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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?

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

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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.

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