# 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.

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 $$\tilde{w}, \tilde{f}, \tilde{m}$$.

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.