# Smooth structure on direct product

Let $$M$$ be the $$E_8$$ manifold. Is there a closed manifold $$N$$ such that $$M\times N$$ is smoothable? What is the smallest possible dimension of $$N$$?

• Worth pointing out that $M\times S^k$ is not smoothable for any $k$; see the lemma on page 219 of The Wild World of 4-Manifolds by Scorpan. – Michael Albanese Sep 29 at 15:53
• @MichaelAlbanese Shouldn't its product with $S^1$ be smoothable? The microbundle of $M$ is stably parallelizable, so the tangent bundle of this product should be trivial, correct? Then smoothing theory shows a parallelizable manifold has at least one smooth structure. – Connor Malin Sep 29 at 17:35
• @ConnorMalin: I am not that familiar with microbundles, but wouldn't there be a non-zero $p_1$? – Michael Albanese Sep 29 at 18:00
• @MichaelAlbanese I definitely could be misunderstanding, but I thought the Milnor manifolds (of which M is the first example) were constructed to show that the surgery obstruction map from the normal invariants of the sphere to $8\mathbb{Z}$ were surjective. Here the surgery obstruction map takes a normal invariant of the sphere to its signature. – Connor Malin Sep 29 at 18:10
• To be the domain of a normal invariant of the sphere is equivalent to have your Spivak normal bundle be trivial, which is weaker than having the microbundle being stably parallelizable (which for a smoothable manifold means the tangent bundle is trivial as an $\mathbb{R}^n$ bundle), so perhaps I have misunderstood and in fact only the Spivak normal bundle is trivial, not the tangent microbundle. – Connor Malin Sep 29 at 18:12

Extending Michael Albanese's answer above, $$M \times N^k$$ will never be smoothable. For if it were then choose a point $$p\in N$$ and a chart U around $$p$$. Then $$M \times U$$ is an open subset of $$M \times N$$, and hence is smoothable. But as argued in Scorpan (p. 219, Lemma), $$M \times \mathbb{R}^k$$ is not smoothable.
Scorpan's argument uses the Kirby-Siebenmann product structure theorem to get down to dimension $$5$$. But I believe (it's been a while) that you could construct an argument based on Novikov's work on signatures.