SYSTEMS OF ORTHOGONAL POLYNOMIALS ARISING FROM THE

Transcription

1 SYSTEMS OF ORTHOGONAL POLYNOMIALS ARISING FROM THE MODULAR -FUNCTION STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH Abstract Let S p x F p [x] be the polynomial whose zeros are the -invariants of supersingular elliptic curves over F p Generalizing a construction of Atkin described in a recent paper by Kaneko and Zagier [5] we define an inner product ψ on R[x] for every ψx Q[x] Suppose a system of orthogonal polynomials {P nψ x} n=0 with respect to ψ exists We prove that if n is sufficiently large and ψxp nψ x is p-integral then S p x ψxp nψ x over F p [x] Further we obtain an interpretation of these orthogonal polynomials as a p-adic limit of polynomials associated to p-adic modular forms Introduction and statement of results Let z denote the usual modular function of weight zero z := q Here and throughout this paper q = e 2πiz Given ψx R[x] we define a symmetric bilinear form on R[x] by letting for fx gx R[x] f g ψ = 6 π π/2 π/3 fe iθ ge iθ ψe iθ dθ For ψ = Kaneko and Zagier investigate this bilinear form which is due to Atkin in great detail [5] Here we generalize some of their results for most ψ Q[x] We begin by making the following definition: Definition A polynomial ψx R[x] is good if there exists a set of monic orthogonal polynomials {P iψ } i=0 with respect to ψ That is the degree of P iψ is i and P nψ P mψ ψ = h mn δ mn where h mn R and δ mn is the Kronecker delta-function Remark For ψ to be good it is sufficient that ψ be a definite bilinear form Since e iθ maps [π/3 π/2] to [0 728] biectively any ψ with no zeros of odd order in the interval will induce a definite bilinear form ψ This observation immediately implies that the vast maority of ψ are good; more precisely if we fix all but the constant term of a prospective ψ Z[x] there is only a finite number of constant terms for which ψ could possibly be bad Date: November The authors would like to thank the following organizations for their generous support of the 2003 Research Experience for Undergraduates that led to this paper: the National Science Foundation the University of Wisconsin at Madison the David and Lucile Packard Foundation the Alfred P Sloan Foundation and the John S Guggenheim Foundation

2 2 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH Let p 5 be prime and let E/F p be an elliptic curve Then E is called supersingular if EF p has no p-torsion The -invariant of such an E is called a supersingular -invariant over F p We fix the notation 2 Ω p := { i : i is a supersingular -invariant over F p } and 3 g p := Ω p We define the supersingular locus S p x as 4 S p x = i Ω p x i F p [x] The fact that g p is finite and that S p x F p [x] is well-known see for example [5] [7] In order to relate a polynomial in Q[x] to S p x we require the notion of p-integrality We will define p-integrality using the p-adic norm Fix a prime p Any nonzero t Q may be written uniquely as t = a b pr where a b r Z b > 0 and gcdp ab = gcda b = We then define the p-adic norm for v Q as { p r if v 0 5 v p := 0 if v = 0 A rational number t is said to be p-integral if t p A polynomial fx Q[x] is p-integral if all of its coefficients are p-integral; in this case the reduction modulo p is well-defined With this in mind we make the following definition Definition Let ψx Q[x] be good and write n P nψ x = a i x i and ψx = If p is a prime then we define the following powers of p i=0 m b i x i i=0 6 α nψ p = max { a i p : 0 i n} β ψ p = max { b i p : 0 i m} Note that α nψ pp nψ x and β ψ pψx are both p-integral We may now state our main theorem Theorem Fix a prime p 5 and suppose ψx Q[x] is good Let over F p [x] If n g p d ψp then in F p [x] d ψp := deggcds p x β ψ pψx S p x β ψ pψxα nψ pp nψ x Remark This theorem in the ψx = case recovers a theorem of Atkin described in a recent paper by M Kaneko and D Zagier [5] Our method offers a new proof of this result

