# Integration of a differential form along a curve

Given the differential form $\alpha = x dy - \frac{1}{2}(s^2+y^2) dt$, I'd like to evaluate $\int_\gamma \alpha$ where $\gamma(s)=(\cos s,\sin s, s)$ and $0\leq s\leq \frac{\pi}{4}$. When attempting to evaluate this in a way similar to a line integral, first I need to find the derivative of each component of $\gamma$ and then take the square root of the sum of their squares: $$dt = \sqrt{x'(s)^2+y'(s)^2+t'(s)^2} = \sqrt{2} ds$$ But when I'm trying to evaluate $$\int_\gamma x dy- \frac{1}{2}(s^2+y^2) dt$$ I'm not sure how to handle the different differentials. The first statement I made about $dt$ doesn't seem to make sense in this case because I don't know how $dy$ and $dt$ compare. I honestly have no idea how to approach this and would appreciate a step-by-step guide to evaluating integrals of this type.

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Differential forms exist to eat vector fields. The only vector field in sight is the velocity field on $\gamma$, so we may as well feed it to $\alpha$:

\begin{eqnarray*}\alpha(\gamma'(s)) &= \cos(s)dy(-\sin(s),\cos(s),1) - \frac{1}{2}(s^2 + \sin^2(s))dt(-\sin(s),\cos(s),1) \\ &= \cos^2(s) - \frac{1}{2}(s^2 + \sin^2(s)).\end{eqnarray*}

This gives us a nice function on $[0,\frac{\pi}{4}]$ which we can integrate. That's the integral of $\alpha$ on $\gamma$: $$\int_\gamma\alpha = \int_0^\frac{\pi}{4} \alpha(\gamma'(s))ds = \int_0^1 \cos^2(s) - \frac{1}{2}(s^2 + \sin^2(s))\ ds.$$ I trust you can evaluate this.

As a general definition, to integrate a differential $k$-form $\omega$ on some oriented smoothly embedded $k$-dimensional submanifold $\Phi: U\hookrightarrow \mathbb{R}^n$, pull the differential form back via $\Phi$ and integrate over $U$: $$\int_U\Phi^*\omega.$$

The pullback form $\Phi^*\omega$ is defined by $\Phi^*\omega(V_1,V_2,\ldots,V_k) = \omega(d\Phi V_1,d\Phi V_2,\ldots, d\Phi V_k)$. (You'll notice that's exactly what we did with $\gamma$ and $\alpha$, since $\gamma'(s) = d\gamma_s(\frac{\partial}{\partial s})$.) Since the pullback is a $k$-form on a $k$-dimensional domain, it's a multiple of the volume form and so it can be integrated as an ordinary function.

In practice, for one-forms, you'll compute the velocity field, feed it to the one-form to get a function, and then integrate that function with respect to the curve parameter.

Here's an exercise to tie integrating one-forms with more familiar line integrals of vector fields. If $\gamma$ is a smooth curve on (say) $[0,1]$ into $\mathbb{R}^3$ and $V(p) = (V_x(p),V_y(p),V_z(p))$ is a smooth vector field on $\mathbb{R}^3$, you have defined $$\int_\gamma V = \int_0^1 \langle V(\gamma(s)), \gamma'(s)\rangle ds.$$ On $\mathbb{R}^3$, to any vector vield $V$, we associate the one-form $\vartheta$ defined by $\vartheta(X) = \langle V,X\rangle$. In terms of differentials, $\vartheta = V_xdx + V_ydy + V_zdz$. Now look at the integrand of $\int_\gamma V$: taking the inner product of $V$ and $\gamma'$ is the same as feeding $\gamma'(s)$ to $\vartheta$. The integral of $V$ over $\gamma$ is nothing but $$\int_\gamma \vartheta.$$

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Thank you very much. That's very illuminating. – chris Oct 18 '12 at 12:18
You're quite welcome :) – Neal Oct 19 '12 at 1:34