# Manifolds admitting CW-structure with single n-cell

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.

[]: 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)

• 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/… Feb 4, 2013 at 21:52
• 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? Feb 4, 2013 at 22:28
• 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. Feb 4, 2013 at 22:35
• @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). Feb 4, 2013 at 23:17
• See also mathoverflow.net/questions/42234/rugged-manifold (especially Greg's answer to a similar question; it works in dimension 4 as well). Feb 8, 2013 at 21:12

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.
• Where do you use $n=4$? Sep 26, 2017 at 4:50
• @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 $h\circ\varphi$. Sep 28, 2017 at 0:29