Let $(\mathcal{M},g)$ be a torsion free compact Riemannian manifold of dimension $n$. Hence from the metric we know there is an associated horizontal sub-bundle $H_u F \mathcal{M}$ of the orthonormal frame bundle $F \mathcal{M}$ at a frame $u:\mathbb{R}^n\to T_p\mathcal{M}$. Furthermore there are theorems that state that for each curve $\gamma:[0,1]\to\mathcal{M}$ and initial frame $u_0$ there is a unique horizontal lift $\tilde{\gamma}:[0,1]\to F \mathcal{M}$.

Question 1

What is the expression for the integrand of the energy functional of a curve $\gamma:[0,1]\to\mathcal{M}$, $$ g_{\gamma}(\dot{\gamma},\dot{\gamma}) $$ in terms of its unique horizontal lift $\tilde{\gamma}:[0,1]\to F \mathcal{M}$? Does the answer depend on $u_0$? Since there is also then a uniquely defined curve in $w:[0,1]\to\mathbb{R}^n$ (which I guess is chart independent), what is the value of $g_{\gamma}(\dot{\gamma},\dot{\gamma})$ via $w$? Could it be just simply $$ g_{\gamma}(\dot{\gamma},\dot{\gamma}) = \langle\dot{w},\dot{w}\rangle_{\mathbb{R}^n} $$?

What is an expression for $\gamma$ in terms of $w$? (I know that $ \dot{\gamma} = \tilde{\gamma}\dot{w}$, but how to solve this for $\gamma$ since there is the lift on the RHS?).

Question 2

It seems like in construction stochastic processes on $\mathcal{M}$ the orthonormal frame bundle is heavily used because one can solve stochastic equations in $\mathbb{R}^n$, move them to $F \mathcal{M}$ in a straight forward manner, and then project them down to $\mathcal{M}$. This is, if I understand correctly, the essence of the Eells–Elworthy–Malliavin construction. My question is then, why not define Brownian motion in $\mathcal{M}$ via charts, first down at the Euclidean space and then pulling back to the manifold using the chart? I guess there would be a way to glue the curve back together.

Is the frame bundle really just a chart-free way to talk about the manifold's Euclidean structure in a way that is compatible with the metric?


On the orthonormal frame bundle we have soldering forms $\omega_i$ and connection forms $\omega_{ij}$. A lift is horizontal just when $\omega_{ij}=0$ on it. So the velocity can be described by its $\omega_i$ components: $v_i(t)=i_{\tilde\gamma'(t)}\omega_i$. The energy is $\sum_i v_i^2$. You don't define $w$, but I guess it is the development, which is coordinate independent but depends on a choice of initial point and frame on Euclidean space. Its energy has the same expression, also equal to $\left<\dot w,\dot w\right>$. To solve for $\gamma$ in terms of $w$, first find the components $v_i=i_{\tilde w'}\omega_i$ on the frame bundle of Euclidean space. Then solve $i_{\tilde\gamma'(t)}\omega_i=v_i(t)$ and $i_{\tilde\gamma'(t)}\omega_{ij}=0$. Then project to find $\gamma(t)$. The frame bundle avoids charts, and allows simpler algebraic expressions for curvature.

Here the map $i_v \xi$ is application of a tangent vector $v$ into a $1$-form $\xi$.

Note that since the frame bundle of Euclidean space is a product of Euclidean space with the rotation group, every lift of a curve $w(t)$ in Euclidean space is a curve $\tilde w(t)=(w(t),E)$ where $E$ is any constant rotation matrix. If the matrix $E$ has columns $E_1,\dots,E_n$, then $v_i(t)=\left<E_i,\dot w(t)\right>$.

Question 2: Brownian motion does not stay inside a single chart for any positive time; it has some chance of escaping. You can see this already in Euclidean space, as the scale invariance of Brownian motion.

  • $\begingroup$ Thanks for your answer. I hesitate to mark it as accepted because it doesn't really address Question 2, but if you think I should accept it anyway I will. Can you say more about the map $i$ which you refer to? What is a good reference for this? $\endgroup$ – PPR May 4 '20 at 21:44

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