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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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    See the book of Hofmann and Morris http://books.google.co.uk/books?id=pOi60a0Th4kC&lpg=PA289&ots=fkLP7uV-gW&dq=%22sphere%20theorem%22%20connected%20compact%20%22lie%20group%22&pg=PA289#v=onepage&q=%22sphere%20theorem%22%20connected%20compact%20%22lie%20group%22&f=false for a proof using the Hopf-Samelson Theorem.2010-11-30
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    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?2010-12-02
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    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.2010-12-02
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    @ 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.)2010-12-03

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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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    Neat! This is actually more elementary than I was expecting.2010-11-30
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    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.2010-11-30
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    To be clear, I don't personally find it *that* elementary. But I was somehow expecting something significantly deeper. I may have been mixing it up with either of the following problems: complex structures on spheres, parallelizability of spheres. I guess these are both harder?2010-11-30
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    Oh, this is a very nice write-up. I may even be able to remember it in the future! Thanks.2010-11-30
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    @Pete: Perhaps. I know parallelizability of spheres is quite hard. As far as I know, whether or not a sphere has a complex structures is nontrivial but not TOO hard, except for $S^6$ where it is still unsolved.2010-11-30
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    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?)2010-12-01
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    BTW, what's it take to prove the (non)existence of complex structures on $S^{2n}$? Or is it too involved to even just mention the major players in the story...2010-12-01
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    And if for nothing else than so that you can drop the right names: one constructs the maximal number of vector fields on spheres using Clifford algebras and proves that they are maximal using the Adams operations on KO-theory. For a quick but readable exposition, check out the first 30 or so pages of http://math.berkeley.edu/~ericp/latex/haynes-notes/haynes-notes.pdf2010-12-01
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    @Aaron: The $\langle \rangle$ on $\mathfrak{g}$ is any $Ad(G)$ invariant inner product. $G$ acts on itself by conjugation and this action fixes the identity element of $G$. Hence, looking at the differential of this action, we get an action of $G$ on it's Lie algebra called the adjoint action or $Ad(G)$ for short. If $G$ is a matrix group (i.e., is embedded in $Gl_n(V)$ for some vector space over $\mathbb{C}$ or $\mathbb{R}$), then $\mathfrak{g}$ can naturally be taken to be a subalgebra of $\mathfrak{gl_n}$. In this picture, $G$ acts on $\mathfrak{g}$ by conjugation (continued)2010-12-01
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    Here, since we're thinking of $G$ and $\mathfrak{g}$ as matrices, I really mean conjugation in that sense - matrix multiplication. Such an inner product always exists (assuming $G$ is compact): Start with any inner product $\langle \rangle_1$ and define $\langle x,y\rangle = \int_G \langle Ad(g)x, Ad(g)y\rangle_1 dg$ where $dg$ is some biinvariant measure (which is easy to constuct).2010-12-01
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    All I know about complex structures on spheres, I learned from http://mathoverflow.net/questions/11664/complex-structure-on-sn2010-12-01
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    Finally, the conclusion that $H^3(G)\neq 0$ for $G$ a Lie group actually doesn't require that $G$ be compact (I think). There is a theorem which states that any (connected) noncompact Lie group $G$ has a maximal compact subgroup $K$ unique up to conjugation and that $G$ is diffeomorphic (but not neccesarily isomorphic) to $K\times\mathbb{R}^n$ for some $n$. This implies that the topology of $G$ is that of $K$. (There is one detail I'm missing: If $G$ is nonabelian, must $K$ be? I think so, but I'm not positive about this)2010-12-01
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    Gotcha. Thanks for the reference. Lastly, this thing you're saying about maximal compact -- can it be anything worse than a semidirect product?2010-12-02
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    Aaron: I'm not sure - I've only really studied compact Lie groups.2010-12-02
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    By the way, this question (to which a comment on the one you gave links, since they overlap) has a somewhat more in-depth discussion about the ins and outs of the question of whether there's a complex structure on $S^6$. http://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere2010-12-03
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    @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.2011-01-26
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    @Jack: To my (very limited) knowledge, it's not published anywhere. I don't even remember where I first heard it. My best guess is that I first heard it in differential geometry class (with Herman Gluck) - we definitely covered the Cartan 3-form in that class. My second best guess is that I heard it while studying for oral exams. If that's the case, then I either heard it from Herman Gluck or Wolfgang Ziller.2011-01-26
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    @JasonDeVito Actually, there are non-abelian groups with abelian maximal compacts. First of all, there are lots of unipotent non-abelian groups, like Heisenbergs, but also $SL_2(\mathbb{R})$ is homotopic to $SO_2(\mathbb{R})$ which is just a circle (rotations of the plane).2013-11-22
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    @Ben: Of course, you're right. That's what I get for never thinking about non-compact groups.2013-11-23
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    @JackLee: I don't know whether you are still looking for a reference (after all you commented on this about 4 years ago), but you can find this proof on page 279 of "An Introduction to Differentiable Manifolds and Riemannian Geometry" by Boothby2015-06-24
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    (of the revised second edition, in the second edition it seems to be on page 288. anyways, it is at the end of the integration on manifolds chapter). @JasonDeVito perhaps your class used this book as textbook. in any case, I will try to find out who first discovered this proof; I guess that it wasn't boothby.2015-06-24
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    Actually this proof seems to be quite old,which is expected I guess. It seems to me that this proof is (albeit somewhat in disguise) already in Chevalley/Eilenberg, Cohomology Theory of Lie Groups and Lie Algebras (Trans. AMS 63), Theorem 21.3. (Note that the paper claims to be of "no great originality", so its probably even older than this paper. The paper refers to Chevalley's Lie groups book.)2015-06-24
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    @Herrmann: Thanks. In the end, I didn't include it in the second edition of my book, because the book was already too long. But it's good to have the references.2015-06-24
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    This proof can also be found in Topology and Geometry by Glen E. Bredon, on page 310. The section including this proof (in a bit more general fashion) is about differential forms on compact lie groups and proofs that their de Rham cohomology is given by Ad-invariant alternating forms on the corresponding lie algebra.2017-04-01