So I was reading through some notes online by William Arveson (https://math.berkeley.edu/~arveson/Dvi/opSpace.pdf) to quickly get up to speed on operator spaces (I'm struggling to find a pdf of Pisier's book). I am experiencing a major brain malfunction on this simple exercise that he gives on page 4.
Example: The row and column operator spaces $\cal{R}, \cal{C}$ are indistinguishable at the level of Banach spaces, but not completely isometric as operator spaces. Arveson shows (pages 3-4) that the natural map which takes a n element $R_z\in\cal{R}$ to $C_z$ for some $z = (z_1, ... z_p)\in \mathbb{C}^p$ cannot be completely isometric: $$\phi : R_z = \begin{pmatrix} z_1 & z_2 &\cdots & z_p\\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0\end{pmatrix} \longmapsto \begin{pmatrix} z_1 & 0 & \cdots & 0\\ z_2 & 0 & \cdots & 0 \\ \vdots & 0 & \ddots & \vdots \\ z_p & 0 & \cdots & 0\end{pmatrix} = C_z$$
He shows this by passing to $\phi_n: M_n(\mathcal{R})\to M_n(\cal{C})$ and noting that for $n\geq p$ we can find partial isometries $A_i\in B(\mathbb{C}^n)= M_n(\cal{C})$ (denoting $A = (A_1, ... A_p)$ the operator in $M_n(\cal{R})$ acting on the direct sum of $p$ copies of $\mathbb{C}^n$ ) such that $A_i^\ast A_i$ all coincide (so $\|R_A\|^2 = \|R_AR_A^\ast\| = \|\sum_i A_i A_i^\ast\|^2 = p$) and $A_iA_i^\ast$ are all mutually orthogonal projections of the same rank - hence with $\|C_A\|^2 = \|C_A^\ast C_A\| = \|\sum_i A_i^\ast A_i\|^2= 1$. This shows that $\phi$ is not completely isometric. I am perfectly happy with this, however it doesnt show that the two operator spaces are not completely isometric in general.
My Question: How to we adapt this approach to a general isometry $\varphi: \cal{R\to C}$. In the notes it is said to be "easy"... I guess I'm just being stupid.
My thinking was to say ok well, given some $R_z\in\cal{R}$ our arbitrary isometry $\varphi$ takes $R_z$ to some $C_{w}$ for some $w = (w_1, ... w_p)\in \mathbb{C}^p$ with $\sqrt{|z_1|^2 + .. + |z_p|^2} = \|R_z\| = \|C_w\| = \sqrt{|w_1|^2 + .. + |w_p|^2}$. When we move to the level of matrices, an $R_A\in M_n(\cal{R})$ for some $A = (A_1, ... A_p)$ must map to $C_B\in M_n(\cal{C})$ for some $B = (B_1, ... B_p)$ with $\|R_A\| = \|C_B\|$. If we use the same C* argument as in the previous case, we can write $\|R_A\|^2 = \|A_1A_1^\ast + \cdots + A_pA_p^\ast\|$ and $\|C_B\|^2 = \|B_1^\ast B_1 + \cdots + B_p^\ast B_p\|$. But now if we make an assumption on the $A_i$'s who's to say that we couldn't have $B_i$'s which sum to the same norm. I have to be forgetting some elementary fact about summing orthogonal projections or something...
Somebody please put me out of my misery. I have other things to think about.
In $\mathcal R$ there are elements $e_1,e_2,\ldots ,e_p$ of norm one (namely the canonical basis elements) such that the matrix $$ \pmatrix{ e_1 & 0 & \ldots & 0 \cr e_2 & 0 & \ldots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr e_p & 0 & \ldots & 0 \cr } \in M_p(\mathcal R) $$ has norm one, while in $\mathcal C$, for any sequence $f_1,f_2,\ldots ,f_p$ of norm one elements, the matrix $$ \pmatrix{ f_1 & 0 & \ldots & 0 \cr f_2 & 0 & \ldots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr f_p & 0 & \ldots & 0 \cr } \in M_p(\mathcal C) $$ has norm $\sqrt p$.