The complex number family.

[@tomkerruish2982]

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

[@goodplacetostop2973]

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

[@ivanklimov7078]

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]

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

[@Tim3.14]

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

[@rajinfootonchuriquen]

​@@Tim3.14 lol 😂

[@freyascats1364]

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]

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

[@MarekMadejski]

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.

[@profkrinkels]

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

[@NicholasPellegrino]

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]

@@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.

[@nosnibor800]

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]

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

[@lawrencedoliveiro9104]

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

[@nHans]

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

[@nHans]

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]

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

[@Alex_Deam]

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]

Wait how? I'm not getting that.

[@rajinfootonchuriquen]

​@@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]

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]

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]

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]

@@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]

@@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]

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]

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]

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

[@anto2593]

i was looking for exactly this comment. Thanks

[@kruksog]

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!

[@pedrocusinato1743]

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

[@taeyeonlover]

that made me so confused

[@schweinmachtbree1013]

​@@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]

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

[@wladwladsnotmyrealname9082]

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]

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.

[@AleksyGrabovski]

Your videos reminded me how I love abstract algebra.

[@rivkahlevi6117]

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

[@austinlincoln3414]

Lmao

[@GeekProdigyGuy]

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

[@TheDannyAwesome]

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]

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

[@angeldude101]

1 + -1 = 0, j² + i² = ε²

[@schweinmachtbree1013]

@@angeldude101 nice!

[@thewhitefalcon8539]

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

[@starsun7455]

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

[@Ricocossa1]

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

[@radadadadee]

for a stationary observer?

[@Skyb0rg]

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]

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

[@thewhitefalcon8539]

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

[@nmmm2000]

@@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)

[@neonsilver1936]

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.

[@AndarManik]

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

[@zemoxian]

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

[@unflexian]

You're making me love abstract algebra:D

[@Zeitgeist9000]

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

[@shalvagang951]

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

[@Zeitgeist9000]

I was wondering the same thing

[@shalvagang951]

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

[@shalvagang951]

@@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]

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

[@camperbbq3026]

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

[@Grassmpl]

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

[@cpiantes]

Can you do an episode on Clifford algebras?

[@rv706]

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.

[@copernic7511]

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

[@apteropith]

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

[@bobdowling6932]

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]

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.

[@CTJ2619]

I like learning about the quaterians

[@Nickle314]

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

[@angeldude101]

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

[@Nickle314]

@@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]

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

[@Noam_.Menashe]

It's also a bit like the cross product.

[@angeldude101]

@@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.

[@brattok]

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

[@stevenwilson5556]

wow so cool. great way to tie this together

[@user-hh5bx8xe5o]

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)

[@neopalm2050]

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

[@sebastiandierks7919]

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!?

[@pajrc1234]

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

[@Dalroc]

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..

[@radupopescu9977]

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

[@drewmandan]

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

[@АлексейТучак-м4ч]

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

[@dtrimm1]

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

[@alexey_burkov]

omg, that preview picture 😂

[@nbooth]

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

[@Grassmpl]

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

[@m1323fj]

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

[@General12th]

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]

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.

[@eschudy]

Very cool!! More please!

[@_P_a_o_l_o_]

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

[@elliottmanley5182]

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]

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]

@@elliottmanley5182 8N = N8

[@arnbrandy]

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]

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.

[@gabrielfrank5142]

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?

[@Grassmpl]

Only in C does the modulus induce a metric.

[@AJ-et3vf]

Great video. Thank you

[@skillerror951]

Looks like curvature of space

[@joel.9543]

Amazing!

[@stephenmorton9789]

Great stuff, thank you

[@deltalima6703]

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]

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

[@ΕχιΜιμζ]

A lovely family indeed.

[@nrrgrdn]

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

[@sciencewithali4916]

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

[@rrr00bb1]

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]

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]

@@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]

@@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]

@@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]

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)

[@catherinebernard3282]

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.

[@djsmeguk]

This must have some sort of relationship to the conics.

[@davidhand9721]

My family is pretty complex, too.

[@NutziHD]

Awesome video!!

[@travisporco]

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

[@angeldude101]

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]

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

[@Fine_Mouche]

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

[@spudhead169]

I think you spilled something down your hoodie there Mike.

[@timseytiger9280]

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

[@williamwright1440]

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.

[@jantarantowicz1306]

as always, great video!

[@aweebthatlovesmath4220]

Numbers are cool!

[@JamesLewis2]

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).

[@marcoottina654]

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"

[@christressler3857]

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.

[@elliottmanley5182]

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]

Interesting if this mixture yields anything

[@angeldude101]

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]

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]

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.

