Preduals for spaces of operators involving Hilbert spaces and trace-class operators

TL;DR

This paper explores the structure of preduals of operator spaces involving Hilbert spaces and trace-class operators, establishing correspondences and uniqueness results for their isometric preduals.

Contribution

It introduces a natural correspondence between isometric preduals of operator spaces and their underlying spaces, extending the understanding of preduals in the context of Hilbert and trace-class operators.

Findings

Isometric preduals of $\\mathcal{L}(H,Y)$ correspond to those of $Y$.

Preduals of $\\mathcal{L}(\mathcal{S}_1)$ relate to preduals of $\mathcal{S}_1$.

The compact operators are the unique predual of $\mathcal{S}_1$ with separately weak* continuous multiplication.

Abstract

Continuing the study of preduals of spaces $\mathcal{L}(H,Y)$ of bounded, linear maps, we consider the situation that $H$ is a Hilbert space. We establish a natural correspondence between isometric preduals of $\mathcal{L}(H,Y)$ and isometric preduals of $Y$. The main ingredient is a Tomiyama-type result which shows that every contractive projection that complements $\mathcal{L}(H,Y)$ in its bidual is automatically a right $\mathcal{L}(H)$-module map. As an application, we show that isometric preduals of $\mathcal{L}(\mathcal{S}_1)$, the algebra of operators on the space of trace-class operators, correspond to isometric preduals of $\mathcal{S}_1$ itself (and there is an abundance of them). On the other hand, the compact operators are the unique predual of $\mathcal{S}_1$ making its multiplication separately weak* continuous.

Equations

$$\Vert af+bg{\Vert }^2\le \Vert af{\Vert }^2+\Vert bg{\Vert }^2.$$

$$\Vert afx+bgx{\Vert }^2=\Vert afx{\Vert }^2+\Vert bgx{\Vert }^2.$$

$$\|af+bg\|^{2}=\sup_{\|x\|\leq 1}\|afx+bgx\|^{2}=\sup_{\|x\|\leq 1}\big{(}\|afx\|^{2}+\|bgx\|^{2}\big{)}\leq\|af\|^{2}+\|bg\|^{2},$$

$$\Vert fa+gb{\Vert }^2\le \Vert fa{\Vert }^2+\Vert gb{\Vert }^2.$$

$$(a^t{)}^{\ast }{b}^t=(a^{\ast }{)}^t{b}^t=(b{a}^{\ast }{)}^t=0.$$

$$\Vert fa+gb{\Vert }^2=\Vert (fa+gb{)}^t{\Vert }^2=\Vert a^t{f}^t+b^t{g}^t{\Vert }^2\le \Vert a^t{f}^t{\Vert }^2+\Vert b^t{g}^t{\Vert }^2=\Vert fa{\Vert }^2+\Vert gb{\Vert }^2.$$

$$\Vert Fa\Vert =\Vert Fap\Vert \le \underset{i}{\underline{\lim }}\Vert f_{i}p\Vert \le \underset{i}{\overline{\lim }}\Vert f_{i}p\Vert \le \Vert Fa\Vert ,$$

$$\Vert Fa+Gb{\Vert }^2\le \underset{i,j}{\underline{\lim }}\Vert f_{i}p+g_{j}q{\Vert }^2\le \underset{i,j}{\underline{\lim }}\left(\Vert f_{i}p{\Vert }^2+\Vert g_{j}q{\Vert }^2\right)=\Vert Fa{\Vert }^2+\Vert Gb{\Vert }^2.\text{qed}$$

$$(1+\lambda {)}^2\Vert \pi (Fp)q{\Vert }^2$$

$$\le \Vert Fp+\lambda \pi (Fp)q{\Vert }^2\le \Vert Fp{\Vert }^2+\Vert \lambda \pi (Fp)q{\Vert }^2$$

$$=\Vert Fp{\Vert }^2+{\lambda }^2\Vert \pi (Fp)q{\Vert }^2.$$

$$(1+2\lambda )\Vert \pi (Fp)q{\Vert }^2\le \Vert Fp{\Vert }^2.$$

$$\pi (Fp)=\pi (Fp)p.$$

$$\pi (Fp)=\pi (Fp)p=\pi (F)p.$$

$${\alpha }_{X,Y}(F)(x)={\mathrm{ev}}_x^{\ast \ast }(F),$$

$$r_{\pi }(F)(x)=\pi ({\mathrm{ev}}_x^{\ast \ast }(F)),$$

$$\mathcal{L}(X,Y)\cong \mathcal{L}(X,F^{\ast })\cong (X\hat{\otimes }F{)}^{\ast }.$$

$$\mathrm{tr}(a):=\sum _j⟨a{e}_j,e_j⟩$$

$$\mathcal{K}(H{)}^{\ast }\cong {\mathcal{S}}_1(H),\text{ and }{\mathcal{S}}_1(H{)}^{\ast }\cong \mathcal{L}(H).$$

$${\mathcal{L}}\big{(}X\hat{\otimes}Y,E\big{)}\cong{\mathcal{L}}\big{(}X,{\mathcal{L}}(Y,E)\big{)}.$$

$$\textstyle{{{\mathcal{L}}\big{(}X\hat{\otimes}Y,E\big{)}^{**}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$

$$\pi(F)(x\otimes y)=\big{[}\pi^{\prime}(\tilde{F})(x)\big{]}(y)=\tau\big{(}{\mathrm{ev}}_{x}^{**}(\tilde{F})\big{)}(y),$$

