Is there an elementary reason $\mathbb{CP}^{n}$ is not homeomorphic to the sphere for $n\ge 2$?

Question

My question is similar to this one, but I am asking this for the complex case.

In the real case, we can use the fact that the fundamental group of $\mathbb{RP}^{n}$ is non-trivial, so the space cannot be homeomorphic to the sphere for $n\ge 2$. We can't really use this fact in the complex case, because $\mathbb{CP}^{n}$ is always simply connected.

Is there any other simple way to argue this? I am hoping to find something similar to the ideas found in the above link that would involve only the simplest of tools from algebraic topology (or as simple as possible if nothing else). Any suggestions are welcome.

Answer 1

Without all the tools of algebraic topology, but with the basics of intersection numbers (e.g., Guillemin & Pollack), you can easily see that, for example, $\Bbb CP^2$ and $S^4$ are not diffeomorphic (or homeomorphic). In $\Bbb CP^2$, any two linear copies of $\Bbb CP^1$ intersect once. However, in $S^4$, since $S^4-\{\text{point}\}$ is contractible, any surfaces have zero intersection number. This argument generalizes immediately to $\Bbb CP^n$ and $S^{2n}$ ($n>2$).

Answer 2

$\mathbb{C}P^0$ is a point and $\mathbb{C}P^1\cong S^2$ is the Riemann sphere. In general $\mathbb{C}P^n$ is the quotient of $S^{2n+1}\subseteq\mathbb{C}^{n+1}$ by the free action of $S^1\subseteq\mathbb{C}\setminus0$. The projection $$S^{2n+1}\rightarrow \mathbb{C}P^n$$ is a locally trivial fibration whose fibre is $S^1$. From this we get a long exact sequence of homotopy groups $$\dots\rightarrow\pi_kS^1\rightarrow \pi_kS^{2n+1}\rightarrow \pi_k\mathbb{C}P^n\rightarrow\pi_{k-1}S^1\rightarrow\dots.$$ Since $$\pi_kS^1\cong\begin{cases}\mathbb{Z}&k=1\\0&\text{otherwise}\end{cases}$$ we get by exactness that $$\pi_k\mathbb{C}P^n\cong\begin{cases}\mathbb{Z}&k=2\\\pi_kS^{2n+1}&\text{otherwise}\end{cases}$$ if $n\geq1$. In particular $$\pi_2\mathbb{C}P^n\cong\mathbb{Z}$$ so if $\mathbb{C}P^n$ is homotopy equivalent to a sphere, then that sphere must be $S^2$. But also $$\pi_{2n+1}\mathbb{C}P^n\cong\mathbb{Z}$$ so if $n\geq2$, then $\mathbb{C}P^n$ cannot be homotopy equivalent to $S^2$. In particular $\mathbb{C}P^n$ cannot be homeomorphic to any sphere if $n\geq2$.

Edit: I suppose I should add that it is known that $\pi_nS^n\cong\mathbb{Z}$ for all $n\geq1$, that $\pi_nS^{n+k}=0$ for all $n$ and all $k\geq1$, and that if $k\geq1$, then $\pi_{n+k}S^n$ is torsion if $n$ is odd, or if $n$ is even and $k\neq 2n-1$.

Answer 3

$H^2(\mathbb{C}P^n,\mathbb{Z})=\mathbb{Z}$ if $p=2$ and $H^2(S^{2n},\mathbb{Z})=0$. (CW homology)

https://topospaces.subwiki.org/wiki/Cohomology_of_complex_projective_space