# Completely Bounded Maps into Certain Hilbertian Operator Spaces

Gilles Pisier

► To cite this version:

Gilles Pisier. Completely Bounded Maps into Certain Hilbertian Operator Spaces. International Mathematics Research Notices, 2004, 74, pp.3983-4018. 10.1155/S1073792804140865. hal-05460873

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# Completely Bounded Maps into Certain Hilbertian Operator Spaces

Gilles Pisier

## 1 Introduction

An operator space is a Banach space given together with an isometric embedding into the space B(H) of all bounded operators on a Hilbert space H. Like the previous papers [23, 24] (to which this one is a natural sequel) this paper mainly studies questions about Hilbertian operator spaces. As is well known, the Hilbert space $ℓ_2$ can be equipped with many different “operator space structures,” that is, there are many inequivalent ways to embed $ℓ_2$ into the space B(H) of all bounded operators on a Hilbert space H. Here, the “inequivalence” is with respect to the operator space theory where the relevant notions of morphism and isomorphism are those of “completely bounded map” and “complete isomorphism.” The basic theory of operator spaces and their duality was developed by Effros-Ruan and Blecher-Paulsen (cf. [5, 22]).

The two basic ways to realize $ℓ_2$ as an operator space are the “row” and “column” ways that are defined, respectively, as follows. Let

$$ 
\begin{align} 
R &= \overline{\operatorname{span}}\{e_{1j} \mid j \ge 1\} \subset B(\ell_2), \tag{1.1} \\ 
C &= \overline{\operatorname{span}}\{e_{i1} \mid i \ge 1\} \subset B(\ell_2). 
\end{align} 
$$

Then R and C are isometric to $ℓ_2$ but are not completely isomorphic (see, e.g., [22, page 21]).

In [20], the author introduced another isometric embedding of $ℓ_2$ into B(H) and denoted by OH the space $ℓ_2$ equipped with this operator space structure. Being self-dual,
the space OH appears as the natural analogue of $\ell_2$ in operator space theory. It also appears as a "midpoint" between $R$ and $C$ in the sense of interpolation theory. More precisely, we have $OH = (C, R)_{1/2}$. The latter formula roughly means that OH is obtained by some sort of "geometric mean" out of the pair $(C, R)$.

In [21], the author introduced more generally a "natural" operator space structure on any $L_p$-space ($1 \le p \le \infty$). When $p = 2$, one recovers the space OH (up to cardinality considerations). This uses the theory developed by Effros-Ruan and Blecher-Paulsen. Roughly, one starts with the obvious (by Gelfand theory) operator space structure on von Neumann or $C^*$-algebras, then by operator space duality one passes to their duals and preduals and, finally, by interpolation one obtains the "natural" operator space structure on noncommutative $L_p$-spaces. In Banach space theory, the fact that $\ell_2$ embeds in $L_p = L_p([0, 1], dt)$ for any $0 < p < \infty$ plays a very important role. The case $p = 1$ is probably the most useful one; it is closely related to Grothendieck's theorem. More precisely, define $J : \ell_2 \to L_p$ by setting $Jx = \sum x_n g_n$, where $(g_n)$ is an independent and identically distributed (i.i.d. in short) sequence of standard (complex-valued) Gaussian random variables. If we normalize $g_n$ so that $\|g_1\|_p = 1$ (note $\|g_n\|_p = \|g_1\|_p$), then $J$ is an isometric embedding. If we replace $(g_n)$ by the Rademacher functions $(r_n)$, then by Khintchine's inequalities, $J$ becomes an isomorphic embedding for any $0 < p < \infty$.

In operator space theory, the question whether OH analogously embeds completely isomorphically (or even hopefully completely isometrically) into a noncommutative $L_p$-space arose immediately after [21]. Curiously, the above embedding $J$ produces a different operator space structure than that of OH. The latter structure actually depends on $p$ (with a sharp distinction between the cases $p < 2$ and $p > 2$) and gives OH only when $p = 2$. Moreover, if we replace $(g_n)$ or $(r_n)$ by its free analogue in Voiculescu's sense (see [34]), the resulting operator space will remain the same (completely isomorphically) for each $p$ as long as $p < \infty$. These facts are closely related to Lust-Piquard's noncommutative Khintchine inequalities (see [21, pages 104–112 and pages 115–121] for full details). Passing to the Fermionic analogue of Gaussian variables (see [9]) also produces the same result; thus neither does this yield a realization of OH in non-commutative $L_p$-spaces. Actually, Junge had observed early on that (by the Lust-Piquard inequalities) such a realization is impossible for $2 < p < \infty$ (see [9] and see also the final remark below), but the case $1 \le p < 2$ and especially the crucial case $p = 1$ resisted all efforts until recently when Junge proved that OH does embed completely isomorphically into $M_*$ for some nonsemifinite von Neumann algebra $M$. Shortly before Junge's paper, the author and Shlyakhtenko [24] already had used free analogues of Gaussian variables but in the *nonsemifinite* case to prove an operator space version of Grothendieck's theorem. With these two papers [10, 24] the *nontracial* or *nonsemifinite* theory (also referred to
as "type III") made its "début" on the operator space scene. Subsequently, the author [23]
showed that no embedding OH ⊂ M* can exist if M is semifinite.

This paper continues this line of research. We first characterize completely bounded maps from a C*-algebra (or an exact operator space) into OH (or into the space that we denote by (C, R)θ), thus refining [24].

Junge actually announced (unpublished work) that his method yields that all subspaces of quotients of the direct sum $R \oplus C$ (of which OH is an example) also embed into $M_*$ for some $M$. This naturally led us to investigate which (infinite-dimensional) quotients of $R \oplus C$ can embed into $M_*$ if $M$ is semifinite. The answer (see **Theorem 2.12** below) is quite satisfactory, there are mainly 3 cases: $R, C$ (these embed into the trace class), and the space (that is denoted by $R+C$ in [20, 21]) corresponding to the range of the above embedding $J$ when $p = 1$. There are 4 other possibilities built out of the first 3 using direct sums and we prove that these 7 possibilities exhaust the list. Finally, in the appendix we show that Junge's embedding for OH (or for subspaces of quotients of $R \oplus C$) can be obtained rather quickly using Shlyakhtenko's generalized free Gaussian random variables. This new proof, based on complex interpolation, is much quicker and yields a better constant, but the completely isometric case remains open. After this proof circulated, Junge and Xu found another proof using real interpolation (see [41] and unpublished work).

We refer either to [5, 17] or to [22] for background on operator spaces and completely bounded maps. As usual, we will abbreviate "completely bounded" as cb and we will say that two operator spaces E, F are completely c-isomorphic if there is an isomorphism $u : E \to F$ with $\|u\|_{cb} \|u^{-1}\|_{cb} \le c$. If E is a subspace of F, we will say that E is completely C-complemented in F if there is a projection $P : F \to E$ with $\|P\|_{cb} \le C$. We recall that, given Hilbert spaces H, K, the "minimal" (or "spatial") tensor product of two operator spaces $E \subset B(H)$ and $F \subset B(K)$ is denoted by $E \otimes_{\min} F$, it is naturally embedded in $B(H \otimes_2 K)$ and its norm is denoted by $\| \cdot \|_{\min}$. The Hilbert-Schmidt norm of a mapping $u : H \to K$ will be denoted by $\|u\|_2$. We will use several times the well-known fact (cf., e.g., [22, page 21]) that for any mappings $u : R \to C$ and $v : C \to R$, we have

$$ \|u : R \to C\|_{cb} = \|u\|_2, \quad \|v : C \to R\|_{cb} = \|v\|_2, \qquad (1.2) $$

while for mappings from C to itself or from R to itself, the cb norm coincides with the operator norm.

More generally, for any Hilbert space H, similar properties hold for the associated "column" (resp., "row") operator spaces denoted by $H_c$ (resp., $H_r$). These are defined by $H_c = B(C, H)$ (resp., $H_r = B(H^*, C)$). When $H = \ell_2$, then $H_c = C$ and $H_r = R$. $H_c$ and $H_r$
are nothing but analogues of C and R relative to general cardinals instead of that of $\mathbb{N}$.
When $H$ is $n$-dimensional, we denote $H_c$ (resp., $H_r$) by $C_n$ (resp., $R_n$).

The letters WEP stand for Lance's "weak expectation property": a $C^*$-algebra $A$ has the WEP if the inclusion $A \to A^{**}$ factors completely contractively through $B(H)$ (see, e.g., [22] for examples of its use in operator space theory).

## 2 Main results

Let $E \subset A$ be an operator space, given as a closed subspace of a $C^*$-algebra $A$. Let $u : E \to \ell_2$ be a linear map. We will identify $\ell_2$ successively with $R$, $C$, and other operator spaces isometric to $\ell_2$.

By [4], we know that $\|u : E \to R\|_{cb} \le 1$ if and only if there is a state $f$ on $A$ such that

$$ \forall x \in E, \quad \|ux\|^2 \le f(xx^*). \qquad (2.1) $$

Equivalently, by [4] this holds if and only if for any finite sequence $x_1, \dots, x_n$ in $E$, we have

$$ \sum \|ux_i\|^2 \le \sum x_i x_i^* \qquad (2.2) $$

Similarly, $\|u : E \to C\|_{cb} \le 1$ if and only if there is a state $g$ on $A$ such that

$$ \forall x \in E, \quad \|ux\|^2 \le g(x^*x). \qquad (2.3) $$

Moreover, this holds if and only if for any finite sequence $x_1, \dots, x_n$ in $E$, we have

$$ \sum \|ux_i\|^2 \le \sum x_i^* x_i \qquad (2.4) $$

In [20], the author introduced a different operator space structure on $\ell_2$, namely, the space OH, an operator space isometric to $\ell_2$ and uniquely characterized among operator spaces by the property that it is (canonically) completely isometric to its antidual.

For any $0 < \theta < 1$, one can extend (see [20]) complex interpolation to the operator space context. Applied to the interpolation pair $(C, R)$ (using the transposition map $x \to {}^t x$ to define "compatibility" in the interpolation sense), this method produces new operator spaces, denoted by $(C, R)_\theta$, each of which is isometric to $\ell_2$. For $\theta = 1/2$, we recover the space OH. To abbreviate we will denote simply $R[\theta] = (C, R)_\theta$. With this notation we have $R[1/2] = (C, R)_{1/2} = (R, C)_{1/2} = \text{OH}$, and also $R[\theta]^* = R[1-\theta]$ completely isometrically. By convention, we set $R[0] = C$ and $R[1] = R$.
The operator space structure on $R[\theta]$ can be described more explicitly as follows. We denote by $\{e_i(\theta) | i = 1, 2, ...\}$ an orthonormal basis in $R[\theta]$ (recall that $R[\theta] \simeq \ell_2$ as a Banach space). Then, for any finite sequence $(a_i)$ in $B(\ell_2)$, we have

$$ \|\Sigma a_i \otimes e_i(\theta)\|_{B(\ell_2) \otimes \min R[\theta]} = \sup \left\{ \left( \sum \|s^\theta a_i t^{1-\theta}\|_2^2 \right)^{1/2} \mid s \ge 0, t \ge 0, \|s\|_2 \le 1, \|t\|_2 \le 1 \right\}, \quad (2.5) $$

where $\|\cdot\|_2$ denotes the Hilbert-Schmidt norm. Equivalently, let $p = (1-\theta)^{-1}$ and $p' = \theta^{-1}$; let $S_p$ and $S_{p'}$ denote the corresponding Schatten classes. Then the left-hand side is equal to the norm of the mapping $x \to \sum a_i^* x a_i$ on $S_{p'}$, and also equal to that of the mapping $y \to \sum a_i y a_i^*$ on $S_p$.

In the extreme case $\theta = 0$ (resp., $\theta = 1$) we recover the space $C$ (resp., $R$) and the above supremum is equal to $\|\Sigma a_i^* a_i\|^{1/2}$ (resp., $\|\Sigma a_i a_i^*\|^{1/2}$). The space $R[\theta]$ can also be described as the space of "row matrices" inside the Schatten class $S_p$ with $p = 1/(1-\theta)$, when the latter is equipped with its "natural" operator space structure defined (by interpolation) in [20]. Similarly, $R[\theta]$ can be described as the space of "column matrices" inside the Schatten class $S_{p'}$.

In [24] it is proved that if $E$ is exact, then $\|u : E \to OH\|_{cb} < \infty$ if and only if there are a constant $C$ and states $f, g$ such that for all $x$ in $E$, we have

$$ \|ux\|^2 \leq Cg(x^*x)^{1/2}f(xx^*)^{1/2}. \quad (2.6) $$

The first goal of this paper is to prove this result with $R[\theta]$ instead of $OH$. Although the ingredients are the same as in [24], our proof is somewhat more direct. Moreover, we are able to recover the extreme cases $\theta = 0$ and $\theta = 1$ (due to Effros and Ruan [4]) described above. Note that no assumption on $E$ is needed in the latter extreme cases, but some assumption (such as exactness) is definitely needed when $0 < \theta < 1$ (see the remark on [24, page 210]).

