# The Operator Hilbert Space OH and Type III Von Neumann Algebras

Gilles Pisier

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Gilles Pisier. The Operator Hilbert Space OH and Type III Von Neumann Algebras. Bulletin of the London Mathematical Society, 2004, 36 (4), pp.455-459. 10.1112/s002460930400311x. hal-05460876

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# The Operator Hilbert Space *OH* and TYPE III von Neumann Algebras*

Gilles Pisier†

## Abstract

We prove that the operator Hilbert space *OH* does not embed completely isomorphically into the predual of a semi-finite von Neumann algebra. This complements Junge's recent result that it admits such an embedding in the non semi-finite case.

In remarkable recent work [5], Marius Junge proves that the operator Hilbert space *OH* (from [8], see also [9]) embeds completely isomorphically (c.i. for short) into the predual $M_*$ of a von Neumann algebra *M* which is of type III; thus this algebra *M* is not semi-finite. In this note, we show that no such embedding can exist when *M* is semi-finite.

The results we just stated all belong to the currently very active field of “operator spaces” for which we refer the reader to the monographs [2, 10]. We merely recall a few basic facts relevant for the present note. An operator space is a Banach space given together with an isometric embedding $E \subset B(H)$ into the algebra $B(H)$ of all bounded operators on a Hilbert space $H$. Using this embedding, we equip the space $M_n(E)$ (consisting of the $n \times n$ matrices with entries in $E$) with the norm induced by the space $M_n(B(H))$, naturally identified isometrically with $B(H \oplus \cdots \oplus H)$. Let $F \subset B(K)$ be another operator space. In operator space theory, the morphisms are the completely bounded (c.b. for short) linear maps: A linear map $u: E \to F$ is called c.b. if the mappings $u_n: M_n(E) \to M_n(F)$ defined by $[x_{ij}] \to [u(x_{ij})]$ are uniformly bounded when $n$ ranges over all integers $\ge 1$, and the cb-norm is defined as $\|u\|_{cb} = \sup_n \|u_n\|$. The resulting normed space of all c.b. maps $u: E \to F$ equipped with the cb-norm is denoted by $CB(E, F)$. If $u$ is invertible with c.b. inverse, then $u$ is called a complete isomorphism. For any operator space $E \subset B(H)$, the Banach dual $E^*$ can be equipped with a specific operator space structure, say $E^* \subset B(H)$, for which the natural identification $M_n(E^*) \simeq CB(E, M_n)$ is isometric. On the other hand, the complex conjugate $\bar{E}$ can obviously be viewed as an operator space using the canonical embeddings $\bar{E} \subset \overline{B(H)} \simeq B(\overline{H})$. Let $I$ be any set. In [8], we exhibited an operator space $E$ which is isometric to $\ell_2(I)$ as a Banach space and such that the canonical isometry (associated to the inner product) $E \simeq \bar{E}^*$ is a complete isometry. The latter operator

<footnote>*MSC 2000: 46L07, 46L54, 47L25, 47L50</footnote>

<footnote>†Partially supported by NSF and Texas Advanced Research Program 010366-163</footnote>
space, which is uniquely characterized by the preceding self-dual property, is denoted by $OH(I)$, or simply by $OH$ when $I = \mathbb{N}$. We call it the operator Hilbert space.

In Banach space theory (and in commutative harmonic analysis) the existence of an isomorphic (actually isometric) embedding of $\ell_2$ (or $\ell_2(I)$) in an $L_1$-space plays a very important role, in connection with the Khintchine and Grothendieck inequalities. However, the non-commutative version of the Khintchine inequality due to Lust-Piquard and the author (see [9, Th. 8.4.1]), when properly interpreted, leads to the embedding of a different (Hilbertian but not OH) operator space into $L_1$, namely the space called $R + C$ in [9]. This essentially implies that $OH$ does not embed (c.i.) into a commutative $L_1$ space (see [6] for details and more general results). Thus the question (raised in [8]) whether $OH$ itself embeds (c.i.) in a non-commutative $L_1$ space remained open, and was only recently solved affirmatively by Junge [5]. By a non-commutative $L_1$-space, we mean the predual $M_*$ of a von Neumann algebra $M$, equipped with the operator space structure induced by the dual $M^*$. Recall that $M$ is called semi-finite if it admits a normal and faithful trace which is also semi-finite, that is, although it is not necessarily finite, it admits sufficiently many elements on which it is finite (see [16] for more).

