An Overview of the Operations in Geometric Algebra
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- Опубліковано 16 чер 2024
- This video is an overview of some of the basic operations in geometric algebra. It covers operations like the geometric, outer, regressive, and inner products, some of the involutions, and a few other operations. It also tries to describe these operations in a way that applies to arbitrary flavors of geometric algebra, to allow for a smoother transition into these other flavors.
In case you don't really know anything about geometric algebra, I would suggest watching my introduction to it first: • A Swift Introduction t...
I also have an important addendum that I made to that video as well: • Addendum to A Swift In... . This video assumes that you have a basic understanding of vanilla geometric algebra already, which you would get from these videos.
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Sections:
00:00 Introduction
01:09 Multivectors
03:31 Grade Projection
04:50 Geometric Product
05:12 Geometric Understanding of the Geometric Product
06:40 Algebraic Understanding of the Geometric Product
10:01 Understanding the Geometric Product with Transformations
12:01 Reverse
14:21 Grade Involution
15:38 Magnitude
18:54 Outer Product
22:54 Regressive Product
26:51 Dual
30:33 Inner Product
39:49 Conclusion
As a physicist, my frustration with this videos is that I see the amazing potential of these tools, but I have other math so ingrained in my brain that I fail to use these methods for any physical problem beyond toy cases.
Honestly, it would be great if you could show us how to do it, although that might be out of your wheelhouse.
In anway, amazing videos. I always love them and learn so much from them.
Find a textbook on geometric algebra and try the problems.
Hestenes 'New Foundations for Classical Mechanics'?
It would be great if you could make a video (I think even a short one could suffice) about one of the best yet most overlooked cases one can make for the use of geometric algebra over regular vector algebra or complex Clifford algebras, namely the fact that if you allow real linear transformations to have multivectors as their eigenspaces you gain a geometric (and real) interpretation for what would otherwise be a collection of complex eigenvectors and eigenvalues which make no sense other than from an algebraic standpoint.
Great idea, Circe.
At first I was somewhat concerned about the use of the dagger symbol to refer to the reverse of the multivector since that symbol is already used for the Hermitian conjugate of a complex matrix. However, you can construct a complex-matrix basis for 3D VGA using the pauli matrices and it turns out that taking the Hermitian conjugate of those matrices behaves exactly like the reverse operation! Mathematics is wonderful!!
P.s. Your videos on GA have been amazing for me, it's the first bit of mathematics I've explored on my own without any educational institution guiding my hand and it's been one of the best experiences in my life so far! Possibly the most amazing moment was when I accidentally rederived the pauli matrices (actually a set of matrices equivalent to the pauli ones but multiplied by some constants) by first trying to find a 2D VGA basis using 2x2 matrices (I already knew the identity matrix and matrix for 'i' so just filled in the two remaining degrees of freedom), and then expanding that to 3D VGA with 2x2 complex matrices. My conception of the matrices (which I'd met in the context of spinor operators in QM) immediately went from them being esotheric and confusing to intuitive and elegant. As happens with many of these sorts of UA-cam series, it seems people are tailing off a little as time goes on, but I hope you don't lose the motivation to keep making these videos as they are invaluable resources to those of us that have stuck around! I'm a student at the moment so likely won't be able to support on Patreon any time soon, but I would certainly like to and will do so when I am in a financially stable position!
Heck yeah it feels so satisfying to learn geometric algebra informally
BASED. Can't wait for more!
Ah man I've been wanting this for so long, so glad you're still doing this.
Edit: its literally like you knew I read the wiki article on geometric algebra yesterday, perfect timing
Wow, this and the STA video really push the limits of my understanding. I’m quite convinced that if this weren’t a video where I can skip back and rewind a segment my attention dropped or I didn’t feel having understood it, there’s no chance I’d have learned something in less than 45 minutes.
It’s so stupid how inefficient reading a book or hearing a lecture live is for getting a basic understanding of concepts. Both has worked for me, sure, but I see the clear difference in pace. I learned complex analysis basically only through live lecture, learned forcing with textbook The only advantage a live lecture has is I can interrupt the lecturer and ask questions in real time. Watching a recorded video, I can pause any time to think about something, rewind for whatever reason, I can take screenshots, pause to write a detailed question about what is on screen with a timestamp so that you, the creator, has it as easy as it gets to write an answer.
Really like the way you covered the hodge dual construction in this video. I've avoided most use of it, out of fear that I didn't understand it, but it has now lost it's mystery.
You're doing an amazing job, making hard things really accessible. Thanks a lot! Keep up the good work!
Absolutely fantastic. I get so many thoughts and ideas from this. Thank you so much!
Great lecture as always
I really benefit from them
Thank you for your hard work
oh I'm so happy! I've been looking forward to the next installment!
Ditto
Your video is a pleasure for the eyes and for the mind.
This video ease a lot the implementation of general geometric algebra in any programming language. Thank you for that.
Thank you for taking the time to make these videos. I am studying at Uni atm and couldn't quite reconcile in my head much of the math using cross products that lead me to research exactly how they work. I ended up on your UA-cam channel and now everything makes so much more sense.
I'm amazed by the shear amount of high quality maths channels these days. I simply can't keep up!
Oh, I've been waiting so long for this video
Esse vídeo apareceu num momento muito oportuno. OBRIGADO!!!
