Thursday, October 10, 2024

The Quirks of Quaternions

The Spark of Curiosity

This post is inspired by a fascinating conversation I had with Igor Bayak and Bjab in the comments of my previous blog. Their input sparked my curiosity to dive deep into quaternions and vector fields on a three-dimensional sphere. 


These concepts aren’t just abstract math—they could be quite handy for my spinor studies. Plus, there’s something aesthetically satisfying about painting these mathematical fields. And let’s be honest, who doesn’t love a little visual beauty in math?

So, here’s a look at the world of quaternions through my curious eyes.


Quaternions 101: Meet i, j, and k

Let’s start with our three quirky quaternion friends: i, j, and k. These are the building blocks of quaternions. A general quaternion, x, can be written like this:

x=x1i+x2j+x3k+x4

Simple enough, right? Now, it’s time to turn these quaternions into something a little more structured—a matrix representation. (Cue dramatic music.)


Quaternion Matrix Magic

We’re going to multiply x by i, j, and k from the left and see what matrices pop out. This is where the magic happens. Let’s start with i.

When we multiply i by x, we get:

ix=x1+x2kx3j+x4i

Now, for the matrix interpretation:

  • At i, we have x^4, which gives us the first row of the matrix: {0, 0, 0, 1}.
  • At j, we have -x^3, giving the second row: {0, 0, -1, 0}.
  • At k, we have x^2, producing the third row: {0, 1, 0, 0}.
  • Finally, at unity, we get -x^1, completing the fourth row: {-1, 0, 0, 0}.

Putting it all together, the matrix that represents multiplication by i on the left is:

L1=(0001001001001000)L1 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}

Matrix magic! (Applause, please.)


Now Multiply by j and k

Using the same process, we get matrices for multiplication by j and k on the left:

For j, we get:

L2=(0010000110000100)L2 = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}

For k, we have:

L3=(0100100000010010)L3 = \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}


What Happens on the Right Side?

Not to be left out (pun intended), we can also multiply quaternions from the right. When we do this, we get the following matrices:

For i on the right:

R1=(0001001001001000)R1 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}

For j:

R2=(0010000110000100)R2 = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}

For k:

R3=(0100100000010010)R3 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}

So now we’ve got both the left and right multiplication matrices. Our quaternion friends are getting quite versatile.


Quaternions, SU(2), and a Three-Dimensional Sphere

Now things get even more interesting. Quaternions with a norm of 1 form a three-dimensional sphere. And not only that, but they form a group that is isomorphic to SU(2)—fancy math speak for “this group behaves like SU(2).”

To make this more concrete, consider the action of this group on the space of all quaternions. Let’s define an action u as:

u:xuxu: x \rightarrow ux

We now have a representation of this group acting on functions, written as:

(T(u)f)(x)=f(u1x)

Are you still with me? Good! Let’s move on.


Vector Fields: Expanding the Fun

Consider a one-parameter subgroup, say exp(ti). This subgroup generates a vector field, which we’ll call X1:

X1(f)=ddtf(exp(ti)x)t=0

When we calculate the action of X1 on the coordinate functions x^i, we get the components of X1:

X1(x)=L1jixjX^i (x) = - L1^i_j x^j

In the same way, we can derive the vector fields X2 and X3

Here are the results:


The Takeaway

Quaternions may sound intimidating at first, but once you break them down, they’re not only manageable—they’re downright fascinating. Their matrix representations, their connection to SU(2), and the rich vector fields they generate all have deep implications in both math and physics. And who knew matrices could be so much fun?

So, the next time you’re pondering the mysteries of the universe (or trying to impress someone at a party), just casually drop some knowledge about quaternion vector fields. It’s sure to be a hit!

47 comments:

  1. "Are you still with me?"

    No.
    But never mind.

    ReplyDelete
    Replies
    1. I lost him already at "i", "j" and "k"; thinking that those are unit vectors of 3d Cartesian coordinate system, making 'ordinary' vector x=(x1, x2, x3) = x1*i + x2*j + x3*k, but no, since those (i, j, k) would be orthonormal. Or perhaps the "multiplication" is not the usual dot product, with kronecker δij when unit vectors are multiplied, but some strange version of Levi-Civita εijk? That is, what's the definition of the "multiplication", what's its neutral and what's the inverse element of such a multiplication?

