2019-01-12 01:02:19 8 Comments

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?

### Related Questions

#### Sponsored Content

#### 0 Answered Questions

### How does quotienting by a finite subgroup act on the framed-cobordism class of a group manifold?

**2019-01-26 00:24:16****Theo Johnson-Freyd****267**View**14**Score**0**Answer- Tags: at.algebraic-topology lie-groups stable-homotopy quantum-field-theory cobordism

#### 1 Answered Questions

### [SOLVED] Embedding spaces and surface knots in high dimensional manifolds

**2017-06-14 21:22:48****Nati****139**View**4**Score**1**Answer- Tags: at.algebraic-topology gt.geometric-topology smooth-manifolds embeddings

#### 1 Answered Questions

### [SOLVED] Is there a geometric construction of hyperbolic Kac-Moody groups?

**2016-03-04 16:48:56****John Baez****888**View**32**Score**1**Answer- Tags: dg.differential-geometry lie-groups lie-algebras

#### 1 Answered Questions

### [SOLVED] Is there an octonionic analog of the K3 surface, with implications for stable homotopy groups of spheres?

**2015-09-14 15:45:10****Chris Schommer-Pries****1370**View**59**Score**1**Answer- Tags: at.algebraic-topology stable-homotopy k3-surfaces octonions homotopy-groups-of-sphere

#### 1 Answered Questions

### [SOLVED] Nilpotence of the stable Hopf map via framed cobordism

**2015-03-31 12:47:45****Noah Snyder****764**View**29**Score**1**Answer- Tags: at.algebraic-topology gt.geometric-topology stable-homotopy cobordism

#### 6 Answered Questions

### [SOLVED] third stable homotopy group of spheres via geometry?

**2010-11-04 19:56:50****Johannes Ebert****6345**View**61**Score**6**Answer- Tags: at.algebraic-topology homotopy-theory stable-homotopy

#### 1 Answered Questions

### [SOLVED] What's a good reference for the following obstruction theory yoga?

**2014-02-01 19:27:56****Theo Johnson-Freyd****408**View**5**Score**1**Answer- Tags: reference-request at.algebraic-topology operads rational-homotopy-theory obstruction-theory

#### 0 Answered Questions

### How to see the quaternionic hopf map generates the stable 3-stem?

**2013-02-19 22:27:02****Chris Schommer-Pries****566**View**14**Score**0**Answer- Tags: stable-homotopy at.algebraic-topology hopf-fibration homotopy-groups-of-sphere

#### 1 Answered Questions

### [SOLVED] The inertia subgroup of `$\Theta_n$` for Lie groups

**2011-12-29 18:42:17****William****203**View**1**Score**1**Answer- Tags: at.algebraic-topology dg.differential-geometry gt.geometric-topology

#### 1 Answered Questions

### [SOLVED] An elementary proof that the degree of a map of spheres determines its homotopy type

**2010-05-21 20:38:21****Charles Staats****1515**View**13**Score**1**Answer- Tags: teaching at.algebraic-topology

## 1 comments

## @Alex Suciu 2019-01-12 03:28:18

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 $\pi_3^s$.## @Theo Johnson-Freyd 2019-01-12 15:41:38

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.

## @Alex Suciu 2019-01-12 20:38:11

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.

## @Theo Johnson-Freyd 2019-01-13 00:30:26

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.