2 Column–row property for noncommutative $L_p$ -spaces

We begin with some preliminaries of operator space theory. For various well-known concepts related to the operator space theory, we refer to [Reference Effros and Ruan6, Reference Pisier16].

Let E be an operator space equipped with a matricial norm structure $(M_n(E), \|.\|_{M_n(E)})_{n\geq 1}$ . Given $x=[x_{ij}]_{i,j=1}^n\in M_n(E)$ , we denote the transpose map by $ t(x):=[x_{ji}]_{i,j=1}^n.$ The transpose maps taking columns to rows or rows to columns are also denoted by $t.$ Sometimes, we also denote it by $t_n$ to specify the dimension of the spaces under consideration. The opposite operator space, $E^{op}$ is defined to be the same space as E, but with the following matricial norm: $\|[x_{ij}]_{i,j=1}^n\|_{M_n(E^{op})}:=\|[x_{ji}]_{i,j=1}^n\|_{M_n(E)}$ , for all $[x_{ij}]_{i,j=1}^n\in M_n(E^{op})$ . For an operator space E, we denote the conjugate operator space by $\overline {E}$ . Any element $x\in E$ corresponds to an element $\overline {x}\in \overline {E}.$ We refer [Reference Pisier16, Section 2.9] for the notion of conjugate operator space. Recall that a linear map $u:E\to F$ between operator spaces is called completely bounded if $\sup \limits _{n\geq 1}\|id_{M_n}\otimes u\|_{M_n(E)\to M_n(F)}<\infty ,$ where $id_E:E\to E$ is the identity map for any vector space E. In this case, one denotes $\|u\|_{cb}:=\sup \limits _{n\geq 1}\|id_{M_n}\otimes u\|_{M_n(E)\to M_n(F)}.$ The map u is called a complete isometry if $id_{M_n}\otimes u$ is an isometry for all $n\geq 1.$ The Hilbert space tensor product of two Hilbert spaces $\mathcal H$ and $\mathcal K$ is denoted by $\mathcal H\otimes _2\mathcal {K}.$ Let A and B be $C^*$ -algebras with faithful representations $\pi _{\mathcal {H}}$ and $\pi _{\mathcal {K}}$ into $B(\mathcal H)$ and $B(\mathcal K)$ , respectively. Note that by natural identification $\pi _{\mathcal H}(A)\otimes \pi _{\mathcal K}(B)$ and hence $A\otimes B$ can be identified as a subalgebra of $B(\mathcal {H}\otimes _2\mathcal K)$ . The minimal tensor product of A and B is defined to be the completion of the algebraic tensor product $A\otimes B$ with norm borrowed from $B(\mathcal {H}\otimes _2\mathcal K)$ and is denoted by $A\otimes _{\text {min}}B$ . It is well known that the $C^*$ -algebra $A\otimes _{\text {min}}B$ thus obtained is independent of the representations $\mathcal {\pi }_{\mathcal H}$ and $\mathcal {\pi }_{\mathcal K}.$ We denote by $I_n$ to be the identity matrix in $M_n.$ We need the following lemma.