      If taking that multiplication above for granted, next question would be what's the meaning or definition of the term "group" in the sentence, "Quaternions with a norm of 1 ... form a group ..."?

      When that's settled, next set of questions is, what's an "action u", what's an "action on functions" (T(u)f)(x), and is that u^(-1) inverse of u, in (T(u)f)(x)=f(u^(−1)x), and if yes how is it defined?

      And only then I can ask about Xi and Yi, that is are those matrices A and B the same as matrices L and R from above, what are those 'partial derivatives' in (1) and (2), and how is the multiplication defined in those commutators [X1, X2] = 2*X3 or [Y1, Y2] = 2*Y3?

      There may be more questions after these above have been answered...

      Delete
    2. Oh, i,j,k are imaginary unit quaternions. So x is a quaternion. Quaternion multiplication is ij=k, jk=i, ki=j, and they anticommute. Moreover i^2=j^2=k^2=-1.

      Delete
    3. Aha, OK, thanks.
      For the rest of the questions and application in physics and real world, I guess I need to dig a bit more into quaternions and group theory and symmetries.
      Any good and useful 'manual' to recommend, perhaps?

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    4. Try this:

      https://cgvr.cs.uni-bremen.de/teaching/cg_literatur/A_Mostly_Linear_Algebraic_Introduction_to_Quaternions.pdf

      It needs some simple linear algebra though.

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    5. O well, according to Wikipedia,
      https://en.wikipedia.org/wiki/Quaternion :

      A quaternion is an expression of the form
      a+bi+cj+dk,
      where a, b, c, d, are real numbers, and i, j, k, are symbols that can be interpreted as unit-vectors pointing along the three spatial axes.

      Hamilton describes a quaternion
      q=a+bi+cj+dk,
      as consisting of a scalar part and a vector part.

      Which means I need to "un-learn" what I thought I knew about vectors and unit vectors to be able to follow quaternions, right?

      Delete
    6. "Try this:
      https://cgvr.cs.uni-bremen.de/teaching/cg_literatur/A_Mostly_Linear_Algebraic_Introduction_to_Quaternions.pdf
      It needs some simple linear algebra though."

      Thank you. Some simple linear algebra I think I have in my tool box. :)

      Delete
    7. You do not have to unlearn anything. You just move from the category of vector spaces to the category of algebras. Quaternions for a 4-dimensional vector space, which is not only vector space, but also an algebra, where you can also multiply vectors, much like the vector product, but now vectors have not only three components, but also the fourth component - the scalar part.

      Delete
    8. You're right, thank you.

      After writing it down and playing a bit, it seems we can work with quaternions pretty much in the same manner as with ordinary complex numbers just where now imaginary part are 3d vectors and we define the product of such imaginary parts as difference between ordinary dot and cross vector products.

      Or more aesthetically pleasing:
      if a = a0 + a1*i and b = b0 + b1*i, where a1 and b1 are ordinary 3d vectors and i is ordinary imaginary unit, we define the product a1*b1 = b1.a1 + (b1×a1)*i, where "." and "×" are ordinary dot and cross products of ordinary 3d vectors.

      In such a way, at least on simple examples I checked, all the usual rules from the 3d vector analysis are still valid, like the norm in general and orthonormality of unit vectors in particular for example, and we can also simply apply ordinary 2d complex number calculus when dealing with quaternions.

      In addition, it's nicely or easily seen that such a product a1*b1 gets us from 3d to 4d, that is from pure quaternions or vector quaternions having only imaginary parts, to real quaternions with both, real and imaginary parts present; and that brackets [a, b] get us from 4d to 3d, that is from real quaternions with both parts present to vector or pure ones with only imaginary parts, as bracket would then simply be [a, b] = 2 * (a1×b1)*i.

      Well, probably old news to you, but to me it's fascinating and super interesting, quaternions are extra fun and ultra cool!
      Thank you very much!

