The complex number family.

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  • Опубліковано 22 лют 2022
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КОМЕНТАРІ • 266

  • @tomkerruish2982
    @tomkerruish2982 2 роки тому +28

    "...associative..." (*sad octonion noises*)

  • @goodplacetostop2973
    @goodplacetostop2973 2 роки тому +29

    17:54 Learning is like rowing upstream: not to advance is to drop back… Have a good day ☀️

  • @ivanklimov7078
    @ivanklimov7078 2 роки тому +138

    i love these recent vids on weird algebras and complex numbers, they're quite helpful too since i'm currently studying complex analysis. great job michael!

    • @jsmdnq
      @jsmdnq 2 роки тому +2

      I wouldn't call these weird... they are quite natural.

    • @Tim3.14
      @Tim3.14 2 роки тому +5

      @@jsmdnq Some of them are natural... Others are rather irrational, and some don't even seem real

    • @rajinfootonchuriquen
      @rajinfootonchuriquen Рік тому

      ​@@Tim3.14 lol 😂

  • @profkrinkels
    @profkrinkels 2 роки тому +35

    10:41 we need to set i = beta / sqrt(abs(beta^2)) so its square will be correct

    • @NicholasPellegrino
      @NicholasPellegrino Рік тому

      I got stuck at that too. Hopefully what you said is what he intended. I guess sqrt should be well defined given beta^2 is real.

    • @danielyuan9862
      @danielyuan9862 14 днів тому +1

      @@NicholasPellegrino Yeah that is what he most likely intended, and it seems necessary that beta^2 is real and, more specifically, abs(beta^2)=-beta^2 is a positive real, so that its square root is a well-defined real number.

  • @freyascats1364
    @freyascats1364 2 роки тому +66

    For those looking for more about this topic, there is a very readable book "Complex Numbers in Geometry" by I.M. Yaglom, Academic Press 1968. The naming of "j" numbers varies in the literature. Wikipedia uses "Split-Complex Number". Yaglom uses "Double Number". In the area of physics research, Walter Greiner uses "Pseudo-Complex Number" ("Pseudo-Complex General Relativity", Springer, FIAS Interdisciplinary Science Series, 2016).

    • @no3339
      @no3339 Рік тому

      I usually just act like the split complex numbers are just the basis vectors of GA

    • @MarekMadejski
      @MarekMadejski 5 місяців тому

      There is also one author, M.E. Irizarry-Gelpí, who uses different terms: „perplex numbers” for numbers with j^2=1 and „nilplex numbers” for numbers with ε^2=0. The latter term is unique to this author. I like these terms more than “double numbers” and “dual numbers”, as those are confusing, whereas “nilplex” is immediately associated with ε^2=0.

  • @rivkahlevi6117
    @rivkahlevi6117 2 роки тому +105

    "a multiplicative identity, which we will generally call one" 🤣🤣🤣 Gotta love mathematicians.

    • @austinlincoln3414
      @austinlincoln3414 2 роки тому +3

      Lmao

    • @hybmnzz2658
      @hybmnzz2658 2 роки тому +11

      It's that or "I" or "e"

    • @GeekProdigyGuy
      @GeekProdigyGuy Рік тому +8

      that is a useful convention for the abstract general setting - after all we do not usually call the identity matrix 1

  • @Alex_Deam
    @Alex_Deam 2 роки тому +62

    There's a nice connection between the hyperbola at the end and light cones in special relativity. You can even show that the equivalent of the Cauchy-Riemann equations for split-complex numbers is a wave equation!

    • @neopalm2050
      @neopalm2050 2 роки тому +2

      Wait how? I'm not getting that.

    • @rajinfootonchuriquen
      @rajinfootonchuriquen Рік тому

      ​@@neopalm2050 it's not hard. I prooved about 3 weeks ago. Now I will see of dual numbers have some kind of CR equations, but just by it's definition it shouldn't.

  • @gudmundurjonsson4357
    @gudmundurjonsson4357 2 роки тому +110

    Fun how the final graphs seem to go from a circle in the negative, to a hyperbola in the positive case, and the zeros case is somehow inbetween them. I wonder if you could some connection to the conic sections if you continuously varied what the squared element was from -1 to 1

    • @PhilBoswell
      @PhilBoswell 2 роки тому +11

      Since those three conic sections are obtained by intersecting a plane with a double-cone at different angles, I imagine the parameter you're looking for would relate to that angle.

    • @elliottmanley5182
      @elliottmanley5182 2 роки тому +9

      The thought that occurred to me is that the hyperbolic trig functions relate to a hyberbola in the same way the standard trigs functions relate to a circle. I wondered if there's a complementary extension of the trig functions that relate to parallel lines. Of course, if there were, it would be completely useless!

    • @angeldude101
      @angeldude101 2 роки тому +26

      @@elliottmanley5182 The circular and hyperbolic trig functions are ultimately just the even and odd parts of the imaginary and real exponential functions respectively. So finding a dual cosine and sine would just be to find the even and odd parts of e^αε = 1 + αε + (αε)²/2 + ... Normally you'd go further, but we already hit ε² so everything past it would be 0. So a cosd(α) = 1 and sind(α) = α. Significantly less interesting than sin/cos or sinh/cosh.

    • @KohuGaly
      @KohuGaly 2 роки тому +9

      @@angeldude101 _less_ interesting? I'd say the fact that sind and cosd are algebraic instead of transcendental is pretty interesting. Especially from computational standpoint, because it means you can do arithmetic with them with fewer approximations.