3 ORTHOGONAL POLYNOMIALS AND MODULAR FORMS 3 Given Theorem it is natural to consider the p-adic properties of these systems of orthogonal polynomials Before we state our result in this direction we require some basic facts from the theory of modular forms For an even integer k 2 define M k to be the finite dimensional vector space of holomorphic modular forms of weight k under the action of the full modular group SL 2 Z We fix the notation σ k n := d k d n and denote by B k the kth Bernoulli number Then the Fourier expansion of the classical Eisenstein series E k z is E k z := 2k σ k nq n B k For k 4 we have E k z M k Although E 2 z is not a modular form it will play an important role later In addition we recall the -function n= z := E 4z 3 z the unique normalized cusp form of weight 2 Remark In terms of z and E 4 z we have z = E 4z 3 z Using these modular forms we define the notion of a divisor polynomial of a modular form If k 4 is even then define Ẽkz by if k 0 mod 2 E 4 z 2 z if k 2 mod 2 E 4 z if k 4 mod 2 7 Ẽ k z := z if k 6 mod 2 E 4 z 2 if k 8 mod 2 E 4 z z if k 0 mod 2 Further define mk by mk := { k/2 if k 2 mod 2 k/2 if k 2 mod 2 With this notation if fz M k and F f x is the unique rational function in x for which 8 fz = z mk Ẽ k z F f z then F f x is a polynomial which we will refer to as the divisor polynomial for f From 7 and 8 the polynomial F f x will have zeros precisely at the -invariant of the zeros of fz with possible exceptions at e 2πi/3 = 0 and i = 728 depending on the weight of fz modulo 2 For a discussion of divisor polynomials see [] Additionally we require the machinery of continued fraction expansions which are of interest in their own right In particular in Section 4 we examine the continued fraction

4 4 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH expansion of ψz E 2zE 4 z Though it may seem most natural to write this as a continued z fraction in terms of q it will be useful to regard it formally in terms of = q 744q 2 + Z[[q]] That is we write ψz E 2zE 4 z z The nth partial convergent of this expansion is 9 ψ ψ = λ λ2 λ3 λ λ2n We show in Section 4 that 9 is equal to Q nψ for certain Q P nψ nψx Q[x] see 42 and P nψ x as above These partial convergents are essentially the p-adic limit of the divisor polynomials of a specific family of p-adic modular forms The order of the error term depends only on n More precisely we have the following result Theorem 2 Let ψx Q[x] be good p 5 be prime and r be a positive integer With α nψ p and β ψ p defined as in 6 and Q nψx P nψ as above we have x F E 4 E φp β ψ pψ r +2 F E p pr β 2n ψpq nψ α nψ pp nψ + O mod p r where φ is Euler s φ-function In the next section we prove some useful general properties of the bilinear form defined by In Section 3 we prove Theorem and in Section 4 we discuss continued fraction expansions in preparation for the proof of Theorem 2 in Section 5 We conclude in Section 6 with an efficient method of computing the P nψ x s and apply this method to illustrate some of our results 2 Preliminary observations on the bilinear form Fix ψx Q[x] and let fx gx Q[x] be polynomials Define rational numbers am and bn as the coefficients of the formal power series in q 2 fgψe 2 = amq m m m 0 and in 22 fgψ E 2E 4 = n n 0 bn n

5 ORTHOGONAL POLYNOMIALS AND MODULAR FORMS 5 Note that am and bn depend on ψ f and g We have the following proposition see also [5] Proposition 3 Let fx gx ψx Q[x] With the notation above we have f g ψ = a0 f g ψ = b0 f g ψ = E 4 fxgxψx 2πx x dx As we will see Proposition 3 simplifies the explicit computation of orthogonal polynomials with respect to ψ In addition it allows us to use the arithmetic of the coefficients of modular forms to deduce properties of these polynomials Before we begin the proof of the proposition we recall the theta operator 26 Θ = q d dq = d 2πi dz and Ramanuan s classical identities see [6] exercise III28 27 ΘE 4 = E 4 E 2 /3 and Θ = E 2 E 2 4/2 Proof of Proposition 3 Let F := {z : 2 } Rez 0 and z {z : 0 < Rez < 2 } and z > be the standard fundamental domain for the action of SL 2 Z on the upper half plane We cut off F by a horizontal line { C := il t : 2 t } 2 where L is a sufficiently large real number Define the contour with counterclockwise orientation around F {Imz L} as in the proof of the k/2 valence formula see for example [6] p 5 where C 2 denotes the arc from e 2πi/3 to i and C 3 the arc from i to e πi/3 We integrate the left hand side of 2 along our contour Breaking the integral into parts we note that the sum of the integrals along the lines given by Rez = ± cancel In addition the residue theorem implies that 2 the entire integral is equal to the sum of the residues This sum is zero because E 2 and are holomorphic on the upper half plane If S := 0 0 then SC 2 traces the same arc as C 3 with opposite orientation These observations imply 28 fgψe 2 dz = fgψe 2 dz + fgψe 2 dz C C 2 C πi q =R fgψe 2 dq q = fgψe 2 dz fgψe 2 dz C 2 SC 2 On the left side of 29 we use the residue theorem to evaluate the integral with respect to q around the circle q = R with negative orientation On the right we use the transformation

