Say we have a Riemannian manifold $(M, g)$ with vector field $Y$, obeying:

  1. $g(Y, Y) = 1$; and
  2. the $1$-form $\varphi(X) = g(X, Y)$ is $d$-closed, $d\varphi = 0$.

I know that the integral curves of $Y$ are geodesics, i.e. $D_Y Y = 0$. Does it follow that these geodesics are locally orthogonal to a family of hypersurfaces $f = k$?

  • $\begingroup$ I would do a coordinate transform so that the orbit of the vector field corresponds to x_0, then the remaining coordinates form a subspace. The condition g(Y,Y)=1 and that fact you specified a manifold insures that I can continuously do this. Since it been some time since I studied this I am putting it in as a comment. $\endgroup$
    – rrogers
    May 18, 2016 at 13:03
  • $\begingroup$ I should have specified that the metric tensor can be made diagonal. $\endgroup$
    – rrogers
    May 18, 2016 at 13:19

1 Answer 1


The condition for the orthogonal distribution $Y^\perp$ to be integrable is given by the Frobenius theorem. In this case the most convenient formulation is in terms of the one-form $\varphi$:

$Y^\perp = \ker \varphi$ is tangent to a foliation by hypersurfaces if and only if $\varphi \wedge d \varphi = 0$.

Since you have assumed $d \varphi = 0$, your answer is yes.


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