      Delete
    9. For the sake of completeness;
      if we denote a quaternion as a = a0 + a1, where a1 is ordinary 3d vector a1 = (a₁, a₂, a₃) (not a best choice of notation for comment section, but hopefully you'll get the idea), then an ordinary product of two quaternions a and b is:
      a*b = a0*a0 - a1.b1 + a0*b1 + b0*a1 + a1×b1,
      where "." and "×" are ordinary dot and cross 3d vector products, respectively, which brings us to that "definition" for multiplication of vector or pure quaternions from the comment above:
      a*b = a1*b1 = a1×b1 - a1.b1,
      as a0 = b0 = 0.
      (details can be found on page 2 of the PDF Ark posted,
      https://cgvr.cs.uni-bremen.de/teaching/cg_literatur/A_Mostly_Linear_Algebraic_Introduction_to_Quaternions.pdf)

      And a "funny" little snippet;
      if a quaternion a is of Pythagorean quadruple type (a₁, a₂, a₃, a0), where a1 = (a₁, a₂, a₃) is a 3d vector, and a0^2 = a1^2 or by components a0^2 = a₁^2 + a₂^2 + a₃^2, that is a0*a0 - a1.a1 = 0, then a product of such Pythagorean quadruple type quaternion with itself is just:
      a*a = 2 * a0*a1,
      because the cross 3d vector product a1×a1 = 0.

      In other words, for Pythagorean quadruple type quaternions their product with themselves also brings us from 4d to 3d, that is from real, complete quaternion we get the pure or vector one, just as the brackets [a, b] do for arbitrary quaternions a and b:
      [a, b] = a*b - b*a = a1×b1 - b1×a1 = 2 * a1×b1 = -2 * b1×a1.

      The main difference between the two being that the brackets result in a pure or vector quaternion that lies in a plane perpendicular to the plane spanned by 3d vectors a1 and b1, while the product of Pythagorean quadruple type quaternions with themselves preserves the original direction of 3d vector part of such a Pythagorean quaternion, that is such product just doubles and squares its magnitude, as a0 = |a1|.

      These quaternions are really fun, thanks!

      Delete
    10. Two small corrections to the comment above.

      In the arbitrary quaternions product formula a*b, it should be a0*b0 instead of a0*a0, that is:

      a*b = a0*b0 - a1.b1 + a0*b1 + b0*a1 + a1×b1,

      and in the last paragraph in the part of the sentence,
      "the product of Pythagorean quadruple type quaternions with themselves preserves the original direction of 3d vector part of such a Pythagorean quaternion",
      maybe more precise term instead of "original direction" would be "colinearity", as in fact a0^2 = a1^2 = |a1|^2 and thus a0 = +/- sqrt(|a1|^2) = +/- |a1|.

      Delete
    11. After playing a bit more with funny quaternions, the conclusion reached is that for all practical purposes so far they can be represented by a "complex" numbers where imaginary part is a 3d vector and so no additional imaginary units would be needed except the usual ordinary "i" where i^2=-1. In such representation the unit 3d vectors (i, j, k) are the usual base vector in R^3, that is (e₁, e₂, e₃) in Cartesian coordinate system, and like all other 3d vectors in this "complex" number representation of quaternions, they just get multiplied with imaginary unit "i". Only thing left sort of unfinished, from the expressions shown on Wikipedia quaternion page that have been checked, is the geodesic norm, that is from expression:
      d_g(a,b) = ||(ln(a^(-1) b))|| = arccos^2(a0*b0 + a1.b1),
      where a=(a0+i*a1) and b=(b0+i*b1) are unit quaternions with scalar parts a0 and b0, and vector parts a1 and b1, respectively, it should be reached:
      d_g(a,b) = arccos(2(a.b)^2 - 1),
      where (a.b) is a quaternion dot product, that is (a.b) = (a0*b0 + a1.b1).
      Regardless of that unfinished crosscheck, the "complex" number representation brought to the surface few interesting things that might, or maybe not, be useful.

