Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

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I will just repeat the title:




Is there a closed non-smoothable 4-manifold with zero Euler
characteristic?




I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.






smooth-manifolds manifolds topological-manifolds







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    up vote
    1
    down vote

    favorite












    I will just repeat the title:




    Is there a closed non-smoothable 4-manifold with zero Euler
    characteristic?




    I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.






    smooth-manifolds manifolds topological-manifolds







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I will just repeat the title:




      Is there a closed non-smoothable 4-manifold with zero Euler
      characteristic?




      I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.






      smooth-manifolds manifolds topological-manifolds







      share|cite|improve this question













      I will just repeat the title:




      Is there a closed non-smoothable 4-manifold with zero Euler
      characteristic?




      I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.






      smooth-manifolds manifolds topological-manifolds




      smooth-manifolds manifolds topological-manifolds






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      Cihan

      34129




      34129




















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          The Kirby-Siebenmann invariant in $H^4(M;Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $chi(M # N) = chi(M) + chi(N) -2$. So one can take eg $F(BbbCP^2) # 3BbbCP^2 #(S^2 times Sigma_2)$, where $F(BbbCP^2)$ is the topological manifold homotopy equivalent to $BbbCP^2$ but with non-trivial Kirby-Siebenmann invariant.






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            The Kirby-Siebenmann invariant in $H^4(M;Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $chi(M # N) = chi(M) + chi(N) -2$. So one can take eg $F(BbbCP^2) # 3BbbCP^2 #(S^2 times Sigma_2)$, where $F(BbbCP^2)$ is the topological manifold homotopy equivalent to $BbbCP^2$ but with non-trivial Kirby-Siebenmann invariant.






            share|cite|improve this answer
























              up vote
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              The Kirby-Siebenmann invariant in $H^4(M;Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $chi(M # N) = chi(M) + chi(N) -2$. So one can take eg $F(BbbCP^2) # 3BbbCP^2 #(S^2 times Sigma_2)$, where $F(BbbCP^2)$ is the topological manifold homotopy equivalent to $BbbCP^2$ but with non-trivial Kirby-Siebenmann invariant.






              share|cite|improve this answer






















                up vote
                5
                down vote










                up vote
                5
                down vote









                The Kirby-Siebenmann invariant in $H^4(M;Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $chi(M # N) = chi(M) + chi(N) -2$. So one can take eg $F(BbbCP^2) # 3BbbCP^2 #(S^2 times Sigma_2)$, where $F(BbbCP^2)$ is the topological manifold homotopy equivalent to $BbbCP^2$ but with non-trivial Kirby-Siebenmann invariant.






                share|cite|improve this answer












                The Kirby-Siebenmann invariant in $H^4(M;Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $chi(M # N) = chi(M) + chi(N) -2$. So one can take eg $F(BbbCP^2) # 3BbbCP^2 #(S^2 times Sigma_2)$, where $F(BbbCP^2)$ is the topological manifold homotopy equivalent to $BbbCP^2$ but with non-trivial Kirby-Siebenmann invariant.







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                answered 3 hours ago









                Mike Miller

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