Clifford Algebras and Time Loops
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- Опубліковано 3 жов 2024
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Hey UA-cam: New episode of mathmajor chat is up on the second channel. I talk with Nate Mankovich about his Ph.D. project on the geometry of big data. We touch on the transition from studying pure mathematics as an undergraduate to this applied field. Here is a link: ua-cam.com/video/WKcTyGMJiz4/v-deo.html
Why does this make no sense? Good topic but just went off the rails. Any chance of a longer version of this video, maybe a short series?
I never knew persons understood spinous from the ground up. Amazing that they like algebraic mathematics with it. Good luck. 👍
I’m not quite following everything in this video. Though, I think I finally understand a common definition I’ve seen for geometric algebras: something like “a vector space over a field with a quadratic form.”
I was wondering if this is part of a playlist? Where the elements are built up? Does the abstract algebra playlist build up to this?
LOVE SEEING MORE ALGEBRAIC STRUCTURES CONTENT! Been trying to learn this to no avail for a while now, but you make the explanation so simple! Thanks for producing this content!
Same! I'm even trying a wee bit too hard to generate engagement for it LoL 😅
The reason r=0 Clifford algebras are so much more common is simply because every Cl[p,q,r](R) algebra can be represented by a Cl[p+r, q+r, 0](R) algebra. If x^2 = 1 and y^2 = -1, then (x+y)^2 = x^2+xy+yx+y^2 = 1 - 1 = 0.
This is also why _The_ Clifford Algebra, is Cl[∞,∞,0](R) rather than Cl[∞,∞,∞](R). One nice property about Clifford Algebras is that if you start with any specific subspace, it's impossible to escape that subspace without introducing an object from outside that subspace. This means that every Clifford Algebra can itself embedded within a higher dimensional version, up to Cl[∞,∞] and it makes no difference to the calculations.
In physics the p,q Clifford algebra is interesting because (3,1) or (1,3) can be used as a coordinate system for spacetime in special relativity (SR). The spacetime equivalent of Pythagoras requires that when you sum the squares you have to _either_ subtract the square of the time difference, _or_ subtract each of the square of the spatial differences.
This leads naturally to the idea of Minkowski space as the natural way to represent spacetime coordinates. This can be defined independently of Clifford Algebra, and usually is in physics classes, but defining Minkowski space as a Clifford Algebra allows the adoption of results already proven by mathematicians.
That the same applies to General Relativity follows from the GR axiom that any "sufficiently small" part of spacetime in free fall approximates "sufficiently well" to the space of SR. The presence of significant masses distorts the metric at larger scales, so that the coefficients used in the metric are no longer all of absolute value of one: in other words space become curved.
Nice video. From a physics perspective the r=0 Clifford algebras are the interesting ones as they are intimately connected with the spin of particles. To be more concrete, spin 1/2 particles are described by spinors and transform under a certain representation of the Lorentz group. The Lie algebra of the Lorentz group in this particular representation can be expressed via the gamma-matrices. These gamma matrices fulfil a certain anticommutation relation that tells us that they form a matrix representation of the Clifford algebra Cl_{1,n-1} if spacetime has dimension n.
I'm gonna pull the undergrad student real hard now (and annoy the heck outta ya) and ask: what's a Clifford Algebra, if you don't mind me asking? 😅😅😅
//in my defense I wasn't gonna, but after watching it a little I decided that this video, person, channel and entire field of content deserve more engagement, and this would probably help a bit... 💁🏻♂️
@@mildlifeisatrisk5727 Sorry not sure if you're confused but the video we're commenting under is entirely about Clifford Algebras
The relationship between Clifford algebras and spinors originated from division algebras. And the commutators of the elements of Clifford algebras gives you the Lie algebras of the Spin groups, which spinors are the certain representation of, O(1,3) is a pseudo orthogonal group. It’s not just Cl(p,q) that gives you a spinor, Cl(10) gut has some nice paper written around it, Cl(3) is your Weyl spinor, with elements being the Pauli matrices
Was going to comment this, but figured it already got posted. I'm not sure why the math community didn't take up r>0 though.
@@InfiniteQuest86 Computer graphics and physics commonly use Cl(3,0,1), or 3D Projective Geometric Algebra, but otherwise any r>0 algebra such as Cl(p, q, r) can be simulated by Cl(p+r, q+r, 0), since by pairing up a positive and a negative basis vector, together they will square to 0. Conformal Geometric Algebra Cl(4, 1) is an example of this since it uses the two extra basis vectors to create a pair of null vectors, one of which can be used as a drop-in replacement for the null basis vector of PGA.
The reason why the degenerate basis is used in computing is largely because it's more efficient since you only have a 16D multivector space rather than a 32D one, along with the fact that the 4D vectors fit very nicely into SIMD registers. A simpler mental model is also a bonus since you can model 2D PGA in 3D space, where as a complete picture would only be able to handle 1D CGA which doesn't tend to be very interesting.
