Gerenuk

I'm a data scientist and physicist who likes dabbling with new ideas for #artificialintelligence 🤖 and #quantumgravity Ψ. Occasionally, I join programming game competitions 🚀 . I am curious about mathematical concepts or novel ideas in general.

#quantum #quantumphysics #science #physics #mathematics #fedi22

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gerenuk_at_mathstodon.xyz@momostr.pink

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Last Notes

2026-02-09 07:07:03 UTC

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Not sure if I had made my question clear enough, but now I've posted a clearer version on https://physics.stackexchange.com/questions/848885/why-do-spinor-bilinear-covariants-satisfy-this-simple-identity .. unfortunately without an answer. The question is about bilinear covariants and Clifford algebras. Interestingly, the AI gave a partial answer to it.

2025-12-08 18:56:09 UTC

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I don't see it explicitely mentioned, but I wish they had told us in physics class, that all of Maxwell's relations follow from $dT dS=dp dV$ https://en.wikipedia.org/wiki/Maxwell_relations#Derivation_based_on_Jacobians

2025-10-23 12:26:59 UTC

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I've watched quite a few hours of lecture on the standard model and also representation theory. Not a full course, bit still quite a bit. For me - from the learners point of view - it would impossible to understand which set of matrices (3,1,-2/3) stands for and why exactly these matrices represent those 3 precise numbers and not others. I could probably parrot some buzz words I heard in lectures, but even the most basic university-style test question on an actual calculation would make me stumble. I could only pretend like a chatbot.
Maybe at least for one example you could show the matrices for people who don't have a grasp on representation theory?

2025-10-23 08:01:48 UTC

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Great to see a lecture about the standard model from you! Looking forward to the next videos.
It would be nice of some of the following would make it into the videos (not sure if I can describe it correctly):
* what do these representation labels ((3,1,-2/3), (1,2,-1), ...) of how particles transform under the groups mean exactly if written in terms as concrete matrices? writing it out as matrices (generators?) with concrete numbers would be enlightening to an undergrad. conversely, if you have a matrix group, how do you find out if it matches the labels.
* where are the free parameters in the standard model exactly (Weinberg angle?) and what are the other numbers in the matrix representations not free parameters.
* a treatment of the weak interaction (just the classical Lagrangian) in terms of concrete matrices would be nice, in order to grasp what the math means. similarly, to what has been presentated about the strong force here: https://youtu.be/-K7P5WQE1X0?si=LMkmrLFr4gxCGd4t&t=3163 it may seem silly to write it all out, but seeing it once makes not hidden behind math terminology.

2025-10-18 06:13:01 UTC

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I haven't thought about it much, but here is some algebra from the book:
https://media.mathstodon.xyz/media_attachments/files/115/393/689/392/951/468/original/c695b6be30f0ab64.png

2025-10-17 13:09:37 UTC

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Ah right. Thanks!
In that sense, $(xx^*)y=x(x^*y)$ would be true if one uses the octonion conjugate on bioctonions.
I asked an AI for more identities. It said the Moufang laws follow from the alternative property (through bumping in the associator). And it suggested that it is conjectured that the possible multilinear identities do not change beyond Sedenions. But not sure if the AI is right.

2025-10-15 11:23:11 UTC

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I can confirm that the Moufang laws work for bioctonions, but my instinct is that any law the works for octonions would work for bioctonions?

2025-10-15 09:20:08 UTC

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Oh, how disappointing. It's a general equality which would work for any bioctonion replacing x*? I forgot. I actually knew about the associator last time I was investigating octonions for new identities with brute computer force. I couldn't find any.
Are there any interesting identities for octonions or bioctonions beyond the alternative property?
I've learned the you can extend most identities to power laws with https://math.stackexchange.com/a/5010791/15588
Maybe there are no other identities for complex conjugate beyond what's already known?
Btw, for the norm you would need to octonion-conjugate (i.e. don't touch the complex numbers)?!

