# Quaternions as rotation generators

The following exercise appears in Geometric Algebra for Physicists by Chris Doran and Anthony Lasenby in section 1.8.

The unit quaternions $$i, j, k$$ are generators of rotations about their respective axes. Are rotations through either $$\pi$$ or $$\pi/2$$ consistent with the equation $$ijk=-1?$$

I am not totally sure what the authors mean here.

The unit quaternions $$i, j, k$$ are generators of rotations about their respective axes.

By "generator," do the authors mean here that they are generators in the sense of a Lie Algebra, i.e. $$e^{it}, e^{jt}, e^{kt}; t \in \mathbb{R},$$ are functions which rotate about the respective axes, or do they mean that $$i^{-1} X i, j^{-1} X j, k^{-1} X k$$ rotate a quaternion $$X$$ about the respective axes?

Are rotations through either $$\pi$$ or $$\pi/2$$ consistent with the equation $$ijk=-1?$$

If we take the Lie Algebra interpretation, is this asking whether or not $$e^{i\pi} = -1,$$ or $$e^{i\pi/2} = -1?$$ Or, is this asking whether or not $$i^{-1} X i$$ rotates $$X$$ by an angle $$\pi$$ or an angle $$\pi/2?$$

• Dec 21, 2022 at 21:18

It is a reference to the section on Hamilton's error in interpreting quaternions discussed at the end of section 1.4 of the book.

The second major difficulty encountered with quaternions was their use in describing rotations. The irony here is that quaternions offer the clearest way of handling rotations in three dimensions, once one realises that they provide a 'spin-1/2' representation of the rotation group. That is, if $$a$$ is a vector (a pure quaternion) and $$R$$ is a unit quaternion, a new vector is obtained by the double-sided transformation law $$a'=RaR^*$$ where the $$^*$$ operation reverses the sign of all three 'imaginary' components. A consequence of this is that each of the basis quaternions $$i$$, $$j$$, and $$k$$ generates rotations through $$\pi$$. Hamilton, however, was led astray by the analogy with complex numbers and tried to impose a single-sided transformation of the form $$a'=Ra$$. This works if the axis of rotation is perpendicular to $$a$$, but otherwise does not return a pure quaternion. More damagingly, it forces one to interpret the basis quaternions as generators of rotations through $$\pi/2$$, which is simply wrong!

If we interpret rotation using Hamilton's single-sided operation, then $$ijk$$ should be equivalent to the result of rotating about each of the three axes in turn. But instead it gives multiplication by $$-1$$, an inversion, which (in 3D) isn't even a rotation!

If instead we use the double-sided operation, then the result of rotating about each of the three basis axes in turn should be $$(-1)a(-1)^*=a$$, the identity. ($$(-1)^*=-1$$ because there are no imaginary components to negate.) Rotations of $$\pi/2$$ about the three axes do not give the identity, but rotating $$\pi$$ about each axis, fortunately, does.

The use of the word "generate" here is unfortunate, particularly as there is actually a direct connection to the Lie algebra interpretation, where pure quaternions $$q$$ (i.e. with no real part) do indeed generate rotations by means of the double-sided operation $$a'=e^{q\theta/2}ae^{-q\theta/2}$$. If $$q$$ is a pure unit quaternion, then this operation rotates the 'vector' $$a$$ through angle $$\theta$$ about axis $$q$$. I suspect that since this is an introductory section of an introductory textbook, the authors are assuming the reader doesn't know about Lie groups and Lie algebras yet, and so is using the word in the non-technical, plain English sense.

This introduction is also talking about the early historical misinterpretation of quaternions, which was also wrong in another way: in interpreting the pure quaternions as 'vectors'. They actually work more like pseudovectors, which for most purposes look a lot like vectors but which reverse sign under reflections. (They are sometimes also called 'axial' vectors, as opposed to 'polar' vectors which are the ordinary sort. Angular velocity, angular momentum, the magnetic field are examples of pseudovectors in physics. The 'vector' cross product of two vectors actually gives a pseudovector, not a vector.) This was another cause of quaternions losing out to Gibbs' vector algebra. (On the other side of the argument, vector algebra identified pseudovectors as being vectors, which is also wrong, and led to some serious conceptual problems.) It was only when Clifford unified the vectors and quaternions as separate parts of a common algebra (which he called geometric algebra) that we got a clear and consistent picture that could be generalised to higher dimensions.

As this chapter seems to be setting the historical scene, describing a situation of confusion that the development of geometric algebra will fix and clarify, I recommend not getting too fixated on the details. Some of it will get corrected later.

• Thanks, this is helpful. So to be clear, $a'=e^{j \theta/2} a e^{-j \theta/2}$ is a rotation by $\theta$ about the $j$ axis, and also $b'= j b j^{*}$ is a rotation by $\pi$ about the $j$ axis? Dec 22, 2022 at 2:05

If we take the Lie Algebra interpretation, is this asking whether or not $$e^{i\pi} = -1,$$ or $$e^{i\pi/2} = -1?$$ Or, is this asking whether or not $$i^{-1} X i$$ rotates $$X$$ by an angle $$\pi$$ or an angle $$\pi/2?$$

The "$$e^{i\theta}$$ vs. $$e^{i\theta/2}$$" dilemma is related to the fact that $$Spin(3)/SU(2)$$ is a double cover of $$SO(3)$$. What do we mean by "double cover"? It has to do with the different ways fermions(spinors) $$\psi$$ and vectors $$X$$ transform under a rotation: $$A \;Spin(3)/SU(2) \;rotation: \quad \psi \rightarrow e^{i\theta/2}\psi, \\ An \;SO(3) \;rotation: \quad X \rightarrow e^{i\theta/2}Xe^{-i\theta/2},$$ Because of the double-sided nature of $$e^{i\theta/2}Xe^{-i\theta/2}$$, a $$\theta/2$$ rotation of a fermion is translated to a $$\theta$$ rotation of a vector. For example, when the $$\theta/2 = \pi$$ $$\psi \rightarrow e^{i\pi}\psi =-\psi\\ X \rightarrow e^{i\pi}Xe^{-i\pi}= X$$ So when a fermion is half way around a circle, a vector has already finished the whole circle rotation. And when a fermion finishes the full circle, a vector has already went around the circle twice. That is what we mean by "$$Spin(3)/SU(2)$$ is a double cover of $$SO(3)$$".