why there is no four dimensional cross product.

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[@MrFtriana]

This is great from the point of view of the physics. We define the angular momentum for a particle as the Cross product of the position and the linear momentum, but when you try to do the same for fields, specially when you study classical fields theory, you can't define a four dimensional Cross product in order to define a four dimensional angular momentum. You use instead a tensor that depends on the position and the momentum-energy tensor to construct the angular momentum tensor and the spacial components of that tensor are the ones that gives the angular momentum vector, that corresponds to the old three dimensional definition.

[@curiosityzero2151]

pretty cool!!

[@JakubS]

why not just make the fields 7 dimensional

[@horsthantelbank293]

@@JakubS you mitght be joking but Look up kazula-klein theory, F theory and Dimensional regularisation (the latter assumes, that the Dimension of spacetime is any Complex number to then Take the Limit as D goes to 4)

[@2fifty533]

or you could just use a bivector as your angular momentum, much simpler solution...

[@horsthantelbank293]

@@MrFtriana that is true, but wdym by „but“. We don’t disagree

[@davidrubel309]

Geometric Algebra be like "Am I a joke to you"?

[@FoughtAgaisntSisera]

please get this to the top, i know MP has looked at clifford algebra but i would love to see him explore the wedge product

[@zemoxian]

@@FoughtAgaisntSisera Yes. Every time I think of the cross product I wonder why the wedge product isn’t more popular. You get the same answers in 1 and 3 dimensions. And you have answers in every other dimension as well. Just a run-of-the-mill bivector. (Which also helps distinguish vectors and pseudo vectors in 3 dimensions.)

[@outofthebots3122]

@@zemoxianwhy in the us do they still use the imperial system instead of the metric system?? Because of legacy reasons. Why is cross product still taught in schools instead of cross product?? Because of legacy reasons.

[@Kram1032]

@@zemoxian it's technically not quite the same answer. With the wedge product it's, in fact, a clearer, better one for 3 dimensions, as it very clearly and single-glance-obviously distinguishes polar and axial vectors. A difference which is obfuscated by the cross product. That said, it also gives you a different answer for 7 dimensions, so the "7D cross product" captures a different thing from rotations there

[@theodorostsilikis4025]

cross is the dual of the wedge ,it is not the same thing.they both exist in geometric algebra.

[@nHans]

Even though there are no cross products in 5D, I still take consolation in the fact that I can nonetheless play *"Rock, paper, scissors, lizard, Spock."*

[@WhattheHectogon]

An interesting interpretation of the cross product in R^1: the resulting vector from a cross product must be orthogonal to the input vectors, and two vectors are orthogonal if and only if their dot product is 0. Every non-zero vector in R^1 is parallel to every other non-zero vector, and so their dot product is never 0. When any vector is dotted with the 0 vector, however, you always get 0, in any dimension. This means two neat things: the 0 vector is orthogonal to all vectors (including itself) in any number of dimensions, and, in R^1 specifically, it is the only vector which is orthogonal to any other vector in the space, which means it must be the result of every cross product in R^1. Thanks for the great video, Michael!

[@wyboo2019]

note: i have not seen the video yet i think of the cross product as the dual of the wedge product in G3. if we use this to define an n-dimensional cross product, then the 'reason' there isn't a 4d cross product is that the dual of a wedge product between two vectors in G4 isn't a vector, its another bivector, which doesn't exist in normal linear algebra

[@BlueGiant69202]

"The vector cross product is largely redundant now that we have the exterior product and duality at our disposal." pg. 37, Geometric Algebra for Physicists by Chris Doran and Anthony Lasenby

[@JoshuaHernandez8a]

Just came to like this comment

[@evandrofilipe1526]

Even though I didn't know, I feel like I already knew

[@ValkyRiver]

Geometric Algebra has this thing called the “wedge product” or “outer product”, which works in any dimension. The cross product is redundant in 3D Geometric Algebra.

[@marcellomarianetti1770]

Lol you are the guy who posts cursed music!! It's nice to find out that we both study maths as well, did you also study composition?

[@Silvermoonmaker]

wtf this has to be a Xenakis moment

[@pacotaco1246]

He has videos on the wedge product on his channel. Some are part of a series on Diffefential Forms if i remember correctly

[@doraemon402]

I've always looked at it from the perspective that you input n vectors of n+1 dimensions and it returns a vector orthogonal to all of the inputs. So, in 4-D, you need 3 vectors, and the cross product is rather a function, not an operation.

[@moonshine7753]

Yesss! That's what I don't understand about this "there doesn't exist any cross product" thing. Of course you need more vectors for more dimensions!

[@askcaralice]

this "common sense" would imply non-existence of the cross product on R^7

[@moonshine7753]

@@askcaralice Why? You'd just need 6 vectors to get the one that's orthogonal to all of them.

