Addendum to A Swift Introduction to Geometric Algebra
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- Опубліковано 10 тра 2024
- This video is an addendum to my most popular video, A Swift Introduction to Geometric Algebra. It clears up some misunderstandings that have arisen from the original video, and then describes two useful ways of understanding the geometric product. This also leads to a discussion of the other flavors of geometric algebra, such as PGA and CGA.
If you want to learn geometric algebra at a much slower pace, check out my video series From Zero to Geo. Here's a link to the playlist for it: • From Zero to Geo
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Sections
00:00 Introduction
01:00 Common Misunderstandings
02:28 Understanding the Geometric Product
03:37 Product of Two Vectors
05:32 Product of Vector With Bivector
07:26 Product of Two Bivectors
09:18 Product of Two Multivectors
09:42 Other Flavors of Geometric Algebra
10:58 Revisiting the Geometric Product
12:43 Geometric Product in VGA
14:44 Problem With 3D Reflections
16:09 2D PGA
19:37 3D PGA
20:47 2D CGA
22:56 Conclusion
Loved this explanation of the geometric product, it feels so natural in a way. And now you've gone and made me excited for different flavours of GA. Zero to Geo will be such an amazing resource if you continue as you're doing. I'm excited for this channel's future, and I hope this type of channel becomes more prevelant in the youtube education sphere.
what is zero to geo?
okay, 0:19
Though I have a background in mathematics, G.A. has remained an elusive sub-branch that one was seemingly already involved in doing PHD related research, or struggling to understand the nomenclature and/or its application beyond "unification" of multiple areas. I tried to read and minimally understand a Grothendieck paper (about tensors relating to topology)... I was pretty overwhelmed to say the least. After doing your series, i still don't really understand that paper- but i did gain a much better grasp on the subject generally speaking, and also why it should be taught more/ added to math cirrocumulus. Anyway i just wanted to say thank you and what a great job you did putting this whole thing together. I tell all who will listen to check out your channel.
This has, so far, been a great series and very informative.
I've done some personal study on Geometric Algebra, even going so far as buying a literal textbook on computer graphics using Geometric algebra. I unfortunately didn't get too far in it because of other projects which I put before it, but your explanation of PGA and CGA have made things so much more clear to me. Thank you.
19:34: it's remarkable and beautiful that a reflection is described by letters that are themselves "reflected".
Last time I watched your GA videos, they inspired me to pick up a book on the topic, but I got a bit overwhelmed by the formal definitions and caveats. Now I'm inclined to give it another go.
Your explanations are incredible. information/sentence is astounding yet somehow very coherent and understandable. I wish there were more resources akin to your brand of teaching when I was an undergrad. Please keep up the phenomenal work.
I just now started studying in university and started with a linear algebra course, your videos on vectors were extremely helpful for me to get a better grasp on vectors! Thanks a lot
I'm mesmerized by the power of geometric algebra. holy cow
patreon'd. your videos have changed my mathematical reasoning considerably. thank you.
My favorite bit about geometric algebra (that is, Clifford algebra) is that tensor algebra = Grassmann algebra + Clifford ideal (where the ideal is simply v tensor v plus its norm squared), so that the geometric multiplication is just the projection of the tensor product on the external algebra, thus recovering in one move both the geometrical insightfulness of the exterior product and the power and simplicity of the tensor, explaining at the same time why the product has both a scalar and a bivector part!
Thanks for this! You answered almost all of the questions I had.
Eventually the intuitive definition. Combination of components! 👏
Incredible job
I knew very little about PGA/CGA, so this was a cool illustration!
i am sooooo grateful for your videos! they are priceless!
Great work Manni
Thanks for making this simple enough for even me to understand.
Thanks! This was clear.
Thank you for this! You should definitely update the video descrption for the Swift Introduction with a prominent link to this video, and maybe add a pinned comment as well.
You are really a legend. I’m gonna dive deep into it
This was a great resource!
I'll really enjoy this video :D
Perfectly done, thank you!
@sudgylacmoe dear sudgy you saved my life and time by these series of video, do you know that ? you are math god to me. I was struggling with clifford algebra so much, but you are the Moses for me. You split the ' sea of difficulties' and help me to pass through from nothing to something. thank you .
So great !!!!
It was freaking awesome
Thks & I likes the way you makes me think and I hates the way you makes my head hurt so-much (keeps-it-up ;).
Well that woke me up! It's a whole new way of thinking (I think). It has been a bit fast but then I have never been to quick on the uptake, I tend to have to 'hand write' all the information in order to process it. I find the idea of multiplying circles hilarious and for there to be consistent mechanisms behind it to be absolutely amazing. In some models of our universe they say that an infinite line eventually loops back, so an infinite vector being akin to a circle isn't so far out as it seems at first glance. OK enough waffling .. I am off to your Geo stream :-)
'infinite line will eventually loop back' but how? by spanning all the universe or by finding a lower dimensional projection (a line or a curve) and how can you guarantee the second outcome ?
@@kutay8421Perhaps a compact universe without boundary...maybe a globally curved universe. Something like a torus or sphere might do it.
