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Let $\mathbb{M}$ be a $n+m$ dimensional manifold. Consider on $\mathbb{M}$ a rank $n$ sub-bundle $\mathcal{H}$ of the tangent bundle. We assume that $\mathcal{H}$ is endowed with a fiber wise inner product $g_\mathcal{H}$.

Under which conditions on $(\mathbb{M}, \mathcal{H}, g_\mathcal{H})$ can we find a totally geodesic Riemannian foliation $(\mathcal{F},g)$ on $\mathbb{M}$ such that:

1) $g$ is bundle-like,

2) $g$ restricts to $g_\mathcal{H}$ on $\mathcal{H}$

3) The leaves of the foliation are orthogonal to $\mathcal{H}$ ?

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  • $\begingroup$ There is a quite recent article on the subject: arxiv.org/abs/1511.05830 (Remark 3.2 a)) $\endgroup$
    – Llohann
    Commented May 27, 2016 at 21:19

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Here is a necessary condition. I have not checked if it is sufficient.

If $\mathcal F$ is totally geodesic, then translation along horizontal paths give local isometries, so you have the condition $$\mathcal L_H g^{\mathcal F}=0\;.$$ Here, $H$ is a horizontal vector field that looks like a horizontal lift of a vector field on the base, so it is parallel along leaves with respect to the Levi-Civita connection, and $g^{\mathcal F}$ is the restriction of the total metric to $\mathcal F$. We have of course used that the flow of $H$ maps local leaves to local leaves because $g$ is bundle-like.

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    $\begingroup$ I admit that this argument implies that you have already found some $g$. You probably need some other way to get the relevant information. Maybe, Kamber-Tondeur classes help. I will add something to the answer above if I can make that more precise. $\endgroup$
    – Sebastian Goette
    Commented May 20, 2016 at 9:04

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