$$\tau(G)(y)=\sigma\big{(}{\mathrm{ev}}_{y}^{**}(G)\big{)},$$

$$\pi(F)(x\otimes y)=\tau\big{(}{\mathrm{ev}}_{x}^{**}(\tilde{F})\big{)}(y)=\sigma\big{(}{\mathrm{ev}}_{y}^{**}({\mathrm{ev}}_{x}^{**}(\tilde{F}))\big{)}=\sigma\big{(}{\mathrm{ev}}_{x\otimes y}^{**}(F)\big{)},$$

$${\mathcal{L}}({\mathcal{S}}_{1}(H))\cong{\mathcal{L}}\big{(}{\mathcal{S}}_{1}(H),{\mathcal{K}}(H)^{*}\big{)}\cong\left({\mathcal{S}}_{1}(H)\hat{\otimes}{\mathcal{K}}(H)\right)^{*},$$

$$⟨{\phi }_{b}a,x⟩=⟨L_a^{\ast }({\phi }_b),x⟩=⟨{\phi }_b,ax⟩=\mathrm{tr}(bax)=⟨{\phi }_{ba},x⟩,$$

$$\underset{D}{\lim }\Vert a{p}_{D}b-ab\Vert =0.$$

Full text

The main ingredient is a Tomiyama-type result which shows that every contractive projection that complements ${\mathcal{L}}(H,Y)$ in its bidual is automatically a right ${\mathcal{L}}(H)$-module map.

As an application, we show that isometric preduals of ${\mathcal{L}}({\mathcal{S}}_{1})$, the algebra of operators on the space of trace-class operators, correspond to isometric preduals of ${\mathcal{S}}_{1}$ itself (and there is an abundance of them). On the other hand, the compact operators are the unique predual of ${\mathcal{S}}_{1}$ making its multiplication separately weak∗ continuous.

The author was partially supported by the Deutsche Forschungsgemeinschaft (SFB 878 Groups, Geometry & Actions).

1. Introduction

An isometric predual of a Banach space $X$ is a Banach space $F$ together with an isometric isomorphism $X\cong F^{*}$. Every predual induces a weak∗ topology. Due to the importance of weak∗ topologies, it is interesting to study the existence and uniqueness of preduals; see the survey [God89] and the references therein.

Given Banach spaces $X$ and $Y$, let ${\mathcal{L}}(X,Y)$ denote the space of operators from $X$ to $Y$. Every isometric predual of $Y$ induces an isometric predual of ${\mathcal{L}}(X,Y)$: If $Y\cong F^{*}$, then ${\mathcal{L}}(X,Y)\cong(X\hat{\otimes}F)^{*}$.

Problem \thepbmCt.

Find conditions on $X$ and $Y$ guaranteeing that every isometric predual of ${\mathcal{L}}(X,Y)$ is induced from an isometric predual of $Y$.

Given reflexive spaces $X$ and $Y$, Godefroy and Saphar show that $X\widehat{\otimes}Y^{*}$ is the strongly unique isometric predual of ${\mathcal{L}}(X,Y)$; see [GS88, Proposition 5.10]. In particular, in this case every isometric predual ${\mathcal{L}}(X,Y)$ is induced from $Y$.

The main result of this paper extends this to the case that $X$ is a Hilbert space $H$ and $Y$ is arbitrary: Every isometric predual of ${\mathcal{L}}(H,Y)$ is induced from an isometric predual of $Y$; see Section 2. In particular, ${\mathcal{L}}(H,Y)$ has a (strongly unique) isometric predual if and only if $Y$ does; see Section 2.

To obtain these results, we use that isometric preduals of $Y$ naturally correspond to contractive projections ${\mathcal{L}}(X,Y)^{**}\to{\mathcal{L}}(X,Y)$ that are right ${\mathcal{L}}(X)$-module maps and have weak∗ closed kernel; see [GT16, Theorem 5.7]. Hence, we are faced with:

Problem \thepbmCt.

Find conditions on $X$ and $Y$ guaranteeing that every contractive projection ${\mathcal{L}}(X,Y)^{**}\to{\mathcal{L}}(X,Y)$ is automatically a right ${\mathcal{L}}(X)$-module map.

It was shown by Tomiyama that every contractive projection from a $C^{*}$-algebra $A$ onto a sub-$C^{*}$-algebra $B$ is automatically a $B$-bimodule map. We therefore consider a positive solution to Section 1 a Tomiyama-type result. Adapting the proof of Tomiyama’s result, we obtain a positive solution to Section 1 whenever $X$ is a Hilbert space; see Section 2.

In Section 3, we show that our results also hold when the Hilbert space $H$ is replaced by the space of trace-class operators ${\mathcal{S}}_{1}(H)$. It follows that isometric preduals of ${\mathcal{L}}({\mathcal{S}}_{1}(H))$ naturally correspond to isometric preduals of ${\mathcal{S}}_{1}(H)$; see Section 3. We note that ${\mathcal{S}}_{1}(H)$ - and consequently ${\mathcal{L}}({\mathcal{S}}_{1}(H))$ - has many different isometric preduals. On the other hand, we show that the ‘standard’ predual of compact operators is the unique predual of ${\mathcal{S}}_{1}(H)$ making its multiplication separately weak∗ continuous; see Section 3.

Acknowledgements

The author would like to thank Eusebio Gardella and Tim de Laat for valuable feedback.

