The n-Category Café

To keep things simple I’ll explain this by focusing, not on the whole Standard Model gauge group, but its subgroup $\text{SU}(2) \times \text{SU}(3)$. Here is a theorem proved by Will Sawin in response to a question of mine on MathOverflow:

Theorem 10. There are exactly two conjugacy classes of subgroups of $\text{Spin}(10)$ that are isomorphic to $\text{SU}(2) \times \text{SU}(3)$. One of them has a representative that is a subgroup of $\text{Spin}(9) \subset \text{Spin}(10)$, while the other does not.

I’ll describe representatives of these two subgroups; then I’ll say a bit about how they show up in physics, and then I’ll show you Sawin’s proof.

We can get both subgroups in a unified way! There’s always an inclusion

$$\text{SO}(m) \times \text{SO}(n) \to \text{SO}(m+n)$$

and taking double covers of each group we get a 2-1 homomorphism

$$\text{Spin}(m) \times \text{Spin}(n) \to \text{Spin}(m+n)$$

In particular we have

$$\text{Spin}(4) \times \text{Spin}(6) \to \text{Spin}(10)$$

so composing with the exceptional isomorphisms:

$$\text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2), \qquad \text{Spin}(6) \cong \text{SU}(4)$$

we get a 2-1 homomorphism

$$k \colon \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) \to \text{Spin}(10)$$

Now, there are three obvious ways to include $\text{SU}(2) \times \text{SU}(3)$ in $\text{SU}(2) \times \text{SU}(2) \times \text{SU}(4)$. There is an obvious inclusion

$$j \colon \text{SU}(3) \hookrightarrow \text{SU}(4)$$

but there are three obvious inclusions

$$\ell, r, \delta \colon \text{SU}(2) \hookrightarrow \text{SU}(2) \times \text{SU}(2)$$

namely the left one:

$$\begin{array}{ccc} \ell \colon \text{SU}(2) &\to& \text{SU}(2) \times \text{SU}(2) \\ g & \mapsto & (g,1) \end{array}$$

the right one:

$$\begin{array}{ccc} r \colon \text{SU}(2) &\to& \text{SU}(2) \times \text{SU}(2) \\ g & \mapsto & (1,g) \end{array}$$

and the diagonal one:

$$\begin{array}{ccc} \delta \colon \text{SU}(2) &\to& \text{SU}(2) \times \text{SU}(2) \\ g & \mapsto & (g,g) \end{array}$$

Combining these with our earlier maps, we actually get a one-to-one map from $\text{SU}(2) \times \text{SU}(3)$ to $\text{Spin}(10)$. So we get three subgroups of $\text{Spin}(10)$, all isomorphic to $\text{SU}(2) \times \text{SU}(3)$:

• There’s the left subgroup $G_\ell$, which is the image of this composite homomorphism:

$$\text{SU}(2) \times \text{SU}(3) \stackrel{\ell \times j}{\longrightarrow} \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) \cong \text{Spin}(4) \times \text{Spin}(6) \stackrel{k}{\longrightarrow} \text{Spin}(10)$$

• There’s the diagonal subgroup $G_\delta$, which is the image of this:

$$\text{SU}(2) \times \text{SU}(3) \stackrel{\delta \times j}{\longrightarrow} \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) \cong \text{Spin}(4) \times \text{Spin}(6) \stackrel{k}{\longrightarrow} \text{Spin}(10)$$

• And there’s the right subgroup $G_r$, which is the image of this:

$$\text{SU}(2) \times \text{SU}(3) \stackrel{r \times j}{\longrightarrow} \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) \cong \text{Spin}(4) \times \text{Spin}(6) \stackrel{k}{\longrightarrow} \text{Spin}(10)$$

The left and right subgroups are actually conjugate, but the diagonal one is truly different! We’ll prove this by taking a certain representation of $\text{Spin}(10)$, called the Weyl spinor representation, and restricting it to those two subgroups. We’ll get inequivalent representations of $\text{SU}(2) \times \text{SU}(3)$. This proves the two subgroups aren’t conjugate.

This argument is also interesting for physics. When restrict to the left subgroup, we get a representation of $\text{SU}(2) \times \text{SU}(3)$ that matches what we actually see for one generation of fermions! This is the basis of the so-called $\text{SO}(10)$ grand unified theory, which should really be called the $\text{Spin}(10)$ grand unified theory.

