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This is a follow up on an earlier question.

In [Lau&Loy, 2008] a Banach algebra $\mathcal{U}$ was called to have the Tomiyama property if any contractive projection $P:\mathcal{U}\to \mathcal{U}$, whose range $P(\mathcal{U})$ is a subalgebra, is a conditional expectation.

Analogously, let's say that $\mathcal{U}$ has the "separable Tomiyama property" if any contractive projection, whose range is a separable subalgebra, is a conditional expectation. Let's also say that $\mathcal{U}$ has property ($T_\omega$) if every $1$-complemented separable subalgebra of $\mathcal{U}$ is contained in a unital separable subalgebra, which is a range of a conditional expectation. Similarly, $\mathcal{U}$ has property left ($T_\omega$) ( resp. right ($T_\omega$) ) if every $1$-complemented separable subalgebra of $\mathcal{U}$ is contained in a unital separable subalgebra $W$, which is a range of a contractive projection $P$ satisfying $P(ux) = uP(x)$ ( resp. $P(xu) = P(x)u$ ) for all $u\in W$, $x\in A$.

Question: Let $A$ be a reflexive, non-separable, simple, unital (not necessarily self-adjoint) operator algebra, which possesses no minimal idempotents, i.e., $socle(A)=\{0\}$. Does $A$ have property ($T_\omega$)?

It would be great if there were an answer for the other properties instead of (or in addition to) property ($T_\omega$).

PS: Could you please share general references in the comments about projections on Banach algebras, specifically about contractive projections whose range is a subalgebra?

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  • $\begingroup$ If $P$ is completely contractive, then $P$ is a conditional expectation by Corollary 4.2.9 in the book by Blecher & le Merdy. books.google.ca/books?id=oBh0EE0HQLQC&pg=PA155 $\endgroup$
    – Onur Oktay
    Commented Apr 11, 2022 at 5:06
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    $\begingroup$ One can prove: for every separable subset $X$ of every operator algebra $A$, there is a separable subalgebra $B$ of $A$ that contains $X$ and that is weakly completely $1$-complemented in $A$, i.e., there is a completely contractive projection $P$ from $A^{**}$ onto $B^{**}$ (which has to be a conditional expectation provided that $A$ is unital and $B$ contains the unit). $\endgroup$
    – Narutaka OZAWA
    Commented Apr 13, 2022 at 8:16
  • $\begingroup$ The proof is as in the earlier question by interlacing. For every separable subspace $X_0$ of every operator space $Y$ and every $n$, there are a separable subspace $X_1\subset Y$ that contains $X_0$ and a linear map $P\colon Y\to X_1^{**}$ with $P|_{X_0}$ canonical inclusion and $\| P\|_n=1$. This fact follows from the same for Banach spaces (or by an argument involving operator space projective tensor product). $\endgroup$
    – Narutaka OZAWA
    Commented Apr 13, 2022 at 8:41

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