Special Values of J-Function Which Are Algebraic

SPECIAL VALUES OF j-FUNCTION WHICH ARE ALGEBRAIC 1. Introduction Let Ek (z) = 1 2 P (c,d)=1 (cz + d)−k be the Eisenstein series of weight k > 2. The E3 1 j-function on the upper half plane is defined by j(z) = ∆4 where ∆(z) = 1728 (E43 − E62 ). For a primitive positive definite quadratic form Q(x, y) = ax2 + bxy + cy 2 , √ −b+ D with D = b2 − 4ac < 0, it is known that j(zQ ) is an algebraic and zQ = 2a integer of degree h(D) by Kronecker and Weber. Here, we show a weaker result that j(zQ ) is an algebraic number of degree at most the class number h(D) using the j-invariant of a complex lattice, orders in an imaginary quadratic field, and complex multiplication. Theorem 1.1. For a primitive positive definite quadratic form Q(x, y) = ax2 + √ D with D = b2 − 4ac < 0, j(zQ ) is an algebraic number bxy + cy 2 , and zQ = −b+ 2a of degree at most the class number h(D). 2. j-invariant of a complex lattice Definition 2.1. A subgroup L of C is called a complex lattice if L = ω1 Z + ω2 Z with ω1 , ω2 ∈ C being linearly independent over R. We simply write L = [ω1 , ω2 ]. We say that two lattices L and L0 are homothetic if there is a nonzero complex number λ such that L0 = λL. Note that homothety is an equivalence relation. Definition 2.2. Weierstrass ℘-function associated to a complex lattice L = [ω1 , ω2 ] is defined by:  X  1 1 1 (2.1) ℘(z; L) = 2 + − 2 . z (z − w)2 w w∈L−{0} We simply write ℘(z) = ℘(z; L). Note that ℘(z + w) = ℘(z) for all w ∈ L. P Lemma 2.1. Let Gk (L) = w∈L−{0} w−k for k > 2. Then, Weierstrass ℘-function for a lattice L has Laurent expansion ∞ X 1 (2.2) ℘(z) = 2 + (2n + 1)G2n+2 (L)z 2n . z n=1 Proof. We have the series expansion ∞ X 1 = 1 + (n + 1)xn (1 − x)2 n=1 for |x| < 1. Thus, if |z| < |w|, we have ∞ X 1 n+1 n 1 − = z . 2 2 (z − w) w wn+2 n=1 1 2 Summing over w, we obtain ∞ ℘(z) = X 1 + (n + 1)Gn+2 (L)z n . 2 z n=1 Since ℘ is an even function, the odd coefficients must vanish and (2) follows.  Lemma 2.2. ℘-function for a lattice L satisfies the differential equation (2.3) ℘0 (z)2 = 4℘(z)3 − g2 ℘(z) − g3 where g2 = 60G4 , and g3 = 140G6 . Proof. Let F (z) = ℘0 (z)2 − 4℘(z)3 + g2 ℘(z) + g3 , then F has possible poles at z = w ∈ L, is holomorphic on C − L, and F (z + w) = F (z) for all w ∈ L. But, Laurent series expansions (followed from Lemma2.1) 9G4 1 ℘(z)3 = 6 + 2 + 15G6 + O(z) z z , and 4 24G4 ℘0 (z)2 = 6 − 2 − 80G6 + O(z) z z imply that F is holomorphic at 0, and F (0) = 0. By Liouville’s theorem, we have F (z) = 0 for all z ∈ C.  Corollary 2.1. ℘0 (z)2 = 4(℘(z) − e1 )(℘(z) − e2 )(℘(z) − e3 ) where e1 = ℘(ω1 /2), e2 = ℘(ω2 /2), and e3 = ℘((ω1 + ω2 )/2). Furthermore, ∆(L) = 16(e1 − e2 )2 (e2 − e3 )2 (e3 − e1 )2 = g23 − 27g32 6= 0. Definition 2.3. The j-invariant j(L) of a lattice L is defined to be the complex number g2 (L)3 g2 (L)3 (2.4) j(L) = 1728 = 1728 . 