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?
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$\begingroup$ They are for Euclidean space, clearly. I think otherwise conformality is rare. $\endgroup$– Ben McKayCommented Aug 8, 2019 at 9:56
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$\begingroup$ Have you checked whether this is true for the standard unit sphere? $\endgroup$– Deane YangCommented Aug 8, 2019 at 12:41
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$\begingroup$ 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 YangCommented Aug 8, 2019 at 12:52
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1$\begingroup$ I have rolled back an edit which seemed needlessly pedantic $\endgroup$– Yemon ChoiCommented Aug 8, 2019 at 15:07
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$\begingroup$ @Yemon Choi:A question is asked, not a topic is explained. The omitted ? makes a grammar mistake. $\endgroup$– user64494Commented Aug 8, 2019 at 16:16
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1 Answer
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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.
