When can we recover a manifold when we attach a $2n$-cell to $S^n$?

I have a question related to this one. In my answer I was going to try and say something about the possible manifolds that might arise in this way, i.e. as mapping cones of elements of $\pi_{2n-1}(S^n)$. Certainly not all of them will be manifolds. At first I felt like there should be a reason why we'd need to be using a torsion-free homotopy generator if we want a manifold (in which case we could appeal to Serre's theorem that the only non-torsion part of the homotopy groups of spheres is $\pi_n(S^n)=\mathbb{Z}$ and $\pi_{4n-1}(S^{2n})=\mathbb{Z}$), but then I realized I couldn't think of any reason why that should be true. Also I was hoping the Hopf invariant would enter into the picture too, since that's a pretty obvious tool at our disposal, but I couldn't get anywhere with that either...and I think at this point I'm more or less all out of tricks. Does anyone have any ideas?

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I believe this is related to the Hopf invariant one problem. Let $X = S^n \cup_f B^{2n}$.

The attaching map from $B^{2n}$ is a map $f$ from $S^{2n-1}$ to $S^n$. I think one way of thinking about the Hopf invariant of this map is as follows.

Let $x\in H^n(X)$. Then $x^2 = k y$ where $y$ generates $H^{2n}(X)$. The number $k$ is the Hopf invariant.

In order for $X$ to be a manifold, it must satisfy Poincare duality (so long as $n>1$ - if $n=1$ different stuff can happen), which implies that $k = \pm 1$. But Adams has shown you can only have Hopf invariant 1 when $n = 2, 4, 8$, getting you something homotopy equivalent to $\mathbb{C}P^2$, $\mathbb{H} P^2$ or $\mathbb{O}P^2$.

The $n=1$ case, you're attaching a disc to a circle and getting a manifold. Here, since $\pi_1(X)$ may be nontrivial, one cannot neccesarily use Poincare duality. However, by a simple analysis, the only manifold this can give rise to is $\mathbb{R}P^2$.

Edit If one wants to avoid cases ($n = 1$ vs $n > 1$) then one can work with cohomology with $\mathbb{Z}/2\mathbb{Z}$ coefficients instead.

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Thanks for fixing my typo. I was thinking $16$ because for some reason I have $\mathbb{O}P^2$ on the brain. –  Jason DeVito Feb 11 '11 at 15:47
This is probably obvious, but why do we need $x^2=y$? The dual statement is that a representative of $H_n$ needs to have self-intersection number 1, and I don't see any reason that that needs to hold either. –  Aaron Mazel-Gee Feb 11 '11 at 20:04
It's Poincare duality. There must be SOME element in $z\in H^n(X)$ such that $xz = y$. Since $x$ generates all of $H^n(X)$, $z = a x$ for some nonzero integer $a$. But then $xz = xax = ax^2 = y$ implies $y$ is a multiple of $x^2$. But $y$ is a generator, so this implies $a = \pm 1$. So, up to rechoosing $-y$ instead of $y$, we must have $x^2 = y$. –  Jason DeVito Feb 11 '11 at 21:27
But why must there exist such a $z$? Sorry if I'm being really dense here. –  Aaron Mazel-Gee Feb 12 '11 at 22:40
It's corollary 3.39 in Hatcher's algebraic topology. The idea is that if $[M]$ is the fundamental class and $x$ is not a multiple of anything, nor is torsion, then $[M]\cap x$ is also not a multiple of anything, nor is torsion. It follows that there is a cocycle $z$ such that $z([M]\cap x) = 1$. But by the formula connecting cap and cup product, we get that $1 = z([M]\cap x) = z \cup x ([M])$, so the $z$ above exists. –  Jason DeVito Feb 13 '11 at 1:25