Nostr notes by Paul Schwahn

There's also the phenomenon in dimension 4 where the ...

2026-07-10T20:59:27+02:00

In reply to nevent1q…8ul9
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There's also the phenomenon in dimension 4 where the SO(4)-representation Λ²ℝ⁴ splits into self-dual and anti-self-dual two-forms, which implies that the Weyl tensor splits into two parts as well. But I don't know if this is true for SO(3,1) as well, and what the physical consequences would be.

I think the derivation from the Einstein-Hilbert action is valid ...

2026-07-09T08:53:49+02:00

In reply to nevent1q…c7c8
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I think the derivation from the Einstein-Hilbert action is valid in any dimension. But I usually assume that the boundary is empty anyway.
At least for Riemannian metrics in 2D, the vacuum Einstein-Hilbert functional is constant because of the Gauß-Bonnet theorem, so every metric is a critical point, so the Einstein tensor is always zero.
So I wouldn't say that the derivation fails, rather that the action itself "degenerates".

On symmetric spaces, each tangent space is a Lie triple system, ...

2026-07-08T09:28:45+02:00

In reply to nevent1q…dtxk
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On symmetric spaces, each tangent space is a Lie triple system, the triple product being the curvature tensor.
Which geometric object does the Jordan triple system correspond to?

Could you give some motivation why we want to study Jordan triple ...

2026-07-07T20:30:42+02:00

In reply to nevent1q…949f
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Could you give some motivation why we want to study Jordan triple systems? What makes them better then Jordan algebras? In particular with respect to complex homogeneous domains.

Regardless of the metric signature, in dimensions other than 2, ...

2026-07-01T22:20:00+02:00

In reply to nevent1q…ujms
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Regardless of the metric signature, in dimensions other than 2, the Einstein tensor carries the same information as the Ricci tensor, so my intuition for the Einstein tensor is the same as for the Ricci tensor. (In particular, a metric satisfies the Einstein equation in vacuum exactly if it is Ricci-flat).

In dimension 2, as John said, the Einstein tensor is already zero, and the only curvature quantity that survives is the scalar curvature (or Gaussian curvature).

But the analogy isn't far fetched: For example, the Schwarzschild metric is Ricci-flat, and all its mass is concentrated in a singularity. Just like a polyhedron is scalar-flat with all its curvature concentrated into singularities.

This reminds me of the Kodaira-Spencer deformation theory for ...

2026-06-25T20:01:31+02:00

In reply to nevent1q…mlpp
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This reminds me of the Kodaira-Spencer deformation theory for complex manifolds (not varieties): the infinitesimal deformations (up to diffeomorphism gauge) of the complex structure can be identified with H¹(TX), while the obstruction space is H²(TX) - in the sense that every integrability obstruction to an infinitesimal deformation is valued in H²(TX), and thus, in case H²(TX) vanishes, all infinitesimal deformations are integrable into actual deformations.

Fun fact: If (X,g,J) is a Kähler-Einstein manifold with negative scalar curvature, then the infinitesimal Einstein deformations of g coincide with the infinitesimal complex deformations of J. In fact there is a correpondence between Einstein metrics near g and complex structures near J (preserving the Kähler-Einstein property), thanks to the Aubin-Yau theorem! Thus, if H¹(TX)≠0 and H²(TX)=0, we suddenly have a (non-explicit) family of solutions to the Einstein equation on X.

Unfortunately this breaks down somewhat for nonnegative scalar curvature.

I have no idea. There is a 72 hour period for Russia to submit an ...

2026-06-14T20:15:14+02:00

In reply to nevent1q…a8gf
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I have no idea.
There is a 72 hour period for Russia to submit an extradition request. If they don't, he can go free.

I must add that I am not completely happy with the petition ...

2026-06-13T20:51:07+02:00

In reply to nevent1q…s2xv
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I must add that I am not completely happy with the petition itself: the text reads somewhat generic, there is no call to action to the Brazilian authorities, and I don't like change.org. But I still found it worth sharing.

My colleague Misha Verbitsky over at IMPA, one of the top ...

2026-06-13T20:42:39+02:00

My colleague Misha Verbitsky over at IMPA, one of the top mathematicians of Brazil, is being detained in Armenia because Russia has designated him a terrorist after his comments on the war in Ukraine.

https://caliber.az/en/post/russian-mathematician-wanted-on-terrorism-charges-detained-in-armenia

This is surreal to me, given that I met him just a month ago.

Please consider signing and sharing the below petition, in the hope that this will generate some media attention:

https://www.change.org/p/free-misha-verbitsky-from-unjust-detention

I know nothing about 2-bundles. What kind of ingredients do you ...

2026-05-31T11:12:11+02:00

In reply to nevent1q…s00d
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I know nothing about 2-bundles. What kind of ingredients do you need?

well, not all of the things! I'm still unsure about how many ...

2026-05-30T21:51:33+02:00

In reply to nevent1q…rtha
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well, not all of the things! I'm still unsure about how many 𝔥₃(ℍ)-subalgebras there are in 𝔥₃(𝕆)...

Plus, there is a number of subtleties going on here, such as whether subplanes of 𝕆ℙ² in the sense of incidence geometry correspond to subalgebras.

But as John said, our findings should be enough to make physicists happy!

I must admit that I never understood the theorem statement of ...

2026-05-25T20:32:03+02:00

I must admit that I never understood the theorem statement of geometrization of 3-manifolds: every (smooth, compact, connected, orientable) 3-manifold (with boundary a disjoint union of tori) "decomposes" into "geometric pieces".

But what exactly does "decompose" mean here, what are the 8 possible "geometric pieces", and how do we decide which of them occur?

This survey of B. Martelli does an excellent job at explaining!

https://arxiv.org/abs/2605.23679

That distilled water is completely safe to drink (contrary to ...

2026-05-07T20:28:15+02:00

In reply to nevent1q…9uxu
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That distilled water is completely safe to drink (contrary to what I learned in school)!

I think in definition 2 G should be required to be discrete.

2026-04-24T21:25:43+02:00

In reply to nevent1q…ue28
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I think in definition 2 G should be required to be discrete.

I'm tempted to support it, but first I need some context. Who ...

2026-03-15T20:09:50+01:00

In reply to nevent1q…7hc5
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I'm tempted to support it, but first I need some context. Who is drafting this letter, and what will happen with the signatures?

It scares me too, right now I am still stuck at trying to answer ...

2026-01-29T22:16:49+01:00

In reply to nevent1q…n638
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It scares me too, right now I am still stuck at trying to answer the question for two octonions (i.e. are there 𝑥,𝑦∈𝕆 such that 𝐿(𝑥)𝐿(𝑦) does the one thing and 𝑅(𝑥)𝑅(𝑦) the other?).

I'll check out those books!

They way I originally stated, before applying the triality ...

2026-01-29T15:10:31+01:00

In reply to nevent1q…fslz
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They way I originally stated, before applying the triality automorphism, was this:

The (-) half-spinor representation h₁₂ should preserve ℍ, and the vector representation h₂₃ should map it to its complement.

We have to be a bit careful when applying outer automorphisms, because they always also involve 𝐶 somewhere, and the explicit realization of each representation matters.

Our original question was: is there an element h∈𝐻 such that ...

2026-01-28T18:30:06+01:00

In reply to nevent1q…kl3j
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Our original question was: is there an element h∈𝐻 such that h₁₂ preserves ℍ, while h₂₃ maps the complement of ℍ to ℍ? We can now try to answer that using the above identification and realization of Spin(8).

To make it even easier, we can interpose the triality automorphism of Spin(8), which (following Harvey) maps

(𝑔₁,𝑔₂,𝑔₃)↦(𝐶𝑔₂𝐶,𝑔₃,𝐶𝑔₁𝐶).

After this automorphism, now 𝐶h₁₂𝐶 and h₂₃ correspond to the half-spinor reps of Spin(8). With h corresponding to a generator 𝑥⋅𝑦, we thus have

h₁₂=−𝐶𝑅(𝑥)𝑅(𝑦*)𝐶=-𝐿(𝑥*)𝐿(𝑦),
h₂₃=−𝑅(𝑥*)𝑅(𝑦),

and specializing to 𝑥=−1 we get

h₁₂=𝐿(𝑦), h₂₃=𝑅(𝑦).

These also generate the group, since the more general form above is a product of two of these. So the question sounds simple enough now:

Do there exist 𝑦₁,…,𝑦ₙ∈𝕆 such that 𝐿(𝑦ₙ)…𝐿(𝑦₁) preserves ℍ, while 𝑅(𝑦ₙ)…𝑅(𝑦₁) maps the complement of ℍ to ℍ?

So far I've been unable to answer that question, and of course it's very possible that I made a calculation mistake there.

(2/2)

Okay, here goes. Let's fix some notation: for any 𝑥∈𝕆 ...

2026-01-28T18:22:19+01:00

In reply to nevent1q…30wl
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Okay, here goes. Let's fix some notation: for any 𝑥∈𝕆 and (i,j)=(1,2),(2,3),(3,1), let ξᵢⱼ(𝑥) be the 3×3-matrix with entry 𝑥 at position (i,j), entry 𝑥* at position (j,i), and 0 elsewhere. So we get linear maps

ξᵢⱼ : 𝕆→𝔥₃(𝕆),

and with the Jordan product ∘ we have the relation

ξᵢⱼ(𝑥)∘ξⱼₖ(𝑦)=ξₖᵢ(𝑦*𝑥*)

for any cyclic permutation (i,j,k) of (1,2,3).

