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Let $A$ be a unital Banach algebra and $C$ be its commutator subspace, i.e., $C$ is the norm-closure of the subspace spanned by the elements of the form $xy-yx$ in $A$.

Notation: Let $C^{\perp}=\{f\in A^{*}: C\subseteq kerf\}$ denote the annihilator of $C$, and let $Z(A)$ denote the center of $A$. Given $a\in A$ and $f\in A^{*}$, $af\in A^{*}$ is defined by $af(x):= f(xa)$ for each $x\in A$.

Q1: What are the necessary and (or) sufficient conditions that $C^{\perp}$ is a complemented subspace in $A^{*}$?

Q2: Suppose $C^{\perp}$ is complemented in $A^{*}$. What are the necessary and sufficient conditions for the existence of a projection $P:A^{*}\to C^{\perp}$ satisfying $P(af) = a(Pf)$ for all $a\in Z(A)$ and $f\in A^{*}$?

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    $\begingroup$ Your space $C^\perp$ seems to be the same as the space of tracial linear functionals on $A$. I don't think there is a single set of necessary and sufficient conditions for Q1, but in some private calculations I was doing about ten years ago I think I found a proof that this holds if $A$ is a Cstar algebra. I was also able to find a Banach algebra for which $C^\perp$ is not complemented in $A^*$ by indirectly using the result in de la Salle's answer to this question mathoverflow.net/questions/81032/… $\endgroup$
    – Yemon Choi
    Sep 24, 2021 at 2:41
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    $\begingroup$ Also, in Q2, do you really want the chosen projection P to be a Z(A)-module map, or do you just want to know if C^\perp being complemented in A implies the existence of some other projection which is a Z(A)-module map? If it is the weaker version that you are interested in, then I think amenablity of Z(A) would suffice, but I have not checked my calculations $\endgroup$
    – Yemon Choi
    Sep 24, 2021 at 2:44
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    $\begingroup$ If you don't mind me asking, what was the original motivation for Q1? The problem that motivated me to look at the same question in 2011 came from trying to prove/disprove things about weak amenability of tensor products of Banach algebras $\endgroup$
    – Yemon Choi
    Sep 24, 2021 at 2:49
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    $\begingroup$ Regarding the link, that MO question doesn't immediately explain how to build the counterexample. I could send you some notes by email if you don't mind waiting a day or two (I never wrote anything up for publication) $\endgroup$
    – Yemon Choi
    Sep 24, 2021 at 2:56
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    $\begingroup$ For Q2, I can't see how to use amenability of $A$, because $C^\perp$ might not be a sub-$A$-module of $A^*$. It is a sub-Z(A)-module, see my answer below. $\endgroup$
    – Yemon Choi
    Sep 24, 2021 at 3:05

2 Answers 2

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Regarding Q2: note that $C$ is a sub-$Z(A)$-module of $A$. So $C^\perp= (A/C)^*$ is a dual $Z(A)$-module, and by assumption $C^\perp$ is complemented as a Banach space in $A^*$.

Now it follows from general results of Helemskii, see also Curtis-Loy JLMS 1989 Theorem 2.3, that whenever B is amenable, M is a B-module and N is a complemented subspace of M such that N is also a dual B-module, then there is a projection of M onto N that is also a B-module map.

Therefore: if $Z(A)$ is amenable and $C^\perp$ is complemented as a Banach space in $A^*$, it is complemented as a Banach $Z(A)$-module.

(I am being a bit sketchy here, if I find time later I can try to fill in details if something is not clear.)

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The purpose of this note is not to answer my own question, but to share a sufficient condition with everyone: If $A$ is amenable, then Q1 & Q2 are answered affirmatively.


Notation: $A\hat{\otimes}_{\pi} A$ is the projective tensor product of $A$ with itself. For an $A$-bimodule $M$, the bimodule center is defined by $\mathcal{Z}(A,M) =\{\beta\in M: a\beta=\beta a\hspace{4mm} \forall a\in A \}$. $A^{*}$ is identified with the image of the map (dual of the product map) $\sim:A^{*}\to (A\hat{\otimes}_{\pi} A)^{*}$, which is defined by $\widetilde{f}(x\otimes y) = f(xy)$ on basic tensors (and extended linearly & continuously afterwards).


$A$ is amenable if and only if $(A\hat{\otimes}_{\pi}A)^{*} = A^{*} \oplus K$ as a direct sum of $A$-bimodules for some sub-$A$-bimodule $K$ of $(A\hat{\otimes}_{\pi}A)^{*}$, see Curtis & Loy . From this, it is not difficult to show that $$\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*}) = \mathcal{Z}(A,A^{*}) \oplus\mathcal{Z}(A,K)$$ as a direct sum of $Z(A)$-modules. It is also not difficult to show that every $\beta\in \mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ is given by $$\beta(x\otimes y) = f(yx) \hspace{8mm} \forall x,y\in A$$ for some $f\in A^{*}$, and every $\beta\in\mathcal{Z}(A,A^{*})$ is similarly given by $f\in A^{*}$ with the extra property that $af=fa$ for every $a\in A$. Hence, one may identify $\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ with $A^{*}$, and $\mathcal{Z}(A,A^{*})$ by $C^{\perp}$ in a natural way via $Z(A)$-module isomorphisms. Similarly, $\mathcal{Z}(A,K)$ is isomorphic to a $Z(A)$-submodule of $A^{*}$, call it $B$. Consequently, $$A^{*} = C^{\perp}\oplus B$$ as a direct sum of $Z(A)$ submodules. In particular, $C^{\perp}$ is a complemented subspace of $A^{*}$.


N.B.: Amenability is a considerably strong condition compared to the ones given by Q1 & Q2. Thus, it is interesting to see weaker sufficient conditions. On the contrary, Q1/Q2 provides relatively easy to check tests for non-amenability. Thus, it is interesting to see a class of Banach algebras where Q1 or Q2 is not satisfied.

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  • $\begingroup$ If $A$ has a Mityagin decomposition, then $A$ has no nonzero traces, e.g. jstor.org/stable/24715582. In the same paper (Example 3.8), for $E=\ell^p, L^p$ for $1\leq p<\infty$, the algebras $A=B(E)$ have no nonzero traces, so $C^{\perp}=\{0\}$. For the James space $E=J_p$ and $A=B(E)$, $C^{\perp}$ is 1-dimensional, so complemented. Clearly $Z(A)$ is trivial (1-dim.) for these $A$. Thus, Q1 and Q2 are answered affirmatively. Notice that none of these algebras are amenable. $\endgroup$
    – Onur Oktay
    Sep 25, 2021 at 3:45

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