For people trying to understand quaternions, the faster way to intuition is understanding axis-angle first. I think axis-angle is totally intuitive. First pick a direction (some 3d unit vector) and then rotate around that direction by some amount (so some scalar). Then convert those four numbers to quaternions
qx = ax * sin(angle/2)
qy = ay * sin(angle/2)
qz = az * sin(angle/2)
qw = cos(angle/2)
where (ax, ay, az) is the unit direction (ax^2 + ay^ + az^2 = 1) and angle is the amount you want to rotate.
For notation, we would often see the basis vectors named (e_1, e_2, e_3) instead of (x, y, z).
The quaternions are the even-ordered subalgebra of the 3D exterior algebra. The exterior algebra has scalars (1), vectors (x, y, z), bivectors (xy, yz, zx), and pseudoscalars (xyz). The even-ordered subalgebra is scalars and bivectors (1, xy, yz, zx). Adding or multiplying two even-ordered multivectors will always give you an even-ordered multivector, and 1 is even-ordered, so the even-ordered multivectors form a subalgebra.
We can also conceive of this subalgebra, the quaternions, as a Clifford algebra. Clifford algebras are generalizations of exterior algebras. Instead of saying v * v = 0, we can put something else on the RHS, and for quaternions we can start with just two basis vectors e_1 and e_2, and then define e_1 * e_1 = e_2 * e_2 = -1. The third basis vector for quaternions is then just e_1 * e_2.
Quaternions are beautiful and natural, not weird, but I'm always happy when they get some attention! This was a very interesting article, thanks. It does seem to confuse quaternions in general with (unit) quaternions as used for rotating vectors. For example:
>So don’t think of quaternions as a 4 dimensional hypersphere of radius 1
This is also a bit weird:
>But nobody would ever suggest that we should think of a rotation matrix as a 9 dimensional hyper-cube with rounded edges of radius 3.
Even weirder when they claim that the axis-angle interpretation of (unit) quaternions "breaks down".
Anyway, back in high-school when I first became fascinated with quaternions I certainly didn't expect to be working with them on a daily basis two decades later. The moral of this is that anything you learn can become crucial to your career...
While we all learned in middle school geometric algebra that the even subalgebra of G3 is isomorphic to the quaternions, what is the relationship between the even subalgebra of G4 and the octonions?
If you write out a multiplication table, it seems that it's isomorphic.
But... Octonions aren't associtive. Does the even subalgebra of G4 somehow lose associativity? Is it equivalent to Octonions with a cannonical multiplication order?
There's a lot of hand waving in that phrase, "it turns out". Sure, "it turns out" that 3D uses four numbers. Why?
Geometric algebra explains that in a succinct way that also appeals to our intuition about geometry. Start by using bivectors to represent reflections, then take the closure of your bivectors and you get the even-ordered subalgebra. This will have dimension 2^(N-1)... so 2 for 2D, 4 for 3D, and 8 for 4D.
This, to me, takes the mystery out of why quaternions can represent rotations, and it places quaternions in a coherent theory of geometry that works in any number of dimensions, not just 3D. Alternatively, we could accept that the math just happens to work out that way, or we could even show that quaternions are a double cover of SO(3), but all that does is analyze why something works, whereas the geometric algebra version is a bit less of a leap and builds quaternions from the ground up.
I think that's a good motivation why we would study quaternions, but it's kinda hiding the big difference between 2D and 3D under the rug. In 2D, we have a nice, global coordinate system for the space of all rotations: what we call the angle. (Actually, it's a coordinate system for the "universal cover" of the space of rotations since angle X and angle X + 2pi give the same rotation, which mostly doesn't really matter.) Meanwhile, in 3D, there is no global coordinate system for the space of rotations! Euler examples uniquely specify a rotation, but the problem of Gimbal lock [1] means that they break down as coordinates at some point (i.e. there's no inverse to go from rotation in 3D to its corresponding Euler angles, which there is in 2D, with the caveat already mentioned).
This is analogous to the problem of finding a coordinate system for the globe: specifying latitude and longitude tells you were you are, but there's a degeneracy at the poles. And no possible coordinate system can solve this problem entirely. Contrast this to the situation of giving a coordinate system for the circle, which we do with it's angle. This isn't quite a coordinate system, due to the problem we already encountered that X and X + 2pi are the same, but that's OK because the these two points are separated from each other. On the sphere, the latitude/longitude pair (pi/2, x) gives the north pole for any value of x, even ones that are arbitrarily close together. That maps not even locally invertible!
You suggest we think of points on the circle as point in 2D space that happen to lie on the circle (i.e. cos and sin of the angle corresponding to that point). Analogously, we can think of points on the sphere as points in 3D space that happen to lie on the sphere (like some point (x,y,z) with x^2 + y^2 + z^2 = 1). And analogously, we can think of rotations of 3D space as a point in 4D space (that happens to satisfy some conditions), and the quaternions give that 4D point. This is fantastic and convenient in both 2D and 3D! But in 2D we didn't need to do this, but could if we wanted to. For 3D rotations, we do need to, or else we have this terrible degeneracy that never rears its head in 2D. In that sense, 2D and 3D are very different!
This is great insight, but it seems a bit silly to act like you don't need a 4D / hypersphere representation when the 4th one is hiding in plain sight. For the not-quaternion to describe a rotation, it needs unit length in 4D, with the two components scaled as a sine/cosine pair.
