Examples of odd-dimensional manifolds that do not admit contact structure












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I'm having an hard time trying to figuring out a concrete example of an odd-dimensional closed manifold that do not admit any contact structure.
Can someone provide me with some examples?






dg.differential-geometry at.algebraic-topology differential-topology contact-geometry







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    Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= non-overtwisted) contact structure.
    $endgroup$
    – KSackel
    yesterday
















10












$begingroup$


I'm having an hard time trying to figuring out a concrete example of an odd-dimensional closed manifold that do not admit any contact structure.
Can someone provide me with some examples?






dg.differential-geometry at.algebraic-topology differential-topology contact-geometry







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= non-overtwisted) contact structure.
    $endgroup$
    – KSackel
    yesterday














10












10








10


3



$begingroup$


I'm having an hard time trying to figuring out a concrete example of an odd-dimensional closed manifold that do not admit any contact structure.
Can someone provide me with some examples?






dg.differential-geometry at.algebraic-topology differential-topology contact-geometry







share|cite|improve this question











$endgroup$




I'm having an hard time trying to figuring out a concrete example of an odd-dimensional closed manifold that do not admit any contact structure.
Can someone provide me with some examples?






dg.differential-geometry at.algebraic-topology differential-topology contact-geometry




dg.differential-geometry at.algebraic-topology differential-topology contact-geometry






share|cite|improve this question















share|cite|improve this question













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edited yesterday









Piotr Hajlasz

9,60843873




9,60843873










asked yesterday









Warlock of Firetop MountainWarlock of Firetop Mountain

28218




28218








  • 2




    $begingroup$
    Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= non-overtwisted) contact structure.
    $endgroup$
    – KSackel
    yesterday














  • 2




    $begingroup$
    Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= non-overtwisted) contact structure.
    $endgroup$
    – KSackel
    yesterday








2




2




$begingroup$
Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= non-overtwisted) contact structure.
$endgroup$
– KSackel
yesterday




$begingroup$
Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= non-overtwisted) contact structure.
$endgroup$
– KSackel
yesterday










2 Answers
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18












$begingroup$

Every compact orientable $3$-dimensional manifold has a contact structure [1]. On the other hand we have




Theorem. For $ngeq 2$ there is a closed oriented connected manifold of dimension $2n+1$ without a contact structure.




For $n=2$, $SU(3)/SO(3)$ has no contact structure and for $n>2$,
$SU(3)/SO(3)timesmathbb{S}^{2n-4}$ has no contact structure, see Proposition 2.4 in [2].



[1] J. Martinet,
Formes de contact sur les variétés de dimension 3. Proceedings of Liverpool Singularities Symposium, II (1969/1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.



[2] R. E. Stong, Contact manifolds. J. Differential Geometry 9 (1974), 219–238.






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    9












    $begingroup$

    Although every closed oriented 3 manifold admits infinitely many contact structures, there isa n obstruction to the existence of contact structure on odd dim manifolds of dim$geq5$.



    Suppose that $n>0$ and $xi= ker alpha$ is a co-oriented contact structure on $Y^{2n+1}$. Then the tangent bundle of $Y$ has a splitting $TY= xi oplus mathbb R$. Then $dalpha$ is symplectic on $xi$ and thus $xi$ admits a compatible complex vector bundle structure[such a splitting called almost complex structure]. This is equivalent to a regudction group to $U(n)times 1 subset SO(2n+1,mathbb R)$. Now lets observe the Chern class of $xi$ viewed as a complex vector bundle over $Y$. Moreover $c_1(xi)$ reduced to Stiefel-Whitney $w_2(Y)in H^2(Y,mathbb Z_2)$. So we can say that if $Y$ admits an almost complex structure then $w_2$ admits an integral lift which is equivalent of third Stiefel-Whitney class $W_3(Y)in H^3(Y,mathbb Z)$ vanishes. [Since $W_3$ is the image of $w_2$ under Bockstein homomorphism $beta$ $to H^2(Y,mathbb Z)to H^2(Y,mathbb Z_2)to_beta H^3(Y,mathbb Z)to$.] Thus $W_3=0$ is a necessary condition.



    Now if $Y= SU(3)/SO(3)$. Then we have a fibration $SO(3)to SU(3)to Y$. By using Steenrod squares, $W_3(Y)$ is the generator of $H^3(Y,mathbb Z)=mathbb Z_2$. So this manifold doesnot admit almost contact structure.



    Theorem[Borman, Eliashberg,Murphy]- There exists a contact structure in every homotopy class of almost contact structure.






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      2 Answers
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      2 Answers
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      active

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      18












      $begingroup$

      Every compact orientable $3$-dimensional manifold has a contact structure [1]. On the other hand we have




      Theorem. For $ngeq 2$ there is a closed oriented connected manifold of dimension $2n+1$ without a contact structure.




      For $n=2$, $SU(3)/SO(3)$ has no contact structure and for $n>2$,
      $SU(3)/SO(3)timesmathbb{S}^{2n-4}$ has no contact structure, see Proposition 2.4 in [2].



