It is clear to me that if $(U,\mathtt{x})$ is a chart of $M$, then $\Omega^{m}(U)$ is a free $\mathcal{C}^{\infty}(U)$-module of rank one. Is $\Omega^{m}(M)$ a free $\mathcal{C}^{\infty}(M)$-module of rank one?

More generally, is ${m}\choose{k}$ the rank of $\Omega^{k}(M)$ as a $\mathcal{C}^{\infty}(M)$- module for $0\leq k \leq m$?

I would also appreciate it if you could give me some reference text where I can look this up.

  • $\begingroup$ Concerning the general case: yes, $\Omega^k(M)$ is a rank $ n \choose{k}$ module, but is in general not free. $\endgroup$
    – Didier
    Jan 6, 2022 at 10:06

1 Answer 1


This answer deals with only $k=m$ (i.e. top forms).

If $M$ is orientable, then $M$ has a non-vanishing top form $\alpha$. Then for any other top form $\beta$, one can write $\beta = f \alpha$ for some smooth functions $f$ on $M$. To see this, note that in each local chart $(U, x)$, one has

$$ \beta = \beta_0 dx^1\wedge \cdots \wedge dx^m, \alpha = \alpha_0 dx^1\wedge \cdots \wedge dx^m$$

for some local functions $\beta_0, \alpha_0$ and $\alpha_0$ is nonzero. Then define $f = \beta_0 /\alpha_0$ on $U$. But this is independent of local charts: given other chart $(\tilde U, \tilde x)$ one has

$$ \beta = \tilde \beta_0 d\tilde x^1 \wedge \cdots \wedge d\tilde x^m = \tilde\beta_0 (\tilde x(x)) \det\left(\frac{\partial \tilde x}{\partial x} \right) dx^1 \wedge \cdots \wedge dx^m$$

and similar for $\alpha$. Thus $$ \frac{\tilde \beta _0(\tilde x (x))}{\tilde \alpha_0 (\tilde x(x))} = \frac{\beta_0(x)}{\alpha _0(x)}$$ and thus $f$ is well-defined. Then $\beta = f\alpha$ and this shows that $\Omega^m (M)$ is a free $C^\infty (M)$-module of rank one.

On the other hand, if $\Omega^m (M)$ is a free module of rank one, then there is a top form $\alpha$ so that for all top forms $\beta$, there is a smooth function $f$ on $M$ so that $\beta = f\alpha$. This implies that $\alpha$ is nowhere zero: if $\alpha (p) = 0$ for some $p\in M$, then $f\alpha $ is also zero at $p$. But one can easily construct a top form $\beta$ so that $\beta (p)$ is nonzero (say, by a bump function).

Thus $\Omega^m (M)$ is a free $C^\infty(M)$-module of rank one if and only if $M$ is orientable.


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