Notation

Given Banach spaces $X$ and $Y$, an operator from $X$ to $Y$ is a bounded, linear map $X\to Y$. The space of such operators is denoted by ${\mathcal{L}}(X,Y)$. The projective tensor product of $X$ and $Y$ is denoted by $X\hat{\otimes}Y$. We identify $X$ with a subspace of its bidual, and we let $\kappa_{X}\colon X\to X^{**}$ denote the inclusion. A projection $\pi\colon X^{**}\to X$ is an operator satisfying $\pi(x)=x$ for all $x\in X$.

2. Preduals involving Hilbert spaces

Throughout, $X$ and $Y$ denote Banach spaces. For conceptual reasons, it is useful to consider preduals of $X$ as subsets of $X^{*}$. More precisely, a closed subspace $F\subseteq X^{*}$ is an (isometric) predual of $X$ if for the inclusion map $\iota_{F}\colon F\to X^{*}$, the transpose map $\iota_{F}^{*}\colon X^{**}\to F^{*}$ restricts to an (isometric) isomorphism $X\to F^{*}$.

The space $X$ is said to have a strongly unique isometric predual if there exists an isometric predual $F\subseteq X^{*}$ and if $F=G$ for every isometric predual $G\subseteq X^{*}$. Every reflexive space $X$ has a strongly unique isometric predual, namely $X^{*}$.

\thepgrCt**.**

The space ${\mathcal{L}}(X,Y)$ has a natural ${\mathcal{L}}(Y)$-${\mathcal{L}}(X)$-bimodule structure. Given $a\in{\mathcal{L}}(X)$, the action of $a$ is given by $R_{a}\colon{\mathcal{L}}(X,Y)\to{\mathcal{L}}(X,Y)$, $R_{a}(f):=f\circ a$, for $f\in{\mathcal{L}}(X,Y)$. Thus, $a$ acts by precomposing on the right of ${\mathcal{L}}(X,Y)$. Similarly, the action of $b\in{\mathcal{L}}(Y)$ is given by postcomposing on the left, that is, by $L_{b}\colon{\mathcal{L}}(X,Y)\to{\mathcal{L}}(X,Y)$, $L_{b}(f):=b\circ f$, for $f\in{\mathcal{L}}(X,Y)$.

We obtain a ${\mathcal{L}}(X)$-${\mathcal{L}}(Y)$-bimodule structure on ${\mathcal{L}}(X,Y)^{*}$. The left action of $a\in{\mathcal{L}}(X)$ on ${\mathcal{L}}(X,Y)^{*}$ is given by $R_{a}^{*}$. The right action of $b\in{\mathcal{L}}(Y)$ on ${\mathcal{L}}(X,Y)^{*}$ is given by $L_{b}^{*}$. Similarly, we obtain a ${\mathcal{L}}(Y)$-${\mathcal{L}}(X)$-bimodule structure on ${\mathcal{L}}(X,Y)^{**}$.

Given a $C^{*}$-algebra $A$ and $a,b,x,y\in A$ with $a^{*}b=0$, we have $\|ax+by\|^{2}\leq\|ax\|^{2}+\|by\|^{2}$, which is an analog of Bessel’s inequality; see [Bla06, II.3.1.12, p.66]. We first prove two versions of this result in a more general context.

Lemma 2.1**.**

Let $H$ be a Hilbert space, let $X$ be a Banach space, let $a,b\in{\mathcal{L}}(H)$ satisfying $a^{*}b=0$, and let $f,g\in{\mathcal{L}}(X,H)$. Then

[TABLE]

Proof.

The equation $a^{*}b=0$ implies that the ranges of $a$ and $b$ are orthogonal. Given $x\in X$, it follows that the elements $afx$ and $bgx$ are orthogonal in $H$, whence

[TABLE]

Using this at the second step, we deduce that

[TABLE]

as desired. ∎

Lemma 2.2**.**

Let $H$ be a Hilbert space, let $Y$ be a Banach space, let $a,b\in{\mathcal{L}}(H)$ satisfying $ab^{*}=0$, and let $f,g\in{\mathcal{L}}(H,Y)$. Then

[TABLE]

Similarly, given $F,G\in{\mathcal{L}}(H,Y)^{**}$, we have $\|Fa+Gb\|^{2}\leq\|Fa\|^{2}+\|Gb\|^{2}$.

Proof.

We denote the transpose of an operator $h$ by $h^{t}$, to distinguish it from the adjoint of an operator in ${\mathcal{L}}(H)$. We have $a^{t},b^{t}\in{\mathcal{L}}(H^{*})$ and $f^{t},g^{t}\in{\mathcal{L}}(Y^{*},H^{*})$. Further, $(fa)^{t}=a^{t}f^{t}$, where $a^{t}f^{t}$ is given by the left action of ${\mathcal{L}}(H^{*})$ on ${\mathcal{L}}(Y^{*},H^{*})$. It follows from $ab^{*}=0$ that $ba^{*}=0$. We have $(a^{*})^{t}=(a^{t})^{*}$ in ${\mathcal{L}}(H^{*})$, and therefore

[TABLE]

Applying Lemma 2.1 at the third step, we compute

[TABLE]

Let us show the second inequality. Let $p$ and $q$ be the right support projections of $a$ and $b$ in ${\mathcal{L}}(H)$, respectively. Then $ap=a$, $bq=b$, and $pq^{*}=0$. It follows that $Fa=Fap$ and $Gb=Gbq$.

