Geometric Algebra in 2D - Fundamentals and Another Look at Complex Numbers
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- Опубліковано 19 чер 2024
- In this video, I introduce some of the concepts of geometric (Clifford) algebra, focusing on two-dimensional space (R^2). We'll talk about the wedge (exterior) product, review the dot product, and introduce the geometric product. We'll see that a new mathematical object, the bivector, arises from considering the wedge product. Furthermore, we're going to see that this bivector behaves like an imaginary unit in that it squares to -1. Since it has this property, it naturally relates to vector rotation in 2D, like the complex numbers.
Recommended video on Clifford algebra:
• Clifford Algebra
...and other video series covering Geometric Algebra (Nick Okamoto): • From Vectors to Multiv... - Наука та технологія
19:30 Very nice to watch the square root of negative -1 comes out of anti-commutativity so cleanly
the way these random algebras map to real things, feels mystical, like it was the reason people chased it for so long.
"I could cook up some vector, 'v.'"
*inhales deeply*
Smells good
Thanks for publishing this series, it's led me down the Geometric Algebra rabbit hole.
Geometric Algebra does away with ideas in mathematics I find aesthetically unpleasing (e.g. cross products, determinants, unmotivated definition of i=sqrt(-1), superfluous coordinate systems) and replaces them with something more unified, beautiful, and intuitive.
I don't understand why the subject is not more popular. My impression from googling is that the subject has been stagnant for the past decade. Books on the subject are mostly aimed at higher-level physics students. Why has no one written an introductory Physics book using Geometric Algebra? If I learned this stuff in high school or early college, I would have been a better student.
Also from googling, my impression is that amateur mathematicians/physicists get more excited about the subject than the experts do. Is Geometric Algebra not as attractive as it appears on the surface, or do amateurs just appreciate good pedagogy more than most experts do?
+Anon Ymous
Like in other aspects in life, people are set in their ways after a certain age and won't change drastically, which is why opinion on large issues only changes as quickly as the sum of the rate of those of the newer opinion populating academia and those of the older opinion retiring. Yes, oftentimes you will find amateurs such as myself more enthusiastic about GA than the experts, and you will often find people on, e.g. blogs or Math Stackexchange, dismissing our appreciation for this branch of math as disproportionate, but there's never any argument backing it up other than _this is what the experts in our lifetime just so happen to be doing_ and invective, neither of which have anything to do with truth or wisdom or proper pedagogy.
Perhaps you could look into the corpus of David Hestenes, since he's written a book entitled _New Foundations for Classical Mechanics_ or if this isn't what you were thinking about, this might be an interesting project.
The problem with Geometric Algebra is a bit more subtle. All the books neatly avoid the simple act of JUST CALCULATE THE DAMN PRODUCT!!! If you spell out every vector/bivector/trivector/scalar... you find that there are 2n components for every multivector. Multiplying two multivectors, using a bit notation for the presence of a unit vector. say that e1 == e_001, e2 == e_010, e2e1 == e_011, e3 == e_100, etc. So, general multivector multiplication has NO pre-requisites, because it's just distribution. (a b) = (a_00 e_00 + a_01 e _01 + a_10 e_10 + a_11 e_11)(b_00 e_00 + b_01 e_01 + b_10 e_10 + b_11 e_11) ... after this is distributed, you cancel out parallel basis vectors and sort orthogonal vectors and negate for every swap required to sort them into a canonical order. When done this way, special products (dot, wedge, cross, reversion, leftcontract, rightcontract, etc) are already CALCULATED for you; and you don't necessarily need to know them.
Start with this CALCULATION of Geometric product first. Only split it into its cos/sin parts later. If this is done, then it doesn't suddenly become complicated when you get beyond R3.
note that (a b) takes O(2^(2n)) operations when you do it in the brute-force definition method, which even ignores the complexity of sorting the orthogonal basis vectors before adding them (to find the sign). This is a big issue with Geometric Algebra. With GA, you really need a smart compiler to do non-trivial things with it.
@Mathoma, is there a closed form for calculating the sign when using the bit notation? It is a simple xor to figure out where the operands are destined for. ... ie: (e_011 e_110) will write into e_101 because (011 xor 110). (e_011 e_110) == (e2 e1 e3 e2) ... i only can figure how how many swaps it takes by literally expanding the orthogonal vectors and counting how many swaps it takes to put it in order.
