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July 2, 2017

The Geometric McKay Correspondence (Part 2)

Posted by John Baez

Last time I sketched how the E 8E_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.

As Kennedy said: ask not what your n-Café can do for you. Ask what you can do for your n-Café!

Remember the basic idea. We start with the rotational symmetry group of the isosahedron and take its double cover, getting a 120-element group Γ\Gamma called the binary icosahedral group. Since this is naturally a subgroup of SU(2)\mathrm{SU}(2) it acts on 2\mathbb{C}^2, and we can form the quotient space

S= 2/Γ S = \mathbb{C}^2/\Gamma

This is a smooth manifold except at the origin — by which I mean the point coming from 0 20 \in \mathbb{C}^2. Luckily we can ‘resolve’ this singularity! This implies that we can find a smooth manifold S˜\widetilde{S} and a smooth map

π:S˜S \pi \colon \widetilde{S} \to S

that’s one-to-one and onto except at the origin. There may be various ways to do this, but there’s one best way, the ‘minimal’ resolution, and that’s what I’ll be talking about.

The origin is where all the fun happens. The map π\pi sends 8 spheres to the origin in 2/Γ\mathbb{C}^2/\Gamma, one for each dot in the E 8\mathrm{E}_8 Dynkin diagram:

Two of these spheres intersect in a point if their dots are connected by an edge; otherwise they’re disjoint.

This is wonderful! So, the question is just how do we really see it? For starters, how do we get our hands on this manifold S˜\widetilde{S} and this map π:S˜S \pi \colon \widetilde{S} \to S?

For this we need some algebraic geometry. Indeed, the whole subject of ‘resolving singularities’ is part of algebraic geometry! However, since I still remember my ignorant youth, I want to avoid flinging around the vocabulary of this subject until we actually need it. So, experts will have to pardon my baby-talk. Nonexperts can repay me in cash, chocolate, bitcoins or beer.

What’s S˜\widetilde{S} like? First I’ll come out and tell you, and then I’ll start explaining what the heck I just said.

Theorem. S˜\widetilde{S} is the space of all Γ\Gamma-invariant ideals I[x,y]I \subseteq \mathbb{C}[x,y] such that [x,y]/I\mathbb{C}[x,y]/I is isomorphic, as a representation of Γ\Gamma, to the regular representation of Γ\Gamma.

If you want a proof, this is Corollary 12.8 in Kirillov’s Quiver Representations and Quiver Varieties. It’s on page 245, so you’ll need to start by reading lots of other stuff. It’s a great book! But it’s not completely self-contained: for example, right before Corollary 12.8 he brings in a crucial fact without proof: “it can be shown that in dimension 2, if a crepant resolution exists, it is minimal”.

I will not try to prove this theorem; instead I will start explaining what it means.

Suppose you have a bunch of points p 1,,p n 2p_1, \dots, p_n \in \mathbb{C}^2. We can look at all the polynomials on 2\mathbb{C}^2 that vanish at these points. What is this collection of polynomials like?

Let’s use xx and yy as names for the standard coordinates on 2\mathbb{C}^2, so polynomials on 2\mathbb{C}^2 are just polynomials in these variables. Let’s call the ring of all such polynomials [x,y]\mathbb{C}[x,y]. And let’s use II to stand for the collection of such polynomials that vanish at our points p 1,,p np_1, \dots, p_n.

Here are two obvious facts about II:

A. If fIf \in I and gIg \in I then f+gIf + g \in I.

B. If fIf \in I and g[x,y]g \in \mathbb{C}[x,y] then fgIf g \in I.

We summarize these by saying II is an ideal, and this is why we called it II. (So clever!)

Here’s a slightly less obvious fact about II:

C. If the points p 1,,p np_1, \dots, p_n are all distinct, then [x,y]/I\mathbb{C}[x,y]/I has dimension nn.

The point is that the value of a function f[x,y]f \in \mathbb{C}[x,y] at a point p ip_i doesn’t change if we add an element of II to ff, so this value defines a linear functional on [x,y]/I\mathbb{C}[x,y]/I . Guys like this form a basis of linear functionals on [x,y]/I\mathbb{C}[x,y]/I, so it’s nn-dimensional.

All this should make you interested in the set of ideals II with dim([x,y]/I)=n\mathrm{dim}(\mathbb{C}[x,y]/I) = n . This set is called the Hilbert scheme Hilb n( 2)\mathrm{Hilb}^n(\mathbb{C}^2).

Why is it called a scheme? Well, Hilbert had a bunch of crazy schemes and this was one. Just kidding: actually Hilbert schemes were invented by Grothendieck in 1961. I don’t know why he named them after Hilbert. The kind of Hilbert scheme I’m using is a very basic one, more precisely called the ‘punctual’ Hilbert scheme.

The Hilbert scheme Hilb n( 2)\mathrm{Hilb}^n(\mathbb{C}^2) is a whole lot like the set of unordered nn-tuples of distinct points in 2\mathbb{C}^2. Indeed, we’ve seen that every such nn-tuple gives a point in the Hilbert scheme. But there are also other points in the Hilbert scheme! And this is where the fun starts!

Imagine nn particles moving in 2\mathbb{C}^2, with their motion described by polynomial functions of time. As long as these particles don’t collide, they define a curve in the Hilbert scheme. But it still works when they collide! When they collide, this curve will hit a point in the Hilbert scheme that doesn’t come from an unordered nn-tuple of distinct points in 2\mathbb{C}^2. This point describes a ‘type of collision’.

More precisely: nn-tuples of distinct points in 2\mathbb{C}^2 give an open dense set in the Hilbert scheme, but there are other points in the Hilbert scheme which can be reached as limits of those in this open dense set! The topology here is very subtle, so let’s look at an example.

