How to prove two curves in the frame bundle to project to the same curve on base manifold? There is a problem about Cartan's development, arising from the paper 'Kinetic Brownian motion on Riemannian manifolds', Subsection 2.4.1. To be precise, let $(M,g)$ be a $d$-dimensional complete Riemannian manifold, $\pi:OM\to M$ be its orthonormal frame bundle with structure group $O(d)$ and equipped with the Riemannian connection. Denote by $H_v$ the standard horizontal vector field on $OM$ corresponding to $v\in\mathbf R^d$, uniquely characterized by the property that $\pi_*(H_v(z)) = e(v)$ for all $z = (x,e) \in OM$. Let $\{\epsilon_1,...,\epsilon_d\}$ be the canonical basis of $\mathbf R^d$, with dual basis $\{\epsilon_1^*,...,\epsilon_d^*\}$. Denote by $V_i, 1\le i\le d$ the vertical vector field induced by $a_i=\epsilon_i \otimes \epsilon_1^* - \epsilon_1 \otimes \epsilon_i^* \in o(d)$.
Given a smooth curve $\{m_t\}_{0\le t\le1}$, define the Cartan's development of $\gamma$ on $OM$ as the solution to the ODE on $OM$,
\begin{equation}\tag{1}
\dot z_t = H_{\dot m_t}(z_t), \quad z_0 = (x_0,e_0)\in OM.
\end{equation}
Now Assume $\{m_t\}_{0\le t\le1}$ is run at unit speed, i.e., $|\dot m_t|\equiv 1$. Then, given an orthonormal basis $f_0$ of $\mathbf R^d$ with $f_0(\epsilon_1) = \dot m_0$, solve the following ODE on $SO(d)$,
\begin{equation}
\dot f_t = \sum_{i=2}^d (f_t(\epsilon_i),\ddot m_t)a_i(f_t),
\end{equation}
started from $f_0$, and define the $\mathbf R^{d-1}$-valued path $\{h_t\}_{0\le t\le1}$, starting from zero, by the ODEs
\begin{equation}
\dot h^i_t = (f_t(\epsilon_i),\ddot m_t), \quad 2\le i\le d.
\end{equation}
Consider the following ODE on $OM$,
\begin{equation}\tag{2}
\dot{\tilde z}_t = H_{\epsilon_1}(\tilde z_t)+ \sum_{i=2}^d V_i(\tilde z_t) \dot h_t^i, \quad \tilde z_0 = (x_0,e_0)\in OM.
\end{equation}
Then the paper, mentioned in the very beginning, has the following claim: 

Claim: $\pi(\tilde z_t) = \pi(z_t)$.

But why?