    • @NutziHD
      @NutziHD 2 роки тому +8

      Visualizing the basis (1,a) of R(a), where a^2 is in R, in a cartesian plane, we can indeed observe that the unit circle, that is all z in R(a) s.t. z•z*=1 (where the * denotes the conjugate) goes from a circle (a^2=-1) to a ellipse (-1

  • @samwinnick4048
    @samwinnick4048 2 роки тому +69

    I think there are a few minor mistakes here. In the second case I think you should instead define i=beta/sqrt(-beta^2) instead of beta/|beta^2|, where for a positive real x, sqrt(x) is its unique nonnegative square root. Similarly in the third case I think you should instead define j=beta/sqrt(beta^2) rather than beta/beta^2.

    • @jkid1134
      @jkid1134 2 роки тому +10

      I haven't seen this stuff before, but something wasn't quite adding up around there.

    • @anto2593
      @anto2593 2 роки тому +9

      i was looking for exactly this comment. Thanks

  • @nosnibor800
    @nosnibor800 2 роки тому +61

    Thanks, this is filling in gaps. I am an Electrical Engineer/Systems Engineer and of course very familiar with complex numbers (except we dont use "i" because "i" is instantaneous current ! - we use j instead). I have also come across quaternions which we use to convert 3-space into 4-space when doing dynamic space transformations in say missile control systems (where trouble is encountered when the angles approach 90 degrees using euler angles). I have also seen your tutorial on duel numbers, which I had not come across before. So the overall picture is emerging. The problem with Engineering is we treat maths like a "tool" in order to do calculations. Whereas you mathematicians look at the "toolbox" as a whole. Thanks Mr Penn, keep it coming !

    • @MrLikon7
      @MrLikon7 2 роки тому +12

      ah yes, the duel numbers :D
      today: pi vs e!
      fight!

    • @lawrencedoliveiro9104
      @lawrencedoliveiro9104 2 роки тому +7

      The “trouble” of which you speak is called “gimbal lock”. We encounter it in CG, too.

    • @nHans
      @nHans 2 роки тому +4

      ​@@MrLikon7 Last I heard, Euler brokered a 5-way peace agreement between _e,_ π, _i,_ 0, and 1.

    • @nHans
      @nHans 2 роки тому +5

      As a fellow engineer, I feel that the tool-in-a-toolbox approach that we take is as it should be. I wouldn't characterize it as a 'problem.'
      There is way too much knowledge out there for any one person to learn in a lifetime. So we have to be very choosy about what we learn, and carefully manage how much time we spend learning. After all, we do need to leave time for ourselves to solve problems in the real world and earn a living!
      I watch these kinds of videos-math, modern physics, biology etc.-because, yes, I do find them very interesting. But I will not be using them in my work-unless I change my line of work 😜. In fact, I don't even use most of the things I learnt in engineering college-they taught us way too many things, hoping to cover all bases. On the other hand, I've had to unlearn and relearn a lot of other things as my work kept evolving.
      No doubt I could study these topics in detail if I wanted to-but that time would have to come from something else that I'm already doing. Priorities!
      Back in college, if I tried to major in all the courses that I was interested in, I'd never have graduated. As it is, 4 years was plenty! 🤣

    • @Grassmpl
      @Grassmpl 2 роки тому

      @@MrLikon7 yes. Epsilon lost to itself. Its square is 0.

  • @pedrocusinato1743
    @pedrocusinato1743 2 роки тому +32

    I think i = beta/sqrt(abs(beta²)), the same with j

    • @taeyeonlover
      @taeyeonlover 2 роки тому +1

      that made me so confused

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 роки тому +4

      ​@@taeyeonlover it's confusing because Michael's notation is confusing; it is better to "write 1/sqrt(abs(beta²)) · beta" to emphasize that the denominator is a scalar multiplication, because if you confuse it for ordinary multiplication then it looks like you can cancel the numerator and denominator ( even more so for "j = beta/sqrt(beta²)" )

    • @taeyeonlover
      @taeyeonlover 2 роки тому +1

      @@schweinmachtbree1013 ah no it was just that the sqrt missing threw me off

  • @kruksog
    @kruksog 2 роки тому +10

    You're awesome Dr. Penn. I'm a guy with a math degree who otherwise doesn't have much of an inroad to mathematics otherwise these days. Your videos bring me that joy of experiencing math I don't get too often in daily life anymore. Thank you!

  • @AleksyGrabovski
    @AleksyGrabovski 2 роки тому +4

    Your videos reminded me how I love abstract algebra.

  • @wladwladsnotmyrealname9082
    @wladwladsnotmyrealname9082 2 роки тому +7

    One arguably nice property of the double numbers (which are defined to be R(j)) is that linear algebra over them is non-trivial. A matrix over the double numbers is a pair of matrices over the real numbers (A,B^T). Addition happens componentwise, and multiplication is sort-of componentwise: (A,B^T) * (C,D^T) = (AC,(DB)^T). The final algebraic operation to define is the "conjugate-transpose" operation: (A,B^T)^* = (B,A^T). Then you have that the Hermitian matrices are pairs (A,A^T) which are a non-trivial embedding of the square matrices, the unitary matrices are pairs (P,(P^-1)^T) which are a non-trivial embedding of the invertible matrices, the "spectral theorem" is just the Jordan canonical form for real matrices, the Cholesky decomposition is effectively the LU decomposition for real matries, the normal matrices are pairs of commuting real matrices, and so on. There is a non-trivial analogue of the Singular Value Decomposition as well.