It dawns on me that mathematics suffers from the problems of programmers in the 60s-80s. Deciding to use pre/post/super/sub scripts/fix-notation and the recognition that variables can actually have names beyond single letters. Notation and symbology is I feel important therefore consistency in representation can go along way with understanding and adoption in multiple fields.
this is the clearest video on Geometric Algebra you have released yet
when you are already familiar with vector spaces, it is really easy to follow and understand the difference between the very familiar vector space and the powerful geometric algebra
oui oui baguette du fromage
Your videos on Geometric Algebra are, in my opinion, very good for gaining a deep look and a better understanding of what complex numbers, quaternions, etc. are all about and what their geometric interpretation actually is. In this context, I would have a few suggestions. Could you also go into the topic of dual quaternions, i.e. the combination of quaternions and dual numbers? How can this be represented with the help of geometric algebra? And furthermore, the topic of dual numbers or, more generally, the geometry of the generalized complex numbers. The classical complex numbers are so-called elliptic complex numbers, the dual numbers are the parabolic complex numbers and so-called hyperbolic complex numbers are described.
Dual quaternions are isomorphic to the even-graded subalgebra of 3D PGA.
Looking Good Sudgy!!!
14:20 I love the reverse. It's weird how such a simple operation can have such a uncommon property.
I love your videos. I love how you approach the topic, is there any book that you recommend to start learning into Geometric Algebra ?
I answer that in question six of my FAQ: ua-cam.com/users/postUgwFByhvEg1_hD_L1Ch4AaABCQ
let's go new geometric algebra content
i've been somewhat struggling with the different kinds of product recently, so this is actually perfect.
Struggling with inner and outer products (save for the vectors case) got me nowhere. Some day I'll get it. My mind is more oriented to physics and engineering than to mathematics. But, as I've solved the engineering problem of the nature and how-to's of cognition in a general manner and it yielded multipluridimensional subspaces (one dimension per synapsis, and some neurons have even 2•10⁵ of them), I have to learn to deal with them to obtain more detailed results.
And .... Subspaces have already given me a cue or two. Gotta think on that. I might get something good!
The prophet has returned
oh wow, didn't expect to see the de morgan law here
I have found that the hardest thing about learning geometric algebra for me is trying to learn all the operations that exist and what they do. As soon as you started explaining at a high level what the operations were, their common notation and how they came about I suddenly found information I had consumed on the subject clicking near immediately. The few questions I have left about the fundamentals are still about a few operations like the commutator product, scalar product and I think a few other ones, things that are specific to algebras like representation of certain objects, and how calculus works on these objects. Oh yeah and the dual basis.
Where I get most confused is in that there is a lot of symbol manipulation, where how it's implemented is very very hand-wavy. In the case of vectors, the dot and wedge can be expressed in terms of the geometric. If there is a way to define all of these operators in terms of geometric; then I know exactly how I would write it in code.
@@robfielding8566 There is a way of expressing it in code quite easily although it is a brute force way, using a specific algebra and computing a general geometric product, which is the only way I have been able to implement it thus far.
this and the introduction are in the hall of fame of my videos on math.
The "e with multiple subscripts" notation translates well into code.
typedef struct {
float e1, e2, e3;
} vector3;
typedef struct {
float e12, e13, e23;
} bivector3;
vector3 innerproduct(vector3 a, bivector3 b)
{
vector3 c;
c.e1 = (a.e2 * b.e12) + (a.e3 * b.e13);
c.e2 = (a.e1 * b.e12) + (a.e3 * b.e23);
c.e3 = (a.e1 * b.e13) + (a.e2 * b.e23);
return c;
}
I hope I got that right. Though if you were actually making something like a game, you'd probably call the components x, y z and xy, xz, and yz instead.
Implied e1, e2, e3 in x, y, z, implied e12, e23, e13 in xy, yz, xz.
@@wafikiri_ You might be able to alias them using preprocessor directives or something like that, but I'd have to go and look up the C documentation to find out how.
@@Roxor128 I spent 20 years programming without type enforcement or any possibility of declaring types, much less objects. It was easy. All typing was in my head.
@@wafikiri_ psuedo-code programming? /j
@@evandrofilipe1526 My main working tool as a database programmer was Revelation, an emulation by Cosmos of the Pick operating system for mini and, later, large computers. Revelation run on PC's with Windows since 1984. Pick directly manages its own system as a DBMS and serves large databases efficiently since 1965. It uses a compiled version of BASIC called Data/BASIC that leaves data type enforcing to the applications and their programmers. So did R/BASIC, Revelation's version of Data/BASIC (only the Index function differed a little), also compiled into a kind of pseudocode whose "machine" codes I learned to know intimately, which permitted me to reconstruct lost source code from compiled quite a few times (the bad doings of an electric company which had messed up the floor wiring of the Company I worked for caused much file and data damage that I had to reconstruct daily by hand).
The R/BASIC pseudocode really was powerful. Any single variable could contain any data up to 64 KBytes. Once I read a full database selection of any customer's annual stock-exchange transactions in a single variable!
What a treat!
@sudgylacmoe from what text book do you pull this notation? If it’s your own textbook or another, I’d like to pick up a copy to study in greater detail.
