# Homotopy type of an oriented, closed, simply connected manifold

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.

Thanks.

• 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... Jul 5, 2016 at 2:58
• 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. Jul 5, 2016 at 3:32
• 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.
– mme
Jul 5, 2016 at 4:30
• 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. Jul 5, 2016 at 11:09

• 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? Jul 6, 2016 at 21:37