2025-10-14 19:19:47 UTC

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yes

2025-10-14 14:18:50 UTC

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But here is a little amendment equation, which I believe holds:
\[(xx^*)y-x(x^*y)=(yx)x^*-y(xx^*)\]
Can someone verify? Or is that trival?

2025-10-14 14:05:00 UTC

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Here is my way to put it as someone who implemented C-D rather than thinking deeply about the math:
C-D has octonion conjugation
\[\begin{aligned}(a,b)\cdot (c,d)&=(ac-d^*b,da+bc*)\\(a,b)^*&=(a^*,-b)\end{aligned}\]
However, once you introduce complex numbers, you also need to get their conjugate right - at least if you naively build bioctonions from the C-D construction complex->bicomplex->biquat->bioct. This means
\[(a,b)^\dagger=(a^\dagger,b^\dagger)\]
And the conjugation used if the OP post is a combination of both
\[x^{*\dagger}\]
This in particular makes a difference for the C-D rule
\[(a,b)^{*\dagger}=(a^{*\dagger},-b^\dagger)\]
All this difficulty arises from my approach building bioctonions with the C-D construction starting from the complex number, rather than thinking about the tensor product.
But I believe now, I got the conjugates right, and I can symbollically check anything you would like to know.

2025-10-14 08:23:18 UTC

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Oh right. Sorry that I missed that. My C-D worked before, but after introducing a new conjugate I made a mistake. Apparently there is a 3rd type of conjugation that is needed: conjugate only the complex numbers. It's quite insightful to implement that to see how the math works. Here is the fixed result. Is that as expected?
https://media.mathstodon.xyz/media_attachments/files/115/371/550/999/835/326/original/724771c02e6d689b.png

2025-10-14 07:16:12 UTC

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I've changed the conjugation the following way: the explicit conjugation affects the complex numbers, the conjugation in the C-D construction does not.
Is the following calculation the result you would expect?
(Xi are complex numbers, e1-e7 is the octonion basis, x.c is the conjugate)
This is now alternative, unless you use conjugation.
https://media.mathstodon.xyz/media_attachments/files/115/371/284/047/563/124/original/70f647b3302a7ef9.png

2025-10-13 22:48:33 UTC

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Hmm. 1) seems to indicate that I have the same construction, but I believe it's not alternative (in any way; except for the one equation I mentioned). Please let me know if you can confirm. But I had the impression that @npub1nf4…nqe4 wrote it should be alternative?!

2025-10-13 22:27:48 UTC

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To understand the convention and check my code, could you please tell me:
* what is ((1+i) e1)* ?
* is the algebra alternative?
* is z z* and/or z+z* always a real number?

2025-10-13 21:46:51 UTC

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What is the conjugate of i e1?

2025-10-13 21:38:11 UTC

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I'm not using pairs. I just put complex numbers at the beginning of CD. But these complex numbers would also be conjugated according to CD.

2025-10-13 21:29:06 UTC

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I've implemented C-D and I conjugate the complex numbers, but you and the AIs tell me that bioctonions are alternative, but my code doesn't show that. Are they really alternative?

2025-10-13 21:00:14 UTC

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Hmm, given that my code does not produce alternative bioctonions, I guess the question is: what is the complex structure? does your conjugate also conjugate the complex numbers?

2025-10-13 20:44:42 UTC

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Is it a non-trivial statement that defining
𝑧=(𝑥𝑥*)𝑦−𝑥(𝑥*𝑦)
yields
𝑧+𝑧*=0 ?

2025-10-13 20:35:12 UTC

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Ah, sorry! I had a typo in my variables (I had y=x). I'm afraid I have to inform you, the equation does not hold:
https://media.mathstodon.xyz/media_attachments/files/115/368/767/156/395/365/original/09667addbb37bbe9.png

2025-10-13 20:08:52 UTC

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To me it looks like bioctonions are just octonions with complex numbers as coefficients? In that case one could just implement the C-D construction and see if the equations works?
I did this as an exercise, and the equation seems to hold. It doesn't even matter on which of the two x you put the conjugate on each side.
Not sure if what I did is correct, but I vote for: the equation holds.

