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\centerline{\BF The Modular Group}
\bigskip
\bigskip
\head 1. The Modular Group
\endhead

\noindent
The {\it modular group\/} $\Gamma$ is the set of all $2\times 2$ matrices
with integral elements and determinant 1.  That is, $\Gamma$ is the 
special linear group of $2\times 2$ matrices over the integers,
$\Gamma = \text{SL}(2, \Bbb Z)$.  It forms a group under
matrix multiplication.  If
$$
M = \bmatrix a & b \\ c & d \endbmatrix \in\Gamma
$$
Then $M$ defines a map
$$
f_M(z) = \frac{az+b}{cz+d}
$$
of the extended complex plane to itself.  Here $f_M(-d/c) = \infty$,
and $f_M(\infty) = a/c$.  Suppose that
$N = \bmatrix \alpha & \beta \\ \gamma & \delta \endbmatrix$.  Then
by direct calculation we find that
$$
f_N(f_M(z)) = \frac{(\alpha a + \beta c)z + (\alpha b + \beta d)}
                   {(\gamma a + \delta c)z + (\gamma b + \delta d)}
            = f_{NM}(z).
$$
While it might seem surprising that the composition of one such rational
function with another would be connected with matrix multiplication,
the mystery can be dispelled by considering how $M$ and $N$ transform
the 2-dimensional vectors $ \Bbb C^2$.  Suppose that
$$
\xalignat2
M\bmatrix z_1\\z_2\endbmatrix &= \bmatrix w_1 \\ w_2\endbmatrix, &
N\bmatrix w_1\\w_2\endbmatrix &= \bmatrix t_1 \\ t_2\endbmatrix.
\endxalignat
$$
Then
$$
NM\bmatrix z_1 \\ z_2 \endbmatrix = \bmatrix t_1 \\ t_2\endbmatrix.
$$
Now suppose we consider these vectors in terms of projective geometry.
Two vectors are then considered to be the same if their coordinates are
proportional (i.e., the vectors are colinear).  In other words, if $c$
is a non-zero complex number, then $\big[{z_1\atop z_2}\big]$ and
$\big[{cz_1\atop cz_2}\big]$ are considered to be the same.  Thus
a projective point $z_1 : z_2$ is determined by the ratio $z = z_1/z_2$ if
its coordinates, and the image $w_1 : w_2$ is determined by the ratio
$w = w_1/w_2$ of its coordinates.  But then
$$
w = \frac{w_1}{w_2} = \frac{az_1 + bz_2}{cz_1 + dz_2}
= \frac{az_1/z_2 + b}{ cz_1/z_2 + d} = \frac{az+b}{cz+d},
$$
so the map from $z$ to $w$ reflects a linear transformation in projective
coordinates.

     Let $\eusm H$ denote the upper half-plane of the complex plane,
$\eusm H = \{ z\in \Bbb C : \Im z > 0\}$.  Write $z = x+iy$.  Then
$$
w = \frac{az+b}{cz+d} = \frac{az+b}{cz+d}\,
\frac{c\overline z + d}{c\overline z + d}
= \frac{ac(x^2+y^2) + (bc+ad)x + bd}{(cx+d)^2 + (cy)^2}
+ i\frac{(ad-bc)y}{(cx+d)^2 + (cy)^2}.
$$
Since $ad-bc = 1$, we deduce that $w\in\eusm H$ if and only if $z\in\eusm H$.
We call two points $z\in\eusm H$ and $w\in\eusm H$ {\it equivalent\/} if
there is an $M\in\Gamma$ such that $w = f_M(z)$.  Since the identity matrix
$I$ takes $z$ to itself, it follows that $z \sim z$. If $M$ takes $z$ to $w$,
then $M^{-1}$ takes $w$ to $z$.  That is, $z \sim w$ implies that $w\sim z$.
Finally, if $z\sim w$ and $w\sim t$, then $NM$ takes $z$ to $t$, so that
$z\sim t$.  Hence this is an equivalence relation, and the upper half-plane
$\eusm H$ is partitioned into equivalence classes.  In analogy with complete
systems of residues (mod $m$), we say that a set $\eusm S \subseteq \eusm H$
is a {\it fundamental region\/} of $\Gamma$ if for every $z\in\eusm H$ there
is exactly one $w\in\eusm S$ such that $z\sim w$.

