TITLE: equivariant stable class of quaternionic Hopf fibration in RO(G)-degrees of ADE-type
QUESTION [11 upvotes]: Does the quaternionic Hopf fibration possibly represent a non-torsion element in the $G$-equivariant stable homotopy groups of spheres, for $G$ a finite subgroup of $SO(3)$ and in RO(G)-degree being its canonical 3d representation?
For the complex Hopf fibration the analog is true, as far as I see: In

Shôrô Araki, Kouyemon Iriye, Equivariant stable homotopy groups of spheres with involutions. I, Osaka J. Math. Volume 19, Number 1 (1982), 1-55. (Euclid)

it is shown (theorem 8.7 i) that the complex Hopf fibration -- canonically regarded as a representative for a $\mathbb{Z}/2$-graded stable homotopy group of spheres in $RO(G)$-degree being the canonical 1-dimensional representation of $\mathbb{Z}/2$ -- is a non-torsion generator.
In this case the construction of the $\mathbb{Z}/2$-equivariant structure is induced, via the Hopf construction, from (on top of their p. 24) the $\mathbb{Z}/2$-equivariance of the product operation on complex numbers with respect to complex conjugation.
Analogously, the product operation on quaternions is of course $SO(3)$-equivariant, again with respect to the canonical action on their imaginary part. So it would seem that restricting this to any finite subgroup $G$ of $SO(3)$ and then applying the Hopf construction to the quaternions will yield a $G$-equivariant quaternionic Hopf fibration that represents an element in the $G$-equivariant stable homotopy groups of spheres in $RO(G)$-degree the corresponding 3d rep. (Right?)
For which choice of $G \hookrightarrow SO(3)$, if any, is this element non-torsion?
Or more generally, what is known about equivariant stable homotopy groups of spheres in what one might call  RO(G)-degrees of ADE type?

REPLY [6 votes]: I think the Hopf construction gives a non-torsion class when G is dihedral or exceptional, but probably not when G is cyclic.
Non equivalently, we can perform Hopf constructions on the 0, 1, 3, and 7 spheres.  Only the 0 sphere gives a non torsion class in stable homotopy.
Suppose $G$ acts on our sphere, preserving the multiplication, with fixed points $(S^n)^G$ also a sphere $S^k$.  If $H: S^{2n+1} \to S^{n+1}$ is the Hopf map, then the restriction of $H$ to $G$ fixed points is also a Hopf map $H': S^{2k+1} \to S^{k+1}$.  If $k=0$, then $H'$ is stably non-torsion (non-equivariantly), and therefore so is $H$ in the  $G$ equivariant stable category.
Conversely, if $k>0$, I think $H$ must be stably torsion $G$ equivariantly: the rational stable equivariant homotopy type of a space $X$ (for finite groups) is (I think) completely determined by the $H^*(X^K, Q)$ as $WK=NK/K$ modules, where $K$ ranges over the conjugacy classes of subgroups of $G$, and rational stable maps are all detected by these groups.