Geodesically convexity and convexity in normal coordinates

differential-geometryglobal-analysismetric-spacesriemannian-geometry

This question regards an assertion in the proof of Theorem 1.3.1 of Douglas Moore's book "Introduction to Global Analysis, Minimal Surfaces in Riemannian Manifolds" (AMS, 2017).

Assumptions/Definitions:

$M$ is a complete, compact Riemannian manifold $M$. Consider $y_1, y_2 \in M$ two points, $\varepsilon_i>0$, $i=1,2.$

Set $D_i := \{v \in T_{y_i}M | \ |v|< \varepsilon_i\} $, and $B_i := \{z\in M \ | d_M(z,y_i) < \varepsilon_i\}$.

We suppose that:

  1. the exponential map gives diffeomorphisms
    $$\exp_{y_i}|_{D_i}:D_i\to B_i$$

  2. $B_1, B_2, $ and $B_1\cap B_2$ are geodesically convex (i.e. any two points can be joined by a unique minimizing geodesic)

Problem

At page 20 of the book there is a claim about convexity which I don't know how to prove, let me state it precisely.

We would like to show that the domain $U$ where the composition
$ (\exp_{y_2}|_{D_2})^{-1}\circ \exp_{y_1}|_{D_1}$
is defined is convex in $T_{y_1}M$.
Notice that $$U := \exp_{y_1}^{-1} (B_2)\cap D_1$$

Question: why is $U$ convex?

My initial idea was to show a property which turned out to be false, as showed by @Kajelad on
Exponential map and convexity (Riemannian geometry)
that is why I decided to add the geodetically convex hypothesis.

We know that $B_1\cap B_2$ is geodetically convex, is not clear how this implies that the preimage of it under the exponential map at $y_1$ is convex.

Best Answer

Even with the added requirement of geodesic convexity, this is false. A similar counterexample is to let $M$ be the unit sphere and choose $\varepsilon_1=\varepsilon_2=\pi/2$. This will always satisfy the convexity requirements, and if $y_1$ and $y_2$ are chosen sufficiently close to antipodal, the resulting $U$ will be a nonconvex "half-crescent" shape.

Glancing at the cited book, the discussion of convexity don't seem to make sense. Unless I'm mistaken, it can be disregarded entirely with a slight change in argument: Fiberwise convexity is only needed in order for the Taylor remainder integral to be well-defined, but even without it, the integral will be well-defined locally (due to compactness), which suffices for the local computations in the proof.

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