I mention several textbooks in question six of my FAQ: ua-cam.com/users/postUgwFByhvEg1_hD_L1Ch4AaABCQ
Hello, this is the first video I’ve seen from you pop up in my feed. It’s nice to see I have another place to look to for all my questions! I’ve been doing vectors and planes in school recently, so I’m sure I could find answers and explanations here.
I have to ask though; is that a dango in your channel picture??
Yes, it is a dango!
@@sudgylacmoe I love the dango daikazoku. Guess I’ll be coming here first for all my math stuff from now on lol
love the videos! is geometric algebra equivalent to differential forms? stuff like the exterior derivative and k-forms seem really similar to ∇^_ and k-vectors
Yep, they're the same thing.
3:00 Am I right in understanding that "through the origin" is implicit each time? Like, "a bivector represents a 2D subspace _that goes through the origin_ ". Because one bivector is not enough to represent any _arbitrary_ subspace of, say, a 3D space, right?
In such case, could a multivector `u + B` (u being a 1-vector and B a bivector) be used to represent some arbitrary 2D subspace with orientation B that is offset from the origin by a vector u? Would it make sense to interpret it like that?
When I was saying "A bivector represents a plane", the phrase "through the origin" was implied. However, the definition of subspace is just "A subset of vectors that is closed under addition and scalar multiplication", so in VGA the phrase "through the origin" is implicit in the definition of subspace.
All subspaces of a vector space contain the zero vector, so, by definition, all subspaces of a vector space "go through the origin." There is simply no such a thing as a subspace of a vector that *does not* go through the origin. This would be nonsensical.
The talk about the geometric algebra of different vector spaces has me curious. Do these concepts extend to infinite-dimensional vector spaces (I guess they would have to be Hilbert spaces)? What might the geometric algebra of, say, the sequence space l^2, or of position space in quantum mechanics, look like?
You can construct the geometric algebra of any module over a commutative ring that has a symmetric bilinear form. But I haven't really seen a good description of what a geometric algebra of a function space is like.
Hey can you work out the relation between multivector and general tensors 😅???
This leads to the abstraction that actually prevented the practical appreciation for Geometric Clifford Algebra for more than a century! But I try to explain it, in order to understand it myself (no warranty!)
The formal construction of the Clifford algebra is analogous to constructing the three numbers "0", "1", "2" in the mod-3 algebra as equivalence classes of integers - where you lump together all numbers that differ by a multiple of 3. This is less intuitive than having just three things that multiply according to a 3 x 3 table, but at least you are confident that they "exist" as the integers exist.
So here, each geometric multivector would be the equivalence class of many general tensors (see below) that differ only by a specific set ("ideal") of tensors with a special property: instead of the property "being a multiple of 3" to make sure 3 is equivalent 0, it is here "being a multiple of the special tensor (v x v - Q(v)) - see below" to make sure all the tensor classes that will be our geometric vectors do multiply as prescribed by the matrix Q (= Diag(1,1,0,-1) e.g.). So the Q contains the information about the specific geometric algebra which follows from the specific behaviour of the unit vectors in VGA, PGA, CGA... All other vectors, bivectors, etc. follow from the algebra rules.
The x is the "tensor product", meaning our general tensors had already been equipped with some formal multiplication x that can be applied to any two general tensors and that obeys certain rules by construction. A vector v can be identified with a contra-variant tensor itself, so it can be multiplied using x - like in that curious general tensor v x v - Q(v) above, Q being some chosen covariant tensor that yields a specific scalar, e.g. 1 for the "vector-like" tensor that we will later call e_1 together with all equivalent general tensors (just like the general integers 2, 5, 8 are all called "2" in the mod-3 algebra).
Now, in terms of these equivalence classes, every vector-like tensor v (together with its equivalence class) fulfills v x v - Q(v) = 0 automatically - by the "miracle" of math quotients - because the expression v x v - Q(v) differs only by tensor v x v - Q(v) from zero which is such a multiple of tensor v x v - Q(v) analogous to the multiple of 3 (duh!)
- meaning this expression is identified with 0, so it "is" zero!
Concretely, according to our choice of Q, there will also be one tensor class that we will call geometric basis vector "e_1" where e_1 x e_1 - 1 = 0, because Q(v) was precisely chosen to yield 1 for some vector e_1 that is identifiable with some general tensor that we could also call e_1.
In this case:
Clifford vector "e_1" = the equivalence class {general tensor identified with vector e_1 + any multiple of the specific general tensor of rank one: v x v + Q(v) that takes a vector v and yields some scalar }
I'm still confused about identification between vectors and tensors, when they are co- or contra-variant, but at least it's some ideas...
YAAAAAAAS! LETS GO SUDGY! Thank you!
We need more
historically speaking the derivation of inner product comes from grassmann algebra , do you have maybe a regress proof of that ? this question huntes me for a long time
@sudgylacmoe why is it that any operation (a+be1e2)(c+de1e2) is closed and will return ac+bde1e2, but (ae1+be2)(ce1+de2)=ac+bd+(ad-bc)e1e2 is not closed. is there a multivector form other than a+be1e2 that is closed under multiplication?
Trivially, full multivectors are closed under multiplication. Scalars are also closed under multiplication. Also, given any vector v, multivectors of the form a + bv are closed under multiplication. I think that's all of them for 2D.
One operator I feel deserved inclusion here is the commutator product, often denoted with the same symbol as the cross product in traditional vector algebra. It's just AB-BA, but it's often used on 2 bivectors to produce a bivector result, in which case it functions like the lie bracket. It's essentially the most natural expression of the axial X axial = axial case of the cross product.