@@robfielding8566 Chris Doran has an interesting talk about implementing GA in Haskell at geometry.mrao.cam.ac.uk/wp-content/uploads/2017/10/HaskellExchangeGA.pptx . His swapping count algorithm is pretty neat though still enumerating all swappings. He encountered the same problems of complexity of computation as you but Haskell heavy lifting hides it.
Awesome series. Every step flows from the previous one in a logical and seamless manner such that the viewer never becomes lost trying to connect missing dots along the way. The fact that your train of thought transports its passengers to the final destination instead of casting them aside is the greatest praise I could bestow on any teacher. Well done!
Your videos on Geometric Algebra really are great. Thanks a lot for making them :)
Read a recommendation to this series in a comment under another video, and I must say, this is very well explained. Thank you for sharing this video-series with us!
Great introduction to the subject which demonstrates clearly the linkage between algebraic and geometric concepts which I previously thought of as quite separate. Excellent and many thanks.
Thank you so much for these videos on Geometric Algebra they have cleared up a lot of doubts I've had in it.
Many thanks for this exhaustive focus on just 2D demonstrating the isomorphism of Geometric product to the complex number. Revealed crucial new connections that help understanding of the Clifford/Geometric Algebra .
Perfect example of video... Nicely explained. Thank you.
I recently saw a blog post about Clifford Algebra and how beautiful it is. Ever since, I've been trying to learn more, but couldn't find a gentle introduction to the topic that still goes deep into the math. Thank you for these great videos--they do exactly that. You are a gifted teacher.
+processing
Yes, it's a very nice branch of math and isn't difficult to learn especially if you have some experience with linear algebra.
Thanks for this series. A few days ago I was trying to find more
information on quaternians and rotating objects in higher dimensions and then *poof* your video introducing rotors came out... but first I had to find out what geometric algebra was all about.
Amazing video! I got especially excited upon seeing cos(theta) + sin(theta)*I, because that reminded me of quaternions. Hyped to watch through the rest of this series, thanks for creating it
I've been waiting for your video on geometric algebra. Finally! Thank you for an awesome introduction. Very clearly explained. Looking forward to more on this topic. Thanks again!
+Clove Cinnamon
You're welcome. If you've been watching my quaternion stuff, you'll see that a lot carries over into the geometric algebra of 3D, really because the quaternions live inside G(R^3) and are bivectors.
I am a Phd (physics) scholar from India and I want to say thank you from core of my heart. Great series indeed!!!👍
Totally agree with @Ryan Sanden. No missing, unexplained or jumped steps. Pace is perfect. Best math instructional vid series that I've seen.
These videos are really great; I haven't binged on a mathematics playlist like this since first discovering 3blue1brown!
As a physics dropout who got super-frustrated with the lack of an intuitive feeling for what the mathematics I was taught was doing (aside from, you know, just sucking at it), this feels like such a simple and intuitive foundation connecting a whole lot of not-quite-as-intuitive mathematics together.
I'm starting to wonder if it wouldn't make much more sense to introduce GA immediately after introducing vectors in high school, even before complex numbers. You know, when vectors are just arrows that you can add and subtract, representing Newtonian forces.
I second that thought. Schools might start teaching GA once teachers have more freedom (eliminate exams I say! Do journal portfolio assessments instead. Free-up both teachers and students.) But this is not even half of it... please check out the work of the Canadian physicist Cohl Furey (arxiv.org/pdf/1611.09182.pdf), who has used Clifford Algebra to retrieve almost all the symmetries of the standard model of particle physics. So, the algebra of spacetime gives us almost for free the standard model. How? Partly it is because spacetime algebra is just an encoding of the geometry of spacetime. But what about all that complexity in gauge theory and Feynman diagrams etc, surely that can't all just be simple geometry? Well, maybe not, but mix in some topology (wormholes) and you probably have ingredients for explaining most of the particles in physics. Also, Nima Arkani-Hamed has shown a simpler geometric structure dubbed the Amplituhedron can replace Feynman diagrams. And Leonard Susskind has shown how quantum entanglement is really just a wormhole bridge. So a lot of physics looks like it is reducing to geometry. If this pans out, then I think physicists will almost be forced to start ditching Gibbs' vector algebra and start working with Clifford's GA.