I try to prove this claim. But I am not able to finish that.
Use the coordinate system $(x^i,e_l^k)$ on $OM$. Then 
\begin{align}
H_v &= v^j e_j^i \partial_{x^i} - v^r \Gamma^k_{ij} e_l^j e_r^i \partial_{e_l^k}, \\
V_i &= e_i^k \partial_{e^k_1} - e_1^k \partial_{e_i^k}.
\end{align}
On the one hand, Eqn. (1) is represented as
\begin{equation}\left\{
\begin{aligned}
\dot{x}^i &= e_j^i \dot m^j, \\
\dot e_l^k &= -\Gamma^k_{ij} e_l^j e_r^i \dot m^r, 
\end{aligned}
\right.
\end{equation}
where $\Gamma^k_{ij}$ are the Christoffel's symbols of the metric $g$. We can obtain
\begin{equation}
\ddot x^i = \dot e_j^i \dot m^j + e_j^i \ddot m^j = -\Gamma^i_{kl} e_j^l e_r^k \dot m^r \dot m^j + e_j^i \ddot m^j = -\Gamma^i_{kl} \dot x^l \dot x^k + e_j^i \ddot m^j,
\end{equation}
that is,
\begin{equation}\tag{1*}
\frac{\nabla \dot x^i}{dt} = e_j^i \ddot m^j.
\end{equation}
On the other hand, Eqn. (2) is represented as
\begin{equation}\left\{
\begin{aligned}
\dot{\tilde x}^i &= \tilde e_1^i, \\
\dot{\tilde e}_1^k &= -\Gamma^k_{ij} \tilde e_1^j \tilde e_1^i + \sum_{i=2}^d \tilde e_i^k \dot h^i, \\
\dot{\tilde e}_l^k &= -\Gamma^k_{ij} \tilde e_l^j \tilde e_1^i - \tilde e_1^k \dot h^l, \quad 2\le l \le d.
\end{aligned}
\right.
\end{equation}
We have
\begin{equation}
\ddot{\tilde x}^i = \dot{\tilde e}_1^i = -\Gamma^i_{kj} \tilde e_1^j \tilde e_1^k + \sum_{k=2}^d \tilde e_k^i \dot h^k = -\Gamma^i_{kj} \dot{\tilde x}^j \dot{\tilde x}^k + \sum_{k=2}^d \tilde e_k^i \dot h^k,
\end{equation}
that is,
\begin{equation}\tag{2*}
\frac{\nabla \dot{\tilde x}^i}{dt} = \sum_{k=2}^d \tilde e_k^i \dot h^k.
\end{equation}
If (1*) and (2*) are the same ODE, then under the same initial condition $x(0) = \tilde x(0) = x_0$, we have $x=\tilde x$, which proves the claim. But I do not know how to compare (1*) and (2*). 
Can anyone give some hints or reference? TIA...
PS: This is a crosspost from math.stackexchange.
 A: If I understand correctly, you're essentially trying to do the following. I've used my own notation, because I don't completely understand yours.
Let $M$ be a smooth Riemannian $d$-manifold and $OM$ its orthonormal frame bundle. Let $I$ be a connected open interval containing $0$.
Given any curve $m: I \rightarrow M$, $A: I \rightarrow so(d)$, and $f_0 \in O_{m(0)}M$, there exists a unique lift $z = (m,f): I \rightarrow OM$ such that
  $$
    f' = fA,\ f(0) = f_0,
  $$
where $f' = \nabla_{m'}f$, and functions $w: I \rightarrow \mathbb{R}^d$ and $h: I \rightarrow \mathbb{R}^d$ satisfying
$$
  m' = fw\text{ and }m'' = fh,
$$
where $m'' = \nabla_{m'}m'$. Therefore,
\begin{align*}
  fh &= m''\\
  &= (fw)'\\
  &= f'w + fw'\\
  &= f(Aw + w'),
\end{align*}
it follows that
$$
  w' + Aw = h.
$$
Conversely, if $h: I \rightarrow \mathbb{R}^d$, $A: I \rightarrow so(d)$, and $f_0 \in O_{m(0)}M$ are the same as above and $w_0 \in \mathbb{R}^d$ satisfies $m'(0) = f_0w_0$, then
there exists a unique lift $\tilde{z} = (\tilde{m},\tilde{f}): I \rightarrow OM$ and $\tilde{w}: I \rightarrow \mathbb{R}^d$ satisfying
\begin{align*}
  \tilde{w}' + A\tilde{w} &= h\\
  \tilde{f}' &= \tilde{f}A\\
  \tilde{m}' &= \tilde{f}\tilde{w},
\end{align*}
where $\tilde{f}' = \nabla_{\tilde{m}'}\tilde{f}$, with the initial conditions
\begin{align*}
  \tilde{w}(0) &= w_0\\
  \tilde{f}(0) &= f_0\\
  \tilde{m}(0) &= m(0).
\end{align*}
Since $w, f, m$ also solve this initial value system, it follows that they are equal to $\tilde{w}, \tilde{f}, \tilde{m}$.
A: Method 1: Let us think about it reversely. 

We want to find a curve $\tilde z = \{z_t\}$ on $OM$ such that $\pi(\tilde z_t) = x_t$ and the horizontal component of the tangent vector field of $\tilde z$ is corresponding to the constant vector $\epsilon_1\in\mathbb R^d$, that is, 
  \begin{equation}\tag{*}
\tilde z_t(\epsilon_1) = \pi_*(H_{\epsilon_1}(\tilde z_t)) = \pi_*(\text{h}(\dot{\tilde z}_t)) = \pi_*(\dot{\tilde z}_t) = \dot x_t,
\end{equation}
  where $\text{h}$ means the horizontal component of vector fields on $OM$. Intuitively, the last equation means that the first basis vector of $\tilde z$ equals to $\dot x$.