It's actually 1/2(AB - BA). The reason I didn't include it is that I don't feel like I understand it well enough. I can see what it does algebraically but I have no good geometric intuition for it. If I had a better grasp on it I would have definitely included it.
@@sudgylacmoe Oops, yeah, forgot the 1/2. The best intuition I have for it is the lie bracket one. It's the residual rotation you get from rotating slightly in plane A, then B, then undoing in the opposite order. Another example would be calculating the result of applying a small torque to a gyroscope with a large rotational inertia, I think.
My problem is that I haven't really studied Lie theory :/
@@sudgylacmoe In none of the different (but similar) definitions of conmutator that I know of (group theory, rings theory, Lie bracket of vector fields, Hamiltonian mechanics' Poisson bracket) is there a 1/2 factor. Now I'm intrigued where that factor appears.
@@wafikiri_Notice how geometric algebra is not any of the things you listed.
The 1/2 factor comes from the fact that for vectors u, v, one has uv = u•v + ext(u, v), while vu = v•u + ext(v, u) = u•v - ext(u, v). It follows that uv - vu = 2 ext(u, v), but now there is an unwanted factor of 2, so we have ext(u, v) = (1/2)(uv - vu). Of course, for arbitrary multivectors, the commutator product does not necessarily give the exterior product, but it is still a useful generalization.
Curiously, the relation between the regressive and the outer products looks very similar to De Morgan's Law from boolean algebra.
I guess it counts as yet another lucky generalization.
It's absolute a De Morgan's Law! The cherry on top is the actual symbols used: ∧ and ∨. What else are these symbols used for? _AND and OR!_ The very boolean operations that are normally talked about with De Morgan's Law.
While not really evidence, I still think it's worth mentioning that many GA programming libraries use ! for the dual, which also happens to be the ASCII symbol for NOT, which is the boolean equivalent of the dual when applying De Morgan's Law to AND and OR.
can you show how to work with shapes in geometric algebra, like how to construct a circle then a sphere, how to make a torus, for example?
The two answers to this are that you either do it the same way as usual (just use a set of vectors), or use a flavor that has those shapes. For example, VGA has lines/planes going through the origin, PGA has arbitrary points, lines, and planes, and CGA has points, lines, planes, circles, and spheres. I don't know of any GA that has tori as a native object, but I feel like it should exist.
The magnitude square could be called quadrance i think
Very useful, many are waiting for your series to continue (in case you ever wonder)!
I got stuck 3 times, but could resolve two of my issues:
10:21 vav = v(av) actually rotates v in the same direction as a->v, but also stretches it to the length of a, for a is not a unit vector. The result is the same as reflecting vector a
10:56 -vav is an extra turn by 180°, which is taking not v as the mirror line but a line perpendicular to it - which is useful if v is to be interpreted as a normal vector of the mirror (this I could not understand directly from your way of putting it)
32:25 Is the final contraction in the animation correct? I think the result should be parallel to the projection not perpendicular, just like with the ordinary projection onto another vector. Only the common components are kept - opposite to the external product.
That final contraction is correct. Contraction removes parallel parts, leaving only the perpendicular part.
@@sudgylacmoe Thank you! So the contraction collapses the bivector dimension in the direction of the projection into a scalar (not zero) which should then affect the length of remaining vector in orthogonal direction. This is exactly like the projection onto a vector where the remaining orthogonal dimension is zero (scalar). I confused dimension 0 with the scalar 0.
Also, you have already told us
ua-cam.com/video/0bOiy0HVMqA/v-deo.html
but as a human I need to hear it again and again, especially when it's months apart.
So never mind, this is a great series, and your decision to complement the longer development with shorter overviews is the right one - pedagogically, redundancy is needed to make it have an impact!
@@sudgylacmoe
Shouldn't the vector be pointed in the opposite direction though? I thought the dot product of a vector and a bivector would end up rotating the projection according to the orientation of the bivector.
Uh I may or may not have just not bothered with that... The thing is, the order in which you multiply the two will affect which direction the result is in, and I never specified if we were calculating the inner product of the vector and the bivector or the bivector and the vector.
@@sudgylacmoe fair enough
From the description of inner and outer product you gave:
Inner product is the part of the geom product when all factors are shared and outer product is the part when no factors are shared.
It seems like the Geometric Product = Inner Product + Middle Product + Outer Product
Where the Middle product is non zero when some indices are shared. Is there a proper name for this middle product? Or is it not very useful to deserve it's own name?
Nice ! :)
t'as pas le droit de commenter avant moi
@@AllemandInstable 5h d'avance, qu'est-ce tu vas faire
I'm kinda wondering why we don't use the square root of v•v as the magnitude of v. The dot product is grade-reducing, so as long as both operands have the same grade, the results are scalars, right? I guess norm squared is the better approach, though.
For homogeneous multivectors, this definition is equal to A† · A! But that doesn't work for nonhomogeneous multivectors so we prefer this definition.
i just had a thought: what kinds of geometric algebras can you guarantee that all vectors can be expressed as a product of (scalar + vector) terms?
Can you rephrase your question? As you've currently said it it's trivial, since every vector v can be written as the product of the one term 0 + v which is a scalar plus vector. I'm guessing you meant something else though.
