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Setup:

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 $X, Y$ hotizontal 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.

UPDATE: I just asked the question on MO, as I haven't received a response yet. If requested, I can certainly delete one of my questions.

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1 Answer 1

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Yes, the horizontal lifts of $N$-geodesics are $M$-geodesics if $\pi: M\to N$ is a Riemannian submersion. And yes, a Riemannian submersion commutes with the geodesic flow in horizontal directions.

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  • $\begingroup$ Thanks for your answers, but my question 1) was slightly different, it was asking for a proof that projection of horizontal $M$-geodesics are $N$-geodesics. If the proof is long, can you at least outline it or cite a resource that prove it? I'd like to see the proof technique. $\endgroup$ Commented Mar 19 at 9:01
  • $\begingroup$ I will try to think of a conceptual proof. In the meantime, you can certainly consult a textbook for such basic facts. @LearningMath Your question 1 is not different, because by uniqueness of the geodesic with given initial conditions, a horizonal $M$-geodesic is exactly the same thing as a horizontal lift of an $N$-geodesic. $\endgroup$ Commented Mar 19 at 9:26

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