A simple argument as to why there are no 3D complex numbers. See the following video by Michael Penn for a more elaborate and rigorous proof. • Why are there no 3 dim...
The mistake in your logic is that you assume that i*j must be a number. But it's not. It's so-called bivector, which is squares to -1 by itself. The whole algebra you've created is the most basic one, containing quaternions. So, the object w + x*i + y*j + z*i*j directly corresponds to classic quaternion. After all, it is 3D complex numbers
i think that assuming ij is a number is the point of the whole video... He shows that a collection of numbers of the form a +bi +cj with i^2=-1, j^2=-1 cannot express ij in terms of such a number without reaching a contradiction or degenerating back to the complex numbers. Hence, you are forced to let ij be its own element. If you make no more assumptions, you still have to deal with iji, jij, (ij)^n and (ji)^n as independent elements (funnily enough (iji)^2 = (jij)^2 = -1). So it leads to an infinite-dimensional real algebra with generators i,j. Of course quaternions arise if you instead require ij=-ji which implies also (ij)^2=(ji)^2=-1. Instead you could require ij=ji which implies (ij)^2 = (ji)^2 =1 and actually corresponds to the commutative real algebra C\otimes_R C. Indeed, just map i \mapsto i \otimes 1 j \mapsto 1 \otimes i then ij = ji \mapsto i \otimes i. An interesting point is that C \otimes_R C is actually not a Clifford algebra/GA, precisely because i and j do not anticommute. Funnily enough the real algebra C \otimes_R C is isomorphic to the real algebra C \oplus_R C and this fact is useful in the classification of Clifford algebras. To me it is more a statement about group presentations, you really have a group G generated by (1,-1,i,j) subject to the relations that (-1)^2 = 1 -1 i = i (-1) -1 j = j (-1) i^2 = -1 j^2 = -1 you cannot state a relation like ij = 1,-1,i,j without introducing dependence between your generators as i=\pm j or contradicting the already imposed relations. As in the real algebra case, if you impose no extra relations iji, jij, (ij)^n, (ji)^n, then they are independent elements of the group. To return to the real algebra case you just take the real group algebra R[G] of G and quotient by the ideal generated by e_{-1} + e_{1} = 0, where e_{g} are basis vectors of R[G].
The problem is the if we look down the real axis the rotation of the axes j relative to I is not normal, worse, it’s undefined. For example any equation in which 0i exists, j is indistinguishable from I. As a result it cannot be a bivector because with a bivector one axis needs to be relatable to the second, otherwise the bivector is undefined.
The whole explanation is based on the fact that there is an assumption that j squared is -1. But that immediately means that j equals i or -i. Extending the rel numbers with i means that we have a new value that can not be written as combination of real numbers. So a real 3D extension of the complex numbers means that we introduce a value( and with that a new dimension for the third axis) that is not a a combination of complex numbers. And that is not the case for j if you introduce that j squared is -1. Then j is just a complex number and the situation for i is different as an extension of the real numbers: i is an extension of the real numbers and j is not an extension of the complex numbers.
Geometric Algebra has 3D “complex” numbers (similar to quaternions) and higher dimensions as well. The key is that i*j is an irreducible bivector in Geometric Algebra. In Geometric Algebra, a “3D complex number” is not expressed as a + bi + cj, it’s expressed as a + bi + cj + dij. Note the introduction of the bivector ij. And “4D complex number” requires the introduction of the trivector ijk. Also, obligatory, since someone will say quaternions are 4D not 3D: ru-vid.comiUvcYNonkaI Geometric Algebra intro: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-60z_hpEAtD8.html
Quaternions are a 4-dimensional real algebra, this is precisely why they are written as a +bi +cj +dk for real a,b,c,d... The unit quaternions on the other hand are not an algebra. They are the 3-dimensional Lie group Spin(3) \cong SU(2). In fact, Spin(3) is diffeomorphic to the 3-sphere.