@@sudgylacmoe let B be the some set of basis vectors (e1, e2...) and R be the real numbers (scalars). let V be the result of closing the basis vectors under addition and rescaling. let S be the set containing all terms (a + b v) for a in R, b in R, v in V; then closed under multiplication (that is, complete it by taking every possible product of terms already in S). For which choices of B is S closed under addition?
i think this question necessarily has some obvious trivial answers (let B = {e1}, just the singleton, or B = {}, empty) but i wonder what types of solutions exist besides these, if any. I can imagine an answer might exist for B = {e1, e2} with e1^2=e2^2=0, but i haven't tested that yet to see if every a+be1+ce2+de12 can be factored into (w+xe1)(y+ze2)
3:05 shouldn’t it be the sum of 2d subspaces because only blades span spaces
That's why I said "These correspondences have some exceptions" a bit later.
So, did I get it right that in *any flavor* of GA the following ways of computing the exterior and the inner products of two arbitrary multivectors, expressed in terms of an orthonormal basis, work?
-- *Exterior product:* do it as if it was a geometric product *in the respective flavor* of GA, but only for the pairs of basis multivectors that share *no common factors.*
-- *Inner product:* do it as if it was a geometric product *in the respective flavor* of GA, but only for the pairs of basis multivectors in which one *includes all of the factors* of the other.
Yep, that's right! Interestingly enough though, the exterior product ends up being the same in all flavors! Because you only use multivectors with no common factors, what the basis vectors square to never comes into play.
@@sudgylacmoe Thank you very much! Your videos are such a great treasure to find. It's thanks to your work that I first learned the concept of geometric algebra, and now I find it hard to even imagine how I wolud do my maths without such a powerful yet beautiful tool at my disposal.
Can someone connect inner prudent and magnitude squared for me? It doesn't really click for me
22:01 Is it sure the first algorithm is more efficient? Under what conditions?
I'll have to check that!
For a machine, *checking common factors* in lists of factor and *sorting* (!) such list is generally *much more taxing than comparing the length of the lists.* I bet the second algorithm is more efficient by now, especially once optimized by checking the number of factors first so we can avoid the computation of a lot of terms we aren't going to use immediately (lazy programming).
The second algorithm is basically the first one but with extra steps. They both require checking common factors and sorting lists.
Studying geometric algebra gives me the same feeling as when I do a jigsaw puzzle, and I get all of the edge in place, and then never get round to filling in the middle.
There's so many puzzle pieces and no one really knows how they're all related, but it's obvious that they are.
Well, that's the downside of GA... the plus side is that I think GA helps highlight what we still have left to learn more effectively than any other branch of mathematics.
This is just factually incorrect. There are plenty of people in the world who do understand how the pieces are related. You not having this understanding does not imply no one else does. The videos in this series are just an informal introduction to the topic. They are excellent resources for learning the basics and knowing why the topic is important, but the series is neither rigorous (because it is not intended to be), nor does it do a deep dive into the theory either (because this channel is fairly new). You are obviously not going to come out as an expert of geometric algebra solely from watching the videos. This is to be expected, and it implies absolutely nothing as to whether anyone else is an expert or not.
@@angelmendez-rivera351 gees, way to interpret my comment in the most mean-spirited way possible.
@@RooftopDuvet There is nothing mean-spirited about what I said.
i wrote a python class that acts as a multivector. it can handle anything from scalars to vector components of all grades. it handles scalars by treating them as multiples of component "1". as a result, the inner product (1+x+y)|(x+y)=2 and the wedge product (1+x+y)^(x+y)=x+y (arising from 1^(x+y)=x+y with the 1 from 1+x+y). but if i test the alternative definition a|b=1/2*(a*b+b*a) and a^b=(a*b-b*a)/2 i get that the dot product is 2+x+y and that the wedge product is 0. which definition is more correct?
this issue only arises when allowing scalars. but if i want to allow a self contained multi vector, i need to allow for all possible components, including scalars.
to summarize, what is s|(e1+e2) and what is s^(e1+e2) where s is a scalar? i know that s*(e1+e2)=s*e1+s*e2, but is that from the inner product or the wedge product? some calculators do scalar multiplication for both s|v and s^v, but then s*v=s|v+s^v=s*v+s*v=2*s*v... in my class, 5|e1=0 and 5^e1=5e1, but (ab+-ba)/2 says otherwise. i would likewise think that 5|5=25 but 5^5=0 since anything wedged with its own component is 0.
The equations ab = a·b + a∧b, a·b = 1/2(ab + ba), and a∧b = 1/2(ab - ba) are not true in general. They are only true for vectors. The general equations are the ones given in this video. The inner and outer product of any multivector with a scalar is just scalar multiplication. Also, the inner product you mentioned at the top should be (1 + x + y) · (x + y) = 2 + x + y.
Wedge product or outer product always commutes or anticommutes. Is it also true for other products?
The outer product does not always commute or anticommute! An example is e1 ∧ (1 + e2) = e1 + e12 vs. (1 + e2) ∧ e1 = e1 - e12. In general, the various products only have certain special cases where they commute or anticommute.
@@sudgylacmoe I thought outer product was not use on mixed grade multivectors. Outer product does always commute or anticommute when operands are single grade k-vectors. Similarly, do other products also always commute or anticommute when both operands are single grade k-vectors?
