P-Adic Theta Functions

P-Adic Theta FunctionsAuthor(s): Peter NormanSource: American Journal of Mathematics, Vol. 107, No. 3 (Jun., 1985), pp. 617-661Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2374372 .

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P-ADIC THETA FUNCTIONS

By PETER NORMAN

Let k be an algebraically closed field of characteristicp > 2; let W(k) be the Witt vectors with entries in k, and let R be the ring of integers in a finite extension of the fraction field of W(k). Let A be an abelian variety over R; S, a line bundle on A, and s e r (A, S). We consider the problem of associating to s a p-adic theta function.

In the case of an elliptic curve with nonintegralj-invariant Tate gave an elegant construction of p-adic theta functions [10] which was subse- quently generalized to abelian varieties.

If A has good reduction there are two constructions of p-adic theta functions. One is due to Barsotti and Cristante ([1], [2]). If A X k is an ordinary abelian variety, they associate to s an element of SR where SR is the completion of the local ring of A X R at the identity and R is the ring of integers of the algebraic closure of W(k). They denote this element by OD

where D is the divisor of s. The element 0 is unique up to a 'linear exponential', that is an elementf e SR such that its first multiplicative difference

W = f (x + y)/f (x)f (y)

is identically one. Here x + y is addition in the formal group of A. The element OD enjoys many nice properties. In particular it satisfies

an analogue of one of the differential equations satisfied by classical theta functions. If A has dimension 1 this comes out as follows: Let X be invariant nonvanishing -differential on A and let d/dz denote the derivation dual to w. Let p be the Wierstrass function of (E, w); let D be the divisor le with e the identity point on A. Then there is a constant C so that

(0.1) (d/dz)2 log 0D = -p + C.

The constant C is determined by (E, co) and the divisor D since OD is unique up to a linear exponential.

Manuscript received January 3, 1983.

617



618 PETER NORMAN

If A does not have ordinary reduction, then OD is not in S, but rather it is in a huge ramified over ring of SR; furthermore, in this case, OD is only unique up to a quadratic exponential in this over ring, i.e., an elementf such that its second multiplicative difference

JD2f = f(x + y)f (x)f (y)f (z)/f (x + y)f (x + z)f (y + z)

is identically one. A second construction is due to Mumford. For an ample, totally sym-

metric line bundle S and s e r (A, S), he gave a technique for canonically evaluatings at torsion points [8]. If we considerA over R/(p'), then points in the formal group are torsion (By "point" I mean an S-valued point where S is a local artin algebra over R/(p')). Using Mumford's technique to evaluate s at these points and taking the limit as n -+ oo we get a theta function O e SR if A X k is ordinary. If A X k is not ordinary, then, after a little struggle, we obtain a theta function O, but 0, is no longer in SR,

rather it is in the same huge over ring that appears in the work of Barsotti/ Cristante. This construction produces an element unique up to a constant.

The aims of this paper are: [1] To carry out Mumford's construction of a p-adic theta function

and deduce its basic properties (Section 1-Section 5). [2] To compare the two theta functions 0, and OD To do this we cast

the Barsotti/Cristante construction in geometric language (Section 6, Sec- tion 7). In the ordinary case we find that 0, is one of the OD's and further, if s is even, by requiring OD to the even, we get OD = 0, up to a constant (Section 8). In the ordinary case we show that OD can also be determined up to a linear exponential and again we can show that O0 = 0D provided 0D iS

even (Section 8). [3] To determine, in the ordinary case, the constant C in the differ-

ential equation (0.1). We can read this question backwards: What is the constant C such that the solution to the differential equation (0.1) has integral coefficients? We give two approaches to this problem: One uses Dwork's theory of thep-adic variation of abelian periods; the other usesp- adic modular forms. Both approaches generalize immediately to abelian varieties.

Acknowledgments. I'd like to thank David Mumford for the suggestion that power series theta could be constructed using the techniques in [8]. The Barsotti/Cristante construction is, of course, due to Barsotti and



P-ADIC THETA FUNCTIONS 619

Cristante ([1], [2]). I've merely translated it. Lastly, I'd like to thank Bernard Dwork for spending many hours explaining his work on p-adic differential equations.

Convention. By a point we mean an S valued point where S is a cer- tain local algebra over R.

Part I. Mumford's Constructions

The aim of this part is: (1) to recapitulate Mumford's technique for evaluating s e r (A, S)

at torsion points of A, (2) to apply this construction to the problem of constructing p-adic

theta functions, and (3) to establish properties of the p-adic theta function with an eye to

comparing it to Barsotti's theta function.

1. Mumford's constructions. The basis idea in both Mumford's and Barsotti's construction is first to pull S back to a "small" group over A:

ir: U -+ A,

so that 7-* ? is trivial and, second, to carefully choose a trivialization

*4 U X A1.

Once this is done we are in a position to evaluate s. This gives our theta function. We express this trivialization in terms of isomorphisms

x-

where Ty denotes translation by an x e U that lies over x e A. Examining the fiber of this map at the identity e e U, we find

S (e) = (r*SC)(e) = TyX*(r*4)(e) = S(x)

Upon choosing, once and for all,

X: R S ?(e)



620 PETER NORMAN

we obtain a coordinate function on S (x) and thus we can evaluate the section s at x. The value obtained will be 0 (x). Of course this depends on the choice of x over x.

We sketch this program following Mumford [8]. We assume that A is defined over a local R-algebra S. For the 'small subgroup' lying over A we use

H(S) = {xEA lo: S + Tx*4}.

For each x there is an isomorphism

X: S IH(2) _+(TX*2) IH(.C)

Our problem is to choose canonically for each x such an isomorphism. To capture the global properties of s we require that any such X we use comes from an isomorphism of S on all of A; we introduce the group

(S ) = {(+I,x)|+: S TX* I

The group law, denoted s * V, is given by composition

0*0 = (Ty*b)oi: ? T TT* C

The group 9 (?) fits into the exact sequence

1 - S* -+ 9(C) - H(L) 0.

Our problem is to produce a canonical section to ir. To construct a section, we rely on the symmetry of S. Let i be the

involution

i:A -+A

x I-+ -x

Choose, arbitrarily, an isomorphism




Define an involution

6: S(G) -S G

by sending (4, x) to the composition

? -~ j*473 -~ i*STX4 = __X(i*2) - T*X S

Since any two choices of a differ by a constant, 6 is independent of the choice of a. Note that if a e S* C g(4), then

b(a *) = a * b().

Definition. An element (0, x) e S (4) satisfies criterion M provided

(1. 1) * 60 = (T*-x 0) o : S -

T*-x S

+

is the identity. If (x, 0) e (4), and

**60 = a eS*,

then ?a 112k satisfies criterion M. Since for each x eHH(2), there are two elements over x in S (4) satisfying criterion M we do not yet have our section. By restricting to 2 * H(SC) we can over come this difficulty: Let x = 2y e 2H(SC); choose (/, y) e !G(2) satisfying (1.1), then

U(X) = * G!(C)

lies over x and is unambiguously defined. This a is our sought-after section.

