## June 19, 2017

### The Geometric McKay Correspondence (Part 1)

#### Posted by John Baez

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.

Posted at June 19, 2017 9:30 PM UTC

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### Re: The Geometric McKay Correspondence (Part 1)

From what I understand, not being an algebraic geometer, resolutions of singularities of surfaces aren’t necessarily unique, although there should be a unique minimal resolution. Is the 8-linked-spheres one you mentioned the minimal resolution? I would assume so, since it’s part of a more general construction, but am not sure. How does this look for E7?

Posted by: Layra Idarani on June 20, 2017 4:24 AM | Permalink | Reply to this

### Re: The Geometric McKay Correspondence (Part 1)

In general there is not a unique minimal resolution (they’re supposed to be related by flips). In this case it is unique.

BTW I think it’s interesting to note that the special fiber of the resolution, this Dynkin curve made of projective lines, is not generically reduced – the multiplicity of a component is the coefficient of the corresponding simple root in the expansion of the highest root (i.e. some number from 2 to 6).

Posted by: Allen Knutson on June 20, 2017 4:52 AM | Permalink | Reply to this

### Re: The Geometric McKay Correspondence (Part 1)

A piece of magic that I have yet to understand connects these singularities directly with the Lie groups of the same name. Something like “consider a string moving in $S$: this physical system has as its automorphism group the Lie group $E_8$, or perhaps the loop group thereof…” The story is supposed to continue that a string, unlike a particle, doesn’t distinguish between $S$ and $\tilde S$, and I guess you are supposed to see the maximal torus of $E_8$ by certain automorphisms corresponding to the spheres in $\tilde S$. Have you yet worked out how to picture that part of the story?

Posted by: Theo Johnson-Freyd on June 20, 2017 9:27 PM | Permalink | Reply to this

### Re: The Geometric McKay Correspondence (Part 1)

One can obtain a 4D quasicrystal by projecting from the $E_8$ lattice, whose cross-section is a 3D quasicrystal with icosahedral symmetry.

Posted by: Metatron on June 23, 2017 5:30 AM | Permalink | Reply to this

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