      [1] J. Martinet,
      Formes de contact sur les variétés de dimension 3. Proceedings of Liverpool Singularities Symposium, II (1969/1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.



      [2] R. E. Stong, Contact manifolds. J. Differential Geometry 9 (1974), 219–238.






      share|cite|improve this answer









      $endgroup$


















        18












        $begingroup$

        Every compact orientable $3$-dimensional manifold has a contact structure [1]. On the other hand we have




        Theorem. For $ngeq 2$ there is a closed oriented connected manifold of dimension $2n+1$ without a contact structure.




        For $n=2$, $SU(3)/SO(3)$ has no contact structure and for $n>2$,
        $SU(3)/SO(3)timesmathbb{S}^{2n-4}$ has no contact structure, see Proposition 2.4 in [2].



        [1] J. Martinet,
        Formes de contact sur les variétés de dimension 3. Proceedings of Liverpool Singularities Symposium, II (1969/1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.



        [2] R. E. Stong, Contact manifolds. J. Differential Geometry 9 (1974), 219–238.






        share|cite|improve this answer









        $endgroup$
















          18












          18








          18





          $begingroup$

          Every compact orientable $3$-dimensional manifold has a contact structure [1]. On the other hand we have




          Theorem. For $ngeq 2$ there is a closed oriented connected manifold of dimension $2n+1$ without a contact structure.




          For $n=2$, $SU(3)/SO(3)$ has no contact structure and for $n>2$,
          $SU(3)/SO(3)timesmathbb{S}^{2n-4}$ has no contact structure, see Proposition 2.4 in [2].



          [1] J. Martinet,
          Formes de contact sur les variétés de dimension 3. Proceedings of Liverpool Singularities Symposium, II (1969/1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.



          [2] R. E. Stong, Contact manifolds. J. Differential Geometry 9 (1974), 219–238.






          share|cite|improve this answer









          $endgroup$



          Every compact orientable $3$-dimensional manifold has a contact structure [1]. On the other hand we have




          Theorem. For $ngeq 2$ there is a closed oriented connected manifold of dimension $2n+1$ without a contact structure.




          For $n=2$, $SU(3)/SO(3)$ has no contact structure and for $n>2$,
          $SU(3)/SO(3)timesmathbb{S}^{2n-4}$ has no contact structure, see Proposition 2.4 in [2].



          [1] J. Martinet,
          Formes de contact sur les variétés de dimension 3. Proceedings of Liverpool Singularities Symposium, II (1969/1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.



          [2] R. E. Stong, Contact manifolds. J. Differential Geometry 9 (1974), 219–238.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Piotr HajlaszPiotr Hajlasz

          9,60843873




          9,60843873























              9












              $begingroup$

              Although every closed oriented 3 manifold admits infinitely many contact structures, there isa n obstruction to the existence of contact structure on odd dim manifolds of dim$geq5$.



              Suppose that $n>0$ and $xi= ker alpha$ is a co-oriented contact structure on $Y^{2n+1}$. Then the tangent bundle of $Y$ has a splitting $TY= xi oplus mathbb R$. Then $dalpha$ is symplectic on $xi$ and thus $xi$ admits a compatible complex vector bundle structure[such a splitting called almost complex structure]. This is equivalent to a regudction group to $U(n)times 1 subset SO(2n+1,mathbb R)$. Now lets observe the Chern class of $xi$ viewed as a complex vector bundle over $Y$. Moreover $c_1(xi)$ reduced to Stiefel-Whitney $w_2(Y)in H^2(Y,mathbb Z_2)$. So we can say that if $Y$ admits an almost complex structure then $w_2$ admits an integral lift which is equivalent of third Stiefel-Whitney class $W_3(Y)in H^3(Y,mathbb Z)$ vanishes. [Since $W_3$ is the image of $w_2$ under Bockstein homomorphism $beta$ $to H^2(Y,mathbb Z)to H^2(Y,mathbb Z_2)to_beta H^3(Y,mathbb Z)to$.] Thus $W_3=0$ is a necessary condition.



              Now if $Y= SU(3)/SO(3)$. Then we have a fibration $SO(3)to SU(3)to Y$. By using Steenrod squares, $W_3(Y)$ is the generator of $H^3(Y,mathbb Z)=mathbb Z_2$. So this manifold doesnot admit almost contact structure.



              Theorem[Borman, Eliashberg,Murphy]- There exists a contact structure in every homotopy class of almost contact structure.






              share|cite|improve this answer











              $endgroup$


















                9












                $begingroup$

                Although every closed oriented 3 manifold admits infinitely many contact structures, there isa n obstruction to the existence of contact structure on odd dim manifolds of dim$geq5$.



