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!

How does this work? We take the hermitian Jordan triple itself to be $\mathbf{k}_1$. The Lie algebra $\mathbf{k}_0$ consists of all linear maps from $\mathbf{k}_1$ to itself that are of this form:

$$x \mapsto [a,b,x] - [b,a,x]$$

for some $a,b \in \mathbf{k}_1$. These maps are called real inner derivations. They form a Lie algebra since the commutator of two such maps is another such map. With a bit more work we can define other operations making all of $\mathbf{k}$ into a $\mathbb{Z}_2$-graded Lie algebra.

So, we get a big Lie algebra $\mathbf{k}$, and a Lie subalgebra $\mathbf{k}_0$ sitting inside it. From this we get two Lie groups: a big one $K$ whose Lie algebra is $\mathbf{k}$, and a subgroup $K_0$, 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. But we’ll start with the bi-Cayley triple.

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, our friend $\mathbb{O}_\mathbb{C}^2$, 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.

I can’t resist explaining some of the quantum physics here. I said I wouldn’t, but I can’t help it. In a hermitian Jordan triple, the triple product $[-,-,-]$ is linear in the first and last slot, but conjugate-linear in the middle slot. So, if you multiply a tripotent by a phase $\alpha$, you get a new tripotent:

$$[\alpha e, \alpha e, \alpha e] = \alpha \overline{\alpha} \alpha e = \alpha e$$

This should remind you of how when you multiply a unit vector in a Hilbert space by a phase, you get a new unit vector. In Jordan triple quantum mechanics, minimal tripotents take the place of these unit vectors. And guess what: the hermitian symmetric space $K/K_0$ that I was talking about earlier is also the space of minimal tripotents mod phase! So, it generalizes the familiar space of ‘pure states’ in quantum mechanics, which are unit vectors mod phase.

But let’s get back to the Standard Model.

A chain of Jordan triples

From here on, the single fact driving everything is this: in any hermitian Jordan triple, any 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 this 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 the bi-Cayley triple colinear with both $e_1$ and $e_2$ is a Jordan triple whose 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 chain taking us from the so-called $\mathrm{SO}(10)$ grand unified theory down to the $\mathrm{SU}(5)$ grand unified theory down to the Standard Model. 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 one generation of Standard Model fermions this way.

The six particles types as Peirce spaces

We have gotten the representation of the Standard Model gauge group on one generation of fermions without any fuss. But it’s also fun to peer into the details, and see how the different kinds of fermions emerge. We can get them using the fact that for any tripotent $e$, we have projections $P_0(e), P_{1/2}(e)$ and $P_1(e)$ onto its three eigenspaces: its so-called Peirce projectors.

Since we get the Standard Model gauge group and its representation on fermions from two minimal tripotents $e_1$ and $e_2$, we have nine Peirce projectors we can apply to our Jordan triple $\mathbb{O}_{\mathbb{C}}^2$. Let’s use these to pick out various kinds of particles!

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
$P_{1/2}(e_2) P_{1/2}(e_1)$	(3, 2, +1/6)	$q_L$
$P_{1/2}(e_2) P_0(e_1)$	($\overline{3}$, 1, +1/3)	$\overline{d_R}$
$P_0(e_2) P_{1/2}(e_1)$	($\overline{3}$, 1, −2/3)	$\overline{u_R}$
$P_0(e_2) P_0(e_1)$	(1, 2, −1/2)	$\ell_L$
$P_1(e_2) P_{1/2}(e_1)$	(1, 1, +1)	$\overline{e_R}$
$P_{1/2}(e_2) P_1(e_1)$	(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!