Garrett Lisi

surfer-physicist

Last Notes

2026-08-05 16:10:24 UTC

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Thanks John. Yes, I suspect at least one of the arxiv moderators may not be the biggest fan of my work. I do hope the appeal process works once it's published -- annoying that has taken over a year. The arxiv date on the paper is April 24, 2025. Citing by that date and URL with the comment "to appear in Advances in Applied Clifford Algebras" would be great, thank you.
From a quick look at Nasmith's paper, his particle assignment looks related to
𝑒7=𝑠𝑢(3)+𝑠𝑢(5)+𝑢(1)+1×5₆+3×10₂+3×5₋₄+1×5̅₋₆+3̅×1̅0₋₂+3̅×5̅₄
He has right-handed neutrinos though, so must be doing something different.
I very much look forward to what you have to say about an e7 model with three generations. I'm almost glad you wrote your paper before seeing mine, as it's unsullied by my ideas. It's always funny when you're searching for an answer to something and the only previous answer you find is your own, which you then have to question.

2026-08-04 23:51:20 UTC

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Much of this money will be going to MU, SKHY, NVDA, and TSM -- good time to buy stock.

2026-08-04 23:49:03 UTC

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Hi John, I tried posting this paper, "Division Algebras, Triality, and Exceptional Magic", to the arxiv over a year ago but they rejected it. I submitted it to Advances in Applied Clifford Algebras and it's been approved for publication. Once it's published I can submit an arxiv appeal. In the mean time, it's here:
https://deferentialgeometry.org/works/DATaEM.6.3.pdf
E7 unification is on p34. Wondering if you're cooking up something similar?

2025-11-25 01:12:51 UTC

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And I think, as long as p>1 or q>1, we have
Cl+(p,q) = Cl(p-1,q)
Cl+(p,q) = Cl(q-1,p)

2025-11-20 22:13:31 UTC

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His equation (8) is very misleading. In his equation (6) you can see he's using Clifford five-vectors for the three spatial directions. Then he presumably uses these, along with a time vector, to make his spin(1,3), that's in Cl(1,11) but is mostly or all six-vectors and not bivectors. His spin(10) is spanned by a collection of Cl(1,11) bivectors.

2025-11-20 17:43:19 UTC

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Nope, he cheated. While everything works great with spin(11,3) plus a 64 spinor of fermions, one cannot fit spin(1,3)+spin(10) in spin(11,1). What he says is that they fit in the "spin(11,1) geometric algebra", which is to say Cl(11,1). That's fine -- I think he has spin(10) as bivectors and spin(1,3) in there as... six-vectors. And that works ok on the spinor. But using a select part of Cl(11,1) for the bosons is pretty ugly. Especially compared to using so(11,3). I don't wish to disparage his work, and I'm happy this paper exists, but it's good to be clear on what was done.

2025-11-20 07:04:44 UTC

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The "six bits" are, I think, easiest to understand by seeing how one generation of fermions (and antifermions) with spin can live in a 64 real chiral spinor representation space of spin(11,3). The seven spinor weight components are then "seven bits", but since the spinor is a chiral half the seventh bit is determined by the other six. The bits partition as you would expect under the spin(1,3)+spin(4)+spin(6) in spin(11,3), and you can see that the left-chiral fermions and right-chiral antifermions interact with the weak su(2). It makes for nice Gravi-GUT unification... though I don't love that name.

2025-09-17 15:44:13 UTC

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I currently only see this as a possible thread connecting inner automorphisms of and by octonions to so(8) automorphisms. If I were to guess at how to firm this up, I would look at when and how so(8) elements, which can be described as ordered products of two octonions, can be described using one octonion for the product.

2025-09-16 22:53:04 UTC

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Ah, if you're interested, I can fill in some more detail. If you take
𝑔=(−½)(𝑒₀+𝑒₁+𝑒₂+𝑒₃)
and compute 𝑔 𝑒ᵢ 𝑔⁻ for the i>3 you will see that these four 𝑒ᵢ are transformed by a 4x4 Hadamard matrix of ±½ entries that is the same as a triality outer-automorphism matrix for a Cartan subalgebra of so(8).

2025-09-16 19:07:03 UTC

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(part two of split comment)
I don't immediately see a connection to inner automorphisms of the octonions, or of w to j, but let's see...
One thing I can say (which will no doubt confuse things further) is that so(2) bivectors can be written as imaginary complex numbers, and so(4) bivectors can be written as imaginary quaternions, and so(8) bivectors cannot always be written as imaginary octonions because of octonionic non-associativity, but can always be written as ordered products of two octonions. (With the multiplication on the right carried out first.)
The inner-automorphisms of the quaternions is so(3), and the inner-automorphisms of the octonions is not so(7) but is a subalgebra, g2, that respects non-associativity.
To address John's question of the connection between so(8) automorphisms and octonion inner-automorphisms, maybe it comes down to when one can and cannot write an ordered octonion bi-product (which is an so(8) element) as a product by a single octonion?
Or, as a specific example that relates to your question, consider the action of the octonion
𝑔=(−½)(𝑒₀+𝑒₁+𝑒₂+𝑒₃)
which satisfies g^3 = 1, permutes the three imaginary quaternionic octonions, and acts on the four remaining imaginary octonions suspiciously like a triality matrix that mixes so(8) roots.
I have a paper on this stuff currently going through the review process.
By the way, John, I don't know if you saw it but I also made a comment on the SM intersection of so(9) with su(3)+su(3) in f4.
https://mathstodon.xyz/@Garrett/115187411739213614