Let’s look at the Hilbert scheme Hilb 2( 2)\mathrm{Hilb}^2(\mathbb{C}^2). Given two distinct points p 1,p 2 2p_1, p_2 \in \mathbb{C}^2, we get an ideal

{f[x,y]:f(p 1)=f(p 2)=0} \{ f \in \mathbb{C}[x,y] \, : \; f(p_1) = f(p_2) = 0 \}

This ideal is a point in our Hilbert scheme, since dim([x,y]/I)=2\mathrm{dim}(\mathbb{C}[x,y]/I) = 2 .

But there are other points in our Hilbert scheme! For example, if we take any point p 2p \in \mathbb{C}^2 and any vector v 2v \in \mathbb{C}^2, there’s an ideal consisting of polynomials that vanish at pp and whose directional derivative in the vv direction also vanishes at pp:

I={f[x,y]:f(p)=lim t0f(p+tv)f(p)t=0} I = \{ f \in \mathbb{C}[x,y] \, : \; f(p) = \lim_{t \to 0} \frac{f(p+t v) - f(p)}{t} = 0 \}

It’s pretty easy to check that this is an ideal and that dim([x,y]/I)=2\mathrm{dim}(\mathbb{C}[x,y]/I) = 2 . We can think of this ideal as describing ‘two particles in 2\mathbb{C}^2 that have collided at pp with relative velocity some multiple of vv’.

For example you could have one particle sitting at pp while another particle smacks into it while moving with velocity vv; as they collide the corresponding curve in the Hilbert scheme would hit II.

This would also work if the velocity were any multiple of vv, since we also have

I={f[x,y]:f(p)=lim t0f(p+ctv)f(p)t=0} I = \{ f \in \mathbb{C}[x,y] \, : \; f(p) = \lim_{t \to 0} \frac{f(p+ c t v) - f(p)}{t} = 0 \}

for any constant c0c \ne 0. And note, this constant can be complex. I’m trying to appeal to your inner physicist, but we’re really doing algebraic geometry over the complex numbers, so we can do weird stuff like multiply velocities by complex numbers.

Or, both particles could be moving and collide at pp while their relative velocity was some complex multiple of vv. As they collide, the corresponding point in the Hilbert scheme would still hit II.

But here’s the cool part: such ‘2-particle collisions with specified position and relative velocity’ give all the points in the Hilbert scheme Hilb 2( 2)\mathrm{Hilb}^2(\mathbb{C}^2), except of course for those points coming from 2 particles with distinct positions.

What happens when we go to the next Hilbert scheme, Hilb 3( 2)\mathrm{Hilb}^3(\mathbb{C}^2)? This Hilbert scheme has an open dense set corresponding to triples of particles with distinct positions. It has other points coming from situations where two particles collide with some specified position and relative velocity while a third ‘bystander’ particle sits somewhere else. But it also has points coming from triple collisions. And these are more fancy! Not only velocities but accelerations play a role!

I could delve into this further, but for now I’ll just point you here:

• John Baez, The Hilbert scheme for 3 points on a surface, MathOverflow, June 7, 2017.

The main thing to keep in mind is this. As nn increases, there are more and more ways we can dream up ideals II with dim([x,y]/I)=n\mathrm{dim}(\mathbb{C}[x,y]/I) = n . But all these ideals consist of functions that vanish at nn or fewer points and also obey other equations saying that various linear combinations of their first, second, and higher derivatives vanish. We can think of these ideals as ways for nn particles to collide, with conditions on their positions, velocities, accelerations, etc. The total number of conditions needs to be nn.

Now let’s revisit that description of the wonderful space we’re seeking to understand, S˜\widetilde{S}:

Theorem. S˜\widetilde{S} is the space of all Γ\Gamma-invariant ideals I[x,y]I \subseteq \mathbb{C}[x,y] such that [x,y]/I\mathbb{C}[x,y]/I is isomorphic, as a representation of Γ\Gamma, to the regular representation of Γ\Gamma.

Since Γ\Gamma has 120 elements, its regular representation — the obvious representation of this group on the space of complex functions on this group — is 120-dimensional. So, points in S˜\widetilde{S} are ideals II with dim([x,y]/I)=120\mathrm{dim}(\mathbb{C}[x,y]/I) = 120 . So, they’re points in the Hilbert scheme Hilb 120( 2)\mathrm{Hilb}^{120}(\mathbb{C}^2).

But they’re not just any old points in this Hilbert scheme! The binary icosahedral group Γ\Gamma acts on 2\mathbb{C}^2 and thus anything associated with it. In particular, it acts on the HIlbert scheme Hilb 120( 2)\mathrm{Hilb}^{120}(\mathbb{C}^2). A point in this Hilbert scheme can lie in S˜\widetilde{S} only if it’s invariant under the action of Γ\Gamma. And given this, it’s in S˜\widetilde{S} if and only if [x,y]/I\mathbb{C}[x,y]/I is isomorphic to the regular representation of Γ\Gamma.

Given all this, there’s an easy way to get your hands on a point IS˜I \in \widetilde{S}. Just take any nonzero element of 2\mathbb{C}^2 and act on it by Γ\Gamma. You’ll get 120 distinct points in 2\mathbb{C}^2 — I promise. Do you see why? Then let II be the set of polynomials that vanish on all these points.

If you don’t see why this works, please ask me.

In fact, we saw last time that your 120 points will be the vertices of a 600-cell centered at the origin of 2\mathbb{C}^2:

By this construction we get enough points to form an open dense subset of S˜\widetilde{S}. These are the points that aren’t mapped to the origin by

π:S˜S \pi \colon \widetilde{S} \to S

Alas, it’s the other points in S˜\widetilde{S} that I’m really interested in. As I hope you see, these are certain ‘limits’ of 600-cells that have ‘shrunk to the origin’… or in other words, highly symmetrical ways for 120 points in 2\mathbb{C}^2 to collide at the origin, with some highly symmetrical conditions on their velocities, accelerations, etc.

