Given a smooth manifold.

Suppose it has an (affine) connection.

How is the exponential map constructed?

  • $\begingroup$ Can you be more explicit about what you are trying for here? $\endgroup$
    – N. Owad
    May 13, 2014 at 18:58
  • $\begingroup$ Yes the point is that I don't know to much yet but need a quick sketchy construction $\endgroup$ May 13, 2014 at 18:59
  • $\begingroup$ In order to define an exponential map, you need either a Riemannian metric or a Lie group structure. $\endgroup$
    – Jack Lee
    May 13, 2014 at 19:43
  • $\begingroup$ hmm and how does that work then ...is it matter of having some sort of connection (affine)? $\endgroup$ May 13, 2014 at 19:44
  • $\begingroup$ Yes, a connection in the tangent bundle would also give you an exponential map. But not a connection in an arbitrary vector bundle. $\endgroup$
    – Jack Lee
    May 14, 2014 at 0:21

1 Answer 1


The usual construction of the exponential map in Riemannian geometry works also for a general affine connection, even if it does not come from any metric, as follows.

Let $\nabla$ be an affine connection on some manifold $M$, ie a connection on the tangent bundle of $M$.

A parametrized curve in $M$ is called a geodesic if its tangent is parallel with respect to $\nabla$. The equation for a geodesic, written in coordinates, is a 2nd order ODE, hence given any initial point $p\in M$ and $v\in T_pM$, there exists a unique geodesic $\gamma (t),$ defined for some open interval around $t=0$, such that $\gamma(0)=p$, $\gamma'(0)=v$. Define $exp(v)=\gamma(1)$ (if $\gamma(t)$ is defined for $t=1$). Then $exp$ maps some open neighborhood of the origin in $T_pM$ to $M$.

For a connection on an arbitrary vector bundle on $M$ I do not know of a definition of an exponential map (and I doubt there is).


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