Geometric Algebra 1

[@danv8718]

I'm starting down the geometric algebra rabbit hole. These videos are incredibly clear and helpful. Thanks a lot for sharing.

[@AlanMacdonald1]

Glad you like them!

[@drandrewsanchez]

Same here! The algorithm is working wonders for me this month.

[@andrewjohnson1378]

Thank you Dr. Macdonald for this. I am an aerospace engineering student interested in the applications of GA to control systems. I am still pretty confused, but will keep with it. By the way, your book is AWESOME

[@BlueGiant69202]

Andrew Johnson You might find the Primer on Geometric Algebra to be helpful. geocalc.clas.asu.edu/GA_Primer/GA_Primer/index.html

[@robfielding8566]

Your book is the clearest introduction to the subject. Something I discovered when trying to actually make a GA calculator in code: I think introduction of M should actually be done first. There are a total of 2^d coefficients for all parts of M, where each on/off combination for the d basis vectors are on/off. Multiplying multivectors is just a distribute over a pair of sums being multiplied. Then you just sort the basis vectors e1,e2,e3 with swaps and cancellations to get the right sign for each part. Using bit notation: e_{bit_e3}{bit_e2}{bit_e1}. Starting with dot and wedge completely obscures how you would just multiply a pair of unknown multivectors (M N). But (M N) = (M_000 e000 + M_001 e001 + M_010 e010 + ...) (N_000 e000 + ... ) .... tells you HOW the low-level multiplication is implemented. Just distribute and sort out swaps and cancels of the basis vectors. From there, you note that the type of a multivector is determined by which coefficients are zero. You can observe the outcome of (M N) and show where the dot and wedge come from. The fact that you need to do an O(2^{2n}) number of low-level operations seems to be the source of what makes this tedious to do literally, even for small numbers like 3 or 5. But when you can see that you are just doing a distribute over (M N) that is mostly zeroes, it is clear that higher level operations are taking advantage of the type determining which fields are zero. with the bit notation, the destination for two parts being multiplied is known to be xor. i was never able to find a simple closed form for taking something like (e011 e110) and calculating the sign without resorting to converting it to (e2 e1 e3 e2) and counting how many swaps are required to sort it.

[@AlanMacdonald1]

Thank you for the appreciation of my book. Yes, if you are going to code a calculator, you only need to know how to add and multiply M's. But if you want an understanding of the geometry and mathematics of geometric algebra, and how to use it, you need more. To me, the best order to present all this is in my book.

[@rrr00bb1]

@@AlanMacdonald1 If you make a simple multivector calculator, you can render in convenient notation by using the parts that are zero to determine the type. But if you don't use an exact arithmetic (rational/symbolic), and use floating point, you get answers like: (0 + 5.3 e1 + 4.0 e2 + 0.0000001 e1 e2). So, it's very easy to get started by just defining a general multivector multiply. But yes, you need to go back and look at the symbolic calculation to figure out which parts had to be exactly zero; to preserve the types. I think real implementations have to do quite a bit of symbolic solving and special cases.

[@robharwood3538]

Very enlightening take on representing multivectors and performing operations on them, from a fellow programmer's perspective. When you can actually code the math into a library/tool/app, that makes things so much more concrete and understandable, in my opinion. Love the idea of using bitstrings as 'types' of GA objects.

[@guslackner9270]

At 5:15 point 6 is wrong. Both objects should be either object P or object Q to properly assert the compatibility of scalar multiplication with field multiplication

[@guslackner9270]

Nevermind, I now see your discussion with James. I, too, will have to investigate your book(s).

[@AlanMacdonald1]

Gus, I had already noted the error at the top of the comments.

[@rohithdsouza8]

beautiful :) one of the few videos(probably most of your videos) which doesn't have even one down vote

[@feraudyh]

at 12:38 you say what about the dimension for G3: it's 1+3+3+1 because vectors have a dimension 3. This contradicts what we were told in the last video. So I have to take this all with a huge grain of salt.

[@MathwithMing]

Crystal clear. Amazing lecture. Thank you very much.

[@AlanMacdonald1]

Thank you Ming.

[@KieranORourke]

Thank you for these clear & concise videos. I am now the proud owner of your second book on Geometric Calculus. GA has allowed me to finally scratch an itch I've had for many years caused by an irritant called an 'imaginary scalar that squares to -1'. (How could such enormous mathematical richness arise from a seemingly trivial requirement to force a solution to the simple polynomial; x^2+1=0)?

[@AlanMacdonald1]

You are welcome -- belatedly.

[@draganostojic6297]

Confusing is that elements of a vector space are called vectors as are oriented geometrical segments. More precisely, an oriented line or plane segment is a member of a corresponding vector space (and thus a vector). Vector or bi-vector in a geometric sense is a vector of an abstract vector space. So we have name vector used in geometric sense (oriented geometrical element: 0D, 1D, 2D ... segment) and in vector space sense. I'm much more clear now.

[@jaimemat3]

¡Gracias! Por fin una idea diferente: los bivectors no tiene shape. Siempre los presentan como paralelogramos ...