2. Properties of 0, We now give several properties of a and the corresponding ones for 0.

Property 1. The map 6 behaves well with respect to base change. Consider the diagram:

7r*A A 1 4

Spec S2 -~Spec S1



622 PETER NORMAN

Let

a: S2 M,

so

'1r*ag: TSiX

from w-*a we can construct an involution

62: g(r*2) ,S(;X

which is just the pullback of 6. From this it follows that

(2.1) 07*S = 7*05?

Property 2. Let

p: A' -- A

be an isogeny of abelian varieties over S; it is easy to see that if

x E 2H(p * 2),

and

p(x) e 2H(SC),

then

(2.2) Op*s(x) = OS(P(x)).

Property 3. Let ? and g2 be totally symmetric ample line bundles onA; let

oti: 9(2S) S(S W ), i= 1, 2

be the involutions corresponding to S1 and S2. Let

6: 9 (-i 0 2) -+ 9(21 0 I2)




be the involution corresponding to 61 0 62. Then

61 0 62 6-

From this it follows that if s1 E r(A, ?Cl), S2 E r(A, ?2), then

(2.3) osl * OS2=0S1(S2

provided

X: S -() (21 0 L2)(e)

is just X = X1 0 X2 where X1, X2 are the evaluation maps for 431(e) and ?2(e). In any case oll S2 differs from OS 1 2 by a constant.

We use the properties above to establish an identity for Mumford's theta functions that will show they are Barsotti theta functions. We begin with some notation. Define maps

P, P1, p2:A XA -A

by

p(al, a2) - as, i = 1, 2,

p(al, a2) f- a1 + a2.

Define the multiplicative difference operator by: O acts on a divisor D by

J)D = p*D - p*D -p*D.

O) acts on a line bundle ? by

05 = p*? 0 (p*?f)-0 (P*2?)1.

OD acts on a function f by

= (P*f)/(Pl*f) * (P2*f).



624 PETER NORMAN

The second multiplicative difference operator, )2, is defined as follows: Using maps

(id) Xp, id XPi, idX P2:A XA X A A X A,

set

2= (id X D) o.

Explicitly, define

7r,x7rij:A XA XAA-A i,j=1,2,3,

by

7r: (a,, a2, a3) - a, + a2 + a3,

7ri: (a1, a2, a3) - ai, i = 1, 2, 3,

7riy: (a,, a2, a3) - ai + aj i,j = 1, 2, 3.

Then

2.4

5D2D = (Wr*D + rl*D + x2rD + r3*D) - (Or*D + 4* D + x*rD),

D2f = (f (x + y + z)f (x)f (y)f (z))(f (x + y)f (x, z)f (y + z)),

D2S = (7r*2 @ 7r*2 4 7r0*4 (D 0r *2) 0 (7*r2C

0 wrl*3 2 4T3)1

To simplify notation, we denote the product of the terms in the nu- merator of (2.4) by n*f, and the product of those in the denominator by d *f, so

(2.5) 5D2f = (n *f )/(d *f)

Analogous notation will be used for divisors and line bundles.




The theorem of the cube asserts that if D is a divisor on A, then there is a rational function F on A X A X A such that

(2.6) 02D = (F).

PROPOSITION 2.1. Let s e r(A, S) and let D = (s), then, up to a constant,

D OS = F12H(2) X 2H(2) x 2H(2)

To prove the proposition we need

Property 4. If s, t e r (A, S), then

(2.7) OS lot = s/t.

Since 0 and Ot are both computed at a point using the same coordinate function we get (2.7).

Proof of proposition. The theorem of the cube gives an isomorphism

n*S -+ d*S

allowing us to identify these line bundles. The ratio of the sections n*s, d*s is

(n*s)/(d*s) = F.

Using properties (2.7), (2.2) and (2.3), we get

F =On*s n*O, = -)2O Q.E.D. Od*s d*Os

The last property of 0 we establish is an analogue of the quasi-periodicity of classical theta functions. Of course we do not have periods, but, instead, examine the behavior of 0 under translation by torsion points. We will use this to pick out Mumford's theta from among Barsot- ti's, and also to compute the constant C in (0.1).

We begin by computing the cocycle of the nontrivial extension



626 PETER NORMAN

1 5-~S* -+S(.0-- H(S) O+

with respect to the section a. Define a bi-multiplicative, skew pairing

e < >:H(SC) X H(S) GGm(S)

by

(X, y) * E*

where (', x), (x, y) are in 9(2). Since S* is in the center of ?(2) this pairing is well defined; furthermore it is nondegenerate ([8]).

PROPOSITION 2.2. Let x = 2x 1 E 2H(0S); let y = 2y 1 EiH(0S). Then

a(x + y) = ec (x, -y/2)a(x) * (y).

Proof. Let a(xl), a(y1) be elements of H(S) sitting overx1,yl satisfying criterion M. Now

(U(x1) * (y1)) * b(uf(xO) * o(y1)) =

ao(x1 ) * r(y1 ) * 6(aY(x1 )) * koa(y1 )) = ec(xl, Y1 );

consequently

a(x ) * a(yl )es<xl,-y,1/2>

satisfies criterion M. By definition and a straightforward calculation we get

a(x + y) = e( <x, -y1 /2) u(x1) * a(y1)

* ec <xj, -yl /2> or(xl) *or(yl)

= e( <Y1, x> a(x)a(y).

We conclude that

a(x + y) = e <x, -y/2> a(x) * (y). Q.E.D.




The group Sg(2) acts on r(4): Let (0, x) e g (4), s e r(A, S), the action of 0 on s is:

UO(s) = r-'(Tx*s).

Since T*(s) e F(A, T*SC), 17'(T*s) e r(A, S). It is straightforward to check that

U

is a right group action. We can rephrase the definition of 0 is terms of this group action. Con-

sider the diagram:

S (X)- 1 T* 2 -M T*4

I a(I I S x-A ?(o) (T*C)(o) = (x)

(T *s)(o) = s (x)

We see that

Os (x) = 'Uo (X) (S)

The action of S(4) on r(A, S) induces an action of S(2) on the associated theta functions-. Let = a * o (y) e (2), a e S*; let t = U? (s). We compute Ot in terms of 0:

Ot(X) = X1(Ua(x)(Uaa(y)(S)))

= a) '(Uor(y) *or(x) (S))

= ae <y, x/2>X V-1 (U,(y+x)(s))

Ot(x) = ae <x, -y/2 > 0 (x + y).

Now s/t f is a rational function with divisor D - Ty*D; by (2.7) we conclude:

(2.8) f = 0s/0t = (ae (<x, y/2>05(x))/05(x + y).



628 PETER NORMAN

3. Explicit Calculation of a Theta Function. Since S is totally symmetric, there is a divisor D associated to S and a covering of A by open sets Ui so that

DIu. = (fi)Iu,

and

i*fi =f.

Transition functions for S are

fij = fi/fi;

hence there is a section s given locally by

s I ui fi I ui

Remark 3.1. If t is another section S, then tis = g is rational on A and at = g * As; many questions about at can be reduced to questions about

Os .

We identify the isomorphism

0: S -

TY*2

that satisfies criterion M. An arbitrary isomorphism is given by multiplication by

Fy (Ty*fi /fi)

where Fy is rational on A and has divisor

(Fy) D - Ty*D.

Since i*fj = fi, we can take S i*S and thus 5(4) is multiplication by

i*(Fy). T*_yfifi-l

Thus q satisfies criterion M provided




T*y(F Y fi) .i*Fy( Y )fi = T* yF i*F = 1.

