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The $E_8$ manifold can be constructed by plumbing together disk bundles of Euler number $2$ over $2$-sphere $S^2$ according to the Dynkin diagram for $E_8$.

In Smooth 4-manifolds with $E_8$ intersection form, it is written that an $E_8$ manifold cannot be smoothed by Donaldson's diagonalization theorem. By Freedman - The topology of four-dimensional manifolds, on the other hand, $E_8$ manifold bounds a contractible topological $4$-manifold. So this is a very interesting example for $4$-dimensional topology.

Saveliev (as well as several mathematicians) wrote in the book Invariants of Homology 3-Spheres that such a plumbing construction gives rise to a simply connected smooth $4$-manifold.

Where is the problem? Are they just typos, or does pasting disk bundles create "corners" which cannot be smoothed so that the resulting $4$-manifold might not be smooth?

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    $\begingroup$ The plumbing along the $E_8$ graph yields a smooth four-manifold with intersection form $E_8$ that bounds the Poincare homology sphere. Freedman constructed the $E_8$ manifold from this by attaching a compact contractible topological four-manifold that bounds the Poincare homology sphere. Donaldson's theorem essentially implies you can't do this smoothly. $\endgroup$
    – Rohil Prasad
    Commented Nov 16, 2020 at 20:54
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    $\begingroup$ Donaldson's diagonalization theorem implies any closed smooth four-manifold cannot have intersection form $E_8$. The plumbing only gives a smooth four-manifold with boundary with intersection form $E_8$. However, Freedman showed that any integral homology three-sphere bounds a compact contractible topological four-manifold, so you can cap off the Poincare sphere boundary to get a topological four-manifold with intersection form $E_8$, which is usually what people call the "$E_8$-manifold". $\endgroup$
    – Rohil Prasad
    Commented Nov 16, 2020 at 21:04
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    $\begingroup$ So, the critical word is "closed", right? $\endgroup$
    – user160180
    Commented Nov 16, 2020 at 21:12
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    $\begingroup$ And the plumbing $4$-manifold with $E_8$ and Freedman's $E_8$ manifold are not same. $\endgroup$
    – user160180
    Commented Nov 16, 2020 at 21:16
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    $\begingroup$ Before Donaldson, there was Rokhlin's theorem, that the signature of the intersection form of a smooth closed spin 4-manifold is divisible by 16 (rather than just 8 as the arithmetic implies). This is enough to show that the $E_8$ manifold is not smoothable. $\endgroup$
    – Ben Wieland
    Commented Nov 17, 2020 at 3:08

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