I was doing Lee's IRM book in page 200,Exercise 7.9 needs to prove the following result:
Let $M$ be a two dimensional Riemannian manifold,pick a local chart
$U_p$ and a tangent vector $v \in T_pM$ then we can construct a vector
field over $U_p$ as follows:First transport $v$ along the $x_1$ axis in $U_p$ then transport them
individually along $x_2$ coordinate line,we get a vector field over
$U_p$,Prove that the given vector field on $U_p$ is smooth.
I try to make it clear ,I do as follows,it's suffices to prove that for each component of the vector field $V^i(x_1,x_2)$ is smooth
If we transport the tangent vector it satifies the equation :$$\dot{V}^k = -\dot{\gamma}^i(t)V^j(t)\Gamma^{k}_{i,j}(\gamma(t))\\V^k(0) = v^k$$
Hence first along each curve,we know vector field is smooth,the hard part is to prove the jointly smooth of $V$ on variable $(x_1,x_2)$.
it seems to rely on the ODE theory which says that solution of $V$ depends smoothly on $t$ and inital data $v^k$,where $v = V(x_1,0)$ hence $V(t,v) $ is smooth function hence $V(t,v) = V(t,V(x_1,0)) = V(t,x_1) = V(x_2,x_1)$ is a smooth function,But the arguement for the jointly smooth above seems not very clear
Best Answer
The idea is to treat the parallel vector field constructed as a integral curve on the tangent bundle.By the fundamental theorem for the flow the integral curve smoothly depend on not only $t$ by also initial condition.
For simplicity let's takeing dim = 2 for example.As the hint says, consider the vector filed $W^{k}_{(p,v)} = \frac{\partial}{\partial x^k} - v^i \Gamma^j_{ki}\frac{\partial}{\partial v^j}$ on the tangent bundle
For $\dim = 2$ first parallel transport along the $x^1$ axis gets the vector field $V(x_1)$(smoothly depend on $x_1$),then take $W^2_{(p,v)}$ as above then consider the integral curve $(x(t),v(t))$ equation :
$$\dot{x}^1(t) = 0 \\\dot{x}^2(t) = 1\\\dot{v}^j(t) = -v^i\Gamma^j_{2 i}$$
with initial condition $(x(0),v(0)) = ((x_1,0),V(x_1,0))$ where $V(x_1,0) = V(x_1)$ which depend smoothly on $x_1$ on both slot.
The integral curve then is the parallel transport vector field.
Since the flow map $\theta (t,(x_0,v_0)) $ depend smoothly on $t$ and $(x_0,v_0)$ while initial condition $(x(0),v(0)) = ((x_1,0),V(x_1,0))$ depend smoothly on $x_1$ by the first construction on the curve.
Therefor the integral curve depend smoothly on $(t,x_1)$ replace $t$ to $x_2$ gives that the parallel vector field on the plane smoothly depend on $(x_1,x_2)$.
For higher dimension, the initial condition will depend smoothly on $(x_1,x_2)$ , and the integral curve will then smoothly depend on $(t,x_1,x_2)$ if we replace $t$ by $x_3$, then we show it smoothly depend on $(x_1,x_2,x_3)$.