1 Introduction
Let
$\ell \ge 2$
be an integer. For
$N\geq 1$
and a Dirichlet character
$\chi $
modulo N, let
$M_{\ell }(N,\chi )$
and
$S_{\ell }(N,\chi )$
be the space of modular forms and cusp forms of weight
$\ell $
, level
$N,$
and nebentypus
$\chi $
, respectively. When
$\chi $
is trivial, we simply write
$M_{\ell }(N)$
and
$S_{\ell }(N)$
. Let
$M_{\ell +1/2}(4N)$
and
$S_{\ell +1/2}(4N)$
be the space of modular forms and the space of cusp forms of weight
$\ell +1/2$
for
$\Gamma _0(4N)$
, respectively. For
$N=1,$
we recall the Kohnen [Reference Kohnen10] plus space as the subspace
$$ \begin{align*} M_{\ell+1/2}^+(4):= \left\{f = \sum_{\substack{n\ge 0}}c_f(n)q^n\in M_{\ell+1/2}(4)~|~ c_f(n) = 0 \text{ if } (-1)^\ell n \equiv 2, 3 \pmod{4}\right\}, \end{align*} $$
and put
$S_{\ell +1/2}^+(4):=M_{\ell +1/2}^+(4)\cap S_{\ell +1/2}(4)$
. Let D be a fundamental discriminant (i.e.,
$D=1$
or is the discriminant of a quadratic field) such that
$(-1)^{\ell }D>0$
. Following Kohnen [Reference Kohnen10, p. 251], for
$f(z)=\sum _{n\geq 0}c_{f}(n)q^n\in M_{\ell +1/2}^{+}(4)$
, we define its D-th Shimura lift as
$$ \begin{align} \mathcal{S}_{D}\left(\sum_{n\ge 0}c_f(n)q^n\right) := \frac{c_f(0)}{2}L_D(1-\ell)+\sum_{n\ge 1}\left(\sum_{d|n}\left(\frac{D}{d}\right)d^{\ell-1}c_f\left(|D|\frac{n^2}{d^2}\right)\right)q^n, \end{align} $$
where
$\left (\hspace {-1pt}\frac {D}{\cdot }\hspace {-1pt}\right )$
is the Kronecker symbol. It is known that
$\mathcal {S}_D$
maps
$M_{\ell +1/2}^+(4)$
to
$M_{2\ell }(1)$
and
$S_{\ell +1/2}^+(4)$
to
$S_{2\ell }(1)$
, and commutes with the action of Hecke operators (see Kohnen [Reference Kohnen10, Theorem 1] and Shimura [Reference Shimura19]).
Now, we recall the Selberg identity on the Shimura lift. Let
$\theta (z)=\sum _{n\in \mathbb {Z}}q^{n^2}\in M_{1/2}(4)$
be the Jacobi theta function. Selberg observed that for a normalized Hecke eigenform
$f(z)\in M_{\ell }(1)$
with
$a_f(1)=1$
, the first Shimura lift provides the identity
$$ \begin{align} \mathcal{S}_1(f(4z)\theta(z))=f(z)^2\in M_{2\ell}(1). \end{align} $$
For a fundamental discriminant D with
$(-1)^{k} D>0$
with
$k\ge 4$
an integer, if one defines
$$ \begin{align*}\mathcal{F}_D(z)&:=\operatorname{\mathrm{Tr}}^D_1(G_{k,D}(z)^2)\in M_{2k}(1)\\ \mathcal{G}_{D}(z)&:=\frac32\left(1-\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)2^{-k}\right)^{-1}\operatorname{\mathrm{pr}}^+\operatorname{\mathrm{Tr}}_4^{4D}(G_{k,4D}(4z)\theta(|D|z))\in M_{k+1/2}^{+}(4),\end{align*} $$
then Kohnen–Zagier [Reference Kohnen and Zagier9, Proposition 3] proved the following generalization of (1.2):
$$ \begin{align}\mathcal{S}_D(\mathcal{G}_D(z))=\mathcal{F}_D(z).\end{align} $$
We must make several definitions for the above to make sense. The Eisenstein series
$G_{k,D}$
and
$G_{k,4D}$
are given by [Reference Kohnen and Zagier9, p. 185]
$$ \begin{align} G_{k,D}(z):=&\frac{L_{D}(1-k)}{2}+\sum_{n=1}^{\infty}\left(\sum_{d\mid n}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)d^{k-1}\right)q^n\in M_{k}\left(|D|,\left(\hspace{-1pt}\frac{D}{\cdot}\hspace{-1pt}\right)\right), \end{align} $$
$$ \begin{align}\kern-18pt G_{k,4D}(z):=&G_{k,D}(4z)-2^{-k}\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)G_{k,D}(2z) \in M_{k}\left(4|D|,\left(\hspace{-1pt}\frac{D}{\cdot}\hspace{-1pt}\right)\right), \end{align} $$
where
$L_D(s)=\sum _{n\ge 1}\left (\hspace {-1pt}\frac {D}{n}\hspace {-1pt}\right )n^{-s}$
. The operator
$\operatorname {\mathrm {pr}}^{+}$
is the projection from
$M_{\ell +1/2}(4)$
to
$M_{\ell +1/2}^{+}(4)$
given by [Reference Kohnen and Zagier9, p. 195]
$$ \begin{align} (\operatorname{\mathrm{pr}}^{+}g)(z)=\frac{1-(-1)^\ell i}{6}(\operatorname{\mathrm{Tr}}^{16}_4Vg)(z)+\frac{1}{3}g(z), \end{align} $$
where
$V(g)(z)=g(z+\frac {1}{4})=g(z)|_{k+1/2}[\begin {smallmatrix} 4 & 1\\ 0& 4 \end {smallmatrix}] $
, using the notation of (1.7) and (1.8). Additionally, for
$N\mid M$
,
$\operatorname {\mathrm {Tr}}_N^M$
is the trace map
$$ \begin{align} \operatorname{\mathrm{Tr}}_N^M: M_{m}(M)\rightarrow M_m(N),\quad g\mapsto \sum_{\gamma\in\Gamma_0(M)\backslash\Gamma_0(N)}g|_m\gamma, \end{align} $$
where for any real number m and
$\gamma =[\begin {smallmatrix} a&b\\c&d \end {smallmatrix}]\in \operatorname {\mathrm {GL}}_2^+({\mathbb {R}}),$
we define the slash operator [Reference Cohen2, Theorem 7.1]
$$ \begin{align} \left(g|_m\gamma\right)(z)=\det(\gamma)^{m/2} (cz+d)^{-m} g\left(\frac{az+b}{cz+d}\right). \end{align} $$
On the other hand, the Selberg identity (1.2) for the first Shimura lift has been generalized to the setting of Rankin–Cohen brackets. Let us first introduce the definition of Rankin–Cohen brackets for modular forms.
Definition 1.1 Let
$f(z)\in M_{a}(\Gamma )$
and
$g(z)\in M_b(\Gamma )$
be modular forms for some congruence subgroup
$\Gamma $
of weights a and b, respectively. For a nonnegative integer e, we define the e-th Rankin–Cohen bracket as
$$ \begin{align} [f(z),g(z)]_e := \sum_{r=0}^e (-1)^r\binom{e+a-1}{e-r}\binom{e+b-1}{r}f(z)^{(r)}g(z)^{(e-r)}, \end{align} $$
where
$f(z)^{(r)}$
is the r-th normalized derivative
$f(z)^{(r)}:=\frac {1}{(2\pi i)^r}\frac {d^r f(z)}{dz^r}$
of f. Here,
$a,b$
can be in
$\frac {1}{2}\mathbb {Z}$
and the binomial coefficients are defined through gamma functions. Moreover,
$[f,g]_e\in M_{a+b+2e}(\Gamma )$
and
$[f,g]_e\in S_{a+b+2e}(\Gamma )$
for
$e>1$
(see [Reference Cohen2, Theorem 7.1]. We remark that the Rankin–Cohen bracket defined in Zagier [Reference Zagier25, (73)] is related to (1.9) through
$F_{e}^{(a,b)}(f(z),g(z))= (-2\pi i)^e e![f(z),g(z)]_e$
(see [Reference Lei, Ni and Xue12, (1.1)].
Choie–Kohnen–Zhang [Reference Choie, Kohnen and Zhang1] and Xue [Reference Xue24] independently showed that if
$k \ge 4$
is an even integer,
$f(z)\in M_{k}(1)$
is a normalized Hecke eigenform, and e is a nonnegative integer, then
$$ \begin{align} \mathcal{S}_1([f(4z), \theta(z)]_e)=\frac{\binom{k+e-1}{e}}{\binom{k+2e-1}{2e}}[f(z),f(z)]_{2e}. \end{align} $$
Note that (1.10) was also proved in [Reference Popa17, Proposition B1] when f is an Eisenstein series. Recently, Wang [Reference Wang21] generalized (1.10) to higher-level forms. Let
$k\geq 4$
and
$e>0$
be integers with
$\ell =k+2e$
and let D be a fundamental discriminant such that
$(-1)^{\ell }D>0$
. We introduce functions
$$ \begin{align} \mathcal{F}_{D,k,e}(z):=&\operatorname{\mathrm{Tr}}^D_1([G_{k,D}(z),G_{k,D}(z)]_{2e})\in S_{2\ell}(1), \end{align} $$
$$ \begin{align} \mathcal{G}_{D, k,e}(z):=&\frac32\left(1-\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)2^{-k}\right)^{-1}\operatorname{\mathrm{pr}}^+\operatorname{\mathrm{Tr}}^{4D}_4\left[G_{k,4D}(z),\theta(|D|z)\right]_e \in S^+_{\ell+1/2}(4). \end{align} $$
Note that both
$\mathcal {F}_{D,k,e}(z)$
and
$\mathcal {G}_{D, k,e}(z)$
are cusp forms, since
$e>0$
. Now, we state our first main result, which can be viewed as a combination of (1.3) and (1.10).
