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Let $M$ denote a manifold admitting a Lorentzian metric $g_{ab}$. Essentially, I would like to know the "minimum domain" on which the exponential map is defined at $p\in M$. To make this concrete, let $(M,g_{ab})$ be time-orientable and let $T\in T_pM$ denote a future-directed timelike vector at $p\in M$. Let $B_T(r)$ denote the open (Euclidean) ball of radius $r$ about the origin of $T_pM$ defined with respect to $T$. Of course, there always exists $r>0$, such that the exponential map on this ball $\text{exp}_p:B_T(r)\to M$ is well defined.

Question: Are there known lower bounds---depending possibly, e.g., on the curvature in a neighborhood of $p$---for the minimum size of $r$? i.e., loosely speaking, for some "observer" at $p$ associated with $T$, what is the minimum (ball-shaped) domain for which the exponential map is well-defined?

Clarification: As Igor Khavkine mentions in his answer, there exists a bound on the "injectivity radius" at $p\in M$ expressed in terms of any $r$ where the exponential map is assumed to be defined. However, I'm interested in a lower bound on $r$ itself.

I'm most interested in the Lorentzian case, but pointers to bounds for ordinary Riemannian manifolds (with no need for the auxiliary vector $T$) may also be enlightening.

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  • $\begingroup$ The question as written makes no sense since the Euclidean metric that we are supposed to use in order to estimate the radius of the domain is unrelated to the Lorentzian metric for which the exponential map is defined. Even if you were dealing with a Riemannian metric, curvature tells you absolutely nothing about lower bounds, just think about open subsets of $R^n$. $\endgroup$
    – Moishe Kohan
    Commented Nov 30, 2023 at 14:45

2 Answers 2

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In Riemannian geometry, the largest such $r$ is the injectivity radius. And there are curvature based bounds for it. To make sense of such a radius in Lorentzian geometry, you need also some reference Riemannian metric. This approach is taken in the following reference, where curvature based bounds are also given:

Injectivity Radius of Lorentzian Manifolds (2006) by Bing-Long Chen, Philippe G. LeFloch arXiv:math/0612860

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  • $\begingroup$ I believe there's some confusion. I think the injectivity radius is the largest $r$ such that $\exp_p$ is injective on $B_T(r)$, but this may be smaller than the largest $r$ such that $\exp_p$ is defined, which OP was asking about: on the unit sphere, as on any geodesically complete manifold, $\exp_p$ is defined on the entire tangent space $T_p M$, but it is injective only up to $r = \pi$. $\endgroup$
    – Gro-Tsen
    Commented Apr 18, 2020 at 9:10
  • $\begingroup$ I grant that your interpretation is a better literal reading of the OP, but I think there is some liberty in interpreting what "well defined" means. I suspect that I answered the intended question, but I may be wrong. $\endgroup$
    – Igor Khavkine
    Commented Apr 18, 2020 at 11:44
  • $\begingroup$ Apologies for any confusion due to imprecise phrasing. However, I'm glad you mention the Chen-LeFloch paper, Igor. Note their injectivity radius bound (Thm 1.1) assumes the exponential map is defined on ball $B_T(r_0)$ for some $r_0>0$. Their bound is then stated in terms of $r_0$. Hence, if one had a lower bound on the (ball-shaped) domain of the exponential map (i.e., a lower bound on $r_0$), Chen-Lefloch would then provide a bound on the minimum possible injectivity radius (for some fixed metric with bounded curvature). So, a bound on $r_0$--which is what I want--would then be very useful. $\endgroup$
    – user143410
    Commented Apr 18, 2020 at 16:04
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    $\begingroup$ @user143410, consider the example of the interior of a Euclidean half-space. The $r_0$ at $p$ cannot be greater than the distance of $p$ to the (open) boundary of the half space. This means that there is no uniform lower bound for $r_0$ on the manifold. Also, whatever the lower bound is at $p$, you cannot get it from local curvature, since the manifold is everywhere flat. The lower bounds that you are looking for somehow depend on the structure of this space "at large". If this example is not what you had in mind, you need some conditions to eliminate it. $\endgroup$
    – Igor Khavkine
    Commented Apr 18, 2020 at 17:43
  • $\begingroup$ Right, I wouldn't expect a uniform bound and, as your example indicates, I should have anticipated that global properties would enter. Essentially, the (ball-shaped) domain of the exponential map is determined by the length of the shortest inextendible geodesic passing through $p$. Perhaps proofs for classic singularity theorems will provide some insight for me. I would think there must be a way to estimate the size of $r_0$ for say a spacetime ($R^D$,$g_{ab}$) with $g_{ab}$ non-flat and satisfying some (physically-reasonable) global constraints. $\endgroup$
    – user143410
    Commented Apr 19, 2020 at 18:28
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You've waited a long time for an answer. And I am surprised that no one has written one.

Let $M=\mathbb{R}\times(0,\infty)$. The exponential map isn't defined at $(x,t)$ for vectors $(u,s)$ with $s<-t$. Since there are points with arbitrarily small $t$ there are points with arbitrarily small injectivity radius. Note that this is independent of curvature. This is a result of incompleteness.

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