@@schmud68 No, it's much more than a fad. All of mathematics comes from geometry, it's in the name, geo-, Earth, and -meter, measure, in Greek. From "taking the measure of the Earth", by which was meant to understand the weights and measures of all things. This comes down to relations and proportions, and we come to concepts like squared numbers and cubed numbers by their geometric origins. Algebraic geometry, and geometric algebra, are continuations past trigonometric identities which fulfill a certain pattern-recognition from the one to the other. I don't think it's merely a fad, at all, but instead is humanity grasping toward a language of a higher intellect, namely, a higher-dimensional comprehension of the Universe alien to our own. What I was noting to our uploader is that his video shows WHY we must bear this other burden of non-numeric values (the bivectors and tricevtors) while attempting to communicate ideas in these other dimensions. It's a necessary constraint, akin to original geometry-proof constraints of only a compass and a straight edge, which I appreciate and helps bridge a gap. I didn't even know about this or consider it before this was recommended to me.
@@quandarkumtanglehairs4743 sure mathematics has historical roots in geometry, and geometry is still a huge part of modern maths, but there are certainly things in maths that are not described geometrically. Maybe because humans are not smart enough, who knows. Geometric algebra is not all of geometry. You can't even do geometric algebra on all Riemannian manifolds (which are the natural higher-dimensional generalisation of surfaces, like a basketball etc). Or if you like physics, you can't always do it on a general spacetime in general relativity. Algebraic geometry, though sounding similar to geometric algebra is a very different thing. I've never used it much so just look at the wiki page to get a taste. Geometric algebra/Clifford algebra, ignoring philosophical viewpoints, is just a mathematical framework to talk about a vector space V with an inner product/dot product. Naturally, when V=R^3 is 3D space, then there are great visualisations which also aid in higher-dimensions when V=R^n. There is also a surprising amount of novel things that come out geometric algebra, like the Pin and Spin groups (I think one calls the Spin group the rotor group in geometric algebra terminology). Like I mentioned above, Spin(3) is essentially the unit quaternions. There is a related fact, in geometric algebra terms, the space of bivectors in 3D is essentially the quaternions themselves. The Pin and Spin groups also have an interesting representation theory, and the geometric algebra, in a way, directly points to so-called pinor and spinor representations. Spinors and pinors are important in quantum physics. The typical example being the use of Dirac spinors in QFT. Pinors and spinors also allow one to get interesting mathematical information about manifolds on which one can do geometric algebra. I am not saying geometric algebra is not useful or insightful, just that it is one part of maths and doesn't really tell you much about a lot of the other parts. I think the fact that you can't always use it on manifolds already says quite a lot. Maybe it is convincing to mention that geometric algebras (Clifford algebras) are completely classified as being isomorphic to matrix algebras Mat_{nxn}(D) or direct sums of matrix algebras Mat_{nxn}(D) + Mat_{nxn}(D) with D the reals, complex numbers or quaternions. Algebraically, this is a very limited selection of possibilities as there are many more types of algebras one could consider.
You started with the axiom/assumption i * i = -1. But what if you don't allow that, and start with the assumption that i * j = -1? It's not going to help solve sqrt(-1) easily, but perhaps it has other properties!
I wondered what the starting assumptions would be and I would have liked them to be at the start. For instance linear independence, associativity and distributivity. However as I am interested in geometric algebra I find this a useful stepping stone and it has filled a gap. Nice video.
I am a physicist, so this is more of a "physicist math" video. If you want the "math math" version, look at the Michael Penn video in the description. Very interesting also, but much more complex, pun intended.
Well, obviously i * j doesn't equal 1. That's not how rotations in 3-d works. We can still have 3-d complex numbers if we keep it clear that multiplication means to rotate into a direction. For example; start with a unit of the real direction - 1. Multiplying ti by i means rotating that vector into the i direction. Then rotate the resulting vector into the j direction. It's still a vector n 3-d space. It's just not the same as doing the multiply by j first and then by i. If you insist on doing it the wrong way, requiring i * j = 1, you're effectively saying rotations in 3-dimensions are impossible.
Hey guys, I might have missed something, but why does j^2 have to be -1? For quaternions for example we have i^2=j^2=k^2=ijk=-1 and I guess that works because it makes sense when doing calculations but I am thinking if it's possible for other values of j in the 3D attempt, I mean we could have j^3=-1 why not...
I am trying to understand this geometrically. If the units not equal to 1 have the property to rotate the vector, it basically means that to keep the space three dimensional, you should not be able to reverse a rotation by i with a rotation by j.