3 2 g2 (L) − 27g3 (L) ∆(L) The remarkable fact is that the j-invariant j(L) characterizes the lattice L up to homothety: Proposition 2.1. If L and L0 are lattices in C, then j(L) = j(L0 ) if and only if L and L0 are homothetic. Proof. It is easy to see that homothetic lattices have the same j-invariant. Namely, if λ ∈ C∗ , then the definition of g2 and g3 implies that g2 (λL) = λ−4 g2 (L) (2.5) g3 (λL) = λ−6 g3 (L), and j(λL) = j(L) follows easily. ω2 For any lattice L = [ω1 , ω2 ], we can assume that z = ω ∈ H = {z ∈ C|Im z > 0} 1 without loss of generality. Then, L and [1, z] become homothetic lattices. Now, we have the connection from j-invariant a lattice L and j-function on the upper half plane: E4 (z)3 j(L) = j([1, z]) = j(z) = . ∆(z) Suppose that L and L0 have the same j-invariant. We first find z, z 0 ∈ H such that L is homothetic to [1, z], and L0 is homothetic to [1, z 0 ]. Then, we have j(z) = SPECIAL VALUES OF j-FUNCTION WHICH ARE ALGEBRAIC 3 j(z 0 ). By the valence formula(See [2] p16, Theorem1.3), we obtain z 0 ≡ z (mod Γ = SL(2, Z)), since j has a simple pole at i∞. This implies that [1, z 0 ] = [1, z]. Hence L and L0 are homothetic.  Lemma 2.3. Let ℘(z) be the ℘-function for the lattice L, and as in Lemma 2.1, let ∞ X 1 (2.6) ℘(z) = 2 + (2n + 1)G2n+2 (L)z 2n . z n=1 be its Laurent expansion. Then for n ≥ 1, the coefficient (2n + 1)G2n+2 (L) of z 2n is a polynomial with rational coefficients, independent of L, in g2 (L) and g3 (L). Proof. We differentiate ℘0 (z)2 = 4℘(z)3 − g2 ℘(z) − g3 to obtain 1 ℘00 (z) = 6℘(z)2 − g2 . 2 Let an = (2n + 1)G2n+2 (L). By substituting in the Laurent expansion for ℘(z) and comparing the coefficients of z 2n−2 , one easily sees that for n ≥ 3, ! n−2 X 2n(2n − 1)an = 6 2an + ai an−1−i , i=1 and hence (2n + 3)(n − 2)an = 3 n−2 X ai an−1−i . i=1 Since g2 (L) = 20a1 and g3 (L) = 28a2 , induction shows that an is a polynomial with rational coefficients in g2 (L) and g3 (L).  3. Orders in quadratic fields Definition 3.1. An order O in a quadratic field K is a subset O ⊂ K such that (i) O is a subring of K containing 1. (ii) O is a finitely generated Z-module. (iii) O contains a Q-basis of K. The ring OK of integers in K is always an order in K. More importantly, (i) and (ii) imply that for any order O of K, we have O ⊂ OK , so that OK is the maximal order of K. Note that the maximal order OK can be written as: √ dK + dK , (3.1) OK = [1, wK ], wK = 2 where dK is the discriminant of K. We can now describe all orders in quadratic fields: Lemma 3.1. Let O be an order in a quadratic field K of discriminant dK . Then O has finite index in OK , and we set f = [OK : O], then (3.2) O = Z + f OK = [1, f wK ]. Proof. Since O and OK are free Z-modules of rank 2, it follows that f = [OK : O] is finite. Since f OK ⊂ O, Z + f OK = [1, f wK ] ⊂ O follows. Thus, O = [1, f wK ].  4 Given an order O in a quadratic field K, discriminant is defined as follows. Let α 7→ α0 be the nontrivial automorphism of K, and suppose O = [α, β]. Then the discriminant of O is the number   2 α β (3.3) D = disc[α, β] = det . α0 β 0 If f = [OK : O], then it follows that D = f 2 dK by Lemma 3.1. Now consider ideals of an order O. Since O may not be a Dedekind domain, we cannot assume that ideals have unique factorization. We introduce the concept of a proper ideal of an order. Definition 3.2. A fractional ideal of O is a subset of K which is a nonzero finitely generated O-module. Then, a fractional O-ideal b is proper provided that (3.4) O = {β ∈ K : βb ⊂ b}. Proposition 3.1. Let O be an order in a quadratic field K, and let a be a fractional O-ideal. Then a is proper if and only if a is invertible. Proof. If a is invertible, then ab = O for some fractional O-ideal b. If β ∈ K and βa ⊂ a, then we have βO = β(ab) = (βa)b ⊂ ab = O, and β ∈ O follows, proving that a is proper.  