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Picture below is from 132th page of do Carmo's Riemannian Geometry. I can feel the red line, but I want to prove it. In my view, it comes down to prove $$ \overline\exp_p|_{T_pM} = \exp_p $$ where $\overline\exp_p$ is the exponential map of $\overline M$, $\exp_p$ is the exponential map of $M$, $T_pM$ is the tangent space of $M$ at $p$.

What I try: for any $v\in T_pM$, there are $$ \overline\exp_p(v) = \overline \gamma(1,p,v),~~~~~~ \exp_p(v)=\gamma(1,p,v) $$ then how to prove $ \overline \gamma(1,p,v)=\gamma(1,p,v) $ ?

enter image description here

Enhao Lan
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1 Answers1

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If $\gamma$ is a geodesic of $\overline{M}$ that lies in a submanifold $M$ of $\overline{M}$, then it is also a geodesic in $M$ with the induced metric. This follows from the fact that the connection in $M$ is the tangent component of the connection in $\overline{M}$, and since $\nabla_\dot{\gamma}^\overline{M} \dot{\gamma} = 0$, then both the tangent and orthogonal components have to be also zero, therefore $\gamma$ is a geodesic in $M$. In particular, this is true for every geodesic brought by $\overline{\exp}$.

Let $\gamma$ be a geodesic in $M$ with $\gamma(0) = p$ and $\dot{\gamma}(0) = v$. Let $\alpha(t) := \overline{\exp}_p(tv)$. Then, $\alpha$ is a geodesic in $\overline{M}$ with $\alpha(0) = p$ and $\dot{\alpha}(0) = v$ , since $d(\overline{\exp}_p)_0 = Id$. Then, by the previous paragraph, $\alpha$ is a geodesic in $M$ with same initial conditions of $\gamma$. Therefore they are the same by the uniqueness of the geodesic.

The equality of the two exponential maps now follows.

Kaitei
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