2
$\begingroup$

Let's consider some particular free group, for instance, the example $G = \langle A, B \rangle < \text{SL}(2, \mathbb{Z})$ in Wikipedia about ping-pong lemma, and some element $g \in G$.

Problem #1 is to compute the reduced word for this $g$, i.e. the corresponding sequence of factors $A$, $A^{-1}$, $B$, $B^{-1}$ (maybe it's the simple problem, but I can't figure out anything better then iterating over possible sequences).

Problem #2 assumes that we have also some $x$, for example $x \in \mathbb{R}^2$, and the action $g. x \in \mathbb{R}^2$, and we need to find reduced word for $g$ using this information (again, maybe it's easy to reduce problem 2 to problem 1, but I'm confused with this a little).

P.S. The application of this will be the following: given the $x, y \in \mathbb{R}^2$ find the $g \in G$ that might give $y$ by acting on $x$. It's like "rounding" to the nearest group element in embedding space $\mathcal{H}$ if we already embedded $G$ in $\mathcal{H}$, for example, using Švarz–Milnor lemma. So the natural requirement for the procedures in answers on problems 1 and 2 is their computational efficiency, because this "rounding" smust be performed as fast as possible.

$\endgroup$
2
  • 1
    $\begingroup$ This paper arxiv.org/abs/2110.02188 does some of what you want for groups similar to the one in Wiki but the parameters they consider don't include that group. $\endgroup$ Mar 26, 2022 at 13:24
  • $\begingroup$ This is one of those "well known" things that many people know how to do but that I don't know a good reference for. My suggestion would be to break this question into two parts. Think of $PSL_2(\mathbb{Z})$ (slightly easier to work with) as $\langle C\rangle*\langle D\rangle$ where $C$ is of order 2 and $D$ is of order 3. Now algorithmically write $g$ as a product of $C$'s and $D$'s. (This is basically just row reduction.) Next write $A$ and $B$ as products of $C$'s and $D$'s, and it will then be easy to write $g$ as a product of $A$'s and $B$'s. $\endgroup$
    – HJRW
    Mar 30, 2022 at 8:29

1 Answer 1

1
$\begingroup$

See the book, Kargapolov-Merzljakov "Fundamentals of group theory", section 14, "Free group". That group is called the "Sanov subgroup" there. The subgroup consists of all $2\times 2$ integral matrices $X$ with $\det(X)=1$, $x_{11}\equiv x_{22}\equiv 1 \pmod 4$, $x_{12}\equiv x_{21}\equiv 0\pmod 2$. On page 94 they give an algorithm to represent such a matrix as a product of generators $e_{12}(2), e_{21}(2)$.That representation is unique because the subgroup is free.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.