That’s what I need to understand.

Posted at July 2, 2017 7:21 PM UTC

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Re: The Geometric McKay Correspondence (Part 2)

I would guess that the Hilbert scheme (of projective space) was so named because on each connected component, the Hilbert polynomial is constant. (Hartshorne’s thesis shows a converse – the Hilbert polynomial separates the connected components.) In your examples, this polynomial is degree 0.

Probably the best way to get started on understanding the E8 Dynkin curve you’re asking about is to consider the resolution as the quiver variety of the Nakajima framing of the E8 diagram. I’m not going to give a full exegesis of Nakajima quiver varieties here so the following is only meant for pointers.

Put the 2-4-6-5-4-3-2 & 3 dimensional spaces on the E8 diagram, with maps in both directions along any edge, and put another eight 1-dimensional spaces mapping to and from those spaces.

To get the quiver variety, impose the preprojective condition along the first set of vertices (the zero level set for the complex moment map; it says that the alternating sum of all products go-out-then-back-in to a vertex v should add to 0, signs given by the original choice of orientation.)

After that you have to do a GIT quotient. If you do the stupid affine GIT quotient, then you get the C^2/Gamma singular variety. Your question is about the extra info in the more interesting quotients.

I suspect the data of a point in the singular variety is given by the maps between the 1-d spaces (go down into the original E8, wander around, come back up into another 1-d space). So I think the extra equations you want, to say (in the Hilbert scheme language) that your points are all at the origin, are that these maps between the 1-d spaces are all 0.

Let’s warm up with the C^2/Z3 case, where the quiver is A2 not E8, whose Nakajima doubling looks like 1-1-1-1. Call the maps (really just numbers) 1-a->1-b->1-c->1 and 1<-x-1<-y-1<-z-1. Then the preprojective conditions are ax = by = cz. But now I also suggest to impose abc = 0 and xyz = 0. I’m guessing that the stability conditions in the GIT quotient say that a,z are not zero, so the equations become bc = xy = 0. Now there are two components, x=b=c=0 and x=y=c=0, which is what I was hoping to get for the A2 case.

Posted by: Allen Knutson on July 3, 2017 6:18 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

To carry on the conversation from last time:

We know what the space SS means in quantum information theory. A point in SS is a quantum measurement—a POVM, possibly a noisy one. The question I hope to answer as this discussion goes on is, “What does the space S˜\widetilde{S} mean in quantum information theory, and what is the significance of the map π:S˜S\pi: \widetilde{S} \to S?”

Posted by: Blake Stacey on July 4, 2017 4:39 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

Blake wrote:

A point in SS is a quantum measurement—a POVM, possibly a noisy one.

Not quite, I think. We have S= 2/ΓS = \mathbb{C}^2/\Gamma, where Γ\Gamma is the binary icosahedral group. In this description we can think of it as the Hilbert space of a spin-1/2 particle, mod the double cover of the icosahedron’s rotational symmetries.

Last time I argued that SS is isomorphic to the space of regular icosahedra of arbitrary size, possibly even zero size, centered at the origin of 3\mathbb{R}^3.

If we eliminate the icosahedra of radius bigger than 1, we can think of these as icosahedra in the unit ball. We can reinterpret this ball as the ‘Bloch ball’: the space of mixed states of a spin-1/2 particle, with the center being the ‘state of maximal ignorance’.

So, we can also think of SS as the Bloch ball modulo the icosahedron’s rotational symmetry group. That is: mixed states of a spin-1/2 particle, but where we count two as the same if they differ by a rotation in the icosahedron’s rotation group.

Can you think of another nice way to think about this?

No matter how we think about it, SS is smooth except at one point in the middle, and S˜\widetilde{S} is SS with this singularity resolved in the nicest possible way.

Posted by: John Baez on July 5, 2017 1:42 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

Here’s what I had in mind. First, to establish some notation and terminology:

Any time we have a vector in 3\mathbb{R}^3 of length 1 or less, we can map it to a 2×22 \times 2 matrix by the formula

ρ=12(I+xσ x+yσ y+zσ z), \rho = \frac{1}{2}\left(I + x\sigma_x + y\sigma_y + z\sigma_z\right),

where (x,y,z)(x,y,z) are the Cartesian components of the vector and (σ x,σ y,σ z)(\sigma_x, \sigma_y, \sigma_z) are the Pauli matrices. This yields a positive semidefinite matrix ρ\rho with trace equal to 1; when the vector has length 1, we have ρ 2=ρ\rho^2 = \rho, and the density matrix is a rank-1 projector that can be written as ρ=|ψψ|\rho = |\psi\rangle\langle\psi| for some vector |ψ|\psi\rangle.

A POVM is a set of “effects” (positive semidefinite operators satisfying 0<E<I0 \lt E \lt I) that furnish a resolution of the identity:

iw iρ i=I, \sum_i w_i \rho_i = I,

for some density operators ρ i\rho_i and weights w iw_i. Note that taking the trace of both sides gives a normalization constraint for the weights w iw_i in terms of the dimension of the Hilbert space (for our purposes today, 2). In this context, the Born rule says that when we perform the measurement described by this POVM, we obtain the ii-th outcome with probability

p(i)=tr(ρw iρ i), p(i) = \mathrm{tr}(\rho w_i \rho_i),

where ρ\rho without a subscript denotes our quantum state for the system. The weighting w iw_i is, up to a constant, the probability we would assign to the ii-th outcome if our state ρ\rho were the maximally mixed state 1dI\frac{1}{d}I, the “state of maximal ignorance.”