6 6 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH formula for E 2 cf [6] p 3 20 E 2 /zz 2 = E 2 z + 2 2πiz and the equalities dz = = izdθ to obtain a0 = a0 = 6 π 2πi dq q fgψe 2 dz 2 C 2 2πi π/2 π/3 fe iθ ge iθ ψe iθ dθ C 2 fgψ dz z C 2 fgψe 2 dz Note that care is required when changing variables in the previous step To prove 24 we calculate the differential d using 26 and 27 d = d E 3 4 E6 2 = 2E6 E 4 3 3E4E 2 4E 6 2 dz dz 728E E4 6 = 2πi 2 E 2 6 E 2 E4E E2 4E 4 E 2 E = 2πi 728 = 2πi E 4 E 4 E E6 2 E = 2πi E 4 Thus we have that dz = E 4 d 2πi Substituting this into the left hand side of 28 and applying the residue theorem we have fgψe 2 dz = fgψ E 2E 4 d = b0 C 2πi γ Because / e 2πiz for L large γ is a simple curve around zero with a counterclockwise orientation To prove 25 recall that on the arc from e iπ/3 to i in F the -invariant takes real values To see this use the invariance of under S and the fact that z can be written as a q-series with real coefficients Since is a continuous biection from F to C e πi/3 = 0 and i = 728 it follows that maps this arc monotonically onto [0 728] Substituting x = e iθ into and noting that d dz = 2 d dz E 3 4 yield the final equivalence Remark Define a measure w ψ x := ψxe 4dx 2πx x Then Proposition 3 tells us that ψ is a real-valued bilinear form on the space of polynomials R[x] with respect to the measure w ψ x Thus we can apply the classical theory of real-valued orthogonal polynomials to our particular obects of study In particular if ψx Q[x] is good then all the zeros of P nψ x Q[x] lie in the interval [0 728] Further if m < n the zeros of P mψ interlace the zeros of P nψ in the sense that any zero of P mψ lies between two zeros of P nψ For a treatment of orthogonal polynomials see [2]

7 ORTHOGONAL POLYNOMIALS AND MODULAR FORMS 7 We will require the following lemma in the next section Lemma 4 Suppose ψx Q[x] is good Write x n = P nψ + c n n P n ψ + + c 0n Then x n P mψ ψ = c mn P mψ P mψ ψ Furthermore c mn = 0 when n < m Proof This follows immediately from the definition of {P nψ } 3 Proof of Theorem Suppose that p 5 is prime We begin by recalling two results Let F E p x if p mod 2 x 3 F E p x := F E p x if p 5 mod 2 x 728 F E p x if p 7 mod 2 xx 728 F E p x if p mod 2 Then we have the following well-known theorem Proposition 5 Deligne [5] If p 5 is prime then S p x F E p x mod p The second result characterizes the logarithmic derivative of a modular form on SL 2 Z and gives an explicit description of Θf Proposition 6 [3] If f M k is a holomorphic modular form and e τ is defined by if τ = i 2 e τ = if τ = e 3 2πi/3 otherwise then we have the following identity: k 2 E 2z = Θfz fz + zz e τ ord τ f E 4 z z τ Remark This formula is equivalent to the famous denominator formula for the Monster Lie algebra For ease of notation we also define 32 K and n n 0 bn 33 K q n m 0 anq n τ F n := b0 to be the constant terms in the respective expansions We are now ready to prove Theorem := a0

