Existence of unit speed geodesics

Question

I am reading Lee's book "Riemannian manifolds: An Introduction To Curvature". Let $(M,g)$ be a Riemannian manifold. Fix $p\in M.$ It is well known that the exponential map $\exp_p$ map is a local diffeomorphism at $0.$ We can use the identification $T_pM$ to be $\mathbb R^n$ by fixing an orthonormal frame of $T_pM.$ Then we can use the exponential map to define a smooth coordinate chart on a neighbourhood of $p\in M.$ A coordinate chart obtained in this fashion is called a normal coordinate chart. Let $(\mathcal U,x^i)$ be a normal coordinate system. Exercise 5.6 of Lee asks to prove that ``At any point $q\in\mathcal U -p,$ $\partial/\partial r$ is he velocity vector of he unit speed geodesic from $p$ to $q$, and therefore has unit length with respect to $g.$"

Lee defines $r(x):=(\sum_i{x^i}^2)^{\frac{1}{2}}$ and $\partial/\partial r:=\frac{x^i}{r}\partial/\partial x^i$. I can show the following. Let $V:=\exp_p^{-1}(q).$ Then consider the maximal geodesic $\gamma_V$ passing through $p$ and initial velocity $V.$ We know that in normal coordinates $\gamma_V(t)=(tV^1,\dots,tV^n)$. Therefore, after easy calculation $\partial/\partial r|_{\gamma_V(t)}=|V|^{-1}\gamma_V(t)$ and if $|V|=1$ probably we have what Lee asks as $\gamma_V$ is also constant speed. But how can we guarantee that there exists $V$ with $|V|=1.$ This can be shown if there exists a $|V|>1$. Then we can scale it by using star-convexity. But what if $|V|<1?$ Also what about uniqueness Lee is mentioning?

Answer 1

As you've written, $\gamma_V(t) = \exp_p(tV)$. This doesn't have unit speed unless $|V| = 1$ so we rescale, defining $\gamma(t) = \exp_p(t|V|^{-1}V)$. By the chain rule this now was constant unit speed (first evaluate $\dot\gamma(0)$ then use that $\lambda \mapsto \exp_p(\lambda v)$ has constant speed). Now, in coordinates, this is given by the curve $$ \tilde{\gamma}(t) = |V|^{-1}(tV^1, \ldots, tV^n) $$ and therefore its tangent vector is $\dot{\tilde{\gamma}}(t) = |V|^{-1}(V^1, \ldots, V^n)$. But, as $\partial/\partial x^i$ is just the $i$th standard basis vector in coordinates, we have $$ \frac{\partial}{\partial r} = \frac{x^i}{r}\frac{\partial}{\partial x^i} = \frac{1}{r}x, $$ where $x = (x^1, \ldots, x^n)$ is the coordinates of a point on the manifold. Thus for instance when $x = \tilde{\gamma}(t) = t|V|^{-1}V$, we have $\frac{\partial}{\partial r} = r^{-1}x = |t|^{-1}t|V|^{-1}V = |V|^{-1}V = \dot{\tilde{\gamma}}(t)$. In this last line I have abused notation, writing $V$ for the coordinate representation $(V^1, \ldots, V^n)$.