      For a quaternion a=(a0+i*a1), its versor U_a is a unit quaternion U_a = a/||a||, where ||a|| is a norm of quaternion a: ||a||^2 = a0^2 + |a1|^2 = aa*, where a* is a conjugate of quaternion a, a*=(a0-i*a1).
      For a Pythagorean quadruple type of quaternion, where a0^2 = |a1|^2 and ||a||^2 = 2*a0^2 = 2*|a1|^2, using the expression for Hamilton square, that is for Hamilton product of a quaternion with itself, from two comments above:
      aa = 2i * a0 * a1 = 2i * a0 * |a1| * a1/|a1| =
      = 2i * a0 * a0 * n_a1 = 2i * |a1| * |a1| * n_a1 =
      = ||a||^2 * n_a1 * i,
      where n_a is a unit vector in the direction of a1, n_a1 = a1/|a1|, for Hamilton square of a versor U_a of such a Pythagorean quadruple type of quaternion, we get simply: U_a U_a = i*n_a1; that is just a unit 3d vector pointing in the direction of a1.

      In its polar form, a quaternion a=(a0+i*a1) can be represented as a = ||a|| * exp(i*n_a1 * ϕ) = ||a|| * (cosϕ + i*n_a1 * sinϕ), where angle ϕ can be obtained from a0 = ||a||*cosϕ or |a1| = ||a||*sinϕ, which for Pythagorean quadruple type of quaternion leads to
      a = a0*√2 * exp(i*n_a1 * pi/4) = |a1|*√2 * exp(i*n_a1 * pi/4),
      but more importantly for its versor it gives
      U_a = exp(i*n_a1 * pi/4),
      which represents spatial rotation for ϕ=pi/4 about n_a1 axis, and for a Hamilton square of such versor we get
      U_a U_a = exp(i*n_a1 * pi/2),
      that is rotation for 90° about n_a1.
      Pretty nice, right?

      Delete
    12. And a few words about the so called Hamilton vector quaternion product as presented several comments above, in "complex" number representation.

      For vector quaternions a=(0+i*a1)=i*a1 and b=(0+i*b1)=i*b1, Hamilton product
      ab = a0*b0 - a1.b1 + i*a0*b1 + i*b0*a1 + i*(a1×b1),
      yields only two terms, that is a dot 3d vector product (a1.b1) and a cross 3d vector product (a1×b1), so we might define Hamilton vector quaternion product as:
      a1*b1 = (a1.b1) - i*(a1×b1),
      which when multiplied by i^2 gives correct signs in the general Hamilton quaternion product above.

      When we exchange places of the factors in this vector quaternion product, that is apply commutation, we get:
      b1*a1 = (b1.a1) - i*(b1×a1) =
      (dot product commutes, cross product anticommutes)
      = (a1.b1) + i*(a1×b1),
      which means that commutation for this vector quaternion product plays the same role as complex conjugation mentioned above, (b1*a1) = (a1*b1)*, which then brings forth few nice properties.

      Lets first check commutator [a,b] = (ab - ba),
      [a1, b1] = (a1*b1) - (b1*a1) =
      = ((a1.b1) - i*(a1×b1)) - ((a1.b1) + i*(a1×b1)) =
      = -2i * (a1×b1),
      meaning that commutator returns sort of imaginary part of the product,
      [a1, b1] = 2i * Im[(a1*b1)] = -2i * Im[(a1*b1)*] = -2i * Im[(b1*a1)].

      Anticommutator {a,b} = (ab + ba) on the other hand returns exactly the real part (up to the factor 2) of a vector quaternion product in "complex" number representation:
      {a1, b1} = (a1*b1) + (b1*a1) =
      = ((a1.b1) - i*(a1×b1)) + ((a1.b1) + i*(a1×b1)) =
      = 2 * (a1.b1),
      {a1, b1} = 2 * Re[(a1*b1)] = 2 * Re[(b1*a1)] = 2 * Re[(a1*b1)*].

      Additionally, when we apply the same procedure, that is check the commutator and anticommutator values of the above defined Hamilton vector quaternion product with itself, we get:
      [(a1*b1) , (b1*a1)] = (a1*b1) * (b1*a1) - (b1*a1) * (a1*b1) =
      = (a1*b1) * ((a1*b1)*) - ((a1*b1)*) * (a1*b1) =
      = ||(a1*b1)|| - ||(a1*b1)|| = 0,
      meaning that this Hamilton vector quaternion product sort of commutes, that is commutes with itself (yay!), but also, when we check the anticommutator:
      {(a1*b1) , (b1*a1)} = (a1*b1) * (b1*a1) + (b1*a1) * (a1*b1) =
      = (a1*b1) * ((a1*b1)*) + ((a1*b1)*) * (a1*b1) =
      = ||(a1*b1)|| + ||(a1*b1)|| =
      = 2 * ||(a1*b1)||,
      we see that the anticommutator of the Hamilton vector quaternion product with itself returns its norm (times 2), which appears like a handy and useful thing.