Theorem 1.1 Let D be an odd fundamental discriminant such that
$(-1)^\ell D>0$
and let
$k\ge 4$
and
$e>0$
be integers such that
$k+2e=\ell $
. Then, we have the identity
$$ \begin{align} \mathcal{S}_D\left(\mathcal{G}_{D,k,e}\right)=|D|^e\frac{\binom{k+e-1}{e}}{\binom{k+2e-1}{2e}}\mathcal{F}_{D,k,e}. \end{align} $$
We have required that
$e>0$
because the case
$e=0$
is exactly (1.3). Our next main result concerns the nonvanishing of twisted central values of L-functions associated with Hecke eigenforms. Before stating the precise result, let us first introduce some notation.
Definition 1.2 Let D be a fundamental discriminant such that
$(-1)^{\ell }D>0$
.
-
(1) Let
$S_{2\ell }^{0,D}(1)$
denote the subspace of
$S_{2\ell }(1)$
generated by normalized Hecke eigenforms f with nonzero central twisted L-values
$L(f,D,\ell )$
, where
$L(f,D,s)=\sum _{n\ge 1}\left (\hspace {-1pt}\frac {D}{n}\hspace {-1pt}\right )a_f(n) n^{-s}$
is the L-function of f twisted by
$\left (\hspace {-1pt}\frac {D}{\cdot }\hspace {-1pt}\right )$
. We write
$S^{-,D}_{2\ell }(1)$
for the orthogonal complement of
$S_{2\ell }^{0,D}(1)$
, which is spanned by Hecke eigenforms with vanishing central twisted L-values. -
(2) Let
$S_{\ell +1/2}^{0,D}(4)$
be the subspace of
$S^+_{\ell +1/2}(4)$
generated by Hecke eigenforms
$g=\sum _{n\geq 1}c_g(n)q^n$
with
$c_g(|D|)\ne 0$
. We write
$S^{-,D}_{\ell +1/2}(4)$
for the orthogonal complement of
$S_{\ell +1/2}^{0,D}(4)$
, which is spanned by Hecke eigenforms
$g=\sum _{n\geq 1}c_g(n)q^n$
with
$c_g(|D|)=0$
.
The twisted L-function
$L(f,D,s)$
, originally defined for
$\operatorname {\mathrm {Re}}(s)\gg 0$
, can be analytically continued to the whole complex plane, and for a Hecke eigenform
$f\in S_{2\ell }(1)$
satisfies [Reference Ono16, Lemma 9.2]:
$$ \begin{align*} \Lambda(f,D,s)=(-1)^{\ell} \left(\hspace{-1pt}\frac{D}{-1}\hspace{-1pt}\right) \Lambda(f,D,2\ell-s), \end{align*} $$
where
$\Lambda (f,D,s)=(2\pi )^{-s}\Gamma (s) L(f,D,s)$
is the completed twisted L-function of f. Since
$\left (\hspace {-1pt}\frac {D}{-1}\hspace {-1pt}\right )$
is the sign of D, the assumption
$(-1)^{\ell }D>0$
implies that the functional equation for
$L(f,D,s)$
has a positive sign. Therefore, the subspace
$S_{2\ell }^{0,D}(1)$
in Definition 1.2 (1) is not trivially zero. It is speculated that the central L-value
$L(f,D,\ell )$
is nonvanishing for every Hecke eigenform
$f\in S_{2\ell }(1)$
. Thus, it is believed that
$S_{2\ell }(1)=S^{0,D}_{2\ell }(1)$
for every fundamental discriminant D. For further discussion, see Section 7.
Our second main result gives an explicit construction of a set of generators for the subspaces
$S_{2\ell }^{0,D}(1)$
and
$S_{\ell +1/2}^{0,D}(4)$
. We hope this result would help investigate the aforementioned speculation on the nonvanishing of twisted central L-values. Furthermore, we prove that the D-th Shimura lift
$\mathcal {S}_D$
gives an isomorphism between
$S_{\ell +1/2}^{0,D}(4)$
and
$S_{2\ell }^{0,D}(1)$
, which generalizes Kohnen’s results [Reference Kohnen10, Theorem 2] and [Reference Xue24, Proposition 3.3].
Theorem 1.2 Let D be an odd fundamental discriminant with
$(-1)^{\ell }D>0$
. Then,
$$ \begin{align*} S_{\ell+1/2}^{0,D}(4)=\operatorname{\mathrm{Span}}\{\mathcal{G}_{D,k,e}\}_{k+2e=\ell},\quad \text{and}\quad S_{2\ell}^{0, D}(1)=\operatorname{\mathrm{Span}}\{\mathcal{F}_{D,k,e}\}_{2k+4e=2\ell}, \end{align*} $$
where
$k\ge 4$
and
$e>0$
. Additionally, the restricted D-th Shimura lift
$$ \begin{align*} S_D: S_{\ell+1/2}^{0,D}(4) \rightarrow S_{2\ell}^{0, D}(1) \end{align*} $$
is an isomorphism.
We assume D to be odd throughout the article in order to avoid the technical complications caused by even D, although we believe our results continue to hold in this case.
This article is organized as follows. Section 2 discusses the main results of this article. The proof of Theorem 1.1 is based on the same idea as the proof of (1.10) (see [Reference Choie, Kohnen and Zhang1, Reference Xue24]), but requires explicit computations of the Fourier coefficients of both sides of (1.13). Most of the technical details required for the proof of Theorem 1.1 are presented in Section 6. Based on the Petersson inner product formulas for
$\mathcal {F}_{D,k,e}$
and
$\mathcal {G}_{D,k,e}$
derived in Section 5, we explicitly construct a spanning set for
$S_{2\ell }^{0,D}(1)$
(Proposition 2.4). We then show that the D-th Shimura lift is an isomorphism from
$S^{0,D}_{\ell +1/2}(4)$
to
$S_{2\ell }^{0,D}(1)$
(Proposition 2.2). Finally, using these results, we prove Proposition 2.6, explicitly constructing a spanning set for
$S_{\ell +1/2}^{0,D}(4)$
and finishing the proof of Theorem 1.2.
The remaining sections are dedicated to proofs of the results needed in Section 2. Section 3 proves an alternate formula for
$\mathcal {G}_{D,k,e}$
, which we use to compute its Fourier coefficients in Section 6. Section 4 recalls the theory of Eisenstein series, which will be useful to the Fourier development of
$\mathcal {F}_{D,k,e}$
and
$\mathcal {G}_{D,k,e}$
in Section 6. Assuming those two sections, Section 5 derives Petersson inner product formulas for
$\mathcal {F}_{D,k,e}$
and
$\mathcal {G}_{D,k,e}$
via the Rankin–Selberg convolution. In Section 6, we carry out the computations of Fourier coefficients for Theorem 1.1. Section 7 discusses the relationship between these results and their potential applications to the nonvanishing of twisted central L-values of Hecke eigenforms in
$S_{2\ell }(1)$
.
2 Selberg identity and spanning sets of subspaces
This section proves our main results, assuming the necessary results to be proved later. We begin by proving Theorem 1.1, a generalization of the Selberg identity.
Proof of Theorem 1.1
Recall that
$\mathcal {G}_{D,k,e}$
(1.12) and
$\mathcal {F}_{D,k,e}$
(1.11) are cusp forms. Write
$$ \begin{align*}\mathcal{S}_D(\mathcal{G}_{D,k,e}(z))=\sum_{n\geq1}g_{D,k,e}(n)q^n\quad\mathrm{ and}\quad\mathcal{F}_{D,k,e}(z)=\sum_{n\geq1}f_{D,k,e}(n)q^n.\end{align*} $$
Comparing the Fourier coefficients
$f_{D,k,e}(n)$
and
$g_{D,k,e}(n)$
that are, respectively, given by Lemmas 6.5 and 6.6, it suffices to show for each nonnegative integer pair
$(a_1, a_2)$
with
$a_1+a_2=n|D_1|$
that
$$ \begin{align} &\binom{k+e-1}{e}\sum_{r=0}^{2e}(-1)^ra_1^ra_2^{2e-r}\binom{2e+k-1}{2e-r}\binom{2e+k-1}{r} \nonumber\\ &\qquad=\binom{k+2e-1}{2e}\sum_{r+s=e}(-1)^r\binom{k+e-1}{s}\binom{e-1/2}{r}4^r(a_2-a_1)^{2s}(a_1a_2)^r. \end{align} $$
Without loss of generality, we may assume that
$R\leq S$
and compare the coefficients of the monomial
$a_1^Ra_2^S$
of the two sides of (2.1). The
$a_1^Ra_2^S$
-coefficient on the left-hand side of (2.1) is
$$ \begin{align*} (-1)^R\binom{k+e-1}{e}\binom{2e+k-1}{2e-R}\binom{2e+k-1}{R}, \end{align*} $$
and the right-hand side of (2.1) has
$a_1^Ra_2^S$
-coefficient
$$ \begin{align*} &\binom{k+2e-1}{2e}\sum_{r=0}^R(-1)^r\binom{k+e-1}{e-r}\binom{e-1/2}{r}4^r\binom{2e-2r}{R-r}(-1)^{R-r}\\ &\qquad=(-1)^R\binom{k+2e-1}{2e}\sum_{r=0}^R\binom{k+e-1}{e-r}\binom{e-1/2}{r}4^r\binom{2e-2r}{R-r}. \end{align*} $$
Lemma 2.1 Let
$R\leq e$
be nonnegative and
$k\geq 4$
. Then, we have the following identity:
$$ \begin{align*} &\binom{k+e-1}{e}\binom{k+2e-1}{2e-R}\binom{k+2e-1}{R}\\&\qquad=\binom{k+2e-1}{2e}\sum_{r=0}^R 4^r\binom{k+e-1}{e-r}\binom{e-1/2}{r}\binom{2e-2r}{R-r}, \end{align*} $$
where fractional binomial coefficients are defined by the
$\Gamma $
function.
