Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following is an abstraction of a more specific problem I've been grappling with, so I might give some extra unnecessary information. (The `bounded approximate left identities' assumption is probably not necessary, but it is definitely the case in the specific problem that this question is abstracted from).

Let $A$ and $B$ be Banach algebras, both having bounded approximate left identities. We assume that we have non-degenerate, isometric representations $\pi:A\to B(X)$ and $\rho:B\to B(Y)$ on some Banach spaces $X$ and $Y$.

We can define the following algebra representation $$ \pi\otimes\rho:A\otimes B\to B(X\hat{\otimes}Y) $$ (where $X\hat{\otimes}Y$ denotes the projective tensor product) in the usual way, for all $a\in A$, $b\in B$, $x\in X$ and $y\in Y$, by $$ \left(\pi\otimes\rho(a\otimes b)\right)x\otimes y:=\pi(a)x\otimes\rho(b)y. $$ Now, since $\pi$ and $\rho$ were assumed to be isometric, we can prove that the map $\pi\otimes\rho$ is injective and satisfies $$ ||\pi\otimes\rho(a\otimes b)||_{op} = ||a||_A ||b||_B $$ for all elementary tensors $a\otimes b\in A\otimes B$, where the norm on the left hand side denotes the operator norm on $B(X\hat{\otimes}Y)$.

My question is the following: Does the map $$ \sum_{i}a_{i}\otimes b_{i}\mapsto\left\Vert \pi\otimes\rho\left(\sum_{i}a_{i}\otimes b_{i}\right)\right\Vert _{\mbox{op}} $$ define a reasonable cross norm on $A\otimes B$ (Ryan's book, Ch. 6)? Seeing that it is dominated by the projective tensor norm on $A\otimes B$ is easy, but is it bounded from below by the injective tensor norm on $A\otimes B$?

Note that we can extend bounded functionals on $A$ to bounded functionals on $B(X)$ by using Hahn Banach and that the representations are isometric; similarly with $B$. Still, using this, a proof still seems just out of reach for want of being able to extend functionals on $B(X)\hat{\otimes}B(Y)$ to functionals on $B(X\hat{\otimes}Y)$ because the norms on these algebras can't be favourably related (projective norm on the one, operator norm on the other).

Any ideas, helpful references or counterexamples showing how this might fail will be greatly appreciated.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Let $D_A\subseteq A^*$ be the functionals of the form $\mu(\pi(\cdot)x)$, for $x\in X, \mu\in X^*, \|x\|\leq 1, \|\mu\|\leq 1$. As $\pi$ is an isometry, Hahn-Banach shows that the convex hull of $X$ is weak$^*$-dense in the closed ball of $A^*$, say $A^*_{[1]}$.

Similarly for $D_B$ using $\rho$. It's clear (*) that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in D_A, \mu_B\in D_B). \] However, this inequality is preserved by taking convex combinations of the elements of $D_A$ and $D_B$, and by taking weak$^*$-closures. But that would then show that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in A^*_{[1]}, \mu_B\in B^*_{[1]}), \] and that's all you need to show, I think?

Why is (*) true? Given $x\in X,\mu\in X^*,y\in Y,\lambda\in Y^*$, we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also $\mu\otimes\lambda \in (X\widehat\otimes Y)^*$ (this is $w\otimes z\mapsto \mu(w)\lambda(z)$) has norm $\|\mu\| \|\lambda\|$. Then $(\mu_A\otimes\mu_B)\tau = (\mu\otimes\lambda)((\pi\otimes\rho)(\tau)(x\otimes y))$. Actually, this shows that the result would remain true if you replaced the projective tensor norm on $X\otimes Y$ be any reasonable cross-norm.

share|improve this answer
    
Ah, thank you so much Matthew! That seems to do the trick, I think. I'll check the details. My intuition was also that it'll work for any reasonable cross-norm, but (for lack of experience) I had trouble even proving it for the projective norm. I'd been going down a similar route as your $D_A$, but on $B(X)$ instead of $A$ and then working with $\pi(A)$ as a subalgebra, but I ran into some annoying problems as things moved a bit too far away from $A\otimes B$ for me to see how to translate back. –  Miek Messerschmidt Dec 9 '11 at 10:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.