Using Goldstine’s theorem, we choose nets $(f_{i})_{i}$ and $(g_{j})_{j}$ in ${\mathcal{L}}(H,Y)$ such that $(f_{i})_{i}$ converges weak∗ to $Fa$, such that $(g_{j})_{j}$ converges weak∗ to $Gb$, and such that $\|f_{i}\|\leq\|Fa\|$ for all $i$ and $\|g_{j}\|\leq\|Gb\|$ for all $j$. Then $(f_{i}p)_{i}$ converges weak∗ to $Fap$. Using this at the second step, we deduce that

[TABLE]

and hence $\lim_{i}\|f_{i}p\|=\|Fa\|$. Analogously, we obtain that $\lim_{j}\|g_{j}q\|=\|Gb\|$.

Using this at the third step, using that the net $(f_{i}p+g_{j}q)_{i,j}$ converges weak∗ to $Fa+Gb$ at the first step, and using the first inequality of this lemma at the second step, we deduce that

[TABLE]

Let $A$ be a $C^{*}$-algebra and let $B\subseteq A$ be a sub-$C^{*}$-algebra. By Tomiyama’s theorem, every contractive projection $\pi\colon A\to B$ is automatically a $B$-bimodule map (called a conditional expectation). The next result is in the same spirit. It provides a partial positive solution to Section 1.

Theorem \thethmCt.

Let $H$ be a Hilbert space, and let $Y$ be a Banach space. Then every contractive projection $\pi\colon{\mathcal{L}}(H,Y)^{**}\to{\mathcal{L}}(H,Y)$ is automatically a right ${\mathcal{L}}(H)$-module map, that is, $\pi(Fa)=\pi(F)a$ for every $F\in{\mathcal{L}}(H,Y)^{**}$ and $a\in{\mathcal{L}}(H)$.

Proof.

First, we show the result for the case that $a$ is a projection. Let $p\in{\mathcal{L}}(H)$ satisfy $p=p^{2}=p^{*}$, and set $q:=1-p$. The following argument is adapted from the proof of Tomiyama’s theorem in [Bla06, Theorem II.6.10.2, p.132]. Let $\lambda>0$. We have $\pi(\pi(Fp)q)q=\pi(Fp)q$. Using this at the first step, using that $\|\pi\|\leq 1$ and $\|q\|\leq 1$ at the third step, and using Lemma 2.2 at the fourth step, we get

[TABLE]

It follows that

[TABLE]

Since this holds for every $\lambda>0$, we deduce that $\pi(Fp)q=0$. Adding $\pi(Fp)p$ to this equation, we obtain

[TABLE]

Switching the place of $p$ and $q$ in the above argument, we get $\pi(Fq)p=0$. Adding $\pi(Fp)p$, we get $\pi(F)p=\pi(Fp)p$. We deduce that

[TABLE]

Finally, we use that every operator on a Hilbert space is a linear combination of finitely many projections; see [PT67, Corollary 2.3]. ∎

\thepgrCt**.**

Given $x\in X$, we let ${\mathrm{ev}}_{x}\colon{\mathcal{L}}(X,Y)\to Y$ denote the evaluation map, given by ${\mathrm{ev}}_{x}(f):=f(x)$, for $f\in{\mathcal{L}}(X,Y)$. Let $\alpha_{X,Y}\colon{\mathcal{L}}(X,Y)^{**}\to{\mathcal{L}}(X,Y^{**})$ be the operator introduced in [GT16, Definition 3.16]. We have

[TABLE]

for all $F\in{\mathcal{L}}(X,Y)^{**}$ and $x\in X$; see [GT16, Lemma 3.18]. The map $\alpha_{X,Y}$ is always a contractive, right ${\mathcal{L}}(X)$-module map.

Let $\pi\colon Y^{**}\to Y$ be a projection. Define $\pi_{*}\colon{\mathcal{L}}(X,Y^{**})\to{\mathcal{L}}(X,Y)$ by $\pi_{*}(f):=\pi\circ f$, for $f\in{\mathcal{L}}(X,Y^{**})$. Set $r_{\pi}:=\pi_{*}\circ\alpha_{X,Y}\colon{\mathcal{L}}(X,Y)^{**}\to{\mathcal{L}}(X,Y)$. Note that

[TABLE]

for $F\in{\mathcal{L}}(X,Y)^{**}$ and $x\in X$. The map $r_{\pi}$ was considered in [GT16, Section 4], where it is shown that $r_{\pi}$ is a right ${\mathcal{L}}(X)$-module projection.

Recall that (concrete) isometric preduals of $Y$ are in natural bijection with contractive projections $Y^{**}\to Y$ that have weak∗ closed kernel; see for example [GT16, Proposition 2.7]. Every predual induces a weak∗ topology. A predual of ${\mathcal{L}}(X,Y)$ makes the right action by each element from ${\mathcal{L}}(X)$ weak∗ continuous if and only if the associated projection ${\mathcal{L}}(X,Y)^{**}\to{\mathcal{L}}(X,Y)$ is a right ${\mathcal{L}}(X)$-module map; see for example [GT16, Proposition B.6]. The following result is contained in [GT16, Theorem 4.7].

Theorem \thethmCt.

Let $X$ and $Y$ be Banach spaces with $X\neq\{0\}$. Given a projection $\pi\colon Y^{**}\to Y$, let $r_{\pi}\colon{\mathcal{L}}(X,Y)^{**}\to{\mathcal{L}}(X,Y)$ be as in Section 2. This assignment defines a natural bijection between the following classes:

1. (1)

Contractive projections $Y^{**}\to Y$; 2. (2)

Contractive, right ${\mathcal{L}}(X)$-module projections ${\mathcal{L}}(X,Y)^{**}\to{\mathcal{L}}(X,Y)$.

