# 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 - Posted by | Differential Topology, Topology

1. Are you going to cover integration of differential forms or de Rham cohomology?

Comment by Andrei | July 20, 2011 | Reply

2. Well, having introduced the exterior derivative and shown that its square is zero, I’m halfway to de Rham cohomology already, so that wouldn’t be a bad guess.

Comment by John Armstrong | July 20, 2011 | Reply

3. [...] The really important thing about the exterior derivative is that it makes the algebra of differential forms into a “differential graded algebra”. We had the structure of a graded algebra before, but now we have a degree-one derivation whose square is zero. And as long as we want it to agree with the differential on functions, there’s only one way to do it. [...]

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