(In fact this works not only for $\text{SU}(2) \times \text{SU}(3)$ but for the whole Standard Model gauge group, which is larger. I’m focusing on $\text{SU}(2) \times \text{SU}(3)$ just because it makes the story simpler.)

When we restrict the Weyl spinor representation to the diagonal subgroup, we get a representation of $\text{SU}(2) \times \text{SU}(3)$ that is not physically correct. Unfortunately, it’s the diagonal subgroup that shows up in several papers connecting the Standard Model gauge group to the octonions. I plan to say a lot more about this later.

The left subgroup

Let’s look at the left subgroup $G_\ell$, the image of this composite:

$$\text{SU}(2) \times \text{SU}(3) \stackrel{\ell \times j}{\longrightarrow} \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) \cong \text{Spin}(4) \times \text{Spin}(6) \stackrel{k}{\longrightarrow} \text{Spin}(10)$$

$\text{Spin}(10)$ has a 32-dimensional unitary representation called the ‘Dirac spinor’ representation. This representation is really on the exterior algebra $\Lambda \mathbb{C}^5$. It’s the direct sum of two irreducible parts, the even grades and the odd grades:

$$\Lambda \mathbb{C}^5 \cong \Lambda^{\text{even}} \mathbb{C}^5 \oplus \Lambda^{\text{odd}} \mathbb{C}^5$$

Physicists call these two irreducible representations ‘right- and left-handed Weyl spinors’, and denote them as $\mathbf{16}$ and $\mathbf{16}\ast$ since they’re 16-dimensional and one is the dual of the other.

Let’s restrict the $\mathbf{16}$ to the left subgroup $G_\ell$ and see what we get.

To do this, first we can restrict the $\mathbf{16}$ along $k$ and get

$$\mathbf{2} \otimes \mathbf{1} \otimes \mathbf{4} \; \oplus \; \mathbf{1} \otimes \mathbf{2} \otimes \mathbf{4}\ast$$

Here $\mathbf{1}$ is the trivial representation of $\text{SU}(2)$, $\mathbf{2}$ is the tautologous representation of $\text{SU}(2)$, and $\mathbf{4}$ is the tautologous rep of $\text{SU}(4)$.

Then let’s finish the job by restricting this representation along $\ell \times j$. Restricting the $\mathbf{4}$ of $\text{SU}(4)$ to $\text{SU}(3)$ gives $\mathbf{3} \oplus \mathbf{1}$: the sum of the tautologous representation of $\text{SU}(3)$ and the trivial representation. Restricting $\mathbf{2} \otimes \mathbf{1}$ to the left copy of $\text{SU}(2)$ gives the tautologous representation $\mathbf{2}$, while restricting $\mathbf{1} \otimes \mathbf{2}$ to this left copy gives $\mathbf{1} \oplus \mathbf{1}$: the sum of two copies of the trivial representation. All in all, we get this representation of $\text{SU}(2) \times \text{SU}(3)$:

$$\mathbf{2} \otimes (\mathbf{3} \oplus \mathbf{1}) \; \oplus \; (\mathbf{1} \oplus \mathbf{1}) \otimes (\mathbf{3}\ast \oplus \mathbf{1})$$

This is what we actually see for one generation of left-handed fermions and antifermions in the Standard Model! The representation $\mathbf{3} \oplus \mathbf{1}$ describes how the left-handed fermions in one generation transform under $\text{SU}(3)$: 3 colors of quark and one ‘white’ lepton. The representation $\mathbf{3}\ast \oplus \mathbf{1}$ does the same for the left-handed antifermions. The left-handed fermions form an isospin doublet, giving us the $\mathbf{2}$, while the left-handed antifermions have no isospin, giving us the $\mathbf{1} \oplus \mathbf{1}$.

This strange lopsidedness is a fundamental feature of the Standard Model.

The right subgroup would work the same way, up to switching the words ‘left-handed’ and ‘right-handed’. And by Theorem 10, the left and right subgroups must be conjugate in $\text{Spin}(10)$, because now we’ll see one that’s not conjugate to either of these.

The diagonal subgroup

Consider the diagonal subgroup $G_\delta$, the image of this composite:

$$\text{SU}(2) \times \text{SU}(3) \stackrel{\delta \times j}{\longrightarrow} \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) \cong \text{Spin}(4) \times \text{Spin}(6) \stackrel{k}{\longrightarrow} \text{Spin}(10)$$

Let’s restrict the $\mathbf{16}$ to $G_\delta$.

