Vector field induced by parallel translation is smooth

connectionsdifferential-geometryordinary differential equationssmooth-manifolds

I'm studying properties of flat manifolds, and I've come across the following lemma (from "Introduction to Riemannian Manifolds" by John M. Lee):

Suppose $M$ is a smooth manifold, and $\nabla$ is any connection on $M$ satisfying
$$
\nabla_X \nabla_Y Z – \nabla_Y \nabla_X Z = \nabla_{[X,Y]}Z, \qquad \forall \: X, Y, Z \in \mathcal{X}(M).
$$

Then given $p \in M$ and any vector $v \in T_pM$, there exists a parallel vector field $V$ on a neighborhood of $p$ such that $V_p = v$.

Lee constructs the vector field in the following way: define a coordinate cube $C_\epsilon = \{(x^1, \ldots x^n) : |x^i| < \epsilon \: \forall i\} \subset M$ centered at $p \in M$. Parallel-transport $v$ along the $x^1$-axis, then from each point $(c^1, 0, \ldots, 0)$ on the $x^1$ axis, parallel-transport the resulting vector along the coordinate curve $t \mapsto (c^1, t, 0, \ldots, 0)$ "parallel" to the $x^2$-axis. Then do the same thing to along the curve $t \mapsto (c^1, c^2, t, 0, \ldots, 0)$, etc. This eventually defines a rough vector field $V$ on the coordinate cube $C_\epsilon$.

Claim: The vector field $V$ defined like so is smooth.

What I've tried: This supposedly follows from the existence and uniqueness theorem for systems of linear differential equations:

Let $I \subset \mathbb R$ bea n open interval, and for $1 \leq j, k \leq n$, let $A_j^k : I \to \mathbb R$ be smooth functions. For all $t_0 \in I$ and every initial vector $c = (c^1, \ldots, c^n) \in \mathbb R^n$, the linear initial value problem
\begin{align*}
\dot V^k(t) &= A_j^k(t) V^j(t) \\ V^k(t_0) &= c^k
\end{align*}

has a unique smooth solution on all of $I$, and the solution depends smoothly on $(t_0,c) \in I \times \mathbb R^n$.

I can show this if $\dim M = 1$. Suppose it's true if $\dim M = n-1$. Let $N = C_{\epsilon} \cap \{x^n = 0\}$. Then $V$ is smooth on $N$. For each curve $\gamma_c(t) = (c^1, \ldots, c^{n-1}, t)$ for $c \in \mathbb R^{n-1}$, parallel-translate $V$ along $\gamma_c(t)$. Since $V$ is parallel along $\gamma_c$, $D_tV \equiv 0$, or
$$
\dot V^k(t) = -\dot\gamma^i_c(t) V^j(t) \Gamma_{ij}^k(\gamma_c(t)) \quad \forall k = 1, \ldots, n,
$$

where $\Gamma_{ij}^k : C_\epsilon \to \mathbb R$ are the Christoffel symbols of $\nabla$ in $C_\epsilon$-coordinates. Since $\dot\gamma_c(t) = \partial_{n}\big|_{\gamma_c(t)}$, we know $\dot\gamma_c^i \equiv \delta_n^i$. So along each curve $t \mapsto (c,t) \in C_\epsilon$, $c \in \mathbb R^{n-1}$, the coordinates of $V$ satisfy the initial value problem
\begin{align*}
\dot V^k(c,t) &= -V^j(c,t) \Gamma_{nj}^k\left(c^1, \ldots, c^{n-1},t\right) \\
V^k(c,0) &= V^k\big|_N(c,0)
\end{align*}

If we could let $A^k_j(t) = -\Gamma_{nj}^k(c,t)$, then we'd be good and the problem would be solved by induction and the existence and uniqueness theorem (since the solution depends smoothly on choice of initial condition).

My problem: I'm not sure the functions $A^k_j$ are allowed to depend on the initial condition. Changing our initial condition in this case also changes the functions $A^k_j$. Does smoothness of $V$ still follow? Alternatively, is there a reason why $\dfrac{\partial}{\partial x^i} \Gamma^k_{nj} \equiv 0$ for $1 \leq i \leq n-1$?

Best Answer

Oy. That's an oversight. The reference should have been to Theorem A.42 in the appendix (the fundamental theorem on flows). You need to apply that theorem to vector fields of the following form on $C_\varepsilon\times \mathbb R^n$: $$ W_k|_{(x,v)} = \frac{\partial}{\partial x^k}-v^i \Gamma^j_{ki}(x)\frac{\partial}{\partial v^j}. $$ I've added a correction to my online list.