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I have some questions about the $n$-sphere:

I know that for $n=0,1,3$, $S^n$ forms a Lie group and I also know why it's true, but why is it not the case for other $n$?

I have the same question for the $n$-spheres admitting an almost complex structures $(n=2,6)$. Is there a general reason to conclude it doesn't have one for any other $n$?

And finally, (this is a homework question):

Can $S^4$ have a Lorentz metric?

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Lorentz metrics come equipped with a notion of `timelike'. This means you can find a continuously-varying 1-dimensional subspace of every tangent space. If you had such a thing, it would tell you something about the Euler Characteristic of $S^4$. This is usually called the Poincare-Hopf Index Theorem. – Ryan Budney Feb 18 '13 at 5:22
Looking into that, Ryan, thank you. – DavidM Feb 18 '13 at 5:37
So I found $\chi (S^4)=2$, does that mean it doesn't admit a Lorentzian metric? Since otherwise it would be $0$, or am I missing something? – DavidM Feb 18 '13 at 5:57

The smooth manifold underlying a Lie group is always parallelizable, and the only spheres which are paralellizable are those of dimension $1$, $3$ and $7$.

On the other hand, the third cohomology group of a compact Lie group is never zero, so that excludes the possibility that $S^7$ be a Lie group.

As for complex structures, see this MO question

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Neither of the two results I mention in the first two paragraphs is obvious, by the way (thaat Lie groups are parallelizable is easy, though). The first one is a theorem of Milnor and Kervaire, and the second one follows from the description of compact Lie groups as (products of tori by semi-simple compact groups) divided by finite abelian subgroups, and the result that semisimple simple compact Lie groups have non-trivial $\pi_3$. – Mariano Suárez-Alvarez Feb 18 '13 at 5:28
I see, I was just curious about it. I also looked online and didn't find many things other than it being known. Thank you for your reply. – DavidM Feb 18 '13 at 5:38
Concerning the complex structures, there is a $K$-theoretic proof for the cases $n > 4$ and $n \neq 6$. Then in the case $n=4$ usually one does another argument. This case I think is actually much easier. Assume $S^4$ where almost complex. Then you can compute the first Pontryagin class in terms of chern classes and get a contradiction (the pontryagin class vanishes, but the second chern class is the euler class, so cannot be evaluated to zero). – mland Feb 18 '13 at 10:56
"On the other hand, the third cohomology group of a compact Lie group is never zero", what about $S^1$? – lee Feb 25 '13 at 15:21
I should have said simply connected. – Mariano Suárez-Alvarez Feb 25 '13 at 21:33

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