Once we have an isomorphism 0 satisfying criterion M we can put a coordinate function on the stalk of S at x = 2y:

TY* T*2*

S(o) S(x)

1 1x

We have

O'(x) = s(x)/lx,

or, up to a constant,

Os (X) = fi (y) /(Ty Fy * Fy * Ty* fi * f i)(o)

(3.1) Os(x) = Fy(y)-'Fy(o)-l fi(O).

4. Power Series Theta Functions. Following a suggestion of Mum- ford we construct power series theta functions. We now assume that A is defined over R; if

f:R -+S

is a map of rings we denote the resulting base change by A X S. We must assume A X k is ordinary. Again assume S is totally symmetric and ample, let s e r(A, S). Let {t1, . . ., tg } be a set of uniformizing parameters at the identity e e A. Let m be the maximal ideal of R and denote R/mn by Rn .

The construction has two steps. First we associate to s an element of Rn, m= Rn [tl ... tg]/(tl ... tg)m for each n and m. Then we patch these truncated power series together to form a power series over R.

Define a point A(n, m) of A X Rn,m by mapping diagonally

A(n, m): Spec Rnm -A XRn,m.



630 PETER NORMAN

If A(n, m) were an element of 2H(S), then we could evaluate s as in Sec- tion 1 and obtain an element of Rn,m; however in general, A(n, m) 0 2H(S). We overcome this difficulty: We pullback ? to a finite cover of A. Since A is ordinary, there exist abelian varieties A(m) and isogenies

Fm :A - A (m)

Vm:A(m) --A

satisfying

(i) Fm o Vm is multiplication byp m on A(m), and Vm oFm is multiplication byp m onA;

(ii) the degree of Fm equals in degree of Vm, and (iii) Vm is etale.

Since Vm is etale, A(n, m) has a distinguished lifting to A(m) X R n which we again denote by A(n, m). We show that for n sufficiently large m the lifting A(n, m) E 2H(V*L). Let A* denote the dual abelian variety to A; then H(G) X Rn is the kernal of the polarization map

q$: X Rn: A X Rn --A* X Rn,

while H(V* ?) X Rn is the kernal of the composition:

A(m) X Rn (A(m)*) X Rn

! Vm XRn t Vm XRn

A X Rn > A* X Rn

We see that the kernal of pn is contained in H(V* ?); in particular since A (n, m) is a point of order pN for N >> 0,

A(n, m) E 2H(V* S) X Rn

for m >? 0. Define

AS (A (n, mi)) = OV*s (A (n, m)).

This is an element of Rn,m.




We patch these elements together. For n' < n and m' < n we have the projection

7r: Rn [tl 9. * tg]/(t, tg)m Rn'[tl *** tg]/(t, tg)m -

We show that

7r(05(A (n, m)) =05(A(n, m).

By property (2.2) we calculate Q,(A(n, m)) and Q,(A (n', m')) by pulling back via the same V* as long as t is sufficiently large. Notice that A(n', m') is the restriction of A(n, m). We apply property (2.1) to conclude that 05(A(n, m)) restricts to 05(A(n', m'). Thus we can patch the various A(n, m) together to obtain a power series associated to s; denote this by P0O, if no confusion is likely, by OQ. Notice P0O e- 8?, where S? is the completion of the local ring at the identity of A.

Remark 3.1. Proposition [1.1] holds for P0O: Let Hm denote H(V* C) and let sm = V*s; then ?20m is the restriction of a rational function, F(m), on A(m) X A(m) X A(m) whose divisor is D2(V*D) where (s) = D. If F is the rational function on A X A X A with divisor D2D, then V* F has divisor 522(V* D); hence, up to a constant, F(m) = V*F. Thus we can insure that V* 22PQS = V* F when restricted to Hm X Hm X

Hm. Taking limits we obtain

(3.1) D2P0S = F

when both functions are viewed as elements of the ring of power series of A.

Remark 3.2. We can also extend the identity [2.8] to PO,, at least modulo powers of m. Lety eA X R/(m)r and assumey E 2H(V*C) X

R/(M)r, then the proof of [2.8] gives

P0O(x) = P0O(x + y)eVt <x, -y/2> f(x) mod mr

wheref is rational on A(n) X R/IMr. Notice that

eV( <x, -y/2> 1 mod(m2)



632 PETER NORMAN

if both x and y are in the connected component of H(Vn*) X R/(mk) since this is of multiplicative type and there are no nontrivial bi- multiplicative pairings on such a group.

5. Theta Functions for General Abelian Varieties. Let A, i, s, m, {t1, . . ., tg} be as in Section 4. If A X k is not ordinary, the above approach does not allow one to associate to s an element of 50 = R [[t1, ..., tg]] since there are not sufficiently many etale covernings of A. We are able to associate to s an element of a huge ramified overring of R [[t1, . ., tg]] that has many of the properties of the theta series for ordinary abelian varieties. Again, we first work over R /mt and take the limit as t goes to infinity.

Fix t e Z, t > 0. For this construction only we denote

A X (R/mt), S X (R/mt), s X (Rimt), ...

by

A, S,s, ....

Let A(n) denote a copy of A = A (o) covering A(o) by multiplication bypn.

Let mA (n) denote the connected component of the kernal of multiplication by pm on A(n); write mA(n) = Spec mR(n); denote the line bundle (pn)*S on A (n) by ?n .

Let An denote the point of A(n) gotten by mapping diagonally

An: 2nA(n) -+ 2nA(n) X A(n).

Let

H(2n) = {xEA(n)lT I ?T n n

Since An e 2H(4%) we can evaluate Sn = pn*y using the procedure in Section 1 at the point An; denote this by s(An ) E 2n R(n). We can patch the

Y(An ) together. We have the commutative diagram

(2n + 2)A(n + 1) -+> (2n + 2)A(n + 1) X A(n + 1)

t T 11 (2n + 1)A(n + 1) -*'- (2n + 1)A(n + 1) XA(n + 1)

IP IP IP

2)nA(n) _5nl 2.nA (n X A (n.)




Evaluating s at these points we obtain:

Y(,An+l) E (2n + 2)R(n + 1) | (inc)*

Y(An*+i)E(2n + 1)R(n + 1)

P*

S(VA) E 2n R(n)

From properties [1.2] and [1.3] we see that p*s(An) = s(Ai*+); furthermore [1.1] implies that (inc)* s7(An+l) = S(A*+ ). We conclude

(5.1) p*s(Am) = (inc)*s(A\n+l)

Let St denote the ring

lim lim -

m (m R(n));

then {2nR(n) } is cofinal and since 37(An) is a compatible sequence we obtain

,s(t) = lim s (An) E 8. -

Remark 5.1. The proof of Remark 3.2 shows that

(5.2) (03-(,)(x + y)e(x 'Y>1/(0(t,(x))

is a rational function in x on A (n) as soon as y E H(2n )). Define 8 to be lim St; similarly let SR be the ring constructed in the

same way with R as base ring. Since the Oy(t) patch together we can define

0 = lim 03-(t) t

Remark 5.2. The proof of remark 3.1 implies that

(5.3) a,20 = F

where F is rational on A X A X A with divisor (F) = 0D2D.



