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Given a Lie group $G$, $e$ is its identity element and $g$ is one element of $G$. I want do find a curve $\gamma(t)$ that satisfies these conditions:

1) passes $g$ and $e$, that is $\gamma(0)=e,\gamma(t_g)=g$;

2) the tagent vector at $e$ is $v\in T_eG$, that is $\gamma'(t)|_{t=0}=\xi$.

Does this curve exist in a left invariant vector field? If yes, how to find it? If no, why? can we find it in other type of vector field?

My try: let $\gamma(t)=\exp(\xi t)$, so $\gamma(0)=e$, then solve $\gamma(t_g)=\exp(\xi t_g)=g$, get $t_g$, then we have the vector at $g$: $\xi_g=\gamma'(t)|_{t=t_g}$. I dont think my solution is right, but I dont know where is the problem, and how to get a better solution.

Thank you.

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In general no for left invariant vector fields, since $\gamma(t_g)=\exp(t_gv)$ may not be equal to the $g$ you want. For general case, of course...say $g=\exp(t_gu)$, then define $\gamma(t)=\exp(tv+t^2(u-v)/t_g)$.

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Note: I interpret "Does this curve exist in a left invariant vector field?" to mean "Can I find a curve satisfying the prescribed conditions that is an integral curve of a left invariant vector field?"


Letting $\mathfrak{g} = T_{e}G$ denote the Lie algebra of $G$, recall that the exponential map $exp : \mathfrak{g} \to G$ is defined by $exp(v) = \gamma(1)$, where $\gamma$ is the integral curve of the left invariant vector field corresponding to $v$ that passes through the identity.

With this in mind, then as it relates to whether or not such a curve exists, your question can be interpreted as

  1. Is $exp : \mathfrak{g} \to G$ onto?
  2. What is the image of the exponential map?

There are a couple of restrictions that prevent you from being able to conclude what you would like. First, the exponential map is continuous and the continuous image of a connected set is connected. Thus, the best that you could hope to find is that the exponential map is onto the connected component of $G$ that contains the identity element. However, even this turns out to be false. The Lie group $SL_{2}\left(\mathbb{R}\right) = \left\{A \in M_{2}\left(\mathbb{R}\right) \vert \det A = 1\right\}$ is connected, but the exponential map is not onto. (I can supply details, if you would like.)

With all of this being said, however, it should be noted that with the correct hypotheses, you can always find an integral curve of a left invariant vector field (or a one-parameter subgroup) satisfying the prescribed conditions. Specifically, it is true that if $G$ is compact and connected than the exponential map $exp : \mathfrak{g} \to G$ is onto. See the following link for more details.

Terry Tao - Surjectivity of Exp

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  • $\begingroup$ Thank you. I read the reference you give but can not understand all the things. If I understand well, the exponential map is onto when $G$ is compact and connected. But how about the image of the exponential map when $G$ is not compact? In fact, in my question, $g$ is an element finite, we dont need $\exp$ be onto; we just need to find a path that connects $g$ and $e$, it doesnt matter that there are no range in other regions. Cant we? $\endgroup$ – Martial Feb 14 '15 at 19:38
  • $\begingroup$ @Martial I am not sure that I understand what you mean by "element finite." If you just want a path that connects $g$ and $e$, then the only requirement that you will need on $G$ is that it be connected (the standard assumptions on manifolds guarantee that connected and path connected are equivalent notions--fix a base point p and show that the set of all points that are path-connected to p is both open and closed.) If you want path connecting $g$ and $e$ to satisfy additional properties, then I believe that you would necessarily have to have some additional assumptions on the Lie group. $\endgroup$ – THW Feb 16 '15 at 1:23

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