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.