Centralizers of elements in $Sl_{2}(\mathbb{Z})$

Question

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 ?

Answer 1

This is a simple computation (write down your matrix, and see what it means for a two-by-two matrix to commute with it, or to normalize the subgroup). When doing the computation, it is useful to remember that two matrices commute if (and only if) they have the same eigenvectors.

EDIT In view of the fancy-pants arguments in the other answers, I do the computation for a parabolic in your notation $(1\ x)(0\ 1).$ Conjugating by $(a,\ b)(c,l\ d)$ gives you

$(1 - a c x,\ a^2 x)(-c^2 x, \ 1+ a c x).$ from the lower left corner we see that $c=0.$ That means (since we are in $SL(2, Z),$ that would not work in $SL(2, R)$) that either $a=d=1$ or $a=d=-1,$ so the normalizer in $PSL(2, Z)$ is just the subgroup itself, in $SL$ you throw in the center. The computation for elliptic (or hyperbolic) element is equally difficult.

Answer 2

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.

Answer 3

I am in a hurry, but I think you can simply work with $GL_2(\mathbb{Q})$ instead, then the rational canonical form (works over any field) gives you four types of conjugacy classes:

• central elements, these are the scalar multiples of the identity, the centralizer is the full group, the normalizer as well.
• hyperbolic elements, i.e. diagonal matrices, but not scalar matrices, their centralizers are just the diagonal matrices, their normalizer is described by the fact that normalizer modulo centralizer is the Weyl group, i.e. has order two.
• elliptic elements, those whose characteristic polynomial is irreducible (these will later correspond to finite order elements), their centralizer is equal to the invertible elements of the $F$-algebra generated by the element $\gamma$, which isomorphic to the splitting field of the characteristic olynomial. The normalizer modulo the centralizer corresponds to the Galois group of this extension (this matrix you will not find in $SL(2)$, since they are conjugated to matrices of the form $\left(\begin{smallmatrix} 1& 0 \newline 0 & -1 \end{smallmatrix} \right)$, i.e. have negative determinant, so in $SL(2)$ normalizer=centralizer for these elements).
• parabolic elements, i.e. a scalar matrix times a strict upper triagnular matrix, there centralizers are prescisely the strict upper triangular matrices, the normalizer is the Borel (upper tringular).

To get the corresponding results for $SL_2(\mathbb{Z})$ simply use the intersection.

Be careful, I am not claiming that any two elements in $SL_2(\mathbb{Z})$ are conjugated, if and only if they are over $GL_2(\mathbb{Q})$. There are more conjugacy classes in $SL_2(\mathbb{Z})$.