# A note on the column-row property

Samya Kumar Ray  Stat-Math Unit, Indian Statistical Institute Kolkata,
Kolkata, 700108, India<EMAIL_ADDRESS>and  Srijan Sarkar  Department
of Mathematics, Indian Institute of Science, Bangalore, 560012, India
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this article, we study the following question asked by Michael Hartz in a
recent paper [8]: which operator spaces satisfy the column-row property? We
provide a complete classification of the column-row property for non-
commutative $L_{p}$-spaces over semifinite von Neumann algebras. We study
other relevant properties of operator spaces that are related to the column-
row property and discuss their existence and non-existence for various natural
examples of operator spaces.

###### Key words and phrases:

Column-row property, Operator spaces

###### 2010 Mathematics Subject Classification:

47L25: Operator spaces (= matricially normed spaces)

##  1\. Introduction

In [24], while studying the Corona problem for Dirichlet spaces on the unit
disc $\mathbb{D}$, the author discovered an important property involving the
corresponding multiplier algebra, which in recent times, is known as the
column-row property. The column-row property has emerged as an important tool
in extending classical results on Hardy spaces to complete Nevanlinna-Pick
(cnp) spaces. In a remarkable recent work [8], Hartz showed that every
normalized cnp space has the column-row property with constant $1$. The notion
of column-row property has led to a plethora of important results for cnp
spaces, to name a few: (a) factorization for weak-product spaces; (b)
interpolating sequences; (c) Corona problem etc. (see [1, 2, 5, 8, 24]).
Motivated by the inherent operator space structure of multiplier algebras and
the immense application of this property, Hartz asked the following question
in [8]:

######  Question.

Which operator spaces satisfy the column-row property?

Our aim in this article is to initiate a study for this question in the
general setting of operator spaces, by looking at several examples. For the
theory of operator spaces and related important results we refer to the
excellent monographs [6, 13, 16]. Let us begin with the description of column-
row property by considering a _concrete_ operator space $E\subseteq
B(\mathcal{H})$. For a sequence $\mathbf{e}:=(e_{1},\dots,e_{n})$ in $E,$ we
define the column operator $C_{\mathbf{e}}:\mathcal{H}\to\mathcal{H}^{n}$ by
$C_{\mathbf{e}}(\zeta):=[e_{1}(\zeta),\dots,e_{n}(\zeta)],\
\zeta\in\mathcal{H}.$ Similarly, let us define the corresponding row operator
$R_{\mathbf{e}}:\mathcal{H}^{n}\to\mathcal{H}$ by
$R_{\mathbf{e}}([\zeta_{1},\dots,\zeta_{n}]):=\sum_{j=1}^{n}e_{j}(\zeta_{j}).$

######  Definition 1.1.

A concrete operator space $E\subseteq B(\mathcal{H})$ is said to have the
column-row property (in short CRP) if there exists a constant $C>0$ such that
for any finite sequence $\mathbf{e}$ in $E$ with
$\|C_{\mathbf{e}}\|_{\mathcal{H}\to\mathcal{H}^{n}}\leq 1$ we have

$\|R_{\mathbf{e}}\|_{\mathcal{H}^{n}\to\mathcal{H}}\leq C.$

In this article we obtain the following characterization for CRP for non-
commutative $L_{p}$-spaces.

######  Theorem 1.2.

Let $\mathcal{M}$ be a semifinite von Neumann algebra. Let $1\leq p\neq
2\leq\infty.$ Then $L_{p}(\mathcal{M})$ has CRP if and only if $\mathcal{M}$
is subhomogeneous.

Moreover, we show that if $L_{p}(\mathcal{M})$ has CRP with constant $1$ for
some $p\in[1,\infty)\setminus\\{2\\}$, then $\mathcal{M}$ must be an abelian
von Neumann algebra. In the case of $p=2$, it is well known that
$L_{2}(\mathcal{M})$ is an operator Hilbert space [16, Page 139] and any
operator Hilbert space is completely isometric to its opposite [16, Exercise
7.1, Page 146] (see also [Footnote 8, Page 246][7]). Moreover, using the
description of the matricial norms, it is a straightforward observation that
any operator Hilbert space satisfies a stronger condition that is, the column-
matrix property (CMP) with constant 1 (see Lemma 2.3). We provide a short
explanation of this fact along with some related results just after Remark
3.2.