Our main theorem below shows that if $OH$ embeds in $M_*$, then $M$ cannot be semi-finite. This proves the need for a type III algebra in Junge's work, and perhaps explains the delay in resolving this embedding problem.

We will use the following easy consequence of [10, Cor. 3.4] (a more direct proof, with a better value of the constant $K$, appears in [12]).

**Lemma 1.** Let $I$ be any set, let $E = OH(I)$ and let $M$ be a von Neumann algebra. Consider a linear map $v: E \to M_*$.

(i) If $v$ is completely bounded (c.b. in short), there is a normal state $\varphi$ on $M$ such that

$$ (1) \qquad \forall x \in M \qquad \|v^*x\|^2 \le K^2(\varphi(xx^*)\varphi(x^*x))^{1/2} $$

where $K = 2^{9/4}\|v\|_{cb}$.

(ii) Conversely if there are $f_1, f_2$ in $M_+^*$ such that

$$ \forall x \in M \qquad \|v^*x\|^2 \le (f_1(xx^*)f_2(x^*x))^{1/2}, $$

then necessarily

$$ (2) \qquad \|v\|_{cb} \le (f_1(1)f_2(1))^{1/4}. $$

*Proof.* Let $K = 2^{9/4}\|v\|_{cb}$. By [13, Cor. 3.4] there is a state $f$ on $M$ such that

$$ (3) \qquad \forall x \in M \qquad \|v^*x\|^2 \le K^2(f(xx^*)f(x^*x))^{1/2}. $$

We will use an argument that can be traced back to [14, p. 352] and, in the non-commutative case, to [3, Prop. 2.3]. Let $f = f_n + f_s$ be the decomposition of $f$ into its normal and singular part. We set $\varphi = f_n$. As explained in the proof of Prop. 2.3 in [3], there is an increasing net $(p_\alpha)$ of projections in $M$ such that $p_\alpha \to 1$, say in the strong operator topology (SOT in short),
and $f_s(p_\alpha) = 0$ for all $\alpha$. Note that (by the SOT-continuity of the product of $M$ on bounded sets) we have $p_\alpha xp_\alpha \to x$, $p_\alpha xp_\alpha x^*p_\alpha \to xx^*$ and $p_\alpha x^*p_\alpha xp_\alpha \to x^*x$ for the SOT. For any $\xi$ in the unit ball of $E^*$, by assumption $x \to \langle \xi, v^*x \rangle$ is in $M_*$ (i.e. is “normal”), hence

$$ \langle \xi, v^*x \rangle = \lim \langle \xi, v^*(p_\alpha xp_\alpha) \rangle $$

hence by (3)

$$ \leq K^2 \lim(f(p_\alpha xp_\alpha x^*p_\alpha)f(p_\alpha x^*p_\alpha xp_\alpha))^{1/2} $$

but since $f_s(p_\alpha) = 0$, a fortiori $f_s(p_\alpha xp_\alpha x^*p_\alpha) = 0 = f_s(p_\alpha x^*p_\alpha xp_\alpha)$, hence we obtain

$$ |\langle \xi, v^*x \rangle| \leq K^2 \lim(\varphi(p_\alpha xp_\alpha x^*p_\alpha)\varphi(p_\alpha x^*p_\alpha xp_\alpha))^{1/2} $$

hence we conclude

$$ |\langle \xi, v^*x \rangle| \leq K^2 (\varphi(xx^*)\varphi(x^*x))^{1/2} $$

which immediately yields (i). The proof of (ii) is identical to that of the last assertion in Corollary 3.4 in [13]. □

Our main result is the following.

**Theorem.** Let $E = OH(I)$ with $I$ an infinite set. Let $F \subset M_*$ be a subspace of the predual of a semi-finite von Neumann algebra $M$. Then for any c.b. maps

$$ u: E \to F \quad \text{and} \quad w: F \to E $$

the composition $wv$ is compact.

