commutator bracketCommutatorBracket2002-04-02 23:43:012004-12-15 09:02:34Definitioncommutator Lie algebracommutator\usepackage{amsmath}
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\newtheorem{proposition}{Proposition}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 {\em 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.
\paragraph{Examples}
\begin{itemize}
\item
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:
$$[X,Y] = XY-YX,\quad X,Y\in \End V.$$
\item 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 $p+q-1$.
\item 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.
\end{itemize}