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Let $A$ be a Banach algebra and let $X$ be an $A$-bimodule. Is there a notion of (relative) injectivity for $X$ which would imply that $\mathcal{H}^n(A,X)$ vanishes for all $n\ge 1$? Here $\mathcal{H}^n(A,X)$ denotes the continuous Hochschild cohomology of $A$ with coefficients in $X$ ($X$ can be assumed to be dual $A$-bimodule).

I expected there is such a notion, but after reading books by Helemskii, Runde and a few other sources on cohomology of Banach algebras I can't seem to find a general statement of this type, even though versions of projectivity and injectivity are discussed there.

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I must be misunderstanding something. Doesn't relative $A$-bi-injectivity of $X$ (a.k.a. relative injectivity as an $A^e$-module) do the job? And if $X$ is a dual module then this is the same as asking for its predual to be (relatively) A-biflat? (This just comes out of Helemskii's version of Ext for Banach modules.) –  Yemon Choi Oct 21 '10 at 18:18
    
@Yemon: why not write that up into an answer?? –  Matthew Daws Oct 21 '10 at 18:25
    
@Yemon: what do you mean by $A^{\epsilon}$? Is it the unitization of $A$? –  Piotr Nowak Oct 21 '10 at 18:51
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@Matt: Because I seem to remember Piotr has written a paper using (or referring to) this kind of stuff, and it is all in Helemskii's pink book (HBTA). @Piotr: $A^e$ is the enveloping algebra of A, this is A proj. tensor A equipped with product $(a\otimes b)(c\otimes d) = (ac\otimes db)$, and its main feature is that every $A$-bimodule can be regarded as a left $A^e$-bimodule, and conversely. –  Yemon Choi Oct 21 '10 at 19:23
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@Piotr: That is correct (I am assuming that A is unital, which if you are working with algebras arising from discrete groups, is usually the case). In practice, what one needs to show is that the natural embedding from $J: X \to {\mathcal L}(A\hat\otimes A, X)$, defined by $J(x)(a\otimes b) = axb$, has a (continuous, linear) left inverse which is an $A$-bimodule map -- this is equivalent to being A-bi-injective, at least in the unital case. –  Yemon Choi Oct 21 '10 at 21:22

1 Answer 1

up vote 4 down vote accepted

A couple of people have encouraged me to post this as an answer, so here goes. I am currently without a copy of Helemskii's Pink Book so I can't give chapter-and-verse references as I would have liked. Everything that follows should be somewhere in there, although perhaps expressed slightly differently, and probably slightly better. Certainly what follows is too wordy, but I haven't had time to work out a condensed version.

To recap: Piotr is asking about the continuous Hochschild cohomology groups ${\mathcal H}^n(A,X)$ where $A$ is a Banach algebra and $X$ a Banach $A$-bimodule. To simplify the discussion slightly, I shall assume that $A$ has an identity element (which is indeed the case if $A$ is one of the usual convolution-type algebras associated to a discrete group) and that $X$ is unit-linked, i.e. that the identity of $A$ acts as the identity operator on $X$.

Conceptual/abstract POV (Helemskian)

As in the classical theory of Cartan-Eilenberg vintage, (continuous) Hochschild cohomology can be expressed in terms of relative Ext. One way to approach this, as Helemskii does, is to introduce the enveloping algebra $A^e$ of a unital Banach algebra $A$. This has underlying Banach space $A\hat{\otimes} A$ (projective tensor product) and has multiplication defined by $(a\otimes b)\cdot (c\otimes d) = (ab\otimes dc)$.

(The definition is slightly different for the non-unital case, and the artificial dichotomy that arises in places in the Pink Book is something that vexes some of us. But I digress...)