In the appendix, we present a simpler proof of Junge's recent remarkable embedding theorem of $OH$ (or $R[\theta]$ for $0 < \theta < 1$) into the predual of a von Neumann algebra $M$. Combined with the results of the present paper, the argument of [23] shows that such an embedding is impossible, for any $0 < \theta < 1$, if $M$ is semifinite.

Let $0 < \theta < 1$. Following [20, Section 2.7], we can view $R[\theta]$ as an operator space such that we have (isometrically)

$$ M_n(R[\theta]) = (M_n(C), M_n(R))_{\theta} \quad (2.7) $$
for any $n \ge 1$. We set

$$ c(\theta) = (\theta^{\theta}(1-\theta)^{1-\theta})^{-1}. \qquad (2.8) $$

**Theorem 2.1.** Let $A$ be a $C^*$-algebra. Then for any complete contraction $u : A \to R[\theta]$, there are states $f, g$ on $A$ such that

$$ \forall a \in E, \quad \|ua\| \le c(\theta)f(a^*a)^{(1-\theta)/2}g(aa^*)^{\theta/2}. \qquad (2.9) $$

We will use the following known fact (see [19, 20] for related results).

**Lemma 2.2.** For any $C^*$-algebra $B$ with the WEP and $0 < \theta < 1$,

$$ B \otimes_{\min} R[\theta] = (B \otimes_{\min} C, B \otimes_{\min} R)_{\theta} \qquad (2.10) $$

with equal norms.

**Proof.** Indeed, this is equivalent to the validity of the following isometric identities for any $n \ge 1$:

$$ B \otimes_{\min} (C_n, R_n)_\theta = (B \otimes_{\min} C_n, B \otimes_{\min} R_n)_\theta. \qquad (2.11) $$

To verify (2.11) we first observe that the case $B = K(H)$ follows from the definition of interpolation. Then, taking the bidual of both sides of (2.11) (still restricted to the case $B = K(H)$), we obtain the case $B = B(H)$. Finally, if $B$ has WEP, the inclusion $B \to B^{**}$ factors completely contractively through $B(H)$, so that (using (2.11) for $B = B(H)$) we have a complete contraction

$$ B \otimes_{\min} (C_n, R_n)_\theta \to (B^{**} \otimes_{\min} C_n, B^{**} \otimes_{\min} R_n)_\theta. \qquad (2.12) $$

But $B^{**} \otimes_{\min} C_n = (B \otimes_{\min} C_n)^{**}$ and $B^{**} \otimes_{\min} R_n = (B \otimes_{\min} R_n)^{**}$, and the norm induced on $B \otimes (C_n, R_n)_\theta$ by the space

$$ ((B \otimes_{\min} C_n)^{**}, (B \otimes_{\min} R_n)^{**})_\theta \qquad (2.13) $$

coincides with the norm of $(B \otimes_{\min} C_n, B \otimes_{\min} R_n)_\theta$. Indeed (see, e.g., [22, page 57] for more details), since the sets appearing on both sides are obviously identical (and each is identical to $(B^*)^n$), we have, isometrically,

$$ (B \otimes_{\min} C_n, B \otimes_{\min} R_n)_\theta^* = ((B \otimes_{\min} C_n)^*, (B \otimes_{\min} R_n)^*)_\theta, \qquad (2.14) $$
and hence, repeating the same argument for the duals, we have, isometrically, $(B \otimes_{\min} C_n, B \otimes_{\min} R_n)^{**} = ((B \otimes_{\min} C_n)^{**}, (B \otimes_{\min} R_n)^{**})_\theta$. Therefore, we find a completely contractive inclusion

$$B \otimes_{\min} (C_n, R_n)_\theta \to (B \otimes_{\min} C_n, B \otimes_{\min} R_n)_\theta. \qquad (2.15)$$

On the other hand, the fact that the converse inclusion is completely contractive is obviously true in general (without any assumption on $B$), as follows easily by considering an embedding of $B$ onto $B(H)$. ■

**Proof of Theorem 2.1.** We will use the following formula valid for any pair $\alpha_0, \alpha_1$ of positive numbers:

$$\alpha_0^{1-\theta} \alpha_1^\theta = \inf_{\lambda > 0} \{(1-\theta)\lambda^\theta \alpha_0 + \theta \lambda^{-(1-\theta)} \alpha_1\}. \qquad (2.16)$$

Using this, (2.9) can be rewritten as

$$\forall \lambda > 0, \forall a \in E, \quad \|ua\|^2 \le (c(\theta))^2 \{(1-\theta)\lambda^\theta f(a^*a) + \theta\lambda^{-(1-\theta)}g(aa^*)\}. \qquad (2.17)$$

By the Hahn-Banach theorem (cf., e.g., [6, Lemma 3.4] for details), it suffices to show that for all finite sequences $(a_i)$ in $E$ and all numbers $\lambda_i > 0$, we have

$$\Sigma \|ua_i\|^2 \le c(\theta)^2 \{(1-\theta) \|\Sigma \lambda_i^\theta a_i^* a_i\| + \theta \|\Sigma \lambda_i^{-(1-\theta)} a_i a_i^*\|\}. \qquad (2.18)$$

We will use the "generalized circular elements" introduced in [29], following Voiculescu's work. Since we closely follow the ideas in [24], we will be brief. Let $H$ be a Hilbert space. We assume given a set $I$ such that $H$ has an orthonormal basis formed of the disjoint union

$$\{e_i | i \in I\} \cup \{e'_i | i \in I\}. \qquad (2.19)$$

Let $\mathcal{F}$ be the full Fock space over $H$, that is,

$$\mathcal{F} = \mathbb{C} \oplus H \oplus H^{\otimes 2} \oplus \dots. \qquad (2.20)$$

Let $\Omega$ be the unit of $\mathbb{C}$, viewed as an element in $\mathcal{F}$. For any $h$ in $H$, we denote by $\ell(h)$ (resp., $r(h)$) the left (resp., right) creation operator on $\mathcal{F}$, that is, $x \to h \otimes x$ (resp., $x \to x \otimes h$). Moreover, we set $\ell_i = \ell(e_i)$ (resp., $r_i = r(e_i)$) and $\ell'_i = \ell(e'_i)$ (resp., $r'_i = r(e'_i)$). We define

$$\begin{aligned} x_i &= (1 - \theta)\lambda_i^{\theta/2} \ell_i + \theta\lambda_i^{-(1-\theta)/2} \ell_i^*, \\ y_i &= (1 - \theta)\lambda_i^{(1-\theta)/2} r_i + \theta\lambda_i^{-\theta/2} r_i^*. \end{aligned} \qquad (2.21)$$
Let $\mathcal{L}$ (resp., $\mathcal{R}$) be the von Neumann algebra generated by $(x_i)_{i \in I}$ (resp., $(y_i)_{i \in I}$) in $B(\mathcal{F})$.
Note that $\mathcal{L}$ and $\mathcal{R}$ commute with each other.

Let $(a_i)$ be a finite sequence in $\mathcal{A}$. By a well-known argument (as in [24, page 202])
we have

$$
\begin{align}
\| \Sigma x_i \otimes a_i \|_{\min} &\le (1-\theta) \| \Sigma \lambda_i^{\theta/2} \ell_i \otimes a_i \|_{\min} + \theta \| \Sigma \lambda_i^{-(1-\theta)/2} \ell_i^{*} \otimes a_i \|_{\min} \nonumber \\
&\le (1-\theta) \| \Sigma a_i^* a_i \lambda_i^{\theta} \|^{1/2} + \theta \| \Sigma a_i a_i^* \lambda_i^{\theta-1} \|^{1/2} \nonumber \\
&\le \{(1-\theta) \| \Sigma a_i^* a_i \lambda_i^{\theta} \| + \theta \| \Sigma a_i a_i^* \lambda_i^{\theta-1} \|\}^{1/2} \quad \text{(by Cauchy-Schwarz).} \tag{2.22}
\end{align}
$$

Therefore the proof of (2.18) (and thus of Theorem 2.1) is reduced to that of the following
sublemma.
■

**Sublemma 2.3.**

$$
\left( \left\| u a_i \right\|^2 \right)^{1/2} \le c(\theta) \left\| \Sigma x_i \otimes a_i \right\|_{\min}. \qquad (2.23) \\
\hfill \square
$$

Proof. By [24] we know that $\mathcal{L}$ is QWEP, that is, there are a $C^*$-algebra $B$ with the WEP and an ideal $\mathcal{J} \subset B$ such that $\mathcal{L} \simeq B/\mathcal{J}$. Since $\|u\|_{cb} \le 1$, $I \otimes u$ defines a contraction from $B \otimes_{\min} A$ to $B \otimes_{\min} R[\theta]$. Since $R[\theta]$ has the completely contractive approximation property, we may clearly assume that $u$ has finite rank. In that case, if we denote by $q: B \otimes_{\min} A \to (B/\mathcal{J}) \otimes_{\min} A$ the canonical map, we must have $(I \otimes u) \ker(q) \subset \mathcal{J} \otimes_{\min} R[\theta]$ (see, e.g., [22, Theorem 15.11] for details). Therefore, $I \otimes u$ defines a contractive map from $(B \otimes_{\min} A)/\ker(q)$ to $(B \otimes_{\min} R[\theta])/\mathcal{J} \otimes_{\min} R[\theta]$. Thus

$$
\| I \otimes u : \mathcal{L} \otimes_{\min} A \to (B \otimes_{\min} R[\theta]) / (\mathcal{J} \otimes_{\min} R[\theta]) \| \le 1. \quad (2.24)
$$

But now, since $B$ has the WEP, by **Lemma 2.2**, the following isometric identity holds:

$$
B \otimes_{\min} R[\theta] = (B \otimes_{\min} C, B \otimes_{\min} R)_{\theta}. \qquad (2.25)
$$

Hence we obtain a natural contractive map

$$
(B \otimes R[\theta]) / (\mathcal{J} \otimes_{\min} R[\theta]) \to (\mathcal{L} \otimes_{\min} C, \mathcal{L} \otimes_{\min} R)_{\theta}. \qquad (2.26)
$$

Thus, to conclude, it suffices to prove the following.

**Claim 2.4.** Let $X_\theta = (\mathcal{L} \otimes_{\min} C, \mathcal{L} \otimes_{\min} R)_\theta$. Then we have

$$
\left( \Sigma \| u a_i \|^2 \right) \le c(\theta) \left\| \Sigma x_i \otimes u a_i \right\|_{X_\theta}. \qquad (2.27)
$$
Let $(z_i)_{i \le n}$ be a finite sequence in $(C, \mathbb{R})_\theta = \mathbb{R}[\theta]$. We will show more generally that

$$ (\Sigma \|z_i\|^2)^{1/2} \le c(\theta) \|\Sigma x_i \otimes z_i\|_{X_\theta}. \qquad (2.28) $$

We denote by $L_c^2$ (resp., $L_r^2$) the completion of $\mathcal{L}$ for the norm $x \to \|x\Omega\|$ (resp., $x \to \|x^*\Omega\|$). Similarly, we denote by $R_c^2$ (resp., $R_r^2$) the completion of $\mathbb{R}$ for the norm $x \to \|x\Omega\|$ (resp., $x \to \|x^*\Omega\|$).

Note that actually $L_c^2$ (resp., $L_r^2$) is clearly isometric to $H$ and the map $x \to x\Omega$ injects $\mathcal{L}$ (resp., $\mathbb{R}$) into $H$. We will denote

$$ \mathcal{L}_{\theta} = (L_c^2, L_r^2)_{\theta}, \qquad \mathbb{R}_{\theta} = (\mathbb{R}_c^2, \mathbb{R}_r^2)_{\theta}. \qquad (2.29) $$

Clearly, for any $(b_1, \dots, b_n)$ in $\mathcal{L}$, we have

$$ (\Sigma \|b_k\|^2_{L_c^2})^{1/2} \le \|\Sigma b_k^* b_k\|^{1/2} = \|\Sigma b_k \otimes e_{k1}\|, \qquad (2.30) $$

$$ (\Sigma \|b_k\|^2_{L_r^2})^{1/2} \le \|\Sigma b_k b_k^*\|^{1/2} = \|\Sigma b_k \otimes e_{1k}\|, $$

hence we have a contractive inclusion from $(\mathcal{L} \otimes_{\min} C, \mathcal{L} \otimes_{\min} \mathbb{R})_\theta$ to $(\ell_2^n(L_c^2), \ell_2^n(L_r^2))_\theta$ and the latter space can be classically identified (see [1]) with $\ell_2^n([L_c^2, L_r^2]_\theta) = \ell_2^n(\mathcal{L}_\theta)$.

