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Does there exist a closed orientable smooth 4-manifold $M$ whose intersection form is the $E_8$-form? Here by the intersection form I mean the $\mathbb{Z}$-valued bilinear form on $H^2(M;\mathbb{Z})/\text{torsion}$ induced by the cup product map $$H^2(M;\mathbb{Z}) \times H^2(M;\mathbb{Z}) \longrightarrow H^4(M;\mathbb{Z}) \cong \mathbb{Z}.$$

By Rochlin's theorem, this is not possible if $M$ is spin. It is almost the case that if the intersection form on a 4-manifold $M$ is even, then $M$ is spin. However, this is not necessarily the case if $H_1(M;\mathbb{Z}/2)$ is nonzero. So we're looking for a manifold with an interesting fundamental group. It's not clear to me whether or not such a thing exists.

I'd also be interested in realizing any other even form whose signature is nonzero modulo $16$.

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    $\begingroup$ "Any other" form, viz. $E_8\oplus2U$, is realized by an Enriques surface. (Starting from that, you can do connected sums.) Of course, you also have $S^2\times S^2$. $\endgroup$
    – Alex Degtyarev
    Commented Jan 12, 2015 at 22:20
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    $\begingroup$ Yes, so you can get $E_8\oplus kU$ for $k\ge2$. $\endgroup$
    – Alex Degtyarev
    Commented Jan 12, 2015 at 22:27
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    $\begingroup$ I don't think the fundamental group matters for Donaldson's diagonalisation theorem. $\endgroup$
    – Johannes Nordström
    Commented Jan 13, 2015 at 1:17
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    $\begingroup$ @AlexDegtyarev: The Enriques surface has intersection form $E_8\oplus U$, not $2U$. Indeed, its second Betti number is 10, since $b_1=0$, and the Euler characteristic is 1/2 the Euler characteristic of the K3 surface. $\endgroup$
    – Alex Suciu
    Commented Jan 13, 2015 at 3:56
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    $\begingroup$ There is a proof of Donaldson's diagonalisation theorem (without the assumption on $\pi_1$) in Ozsváth-Szabó's Absolutely graded Floer homologies and intersection forms for four-manifolds with boundary, which is an adaptation of Frøyshov's proof (for which I do not have a reference). $\endgroup$
    – Marco Golla
    Commented Jan 13, 2015 at 13:09

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This post is a summary of the comments above.

No, such a manifold doesn't exist. Donaldson proved in 1982 that if the intersection form of a simply connected, closed, orientable, smooth 4-manifold is definite, then it is diagonal.

Since then, the theorem has been extended to arbitrary fundamental groups: see (for example) Ozsváth and Szabó's 2003 paper Absolutely graded Floer homologies and intersection forms for four-manifolds with boundary in Adv. Math.

In particular, $E_8$ can't be the intersection form of a closed, orientable, smooth 4-manifold.

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    $\begingroup$ The result actually is due to Donaldson; see The orientation of Yang-Mills moduli spaces and 4-manifold topology, JDG 26 (1987) 397-428. $\endgroup$
    – Danny Ruberman
    Commented Jul 27, 2015 at 13:40
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    $\begingroup$ It occurred to me that I should point out a gap the proof of lemma 2.9 of Donaldson's paper; the "simple algebraic fact" left to the reader is actually incorrect! This was noted by Chris Herald, and in fact he gave a counterexample to that "fact". On the other hand, the result (that there exist holonomy perturbations to get rid of flat connections) still holds. One reference for the details (the idea basically follows Donaldson) is section 4 of my paper with Saveliev, Geom. & Top. 9 (2005) 2079–2127. $\endgroup$
    – Danny Ruberman
    Commented Jul 27, 2015 at 19:09

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