# Does a Riemannian submersion map geodesics to geodesics?

Let's suppose I have a Riemannian submersion (projection) $$\pi: G \rightarrow G/K$$, with $$G$$ a matrix Lie group and where $$G/K$$ is a reductive homogeneous space. I have a decomposition of Lie algebras $$\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{m}$$.

I know how geodesics look like in a reductive homogeneous space. This is in the book of O'Neill. If we write $$o := eK$$, they are of the form $$\gamma_{d\pi X} (t) = \alpha(t) o = \pi (\alpha(t))$$ where $$\alpha(t)$$ is the one-parameter subgroup of $$X \in \mathfrak{m}$$.

I also know that horizontal geodesics map under $$\pi$$ to geodesics.

But let's say I have a geodesic in $$G$$, this is of the form $$e^{tX}$$, where $$X \in \mathfrak{g}$$. Is the image of this geodesic under $$\pi$$ also a geodesic? Or do I have to figure out what $$\pi$$ does with the vertical component?

• Be careful. In general, one-parameter subgroups are not geodesics. Dec 2, 2020 at 19:05

There is no reason for this to be true. If you are looking for a counter example, consider $$G=SL(2,\mathbb{R})$$ and $$X=\begin{pmatrix}1&1\\-1&-1\end{pmatrix}\in \mathfrak{sl}(2,\mathbb{R})$$. Then the Cartan decomposition of $$X$$ is $$X=\begin{pmatrix}0&1\\-1&0\end{pmatrix}+\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ Note that $$\pi_\ast(Id)$$ has kernel $$\mathfrak{l}$$. So the geodesic passing through $$eK$$ in the direction $$\pi_\ast(Id)(X)$$ is $$t\to \pi\begin{pmatrix}e^t&0\\0&e^{-t}\end{pmatrix}$$. Also note that $$t\to \pi(e^{tX})$$, is not this geodesic, as at time $$t=1$$ it can be checked that their difference is not in $$K$$. For this, note that $$e^X=Id+X$$, as $$X$$ is nilpotent.