Spinors for Beginners 12: How the Spin Group Generalizes Quaternions to any Dimension
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- Опубликовано: 9 июл 2024
- Full spinors playlist: • Spinors for Beginners
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Powerpoint slide files + Exercise answers: github.com/eigenchris/MathNot...
0:00 - Introduction
2:45 - Terminology overview
4:00 - Reflections in 3D space
9:00 - Reflections in 4D spacetime
13:20 - Rotations in 3D space
22:07- Exponentials
26:37 - Rotations + Boosts in 4D spacetime
32:07 - Galilean Boosts
32:42 - Spin(n) Groups
40:27 - Grade Involution
42:07 - Spin(p,q) Groups
44:15 - Transforming Multi-vectors
46:34 - Hestenes Definition of "spinor"
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Chris. I love your site. As a math and physics geek from the 70s, I often ask "where was eigen Chris back then". I love your point of view. Thanks. MetricTensor!
the pin group being derived from crossing s in spin group is just perfect
we must have more tomfoolery like that
On the other hand, rotations do keep lengths and the position of one point (or axis/higher dimensional thing) “pinned”
I've been thinking about this video dropping for weeks, so glad it finally did!
Truly a blessing to the world
Great video as always, keep the good work, we appreciate it!
Enjoyed the video Chris. Great work!
Excellent work. Thank you.
Thank you. I'm preferring your take on the terminology. We all headed straight into confusion a few decades ago when we couldn't agree on what 'spinor' meant to those of us using clifford algebras in physics. Hestenes had his way, but I'm come to prefer 'rotor' for the defining element of 'rotation' operations. Spinors should be what we need them them to be for quantum and polarization. 8)
11:23 The hyperbolic angle is defined as the half area according to the arc length along the hyperbolic curve for my knowledge. So I think that the end point of Vreflected should be positioned such that the reflection axis (cyan line) divide the area of the diagram composed of V, Vreflected and hyperbolic curve exactly in half (or divide the length of arc from V to Vreflected along the hyperbolic curve exactly in half).
Performing the inverse of the flim-flam operator is what always trips me up, help! Kind request to cover that next video!
I don't get the diagram at 32:00. Reflecting the mauve line (in -t direction) across the mirror (pale red) would give a line near the -x/+t lightline but the diagram shows the purple line near the +x/+t lightline. What am I missing or is this a mistake?
"I'm in pain without the a"
"Wait, you're in pin?"
lets gooOooOo new videooo
Great video, as always! Thank you :) When you say that V is outside of the plane in 19:22, does it mean perpendicular to the plane?
Yes. I didn't phrase that very well.
I find it a bit disturbing to use "length" for the number of vectors in a multivector, as it has little to do with the usual concept of length. "multiplicity" would have been better IMHO.
As usual, thanks for the great contents.
"Grade" is the more common term for k-vectors. Alternatively, you can call products of 2, 3, or 4 1-vectors respectively "bireflections", "trireflections", and "quadreflections", though this makes more sense when visualizing 1-vectors as mirrors rather than arrows.
I initially used the word "grade" but I wasn't so sure if I should use it. What I call a "length 2 versor" ends up being a scalar (grade-0) + a bivector (grace-2). So I'm not sure what the proper terminology is.
3:01
This only gives k-blades, not k-vectors generally.
x1^x2+x3^x4 is a 2-vector, but not a 2-blade, so it can't be made by wedging vectors. Right?
Great Video! As always! Between 31:26 and 32:06 should it be Gamma_x instead of Gamma_z in front of sinh?
Yeah, whoops. My fault.
Have you been focusing on Quantum field ?
Hi Chris, part way through another good video. I couldn't understand the formula for tau at 25:29 in the last line (is tau here a unit vector?), or where the alpha squared comes from in the equation.
I don't think Chirs have seen your question yet, so may I leave my answer first? Here, Chris just replaced the vector αsin(𝜃/2)−𝛽sin(𝜃/2) with the vector 𝜏. And we can see that 𝜏 is also a unit vector, i.e., 𝜏^2=(αsin(𝜃/2)−𝛽sin(𝜃/2))^2=1. Therefore α𝜏 is a geometric product of two unit vectors. In mathematics, such a quantity that is defined as a geometric product of two unit vectors is called a versor. It is represented by a scalar plus a bivector (a rotor in geometric algebra) as shown in video. It is indeed isomorphic to the unit quaternion.
