# The Unapologetic Mathematician

## The Exterior Derivative is Nilpotent

One extremely important property of the exterior derivative is that $d(d\omega)=0$ for all exterior forms $\omega$. This is only slightly less messy to prove than the fact that $d$ is a derivation. But since it’s so extremely important, we soldier onward! If $\omega$ is a $p$-form we calculate

$\displaystyle\begin{aligned}\left[d(d\omega)\right](X_0,\dots,X_{p+1})=&\sum\limits_{i=0}^k(-1)^iX_i\left(d\omega(X_0,\dots,\hat{X}_i,\dots,X_{p+1})\right)\\&+\sum\limits_{0\leq i<j\leq k}(-1)^{i+j}d\omega\left([X_i,X_j],X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1}\right)\end{aligned}$

We now expand out the $d\omega$ on the first line. First we extract an $X_j$ from the list of vector fields. If $j<i$, then we get a term like

$\displaystyle(-1)^{i+j}X_iX_j\omega(X_0,\dots,\hat{X}_j,\dots,\hat{X}_i,\dots,X_{p+1})$

while if $j>i$ then we get a term like

$\displaystyle(-1)^{i+j-1}X_iX_j\omega(X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1})=(-1)^{i+j-1}X_jX_i\omega(X_0,\dots,\hat{X}_j,\dots,\hat{X}_i,\dots,X_{p+1})$

If we put these together, we get the sum over all $i<j$ of

$\displaystyle(-1)^{i+j}[X_j,X_i]\omega(X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1})$

We continue expanding the first line by picking out two vector fields. There are three ways of doing this, which give us terms like

$\displaystyle\begin{aligned}(-1)^{i+j+k}&X_i\omega([X_j,X_k],X_0,\dots,\hat{X}_j,\dots,\hat{X}_k,\dots,\hat{X}_i,\dots,X_{p+1})\\(-1)^{i+j+k-1}&X_i\omega([X_j,X_k],X_0,\dots,\hat{X}_j,\dots,\hat{X}_i,\dots,\hat{X}_k,\dots,X_{p+1})\\(-1)^{i+j+k-2}&X_i\omega([X_j,X_k],X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,\hat{X}_k,\dots,X_{p+1})\end{aligned}$

Next we can start expanding the second line. First we pull out the first vector field to get

$\displaystyle(-1)^{i+j}[X_i,X_j]\omega(X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1})$

which cancels out against the first group of terms from the expansion of the first line! Progress!

We continue by pulling out an extra vector field from the second line, getting three collections of terms:

$\displaystyle\begin{aligned}(-1)^{i+j+k+1}&X_k\omega([X_i,X_j],X_0,\dots,\hat{X}_k,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1})\\(-1)^{i+j+k}&X_k\omega([X_i,X_j],X_0,\dots,\hat{X}_i,\dots,\hat{X}_k,\dots,\hat{X}_j,\dots,X_{p+1})\\(-1)^{i+j+k-1}&X_k\omega([X_i,X_j],X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,\hat{X}_k,\dots,X_{p+1})\end{aligned}$

It’s less obvious, but each of these groups of terms cancels out one of the groups from the second half of the expansion of the first line! Our sum has reached zero!

Unfortunately, we’re not quite done. We have to finish expanding the second line, and this is where things will get really ugly. We have to pull two more vector fields out; first we’ll handle the easy case where we avoid the $[X_i,X_j]$ term, and we get a whopping six cases:

