gerenuk - Network

Not an answer to my question, but related:It's a bit hard to google what I describe (and read in Lounesto), but Perplexity Deepresearch...

Not an answer to my question, but related:
It's a bit hard to google what I describe (and read in Lounesto), but Perplexity Deepresearch did quite a good job and even gave me a better name for the identities:

**How spinors square to a real number**Spinors can be mapped to their bilinear...

**How spinors square to a real number**

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 exactly what you need for a general element from the Clifford algebra to "square" to a real number:\[\begin{aligned}
A&=\Omega_1+J+S+I K-I\Omega_2\\
A\tilde A&\in\mathbb{R}
\end{aligned}\](using the reverse)

Therefore there is a 1-to-1 correspondence between spinors and multivectors"squaring" to a real number.
Has anyone seen that discussed somewhere?
(I did check the algebra, but don't mind a confirmation)

**Beyond complex numbers, quaternions and octonions**The Cayley-Dickson construction provides a nice way to create more and more...

**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

Next time you order a cocktail in a bar, check how full it is: https://www.youtube.com/watch?v=Mkn3PzdaByY

Next time you order a cocktail in a bar, check how full it is: https://www.youtube.com/watch?v=Mkn3PzdaByY

My theory: they were late for deciding who should win the #Nobel prize for #physics, and then they asked ChatGpt the night before.

My theory: they were late for deciding who should win the #Nobel prize for #physics, and then they asked ChatGpt the night before.

An insightful video about trying to guess closed-form solutions from empirical analysis and recent...

An insightful video about trying to guess closed-form solutions from empirical analysis and recent results:
https://www.youtube.com/watch?v=-uIwboK4nwE
(related to Sinkhorn limit and iterative calculation of doubly stochastic matrices)

Somehow it is rarely found in introductory material to Clifford Algebras how to determine when basis blades commute. Fortunately, that's...

Somehow it is rarely found in introductory material to Clifford Algebras how to determine when basis blades commute. Fortunately, that's pretty easy:

Basis blades \(\mathtt{e}_A\) and \(\mathtt{e}_B\) commute for some set of indices \(A\) and \(B\) iff
\[|A|\cdot |B|=|A\cap B|\mod 2\]

Else, they anti-commute.

If you want to multiply basis blades \(e_{ij\ldots}\) in Clifford algebra, it is fairly easy to see what the resulting basis blade will be....

If you want to multiply basis blades \(e_{ij\ldots}\) in Clifford algebra, it is fairly easy to see what the resulting basis blade will be. However, seeing the sign of the product is trickier.

Interestingly, there is a highly efficient way to get the sign if you use bitfields to encode your basis blades. The resulting bitfield (without sign) is `a ^ b` (XOR).

Now, you can get the required sign with `parity((a ^ (a>>1)) & b & mask_high_bit)` which is extremely efficient on computers. The only thing you have to do is to not use plain bitfields, but rather do a bitwise accumulative XOR from the lowest bit to "encode" them beforehand.

This nice video shows how to invert any function with SAT and how it is related to P vs NP.https://www.youtube.com/watch?v=6OPsH8PK7xM

This nice video shows how to invert any function with SAT and how it is related to P vs NP.
https://www.youtube.com/watch?v=6OPsH8PK7xM

A nicely made video about gauge theory https://www.youtube.com/watch?v=Sj_GSBaUE1oIf you watch this, you will have a "positive...

A nicely made video about gauge theory https://www.youtube.com/watch?v=Sj_GSBaUE1o
If you watch this, you will have a "positive time" 🙂

If you look at the Born rule \[P=|\langle \psi_1|\psi_2\rangle|^2\]it seems to have something non-linear, but density matrices show that...

