PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: High Entry average rating: No information on entry rating
commutator bracket (Definition)

Let $A$ be an associative algebra over a field $K$ . For $a,b \in A$ , the element of $A$ defined by $$[a,b]=ab-ba$$ is called the commutator of $a$ and $b$ . The corresponding bilinear operation $$[-,-]: A\times A\rightarrow A$$ is called the commutator bracket.

The commutator bracket is bilinear, skew-symmetric, 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 left-hand 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 right-adjoint to the universal enveloping algebra functor.

Examples

  • Let $V$ be a vector space. Composition endows the vector space of endomorphisms $\End V$ with the structure of an associative algebra. However, we could also regard $\End V$ as a Lie algebra relative to the commutator bracket: $$[X,Y] = XY-YX,\quad X,Y\in \End V.$$
  • 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 non-commutative 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 $p+q-1$ .
  • In light of the preceding comments, it is evident that the vector space of first-order differential operators is closed with respect to the commutator bracket. Specializing even further we remark that, a vector field is just a homogeneous first-order differential operator, and that the commutator bracket for vector fields, when viewed as first-order operators, coincides with the usual, geometrically motivated vector field bracket.




"commutator bracket" is owned by rmilson.
(view preamble | get metadata)

View style:

See Also: Lie algebra

Also defines:  commutator Lie algebra, commutator
Log in to rate this entry.
(view current ratings)

Cross-references: homogeneous, vector field, closed, operators, order, commutative, non-commutative operation, even, properties, differential operators, structure, endomorphisms, composition, universal enveloping algebra, operation, multiplication, vector space, action, fixed field, Lie algebras, algebras, category, functor, categorical, terms, expands, side, proof, Jacobi identity, skew-symmetric, bilinear, bilinear operation, element, field, algebra, associative
There are 23 references to this entry.

This is version 5 of commutator bracket, born on 2002-04-02, modified 2004-12-15.
Object id is 2811, canonical name is CommutatorBracket.
Accessed 14048 times total.

Classification:
AMS MSC17A01 (Nonassociative rings and algebras :: General nonassociative rings :: General theory)
 17B05 (Nonassociative rings and algebras :: Lie algebras and Lie superalgebras :: Structure theory)
 18A40 (Category theory; homological algebra :: General theory of categories and functors :: Adjoint functors )

Pending Errata and Addenda
None.
[ View all 1 ]
Discussion
Style: Expand: Order:
forum policy
algebra or ring by HkBst on 2004-10-16 12:29:39
what about defining the commutator for rings?
[ reply | up ]

Interact
post | correct | update request | add derivation | add example | add (any)