Milne. .Elliptic.Curves.And.Algebraic.Geometry.(1996).[sharethefiles.com]

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can give it a complex structure as follows: let � : C �! C/Z be the quotient map; for any
P " C/Z and Q " f-1(P ) we can find open neighbourhoods U of P and V of Q such that
� : U �! V is a homeomorphism; take any such pair (U, �-1 : U �! V ) to be a coordinate
function.
For any open U �" C/Z, a function f : U �! C is holomorphic for this complex structure
if and only if f �% � is holomorphic. Thus the holomorphic functions f on U �" C/Z can
be identified with the holomorphic functions g on �-1(U) invariant under Z, i.e., such that
g(z + 1) = g(z).
For example, q(z) = e2�iz defines a holomorphic function on C/Z. In fact, it gives an
isomorphism C/Z �! C� whose in inverse C� �! C/Z is (by definition) (2�i)-1 � log.
114 J.S. MILNE
Example 23.5. Let D be the open unit disk {z | |z|
on D. The Schwarz lemma implies that Aut(D) = {z " C | |z| = 1} H" R/Z, and it follows
that " is a finite cyclic group. Let z �! �z be its generator and suppose that � has order m,
i.e., �m = 1. Then zm is invariant under ", and so defines a function on "\D, which in fact
is a homeomorphism "\D �! D, and therefore defines a complex structure on "\D.
Let � : D �! "\D be the quotient map. Then f �! f �%� identifies the space of holomorphic
functions on U �" "\D with the space of holomorphic functions on �-1(U) such that f(�z) =
f(z), i.e., which are of the form f(z) = h(zm) with h holomorphic. Note that if �(Q) = P ,
1
then ordP (f) = ordQ(f �% �).
m
Let � be a group acting on a Riemann surface X. A fundamental domain for � is a
connected open subset D of X such that
(a) no two points of D lie in the same orbit of �;
�
(b) the closure D of D contains at least one element from each orbit.
For example,
D = {z " C | 0
is a fundamental domain for Z acting on C (as in 23.4), and
D0 = {z " D | 0
is a fundamental domain for Z/nZ acting on the unit disk (as in 23.5).
The Riemann surfaces X(�). Let � be a subgroup of finite index in SL2(Z). We want
to define the structure of a Riemann surface on the quotient �\H. This we can do, but the
resulting surface will not be compact. Instead, we need to form a quotient �\H" where H"
properly contains H.
The action of SL2(Z) on the upper half plane. Recall that SL2(Z) acts on H = {z | (z) > 0}
according to
az + b
a b
z = .
c d
cz + d
-1 0
Note that -I = acts trivially on H, and so the action factors through
0 -1
PSL2(Z) = SL2(Z)/{�I}.
Let
-1
0 -1
S = , so Sz = ,
1 0
z
and
1 1
T = , so T z = z + 1.
0 1
Then
S2 = 1, (ST )3 = 1 in PSL2(Z).
Proposition 23.6. Let
1 1
D = z " H | |z| > 1, -
2 2
ELLIPTIC CURVES 115
�
(a) D is a fundamental domain for SL2(Z); moreover, two elements z and z of D are in
the same orbit if and only if
(i) (z) = �1 and z = z � 1 (so z = T z or z = T z );
2
(ii) |z| = 1 and z = -1/z (= Sz).
�
(b) For z " D, the stabilizer of z is = {�I} if and only if z = i, in which case the
stabilizer is , or � = e2�i/6, in which case the stabilizer is , or �2, in which
case it is .
(c) The group PSL2(Z) is generated by S and T .
�
Proof. Let � =. One first shows that � D = H, from which (a) and (b) follow
easily. For (c), let � " SL2(Z), and choose a point z0 in D. There exists a � in � such that
�z0 = � z0, and it follows from (b) that � �-1 = �I. For the details, see (Serre, Course on
Arithmetic, VII.1.2).
Remark 23.7. Let � be a subgroup of finite index in SL2(Z), and write
SL2(Z) = ��1 *" . . . *" ��m (disjoint union).
