commutator bracket
Let $A$ be an associative algebra over a field $K$. For $a,b\in A$, the element of $A$ defined by
$$[a,b]=abba$$ 
is called the commutator^{} of $a$ and $b$. The corresponding bilinear operation
$$[,]:A\times A\to A$$ 
is called the commutator bracket.
The commutator bracket is bilinear, skewsymmetric, and also satisfies the Jacobi identity^{}. To wit, for $a,b,c\in A$ we have
$$[a,[b,c]]+[b,[c,a]]+[c,[a,b]]=0.$$ 
The proof of this assertion is straightforward. Each of the brackets in the lefthand side expands to 4 terms, and then everything cancels.
In categorical terms, what we have here is a functor^{} from the category^{} of associative algebras to the category of Lie algebras over a fixed field. The action of this functor is to turn an associative algebra $A$ into a Lie algebra that has the same underlying vector space as $A$, but whose multiplication^{} operation^{} is given by the commutator bracket. It must be noted that this functor is rightadjoint to the universal enveloping algebra functor.
Examples

•
Let $V$ be a vector space. Composition endows the vector space of endomorphisms^{} $\mathrm{End}V$ with the structure^{} of an associative algebra. However, we could also regard $\mathrm{End}V$ as a Lie algebra relative to the commutator bracket:
$$[X,Y]=XYYX,X,Y\in \mathrm{End}V.$$ 
•
The algebra^{} of differential operators has some interesting properties when viewed as a Lie algebra. The fact is that even though the composition of differential operators is a noncommutative operation, it is commutative^{} when restricted to the highest order terms of the involved operators. Thus, if $X,Y$ are differential operators of order $p$ and $q$, respectively, the compositions $XY$ and $YX$ have order $p+q$. Their highest order term coincides, and hence the commutator $[X,Y]$ has order $p+q1$.

•
In light of the preceding comments, it is evident that the vector space of firstorder differential operators is closed with respect to the commutator bracket. Specializing even further we remark that, a vector field is just a homogeneous^{} firstorder differential operator, and that the commutator bracket for vector fields, when viewed as firstorder operators, coincides with the usual, geometrically motivated vector field bracket.
Title  commutator bracket 

Canonical name  CommutatorBracket 
Date of creation  20130322 12:33:51 
Last modified on  20130322 12:33:51 
Owner  rmilson (146) 
Last modified by  rmilson (146) 
Numerical id  8 
Author  rmilson (146) 
Entry type  Definition 
Classification  msc 17A01 
Classification  msc 17B05 
Classification  msc 18A40 
Related topic  LieAlgebra 
Defines  commutator Lie algebra 
Defines  commutator 