To do this, first let’s restrict the $\mathbf{16}$ along $k \colon \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) \to \text{Spin}(10)$ and get

$$\mathbf{2} \otimes \mathbf{1} \otimes \mathbf{4} \; \oplus \; \mathbf{1} \otimes \mathbf{2} \otimes \mathbf{4}\ast$$

as before. Then let’s restrict this representation along $\delta \times j$. The $\SU(3)$ part works as before, but what happens when we restrict $\mathbf{2} \otimes \mathbf{1}$ or $\mathbf{1} \otimes \mathbf{2}$ along the diagonal map $\delta \colon \text{SU}(2) \to \text{SU}(2) \times \text{SU}(2)$? We get $\mathbf{2}$. So, this is the representation of $G_\delta$ that we get:

$$\mathbf{2} \otimes (\mathbf{3} \oplus \mathbf{1}) \; \oplus \; \mathbf{2} \otimes (\mathbf{3}\ast \oplus \mathbf{1})$$

This is not good for the Standard Model. It describes a more symmetrical universe than ours, where both left-handed fermions and antifermions transform as doublets under $\text{SU}(2)$.

The fact that we got a different answer this time proves that $G_\ell$ and $G_\delta$ are not conjugate in $\text{Spin}(10)$. So to complete the proof of Theorem 10, we only need to prove

1. Every subgroup of $\text{Spin}(10)$ isomorphic to $\text{SU}(2) \times \text{SU}(3)$ is conjugate to $G_\ell$ or $G_\delta$.

2. $G_\delta$ is conjugate to a subgroup of $\text{Spin}(9) \subset \text{Spin}(10)$, but $G_\ell$ is not.

I’ll prove 2, and then I’ll turn you over to Will Sawin to do the rest.

Why the diagonal subgroup fits in $\text{Spin}(9)$

Every rotation of $\mathbb{R}^n$ extends to a rotation of $\mathbb{R}^{n+1}$ that leaves the last coordinate fixed, so we get an inclusion $\text{SO}(n) \hookrightarrow \text{SO}(n+1)$, which lifts to an inclusion of the double covers, $\text{Spin}(n) \hookrightarrow \text{Spin}(n+1)$. Since we have exceptional isomorphisms

$$\text{Spin}(3) \cong \text{SU}(2), \qquad \text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2)$$

it’s natural to ask how the inclusion $\text{Spin}(3) \hookrightarrow \text{Spin}(4)$ looks in these terms. And the answer is: it’s the diagonal map! In other words, we have a commutative diagram

$$\begin{array}{ccc} \text{SU}(2) & \xrightarrow{\sim} & \text{Spin}(3) \\ \delta \downarrow & & \downarrow \\ \text{SU}(2) \times \text{SU}(2) & \xrightarrow{\sim} & \text{Spin}(4) \end{array}$$

Now, we can easily fit this into a larger commutative diagram involving some natural maps $\text{Spin}(m) \times \text{Spin}(n) \to \text{Spin}(m+n)$ and $\text{Spin}(n) \to \text{Spin}(n+1)$:

$$\begin{array}{ccccccc} \text{SU}(2) & \xrightarrow{\sim} & \text{Spin}(3) & \to & \text{Spin}(3) \times \text{Spin}(6) & \to & \text{Spin}(9) \\ \delta \downarrow & & \downarrow & & \downarrow & & \downarrow \\ \text{SU}(2) \times \text{SU}(2) & \xrightarrow{\sim} & \text{Spin}(4) & \to & \text{Spin}(4) \times \text{Spin}(6) & \to & \text{Spin}(10) \end{array}$$

We can simplify this diagram using the isomorphism $\text{Spin}(6) \cong \text{SU}(4)$:

$$\begin{array}{ccccccc} \text{SU}(2) \times \text{SU}(4) & \to & \text{Spin}(9) \\ \delta \times 1 \downarrow & & \downarrow \\ \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) & \to & \text{Spin}(10) \end{array}$$

and then we can use our friend the inclusion $j \colon \text{SU}(3) \to \text{SU}(4)$:

$$\begin{array}{ccccccc} \text{SU}(2) \times \text{SU}(3) & \xrightarrow{1 \times j} & \text{SU}(2) \times \text{SU}(4) & \to & \text{Spin}(9) \\ & & \delta \times 1 \downarrow & & \downarrow \\ & & \text{SU}(2) \times \text{SU}(2) \times \text{SU}(4) & \to & \text{Spin}(10) \end{array}$$

This shows that the diagonal subgroup $G_\delta$ of $\text{Spin}(10)$ is actually a subgroup of $\text{Spin}(9)$!

