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.)

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    $\begingroup$ See the book of Hofmann and Morris books.google.co.uk/… for a proof using the Hopf-Samelson Theorem. $\endgroup$ Nov 30 '10 at 11:51
  • $\begingroup$ Is the smoothness such a big condition that there is no way to get from the below proof to the $\mathbb{R}$ divison algebra result of Adams? $\endgroup$ Dec 2 '10 at 3:58
  • $\begingroup$ Well, $\mathbb{R}^n$ is a division algebra iff $S^{n-1}$ is parallelizable. That's a fundamentally topological constraint, which is weaker than being an H-space, which is weaker than being a Lie group. So proving that most spheres aren't Lie groups doesn't seem (to me) to make too big of a step in that direction. $\endgroup$ Dec 2 '10 at 17:31
  • $\begingroup$ @ Willie: Thanks for the reference. This book looks like exactly what I've been trying to find. It feels like a shame that there's no differential geometry/topology prerequisite for my Lie theory class, and so as a result we miss out on some really great stuff. (Or rather, I guess it's a shame that we don't have a Lie theory class that does have such a prerequisite.) $\endgroup$ Dec 3 '10 at 8:07
  • $\begingroup$ maybe of interests math.stackexchange.com/q/4220631/955245 please help/comment if you can $\endgroup$ Aug 9 at 16:38

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 $n=3$.

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    $\begingroup$ Neat! This is actually more elementary than I was expecting. $\endgroup$ Nov 30 '10 at 13:34
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    $\begingroup$ Well, I'm sweeping MANY details under the rug. For example, the fact that simply connected Lie groups are classified by their Lie algebras is somewhat nontrivial. (Though I think this is the most nontrivial part). The rest of details aren't too hard, but some of them are tedious. $\endgroup$ Nov 30 '10 at 14:03
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    $\begingroup$ Oh, this is a very nice write-up. I may even be able to remember it in the future! Thanks. $\endgroup$ Nov 30 '10 at 14:58
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    $\begingroup$ Whaaaaa. Too good. Accepted + upvoted. What's this langle-rangle product on $\mathfrak{g}$, though? (Also, at the end it's probably worth re-stating the hypothesis that $G$ be compact. Or is that not necessary?) $\endgroup$ Dec 1 '10 at 7:07
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    $\begingroup$ @Jason: Is this simple proof published anywhere? And do you know who first discovered it? I'd like to include it in the next edition of my Smooth Manifolds book, but I want to make sure to give credit where it's due. $\endgroup$
    – Jack Lee
    Jan 26 '11 at 1:05

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