Then D = *"�iD satisfies the conditions to be a fundamental domain for �, except that it
won t be connected. However, it is possible to choose the �i so that the closure of D is
connected, in which case the interior of the closure will be a fundamental domain. for �.
The extended upper half plane. The elements of SL2(Z) act on P1(C) by projective linear
transformations,
a b
(z1 : z2) = (az1 + bz2 : cz1 + dz2).
c d
Identify H, Q, and {"} with subsets of P1(C) according to
z �! (z : 1) z " H
r �! (r : 1) r " Q .
" �! (1 : 0)
The action of SL2(Z) stabilizes H" =df H *" Q *" {"}. For example, for z " H,
az + b
a b
(z : 1) = (az + b : cz + d) = ( : 1)
c d
cz + d
as usual, and for r " Q,
a b (ar+b : 1) r = -d
cr+d c
(r : 1) = (ar + b : cr + d) = ,
c d
" r = -d
c
and, finally,
a b (a : 1) c = 0,
c
" = (a : c) = .
c d " c = 0
Thus in passing from H to H", we have added one additional SL2(Z) orbit. The points in
H" not in H are often called the cusps.
We make H" into a topological space as follows: the topology on H is that inherited from
C; the sets
{z | (z) > M}, M > 0
116 J.S. MILNE
form a fundamental system of neighbourhoods of "; the sets
{z | |z - (a + ir)|
form a fundamental system of neighbourhoods of a " Q. One shows that H" is Hausdorff,
and that the action of SL2(Z) is continuous.
The topology on �\H". Recall that if � : X �! Y is a surjective map and X is a topological
space, then the quotient topology on Y is that for which a set U is open if and only of
�-1(U) is open. In general the quotient of a Hausdorff space by a group action will not be
Hausdorff, even if the orbits are closed one needs that distinct orbits have disjoint open
neighbourhoods.
Let � be a subgroup of finite index in SL2(Z). One can show that such a � acts properly
discontinuously on H, i.e., that for any pair of points x, y " H, there exist neighbourhoods
U of x and V of y such that
{� " � | �U )" V = "}
is finite. In particular, this implies that the stabilizer of any point in H is finite (which we
knew anyway).
Proposition 23.8. (a) For any compact sets A and B of H, {� " � | �A )" B = "} is
finite.
(b) Any z " H has a neighbourhood U such that
�U )" U = "
only if �z = z.
(c) For any points x, y of H not in the same �-orbit, there exist neighbourhoods U of x
and V of y such that �U )" V = " for all � " �
Proof. (a) This follows easily from the fact that � acts properly discontinuously.
(b) Let V be compact neighbourhood of z. From (a) we know that there is only a finite
set {�1, . . . , �n} of � such that V )" �iV = ". Let �1, . . . , �s be the �i s fixing z, and for each
i > s, choose disjoint neighbourhoods Vi of z and Wi of �iz, and set
-1
U = V )" ()"i>sVi )" �i Wi).
For i > s, �iU �" Wi, which is disjoint from Vi, which contains U.
(c) Choose compact neighbourhoods A of x and B of y, and let �1, . . . , �n be the elements
of � such that �iA )" B = ". We know �ix = y, and so we can find disjoint neighbourhoods
Ui and Vi of �ix and y. Take
-1 -1
U = A )" �1 U1 )" . . . )" �n Un, V = B )" V1 )" . . . )" Vn.
Corollary 23.9. The space �\H is Hausdorff.
Proof. Let x and y be points of H not in the same �-orbit, and choose neighbourhoods U and
V of x and y as in (c) of the last proposition. Then �U and �V are disjoint neighbourhoods
of �x and �y.
In fact, �\H" will be Hausdorff, and compact.
ELLIPTIC CURVES 117
The complex structure on �0(N)\H". The subgroups of SL2(Z) that we shall be espe-
cially interested in are
a b
�0(N) = c a" 0 mod N .
c d
We let �0(1) = SL2(Z).
For z0 " H, choose a neighbourhood V of z0 such that
�V )" V = " =�! �z0 = z0,
and let U = �(V ) it is open because �-1U = *"�V is open. [ Pobierz całość w formacie PDF ]

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