Let $(e_k)$ be the canonical basis of $\mathbb{R}[\theta]$ (corresponding to $(e_{k1})$ or $(e_{1k})$). Let $z_i = \Sigma z_i(k)e_k$ and let

$$ x(k) = \Sigma_i z_i(k) x_i. \qquad (2.31) $$

By the preceding discussion, to prove our claim (2.28), it suffices to show that

$$ \Sigma \|z_i\|^2 \le c(\theta)^2 \Sigma \|x(k)\|_{\mathcal{L}_{\theta}}^2. \qquad (2.32) $$

Actually we will show that there is equality in (2.32). To verify this, we now introduce

$$ y(k) = \Sigma_i \overline{z_i(k)} y_i \qquad (2.33) $$

so that

$$ \Sigma \langle y(k) x(k) \Omega, \Omega \rangle = \Sigma_k \Sigma_i |z_i(k)|^2 \theta(1-\theta) = \theta(1-\theta) \Sigma \|z_i\|^2. \qquad (2.34) $$
We will use the fact that for any $(X_k)$ in $\mathcal{L}$ and any $(Y_k)$ in $\mathcal{R}$ we have

$$|\Sigma\langle Y_k X_k \Omega, \Omega \rangle| \leq (\Sigma\|X_k\|_{\mathcal{L}_{\theta}}^2)^{1/2} (\Sigma\|Y_k\|_{\mathcal{R}_{1-\theta}}^2)^{1/2}. \qquad (2.35)$$

Indeed, we have

$$\begin{aligned} |\Sigma\langle Y_k X_k \Omega, \Omega \rangle| &\leq (\Sigma\|X_k\Omega\|^2)^{1/2} (\Sigma\|Y_k\Omega\|^2)^{1/2} \\ &= (\Sigma\|X_k\|_{L_c^2}^2)^{1/2} (\Sigma\|Y_k\|_{R_c^2}^2)^{1/2}, \end{aligned} \qquad (2.36)$$

and also since $Y_k X_k = X_k Y_k$,

$$|\Sigma\langle Y_k X_k \Omega, \Omega \rangle| \leq (\Sigma\|X_k\|_{L_r^2}^2)^{1/2} (\Sigma\|Y_k\|_{R_c^2}^2)^{1/2}, \qquad (2.37)$$

hence by the bilinear interpolation theorem (see [1]), (2.35) follows (since $(R_r^2, R_c^2)_\theta = (R_c^2, R_r^2)_{1-\theta} = \mathcal{R}_{1-\theta}$).

We will show that

$$(\Sigma\|y(k)\|_{\mathcal{R}_{1-\theta}}^2)^{1/2} \leq (1-\theta)^{\theta} \theta^{1-\theta} (\Sigma\|z_i\|^2)^{1/2}. \qquad (2.38)$$

Let $f_i$ be the $\mathcal{R}$-valued analytic function defined on $\mathbb{C}$ by

$$f_i(z) = ((1-\theta)^{1-z} \theta^z)^{-1} ((1-\theta) \lambda_i^{z/2} r_i' + \theta \lambda_i^{-(1-z)/2} r_i^*). \qquad (2.39)$$

Note that $f_i(1-\theta) = ((1-\theta)^\theta \theta^{1-\theta})^{-1} y_i$. Moreover, for any $(\alpha_i)$ in $\ell_2(\mathbb{I})$, we have

$$\begin{aligned} \|\Sigma \alpha_i f_i(z)\|_{R_c^2} &= (\Sigma |\alpha_i|^2)^{1/2} && \text{if } \mathrm{Re}(z) = 0, \\ \|\Sigma \alpha_i f_i(z)\|_{R_r^2} &= (\Sigma |\alpha_i|^2)^{1/2} && \text{if } \mathrm{Re}(z) = 1. \end{aligned} \qquad (2.40)$$

Hence

$$\|\Sigma \alpha_i f_i(1-\theta)\|_{\mathcal{R}_{1-\theta}} \leq (\Sigma |\alpha_i|^2)^{1/2}. \qquad (2.41)$$

But now

$$\Sigma \overline{z_i(k)} f_i(1-\theta) = ((1-\theta)^\theta \theta^{1-\theta})^{-1} y(k), \qquad (2.42)$$
hence (2.38) follows from (2.41) and (2.42). Now, combining (2.38) with (2.35) and (2.34),
we find

$$
\begin{align}
(1 - \theta)\theta \Sigma \|z_i\|^2 &= \Sigma \langle y(k)x(k)\Omega, \Omega \rangle \nonumber \\
&\le (\Sigma \|x(k)\|_{\mathcal{L}_{\theta}}^2)^{1/2} (\Sigma \|y(k)\|_{\mathbb{R}_{1-\theta}}^2)^{1/2} \tag{2.43} \\
&\le (\Sigma \|x(k)\|_{\mathcal{L}_{\theta}}^2)^{1/2} (1 - \theta)^{\theta} \theta^{1-\theta} (\Sigma \|z_i\|^2)^{1/2}, \nonumber
\end{align}
$$

which, after a suitable division, yields (2.32). This completes the proof of (2.28), of the
above claim, and of the sublemma. Note that an obvious modification of the proof of
(2.38) shows that the converse of (2.32) also holds, so (2.32) is indeed an equality. ■

Remark 2.5. Arguing as in [24], it is easy to check that Theorem 2.1 remains valid when A is replaced by an exact subspace E ⊂ A with exactness constant less than or equal to c, provided the constant c(θ) in (2.9) is replaced by c(θ)c.

Remark 2.6. Note that $c(\theta) \to 1$ when either $\theta \to 0$ or $\theta \to 1$. In the cases $\theta = 0$ and $\theta = 1$, Theorem 2.1 is well known (cf. [4]). In that case, Theorem 2.1 still holds when A is replaced by an arbitrary subspace $E \subset A$. However, when $0 < \theta < 1$, some extra assumption (such as exactness) is necessary. Indeed, if we take $\theta = 1/2$, let $(a_i)$ be the orthonormal basis of OH = $(C, \mathbb{R})_{1/2}$ and let $u$ be the identity map, we have $\| \sum_1^n a_i^* a_i \|^{1/2} = \| \sum_1^n a_i a_i^* \|^{1/2} = n^{1/4}$, but $\sum_1^n \|u a_i\|^2 = \sum_1^n \|a_i\|^2 = n$, which shows that (2.18) fails. Similarly, the extension property valid when either $\theta = 0$ or $\theta = 1$ is no longer true in general; indeed this is closely related to the fact that $\mathbb{R}$ or $C$ is an injective operator space, while $R[\theta]$ is not when $0 < \theta < 1$.

Remark 2.7. In the preceding argument, the only delicate point is (2.32). Note that, actually, it is easy to show that equality holds in (2.32). We chose to base the above proof of (2.32) solely on complex interpolation to make it accessible to a reader unfamiliar with the Tomita-Takesaki theory. However, if one uses the latter theory, in the form made explicit by Shlyakhtenko in [29], it is very easy to explain why (2.32) should be true. We now review this alternate approach.

Alternate proof of (2.32). Let $\varphi$ be the vacuum state, defined on $B(\mathcal{F})$ by $\varphi(T) = \langle T\Omega, \Omega \rangle$. Let $\xi_i = \theta(1-\theta)^{-1}\lambda_i^{-1/2}$. Note that

$$
x_i = (1 - \theta) \lambda_i^{\theta/2} (\ell_i + \xi_i \ell_i^{*}); \tag{2.44}
$$
therefore $\mathcal{L}$ can be viewed as generated by $\{\ell_i + \xi_i \ell_i^*\}$. We define a one-parameter group of unitary operators $u_t$ on $\mathcal{H}$ by setting for any $t$ in $\mathbb{R}$,

$$ \forall j \in \mathcal{I}, \quad u_t e_j = (\xi_j)^{2it} e_j, \quad u_t e'_j = (\xi_j)^{-2it} e'_j. \qquad (2.45) $$

We extend $u_t$ (by the so-called second quantization) to a unitary operator $U_t$ on $\mathcal{F}$ such that $U_t \Omega = \Omega$ and $U_t = u_t \otimes \cdots \otimes u_t$ on $\mathcal{H} \otimes \cdots \otimes \mathcal{H}$. For any $x$ in $\mathcal{L}$, we denote

$$ \sigma_t(x) = U_t x U_t^{-1}. \qquad (2.46) $$

Note that $\sigma_t(\ell_j) = \xi_j^{2it}\ell_j$ and $\sigma_t(\ell'_j) = \xi_j^{-2it}\ell'_j$ so that we have for all $j$,

$$ \sigma_t(x_j) = \xi_j^{2it} x_j, \quad \sigma_t(x_j^*) = \xi_j^{-2it} x_j^*. \qquad (2.47) $$

Therefore, $\sigma_t$ is a one parameter group of automorphisms of $\mathcal{L}$ that is nothing but the classical modular automorphism group of $\mathcal{L}$ relative to the state $\varphi$. In particular, $(\sigma_t)$ satisfies the KMS condition: for any polynomials $x$, $y$ in the generators $\{x_j\}$, we have

$$ \varphi(\sigma_t(x)y) = \varphi(yx), \qquad (2.48) $$

where $z \to \sigma_z$ is the obvious analytic extension of $t \to \sigma_t$. As is well known (cf., e.g., [25, 26]), we have in this situation,

$$ \|x\|_{\mathcal{L}_{\theta}} = \|x\|_{(L_c^2, L_r^2)_{\theta}} = \|\sigma_{-i\theta/2}(x)\|_{L_c^2} = \varphi(\sigma_{-i\theta/2}(x)^* \sigma_{-i\theta/2}(x))^{1/2}. \qquad (2.49) $$

Hence, we can write, since $\xi_j \lambda_j^{1/2} = \theta(1 - \theta)^{-1}$,

$$
\begin{align*}
\Sigma \|x(k)\|_{\mathcal{L}_{\theta}}^2 &= \Sigma \varphi (\sigma_{-i\theta/2}(x(k))^* \sigma_{-i\theta/2}(x(k))) \\
&= \Sigma \|\sigma_{-i\theta/2}(x(k))\Omega\|^2 \\
&= \sum_k \left\| \sum_j z_j(k) \sigma_{-i\theta/2}(x_j) \Omega \right\|^2 \\
&= \sum_k \left\| \sum_j z_j(k) \xi_j^\theta x_j \Omega \right\|^2 \tag{2.50} \\
&= \sum_k \left\| \sum_j z_j(k) \xi_j^\theta (1-\theta) \lambda_j^{\theta/2} e_j \right\|^2 \\
&= \sum_j \|z_j\|^2 (c(\theta))^{-2}.
\end{align*}
$$
Hence, we obtain the announced equality

$$ (\Sigma\|z_j\|^2)^{1/2} = c(\theta) (\Sigma\|x(k)\|_{\mathcal{L}_{\theta}}^2)^{1/2}. \qquad (2.51) $$

The converse of Theorem 2.1 also holds as follows. In the case $\theta = 1/2$, this was proved in [24]. We give a more direct argument.

**Proposition 2.8.** Let $E \subset A$ be an operator space embedded in a $C^*$-algebra $A$. Then any linear map $u: E \to \ell_2$ for which there are states $f, g$ on $A$ and a constant $C$ such that

$$ \forall a \in E, \quad \|ua\| \le C f(a^*a)^{(1-\theta)/2} g(aa^*)^{\theta/2} \qquad (2.52) $$

is completely bounded, with $\|u\|_{cb} \le C$, as a mapping into $\ell_2$ equipped with the operator space structure of $(C, R)_\theta$. $\square$

**Proof.** Let $F = (C, R)_\theta$ (recall that this is isometric to $\ell_2$). We will show that for any $a = (a_{ij})$ in the unit ball of $M_n(E)$ we have $\|(u(a_{ij}))\|_{M_n(F)} \le C$. Let $(T_k)$ be an orthonormal basis of $F = (C, R)_\theta$. Let $u_k: E \to C$ be defined by $u(a) = \sum_k u_k(a)T_k$. We denote $\gamma_k = \sum_{ij} u_k(a_{ij})e_{ij} \in M_n$. Using $M_n(F) \simeq M_n \otimes F$, the matrix $(u(a_{ij}))$ can then be rewritten as

$$ \Sigma e_{ij} \otimes u(a_{ij}) = \sum_k \gamma_k \otimes T_k. \qquad (2.53) $$

Let $p, p'$ be defined by $1 - \theta = 1/p$ and $\theta = 1/p'$. By the definition of $(C, R)_\theta$, we have (see [20, identity (8.5), page 83], but note that our space $R[\theta]$ corresponds to the space denoted by $R(1 - \theta)$ in [20])

$$ \|(u(a_{ij}))\|_{M_n(F)} = \sup \left\{ \left( \sum_k \|s\gamma_k t\|_2^2 \right)^{1/2} \right\}, \qquad (2.54) $$

where $\| \cdot \|_2$ denotes the Hilbert-Schmidt norm on $M_n$ and where the supremum runs over all pairs $(s, t)$ in $(M_n)_+ \times (M_n)_+$ such that $\text{tr} s^{2p'} \le 1$ and $\text{tr} t^{2p} \le 1$.

