2
$\begingroup$

Does a finite set of non-coaxial irrational rotations in 3d form a free group? That is, if I have $n$ irrational rotations all around different axes and I take non-trivial combinations of these rotations and their inverses is it possible to arrive at the identity?

In 2 dimensions the answer is no, because all the rotations commute the sequence $R(\alpha_1)R(\alpha_2)R(-\alpha_1)R(-\alpha_2)=Id$. Moreover if we have two non-coaxial irrational rotations as the generators we will also have a free group.

1 Answers 1

1

Given a pair of non-coaxial irrational rotations, $a,b$, let $c=ab$. $c$ is a rotation, which is also not coaxial with $a$ or $b$. If $c$ is a rational rotation, then $(ab)^n=1$. If $c$ is an irrational rotation, then the trio of irrational rotations $a,b,c$ satisfies $abc^{-1}=1$.

[As commenter Tobias notes below, this second case does not show that the group generated by $a,b,c$ is not free, only that the group is not free on $a,b,c$. I took from the language of the question that the latter is what was wanted. ]

There might be a way to prove that, given $n$, there exists a set of $n$ rotations with no relations between them.

Actually, you only need to show $n=2$.

Once you have a free group on two elements, $x,y$, you have a free group on $n$ elements for any $n$ by taking a subgroup, so you really just need to find two non-coaxial irrational rotations with no relations, $a,b$, to find $n$ such, by taking $a,bab^{-1},b^2ab^{-2},\dots,b^{n-1}ab^{-(n-1)}$.

  • 0
    The relation $abc^{-1} = 1$ does not rule out the group being free. – 2017-01-19
  • 0
    yeah, I took the later language "... and I take non-trivial combinations of these rotations and their inverses is it possible to arrive at the identity?" to mean the question wanted to know if the group was free on the $n$ elements, not just free. But it is possible for this group to be free, yes. I do think we can easily construct $a,b$ so that $ab$ is rational, however. @TobiasKildetoft – 2017-01-19