2025-09-16 19:02:22 UTC

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Sorry if the LaTeX to unicode is wonky -- it seems like the best of available options. John, the context wasn't directly your question on inner automorphisms of octonions (which I think mostly relates to g2) but (I think) a related description of inner automorphisms of f4. Sorry for the lack of clarity on my part.
To address specific questions you and @npub1d60…mhwm raised: For f4, one has the decomposition
𝑓₄ = 𝑠𝑜(9) + 16ˢ
which implies we build a matrix representation of the Clifford algebra Cl(9), which consists of scalars, vectors, bivectors, etc. The Clifford bivectors, under their anti-symmetric product, are so(9), and these bivectors, as representative matrices, act on the spinor representation space, 16ˢ. To simple-minded physicists, this is what spinors are.
One can choose any bivector, let's call it w instead of j, inside so(9) in f4, and exponentiate it
𝑆 = 𝑒𝑥𝑝(𝑤 θ)
to get an automorphism of f4, with S acting on bivector and spinor elements as a "rotation",
𝑆 𝑠𝑜(9) 𝑆⁻ 𝑆 16ˢ
What is non-obvious is that for θ = 2π/3 this corresponds to a canonical triality automorphism of f4, relating to a decomposition
𝑓₄ = 𝑠𝑜(8) + 8ᵛ + 8ˢ⁻ + 8ˢ⁺
This correspondence is established via the eigenspace decomposition of f4 under the canonical triality automorphism, which ends up giving the 2π/3 rotation by some w (that as John pointed out is not in the so(8) above).
(hit length limit -- splitting this comment in two)

2025-09-12 23:56:59 UTC

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I'm less familiar with 𝑠𝑢(3)+𝑠𝑢(3) as a subalgebra of 𝑓₄ fixed by an automorphism of order 3, but the canonical triality automorphism of 𝑓₄ fixes a 𝑠𝑜(7)+𝑢(1) subalgebra. How this works is, we have
𝑓₄ = 𝑠𝑜(8) + 8ᵛ + 8ˢ⁻ + 8ˢ⁺
and a triality automorphism permutes the 8's and is an outer-automorphism of 𝑠𝑜(8). The fixed subalgebra in 𝑠𝑜(8) is 𝑔₂, and the average of the permuted 8's is an 8 which is acted on by the 𝑔₂ as a 7+1. The 𝑔2+7 = 𝑠𝑜(7) and the remaining 1 generates a 𝑢(1). If one looks at this inside
𝑓4 = 𝑠𝑜(9) + 16𝑠
f4 = so(9) + 16s
then the u(1) is generated by any so(9) bivector, which we can call j, and the triality automorphism of f4 is indeed a Clifford rotation
S = e^{2 pi j/3} = -½ + \frac{\sqrt{3}}{2} acting on the so(9) bivectors and 16s spinor as
S so(9) S^- S 16s
As eigenspaces of this triality automorphism of f4, the invariant eigenspace is
so(7) + u(1)
and the +-1 eigenspaces are
7v+-1 + 8s-+½

2025-09-11 19:53:38 UTC

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I don't know how much it will help, but I think I see how this works algebraically in 𝑓₄. One has the decompositions
𝑓₄ = 𝑠𝑜(9) + 16ˢ
and
𝑓₄ = 𝑠𝑢(3) + 𝑠𝑢(3) + 3×6 + 3̅×6̅
The 𝑠𝑢(3) + 𝑠𝑢(3) almost but doesn't quite fit in the 𝑠𝑜(9). The 𝑠𝑜(9) is
𝑠𝑜(9) = 𝑠𝑜(3) + 𝑠𝑜(6) + 3×6
= 𝑠𝑢(2) + 𝑠𝑢(4) + 3×6
= 𝑠𝑢(2) + 𝑠𝑢(3) + 𝑢(1) + 3 + 3̅ + 3×6
So you can see the ''intersection'' 𝑠𝑢(2)+𝑠𝑢(3)+𝑢(1) in 𝑠𝑜(9), and the related 𝑠𝑢(3)+𝑠𝑢(3) in 𝑓₄ is found by pulling a 2+2̅ out of the 16ˢ to make the
𝑠𝑢(3) = 𝑠𝑢(2) + 𝑢(1) + 2 + 2̅
with the other 𝑠𝑢(3) the same as in 𝑠𝑜(9).
Now, my thinking as to why this isn't especially great for Standard Model building is that 𝑢(1) in the 𝑠𝑜(9) acting on the 16ˢ is B-L and not hypercharge, Y. What does work well with the SM is this same game in 𝑒₆, with 𝑠𝑜(10) -- as you and JH figured out.