The n-Category Café

Here’s how the Standard Model gauge group, together with its representation on one generation of fermions, drops out of a Jordan triple.

The bi-Cayley triple

Let

$$\mathbb{O}_\mathbb{C} = \mathbb{C} \textstyle{\otimes}_\mathbb{R} \mathbb{O}$$

be the bioctonions: octonions with complex coefficients. Write $\mathbb{O}_\mathbb{C}^2$ for the space of column vectors with two bioctonion entries.

$\mathbb{O}_\mathbb{C}^2$ has a certain triple product

$$[x,y,z]=\frac{1}{2}(x(y^{\dagger}z)+z(y^{\dagger}x))$$

which obey the axioms of a gadget called a ‘positive hermitian Jordan triple’. It’s called the bi-Cayley triple.

Now, every positive hermitian Jordan triple gives rise to a $\mathbb{Z}_2$-graded real Lie algebra

$$\mathbf{k} = \mathbf{k}_0 \textstyle{\oplus} \mathbf{k}_1$$

Not a Lie superalgebra: a plain old-fashioned Lie algebra with a $\mathbb{Z}_2$-grading! The hermitian Jordan triple itself is $\mathbf{k}_1$. The Lie algebra $\mathbf{k}_0$, cooked up from the Jordan triple in a certain way, acts on it.

From this situation we get two Lie groups: a big one $K$ whose Lie algebra is $\mathbf{k}$, and a little one $K_0$, called the whose Lie algebra is $\mathbf{k}_0$. The quotient is $K/K_0$ is a nice kind of manifold called a hermitian symmetric space. Conversely, any compact hermitian symmetric space give rise to a positive hermitian Jordan triple!

This geometric picture is revealing. The group $K$ acts transitively as symmetries of our hermitian symmetric space, while the stabilizer of any point is isomorphic to $K_0$. Our original Jordan triple, $\mathbf{k}_1$, is then the tangent space of that point. So, $K_0$ acts on the Jordan triple. This action preserves the triple product, and we call $K_0$ the real inner automorphism group of our Jordan triple.

Here’s another great thing about the geometric picture: hermitian symmetric spaces were classified by Eli Cartan (who seems to have spent his life classifying things). As a result we also know the classification of positive hermitian Jordan triples. They come in four infinite series together with two exceptions. One is the bi-Cayley triple, and other is the Albert triple, which is the complexification of the exceptional Jordan algebra. The bi-Cayley triple is a subtriple of the Albert triple. It’s these two exceptions that are connected to the Standard Model.

The 3-graded Lie algebra coming from the bi-Cayley triple is the compact real form of $\mathfrak{e}_6$:

$$\mathfrak{e}_6 = \big[\mathfrak{so}(10) \textstyle{\oplus} \mathfrak{u}(1)\big] \textstyle{\oplus} \mathbb{O}_\mathbb{C}^2.$$

The even part of this Lie algebra is in brackets. The corresponding hermitian symmetric space is called the bioctonionic plane $(\mathbb{C}\otimes\mathbb{O})P^2$. I explained it in Part 12. The even part of our 3-graded Lie algebra, $\mathfrak{so}(10)\oplus \mathfrak{u}(1)$, generates the stabilizer of a point in the bioctonionic plane. The odd part, $\mathbb{O}_\mathbb{C}^2$, is the tangent space of that point.

Here’s the first big surprise. The even part transforms as the adjoint representation of $\mathrm{Spin}(10)$, while the odd part itself transforms as the 16-dimensional complex spinor representation of $\mathrm{Spin}(10)$. Ignoring the extra $\mathrm{U}(1)$ for a moment, this is exactly what we see in a $\mathrm{SO}(10)$ grand unified theory: gauge bosons in the adjoint representation, and one generation of fermions in the 16-dimensional spinor representation.

So before we do anything, the bi-Cayley triple already smells like it contains the ingredients of an $\mathrm{SO}(10)$ grand unified theory.

Tripotents

In a Jordan algebra the important elements are the idempotents, $e^2 = e$. In a Jordan triple $W$ their role is played by tripotents: elements $e$ with

$$[e,e,e] = e.$$

A tripotent always lets us split $W$ into three parts via something called its Peirce decomposition. The operator $w \mapsto [e,e,w]$ has eigenvalues $0, \tfrac{1}{2}, 1$, and $W$ splits into the corresponding eigenspaces

$$W = W_0(e) \textstyle{\oplus} W_{1/2}(e) \textstyle{\oplus} W_1(e),$$

which are called the Peirce 0-space, Peirce $\tfrac{1}{2}$-space and Peirce 1-space of $e$. A tripotent is called minimal when its Peirce $1$-space is one-dimensional: minimal tripotents are the analogues of unit vectors in ordinary quantum theory. Two tripotents $e_1, e_2$ are called colinear when each lies in the other’s Peirce $\tfrac{1}{2}$-space.

The single fact driving everything below is this: a minimal tripotent’s Peirce $\tfrac{1}{2}$-space is itself a hermitian Jordan triple. If we run this starting from the bi-Cayley triple, we get a chain of hermitian Jordan triples, where each row’s $\tfrac{1}{2}$-space is the next row’s triple:

Jordan triple	Lie algebra $\mathbf{k}_0 \oplus \mathbf{k}_1$ (even part in brackets)	real inner automorphism group
$W = \mathbb{O}_\mathbb{C}^2$	$\mathfrak{e}_6 = [\mathfrak{so}(10) \oplus \mathfrak{u}(1)] \oplus \mathbb{O}_\mathbb{C}^2$	$(\mathrm{Spin}(10) \times \mathrm{U}(1)) / \mathbb{Z}_4$
$W' = \mathfrak{a}_5(\mathbb{C})$	$\mathfrak{so}(10) = [\mathfrak{su}(5) \oplus \mathfrak{u}(1)] \oplus \mathfrak{a}_5(\mathbb{C})$	$\mathrm{SU}(5) \times \mathrm{U}(1)$
$W'' = \mathrm{M}_{3,2}(\mathbb{C})$	$\mathfrak{su}(5) = [\mathfrak{g}_{\mathrm{SM}}] \oplus \mathrm{M}_{3,2}(\mathbb{C})$	$G_{\mathrm{SM}}$