Now, let 𝐻 be the stabilizer group of the standard Jordan frame. This group is isomorphic to Spin(8), and it preserves the decomposition

𝔥₃(𝕆)=ℝ𝑒₁⊕ℝ𝑒₂⊕ℝ𝑒₃⊕ξ₁₂(𝕆)⊕ξ₂₃(𝕆)⊕ξ₃₁(𝕆).

For any group element h∈𝐻, let hᵢⱼ∈SO(𝕆) be determined by

h⋅ξᵢⱼ(𝑥)=ξᵢⱼ(hᵢⱼ⋅𝑥).

Then the multiplicative relation between the ξ's above tells us that

(h₁₂⋅𝑥)⋅(h₂₃⋅𝑦)=𝐶h₃₁𝐶⋅(𝑥𝑦),

where 𝐶 : 𝕆→𝕆 denotes conjugation, 𝐶𝑥=𝑥*.

Now we can apply the triality theorem with 𝑔₁=𝐶h₃₁𝐶, 𝑔₂=h₁₂ and 𝑔₃=h₂₃.

But before we do that, consider the realization of Spin(8) inside Cl(𝕆) ≅ End(𝕆⊕𝕆), where the isomorphism is generated by the map 𝐴 : 𝕆→End(𝕆⊕𝕆),

$x\mapsto\begin{pmatrix}0&R(x)\\-R(x^\ast)&0\end{pmatrix}.$

The spin group is generated by 𝑥⋅𝑦∈Cl(𝕆) for unit octonions 𝑥,𝑦, whose image in End(𝕆⊕𝕆) is

𝐴(𝑥)𝐴(𝑦)=diag(−𝑅(𝑥)𝑅(𝑦*),−𝑅(𝑥*)𝑅(𝑦)).

This gives us the left/right-handed half-spin representations of Spin(8).

Returning to our group 𝐻 and the triality theorem, we see now that there is an isomorphism Spin(8)≅𝐻 under which 𝐶h₃₁𝐶 and h₁₂ are the half-spinor representations, and h₂₃ the vector representation of the element h.

(1/2)

@npub1nf4…nqe4 Today I revisited our project on the subgroups ...

2026-01-27T01:53:29+01:00

John Carlos Baez (npub1nf4…nqe4)

Today I revisited our project on the subgroups of F₄. Recall, we were hoping to show that for each 𝕂=ℝ,ℂ,ℍ, all subalgebras of 𝔥₃(𝕆) which are isomorphic to 𝔥₃(𝕂) are F₄-related.

We were almost done with this; in fact, I think I have shown that a subalgebra isomorphic to 𝔥₃(ℍ) is F₄-related to either the standard 𝔥₃(ℍ)⊂𝔥₃(𝕆) or a "twisted" one, where two of the off-diagonal elements are in the orthogonal complement of ℍ.

It remained to see if Spin(8), i.e. the subgroup of F₄ preserving each off-diagonal space, has an element mapping ℍ to ℍ in one of its 8-dimensional representations and to its complement in one of the others. There was the issue of how exactly Spin(8) acts on each of these copies of 𝕆. Both the note of Bryant and the book of Harvey have different ways of writing things down, and the result depends delicately on which representation you have on which copy of 𝕆.

I did a back-of-the-envelope calculation, and now it seems to me that there are actually two different F₄-orbits of 𝔥₃(ℍ)-subalgebras...

The idea was to use the "Triality Theorem" (quoted from Harvey), which states that under the right isomorphism Cl(𝕆) ≅ End(𝕆⊕𝕆), Spin(8) gets mapped to the set

{diag(g₁,g₂) | g₁,g₂∈SO(𝕆), ∃g₃∈SO(𝕆), ∀x,y∈𝕆, g₁(xy)=g₂(x)g₃(y)}.

Then g₁,g₂ are the half-spinor representation, and g₃ the vector representation. Together with the multiplicative relations between the off-diagonal elements in 𝔥₃(𝕆), this suffices to determine (up to automorphism) how Spin(8) acts on those copies of 𝕆.

I will try to explain the calculation later; for now I urgently have to get some food.

I think the question should rather be: why are these particular ...

2025-12-25T15:29:10+01:00

In reply to nevent1q…fz9c
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I think the question should rather be: why are these particular scales called "major" and "minor" in English? This is not the case in every language.

it's an internet meme referring to similarly shaped objects ...

2025-12-23T19:04:22+01:00

In reply to nevent1q…2rsk
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it's an internet meme referring to similarly shaped objects (https://knowyourmeme.com/memes/things-that-look-like-among-us-crewmates), so not a mathematical term - yet.

This amogus surface in ℝ³ has principal curvatures contained ...

2025-12-23T16:40:11+01:00

This amogus surface in ℝ³ has principal curvatures contained in [-1, 1] and is homeomorphic to a sphere, but its enclosed volume is less than that of the unit ball.

https://arxiv.org/abs/2512.19659

Exactly, 𝕆⊕𝕆 is the spinor module, and the action of g(x) ...

2025-12-09T19:27:50+01:00

In reply to nevent1q…khy6
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Exactly, 𝕆⊕𝕆 is the spinor module, and the action of g(x) is by L(x) on the left summand, resp. R(x) on the right summand.

For Spin(9), Bryant writes that it is the subgroup of GL(𝕆⊕𝕆) generated by
\[p(r,x)=\begin{pmatrix}rI_8&CR(x)\\-CL(x)&rI_8\end{pmatrix},\]
where 𝑟∈ℝ and 𝑥∈𝕆 with 𝑟²+|𝑥|²=1, and 𝐶 denotes conjugation in the octonions.

I should also look into Krasnov's calculation (or the ones from Yokota), because I haven't yet fully understood how to arrive at these precise subgroups. But it looks like it's the same as what's going on inside F₄: we fix a point 𝑒∈𝕆ℙ²⊂𝔥₃(𝕆), giving as isotropy group Spin(9), and obtain 𝕆⊕𝕆 as one of the spaces in the Peirce decomposition w.r.t 𝑒. Then we intersect everything with the right 𝔥₃(ℂ)-subalgebra.

What do I mean by "right"? I'm not sure. Is the stabilizer of both a 𝔥₂(𝕆)-subalgebra and a 𝔥₃(ℂ)-subalgebra always conjugate to S(U(2) × U(3))? Or do they have to be "compatible" in some way?

Thanks for reminding me of this! I've tried playing around ...

2025-12-08T23:03:39+01:00

In reply to nevent1q…2ce9
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Thanks for reminding me of this! I've tried playing around with this, but it's kind of hard seeing subalgebras in both the vector and the spinor representations at the same time.

However I think I found the solution in a note of Bryant: https://arxiv.org/abs/2011.05568

He explains that the Clifford algebra Cl(8) is isomorphic to End(𝕆⊕𝕆), and that the subgroup Spin(8) is generated by elements of the form

g(x) := diag(L(x), R(x)),

where x is a unit octonion and L(x), R(x) : 𝕆 → 𝕆 are left-/right-multiplication by x.

Now comes the kicker: If I read it correctly, in the vector representation on 𝕆, the element g(x) gets mapped to L(x)R(x)! So we know exactly how these representations interact with octonion multiplication.

As for the question whether there is an element of Spin(8) preserving ℍ in the vector representation, but swapping it with its complement in a half-spinor representation: we can just take g(ℓ), where ℓ is a unit octonion ortogonal to ℍ, because

L(ℓ)R(ℓ)ℍ = ℍ,
L(ℓ)ℍ ⊥ ℍ.

It's me who is confused. What I wrote earlier didn't make ...

2025-12-05T17:42:24+01:00

In reply to nevent1q…rrsf
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It's me who is confused. What I wrote earlier didn't make sense. (Sorry! It was also late for me.)

I guess what I am really asking is: how are the three 8-dimensions irreps of Spin(8) related to each other and to octonion multiplication? I'm guessing we will have to think about the Clifford algebra Cl(8).

If I recall correctly, Cl(7) is generated by left-multiplications with imaginary octonions. Was there something similar for Cl(8)?

I'm not sure. There are many cube roots of 1, right? Each of ...

2025-12-05T03:33:14+01:00

In reply to nevent1q…3c8g
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I'm not sure. There are many cube roots of 1, right? Each of them would give a different such intertwining map.

Then applying one such map and going back with another would give many different Spin(8)-equivariant automorphisms of an irrep. But by Schur's Lemma, there is only one (up to a constant).

Actually, I think I have a proof of this now. (It's a bit ...

2025-12-04T21:28:47+01:00

In reply to nevent1q…dchx
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Actually, I think I have a proof of this now. (It's a bit technical, but I would be glad if you could check it!)

If you agree with me, that would leave us with two types of 𝔥₃(ℍ)−subalgebras to consider: the standard one, in which case we are done, and the one for which 𝑉₂₃=𝑉₃₁ is the orthogonal complement of ℍ in 𝕆.

I think we are in some sense free to choose which of the three off-diagonal 𝕆s gets identified with which of the three 8-dimensional irreps of Spin(8), right? So lets assume that 𝑉₁₂=ℍ lies in the vector rep, and that 𝑉₂₃ lies in one of the half-spinor reps.