If A is a $C^*$ -algebra, then let us fix a faithful $*$ -representation of A on ${B}(\mathcal {H})$ . The canonical operator space structure on A is obtained by borrowing the matricial structure from ${B}(\mathcal {H})$ , and we will be using this canonical structure in the sequel. A $C^*$ -algebra A is called n-subhomogeneous if all the irreducible representations of A have dimensions at most $n.$ A $C^*$ -algebra is called subhomogeneous if it is n-subhomogeneous for some $n\in \mathbb N.$ Given a von Neumann algebra $({\mathcal M},\tau )$ with normal faithful semifinite trace $\tau $ , let $L_p({\mathcal M},\tau )$ be the corresponding noncommutative $L_p$ -space for $0<p<\infty $ . One denotes $L_{\infty }({\mathcal M})={\mathcal M}.$ When $({\mathcal M},\tau )=(B(\mathcal H),Tr)$ with $Tr$ being the natural trace on $B(\mathcal H),$ the corresponding noncommutative $L_p$ -spaces are called the Schatten-p classes and are denoted by $S_p(\mathcal H)$ for $1\leq p<\infty .$ The space $S_{\infty }(\mathcal H)$ is the space of all compact operators on $\mathcal H.$ These spaces are denoted by $S_p$ and $S_p^n$ when $\mathcal H$ is $\ell _2$ or $\ell _2^n$ , respectively, for $1\leq p\leq \infty .$ We often identify ${B}(\ell _2^n)$ with $M_n$ . Note that if ${\mathcal M}$ is a von Neumann algebra with a normal faithful semifinite trace $\tau $ , $M_n({\mathcal M})$ is again a von Neumann algebra equipped with the canonical tensor trace $Tr\otimes \tau $ . Indeed, we have a canonical identification of $M_n({\mathcal M})$ with the von Neumann algebra $M_n\overline {\otimes }{\mathcal M}$ , where $M_n\overline {\otimes }{\mathcal M}$ is the von Neumann algebra tensor product of $M_n$ and ${\mathcal M}.$ Then $Tr\otimes \tau (\sum _{i=1}^Na_i\otimes x_i):=\sum _{i=1}^NTr(a_i)\tau (x_i)$ extends to a normal faithful semifinite trace on $M_n\overline {\otimes }{\mathcal M}$ , where $a_i\in M_n$ and $x_i\in {\mathcal M}$ are positive elements for all $1\leq i\leq N$ and $N\geq 1.$ Equivalently, for any positive $x\in M_n({\mathcal M})$ , we have $Tr\otimes \tau (x)=\sum _{i=1}^n\tau (x_{ii}).$ Equipped with the canonical operator space structures $({\mathcal M}_{*}, {\mathcal M})$ becomes an interpolation couple (see [Reference Fidaleo7] and [Reference Pisier16, Section 2.7 and Chapter 7]), where $\mathcal {M}_{*}$ is the predual of ${\mathcal M}.$ We identify the predual ${{\mathcal M}}_{*}$ with $L_1({\mathcal M})$ via the map $\phi :L_1({\mathcal M})\to {{\mathcal M}}_{*}$ as $\phi (y)(x):=\tau (xy)$ for all $y\in L_1({\mathcal M})$ and $x\in {\mathcal M}.$ Moreover, $L_1({\mathcal M})$ has a natural operator space structure induced by ${\mathcal M}_{*}$ (see [Reference Pisier16, p. 139]). We have the following description of the operator space structure of $L_p({\mathcal M},\tau )$ from the convention established in [Reference Pisier16]. For $x\in M_n(L_p({\mathcal M},\tau )),$ we have, for all $1\leq p\leq \infty $ ,

(2.1) $$ \begin{align} \|x\|_{M_n(L_p({\mathcal M},\tau))}=\sup\{\|axb\|_{L_p(M_n({\mathcal M} ))}:\|a\|_{S_{2p}^n}\leq 1 \ \|b\|_{S_{2p}^n}\leq 1\}, \end{align} $$

where $axb$ is the usual product of matrices. It follows from [Reference Pisier15, Lemma 1.7] that, for a map $u:L_{p}({\mathcal M})\to L_p(\mathcal N)$ , we have $\|u\|_{cb, L_p({\mathcal M})\to L_p(\mathcal N)}=\|id_{S_p^n}\otimes u\|_{L_p(M_n\overline {\otimes }{\mathcal M})\to L_p(M_n\overline {\otimes }\mathcal N)}.$ Moreover, u is a complete isometry iff $id_{S_p^n}\otimes u$ is an isometry for all $n\geq 1.$ We refer [Reference Takesaki21] for the general theory of von Neumann algebras along with the notions of trace and tensor products of von Neumann algebras. We refer [Reference Pisier and Xu17, Reference Terp22] for noncommutative $L_p$ -spaces.

We will now introduce a class of operator spaces which is important for this article. Following [Reference Pisier14, Theorem 1.1] (see also [Reference Pisier16, Theorem 7.1]), we can give the following definition of an operator Hilbert space denoted by $OH(I)$ .

Definition 2.1 For any index set I, there always exists an unique operator space $OH(I)$ (up to complete isometry) such that the following properties are satisfied:

1. (i) $OH(I)$ is isometric to $\ell _2(I)$ as a Banach space.

