When is the annihilator of the commutator subspace a complemented subspace?

Question

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 $f\in A^{*}$?

Answer 1

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.)

Answer 2

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.