Proof We reproduce the proof of [Reference Xue24, Proposition 2.1]. By definition, we have
$$ \begin{align*} \binom{e-1/2}{r}=\frac{\Gamma(e+1/2)}{\Gamma(r+1)\Gamma(e+1/2-r)}. \end{align*} $$
By Legendre’s duplication formulas, we have
$$ \begin{align*} \Gamma(e+1/2)=\frac{(2e)!}{4^ee!}\sqrt{\pi}, \qquad \Gamma(e-r+1/2)=\frac{(2(e-r))!}{4^{e-r}(e-r)!}\sqrt{\pi}. \end{align*} $$
These together yield
$$ \begin{align*} \binom{e-1/2}{r}=\frac{(2e)!4^{e-r}(e-r)!\sqrt{\pi}}{r!4^ee!(2(e-r))!\sqrt{\pi}}=\frac{(2e)!4^{-r}(e-r)!}{r!e!(2(e-r))!}, \end{align*} $$
which yields the following formula for each term on the right-hand side:
$$ \begin{align*} &4^r\binom{k+2e-1}{2e}\binom{k+e-1}{e-r}\binom{e-1/2}{r}\binom{2e-2r}{R-r}\hspace{-1pt}\\&\qquad=\hspace{-1pt}\frac{(k+2e-1)!(k+e-1)!}{(k-1)!\hspace{-1pt}(k+r-1)!\hspace{-1pt}(R-r)!\hspace{-1pt}(2e-R-r)!e!r!}. \end{align*} $$
The left-hand side expands into
$$ \begin{align*} &\binom{k+e-1}{e}\binom{k+2e-1}{2e-R}\binom{k+2e-1}{R}\\&\qquad=\frac{(k+e-1)!(k+2e-1)!(k+2e-1)!}{e!(k-1)!(2e-R)!(k+R-1)!R!(k+2e-1-R)!}. \end{align*} $$
If we cancel
$(k+e-1)!(k+2e-1)!$
from both sides, and multiply by
$R!(k+2e-R-1)$
, we see that it suffices to show
$$ \begin{align*} \sum_{r=0}^R \binom{R}{R-r}\binom{k+2e-R-1}{k+r-1}=\binom{k+2e-1}{k+R-1}. \end{align*} $$
After applying the involution
$r\mapsto R-r$
, this is then Vandermonde’s identity [Reference Riordan18, p. 11]
$$ \begin{align*} \sum_{j=0}^t \binom{n}{j}\binom{m}{t-j}=\binom{n+m}{t} \end{align*} $$
for the case of
$n=R$
,
$m=k+2e-R-1$
, and
$t=k+R-1$
.
We now build toward the proof of Theorem 1.2. We begin by showing that the D-th Shimura lift gives rise to an isomorphism between
$S_{\ell +1/2}^{0,D}(4)$
and
$S_{2\ell }^{0,D}(1)$
, which is a generalization of [Reference Kohnen10, Theorem 2] for
$D=1$
.
Proposition 2.2 Let D be an odd fundamental discriminant with
$(-1)^{\ell }D>0$
. Then, the D-th Shimura lift
$\mathcal {S}_D$
restricts to an isomorphism
$S_{\ell +1/2}^{0,D}(4) \rightarrow S_{2\ell }^{0,D}(1)$
for all
$\ell \geq 6$
.
Proof Recall that by [Reference Kohnen10, Theorem 1] or [Reference Kohnen and Zagier9, p. 182], if
$g=\sum _{n\ge 1} c_g(n)q^n\in S^+_{\ell +1/2}(4)$
is a Hecke eigenform and
$f\in S_{2\ell }(1)$
is the normalized Hecke eigenform corresponding to g, then
$\mathcal {S}_D(g)=c_g(|D|) f$
. This means that
$\mathcal {S}_D$
is a monomorphism when restricted to
$S^{0,D}_{\ell +1/2}(4)$
. Thus, in order to show
$\mathcal {S}_D$
restricts to an isomorphism from
$S_{\ell +1/2}^{0,D}(4)$
to
$ S_{2\ell }^{0,D}(1)$
it suffices to show that
$\dim S_{\ell +1/2}^{0,D}(4)=\dim S_{2\ell }^{0,D}(1)$
since
$c_g(|D|)=0$
if and only if
$L_D(f,\ell )=0$
by [Reference Kohnen and Zagier9, Theorem 1].
Note that
$\dim S_{2\ell }^{0,D}(1)$
is the number of Hecke eigenforms in
$S_{2\ell }(1)$
with nonzero central twisted L-value, and
$\dim S_{\ell +1/2}^{0,D}(4)$
is the number of Hecke eigenforms in
$S^+_{\ell +1/2}(4)$
with nonzero
$|D|$
-th Fourier coefficient. According to [Reference Kohnen and Zagier9, Theorem 1], these two nonvanishing conditions are the same under the Shimura correspondence, thus we conclude that
$\dim S_{\ell +1/2}^{0,D}(4)=\dim S_{2\ell }^{0,D}(1)$
.
Remark 2.3 In the
$\ell = 5,7$
case, the space of cusp forms
$S_{2\ell }(1)$
is zero, and so is the space
$S^+_{\ell +1/2}(4)$
. So this proposition is trivially true.
We now construct an explicit spanning set for
$S_{2\ell }^{0,D}(1)$
. Before doing so, we need to introduce the period of a modular form. For
$f\in S_{2\ell }(1)$
and
$0\leq t\leq 2\ell -2$
, the t-th period of f is given by
$$ \begin{align}r_t(f):=\frac{t!}{(-2\pi i)^{t+1}}L(f,t+1).\end{align} $$
Here, the L-series of
$f(z)=\sum _{n\geq 1}a_nq^n$
is
$L(f,s)=\sum _{n\geq 1}a_nn^{-s}$
, which converges for
$\operatorname {\mathrm {Re}}(s)\gg 0$
and can be extended analytically to the whole complex plane; for details, see [Reference Manin14].
Proposition 2.4 The set
$\{\mathcal {F}_{D, k, e}\}_{2k+4e=2\ell }$
for
$1\leq e \leq \lfloor \frac {\ell -4}{2}\rfloor $
spans
$S_{2\ell }^{0,D}(1)$
, for all
$\ell \geq 6$
.
Proof By Proposition 5.6, we know that if
$g\in S_{2\ell }^{-, D}(1)$
then g is orthogonal to the subspace of
$S_{2\ell }^D(1)$
spanned by
$\{\mathcal {F}_{D, k, e}\}_{2k+4e=2\ell }$
. So it suffices to show that the orthogonal complement of the span of
$\{\mathcal {F}_{D, k, e}\}_{2k+4e=2\ell }$
is contained in
$S_{2\ell }^{-, D}(1)$
.
We will show that any modular form
$G=\sum _j c_jg_j$
, which is a linear combination of normalized Hecke eigenforms in
$g_j \in S_{2\ell }^{0, D}(1)$
such that
$\langle G, \mathcal {F}_{D,k,e} \rangle =0$
for all
$\mathcal {F}_{D, k, e}$
must be zero.
Note that Proposition 5.6 and (2.2) imply that
$$ \begin{align*} \langle \mathcal{F}_{D,k,e},g_j\rangle &=\frac{1}{2}\frac{\Gamma(2k+4e-1)\Gamma(k+2e)}{(2e)!(4\pi)^{2k+4e-1}\Gamma(k)}\\&\quad\frac{L_{D}(1-k)}{L_D(k)}\frac{(-2\pi i)^{2k+2e-1}}{(2k+2e-2)!}L(g_j,D,k+2e)r_{2k+2e-2}(g_j). \end{align*} $$
Thus, the orthogonality condition
$ \langle G,\mathcal {F}_{D,k,e}\rangle =0$
is equivalent to
$$ \begin{align} \sum_{j}c_jL(g_j,D,k+2e)r_{2k+2e-2}(g_j)=0. \end{align} $$
Following an idea from the proof of [Reference Imamolu and Kohnen6, Theorem 1], we define another form in
$S_{2\ell }(1)$
by
$$ \begin{align*}F=\sum_j c_j L(g_j,D,k+2e)g_j.\end{align*} $$
Hence, (2.3) implies that
$$ \begin{align*} r_{2k+2e-2}(F)=\sum_jc_jL(g_j,D,k+2e)r_{2k+2e-2}(g_j)=0. \end{align*} $$
As
$1\leq e \leq \lfloor \frac {\ell -4}{2}\rfloor $
and
$k+2e = \ell $
, we have
$\ell -2 \geq k \geq 4$
. Then,
$t=2k+2e-2$
ranges through all even values
$\ell +2\leq t\leq 2\ell -4$
, so
$r_{t}(F)=0$
for all even
$\ell +2\leq t\leq 2\ell -4$
. As a result of the following lemma, we have
$F=0$
. Since
$L(g_j,D,k+2e)\neq 0$
as
$g_j\in S_{2\ell }^{0,D}(1)$
, we must have
$c_j=0$
for all j, and thus
$G=0$
.
Lemma 2.5 Let
$F\in S_{2\ell }(1)$
and
$\ell \geq 6$
, and let
$r_t(F)$
be the t-th period of F. If
$r_t(F)=0$
for all even t such that
$\ell +2\leq t \leq 2\ell -4$
, then
$F=0$
.
Proof We follow the idea of [Reference Xue23]. By the Eichler–Shimura theory [Reference Manin14, Proposition 2.3(b)] and [Reference Xue23, Remark 2.4], we know that
$F=0$
if and only if
$r_t(F)=0$
for all even
$2\leq t\leq 2\ell -4$
. By the Eichler–Shimura relation
$$ \begin{align} r_t(F)+(-1)^{t}r_{2\ell-2-t}(F)=0, \end{align} $$
and the assumption that
$r_t(F)=0$
for all even
$\ell +2\leq t \leq 2\ell -4$
, we know that
$r_t(F)=0$
also for all even
$2\leq t\leq \ell -4$
. To show that the periods
$\ell -4 < t < \ell +2$
are zero, we split into cases based on the parity of
$\ell $
.
