# Lie group structure on the complex projective space

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)?

• It is an amusing observation that a polynomial $f$ over $\Bbb C$ that does not split into linear factors gives a non-trivial finite field extension $F/\Bbb C$, and $\Bbb P(F) = \Bbb CP^{\dim F - 1}$ is then a commutative positive-dimensional Lie group, and so the proofs below give as a corollary the fundamental theorem of algebra. – user98602 Dec 17 '18 at 3:21
• @MikeMiller: Could I ask what makes it amusing?! (I'm not a mathematician...) – Mehrdad Dec 17 '18 at 9:02
• @Mehrdad I'll let Mike answer for himself, but personally, I think that such a sophisticated and conceptual proof (it involves field extensions, Lie groups, algebraic topology) for such a simple-looking result (any non-constant complex polynomial has a root) is amusing. And it doesn't really use any analysis! – Najib Idrissi Dec 18 '18 at 14:50
• @Mehrdad There are many proofs of FTA, most of which are much faster routes to the actual theorem. This is one that requires the technical input of algebraic topology (or by a different proof, the Lie-theoretic notion of exponential map and some easier algebraic topology) for a concise and fully topological argument. – user98602 Dec 19 '18 at 16:48
• @MikeMiller what does $\mathbb{P}(F)$ stand for? – doetoe Jan 11 at 14:26

## 4 Answers

$$\mathbb{CP}^n$$ 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.

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.

• Can you talk a bit about the Hopf structure on the cohomology ring? The "standard" way I've seen to get a Hopf algebra out of a Lie group is to consider the universal enveloping algebra of its Lie algebra, but this is obviously coarser. Are the two related in any way? – Ashwin Trisal Dec 17 '18 at 3:48
• It's very simple: if $G$ is a topological group, the multiplication $\mu:G\times G\to G$ gives a map $\mu^*: H^*(G)\to H^*(G\times G)\cong H^*(G)\otimes H^*(G)$ (here cohomology is with coefficients in a field to get the latter isomorphism), and the group axioms for $\mu$ say exactly that $\mu^*$ is the comultiplication of a Hopf algebra structure on $H^*(G)$. I don't know of any connection to the universal enveloping algebra (there may be one but it would have to be fairly indirect). – Eric Wofsey Dec 17 '18 at 3:54
• $H^*(G;\mathbb{Q})$ is the universal enveloping algebra of the Lie algebra given by $\pi_*(\Omega G) \otimes_\mathbb{Z} \mathbb{Q}$ (with Whitehead product, which is trivial here). But of course the Lie algebra involved isn't $\mathfrak{g} = T_e G$... – Najib Idrissi Dec 18 '18 at 15:01
• @Najib: I think you want $H_{\bullet}(\Omega G, \mathbb{Q})$ there. – Qiaochu Yuan Dec 18 '18 at 22:46

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.

$$\pi_2(\mathbb{C}P^n)$$ 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