Well, this was a very nice objection. Geometric algebra unifies mathematics, brings clarity and geometric intuition. It should be introduced in high-school programs, but we have a problem that it is hard to bring geometric algebra to universities. But there is no doubt, it will prevail, due to its power.
At 11:24 you mentioned that it was Clifford who came up with the geometric product of vectors as the sum of Grassmann's inner and outer (dot and wedge) products. That's what many GA practitioners think, and it is correct, but I found it fascinating to learn that Grassmann independently came up with that idea near the end of his life. In the paper "Grassmann's Vision," Hestenes points out that Grassmann (influenced by Hamilton's quaternions) defined what he called the "central product" in a paper that was actually published before Clifford's paper introducing the geometric product. Fascinating stuff!
+Nick Okamoto
I actually read that paper just a few days ago, it's funny you mentioned it. Hestenes is a lot of fun to read, as there's a polemical quality to the writing in addition to straight-up math. Another fun paper is the "Mathematical viruses" one because I think I've unfortunately been infected with the "coordinate virus". But, what can I do? I may as well let the virus take its course and build my immunity.
Ah yes, I really enjoyed that paper too. I asked Hestenes about it at a GA conference a few years ago, and he said that article started as a tongue-in-cheek response to critical remarks made by another mathematician (as mentioned in the acknowledgement section at the end of the paper), but I don't recall what the criticism was about. Though it's presented in a humorous tone, I also found the "Mathematical Viruses" paper to be very enlightening.
Mathoma What do you mean by “coordinate virus”? I understand it’s probably meant as a joke, I’m just not sure what you’re getting at specifically
@@raydencreed1524 Well, according to Hestenes, the "coordinate virus" is the fact that many students (and others) imagine vectors as lists of coordinates. In geometric algebra, vectors are treated as geometric objects, as well as bivectors and many other elements. Coordinates are not important. Here is a nice example. We use the Pauli matrices in physics, but they are just 2x2 matrix representation of 3D Euclidean geometric algebra. This means that we can formulate quantum mechanics using vectors, without need for the imaginary unit.
Miroslav Josipović That’s what I was thinking they meant, but it’s good to be sure. Thanks for taking the time to explain that.
This video is great! I found it extremely helpful. Thank you!
This is a great resource, thank you!
Thank you for this series! This along with a conference paper I found on arxiv are helping me learn this cool subject.
Which conference paper was it?
This video is so much more than it's title implies. It opens the door to Geometric Algebra, Cartan's Exterior Algebra, Modern Physics and a way of thinking about deep and beautiful correspondences and analogies.
+anthonysegers01
I try to be subtle.
Cool. I'm not a mathematician, but I'm finding generalized geometric algebra very cool as an alternative to other approaches to computing projections. Keep it up!
Very Good channel! I usually don't understand everything, but where is the fun if I do get it all! everything is so efficiently explained and the quality is very high.
+Mrbillybobcorndog
Well, if you have any questions, feel free to ask. I may not have all the answers but it's fun talking about these topics. Hopefully my videos occasionally make some sense.
you are a great teacher!
A model of clarity my man! You have a new sub.
+Bob Kelly
Thanks!
Thank you so much for your video!
Wonderfully explained!
+oberon the first
Thanks!
23:55 - "This set forms a basis for..."
Ah, crap, my Senior-level Linear Algebra did not prepare me as well as I hoped it would.
That dude made up his own notation, and I didn't see why. Turns out, when you research new math...
These are really good.
Wow, that was amazing
Thanks for this series, this is amazing. Would you have any simple and practical textbook to suggest?
29:50 if you instead view a vector (ae1+be2) as being e1(a+be1e2), then it makes sense why left and right multiplication by e1e2 is different. in fact, the e1 factor can be thought of as a "mapping" from the true complex numbers (a+be1e2) to the vectors (ae1+be2) for the purposes of graphing
Thank you so much for these incredible videos.
Will you introduce geometric calculus?
Great tutorial.
Much better video on Cliff Alg then the other video discourses that drown you in notation
Brilliant!
This is so cool I'm losing my mind
Muchas gracias por compatir tu video. ¡relike!
i surely did enjoy watching this, since I've never heard of bivectors and geometric product before
+no name
The wedge product is closely connected to both matrix determinants and the cross product, so you may have implicitly learned about it. The geometric product, on the other hand, isn't commonly known.