Now the question is: can we find an element of Spin(8) that preserves ℍ in the vector rep, but maps ℍ⊥ to ℍ in the half-spinor rep?

For this it would be useful to write down the linear isomorphism 𝕆 ≅ 𝕆 that intertwines these two representations. Do you happen to know it?

"The group of all rotations of a ball in space, known to ...

2025-12-04T15:13:00+01:00

In reply to nevent1q…4x2v
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"The group of all rotations of a ball in space, known to mathematicians as SO(3), is a six-dimensional tangle of spheres and circles."

🤨

I think it also holds when 𝑎=ℓ, some unit octonion ...

2025-12-03T16:42:39+01:00

In reply to nevent1q…42m9
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I think it also holds when 𝑎=ℓ, some unit octonion orthogonal to ℍ.

Don't we have ℍℓ=ℓℍ and (ℍℓ)⋅(ℍℓ)⊂ℍ?

@npub1nf4…nqe4 Coming back to our discussion about ...

2025-12-03T00:57:22+01:00

John Carlos Baez (npub1nf4…nqe4)
Coming back to our discussion about 𝔥₃(𝕂)-subalgebras of 𝔥₃(𝕆), and whether F₄ acts transitively on them:

I've written up a unified argument why one can always assume, up to the action of F₄, that the subalgebra consists of elements of the form
\[\begin{pmatrix}
\alpha_1&x&z^\ast\\
x^\ast&\alpha_2&y\\
z&y^\ast&\alpha_3\end{pmatrix}\]
with αᵢ∈ℝ, 𝑥∈𝑉₁₂, 𝑦∈𝑉₂₃, 𝑧∈𝑉₃₁. And recall, we still have a freedom of Spin(8) to act.

The linear subspaces 𝑉ᵢⱼ⊂𝕆 satisfy some octonionic relations like 𝑉₁₂⋅𝑉₂₃⊂𝑉̅₃₁, and have the same dimension as 𝕂. We can use the action of Spin(8) to map one of them (say, 𝑉₁₂) to the actual 𝕂⊂𝕆. I worked this out and found that 𝑉₂₃ must be of the form 𝕂⋅𝑎 for some 𝑎∈𝕆 satisfying the following mysterious condition:

(𝕂⋅𝑎)⋅(𝑎*⋅𝕂)⊂𝕂.

For 𝕂∈{ℝ,ℂ}, this is always the case (because then everything associates). But for 𝕂=ℍ it isn't! This might be our saving grace: the stabilizer of ℍ⊂𝕆 in Spin(8) is not transitive on the unit sphere, but we're only considering a subset of the unit sphere!

But what is this subset?

"total bottleneck formations & ℝ⁴" - who let the ...

2025-11-28T08:13:53+01:00

In reply to nevent1q…spa0
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"total bottleneck formations & ℝ⁴" - who let the geometric topologist out??

W from the Chair of Representation Theory at EPFL. ...

2025-11-27T22:09:36+01:00

W from the Chair of Representation Theory at EPFL.

Let me see if my rudimentary understanding of particle physics is ...

2025-11-18T19:52:17+01:00

In reply to nevent1q…r5nh
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Let me see if my rudimentary understanding of particle physics is correct: a "state of a particle" corresponds to a vector in some representation of the gauge group, right?

Particles with different charges are in different irreducible representations of U(1). And particles with different total (iso-?)spins are in different representations of SU(2) (I guess it depends on which SU(2) you pick). But the direction of their spin can be different! E.g. there are several distinct ways a particle can have spin 1/2.

U(1)-irreps are complex 1-dimensional, so there is only one "way" a particle can have a certain charge, but the half-spin irrep of SU(2) is complex 2-dimensional, giving (after projectivizing) a sphere of possible directions for the spin (and the two states of spin up/down are a basis for this irrep).

To use the lingo of a recent video of John, U(1) acts on "quunits", while SU(2) acts on "qubits". Similar, SU(3) acts on "qutrits", corresponding to color charge.

When you enlarge the gauge group, then two particle states that have previously been in different irreps may now lie in the same irrep, and thus be viewed as two states of the same particle!

At least that's how I understand it currently; please correct me in case I'm spreading misinformation!

What was the representation-theoretic interpretation for the six quark flavors?

There is another way to view the sum of all irreps of G₂ as an ...

2025-11-07T05:55:28+01:00

In reply to nevent1q…lsv6
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There is another way to view the sum of all irreps of G₂ as an algebra.

By Peter-Weyl, the space of smooth functions (complex-valued of finite type) on a compact Lie group G decomposes (under the left- and right-translation action) into a direct sum ⨁V*⊗V, where V runs through all complex irreps of G.

If we choose a maximal nilpotent subalgebra 𝔫 ⊂ 𝔤⊗ℂ of the complexified Lie algebra (e.g. the sum of positive root spaces after making the necessary choices), then the subspace of V annihilated by 𝔫 is one-dimensional (precisely the highest weight space). So the space of functions on G which are annihilated by the infinitesimal left-translation action of 𝔫 is pretty much ⨁V* (≅ ⨁V), the sum of all irreps of G!

The algebra multiplication is just pointwise multiplication of functions; and it follows from the construction that the multiplication maps any two irreps into their Cartan product!

What's more, you can also write this algebra as a quotient of the symmetric algebra over {the sum of all fundamental representations of G}.

In particular this works for G = G₂, and we obtain the same algebra as before: 𝔰𝔬(7) ≅ ℝ⁷⊕𝔤₂ is indeed the sum of the two fundamental reps of G₂, and we also quotient out by the same ideal!

So to summarize: some algebra of special functions on the group G₂ is isomorphic to the coordinate ring of the adjoint variety of 𝔰𝔬(7).

What does this mean?

(3/3)

The space ⨁ₖV(k) has a geometric meaning. Think back to V(k) ...

2025-11-07T05:34:58+01:00

In reply to nevent1q…tzax
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The space ⨁ₖV(k) has a geometric meaning. Think back to V(k) as a quotient of the symmetric power Symᵏ𝔰𝔬(7).

It turns out that the direct sum ⨁ₖV(k) can be realized as a quotient of the symmetric algebra over 𝔰𝔬(7), which (after dualizing with an invariant inner product) means a quotient of the ring of polynomials on 𝔰𝔬(7).

The (homogeneous) ideal by which we have to quotient cuts out a projective variety, which is the orbit of a highest weight vector in 𝔰𝔬(7) and is called its 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝑣𝑎𝑟𝑖𝑒𝑡𝑦. This is a consequence of a theorem of Kostant.

As a homogeneous space, we can write this variety as SO(7)/(U(2)×SO(3)).

So ⨁ₖV(k) is the coordinate ring of this adjoint variety.

What do the irreducible representations of G₂ have to do with this particular homogeneous variety?

(2/3)

Take the adjoint representation of the Lie algebra 𝔰𝔬(7) ...

2025-11-07T05:27:23+01:00

Take the adjoint representation of the Lie algebra 𝔰𝔬(7) (i.e. the representation of 𝔰𝔬(7) on itself). This is one of the fundamental representations of 𝔰𝔬(7) (together with ℝ⁷ and the spinor module) - all irreducible representations of 𝔰𝔬(7) can be obtained by taking Cartan products of these three representations.

(The Cartan product of two irreps is the "largest" irreducible subrepresentation of their tensor product.)

Now notoriously the exceptional Lie group G₂ is a subgroup of SO(7). When we restrict an irrep of SO(7) to G₂, it usually decomposes into several irreps of G₂.

Here's a crazy fact that my collaborator Gregor Weingart told me:

Consider the Cartan powers of the adjoint representation 𝔰𝔬(7). Let's denote them with V(k), so that V(0) is the trivial representation, V(1) is 𝔰𝔬(7), V(2) is the representation of Weyl tensors in 7 dimensions and so on. We can define V(k) as a subspace (or quotient) of the symmetric power Symᵏ𝔰𝔬(7).

Fact: Every irreducible representation of G₂ occurs inside exactly one of the V(k), with multiplicity exactly one.

What??

That means that the direct sum ⨁ₖV(k), where k≥0, contains every irrep of G₂ exactly once.

But there's more...

(1/3)

I know U(1), but what's U₁(1)?

2025-10-20T18:39:14+02:00

In reply to nevent1q…3cx7
_________________________

I know U(1), but what's U₁(1)?

Also, a similar argument may work for subalgebras isomorphic to ...

2025-10-19T07:10:10+02:00

In reply to nevent1q…dyey
_________________________

Also, a similar argument may work for subalgebras isomorphic to 𝔥₃(ℝ): now V₁, V₂, V₃ are one-dimensional. With Spin(8) we can map a unit vector in V₁ to 1∈𝕆, leaving us with a Spin(7) - and Spin(7) still acts transitively on 𝑆⁷!

For 𝔥₃(ℍ) is might get more complicated: we have V₁, V₂, V₃ four-dimensional, we use Spin(8) to map V₁ to ℍ⊂𝕆, and then are left with a Spin(4)×Spin(4) ≅ SU(2)⁴.

I think under restriction to this subgroup, the left-/right-handed spinor modules of Spin(8) also decompose into two ℝ⁴s, but corresponding to different pairs of SU(2)s inside SU(2)⁴. And this time Spin(4)×Spin(4) does not act transitively on 𝑆⁷, but just with cohomogeneity one.

