Generic CR submanifolds

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So the circle $S^1$ is a hypersurface of $mathbb C$ and hence a generic CR submanifold. But $mathbb Csetminus0$ is the complexification of $S^1$ and $mathbb Csetminus0subset mathbb C$. Can $S^1$ be also a generic CR submanifold of $mathbb Csetminus0$ ?



And in general, $mathbb Rtimes S^2$ is a generic submanifod of $mathbb C^2$, $(mathbb Csetminus0)^2$, or $mathbb Ctimes( mathbb Csetminus0)$?










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    So the circle $S^1$ is a hypersurface of $mathbb C$ and hence a generic CR submanifold. But $mathbb Csetminus0$ is the complexification of $S^1$ and $mathbb Csetminus0subset mathbb C$. Can $S^1$ be also a generic CR submanifold of $mathbb Csetminus0$ ?



    And in general, $mathbb Rtimes S^2$ is a generic submanifod of $mathbb C^2$, $(mathbb Csetminus0)^2$, or $mathbb Ctimes( mathbb Csetminus0)$?










    share|cite|improve this question























      up vote
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      up vote
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      down vote

      favorite











      So the circle $S^1$ is a hypersurface of $mathbb C$ and hence a generic CR submanifold. But $mathbb Csetminus0$ is the complexification of $S^1$ and $mathbb Csetminus0subset mathbb C$. Can $S^1$ be also a generic CR submanifold of $mathbb Csetminus0$ ?



      And in general, $mathbb Rtimes S^2$ is a generic submanifod of $mathbb C^2$, $(mathbb Csetminus0)^2$, or $mathbb Ctimes( mathbb Csetminus0)$?










      share|cite|improve this question













      So the circle $S^1$ is a hypersurface of $mathbb C$ and hence a generic CR submanifold. But $mathbb Csetminus0$ is the complexification of $S^1$ and $mathbb Csetminus0subset mathbb C$. Can $S^1$ be also a generic CR submanifold of $mathbb Csetminus0$ ?



      And in general, $mathbb Rtimes S^2$ is a generic submanifod of $mathbb C^2$, $(mathbb Csetminus0)^2$, or $mathbb Ctimes( mathbb Csetminus0)$?







      differential-geometry smooth-manifolds several-complex-variables






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      asked Aug 30 at 2:53









      Amrat A

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          Yes, $S^1$ is a generic submaifold of $mathbb C setminus 0 $ as being generic is a local condition, so it is a generic submanifold of any neighborhood of $S^1$ in $mathbb C$.



          Similarly $mathbb R times S^2$ is going to be a generic submanifold in any complex dimension 2 mmanifold, since however you embed it, it will be a hypersurface, and a hypersurface is always generic.



          Again, the main thing is that being generic is a local condition, that is, $M$ is generic at $p$ if $T_p M + J T_p M$ ($J$ is the complex structure) is the entire tangent space of the ambient manifold.






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Yes, $S^1$ is a generic submaifold of $mathbb C setminus 0 $ as being generic is a local condition, so it is a generic submanifold of any neighborhood of $S^1$ in $mathbb C$.



            Similarly $mathbb R times S^2$ is going to be a generic submanifold in any complex dimension 2 mmanifold, since however you embed it, it will be a hypersurface, and a hypersurface is always generic.



            Again, the main thing is that being generic is a local condition, that is, $M$ is generic at $p$ if $T_p M + J T_p M$ ($J$ is the complex structure) is the entire tangent space of the ambient manifold.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              Yes, $S^1$ is a generic submaifold of $mathbb C setminus 0 $ as being generic is a local condition, so it is a generic submanifold of any neighborhood of $S^1$ in $mathbb C$.



              Similarly $mathbb R times S^2$ is going to be a generic submanifold in any complex dimension 2 mmanifold, since however you embed it, it will be a hypersurface, and a hypersurface is always generic.



              Again, the main thing is that being generic is a local condition, that is, $M$ is generic at $p$ if $T_p M + J T_p M$ ($J$ is the complex structure) is the entire tangent space of the ambient manifold.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Yes, $S^1$ is a generic submaifold of $mathbb C setminus 0 $ as being generic is a local condition, so it is a generic submanifold of any neighborhood of $S^1$ in $mathbb C$.



                Similarly $mathbb R times S^2$ is going to be a generic submanifold in any complex dimension 2 mmanifold, since however you embed it, it will be a hypersurface, and a hypersurface is always generic.



                Again, the main thing is that being generic is a local condition, that is, $M$ is generic at $p$ if $T_p M + J T_p M$ ($J$ is the complex structure) is the entire tangent space of the ambient manifold.






                share|cite|improve this answer












                Yes, $S^1$ is a generic submaifold of $mathbb C setminus 0 $ as being generic is a local condition, so it is a generic submanifold of any neighborhood of $S^1$ in $mathbb C$.



                Similarly $mathbb R times S^2$ is going to be a generic submanifold in any complex dimension 2 mmanifold, since however you embed it, it will be a hypersurface, and a hypersurface is always generic.



                Again, the main thing is that being generic is a local condition, that is, $M$ is generic at $p$ if $T_p M + J T_p M$ ($J$ is the complex structure) is the entire tangent space of the ambient manifold.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Aug 31 at 3:21









                Jiri Lebl

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