# Are the exp and log maps of Riemannian geometry conformal

For any Riemannian manifold are the exp and log maps (from a predetermined base) conformal? If not, are there some manifolds where they are and others where they aren't?

• They are for Euclidean space, clearly. I think otherwise conformality is rare. Aug 8 '19 at 9:56
• Have you checked whether this is true for the standard unit sphere? Aug 8 '19 at 12:41
• The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point. Aug 8 '19 at 12:52
• I have rolled back an edit which seemed needlessly pedantic Aug 8 '19 at 15:07
• @Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake. Aug 8 '19 at 16:16

If the exponential map is conformal on some Riemannian manifold of dimension $$\ge 3$$, it is a conformal map on the intersection $$P \cap B$$ of any 2-plane $$P\subset T_m M$$ in any tangent space with the ball $$B\subset T_m M$$ of radius equal to the injectivity radius. So the curvature vanishes on $$e^{P\cap B}$$, i.e. the sectional curvature vanishes, so the Riemannian manifold is flat.