Ok, I am too excited about this. This mirrors my first experience with geometric algebra on your channel, where suddenly everything made sense But now, i'm like a sports fan cheering whenever some new math operation appears out of thin air I can't wait to see the other video!
@purplenanite
Жыл бұрын
I was just learning about Projective geometry, and also PGA - what a coincidence!
@imnimbusy2885
Жыл бұрын
Be wary for it was not a coincidence. The algorithm is evolving. It knew you were interested in PGA. Just this second it’s tracking your brain waves and chugging it full of advertisements! B7. IM RUNNIN OUT OF- &£56*- Don’t trust any- BUY GOLD BUYYYYYY!!…..
@caspermadlener4191
Жыл бұрын
I normally don't watch these videos, but when the video is a "bonus video", you already know it is gonna be good.
@HiddenPowerIce
8 сағат бұрын
Pov the thumbnail Mathematicians: 😁 Physicists: 😡
@HotakaPeter
Жыл бұрын
This is an excellent series. I am interested in using Clifford algebra for nautical applications such as autopilot construction. Can't wait to see more.
@fibonaccicats7179
Жыл бұрын
2 videos on one day?Unprecedented.
@Cobblestoned100
Жыл бұрын
Hey, thanks for making this series! As I'm not so much of a math person, but like to use math in my work as a software engineer, it would be great if you could also cover some more practical examples. Something like: how to transform objects in 3D space using geometric algebra as opposed to the linear algebra ways of doing it. Or how to solve some simple introductury physics problems we learned in school more intuitively with GA. Sometimes it's hard for me to follow the abstract formulations. Or maybe I should just learn more math to understand it better :D
@BboyKeny
Жыл бұрын
I have a similar background. I've been going deeper into math and knowing more math does help understand some of the explanation better because some use other subjects as reference. But you will be putting off the "trying to understand" part to that other math subject. Do take my opinion with a grain of salt since I'm also still learning this all. It's like learning a language. The first language you need to babble through, the second you can translate to the first. I think for GA knowing algebra, trig and complex numbers helps. Also (anti-)commutativity is very important to understand. You can also do calculus on GA which I imagine is very powerful, so learning that is useful I think. GA is a superset of Linear Algebra and Quaternions so you don't have to learn that. But learning materials are more abundant for them so that's a trade-off. I think for a programmer you should check out Clifford library for Python. There is also Ganja for JS. It's easier to play around with the different operations and objects with direct visual feedback. Once you build up an intuition you can delve into the math to put words onto your intuition. So instead of trying to understand the symbols through the algebra, you can understand the symbols through the geometry by fiddling around and then understand the algebra. Since now it's weird algebra done with weird symbols. Then it's just weird algebra with familiar symbols. My plan is: Start with making a square with 2D PGA, then make rotate and move. Make it 3d and project it to 2d for visualisation. At this point you're are pretty deep into linear algebra if you weren't using GA. In graphics programming you would want to load in a 3d model and build a bunch of triangles based on the vertices in the 3d model file. The Utah teapot is an example for this. You could also explore how to interpolate between 2 points by moving an object from 1 point to another and back. This could serve as a foundation for Bezier Curve's which enables smooth animations. At this point I think physics is also available to play around with. Since this is done on the CPU, I don't think you can make a high res games like this. You would need to run it on the GPU and I got no clue how nice GA and the GPU play together. Hope this wall of text is useful.
@vivekdabholkar5965
7 ай бұрын
Awesome presentation!
@yiheyao2859
Жыл бұрын
I was just learning about spinors and some introductory videos to Clifford algebra. Your videos are super helpful to physics students learning special relativity.
@1p6Gaming
Жыл бұрын
Huh, same trick works for Conformal Geometric Algebra. Consider the equation of a hypersphere aX^2 + B*X + c = 0, where a and c are scalars, B and X are vectors from some vector space, and B*X is the dot product between vectors, and X^2 is the squared magnitude of the vector. In this case, it was helpful to consider when does this equation have solutions as the starting point, which led to figuring out the radius of the hypersphere. One gets B^2 / (4a^2) - c/a as the radius squared, and when this expression is negative, the sphere's equation has no solutions. Multiplying by 4a^2 does not change the sign, so B^2 - 4ac can potentially be the quadratic form for this linear space of hypersphere equations. One can then get the inner product by considering the form applied to the sum of two sphere equations, and subtracting the forms of the two equations considered separately. This gives B1*B2 - 2*a_1*c_2 - 2*a_2*c_1. Though hm, the inner product has different coordinates multiplied together, and as such can't quite match up as is with a geometric algebra. Perhaps a change of basis is needed...