If you look at the Born rule \[P=|\langle \psi_1|\psi_2\rangle|^2\]it seems to have something non-linear, but density matrices show that it's just an inner product between states.
To reproduce this in an algebra way, one could use complex Clifford algebra and write
\[\Psi=\frac1{\sqrt2}\sum_k (\mathtt{e}_{2k}+I\mathtt{e}_{2k+1})(\Re z_k+iI\,\Im z_k)\]where \(z_k\) are components of \(\psi\), \(I\) the pseudo-scalar and all \(\mathtt{e}^2=1\) .
Then the "state" for the inner product becomes just \[\Omega=\Psi^2\] and the Born rule the inner product \[P=\langle\Omega_1\Omega_2^\dagger\rangle\] with no pesky square.
Maybe this helps understanding multi-spin systems in #quantumcomputing #quantum

Why gears always slip and finding the envelope curve algebraically in the complex plane https://youtu.be/eG-z-791_ak?si=9vlfblAWYFvvDn65

Why gears always slip and finding the envelope curve algebraically in the complex plane
https://youtu.be/eG-z-791_ak?si=9vlfblAWYFvvDn65

Does someone know about matrices and groups?If you take a set of matrices which square to zero \(A^2=0\) and which are all related by a...

Does someone know about matrices and groups?

If you take a set of matrices which square to zero \(A^2=0\) and which are all related by a similarity transform \(A'=TAT^{-1}\), do these similarity transformations form a group? If so, which group (for 4x4 matrices)? If not, why not?

I'm looking for an answer which would identify matrices \(T_1\equiv T_2\) having the same action (i.e. same \(A'\)) as equivalent. Therefore the group is not all of \(GL\). Is it still a group? If not, can you show an example of what goes wrong?

With the Jordan normal form it's easy to see that there are only a few sets of such matrices which are similar within a set.

A nice presentation of basic techniques for non-linear ODEs:https://www.youtube.com/playlist?list=PLw5b-jhi9cCD_YQgijiHnodoD0G91lu4c

A nice presentation of basic techniques for non-linear ODEs:
https://www.youtube.com/playlist?list=PLw5b-jhi9cCD_YQgijiHnodoD0G91lu4c

One of the basic things you learn in #quantum mechanics is that 2 spins can be in a state of definite total angular momentum J^2.In...

One of the basic things you learn in #quantum mechanics is that 2 spins can be in a state of definite total angular momentum J^2.

In elementary textbooks, this is usually presented as a spin singlet and triplet (see image). It seems simple, but how do you know that you understand what has been done and are not just reciting what you have seen? A good way to test your knowledge is to recreate the same approach for 3 spins which some researchers may struggle with.

It is true that you will have a total spin 3/2 state with 4 J_z eigenstates and a spin 1/2 state with 2 J_z eigenstates. But this is only 6 states in total, whereas you need 8 to label the 8-dim. 3-spin-basis.

It turns out the J^2 spin 1/2 state is doubly degenerate. However, this means you need to invent a new operator to create a new quantum number to resolve this degeneracy.

So, to really recreate the approach for 3-spins you need to provide 8 eigenstates of spin state superpositions (RHS of image) and 8 triples of quantum numbers to label them (LHS of image). You would also need to explicitly write down the 3rd operator in terms of individual spin operators. The result is a complete set of commuting operators J^2, J_z and your new operator.

To test my understanding I've tried this exercise as it requires no more that some linear algebra. I could find a simple operator which can fulfill the role of lifting the degeneracy.
\[\begin{align*}
C&=S_1^xS_2^yS_3^z+S_1^zS_2^xS_3^y+S_1^yS_2^zS_3^x\\
&\quad -S_1^xS_2^zS_3^y-S_1^yS_2^xS_3^z-S_1^zS_2^yS_3^x
\end{align*}\]
It is a simple anti-symmetrisation.
Weirdly, I didn't find this solved anywhere else.

If you want to play with simple algebraic expressions and would like to benefit from an extensible Python package, you could try my Python...

If you want to play with simple algebraic expressions and would like to benefit from an extensible Python package, you could try my Python scripts. The Github repo and an example notebook can be found on https://github.com/Gerenuk/Algebrant/blob/main/examples/Example.ipynb

It supports basic symbol algebra, Clifford algebra, non-commutative symbols, Clifford algebra matrix representations and a bit more.

I've watched a debate about gravity and the universe https://www.youtube.com/watch?v=1j0Xh9XM34M and I'm not sure how some...

I've watched a debate about gravity and the universe https://www.youtube.com/watch?v=1j0Xh9XM34M and I'm not sure how some approaches make sense. If you have seen such discussions before, there is probably nothing new for you in that video.

