11
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Just a remark: to detect non-torsion classes in equivariant stable homotopy theory, you need rational e.s.h.t. For finite groups, 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
  • 1
    $\begingroup$ Thanks. Apparently I need to get into contact with David Barnes ncatlab.org/nlab/show/… $\endgroup$ – Urs Schreiber Nov 20 '15 at 18:43
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks! I get it. Just to amplify: your argument says that a Hopf generator should become non-torsion equivariantly whenever the fixed points of the action reduce it to a 1-sphere; and for n = 3 this is case for all the ADE finite groups except the cyclic ones. Incidentally, that is consistent with Araki-Iriye'82 cited above: their prop. 10.1 and theorem 10.11 says that Z/2-equivariantly the quaternionic Hopf fibration is still torsion of order 24. $\endgroup$ – Urs Schreiber Nov 22 '15 at 20:15
  • $\begingroup$ For what it's worth, the exclusion of the cyclic ADE groups actually matches the situation in the application that motivates me here: one sees the same exclusion on p. 3 of arxiv.org/abs/hep-th/9812205 (ncatlab.org/nlab/show/G2+manifold#WithADEOrbifoldStructure). $\endgroup$ – Urs Schreiber Nov 23 '15 at 8:22
  • $\begingroup$ I have now written out a proof, using the above argument and appealing to Greenlees-May decomposition and tom Dieck splitting: ncatlab.org/nlab/show/… $\endgroup$ – Urs Schreiber Nov 26 '15 at 8:25

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.