Let $M$ be a topological $n$-manifold, closed and connected (not necessarily oriented):
When does $M$ not admit (up to homotopy-type) a CW-structure with a single $n$-cell?

By classification of surfaces we assume $n>2$. By existence of smooth structures we assume $n>3$. In particular, if $M$ is smoothable then Morse theory provides us the desired structure.

[[Edit]]: To put this question into context, we have various ways of showing that $H_{n-1}(M)$ has either $0$ or $\mathbb{Z}_2$ as its torsion subgroup depending on orientability. One way, when $M$ is such a CW-complex, is to quickly look at the chain-complex differential $d:C_n(M)\cong\mathbb{Z}\to C_{n-1}(M)$ and note that $H_n(M)\cong\mathbb{Z}$ for $M$ orientable and $H_n(M;\mathbb{Z}_2)\cong\mathbb{Z}_2$ otherwise. So I would like to see how large of a class of manifolds this argument holds for.

[[Addendum]]: After chatting with Allen Hatcher and Rob Kirby, who reaffirm the comments below, here are their resulting thoughts:
1) We should be careful with the Kirby theorem of $M$ being homotopy-equivalent to a finite complex, because this complex is obtained by first embedding $M$ into $\mathbb{R}^N$ and then wiggling the boundary of a tubular neighborhood ($M\times D^{N-n}$) of $M$ to be PL, and so the resulting complex could have $i$-cells with $i>n$.
2) When $\dim M\ne 4$ there is a handlebody-decomposition, and this can be arranged to have a single 0-handle (canceling the other 0-handles with available 1-handles -- we can do this because there are no smoothing obstructions in a neighborhood of the 3-skeleton). Taking the dual handlebody, we have a decomposition with a single n-handle. Passing from the handlebody-decomposition to the CW-decomposition (shrinking everything to their cores), we obtain the desired CW-complex with a single n-cell.
3) When $\dim M=4$ then a handlebody-decomposition exists if and only if $M$ is smoothable. So when $M$ is smoothable we can apply the argument in (2).
4) But even when $M$ is not smooth we get some positive results, in particular for the $E_8$ manifold. We build $E_8$ using Kirby calculus on an 8-link diagram, giving a decomposition of $E_8$ into a 0-handle plus eight 2-handles plus a contractible piece (without the contractible piece we get a space with boundary being a homology 3-sphere, namely the Poincare-sphere $S^3/G$ with $G=$ binary icosahedral group). In particular, flipping this structure over we see that $E_8$ is homotopy-equivalent to a CW-complex with a single 4-cell. Furthermore, Lennart Meier's remark gets us all other simply-connected 4-manifolds.

We are thus left with the scenario that $M$ (up to homotopy) is a closed connected non-simply-connected non-smoothable 4-manifold. (which the comments below assert)

  • 2
    $\begingroup$ Since not all topological manifolds admit CW structures, your question must be about homotopy type. So any counter-examples would have to fail to be homotopy equivalent to a smooth manifold. The examples here might be candidates: mathoverflow.net/questions/34848/… $\endgroup$
    – Mark Grant
    Feb 4, 2013 at 21:52
  • 1
    $\begingroup$ Isn't the $E_8$ manifold homotopy equivalent to a CW complex with one 0-cell, 8 2-cells, 0 3-cells, and 1 4-cell? $\endgroup$
    – Will Sawin
    Feb 4, 2013 at 22:28
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    $\begingroup$ I just want to remark: If you care only about homotopy type, then every simply-connected closed, connected manifold has a CW-structure with just one 1-cell. This follows from Proposition 4C.1 of Hatcher's Algebraic Topology (about minimal cell structures) and Poincare duality. This does not exclude the aspherical manifold mentioned by Misha in the other Question. $\endgroup$ Feb 4, 2013 at 22:35
  • 4
    $\begingroup$ @Mark: Nevertheless, every compact manifold of dimension other than four admits a CW-structure (see mathoverflow.net/questions/36838). The question posed by Chris Gerig then makes sense precisely as stated. In fact, it is still interesting in dimension four, even if some 4-manifolds may not admit a CW-structure. Note: the question of CW-structures on 4-manifolds seems to be fairly open (see mathoverflow.net/questions/73428). $\endgroup$ Feb 4, 2013 at 23:17
  • 1
    $\begingroup$ See also mathoverflow.net/questions/42234/rugged-manifold (especially Greg's answer to a similar question; it works in dimension 4 as well). $\endgroup$
    – Misha
    Feb 8, 2013 at 21:12

1 Answer 1


Let $M$ be a closed topological four-manifold and $D$ an embedded closed four-disc, then $M\setminus D^{\circ}$ is a four-manifold with boundary $\partial D = S^3$. As I learnt from this answer, the boundary has a collar neighbourhood and hence $M\setminus D^{\circ}$ is homotopy equivalent to its interior, namely $M\setminus D$. As $M\setminus D$ is an open four-manifold, it is smoothable (a fact I learnt from this answer). A smooth open $n$-manifold is homotopy equivalent to an $(n-1)$-dimensional CW complex (see this answer), so $M\setminus D$ is homotopy equivalent to a three-dimensional CW complex $X$.

As $M$ is homeomorphic to $M\setminus D^{\circ}$ with a $4$-disc attached, and $M\setminus D^{\circ}$ is homotopy equivalent to $X$, $M$ is homotopy equivalent to $X$ with a $4$-disc attached, i.e. a CW complex with one top-cell.

  • $\begingroup$ Where do you use $n=4$? $\endgroup$ Sep 26, 2017 at 4:50
  • $\begingroup$ An open four-manifold is smoothable. I don't know if this is true in higher dimensions (I suspect it isn't). $\endgroup$ Sep 26, 2017 at 12:29
  • $\begingroup$ My hesitation is in the last sentence: How is the original 4-disc now attached to X? Does the attaching map "follow" the homotopy somehow, and does that give an honest CW decomposition? (I'll think more later) $\endgroup$ Sep 28, 2017 at 0:26
  • $\begingroup$ @ChrisGerig: I understand your hesitation and it's not completely clear to me that it works. If $\varphi : S^3 \to M\setminus D^{\circ}$ is the original attaching map and $h : M\setminus D^{\circ} \to X$ is the homotopy equivalence, then the new attaching map should be $h\circ\varphi$. $\endgroup$ Sep 28, 2017 at 0:29

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