Finite subgroups of SL_2(Z) (reference request) - MathOverflow

Finite subgroups of SL_2(Z) (reference request)

2010-06-06T18:05:24Z

I recently read the statement "up to conjugacy there are 4 nontrivial finite subgroups of $SL_2(\mathbb{Z})$." They are generated by

$\left(\begin{array}{cc} -1&0 \\ 0&-1\end{array}\right), \left(\begin{array}{cc} -1&-1 \\ 1&0\end{array}\right), \left(\begin{array}{cc} 0&-1 \\ 1&0\end{array}\right), \left(\begin{array}{cc} 0&-1 \\ 1&1\end{array}\right) $ and are isomorphic to Z_2, Z_3, Z_4, and Z_6, respectively. Does someone know a reference for this statement? (Or, is it easy to see?) My attempt at a google search turned up this statement, but I wasn't able to find a reference.

Answer by Robin Chapman for Finite subgroups of SL_2(Z) (reference request)

2010-06-06T18:19:09Z

Each finite subgroup $G$ of $GL_2(\mathbb{Z})$ is a finite subgroup of $GL_2(\mathbb{R})$. Taking a positive definite form on $\mathbb{R}^2$ and averaging by the action of $G$ gives a positive definite form invariant under $G$. This implies that $G$ is conjugate in $GL_2(\mathbb{Z})$ to a subgroup of $O(2)$. All such groups are cyclic or dihedral.

A finite subgroup of $GL_2(\mathbb{Z})$ corresponds to a finite subgroup of $O(2)$ leaving a lattice invariant. With respect to generators of this lattice the matrix of a group element has integer coefficients. As it is a rotation or reflection matrix, its trace is $2\cos\theta$ (where $\theta$ is the angle of rotation) or zero (if it is a reflection). The trace must be an integer so it is $2$, $1$, $0$, $-1$ or $-2$, corresponding to rotations of order $1$, $6$, $4$, $3$ or $2$. These are the only possible orders of rotation possible (the crystallographic restriction) and this leads to the classification you cite.

See http://en.wikipedia.org/wiki/Crystallographic_restriction_theorem .

Answer by Torsten Ekedahl for Finite subgroups of SL_2(Z) (reference request)

2010-06-06T19:17:23Z

I interpret your statement as being concerned with conjugation in $\mathrm{SL}_2(\mathbb Z)$. In that case I think that the arguments given only give that the groups are cyclic of order $1$, $2$, $3$, $4$ or $6$ not that they are unique up to conjugacy. For this latter fact one need only consider order $1$, $3$ or $4$ as the others are obtained by multplying a generator by $-E$. The case of order $1$ is trivial and that of order $3$ or $4$ gives a module of rank $1$ over the ring of $3$'rd and $4$'th roots of unity and then the statement is equivalent to these rings having class number $1$.

Addendum: I was a little it sketchy as Victor pointed out. In the present case all the relevant representations of $\mathbb Z[\mathbb Z/p]$factor through $\mathbb Z[\zeta_p]$ which is seen by looking at the characteristic polynomial and hence one doesn't have to look at the more general representations. In higher ranks I certainly agree with Victor about the need for the full ring. As for $\mathrm{GL}_2(\mathbb Z)$ versus $\mathrm{SL}_2(\mathbb Z)$-conjugacy I think that is taken care of by the fact that complex conjugation acts trivially on the class groups (which are trivial) and hence there is an automorphism of determinant $-1$ of the modules in question.

Answer by Allen Knutson for Finite subgroups of SL_2(Z) (reference request)

2010-06-06T23:40:09Z

A finite subgroup will map to a finite subgroup of $PSL_2(\mathbb Z)$, which is a free product $Z_2 * Z_3$. I believe I have been told that finite subgroups of free products of finite groups are conjugate to subgroups of the factors being free-producted together.

Answer by Roland Bacher for Finite subgroups of SL_2(Z) (reference request)

2010-06-07T07:43:50Z

A slightly different argument showing that every finite subgroup of $SL_2(\mathbb Z)$ is of cardinality a divisor of $24$ goes as follows: Consider such a finite subgroup $H$. Since the coefficients of all elements of $H$ involve only a finite number of prime divisors, the obvious group homomorphism from $SL_2(\mathbb Z)$ into $SL_2(\mathbb F_p)$ where $\mathbb F_p$ is the finite field with cardinality a prime number $p$ is injective for almost all primes. Since $SL_2(\mathbb F_p)$ has $p(p^2-1)$ elements, the cardinality $h$ of the finite group $H$ divides $p(p^2-1)$ for almost all prime numbers. This implies that $h$ divides $24$. Indeed, quadratic reciprocity shows that $2$ and $3$ are the only possible prime divisors of $h$ and gives upper bounds on the maximal exponents $\alpha,\beta$ such that $2^\alpha\cdot 3^\beta$ divides $p(p^2-1)$ almost all primes.

