Definition: Let $\pi:V \rightarrow M$ be a vector bundle. $\Omega^p_{cv}(V)$ is the sections of $p$ form on $V$, such that $\pi^{-1}(K) \cap supp \, (w) $ for all $K \subseteq M$ compact.

Definition: $\Omega^p_c(V)$ is the set of sections $p$ forms on $V$ such that its support is compact.

It is claimed that

$\Omega^p_c(V) = \Omega^p_{cv}(V)$ when $M$ is compact.

How is this true? $\subseteq $ is clear.

Source, Pg 84, Proof of Theorem 6.2.10, line +3


Let $\omega$ be a form on $V$ which is compactly supported on each fiber $V_x$. Consider $\text{supp}(\omega) \subset V$; this is a closed subset of $V$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.

Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $X \subset V$ is closed, with the property that each $X \cap V_x$ is compact, then we may define a function $\sigma: M \to \Bbb R_{\geq 0}$ given by sending $\sigma(x)$ to the largest $\|v\|$ of any $v \in X \cap V_x$. This is well-defined because $X \cap V_x$ is compact. The map $\sigma$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $X \cap V_x$ are 'close', and hence should have 'nearby' size bounds.)

Thus $\sigma$ is a continuous map from a compact space to $\Bbb R$, and hence has an absolute maximum, say $L$. Thus the set $X$ is contained in the closed $L$-disc bundle of $V$, which is a compact space (given any sequence $(m_n, v_n)$ of norm bounded by $L$, choose a subsequence so that $m_n$ is convergent to $m$; working on a chart to identify all nearby fibers with $V_x$, use compactness of the $L$-disc ball to see that a further subsequence may be chosen so that $v_n$ converges). $X$, being a closed subspace of a compact Hausdorff space, is compact.

So any form with compact vertical support on a vector bundle $V$ over compact base $M$ has compact support, as desired.


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