Let $M$ be a closed orientable Riemannian manifold. Recall that a plane field on a Riemannian manifold is said to be geodesic if any geodesic tangent to the plane field at one point is tangent to it at every point. Is it true that if $E\subset TM$ a one-dimensional geodesic plane field, then there exists $X\in \Gamma(E)$ such that $X$ is a Killing vector field? where $\Gamma$ means the section of a bundle.

How about $E$ has dimension $\ge 2$?


(I assume by a plane field you mean a distribution.)

No, take $\mathbb{T}^3$ parametrized as $(x,y,z) \in [0,2\pi)^3$. The field $v = \partial_x + \sin(z) \partial_y$ is geodesic. But for any $\phi(x,y,z)$ the deformation tensor $\mathcal{L}_{\phi v} g$ can be computed to be $$ \begin{pmatrix} 2 \phi_x & \phi_y + \phi_x \sin(z) & \phi_z \newline \phi_y + \phi_x \sin(z) & 2\phi_y \sin(z) & \phi_z \sin(z) + \phi \cos(z) \newline \phi_z & \phi_z \sin(z) + \phi \cos(z) & 0 \end{pmatrix} $$ which can only vanish if $\phi_x = \phi_y = \phi_z = 0$ (from the (1,1), (2,2) and (1,3) components) and $\phi \equiv 0$ (from the (2,3) component).

For distributions of higher dimension, you can observe that for any complete Riemannian manifold $TM$ as a top-dimensional distribution is geodesic. So it suffices to find a Riemannian manifold of the appropriate dimension that does not admit any Killing vector field to get a counterexample. Alternatively you can also modify the above one dimensional construction in the obvious way.

| cite | improve this answer | |

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.