Operator space

Remarks

Example: $B(H)$ itself, and every norm-closed subalgebra of it, are operator spaces. Example: the set of matrices $\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$ with $a, b \in \C$ is a closed subspace of $M_2(\C) = B(\C^2)$, since every subspace of a finite-dimensional space is closed. Non-example: if $H$ is infinite-dimensional, the finite-rank operators form a subspace of $B(H)$ that is not closed, so it is not an operator space. Its closure, the space of compact operators, is one. Abstractly, an operator space is a complex vector space $E$ with a complete norm on $E$ and norms $\norm{\cdot}_n$ on each $M_n(E)$ with two properties. First, $\norm{x \oplus y}_{m+n} = \max(\norm{x}_m, \norm{y}_n)$ for $x \in M_m(E)$ and $y \in M_n(E)$. Second, $\norm{\alpha x \beta}_n \le \norm{\alpha} \norm{x}_m \norm{\beta}$ for $x \in M_m(E)$ and scalar matrices $\alpha$ of size $n \times m$ and $\beta$ of size $m \times n$. Ruan's theorem states that every such abstract operator space is completely isometric to a concrete one.