Examples of smooth manifolds admitting inbetween one and a continuum of complex structures












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For smooth manifolds it is known that they can admit a unique, finitely many, or a continuum of distinct smooth structures (I don't know whether there are any examples admitting precisely a countably infinite number).



For complex manifolds there are examples of smooth manifolds admitting a unique complex structure ($mathbb{CP}^1$) or a continuum (compact Riemann surfaces, K3 surfaces, etc.) but are there examples admitting only finitely many or a countably infinite number?



By deformation theory the tangent space to the moduli space of complex structures on $X$ should be given by $H^1(X, TX)$ (at least morally) so it must be necessary for this to vanish for every possible complex structure on $X$ to have any hope.






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    if you allow non-closed manifolds, smooth unit disc admits exactly two complex structures I believe (one its own, the other coming from the diffeomorphism with $mathbb{R}^2$)
    $endgroup$
    – Aknazar Kazhymurat
    8 hours ago
















12












$begingroup$


For smooth manifolds it is known that they can admit a unique, finitely many, or a continuum of distinct smooth structures (I don't know whether there are any examples admitting precisely a countably infinite number).



For complex manifolds there are examples of smooth manifolds admitting a unique complex structure ($mathbb{CP}^1$) or a continuum (compact Riemann surfaces, K3 surfaces, etc.) but are there examples admitting only finitely many or a countably infinite number?



By deformation theory the tangent space to the moduli space of complex structures on $X$ should be given by $H^1(X, TX)$ (at least morally) so it must be necessary for this to vanish for every possible complex structure on $X$ to have any hope.






dg.differential-geometry complex-geometry differential-topology







share|cite|improve this question







New contributor




John McCarthy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    if you allow non-closed manifolds, smooth unit disc admits exactly two complex structures I believe (one its own, the other coming from the diffeomorphism with $mathbb{R}^2$)
    $endgroup$
    – Aknazar Kazhymurat
    8 hours ago














12












12








12


2



$begingroup$


For smooth manifolds it is known that they can admit a unique, finitely many, or a continuum of distinct smooth structures (I don't know whether there are any examples admitting precisely a countably infinite number).



For complex manifolds there are examples of smooth manifolds admitting a unique complex structure ($mathbb{CP}^1$) or a continuum (compact Riemann surfaces, K3 surfaces, etc.) but are there examples admitting only finitely many or a countably infinite number?



By deformation theory the tangent space to the moduli space of complex structures on $X$ should be given by $H^1(X, TX)$ (at least morally) so it must be necessary for this to vanish for every possible complex structure on $X$ to have any hope.






dg.differential-geometry complex-geometry differential-topology







share|cite|improve this question







New contributor




John McCarthy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




For smooth manifolds it is known that they can admit a unique, finitely many, or a continuum of distinct smooth structures (I don't know whether there are any examples admitting precisely a countably infinite number).



For complex manifolds there are examples of smooth manifolds admitting a unique complex structure ($mathbb{CP}^1$) or a continuum (compact Riemann surfaces, K3 surfaces, etc.) but are there examples admitting only finitely many or a countably infinite number?



By deformation theory the tangent space to the moduli space of complex structures on $X$ should be given by $H^1(X, TX)$ (at least morally) so it must be necessary for this to vanish for every possible complex structure on $X$ to have any hope.






dg.differential-geometry complex-geometry differential-topology




dg.differential-geometry complex-geometry differential-topology






share|cite|improve this question







New contributor




John McCarthy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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John McCarthy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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asked 13 hours ago









John McCarthyJohn McCarthy

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New contributor





John McCarthy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 2




    $begingroup$
    if you allow non-closed manifolds, smooth unit disc admits exactly two complex structures I believe (one its own, the other coming from the diffeomorphism with $mathbb{R}^2$)
    $endgroup$
    – Aknazar Kazhymurat
    8 hours ago














  • 2




    $begingroup$
    if you allow non-closed manifolds, smooth unit disc admits exactly two complex structures I believe (one its own, the other coming from the diffeomorphism with $mathbb{R}^2$)
    $endgroup$
    – Aknazar Kazhymurat
    8 hours ago








2




2




$begingroup$
if you allow non-closed manifolds, smooth unit disc admits exactly two complex structures I believe (one its own, the other coming from the diffeomorphism with $mathbb{R}^2$)
$endgroup$
– Aknazar Kazhymurat
8 hours ago




$begingroup$
if you allow non-closed manifolds, smooth unit disc admits exactly two complex structures I believe (one its own, the other coming from the diffeomorphism with $mathbb{R}^2$)
$endgroup$
– Aknazar Kazhymurat
8 hours ago










2 Answers
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15












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Examples of smooth manifolds admitting finitely many complex structures are provided by some holomorphically rigid complex varieties.



For instance, considering fake projective planes (i.e., smooth compact complex surfaces which are not the complex projective plane but have the same Betti numbers as the
complex projective plane), there are $100$ examples determined up to biholomorphism, but only $50$ examples determined up to isometry.



In fact, the real $4$-manifold underlying any fake projective plane admits precisely two inequivalent complex structures, that are exchanged by complex conjugation.



Moreover, by standard results on rigidity, any Kähler surface with the same fundamental group as a fake projective plane is actually biholomorphic or conjugate biholomorphic to it.






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    $begingroup$

    There are countably many complex structures on $S^2 times S^2$ up to isomorphism. Specifically, the even Hirzebruch surfaces $F_{2k}$ are the only options. This is the main result of



    Qin, Zhenbo, Complex structures on certain differentiable 4-manifolds, Topology 32, No. 3, 551-566 (1993). ZBL0796.57010.