-
(1) If
$\ell $
is even, it suffices to show that
$r_\ell (F)=r_{\ell -2}(F)=0$
. Since
$\ell $
is even, by (2.4), (2.5)
$$ \begin{align} r_\ell(F)+r_{\ell-2}(F)=0. \end{align} $$
Substituting
$t=\ell -2$
into the Eichler–Shimura relation (2.6)
$$ \begin{align} (-1)^tr_t(F)&+\sum_{\substack{0\leq m\leq t\\ m\equiv 0\pmod2}}\binom{t}{m}r_{2\ell-2-t+m}(F)+\sum_{\substack{0\leq m\leq 2\ell-2-t\\ m\equiv t\pmod2}}\binom{2\ell-2-t}{m}r_{m}(F)=0 \end{align} $$
and noting that
$r_0(F)+r_{2\ell -2}(F)=0$
, we obtain
$$ \begin{align*} \left(\binom{\ell}{2}+1\right)r_{\ell-2}(F)+2r_{\ell}(F) = 0. \end{align*} $$
This equation, along with (2.5), implies that
$r_\ell (F)=r_{\ell -2}(F)=0$
for
$\ell \ge 6$
. -
(2) If
$\ell $
is odd, it suffices to show that
$r_{\ell -3}(F)=r_{\ell -1}(F)=r_{\ell +1}(F)=0$
. Substituting
$t = \ell -1$
into (2.6), we get
$$\begin{align*}3r_{\ell-1}(F) + \binom{\ell-1}{2} r_{\ell+1}(F) + \binom{\ell-1}{\ell-3} r_{\ell-3}(F) = 0.\end{align*}$$
Since
$\binom {\ell -1}{2} = \binom {\ell -1}{\ell -3}$
, and we know by (2.4) that
$ r_{\ell -3}(F)+r_{\ell +1}(F)=0$
, we conclude that
$r_{\ell -1}(F) = 0$
. Substituting
$t=\ell +1$
into (2.6) yields
$$ \begin{align*} 2r_{\ell-3}(F)+\binom{\ell+1}{2}r_{\ell-1}(F)+\left(1+\binom{\ell+1}{4}\right)r_{\ell+1}(F)=0, \end{align*} $$
and noting that
$r_{\ell -1}(F)=0$
, we conclude by (2.4) that
$$ \begin{align*} r_{\ell-3}(F)=r_{\ell+1}(F)=0. \end{align*} $$
This finishes the proof.
Finally, we construct a spanning set for
$S_{\ell +1/2}^{0,D}(4)$
and finish the proof of Theorem 1.2.
Proposition 2.6 The set
$\{\mathcal {G}_{D,k,e}\}_{k+2e=\ell }$
for
$1\leq e \leq \lfloor \frac {\ell -4}{2}\rfloor $
spans the subspace
$S_{\ell +1/2}^{0,D}(4)$
, for all
$\ell \geq 6$
.
Proof For a Hecke eigenform
$g \in S_{\ell +1/2}^{-, D}(4)$
, we have
$\langle g,\mathcal {G}_{D,k,e}\rangle =0$
by Proposition 5.7. So g is orthogonal to
$\operatorname {\mathrm {Span}}\{\mathcal {G}_{D,k,e}\}_{k+2e=\ell }$
and thus
$$ \begin{align}\operatorname{\mathrm{Span}}\{\mathcal{G}_{D,k,e}\}_{k+2e=\ell}\subseteq S_{\ell+1/2}^{0,D}(4).\end{align} $$
Note that Theorem 1.1 implies that
$$ \begin{align*} \dim \operatorname{\mathrm{Span}}\{\mathcal{G}_{D,k,e}\}_{k+2e=\ell}\geq \dim\operatorname{\mathrm{Span}}\{\mathcal{F}_{D,k,e}\}_{k+2e=\ell}. \end{align*} $$
By Propositions 2.2 and 2.4, we have
$$ \begin{align} \quad \operatorname{\mathrm{Span}}\{\mathcal{F}_{D,k,e}\}_{k+2e=\ell}=S_{2\ell}^{0,D}(1) \quad\mathrm{and}\dim S_{2\ell}^{0,D}(1)=\dim S_{\ell+1/2}^{0,D}(4). \end{align} $$
Now, (2.7)–(2.8) together imply that
$\dim \operatorname {\mathrm {Span}}\{\mathcal {G}_{D,k,e}\}_{k+2e=\ell } \geq \dim S_{\ell +1/2}^{0,D}(4)$
. So we conclude that
$\operatorname {\mathrm {Span}}\{\mathcal {G}_{D,k,e}\}_{k+2e=\ell } = S_{\ell +1/2}^{0,D}(4)$
.
Combining Propositions 2.2, 2.4, and 2.6, we complete the proof of Theorem 1.2.
3 Projection
In this section, we prove an alternate formula for
$\mathcal {G}_{D,k,e}$
(1.12):
$$ \begin{align*} \mathcal{G}_{D, k, e}(z)=\operatorname{\mathrm{Tr}}^{4D}_4 [G_{k,D}(4z), \theta(|D|z)]_e. \end{align*} $$
A similar formula is implicit in equations (6) and (7) in [Reference Kohnen and Zagier9]. This formula allows us to compute the Fourier coefficients (Proposition 6.4).
We need to introduce some notation and facts needed for the proof of Lemma 3.2. Let
$$ \begin{align*} \mathbb{P}^1(\mathbb{Z}/N\mathbb{Z})=\{(a:b):a,b\in\mathbb{Z}/N\mathbb{Z},\,\gcd(a,b,N)=1\}/\sim \end{align*} $$
be the projective line over
$\mathbb {Z}/N\mathbb {Z}$
, where
$(a:b)\sim (a':b')$
if there exists
$u\in (\mathbb {Z}/N\mathbb {Z})^{\ast }$
such that
$a=ua', b=ub'$
. It is known that there is a bijection between
$\Gamma _0(N)\backslash \operatorname {\mathrm {SL}}_2(\mathbb {Z})$
and
$\mathbb {P}^1(\mathbb {Z}/N\mathbb {Z})$
, which sends a coset representative
$[\begin {smallmatrix} a &b\\c&d \end {smallmatrix}]$
to the class
$(c:d)$
in
$\mathbb {P}^1({\mathbb {Z}}/N\mathbb {Z})$
(see [Reference Stein20, Proposition 3.10]. For future reference, we prove a result on coset representatives of certain quotients of congruence subgroups.
Lemma 3.1 Let
$N\in \mathbb {N}$
and let
$S\in \mathbb {N}$
be squarefree with
$(N,S)=1$
. Then,
$$ \begin{align*} \left\{ \begin{bmatrix} 1 & 0 \\ NS_1 & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix}\quad :\quad S_1\mid S,\quad \mu\ \mod \frac{S}{S_1}\right\} \end{align*} $$
is a set of coset representatives for
$\Gamma _0(NS)\backslash \Gamma _0(N)$
.
Proof The statement follows easily from the description of the cosets given in [Reference Gross and Zagier5, p. 276].
Lemma 3.2 Let
$\ell \ge 1$
be an integer and D be odd. We have
$V\operatorname {\mathrm {Tr}}^{4D}_4g=\operatorname {\mathrm {Tr}}^{16D}_{16}Vg$
for all
$g\in M_{\ell +1/2}(4|D|)$
.
Proof The statement was mentioned in [Reference Kohnen and Zagier9, p. 195], we only sketch it. We first remark that by direct calculation,
$Vg\in M_{\ell + \frac {1}{2}}(16|D|)$
, so
$\operatorname {\mathrm {Tr}}^{16D}_{16}Vg$
is well-defined. Note that applying the fixed set of cosets for
$\Gamma _0(4D)\backslash \Gamma _0(4)$
and
$\Gamma _0(16D)\backslash \Gamma _0(16)$
given by Lemma 3.1 to
$N=4,16$
and
$S=|D|$
, we have the following explicit formulas (see (1.8) for the definition of slash operators):
$$ \begin{align*} V\operatorname{\mathrm{Tr}}^{4D}_4g(z)&=\sum_{D_1D_2=D}\sum_{\mu~\text{mod}~|D_2|} g(z)|_\ell \gamma_{D_1,\mu}[\begin{smallmatrix} 1 & \frac{1}{4} \\ 0 & 1 \end{smallmatrix}], \\ \operatorname{\mathrm{Tr}}^{16D}_{16}Vg(z)&=\sum_{D_1D_2=D}\sum_{\mu~\text{mod}~|D_2|} g(z)|_\ell [\begin{smallmatrix} 1 & \frac{1}{4} \\ 0 & 1 \end{smallmatrix}]\gamma^{\prime}_{D_1,\mu}, \end{align*} $$
where
$$ \begin{align*} \gamma_{D_1,\mu}=\begin{bmatrix} 1 & 0 \\ 4|D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix}\quad\mathrm{and}\quad\gamma^{\prime}_{D_1,\mu}=\begin{bmatrix} 1 & 0 \\ 16|D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix}. \end{align*} $$
And the outer sums are over all factorizations of D into a product of fundamental discriminants
$D_1, D_2$
. Therefore, to prove the desired equality, it suffices to show that the set of cosets
$$ \begin{align*} \left\{\Gamma_0(4|D|)\begin{bmatrix} 1 & 1/4 \\ 0 & 1 \end{bmatrix}\gamma^{\prime}_{D_1,\mu}\begin{bmatrix} 1 &1/4\\0 &1 \end{bmatrix}^{-1}:\,D_1D_2=D,\quad\mu~\text{mod}~{|D_2|}\right\} \end{align*} $$
is a system of representatives of
$\Gamma _0(4|D|)\backslash \Gamma _0(4)$
, which can be easily checked.
Definition 3.1 For
$m\in \mathbb {N}$
and
$f(z)=\sum _{n\geq 0}a_f(n)q^n\in S_{k}(N,\chi ),$
we define
$U_mf$
by
$$ \begin{align} U_mf(z)=\frac{1}{m}\sum_{v~\mathrm{mod}~m}f\left(\frac{z+v}{m}\right)=\sum_{n\geq0}a_f(mn)q^n. \end{align} $$
Equivalently, we may write via (1.8)
$$ \begin{align}U_mf(z)=m^{k/2-1}\sum_{v\text{ mod }m}f(z)\bigg|_k\begin{bmatrix} 1 & v\\ 0& m \end{bmatrix}.\end{align} $$
We need the following two simple observations. Note that Lemma 3.3 follows from (3.1) and it implies Lemma 3.4.