Why the left subgroup does not fit in $\text{Spin}(9)$

The three-fold way is a coarse classification of irreducible representations of compact Lie group. Every such representation is of one and only one of these three kinds:

1) not self-dual: not isomorphic to its dual,

2a) Self-dual and orthogonal: isomorphic to its dual via an invariant nondegenerate symmetric bilinear form, also called an orthogonal structure,

2b) Self-dual and symplectic: isomorphic to its dual via an invariant nondegenerate antisymmetric bilinear form, also called a symplectic structure.

I’ve written about how these three cases are related to the division algebras $\mathbb{C}, \mathbb{R}$ and $\mathbb{H}$, respectively:

• John Baez, Division algebras and quantum theory.

But we don’t need most of this here. We just need to know one fact: when $n$ is odd, every irreducible representation of $\text{Spin}(n)$, and thus every representation of this Lie group, is self-dual! In particular this is true of $\text{Spin}(9)$.

Why does this matter? Assume the left subgroup $G_\ell \subset \text{Spin}(10)$ is a subgroup of $\text{Spin}(9)$. When we restrict the Weyl spinor representation of $\text{Spin}(10)$ to $\text{Spin}(9)$ it will be self-dual, like every representation of $\text{Spin}(9)$. Then when we restrict this representation further to $\text{SU}(2) \times \text{SU}(3)$ it must still be self-dual, since the restriction of a self-dual representation is clearly self-dual.

However, we know this representation is

$$\mathbf{2} \otimes (\mathbf{3} \oplus \mathbf{1}) \; \oplus \; (\mathbf{1} \oplus \mathbf{1}) \otimes (\mathbf{3}\ast \oplus \mathbf{1})$$

and this is not self-dual, since $\mathbf{1}\ast \cong \mathbf{1}$ and $\mathbf{2}\ast \cong \mathbf{2}$ but $\mathbf{3}\ast \ncong \mathbf{3}$.

So, it must be that $G_\ell$ is not a subgroup of $\text{Spin}(9)$.

Proof of Theorem 10

To complete the proof of Theorem 10 we just need to see why there are just two conjugacy classes of subgroups of $\text{Spin}(10)$ isomorphic to $\text{SU}(2) \times \text{SU}(3)$. But in fact Will Sawin proved a stronger result! He was answering this question of mine:

Define the Standard Model gauge group to be $\text{S}(\text{U}(2) \times \text{U}(3))$, the subgroup of $\text{SU}(5)$ consisting of block diagonal matrices with a $2 \times 2$ block and then a $3 \times 3$ block. (This is isomorphic to the quotient of $\text{U}(1) \times \text{SU}(2) \times \text{SU}(3)$ by the subgroup of elements $(\alpha, \alpha^{-3}, \alpha^2$) where $\alpha$ is a 6th root of unity.)

Up to conjugacy, how many subgroups isomorphic to the Standard Model gauge group does $\text{Spin}(10)$ have?

This question is relevant to grand unified theories of particle physics, as explained here:

• John C. Baez and John Huerta, The algebra of grand unified theories, Bull. Amer. Math. Soc. 47 (2010), 483-552.

This article focuses on one particular copy of $\text{S}(\text{U}(2) \times \text{U}(3))$ in $\text{Spin}(10)$, given as follows. By definition we have an inclusion $\text{S}(\text{U}(2) \times \text{U}(3)) \hookrightarrow \text{SU}(5)$, and we also have an inclusion $\text{SU}(5) \hookrightarrow \text{Spin}(10)$ because for any $n$ we have an inclusion $\text{SU}(n) \hookrightarrow \text{SO}(2n)$, and $\text{SU}(n)$ is simply connected so this gives a homomorphism $\text{SU}(n) \hookrightarrow \text{Spin}(2n)$.

However I think there is also an inclusion $\text{S}(\text{U}(2) \times \text{U}(3)) \hookrightarrow \text{Spin}(9)$, studied by Krasnov:

• Kirill Krasnov, SO(9) characterisation of the Standard Model gauge group.

