What information is contained in the 'list of cells' of CW-complexes

In Milnor's book on Morse theory there are many statements of the form '[...] has the homotopy type of a CW-complex with these kind of cells'. For example, Theorem 17.3 (fundamental theorem of Morse theory) states that the path-space $$\Omega(p,q)$$ (between two points $$p,q$$ on a complete manifold $$M$$) has one cell of dimension $$\lambda$$ for each geodesic between $$p$$ and $$q$$ of index $$\lambda$$.

I would like to understand how much information this gives about the space under consideration. Certainly, to describe the whole CW-structure, one needs to describe the attaching maps as well. But of course, after classifying by homotopy type some information is lost.

Question. Suppose I know that a space $$X$$ has the homotopy type of a CW-complex with $$k_0$$ zero-cells, $$k_1$$ one-cells and so on. What information about $$X$$ can I deduce from the list $$(k_0,k_1,\dots)$$?

Surely in some special cases one can compute homology groups: Just because $$\mathbb{C}P^n$$ has exactly one cell of dimension $$2j$$ for $$j=0,\dots,n$$, we know how its homology groups look like. I would also not be suprised (but cannot efficiently verify this myself, because my algebraic topology is quite rusty), that any any CW-complex with the same list as $$\mathbb{C}P^n$$ is homotopy equivalent to it. (wrong, see comments)

In general however, not the full homotopy type is captured by the list of cells: If there is exactly one cell of dimension zero, one and two, then everything depends on the degree $$d$$ of the last attaching map. If we call the resulting space $$X_d$$, then its fundamental group is $$\mathbb{Z}/d$$ and thus $$X_{d}$$ and $$X_{d'}$$ are homotopy equivalent if and only if $$d=d'$$.

Addendum 1: I am happy about all concrete examples, like e.g. in Lee Mosher's comment. In a slightly different direction, I would also appreciate if someone can comment on the situation in Morse theory: Is the list of cells just all we get from Morse theory, is it because the attaching maps do not have a nice geometric interpretation in terms of the underlying manifold $$M$$, etc...

Addendum 2: If it turns out that (besides Euler characteristic and the case of cells only in non-adjacent dimension) little can be said from the list of cells, then what is the main message that Milnor wants to send with this phrasing? (Below the theorem quoted above, he does mention an application, where $$X=\Omega(p,q)$$ is the path-space of $$S^n$$ (round sphere). For $$n\ge 3$$ one is again in the case that cells don't live in adjacent dimensions and there are some nice geometric consequences for Riemannian manifolds that are homotopy equivalent to $$S^n$$, but again, here we are in this special case.)

• For starters, you can deduce the Euler characteristic $$\sum_i (-1)^i k_i = \sum_i (-1)^i \text{Rank}(H_i(X;\mathbb Z))$$ – Lee Mosher Jul 24 at 13:25
• It is definitely not true that any complex with the same cells as $\mathbb{C}P^n$ is equivalent to it. This would imply it is a wedge of spheres, but cohomology ring implies this can't be. I'd be surprised if there was anything you can say besides Euler characteristic and homology in cases where cells don't live in adjacent dimensions. This is with the exception of one cell complexes (obvious), complexes of dimension 1, and complexes with cells in only dimensions 0, 1, and n>2. – Connor Malin Jul 24 at 15:30
• There are analogous results for other homology theories when we have cells only in nonadjacent dimensions, but they need more restriction. The study of two cell complexes with information about the attaching map is very difficult. One small portion of it is given by the Hopf Invariant 1 problem. This is a difficult problem that took many years to solve, and it only cares about the ring structure on cohomology. – Connor Malin Jul 24 at 15:32