2. (ii) The canonical identification between $\ell _2(I)$ and $\overline {\ell _2(I)^*}$ induces a complete isometry from $OH(I)$ to $\overline {OH(I)^*}$ .

This unique operator space $OH(I)$ is known as the operator Hilbert space. Moreover, if $\mathcal K$ is a Hilbert space and $(T_i)_{i\in I}$ is any orthonormal basis of $OH(I)$ , then for any finitely supported family $(x_i)_{i\in I}$ in $B(\mathcal K)$ , we have that

$$\begin{align*}\Big\|\sum_{i\in I}x_i\otimes T_i\Big\|_{\text{min}}=\Big\|\sum_{i\in I}x_i\otimes \overline{x_i}\Big\|_{\text{min}}^{\frac{1}{2}},\end{align*}$$

where $\|.\|_{\text {min}}$ denotes the norm of minimal tensor product between $C^*$ -algebras.

Thus, for any $[x_{ij}]_{i,j=1}^n\in M_n(OH(I))$ , we have

(2.2) $$ \begin{align} \Big\|[x_{ij}]_{i,j=1}^n\Big\|_{M_n(OH(I))}=\Big\|\Big[ [\langle x_{ij}, x_{kl}\rangle]_{k,l=1}^n \Big]_{i,j=1}^n\Big\|^{\frac{1}{2}}_{M_{n^2}} \end{align} $$

(see, for instance, [Reference Fidaleo7, Proposition 1] and [Reference Pisier16, Exercise 7.5] for a proof). Since $L_2({\mathcal M})$ is completely isometrically isomorphic to an operator Hilbert space [Reference Pisier16, p. 139] (also see [Reference Pisier16, p. 125]), we can describe the matricial structure of $L_2({\mathcal M})$ by the same formula as (2.2) for any $[x_{ij}]_{i,j=1}^n\in M_n(L_2({\mathcal M}))$ with the inner product $\langle x,y\rangle :=\tau (xy^*)$ for all $x,y\in L_2({\mathcal M}).$ It is well known that $\|id_{OH(I)}\|_{cb,OH(I)\to OH(I)^{op}}=1$ (see [Reference Fidaleo7]). We state the following lemma without giving the straightforward proof.

In the sequel, we shall frequently use the following block matrices for proving our results.

$$\begin{align*}A_n: = \begin{bmatrix} E_{11} & E_{12} & \dots & E_{1n}\\ 0 & 0 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \dots & 0 \end{bmatrix}_{n \times n}; \quad B_n: = \begin{bmatrix} E_{11} & E_{21} & \dots & E_{n1}\\ 0 & 0 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \dots & 0 \end{bmatrix}_{n \times n}. \end{align*}$$

Now, let us record a straightforward observation that will be useful in the sequel.

Lemma 2.5 Let $y=[y_{kl}]_{k,l=1}^n\in M_n(S_{\infty }^n)$ such that $\|y\|_{M_n(S_{\infty }^n)}\leq 1.$ Then, for all $a\in S_2^n$ with $\|a\|_2\leq 1,$ we have

$$\begin{align*}\sum_{l=1}^n\Big|\sum_{j,k=1}^na_{jk}y_{j1}^{kl}\Big|^2\leq 1, \end{align*}$$

where $y_{kl}=[y_{ij}^{kl}]_{i,j=1}^n.$

Proof Note that $z=[z_{lk}]_{l,k=1}^n$ has norm $\leq 1,$ where $z_{lk}=y_{kl}^*.$ For all $1\leq k \leq n$ , consider $v_k:=(v_{1k}, \ldots , v_{nk})\in \ell _2^n$ such that $\|v\|_{\ell _2^{n^2}}\leq 1$ , where $v:=(v_1,\ldots ,v_n).$ Note that we must have $\|zv\|_{\ell _2^{n^2}}^2\leq 1$ as well. Hence, we get

(2.3) $$ \begin{align} \sum_{l=1}^n\Big\|\sum_{k=1}^nz_{lk}v_k\Big\|^2\leq 1. \end{align} $$

