Cartan calculusCartanCalculus2005-11-30 17:28:362008-09-04 14:09:57Definitionanticommutator bracketCartan's magic formulasupercommutation relationgraded derivationCartan's magic formulasupercommutation relationgraded derivationLie superalgebra% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\lie}{\mathcal{L}}Suppose $M$ is a smooth manifold, and denote by $\Omega(M)$ the algebra of differential forms on $M$. The \emph{Cartan calculus} consists of the following three types of linear operators on $\Omega(M)$:
\begin{enumerate}
\item the exterior derivative $d$,
\item the space of Lie derivative operators $\lie_X$, where $X$ is a vector field on $M$, and
\item the space of contraction operators $\iota_X$, where $X$ is a vector field on $M$.
\end{enumerate}
The above operators satisfy the following identities for any vector fields $X$ and $Y$ on $M$:
\begin{align}
d^2 &= 0, \label{cartfirst}\\
d \lie_X - \lie_X d &= 0, \\
d \iota_X + \iota_X d &= \lie_X, \label{magic}\\
\lie_X \lie_Y - \lie_Y \lie_X &= \lie_{[X,Y]}, \\
\lie_X \iota_Y - \iota_Y \lie_X &= \iota_{[X,Y]},\\
\iota_X \iota_Y + \iota_Y \iota_X &= 0, \label{cartlast}
\end{align}
where the brackets on the right hand side denote the Lie bracket of vector fields.
The identity (\ref{magic}) is known as \emph{Cartan's magic formula} or \emph{Cartan's identity}
\subsection*{Interpretation as a Lie Superalgebra}
Since $\Omega(M)$ is a graded algebra, there is a natural grading on the space of linear operators on $\Omega(M)$. Under this grading, the exterior derivative $d$ is degree $1$, the Lie derivative operators $\lie_X$ are degree $0$, and the contraction operators $\iota_X$ are degree $-1$.
The identities (\ref{cartfirst})-(\ref{cartlast}) may each be written in the form
\begin{equation}
AB \pm BA = C,
\end{equation}
where a plus sign is used if $A$ and $B$ are both of odd degree, and a minus sign is used otherwise. Equations of this form are called \emph{supercommutation relations} and are usually written in the form
\begin{equation}\label{supercom}
[A,B] = C,
\end{equation}
where the bracket in (\ref{supercom}) is a \emph{Lie superbracket}. A Lie superbracket is a generalization of a Lie bracket.
Since the Cartan Calculus operators are closed under the Lie superbracket, the vector space spanned by the Cartan Calculus operators has the structure of a \emph{Lie superalgebra}.
\subsection*{Graded derivations of $\Omega(M)$}
\begin{dfn}
A degree $k$ linear operator $A$ on $\Omega(M)$ is a \emph{graded derivation} if it satisfies the following property for any $p$-form $\omega$ and any differential form $\eta$:
\begin{equation}
A (\omega \wedge \eta) = A(\omega) \wedge \eta + (-1)^{kp} \omega \wedge A(\eta).
\end{equation}
\end{dfn}
All of the Calculus operators are graded derivations of $\Omega(M)$.