[@henrikljungstrand2036]

It was implicit that we asked for a binary cross product V×V -> V in V = R^n. And this only exists in R⁰, R¹, R³ and R⁷, where the first two are trivial. However, there is a ternary cross product V × V × V -> V in both R⁴ and R⁷ (and trivially in R⁰, R¹ and R²). All cross products are to be non-degenerate alternating (totally antisymmetrical) multilinear functions. In char(K) ≠ 2, alternation a×a =0 (and a×a×b = a×b×a = b×a×a = 0 etc) is the same as antisymmetry a×b = - b×a (and a×b×c = - b×a×c = - c×b×a = - a×c×b etc), and this is true for char(K) = 0 when e.g. K = R or K = Q. But using e.g. K = Z/pZ (or some field extension of Z/pZ) for primes p, which makes char(K) = p, this equivalence doesn't hold for p = 2, even if it holds for odd p. Because a×b = - b×a means a×b + b×a = 0, and if 2 = 0 (as is the case for char(K) = 2), this might hold for a×b = b×a ≠ 0, and always holds for a = b, giving 2 * a×a = 0. However the condition a×a = 0 (always) leads to a×b = - b×a because 0 = (a+b)×(a+b) = a×a + b×a + a×b + b×b = b×a + a×b. Thus i expect the argument K^n has a binary cross product only when n is 0, 1, 3 or 7, actually holds for char(K) = p an odd prime (since then we can divide by 2), but i am unsure if it holds for char(K) = 2.

[@joshuagenes]

I think I like Geometric Algebra better. Wedge Product Baby!

[@Kram1032]

For rotations definitely Though the 7D cross product is different from the wedge product as it corresponds to Octonions and not Cl(7, 0)

[@jozsefgurzo8777]

That was very interesting! When we defined cross product in an abstract way, we said it's the Lie-bracket of a Lie-algebra on R^3, so I wonder how it would change the result if you assumed the Jacobian identity instead of |a×b|^2 = |a|^2|b|^2 - (a "dot" b)^2 . Also, fun fact: the reciprocal of the fine-structure constant is approximatly 137, whose digits are the solution of the problem :)

[@henrikljungstrand2036]

The cross product on R⁷ fails to be a Lie algebra, and must do so in order to be non-trivial. It fails the Jacobi identity, and in fact its Jacobi triple product is (more or less) a "Pfister 3-form" related to the Octonion associator (a*b)*c - a*(b*c). This R⁷ cross product is however a Malcev algebra, which is kind of like a Lie algebra, but tangent (at identity) to a non-associative differential Moufang loop (Malcev loop?) rather than to a differential group i.e. a Lie group. This differential loop is usually called M_7 iirc. It is simple, and probably is the only simple non-associative differential Moufang loop. Non-simple such loops may be built from M_7 together with the simple Lie groups.

[@henrikljungstrand2036]

Analytic loop is the proper name. Or analytic Moufang loop. An analytic group is the same as a Lie group.

[@vassillenchizhov290]

I feel like there should have been a disclaimer that one can define a generalisation of the 3D cross product using the hodge star of the wedge product/the Levi-Civita contraction of n-1 vectors in R^n. In a lot of applications this is precisely what one wants.

[@stephenhamer8192]

The video establishes that the allowed values of n for an n-dimensional cross-product are 1, 3, 7 (very surprised to find that there's a 7-dimensional cross-product!) I note: 1 = the dimension of the "imaginary part" of the complex numbers 3 = the dimension of the "vector part" of the quaternions 7 = the dimension of the "non-real part" of the octonions (is there such a thing as the "non-real part" of the octonions?) And, as shown in an earlier video, the complex nos, quaternions and octonions are the only normed division algebras of dimension > 1 Does the existence of a cross-product of order n imply the existence of a division-algebra of order n+1 (1

[@stephenhamer8192]

I'm thinking that the cross-product of the 7-vectors a, b is the non-real part of the octonion product (0, a).(0, b)

[@soyokou.2810]

This was my thought as well. You can find this in older advanced calculus books too

[@TheEternalVortex42]

Yes, each cross product can be used to define a division algebra.

[@user-bz3kd2mt3u]

yeah the octonions are a superset of the real numbers, they have a real part. In fact what you just described is that the cross product is defined for the non-real part of the complex numbers, the non-real part of the quaternions, and the non-real part of the octonions. I wonder why there isn't a 15-dimensional product for the sedenions. Is the alternative(?) property (weak form of associative property the octonions retain which the quaternions lack) the dealbreaker for cross productizing?

[@Kram1032]

@@user-bz3kd2mt3u one of the axioms in the video involved a x (a x b) right? I'd imagine the sedenions fail that axiom

[@GreenDayFanMT]

Great to see you getting recognized for your great videos by Brilliant. Your work is more than worthy of a sponsorship.

