Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My advanced calculus class is currently doing differential forms and I have a hard time really understanding what they are all about. I can read the proofs of the theorems given in Rudin's PMA chapter 10 and the proofs and I can follow the logic and verify that they are true. However, I don't see why someone would come up with their definition and what makes them useful for building a theory of integration.

To rephrase this: What information exactly is encapsulated by the definition of differential forms and what makes them work out so nicely with respect to wedge products? Why is this the "right" formulation for an integration theory?

share|cite|improve this question
I answered a similar question on MathSE's 'sister site' mathoverflow a while back- you may be after something more in depth, but you may find the second and third paragraphs useful:‌​e-space-of-1-forms-on-a-manifold-c/25393#25393 – Tom Boardman Mar 27 '12 at 21:20
When integrating a function over a surface in R^3, a change of variables leads you to integrate a different function: the function multiplied by the abs. value of the Jacobian of the change of coordinates. This is even apparent in one-variable integration (u-substitution). So if you want to do integration in a coordinate-free way (say, on a manifold without natural coordinates), integration of functions is not well-defined. You want to integrate objects which change under a change of coordinates by mult. by the abs. value of the Jacobian. Those are differential forms (on oriented manifold). – KCd Mar 28 '12 at 1:29

Differential 1-forms are the dual of vector fields on a manifold. Thus, if you are interested in vector fields, you should also be interested in their ``mirror image,'' differential forms. A good intuition regarding their multiplication is that $\mathrm{d}x\wedge\mathrm{d}y$ is an oriented area. Thus, $\mathrm{d}y\wedge\mathrm{d}x = -\mathrm{d}x\wedge\mathrm{d}y$ since its orientation is opposite that of $\mathrm{d}x\wedge\mathrm{d}y$. The exterior derivative unifies and generalizes the gradient, curl, divergence, and Laplacian. Some special cases of the nilpotency of $\mathrm{d}$ (i.e., $\mathrm{d}^2 = 0$) are $\nabla\cdot\nabla\times{\bf F} = 0$ and $\nabla\times\nabla f = 0$.

In terms of integration, notice first that the volume form transforms the right way under coordinate transformations, $$\begin{eqnarray} \omega &=& h(p) \mathrm{d}x^1\wedge\cdots\wedge \mathrm{d}x^n \\ &=& h(p) \frac{\partial x^1}{\partial y^{\mu_1}} \mathrm{d}x^{\mu_1}\wedge\cdots\wedge \frac{\partial x^n}{\partial y^{\mu_n}} \mathrm{d}x^{\mu_n} \\ &=& h(p) \det\left(\frac{\partial x^\mu}{\partial y^\nu}\right) \mathrm{d}y^1\wedge\cdots\wedge \mathrm{d}y^n, \end{eqnarray}$$ where $h(p)$ is positive definite and $\det\left(\frac{\partial x^\mu}{\partial y^\nu}\right)$ is the Jacobian of the coordinate transformation. It also ``knows'' about the orientability of the space. In fact, a space is orientable if and only if it has a volume form.

With differential forms, many important results from multivariable calculus can be unified into a single formula, Stokes' theorem: $$\int_M \mathrm{d}\omega = \int_{\partial M} \omega.$$ For example, some special cases of Stokes' theorem are the curl and divergence theorems in 3-space: $$\begin{eqnarray} \int_S \nabla\times {\bf F}\cdot d{\bf S} &=& \oint_{\partial S} {\bf F}\cdot d{\bf r} \\ \int_V \nabla\cdot {\bf F} dV &=& \oint_{\partial V} {\bf F}\cdot d {\bf S} \end{eqnarray}$$ Of course, Stokes' theorem also generalizes these theorems to higher dimensions.

The utility of differential forms extends far beyond this. For example, de Rham cohomology depends fundamentally on differential forms and can be used to classify topological spaces.

share|cite|improve this answer
The man's name was Stokes, not Stoke, so it's Stokes's theorem (or Stokes' theorem in an older style). – Robert Israel Mar 28 '12 at 1:51
@RobertIsrael: Of course. Thanks for catching that. – user26872 Mar 28 '12 at 1:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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