To prove the converse, we need the following lemma: Lemma 3.2. Let K = Q(τ ) be a quadratic field, and let ax2 + bx + c be the minimal polynomial of τ , where a, b and c are relatively prime integers. Then [1, τ ] is a proper fractional ideal for the order [1, aτ ] of K. Proof. First, [1, aτ ] is an order since aτ is an algebraic integer. Then, given β ∈ K, 2 note that β[1, τ ] ⊂
[1, τ ] is equivalent to β = m+nτ , m, n ∈ Z, and βτ = mτ +nτ = −bn −cn a + a + m τ ∈ [1, τ ]. But, this is also equivalent to a|n, since (a, b, c) = 1. Thus, [1, τ ] is a proper fractional ideal for the order [1, aτ ]. Now, we can prove that proper fractional ideals are invertible. First note that a is a Z-module of rank 2, so that a = [α, β] for some α, β ∈ K. Then a = α[1, τ ], where τ = β/α. If ax2 + bx + c, (a, b, c) = 1, is the minimal polynomial of τ , then Lemma 3.2 implies that O = [1, aτ ]. Let β 7→ β 0 denote the nontrivial automorphism of K. Since τ 0 is the other root of ax2 + bx + c, using Lemma 3.2 again shows a0 = α0 [1, τ 0 ] is a fractional ideal for [1, aτ ] = [1, aτ 0 ] = O. To see why a is invertible, note that aaa0 = aαα0 [1, τ ][1, τ 0 ] = N (α)[a, aτ, aτ 0 , aτ τ 0 ]. Since τ + τ 0 = −b/a and τ τ 0 = c/a, this becomes aaa0 = N (α)[a, aτ, −b, c] = N (α)[1, aτ ] = N (α)O since (a, b, c) = 1. This proves that a is invertible.  Definition 3.3. Given an order O, let I(O) denote the set of proper fractional O-ideals. By Proposition2, I(O) forms a group. The principal O-ideals give a subgroup P (O) ⊂ I(O), and thus we can form the quotient C(O) = I(O)/P (O), which is the ideal class group of the order O. SPECIAL VALUES OF j-FUNCTION WHICH ARE ALGEBRAIC 5 Let C(D) be the set of proper-equivalence classes of primitive quadratic forms with discriminant D. Denote h(D) = |C(D)|. Theorem 3.1. Let O be the order of discriminant D in an imaginary quadratic field K. 2 (i) If f (x, y) = ax2 + bxy + cy√ is a primitive positive definite quadratic form of discriminant D, then [a, (−b + D)/2] is√a proper ideal of O. (ii) The map sending f (x, y) to [a, (−b + D)/2] induces a bijection between C(D) and the ideal class group C(O). Remark that h(D) = |C(D)| = |C(O)|. √ √ Proof. (i) Let τ = (−b + D)/2a. Then [a, (−b + D)/2] = [a, aτ ] = a[1, τ ]. Note that by Lemma 3.2, a[1, τ ] is a proper ideal for the order [1, aτ ]. However, if f = [OK : O], then D = f 2 dK , and thus b + f dK + f wK ∈ [1, f wK ]. 