Fundamental confusion with the explanation is assuming complex plane and 2d x, y plane is same. But it is not.. To understand complex number properties we need to understand diferrence between 2 axioms.. If we rotate complex axis 90degree we will get real in negative but 2d we cannot rotate and even rotate xy coordinates will stay same it can never be converted between x and y For example a+ib if we rotate then we get ia-b Straight line x+y if we rotate -x+y Therefore both are not equal.. It just coincide betwen our thinking.. Therefore complex number 3d is qual to 1st or 2nd dimension.. 11:36
It depends on what we mean by "3D complex numbers." If we are referring to objects that describe affine transformations in 3D space, then quaternions can be considered as such numbers.
Anything divided by 0 would be not a number, but it can be a mathematical abstraction, convertible to numbers in expressions, also you have to define non-negative 0 to avoid negative/positive option.
What would happen if we made also “scalar” multiplication by a real number also not commutative? Is there in that a similar argument? Do we have to be more ingenious? Or is it actually possible in that case?
If you lose commutativity, you lose a lot of nice properties. It's been a while since I recorded the video, but I think you can make the proof work without commutativity with real numbers. Note that when you get to the next level, which are the quaternions, you lose commutativity between the non-real component (numbers of the form A + B i + C j + D k, capital letters real numbers, i^2=j^2=k^2=-1, you have for example i j = - j i.
@@DeeperScience I see that you have maintained a strict constraint in your starting premises, and retained yourself from arguing algebraic geometry using ij bivector or ijk trivector. I like your maintenance of this constraint, as it shows us exactly WHY we need a non-numeric vector in expressing imaginary values in higher dimensions. I think your presentation here, exactly with this constraint, is perfect. Thank you! I had never considered it, before...
one slight objection: in your derivation you used commutativity of multiplication. That does not hold for quaternions, so it is a big step to assume it for 3D numbers.
@@tomgavelin2879 when multiplying an equation with j he freely distributed the j around. To begin with, when multiplication is not (necessarily) commutative you have to clarify whether you multiply on the left or on the right.
I am aware of commutativity and distributivity. I just do not see anywhere in this video where commutativity was assumed where it didn't apply. Do you have a time stamp of what you're talking about? I am curious because I feel like I'm missing something obvious.
It looks like the only commutative assumption in the last step is the commutativity of scalar multiplication (ie a * z = z*a if a is a real number and x is a 3-d complex number). Any multiplication of i and j are not assumed to be commutative here in any step, as far as I can see.
by 5:58 I think I already get it. You would have two numbers to go to -1 but that leaves us with an incomplete 3d. So Im assuming that if j was complex that would fix it bringing us to 4d numbers.
Good question. The first issue is that you then no longer have the relationship "multiplying by j is equivalent to rotating 90 degree", which makes the complex number useful in calculations. Also, this will not work. You can look at the video linked in the description for a deeper understanding of the subject. If you want, you can try to come up with a simple proof using a contradiction similar to the one in my video yourself. Suppose j^2=a+b i + c j, i j = d +e i + f j, and do manipulations until you get to a contradiction. Since you have 6 parameters a to f to play with, the algebra may be a bit ugly.
I know the channel has been inactive for a while. I have a nice one in preparation about the movement of spinning tops and similar objects. I just have to find some time to record and edit it.
Yes. You then get standard 3D vector analysis which is quaternions with the real parts always equal to zero. There was a big argument in the 19th century about which was better, vectors or quaternions. Most physicists, particularly Heaviside and Gibbs, thought quaternions cumbersome and decided to split off the vector part and just worked with that.
why you defined "j" as the same as "i" ? What if "j" is another "impossible" solution? i is √-1 what if j is -1! or some else, but not the same as √-1 ?
well you want j to not be a real number, and you want it to have simple algebraic properties. You can’t have j=infinity, for example, because you would have no useful multiplication rule for it
Nice try. First, x+iy alone goes into three dimensions by rotating into z. Second, for your thinking, you have to use Heaviside's selective j where ij=-1, not the tessarine ij=+1. Then (jx+iy)^2=-(x^2+y^2) and you can then step out of this arithmetic box and define the other imaginary numbers to expand from two to three dimensions.
the cube root of -1 is represented in the two dimensional complex numbers. It is a particular linear combination of 1 and i. Look up “roots of unity” on Wikipedia for more info. You can use them to get the n-th roots of any complex number
I think you should try to start with i, j and then show that no, matter what the rules are for multiplication using i and j, that not all numbers can have inverses. That may have a bigger scope (lack of a division algebra), but I don't quite understand clearly the assumptions you have made here. Sincerely .....