    • @wladwladsnotmyrealname9082
      @wladwladsnotmyrealname9082 2 роки тому +3

      Hopefully, this won't be perceived as a rant by most.
      I think that the analogue of complex analysis over the double numbers is not as nice as their linear algebra. This is because linear algebra uses the conjugation operation (a + bj)^* = a - bj. Linear algebra over the double numbers depends strongly on this operation. Complex analysis studies the analytic functions, which do not include conjugation. I think that because of the isomorphism R(j) = RxR, a holomorphic function over the double numbers is just an arbitrary pair of differentiable real functions, so double-number analysis is effectively real analysis AFAICT.

  • @TheDannyAwesome
    @TheDannyAwesome 2 роки тому +7

    I also like to consider these number systems as the ring of polynomials over the real numbers quotiented out by a quadratic polynomial. In the case of the dual numbers, x^2, in the case of the complex numbers, x^2+1, and in the case of the split complex numbers, x^2-1.

    • @adityajain16701
      @adityajain16701 2 роки тому

      Those are indeed all isomorphic to their respective 2D R-algebras

    • @angeldude101
      @angeldude101 2 роки тому +3

      1 + -1 = 0, j² + i² = ε²

    • @schweinmachtbree1013
      @schweinmachtbree1013 2 роки тому

      @@angeldude101 nice!

    • @thewhitefalcon8539
      @thewhitefalcon8539 4 місяці тому

      This raises the question of other quotient rings such as real polynomials/x³+x²+1. Maybe they end up just being complex numbers with a weird choice of I or maybe not

  • @cpiantes
    @cpiantes 2 роки тому +6

    Can you do an episode on Clifford algebras?

  • @Skyb0rg
    @Skyb0rg 2 роки тому +3

    You should explain Clifford algebras! Many of those "families" can co-exist in those algebras, and some examples (such as R_3; R_3,2; and R_3,0,1) model 3D spaces really well.

  • @nmmm2000
    @nmmm2000 2 роки тому +8

    Woaw :) Definitely need to render Mandelbrot in Split complex numbers too...

    • @thewhitefalcon8539
      @thewhitefalcon8539 4 місяці тому

      There's a fractal called the Burning Ship Fractal that you get if you mess up the Mandelbrot equations slightly.

    • @nmmm2000
      @nmmm2000 4 місяці тому

      @@thewhitefalcon8539 burning ship utilize standard complex numbers, but get absolute value (modul) from the complex part (is not that easy but this is the basic idea)

  • @Ricocossa1
    @Ricocossa1 2 роки тому +3

    Circles in R(j) and R(\epsilon) are like lines of constant proper time in Mikowskian and Gallilean space-times.

  • @starsun7455
    @starsun7455 2 роки тому +1

    I've been studying hyperbolic for several months, so it was really good that Professor Penn dealt with 'split complex numbers' called hyperbolic numbers.

  • @neonsilver1936
    @neonsilver1936 Рік тому +1

    I've watched 2 of your videos now, and I can already tell that I'm going to learn a LOT from you. Thank you for showing me that I need to improve my mathematics.

  • @camperbbq3026
    @camperbbq3026 2 роки тому +2

    Please do more content on the split complex numbers! I love your videos!

  • @dtrimm1
    @dtrimm1 2 роки тому

    Great videos - I’m loving them, and learning a ton - thanks for doing them!

  • @SiiKiiN
    @SiiKiiN 2 роки тому +1

    I like these style of videos which helps build a landscape for motivation behind algebras

  • @zemoxian
    @zemoxian 2 роки тому +4

    Reminds me of the various geometric algebras. The multivectors have similar bases and can even reproduce quaternions and Minkowski spacetime.

  • @m1323fj
    @m1323fj 2 роки тому

    Really nice video. I thoroughly enjoyed it and would like to see more content like this.

  • @stephenmorton9789
    @stephenmorton9789 Рік тому

    Great stuff, thank you

  • @AJ-et3vf
    @AJ-et3vf Рік тому

    Great video. Thank you

  • @eschudy
    @eschudy 2 роки тому

    Very cool!! More please!

  • @unflexian
    @unflexian 2 роки тому +4

    You're making me love abstract algebra:D

  • @stevenwilson5556
    @stevenwilson5556 2 роки тому +1

    wow so cool. great way to tie this together

  • @neopalm2050
    @neopalm2050 2 роки тому +2

    I predict the clifford algebras and perhaps the weyl algebras at some point in the near-ish future

  • @NutziHD
    @NutziHD 2 роки тому

    Awesome video!!

  • @Grassmpl
    @Grassmpl 2 роки тому +1

    You should never use round parenthesis when the ring extension does not form a field. Use square parenthesis.
    R[epsilon], R[j], R(i) this last one forms a field

  • @jantarantowicz1306
    @jantarantowicz1306 2 роки тому

    as always, great video!

  • @bobdowling6932
    @bobdowling6932 2 роки тому +12

    Typo at 11:05-ish and again a little later. I think you are missing a square root in your normalizations to get i and j. The denominator needs to be sqrt(-beta^2) and sqrt(beta^2) respectively.

    • @badlydrawnturtle8484
      @badlydrawnturtle8484 2 роки тому +2

      Thank you. That makes sense. I had to pause the video and look through the comments for an explanation when that bit didn't follow.

  • @joel.9543
    @joel.9543 2 роки тому +2

    Amazing!