To be precise, if we write ℝ⁸=ℝ⁴₁⊕ℝ⁴₂ with 𝑆³ᵢ the unit sphere in ℝ⁴ᵢ, then
𝑆⁷={ax+by | x∈ 𝑆³₁, y∈𝑆³₂, a²+b²=1},
and the action of SO(4)×SO(4) preserves (a,b).

So we need to write down how the splitting interacts with the multiplicative relation between the Vᵢ.

Let's arrange it so that V₁ lies in the vector ...

2025-10-17T23:58:29+02:00

In reply to nevent1q…duqz
_________________________

Let's arrange it so that V₁ lies in the vector representation, and V₂, V₃ in the left-/right-handed spinor modules of Spin(8).

In our recent chat, you had the following idea: use the Spin(8)-action to map the two-plane V₁ to the standard ℂ⊂𝕆, and then see what the relations V₁V₂ ⊆ V₃ give us. Working this leads us to the conclusion that V₂ = V₃ = span{x, ix} for some octonion x with x²∈ℂ.

This means that either x∈ℂ, or x is some purely imaginary octonion. (We can also assume that x has unit norm.) In the first case we are done, but we got stuck at the second.

In fact I think x∉ℂ can happen! But we still have some unused freedom: we can still act by an element of the subgroup of Spin(8) which fix V₁ in the vector representation. This group, at least infinitesimally, is Spin(6)Spin(2), or U(4), and it acts irreducibly on the left-/right-handed spinor representations.

Notably, U(4) acts transitively on the unit sphere 𝑆⁷⊂ℂ⁴. So I think we can use the action of U(4) to map x to 1∈𝕆. And then we are done, since V₁, V₂, V₃ are all mapped to the standard copy of ℂ⊂𝕆, so our subalgebra is mapped to the standard 𝔥₃(ℂ).

Right now I don't have an answer for the exact you posed, but ...

2025-10-14T02:46:30+02:00

In reply to nevent1q…ksu5
_________________________

Right now I don't have an answer for the exact you posed, but are you sure this is the definition of conjugation you want?

Viewing ℂ⊗𝕆 as a composition algebra over ℂ (composition algebras satisfy a bunch of "nice" identities even if they're not associative), it has a canonically defined conjugation given by

(a ⊗ b)* = a ⊗ b*

for a ∈ ℂ, b ∈ 𝕆. With this we have xx*∈ ℂ, and the identity (xx*)y = x(x*y) holds true (see e.g. the book "Octonions, Jordan Algebras and Exceptional Groups" by Springer & Veldkamp).

The characterization through the Pappian property is good to ...

2025-10-08T04:09:35+02:00

In reply to nevent1q…jghz
_________________________

The characterization through the Pappian property is good to know. Reminds me to get on with my projective geometry formalization project...

Writing Pappian property in Jordan-algebraic terms seems involved, but maybe it could be worthwhile creating a "dictionary" between statements in the incidence geometry and statements in the Jordan algebra.

I didn't know about the transitive action of F₄ on Jordan ...

2025-10-07T03:46:57+02:00

In reply to nevent1q…duqz
_________________________

I didn't know about the transitive action of F₄ on Jordan frames, but that seems very useful! I also think the subgroup fixing each of e₁, e₂, e₃ is Spin(8). Here's a sketch of the argument:

The stabilizer of e₁ is Spin(9), and this also preserves the 𝔥₂(𝕆)-subalgebra containing e₂+e₃. I think you have shown somewhere else that this Spin(9) acts on this 𝔥₂(𝕆) by automorphisms, via the double covering Spin(9) → SO(9) = Aut 𝔥₂(𝕆).

Now we are one level lower: the set of points (trace 1 idempotents) in 𝔥₂(𝕆) is 𝕆ℙ¹ ≅ SO(9)/SO(8) ≅ Spin(9)/Spin(8), so the subgroup of Spin(9) fixing in addition the point e₂ (and therefore its complement e₃) must be Spin(8).

There is another thing I would like to understand: whether there is an incidence-geometric way of seeing what (SU(3) × SU(3))/ℤ₃ stabilizes. As before, Spin(9) is the stabilizer in F₄ of a trace 1 idempotent, so the set of points of 𝕆ℙ² can be identified with F₄/Spin(9).

We mused earlier that F₄/(SU(3) × SU(3))/ℤ₃ may be the set of "ℂℙ²s in 𝕆ℙ²" - but what even is a ℂℙ² in 𝕆ℙ²? I want to have a definition like this:

"An 𝕆ℙ¹⊂𝕆ℙ² is the set of all trace 1 idempotents in 𝔥₃(𝕆) incident with a given trace 2 idempotent."
"A ℂℙ²⊂𝕆ℙ² is the set of all trace 1 idempotents in 𝔥₃(𝕆) satisfying [condition]".

Sure, thanks for pointing it out! I've been thinking a bit ...

2025-10-05T22:03:49+02:00

In reply to nevent1q…60vq
_________________________

Sure, thanks for pointing it out!

I've been thinking a bit about 𝔥₃(ℂ) and whether its multiplication table gives some hint about how to give an analogous treatment, but so far unsuccessfully.

Yes, my mistake - 𝑋 and 𝑌 don't need to be idempotents, ...

2025-10-05T21:56:01+02:00

In reply to nevent1q…fxqd
_________________________

Yes, my mistake - 𝑋 and 𝑌 don't need to be idempotents, so you can ignore the 𝑋²,𝑌².

Is there something that makes you think there is such a bundle? ...

2025-10-04T03:54:48+02:00

In reply to nevent1q…man3
_________________________

Is there something that makes you think there is such a bundle?

We could also look for S⁷-bundles over 𝕆ℙ²=F₄/Spin(9) - but I don't immediately see any viable candidate.

There are, however, candidates of rank-8 bundles over F₄/Spin(8), namely the homogeneous vector bundles associated to the 3 irreps of Spin(8). One of these should come from the octonionic Hopf fibration

𝑆⁷=Spin(8)/Spin(7) → 𝑆¹⁵=Spin(9)/Spin(7) ↠ 𝕆ℙ¹=Spin(9)/Spin(8).

Note that there is no Hopf fibration over the Cayley plane, for some reasons.

John Carlos Baez (npub1nf4…nqe4)

Hmm, I don't even know what the Dynkin index would mean for a ...

2025-09-21T09:57:35+02:00

In reply to nevent1q…gths
_________________________

Hmm, I don't even know what the Dynkin index would mean for a non-simple subalgebra - the Killing forms of 𝔣₄ and the subalgebra will certainly not be proportional. There's also this interpretation in terms of π₃ - but I am a lower geometer and have no understanding of π₃ ;)
So, I am sorry to be of no help here...

> What is not so clear to me now is that 𝑎𝑛𝑦 ...

2025-09-20T00:28:07+02:00

In reply to nevent1q…hu3j
_________________________

> What is not so clear to me now is that 𝑎𝑛𝑦 subalgebra isomorphic to 𝔥₂(𝕆) is of the form 𝐴(ℓ).

I have finally written down a rigorous proof. I couldn't find abstract arguments why the trace should be preserved by the embedding, so I use explicit matrix calculations; but the proof is still very short:

https://pschwahn.github.io/assets/op2groups.pdf

There's also a classification of subalgebras inside ...

2025-09-17T18:02:16+02:00

In reply to nevent1q…neu7
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There's also a classification of subalgebras inside exceptional Jordan algebras by Racine (Journal of Algebra 46, pp. 12-21, 1977, https://doi.org/10.1016/0021-8693(77)90391-x). But I don't understand what's happening there. In particular I don't understand whether with unital subalgebra they mean that the subalgebra has a unit, or that the unit of the larger algebra also lies in the subalgebra - the latter is something I would like to rule out for subalgebras of type 𝔥₂(𝕆).

Pursuing the question "what structures in the Cayley plane ...

2025-09-17T17:56:11+02:00

In reply to nevent1q…4eqa
_________________________

Pursuing the question "what structures in the Cayley plane are preserved by (SU(3)×SU(3))/ℤ₃ or the standard model gauge group" is the same as giving a geometric interpretation to the homogeneous space F₄/SU(3)SU(3) and F₄/S(U(3)U(2)) (I'll be a bit lax with notation here and treat groups with the same Lie algebra as the same). Now these spaces show up in several fibrations:

First of all, F₄/S(U(3)U(2)) fibers over both F₄/SU(3)SU(3) and 𝕆ℙ²=F₄/Spin(9), with fiber ℂℙ² and Spin(9)/S(U(3)U(2)), respectively. (What does the latter mean?)

F₄/S(U(3)U(2)) is also a flag manifold of F₄ (i.e. a homogeneous projective variety), and can be viewed as the space of complex two-dimensional isotropic subspaces of the tracesless part 𝔥₃⁰(𝕆⊗ℂ) ⊂ 𝔥₃(𝕆⊗ℂ). See Prop. 3.3 in [Correia, Pacheco, Svensson: Harmonic surfaces in the Cayley plane, https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms.12376] or Prop. 6.7 in [Landsberg, Manivel: On the projective geometry of rational homogeneous varieties, https://ems.press/journals/cmh/articles/382]. (How is this related to structures inside the 𝑟𝑒𝑎𝑙 Cayley plane?)