@angeldude101
10 ай бұрын
There are 2 different bases for CGA. One of them is e₁ -> x = 0 e₂ -> y = 0 o -> x² + y² = 0 ∞ -> 1 = 0 (analogous to PGA's e₀) The other has the same two first vectors, but the third and fourth are replaced with e₋ -> ½(x² + y²) + ½ = 0 e₊ -> -½(x² + y²) + ½ = 0 The mapping between the two forms are o = e₋ - e₊ and ∞ = e₋ + e₊ (up to a scaling factor that I forget) Usually, CGA will be implemented with the latter version, but described using the former, since the former has a more clear geometric interpretation, but the latter keeps the two new basis vectors fully orthogonal, where as o · ∞ ≠ 0.
@smolboi9659
7 ай бұрын
I would argue that lines in 2 Dimensions live in a 2 Dimensional Space. We have 3 coordinates but scalar multiples of the coordinates are the same line. Hence we have reduced 1 degree of freedom to this scaling constant. Another way to think about it is y = mx + c has m and c as 2 free parameters. Consider the 2D surface of a 3D ball xx + yy + zz = 1. We have 3 coordinates but we reduce one degree of freedom to obey the equation. This 2D space is not a linear space but a 'curvy' space called the 2-sphere. Now what sort of 2D space is the space of lines in 2D (including the line at infinity)? First the coordinates belong to R^3 without the origin where all coefficients are 0. Then, we need to identify all scalar multiples as the same point cause they represent the same line. We do this by choosing a representative vector. For convenience let the vector norm be 1. Then we 'glue' together or collapse all vectors whose norm is not 1 into the scaled version of itself where the norm is 1. You can visualize this in 3D space where all non zero vectors collapse into the 2-sphere. We are not done yet though. Directly opposite ends of the sphere differ by a scale of -1 and represent the same line. Hence all antipodal points on the 2D surface of this 3D ball need to undergo this glueing as well but this is hard to visualize. Another way to do this is to realize we double counted the lines on this ball and cut the ball in half and only keep the northern hemisphere. Then on the circular boundary caused by the cut, you also double counted so u peel of half of this circle and then on the remaining half circle, the 2 exposed points from this peeling is also double counted so get rid of one of them. This space is precisely the real projective plane RP2. It is not a vector space or linear space but a curvy space or manifold. The dimension of this manifold is 2 just like how the dimension of the surface of a ball is 2. TLDR the space of all lines in 2D including the line at infinity is the 2D manifold called the real projective plane.
@vwcanter
Жыл бұрын
Is there another longer playlist with more videos? This is the last one I see
@sudgylacmoe
Жыл бұрын
No, this is the most recent video made in the series. I'm still working on more!
@rarebeeph1783
Жыл бұрын
considering e0 to be the line at infinity is interesting. i was thinking of the lines as being cross-sections of planes in 3d euclidean space, so e0 would just be a parallel plane to the cutting plane. but projectively, that's the same as a plane that intersects the cutting plane along the line at infinity. this comparison probably has something to do with homogeneous coordinates. even further into the video now, keeping track of a corresponding vector perpendicular to the line is really reminding me of covectors, which can also be visualized with cross sections of planes or vectors perpendicular to those cross sections. but in terms of the video title, really the necessity of that vector is an indication that our chosen representation for lines contains redundant information when specifying lines, i.e. the lines themselves are 2 dimensional, because the 3d basis has a free parameter that preserves the set in the line.
@jsrada29
Жыл бұрын
Great video, thank you
@Hector-bj3ls
Жыл бұрын
We have to wait until chapter 5?! Man, I'm going to be old when that comes out haha
@evandrofilipe1526
Жыл бұрын
Can't wait to watch the swift introduction
@BenGeorge77
Жыл бұрын
I have so many poorly formed questions. What an interesting video.