2025-10-10 14:41:57 UTC

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That's interesting, too. Thanks!
For Clifford algebras, I'd be most interested in how to do differential geometry on curved manifolds with CA. I think there is no source which explains this as clearly as conventional books explain DiffGeom. A coordinate-independent treatment would be important, where coordinates are not a vector space. And also an expression for curvature would be nice to see. Ideally, with pure CA and without a mix with tensor algebra where CA plays just a minor role. I couldn't find anything like that.
For exterior algebra a set of theorems which help solving algebra would be nice to see. For example how to see that in a 4-dim CA the expression $A\wedge X+(A\wedge Y)^\star$ with vectors $A,X,Y$ for constant $A$ and freely chosen $X,Y$ can span all bi-vectors.

2025-10-10 13:38:44 UTC

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If you decide to talk about SU(5) or Spin(10), it would be great if there is a way to do some part down to the very basics so that an undergrad without much knowledge about representations can verify. For example writing something down concretely with matrices. I'd imagine the must be a way to demonstrate things concretely with matrices.

2025-10-07 15:44:51 UTC

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I'd like to learn anything which gets as close as possible to what I describe. Since I don't know about it yet, I wasn't able to describe it better. Sorry if it was vague.

2025-10-07 14:42:22 UTC

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Most interesting would be: Successful simplifications of the Standard model of particle physics. Therefore, any attempt which would be able to reproduce some(?) predictions of the SM, but with more concise/elegant math.
Ideally, using Clifford algebras.
And if that doesn't exist then: if a new, simpler model replacing the SM were found, what would be the most convincing way to prove a success? Predict some open questions (strong CP-violation, 3 generations,....)? Or maybe even predict some parameters of the SM?

2025-02-02 14:28:02 UTC

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I've found a way to put it more concretely:
For any spinors and it's conventional bilinear covariants you can check that
\[(\Omega_1+J+i\,S+i\,IK-\Omega_2 I)^2\in\mathbb{R}\]
Therefore all non-scalar parts cancel.
Not only the aggregate seems more "natural" than the Fierz aggregate, because it contains all grades, it also has simpler conditions that $Z$ (compare Lounest Ch 11.4).
I'm surprised this does not get mentioned in Lounesto.
My main question is now: If I drop the imaginary units (in order to stay in the real Clifford algebra), do I retain the important algebraic/geometric/topological properties which are necessary for spinors?

2025-02-01 18:27:57 UTC

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It plays a similar role to the Fierz aggregate $Z$, except that $A$ is from the real CA and contains all grades. Being based on the bilinears, it's rather quadratic in dependence. Squaring to 1 may have been a confusing statement. Squaring to a scalar (grade 0) is more what I meant.
The interesting part is that the condition $\operatorname{grade}(A^2)=0$ somehow yields all Fierz identities (except for this one sign).
Maybe "isomorphic" is not the right word, but I mean $A$ with the condition $A^2\in\mathbb{R}$ seem to relate to spinors just like $Z$ does.
I'm looking for a way to determine if it has similar (topological? algebraic?) properties, just like the Fierz aggregate is 1-to-1 with spinors (except for the phase).

2025-02-01 13:10:13 UTC

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Thanks, I'd appreciate that a lot! I read Lounesto and was surprised that I found an alternative way to derive the Fierz identities.
Here is the details to reproduce. If you examine the conditions for \[(\Omega_1+J-S+IK+I\Omega_2)^2\in\mathbb{R}\], then you get a set of identities compatible with the many Fierz identities that Lounesto mentions. For the case $\Omega_1\neq 0$ (only possible in Cl(3,1)?) you recover the 4 identities for $J,K$ exactly except that you get $J^2=K^2$. I guess, having vectors with the same sign of the square in spacetime is a good thing though.
This makes me think that spinors could be somehow isomorphic to elements from Cl(3,1) squaring to 1. Is that possible despite the sign difference of $K^2$?
Is there any simple proof starting from the definition of bilinear covariants and does that work in any dimension?
Do spinors "behave" the same as general elements which square to 1?
Looking at elements from Cl(3,1) that square to 1, one can find that there is one class with $\Omega_1\neq 0$ and two classes with $\Omega_1=0$. It's somehow similar to Lounesto's classifcation, but I haven't compared that yet.

