Geometric Algebra in 3D - Fundamentals

[@skeletonrowdie1768]

when i saw 1,2,1 as being scalar, vector and bivector, i just knew G(3) would have 1,3,3,1 objects and G(4) will have 1,4,6,4,1 objects :P pascal's triangle for the win. Now to prove.. xO

[@bonbonpony]

The reason is how you pick stuff to combine, because that's basically 1 choose N, 2 choose N, 3 choose N etc., so it's pretty obvious why the binomial coefficients appear here. You have N basis vectors. In how many ways you can pick two of them? In how many different ways you can pick three? etc.

[@wreksy7522]

I am currently doing a research project on Geometric Algebra, and these videos have been a blessing. I appreciate all you do sir.

[@oberonthefirst8886]

These videos are fantastic, they really elucidate the applicability of geometric algebra.

[@Math_oma]

+oberon the first Thanks! I too like this topic for its theoretical elegance and physical applicability.

[@aleclarsen7434]

I love your explanatory style, and I'm really looking forward to the video on gyroscopic procession. Thanks for taking the time to make such great content!

[@Math_oma]

+Alec Larsen My pleasure.

[@bonbonpony]

What if one takes a wedge product of a vector and a scalar? Is it possible at all? (I mean, it _should_ be, if bivector wedge vector is OK.) How about the other combinations? Like trivector wedged with vector?

[@1495978707]

Well remember, the wedge part of the geometric part. So what product of a vector and scalar is antisymmetric? Remember that a wedge product with a vector and another one is only nonzero when the vectors aren’t parallel. It would just be zero. A vector geometric producted with a scalar is still a vector

[@1495978707]

Remember, *any* object in the G(3) algebra is a linear combination of the 8 basis elements

[@angelmendez-rivera351]

Your logic is flawed. The wedge product is associative, so bivector wedge vector is only okay, because a bivector is itself a wedge product of two vectors. The wedge product is ultimately only applied to grade 1 objects to build higher grade objects. You cannot build grade 0 objects out of it, so you cannot wedge a scalar with a vector.

[@bashirabdel-fattah9499]

It is indeed possible to take the wedge product of a scalar with a vector, and the result is the same as just the ordinary scalar multiplication. There are a couple of ways to explain why, but really it has to do with the properties of the exterior algebra of a vector space.

[@mahmoudmaher00]

Man, your are amazing!

[@zlatanbrekke6538]

How do we justify that e₁e₂e₃ = e₁∧e₂∧e₃ by the definition of the geometric product? If you write it out you could write e₁e₂e₃ = (e₁⋅e₂ + e₁∧e₂)e₃ = (e₁∧e₂)e₃ But this is as far I reach with my knowledge of geometric algebra

[@animowany111]

The dot product of two separate basis vectors is zero, because they are orthogonal - they have no common facing direction, so all the dot products end up being zero. All that's left is wedge products. You don't even need to perform the calculation.

[@jamesedwards6173]

What animowany111 says is true, but it's also not difficult to finish that computation: ... = (e₁᭴e₂)·e₃ + (e₁᭴e₂)᭴e₃ = 0 + e₁᭴e₂᭴e₃ = e₁᭴e₂᭴e₃ (e₁᭴e₂)·e₃ is 0 because e₁᭴e₂ is orthogonal to e₃. (e₁᭴e₂ is in the "xy plane", and e₃---or "z"---is perpendicular to that.)

[@metuphys5611]

@@jamesedwards6173 i feel the guy who first asked the question knows that if you treat the vectors e_i as our usual cartesian base vectors the target result is obvious. Also, if we're going to consider the bivector as a unit vector normal to a plane, why do even bother with the directed area stuff? Using usual cross product relations between cartesians would do just as well

[@jamesedwards6173]

@@metuphys5611 They're not treated as "the usual Cartesian base vectors"; they're _any_ orthonormal basis. We are also not considering "the bivector as a unit vector normal to a plane"; we can't do that, because it isn't that; it's an oriented area _in_ the plane. And, the reason we would use geometric algebra instead is because it generalizes and more naturally represents the elements involved. (The cross product does not; it exists in three dimensions, and _only_ in three dimensions, and is something of a "hack", in some senses.) Note: I did use x, y, and z terminology, but it was only for communications' effect. The e vectors do not need to map directly to x, y, and z.

[@pronounjow]

Now I understand why you called torque a pseudovector on Veritasium's Gyroscopic Precession video. I thought you were referring to the cross product as a fake vector, but I misunderstood. Could the cross product be called an axial vector then, like an axis of rotation?

[@Math_oma]

+Jo Reven It depends on what kinds of vectors are getting crossed. Torque, for example, is a cross product between two polar vectors (position and force) and that will generate a pseudovector. In GA, that corresponds to wedging two vectors to get a bivector. However, you can cross a polar vector with a pseudovector to get a polar vector. For example, if you have a charged particle moving in a magnetic field, the direction of the magnetic force (a polar vector) is given by crossing the velocity vector (polar) with the magnetic field pseudovector. In standard physics using _arrows_ for both those quantities seems to lead to confusion and an over-reliance on the _right hand_ whereas in GA it seems you cannot confuse them because one is grade-1 (true vectors) and the other is grade-2 (pseudovector).

