Intrinsic distance is symmetric

Question

Define the intrinsic distance between points $p, q\in M$ to be $\rho(p,q)=\inf (\int_{t_0}^{t_1}{\lvert\lvert\alpha^{\prime}(t)\lvert\lvert dt})$ where $\alpha$ is a curve segment on $M$ such that $\alpha(t_0)=p$ and $\alpha(t_1)=q$.

I need to show that $\rho(p,q)=\rho(q,p)$. Please help me finish this proof.

Define $A$ to be the set of curves $k$ such that $k(t_0)=p$ and $k(t_1)=q$. Let $h(s)=t_0+t_1-s$. Then $h(t_1)=t_0$ and $h(t_0)=t_1$. And we have, if $\alpha=\inf(L(A))$, that $$ \int_{t_0}^{t_1}{\lvert\lvert\alpha^\prime{t}\lvert\lvert dt}=\int_{t_1}^{t_0}{\lvert\lvert\alpha^\prime(h(s))\lvert\lvert dh(s)}=-\int_{t_0}^{t_1}{\lvert\lvert\alpha^\prime(h(s))\lvert\lvert h^\prime(s)ds} $$ Because $t_0<t_1$ we have $h(s)$ to be monotonically decreasing and $h'(s)<0$ for any $s\in [t_0,t_1]$. Hence, $-\lvert h^\prime(s)\lvert=h^\prime(s)$ and

$$ -\int_{t_0}^{t_1}{\lvert\lvert\alpha^\prime(h(s))\lvert\lvert h^\prime(s)ds} = \int_{t_0}^{t_1}{\lvert\lvert\alpha^\prime(h(s))\lvert\lvert \lvert h^\prime(s)\lvert ds = \int_{t_0}^{t_1}{\lvert\lvert\alpha^\prime(h(s))h^\prime(s)\lvert\lvert ds} = \int_{t_0}^{t_1}{\lvert\lvert \beta^\prime(s) \lvert\lvert ds}} $$ Where $\beta(s)=(\alpha\circ h)(s)$. If $\beta$ is not the curve with shortest arclength from $q$ to $p$ then we can convert this to some curve $\gamma$ from $p$ to $q$ that has shorter arclength than $\alpha$, which is a contradiction.

Is this even true? If so, is there a more elegant way? Do I need to do this much work?

Answer 1

More elegant way: take any curve segment $\alpha$ as in your first line. Define $\bar{\alpha}(t) = \alpha(t_1+t_0-t)$. Then $\bar{\alpha}$ is also a curve segment on $M$, and further, $\bar{\alpha}(t_0) = q$, and $\bar{\alpha}(t_1) = p$. Then, clearly $$\int_{t_0}^{t_1}\|\bar{\alpha}'(t)\|dt = \int_{t_0}^{t_1}\|\frac{d}{dt}\alpha(t_1+t_0-t)\|dt = \int_{t_1}^{t_0}-\|\alpha'(t)\|dt = \int_{t_0}^{t_1}\|\alpha'(t)\|dt \leq \rho(p,q).$$

But this holds for any $\alpha$, so we have $\rho(q,p)\leq\rho(p,q)$. But this is symmetric in $p$ and $q$, so the exact same argument gives that $\rho(p,q)\leq\rho(q,p)$, so indeed, $\rho(p,q) = \rho(q,p)$.

Answer 2

You have already shown $$\int_{t_0}^{t_1}{\lVert\alpha^\prime(t)\rVert dt}=\int_{t_0}^{t_1}{\lVert\beta^\prime(t)\rVert dt}$$ for a certain curve $\beta$ from $q$ to $p$. This is already enough. Here is why:

You are trying the show the infimums of the two sets $\left\{\int_{t_0}^{t_1}{\lvert\lvert\alpha^{\prime}(t)\lvert\lvert dt}:\alpha(t_0)=p,\alpha(t_1)=q\right\}$ and $\left\{\int_{t_0}^{t_1}{\lvert\lvert\gamma^{\prime}(t)\lvert\lvert dt}:\gamma(t_0)=q,\gamma(t_1)=p\right\}$ are equal. Your proof already tells you that $$\left\{\int_{t_0}^{t_1}{\lvert\lvert\alpha^{\prime}(t)\lvert\lvert dt}:\alpha(t_0)=p,\alpha(t_1)=q\right\}\subseteq\left\{\int_{t_0}^{t_1}{\lvert\lvert\gamma^{\prime}(t)\lvert\lvert dt}:\gamma(t_0)=q,\gamma(t_1)=p\right\}.$$ The same proof applies to tell you that the reverse inclusion is also true. So the two sets are equal, and so the infimum must be equal.