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 $N$?
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6$\begingroup$ 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. $\endgroup$– Michael AlbaneseCommented Sep 29, 2020 at 15:53
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$\begingroup$ @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. $\endgroup$– Connor MalinCommented Sep 29, 2020 at 17:35
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$\begingroup$ @ConnorMalin: I am not that familiar with microbundles, but wouldn't there be a non-zero $p_1$? $\endgroup$– Michael AlbaneseCommented Sep 29, 2020 at 18:00
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$\begingroup$ @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. $\endgroup$– Connor MalinCommented Sep 29, 2020 at 18:10
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$\begingroup$ 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. $\endgroup$– Connor MalinCommented Sep 29, 2020 at 18:12
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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.
answered Sep 29, 2020 at 17:40