# If $M$ is a $m$-dimensional smooth manifold, what is the rank of $\Omega^{m}(M)$ as a $\mathcal{C}^{\infty}(M)$-module?

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.

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

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.