6
$\begingroup$

Is it possible to find two matrices $A$ and $B$, so that there does not exists a product of matrices $A$,$A^{-1}$,$B$,$B^{-1}$ that is equal to $Id$, under the condition that the product is irreducible, that is, it is is not trivial as a word (e.g. $AA^ {-1}B^{-1}B$). If there is no simple answer I would be happy with some references to where I should start.

Improve this question
$\endgroup$

3 Answers 3

Reset to default
14
$\begingroup$

If I understand correctly, you will find the answer in a good exposition of the Banach-Tarski paradox. Finding two rotation matrices in $\mathbb{R}^3$ that generate the free group in two generators is a usually the biggest part of the construction.

Improve this answer
$\endgroup$
2
11
$\begingroup$

As far as I understand, the OP asks about existence of free subgroups with 2 generators in matrix groups. The simplest example (if one does not require the matrices to be orthogonal which the OP did not) is probably the free group generated by the matrices $$\left[ \begin{array}{cc} 1 & 2 \\\ 0 & 1 \end{array} \right]$$ and $$\left[ \begin{array}{cc} 1 & 0 \\\ 2 & 1 \end{array} \right]\;,$$ see The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$)

Improve this answer
$\endgroup$
7
$\begingroup$

A brief comment, really, on R W's answer. Sanov's theorem states that the matrices $$\begin{pmatrix} 1 & a \\ 0 & 1\end{pmatrix}$$ and $$\begin{pmatrix}1 & 0 \\ a & 0\end{pmatrix}$$ generate a free group as long as $a\geq 2,$ though $a>2, a \in \mathbb{N}$ gives a free group of infinite index in $SL(2, \mathbb{Z}).$

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Actually, it is an interesting question (known as the "eye-problem'') which goes back to Lyndon and Ullman (1969) to describe those complex $a$ for which the group generated by these two parabolic elements is free. $\endgroup$
    – R W
    Commented Jan 18, 2018 at 18:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .