7
$\begingroup$

Is there a non integrable $2$ dimensional distribution $D$ of a $3$ dimensional Riemannian manifold such that the distribution is totally geodesic in the following sense:

Every geodesic whose tangent vector of its intitial point is tangent to the distribution then the tangent vector at all its points is tangent to $D$, too.

$\endgroup$
9
$\begingroup$

Yes, the standard contact structure on the unit three-sphere in $\mathbb{R}^4 = \mathbb{C}^2$, for instance. The Legendrian great circles are the intersections of the sphere with the Lagrangian two-planes.

$\endgroup$
  • $\begingroup$ Thank you and +1 for your answer. To be honest I need some background to understand your answer. I think that the wikipedia link in the other answer can help me to understand the details. I would appreciate if you read my comment to "mathematician", the other answer er. Thanks again for your attention to my question. $\endgroup$ – Ali Taghavi Jul 25 '18 at 7:29
  • $\begingroup$ I am sorry if my question is elementary. $\endgroup$ – Ali Taghavi Jul 25 '18 at 7:39
  • 1
    $\begingroup$ @AliTaghavi. it is a good question. Some years ago Patrick Massot, a student of Giroux, did some nice work on this: projecteuclid.org/euclid.gt/1513800108 $\endgroup$ – alvarezpaiva Jul 26 '18 at 14:37
  • $\begingroup$ Thanks again for your attention to my question and very interesting answer and your reference to the paper of Patrick Massot. $\endgroup$ – Ali Taghavi Jul 27 '18 at 22:10
2
$\begingroup$

Take $\mathbb{R}^3$ with the distribution which is the kernel of the one-form $dz - y dx$. This is the standard example of a contact structure on $\mathbb{R}^3$. See https://en.wikipedia.org/wiki/Contact_geometry .

$\endgroup$
  • $\begingroup$ Thank you and +1 for your answer. Did I understand your answer correctly: Let $\alpha$ be a contact structure with the unique Reeb vector field $R$(as I learn from the link in your answer). Let $X$ be the tangent vector field to our geodesic. We consider a Riemannian metric whose frame is $\{RD,\}$ where $D=\ker \alpha$. Then we have to prove $X.R$ is identically zero provided it is zero at the initial point. But I can not prove this Sould I understand from your answer the following: If we have a contact structure $\alpha$ with the Reeb field R $\endgroup$ – Ali Taghavi Jul 25 '18 at 7:37
  • $\begingroup$ and a Riemannian metric $g$ such that $R$ is $g-$ perpendicular to $\ker \alpha$ then the later distribution is totally geodesic in the sense of my question? $\endgroup$ – Ali Taghavi Jul 25 '18 at 7:38
  • $\begingroup$ I am sorry if my question is elementary. $\endgroup$ – Ali Taghavi Jul 25 '18 at 7:39
  • $\begingroup$ In fact I computed $X.(<X.R>)=<\nabla_X X,.R>+<\nabla_X R,X>=<\nabla_X R,X>$ but why the later is zero? As I learned the Reeb vector field from the link of your answer I realize that the Reeb vector field satisfies the conditions of propostion 6.8 mentioned in this answer mathoverflow.net/questions/273635/… $\endgroup$ – Ali Taghavi Jul 25 '18 at 8:02
  • 1
    $\begingroup$ I'm not a Riemannian geometer, so I don't know if I can respond directly to your comment. But you might try looking up first the Chow-Rashevskii theorem (en.wikipedia.org/wiki/Chow%E2%80%93Rashevskii_theorem ). Existence of geodesics comes from the Hopf-Rinow theorem. You can argue that a geodesic must be a.e. tangent to the horizontal distribution. $\endgroup$ – mdr Jul 29 '18 at 7:30

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.