for a function fieldK of genusg, and for any divisorD,
deg(D) + 1−g≤l(D)≤deg(D) + 1.
4.5.4 Corollary:
For any divisor D such that deg(D) ≥ c, where c is the constant from Corollary 4.5.3, we have
AK(D) +K =AK.
Proof.
By the claim in Lemma 4.4.5, dimF
AK(D2) +K
AK(D1) +K
=r(D2)−r(D1)
for any divisorsD2 ≥D1. By Corollary 4.5.3, deg(D)≥cimplies
r(D) =g−1. Thus if deg(D2),deg(D1)≥c, then
AK(D2) +K=AK(D1) +K.
For any divisor D= ΣP nPP with deg(D)≥cand for any adele
(φP), define
E = max(D,−div((φP))).
Therefore E ≥ D, deg(E) ≥ deg(D) ≥ c thus AK(E) +K =
AK(D) +K, we have that
(φP)∈AK(−div((φP)))⊂AK(E)⊂AK(E) +K=AK(D) +K.
If deg(D) is large enough, then any adele is in AK(D) +K,
AK ⊂AK(D) +K. And we have that AK ⊃AK(D) +K since
AK(D) and K are both subsets of the adeles under the diagonal
embedding. Thus AK(D) +K=AK.
4.6
Weil Differentials
So far, we have a precise bound for l(D) based on deg(D), then to cal- culate l(D) precisely, we need to introduce another object called the Weil differentials.
4.6.1 Motivation:
Before we define a Weil differential, here is some motivation following [5][Chapter 6].
Recall that a Riemann surface X is a connected complex manifold of com- plex dimension one. And in complex analysis, a meromorphic function on an open subsetU is a function that is holomorphic everywhere onU except for a set of isolated points, which are so called poles of the function. Let X be a compact Riemann surface of genus g, and M is the field of meromorphic functions onXand Ω is the space of meromorphic differential forms on X.
Fix someω∈Ω, a pointP ∈X, picktsuch thattvanishes to order one at the point P. If we pick some derivation donX, there is
ω= Σk∈Zaktkdt
Sinceωis meromorphic, there is some least integerNsuch thatakis nonzero
atP. Let N be the order ofω atP, ordP(ω) = 0 for all but finitely many
P, thus we could define
div(ω) = ΣPordP(ω)P
as a divisor of the points of X. For a function f ∈ M, we could write
f = Σ∞j=−1bjtj when look at φ locally. By integrating over a small simple
closed path around the point P, the residue off ω atP is ResP (f ω) =c−1 = Σi+j=−1aibj.
Then we can define a mapωP :M →Cby defining
f 7→ResP (f ω).
Then by the Residue theorem [6][Section 8.1], we have ΣP∈X ResP (ω) = 0 on X.
Then for all f ∈M we have
ΣP∈X ωP(f) = 0.
Let HP be the set of functions in M which are holomorphic at P, let AX
be the vector space over Cinside ΠP M which satisfies
φ= (φP)∈AX ⇒ φP ∈HP for all but finitely manyP.
For any divisor D= ΣP nPP, letAX(D) be defined by
AX(D) ={φ∈AX : ordP(φP) +nP ≥0 or φP = 0 for allP}.
Then define a function ˆω:AX →Cby ˆ
ω((φP)) := ΣPωP(φP) = ΣPResP(φPω).
If φP ∈HP, ordP(ω)≥0, then ResP(φPω) = 0, thus ˆω((φP)) is defined by
a finite sum, therefore the function ˆω((φP)) is well-defined.
4.6. WEIL DIFFERENTIALS 51
4.6.2 Definition: Weil Differential
A Weil differentialω on a function fieldK over an algebraically closed field
F, is an F-linear map from AK to F such that there is some divisor D of
K whereω vanishes both on K and on AK(D).
We denote the space of differentials onKby ΩKand the space of differential
vanishes onAK(D) for some divisor Ddenote by ΩK(D).
4.6.3 Theorem:
For any divisor D, ΩK(D) is finite dimensional overF,
l(D) = deg(D)−g+ 1 + dimF ΩK(D).
. Proof.
For all a∈F,ω ∈ΩK,aω is also an F- linear map from AK to
F, thus ΩK and ΩK(D) can be both viewed as F- vector spaces.
And since ω ∈ΩK(D) if and only if when ω is from an F-linear
map fromAK/(AK(D) +K) to F.