Note that $z_{lk}v_k=(\sum _{j=1}^nz_{ij}^{lk}v_{jk})_{i=1}^n$ , where $z_{lk} = [z_{ij}^{lk}]_{i,j=1}^n$ . Therefore, we get $\sum _{k=1}^nz_{lk}v_k=(\sum _{k=1}^n\sum _{j=1}^nz_{ij}^{lk}v_{jk})_{i=1}^n$ . Thus, from equation 2.3, we obtain that

$$\begin{align*}\sum_{l=1}^n\sum_{i=1}^n\Big|\sum_{k=1}^n\sum_{j=1}^nz_{ij}^{lk}v_{jk}\Big|^2\leq 1. \end{align*}$$

Fixing $i=1$ in the above inequality, we get $ \sum _{l=1}^n\Big |\sum _{k=1}^n\sum _{j=1}^nz_{1j}^{lk}v_{jk}\Big |^2\leq 1. $ Rewriting we get $\sum _{l=1}^n\Big |\sum _{k=1}^n\sum _{j=1}^ny_{j1}^{kl}\overline {v_{jk}}\Big |^2\leq 1.$ Putting $\overline {v_{jk}}=a_{jk}$ for $1\leq j,k\leq n$ , we obtain the desired inequality.

Proof Let us fix $2<p<\infty $ and consider $A_n \in M_n(S_p^n).$ Then by choosing

(2.4) $$ \begin{align} a=E_{11}; \quad \ b=n^{-\frac{1}{2p}}I_n, \end{align} $$

in the formula (2.1), we have the estimate $\|A_n\|_{M_n(S_p^n)}\geq \|a A_n b\|_{L_p(M_n(B(\ell _2^n)))}.$ By this and Lemma 2.4, we obtain that $\|A_n\|_{M_n(S_p^n)}\geq n^{-\frac {1}{2p}}\|A_n\|_{S_p^{n^2}}=n^{\frac {1}{2}-\frac {1}{2p}}=n^{\frac {1}{2p^{\prime }}},$ where $\frac {1}{p}+\frac {1}{p^{\prime }}=1.$

Now, we shall estimate $\|t(A_n)\|_{M_n(S_p^n)}$ . We know that $\|t(A_n)\|_{M_n(B(\ell _2^n))}=1.$ We start with calculating $\|t(A_n)\|_{M_n(S_2^n)}$ . Note that the operator space structure of $S_2^n$ agrees with the operator Hilbert space structure [Reference Fidaleo7]. Therefore, by [Reference Pisier16, Theorem 7.1] and Lemma 2.3, we get

(2.5) $$ \begin{align} \|t(A_n)\|_{M_n(S_2^n)}=\left(\sum_{i=1}^n\sum_{j=1}^n|\langle E_{1i},E_{1j}\rangle|^2\right)^{\frac{1}{4}}=\left(\vphantom{\sum_{j=1}^n}\sum_{i=1}^n|\langle E_{1i},E_{1i}\rangle|^2\right)^{\frac{1}{4}}=n^{\frac{1}{4}}. \end{align} $$

Now, we shall use the method of complex interpolation (for reference, see [Reference Bergh and Löfström3, Chapter 4, p. 87] and [Reference Pisier16, Section 2.7, p. 52]). Consider $\theta \in (0,1)$ be such that $\frac {1}{p} = \frac {\theta }{2}+0$ . Thus, we get

$$\begin{align*}\|t(A_n)\|_{M_n(S_p^n)}\leq (n^{\frac{1}{4}})^{\frac{2}{p}} = n^{\frac{1}{2 p}}. \end{align*}$$

Hence, we obtain the estimate

$$\begin{align*}\frac{\|A_n\|}{\|t(A_n)\|}\geq\frac{n^{\frac{1}{2p^{\prime}}}}{n^{\frac{1}{2p}}}=n^{\frac{p-2}{2p}}.\end{align*}$$