634 PETER NORMAN

Remark 5.3. St contains the ring R[[t1, * .* tg]]. We establish some conventions concerning the rings St and 8 =

lim St. Although there is no formal group corresponding to 8 it is useful to think in terms of some sort of geometric object corresponding to S. We denote this "object" by A. Statements about A are to be interpreted in terms of statements about a sequence of finite objects. For example, a bilinear pairing

B:A XA Gm

is a set of compatible bilinear maps for all t and n:

- ~~B(n+l) -

(5.4) (2n + 2)A(n + 1) X (2n + 2)A(n + 1) Gm

(2n + l)A(n + 1) X (2n + l)A(n + 1) P P

B(n) -

2nA(n) X 2nA(n) B GM

An extension

1 Gm E A 0

is a group law on Gm X A, that is, a diagram like (5.4) except the maps B(n) have range A' and satisfy the cocyle condition. By a point of A we mean a compatible family of points Pn E 2nA (n) for all n.

Part I. Barsotti's Theta Functions.

Let A, R, k R/m, be as above. Let D be a divisor on A and letL be its associated line bundle. Let { t1, . . ., tg } be a set of uniformizing parameters at e, the identity of A. Following Barsotti & Cristante ([1], [2]) we associate to D an element 0D of R?IJtl, .. ., tg]] if A X k is ordinary, and if

A X k is not ordinary 0D iS in SR. We carry out their program using the geometric language of Part I; this allows us to compare the approach in Part I with that of Barsotti and Cristante.

Let A be the formal group of A, and ?C the restriction of S to A. The underlying idea of the construction is to first distinguish a subgroup, ?(2)B, of the group

{(4, x) Ix EA, 0: OS T *3}.




The second step is to distinguish a section A to the map

: S(SB A.

(+, x) x.-

Once we have A, we obtain coordinates on the stalk of S over x:

A(x): S(e) -+ (Tx*4)(e) = S(x).

Define ODto be the function gotten by evaluating a section of S with divisor D using these coordinates.

6. The Divisor is Algebraically Equivalent to Zero. In this section we assume D is algebraically equivalent to zero; furthermore, we assume, for the moment, that the support of D does not contain e. We construct OD

in two steps:

STEP ONE. We construct an isomorphism

A: S(4s) S *= S-{O}.

This gives S* a group structure (and conversely it gives S (4) a scheme structure). Fix P e S (0), P * 0. Define

A: (+,x) _ 0 (P) e SC(x).

For details see [6] page 225ff. We obtain an extension of group schemes

S (X) r [6.1] 1 Gm 11 A -+0.

STEP TWO. We take the formal completion of (6.1) and obtain the extension of formal groups

(6.2) 1 Gm S * A O0.

If A X k is ordinary, then the extension (6.2) splits. If A X k is not ordinary, we must enlarge R to R, the integral closure of R is the algebraic



636 PETER NORMAN

closure of the quotient field of R. Once we have split (6.2) we can evaluate a section of r (A, S) with divisor D. The resulting power series is OD. The point is that we have chosen a section of 2* that does not merely split the line bundle, but one that splits the bundle plus a globally defined group structure. Notice that the splitting, and hence 0D, is unique up to a linear exponential.

We explicitly compute the group law on 2* and the resulting 0D* We must first find a section to ir. Since D is algebraically equivalent to zero, there is a rational function F on A X A such that

(F) = SD.

Normalize F by requiring

F(o,y) F(x, o) = 1.

If x is a point outside the support of D, then

(FJ {x}XA) = T*D - D

and F', using the conventions of Section 3, determines an isomorphism

A(x): S - T *C.

By comparing V'(x) * f (y) with V(x + y) over the identity point of A we obtain

(6.3) V(x) * V(y) = F-l(x, y)V(x + y).

In order to split (6.2), we need to find a map

X: A Gm.

so that

WKWA(x * X(yWAY) = (x + y)A(x + y)

In terms of the power series representation of X, this means that

X(x)X(y)/X(x + y) = F(x, y)




We explicitly characterize 0D. Using the section Xb with the method and notation of Section 3 we put coordinates on each stalk:

k(x) Q(x): S (o) (T *2)(o) = S

(x)

1 -a X(x)F'l(x, o)(T*f )(o)/fi(o) = X(x)f1(x)/fj(o).

Now s (x) = fi(x) in terms of trivializations, so

GD(x) = fi(x)/X(x)fi(x)fi(o)')

= XW-1 * (constant)

We conclude that, up to a constant,

(6.4) =D = F.

Remark 1. Notice that 0D iS unique up to a linear exponential, i.e., a homomorphism of formal groups

A -+ GM .

Also note that two splittings of the extension (6.2) differ by a linear exponential.

Remark 2. If D is totally symmetric we can define OD uniquely up to a constant by requiring 0D to be even.

Remark 3. If D is linearly equivalent to zero, say D = (f ), then we can find a preferred splitting of (6.2), namely the one which extends to a splitting of (6.1). This gives

OD f

Remark 4. If e is in the support of D, then we can find a rational function g so that e is not in the support of

E = (g) + D.

Then we can proceed as above with E. We obtain

OD =

OE g



638 PETER NORMAN

7. Barsotti's Theta Function for General Divisor. We associate to a divisor D an element OD, of SR. If A X k is ordinary, then OD is a power series with coefficients in R. We outline the construction:

STEP ONE. The theorem of the cube implies that D2L is trivial; this allows us to construct the structure of biextension on

O*S = OS3-{o} -+A X>A.

The biextension structure is given by two compatible families of group operations. If we consider A X A as an abelian scheme over A via the first projection (write AA in this case), then one of the group operations gives an extension of group schemes

(7.1a) 1 +AGm - * AA O;

Analogously, the second group operation gives rise toQ a second extension:

(7.lb) 1 GmA -S* AA ? ;

STEP TWO. We restrict the biextension toA X A (orA 'X A if A X k is ordinary) and split the extension in (7.la). This distinguishes a family of isomorphisms

S(S)B C { (4, X) I 0: S TX*, x EA}.

The group S (B) sits in an exact sequence

(7.2) 1 Gm S ) B A O.

(Again A is replaced by A is A X k if ordinary).

STEP THREE. We show that the partial splitting of the biextension in step two actually splits the entire biextension structure. This leads to a splitting of (7.2) and thence to OD is the usual way.

We now carry out these steps in detail.

STEP ONE. We give (05)* the structure of biextension. The line bundle

Jr2* -+A




is algebraically equivalent to zero; indeed, upon applying the operator (Id X D) we obtain

(Id X )) 9 = V92

which is trivial by the theorem of the cube. Thus

3)2* -AA

has a group structure which fits into the exact sequence (7.1a). To compute the group law we proceed as in Section 4: Apply Id X O to O)D; this gives the divisor 0iT2D on A(A X A). Let G be a rational function on

A(A X A) whose divisor is 02D. Normalize G by requiring

G(o, y, z)=1

Then G-1 (x, y, z) gives a rational scheme (but not group) theoretic splitting of (7.1a); with respect to this section the group law is G' (x, y, z) where x is considered as the parameter on the base space.

By considering the second copy of A in A X A as the base space, we get a second group law on 052*. Since G is symmetric under permutation of variables, the second group law is also G-1 (x, y, z) except z is now the parameter on the base space. It is easy to check that these two group laws are compatible and give a biextension.

