Existence of a type of simplicial complex












1












$begingroup$


I want to prove that the following proposition is false:



There exists a homologically trivial finite 2 dimensional simplicial complex $mathcal K$ such that every edge (1 dimensional simplex) has at least 3 adjacent faces (2 dimensional simplex)



Testing with some examples I notice that if every edge has 3 adjacent faces the complex is not trivial but I do not know how to prove the proposition.






algebraic-topology homology-cohomology simplicial-complex







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you require the complex to be finite/compact ?
    $endgroup$
    – Card_Trick
    Dec 22 '18 at 21:12










  • $begingroup$
    The complex is finite
    $endgroup$
    – giovanni gajac
    Dec 22 '18 at 21:14
















1












$begingroup$


I want to prove that the following proposition is false:



There exists a homologically trivial finite 2 dimensional simplicial complex $mathcal K$ such that every edge (1 dimensional simplex) has at least 3 adjacent faces (2 dimensional simplex)



Testing with some examples I notice that if every edge has 3 adjacent faces the complex is not trivial but I do not know how to prove the proposition.






algebraic-topology homology-cohomology simplicial-complex







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you require the complex to be finite/compact ?
    $endgroup$
    – Card_Trick
    Dec 22 '18 at 21:12










  • $begingroup$
    The complex is finite
    $endgroup$
    – giovanni gajac
    Dec 22 '18 at 21:14














1












1








1





$begingroup$


I want to prove that the following proposition is false:



There exists a homologically trivial finite 2 dimensional simplicial complex $mathcal K$ such that every edge (1 dimensional simplex) has at least 3 adjacent faces (2 dimensional simplex)



Testing with some examples I notice that if every edge has 3 adjacent faces the complex is not trivial but I do not know how to prove the proposition.






algebraic-topology homology-cohomology simplicial-complex







share|cite|improve this question











$endgroup$




I want to prove that the following proposition is false:



There exists a homologically trivial finite 2 dimensional simplicial complex $mathcal K$ such that every edge (1 dimensional simplex) has at least 3 adjacent faces (2 dimensional simplex)



Testing with some examples I notice that if every edge has 3 adjacent faces the complex is not trivial but I do not know how to prove the proposition.






algebraic-topology homology-cohomology simplicial-complex




algebraic-topology homology-cohomology simplicial-complex






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 21:23









Eric Wofsey

188k14216346




188k14216346










asked Dec 22 '18 at 21:05









giovanni gajacgiovanni gajac

133




133












  • $begingroup$
    Do you require the complex to be finite/compact ?
    $endgroup$
    – Card_Trick
    Dec 22 '18 at 21:12










  • $begingroup$
    The complex is finite
    $endgroup$
    – giovanni gajac
    Dec 22 '18 at 21:14


















  • $begingroup$
    Do you require the complex to be finite/compact ?
    $endgroup$
    – Card_Trick
    Dec 22 '18 at 21:12










  • $begingroup$
    The complex is finite
    $endgroup$
    – giovanni gajac
    Dec 22 '18 at 21:14
















$begingroup$
Do you require the complex to be finite/compact ?
$endgroup$
– Card_Trick
Dec 22 '18 at 21:12




$begingroup$
Do you require the complex to be finite/compact ?
$endgroup$
– Card_Trick
Dec 22 '18 at 21:12












$begingroup$
The complex is finite
$endgroup$
– giovanni gajac
Dec 22 '18 at 21:14




$begingroup$
The complex is finite
$endgroup$
– giovanni gajac
Dec 22 '18 at 21:14










2 Answers
2






active

oldest

votes


















1












$begingroup$

Observe that the given condition implies that the number of edges $e$ is less than or equal to the number of faces $f$.



The chain $mathbb{Z}^frightarrowmathbb{Z}^erightarrowmathbb{Z}^v$ gives trivial homology if the first map is injective and the image of the first map is the kernel of the second map. Since $eleq f$ and the first map is injective, the image is whole $mathbb{Z}^e$ so the second map must be the zero map. Can you see why this is impossible?






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint: Think about what you can say about the Euler characteristic of $mathcal{K}$.



    A full proof is hidden below.




    Say $mathcal{K}$ has $V$ vertices, $E$ edges, and $F$ faces. Consider the set $S$ of pairs $(a,b)$ where $a$ is a face and $b$ is a boundary edge of $a$. Since every face has three boundary edges, $|S|=3F$. But since every edge is in the boundary of at least $3$ faces, $|S|geq 3E$. Thus $Fgeq E$. Now the Euler characteristic of $mathcal{K}$ is $$V-E+Fgeq V.$$ Since $mathcal{K}$ is $2$-dimensional, it has at least one face, and thus at least $3$ vertices. So the Euler characteristic of $mathcal{K}$ is at least $3$, and in particular greater than $1$, so the homology of $mathcal{K}$ cannot be trivial.







    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      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%2f3049848%2fexistence-of-a-type-of-simplicial-complex%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









      1












      $begingroup$

      Observe that the given condition implies that the number of edges $e$ is less than or equal to the number of faces $f$.



      The chain $mathbb{Z}^frightarrowmathbb{Z}^erightarrowmathbb{Z}^v$ gives trivial homology if the first map is injective and the image of the first map is the kernel of the second map. Since $eleq f$ and the first map is injective, the image is whole $mathbb{Z}^e$ so the second map must be the zero map. Can you see why this is impossible?






