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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.