$\displaystyle\begin{aligned}(-1)^{i+j+k+l+2}&\omega([X_k,X_l],[X_i,X_j],X_0,\dots,\hat{X}_k,\dots,\hat{X}_l,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1})\\(-1)^{i+j+k+l+1}&\omega([X_k,X_l],[X_i,X_j],X_0,\dots,\hat{X}_k,\dots,\hat{X}_i\dots,\hat{X}_l,\dots,\hat{X}_j,\dots,X_{p+1})\\(-1)^{i+j+k+l}&\omega([X_k,X_l],[X_i,X_j],X_0,\dots,\hat{X}_k,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,\hat{X}_l,\dots,X_{p+1})\\(-1)^{i+j+k+l}&\omega([X_k,X_l],[X_i,X_j],X_0,\dots,\hat{X}_i,\dots,\hat{X}_k,\dots,\hat{X}_l,\dots,\hat{X}_j,\dots,X_{p+1})\\(-1)^{i+j+k+l-1}&\omega([X_k,X_l],[X_i,X_j],X_0,\dots,\hat{X}_i,\dots,\hat{X}_k,\dots,\hat{X}_j,\dots,\hat{X}_l,\dots,X_{p+1})\\(-1)^{i+j+k+l-2}&\omega([X_k,X_l],[X_i,X_j],X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,\hat{X}_k,\dots,\hat{X}_l,\dots,X_{p+1})\end{aligned}$

In each group, we can swap the $[X_k,X_l]$ term with the $[X_i,X_j]$ term to get a different group. These two groups always have the same leading sign, but the antisymmetry of $\omega$ means that this swap brings another negative sign with it, and thus all these terms cancel out with each other!

Finally, we have the dreaded case where we pull the $[X_i,X_j]$ term and one other vector field. Here we mercifully have only three cases:

$\displaystyle\begin{aligned}(-1)^{i+j+k+1}&\omega([[X_i,X_j],X_k],X_0,\dots,\hat{X}_k,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1})\\(-1)^{i+j+k}&\omega([[X_i,X_j],X_k],X_0,\dots,\hat{X}_i,\dots,\hat{X}_k,\dots,\hat{X}_j,\dots,X_{p+1})\\(-1)^{i+j+k-1}&\omega([[X_i,X_j],X_k],X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,\hat{X}_k,\dots,X_{p+1})\end{aligned}$

Here we can choose to re-index the three vector fields so we always have $0\leq i<j<k\leq p+1$. Adding all three terms up we get

$\displaystyle(-1)^{i+j+k}\omega(-[[X_i,X_j],X_k]+[[X_i,X_k],X_j]-[[X_j,X_k],X_i],X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,\hat{X}_k,\dots,X_{p+1})$

Taking the linear combination of double brackets out to examine it on its own we find

$\displaystyle-[[X_i,X_j],X_k]+[[X_i,X_k],X_j]-[[X_j,X_k],X_i]=[X_k,[X_i,X_j]]-\left([[X_k,X_i],X_j]+[X_i,[X_k,X_j]]\right)$

Which is zero because of the Jacobi identity!

And so it all comes together: some of the terms from the second row work to cancel out the terms from the first row; the antisymmetry of the exterior form $\omega$ takes care some remaining terms from the second row; the Jacobi identity mops up the rest.

Now I say again that the reason we’re doing all this messy juggling is that nowhere in here have we had to pick any local coordinates on our manifold. The identity $d(d\omega)=0$ is purely geometric, even though we will see later that it actually boils down to something that looks a lot simpler — but more analytic — when we write it out in local coordinates.

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Posted by John Armstrong | Differential Topology, Topology

## 3 Comments	»

1. [...] we consider the differential of a function. If we apply to it, the nilpotency of the exterior derivative tells us that we automatically get zero. On the other hand, if we apply [...]

Pingback by | July 26, 2011 | Reply

2. [...] Even so, the differential is going to be identically zero. Away from the “branch curve” on which we cut in our setup — the negative real axis here — this should be obvious, because here we have since the square of the exterior derivative is automatically zero. [...]

Pingback by | August 22, 2011 | Reply

3. [...] operator is that the curl of a gradient is zero: . In our terms, this is a simple consequence of the nilpotence of the exterior derivative. Indeed, when we work in terms of -forms instead of vector fields, the composition of the two [...]

Pingback by | October 13, 2011 | Reply

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