Quaternions, Rotations and Real numbers

Question

I haven't really formally studied Algebra at anywhere near this level, but I was told about the existence of Quaternions a few years ago and I find them really cool. I also like how pure quaternions are analogous to cross products in $R^3$, and that it gives me a way of doing it algebraically rather than relying on hand rules.

Over the course of using them it's my understanding that multiplying an element by another, both imaginary numbers, rotates it in some manner so as to be orthogonal to both, similar to rotations of $\frac{\pi}{2}$ in the space with one real and one imaginary axis when multiplied by $i$, which is why it's useful for cross products as it will rotate it to be orthogonal to a plane spanned by linear combinations of those two vectors [afaik].

What I can't intuit though is why for any pure quaternion multiplication, there's a negative real part [the scalar product] if it's meant to correspond to a rotation within $R^3$ and the space of pure Quaternions is Isomorphic to $R^3$. Similarly I don't understand why $i^2 = j^2 = k^2 = $ A negative real.

Answer 1

Don't try interpret quaternion multiplication $q_1 q_2$ as immediatedly connected to rotations; it's more indirect.

The operation which performs a rotation is $q u q^*$, where $u$ is a pure vector quaternion (to be rotated), and $q$ is a unit quaternion $$ q = \cos(\alpha/2) + \sin(\alpha/2) \, n , $$ where $n$ is a unit vector giving the axis of rotation, and $\alpha$ is the angle of rotation. (And $q^* = \cos(\alpha/2) - \sin(\alpha/2) \, n$ is the conjugate.)

Then a general quaternion (a constant times a unit quaternion) can be thought of as representing a linear transformation which is a combination of a rotation and a scaling, and the multiplication of two quaternions is an algebraic way of computing the composition of two such linear transformations: $$ q_1 \bigl( q_2 u q_2^* \bigr) q_1^* = (q_1 q_2) u (q_1 q_2)^* . $$ For example, a pure vector (unit) quaternion performs a 180 degree rotation around the corresponding axis, but when you compose two 180 degree rotations you don't get a new 180 degree rotation, unless the two axes of rotation happen to be orthogonal. So it makes perfect sense to have a scalar part (nonzero in general) in the quaternion product; it is related to the angle of the composed rotation.

Answer 2

Let's start by considering the uses of multiplication of complex numbers, since that is how you tried to derive an intuition about quaternions being suitable for rotation in $\mathbb R^3.$

First, it might be worth noting that to use complex numbers for rotation, we use multiplication, and that although $\mathbb C$ is in some ways isomorphic to $\mathbb R^2,$ the field of $\mathbb C$ with multiplication is not isomorphic to $\mathbb R^2,$ at least not in the sense that we would have "guessed" the rules for complex multiplication just by looking at obvious operations to perform on $\mathbb R^2.$ See an answer to an earlier question for further discussion.

Now let's consider the function $f(x) = \cos x.$ This is a transformation on $\mathbb R,$ that is, a function $\mathbb R\to\mathbb R$; its input and output are strictly one-dimensional. But we can also write $f(x) = \Re[e^{ix}],$ that is, we can use complex numbers (which are two-dimensional in this context) to express a one-dimensional operation. This seems to add a lot of unnecessary complexity (pun intended) to something that really only needs one dimension, not two, but electrical engineers and physicists find that this "unnecessary complexity" actually simplifies a lot of calculations.

You could think of rotation by quaternions in a similar way: as an "unnecessary" extra dimension added to the three-dimensional space of the objects you want to rotate, which happens to be handy for calculation. We allow intermediate steps of the calculation to have this fourth dimension, but in the end we will be back in three dimensions again.

As for how to make sense of the fact that $i^2 = j^2 = k^2 = -1,$ we need to review some details of how calculations with quaternions relate to rotation in $\mathbb R^2.$ Referring to an answer to another previous question, we can write a quaternion in the form $$ q = w + ix + jy + kz,$$ and define the conjugate of the same quaternion as $$ q^* = w - ix - jy - kz,$$ that is, the conjugate has the same real part but opposite imaginary parts. We can also view the imaginary parts of a quaternion as a vector in $\mathbb R^3,$ so $\mathbb R^3$ is (in this sense) isomorphic to the set of purely imaginary quaternions.

Then if we take a purely imaginary quaternion $r$ and a unit quaternion $q$ (such that $qq^* = 1$), the product $$ r' = qrq^* $$ gives the same result $r'$ that we would get if we made the usual identification of $r$ with a vector in $\mathbb R^3$ and rotated it by an angle $\theta$ around an axis $\hat n = (n_x, n_y, n_z),$ where $$ q = \cos\frac\theta2 + (in_x + jn_y + kn_z)\sin\frac\theta2. $$

Notice that $q$ can be purely imaginary only if $\cos\frac\theta2 = 0.$ A purely imaginary quaternion corresponds to a rotation by $\pi$ radians ($180$ degrees). Perform the same rotation twice, and you come back to where you started. So it works out nicely that $i^2 = -1,$ because if we perform the "$i$" rotation twice, the result is $$r' = i^2 r (i^2)^* = (-1)r(-1) = r,$$ as it should be.

Answer 3

That $i^2 = j^2 = k^2 = -1$ is related to that two rotations à $90^\circ$ give a rotation of $180^\circ$ which reflects (two) axes, i.e. multiply the $x$, $y$ or $z$ values with $-1$.

Answer 4

It could be added that when you represent vectors of a 4D euclidean space $E^4$ by (all) quaternions, one-sided multiplication by a quaternion is a special kind of rotation in $E^4$, an isoclinic rotation. Other than $±1$, there are two kinds: left and right, which you get by multiplying from the left or from the right.

These rotations are nice (and they’re the first glance at higher-dimensional rotation groups). Under these, each vector rotates by the same angle, and there are infinitely many invariant planes spanned by any vector and its rotated image. A way to get such kind of rotation is to pick two orthogonal planes (say, $\langle 1, j\rangle$ and $\langle i, k\rangle$) and rotate by the same angle in the first one then in the other. You have four ways to do so, two pairs of inverses; and we just call one pair left and the other right; and all isoclinic rotations get called left and right in such a way similar rotations get called the same. Also note that the planes used in the construction aren’t special: any pair of orthogonal invariant planes will suffice.

And here is the interesting trivia:

• First, by combining both kinds by mapping $v$ to $q_1 v q_2$, you can get any 4D rotation whatsoever (like, not only these special isoclinic ones, but also single-plane rotations and double-plane rotations with their two angles unequal; and of course $±1$ in case one treats them differently).

• Second, if you choose $q_1 = q_2^*$, you now get a rotation for which the subspace spanned by $i, j, k$ is invariant — you get a 3D rotation. (Likewise, if $q_1$ coincides with $q_2$ under a reflection in another 3-subspace, you get a rotation of that subspace alone.)

Why is the latter? Suppose we found two orthogonal planes $A$ and $B$ which are invariant for both a left-isoclinic rotation and a right-isoclinic rotation by the same angle. Then if they both do rotate $A$ the same way, they need to rotate $B$ in opposite ways, or vice versa. So composing both, you get a rotation by the double angle in one plane and a cancelled zero rotation in another. This isn’s complete but it shows the ideas behind $q^* v q$.