Now, we will study the case for $p \in [1,2)$ , for which let us consider $B_n \in M_n(S_p^n).$ Note that by choosing a and b as in (2.4) in the formula (2.1), we have the estimate $\|B_n\|_{M_n(S_p^n)}\geq \|aB_nb\|_{L_p(M_n(B(\ell _2^n)))}= n^{\frac {1}{2p}}.$ Now, let us estimate $\|t(B_n)\|_{M_n(S_p^n)}.$ To do this, we use the method of complex interpolation again. It is easy to see that $ \| t(B_n) \|_{M_n(S_1^n)}=\|A_n\|_{M_n(S_1^n)}$ . For the sake of computation, let us denote the latter block matrix by $\boldsymbol {z}:=(z_{ij})_{i,j=1}^n$ , that is, $z_{1i}=E_{1i}$ fo $1\leq i\leq n$ and $z_{ij}=0$ for $i\geq 2.$

Now, we will estimate $\|\boldsymbol {z}\|_{M_n(S_1^n)}.$ Following the method of [Reference Pisier15, Lemma 1.7], we get that $ \|\boldsymbol {z}\|_{M_n[S_1^n]} = \sup \{ \|a \cdot [z_{ij}] \cdot b \|_{S_1^n[S_1^n]}: a,b \in S_{2}^n, \|a\|_{S_2^n}, \|b\|_{S_2^n} \leq 1\}. $ Using the fact, $M_n[S_1^n]^* = S_1^n[S_{\infty }^n]$ and [Reference Pisier15, Theorem 1.5], we observe that

$$\begin{align*}\|\boldsymbol{z}\|_{M_n(S_1^n)} = \sup \left\{ \big | \langle [z_{ij}], a \cdot [y_{ij}] \cdot b \rangle \big | : a,b \in S_{2}^n, \|a\|_{S_2^n}, \|b\|_{S_2^n}, \|[y_{ij}]\|_{M_n[S_{\infty}^n]} \leq 1\right\}. \end{align*}$$

Now, $ a\cdot [y_{ij}] \cdot b = a \cdot [\sum _{l=1}^n y_{il}b_{lj}]_{i,j=1}^n = [\sum _{k=1}^n a_{ik} (\sum _{l=1}^n y_{kl} b_{lj})]_{i,j=1}^n = [\sum _{k,l=1}^n a_{ik} y_{kl} b_{lj}]_{i,j=1}^n $ . By definition,

$$ \begin{align*} \langle [z_{ij}], a \cdot [y_{ij}] \cdot b \rangle = \sum_{i,j=1}^n \left\langle z_{ij}, \sum_{l,k=1}^n a_{jk} y_{kl} b_{li} \right\rangle = \sum_{i,j,k,l=1}^n a_{jk} \langle z_{ij}, y_{kl} \rangle b_{li}. \end{align*} $$

Hence,

$$\begin{align*}\|\boldsymbol{z}\|_{M_n(S_1^n)} = \sup \left\{ \left | \sum_{i,j,k,l=1}^n a_{jk} \langle z_{ij}, y_{kl} \rangle b_{li}\right | : a,b \in S_{2}^n, \|a\|_{S_2^n}, \|b\|_{S_2^n}, \|[y_{ij}]\|_{M_n[S_{\infty}^n]} \leq 1\right\}. \end{align*}$$

Therefore, from the definition of $\boldsymbol {z}$ , we get

$$\begin{align*}\|\boldsymbol{z}\|_{M_n(S_1^n)} = \sup \left\{\left | \sum_{j,k,l=1}^n a_{jk} \langle E_{1j}, y_{kl} \rangle b_{l1} \right | : a,b \in S_{2}^n, \|a\|_{S_2^n}, \|b\|_{S_2^n}, \|[y_{ij}]\|_{M_n[S_{\infty}^n]} \leq 1\right\}. \end{align*}$$

