Cfrgtkky

Define the intrinsic distance between points $p, qin M$ to be $rho(p,q)=inf (int_{t_0}^{t_1}{lvertlvertalpha^{prime}(t)lvertlvert dt})$ 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 $$
int_{t_0}^{t_1}{lvertlvertalpha^prime{t}lvertlvert dt}=int_{t_1}^{t_0}{lvertlvertalpha^prime(h(s))lvertlvert dh(s)}=-int_{t_0}^{t_1}{lvertlvertalpha^prime(h(s))lvertlvert h^prime(s)ds}
$$
Because $t_0<t_1$ we have $h(s)$ to be monotonically decreasing and $h'(s)<0$ for any $sin [t_0,t_1]$. Hence, $-lvert h^prime(s)lvert=h^prime(s)$ and

$$
-int_{t_0}^{t_1}{lvertlvertalpha^prime(h(s))lvertlvert h^prime(s)ds} = int_{t_0}^{t_1}{lvertlvertalpha^prime(h(s))lvertlvert lvert h^prime(s)lvert ds = int_{t_0}^{t_1}{lvertlvertalpha^prime(h(s))h^prime(s)lvertlvert ds} = int_{t_0}^{t_1}{lvertlvert beta^prime(s) lvertlvert ds}}
$$
Where $beta(s)=(alphacirc h)(s)$. 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?

You have already shown $$int_{t_0}^{t_1}{lVertalpha^prime(t)rVert dt}=int_{t_0}^{t_1}{lVertbeta^prime(t)rVert dt}$$ 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 $left{int_{t_0}^{t_1}{lvertlvertalpha^{prime}(t)lvertlvert dt}:alpha(t_0)=p,alpha(t_1)=qright}$ and $left{int_{t_0}^{t_1}{lvertlvertgamma^{prime}(t)lvertlvert dt}:gamma(t_0)=q,gamma(t_1)=pright}$ are equal. Your proof already tells you that $$left{int_{t_0}^{t_1}{lvertlvertalpha^{prime}(t)lvertlvert dt}:alpha(t_0)=p,alpha(t_1)=qright}subseteqleft{int_{t_0}^{t_1}{lvertlvertgamma^{prime}(t)lvertlvert dt}:gamma(t_0)=q,gamma(t_1)=pright}.$$ 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.

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 $rho(q,p)leqrho(p,q)$. But this is symmetric in $p$ and $q$, so the exact same argument gives that $rho(p,q)leqrho(q,p)$, so indeed, $rho(p,q) = rho(q,p)$.