It's definitely an alternative in the sense that it gives you an alternative framework for concepts that are taught under the banner of linear algebra in school. For example, it gives an alternative construction for quaternions as a subalgebra, and it gives the exterior product as an alternative to the cross product.
> Redundant information. The latter follows from the former.
It's a blog post, not a paper. If I give you a brand new operator ^ that takes an arbitrary object a and another object b and tell you that it has anti-commutativity, it's not a given that a ^ a = 0. So it's useful to highlight that this IS the case when a and b are the vectors of GA, for intuition's sake. a = -a iff a is the zero vector. What if our objects are rotations or something though? R = -R does not imply R = the zero rotation.
arnioxux|8 years ago
http://www.euclideanspace.com/maths/geometry/rotations/conve...
http://www.euclideanspace.com/maths/geometry/rotations/conve...
klodolph|8 years ago
For notation, we would often see the basis vectors named (e_1, e_2, e_3) instead of (x, y, z).
The quaternions are the even-ordered subalgebra of the 3D exterior algebra. The exterior algebra has scalars (1), vectors (x, y, z), bivectors (xy, yz, zx), and pseudoscalars (xyz). The even-ordered subalgebra is scalars and bivectors (1, xy, yz, zx). Adding or multiplying two even-ordered multivectors will always give you an even-ordered multivector, and 1 is even-ordered, so the even-ordered multivectors form a subalgebra.
We can also conceive of this subalgebra, the quaternions, as a Clifford algebra. Clifford algebras are generalizations of exterior algebras. Instead of saying v * v = 0, we can put something else on the RHS, and for quaternions we can start with just two basis vectors e_1 and e_2, and then define e_1 * e_1 = e_2 * e_2 = -1. The third basis vector for quaternions is then just e_1 * e_2.
logfromblammo|8 years ago
Grustaf|8 years ago
>So don’t think of quaternions as a 4 dimensional hypersphere of radius 1
This is also a bit weird:
>But nobody would ever suggest that we should think of a rotation matrix as a 9 dimensional hyper-cube with rounded edges of radius 3.
Even weirder when they claim that the axis-angle interpretation of (unit) quaternions "breaks down".
Anyway, back in high-school when I first became fascinated with quaternions I certainly didn't expect to be working with them on a daily basis two decades later. The moral of this is that anything you learn can become crucial to your career...
Govindae|8 years ago
If you write out a multiplication table, it seems that it's isomorphic. But... Octonions aren't associtive. Does the even subalgebra of G4 somehow lose associativity? Is it equivalent to Octonions with a cannonical multiplication order?
marcv81|8 years ago
klodolph|8 years ago
Geometric algebra explains that in a succinct way that also appeals to our intuition about geometry. Start by using bivectors to represent reflections, then take the closure of your bivectors and you get the even-ordered subalgebra. This will have dimension 2^(N-1)... so 2 for 2D, 4 for 3D, and 8 for 4D.
This, to me, takes the mystery out of why quaternions can represent rotations, and it places quaternions in a coherent theory of geometry that works in any number of dimensions, not just 3D. Alternatively, we could accept that the math just happens to work out that way, or we could even show that quaternions are a double cover of SO(3), but all that does is analyze why something works, whereas the geometric algebra version is a bit less of a leap and builds quaternions from the ground up.
tgb|8 years ago
This is analogous to the problem of finding a coordinate system for the globe: specifying latitude and longitude tells you were you are, but there's a degeneracy at the poles. And no possible coordinate system can solve this problem entirely. Contrast this to the situation of giving a coordinate system for the circle, which we do with it's angle. This isn't quite a coordinate system, due to the problem we already encountered that X and X + 2pi are the same, but that's OK because the these two points are separated from each other. On the sphere, the latitude/longitude pair (pi/2, x) gives the north pole for any value of x, even ones that are arbitrarily close together. That maps not even locally invertible!
You suggest we think of points on the circle as point in 2D space that happen to lie on the circle (i.e. cos and sin of the angle corresponding to that point). Analogously, we can think of points on the sphere as points in 3D space that happen to lie on the sphere (like some point (x,y,z) with x^2 + y^2 + z^2 = 1). And analogously, we can think of rotations of 3D space as a point in 4D space (that happens to satisfy some conditions), and the quaternions give that 4D point. This is fantastic and convenient in both 2D and 3D! But in 2D we didn't need to do this, but could if we wanted to. For 3D rotations, we do need to, or else we have this terrible degeneracy that never rears its head in 2D. In that sense, 2D and 3D are very different!
[1] https://en.wikipedia.org/wiki/Gimbal_lock
amai|8 years ago
tnone|8 years ago
mwkaufma|8 years ago
catnaroek|8 years ago
No. Geometric algebra is a use case of linear algebra. How can it be an alternative?
> Before I tell you how to actually evaluate the wedge product, I first have to tell you the properties that it has:
> 1. It’s anti-commutative: a \wedge b = -b \wedge a
> 2. The wedge product of a vector with itself is 0: a \wedge a = 0
Redundant information. The latter follows from the former.
klodolph|8 years ago
sebastos|8 years ago
It's a blog post, not a paper. If I give you a brand new operator ^ that takes an arbitrary object a and another object b and tell you that it has anti-commutativity, it's not a given that a ^ a = 0. So it's useful to highlight that this IS the case when a and b are the vectors of GA, for intuition's sake. a = -a iff a is the zero vector. What if our objects are rotations or something though? R = -R does not imply R = the zero rotation.