Let $x_{ij}$ denote the $(ij)$-entry of $s\hat{t}$ (so that $x_{ij} = \sum_{k\ell} s_{ik} a_{k\ell} t_{\ell j}$). We have

$$ \sum_k \|s\gamma_k t\|_2^2 = \sum_{ij} \|u(x_{ij})\|_F^2. \qquad (2.55) $$
We claim that (2.52) implies

$$ \sum_{ij} \|u(x_{ij})\|_{F}^{2} \le C^{2}. \qquad (2.56) $$

By (2.54), this claim implies that $\|u\|_{cb} \le C$, thus completing the proof. To show this claim, we may as well assume (replacing $\alpha$ by $v_1 a v_2$ for suitable unitaries $v_1, v_2$ in $M_n$, and using (2.54) again) that $s$ and $t$ are diagonal matrices. We then have

$$ \sum_{ij} \|u(x_{ij})\|^2 = \sum_{ij} s_{ii}^2 \|u(a_{ij})\|^2 t_{jj}^2, \qquad (2.57) $$

hence, since $\sum_{ii} s_{ii}^{2p'} \le 1$ and $\sum_{jj} t_{jj}^{2p} \le 1$, we have, using (2.52),

$$ \sum_{ij} \|u(x_{ij})\|^2 \le C^2 \left\| \sum f(a_{ij}^* a_{ij})^{1-\theta} g(a_{ij} a_{ij}^*)^{\theta} e_{ij} \right\|_{B(\ell_p^n, \ell_p^n)}. \qquad (2.58) $$

Thus, the above claim follows from the next lemma.

<figure><img src="image_2.png" /></figure>

**Lemma 2.9.** Let $f, g$ be states on a $C^*$-algebra $\mathcal{A}$. Then, for any $n \ge 1$ and any $\alpha$ in $M_n(\mathcal{A})$,

$$ \left\| \sum_{ij} f(a_{ij}^* a_{ij})^{1-\theta} g(a_{ij} a_{ij}^*)^{\theta} e_{ij} \right\|_{B(\ell_p^n, \ell_p^n)} \le \|a\|_{M_n(\mathcal{A})}. \qquad (2.59) $$

<figure><img src="image_3.png" /></figure>

**Proof.** To prove Lemma 2.9, we first check that if $\|a\|_{M_n(\mathcal{A})} \le 1$, then $\alpha_0(i, j) = f(a_{ij}^* a_{ij})$ and $\alpha_1(i, j) = g(a_{ij} a_{ij}^*)$ satisfy

$$ \sup_j \sum_i \alpha_0(i, j) \le 1, \qquad \sup_i \sum_j \alpha_1(i, j) \le 1. \qquad (2.60) $$

Indeed, we have, for any fixed $j$,

$$ \sum_i f(a_{ij}^* a_{ij}) \le \left\| \sum_i a_{ij}^* a_{ij} \right\| = \left\| \sum_i a_{ij} \otimes e_{ij} \right\| \le \|a\|_{M_n(\mathcal{A})}, \qquad (2.61) $$

and similarly for the other sum. Now let $\alpha_0(i, j)$, $\alpha_1(i, j)$ be any $n \times n$ matrices with non-negative entries satisfying (2.60). Then it is well known and easy to check by interpolation (see, e.g., [18] for more on this topic) that

$$ \left\| \sum_{ij} \alpha_0(i, j)^{1-\theta} \alpha_1(i, j)^{\theta} e_{ij} \right\|_{B(\ell_p^n, \ell_p^n)} \le 1, \qquad (2.62) $$
or equivalently, for any $s_i, t_j \ge 0$ with $\sum s_i \le 1$, $\sum t_j \le 1$, we have

$$ \sum_{ij} \alpha_0(i, j)^{1-\theta} \alpha_1(i, j)^{\theta} s_i^{1/p'} t_j^{1/p} \le 1. \qquad (2.63) $$

Indeed, by Hölder's inequality (recall that $1/p = 1 - \theta$ and $1/p' = \theta$), this is

$$ \begin{aligned} & \le \left( \sum_{ij} \alpha_0(i, j) t_j \right)^{1-\theta} \left( \sum_{ij} \alpha_1(i, j) s_i \right)^{\theta} \\ & \le \left( \sum_j t_j \right)^{1-\theta} \left( \sum_i s_i \right)^{\theta} \\ & \le 1. \end{aligned} \qquad (2.64) $$

Remark 2.10. When $\lambda_i = 1$ for all $i$, and $\theta = 1/2$, we have $x_i = (1/2)x_i'$, where

$$ x_i' = \ell_i + \ell_i^*. \qquad (2.65) $$

Then $(x_i')$ is a free circular (i.e., free analogue of complex Gaussian) family in Voiculescu's sense (cf. [34]). It is easy to see in this case that for any finite sequence $(a_i)$ in $B(\ell_2)$ we have

$$ \frac{1}{2} \left\| \sum a_i \otimes x_i' \right\| \le \max \left\{ \left\| \sum a_i^* a_i \right\|^{1/2}, \left\| \sum a_i a_i^* \right\|^{1/2} \right\} \le \left\| \sum a_i \otimes x_i' \right\|. \qquad (2.66) $$

Thus, $\overline{span}[x_i']$ is completely isomorphic to the space $R \cap C$ studied in [7] (see also [22, page 209]). The notation $R \cap C$ comes from the fact that if we consider $\delta_i = e_{1i} \oplus e_{i1}$ in $R \oplus C$ then we have for $(a_i)$, as above,

$$ \left\| \sum a_i \otimes \delta_i \right\| = \max \left\{ \left\| \sum a_i a_i^* \right\|^{1/2}, \left\| \sum a_i^* a_i \right\|^{1/2} \right\}, \qquad (2.67) $$

so that

$$ \overline{span}[\delta_i] = \{ (x, t)x \mid x \in \mathbb{R} \} \subset R \oplus C \qquad (2.68) $$

appears as the diagonal in $R \oplus C$. Let $\mathcal{L}$ be again the von Neumann algebra generated by $\{x_i'\}$. We claim (see [7] and also [22, page 209]) that there is a normal cb projection $P: \mathcal{L} \to \overline{span}[x_i']$ with $\|P\|_{cb} \le 2$. Indeed, let $Q: \mathcal{F} \to \mathcal{F}$ (resp., $Q': \mathcal{F} \to \mathcal{F}$) be the orthogonal projection onto $\overline{span}[e_i \mid i \in I]$ (resp., $\overline{span}[e_i' \mid i \in I]$) viewed as a subspace of $H$, itself embedded into $\mathcal{F}$ via tensors of degree 1. Then the map $P$ defined by

$$ \forall T \in \mathcal{L}, \quad P(T) = \ell(Q(T\Omega)) + \ell(Q'(T^*\Omega))^* \qquad (2.69) $$
is the announced projection (see more generally Lemma 2.11 below). Therefore, $(\mathbb{R} \cap \mathbb{C})^*$ embeds completely isomorphically into $\mathcal{L}_*$ and in the present special case, $\mathcal{L}$ is semifinite (and actually finite).

Consider now a family $\xi = (\xi_i)$ with $\xi_i > 0$. Let $\delta_i^\xi = e_{1i} \oplus \xi_i e_{i1} \in \mathbb{R} \oplus \mathbb{C}$. Thus, if $(a_i)$ is as before, we have

$$ \left\| \sum a_i \otimes \delta_i^\xi \right\| = \max \left\{ \left\| \sum a_i a_i^* \right\|^{1/2}, \left\| \sum \xi_i^2 a_i^* a_i \right\|^{1/2} \right\}, \qquad (2.70) $$

and also

$$ (1/2) \left\| \sum a_i \otimes (\ell_i + \xi_i \ell_i^*) \right\| \le \left\| \sum a_i \otimes \delta_i^\xi \right\| \le \left\| \sum a_i \otimes (\ell_i + \xi_i \ell_i^*) \right\|. \qquad (2.71) $$

Thus we have, completely isomorphically,

$$ \overline{\mathrm{span}}[\ell_i + \xi_i \ell_i^*] \simeq \overline{\mathrm{span}}[\delta_i^\xi]. \qquad (2.72) $$

In particular, if $(\xi_i)$ is as in (2.44), then $\overline{\mathrm{span}}[x_i] \simeq \overline{\mathrm{span}}[\delta_i^\xi]$ (completely isomorphically). Note that $\overline{\mathrm{span}}[\delta_i^\xi]$ can also be viewed as the graph of the unbounded operator $\Lambda : \mathbb{R} \to \mathbb{C}$ taking $e_{1i}$ to $\xi_i e_{i1}$, with $\mathrm{Dom}(\Lambda) = \{x = \sum x_i e_{1i} \in \mathbb{R} \mid \sum |\xi_i x_i|^2 < \infty\}$. More precisely, if we denote

$$ G(\Lambda) = \{(x, \Lambda x) \mid x \in \mathrm{Dom}(\Lambda)\}, \qquad (2.73) $$

then we have $\overline{\mathrm{span}}\delta_i^\xi \simeq G(\Lambda)$ completely isometrically. (Note that in analogy with $\mathbb{R} \cap \mathbb{C}$, it would be natural to denote $G(\Lambda)$ by $\mathbb{R} \cap \Lambda^{-1}(\mathbb{C})$, but we prefer not to use this notation.)

The next result (extending the case $\xi_i = 1$ for all $i$) is easy to deduce from Shlyakhtenko's [29].

**Lemma 2.11.** Let $\mathcal{L}$ be the von Neumann algebra generated by the family $(x_i)$ defined by (2.44). Then the mapping $P : \mathcal{L} \to \mathcal{L}$ defined by (2.69) is a normal cb projection from $\mathcal{L}$ onto $\overline{\mathrm{span}}[x_i]$ with $\|P\|_{\mathrm{cb}} \le 2$. In particular, $\overline{\mathrm{span}}[x_i]$ embeds completely isomorphically into $\mathcal{L}_*$.

**Proof.** We first claim that $T \to P(T)$ is cb on $B(\mathcal{F})$ with cb norm less than or equal to 2. This is easy to see. Indeed, consider $P_1 : B(\mathcal{F}) \to B(\mathcal{F})$ and $P_2 : B(\mathcal{F}) \to B(\mathcal{F})$ defined by

$$ P_1(T) = \ell(Q(T\Omega)), \qquad P_2(T) = \ell(Q'(T^*\Omega))^*, \qquad (2.74) $$
so that $P(T) = P_1(T) + P_2(T)$. We will show that $\|P_1\|_{cb} \le 1$ and $\|P_2\|_{cb} \le 1$. Indeed, the ranges of $P_1$ and $P_2$ are, respectively, $\overline{\mathrm{span}}[\ell_i]$ and $\overline{\mathrm{span}}[\ell_i^*]$. Assuming $I = \mathbb{N}$ for simplicity, we have (see, e.g., [22, pages 176–177])

$$ \overline{\mathrm{span}}[\ell_i] \simeq C, \quad \overline{\mathrm{span}}[\ell_i^*] \simeq \mathbb{R}. \qquad (2.75) $$

Note that we obviously have (recall that $\varphi$ is the vacuum state)

$$ \begin{aligned} \|P_1(T)\| &= \|Q(T\Omega)\| \le \|T\Omega\| = \langle T^*T\Omega, \Omega \rangle^{1/2} = \varphi(T^*T)^{1/2}, \\ \|P_2(T)\| &= \|Q'(T^*\Omega)\| \le \|T^*\Omega\| = \langle T^*T^*\Omega, \Omega \rangle^{1/2} = \varphi(T^*T)^{1/2}. \end{aligned} \qquad (2.76) $$

Hence, $\|P_1\|_{cb} \le 1$ and $\|P_2\|_{cb} \le 1$ (and a fortiori $\|P\|_{cb} \le 1$) follow using (2.3) and (2.1). Thus it suffices to prove that the restriction of $P$ to $\mathcal{L}$ is a projection onto $\overline{\mathrm{span}}[x_i]$. We know (see [29]) that the map $T \to T\Omega$ is faithful (i.e., injective) on $\mathcal{L}$. Let $T$ be a polynomial in $x_i$, $x_i^*(i \in I)$. We can a priori write

$$ T\Omega = \sum t_i e_i + \sum t'_i e'_i + r, \qquad (2.77) $$

where $r$ is a sum of tensors of degree greater than 1. By [29, Lemma 3.2], we know that the (antilinear) map $S$ taking $T\Omega$ to $T^*\Omega$ takes $r$ to another sum $r'$ of tensors of degree greater than 1 in $\mathcal{F}$. Moreover, since $(\ell_i + \xi_i \ell_i^*)\Omega = e_i$ and $(\ell_i + \xi_i \ell_i^*)^*\Omega = \xi_i e_i'$, we have

$$ Se_i = \xi_i e'_i, \quad Se'_i = \xi_i^{-1} e_i, \qquad (2.78) $$

and hence

$$ T^*\Omega = \sum \bar{t}_i \xi_i e'_i + \sum \overline{t'_i \xi_i^{-1}} e_i + r'. \qquad (2.79) $$

Therefore we have $Q(T\Omega) = \sum t_i e_i$, $Q'(T^*\Omega) = \sum \bar{t}_i \xi_i e_i'$, and we finally obtain

$$ P(T) = \sum t_i \ell(e_i) + \sum t_i \xi_i \ell(e_i')^* = \sum t_i (\ell_i + \xi_i \ell_i^*) \in \mathrm{span}[x_i]. \qquad (2.80) $$

