Integral Curves by Right-Translation of Lie-Algebraic Elements

[Kreizhn]

Homework Statement

Let G be a smooth Lie group with Lie algebra $\mathfrak g$. Consider a curve $X: [t_0,t_1] \to G$ whose dynamics are given by
$$\frac{dX}{dt}(t) = H X(t)$$
for some $H \in \mathfrak g$.

Show that this equation is well defined for all time.

The Attempt at a Solution

Okay, so we know that the Lie algebra can be view as either the tangent space at the group identity or the set of right(left) invariant vector fields. Let the group action of right translation be given by $R_X:G\to G$ by $R_X(Y) = YX$. Then $H \in \mathfrak g$ is right-invariant means that we can write
$$dR_X(H) = H(X)$$
where H(X) is the vector field H evaluated at the point X.

Okay, so now here is where the confusion starts to come in. Firstly, if we are given a fixed $H \in T_I G$ I do not see how we can naturally extend this to a right-invariant vector field.

Secondly, using the definition of the pushforward/differential, it must follow that for any function $f: M \to \mathbb R$ that
$$H(f\circ R_X) = (H X)f$$

Now, is this question just trivial? In particular, do we define the extension of the tangent $H \in T_I G$ to a vector-field on G by $H(X) = HX \in T_XG$? Secondly, does this imply that the pushforward of the right-translation operator $dR_X$ just acts as $dR_X(H) = HX$? I guess both these first and second points are essentially the same, but an answer to either should give an answer to both.

Any help would be appreciated.

 

[Dick]

If you define Y(t)=X(t0)^(-1)*X(t), then isn't Y(t) a curve in the Lie group such that Y(0)=I, so Y'(0)=H is a member of the tangent space at the group identity? I don't pretend to be an expert on these things these days, but can't you just define Y(t) using the exponential map?

 

[Kreizhn]

I think it seems close.

I think we have to switch the order of X(t) and X(t_0) though, so that $Y(t) = X(t) X(t_0)^{-1}$ since otherwise we'll have $Y'(t_0) = X(t_0)^{-1} H X(t_0)$ and we cannot be guaranteed that H and $X(t_0)$.

Anyway, I'm not worried so much about solving this. That I can do . I'm just trying to justify why $HX(t) \in T_{X(t)} G$ because I can't immediately see it to be true unless it follows from the definition of H being right-invariant.

 

[Dick]

H is in the Lie algebra, so it's in the tangent space at the identity. So it's the derivative of a curve through the identity, say H=c'(s) evaluated at s=0 where c(0)=I, or whatever you call the identity. So HX(t) is in the tangent space at X(t). It's the derivative with respect to s of c(s)X(t) which is a curve passing through X(t) at s=0. How's that?

 

[Kreizhn]

Hmm...I think that might do it. Though by "H=c'(s) evaluated at s=0" do you mean that c(s) is some arbitrary curve with c'(0) = H?

 

[Dick]

Hmm...I think that might do it. Though by "H=c'(s) evaluated at s=0" do you mean that c(s) is some arbitrary curve with c'(0) = H?

Sure. That's one way of defining what a tangent space is, right? You can think of it as the derivatives of curves passing through that point, evaluated at that point. Sorry about the confusing phrasing.

 

[Kreizhn]

Yeah, I was just making sure that was

$$\left. H = c'(s) \right|_{s=0}$$

rather than everywhere, since otherwise we're back in the same problem. I think it all works. Thanks.