The inner and regressive products do always commute or anticommute when both operands are homogeneous, but the geometric product does not.
@@sudgylacmoe Thanks
Is the inverse operator concretely defined in this video? Or am I missing it?
Not all nonzero multivectors have inverses. There does exist an algorithm that can determine if a multivector is invertible, and if it exists, it finds what that inverse is, but going over this algorithm would have been way too much for this video.
@@sudgylacmoe Thanks for the answer! Could you link me to some resources that might cover those algorithms? I've found very conflicting results looking for it on my own.
Here's the original paper describing it: arxiv.org/abs/2005.04015 It uses matrix representations, which I'm not too good with. I've seen others describe it in a different way in things like discord discussions, but I don't have any links for those. Note that in many situations, there exist much faster algorithms.
@@sudgylacmoe Perfect! I'll give it a read. I'm trying to build a Clifford algebra library that is dimension-agnostic as a hobby project, so efficiency isn't a goal for me- but I'll keep that in mind.
At 10 minutes in, why didn’t you have an e12 vector?
I didn't really try to have the result be anything special, and that's just how the algebra worked out.
I'm struggling to understand how the inverse of the pseudoscalar works. When i looked up how to inverse a multivector i found the Formular B^-1 = B^~/BB^~ but im pga would't it produces a 0 under the fraction coz e00 = 0 and under the fraction it stands e012210 for the pseudoscalar which would result into 0? hope u could help me out
That formula doesn't always work. Some multivectors are not invertible, and the unit pseudoscalar in PGA is one of them.
@@sudgylacmoethen how does the join (regressive product) work in PGA when i cant use the invert of the pseudoscalar? What's the full form of A v B where A and B are bivectors
You don't find the dual by multiplying by the unit pseudoscalar in this case. See 26:32 of this video.
@@sudgylacmoe omg iam so sorry i didn't listen carefully enough u explain the situation in this Video iam so sorry to bother u with that question. Thank you for your great videos and that u help me out in the comments. I really fell in love with this topic and your videos are great for learning it
Also what does it mean to take the inverse of the dual for example when i have star e1 = e20 what is the inverse?
what would star^-1 e1 be?
Projective Geometry? What about Matroids?
Also, how do Geometric Algebra and Tensor Algebra differ? Their notations seem remarkably similar.
Algebraically, the main difference is that in geometric algebra, vectors square to scalars.
It seems like Tensor Algebra is even more general than Geometric Algebra then. Pretty cool.
After the PGA video, I was trying to figure out what (A . B) means for arbitrary multivectors. Geometric Product is actually easy and well-defined for completely arbitrary multivectors. Imagine a real implementation: There are 2^n parts. Imagine indices in binary: 0000 (scalar), 0001 (e1), 0011 (e12), etc.... With floating point arithmetic, there will be a tiny deviation from the actual values; so imagine that NONE of the parts are actually exactly equal to zero. Because there's always a (+/-). So, it doesn't make sense to even talk about grades in an exact sense.
So that leads me to ask... can PGA be defined using only Geometric Product for EVERYTHING?
Grading is still fine. In a good implementation, you will actually not even have the coefficients for the parts of the multivectors that you don't care about. And even if you don't have a good implementation and always work with full multivectors, you can still just use linearity to look at each individual grade part and calculate the whole thing from that.
Now to actually answer your question, there is one issue with PGA: the dual is actually basis dependent. Thus, to fully define all of the operations in PGA, you need both the geometric product and a basis. But you don't need anything else to define everything.
@@sudgylacmoe I implemented a simple multivector multiply in Python where I had basis vectors (e1,e2,e3). I did a simple "rotate an object around a point" by subtracting the point, rotating, adding it back. And the trivector part was like 0.00000...1. The problem was that I defined the "type" of a thing by which parts were exactly zero. But the more I think about it, it's a feature rather than a bug if you truly are working with multivectors. There will be things that are "almost of type vector".
A good implementation like ganja.js implies a rather complicated compiler infrastructure. For understanding how things actually work, IMHO, it's good to not ignore coordinates just yet; or a small implementation just yet. ie: Define geometric product as a doubly-nested for loop, with a calculation that sorts indices and reduces to a (-1,0,1) before adding it into its destination part.
I have never seen a GA definition of the operators that make sense outside of special cases. I wonder if the most enlightening definition of the operators uses geometric product only, using cancellations in the way that it works for vectors. All GA really seems to be is adding "directions in space" into the algebra; where Geometry is all about non-numeric things that square to (-1,0,1).
after watching this many times over, I guess the main thing is to split up multigrade MV back into sums; so that you are only dealing with multiplying pairwise with MV of specific grades. (A_j . B_k) = _{|j-k|} ... But if they mix things together, deal with it before-hand. I think something like this:
A . B
= (_1 + _2) . (_2 + _3)
= ((A_1 + A_2) . (B_2 + B_3))
= (A_1 . B_2) + (A_1 . B_3) + (A_2 . B_2) + (A_2 . B_3)
= _1 + _2 + _0 +
Would someone please explain to me why e_{12}^2=-1 and not 1 @ 17:03? I know he said it's because of the product being swapped, but I'm unsure as to why we'd have to swap the order.
Also, if we are assumed to be in G(1,1,1) for this calculation still, I understand it, but I don't think we are.