Moreover, the kernel of $\pi$ is weak∗ closed if and only if the kernel of $r_{\pi}$ is. Thus, the above correspondence restricts to a natural bijection between isometric preduals of $Y$ and isometric preduals of ${\mathcal{L}}(X,Y)$ that make the right action by ${\mathcal{L}}(X)$ weak∗ continuous.

Combining Theorems 2 and 2, we obtain the main result of this paper:

Theorem \thethmCt.

Let $H$ be a Hilbert space with $H\neq\{0\}$, and let $Y$ be a Banach space. Assigning to a projection $\pi\colon Y^{**}\to Y$ the projection $r_{\pi}\colon{\mathcal{L}}(H,Y)^{**}\to{\mathcal{L}}(H,Y)$, as in Section 2, establishes a natural bijection between the following classes.

1. (1)

Contractive projections $Y^{**}\to Y$; 2. (2)

Contractive projections ${\mathcal{L}}(H,Y)^{**}\to{\mathcal{L}}(H,Y)$.

Restricted to projections with weak∗ closed kernel, we obtain a natural bijection between isometric preduals of $Y$ and isometric preduals of ${\mathcal{L}}(H,Y)$.

Corollary \thecorCt.

Let $H$ be a Hilbert space with $H\neq\{0\}$, and let $Y$ be a Banach space. Then ${\mathcal{L}}(H,Y)$ is $1$-complemented in its bidual if and only if $Y$ is. Further, ${\mathcal{L}}(H,Y)$ has an isometric predual if and only if $Y$ does. Moreover, if ${\mathcal{L}}(H,Y)$ has a strongly unique isometric predual if and only if $Y$ does.

Corollary \thecorCt.

Let $H$ be a Hilbert space, let $Y$ be a Banach space, and let $F\subseteq{\mathcal{L}}(H,Y)^{*}$ be an isometric predual. Then for each $a\in{\mathcal{L}}(H)$, the right action of $a$ on ${\mathcal{L}}(H,Y)$ is continuous for the weak∗ topology induced by $F$.

Remark \thermkCt.

There is a canonical isometric isomorphism $(X\hat{\otimes}Y)^{*}\cong{\mathcal{L}}(X,Y^{*})$. Given an isometric isomorphism $Y\cong F^{*}$, we obtain isometric isomorphisms

[TABLE]

Hence, $X\widehat{\otimes}F$ is an isometric predual of ${\mathcal{L}}(X,Y)$. Given a Hilbert space $H$, Section 2 states that every isometric predual of ${\mathcal{L}}(H,Y)$ occurs this way. In particular, given an isometric isomorphism ${\mathcal{L}}(H,Y)\cong G^{*}$ for some Banach space $G$, there is an isometric isomorphism $H\hat{\otimes}F\cong G$, for some isometric predual $F$ of $Y$.

Remark \thermkCt.

By [GS88, Proposition 5.10], $X\hat{\otimes}F$ is the strongly unique isometric predual of ${\mathcal{L}}(X,F^{*})$ if $X$ and $F$ satisfy the Radon Nikodým property (RNP). Every reflexive space (in particular, every Hilbert space) satisfies the RNP. Thus, if $Y$ has an isometric predual satisfying the RNP, then Section 2 follows from [GS88]. However, not every Banach spaces with strongly unique isometric predual occurs as the dual of a space satisfying the RNP; see Section 2.

Example \theexaCt.

Let $M$ be a von Neumann algebra. By Sakai’s theorem, $M$ has a strongly unique isometric predual, denoted by $M_{*}$. It follows from Section 2 that $H\hat{\otimes}M_{*}$ is the strongly unique isometric predual of ${\mathcal{L}}(H,M)$. By [Chu81, Theorem 4], $M_{*}$ satisfies the RNP if and only if $M$ is a direct sum of type $\mathrm{I}$ factors.

Let $\mathcal{R}$ denote the hyperfinite $\mathrm{II}_{1}$-factor. Then $\mathcal{R}_{*}$ does not have the RNP, yet $H\hat{\otimes}\mathcal{R}_{*}$ is the strongly unique isometric predual of ${\mathcal{L}}(H,\mathcal{R})$.

Question \theqstCt.

Does Section 2 hold when the Hilbert space is replaced by a general Banach space satisfying the RNP? More modestly, if $X$ is an $L^{p}$-space, do isometric preduals of ${\mathcal{L}}(X,Y)$ correspond to isometric preduals of $Y$?

Remark \thermkCt.

Note that every positive solution of Section 1 leads to an analog of Section 2. Therefore, Section 2 has a positive answer if the following instance of Section 1 has a positive solution: Given a measure space $\mu$ and a Banach space $Y$, is every contractive projection ${\mathcal{L}}(L^{p}(\mu),Y)^{**}\to{\mathcal{L}}(L^{p}(\mu),Y)$ automatically a right ${\mathcal{L}}(L^{p}(\mu))$-module map?

