A Banach algebra $A$ is contractible if $H^1(A, X)=0$ for all Banach $A$-bimodules $X$. Now to my question

Let $A$ be Banach algebra and $I$ be closed ideal of $A$. If $I$ and $A/I$ are both contractible, then is $A$ contractible?

My attempt: Let $X$ be a Banach $A$-bimodules, and let $D\in \mathcal{Z^{1}}(A, X)$, then $D_{|_I}\in \mathcal{Z^{1}}(I, X)$ since $I$ contractible, there is $x\in X$ such that $$Da= a.x-x.a=ad_x(a) $$ for all $a\in I$. Let $\bar{D}=D-ad_x$, then $\bar{D}_{|_I}=0$.

I'm stuck here, Does anyone have a recommendation of how to continue?

I am grateful for anyone to have a good recommendation of books//lectures/resources/etc.? for contractible Banach algebra Thanks.

  • $\begingroup$ If you have access to Johnson's 1972 memoir, or Runde's book, then they should have complete proofs of the corresponding result for amenability. You should then be able (this is my intuition and not a certain prediction) to adapt those proofs $\endgroup$
    – Yemon Choi
    May 20 '19 at 16:19
  • $\begingroup$ That said, note that $\overline{D}$ is a derivation on $A$ which vanishes on $I$. So it factors through the quotient map $q:A\to A/I$ as $\overline{D}= D_2 \circ q$ where $D_2: A/I \to X$. Can you show that $D_2$ is a derivation? $\endgroup$
    – Yemon Choi
    May 20 '19 at 16:21
  • $\begingroup$ @Yemon Choi, If I show $D_2$ is a derivation, then there is an $x_2 \in X$ such that $D_2=ad_{x_2}$ but I'm stuck here again $\endgroup$
    – user62498
    May 20 '19 at 16:45
  • $\begingroup$ Since any contractible Banach algebra has a unit (as any amenable Banach algebra has b.a.i.), an ideal of it is nothing but a direct summand. $\endgroup$ May 20 '19 at 22:24
  • 1
    $\begingroup$ @Yemon Choi: (0) The way I wrote was confusing. I meant $I$ is unital because it is contractible. (1) If $B$ is contractible and non-unital, then $B$ is a $B$-bimodule w.r.t. the usual left multiplication and zero right multiplication. Then the formal identity map on $B$ is a derivation, which must be inner by contractibility. This means $B$ has a right unit. Likewise for the left. (2) Contractibility of $B(E)$ doesn't imply that of $K(E)$. $\endgroup$ May 23 '19 at 3:45

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