Timeline for A non integrable distribution which is totally geodesic
Current License: CC BY-SA 4.0
7 events
when toggle format | what | by | license | comment | |
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Jul 30, 2018 at 4:37 | comment | added | Ali Taghavi | Thanks a lot for your attention and your very helpful comment. | |
Jul 29, 2018 at 7:30 | comment | added | mdr | 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. | |
Jul 25, 2018 at 8:02 | comment | added | Ali Taghavi | 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/… | |
Jul 25, 2018 at 7:39 | comment | added | Ali Taghavi | I am sorry if my question is elementary. | |
Jul 25, 2018 at 7:38 | comment | added | Ali Taghavi | 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? | |
Jul 25, 2018 at 7:37 | comment | added | Ali Taghavi | 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 | |
Jul 23, 2018 at 14:11 | history | answered | mdr | CC BY-SA 4.0 |