Remark 7.1. Since G(o, y, z) 1, the fiber of 0D52* over e X {A } considered as extensions of A by Gm is trivial; moreover the section of this fiber provided by G(o, y, z) is the unique one splitting this group extension. The analogous statements are true for A X {e }.

STEP TWO. We restrict the biextension to a small neighborhood of the identity and split the extension (7. la). For the moment, assume A X k is ordinary; we postpone the general case. Let 052* denote the formal completion 052* at the identity. We obtain an extension of formal groups from (7.la):

(7.3) 1 AGm 05* AA o

LEMMA 7. 1. The extension (7.3) splits.



640 PETER NORMAN

Proof (sketch): Let p'H denote the kernal of multiplication byp on a formal group scheme H. Since Gm is p-divisible, (7.3) gives the exact sequences:

(7.4) 1 /,pn _lpn j)7 * -,pnAA - 0.

Since the base space is a complete local ring with algebraically closed resi- due field we can split the Cartier dual of (7.4) and hence we can split (7.4). Q.E.D.

The splitting of (7.3) is given by a map

: :AA -+Gm ;

the power series associated to X (again denoted 4) satisfies

(Id X D)) = ?(x, y + z)/c(x, y)ck(x, zY = G'.

We now treat the case of A X k not ordinary. Restrict 05* toA X A; denote this by D 2 * and consider it as a formal group scheme over A.

LEMMA 7.2. The exact sequence

(7.5) O -Gm *S* -AA O

splits.

Proof. The proof follows Proposition 4 in [7] page 319 exactly. Let nA denote the connected component of the kernal of multiplication by p' on A; let ?n denote the restriction of D2* to nA X nA. Consider nA X nA as group scheme over nA via the first projection. To split (7.5) it suffices to construct a sequence of compatible maps:

on

On l rn

2nA X nA nA X nA (pn, 1)

We define On. Let (x, y) E 2nA X nA; choose z1 E ?2n so that

1r2n (z1) = (x, y). Denote byp pn (resp. p n) the operation of adding an element of D4 to itself pn times in the biextension using the first (resp. sec-




ond) group operation. Use the canonical splitting of DIC* along the fibers {e } X A and A X {e} to give coordinates on these fibers. Since

rn(pn(zl )) =x (x,) pO,(zl) = (X, X, O) with X E Gm, and we can find an element Il E G, so thatp 2(z1) = (1, x, o). SetZ2 = jAz, and define

on (X, Y) = p1(n2

Notice that Z2 is characterized up to a (ppn )th root of unity by

(7.6a) 1rn (Z 2) =(X, Y)

(7.6b) pn2(z2) = (1, x, );

furthermore 'On (X, y) is independent of this choice of root of unity. It is clear that n is additive. To show that the 4n's patch together we

need to check that the diagram below commutes:

(2n + 2)A X (n + 1)A - ?n+1

t 11

(2n + 1)A X (n + 1)A

~P t t 2nA X nA -I-4 n

Let (x, y), z1 andz2 be as above. Let (x', y') e (2n + 1)A X (n + 1)A satisfy y' = y and px' = x. Choose z' so that 7r(z?) = (x', y) and

p 2 l(Z2') =(1, x', o);

then pnI +I (Z2') = 'On+ (X ', y'). To show that On+ (X', Y') = On (X, y), it

suffices to show that

(7.7) P1Z2 = Z2

up to a (pn)th root of unity; for if (7.7) holds,

n (x, y) pZ2 = P(pZ2) = n+l(Xy)

To show (7.7), we need to show that p1z' satisfies



642 PETER NORMAN

(a) o(PIZ2) = (x, y),

(b) P2n (PIlZ2 ) =(1 , o, y).

Part (a) is obvious. Notice thatp'z= (X', x', o) sincey' E nA; furthermore

(,x, ?) =P2+ Z2 = P2(, x', 0) = (Xp, x', O).

We see that XP = 1. Now

p (P1(Z9)) = pl(p2(Z2)) = p1(X, x, o) = (1, x, o).

Q.E.D.

With respect to the scheme splitting of D42* induced by G-1, the splitting of DS* is given by an element of SR 9 SR that satisfies

(7.8) (Id X )) = G-'.

We normalize 0. If we set x = 0 in (7.8), we get

(Id X D) (X (o, y)) 1.

Since we can multiply 0 by a multiplicative power series in y, we can replace 0 by ?(x, y)>(o, y)-; hence we may assume

(7.9) 4(o,y) 1.

Substitutingy = o in (7.8) shows that +(x, o) 1. We show how the splitting OG-1 constructed above gives rise to a

distinguished group of isomorphisms

with x E A if A X k is ordinary and x E A otherwise. The fiber of D43 over {x } X A is an algebraically trivial bundle, S 0) T*SC which we denote by ?X. The splitting X G' trivializes SC = 0 T *S; but to do this is equivalent to giving an isomorphism from S to T *.. The isomorphisms




we wish to distinguish are the ones induced by multiples of + G-1. We work out explicitly what these isomorphisms are. One way of giving a trivialization of Sx is to give, for each y E A, an isomorphism

SX- > Ty*2x

and by evaluating this map over z = e we get an identification of ?tx(e) with 2x(y). This is what the isomorphism G-1 does. We look more carefully. Specialize to z = e; to give a point in ^x (e) is to give an isomorphism

?(e) -(x);

furthermore points in ?x(y) correspond to isomorphisms

S(y) S(x +y).

In this way, asy varies, we get a map from S to T*S. Since G(x,y, o) 1, this map is given by multiplication by + (x, y) (we use the conventions in the calculation in Section 3). Let k denote the induced isomorphism

Let

2--+ T IeG ,XA} S(SB = : {X

' XI S E TS GM, X A}

The group 9 ()B sits in the exact sequence

(7.11) 1 G S ( S)B A 0.

If A X k is ordinary, we replace A by A.

STEP THREE. We show that 4 splits the biextension when restriction toA X A and later use this to split (7.11). In addition to (7.8) we need to show

(7.12) (JD X 1)4 = G-1.



644 PETER NORMAN

Define maps zr1, -r2, 12 fromA X A to A by

-XI(X,y) = x, -r2(x,y) Y, r12(X, y) = X + y

Now the extension (7. la) has base space A. Pull 7.1a back toA X A via -rl, i2, and ir12. This gives extensions E1, E2 and E12. Let E denote the extension with base space A X A gotten by adding, as extensions, E12 - E-

E2. The extension E has group law

(D X Id X Id)G'l.

Since 0 splits 7.5 (or 7.2 if A X k is ordinary), (D X 1) splits E with respect to this group law; hence

(Id X Id X D)(D X 1)0 = (D X Id X Id)G'l.

This implies

(D X Id)(0)(x, y, z) = (z)G (x, y, z)

for a linear exponential it (z). By substituting x o and using X (o, y) 1 in the above equation, we see that ,u(z) 1. We conclude that (7.12) holds.