Maths begins at 3:16.
I expected an introduction to Clifford algebras via the route of Hamilton's quaternions and their generalizations to be explained in a future video. What I got was John Titor. Well played, Professor!
You might like to do a video on geometric algebra, which is Clifford algebra but with a little twist.
@23:28. You called those vectors with x_i^2 = 1 "even", but said that you weren't sure if that was a good name. In physics, we call the vectors with +1, 0, and -1 squares EITHER: spacelike, lightlike and timelike respectively, OR timelike, lightlike, or spacelike respectively. The two different possibilities depend on the quadratic that is being used by the author (i.e.: the metric convention). There is no general agreement, but I understand that in QFT, +1 usually means timelike, whereas in GR, -1 usually means timelike.
In addition, "even" is already used for something else in the context of Clifford Algebras, namely the elements that only contain components of even grade. This forms the "even subalgebra" of a larger Clifford algebra. Ie: grade 0 scalars, grade 2 bivectors, grade 4 vectors, grade 6... etc. An interesting thing to note is that from what I can tell, the even subalgebra is closed under the geometric product, and it's mainly used to represent various transformations.
@@angeldude101 Indeed. In fact, the even subalgebra is the complex numbers for 2d Euclidean vectors, and the quaternions for the 3d Euclidean vectors.
I believe the original Clifford algebra was a set of symbols e1, e2, e3 etc with ek^2=1, ei ej + ej ei = 0 (ij) [1]. They have got something to do with Euclidean translations and rotations but even that much curdles my brain! I should like to see the connection between this (doubtless specialised) case and the development with quadratics.
[1] George Temple, 100 Years of Mathematics; London:Duckworth 1981.
To answer the last question, notice that having a basis z with z*z=0 is almost equivalent to having extra bases p,q with p*p=-1 and q*q=1. This is because (p+q)(p+q) = p*p + pq + qp + q*q = -1 + pq - pq + 1 = 0.
So if you want a Clifford algebra with null vectors (ie v such that v*v=0) you can do so without an explicit basis vector that does so.
Using spaces with null vectors directly is useful in computer visualizations using geometric algebra, since an extra bases are a lot of extra dimensions (more memory usage + computation). There, bases that square to 1 represent reflections over a Euclidean plane, and the base that squares to 0 represents a reflection over the “plane at infinity”. Then build from there (rotation = 2 Euclidean reflections, translation = Euclidean * projective reflections, etc)
Well it’s not. The z’s you’re talking about pair with EVERYTHING to be 0, not just themselves, since the basis gives a diagonal pairing. The p+q you created only pairs with itself to be 0, but the basis isn’t diagonal anymore since pairing with p or q gives something nonzero. In fact, the set of vectors v for which (v,w)=0 for all w is a well-defined subspace Null(q), and gives a SES 0->Null(q)->V->Reg(V)->0. Reg(V) is either the quotient as described by this SES, or a non-canonical splitting of this SES (that is, a choice of lift V’ of this quotient to the original space), and is such that the pairing is perfect (gives a bijection between V’ and V’*). I think the point is that there just isn’t anything interesting to say about Null(q), which is why we assume it doesn’t exist.
@@noahtaul It only pairs with itself because all basis vectors in Clifford algebras pair with themselves. You are correct though in that it isn't diagonal anymore unless you ignore instances where the two are not seen in the form x(p+q). Luckily, I've found that Clifford algebras don't seem to let you escape the subspaces that you started with, so unless you somehow get a lone p, the algebra itself doesn't seem to separate the two by itself.
Too early to think about intro, really enjoyed waking up to this video, infinite respect to you Michael Penn.
That's an amusing intro!
Maybe also looking into the mother algebra? Infinite dimensional Clifford algebra. Great video!
maths lore ???
This video helps a lot although I am far from complete understanding. A group that I am following uses Clifford algebras for their version of a unified field theory. One video from Dr Penn helps me understand equivalent to hours of independent study of this material. These particular examples used helped a lot. Thanks
yes! more algebraic structure expositions!!
I don't know why mathematicians don't Clifford Algebra's with vectors that square to zero, but I will point out that the Clifford algebra where all vectors square to zero is just the fermion algebra from many particle QM, since all creation/annihilation operators square to zero and anticommute.
The most wonderful thing is that once you can build the original foundation it’s based on then how you want it is what you can pursue.
Unless I've missed a nuance in the definition, the Clifford algebra with vectors that square to zero is just isomorphic to the exterior algebra, which mathematicians do use. They just usually won't call it a Clifford algebra.
@@jaredbrown2185 Yeah, that true it’s equivalent to the exterior algebra
Inspired by the word “nilpotent” for 0-squaring elements, I tend to refer to +1-squaring elements as “unipotent” and -1-squaring elements as “antipotent”.