Consider the space $\ell^{\infty}$ of bounded sequences. Since $\ell^{\infty}$ is a von Neumann algebra, it has a strongly unique isometric predual. Thus, if Section 1 had a positive solution for $X=Y=\ell^{\infty}$, then ${\mathcal{L}}(\ell^{\infty})$ would have a strongly unique isometric predual. However, it was noted in [GS88, Remark 5.12] that ${\mathcal{L}}(\ell^{\infty})$ has nonisomorphic isometric preduals. It follows in particular that there exists a contractive projection ${\mathcal{L}}(\ell^{\infty})^{**}\to{\mathcal{L}}(\ell^{\infty})$ that is not a right ${\mathcal{L}}(\ell^{\infty})$-module map.

3. Preduals involving trace-class operators

Throughout this section $H$ denotes a Hilbert space with $H\neq\{0\}$. We let ${\mathcal{K}}(H)$ and ${\mathcal{S}}_{1}(H)$ denote the compact and trace-class operators on $H$, respectively.

An operator $a\in{\mathcal{L}}(H)$ belongs to ${\mathcal{S}}_{1}(H)$ if and only if for some (equivalently, every) orthonormal basis $(e_{j})_{j}$ of $H$ the sum $\sum_{j}\langle|a|e_{j},e_{j}\rangle$ is finite. Given $a\in{\mathcal{S}}_{1}(H)$ and an orthonormal basis $(e_{j})_{j}$ of $H$, the sum $\sum_{j}\langle ae_{j},e_{j}\rangle$ converges absolutely. Moreover, it is independent of the choice of a orthonormal basis and we call

[TABLE]

the trace of $a$. We set $\|a\|_{1}:=\operatorname{tr}(|a|)$. This defines a norm on ${\mathcal{S}}_{1}(H)$, turning the trace-class operators into a Banach space. Note that ${\mathcal{S}}_{1}(H)$ is a (non-closed) two-sided ideal in ${\mathcal{L}}(H)$. Moreover, we have ${\mathcal{S}}_{1}(H)\subseteq{\mathcal{K}}(H)$.

\thepgrCt**.**

Given $a\in{\mathcal{L}}(H)$, the map ${\mathcal{S}}_{1}(H)\to{\mathbb{C}}$, $x\mapsto\operatorname{tr}(ax)$, is a bounded, linear functional on ${\mathcal{S}}_{1}(H)$. This induces an isometric isomorphism ${\mathcal{L}}(H)\cong{\mathcal{S}}_{1}(H)^{*}$. It is also well known that ${\mathcal{K}}(H)\subseteq{\mathcal{L}}(H)={\mathcal{S}}_{1}(H)^{*}$ is an isometric predual of ${\mathcal{S}}_{1}(H)$. Thus, we have isometric isomorphisms

[TABLE]

Let us say that a Banach space $X$ satisfies ($*$) if for every Banach space $E$, every contractive projection ${\mathcal{L}}(X,E)^{**}\to{\mathcal{L}}(X,E)$ is automatically a right ${\mathcal{L}}(X)$-module map. If $X$ satisfies ($*$), then it follows from Section 2 that every contractive projection ${\mathcal{L}}(X,E)^{**}\to{\mathcal{L}}(X,E)$ is of the form $\pi_{\ast}\circ\alpha_{X,E}$ for a unique contractive projection $\pi\colon E^{**}\to E$. Section 2 states that Hilbert spaces satisfy ($*$).

Lemma 3.1**.**

If two Banach spaces $X$ and $Y$ satisfy ($*$), then so does $X\hat{\otimes}Y$.

Proof.

Assume that $X$ and $Y$ satisfy ($*$). Let $E$ be another Banach space, and let $\pi\colon{\mathcal{L}}(X\hat{\otimes}Y,E)^{**}\to{\mathcal{L}}(X\hat{\otimes}Y,E)$ be a contractive projection.

Banach spaces form a closed monoidal category for the projective tensor product and with ${\mathcal{L}}(Y,\_\,)$ adjoint to $\_\,\hat{\otimes}Y$. Thus, there is a natural isometric isomorphism

[TABLE]

An operator $f\colon X\hat{\otimes}Y\to E$ is identified with the operator $\tilde{f}\colon X\to{\mathcal{L}}(Y,E)$ that sends $x\in X$ to the operator $\tilde{f}(x)\colon Y\to E$ given $\tilde{f}(x)(y):=f(x\otimes y)$, for $y\in Y$.

The projection $\pi$ corresponds to a contractive projection $\pi^{\prime}\colon{\mathcal{L}}(X,{\mathcal{L}}(Y,E))^{**}\to{\mathcal{L}}(X,{\mathcal{L}}(Y,E))$. Applying that $X$ satisfies $(*)$ to the projection $\pi^{\prime}$, there exists a unique contractive projection $\tau\colon{\mathcal{L}}(Y,E)^{**}\to{\mathcal{L}}(Y,E)$ such that $\pi^{\prime}=\tau_{*}\circ\alpha_{X,{\mathcal{L}}(Y,E)}$. The situation is shown in the following diagram:

[TABLE]

Let F\in{\mathcal{L}}\big{(}X\hat{\otimes}Y,E\big{)}^{**}, which corresponds to \tilde{F}\in{\mathcal{L}}\big{(}X,{\mathcal{L}}(Y,E)\big{)}^{**}. Then

[TABLE]

for $x\in X$ and $y\in Y$.

Applying that $Y$ satisfies $(*)$ to the projection $\tau$, there exists a unique contractive projection $\sigma\colon E^{**}\to E$ such that

[TABLE]

for every $G\in{\mathcal{L}}(Y,E)^{**}$ and $y\in Y$. We claim that $\sigma$ is the desired projection to verify that $X\hat{\otimes}Y$ satisfies $(*)$.

