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.


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.


1 Answer 1


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 $\tilde{Y}(\tilde{\gamma}(t)) = \tilde{V}(t)$.

I recommend taking a look at this paper for more insight.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .