Mock-modular forms of weight one - UCLA Department of Mathematics

52 W. DUKE AND Y. LI
for any d. By the same argument in the proof of Theorem 9.1, one can show that equation
(12) holds when d ≠ pm 2 for any integer m. So in that case, the equation above becomes
∣
− 1 ∑
( ) pd − k
r + 2
1 ∑
B
δ(k) = ψ(B −2 ) ∑ ∣∣∣∣∣
∏
ψ 2 (A) log
4
2 4
H(p)
− j(τ q ))
k∈Z
ψ
A ∣q∈Q −d /Γ(j(A) 1/wq ∣ ∣∣∣∣∣ ∏
= log
(j(B) − j(τ q )) 1/wq .
∣
q∈Q −d /Γ
10. Case p = 283: Some numerical calculations
Finally, we will present another numerical example to demonstrate that statements similar
to the Conjecture in §1 should be true for octahedral newforms. These calculations were
conducted in SAGE [41]. When p = 283, Cl(F ) has order 3 and the space S 1 (p) is 3-
dimensional spanned by dihedral newform h(z) and octahedral newforms f ± (z) = f 1 (z) ±
√ −2f2 (z), where
h(z) = q + q 4 − q 7 + q 9 + O(q 10 ),
f 1 (z) = q − q 4 + 2q 6 − q 7 − q 9 + O(q 10 ),
f 2 (z) = q 2 − q 3 − q 5 + O(q 10 ).
By Lemma 3.3, h(z), f 1 (z) ∈ S + 1 (p) and f 2 (z) ∈ S − 1 (p). By Deligne-Serre Theorem, f ± (z)
arises from ρ : Gal(K/Q) → GL 2 (C), where K is the degree 2 extension of the normal closure
of Q[X]/(X 4 − X − 1) such that Gal(K/Q) ∼ = GL 2 (F 3 ).
Let F 6 be the subfield of K fixed by a subgroup of Gal(K/Q) isomorphic to Z/8Z, and H 3
the cubic subfield of F 6 . Explicitly, we can write
H 3 = Q[Y ]/(Y 3 + 4Y + 1),
F 6 = Q[X]/(X 6 − 3X 5 + 6X 4 − 7X 3 + 10X 2 − 7X + 6),
□
with the embedding
H 3 ↩→ F 6
Y ↦→ X 2 − X + 1.
Fix a complex embedding of F 6 ↩→ C such that X = t, Y = θ with t, θ ∈ C having positive
imaginary part. Since F 6 is totally complex, the unit group of F 6 has rank 2 and is generated
by
u 1 = t 2 − t + 1, u 2 = t 3 + t − 1.
As in the case of dihedral newforms, one can express the Petersson norms of f + (z) and f − (z)
as
〈f + , f + 〉 = 〈f − , f − 〉 = p + 1
12
Res s=1
ζ(s)L(s, F 6 )
L(s, H 3 )ζ(2s)(1 + p −s ) .
Notice the ratio L(s, F 6 )/L(s, H 3 ) is the L-function of the quadratic Hecke character of H 3
associated to the Hilbert quadratic norm residue symbol of F 6 /H 3 . So it is holomorphic at
s = 1 and the right hand side can be evaluated to be 8 log |u 1 u 2 2|.