# Geodesics and Riemannian submersions

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.)

• 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. – Marco Golla Jan 25 '15 at 12:51
• @MarcoGolla, but if a geodesic is horizontal at one point then it is horizontal at all points, no? – Paul Reynolds Jan 25 '15 at 13:14
• @MarcoGolla: Riemannian submersion is not so straightforward/easy to perturb. – Alex Degtyarev Jan 25 '15 at 13:16
• Ah, I see.. I hadn't noticed that "Riemannian". Sorry about that. – Marco Golla Jan 25 '15 at 13:49

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".