Lemma 3.3 Let
$U_2$
be the operator defined in (3.1). Then,
$$ \begin{align*} U_2G_{k,D}(z)=\left(1+2^{k-1}\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)\right)G_{k,D}(z)-2^{k-1}\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)G_{k,D}(2z). \end{align*} $$
Lemma 3.4 The following identity holds:
$$\begin{align*}&G_{k,D}(4z) -G_{k,D}(8z)-2^{-k}\left(\frac{D}{2}\right) \left(G_{k,D}\left(2z+\frac{1}{2}\right) + G_{k,D}(2z)\right)\\&\quad=-\left(\frac{D}{2}\right)2^{-k+1}G_{k,D}(4z).\end{align*}$$
Note that
$\gamma _v=[\begin {smallmatrix}1 & 0\\4|D|v & 1\end {smallmatrix}]$
for
$v=0,1,2,3$
form a system of representatives of
$\Gamma _0(16|D|)\backslash \Gamma _0(4|D|)$
[Reference Kohnen and Zagier9, p. 195]. The following lemma explicitly computes each term in
$\operatorname {\mathrm {Tr}}^{16D}_{4D}(VG_{k,D}(2z))$
.
Lemma 3.5 For
$\gamma _v=[\begin {smallmatrix}1 & 0\\4|D|v & 1\end {smallmatrix}]$
, we have
$$ \begin{align*} V(G_{k,D}(2z))\Big|_{k}\gamma_v&=\begin{cases} G_{k,D}\left(2z+\frac{1}{2}\right) & v\equiv0,2\pmod4 ,\\ \left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)2^{k}G_{k,D}(8z) & v\equiv1,3\pmod4. \end{cases} \end{align*} $$
Proof First,
$$ \begin{align*} V(G_{k,D}(2z))\Bigg|_{k}\begin{bmatrix}1 & 0\\ 4|D|v & 1 \end{bmatrix}&=2^{-k/2}G_{k,D}(z)\Bigg|_{k}\begin{bmatrix}2 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix}4 & 1\\ 0 & 4 \end{bmatrix}\begin{bmatrix}1 & 0\\ 4|D|v & 1 \end{bmatrix}\\&=2^{-k/2}G_{k,D}(z)\Bigg|_{k}\begin{bmatrix}8(|D|v+1) & 2\\ 16|D|v & 4 \end{bmatrix}.\end{align*} $$
Now, we do some casework.
-
(1)
$v=0$
: We have
$$\begin{align*}V(G_{k,D}(2z))\Bigg|_{k}\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}=V(G_{k,D}(2z))=G_{k,D}\left(2\left(z+\frac14\right)\right)=G_{k,D}\left(2z+\frac{1}{2}\right).\end{align*}$$
-
(2)
$v=1, 3$
: Since v and
$|D|$
are odd,
$v|D|+1$
must be even,
$\gcd (\frac {|D|v+1}{2}, |D|v)=1$
, and there exist some
$x, y \in \mathbb {Z}$
such that
$\frac {|D|v+1}{2}x+|D|vy=1$
. Note also that
$x\equiv 2\pmod D$
and
$\left (\hspace {-1pt}\frac {D}{x}\hspace {-1pt}\right )=\left (\hspace {-1pt}\frac {D}{2}\hspace {-1pt}\right )$
. Thus,
$$ \begin{align*} V(G_{k,D}(2z))\Bigg|_{k}\begin{bmatrix}1 & 0\\ 4|D|v & 1 \end{bmatrix}&=2^{-k/2}G_{k,D}(z)\Bigg|_{k}\begin{bmatrix}8(|D|v+1) & 2\\ 16|D|v & 4 \end{bmatrix}\\ &=2^{-k/2}G_{k,D}(z)\Bigg|_{k}\hspace{-3pt}\begin{bmatrix}\frac{|D|v+1}{2} & -y\\ |D|v & x \end{bmatrix}\hspace{-5pt}\begin{bmatrix}16 & 2x+4y\\ 0 & 2 \end{bmatrix}\\ &=2^{-k/2}\left(\frac{D}{x}\right)G_{k,D}(z)\Bigg|_{k}\begin{bmatrix}16 & 2x+4y\\ 0 & 2 \end{bmatrix}\\ &=2^{k}\left(\frac{D}{x}\right)G_{k,D}(8z+x+2y)\\&=\left(\frac{D}{2}\right)2^{k}G_{k,D}(8z). \end{align*} $$
-
(3)
$v=2$
: Since
$\text {gcd}(2|D|+1, 4|D|)=1$
, we can pick
$x,y\in \mathbb {Z}$
such that
$ (2|D|+1)x+4|D|y=1$
. As
$4|D|y$
is even, x must be odd, so
$G_{k,D}(2z+\frac {x}{2})=G_{k,D}(2z+\frac {1}{2})$
, and further
$\left (\frac {D}{x}\right )=1$
since
$x\equiv 1\pmod D$
. Hence,
$$ \begin{align*} V(G_{k,D}(2z))\Bigg|_{k}\begin{bmatrix}1 & 0\\ 8|D| & 1 \end{bmatrix}&=2^{-k/2}G_{k,D}(z)\Bigg|_{k}\begin{bmatrix}8(2|D|+1) & 2\\ 32|D| & 4 \end{bmatrix}\\&=2^{-k/2}G_{k,D}(z)\Bigg|_{k}\begin{bmatrix}2|D|+1 & -y\\ 4|D| & x \end{bmatrix}\begin{bmatrix}8 & 2x+4y\\ 0 & 4 \end{bmatrix}\\ &=\left(\frac{D}{x}\right)2^{-k/2}G_{k,D}(z)\Bigg|_{k}\begin{bmatrix}8 & 2x+4y\\ 0 & 4 \end{bmatrix}\\&=G_{k,D}\left(2z+\frac{x}{2}\right)\\&=G_{k,D}\left(2z+\frac{1}{2}\right). \end{align*} $$
Thus, the proof is complete.
The following lemma explicitly computes each term in
$\operatorname {\mathrm {Tr}}^{16D}_{4D}(V\theta (|D|z))$
.
Lemma 3.6 Let D be an odd fundamental discriminant. Then,
$$ \begin{align*} V(\theta(|D|z))\Big|_{\frac{1}{2}}\gamma_0&=\theta\left(|D|z+\frac{|D|}{4}\right),\\ V(\theta(|D|z))\Big|_{\frac{1}{2}}\gamma_1&=\begin{cases}(2i)^{1/2}(\theta(|D|z)-\theta(4|D|z)) & D>0,\\ -i(2i)^{1/2}\theta(4|D|z) &D<0, \end{cases} \\ V(\theta(|D|z))\Big|_{\frac{1}{2}}\gamma_2&=\mathrm{sgn}(D)i\theta\left(|D|z-\frac{|D|}{4}\right),\\ V(\theta(|D|z))\Big|_{\frac{1}{2}}\gamma_3&=\begin{cases}(2i)^{1/2}\theta(4|D|z) & D>0,\\ -i(2i)^{1/2}(\theta(|D|z)-\theta(4|D|z)) & D<0, \end{cases} \end{align*} $$
taking the principal branch of every square root.
Proof Recall that for
$W_4:=[\begin {smallmatrix} 0 & -1 \\ 4 & 0 \end {smallmatrix}]$
, we have
$$ \begin{align*}\theta(z)|_{\frac{1}{2}}W_4=i^{-1/2}\theta(z),\end{align*} $$
see, e.g., [Reference Cohen and Strömberg4, Proposition 15.1.1]. Note that
$$ \begin{align*} V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 4|D|v & 1 \end{bmatrix}&=|D|^{-1/4}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix}4 & 1\\ 0 & 4 \end{bmatrix}\begin{bmatrix}1 & 0\\ 4|D|v & 1 \end{bmatrix}\\ &=|D|^{-1/4}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}|D|v+1 & |D|\\ 4v & 4 \end{bmatrix}\begin{bmatrix}4|D| & 0\\ 0 & 1\\ \end{bmatrix}. \end{align*} $$
In the following, we only give detailed proofs for
$v=0,1,2$
, and leave out details for
$v=3$
because it follows a similar argument to
$v=1$
.