Given $f\in{\mathcal{L}}(X\hat{\otimes}Y,E)$ with corresponding element $\tilde{f}\in{\mathcal{L}}(X,{\mathcal{L}}(Y,E))$, note that ${\mathrm{ev}}_{x\otimes y}(f)={\mathrm{ev}}_{y}({\mathrm{ev}}_{x}(\tilde{f}))$. It follows that ${\mathrm{ev}}_{x\otimes y}^{**}(F)={\mathrm{ev}}_{y}^{**}({\mathrm{ev}}_{x}^{**}(\tilde{F}))$. Using this at the last step, we deduce that

[TABLE]

for every $x\in X$ and $y\in Y$. Thus, we have $\pi(F)(t)=\sigma({\mathrm{ev}}_{t}^{**}(F))$, for every simple tensor $t$ in $X\hat{\otimes}Y$. It follows from linearity and continuity of the involved maps that the same equation holds for every $t\in X\hat{\otimes}Y$, as desired. ∎

Lemma 3.2**.**

Let $Y$ be a Banach space. Then every contractive projection from ${\mathcal{L}}({\mathcal{S}}_{1}(H),Y)^{**}$ to ${\mathcal{L}}({\mathcal{S}}_{1}(H),Y)$ is automatically a right ${\mathcal{L}}({\mathcal{S}}_{1}(H))$-module map.

Proof.

It is well known that the trace-class operators on $H$ are isometrically isomorphic to $H\hat{\otimes}H$. Therefore, the statement follows from Lemma 3.1. ∎

Using Lemma 3.2 instead of Section 2, we obtain the analog of Section 2 with the space of trace-class operators in place of the Hilbert space.

Proposition \theprpCt.

Let $Y$ be a Banach space. Assigning to a projection $\pi\colon Y^{**}\to Y$ the projection $r_{\pi}\colon{\mathcal{L}}({\mathcal{S}}_{1},Y)^{**}\to{\mathcal{L}}({\mathcal{S}}_{1},Y)$, as in Section 2, established a natural bijection between the following classes.

1. (1)

Contractive projections $Y^{**}\to Y$; 2. (2)

Contractive projections ${\mathcal{L}}({\mathcal{S}}_{1}(H),Y)^{**}\to{\mathcal{L}}({\mathcal{S}}_{1}(H),Y)$.

Restricted to projections with weak∗ closed kernel, we obtain a natural bijection between isometric preduals of $Y$ and isometric preduals of ${\mathcal{L}}({\mathcal{S}}_{1}(H),Y)$.

Corollary \thecorCt.

Let $Y$ be a Banach space, and let $F\subseteq{\mathcal{L}}({\mathcal{S}}_{1}(H),Y)^{*}$ be an isometric predual. Then for each $a\in{\mathcal{L}}({\mathcal{S}}_{1}(H))$, the right action of $a$ on ${\mathcal{L}}({\mathcal{S}}_{1}(H),Y)$ is continuous for the weak∗ topology induced by $F$.

Remark \thermkCt.

Using Lemma 3.1 inductively, Section 3 and Section 3 hold for any projective tensor power $H\hat{\otimes}\ldots\hat{\otimes}H$ in place of ${\mathcal{S}}_{1}(H)$.

Remark \thermkCt.

We have isometric isomorphisms

[TABLE]

and we consider ${\mathcal{S}}_{1}(H)\hat{\otimes}{\mathcal{K}}(H)$ as the ‘standard’ predual of ${\mathcal{L}}({\mathcal{S}}_{1}(H))$. Right multiplication by an element from ${\mathcal{L}}({\mathcal{S}}_{1}(H))$ is continuous for the induced weak∗ topology; see Section 3. However, this is not the case for left multiplication. Indeed, a Banach space $X$ is reflexive if and only if left multiplication by elements from ${\mathcal{L}}(X)$ is weak∗ continuous (for any predual); see [GT16, Corollary 6.4].

Example \theexaCt.

By Section 3, there is a natural correspondence between isometric preduals of ${\mathcal{L}}({\mathcal{S}}_{1}(H))$ and isometric preduals of ${\mathcal{S}}_{1}(H)$. The compact operators on $H$ form the canonical isometric predual of ${\mathcal{S}}_{1}(H)$.

However, if $H=\ell_{2}$, then ${\mathcal{S}}_{1}(H)$ has also many other isometric preduals. Indeed, the diagonal operators in ${\mathcal{S}}_{1}(H)$ form an isometric copy of $\ell_{1}$ that is closed for the ‘standard’ weak∗ topology induced by the compact operators. Since $\ell_{1}$ does not have a strongly unique isometric predual, neither does ${\mathcal{S}}_{1}(H)$.

Theorem \thethmCt.

The ‘standard’ predual ${\mathcal{K}}(H)$ of ${\mathcal{S}}_{1}(H)$ makes multiplication in ${\mathcal{S}}_{1}(H)$ separately weak∗ continuous. Moreover, ${\mathcal{K}}(H)$ is the only such predual: If $F\subseteq{\mathcal{S}}_{1}(H)^{*}={\mathcal{L}}(H)$ is a (not necessarily isometric) predual that makes multiplication in ${\mathcal{S}}_{1}(H)$ separately weak∗ continuous, then $F={\mathcal{K}}(H)$. In particular, every predual of ${\mathcal{S}}_{1}(H)$ making multiplication separately weak∗ continuous is automatically an isometric predual.

