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November 6, 2020

Octonions and the Standard Model (Part 7)

Posted by John Baez

Last time I explained the connection between the exceptional Jordan algebra and 10-dimensional Minkowski spacetime. Today I want to report on some work that Greg Egan, John Huerta and I did in November 2015. We figured out how to describe the big 78-dimensional symmetry group of the exceptional Jordan algebra in terms of 10d spacetime geometry!

This group is called E 6\mathrm{E}_6, and we saw how it’s built up from:

  • the double cover of the Lorentz group of 10d spacetime, Spin(9,1)\mathrm{Spin}(9,1)
  • the right-handed spinors in 10d spacetime, S +S_+
  • the left-handed spinors in 10d spacetime, S S_-
  • scalars, \mathbb{R}.

This lets us chop up the Lie algebra of E 6\mathrm{E}_6 as follows:

𝔢 6𝔰𝔬(9,1)S +S \mathfrak{e}_6 \cong \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R}

or counting dimensions,

78=45+16+16+1 78 = 45 + 16 + 16 + 1

Remember from last time: the exceptional Jordan algebra 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) consists of 3×33 \times 3 self-adjoint matrices of octonions, while 10d Minkowski spacetime can be seen as 𝔥 2(𝕆)\mathfrak{h}_2(\mathbb{O}) — the 2×22 \times 2 self-adjoint matrices of octonions.

So, 10d Minkowski spacetime sits inside the exceptional Jordan algebra! It does so in many ways, and there’s no ‘best’ one. But if we pick any one, then the exceptional Jordan algebra, which is 27-dimensional, splits into three parts, each with a nice interpretation in terms of 10d spacetime geometry:

𝔥 3(𝕆)VS \mathfrak{h}_3(\mathbb{O}) \cong V \oplus S_- \oplus \mathbb{R}

or in terms of dimensions:

27=10+16+1 27 = 10 + 16 + 1

The 10-dimensional part VV consists of vectors in 10d Minkowski spacetime, the 16-dimensional part S +S_+ consists of left-handed spinors, and the 1-dimensional part \mathbb{R} consists of scalars. Each part is an irreducible representation of Spin(9,1)\mathrm{Spin}(9,1) — and the key structure on the exceptional Jordan algebra, the determinant, is invariant under Spin(9,1)\mathrm{Spin}(9,1).

This is cute, because the three most important kinds of particles in physics are spin-1 (vectors), spin-1/2 (spinors), and spin-0 (scalars). But I’ll again remind everyone that I’m not proposing or advocating a theory of physics here. I’m just doing math.

And here’s what Greg Egan, John Huerta and I showed:

Theorem 9. The Lie group Spin(9,1)\mathrm{Spin}(9,1) and abelian Lie groups isomorphic to S +S_+, S S_- and \mathbb{R} show up as Lie subgroups of E 6\mathrm{E}_6, in such a way that the direct sum of their Lie algebras is — as a vector space — all of 𝔢 6\mathfrak{e}_6.

Proof. First, remember that we can write any element of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) in this way:

(α z y * z * β x y x * γ)=(α ψ ψ v) \left( \begin{array}{ccc} \alpha & z & y^\ast \\ z^\ast & \beta & x \\ y & x^\ast & \gamma \end{array} \right) = \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right)

where:

v=(β x x * γ) v = \left( \begin{array}{cc} \beta & x \\ x^\ast & \gamma \end{array} \right)

is a vector in 10d Minkowski spacetime: an element of the space V=𝔥 2(𝕆)V = \mathfrak{h}_2(\mathbb{O}) of 2×22\times 2 self-adjoint octonionic matrices,

ψ=(z * y) \psi = \left( \begin{array}{cc} z^\ast \\ y \end{array} \right)

is a left-handed spinor: an element of S =𝕆 2S_- = \mathbb{O}^2, and aa is a scalar: an element of \mathbb{R}.

