Coordinate Free Proof of $ d^2 = 0$

Question

We all know that when the exterior derivative is applied twice to a $k$ form, it always yields the null $k+2$ form (i.e. $d^2 = 0$). However, the only proofs I've seen of this does it in some chart and using coordinate notation computes this. I was wondering if anyone has seen or knows of a proof of $d^2 = 0$ that doesn't use coordinates and instead proves this in a coordinate free way.

Answer 1

I found a self-contained answer that really requires no real computation. I'll reproduce a sketch of it below but it can be found in Natural Operations in Differential Geometry on pg. 66. I think the key is recognizing what generates the algebra $ \Omega(M) $ locally at least.

Since $d$ is a graded derivation of degree one, $ d^2 = \frac12[d,d] $ is a graded derivation of degree two and this is local. And since $\Omega(M) $ is locally generated by $ C^\infty(M,\mathbb{R} )$ and $ \{ df: f\in C^\infty(M,\mathbb{R})\} $ you can just show $ d^2f=0$. By simple computation $$ d^2f(X,Y)= Xdf(Y)-Ydf(X)-df([X,Y])=XYf-YXf-[X,Y]f = 0 $$

Therefore $d^2 = 0$ and $ d^l=0$ for all $ l \geq 2$ for that matter.

Answer 2

I think it is not difficult to get a proof from the "global" formula, which expresses the action of an exterior derivative on vector fields, i.e. $$d\phi (\xi_0,\dots,\xi_k)=\textstyle\sum_i(-1)^i\xi_i\cdot \phi(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)\\ +\textstyle\sum_{i<j}(-1)^{i+j}\phi([\xi_i,\xi_j],\xi_0,\dots,\widehat{\xi_i},\dots,\widehat{\xi_j},\dots,\xi_k),$$ with the hats denoting omission. It is just tedious/annoying to write out these things nicely and get all signs right. Basically this just reflects the general algebraic fact that the differential in the standard complex computing Lie algebra cohomology (with values in some representation) squares to zero.

Answer 3

Let $M$ be a nice smooth manifold, $\alpha\in\Omega^k(M;\mathbb{R})$ be a differential $k$-form, and $X,X_1,\dots,X_{k+2}\in\mathfrak{X}(M;\mathbb{R})$ any collection of vector fields. A definition for the exterior derivative is

$\mathrm{d}\alpha(X_1,\dots,X_k,X_{k+1}):=\sum\limits_{i=1}^{k+1}(-1)^{i+1}X_i(\alpha(X_1,\dots,\hat{X_i},\dots,X_{k+1}))+\sum\limits_{i<j}^{}(-1)^{i+j}\alpha([X_i,X_j],X_1,\dots,\hat{X_i},\dots,\hat{X_j},\dots,X_{k+1})$

As for smooth functions $f\in C^\infty(M;\mathbb{R})$,

$\mathrm{d}f(X)=X(f)$

I think that this is Koszul's formula for the exterior derivative, and, as usual, the vector fields with a hat, $\hat{X_i}$, are omitted from the evaluation on $\alpha$.

Let me show what happens to functions when the exterior derivative is applied twice.

$\mathrm{d}\circ\mathrm{d}f(X_1,X_2)=X_1(\mathrm{d}f(X_2))-X_2(\mathrm{d}f(X_1))-\mathrm{d}f([X_1,X_2])=X_1\circ X_2(f)-X_2\circ X_1(f)-[X_1,X_2](f)=0$

Now for the general case, a two page of tedious manipulations of summations proves the result. This also works for the \v{C}ech cohomology.

Although the beauty of proving this result using local coordinates is that it is clear that "the exterior derivative is a coboundary operator" is a fancy way of saying that mix derivatives commute for smooth functions, $\frac{\partial^2}{\partial x\partial y}f(x,y)=\frac{\partial^2}{\partial y\partial x}f(x,y)$.