The aim is to find the ODE for $\tilde z$. Assume 
\begin{equation}\tag{a}
\tilde z = zf
\end{equation}
for some curve $f=\{f_t\}\subset O(d)$. Then using $z_t(\dot m_t) = \pi_*(H_{\dot m_t}(z_t)) = \pi_*(\dot z_t) = \dot x_t$, it's easy to see that (*) holds if 
\begin{equation}\tag{b}
f_t(\epsilon_1) = \dot m_t.
\end{equation}
Write $f$ in component as $\{f_i^j\}$. Then in local coordinates, (*), (a) and (b) are equivalent to
\begin{align}
\dot x^i &= \tilde e_1^i, \tag{c}\\
\tilde e_l^k &= e_n^k f_l^n, \tag{d} \\
f_1^r &= \dot m^r. \tag{e}
\end{align}
Suppose $f$ is the integral curve of the left-invariant vector field corresponding to some $A\in o(d)$. Then 
$$\dot f = A_f = d(L_f)_{I_d}(A) = f_i^j A_k^i \frac{\partial}{\partial f_k^j},$$
or in local coordinates,
$$\dot f_k^j = f_i^j A_k^i.\tag{f}$$
Take derivative on both sides of (d),
\begin{equation}
\begin{split}
\dot{\tilde e}_l^k = \dot e_n^k f_l^n + e_n^k \dot f_l^n &= -\Gamma^k_{ij} (e_n^j f_l^n) (e_r^i \dot m^r) + e_n^k f_j^n A_l^j  \\
&= -\Gamma^k_{ij} \tilde e_l^j \dot x^i + \tilde e_j^k A_l^j \\
&= -\Gamma^k_{ij} \tilde e_l^j \tilde e_1^i + \tilde e_j^k A_l^j.
\end{split}
\end{equation}
Then combining this with (c), it's easy to see that the curve $\tilde z=(x,\tilde e)$ satisfies
$$\dot{\tilde z}_t = H_{\epsilon_1}(\tilde z_t)+ A^*(\tilde z_t),\tag{g}$$
where we denote by $A^*$ the fundamental vertical vector field corresponding to $A$, it has the following representation in local coordinates
$$A^* = A_l^j e_j^k \partial_{e^k_l}.$$
Now we figure out the ODE that $A$ and $f$ satisfies. By (e) and (f), we have
$$\ddot m^r = \dot f_1^r = f_j^r A_1^j.$$
Using the fact that $f\in O(d)$, we get
$$A_1^i = \sum_{r=1}^d f_i^r \ddot m^r = (f(\epsilon_i),\ddot m).$$
Note that $A_1^1 = \sum_{r=1}^d f_1^r \ddot m^r = \sum_{r=1}^d \dot m^r \ddot m^r = \frac{d}{dt}|\dot m|^2 = 0$ by virtue of the assumption $|\dot m|\equiv1$. This agrees with the fact that $A\in o(d)$.
Using again (f),
$$\dot f_1^j = f^j_i A_1^i = \sum_{i=2}^d f^j_i A_1^i = \sum_{i=2}^d (f(\epsilon_i),\ddot m)f^j_i.\tag{h}$$
An important observation is that we can only determine the entries $A_1^i$ and $f_1^j$, hence the choosing of $A$ and $f$ is not unique, and consequently, $\tilde z$ is not unique. The simplest way of choosing $A$ is that $A_i^1 = - A_1^i$ for $i=2,...,d$ and all other $A_i^j$'s are zero, that is, $A = A_1^i a_i$, where $a_i=\epsilon_i \otimes \epsilon_1^* - \epsilon_1 \otimes \epsilon_i^* \in o(d)$. In this case, $A^* = A_1^i V_i$ where $V_i$ is the vertical vector field induced by $a_i$, and for $k=2,...,d$,
$$\dot f_k^j = f_i^j A_k^i = f_1^j A_k^1 = -f_1^j A^k_1 = (f(\epsilon_k),\ddot m)f^j_1.\tag{i}$$
To summarise up (h) and (i), we get
\begin{equation}
\dot f_t = \sum_{i=2}^d (f_t(\epsilon_i),\ddot m_t)a_i(f_t).
\end{equation}
Note that $\dot h^i$ is nothing but $A_1^i$.

Method 2: First assume (a) and (g). Then (cf. Kobayashi & Nomizu, Proposition II.3.1)
$$\dot{\tilde z} = \dot z f + z \dot f.$$
Let $\omega$ be the connection form of the given Riemannian connection on $OM$. Then,
$$\omega(\dot{\tilde z}) = \mathrm{Ad}(f^{-1})\omega(\dot z) + d(L_{f^{-1}})(\dot f) = d(L_{f^{-1}})(\dot f).$$
On the other hand, by (g)
$$\omega(\dot{\tilde z}) = \omega(A^*) = A.$$
Hence,
$$\dot f = d(L_{f})(A),$$
and we get (f) again.

PS: For the converse, i.e., deriving (1) from (2), see Xue-Mei Li's paper 'Random perturbation to the geodesic equation', Lemma 3.1, which is similar to Method 2.
