1
$\begingroup$

I'm working on the amenability of some Banach algebras, and I'm wondering why $l^\infty$ is amenable ? Does any one has any idea how to start ?

Thank you in advance.

$\endgroup$

closed as off-topic by Nik Weaver, Chris Godsil, Yemon Choi, Andrey Rekalo, Ricardo Andrade Apr 3 '14 at 1:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Nik Weaver, Chris Godsil, Andrey Rekalo
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ As indicated in the answer below: see Johnson's 1972 monograph, or the proof in Bonsall and Duncan's book $\endgroup$ – Yemon Choi Mar 30 '14 at 4:46
  • 2
    $\begingroup$ This question admits an easy answer if you know the right references, so I am voting to close $\endgroup$ – Yemon Choi Mar 30 '14 at 17:15
3
$\begingroup$

In general, commutative $C^\ast$-algebras are amenable. See A new proof of the amenability of $C(X)$ by Mortaza Abtahi and Yong Zhang (Bulletin of the Australian Mathematical Society, Volume 81, Issue 3, June 2010, pages 414-417).

$\endgroup$
  • $\begingroup$ Can we prove it using the definition only ? Can we prove the existence of a bouded approximate diagonal for $l^\infty$ $\endgroup$ – user128591 Mar 30 '14 at 4:48
  • 1
    $\begingroup$ @user128591 Please see the references mentioned in my comment above $\endgroup$ – Yemon Choi Mar 30 '14 at 17:14
0
$\begingroup$

$l^\infty$ is a $C^*$-algebra, in fact a von Neumann algebra. For these algebras, there are many equivalent definitions of amenability. For example, for finite von Neumann algebras one such example is the existence of a Hilbert bi-module $\mathcal{H}$ and a sequence of vector $\{\xi_i\}$ such that $\|a\xi_i-\xi_ia\|_2\rightarrow 0$, for all $a$ in the algebra. You can take the usual representation of $l^\infty$ on $l^2$ by multiplication (since $l^\infty$ is a abelian this actually make it into a bimodule and the above convergence is trivial since it is abelian.

$\endgroup$
  • 2
    $\begingroup$ With all due respect, this is massive overkill. Also the "correct" notion of amenability for a von Neumann algebra is not the same as saying its underlying Cstar-algebra is amenable in the Banach-algebraic sense $\endgroup$ – Yemon Choi Mar 30 '14 at 17:14
  • $\begingroup$ E.g. a two-one factor is never amenable as a Banach algebra $\endgroup$ – Yemon Choi Mar 30 '14 at 19:50

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