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$\newcommand{\R}{\mathbf R}$ Consider a smooth function $f:\R^n\to\R$ and let $Df:\R^n\to \R^{n*}$ be the map which takes a point $\mathbf a\in R^n$ to the linear map $Df_{\mathbf a}:\R^n\to \R$.

This itself is a smooth map whose derivative at a point can be thought of as a multilinear map from $\R^n\times \R^n\to \R$, the matrix representation of which (with respect to, say, the standard basis) will have second partial derivatives.

On the other hand, the map $Df$ can be thought of as a $1$-form on $\R^n$, for at each point $\mathbf a\in \R^n$, we have a functional $Df_{\mathbf a}$. Explicitly, this $1$-form can be written as $$\sum_{i=1}^n\frac{\partial f}{\partial x^i}dx^i$$

The exterior derivative of this is $0$.

So there is a big difference between the notion of the (usual?) derivative and that of the exterior derivative.

If I were asked to say something about what the usual derivative is trying to capture, I would say that the derivative is the natural generalization of the derivative of a map from $\R$ to $\R$. A smooth function $f:\R^n\to \R^m$ at any given point in the domain behaves as closely as we please to a linear map, the derivative, in a sufficiently small neighborhood of the given point.

But I do not have any intuition as to what is the idea behind the exterior derivative.

Can somebody say something about this and about how does the idea of the exterior derivative help us formalize the truth of stokes theorem?

Thanks.

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  • $\begingroup$ the exterior derivatives of a scalar function (0-form) gives you a 1-form, the exterior derivative of a 1-form gives you a 2-form, etc these "devices" allow to express electromagnetism phenomena in a dressing that physicists like to talk, for example $\endgroup$ – janmarqz Jul 7 '15 at 14:17
  • $\begingroup$ in other hand, had you tried to differentiate something like $$\sum_{i=1}^nG_idx^i$$ where the $G_i$ are arbitrary functions in the variables $x^i$? $\endgroup$ – janmarqz Jul 7 '15 at 16:49
  • $\begingroup$ Yes. The derivative is $$\sum_{i, j=1}^n \frac{\partial G_i}{\partial x_j}dx^i\otimes dx^j$$ $\endgroup$ – caffeinemachine Jul 7 '15 at 19:09
  • $\begingroup$ No, it is better to consider $$\sum_{i,j=1}^n\frac{\partial G_i}{\partial x^j}dx^j\wedge dx^i$$ and the point is that it isn't zero always $\endgroup$ – janmarqz Jul 7 '15 at 20:14
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The exterior derivative is exactly what you need in the fundamental theorem of calculus, once you start pushing it to higher dimensions. Let's review the 1-dimensional form: $$f(b)-f(a) = \int_a^b f'(x)\,dx$$ The expression that you see on the left is secretly the integral of $f$ over the outward-oriented boundary of the interval $(a,b)$. Indeed, this boundary consists of two points $\{a,b\}$ of which $b$ gets the plus sign (outward direction is positive there) and $a$ gets the minus sign (outward direction is negative). The FTC says that the boundary integral of $f$ is equal to the interior integral of $f'$.

Pushing this to higher dimensions means replacing the interval $(a,b)$ by an open set $\Omega$ (or more generally, by a manifold), and $\{a,b\}$ by its oriented boundary $\partial \Omega$.

For simplicity, consider two-dimensional $\Omega$, say, a disk in $\mathbb{R}^2$. Then $\partial \Omega$ is an oriented circle. We can integrate a 1-form $f=pdx+qdy$ over this thing. What should be on the other side, in the integral over $\Omega$ itself? The exterior derivative $df = (q_x-p_y)dx\wedge dy$ — this is what Green's theorem says.

The fact that $df=0$ when $f=dg$ is quite natural, since $dg$ integrates to zero over closed curves. Also, the fact that exterior derivative is a counterpart of the boundary operator $\Omega\to\partial \Omega$ brings forward this parallel:

  • the exterior derivative of any form has zero exterior derivative
  • the boundary of any manifold (with boundary) is a manifold without boundary
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