I'm self-learning differential forms. I've been happily integrating 1-forms over parameterised curves, and 2-forms over parameterised surfaces, both in $\mathbb{R}^{3}$. Now I've just found out that integrating an n-form $\omega=f\left(x_{1},\ldots,x_{n}\right)dx_{1}\wedge dx_{2}\cdots\wedge dx_{n}$ over an n-dimensional manifold M in $\mathbb{R}^{n}$ is defined by$$\intop_{M}\omega=\pm\intop_{M}f\left(x_{1},\ldots,x_{n}\right)dx_{1}\cdots dx_{n}.$$

Am I correct in thinking that this definition describes what's going on with an ordinary calculus definite integral$$\int_{b}^{a}f\left(x\right)dx.$$So $f\left(x\right)dx$ would be a 1-form and the one-dimensional manifold it is integrated over is the interval $\left(a,b\right)$?

  • $\begingroup$ That is exactly correct. $\endgroup$ – Lee Mosher Sep 10 at 15:42
  • $\begingroup$ To be extra careful, manifolds should be oriented for purposes of integrating forms, so you should specify the orientation on $(a,b)$ as being induced by restriction from the "basic" orientation on $\mathbb R$. Reversing that orientation means integrating backwards from $b$ to $a$, which changes the sign. $\endgroup$ – Lee Mosher Sep 10 at 15:44
  • $\begingroup$ Actually, to be extra extra careful, the "ordinary calculus definite integral" is ambiguous. It can be interpreted either as (1) the integral of a density on an unoriented smooth manifold with boundary, namely the interval $[a,b]$, in which case you don't need an orientation; or as (2) the integral of a form on an oriented smooth manifold, namely the oriented interval $[a,b]$ with the standard orientation. We can integrate functions even on unoriented smooth manifolds, because of densities. And indeed the whole point of an orientation here is to convert a form into a density. $\endgroup$ – symplectomorphic Sep 10 at 19:00
  • $\begingroup$ @symplectomorphic - Tau, in “Differential Forms and Integration”, distinguishes between the “unsigned definite integral $\int_{\left[a,b\right]}f\left(x\right)dx$ (which one would use to find area under a curve, or the mass of a one-dimensional object of varying density), and the signed definite integral $\int_{a}^{b}f\left(x\right)dx$ (which one would use for instance to compute the work required to move a particle from $a$ to $b$).” Is that what you mean? Thanks $\endgroup$ – Peter4075 Sep 11 at 7:03
  • $\begingroup$ Sorry, that should be Tao not Tau. $\endgroup$ – Peter4075 Sep 11 at 8:09


Moreover, you can think of 0-forms which are just scalars. Then generalized Stokes' theorem $$ \int_{d\Omega} \omega=\int_\Omega d\omega $$ in case of 0-form $\omega$ (and 1-form $d\omega$) becomes the fundamental theorem of Calculus: $$ \left.F(x)\right|_a^b =\int_a^b\frac{dF}{dx}dx $$


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.