7
$\begingroup$

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 $\mathcal{H}$ ?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 May 20 '16 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.