      Well, that's it for tonight, will be looking forward to feedback and/or if something's been messed up in this longish comment, which had to be split in 2 parts as it was too long for just one go.
      Nice regards!

      Delete
    13. Combining things presented in previous two-part comment with some ordinary usual geometry relations, we can reach rather useful and practical interpretation of Hamilton vector quaternion product and perhaps general Hamilton quaternion product as well.

      So, for two arbitrary vector quaternions, a = i*a1 and b = i*b1, we defined their Hamilton vector product as:
      a * b = a1 * b1 = a1.b1 - i*(a1×b1),
      where a1.b1 is a dot 3d vector product, which can be evaluated as a1.b1 = |a1|*|b1|*cosϕ with ϕ being the angle between a1 and b1 (or vice versa), and where a1×b1 is a cross 3d vector product, which can be written as
      a1×b1 = |a1|*|b1|*sinϕ * n_⊥(a1,b1),
      with n_⊥(a1,b1) being unit 3d vector orthogonal to the plane spanned by vectors a1 and b1 (where the angle ϕ also lies). For the sake of clarity through the rest of the comment unit vector n_⊥(a1,b1) will be denoted simply as n, if not stated otherwise. Importing these sine and cosine relations into our Hamilton vector quaternion product yields:
      a1 * b1 = |a1|*|b1|*cosϕ - i*n * |a1|*|b1|*sinϕ =
      = |a1|*|b1| * (cosϕ - i*n*sinϕ) = |a1|*|b1| * exp(-i*n*ϕ).

      Comparing the quaternion polar form from previous comment, q = ||q|| * exp(i*n_q*phi), with this result, we see that the norm of Hamilton vector quaternion product is simply the product of norms of vectors a1 and b1, the same as for general Hamilton quaternion product, ||(a*b)|| = ||(a1*b1)|| = |a1|*|b1|.
      Comparing the result to the unique quaternion polar decomposition, q = ||q|| * U_q, where U_q is a unit quaternion U_q = q/||q|| called versor, also mentioned in previous comment, we see that the versor for Hamilton vector quaternion product is U_(a1*b1) = exp(-i*n*ϕ), which can be interpreted as spatial or planar rotation for the angle between vectors a1 and b1 about the axis n, that is in the plane spanned by those two vectors in the direction "from b1 to a1" as for the minus sign.

      Using the expression for Hamilton square of a general quaternion a=(a0+i*a1):
      a * a = a0^2 - |a1|^2 + 2*a0*|a1|* n_a * i,
      for Hamilton square of a Hamilton vector quaternion product we get after some regrouping of the elements:
      (a1 * b1) * (a1 * b1) = |a1|^2 * |b1|^2 * (cos^2(ϕ) - sin^2(ϕ) - i*n * 2*sinϕ*cosϕ) =
      = (|a1|*|b1|)^2 * (cos(2ϕ) - i*n *sin(2ϕ)) =
      = ||(a1*b1)||^2 * exp(-i*n*2ϕ),
      that is exactly the square of its polar form, meaning that in general Hamilton square of a quaternion squares the norm of the quaternion and sort of doubles the angle of rotation about the original axis of the vector quaternion part.

      Similar reasoning can then be applied to interpret the equivalence of commutation and complex conjugation for our Hamilton vector quaternion product.
      Since the product (a1*b1) * (b1*a1) gives the norm squared, that is
      (a1*b1) * (b1*a1) = (a1*b1) * ((a1*b1)*) = ||(a1*b1)||^2,
      (erroneously written as only the norm, ||(a1*b1)||, in last paragraph of the second part of previous post when discussing commutator and anticommutator values for Hamilton vector quaternion product with itself), indicating that the overall ϕ is effectively 0, we can deduce that while (a1*b1) rotates in the direction "from b1 to a1", its complex conjugate and commutative partner (b1*a1) rotates in the opposite direction about the same axis n, that is "from a1 to b1", leading to effectively no rotation at all at the end of such action or vector multiplication.