2 It follows that [1, aτ ] = [1, f wK ] = O by Lemma 3.1. This proves that a[1, τ ] is a proper O-ideal. (ii) Let f (x, y) and g(x, y) be forms of discriminant D, and let τ and τ 0 be their respective roots. We will prove: aτ = − f (x, y), g(x, y) are properly equivalent   pτ + q p q , ∈ SL(2, Z) ⇐⇒τ 0 = r s rτ + s ⇐⇒[1, τ ] = λ[1, τ 0 ], λ ∈ K ∗ . To seethe first equivalence, assume that f (x, y) = g(px + qy, rx + sy), where  p q ∈ SL(2, Z). Then r s   pτ + q 2 0 = f (τ, 1) = g(pτ + q, rτ + s) = (rτ + t) g ,1 , rτ + s so that g((pτ + q)/(rτ + s), 1) = 0. However, if τ ∈ H, then (pτ + q)/(rτ + s) ∈ H, thus τ 0 = (pτ + q)/(rτ + s). Conversely, if τ 0 = (pτ + q)/(rτ + s), then we have f (x, y) and g(px + qy, rx + sy) have the same root, hence they are equal. Next, if τ 0 = (pτ + q)/(rτ + s), let λ = rτ + s ∈ K ∗ . Then   pτ + q λ[1, τ 0 ] = (rτ + s) 1, rτ + s = [rτ + s, pτ + q] = [1, τ ]   p q since ∈ SL(2, Z). Conversely, if [1, τ ] = λ[1, τ 0 ] for some λ ∈ K ∗ , then r s [1, τ ] = [λ, λτ 0 ], which implies λτ 0 = pτ + q λ = rτ + s   p q ∈ GL(2, Z). This gives us τ 0 = r s   p q we have ∈ SL(2, Z). r s for some pτ +q rτ +s . Since τ, τ 0 are both in H, 6 These equivalences show that the map sending f (x, y) to a[1, τ ] induces an injection C(D) −→ C(O). To show that the map is surjective, let a be a proper fractional O-ideal. We write a = [α, β] for some α, β ∈ K with τ = β/α lies in H. Let ax2 + bx + c be the minimal polynomial of τ . We may assume that (a, b, c) = 1 and a > 0. Then f (x, y) = ax2 + bxy + cy 2 is positive definite of discriminant D, and maps to a[1, τ ] which is in the class of a. We thus have a bijection of sets (3.5) C(D) −→ C(O).  4. Complex Multiplication First, we observe that orders in imaginary quadratic fields give rise to a natural class of lattices. If O is an order in a quadratic field K and a = [α, β] is a proper fractional O-ideal, then α and β are linearly independent over R. Thus a ⊂ C is a lattice. Conversely, let L ⊂ C be a lattice which is contained in K. Then L is a proper fractional O-ideal for some order O of K. As a consequence, we have that a and b determine the same class in the ideal class group C(O) if and only if they are homothetic as lattices in C. Moreover, this enables us to define j(a) for a proper fractional O-ideal. We defined ℘-function for a lattice L ⊂ C. In fact, any elliptic function for L is a rational function of ℘ and ℘0 . Lemma 4.1. Any even elliptic function for L is a rational function in ℘(z). Proof. (a) Let f (z) be an even elliptic function which is holomorphic on C − L. Then there is a polynomial A(x) such that the Laurent expansion of f (z) − A(℘(z)) is holomorphic on C. By Liouville’s theorem, f (z) − A(℘(z)) is a constant. Thus, f (z) is a polynomial in ℘(z). (b) Let f (z) be an even elliptic function that has a pole of order m at w ∈ C − L. If 2w ∈ / L, then (℘(z) − ℘(w))m f (z) is holomorphic at w, since (℘(z) − ℘(w)) has a zero at z = w. If 2w ∈ L, then m is even, since the Laurent expansion for f (z) and f (2w − z) at z = w must be equal. In this case, (℘(z) − ℘(w))m/2 f (z) is holomorphic at w, since (℘(z) − ℘(w)) has double zero at z = w. (c) Now we can show that for an even elliptic function f (z), there is a polynomial B(x) such that B(℘(z))f (z) is holomorphic on C − L. Then the lemma follows by part (a).  