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?

rationale.s.h.t. Forfinitegroups, r.e.s.h.t is supposed to be equivalent to the homotopy theory of chain complexes of rational Mackey functors on the group; and its homotopy category is just equivalent to graded rational Mackey functors. That is: it's all just algebra. So the answer to your question is very amenable to solution by calculation. I have no idea who (if anyone) has done these calculations, or what they look like ... $\endgroup$ – Charles Rezk Nov 20 '15 at 15:30