  • @copernic7511
    @copernic7511 Рік тому +1

    That merch is great! No way I'm forgetting that formula now.

  • @sciencewithali4916
    @sciencewithali4916 2 роки тому

    Amazing topic 👏 thank you very much. You second UA-cam channel truly is very ins8ghtful and helpful

  • @nbooth
    @nbooth 2 роки тому +4

    Excellent, thank you. Any chance you'd do a video/series on multivectors from geometric algebra?

  • @sebastiandierks7919
    @sebastiandierks7919 2 роки тому +4

    Maybe a follow-up video on Grassmann numbers, i.e. the the exterior algebra of a vector space, and their differentiation and integration (Berezin integral) would be interesting, as they are a generalisation of the dual numbers! Would love to see that video :) Basically, I don't understand how you integrate over a Grassmann variable. In the context of this video, how would you integrate over the dual number epsilon? It's just a single basis element of the algebra!?

  • @apteropith
    @apteropith Рік тому +1

    these can all be understood as planar rotational algebras for vector spaces with particular metrics, as well
    with the right choice of setup you can get some fairly impressive use out of them (and their higher dimensional analogues)
    such as representing arbitrary translations as dual-number rotations around a paraboloid, which allows the composing of translations with (regular) rotations much more effectively
    ... i don't have much opportunity to make use of this knowledge myself, but it is a subject that fascinates me

  • @Zeitgeist9000
    @Zeitgeist9000 2 роки тому +12

    Love this area of math, thanks for the cool content!

    • @shalvagang951
      @shalvagang951 2 роки тому +1

      which area of math is this I would really love to know

    • @Zeitgeist9000
      @Zeitgeist9000 2 роки тому +1

      I was wondering the same thing

    • @shalvagang951
      @shalvagang951 2 роки тому +1

      @@Zeitgeist9000 I have searched for it and I got the result something about hypercomplex number in group representation theory

    • @shalvagang951
      @shalvagang951 2 роки тому +1

      @@Zeitgeist9000 I just wanna ask you one thing that anytime you hear about any new math topic that has recently been discovered what would be your thoughts about it pls answer

    • @shalvagang951
      @shalvagang951 2 роки тому +1

      like you wanna learn it or just "ahh heck with it I am not doing this"

  • @Nickle314
    @Nickle314 2 роки тому +5

    On the quaternions, the more human understandable approach is bivectors from a geometric algebra, and how they multiply.

    • @angeldude101
      @angeldude101 2 роки тому +1

      It's definitely much easier to see (e1e2)(e2e3) = e1e2e2e3 = e1e3 = -e3e1 than to try and remember ij = k.

    • @Nickle314
      @Nickle314 2 роки тому +3

      @@angeldude101 Agreed. GA to me has been a revelation. Covers large parts of Tensor calculus, Quaternions, Complex numbers, Dirac algebra .... Then you have the physical equations.
      ∇F=J/(ϵ0c)+MI
      One equation, the whole lot.
      On complex numbers, I think there is a case for teaching the GA approach first at a very basic level, to show that things can square to -1 in a real sense.

    • @Jono98806
      @Jono98806 2 роки тому

      Quaternions are actually a subset of the geometric algebra (as are complex numbers and split complex numbers).

    • @Noam_.Menashe
      @Noam_.Menashe 2 роки тому

      It's also a bit like the cross product.

    • @angeldude101
      @angeldude101 2 роки тому +3

      @@Noam_.Menashe But the main difference is that the cross product is evil and needs to die in a fire because it's completely unintuitive both geometrically and algebraically. The wedge product by contrast is very intuitive both geometrically and algebraically.

  • @General12th
    @General12th 2 роки тому +2

    This is a really neat video!
    I don't think I would have been able to follow along if I was just reading through a textbook, even if the book had all the same arguments laid out in the same order. I wonder what makes lectures different and easier for me to digest.

    • @xrhsthsuserxrhsths
      @xrhsthsuserxrhsths 2 роки тому

      I had this problem too, I still do to some extent, but it gets better. It also depends on how familiar is the subject you read.

  • @brattok
    @brattok Рік тому

    In 13:45 we could also say that R(j) is not a field because polynomial x²-1 has more than 2 roots

  • @rv706
    @rv706 Рік тому +2

    Please denote the dual numbers by R[ε], not R(ε), cause it's just a ring not a field; generally the notation with round parentheses is for field extensions, as opposed to ring (or algebra) extensions.

  • @CTJ2619
    @CTJ2619 2 роки тому +3

    I like learning about the quaterians

  • @travisporco
    @travisporco 2 роки тому

    fascinating stuff. Would love to see more applications of the dual numbers

    • @angeldude101
      @angeldude101 2 роки тому +1

      Dual numbers can be used for automatic differentiation since the "imaginary" term acts like the derivative in many ways. While it's not specifically the dual numbers, a basis squaring to 0 is also useful in representing translations of objects, much like complex numbers are good at rotating, and split-complex numbers are good at hyperbolic boosts (more niche than the other two, but very useful in relativity).

    • @drewmandan
      @drewmandan 2 роки тому

      Probably has applications for things involving perturbations. Maybe even stochastic calculus.

