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.)



Geodesics and Riemannian submersions
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".


