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i=1
A quadratic form is integral if its coefficients as a polynomial are integers.
We will be concerned in the sequel with integral quadratic forms of two variables, such
as 3x2 + 2xy - 7y2 or -x2 - 3xy.
Notice for every quadratic form f(, ) there corresponds uniquely a symmetric matrix
x
Mf such that f(x, y) = x y Mf . In particular, if f(x, y) = ax2 + bxy + cy2 then
y
b
a
2
Mf =
b
c
2
Definition 3.35. We say that a quadratic form f(x, y) represents an integer n if there is
some assignment of integer values n1, n2 to x, y for which
f(n1, n2) = n
Given an integral quadratic form, it is a natural question to ask what interger values
it represents. Observe that there are some elementary transformations on quadratic forms
which do not change the set of values represented. If we substitute x ! xy or y ! y x
we get a new quadratic form. Since this substitution is invertible, the new quadratic form
obtained represents exactly the same set of values as the original. For instance,
x2 + y2 !! x2 + 2xy + 2y2
This substitution defines an equivalence relation on quadratic forms.
CLASSICAL GEOMETRY LECTURE NOTES 27
The most general form of substitution allowed is a transformation of the form
x ! px + qy, y ! rx + sy
For integers p, q, r, s. Again, since this substitution must be invertible, we should have
p q
ps - qr = 1. That is, the matrix is in SL(2, Z). Since these forms are
r s
homogeneous of degree 2, the substitution x ! -x, y ! -y does nothing. One can
easily check that every other substitution has a nontrivial effect on some quadratic form.
In particular, we have the following theorem:
Theorem 3.36. Integral quadratic forms up to equivalence are parameterized by equiva-
b
a
2
lence classes of matrices of the form for integers a, b, c modulo the conjugation
b
c
2
action of P SL(2, Z).
Observe that what is really going on here is that we are evaluating the quadratic form f
on the integral lattice Z " Z. The group SL(2, Z) acts by automorphisms of this lattice,
and therefore permutes the set of values attained by the form f.
There is nothing special about the integral lattice here; it is obvious that SL(2, Z) acts
by automorphisms of any lattice L. In particular, if L = e1, e2 then M = "
SL(2, Z) acts on elements of this lattice by
M re1 + se2 ! (r + s)e1 + (r + s)e2
3.2.3. Moduli of tori, continued fractions and P SL(2, Z).
Definition 3.37. Let r be a real number. A continued fraction expansion of r is an expres-
sion of r as a limit of a (possibly terminating) sequence
1 1 1
n1, n1 + , n1 + , n1 + , . . .
1 1
m1 m1 + m1 +
1
n2
n2+
m2
where each of the ni, mi is a positive integer.
A continued fraction expansion of r can be obtained inductively by Euclid s algorithm.
First, n1 is the biggest integer d" r. So 0 d" r - n1
1
Otherwise, r = > 1 and we can define m1 as the biggest integer d" r . So 0 d"
r-n1
r - m1
which is the continued fraction expansion of r. If r is rational, this procedure terminates at
a finite stage. The usual notation for the continued fraction expansion of a real number r is
1 1 1 1
r = n1 + . . .
m1+ n2+ m2+ n3+
The following theorem is quite easy to verify:
Theorem 3.38. If n1, m1, . . . is a continued fraction expansion of r, then the successive
approximations
1 1
n1, n1 + , n1 + , . . .
1
m1 m1 +
n2
denoted r1, r2, r3, . . . satisfy
|r - ri| d" |r - p/q|
for any integers (p, q), where q
28 DANNY CALEGARI
Thus, the continued fraction approximations of r are the best rational approximations
to r for a given bound on the denominator.
Let T be a flat torus. Then the isometry group of T is transitive (this is not too hard
to show). Pick a point p, and cut T up along the two shortest simple closed curves which
start and end at p. This produces a Euclidean parallelogram P . After rescaling T , we can
assume that the shortest side has length 1. We place P in E2 so that this short side is the
segment from 0 to 1, and the other side runs from 0 to z where Im(z) > 0. By hypothesis,
1
|z| e" 1. Moreover, if |Re(z)| e" we can replace z by z + 1 or z - 1 with smaller norm,
2
contradicting the choice of curves used for the decomposition. Let D be the region in the
1 1
upper half plane bounded by the two vertical lines Re(z) = , Re(z) = - and the circle
2 2
|z| = 1.
The group P SL(2, Z) acts naturally on H as a subgroup of P SL(2, R). The action there
is properly discontinuous. The action of P SL(2, Z) permutes the sides of D. The element
1 1 0 1
pairs the two vertical sides, and the element preserves the bottom side,
0 1 -1 0
interchanging the left and right pieces of it. In particular, D is a fundamental domain for
P SL(2, Z). The quotient is topologically a disk, but with two cone points of order 2 and
"
1+i 3
3 respectively, which correspond to the points i and respectively, whose stabilizers
2
are Z/2Z and Z/3Z respectively.
This quotient is an example of an orbifold.
Definition 3.39. An n dimensional orbifold is a space which is locally modelled on Rn
modulo some finite group.
A 2 dimensional orbifold looks like a surface except at a collection of isolated points
pi where it looks like the quotient of a disk by the action of Z/niZ, a group of rotations
centered at pi. The point pi is a cone point, sometimes also called an orbifold point. The
finite group is part of the data of the orbifold. One can think of the orbifold combinatorially
as a surface (in the usual sense) with a finite number of distinguished points, each of which
has an integer attached to it. Geometrically, this point looks like a cone made from a
wedge of angle 2/ni.
o
We can define an orbifold fundamental group 1() for a surface orbifold. Thinking
of our orbifold as X/ for the moment where acts properly discontinuously but not
freely, the orbifold fundamental group of should be exactly . This means that a small
o
loop around an orbifold point pi should have order ni in 1(). Note that we are being
casual about basepoints here, so we are only thinking of these groups up to isomorphism.
In any case, the orbifold H2/P SL(2, Z) should have orbifold fundamental group iso-
morphic to P SL(2, Z). There is an element of order 2 corresponding to the loop around
0 1
the order 2 point; a representative of this element in P SL(2, Z) is . There is an
-1 0
element of order 3 corresponding to the loop around the order 3 point; a representative
1 1
of this element in P SL(2, Z) is . Note that these elements have order 2 and 3
-1 0
respectively in P SL(2, Z), even though the corresponding matrices have orders 4 and 6
respectively in SL(2, Z).
Every loop in the disk can be shrunk to a point; it follows that every loop in the orbifold
H2/P SL(2, Z) can be shrunk down to a collection of small loops around the two cone
points in some order. That is, the group P SL(2, Z) is generated by these two elements.
CLASSICAL GEOMETRY LECTURE NOTES 29
Theorem 3.40. A presentation for P SL(2, Z) is given by
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