## March 14, 2019

### Sporadic SICs and Exceptional Lie Algebras III

#### Posted by John Baez

guest post by Blake C. Stacey

On the fourteenth of March, 2016 — so, exactly three years ago — Maryna Viazovska published a proof that the $\mathrm{E}_8$ lattice is the best way to pack hyperspheres in eight dimensions. Today, we’ll see how this relates to another packing problem that seems quite different: how to fit as many equiangular lines as possible into the complex space $\mathbb{C}^8$. The answer to this puzzle is an example of a SIC: a Symmetric Informationally Complete quantum measurement.

Last time, we saw how to build SICs by starting with a fiducial vector and taking the orbit of that vector under the action of a group, turning one line into $d^2$. We said earlier that the Weyl–Heisenberg group was the group we use in call cases but one. Now, we take on that exception. It will lead us to the exceptional root systems $\mathrm{E}_7$ and $\mathrm{E}_8$. Actually, it will be a bit easier to tackle the latter first. Whence $\mathrm{E}_8$ in the world of SICs?

We saw how to generate the Hesse SIC by taking the orbit of a fiducial state under the action of the $d = 3$ Weyl–Heisenberg group. Today, we will do something similar in $d = 8$. We start by defining the two states

$|\psi_0^\pm \rangle \propto (-1 \pm 2i, 1, 1, 1, 1, 1, 1, 1)^{\mathrm{T}}.$

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 $\langle \psi_0^+ | \psi_0^+ \rangle = \langle \psi_0^- | \psi_0^- \rangle = 1$.

First, we focus on $|\psi_0^+\rangle$. To create a SIC from the fiducial vector $|\psi_0^+\rangle$, we take the set of Pauli matrices, including the identity as an honorary member: $\{ I, \sigma_x, \sigma_y, \sigma_z \}$. 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 $|\psi_0^+\rangle$ under multiplication (equivalently, the orbit of $\Pi_0^+ = |\psi_0^+\rangle \langle \psi_0^+|$ under conjugation), we find a set of 64 states that together form a SIC set.

The same construction we used to make the $|\psi_0^+\rangle$ SIC works for the other choice of sign, $|\psi_0^- \rangle$, creating another SIC with the same symmetry group. We can call both of them SICs of Hoggar type, in honor of Stuart Hoggar.

To make the link with $\mathrm{E}_8$, 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 $d^2 - 1$ vectors. Huangjun Zhu observed that the stabilizer of any fiducial for a Hoggar-type SIC is isomorphic to the group of $3\times 3$ unitary matrices over the finite field of order 9. This group is sometimes written $U_3(3)$ or $PSU(3,3)$. In turn, this group is up to a factor $\mathbb{Z}_2$ isomorphic to $G_2(2)$, the automorphism group of the Cayley integers, a subset of the octonions also known as the octavians. Up to an overall scaling, the lattice of octavians is also the lattice known as $\mathrm{E}_8$.

The octavian lattice contains a great deal of arithmetic structure. Of particular note is that it contains 240 elements of norm 1. In addition to the familiar $+1$ and $-1$, which have order 1 and 2 respectively, there are 56 units of order 3, 56 units of order 6 and 126 units of order 4. The odd-order units generate subrings of the octavians that are isomorphic to the Eisenstein integers and the Hurwitz integers, lattices in the complex numbers and the quaternions (as Conway and Smith describe in their book). From the symmetries of these lattices, we can in fact read off the stabilizer groups for fiducials of the qubit and Hesse SICs. It is as if the sporadic SICs are drawing their strength from the octonions.

Before moving on, we pause to note how peculiar it is that by trying to find a nice packing of complex unit vectors, we ended up talking about an optimal packing of Euclidean hyperspheres!

Now that we’ve met $\mathrm{E}_8$, it’s time to visit the root system we skipped: Where does $\mathrm{E}_7$ fit in?