Given any polyhedron of unit radius or less in 3\mathbb{R}^3, we can feed its vertices into the Bloch representation and obtain a set of density operators (which are pure states if they lie on the surface of the Bloch sphere). For a simple example, we can do a regular tetrahedron. Let ss and rr take the values ±1\pm 1, and define

ρ s,r=12(I+13(sσ x+rσ y+srσ z)). \rho_{s,r} = \frac{1}{2}\left(I + \frac{1}{\sqrt{3}}(s\sigma_x + r\sigma_y + sr\sigma_z)\right).

(The blog software is cutting my comment off here in the preview, so I’ll continue below.)

Posted by: Blake Stacey on July 5, 2017 3:33 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

Blake wrote:

So, we can think of SS in two ways: either as the Bloch ball modulo the icosahedron’s rotational symmetry group, or as the set of icosahedral POVMs.

Okay, you’re right about this. Great! In my previous skeptical comment I had the wrong idea about what a POVM was.

Posted by: John Baez on July 7, 2017 1:17 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

No problem! Learning how to communicate across disciplinary boundaries is a big reason why I hang out around here.

Posted by: Blake Stacey on July 7, 2017 1:28 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

Continuing from previous comment:

To make the POVM effects, weight the rank-1 projectors ρ s,r\rho_{s,r} equally. That is, take

E s,r=12ρ s,r. E_{s,r} = \frac{1}{2} \rho_{s,r}.

Then, the four operators E s,rE_{s,r} will sum to the identity. The icosahedron will work out basically the same way, but I don’t already have it in a notebook. Because I’m lazy tonight, I looked it up; the most detailed description I found in a couple minutes of checking is in section 10 of T. Decker, D. Janzing and T. Beth, “Quantum circuits for single-qubit measurements corresponding to Platonic solids,” Int. J. Quant. Info. 2 (2004), 353–77, arXiv:quant-ph/0308098.

So, we can think of SS in two ways: either as the Bloch ball modulo the icosahedron’s rotational symmetry group, or as the set of icosahedral POVMs.

(I figured out why the preview of my comment was stopping partway through: I had a square bracket that should have been a curly bracket.)

Posted by: Blake Stacey on July 5, 2017 3:40 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

One more bit of terminology I will probably use later:

Every known SIC has a group covariance property. Talking in terms of projectors, a SIC is a set of d 2d^2 rank-1 projectors {Π j}\{\Pi_j\} on a dd-dimensional Hilbert space that satisfy the Hilbert–Schmidt inner product condition

tr(Π jΠ k)=dδ jk+1d+1. \mathrm{tr} (\Pi_j \Pi_k) = \frac{d\delta_{jk} + 1}{d+1}.

These form a POVM if we rescale them by 1/d1/d. In every known case, we can compute all the projectors {Π j}\{\Pi_j\} by starting with one projector, call it Π 0\Pi_0, and then taking the orbit of Π 0\Pi_0 under the action of some group. The projector Π 0\Pi_0 is known as the fiducial state. (I don’t know who picked the word “fiducial”; I think it was something Carl Caves decided on, way back.)

In all known cases but one, the group is the Weyl–Heisenberg group in dimension dd. To define this group, fix an orthonormal basis {|n}\{|n\rangle\} and define the operators XX and ZZ such that

X|n=|n+1, X|n\rangle = |n+1\rangle,

interpreting addition modulo dd, and

Z|n=e 2πin/d|n. Z|n\rangle = e^{2\pi i n / d} |n\rangle.

The Weyl–Heisenberg displacement operators are

D lα=(e iπ/d) lαX lZ α. D_{l\alpha} = (-e^{i\pi / d})^{l\alpha} X^l Z^\alpha.

Because the product of two displacement operators is another displacement operator, up to a phase factor, we can make them into a group by inventing group elements that are displacement operators multiplied by phase factors. This group has Weyl’s name attached to it, because he invented XX and ZZ back in 1925, while trying to figure out what the analogue of the canonical commutation relation would be for quantum mechanics on finite-dimensional Hilbert spaces. It is also called the generalized Pauli group, because XX and ZZ generalize the Pauli matrices σ x\sigma_x and σ z\sigma_z to higher dimensions.

In many cases, we have both a displacement operator and its adjoint, e.g., when we conjugate Π 0\Pi_0:

Π l,α=D lαΠ 0D lα . \Pi_{l,\alpha} = D_{l\alpha} \Pi_0 D_{l\alpha}^\dagger.

So, we can often drop the phase factors and think of our SIC as being the orbit of a state under a group isomorphic to d× d\mathbb{Z}_d \times \mathbb{Z}_d.

The only known exception to Weyl–Heisenberg covariance is a SIC that was discovered early on, by Stuart Hoggar. It lives in dimension 8, it is covariant under the tensor product of three copies of the Pauli group, and of course, it is secretly related to the octonions.

Posted by: Blake Stacey on July 5, 2017 4:56 AM | Permalink | Reply to this

Symmetric Quantum Measurements and Exceptional Structures

I wrote the following notes a few months ago. Looking at them again, I’m not sure why I didn’t post them here; no doubt I felt they were unfinished in some important way, but now they just look incomplete in an inevitable way.

Part 1: E 6\mathrm{E}_6

In what follows, I will refer to H. S. M. Coxeter’s Regular Complex Polytopes, second edition (Cambridge University Press, 1991). Coxeter devotes a goodly portion of chapter 12 to the Hessian polyhedron, which lives in 3\mathbb{C}^3 and has 27 vertices. These 27 vertices lie on nine diameters in sets of three apiece. He calls the polyhedron “Hessian” because its nine diameters and twelve planes of symmetry interlock in a particular way. Their incidences reproduce the Hessian configuration, a set of nine points on twelve lines such that four lines pass through each point and three points lie on each line. (This configuration is also known as the discrete affine plane on nine points, and as the Steiner triple system of order 3. It is the solution to this interactive puzzle at Quanta Magazine.)

