# Limit definition of the exterior derivative

In most treatments of differential geometry, the exterior derivative is defined in a purely algebraic fashion, as map on the exterior algebra of a manifold satisfying certain properties, from which one can show it is well defined and unique.

I am curious if there is any way to define the exterior derivative as a limit. I am mainly interested in this, since "limit definitions" usually carry more geometric meaning than algebraic definitions. Case in point, one could define the Lie derivative $\mathcal{L}_X$ as the unique derivation of the tensor algebra satisfying $\mathcal{L}_Xf=Xf$ and $\mathcal{L}_XY=[X,Y]$ and that it commutes with contractions, however the definition $$\mathcal{L}_XT=\lim_{t\rightarrow 0}\frac{(\phi^X_{t})^*T-T}{t}$$ is far more enlightening in terms of geometric meaning.

Question: Is there any way to define the exterior derivative of a differential form as a limit?

• In local coordinates the exterior derivative can be defined in terms of partial derivatives, which are limits.
– ಠ_ಠ
Nov 7 '16 at 23:11
• math.stackexchange.com/a/2395095/297841
– user297841
Feb 7 '19 at 10:07
• The exterior derivative can be motivated via integration. This is explained in, for example, C. H. Edwards, Advanced Calculus of Several Variables.Here is my write-up o this: math.nyu.edu/~yangd/papers/exterior_derivative.pdf for how the exterior derivative arises naturally from integration. May 8 at 23:19

I don't have time to write out the details, but if $\omega$ is a $k$-form, you can write out a formula for $d\omega(v_1,\dots,v_{k+1})$ as a limit of integrals of $\omega$ over boundaries of smaller and smaller $(k+1)$-dimensional parallelepipeds determined by the $v_i$'s. For example, since $d\omega(v_1,\dots,v_{k+1})$ depends only on the values of the $v_i$'s at one point, you could extend the $v_i$'s to commuting vector fields and then define the parallelpiped using their flows.
Here's Jack Lee's argument applied to the exterior derivative of a $$1$$-form. The argument extends easily to $$k$$-forms.
First, recall that the fundamental theorem of calculus (or line integrals) says $$\int_{t=a}^{t=b} \langle c'(t), df(c(t))\rangle\,dt = f(c(b)) - f(c(a)).$$
Given $$p \in M$$ and $$v_1, v_2 \in T_pM$$, Let $$\Phi: E \rightarrow M$$ satisfy $$\Phi(0,0) = p$$, where $$R = [0,\delta]\times [0,\delta]$$, $$\partial_1\Phi(0,0) = v_1$$, and $$\partial\Phi_2(0,0) = v_2$$. Given a $$1$$-form $$\theta$$, consider the line integral \begin{align*} \int_{\partial R} \theta &= \int_{x^1=0}^{x^1=\delta} \langle \partial_1, \theta(x^1,0)\rangle\,dx^1 + \int_{x^2=\delta}^{x^2=0} \langle \partial_2, \theta(\delta,x^2)\rangle\,dx^2\\ &\quad+ \int_{x^1=\delta}^{x^1=0} \langle \partial_1, \theta(x^1,\delta)\rangle\,dx^1 + \int_{x^2=\delta}^{x^2=0} \langle \partial_2, \theta(0,x^2)\rangle\,dx^2\\ &=\int_{x^2=0}^{x^2=\delta} \langle \partial_2,\theta(\delta,x^2) - \theta(0,x^2)\rangle\,dx^2\\ &\quad- \int_{x^1=0}^{x^1=\delta} \langle \partial_1, \theta(x^1,\delta) - \theta(x^1,0)\rangle\,dx^1\\ &=\int_{x^2=0}^{x^2=\delta} \int_{x^1=0}^{x^1=\delta} \langle \partial_1, d\langle \partial_2, \theta(x^1,x^2)\rangle\rangle\,dx^1\,dx^2\\ &\quad - \int_{x^1=0}^{x^1=\delta} \int_{x^2=0}^{x=\delta} \langle \partial_2, d\langle \partial_1, \theta(x^1,x^2)\rangle\rangle\,dx^2\,dx^1\\ &=\int_{x^2=0}^{x^2=\delta} \int_{x^1=0}^{x^1=\delta} \langle \partial_1, d\langle \partial_2, \theta(x^1,x^2)\rangle\rangle - \langle \partial_2, d\langle \partial_1, \theta(x^1,x^2)\rangle\rangle\,dx^1\,dx^2. \end{align*} Therefore, \begin{align*} \lim_{\delta\rightarrow 0} \frac{1}{\delta^2}\int_{\partial R} \theta &= \langle \partial_1, d\langle \partial_2, \theta\rangle\rangle - \langle \partial_2, d\langle \partial_1, \theta\rangle\rangle\\ & = \langle \partial_1\otimes\partial_1,d\theta\rangle\\ &= \langle v_1\otimes v_2,d\theta\rangle. \end{align*}