$\int_{\gamma\circ\theta}\alpha=\sigma\int_{\gamma}\alpha$ proof

Question

Let $A$ be an open set of $\mathbb{R}^m$ and $\gamma:[a,b]\to A,t\to\gamma(t)$ a curve of $C^1$ class. and let $\alpha\in\Omega^1(A)$. If $\theta:[c,d]\to[a,b]$ is a diffeomorphism of $C^1$ class, prove

$\int_{\gamma\circ\theta}\alpha=\sigma\int_{\gamma}\alpha$
where $\sigma={\rm sgn}(\theta')$

Since we have a diffeomorphism, $(\theta^{-1})::[a,b]\to [c,d]$. and $\frac{d\theta}{dt}$ is a constant, gives me an intuition why $\int_{\gamma\circ\theta}\alpha=\sigma\int_{\gamma}\alpha$.

Question:

However I have no idea on hoe to prove the statement. Could someone provide me a proof?

Answer 1

Let's write $t$ for the $[c,d]$ coordinate and $s = \theta(t)$ for the $[a,b]$ coordinate. By definition, $\int_{\gamma \circ \theta}\alpha = \int_c^d(\gamma \circ \theta)^*\alpha$. If $\alpha = \alpha_idx^i$ in coordinates, we find for $\partial / \partial t|_{t=t_0} \in T_{t_0}[c,d]$, $$ (\gamma \circ \theta)^*\alpha(\partial/\partial t) = \alpha_i(\gamma(\theta(t_0))d(x^i \circ \gamma \circ \theta)(\partial / \partial t). $$ Writing $\gamma(s) = (\gamma^1(s), \ldots, \gamma^m(s)) \in \mathbb{R}^m$, we have $$ d(x^i \circ \gamma \circ \theta)(\partial / \partial t) = d(\gamma^i) \circ d\theta (\partial / \partial t) = d(\gamma^i)(\theta'(t_0)\partial / \partial s) = \frac{d\gamma^i}{ds}(\theta(t_0))\theta'(t_0) $$ so that $$ \int_{\gamma \circ \theta} \alpha = \int_c^d \alpha_i(\gamma(\theta(t)))\frac{d \gamma^i}{d s}(\theta(t))\theta'(t)dt. $$ Now, doing a change of variables as in Calc 1, i.e. rewriting this in the $s = \theta(t)$ coordinate (which satisfies $ds = \theta'(t)dt$), we find $$ \int_{\gamma \circ \theta}\alpha = \int_{\theta(c)}^{\theta(d)}\alpha_i(\gamma(s))\frac{d\gamma^i}{ds}(s)ds = \pm \int_a^b \alpha_i(\gamma(s))\frac{d\gamma^i}{ds}(s)ds, $$ the sign being negative if $\theta$ reverses orientation. But, same as before, we have $$ \gamma^*\alpha(\partial / \partial s) = \alpha_i(\gamma(s))\frac{d\gamma^i}{ds}(s) $$ so that the integrand in that last term is simply $\gamma^*\alpha$, hence $$ \int_{\gamma \circ \theta}\alpha = \text{sgn}(\theta')\int_{\gamma}\alpha. $$ By the way, you wrote in your question that $d\theta/dt$ is a constant but that's not necessarily true. What's true is that $\theta'(t)$ is either always positive or always negative, since it's continuous and can never be $0$ if $\theta$ is a diffeomorphism.

Answer 2

In advanced calculus you learned that an increasing reparametrization of $\gamma$ does not change the value of the line integral $\int_\gamma {\bf F}\cdot d{\bf x}$, and similarly, that changing the "sense of direction" of $\gamma$ multiplies the value of this integral by $-1$. Your problem is just a fancy expression of these facts.