  • @drewmandan
    @drewmandan 2 роки тому +1

    This looks so much like general relativity and closed/open/flat space! Cool

  • @wilderuhl3450
    @wilderuhl3450 2 роки тому

    Fascinating

  • @user-hh5bx8xe5o
    @user-hh5bx8xe5o 2 роки тому

    Geometrically, the multiplication of the algebra is a composition of a dilatation and a rotation for C, a reflection with y=x for R(j) and a projection along x=0 for R(epsilon)

  • @radupopescu9977
    @radupopescu9977 2 роки тому +1

    In fact, complex number and split complex numbers, are subalgebras of bicomplex numbers.

  • @Dalroc
    @Dalroc 11 місяців тому

    Love it that Michael can't differentiate between his a's and his u's.
    Makes me feel better about my own a's and u's!... and my z's and 2's. and my I's and 1's..

  • @Khwartz
    @Khwartz 2 роки тому

    Excellent! Very Clear I think thanks of a Very Nice Progression in the Lecture :) (y) (y)

  • @pajrc1234
    @pajrc1234 Рік тому +1

    So when I saw the video on the dual numbers I tried writing down a bunch of relations between it and i and the matrix stuff. I discovered that e^(+-b epsilon)=1+-b*epsilon--the same curve that represents the "unit circle." I figure the same is true with j. Also, (ep*i)^2+(i*ep)^2=1 and (ep*i)+(i*ep)=-1. It's weird because these are not commutative. I'll have to add j to the mix to find other other things. I'm going to see if f(x+b*ep)=f(x)+b*ep*f'(x) works even when b is complex, and complex algebra treats dual numbers the same way

  • @Grassmpl
    @Grassmpl 2 роки тому

    An associative algebra is a ring and a real vector space, with the compatibility of scalar multiplication.

  • @user-fi6if8gx3g
    @user-fi6if8gx3g 2 роки тому +1

    A lovely family indeed.

  • @beatadalhagen
    @beatadalhagen 2 роки тому +1

    Will there be a follow-up involving combining these?

  • @nrrgrdn
    @nrrgrdn 2 роки тому +1

    Thanks, why don't we hear of this in a standard abstract algebra course

  • @gabrielfrank5142
    @gabrielfrank5142 Рік тому

    Some questions:
    1. Is the fact that the dual and split cases are not a field is related (i.e. iff) to the fact there unit circles are not continuous?
    2. Can we define other types of numbers based on arbitrary "unit circle" curve geometry?

  • @_P_a_o_l_o_
    @_P_a_o_l_o_ 2 роки тому +3

    Wonderful content! Will you consider making a video on quaternions?

    • @elliottmanley5182
      @elliottmanley5182 2 роки тому

      If so, how about taking a similar "family" approach and explaining how constraints of addition and multiplication drop away as you move from complex to quaternions to octonions to sedenions?

    • @elliottmanley5182
      @elliottmanley5182 2 роки тому

      A bit more background reading today: I started to get a handle on Clifford algebras and then got lost completely when I got to Bott Periodicity.

    • @cpiantes
      @cpiantes 2 роки тому

      @@elliottmanley5182 8N = N8

  • @aweebthatlovesmath4220
    @aweebthatlovesmath4220 2 роки тому +1

    Numbers are cool!

  • @arnbrandy
    @arnbrandy 2 роки тому +1

    IIUC he worked out the modulus of dual and split-complex numbers by multiplying them by their conjugate, very much like with complex numbers.
    I don't understand why would we use the same definition here. I understand the motivation behind using this definition for complex numbers, it gives the distance from the origin. Why don't we define modulus differently for the other algebras so we get the same property.
    I found it fascinating how the modulus definition leads to different conic curves but I wonder why would it be especially important.

    • @angeldude101
      @angeldude101 2 роки тому

      If you were to just square a split-complex number, then you'd get something like (x+yj)^2 = x^2 + 2xyj + y^2; not quite what we're looking for. The conjugate is needed to eliminate the middle term of the expansion and get a pure difference of squares.

  • @deltalima6703
    @deltalima6703 2 роки тому +1

    Cohl Furey over at perimeter institute did a series relating octonions to QM and GR. Perimeter is legit, as I am sure most people know, having been a phd and nobel prize factory for awhile. It might be worthwhile adding octonions in as well, they are not as useless as you would suspect.

    • @deprivedoftrance
      @deprivedoftrance 2 роки тому

      Now I might need to go re-watch them as it's been a while.
      Those were good ones.

  • @alexey_burkov
    @alexey_burkov 2 роки тому +2

    omg, that preview picture 😂

  • @Grassmpl
    @Grassmpl 2 роки тому +1

    Only in C does the modulus induce a metric.

  • @user-qu2oz2ut2h
    @user-qu2oz2ut2h Рік тому +1

    hmm
    dual and split-complex Mandelbrot set analogs... Sweeps of "i" from minus infinity up to infinity...

  • @CesarMaglione
    @CesarMaglione 2 роки тому +2

    Buena pizarra Michel 👍

  • @catherinebernard3282
    @catherinebernard3282 Рік тому

    Really silly thing to point out, but if you start at 17:23 or so you can hear the chalk break, and if you go frame by frame you can actually see it fall... which for some reason I found amusing.

  • @spudhead169
    @spudhead169 Рік тому

    I think you spilled something down your hoodie there Mike.

