Geometric Algebra in 2D - Two Reflections is a Rotation

[@cristian-bull]

I dont understand why students are not taught this. It's hard to find quality beginner content of this topic. This is exactly what I needed. These videos are so good, that I'm neglecting my studies for study.

[@kasufert]

Students aren't taught it because the teachers don't know it. If teachers were taught this, they would gladly teach this rather than slog through trying to teach the roundabout and over-complicated concepts such as cross products, "into the page", and complex numbers/quaternions for rotations, which would be much more easily explained via GA.

[@Bubatu7]

I am consuming these GA videos by bulk. They greatly simplify the learning process of GA for me. Thanks for this high quality content!

[@robertmcmorran6680]

This whole series so far can be described with one word awesome! Thank you so much for making such a complicated subject, so easy to understand.

[@MultivectorAnalysis]

YAAS! I love the rotor form of the geometric product. I'm teaching a Calc 2 class this semester, and we're about to get into Taylor expansions. That means I'll be deriving Euler's formula soon. Feynman called Euler's formula "the most remarkable formula in mathematics," and my appreciation of it went through the roof when I learned about its intimate connection to the geometric product of vectors. Looking back at my comments on your earlier GA videos, they might come across as sounding impatient for you to move up to G(3), but I think your approach of really focusing in on understanding G(2) first is wise. I did notice (and appreciate) the subtle foreshadowing @15:17 when you mentioned that the lower formula applies only "in the special case where the bivector and the vector of interest are in the same plane." Hopefully viewers learning GA for the first time here will catch that, preempting erroneous hasty generalizations to rotations in higher dimensions. If the vector is not in the same plane as the bivector, that one-sided product would rotate the coplanar component as expected, but the orthogonal component would produce a trivector. That "sandwiching product" slickly avoids the trivector issue, taking advantage of the anticommutativity of bivectors with coplanar vectors, paired with the commutativity of bivectors with orthogonal vectors. Beautiful!

[@Math_oma]

+Nick Okamoto Yeah, you know Euler's formula is golden when it just keeps appearing again and again. An additional way in which it pops up that I recently rediscovered is in differential equations. If you consider y = e^(ix), it's easy to see this is a solution to the differential equation y''=-y. Now, a basis for the solution space is {sin(x),cos(x)} so I should be able to write e^(ix) as some linear combination of the basis vectors Asin(x)+Bcos(x) and the correct one is seen by considering the initial conditions y(0)=1 and y'(0)=i which gives A=i and B =1 so e^(ix)=cos(x)+i*sin(x). Of course this is but one of many arguments but this one is nice because you get to exercise some linear algebra concepts.

[@MultivectorAnalysis]

Ooh, that's a fun one! I'll remember to share that next time I'm assigned to teach an differential equations course. Well, now that I think about it, it's a simple enough ODE example to share with my Calc 2 students. Thanks! I just thought of another neat way to think about the geometric product of vectors in G(2); this one related to the graded form (i.e. the sum of inner and outer products) rather than the rotor form. Choosing your unit bivector I (the pseudoscalar basis element in G(2)) defines which angular direction is "positive." So the angle from a vector u to another vector v is positive or negative depending on whether the angular direction from u to v matches the bivector direction of I. That means the angle measure from v to u is opposite the angle measure from u to v, and the commutativity/anticommutativity of the inner/outer product is seen to be resulting from the even/odd nature of the cosine/sine function. Hence Euler's formula decomposing e^{I\theta} into even and odd components (cosine and sine fcns) is effectively the geometric product of vectors being decomposed into symmetric and antisymmetric components (inner and outer products).

[@cristian-bull]

Actually I am more scared now of generalising to 3D things that I shouldn't.

[@NoNameEntered]

Awesome video 👍. I had to watch this a couple times before everything suddenly really clicked. The sandwich operator is so elegant. It's interesting to note that the sandwich operator can be used both to reflect a vector around another vector (v-1uv) or to rotate a vector using a rotor (r-1ur)

[@davidhand9721]

The more I learn about GA and PGA, the more I think PGA has important advantages. For example, in PGA it is quite clear that the operation you are doing here is a _plane_ reflection, and how it's different from a point reflection. For a point reflection, the lengths of *u* and *v* matter, whereas reflecting *u* on *v* here discards the length of *v*. If the magnitude is meaningless, then it isn't truly the vector that matters, but the plane that it defines in 2D. PGA makes this perfectly clear in that the first order objects always bisect the space linearly, so in 2D it's a line. I guess they still have magnitudes that get ignored, though, so maybe I'm talking nonsense.