When we do this, the determinant takes on 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) takes on a simpler form. By definition we have

det(α z y * z * β x y x * γ)=αβγ(α|x| 2+β|y| 2+γ|z| 2)+2Re(xyz) \mathrm{det} \left( \begin{array}{ccc} \alpha & z & y^\ast \\ z^\ast & \beta & x \\ y & x^\ast & \gamma \end{array} \right) = \alpha \beta \gamma - (\alpha |x|^2 + \beta |y|^2 + \gamma |z|^2) + 2 Re(x y z)

but in Theorem 8 we saw that

det(α ψ ψ v)=αg(v,v)+g(v,[ψ,ψ]) \mathrm{det} \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) = - \alpha g(v,v) + g(v, [\psi, \psi])

Here gg is the Minkowski metric and the bracket is a way of turning two left-handed spinors into a vector, explained last time.

Next, remember that whenever I say E 6\mathrm{E}_6, I mean the group of linear transformations of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) preserving the determinant. To be precise, this is the noncompact form of E 6\mathrm{E}_6 called E 6(26)\mathrm{E}_{6(-26)}.

Now, here are four subgroups of E 6\mathrm{E}_6:

1) First, Spin(9,1)\mathrm{Spin}(9,1) can be seen as a Lie subgroup of E 6\mathrm{E}_6. If we use our identification

𝔥 3(𝕆)VS \mathfrak{h}_3(\mathbb{O}) \cong V \oplus S_- \oplus \mathbb{R}

then Spin(9,1)\mathrm{Spin}(9,1) acts on each summand, hence the whole space — and its action preserves the determinant

det(α ψ ψ v)=αg(v,v)+g(v,[ψ,ψ]) \mathrm{det} \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) = - \alpha g(v,v) + g(v, [\psi, \psi])

since this is defined using operations that we saw last time are Spin(9,1)\mathrm{Spin}(9,1)-equivariant. So, Spin(9,1)E 6\mathrm{Spin}(9,1) \subseteq \mathrm{E}_6.

2) Second, the vector space S +=𝕆 2S_+ = \mathbb{O}^2, regarded as an abelian Lie group, can be seen as a Lie subgroup of E 6\mathrm{E}_6. For this, let ϕS +\phi \in S_+ act on

(α ψ ψ v)𝔥 3(𝕆) \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})

as follows:

v v ψ ψ+γ(vϕ) α α+g(v,[ϕ,ϕ])+2ϕ,ψ \begin{array}{ccl} v & \mapsto & v \\ \psi &\mapsto& \psi + \gamma(v \otimes \phi) \\ \alpha & \mapsto & \alpha + g(v, [\phi, \phi]) + 2 \langle \phi, \psi \rangle \end{array}

The bracket here describes a way to turn two right-handed spinors into a vector. The angle brackets denote the dual pairing between S +S_+ and S S_-, which produces a scalar. And γ\gamma is a way for any vector to act on a right-handed spinor to give a left-handed spinor. I described all these operations last time — and by the way, all these too are Spin(9,1)\mathrm{Spin}(9,1)-equivariant.

We can check that these formulas define a transformation of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) that preserves the determinant αdet(v)+g(v,[ψ,ψ])\alpha \mathrm{det}(v) + g(v, [\psi, \psi]). By adding γ(vϕ)\gamma(v \otimes \phi) to ψ\psi, g(v,[ψ,ψ])g(v, [\psi, \psi]) gains these three extra terms:

g(v,[ψ,γ(vϕ)]) = det(v)ϕ,ψ g(v,[γ(vψ),ψ]) = det(v)ϕ,ψ g(v,[γ(vϕ),γ(vϕ)]) = det(v)g(v,[ϕ,ϕ]) \begin{array}{ccl} g(v, [\psi, \gamma(v \otimes \phi)]) &=& -det(v) \langle \phi,\psi \rangle \\ g(v, [\gamma(v \otimes \psi), \psi]) &=& -det(v) \langle \phi,\psi \rangle \\ g(v, [\gamma(v \otimes \phi), \gamma(v \otimes \phi)]) &=& -det(v) g(v, [\phi, \phi]) \end{array}

To prove these equations, just play around with the explicit formulas given last time. These three extra terms cancel out the extra terms that appear in αdet(v)\alpha \mathrm{det}(v), so the determinant is unchanged.

