C*-cross norm for commutative C*-algebras

Not. Let $A$ be a commutative C*-algebra, $B$ a C*-algebra.

Note that we can consider a left action of the unitary group $U\left(B^{+}\right)$ of the unitization $B^{+}$ of $B$ by $u.\phi \left(b\right)=\phi \left(u^{*}\mathrm{bu}\right)$.

Fact (Noncommutative Lusin's theorem) Let $\tau :B\to B\left(K\right)$ be an irreducible representation of $B$, $x$ and $y$ unit vectors of $K$. There exists a unitary element $b$ of $B^{+}$ satisfying $\tau \left(b\right)x=y$.

Dfn. A seminorm $\alpha$ on the algebraic tensor product $A\otimes B$ is said to be a C*-cross norm when (i) the completion $A{\otimes }_{\alpha }B$ is a C*-algebra compatible with the *-algebra structure on $A\otimes B$, (ii) $\alpha (a\otimes b)=\parallel a\parallel .\parallel b\parallel$ for any $a\in A$ and $b\in B$.

Rem. The injective C*-norm is a C*-cross norm.

We have seen that any C*-cross norm $\alpha$ on $A\otimes B$ is majored by the injective C*-norm because any irreducible representation of $A\otimes B$ is decomposed as a tensor product of irreducible representations of $A$ and $B$. In fact we have the converse inequality, C*-cross norm on $A\otimes B$ is unique.

What we have to check out is that we have an extension $A{\otimes }_{\alpha }B\to B\left(H\bar{\otimes }K\right)$ for any pair of irreducible representations $\sigma :A\to B\left(H\right)$ and $\tau :B\to B\left(K\right)$. Taking (arbitrary) vectors $h\in H$ and $k\in K$, we can extend $\sigma \otimes \tau$ on $A{\otimes }_{\alpha }B$ if and only if the product functional ${\omega }_h\cdot {\omega }_k$ is continuous with respect to $\alpha$ where ${\omega }_h$ is the pure functional defined by $\sigma$ and $h$.

Put $$S_{\alpha }=\left\{\right(\phi ,\psi )\in P(A)\times P(B):\phi \cdot \psi \text{is continuous with respect to}\alpha \}$$ and suppose it was not the whole $P\left(A\right)\times P\left(B\right)$. Note that $S_{\alpha }$ is w*-closed in $P\left(A\right)\times P\left(B\right)$ and closed under the action of $U\left(A^{+}\right)\times U\left(B^{+}\right)$. Hence there exist w*-closed unitary invariant subsets $F$ of $P\left(A\right)$ and $G$ of $P\left(B\right)$ such that $S_{\alpha }\subset F\times P\left(B\right)\cup P\left(A\right)\times G$.

Fact. ``Polar set'' operation gives a bijective (and order-reversing) correspondence between the closed ideals of $B$ and the w*-closed subsets of $P\left(B\right)$ which are closed under the action of $U\left(B^{+}\right)$.

Thus we get nonzero positive elements $a\in F^{ˆ}$ and $b\in G^{ˆ}$. This means tensor $a\otimes b$ is mapped to zero under any irreducible representation of $A{\otimes }_{\alpha }B$ which is a contradiction.