# The exterior derivative of a 1-form

The Lie derivative $${L_{v}\varphi}$$ is defined in terms of a vector field $${v}$$, and its value as a “change in $${\varphi}$$” is computed by using $${v}$$ to transport the arguments of $${\varphi}$$. In contrast, recall that the differential $${\mathrm{d}}$$ takes a 0-form $${f\colon M\to\mathbb{R}}$$ to a 1-form $${\mathrm{d}f\colon TM\to\mathbb{R}}$$ with $${\mathrm{d}f(v)=v(f)}$$. Thus $${\mathrm{d}}$$ is a derivation of degree +1 on 0-forms, whose value as a “change in $${f}$$” is computed using the vector field argument of the resulting 1-form.

We would like to generalize $${\mathrm{d}}$$ to $${k}$$-forms by extending this idea of including the “direction argument” by increasing the degree of the form. It turns out that if we also require the property $${\mathrm{d}\left(\mathrm{d}\left(\varphi\right)\right)=0}$$ (or “$${\mathrm{d}^{2}=0}$$”), there is a unique graded derivation of degree +1 that extends $${\mathrm{d}}$$ to general $${k}$$-forms; this derivation is called the exterior derivative. We first explore the exterior derivative of a 1-form.

The exterior derivative of a 1-form is defined by

$$\displaystyle \mathrm{d}\varphi\left(v,w\right)\equiv v\left(\varphi\left(w\right)\right)-w\left(\varphi\left(v\right)\right)-\varphi\left(\left[v,w\right]\right),$$

where e.g.

$$\displaystyle v\left(\varphi\left(w\right)\right)=\underset{\varepsilon\rightarrow0}{\textrm{lim}}\frac{1}{\varepsilon}\left[\varphi\left(w\left|_{v_{p}\left(\varepsilon\right)}\right.\right)-\varphi\left(w\left|_{p}\right.\right)\right]$$

measures the change of $${\varphi\left(w\right)}$$ in the direction $${v}$$, so that

\begin{aligned}\mathrm{d}\varphi\left(v,w\right) & =\underset{\varepsilon\rightarrow0}{\textrm{lim}}\frac{1}{\varepsilon^{2}}\left[\left(\varphi\left(\varepsilon w\left|_{v_{p}\left(\varepsilon\right)}\right.\right)-\varphi\left(\varepsilon w\left|_{p}\right.\right)\right)\right.\\
& \phantom{{=\underset{\varepsilon\rightarrow0}{\textrm{lim}}\frac{1}{\varepsilon^{2}}\left[-\right.}}-\left(\varphi\left(\varepsilon v\left|_{w_{p}\left(\varepsilon\right)}\right.\right)-\varphi\left(\varepsilon v\left|_{p}\right.\right)\right)\\
& \phantom{{=\underset{\varepsilon\rightarrow0}{\textrm{lim}}\frac{1}{\varepsilon^{2}}\left[-\right.}}\left.-\varphi\left(\varepsilon^{2}\left[v,w\right]\right)\right].
\end{aligned}

The term involving the Lie bracket “completes the parallelogram” formed by $${v}$$ and $${w}$$, so that $${\mathrm{d}\varphi\left(v,w\right)}$$ can be viewed as the “sum of $${\varphi}$$ on the boundary of the surface defined by its arguments.”

The above depicts the exterior derivative of a 1-form $${\mathrm{d}\varphi\left(v,w\right)}$$, which is the sum of $${\varphi}$$ along the boundary of the completed parallelogram defined by $${v}$$ and $${w}$$. So if in the diagram $${\varepsilon=1}$$, we have $${\mathrm{d}\varphi\left(v,w\right)=\left(2-1\right)-\left(0-0\right)+2=3}$$. This value is valid in the limit $${\varepsilon\rightarrow0}$$ if the sum varies like $${\varepsilon^2}$$ as depicted in the figure.

