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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 $\mathrm{SL}_2(\mathbb{R})$.

Thank you for your help.

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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 $$R^{j_1} L^{k_1} R^{j_2} L^{k_2} \cdots R^{j_I} L^{k_I} $$ 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 $$R=\pmatrix{1 & 1 \\ 0 & 1}, \quad L=\pmatrix{1 & 0 \\ 1 & 1} $$ These 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 $\sqrt{t^2-4}$.

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    $\begingroup$ Not the square root of the trace, but rather $\sqrt{t^2-4}$, where $t$ is the trace. $\endgroup$ – David E Speyer Apr 14 '16 at 2:52
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    $\begingroup$ @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 $R,L$. $\endgroup$ – Jérémy Blanc Mar 11 '17 at 6:28
  • $\begingroup$ I don't have a specific reference. I learned it over a long process of ingestion, maybe starting with my first book on continued fractions as a kid, and continuing with my education in mapping class groups ($PSL_2(\mathbb{Z})$ is the mapping class group of the torus). The answer of @IgorRivin answer gives some references. $\endgroup$ – Lee Mosher Mar 11 '17 at 13:57
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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 $SL(n, \mathbb{Z}).$

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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.

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

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