Next, there's the symmetric space F₄/Sp(3)SU(2) of all ℍℙ²s in 𝕆ℙ². Over it fibers the space F₄/SU(3)SU(2) of all chains ℂℙ²⊂ℍℙ² in 𝕆ℙ² (with fiber Sp(3)/SU(3), the space of all ℂℙ²s in ℍℙ²). But F₄/SU(3)SU(2) is also a circle bundle over the flag manifold F₄/S(U(3)U(2)). (What does the circle mean here?)

So instead of providing answers, I just multiplied the questions. Oops...

Yes, indeed the identiy element should do it - I will meditate on ...

2025-09-16T00:44:53+02:00

In reply to nevent1q…c6ch
_________________________

Yes, indeed the identiy element should do it - I will meditate on this.

In a similar vein, is it clear that every subalgebra isomorphic to 𝔥₃(ℂ) is an F₄-translate of the standard 𝔥₃(ℂ) in 𝔥₃(𝕆)?

We need a way to translate between the geometric picture and the ...

2025-09-15T22:12:44+02:00

In reply to nevent1q…lxjm
_________________________

We need a way to translate between the geometric picture and the Jordan algebra picture.

Let's start with 𝕆ℙ¹⊂𝕆ℙ². Let 𝐴=𝔥₃(𝕆) be the Jordan algebra of octonionic 3×3 matrices. A 𝑝𝑜𝑖𝑛𝑡 in 𝕆ℙ² is a trace 1 idempotent, i.e. 𝑝∈𝐴 with 𝑝²=𝑝 and tr(𝑝)=1. A 𝑙𝑖𝑛𝑒 in 𝕆ℙ², on the other hand, is a trace 2 idempotent. We say a point 𝑝 lies on a line ℓ if 𝑝∘ℓ=𝑝.

The automorphism group F₄ acts transitively on the set of points as well as on the set of lines, and the stabilizer of a given line is isomorphic to Spin(9).

For a given idempotent ℓ, define
𝐴(ℓ) := {𝑋∈𝐴 | 𝑋∘ℓ=𝑋}.
I claim:
1) 𝐴(ℓ) is a subalgebra of 𝐴. (This can be shown using the linearized Jordan identity.)
2) The map ℓ↦𝐴(ℓ) is injective.
3) An automorphism of 𝐴 preserves 𝐴(ℓ) if and only if it fixes ℓ.
4) If ℓ is a line, 𝐴(ℓ) is isomorphic to 𝔥₂(𝕆).

What is not so clear to me now is that 𝑎𝑛𝑦 subalgebra isomorphic to 𝔥₂(𝕆) is of the form 𝐴(ℓ). But already, this works pretty well.

Now we need to play the same game for ℂℙ²s, resp. subalgebras isomorphic to 𝔥₃(ℂ), and relate that with our order 3 inner automorphism. What even 𝑖𝑠 a ℂℙ² inside 𝕆ℙ² - can we express it Jordan-algebraically, like we just did for a line inside 𝕆ℙ²?

Can't wait to see what you find out!

2025-09-14T22:45:33+02:00

In reply to nevent1q…32a9
_________________________

Can't wait to see what you find out!

This description as an inner automorphism is very nice - it makes ...

2025-09-12T21:04:41+02:00

In reply to nevent1q…v269
_________________________

This description as an inner automorphism is very nice - it makes it plausible that an element of G₂ fixes some imaginary octonion j if and only if it commutes with the "rotation" which is conjugation by exp(2πj/3) (or exp(2πj/6), I guess).

Note that this is stronger than just preserving a subalgebra ℂ⊂𝕆 - elements of G₂ can map j ↦ ±j. These form a subgroup of type SU(3)×ℤ₂, and don't all commute with this rotation!

So since (SU(3)×SU(3))/ℤ₃ is the group of Jordan algebra automorphisms that commute with conjugation by exp(2πj/3), does that mean we can think of F₄/((SU(3)×SU(3))/ℤ₃) as the set of "positively oriented" ℂℙ²s in 𝕆ℙ²? It would be nice to understand what exactly each SU(3) factor does.

Indeed, Besse's book on Einstein manifolds lists the ...

2025-09-11T22:33:28+02:00

In reply to nevent1q…dnah
_________________________

Indeed, Besse's book on Einstein manifolds lists the symmetric space F₄/(Sp(3)×SU(2)) as the "set of ℍℙ²s in 𝕆ℙ²".

The subgroup (SU(3)×SU(3))/ℤ₃ is not mentioned there, because this does not define a symmetric space. When you say the answer is "sort of known", what do you mean by that? At a glance, I can't see what structure is stabilized by (SU(3)×SU(3))/ℤ₃.

You say the Standard Model gauge group are those symmetries of an ...

2025-09-09T19:32:08+02:00

In reply to nevent1q…t4vk
_________________________

You say the Standard Model gauge group are those symmetries of an octonionic qutrit that preserve a choice of ℂ⊂𝕆 and a choice of a complex qubit (ℂℙ¹) in the octonionic qutrit (𝕆ℙ²). But in the video you say that we take those symmetries which preserve a ℂ⊂𝕆 and which restrict to symmetries of an octonionic qubit (𝕆ℙ¹⊂𝕆ℙ² resp. 𝔥₂(𝕆)⊂𝔥₃(𝕆)). Are these descriptions equivalent?

Also, now I'm curious what these new avenues are!

Maybe a more general question: there are certainly many ways to cook up the gauge group S(U(2)×U(3)) as the symmetries of some object - the easiest probably being the isotropy group of a 2-dimensional subspace in ℂ⁵ under SU(5). But how do we proceed from there - how do we decide which of these constructions is the best/most interesting/most promising for physics?

Anytime I try to read copyright agreements, my mind just goes ...

2025-09-07T09:26:42+02:00

In reply to nevent1q…h27f
_________________________

Anytime I try to read copyright agreements, my mind just goes blank. Even when I genuinely want to understand them! It's like back in school when I tried (and failed) to study for history exams.

As a consequence, I must admit I usually don't really read nor understand the publishing agreements, and I suspect that it is similar for many colleagues: no matter what the copyright/licensing agreement says, we put our article on arXiv, and update it after it has been peer-reviewed.

So far, I haven't witnessed anyone suffer negative consequences from that.

I think it's a standard fact that Λ²ℂⁿ is irreducible ...

2025-07-28T07:20:46+02:00

In reply to nevent1q…w9fy
_________________________

I think it's a standard fact that Λ²ℂⁿ is irreducible over SU(n), while Λ²ℂ²ⁿ=ℂω⊕Λ²₀ℂ²ⁿ over Sp(n).

If you need a reference for specific branchings anyway, try
"Tables of dimensions, indices, and branching rules for representations of simple Lie algebras" by McKay and Patera. For dimension-independent results, "Representation Theory: A First Course" by Fulton and Harris has many such statements scattered across the book, often as exercises.

Oh yes, thanks! Somehow I messed up in the middle and assumed ...

2025-06-20T07:03:20+02:00

In reply to nevent1q…f7fq
_________________________

Oh yes, thanks! Somehow I messed up in the middle and assumed that |p| is constant. Edited the post - I hope it is correct now.

Excellent informal explanation! This was fun to reverse-engineer, ...

2025-06-19T19:19:32+02:00

In reply to nevent1q…n4wc
_________________________

Excellent informal explanation! This was fun to reverse-engineer, as I wasn't able to look up Killing's Theorem:

Let 𝑔 be the metric, 𝑤 be a Killing vector field, γ a unit speed geodesic. Then I interpret Killing's theorem as stating that 𝑔(𝑤,γ') is constant along γ.

In our case, we need to know that there is a (spacelike) Killing field 𝑤 pointing always in the same direction as the projection 𝑝 of γ'. to the "spatial" part (without restriction, assume 𝑤(γ(0))=𝑝(0)). To me this is not obvious - I assume we are talking about a FLRW spacetime ℝ×Σ with metric

𝑔=−𝑐²𝑑𝑡²+𝑎(𝑡)²𝑔ₛ,

where 𝑔ₛ is some Riemannian metric on Σ with constant curvature. Is is true that geodesics of 𝑔 always project to geodesics of 𝑔ₛ?

Anyway, if yes, then I think we can find such a Killing field 𝑤, which is the lift of a Killing field of 𝑔ₛ times the scale factor 𝑎(𝑡) (?). It remains to observe that

|𝑤(γ(τ))|=𝑎(γ(τ))|𝑤(γ(0))|=𝑎(γ(τ))|𝑝(0)|=𝑎(γ(τ))|𝑝(τ)|

along the geodesic (abusing notation here; 𝑎 is a function on the spacetime which only depends on the 𝑡 coordinate), and thus

𝑤(γ(τ))=𝑎(γ(τ))𝑝(τ)

since 𝑤 is pointwise parallel to 𝑝. So we finally obtain that

𝑔(𝑤,γ')=𝑔(𝑤,𝑝)=𝑔(𝑎²𝑝,𝑝)=𝑎|𝑝|²

is constant along γ. (Isn't there a square missing? Hmm.)

For me, conceptually v ⋅ ∇ is the easiest of the terms - ...

2025-05-20T03:21:32+02:00

In reply to nevent1q…983d
_________________________

For me, conceptually v ⋅ ∇ is the easiest of the terms - it's a directional derivative in the direction of v! On curved spaces, it gets replaced by the covariant derivative.