Having constructed a splitting of the biextension, we are now able to produce a distinguished section for (7.11). First notice that the group law for ?(O)B is +(x, y); next notice that (7.8) and (7.12) together imply

+k(x, y)/(y, x) = X(X, Y)

is bi-multiplicative. We now treat the ordinary case. Since 0 (x, y) is a power series in both

variables, so is X. Since there are no non-trivial bilinear maps

A XA-Gm

if A is ordinary, we conclude that X 1. This means 9 (S)B is a commutative formal group and, extending byR if necessary, we can split (7.11). The splitting is given by a power series X E SR which satisfies

DX = q




We calculate AD by evaluating a section s (possible rational) of S whose divisor is 0). The evaluation is done using the splitting X(x)O(x, y) of ?(OC)B. We find, just as in Section 3, that, up to a constant,

O = XX

and hence

(7.13) DOD O

Combining this with (7.8) we see that

(7.14) D20D = G.

This is the equation Barsotti and Cristante take as a definition of OD.

Clearly AD is unique up to a quadratic exponential.

Remark 7.2. It A is a quadratic exponential in So then DA is bilinear. If A X k is ordinary, then DA 1. This means A is actually a linear exponential, hence (7.14) determines OD, in the ordinary case, up to a linear exponential.

We now treat the general case. The pairing X is no longer a power series, rather it is in SR 0) Sj. In general X is not identically one and hence (7.11) does not split. Instead of splitting (7.11), we find a section so that the group law assumes a special form. Consider the group law

X- 1/2.X4;

this defines an extension of Gm by A. Let X E S split this extension; hence

X=x -

-1/2o.

The group law of 9 (S)B with respect to the section X * is the bimultiplica- tive element X/2; furthermore, evaluating a section s of r (A, S) such that (s) = D using X gives, up to a constant,

OD = XA

Again OD satisfies

D20D = G.



646 PETER NORMAN

Remark 7.3. Again 0D is unique up to a quadratic exponential; moreover there are non-trivial quadratic exponentials in the ring S?. We can characterize 0D up to a linear exponential by examining the construction in more detail. We begin with the

LEMMA 7.3. There is a unique element X in SR 0 SR such that

(D X Id)q$ = (Id X D) = G-1.

Proof. We only need to show uniqueness. This will follow once we show that there are no nontrivial bilinear pairings.

t:A XA G

This follows from the two claims.

Claim L Let B be ap-divisible formal group over R and assume B X k is of local-local type; then any bilinear pairing

AX:A XB Gm

is trivial.

Proof. By viewing A X B as a formal group over A via the projection map, we can view A as inducing a morphism of formal groups over A:

A X BA X Gm.

Since B is of local-local type there are no non-trivial homomorphisms of this kind.

Claim I. Assume B is a multiplicative type. There are no bilinear pairings.

AXB Gm.

Proof. On the finite layers the pairings are necessary trivial. Q.E.D.

The lemma shows that if we define 0D by

(7.15i) 5)OD = 0-1 X1/2




(7.15ii) 0 Ei ~SO,

(7.15iii) (D X Id)) = (Id X D))4O G

(7.15iv) 4(x, o) 0(o,y) 1

then AD is unique up to a linear exponential. We say a theta function AD is special provided it satisfies (7.15i-iv). In addition notice that Lemma 7.3 implies that x is unique also.

8. Comparison of different thetas. Assume S is an ample, totally symmetric line bundle; lets E r(A, S) be such that (s) = D with D = (fi) locally and i*fi fi. From this data we obtain 6D and E, both elements of

SN. We compare these as explicitly as possible. The element AD is characterized by: 02OD is a rational function on A X A X A with divisor 5D2D. This determines 0D only up to a quadratic exponential. By Proposition 2.1 6s also satisfies this property, hence As is one of the OD's. This raises the question: Among the elements of Si? characterized as a OD, how do you pick out As ? The answer is given in

PROPOSITION 8.1. For an isogeny xr0: A0 - A, let S, denote 7r*iS. If A X k is ordinary, assume r is etale. Assume 0 is an element of SR such that for any y in the connected component of H(2,C0)

fy(x) = 0(x)eCa<x,y/2>/0(x +y)

is rational on A. with divisor

7r*D - TY*7r*D.

Then 0 = A, up to a constant.

Proof. The explicit calculation in Section 3 shows that

Os (Y)-=G - (o) G- 1 (z)

where 2z = y, G is rational on A., and G satisfies

(i) i*G T*G = 1 (ii) (G) = r*D - T*7r*D.



648 PETER NORMAN

It is easy to check thatf,(x) satisfies properties (i) and (ii). Now a simple calculation gives

0s(Y) = G(y)0(o')-. Q.E.D.

We come to the question: What is the relation between a special Bar- sotti theta, OD, and As ? The answer is given in

THEOREM 8.1. Up to a linear exponential As = OD.

Proof. The first step in establishing the relation between As and 6D is

to construct elements 4?m and e2 which are Mumford analogues of Barsot- ti's 0 and X. Let A (n) denote a copy of A covering A by multiplication by pn; let mA (n) denote the connected component of the kernal of multiplication bypm on A (n). To define an element of SR ( SR we need to define a compatible family of maps

2nA(n) X 2nA(n) - A'.

Let Sn denote the line bundle pn*2 on A (n); let

A (n): 2nA (n) -+ A (n) X 2nA (n)

be the diagonal map. Using criterion M we construct a distinguished isomorphism

A +A(n) ?n -

Using the conventions in Section 3, we see that kA(n) is induced by a unique rational function 0 (-, A(n)) on A (n) X 2nA (n). Restricting the first variable to 2nA (n) we obtain

On 2nA(n) X 2nA(n) -+ A'

Properties 2.1 and 2.2 insure that the diagram below commutes:

(2n + 2)A(n + 1) X (2n + 2)A(n + 1) A

(2n + 1)A(n + 1) X (2n + 1)A(n + 1) P P

2nA (n) X 2nA (n) A

Let OM denote the limit of the On's.




We can also patch the pairings e n together to obtain a bilinear element of SR 0 SR. To do this it is necessary that the diagram below com- mute:

ekn+1 A

(2n + 2)A(n + 1) X (2n + 2)A(n + 1) Gm

(2n + 1)A(n + 1) X (2n + 1)A(n + 1)

eA 2nA (n) X 2nA (n) G

Since e"n is constructed from <tn and the 4,, patch together, so do the e We express Barsotti's 4 in terms of OM. From the proof of Prop. 8.1

we have

'kM(Y, x) (6,(y)egn <y, x/2>)/1O(x + y);

hence OkM(o, x) = ,(x)f. We replace OkM by

'1'M(Y, x) = 'M(Y, x)Os(x) = 0DO7'(y, x)ec(y, x/2).

Since D2Os = G and es is bilinear, (kM satisfies 7.15(iii); furthermore a direct check shows '?M satisfies 7.15(iv). In general 7.15(ii) does not hold for (DM.

We proceed by identifying the obstruction to 4)M being an element of so S. This leads to the construction of a bilinear element B E 8 0 8

such that

4,m(y, x)B(y, x) E SiR () S;R

This implies that 4M(y, x) * B(y, x) = ?(x, y). The element "DM would be in S'R ( R provided, for each n, the map

On in the diagram below descended to a mapfn:

2nA(n) X 2nA(n) A pn fn

nA(o) X 2nA(n)

The obstruction to this descent is the pairing



650 PETER NORMAN

(y, x) -+ <y, X>n* = (t(Z, x)/4"(z + y, x)

nA (n) X 2nA (n) GmM.