33:00 is a "plane" mistake (pun intended 🙂). There are misconceptions here about rotation planes in more than 3 spatial dimensions.
The error is already happening in 4D, so I will focus on that case. Starting in 4D, the 2-vectors (multivectors of grade 2, that is, arbitrary linear combinations of e_i e_j bivectors) are actually distinct from the 2-blades (exterior product of two vectors). In 4D, the 2-vectors have 1 more degree of freedom than the 2-blades! A 2-blade represents an oriented plane with magnitude (a 'simple bivector') but this only has 5 degrees of freedom. A 2-vector has 6 degrees of freedom, so it is something greater than a mere rotation in a plane.
But 4D rotations (SO4) have 6 degrees of freedom, and indeed they are generated by exp() of 2-vectors. While some 4D rotations may happen in a single plane (as generated by exp()ing a 2-blade), those are a special case. In general 4D rotations are 'double rotations' that happen in two orthogonal planes at the same time, and the rates of rotation in the two planes can be totally different. As an example, you can rotate a bit in x-y plane, then rotate a different amount in z-w plane. The x-y rotation and z-w rotations commute by the way.
(By the way since the two rates of rotation can be anything, this means a spinning object in 4D can, in a sense, take several turns to return to close to its original orientation. It may never return exactly if the rates of rotation have an irrational ratio.)
Anyway, this also means that discussing exp(B*theta/2), i.e., decomposing the 2-vector into B*theta, is mathematically valid but misleading. First, there are conceptually two rotation angles, not one. We can use exp(B*theta/2) anyway but then that 'theta' does not quite have the meaning of an angle which we can see by the fact that adding 4pi to theta produces a completely different rotation in general. I'd say that normalizing the bivector B is conceptually not helpful since we can't separate out a clean angle this way.
It's not just rotations that aren't 2-blades. Angular momentum is also a full 2-vector. You could take a sphere in 4D and torque it in the x-y plane, then torque it in the z-w plane, and then every point on the surface of that sphere could be in motion (no stationary 'poles'!).
In D>=2 dimensions, the 2-blades have 2D-3 degrees of freedom, and 2-vectors have D(D-1)/2. Rotations can be decomposed to rotations on floor(D/2) orthogonal planes, where for odd D the leftover axis is the fixed point of rotation (e.g. the fixed axis in 3D rotation, the bald spot of the hairy ball theorem).
(Thanks for the wonderful video series by the way. I used to be a condensed matter experimental physicist and I always found spin to be a bit mystifying even though my work directly involved spin and spinors at times. I just shut up and calculated.)
23:41 to complete the proof you also need to prove absolute convergence (so you can reorder the terms), which amounts to proving e^θ is finite
It shouldn't be difficult to do. It should be clear that n! grows much quicker than θⁿ so the series must converge for all values of θ. I think it's fine to skip some of the rigour for the sake of ease and clarity.
looking forward for lie groups!
Its a bit unrelated to this point in the series, but how does ths spinor(n) spaces relate to the spin ½, 1, 1½...? If i remember correctly all those spin systems like the spin ½ system are the unitary representations of SO(3) in various dimensions. But now i am a bit confused about how all of this relates to the physicists and chemists point of view. Im guess you are going to go over it in the section about lie groups
Vectors transform with 2 members from Spin(n) (i.e. two SU(2) matrices). Spinors transform with 1 member from Spin(n) (i.e. one SU(2) matrix). The "spin" number counts the number of Spin(n) elements are needed to transform an object when we change coordinates.
@@eigenchris thank you for your reply and for this amazing video series! 😁
So "spin" number 1½ would need 3 elements of SU(2)?
@@Duskull666Yes. It would have 3 spinor indices, requiring 3 SU(2) matrices. Just as a vector has 2 spin indices, requiring 2 SU(2) matrices. Usually for anything that's 3 indices or greater, you need to ditch array notation and stick to summation notation.
Amazing as always.
I wonder, are Cl(m,n) and Cl(n,m) homomorphic for any natural n and m? I know that it's true for 1,3 since the n special relativity the difference between spaces with signatures +--- vs. -+++ is only up to a sign.