@@umbraemilitos the answer I got from ChatGPT:
The question: You are on a 1 meter radius Torus. The thickness radius is 10 cm. You choose a random direction and start walking. What is the average path in meters you take before hitting the same starting position again?
The answer: A = ∞
The average path you take before hitting the same starting position again is infinite. This result might seem counterintuitive, but it is a consequence of the particular geometry of the torus and the random nature of the walk. It means that, on average, you would wander infinitely far before returning to your starting point. This does not mean you will never return; it just means that there is no finite average distance for the path you take.
Just thought I would comment that while the outer/tensor product and wedge product may differ in general, b tensor a is also a linear transformation that takes a to ||a||^2 b ! So there is some relationship and I suspect you could make an identification in certain dimensions.
Please more!
I feel like this was more than just an addendum. But whatever; it was very super interesting and I learned a ton of stuff so I'm definitely not complaining! 🤩
The moment you pointed out that ab≠a·b+a∧b I suddenly realized what we've been doing all is just adding elements { (e0), e1, ..., en } to the damn real space and treat them like normal polynomials! Except that we have to simplify their products by the multiplication table, and that they are anti-commutative.
I didn't understand your comment. Can you explicit it a little ?
Thanks for these video series, the animations are great and this one in particular shows the best intuitive explanation I have seen for what the geometric product is.
That led me to the following intuition, which is probably tainted by the fact I mostly work with vector spaces in machine learning and natural language processing [1]:
the geometric product ab "reshapes" (or "splits") the (a,b) pair into two new components: their similarity and their dissimilarity, ie. what they have in common vs. what they don't have in common. This intuition works if a and b are vectors, but as you showed it fails in higher dimensions where geom products can generate more than 2 terms. Do you think that intuition is still somewhat in the right direction, and would there be a way to generalize it so it applies whatever the dimension and whatever flavour of GA you are using?
[1] In these fields, using vector encodings for bits of data (words for instance) is very common so you can use things like the dot product to measure the similarity between them.
Eg. if you have a body of text containing documents labeled from 1 to N, some naïve approach would be to consider that you have a N-dim vector space. Each word occurring in this body of text would be a vector, and each component x_i of that vector just counts how many times that word occurs in document i. This approach therefore considers similar words that tend to frequently occur in the same documents (high cross product). Less naïve approaches would try to learn better vector encodings, do dimensionality reduction, etc, but the general logic remains valid.
Man I wasn't expecting to see the stuff we covered in finite geometries to pop up here (e.g. using reflections to get all isometries), always neat to see that stuff pop up
At least this was in English instead of German
Your videos have given many an introduction to the wonderful concepts behind geometric algebra, I think we should all thank you! I am wondering where you learned these concepts and intuitions from, and if you could link those sources for us to take a look at.
Also, after seeing your videos on GA, I tried learning about geometric calculus. Do you plan on making videos on that in your Geo to Hero series? And do you have any recommendations as to where we can learn it?
I've covered introductory textbooks and my journey in learning geometric algebra in questions six and seven of my FAQ: ua-cam.com/users/postUgwFByhvEg1_hD_L1Ch4AaABCQ
As for the concepts and intuitions that I presented in this video, I had to come up with most of them myself. Thus, I am unable to provide you with resources other than this video. Sorry.
For geometric calculus, I am not planning on covering it in From Zero to Geo. Even back in college I struggled with vector calculus, and that has transferred somewhat over to geometric calculus. I have gotten to the point that I understand the basic concepts, but I am not at the point that I would be comfortable teaching the concepts to others. However, it is definitely a topic that I would like to be able to cover, and I hope that I might be able to learn the concepts well enough to do a series on it after From Zero to Geo is done.
I forgot to mention that if you want a textbook for geometric calculus, MacDonald's Vector and Geometric Calculus is a good start. If you want to go hardcore, you can go for Hestenes and Sobczyk's Clifford Algebra to Geometric Calculus.
Great video. A few dummy questions:
4:20 why would we lose associativity?
Consider three vectors u, v, and w, such that u and w point in different directions. If the product of two vectors was always the product of their lengths, (uv)w would point in the direction of w and u(vw) would point in the direction of u, meaning that they are not equal and associativity is lost.
So, it seems like the complex number plane is some kind of weird variation on a 2D vector space, where we only keep the scalar and the bivector parts? It's weird that a bivector would show up in the solution to polynomials.
Complex numbers and quaternions are actually the same as even subalgebras of geometric algebras for 2D and 3D: i, j, k are bivectors, 1 is a scalar. (An even subalgebra contains just elements of even degree and their sums etc..)
And complex numbers (or quaternions or any algebra like geometric algebras of spaces) *are* a vector space, if you strip multiplication and conjugation from them and leave just addition (and scaling). It’s an extra structure that makes them complex numbers and not just points on a plane. 🙂
I have been thinking about the connection between geometric product and matrix since the swift introduction video.
At 12:30, "the geometric product AB is a linear transformation that transforms A to B"
I think this would be equivalent to this description, if a 2d vector A(a1,a2),is timed by a 2x2 matrix M[m1,m2,m3,m4], we can get a new vector B(b1,b2)=MA, so M is a linear transformation that transforms A to B.