\proclaim{Theorem 1}Let $\eusm R$ be the set of those $z = x+iy\in\eusm H$
such that either $-1/2 \le x < 1/2$ and $|z| > 1$ or else $-1/2 \le x \le 0$
and $|z| = 1$.  Then $\eusm R$ is a fundamental region for $\Gamma$.
\endproclaim
\headline{\hfil\bf The modular group\hfil}

\demo{Proof}We show first that if $z\in \eusm H$, then there is a
$w\in \eusm R$ that is equivalent to $z$.
Let $S = \bmatrix 0 & -1 \\ 1 & 0\endbmatrix$ and
$T = \bmatrix 1 & 1 \\ 0 & 1\endbmatrix$, and note that
$$
\align
S\eusm R &= \{z\in\eusm H : |z+1| > 1, |z - 1| \ge 1, |z| < 1\}
\cup \{z\in\eusm H : |z| = 1, 0\le x \le 1/2\}, \\
ST^{-1}\eusm R &= \{z\in\eusm H : |z-1| < 1, |z - 1/3| \ge 1/3, x < 1/2\} \\
&\qquad\cup \{z\in\eusm H : x = 1/2, \sqrt 3/6\le y \le 1/2\}, \\
ST \eusm R &= \{z\in\eusm H : |z+1|\le 1, |z + 1/3| > 1/3, x > -1/2\} \\
&\qquad\cup \{z\in\eusm H : x = -1/2, 1/2 \le y \le \sqrt 3/2\}.
\endalign
$$
Suppose that $z\in\eusm H$.  Choose an integer $m$ so that $w = z + m$
satisfies $-1/2 \le \Re w < 1/2$.  If $w\in \eusm R$, then we are done.
If $w \in ST^{-1}\eusm R$, then $TSw\in \eusm R$, and we are done.  If
$w\in ST\eusm R$, then $T^{-1}Sw\in\eusm R$, and we are done.  Otherwise,
$|w|\le 1/\sqrt 3$, so that $|Sw| \ge 3|w|$.  In this case we begin again
with $Sw$.  Since the imaginary part increases by a factor of at least 3
upon each repetition, the process eventually terminates with an equivalent
member of $\eusm R$.

     To complete the proof we show that no member of $\eusm R$ is equivalent
to a different member of $\eusm R$.  Suppose that $z\in\eusm R$, and that
$M$ takes $z$ to $w \in\eusm R$.  If $c = 0$, then the condition
det($M$) = 1 implies that $ad = 1$, so that without loss of generality
$a = c = 1$.  Then $w = z + b$, and so the condition
$-1/2 \le \Re w < 1/2$ implies that $b = 0$.  That is, $M = I$, $z = w$.
Now suppose that $c \ne 0$.  By direct calculation we find that
$$
cw - a = \frac{-1}{cz + d}.
$$
Thus if $|cz+d| > 1$, then $|cw - a| < 1$.
Suppose that $|c| > 1$.  Then $z\in\eusm R$ implies that
$|cz + d| > 1$, which implies that $|cw - a| < 1$, which implies that
$w\notin \eusm R$.  Suppose that $|c| = 1$.  By replacing $M$ by $-M$ we may 
suppose that $c = 1$.  The disc $|z+d|< 1$ does not intersect $\eusm R$,
and the closed disc $|z+d|\le 1$ has non-empty intersection with
$\eusm R$ only when $d = 0$ or $d = 1$.  Similarly, the disc $|w-a| < 1$ does not
intersect $\eusm R$, and the closed disc $|w-a|\le 1$ has
non-empty intersection with $\eusm R$ only when $a = 0$ or $a = -1$.  Hence if
$z\in \eusm R$ and $w\in\eusm R$, then $|z+d|=1$ and $|w-a|=1$, and we have
four cases:\hfil\break
1. $a = d = 0$.  In order that $\operatorname{det}(M)=1$, we must have $b=-1$, which
is to say that $M=S$ and $f(z)=-1/z$.  But $\Re z < 0$ implies $\Re w > 0$, and then
$w\notin \eusm R$, since $|w|=1$.  If $z = i$, then $w = i$, so $z=w$, even though
$M\not= \pm I$.\hfil\break
2. $a = 0$, $d = 1$.  Since $\operatorname{det}(M)=1$, we have $b = -1$.  The only
point $z\in\eusm R$ such that $|z+1|=1$ is $z = \rho = -1/2+i\sqrt 3/2$.  But
$f(\rho)= -1/(\rho+1)= \rho$, so again we have $w = z$.\hfil\break
3. $a = -1$, $d = 0$.  Since $\operatorname{det}(M)=1$, we must have $b = -1$.  
This determines $M$, and we note that $M^{-1}$ is the matrix treated in the
preceding case, so again $z\in \eusm R$ and $w\in\eusm R$ only when $z = w = \rho$.
\hfil\break
4. $a = -1$, $d = 1$.  In order that $\operatorname{det}(M)=1$ we must have $b = -2$.
Now $|z+1|=1$ implies that $z = \rho$.  But 
$f(\rho)= (-\rho-2)/(\rho+1)= -3/2+i\sqrt3/2\notin\eusm R$.