[@outofthebots3122]

Understanding the relationship between cross product and wedge product is an easier way to explain this. The cross product is the duel of the wedge product in 3D space. More correctly it is the complement of the wedge product in 3D space i.e axb = *a^b where * is the Hodge star operator. The *a^b is only a vector in 3D space, in 4D space it would be a Bi-vector.

[@pwmiles56]

Surely the cross product is most clearly understood as a determinant x^y = | i j k | | x1 x2 x3 | | y1 y2 y3 | in which i, j and k are the three orthogonal unit vectors for the coordinates. This has an application in the geometry of the projective plane m = x^y where x and y are the homogeneous coordinates of two points, and m is the homogeneous coordinates (known as tangential coordinates) of their line join. It is easily seen that m.x=0, m.y=0 so the points both fall on the line. Here i, j, and k are placeholders for the so-called triangle of reference i.e. the points (1,0,0); (0;1;0); (0,0,1). The dual relationship is x = m^n where x is the point intersection of lines m and n. The generalisation to 4 coordinates is a = | i j k l | | x1 x2 x3 x4 | | y1 y2 y3 y4 | ' | z1 z2 z3 z4 | a is the coordinate set for the plane holding x, y, and z, Again a.x=0, etc. The construction dualises to the point intersection of three planes.

[@lawrencejelsma8118]

That is the way I was taught also in Electricity and Magnetism 3D space "lines of force" and "flux" calculations of current first of all from a 2D orientation such as a circle loop (your first half information) then 3D extending out as in a coil etc. (your second half information now needing a 4th dimension cross product retainer)! Good thoughts on helping us understand better all our electrical x magnetic fields in Maxwell's Equations better of vector E cross vector B for Electrical Engineering students!! 👍

[@clickaccept]

Kinda the other way round, mate. Just ditch cross product, its a leftover from ages past, when we didn't know exterior algebra.

[@BlueGiant69202]

@@clickaccept Now we know GEOMETRIC ALGEBRA (Hestenes unified language for math and physics/formalism/notation not just another name for Clifford Algebra), too!

[@clickaccept]

​@@BlueGiant69202 Hestnes is a con artist.

[@clickaccept]

​@@BlueGiant69202 May I ask what precisely you have in mind by "not just another name for Clifford Algebra"? Can you offer a definition? For example, the exterior algebra of a module X is the quotient of the free algebra on X, by the ideal generated by x*x over x in X.

[@iqtrainer]

This video is pretty nice to explain the stuff

[@thetrueshadowfox]

I'm really enjoying the video, and having just taken a vector calculus course, some of the techniques learned definitely helped me keep up with what was presented. It got me to try doing some of the proofs you were doing in ways that my class taught me, and I think it would be interesting for you to look into or make a video on one technique in particular if you haven't already. It's called "suffix notation". I can definitely give it a good recommendation to look into due to how it can help deal with vector operators with more ease than standard vector proofs. The terminology can be a bit daunting at first (but this is math, so what else is new? XD), though personally it could make for something very interesting. For the proof starting at 15:30 for example, suffix notation can make the proof - in my opinion - more straight forward and easy to deal with since there isn't the dummy vector d, nor the a -> a+c substitute later. I'd like to write the proof that I did here, but KZbin's comment section formatting isn't kind to the usage of suffixes, so all I can say is give it a look if you're interested, and you may find something interesting. Best of luck with future endeavors!

[@mrengstad]

Would be great to see what you think about Geometric (Clifford) Algebra.

[@gasparliboreiro4572]

remember about the way to compute the cross product with a determinant? that if you have a x b it would be | i j k | | a1 a2 a3 | | b1 b2 b3 | well I tested it and if you try this same method for a 2D vector, you also get one that is perpendicular | i j | | a1 a2 | if a = (a1, a2) that determinant gives you (a2, -a1), and that vector is always perpendicular to a in 2D if with the same pattern we make this determinant for 4D it would be | i j k w | | a1 a2 a3 a4 | | b1 b2 b3 b4 | | c1 c2 c3 c4 | of course you need a third vector, because in 2D with one vector you can only have 1 direcction that is perpendicular, but in 3D you have infinite directions that are perpendicular to 1 vector, but if you use 2 there is only one direction (positive and negative directions I'm counting as the same) and as might seem logical, in 4D space there must be infinite directions that are perpendicular to 2 vectors, so to get just one you need 3 I don't know how to prove that the vector resultant is always perpendicular to the others, but I tried using a=(1,0,0,0) b=(0,1,0,0) c=(0,0,1,0) and I got the vector (0,0,0,1), so at least there it works also I don't recommend to get stuck in the notation of this kind of stuff, because this being a 3 object operation, you cant really type it out in a simple way like with multiplication or something like that, and to try to derive to a 4dimensional equation using only our 2 dimensional expressions seems kindof weird to me with this I'm reffering to the non comuntative properties of the cross product for example, yes with 2 is simple enough, there are just 2 ways of doing the operation, but with 3 is a whole new thing that should be studied apart from what we know

[@robshaw2639]

it looks like the axioms don't uniquely define the cross product. for instance in R3, I could have e1xe2 = -e3, e2xe3 = -e1 and e3xe1 = -e2 and all the axioms are satisfied i think.... There is no chirality given by the axioms.