They're not homomorphic. By its very nature, Cl(1,3) has 1 vector that squares to +1, and 3 vectors that square to -1 (same for the trivectors). And Cl(3,1) has 3 vectors that square to +1 and 1 vector that squares to -1 (same for the trivectors). However, their "even subalgebras" are homomorphic: they each have a scalar that squares to +1, a quadvector that squares to -1, and 3 bivectors that square to +1 and another 3 bivectors that square to -1. So you can switch back and forth between Cl(1,3) and Cl(3,1) if you restrict yourself to the even subalgebra only. When it comes to doing rotations and boosts, only the even sub-algebra is used, so this is why we can get away with doing special relativity using both + - - - in Cl(1,3) and - + + + in Cl(3,1).
@@eigenchris Right! I missed this obvious distinction. What I meant is that they are somehow "equivalent" in the sense that they can be used interchangeably for flat space-time. Thanks for the correction and the insight :)
Cl(4,0) and Cl(0,4) are isomorphic to eachother (and Cl(1,3)!!)
34:25 I have a question. When you're doing rotations in 3D using quaternions, you can instead of using the inverse have just a conjugate of the quaternion on the right side. Done like this non-unit quaternions carry out rotation with (isotropic) scaling by a non-negative factor. What if you wanted to do the same using Clifford algebras in higher dimensions? What would be the analogue of conjugation? Would that also be just rotation+scaling or would there be more degrees of freedom?
The "Clifford Algebra" version of quaternion conjugation is the reversal operator. If we take "k = -σxσy", conjugating "k" gives us "-k" and reversing "-σxσy" gives us "-σyσx = +σxσy". This is a special case of the more general "inverse" operation, where we don't need to divide by the squared length, because the length is 1.
@@eigenchris Thanks. So what about wedge products of more than 2 vectors? Is their order totally reversed? So something like σ₁^σ₂^σ₃^σ₄ becomes σ₄^σ₃^σ₂^σ₁? And what happens if you use this, rather than inverse, for rotations in higher dimensions than 3D? In 3D using quaternions like this produces rotation with scaling and that makes sense because rotation in 3D has 3 degrees of freedom whereas quaternions have 4 so the one extra degree of freedom can describe scaling (if quaternions aren't normalized). If I understand correctly, rotations in e.g. 4D would need 6 degrees of freedom whereas the even degree subalgebra of Cl(4,0) seems to have 8 degrees of freedom. So there seems to 2 extra degrees of freedom so it might be able to describe more than just rotation with isotropic scaling, but it's not clear to me what.
@@seneca983 Yes, that's how reversal of 4 elements works. If you multiply a k-vector and its reverse together, you just get the products of squares, so you'll get an extra scaling factor proportional to the squares of the vectors involved in the versor.
As for your question, I'm not sure. There are only as many rotations in a given space as there are bivectors (the number of pairs of axes). But in larger and larger spaces, the even subalgebra will get bigger and bigger, and include all possible combinations of bivectors as well. Spin(10) would have 45 bivectors but the even subalgebra would be 512-dimensional. I'm not sure what the extra degrees of freedom mean, other than the fact that they include combinations of rotations along mutually exclusive bivectors (e.g. a rotation along xy and another along tz).
@@eigenchris Thanks. I guess I'll have to try it in e.g. 4D when I have the time and energy.
hey can u make a video on a map of math and physics and how they intertwine together at each level(from school to uni to research)???
@eigenchris
Sorry, I'm focusing on this series on spinors right now.
Knowing that γ matrices can be written in terms of σ matrices in the Weyl representation
ex: γx = | 0 σx |
| -σx 0 |
does this imply a deeper connection between CL(3,0) and CL(1,3) ?
I'm not sure if it says anything specific. It's common to write "larger" clifford algebras using 2x2 matrices of "smaller" clifford algebras. For example, the complex numbers are Cl(0,1). We write the clifford algebra Cl(3,0) as 2x2 matrices with entries from Cl(0,1). What you've shown above is writing Cl(1,3) as 2x2 matrices with entries from Cl(3,0). I'm not sure if there's a "formal" way to figure out this process... there probably is. I just don't know it.
As an aside, specifically for Cl(3,0) and Cl(1,3), it's common to re-write γt*γi as σi. This is called a "spacetime split", and it depends on a reference frame defined by γt. Sudgylacmoe has a video called "A Swift Introduction to Spacetime Algebra" where he goes over it.