    In all cases, if $z\in \eusm R$, $w\in\eusm R$, and $z\sim w$, then $z = w$,
so the proof is complete.
\enddemo
\bigskip
\line{\hfil}
\psfig{file=Figmodgp.ps,rheight=3truein}
\vskip1.3truein
\centerline{\smc Figure 1. The Fundamental Domain and some of its images.}
\vfil
\eject
    By examining the above proof, we may observe that $z \in \eusm R$ and $f(z)=z$
only when $M = \pm I$, with the exceptions that $f(i)= i$ when $M$ is one of the
four matrices $\pm I$, $\pm S$, and $f(\rho)= \rho$ when $M$ is one of the six matrices
$\pm I$, $\pm\bmatrix 0&-1\\ 1&1\endbmatrix$, $\pm\bmatrix -1&-1\\ 1&0\endbmatrix$.
\bigskip
\bigskip
\head
2. Positive definite binary quadratic forms
\endhead

\noindent
Write
$$
\align
f(x,y) &= ax^2 + bxy + cy^2 = a(x - ry)(x - \overline ry), \\
g(x,y) &= Ax^2 + Bxy + Cy^2 = A(x - Ry)(x - \overline Ry)
\endalign
$$
where $r$ and $R$ are in the upper half-plane $\eusm H$.  Since $b = -2a\Re r$, to
say that $-a<b\le a$ is equivalent to $-1/2\le \Re r<1/2$.  Since $c = a|r|^2$,
the inequality $c > a$ is equivalent to $|r|>1$.  Similarly, the
inequalities $0\le b\le a = c$ are equivalent to $-1/2\le \Re r\le 0$, $|r|=1$.
Thus we see that $f$ is reduced if and only if $r\in\eusm R$.  Moreover, if
$g(x,y)=f(m_{11}x+m_{12}y,m_{21}x+m_{22}y)$, then 
$$
\align
g(x,y) &=a(m_{11}x+m_{12}y-r(m_{21}x+m_{22}y))
(m_{11}x+m_{12}y-\overline r(m_{21}x+m_{22}y)) \\
&=a((-m_{21}r+m_{11})x - (m_{22}r-m_{12})y)
((-m_{21}\overline r+m_{11})x - (m_{22}\overline r-m_{12})y) \\
&= a(m_{11}-rm_{21})(m_{11}-\overline r m_{21})
\Big(x-\frac{m_{22}r-m_{12}}{-m_{21}r+m_{11}}y\Big)
\Big(x - \frac{(m_{22}\overline r-m_{12}}{-m_{21}\overline r + m_{11}}y\Big)\,.
\endalign 
$$
Here we see that $A = a(m_{11}-rm_{21})(m_{11}-\overline r m_{21})=f(m_{11},m_{21})$,
which we already knew.  More importantly, we see that
$$
R = \frac{m_{22}r-m_{12}}{-m_{21}r+m_{11}},
$$
which is to say that
$$
r = \frac{m_{11}R+m_{12}}{m_{21}R+m_{22}}\,.
$$
That is, $M$ takes $R$ to $r$, so that $r\sim R$.  Since each $r\in \eusm H$ is
equivalent to a unique $R\in\eusm R$, every positive definite binary quadratic 
form is equivalent to a unique reduced quadratic form.

     The coefficients $a, b, c$ of $f$ determine a unique $r\in \eusm H$, provided
that $d = b^2 - 4ac < 0$.  Conversely, once $d<0$ is fixed, the coefficients $a, b, c$
can be recovered from $r$ by the relations
$$
a^2(r-\overline r)^2 = d, \qquad a > 0, \qquad b = -a(r+\overline r), 
\qquad c = ar\overline r\,.
$$	 
Thus if $r\sim R$, then we have associated $f$ and $g$ with $f\sim g$.
\enddocument