It's a tessarine. He isn't looking at how the operators actually work or following the vector analysis or anything. Give him credit for trying. He erased my comment where I pointed him to how j works, or the documentation showing it's been known and used that way for over a century by the guy who invented vector calculus. Using it right solves his problem. Using the imaginary numbers wrong... clearly does not.
I think we can also answer as: since there were a problem in R ( it was √(-1) ) that led to birth of "i" (complex numbers) , so what problem is there ( in R or C ) which leads to birth of "j" ??? so just putting a random letter which based on no problem makes no sense at all.
There are at least two motivations for the video. First, there is something called the quaternions, that are 4-D. One may wonder if there is something between the complex numbers and the quaternions. Second, as mentioned in the videos, complex numbers are very useful for making 2-D calculations. One may want to do the same in 3D, since our physical space is 3D.
Somebody claims he has invented that 3d complex numbers you refers on the video, please check it out on: Italian journal of pure and applied mathematics (IJPAM) n. 29 year 2012, pages from 187 to 300. I'm not expert in math so I can only suggest you to see it personnally.
C is isomorphic as a set to R² so points in C are like points in R². Similarly C² is isomorphic to R⁴ so points in C² would be equivalent to points in 4-dimensional real euclidean space. Iirc for all n in N we have that C^n is isomoprhic to R^(n/2) so you can make connections between complex spaces and real spaces for any dimension. TL;DR: points in C² are like quaternions
I tried to have symmetrical components x,y,z, with x²=x, y²=y, z²=z, xy=z, xz=y, yz=x. which works fine except for associativity, so for three numbers in this system a(bc) != (ab)c, which makes them useless. There seems to be something deeper that prevents useful numbers with three components to exist, even if you don't limit yourself to "complex-like" numbers.
@@theupson Numbers that follow the usual laws of reflexivity, associativity etc. Of course Vectors work, but the multiplication is either trivial (scalar product, the three dimensions don't interact) or anti-commutative (cross product), which isn't really nice, e.g. there are two ways to define it (left- and right-handed)
Thank you for this. This is so simple that even my 14 year old children can understand. I love Michael Penn's channel, but he is often rigorous to the exclusion of making it easy.
The video itself is fine but I disagree with the assumptions made. Also, we could have ai + bj + ck = T and then define a braid like relationship between the three. Or, we could look for solutions to T^3 = -1 other than -1.
Why even allow i and j to mix AT ALL? i doesn’t mix with the reals as is. So, j shouldn’t be allowed to mix with i except as another complex term. That would work! The whole thing wouldn’t end up becoming a pure real unless BOTH i and j were multiplied by themselves and even number of times become either -1 or 1 real to remove its imaginary contribution.
i, j, & k would be orthogonal vectors, just as x, y, & z; values, and therefore would have to be handled by vector algebra (?) - The possibility of this approach seems not to be addressed in this video.
I think (havent actually looke into it though) is that you can avoid the problems faced in the video for trying to define multiplication. This is done by essentially "escaping" the new dimension, so for i,j and k if we were to define i*j it was already showed in three dimensions that none of the options with 1,i and j would work so we choose i*j=k and same for the others i*k=j and so on. We essentially just have room using the extra dimension to define our multiplication in a nice way. Again, i did not actualky do any calculations so the above just gives the jist of why i think it might work in 4 but not 3
I think this is like another question below. It seems like having a complex number and taking its first derivative with respect to time would accomblish making a 3-d number?
