TITLE: Geometric algebra: Rotation of a rotor
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.