[@randomgamevideos241]

I see a lot of comments saying that "Why do we need two reflections, when one reflection can get us the rotation we need" : Well, when you are applying geometric products, you wont be doing it on a single vector, as show in this video, you will be doing it on more complex shapes, like a triangle, so lets look at a triangle, its made of 3 points ( 3 position vectors ), so now to rotate the triangle we need to rotate each of those points about the origin of rotation, by the same angle, ok, now what do you think will happen if you reflected this triangle once ? you will get a reflected triangle, the reflected triangle looks like the original triangle but it not the same, its like your right hand and left hand, they look the same but if you put one hand over the other you realize that they are not the same, they are just a mirror of each other, that is what one reflection gets you, a mirror object (with some rotation), so the second reflection undo's the mirror effect, leaving you with pure rotation of the shape, that's why you need two reflections, the fact that U and V could be any obituary vectors as long as they have the right angle between them, is just cherry on top of the need for a double reflection. For those of you who still need to draw things or do things by hand, i will give you a workable example. 1- Draw your x-axis and y-axis on paper 2- Pick three points to represent a triangle, the points you pick dont matter as long as the x and y values for each point is positive ( this makes it easy to visualize) 3- Do the above 3 times, so that you have three graphs, graph1 graph2 and graph3 4- in graph1, rotate your triangle 180 degrees anti-clockwise, use your normal protractor means to do it (this will be the correct answer we will use to judge graph2 and graph3) 5 - in graph2, reflect about ( x= 0, y=1) and then (x=-1, y=0) (this is the double reflection result) 6 - in graph3, this is the one reflection example, now i will say that, there is no 1 reflection that will give you the same result as graph1 or graph2, but you are welcome to try, if you find a vector that does that, let me know. I hope this comment gets pinned.

[@angeldude101]

Something interesting I've noticed about the rotation being twice the angle between the vectors is that, well, the wedge product is the _area_ of the parallelogram between the vectors. What's the area of a sector of a unit circle that covers 2θ radians? θ! The area of the sector is always half the angle, so the formula e^θI could be seen as rotating an angle of θ radians, or it could be seen as sweeping out a sector with area θ. Given that θI itself is a planar object with area θ as well... Worth noting is that in hyperbolic geometry, angles are defined as twice the area of the sector above the unit hyperbola. There's a similar flat equivalent that satisfies the same properties, but also has the length itself equal the "angle" like with circular angles. Either way, this idea of e^φB sweeping out a sector with area of φ for any "unit" bivector B holds for all 3 types of transformations.

[@davidste60]

Does anyone know the best way to do non-uniform scaling in G(R2) ? There doesn't seem to be an operator for component-wise multiplication with respect to e1 and e2.

[@angeldude101]

As far as I know, non-uniform scaling is very hard to do with geometric algebra. It really seems to be the main area where matrices win out. Since working with components tends to be discouraged, the proper way to extract them is with the dot product with the basis vectors. a(v . e1)e1 + b(v . e2)e2 seems to be the best I can give.

[@davidste60]

@@angeldude101 Thanks for the reply. It doesn't seem so hard from where I'm standing. If that's the main weakness of GA I'd say it has a lot to recommend it. One thing I've noticed with all types of 3D math is the emphasis on composing rotations versus "combining" them, if that's even the right term. Like multiple forces acting simultaneously, I mean. I've only really seen it covered once, by converting them to rotation vectors and then componentwise adding them, then converting back, I think it was. But that wouldn't really fit into GA very well either AFAIK...

