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







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$endgroup$












  • $begingroup$
    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






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oldest

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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









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      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






























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