ΩK(D) = HomF(AK/AK(D) +K) = ( AK
AK(D) +K
)∗ (Recall the definition of dual vector space in Definition 1.6.1.) Next, by Corollary 4.5.4 We have
ΩK(D) = ( A K AK(D) +K )∗= (AK(E) +K AK(D) +K )∗
for any divisorDfor any divisorE≥Dwith large enough degree. Then by the claim from Lemma 4.4.5, for any divisorD0 ≥D,
dimF(
AK(D0) +K
AK(D) +K
) =r(D0)−r(D)
which is of finite dimension. Thus ΩK(D) is finite-dimensional,
and by duality
dimF(ΩK(D)) = dimF( AK
AK(D) +K
).
Next, sincer(D0) =g−1 andAK(D0) +K=AK for any divisor
D’ with large enough degree,
dimFΩK(D) =g−1−(deg(D)−l(D)).
Thus the results hold. .
4.6.4 Corollary:
The genus
Proof. By Theorem 4.6.2 above, l(0) = deg(0)−g+ 1 + dimFΩK(D) 1 = 0−g+ 1 + dimFΩK(D), thus dimFΩK(D) =g. (4.9) 4.6.5 Theorem:
Let ω be any nonzero differential, there exists a greatest divisor D such that, for any other divisorD0,D0 ≤Dif and only ifω vanishes onAK(D0).
Proof.
Let Sω be the set of divisors such thatω vanishes on AK(E) By
Corollary 4.5.4,
deg(D0)≥c ⇒ AK(D0) +K =AK.
Or, equivalently,
AK 6=AK(E) +K ⇒ deg(D0)< c.
Next, there is some adele φ on which ω does not vanish since ω
is nonzero. ThusAK(D0) +K is not all of the adeles, so we have
a bound on the degree of divisors in the setSω.
Now, let’s fix some divisor Dof maximal degree in Sω. To show
that this divisor of maximal degree is unique, if we pick any other divisor, sayE inSω,ω vanishes on bothAK(E) andAK(D) thus
ω will also vanishes on the union
AK(E) +AK(D) =AK(max(D, E)).
Thus max(D, E)∈Sω. By the definition of max(D, E), deg(max(D, E))≥
deg(D). Since D is of maximal degree in Sω, we have that
D=max(D, E), D≥E, thus the divisor inSω which has max-
imal degree is indeed unique. And the statement of the theorem follows.
4.6.6 Definition: div(ω)
div(ω) is defined as the divisor D of greatest degree such that ω vanishes on AK(D).
4.6. WEIL DIFFERENTIALS 53
4.6.7 Remark:
By Theorem 4.6.5 above, for all φ ∈ AK(divω), ω(φ) = 0, and if for all
φ∈AK(D0) we have ω(φ) = 0, then
D0 ≤div(ω).
4.6.8 Lemma:
For anyω ∈ΩK ,α∈K×, we have
div(αω) = div(α) + div(ω).
Proof.
Letφ∈AK,D= ΣP nPP,α∈K×.
αφ∈AK(D)⇐⇒ordP(αφP) +nP ≥0 orφP = 0 for allP
⇐⇒ordP(φ) + (ordP(α) +nP)≥0 orφP = 0 for allP
⇐⇒φ∈AK(div(α) +D)
(4.10) Thus ifω vanishes on AK(D), thenαω vanishes onAK(div(α) +
D), and the reverse is also true, thus we have
ω∈ΩK(D)⇐⇒αω∈ΩK(div(α) +D).
Now, obviously
ω ∈ΩK(div(ω)),
let Sαω be defined as in Theorem 4.6.4, thus
div(α) + div(ω)∈Sαω.
Thus
div(αω)≥div(α) + div(ω).
By the relation
ω ∈ΩK(D)⇐⇒αω∈ΩK(div(α) +αω),
we know thatαω∈ΩK(div(α) + (div(αω)−div(α))) implies
ω ∈ΩK(div(αω)−div(α)).
Then by Definition,
div(αω)≤div(α) + div(ω).
since div(ω) ≥ div(αω)−div(α), Combining with the previous result div(αω)≥div(α) + div(ω),the statements hold.
4.6.9 Lemma:
Let ω∈ΩK be any nonzero differential,D be any divisor, then
L(div(ω)−D)ω ⊂ΩK(D).
Proof.
For anyα ∈K×, we have
α∈L(div(ω)−D) if and only if div(α)+div(ω) = div(αω)≥D.
Therefore
AK(div(αω))⊃AK(D).
And αω vanishes on ΩK(D) since it vanishes on ΩK(αω).