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Classification of surfaces


connected sum of torus with projective planeWhy can all surfaces with boundary be realized in $mathbb{R}^3$?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













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$begingroup$


The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_{g}T^2)# (#_{b} D^2)# (#_{c} mathbb{R}P^2),$$ 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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  • $begingroup$
    This can help you math.stackexchange.com/q/358724/654562
    $endgroup$
    – dcolazin
    2 hours ago
















4












$begingroup$


The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_{g}T^2)# (#_{b} D^2)# (#_{c} mathbb{R}P^2),$$ 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









$endgroup$












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














4












4








4





$begingroup$


The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_{g}T^2)# (#_{b} D^2)# (#_{c} mathbb{R}P^2),$$ 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









$endgroup$




The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_{g}T^2)# (#_{b} D^2)# (#_{c} mathbb{R}P^2),$$ 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




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






share|cite|improve this question













share|cite|improve this question











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asked 3 hours ago









KarenKaren

1336




1336












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


















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
















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




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










2 Answers
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$begingroup$

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









$endgroup$





















    2












    $begingroup$

    Since $mathbb{R}P^2# mathbb{R}P^2#mathbb{R}P^2cong mathbb{R}P^2 # T^2$, $c$ 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









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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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









      $endgroup$


















        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









        $endgroup$
















          2












          2








          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









          $endgroup$



          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












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Adam ChalumeauAdam Chalumeau

          48010




          48010























              2












              $begingroup$

              Since $mathbb{R}P^2# mathbb{R}P^2#mathbb{R}P^2cong mathbb{R}P^2 # T^2$, $c$ 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









              $endgroup$


















                2












                $begingroup$

                Since $mathbb{R}P^2# mathbb{R}P^2#mathbb{R}P^2cong mathbb{R}P^2 # T^2$, $c$ 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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Since $mathbb{R}P^2# mathbb{R}P^2#mathbb{R}P^2cong mathbb{R}P^2 # T^2$, $c$ 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









                  $endgroup$



                  Since $mathbb{R}P^2# mathbb{R}P^2#mathbb{R}P^2cong mathbb{R}P^2 # T^2$, $c$ 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












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Eric WofseyEric Wofsey

                  194k14223354




                  194k14223354






























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