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

Solution 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 $n=3$.