Coxeter writes the 27 vertices of the Hessian polyhedron explicitly, in the following way. First, let ω\omega be a cube root of unity, ω=e 2πi/3\omega = e^{2\pi i / 3}. Then, construct the complex vectors

(0,ω μ,ω ν),(ω ν,0,ω μ),(ω μ,ω ν,0), (0, \omega^\mu, -\omega^\nu),\ (-\omega^\nu, 0, \omega^\mu),\ (\omega^\mu, -\omega^\nu, 0),

where μ\mu and ν\nu range over the values 0, 1 and 2. As Coxeter notes, we could just as well let μ\mu and ν\nu range over 1, 2 and 3. He prefers this latter choice, because it invites a nice notation: We can write the vectors above as

0μν,ν0μ,μν0. 0\mu\nu,\ \nu0\mu,\ \mu\nu0.

For example,

230=(ω 2,1,0), 230 = (\omega^2, -1, 0),


103=(ω,0,1). 103 = (-\omega, 0, 1).

Coxeter then points out that this notation was first introduced by Beniamino Segre, “as a notation for the 27 lines on a general cubic surface in complex projective 3-space. In that notation, two of the lines intersect if their symbols agree in just one place, but two of the lines are skew if their symbols agree in two places or nowhere.” Consequently, the 27 vertices of the Hessian polyhedron correspond to the 27 lines on a cubic surface “in such a way that two of the lines are intersecting or skew according as the corresponding vertices are non-adjacent or adjacent.”

Casting the Hessian polyhedron into the real space 6\mathbb{R}^6, we obtain the polytope known as 2 212_{21}, which is related to E 6\mathrm{E}_6, since the Coxeter group of 2 212_{21} is the Weyl group of E 6\mathrm{E}_6.

We make the connection to symmetric quantum measurements by following the trick that Coxeter uses in his Eq. (12.39). We transition from the space 3\mathbb{C}^3 to the complex projective plane by collecting the 27 vertices into equivalence classes, which we can write in homogeneous coordinates as follows:

(0,1,1),(1,0,1),(1,1,0) (0, 1, -1),\ (-1, 0, 1),\ (1, -1, 0) (0,1,ω),(ω,0,1),(1,ω,0) (0, 1, -\omega),\ (-\omega, 0, 1),\ (1, -\omega, 0) (0,1,ω 2),(ω 2,0,1),(1,ω 2,0) (0, 1, -\omega^2),\ (-\omega^2, 0, 1),\ (1, -\omega^2, 0)

Let uu and vv be any two of these vectors. We find that

|u,u| 2=4|\langle u, u\rangle|^2 = 4

when the vectors coincide, and

|u,v| 2=1 |\langle u, v\rangle|^2 = 1

when uu and vv are distinct. We can normalize these vectors to be quantum states on a three-dimensional Hilbert space by dividing each vector by 2\sqrt{2}.

We have found a SIC for d=3d = 3. When properly normalized, Coxeter’s vectors furnish a set of d 2=9d^2 = 9 pure quantum states, such that the magnitude squared of the inner product between any two distinct states is 1/(d+1)=1/41/(d+1) = 1/4.

To relate this with the Weyl–Heisenberg business we discussed earlier, turn the first of Coxeter’s vectors into a column vector:

(0 1 1). \left(\begin{array}{c} 0 \\ 1 \\ -1\end{array}\right).

Apply the XX operator twice in succession to get the other two vectors in Coxeter’s table (converted to column-vector format). Then, apply ZZ twice in succession to recover the right-hand column of Coxeter’s table. Finally, apply XX to these vectors again to effect cyclic shifts and fill out the table. This set of nine states is known as the Hesse SIC.

Part 2: E 7\mathrm{E}_7

We saw how to generate the Hesse SIC by taking the orbit of a fiducial state under the action of the d=3d = 3 Weyl–Heisenberg group. Next, we will do something similar in d=8d = 8. We start by defining the two states |ψ 0 ±(1±2i,1,1,1,1,1,1,1) T. |\psi_0^\pm\rangle \propto (-1 \pm 2i, 1, 1, 1, 1, 1, 1, 1)^{\mathrm{T}}. Here, we are taking the transpose to make our states column vectors, and we are leaving out the dull part, in which we normalize the states to satisfy ψ 0 +|ψ 0 +=ψ 0 |ψ 0 =1. \langle \psi_0^+ | \psi_0^+ \rangle = \langle \psi_0^- | \psi_0^- \rangle = 1. First, we focus on |ψ 0 +|\psi_0^+\rangle. To create a SIC from the fiducial vector |ψ 0 +|\psi_0^+\rangle, we take the set of Pauli matrices, including the identity as an honorary member: {I,σ x,σ y,σ z}. \{ I, \sigma_x, \sigma_y, \sigma_z \}. We turn this set of four elements into a set of sixty-four elements by taking all tensor products of three elements. This creates the Pauli operators on three qubits. By computing the orbit of |ψ 0 +|\psi_0^+\rangle under multiplication (equivalently, the orbit of Π 0 +=|ψ 0 +ψ 0 +|\Pi_0^+ = |\psi_0^+\rangle\langle \psi_0^+| under conjugation), we find a set of 64 states that together form a SIC set.

The same construction works for the other choice of sign, |ψ 0 |\psi_0^-\rangle, creating another SIC with the same symmetry group. We can call both of them SICs of Hoggar type.

Recall that when we invented SICs for a single qubit, they were tetrahedra in the Bloch ball, and we could fit together two tetrahedral SICs such that each vector in one SIC was orthogonal (in the Bloch picture, antipodal) to exactly one vector in the other. The two Hoggar-type SICs made from the fiducial states Π 0 +\Pi_0^+ and Π 0 \Pi_0^- satisfy the grown-up version of this relation: Each state in one is orthogonal to exactly twenty-eight states of the other.

