It is well known that every closed, oriented, simply-connected four-manifold $M$ is homotopy equivalent to a CW-complex consisting on a 0-cell, a wedge of two spheres and a 4-cell.

I was wondering if similar results hold for higher dimensional manifolds, in particular for closed, oriented, simply-connected and spin manifolds in dimension eight. In particular, I would like to know if a closed, oriented, simply-connected and spin 8-manifold admits a "simple" type of cell decomposition.


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    $\begingroup$ I think that a 4-manifold being simply connected, and the results that flow from that, could really a manifestation of the fact it is 4-3=1-connected. Hence for an 8-manifold you might find a nice result for 5-connected examples... $\endgroup$
    – David Roberts
    Commented Jul 5, 2016 at 2:58
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    $\begingroup$ I think you get an analogous result if you ask for it to be 3-connected. math.stanford.edu/~ksiegel/N-1Connected2NManifolds.pdf discusses a result of this form. $\endgroup$ Commented Jul 5, 2016 at 3:32
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    $\begingroup$ You don't need to demand that things be so highly connected to have classification theorems; there is a (very difficult) classification theorem for simply connected 6-manifolds; see the manifold atlas. But this is already quite complicated, and if one for simply connected 8-manifolds is possible, it would be really very complicated indeed. Perhaps with some bravery one might dare to have a classification of 2-connected 8-manifolds. $\endgroup$
    – mme
    Commented Jul 5, 2016 at 4:30
  • $\begingroup$ Maybe the following paper is interesting to you: Schmitt, Alexander, On the classification of certain piecewise linear and differentiable manifolds in dimension eight and automorphisms of connected sums of (S2×S5). Enseign. Math. (2) 48 (2002), no. 3-4, 263–289. $\endgroup$
    – user83633
    Commented Jul 5, 2016 at 11:09

1 Answer 1


Essentially the same question has been answered; see Generalizations of the handle trading techniques. For simply-connected manifolds, Smale showed that you can get a handle decomposition that is as simple as required by the homology groups. (The spin condition is not relevant at all.) A handle decomposition gives a homotopy equivalent cell complex; you won't get any simpler than this.

  • $\begingroup$ Thanks for the reference. However, the same topological space, in this case manifold, can admit many different cell decomposition. What would be the simplest one in this case? $\endgroup$
    – Bilateral
    Commented Jul 5, 2016 at 19:49
  • $\begingroup$ What do you mean by simplest? Can you give a non-trivial example of what this might mean? $\endgroup$ Commented Jul 5, 2016 at 21:13
  • $\begingroup$ For example, from the fact that $M$ is closed and oriented we deduce that it is homotopy equivalent to a complex having a unique 0-cell and a unique 8-cell. Is there a way to deduce that other k-cells should be absent, or there should be a given number of them? $\endgroup$
    – Bilateral
    Commented Jul 6, 2016 at 21:37
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    $\begingroup$ It's essentially an algebraic problem. Look at the homology groups, and figure out a chain complex with the fewest of generators in each dimension that could give those groups. Then Smale says you can find a cell decomposition with exactly the number of cells in each dimension as there are generators of the chain groups. For instance, if the homology groups are free (as in your example) then you'd have one cell per homology generator and boundary operators 0. If there's torsion in the homology, you'd need some additional cells because there must be some non-trivial boundary operators. $\endgroup$ Commented Jul 7, 2016 at 15:30
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    $\begingroup$ The vanishing of Betti numbers doesn't mean that the homology groups vanish! So you'd need cells to account for the torsion in the homology, which is not ruled out by the Kaehler condition. If there's a lot of torsion, you need a lot of cells. So you need to know the integral homology, not just what you can deduce from Hodge theory. $\endgroup$ Commented Jul 7, 2016 at 19:37

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