# Completely co-bounded Schur multipliers

Gilles Pisier

► To cite this version:

Gilles Pisier. Completely co-bounded Schur multipliers. Operators and Matrices, 2012, 6, pp.263-270.
10.7153/oam-06-18 . hal-05460850

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Completely co-bounded Schur multipliers

by

Gilles Pisier*

Texas A&M University

College Station, TX 77843, U. S. A.

and

Université Paris VI

Equipe d'Analyse, Case 186, 75252

Paris Cedex 05, France

Abstract

A linear map $u: E \to F$ between operator spaces is called completely co-bounded if it is completely bounded as a map from $E$ to the opposite of $F$. We give several simple results about completely co-bounded Schur multipliers on $B(\ell_2)$ and the Schatten class $S_p$. We also consider Herz-Schur multipliers on groups.

In this short note, we wish to draw attention to the notion of “completely co-bounded’ mapping between two operator spaces. Recall that an operator space can be defined as a Banach space $E$ given together with an isometric embedding $E \subset B(H)$ into the space $B(H)$ of all bounded operators on a Hilbert space $H$. The theory of operator spaces started around 1987 with Ruan’s thesis and has been considerably developed after that (notably by Effros-Ruan and Blecher-Paulsen, see [2, 6]), with applications mainly to Operator Algebra Theory. In this theory, the morphisms between operator spaces are the completely bounded maps (c.b. in short), defined as follows. First note that if $E \subset B(H)$ is any subspace, then the space $M_n(E)$ of $n \times n$ matrices with entries in $E$ inherits the norm induced by $M_n(B(H))$. The latter space is of course itself equipped with the norm of single operators acting naturally on $H \oplus \cdots \oplus H$ ($n$ times). Then, a linear map $u: E \to F$ is called completely bounded (c.b. in short) if

$$ (1) \qquad \|u\|_{cb} \stackrel{\text{def}}{=} \sup_{n \ge 1} \|u_n: M_n(E) \to M_n(F)\| < \infty $$

where, for each $n \ge 1$, $u_n$ is defined by $u_n([a_{ij}]) = [u(a_{ij})]$. One denotes by $CB(E, F)$ the space of all such maps.

Given an operator space $E$, the opposite $E^{op}$ is the same Banach space as $E$, but equipped with the operator space structure (o.s.s. in short) associated to any embedding $E \subset B(H)$ such that for any $a = [a_{ij}] \in M_n(E)$, we have $\|a\|_{M_n(B(H))} = \|[a_{ji}]\|_{M_n(E)}$. Thus $\|a\|_{M_n(E^{op})} = \|[a_{ji}]\|_{M_n(E)}$. It is easy to check that the (isometric linear) mapping $T \mapsto {}^tT \in B(H^*)$ realizes such an embedding (warning: here ${}^tT: H^* \to H^*$ designates the adjoint of $T$ in the Banach space sense).

We call a map $u: E \to F$ between operator spaces completely co-bounded if the same map is c.b. from $E$ to $F^{op}$. This definition is inspired by existing work on completely co-positive maps (cf.

<footnote>*Partially supported by NSF grant 0503688.</footnote>
e.g. [MM] and references there). I started to think about this notion after hearing Marciniak's lecture on co-positive multipliers at the 2004 Quantum probability conference in Bedlewo.

While this definition seems at first glance a pointless variation, easy to reduce to the usual case, we hope in what follows to convince the reader that it has a natural place in operator space theory and that it suggests many interesting questions. As a first motivation for this notion, we should mention that the non-commutative Grothendieck theorem, that came out of work by the author and Haagerup, can be rephrased as saying that, if $A, B$ are $C^*$-algebras, any bounded linear mapping $u: A \to B^*$ is the sum of a c.b. mapping and a co-c.b. one, see [7, p. 189] for details and more references.

**Definition 1.** A linear map $u: E \to F$ between operator spaces will be called completely co-bounded if it is completely bounded as a mapping from $E$ into $F^{op}$ the opposite operator space. We then denote

$$ \|u\|_{cob} = \|u: E \to F^{op}\|_{cb}. $$

*Remark 2.* Obviously, $\|u: E \to F^{op}\|_{cb} = \|u: E^{op} \to F\|_{cb}$ and $(F^{op})^* = F^{*op}$ completely isometrically. Therefore, $u: E \to F$ is completely co-bounded iff the same is true for $u^*$ and $\|u\|_{cob} = \|u^*\|_{cob}$, since this is valid for c.b. maps (cf. e.g. [2, 6]).