Moreover, 28 = 28.

By that, I mean that these orthogonalities correspond to the antisymmetric elements of the three-qubit Pauli group. It is simplest to see why when we look for those elements of the Π 0 \Pi_0^- SIC that are orthogonal to the projector Π 0 +\Pi_0^+. These satisfy

tr(Π 0 +DΠ 0 D )=0 \mathrm{tr}(\Pi_0^+ D \Pi_0^- D^\dagger) = 0

for some operator DD that is the tensor product of three Pauli matrices. Intuitively speaking, the product Π 0 +Π 0 \Pi_0^+ \Pi_0^- is a symmetric matrix, so if we want the trace to vanish, we ought to try introducing an asymmetry, but if we introduce too much, it will cancel out, on the “minus times a minus is a plus” principle. So, we want DD to comprise an odd number of factors of σ y\sigma_y.

As explained in more detail in this old comment, these 28 antisymmetric matrices correspond exactly to the 28 bitangents of a quartic curve, and to pairs of opposite vertices of the Gosset polytope 3 213_{21}.

Two fun things have happened here: First, we started with complex equiangular lines. By carefully considering the orthogonalities between two sets of complex equiangular lines, we arrived at the Gosset polytope 3 213_{21}, which yields us a maximal set of real equiangular lines in 7\mathbb{R}^7. And since one cannot actually fit more equiangular lines into 8\mathbb{R}^8 than into 7\mathbb{R}^7, we have a connection between a maximal set of equiangular lines in 8\mathbb{C}^8 and a maximal set of them in 8\mathbb{R}^8.

Second, because we have made our way to the polytope 3 213_{21}, we have arrived at E 7\mathrm{E}_7.

Part 3: E 8\mathrm{E}_8

I have already “spoiled” this part of the story, by hinting at a connection with the octonions. To spell it out more explicitly, we consider the stabilizer of the fiducial vector, i.e., the group of unitaries that map the SIC set to itself, leaving the fiducial where it is and permuting the other d 21d^2 - 1 vectors. Huangjun Zhu observed that the stabilizer of any fiducial for a Hoggar-type SIC is isomorphic to the group PSU(3,3)\mathrm{PSU}(3,3). In turn, this is up to a factor 2\mathbb{Z}_2 isomorphic to G 2(2)G_2(2), the automorphism group of the octavians. And the unit octavians furnish the E 8\mathrm{E}_8 lattice.

Posted by: Blake Stacey on November 25, 2017 7:42 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

No need to go to E 8E_8 immediately; let’s start with A n1A_{n-1}. This acts on 2\mathbb{C}^2 by (x,y)(ζx,ζ 1y)(x,y) \mapsto (\zeta x, \zeta^{-1} y) where ζ\zeta is a primitive nn-th root of unity. Here is the answer, proofs below:

The kk-th 1\mathbb{P}^1 (for 1kn11 \leq k \leq n-1) corresponds to ideals of the form px k+qy nk,xy,x k+1,y nk+1\langle p x^k + q y^{n-k}, x y, x^{k+1}, y^{n-k+1} \rangle, where (p:q)(p:q) are homogenous coordinates on the 1\mathbb{P}^1. A family of ideals which lands on this sphere, for pq0p q \neq 0, is xyt,px k+qy nk\langle x y-t, p x^k + q y^{n-k} \rangle, which corresponds to the nn points (x,y)=((q/p) 1/nt (nk)/n,(q/p) 1/nt k/n)(x,y) = ((-q/p)^{1/n} t^{(n-k)/n}, (-q/p)^{1/n} t^{k/n}).

Proof that all ideals lying over the singular point are of this form: Note that xyx y, x nx^n and y ny^n are all invariant polynomials, so they must take a constant value on any ideal where Γ\Gamma acts by the regular representation, and this value must be 00 to lie over the singular point. So we are looking at ideals which contain xy,x n,y n\langle x y, x^n, y^n \rangle. Since x kx^k and y nky^{n-k} transform by the same character, they must be proportional in any such ideal. If px k+qy nkp x^k + q y^{n-k} is in the ideal for pq0p q \neq 0, then x k+1=(1/p)(x(px k+qy nk)xy(qy nk1)x^{k+1} = (1/p) (x (p x^k + q y^{n-k}) - x y (q y^{n-k-1}) is and similarly so is y nk+1y^{n-k+1}; a similar analysis covers the cases where pq=0p q=0.

Posted by: David Speyer on July 7, 2017 8:42 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)


Now I’d like to transform this algebra into geometry. I’m less interested in proving things here than in ‘seeing’ them.

We’ve got the nn-element cyclic group Γ= n\Gamma = \mathbb{Z}_n acting on the plane. A ‘typical’ point in the minimal resolution of 2/Γ\mathbb{C}^2/\Gamma consists of nn distinct particles lying at the corners of a regular nn-gon. Somehow there is a way for such a typical point to approach any point on any of the 1\mathbb{P}^1’s you describe. To do so, all nn particles need to approach the origin… but they need to do so in a carefully specified way!

The simplest thing is for the nn-gon to shrink to a point without rotating at all. Which 1\mathbb{P}^1 does it approach then, and why?

Then: what’s the ‘second simplest thing’?

(Note: I’m using ‘particle’ as a name for a point in 2\mathbb{C}^2, and ‘point’ mainly as a name for a point in S= 2/ΓS = \mathbb{C}^2/\Gamma or its minimal resolution S˜\widetilde{S}. Also, where I’m saying ‘typical point’ above, the differential geometer in me wants to say ‘generic point’, but I’m afraid that could be confusing in algebraic geometry.)

