9
$\begingroup$

Let $X,Y$ be Riemannian manifolds, and $f\colon X\to Y$ be a Riemannian submersion. Let $\gamma$ be a geodesic on $X$ starting at a point $x\in X$ and which is orthogonal to the fiber $f^{-1}(f(x))$.

Questions. (1) Is it true that $\gamma$ is orthogonal to any fiber of $f$ which it intersects?

(2) If this is not true in general, is it true under the extra assumption that $f$ has totally geodesic fibers?

Remark. I think this is true in the special case when $X=G$ is a compact Lie group with a bi-invariant metric, and $Y=G/H$ is its homogeneous space with the only metric such that the canonical map $f\colon G\to G/H$ is a Riemannian submersion. (In this case $f$ does have totally geodesic fibers.)

$\endgroup$
  • $\begingroup$ The answer to (1) is definitely "no"; morally, orthogonality is a closed condition, while being a submersion is an open condition. If you take any submersion for which it is true and perturb it a little (along a geodesic starting at $x$, but away from $x$), you get a counterexample. $\endgroup$ – Marco Golla Jan 25 '15 at 12:51
  • $\begingroup$ @MarcoGolla, but if a geodesic is horizontal at one point then it is horizontal at all points, no? $\endgroup$ – Paul Reynolds Jan 25 '15 at 13:14
  • $\begingroup$ @MarcoGolla: Riemannian submersion is not so straightforward/easy to perturb. $\endgroup$ – Alex Degtyarev Jan 25 '15 at 13:16
  • $\begingroup$ Ah, I see.. I hadn't noticed that "Riemannian". Sorry about that. $\endgroup$ – Marco Golla Jan 25 '15 at 13:49
6
$\begingroup$

Take horizontal lift $\gamma$ of the minimal geodesic $\bar\gamma$ in Y, connecting two points $p$ and $q$ close to each other. It is (minimal) geodesic, since any other curve connecting them is not shorter than its horizontal projection, which in turn not shorter than minimal $\bar\gamma$. Since in each direction we can issue only one geodesic - any geodesic normal to the fiber is a horizontal lift. Which in turn implies that such geodesics stay normal to fibers. So, the answer is "yes".

$\endgroup$
4
$\begingroup$

Yes. For a printed proof see 26.12 of here.

$\endgroup$
  • $\begingroup$ Haha, the book link comes two days late for me! I have ordered this treasure at amazon. :-) $\endgroup$ – Martin Gisser Jan 27 '15 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.