Classification of surfaces












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
    3 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
    3 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











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









KarenKaren

1336




1336












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


















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
















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




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










2 Answers
2






active

oldest

votes


















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












    $begingroup$

    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









    $endgroup$














      Your Answer








      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203841%2fclassification-of-surfaces%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203841%2fclassification-of-surfaces%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      What other Star Trek series did the main TNG cast show up in?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75
                      Read more

                      Berlina muro

                      Image
                      La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l
                      Read more

                      Berlina aerponto

                      Image
                      Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa
                      Read more