Note that Hartz proved a stronger result by showing that cnp spaces satisfy
the column-matrix property [8, Corollary 3.6]. Motivated by these results, we
study the column-matrix property (in short CMP) for operators spaces via some
naturally occuring examples. We also introduce the completely bounded version
of CRP and CMP. It was shown in [8, Section 5.1] that normalized cnp spaces do
not have completely bounded CRP with constant $1.$ However, it turns out that
no non-trivial operator space can have completely bounded CRP. We also discuss
CRP for many naturally occurring operator spaces including $C^{*}$-algebras.

Here, we would like to point out that the operator spaces we have mainly
considered are self-adjoint. Thus, this situation is different from the study
of the column-row property of multiplier algebras of rkHs, which are very much
asymmetrical in nature. We thank the anonymous reviewer for this remark. The
main advantage that self-adjointness brings into the problem is the concept of
subhomogeneity. It allows us to form a line of thought where if subhomogeneity
fails then the operator space contains certain matrix algebras, where we can
study the column-row property by means of computation of norms. However, these
computations are not straightforward and are obtained via non-trivial estimate
of the norm of certain matrix-valued rows and columns. The main difficulty for
establishing Theorem 1.2 lies in the fact that the operator space structure of
non-commutative $L_{p}$-spaces are difficult to work with. To overcome this
difficulty, we first compute the norm of a column with certain entries coming
from $S_{2}.$ Here we use the operator Hilbert space structure of $S_{2}$ and
the description of the norm given by Pisier in [14]. Following this, we get an
upper bound for the norm of the column by using the method of complex
interpolation. Then by exhibiting lower bounds of the rows, we obtain the
desired estimates for $p>2.$ The approach for the case of $p<2$ differs in the
following manner: here we estimate norms of columns with entries from $S_{1}$.
This is done using certain duality relationship between non-commutative vector
valued $L_{p}$-spaces which was developed by Pisier in [15].

Let us now briefly discuss the manner in which the rest of the article has
been organised. Section 2 contains the characterization of CRP for non-
commutative $L_{p}$-spaces. In section 3, we have studied CRP and CMP and
their completely bounded versions for some natural examples of operator
spaces.

##  2\. Column-Row property for non-commutative $L_{p}$-spaces

We begin with some preliminaries of operator space theory. For various well-
known concepts related to the operator space theory we refer to [6] and [16].

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 [16, 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.

######  Lemma 2.1.

Let $E$ be an operator space. Then $E$ has CRP if and only if there exists a
constant $C>0$ such that for all $n\geq 1$, $\|t_{n}\otimes
id_{E}\|_{M_{n,1}\check{\otimes}E\to M_{1,n}\check{\otimes}E}\leq C,$ where
$\check{\otimes}$ denotes the operator space injective tensor product.

###### Proof.

From [6, Corollary 8.1.3] we note that $M_{m,n}(E)=M_{m,n}\check{{\otimes}}E$,
where $\check{\otimes}$ denotes the operator space injective tensor product of
operator spaces. ∎

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 non-commutative $L_{p}$-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 non-commutative $L_{p}$-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 $Tr\otimes\tau(\sum_{i=1}^{N}a_{i}\otimes
x_{i}):=\sum_{i=1}^{N}Tr(a_{i})\tau(x_{i})$ 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 [16, Section 2.7 and Chapter 7] and [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 [16, Page 139]). We have the following description of
the operator space structure of $L_{p}(\mathcal{M},\tau)$ from the convention
established in [16]. For $x\in M_{n}(L_{p}(\mathcal{M},\tau)),$ we have for
all $1\leq p\leq\infty$

(2.1)
$\displaystyle\|x\|_{M_{n}(L_{p}(\mathcal{M},\tau))}=\sup\\{\|axb\|_{L_{p}(M_{n}(\mathcal{M}))}:\|a\|_{S_{2p}^{n}}\leq
1\ \|b\|_{S_{2p}^{n}}\leq 1\\},$

where $axb$ is the usual product of matrices. It follows from [15, 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 [21] for the general theory of von Neumann algebras along with the
notions of trace and tensor products of von Neumann algebras. We refer [17,
23] for non-commutative $L_{p}$-spaces.

######  Remark 2.1.

Note that the norm on $M_{n}(L_{p}(\mathcal{M}))$ is very different from
$L_{p}(M_{n}\overline{\otimes}\mathcal{M})$ as can be seen from our
computations in the proof of Theorem 2.6.

We will now introduce a class of operator spaces which is important for this
article. Following [14, Theorem 1.1] (see also [16, Theorem 7.1]), we can give
the following definition of an operator Hilbert space denoted by $OH(I)$.