-
(1)
$v=0$
:
$$\begin{align*}V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}=\theta\left(|D|z+\frac{|D|}{4}\right).\end{align*}$$
-
(2)
$v=1$
: We have
$$ \begin{align*} V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 4|D| & 1 \end{bmatrix}&=|D|^{-1/4}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}|D|+1 & |D|\\ 4 & 4 \end{bmatrix}\begin{bmatrix}4|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=|D|^{-1/4}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}-|D| & |D|+1\\ -4 & 4 \end{bmatrix} W_4 \begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}. \end{align*} $$
Since
$|D|$
is odd, we can choose
$x,y\in \mathbb {Z}$
such that
$-|D|x-4y=1$
. This gives us
$$ \begin{align*} \begin{bmatrix}-|D| & |D|+1\\ -4 & 4 \end{bmatrix} =\begin{bmatrix}-|D| & -y\\ -4 & x \end{bmatrix}\begin{bmatrix}1 & x-1\\ 0 & 4 \end{bmatrix}. \end{align*} $$
Note that
$\begin {bmatrix}-|D| & -y\\ -4 & x \end {bmatrix}\in \Gamma _0(4)$
, so we have
$$ \begin{align*} V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 4|D| & 1 \end{bmatrix}&=|D|^{-1/4}\left(\frac{-4}{x}\right)\varepsilon_{x}^{-1}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & x-1\\ 0 & 4 \end{bmatrix} W_4 \begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=2^{-1/2}|D|^{-1/4}\left(\frac{-4}{x}\right)\varepsilon_{x}^{-1}\theta\left(\frac{z+x-1}{4}\right)\Bigg|_{\frac{1}{2}} W_4 \begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}, \end{align*} $$
where
$$ \begin{align*}\varepsilon_x=\begin{cases} 1 &x\equiv 1\pmod4,\\ i &x\equiv 3\pmod 4. \end{cases}\end{align*} $$
Now, we have two cases since
$-|D|x-4y=1$
and the sign of D determines
$\varepsilon _x$
.-
(a) If
$D>0$
, then
$|D|\equiv 1\pmod 4$
, so
$x\equiv 3\pmod 4$
,
$\theta \left (\frac {z+x-1}{4}\right )=\theta \left (\frac {z}{4}+\frac {1}{2}\right )=2\theta (z)-\theta (\frac {z}{4})$
,
$\varepsilon _{x}=i$
, and
$\left (\frac {-4}{x}\right )=-1$
. So we have
$$ \begin{align*} V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 4|D| & 1 \end{bmatrix}&=2^{-1/2}|D|^{-1/4}\left(\frac{-4}{x}\right)\varepsilon_{x}^{-1}\theta\left(\frac{z+x-1}{4}\right)\!\Bigg|_{\frac{1}{2}}\! W_4\begin{bmatrix}|D| &\! 0\\ 0 &\! 1 \end{bmatrix}\\ &=i2^{-1/2}|D|^{-1/4}\left(2\theta(z)-\theta\left(\frac{z}{4}\right)\right)\Bigg|_{\frac{1}{2}} W_4 \begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}. \end{align*} $$
Explicitly computing these, we get
$$ \begin{align*} \theta(z)\Bigg|_{\frac{1}{2}} W_4\begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}&=i^{-1/2}|D|^{1/4}\theta(|D|z),\\ \theta\left(\frac{z}{4}\right)\Bigg|_{\frac{1}{2}} W_4\begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}&=2^{1/2}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 0 & 4 \end{bmatrix} W_4\begin{bmatrix}|D| & 0\\ 0 & 1\\ \end{bmatrix}\\ &=2^{1/2}\theta(z)\Bigg|_{\frac{1}{2}} W_4 \begin{bmatrix}4 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=2^{1/2}i^{-1/2}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}4 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=2i^{-1/2}|D|^{1/4}\theta(4|D|z). \end{align*} $$
So our expression simplifies to
$$ \begin{align*} i2^{-1/2}|D|^{-1/4}\left(2\theta(z)-\theta\left(\frac{z}{4}\right)\right)\Bigg|_{\frac{1}{2}} W_4\begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}=(2i)^{1/2}(\theta(|D|z)-\theta(4|D|z)). \end{align*} $$
-
(b) If
$D<0$
, then
$|D|\equiv 3\pmod 4$
,
$x\equiv 1\pmod 4$
,
$\theta \left (\frac {z+x-1}{4}\right )=\theta (\frac {z}{4}),\varepsilon _{x}=1,$
and
$\left (\frac {-4}{x}\right )=1$
. So we have
$$ \begin{align*} V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 4|D| & 1 \end{bmatrix}&=2^{-1/2}|D|^{-1/4}\left(\frac{-4}{x}\right)\varepsilon_{x}^{-1}\theta\left(\frac{z+x-1}{4}\right)\!\Bigg|_{\frac{1}{2}}\!\! W_4\begin{bmatrix}|D| &\! 0\\ 0 &\! 1 \end{bmatrix}\\ &=2^{-1/2}|D|^{-1/4}\theta\left(\frac{z}{4}\right)\Bigg|_{\frac{1}{2}} W_4\begin{bmatrix}|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=-i(2i)^{1/2}\theta(4|D|z). \end{align*} $$
-
-
(3)
$v=2$
: Since
$2|D|+1$
is coprime to
$8$
, we can find
$x,y\in \mathbb {Z}$
such that
$(2|D|+1)x+8y=1$
. Now that
$$ \begin{align*} V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 8|D| & 1 \end{bmatrix}&=|D|^{-1/4}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}2|D|+1 & |D|\\ 8 & 4 \end{bmatrix}\begin{bmatrix}4|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=|D|^{-1/4}\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}2|D|+1 & -y\\ 8 & x \end{bmatrix}\\&\quad\begin{bmatrix}1 & |D|x+4y\\ 0 & 4 \end{bmatrix}\begin{bmatrix}4|D| & 0\\ 0 & 1 \end{bmatrix}. \end{align*} $$
$\begin {bmatrix}2|D|+1 & -y\\ 8 & x \end {bmatrix}$
is in
$\Gamma _{0}(4)$
, we get As
$$ \begin{align*} V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 8|D| & 1 \end{bmatrix}&=|D|^{-1/4}\varepsilon_{x}^{-1}\left(\frac{8}{x}\right)\theta(z)\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & \frac{1-x}{2}\\ 0 & 4 \end{bmatrix}\begin{bmatrix}4|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=|D|^{-1/4}\varepsilon_{x}^{-1}\left(\frac{8}{x}\right)2^{-1/2}\theta\left(\frac{z}{4}+\frac{1-x}{8}\right)\Bigg|_{\frac{1}{2}}\begin{bmatrix}4|D| & 0\\ 0 & 1 \end{bmatrix}\\ &=\varepsilon_{x}^{-1}\left(\frac{8}{x}\right)\theta\left(|D|z+\frac{1-x}{8}\right). \end{align*} $$
$(2|D|+1)x+8y=1$
and the sign of D determines
$\varepsilon _x$
, we do casework again.
-
(a) If
$D>0$
, then
$|D|\equiv 1\pmod 4$
, which implies that
$3x\equiv 1\pmod 8$
,
$x\equiv 3\pmod 8$
,
$\varepsilon _x=i$
and
$\left (\hspace {-1pt}\frac {8}{x}\hspace {-1pt}\right )=\left (\hspace {-1pt}\frac {8}{3}\hspace {-1pt}\right )=-1$
. Note also that
$\theta (|D|z+\frac {1-x}{8})=\theta (|D|z-\frac {1}{4})$
. -
(b) If
$D<0$
, then
$|D|\equiv 3\pmod 4$
, which gives
$7x\equiv 1\pmod 8, x\equiv 7\pmod 8$
,
$\left (\hspace {-1pt}\frac {8}{x}\hspace {-1pt}\right )=\left (\hspace {-1pt}\frac {8}{7}\hspace {-1pt}\right )=1$
,
$\varepsilon _x=i$
and
$\theta (|D|z+\frac {1-x}{8})=\theta (|D|z-\frac {3}{4})$
.
Combining these two cases, we can write
$$\begin{align*}V(\theta(|D|z))\Bigg|_{\frac{1}{2}}\begin{bmatrix}1 & 0\\ 4|D|v & 1 \end{bmatrix}=\text{sgn}(D)i\theta\left(|D|z-\frac{|D|}{4}\right).\end{align*}$$
-
-
(4)
$v=3$
: The argument in this case is similar to that of
$v=1$
, and is omitted.
The above arguments complete the proof.
We also need the following two lemmas.
Lemma 3.7 We have that
$$\begin{align*}V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_1+V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_3 = \varepsilon_{|D|}^{-1}(2i)^{1/2}\theta(|D|z).\end{align*}$$
Proof It is a trivial consequence of Lemma 3.6.
Lemma 3.8 We have that
$$\begin{align*}V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_0+V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_2=(1+i\mathrm{sgn}(D))\theta(|D|z).\end{align*}$$
Proof By Lemma 3.6, we have
$$ \begin{align*}V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_0+V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_2=\theta\left(|D|z+\frac{|D|}{4}\right)+i\text{sgn}(D)\theta\left(|D|z-\frac{|D|}{4}\right).\end{align*} $$
Note that
$$\begin{align*}\theta\left(|D|z+\frac{|D|}{4}\right) = \sum\limits_{n\in\mathbb{Z}}e^{2\pi i \frac{n^2|D|}{4}}e^{2\pi i n^2|D|z}=\sum\limits_{n\in\mathbb{Z}}a(n)e^{2\pi i n^2|D|z},\end{align*}$$
where
$a(n)=i\text {sgn}(D)$
if n is odd and
$a(n)=1$
if n is even. On the other hand,
$$\begin{align*}\theta\left(|D|z-\frac{|D|}{4}\right) = \sum\limits_{n\in\mathbb{Z}}e^{2\pi i \frac{-n^2|D|}{4}}e^{2\pi i n^2|D|z}=\sum\limits_{n\in\mathbb{Z}}b(n)e^{2\pi i n^2|D|z},\end{align*}$$
where
$b(n)=-i\text {sgn}(D)$
if n is odd and
$b(n)=1$
if n is even. Hence,
$$ \begin{align*} \theta\left(|D|z+\frac{|D|}{4}\right)+\mathrm{sgn}(D)i\theta\left(|D|z-\frac{|D|}{4}\right)&= \sum\limits_{n\in\mathbb{Z}}(a(n)+i\text{sgn}(D)(b(n))e^{2\pi i n^2|D|z}\\ &= (1+i\text{sgn}(D))\theta(|D|z), \end{align*} $$
as desired.
Now, we are ready to prove the alternate formula (1.12) for
$\mathcal {G}_{D,k,e}$
promised at the beginning of this section.