The eigenvalue of M, λ, can be found by solving the equation,(m1-λ)*(m4-λ)-(m2)*(m3)=0.
I think, the eigenvalue of M, λ, is equal to the geometric product of A and B.
For example, the matrix M[0,-1,1,0], it's eigenvalue λ can be found by solving (0-λ)*(0-λ)-(-1)*(1)=0, λ=i
Given a vector A(1,0), vector B=MA=[0,-1,1,0]*(1,0)=(0,1), the geometric product AB=A⬤B+A∧B=0+i, they are equal.
I'm quite excited when I came up with this idea, but I don't have many friends in real life who have interest in this topic to discuss with, so I hope somebody can give me some comment.
An n-dimensional square matrix is diagonalisable over C with algebraic multiplicity n, so there can be in general n different eigenvalues for M, how does this fit with your analogy?
Also, the eigenspace has geometric multiplicity n, so there could be in general n different eigenvectors too
We could think of that last fact as saying that A and B are related to the eigenvectors of M, but this doesn't make sense once we increase in dimension because the argument of the 2-d linear subspace needs more treatment
@@MagicGonads
I've forgot the details about matrix diagonalization and geometric multiplicity,I've searched a little bit ,but still have not fully understood ,thus can't fully understand your meaning. But I think the rough meaning you want to express is like, for an eigenvalue, there are many matrices correspond to this value, so we can't know which one is right. But I'll talk about that later.
I want to add two more idea to my previous idea.
First, I should divide the result of the product AB by |A|^2, to do some kind of "normalization".
Second, given two vectors A and B, B=MA, there are many M can do this, but I want to focus on the type of M which can be considered as the summation of "scaling matrix",[λ,0,0,λ], and "rotation+scaling matrix",[λcosθ,λ-sinθ,λsinθ,λcosθ].
Because the eigenvalue of [a,0,0,a] is a, and the eigenvalue of [bcosθ,-bsinθ,bsinθ,bcosθ] is bcosθ+ibsinθ, their summation is a+bcosθ+ibsinθ.There are many possible combinations of a and b+θ, but it doesn't matter, anyone is OK.
More simply, if a matrix has eigenvalue a+bi, then it might be [a , -b , b , a], as this one
www.reddit.com/r/math/comments/d7bxdu/matrix_representation_of_complex_numbers/
So the my conclusion is, given A and B, B=MA, to find M , first calculate AB, then divide it by |A|^2, and express that complex number in matrix form, and that is a working M.
I've tried a few, and it seems to be working, though I'm not quite sure how it's working yet, though it might be able to explained by simple algebra.
For example, vector A(1,3) and vector B(5,7)
AB=26-8i,|A|^2=10,
AB/|A^2|=2.6-0.8i
M=[2.6 , 0.8 , -0.8 , 2.6]
B1=2.6*1+0.8*3=5
B2=-0.8*1+2.6*3=7
Another example,vector A(1,2) and vector B(3,4)
AB=11-2i,|A|^2=5,
AB/|A^2|=2.2-0.4i
M=[2.2 , 0.4 , -0.4 , 2.2]
B1=2.2*1+0.4*2=3
B2=-0.4*1+2.2*2=4
And I agree I don't know how to do this trick on higher dimension, I don't know how to express a+bi in a 3*3 matrix.
@@user-fr2jr6hd4i Like in the video, we ignore the magnitudes involved, only care about the orientation, so no need to correct with dividing by |A| or whatever
@@user-fr2jr6hd4i My point is that a 2x2 matrix has two eigenvalues, for example [0, -1, 1, 0] has eigenvalues i and - i, so there are *two* values to consider, so which of those is ab and how do we assign that, and for 3x3 matrices this becomes three eigenvalues etc.
@@MagicGonads As you said, in the video 12:05 , he dropped the A squared, it seems to be a tiny unobtrusive step for just a few seconds in the video, but it is a very important step for me to make sure the result of the calculation is right, so I still want to emphasize it.
M1= [0 , -1 , 1 , 0] has two eigenvalues, i and -i
M2= [0 , 1 , -1 , 0] also has two eigenvalues, i and -i
We'll choose i for M1, and -i for M2, because M1 is CCW 90 degree rotation and M2 is CW 90 degree rotation. Although the calculation pop out two answers, we have to decide on our own, which one to use.
I want to know about CGA more, too!
And also how do we derive, erm, abstractly why in PGA a vector represents a hyperplane? Naively when one had worked before with projective spaces as is, without geometric product added, one might think that vectors in PGA would represent points (and then of course get stuck in things not working at all). I mean if one uses just an exterior algebra, then it’s pretty natural that vectors, bivectors and so on would represent points, lines and so on. But we couldn’t add the scalar part to this, we somehow need to projective-dualize first for things to work. Why that’s so eludes me... 😑
Every geometric algebra has a "dual algebra" where the role of scalars and pseudoscalars are swapped, vectors and pseudovectors are swapped, etc. So thinking of these dual algebras is not a strange thing in and of itself. But that still leaves the question of why we should think to look at the dual algebra for PGA in the first place. Looking into it myself, I couldn't find any direct historical motivation. The oldest thing I could find was somebody just saying "Hey let's use planes". For some more motivation from something more recent, you can check out this video: ua-cam.com/video/ichOiuBoBoQ/v-deo.html. They also did a longer series on PGA here: ua-cam.com/play/PLsSPBzvBkYjxrsTOr0KLDilkZaw7UE2Vc.html
Thanks!
