For which positive integers $n$ can $\mathbb R P^n$ be given the structure of a topological group?

I believe that $\mathbb R P^n$ cannot be given a Lie group structure for even $n$, since then it is not orientable. But, this doesn't necessarily imply it doesn't have a topological group structure (which is not smooth); moreover, it tells us nothing about odd $n$. And ideas?

  • $\begingroup$ $S^3$ can be equated with the group of unit quaternions. Taking the quotient by $\langle -1\rangle$ equates the antipodal points, so that should work. I'm not very optimistic about most of the others, because having a differentiable group structure should allow us to build a lot of nowhere vanishing vectorfields (by translation), and these can be lifted to the covering space. Even with just topological structure you may need to have exotic spheres, but I know nothing about those. $\endgroup$ – Jyrki Lahtonen Jun 21 '12 at 21:41
  • $\begingroup$ There might be a group structure that has little to do with the sphere before taking quotient, though, right? $\endgroup$ – tomasz Jun 21 '12 at 21:45
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    $\begingroup$ for $n=1$ the projective line is a circle, so it is a group, and for $n=3$ it is isomorphic to the special orthogonal group in $SO(3)$, so it is a group again. Whenever a projective space is a topological group, its universal cover, the sphere of same dimension, is a topological group too in such a way that the cover is a homomorphism of groups. So the question for projective space is answered by that for spheres. The question for spheres is settled, but somebody more qualified than me will have to tell you the answer... $\endgroup$ – Olivier Bégassat Jun 21 '12 at 21:46
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    $\begingroup$ Following Olivier Bégassat's argument, you can take a look at unizar.es/acz/05Publicaciones/Revistas/Revista62/p075.pdf $\endgroup$ – Ehsan M. Kermani Jun 21 '12 at 21:53
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    $\begingroup$ It does for $n=\infty$. $\endgroup$ – Sean Tilson Jun 23 '12 at 19:44

As Olivier has soon this can be reduced to the question for spheres.

One way to prove this is to note that for a topological group $G$ we have that $G$ is homotopy equivalent to $\Omega BG$, the loopsapce of the classifying space. For example $\Omega BS^1 = \Omega \mathbb{C}P^\infty = \Omega K(\mathbb{Z},2) = K(\mathbb{Z},1) = S^1$. Thus the question is which spheres are loop spaces of classifying spaces. Adams' work showed this is true only for $n=0,1,3$.

  • $\begingroup$ It should be noted that Adams was concerned with when the spheres are $H$-spaces (amongst a raft of other equivalent conditions). He showed that this is only true for $n=0,1,3,7$. Note a topological group is always a $H$-space, but the converse can fail to be true. The $H$-space structure on $S^7$ does not give a group structure since the octonion's are not associative under multiplication $\endgroup$ – Juan S Jun 22 '12 at 3:40

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