*Proof.* Clearly this reduces to $I = \mathbb{N}$ if we wish. Let $\tau$ be a normal faithful semi-finite trace on $M$.

We will argue by contradiction. Assume that $wv$ is not compact. Then by the homogeneity of $OH$ (see [8, p. 18]) we may assume that $wv$ is diagonal or even furthermore that $wv$ is the identity on $E$. In other words, we may as well assume $u$ invertible and $w = u^{-1}$.

Let $v: E \to M_*$ be the same map as $u$ but viewed as acting into $M_*$. By Lemma 1, there is a normal state $\varphi$ such that (1) holds. Let $e$ be the support projection of $\varphi$ (i.e. we have $\varphi(1 - e) = 0$ and $\varphi(q) > 0$ for any non zero projection $q$ in $M$ with $q \leq e$). Then (1) implies that for any $x$ in $M$, we have $v^*(x(1 - e)) = 0 = v^*((1 - e)x)$ hence

$$ v^*(x) = v^*(exe). $$

Thus if we replace $M$ by $eMe$ and $\varphi$ by $e\varphi$, we may assume in addition that $\varphi$ is faithful.

Since $M_* \simeq L_1(\tau)$, we may assume $\varphi = \psi \cdot \tau$. Fix $0 < \delta < 1$. Let $p$ be the spectral projection associated to $\psi$ with respect to the set $[\delta, \delta^{-1}]$ (for more details, see e.g. [15, p. 338] or also [7]). Note that $\delta\tau(p) \leq \tau(\psi) = \varphi(1) \leq 1$, so that in particular $\tau(p) < \infty$ and moreover

$$ (4) \qquad p \cdot \psi = \psi \cdot p = p \cdot \psi \cdot p \leq \delta^{-1}p. $$
On the other hand, let $\varepsilon(\delta) = \tau((1-p)\psi) = \varphi(1-p)$. Clearly (since $\varphi$ is faithful) we have $\varepsilon(\delta) \to 0$ when $\delta \to 0$. Thus if we set, for all $y$ in $E$,

$$v_{\delta}(y) = p \cdot v(y) \cdot p \text{ and } T_{1}(y) = v(y)(1-p), T_{2}(y) = (1-p)v(y)p,$$

we have $v = v_{\delta} + T_1 + T_2$. We will show that $T_1 + T_2$ is small when $\delta \to 0$ so that $v_{\delta}$ can be viewed as a perturbation of $v$. Indeed, for $x$ in $M$, we have $T_1^*(x) = v^*((1-p)x)$ and $x^*(1-p)x \le x^*x$ hence by (1) we have

$$\|T_1^*(x)\|^2 \le K^2(\varphi((1-p)xx^*(1-p))\varphi(x^*x))^{1/2}$$

hence by (2)

$$\begin{aligned} \|T_1\|_{cb} &\le K(\varphi(1-p))^{1/4} \\ &\le K(\varepsilon(\delta))^{1/4}. \end{aligned}$$

Similarly $\|T_2\|_{cb} \le K(\varepsilon(\delta))^{1/4}$, hence $\|v - v_{\delta}\|_{cb} \le \|T_1\|_{cb} + \|T_2\|_{cb} \le 2K(\varepsilon(\delta))^{1/4}$. Let $\varepsilon_0 = \|u^{-1}\|_{cb}^{-1}$. Clearly, if we choose $\delta$ small enough so that $2K(\varepsilon(\delta))^{1/4} < \varepsilon_0$ we have $\|v - v_{\delta}\|_{cb} < \varepsilon_0$, hence, by an elementary reasoning (based solely on the triangle inequality in $M_n(M_*)$), the map $v_{\delta}: E \to pM_*p \subset M_*$ is a completely isomorphic embedding. But now by (4) we have

$$\|v_{\delta}^*(x)\|^2 = \|v^*(pxp)\|^2 \le K^2\delta^{-1}\|pxp\|_{L^2(\tau)}\|px^*p\|_{L^2(\tau)}$$

hence since $\tau$ is tracial (this is where we make crucial use of the semi-finiteness assumption)

$$\|v_{\delta}^*(x)\|^2 \le K^2\delta^{-1}\|pxp\|_{L^2(\tau)}^2 \le K^2\delta^{-1}\min\{\|px\|_{L^2(\tau)}^2, \|xp\|_{L^2(\tau)}^2\} \le K^2\delta^{-1}\|px\|_{L^2(\tau)}^2.$$