I lament this symmetry in notation where there should be none: v ⋅ ∇ is a differential operator, while ∇ ⋅ v is a function. (Personally I'd write ∇ᵥ resp. div(v).)

Of course the real pain starts when the cross product and rotation operator start to show up. One could try to reformulate it all in terms of the wedge product/Hodge star operator and codifferential, eliminating the need for entrywise calculation.

For example, ∇×(𝐴×𝐵) can be recast as:

δ(𝐴∧𝐵)=−∑ᵢ𝑒ᵢ⌟∇ᵢ(𝐴∧𝐵)=−∑ᵢ𝑒ᵢ⌟(∇ᵢ𝐴∧𝐵+𝐴∧∇ᵢ𝐵)
=−(∑ᵢ𝑒ᵢ⌟∇ᵢ𝐴)⋅𝐵+∑ᵢ∇ᵢ𝐴⋅𝐵ᵢ−∑ᵢ𝐴ᵢ⋅∇ᵢ𝐵+𝐴⋅(∑ᵢ𝑒ᵢ⌟∇ᵢ𝐵)
=−(∇⋅𝐴)𝐵+(𝐵⋅∇)𝐴−(𝐴⋅∇)𝐵+(∇⋅𝐵)𝐴.

wait, will you actually be in São Paulo/São Carlos?

2025-05-14T01:57:57+02:00

In reply to nevent1q…mylt
_________________________

wait, will you actually be in São Paulo/São Carlos?

Haven't seen this book before. It looks interesting. Seems ...

2025-05-12T00:46:47+02:00

In reply to nevent1q…92r5
_________________________

Haven't seen this book before. It looks interesting. Seems more aimed at analysts than geometers, doesn't seem to talk much about spectra of Dirac operators, and doesn't touch on the Peter-Weyl theorem, connections with torsion, Killing spinors...

What confuses me, and I think I might be misunderstanding ...

2025-05-05T00:21:07+02:00

In reply to nevent1q…36u6
_________________________

What confuses me, and I think I might be misunderstanding something here, is the following: the "problem" of the "never-ending, incomplete ∛7" is supposed to be a justification for forgetting about radicals and instead replacing them by power-series solutions - but aren't they also just as never-ending and incomplete?

Wildberger seems to have no problems with reasoning about "non-existent" objects (after all, the actual paper is solid and interesting math), but then why does he start out by trashing radicals?

Glad I could help. Can't wait to find out what you're up ...

2025-04-12T03:52:31+02:00

In reply to nevent1q…ucqx
_________________________

Glad I could help. Can't wait to find out what you're up to that involves the Dirac operator!

The relation between Levi-Civita and Cartan connection is ...

2025-04-11T18:02:23+02:00

In reply to nevent1q…se62
_________________________

The relation between Levi-Civita and Cartan connection is essentially contained in Kobayashi&Nomizu's "Foundations of Differential Geometry", vol.2, the chapter on homogeneous spaces. But it is a bit difficult to get the information out of there.

The Cartan connection is a special case of the "canonical connection" of a reductive homogeneous space (i.e. a homogeneous space G/H with a choice of Ad(H)-invariant decomposition 𝔤=𝔥⊕𝔪; in our case, 𝔤=𝔰𝔲(2) and 𝔥=0). See for example §2.2 of https://arxiv.org/abs/math/0606705.

The canonical and LC connections are both invariant connections and can be characterized by something called the Nomizu map Λ: 𝔪→𝔤𝔩(𝔪). For the canonical connection Λ=0, so the Nomizu map of the Levi-Civita connection is exactly the difference ∇ᵍ-∇.

The upshot, i.e. the formula for the difference, is also contained in this article of mine, §2.2: https://arxiv.org/abs/2304.10607
(It was +, not -. Oops.)
(Since 𝔥=0, we can ignore the projection to 𝔪.)
And then it remains to lift this to the spinor bundle.

Here's an idea. We can only identify the sections of the ...

2025-04-11T04:04:39+02:00

In reply to nevent1q…jfm2
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Here's an idea. We can only identify the sections of the spinor bundle on S³ with smooth ℍ-valued functions by using the Lie group structure. The fibres get identified by means of the a flat connection on S³ sometimes called the 𝐶𝑎𝑟𝑡𝑎𝑛 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛 (let's denote it by ∇). And the directional derivative on $C^\infty(S^3,\mathbb{H})$ corresponds to covariant differentiation under this Cartan connection.

But this is not the same as the Levi-Civita connection ∇ᵍ, which features in the definition of the Riemannian Dirac operator:
\[D^g\psi=\sum_ie_i\cdot\nabla^g_{e_i}\psi.\]
Here ⋅ is Clifford multiplication, and (𝑒ᵢ) some orthonormal frame - e.g. i,j,k. Kronheimer's Dirac-like operator is more like
\[D\psi=\sum_ie_i\cdot\nabla_{e_i}\psi.\]
(I don't know why he multiplies on the right, maybe this in the definition of Clifford multiplication.)

Anyway, the difference between these connection should be (on vectors, at the base point, up to sign error)
\[\nabla^g_XY=\nabla_XY-\frac12\mathrm{ad}(X)Y.\]
In the spinor representation this becomes
\[\nabla^g_X\psi=\nabla_X\psi-\frac12X\cdot\psi.\]
So the difference of the Dirac operators is
\[D^g\psi-D\psi=-\frac12\sum_ie_i\cdot e_i\cdot\psi=-\frac12\sum_ie_i^2\cdot\psi=\frac32\psi.\]
There we have our 3/2!

Ah, got it, thanks! I remember, I've seen the quantum ...

2025-04-10T06:27:18+02:00

In reply to nevent1q…2va0
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Ah, got it, thanks! I remember, I've seen the quantum harmonic oscillator before.

In analogy to that, we could take the potential function to be the square of the geodesic distance to the "origin" 𝑜, which would be the point in M representing the unit sphere, i.e. the identity matrix. So
𝑉(𝑞)=𝑑(𝑞,𝑜)².

We actually have some free parameters here - there is a 2-dimensional space of invariant Riemannian metrics on GL(3,ℝ)/O(3), allowing us to control the volume-preserving and the shape-preserving directions separately.

To construct eigenfunctions for the Hamiltonian |p|²+V, maybe one could foliate M by geodesic spheres around 𝑜, and then use the harmonics of those 5-spheres. But these will in general not be round... they only have a SO(3)-invariant metric (which is not homogeneous, but of cohomogeneity two I think). That'll yield some complicated PDE system... yuck.

I'm fairly new to this quantization thing. The points (q,p) ...

2025-04-10T02:23:18+02:00

In reply to nevent1q…4la3
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I'm fairly new to this quantization thing. The points (q,p) in T*M should correspond to operators on L²(M)? (Just replacing ℝ⁶ in your blog article with the manifold M=GL(3,ℝ)/O(3).)

So if the Hamiltonian happens to be something like |p|², we're looking at eigenfunctions of the Laplacian on M, right?

I've yet to properly learn how harmonic analysis works on noncompact symmetric spaces. But I know there is something like a Plancherel formula, and one can have both discrete and continuous spectrum.

Ah yes, the symmetric space GL(3,ℝ)/SO(3).

2025-04-10T01:50:28+02:00

In reply to nevent1q…0pn3
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Ah yes, the symmetric space GL(3,ℝ)/SO(3).

Thanks for sharing this.

2025-04-10T01:34:31+02:00

In reply to nevent1q…ms7f
_________________________

Thanks for sharing this.

Just learned that the Mathematical Congress of the Americas 2025 ...

2025-04-10T01:13:22+02:00

Just learned that the Mathematical Congress of the Americas 2025 in taking place in Miami, Florida, of all places. Perhaps not the most auspicious location nowadays.

Ah very nice! That's in line with what I figured. So for ...

2025-04-07T00:17:12+02:00

In reply to nevent1q…ctsh
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Ah very nice! That's in line with what I figured.

So for people whose minds ARE corrupted by representation theory (like me), we have a tensor product of SU(2)-representations 𝐿²(ℝ³)⊗𝑉. If we're looking only at functions which are determined by their restriction to the unit sphere, we get the representation 𝐿²(𝑆²)⊗𝑉, which by Peter-Weyl is completely reducible into finite-dimensional pieces. And for "spin 0"-particles we can take 𝑉 to be the trivial representation, and then it indeed descends to a representation of the group SO(3).

I was confused about the difference between quantum numbers and the magnitude of the angular momentum, thanks for clearing that up.

The problem is that protons and neutrons are both fermions, so ...

2025-04-06T22:44:33+02:00

In reply to nevent1q…sm9l
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The problem is that protons and neutrons are both fermions, so they have spin 1/2. But we already assume the spin of the pairs to cancel out.

The number 0 or 2 apparently refers to some other kind of spin (John said "orbital angular momentum")? So I assume it is referring to the spin of the SO(3)-rotation representation.

In that case, even spin would refer to wave functions which are (point-)symmetric, or even, around the origin. So I am confused by the phrase "antisymmetric state".

Just to make sure, the "total angular momentum" is the Casimir operator of the SO(3)-representation, right? If the spin is 𝑠, then the total angular momentum should be 𝑠(𝑠+1) iirc. So for bosons, the total angular momentum is always even.

Phase transitions in the atomic nucleus? That sounds funky! ...

