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

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 $$f\in A^{*}$$?

• 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/… Sep 24, 2021 at 2:41
• 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 Sep 24, 2021 at 2:44
• 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 Sep 24, 2021 at 2:49
• 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) Sep 24, 2021 at 2:56
• 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. Sep 24, 2021 at 3:05

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.)

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.

• 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. Sep 25, 2021 at 3:45