[@jamesedwards6173]

A nice series so far. Just a couple observations. Rule #6 at 4:55 reads "a(bQ) = (ab)P." but should instead read: a(bP) = (ab)P. At 11:49, the third bivector's description ought to reference the yz-plane (just for consistency with the preceding two bivectors' descriptions).

[@AlanMacdonald1]

Thank you for your comment, James. When I first posted my videos it was possible to add "cards" to them to correct errors. KZbin then stopped offering this capability. So the only option for me would be to replace the video. This would mean gathering all the tools I used to create them, which I don't want to do. Even if I were to recreate and post a new video, I would lose all the positive reviews (KZbin rules).

[@jamesedwards6173]

@@AlanMacdonald1 Yep. No problem. I'd probably do the same. I suppose one thing that should be possible, though, is to update the video's Description field to include any known errata details, perhaps? Or maybe a pinned comment? Btw, I ordered your GA book. :)

[@AlanMacdonald1]

Thank you for the suggestion, James. Done. Since you seem to be an eagle eyed reader, I invite you to send me comments as you read my book. By email, not here.

[@AlanMacdonald1]

Yevgeniy Terekhin: When replying to your question I accidentally deleted it. Here is my reply: I don't see a contradiction. Please elaborate on why you think that my statement at 12:48 contradicts my earlier statement at 1:36 that vectors are one dimensional objects. Also, my statement was "They may *live* [not leave] in 3D space."

[@geertdejonge4194]

Thank you this is great.

[@olbluelips]

Binomial theorem at 12:27? :) the idea I get is that some lower-dimensional multivectors have more freedom to be oriented in the higher space than others, and they each contribute an amount of dimensionality according to Pascal’s triangle. A trivector in G^3 has almost no “room” to orient and becomes a psuedoscalar So we’d have a k-dimensional multivector in G^n forms an (n choose k)-dimensional vector space

[@AlanMacdonald1]

Thank you for your question. Yes, the dimension of k-vectors in G^n is kth term in the binomial expansion of (1 + 1)^n -- (n choose k as you wrote). it is the the dimension of the ways one can orient k-vectors, for n-vectors it is 1 = n choose 1, only one way to orient a pseudoscalar. I think that your intuition is good here. The entries in Pascal's triangle are different, getting larger as one moves down its rows. The entries in the binomial expansion are two triangles. one upside down.

[@pi5549]

6:26 I can't see visually/intuitively (or by any logic) that only 2 orientations are possible for a 3D component. Moreover, I sense an implication that this extends to an N-dimensional component, and I can't even imagine how to go about proving that for such a beast. I'm guessing e1.(-e2).e3.(-e4).(-e5) ends up with a single - sign... something down this line...? Looking at the 2D case, I suppose we'd need ei.ej = -ej.ei... EDIT: It's explained in the next vid.

[@AlanMacdonald1]

Permutations of e1 e2 ... en fall into two classes -- even or odd: en.wikipedia.org/wiki/Parity_of_a_permutation . One can choose a permutation and assign an n-dimensional object in R^n one or the other orientation. See also p. 81 of my book linear and geometric algebra.

[@pi5549]

Typo @ 5:04, Vector Space Axiom 6 should be a(bP) not a(bQ)

[@AlanMacdonald1]

Thank you for pointing this out. I have added a note to the video @5:06 giving the correction.

[@adamsniffen5187]

@@AlanMacdonald1 you should out it at first appearance, @4:54, I paused to read, found the error, then needed to scroll to find this.....if only that correction was there when it first appeared, most people have a little faster reaction time than 11 seconds....if you are paying attention.

[@mr.es1857]

In which type, of course, do we have to study outer product and inner products for example? I know that the inner product is some kind of generalization of the dot product but never learned in linear algebra!

[@jonnymahony9402]

The inner product has applications in quantum mechanics, where you can calculate the probability of a state.

[@garethwilliams2173]

You use the e12,e13,e23 bivector basis, others use e12,e23, e31 which is consistent with standard vector x product and quaternions. This is confusing, does it matter?

[@AlanMacdonald1]

Each is a basis: negating one or more vectors in a basis results in a basis. (e1e2 = -e2e1, etc.) Sometimes e12,e13,e23 is more convenient. Sometimes e12,e23, e31 is more convenient.

[@feraudyh]

at 9:19 you could just write M=(s,v,B,T) using tuple notation.

[@AlanMacdonald1]

Yes I could. But this would not be as useful, for example when multiplying two multivectors. For the same reason one does not denote a complex number as (x,y).

[@feraudyh]

At 1:32 we are told that vectors are 1D objects. This is of course in contradiction with the usual knowledge of linear algebra etc... So what has changed? The word dimension is being used in an entirely different way. Why does the author say "Note that"?

[@rwharrington87]

I think what he’s getting at is the endpoint of the vector is just a 1D point in 3D space.

[@ricardowilsonaguiardacruz4554]

Excuse me, i mean, Dr. Macdonald.

[@AlanMacdonald1]

Amazon.

[@HISEROD]

Shouldn't rule 6 in the definition of a vector space be a(bP) = (ab)P rather than a(bQ) = (ab)P?

