Let $M$ be a smooth manifold. One $k$-form is a section of the bundle $\bigwedge^k T^\ast M$, that is, if $p\in M$ and $\omega$ is a $k$-form then $\omega_p$ is one $k$-linear alternating real function defined in the cartesian product of $k$ copies of the tangent space at $p$.

Given $\omega$ one can then build a new differential $k+1$ form called the exterior derivative $d\omega$. This can be easily defined in a coordinate system $(x,U)$ as

$$d\omega = \sum_{}d\omega_{i_1\dots i_k}\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}$$

where $\omega_{i_1\dots i_k}$ are the components of $\omega$ in the coordinate system $(x,U)$. One then can easily show this definition doesn't depend on the coordinate system chosen. Based on that it is possible to show many properties of the exterior derivative then.

Now, the problem is that I simply can't get one geometrical intuition on what the exterior derivative is. As I know, a $k$-form can be thought of as an object which performs measurements on $k$-vectors, that is, an object capable of measuring projections of $k$-vectors. This point of view comes from the fact that $\bigwedge^k V^\ast$ is isomorphic to the space of linear functions on $\bigwedge^k V$.

In that setting, what is the geometrical intuition behind the exterior derivative? Given one $k$-form how can one understand from one intuitive point of view what its exterior derivative is?

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    $\begingroup$ Just asking: Does thinking about grad, curl, and div in three dimensions help? Have you encountered Cartan's formula? Does it help to interpret Stokes' theorem as asserting that $d$ and the boundary operator for manifolds are adjoint with respect to the pairing given by integration? (I expect you won't get much better than "the picture is analogous in higher dimensions", though I'd be delighted to be proven wrong.) $\endgroup$ – Andrew D. Hwang May 7 '15 at 1:24

I want to start with a simple analogy to the (ordinary) derivative. So suppose that $\omega$ is a $k$-form, and $X_1, \ldots, X_k$ are vector fields. And for the moment, I want you to imagine that the $X_i$ fields are all "constant near some point $p$. Now that doesn't really make sense (unless you're in $\Bbb R^n$) but bear with me. If $p$ is the north pole, and $X_1(p)$ is a vector field that points towards, say, London, then it makes sense to define $X_1$ near $p$ to also point towards London, and those vectors will (in 3-space) all be pretty close to $X_1(p)$.

Then we can define a function $$ f(q) = \omega(q)[X_1(q), \ldots, X_k(q)] $$ defined for $q$ near $p$.

How does $f(q)$ vary as $q$ moves away from $p$? Well, it depends on the direction that $q$ moves. So we can ask: What is $$ f(p + tv ) - f(p)? $$ Or better still, what is $$ \frac{f(p + tv ) - f(p)}{t}? $$ especially as $t$ gets close to zero?

That "derivative" is almost the definition of $$ d\omega(p)[X_1(p), \ldots, X_k(p), v]. $$

There are a couple of problems with that "definition" as it stands:

  1. What if there are multiple ways to extend $X_i(p)$, i.e., what if "constant" doesn't really make sense? Will the answer be the same regardless of the values of $X_i$ near $p$ (as opposed to at $p$)?
  2. How do we know that $d\omega$ has all those nice properties like being antisymmetric, etc.?
  3. How does this fit in with div, grad, curl, and all that?

Problems 1 and 2 are why we have fancy definitions of $d$ that make theorems easy to prove, but hide the insight. Let me just briefly attack item 3.

For a 0-form, $g$, the informal definition I gave above is exactly the definition of the gradient. You have to do some stuff with mixed partials (I think) to verify that the gradient, as a function of the vector $v$ is actually linear in $v$, and therefore can be write $dg(p)[v] = w \cdot v$ for some vector $w$, which we call the "gradient of $g$ at $p$."

So that case is pretty nice.

What about the curl? That one's messier, and it involves the identification of every alternating 2-form with a 1-form (because $2 + 1 = 3$), so I'm going to skip it.

What about div? For the most basic kind of 2-form, something like $$ \omega(p) = h(x, y, z) dx \wedge dy $$ and the point $p = (0,0,0)$ and the vector $v = (0,0,1)$, and the two "vector fields" $X_1(x,y,z) = (1,0,0)$ and $X_2(x, y, z) = (0, 1, 0)$, we end up looking at

\begin{align} f(p + tv) &= h(0, 0, t) dx \wedge dy[ (1,0,0), (0, 1, 0)]\\ &= h(0,0, t) \end{align} and the difference quotient ends up being just $$ \frac{\partial h}{\partial z}(0,0,0) $$ That number tells you how $\omega's$ "response" to area in the $xy$-plane changes as you move in the $z$ direction.

What's that have to do with the divergence of a vector field? Well, that vector field is really a 2-form-field, and duality has been applied again. But in coordinates, it looks like $(0,0,h)$, and its divergence is exactly the $z$-derivative of $h$. So the two notions match up again in this case.

I apologize for not drawing out every detail; I think that the main insight comes from recognizing the idea that the exterior derivative is really just a directional derivative with respect to its last argument...and then doing the algebra to see that it's also a directional derivative with respect to the OTHER arguments as well, which is pretty cool and leads to cool things like Stokes' theorem.


You can visualize the exterior derivative in terms of a boundary operation. The fact that $d \circ d = 0$ can then be understood as following from the fact that the boundary of a boundary is empty. The idea is difficult to explain without pictures, so here's a short PDF that explains the idea and gives some nice diagrams to help you visualize differential forms and the exterior derivative operation in the the Euclidean setting.

There's also a good discussion on mathoverflow on this topic which you might find useful.

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    $\begingroup$ Dear ಠ_ಠ, the pdf link no longer works. Could you please update it? $\endgroup$ – Arrow May 22 '19 at 22:24
  • $\begingroup$ any chance you could update the link to the pdf? $\endgroup$ – ziggurism Dec 18 '20 at 21:48
  • $\begingroup$ @ziggurism Sorry I don't remember the author anymore. I'll see if I can find it, though. $\endgroup$ – ಠ_ಠ Dec 19 '20 at 10:58

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