# The Exterior Derivative

I need to stop putting blogging off until night, because then I just say I’ll do it the next morning, and then I just say I’ll put it off until night, etc. This is the main source of why I haven’t posted anything for awhile.

I think we only need one more tool out of our differential forms bag. Define $d:\Omega^k(M)\to \Omega^{k+1}(M)$ to be the unique $\mathbb{R}$-linear operator satisfying

1) If $\omega\in\Omega^k(M)$ and $\eta\in\Omega^l(M)$, then $d(\omega\wedge\eta)=d\omega\wedge\eta + (-1)^k\omega\wedge d\eta$.
2) $d\circ d\equiv 0$
3) If $f\in\Omega^0(M)=C^\infty(M)$ and $X$ is a smooth vector field, then $df(X)=Xf$.

Note that this last part is sort of the motivation for this operator. We have that 0-forms are just smooth functions, and that this operator gives the differential of the function. So it is a generalization of the differential to all forms. It is non-trivial to show such an operator exists and is unique, but I’ll define an operator that does the trick next, and leave the checking that it actually works as an exercise.

The above definition tells me almost nothing about how to actually compute what the operator does. Let’s look at it in coordinates. Let $\omega=\sum_J \omega_Jdx^J$, where $\omega_J$ are smooth functions, and the multi-indexes are in some natural order. Then $\displaystyle d\left(\sum \omega_Jdx^J\right)=\sum d\omega_J\wedge dx^J$, where we already said that the operator on 0-forms just gives the differential.

Thus we get $\displaystyle d \left( \sum \omega_Jdx^{j_1}\wedge \cdots \wedge dx^{j_k}\right) =\sum_{J}\sum_{i}\frac{\partial \omega_J}{\partial x^i} dx^i\wedge dx^{j_1}\wedge \cdots \wedge dx^{j_k}$.

There are so many interesting things we can go on to say about this, but none are relevant to near future posts.

I’ll just briefly mention a few. The condition about $d\circ d\equiv 0$ actually gives us a cochain complex $\Omega^0(M)\to \Omega^1(M)\to \Omega^2(M)\to \cdots$. If we look at the cohomology of this, we have what is known as de Rham cohomology.

Another interesting bit is that if we take our manifold to be $\mathbb{R}^3$, then this operator gives us lots of our familiar calc 3 operators (in slightly modified form). For example, the components of the exterior derivative of a 1-form, gives the curl of the components of the 1-form treated as a vector field. Also, the 2-form to 3-form computation similarly gives you the divergence.

One thing that might come up later is that the exterior derivative commutes with pullbacks. By this I mean that if $F:M\to N$ is a smooth map, then $F^*(d\omega)=d(F^*\omega)$. One way to prove this is to show it holds for smooth functions, and then induct and use the formula $d(\omega\wedge\eta)=d\omega\wedge (-1)^k\omega\wedge d\eta$.

The last sort of interesting thing to mention is that there is a coordinate independent form of the exterior derivative, but I’m pretty sure I’ve never actually used it.