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Let $G\subset \mathrm{SU}(2)$ be a finite group. (These are famously classified through the McKay correspondence.) The Lie group framing of $\mathrm{SU}(2) = S^3$ descends to the quotient manifold $S^3 / G$, at least after getting some "left"s and "right"s in the correct places.

Every framed $k$-manifold determines a class in the $k$th stable homotopy group of spheres $\pi_k(\text{Sphere})$. For example, $\mathrm{SU}(2)$ with its Lie group framing provides a generator of $\pi_3(\text{Sphere}) = \mathbb{Z}/24$.

What are the values of the homotopy-group classes determined by the framed 3-manifolds $S^3/G$? How do these values relate to other Lie theoretic data like the rank or (dual) Coxeter number of the ADE Dynkin diagram?

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    $\begingroup$ The finite subgoups of $\operatorname{SU}(2) $ have been classified by Felix Klein, slightly before McKay... $\endgroup$
    – abx
    Commented Jan 12, 2019 at 5:33
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    $\begingroup$ @abx Thanks for the correction. I hadn't meant that the classification was due to McKay, only that I wanted to remind that it is convenient to sort them in terms of ADE data. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 12, 2019 at 5:36
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    $\begingroup$ Sorry to be pedantic again, but the correspondence in terms of ADE goes back (at least) to P. Duval (around 1930). $\endgroup$
    – abx
    Commented Jan 12, 2019 at 7:14
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    $\begingroup$ @abx Good to know. It is a fairly common phenomenon that mathematics is not named after its inventor... $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 12, 2019 at 15:32
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    $\begingroup$ @aglearner It is the type of fact I sort of grew up with. In deed, the statement is that the group $\Omega_k^{fr}$ of stably-framed $k$-manifolds up to stably-framed cobordism is isomorphic to the stable homotopy group $\pi_k^s$ of spheres. In his answer below, Alex Suciu cites a paper of Seade and Steer, which calls the map $\Omega_k^{fr} \to \pi_k^s$ the "Pontryagin construction" and cites Milnor's book Topology from the differential viewpoint. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 12, 2019 at 15:38

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The answer can be found in Theorem 2.1 from a paper of José Seade and Brian Steer (Complex singularities and the framed cobordism class of compact quotients of $3$-dimensional Lie groups by discrete subgroups, Comment. Math. Helv. 65 (1990), no. 3, 349–374, available here): If $G$ is the cyclic group of order $r$, then $S^3/G$ represents $r$ times a generator of $\pi_3^s=\mathbb{Z}/24\mathbb{Z}$, whereas if $G$ is the $\langle p,q,r\rangle$ triangle group, then $S^3/G$ represents $(p+q+r-1)$ times a generator of $\pi_3^s$.

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    $\begingroup$ If I am not mistaken, this number $r$ or $(p+q+r-1)$ is the number of complex irreps of $G$, which in terms of the ADE classification is one more than the rank of the compact Dynkin diagram. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 12, 2019 at 15:41
  • $\begingroup$ Quick reality check: Take $G=I^*$ (the binary icosahedral group of order 120), so that $S^3/G=\Sigma(2,3,5)$ is the Poincaré homology sphere. Then the corresponding Dynkin diagram is $E_8$, and $2+3+5-1=9=8+1$. Yep, that works. $\endgroup$
    – Alex Suciu
    Commented Jan 12, 2019 at 20:38
  • $\begingroup$ Incidentally, the reason I wanted to say things in terms of the Dynkin diagram is that there is a moonshine-type construction that takes in certain Dynkin data and produces certain modular objects, and which might, in some yet-to-be understood way, factor through some geometric objects e.g. 3-manifolds. I had hoped to go (Dynkin data) -> (S^3/G) -> (modular object), but this seems not to work, or at least I need to think more. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 13, 2019 at 0:30

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