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Observe that the given condition implies that the number of edges $e$ is less than or equal to the number of faces $f$.



        The chain $mathbb{Z}^frightarrowmathbb{Z}^erightarrowmathbb{Z}^v$ gives trivial homology if the first map is injective and the image of the first map is the kernel of the second map. Since $eleq f$ and the first map is injective, the image is whole $mathbb{Z}^e$ so the second map must be the zero map. Can you see why this is impossible?






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Observe that the given condition implies that the number of edges $e$ is less than or equal to the number of faces $f$.



          The chain $mathbb{Z}^frightarrowmathbb{Z}^erightarrowmathbb{Z}^v$ gives trivial homology if the first map is injective and the image of the first map is the kernel of the second map. Since $eleq f$ and the first map is injective, the image is whole $mathbb{Z}^e$ so the second map must be the zero map. Can you see why this is impossible?






          share|cite|improve this answer









          $endgroup$



          Observe that the given condition implies that the number of edges $e$ is less than or equal to the number of faces $f$.



          The chain $mathbb{Z}^frightarrowmathbb{Z}^erightarrowmathbb{Z}^v$ gives trivial homology if the first map is injective and the image of the first map is the kernel of the second map. Since $eleq f$ and the first map is injective, the image is whole $mathbb{Z}^e$ so the second map must be the zero map. Can you see why this is impossible?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 21:21









          LeventLevent

          2,729925




          2,729925























              1












              $begingroup$

              Hint: Think about what you can say about the Euler characteristic of $mathcal{K}$.



              A full proof is hidden below.




              Say $mathcal{K}$ has $V$ vertices, $E$ edges, and $F$ faces. Consider the set $S$ of pairs $(a,b)$ where $a$ is a face and $b$ is a boundary edge of $a$. Since every face has three boundary edges, $|S|=3F$. But since every edge is in the boundary of at least $3$ faces, $|S|geq 3E$. Thus $Fgeq E$. Now the Euler characteristic of $mathcal{K}$ is $$V-E+Fgeq V.$$ Since $mathcal{K}$ is $2$-dimensional, it has at least one face, and thus at least $3$ vertices. So the Euler characteristic of $mathcal{K}$ is at least $3$, and in particular greater than $1$, so the homology of $mathcal{K}$ cannot be trivial.







              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint: Think about what you can say about the Euler characteristic of $mathcal{K}$.



                A full proof is hidden below.




                Say $mathcal{K}$ has $V$ vertices, $E$ edges, and $F$ faces. Consider the set $S$ of pairs $(a,b)$ where $a$ is a face and $b$ is a boundary edge of $a$. Since every face has three boundary edges, $|S|=3F$. But since every edge is in the boundary of at least $3$ faces, $|S|geq 3E$. Thus $Fgeq E$. Now the Euler characteristic of $mathcal{K}$ is $$V-E+Fgeq V.$$ Since $mathcal{K}$ is $2$-dimensional, it has at least one face, and thus at least $3$ vertices. So the Euler characteristic of $mathcal{K}$ is at least $3$, and in particular greater than $1$, so the homology of $mathcal{K}$ cannot be trivial.







                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint: Think about what you can say about the Euler characteristic of $mathcal{K}$.



                  A full proof is hidden below.




                  Say $mathcal{K}$ has $V$ vertices, $E$ edges, and $F$ faces. Consider the set $S$ of pairs $(a,b)$ where $a$ is a face and $b$ is a boundary edge of $a$. Since every face has three boundary edges, $|S|=3F$. But since every edge is in the boundary of at least $3$ faces, $|S|geq 3E$. Thus $Fgeq E$. Now the Euler characteristic of $mathcal{K}$ is $$V-E+Fgeq V.$$ Since $mathcal{K}$ is $2$-dimensional, it has at least one face, and thus at least $3$ vertices. So the Euler characteristic of $mathcal{K}$ is at least $3$, and in particular greater than $1$, so the homology of $mathcal{K}$ cannot be trivial.







                  share|cite|improve this answer









                  $endgroup$



                  Hint: Think about what you can say about the Euler characteristic of $mathcal{K}$.



                  A full proof is hidden below.




                  Say $mathcal{K}$ has $V$ vertices, $E$ edges, and $F$ faces. Consider the set $S$ of pairs $(a,b)$ where $a$ is a face and $b$ is a boundary edge of $a$. Since every face has three boundary edges, $|S|=3F$. But since every edge is in the boundary of at least $3$ faces, $|S|geq 3E$. Thus $Fgeq E$. Now the Euler characteristic of $mathcal{K}$ is $$V-E+Fgeq V.$$ Since $mathcal{K}$ is $2$-dimensional, it has at least one face, and thus at least $3$ vertices. So the Euler characteristic of $mathcal{K}$ is at least $3$, and in particular greater than $1$, so the homology of $mathcal{K}$ cannot be trivial.








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 '18 at 21:21









                  Eric WofseyEric Wofsey

                  188k14216346




                  188k14216346






























                      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%2f3049848%2fexistence-of-a-type-of-simplicial-complex%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

                      To store a contact into the json file from server.js file using a class in NodeJS

                      Image
                      0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
                      Read more

                      Redirect URL with Chrome Remote Debugging Android Devices

                      Image
                      1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
                      Read more

                      Dieringhausen

                      Image
                      Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                      Read more