Let $G \subset \mathrm{SL}_2(\mathbf{C}^2)$ be a finite subgroup isomorphic to the binary icosahedral group. Let $Y$ be the minimal resolution of $\mathbf{C}^2/G$. The irreducible components of the exceptional fiber of $Y$ are naturally in correspondence with nodes of the Dynkin diagram of $\mathrm{E}_8$.

Each of these components has a linking circle in $(\mathbf{C}^2 - \{0\})/G$, whose fundamental group is $G$. Thus, each component determines a nontrivial conjugacy class in $G$.

There are 8 nontrivial conjugacy classes in $G$, of orders 2,3,4,5,5,6,10,10. For each node of $\mathrm{E}_8$, which conjugacy class is it labeled by? In particular, what is the order of this class?

Later: It sounds like Hugh is proposing $$ \begin{array}{rrrrrrr} & & 4\\ 6 & 3 & 2 & 5 & 10 & 5 & 10 \end{array}$$

but the locations of the (6,3), and of (5,10,5,10) are just guesses.

  • 1
    $\begingroup$ I was under the impression that the bijection with characters was more natural, since the tensor product with the 2-dimensional representation yields the extended Dynkin graph adjacency relation. $\endgroup$
    – S. Carnahan
    Jun 24, 2014 at 8:20
  • $\begingroup$ S., I agree that's a natural and beautiful bijection, but it is not the map I am asking about. $\endgroup$ Jun 24, 2014 at 15:10
  • $\begingroup$ The sizes of the conjugacy classes given by Suter seem to differ from yours. Can you explain? Are we talking about different things somehow? $\endgroup$ Jun 24, 2014 at 17:58
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    $\begingroup$ By "order of a conjugacy class C" I suppose you could mean either the cardinality of C, or order of an element of C. I meant the second thing. $\endgroup$ Jun 24, 2014 at 18:12
  • $\begingroup$ In principle, this could be determined from the paper of Kirby and Scharlemann: math.berkeley.edu/~kirby/papers/… If I get some time, I'll try to have a look to see if there's a simple way to see it. $\endgroup$
    – Ian Agol
    Jun 25, 2014 at 4:34

2 Answers 2


I found a paper of Kirby and Scharlemann which does the computations. In figure 2 of the paper, they give the surgery diagram for the Poincar\'e homology sphere:

enter image description here

One obtains a presentation for the fundamental group of the link complement by the usual Wirtinger method. Moreover, 2-handles are attached to each loop with framing 2 to give the 4-manifold, giving 8 more relations, appearing in the 3rd paragraph on p. 116. As indicated there, the group is generated by $a, g$ with presentation $\langle a, g | a^5 = g^3 =(ag)^2 \rangle$. We have $h=(ag)^{-1}$, and the central involution is given by $e= a^5 = g^3 = h^{-2}$ (notice $a,g,h$ are the generators of the outermost vertices of the Dynkin diagram, and $e$ represents the trivalent vertex). From their presentation, one computes $b=a^{-2}, c= a^3, d= a^{-4}, f=g^{-2}$. So indeed, the orders of these elements agree with the orders in your diagram, and essentially agrees with Hugh Thomas' answer.


According to arXiv:math/0503542 by Suter (Fact 5.1) the sizes of the conjugacy classes of $G$ other than the trivial one are 1, 30, 20, 20, 12, 12, 12, 12. These numbers (plus 1 for the trivial) sum to 120, which is the size of $G$, providing a sanity check.

In general (also from the same source), if the branch lengths of the finite Dynkin diagram are $p_1$, $p_2$, $p_3$, there are $p_i-1$ conjugacy classes of size $|G|/2p_i$, generated by powers of some element $\gamma_i$ satisfying $\gamma_i^{p_i}$ all equalling the central non-identity element, plus a conjugacy class for it.

One would therefore expect the $\gamma_i^j$ conjugacy classes to correspond to vertices along the branches, and the central element to correspond to the branch vertex. Suter actually labels an $E_8$ diagram like this (with powers increasing towards the branch point, which makes sense since that way the branch point is $\gamma_i^{p_i}$ for all $i$) but I don't see any explanation in the paper of what that labelling means for him.

And, of course, this doesn't begin to answer the question the OP asked...

Later: it seems like we would be most of the way there if one could show that the linking circle for a node, squared, should equal the product of the linking circles for the adjacent nodes. (Say, assuming the degree of the node is at most two -- this formula doesn't seem to hold for the central node, where the order of the product also potentially enters.)


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