# Simultaneous rotations

I've been puzzled by this question for a while now (something I stumbled upon myself):

Say I have two 2-spheres $S^2$. For the $k$-th sphere ($k \in \{ 0,1 \}$), I define two axes: $$a_k = \hat{z} \\ b_k = \sin\left(\frac{\pi(k+1/2)}{2}\right) \hat{y} + \cos\left(\frac{\pi(k+1/2)}{2}\right)\hat{z}$$ Denote an $SO(3)$ rotation around $a_k$ by angle $x$ (with some convention about handedness) by $R_{a_k}(x)$, and similarly for $b_k$.

For each sphere, I rotate a point $v_i^{(k)} = (0,0,1)$ to a new point $v_f$ according to $$v_f^{(k)}(x_1,x_2,x_3,x_4) = R_{a_k}(x_4) R_{b_k}(x_3) R_{a_k}(x_2) R_{b_k}(x_1) v_i^{(k)}$$ where the angles $(x_1,x_2,x_3,x_4) \in [0,2\pi)^4$ are the same for both sphere (hence the title simultaneous rotations, just that the rotations are around different fixed axes for each sphere)

Now, after playing around with this a bit, I feel like under this map, there always exists a (non-unique) set of angles such that I can get to any point on $(S^2)^2$.. that is, the map $( v_j^{(1)}, v_j^{(2)})(x_1,x_2,x_3,x_4)$ is surjective.

How do I go about showing that? Intuitively by drawing some pictures, it seems true but I struggle to wrap my head around showing it properly..

Consider the pair $(-\hat{z}, \hat{z})$ in $(S^2)^2$. I claim that no choice of $(x_1,x_2,x_3,x_4)$ will get you there. (Observe that $R_{a_k}(x)$ fixes both $-\hat{z}$ and $\hat{z}$, so we may ignore the final of the four rotations: $R_{a_k}(x_4)R_{b_k}(x_3)R_{a_k}(x_2)R_{b_k}(x_1)v^{(k)}_i = \pm \hat{z} = \pm R_{a_k}(-x_4)\hat{z} = R_{b_k}(x_3)R_{a_k}(x_2)R_{b_k}(x_1)v^{(k)}_i$)
Up to the irrelevant $x_4$, there is a unique choice, $(x_1,x_2,x_3,x_4) = (\pi,\pi,\pi,x_4)$ so that $v^{(1)}(x_1,x_2,x_3,x_4) = -\hat{z}$, which is seen by working backwards (you'll probably want to draw yourself the picture to make sense of following): the possible vectors that could rotate through $x_3$ around $b_1$ to land on $-\hat{z}$ comprise a circle in the lower hemisphere of $S^2$. Then the vectors that could rotate around $a_1$ to land on this circle comprise the entire lower hemisphere (including the equator). Finally, the only rotation of $(0,0,1)$ about $b_1$ that lands in the lower equator is by the angle of $\pi$. Working forward now, it follows that $x_2 = \pi$ and $x_3 = \pi$.
On the other hand, clearly $v^{(1)}(\pi,\pi,\pi,x_4) = -\hat{z}$ for all choices of $x_4$, so we see that $(-\hat{z}, \hat{z})$ is not in the image.