8 8 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH Proof of Theorem The von Staudt-Clausen Bernoulli number congruences imply the q- series congruence E p z mod p see Section 5 for details In particular the Fourier coefficients of E p z are p-integral Since and F E p = E p Ẽ k mk Ẽ k mk Z[[q]] it follows that F E p is p-integral when written as a polynomial in Use the Euclidean algorithm to find hx Q[x] and ai Q such that 34 α nψ pβ ψ pp mψ xψx = hxf E p x + ag p x g p + + a0 As the set of p-integral elements of Q[x] forms a ring the fact that F E p x and α nψ pβ ψ pp nψ xψx are p-integral implies that h and the ai s will also be p-integral By Lemma 4 we have c mn P mψ α nψ pβ ψ pp mψ ψ = x n α nψ pβ ψ pp mψ ψ By Proposition 3 35 x n α nψ pβ ψ pp mψ ψ = K n α nψ pβ ψ pp mψ ψ E 2E 4 This constant term is p-integral because E 2E 4 is p-integral as a Applying Proposition 6 with f = E p and k = p we derive the relation 36 E 2 = 2 p Combining 35 with 36 and letting ΘE p E p + z E 4 37 P := α nψ pβ ψ pp mψ ψ we have p 2 c mn P mψ P mψ ψ = K = K n P E 4 τ F e τ ord τ E p z τ -series [ ΘE p E p + z E 4 [ n P E 4 ΘE p + E p τ F τ F ] e τ ord τ E p z τ ] e τ ord τ E p z z τ We expand 34 with x = and rewrite the above equality as p 2 c mn P mψ P mψ ψ K n P E 4 ΘE p + n hf E p E p τ F = K n ag p gp + + a0 e τ ord τ E p z z τ τ F e τ ord τ E p z z τ

9 ORTHOGONAL POLYNOMIALS AND MODULAR FORMS 9 Rewriting the sums as geometric series we have p 2 c mn P mψ P mψ ψ K n P E 4 ΘE p τ F + n hf E p e τord τ E p τ k 38 E p z k k=0 = K n ag p gp + + a0 τ F e τord τ E p τ k z k Equivalently we may write this equality as the following system of linear equations: τ F e τord τ E p τ 0 τ F e τord τ E p τ gp a0 39 = V τ F e τord τ E p τ g p τ F e τord τ E p τ 2g p ag p where V is a vector with g p coordinates each of which is p-integral Reducing the determinant of this matrix modulo p and applying Proposition 5 we have i Ω p i 0 i Ω p gp i D := i Ω p g p i i Ω p 2g p i We use multilinearity in the rows to simplify this determinant Let S gp be the symmetric group on g p elements Then noting that all matrices with repeated rows have zero determinant a straightforward calculation gives D = σ 0 g p σ 0 gp 2 = σ S g p g p σg p 2g p g 0 p g p g p σg p This is the square of a Vandermonde determinant which is nonzero because the i are distinct Thus there is a unique vector a 0 a gp F gp p that is the solution to the reduction of 39 modulo p To find this unique solution denote the reduction modulo p of h by h and use the fact that ΘE p z 0 mod p to reduce 38 0 = K n hsp z τ τ S p = K n ag p gp + + a0 z τ k=0 τ S p It is clear that a 0 a gp = is a solution to this equation over F p But then we have α nψ pβ ψ pp mψ xψx hxs p x + ag p x g p + + a0 hxs p x mod p which immediately implies the theorem

10 0 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH Remark Notice that this proof allows us to weaken the hypotheses regarding the existence of {P nψ } In particular if Rx Q[x] is a monic polynomial of degree n g p d ψp and x r Rx ψ = 0 for all r < n then the same conclusion holds Remark With slight modification the proof of Theorem can be extended to the number field case In particular fix p 5 K a number field O K its ring of integers and p O K a prime ideal above p Suppose ψx O K [x] is good Then under certain conditions over O K /p [x] S p x ψxp nψ x 4 Continued fraction expansions Orthogonal polynomials are closely related to a certain type of continued fraction expansion see [2] Consider the -expansion of 22 with f = P nψ 4 P nψ gψ E 2E 4 = g i n 0 b ψ i i The q-series expansion of E 2E 4 begins with a constant while the order of the lowest order term of the q-series expansion of P nψ ψ is n + m where m is the degree of ψx Therefore as a -series n 0 = n + m in 4 Since P nψ is an orthogonal polynomial gb ψ i = 0 for all gx of degree lower than n which in turn implies that b ψ i = 0 for 0 i n Hence let us define the polynomial i 42 Q nψ = b ψ i i= m+n Dividing both sides of 4 by P nψ g we obtain 43 ψ E 2E 4 = Q 2n nψ P nψ + O Let y = then 43 has a unique continued fraction expansion in terms of λk as 44 ψ E 2E 4 = ψ y λy λ2y λ3y with partial convergents given by 45 where Q nψ P nψ = ψ y λy λ2n y = ψ y + C n + D n