*Remark 3.* Clearly if $B$ is a $C^*$-algebra with $F \subset B$ and if $\alpha: B \to B$ is an anti-automorphism (for instance transposition on $B(\ell_2)$), then $u: E \to F$ is completely co-bounded iff $\alpha u$ is c.b. and $\|u\|_{cob} = \|\alpha u\|_{cb}$. It is well known that the transposition on $M_n$ has c.b. norm equal to $n$ (cf. e.g. [6, p. 418-419]). Therefore, the identity map on $B(H)$ is not completely co-bounded unless $H$ is finite dimensional. More generally, the identity map on a von Neumann algebra $B$ is completely co-bounded iff $B$ is of type $I_n$ for some finite $n$, i.e. iff $B$ is a direct sum of finitely many algebras of the form $M_n \otimes A_n$, with $A_n$ commutative.

At first glance, the reader may have serious doubts for the need of the preceding notion! but hopefully the next result will provide some justification.

**Theorem 4.** A Schur multiplier $M_\varphi$: $[x_{ij}] \to [\varphi_{ij}x_{ij}]$ is completely co-bounded on $B(\ell_2)$ iff the matrix $[|\varphi_{ij}|]$ defines a bounded operator on $\ell_2$ and we have

$$ (2) \qquad \|M_\varphi\|_{cob} = \|TM_\varphi\|_{cb} = \|[\varphi_{ij}]\|_{B(\ell_2)} $$

where $T: B(\ell_2) \to B(\ell_2)$ denotes the transposition. Moreover, if $\|M_\varphi\|_{cob} \le 1$, then $M_\varphi$ admits a factorization

$$ B(\ell_2) \xrightarrow{J} \ell_\infty(\mathbb{N} \times \mathbb{N}) \xrightarrow{M_\varphi} B(\ell_2) $$

where $J$ is the natural inclusion map and where $\|M_\varphi\|_{cb} \le 1$.

*Proof.* Assume that $[|\varphi_{ij}|]$ is in $B(\ell_2)$. Then the mapping

$$ \mathcal{M}_\varphi: \ell_\infty(\mathbb{N} \times \mathbb{N}) \to B(\ell_2) \\ [x_{ij}] \to [x_{ij}\varphi_{ij}] $$

is obviously bounded with $\|\mathcal{M}_\varphi\| = \|[\varphi_{ij}]\|_{B(\ell_2)}$. Actually, more generally, if $x_{ij} \in B(H)$ with $\|x_{ij}\| \le 1$, then the matrix $[\varphi_{ij}x_{ij}]$ defines a bounded operator on $\ell_2(H)$ with norm easily seen to be majorized by $\|[\varphi_{ij}||x_{ij}]\|_{B(\ell_2)} \le \|[\varphi_{ij}]\|_{B(\ell_2)}$. This shows that $\|\mathcal{M}_\varphi\|_{cb} \le \|[\varphi_{ij}]\|_{B(\ell_2)}$. Let $J$ be as above. Clearly we have

$$ \|J\|_{cb} = 1 $$
and hence

$$M_{\varphi} = M_{\varphi}J$$

with $||M_{\varphi}||_{cb} = ||[\varphi_{ij}]||_{B(\ell_2)}$. But since $\ell_{\infty}(\mathbb{N} \times \mathbb{N})^{op}$ and $\ell_{\infty}(\mathbb{N} \times \mathbb{N})$ are identical we can factorize $M_{\varphi}$ as follows

$$M_{\varphi}: B(\ell_2) \xrightarrow{J} \ell_{\infty}(\mathbb{N} \times \mathbb{N}) = \ell_{\infty}(\mathbb{N} \times \mathbb{N})^{op} \xrightarrow{M_{\varphi}} B(\ell_2)^{op}$$

it follows that

$$||M_{\varphi}: B(\ell_2) \to B(\ell_2)^{op}||_{cb} \le ||M_{\varphi}||_{cb} \le ||[\varphi_{ij}]||_{B(\ell_2)}.$$

This proves the “if” part.