+Mathoma that's true, cross and wedge product have a resemblance. I also liked how by using geometric product you can show that complex numbers or "i" naturally comes up in geometry of 2d that was an "aha" moment
+no name
In fact, the cross product is dual to an associated bivector. Also, the quaternions will show up in 3D as naturally as the complex numbers showed up in 2D, and by similar means. Both of these thoughts will be covered in future videos.
Thanks a lot sir💗💗💗
Thanks very much....
especially for 14:15
Thank you. Now I am able to make sense and follow along with David Hestens' 2002 paper.
amazing thank you
this is excellenr -- thank you!
+Ralph Dratman
My pleasure.
Thanks!
19:20 seeing (e1e2)^2 I would be really tempted to write '= (e1^2)(e2^2) = 1', and it took me a while to realise why that's wrong, and that's because there's a secret commution in breaking down a power like that.
(ab)^2 = abab, and not aabb
well done
@Mathoma, at time stamp 31:17 (ae1 + be2) is multiplied with (cos{theta} + sin(theta) I) why isn't the geometric product (dot and wedge product) calculated between these (multi)vectors?
Geometric Algebra is so beautiful. It reveals how interconnected math really is.
Please , demonstrate how I would use the wedge product for calculus 3
@20:33 OK, I'm hooked!
You have made the world a better place.
You are criminally undersubbed
So when summarizing G(R^2), we see we have 1 as unit scalar, e1 and e2 as unit vectors and e1\wedge e2 as unit bivector?
In 33:20 I see how you arrived at u' expression n the right but I don't understand how that expression is really a rotation of theta degrees
Bravo, tome otro like buen hombre!!
Please suggest me books for Geometric Algebra with insights on Tensor calculus. Thanks!
+Jonathan Salazar
I forgot to include references in this particular video (I'll get on it) although I have references in most videos in the geometric algebra series. My main text is Lasenby and Doran's _Geometric Algebra for Physicists_. I think it's a nice gentle introduction to the subject and if you're interested in particular aspects of the subject you can always start there and follow the references.
Hi Mathoma, would you possibly know if there's anything I should learn before learning this topic? I'm a freshman in high school looking to learn this kind of stuff for game development and physics. Cheers!
i would use "x hat" and "y hat" instead of "e1" and "e2" to represent the two unit vectors because it's more elegant and less confusing (with the constant e we've got 3 e's on the board). But still this is a very good video
Do you have a Patreon? As of now I only support 3Blue1Brown, but you produce the kind and quality of content I would be happy to support!
+VFB1210
That's very generous of you. Perhaps in a half-year to year's time I'll make an account because I should increase the production quality and series coherence before I feel good about accepting donations. I have a long-term project in mind on nonstandard analysis (infinitesimals and all that) and perhaps only at that point and for similar projects would I accept monetary donations. I genuinely appreciate the enthusiasm but for the time being, simply sharing videos is enough of a donation, as often as you might hear that from UA-cam content creators.
I lost my notebook, now I have to start the series all over :(
:feelsbadman
Lucky!
But if e_1 and e_2 are both orthogonal and unit vectors, isn't their wedge product just equal to +/- 1 ?
End of this video gets pretty weird. Maybe I was lost somewhere, but there's weird analogy between multiplying two complex numbers and multiplying vector by scalar+bivector and getting vector as a result. That's weird, because in the first case vectors and complex number are used sort of like the same thing.
Anyways, even complex numbers are somewhat weird for me. I never understood how to interpret complex solutions to quadratic equation and I used them a lot at university to represent phase and amplitude of electric current, analog signals or basically any wave.
Man this stuff is absolutely fascinating. I'm hooked! Thanks for this great content, big fan of your channel. Do you know of a good textbook to get going with this stuff?
Check out the description boxes under the videos for recommendations.
Cool stuff! What's the purpose of writing down rotation in Clifford Algebra though? seems rather odd. I'm more used to regular complex numbers as supposed to this stuff.
+DonutKop
In short, it's nice because it's the same formula in all dimensions (which actually isn't the formula in this video) and it makes it clear that rotation should be thought of as being done by a two-sided operator, which only reduces to a one-sided operator in special cases (when the vector being rotated is completely in the plane of rotation).