In particular, we find $P(\ell_i + \xi_i \ell_i^*) = \ell_i + \xi_i \ell_i^*$, and hence $P(x_i) = x_i$ for all $i$. This proves that $P|_{\mathcal{L}}$ is a projection from $\mathcal{L}$ onto $\overline{\mathrm{span}}[x_i]$. ■

Note that, by (2.71), if

$$ 0 < \inf \xi_i \le \sup \xi_i < \infty \qquad (2.81) $$
then $\overline{\text{span}}[x_i]$ (or equivalently $G(\Lambda)$) is again completely isomorphic to $R \cap C$, and hence its dual embeds in $M_*$ for some semifinite $M$. We now show that if either $\inf \xi_i = 0$ or $\sup \tilde{\xi}_i = \infty$, then such an embedding $G(\Lambda)^* \subset M_*$ with $M$ semifinite exists if and only if we have for some $\varepsilon > 0$,

$$ \sum_{i: \xi_i < \varepsilon} \xi_i^2 + \sum_{i: \xi_i^{-1} < \varepsilon} \xi_i^{-2} < \infty. \qquad (2.82) $$

Let $M$ be a von Neumann algebra with predual $M_*$. As already mentioned at the end of [23], Theorem 2.1 and its converse (Proposition 2.8) admit the following corollary: for any $0 < \theta < 1$, the space $R[\theta]$ does not embed completely isomorphically into $M_*$ when $M$ is semifinite. It can be shown (see the appendix) that $R[\theta]$ is completely isometric to a quotient of a subspace $S$ of $R \oplus C$. Thus, $R[\theta]^*$ embeds in $S^*$, and hence to embed $R[\theta]^*$ into $M_*$, it suffices to embed $S^*$ into $M_*$. Indeed, Marius Junge announced that if $S$ is any subspace of $R \oplus C$, then $S^*$ embeds completely isomorphically into $M_*$ for some von Neumann algebra $M$. Let $S \subset R \oplus C$ be such a subspace. For convenience, we assume that $S$ is not completely isomorphic to either $R$, $C$, or $R \oplus C$. Then Xu [38] observed the fact (presumably known to Junge) that $S$ can be rewritten (up to complete isomorphism) as a direct sum $H_r \oplus \tilde{S} \oplus K_c$, where $H_r$, $K_c$ are suitable Hilbert spaces equipped, respectively, with the row and column operator space structures, and where $\tilde{S} \subset R \oplus C$ is the (closed) graph of a (closed) densely defined operator $\Lambda : R \to C$, injective (on its domain) and with dense range. As explained in the appendix, the fact that $(\tilde{S})^*$ embeds into $M_*$ for some suitable $M$ can be deduced from the basic properties of Shlyakhtenko's generalized free circular elements, already used in [24]. The typical $M$ is then not semifinite. The next result shows that this cannot be avoided.

**Theorem 2.12.** Let $S \subset R \oplus C$ be an arbitrary infinite-dimensional subspace. Then there is a semifinite von Neumann algebra $M$ such that $S^*$ embeds completely isomorphically into $M_*$ if and only if $S$ is completely isomorphic to one of the spaces

$$ R, C, R \oplus C, R \cap C, R \oplus (R \cap C), C \oplus (R \cap C), R \oplus (R \cap C) \oplus C. \qquad (2.83) $$

□

**Remark 2.13.** As is well known, we have $R^* \simeq C$ and $C^* \simeq R$, so that $R^*$ and $C^*$ embed (completely isometrically) in $K^* \simeq S_1$ (the trace class). Consequently, $(R \oplus C)^* \simeq R \oplus C$ embeds completely isomorphically into $S_1 \oplus S_1 \simeq S_1$. The space $R \cap C$ is less trivial, but it was shown by Lust-Piquard and the author (see [15] or [22, page 194]) that $(R \cap C)^*$ embeds completely isomorphically into the most classical $L_1$-space, namely, $L_1([0, 1], dt)$. Therefore, we do have $S^* \subset M_*$ with $M$ semifinite for any of the 7 spaces in the above list.
Our task will now be to show that the latter list is complete.

Remark 2.14. Consider a (closed) subspace $S \subset R \oplus C$. As explained above, we can write

$$S \simeq H_r \oplus G(\Lambda) \oplus K_c \qquad (2.84)$$

with

$$G(\Lambda) = \{(x, \Lambda x) | x \in \text{Dom}(\Lambda)\} \subset R \oplus C, \qquad (2.85)$$

where $\text{Dom}(\Lambda) \subset R$ is a dense subspace and $\Lambda : \text{Dom}(\Lambda) \to C$ is a closed unbounded operator with zero kernel and dense range.

By the polar decomposition of $\Lambda$ and the “homogeneity” of $R$ and $C$ (in the sense of [22, page 172]), we may assume that $\Lambda > 0$. Using the spectral theory of Hermitian operators, we can then decompose $\Lambda$ as $\Lambda = \Lambda_1 + \Lambda_2$ with $0 < \Lambda_1 \le 1$ and $\Lambda_2 \ge 1$, and consequently we may decompose

$$G(\Lambda) \simeq G(\Lambda_1) \oplus G(\Lambda_2), \qquad (2.86)$$

where $\Lambda_1, \Lambda_2$ are unbounded selfadjoints of the same form as $\Lambda$ but, in addition, such that $\Lambda_1$ and $\Lambda_2^{-1}$ are bounded with norm less than or equal to 1. The key to the preceding theorem then lies in the next statement.

**Lemma 2.15.** Consider $\Lambda > 0$ with $\|\Lambda\| \le 1$ and $\Lambda^{-1}$ unbounded. Let $E(\varepsilon)$ be the spectral projection of $\Lambda$ relative to $(0, \varepsilon)$, so that $0 \ne \|\Lambda E(\varepsilon)\| \le \varepsilon$ for any $\varepsilon > 0$. Assume that there is a semifinite $M$ such that $G(\Lambda)^*$ embeds completely isomorphically into $M_*$. Then, for $\varepsilon > 0$ small enough, $\Lambda E(\varepsilon)$ must be Hilbert-Schmidt. $\square$

**Proof.** The basic idea is similar to the one in [23] but the details are more complicated. By assumption, we have an embedding $j : G(\Lambda)^* \subset M_*$. Let $u = j^* : M \to G(\Lambda)$. We may assume that $\|u\|_{cb} \le 1$ and that there is a constant $c$ such that for any $n$ and any $a$ in $M_n(G(\Lambda))$ with $\|a\| < 1$, there is $\tilde{a}$ in $M_n(M)$ with

$$\|\tilde{a}\| < c \qquad (2.87)$$

such that $(I \otimes u)(\tilde{a}) = a$. Note that $u$ is “normal,” that is, $(\sigma(M, M_*), \sigma(G(\Lambda), G(\Lambda)^*))$ continuous. The map $u$ can clearly be rewritten as $ux = (vx, \Lambda vx)$ with

$$\|v : M \to R\|_{cb} \le 1, \qquad \|\Lambda v : M \to C\|_{cb} \le 1. \qquad (2.88)$$
Let $\tau$ be a semifinite faithful normal trace on M. Since $v$ and $\Lambda v$ are normal, arguing as in [23], we find normal states $f, g$ on M such that

$$ \begin{aligned} \|vx\| &\le f(xx^*)^{1/2}, \\ \|\Lambda vx\| &\le g(x^*x)^{1/2}, \end{aligned} \qquad (2.89) $$

for all $x$ in M.

We may view $f, g$ as elements of $L_1(\tau)$, that is, positive unbounded operators affiliated to M such that $\tau(f) = \tau(g) = 1$, and consequently we will write $f(\cdot) = \tau(f\cdot)$ and $g(\cdot) = \tau(g\cdot)$. Fix $\alpha > 0$. Let $p$ (resp., $q$) be the spectral projection of $f$ (resp., $g$) associated to $(\alpha^{-1}, \alpha]$, so that in $M_*$, we have $\alpha^{-1}p \le pfp \le \alpha p$ and $\alpha^{-1}q \le qgq \le \alpha q$. Choosing $\alpha = \alpha(\delta)$ large enough, we can ensure that moreover,

$$ \|f(1-p)\|_{M_*} < \delta^2, \quad \|g(1-q)\|_{M_*} < \delta^2. \qquad (2.90) $$

Moreover, we have $\alpha^{-1}\tau(p) \le \tau(f) = 1$ and $\alpha^{-1}\tau(q) \le \tau(g) = 1$, and hence

$$ \tau(p) \le \alpha, \quad \tau(q) \le \alpha. \qquad (2.91) $$

We then define

$$ v_{\delta}x = v(pxq). \qquad (2.92) $$

Note that

$$ \|v_{\delta}x\| \le f(pxqx^*p)^{1/2} \le \sqrt{\alpha\tau(pxqx^*)^{1/2}} = \sqrt{\alpha\tau(qx^*px)^{1/2}} \le \sqrt{\alpha\tau(qx^*x)^{1/2}}, \qquad (2.93) $$

where the last equality holds by the trace property, so that by (2.3),

$$ \|v_{\delta} : M \to C\|_{cb} \le (\alpha\tau(q))^{1/2} \le \alpha. \qquad (2.94) $$

On the other hand, we have

$$ v_x - v_{\delta}x = v_1x + v_2x \qquad (2.95) $$

with $v_1x = v(x(1-q))$ and $v_2x = v((1-p)xq)$. Note that

$$ \|\Lambda v_1 x\| \le g((1-q)x^*x(1-q))^{1/2} = \tau(g(1-q)x^*x)^{1/2}, \qquad (2.96) $$
hence by (2.3),

$$ \|\Lambda v_1 : M \to C\|_{cb} \le \delta. \qquad (2.97) $$

Similarly, we have

$$ \|v_2 x\| \le \tau (f(1-p)xx^*)^{1/2}, \qquad (2.98) $$

hence by (2.1),

$$ \|v_2\| \le \|v_2 : M \to R\|_{cb} \le \delta. \qquad (2.99) $$

We now turn to the following.

**Claim 2.16.** $\varepsilon > 0$ can be chosen so that $\|\Lambda E(\varepsilon)\|_2 \le 1$. $\square$

For each integer $n \ge 1$, let

$$ \pi_2^n = \sup \left\{ \left( \sum_{i=1}^{n} \| \Lambda e_i \|^2 \right)^{1/2} \right\}, \qquad (2.100) $$

where the supremum runs over all possible orthonormal $n$-tuples $(e_1, \dots, e_n)$ in the range of the projector $E(\varepsilon)$. Note that for any operator $w : M \to E(\varepsilon)$ and for any $a_1, \dots, a_n$ in $M$, we have

$$ \left( \sum_{i=1}^{n} \| \Lambda w a_i \|^2 \right)^{1/2} \le \pi_2^n \| w \| \left\| \sum a_i^* a_i \right\|^{1/2}. \qquad (2.101) $$

Indeed, let $T : [e_1, \dots, e_n] \mapsto M$ be the map defined by $Te_i = a_i$. Note that $\|T\| \le \|\Sigma a_i^* a_i\|^{1/2}$. Let $F = \text{span}[wa_i]$,

$$ \sum \| \Lambda w a_i \|^2 = \sum_{i=1}^{n} \| \Lambda w Te_i \|^2 = \| \Lambda w T \|_2^2 \le \| \Lambda_F \|_2^2 \| w \|^2 \| T \|^2, \qquad (2.102) $$

and since $\dim F \le n$, we have $\|\Lambda_F\|_2 \le \pi_2^n$, hence $\Sigma \| \Lambda w a_i \|^2 \le (\pi_2^n)^2 \|w\|^2 \|\Sigma a_i^* a_i\|$, which establishes (2.101).