[@CesarMaglione]

Buena pizarra Michel 👍

[@przemekmajewski1]

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]

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]

Case 3 is also known as hyperbolic numbers

[@tothm129]

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

[@linuxophile]

There are two mistakes (really the same twice). If beta^2 is positive, then the normalized is beta/sqrt (beta^2 ). Ditto for the other case

[@profapotema]

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

[@HoSza1]

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]

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

[@weonlygoupfromhere7369]

My brain is too small for this

[@tiborgrun6963]

H : [0,1] ∋ t ↦ (1-t)*(x^2 + y^2 - 1) + (t)*(x^2 - y^2 - 1) is a homotopy between all three of these circles. H(0) = x^2 + y^2 - 1, H(0.5) = x^2 - 1, H(1) = x^2 - y^2 - 1. Does that mean, there are other 2-dimensional associative algebras for the in-between values of t?

[@schweinmachtbree1013]

Michael proved that the only two-dimensional unital (associative) *R*-algebras are *R* [j], *R* [epsilon], and *R* [i] = *C* , so I don't think the values of t correspond to two-dimensional *R*-algebras.

[@japanada11]

Every value of t does correspond to a valid 2d associative algebra! It's just that they're all isomorphic to the complex numbers when t

[@japanada11]

More precisely, you can say that t corresponds to the algebra R[b]/(b^2-2t+1). The "unit circle" is defined by the equation (x+yb)(x-yb)=1; using the identity b^2=2t-1, this comes out to x^2 - (2t-1)y^2 -1 = 0 which is exactly the equation H(t)=0 using your definition of H. When 2t-1≠0, you can scale b by sqrt(|2t-1|) to get a new element c in the algebra that satisfies either c^2+1=0 (if t1/2), which is why all these algebras still fit in one of the three categories.

[@schweinmachtbree1013]

@@japanada11 I thought something similar at first: I thought that all the *R*-algebras for t1/2 would be isomorphic by scaling the split-imaginary unit. However scaling the (split or non-split) imaginary unit, e.g. the map phi: *C* -> *C* that sends 1 to 1 and i to 2i (extended linearly, i.e. phi(a+bi) = a+2bi), is a *R*-vectorspace isomorphism but not an *R*-algebra isomorphism, since it does not respect multiplication. You also don't get an *R*-algebra isomorphism if you scale both the real and imaginary units; indeed the only *R*-algebra automorphisms of *C* are the identity map and complex conjugation. I think your approach will run into the same issue, but I am not sure. What does your isomorphism from R[b]/(b^2+1) (when t=0) to R[b]/(b^2+1/4) (when t=3/8) look like? - then we can see whether it is multiplicative. So we don't get confused with notation let's write it as phi: R[x]/(x^2+1) -> R[y]/(y^2+1/4).

[@japanada11]

@@schweinmachtbree1013 You're right that it is a subtle point that needs to be checked! In this case the map is defined by sending a+bx to a+2by. Then f((a+bx)(c+dx)) = f(ac + (ad+bc)x + bdx^2) = f((ac-bd) + (ad+bc)x) = (ac-bd) + 2(ad+bc)y, f(a+bx)f(c+dx) = (a+2by)(c+2dy) = ac + 2(ad+bc)y + 4bdy^2 = (ac-bd) + 2(ad+bc)y, using x^2=-1 in the first line and y^2=-1/4 in the second.

[@geelaw]

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.

[@roberttelarket4934]

Michael will not be at Franklin College much longer! With his abilities he’ll be at Stanford.

[@simeonsurfer5868]

hum, case 2: (B/|B^2|)^2=B^2/|B^2|^2=-1/|B^2|, not -1. same for case 3

[@tomasstana5423]

Wow, a whole second of video after "this is a good place to stop". Michael, you are slippin'

[@KitagumaIgen]

I had to check that the geometric-algebra Clifford was the same person as the dual-numbers Clifford - imagine if it wasn't - and how grumpy that would make G-A Clifford in the mathematician's heaven "quaternions and dual numbers - BAH!"...

[@argonwheatbelly637]

Sorry, associative algebra. No Octonions or Sedenions.

[@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 😅

[@BigBoy6741]

When you start to Suppose A is a 2D unital algebra, shouldn't you also be supposing that A is commutative? Otherwise you don't exactly get the dual numbers or the complex numbers and what not.

[@beatadalhagen]

Will there be a follow-up involving combining these?

[@VideoFusco]

Must be i=beta/sqrt(|beta|^2), and the same for j

[@JakobVirgil]

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?

[@viniciusmoretti]

Engineers are confused by case 3.

[@wilderuhl3450]

Fascinating