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    14. With these remarks above in mind, it's then natural to interpret general Hamilton quaternion product (a*b) as being composed of three elements with their distinct actions:
      (a0+i*a1) * (b0+i*b1) =
      1) a0 * b0 --> compact : straightforward multiplication of reals;
      2) i*(b0*a1 + a0*b1) --> composite : scaling the vectors with reals and their subsequent addition;
      3) i^2 * (a1 * b1) --> complexive : Hamilton vector product with multiplication of vector norms and spatial or planar rotation in the plane spanned by those vectors for the angle value between them.

      Interpreted like this, it's suggestive that the general Hamilton quaternion product represents sort of planar actions more than pure spatial ones, as both 2) and 3) result and concern primarily in movements in the plane defined by vectors a1 and b1, while the axis n in 3), perpendicular to that plane and so sort of complicating things, apparently serves for orientation purposes vis-a-vis the informational exchange with other "actors" with their Hamilton quaternion products around.

      Viewed on its own, the quaternion q=(q0+i*q1) in its polar form q = ||q||*exp(i*n_q*phi) = ||q||*U_q again sort of speaks about rotation in a plane, only this time about the axis n_q = q1/|q1|, that is in the plane perpendicular to the direction of its vector part.

      Well, things can start to be rather complicated to describe analytically when we also consider simultaneous movement just up and down that axis while spinning about it. But that's a topic for perhaps some other comment.

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    15. There will be little bit of quaternions in my new post tomorrow.

      Delete
  2. Funny enough there is a typo in (1) X1.

    Funny enough there is also a typo in (2) Y1.

    ReplyDelete
  3. 𝑋𝑖(𝑥)=−𝐿1𝑗𝑖𝑥𝑗 ->
    𝑋1𝑖(𝑥)=−𝐿1𝑗𝑖𝑥𝑗

    ReplyDelete
    Replies
    1. "Indeed there was."

      I can't see corrections of (1) and (2).

      Delete
    2. "Now you should see them"
      Yes, thank you.

      Delete
  4. Если уж эксплуатировать тему векторных полей, то на полную катушку. Почему бы вам не построить из этих векторных полей трёхмерное слоение с типичным слоем S^3. По крайней мере, евклидова ортогональность трёх полей это как ог в помощь. Главное не забыть добавить к этим компактным полям некомпактное поле эволюции.

    ReplyDelete
    Replies
    1. "Почему бы вам не построить из этих векторных полей трёхмерное слоение с типичным слоем S^3"
      You will need to be less cryptic, Igor!

      Delete
    2. У вас же уже есть опыт рисования предельных циклов. А тут слоение без особенностей, это даже проще. Нарисуйте сначала логарифмические спирали линий тока векторного поля ∂ρ+∂φ, а потом уже надо думать как изобразить поля ∂ρ+∂φ_x, ∂ρ+∂φ_y, ∂ρ+∂φ_z и образуемое ими слоение.

      Delete
    3. ∂ρ+∂φ_x? Or ∂ρ+φ_x? But these vector fields are not on the sphere S^3. They need 4d space. Projecting from 4d to 2d is of doubtful use. And you should be less cryptic. You say "слоение", but you do not say what is the basis, what is the fiber, what is the projection, and how it relates to your vector fields. Moreover, Bjab asked you about your probable mistake with ∂ρ in your paper.

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    4. Я имел в виду теорему Фробениуса, согласно которой три интегральные кривые образуют трёхмерное слоение, если попарные скобки Ли векторных полей этих кривых принадлежат множеству алгебры векторных полей.

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    5. I don't know such a theorem. Can you provide a reference?

      Delete
    6. It's mentioned on Wikipedia page about quaternions:

      "The set H of all quaternions is a vector space over the real numbers with dimension 4. Multiplication of quaternions is associative and distributes over vector addition, but with the exception of the scalar subset, it is not commutative. Therefore, the quaternions H are a non-commutative, associative algebra over the real numbers. Even though
      H contains copies of the complex numbers, it is not an associative algebra over the complex numbers.

