Any two dimensional orientable manifold is complex [duplicate]












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This question already has an answer here:




  • existence of complex structure in dimension 2

    1 answer



  • Why does every noncompact orientable surface have a complex structure?

    1 answer




Let $M$ be a $2$-dimensional orientable manifold. Is $M$ a Riemann surface? If it is true, how can I show it? Thank you very much.






riemann-surfaces







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marked as duplicate by Dietrich Burde, Eric Wofsey, Lord Shark the Unknown, amWhy, Robert Soupe Dec 3 '18 at 3:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    It is a $1$-dimensional complex manifold (in complex sense).
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:27








  • 1




    $begingroup$
    See: math.stackexchange.com/a/907476/534616
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:27










  • $begingroup$
    Not quite a duplicate of that one because the surface isn't required to be compact
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:29










  • $begingroup$
    I do not see why they answer my question.
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:40
















0












$begingroup$



This question already has an answer here:




  • existence of complex structure in dimension 2

    1 answer



  • Why does every noncompact orientable surface have a complex structure?

    1 answer




Let $M$ be a $2$-dimensional orientable manifold. Is $M$ a Riemann surface? If it is true, how can I show it? Thank you very much.






riemann-surfaces







share|cite|improve this question









$endgroup$



marked as duplicate by Dietrich Burde, Eric Wofsey, Lord Shark the Unknown, amWhy, Robert Soupe Dec 3 '18 at 3:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    It is a $1$-dimensional complex manifold (in complex sense).
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:27








  • 1




    $begingroup$
    See: math.stackexchange.com/a/907476/534616
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:27










  • $begingroup$
    Not quite a duplicate of that one because the surface isn't required to be compact
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:29










  • $begingroup$
    I do not see why they answer my question.
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:40














0












0








0





$begingroup$



This question already has an answer here:




  • existence of complex structure in dimension 2

    1 answer



  • Why does every noncompact orientable surface have a complex structure?

    1 answer




Let $M$ be a $2$-dimensional orientable manifold. Is $M$ a Riemann surface? If it is true, how can I show it? Thank you very much.






riemann-surfaces







share|cite|improve this question









$endgroup$





This question already has an answer here:




  • existence of complex structure in dimension 2

    1 answer



  • Why does every noncompact orientable surface have a complex structure?

    1 answer




Let $M$ be a $2$-dimensional orientable manifold. Is $M$ a Riemann surface? If it is true, how can I show it? Thank you very much.





This question already has an answer here:




  • existence of complex structure in dimension 2

    1 answer



  • Why does every noncompact orientable surface have a complex structure?

    1 answer







riemann-surfaces




riemann-surfaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 19:15









joseabp91joseabp91

1,243411




1,243411




marked as duplicate by Dietrich Burde, Eric Wofsey, Lord Shark the Unknown, amWhy, Robert Soupe Dec 3 '18 at 3:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Dietrich Burde, Eric Wofsey, Lord Shark the Unknown, amWhy, Robert Soupe Dec 3 '18 at 3:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    It is a $1$-dimensional complex manifold (in complex sense).
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:27








  • 1




    $begingroup$
    See: math.stackexchange.com/a/907476/534616
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:27










  • $begingroup$
    Not quite a duplicate of that one because the surface isn't required to be compact
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:29










  • $begingroup$
    I do not see why they answer my question.
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:40


















  • $begingroup$
    It is a $1$-dimensional complex manifold (in complex sense).
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:27








  • 1




    $begingroup$
    See: math.stackexchange.com/a/907476/534616
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:27










  • $begingroup$
    Not quite a duplicate of that one because the surface isn't required to be compact
    $endgroup$
    – K B Dave
    Dec 2 '18 at 19:29










  • $begingroup$
    I do not see why they answer my question.
    $endgroup$
    – joseabp91
    Dec 2 '18 at 19:40
















$begingroup$
It is a $1$-dimensional complex manifold (in complex sense).
$endgroup$
– joseabp91
Dec 2 '18 at 19:27






$begingroup$
It is a $1$-dimensional complex manifold (in complex sense).
$endgroup$
– joseabp91
Dec 2 '18 at 19:27






1




1




$begingroup$
See: math.stackexchange.com/a/907476/534616
$endgroup$
– K B Dave
Dec 2 '18 at 19:27




$begingroup$
See: math.stackexchange.com/a/907476/534616
$endgroup$
– K B Dave
Dec 2 '18 at 19:27












$begingroup$
Not quite a duplicate of that one because the surface isn't required to be compact
$endgroup$
– K B Dave
Dec 2 '18 at 19:29




$begingroup$
Not quite a duplicate of that one because the surface isn't required to be compact
$endgroup$
– K B Dave
Dec 2 '18 at 19:29












$begingroup$
I do not see why they answer my question.
$endgroup$
– joseabp91
Dec 2 '18 at 19:40




$begingroup$
I do not see why they answer my question.
$endgroup$
– joseabp91
Dec 2 '18 at 19:40










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