[@dmytryk7887]

There is a pretty common mnemonic diagram used in introductory courses when the cross product is introduced. You have the vectors i, j, and k placed around a circle with clockwise circular arrows from I to j , from j to k, and from k to i, etc. What you are suggesting is just reversing the direction of the arrows. So aren't you just replacing the right-hand rule by the left-hand rule? That is, the usual cross product with a mirrored version.

[@fw15str11]

There are two equally valid definitions of a cross product in R^3, a right-handed (R) and a left-handed (L) version (agreeing with @dmytryk7887). In a more physical sense, this could be seen as a manifestation of defining the (R vs. L) cross products through the magnetic forces that act on (+ vs. -) charges moving with some specified velocity through an ambient magnetic field. That is to say, we could flip our definition of the cross product by a sign but then we would also need to change the Lorentz force law accordingly.

[@robshaw2639]

@@dmytryk7887 Correct - the given axioms don't enforce the handedness (in math contexts I tend to say "chirality" instead of "handedness") both a left or right hand rule would satisfy the axioms, so I assume in the 7 dimensional case there is maybe 5-choose-2 chiralities. Fix e1 and e2 and you can choose either way for the other 5 axes...

[@bjornfeuerbacher5514]

@@dmytryk7887 You mean exactly the circle shown at 11:00 in the video? ;)

[@Kram1032]

yeah, the axioms as given only specify the outcome up to a chosen ordering of the unit vectors. Technically you could have six of them, but some of that choice is simply down to which unit vector you consider "the first" (so normally you'd just call it e1 and then the other two e2 and e3) and the rest of the choice comes down to chirality. The 7dimensional version is consequently going to have like 5040 such variations I think (but much less once you fix a first element etc.)

[@yoav613]

45 min video,michael penn started competing netflix now.

[@karthikeyank132010]

Never mind Netflix, this is an entire length of a regular class

[@AnrewDGK]

Everybody talks about the wedge product, but I personally enjoy another explanation - by the convolution with structure constant of appropriate algebra

[@Calcprof]

I like the connection between the problem and the problem of finding n mutually orthogonal tangent fields on the n - dimensional unit sphere in R^(n+1). The non-existence is then just algebraic topology. Existence is by construction (from the quaternions and octonians).

[@uselesscommon7761]

Cross product is awkward anyway. Outer product is the way, and is generalizable.

[@dougborter]

I should add that you can extend the quaternion concept to the octonions that consist of a scaler and seven imaginaries! There's another video on this channel that describes the octonions: kzbin.info/www/bejne/kHSaiqKcj9Vgmdk&pp=ygURb2N0b25pb25zIG51bWJlcnM%3D.

[@douglasmagowan2709]

We can define the cross product as the dual of the wedge product in a geometric algebra. This generalizes to any dimension. However, it is an operation on two vectors that returns a multivector outside of three dimensions.

[@JakubS]

another cheeky michael penn video to watch before going to sleep

[@burrbonus]

2:56 -- Does that property have a name? It seems to be related to the simplest of the Pythagorean identities from trigonometry because |a×b| involves the sine of the angle between vectors a and b while |a·b| involves the cosine.

[@VeteranVandal]

There's an arbitrary dimension external product, thankfully. But you have to embrace multivectors for it.

[@agranero6]

12:36 I made a question but removed you answered it later...(I misread the blue dot expression (a·b)c as (a·b)·c but when you deduce it the context made me I realized it as a dot product time a vector). So if you read it ignore it.

[@angeldude101]

It's possible to define a cross-product-like family of Lie algebras that act on (n choose 2)-dimensional objects for any n. The big thing that this sacrifices is the product of any two orthonormal elements having a length of 1, which is exactly the condition that was proven at the end. The video however clung to the axiom and discarded any potential products that accepting the result would bring. For n = 2 and n = 3, this family gives products that are isomorphic to the 1 and 3 dimensional cross products shown in the video. There is however no n that gives the 7 dimensional cross product since there is no n for which n choose 2 = 7.

[@kaay8983]

So we have a situation when skew-symmetric bilinear map VxV \to V can't simultaneously send the result to the orthogonal completion of input span and have a norm fixed subspace at |v|=1. I wonder if we can deduce it from the properties of orthogonal completion and norm alone, implicitly?

[@dougborter]

It might be of some interest to viewers that the cross product and dot product were discovered in the product of quaternions discovered by a mathematician, Hamilton.

