This preview shows page 1. Sign up to view the full content.
90
LECTURE 8. THE MODULAR CURVE
Proof.
We know this already when Γ = Γ(1). So we may assume that
μ
Γ
>
1. By
Theorem 8.5 we can identify the space
M
6
(Γ) with
L
(
D
), where
deg
D
= 6deg
K
X
+ 6
r
∞
+ 3
r
2
+ 4
r
3
= 12
g

12 + 6
r
∞
+ 3
r
2
+ 4
r
3
.
I claim that deg
D >
2
g
+ 1
.
If
g
≥
0 this is obvious. If
g
= 0 we use the formula for
the genus from Theorem 8.4. It easily gives that

12 + 6
r
∞
+ 3
r
2
+ 4
r
3
=
μ >
2
g
+ 1 = 1
.
It follows from the proof of Theorem 8.8 that
ν
x
(
f
i
) =
ν
x
(
f
i
/j
0
6
) +
D
(
x
)
.
(8.28)
Now we use the standard argument from the theory of algebraic curves. First of all the
map is welldeﬁned. In fact, if all functions
f
i
vanish at the same point
x
, we obtain
ν
x
(
f
i
)
>
0 for all
i
= 0
,...,N
, and hence
ν
x
(
f
i
/j
0
6
) +
D
(
x
)

1
≥
0 for
i
= 0
,...,N
.
This implies that
L
(
D
) =
L
(
D

x
)
.
However, this contradicts the RiemannRoch
theorem: since
deg(
K
X

D
)
<
deg(
K
X

(
D

x
)) = 2
g

2

deg
D
+ 1
<
0
,
it gives dim
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.
 Fall '09
 ONTONKONG

Click to edit the document details