Now, $ \langle E_{1j}, y_{kl} \rangle = y_{j1}^{kl}$ , where $y_{kl}=[y_{ij}^{kl}]_{i,j=1}^n$ and, therefore,

$$ \begin{align*} \|\boldsymbol{z}\|_{M_n(S_1^n)} &= \sup \left\{ \left | \sum_{j,k,l=1}^n a_{jk} y_{j1}^{kl} b_{l1} \right | : a,b \in S_{2}^n, \|a\|_{S_2^n}, \|b\|_{S_2^n}, \|[y_{ij}]\|_{M_n[S_{\infty}^n]} \leq 1\right\}\\ &= \sup \left\{ \left(\sum_{l=1}^n \left| \sum_{j,k=1}^n a_{jk}y_{j1}^{kl} \right|^2\right)^{\frac{1}{2}} : a \in S_{2}^n, \|a\|_{S_2^n}, \|[y_{ij}]\|_{M_n[S_{\infty}^n]} \leq 1\right\}, \end{align*} $$

where the last equality is obtained by taking supremum over $\|b\|_{S_2^n} \leq 1$ . Now using Lemma 2.5, we obtain $\|\boldsymbol {z}\|_{M_n[S_1^n]}\leq 1$ and thus, $\|t(B_n)\|_{M_n[S_1^n]} \leq 1$ . If we do computations in a manner similar to condition (2.5), we get $\|t(B_n)\|_{M_n[S_2^n]} = n^{\frac {1}{4}}$ . Therefore, by the method of complex interpolation, we get $ \|t(B_n)\|_{M_n[S_p^n]}\leq n^{\frac {1}{2p^{\prime }}} \quad (1\leq p<2). $ Thus, by combining all the facts obtained above for any $p \in [1,2)$ , we get

(2.6) $$ \begin{align} \frac{\|B_n\|}{\|t(B_n)\|}\geq\frac{n^{\frac{1}{2p}}}{n^{\frac{1}{2p^{\prime}}}}=n^{\frac{2-p}{2p}}. \end{align} $$

This completes the proof.

Remark 2.8. Let E be an operator space. Then we have the following general estimate:

$$ \begin{align*} \|t_n\otimes id_{E}\|_{M_{n,1}\check{\otimes}E\to M_{1,n}\check{\otimes}E} &\leq \|t_n\otimes id_{E}\|_{cb,M_{n,1}\check{\otimes}E\to M_{1,n}\check{\otimes}E} \\ &\leq\|t_n\|_{cb,M_{n,1}\to M_{1,n}}\|id_{E}\|_{cb,E\to E}\leq\sqrt{n}. \end{align*} $$

In the above, we have used the estimate $\|t_n\|_{cb,M_{n,1}\to M_{1,n}}=\sqrt {n}$ (see [Reference Pisier16, p. 22]).

We are now ready to prove Theorem 1.1 in the following manner.

3 Column–row property for other operator spaces

We shall now study the completely bounded versions of the notions CRP and CMP. To state our result, we introduce the following definitions.

Definition 3.1 Let E be an operator space. We say E has the CMP if there exists a constant $C>0$ such that, for all $n\in \mathbb N$ and $[x_{ij}]_{i,j=1}^n\in M_n(E),$ we have

$$\begin{align*}\|[x_{ij}]_{i,j=1}^n\|_{M_n(E)}\leq C\| \begin{bmatrix} x_{11} \\ \vdots\\ x_{n1}\\ \vdots\\ x_{nn} \end{bmatrix}\|_{M_n(E)}. \end{align*}$$

Definition 3.2 Let E be an operator space. E is said to satisfy the completely bounded CRP if there exists a constant $C>0$ such that, for all $n\geq 1$ , we have

$$\begin{align*}\|[x_1\dots x_n]^t\mapsto [x_1\dots x_n]\|_{cb, M_{n,1}(E)\to M_{1,n}(E)}<C.\end{align*}$$

We now show that there is no nontrivial operator space with the completely bounded CRP.

We present a characterization of CRP for $C^*$ -algebras. First, we need the following lemma.

Remark 3.4. One can also have the following alternative approach. Recall that the Haagerup tensor product norm on the algebraic tensor product is defined by

$$\begin{align*}\|z\|_h:=\inf\left\{\left\|\sum_{i=1}^n x_ix_i^{*}\left\|^{\frac{1}{2}}\right\|\sum_{i=1}^n y_i^*y_i\right\|^{\frac{1}{2}}:z=\sum_{i=1}^n x_i\otimes y_i\right\} \quad (z \in A\otimes A). \end{align*}$$

Let us consider $\text {inv}:A\otimes A\to A\otimes A$ , defined on the elementary tensors by $\text {inv}(x\otimes y ):=x^*\otimes y^*.$ If A has CRP it follows that for some $C>0$ and for all $z\in A\otimes A$ , $ \|\text {inv}(z)\|\leq C\|z\|. $ The result follows from [Reference Kumar9, Part (iii) Theorem 2.2.]. If $C=1,$ then A is abelian follows from [Reference Kumar9, Theorem 2.3].