Also, we can check that this map from S +S_+ to determinant-preserving linear transformations of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) is a Lie group homomorphism, where we make S +S_+ into an abelian Lie group using addition. We can also check that it’s one-to-one. All this is a calculation: the only ‘trick’ we need is the identity

g(v,[ϕ,ϕ])=ϕ,γ(vϕ) g(v, [\phi, \phi']) = \langle \phi' , \gamma(v \otimes \phi) \rangle

for all vVv \in V and ϕ,ϕS +\phi, \phi' \in S_+. This follows from the definitions of these operations. So, S +S_+ becomes a Lie subgroup of E 6\mathrm{E}_6.

3) Similarly, S =𝕆 2S_- = \mathbb{O}^2 can be seen as an abelian Lie subgroup of E 6\mathrm{E}_6. For this, let ηS \eta \in S_- act on

(α ψ ψ v)𝔥 3(𝕆) \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})

as follows:

v v+12α[η,η]+[ψ,η] ψ ψ+αη α α \begin{array}{ccl} v &\mapsto& v + \frac{1}{2} \alpha [\eta, \eta] + [\psi, \eta] \\ \psi & \mapsto& \psi + \alpha \eta \\ \alpha &\mapsto& \alpha \end{array}

Again we can show that this preserves the determinant. And again, we can check that this map from S +S_+ to determinant-preserving linear transformations of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) is a Lie group homomorphism, which is one-to-one. Again these are just calculations, and this time no tricks at all are required. So the additive group of S +S_+ becomes a Lie subgroup of E 6\mathrm{E}_6.

4) Similarly, *\mathbb{R}^\ast, the multiplicative group of nonzero real numbers, can be seen as an abelian Lie subgroup of E 6\mathrm{E}_6. For this, let any nonzero real number tt act on

(α ψ ψ v)𝔥 3(𝕆) \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})

as follows:

v t 2v ψ tψ α t 4α \begin{array}{ccl} v &\mapsto& t^{-2} v \\ \psi &\mapsto& t \psi \\ \alpha &\mapsto& t^4 \alpha \end{array}

These rescalings clearly preserve the determinant αdet(v)+g(v,[ψ,ψ])\alpha \det(v) + g(v, [\psi, \psi]).

As a consequence, we get inclusions of Lie algebras

𝔰𝔬(9,1) 𝔢 6 S + 𝔢 6 S 𝔢 6 𝔢 6 \begin{array}{ccl} \mathfrak{so}(9,1) &\hookrightarrow& \mathfrak{e}_6 \\ S_+ &\hookrightarrow& \mathfrak{e}_6 \\ S_- &\hookrightarrow& \mathfrak{e}_6 \\ \mathbb{R} &\hookrightarrow& \mathfrak{e}_6 \end{array}

where the last three are treated as abelian Lie algebras. We thus get a linear map (not a Lie algebra homomorphism!)

f:𝔰𝔬(9,1)S +S 𝔢 6 f: \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R} \to \mathfrak{e}_6

I will take it for granted that the dimension of 𝔢 6\mathfrak{e}_6 is 78: this is well-known. The dimensions of the four summands add up to this number: 45+16+16+1=7845 + 16 + 16 + 1 = 78. So, to show that ff is an isomorphism of vector spaces, we just need to check that it is one-to-one.

When we differentiate the above formulas and get actions of the Lie algebras 𝔰𝔬(9,1)\mathfrak{so}(9,1), S +S_+, S S_-, and \mathbb{R} on 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}), we can direct sum them and get a map from 𝔰𝔬(9,1)S +S \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R} to linear transformations of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}). Here’s what we get. Write L𝔰𝔬(9,1)L \in \mathfrak{so}(9,1), ϕS +\phi \in S_+, ηS \eta \in S_- and tRt \in R. Then (L,ϕ,η,t)(L, \phi, \eta, t) gives this transformation of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}):

(L,ϕ,η,t):(α ψ ψ v)(2ϕ,ψ+4tα (Lψ+γ(vϕ)+tψ) Lψ+γ(vϕ)+tψ Lv+[ψ,η]2sv) (L, \phi, \eta, t) \colon \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \mapsto \left( \begin{array}{cc} 2\langle \phi, \psi \rangle + 4t \alpha & (L \psi + \gamma(v \otimes \phi) + t \psi)^\dagger \\ L \psi + \gamma(v \otimes \phi) + t \psi & L v + [\psi,\eta] - 2s v \end{array} \right)

