I know very little about the fancy equivariant stable homotopy category, so I apologize if this question is silly for one reason or another, but:


I think that stable homotopy, in the non-equivariant case, corresponds to the homology theory associated to the sphere spectrum. And indeed, in the equivariant context we have $RO(G)$-graded homology and cohomology theories for every spectrum. On the other hand, equivariant stable homotopy groups are defined by looking at maps, not from representation spheres, but from spheres smashed with homogeneous spaces. The resulting object has a sort of quirky bigrading, it seems to me, over the integers and subgroups of $G$. 


Is there some reason that this choice is more natural than defining homotopy groups as $\pi\_{\nu}(E) = [S^{\nu}, E]\_G$ for $\nu \in RO(G)$? Is this object somehow trivial or something? 


If the answer to this question is that I'm missing something easy or that the objects I gave above are trivial for some reason- then let me know but don't tell me the answer! It'll probably be good for me to figure that out on my own...



Why are equivariant homotopy groups not RO(G)-graded?
If $G$ acts linearly and isometrically on $V$ then the unit disk $D(V)$ and its boundary $S(V)$ are an appealing kind of equivariant disk and sphere. You can try taking the pairs $(D(V),S(V))$ as the building blocks for equivariant complexes, but you won't get all the equivariant homotopy types that you want, not even the manifolds with smooth $G$-action.


You will get them all if you consider also representations of subgroups. If $H$ acts on $V$ then look at $(G\times\_HD(V),G\times\_HS(V))$, where $G\times\_HX$ means the quotient of $G\times X$ with $(g,hx)$ identified with $(gh,x)$. (This converts an $H$-action into a $G$-action.)


On the other hand, you also get everything you want if you only use trivial representations of arbitrary subgroups rather than arbitrary representations of arbitrary subgroups.


