There is a problem about Cartan's development, arising from the paper ['Kinetic Brownian motion on Riemannian manifolds', Subsection 2.4.1](https://projecteuclid.org/euclid.ejp/1465067216). 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](https://math.stackexchange.com/questions/3218388/how-to-prove-two-curves-in-the-frame-bundle-to-project-to-the-same-curve-on-base).



How to prove two curves in the frame bundle to project to the same curve on base manifold?
**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](https://books.google.com.tw/books?id=BHp4wPTwaVwC&dq=Kobayashi+Nomizu&hl=zh-CN&sa=X&redir_esc=y))
$$\dot{\tilde z} = \dot z f + z \dot f.$$
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.




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**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](https://projecteuclid.org/euclid.aop/1454423049), which is similar to Method 2.