Posted by: John Baez on July 7, 2017 9:15 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

Here are a few realizations. I’m going to continue talking about the nn points in 2\mathbb{C}^2 version for a while, before I try to work with the analogue of the dodecahedron picture.

Key realization — this is a complex geometry problem, not just a differential geometry problem. Let ζ\zeta be a primitive nn-th root of unity. There are two very different ways for the cyclic group of order nn to act on 2\mathbb{C}^2 – by (x,y)(ζx,ζ 1y)(x,y) \mapsto (\zeta x, \zeta^{-1} y) and by (x,y)(ζx,ζy)(x,y) \mapsto (\zeta x, \zeta y). I’ll call these the (ζ,ζ 1)(\zeta, \zeta^{-1}) action and the (ζ,ζ)(\zeta, \zeta) action. The former is the one whose minimal resolution corresponds to A n1A_{n-1}. The latter’s minimal resolution is just to blow up the origin, creating a single sphere. In terms of ideals, the sphere of ideals at the origin for the (ζ,ζ)(\zeta, \zeta)-action is those of the form pxqy,x n,y n\langle p x-q y, x^n, y^n \rangle, consisting of nn points on a line with some slope p/qp/q through the origin and all on top of each other.

Yet, as actions on 4\mathbb{R}^4, (ζ,ζ)(\zeta, \zeta) and (ζ,ζ 1)(\zeta, \zeta^{-1}) are isomorphic! They are conjugate by (x,y)(x,y¯)(x,y) \mapsto (x, \bar{y}). So we need to make use of the complex structure on 2\mathbb{C}^2 in order to understand things. (I suspect there might also be an alternate perspective using symplectic geometry; I’d love a symplectic geometer to comment.)

When you speak of a ring of nn points shrinking linearly toward the origin, you mean a family of the form (ruζ k,rvζ k)(r u \zeta^k, r v \zeta^{-k}) for uu and vv fixed constants and r0r \to 0. If n=2mn=2m is even, then this ideal is v mx mu my m,xyr 2\langle v^{m} x^{m} - u^{m} y^{m}, x y - r^2 \rangle and the limit is the ideal v mx mu my m,xy\langle v^{m} x^{m} - u^{m} y^{m}, x y \rangle, on the middle sphere of the 2m12m-1 spheres. If n=2m+1n=2m+1 is odd and uu, v0v \neq 0, the ideal is v mx m+1ru m+1y m,u my m+1rv m+1x m,xyr 2\langle v^m x^{m+1} - r u^{m+1} y^m, u^m y^{m+1} - r v^{m+1} x^m, x y-r^2 \rangle, with limit x m+1,y m+1,xy\langle x^{m+1}, y^{m+1}, x y \rangle, at the node at the center of the line of 2m2m spheres. The less simple ways to shrink to the origin are (r auζ k,r navζ k)(r^a u \zeta^k, r^{n-a} v \zeta^{-k}) for 1an11 \leq a \leq n-1. In general, if we look at (r cuζ k,r dvζ k)(r^c u \zeta^k, r^d v \zeta^{-k}), its limit is determined by which of the intervals (0,1/(n1))(0, 1/(n-1)), (1/(n1),2/(n2))(1/(n-1), 2/(n-2)), …, (n1,)(n-1, \infty) the ratio c/dc/d lies in — the resonances at a/(na)a/(n-a) land on the n1n-1 spheres.

Posted by: David Speyer on July 9, 2017 3:03 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

This is very enlightening! It’ll take me a while to muster a good response, but it’s very nice to see the various ways a /n\mathbb{Z}/n-orbit in 2\mathbb{C}^2 can collapse to the origin.

What, in words, are the special geometrical features of those collapse processes that converge to nodes—points where two Riemann spheres in the exceptional fiber of π:S˜S\pi : \widetilde{S} \to S intersect? This is one thing I’ve gotta know.

Posted by: John Baez on July 9, 2017 7:47 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

Now, let me try to get to the quaternions, with the goal of eventually getting to the analogue of the dodecahedron picture.

Let HH be the ring of quaternions and Γ\Gamma the cyclic group of order nn. Let Γ\Gamma act on HH by right multiplication by cos(2π/n)+isin(2π/n)\cos (2 \pi/n) + i \sin(2 \pi/n). We need to make HH into a \mathbb{C}-vector space by left multiplication, because that way we will get the (ζ,ζ 1)(\zeta, \zeta^{-1}) action. But there are many copies of \mathbb{C} inside HH, and I am going to show you that the choice of which embedding of \mathbb{C} into HH we use to make HH a \mathbb{C} vector space matters.

First, embed \mathbb{C} by 1\sqrt{-1} goes to ii, so (x 1+x 21,y 1+y 21) 2(x_1+x_2 \sqrt{-1}, y_1+y_2\sqrt{-1}) \in \mathbb{C}^2 corresponds to (x 1+x 2i)+(y 1+y 2i)j=x 1+x 2i+y 1j+y 2k(x_1+x_2 i) + (y_1+y_2 i) j = x_1 + x_2 i + y_1 j + y_2 k. Then the ζ\zeta and ζ 1\zeta^{-1} eigenspaces for the action of Γ\Gamma are (1,0)(1,0) and (0,1)(0,1). In real terms, these eigenspaces are Span (1,i)\mathrm{Span}_{\mathbb{R}}(1,i) and Span (j,k)\mathrm{Span}_{\mathbb{R}}(j,k).