Conversely, assume that $M_{\varphi}$ is completely co-bounded with $||M_{\varphi}||_{cob} \le 1$. Let $x = [x_{ij}]$ be an $n \times n$ matrix viewed as sitting in $B(\ell_2)$. Let $B = B(\ell_2)$. Note that

$$\left\| \sum_{ij=1}^{n} e_{ij} \otimes e_{ij} x_{ij} \right\|_{M_n(B)} = \|x\|_B$$

while

$$\left\| \sum_{ij=1}^{n} e_{ij} \otimes e_{ij} x_{ij} \right\|_{M_n(B^{op})} = \left\| \sum_{ij=1}^{n} e_{ji} \otimes e_{ij} x_{ij} \right\|_{M_n(B)} = \sup |x_{ij}|.$$

By definition of $||M_{\varphi}||_{cob} \le 1$, we have

$$\left\| \sum e_{ij} \otimes e_{ij} \varphi_{ij} x_{ij} \right\|_{M_n(B)} = \left\| \sum e_{ji} \otimes e_{ij} \varphi_{ij} x_{ij} \right\|_{M_n(B^{op})} \le \left\| \sum e_{ji} \otimes e_{ij} x_{ij} \right\|_{M_n(B)}$$

which yields

$$||[\varphi_{ij}x_{ij}1_{\{i,j \le n\}}]||_B \le \sup_{ij} |x_{ij}| \le 1.$$

This implies

$$||[\varphi_{ij}]_{ij \le n}|| \le 1$$

and since $n$ is arbitrary we obtain

$$||[\varphi_{ij}]||_B \le 1.$$

This proves the “only if” part. The proof also yields (2). □

**Corollary 5.** A Schur multiplier $M_{\varphi}$ is completely co-bounded on $B(\ell_2)$ iff it factors through a commutative $C^*$-algebra or iff it factors through a minimal operator space and the corresponding factorization norm coincides with $||M_{\varphi}||_{cob}$.

*Proof.* For any commutative $C^*$-algebra $C$ or for any $E \subset C$, we have clearly $E = E^{op}$, so a cb-factorization $M_{\varphi}: B \xrightarrow{u_1} E \xrightarrow{u_2} B$ yields

$$||M_{\varphi}: B \to B^{op}||_{cb} \le \inf\{||u_1||_{cb}||u_2||_{cb}\}$$

where the infimum runs over all possible factorizations. Conversely, the preceding shows the converse with a factorization through $\ell_{\infty}(\mathbb{N} \times \mathbb{N})$. □
Remark 6. Let $G$ be an infinite discrete group. Consider a function $f: G \to \mathbb{C}$, and the function $\hat{f}$ defined on $G \times G$ by $\hat{f}(s, t) = f(st^{-1})$. By the well known Kesten-Hulanicki criterion (cf. e.g. [5, Th. 2.4]) $G$ is amenable iff there is a constant $C$ such that for any finitely supported $f$ we have $\sum_{t \in G} |f(t)| \le C \|[\hat{f}(s, t)]\|_{B(\ell_2(G))}$ and when $G$ is amenable this holds with $C = 1$. Thus, by Theorem 4, the inequality

$$ \sum_{t \in G} |f(t)| \le C \|M_{\hat{f}}\|_{cob} $$

characterizes amenable groups. This should be compared with Bożejko's and Wysoczanski's criteria described in [5, p. 54] and [5, p. 38].

We now generalize Theorem 4 to the Schur multipliers that are bounded on the Schatten $p$-class $S_p$. We assume $S_p$ equipped with the “natural” operator space structure introduced in [4] using the complex interpolation method. We will use freely the notation and results from [4].

**Theorem 7.** Let $2 \le p \le \infty$. Let $T: S_p \to S_p$ denote again the transposition mapping $x \to {}^t x$. Then a bounded Schur multiplier $M_\varphi: S_p \to S_p$ is completely co-bounded iff it admits a factorization as follows:

$$ S_p \xrightarrow{J_p} \ell_p(\mathbb{N} \times \mathbb{N}) \xrightarrow{M_\varphi} S_p $$

where $J_p$ is the natural (completely contractive) inclusion and where $\|M_\varphi\|_{cob} = \|M_\varphi\|_{cb}$.

*Proof.* Note that the fact that $J_p: S_p \to \ell_p(\mathbb{N} \times \mathbb{N})$ is completely contractive is immediate by interpolation between the cases $p = 2$ and $p = \infty$.

