# Exponential maps depends on Riemannian metric?

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?

• Apr 9, 2020 at 22:25

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)

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