[@angeldude101]

@@davidste60 I'm not sure exactly what you mean by "combining" transformations, but the way I read it does happen sometimes in GA with rotations interacting with translations. In projective GA and its supersets, you can compose a rotation and a translation to perform them one after another, but you can also apply the to each other. Rotating a translator makes it translate in a different direction, while translating a rotor effectively changes the center of rotation. Compose: RT or TR Combine: RTR^-1 or TRT^-1 Doing any of these no longer results in a pure rotation or pure translation, but it still acts as a transformation, usually called a "motor," that can be applied to various objects. Attempting to translate a translation doesn't really do anything since it doesn't have a position, while rotating a rotation doesn't do anything beyond changing the plane of rotation. So you can combine two rotations as long as they're not parallel. (I actually worked this through after writing most of this comment.)

[@angeldude101]

@@davidste60 Looking at the forces example, the best way to do that is probably either some form of composition, or just adding the forces into one total force, though the latter option might need some form of normalization? GA is perfectly capable of adding two multivectors component-wise. It's just accessing the individual components and operating on them separately that's hard.

[@davidste60]

@@angeldude101 I meant combining rotations as in if I pushed your shoulder towards the ground and your hip towards the Sun at the same time. Most math resources seem obsessed with composing one rotation after another, but don't seem to care about what happens if they coincide.

[@paulbloemen7256]

Hi Mathoma, I recently discovered your videos on GA, me not being a mathematician but recently interested in GA, I think they are excellent. I saw most of them, take notes, I intend to see all of them to get a clear total picture. I have a blindspot, couldn’t find the answer also in other publications: my mistake no doubt, but still, my blindspot. It’s about the inverse of a geometric product. As the latter is a non-trivial operation, adding the dot and wedge product of two vectors, such an inverse could be tricky. Based on uu = |u|2, you arrive on (u/|u|2)u = 1, which seems OK to me, then go to u/|u|2 = (1/u) = u-1, which seems troublesome to me: why are you allowed to truncate a geometrical product (from (u/|u|2)u to (u/|u|2)), and why are you allowed to treat 1/u as a normal fraction, u being a vector? (1/u) = u-1 by definition is the inverse of u, thus uu-1 = 1. But what kind of product is uu-1? If u-1 is a vector too, one might think uu-1 is a geometric product too, but if so, then uu-1 = 1 doesn’t make sense, or does it? This all the more as you may split uu-1, as in the neat trick I saw elsewhere (splitting vv-1): u = u(vv-1) = (uv)v-1 = ((u.v)+(u^v))v-1 = (u.v)v-1+(u^v)v-1 = uproj+urej. Finally, there seems to be an inverse of a geometric product, (vw)-1 to vw, where thus by definition (vw)(vw)-1 = 1. But what is this (vw)-1? I appreciated your proof (w-1v-1)(vw) = (vw)(w-1v-1) = 1, thus indicating (w-1v-1) = (vw)-1. Rereading this text, I am confused, somehow I seem to fail to pinpoint exactly what I don’t get, sorry. My question is: do you have an answer for me, can you direct me to a video or a text at my level to make me see the, quite probably, simple truth about these inverses and the permitted operations with them? Thank you very much in advance for your answer. Hmm, I prepared the text elsewhere, I see all the superscripts and subscripts flattened out. Hopefully, the above still makes sense. Kind regards, Paul

[@hfyaer]

Ok so the important part is that the pair formed by v and w can be oriented in any direction and this would still work. Otherwise 1 reflection is also a rotation but it's dependant on the angle between v and u. Is that correct?

[@randomgamevideos241]

No, one reflection is not rotation, two reflection is a rotation, the reason you can't see it here is that, you are reflecting only one vector, so you can't see the issue that rises with one reflection, so humor me for a bit, take three vectors ( how you arrange them dont matters) now assume that the tip of these vectors are the points/vertices of a triangle (draw it in your head ... so essentially you have a triangle that is made of three vectors that we are going to reflect), now perform 1 reflection on these three vectors (pick any reflection vector you want), you will realize that you will have a new triangle that is just the mirror image of the original, that is not a rotation, but now do a second reflection, ....... did you get the original triangle? the double reflection is needed to undo the effect of reflections and leave only rotation, that is why the double reflection is needed? so 1 reflection is a rotation if you are dealing with only one vector or point but if you are dealing with a 2d shape, one reflection will lead to a shape that is the mirror image of the original, you need two reflections to get the original shape back.

