## The Geometric McKay Correspondence (Part 1)

The ‘geometric McKay correspondence’, actually discovered by Patrick du Val in 1934, is a wonderful relation between the Platonic solids and the ADE Dynkin diagrams. In particular, it sets up a connection between two of my favorite things, the icosahedron: and the $\mathrm{E}_8$ Dynkin diagram: When I recently gave a talk on this topic, I realized I didn’t understand it as well as I’d like. Since then I’ve been making progress with the help of this book:

• Alexander Kirillov Jr., Quiver Representations and Quiver Varieties, AMS, Providence, Rhode Island, 2016.

I now think I glimpse a way forward to a very concrete and vivid understanding of the relation between the icosahedron and E8. It’s really just a matter of taking the ideas in this book and working them out concretely in this case. But it takes some thought, at least for me. I’d like to enlist your help.

The rotational symmetry group of the icosahedron is a subgroup of $\mathrm{SO}(3)$ with 60 elements, so its double cover up in $\mathrm{SU}(2)$ has 120. This double cover is called the binary icosahedral group, but I’ll call it $\Gamma$ for short.

This group $\Gamma$ is the star of the show, the link between the icosahedron and E8. To visualize this group, it’s good to think of $\mathrm{SU}(2)$ as the unit quaternions. This lets us think of the elements of $\Gamma$ as 120 points in the unit sphere in 4 dimensions. They are in fact the vertices of a 4-dimensional regular polytope, which looks like this:

It’s called the 600-cell.

Since $\Gamma$ is a subgroup of $\mathrm{SU}(2)$ it acts on $\mathbb{C}^2,$ and we can form the quotient space $S = \mathbb{C}^2/\Gamma$

This is a smooth manifold except at the origin—that is, the point coming from $0 \in \mathbb{C}^2.$ There’s a singularity at the origin, and this where $\mathrm{E}_8$ is hiding! The reason is that there’s a smooth manifold $\widetilde{S}$ and a map $\pi : \widetilde{S} \to S$

that’s one-to-one and onto except at the origin. It maps 8 spheres to the origin! There’s one of these spheres for each dot here: Two of these spheres intersect in a point if their dots are connected by an edge; otherwise they’re disjoint.

The challenge is to find a nice concrete description of $\widetilde{S},$ the map $\pi : \widetilde{S} \to S,$ and these 8 spheres.

But first it’s good to get a mental image of $S.$ Each point in this space is a $\Gamma$ orbit in $\mathbb{C}^2,$ meaning a set like this: $\{g x : \; g \in \Gamma \}$

for some $x \in \mathbb{C}^2.$ For $x = 0$ this set is a single point, and that’s what I’ve been calling the ‘origin’. In all other cases it’s 120 points, the vertices of a 600-cell in $\mathbb{C}^2.$ This 600-cell is centered at the point $0 \in \mathbb{C}^2,$ but it can be big or small, depending on the magnitude of $x.$

So, as we take a journey starting at the origin in $S,$ we see a point explode into a 600-cell, which grows and perhaps also rotates as we go. The origin, the singularity in $S,$ is a bit like the Big Bang.

Unfortunately not every 600-cell centered at the origin is of the form I’ve shown: $\{g x : \; g \in \Gamma \}$

It’s easiest to see this by thinking of points in 4d space as quaternions rather than elements of $\mathbb{C}^2.$ Then the points $g \in \Gamma$ are unit quaternions forming the vertices of a 600-cell, and multiplying $g$ on the right by $x$ dilates this 600-cell and also rotates it… but we don’t get arbitrary rotations this way. To get an arbitrarily rotated 600-cell we’d have to use both a left and right multiplication, and consider $\{x g y : \; g \in \Gamma \}$

for a pair of quaternions $x, y.$

Luckily, there’s a simpler picture of the space $S.$ It’s the space of all regular icosahedra centered at the origin in 3d space!

To see this, we start by switching to the quaternion description, which says $S = \mathbb{H}/\Gamma$

Specifying a point $x \in \mathbb{H}$ amounts to specifying the magnitude $\|x\|$ together with $x/\|x\|,$ which is a unit quaternion, or equivalently an element of $\mathrm{SU}(2).$ So, specifying a point in $\{g x : \; g \in \Gamma \} \in \mathbb{H}/\Gamma$

amounts to specifying the magnitude $\|x\|$ together with a point in $\mathrm{SU}(2)/\Gamma$. But $\mathrm{SU}(2)$ modulo the binary icosahedral group $\Gamma$ is the same as $\mathrm{SO(3)}$ modulo the icosahedral group (the rotational symmetry group of an icosahedron). Furthermore, $\mathrm{SO(3)}$ modulo the icosahedral group is just the space of unit-sized icosahedra centered at the origin of $\mathbb{R}^3.$

So, specifying a point $\{g x : \; g \in \Gamma \} \in \mathbb{H}/\Gamma$

amounts to specifying a nonnegative number $\|x\|$ together with a unit-sized icosahedron centered at the origin of $\mathbb{R}^3.$ But this is the same as specifying an icosahedron of arbitrary size centered at the origin of $\mathbb{R}^3.$ There’s just one subtlety: we allow the size of this icosahedron to be zero, but then the way it’s rotated no longer matters.

