Given a smooth manifold.
Suppose it has an (affine) connection.
How is the exponential map constructed?
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asked May 13, 2014 at 18:53
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$\begingroup$ Can you be more explicit about what you are trying for here? $\endgroup$– N. OwadMay 13, 2014 at 18:58
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$\begingroup$ Yes the point is that I don't know to much yet but need a quick sketchy construction $\endgroup$– C-star-W-starMay 13, 2014 at 18:59
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$\begingroup$ In order to define an exponential map, you need either a Riemannian metric or a Lie group structure. $\endgroup$– Jack LeeMay 13, 2014 at 19:43
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$\begingroup$ hmm and how does that work then ...is it matter of having some sort of connection (affine)? $\endgroup$– C-star-W-starMay 13, 2014 at 19:44
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$\begingroup$ 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 LeeMay 14, 2014 at 0:21
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1 Answer
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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).