  • @elliottmanley5182
    @elliottmanley5182 2 роки тому +2

    Really nice survey that made these (previously, to me) very abstract concepts accessible.
    Wondering if there's an algebra that joins more than one of these imaginary bases: e.g. a four-dimensional R(i,j,epsilon)

    • @elihowitt4107
      @elihowitt4107 2 роки тому +1

      Interesting if this mixture yields anything

    • @angeldude101
      @angeldude101 2 роки тому +2

      Geometric algebra lets you specify the signature of the algebra as how many bases square to 1, -1, and 0. You can have G_1,1,1(R) which has {1, i, j, ε, ij, iε, jε, ijε}. In practice you don't need this since two positive bases already have an imaginary product, and it's possible to make a dual basis vector from a positive and a negative basis.

    • @opfromthestart3645
      @opfromthestart3645 2 роки тому

      I believe that this 4d structure would just be the set of all 2x2 matrices. If you represent i,j, and epsilon as matrices it makes sense

    • @opfromthestart3645
      @opfromthestart3645 2 роки тому

      You could also make them as 8x8 matrices (technically in 3 ways, but all will act the same). You can also make them into 4×4 matrices in three distinct ways, following each pair, which won't commute with the other one in the pair, but they should both commute with the one left out.

  • @Fine_Mouche
    @Fine_Mouche 2 роки тому

    17:54 : "ok that's a good place to start" So it will have a next episode of this series about "hypercomplexe family" ?

  • @JakobVirgil
    @JakobVirgil 2 роки тому +1

    I have noticed that if I want to use dual and complex numbers together that it sort of forces me into quartics but I am having a hard time expressing why. Could you shed some light on this?

  • @tothm129
    @tothm129 2 роки тому +1

    can I give you a closed curve and then say its a unit circle and then you work backwards

  • @timseytiger9280
    @timseytiger9280 Рік тому +1

    Nice, how about redefining the absolute value so we always get the "same" picture?

  • @DirkDanckaert
    @DirkDanckaert 2 роки тому

    Small error in the normalisation op beta: i,j := beta/|beta| and ot beta/|beta|^2

  • @EMAngel2718
    @EMAngel2718 2 роки тому +1

    I wonder if the very similar correlation between basis sign and shape to that of spatial curvature has any meaning

    • @EMAngel2718
      @EMAngel2718 2 роки тому

      I just realized it's actually the reverse but still

  • @marcoottina654
    @marcoottina654 Рік тому

    13:20 is R(j) just the negated complex of C ?
    17:55 also, it seems that:
    -) C is something like "two dimensions swirling all together in a coupled, attractive way, with the "balancing" fact that the radius always stays the same (1)"
    -) R(j) are two "divergent" dimensions, like a "swirling where the interaction leads to divergence, like particle with the same electrical charge"
    -) R(epsilon) seems two completely unrelated, never-interacting "dimensions", which closely resembles the very concept of "parallelism"

  • @geelaw
    @geelaw 2 роки тому

    It's a bit confusing why something in A in the form of x*1+0*alpha is considered to be in R (as you mentioned 1 means the identity in A) --- it would be instructive to first show that R is naturally embedded into A so that x in R can be regarded as x*1 in A.

  • @skillerror951
    @skillerror951 2 роки тому +2

    Looks like curvature of space

  • @przemekmajewski1
    @przemekmajewski1 2 роки тому +1

    Nice video! I have a remark: You've been talking about associative algebras, then suddenly you've switched to "normed algebras", IMO it was too quick a jump. The notion of an adjoint and how to introduce norm are important stuff

    • @przemekmajewski1
      @przemekmajewski1 2 роки тому

      In fact in your very nice proof of the fact how many "2D algs" there are you've shown, that any such algebra is in fact a normed *-algebra, as you nicely constructed the element "beta" that squared to a central element.

  • @VideoFusco
    @VideoFusco 2 роки тому

    Case 3 is also known as hyperbolic numbers

  • @rrr00bb1
    @rrr00bb1 2 роки тому +1

    Look at how Clifford Algebra (aka: Geometric Algebra) is formulated. The Quaternions can be thought of as "wrong" in a way, as they are half of R^3 in Clifford Algebra; which loses information required to perform division, among other difficulties. But GA is so weirdly notated that its obviousness is totally obscured. In it, you get these basis objects that square to -1,0,1.
    // directions in space square to 1
    east * east = 1
    north * north = 1
    // extra names
    east = -west
    north = -south
    // orthogonal directions create a plane of rotation
    // counter-clockwise in (east north) plane
    east * north = north * west = west * south = south * east
    // from this, you can deduce anti-commutativity:
    east * north = -north * east
    // Let's define (east north) = i
    // from this, you can see that such a plane squares to -1
    (east north)(east north)
    =
    (-north east)(east north)
    =
    -(north (east east) north) = -(north 1 north) = -1
    So, "i" can be factored into a pair of orthogonal directions.
    R^3 has: 2^d components. It has directions {east,north,up}.
    (4 east + 3 north)(5 east + 2 north)
    =
    20 east^2 + 6 north^2 + 8(east north) + 15(north east)
    =
    26 + (8 - 15)(east north)
    =
    26 + (8 - 15)i
    When you say "i", it is ambiguous which plane you mean. It might mean "(east up)" or "east north". A pair of multiplied directions creates a thing that squares to "i" in a complex number.
    There is NOT a single square root of -1 called "i"!!! "i" is ambiguous about what plane it is in. That's why Quaternions have the i,j,k bivectors. But they are missing the vectors and pseudoscalar that they should have. If you just cross-multiply out two R^3 multivectors, the Quaternions just contain the rotation part only. Use full R^3 and division works fine. And generalizations to R^4 etc are very straight-forward.

