Cryptography Reference
In-Depth Information
If z
C , recall that ord z f is the order of f at z .Thatis,
z ) ord z ( f ) g ( τ ) ,
f ( τ )=( τ
with g ( z ) =0 ,∞ . We can also define the order of f at i∞ . Suppose
f ( τ )= a n q n + a n +1 q n +1 + ··· ,
(9.15)
with n
Z and a n = 0, and assume that this series converges for all q close
to 0 (with q =0when n< 0). Then
ord i∞ ( f )= n.
Note that q → 0as τ → i∞ ,soord i∞ ( f ) expresses whether f vanishes ( n> 0)
or blows up ( n< 0) as τ → i∞ .
PROPOSITION 9.16
Let f be a function m erom orphicin H su ch that f isnotidenticallyzeroand
su ch that
f + b
+ d
= f ( τ )
for all ab
∈ SL 2 ( Z ) .
cd
Then
2 ord i ( f )+
z
ord i∞ ( f )+ 1
3 ord ρ ( f )+ 1
ord z ( f )=0 .
= i,ρ,i
REMARK 9.17 The function f can be regarded as a function on the
surface obtained as follows. Identify the left and right sides on F to get a
tube, then fold the part with
to
a point. This gives a surface that is topologically a sphere. The proposition
expresses the fact that the number of poles of f equals the number of zeros on
such a surface, just as occurred for doubly periodic functions in Theorem 9.1.
The point i is special since a small neighborhood around i contains only half
of a disc inside F . Similarly, a small neighborhood around ρ includes only
1/3 of a disc from F (namely, 1/6 near ρ and 1/6 near 1 + ρ , which is folded
over to meet ρ ). This explains the factors 1/2 and 1/3 in the proposition. For
a related phenomenon, see Exercise 9.3.
|
z
|
=1at i . Thenpinchtheopenendat i
PROOF Let C be the path shown in Figure 9.4. Essentially, C goes around
theedgeof F . However, it consists of a small circular arc past each of ρ ,1+ ρ ,
and i . If there is a pole or zero of f at a point on the path, we make a small
detour around it and a corresponding detour at the corresponding point on
the other side of F . The arcs near ρ ,1+ ρ ,and i have radius ,where is
chosen small enough that there are no zeros or poles of f inside the circles,
except possibly at ρ ,1+ ρ ,or i . Similarly, the top part of C is chosen to have
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