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Suppose that $A,B$ and $C$ are $C^*$-algebras. If $\pi: A \otimes B \to C$ is a $*$-homomorphism, then there exists a unique $*$-homomorphism from $A\otimes_{\text{max}}$B to $C$ which extends $\pi$. In particular, any pair of $*$-homomorphisms with commuting ranges $\pi_A: A \to C$ and $\pi_B:B \to C$ induces a unique $*$-homomorphism $$\pi_A \times \pi_B:A\otimes_{\text{max}}B \to C$$

Given $A$ and $B$, we define the maximal $C^*$-norm on $A\otimes B$ to be $$\|x\|=\sup\{\|\pi(x)\|: \pi:A \otimes B \to \mathcal{B}(H) \text{ a cyclic $*$-homomorphism}\}$$

For the second statement: Since the map from $A \times B \to C$ defined by sending $(a,b)$ to $\pi_A(a)\pi_B(b)$ is bilinear, by universality there exists a unique map between $A \otimes B$ to $C$ which send $a \otimes b$ to $\pi_A(a)\pi_B(b)$. This happens to be a $*$-homomorphism because of the commuting ranges. Hence by the first statement, this map extends to a map from $A\otimes_{\text{max}}B$ to $C$.

For the first Statement: Let $\pi_C$ be a faithful representation of $C$ inside $\mathcal{B}(H)$. Then for $x \in A \otimes_{\text{max}} B$, define $$\tilde{\pi}(x)=\lim_n\pi_c \circ \pi(x_n)$$ where $x_n \in A\otimes B$ is a sequence converging to $x$ in $\|.\|_{\text{max}}$. This definition makes sense because $\|\pi_c \circ \pi(x_n)-\pi_c \circ \pi(x_m)\| \le ||x_n-x_m||_{\text{max}}$ and $\{\pi_c \circ \pi(x_n)\}$ being a cauchy sequence in $\mathcal{B}(H)$ converges.

I am unable to show precisely why the map has to end up in $C$. I think the idea is to think of $C$ as already being faithfully represented in some Hilbert space. I am not able to show it mathematically though.

Thanks for the help!!

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Let $\psi : C \hookrightarrow B(H)$ be a faithful non-degenerate representation of $C $ (e.g. the universal representation coming from the GNS construction). Then $$ \lVert \pi(x) \rVert_C = \lVert (\psi \circ \pi)(x)\rVert_{B(H)} \leq \lVert x \rVert_{\mathrm{max}}. $$

The first equality holds true, since $\pi$ is injective and hence isometric. The second inequality holds by definition of the maximal norm. Since each non-degenerate representation decomposes as a direct sum of cyclic representations, it is easy to check that $$ \lVert x \rVert_{\mathrm{max}} = \sup \{ \lVert \pi(x) \rVert : \pi \text{ is a non-degenerate $*$-rep. of } A \otimes B \}. $$

This shows that $\pi$ extends to the completion $A \otimes_{\mathrm{max}} B$.

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  • $\begingroup$ S. What about my arguement?? $\endgroup$ – tattwamasi amrutam Jul 4 '17 at 2:08
  • $\begingroup$ You show how to extend $\pi$ to the closure. You should replace $\pi_C \circ \pi (x_n)$ by $\pi(x_n)$ everywhere, because (as you say) you want to map into $C$. Then use everywhere that $\pi_C$ is isometric. $\endgroup$ – user42761 Jul 4 '17 at 7:06

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