######  Definition 2.2.

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:

  * (i)

$OH(I)$ is isometric to $\ell_{2}(I)$ as a Banach space;

  * (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

$\Big{\|}\sum_{i\in I}x_{i}\otimes
T_{i}\Big{\|}_{\text{min}}=\Big{\|}\sum_{i\in
I}x_{i}\otimes\overline{x_{i}}\Big{\|}_{\text{min}}^{\frac{1}{2}},$

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

(see for instance, [16, Exercise 7.5] and [7, Proposition 1] for a proof).
Since $L_{2}(\mathcal{M})$ is completely isometrically isomorphic to an
operator Hilbert space [16, Page 139] (also see [16, Page 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 [7]). We state the following lemma without giving the
straightforward proof.

######  Lemma 2.3.

Let $a_{i}\in OH(I),$ $1\leq i\leq n.$ Then $\|[a_{1}\dots
a_{n}]^{t}\|_{M_{n,1}(OH(I))}=\Big{(}\sum_{i,j=1}^{n}|\langle
a_{i},a_{j}\rangle|^{2}\Big{)}^{\frac{1}{4}}.$

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

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

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

######  Lemma 2.4.

Let $1\leq p<\infty.$ Then $\|A_{n}\|_{S_{p}^{n^{2}}}=\sqrt{n}\ \text{and}\
\|B_{n}\|_{S_{p}^{n^{2}}}=n^{\frac{1}{p}}.$

######  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

$\sum_{l=1}^{n}\Big{|}\sum_{j,k=1}^{n}a_{jk}y_{j1}^{kl}\Big{|}^{2}\leq 1,$

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) $\sum_{l=1}^{n}\Big{\|}\sum_{k=1}^{n}z_{lk}v_{k}\Big{\|}^{2}\leq 1.$

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

$\sum_{l=1}^{n}\sum_{i=1}^{n}\Big{|}\sum_{k=1}^{n}\sum_{j=1}^{n}z_{ij}^{lk}v_{jk}\Big{|}^{2}\leq
1.$

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

######  Lemma 2.6.

Let $1\leq p\neq 2\leq\infty.$ Then $\|t_{n}\otimes
id_{S_{p}^{n}}\|_{M_{n,1}\check{\otimes}S_{p}^{n}\to
M_{1,n}\check{\otimes}S_{p}^{n}}\geq n^{\frac{|p-2|}{2p}}.$

###### Proof.

Let us fix $2<p<\infty$ and consider $A_{n}\in M_{n}(S_{p}^{n}).$ Then by
choosing

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

in the formula (2.1), we have the estimate
$\|A_{n}\|_{M_{n}(S_{p}^{n})}\geq\|aA_{n}b\|_{L_{p}(M_{n}(B(\ell_{2}^{n})))}.$
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 [7]. Therefore,
by [16, Theorem 7.1] and Lemma 2.3, we get

(2.5) $\|t(A_{n})\|_{M_{n}(S_{2}^{n})}=(\sum_{i=1}^{n}\sum_{j=1}^{n}|\langle
E_{1i},E_{1j}\rangle|^{2})^{\frac{1}{4}}=(\sum_{i=1}^{n}|\langle
E_{1i},E_{1i}\rangle|^{2})^{\frac{1}{4}}=n^{\frac{1}{4}}.$

Now we shall use the method of complex interpolation (for reference, see [16,
Section 2.7, Page 52] and [3, Chapter 4, Page 87]). Consider $\theta\in(0,1)$
be such that $\frac{1}{p}=\frac{\theta}{2}+0$. Thus, we get

$\|t(A_{n})\|_{M_{n}(S_{p}^{n})}\leq(n^{\frac{1}{4}})^{\frac{2}{p}}=n^{\frac{1}{2p}}.$

Hence we obtain the estimate

$\frac{\|A_{n}\|}{\|t(A_{n})\|}\geq\frac{n^{\frac{1}{2p^{\prime}}}}{n^{\frac{1}{2p}}}=n^{\frac{p-2}{2p}}.$

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
$\|B_{n}\|_{M_{n}(S_{p}^{n})}\geq\|aB_{n}b\|_{L_{p}(M_{n}(B(\ell_{2}^{n})))}=n^{\frac{1}{2p}}.$
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
$\bm{z}:=(z_{ij})_{i,j=1}^{n}$ that is, $z_{1i}=E_{1i}$ fo $1\leq i\leq n$ and
$z_{ij}=0$ for $i\geq 2.$