For any elliptic function f (z) for L, we have   f (z) − f (−z) f (z) + f (−z) + ℘0 (z). f (z) = 2 2℘0 (z) Hence, any elliptic function for L is a rational function of ℘ and ℘0 . We turn into an important proposition about complex multiplication: Proposition 4.1. Let L be a lattice, and let ℘(z) be the ℘-function for L. Then, for a number α ∈ C − Z, the following statements are equivalent: (i) ℘(αz) is a rational function in ℘(z). (ii) αL ⊂ L. (iii) There is an order O in an imaginary quadratic field K such that α ∈ O and L SPECIAL VALUES OF j-FUNCTION WHICH ARE ALGEBRAIC 7 is homothetic to a proper fractional O-ideal. Furthermore, if these conditions are satisfied, then ℘(αz) can be written in the form (4.1) ℘(αz) = A(℘(z)) B(℘(z)) where A(x) and B(x) are relatively prime polynomials such that deg(A(x)) = deg(B(x)) + 1 = [L : αL] = N (α). Proof. (i)⇒(ii). If ℘(αz) is a rational function in ℘(z), then there are polynomials A(x) and B(x) such that B(℘(z))℘(αz) = A(℘(z)). Comparing the order or pole at z = 0, we have deg(A(x)) = deg(B(x)) + 1. Now, let ω ∈ L. Then the above show that ℘(αz) has a pole at ω, which means that ℘(z) has a pole at αω. Since the poles of ℘(z) are exactly at members in L, this implies αω ∈ L, and αL ⊂ L follows. (ii)⇒(i). If αL ⊂ L, it follows that ℘(αz) is meromorphic and has L as a lattice of periods. Furthermore, ℘(αz) is an even function. By Lemma 4.1, we have ℘(αz) is a rational function in ℘(z). (ii)⇒(iii). Suppose that αL ⊂ L. Replacing L by λL for suitable λ, we can assume that L = [1, τ ] for some τ ∈ C − R. Then αL ⊂ L means that α = a + bτ and ατ = c + dτ for some integers a, b, c and d. Then we obtain, τ= c + dτ , a + bτ which implies bτ 2 + (a − d)τ − c = 0. Since τ is not real, we have b 6= 0, and K = Q(τ ) is an imaginary quadratic field. Thus, O = {β ∈ K|βL ⊂ L} = {β ∈ K|β[1, τ ] ⊂ [1, τ ]} = [1, bτ ] is an order of K for which L is a proper fractional O-ideal(by Lemma 3.2), and since α is obviously in O, we are done. (iii)⇒(ii) is trivial. Suppose αL ⊂ L = [1, τ ]. From the definition of discriminant (3.3), we obtain N (α)2 disc[1, τ ] = disc[α, ατ ] = [L : αL]2 disc[1, τ ]. Thus, [L : αL] = N (α). It remains to prove that degree of A(x) is the index [L : αL]. Fix z ∈ C such that 2z ∈ / (1/α)L, and consider the polynomial A(x)−℘(αz)B(x). This polynomial has the same degree as A(x), and z can be chosen so that it has distinct roots(multiple root of A(x)−℘(αz)B(x) is a root of A(x)B 0 (x)−A0 (x)B(x).) Then consider the lattice L ⊂ (1/α)L, and let {wi } be coset representatives of L in (1/α)L. Our assumption on z implies that ℘(z +wi ) are distinct. From (4.1), we see that A(℘(z + wi )) = ℘(α(z + wi ))B(℘(z + wi )). But αwi ∈ L, hence ℘(α(z + wi )) = ℘(αz). This shows that ℘(z + wi ) are distinct roots of A(x) − ℘(αz)B(x). Let u be another root. Then we see that u = ℘(w) for some complex number w. Then, ℘(αz) = ℘(αw), and w ≡ z + wi mod L for some i. Hence ℘(z + wi ) are all roots of A(x) − ℘(αz)B(x), giving that degA(x) = [(1/α)L : L] = [L : αL].  8 Together with Laurent series of ℘(z), this theorem allows us to compute some special values of j-function for example: √   √ 1 + −7 j( −2) = 8000 and j = −3375. 2 To complete the proof of Theorem1.1, we need a lemma involving the invariants g2 (L) and g3 (L). Lemma 4.2. Let g2 and g3 be given complex numbers satisfying g23 − 27g32 6= 0. Then there is a unique lattice L ⊂ C such that g2 (L) = g2 and g3 (L) = g3 . Proof. We can find z ∈ H such that j(z) = E4 (z)3 /∆(z) = 1728g23 /(g23 − 27g32 ). By valence formula, this z is uniquely determined modulo Γ. If g2 6= 0, then we find w1 ∈ C such that (4.2) g2 = 4π 4 E4 (z). 