    The author also proves that the odd Hirzebruch surfaces are the only complex structures on the $4$-fold $mathbb{CP}^2 # overline{mathbb{CP}^2}$.






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      2 Answers
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      active

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      2 Answers
      2






      active

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      active

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      active

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      15












      $begingroup$

      Examples of smooth manifolds admitting finitely many complex structures are provided by some holomorphically rigid complex varieties.



      For instance, considering fake projective planes (i.e., smooth compact complex surfaces which are not the complex projective plane but have the same Betti numbers as the
      complex projective plane), there are $100$ examples determined up to biholomorphism, but only $50$ examples determined up to isometry.



      In fact, the real $4$-manifold underlying any fake projective plane admits precisely two inequivalent complex structures, that are exchanged by complex conjugation.



      Moreover, by standard results on rigidity, any Kähler surface with the same fundamental group as a fake projective plane is actually biholomorphic or conjugate biholomorphic to it.






      share|cite|improve this answer











      $endgroup$


















        15












        $begingroup$

        Examples of smooth manifolds admitting finitely many complex structures are provided by some holomorphically rigid complex varieties.



        For instance, considering fake projective planes (i.e., smooth compact complex surfaces which are not the complex projective plane but have the same Betti numbers as the
        complex projective plane), there are $100$ examples determined up to biholomorphism, but only $50$ examples determined up to isometry.



        In fact, the real $4$-manifold underlying any fake projective plane admits precisely two inequivalent complex structures, that are exchanged by complex conjugation.



        Moreover, by standard results on rigidity, any Kähler surface with the same fundamental group as a fake projective plane is actually biholomorphic or conjugate biholomorphic to it.






        share|cite|improve this answer











        $endgroup$
















          15












          15








          15





          $begingroup$

          Examples of smooth manifolds admitting finitely many complex structures are provided by some holomorphically rigid complex varieties.



          For instance, considering fake projective planes (i.e., smooth compact complex surfaces which are not the complex projective plane but have the same Betti numbers as the
          complex projective plane), there are $100$ examples determined up to biholomorphism, but only $50$ examples determined up to isometry.



          In fact, the real $4$-manifold underlying any fake projective plane admits precisely two inequivalent complex structures, that are exchanged by complex conjugation.



          Moreover, by standard results on rigidity, any Kähler surface with the same fundamental group as a fake projective plane is actually biholomorphic or conjugate biholomorphic to it.






          share|cite|improve this answer











          $endgroup$



          Examples of smooth manifolds admitting finitely many complex structures are provided by some holomorphically rigid complex varieties.



          For instance, considering fake projective planes (i.e., smooth compact complex surfaces which are not the complex projective plane but have the same Betti numbers as the
          complex projective plane), there are $100$ examples determined up to biholomorphism, but only $50$ examples determined up to isometry.



          In fact, the real $4$-manifold underlying any fake projective plane admits precisely two inequivalent complex structures, that are exchanged by complex conjugation.



          Moreover, by standard results on rigidity, any Kähler surface with the same fundamental group as a fake projective plane is actually biholomorphic or conjugate biholomorphic to it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 13 hours ago









          Francesco PolizziFrancesco Polizzi

          48.6k3130212




          48.6k3130212























              4












              $begingroup$

              There are countably many complex structures on $S^2 times S^2$ up to isomorphism. Specifically, the even Hirzebruch surfaces $F_{2k}$ are the only options. This is the main result of



              Qin, Zhenbo, Complex structures on certain differentiable 4-manifolds, Topology 32, No. 3, 551-566 (1993). ZBL0796.57010.



              The author also proves that the odd Hirzebruch surfaces are the only complex structures on the $4$-fold $mathbb{CP}^2 # overline{mathbb{CP}^2}$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                There are countably many complex structures on $S^2 times S^2$ up to isomorphism. Specifically, the even Hirzebruch surfaces $F_{2k}$ are the only options. This is the main result of



                Qin, Zhenbo, Complex structures on certain differentiable 4-manifolds, Topology 32, No. 3, 551-566 (1993). ZBL0796.57010.



                The author also proves that the odd Hirzebruch surfaces are the only complex structures on the $4$-fold $mathbb{CP}^2 # overline{mathbb{CP}^2}$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  There are countably many complex structures on $S^2 times S^2$ up to isomorphism. Specifically, the even Hirzebruch surfaces $F_{2k}$ are the only options. This is the main result of



                  Qin, Zhenbo, Complex structures on certain differentiable 4-manifolds, Topology 32, No. 3, 551-566 (1993). ZBL0796.57010.



                  The author also proves that the odd Hirzebruch surfaces are the only complex structures on the $4$-fold $mathbb{CP}^2 # overline{mathbb{CP}^2}$.






                  share|cite|improve this answer









                  $endgroup$



                  There are countably many complex structures on $S^2 times S^2$ up to isomorphism. Specifically, the even Hirzebruch surfaces $F_{2k}$ are the only options. This is the main result of



                  Qin, Zhenbo, Complex structures on certain differentiable 4-manifolds, Topology 32, No. 3, 551-566 (1993). ZBL0796.57010.



                  The author also proves that the odd Hirzebruch surfaces are the only complex structures on the $4$-fold $mathbb{CP}^2 # overline{mathbb{CP}^2}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 46 mins ago









                  David E SpeyerDavid E Speyer

                  107k9282539




                  107k9282539






















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