Suppose, on the other hand, we send 1\sqrt{-1} to jj. We’ll take (1,i)(1,i) as our \mathbb{C}-basis for HH, so (x 1+x 21,y 1+y 21)(x_1+x_2 \sqrt{-1}, y_1+y_2 \sqrt{-1}) goes to (x 1+x 2j)+(y 1+y 2j)i=x 1+y 1i+x 2jy 2k(x_1+x_2 j) + (y_1+y_2 j) i = x_1 + y_1 i + x_2 j - y_2 k. The right action of ii sends this to y 1+x 1iy 2jx 2k=(y 1y 2j)+(x 1+x 2j)i-y_1 + x_1 i - y_2 j - x_2 k = (-y_1 - y_2 j) + (x_1 + x_2 j) i which is to say it acts on (x 1+x 21,y 1+y 21(x_1+x_2 \sqrt{-1}, y_1+y_2 \sqrt{-1} by (0 1 1 0)\left( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right). So right mulitplication by cos(2π/n)+isin(2π/n)\cos (2 \pi/n) + i \sin (2 \pi/n) acts by (cos(2π/n) sin(2π/n) sin(2π/n) cos(2π/n))\left( \begin{smallmatrix} \cos(2 \pi/n) & \sin(2 \pi/n) \\ -\sin(2 \pi/n) & \cos(2 \pi/n) \end{smallmatrix} \right). The eigenvectors of this matrix are (1 ±1)\left( \begin{smallmatrix} 1\\ \pm \sqrt{-1} \end{smallmatrix} \right), which correspond to 1±k1 \pm k in HH. So, with this \mathbb{C}-vector space structure, the eigenspaces of the Γ\Gamma action are Span (1±k,j(1±k))\mathrm{Span}_{\mathbb{R}}(1 \pm k, j(1 \pm k)).

This matters! A ring of points lying in one of the eigenspaces and shrinking to 00 lands on the end of the chain of spheres. So different complex structures on HH give different resolutions of singularities.

Posted by: David Speyer on July 9, 2017 3:06 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

It occurred to me last night while I was trying to fall asleep that there is another way to bring E 8\mathrm{E}_8 into the story of a qubit, which might be illuminating by contrast. Use the Pauli matrices as an orthonormal basis for the space of Hermitian operators on the qubit state space. Then we can expand any Hermitian operator AA as

A= i=0 3a iσ i, A = \sum_{i=0}^3 a_i \sigma_i,

where σ 0=I\sigma_0 = I and a ia_i \in \mathbb{R}. Use the correspondence between the Pauli matrices and the quaternion units {1,i,j,k}\{1,i,j,k\} to define a map from Hermitian operators to quaternions:

Aa=(a 0+ia 1+ja 2+ka 3). A \mapsto a = (a_0 + i a_1 + j a_2 + k a_3).

The quaternion norm of aa is

||a||= ia i 2=trA 2. ||a|| = \sqrt{\sum_i a_i^2} = \sqrt{\mathrm{tr} A^2}.

Now, as in week20, let aa range over the icosians, which we generate from the double cover of A 5\mathrm{A}_5. When aa is an icosian, each of the a ia_i belongs to the “golden field” (5)\mathbb{Q}(\sqrt{5}), and so we can write

a i=r i+s i5, a_i = r_i + s_i\sqrt{5},

with r i,s ir_i, s_i \in \mathbb{Q}, making the quaternion aa into an 8-tuple of rationals. The ordinary quaternion norm of aa will also be of the form r+s5r + s\sqrt{5}. We can define a new norm, rather unnatural in the context of quantum information but simple enough as a matter of quaternion arithmetic, as the sum of rr and ss. With respect to this norm, the lattice of icosians is isomorphic to E 8\mathrm{E}_8.

What does this mean in terms of quantum observables and measurements on qubits? I don’t know—perhaps nothing! As I said, it seems rather contrived in that context to add the “real” part rr and the “surd” part ss of a quaternion norm, even though the quaternion norm itself has a plain enough meaning. The imagery of contracting the vertices of a POVM closer and closer to the origin is more vivid and more natural to me.

But since we can construct the Leech lattice from three copies of E 8\mathrm{E}_8, this hints that the “icosian Leech lattice” might in some way pertain to the quantum state space of three qubits. And poking into the literature, I noticed that people have drawn a connection between this construction of the Leech lattice and the sporadic Hall–Janko group:

Which is intriguing to me (and perhaps nobody else) because I had found a sketchy sort of path from Hoggar’s SIC on three qubits to the Hall–Janko group. So, I have a new thing I need to learn about and get sorted out.

Posted by: Blake Stacey on July 17, 2017 8:49 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

Edit: better to write the third equation as

|a| 2= ia i 2=trA 2. |a|^2 = \sum_i a_i^2 = \mathrm{tr} A^2.

Posted by: Blake Stacey on July 17, 2017 8:57 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

While eagerly awaiting the next installment of this series, I’ve been working my way through this article: The Birth of E8 out of the Spinors on the Icosahedron by Pierre-Philippe Dechant. I don’t believe this approach to E8 has been discussed previously at the Cafe, and it seems that it may be relevant to the topic at hand.

Posted by: Charles G Waldman on August 16, 2017 10:41 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

For some reason I’m not getting emails about comments on this post—sorry, Blake. It’s just sheer luck that I tuned back in shortly after Charles’ comment.

I think of the ‘birth of E8 out of spinors on the icosahedron’ as a slight variant of the work of Fring and Korff discussed in “week270” and “week271” of This Week’s Finds, and also here:

Icosidodecahedron from D6, Visual Insight, January 1, 2015.

Posted by: John Baez on August 18, 2017 4:14 AM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

No problem—software will be software, it seems. I’ve been doing calculations relating to this off and on since I made my previous post, and I need to gather the results up in an intelligible form.

Posted by: Blake Stacey on August 18, 2017 9:40 PM | Permalink | Reply to this

Re: The Geometric McKay Correspondence (Part 2)

I finally got organized enough to gather these notes together, incorporating edits for clarity and recording one construction I haven’t found written in the literature anywhere.

Posted by: Blake Stacey on December 30, 2018 9:45 PM | Permalink | Reply to this

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