So, $S$ is the space of icosahedra centered at the origin, with the ‘icosahedron of zero size’ being a singularity in this space. When we pass to the smooth manifold $\widetilde{S},$ we replace this singularity with 8 spheres, intersecting in a pattern described by the $\mathrm{E}_8$ Dynkin diagram.

Points on these spheres are limiting cases of icosahedra centered at the origin. We can approach these points by letting an icosahedron centered at the origin shrink to zero size in a clever way, perhaps spinning about wildly as it does.

I don’t understand this last paragraph nearly as well as I’d like! I’m quite sure it’s true, and I know a lot of relevant information, but I don’t see it. There should be a vivid picture of how this works, not just an abstract argument. Next time I’ll start trying to assemble the material that I think needs to go into building this vivid picture.

### 7 Responses to The Geometric McKay Correspondence (Part 1)

1. Bruce Smith says:

I don’t understand this last paragraph nearly as well as I’d like!

However, I bet you understand it a lot better than we do! So unless you want to save everything for “next time”, can you give a hint about why these limits are anything but the “obvious one”? (To me, I think the only obvious one is shrinking to zero without spinning around significantly, which I presume corresponds to a single point on one or more of those 8 spheres. Can you give at least one other example of a different kind of limiting point (and the corresponding sequence with a limit), and explain why it’s different?)

• John Baez says:

Giving a nice explanation of in what sense there are 8 spheres’ worth of ways for an icosahedron to shrink to zero size is the goal of this series of posts! It’s utterly non-obvious, and the only way I understand it so far is by following along the proofs in some books. So, I can’t explain it here.

But we certainly don’t want there to be just one way for an icosahedron to shrink to zero size. There’s just one way in the space $S = \mathbb{C}^2/\Gamma$

but this space has a singularity at the point coming from the origin, which represents the zero-sized icosahedron. We’re going to replace this space $S$ by a subtler space $\widetilde{S}$ which is smooth everywhere, and has a map $\pi : \widetilde{S} \to S$ sending 8 spheres to the singular point in $S.$ This trick, standard in algebraic geometry, is called a resolution of singularities. But understanding how points on those 8 spheres can be seen as limits of icosahedra shrinking to zero size—that’s the hard part.

Naively, I’d expect $\widetilde{S}$ to have a bunch of points that are limits of icosahedra that don’t rotate as they shrink to zero size. If there were one such point for each icosahedron of given size, that would give one point for each point in $\mathrm{SO}(3)/\mathrm{A}_5$ where $\mathrm{A}_5$ is the rotational symmetry group of the icosahedron. $\mathrm{SO}(3)/\mathrm{A}_5$ is the space of ‘visibly distinct icosahedra of given size centered at the origin’.

But this answer is not right so it must be too naive!

There is a lot more to say, but that’s why this is just Part 1.

• Bruce Smith says:

Thanks, that makes it clearer. I’m looking forward to the other Parts!

2. John Baez says:

In an experiment, I also posted this article here:

The Geometric McKay Correspondence (Part 1), The n-Category Café.

I got more comments from professional pure mathematicians there… not surprising, since Azimuth is more focused on applied math. Check out the comments there, and my replies!

• Blake Stacey says:

I commented over there, but I’m not a pure mathematician (I don’t even play one on TV).

• John Baez says:

You should start by playing one on TV, get the hang of it that way, and then parlay it into a full-time job. Nobody understands math anyway; I’m sure if you string together impressive words in a plausible way you’ll get tenure.

For some reason I’m not getting notifications of comments on that particular n-Café entry, so I just noticed and replied to your comments there today. They’re really exciting to me!

So exciting, in fact, that I finally wrote Part 2 of this series, both here and on the n-Café.

3. Last time I sketched how the $E_8$ Dynkin diagram arises from the icosahedron. This time I’m fill in some details. I won’t fill in all the details, because I don’t know how! Working them out is the goal of this series, and I’d like to enlist your help.

(In fact, I’m running this series of posts both here and at the n-Category Café. So far I’m getting more comments over there, so I suspect the serious helping will happen there. You may want to pop over there to see what people are talking about.)

This site uses Akismet to reduce spam. Learn how your comment data is processed.