[@pronounjow]

I didn't know you could cross those together. Is the idea of "fake vectors" a thing, though?

[@Math_oma]

+Jo Reven I'm not sure if I've ever heard of pseudovectors referred to as _fake vectors_ but the prefix pseudo- does mean _false_. There is indeed something fishy about the way pseudovectors transform when looking in a mirror, which is why you have to distinguish them from true vectors at all.

[@v8poweredpony]

Thank you so much for your videos on Geometric Algebra...I have been self learning this subject to aid me in Python coding . I wish I took Geometric Algebra while I was in school in my younger years. I very much appreciate your lessons and style of teaching. Have an outstanding day.

[@maxwibert]

Just realized a cool thing: a geometric algebra generated by n-dimensional space is going to be 2^n dimensional as an abstract vector space. This is because a k-grade basis element will be the multiple of any k distinct 1-grade basis element, so you have n choose k distinct k-grade basis elements. But the sum from k = 0 to k = n of n choose k is just 2^n.

[@bonbonpony]

Yup, which by the way is the sum of the binomial coefficients of order N (or Pascal triangle's row). That's also the reason behind the algebras of space (Clifford algebras) getting closed for every 2^n dimension: complex numbers getting closed at dimension 2, quaternions getting closed at dimension 4, octonions at dimension 8, sedenions at dimension 16, etc.

[@mortelli7]

Loving this series of videos! Would you also consider doing a series on differential forms and dual vector fields? I would really like to hear another perspective on this, and it develops pretty naturally from geometric algebra.

[@Math_oma]

+mortelli7 I suspect these topics will come up naturally in this video series. I also like these two topics.

[@joshuabaker1616]

This is great stuff, you are really doing such a great job explaining, and helping to demystify an area/ branch of mathematics... Which even to this day, there doesn't exist many prerequisites or introduction type material. I just wanted to say thank you and great job!!

[@alfonshomac]

take a very christmasy high five from me. thanks for your work.

[@Math_oma]

+and then i said you're welcome.

[@NicholsonNeisler-fz3gi]

Mathoma? More like a Mathanoma.

[@Timo4eus]

That's very elegant way to understand those things that were explained (actually not explained) as something given as is.

[@NathanPK]

Thank you

[@ExCyberino]

Ty bro

[@henrmota]

This volume trivector smells like integration

[@Ambagaye]

@Mathoma why the -1 scalar is not in the set to complete the closure property?

[@bonbonpony]

Because this is only the set of units. -1 is just the opposite of 1. Same reason why 2, 3, 4, etc. (and -2, -3, -4 etc.) are not in this basis set: they can be obtained from the unit by multiplying it by the elements from the underlying field (usually real numbers).

[@leokovacic707]

Can it be done without basis vectors?

[@angelmendez-rivera351]

Yes, but it gets quite difficult

[@Cihak200]

18:24

[@CephalopodParty]

I've taken it up as a personal project to write a program for rotating vectors in higher dimensions. Are you going to talk about rotating vectors in R^n with the geometric algebra? PS I realized this is the clifford algebra of R^3 and I feel silly for saying it was the tensor algebra. I think I also have a better understanding of the exterior algebra as well now, although I still don't quite understand how the exterior or clifford algebras are constructed as a quotient space of the tensor algebra.

[@Math_oma]

+Anon Ymous That sounds like an interesting project, as I'll likely end up making a video on higher dimensional rotations - people seem to like high-dimensional stuff too. Unfortunately, I'll be reaching the end of my drawing abilities, so I'll have to do some thinking on how we're going to cover that material in addition to symbols.

[@CephalopodParty]

Mathoma Personally I don't mind looking at symbolic abstractions for higher dimensions as opposed to diagrams, but on the topic of visualizing higher dimensions-is it possible to represent, say, a 10 dimensional polyhedra and its rotations with 5 2D cross sections the same way you could represent a 2D figure with 2 1D cross sections or a 3D figure with a 2D and 1D cross section? It seems intuitive but I feel like I read somewhere that multiple 2D subspaces are preserved in higher dimensional rotations and my intuition tells me that this would get in the way of representing some 2n-dimensional figure with n 2D cross sections. I was going to ask this on stackexchange but I figured I might as well ask you too. Thanks!

[@CephalopodParty]

Mathoma I posted a revision of my question here math.stackexchange.com/questions/2076888/how-does-one-explicitly-compute-the-result-of-a-rotation-on-a-higher-dimensional

[@Math_oma]

+Anon Ymous I think the answer to your question is you wedge n_1 and n_2 to form a bivector, B, then form the rotor, R=exp(-B*theta/2) where theta is your angle of rotation. Then to rotate some vector, v, you perform RvR^(-1), where R^(-1) is the inverse rotor exp(B*theta/2).

[@CephalopodParty]

Mathoma Ah hah! That makes far more sense and its way easier to understand why the quaternions are the clifford algebra of R3 now, and the rotation formula follows from the fact that the wedge product of the two basis vectors in 3 dimensions can be equated, after multiplication of the trivector, to the cross product which is normal to the surface of rotation! That's way simpler than the proof for that formula that I learnt a week ago!