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Let $\alpha: S^1 \to \mathbb{R}^2$ be a smooth immersion. Let $\Gamma: D^2 \to \mathbb{R}^2$ be a smooth map that restricts to $\alpha$ on the boundary. After messing around a bit with Mathematica, I'm convinced that

$$\int_{D^2} \Gamma^*\,dx_1 \wedge dx_2 = \int_\alpha x_1\,dx_2.$$

How do I see this? I know we want to use Stokes' theorem somehow, could anybody help?

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2 Answers 2

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This is precisely Stokes's Theorem. $d(x_1\,dx_2) = dx_1\wedge dx_2$ and $d(\Gamma^*\phi) = \Gamma^*(d\phi)$ for any form $\phi$, so $$\int_{D^2} \Gamma^* d(x_1\,dx_2) = \int_{D^2} d\big(\Gamma^*(x_1\,dx_2)\big) = \int_{\partial D^2}\Gamma^*(x_1\,dx_2) = \int_{S^1} \alpha^*(x_1\,dx_2),$$ as required.

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  • $\begingroup$ Beat me to it. Do you know what the OP means by $\int_{\alpha}x_1dx_2$? $\endgroup$ Mar 13, 2016 at 16:29
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    $\begingroup$ @MichaelA: Sure, that's somewhat standard notation for integrating over a (smooth) singular chain. $\endgroup$ Mar 13, 2016 at 16:35
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Note that

\begin{align*} \int_{D^2}\Gamma^*(dx_1\wedge dx_2) &= \int_{D^2}\Gamma^*(d(x_1 dx_2))\\ &= \int_{D^2}d(\Gamma^*(x_1dx_2))\\ &= \int_{\partial D^2}\Gamma^*(x_1dx_2)\\ &= \int_{S^1}\Gamma^*(x_1dx_2)\\ &= \int_{S^1}\alpha^*(x_1dx_2). \end{align*}

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