@@eigenchris I see. my guess is that whatever the connection is, it can explain the sequence n,m of CL(n,m)
why is "smaller" to "bigger" clifford algebras CL(0,1) -> CL(3,0) -> CL(1,3) -> ... instead of n or m = 0,1,2,... ? why can't we write the members of CL(1,3) as 2x2 or 3x3 matrices instead?
@@GeoffryGifari Cl(1,3) is 16-dimensional (16 = 2^(1+3)), and 2x2 complex matrices only give us 8 dimensions. So that's one answer. I don't actually know if it's possible to write Cl(1,3) in terms of 2x2 quaternion matrices. I also don't know why we can't make 3x3 matrices work. Sorry! I assume these types of questions have been investigated before by someone, but I don't know where to look. "Representation theory of clifford algebras" might be phrase to search.
@@eigenchrisFor the case of Dirac matrices, only even dimensional matrices are allowed, this is because the square must be 1 (or -1), so the eigenvalues must be 1,-1 (or i, - i). Also, at least for the case of Dirac matrices, one can prove they must be traceless, probably this can be generalised to other Clifford algebras, it is impossible to have an odd dimensional traceless matrix with eigenvales 1,-1
@@eigenchris Issok. this got me thinking... pauli spin matrices (happens to be SU(2) generator) can have different numbers of row/column (n,n) dependending on the spin. 2x2 matrices is for spin-1/2, while 3x3 pauli matrices work for spin-1. 4x4 γ matrices (which contain pauli matrices) is essentially used for spin-1/2 particle and antiparticle pair. Can CL(3,0) and CL(1,3) be related that way because they deal with half integer spin (fermions)?
yet, in the study of QCD, SU(3) color symmetry has as its generator eight 3x3 Gell-mann matrices. Is it possible for this SU(3) algebra of 3x3 matrices to have an associated clifford algebra (like CL(3,0) and SU(2))?
fascinating.....
Question: can a hyperbolic rotation be seen as a spinor? The hyperbolic angle is in the range from minus infinity to plus infinity has no characteristic half Euclidean angle relationship.
I'm nit sure what yoy mean by "spinor" in this case. I normally think of a spinor as an object that can be boosted or rotated, similar to the way a vector can be boosted or rotated. When a spinor is boosted, is is boosted by half the hyperbolic angle compared to when a vector is boosted.
Thanks! What I mean: with a hyperbolic rotation the picking up of a minus sign at a 2pi rotation and being back at a 4pi rotation is not possible. This is a characteristic of a spatial spinor (Pauli spinor), but not of a hyperbolic spinor. So, can we speak of a hyperbolic spinor?
@@AMADEOSAM The "negative sign after 2pi rotation" is specifically with circular rotations, not hyperbolic rotations. Pauli spinors in 3D can only be rotated in space. But Weyl spinors in 1+3D spacetime can be both spatially rotated and boosted (hyperbolic spacetime rotation).
Thanks for your excellent work! The division in hyperbolic and Euclidean rotation parameters of the Lorentz group can be broken by the introduction of a hyperbolic rotation with Euclidean rotation parameter. This demonstrates the existence of a temporal spinor with the typical minus sign after a 2pi probation and being back original after a 4pi rotation. The paper about this subject is under review at the journal “Foundations of Physics”. Will keep you updated.
What software do you use to create a display like this?
I use Microsoft Powerpoint. It lets you do animations and record slides to video.
@@eigenchris Thank you very much
45:49 … raised to the power of the length of M?
Sorry, I don't follow. What are you trying to say?
47:05 You need to pick a specific ideal don't you? For instance Juvet, Fritz and Sauter 1930 could've chosen the ideal of Matrixes with the first column being zero instead of the second column. You get the same behaviour for spinors but you are *not* allowed to mix elements of both ideals, in fact that would give you the whole L(2,C) in this example.
I still dont get it.. why are the Sigmas matrices instead of vectors. Where is the link between normal and clifford basevectors?? I cant find a formula.
The sigmas are matrices so that they obey the anti-commutation properties. Which formula are you looking for? I don't understand.
Hi! I understand that they are basevectors in clifford algebra. But i cant find an explanation why. Can they be somehow derived from or linked to normal basevectors (ex,ey,ez)? It could be important for my work to understand the connection. "Shut up and calculate" is not my stile.
