exp${}_p v = p$ $\stackrel{w ?}{\Longrightarrow}$ exp${}_q w = p$

Question

Let us define a manifold $(g, \mathcal{X})$ in a coordinate chart $(U, \varphi)$, then the exponential map exp${}_p v$, for some coordinate $p \in \mathcal{X}$, vector $v \in T_p \mathcal{X}$ and coordinate $q \in \mathcal{X}$ is given by exp${}_p v = q$.

How can we define $w$ such that exp${}_q w = p$?

Answer 1

Let $c(t)$ be geodesic generated by $v$, then $exp_q(v)=c(1)$. Now its reverse $c(1-t)$ is another geodesic curve from $q$ to $p$, so $w=\frac{d}{dt}|_{t=0}c(1-t)=-c'(1)$ is the tangent vector you wish to find.