The way I picture it is that e₁ means x = 0, e₂ means y = 0, and e₃ means z = 0. PGA also has e₀ which could be interpreted as the equation 1 = 0, which is obviously a contradiction, but it can be added to the other equations giving results that actually have solutions. Taking the wedge product between two vectors is akin to making a system of linear equations, so e₁e₂ means the system of equations {x = 0, y = 0}.
The other way of interpreting it is that mirror reflection is the simplest transformation and all others can be built up from multiple mirrors. 2 mirrors gives a rotation around their shared intersection line. 3 mirrors gives a reflection around the point where all three intersect. This is often called the "meet" operator, and it basically takes the intersection of two subspaces. It's opposite, the join operator is actually a little more complex conceptually since it also fills in other points between the two, and how it does so can different depending on the geometry. In CGA, the join of two points is a point pair, where as in PGA it's a line. Why a point pair even exists as a primitive object can be unclear until you see how it's simply the intersection/meet between a circle and either a plane or a sphere, or in 2D: the intersection of two circles.
I need a recommendation for a good book on geometric algebra. I prefer a more rigorous style, rather than sketched proofs. On the other hand I appreciate insightful examples and applications of theory as well. Do you have any suggestions?
I answer this in question six of my FAQ: ua-cam.com/users/postUgwFByhvEg1_hD_L1Ch4AaABCQ
@@sudgylacmoe cool! Thx!
The reason why it seems that (equation in the thumbnail) doesn't hold universally could be the way the inner and outer product were introduced in the original video.
According to their usual definitions (when the geometric product is defined axiomatically)
a.b := 1/2*(ab + ba) (symmetric part of the product),
a^b := 1/2*(ab - ba) (antisymmetric part of the product),
it must hold due to the axioms of linear space that
a.b + a^b =(definitions)= 1/2*(ab + ba) + 1/2*(ab - ba) =(distributivity over vectors)= 1/2*((ab + ba) + (ab - ba)) =(associativity, commutativity)= 1/2*(ab + ab + (ba - ba)) =(additive inverse)= 1/2*(ab + ab) =(distributivity over vectors)= 1/2*ab + 1/2*ab =(distributivity over scalars)= (1/2+1/2)*ab = 1*ab =(compatibility)= ab.
Even though the algebra consists of elements of different grades, it still forms a linear space over the field (of real numbers) as a whole.
In case of a product of two bivectors, the resulting bivector component is antisymmetric and thus part of the outer product.
I'm not sure how this could be explained intuitively, though.
The inner and outer products are not the symmetric and antisymmetric parts of the geometric product. That's only true for vectors (well and a few other cases). Really I think the focus on symmetry and antisymmetry obfuscates the geometric, inner, and outer products, and it's what confused me when I made my first video.
Really, the geometric product should be defined first (in a few different possible ways), and then the inner product is the lowest-grade part of the geometric product and the outer product is the highest-grade part of the geometric product. Everything else is just a special case of this fundamental idea.
The VGA case is intuitive because I already understand the relationship between the dot product and the cosine of an angle between two vectors. The PGA case where you’re moving lines around is just taking your word. Guess I need to read a text to see an actual proof.
Interestingly enough, with e0 · e0 = 0, the inner product of two lines in PGA is precisely the magnitude of the two lines times the cosine of the angle between them, just like in VGA!
@@sudgylacmoe The thing I’m not understanding is what you are talking about when you put points and vectors in space. The vectors in terms of the basis e0, e1, e2, e3 are planes in space now. But the planes are overdetermined as there are many ways to have a0 + a1X + a2Y + a3Z = 0 ( I’m using “Vector” = a0e0 + a1e1 + a2e2 + a3e3 ). Also, if vectors are now representing planes, what is representing the single “points” you are drawing and reflecting over the “planes” that are actually vectors. This is what is confusing.
Planes being overdetermined isn't really an issue. There isn't really anything stopping you from having multiple vectors representing the same object. It's pretty much just like homogeneous coordinates. Also, in n-dimensional PGA, points are n-1 vectors, and it turns out that exactly the same formula for reflecting planes works for reflecting points.
In general I'll admit that I didn't really explain things in detail here. The idea was just to get a rough overview of these ideas, and not to enable you to actually do PGA yourself.
@@sudgylacmoe Okay. So the vector in "world space" is the hodge star of the vector in PGA space. The world-vectors are combinations of the pairs e1e2, e2e3, e1e3 in the PGA space. I have yet to see the math, but its at least making some sense now.