With respect to the probabilistic representation furnished by the $\Pi_0^+$ SIC, the states of the $\Pi_0^-$ SIC minimize the Shannon entropy, and vice versa. Here are four of the states from the $\{\Pi_i^-\}$ Hoggar-type SIC, written in the probabilistic representation of three-qubit state space provided by the $\{\Pi_i^+\}$ SIC. Up to an overall normalization by $1/36$, these states are all binary sequences, i.e., they are uniform over their supports.

 1110111011100001111011101110000111101110111000010001000100011110 1101110111010010110111011101001011011101110100100010001000101101 1011101110110100101110111011010010111011101101000100010001001011 0111011101111000011101110111100001110111011110001000100010000111 

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 $\Pi_0^+$ and $\Pi_0^-$ satisfy the grown-up version of this relation: Each state in one is orthogonal to exactly twenty-eight states of the other.

We can understand these orthogonalities as corresponding to the antisymmetric elements of the three-qubit Pauli group. It is simplest to see why when we look for those elements of the $\Pi_0^-$ SIC that are orthogonal to the projector $\Pi_0^+$. These satisfy

$tr (\Pi_0^+ D \Pi_0^- D^\dagger) = 0$

for some operator $D$ that is the tensor product of three Pauli matrices. For which such tensor-product operators will this expression vanish? Intuitively speaking, the product $\Pi_0^+ \Pi_0^-$ 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. Recall the Pauli matrices:

$\sigma_x = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right),\qquad \sigma_y = \left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right),\qquad \sigma_z = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right).$

Note that of these three matrices, only $\sigma_y$ is antisymmetric, and also note that we have

$\sigma_z \sigma_x = -\sigma_x\sigma_z = i\sigma_y.$

This much is familiar, though that minus sign gets around. For example, it is the fuel that makes the GHZ thought-experiment go, because it means that

$\sigma_x \otimes \sigma_x \otimes \sigma_x = -(\sigma_x \otimes \sigma_z \otimes \sigma_z)(\sigma_z \otimes \sigma_x \otimes \sigma_z)(\sigma_z \otimes \sigma_z \otimes \sigma_x).$

Let’s consider the finite-dimensional Hilbert space made by composing three qubits. This state space is eight-dimensional, and we build the three-qubit Pauli group by taking tensor products of the Pauli matrices, considering the $2 \times 2$ identity matrix to be the zeroth Pauli operator. There are 64 matrices in the three-qubit Pauligroup, and we can label them by six bits. The notation

$\left(\begin{array}{ccc} m_1 & m_3 & m_5 \\ m_2 & m_4 & m_6\end{array}\right)$

means to take the tensor product

$\sigma_x^{m_1} \sigma_z^{m_2} \otimes (-i)^{m_3m_4} \sigma_x^{m_3} \sigma_z^{m_4} \otimes (-i)^{m_5m_6} \sigma_x^{m_5} \sigma_z^{m_6}.$

Now, we ask: Of these 64 matrices, how many are symmetric and how many are antisymmetric? We can only get antisymmetry from $\sigma_y$, and (speaking heuristically) if we include too much antisymmetry, it will cancel out. More carefully put: We need an odd number of factors of $\sigma_y$ in the tensor product to have the result be an antisymmetric matrix. Otherwise, it will come out symmetric. Consider the case where the first factor in the triple tensor product is $\sigma_y$. Then we have $(4-1)^2 = 9$ possibilities for the other two slots. The same holds true if we put the $\sigma_y$ in the second or the third position. Finally, $\sigma_y \otimes \sigma_y \otimes \sigma_y$ is antisymmetric, meaning that we have $9 \cdot 3 + 1 = 28$ antisymmetric matrices in the three-qubit Pauli group. In the notation established above, they are the elements for which

$m_1m_2 +m_3m_4 + m_5m_6 = 1 \mod 2.$

Moreover, these 28 antisymmetric matrices correspond exactly to the 28 bitangents of a quartic curve, and to pairs of opposite vertices of the Gosset polytope $3_{21}$. In order to make this connection, we need to dig into the octonions.

