Any two dimensional orientable manifold is complex [duplicate]
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
asked Dec 2 '18 at 19:15
joseabp91
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.
add a comment |
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
asked Dec 2 '18 at 19:15
joseabp91
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.
It is a $1$-dimensional complex manifold (in complex sense).
– joseabp91
Dec 2 '18 at 19:27
1
See: math.stackexchange.com/a/907476/534616
– K B Dave
Dec 2 '18 at 19:27
Not quite a duplicate of that one because the surface isn't required to be compact
– K B Dave
Dec 2 '18 at 19:29
I do not see why they answer my question.
– joseabp91
Dec 2 '18 at 19:40
add a comment |
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
asked Dec 2 '18 at 19:15
joseabp91
1,243411
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
asked Dec 2 '18 at 19:15
joseabp91
1,243411
asked Dec 2 '18 at 19:15
joseabp91
1,243411
asked Dec 2 '18 at 19:15
joseabp91
1,243411
asked Dec 2 '18 at 19:15
joseabp91
1,243411
asked Dec 2 '18 at 19:15
joseabp91
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.
It is a $1$-dimensional complex manifold (in complex sense).
– joseabp91
Dec 2 '18 at 19:27
1
See: math.stackexchange.com/a/907476/534616
– K B Dave
Dec 2 '18 at 19:27
Not quite a duplicate of that one because the surface isn't required to be compact
– K B Dave
Dec 2 '18 at 19:29
I do not see why they answer my question.
– joseabp91
Dec 2 '18 at 19:40
add a comment |
It is a $1$-dimensional complex manifold (in complex sense).
– joseabp91
Dec 2 '18 at 19:27
1
See: math.stackexchange.com/a/907476/534616
– K B Dave
Dec 2 '18 at 19:27
Not quite a duplicate of that one because the surface isn't required to be compact
– K B Dave
Dec 2 '18 at 19:29
I do not see why they answer my question.
– joseabp91
Dec 2 '18 at 19:40
It is a $1$-dimensional complex manifold (in complex sense).
– joseabp91
Dec 2 '18 at 19:27
It is a $1$-dimensional complex manifold (in complex sense).
– joseabp91
Dec 2 '18 at 19:27
1
1
See: math.stackexchange.com/a/907476/534616
– K B Dave
Dec 2 '18 at 19:27
See: math.stackexchange.com/a/907476/534616
– K B Dave
Dec 2 '18 at 19:27
Not quite a duplicate of that one because the surface isn't required to be compact
– K B Dave
Dec 2 '18 at 19:29
Not quite a duplicate of that one because the surface isn't required to be compact
– K B Dave
Dec 2 '18 at 19:29
I do not see why they answer my question.
– joseabp91
Dec 2 '18 at 19:40
I do not see why they answer my question.
– joseabp91
Dec 2 '18 at 19:40
add a comment |
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It is a $1$-dimensional complex manifold (in complex sense).
– joseabp91
Dec 2 '18 at 19:27
1
See: math.stackexchange.com/a/907476/534616
– K B Dave
Dec 2 '18 at 19:27
Not quite a duplicate of that one because the surface isn't required to be compact
– K B Dave
Dec 2 '18 at 19:29
I do not see why they answer my question.
– joseabp91
Dec 2 '18 at 19:40