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

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.

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) $$F(\Bbb{CP}^2) \# 3\Bbb{CP}^2 \#(S^2 \times \Sigma_2).$$

• Is there any easy explanation on why $F(CP^2)$ exists? – Anubhav Mukherjee Sep 27 '18 at 11:36
• @AnubhavMukherjee This is Freedman's theorem, stated here. I do not think there is an easy explanation for the existence of any non-smoothable 4-manifold. – Mike Miller Sep 27 '18 at 13:04
• yes you are right. All of a sudden I forgot about that theorem. Thanks :) – Anubhav Mukherjee Sep 27 '18 at 13:07
• It is not stated at the link I provided, sorry. I edited a statement into my answer. – Mike Miller Sep 27 '18 at 13:15
• This statement is there in the Kirby calculus book by Gompf and Stipshicz – Anubhav Mukherjee Sep 27 '18 at 13:20

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.

• Nice answer, Marco--I simultaneously wrote the same thing! – Danny Ruberman Sep 27 '18 at 13:27
• Likewise, Danny! :) – Marco Golla Sep 27 '18 at 13:27
• 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? – Mike Miller Sep 27 '18 at 13:30
• 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. – Danny Ruberman Sep 28 '18 at 0:50

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.