# Have I been using differential forms without knowing it?

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 $$\left(a,b\right)$$?

• That is exactly correct. – Lee Mosher Sep 10 at 15:42
• 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. – Lee Mosher Sep 10 at 15:44
• 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. – symplectomorphic Sep 10 at 19:00
• @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 – Peter4075 Sep 11 at 7:03
• Sorry, that should be Tao not Tau. – 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: $$\left.F(x)\right|_a^b =\int_a^b\frac{dF}{dx}dx$$