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It is known that for any compact Lie group $G$ with maximal torus $T$, that any other maximal torus $T'$ is conjugate to $T$. This might be a bit of a stretch, but I was wondering if it is possible to use this result to deduce which positive dimensional spheres can have a group structure (or maybe just a Lie group structure). The program I had in mind was this:

1) Show that if $S^n$ has a Lie group structure that its maximal torus would be $S^1$

2) Look at the set of all maximal tori (each one being a copy of $S^1$) and somehow turn each of these into the fiber of some bundle $S^n\rightarrow B$.

3) Hopefully be able to conclude that $n$ must be $1$ or $3$ (my idea was that perhaps we could show that $B= S^{n-1}$

Any suggestions would be greatly appreciated. Also, I understand that this is using much more machinery than necessary to attack the problem of which spheres admit group structures but the idea still makes me curious if there is something like this.

Edit: I don't think you need this much to deduce which spheres admit a Lie group structure; I'm not sure if this argument provides a path to showing which spheres admit a topological group structure.

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(2) and (3) look reasonable provided you have (1). The bundle (2) could be obtained by looking at a unit tangent vector at the neutral element in the maximal torus, giving $B=\mathbb RP^{n-1}$ or $\mathbb S^{n-1}$. If you argue $S^n$ acts transitively on oriented maximal tori, modulo (1) that would finish your problem provided $S^n$ acted transitively on oriented maximal tori. It's not clear to me how to argue point (1). –  Ryan Budney Dec 28 '11 at 23:34
    
I am curious about the statement that this uses "much more machinery than necessary". I am not aware of any elementary arguments (are cohomology operations elementary?) –  Igor Rivin Dec 28 '11 at 23:42
    
I suppose the method I suggested isn't necessarily more advanced, but I think if all we are interested in is showing that there isn't a Lie group structure on spheres of dimension not 1 or 3, doesn't it suffice to just look at some results about cohomology of general Lie groups (i.e. simply connected Lie groups have nonzero 3rd cohomology groups)? I don't think this other approach involves cohomology operations. –  Geoffrey Dec 28 '11 at 23:53
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@ Ryan: 1) is actually quite easy. Given a torus $T^k$ in $G=S^n$ we have a bundle $T\to G\to G/T$. Next, $n$ must obviously be odd. Now, given a bundle $S^1\to G\to G/S^1$ it's easy from the Gysin sequence that $G/S^1$ has evenly graded cohomology and hence a positive Euler characteristic. That means that it can not admit a free $S^1$ action since otherwise its Euler charactersitic would be zero. that means that $k\le 1$. Together with your argument for 2) and 3) that finishes the problem. @emiliocba: $S^1\times S^1$ certainly embeds into any $S^n$ with $n\ge 3$. –  Vitali Kapovitch Dec 29 '11 at 1:37
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Maximal tori are all conjugate. So the idea is that the group acts on the space of maximal tori by conjugation. Since each maximal torus is 1-dimensional, you can think of parametrizing the space of them by the 1-dimensional subspaces of the tangent space of the neutral element, i.e. $\mathbb RP^{n-1}$. So the map from $S^n$ to this projective space is given by conjugation of a prescribed point in $\mathbb RP^{n-1}$ with any element of $S^n$. –  Ryan Budney Dec 29 '11 at 9:51
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