4
$\begingroup$

If $A$ is a Banach agebra and $M$ is a Banach $A$-bimodule then a linear map $T:A\to M$ is called an $A$-module homomorphism if $$T(ab)=aT(b),\quad T(ab)=T(a)b,\qquad a,b\in A.$$ Also $A\hat{\otimes} A$ is a two sided $A$-module with the following actions $$c(a\otimes b)=ca\otimes b,\quad (a\otimes b)c=a\otimes bc.$$ (Here $\hat{\otimes}$ deotes the projective tensor product.)

Does there exist a Banach algebra $A$ and a net of linear $A$-module homomorphisms $\{\rho_\alpha:A\to A\hat{\otimes} A\}_{\alpha\in I}$ such that for the linear product map $\pi:A\hat{\otimes} A\to A,$ $\pi(a\otimes b)=ab$ we have:

  1. $\pi\circ\rho_\alpha(a)\to a$ in norm for all $a\in A$;

  2. the set $\left\{\frac{\|\pi\circ\rho_\alpha(a)b-ab\|}{\|ab\|}:a,b\in A,\alpha\in I\right\}$ is bounded;

  3. the set $\left\{\frac{\|\pi\circ\rho_\alpha(a)-a\|}{\|a\|}:a\in A,\alpha\in I\right\}$ is not bounded.

$\endgroup$
3
  • 1
    $\begingroup$ (I just rolled back an edit which replaced imperfect English with slightly worse English) $\endgroup$
    – Yemon Choi
    Oct 9, 2016 at 2:56
  • $\begingroup$ Just to get the background straight in my head: is this related to questions about approximate biprojectivity? $\endgroup$
    – Yemon Choi
    Oct 9, 2016 at 2:57
  • $\begingroup$ Yes! It is related to approximate biprojectivity. Thanks for edit. $\endgroup$ Oct 9, 2016 at 15:11

1 Answer 1

5
$\begingroup$

The following is not a complete answer, but suggests that it will be difficult to find such an $A$ (and it leads me to conjecture that no such $A$ exists).

Let $A^2$ denote $\{ ab \colon a,b\in A\}$. (This notation is not entirely standard: many authors use $A^2$ to denote the linear span of this set, and some use $A^2$ to denote the closed linear span of this set. If $A$ has a bounded approximate identity then all of these sets are the same, by Cohen's factorization theorem.) Then your condition 2 is equivalent to

2'. the set $\left\{\frac{\|\pi\circ\rho_\alpha(x)-x\|}{\|x\|} \colon x\in A^2,\alpha\in I\right\}$ is bounded.

So in the case where $A^2$ is dense in $A$, a straightforward argument shows that condition 2' contradicts condition 3.

On the other hand, condition 1 clearly implies that ${\rm lin} A^2$ is dense in $A$.

Therefore, if a Banach algebra $A$ exists which satisfies 1, 2 and 3, it must have the property that $A^2$ is not dense in $A$ but ${\rm lin} A^2$ is dense in $A$. This seems quite difficult to me; most examples I can think of where ${\rm lin} A^2$ is dense in $A$ have the property that $A^2$ is dense in $A$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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