TITLE: Centralizers  of  elements  in $Sl_{2}(\mathbb{Z})$
QUESTION [5 upvotes]: Conjugacy  classes  Of  elements  of  $SL_{2}(\mathbb{Z})$  can be  characterized  by  its  trace. If  $\mid T(m)\mid=1 $, then it  is  an  element  of  finite  order,  if  $\mid T(m)\mid =2$, then m is  conjugate  to  a  matrix up to  sign,$ (1 n)  (0  1)$ (  row  notation)  where $ n $ a  natural  number.  in other  case, there  is  an  expansion   in words $ML^{n_{i}} MR^{n_{i}}$,   where  ML is  the  matrix $(10)(11)$,  and  MR is the  matrix $(01)(11)$ (Again,  row  notation).  Is  there  a  Classification or  method for  the  determination of the  centralizers  of  these  elements, or  the  Normailizers of  the  subgroups generated  by  them ?

REPLY [3 votes]: A good way to visualize this issue is by looking at the fractional linear action of $SL(2,\bf Z)$ on the upper half plane model of $\bf H^2$; see section 4.2 of Serre's 
"Trees". The kernel of the action is the center of the group, ${\bf Z}/2$ generated by minus the identity matrix. I'll ignore the center henceforth and describe what happens in its quotient $PSL(2,{\bf Z})$. 
The action of $PSL(2,{\bf Z})$ preserves the Farey triangulation (a vertex at each rational point at infinity, an edge connecting $p/q$ to $r/s$ if $ps-rq=\pm 1$). The dual graph of the Farey triangulation is an infinite trivalent tree $T$ embedded in $\bf H^2$. The action of $PSL(2,\bf Z)$ on $T$ is the same as the Bass-Serre tree for the free product decomposition $PSL(2,{\bf Z}) = {\bf Z}/3 * {\bf Z}/2$. An element is elliptic if and only if it fixes a vertex or an edge midpoint (where an edge of $T$ crosses an edge of the Farey triangulation) and its normalizer and centralizer are the stabilizer of that point, ${\bf Z}/3$ for a vertex and ${\bf Z}/2$ for an edge midpoint. An element is parabolic if and only if it fixes a rational point at infinity and its normalizer and centralizer are the stabilizer of that point, isomorphic to ${\bf Z}$; e.g. for the rational point $1/0$ the stabilizer is the cyclic group generated by $(1 ,n)(0 ,1)$ (row notation). 
A hyperbolic element translates along its axis, a line $A$ in $T$. The circular word $ML^{n_1} MR^{n_2} ... ML^{n_{2k-1}} MR^{n_{2k}}$ describes the behavior of a fundamental domain of $A$, making choices of Left and Right at each valence 3 vertex in $T$. The centralizer is the infinite cyclic group generated by the primitive translation of $A$, represented by the primitive cycle of the circular word $ML^{n_1} MR^{n_2} .. ML^{n_{2k-1}} MR^{n_{2k}}$. The normalizer may have one more order 2 generator, and here's how you tell: are the choices made by $A$ going in the forward direction the same, up to cyclic permutation, as the choices made going in the backward direction? To put it another way, reverse the cyclic word and interchange $L$ with $R$, obtaining $ML^{n_{2k}} MR^{n_{2k-1}} ... ML^{n_2} MR^{n_1}$ and ask: is the result a cyclic permutation of the original word? If not, the normalizer is the centralizer; if so, the centralizer has index 2 in the normalizer.