Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
Should be Dranishnikov? –  Mark Sapir May 21 '12 at 21:06
3  
@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 arxiv.org/abs/math/0010141 would change if one defines it using smooth (!) properly discontinuous actions. –  Igor Belegradek May 21 '12 at 22:03
1  
@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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.