Indeed, by [9], if <y, x >n* is identically 1 for ally E nA (n), the mapfn does exist. Writing<, >C, n (instead of < >Cpn) for the Riemann pairing we see ([6] page 184) that

<y, pnX> oSln = <y, X>n

(the exponential -1 arises because On comes from a function with divisor (T*D - D) while the Riemann pairing uses D - T*D).

Claim. If there are bilinear pairings

Bn (y, x): 2nA (n) X 2nA (n) Gm

extending <y, x>*, and if the Bn are compatible and hence give an element B of SR 0 SR then

4IM(Y, x)B(y, x) E S'R 0 SR

Proof of Claim. It suffices to show that 4(y, x)B(y, x) descends to a map

(2nA (n))/(nA (n)) X 2nA (n)) -- A

Lety, x e 2nA (n), z e nA (n); then

4IM(Y + z, x)B(y + z, x) = DM(Y + z, x)B(y, x)B(z, x)

= (DM(Y + z, x)B(y, x)4M(y, x)/4M(y + z, x)

= -M(Y, x)B(y, x).

This proves the claim. We extend the pairing <y, x >* by defining B (y, x) via the diagram:




nA (n) X 2nA (n) Gm

nA (o) X nA (o) Gm

t t < -1 11 2nA(o) X 2nA(o) Gm

2nA(2n) X 2nA(2n) BGm)

To check that the diagram commutes, we merely need to check that the middle square does since the top and bottom squares do so by definition. To do this we use the property of the Riemann pairing:

(8.1) <y, X>IS,n+l = <PY, PX>,Cn

(see [6] page 185). Letp'x, pny E nA(o); by (8.1)

< pny, p nX > pit2 (pfy ">2,n (Y, X>2,2,,

bilinearity gives

<py,Y X >SC 2n -<Y' X >Pn

Thus the diagram commutes. We now check that the various Bn form a compatible system. We

check that the diagram below commutes:

(2n + 2)A(n + 1) X (2n + 2)A(n + 1) n+ Gm

it 1 (2n + 1)A(n + 1) X (2n + 1)A(n + 1)

iP l~~~~P

2nA (n) X 2nA (n) B n Gm

Letpx1 = x andpy1 = y E(2n + 1)A (n + 1), then

Bn+1(Y, x) = <Y' X> S,2n+2 = <Y1 x >15n+2



652 PETER NORMAN

On the other hand, using (8.1), we have

<Y1, x1> 2,2n+2 = <y, x>7,21 = <pypx>'2n = Bn(py, px).

We conclude

(8.2) 4'M(Y, x)B(y, x) = (x, y).

The final step in relating 4?M(y, x) and +(x, y), and hence As and OD ,

is to identify B(y, x)- as es (y, x). We have ([6] page 228)

(8.3) <y, X>s n = e <y, x>.

We wish to compare

en(< >: 2nA(n) X 2nA(n) Gm

and

(Bn)-': 2nA(n) X 2nA(n) 31 Gm

< >2, 2n: 2nA (n) X 2nA (o) 31 Gm

LEMMA 8.1. e n _ Bn7

Proof of Lemma. We have

Bn (Y, x)1 = <Y, x >C, 2n

= e Cp2n

(y, X> (by 8.3)

epn*s(y,x)

= e n<y, X>.

The third equality results from the fact that S is symmetric and thus SP

is algebraically equivalent to pn*?. This proves the lemma. From [3] and Lemma 8.1 we have




furthermore, from the definition of 4DM we have

-tm(y, x)e' (x, y) = DO-1 (y, x)e2 (y, xl2)e2 (x, y)

= DO- 1 (y, x)e2 (x, y/2)

This enables us to prove

(8.5) x(x,y) = e (y, x).

We have, by definition,

x = X (x, y)/c (y, x)

which by the equations above is

X = e2 (y/2, x)e2 (x/2, y)-

= e"(y, x).

We now finish the proof of the theorem. Since 0D is determined by its functional equations and the requirements 7.15(i)-(iv), we only have to show that DOS = DOD. We have, by definition

DOD(X, Y) = -

(X, y)x (X, y);

now using (8.4) and (8.5) we obtain

DOD(X, y) = SDO(y, x)e2(y, x/2)-1' e2(y, x/2)

= 5)0, (y, x). Q.E.D.

Part IH.

9. Differential equations. Let R = W(k), the ring of Witt vectors with entries in k; let E be an elliptic curve over R and assume E X k is ordinary. Let D be the divisor 2e where e is the identity point on E; let 0D be a Barsotti theta function associated to D. Let X be a non-vanishing invari-



654 PETER NORMAN

ant differential on E and let D be the derivation dual to c; then there is a constant C which is determined by E and D such that ([1])

02 log OD = -2p + 2C.

Here p is the Wierstrass function associated to E and w.

Problem. What is C?

Another way of looking at this is: Find a constant C so that the differential equation above has a power series solution with integral coefficients. The aim of this section is to solve this problem.

We begin with some classical motivation. Let z E C; let T be in the upper half plane, and set q = e2'iT. The theta function associated to the lattice L = Z + Z * r and divisor 1 * e is

a(Z, T) = -i S (- (nq+(112)) 27xi(n+(112))z. 6(z, r) = -i E -)'

furthermore a classical calculation shows

( d )2log0(z, r) dz = (-p(z,L) + 771)dz

where 1q is the period

jApdz.

Notice that the differential

(-P + -,l)dz

has periods independent of r; indeed they are o and zxi. This suggests that we study the p-adic variation in the periods of differentials on families of elliptic curves.

The study of the p-adic variation of periods was initiated by Dwork (e.g. [3]). We formulate a part of this work following Katz ([5]). Let S =

Spec(B) be a smooth affine variety and let

90 -+Spec(B) = S




be a family of elliptic curves; let S' be ap-adic completion of S, i.e.,

So = Spec(lim(B/p'B)).

Let ? be the restriction of &; to S'. Let 0 be a lifting of the Frobenius on W(k) to B. Then HDJR (E) is equipped with

(a) the Gauss-Manin connection,

V: HDRR(&) --+HDR(&) 0 Q, and

(b) an action of Frobenius

F: 4*HDR (E) -+HDRj(E)E

Together (HDJR (E), F, V) gives an F-crystal. In addition, we have a locally free submodule of rank 1 with locally

free cokernal

HO?(Q) HDR()

This filtration satisfies

F: q*(HO(Q) -+pHD R (&)

We say ? is ordinary provided the Hasse invariant of ? is ap-adic unit. In this case we have a basic theorem due to Dwork ([3], [5]):

THEOREM. There exists a unique suberystal (hence stable under V and F) U C HDJR (E), locally free of rank 1, that satisifies

(i) F is an isomorphism on U, and (ii) HDR(&) = H?(Q) ? U.

THEOREM 9.1. Assume ? is ordinary; let D be the divisor 2 * e, let S be the line bundle associated to D and let s E I(&, S) be a section with divisor 2 * e. Let X be a non-vanishing invariant differential on ? and D, the dual derivation. Then

(jD2log 0) C=o

spans the suberystal U.