@@marshallsweatherhiking1820 PGA generally works with normalized subspaces. Multiplying an object by a scalar doesn't change its geometry, but does change its "weight" in a sense, which only affects how it adds with other objects. As for how points are represented, points are treated as point reflections, which in 3D are compositions of 3 orthogonal reflections, or equivalently, systems of 3 linear equations. Hence, they are the outer product of 3 mirrors, ie trivectors.
Aren't multivectors tensors tho? They just don't allow elements, that have 2 or more equal indices.
So outer product isn't all that different between the two.
Multivectors are not tensors. While one of the constructions of geometric algebra involves defining the geometric algebra as a suitable quotient of the tensor algebra, it's best not to think about multivectors in this way, just like how we shouldn't think of the real numbers as being equivalence classes of Cauchy sequences.
From what I can tell, the GA outer product is like the tenser outer product but with the diagonal zeroed out since those components correspond to the inner product. In this sense the full tensor outer product is more closely related to the full geometric product than just the GA outer product.
And yes, it would be true to say that multivectors, or at least k-vectors are tensors in the same way it's true to say that linear transformations are matrices. Matrices are just one extremely powerful way to express linear transformations, but they're also something of a sledgehammer solution. Geometric Algebra has a narrowed focus, so it can't represent everything a matrix can, but what it can represent tends to be much clearer than the matrix form.
@@angeldude101 If you want to define multivectors in terms of tensors, I believe the correct way is to define the GA outer product to be the antisymmetrized tensor product.
I.E. If we write the tensor product of two tensors A and B as T(A,B) (since I can't put the standard notation in a youtube comment), then the wedge product of A and B is = T(A,B) - T(B,A)
There is one interesting thing that antisymmetric tensors taken as is (from the vanilla tensor algebra) aren’t naturally isomorphic to elements of the antisymmetrized tensor algebra. It appears in concrete terms when people define antisymmetrization of tensors with or without dividing by n!, and that n! then appearing elsewhere, with no natural way to decide once and for all where does one hide it and where it’s a bad taste. So for the sake of representational gods we need not conflate elements of the exterior algebra and antisymmetric tensors, and we even do need to be careful with thinking p-vectors from exterior algebra are the same as p-vectors from geometric algebra (or any Clifford algebra): there are details in the dark too! I don’t particularly remember what’s up right now though.
But our salvation is that we can define all these algebras by factorizing the tensor algebra and this factorization will make things in a right manner. To make the exterior algebra, we treat all tensors v ⊗ v (where v is a vector) equivalent to 0 (this is the “antisymmetrization” of the whole algebra). Now a ∧ b is represented by any tensor equivalent to a ⊗ b, like a ⊗ b or for example a ⊗ (b + a) − 2 b ⊗ b ⊗ a. Later we just use ∧ and its emerged properties as primitives and forget about all infinitely many equivalent tensors. Likewise, for geometric algebra or any Clifford algebra at all, we treat all tensors v ⊗ v as equivalent to the length of v squared, a scalar, and not just zero. Then we get from this all niceties we love. Again each thing has infinitely many representatives but we can just forget about underlying tensors and use our new notions and their properties as atomic.
I think I see a mistake. At around 11:28 it's stated that the product of three vectors in two dimensions is always another vector. However, this seems fishy. The product of two vectors a and b is the sum of a scalar and a bivector. If you take the product of this sum with a third vector, we get a vector from the scalar part, ie (a /dot b)c, but you get a scalar from the bivector part since we are in two dimensions meaning the third vector lies in the plane spanned by a and b. Therefore, in general the product of three vectors in two dimensions is the sum between a vector and a scalar. This scalar part is only zero if two of the three vectors are in the same direction.
Let me know if you agree with me or if you see a flaw in my reasoning!
The problem with your reasoning is that a bivector times a vector inside that bivector is a vector, not a scalar. As an example, e1 * e1e2 = e2. If you want to verify that the statement is correct for any three two-dimensional vectors, just calculate the general product of three vectors: (a e1 + b e2) * (c e1 + d e2) * (f e1 + g e2). If you fully distribute and simplify, you'll see that you are left with a vector.
@@sudgylacmoe Ahh, I see! Thanks for the explanation!
@@saudmolaib2764 It is true in any number of dimensions. As long as v3 is in the same plane as v1 and v2. If v1 and v2 are parallel, then their product is only their dot product, i.e. a scalar, which makes a vector when you multiply it by the v3. If v1 and v2 are not parallel, then the v3 can be written as a linear combo of the v1 and v2, since it is in the same plane. Substitute the linear combo and distribute an you get a*v1*v2*v1 + b*v1*v2*v2. Both terms are vectors.
niiiice
What software are you using to make these?
This is covered in question 2 of my FAQ: ua-cam.com/users/postUgwFByhvEg1_hD_L1Ch4AaABCQ
@@sudgylacmoe thanks!
Regarding reflections in VGA, I think you've actually got it backwards when you say you want to avoid "thinking about normal vectors again". The most correct way to specify a reflection is to specify the thing you want to reflect *along*, not *across*.
