Conjugacy classes of $\mathrm{SL}_2(\mathbb{Z})$

I was wondering if there is some description known for the conjugacy classes of $$\mathrm{SL}_2(\mathbb{Z})=\{A\in \mathrm{GL}_2(\mathbb{Z})|\;\;|\det(A)|=1\}.$$ I was not able to find anything about this. Most references only give solutions for $$\mathrm{SL}_2(\mathbb{R})$$.

• There is plenty of literature about conjugacy within ${\rm GL}(2,\mathbb{Q}),$ which is essentially the theory of the rational canonical form (in a special case). – Geoff Robinson Apr 13 '16 at 21:40
• – Steve Huntsman Apr 14 '16 at 12:20
• A reasonable follow-up question might be whether there's any relation to geometry, e.g. via the action of $\text{SL}_2(\mathbf{Z})$ on the upper half plane. – Meow 23 hours ago

One can proceed as follows for $$SL_2(\mathbb{Z})$$.

1. First, the trace is a conjugacy invariant.
2. For trace $$0$$ there are two conjugacy classes represented by $$\pmatrix{0 & 1 \\ -1 & 0}$$ and $$\pmatrix{0 & -1 \\ 1 & 0}$$. These representatives can be thought of as $$90^\circ$$ and $$270^{\circ}$$ degree rotations of a lattice generated by the corners of a square centered on the origin.
3. For trace $$1$$ and $$-1$$ there are two conjugacy classes each, represented by the matrices $$M=\pmatrix{1 & -1 \\ 1 & 0}, M^2=\pmatrix{0 & -1 \\ 1 & -1}, M^4=\pmatrix{-1 & 1 \\ -1 & 0}, M^5 = \pmatrix{0 & 1 \\ -1 & 1}$$ These representatives can be thought of as $$60^\circ$$, $$120^\circ$$, $$240^\circ$$, and $$300^\circ$$ degree rotations of a lattice generated by the vertices of a regular hexagon centered at the origin.
4. For trace $$2$$ there is a $$\mathbb{Z}$$-indexed family of conjugacy classes, represented by $$\pmatrix{1 & n \\ 0 & 1}$$; these are all "shear" transformations except for the identity. For trace $$-2$$ there is a similar $$\mathbb{Z}$$-indexed family of conjugacy classes represented by $$\pmatrix{-1 & n \\ 0 & -1}$$.
5. In general, for nonzero trace the conjugacy classes come in opposite pairs, represented by a matrix $$M$$ with trace $$t>0$$ and an opposite representative $$-M$$ with trace $$-t<0$$.
6. For trace of absolute value $$> 2$$, there is one conjugacy class for each word of the form $$\pm R^{j_1} L^{k_1} R^{j_2} L^{k_2} \cdots R^{j_I} L^{k_I}$$ up to cyclic conjugacy, where $$I \ge 1$$ and all the exponents are positive integers. A matrix representing this form is obtained from the above word by making the replacements $$R=\pmatrix{1 & 1 \\ 0 & 1}, \quad L=\pmatrix{1 & 0 \\ 1 & 1}$$ The transformations represented by such words are all "hyperbolic" transformations, having an independent pair of real eigenvectors. The slope of the expanding eigenvector is a quadratic irrational, and hence has eventually repeating continued fraction expansion. The cyclic sequence $$(j_1,k_1,j_2,k_2,\ldots,j_I,k_I)$$ can be thought of as the fundamental repeating portion of the continued fraction expansion of the slope of the expanding eigenvector, or, better, as an appropriate power of the fundamental repeating portion where the power is equal to the exponent of the given matrix.

Number theorists will tell you that the number of conjugacy classes of each trace $$t>2$$ is closely related to the class number of the number field generated by $$\sqrt{t^2-4}$$.

• Not the square root of the trace, but rather $\sqrt{t^2-4}$, where $t$ is the trace. – David E Speyer Apr 14 '16 at 2:52
• @LeeMosher Nice answer. Do you have reference for this? For the case 6, you probably have to say that the conjugacy class is of the form $\pm R^{j_1}L^{k_1}\cdots R^{j_I}L^{k_I}$. Indeed, if the trace is negative, you cannot be conjugate to a product of $R,L$. – Jérémy Blanc Mar 11 '17 at 6:28
• Your calculations of $M^2$ and $M^5$ in step 3 are incorrect: the non-diagonal entries have the wrong signs. Your (incorrect) $M^2$ can be conjugated by $U = (\begin{smallmatrix}1&0\\1&1\end{smallmatrix})$ to your (correct) $M^3$ and your (incorrect) $M^5$ can be conjugated by $U$ to your (correct) $M$. – KConrad yesterday
• Ah, how embarassing. Thanks for catching that. – Lee Mosher yesterday
• The $M^4$ had a sign error in one entry (giving the matrix determinant $-1$ instead of $1$) so I fixed that and rechecked by hand that the powers of $M$ are all as indicated. – KConrad 23 hours ago

This is the subject of Gauss' reduction theory, as discussed in Karpenkov's book (among many other places). In this 2007 paper, Karpenkov also extends the method to $SL(n, \mathbb{Z}).$

The conjugacy classes of elements of ${\rm SL}(2,\mathbb{Z})$ with given trace are counted in:

S. Chowla, J. Cowles and M. Cowles: On the number of conjugacy classes in SL(2,Z). Journal of Number Theory 12(1980), Issue 3, Pages 372-377.

• Extraordinary: the paper which has Chowla as an author, was also communicated by Chowla! – Lucia Apr 13 '16 at 22:29