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 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.

Kaloshin and Rodnianski suggest this is always the case for pretty much any collecton of rotations in Diophantine Properties of $SO(3)$. 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.

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