I am currently trying to understand the proof of the following statement from the Riemannian Geometry book from Sylvestre Gallot, Dominique Hulin, Jacques Lafontaine. The statement is:
Let $m_0 \in M$. There exist a neighborhod $U$ of $m_0$ and $\epsilon >0$ sucht that, for any $m,p \in U$, there is a unique geodesic $c$ of length less than $\epsilon$ from $m$ to $p$. More, $L(c)= d(m,p)$.
The proof of the statement is given as follow:
There are several points, which are not clear for me in this proof:
- Why is for a fixed $v \in S^{n-1}$, the curve $c_v$ a geodesic?
- How is the function $r(t)$ defined? Does it goes from $(0, \epsilon)$? But why is $\int_0^s|r'(t)|dt \geq \epsilon$?
- Finally this proof is strongly correlated with the Gauss lemma. Now the statement of the Gauss lemma, is that the map $exp_p$ is a radial isometry... How can we with it conclude that, we have the following inequality: $$ \int_0^s [g_{\gamma(t)}(\gamma'(t),\gamma'(t)]^{\frac{1}{2}} \geq \int_0^s |r'(t)|dt$$
Many thanks for some help!
It is a geodesic by definition. Since the exponential map is a local diffeomorphism, then for $t \neq 0$ the curve $t \mapsto \gamma(t)$ can be uniquely written as $t \mapsto \exp_p(r(t) v(t))$ for some positive piecewise differentiable function $r: (0, 1] \to \mathbb{R}$ and some curve $t \mapsto v(t)$ such that $\| v(t) \| = 1$. $r(t)$ and $v(t)$ are both determined by $\gamma(t) = \exp_p(r(t) v(t)) $, that's how they are defined. And the exponential map by definition travels only along geodesics (i.e, $\exp_p(w)$ is the point in $M$ we get to after traveling along the unique geodesic with initial tangent vector $w$ for a single unit of time). So, of course if you fix $t_0$, then the curve $t \mapsto \exp_p(r(t) v(t_0)) = \gamma_{v(t_0)}(t)$ is a geodesic, because the exponential map is born from traveling along geodesics. Mutatis mutandis, the same is true for $c$.
The function $r(t)$ is defined as I just mentioned in 1. Notice that $r(1) = \ell(\gamma)$ by the definition of the exponential map. Now, for any $\varepsilon > 0$, we have: $$\int_{\varepsilon}^1 \| r'(t) \| \ \mathrm{d} t \geq \int_{\varepsilon}^1 r'(t) \ \mathrm{d} t = r(1) - r(\varepsilon)$$ Taking $\varepsilon \to 0$, we get $$\int_{0}^{1} \|r'(t) \| \ \mathrm{d}t \geq r(1) = \ell(\gamma) \geq \varepsilon$$ where the last inequality follows from the hypothesis that $\gamma$ leaves the ball of radius $\varepsilon$.
Finally, notice that
\begin{aligned} \gamma'(t) &= \mathrm{d}(\exp_p)_{r(t) v(t)} (r'(t) v(t) + r(t) v'(t)) \\ &= \mathrm{d}(\exp_p)_{r(t) v(t)}(r'(t) v(t)) + \mathrm{d}(\exp_p)_{r(t) v(t)}(r(t) v'(t)) \end{aligned}
And therefore
\begin{aligned} \langle \gamma', \gamma' \rangle &= \left\langle\mathrm{d}(\exp_p)_{r(t) v(t)}(r'(t) v(t)), \mathrm{d}(\exp_p)_{r(t) v(t)}(r'(t) v(t)) \right\rangle + 2 \left\langle \mathrm{d}(\exp_p)_{r(t) v(t)}(r'(t) v(t)), \mathrm{d}(\exp_p)_{r(t) v(t)}(r(t) v'(t))\right\rangle \\ &+ \|\mathrm{d}(\exp_p)_{r(t) v(t)}(r(t) v'(t)) \|^2 \\ &\geq \frac{r'}{r} \left\langle \mathrm{d}(\exp_p)_{r(t) v(t)}(r(t) v(t)), \mathrm{d}(\exp_p)_{r(t) v(t)}(r'(t) v(t))\right\rangle + 2 \frac{r'(t)}{r(t)} \left\langle \mathrm{d}(\exp_p)_{r(t) v(t)}(r(t) v(t)), \mathrm{d}(\exp_p)_{r(t) v(t)}(r(t) v'(t)) \right\rangle \\ &= \frac{r'(t)}{r(t)} \langle r(t) v(t), r'(t)v(t) \rangle + 2 \frac{r'(t)}{r(t)} \langle r(t) v(t), r(t) v'(t) \rangle \\ & = (r'(t))^2 \end{aligned}
where in the last equality we used the Gauss lema and the fact that $\langle v(t), v(t) \rangle$ (so that $\langle v'(t), v(t) \rangle = 0$).
Hence
$$\| \gamma'(t) \|^2 \geq \| r'(t) \|^2$$ which implies the inequality you want.