[@brod515]

one thing that I'm not sure why is the single reflection itself not a rotation. isn't it already clear that the reflection itself is a rotation. is the double reflection meant to be a general way to say given any to vectors we can rotate another vector about the angle that those two vectors form?

[@DanieleMortari]

I try to compute the rotor satisfying the equation, b1 \wedge b2 = R (r1 \wedge r2) R^(-1), where all vectors are unit-vectors satisfying, b1 \cdot b2 = r1 \cdot r2 .... but I get lost. Do you have any idea how to find R from that equation?

[@ufh3478629tegryu]

Why do you need to reflect twice for a rotation? If v is at an angle of theta to u, then isn't u' a rotation of 2*theta?

[@Math_oma]

+ufh3478629tegryu But this operation doesn't rotate any given vector in the plane by the same angle, which is something we're looking for.

[@tomwilliam7299]

@@Math_oma still I did not get it man?!! in your example, the angle between u and v is PI/4 and and after we reflect u around v, the angle between u and u' is PI/2, so this is a rotation, why we need w to reflect around it ?!! Assume we want to rotate u by PI/2, all we need to find a vector v that make angle between v and u by PI/4, am i right ?!! so why we need w ?

[@jakman2179]

@@tomwilliam7299 If I'm understanding correctly it. Reflecting by 2 vectors is capable of a full rotation and, more importantly, does not depend on the angle between u and v, meaning we can rotate by selecting the angle between v and w and it will work for any starting vector u. Additionally, if thinking in terms of a group of vectors we want to rotate them together, a single reflection would invert the angles between our group of vectors, so it would not be a rotation with just one vector. So, while a single reflection by v 'may' work for a particular case, it does 'not' work for all cases.

[@davidste60]

@@tomwilliam7299 To make it simple, if you reflect two vectors, A and B, through vector C, and A is clockwise compared to B when you begin. Then, after the reflection, A will be counter-clockwise compared to B. That isn't a rotation.

[@tomwilliam7299]

in this video why we do not consider u' is a rotation of u when we reflect it around v? why we need w also? 🤔🤔🤔🤔🤔🤔@Mathoma

[@ToranSharma]

Only doing the first reflection, does rotate the vector by the angle between them. But this action is dependent on the original vector u that we are reflecting/rotating. When we do the second reflection, the result of this action is no longer dependent on our choice of u, it will rotate any vector by the same amount, given by twice the angle between the two reflection 'axis' vectors. Say you have a number of vectors that you want to rotate by a desired angle. There is no one vector that if we reflect across will give the same rotation for each of our starting vector. It will just reflect all of them. Their relative angles will all be inverted, like looking in a mirror. Introducing the second reflection corrects for this orientation inversion, and the result is a true rotation but the same angle for every vector. The initial confusion is caused by only thinking about the result of the reflection on a single vector. If an empty universe a reflection of a single arrow would look like a rotation, but if you think about reflecting multiple things it is clear that reflections and rotations are not the same. Think how text reflects in a mirror, it isn't just a rotation.

[@5hape5hift3r]

A reflection is just a 180 degree rotation in a higher dimention.

[@blackmephistopheles2273]

Now, are two reflections a rotation...or is a reflection, two rotations? YES!

[@angeldude101]

Not quite on the latter. Two rotations is still just a rotation. One fun thing that you can do though, particularly in 2D is attempt to rotate a rotation, as in apply a rotation to a rotor rather than composing them. The result is the same as doing absolutely nothing since rotating the two vectors in the rotation doesn't actually change the angle between them, and it can only change their orientation in 3D and above.

[@gabitheancient7664]

16:14 "that vector could be u"... so... am I a vector??????

[@poisonketchup8098]

this proof doesnt need geometric algebra. actually, this can be proven using the mathematical tools of ancient greece, given you simplify the defenitions of vectors.

[@angeldude101]

Correct. Geometric Algebra merely lets us represent this simple fact as an equally simple expression.

[@tomwilliam7299]

please make a discord server and invite us and put the link here to ask you some questions 😘😘😘😘😘😘😘😘😘 @Mathoma