- #1

- 192

- 17

Let ##M## be a smooth n-manifold, let ##\gamma: [a,b] \rightarrow M## be a smooth curve in M, let ##\omega## be a smooth 1-form on M. Then by definition, $$\int_{\gamma} \omega = \int_{[a,b]} \gamma^{*} \omega$$

Choosing smooth coordinates ##(x^i)## on ##M##, we can write ##\omega = w_i dx^i##. So then by definition, $$\int_{[a,b]} \gamma^{*} \omega = \int_a^b (w_i \circ \gamma) d\gamma^i = \int_a^b (w_i \circ \gamma)\frac{d\gamma^i}{dt} dt$$

From an elementary calculus point of view, this makes sense to me. We take the value of ## (w_i \circ \gamma)\frac{d\gamma^i}{dt} ## at a point on some rectangle of some partition, multiply it by the length of the subrectangle (which is approximated by ##dt##), then we let the mesh of the partition tend to ##0##. The summation makes sense to me because it's a summation of

**real numbers**(the product I just mentioned is a product of real numbers).

Now when I think about this from a differential forms perspective, well ## (w_i \circ \gamma)\frac{d\gamma^i}{dt} ## is a function that is being multiplied by the 1-form ##dt##. The value of ## (w_i \circ \gamma)\frac{d\gamma^i}{dt} ## at a point of a subrectangle is a real number, but the value of ##dt## at a point is a covector,

**not**a real number. So what we really have is a summation of a product of a real number with a covector, which is itself a covector, so how can the result of the integral be a real number?