QED Prerequisites Geometric Algebra 6 - Multivector Products

This lesson begins by clearing up the definition of a "k-blade". Then we proceed to write down explicit expressions for most of the multivector products between vectors and bivectors, vectors and trivectors, and two bivectors. We do this using the "relative" formalism with a basis set given for M_1_3.

The software I usually use to produce the lectures is:

[drarp]

Hey. Long-time viewer, first-time commenter here. First of all, thank you for all the work you've put into these lectures over the years. They are an amazing resource that's helped me see the light where I had previously repeatedly failed. Second, with regard to this project on geometric algebra, you are doing an entertaining, engaging, and easy to follow work, for which I am again thankful. I wasn't expecting to take this subject up and go along with you, but here I am, hooked to it. And third, good call writing some expressions before expounding on them; it's surely made the past couple of lectures easier to follow, not to mention shorter than they otherwise would have been. Small changes such as this and your use of different colors make for big improvements on my understanding.

Hope you continue doing an amazing work and have a healthy and happy life. See you around!

[maccychannel]

Really enjoying this series so far. I have been studying GA for a while now and I just had a thought regarding part of this video. Where you are expalining about blade as and that not all 2-vectors are 2-blades etc. I think in the paper they are saying that the grade-k part of a multivectors is a k-blade which is true. A multivector in STA is written as A=a(A_0)+b(A_1)+c(A_2)+d(A_3) here each different k-vector. Within the multivector each of these A_i is a simple k-vector so could be said to be a k-blade.

[jackozeehakkjuz]

At 34:00, the B^ij are already antisymmetric, right? so you can simplify that to 2B^ijv_j I think.

[PeeterJoot]

@45:49. You've used a dot product notation for the grade-2 selection. That would normally only be used for the lowest grade part. i.e.: you should write: AB = A . B + <AB>_2 + A /\ B

where the dot and wedges are defined as:

A . B \equiv <AB>_{|p -q|}
A /\ B \equiv <AB>_{p + q}

assuming that p and q are the grades of A, B respectively.

[riakm]

I’m definitely enjoying your exploration of the paper! I find myself wondering how much more complicated the GA method becomes when applied to GR... I guess rather than having a single graded algebra, it would be a graded algebra field (in the physics sense)?

[davidhand]

I don't know why you're doing this to yourself with the tensor notation and index fiddling. If my basis is X, Y, and Z, each squaring to -1, then for bivectors A and B, the geometric product is as follows:
AB = (AxyXY + ...)(BxyXY + ByzYZ + BzxZX)
= AxyBxyXYXY + AxyByzXYYZ + AxyBzxXYZX + ...

Then work out what the products of the basis vectors are using anti commutation and definitions of squares, also very very simple:

XYXY = -XXYY = -1
XYYZ = -XZ = ZX
XYZX = XXYZ = -YZ
...

And then group your coefficients by basis:

(-AxyBxy + ...)1 + (AxyByz + ...)ZX + (-AxyBzx + ...)YZ + ...

I get the appeal of condensing the equations with the dummy index contraction, but it seems like maybe at least once you could mention that the geometric product is just multiplication of sums, and you can always work it out exhaustively just like that. Honestly it doesn't seem like you're saving a lot of time doing it with index contraction, and it often looks like a risk of double counting. One of the best parts of GA is that any idiot that graduated from middle school can do the geometric product, i.e. it's highly axiomatic.

I get the desire for mathematical pedantry when you introduce concepts like the direct sum of vector spaces, and that's a fine diversion. You seem to want to present it as something "going on under the hood", but that's subjective. You may have learned all about traditional mathematical vector spaces _before_ ever seeing GA, and in that case, it seems appropriate to see that as a technical underpinning of GA. However, GA itself is axiomatic. It doesn't require the metric tensor to work - just definitions of each basis squared is all that's needed. I have a feeling the author uses the metric eta only because it's what people who don't know GA as an independent topic are expecting to see in relativity, because I've seen many discussions on STA, and this is the first time I've seen eta shoehorned in.