Now we will estimate $\|\bm{z}\|_{M_{n}(S_{1}^{n})}.$ Following the method of
[15, Lemma 1.7], we get that
$\|\bm{z}\|_{M_{n}[S_{1}^{n}]}=\sup\\{\|a\cdot[z_{ij}]\cdot
b\|_{S_{1}^{n}[S_{1}^{n}]}:a,b\in
S_{2}^{n},\|a\|_{S_{2}^{n}},\|b\|_{S_{2}^{n}}\leq 1\\}.$ Using the fact,
$M_{n}[S_{1}^{n}]^{*}=S_{1}^{n}[S_{\infty}^{n}]$ and [15, Theorem 1.5], we
observe that

$\|\bm{z}\|_{M_{n}(S_{1}^{n})}=\sup\\{\big{|}\langle[z_{ij}],a\cdot[y_{ij}]\cdot
b\rangle\big{|}:a,b\in
S_{2}^{n},\|a\|_{S_{2}^{n}},\|b\|_{S_{2}^{n}},\|[y_{ij}]\|_{M_{n}[S_{\infty}^{n}]}\leq
1\\}.$

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,

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

Hence,

$\|\bm{z}\|_{M_{n}(S_{1}^{n})}=\sup\\{\big{|}\sum_{i,j,k,l=1}^{n}a_{jk}\langle
z_{ij},y_{kl}\rangle b_{li}\big{|}:a,b\in
S_{2}^{n},\|a\|_{S_{2}^{n}},\|b\|_{S_{2}^{n}},\|[y_{ij}]\|_{M_{n}[S_{\infty}^{n}]}\leq
1\\}.$

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

$\|\bm{z}\|_{M_{n}(S_{1}^{n})}=\sup\\{\big{|}\sum_{j,k,l=1}^{n}a_{jk}\langle
E_{1j},y_{kl}\rangle b_{l1}\big{|}:a,b\in
S_{2}^{n},\|a\|_{S_{2}^{n}},\|b\|_{S_{2}^{n}},\|[y_{ij}]\|_{M_{n}[S_{\infty}^{n}]}\leq
1\\}.$

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

$\displaystyle\|\bm{z}\|_{M_{n}(S_{1}^{n})}$
$\displaystyle=\sup\\{\big{|}\sum_{j,k,l=1}^{n}a_{jk}y_{j1}^{kl}b_{l1}\big{|}:a,b\in
S_{2}^{n},\|a\|_{S_{2}^{n}},\|b\|_{S_{2}^{n}},\|[y_{ij}]\|_{M_{n}[S_{\infty}^{n}]}\leq
1\\}$
$\displaystyle=\sup\\{(\sum_{l=1}^{n}|\sum_{j,k=1}^{n}a_{jk}y_{j1}^{kl}|^{2})^{\frac{1}{2}}:a\in
S_{2}^{n},\|a\|_{S_{2}^{n}},\|[y_{ij}]\|_{M_{n}[S_{\infty}^{n}]}\leq 1\\},$

where the last equality is obtained by taking supremum over
$\|b\|_{S_{2}^{n}}\leq 1$. Now using Lemma 2.5, we obtain
$\|\bm{z}\|_{M_{n}[S_{1}^{n}]}\leq 1$ 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)
$\displaystyle\frac{\|B_{n}\|}{\|t(B_{n})\|}\geq\frac{n^{\frac{1}{2p}}}{n^{\frac{1}{2p^{\prime}}}}=n^{\frac{2-p}{2p}}.$

This completes the proof. ∎

######  Remark 2.2.

Let $1\leq p\neq 2\leq\infty.$ In view of [11, Lemma 5.3] and Lemma 2.6 we
have the estimate $n^{\frac{|p-2|}{2p}}\leq\|t_{n}\otimes
id_{S_{p}^{n}}\|_{M_{n,1}\check{\otimes}S_{p}^{n}\to
M_{1,n}\check{\otimes}S_{p}^{n}}\leq n^{\frac{|p-2|}{p}}.$ We believe that the
lower bound is actually sharp.

######  Remark 2.3.

Let $E$ be an operator space. Then we have the following general estimate.

$\displaystyle\|t_{n}\otimes id_{E}\|_{M_{n,1}\check{\otimes}E\to
M_{1,n}\check{\otimes}E}$ $\displaystyle\leq\|t_{n}\otimes
id_{E}\|_{cb,M_{n,1}\check{\otimes}E\to M_{1,n}\check{\otimes}E}$
$\displaystyle\leq\|t_{n}\|_{cb,M_{n,1}\to M_{1,n}}\|id_{E}\|_{cb,E\to
E}\leq\sqrt{n}.$

In the above we have used the estimate $\|t_{n}\|_{cb,M_{n,1}\to
M_{1,n}}=\sqrt{n}$ (see [16, Page 22]).