The purpose of doing this is as follows: every Banach $A$-bimodule $X$ becomes a left Banach $A^e$-module via $(a\otimes b)\cdot x = axb$; and conversely, every left Banach $A^e$-module becomes a Banach $A$-bimodule via the same formula. Now, taking as read the definition of relative Ext that is given in the Pink Book, we have $$ {\mathcal H}^n(A,X) \cong \operatorname{Ext}_{A^e}^n (A,X) $$ This is an isomorphism of seminormed spaces for each $n$ (I guess it would be more precise to say an isomorphism of seminormed-space-valued $\delta$-functors or some such high-falutin' phrase)

Now, recall that if $B$ is a Banach algebra, then a left Banach $B$-module $X$ is said to be (relatively) $B$-injective if it satisfies the following:

whenever $N$ is a left Banach $B$-module and $M$ is a closed $B$-submodule of $N$ which is complemented as a Banach subspace, then each continuous linear $B$-module map $M\to X$ has a continuous linear extension to a $B$-module map $N\to X$.

Moreover, if $X$ is relatively $B$-injective then $\operatorname{Ext}_B^n(\cdot,X)=0$ for each $n\geq 1$. (The converse also holds, in fact.) Therefore:

if $X$ is relatively $A^e$-injective, then ${\mathcal H}^n(A,X)=0$ for all $n\geq 1$.

The get-your-hands-dirty approach (Johnsonite)

We continue to suppose that $A$ has an identity element. Now let $E$ be any Banach space and equip $V_E :={\mathcal L}(A\hat\otimes A, E)$ with the following natural $A$-bimodule structure: $$ (b\cdot T \cdot a)(c\otimes d) = T(ac\otimes db) \quad\quad(T\in V_E). $$

Claim: ${\mathcal H}^n(A,V_E)=0$ for all $n\geq 1$.

This is most easily proved by proving something stronger:

Exercise: Let $\delta: {\mathcal C}^n(A,V_E)\to {\mathcal C}^{n+1}(A,V_E)$ denote the Hochschild coboundary operator. Define $\sigma: {\mathcal C}^{n+1}(A,V_E) \to {\mathcal C}^n(A,V_E)$ by $$ [\sigma\psi(a_1,\dots,a_n)]{(c\otimes d)} = [\psi(d,a_1,\dots, a_n)]{(c\otimes 1)}. $$ Then $\delta\sigma(\psi)+\sigma\delta(\psi) =\psi$ for every $\psi\in\mathcal C^k(A,V_E)$.

We now observe the following: if $V$ is any Banach $A$-bimodule, and it can be written as $V\cong X\oplus Y$ where $X$ and $Y$ are closed $A$-sub-bimodules of $V$, then $$ {\mathcal H}^n(A,V) \cong {\mathcal H}^n(A,X) \oplus {\mathcal H}^n(A,Y) \quad\hbox{for all $n$.} $$ One can check this directly or appeal to the long exact sequence of Hochschild cohomology (which is a special case of the one for relative Ext).

Finally, for each Banach $A$-bimodule $X$ there is a canonical $A$-bimodule map $J:X\to V_X$ which is defined by $[J(x)](c\otimes d) = dxc$. Therefore:

If there exists an $A$-bimodule map $P:V_X\to X$ such that $PJ$ is the identity, then ${\mathcal H}^n(A, X)=0$ for all $n\geq 1$.

Clowns to the left of me, jokers to the right

As may be apparent to anyone who's read this far: the two conditions we have obtained on $X$, each of which implies that Hochschild cohomology with coefficients in $X$ vanishes, are one and the same condition. [The calculations in the second version actually show that $V_E$ is $A$-bi-injective - meaning the same as $A^e$-injective. This relied on $A$ having an identity element! Then, knowing that a complemented submodule of an injective module is injective, we see that the second condition implies $X$ is $A$-bi-injective.] The nice thing about the direct approach is that it gives one explicit formulas one can try even in settings where the coefficient module is not bi-injective (see for instance my first excuse for a paper ). Personally I think it is good to have both points of view.

It should lastly be noted that almost none of the above actually used analysis - everything is taken care of by working in a particular category with a particular tensor product. So what I have just written out is no more than was known at the time of Cartan-Eilenberg.

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Having typed all that out, I guess I should really have written a blog post and left a link here. Ah well. –  Yemon Choi Oct 25 '10 at 1:26
    
This is great, thanks for such a complete answer! –  Piotr Nowak Oct 25 '10 at 2:06

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