# Differential Forms - Algebraic Point of View

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.

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

$$\Omega^n_M = \bigwedge^n \Omega^1_M$$

• Thank you, that is certainly algebraic. I assume that if $f$ is a smooth function then you define $df(X) = X(f)$ where $X$ is a derivation? Defining $d(\omega)$ is trickier though, I guess sorta like a co-boundary operator. Is this "algebraic approach" in any way common in the subject of differential forms? I always seem then treated a lot more geometrically. Oct 25 '19 at 1:51
• Yes, precisely that is how you define exterior derivative for smooth functions. This approach is less common than the standard one of working with local coordinates. There are some books that take this point of view, but to be honest I don't know if any extensive treatment of this topic along this lines exists in english. You may c.f. Smooth manifolds and observables by Nestuev or Models for Smooth Infinitesimal Analysis by Moerdijk and Reyes.
– Slup
Oct 25 '19 at 7:20
• Thank you. There is one more, to-be-expected question, on the matter of "coordinates" from this algebraic-view. Let us say $U$ is an open subset of $M$, and $(f_1,...,f_n)$ is a system of local coordinates, where $f_i:U\to \mathbb{R}$ are smooth. Then, I am assuming, if $\omega$ is a differential form on $M$, then over the open set $U$, we can write $\omega$ as a sum of differentials $d f_i$? Oct 25 '19 at 15:24
• @Slup How is $Vect(\mathbb R)$ defined? Isn't $\Theta_M\cong \mathfrak X(M)$? Sep 15 '21 at 17:29
• @Alchemist The category of vector spaces over $\mathbb{R}$. Yes, $\Theta_M$ can be identified with $\mathfrak{X}(M)$
– Slup
Nov 16 '21 at 12:11