If $U$ is an open subset of $\mathbb{R}^n$ we can define "differential forms on $U$" to simply be the exterior algebra generated by smooth maps $\mathcal{C}^{\infty}(U,\mathbb{R})$ and the symbols $d_1,d_2,…,d_n$ which satisfy the anti-commutative properties.
Is there an analogue that works for an $n$-dimensional manifold? I am familiar with the notion of a differential form as being an $n$-covector on the tangent space, … ect. But this is a partially geometric definition, it invokes thinking of tangent vectors, and requires us to think of a differential form as acting on those tangent vectors. However, in the above paragraph, we could define differential forms from a purely algebraic way.
If $M$ is a manifold which has an atlas consisting of just one chart then we can define, likewise, the differential forms of $M$ to simply be the same exterior algebra but generated by smooth maps $\mathcal{C}^{\infty}(M,\mathbb{R})$ instead. What approach would one take for a more general manifold? Is there even such an approach? I looked at the book by Bott and Tu, and it appears they are trying to do something like that in the early pages, but I do not quite see how they define it algebraically.
Best Answer
I depends on what you mean by algebraic. Suppose that $M$ is a manifold (but you can also pick something more general - for instance differentiable space in the sense of R.Sikorski). Now consider the $\mathbb{R}$-algebra $C^{\infty}(M)$ of smooth functions on $M$ and define
$$\Theta_M = \big\{d:C^{\infty}(M)\rightarrow C^{\infty}(M)\,\big|\,d\mbox{ is }\mathbb{R}\mbox{-linear and satisfies Leibniz rule}\big\}\subseteq \mathrm{Hom}_{{Vect}(\mathbb{R})}\left(C^{\infty}(M),C^{\infty}(M)\right)$$
Recall that the Leibniz rule is $$d(f\cdot g) = f\cdot d(g) + g\cdot d(f)$$ for every $f, g\in C^{\infty}(M)$. Now you can check that $\Theta_M$ is a $C^{\infty}(M)$-submodule of $\mathbb{R}$-linear maps $\mathrm{Hom}_{Vect(\mathbb{R})}\left(C^{\infty}(M),C^{\infty}(M)\right)$ and it can be identified with global vector fields on $M$, because smooth functions can be "differentiated" along vector fields. Now define $C^{\infty}(M)$-module
$$\Omega^1_M = \mathrm{Hom}_{C^{\infty}(M)}\left(\Theta_M, C^{\infty}(M)\right)$$
of 1-differential forms and finally
$$\Omega^n_M = \bigwedge^n \Omega^1_M$$