What manifolds $M$ have a $CW-$structure so that the $n-$skeleton, $M_n$, is a manifold for all $n$ aswell?

Solution 1:

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.