The proof can then be completed following the same idea as for Theorem 4. We have for any $[x_{ij}]$ in $M_n(S_p)$

$$ \left\| \sum e_{ij} \otimes e_{ji} \otimes x_{ij} \right\|_{S_p^n [S_p^n [S_p]]} = \left( \sum_{ij} \|x_{ij}\|_{S_p}^p \right)^{1/p} $$

while

$$ \left\| \sum e_{ij} \otimes e_{ij} \otimes x_{ij} \right\|_{S_p^n [S_p^n [S_p]]} = \|[x_{ij}]\|_{S_p^n [S_p]} $$

Both of these identities can be proved by routine interpolation arguments starting from $p = \infty$ and $p = 2$. This gives us

$$ \|[\varphi_{ij}x_{ij}]\|_{S_p^n[S_p]} \le \|M_\varphi\|_{cob} \left( \sum \|x_{ij}\|_{S_p}^p \right)^{1/p} $$

which means (cf. [4]) that

$$ \|\mathcal{M}_\varphi\|_{cb} \le \|M_\varphi\|_{cob}. $$

To prove the converse, it suffices to notice again that

$$ \ell_p(\mathbb{N} \times \mathbb{N})^{op} = \ell_p(\mathbb{N} \times \mathbb{N}). \quad \square $$

Remark 8. Consider Schur multipliers from $B(\ell_2)$ (or the subalgebra of compact operators $K$) into the trace class $S_1$. We refer to [7] for a detailed discussion of when such a multiplier is bounded and when it is c.b. From that discussion follows easily that such a multiplier is completely co-bounded iff it is bounded. Indeed, more generally, if $A, B$ are $C^*$-algebras a linear map $u: A \to B^*$ is completely co-bounded iff there are a constant $c$ and states $f_1, f_2, g_1, g_2$ on $A, B$ respectively, such that for any $(a, b) \in A \times B$

$$ (3) \qquad |\langle u(a), b \rangle| \le c \left( (f_1(a^*a)g_1(b^*b))^{1/2} + (f_2(aa^*)g_2(bb^*))^{1/2} \right). $$
Consider a bounded Schur multiplier $\varphi = [\varphi_{ij}]$ from $B(\ell_2)$ (or $K$) to $S_1$, where $S_1$ is equipped with its natural o.s.s. as the dual of $K$. By this we mean that $\langle M_\varphi(a), b \rangle = \sum \varphi_{ij} a_{ij} b_{ij}$. Then (see [7]) there are two nonnegative summable sequences $(\lambda_i)$ and $(\mu_i)$ such that for any $i, j$

$$|\varphi_{ij}| \le \lambda_i + \mu_j.$$

Then, using the Cauchy-Schwarz inequality, we obtain (3) with the states $f_1 = g_1 = (\sum \lambda_j)^{-1} \sum \lambda_j e_{jj}$ and $f_2 = g_2 = (\sum \mu_j)^{-1} \sum \mu_j e_{jj}$. This shows that $M_\varphi$ is automatically completely co-bounded. See [8] for an extension to Schur multipliers from $S_p$ to $S_{p'}$ with $2 < p < \infty$.

**Remark 9.** The preceding two theorems illustrate the following simple observation: Assume that a linear map $u: E \to F$ is both c.b. and completely co-bounded, then $u$ can be completely boundedly factorized through an operator space $G$ for which the identity map $I_G$ is completely co-bounded with $\|I_G\|_{cob} = 1$. Indeed, we just consider $G = F \cap F^{op}$ in the sense of [4] (this means $G = F$ equipped with the o.s.s. induced by the diagonal embedding $F \subset F \oplus F^{op}$), then $u$ can be viewed as $u: E \to G \to F$, with $\|u: E \to G\|_{cb} = \max\{\|u\|_{cb}, \|u\|_{cob}\}$ and $\|G \to F\|_{cb} = 1$. Conversely, any mapping of the form $u: E \xrightarrow{v} G \xrightarrow{w} F$, with c.b. maps $v, w$ and $G$ such that the identity $I = I_G$ on $G$ is completely co-bounded, must be both c.b. and completely co-bounded (since $u: E \xrightarrow{v} G \xrightarrow{w} G^{op} \xrightarrow{w} F^{op}$ is c.b.).

Let us say that an operator space $G$ is self-transposed if $I_G$ is completely co-bounded. This property passes obviously to subspaces, quotients (and hence subquotients) and dual spaces. It is also stable under ultraproducts. Examples include any commutative $C^*$-algebra (or any minimal operator space), by duality any $L_1$-space (or any maximal operator space) and by interpolation any $L_p$-space ($1 \le p \le \infty$). Perhaps there is a nice characterization of self-transposed operator spaces?

Let $G$ be a finite group. Let $f, g: G \to \mathbb{C}$ be functions on $G$ and let $\lambda(f): x \to f * x$ and $\rho(g): x \to x * g$ be the associated convolutors on $\ell_2(G)$.