3w14 Using −w1 if necessary, we obtain (4.3) g3 = 8π 6 E6 (z). 27w16 If g2 = 0, then we have g3 6= 0 and we find w1 ∈ C using (4.3). Let w2 = zw1 , then L = [w1 , w2 ] is the desired lattice. Suppose we have two lattices L and L0 with g2 (L) = g2 (L0 ) and g3 (L) = g3 (L0 ). Then, j(L) = j(L0 ) implies that there exists λ ∈ C − {0} such that L0 = λL by Proposition2.1. If g2 (L) 6= 0 and g3 (L) 6= 0, then (2.5) give λ2 = 1. Thus, L = L0 . If g2 (L) = 0, then L = λ1 [1, w] and L0 = λ2 [1, w] for some λ1 , λ2 ∈ C − {0} with w = exp(2πi/3). But, (2,5) gives λ61 = λ62 . Thus, L = L0 If g3 (L) = 0, then L = λ1 [1, i] and L0 = λ2 [1, i] for some λ1 , λ2 ∈ C − {0}. In this case, (2.5) gives λ41 = λ42 . Hence, L = L0 .  Now, we are ready to prove Theorem1.1. In fact, the following theorem will imply Theorem1.1. Theorem 4.1. Let O be an order in an imaginary quadratic field, and let a be a proper fractional O-ideal. Then j(a) is an algebraic number of degree at most h(O). Proof. By Lemma2.3, we can write the Laurent expansion of ℘(z) for the lattice a as ∞ X 1 ℘(z) = 2 + an (g2 , g3 )z 2n , z n=1 where an is a polynomial in g2 and g3 with rational coefficients. To emphasize the dependence on g2 and g3 , we write ℘(z) as ℘(z; g2 , g3 ). By assumption, for any α ∈ O, we have αa ⊂ a. Thus, by Proposition4.1, ℘(αz) is a rational function in ℘(z). ∞ ℘(αz; g2 , g3 ) = = X 1 + an (g2 , g3 )α2n z 2n 2 2 α z n=1 A(℘(z; g2 , g3 )) B(℘(z; g2 , g3 )) SPECIAL VALUES OF j-FUNCTION WHICH ARE ALGEBRAIC 9 for some polynomials A(x) and B(x). We can regard this as an identity in the field of meromorphic Laurent series C((z)). Now let σ be any automorphism of C. Then σ induces an automorphism of C((z)). Applying σ, we obtain (4.4) ℘(σ(α)z; σ(g2 ), σ(g3 )) = Aσ (℘(z; σ(g2 ), σ(g3 ))) . B σ (℘(z; σ(g2 ), σ(g3 ))) We observe that g23 − 27g32 6= 0 implies σ(g2 )3 − 27σ(g3 )2 6= 0. By Lemma4.2, there exists a lattice L such that g2 (L) = σ(g2 ) g3 (L) = σ(g3 ). Since ℘(z; σ(g2 ), σ(g3 )) = ℘(z; L), (4.4) implies that ℘(z; L) has complex multiplication by σ(α). Let O0 be the ring of all complex multiplications of L, then we have proved that O = σ(O) ⊂ O0 . Applying σ −1 and we interchange a and L, then above argument shows O0 ⊂ O, which shows that O is the ring of all complex multiplications of both a and L. Now consider j-invariants. Above formulas for g2 (L) and g3 (L) imply that j(L) = σ(j(a)). Since L has O as its ring of complex multiplications, there are only h(O) possibilities for j(L). It follows that j(a) must be an algebraic number, and the degree is at most h(O).  Suppose that O is an order of discriminant D in an imaginary quadratic field, and ax2 + bxy + cy 2 is a √ primitive positive definite quadratic form with discriminant D. D , a = [1, zQ ] is a proper fractional ideal in O by Lemma3.2. Then, for zQ = −b+ 2a Now, Theorem4.1 implies that j(a) = j([1, zQ ]) = j(zQ ) is an algebraic number of degree at most h(O) = h(D) (by Theorem3.1). This completes the proof of Theorem1.1.