> Yes, sometimes you just need the dot product, and sometimes you just need the exterior product. If you are coding, or giving the final form of some formula, you don't have to always put both of them in your code or paper.
In my experience 99% of the time you just want the dot product or the exterior product. Even when you want both it is rare that you want to combine them linearly except in some niche physics/mathematics.
> But neither the dot product nor the wedge product are investable by themselves. Having an investable product on vectors is endlessly useful while you are deriving the formulas.
Do you mean invertible? Why is invertibility is so useful?
First of all it is only invertible for some non-zero elements, especially if `a` is a linear combination of multivectors or we work in PGA that explicitly adds a basis vector of norm 0. Yes sometimes it is useful but that doesn't automatically makes it more fundamental than the inner product and exterior product.
You make it sound as though multivectors being invertible is a special case, when the opposite is true. In 2D and 3D GA, every non-zero k-vector and versor has an inverse. In PGA, every non-zero, non-ideal plane, line, and point, and versor has an inverse. The inverse is used all the damn time when composing and applying transformations and performing projections and rejections.
As to which is more fundamental, I don't think it matters. You could argue that the dot and exterior products are more fundamental because the geometric product is their sum (for vectors). You could also argue that the geometric product is more fundamental because it is simply the Cartesian product of two multivectors, and you derive the dot, exterior and commutator products by filtering that product by grade. Both definitions are true, and "fundamental" is both a matter of perspective and irrelevant to any practical concern.
> In 2D and 3D GA, every non-zero k-vector and versor has an inverse.
Of course by definition every versor has an inverse. The invertibility of k-vector gets hairier for higher dimensions though. Even in 3D GA, some mixed-grade elements are not invertible.
> As to which is more fundamental, I don't think it matters.
It doesn't matter mathematically but it matters pedagogically. GA enthusiasts seem to advocate teaching GA to anyone that has learnt linear algebra. I believe it is more appropriate to stick to teaching tensor algebra and its quotient exterior algebra. Then it is up to you to learn Clifford algebra as a generalization of exterior algebra; especially if you are a game dev, a physicist, or a topological K-theorist.
I'm curious about your perspective on exterior algebra. So far I've mostly seen it as a special case of GA so my view is probably GA-tinted. I'm just curious how you even do any sort of transformations since the exterior product doesn't allow these sorts of things. It seems like it only gives you the "things" in your algebra with few ways to do anything with them. You seem to agree with the author of the article you linked to in that the most useful aspects of GA come from EA. I find this very hard to see, so maybe you can shed some light on it? E.g. how do you rotate a bivector in EA?
There is no equivalent notion of rotor in EA. I would say its most important use is that you need it to define differential form in which we can define exterior derivative and integeration in arbitrary differentiable manifolds. It also enables to define determinant elegantly.
Sorry I forgot to answer your question: To rotate a bivector of the form v ^ w, just do R v ^ R w where R is the rotation matrix. This is a linear map so you can extend the operation linearly to arbitrary bivector.
>Even in 3D GA, some mixed-grade elements are not invertible.
This paper [1] claims to have inverses for general multivectors up to a certain dimension, but I've never needed them and haven't dived into it. I'm curious what the applications would be for general multivectors, I've never come across them in practice.
we first establish algebraic product formulas for the direct computation of the Clifford product inverses of multivectors in Clifford algebras Cl(p, q), n = p + q \le 5, excluding the case of divisors of zero.
In my experience 99% of the time you just want the dot product or the exterior product. Even when you want both it is rare that you want to combine them linearly except in some niche physics/mathematics.
> But neither the dot product nor the wedge product are investable by themselves. Having an investable product on vectors is endlessly useful while you are deriving the formulas.
Do you mean invertible? Why is invertibility is so useful?