I intended to post the following as a comment but it is too long:

A very easy argument showing that every prime $p>n+1$ works for the injectivity of the reduction modulo $p$ of a finite subgroup $H$ of $SL_n(\mathbb Z)$ is as follows: Since every element of $H$ is finite, its characteristic polynomial is a product of cyclotomic polynomials. Reductions of cyclotomic polynomials modulo $p$ with order (defined as the order of an underlying root of unity in the multiplicative group of invertible elements in $\mathbb C$) prime to $p$ are never congruent to a power of $(1-x)$. Cyclotomic polynomials of order divisible by $p$ are of degree at least $p-1$. Since a non-trivial element $h\in H$ is diagonalisable, its characteristic polynomial is not a power of $(1-x)$. The reduction of this characteristic polynomial modulo $p$ is thus also distinct from a power of $(1-x)$. This implies that $h$ modulo $p$ is non-trivial in $SL_n(\mathbb F_p)$.

Answer by james-parson for Finite subgroups of SL_2(Z) (reference request)

2010-06-07T15:04:50Z

Perhaps the following explicit calculations clarify Ekedahl's argument: consider, for example, the case where the group is cyclic of order 3. Let $w$ be a generator of the group. The characteristic polynomial of w is then necessarily $X^2 + X + 1$. Thus the action of $w$ on $M = \mathbf{Z}\oplus\mathbf{Z}$ makes $M$ into a $\mathbf{Z}[X]/(X^2 + X + 1)$-module. This module is torsion free (since it is torsion-free as a $\mathbf{Z}$-module). Consequently, it is locally free (necessarily of rank $1$) and hence free since $\mathbf{Z}[X]/(X^2+X+1)$ is a PID.

The freeness means that one can find a module generator $m\in M$. Since $(-X,1)$ is a $\mathbf{Z}$-module basis for $\mathbf{Z}[X]/(X^2+X+1)$, one finds that that $(-wm,m)$ is a $\mathbf{Z}$-module basis for $M$.

If the basis $(-wm,m)$ has the same orientation as the standard basis for $M$, then the change-of-basis matrix to $(-wm,m)$ is in ${\rm SL}_2(\mathbf{Z})$. Since the matrix of $w$ with respect to $(-wm,m)$ is the standard matrix of the original problem statement, this construction finishes the argument in this case.

If, on the other hand, the basis $(-wm,m)$ has the opposite orientation from the standard basis for $M$, then $(-w^2 m, m)$ has the same orientation as the standard basis of $M$ (algebra omitted). The change-of-basis matrix to $(-w^2 m, m)$ is thus in ${\rm SL}_2(\mathbf{Z})$. The matrix of $w^2$ with respect to $(-w^2 m, m)$ is the standard matrix of the original problem statement, which finishes the argument in this case.

Let me state this last part of the calculation abstractly: since the question involves ${\rm SL}_2(\mathbf{Z})$ conjugacy, one should consider locally free $\mathbf{Z}[X]/(X^2+X+1)$-modules of rank $1$ equipped with an orientation. We use the orientation $(-X,1)$ on $\mathbf{Z}[X]/(X^2+X+1)$. Given such a module $N$, we can obtain a new conjugate module $N'$ by changing the $\mathbf{Z}[X]/(X^2+X+1)$ action by the (orientation-reversing) automorphism $X\mapsto X^2$. For any oriented $N$, either $N$ or $N'$ is module-isomorphic to $\mathbf{Z}[X]/(X^2+X+1)$ (with orientation). Corresponding to these two cases, one can conjugate (in ${\rm SL}_2(\mathbf{Z})$) either $w$ or $w^2$ to the desired standard form.

Note that it is straightforward to find a module generator $m$ (and hence to find a standardizing basis for $M$): the function $$ q(m):m\mapsto \det(-wm \; m) $$ is a quadratic form on $\mathbf{Z}\oplus\mathbf{Z}$. It is either positive definite or negative definite. (One can see that easily in the abstract using a module generator for $M$.) An $m\in M$ is a module generator precisely when $q(m) = \pm 1$. We know that such an $m$ exists, and since $q$ is definite, it is easy to search for one.