# For a $1$-form $h$, why does $\int_\Gamma \varphi^*h=\int_{\varphi\circ\Gamma}h$?

I'm trying to understand why for a differentiable arc $\Gamma:[a,b]\to\Omega$ and a $1$-form $h=fdx+gdy$, then $$\int_\Gamma\varphi^*h=\int_{\varphi\circ\Gamma}h?$$

For background, $\Omega$ is an open set in $\mathbb{C}$, and $\varphi:\Omega\to\mathbb{C}$ a smooth map. For a function $f$, I have the definition $\varphi^*f=f\circ\phi$, (when this makes sense for $f$ of course).

I also have the definitions $$\varphi^*\,dx=\frac{\partial u}{\partial x}\,dx+\frac{\partial u}{\partial y}\,dy, \qquad \varphi^*dy=\frac{\partial v}{\partial x}\,dx+\frac{\partial v}{\partial y}\,dy,$$ where $u$ is the $x$ component of $\varphi$ and $v$ is the $y$ component. For a $1$-form $h=f\,dx+g\,dy$, $$\varphi^*h=(\varphi^*f)\varphi^*\,dx+(\varphi^*g)\varphi^*\,dy.$$

I calculate \begin{align*} \int_\Gamma \varphi^*h &= \int_\Gamma(\varphi^*f)\varphi^*dx+\int_\Gamma (\varphi^*g)\varphi^*dy\\ &= \int_\Gamma(f\circ\varphi)\frac{\partial u}{\partial x}dx+ \int_\Gamma(f\circ\varphi)\frac{\partial u}{\partial y}dy+ \int_\Gamma(g\circ\varphi)\frac{\partial v}{\partial x}dx+ \int_\Gamma(f\circ\varphi)\frac{\partial v}{\partial y}dy\\ &= \int_\Gamma\left((f\circ\varphi)\frac{\partial u}{\partial x}+(g\circ\varphi)\frac{\partial v}{\partial x}\right)dx+\int_\Gamma\left((f\circ\varphi)\frac{\partial u}{\partial y}+(g\circ\varphi)\frac{\partial v}{\partial y}\right)dy \end{align*}

but I don't see if this fits into the form $\int_{\varphi\circ\Gamma}h=\int_{\varphi\circ\Gamma}fdx+\int_{\varphi\circ\Gamma}gdy$? Can it be made to fit? Thanks.

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Hint: Think of the way you compute an integral of a form in terms of a parametrization to turn it into an ordinary definite integral. – Matt Apr 19 '12 at 19:03

## 2 Answers

Things become much clearer when you use different names for the variables: You have a curve $\Gamma:\ t\mapsto \bigl(x(t),y(t)\bigr)$ in the $(x,y)$-plane, a map $\phi:\ (x,y)\mapsto (u,v)$ and a $1$-form $\omega$ in the $(u,v)$-plane given as $$\omega=f(u,v)du + g(u,v)dv\ .$$ Then $$\phi^*f=f\circ\phi,\quad \phi^*g=g\circ\phi,\quad \phi^*du=u_xdx + u_y dy,\quad \phi^*dv=v_x dx+v_y dy\ ,$$ and $\phi^*\omega$ becomes $$\phi^*\omega=\bigl((f\circ\phi)u_x+(g\circ\phi)v_x\bigr)dx + \bigl((f\circ\phi)u_y+(g\circ\phi)v_y\bigr)dy=:p(x,y)dx + q(x,y)dy\ .$$ Therefore $$\int_\Gamma \phi^*\omega=\int_a^b\bigl(p(x(t),y(t))\dot x(t) + q(x(t),y(t))\dot y(t)\bigr)dt=\ldots$$ and writing it all out allows you to interpret the resulting integral as $\int_{\phi(\Gamma)} \omega$.

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Thanks, it helps to see it spelled out like that. – Hana Bailey Apr 28 '12 at 6:17

Regarding your first question, we can note that $\int_{\Gamma} \phi ^\ast h=\int_{a}^{b}\Gamma ^\ast \phi ^\ast h =\int_{a}^{b}h\circ\phi \circ \Gamma =\int_{a}^{b}(\phi\circ\Gamma)^\ast h=\int_{\phi \circ\Gamma}h$. Also note that $\phi^\ast dx=d\phi_{x}$, and likewise for $d\phi_y$, which implies that $\phi \ : (x,y) \rightarrow(u,v)$ i.e. is a change of variables. Hopefully this helps!

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Thanks, that is helpful. – Hana Bailey Apr 28 '12 at 6:17