representation of $C$. In particular $\sigma$ is isometric.

Then $\lVert \pi(x) \rVert = \lVert \sigma(\pi(x)) \rVert \leq \lVert x \rVert_{\mathrm{max}}$, since $\sigma \circ \pi$ is a $*$-rep. of $A \odot B$.