We are now ready to prove Theorem 1.2 in the following manner.

###### Proof.

See Theorem 3.5 for $p=\infty.$ Let $p\neq\infty.$ Let $\mathcal{M}$ be
subhomogeneous. Then it follows from [10, Lemma 4.4] that $L_{p}(\mathcal{M})$
has CRP. For the converse, let us assume that $\mathcal{M}$ is not
subhomogeneous. Then by [10, Lemma 2.1] it follows that for all $n\geq 1$
there is a complete isometry from $S_{p}^{n}$ into $L_{p}(\mathcal{M}).$ Hence
by Lemma 2.6 and Lemma 2.1, $L_{p}(\mathcal{M})$ cannot have CRP for $p\neq
2.$ This completes the proof of the theorem. ∎

######  Proposition 2.7.

Let $\mathcal{M}$ be a semifinite von Neumann algebra. Let $1\leq p\neq
2\leq\infty$. If $L_{p}(\mathcal{M})$ has CRP with constant $1$ then
$\mathcal{M}$ must be an abelian von Neumann algebra.

###### Proof.

See Theorem 3.5 for $p=\infty.$ Note that by Theorem 1.2, it follows that
$\mathcal{M}$ must be subhomogeneous. By [10, Lemma 2.1] if $\mathcal{M}$ is
not subhomogeneous of degree $1,$ it contains $S_{p}^{2}$ completely
isometrically. Therefore, by Lemma 2.6 we have the estimate $\|t_{n}\otimes
id_{S_{p}^{2}}\|_{M_{2,1}\check{\otimes}S_{p}^{2}\to
M_{1,2}\check{\otimes}S_{p}^{n}}\geq 2^{\frac{|p-2|}{2p}}>1.$ This shows that
$\mathcal{M}$ must be subhomogeneous of degree $1.$ Hence by [20, Theorem
7.1.1], we get that $\mathcal{M}$ is an abelian von Neumann algebra. ∎

##  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 column-matrix property (in
short 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

$\|[x_{ij}]_{i,j=1}^{n}\|_{M_{n}(E)}\leq
C\|\begin{bmatrix}\vspace{-2mm}x_{11}\\\ \vspace{-2mm}\vdots\\\
\vspace{-3mm}x_{n1}\\\ \vspace{-2mm}\vdots\\\
x_{nn}\end{bmatrix}\|_{M_{n}(E)}.$

######  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

$\|[x_{1}\dots x_{n}]^{t}\mapsto[x_{1}\dots x_{n}]\|_{cb,M_{n,1}(E)\to
M_{1,n}(E)}<C.$

We now show that there is no non-trivial operator space with the completely
bounded CRP.

######  Proposition 3.3.

Let $E$ be an operator space which is not equal to the zero vector space. Then
$E$ does not have the completely bounded CRP.

###### Proof.

Let $e\in E$ be such that $\|e\|=1.$ Then it follows from the property of the
operator space injective tensor product [6, Page 142] that the map
$i_{n}:M_{n,1}\to M_{n,1}\check{\otimes}E$ given by $x\mapsto x\otimes e$ is a
complete isometry. Also $t_{n}(x\otimes e)=t_{n}(x)\otimes e$ for all $x\in
M_{n,1}.$ We have $\|t_{n}\|_{cb,M_{n,1}\check{\otimes}E\to
M_{1,n}\check{\otimes}E}\geq\sup_{N\geq 1,[x_{ij}]_{i,j=1}^{N}\neq
0}\frac{\|[t_{n}(x_{ij})]_{i,j=1}^{N}\|_{M_{N}(M_{1,n})}}{\|[x_{ij}]_{i,j=1}^{N}\|_{M_{N}(M_{n,1})}}=\sqrt{n}.$
The last estimate in the above inequality follows from [16, Page 22]. This
shows that $E$ does not have completely bounded CRP. This completes the proof
of the theorem. ∎

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

######  Lemma 3.4.

Let $E$ be an operator space. Then $E$ has CRP iff $E^{**}$ has CRP.