By Lemma 2 below, since $\tau(p) < \infty$, this is impossible. $\square$

**Lemma 2.** With the above notation. Let $V: OH \to L_1(\tau)$ be a linear map for which there is a in the unit ball of $L_2(\tau)$ and a constant $B$ such that, for any $x$ in $M$, we have

$$(5) \qquad \|V^*(x)\| \le B\|ax\|_{L^2(\tau)}.$$

Then, for any isometry $J: C \to OH$, $VJ$ is completely bounded from $C$ to $L_1(\tau)$. In particular $V$ cannot be a completely isomorphic embedding.

*Proof.* By (5), for any finite sequence $(x_i)$ in $M$

$$(6) \qquad \sum \|(VJ)^*(x_i)\|^2 = \sum \|V^*(x_i)\|^2 \le B^2 \left\|\sum x_i x_i^*\right\|.$$

By a well known argument from [1], it follows that $\|(VJ)^*: M \to C^*\|_{cb} \le B$, and hence $VJ$ is c.b. . Finally, if $V$ were a completely isomorphic embedding, then $VJ$ (when viewed as acting into the range of $V$) would be a completely bounded map from $C$ to $OH$, hence would be (by Remark 2.11 in [8]) in the Schatten class $S_4$, and a fortiori would be compact. But then $J$ itself would have to be compact which is absurd. $\square$

*Remarks.*
(i) Junge [5] proves that $OH_n$ embeds c.i. (with uniform constants) into the predual of a finite dimensional (hence semi-finite!) von Neumann algebra. More precisely, he proves that there is $C > 0$ so that for any $n$, there is an integer $N$, a subspace $F_n \subset M_N^*$ and a (complete) isomorphism $u_n: OH_n \to F_n$ such that $\sup_n \|u_n\|_{cb} \|u_n^{-1}\|_{cb} \le C$. It would be interesting to estimate $N$ as a function of $n$.

(ii) The non-existence of embeddings of $OH$ into $M_*$ when $M$ is commutative is rather easy. In that case, even the finite dimensional case (as in the preceding point) is ruled out (see [11] for related facts). The paper [6] contains stronger results in the same direction.

(iii) The above theorem remains valid with essentially the same proof for $E = (R, C)_\theta$ (with $0 < \theta < 1$) in the sense of [9], but this requires the generalized version of Lemma 1 proved in [12]. This implies that, for any $1 < p < 2$, the Schatten classes $S_p$ (and hence most non-commutative $L_p$-spaces) do not embed (c.i.) into the predual of any semi-finite von Neumann algebra.

(iv) Let $N_*$ be the predual of the injective factor of type III$_\lambda$ with $0 < \lambda \le 1$, and let $M$ be a (semi-finite) von Neumann algebra of type II$_\infty$. Junge proved that $OH$ embeds completely isomorphically into $N_*$. Hence Theorem 1 implies that $N_*$ does not embed c.i. into $M_*$. This gives a somewhat partial answer to the (still open) question raised in [4] of the existence of a (Banach space sense) isomorphic embedding of $N_*$ into $M_*$.

## References

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[10] G. Pisier, *Introduction to Operator Space Theory*, London Math. Soc. Lecture Notes **294**, Cambridge Univ. Press, (2003).
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[12] G. Pisier, Completely bounded maps into certain Hilbertian operator spaces. To appear.

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[14] H.P. Rosenthal, On subspaces of $L^p$, *Ann. of Math.* **97** (1973), 344-373.

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Texas A&M University
College Station, TX 77843, USA

and

Université Paris VI
Equipe d'Analyse, Case 186, 75252
Paris Cedex 05, France,

pisier@math.tamu.edu