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Commutator

Let A^~, B^~, ... be operators. Then the commutator of A^~ and B^~ is defined as

[A^~,B^~]=A^~B^~-B^~A^~.
(1)

Let a, b, ... be constants, then identities include

[f(x),x]=0
(2)
[A^~,A^~]=0
(3)
[A^~,B^~]=-[B^~,A^~]
(4)
[A^~,B^~C^~]=[A^~,B^~]C^~+B^~[A^~,C^~]
(5)
[A^~B^~,C^~]=[A^~,C^~]B^~+A^~[B^~,C^~]
(6)
[a+A^~,b+B^~]=[A^~,B^~]
(7)
[A^~+B^~,C^~+D^~]=[A^~,C^~]+[A^~,D^~]+[B^~,C^~]+[B^~,D^~].
(8)

Let A and B be tensors. Then

[A,B]=del _AB-del _BA.
(9)

There is a related notion of commutator in the theory of groups. The commutator of two group elements A and B is ABA^(-1)B^(-1), and two elements A and B are said to commute when their commutator is the identity element. When the group is a Lie group, the Lie bracket in its Lie algebra is an infinitesimal version of the group commutator. For instance, let A and B be square matrices, and let alpha(s) and beta(t) be paths in the Lie group of nonsingular matrices which satisfy

alpha(0)=beta(0)=I
(10)
(partialalpha)/(partials)|_(s=0)=A
(11)
(partialbeta)/(partials)|_(s=0)=B,
(12)

then

partial/(partials)partial/(partialt)alpha(s)beta(t)alpha^(-1)(s)beta^(-1)(t)|_((s=0,t=0))=2[A,B].
(13)

See also

Adjoint Representation, Anticommutator, Commutator Subgroup, Jacobi Identities

Portions of this entry contributed by Todd Rowland

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Cite this as:

Rowland, Todd and Weisstein, Eric W. "Commutator." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/Commutator.html

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