pullback of a $k$-formPullbackOfAKForm2003-10-15 01:25:022006-08-22 11:09:11Definition%fancy typeface/symbols
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%% \include{argawarga}If $X$ is a manifold, let $\Omega^k(X)$ be the vector space of $k$-forms on $X$.
\PMlinkescapeword{pullback}
{\bf Definition} Suppose $X$ and $Y$ are smooth manifolds, and suppose
$f$ is a smooth mapping $f:X\to Y$. Then the {\bf pullback}
induced by $f$ is the mapping $f^\ast:\Omega^k(Y)\to\Omega^k(X)$ defined as
follows: If $\omega\in \Omega^k(Y)$, then $f^\ast(\omega)$ is the $k$-form on $X$
defined by the formula
$$(f^*\omega)_x(X_1,\ldots ,X_k)=\omega_{f(x)}\big((Df)_x(X_1),\ldots ,(Df)_x(X_k))$$
where $x\in X$, $X_1,\ldots, X_k\in T_x(X)$, and $Df$ is the
tangent map $Df:TX\to TY$.
\subsubsection{Properties}
Suppose $X$ and $Y$ are manifolds.
\begin{itemize}
\item If $\mbox{id}_X$ is the identity map on $X$, then $(\mbox{id}_X)^\ast$
is the identity map on $\Omega^k(X)$.
\item If $X,Y,Z$ are manifolds, and $f,g$ are mappings
$f:X\to Y$ and $g:Y\to Z$, then
$$(g\circ f)^\ast = f^\ast\circ g^\ast.$$
\item If $f$ is a diffeomorphism $f:X\to Y$, then $f^\ast$ is a diffeomorphism
with inverse
$$(f^{-1})^\ast = (f^\ast)^\ast.$$
\item If $f$ is a mapping $f:X\to Y$, and $\omega\in \Omega^k(Y)$, then
$$df^\ast \omega = f^\ast d\omega,$$
where $d$ is the exterior derivative.
\item Suppose $f$ is a mapping $f:X\to Y$, $\omega\in \Omega^k(Y)$, and $\eta\in \Omega^l(Y)$. Then
$$ f^\ast(\omega\wedge \eta)= f^\ast(\omega)\wedge f^\ast(\eta).$$
\item If $g$ is a $0$-form on $Y$, that is, $g$ is a real valued function $g:Y\to \mathbb{R}$, and $f$ is a mapping $f:X\to Y$,
then $f^\ast(g) = f\circ g$.
\item Suppose $U$ is a submanifold (or an open set) in an manifold $X$, and
$\iota:U\hookrightarrow X$ is the inclusion mapping. Then $\iota^\ast$
restricts $k$-forms on $X$ to $k$-forms on $U$.
\end{itemize}