Looking at it naively it seems the problem statement of quantum gravity is easy to express, but yet it is rarely summarized this way. Gravity is governed by the Einstein field equations. Particle physics is successfully modeled by the Standard Model Lagrangian. The issue is that this is 2 equations and not just 1. If we could apply one equation to the scale of the other, they would make conflicting predictions and obviously only one can be right. So the task of quantum gravity is "just" to find a *single equation* reproducing results at all scales, right? Why even debate this?

That's a purely mathematical task. Why would we need to wait for more experiments to do this? Why do people say we cannot test quantum gravity, when all you have to do is to present a single equation giving correct results at both scales? Not saying it's simple, though.

If you cannot present such a equation and actually do the calculation for real on a computer and get the numeric values of the anomalous magnetic moment and the perihelion precession of mercury correctly, it's a "fail"?! Why do theorists speak of proposals to make one or the other more classical/quantum and believe in theories which "successfully" produce quantum gravity when (to my knowledge) all these approach fail the initial request to provide a single equation of the world. Some theories say it's all consistent and solved, then say "we cannot really get the standard model out, because there are too many ways". Does this makes sense?

When you want to measure the area of a curved surface by triangulation, it is not enough to just shrink the triangle sides to zero. The...

When you want to measure the area of a curved surface by triangulation, it is not enough to just shrink the triangle sides to zero. The summed triangle area could still diverge, unless you also ensure that the surface normals are parallel.
https://www.youtube.com/watch?v=yAEveAH2KwI

Algebras have multiple equivalent ways to see them. Maybe some relations are less known. Here, by algebra I mean a \(2^{2n}\)-dimensional...

Algebras have multiple equivalent ways to see them. Maybe some relations are less known. Here, by algebra I mean a \(2^{2n}\)-dimensional complex vector-space with an associative multiplication between elements.

**Matrix**
Obviously, the algebra of \(2^n\times 2^n\) matrices is such an algebra.

**2-generator**
The whole algebra can be generated from just two abstract elements with \[\begin{align*}P^2&=P\\C^n&=1\\PC^kP&=0\quad n\nmid k\end{align*}\] \(P\) is a matrix with a single 1 and \(C\) is a matrix which permutes rows/columns (see https://mathstodon.xyz/@gerenuk/111539875588189260). All elements will be of the form ∑αₖₗ𝐶ᵏ𝑃𝐶ˡ .

**Clifford**
You can always find \(2n+1\) anti-commuting matrices which as abstract elements can generate the whole algebra:
\[\begin{align*}e_i^2&=1 & i&\in\{0,\ldots 2n-1\}\\e_ie_j&=-e_je_i&i&\neq j\end{align*}\]
(for example https://mathstodon.xyz/@gerenuk/111488226649579147)
For even \(n\), in order to highlight a symmetry, you could add in the pseudo-scalar \(I=e_{2n}\) which obeys the same rules and then \(\prod_{i=0}^{2n} e_i=1\).

**Creation**
Pairs of basis elements from the Clifford algebra could be combined to form a new basis resembling creation and annihilation operators in quantum mechanics:
\[a_k=\frac12(e_{2k}+i e_{2k+1})\]with the anti-commutation rules
\[\begin{align*}\{a^\dagger_k,a_l\}&=\delta_{kl}\\\{a_k,a_l\}&=0\\\{a^\dagger_k,a^\dagger_l\}&=0\end{align*}\]

All of these are equivalent ways to see the same algebra!

All \(n\times n\) matrices can always be generated from just 2 matrices with addition, multiplication and scalar multiplication. The...

All \(n\times n\) matrices can always be generated from just 2 matrices with addition, multiplication and scalar multiplication. The following example makes it obvious:
\[P=\begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}
\] and \[C=\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{pmatrix}
\]will generate all \(4\times 4\) matrices from sums of \(\alpha\cdot C^kPC^l\) where \(C\) cycles rows or columns if pre- or post-multiplied and essentially moves the single 1 around.

It seems that the \(n\times n\) matrix algebra can be defined and generated from \[\begin{align*}P^2&=P\\C^n&=1\\PC^kP&=0\quad n\nmid k\end{align*}\] (with some additional details).