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Updates on last chapter
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Signed-off-by: Marcello Seri <[email protected]>
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mseri committed Jan 12, 2021
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2 changes: 1 addition & 1 deletion 4-cotangentbdl.tex
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Expand Up @@ -368,7 +368,7 @@ \section{One-forms and the cotangent bundle}
In this case you often say that the vector fields are $F$-related\footnote{This is a definition that can be properly formalized, but we will not spend any time on it in during the course.} or that they behave naturally: you can either pull back the function $f$ to $M$ or push forward the vector field $X$ to $N$.

\begin{exercise}
Let $\{\rho\alpha\}$ denote a partition of unity on a manifold $M$ subordinate to an open cover $\{U_\alpha\}$.
Let $\{\rho_\alpha\}$ denote a partition of unity on a manifold $M$ subordinate to an open cover $\{U_\alpha\}$.
Let $F:N\to M$ denote a smooth map between smooth manifolds.
With the definition of pullback of functions given above, prove that
\begin{enumerate}
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6 changes: 3 additions & 3 deletions 6-differentiaforms.tex
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Expand Up @@ -236,14 +236,14 @@ \section{The interior product}
\end{definition}

In other words, $\iota_v\omega$ is obtained from $\omega$ by inserting $v$ into ``the first slot''.
By convention $\iota_v\omega = 0$ if $\omega\in\Lambda^0$.
By convention $\iota_v\omega = 0$ if $\omega\in\Lambda^0(V)$.

\begin{lemma}\label{lemma:intprod}
Let $V$ be a real $n$-dimensional vector space and $v\in V$.
Then the following hold.
\begin{enumerate}
\item $\iota_v\circ \iota_v = 0$ and, thus, $\iota_v\circ\iota_u = -\iota_u\circ\iota_v$;
\item if $\omega\in\Lambda^k$ and $\eta\in\Lambda^h$,
\item $\iota_v\circ \iota_v = 0$ and $\iota_v\circ\iota_u = -\iota_u\circ\iota_v$;
\item if $\omega\in\Lambda^k(V)$ and $\eta\in\Lambda^h(V)$,
\begin{equation}
\iota_v(\omega\wedge\eta) = (\iota_v\omega)\wedge\eta + {(-1)}^k\omega\wedge(\iota_v\eta).
\end{equation}
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