# Existence of unit speed geodesics

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?

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 $$(V^1, \ldots, V^n)$$.
• Lee has defines the domain of $\exp_p$ to be $T_pM\cap\mathcal E$ where $\mathcal\subseteq TM$ is the set of all $(p,V)$ such that domain of $\gamma_V$ contains $[0,1]$. How can we guarantee this condition after rescaling? Jan 19, 2021 at 18:04
• We don't know that $\exp_p(|V|^{-1}V)$ is defined. However, we don't need to follow the curve that long; we just need to go until we hit the point $q$, which occurs at $t = |V|^{1}$. Jan 19, 2021 at 18:16