@brendawilliams8062
Жыл бұрын
Don’t take 3.13333 cubes there
@nicetrybut7423
6 ай бұрын
Awesome video! I'm relatively new to PGA but I like a more grassmannian approach better. In that version you have vectors and points as grade 1 objects and a line/ plane/ hyperplane is the wegde between these. I think it's better because you don't have to use a sort of dual to represent these so called bound vectors(lines/planes...) For example let the basis vectors be e1, e2 and an origin point e0 (vectors are "carriers of points" so e0 + e1 would be the point you get if you move e0 the origin point along e1). What you would call - e1+2 e2 I would call e0 ^ (2 e1 + e2). (Notice 2 e1 + e2 represents the subspace in which the line lies and the magnitude and orientation of the line is more clear imo) What you would call e2 + 2 e0 I would call (e0 ^ e1) + 2(e1 ^ e2) = (e0 ^ e1) - (2 e2 ^ e1) = (e0 - 2 e2) ^ (e1) and so what you call e0 is to me a bivector, a "carrier of lines" and you are right in that it is a line at infinity just as a vector is a point at infinity. If you are interested I really recommend John Browne's Grassmann algebra book... it has a bit different approach and does not really focus on the geometric product but it has fascinating perspectives on the so called exterior algebra
@sudgylacmoe
6 ай бұрын
The issue with this approach is that the geometric product just doesn't work in the same way, and it's much less useful. Using the line/plane-based version, the geometric product lets you encode arbitrary rigid transformations, whereas in the point-based version, I haven't heard of much useful that you can do with the geometric product.
@nicetrybut7423
6 ай бұрын
@@sudgylacmoe I haven't heard much about the geometric product in this space either... The problem is there are only very few materials on the topic. But from what I found all the other products (exterior, regressive, inner, +complement operation) work pretty neatly. So maybe there lurks an interpretaion of the geometric product out there for this space too? Well whatever the answear is I'll keep coming back to your videos :D
@sudgylacmoe
6 ай бұрын
@@nicetrybut7423 I know that it does have an interpretation in this space, and that it's not nearly as useful as in the line/plane-based view. All non-metric operations work equally between the two views by dualization so you can't really use them to argue which viewpoint is better. It's only when you start looking at the geometric product that it starts to make a difference, and in this case the line/plane-based view gives you rigid transformations while the point-based view just doesn't.
@nicetrybut7423
6 ай бұрын
@@sudgylacmoe I see... so you prioritize the geometric product. I'm fine with that. What bothers me is the use of duals to represent objects. Using a vector perpendicular to the line it represents surely works in 2D but you need different objects in different dimensions (while the line itself is independent of the dimension of the underlying space). This also results in the switching of the wedge and regressive products (in plain GA wedge works like the join and the regressive product works as a meet) Do you think that there's a construction of PGA that avoids the use of duals?
@sudgylacmoe
6 ай бұрын
To me, one of the fundamental insights you need to understand PGA is to realize that it's not a "dual" representation. A priori, we have no reason to think of vectors as points, or vectors as lines/planes. It's just a historical accident that we picked points. Neither view is more fundamental than the other, and neither is "the dual view". We should just pick whichever one works out better. And, given the results in PGA, I personally think the line/plane-based view is superior. To actually respond to your points, I don't like thinking of the objects in terms of duals. In PGA, a vector a1 e1 + a2 e2 + ... + an en + a0 e0 corresponds to the object that's the solution set of the equation a1 x1 + a2 x2 + ... + an xn + a0 = 0. This is the view I presented in this video and in my swift introduction to PGA. This way you don't need to consider duals or perpendicular arrows or anything. For the join and the meet, I'm perfectly fine with ∧ being the meet and ∨ being the join. It's just what geometric operations the algebraic operations happen to map to. As for a construction of PGA that avoids the use of duals, it depends on what exactly you mean. If you are saying "one that avoids using the dual operator", I would say what I just specified above suffices. There are a couple of things that still require the dual (notably, measuring distance), but those aren't required for the fundamental definitions. If you are saying "one that defines vectors as points", that's a no. It's impossible to get the things we want from PGA, like rigid transformations, without duals in that case. It is possible to do that kind of stuff with a ridiculous amount of duals, but it's more complicated than just using the plane-based approach, and it involves defining operations that aren't preserved by orthogonal transformations, and thus are not natural.
@steffejohnson3193
11 ай бұрын
Thank you for existing!! Just one question why is the e1(x) vertical and e2(y) horizontal? Im used to have x being horizontal and y being vertical. Big Love!!
@sudgylacmoe
11 ай бұрын
This is just a side effect of the way we represent lines. e1 represents the line x = 0 and e2 represents the line y = 0. Thus, e1/x and e2/y are still related to each other in this way, but it ends up producing lines going in different directions than you would expect.