UA-cam autocorrect seems to think “unipotent” is a word, but “antipotent” isn’t. If someone knows better terminology please do tell!
Definitely ,MORE PLEASE!!!!
love that intro
It would be interesting to continue looking into these Clifford algebras, maybe looking into the Clifford calculus a bit. Relating to the Cl with r>0, I think it's mostly used in computer science because it allows for efficient geometric transformations, so it can be used in 3D engines (I'm pretty sure its possible to replace each basis vector in the r case by 2 basis vectors 1 in the p case and another in the q case, but that would imply computational overhead so it is still more efficient to have vectors in the r case)
Yes. I think each additional dimension doubles the dimensions of the multi-vectors produced.
that intro was great
great intro
24:59
On 21:57, there is a mistake on the quadratic formula, because is missing squared at each alpha and beta coefficients
I mean, on the greenboard, the wrong equality is
Q(a1x1+...apxp+b1y1+… +bqyq+g1z1+…+grzr) = (a1+… +ap) - (b1 +… + bq)
So, the right equality is
Q(a1x1+...apxp+b1y1+… +bqyq+g1z1+…+grzr) = (a1²+…ap²) - (b1²+… +bq²)
Was wondering about this
Clifford, my favorite torus / Kaiju canine! ^.^
The Clifford algebras with elements squaring to zero form the basis of projective geometric algebra, and as other commenters have noted it's applicable to computer graphics - but also, to space groups in crystallography, as you can describe the group operations in terms of flectors, rotors, and motors (corresponding to reflections, rotations, and translations).
I'd love to see a video where you take the purely geometric, intuitive perspective of Clifford algebras and build up the formalism piece by piece. It's one thing to start with a tensor algebra, but I think a process of discovering the formalism as you go along would be great for helping others realize how these definitions are constructed.
at 21:00 (x_i x_j = -x_j x_i), is it true for all i and j or just i != j, since if i=j, x_i^2 = -x_i^2 => lambda_i = -lambda_i => for all i, lambda_i = 0?
Do the Quaternions form a Clifford algebra from a 3D vector space over the reals, where each lambda is -1?
Yes it's the (3, 0) Clifford algebra over the reals
Quaternions can be generated from Cl(0,2), which gives a 4D vector space where the non-real bases all square to -1. {1, x, y, xy} In practice, it seems to be more common to see the quaternions used as Cl+(3,0), which is the subalgebra of Cl(3,0) of terms of even grade, so the quaternion bases are {1, xy, yz, zx} Cl(3,0) as a whole would have the bases {1, x, y, z, xy, xz, yz, xyz}.
In the summary the coefficients in the quadratic form should be squared:
Q(.. alpha_i x_i .. + .. beta_j y_j + .. + .. gamma_k z_k ..) = (.. + alpha_i^2 + ..) - (.. + beta_j^2 + ..)
Love the intro
Why starting from tensors to define a Clifford algebra?
@21:05. You said x_i x_j = - x_j x_i for all i, j. I think that should have been for all i not equal to j, since if it held for all i == j, it would mean that x_i^2 = 0 for all i.
right, but the argument seems to still go through when i = j (since i =/= j isn't used in the argument), so I'm not sure what's going on there...
@@schweinmachtbree1013 There's an implicit assumption of not equals hiding there. Suppose that x_i = a, x_j = b (to make typing easier). Since (a+b)^2 - a^2 - b^2 = a b + ba, he gave meaning to this symmetric sum by calculating Q(a+b) - Q(a) - Q(b) = (lambda_i + lambda_j) - lambda_i - lambda_j = 0. However, for a = b, we have (a+b)^2 - a^2 - b^2 = 2 a^2 -- i.e.: Q(\sqrt(2) a) = 2 lambda_i
e 0 (unless lambda_i = 0.)
Or more simply, if a = b, Q(a+b) - Q(a) - Q(b) = Q(2 a) - 2 Q(a) = 2 Q(a) -- which isn't zero unless Q(a) = 0.
@@PeeterJoot Ah yes of course, I must have been very tired last night not to see that xD
I am commenting just to generate engagement for the video, because it's so well deserved and the platform only measures it through engagement. So here we go! 💁🏻♂️
Engagement includes liking as well as commenting (if I understand correctly). So if your motive is to help Michael become more visible to the algorithm it is worth doing both on every video offered :)
What is the connection to "time-loops"?
I believe it relates to the opening skit by the interns, the sub-plot is that they are looping in time
I was wondering the same. But what do the interns looping in time have to do with Clifford Algebras?
As a physics tutor who taught General Relativity before I retired, the connection I made was that you can represent spacetime as a four-tensor from a Clifford algebra where (p,q) is either (3,1) or (1,3) depending on your sign convention.