To recap: Each of the 64 vectors (or, equivalently, projectors) in the Hoggar SIC is naturally labeled by a displacement operator, which up to an overall phase is the tensor product of three Pauli operators. Recall that we can write the Pauli operator $\sigma_y$ as the product of $\sigma_x$ and $\sigma_z$, up to a phase. Therefore, we can label each Hoggar-SIC vector by a pair of binary strings, each three bits in length. The bits indicate the power to which we raise the $\sigma_x$ and $\sigma_z$ generators on the respective qubits. The pair $(010,101)$, for example, means that on the three qubits, we act with $\sigma_x$ on the second, and we act with $\sigma_z$ on the first and third. Likewise, $(000, 111)$ stands for the displacement operator which has a factor of $\sigma_z$ on each qubit and no factors of $\sigma_x$ at all.

There is a natural mapping from pairs of this form to pairs of unit octonions. Simply turn each triplet of bits into an integer and pick the corresponding unit from the set $\{1, e_1, e_2, e_3, e_4, e_5, e_6, e_7\}$, where each of the $e_j$ square to $-1$.

We can choose the labeling of the unit imaginary octonions so that the following nice property holds. Up to a sign, the product of two imaginary unit octonions is a third, whose index is the XOR of the indices of the units being multiplied. For example, in binary, $1 = 001$ and $4 = 100$; the XOR of these is $101 = 5$, and $e_1$ times $e_4$ is $e_5$.

Translate the Cayley–Graves multiplication table here into binary if enlightenment has not yet struck:

$\begin{array}{c|cccccccc} e_i e_j & 1 & e_1 & e_2 & e_3 & e_4 & e_5 & e_6 & e_7 \\ 1 & 1 & e_1 & e_2 & e_3 & e_4 & e_5 & e_6 & e_7 \\ e_1 & e_1 & -1 & e_3 & -e_2 & e_5 & -e_4 & -e_7 & e_6 \\ e_2 & e_2 & -e_3 & -1 & e_1 & e_6 & e_7 & -e_4 & -e_5 \\ e_3 & e_3 & e_2 & -e_1 & -1 & e_7 & -e_6 & e_5 & -e_4 \\ e_4 & e_4 & -e_5 & -e_6 & -e_7 & -1 & e_1 & e_2 & e_3 \\ e_5 & e_5 & e_4 & -e_7 & e_6 & -e_1 & -1 & -e_3 & e_2 \\ e_6 & e_6 & e_7 & e_4 & -e_5 & -e_2 & e_3 & -1 & -e_1 \\ e_7 & e_7 & -e_6 & e_5 & e_4 & -e_3 & -e_2 & e_1 & -1 \end{array}$

So, each projector in the Hoggar SIC is labeled by a pair of octonions, and the group structure of the displacement operators is, almost, octonion multiplication. I say “almost” because we are neglecting the sign factors and only caring about which imaginary octonion a multiplication yields: All we need is the label. Rest assured, the sign factors will return soon.

Another way to express the Cayley–Graves multiplication table is with the Fano plane, a set of seven points grouped into seven lines that has been called “the combinatorialist’s coat of arms”. We can label the seven points with the imaginary octonions $e_1$ through $e_7$. When drawn on the page, a useful presentation of the Fano plane has the point $e_4$ in the middle and, reading clockwise, the points $e_1$, $e_7$, $e_2$, $e_5$, $e_3$ and $e_6$ around it in a regular triangle.

The three sides and three altitudes of this triangle, along with the inscribed circle, provide the seven lines: $(e_1,e_2,e_3)$, $(e_1,e_4,e_5)$, $(e_1,e_7,e_6)$, $(e_2,e_4,e_6)$, $(e_2,e_5,e_7)$, $(e_3,e_4,e_7)$, $(e_3,e_6,e_5)$. It is apparent that each line contains three points, and it is easy to check that each point lies within three distinct lines, and that each pair of lines intersect at a single point. One consequence of this is that if we take the incidence matrix of the Fano plane, writing a 1 in the $i j$-th entry if line $i$ contains point $j$, then every two rows of the matrix have exactly the same overlap:

