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Classification of surfaces

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Classification of surfaces

connected sum of torus with projective planeWhy can all surfaces with boundary be realized in $mathbbR^3$?2-cell embeddings of graphs in surfaces and Euler formulaAbout zeros of vector fields in compact surfacesClassification of orientable non-closed surfacesClosed, orientable surface whose genus is very hard to find intuitivelyNonorientable surfaces: genus or demigenus?Surface has Euler characteristic 2 iff equal to sphereClassification of surfaces theoremClassification of 2-dim topological manifold (not necessarily second countable)Connected sum of two non homeomorphic surfaces

4

$begingroup$

The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_gT^2)# (#_b D^2)# (#_c mathbbRP^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.

From there, it is easy to compute $chi(M)=2-2g-b-c$.

Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.

I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $chi(M)$, and orientability only tells us if $c=0$ or $cgeq 1$. Can someone help me, please?

manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces

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asked 3 hours ago

KarenKaren

1336

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can help you math.stackexchange.com/q/358724/654562
  $endgroup$
  – dcolazin
  2 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_gT^2)# (#_b D^2)# (#_c mathbbRP^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.

From there, it is easy to compute $chi(M)=2-2g-b-c$.

Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.

I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $chi(M)$, and orientability only tells us if $c=0$ or $cgeq 1$. Can someone help me, please?

manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces

share|cite|improve this question

asked 3 hours ago

KarenKaren

1336

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can help you math.stackexchange.com/q/358724/654562
  $endgroup$
  – dcolazin
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_gT^2)# (#_b D^2)# (#_c mathbbRP^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.

From there, it is easy to compute $chi(M)=2-2g-b-c$.

Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.

I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $chi(M)$, and orientability only tells us if $c=0$ or $cgeq 1$. Can someone help me, please?

manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces

share|cite|improve this question

asked 3 hours ago

KarenKaren

1336

$endgroup$

share|cite|improve this question

asked 3 hours ago

KarenKaren

1336

share|cite|improve this question

asked 3 hours ago

KarenKaren

1336

asked 3 hours ago

KarenKaren

1336

asked 3 hours ago

KarenKaren

1336

2 Answers
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2

$begingroup$

If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.

This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.

Please correct me if I misunderstood your question.

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

$endgroup$

add a comment | 

2

$begingroup$

Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.

Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

add a comment | 

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2

$begingroup$

If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.

This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.

Please correct me if I misunderstood your question.

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

$endgroup$

add a comment | 

2

$begingroup$

If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.

This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.

Please correct me if I misunderstood your question.

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

$endgroup$

add a comment | 

$begingroup$

If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.

This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.

Please correct me if I misunderstood your question.

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

2

$begingroup$

Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.

Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

add a comment | 

2

$begingroup$

Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.

Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

add a comment | 

$begingroup$

Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.

Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354