**Claim 2.17.** If $\pi_2^n \le 1$, then we have

$$ \pi_2^n \le (\varepsilon\alpha + \delta + \delta\pi_2^n)c. \qquad (2.103) $$

$\square$
To prove this, consider $(e_1, \dots, e_n)$ in $E(\varepsilon)$ and let

$$a = \sum_{i=1}^{n} e_{i1} \otimes (e_i, \Lambda e_i) \in M_n(G(\Lambda)). \qquad (2.104)$$

We have

$$\|a\| = \max \left\{ \left\| \sum e_{i1} \otimes e_i \right\|_{C_n \otimes_{\min R}} , \left\| \sum e_{i1} \otimes \Lambda e_i \right\|_{C_n \otimes_{\min C}} \right\}, \qquad (2.105)$$

hence (since $\pi_2^n \le 1$)

$$\|a\| = \max \left\{ 1, \left( \sum \|\Lambda e_i\|^2 \right)^{1/2} \right\} \le 1. \qquad (2.106)$$

By (2.87), there is $\tilde{a}$ in $M_n(M)$ with $\|\tilde{a}\| \le c$, such that $(I \otimes u)(\tilde{a}) = a$. Clearly, we may assume $\tilde{a} = \sum_1^n e_{i1} \otimes a_i$ with $a_i \in M$ such that $e_i = va_i$. Note that

$$\left\| \sum a_i^* a_i \right\|^{1/2} = \|\tilde{a}\| \le c. \qquad (2.107)$$

Note that since $v = v_\delta + v_1 + v_2$, we have $e_i = v_\delta a_i + v_1 a_i + v_2 a_i$, hence if we let $\Lambda_\varepsilon = E(\varepsilon)\Lambda = \Lambda E(\varepsilon)$, we have

$$(\sum \|\Lambda e_i\|^2)^{1/2} \le (\sum \|\Lambda_\varepsilon v_\delta a_i\|^2)^{1/2} + (\sum \|\Lambda_\varepsilon v_1 a_i\|^2)^{1/2} \\ + (\sum \|\Lambda_\varepsilon v_2 a_i\|^2)^{1/2}. \qquad (2.108)$$

By (2.94), (2.107), and (2.4), we have

$$(\sum \|\Lambda_\varepsilon v_\delta a_i\|^2)^{1/2} \le \varepsilon (\sum \|v_\delta a_i\|^2)^{1/2} \le \varepsilon (\alpha \tau(q))^{1/2} c, \qquad (2.109)$$

and also by (2.97),

$$(\sum \|\Lambda_\varepsilon v_1 a_i\|^2)^{1/2} \le (\sum \|\Lambda v_1 a_i\|^2)^{1/2} \le \delta c. \qquad (2.110)$$

Finally, by (2.99), (2.101), and (2.107), we have (recall that $\Lambda_\varepsilon = \Lambda E(\varepsilon)$)

$$(\sum \|\Lambda_\varepsilon v_2 a_i\|^2)^{1/2} \le \pi_2^n \|E(\varepsilon)v_2\|c \le \delta c \pi_2^n. \qquad (2.111)$$

Recapitulating, we can now deduce (2.103) from (2.108), and Claim 2.17 follows.
We can now prove Claim 2.16.

We assume $\epsilon < 1/2$. We will argue by contradiction. Assume that $\sup_m \pi_2^m > 1$. We will show that this is impossible. Let $n+1$ be the smallest integer such that $\pi_2^{n+1} > 1$. Note that $\pi_2^n \le 1$ and $n \ge 1$ (because $\pi_2^1 \le \epsilon < 1$). Moreover, we obviously have $\pi_2^{n+1} \le \pi_2^n + \epsilon \le \pi_2^n + 1/2$, hence $\pi_2^n > 1/2$. But now, if we choose $\delta$ so that $\delta c < 1/2$, (2.103) implies

$$ \pi_2^n \le c(\epsilon\alpha + \delta) + \frac{\pi_2^n}{2}, \qquad (2.112) $$

hence

$$ \pi_2^n \le 2c(\epsilon\alpha + \delta), \qquad (2.113) $$

so that since $\pi_2^n > 1/2$, we obtain

$$ \frac{1}{2} \le 2c(\epsilon\alpha + \delta). \qquad (2.114) $$

But now, if we choose $\delta = 1/8c$, this implies

$$ \frac{1}{4} \le 2c\epsilon\alpha, \qquad (2.115) $$

and here $\alpha = \alpha(\delta)$ is determined by $\delta$ but $\epsilon$ can still be made arbitrarily small. Thus we reach a contradiction, proving that $\sup_m \pi_2^m \le 1$ for any $\epsilon < 1/2$ for which (2.115) fails. This proves Claim 2.16 and completes the proof of Lemma 2.9. ■

**Proof of Theorem 2.12.** By Xu's result (2.84), we are reduced to S of the form $S = G(\Lambda)$ for $\Lambda > 0$ with dense range. By (2.86), we may assume that either $\Lambda$ or $\Lambda^{-1}$ has norm less than or equal to 1. But observe that if $\|\Lambda^{-1}\| \le 1$,

$$ G(\Lambda^{-1}) = \{(x, \Lambda^{-1}x) \mid x \in C\} = \{(\Lambda y, y) \mid y \in \text{Dom}(\Lambda)\} \subset C \oplus \mathbb{R}, \qquad (2.116) $$

hence $G(\Lambda^{-1}) \simeq G(\Lambda)$ since the first is obtained from the second via the mapping $(x, y) \to (y, x)$ which is obviously a complete isometry from $C \oplus R$ to $R \oplus C$. In particular, $G(\Lambda^{-1})$ embeds in $M_*$ if and only if $G(\Lambda)$ embeds in $M_*$. Thus, to conclude, we may as well assume that $\|\Lambda\| \le 1$. But then Lemma 2.15 shows that for $\epsilon$ small enough we have a decomposition $R = H_\epsilon \oplus H_\epsilon^\perp$ and $\Lambda = \Lambda_\epsilon \oplus \Lambda'_\epsilon$ with $\|\Lambda_\epsilon\|_2 < \infty$. Clearly, this implies $G(\Lambda) \simeq G(\Lambda_\epsilon) \oplus G(\Lambda'_\epsilon)$, but since $\|\Lambda_\epsilon : (H_\epsilon)_r \to C\|_{cb} = \|\Lambda_\epsilon\|_2 < \infty$, we have $G(\Lambda_\epsilon) \simeq (H_\epsilon)_r$, and since $\epsilon \le \Lambda'_\epsilon \le 1$, we obviously have (arguing as in the case when (2.81) holds) $G(\Lambda'_\epsilon) \simeq (H_\epsilon^\perp)_r \cap (H_\epsilon^\perp)_c$. This completes the proof of Theorem 2.12. ■
Remark 2.18. It may be worthwhile to point out that in Lemma 2.15, even if we know that $G(\Lambda)^*$ is completely c-isomorphic to a subspace of $M_*$ with a fixed c, the $\varepsilon$ given by Lemma 2.15 may be arbitrarily small, and this happens even for M finite. Indeed, for the relevant examples, consider a free circular sequence $(x'_i)$ on $(M, \tau)$ (with $\tau$ a normalized trace) and a projection $p$ that is free from that family and such that $\tau(p) = \varepsilon$ (see [34]). Then $\overline{\mathrm{span}[px'_i]}$ provides the required phenomenon.

In the next statement, we observe that Xu's decomposition for subspaces of $R \oplus C$ leads to an easy proof of a result due to Oikhberg [16] (with an improved bound).

**Theorem 2.19.** Let $S \subset R \oplus C$ be a closed subspace. If there is a cb projection $P: R \oplus C \to S$, then there are Hilbert spaces $H$, $K$ such that $S \simeq H_r \oplus K_c$. Moreover, there is a numerical constant $C$ such that $d_{cb}(S, H_r \oplus K_c) \le C\|P\|_{cb}$. $\square$

**Proof.** By Xu's decomposition and the above remarks, it suffices to prove this for $S = G(\Lambda)$ with $0 < \Lambda$ and $\|\Lambda\| \le 1$. Then the projection $P$ can be written as

$$ \forall (x, y) \in R \oplus C, \quad P(x, y) = (\alpha x + \beta y, \Lambda(\alpha x + \beta y)), \qquad (2.117) $$

where $\alpha \in CB(R, R)$ and $\beta \in CB(C, R)$. By restricting $P$, we find

$$ \max \{\|\alpha\|_{CB(R,R)}, \|\Lambda\alpha\|_{CB(R,C)}, \|\beta\|_{CB(C,R)}, \|\Lambda\beta\|_{CB(C,C)}\} \le \|P\|_{cb}. \qquad (2.118) $$

Moreover, since $P$ is a projection onto $G(\Lambda)$, we have for any $x$ in $R$,

$$ \alpha x + \beta \Lambda x = x, \qquad (2.119) $$

hence

$$ \Lambda\alpha + \Lambda\beta\Lambda = \Lambda, \qquad (2.120) $$

which implies by (1.2) and (2.118) (since we assume $\Lambda \le 1$),

$$ \|\Lambda\|_{CB(R,C)} = \|\Lambda\|_2 \le \|\Lambda\alpha\|_2 + \|\Lambda\beta\Lambda\|_2 = \|\Lambda\alpha\|_{CB(R,C)} + \|\Lambda\beta\Lambda\|_{CB(C,R)} \le 2\|P\|_{cb}. \qquad (2.121) $$

Thus we conclude that

$$ \|\Lambda\|_{CB(R,C)} \le 2\|P\|_{cb}, \qquad (2.122) $$
and hence the map $u : x \to (x, \Lambda x)$ is a complete isomorphism between $\mathbb{R}$ and $G(\Lambda)$ with

$$d_{cb}(\mathbb{R}, G(\Lambda)) \le \|u\|_{cb} \|u^{-1}\|_{cb} \le 2 \|P\|_{cb}. \qquad (2.123)$$

<figure><img src="image_4.png" /></figure>

Remark 2.20. The preceding statement yields a rather satisfactory estimate in the following result from [24]: if an operator space $E$ is exact as well as its dual, then there are Hilbert spaces $H$, $K$ such that $E \simeq H_r \oplus K_c$, and moreover,

$$d_{cb}(E, H_r \oplus K_c) \le 2^{5/2} \operatorname{ex}(E) \operatorname{ex}(E^*), \qquad (2.124)$$

where $\operatorname{ex}(E)$ denotes the exactness constant of $E$. This seems rather sharp when $\operatorname{ex}(E) \operatorname{ex}(E^*)$ is large: consider, for instance, the case $E = \operatorname{OH}_n$; we then have (cf. [23, page 336]) $\operatorname{ex}(E) = \operatorname{ex}(E^*) \simeq n^{1/4}$ but, on the other hand, it is easily checked that

$$d_{cb}(\operatorname{OH}_n, H_r \oplus K_c) \simeq n^{1/2} \simeq \operatorname{ex}(E) \operatorname{ex}(E^*). \qquad (2.125)$$

## Appendix

In this appendix, we will reprove Junge's result [10] that OH embeds completely isomorphically into a noncommutative $L_1$-space. The main idea is the same as his, but our exposition is shorter and makes more transparent the relationship between the methods from [10, 24]. We base the argument on the complex interpolation method instead of the Pusz-Woronowicz formula. Actually, there is nothing mysterious there: indeed, the "purification of states" associated in [26] (see also [25, 35, 36, 37]) to a pair of faithful states $(\varphi, \psi)$ on a $C^*$-algebra $A$ is known to be very closely related to the complex interpolation space $(A_0, A_1)_{1/2}$, where the Hilbert spaces $A_0, A_1$ are obtained by completing $A$ for the norms

$$\|x\|_{A_0} = (\varphi(x^*x))^{1/2}, \qquad \|x\|_{A_1} = (\psi(xx^*))^{1/2}. \qquad (A.1)$$

This close connection has been explored in depth notably by B. Simon, Uhlman, Peetre and probably others, besides Pusz and Woronowicz.

The proof rests on the following basic fact which had been known to the author (and probably also to Junge) for some time, before Junge proved his embedding result for OH. A detailed proof is included as the solution to [22, Exercise 7.8]. We reproduce it here for the convenience of the reader.

**Proposition A.1.** OH is completely isometric to a quotient of a subspace of $\mathbb{R} \oplus C$. $\square$
Proof. Let $\mu$ be the harmonic measure of the point $z = 1/2$ in the strip $S = \{z \in \mathbb{C} | 0 < \text{Re}(z) < 1\}$. Recall that $\mu$ is a probability measure on $\partial S$ such that $f(1/2) = \int f d\mu$ whenever $f$ is a bounded harmonic function on $S$ extended nontangentially to $\bar{S}$. Obviously, $\mu$ can be written as $\mu = 2^{-1}(\mu_0 + \mu_1)$, where $\mu_0$ and $\mu_1$ are probability measures supported, respectively, by

$$ \partial_0 = \{z | \text{Re}(z) = 0\}, \quad \partial_1 = \{z | \text{Re}(z) = 1\}. \qquad (A.2) $$

Let $(A_0, A_1)$ be a compatible pair of Banach spaces. We first need to describe $(A_0, A_1)_{1/2}$ as a quotient of a subspace of $L_2(\mu_0; A_0) \oplus L_2(\mu_1; A_1)$. The classical argument for this is as follows.

We denote by $\mathcal{F}(E_0, E_1)$ the set of all bounded continuous functions $f: \bar{S} \to E_0 + E_1$ which are holomorphic on $S$ and such that $f|_{\partial_0}$ and $f|_{\partial_1}$ are bounded continuous functions with values, respectively, in $E_0$ and $E_1$.

We start by showing that for any $x$ in $(A_0, A_1)_{1/2}$ we have

$$ \|x\|_{(A_0, A_1)_{1/2}} = \inf \left\{ \max \left\{ \|f\|_{L_2(\mu_0; A_0)}, \|f\|_{L_2(\mu_1; A_1)} \right\} \right\}, \qquad (A.3) $$

where the infimum runs over all $f$ in $\mathcal{F}(A_0, A_1)$ such that $f(1/2) = x$. For a proof, see [14, page 224]. Let then $E = L_2(\mu_0; A_0) \oplus \infty L_2(\mu_1, A_1)$ and let $G \subset E$ be the closure of the subspace $\{f|_{\partial_0} \oplus f|_{\partial_1} | f \in \mathcal{F}(A_0, A_1)\}$. The preceding equality shows that the mapping $f \to f(1/2)$ defines a metric surjection $Q: G \to (A_0, A_1)_{1/2}$. We now consider the couple $(A_0, A_1) = (\mathbb{R}, \mathbb{C})$, where we think of $\mathbb{R}$ and $\mathbb{C}$ as operator space structures on the "same" underlying vector space, identified with $\ell_2$. We introduce the operator space $E = L_2(\mu_0; \ell_2)_r \oplus L_2(\mu_1; \ell_2)_c$. Let $G$ and $Q: G \to \ell_2$ be the same as before. Note that $G$ is nothing but the $\ell_2$-valued version of the Hardy space $H^2$ on the strip $S$, so that if we assume $f$ analytically extended inside $S$, we have $Q(f) = f(1/2)$.

