Dearh math.stackeschange-community,
I'm at a loss with the following problem:
Let $I=(a,b)$, then the inverse of the parallel transport $P_{s,t}^\gamma$ from $s \in I$ to $t \in I$ along the curve $\gamma: I \rightarrow M$, then its inverse is given by the transport $P_{t,s}^\delta$ from $t$ to $s$ along the "backwards" curve $\delta: I \rightarrow M, \delta(r) = \gamma(a+b-r)$ for all $r \in I$.
I've seen this proven in several books and I'm also aware of this post, so I'm convinced that it is correct. However, why is the inverse not simply given by $P_{t,s}^\gamma$, i.e. by the transport from t to s along the same curve?
To be honest, this seems like a stupid question - shouldn't those two things intuitively be exactly the same thing? I can imagine three different outcomes: Both approaches indeed define the same parallel transport, the second transport it somehow non-sensical and can't be defined or the second one is simply not the correct inverse.
However, here is my proof that $P_{t,s}^\gamma P_{s,t}^\gamma = Id$:
Let $X_0 \in T_{\gamma(s)}M$, then $P_{s,t}(X_0) = X(t)$, where $X: I \rightarrow TM$ is the unique vector field along $\gamma$ (i.e. $X(r) \in T_{\gamma(r)}M $ holds for all $r \in I$) with
$$D_t X (r):= \nabla_{\dot{\gamma}(r)} X = 0$$
for all $r \in I$, and
$$X(s) = X_0.$$
Now, since X fulfills those properties, it is also the unique vector field used in the second parallel transport. It follows that
$$P_{t,s}^\gamma P_{s,t}^\gamma(X_0) = P_{t,s}^\gamma(X(t)) = X(s) = X_0$$
which was the claim.
I'd be very thankful for any comments, explanations or hints!