# Covariant derivative of horizontal lift

Suppose $$\pi\colon (\tilde{M},\tilde{g})\to(M,g)$$ is a Riemannian submersion. If $$Z$$ is a vector field on $$M$$, denote its horizontal lift by $$\tilde{Z}$$. Now, consider a curve $$\gamma$$ on $$M$$ starting at $$p$$. We can lift this curve to a horizontal curve $$\tilde\gamma$$ starting at some $$\tilde{p}\in\pi^{-1}(p)$$ - essentially by lifting the velocity vector field of $$\gamma$$ pointwise via the isomorphism $$\mathrm{d}\pi_{\tilde{p}}|_{H_{\tilde{p}}}\colon H_{\tilde{p}}\to T_{\pi(\tilde{p})}M$$ and then solving the flow equation by Picard-Lindelöff.

If V is a vector (tensor) field along $$\gamma$$ then we can also lift it to a vector (tensor) field $$\tilde{V}$$ along $$\tilde{\gamma}$$ by attaching to each $$\tilde{\gamma}(t)$$ the vector (tensor) $$\mathrm{d}\pi_{\tilde{\gamma}(t)}|_{H_{\tilde{\gamma}(t)}}^{-1}(V_{\gamma(t)})$$.

Exercise 5.6 (b) in Lee's Riemannian manifolds shows that for lifts of vector fields $$\tilde{\nabla}_{\tilde{X}}{\tilde{Y}} = \widetilde{\nabla_{X}{Y}} + \tfrac12[\tilde{X},\tilde{Y}]^{V}$$ where the superscript $$V$$ denotes the projection onto the vertical tangent bundle.

## Question 1:

Is there a corresponding statement for the covariant derivative along $$\gamma$$ and its lift?
If I can locally extend the vector field $$\gamma^{\prime}$$ and $$V$$ downstairs then I can use Lee's formula for the lifts. Of course, that's not always possible. So my guess would be $$\tilde{D_t}{\tilde{V}} = \widetilde{D_t{V}} + \tfrac12[\tilde{\gamma}^{\prime},\tilde{V}]^{V}$$ but I don't know if a commutator of vector fields along a curve makes sense or how it acts on functions.

## Question 2:

Is the following true at least for geodesics $$\gamma$$ downstairs?
For those I would guess that the 'commutator' disappears and it should simply be: $$\tilde{D_t}{\tilde{\gamma}^{\prime}} = \widetilde{D_t{\gamma^{\prime}}} = 0.$$ Is this true? That is, the lifted curve only has vertical curvature. Or in other words: Horizontal geodesics upstairs are horizontal lifts of geodesics downstairs.

## Edit:

For what it's worth I can show that the horizontal part of $$\tilde{D}_{t}\tilde{V}$$ is the lift of $$D_{t}V$$ by showing that $$\mathrm{d}\pi\circ\tilde{D}_{t}\tilde{V}$$ fulfils the three defining properties of the covariant derivative along the curve $$\gamma$$ downstairs. Uniqueness of the covariant derivative operator then tells me that $$\mathrm{d}\pi\circ\tilde{D}_{t}\tilde{V} = D_{t}V$$ for all vector fields $$V\in\mathfrak{X}(\gamma)$$. Or in other words: $$(\tilde{D_t}{\tilde{V}})^{H} = \widetilde{D_t{V}}.$$ This also shows that horizontal geodesics upstairs descend to geodesics downstairs. Question 1 about the vertical part of $$\tilde{D_t}{\tilde{V}}$$ still remains open though.

It can be seen that $$[\tilde{X}, \tilde{Y}]^V$$ is tensorial (i.e. $$[\tilde{X}, \tilde{Y}]^V$$ depends only on $$\tilde{X}(p)$$ and $$\tilde{Y}(p)$$, not on the local behavior of the vector fields). Because of this, it follows that $$[\tilde{\gamma}', \tilde{V}]^V$$ is well-defined, and in particular $$[\tilde{\gamma}', \tilde{V}]^V(t) = [\tilde{X}, \tilde{Y}]^V(\tilde{\gamma}(t))$$ for any horizontal vector fields $$\tilde{X}, \tilde{Y}$$ such that $$\tilde{X}(\tilde{\gamma}(t)) = \tilde{\gamma}'(t)$$ and $$\tilde{Y}(\tilde{\gamma}(t)) = \tilde{V}(t)$$.