The Dodecahedron, the Icosahedron and E8

Here you can see the slides of a talk I’m giving:

The dodecahedron, the icosahedron and E8, Annual General Meeting of the Hong Kong Mathematical Society, Hong Kong University of Science and Technology.

It’ll take place on 10:50 am Saturday May 20th in Lecture Theatre G. You can see the program for the whole meeting here.

The slides are in the form of webpages, and you can see references and some other information tucked away at the bottom of each page.

In preparing this talk I learned more about the geometric McKay correspondence, which is a correspondence between the simply-laced Dynkin diagrams (also known as ADE Dynkin diagrams) and the finite subgroups of \mathrm{SU}(2).

There are different ways to get your hands on this correspondence, but the geometric way is to resolve the singularity in \mathbb{C}^2/\Gamma where \Gamma \subset \mathrm{SU}(2) is such a finite subgroup. The variety \mathbb{C}^2/\Gamma has a singularity at the origin–or more precisely, the point coming from the origin in \mathbb{C}^2. To make singularities go away, we ‘resolve’ them. And when you take the ‘minimal resolution’ of this variety (a concept I explain here), you get a smooth variety S with a map

\pi \colon S \to \mathbb{C}^2/\Gamma

which is one-to-one except at the origin. The points that map to the origin lie on a bunch of Riemann spheres. There’s one of these spheres for each dot in some Dynkin diagram—and two of these spheres intersect iff their two dots are connected by an edge!

In particular, if \Gamma is the double cover of the rotational symmetry group of the dodecahedron, the Dynkin diagram we get this way is E_8:

The basic reason \mathrm{E}_8 is connected to the icosahedron is that the icosahedral group is generated by rotations of orders 2, 3 and 5 while the \mathrm{E}_8 Dynkin diagram has ‘legs’ of length 2, 3, and 5 if you count right:

In general, whenever you have a triple of natural numbers a,b,c obeying

\displaystyle{ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} > 1}

you get a finite subgroup of \mathrm{SU}(2) that contains rotations of orders a,b,c, and a simply-laced Dynkin diagram with legs of length a,b,c. The three most exciting cases are:

(a,b,c) = (2,3,3): the tetrahedron, and E_6,

(a,b,c) = (2,3,4): the octahedron, and E_7,

(a,b,c) = (2,3,5): the icosahedron, and E_8.

But the puzzle is this: why does resolving the singular variety \mathbb{C}^2/\Gamma gives a smooth variety with a bunch of copies of the Riemann sphere \mathbb{C}\mathrm{P}^1 sitting over the singular point at the origin, with these copies intersecting in a pattern given by a Dynkin diagram?

It turns out the best explanation is in here:

• Klaus Lamotke, Regular Solids and Isolated Singularities, Vieweg & Sohn, Braunschweig, 1986.

In a nutshell, you need to start by blowing up \mathbb{C}^2 at the origin, getting a space X containing a copy of \mathbb{C}\mathrm{P}^1 on which \Gamma acts. The space X/\Gamma has further singularities coming from the rotations of orders a, b and c in \Gamma. When you resolve these, you get more copies of \mathbb{C}\mathrm{P}^1, which intersect in the pattern given by a Dynkin diagram with legs of length a,b and c.

I would like to understand this better, and more vividly. I want a really clear understanding of the minimal resolution S. For this I should keep rereading Lamotke’s book, and doing more calculations.

I do, however, have a nice vivid picture of the singular space \mathbb{C}^2/\Gamma. For that, read my talk! I’m hoping this will lead, someday, to an equally appealing picture of its minimal resolution.

18 Responses to The Dodecahedron, the Icosahedron and E8

  1. jim stasheff says:

    only the animated opening slide

    is available at that click


  2. John Baez says:

    I just ran into the mathematician Roger Howe, who is visiting the Insititute of Advanced Studies here at HKUST. He was a professor at Yale back when I was a postdoc there… and still is. He argued that there are insufficiently exploited connections between the Platonic solids and the number 12.

    Puzzle 1. What does the regular tetrahedron have 12 of?

