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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 Mar 30 '16 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 Mar 30 '16 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 Mar 30 '16 at 21:25
  • $\begingroup$ 1) I'm puzzled by the fact you say that the density of the subgroup $\langle A,B\rangle$ is obvious in your message, and that is seems hard in your comment. It's not trivial but it's not hard (each generates a dense subgroup of two "orthogonal" copies of $SO(3)$ and it's easy to conclude. $\endgroup$ – YCor Mar 30 '16 at 21:39
  • $\begingroup$ 2) "find product of $A,B$ that are very close to identity is vague: possibly you mean an element of the subgroup generated by $A$ and $B$ (then you should require a non-trivial element of the free group, or a nontrivial element, which is the same is $(A,B)$ is a free family. Or you mean, in the semigroup generated by $A$ and $B$, and the latter is also dense (by the same argument, since the positive powers of $A$ are dense in the corresponding copy of $SO(2)$, etc, so this is a reasonable question too. $\endgroup$ – YCor Mar 30 '16 at 21:43
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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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