TITLE: Smooth 4-manifolds with $E_8$ intersection form
QUESTION [16 upvotes]: 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 $16$.

REPLY [5 votes]: 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.