I've heard a little recently about equivariant homotopy theory, and so I decided to try out some baby examples just to get a feel for it. I'm not even sure if these are the right thing to look at, and I'm sure I'm butchering the notation, but I've attempted to compute $\pi_2^{C_n}(S^2)$ and $\pi_3^{C_3}(S^2)$; here, $C_n$ acts by rotations on the plane and $C_3$ acts by the standard representation on 3-space.

For the first one, there's the obvious equivariant cell structure on $S^2$, which has fixed vertices at 0 and at the basepoint $\infty$, one orbit of $n$ edges, and one orbit of $n$ faces. My maps are based and I need to map fixed points to fixed points, so either both vertices go to $\infty$ or else my map fixes them.

  • Suppose both go to $\infty$. The image of any one edge is an arbitrary based loop, and of course the other edges' images are determined by this. Assuming my faces are going to able to be mapped in at all, the inclusion of an edge into a face is a cofibration, so I may as well just homotope the edges to $\infty$ now. From here, the map is determined by the image of one face, i.e. an element of (nonequivariant) $\pi_2(S^2)$.
  • In the case where $0$ and $\infty$ are both fixed, the image of an edge is an arbitrary path between $0$ and $\infty$, and then we just get to decide where a face goes, which I think is just an element of the preimage of 1 in the connecting map $\pi_2(S^2,S^1)\rightarrow \pi_1(S^1)$. There's no extension problem in the lexseq, so this is just a projection onto one factor $\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}$, so this looks like $\mathbb{Z}\times \{1\}$.

So as a set, $\pi_2^{C_n}(S^2)=\mathbb{Z} \sqcup \mathbb{Z}$.

For the second one, I have a 1-cell of fixed points at $(x,y,z)=(t,t,t)$, my three 2-cells are the half-planes through that 1-cell and one of the three axes, and my three 3-cells fill in the rest. The 1-cell needs to be sent to fixed points, and these are still just $0$ and $\infty$, so by continuity it's all sent to $\infty$. So the image of a 2-cell is just an element of $\pi_2(S^2)$. Once we choose that (which determines the images of the other 2-cells, of course) we can always extend to the 3-cells, since these are just homotopies from a map to itself. Once we've chosen one, any other gives an element of $\pi_3(S^2)=\mathbb{Z}$. So as a set, $\pi_3^{C_3}(S^2)=\mathbb{Z} \times \mathbb{Z}$.

So, my first question is: Are these right? Also, I've learned to compute (usual) homotopy groups of spheres by making a Postnikov tower, and I'm wondering if there's a sufficiently easy example for the equivariant case where I can do the analogous calculations by hand without the full generality of slice cells or whatever's going on (those could be the wrong words -- I don't think I understand what these are well enough to know whether this is a decent request, either). In any case, I'd love suggestions of better/more instructive examples. Lastly, I'm wondering if there are actually group structures here. In the first example, it looks like I can reasonably hope to add guys that are both in the same copy of $\mathbb{Z}$, but not otherwise. I think it's easy to show that there's no $C_n$-equivariant coproduct on $S^2$. On the other hand, quotienting by the plane $x+y+z=0$ in $\mathbb{R}^3$ looks like it gives a $C_3$-equivariant coproduct on $S^3$. If two elements came from the same choice of $\pi_2(S^2)$, then there's an obvious origin for the $\pi_3(S^2)$-torsor, namely using the trivial homotopy to extend the map over the 3-cells. I think this agrees with my equivariant coproduct, just because it matches up with the usual picture you draw of how to add elements of $\pi_2$ (two squares sitting on top of each other). However, I can't tell whether this makes any sense when the elements came from different choices of $\pi_2(S^2)$. It seems like if it does work at all, there might be something funnier than the obvious group structure...

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    $\begingroup$ Something is messed up here. If you want (t,t,t) to give a 1-cell of fixed points you cannot be thinking of the unit S^2 in R^3. If you are instead thinking of the one-point compactification of R^3, that would be \pi_3^{C_3}(S^3), not \pi_3^{C_3}(S^2). If you really are thinking of the unit S^2, then that can be regarded as the one-point compactification of the plane x+y+z=0, and the cyclic permutation acts as a 1/3-twist on that plane so you essentially have the general C_n case in disguise. $\endgroup$ – Neil Strickland Nov 10 '10 at 23:29

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