The following list summarizes the results which establishes CRP and other related properties for some well-known examples of operator spaces. Checking the properties are easy in the most of the cases and some of it maybe present in the literature.

• • $MIN(E), MAX(E)$ , and $L_p(\Omega )$ , for $1\leq p\leq \infty ,$ have CRP with constant $1$ , where $MIN(E)$ and $MAX(E)$ denote the so called minimal and maximal quantization of a Banach space E. For $MIN(E)$ and $MAX(E)$ , this follows from their definitions. By duality and interpolation one sees that the result for $L_p(\Omega )$ follows from complex interpolation. Moreover, $MIN(E)$ has CMP with constant $1.$ This is because $\text {MIN}(E)$ embeds completely isometrically into $C(K)$ for some compact Hausdroff topological space K [Reference Effros and Ruan6, Proposition 3.3.1] and for all $[f_{ij}]_{i,j=1}^n\in M_n(C(K)),$ we have $\sup _{s\in K}\|[f_{ij}(s)]\|_{op}\leq \sup _{s\in K}(\sum _{i,j=1}^n|f_{ij}(s)|^2)^{1/2}.$

• • The operator Hilbert space over the indexing set I, denoted by $OH(I)$ , have CMP with constant $1$ , which further implies that $S_2$ (the class of Hilbert–Schmidt operators on $\ell _2$ ) and, more generally, $L_2({\mathcal M})$ have this property as well with constant $1$ , where ${\mathcal M}$ is a semifinite von Neumann algebra. Note that, by the description of the norm of $OH(I)$ , we have the formula

  $$\begin{align*}\|[x_{ij}]_{i,j=1}^n\|_{M_n(OH(I))}=\|\big [[\langle x_{ij}, x_{kl}\rangle]_{k,l=1}^n \big ]_{i,j=1}^n\|^{\frac{1}{2}}_{M_{n^2}}.\end{align*}$$
  From Lemma 2.3, we get $\|[a_1 \ldots a_n]^t\|_{M_{n,1}(OH(I))}=(\sum _{i,j=1}^n|\langle a_i, a_j\rangle |^2)^{\frac {1}{4}}$ , and thus, we get the trivial inequality
  $$\begin{align*}\|[x_{ij}]_{i,j=1}^n\|_{M_n(OH(I))}\leq \left(\sum_{i,j=1}^n\sum_{k,l=1}^n|\langle x_{ij},x_{kl}\rangle|^2\right)^{\frac{1}{2}}. \end{align*}$$

• • The column Hilbert space C has the CRP with constant $1$ . The row Hilbert space does not have the CRP.

4 Concluding remarks

In the literature, there are several important studies on properties which have a certain resemblance to CRP. An operator space is called symmetric if it is completely isometric to its opposite [Reference Blecher4] via the identity map. Okayusu [Reference Okayusu12] showed that a $C^*$ -algebra is symmetric if and only if it is commutative. Thus Theorem 3.3, can be considered as a generalization of the aforementioned result. Using Okayusu’s characterization, Tomiyama [Reference Tomiyama23] studied the positivity of the transpose map on $C^*$ -algebras. Furthermore, Blecher [Reference Blecher4] extended Okayusu’s result and showed that any unital operator algebra which is symmetric must be commutative. We refer the reader to [Reference Roydor18] for some related results. Our characterizations on $L_p$ -spaces with CRP can be observed as a contribution toward results belonging to the theme of the above direction. Moreover, estimates on the completely bounded norm of the transpose maps on noncommutative $L_p$ -spaces were crucial in the recent work of Le Merdy and Zadeh for studying separating maps on noncommutative $L_p$ -spaces (see [Reference Le Merdy and Zadeh10, Reference Le Merdy and Zadeh11]).