    Puzzle 2. What does the cube have 12 of?

    Puzzle 3. What does the dodecahedron have 12 of?

    Puzzle 4. How can you set up a bijection between the 12 things of the tetrahedron and the 12 things of the cube?

    Puzzle 5. How can you set up a bijection between the 12 things of the cube and the 12 things of the dodecahedron?

    Some of these are easier than others.

    • Scott Hotton says:

      These seem like the easy ones.

      Puzzle 2. A cube has 12 edges.

      Puzzle 3. A dodecahedron has 12 faces.

      Puzzle 5. A cube can be inscribed in a dodecahedron so that each edge of the cube is a diagonal of a face of the
      dodecahedron. This is shown here:

      These seem a little harder.

      Puzzle 1. A regular tetrahedron has 12 face medians (the medians of each triangular face).

      Puzzle 4. A regular tetrahedron can be inscribed in a cube so that each edge of the tetrahedron is a diagonal of a face of the cube. The tetrahedron and cube will be concentric. We can radially project the tetrahedron to the cube from their common center. The 4 face centers of the tetrahedron get mapped to the 4 vertices of the cube which are not also vertices of the tetrahedron. The segment of the tetrahedron’s face medians which connects the face center to a vertex will be mapped to an edge of the cube.

      This image shows a tetrahedron inscribed in a cube. It shows segments of the tetrahedron’s 4 medians rather than segments of the 12 face medians but it might be helpful for visualizing the bijection:

      • John Baez says:

        Great! Well done!

        Roger Howe actually told me about another thing that the tetrahedron has 12 of, and he managed to elegantly relate those to the 12 edges of a cube. So, we get more puzzles:

        Puzzle 1′: What are these other 12 things that the tetrahedron has?

        Puzzle 4′: How can you set up a bijection between these other 12 things and the 12 edges of the cube?

        Puzzle 6: How are these other 12 things related to the 12 medians of the tetrahedron’s faces?

        I suppose I should warn everyone that these other 12 things are considerably more abstract, and not just another way of talking about its 12 face medians.

        • Philip Gibbs says:

          Is it rotations that map the tetrahedron onto itself?

          Chose two medians (not necessarily distinct) and there is a unique rotation that maps one to the other. Fixing one of the two gives a bijection between the rotations and the other median.

          Is it also of interest that there is more than one way of constructing each bijection?

        • John Baez says:

          Philip wrote:

          Is it rotations that map the tetrahedron onto itself?

          Yes, that’s what Roger Howe had in mind. We’ve been discussing this stuff over on G+, and here’s what I said:

          I think I’ll give away Roger Howe’s preferred answers to puzzles 1 and 4.

          Puzzle 1. A tetrahedron has 12 rotational symmetries: we can rotate our favorite vertex to any of the 4 vertices in 3 different ways.

          Puzzle 4. If we inscribe a tetrahedron in either of the ways shown below (thanks again to Greg Egan), and pick a ‘favorite’ edge of the cube, each of the 12 rotational symmetries of the tetrahedron will carry our favorite edge to one of the 12 edges. This gives a bijection between rotational symmetries of the tetrahedron and edges of the cube.

          This is not perfectly natural since it involves a choice – a choice of a cube edge. We can eliminate this unnaturalness if we follow Allen Knutson’s suggestion: instead of using rotational symmetries of the tetrahedron, use vertex-edge flags. A vertex-edge flag for the tetrahedron is a vertex together with an edge it lies on.

          There is a natural bijection between edges of the cube and vertex-edge flags of the tetrahedron. Here I’m using ‘natural’ to mean preserved by the rotational symmetries of the tetrahedron.

          For experts: vertex-edge flags form a torsor for the group of rotational symmetries of the tetrahedron: i.e., a set on which this group acts freely and transitively. Identifying a torsor with the group itself requires choosing an arbitrary point of the torsor, which we identify with the identity in the group.

          With less jargon: if we choose any vertex-edge flag as our ‘favorite’, each rotational symmetry carries it to another vertex-edge flag, and this sets up a bijection between rotational symmetries and vertex-edge flags.