                Suppose that $n>0$ and $xi= ker alpha$ is a co-oriented contact structure on $Y^{2n+1}$. Then the tangent bundle of $Y$ has a splitting $TY= xi oplus mathbb R$. Then $dalpha$ is symplectic on $xi$ and thus $xi$ admits a compatible complex vector bundle structure[such a splitting called almost complex structure]. This is equivalent to a regudction group to $U(n)times 1 subset SO(2n+1,mathbb R)$. Now lets observe the Chern class of $xi$ viewed as a complex vector bundle over $Y$. Moreover $c_1(xi)$ reduced to Stiefel-Whitney $w_2(Y)in H^2(Y,mathbb Z_2)$. So we can say that if $Y$ admits an almost complex structure then $w_2$ admits an integral lift which is equivalent of third Stiefel-Whitney class $W_3(Y)in H^3(Y,mathbb Z)$ vanishes. [Since $W_3$ is the image of $w_2$ under Bockstein homomorphism $beta$ $to H^2(Y,mathbb Z)to H^2(Y,mathbb Z_2)to_beta H^3(Y,mathbb Z)to$.] Thus $W_3=0$ is a necessary condition.



                Now if $Y= SU(3)/SO(3)$. Then we have a fibration $SO(3)to SU(3)to Y$. By using Steenrod squares, $W_3(Y)$ is the generator of $H^3(Y,mathbb Z)=mathbb Z_2$. So this manifold doesnot admit almost contact structure.



                Theorem[Borman, Eliashberg,Murphy]- There exists a contact structure in every homotopy class of almost contact structure.






                share|cite|improve this answer











                $endgroup$
















                  9












                  9








                  9





                  $begingroup$

                  Although every closed oriented 3 manifold admits infinitely many contact structures, there isa n obstruction to the existence of contact structure on odd dim manifolds of dim$geq5$.



                  Suppose that $n>0$ and $xi= ker alpha$ is a co-oriented contact structure on $Y^{2n+1}$. Then the tangent bundle of $Y$ has a splitting $TY= xi oplus mathbb R$. Then $dalpha$ is symplectic on $xi$ and thus $xi$ admits a compatible complex vector bundle structure[such a splitting called almost complex structure]. This is equivalent to a regudction group to $U(n)times 1 subset SO(2n+1,mathbb R)$. Now lets observe the Chern class of $xi$ viewed as a complex vector bundle over $Y$. Moreover $c_1(xi)$ reduced to Stiefel-Whitney $w_2(Y)in H^2(Y,mathbb Z_2)$. So we can say that if $Y$ admits an almost complex structure then $w_2$ admits an integral lift which is equivalent of third Stiefel-Whitney class $W_3(Y)in H^3(Y,mathbb Z)$ vanishes. [Since $W_3$ is the image of $w_2$ under Bockstein homomorphism $beta$ $to H^2(Y,mathbb Z)to H^2(Y,mathbb Z_2)to_beta H^3(Y,mathbb Z)to$.] Thus $W_3=0$ is a necessary condition.



                  Now if $Y= SU(3)/SO(3)$. Then we have a fibration $SO(3)to SU(3)to Y$. By using Steenrod squares, $W_3(Y)$ is the generator of $H^3(Y,mathbb Z)=mathbb Z_2$. So this manifold doesnot admit almost contact structure.



                  Theorem[Borman, Eliashberg,Murphy]- There exists a contact structure in every homotopy class of almost contact structure.






                  share|cite|improve this answer











                  $endgroup$



                  Although every closed oriented 3 manifold admits infinitely many contact structures, there isa n obstruction to the existence of contact structure on odd dim manifolds of dim$geq5$.



                  Suppose that $n>0$ and $xi= ker alpha$ is a co-oriented contact structure on $Y^{2n+1}$. Then the tangent bundle of $Y$ has a splitting $TY= xi oplus mathbb R$. Then $dalpha$ is symplectic on $xi$ and thus $xi$ admits a compatible complex vector bundle structure[such a splitting called almost complex structure]. This is equivalent to a regudction group to $U(n)times 1 subset SO(2n+1,mathbb R)$. Now lets observe the Chern class of $xi$ viewed as a complex vector bundle over $Y$. Moreover $c_1(xi)$ reduced to Stiefel-Whitney $w_2(Y)in H^2(Y,mathbb Z_2)$. So we can say that if $Y$ admits an almost complex structure then $w_2$ admits an integral lift which is equivalent of third Stiefel-Whitney class $W_3(Y)in H^3(Y,mathbb Z)$ vanishes. [Since $W_3$ is the image of $w_2$ under Bockstein homomorphism $beta$ $to H^2(Y,mathbb Z)to H^2(Y,mathbb Z_2)to_beta H^3(Y,mathbb Z)to$.] Thus $W_3=0$ is a necessary condition.



                  Now if $Y= SU(3)/SO(3)$. Then we have a fibration $SO(3)to SU(3)to Y$. By using Steenrod squares, $W_3(Y)$ is the generator of $H^3(Y,mathbb Z)=mathbb Z_2$. So this manifold doesnot admit almost contact structure.



                  Theorem[Borman, Eliashberg,Murphy]- There exists a contact structure in every homotopy class of almost contact structure.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 4 hours ago

























                  answered 13 hours ago









                  Anubhav MukherjeeAnubhav Mukherjee

                  1,0341019




                  1,0341019






























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