@@ThurVal I'm not sure what you mean by "why"? All Clifford Algebras above dimension 1 will have bivectors. I didn't think this required further explanation.
@@eigenchris Of course, but doesnt explain a thing. In Clifford-Algebra and related (Base)-Bivectors are products of the Sigmas as Base-Vectors! At least in all docu I know of.
Just one idea: it seems to me that Clifford-Base-Vectors somehow doubles the content of a normal Base-Vector. Their anti-commutation leads to a scalar, the content of the correlated normal Base-Vector.
Ps: is there any docu how Clifford developed his algebra?
Hello! I already tried the matrix-representation of geometric product. It doesnt work for every Pauli-Matrix, because only two are antidiagonal. I could add one diagonal matrix for the scalar produkt and link it to the NUL-Pauli (you know,that fourth one in dirac algebra). But what about Sigma Z? Its diagonal and doesnt fit a antidiagonal matrix-representation of a Bi-Vector. Found one more problem: sigmaX. A Bi-Vector is antidiagonal(-1,1). But simaX is antidiagonal (1,1).
I find it very confusing that when talking about just space, we use Cl(3,0), while when taking about spacetime, we use Cl(1,3), so the σ_x, σ_y, σ_z change behaviour. That seems like a quite unnecessary cognitive load
Yeah, people argue about this a lot. Maybe Cl(3,1) would be better. Most sources I see use Cl(1,3), so just got used to using that one.
Usually in STA (Cl(1,3)) we use the sigmas for bivectors that involve one space basis vector and the time basis vector. We call the basis vectors gammas. I think you mentioned the gammas at some point, but it looks a lot like the rest of the video is still using sigmas as basis vectors. Confused the crap out of me for a minute.
I tried to use sigmas for Cl(n) and gammas for Cl(p,q). Sorry for the confusion.
I think some people use sigmas for STA when they do a "spacetime split" (multiplying all gamma elements by the gamma0 time basis vector for a given frame).
@@eigenchris I think I might have gotten confused about which algebra you were using at some point, then. Yes, as I said, the sigmas are the bivectors produced by the spacetime split. Sigma i = gamma i0. The subalgebra formed by the sigmas and their products is identical to that of Cl(3).
You keep talking about the mirror original to an arrow, but not as much about the arrow itself. Why not ditch the arrow completely and just talk about the mirror?
Rather than reflecting across the mirror original to the z axis, why not just reflect across the z = 0 mirror? Reflecting again across the x = 0 mirror then gives a rotation around the zx axis, right where the two mirrors intersect.
Also there is a subtle difference between k-vectors and k-blades, namely that k-blades are the wedge product of k 1-vectors, while k-vectors are the sums of k-blades. These are only actually different in 4D and higher, and in 4D it's only relevant for bivectors. The exponential function is defined for bivectors that aren't 2-blades, but the formula is more complicated since the bivector might not square to a scalar. If broken into a sun of commuting 2-blades, then Euler's formula can be applied to the two blades independently before multiplying them afterwards as per the power law (which only applies if the two terms commute).
Also, there _aren't_ two versors that do the same rotation. There _are_ two versors that rotate to the same _orientation._ Rotations can go clockwise or counterclockwise, giving two _different rotations_ that end in the same _orientation._ "Rotation matrices" aren't rotations, but are _orientations._
Is that you boomshroom?
I would argue that if two rotations lead to the same orientation, then they would really be the same rotation. Sure, you can interpolate it differently between the identity rotation and the final rotation, but multiplication by the full thing just does the rotation in one go. At least for rotating vectors, there's no way to tell them apart.
@@IronLotus15 A "spinning" particle doesn't have an orientation, only an angular momentum. Angular momentum in macroscopic objects is different between whether they're oriented clockwise or counterclockwise. Why should intrinsic angular momentum be any different? From what I can tell "spin-up" and "spin-down" are really just fancy words for clockwise and counter-clockwise (not necessarily in that order because nothing, quantum or not, spins along a line, only around one).
@@angeldude101 Perhaps spin 1/2 is more fundamental, but it shouldn't it still be the case that the positive and negative versions of the rotation/transformation are indistinguishable for integer spins?
whew... this is *dense*