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For a Riemannian manifold $M$, usually, the injectivity radius of $p \in M$ is defined by the supremum of positive numbers $r$ such that $\exp_p|_{B(0, r)}$ is a diffeomorphism onto the image. Here, $B(0,r)$ is the geodesic ball of radius $r$. However, in Klingenberg's book, the injectivity radius is defined by by the supremum of positive numbers $r$ such that $\exp_p|_{B(0, r)}$ is injective. Are these two definitions equivalent?

Considering the global rank theorem, I tried to prove the exponential map at $p$ is of constant rank, but it seems that this try does not work. In fact, Klingenberg frequently says that the exponential map is just injective instead of saying the map is a diffeomorphism. To guarantee that the exponential map at $p$ is a diffeomorphism on a geodesic ball, is it enough to show that it is injective?

Thanks!

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2 Answers 2

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It's not a completely straightforward consequence of the inverse function theorem. Here's a proof.

Theorem. Suppose $(M,g)$ is a Riemannian manifold, $p\in M$, and $B_c(0)\subset T_pM$ is a ball on which $\exp_p$ is defined. Then the restriction of $\exp_p$ to $B_c(0)$ is injective if and only if it is a diffeomorphism onto its image.

Proof. If the restriction of $\exp_p$ is a diffeomorphism onto its image, then clearly it's injective. For the converse, assume it's injective. For each vector $v\in B_c(0)$, let $\gamma_v(t)= \exp_p(tv)$, which is a geodesic defined for all $t$ such that $|tv|<c$.

There are two key facts needed for the proof. Hopefully the following facts are proved in Klingenberg's book, but I don't have a copy of the book here, so I'll give references to my book Introduction to Riemannian Manifolds (2nd ed.) [IRM].

  1. A vector $v\in B_c(0)$ is a critical point of $\exp_p$ if and only if $\gamma_v(1)$ is conjugate to $p$ along $\gamma_v$ [IRM, Proposition 10.20].
  2. No geodesic is minimizing past its first conjugate point [IRM, Theorem 10.26].

Using these facts, the argument goes as follows. It suffices to show that $\exp_p$ has no critical points in $B_c(0)$, for then it's a local diffeomorphism by the inverse function theorem, and an injective local diffeomorphism is a diffeomorphism onto its image.

Suppose for the sake of contradiction that $v\in B_c(0)$ is a critical point. Then by fact 1 above, $\gamma_v(1)$ is conjugate to $p$ along $\gamma_v$. Thus by fact 2, for any number $r$ such that $|v| < |rv| < c$, it follows that $\gamma_v$ restricted to $[0,r]$ not minimizing. Thus there is a shorter geodesic $\sigma: [0,b]\to M$ with $\sigma(0)=p$ and $\sigma(b) = \gamma_v(r)$. Since every geodesic starting at $p$ is the image of a radial line under $\exp_p$, we can parametrize $\sigma$ so that it has parameter domain $[0,1]$ and is of the form $\sigma(t) = \exp_p(tw)$ for some $w\in T_pM$ with $|w| = \text{Len}(\sigma)$, and the fact that $\sigma$ is shorter implies that $|w|<|rv|<c$. Thus $\exp_p(w) = \exp_p(rv)$, contradicting the assumption that $\exp_p$ is injective on $B_c(0)$. $\square$

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Edit: According to this answer, Klingenberg's Riemannian Geometry shows that $\exp: TM \to M$ is non-injective in an neighborhood of any critical point in Chapter 2.1, "Completeness and Cut Locus:"

The following result is from Klingenberg's Riemannian geometry book (Theorem 2.1.12):

Theorem. On a complete Riemannian manifold $(M,g)$, if $\mathrm{ker}(d\exp_p\vert_v)\neq 0$ for some $(p,v)\in TM$, then $\exp_p$ fails to be injective in every neighbourhood of $v$.

(Thus if injectivity fails infinitesimally, it also fails in every neighbourhood. This is of course false for a general smooth map.)

That being proven, Klingenberg is then free to give a weaker-looking, yet equivalent, alternative definition of injectivity radius, and apply it for the remainder of the text.

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    $\begingroup$ It's not true that "locally injective" is equivalent to "local diffeomorphism" for differentiable functions. For example, $f\colon \mathbb R\to \mathbb R$ given by $f(x) = x^3$ is locally (in fact, globally) injective, but it's not a local diffeomorphism. $\endgroup$
    – Jack Lee
    Jan 11, 2021 at 1:30
  • $\begingroup$ @JackLee True, I shouldn't have been sloppy. I fixed it to say the exponential map specifically. $\endgroup$ Jan 11, 2021 at 1:38
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    $\begingroup$ OK, I think it's true for the exponential map. But I don't think it follows just from the inverse function theorem -- you have to use some information about conjugate points and the second variation formula. If you have an argument in mind that uses only the inverse function theorem, it would be worth posting it here. $\endgroup$
    – Jack Lee
    Jan 11, 2021 at 1:50
  • $\begingroup$ @JackLee: I fall in this seduction of functions always. E.g $x^{1/3}$ is not smooth but its graph $(x,x^{1/3})$ is!!. Most of times I recognize functions by their graphs. Could this be misleading? $\endgroup$
    – C.F.G
    Jan 11, 2021 at 7:45

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