[@zh84]

didn't know that, and it is interesting. Hamilton's quaternions were, IIRC, used in the original formulation of Maxwell's equations. Oliver Heaviside recast them in terms of vector analysis (using cross and dot products) which is how they are now usually written.

[@BlueGiant69202]

@@zh84 The geometric product of Geometric Algebra and Geometric Calculus (the sum of the dot product and the outer product ( - the dual of the cross product - )) now allows Maxwell's equation to be written as a single equation. "∇F = J (7.14)" "This unites all four Maxwell equations into a single spacetime equation based on the geometric product with the vector derivative." pp. 230-231, Geometric Algebra for Physicists by Chris Doran and Anthony Lasenby

[@zh84]

​@@BlueGiant69202That's way beyond my level of edulcation, I'm afraid!

[@bjornfeuerbacher5514]

Didn't Grassmann invent the dot product and the cross product independently of Hamiltonian's ideas about quaternions?

[@eartphoze]

Cross products glitched by a factor of Log in neural network in unit vectors for impedance in Pythagorean triangle as analog for distance in joules to horses in sidereal time in hertz

[@KipIngram]

This can be explained in a single sentence. In 4D, vectors have four degrees of freedom, but bivectors have six degrees of freedom - therefore there can be no one-to-one correspondence established between them. This is what we're really doing with the cross product in 3D - the cross product is a vector which is the DUAL of the bivector that properly represents the quantity under study. We do our kids a disservice teaching them the cross product approach to things in the first place - we really should just teach them proper geometric algebra starting in the ninth grade or so. It's not like it's hard, and it unifies so many things that otherwise get taught as separate things. For example, all of complex number theory arises from geometric algebra - complex arithmetic is the "even order sub-algebra of 2D geometric algebra." Quaternion theory? Built in. Etc. It's a long list of things that you just "take care of" once you learn geometric algebra. Those two vectors that you feed into a cross product? They actually define a PLANAR AREA. In 3D we can "cheat" and use the vector normal to that plane as a "stand-in" for the planar quantity itself. But this trick does not extend to other dimensions. It just HAPPENS TO BE TRUE in 3D that vectors have three degrees of freedom and bivectors have three degrees of freedom. It's an accident. If we didn't just happen to live in three spatial dimensions, cross products "wouldn't work" in the way that we learn them.

[@henrikljungstrand2036]

You never proved that the cross product in R⁷ actually IS a cross product! I know it is since earlier computations of my own, but others may be interested in this. Anyways, thanks for the rest of the lecture, it was good and possible to follow!

[@3snoW_]

Without watching the video, the most relevant property of a cross product is that its output is a vector orthogonal to both inputs, and therefore linearly independent to both. If both inputs are linearly independent, in 3D space there is only 1 direction where this is true, so it makes sense for the output to be a vector. Below 3D this is impossible, since you can't have 3 linearly independent vectors. Above 3D, the orthogonal direction of 2 independent vectors is not a single vector but a hyperplane, so my guess is that you need to give up the output being a vector for the operation to still be meaningful, and at that point it's no longer something you could call a vector product since the output is no longer a vector.

[@kennethforeman6164]

The proof at the end 34:16 ff strongly depends on A being closed under the cross product. But should we expect that the embedded R3 in the higher dimensional space be closed under the cross product? It is only one possible way to define the relations between e1-3. Perhaps we could expect that e2xe3 gives e4 instead of e1. Then none of the conclusions at the end hold because A is not closed under the cross product. Why is this circumstance excluded?

[@mathunt1130]

Having learnt geometric algebra, I think you can get cross-products in higher dimensions. It's very easy to get orthogonal vectors using geometric algebra. It warms my heart to see so many comments about geometric algebra.

[@evandrofilipe1526]

Yeah, it's at least some way to gauge popularity. It seems to be getting more popular, at least looking at this guy's previous comment sections

[@mathunt1130]

@@evandrofilipe1526 When I was working in GA around 5 years ago it seemed to be very niche but now, I see it lots of places.

[@baruchba7503]

What about complex cross product as in quaternions.

[@2fifty533]

that's 3d

[@PRIYANSH_SUTHAR]

They are different as they are of the form ai + bj + ck + d where d is a scalar which is not associated with any of the dimensions unlike in 3D vectors.

[@henrikljungstrand2036]

The "cross product" of quaternions is somewhat degenerate (scalars commute with everything), but becomes non-degenerate when restricted to "imaginary" i.e. vector part quaternions only. Which restricts R⁴ into R³.

[@davidawakim5473]

I wonder how the result would change if the base field changes. The lemmas fail in the quaternions for example

[@samuelbevins9151]

Can there be an L_\infty product on R^n which “mimics” the operation of the cross product?

[@ramsey121kamar]

At 34:56, how do we know that n = dim(A) + dim(A_orthogonal_complement) ?