For example, when you reflect across a line in 2D, you don't actually care about the direction along the line at all, the reflection completely happens within the dimensions perpendicular to it, so it's actually *more natural* to describe it as reflection *along* the normal vector to that line, and this is exactly what VGA guides us toward doing.
Another way to look at it is to consider how a transformation changes due to the mere addition of an uninvolved dimension. If you're reflecting across a line in 2 dimensions, and someone throws a new dimension into the space, suddenly your reflection has to be across a plane to still be the same reflection. If you think in terms of reflecting *along* things, this doesn't happen. The reflection is still along the exact same vector as before. Uninvolved dimensions are actually uninvolved.
But that's just reflection across a plane. In PGA, it doesn't actually care about the objects used and instead only cares about the transformations. In PGA, a vector is a single reflection; a mirror. By convention, interpreting this mirror as a distinct object is to use the eigenspace. The points that remain unchanged by this reflection are the ones in the plane the mirror lies in, so a vector is interpreted as a plane, at least in 3D. In 2D, this is a line. In CGA, you're also able to use spherical mirrors for inversions, so there vectors can be interpreted as spheres.
There is no need to invoke "normal vectors" when e₁ isn't a line segment in the x direction, but rather the equation x = 0. The equation is a point in 1D, a line in 2D, and a plane in 3D, but ultimately it is the same vector and same reflection.
@@angeldude101 My point was only that the representation of reflections in VGA isn't unnatural at all, contrary to 15:47. PGA and CGA are a whole other matter, and I'm not discounting their usefulness :-)
@@tejing2001 Something about "mirror space" as I like to call it is that it appears to work regardless of the geometry you're working in. Even in VGA, you can represent vectors as mirrors and then reflect across them using the sandwich product. Your only restriction is that the mirrors can only be planes through the origin.
Reflecting _along_ a line on the other hand seems very unnatural since the reflection has little to do with that line beyond just happening to be perpendicular to it, and that fact _only_ seems meaningful in VGA rather than other spaces and that's entirely because the origin is treated as special.
If you have a line, the most natural reflection is across it. Such a reflection also acts as a 180° rotation, and pure rotations are bivectors. Ergo, lines should be bivectors. Said reflection is also much less commonly used than a mirror reflection, so it would make more sense for the mirror to be the primitive transformation rather than the line orthogonal to it.
@@angeldude101 A reflection most fundamentally involves inverting dimensions, so it's natural to specify it by the dimensions that are inverted, not the dimensions that are untouched.
@@tejing2001 You're thinking in terms of dimensions? So like directions? Generally geometric algebra focuses more on subspaces than directions. Not what directions are inverted, but what points are inverted. The points that are inverted are everything not lying on the plane, which includes all but 1 points on the line, as well as many points not on the line, but that doesn't form a subspace. The opposite, the points not inverted, do however form a connected subspace.
Specifying the dimensions that get inverted feels like that fits well with a matrix though. [[1, 0], [0, -1]] does flip the y-coordinate specifically and not the x-coordinate, but geometric algebra generally doesn't concern itself with the actual coordinates of something.
Love your videos on GA, but 1 problem with your bit about the formula being wrong. Dot product is only defined initially for vectors (and numbers also work if you want) as inputs (unsure about outer product). For multivectors a dot b = 1/2 (ab+ba) and a outer b = 1/2 (ab-ba). Which yes is kind of a circular definition, but saying that ab != a.b + a^b is wrong afaik. It is built up by defining dot product and outer product and multiplying by scalars on simple k-vectors (simple meaning with no pluses and all basis vectors are perpendicular, ie. (e_i).(e_i)=1 and e_i^e_j=-e_j^e_i and otherwise 0) - the distribution law with geometric multiplication can then extend it to all multivectors using the simple product of k-vectors and ab=a.b+a^b (ie. giving (e_n)^2=1 and e_i*e_j=-e_i*e_j) and how we multiply by a scalar - this then allows the generic redefinitions of dot and outer product. For example (e1e2+e2e3+e3e4)(e3e4) = e1e2e3e4+e2e4-1 and (e3e4)(e1e2+e2e3+e3e4) = e1e2e3e4-e2e4-1: therefore dot (the symmetric part) = e1e2e3e4-1 and outer (the asymmetry part) = e2e4. Note: I'm not an expert (all info from the internet) on this so if I'm wrong, please tell me where I'm wrong, thanks.
You just reminded me that there's another formula I should have said was not always true: a · b = 1/2 (ab + ba) and a ∧ b = 1/2 (ab - ba). Those are not always true as well. The standard definition of the inner and outer products is the lowest and highest grade part of the geometric product as I described here.
@@sudgylacmoe okay, that makes more sense. Thanks :)
I think the previous introducción is correct in G³.
Why do we define the product of two parallel vectors to be the product of their lenghts? In my head it would make more sense to get a vector with the length of the product of the two lengths. So 5e_x * 6e_x =* 30e_x and not 5e_x * 6e_x = 30. There is propably a good reason why we would define it as a scalar, but what is the reason? 3:40 (and thanks for the great video)
Which vector would e_x * -e_x be? On one hand you might think it's e_x * e_x * -1 = e_x * -1 = -e_x, but then on the other hand e_x * -e_x = --e_x * e_x = -1 * -e_x * -e_x = -1 * -e_x = e_x, so we have both e_x * -e_x = -e_x and e_x * -e_x = e_x, which is impossible. We would have to lose associativity and/or distributivity to make this work, and we don't want that.
