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 $X^{\*\*}$.



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> **Question:** Is $J$ always algebraically closed?
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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.



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

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> The answer is **no**.
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*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.



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> Cf. [Here](https://mathoverflow.net/questions/299759/open-projections-and-murray-von-neumann-equivalence) is my related question.
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