Relevance of Differential Forms

Question

I recently started reading about differential forms, and I am trying to figure out their purpose. Lets say $\omega=y\,dx+x\,dy$, and we want to evaluate $\int_C \omega$ over the curve parametrized by $\phi(t)=(t^2,t^3)$ from 0 to 1. So we have $\int_C \omega=\int_C y\,dx+x\,dy$...now I am trying to figure out what the purpose in defining differential forms as functions that send points to $T^*_p C$...when I parametrize the integral, am I suppose to evaluate $\omega$ at $\omega_{\phi(t)}(\phi(t))\,dt$, so that $\int_C \omega=\int_0^1 \omega_{\phi(t)}(\phi(t))\,dt$, where $\omega_{\phi(t)}=t^3\,dx_\phi+t^2\,dy_\phi$, and so $\omega_{\phi(t)}(\phi(t))=\omega_{\phi(t)}(t^2\partial_x,t^3\partial_y)=2t^5$, where I used that $dx\partial_x=1, dx\partial_y=0$. I know that this is wrong though.

Answer 1

The problem is that you need to look at $\omega_{\phi(t)}(\phi'(t))$, and not $\omega_{\phi(t)}(\phi(t))$ as you were doing.

Remember, $\omega_{\phi(t)}$ is an element of $T_{\phi(t)}^*C$, so $\omega_{\phi(t)}\colon \ T_{\phi(t)}C \to \mathbb{R}.$ This means that $\omega_{\phi(t)}$ has to take tangent vectors as inputs.

Since $\omega_{\phi(t)}(\phi'(t)) = (t^3\,dx_\phi + t^2\,dy_\phi)(2t\,\partial_x + 3t^2\partial_y) = 2t^4 + 3t^4 = 5t^4,$ we have $$\int_C \omega = \int_{[0,1]}\phi^*\omega = \int_0^1 \omega_{\phi(t)}(\phi'(t))\,dt = \int_0^1 5t^4\,dt = 1,$$ where $\phi^*\omega$ is the pullback of $\omega$.

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When you ask why differential forms are "relevant," I take your question to mean, "Why are differential forms the objects that we integrate?" Or maybe, "Why is the definition of differential form so technical?"

I don't have time to answer that now, but I do know that there are a number of good answers on this very website (and possibly MathOverflow) already. Perhaps another user can provide the links.