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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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
    Commented Jul 25, 2018 at 7:29
  • $\begingroup$ I am sorry if my question is elementary. $\endgroup$
    – Ali Taghavi
    Commented Jul 25, 2018 at 7:39
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    $\begingroup$ @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
    Commented Jul 26, 2018 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
    Commented Jul 27, 2018 at 22:10
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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 .

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  • $\begingroup$ 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
    Commented Jul 25, 2018 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
    Commented Jul 25, 2018 at 7:38
  • $\begingroup$ I am sorry if my question is elementary. $\endgroup$
    – Ali Taghavi
    Commented Jul 25, 2018 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
    Commented Jul 25, 2018 at 8:02
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    $\begingroup$ 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
    Commented Jul 29, 2018 at 7:30

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