    • @angeldude101
      @angeldude101 2 роки тому +1

      Interesting writing out the bases as full words. (east, north, west, south, up, down) There's certainly nothing wrong with it, it's mainly just the space it takes up, so they're usually shortened to x, y, -x, -y, z, and -z, but ultimately they mean the same thing. A lot of GA literature uses e1, e2, and e3, which _definitely_ makes things harder to follow for someone new to the subject, but they also tend to make things easier to extend to higher dimensions.
      Something that's honestly kind of hilarious is that GA shows that the quaternions are a _left_ handed basis rather than a right handed one like people thought it was. Honestly though, I'd say to get rid of i, j, and k for good just because of how unhelpful the names are. Call them what they are: zy, xz, and yx. (To be clear, zy*xz = yx, zy*xz*yx = -1, unlike a silly right-handed basis like yz*zx*xy = 1)

    • @rrr00bb1
      @rrr00bb1 2 роки тому

      @@angeldude101 I see \hat x, used to mean the normal in the direction of the x axis sometimes. But x,y,z usually mean real, complex; and they are so overloaded that I would rather use full words to get the idea across, like well-documented code. I am bothered by how clear Geometric Algebra CAN be, but every book you ever see on it is impossibly obtuse. If you spell out all of the coordinates, you can see the power of adding directions in space into the algebra. This should be the primary task. Taking lots of shortcuts with dot and wedge should be done long after we are comfortable just multiplying two general multivectors together; and naming the important structure that results from that. This is because the shortcut notation that tries to be coordinate-free is full of special cases when it's not just vectors and scalars being multiplied. It wasn't until I read 2 books, and started a book on how to implement it in a computer that I realized that general multivector multiplication is just cross-multiplying, where the directions in space follow algebraic rules.
      What I really love about it is that "i" is shown to be factorable into a pair of orthogonal directions. It exposes the ambiguity of saying "square root of -1", as if there is just one. The greatest of all is that you can produce all of trigonometry by starting with multiplying a pair of directions in space together. It is then not mysterious why rotations come from complex numbers; and it tells you that when you see it in physical applications, that you need to identify WHICH planes are producing the complex numbers. If forces you to ask whether the orientation is arbitrary as well.

    • @angeldude101
      @angeldude101 2 роки тому

      @@rrr00bb1 I'm not sure what you mean by "x,y,z usually mean real, complex". From what I've seen, x, y, and z either just mean unknown variables, or the 3D coordinate axes.
      One thing that I really like about multivector multiplication is that it's very easily derived from the standard properties like linearity, plus 1 axiom, which is the contraction axiom. The contraction axiom just says that v² = v*v is its magnitude squared (or more generally, any scalar. With that alone, and two arbitrary vectors that you just define as orthonormal, anticommutitivity becomes very easy to prove. x̂ + ŷ clearly has a magnitude of sqrt(2), so (x̂ + ŷ)² = 2. Expand it out: (x̂ + ŷ)(x̂ + ŷ) = x̂² + x̂ŷ + ŷx̂ + ŷ² = 2. Since x̂ and ŷ are both unit vectors: x̂² + x̂ŷ + ŷx̂ + ŷ² = x̂ŷ + ŷx̂ + 2 = 2; x̂ŷ + ŷx̂ = 0, x̂ŷ = -ŷx̂. From there, it's not hard to prove that (x̂ŷ)² = (ŷx̂)² = -1. With only this, it's very easy to generalize to any number of dimensions. All the rules stay exactly the same. At most, the contraction axiom gets relaxed to an arbitrary scalar allowing for negative and null squares of basis vectors.
      Also, sqrt in general is dubiously a function, since it's an inverse of a function that isn't 1-1 even within the reals. -i is just as much a square root of -1 as i is. We just arbitrarily choose one for convenience.

    • @rrr00bb1
      @rrr00bb1 2 роки тому

      @@angeldude101 i would not bother with a contraction theorem; because it's another one of those formulas that makes people get confused by the algebra; when v is not of type vector. In a typical GA book, it's extremely unclear what bivector times trivector mean, etc. "is of type vector" literally means that there are zero co-efficients for everything not in the vector. If you add a vector to a bivector, there isn't really even a name for the type of that multivector. That's why I like the idea of: parallel unit directions in space multiply to 1, perpendicular unit directions in space multiplied will anti-commute. Contraction for vectors falls out of the definition; and you don't get confusing issues like "what's a trivector squared?"
      (a1 e1 + a2 e2 + a3 e3)(b1 e1 + b2 e2 + b3 e3)
      = // cross-multyply it all out
      a1 b1 e1 e1 + a1 b2 e1 e2 + a1 b3 e1 e3 + ....
      if it's "of type scalar" when all coefficients are zero except for 1. "of type vector" is when all zero except for (e1,e2,e3).
      When you write code to do this, you do have issues with approximations though; where you implement rotation of a vector in space with rotors; and have a trivector that's almost-zero due to floating point. So in theory, the calculations are trivial; but in practice, they use complicated compilers so that the approximations at least give the exact type.
      The only real problem I have with GA is that for a long time, I have been unable to figure out what is going on in Geometric Calculus. The gradient operator on multivector inputs doesn't seem to be as obviously defined as for multiplication. Given that (unit) directions in euclidean space square to 1, (unit) rotations square to -1, objects that square to 0 may play a role in getting gradients defined in an obvious way; that fits well with writing computer code. I ran into these things trying to use GA in the context of AutoDiff. ie: use gradient descent to fit parameters.

