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?


1 Answer 1


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)

  • $\begingroup$ Thank you very much Thomas :) $\endgroup$
    – ABIM
    Oct 28, 2015 at 6:15
  • $\begingroup$ The example is also extremely helpful! $\endgroup$
    – ABIM
    Oct 28, 2015 at 6:29
  • $\begingroup$ @thomas Do you know something about converse? I mean, if the exponential maps are equal than are the metrics isometric? $\endgroup$
    – melomm
    May 29, 2017 at 19:13
  • $\begingroup$ @melomm this is way more tricky. See, e.g., here: mathoverflow.net/questions/132244/… $\endgroup$
    – Thomas
    May 30, 2017 at 5:36

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