Opposite $C^*$ algebras $\DeclareMathOperator\op{op}$Let $A$ be a $C^*$-algebra. We know that $A$ admits a natural operator space structure, namely the operator space structure induced by any faithful $*$-representation of $A$ into some $B(\mathcal H)$ for some Hilbert space $\mathcal H.$ Now consider the opposite $C^*$ algebra $A^{\op}$ which is the same as $A$ but multiplication reversed. Now $A^{\op}$ has a natural operator space structure as well.

*

*Is it true that the matrix norms are given by $\|[x_{ij}]_{i,j=1}^n\|_{M_n(A^{\op})}=\|[x_{ji}]_{i,j=1}^n\|_{M_n(A)}$?

*Suppose $A$ and $A^{\op}$ is isomorphic as $C^*$-algebras then there exists some constant $C>0$ such that for all $n\geq 1$ and $[x_{ij}]_{i,j=1}^n\in M_n(A)$ we have $\|[x_{ij}]_{i,j=1}^n\|_{M_n(A)}\leq C \|[x_{ji}]_{i,j=1}^n\|_{M_n(A)}$?

 A: *

*Yes.  You could consult Pisier's Operator Space Theory book sections 2.9 and 2.10.

To give details: let $H$ be a Hibert space and $\overline H$ be the complex conjugate space.  This has vectors $\{ \overline\xi : \xi\in H \}$ with the same addition, the complex conjugate multiplication, and the inner product $(\overline\xi|\overline\eta)_{\overline H} = (\eta|\xi)_H$.  For $x\in\mathcal{B}(H)$ define $x^\top\in\mathcal{B}(\overline H)$ by $x^\top{\overline\xi} = \overline{x^*(\xi)}$.  Then $x\mapsto x^\top$ is an anti-homomoprhism, and we have a coordinate-free notion of the transpose.
Given a (faithful) $*$-representation $\pi:A\rightarrow\mathcal B(H)$ define $\pi^\top:A^{op}\rightarrow\mathcal B(\overline H)$ by $\pi^\top(x) = \pi(x)^\top$.  Then $\pi^\top$ is a faithful) $*$-representation of $A^{op}$.  Let $\pi$ be faithful.  Given $x = (x_{ij})\in M_n(A^{op})$ and $\xi=(\overline{\xi_j})$ and $\eta=(\overline{\eta_i})$ in $\overline{H}^n$,
$$ (\eta|(\operatorname{id}\otimes\pi^\top)(x)\xi)_{\overline{H}^n}
= \sum_{i,j} (\overline{\eta_i}|\pi(x_{ij})^\top \overline{\xi_j}  )_{\overline{H}}
= \sum_{i,j} (\overline{\eta_i}|\overline{\pi(x_{ij}^*)\xi_j}  )_{\overline{H}}
= \sum_{i,j} (\pi(x_{ij}^*)\xi_j | \eta_i )_H
= \sum_{i,j} (\xi_j | \pi(x_{ij})\eta_i )_H. $$
As $\operatorname{id}\otimes\pi^\top$ induces the norn on $M_n(A^{op})$, and $\operatorname{id}\otimes\pi$ induces the norn on $M_n(A)$, it follows that $\|x\|_{M_n(A^{op})} = \| (x_{ji}) \|_{M_n(A)}$.


*I don't think so.  Fix $n$, and consider the matrix units $e_{ij}\in M_n$.  Consider $x = \sum_{i,j} e_{ij} \otimes e_{ji} \in M_n\otimes M_n$.  Being careful, if $A=M_n$, then $x = (x_{ij}) \in M_n(A)$ has $x_{ij} = e_{ji}$.  Thus $(x_{ji}) = \sum_{i,j} e_{ij} \otimes e_{ij} \in M_n\otimes M_n$.  Using that $M_n\otimes M_n \cong \mathcal{B}(C^{n^2})$, for $\xi = \xi_{a,b} \in \mathbb C^n\otimes \mathbb C^n \cong \mathbb C^{n^2}$ we see that
$$ x(\xi) = \sum_{i,j,a,b} \xi_{a,b} e_{ij}(e_a) \otimes e_{ji}(e_b)
= \sum_{i,j} \xi_{j,i} e_i \otimes e_j
\quad\implies \|x(\xi)\| = \|\xi\|, $$
while
$$ (x_{ji})(\xi) = \sum_{i,j,a,b} \xi_{a,b} e_{ij}(e_a) \otimes e_{ij}(e_b)
= \sum_{i,j} \xi_{j,j} e_i \otimes e_i, $$
from which it follows that $\|x\| = n$.

However, $M_n$ and $M_n^\op$ are isomorphic via the transpose map $\theta(x) = x^t$.  Check: $\theta$ is linear, a homomorphism, and $\theta(x^*) = (x^*)^t = \overline{x} = (x^t)^* = \theta(x)^*$.
So a counter-example comes from letting $A$ be the $c_0$-direct sum $\bigoplus_n M_n$.
