While attempting to introduce the $\exp$ map, one is often lead through integral curves (say $\gamma:\mathbb R \times \mathfrak g \times G \rightarrow G$ ) on a Lie group $G$, with initial conditions at $t=0$ being $\dot\gamma(0)=X \in \mathfrak g$ and the initial point at the identity $e \in G$.

To obtain an integral curve, we use left invariance to send the single vector $X$ to any other required point, thus obtaining a vector field by using the group structure (see for ex: Fulton-Harris, beginning of $\S$ 8.3)

$$v_X(g)={L_g}_* X \tag{1}\label{vecf}$$

Finally, existence and uniqueness from theory of differential equations is used to declare $\gamma(t)$ as solution of

$$\dot\gamma = v_X(\gamma(t)) \tag{2}\label{intcurv}$$

My confusions are the following:

  1. To solve \eqref{intcurv}, it seems that i would need to know $\gamma(t)$ beforehand to be able to use $\eqref{vecf}$.
  2. At a formal level, is it allowed to deduce existence of $\gamma(t)$ from \eqref{intcurv} with $v_X$ being defined only implicitly?

I wanted to add that usually the presentation of $\exp$ map comes after this, defined through $\gamma $ (for ex: Fulton-Harris). Hence i would not like to use $\exp$ here.

  • $\begingroup$ $v_X$ is explicitly defined... you just defined it. You need to verify that it's smooth, but it is. 2) One picks charts on the manifold and writes this as a linear ODE in $\text{dim } G$ variables with smooth coefficients (for $t$ sufficiently close to some point $t_0$). One then applies existence and uniqueness theorems for solutions to such ODEs to see that $\gamma$ is uniquely defined for $t$ near $t_0$, given a value of $\gamma(t_0)$. This plus the group structure is enough to define $\gamma$ for all time. This is probably done very explicitly in Lee's smooth manifolds book. $\endgroup$ – user98602 Oct 5 '17 at 19:53
  • $\begingroup$ @MikeMiller Thanks for your comment. Now that you say it, i realize that it's kind of obvious :) There is no need to search for a special path to define $v_X$. We define it over the whole manifold, thereby getting a vector field, and then the usual. By the way, while searching for answers, i encountered a nice numerical technique that i wanted to add: arxiv.org/abs/1207.0069. $\endgroup$ – mildmay Oct 6 '17 at 0:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.