Well, the rotor formalism in geometric algebra is the most powerful formalism that I know. It is universal, simple, intuitive, works without exceptions (ever), etc. It is directly connected to spinors (which we can imagine as rotors multiplied by a scalar).
If the definition of the geometric product seems arbitrary, I recommend this series, which shows that all of these definitions can be derived from a single assumption that a vector squared gives its length squared!: ua-cam.com/video/Fc5rlZ3Malw/v-deo.html
07:54 What is the justification for that? How can we know that the distributive law indeed works in this case, other than just "checking and seeing that it seems to work"? (because there might be some edge cases in which it breaks, e.g. for certain configurations of vectors, or at certain number of dimensions).
19:35 Again, what is the justification for associativity here?
"Check it and see if it works" is a completely valid concept here. The associativities you mention are generalized formulas; they use placeholder variables and you can substitue whichever numbers you like in there to try and break it. Since the operations involved are all linear, there's no danger of singularities occuring (as there would be if we used a quotient, for example). You're never going to get an undefined division by zero or infinity unless you started with one to begin with.
I'm a little hazy in the proof that the square of the bivector, e1e2e1e2 , yields -1. It seems to require that the wedge product has to be
associative. I don't know if this is true. The square of the bivector reduces to (e1^e2)^(e1^e2). To get -1 out of this requires that
(e1^e2)^(e1^e2) can be written as e1^(e2^e1)^e2. Is this correct?
The wedge (outer) product is associative, it is something that is easy to prove.
11:00 and the other product (e1 /\ e2) denotes the orientation of the bivector then?
+Procrastinator7
The order in which the product is written tells us the orientation. The product e2 ^ e1 has reversed orientation. The scalar sitting in front gives us information on the magnitude of the bivector, how much area it contains.
Yes, I meant that this second product of the result of the calculation describes the other property of the bivector, the rotation/orientation of it.
Great video by the way!
This video is 7years old and still many universities didn’t add this important course in their curricula in engineering.
God I wish I could see this in relation to linear transformations :P
Now that I am have been studying these I have stumbled onto an actual problem. It seems natural to me how two vectors product for a bi-vector described with a signed area and scalar to the perpendicular projections relative to vector magnitudes squared. But the algebra multiplies several vectors in sequence to make its proofs. Once more than two vectors are multiplied in 2D the immediate geometric sense of what these equations mean starts to break down.
vector (geometric product) vector = bivector
but what does bivector (geometric product) vector =
how about bivector (geometric product) bivector = ??
I ask because (e1e2)^2 is suppose to = -1 (a scaler of minus one and an area of zero, because no 'I' part) that is taking (a vector??) and moving it 180 degrees. but it is composed of 4 vectors so which vector is it moving 180? The scalars and the wedge products are well behaved under the complex plane in making the algebraic steps from (e1e2)^2 to e1e2e1e1 to -e1e2e2e1 to -e1e1 to -1. But geometricaly e1e2 is a bivector and we are taking an operation that takes two vectors and geometrically produces a bivector and applying it to two bivectors. If we have e1e2e1 we are taking a bivector and a vector as inputs, what is the geometric interpretation that is happening, not just the scalar and oriented areas because those only describe bivectors. According to the algebra each new vector seems to be producing yet another bivector which implies there is an inherent vector a bivector collapses to when a new vector is applied to as a geometric product.
Can i have your textbook's name pleaseeee !!
Why can we use Euler's formula with the bivector I ?
+seen0it
Because it squares to -1.
that's exactly what happens. normal multiplication is actually the same as the complex exponential. this is not possible to see if multiplication is not defined with Geometric Algebra. (u v) = (u . v) + (u ^ v) = |u| |v| ( cos_uv[ theta ] + I_uv sin_uv[ theta ] ) = |u| |v| E^(2*Pi* I_uv *theta) ... the standard notation is ambiguous about which plane I is in, and what plane sin/cos are over. the standard notation is basically wrong. it lacks important information.
Taking the wedge product of two vectors yields a bivector, so why is it correct to assert that u∧u=0? Is it more correct to state that the magnitude of the resulting bivector is 0?
It follows from antisymmetry of the wedge product, i.e. that u∧v = -v∧u but more intuitively it is because the oriented area swept out from a vector to itself is 0.