Here $\mathfrak{a}_5(\mathbb{C})$ is the Jordan triple of antisymmetric $5\times 5$ complex matrices, $\mathrm{M}_{3,2}(\mathbb{C})$ is the Jordan triple of $3\times 2$ complex matrices, $\mathfrak{g}_{\mathrm{SM}} = \mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus \mathfrak{u}(1)$, and

$$G_{\mathrm{SM}} = \mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3)) \cong (\mathrm{SU}(3)\times\mathrm{SU}(2)\times\mathrm{U}(1))/\mathbb{Z}_6$$

is the true Standard Model gauge group.

The gauge group from two tripotents

Now pick two colinear minimal tripotents $e_1, e_2 \in W$. Descend the table twice:

• Start with $W = \mathbb{O}_\mathbb{C}^2$, which has real inner automorphism group $(\mathrm{Spin}(10)\times\mathrm{U}(1))/\mathbb{Z}_4$.
• Fix $e_1$. Its Peirce $\tfrac{1}{2}$-space is $W' = \mathfrak{a}_5(\mathbb{C})$, with real inner automorphism group $\mathrm{SU}(5)\times\mathrm{U}(1)$.
• Fix $e_2$ (colinear with $e_1$, so living in $W'$). Its Peirce $\tfrac{1}{2}$-space in $W'$ is $W'' = \mathrm{M}_{3,2}(\mathbb{C})$, with real inner automorphism group exactly $G_{\mathrm{SM}}$.

In other words, the subspace of $W$ colinear with both $e_1$ and $e_2$ is $\mathrm{M}_{3,2}(\mathbb{C})$, and its real inner automorphism group is the Standard Model gauge group.

The choice of $e_1$ and $e_2$ also pins down how $G_{\mathrm{SM}}$ sits inside the original group $\mathrm{E}_6$. At each we step take the subgroup that acts with determinant $1$ and preserves the chosen tripotent up to a phase; this gives a chain of subgroups whose members are $\mathrm{Spin}(10)$, $\mathrm{U}(5)$, and $G_{\mathrm{SM}}$, so we get the embedding

$$G_{\mathrm{SM}} \subset \mathrm{SU}(5) \subset \mathrm{Spin}(10).$$

In particle physics, this is the classic Georgi–Glashow chain. And it’s well known that restricting the 16-dimensional complex spinor representation of $\mathrm{Spin}(10)$ along this chain gives precisely the Standard Model representation $\rho_{\mathrm{SM}}$ on one generation of fermions! So we get this representation too.

The six particles types as Peirce spaces

Here’s the part I find most charming. As a representation of the Standard Model Lie algebra

$$\mathfrak{g}_{\mathrm{SM}} = \mathfrak{su}(3) \textstyle{\oplus} \mathfrak{su}(2) \textstyle{\oplus} \mathfrak{u}(1) ,$$

any generation of Standard Model fermions transforms as the direct sum of six irreducible representations:

$$\rho_{\mathrm{SM}} = (3,2,\tfrac{1}{6}) \textstyle{\oplus} (\bar 3,1,\tfrac{1}{3}) \textstyle{\oplus} (\bar 3,1,-\tfrac{2}{3}) \textstyle{\oplus} (1,2,-\tfrac{1}{2}) \textstyle{\oplus} (1,1,1) \textstyle{\oplus} (1,1,0),$$

These correspond to the six types of left-handed fermion: $q_L, \overline{d_R}, \overline{u_R}, \ell_L, \overline{e_R}, \overline{\nu_R}$. Six irreducible pieces, six particle types.

It turns out these are exactly the six nonzero components of the Peirce decomposition of $\mathbb{O}_\mathbb{C}^2$ with respect to both $e_1$ and $e_2$. Those six match up one-to-one with the particle types:

Peirce projector	representation of $G_{\text{SM}}$	particle type
P1/2(e2) P1/2(e1)	(3, 2, +1/6)	$q_L$
P1/2(e2) P0(e1)	($\overline{3}$, 1, +1/3)	$\overline{d_R}$
P0(e2) P1/2(e1)	($\overline{3}$, 1, −2/3)	$\overline{u_R}$
P0(e2) P0(e1)	(1, 2, −1/2)	$\ell_L$
P1(e2) P1/2(e1)	(1, 1, +1)	$\overline{e_R}$
P1/2(e2) P1(e1)	(1, 1, 0)	$\overline{\nu_R}$

The remaining three combinations — $P_1(e_2)P_1(e_1)$, $P_1(e_2)P_0(e_1)$, and $P_0(e_2)P_1(e_1)$ — all vanish, which is why we land on six pieces and not nine.

So the whole package — the gauge group $G_{\mathrm{SM}}$, the embedding $G_{\mathrm{SM}} \subset \mathrm{Spin}(10)$, the representation $\rho_{\mathrm{SM}}$, and even the split of one generation into its six particle multiplets as distinct Peirce components — all comes out of the single object $\mathbb{O}_\mathbb{C}^2$ once you choose two colinear minimal tripotents.

And if you prefer to start one level up, with the Albert triple $\mathfrak{h}_3(\mathbb{O}) \otimes \mathbb{C}$, you get the same result by choosing three mutually colinear tripotents instead of two — but for that, read our paper!