Composition of rotations is a new rotation. Pure geometric idea


Two ways to see this in a visual mode:

First way.

First of all, two reflections give rise to a rotation, the angle of the rotation being the double of the angle formed by the planes defining the reflections. Reciprocally, a rotation can be seen as the product of two reflection, and we have the freedom to choose one of the reflection, except for the fact that its plane must contain the axis of the rotation.

Once you are convinced of this, if you have two rotations in 3D, \(R_1=\{e_1,\phi_1\}\) and \(R_2=\{e_2,\phi_2\}\), you can fix a reflection along the plane defined by the axis of \(e_1\) and \(e_2\), call it \(T\). So we have that
\[
R_1=TU
\]
and
\[
R_2=VT
\]
and therefore
\[
R_2 R_1=VTTU=VU
\]
is the product of two reflection and hence a rotation.


In the picture, we have expressed reflections by means of the intersection of their defining planes with a unitary sphere.


Second way.

We can express rotations by giving two points in a unitary sphere. This way, two points \(A\) and \(B\) determine the rotation \(R_{AB}\) whose axis is the orthogonal line to the plane \(OAB\) and whose angle is twice the distance from \(A\) to \(B\) (the arc of the great circle resulting from the intersection of plane \(OAB\) with the sphere). Observe that giving a rotation, we have some freedom to choose one of the points.

With this convention to define rotations, suppose you have two rotations \(R_1\) and \(R_2\). We can choose the point \(B\) of intersection of the two great circles defined by both rotations in the unitary sphere. Then we have two points \(A\) and \(C\) such that
\[
R_1=R_{AB}
\]
and
\[
R_2=R_{BC}
\]

This way, we get a spherical triangle \(ABC\). Now, we are going to construct three more triangles congruent with the original one. We take symmetric point along the great circles and give them names the way you see in the figure:


Triangles \(T_1=AB'C'\), \(T_2=A''BC''\) and \(T_3=A'''B'''C\) are all congruent with \(ABC\) since they have in common with it an angle and two sides. But observe that \(R_{BC}R_{AB}\) take \(T_1\) to \(T_3\), the same that it does \(R_{AC}\). And since three points determine any symmetry of the sphere, we are done:
\[
R_{BC}R_{AB}=R_{AC}
\]


These two geometric ideas are of great importance in Clifford algebras and spinors. I will tackle this issues next days.

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