Mock-modular forms of weight one - UCLA Department of Mathematics
Mock-modular forms of weight one - UCLA Department of Mathematics
Mock-modular forms of weight one - UCLA Department of Mathematics
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26 W. DUKE AND Y. LI<br />
By (49) and the substitution A = A ′2 , we have<br />
lim<br />
s→1 4π ∑ A<br />
ψ(A)(E(τ √ AB −1 , s)+E(τ √ AB , s)) = −24(ψ(B)+ψ(B)) ∑ A ′ ψ 2 (A ′ ) log | √ y A ′η 2 (τ A ′)|.<br />
From (52) and<br />
H(p)ϑ B (z) = ∑ ψ ′<br />
ψ ′ (B)g ψ ′(z),<br />
it is easy to see that<br />
〈ϑ B , g ψ 〉 = −(ψ(B) + ψ(B)) ∑ A ′ ψ 2 (A ′ ) log | √ y A ′η 2 (τ A ′)|.<br />
After substituting these into equation (64) and changing A, B to A ′2 , B ′2 respectively, we<br />
obtain equation (53).<br />
□<br />
For any m ≥ 1, r ≥ 0 and τ, τ ′ ∈ H, define the level <strong>one</strong> <strong>modular</strong> function Ψ ∗ m(z; τ, τ ′ , r)<br />
by<br />
Ψ ∗ m(z; τ, τ ′ Ψ m (τ, z)<br />
, r) :=<br />
(j(τ ′ ) − j(z)) r<br />
In terms <strong>of</strong> Ψ ∗ m(z; τ, τ ′ , r), we can re-write the right hand side <strong>of</strong> (53) as<br />
−2<br />
∑<br />
ψ 2 (A ′ ) (log |Ψ ∗ m(τ A ′ B ′; τ A ′ B ′−1, τ A ′ B ′, r B(m))| + r B (m) log |y A ′ B ′j′ (τ A ′ B ′)|)<br />
A ′ ∈Cl(F )<br />
The quantity Ψ ∗ −1<br />
m(τ A ′ B ′; τA ′ B<br />
, τ ′ A ′ B ′, r B(m)) is well-defined and non-zero, since the order <strong>of</strong><br />
Ψ m (τ A ′ B ′−1, z) at z = τ A ′ B ′ is exactly r B(m) from the pro<strong>of</strong> above. By equation (47) and the<br />
fact that<br />
(j ′ (z)) 6 = j(z) 4 (j(z) − 1728) 3 ∆(z),<br />
we have<br />
σ C<br />
⎛<br />
⎝<br />
∏<br />
C ′ ∈Cl(F )<br />
y 6 A ′ B ′j′ (τ A ′ B ′)6<br />
y 6 C ′ ∆(τ C ′)<br />
⎞<br />
⎛<br />
⎠ = ⎝<br />
∏<br />
C ′ ∈Cl(F )<br />
y 6 C −1 A ′ B ′ j ′ (τ C −1 A ′ B ′)6<br />
y 6 C ′ ∆(τ C ′)<br />
⎞<br />
⎠ ∈ H.<br />
Similarly, the <strong>modular</strong> function Ψ ∗ m(z; τ A ′ B ′−1, τ A ′ B ′, r B(m)), which is defined over H, is sent<br />
to Ψ ∗ m(z; τ C −1 A ′ B ′−1, τ C −1 A ′ B ′, r B(m)) under σ C . So if we let<br />
⎛<br />
u m,B (A ′ ) := Ψ ∗ m(τ A ′ B ′; τ A ′ B ′−1, τ A ′ B ′, r B(m)) 6H(p) · ⎝<br />
⎞<br />
∏<br />
r B (m)<br />
yA 6 ′ B ′j′ (τ A ′ B ′)6 ⎠<br />
yC 6 ∆(τ ′ C ′)<br />
,<br />
then we can write<br />
with u m,B (A ′ ) ∈ H satisfying<br />
〈j lift,B<br />
m , g ψ 〉 reg = − 1<br />
3H(p)<br />
∑<br />
A ′ ∈Cl(F )<br />
C ′ ∈Cl(F )<br />
(65) σ C (u m,B (A ′ )) = u m,B (A ′ C −1 ).<br />
ψ 2 (A ′ ) log |u m,B (A ′ )|<br />
When m = 0, we have j lift,B<br />
0 (z) = ϑ B (z) and<br />
∑<br />
〈ϑ B , g ψ 〉 = − 1 ψ 2 (A ′ ) log |u<br />
12H(p)<br />
A ′ B ′u A ′ B ′−1|,<br />
A ′