2025-04-06T22:22:47+02:00

In reply to nevent1q…x6j8
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Phase transitions in the atomic nucleus? That sounds funky!
Thanks for digging this one up, even though I understand virtually nothing.

I recognize some of the symbols here - O(6), U(5), SU(3) - as ...

2025-04-06T02:29:12+02:00

In reply to nevent1q…yvls
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I recognize some of the symbols here - O(6), U(5), SU(3) - as subgroups of U(6). But what in the world are E(5) and X(5)?

It might become much worse than under Bolsonaro. As far as I ...

2025-03-18T19:16:24+01:00

In reply to nevent1q…ywuk
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It might become much worse than under Bolsonaro. As far as I know, in Brazil a large part of the funding for science is state-based, not federal. For example, in São Paulo, a certain portion of tax revenue is guaranteed to go into research funding by the state constitution. No Bolsonaro can shut this down!

What an icon. I didn't know you were connected to him in this ...

2025-03-13T05:54:52+01:00

In reply to nevent1q…d30p
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What an icon. I didn't know you were connected to him in this way! The world really is small.

Everywhere I go researchwise, it somehow seems that Bertram ...

2025-03-13T01:23:13+01:00

Everywhere I go researchwise, it somehow seems that Bertram Kostant has already been there and done that... maybe I should read his work more closely.

Regarding quantum field theory, would you say this is getting ...

2025-01-03T23:17:41+01:00

In reply to nevent1q…s5n3
_________________________

Regarding quantum field theory, would you say this is getting better? Like, is there a community effort to lay out and teach the foundations in a rigorous way?

I'm asking because I've also recently tried to get into it, and I'm a bit dissatisfied with the literature - so I wonder whether the problem is that no one KNOWS how to make things rigorous, or that no one CARES to do so.

From what I've gathered, the level of mathematical rigor there has stagnated since Wigner's time, but I'd love to be corrected...

Sorry for the imprecision - I didn't mean "bound" in ...

2024-12-28T03:37:17+01:00

In reply to nevent1q…yy28
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Sorry for the imprecision - I didn't mean "bound" in the traditional sense. I was thinking in analogy to a metallic conductor that we know from school, where the (valence) electrons are not bound to individual atoms, but idle around inside the metal for all practical purposes - until a current happens. Sorry again, I tend to have rather naive mental models for some physical processes when I don't understand the math behind it.

With "ripped apart" I guess I meant a situation in which we have to drop the assumption of 𝐄=0, because the Lorentz force separates charges faster than the fluid/plasma can conduct. I was wondering if this ever happens macroscopically.

thanks for the answers! The lecture notes were also a bit ...

2024-12-27T19:10:50+01:00

In reply to nevent1q…47rx
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thanks for the answers! The lecture notes were also a bit enlightening. So what I've gathered from this is that the electrons should be thought of as bound to the fluid/plasma, but being able to freely move inside it. I was worried about the changes in momentum by the Lorentz forces on ions and electrons cancelling out, but now I see that the surviving part, which pushes on the fluid as a whole, comes from the motion of electrons *within/relative to* the (electrically neutral) fluid.

Still, I am wondering if there are scenarios where a separation of charges *does* become important - i.e. the charges get ripped apart faster than they can recombine? Maybe at very high velocities/magnetic fields? Although I guess one would have to take relativistic effects into account then...

maybe a stupid question: when we talk about plasma, we mean a ...

2024-12-27T07:55:03+01:00

In reply to nevent1q…sz7d
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maybe a stupid question: when we talk about plasma, we mean a bunch of atomic nuclei and unbound electrons, together electrically neutral, correct?

So, when one of the terms describing its motion comes from the Lorentz force - what do we mean exactly? Does it act just on the positively charged part (ions)? Wouldn't it act in the opposite direction, with much greater acceleration, on the free electrons, somehow separating them from the ions?

On the other hand the electric field is assumed to be zero everywhere, so this means no separation of charge - does that mean this effect can be ignored somehow, and why? Can we show that the charge cancels out much faster than the plasma moves? But when the ions and electrons come back together and the charge cancels out, shouldn't the same happen for the (opposite) momenta imparted by the Lorentz force on both the ions and the electrons, thereby nullifying the motion?

I'm probably confusing a lot of things here, but I don't even know where to start untangling this.

I wonder if there exists a fluid-dynamical framework for Alfvén ...

2024-12-25T06:25:07+01:00

In reply to nevent1q…t4wa
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I wonder if there exists a fluid-dynamical framework for Alfvén waves. Like, is there a wave equation (PDE) associated to them? The wikipedia article doesn't say anything about it.

Confusingly, there's also 2-connected, which means that the ...

2024-12-15T15:13:47+01:00

In reply to nevent1q…455j
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Confusingly, there's also 2-connected, which means that the homotopy groups πₖ vanish up to k=2. (But that's certainly not what's meant here.)

It's finally here! Sandwich operators and Einstein ...

2024-12-13T16:41:27+01:00

It's finally here!

Sandwich operators and Einstein deformations of compact symmetric spaces related to Jordan algebras, with Stuart J. Hall and Uwe Semmelmann.

https://arxiv.org/abs/2412.08770

Many of my posts in the past months actually came from working on this project; for example, the one about Jordan algebras and projective spaces: https://mathstodon.xyz/@pschwahn/113388126188923908

In the end, we did not address the question of what sort of incidence geometries one gets starting from different power-associative algebras, as the paper was already getting too long; we only reviewed the formally real Jordan algebras. Still, this didn't stop me from cross-posting it in math.RA, and I want to pursue this further at some point.

John Carlos Baez (npub1nf4…nqe4)

That's true at least! Is there any publication you could ...

2024-11-14T01:16:20+01:00

In reply to nevent1q…6mwm
_________________________

That's true at least! Is there any publication you could point me to where this expression appears with idempotents?

I need suggestions on how to name this operator! In an upcoming ...

2024-11-13T18:31:38+01:00

I need suggestions on how to name this operator!

In an upcoming paper, we take a matrix/a vector space endomorphism 𝐴∈End(𝑉), together with a subspace of End(𝑉) closed under commutators with some orthonormal basis (𝑋ᵢ), and create the mapping

End(𝑉)→End(𝑉), 𝐴↦𝑋ᵢ𝐴𝑋ᵢ.

(More generally, we can map 𝐴↦π(𝑋ᵢ)𝐴π(𝑋ᵢ) where π: 𝔤→𝔤𝔩(𝑉) is any Lie algebra representation and (𝑋ᵢ) is an ONB of 𝔤. This map is then related to quadratic Casimir operators of 𝔤.)

Since the endomorphism 𝐴 gets 'sandwiched' between the 𝑋ᵢ, we have semi-creatively taken to call this mapping 𝑠𝑎𝑛𝑑𝑤𝑖𝑐h 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟. Can anyone think of a better name? Or has anyone seen this thing in action before?

Reading the article of Viro, it seems that hyperfields are ...

2024-11-06T06:16:27+01:00

In reply to nevent1q…ul6z
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Reading the article of Viro, it seems that hyperfields are somehow more flexible than fields - they allow for more constructions. As a category theory amateur this makes me wonder - how do the categories of hyperrings/-fields behave? Do they have limits and colimits, for example?

Sure, I will! Although at this point, Jordan algebras appear only ...

2024-10-31T01:28:52+01:00

In reply to nevent1q…efax
_________________________

Sure, I will! Although at this point, Jordan algebras appear only tangentially - in both a figurative and a literal sense ;)

For example, the automorphism group of a Jordan algebra $J$ ...

2024-10-29T02:50:29+01:00

In reply to nevent1q…0rmj
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For example, the automorphism group of a Jordan algebra $J$ acts transitively and by isometries on the projective space $\mathbb{P}_J$ (and we get all isometries in this way).

Now, for the examples we've seen, there is a sneaky way to let a larger group act on $\mathbb{P}_J$, not by isometries but at least by 𝑐𝑜𝑛𝑓𝑜𝑟𝑚𝑎𝑙 (i.e. angle-preserving) transformations. And there is an even sneakier way to extend this group action to an action of the complexified group on the complexified $J^\mathbb{C}$. The orbit of $\mathbb{P}_J$ under this action is - you guessed it - a "complexified" version of $\mathbb{P}_J$. This is the way to define the projective spaces $(\mathbb{K}\otimes\mathbb{C})\mathbb{P}^{n-1}$ for $\mathbb{K}=\mathbb{R,C,H,O}$!

Having done this, we may consider the space of "$\mathbb{KP}^{n-1}$'s inside $(\mathbb{K}\otimes\mathbb{C})\mathbb{P}^{n-1}$" (made rigorous by writing it as a quotient of groups). I am currently working on a project involving these, and exploring these relations is actually quite fun and satisfying. In fact, we started from the list of these spaces written as quotient because they share a certain other property, and I only recently recognized the relation to Jordan algebras!

(3/3)

This allows one to define generalized "spaces of lines ...

2024-10-29T02:37:52+01:00

In reply to nevent1q…um7t
_________________________

This allows one to define generalized "spaces of lines through the origin" (projective spaces): We identify a one-dimensional subspace of a vector space with a projection (=idempotent) of trace one. Then we forget about the subspace, and consider for any Jordan algebra $J$ the subset
\[\mathbb{P}_J=\{x\in J\,|\,x^2=x,\ \operatorname{tr}x=1\}.\]
These will be the set of points of our "projective space" $\mathbb{P}_J$, and the space of lines can easily be defined as the set of idempotents of trace $2$. Just as it should be for projections in vector spaces, we say that the point (one-dim. subspace) $p\in\mathbb{P}_J$ lies on the line $\ell$ if $p\cdot\ell=p$.

