Quaternion-forms: why you should probably stop doing differential geometry with just the wedge product
First let’s have a motivating problem. You're working with a robot arm or animated character that has two universal joints. For both joints, you have a quaternion telling you its state. A quaternion is, I assure you, a nicer thing to have than euler angles, because with those you won’t be able to tell whether the thing is twisted up (and about to break!) or not.
You're considering the situation where the end of the arm is in a very specific position with a specific orientation, and is such that there is not a unique pair of quaternions defining it (like how you can grab the corner of a table and still waggle your elbow without moving your shoulder). In fact, the arm is such that the first joint could be any quaternion you like, and the end of the arm will be in the correct place, so long as you put the second quaternion in the correct pose. Call joint 1's quaternion q1 and joint 2's quaternion q2. You have a function F:H→H (H is the quaternions) such that F(q1) gives the unique quaternion you need to assign q2 to in order for the end of the arm to be in the position and orientation you desire.
And now, you’re finding yourself with extra constraints, including using less battery, and you’re trying to figure out how to modify q1 and q2 to meet those constraints. Consequently, you’re thinking about about F as a manifold to navigate - finding shortest paths, maybe even taking integrals along them. Serious stuff, for which you want to use all the tools of differential geometry.
For this, and related problems, I claim you would like to have a quaternion-form! This blends differential geometry and geometric algebra. But, there’s a bit of a gap between these two fields; let’s review that gap.
Differential Geometry versus Geometric Algebra
Here’s the state of play, in my perhaps-limited understanding:
One thing everyone can agree on is that Grassmann algebra (the wedge product) is the bee’s knees; both sides use it in some form, and for closely related if not identical things.
If you’re in differential geometry, you’ll be using the wedge product on k-vectors and k-forms. By definition, k-forms measure k-vectors. So if you have a 2-vector, you measure it using a 2-form. In geometric algebra, you rarely see k-forms.
Again in DG, you impose the rule that, say, 2-vectors (like the parallelograms above) can be added to other 2-vectors, but a 2-vector would not be added to a 4-vector. Same for forms; you can add two 2-forms; you can’t add a 2-form to a 4-form.
On the other hand, if you’re in geometric algebra, you have no problem adding vectors with different k’s. In fact, a quaternion is a 2-vector added to a 0-vector. A quite general word for sums-of-vectors-with-different-k’s is “k-reflection” - a composition of k reflections (duh). A quaternion is a 2-reflection (here’s a nice explanation of that). To say it again: a general quaternion is not a 2-vector, or a 0-vector, but a mixture of the two; intuitively, this is because a general rotation is a mixture of a 180 rotation and a 0-degree rotation.
Still in geometric algebra, you’ll view the wedge product as a limited form of the real badass, which is the geometric product. The geometric product is the precise thing that composes 1-reflections (planar reflections) into k-reflections (quaternions, rotoreflections, translations, screw motions). When the geometric product is taken, the wedge product is happening as a part of it. In some rare cases, the geometric product is the wedge product!
And here’s the claim: for the problem above, you might find yourself wanting a quaternion-form. I’m not positive of this; but, it makes logical sense to me, so if I’m wrong in this particular case, plausibly there is some other problem for which quaternion forms are the solution. I’ve no more to say on that!
Assuming I’m correct now, there’s bad news for people who want to only use the exterior product and not the geometric product: the exterior product only allows you to create k-forms for a single integer k. That’s a problem. And alright, maybe you’ll have a separate 0-form and 2-form, and measure you’re quaternions by measuring them first with the 0-form, then with the 2-form, then, I guess, adding the results?
Here you are being motivated by the naieve desire for what you think will be simpler that motivated Gibbs and Heaviside to split quaternions into scalars and “vectors”. This was one of the worst ideas in history. Philosophically, the problem is that one man's 10 degree clockwise turn around the x axis is another man's 5 degree counter clockwise turn around the axis between the x axis and the y axis. What this means in terms of the coordinates is that as you change things, your 0-form and your 2-form will trade pieces with each other. Almost as if, you know, they were one object. So, you’ll just be working with a quaternion-form by another name. So don’t do that.
Quaternion-forms are not quaternions - they are something cooler!
Here’s a problem: what sort of object is your quaternion-form?
(Warning: while I am pretty pleased with my potential solution, you’ll only understand it if you understand Projective Geometric Algebra. Otherwise, maybe get out now while you can pretend this problem doesn’t exist. If you’d like to learn it, I recommend Sudgy’s introduction (my lectures are here and here, but I don’t cover the regressive product, which is the main thing that this is about!))
You might have thought it would be another quaternion. Your plan might have been to use the quaternion inner product as a way of measuring a given quaternion using a given quaternion-form.
Alas, no. The inner product of the j quaternion with itself is -1. But clearly the linear function that measures how much like the j quaternion you are should give +1 as output if you give it the j quaternion. So, inner product is not an option!
Here’s an idea based on something suggested to me by Steven De Keninck. There’s another product involving k-reflections that should work. It is the regressive product, which is defined using the wedge product and the hodge dual, so it shouldn’t be too unfamiliar to DG people. It can be visualized as the join, although bear in mind that it goes very far: it turns out that the join of a line with the line at infinity that “encircles” it is the scalar.
So, it turns out that the appropriate space to get my quaternion-forms basis from is: the linear space spanned by e01, e02, e03, e0123 (e0123 is involved in screw motions). I take some basis of that subspace, and if I want to measure a quaternion, I take its regressive product with my basis - and there’s my measurement. So, these are the objects you can use to do calculus on the surface of a manifold where the points are quaternions. I think! If I’m right, I guess the general concept is k-reflection-forms. Krorms?