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Suppose $M$ is a simply-connected and complete Riemannian manifold, with Gaussian Curvature $K\leq \beta$,where $\beta>0$ is a positive real number. Choose a point $O\in M$, using the method of Jacobi field we can know that $|dexp_O(\frac{d}{d\theta}(r,\theta))|\geq \frac{1}{\sqrt{\beta}}sin\sqrt{\beta}r$,thus the exponetial map $exp_O$ is a local diffeomorphism restricted on $B=\{v\in T_OM| |v|< \frac{\pi}{\sqrt{\beta}}\}$ since exponential map is nondegenerate in radial direction and M is 2-dimensional. I want to show that $exp_O$ is actually a diffeomorphism from $B$ to $exp_O(B)$(equivalently, $exp_O$ is injective on $B$)，but I can't find a way.

Tsoshamry
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  • Set $r = \sup{\varepsilon~|~ \exp_O \text{ is injective on } B_O(\varepsilon)}$. Show that if $r <\frac{\pi}{\sqrt{\beta}}$ then there exists a point $p$ at distance $r$ from $O$ with two minimizing geodesics from $O$ to $p$. Then there exists a Jacobi field along these segments with $J(O) = J(p) = 0$ and which is not zero. This will be a contradiction with the inequality you proved. – Didier Dec 12 '20 at 12:54
  • If it is not injective, then there are two points in the tangent plane mapping to one point in the manifold under the exponential map, how can this lead to the result that there are two minimizing geodesics from O to p? – Tsoshamry Dec 13 '20 at 10:53

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The reason you can't find a way to show this is because it's not true in general. Here's a counterexample. Let $M$ be a dumbbell-shaped surface of revolution in $\mathbb R^3$ obtained by revolving a curve like this around the $z$-axis:

enter image description here

Using bump functions, it's easy to create a profile curve that has arcs of unit circles at the top and bottom, and then transitions smoothly to a straight line in the middle part. The corresponding surface of revolution will have Gaussian curvature equal to $1$ at the bottom and top, equal to $0$ in the cylindrical part, and bounded above by $1$ everywhere. But if $O$ is a point in the cylindrical part, the injectivity radius at $O$ is only $L/2$, where $L$ is the circumference of the cylinder.

In general, lower bounds on the injectivity radius are not easy to come by. As this example shows, just bounding the curvature above isn't enough, even if you assume the manifold is simply connected. (To be sure, if the curvature is bounded above by $0$ and the manifold is complete and simply connected, then the injectivity radius is infinite; but a positive upper bound won't do.)

One useful estimate due to Klingenberg is the following: If $(M,g)$ is a compact Riemannian manifold such that all sectional curvatures are bounded above by $\beta>0$ and all closed geodesics have length at most $L$, then $$ \operatorname{inj}(M) \ge \min \left( \frac{\pi}{\sqrt{\beta}},\ \frac{L}{2}\right).$$ (See, for example, Petersen's Riemannian Geometry (3rd ed.), Lemma 6.4.7.)

Jack Lee
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