Proposition 3.9 Let
$k\ge 4$
and
$e>0$
be integers such that
$k+2e=\ell $
and let D be an odd fundamental discriminant such that
$(-1)^\ell D>0$
. Then,
$$ \begin{align*} \mathcal{G}_{D, k, e}(z)=\operatorname{\mathrm{Tr}}^{4D}_4 [G_{k,D}(4z), \theta(|D|z)]_e. \end{align*} $$
Proof We closely follow [Reference Kohnen and Zagier9, p. 195], where a similar result is implicit in the proof of formulas [Reference Kohnen and Zagier9, (6) and (7)]. Write
$h=[G_{k,4D}(z),\theta (|D|z)]_e$
. By Lemma 3.2, we get
$$ \begin{align} \mathcal{G}_{D,k,e}&=\frac32\left(1-\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)2^{-k}\right)^{-1}\operatorname{\mathrm{pr}}^+\operatorname{\mathrm{Tr}}_4^{4D}(h) \nonumber\\ &=\frac32\left(1-\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)2^{-k}\right)^{-1}\left(\frac{1-(-1)^\ell i}{6}\operatorname{\mathrm{Tr}}_4^{16}V(\operatorname{\mathrm{Tr}}^{4D}_{4}(h)) + \frac{1}{3}\operatorname{\mathrm{Tr}}^{4D}_{4}(h)\right) \nonumber\\ &=\frac32\left(1-\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)2^{-k}\right)^{-1}\operatorname{\mathrm{Tr}}_4^{4D}\left( \frac{1-(-1)^\ell i}{6}\operatorname{\mathrm{Tr}}^{16D}_{4D}(V(h)) + \frac{1}{3}h\right) \nonumber\\ &=\frac32\left(1-\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)2^{-k}\right)^{-1}\operatorname{\mathrm{Tr}}_4^{4D}g_D, \end{align} $$
with
$$ \begin{align*}g_D = \frac{1-(-1)^k i}{6}\operatorname{\mathrm{Tr}}^{16D}_{4D}(V(h)) + \frac{1}{3}h.\end{align*} $$
Note that
$k\equiv \ell $
mod 2, so we can substitute in
$(-1)^k$
for
$(-1)^\ell $
above. We now compute
$g_D$
. The matrices
$\gamma _v = [\begin {smallmatrix}1 & 0\\4|D|v & 1\end {smallmatrix}]$
, where
$v=0,1,2,3$
, form a set of coset representatives for
$\Gamma _0(16|D|)\backslash \Gamma _0(4|D|)$
[Reference Kohnen and Zagier9, p. 195]. Then, we have
$$ \begin{align*} \operatorname{\mathrm{Tr}}^{16D}_{4D}(V(h))&=\operatorname{\mathrm{Tr}}^{16D}_{4D}(V[G_{k,4D}(z),\theta(|D|z)]_e)\\ &=\operatorname{\mathrm{Tr}}^{16D}_{4D}\left[VG_{k,4D}(z),V\theta(|D|z)\right]_e\\ &=\operatorname{\mathrm{Tr}}^{16D}_{4D}\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)V(G_{k,D}(2z)),V\theta(|D|z)\right]_e\\ &=\sum\limits_{\gamma_v}\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)V(G_{k,D}(2z)),V\theta(|D|z)\right]_e\Bigg|_{k+\frac{1}{2}+2e}\gamma_v. \end{align*} $$
Since
$\gamma _v\in \Gamma _0(4|D|)$
,
$G_{k,D}(4z)\Big |_{k}\gamma _v=G_{k,D}(4z)$
. By Lemma 3.5, we get
$$ \begin{align*} \operatorname{\mathrm{Tr}}^{16D}_{4D}(V(h))&=\sum\limits_{\gamma_v}\left[G_{k,D}(4z)\Big|_{k}\gamma_v-2^{-k}\left(\frac{D}{2}\right)V(G_{k,D}(2z))\Big|_{k}\gamma_v,V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_v\right]_e\\ &=\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)G_{k,D}\left(2z+\frac{1}{2}\right),V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_0+V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_2\right]_e\\ &\hspace{20pt}+\left[G_{k,D}(4z)- G_{k,D}(8z),V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_1+V\theta(|D|z)\Big|_{\frac{1}{2}}\gamma_3\right]_e. \end{align*} $$
Using Lemmas 3.7 and 3.8 and noting that sgn
$(D)=(-1)^k$
by our assumption, we can simplify this to
$$ \begin{align*} \operatorname{\mathrm{Tr}}^{16D}_{4D}(V(h)) &=\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)G_{k,D}\left(2z+\frac{1}{2}\right),(1+i(-1)^k)\theta(|D|z)\right]_e\\ &\hspace{20pt}+\left[G_{k,D}(4z)- G_{k,D}(8z),\varepsilon_{|D|}^{-1}(2i)^{1/2}\theta(|D|z)\right]_e. \end{align*} $$
Now, we can finally compute the projection
$$ \begin{align*} g_D(z)&= \frac{1-i(-1)^k }{6}\operatorname{\mathrm{Tr}}^{16D}_{4D}(V(h(z))) + \frac{1}{3}h(z)\\&= \frac{1-i(-1)^k}{6}\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)G_{k,D}\left(2z+\frac{1}{2}\right),(1+i(-1)^k)\theta(|D|z)\right]_e\\&\hspace{20pt}+ \frac{1-i(-1)^k}{6}\left[G_{k,D}(4z)- G_{k,D}(8z),\varepsilon_{|D|}^{-1}(2i)^{1/2}\theta(|D|z)\right]_e\\&\hspace{20pt}+ \frac{1}{3}\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)G_{k,D}(2z),\theta(|D|z)\right]_e\\&=\frac{1}{3}\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)G_{k,D}\left(2z+\frac{1}{2}\right),\theta(|D|z)\right]_e\\&\hspace{20pt}+\frac{1}{3}\left[G_{k,D}(4z)- G_{k,D}(8z),\theta(|D|z)\right]_e\\&\hspace{20pt}+ \frac{1}{3}\left[G_{k,D}(4z)-2^{-k}\left(\frac{D}{2}\right)G_{k,D}(2z),\theta(|D|z)\right]_e\\&=\frac{1}{3}\left[\!G_{k,D}(4z) -G_{k,D}(8z)-2^{-k}\left(\frac{D}{2}\right) \left(\!G_{k,D}\left(2z+\frac{1}{2}\right) + G_{k,D}(2z)\!\right), \theta(|D|z)\!\right]_e\\&\hspace{20pt}+\frac{2}{3}[G_{k,D}(4z),\theta(|D|z)]_e. \end{align*} $$
Using Lemma 3.4, we can simplify the first term to get
$$ \begin{align} g_D(z)&=\frac{1}{3}\left[-\left(\frac{D}{2}\right)2^{-k+1}G_{k,D}(4z), \theta(|D|z)\right]_e+\frac{2}{3}[G_{k,D}(4z),\theta(|D|z)]_e \nonumber\\ &=\frac{2}{3}\left(1-\left(\frac{D}{2}\right)2^{-k}\right)\left[G_{k,D}(4z), \theta(|D|z)\right]_e. \end{align} $$
4 Eisenstein Series
In this section, we define various Eisenstein series and show that
$G_{k,4D}(z)$
(1.5) is an Eisenstein series for the cusp at infinity of level
$4|D|$
. We recall the theory of Eisenstein series as developed in Miyake [Reference Miyake15, Section 7]. Let
$\chi $
and
$\psi $
be Dirichlet characters mod L and mod M, respectively. For
$k\geq 3$
, we put
$$ \begin{align*} E_{k}(z;\chi,\psi)=\sideset{}{'}\sum\limits_{m,n\in\mathbb{Z}}\chi(m)\psi(n)(mz+n)^{-k}. \end{align*} $$
Here,
$\sum $
is the summation over all pairs of integers
$(m,n)$
except
$(0,0)$
. In particular,
$E_{k}(Mz;\chi ,\psi )$
is a modular form in
$M_k(LM,\chi \overline {\psi })$
(see [Reference Miyake15, pp. 269–271] for details).
Lemma 4.1 [Reference Miyake15, Theorem 7.1.3]
Assume
$k\geq 3$
. Let
$\chi $
and
$\psi $
be Dirichlet characters mod L and mod M, respectively, satisfying
$\chi (-1)\psi (-1)=(-1)^k$
. Let
$m_{\psi }$
be the conductor of
$\psi $
, and
$\psi ^0$
be the primitive character associated with
$\psi $
. Then,
$$ \begin{align*} E_{k}(z;\chi,\psi)=C+A\sum_{n=1}^{\infty}a(n)e^{2\pi inz/M}, \end{align*} $$
where
$$ \begin{align*} A&=2(-2\pi i)^kG(\psi^0)/M^{k}(k-1)!,\\ C&=\begin{cases}2L_M(k,\psi) &\chi:\mathrm{the~principal~character},\\ 0 &\mathrm{otherwise}, \end{cases}\\ a(n)&=\sum_{0<c\mid n}\chi(n/c)c^{k-1}\sum_{0<d\mid(l,c)}d\mu(l/d)\psi^0(l/d)\overline{\psi^0}(c/d). \end{align*} $$
Here,
$l=M/m_{\psi }$
,
$\mu $
is the Möbius function,
$L_M(k,\psi )=\sum _{n=1}^{\infty }\psi (n)n^{-k}$
is the Dirichlet series, and
$G(\psi ^0)$
is the Gauss sum of
$\psi ^0$
.
For a fundamental discriminant D, we write
$\chi _D(\cdot )=\left (\hspace {-1pt}\frac {D}{\cdot }\hspace {-1pt}\right )$
and
$L_D(k)=\sum _{n=1}^{\infty }\chi _D(n)n^{-k}$
.
Example 4.2 Let D be a fundamental discriminant and
$\mathbf {1}$
be the principal character. Then,
$$ \begin{align*} E_{k}(z;\mathbf{1},\chi_D)=2L_{D}(k)+\frac{2(-2\pi i)^kG(\chi_D)}{(k-1)!|D|^k}\sum_{n=1}^{\infty}\left(\sum_{d\mid n}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)d^{k-1}\right)q^{2\pi i n z/|D|}. \end{align*} $$
Example 4.3 If
$D=D_1D_2$
is a product of relatively prime fundamental discriminants then
$$ \begin{align*} E_{k}(z;\chi_{D_2},\chi_{D_1}):=C+\frac{2(-2\pi i)^kG(\chi_{D_1})}{|D_1|^k(k-1)!}\sum_{n=1}^{\infty}\left(\sum_{\substack{d_1,d_2>0\\ d_1d_2=n}}\left(\hspace{-1pt}\frac{D_1}{d_1}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_2}{d_2}\hspace{-1pt}\right)d_1^{k-1}\right)e^{2\pi inz/|D_1|}, \end{align*} $$
where C is zero unless
$D_2=1$
.
