# Geometric algebra: Rotation of a rotor

In short my question is: Why is the rotation of a rotor in geometric algebra implemented by a single-sided rotation?

To elaborate:

In geometric algebra, rotating an object is done by multiplying it double-sidedly by a rotor and its inverse:

$$V_{\text{rotated}}=RVR^\dagger\qquad(1)$$

A rotor is defined as a normalized element consisting of scalar part and a bivector part:

$$R(\theta) = \cos(\theta/2) + \sin(\theta/2)B = e^{\theta/2 B}$$ $$RR^\dagger=1,$$

where $B$ is a normalized bivector.

However it is claimed, in e.g. Doran & Lasenby's Geometric algebra for physicists, that the rotation of a rotor itself is obtained by multiplying it single-sidedly by a rotor:

$$R_{1,\text{rotated by }R2} = R_2R_1$$

As a clarifying example the rotation of rotated vector is then shown:

$$V_{\text{rotated by }R1\text{ and }R2}=R_2R_1VR_1^\dagger R_2^\dagger=R_{\text{tot}}VR_{\text{tot}}^\dagger$$ $$R_{\text{tot}}=R_2R_1,$$

claiming that this shows the rotor $R_1$ is rotated by left multiplication by $R_2$.

I do not understand this claim. I would expect that to know how $R_1$ behaves under rotation by $R_2$, one should check that equation $(1)$, with $R=R_1$, still holds in a frame rotated by $R_2$. One should therefore rotate all elements in equation $(1)$:

$$V_{\text{rotated by }R1\text{ and }R2}=R_{1,\text{rotated by }R2}V_{\text{rotated by }R2}R_{1,\text{rotated by }R2}^\dagger$$

This leads to a double-sided multiplication rule for rotors:

$$R_{1,\text{rotated by }R2}=R_2R_1R_2^\dagger$$

Which is clearly different from the single sided form.

Why is this incorrect?

• What page do you see this on? I see something very different. – rschwieb Aug 17 '17 at 0:08
• @rs Upon rereading the passage for the twentiest time it is actually not as clearcut as I thought. Still, on page 49 it is stated: "Rotors are the exception to the rule that all multivectors are subject to a double-sided transformation law." Later – Crimson Aug 17 '17 at 14:15
• what page are you pulling from in the body of the post? – rschwieb Aug 17 '17 at 14:17
• @rschwieb Upon rereading the passage for the twentiest time it is actually not as clearcut as I thought. Still, on page 49 it is stated: "Rotors are the exception to the rule that all multivectors are subject to a double-sided transformation law." At the bottom of the same page it is mentioned that a rotor has changed sign under a $2\pi$ rotation. The combination of these two statements made me interpret the section as described in my question. I am now starting to realize that this last statement is just trying to say that the rotor corresponding to a $2\pi$ rotation equals -1. – Crimson Aug 17 '17 at 14:25
• @rschwieb Then my question remains, what is meaning of the first statement and how does a rotor behave under a rotation of the reference frame? – Crimson Aug 17 '17 at 14:25

This is the text you're referring to:

Rotations for a group, as the result of combining two rotations is a third rotation. The same must therefore be true of rotors. Suppose that $R_1$ and $R_2$ generate two distinct rotations. The combined rotations take $a$ to

$$a\mapsto R_2(R_1(a)R_1^\dagger)R_2^\dagger=R_2R_1aR_1^\dagger R_2^\dagger$$

We therefore define the product rotor $R=R_2R_1$

so that the result of the composite rotation is described by $RaR^\dagger$, as usual. The product $R$ is a new rotor, and in general it will consist of geometric products of an even number of unit vectors[...]

There isn't anything mentioned here about "rotating rotors," just compositions of rotations, and the meaning of the passage seems completely clear: to compute the composite of rotations, you just multiply them.

how does a rotor behave under a rotation of the reference frame?

Well, that is a completely different topic from composition of rotations!

If $C$ is the rotor that converts from the old basis $\{e_1,\ldots, e_n\}$ onto the new basis $\{f_1,\ldots f_n\}$, then

$CRe_iR^\dagger C^\dagger = CRC^\dagger f_iCR^\dagger C^\dagger$, and so $CRC^\dagger$ expresses the rotation in the new basis of $f_i$'s.

Personally I don't like the idea of using the phrase "rotating a rotor." Conceptually there is supposed to be a divide here between "space" and "operators." That is, the space is the set that models what is being transformed, and the operators are what do the transforming.

"Rotating a rotor" makes me think the two things are being muddled together. (Although, it might be a good mnemonic for remembering how an orthogonal change of basis can be used.) Maybe something like this is what led to your question.

• Thanks a lot! Somehow my preconceived notions of a geometrical interpretation of quantum mechanical spin made me completely misinterpret this section. A lot of stuff is falling into place now. Thanks! – Crimson Aug 18 '17 at 9:17
• @Crimson glad to have helped! – rschwieb Aug 18 '17 at 10:17
• I'd just like to add that what you call rotors, have precisely the same transformation properties as spinors, for they are the same thing, and are used extensively in quantum theory in physics – R. Rankin Jul 9 '18 at 5:45
• @R.Rankin Thanks for the comment. I hope to understand that someday. For years now I have been slowly making efforts to decipher texts on the subjects, but it is slow going for a mathematician on his own. – rschwieb Jul 9 '18 at 13:51

From the definition of a rotator, it is natural that the concatenation of rotators will follow the rules

$$R_2R_1VR_1^\dagger R_2^\dagger=R_{\text{tot}}VR_{\text{tot}}^\dagger$$ where $$R_{\text{tot}}=R_2R_1.$$

But you also need

$$R_{\text{tot}}^\dagger=R_1^\dagger R_2^\dagger,$$ which is not necessarily implied by the previous.

And this says nothing about the way to compute the coefficients of the total rotator and its dagger counterpart.

Now, it may very well be that "rotating a rotator" is a meningless operation, or one that doesn't yield another rotator. What matters is the ability to concatenete rotators, not to rotate them.