Geometric Algebra 1

Errata. Rule #6 at 4:55 should read a(bP) = (ab)P.
At 11:49, the third bivector's description should reference the yz-plane (for consistency with the preceding two bivectors' descriptions). (James Edwards)

[danv]

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

[andrewjohnson]

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

[robfielding]

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.

[mingmiao]

Crystal clear. Amazing lecture. Thank you very much.

[guslackner]

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

[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.

[rohithdsouza]

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

[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)?

[jaimemat]

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

[AlanMacdonald]

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."

[garethwilliams]

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?

[draganostojic]

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.

[ricardowilsonaguiardacruz]

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

[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

[HISEROD]

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

[saschabaer]

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)

[mr.es]

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!

[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"?

[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?

[yevgeniyterekhin]

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.