$M = \begin{pmatrix} 1 &1 &1 &0 &0 &0 &0 \\ 1 &0 &0 &1 &1 &0 &0 \\ 1 &0 &0 &0 &0 &1 &1 \\ 0 &1 &0 &1 &0 &1 &0 \\ 0 &1 &0 &0 &1 &0 &1 \\ 0 &0 &1 &1 &0 &0 &1 \\ 0 &0 &1 &0 &1 &1 &0 \end{pmatrix}.$

The rows of the incidence matrix furnish us with seven equiangular lines in $\mathbb{R}^7$. We can build upon this by considering the signs in the Cayley–Graves multiplication table, which we can represent by adding orientations to the lines of the Fano plane. Start by taking the first row of the incidence matrix $M$, which corresponds to the line $(e_1, e_2, e_3)$, and give it all possible choices of sign by multiplying by the elements not on that line. Multiplying by $e_4$, $e_5$, $e_6$ and $e_7$ respectively, we get

$\begin{pmatrix} + &+ &+ &0 &0 &0 &0 \\ - &+ &- &0 &0 &0 &0 \\ - &- &+ &0 &0 &0 &0 \\ + &- &- &0 &0 &0 &0 \end{pmatrix}.$

The sign we record here is simply the sign we find in the corresponding entry of the Cayley–Graves table. Doing this with all seven lines of the Fano plane, we obtain a set of 28 vectors, each one given by a choice of a line and a point not on that line. Moreover, all of these vectors are equiangular. This is easily checked: For any two vectors derived from the same Fano line, two of the terms in the inner product will cancel, leaving an overlap of magnitude 1. And for any two vectors derived from different Fano lines, the overlap always has magnitude 1 because each pair of lines always meets at exactly one point. Van Lint and Seidel noted that the incidence matrix of the Fano plane could be augmented into a full set of 28 equiangular lines, but to my knowledge, extracting the necessary choices of sign from octonion multiplication is not reported in the literature.

So, the 28 lines in a maximal equiangular set in $\mathbb{R}^7$ correspond to point-line pairs in the Fano plane, where the point and the line are not coincident. In discrete geometry, the combination of a line and a point not on that line is known as an anti-flag. It is straightforward to show from the above that the antisymmetric matrices in the three-qubit Pauli group also correspond to the anti-flags of the Fano plane: Simply take the powers of the $\sigma_x$ operators to specify a point and the powers of the $\sigma_z$ operators to specify a line.

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 a maximal set of real equiangular lines in $\mathbb{R}^7$. And it turns out that one cannot actually fit more equiangular lines into $\mathbb{R}^8$ than into $\mathbb{R}^7$, meaning that we have a connection between a maximal set of equiangular lines in $\mathbb{C}^8$ and a maximal set of them in $\mathbb{R}^8$.

Second, our equiangular lines in $\mathbb{R}^7$ are the diameters of the Gosset polytope $3_{21}$. And because we have made our way to the polytope $3_{21}$, we have arrived at $\mathrm{E}_7$. To quote a fascinating paper by Manivel,

Gosset seems to have been the first, at the very beginning of the 20th century, to understand that the lines on the cubic surface can be interpreted as the vertices of a polytope, whose symmetry group is precisely the automorphism group of the configuration. Coxeter extended this observation to the 28 bitangents, and Todd to the 120 tritangent planes. Du Val and Coxeter provided systematic ways to construct the polytopes, which are denoted $n_{21}$ for $n = 2, 3, 4$ and live in $n + 4$ dimensions. They have the characteristic property of being semiregular, which means that the automorphism group acts transitively on the vertices, and the faces are regular polytopes. In terms of Lie theory they are best understood as the polytopes in the weight lattices of the exceptional simple Lie algebras $\mathfrak{e}_{n+4}$, whose vertices are the weights of the minimal representations.

When we studied the Hesse SIC, we met the case $n = 2$ and $\mathfrak{e}_6$. The intricate orthogonalities between two conjugate SICs of Hoggar type have led us to the case $n = 3$ and $\mathfrak{e}_7$.

Puzzle: Another place that the number 28 arises when considering an 8-dimensional space is the free modular lattice on 3 generators. Is that 28 the same as this 28?

Posted at March 14, 2019 1:33 AM UTC

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