@@sudgylacmoe thank you! That was the simple contradiction I was looking for
@@sudgylacmoe after looking at it for a while, I have a question about the second statement.
Wouldn't -e_x * -e_x result in +e_x?
In longer: -e_x * - e_x = -1 * e_x * -1 * e_x = -1 * -1 * e_x * e_x = e_x * e_x = e_x
But you yourself said that the product of a vector with itself should be another vector pointing in the same direction.
@@sudgylacmoe ah i dont want to waste your time if I cant explain properly what I mean.
I meant as an equation:
(a * e_x) * (b * e_x) = a * b * e_x
a,b are scalars. Does this make more sense?
oh my god. :-). This is why I prefer using explicit coordinates for as long as possible.
If you prefer using explicit coordinates, then GA is probably not for you. One of the biggest focuses in GA is representing various transformations abstractly with pure algebra without specifying the structure of the objects you're transforming or even transforming with. Pretty much all formulae that aren't constructing a specific vector from coordinates in the first place are done without mentioning coordinates, or even dimension. a(b) = aba¯¹ is some number of reflections with the number of reflections equal to the highest grade of a. I have no idea what a looks like internally, the dimension the objects are in, or even the shape of the reflections, but I can say with relative certainty that they are reflections.
@@angeldude101 I used to think this until I started to use it in practice; in code. You can fool yourself into thinking you got it when you write it on paper. When you implement it in code, you discover that you can just implement a function to GeometricProduct(a,b) once in code, and there are no special cases at all to implement when it's done internally with the coordinates. Then the definition of "is this is a vector" is all a matter of which values are non-zero.
When you read GA books and stop multiplying just vectors, for formulas are actually kind of complicated. Where you benefit from using special cases is when the floating point precision is an issue, and you use all the special cases to nail down when you KNOW which components must be exactly zero, and therefore the types are exact.
For example, multiplication of two 3D multivectors can leave you with a trivector that is very slightly non-zero, because of floating point. It is kind of complicated to do the code such that round-off errors ONLY happen in the values that are not guaranteed to be zero.
Implement GeometricProduct of two arbitrary multivectors in code to see what I mean. If you want your CODE to never mention coordinates, you implement this one function that multiplies two multivectors internally with the coordinates. It's so gloriously simple because all of GA is in this one tiny function. Then, you define two vectors a and b in 3D...
// You only see coordinates when
// when you actually specify WHERE a vector is,
// and you are not just playing around on paper
// the operator * is the general Multivector
GA.Dimensions(3)
a := vec3(a1,a2,a3)
b := vec3(b1,b2,b3)
p := vec3(b1,b2,b3)
// do a rotation
r := a * b * p * Inv(b) * Inv(a)
What do you think happens? The scalar part comes out to be like 0.0000000001, and the trivector part is like -0.0000000001.
It's really only IsVector(r) when:
- scalar == 0
- trivector == 0
- all 3 coefficients of bivectors are == 0
How do you fix this? You basically have to write code that doesn't just have one GAMultiply function. You need a complicated compiler that looks at which coefficients are already exactly zero, and pick the best special case way to multiply the two multivectors so that the parts that MUST be exactly zero STAY exactly zero. Unless you want to demand completely symbolic calculations (ie: rational numbers, with symbolic roots and symbolic special numbers... answers like .... (235/3939 - pi^-2*e) .... with any kind of error in the result... some of the "not necessarily zero" values will end up being zero. Even tiny errors will make a big difference when you are dividing.
This whole thing about coordinate-free is only true when you are doing fictional calculations on paper. As soon as you implement code to actually manipulate vectors with realized values; you dont go all the way back to matrices, but the implementation gets complicated very quickly to deal with the fact that just converting everything to approximate numbers is an issue.
For example: at the end of a GA calculation that yields two vectors; how do you define "these two things are parallel", or "these two things are orthogonal"? in the face of approximate answers? When you try to do it symbolically, the number representations can get huge (ie: like when you sit in a loop collecting statistics for an hour and a pair of vectors come out of it... naive stddev calculation can be catastrophically wrong with naive floating point, or numbers too huge to compute with symbolically).
@@angeldude101 You DO need to know the dimension, or else answers will come out wrong; symbolically while avoiding coordinates as well. GA is straightforward when you are talking about multiplying vectors. But while you are multiplying arbitrary multivectors, every possible combination is possible.
a * b * p * inv(b) * inv(a).
What drives me most crazy about standard GA books is that they talk about it being "object-oriented", and it's almost never clear how you would take a GA expression that is a function of a bunch of inputs, and compile it down to a function that computes the answer when given realized values as input.
Learns VGA and CGA
Bookmarks chapters to study EGA and sVGA.
lol
The main reason I see for the discrepancy in views between "From 0 to geo" and "A swift introduction" it's precisely the "from zero" part.