The Fourier transform of $f$ is defined as follows: for any irreducible representation $\pi$ on $G$ (i.e. $\pi \in \hat{G}$)

$$\hat{f}(\pi) = \int f(t)\pi(t)^*dm(t)$$

where $m$ is the normalized Haar measure on $G$. We have then

**Proposition.** With the above notation, we have

$$\|\lambda(f)\rho(g)\|_{cob} = \sup_{\pi \in \hat{G}} \|\hat{f}(\pi)\|_2 \|\hat{g}(\pi)\|_2$$

where $\| \|_2$ denotes the Hilbert-Schmidt norm on $H_\pi$. In particular,

$$\| \mathrm{Id}: C^*(G) \to C^*(G) \|_{cob} = \sup_{\pi \in \hat{G}} \dim(\pi).$$

*Proof.* Passing to Fourier transforms, we see that $\lambda(f)\rho(g)$ coincides with $\bigoplus_{\pi \in \hat{G}} L(\hat{f}(\pi))R(\hat{g}(\pi))$ where $L(a)$ (resp. $R(a)$) denotes left (resp. right) multiplication by $a$ on $H_\pi$. Thus the result follows from the next lemma. $\square$

**Lemma.** Let $H$ be a Hilbert space and let $u: B(H) \to B(H)$ be defined by $u(x) = axb$. Then $u$ is completely co-bounded iff $a, b$ are both Hilbert-Schmidt operators and $\|u\|_{cob} = \|a\|_2 \|b\|_2$.

*Proof.* We may easily reduce this to the finite dimensional case. So we assume $B(H) = M_n$. Then again we can write

$$\left\| \sum e_{ij} \otimes a e_{ij} b \right\| \le \|u\|_{cob} \left\| \sum e_{ji} \otimes e_{ij} \right\| = \|u\|_{cob}.$$
But now, $T = \sum e_{ij} \otimes ae_{ij}b$ is a rank one operator on $\ell_2^n \otimes \ell_2^n = \ell_2^n(H)$ with $H = \ell_2^n$. Indeed, for any $h = (h_i)$, $k = (k_j) \in \ell_2^n(H)$ we have:

$$ \begin{aligned} \langle Th, k \rangle &= \sum_{ij} \langle ae_{ij}bh_j, k_i \rangle \\ &= \sum_i \langle ae_i, k_i \rangle \sum_j \langle h_j, b^*e_j \rangle \\ &= \langle (ae_i), k \rangle \langle h, (b^*e_j) \rangle \end{aligned} $$

and hence

$$ \|T\| = \left(\sum \|ae_i\|^2\right)^{1/2} \left(\sum \|b^*e_j\|^2\right)^{1/2} = \|a\|_2 \|b\|_2. \quad \square $$

Let us denote by $C_\lambda^*(G)$ the reduced $C^*$-algebra of a discrete group $G$, i.e. the $C^*$-algebra generated by the left regular representation $\lambda: G \to B(\ell_2(G))$. By a Herz-Schur multiplier on $C_\lambda^*(G)$, we mean a bounded linear $T$ map on $C_\lambda^*(G)$ for which there is a function $f: G \to \mathbb{C}$ such that, for any $t \in G$

$$ T(\lambda(t)) = f(t)\lambda(t). $$

We then denote $T_f = T$.

**Corollary 10.** If $G$ is a finite group and $f$ is the function constantly equal to 1, then $\|T_f\|_{\text{cob}}$ (i.e. the identity map on $C_\lambda^*(G)$) is equal to the supremum of the dimensions of the irreducible representations of $G$.

*Remark 11.* By Theorem 4, if $f \equiv 1$ as above, then the function defined on $G \times G$ by $\hat{f}(s, t) = f(st^{-1})$ (that is also constantly equal to 1), satisfies (when viewed as a Schur multiplier on $B(\ell_2(G))$) $\|M_f\|_{\text{cob}} = |G|$ and in general this is different from $\|T_f\|_{\text{cob}} = \sup\{\dim(\pi) \mid \pi \in \hat{G}\}$.

Now let $G$ be an infinite discrete group. By Remarks 2 and 3, the identity map on $A = C_\lambda^*(G)$ is completely co-bounded iff the same is true for $A^{**}$, and this implies that the latter is a finite direct sum of algebras of the form $M_n \otimes A_n$, with $A_n$ commutative. In particular, of course $A^{**}$ is injective, $A$ is nuclear and hence $G$ is amenable.

It would be interesting to describe the completely co-bounded Herz-Schur multipliers on the reduced $C^*$-algebra of a discrete group $G$ in analogy with what is known for the c.b. ones (see [1]).

## References

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