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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.

Maximal Ideal
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    The sort of point-set topology usually covered before the first course in algebraic topology has very few tools for proving that two spaces are distinct (in whatever category). On the other hand, $S^{2n}$ and $\mathbb{CP}^n$ have different cohomology groups (let alone rings) for $n > 2$, which seems like the easiest way to distinguish them. Besides, this is exactly the sort of problem algebraic topology is designed to solve. – anomaly Jul 02 '20 at 21:01
  • (I meant $n > 1$ above. All the $\mathbb{CP^n}$ have $H^2$ nonzero.) – anomaly Jul 02 '20 at 21:10
  • If $V$ is any vector space, any complete flag for $V$ induces a cellular decomposition for the projective space ${\rm P}V$. Since we're dealing with complex spaces, the complex consists of $0$'s and $\Bbb Z$'s. – Ivo Terek Jul 02 '20 at 23:34
  • Probably by now you should accept one of the answers or follow up with some specific questions. – Ted Shifrin Jan 09 '22 at 00:40
  • @TedShifrin Fair enough, I wasn't sure which one to accept honestly. – Maximal Ideal Jan 09 '22 at 18:40

3 Answers3

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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 $S^{2n}$ ($n>2$).

Ted Shifrin
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  • I'm not familiar with intersection numbers, so I have just two questions. I would gladly appreciate either a direct response or some references. The first is, what is meant by "linear copy" of $\mathbb{C}P^{1}$? The second is, why does contractibility of $S^{4}-{\text{point}}$ imply that any surfaces have zero intersection number in $S^{4}$? – Maximal Ideal Jan 09 '22 at 18:40
  • The differential topological notion of intersection numbers of compact oriented submanifolds corresponds to the cup product structure on cohomology, but generally the latter is more advanced. Guillemin and Pollack is a good readable reference for intersection numbers and degree; Hirsch's Differential Topology is a far more sophisticated treatment. By a linear copy of $\Bbb CP^1$ I mean a literal line, given by $a_0z_0+a_1z_1+a_2z_2=0$ in $\Bbb CP^2$ (there are also smooth conics, which are holomorphically equivalent to $\Bbb CP^1$). – Ted Shifrin Jan 09 '22 at 20:20
  • In any contractible space, intersection numbers vanish (because intersection numbers are a homotopy invariant and you can homotope any compact submanifold off any given closed submanifold). – Ted Shifrin Jan 09 '22 at 20:22
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$\mathbb{C}P^0$ 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 $k\neq 2n-1$.

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

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

Tsemo Aristide
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