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?

  • $\begingroup$ They are for Euclidean space, clearly. I think otherwise conformality is rare. $\endgroup$
    – Ben McKay
    Aug 8 '19 at 9:56
  • $\begingroup$ Have you checked whether this is true for the standard unit sphere? $\endgroup$
    – Deane Yang
    Aug 8 '19 at 12:41
  • $\begingroup$ The Taylor series expansion of the exponential map shows that, at the very least, the curvature has to vanish at the base point. $\endgroup$
    – Deane Yang
    Aug 8 '19 at 12:52
  • 1
    $\begingroup$ I have rolled back an edit which seemed needlessly pedantic $\endgroup$
    – Yemon Choi
    Aug 8 '19 at 15:07
  • $\begingroup$ @Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake. $\endgroup$
    – user64494
    Aug 8 '19 at 16:16

See Robert Bryant's answer to Complex manifolds in which the exponential map is holomorphic in which he proves that the surfaces for which the exponential map is conformal are precisely the flat ones.

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.


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