Proof.

Given $a\in{\mathcal{L}}(H)$, we let $\varphi_{a}$ denote the functional on ${\mathcal{S}}_{1}(H)$ given by $\langle\varphi_{a},x\rangle:=\operatorname{tr}(ax)$, for $x\in{\mathcal{S}}_{1}(H)$. This identifies ${\mathcal{S}}_{1}(H)^{*}$ with ${\mathcal{L}}(H)$.

The multiplication on ${\mathcal{S}}_{1}(H)$ induces a ${\mathcal{S}}_{1}(H)$-bimodule structure on its dual; see [GT16, Paragraph A.3]. Let us recall some details. Given $a\in{\mathcal{S}}_{1}(H)$, let $L_{a},R_{a}\colon{\mathcal{S}}_{1}(H)\to{\mathcal{S}}_{1}(H)$ be given by $L_{a}(x):=ax$ and $R_{a}(x):=xa$, for $x\in{\mathcal{S}}_{1}(H)$. Then the left action of $a$ on ${\mathcal{S}}_{1}(H)^{*}$ is given by $R_{a}^{*}$, and the right action is given by $L_{a}^{*}$. Thus, given $a\in{\mathcal{S}}_{1}(H)$ and $b\in{\mathcal{L}}(H)$, we have

[TABLE]

for $x\in{\mathcal{S}}_{1}(H)$, and therefore $\varphi_{b}a=\varphi_{ba}$. Similarly, we obtain $a\varphi_{b}=\varphi_{ab}$.

Let $F\subseteq{\mathcal{S}}_{1}(H)^{*}={\mathcal{L}}(H)$ be a predual. Then left (right) multiplication on ${\mathcal{S}}_{1}(H)$ is $\sigma({\mathcal{S}}_{1}(H),F)$-continuous if and only if $F$ is a right (left) ${\mathcal{S}}_{1}(H)$-submodule of ${\mathcal{L}}(H)$; see [GT16, Corollary B.7]. Given $a\in{\mathcal{K}}(H)$ and $b\in{\mathcal{S}}_{1}(H)$, we have $ab,ba\in{\mathcal{K}}(H)$. Thus, the predual ${\mathcal{K}}(H)$ is a ${\mathcal{S}}_{1}(H)$-sub-bimodule of ${\mathcal{L}}(H)$, which shows that it makes multiplication in ${\mathcal{S}}_{1}(H)$ separately weak∗ continuous.

Conversely, let $F\subseteq{\mathcal{S}}_{1}(H)^{*}={\mathcal{L}}(H)$ be a predual making multiplication in ${\mathcal{S}}_{1}(H)$ separately weak∗ continuous. Then $F$ is invariant under the left and right action of ${\mathcal{S}}_{1}(H)$ on ${\mathcal{L}}(H)$. We have shown above that the left (right) action of $a\in{\mathcal{S}}_{1}(H)$ on ${\mathcal{L}}(H)$ is simply given by right (left) multiplication with $a$.

Claim: The set $F\cap{\mathcal{K}}(H)$ is a closed, two-sided ideal in ${\mathcal{K}}(H)$. To verify the claim, let $a\in F\cap{\mathcal{K}}(H)$, and let $b\in{\mathcal{K}}(H)$. Given a finite-dimensional subspace $D\subseteq H$, let $p_{D}$ be the orthogonal projection onto $D$. We order the finite-dimensional subspaces of $H$ by inclusion. Since $b\in{\mathcal{K}}(H)$, we have $\lim_{D}\|p_{D}b-b\|=0$ and therefore

[TABLE]

For each $D$, we have $p_{D}b\in{\mathcal{S}}_{1}(H)$. Since $F$ is invariant under right multiplication by ${\mathcal{S}}_{1}(H)$, it follows that $ap_{D}b\in F\cap{\mathcal{K}}(H)$. Since $F$ is norm-closed, we deduce that $ab\in F\cap{\mathcal{K}}(H)$. Analogously, one shows that $F\cap{\mathcal{K}}(H)$ is a left ideal in ${\mathcal{K}}(H)$, which proves the claim.

The only closed, two-sided ideals of ${\mathcal{K}}(H)$ are $\{0\}$ and ${\mathcal{K}}(H)$. It is easy to see that $F\cap{\mathcal{K}}(H)\neq\{0\}$. Thus, ${\mathcal{K}}(H)\subseteq F$. Since both ${\mathcal{K}}(H)$ and $F$ are preduals of ${\mathcal{S}}_{1}(H)$, it follows that ${\mathcal{K}}(H)=F$, as desired. ∎

Corollary \thecorCt.

Let ${\mathcal{S}}_{1}(H)$ be the trace-class operators on a Hilbert space $H$. Then every Banach algebra isomorphism ${\mathcal{S}}_{1}(H)\to{\mathcal{S}}_{1}(H)$ is weak∗ continuous (for the ‘standard’ predual ${\mathcal{K}}(H)$.)

Remark \thermkCt.

A dual Banach algebra is a Banach algebra $A$ together with a predual $F\subseteq A^{*}$ making the multiplication in $A$ separately weak∗ continuous. This concept was introduced by Runde, [Run02, Definition 4.4.1, p.108], and extensively studied by Daws; see [Daw11] and the references therein. Section 3 states that the trace-class operators with their ‘standard’ predual of compact operators form a dual Banach algebra. Moreover, the compact operators are the only predual making the trace-class operators into a dual Banach algebra.