In some deliberately artificial scenarios you can create a spacetime in GR where by traversing a special path you can end up arriving back at the space and time you started from (ie colloquially a loop in time. If you reach your starting place before you left then in the jargon it's called a "global causality violation")
In 1974 a paper to demonstrate such a closed loop proposed having an infinitely long cylinder of significant mass per unit length rotating about its long axis. Tipler showed that by traversing around the cylinder you could arrive back where you started at the time you started out despite taking time to go round the loop, or even (if I remember correctly) to arrive before you left. The paper was therefore entitled "Rotating Cylinders and the Possibility of Global Causality Violation".
The Tipler cylinder time machine has also been used by science fiction writers, and notably Larry Niven called his short story about a project to build one "Rotating Cylinders and the Possibility of Global Causality Violation", borrowing the title as well as the principle from Tipler.
See en.m.wikipedia.org/wiki/Tipler_cylinder
I have found myself repeatedly wondering if Tipler actually wrote that paper. Perhaps he just copied it from the journal that printed it, and then traversed such a path in order to submit the plagiarised item to the journal's editor...
Genuine question: can't we split Cl(V,Q) into the direct sum of a Clifford p,q-algebra and "the rest"? I haven't worked it out, but it sounds plausible, no?
Kind of: taking the Clifford algebra transforms direct sum to graded tensor products. Over the real numbers, this gives that a C(p,q) is always a matrix algebra over a C(p-q,0) or a C(0,q-p) depending on the sign of p-q. You can look it up in Karoubi's book on K-theory for instance.
@@clementdellaiera3144 Super merci beaucoup! Ca me donne enfin une excuse d'ouvrir ce monument.
I dig it.
I jumped in the middle of something. How about a list in order from ignorance to enlightenment (algebra-wise, anyhow) which leads up to this?
20:30 how do they anticommute? why didn't the algebra in the last example anticommute like that?
I wondered about it too, but It is because in the algebra before, the relations were not given in the eigen-vector basis. You need to diagonalize the matrix of the quadractic form first and then find the basis of eigen-vectors (the matrix U in the video). The new basis vectors (that you then have found) anti-commute.
I've been waiting for you to cover this!
22:50 - I think there's a typo: all those alphas and betas in the RHS of Q(...)=... should be squared.
Bet you can't simulate a Hexa-flexagon in JavaScript tho
There's a lot of squares missing on that last board!
You guys are so silly 🤣 Good Job 👌
I've often wondered if there was any way to generalize tensors to transform arbitrary Geometric Algebra objects instead of just the 1-vector subspace. It's pretty clear how this would work for the M(p,p) × Cl(p,0) -> Cl(p,0) case, at least: it's a change in basis. Except then you have (sI)(a + xe1 + ye2 + be1e2) = a + sxe1 + sye2 + s²be1e2 != s(a + xe1 + ye2 + be1e2) = sa + sxe1 + sxe2 + sbe1e2...
r=/=0 gives you projective geometry which is used in graphics rendering all the time. Cl(p,q) is more interesting because that gets you spacetime algebra which is a unifying language for physics
What is that song in the outro?
Huh, a couple of days ago I drew almost that exact image in your thumbnail with clocks in an infinity symbol and it wasn't about clifford algebras, though I am a big fan of those.
I was looking at modular arithmetic with odd prime power modulii where half the modulus, p, is zero since ceil(p/2) is always 1/2 and adding 1/2 to floor(p/2) always means you are adding its negation. There are many more (infinite?) zero crossings if you start considering the modular rationals, however, so it's really just the first twist of many.
Nice!
I'm confused as to why in the general case we found that xy + yx always equals 0, but in the particular case where Q(ax + by) = 2a² - 3ab - b² you derived that xy + yx = -3.
I see the -3 comes from the -3ab term in the definition of the quadratic form, but it seems like a contradiction, no?
I think this is because xy+yx = 0 when the quadratic form is diagonalized
xy+yx = 0 only if x and y are orthogonal to each other. In general for grade-1 vectors x and y, xy+yx = x . y = |x||y|cos(phi), where phi is the angle between them. If they're orthogonal, phi is 90 degrees, and cos(90 degrees) = 0.
Similarly, xy-yx = 0 only if x and y are colinear, or pointing in the same or opposite directions.
I think that can be connected with the Gr¨obner basis...
the number theory/geometry/integration trick videos on this channel are nice and all but the algebra videos are definitely the ones that get me excited
Literally an algebra video
This rigorous definition is completely useless as it doesn't give you any insight on why this would even matter at all. It is such an arbitrary concept. It makes a lot more sense when presented from the point of view of geometric algebra.
the 3 minute intro was annoying and i skipped it.
Most people liked it.
How do you know it was annoying if you skipped it? Alternate timeline?
Fortunately, I was able to skip the intro (3:16), and end early.