Extension of a morphism from algebraic tensor tensor to algebraic tensor product of double duals

Question

Let $A$ and $B$ be $C^{\ast}$-algebras and $A \otimes B$ be their algebraic tensor product. Let $\pi: A \otimes B \to B(\mathcal{H}) $ be a $*$-homomorphism.

Does there exist a $*$-homomorphism $\tilde{\pi} : A^{\ast \ast} \otimes B^{\ast \ast} \to B(\mathcal{H})$such that restriction of $\tilde{\pi}$ to $A \otimes B$ is $\pi$

Since arbitrary elements of second dual will not come from $A$ so i am facing difficulty in extending it. Any references or ideas?

Answer 1

I will denote the algebraic tensor product with the $\odot$ symbol.

Let's assume that $A,B,\pi$ are all unital. Otherwise, you can first extend $A\odot B\to B(H)$ to a unital $*$-homomorphism $\tilde{A}\odot\tilde{B}\to B(H)$, then apply the below argument to extend to a $*$-homomorphism $(\tilde{A})^{**}\odot(\tilde{B})^{**}\to B(H)$, and finally restrict this to $A^{**}\odot B^{**}\subset (\tilde{A})^{**}\odot(\tilde{B})^{**}$.

In that case, set $\pi_A:A\to B(H)$ by $\pi_A(a)=\pi(a\otimes 1)$ and $\pi_B:B\to B(H)$ by $\pi_B(b)=\pi(1\otimes b)$. These are unital $*$-homomorphisms and it is not hard to check that $\pi_A(a)\pi_B(b)=\pi_B(b)\pi_A(a)=\pi(a\otimes b)$.

Now it is a standard fact of representation theory (see for example Pederesen's book) that a non-degenerate $*$-representation of a $C^*$-algebra extends uniquely to a unital, ultraweakly continuous representation of the enveloping von Neumann algebra. In particular, we obtain $*$-homomorphisms $\bar{\pi}_A:A^{**}\to B(H)$ and $\bar{\pi}_B:B\to B(H)$ extending $\pi_A$ and $\pi_B$ respectively. Moreover, by Goldstine's theorem, we have that $(A)_1$ is ultraweakly dense in $(A^{**})_1$ and similarly for $B$, and since $\bar{\pi}_A,\bar{\pi}_B$ are ultraweakly continuous, we deduce that $\bar{\pi}_A,\bar{\pi}_B$ have commuting ranges in $B(H)$ (since $\pi_A$ and $\pi_B$ do so).

Now by the universal property of the maximal tensor product, we obtain a $*$-homomorphism $\bar{\pi}:A^{**}\otimes_\max B^{**}\to B(H)$ such that $\bar{\pi}(x\otimes y)=\bar{\pi}_A(x)\bar{\pi}_B(y)=\bar{\pi}_B(y)\bar{\pi}_A(x)$ on elementary tensors. Note that, if $x\in A$, and $y\in B$ to begin with, then $\bar{\pi}(x\otimes y)=\pi_A(x)\pi_B(y)=\pi(x\otimes y)$, i.e. $\bar{\pi}$ extends $\pi$.