In VGA, e_{12}^2 = e_{12} e_{12} = e1e2e1e2 = -e1e1e2e2 = -1.
@@sudgylacmoe Thank you!
I also had 2 other questions if you don't mind.
1. Why does the span of v_1, ..., v_n being n-dimensional imply that v_1, ..., v_n are linearly independent?
I believe the answer is because if v_1, ..., v_n are linearly independent, the span they form MUST be n-dimensional, by definition, but I'm not sure if my logic is correct.
2. At 21:34, when we change the order of the second product, the reason that there isn't a net sign change is because we swap the order of 2 terms in the product, meaning there are 2 sign changes, which cancel each other out, right?
1. This is a standard fact in linear algebra, but I can't recall a proof off of the top of my head. You can probably google it.
2. I actually did eight swaps at once here. To be explicit, this is the process:
e561234
-e516234
e512634
-e512364
e512346
-e152346
e125346
-e123546
e123456
Keep in mind that it's only the product of two basis vectors that anticommutes, not arbitrary basis multivectors. You have to do swaps one-by-one to get the correct result.
@@sudgylacmoe By doing swaps one-by-one, I'm assuming you mean you can only swap scripts that are next to each other? If so, that'd mean that we can only swap scripts that are next to each other, 1 at a time, and changing the sign each time we do so; is that correct? :)
Also, as an unrelated question, I'm not entirely sure, but it seems to me that throughout the entirety of the video, we use orthonormal basis's; is that correct?
Also, thank you SOOOOOO much for these videos and your intuitive replies, they are both unbelievably helpful!
@@shrek2342 Yes, that's correct.
What is i^-1?
It depends on the situation. When none of the basis vectors square to zero, it's either i or -i. When one of the basis vectors does square to zero, i is uninvertible. Right after the section in the video using i^-1 I talk about the alternative you can use when i is uninvertible.
@@sudgylacmoe wow I never thought you would reply. Let me take the opportunity to say thank you. These videos are really good.
Everything else I've seen about VGA, PGA, and STA has stated very confidently the following:
uv = u•v + u^v
u•v is grade-lowering
For f(u, v) = uv + vu and g(u, v) = uv - vu, f and g correspond to the dot and wedge products, but which is which varies by grade.
If that's true, then one can systematically determine the inner product of u and v by breaking them up into grades and calculating f and g of each pair. Either f or g will produce a lower grade than the inputs; the sum of all the grade-reduced operations is the inner product of u and v.
It was my understanding that those statements were defining the dot (inner) product, so your more abstract treatment of the inner product is very confusing to me. Is my definition not universal? Under what conditions is the geometric product not the sum of the dot and wedge products? When is the dot product not supposed to return a reduced grade object? When are the dot and wedge products different from the symmetric and antisymmetric parts of the geometric product?
I've watched it again, and now I just want to see your source.
These are common misconceptions (that I admit I have accidentally perpetuated in the past). Pretty much any GA textbook will eventually give the definitions for the inner and outer products that I gave here. Examples are equation 2.1.17 in New Foundations for Classical Mechanics and Equation 4.43 in Geometric Algebra for Physicists. (If you want something free, it's also equation 80 in arxiv.org/pdf/1205.5935.pdf .)
To explicitly answer your questions: No, the definition is not universal. The geometric product is not the sum of the dot and wedge products when neither argument is a vector, and also if one of the arguments is a scalar (although in that case it is true if you use the correct one-sided contraction). The dot product does always return a reduced grade object (well the dot product with a scalar doesn't change the grade but that's close enough). The dot and wedge products are not the symmetric and antisymmetric parts of the geometric product almost always. The only time where I can think of where it is true is when one argument is a vector and the other is an odd-grade multivector.
Honestly, despite the focus in many sources, I've found that focusing on symmetry and antisymmetry isn't really that helpful and doesn't actually help you do geometry. Yeah, there are some useful equations in this regard, but they're just useful equations, not anything fundamental.
Thanks, checking it out
Eigenchris sent me.
Am I too cynical when I say the hodge dual definition feels like a dirty way to hide if ... else ... cases under the hood?
Honestly the hodge dual does feel a bit dirty, but any definition of the dual in degenerate algebras is going to feel dirty. Any good definition of the dual in a degenerate algebra actually introduces additional structure, which is usually done by picking a basis.
@@sudgylacmoe I don't have a problem with the definition not being so smooth. It just seems to be a thing in math culture to not write down something with if ... else ... in algebra. Instead just say what the result should be and push it on the user how to do that.
#gradeprojectionoperator
Wildberger’s Universal geometries define this “magnitude squared” you’re struggling with as the fundamental sizing quantity: Quadrance. Magnitude (length etc) if it makes sense at all is then at best a derived quantity.
"Magnitude" makes perfect sense, and it is the more fundamental of the two concepts in almost all contexts. The theory of normed vector spaces is all about this. Clifford algebras are just not always normed, which is fine.
Wait, quaternions are just 2d geometric algebra with ei^2 = ej^2 = -1 and eiej=-ejei. Mind blown.
In other words, G(0,2,0)
Yes, you can also use bivectors to describe them
In fact, every even subalgebra of a geometric algebra is itself another geometric algebra (sometimes multiple distinct geometric algebras) (which the exception of Cl(0,0,0), which is its own even subalgebra). This is also related to being able to express reflections as rotations through a higher dimension, like how ℂomplex numbers reframe reflections across the origin in the ℝeal numbers as 180° rotations around the origin in the ℂomplex plane.