@@Math_oma Applying the wedge product on two 3-dimensional vectors 𝒖 and 𝒗, where:
𝒖=a𝒙+b𝒚+c𝒛 and 𝒗=d𝒙+e𝒚+f𝒛
𝒖∧𝒗=(ae-bd)(𝒙∧𝒚)+(af-cd)(𝒙∧𝒛)+(bf-ce)(𝒚∧𝒛)
Is this correct?
I ask because I'm struggling to rationalize how to programmatically represent a bivector given any two arbitrary N-dimensional vectors. I feel there is no greater way to understand the underlying mathematical concepts than to create a library capable of performing such calculations.
@@Tannz0rz Yes. You'll notice the coefficients are the same as those of the cross product of u and v.
14:30 how are 2-dimensional vectors suddenly 1 dimensional. And how is the scalar 0 dimensional and not 1 dimensional?
+TheMadmaurice
For your first question, could you point out which statement you're asking about? I suspect the answer to that question won't be any deeper than it's just the way the geometric product is set up. I'm also not sure if I have a good answer for why scalars are 0-dimensional other than saying there's not much else for them to be, seeing as how they don't have direction and don't require any particular number of vectors to be described. When we formed those four objects (one scalar, two vectors, one bivector) we were choosing a certain number of basis elements at a time. For the vectors, of the 2 we pick only 1, and so the number of vectors is 2 choose 1 = 2. For the bivectors, we choose both, so there's 1 bivector. By abstraction, we choose 0 elements and those are the scalars in the geometric algebra.
+TheMadmaurice
Perhaps I should have used the word "grade" instead of "dimension". Probably a blunder on my part.
"We have these one dimensional objects that we're inputting into the geometric product". These "one dimensional objects" apparently being the two 2D vectors u and v.
I'd say scalars are 1 dimensional. The graph for a 1 dimensional vector would be the number line and thus showing the real numbers which are scalars. The direction of a 1d vector would be negative or positive or none at all for 0 (like there wouldn't be a direction of a vector if the vector's length is 0 in pretty much any amount of dimensions). Their describing "vector" would be e_1 = 1. Another way to describe it would be: When a 2 dimensional vector is an element of R^2 (just for the example) then a 1 dimensional vector would be an element of R^1 or R and thus being a scalar.
I didn't really understand what bivectors are yet though which might because of the language barrier. I sadly don't know all mathematical terms in English that I learned in German.
Alright i just looked up bivectors now and i guess this whole thing kinda succeeds my mathematical horizon. But at least on wikipedia they talk about scalars being a "order zero quantity". Maybe that's where I misunderstood.
+TheMadmaurice
As I said above, I probably should have stuck to the word "grade" instead of "dimension". That bivector is just an area element that has magnitude equal to the area swept out and some orientation.
If a scalar + bivector couple is isomorphic to a complex number/2D vector, why is G(R²) of dimension 4 instead of just dimension 2?
Why don't we just consider 1 and 𝑰 the only objects we need to define G(R²)?
Both the scalar 1 and the bivector 𝑰 are defined using the basis vectors e_1 and e_2, so together, all 4 objects define G(R²).
This is very interesting when applied to the laws of physics, as suggested by Gauss, to integrate the time of transformation of information transmission into the the electrodynamics. I am now a big fan of your videos.
note: In his famous letter on non-Euclidean geometry to Farkas Bolyai of 6 March 1832, Gauss suggested to Farkas that he should study complex numbers, thus relating non-Euclidean geometry and complex numbers (see Section 7, Letter 8).
I am curious how the reformation of the laws of physics look in this language. It appears to me on the surface that the reformations of Maxwell's equations by Gibbs and Heavyside may have been a step in the wrong direction.
Are you aware of any attempts to formulate the laws of physics into this symbology?
+Richard Alsenz
Yes, geometric algebra appears to be a very natural system for doing physics and one of the big selling points for geometric algebra is that the four Maxwell equations can be compressed and written as a single equation. This is briefly explained here:
en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Geometric_algebra_formulations
However, I'd encourage you to check out the literature to learn more, particularly the work of Doran and Hestenes. The main reference I use for this series is _Geometric Algebra for Physicists_ and this would likely answer your question in the greatest detail.
