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It is known that any closed topological manifold is homotopy equivalent to an open smooth manifold.

Question 1. What can be said about the smallest dimension of a smooth manifold that is homotopy equivalent to a given closed topological manifold?

The following somewhat heavy-handed argument yields a smooth manifold of roughly twice the dimension, namely, use

  • West's solution of Borsuk's conjecture that any compact ANR of dimension $n$, and in particular any closed $n$-manifold, is homotopy equivalent to a finite polyhedron of dimension $\max\{n, 3\}$;
  • Stallings-Dranishnikov-Repovs's embedding up to homotopy type theorem that any finite $n$-dimensional polyhedron is homotopy equivalent to a finite $n$-dimensional subpolyhedron of $\mathbb R^{2n}$, so its regular neighborhood is the desired open smooth manifold.

Here is a specific question that shows the state of my ignorance on this matter:

Question 2. Is there a closed $n$-manifold which is not homotopy equivalent to a smooth $(n+1)$-manifold?

The naive idea to look at the product of a non-smoothable manifold of dimension $\ge 5$ with $\mathbb R$ fails, because such a product is also non-smoothable (by the topological product structure theorem of Kirby-Siebenmann).

Edit: Misha kindly corrects me that a $5$-manifold is smoothable if and only if its Kirby-Siebenmann invariant vanishes; in particular, this apples to products of a $\mathbb R$ and a closed $4$-manifold $M$. Thus $M\times\mathbb R$ is smoothable iff $M$ has zero KS invariant. Smooth $4$-manifolds have zero KS invariant, but amazingly so do some non-smoothable ones.

Igor Belegradek
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  • Should be Dranishnikov? –  May 21 '12 at 21:06
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    @Igor: 1. Sometimes, the product with ${\mathbb R}$ is smoothable. 2. You can avoid Stallings' theorem if you use immersion to ${\mathbb R}^{2n}$ instead of an embedding. 3. Try Davis-Januszkiewicz argument, see the reference here and see if this gives a counter-example at least for a tame smoothing. (Try to argue that your open smooth (n+1)-manifold is properly homotopy-equivalent to an open interval bundle over a non-smoothable n-manifold.) – Misha May 21 '12 at 21:08
  • @Mark: corrected; I hope you won't insists on accents in Repovs. @Misha: Do you know a $4$-dimensional example? In higher dimensions there is an old preprint of Galewski-Hollingworth that implies that there is no $(n+1)$-dimensional compact smooth thickening $V$ of an $n$-dimensional Poincare complex $X$, provided both of them are orientable; compactness of $V$ can be replaced by the assumption that an image of $X$ separates a neighborhood in $V$. Showing that $V$ is properly h.e. to an $\mathbb R$-bundle seems tricky. – Igor Belegradek May 21 '12 at 21:36
  • @Misha: I am familiar with examples of closed non-smoothable aspherical manifolds by Davis-Januszkiewicz and Davis-Hausmann, in fact, my question was motivated by the desire to see how much the action dimension of http://arxiv.org/abs/math/0010141 would change if one defines it using smooth (!) properly discontinuous actions. – Igor Belegradek May 21 '12 at 22:03
  • @Igor: Actually, I realized that even for n=3, a tame 4-dimensional thickening of a hyperbolic 3-manifold need not be properly homotopy-equivalent to an interval bundle. – Misha May 23 '12 at 15:06
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    @Igor: Concerning KS invariant for 4-manifolds: It vanishes iff $M^4\times {\mathbb R}$ is smoothable. In dimension 4 there are non-smoothable manifolds with both zero and non-zero KS invariants since there are other, gauge-theoretic, obstructions to smoothability. – Misha May 23 '12 at 15:12
  • @Misha: certainly, not all $(n+1)$-dimensional thickenings of closed $n$-manifolds are proper homotopic to $I$-bundles. One can start with an $I$-bundle, and then attach a one-sided $h$-cobordism along the boundary; there are general methods to construct those in higher dimensions. One can also take boundary connected sums with (possibly noncompact) contractible manifolds with boundary, and this way one gets tons of examples when $n\ge 3$ whose ends are quite complicated. – Igor Belegradek May 23 '12 at 15:33

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