@steffejohnson3193
11 ай бұрын
Thank you!!@@sudgylacmoe
@linuxp00
10 ай бұрын
As GA/PGA are supersets of other algebraic representations of geometry, it would be reasonable to revisit some concepts from tensor algebra (TA), topology and differential geometry for better clarity. In TA, lines are not actually vectors, but rather covectors or dual vectors, because they exist in the complementary space of the vector that governs their sliding and orientation. In this framework, lines (or level curves) are dual to 2D vectors, and planes (or level surfaces) are dual to 3D vectors, and so on. Now, to illustrate the concepts in a topological view, one could apply a parallel transport of directing vectors noted as e_0, "the line at infinity". Imagine that the entire 2D plane is rolled up into a doughnut-like shape with infinite radii. A great circle drawn around the 'equator' (the circle with the largest radius) of that surface could represent the x-axis, while another perpendicular circle, drawn around its 'handle' (one of the smallest circles), could represent the y-axis. In this scenario, a translation of lines in the x-direction would equate to an even rotation of the doughnut around its equator. Similarly, a translation in the y-direction would result in an even rotation around its handle, moving circles near the hole further away and vice versa. A combination of lines could be represented by a twist of the handle, fixing the x and y axes in place. As viewed extrinsically of the space, i.e. in the 3rd dimension, by us.
@jacborne614
5 ай бұрын
when you associated the line with the arrows, i noticed that e1(x=0) was implicitly associated with the unit arrow pointing right, which is perpendicular to the line x=0 that e1 was defined as, and the e2(y=0) was implicitly associated with the unit arrow pointing up, which was also perpendicular to e2(y=0), so...the perpendicularity of the arrows to the lines, which derived from that implicit choice of association, isn't it kinda arbitrary? cuz if e1 was to associate unit arrow pointing up along e1(x=0), and e2 was ......etc, what are your thoughts, i kinda lose my thread from here, dunno what to think of it, is it less useful this way?
@sudgylacmoe
5 ай бұрын
This doesn't work in higher dimensions. In three dimensions, the vectors are planes, and there the only good vector associated with the planes is its normal vector.
@That_One_Guy...
Жыл бұрын
I have some question : 1) Why do you represent one vector as the whole LINE ? line is infinitely many vector added together meanwhile vector is a directed line segments ? extended question of this : why do you simplify cartesian equation of line which doesn't go through (0,0) (ax+by+c = 0, where c != 0) into very simple linear combination equation (a*i + b*j + c*e = 0, e = line at infinity) ? i don't think that's very accurate as you need to specify the starting point of that line, which basically what parametric equation does. 2) What's the role of this line at infinity ? how do you even determine the coefficients of this line at infinity.
@sudgylacmoe
Жыл бұрын
1. Vectors are not directed line segments. See kzitem.info/news/bejne/qpxu2GusqGamooo. I'm just making a new linear space out of lines. If you want to, just think of it as being an abstract 3D linear space with a basis given by e1, e2, and e0, which we then add a geometric description to (where a e1 + b e2 + c e0 corresponds to the line given by ax + by + c = 0). 2. The line at infinity shows up when translating lines, for one. In more advanced contexts like PGA it often appears as the result of certain geometric operations.
@2fifty533
Жыл бұрын
is this the last video of this series
@sudgylacmoe
Жыл бұрын
At the current moment, but much more is planned!
@thomasolson7447
Жыл бұрын
tan(pi/4-θ/2)= sec(θ)-tan(θ) sec(θ)^2-tan(θ)^2=1 sec(arctan(b/(2a))-tan(arctan(b/(2a))) sqrt(b^2+(2*a)^2)/(2a)-b/(2a), if c=-a sqrt(b^2+(4*a*(-c)))/(2a)-b/(2a) (sqrt(b^2-4*a*c)-b)/(2a) is the quadratic formula. (sqrt(b^2+(4*a*(-c)))/(2a))^2-(b/(2a))^2=-c/a (b^2+(4a(-c)))/(4a^2)-(b/(2a))^2=-c/a
@thomasolson7447
Жыл бұрын
Google this: "Question Video: Using the Divergence Theorem to Find the Flux of a Vector Field over a Surface" I'm just trying to figure out if that video is this. arctan(y/x) So, if b=y and y=0, it is on the x-axis and the slope is also 0. So, the vertex must be on that axis. Which is the same as the video. I think they are the same.
@wipetywipe
Жыл бұрын
Which line does the zero vector represent?
@sudgylacmoe
Жыл бұрын
That is an excellent question! I don't know. Part of me wants to say it's in a sense both no line at all and all lines at once.
@Manisphesto
Жыл бұрын
I would like to use the unit vectors written as i1, i2, and i0 because I want to make it connection with the imaginary number since i1·i2 = i in 2D vector space. Probably also because it feels like they'res Eulers Number even though it's not.