We first claim that

$$ \|Q: G \to \text{OH}\|_{cb} \le 1. \qquad (A.4) $$

To verify this, consider $x$ in $M_n(G)$ with $\|x\|_{M_n(G)} \le 1$. We claim that $\|x(1/2)\|_{M_n(\text{OH})} \le 1$. We may view $x$ as a sequence $(x_k)$ of $M_n$-valued functions on $\partial S$ extended analytically inside $S$, so that

$$ \|x\|_{M_n(G)} = \max \left\{ \left\| \left( \int \sum x_k x_k^* d\mu_0 \right)^{1/2} \right\|_{M_n}, \left\| \left( \int \sum x_k^* x_k d\mu_1 \right)^{1/2} \right\|_{M_n} \right\}, \qquad (A.5) $$
and by [22, (7.3)'],

$$
\begin{align}
\left\| x\left(\frac{1}{2}\right) \right\|_{M_n(\text{OH})}^2 &= \left\| \sum_k x_k\left(\frac{1}{2}\right) \otimes \overline{x_k\left(\frac{1}{2}\right)} \right\|_{\min} \tag{A.6} \\
&= \sup \left\{ \left| \operatorname{tr}\left( \sum_k x_k\left(\frac{1}{2}\right) \alpha x_k\left(\frac{1}{2}\right)^* b \right) \right| \right\}, \nonumber
\end{align}
$$

where the supremum runs over all $\alpha, \beta \ge 0$ in $M_n$ such that $\operatorname{tr}|\alpha|^2 \le 1$ and $\operatorname{tr}|\beta|^2 \le 1$. Fix $\alpha, \beta$ satisfying these conditions. Consider then the analytic function

$$
F(z) = \operatorname{tr} \left( \sum_k x_k(z) a^{2z} x_k(\bar{z})^* b^{2(1-z)} \right) \quad (A.7)
$$

on S. Note that

$$
F\left(\frac{1}{2}\right) = \operatorname{tr}\left(\sum_k x_k\left(\frac{1}{2}\right) \alpha x_k\left(\frac{1}{2}\right)^* b\right) = 2^{-1}\left(\int_{\partial_0} F d\mu_0 + \int_{\partial_1} F d\mu_1\right). \quad (A.8)
$$

But, for all $z = it$ in $\partial_0$, we have

$$
F(it) = \sum_k \operatorname{tr} (b^{1-it} x_k(it) a^{2it} x_k(-it)^* b^{1-it}), \quad (A.9)
$$

hence by Cauchy-Schwarz for any $z$ in $\partial_0$,

$$
|F(z)| \le \left( \sum_k \operatorname{tr} (bx_k(z)x_k(z)^* b) \right)^{1/2} \left( \sum_k \operatorname{tr} (bx_k(\bar{z})x_k(\bar{z})^* b) \right)^{1/2} . \quad (A.10)
$$

A similar verification shows that for any $z$ in $\partial_1$, we have

$$
|F(z)| \le \left( \sum_k \operatorname{tr} (\alpha x_k(z)^* x_k(z) a) \right)^{1/2} \left( \sum_k \operatorname{tr} (\alpha x_k(\bar{z})^* x_k(\bar{z}) a) \right)^{1/2} . \quad (A.11)
$$

Thus we obtain by Cauchy-Schwarz,

$$
\begin{align}
\left| F\left(\frac{1}{2}\right) \right| &= \left| \int F d\mu \right| \le 2^{-1} \left( \int_{\partial_0} |F| d\mu_0 + \int_{\partial_1} |F| d\mu_1 \right) \nonumber \\
&\le 2^{-1} \left\{ \operatorname{tr} \left( b^2 \int \sum_k x_k x_k^* d\mu_0 \right) + \operatorname{tr} \left( a^2 \int \sum_k x_k^* x_k d\mu_1 \right) \right\} \le \|x\|_{M_n(G)} \le 1, \tag{A.12}
\end{align}
$$

which proves our claim.
It is now easy to show that $\mathbb{Q}$ is actually a complete metric surjection, or equivalently, that $\mathbb{I} \otimes \mathbb{Q} : \mathbb{M}_n(\mathbb{G}) \to \mathbb{M}_n(\mathbb{OH})$ is a metric surjection for any $n \ge 1$. Indeed, consider $x \in \mathbb{M}_n(\mathbb{OH})$ with $\|x\|_{\mathbb{M}_n(\mathbb{OH})} < 1$. Since $\mathbb{M}_n(\mathbb{OH}) = (\mathbb{M}_n(\mathbb{R}), \mathbb{M}_n(\mathbb{C}))_{1/2}$ (isometrically) by [22, Corollary 5.9], there is a bounded continuous analytic function $f$ on $\bar{\mathbb{S}}$ with values in $\mathbb{M}_n(\mathbb{R}) + \mathbb{M}_n(\mathbb{C})$ such that

$$ 
\begin{align} 
\alpha_0 &= \sup \{ \|f(z)\|_{\mathbb{M}_n(\mathbb{R})} \mid z \in \partial_0 \} < 1, \\ 
\alpha_1 &= \sup \{ \|f(z)\|_{\mathbb{M}_n(\mathbb{C})} \mid z \in \partial_1 \} < 1, \tag{A.13} \\ 
f\left(\frac{1}{2}\right) &= x. 
\end{align} 
$$

We write $f(z) = (f_k(z))_k$, where $f_k$ is an $\mathbb{M}_n$-valued function on $\bar{\mathbb{S}}$. We trivially have

$$ 
\left\| \left( \int \sum f_k(z) f_k(z)^* d\mu_0(z) \right)^{1/2} \right\|_{\mathbb{M}_n} \le \alpha_0 < 1, \tag{A.14} \\ 
\left\| \left( \int \sum f_k(z)^* f_k(z) d\mu_1(z) \right)^{1/2} \right\|_{\mathbb{M}_n} \le \alpha_1 < 1, 
$$

hence $\|f\|_{\mathbb{M}_n(\mathbb{G})} < 1$. Since clearly $(\mathbb{I} \otimes \mathbb{Q})(f) = x$, this shows that $\mathbb{I} \otimes \mathbb{Q} : \mathbb{M}_n(\mathbb{G}) \to \mathbb{M}_n(\mathbb{OH})$ is a metric surjection. Thus we have, completely isometrically, $\mathbb{OH} \simeq \mathbb{G}/\ker(\mathbb{Q})$. Finally, since $G \subset \mathbb{R} \oplus \mathbb{C}$, this completes the proof. ■

$$ 
\begin{gather} 
\text{Let } E = L_2(\mu_0; \ell_2)_r \oplus L_2(\mu_1; \ell_2)_c. \text{ The preceding argument shows that} \nonumber \\ 
\mathbb{OH} \simeq \mathbb{G}/\mathbb{N}, \tag{A.15} 
\end{gather} 
$$

where $G \subset E$ is the subspace of boundary values of analytic functions on the strip $S = \{0 < \operatorname{Re} z < 1\}$, and where $N$ is the subspace of $G$ formed of all $f$ in $G$ such that $f(1/2) = 0$. Thus, $\mathbb{OH}$ appears as a quotient, namely $\mathbb{G}/\mathbb{N}$, of a subspace, namely $G$, of $\mathbb{R} \oplus \mathbb{C}$ since obviously $E \simeq \mathbb{R} \oplus \mathbb{C}$. Moreover, the subspace $G \subset E$ is the *graph* of a (necessarily closed) unbounded operator $T : \operatorname{Dom}(T) \to L_2(\mu_1; \ell_2)_c$, where $\operatorname{Dom}(T) \subset L_2(\mu_0, \ell_2)_r$ is the dense subspace formed of all the restrictions $f|_{\partial_0}$ when $f$ runs over $G$. Since $G$ is formed of *analytic* functions, the restriction of $f$ to $\partial_0$ (or $\partial_1$) entirely determines $f$, therefore $f \in G \to f|_{\partial_0}$ and $f \in G \to f|_{\partial_1}$ are one-to-one, so that the definition of $T$ is clear: we simply have

$$ T(f|_{\partial_0}) = f|_{\partial_1}. \tag{A.16} $$

Note that $T$ has dense range. By the polar decomposition of $T$ (cf. [2, page 1249]), we have $T = U|T|$, where $U : L_2(\mu_0; \ell_2) \to L_2(\mu_1; \ell_2)$ is unitary and where $|T| : L_2(\mu_0; \ell_2) \to L_2(\mu_1; \ell_2)$ is an unbounded, positive operator.
Clearly, since $L_2(\mu_0; \ell_2)_r \simeq \mathbb{R}$ and $L_2(\mu_1; \ell_2)_c \simeq \mathbb{C}$ are “homogeneous” operator spaces (i.e., for any $u: \mathbb{R} \to \mathbb{R}$ or $u: \mathbb{C} \to \mathbb{C}$, we have $\|u\|_{cb} = \|u\|$), $\mathcal{U}$ (or its inverse) is completely isometric from $L_2(\mu_0; \ell_2)_c$ to $L_2(\mu_1; \ell_2)_c$, and hence $\mathbb{I} \oplus \mathcal{U}^{-1}$ is completely isometric on $L_2(\mu_0, \ell_2)_r \oplus L_2(\mu_1, \ell_2)_c$. Let $\Lambda: L_2(\mu_1; \ell_2)_r \to L_2(\mu_1; \ell_2)_c$ be the same map as $|T|$ but viewed as acting between the indicated operator spaces (so that $T = \mathcal{U}\Lambda$).

Then we trivially have

$$G \simeq (\mathbb{I} \oplus \mathcal{U}^{-1})(G), \qquad (A.17)$$

but

$$(\mathbb{I} \oplus \mathcal{U}^{-1})(G) = \{(x, \Lambda x) \mid x \in \mathrm{Dom}(\Lambda)\}. \qquad (A.18)$$

So we are reduced to the following.

**Proposition A.2.** Let $\Lambda: \mathbb{R} \to \mathbb{C}$ be a closed selfadjoint densely defined unbounded operator with $\Lambda \ge 0$. Let

$$G(\Lambda) = \{(x, \Lambda x) \mid x \in \mathrm{Dom}(\Lambda)\} \subset \mathbb{R} \oplus \mathbb{C} \qquad (A.19)$$

be the graph of $\Lambda$. Then the dual $G(\Lambda)^*$ is completely 2-isomorphic to a subspace of a noncommutative $L_1$-space. In particular, OH embeds completely 2-isomorphically in $M_*$ for some von Neumann algebra $M$. $\square$

**Proof.** Let $\{E_\alpha\}$ be a net of finite-dimensional subspaces of $\mathrm{Dom}(\Lambda)$ directed by inclusion and such that $\cup E_\alpha = \mathrm{Dom}(\Lambda)$. Let $G_\alpha = \{(x, \Lambda x) \mid x \in E_\alpha\}$. Then $G(\Lambda) = \cup G_\alpha$ (directed union), and hence for any cb map $u: G(\Lambda) \to M_n$, we have

$$\|u\|_{cb} = \lim \uparrow \|u|_{G_\alpha} : G_\alpha \to M_n\|_{cb}. \qquad (A.20)$$

It follows that $G(\Lambda)^*$ embeds completely isometrically into an ultraproduct of the spaces $G_\alpha^*$. Since by [27], ultraproducts preserve the class of subspaces of noncommutative $L_1$-spaces (the operator space version of this is easy to derive from [27]), we are reduced to proving this with $G(\Lambda)$ replaced by $G_\alpha$. In that case we may as well replace $\mathbb{C}$ by $\mathbb{C}_n$ (where $n = \dim(E_\alpha)$) and replace $\mathbb{R}$ by $\mathbb{R}_n$.