          If we inscribe the tetrahedron in a cube, the edges of that cube form a torsor for the group of rotational symmetries of the tetrahedron. If we then inscribe the cube in a dodecahedron, so do the faces of the dodecahedron!

    • Bob says:

      Puzzle 1: There are 12 unique directions along the edges. Each can be traversed only once in making a complete circuit.

  3. Bruce Smith says:

    In fact, there are several kinds of things that the tetrahedron has 12 of (even staying no more abstract than “face median”). For two answers to Puzzle 1 that I have in mind, Puzzle 4 is easy for one of them, but I don’t know whether it’s solvable for the other one.

    • John Baez says:

      Over on G+, Allen Knutson suggested something besides ‘face medians’ that a tetrahedron has 12 of: ‘vertex-edge flags’. A vertex-edge flag is a vertex together with a face it lies on.

      Whenever we’ve got a set S of things that a tetrahedron has 12 of, we should get an action of the tetrahedron’s rotational symmetry group A4 on this 12-element set S. A set on which a group G acts is called a G-set, and an isomorphism between G-sets S and S' is a bijection

      f : S \to S'

      such that

      g f(s) = f(gs)

      for all g \in G, s \in S.

      Puzzle 6. How many isomorphism classes of 12-element A4-sets are there?

      In simple rough terms: how many ‘fundamentally different ways’ are there to find 12 things in a tetrahedron?

      For example, the set of face medians is a 12-element A4-set, and so is the set of ‘vertex-face flags’ (defined in the obvious way). But these are isomorphic as A4-sets, since a face median touches one vertex and one face, and these give a vertex-face flag.

      • Scott Hotton says:

        This looks a little bit like the math problem for periodic patterns in peptides

        except that it seems to come down to counting unordered sums instead of ordered sums.

        Two actions of a finite group on a finite set are isomorphic if and only if they have the same number of orbits of the same length. The length of an orbit is the index of a point stabilizer in the full group.

        The subgroups of A_4 have orders 1,2,3,4,12 so the orbits can have lengths 12, 6, 4, 3, 1. The number of equivalence classes of actions of A_4 on a set of 12 elements comes down to the number of ways 12 can be written as an unordered sum of 1, 3, 4, 6, 12. I wrote the cases out as Young diagrams here:

        There appear to be 17 equivalence classes of actions of A_4 on a 12 element set.

        In a solid tetrahedron there are no points with orbits of length 3 under the action of A_4. So it appears that for any 12 point subset of a solid tetrahedron there are only 7 equivalence classes of actions by A_4.

        But if A_4 acts on collections of subsets of the tetrahedron then we can get an orbit of length 3. For instance there are 3 pairs of opposite edges.

        We can reduce this to finite sets. Denote the vertices by 1,2,3,4. A_4 acts transitively on the 3 element “class of collections of finite subsets”:

        {{{1,2},{3,4}}, {{1,3},{2,4}}, {{1,4},{2,3}}}

        Other orbits of length 3 could be obtained this way by selecting other points from the tetrahedron’s medians. So it seems that all possible equivalence classes of actions of A_4 on a 12 element set would be possible with a regular tetrahedron.

      • Philip Gibbs says:

        I haven’t been following the discussion in G+ so excuse me if this has already been discussed. You can count vertex-face flags, vertex-edge flags or face-edge flags. All three give the same number for any polyhedron. They are also the same for dual polyhedra. Is there a name for this number? The numbers are 12,24,60 for tetrahedron, cube/octahedron and dodecahedron/icosahedron. These numbers come up a lot in exceptional structures, especially 24 of course. For example

        • John Baez says:

          For the regular polyhedra someone on G+, probably Layra Idarani, pointed out that these numbers are called “the order of the rotational symmetry group of the polyhedron”. There’s a nice generalization to the polytopes associated to Coxeter groups.

          For other, non-regular polyhedra these numbers seem a bit more mysterious to me.

