## Gauss lemma and geodesics

From the Gauss lemma to the minimality of geodesics

Lemma:(Gauss) Suppose ${p\in M}$, ${v\in T_u(T_pM)}$

$\displaystyle \langle u,v\rangle=\langle(d\exp_p)_uu,(d\exp_p)_uv\rangle$

Here we make natural equivalence of ${T_u(T_pM)}$ and ${T_pM}$. This lemma means ${d\exp}$ presvers the inner product.

For any ${p\in M}$ there exists a neighborhood ${U}$ of ${p}$ such that the exponential map is diffeomorphism from ${B_\epsilon(0)\subset T_p(M)}$ to ${U_p}$.

Thm: ${p\in M}$, for such ${U_p}$, suppose ${B=\{\exp_pv|||v||<\epsilon\}}$ is any geodesic ball contained in ${U_p}$. Then the shortest length connecting a point on ${\partial B}$ and ${p}$ is the geodesic ray from ${p}$.

Proof: Suppose ${\gamma(t)=\exp_p\sigma(t):[0,1]\rightarrow M}$ connects ${p}$ and ${q\in \partial B}$, where ${\sigma(t)}$ is a curve in ${T_pM}$. Suppoe ${\displaystyle \alpha(t)=\frac{\sigma(t)}{|\sigma(t)|}}$ is the unit normal along ${\sigma(t)}$, then ${\dot{\sigma}(t)=\langle \dot{\sigma}(t),\alpha(t)\rangle \alpha(t)+\beta(t)}$ with ${\beta(t)}$ is perpendicular to ${\alpha(t)}$. Then by Gauss lemma

$\displaystyle \dot{\gamma}(t)=(d\exp_p)_{\sigma(t)}\dot{\sigma}(t)=(d\exp_p)_{\sigma(t)}\left[\langle\dot{\sigma}(t),\alpha(t)\rangle \alpha(t)+\beta(t)\right]$

$\displaystyle =\langle\dot{\sigma}(t),\alpha(t)\rangle(d\exp_p)_{\sigma(t)}\alpha(t)+(d\exp_p)_{\sigma(t)}\beta(t)$

$\displaystyle l(\gamma)=\int_0^1 |\dot{\gamma}(t)|dt\geq \int_0^1|\langle\dot{\sigma}(t),\alpha(t)\rangle(d\exp_p)_{\sigma(t)}\alpha(t)|dt$

$\displaystyle =\int_0^1 |\langle\dot{\sigma}(t),\alpha(t)\rangle|dt(\text{ since }||\alpha(t)||=1)$

$\displaystyle \geq \int_0^1 \langle\dot{\sigma}(t),\alpha(t)\rangle dt=\int_0^1 \langle {\sigma}(t),\alpha(t)\rangle'dt=\langle {\sigma}(1),\alpha(1)\rangle$

$\displaystyle =||\sigma(1)||=\text{length of }\exp t\sigma(1)= \text{geodesic ray from }p \text{ to } q$

We use the fact ${\langle{\sigma}(t),\dot{\alpha}(t)\rangle=0}$, because ${||\alpha(t)||=1}$ which implies ${\langle{\alpha}(t),\dot{\alpha}(t)\rangle=0}$.

Remark: I own this proof to Prof. Xiaochun Rong.