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.
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3 Answers
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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, $\chi(M \# N) = \chi(M) + \chi(N) -2$.
To construct manifolds with nontrivial Kirby-Siebenmann invariant we should apply Freedman's theorem: simply connected topological 4-manifolds are determined by their intersection form and Kirby-Siebenmann invariant; if the intersection form is odd, both Kirby-Siebenmann invariants are realized, and if the intersection form is even, only one KS-invariant is realized. So there is a topological manifold $F(\Bbb{CP}^2)$, homotopy equivalent to $\Bbb{CP}^2$ but with nontrivial Kirby-Siebenmann invariant.
So for the desired non-smoothable manifold with zero Euler characteristic, one can take (for instance) $$F(\Bbb{CP}^2) \# 3\Bbb{CP}^2 \#(S^2 \times \Sigma_2).$$
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$\begingroup$ Is there any easy explanation on why $F(CP^2)$ exists? $\endgroup$ Commented Sep 27, 2018 at 11:36
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$\begingroup$ yes you are right. All of a sudden I forgot about that theorem. Thanks :) $\endgroup$ Commented Sep 27, 2018 at 13:07
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$\begingroup$ It is not stated at the link I provided, sorry. I edited a statement into my answer. $\endgroup$– mmeCommented Sep 27, 2018 at 13:15
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$\begingroup$ This statement is there in the Kirby calculus book by Gompf and Stipshicz $\endgroup$ Commented Sep 27, 2018 at 13:20
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Alternatively, one can start with the $E_8$-manifold and connect sum with (five copies of) $S^1\times S^3$, and appeal to Donaldson's diagonalisation theorem instead.
More precisely, the (negative) $E_8$-plumbing $P$ bounds the Poincaré homology sphere $Y$; by Freedman's theorem, $Y$ is also the boundary of a contractible topological 4-manifold $W$, and gluing $P\cup_Y -W$ yields a closed, simply connected topological 4-manifold $X$ with intersection form $-E_8$; $\chi(X) = 10$.
The connected sum $X\#5(S^1\times S^3)$ has Euler characteristic 0, by Mike's computation, and the intersection form is still $-E_8$. Hence, by Donaldson's theorem, $X\#5(S^1\times S^3)$ does not have a smooth structure, since its intersection form is negative definite but not diagonal.
answered Sep 27, 2018 at 13:25
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$\begingroup$ Nice answer, Marco--I simultaneously wrote the same thing! $\endgroup$ Commented Sep 27, 2018 at 13:27
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1$\begingroup$ Speaking of ignorance from Freedman's work (something I should one day remedy), is the construction of the contractible manifold bounding the Poincare sphere significantly easier than the full power of his 1982 JDG paper? $\endgroup$– mmeCommented Sep 27, 2018 at 13:30
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3$\begingroup$ The hardest part of all of those theorems is finding a disk inside a Casson handle (plus Casson's original work putting Casson handles where you want disks.) After that, the other theorems about simply-connected manifolds require work, but the machinery was pretty well understood. $\endgroup$ Commented Sep 28, 2018 at 0:50
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You can also get this from Donaldson's theorem by a similar device. Take a non-diagonalizable definite form with even rank $2n$, and realize it (Freedman again) by a simply connected manifold. Then $W\ \#\ (n+1) (S^1 \times S^3)$ is not smoothable and has Euler characteristic $0$. The argument is that if it were smoothable, then you could surger away the fundamental group and realize that intersection form by a simply connected manifold.
answered Sep 27, 2018 at 13:26