[@krisbrandenberger544]

@ 29:00 The a and a_i should be reversed. Interchanging the b and a_i on the LHS would give the correct result on the RHS.

[@Songfugel]

Sort of, but with reservations. Since for example you can step out of the realm of real numbers into the complex plane and there are several ways to generalize the cross product to higher dimensions.

[@zemoxian]

That sounds interesting. Is there a term I could use to search for more information on this? Would it be related to Clifford algebras or something different?

[@Kram1032]

You're probably talking about the wedge product but, while almost identical to the cross product, it's not actually a generalized cross product. To see this, compare the 7D case where the cross product ends up being related to the octonions, but the wedge product is associative In fact, the 3D cross and wedge products are related by a factor -i (where i is the pseudo scalar in Cl(3,0)) That said, the things the cross product normally is being used for, the wedge product is typically straight up better at. I.e. describing rotations. The 7D cross product describes something else I think

[@Kram1032]

@@zemoxian Wedge Product, Outer Product, Hodge Star are all related to the Cross Product Clifford Algebra, Geometric Algebra are some broader areas (pretty much synonymous) where some of those are the chosen take on the Cross Product

[@zemoxian]

@@Kram1032 It’s my understanding that Geometric Algebras focus on real Clifford Algebraic. But Clifford algebras can also include complex numbers as well. I’m not really familiar with those. The wedge product in a real Cl(3, 0) algebra would be the dual of the cross product. Is the behavior of the wedge product in a complex Clifford algebra different in a way that gives rise to a more general definition of the cross product? Or was the OP describing a different way to generalize vector spaces using complex numbers that’s not covered by Clifford algebras at all?

[@Songfugel]

Look wikipedia for cross product and generalizations for an exhaustive list of the multiple ways of generalizations. Sorry, I am too lazy to write a serious academic answer in YT comment when the wiki page describes it beautifully, I am also not a native English speaker The complex realm and generalizations, while related in some generalizations and required in 2D (since 2d in real space lacks the information capacity required, expanding to complex realm allows you to keep track of the required extra information without expanding to 3rd dimensions to calculate the cross product for a 2d vector), was ultimately two separate statements A and B Lie algebra, multilinear algebra and quaternions are examples I recommend checking out

[@kono152]

ooh we get a big video today yipee

[@BenAlternate-zf9nr]

For n=1 when you said the cross product is always zero, doesn't that violate the last axiom since a cross product of unit vectors won't be a unit vector?

[@TheEternalVortex42]

Technically it does still satisfy the last axiom as written, just not the interpretation of it as "cross product of unit vectors is a unit vector".

[@fullfungo]

Yeah, the interpretation was not very precise. It is only correct for *distinct* unit vectors.

[@DavidRybka]

doesn't this also assume the cross product is a binary operation. for instance if you allow for three vectors you can create something that fits most of those axioms using the determinant of a 4x4 matrix with the 4d orthogonal direction vectors in the top row. changing order negates the determinant. so does adding two vectors and constant multiple. the biggest issue is the bottom axiom I think.

[@renanmonteirobarbosa8129]

You forgot special cases like quasi-orthogonal space which allows for n-dimensional cross product as well n=4 is the highest dimension for proving orthogonality.

[@azai.mp4]

n=0 also has a cross product by this definition: R^0 only has one vector, the zero vector, so you can just use 0×0=0

[@Frisbieinstein]

The cross product is a convenient mathematical trick for magnetism but only works in 3D. Elsewhere I use geometric algebra.

[@emwave100]

Are you the host from Hot ones?

[@flavioxy]

i remember we were given a question at math 1 for scientists. it was defined in R2 but i expanded to R3 to get the right results using the cross product. got minus points for that, awww :(

[@goodplacetostop2973]

45:05

[@Milan_Openfeint]

More like 0:41 honestly. If this was the nicest proof MP could find, I guess he should've kept looking.

[@crehenge2386]

I thought that this is what differential forms where made to generalize?

[@Noam_.Menashe]

Do you mean exterior algebra

[@TheLethalDomain]

​​​@@Noam_.MenasheExterior algebras are baked into the nature of fundamental forms and associated geometric algebras. In these languages, the cross product of R^n is a limiting case where you obtain a Lie bracket for the Lie algebra so(3). This means that the cross product is an operator related to orthonormal matrices of dimension 3 with a determinant of 1.

[@dougborter]

You could say that the cross product and dot product were "discovered" but I think "invented" isn't accurate. When you do a multiplication of two quaterions and group the terms, the dot product and the cross product are in the result. That's not necessarily all you get; I think there are some left over terms. It's been a while since I did it. You should try it: (a+bi+cj+dk)(r+si+tj+uk) where a,r are scalers and b, c, d, s, t, u are imaginary. When multiplying the imaginaries: ij = k; jk = i; ki = j; ji = -k; kj = -i; ik = -j; ijk = -1 where i, j, and k are unit vectors. There was a rift between two groups: some sided with Hamilton about using the quaterions only; the other group felt that the quaternions were needlessly complicated and wanted to use (and teach) just the cross product and dot product. And for the most part the second group prevailed. I myself was taught just the dot product and cross product in college. I learned of the quaternions decades later when reading some histories of mathematics. There's a lot of good material on the subject that can be found by doing a search on your favorite search engine.

