I've been trying to find a clear explanation on the Internet but failed unfortunately, so I'm asking here. How does the exponential map relate to parallel transport and geodesics for Lie groups. If it makes things simpler we can assume the Lie group to be compact and/or linear. Also if there is a choice between left and right action/invariance etc. I'd prefer the left one.

There are several bits of information that I picked up here and there, but I'm struggling to put them together consistently:

  • $\exp_p(v):=\gamma_v(1)$, where $\gamma$ is the unique geodesic with $\gamma(0)=p$ and $\dot\gamma(0)=v$.
  • $\exp(tv)$ is in general not a geodesic
  • $\exp(tv)$ is the integral curve of the left-invariant vector field associated to $v\in\mathfrak g$
  • The canonical connection $\nabla$ on $G$ is given by $\nabla_XY=0$ for all left-invariant vector fields $X$ and $Y$
  • The canonical parallel transport is given by left-multiplication.

Can the above statements hold consistently for a Lie group $G$? What is the precise relation between $$ \exp \quad\Leftrightarrow\quad\text{geodesics} \quad\Leftrightarrow\quad\text{connection} \quad\Leftrightarrow\quad\text{parallel transport} $$


1 Answer 1


There are two separate notions of exponential map: (1) a riemannian-geometric notion, for which your first item holds; and (2) a Lie group-theoretic notion, for which your second and third items hold.

In certain special situations these two coincide, for example a biinvariant metric on a compact Lie group.

A good reference for this is an appendix in Buser and Karcher:

Buser, Peter; Karcher, Hermann Gromov's almost flat manifolds. Astérisque, 81. Société Mathématique de France, Paris, 1981. 148 pp.


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