Parameterizations for quintic Eisenstein series

Balanced Modular Parameterizations

“…, a r ; q) n = r j−1 (a j ; q) n for n ∈ N ∪ ∞. These are special cases of more general symmetric parameterizations in A 5 , B 5 from [8]. At level seven, certain modular forms are symmetric functions of the parameters x = q (q 2 , q 5 , q 7 , q 7 ; q 7 ) ∞ (q 3 , q 4 ; q 7 ) 2 ∞ , y = −q (q, q 6 , q 7 , q 7 ; q 7 ) ∞ (q 2 , q 5 ; q 7 ) 2 ∞ , z = (q 3 , q 4 , q 7 , q 7 ; q 7 ) ∞ (q, q 6 ; q 7 ) 2 ∞ .…”

A theory of theta functions to the quintic base

“…These series are among the classes of Eisenstein series considered in [14] that are expressed in terms of symmetric homogeneous polynomials in A 5 (q) and B 5 (q). Corresponding parameterizations exist for the Eisenstein series on the full modular group, defined by (1.32).…”

“…In particular, Theorem 3.6 may be applied to parameterizations from [14,Lemma 3.9] We may iteratively differentiate (4.45) with respect to θ to obtain…”

“…To prove the coupled system of differential equations for A(q) and B(q) given in Theorem 1.6, we employ parameterizations for the elliptic parameters from (1.33), namely e α (q), P α (q), and Q α (q), α = 1/5, 2/5 in terms of Eisenstein series. We will use formulas from [14] to write the Eisenstein series of weight k in terms of homogeneous polynomials of degree k in A 5 (q) and B 5 (q). These calculations allow us to translate Theorem 1.7 into Theorem 1.6.…”

Differential equations for septic theta functions

“…(q) = x 4 − 116x 3 y + 116xy 3 + y 4 − 116x 3 z + 848xyz 2 + 848x 2 yz − 848xy 2 z − 116y 3 z + 116xz3 + 116yz 3 + z 4 , (4.27) E 4 (q 7 ) = x 4 + 4x 3 y − 4xy 3 + y 4 + 4x 3 z + 8xyz + 8x 2 yz − 8xy 2 z + 4y 3 z − 4xz 3 − 4yz 3 + z 4 , (4.28) E 6 (q) = x 6 + 258x 5 y − 5904x 4 y 2 − 5904x 2 y 4 − 258xy 5 + y 6 + 258x 5 z (4.29) + 7310x 3 y 2 z + 7310x 2 y 3 z + 258y 5 z − 5904x 4 z 5904x 2 z 4 − 5904y 2 z 4 − 258xz 5 − 258yz 5 + z 6 , (4.30) E 6 (q 7 ) = x 6 + 6x 5 y + 18x 4 y 2 + 18x 2 y 4 − 6xy 5 + y 6 + 6x 5 z + 2x 3 y 2 z + 2x 2 y 3 z + 6y 5 z + 18x 4 z 2 + 2x 3 yz 2 − 57x 2 y 2 z 2 − 2xy 3 z 2x 2 yz 3 − 2xy 2 z 3 + 18x 2 z 4 + 18y 2 z 4 − 6xz (4.31)Proof. To prove (4.28), apply (4.24) and the formulas fromLemmas 4.2-4.3 to derive Zσ(1 + 5X + X 2 ) (4.32) − (x 4 + 4x 3 y + 4x 3 z + 8x 2 yz − 4xy 3 − 8xy 2 z + 8xyz 2 − 4xz 3 + y 4 + 4y 3 z − 4yz 3 + z 4 ) = (xy − xz + yz) −42x 4 z 2 − 79x 3 y 2 z − 258x 3 yz 2 + 52x 3 z 3 − 37x 2 y 4 (zx − z 2 + y 2 + 6yz) (xy − xz + yz) −184x 2 y 3 z + 3x 2 y 2 z 2 − 6x 2 yz 3 + 30x 2 z 4 + 45xy 5 (zx − z 2 + y 2 + 6yz) 2 + (xy − xz + yz) +292xy 4 z + 90xy 3 z 2 − 449xy 2 z 3 + 334xyz 4 − 48xz 5 (zx − z 2 + y 2 + 6yz) 2 + (xy − xz + yz) −10y 6 − 84y 5 z − 133y 4 z 2 + 96y 3 z 3 + 121y 2 z 4 − 70yz 5 + 8z 6 (zx − z 2 + y 2 + 6yz) 2 .…”