[@AlanMacdonald1]

You are of course right. Unfortunately, KZbin doesn't allow me to make a correction (they used to). If you comment again please give the time in the video so I don't have to search for it.

[@HISEROD]

@@AlanMacdonald1 Ah ok, I wasn't 100% sure since I'm new to GA. I'll be sure to include the timestamp next time :)

[@alphalunamare]

2:40 come on dude! squaring the circle is a joke ... you could have at least sniggered. 7:57 I was at the end of my seat and it all fell flat with just another bloody definition .. why not explain for why it is considered a scalar and then that it necessitates a new definition? Why do trivectors form a space of dimension 1???? 9:50 ..A Theorem??? you have totally bypassed for why bivectors form a 1 dimensional space, isn't this slight of hand?

[@kx4532]

Woahhhh!

[@ricardowilsonaguiardacruz4554]

By the way, Dr. Andrew, where can i find your book for sale?

[@AlanMacdonald1]

At Amazon. My name is Alan Macdonald.

[@yevgeniyterekhin5523]

Slide 9 at approximately 1:36. It is said that “the vectors are 1D objects. They may leave in 3D space but the are one dimensional.” Doesn’t it contradict to slide 15 (at approximately 12:48) where you defined “vector basis”. Based on the “Dimension: 1+3+3+1=8” equation, it looks like vectors are three dimensional objects.

[@Fauxmar]

I'm also slightly confused about this.

[@AlanMacdonald1]

I don't see a contradiction. Please elaborate on why you think that my statement at 12:48 contradicts my earlier statement at 1:36 that vectors are one dimensional objects. Also, my statement was "They may *live* [not leave] in 3D space."

[@saschabaer3327]

When you state that two trivectors are considered identical if |T| does not change, I assume the orientation must also remain the same, right? (Otherwise T = -T which I think would make the vector space somewhat degenerate)

[@AlanMacdonald1]

I didn't quite say "trivectors are considered identical if |T| does not change"; I said "reshaped, retaining |T|". I guess that there is an implicit assumption that orientation is preserved when reshaping.

[@LazySundayBook]

Could you elaborate on the basis of the Bivectors and Trivectors near the end of the video and how they form a basis of the Bivector (e1 ^ e2) and Trivector (e1 ^ e2 ^ e3)? You said the '^' symbol represented the outer product. I looked up what the definition of outer product is and it seems to be doing vector multiplication that results in a matrix. How does this outer product generate a parallelogram. I learned somewhere that the cross product could be used to find the area of a parallelogram. Is the outer product in some way related to the cross product? How does the outer product give you the unit square or the unit cube?

[@AlanMacdonald1]

+Scott Munro I defined the outer product on Slide 18. Your question about the cross product is answered on Slide 34. Your question about bases tells me, I think, that you are not familiar with the concept from linear algebra. It that right? In the video Geometric Algebra 0 I say that some knowledge of linear algebra is required to fully understand the videos in my playlist.

[@LazySundayBook]

+Alan Macdonald I am familiar with Linear Algebra and I heard your warning that a working knowledge of linear algebra (which I have) would be required. Neither of those slides are in this video which runs from slides 6 to 15. If those concepts are tough to understand until later videos it might be worth putting a footnote within the video describing where further explanations of those concepts could be found.

[@AlanMacdonald1]

+Scott Munro You are right: I used the outer product /\ before I defined it. Now I have to figure out what do do. If I replace a video with a new version I lose comments on the video, number of views, and thumbs up/down.

[@LazySundayBook]

+Alan Macdonald I don't think taking the entire video down is necessary. I think KZbin has an annotations system that you can use to embed notes within the KZbin videos

[@AlanMacdonald1]

+Scott Munro I had known about annotations, but wanted to use them only as a last resort, hoping to find some way, perhaps using the video editor, to change the videos. No dice. So I used an annotation. Thank you for pointing out my error.

[@jrkirby93]

When you say in slide 8, "scalars are round in R3" what does that mean?

[@AlanMacdonald1]

+jrkirby My enunciation was not clear enough: "around", not "round". Scalars are around in R3; there is scalar multiplication. But they are not members of R3. They are members of G3.

[@jrkirby93]

Ah, that makes sense. Also thanks for making these. They're exceedingly concise and clear. And geometric algebra is a really cool way of looking at mathematical spaces I'd never thought of before.

[@johnholmes912]

bivectors have no shape? hmmph, they are topologically equivalent...a doughnut is the same as a coffee cup in topology, but they DO have different shapes

[@AlanMacdonald1]

Rectangles of different areas are topologically equivalent but not considered equal in geometric algebra. Squares in 2D with the same area but positioned differently are considered different in Euclidean geometry but considered the same in geometric algebra. Each geometry gets to specify which geometric objects are considered equal. So I don't understand what you are getting at.

[@account1307]

Is this grassman algebra ?

[@AlanMacdonald1]

The Grassmann algebra as usually understood is only a part of geometric algebra, the part involving the outer product. Grassmann was aware of the geometric product, but only with the work of Clifford that it became more generally known.