# Orthogonal geodesics to hypersurfaces

+ 6 like - 0 dislike
184 views

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 $f = k$?

This post imported from StackExchange Mathematics at 2016-06-20 15:05 (UTC), posted by SE-user user338358

edited Jun 20, 2016
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.

This post imported from StackExchange Mathematics at 2016-06-20 15:05 (UTC), posted by SE-user rrogers
I should have specified that the metric tensor can be made diagonal.

This post imported from StackExchange Mathematics at 2016-06-20 15:05 (UTC), posted by SE-user rrogers

+ 2 like - 0 dislike

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.

This post imported from StackExchange Mathematics at 2016-06-20 15:06 (UTC), posted by SE-user Anthony Carapetis
answered May 18, 2016 by (20 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.