exterior derivative

The exterior derivative is a map that mimics/ generalizes the total derivative from multivariate calculus. Instead of just functions, it uses differential forms. Its definition is motivated by the idea of finding a “closest linear approximation”.

Definition

Let $M$ be a smooth manifold. The exterior derivative is the unique linear map

$$d:{\Omega }_m⟶{\Omega }_m$$

that satisfies the following conditions: 1.

$$d\circ d=0$$

Doing it twice gives the zero map.

$$d({\Omega }_M^k)\subseteq {\Omega }_M^{k+1}$$

It doesn’t respect the grading of the graded algebra structure of ${\Omega }_M$

3. For $\alpha$ a $k$-form and $\beta$ a $l$-form

$$d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^k\alpha \wedge d\beta$$

Follows the Liebsnitz rule (with a graded enhancement).

4. If $f\in {\Omega }_M^0={\mathcal{O}}_M$, then $df\in {\Omega }_M^1$ is defined by $d{f}_p\in T_p^{\ast }M$ by

$$\begin{array}{rl}d{f}_p:T_{p}M& ⟶\mathbb{R}\\ D& ⟼D([(M,f)])\end{array}$$

In other words, given any derivation of local behavior, $d{f}_p$ evaluates the derivation on $f$.

Explicit formula

The exterior derivative can be calculated using the formula

$$\begin{array}{r}d\omega (X_1,\dots ,X_{k+1})=\sum _{1\le i\le k+1}(-1{)}^{i-1}(X_i(\omega (X_1,\dots ,\widehat{X_i},\dots ,X_{k+1}))\\ +\sum _{1\le i<j\le k+1}(-1{)}^{i+j}\omega ([X_i,X_j],X_1,\dots ,\widehat{X_i},\dots ,\widehat{X_j},\dots ,X_{k+1})\end{array}$$

where $\widehat{X_i}$ denotes omitting those arguments.

Notes

Derivative on smooth functions

For a smooth function $f\in {\Omega }^0(M)$ then

$$d{f}_p=\sum \frac{\partial }{\partial x_i}{|}_p(d{x}_i{)}_p$$

Proof

For $p\in M$, let

$$\left\{\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_n}\right\}$$

be a basis for $T_{p}M$.

For $f\in {\Omega }^0(M)={\mathcal{O}}_M$, $d{f}_p\in T_P^{\ast }M$. Therefore,

$$d{f}_p=\sum a_i(d{x}_i{)}_p{a}_i\in \mathbb{R}$$

where $(d{x}_i{)}_p$ is the dual basis of $T_{p}M$, and $a_i$ are the unique coefficients from basis vectors.

However, if we evaluate $d{f}_p$ on the basis for $T_{p}M$ we get

$$d{f}_p\left(\frac{\partial }{\partial x_i}{|}_p\right)=a_i$$

because of the the dual basis, and linearity.

By the definition, and 4 above

$$d{f}_p\left(\frac{\partial }{\partial x_i}{|}_p\right)=\frac{\partial }{\partial x_i}{|}_p([f])$$