People who would care about and watch an accesible UA-cam intro to geometric algebra probably aren't at "zero".
If you don't know some linear algebra, use vectors or at least know how they work then you most definitely won't be into watching a video about a novel product between them.
TL DR: The views come from the "geo" part. Not the "zero* part.
That's totally true! I believe the majority of viewers has a decent understanding of linear algebra, they are simply bored by studying things they already know. We don't watch these videos not because the series are dull at their core, no, we're simply waiting for the interesting part to begin. As for me, I'm really looking forward to this moment!
My thoughts exactly.
plus there are already so many videos on youtube about linear algebra
I am still watching them as they come out since they are still somewhat interesting, but I won't go back to rewatch them nearly as much as I do for the Swift Introduction.
I didn't understand why "things that are parallel"cancel. What happened to the inner product? When to parallel vector are multipied it gives a scaler not 0...
By cancel, I mean their directions cancel. The magnitudes of those directions multiply.
@@sudgylacmoe but shouldn't it be a multivector? Because the scaler is still there
I'm not sure what your question is. When you multiply two parallel vectors, their magnitudes multiply and their directions cancel, leaving a directionless magnitude, which is a scalar. Scalars are also multivectors, so the result of the product is a multivector.
@@sudgylacmoe Ok I think I understand
@15:10 you show that a reflection around a vector is a half-way rotation. Why is this an issue? Then you go on to show "a few ways to fix this". It's not very clear to me what needs fixing. Thank you.
The problem is that a half-way rotation is not a three-dimensional reflection. Three-dimensional reflections go through planes, not lines. So the issue is that the reflection formula that worked in two dimensions did not produce a reflection in three dimensions.
It's less of a problem with geometric algebra and more a problem with generalizing reflections to higher dimensions. Just looking at the complex plane shows why this is the case. Traditionally, -1 is a reflection around the origin, but since there's no halfway point to represent a transformation that when done twice gives a reflection, a new axis had to be added and suddenly the reflection has become a rotation, and the halfway point is a 90 degree rotation. The operation to reflect a real number is completely different from the operation to reflect a higher dimensional object and is also indistinguishable from a rotation in that higher dimension.
With PGA, people insisted that reflection across a "vector" stay as a reflection in higher dimensions rather than becoming a rotation, even if every reflection is indistinguishable from a rotation through a higher dimension.
Edit: I have since changed my stance on this. I now accept vectors as mirrors rather than directions, and that a rotation around a direction is to be a bivector (or composition of 2 mirrors) regardless of what that rotation is. The part about reflection being rotations through a higher dimension is however something I still hold and it relates to the even subalgebra of a geometric algebra. When going up a dimension, rather than simply adding a new basis vector, it's possible to encode the old algebra as the even subalgebra of the higher dimensional algebra. This explicitly turns reflections into rotations through the higher dimension.
5:41 "cancel each other" 🙄... oh, the direction
Eh, I don't think you can produce any conformal transformation in 2D using a finite number of generators, because the 2D conformal group is infinite-dimensional, unlike in higher dimensions, where the dimension of the conformal group is finite.
5:20 inside who?
When will you stop blowing my mind
Aw c'mon man, you're just makin' this shit up as ypu go along! 🙂
Hnnng thanks
hi
Why DON'T you make a GEOMETRIC CALCULATOR ?
So we may play and tweak with it and observe the results
Two reasons:
1. A geometric calculator is highly non-trivial to make.
2. People have already done similar things, like here: bivector.net/tools.html
Dango
Its traditionally called the exterior product. Not the outer product.
Textbook?!! Bite your 19th century tongue! Where is the interactive software for my Physics as Modeling Workstation?? How about a Jupyter Notebook to start rather than a textbook? How about a suite of Brilliant Geometric Algebra exercises?
😜 𝓟Ř𝔬𝓂𝔬𝐒ϻ
acab at 11:48 ✊
This video makes me very sad. I wish all videos went at this pace, but sadly they all spend forever on trivial junk, then throw in a bunch of worthless jargon and turn into a mess.
Does anyone know of any other math/physics topics in other videos shown at this same super fast pace?
You could try 'Physics with Elliot' .. that's always good for a binge :-)
I have found the last few episodes of Zero to Geo too basic. I realize you are going for rigor, but spending entire videos on defining basis and spaces is too much. This is probably a bias on my part since I have gone all the way through math courses and so already know this stuff from calculus and linear algebra classes, but I keep sitting there going 'come on get to the new stuff'.
As you made a video about spacetime algebra (STA), do you plan making videos about projective (PGA) and conformal geometric algebra (CGA) as well? The basis vector of PGA that squares to 0 is especially interesting because the zero vector already does that. This is in contrast to, e.g., the complex numbers that extend the real numbers by i, which has a property not found in the real numbers and thus does not need that much explanation _why_ we do that, adding an object with a property already present likely has some interesting reasoning behind it. I’ve looked into en.wikipedia.org/wiki/Dual_number but it did not really make me feel I’ve learned something.
I'll be covering them eventually in From Zero to Geo.