Which pairs of conjugates of $\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$ generate $\operatorname{SL}(2,\mathbb{Z})$?

Question

When do two distinct conjugates of $U := \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ generate $\DeclareMathOperator\SL{SL}\SL(2,\mathbb{Z})$? The classic example is $U,L^{-1}$, where $L = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.

Here I mean conjugate in $\SL(2,\mathbb{Z})$. Note that by Lee Mosher's answer to Conjugacy classes of $\operatorname{SL}_2(\mathbb Z)$, $U$ is not conjugate to its inverse in $\SL(2,\mathbb{Z})$.

I checked in GAP that there are examples of distinct conjugates which fail to generate. Is there a nice characterization of the pairs of conjugates which generate $\SL(2,\mathbb{Z})$?

This is related to the action of $\text{Aut}(\operatorname{SL}(2,\mathbb{Z}))$ on the set of conjugates of $U$. The above observation implies that this action is not quite 2-transitive. Is there a nice description of the orbits of $\text{Aut}(\operatorname{SL}(2,\mathbb{Z}))$ on the set of pairs of distinct conjugates of $U$, and the collection of such orbits which consist of pairs which generate?

Answer 1

$U$ and $L$ generate. Thus so do $AUA^{-1}$ and $ALA^{-1}$, for any element $A$ in $\mathrm{SL}(2, \mathbb{Z})$. These are the only pairs of conjugates which generate. The shortest proof I know is "read the discussion of groups generated by a pair of Dehn twists in the Primer on Mapping Class Groups".

Answer 2

Here's a proof of the criterion in the answer of Sam Nead.

Let $T$ be the Bass-Serre tree of the decomposition $SL(2,\mathbb Z) = \mathbb Z/3\mathbb Z * \mathbb Z / 2 \mathbb Z$. Let $U',L'$ be conjugates of $U,L$, and let $\mathcal A_{U'}$, $\mathcal A_{L'}$ be their axes in $T$.

There are three cases.

Case 1: $\mathcal A_{U'}=\mathcal A_{L'}$. Then $U'=L'$ and they don't generate.

Case 2: $\mathcal A_{U'} \cap \mathcal A_{L'}$ is the union of two edges of $T$ meeting at a $\mathbb Z / 2 \mathbb Z$ vertex. Then there exists $A$ such that $U'=AUA^{-1}$ and $L'=ALA^{-1}$ and $U',L'$ generate.

Case 3: $\mathcal A_{U'} \cap \mathcal A_{L'} = \emptyset$. Let $\sigma$ be the shortest segment in $T$ connecting $\mathcal A_{U'}$ to $\mathcal A_{L'}$. This segment $\sigma$ is a concatenation of some number $k \ge 1$ of paths each of which is a union of two edges meeting at a $\mathbb Z/2\mathbb Z$ vertex.

One can construct a fundamental domain $D' \subset T$ for the action of $\langle U',L' \rangle$ on its minimal subtree $T' \subset T$, consisting of the union of $\sigma$ with a fundamental domain of $U'$ in $\mathcal A_{U'}$ and a fundamental domain of $L'$ in $\mathcal A_{L'}$.

If $k \ge 2$ then using $D'$ one can see that $T'$ is a proper subtree of $T$ and so $\langle U',L' \rangle$ has infinite index in $SL(2,\mathbb Z)$.

If $k=1$ then $T'=T$, and $D'$ is a union of two fundamental domains of the action of $SL(2,\mathbb Z)$ on $T$, and so $\langle U',L' \rangle$ has index 2 in $SL(2,\mathbb Z)$