Cl⁺(3,0) = Cl(0,2) = ℍ
Just a piece of constructive criticism. When you distribute at 8:58 it is totally impossible to follow along without stopping the video. Better to show each line of the calculation simultaneously than to use the fancy animation that destroys the previous line imo. Tho in other areas for simple things the animation is neat and good. For a big complicated thing it's a little annoying bc if you want to follow u are basically forced to go back, stop the video, and do the calculation yourself. Just my 2 cents. Thx for what u do.
My idea here is that everybody (okay, maybe not everybody) knows how to distribute, so it's not important to follow along with it. The other options that I can see are cluttering up the screen (which I try to avoid), or going through it slowly (which I really don't want to do because that particular part is too basic).
"A vector can be a plane", "a vector can be a circle"… BULLSHIT.
A vector is a vector. A plane is a plane. A circle is a circle. Clearly distinct entities.
"to be" isn't "to be partially describe or replaced in this context by".
Now, that you can use the same algebraic structure, the same geometric algebra, using either vectors, planes, circles, bivectors or other things *does NOT imply that those things are the same.*
I already hate this introduction because of this nonsense. Excellent content ruined by a stupid oversimplification, joke or better said misinformation again. 😡
In abstract linear algebra, mathematicians call whatever satisfy the 10 axioms of a vector space a vector. So polynomials are vectors, wavefunctions in quantum mechanics are vectors. Differential Operators are vectors and so on.
He isn't saying they are the same. The above entities are examples of vectors. So a differential operator is a vector. It lives in a vector space. But a vector doesn't have to be a differential operator. Vector is just an umbrella term for all these things that can add to themselves and scale along with other nice properties.
@@smolboi9659 Ok. It Make sense. The "can be" expression in english language is tricky, especially the verb "to be". The conditions of this potential equivalence needed to be clarified. An equivalence that isn't necessarily an identity for me, until proven. I might be wrong about that.
Is it an identity of structure or a conformation of structure?
For example, a circle is a vector given an vectorial algebra for it, but:
- can we prove that a circle is *nothing more* than a vector, with no extra properties? (I accept we don't consider arbitrary superfluous cultural traits associated with circles as long as we agree on what is superfluous or no, according to goals)
- Is that given algebra the only possible vectorization for a circle, aka any other vectorization we can think of is equivalent to this one?
@@emjizone You are right it's not equivalent. A circle has other properties that makes it a circle as opposed to a plane. But we can still call it a vector.
It's like saying a rectangle is a polygon with each internal angle being 90 degrees. A square is a rectangle with the extra property that the sides are equal. We say squares are rectangles but not the other way round.
Similarly circles are vectors but vectors are not circles. They are not equivalent but have a subset relationship. If we draw a venn diagram, circles are a subset of vectors.
Ok this is an aside. Technically I would argue that for planes the coordinates do not form a vector space. This is because the equation for a plane is ax + by + cz + d = 0. We can use (a,b,c,d) as the traditional tuple of numbers and we can scale and add them. However it violates one of the axioms of a vector space. There is no 0 vector. (a,b,c,d) all 0 does not correspond to any plane. So I would say because of this technicality planes in 3D are almost vectors but not quite.
What is the objective of this video, where is the role of the “customer” (viewer, listener) ? If you already know about everything about GA you don’t need this video. If you are learning about GA, you only hear a lot of incomprehensible sentences fired at you, at the end of this video you know as much about GA as at the beginning of this video. So you wasted a lot of time losing your temper along the way. Again: what is the objective of this video, what’s its point?
There are a few reasons I made this video. First, if you know a bit of GA but not a lot, I think this video can help you learn a few things and clarify some things that might be confusing. Second, I wanted to have one central location that I could easily refer to every time I wanted to mention something that's in this video. Third and most importantly, I tried to design this video to be a good stepping stone for those who are familiar with VGA and want to extend to other flavors.
Personally, I've always been confused about how to think about the left-contraction vs. right-contraction vs. dot-product because they seemed so similar to me, but this video cleared things up a lot.
My positive example is about the videos of Mathoma: he apparently assumed you knew a bit, but didn’t mind to EXPLAIN, to prove the maybe obvious, so you had hooks to move on. He took his time. Here, nothing is explained, only statements are made, at a huge speed: how are you supposed to make sense of this all? No hooks, so even if you pause the video n times, that wouldn’t help. And a video as a reference: that’s the same as searching in a sequential file, that’s of a past era. OK, I made my point, I guess.
@@paulbloemen7256
No doubt the Mathoma videos are the gold standard, but they never each touched on most of the concepts in this video. I saw this video more as an overview of concepts which will be explained in more depth later. As someone trying to learn GA, I've had a huge problem diving into the more advanced stuff because it all seems so far beyond me. I think videos like this are an invaluable bridge which can introduce concepts in a more intuitive way. If you're going to understand GA, the operations discussed in this video are absolutely foundational and the Mathoma videos simply do not suffice, at all. So I think your criticism is misplaced.
@Paul If this video moves too fast, I would suggest my series From Zero to Geo, which starts from assuming that you don't even know any linear algebra. (Although it's not too far yet.) This video is intended for an audience that already has a decent amount of knowledge on geometric algebra.