    • @rrr00bb1
      @rrr00bb1 2 роки тому

      example where square to zero may be useful if also integrated with directions in space:
      // "x is scalar" literally means that it's a multivector where every coefficient is zero, except for component 1
      f[x] := 3x^2
      // define (dt * dt) = 0 ?
      f[t + dt] = 3(t + dt)^2
      try an x that is a multivector:
      dx = da1 e1 + da2 e2 + da3 e3
      x = a1 e1 + a2 e2 + a3 e3.
      da1^2 = 0.
      da2^2 = 0.
      da3^2 = 0
      for that which is "held constant", such as da2 and da3, it would mean that "da2 = 0. da3 = 0".
      Dual numbers to perform the differentiation. Something like...
      // implicit differential
      d[f] = f[x + da1 e1] - f[x]
      // derivative with respect to (da1 e1), where da1 is undefined, but squares to 0
      d[f]/ (da1 e1)

  • @christressler3857
    @christressler3857 2 роки тому

    Using ß for beta..
    I don't think you want,
    i = ß/|ß^2|, unless by |•| you mean something like the *norm*, iow a square root value.
    It might've made more sense to put,
    i = ß/|ß| which looks more like a unit vector than your notation does.

  • @MrRyanroberson1
    @MrRyanroberson1 2 роки тому

    How about the triplex numbers?

  • @antoniusnies-komponistpian2172

    Though j just has the properties that 1 and -1 have in common. j is a kind of Schrödinger's 1 that you give a letter just because you don't know if it's positive or negative 😅

  • @HoSza1
    @HoSza1 Рік тому

    About quaternions: I understand that i²=j²=k²=-1 rules may naturally come as an extension of the complex numbers, but where do the last 3 rules come from? I was possible to derive jk=i and ki=j from the other rules, but I had to use ij=k, and I don't know if this last one can be derived too, or it is a free choice itself.

    • @dantebroggi3734
      @dantebroggi3734 Рік тому

      I believe the `jk=i` form rules are generally derived from a rule `ijk = -1`.

  • @littlecousin5630
    @littlecousin5630 Рік тому

    10:44
    Doesn’t that yield i^2=B^2/|B^2|^2=|B^2|(-1)/|B^2|^2=-1/|B^2|. How do we get to -1 from there?

  • @HoSza1
    @HoSza1 Рік тому

    Well after digging around a bit, it looaks like the defining properties of H are i²=j²=k²=ijk=-1 and the other three "rules" in the video are in fact theorems that are easy consequences of these four. I wonder what happens if we define ijk=1 (or zero) instead, is the resulting structure still useful in some ways?

    • @James2210
      @James2210 11 місяців тому

      Split quaternions? Only j^2 and k^2 are 1

  • @williamwright1440
    @williamwright1440 Рік тому

    Has anyone seen any papers/videos/textbooks on algebras where the products of the basis vectors with themselves are not necessarily real? E.g. an algebra over the reals where the basis vectors are 1 (the real number) and x (not a real number, or anything in particular, for that matter), with 1*1=1 (of course), 1*x=x, x*1=x, and x*x=x.

  • @mvsh
    @mvsh 2 роки тому

    I expected the "circle" in the dual-numbers plane to be a parabola after seeing the circle and hyperbola because of the conic sections… Is there anything that will give us it?

  • @SoonRaccoon
    @SoonRaccoon 2 роки тому

    Now I'm curious if there are values analogous to pi in the other two number systems.

    • @angeldude101
      @angeldude101 2 роки тому

      The hyperbola and straight lines don't converge to anything, so the closest thing they each have to pi would probably just be infinity.

  • @Raptormind1
    @Raptormind1 2 роки тому

    I feel like I'm missing something. at 5:50, x+y(alpha) is just shorthand for x(1)+y(alpha) where 1 is the multiplicative identity of the original vector field, right? So at 7:31, isn't 4a+b^2 shorthand for 4a+b^2 times the 1 from the vector field? So shouldn't beta squared be a real number times the 1 from the vector field, not just a real number? What am I missing?

  • @JamesLewis2
    @JamesLewis2 Рік тому

    You have a mistake in your isomorphisms with the complex and split-complex numbers: In fact, both i and j are equal to β/√|β²| in their respective algebras. or you could say that i=β/√(−β²) and j=β/√(β²).

  • @jercki72
    @jercki72 2 роки тому

    so those are the only associative R algebra with dimension 2 in terms of vector field

  • @brian554xx
    @brian554xx Рік тому

    Just a stray thought that I hope will be picked up by someone with more talent than I have: Is there a way to use dual numbers in string theory for the collapsed dimensions? Is there such a thing as a hybrid of quaternions and duals? I don't see any conflict. Can one make something with multiple epsilons?
    Perhaps this has been tried. If so, could you direct me to search terms that will help me find it?

  • @MisterPenguin42
    @MisterPenguin42 26 днів тому

    What should I have studied before I can begin to understand this video?

  • @profapotema
    @profapotema 10 місяців тому

    I think you forgot to take the square root in the denominator in normalizing beta to get i and j.

  • @Kris_not_Chris
    @Kris_not_Chris 2 роки тому

    does this mean there's a connection between the hyperbolic functions and split complex numbers like there is between the trigonometric functions and complex numbers?

    • @angeldude101
      @angeldude101 2 роки тому +1

      Yup! e^ix = cos(x) + isin(x), e^jx = cosh(x) + jsinh(x). Dual numbers have an odd variant: e^iε = 1 + εx.