Let $M$ be a Riemannian $n$-manifold and $x$ be a point on $M$. Let $\varepsilon>0$ and $B(0, \varepsilon)$ be the $\varepsilon$-ball centered at the origin in $\mathbf R^n$. Suppose that there is a diffeomorphism $e:B(0, \varepsilon)\to U$, where $U$ is a an open subset of $M$, with $e(0)=x$. Further assume that the following two conditions are satisfied.

a) For all $y\in B(0, \varepsilon)$, and for all $v\in \mathbf R^n$, we have $|de_y(v)|\geq |v|$.

b) $|de_y(v)|=|v|$ whenever $v$ is along the line joining $0$ and $y$.

Then the following implications hold:
1) The image under $e$ of the line joining the origin and a point $y\in B(0, \varepsilon)$ is a geodesic in $M$.
2) The geodesic segment in (1) is the unique geodesic segment joining $e(0)$ and $e(y)$.

I do not see how $(2)$ holds.

(1) can be easily proved by noting that if $\gamma$ is the unit speed straight line segment in $\mathbf R^n$ joining $0$ and $y$, then $e\circ \gamma$ is a minimizing curve joining $e(0)$ and $e(y)$ in $M$. Since minimizing curves are geodesics, we have $e\circ \gamma$ is a geodesic.

But I am unable to see why the image of $e\circ \gamma$ is the unique geodesic segment in $M$ joining $e(0)$ and $e(y)$.

  • 2
    $\begingroup$ I presume that the intended interpretation of (2) is that it’s the unique geodesic segment joining the points and contained in $U$. $\endgroup$ – Ted Shifrin Oct 24 '17 at 17:43
  • $\begingroup$ @TedShifrin I suppose you are right, Though the book doesn't explicitly mention it. At any rate. I am still stuck. $\endgroup$ – caffeinemachine Oct 25 '17 at 5:04

First, to see why you need to say unique geodesic segment joining the two points and contained in $U$, consider two nearby points $P=e(0)$ and $Q=e(y)$ on a great circle on a sphere; obviously you could also take the long way around the sphere, starting at $P$ and going through $-P$ before going back to $Q$.

For the uniqueness of small geodesics starting at $P$ and going to $Q\in U$, this follows from the fact that there is a unique geodesic with any given tangent vector at $P$. Because $e$ is a diffeomorphism, if you follow a geodesic starting at $P$ in a different tangent direction, the segment contained in $U$ cannot also pass through $Q$.


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