# Orthonormal frame bundles on a manifold

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 $$v_i(t)=\left$$.
• 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? – PPR May 4 '20 at 21:44