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I asked this question on MSE, but I didn't receive a response yet, so I'm asking here. Apologies if the question is not exactly a research level question, but I'm having some trouble in figuring them out myself. Hints or solutions would be appreciated, as well as resources (book chapter etc.). I did an internet search on this one, but didn't see any affirmative answer.

Let $\pi:(M,g)\to (N,h)$ be a surjective Riemannian submersion,

  • i.e. $\forall p\in M, D\pi_p$ is surjective between the respective tangent spaces and that .
  • $T_pM=H_pM \oplus V_pM$ ($g_p$-orthogonal direct sum), and
  • $H_pM$ is isometric to $T_qN, q:=\pi(p)$ via $D\pi_p, i.e. h_q(D\pi_p(v), D\pi_p(w)=g_p(v,w)\forall v,w \in H_pM.$ (this defines the Riemannian submersion part, isometry of the horizontal part of the tangent space with the tangent space of the image/quotient).

Questions:

  1. Is it true that $\pi$ maps horizontal geodesics in $M$ to geodesics in $N?$ Perhaps relevant is this MO question: initially horizontal geodesics are always horizontal. I can't help thinking the way we show that an isometry $\phi$ maps geodesics to geodesics: the idea is to show for vectore fields $X,Y$ that $\phi_{*}(\nabla^M_X {Y})= \nabla^N_{\phi_{*}X}{\phi_{*}Y}.$ (John Lee: Riemannian manifolds: an introduction to curvature, P.71). Should we show the same here for $X, Y$ horizontal vector fields?

  2. Is it true that $\forall v \in H_pM, exp^N_q(D\pi_p(v))= \pi(exp_p^M(v)),$ where $exp^M, exp^N$ represent the corresponding exponential maps? I think this can be proven if 1) is proved, and noting that the initial velocity of the geodesic (if 1) is proven) $t\mapsto \pi(exp_p^M(tv))$ is indeed $D\pi_{p}(v),$ and then using the uniqueness of geodesic with initial point and initial velocity.

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Yes, this is true. There is a DG-style proof, but let me do it metrically. (The following proof works only in the Riemannian world; the DG-style proof should work in pseudo-Riemannian as well.)

Choose a geodesic $\gamma$ in $N$; let $\tilde \gamma$ be its horizontal lifting in $M$. Passing to a subinterval, we can assume that $\gamma$ is length-minimizing. Note that in this case so is $\tilde \gamma$; in particular $\tilde \gamma$ is a geodesic.

Now, choose a point $p\in M$ and a horizontal direction $ \xi$ at $p$. We may choose $\gamma$ running from $\pi(p)$ in the direction of $d\pi(\xi)$. Since geodesics cannot bifurcate, the above argument implies that if $\tilde \gamma$ is an $M$-geodesic in the direction $\xi$, then it is horizontal. Moreover $\gamma=\pi\circ\tilde\gamma$.

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  • $\begingroup$ @Prof. Petrunin: thank you for your answer! I'd appreciate if you could please answer this follow up questions, related to better understanding of your answer. My first question is: which theorem/result guarantees the existence of a lift since $\tilde{M}$ isn't covering space (intervals are simply connected, so such a lift exists for a covering space)? Next, assuming (upon proving) such a lift $\tilde{\gamma}$ of $\gamma$ does exists, I guess your second paragraph shows that this lift must be i) horizontal and ii) locally length minimizing? (continued to the next paragraph). $\endgroup$
    – Learning math
    Commented Mar 20 at 12:47
  • $\begingroup$ (contd.) To show $\tilde{\gamma}$ is locally length minimizing, and to conclude the part "note that in this case, so (locally length minimizing) is $\tilde{\gamma},$" you use the fact that $D\pi_{\tilde{\gamma(t)}}:T_{\tilde{\gamma(t)}}M\to T_{\gamma(t)}N$ is a linear isometry, correct? I ask this because to calculate the length of a horizontal lift $\tilde{\gamma}$ in $M$ and to show that the length remains the same after projecting it to $\gamma \subset N,$ we need to integrate $||\tilde{\gamma(t)}||=||\gamma'(t)||,$ where the last equality follows from $D\pi$ being a linear isometry. $\endgroup$
    – Learning math
    Commented Mar 20 at 13:41
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    $\begingroup$ For the first part: by definition of submersion, any vector field on $N$ can be lifted to a horizontal field on $M$ --- it remains to extend tangent field along $\gamma$ to a field on $N$, lift it and take integral curve. For the second part: note and use that Riemannian submersion is 1-Lipschitz. $\endgroup$
    – Anton Petrunin
    Commented Mar 24 at 21:18

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