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My definition of a left invariant vector field is one for which the equation $dL_xX = X$ is satisfied, where $X$ is the differentiable vector field and $x \in G$ where G is a Lie group.

Consider the two sides of the equation:

$L_y \circ x_t (p) = L_y(x_t(p)) = yx_t(p)$

$x_t \circ L_y (p) = x_t(L_y(p)) = x_t(yp))$

Why are these equal?

user127890
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1 Answers1

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Hint: Let $\theta^{(p)}(t) = yx_t(p)$ and $\psi^{(p)}(t) = x_t(yp)$. Compute both ${\theta^{(p)}}'(0)$ and ${\psi^{(p)}}'(0)$. Then use uniqueness of integral curves.

Jesse Madnick
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  • In my calculation, ${\theta^{(p)}}'(0) =\frac{d}{dt}|{t=0} ~yx_t(p)=yX(p)$, but I don't know what is the $yX(p)$, since $y$ is a element of $G$, and $X(p)$ is a tangent vector at $p$, I don't know what $yX(p)$ is ? Namely, how do they multiply ? Besides, ${\psi^{(p)}}'(0)= \frac{d}{dt}|{t=0} x_t(yp) = X(yp)$. Seemly, I can't get $L_y \circ x_t = x_t \circ L_y$. Could you talk it detail ? I meet this problem when I read the section 2 of chapter 1 of do Carmo's Riemannian Geometry. My knowledge background is very weak. Could you talk it detail ? Thanks very much. – Enhao Lan Apr 07 '21 at 05:59