Composing this with $\text{Spin}(9) \hookrightarrow \text{Spin}(10)$, this should give another inclusion $\text{S}(\text{U}(2) \times \text{U}(3)) \hookrightarrow \text{Spin}(10)$, and I believe this one is ‘truly different from’ — i.e., not conjugate to — the first one I mentioned.

So I believe my current answer to my question is “at least two”. But that’s not good enough.

Sawin’s answer relies on the 3-fold way. we embed $\text{SU}(2) \times \text{SU}(3)$ in $\text{Spin}(10)$, we are automatically giving this group an orthogonal 10-dimensional representation, thanks to the map $\text{Spin}(10) \to \text{SO}(10)$. We can classify the possibilities.
He writes:

There are infinitely many embeddings. However, all but one of them is “essentially the same as” the one you studied as they become equal to the one you studied on restriction to $\text{SU}(2)\times \text{SU}(3)$. The remaining one is the one studied by Krasnov.

I follow the strategy suggested by Kenta Suzuki.

$\text{SU}(3)$ has irreducible representations of dimensions $1,3,3,6,8,6, 10, 10$, and higher dimensions. The $10$-dimensional ones are dual to each other, as are the $6$-dimensional ones, so they can’t appear. The $3$-dimensional ones are dual to each other and can only appear together. So the only $10$-dimensional self-dual representations of $\text{SU}(3)$ decompose as irreducibles as $8+1+1$, $3+3+1+1+1+1$, or ten $1$s. All of these are orthogonal because the 8-dimensional representation is orthogonal. However, the ten $1$s cannot appear because then $\text{SU}(3)$ would act trivially.

A representation of $\text{SU}(3) \times \text{SU}(2)$ is a sum of tensor products of irreducible representations of $\text{SU}(3)$ and irreducible representations of $\text{SU}(2)$. Restricted to $\text{SU}(3)$, each tensor product splits into a sum of copies of the same irreducible representation. So $\text{SU}(2)$ can only act nontrivially when the same representation appears multiple times. Since the $3+3$ is two different $3$-dimensional representation, only the $1$-dimensional representation can occur twice. Thus, our 10-dimensional orthogonal representation of $\text{SU}(3) \times \text{SU}(2)$ necessarily splits as either the $8$-dimensional adjoint repsentation of $\text{SU}(3)$ plus a $2$-dimensional orthogonal representation of $\text{SU}(2)$ or the $6$-dimensional sum of standard and conjugate [i.e., dual] representations of $\text{SU}(3)$ plus a $4$-dimensional orthogonal representation of $\text{SU}(2)$. However, $\text{SU}(2)$ has a unique nontrivial representation of dimension $2$ and it isn’t orthgonal, so only the second case can appear. $\text{SU}(2)$ has representations of dimension $1,2,3,4$ of which the $2$ and $4$-dimensional ones are symplectic and so must appear with even multiplicity in any orthogonal representation, so the only nontrivial $4$-dimensional orthogonal ones are $2+2$ or $3+1$.

So there are two ten-dimensional orthogonal representations of $\text{SU}(2) \times \text{SU}(3)$ that are nontrivial on both factors, those being the sum of two different $3$-dimensional irreducible representations of $\text{SU}(3)$ with either two copies of the two-dimensional irreducible representation of $\text{SU}(2)$ or the three-dimensional and the one-dimensional irreducible representation of $\text{SU}(2)$. The orthogonal structure is unique up to isomorphisms, so these give two conjugacy classes of homomorphisms $\text{SU}(2) \times \text{SU}(3) \to SO(10)$ and thus two conjugacy classes of homomorphisms $\text{SU}(2) \times \text{SU}(3) \to \text{Spin}(10)$. The first one corrresponds to the embedding you studied while only the second one restricts to $\text{Spin}(9)$ so indeed these are different.

To understand how to extend these to $\text{S}(\text{U}(2) \times \text{U}(3))$, I consider the centralizer of the representation within $\text{Spin}(10)$. Since the group is connected, this is the same as the centralizer of its Lie algebra, which is therefore the inverse image of the centralizer in $\text{SO}(10)$. Now there is a distinction between the two examples because the example with irrep dimensions $3+3+2+2$ has centralizer with identity component $\text{U}(1) \times \text{SU}(2)$ while the example with irrep dimensions $3+3+3+1$ has centralizer with identity component $\text{U}(1)$. In the second case, the image of $\text{U}(2) \times \text{U}(3)$ must be the image of $\text{SU}(2) \times \text{SU}(3)$ times the centralizer of the image of $\text{SU}(2) \times \text{SU}(3)$, so this gives a unique example, which must be the one considered by Krasnov.

