$k$-forms as modules over $C^\infty(M)$ Let $M$ be an $n$-dimension manifold. Is $E^k(M)$ a finite dimension module over $C^\infty(M)$? Here $E^k(M)$ is the space of $k$-form on $M$, $k<n$.
 A: I think what you want to ask is whether $E^k(M)$ is finitely generated as a module over $C^\infty(M)$.  The answer is yes (at least if your manifold is second countable); in fact, it's true for the space of smooth sections of any finite-rank smooth vector bundle over $M$.
Let $E\to M$ be a smooth vector bundle of rank $m$.  If $M$ has a finite trivializing cover $\lbrace W_1,\dots,W_k\rbrace$ (i.e. a cover by open sets over which $E$ is trivial), then the result is easy to prove: Let $\lbrace \phi_1,\dots,\phi_k\rbrace$ be a smooth partition of unity subordinate to $\lbrace W_1,\dots,W_k\rbrace$, and for each $j=1,\dots,k$, let $(\sigma_j^1,\dots,\sigma_j^m)$ be a smooth local frame for $E$ over $W_j$.  For each $i=1,\dots,m$ and each $j=1,\dots,k$, the section $\tilde\sigma_j^i = \phi_j\sigma_j^i$ extends to a smooth global section of $E$ by defining it to be zero outside the support of $\phi_j$.  The $m\times k$ sections $\lbrace\tilde\sigma^i_j\rbrace$ generate $\Gamma(E)$ as a $C^\infty(M)$-module.
The fact is that $M$ always has a finite trivializing cover.  If $M$ is compact, this is obvious.  For noncompact $M$, it's not so easy to prove.  It depends on the fact that a finite-dimensional manifold has finite topological dimension, meaning that every open cover has an open refinement with the property that there is some integer $K$ such that each point lies in at most $K$ of the sets of the refinement.  One place to find the proof is in Connections, Curvature, and Cohomology, Vol. I, by Greub, Halperin, and Vanstone, pp. 17-21.  (Maybe someone else can give a more convenient reference.)