2025-02-01 08:13:37 UTC

**How spinors square to 1**
Spinors can be mapped to their bilinear covariants
\[\begin{aligned}
\Omega_1&=\psi^\dagger\gamma_0\psi\\J^\mu&=\psi^\dagger\gamma_0\gamma^\mu\psi\\
S^{\mu\nu}&=\psi^\dagger \gamma_0 i \gamma^{\mu\nu}\psi\\
K^\mu&=\psi^\dagger \gamma_0 u \gamma^{0123}\gamma_\mu\psi\\
\Omega_2&=\psi^\dagger \gamma_0 \gamma^{0123}\psi
\end{aligned}\]
These covariants can be collected into multivectors of the Clifford algebra Cl(1,3) (Lounesto Ch 10.5)
The spinor can be recovered from the covariants uniquely up to a complex phase (Lounesto Ch 11.2)
Since the spinor had only 8 parameters, but the covariants have 16 all together, the covariants satisfy some identities known as Fierz identities (in Clifford algebra)
\[\begin{aligned}
J^2&=\Omega_1^2+\Omega_2^2\\
&=-K^2\\
J\cdot K&=0\\
J\wedge K&=-(\Omega_2+\Omega_1\gamma_{0123})S
\end{aligned}\]
But what is surprising now and what I haven't seen in any other source is, that these identities are (almost) exactly what you need for a general element from the Clifford algebra to square to a real number:\[\begin{aligned}
A&=\Omega_1+J+S+I K+I\Omega_2\\
A^2&\in\mathbb{R}
\end{aligned}
\]
Therefore there is a 1-to-1 correspondence between spinors and multivectors (or equally real 4x4 matrices) squaring to a real number.
Has anyone seen that discussed somewhere?
There is a small difference in one sign and the Fierz identities with this choice covers only one type of element squaring to 1, but elements squaring to 1 thus describe the Fierz identities more generally.

2024-12-10 05:39:59 UTC

**Beyond complex numbers, quaternions and octonions**
The Cayley-Dickson construction provides a nice way to create more and more anti-commuting imaginary units which form an algebra. Unfortunately, along the way familiar algebraic properties are lost (like commutativity in quaternions and even associativity in octonions).
I read that you always have power associativity and the flexible property - i.e. parenthesis around $ xx\cdots x $ and $ xyx $ do not matter. But are there more identities which could be useful?
I tried it with a small Python program and empirically it seems that the following identities hold in higher Cayley-Dickson algebras, too.
\[\begin{align*}x^n(x^my)&=x^m(x^ny)\\
(yx^n)x^m&=(yx^m)x^n\\
(x^ny)x^m&=x^n(yx^m)=x^nyx^m\\
(xy)z-x(yz)&=z(yx)-(zy)x\end{align*} \]
With a "reversing associator" (any other name?) $\{x,y,z\}=(xy)z-z(yx)$, the last identity could be written as
\[\begin{align*}
\{x,y,z\}&=-\{z,y,x\}
\end{align*}\]
Has someone listed these identities or named them? And are there any others that would work in all C-D-algebras?
Some derived identity is
\[(xy)(yx)-(yx)(xy)=[x^2,y]y+x[y^2,x]\]
Someone notes that all of this can be derived from flexibility and $x^2=(x+\overline x)x- \overline x x$ and it makes sense.
#math #algebra

2024-10-08 17:29:05 UTC

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That's the causation the wrong way round. Non-physics inspiring breakthroughs in physics, would be physics. But physics inspiring breakthroughs in non-physics is not physics. Or should spin networks win the Nobel prize then? Neural networks have not produced anything meaningful for physics yet, apart from findings without any interest for physicists.