11 ORTHOGONAL POLYNOMIALS AND MODULAR FORMS C mr = D mr = r s s=m r s s=m 2 k <<ks 2r k i+ k i 2 k <<ks 2r k i+ k i 2 s λk i y i= s λk i y i= Proposition 7 Given a continued fraction expansion in λk as in 44 the recursion relation for P n+ψ is 46 P n+ψ = λ2n + λ2n + P nψ λ2nλ2n P n ψ Proof Recall that y = and denote P nψ by P nψ By the partial convergents in 45 P 0ψ = and P ψ = λy Assume that 46 generates the orthogonal polynomials up to P nψ Let P n+ be the next term in the recurrence relation It is sufficient to show that P n+ = P n+ψ ; observe that P n+ = λ2n + λ2n + P nψ λ2nλ2n P n ψ = n+ { λ2n + λ2n + yd n λ2nλ2n y 2 D n } = n+ D n+ + λ2nλ2n n D n λ2nλ2n n D n = P n+ψ We are done by induction The relation given later as 46 matches the relation given in 6 for the orthogonal polynomials and the initial conditions P ψ = 0 P 0ψ = match; hence the orthogonal polynomials generated are identical When ψx = Kaneko and Zagier proved the following additional forms of the recurrence relation in [5] P n+ = 44n P n 2n + 2n 2n 32n 72n 52n + 36 P nn 2n 2 n 47 P n = [ n i 2 3i m=0 i= /2 i m 5/2 i m ] n + 2 n 7/2 2n n m m m Proposition 8 The orthogonal polynomials P n are p-integral for all n < p+ 2 Proof This follows from 47 since 2n m is the only factor contributing an n to the denominator so n = p+ is the first time this expression for P 2 n is possibly not p-integral

12 2 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH 5 Proof of Theorem 2 We first recall two classical Bernoulli number congruences see [4] p Let D n be the denominator of the nth Bernoulli number written in lowest terms The von Staudt- Clausen congruences state 5 D n = 6 p i n where the p i s are prime Let p 5 be prime Now suppose m 2 is even and m m mod φp r where φ is the Euler φ-function Then the Kummer congruences state p m B m 52 pm B m m m Using these congruences we prove the following lemma Lemma 9 For r the following q-series congruences hold 53 E p z pr mod p r and 54 E φp r +2z E 2 z mod p r Proof For 53 we have 2p E p z pr = B p p i mod p r pr σ p nq n = 2p D p U p n= pr σ p nq n where U p is an integer coprime to D p From 5 we have p D p which implies 53 after an application of the binomial theorem To prove 54 if we let m = 2 m = φp r +2 in 52 and note that p φpr + p mod p r we obtain B 2 2 Bφpr +2 mod p r φp r + 2 Also by Euler s theorem σ n σ φp r +n mod p r With these two observations we have E 2 z = 22 B 2 n= Proof of Theorem 2 From Lemma 9 σ nq n 2φpr + 2 B φp r +2 n= σ φp r +nq n mod p r 55 β ψ pψ E 2E 4 β ψ pψ E φpr +2E 4 E p pr mod p r Note that the definition of β ψ p and Lemma 9 imply that the left hand side of 55 is p-integral The right hand side of 55 is a weight zero modular function; hence it is a rational function in z By examining its zeros we obtain 56 β ψ pψ E 4E φp r +2 E p pr = β ψ pψ n= F E 4 E φp r +2 F E p pr

13 ORTHOGONAL POLYNOMIALS AND MODULAR FORMS 3 This together with 43 implies that β ψ pψ F E 4 E φp r +2 F E p pr = β ψpq nψ α nψ pp nψ + O 2n 6 Examples In this section we carry out explicit numerical examples illustrating properties of ψ Proposition 3 and Theorem Let us first compute x ψ for ψ = x Proposition 3 allows us to forgo the computation of integrals Using 23 we compute that x ψ = K q 2 + 5E 2 = K q q q q 2 + q q q q 2 + E 2 = K q q q q q 72q 2 96q 3 + = K q q q q = This method is simpler than the integral methods but the number of terms to which each power of must be computed in a given polynomial depends upon the degrees of the other polynomials involved Further computing powers of as a q-series can be computationally inefficient With this in mind we investigate 24 First we compute that 2 3 E 2 E 4 = Thus x ψ = K = K = K E 2E = which is of course the same result as that obtained from the q-expansion method This method benefits from the fact that the -expansion of E 2E 4 only needs to be computed