###### Proof.

Note that if $E^{**}$ has CRP, then $E$ has CRP follows from the fact that the
natural inclusion of $E$ inside $E^{**}$ is a complete isometry [6,
Proposition 3.2.1]. To prove the converse, note by the duality between the
projective and injective tensor product of operator spaces (see [16, Theorem
4.1]) and the fact that $M_{n,1}^{*}=M_{1,n}$ completely isometrically (see
[16, Exercise 2.3.5]), we have $\|t_{n}\otimes
id_{E^{**}}\|_{M_{n,1}\check{\otimes}E^{**}\to
M_{1,n}\check{\otimes}E^{**}}=\|t_{n}\otimes
id_{E}\|_{M_{n,1}\check{\otimes}E\to M_{1,n}\check{\otimes}E}.$ The proof
follows by using Lemma 2.1. ∎

######  Theorem 3.5.

Let $A$ be a $C^{*}$-algebra. Then $A$ has CRP iff $A$ is subhomogeneous.
Moreover, $A$ has CRP with constant $1$ iff $A$ is abelian.

###### Proof.

One direction follows from [18]. Note that if $A$ has CRP, then by Lemma 3.4
$A^{**}$ is a von Neumann algebra which has CRP. However, it is folklore that
if $A^{**}$ is not subhomogeneous then it contains $M_{n}$ for all $n\geq 1.$
We get a contradiction by looking at $A_{n}$ and $B_{n}$ defined in Section 2.
∎

######  Remark 3.1.

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

$\|z\|_{h}:=\inf\Big{\\{}\Big{\|}\sum_{i=1}^{n}x_{i}x_{i}^{*}\Big{\|}^{\frac{1}{2}}\Big{\|}\sum_{i=1}^{n}y_{i}^{*}y_{i}\Big{\|}^{\frac{1}{2}}:z=\sum_{i=1}^{n}x_{i}\otimes
y_{i}\Big{\\}}\quad(z\in A\otimes A).$

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 [9, Part (iii) Theorem
2.2.]. If $C=1,$ then $A$ is abelian follows from [9, Theorem 2.3].

######  Remark 3.2.

Operator spaces for which $\|id_{E}\|_{cb,E\to E^{op}}=1$, have been studied
for $C^{*}$-algebras and operator algebras in [12] and [4].

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.

  1. $\bullet$

$MIN(E),MAX(E)$ 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 $K$ [6, Proposition 3.3.1] and for all $[f_{ij}]_{i,j=1}^{n}\in
M_{n}(C(K)),$ we have $\sup_{s\in K}\|[f_{ij}(s)]\|_{op}\leq\sup_{s\in
K}(\sum_{i,j=1}^{n}|f_{ij}(s)|^{2})^{1/2}.$

  2. $\bullet$

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

$\|[x_{ij}]_{i,j=1}^{n}\|_{M_{n}(OH(I))}=\|\big{[}[\langle
x_{ij},x_{kl}\rangle]_{k,l=1}^{n}\big{]}_{i,j=1}^{n}\|^{\frac{1}{2}}_{M_{n^{2}}}.$

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

$\|[x_{ij}]_{i,j=1}^{n}\|_{M_{n}(OH(I))}\leq(\sum_{i,j=1}^{n}\sum_{k,l=1}^{n}|\langle
x_{ij},x_{kl}\rangle|^{2})^{\frac{1}{2}}.$

  3. $\bullet$

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

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 [4] via the
identity map. Okayusu [12] showed that a $C^{*}$-algebra is symmetric if and
only if it is commutative. Thus Theorem 3.5, can be considered as a
generalization of the afore-mentioned result. Using Okayusu’s
characterization, Tomiyama [22] studied the positivity of the transpose map on
$C^{*}$-algebras. Furthermore, Blecher [4] extended Okayusu’s result and
showed that any unital operator algebra which is symmetric must be
commutative. We refer the reader to [18] for some related results. Our
characterizations on $L_{p}$-spaces with CRP can be observed as a contribution
towards results belonging to the theme of the above direction. Moreover,
estimates on the completely bounded norm of the transpose maps on non-
commutative $L_{p}$-spaces were crucial in the recent work of Le Merdy and
Zadeh for studying separating maps on non-commutative $L_{p}$-spaces (see [10]
and [11]).

## Acknowledgement

The authors are thankful to Michael Hartz for some useful communications
throughout the work. They would also like to thank the anonymous referee for
reading the manuscript intricately and giving important comments and revisions
which have improved the presentation of this article. The first and second
author acknowledges the DST-INSPIRE Faculty Fellowship
DST/INSPIRE/04/2020/001132 and DST/INSPIRE/04/2019/000769, respectively.

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