Is there an easy way to show which spheres can be Lie groups?

Question

I heard that using some relatively basic differential geometry, you can show that the only spheres which are Lie groups are $S^0$, $S^1$, and $S^3$. My friend who told me this thought that it involved de Rham cohomology, but I don't really know anything about the cohomology of Lie groups so this doesn't help me much. Presumably there are some pretty strict conditions we can get from talking about invariant differential forms -- if you can tell me anything about this it will be a much-appreciated bonus :)

(A necessary condition for a manifold to be a Lie group is that is must be parallelizable, since any Lie group is parallelized (?) by the left-invariant vector fields generated by a basis of the Lie algebra. Which happens to mean, by some pretty fancy tricks, that the only spheres that even have a chance are the ones listed above plus $S^7$. The usual parallelization of this last one comes from viewing it as the set of unit octonions, which don't form a group since their multiplication isn't associative; of course this doesn't immediately preclude $S^7$ from admitting the structure of a Lie group. Whatever. I'd like to avoid having to appeal to this whole parallelizability business, if possible.)

Answer 1

Here is the sketch of the proof.

Start with a compact connected Lie group G. Let's break into 2 cases - either $G$ is abelian or not.

If $G$ is abelian, then one can easily show the Lie algebra is abelian, i.e., $[x,y]=0$ for any $x$ and $y$ in $\mathfrak{g}$. Since $\mathbb{R}^n$ is simply connected and has the same Lie algebra as $G$, it must be the universal cover of $G$.

So, if $G$ is a sphere, it's $S^1$, since all the others are simply connected, and hence are their own universal covers.

Next, we move onto the case where $G$ is nonabelian. For $x,y,$ and $z$ in the Lie algebra, consider the map $t(x,y,z) = \langle [x,y], z\rangle$. This map is clearly multilinear. It obviously changes sign if we swap $x$ and $y$. What's a bit more surprising is that it changes sign if we swap $y$ and $z$ or $x$ and $z$. Said another way, $t$ is a 3 form! I believe $t$ is called the Cartan 3-form. Since $G$ is nonabelian, there are some $x$ and $y$ with $[x,y]\neq 0$. Then $t(x,y,[x,y]) = ||[x,y]||^2 \neq 0$ so $t$ is not the 0 form.

Next, use left translation on $G$ to move $t$ around: define $t$ at the point $g\in G$ to be $L_{g^{-1}}^*t$, where $L_{g^{-1}}:G\rightarrow G$ is given by $L_{g^{-1}}(h) = g^{-1}h$.

This differential 3-form is automatically left invariant from the way you've defined it everywhere. It takes a bit more work (but is not too hard) to show that it's also right invariant as well.

Next one argues that a biinvariant form is automatically closed. This means $t$ defines an element in the 3rd de Rham cohomology of $G$. It must be nonzero, for if $ds = t$, then we may assume wlog that $s$ is biinvariant in which case $ds = 0 = t$, but $t$ is not $0$ as we argued above.

Thus, for a nonabelian Lie group, $H^3_{\text{de Rham}}(G)\neq 0$. But this is isomorphic to singular homology. Hence, for a sphere to have a nonabelian Lie group structure, it must satisfy $H^3(S^n)\neq 0$. This tells you $n=3$.