        • Philip Gibbs says:

          It must be more general than the order of the rotation group if it applies to all polyhedra. Perhaps it is the vertex-edge-face flags that count. There would be twice as many of those. This would generalise to higher dimensions. Is that something from cohomology?

        • John Baez says:

          Certainly the vertex-edge-face flags are more fundamental—or in general, for any polytope, the ‘complete’ flags.

          I don’t know the meaning of the number of these flags. But there’s an excellent group that acts on the complete flags regardless of whether the polytope is regular or not! It’s called the cartographic group. It has one generator for each dimension.

          In the case at hand, when we’re starting with a polyhedron, these generators could be called V, E, and F. Their meaning is ‘change the vertex’, ‘change the edge’ and ‘change the face’.

          You see, if you have a vertex-edge-face flag, there’s exactly one way to change the vertex and get a new such flag. Similarly there’s one way to change the edge, and one way to change the face.

          In the cartographic group, these generators obey only the relations that hold regardless of which polyhedron we’re dealing with:

          V^2 = E^2 = F^2 = 1, VF = FV

          This presentation defines the cartographic group.

          This story extends to vertex-edge flags, or edge-face flags, or vertex-face flags. Various quotient groups of the cartographic group act on these.

          All this generalizes to arbitrary dimensions. In general, each generator of the cartographic group. squares to 1, and generators that aren’t of neighboring dimensions commute.

  4. John Baez says:

    Today at breakfast Roger Howe suggested trying to construct the Leech lattice (a 24-dimensional lattice) or Golay code (a linear code consisting of 24-bit strings) starting the tetrahedron, roughly as follows. There are lots of significant problems left to solve, but it’s a cute idea.

    The tetrahedron has 12 rotational symmetries, giving

    \mathrm{A}_4 \subseteq \mathrm{SO}(3)

    Using the double cover

    \pi : \mathrm{SU}(2) \to \mathrm{SO}(3)

    we get a 24-element subgroup of \mathrm{SU}(2). This is quite famous:

    • Wikipedia, Binary tetrahedral group.

    Since \mathrm{SU}(2) is the unit sphere in the quaternions the binary tetrahedral group forms the vertices of a 4d polytope, called the 24-cell:

    • Wikipedia, 24-cell.

    The 24-cell is a 4-dimensional regular polytope. The vertices of the 24-cell can be broken up into 3 sets of 8, each set being the vertices of another 4-dimensional polytope, which is called the ‘hyperoctahedron’ or ‘cross-polytope’ or ‘4-dimensional orthoplex’ or ’16-cell’:

    • Wikipedia, 16-cell.

    If you set up things nicely, these 3 cross-polytopes inside the 24-cell get permuted when we permute the quaternions i, j, and k. This is called ‘triality’.

    Now the idea is this: since the 3 cross-polytopes inside the 24-cell are a very nice way to think about the identity

    24 = 8 + 8 + 8,

    perhaps they can be used to help understand how we build the Leech lattice out of 3 copies of the E8 lattice, or build the Golay code out of 3 copies of the 8-bit Hamming code!

    We (in some sense of ‘we’) already know how to build the Leech lattice out of 3 copies of the E8 lattice, or build the Golay code out of 3 copies of the 8-bit Hamming code. It’s called the ‘Turyn construction’. In the case of building the Leech from 3 E8‘s, Greg Egan and I described it here:

    Integral octonions (part 9).

    The question is whether this is at all illuminated by the geometry I just described.

    It’s worth noting that Greg and I already did connect triality to the Turyn construction of the Leech lattice. The three E8 lattices used to build the Leech lattice can be thought of as lying in the vector, left-handed spinor, and right-handed spinor representations of \mathrm{Spin}(8). These representations are permuted by the triality automorphisms of \mathrm{Spin}(8). Furthermore, the weight lattice of \mathrm{Spin}(8) is a 4-dimensional lattice generated by the points in the 24-cell, and the weights of the vector, left-handed spinor and right-handed spinor representations lie in the 3 cross-polytopes I mentioned!

    So, all this stuff does fit together. I guess the question is whether the Turyn construction makes a bit more sense if we examine it in this larger context.

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