Applying this construction to $J=\{\text{Hermitian $n\times n$-matrices}\}$ yields exactly the projective spaces $\mathbb{RP}^{n-1},\mathbb{CP}^{n-1},\mathbb{HP}^{n-1}$ - and even the Cayley plane $\mathbb{OP}^2$, for which the usual way to define projective planes fails!

These examples are neatly documented in John Carlos Baez (npub1nf4…nqe4) 's article "The Octonions". There's also a Jordan algebra structure on $\mathbb{R}^n\times\mathbb{R}$, called the 𝑠𝑝𝑖𝑛 𝑓𝑎𝑐𝑡𝑜𝑟, and there the construction gives the so-called h𝑒𝑎𝑣𝑒𝑛𝑙𝑦 𝑠𝑝h𝑒𝑟𝑒 $S^{n-1}$ in Minkowski space $\mathbb{R}^{n,1}$.

Now I'm wondering what happens when we plug other power-associative algebras into the constructions? What is needed to prove the axioms of a projective space, and what kind of geometric objects do we get? There is also a lot to be said about actions of certain symmetry groups here...

(2/n)

Today I learned that ANY finite-dimensional unital algebra which ...

2024-10-29T02:31:39+01:00

Today I learned that ANY finite-dimensional unital algebra which is power-associative (that is, powers $x^k$ are well-defined) carries a polynomial $\det$, called the 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡, and a linear form $\operatorname{tr}$, called the 𝑡𝑟𝑎𝑐𝑒, with the property that for all "regular" elements $x$ (so "almost all" in a sense),
\[\lambda\mapsto\det(x-\lambda\cdot 1)\]
is the minimal polynomial of $x$ -- and $-(\operatorname{tr}x)\lambda$ is its linear term!

This way of viewing the determinant is conceptually quite different from the usual geometric way (via exterior powers of linear maps). But it pays to have it: in particular, this statement applies to all 𝐽𝑜𝑟𝑑𝑎𝑛 𝑎𝑙𝑔𝑒𝑏𝑟𝑎𝑠 (named after Pascual Jordan), and in particular particular to the algebras of Hermitian $n\times n$-matrices over $\mathbb{R,C,H}$ or $\mathbb{O}$ (the latter only works for $n=3$) with the symmetrized matrix product as multiplication.

(1/n)

On (differentiable) manifolds, lots of things that we know and ...

2024-09-26T15:51:44+02:00

In reply to nevent1q…qnum
_________________________

On (differentiable) manifolds, lots of things that we know and love from analysis come for free: smooth functions, differentials, coordinates, vector fields, exterior calculus, integration and Stokes' theorem.

But there is no canonical way of identifying tangent spaces at different points! People like Einstein thought long and hard about this - he called it 𝑡𝑒𝑙𝑒𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑖𝑠𝑚.

The thing that may do the job is an additional structure called a 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛. A connection allows you to "parallel transport" vectors along a curve (and thus identify the tangent spaces across the curve), or equivalently, to differentiate vector fields - this aspect is called a 𝑐𝑜𝑣𝑎𝑟𝑖𝑎𝑛𝑡 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 and usually denoted by something like ∇ (nabla).

However, there is an important subtlety here: How different tangent spaces are identified using parallel transport depends on the chosen path! This gives rise to concepts like torsion, curvature and holonomy.

The 𝑡𝑜𝑟𝑠𝑖𝑜𝑛 of a connection may be thought of as measuring the amount of "twisting around" when you parallel transport a vector in a given direction.

The 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒, on the other hand, measures what happens to a vector when you parallel transport it along an infinitesimal CLOSED curve. The h𝑜𝑙𝑜𝑛𝑜𝑚𝑦 𝑔𝑟𝑜𝑢𝑝 is defined as the group of linear transformations of the tangent space that are obtained from parallel transporting along closed curves.

The relation between curvature and holonomy is made explicit in the famous Ambrose-Singer Holonomy Theorem. As it turns out, Einstein's dream of teleparallelism - i.e. no dependence on the chosen curve - can only come true if you have a connection which is 𝑓𝑙𝑎𝑡, i.e. with zero curvature.

(2/n)

Turns out you were right: I just learned that conjugacy classes ...

2024-09-25T15:57:41+02:00

In reply to nevent1q…n9zm
_________________________

Turns out you were right: I just learned that conjugacy classes of three-dimensional simple subalgebras (TDS) of a simple Lie algebra 𝔤 are in 1-1 correspondence with orbits of nilpotent elements in 𝔤.
[Cor. 3.7 in B. Kostant, Amer. J. Math. 81, No. 4 (1959), pp. 973-1032]

There is always an open such orbit called the "principal orbit", and the corresponding subalgebras are also called "principal". In the case of 𝔤₂, the principal TDS are precisely those which act irreducibly on ℝ⁷!

So 𝔰𝔬(3)ᵢᵣᵣ is indeed generic (i.e. principal). The principal orbit is the 11-dimensional homogeneous space 𝐺₂/SO(3)ᵢᵣᵣ. And this might indeed be the best way of thinking about this subalgebra.

Addendum: The conjugacy class of 𝔰𝔬(3)ᵢᵣᵣ itself is 𝐺₂/N(SO(3)ᵢᵣᵣ), where N is the normalizer of a subgroup. This has to have the same dimension, but I'm not sure if N(SO(3)ᵢᵣᵣ)=SO(3)ᵢᵣᵣ - because I'm not aware of a study of maximal sub𝑔𝑟𝑜𝑢𝑝𝑠 of 𝐺₂ (which may be disconnected). If you happen to stumble across such a reference, please let me know!

I'm reluctant to make a habit out of advertising my papers ...

2024-09-24T13:15:05+02:00

I'm reluctant to make a habit out of advertising my papers here. But I'm incredibly happy to finally have this preprint out:

Submersion constructions for geometries with parallel skew torsion (with Andrei Moroianu),
https://arxiv.org/abs/2409.14421

I might toot something explaining the background later!

that's a nice way of thinking about them that I hadn't ...

2024-09-09T13:55:13+02:00

In reply to nevent1q…esx5
_________________________

that's a nice way of thinking about them that I hadn't considered before!

great post, great explanation! I must add, the space of bivectors ...

2024-09-08T10:09:30+02:00

In reply to nevent1q…d933
_________________________

great post, great explanation! I must add, the space of bivectors becomes isomorphic to 𝔰𝔬(p,q), not End(V).

Another comment: you make the argument that multivectors have more geometric intuition than k-forms. But I think this intuition cannot be stretched too far (I'm sure you know all this, but I gotta be pedantic): a (simple) bivector 𝑎∧𝑏 is NOT the parallelogram spanned by 𝑎 and 𝑏, but rather some sort of "oriented amount" of the plane spanned by the two vectors. The parallelogram is only one way to represent 𝑎∧𝑏, but there are many other parallelograms (with same signed area) that represent the same bivector. The bivector 𝑎∧𝑏 does not know anything about the angles and lengths of 𝑎,𝑏.

It becomes much more unintuitive when you ADD different bivectors - and you'll have to, because not all bivectors are simple!

Passing to 2-forms is not very hard conceptually: think of them as being "measuring devices" for bivectors in the sense that they are dual to them. In fact, in order to define a 2-form it suffices to define it on simple bivectors, so in my mind this makes them actually MORE geometrically intuitive than bivectors!

However I strongly agree with the point that we should encourage to identify bivectors, 2-forms and infinitesimal rotation once we have a metric available - this really does help! But as a Riemannian geometer I might be biased here.

It seems this maximal SO(3) is more mysterious than anticipated! ...

2024-05-15T19:28:20+02:00

In reply to nevent1q…2ynd
_________________________

It seems this maximal SO(3) is more mysterious than anticipated!

Looking at Dynkin's paper (Table 16), there are four conjugacy classes of 3-dimensional subalgebras of 𝔤₂. Two of those are the 𝔰𝔲(2) factors of the 𝔰𝔬(4) subalgebra, one belongs to the SO(3)⊂SO(4) acting reducibly on Im 𝕆, and one belongs to the maximal subgroup SO(3)ᵢᵣᵣ.

Each 𝔰𝔬(3) subalgebra contains what Dynkin calls a "defining vector", that is, an element in a fixed maximal torus of 𝔤₂. Not just any element of the torus can be a defining vector: Dynkin gives the possible coordinates in the basis of simple roots.

From this point of view all 𝔰𝔬(3) subalgebras are on equal footing, so I'm not sure whether one can speak of "generic" objects here. At a glance it looks like they all have the same degrees of freedom, namely a choice of maximal torus in 𝔤₂ plus a choice of simple roots.

Yes, but it this the right representation? Perhaps you missed my ...

2024-05-15T17:18:13+02:00

In reply to nevent1q…p3h7
_________________________

Yes, but it this the right representation? Perhaps you missed my earlier reply, but I think this representation is not irreducible. Aren't span{i,j,ℓ}, span{k,iℓ,jℓ} and span{kℓ} invariant subspaces?