Composition of two rotations of the same angle $\alpha$ fixing points $a,b \in{S^2}$

Question

Let $g,h$ be rotations of the same angle $\alpha$ around fixed points $a,b \in{S^2}$. Show $gh$ fixes $c \in{S^2}$ on the great circle that forms the perpendicular bisector of the segment $ab$, such that $ac$ and $bc$ make an angle $\alpha/2$ with $ab$.

All I know is that two rotations of the same angle conjugate, but othe than that i don't know a lot. $gh$ is the composition of two rotations and hence is a rotation, so by definition must fix a point (technically antipodal points), but i can't see how to show it lies n this great circle. It makes intuitive sense that the point is in some way an 'average' of points $a$ and $b$, but i cannot see how one would show this.

Any pointers in the right direction are greatly appreciated.

Answer 1

I woke this morning with the answer in my head. Look at this picture in the plane:

Consider the point $P$. Under a rotation of angle $\alpha$ about $A$ (clockwise), $P$ will move to $Q$. When we then rotate about $B$ clockwise by $\alpha$, $Q$ will move back to $P$. The net result? The point $P$ ends up fixed by the sequence of two rotations.

Now imagine that this is really an overhead shot of a sphere. We're looking down on a portion of the great circle between $A$ and $B$; that portion is drawn in blue. The orthogonal great circle, through the midpoint of the blue segment, is drawn in orange (or at least a portion of it is!). Rotation of the sphere about $A$ by angle $\alpha$ will again take $P$ to $Q$, and the same argument applies to $B$, and we're done.

You might say "But what if $A$ and $B$ are far apart, like $3\pi/2$ apart? Then we can look from the other side, where they'll only be $\pi/2$ apart. For any two points that are not antipodal, there's some hemisphere whose central great circle contains both, with the center point of the hemisphere being the midpoint of $A$ and $B$ along the great circle. Once you've got that, this picture is all you need.

(More formal argument: under stereographic projection from the sphere to the plane tangent to the (spherical) midpoint of $A$ and $B$, the figure described in the problem becomes the figure drawn in this plane. Since stereographic projection is conformal, angles are preserved, and great circles map to either great circles or lines, etc.)

Answer 2

Let $$ \mathbf{K_a}= \left[\begin{array}{ccc} 0& -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{array}\right] $$ where $a = [a_1, a_2, a_3]$ Then $$ \mathbf{R_a} = \mathbf{I} + (\sin\alpha) \mathbf{K_a} + (1-\cos\alpha)\mathbf{K_a}^2 $$ is the rotation by the vector $a$. (This is called Rodrigues' Formula.)

You can, with this, write out the formula for $K_b$ as well, and multiply $K_b K_a$ by $a+b$ (which is just twice the average) to see that it fixes the correct vector. It helps to remember that $K_x x = 0$ for any vector $x$. Let's just do it. I'm going to let $c$ and $s$ denote the cosine and sine of $\alpha$. \begin{align} \mathbf{R_b}\mathbf{R_a}(\mathbf{a}+\mathbf{b}) &= (\mathbf{I} + (\sin\alpha) \mathbf{K_b} + (1-\cos\alpha)\mathbf{K_b}^2)(\mathbf{I} + (\sin\alpha) \mathbf{K_a} + (1-\cos\alpha)\mathbf{K_a}^2)(\mathbf{a}+\mathbf{b} )\\ &= (\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)(\mathbf{I}(\mathbf{a}+\mathbf{b} ) + s \mathbf{K_a}(\mathbf{a}+\mathbf{b} ) + (1-c)\mathbf{K_a}^2 (\mathbf{a}+\mathbf{b} )\\ &= (\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)((\mathbf{a}+\mathbf{b} ) + s \mathbf{K_a}\mathbf{b} + (1-c)\mathbf{K_a}^2 \mathbf{b} ) \end{align} Now we use the fact that $K_a \mathbf{u} = \mathbf{a} \times \mathbf{u} $ for any $\mathbf{u}$, and similarly for $K_b$ to get $$ \newcommand{\a}{\mathbf{a}} \newcommand{\b}{\mathbf{b}} \newcommand{\u}{\mathbf{u}} \newcommand{\v}{\mathbf{v}} $$ \begin{align} \mathbf{R_b}\mathbf{R_a}(\a+\b) &=(\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)((\a+\b) + s \mathbf{K_a}\b + (1-c)\mathbf{K_a}^2 \b)\\ &= (\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b))\\ &= \left[ ((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b)) \right] + s \b \times \left[ ((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b)) \right] + (1-c) \b \times \left( \b \times \left[ ((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b)) \right]\right ) \end{align} Ugh. Pretty ugly. Let's let $\u$ denote $\a \times (\a \times \b)$, and $\v$ denote $\b \times (\b \times \a)$. Then use the fact that $\mathbf{w \times w} = 0$ for any vector $\mathbf{w}$, repeatedly, and we get \begin{align} \mathbf{R_b}\mathbf{R_a}(\a+\b) &= \left[ (\a+\b) + s \a \times \b + (1-c)\u \right] + s \b \times \left[ (\a+\b) + s \a \times \b + (1-c)\u \right] + (1-c) \b \times \left( \b \times \left[ (\a+\b\ ) + s \a \times \b + (1-c)\u \right]\right ) \\ &= \left[ (\a+\b) + s \a \times \b + (1-c)\u \right] + \left[ (s \b \times (\a+\b) + s^2 (\b \times \a) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left( \b \times \left[ ((\a+\b) + s \a \times \b + (1-c)\u \right]\right ) \\ &= \left[\a+\b + s \a \times \b + (1-c)\u \right] + \left[ -s \a \times \b - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left( \b \times \left[ (\a+\b) + s \a \times \b + (1-c)\u \right]\right ) \end{align} And at this point, we have our first cancellation -- the second and fourth terms -- so we cancel them and further expand cross products in the latter terms: \begin{align} \mathbf{R_b}\mathbf{R_a}(\a+\b) &= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left( \b \times \left[ (\a+\b) + s \a \times \b + (1-c)\u \right]\right ) \\ &= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) + s \b \times (\a \times \b) + (1-c)\b \times\u \right] \\ &= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \b \times (\b \times \a) + (1-c)\b \times\u \right] \\ &= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \v + (1-c)\b \times\u \right] \\ &= \left[\a+\b + (1-c)\u \right] + \left[ + s^2 (\b \times \a) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \v + (1-c)\b \times\u \right] \\ &= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 \b \times (\b \times \a) + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \v + (1-c)\b \times\u \right] \\ &= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 \v + (1-c)s \b \times \u \right] + (1-c) \left[ \b \times(\b \times\a) - s \b \times\v + (1-c)\b \times (\b \times\u )\right] \\ &= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 \v + (1-c)s \b \times \u \right] + (1-c) \left[ \v - s \b \times\v + (1-c)\b \times (\b \times\u )\right] \\ \end{align}

Now it's a matter of figuring out simpler expressions for $\b \times \u$ and things like that. I suggest that you define $\a^\perp$ to be $\a \times (\a \times \b)$, which is a vector perpendicular to $\a$ in the plane spanned by $\a$ and $\b$, and define $\b^\perp$ similarly. You can then use these to simplify things like $\b \times \u$.

Processing each equation I write is taking longer and longer, and I can't bring myself to write out any more...but since this is your problem, I'll bet that you can do so.

This isn't a very elegant solution, I admit...but it'll get you there.