---
tag: S805N6Z8G7
kind: definition
name: Operator space
msc: [47L25, 46L07]
defines: VB1J9AA4WX
uses: [HM2HHE3J51, BMDG4JQPEH, 06EKW1AKXF]
sources:
  - url: https://en.wikipedia.org/wiki/Operator_space
---

## Definition

Let $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$.
A *concrete operator space* is a linear subspace $E \subseteq B(H)$ that is closed in the [[operator norm|06EKW1AKXF]].
For each $n \ge 1$, let $M_n(E)$ be the space of $n \times n$ matrices with entries in $E$.
A matrix $[x_{ij}] \in M_n(E)$ acts on the direct sum $H^n$ of $n$ copies of $H$, and its norm is the operator norm of this action:
$$[x_{ij}] (h_1, \ldots, h_n) = \Big( \sum_{j} x_{1j} h_j, \ldots, \sum_{j} x_{nj} h_j \Big).$$
The operator space is $E$ together with this sequence of norms $(\norm{\cdot}_n)_{n \ge 1}$.

## 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.