[@keeperofthelight9681]

Wait cross products can be defined for non euclidean spaces too!!?

[@liamrapanan9434]

1D Being be like: omg how to understand whatever 2D this is - Us: bruh- Us: How to understand 4D??!?!?!!!?!?! 5D Being: lmao small brain

[@ronwittmann8400]

At 41:40 shouldn't (ei x b) x b = 0?

[@APaleDot]

No, because e_i × b is orthogonal to b

[@theupson]

there absolutely IS a four dimensional cross product, it just requires three input vectors instead of two. which should be your basic intuition from the cross product afinding the uniquely defined mutual perpendicular.

[@bjornfeuerbacher5514]

The product in four dimensions which takes three input vectors isn't called a "cross product", as far as I know.

[@denielalain5701]

Dear Michael Penn! I have worked out an algorithm for calculating a cross product in semi-arbitrary dimensions. It works not only in 3, and 7 dimension, but any dimension greater than 2 (yet). I can not actualy prove that the results of such calculations are the cross product (yet), but i think you can actualy help me with it. I am firmly believing that it is the cross product due to the way it works, and if you reach out to me, i will share it with you. If i show you the algorithm, you might be able to help to prove it, and possibly figuring out a way to use it too. My greatest problem is that i can not use it. What usage do you suggest of such algorithm?

[@Archik4]

what happen if you define it like that axbxc = determinant M | i j k r | | a1 a2 a3 a4| | b1 b2 b3 b4| | c1 c2 c3 c4|

[@Archik4]

For R2 | i j | | a1 a2 | We need only one vector. (a1, a2) => (a2, -a1) We get a rotation of 90 degrees clockwise. For R1 we need 0 vectors and always receive just i.

[@pasavant]

If this is the " new math" I am glad my kids are being home schooled.

[@ed.puckett]

[22:00] Am I being dense or is the argument a little subtle that (x.d = y.d for all d) implies x = y ?

[@amankarunakaran5080]

Yes I had the same thought - that statement is generally incorrect (e.g. in R^3 $e_1 \cdot e_2 = e_3 \cdot e_2$ but $e_1 eq e_3$) so I think there must be some other way to show this

[@Jester01]

Probably not the simplest proof, but I think you can choose d = x - y and then x.(x - y) = y.(x-y) => x.(x - y) - y.(x - y) => (x - y).(x - y) = 0 => x - y = 0 => x = y

[@kennethforeman6164]

Equivalent to (x-y).d=0, meaning x-y and d are orthogonal. But x-y is an arbitrary vector and d is a constant given vector. Hence, at least one of x-y and d is the 0 vector. We assumed d was more than just the 0 vector, so he concluded x must be y. But that contradicts the statement that x-y is an arbitrary vector! I think the resolution is that we assume d to be the arbitrary vector and hence x must equal y to make the dot product always true. Agreed that this step is problematic when d is the 0 vector since the constraint on x and y is no longer requisite, though it still holds true if the constraint applies.

[@Handelsbilanzdefizit]

Please, can you make a video about the "Gaussian process" in stochastics. I've some trouble with understanding. Thank you.

[@fernandogaray1681]

This is top tier content

[@BlueGiant69202]

??? This is not the blackboard and chalk years of the 18880's, 1950's or 1960's. It's 2024! See Clifford Algebra to Geometric Calculus (1984), New Foundations for Classical Mechanics(1986, 1999), Geometric Algebra for Physicists (2003).

[@fernandogaray1681]

@@BlueGiant69202 I mean KZbin content

[@user-qb2jn9zh9i]

It seems to me that the key condition of all the evidence in the video was the requirement of anticommutability. But why exactly should it be preserved in another number of dimensions? There was surprisingly so little justification for this in the video. What will the author call an object with the same system of axioms, but without the axiom of anticomutativity? With the axiom of commutativity?

[@TheEternalVortex42]

It wouldn't make sense to call it a "cross product" if it doesn't obey the same axioms in higher dimensions.

[@user-qb2jn9zh9i]

You see some meaning that eludes me. Wikipedia says that this operation was introduced by Hamilton. He multiplied the quaternions and divided the answer into a commutative and an anticomutative part. This turned out to be useful. I could do the same thing with octaves, by analogy. I guess I'll get the same R7 from this video, just in a different way? Does this make sense, some kind of generalization? Unfortunately, it doesn't work. An additional generalization clarifies something in particular; if we remove the requirement of commutativity, will it lead to a more general concept? Someone must have researched this, what is it called? I'm trying to fight my ignorance and my misunderstanding. I appreciate any clarification in advance.