One can check that this map from 𝔰𝔬(9,1)S +S \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R} to linear transformations of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) is one-to-one. Thus, we get an isomorphism of vector spaces

𝔢 6𝔰𝔬(9,1)S +S \mathfrak{e}_6 \cong \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R}

as desired.   █

This isomorphism of vector spaces is not an isomorphism of Lie algebras, since for example the abelian Lie subalgebras S +S_+ and S S_- don’t commute with each other in 𝔢 6\mathfrak{e}_6. However, each summand is a Lie subalgebra of 𝔢 6\mathfrak{e}_6, and we can use this isomorphism of vector spaces to describe the bracket in 𝔢 6\mathfrak{e}_6 in terms of operations on 𝔰𝔬(9,1)\mathfrak{so}(9,1), S +S_+, S S_- and \mathbb{R}. Maybe I’ll do that next time; I think you’re probably tired by now, if you’re reading this at all.

Also — and perhaps more importantly — we can get a geometrical understanding of what’s going on here, using the octonionic projective plane 𝕆P 2\mathbb{O}\mathrm{P}^2. But again, I’ll wait and talk about that later.


  • 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 SU(3)\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 SU(3)\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 SU(3)\mathrm{SU}(3).
  • Part 4. Introducing the exceptional Jordan algebra 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}): the 3×33 \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 2×22 \times 2 adjoint octonionic matrices form precisely the Standard Model gauge group.
  • Part 5. How to think of 2×22 \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 E 6\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 E 6\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 E 6\mathrm{E}_6 is connected to 10d spacetime, based on the octonionic projective plane.
Posted at November 6, 2020 8:26 PM UTC

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Re: Octonions and the Standard Model (Part 7)

Each part is an irreducible representation of Spin(9,1)\mathrm{Spin}(9,1). So is 𝕙 3(𝕆)\mathbb{h}_3(\mathbb{O}) — and this is how it decomposes into irreducible representations.

typo? Surely 𝕙 3(𝕆)\mathbb{h}_3(\mathbb{O}) isn’t irreducible if it breaks into irreps.

Posted by: David Roberts on November 8, 2020 12:39 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 7)

Sorry, I was just trying to remind people that 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) is a representation of Spin(9,1)Spin(9,1). It sure ain’t irreducible! I’ll fix that sentence.

Posted by: John Baez on November 8, 2020 12:42 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 7)

The Peirce decomposition can be generalized to a (Vinberg) T-algebra of dimension D=275. This is a nonassociative algebra of 3×33\times 3 matrices as well, with Peirce decomposition 275=1+18+256. The 18 part is a spin factor i.e., vector of SO(17,1)SO(17,1) with 256 spinor that, under SO(16)SO(16) decomposes as 128 s128_s + 128 c128_c. The higher analog of 𝔢 6(26)\mathfrak{e}_{6(-26)} here is an algebra 𝔤=256(so(17,1)+1)256\mathfrak{g}=256\oplus (so(17,1)+1)\oplus 256.

In fact, the exceptional Jordan algebra 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) can be seen as part of an infinite family of such 3×33\times 3 T-algebras. Each such T-algebra admits a Peirce decomposition, under a fixed point (rank one idempotent). There is even a T-algebra with 26-dimensional spin factor with SO(25,1)SO(25,1) affine symmetry. One can use these three T-algebras to formulate (matrix) string theories in D=9+1D=9+1, D=17+1D=17+1 and D=25+1D=25+1.

Posted by: Metatron on November 8, 2020 6:14 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 7)

For some background see

Posted by: jackjohnson on November 9, 2020 12:07 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 7)

Nice paper! Cartier seems to be talking about an unrelated concept of ‘Vinberg algebra’: these are approximately the same as pre-Lie algebras, which are vector spaces with a binary operation *\ast such that the bracket [a,b]=a*bb*a[a,b] = a\ast b - b \ast a gives a Lie algebra.

Posted by: John Baez on November 12, 2020 2:36 AM | Permalink | Reply to this

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