In the first case, we can restrict attention to a torus $\text{U}(1) \times \text{U}(1)$ in $\text{SU}(2) \times \text{SU}(2)$. The center of $\text{S}(\text{U}(2) \times \text{U}(3))$ maps to a one-dimensional subgroup of this torus, which can be described by a pair of integers. Explicitly, given a two-by-two-unitary matrix $A$ and a three-by-three unitary matrix $B$ with $\det(A) \det(B) =1$, we can map to $\text{U}(5)$ by sending $(A,B)$ to $A \gamma^a \oplus B \gamma^b$ where $\gamma = \det (A) = \det(B)^{-1}$, and then map from $\text{U}(5)$ to $SO(10)$. This lifts to the spin group if and only if the determinant in $\text{U}(5)$ is a perfect square. The determinant is $\gamma^{ 1 + 2a - 1 + 3b} = \gamma^{2a+3b}$ so a lift exists if and only if $b$ is even.

The only possible kernel of this embedding is the scalars. The scalar $A = \lambda^3 I_2, B = \lambda^{-2} I_3$ maps to $\lambda^{3+ 6a} I_2 \oplus \lambda^{-2 + 6b} I_3$ and so the kernel is trivial if and only if $\gcd(3+6a,-2 + 6b)=1$.

However, there are infinitely many integer solutions $a,b$ to $\gcd(3+6a,-2a+6b)=1$ with $b$ even (in fact, a random $a$ and even $b$ works with probability $9/\pi^2$), so this gives infinitely many examples.

---

• Part 1. How to define octonion multiplication using complex scalars and vectors, much as quaternion multiplication can be defined using real scalars and vectors. This description requires singling out a specific unit imaginary octonion, and it shows that octonion multiplication is invariant under $\mathrm{SU}(3)$.
• Part 2. A more polished way to think about octonion multiplication in terms of complex scalars and vectors, and a similar-looking way to describe it using the cross product in 7 dimensions.
• Part 3. How a lepton and a quark fit together into an octonion — at least if we only consider them as representations of $\mathrm{SU}(3)$, the gauge group of the strong force. Proof that the symmetries of the octonions fixing an imaginary octonion form precisely the group $\mathrm{SU}(3)$.
• Part 4. Introducing the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$: the $3 \times 3$ self-adjoint octonionic matrices. A result of Dubois-Violette and Todorov: the symmetries of the exceptional Jordan algebra preserving their splitting into complex scalar and vector parts and preserving a copy of the $2 \times 2$ adjoint octonionic matrices form precisely the Standard Model gauge group.
• Part 5. How to think of $2 \times 2$ self-adjoint octonionic matrices as vectors in 10d Minkowski spacetime, and pairs of octonions as left- or right-handed spinors.
• Part 6. The linear transformations of the exceptional Jordan algebra that preserve the determinant form the exceptional Lie group $\mathrm{E}_6$. How to compute this determinant in terms of 10-dimensional spacetime geometry: that is, scalars, vectors and left-handed spinors in 10d Minkowski spacetime.
• Part 7. How to describe the Lie group $\mathrm{E}_6$ using 10-dimensional spacetime geometry. This group is built from the double cover of the Lorentz group, left-handed and right-handed spinors, and scalars in 10d Minkowski spacetime.
• Part 8. A geometrical way to see how $\mathrm{E}_6$ is connected to 10d spacetime, based on the octonionic projective plane.
• Part 9. Duality in projective plane geometry, and how it lets us break the Lie group $\mathrm{E}_6$ into the Lorentz group, left-handed and right-handed spinors, and scalars in 10d Minkowski spacetime.
• Part 10. Jordan algebras, their symmetry groups, their invariant structures — and how they connect quantum mechanics, special relativity and projective geometry.
• Part 11. Particle physics on the spacetime given by the exceptional Jordan algebra: a summary of work with Greg Egan and John Huerta.
• Part 12. The bioctonionic projective plane and its connections to algebra, geometry and physics.
• Part 13. Two ways to embed $\text{SU}(2) \times \text{SU}(3)$ in $\text{Spin}(10)$, and their consequences for particle physics.