@angeldude101
Жыл бұрын
i squares to -1, but none of e₁, e₂, or e₃ do. They're also geometric lines rather than an abstract "imaginary" quantity that is exactly as imaginary as any other number. The origin in 2D PGA behave a lot more like i, but is also very much not imaginary since it's an actual geometric point in space, and its role can be performed just as easily by any other point. Aside from e₁, e₂, and e₀, I occasionally call them instead x̂, ŷ, and d̂, since the "hat" notation for orthonormal basis vectors feels more familiar and less abstract than the initial of the German word for "unit." Alternatively, just call them what they are: x = 0, y = 0, and 1 = 0 (or "w = 0" if you'd rather talk about the embedding space or have it actually have solutions). The origin is then described as either e₁₂, x̂ŷ, or "x = y = 0" depending on which notation you're using. "I" is used as an object in PGA, but usually refers to the pseudoscalar, which _could_ technically be called "imaginary" since it's outside the euclidean modelling space, but more often I see it explained as "eye," since it acts as the focal point for the "projective" part of "projective geometric algebra."
@Manisphesto
Жыл бұрын
@@angeldude101 ok yes i² = -1, but that's because the units are anti communicative, as ab = -ba. Here's this: here's (e₁e₂)², and (e₁²)(e₂²). (e₁e₂)² = e₁e₂e₁e₂ = -e₁e₁e₂e₂ = -(e₁²)(e₂²). As we know that unit vectors square to one, this means (e₁e₂)² = -1. So I call them i₁ and i₂ because of how they multiply to i and not being because they not square to -1.
@angeldude101
Жыл бұрын
@@Manisphesto Ok, so then what's e₃ from 3D PGA then? It gives other negative square objects when multiplied by the other basis elements, but they're not the same. You could probably say that e₃ * e₁ = j if you want to confuse them with quaternions. But that would mean "i₁i₂ = i" but "i₂i₃ = k", which doesn't make a whole lot of sense.
@Manisphesto
Жыл бұрын
@@angeldude101 The "i" in complex numbers is not the same as the "i" in quaternions. the complex i in here = e₁₂₃ but in quaternions, it's e₁₂, and also, the quaternion i is actually (or at least isomorphic to) "ie₃" because e₁e₂e₃ e₃ = e₁e₂ (e₃)² = e₁e₂. Also, they're many more things in 3D than 2D like the owner said, so 2D is not the same as 3D.
@davidgillies620
4 ай бұрын
So the perpendicular arrows are covectors (dual vectors), right?
@sudgylacmoe
4 ай бұрын
Covectors are linear functionals, so no. While these perpendicular arrows have some similarities to the dual in GA, there are some differences, such as ignoring the e0 part.
@davidgillies620
4 ай бұрын
@@sudgylacmoe So what is the formal term for them?
@sudgylacmoe
4 ай бұрын
In more general contexts you would call them normal vectors, but I haven't heard of a name for their use in this particular context.
@davidgillies620
4 ай бұрын
@@sudgylacmoe Thanks. Really looking forward to Chapter 2.
@housamkak8005
Жыл бұрын
hey sudgy, how do u make your videos? manim?
@sudgylacmoe
Жыл бұрын
Yep
@Pluralist
Жыл бұрын
@zTJq40sl
7 ай бұрын
Can we really conclude that lines in 2D space are vectors? In kzitem.info/news/bejne/qpxu2GusqGamooo you've shown that equations of the form a x + b y + c = 0 can be used to _represent_ lines and that _these equations_ are vectors. As far as I can tell, it doesn't follow that _the lines themselves_ are also vectors. Wouldn't that be like saying that lists of numbers (which can be used to _represent_ arrows in space) _are_ arrows in space? Huh. Let's take a step back. The way you've shown that equations of this type are vectors is by showing that we can (trivially) define scaling and multiplication for them and that the results are again equations of this type and that (less trivially) these operations fulfill all the rules of a linear space. If each line can be represented by such an equation, doesn't it follow that we can define the two operations analogously (via the representation, if you will) as you (implicitly) did _and_ that then all the arguments/proofs apply to them for lines in 2D analogously? If I'm not mistaken, that only follows if the mapping to the representation is a bijection, which it clearly isn't, as each line can be represented by infinitely many equations of said form. This wouldn't matter if the choice of which equation we use to represent a line wouldn't matter, but as you've shown that choice _does_ influence the results of the operations, even when mapped back to lines. So while the objects each of which is a combination of a line and a "magnitude" seem to form a linear space (and thus by definition to be vectors), lines themselves don't seem to.
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