Thus we are reduced to proving the result for $G(\Lambda) \subset \mathbb{R}_n \oplus \mathbb{C}_n$ for some invertible operator $\Lambda \ge 0$ from $\mathbb{R}_n$ to $\mathbb{C}_n$. In that case, we may as well assume (by homogeneity) that $\Lambda e_{1i} = \lambda_i e_{i1}$ for some $\lambda_i > 0$. But then this follows from the next result which is somewhat implicit in Shlyakhtenko's work [29], and in any case is included in Lemma 2.11. $\blacksquare$

<figure><img src="image_5.png" /></figure>
**Proposition A.3.** With the notation as in the first part, let $I = \{1, \dots, n\}$ and

$$a_i = l_i + \lambda_i l_i^*. \qquad (A.21)$$

Let $G_n = \text{span}[a_1, \dots, a_n]$ and let $W_n$ be the von Neumann algebra generated by $G_n$. Then $G_n$ is completely 2-isomorphic to $G(\Lambda_n) \subset R_n \oplus C_n$ and $G_n$ is completely 2-complemented in $W_n$. More precisely, there is a surjective mapping $P_n : W_n \to G(\Lambda_n)$ with $\|P_n\|_{cb} \le 1$ such that $G(\Lambda_n)$ is completely 2-isomorphic to the quotient $W_n/\ker(P_n)$. Therefore, $G(\Lambda_n)^*$ is completely 2-isomorphic to a subspace of a noncommutative $L_1$-space, namely, the predual of $W_n$. $\square$

**Proof.** We let $P$ be as in the proof of **Lemma 2.11**. Let $V : G_n \to G(\Lambda_n)$ be defined by $V(l_i + \lambda_i l_i^*) = e_{1i} \oplus \lambda_i e_{i1}$. Finally, let $P_n = VP : W_n \to G(\Lambda_n)$. The proof of **Lemma 2.11** shows that $\|P_n\|_{cb} \le 1$ and by the triangle inequality we have $\|V^{-1}\|_{cb} \le 2$. Therefore, $G(\Lambda_n)$ is completely 2-isomorphic to $W_n/\ker(P_n)$. $\blacksquare$

**Note.** In the above we used a discretization of $\Lambda$ to make the proof as elementary as possible, but this is not really necessary if one uses the general picture described in [29]. This alternate route is much more elegant but perhaps a bit more “abstract.” We will merely outline it. We consider the (complex) Hilbert space $H = L_2(\mu_0; l_2) \oplus L_2(\mu_1, l_2)$ equipped with the norm

$$\|(x_0, x_1)\| = (2^{-1} (\|x_0\|^2 + \|x_1\|^2))^{1/2}. \qquad (A.22)$$

As is classical, any $x = (x_0, x_1)$ admits (via Poisson integrals) a harmonic extension inside $S$, that is, there is a harmonic function $\tilde{x} : S \to l_2$ such that $\|\tilde{x}(\cdot)\|^2$ admits a harmonic majorant $u$ on $S$ and admitting $x_0$ and $x_1$ as its nontangential boundary values, respectively, on $\partial_0$ and on $\partial_1$. Note that $\|x\| = \inf\{u(1/2)\}$, where the infimum runs over all such majorants and the Poisson integral of the function equal to $\|x_0(\cdot)\|^2$ on $\partial_0$ and $\|x_1(\cdot)\|^2$ on $\partial_1$ produces the minimal $u$.

We will denote by $h^2(l_2)$ the space of all harmonic functions $\tilde{x}$ obtained in this way. All such functions are implicitly extended nontangentially to the closure of $S$. Thus $h^2(l_2)$ can be identified with $H$. We denote by $H^2(l_2)$ the subspace of $h^2(l_2)$ formed of all the analytic functions. The spaces $h^2(l_2)$ and $H^2(l_2)$ may be viewed as conformally equivalent copies of the usual spaces on the unit disc.

For any $f = (f_k)$ in $h^2(l_2)$, we set $\bar{f} = (\bar{f}_k)$.
We denote by $H_R$ the real linear subspace of $H$ of all elements of the form $(\bar{f}_{|\partial_0}, f_{|\partial_1})$ when $f$ runs over all functions in $H^2(\ell_2)$. Note that the map

$$j: H^2(\ell_2) \to (\bar{f}_{|\partial_0}, f_{|\partial_1}) \in H \qquad (A.23)$$

is a real linear isometry, with range $H_R$. It is easy to check that $H_R \cap iH_R = \{0\}$ (because an analytic function in $H^2(\ell_2)$ that vanishes on $\partial_1$ must vanish everywhere) and that $H_R + iH_R$ is dense in $H$ (because if an element of $h^2(\ell_2)$ is supported on $\partial_1$ or $\partial_0$ and is orthogonal to $H^2(\ell_2)$, it must be antianalytic, and hence must vanish identically; therefore the restrictions $\{f_{|\partial_1} | f \in H^2(\ell_2)\}$ are dense in $L_2(\mu_1; \ell_2)$, and similarly for $\partial_0$).

As pointed out in [29, Remark 2.6], the basic construction of [29] can be carried out starting from the data of the embedding $j: H_R \hookrightarrow H$, using [28] to obtain a group of orthogonal transformations of $H$ satisfying the KMS condition relative to this embedding. Let $\mathcal{F}$ be the full Fock space over $H$. We will identify $H$ with $L_2(\partial_0 \cup \partial_1; \mu)$. With the previous notation we set for any $f$ in $H^2(\ell_2)$,

$$t(f) = (\ell(\bar{f}1_{\partial_0}))^* + \ell(f1_{\partial_1}). \qquad (A.24)$$

Observe that $f \to t(f)$ is now a complex linear isomorphic embedding of $H^2(\ell_2)$ into $B(\mathcal{F})$. Note that this "quantization" of $H^2(\ell_2)$ seems to be of independent interest (even for scalar-valued Hardy spaces, when $\ell_2$ is replaced by $\mathbb{C}$). More generally, (A.24) makes sense for any $f$ in $h^2(\ell_2)$; the resulting mapping is then a completely isomorphic embedding of $L_2(\mu_0; \ell_2)_r \oplus L_2(\mu_1; \ell_2)_c$ into $B(\mathcal{F})$.

Shlyakhtenko [29] made an extensive study on the von Neumann algebra $M$ generated by the operators $\{s(h) = \ell(h) + \ell(h)^*, h \in H_R\}$. Since for any $f$ in $H^2(\ell_2)$,

$$2t(f) = s(j(f)) - is(if), \qquad (A.25)$$

$M$ is equivalently generated by the family $\{t(f), f \in H^2(\ell_2)\}$. Finally, arguing as for the above Lemma 2.11, we see that there is a projection $P: M \to t(H^2(\ell_2))$ with $\|P\|_{cb} \le 2$, and $t(H^2(\ell_2))$ is completely isomorphic to the space $G$ appearing in (A.15). Thus we can conclude, as in Proposition A.3, that $G$ (and a fortiori $OH$) is completely 2-isomorphic to a quotient of $M$, via a normal surjection $M \to OH$. Thus, taking adjoints, we find that $OH$ embeds completely isomorphically into $M_*$.

Remark A.4. The constant 2 appearing in Proposition A.2 is better than that in Junge's original work. This is not too significant but the question whether $OH$ embeds completely isometrically into $M_*$ for some von Neumann algebra $M$ remains open.
**Remark A.5.** Our proof does not yield the fact announced by Junge (yet unpublished) that, in the above Proposition A.2, M can be chosen hyperfinite. Note, however, that since by [24] we obtain an M that is a quotient of a C\*-algebra with the WEP (QWEP), one can deduce from the strong local reflexivity of noncommutative L₁-spaces (see [3]) Junge's result that for each n and c > 2 there are an integer N and a subspace E_n ⊂ S₁N such that d<sub>cb</sub>(E_n, OH_n) ≤ c.

**Remark A.6.** The same proof suitably modified shows that OH embeds completely isomorphically in a noncommutative L<sub>p</sub>-space for any p with 1 < p < 2. (The case p = 2 is of course trivial.) That result was known to Junge and Xu. Indeed, for any 0 < θ < 1, we have, by [20], OH = (R[θ], R[1 − θ])<sub>1/2</sub>, hence (arguing as for Proposition A.1) we find that OH is a quotient of a subspace of R[θ] ⊕ R[1 − θ]. Now, let p = (1 − θ)<sup>-1</sup> as before. In that case we claim that (R[θ] ⊕ R[1 − θ])<sup>*</sup> embeds completely isomorphically into S<sub>p</sub>. Indeed, as we already mentioned, R[θ]<sup>*</sup> = R[1 − θ] (resp., R[1 − θ]<sup>*</sup> = R[θ]) can be identified with the subspace of column (resp., row) matrices in S<sub>p</sub>. This proves our claim.

More generally, it follows from Xu's results in [39, 40] (see also [13] for related facts) that for any closed unbounded positive operator Λ : R[θ] → R[1 − θ] with dense domain, dense range, and zero kernel, the graph G(Λ) ⊂ R[θ] ⊕ R[1 − θ] is such that G(Λ)* embeds in a noncommutative L<sub>p</sub>-space. Thus, by the same principle as above for p = 1, we can show that OH embeds completely isomorphically in a noncommutative L<sub>p</sub>-space for all 1 < p < 2. See [39] for more on this theme.

**Remark A.7.** Junge already observed in [11] that OH does not embed completely isomorphically into a noncommutative L<sub>q</sub>-space for 2 < q < ∞. Actually, in that case it is even impossible to embed OH<sub>n</sub> uniformly over n into such a space. For the reader's convenience, we now sketch Junge's argument for this fact. We will use the noncommutative Khintchine inequalities due to Lust-Piquard (cf. [22, page 193]). For our present purpose, it is convenient to state them using the "vector-valued" version of the Schatten classes introduced in [21] and denoted by S<sub>p</sub>[E], where 1 ≤ p < ∞ and E is an arbitrary operator space. Let (ε<sub>k</sub>) denote the Rademacher functions on (Ω, m), where (say) Ω = [0, 1] and m is a normalized Lebesgue measure. Then, if the operator space E is assumed to be (completely isometrically) a subspace of a noncommutative L<sub>q</sub>-space and if 2 ≤ q < ∞, then for any finite sequence a<sub>1</sub>, ..., a<sub>n</sub> in E, we have

$$ \left( \left( \int \left\| \sum_{k=1}^{n} \varepsilon_k a_k \right\|^q dm \right)^{1/q} \le B_q \left( \left\| \sum_{k=1}^{n} e_{1k} \otimes a_k \right\|_{S_q[E]} + \left\| \sum_{k=1}^{n} e_{k1} \otimes a_k \right\|_{S_q[E]} \right) \right), \quad (A.26) $$
where $B_q$ is a constant depending only on $q$. See [13] for the extension of (A.26) to the case of general noncommutative $L_q$-spaces, including the nonsemifinite case.

Now, let $u : OH_n \to E$ be a linear isomorphism and let $(e_1, \dots, e_n)$ denote an orthonormal basis in $OH_n$. We clearly have

$$ \sqrt{n} \le \|u^{-1}\| \left( \int \left\| \sum_{k=1}^{n} \varepsilon_k u e_k \right\|^q dm \right)^{1/q} . \quad (A.27) $$

On the other hand, applying (A.26) to $\alpha_k = u e_k$ and using [21, Corollary 1.2], we find

$$ \left( \int \left\| \sum_{k=1}^{n} \varepsilon_k u e_k \right\|^q dm \right)^{1/q} \le \|u\|_{cb} B_q \left( \left\| \sum_{k=1}^{n} e_{1k} \otimes e_k \right\|_{S_q[OH_n]} + \left\| \sum_{k=1}^{n} e_{k1} \otimes e_k \right\|_{S_q[OH_n]} \right). \quad (A.28) $$

Finally, by easy interpolation arguments (based on [21, Corollary 1.4]), we find

$$ \left\| \sum_{k=1}^{n} e_{1k} \otimes e_k \right\|_{S_q[OH_n]} \le n^{1/2q+1/4}, \quad (A.29) $$

and similarly for $\left\| \sum_{k=1}^{n} e_{k1} \otimes e_k \right\|_{S_q[OH_n]}$. Thus we conclude that

$$ n^{1/2} \le \|u^{-1}\| \|u\|_{cb} 2B_q n^{1/2q+1/4} \quad (A.30) $$

and a fortiori we find

$$ d_{cb}(E, OH_n) = \inf \|u\|_{cb} \|u^{-1}\|_{cb} \ge (2B_q)^{-1} n^{1/4 - 1/2q}. \quad (A.31) $$

A similar argument can be applied with $(C_n, R_n)_\theta$ instead of $OH_n$. The same calculations yield that for any $q > \max\{p, p'\}$ with $p = (1 - \theta)^{-1}$ and $p' = \theta^{-1}$, we have

$$ d_{cb}(E, (C_n, R_n)_\theta) \ge (2B_q)^{-1} n^{\beta/2}, \quad (A.32) $$

where $\beta = \min\{p^{-1} - q^{-1}, p'^{-1} - q'^{-1}\}$. Note however that if $p = (1 - \theta)^{-1}$, then $(C_n, R_n)_\theta$ embeds completely isometrically in both $S_p$ and $S_{p'}$; indeed it can be identified with $\operatorname{span}[e_{1k} | 1 \le k \le n]$ in $S_p$ and with $\operatorname{span}[e_{k1} | 1 \le k \le n]$ in $S_{p'}$.
Acknowledgments

I am grateful to Quanhua Xu for fruitful conversations, useful correspondence, and for reminding me
of some results that I should not have forgotten. Thanks also to the referee for remarks that led to
an improved presentation. The author was partially supported by NSF Grant no. 0200690 and Texas
Advanced Research Program 010366-163.

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Gilles Pisier: Department of Mathematics, Texas A&M University, College Station, TX 77843-3368, USA; Equipe d'Analyse, Université Paris VI, Case 186, 75252 Paris Cedex 05, France
E-mail address: pisier@math.tamu.edu