Is the ideal property of $X^{**}$ inheritable to $X$?

Question

Let $X$ be an operator space such that there is a weak$^*$-continuous complete isometry $\phi$ from its second dual $X^{**}$ into a $W^*$-algebra $M$ in which $\phi(X^{**})$ is a (necessarily weak$^*$-closed) left ideal. Put $J:=\phi(\hat{X})$, where $\hat{X}$ is the canonical copy of $X$ in $X^{**}$.

Question: Is $J$ always algebraically closed?

I have observed that $J$ is a subtriple of $M$, that is, $ab^*c\in J$ whenever $a,b,c\in J$. However, I was not able to see that $J$ is closed under the product on $M$, i.e., $ab\in J$ whenever $a,b\in J$. If the answer is yes, then I can easily see that $J\subseteq JJ^*$, and hence $J$ is a left ideal in the $C^*$-algebra $JJ^*$, which I desire.

Answer 1

What about taking $J$ to be the set of bounded bi-infinite sequences satisfying $\lim_{n \to \infty} a_n = 2\lim_{n \to -\infty} a_n$ (with both limits existing)? That is a norm closed subspace of $l^\infty(\mathbb{Z})$, and it contains $c_0(\mathbb{Z})$, so it is weak * dense in $l^\infty(\mathbb{Z})$. But $J$ is not closed under products, nor under triple products, so something is wrong ...

Answer 2

The answer is no.

Proof. Theorem 9 in the reference [3] below implies that there exist a $C^{\star}$-algebra $\mathcal{A}$ and open projections $p,q\in\mathcal{A}^{**}$ such that $p$ and $q$ are Murray-von Neumann equivalent, but the $C^{\star}$-algebras $\mathcal{A}_p\,(:=p\mathcal{A}^{**}p\cap\mathcal{A})$ and $\mathcal{A}_q$ are not $^{\star}$-isomorphic. (In his example, actually $p,q\in\mathcal{A}$ ($q$ is the identity of $\mathcal{A}$), but here we do not need to assume this.) Let $e\in\mathcal{A}^{**}$ be the partial isometry that implements the equivalence, that is, $p=ee^{\star}$ and $q=e^{\star}e$. Put $X:=\mathcal{A}^{**}q\cap\mathcal{A}$, then $X$ is a left ideal, in particular, a TRO in $\mathcal{A}$. Since $q$ is open, $\mathcal{A}^{**}q=X^{\perp\perp}$ ($\cong X^{**}$ completely isometrically and weak$^*$-homeomorphically). Also note that $e\in X^{\perp\perp}$. (Indeed, $e$ is an extreme point of $\operatorname{Ball}(X^{\perp\perp})$.) Define a weak$^*$-continuous complete isometry $\phi$ from $X^{\perp\perp}$ into the $W^{\star}$-algebra $(XX^{\star})^{\perp\perp}$ by $\phi(x):=xe^{\star}\,\forall x\in X^{\perp\perp}$. (Note that in this case the inverse $\phi(X^{\perp\perp})\to X^{\perp\perp}$ is also weak$^*$ continuous.) Clearly $\phi(X^{\perp\perp})\,(=X^{\perp\perp}e^{\star})$ is a weak$^*$-closed left ideal in $(XX^{\star})^{\perp\perp}$. Put $J:=\phi(X)\,(=Xe^{\star})$, and suppose that $J$ is closed under the product on $(XX^{\star})^{\perp\perp}$, which is a restriction of the product on $\mathcal{A}^{**}$. Let $(e_{\alpha})$ be a net in $\operatorname{Ball}(X)$ converging to $e$ in weak$^*$. Since $J$ is algebraically closed, $Xe^{\star}e_{\alpha}e^{\star}\subseteq J$ for every $\alpha$, and $xe^{\star}e_{\alpha}e^{\star}\to xe^{\star}$ for each $x\in X$ in weak$^*$, hence in the weak topology on $J$. (If necessary, consider the pullback of $J=Xe^{\star}$ to $\hat{X}\,(\subseteq X^{**})$.) Thus by a well-known argument from the proof of Theorem 2.2 in [2], one can take a net $(\tilde{e}_{\alpha})\subseteq\operatorname{Ball}(X)$ such that $xe^{\star}\tilde{e}_{\alpha}e^{\star}\to xe^{\star}\,\forall x\in X$ in norm. That is, $J$ has a contractive approximate right identity $(\tilde{e}_{\alpha}e^{\star})$. Note that $ee^{\star}$ is a weak$^*$ accumulation point of the net $(\tilde{e}_{\alpha}e^{\star})$ since a weak$^*$-accumulation point of $(\tilde{e}_{\alpha}e^{\star})$ must serve as a right identity of $X^{\perp\perp}e^{\star}$ which is $ee^{\star}$. Thus by considering the adjoints, $ee^{\star}$ is a weak$^*$ accumulation point of $(e\tilde{e}_{\alpha}^{\star})$ as well. By Lemma 2.2 (1) in [1], the adjoint of $(\tilde{e}_{\alpha}e^{\star})$ also serves as a right approximate identity of $J$ (though it may not be in $J$), that is, $xe^{\star}e\tilde{e}_{\alpha}^{\star}\to xe^{\star}\,\forall x\in X$ in norm. But $xe^{\star}e\tilde{e}_{\alpha}^{\star}=x\tilde{e}_{\alpha}^{\star}\in XX^{\star}$, so that $J=Xe^{\star}\subseteq XX^{\star}$. This implies that $eX^{\star}\subseteq XX^{\star}$ and $eX^{\star}X\subseteq X$ noting that $X$ is a TRO. Thus $e$, thought as in the second dual of the linking $C^{\star}$-algebra $\mathcal{L}(X)$ of $X$, implements a Peligrad-Zsidó equivalence (Definition 1.1 in [4]) between $p\oplus0$ and $0\oplus q$ noting Lemma 1.3 in [4]. Hence by the last sentence of Definition 1.1 in [4] the $C^{\star}$-algebras $\mathcal{L}(X)_{p\oplus0}\,(:=(p\oplus0)\mathcal{L}(X)^{**}(p\oplus0)\cap\mathcal{L}(X))$ and $\mathcal{L}(X)_{0\oplus q}$ are $^{\star}$-isomorphic. But it is easy to see that $\mathcal{L}(X)_{p\oplus0}\cong\mathcal{A}_{p}$ and $\mathcal{L}(X)_{0\oplus q}\cong\mathcal{A}_{q}$ ($^{\star}$-isomorphically), so that $\mathcal{A}_{p}$ and $\mathcal{A}_{q}$ are $^{\star}$-isomorphic, a contradiction. $\square$

References

[1] D. P. Blecher, One-sided ideals and approximate identities in operator algebras, Journal of the Australian Mathematical Society 76(3) (2004), 425-447.

[2] E. G. Effros and Z.-J. Ruan, On non-self-adjoint operator algebras, Proceedings of the American Mathematical Society 110(4) (1990), 915-922.

[3] H. Lin, Equivalent open projections and corresponding hereditary $C^*$-subalgebras, Journal of the London Mathematical Society, Second Series 41(2) (1990), 295-301.

[4] C. Peligrad and L. Zsidó, Open projections of $C^*$-algebras: comparison and regularity, Operator theoretical methods (Timișoara, 1998), 285-300.

Cf. Here is my related question.