Exponential maps depends on Riemannian metric?

Question

This may be a silly question but I am new to Riemannian Geometry.

If I have two different Riemannian Metrics $g_1,g_2$ on a smooth manifold $M$, then do the geodesics on the Riemannian manifolds $(M,g_1)$ and $(M,g_2)$ differ? That is do the Exponential maps depend on our choice of Riemannian metric?

See also: https://math.stackexchange.com/questions/2301858/could-a-riemannian-metric-be-uniquely-determined-by-its-exponential-map/3614249#3614249 — Moishe Kohan, Apr 09 '20 at 22:25

Thomas · Accepted Answer · 2015-10-28T06:21:47.013

2

Yes, sure. You change the definition of length, so you change the shortest paths between two given points, so change $\exp$.

(Edit: A nice visualization you get by looking at the upper half plane in $\mathbb{R}^2$. If this is equipped with the standard Euclidean metric, a geodesic is a straight line. If you use $g_{ij} = \frac{1}{y^2}\delta_{ij}$ you get one standard model of hyperbolic space, and the geodesics are half circles with center on the line $y=0$. See here)

edited Oct 28 '15 at 06:21

answered Oct 28 '15 at 06:13

Thomas

22,361

Thank you very much Thomas :) – ABIM Oct 28 '15 at 06:15
The example is also extremely helpful! – ABIM Oct 28 '15 at 06:29
@thomas Do you know something about converse? I mean, if the exponential maps are equal than are the metrics isometric? – melomm May 29 '17 at 19:13
@melomm this is way more tricky. See, e.g., here: https://mathoverflow.net/questions/132244/can-one-recover-a-metric-from-geodesics – Thomas May 30 '17 at 05:36

Exponential maps depends on Riemannian metric?

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