656 PETER NORMAN

Proof. It suffices to show that 7 is in the unit eigenspace for the action of Frobenius on HDJR (6). Since ? is ordinary, for each integer n 2

o, there is an elliptic curve ?(n) and maps

Fn: 6 -8 (n)

V: 8(n) ?

satisfying

(i) Vn oFn = Fn o Vn = multiplication by pn, and (ii) Vn is etale and of degree pn.

To show - is in the unit eigenspace it suffices to show that for all n > 0

n*1 0 mod pn;

Let Wn = W(k)/pn W(k); we must show that V"*(8) considered as an element H)DR (E;(n) X Wn ) is zero. Let 6/n denote

6 X WnIV[]/(eC);

let

e: Spec W, [dA/(E2) 6 U/n

be the map that maps the closed point to the identity and induces an iso-

morphism on tangent spaces. We may assume c maps to D. Then

VOS _ 1 OS(z + e) -As() s-- Os (z)

1 (O (z+e) _ e O\ a(z) /

Since e is a point of the connected component of points of order n on 6/n,

by Remark 3.2, we have that Os(z + c)/0s(z) is rational when pulled back to 6(n)/In by V,,; hence

d (Vn*'QOs) = d (rational function on En)1n),




i.e., it is exact on g(n)/n. Since V,, is etale, we have

d(Vn*D)Os) = Vn*(dQOs).

Thus we have shown Vn*(d(Qs)) is exact modulo pn. This completes the proof. Q.E.D

Given a family of elliptic curves ? and a nowhere vanishing differential X we can write

?: y2 = 4x3 - g2X -

with

dx

y

furthermore, { (dx/y), x (dx/y) } provides a basis for HDJR (g). We characterize a constant C = C(&, w) by

x dx dx ~~+ C-

y y

spans the submodule U. Katz [4] in the case of the Tate curve over W(k)((q)) computed C. He found that

(9.1) C(q) = 1 -24 E orl(n)qn. n21

Theorem 9.1 asserts that

xdx dx (D020S)w= + C

y y

The above remark can be taken as the starting point of a second proof of Theorem 9.1 using modular forms. Recall that ap-adic modular form is a rulef that associates to every pair, (E, ), consisting of an ordinary elliptic curve ? over a W(k)-algebra S and a non-vanishing differential w on ?, an element of S. This rule must satisfy:



658 PETER NORMAN

(1) f depends only on the isomorphism class of (E, co), (2) f commutes with base change, and

(3) f(6, X@) = Xk f(, c)

We say k is the weight of f. We now define two p-adic modular forms. One is

(E, I ) i- C( , w).

The second one, P, is given by the formula

(Q2log & C,o =-2x dx + 2P-. y y

Here w = (dx/y) and D is the derivation dual to w. It is easy to check that both P and C are weight 2.

To prove Theorem 9.1 we must show P = C; to do this, by the q- expansion principle, we must merely show P agrees -with C on the Tate curve with its canonical differential. For C, we have (9.1). To calculate P we must first calculate As on the Tate curve; this is done in the appendix. We find

OS = X( II (1 - qnX'1) II (1 - q-nx))2 n -- n<O

where X is a multiplicative parameter on the Tate curve. Now a classical calculation shows that, indeed, P = C.

Appendix: Calculation of Os on the Tate curve. Let E be the Tate curve over W(k)((q)). Let D be the divisorD = 2 * e and let S be the corresponding line bundle. Let s E F(E, S) be a section with divisor (s) = D. We calculate Os in terms of Tate's theta:

OT(X) = II (1 - qnX-1) II (1 - qnX). n?O n<O

The variable X is multiplicative on E. The rational functions on E are all obtained as products and quotients of translation of 6T with possibly a fac- tor Xn thrown in. Any such expression that is q-periodic is rational on E.

We use the following basic formulas:

f A-1) Of q _-X) = _ _

T(X)




We have

(A-2) i*OT(X) = -XOT(X).

From this we see that

X02 (X)

is even. We now outline the calculation of Os at a torsion point x,. Let x0 = 2yO

withyo a point of order n. Of course, in general, S * Ty* ?; we must work on the covering of E by itself by multiplication by n. Let x, y E E so that zy

x and ny = y,. We then have

n*y = Ty(n*J0

We begin by constructing a rational function on EKwith divisor

n*D - Ty*n*D

Let

F =n*(XO2)

and

Go = F/Ty*F.

Although Go is not q-periodic, we can obtain a q-periodic function by mul- tiplying Go by a suitable power of X. We sketch this calculation. First from [1] we get

F(q'X) q n2X2n2 F;

second we have

(TyF)(q-'E) - qn2X-2n2y2n2F.

Thus

G.(q X) =y2n2 Go(X)



660 PETER NORMAN

Since y is a point of order n2 on E, y-2n2 = q2r for some integer r. Set

G =X-2rG;

then G is q-periodic. We adjust G so that it satisfies criterion M. First note that since

i*F = F,

Ty Goi*Go= Ty*- (F/Ty*F)i*(F/Ty*F)

- 1.

From this we see that

TY*-ai G i*G= (Ty*- X-2r)(i*X-2r) =y2r

We conclude that

G = y -rG = y -rX2rn* (XOf2)/Ty*n * (XO2)

gives an isomorphism from ?n to Ty*Cn that satisfies criterion M. We calculate AS. From (3.1) we have, up to s constant,

0,(xo) = Gy(y)-'G-'(y).

Thus, up to a constant

05(Xo) = [y-ry2r(n*(X02) jIX=y 1 [Ty*n*(X02) X=Y

[y-rn*(XO2) Jx=1 ]1 [Ty*n*(Xf42) IX=- I.

Now using yn = yo and y2 = x0 we obtain

0 (xO) = y2nx O 2 (X0) 02 (1).

Notice that if xo is a root of unity we may choose n = 0; this is the canonical lifting of xo to the covering of E by itself by multiplication by n. In addition the power series theta function is determined by these xo; thus we conclude that

P0O (X) = X02 (X).




The discrepancy between As and 0T on the torsion points that are not roots of unity should be expected since As does not argue with the classical theta function ([8]).

UNIVERSITY OF MASSACHUSETTS

REFERENCES

[1] I. Barsotti, Considerazioni sulle funzioni theta, Symp. Math., 3 (1970), 247-277. [2] V. Cristante, Theta functions and Barsotti-Tate groups, Ann. Scuola Norm. Sup. Pisa,

7 (1980), 181-225. [3] B. Dwork, p-adic cycles, Pub. Math IHES, 37 (1968), 327-415. [4] N. Katz, p-adic properties of modular schemes and modular forms, Modular Functions

of One variable III, Lect. Notes in Math. 350, Springer (1973), 89-190. [5] N. Katz, Travaux de Dwork, Expose 409, Seminaire Bourbaki 1971/72, Lect. Notes in

Math. 317, Springer, (1973), 167-200. [6] D. Mumford, Abelian Varieties, Oxford University Press, 1970. [7] D. Mumford, Biextensions of formal groups, Algebraic Geometry, Oxford, 1968. [8] D. Mumford, Equations defining abelian varieties I-ILL, Inventiones I (1966) 287-354;

III (1967) 75-135, 179-214. [9] M. Raynaud, Passage au quotient par une relation d'equivalence plate, Local Fields,

Springer, 1967. [10] P. Roquette, Analytic Theory of Elliptic Functions over Local Fields, Vandenhoeck u.

Ruprecht (1970).



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