Lie group structure on the complex projective space

Question

There is a famous theorem about when $S^n$ has the structure of a Lie group. What about the complex projective space $\mathbb CP^n$? For example, why $\mathbb CP^2$ is not a Lie group (without using classification for low dimension compact Lie groups)?

Answer 1

$\mathbb{CP}^n$ has Euler characteristic $n+1$, but a compact (positive-dimensional) Lie group has Euler characteristic $0$, for example by the Lefschetz fixed point theorem.

Alternatively, you can show in various ways that the rational cohomology ring of a compact Lie group must be an exterior algebra on a finite number of odd generators. But $H^{\bullet}(\mathbb{CP}^n, \mathbb{Q})$ is concentrated in even degrees, so doesn't admit odd generators.

Answer 2

$\pi_2(\mathbb{C}P^n)$ is $\mathbb{Z}$ and $\pi_2(G)$ is trivial where $G$ is a connected Lie group.

https://en.wikipedia.org/wiki/Complex_projective_space#Homotopy_groups

https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups

Answer 3

A complex projective space (of positive dimension) never admits a Lie group structure. There are lots of ways to prove this. For instance, the rational cohomology ring of any Lie group is a graded Hopf algebra (the comultiplication coming from the group operation) but the cohomology ring $\mathbb{Q}[x]/(x^{n+1})$ of $\mathbb{CP}^n$ does not admit a Hopf algebra structure. Indeed, for reasons of degree, $\Delta(x)$ would have to be $x\otimes 1+1\otimes x$ but then $\Delta(x^n)=\Delta(x)^n=\sum_{k=0}^n \binom{n}{k} x^k\otimes x^{n-k}$ would be nonzero (all the terms except $k=0$ and $k=n$ are nonzero), which is a contradiction.

Answer 4

You can actually generalise everything in the question to show that $\mathbb{C}P^n$ ($1\leq n<\infty$) cannot even admit a Hopf structure (https://en.wikipedia.org/wiki/H-space). And one way to see this is to demonstrate the existence of a non-trivial Whitehead product in $\pi_*\mathbb{C}P^n$. I'll point out that you already have fantastic answers, and most of them can be generalised directly to cover this case. This answer is only supposed to add another perspective.

Recall the quotient map $\gamma_n:S^{2n+1}\rightarrow \mathbb{C}P^n$ and the fact that it induces isomorphisms on $\pi_*$ for $*>2$. In particular $\gamma_{n*}:\pi_{4n+1}S^{2n+1}\xrightarrow{\cong}\pi_{4n+1}\mathbb{C}P^n$ is an isomorphism that takes the Whitehead square $\omega_{2n+1}=[\iota_{2n+1},\iota_{2n+1}]\in\pi_{4n+1}S^{2n+1}$ to the Whitehead product

$$\gamma_{n*}\omega_{2n+1}=[\gamma_n,\gamma_n]\in\pi_{4n+1}\mathbb{C}P^n.$$

It is classical fact related to the Hopf invariant one problem that for odd $k$, $\omega_k$ vanishes exactly when $k=1,3$ or $7$. Therefore, since $\gamma_{n*}$ is an isomorphism, $[\gamma_n,\gamma_n]$ is non-zero in $\pi_{4n+1}\mathbb{C}P^n$ as long as $n\neq 1,3$. Now $\mathbb{C}P^1\cong S^2$ is not an $H$-space exactly because of Adam's solution to the Hopf invariant one problem (and more in line with the current trail of thought, $[\iota_2,\iota_2]=-2\eta\in\pi_3S^2$).

Therefore we single out $\mathbb{C}P^3$ as the interesting case for which this line of reasoning does not apply. In fact all Whitehead products vanish in $\mathbb{C}P^3$ (Stasheff: "On homotopy Abelian H-spaces") and $\Omega \mathbb{C}P^3$ is homotopy commutative. I assure you still that $\mathbb{C}P^3$ is not an $H$-space, since either Qiaochu's or Eric's answers for compact Lie groups apply more or less verbatim to finite H-spaces.