Let $A,B$ be two rational rotations:


$$ A = \left[\begin{array}{rcc} \frac{3}{5} & \frac{4}{5} & 0 \\
-\frac{4}{5} & \frac{3}{5} & 0 \\
0 & 0 & 1 \end{array}\right] 
\quad\text{ and }\quad
B = \left[\begin{array}{crc} 
1 & 0 & 0 \\ 
0 & \frac{3}{5} & \frac{4}{5} \\
0 & -\frac{4}{5} & \frac{3}{5} \end{array}\right]
$$


Then we build the free group $\mathbb{F}\_2 = \langle A, B \rangle$ of all possible products of $A, B, A^{-1}, B^{-1}$. Or I think since we chose $A,B$ to be rotations, we have a **homomorphism**


$$ \phi: \mathbb{F}\_2 \to SO(3)$$


just by evaluating the product. The image $\langle A,B \rangle$ is obviously dense in $SO(3)$, but how much work is it to find products of $A$ and $B$ that are very close to the identity as a function of the word length $n$?


$SO(3)$ only has [a few](http://www.cs.cornell.edu/~ginsparg/physics/phys446-546/disc3d.pdf) discrete subgroups. Proving that $\overline{\langle A,B \rangle} = SO(3)$ seems to be a matter of the pigeonhole principle or maybe showing this group is dense [Zariski topology](https://ncatlab.org/nlab/show/Zariski+site).


Kaloshin and Rodnianski suggest this is always the case for pretty much any collecton of rotations in [Diophantine Properties of $SO(3)$](http://www.its.caltech.edu/~kaloshin/research/dioph-SO3.pdf). I am wondering if the argument simplifies for any specific choice of group elements.


On my computer this morning I was able to find:


$$ AB^{-1}A^{-3}B^{-1}A^2 =
\left[\begin{array}{rrr}
 0.98189568 & -0.12261376 & 0.144384 \\
 0.15538176 & 0.95731968 & -0.243712 \\
 -0.10833920 & 0.26173440 & 0.959040 \end{array}\right]$$


which is not that close to the identity matrix, but it's the best one I can find with 8 letters.


Maybe there's a "cheap trick" that works here. I would especially love that. This seemed like such an easy problem, but all the solutions I found are quite hard.



Can two rational rotations $F_2 = \langle A, B \rangle \to SO(3)$ efficiently approximate the $3 \times 3$ identity matrix?
Setting $a\_0=A^7, b\_0=B^7$ and then $a\_{n+1}=[b\_n^{-1},a\_n], b\_{n+1}=[a\_n,b\_n]$ seems very efficient. The length grows like $C \alpha^n$ with $\alpha= \frac{3 + \sqrt{17}}2$ and the operator norm distance to the identity matrix gets squared in each step, since
$$\|1 - [a,b]\| \leq 2 \|1-a\| \cdot \|1-b\|.$$ I used a similar construction in [On the length of the shortest non-trivial element in the derived and the lower central series](http://arxiv.org/abs/1311.0138). I do not think this is best possible, but it is very explicit.