14 4 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH once All remaining computations are simple polynomial manipulations where can be treated as an indeterminate Having found an efficient method for the computation of these bilinear forms we now describe an efficient method of calculating {P nψ } for arbitrary good ψ If ψ is good then the Gram-Schmidt process will find the orthogonal polynomials However we use the simpler three-term recurrence relation 6 P n+ψ = x xp nψ P nψ ψ P nψ P nψ P nψ ψ P n ψ P nψ P nψ ψ P n ψ P n ψ ψ with P 0ψ = and P ψ = x x ψ ψ This relation produces monic orthogonal polynomials as is easily verified In particular note that due to the orthogonality of the P nψ s and the properties of the bilinear form xp nψ P kψ ψ = P nψ P k+ψ ψ for k < n We now consider some examples of {P nψ } for various ψ For ψ = the case considered by Kaneko and Zagier in [5] we find that P 0 = P = x 720 P 2 = x 2 640x P 3 = x x x P 4 = x x x x in agreement with their results We pause here to emphasize that by Theorem a particular polynomial may directly represent as many as four different supersingular loci simultaneously It will also give information about other supersingular loci by containing them as factors We illustrate this with our next example For ψx = x we find that P 0ψ = P ψ = x P 2ψ = x x P 3ψ = x x P 4ψ = x x x x x

15 ORTHOGONAL POLYNOMIALS AND MODULAR FORMS 5 Factoring P 3ψ over F p for p { } we see that P 3ψ xx + 4x + 20 mod 23 P 3ψ xx + 4x + 27 mod 29 P 3ψ x + 8x + 27x + 29 mod 3 P 3ψ x + 29x 2 + 3x + 3 mod 37 Further while P 3ψ is not 7-integral we have that 7 2 P 3ψ 5x + 5x + mod 7 Alternately using Proposition 5 we may find the supersingular locus for p by computing F E p x We thus compute that S 23 x = xx + 4x + 20 S 29 x = xx + 4x + 27 S 3 x = x + 8x + 27x + 29 S 37 x = x + 29x 2 + 3x + 3 S 7 x = x + which agrees with our main result Thus P 3ψ is a lift of four different supersingular loci It also contains as factors the lifts of the supersingular loci of all the primes from 5 to 9 once the issue of p-integrality is properly taken into account Both of the ψ above are good as they have no zeros in the interval What are some examples of bad ψ? A particular ψ is bad if and only if the Gram-Schmidt process fails which happens exactly when there exists a P nψ such that P nψ P nψ ψ = 0 In this case P n+ψ does not exist Thus both x 720 and x 2 266x are bad as in both cases ψ = 0 Also x is bad because although ψ = P ψ P ψ ψ = 0 There are two natural questions: first are there bad rational ψ such that ψ is non-zero? Second are there real ψ which are bad but for which both ψ and P ψ P ψ ψ are non-zero? In fact we were unable to find ψ which met either criterion Acknowledgements The authors sincerely appreciate the opportunity provided by K Ono and W McGraw who organized the 2003 Research Experience for Undergraduates REU during which this research was conducted J Anderson N Boston R Chatteree G Coogan J Loveoy J Rouse and K Zuhr also deserve thanks for their lectures and assistance References [] S Ahlgren and K Ono Weierstrass points on X 0 p and supersingular -invariants Mathematische Annalen [2] G Andrews R Askey and R Roy Special Functions Cambridge University Press Cambridge 999 [3] J Bruinier W Kohnen and K Ono The arithmetic of the values of modular functions and the divisors of modular forms Compositio Math to appear [4] K Ireland and M Rosen A Classical Introduction to Modern Number Theory Springer Verlag 99 [5] M Kaneko and D Zagier Supersingular -invariants hypergeometric series and Atkin s orthogonal polynomials Computational Perspectives on Number Theory Chicago IL 995 AMS/IP [6] N Koblitz Introduction to Elliptic Curves and Modular Forms Springer-Verlag New York 993

16 6 STEPHANIE BASHA JAYCE GETZ HARRIS NOVER AND EMMA SMITH [7] J Silverman The Arithmetic of Elliptic Curves Springer-Verlag New York 986 Department of Mathematics Santa Clara University Santa Clara CA address: sbasha@scuedu 4404 South Ave W Missoula MT address: getz@fasharvardedu Department of Mathematics California Institute of Technology Pasadena CA 925 address: nover@itscaltechedu 5849 Sand Rd Bellingham WA address: emmas@mitedu