6
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ In local coordinates the exterior derivative can be defined in terms of partial derivatives, which are limits. $\endgroup$
    – ಠ_ಠ
    Nov 7 '16 at 23:11
  • 1
    $\begingroup$ math.stackexchange.com/a/2395095/297841 $\endgroup$
    – user297841
    Feb 7 '19 at 10:07
  • $\begingroup$ 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. $\endgroup$
    – Deane
    May 8 at 23:19
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.