[@gsestream]

oh but there is, a complex cross product, note how those axis naming stuff goes, well the non-counted dimensions are imaginary. well insert some imaginary dimensions to get up to the dimension number that has real cross products. to get real products. but imaginary.

[@roger7341]

The space we live in has six orthogonal directions, three translations, tx, ty, tz, and three rotations, rx, ry, rz. There are 21 valid subspaces. tx, ty, tz, rx, ry, rz, (tx,ty), (ty,tz), (tz,tx), (tx,rx), (ty,ry), (tz,rz), (tx,ty,tz), (tx,ty,rz), (ty,tz,rx), (tz,tx,ry), (rx,ry,rz), (tx,ty,tz,rx), (tx,ty,tz,ry), (tx,ty,tz,rz), (tx,ty,tz,rx,ty,rz)

[@willemesterhuyse2547]

There is a 4-dimensional cross product of two vectors, using the determinant of a non-square matrix. I just don't know if it satisfies all the properties of the 3-dimensional cross product.

[@remember25october]

Because 4-dim is the true world, from where ball lightnings come to us?

[@GreenDayFanMT]

What about Minkowski space?

[@chillfill4866]

I claim that the 2d cross product does not exist because if I consider 2d vectors to be 3d vectors in the same plane, the cross product will not be in the same plane. Which axioms of Euclidean geometry can I strip away to make me wrong?

[@angeldude101]

The axiom that all elements are 1-dimensional arrows. Get rid of that and the cross product between two lines in 2D gives a scaled copy of the 2D origin.

[@ChaoticNeutral6]

Coincidentally, I was just looking this up. I thought the generalised version of the cross product was the lie bracket?

[@FlaminTubbyToast]

To my knowledge the general perpendicular product is the wedge product.

[@angeldude101]

It is. The wedge/exterior product others have been talking about is a generalization of the 1D and 3D cross products shown in the video, but the Lie bracket generalizes the 7D cross product as well as the Geometric Algebra commutator product (which operates on the elements output by the exterior product).

[@meofamily4]

I'm a fan of the American Mathematical Monthly. Go, MAA.

[@tomctutor]

@2:43 axiom |a☓b|² = |a|²|b|² -(a.b)² comes simply from the Pythagarian [absin(θ)]² = (ab)² - [abcos(θ)]² , where θ is a scalar function, defined as the argument θ(a,b). Not too sure what θ(a,b) actually is in n>3 dimensions however.🤔

[@icew0lf98]

i didnt have the patience to watch the entire video just to find out if my question is answered or not: would a product that takes three inputs work in some analogous way as a cross product, for example the first rule could be abc=bca=cab=-acb=-bac=-cba

[@TheEternalVortex42]

There is a 4-D "cross product" that works with three vectors in an analogous way

[@APaleDot]

Look up the wedge product

[@isabelshurmanfeitoza9897]

Quase existe uashua quaternios usa uashia só que é associativo

[@wcdeich4]

You can make an n dimensional analog.

[@theodorostsilikis4025]

The step at 22:00 is totally unjustified.

[@System.Error.]

7th dim cross product is possible from what i heard

[@talastra]

Lengthy today.

[@ILSCDF]

And that's why I hate the cross product

[@boffo25]

What about sedonions?

[@camilocagliolo]

Oh my god, KZbin peaked here.

[@kyaume21]

There is a 4-dimensional cross product, but it is not binary but ternary.

[@kyaume21]

If you pose the wrong conditions, you get an empty or wrong answer.

[@denielalain5701]

-Hello there! I have figured out a way to do cross product in any dimensions, not only 3, and 7 dimensions. If i have made it, you will be able to do it too. Do not search for reasons, just do it instead!- Never mind, see my latest comment.

[@annaclarafenyo8185]

This is not the right way to teach this. You should explain the epsilon tensor, and that's it, it gives the "cross product" in any number of dimensions. In 4 dimensions, it takes 3 vectors and gives one new one.

[@bjornfeuerbacher5514]

First, he does not "teach" here, he explores and plays around. Second, there is not one single "right" way, there are several ways in which one could do thi.s

[@annaclarafenyo8185]

@@bjornfeuerbacher5514He does teach here, often well, sometimes he misses. The idea of a "cross product" in 4d taking THREE vectors to one makes perfect sense. In 5d it would take four vectors to 1, in 2d it takes one vector to one. This is the epsilon tensor.

[@bjornfeuerbacher5514]

@@annaclarafenyo8185 Again, there is not one single right way, there are several ways in which one could do this. He explores one of the ways here, the way you suggest is a different one. Nowhere did I say that your suggestions makes no sense. I only stressed that there are different ways to approach this.