1-parameter group of a vector field

Question

Let $(M,g)$ be a Riemannian manifold and $\nabla$ be the Levi-Civita connection of $g$ and let $X,Y$ be vector fields on $M$. If $\lbrace \phi _t \rbrace $ is the 1-parameter group of $X$ then what is the relation between $\nabla _YX$ and $\phi _{t*}Y$($\phi _{t*}$ is the differential of $\phi _t$)?

Answer 1

The Levi-Civita connection has torsion zero so that

$$ \nabla_XY- \nabla_Y X= [X,Y] . $$

On the other hand, $[X,Y]$ is the Lie derivative of $Y$ along $X$. Arnold called this the ``fisherman's derivative'' due to its description in terms of the action of the flow $\Phi^t$. (Note that I use the time $t$ as a superscript for typographical reasons, and not only.) More precisely

$$ [X, Y]_{p_0}=- \lim_{t\to 0} \frac{1}{t}\Bigl(\,\Phi^t_* Y(p_{-t})- Y(p_0)\,\Bigr),\;\;p_0\in M,\;\;p_t= \Phi^t(p_0),\;\;\forall t\in\mathbb{R}. $$

For a proof of the last equality see section 3.1.1 of these notes.

Answer 2

Let $C(X)$ be the horizontal lift (a vector field on $TM$) of the vector field $X$ with respect to the the Levi Civita connection. Then the flow $Fl^{C(X)}_t$ is parallel transport along each flow line of $Fl^X_t$, and we have $$ \nabla_ XY = \frac{d}{dt}|_0 (Fl^{C(X)}_{-t}\circ Y \circ Fl^X_t). $$ For a proof see 24.2 of here

Answer 3

Denote $\phi_t = e^{t X}$. By definition of Lie derivative of a vector field

$$ \left.\frac{d}{dt}\right|_{t=0}e^{-t X}_* Y = [X,Y] = \nabla_X Y - \nabla_Y X.$$

The second equality holds since the Levi-Civita connection has zero torsion.