# Completely bounded operators

Continuing with the linear algebra theme, let’s consider the space $\mathcal M$ of all $2\times 2$ matrices with real entries. This is a 4-dimensional vector space; we can treat a matrix $(a_{ij})$ as a vector $(a_{11},a_{12},a_{21},a_{22})$. So a general linear map $T\colon \mathcal M\to \mathcal M$ can be described by a $4\times 4$ matrix. That is, the space of linear maps $L(\mathcal M)$ is 16-dimensional.

However, some linear maps $T\colon \mathcal M\to \mathcal M$ arise more naturally than others; they are somehow more “matricial”. For example, left multiplication by a fixed matrix $A\in\mathcal M$. Or right multiplication by a fixed $B\in\mathcal M$. Or, to hit both with one stone, the two-sided multiplication map $T_{A,B}\in L(\mathcal M)$ defined by $X\mapsto AXB$. The matrix of this linear map is the Kronecker product $A\otimes B^T$:

$\displaystyle A\otimes B^T = \begin{pmatrix} a_{11}b_{11} & a_{11}b_{21} & a_{12}b_{11} & a_{12}b_{21} \\ a_{11}b_{12} & a_{11}b_{22} & a_{12}b_{12} & a_{12}b_{22} \\ a_{21}b_{11} & a_{21}b_{21} & a_{22}b_{11} & a_{22}b_{21} \\ a_{21}b_{12} & a_{21}b_{22} & a_{22}b_{12} & a_{22}b_{22} \end{pmatrix}$

Observe that the set of Kronecker products is not closed under addition. This is somehow unsatisfactory. It would be nice if “matricial” transformations, whatever they are, formed a linear subspace of $L(\mathcal M)$. This can be fixed by generalizing the two-sided multiplication map: any representation $\pi$ of $\mathcal M$ on some vector space $V$ and any pair of operators $A\colon V\to \mathcal M$, $B\colon \mathcal M\to B$, give rise to a linear map $X\mapsto A\pi(X)B$. Given another triple $(\tilde \pi, \tilde A, \tilde B)$, we can take the direct product of representations and arrange the operators accordingly:

$\displaystyle \begin{pmatrix} A & \tilde A\end{pmatrix} \begin{pmatrix} \pi(X) & 0 \\ 0 & \tilde \pi(X) \end{pmatrix} \begin{pmatrix} B \\ \tilde B\end{pmatrix} = A\pi(X)B+\tilde A\tilde\pi(X)\tilde B$

Hurray, we have a linear subspace of $L(\mathcal M)$! Now, what is this subspace? Maybe all $L(\mathcal M)$? Hm.

Let $\mathcal M_{2n}$ be the space of square matrices of size $2n$. We can think of them as block matrices with $n\times n$ blocks of size $2\times 2$. A linear map $T\in L(\mathcal M)$ induces $T_n\in L(\mathcal M_{2n})$: just apply $T$ to every block. If $T$ comes from a triple $(\pi,A,B)$ as above, then $T_n$ can be obtained if we let $\mathcal M_{2n}$ act on $\bigoplus^n V$ blockwise and multiply on left and right by $A\otimes I_n$ and $B\otimes I_n$. In this case the norm of $T_n$ is bounded by $\|A\|\|B\|$the bound is independent of $n$. Note that the norm of $T_n$ is taken with respect to the matrix norm on $\mathcal M_{2n}$

An operator $T\in L(\mathcal M)$ such that $\sup_n \|T_n\|<\infty$ is called completely bounded. We just saw that any operator of the form $X\mapsto A\pi(X)B$ is completely bounded. The converse is also true; it was proved independently by Haagerup, Paulsen and Wittstock (and not just for $2\times 2$ matrices, of course).

This certainly helps. Still, which operators on $\mathcal M$ are completely bounded? So far we saw that they form a linear space of dimension between 8 and 16…