Your video is superficial because you miss the starting crucial point of giving a clear definition or even try to address the question : « what is a number ». So your « proof » is mainly empty because it stands on the sand ground of an undefined definition. And indeed, « i » for instance is actually less a « number » than an « oriented area », despites its arithmetic and algebraic apparent behavior « as a (almost) usual number ». And moreover, by the way, there is not only one « complex number structure » in 2D, but THREE, directly related to the three subalgebras of 2D real matrix algebra. And thus in this deeper regard, there is somehow no « (complex) number » in 2D, but instead, more correctly : ONE SCALAR, TWO VECTORS, and ONE PSEUDO-SCALAR (UNIT ORIENTED AREA). Which leads to three different versions, depending on the precise algebraic structure chosen. That’s the actual unmasked picture of such « three-fold » ALGEBRAICO-GEOMETRIC STRUCTURE. The « purely number » question is thus too naive and narrow minded. It misses the key point of the fundamental inseparability of Algebra and Geometry, despites the usual mainstream misleading perroting blabla. And moreover, in the same line of thoughts, there IS actually « generalised complex numbers » in 3D : ONE SCALAR, THREE VECTORS, THREE BIVECTORS (ORIENTED AREA), and ONE PSEUDO SCALAR (ORIENTED VOLUME = DETERMINANT FORM). You’re thus just proving that there is no coherent structure of the (obvious naive and incomplete) form that you « historically » (pre Clifford and Grassman) suppose. But by no way you’re proving what you pretend to prove : that there is no coherent algebraic complex number structure in 3D. On the contrary, THERE IS! You are making a similar mistakes than the usual superficial blabla of false beliefs about the uselessness of divergent series. They are often on the contrary more rich and useful than boring convergent ones. In brief here, you urgently need to update two centuries of intensive developpements on Clifford Algebras.
Complex numbers are often used specifically (solving wace equation for example) to not have to do cumbersome sine and cosine stuff, sure it could be done that way, its just much easier not too.
@@mattias2576 As a game programmer, I prefer working with sine and cosine trig, and if a language features is optional, I try to avoid it for consistency and simplicity. I feel like imaginary complex numbers, breaks the rules of algebra too much to accept, redefining multiplication is a serious thing, that should be justified by it doing something that can't be done otherwise. it takes syntax that math already used, and redefined it, so now nobody can use i as a variable or it gets confusing. I am glad every use of complex numbers can be turned into sine and cosine. i j and k should be used for iterators.
@@ScottSpadea i cannot speak to it in terms of programmikg but as a physicist, complex numbers are a very good tool for waves and similar things and also are completely necessary for quantum mechanics. Complex numbers are not always just equivalent to a different way of writing sine and cosine, QM is the best example i have for when complex numbers are fundamentally required, the schroedinger equation always gives complex solutions due to its structure. Also how does it redefine multiplication? It is associative, commutative, distributive with respect to addition etc. I can see why they might be overkill in some places but that does not mean they are not practical in other areas, electrical engineering use them a lot for describing electricity for example
@@mattias2576 You claimed "complex numbers are completely necessary for quantum mechanics." but they aren't. you can use cos and sin to solve the same problems, every time. The only reason complex numbers are ever used, are dealing with circles on the complex plane, and you can always just use sin and cos. And if you say it takes more writing, I don't care. All computer code can be translated to bytes, if you can solve the equations on a computer, you can solve it with nothing but an array of bytes and a small list of op codes, for add, get, set, jump, jump if zero, bitwise not, shift left and shift right. Simulating spherical waves that interfere just doesn't require any weird math, you can do it with scaled and shifted sine waves to get the same interference patterns. scale sine waves by their EMWaveLength, Shift them by the measured points relative distance from each source, add the waves, and you have interfering spherical waves. If you complain that doesn't take into account the phase difference, like Gemini AI keeps arguing, it does if you wobble the wave source locations with sine waves, toward and away from the measurement point. Those sine waves are offset by the difference in time each wave peaks.
@mattias2576 Simulating spherical waves that interfere doesn't require complex math, you can do it with scaled and shifted sine waves to get the same interference patterns. scale sine waves by their EM Wave Length, Shift them by the measured point's relative distance from each source, add the waves, and you have interfering spherical waves. If you want a phase difference, you can wobble the wave source locations with sine waves, toward and away from the measurement point. Those sine waves are offset by the difference in time each wave peaks. complex numbers are never actually needed, they don't provide more accurate answers, and if a computer can solve the equations, so can an array of bytes and basic op codes.
Complex numbers only deal with 2 orthogonal axes and are easily handled with planar math. In my 89 years (over 30 as an electrical engineer) I have never encountered a need for 3D complex number representation or notation. If somebody comes up with a 3 dimensional Smith Chart (Impedance Globe ?) maybe I'll change my mind.
Most people have heard of complex numbers but rarely ever hear about quaternions. The naive would assume you could just "add j" and make it work, I certainly did when I thought about extending complex numbers to 3D.
Uh ... I do not know what you mean by "3D complex numbers" ... but there are quaternions that handle 3D rotations beautifully (better than Euler matricies), and then octonions ... and that is the end of the division algebras.
