# The Unapologetic Mathematician

## The Uniqueness of the Exterior Derivative

It turns out that our exterior derivative is uniquely characterized by some of its properties; it is the only derivation of the algebra $\Omega(M)$ of degree $1$ whose square is zero and which gives the differential on functions. That is, once we specify that $d:\Omega^k(M)\to\Omega^{k+1}(M)$, that $d(\alpha+\beta)=d\alpha+d\beta$, that $d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^p\wedge d\beta$ if $\alpha$ is a $p$-form, that $d(d\omega)=0$, and that $df(X)=Xf$ for functions $f\in\Omega^0(M)$, then there is no other choice but the exterior derivative we already defined.

First, we want to show that these properties imply another one that’s sort of analytic in character: if $\alpha=\beta$ in a neighborhood of $p$ then $d\alpha(p)=d\beta(p)$. Equivalently (given linearity), if $\alpha=0$ in a neighborhood $U$ of $p$ then $d\alpha(p)=0$. But then we can pick a bump function $\phi$ which is $0$ on a neighborhood of $p$ and $1$ outside of $U$. Then we have $\phi\alpha=\alpha$ and

\displaystyle\begin{aligned}d\alpha(p)&=\left[d(\phi\alpha)\right](p)\\&=d\phi(p)\wedge\alpha(p)+\phi(p)d\alpha(p)\\&=0+0=0\end{aligned}

And so we may as well throw this property onto the pile. Notice, though, how this condition is different from the way we said that tensor fields live locally. In this case we need to know that $\alpha$ vanishes in a whole neighborhood, not just at $p$ itself.

Next, we show that these conditions are sufficient for determining a value of $d\omega$ for any $k$-form $\omega$. It will helps us to pick a local coordinate patch $(U,x)$ around a point $p$, and then we’ll show that the result doesn’t actually depend on this choice. Picking a coordinate patch gives us a canonical basis of the space $\Omega^k(U)$ of $k$-forms over $U$, indexed by “multisets” $I=\{0\leq i_1<\dots. Any $k$-form $\omega$ over $U$ can be written as

$\displaystyle\omega(q)=\sum\limits_I\omega_I(q)dx^{i_1}(q)\wedge\dots\wedge dx^{i_k}(q)$

and so we can calculate

\displaystyle\begin{aligned}d\omega(p)=&d\left(\sum\limits_I\omega_I(p)dx^{i_1}(p)\wedge\dots\wedge dx^{i_k}(p)\right)\\=&\sum\limits_Id\left(\omega_I(p)dx^{i_1}(p)\wedge\dots\wedge dx^{i_k}(p)\right)\\=&\sum\limits_I\Bigl(d\omega_I(p)\wedge dx^{i_1}(p)\wedge\dots\wedge dx^{i_k}(p)\\&+\omega_I(p)d\left(dx^{i_1}(p)\wedge\dots\wedge dx^{i_k}(p)\right)\Bigr)\\=&\sum\limits_I\Biggl(d\omega_I(p)\wedge dx^{i_1}(p)\wedge\dots\wedge dx^{i_k}(p)\\&+\omega_I(p)\sum\limits_{j=1}^k(-1)^jd\left(dx^{i_1}(p)\wedge\dots\wedge d\left(dx^{i_j}\right)\wedge\dots\wedge dx^{i_k}(p)\right)\Biggr)\\=&\sum\limits_Id\omega_I(p)\wedge dx^{i_1}(p)\wedge\dots\wedge dx^{i_k}(p)\end{aligned}

where we use the fact that $d\left(dx^i\right)=0$.

Now if $(V,y)$ is a different coordinate patch around $p$ then we get a different decomposition

$\displaystyle\omega(q)=\sum\limits_J\omega_J(q)dy^{i_1}(q)\wedge\dots\wedge dy^{i_k}(q)$

but both decompositions agree on the intersection $U\cap V$, which is a neighborhood of $p$, and thus when we apply $d$ to them we get the same value at $p$, by the “analytic” property we showed above. Thus the value only depends on $\omega$ itself (and the point $p$), and not on the choice of coordinates we used to help with the evaluation. And so the exterior derivative $d\omega$ is uniquely determined by the four given properties.

July 19, 2011

## 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

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, 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 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. 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.

July 19, 2011