      Because it is possible to divide quaternions, they form a division algebra. This is a structure similar to a field except for the non-commutativity of multiplication. Finite-dimensional associative division algebras over the real numbers are very rare. The Frobenius theorem states that there are exactly three: R, C, and H. ..."

      https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)

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    7. Saša, это другая его теорема.

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  5. Скорее всего знаете, а я сказал какую-то глупость. Однако, ссылку дам
    https://ru.wikipedia.org/wiki/%D0%A0%D0%B0%D1%81%D0%BF%D1%80%D0%B5%D0%B4%D0%B5%D0%BB%D0%B5%D0%BD%D0%B8%D0%B5_(%D0%B4%D0%B8%D1%84%D1%84%D0%B5%D1%80%D0%B5%D0%BD%D1%86%D0%B8%D0%B0%D0%BB%D1%8C%D0%BD%D0%B0%D1%8F_%D0%B3%D0%B5%D0%BE%D0%BC%D0%B5%D1%82%D1%80%D0%B8%D1%8F)

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    Replies
    1. Nu, da. распределение. Here it is integrable, and the surfaces (интегральная поверхность) are 3-spheres with different radii. You know it, we know it.

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    2. Поверхность не компактна и на ней определено евклидово скалярное произведение. Может быть это и есть евклидово пространство Вселенной?

      Delete
    3. Here these distributions (распределения) are compact. They are spheres S^3. If you also want to add ∂ρ, first fix your formula for ∂ρ, before we can discuss the subject further.

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    4. Your sentence "векторное поле ∂ϕ ..." makes no sense for me. You need to fix it too. Better remove your paper, right now is too bad.

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    5. Что там не ладно с \partial\rho? Это просто еденичное линейное векторное поле. Чем можно заменить еденицу?

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    6. Read what Bjab wrote:
      https://ark-jadczyk.blogspot.com/2024/10/navigating-quantum-maze-spin-state.html?showComment=1728495587440#c8006803352819991078

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    7. Так как предлагает Бьяб получится нормированное векторное поле, а нам в алгебре надо использовать единичное линейное векторное поле.

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    8. Bjab is not proposing. He is calculating from the formula for rho. If you want your formula, give it a different name, otherwise it is misleading.
      And again: Your sentence "векторное поле ∂ϕ ..." makes no sense for me. You have three vector fields. How do you get from them just one field? You understand Igor?

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    9. "Bjab is not proposing. He is calculating from the formula for rho. If you want your formula, give it a different name, otherwise it is misleading."

      You took those words right out of my keyboard.

      Igor could use instead of
      ρ=√(x²+y²)
      equation:
      r=√(x²+y²)

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    10. It would be the same, r or ρ, if either of them would be √(x²+y²), its derivative or variation is incorrectly calculated. And there's also a factor 2 missing on the right hand side.
      It looks like the same error happened for ∂ϕ, otherwise ϕ or its variation or derivative would have a dimension of area or surface as x∂x or y∂x is basically of the same dimension as x*x. Right?

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    11. Saša
      "It would be the same, r or ρ, if either of them would be √(x²+y²)"

      No, only r would be √(x²+y²). Then letter ρ would be free to be defined as anything (for example as ∂ρ = x∂x + y∂y).

      "And there's also a factor 2 missing on the right hand side."

      In which equation?

      "It looks like the same error happened for ∂ϕ"

      I haven't analise ϕ.





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    12. Бьяб прав, в его нотации лучше. Тогда можно было бы ещё указать, что \rho=\log(r), а \varphi=\arctan(y/x)

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    13. "You have three vector fields. How do you get from them just one field? You understand Igor?"

      Вы спрашиваете, как из трёх векторных полей получить дифференциальную 1-форму, чтобы задать слоение?

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    14. Simply: Your sentence "векторное поле ∂ϕ ..." makes no sense for me.
      So, please, explain or fix.

      Delete

Thank you for your comment..

The Spin Chronicles (Part 3): Spin frames

What’s a Spin Axis, Anyway? This post was inspired by a question from Bjab, who asked: "What do you mean by ‘spin axis’?" Well, bu...