The identity $${\mathrm{d}^{2}=0}$$ can then be seen as stating the intuitive fact that the boundary of a boundary is zero. If $${\varphi=\mathrm{d}f}$$, then $${\varphi\left(v\right)=\mathrm{d}f\left(v\right)=v\left(f\right)}$$, the change in $${f}$$ along $${v}$$. Thus e.g. $${\varepsilon\varphi\left(v\left|_{p}\right.\right)=f\left(v_{p}\left(\varepsilon\right)\right)-f\left(p\right)}$$, so that the value of $${\varphi}$$ on $${v}$$ is the difference between the values of $${f}$$ on the two points which are the boundary of $${v}$$. Each endpoint will be cancelled by a starting point as we add up values of $${\varphi}$$ along a sequence of vectors, resulting in the difference between the values of $${f}$$ at the boundary of the total path defined by these vectors. $${\mathrm{d}\varphi}$$ is the value of $${\varphi}$$ over the boundary path of the surface defined by its arguments, which has no boundary points and so vanishes.

The above depicts how $${\mathrm{d}^{2}=0}$$ corresponds to the boundary of a boundary is zero: each term $${\varphi(v)=\mathrm{d}f(v)}$$ is the difference between the values of $${f}$$ on the boundary points of $${v}$$, which cancel as we traverse the boundary of the surface defined by the arguments of $${\mathrm{d}\varphi(v,w)}$$. In the figure we assume a vanishing Lie bracket for simplicity.

Note that $${\mathrm{d}\varphi\left(v,w\right)}$$ measures the interaction between $${\varphi}$$ and the vector fields $${v}$$ and $${w}$$, thus avoiding the need to “transport” vectors. In particular, a non-zero exterior derivative can be pictured as resulting from either the vector fields or $${\varphi}$$ “changing,” i.e. changing with regard to the implied coordinates of our pictures.

The above depicts a non-zero exterior derivative $${\mathrm{d}\varphi\left(v,w\right)}$$, which results from changes in $${\varphi\left(v\right)}$$ or $${\varphi\left(w\right)}$$, not changes in either $${\varphi}$$ or the vector fields alone as compared to some transport.

If we calculate $${\mathrm{d}\varphi\left(e_{1},e_{2}\right)}$$ explicitly in a holonomic frame in two dimensions, $${\mathrm{d}\left(\varphi_{1}\mathrm{d}x^{1}+\varphi_{2}\mathrm{d}x^{2}\right)=\mathrm{d}\varphi_{1}\wedge \mathrm{d}x^{1}+\mathrm{d}\varphi_{2}\wedge \mathrm{d}x^{2}}$$, so applying this to the basis vector fields $${e_{1}}$$ and $${e_{2}}$$ we have

\begin{aligned}\mathrm{d}\varphi\left(e_{1},e_{2}\right) & =\mathrm{d}\varphi_{1}\left(e_{1}\right)\cdot \mathrm{d}x^{1}\left(e_{2}\right)-\mathrm{d}\varphi_{1}\left(e_{2}\right)\cdot \mathrm{d}x^{1}\left(e_{1}\right)\\
& \phantom{{}=}+\mathrm{d}\varphi_{2}\left(e_{1}\right)\cdot \mathrm{d}x^{2}\left(e_{2}\right)-\mathrm{d}\varphi_{2}\left(e_{2}\right)\cdot \mathrm{d}x^{2}\left(e_{1}\right)\\
& =e_{1}\left(\varphi_{2}\right)-e_{2}\left(\varphi_{1}\right)\\
& =\frac{\partial\varphi_{2}}{\partial x^{1}}-\frac{\partial\varphi_{1}}{\partial x^{2}}.
\end{aligned}

Note that a holonomic dual frame $${\beta^{\mu}=\mathrm{d}x^{\mu}}$$ satisfies $${\mathrm{d}\beta^{\mu}=\mathrm{dd}x^{\mu}=0}$$.