Using 1-forms to integrate along curves

Question

The following is written in my lecture notes: 'Suppose that $\gamma:[a,b]\rightarrow\mathcal{M}$ is a smooth curve and $\omega$ is a 1-form on $\mathcal{M}$. Then we get a smooth function $[t\rightarrow(\omega(\gamma(t)))(\gamma'(t))]$, evaluating $\omega(\gamma(t)))\in{T}^*_{\gamma(t)}\mathcal{M}$ at the vector $\gamma'(t)\in{T}_{\gamma(t)}\mathcal{M}$. We write $\int_\gamma\omega=\int_a^b{(\omega(\gamma(t)))(\gamma'(t))}dt$.'

As an example, I'm trying to show that for a smooth function $f\in{C^{\infty}}(\mathcal{M})$, $\int_\gamma{df}=f(\gamma(b))-f(\gamma(a))$. This is my first attempt at using 1-forms in integration, and whilst I understand the definition above to the extent of what is being mapped where, I don't have much intuition yet and so am quite lost.

Anyway, I have the fact that $df=\sum_{i=1}^{m}\frac{\partial{f}}{\partial{x_i}}dx_i$ and I've substituted this into the expression above to get: \begin{equation} \int_\gamma{df}=\int_a^b\bigg{(}\sum_{i=1}^m{\frac{\partial{f}}{\partial{x_i}}dx_i}(\gamma(t))\bigg{)}(\gamma'(t))dt \end{equation}

I'm aware there are a few similar questions that have been asked, but they all use slightly different notation and I hope that seeing what do with this specific example would clear up the general theory for me as well. Any advice on how to proceed would be much appreciated!

Answer 1

Two comments:

1. You expression $df = \Sigma_{i = 1}^m \frac{\partial f}{\partial x_i} dx_i$ only makes sense in a coordinate chart. That is okay, because...

2. One way to proceed is to observe that the definition of integration of differential forms is set up so that it doesn't matter what coordinates one uses to evaluate the integral (this is important and subtle - you should try to verify it). So you can solve your question by breaking up the path into pieces that lie each in one coordinate chart (this can be done with finitely many charts - why?), prove the proposition there using calculus (Andrew D. Hwang's hint), and then figure out how to paste your computations back together.

Answer 2

You have set up a very reasonable problem for yourself.

Your general setup, applied to $\omega:=df$, gives $$\int_\gamma df=\int_a^b df\bigl(\gamma(t)\bigr).\gamma'(t)\>dt\ .$$ Now this $\gamma$ is given in terms of local coordinates $(x_1,\ldots, x_m)$, and for simplicity we shall assume that all of $\gamma$ fits into one single coordinate patch. Then $df\bigl(\gamma(t)\bigr)$ is encoded in the $(1\times m)$-matrix $\bigl[{\partial f\over\partial x_1},\ldots,{\partial f\over\partial x_m}\bigr]_{\gamma(t)}\>$, and $\gamma'(t)$ is encoded in the $(m\times 1)$-matrix $\bigl[\gamma_1'(t),\ldots,\gamma_m'(t)\bigr]^\top$. It follows that $$\int_a^b df\bigl(\gamma(t)\bigr).\gamma'(t)\>dt=\int_a^b \sum_{i=1}^m{\partial f\over\partial x_i}\biggr|_{\gamma(t)}\>\gamma_i'(t)\ dt\ .$$ Introducing the function $g(t):=f\bigl(\gamma(t)\bigr)$ we see by the chain rule, that the last integral is $$=\int_a^b g'(t)\>dt=g(b)-g(a)=f\bigl(\gamma(b)\bigr)-f\bigl(\gamma(a)\bigr)\ ,$$ as expected.