<s>[INST] Composition of two rotations of the same angle $\alpha$ fixing points $a,b \in{S^2}$ 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.
 [/INST] 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. 
</s>