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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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  • $\begingroup$ proving density $\langle A,B \rangle = SO(3)$ seems hard $\endgroup$
    – john mangual
    Commented Mar 30, 2016 at 15:54
  • $\begingroup$ also... despite floating point error using decimals seems correct since if I just use rationals, I am saying implicitly $\mathbb{Q} \subset \mathbb{R}$ $\endgroup$
    – john mangual
    Commented Mar 30, 2016 at 15:56
  • $\begingroup$ Maybe I'm missing something here, but what's the problem with the implicit statement that $\mathbb{Q}\subset\mathbb{R}$? $\endgroup$
    – Steven Stadnicki
    Commented Mar 30, 2016 at 21:25
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    $\begingroup$ 3) Related: here's a survey (in French) by Sevennec about equirepartition of subgroups in $SO(3)$: umpa.ens-lyon.fr/sevennec/txt-lebesgue.ps $\endgroup$
    – YCor
    Commented Mar 30, 2016 at 22:40
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    $\begingroup$ Proving density isn't hard. The closure $\overline{\langle A,B \rangle}$ is a closed subgroup of $SO(3)$, and the complete listing of those is known; it is not too hard to rule out all options other than $SO(3)$ itself. $\endgroup$
    – David E Speyer
    Commented Mar 30, 2016 at 23:57

2 Answers 2

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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. I do not think this is best possible, but it is very explicit.

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Have you tried b^{-34}*a^{34}

It gets pretty close...
$b^{-34}*a^{34} = \left[ \begin{matrix} 0.9937 & 0.1119 & 0 \\ -0.1112 & 0.9875 & -0.1119 \\ -0.0125 & 0.1112 & 0.9937 \end{matrix} \right]$

in fact chaining b^-17*a^17*...*b^-17*a^17 an even number of times will get you closer while maintaining unit determinant...

$b^{-17}a^{17}b^{-17}a^{17}b^{-17}a^{17}b^{-17}a^{17}=$ $\left[ \begin{matrix} 1.0000 & 0.0002 & -0.0063 \\ -0.0002 & 1.0000 &-0.0002 \\ 0.0063 & 0.0002 & 1.0000 \end{matrix} \right]$

Are these results product of rounding errors?

No!

The geometric meaning is that $17 \cdot \arccos(3/5) \approx 5\pi$, so $a^{17}$ is like rotating two and a half times around the $\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right]$ axis, and $a^{34}$ is 5 way round, so you obtain the identity (approximately).

You could chase down the decimals after $\arccos(3/5)$ and get so close to identity as you wish. And the same goes with b. The question about combinations of powers of multiples other than 17 is far more difficult.

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