---
tag: YQ626FK82G
kind: definition
name: Commutant
msc: [46L10, 47L05]
defines: 1VC7EJ1ZBZ
uses: [HM2HHE3J51, BMDG4JQPEH]
sources:
  - url: https://en.wikipedia.org/wiki/Centralizer_and_normalizer
---

## Definition

Let $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$.
The *commutant* of a subset $S \subseteq B(H)$ is
$$S' = \set{T \in B(H) : TA = AT \text{ for all } A \in S}.$$
The *bicommutant*, or double commutant, of $S$ is $S'' = (S')'$.

## Remarks

Example: let $D \subseteq M_n(\C) = B(\C^n)$ be the algebra of diagonal matrices.
If $T$ commutes with $E = \operatorname{diag}(e_1, \dots, e_n)$ for distinct $e_k$, then $(e_i - e_j) T_{ij} = 0$, so $T$ is diagonal; hence $D' = D$.
Likewise, $M_n(\C)' = \C I$, since commuting with all matrix units forces $T$ to be scalar.

Non-example: the matrix unit $E_{12}$ does not lie in $D'$, because $E_{12} \operatorname{diag}(1, 2) = 2E_{12}$ while $\operatorname{diag}(1, 2) E_{12} = E_{12}$.

For all subsets, $S \subseteq S''$, $S_1 \subseteq S_2$ implies $S_2' \subseteq S_1'$, and $S''' = S'$.
The commutant $S'$ is always a unital subalgebra of $B(H)$ that is closed in the weak operator topology.
If $S$ is closed under adjoints, then $S'$ is also closed under adjoints, so $S'$ is a von Neumann algebra.
The same definition in any ring or algebra gives the centralizer of a subset.