If you have a $CW-$structure on a connected manifold $M$ we obtain a filtration $M_n$ of $M$ where $M_n$ is the $n-$skeleton. If in addition we have that the $M_n$ are also all manifolds (like with the standard $CW-$decomposition of $\mathbb RP^n$ or $\mathbb CP^n$) what can we say about the manifold $M$?

Not all manifolds $M$ have such a CW structure.

Like $S^1 \times S^1 = M$ for example, $M_1$ has to be either a point or a circle since it can't be a wedge of circles since that's not a manifold, it also has to be connected since if $M_1$ is disconnected $M_2$ will always be disconnected but that's impossible since $M_2$ is supposed to be $S^1 \times S^1$, and if we denote by $m_i$ the number of $i-$cells in $M$ we have that

$\text{rank } H_i(M) \leq m_i$ (by a $CW-$homology argument) which is not true for $i = 1$ in $S^1 \times S^1$ since $H_1(S^1 \times S^1) = \mathbb Z^2$ and $m_1 = 1$ or $0$.

How far can we go with this? Is there some easy way to see through invariants like homology or cohomology wether such a filtration $M_n$ exists or not? All of the spaces $S^n, \mathbb RP^n, \mathbb CP^n, \mathbb HP^n$ are examples of such manifolds. Don't know of any else from the top of my head.


I believe you gave a homological classification.

Let's work with $\mathbb{Z}/2$ coefficients; then Poincare duality tells us that if our lowest homology is in degree $k$, to obtain a manifold from $M^n$ by attaching higher degree cells we must be attaching degree $(n+k)$-cells. Again by Poincare duality (obvious in the case $k>1$) is that we can only attach a single $(n+k)$-cell because the differential from the $(n+k)$-cells is homologically trivial since otherwise it would destroy Poincare duality.

Let us call the generator of $H^k(M)$ the element $x$, we first show that there is a m so $x^m=g$ generates $H^n (M)$. We know that there is a $y$ so $xy=g$ and by Poincare duality $x^2 y=g'$ where $g'$ generates $H^{n+k}(M')$. Hence $x^2$ is nontrivial. Repeating this, we know there is a $y'$ so $x^2 y'=g$ and $x^3 y'=g'$, so $x^3$ is nontrivial. Repeating this we conclude that if $|x^{m'}| \leq n$ then $x^{m'+1}$ is nontrivial, so by degree considerations and Poincare duality we know there is a $m$ so $x^m =g$, the generator of $H^n (M)$.

Hence, the cohomology of $M'$ contains a truncated polynomial algebra on $x$ with highest degree element $n+k$. Could there be other elements? No, since for any element $y$, we have some equation $yy'=g=x^m$ and so $yy'x=g'$, but by Poincare duality $yx=0$ if $y$ is not a power of $x$. Of course this is rather sloppy, to write this down formally we need to pick a basis of the cohomology and talk about basis elements $y \neq x^l$, but this is possible because the pairing by the cup product is nonsingular.

Hence the cohomology algebra of $M'$ is a truncated polynomial algebra. Hopf classified up to $\mathbb{Z}/2$ homology equivalence all spaces which have truncated polynomial $\mathbb{Z}/2$ cohomology algebras and they are exactly what you listed. It is also known that there are many manifolds homotopy equivalent but not homeomorphic to projective spaces. Obviously this is only the algebraic obstructions, but I suspect it will not be possible to give an actual local analysis. There are many types of these questions where giving a homological answer is rather easy, but strengthening this to homeomorphism is almost impossible. For example, the topological Poincare conjecture is equivalent to the statement "The suspension of the manifold $M$ is a manifold, if and only if, $M$ is a sphere." Whereas it is extremely easy to see $M$ must be a homology sphere.

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    $\begingroup$ Oh and by the way I only used that its largest proper skeleton was a manifold, not that all were. $\endgroup$ – Connor Malin Feb 15 at 19:39
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    $\begingroup$ One small comment: $\mathbb{O}P^2$ should also be on the list. $\endgroup$ – Jason DeVito Feb 22 at 15:30

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