A Special Module in the Field of Modular Functions

Transcription

1 A Special Module in the Field of Modular Functions Research Institute for Symbolic Computation (RISC), Johannes Kepler University, A-4040 Linz, Austria Dedicated to Peter Paule at the Occasion of his 60th Birthday May 17, 2018

2 Modular Functions Let M be a fixed positive integer and r = (r δ1,r δ2,,r δs ) a sequence indexed by the positive divisors δ i of M with e(r) := δ M δr δ 0 (mod 24). Let k=e(r) a r (k)q k := q e(r) (1 q δ1n ) r δ 1 (1 q δ2n ) r δ 2 (1 q δsn ) r δs. n=1 The reason for the q e(r) factor is that k=e(r)a r (k)e 2πikτ = δ M η(δτ) r δ.

3 The Dedekind eta function is defined as η(τ) := e πiτ 12 (1 e 2πiτn ). n=1 η( 1/τ) = ( iτ) 1/2 η(τ) More general for a,b,c,d Z with ad bc = 1 we have the following nice transformation formula: ( aτ +b ) η = ǫ(a,b,c,d)( i(cτ +d)) cτ +d where ǫ(a,b,c,d) = e πia 12 (c b 3) when a,c > 0 and gcd(a,6) = 1

4 η(2τ) 24 η(τ) 16 η(4τ) 8 16η(τ) 8 η(4τ) 16 = 0 η(4τ) 12 η(τ) 4 η(2τ) 2 η(4τ) 2 η(8τ) 4 4η(2τ) 4 η(8τ) 8 = 0 η(3τ) 12 η(τ) 9 η(9τ) 3 9η(τ) 6 η(9τ) 6 27η(τ) 3 η(9τ) 9 = 0 Replacing τ = 1/8t we obtain: η( 1 2t )12 η( 1 8t )4 η( 1 4t )2 η( 1 2t )2 η( 1 t )4 4η( 1 4t )4 η( 1 t )8 = 0 Applying the eta transformation formula we obtain: ( i2t) 6 η(2t) 12 ( i8t) 2 η(8t) 4 ( i4t)η(4t) 2 ( i2t)η(2t) 2 ( it) 2 η(t) 4 4( i4t) 2 η(4t) 4 ( it) 4 η(t) 8 = 0

5 Now dividing out ( it) 6 gives 2 6 η(2t) η(8t) 4 η(4t) 2 η(2t) 2 η(t) η(4t) 4 η(t) 8 = 0 Finally dividing the whole identity by 2 6 gives η(2t) 12 8η(8t) 4 η(4t) 2 η(2t) 2 η(t) 4 η(4t) 4 η(t) 8 = 0 This shows at least that the transformation formulas for the eta function can be used to transform a q-identity into another q-identity in a non trivial way.

6 Now we turn to the problem of proving and finding such eta identities. Let R(N) be the set of all functions f such that f holomorphic in the upper half plane ) = f(τ),a,b,c,d Z,N c f( aτ+b cτ+d f(τ) is a Laurent series in powers of q with finite principal part ) is a Taylor series in powers of q n/gcd(c2,n) for all f( aτ+b cτ+d a,b,c,d Z with ad bc = 1 and N c.

7 ( η(τ) ) 6 R(5) η(5τ) ( η(τ) ) 4 R(7) η(7τ) ( η(τ) ) 12 R(11) η(11τ) η(τ) 3 η(5τ) η(2τ)4 η(5τ) 2 η(2τ)η(5τ)5 η(2τ)η(10τ) 3, η(τ) 2 η(10τ) 4, η(τ)η(10τ) 5 R(10).

8 For each positive integer N the set R(N) has the structure of a C-algebra. The most important property of the set R(N) is that if a q-series has zero principal part then it is zero. Once we know that two q-series F(q),G(q) are in R(N), to prove an identity of the form F(q) G(q) = 0 amounts to proving that F(q) G(q) is a Taylor series with zero constant term. Since F(q) and G(q) are expressions involving the eta function we can compute any coefficient in the q-expansion of F(q) and G(q) if we have enough computational power.

9 Representing R(N) as a module Take a non constant element t R(N). t = q n +O(q) Let I :={i {0,..,n 1} : exists g R(N) with ord(g) i (mod n)} For i I, let O i := {ord(g) : g R(N),ord(g) i (mod n)}. o i := mino i G i := {g R(N) : ord(g) = o i } Take any (g 0,g 1,..,g m 1 ) G i0 G i1 G im 1. Then R(N) = g 0,g 1,...,g m 1 C[t].

10 Proof Let S be the set of all g R(N) with g g 0,g 1,...,g m 1 C[t]. Take g S such that ord(g) is minimal. Take g i with ord(g i ) ord(g) mod n. Then g := g cg i t ord(g i ) ord(g) n g S. Contradiction. has smaller order then g and

11 This proof also works for any C-subalgebra R of R(N). This means that R = h 1,...,h m C[u] Once we have such a presentation of R we can easily check if f R(N) is in R which also results in an identity: where c i (X) C[X]. f = c 1 (u)h 1 + +c m (u)h m,

12 An idea by Peter Paule Choose and ( 1 q t := q 5 k ) 12 1 q 11k k=1 f := qt (1 q 11k ) p(11n+6)q n. k=1 n=0 Then f,t R(11) and C[t,f] is a C-subalgebra of R(11).

13 One verifies that C[t,f] = 1,f,f 2,f 3,f 4 C[t] Let us check membership of f 5, if yes this gives us an expression of the form f 5 = c 0 (t)+c 1 (t)f + +c 4 (t)f 4.

14 Indeed we have membership! f 5 = f ( t)f 3 (1) ( t +4093t 2 )f ( t t t 3 )f (11 4 +t)( t t 2 +t 3 ). Recall f := qt (1 q 11k ) p(11n+6)q n. k=1 n=0

15 ( qt (1 q 11k ) p(11n+6)q n) 5 0 (mod 11). k=1 This implies k=1 n=0 qt (1 q 11k ) p(11n+6)q n 0 (mod 11). This further implies n=0 p(11n+6)q n 0 (mod 11). n=0

16 This identity was appreciated by Michael Hirschhorn who congratulated us for the discovery of this identity. The module has been generalized by Ralf Hemmecke in the sense that he proved that all q-series in R(N) with integer coefficients is equal to h 1,...,h m Z[t] This also holds for any Z-subalgebra. This fantastic discovery resulted from a problem posed by Peter Paule which Hemmecke was able to solve, this problem was also related to the module, which now needed to be generalised.

17 I must confess that I personally never saw any potential in this module other then using it for describing the structure of R(N). To my surprise Peter always said that this module has potential and at least three joint papers where dedicated to this module. Now I think I understand why. I say that I think because I clearly see that there is a lot more potential to this module then I thought but perhaps even more than I think.

18 Happy Birthday, Peter! Thank you for you attention!