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commutator bracket (Definition)

Let $ A$ be an associative algebra over a field $ K$. For $ a,b \in A$, the element of $ A$ defined by

$\displaystyle [a,b]=ab-ba$
is called the commutator of $ a$ and $ b$. The corresponding bilinear operation
$\displaystyle [-,-]: A\times A\rightarrow A$
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

$\displaystyle [a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0.$
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 $ \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:
    $\displaystyle [X,Y] = XY-YX,\quad X,Y\in \mathrm{End}V.$
  • 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 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.
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See Also: Lie algebra

Also defines:  commutator Lie algebra, commutator
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Cross-references: homogeneous, vector field, closed, operators, order, restricted, commutative, non-commutative, even, properties, differential operators, structure, endomorphisms, composition, universal enveloping algebra, operation, multiplication, vector space, action, fixed field, Lie algebras, category, functor, categorical, terms, expands, side, Jacobi identity, skew-symmetric, bilinear, bilinear operation, field, algebra, associative
There are 19 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 11375 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 )

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Discussion
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algebra or ring by HkBst on 2004-10-16 12:29:39
what about defining the commutator for rings?
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