Thanks.
I also just discovered "Physical Applications of Geometric Algebra" to be very helpful.
geometry.mrao.cam.ac.uk/wp-content/uploads/2015/02/hout01.pdf
I've prepared some materials that might also be of interest: ua-cam.com/play/PL4P20REbUHvwZtd1tpuHkziU9rfgY2xOu.html . Most of those videos reference GeoGebra worksheets and pdfs of written versions. Your participation would be very welcome in the LinkedIn group, "Pre-University Geometric Algebra": www.linkedin.com/groups/8278281 .
Very good intuitive explaining (Cartesian-system)^(Rotation-system) ⇔ Quarternion-4D ≡ 2dimention-xy MAPPING onto 2dimention-R(theta) ⇔ 4D ⇒ 5 DIMENTION world ! This geometric-algebra is much simpler then complicated-METRIC-TENSOR ! Thus the MATHS-topology can explains the Physics-Einstein-cosmology .Beginning-bigBang-theory and Ending-blackHoleTheory MirrorReflexion of 5DimentionWorld . The 10+1 dimention stringTheory can reduced to 3space1time/4Dspacetime .
thank you for the Free Gift .. ( i am far from this ) but I SEE the value.. Freely give what was freely given. Amen
+jennifer siagian
Thanks! A little at a time and you'll successfully learn this. Non nobis, Domine.
where there is Life there is hope (circle me google+ SEE what I am saying.) Ps Math IS the Universal language and sooner than later WE will all be speakingi (( again)) g nite. Non nobis - Wikipedia
en.wikipedia.org/wiki/Non_nobis
Non nobis is a short Latin hymn used as a prayer of thanksgiving and expression of humility. ... Non nobis Domine is now known in the form of a 16th-century canon derived from two passages in the motet Aspice Domine (a5) by the South ...
Make more videos dude.
26:15 Oh no, a bee
34:18 This is a mistake, isn't it?
“Cook up” = suppose?
Yes
At 14:36 you say that there are one dimensional objects that we are inputing into the geometrical product and what we are getting out is a 0 dimensional object ( a scalar) and a two dimensional object (a bivector). Now vectors on the plane are two dimensional objects in my book and scalars are one dimensional objects. So clearly you are using "dimension" with a completely different meaning. Are you aware of the confusion that you are creating?
I suspect that you are using "dimension" for "grade".
Geometric dimension ≠ dimension of a linear space.
Grade 1 objects are 1-dimensional subspaces of a larger N-dimensional space, but to describe any given 1-dimensional subspace in an N-dimensional space can require N components, so the linear space of 1-dimensional supspaces in N-dimensions itself has a dimension of N. In any number of dimensions, scalars represent 0-dimensional subspaces, and independently act as a 1-dimensional linear space.
The video seems to be saying "dimension" to refer to the geometric dimension of a given subspace, where as you seem to be saying "dimension" to mean the dimension of the linear spaces that describe said subspaces.
everytime I see the (e1e2)^2 = -1 I just think "top10 anime plot twists"
It's equal to -1 not i btw
@@officiallyaninja shhhhh
u wdj u = 0 *bi-vector*
You can't just erase it
It may be ok but you need to show how
You never really show how associativity is possible with bivectors, or how you can multiply a vector by a bivector. It seems like you would need to prove these properties by demonstrating how to multiply a vector by a wedge product but you never address that.
You may want to look at the "abstract algebra" definition of the geometric product then, which defines the Geometric product as having the properties you describe (everything can be multiplied with anything), and defines the wedge and inner product with respect to the geometric product. It's basically a product that satisfies
1) AB is in the same space (e.g. closure)
2) A(BC) = (AB)C (associativity)
3) 1A = A1 = A (the existence of an identity element)
4) A(B+C) = AB + AC, and also (B + C)A = BA + CA (distributivity)
5) For every vector a, a^2 = a dot a (contraction)
The reason (that I support) for not introducing the geometric product as such is that 1, 2, 3 and 4 are all properties of a 'normal' product and thus introducing them again would be a waste of time and confuses the main purpose: geometric intuition. Moreover, as far as I can tell 5 is equivalent to the Fundamental Identity (e.g. ab = a wedge b + a dot b), which he focused on instead.
My feet are all wet
Thanks!
Thank you very much