The exponential map on a manifold $M$ is defined at a point $ p\in T_p(M)$ as

$$exp_p:T_p(M)\rightarrow M \\ exp_p(v)=\gamma_v(1) $$ where $\gamma_v$ is the constant speed geodesic with initial velocity as $v$

Can anyone please tell me what the existence of an affine connection has to do with this definition and also why is the curve's value taken at the point 1? The latter probably has a simple answer, if so a hint would suffice.


1 Answer 1


Well, what is $\gamma_v(t)$? It is a solution to the equation $$\nabla_{\gamma'_v}\gamma'_v = 0,$$ where $\nabla$ is the unique affine connection satisfying the torsion free and metric conditions. So this provides a relationship between the exponential map and the Riemannian connection. As far as why you use the point $1$, the reason is due to a so-called "homogeneity" condition, i.e. you may equivalently consider $\gamma_v(s)$ or $\gamma_{sv}(1)$, see do Carmo's book ``Riemannian Geometry'' for details on this.

  • $\begingroup$ Thanks. I am new to this and therefore a bit iffy . Are Affine connections always easily available though? $\endgroup$
    – Vishesh
    Aug 28, 2013 at 18:04
  • 2
    $\begingroup$ Yes, in fact the set of affine connections is an affine space :) Adding the requirements that the connection is torsion free and compatible with the metric uniquely determines an affince connection within the space of all connections. Details are clearly written out in one of the early chapters of do Carmo mentioned above. $\endgroup$
    – treble
    Aug 28, 2013 at 18:24

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