We shall compare our Eisenstein series
$G_{k,D}(z)$
(1.4) and
$G_{k,D_1,D_2}(z)$
, defined below in (4.2) [Reference Kohnen and Zagier9, p. 193] with the ones above given in Miyake [Reference Miyake15]. Comparing the Fourier coefficients of
$G_{k,D}(z)$
and
$E_{k}(z;\mathbf {1},\chi _D)$
gives
$$ \begin{align} G_{k,D}(z)=\frac{(k-1)!|D|^k}{2(-2\pi i)^kG(\chi_D)}E_{k}(|D|z,\mathbf{1},\chi_{D})\in M_k(|D|,\chi_D). \end{align} $$
Recall that [Reference Kohnen and Zagier9, p. 193] for
$D_1,D_2$
relatively prime fundamental discriminants with
$(-1)^kD_1D_2>0$
:
$$ \begin{align} G_{k,D_1,D_2}(z)&=\sum_{n\geq0}\sigma_{k-1,D_1,D_2}(n)q^n, \\ \sigma_{k-1,D_1,D_2}(n)&=\begin{cases} -L_{D_1}(1-k)L_{D_2}(0) &n=0,\\\sum\limits_{\substack{d_1,d_2>0\\ d_1d_2=n}}\left(\hspace{-1pt}\frac{D_1}{d_1}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_2}{d_2}\hspace{-1pt}\right)d_1^{k-1} &n>0,\nonumber \end{cases} \end{align} $$
where the constant term is zero unless
$D_2=1$
. Hence, by comparing the Fourier coefficients of
$G_{k,D_1,D_2}(z)$
and
$E_{k}(z;\chi _{D_2},\chi _{D_1})$
, we get
$$ \begin{align} G_{k,D_1,D_2}(z)=\frac{|D_1|^k(k-1)!}{2(-2\pi i)^k G(\chi_{D_1})}E_{k}(|D_1|z;\chi_{D_2},\chi_{D_1})\in M_k(|D_1D_2|,\chi_{D_1}\chi_{D_2}). \end{align} $$
The following expression of
$G_{k,D_1,D_2}(z)$
is useful for Lemma 6.1.
Lemma 4.4 Let
$k\ge 3$
and
$D=D_1D_2$
be a product of coprime fundamental discriminants. Then,
$$ \begin{align} G_{k,D_1,D_2}(z)=\frac{|D_1|^k(k-1)!}{2(-2\pi i)^k G(\chi_{D_1})}\chi_{D_2}(|D_1|)\sideset{}{'}\sum_{\substack{m,n\in\mathbb{Z}\\ D_1\mid m}}\frac{\chi_{D_2}(m)\chi_{D_1}(n)}{(mz+n)^k}. \end{align} $$
Proof Note that
$$ \begin{align*} E_{k}(|D_1|z;\chi_{D_2},\chi_{D_1})&=\sideset{}{'}\sum\limits_{m,n\in\mathbb{Z}}\chi_{D_2}(m)\chi_{D_1}(n)(m|D_1|z+n)^{-k}\\ &=\chi_{D_2}(|D_1|)\sideset{}{'}\sum_{\substack{\ell,n\in\mathbb{Z}\\D_1\mid\ell}}\chi_{D_2}(\ell)\chi_{D_1}(n)(\ell z+n)^{-k}. \end{align*} $$
Thus, the result follows from (4.3).
Let
$k\ge 3$
and
$\chi $
be a Dirichlet character mod N. We define the Eisenstein series for the cusp at infinity [Reference Miyake15, p. 272] as
$$ \begin{align*} E_{k,N}^{\ast}(z;\chi)=\sum_{[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]\in\Gamma_{\infty}\backslash\Gamma_0(N)}\frac{\chi(d)}{(cz+d)^k}, \end{align*} $$
where
$\Gamma _{\infty }=\{\pm [\begin {smallmatrix} 1 & n \\ 0 &1 \end {smallmatrix}]:n\in \mathbb {Z}\}$
.
Now, we are ready to prove that
$G_{k,4D}$
is an Eisenstein series for the cusp at infinity of level
$4|D|$
.
Lemma 4.5 [Reference Miyake15, (7.1.30)]
Let
$\mathbf {1}$
denote the principal Dirichlet character. Then,
$$ \begin{align*} 2L_{N}(k,\chi)E_{k,N}^{\ast}(z;\chi)=E_{k}(Nz;\mathbf{1},\chi). \end{align*} $$
From (4.1) and Lemma 4.5, we know that
$G_{k,D}(z)$
is an Eisenstein series at infinity. We have
$$ \begin{align} G_{k,D}(z)=\frac{L_D(1-k)}{2}E_{k,|D|}^{\ast}(z;\chi_D). \end{align} $$
Note also that (4.1) and the proof of Lemma 4.5 imply that
$$ \begin{align} G_{k,D}(z)=\frac{(k-1)!|D|^k}{2(-2\pi i)^kG(\chi_D)}\sideset{}{'}\sum_{\substack{c,d\in{\mathbb{Z}}\\D\mid c}}\frac{\chi_D(d)}{(cz+d)^k}. \end{align} $$
In fact, equation (4.6) will be more convenient for us to compute the Fourier expansion of
$G_{k,D}(z)$
at different cusps. We need the following lemma (see also [Reference Gross and Zagier5, p. 271].
Lemma 4.6 Let
$L_D^{(4)}(k)=\sum \limits _{\substack {(n,4)=1\\n\geq 1}}\chi _D(n)n^{-k}$
. Then,
$$ \begin{align*} E_{k,4|D|}^{\ast}(z;\chi_D)=\frac{L_D(k)}{L_{D}^{(4)}(k)}\left(E_{k,|D|}^{\ast}(4z;\chi_D)-2^{-k}\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)E_{k,|D|}^{\ast}(2z;\chi_D)\right). \end{align*} $$
Proof Observe that
$$ \begin{align*} 2L_D^{(4)}(k)E_{k,4|D|}^{\ast}(z; \chi_D) &=2\sum_{\substack{n\ge 1\\(4,n)=1}}\frac{\chi_D(n)}{n^k}\left(\frac{1}{2}\sum_{\substack{(c,d)=1 \\ 4D\mid c}}\frac{\chi_D(d)}{(cz+d)^{k}}\right) \\[4pt] &=\sum_{\substack{n\ge 1\\(4,n)=1}} \sum_{\substack{(c,d)=1\\4D\mid c}}\frac{\chi_D(nd)}{(ncz+nd)^k} \\[4pt] &=\sum_{\substack{(d',4D)=1\\4D\mid c'}}\frac{\chi_D(d')}{(c'z+d')^k}, \end{align*} $$
where
$nc=c'$
and
$nd=d'$
. Note that we can replace
$(d',4D)=1$
by
$(d',4)$
since
$\chi _D(d')=0$
otherwise. It follows that
$$ \begin{align*} 2L_D^{(4)}(k)E_{k,4|D|}^{\ast}(z; \chi_D) &=\sum_{\substack{(d, 4)=1\\4D\mid c}} \frac{\chi_D(d)}{(cz+d)^k}\\[4pt] &=\sideset{}{'}\sum_{\substack{c,d\in\mathbb{Z}\\4D|c}} \left(\sum_{e|(d,4),e>0}\mu(e)\right)\frac{\chi_D(d)}{(cz+d)^k}\\[4pt] &=\sum_{e\mid4,e>0}\mu(e)\sideset{}{'}\sum_{\substack{c,d\in \mathbb{Z}\\4D\mid c, e\mid d}}\frac{\chi_D(d)}{(cz+d)^k}, \end{align*} $$
where we used
$\sum \limits _{e\mid (d,4)}\mu (e)=0$
for
$(d,4)>1$
in the second equality. Substituting
$d=ey$
and
$c=4x$
,
$$ \begin{align*} 2L_D^{(4)}(k)E_{k,4|D|}^{\ast}(z; \chi_D)&=\sum_{e\mid 4,e>0}\mu(e) \sideset{}{'}\sum_{\substack{x,y\in\mathbb{Z}\\D\mid x}}\frac{\chi_D(ey)}{(4xz+ey)^k}\\&=\sum_{e\mid 4,e>0}\mu(e)e^{-k}\chi_D(e) \sideset{}{'}\sum_{\substack{x,y\in\mathbb{Z}\\D|x}}\frac{\chi_D(y)}{(x4z/e+y)^k} \\ &=\sum_{e\mid 4,e>0}\mu(e)e^{-k}\chi_D(e)2L_D(k)E_{k,D}^{\ast}(4z/e,\chi_D)\\&=2L_D(k)\left(E_{k,|D|}^{\ast}(4z;\chi_D)-2^{-k}\left(\hspace{-1pt}\frac{D}{2}\hspace{-1pt}\right)E_{k,|D|}^{\ast}(2z;\chi_D)\right), \end{align*} $$
where the second to last equality is from the proof of Lemma 4.5.
From Lemma 4.6 and (4.5), we know that
$G_{k,4D}(z)$
is an Eisenstein series for the cusp at infinity in
$M_{k}(4|D|,\chi _D)$
. We have
$$ \begin{align} G_{k,4D}(z)=\frac{L_{D}(1-k)}{2}\left(1-2^{-k}\left(\frac{D}{2}\right)\right)E_{k,4D}^{\ast}(z;\chi_D). \end{align} $$
5 The Rankin–Selberg convolution
The purpose of this section is to prove Propositions 5.6 and 5.7. For two elements f and g of
$M_k(N)$
such that
$fg$
is a cusp form, the Petersson inner product is given by
$$ \begin{align} \langle f,g\rangle_{\Gamma_0(N)}=\int_{\Gamma_{0}(N)\backslash\mathbb{H}}f(z)\overline{g(z)}\operatorname{\mathrm{Im}}(z)^kd\mu, \end{align} $$
where
$z=x+iy$
and
$d\mu =dxdy/y^2$
. We use
$\langle \cdot ,\cdot \rangle $
to denote
$\langle \cdot ,\cdot \rangle _{\Gamma _0(N)}$
if the level is clear from the context. For
$f(z)=\sum _{n\geq 1}a_f(n)q^n\in S_{k}(N,\chi )$
, we put
$f_{\rho }(z):=\sum _{n\geq 1}\overline {a_f(n)}q^n$
. Note that
$f_{\rho }(z)=f(z)$
if f is a newform and
$\chi $
is trivial.
We now review the classical result on the Rankin–Selberg convolution, which was reformulated and generalized in Zagier [Reference Zagier25], keeping in mind the difference between our definition of the Rankin–Cohen bracket and the one used therein.
Lemma 5.1 [Reference Zagier25, Proposition 6]
Let
$k_1$
and
$k_2$
be real numbers with
$k_2\geq k_1+2>2$
. Let
$f(z)=\sum _{n=1}^{\infty }a(n)q^n$
and
$g(z)=\sum _{n=0}^{\infty }b(n)q^n$
be modular forms in
$S_k(N,\chi )$
and
$M_{k_1}(N,\chi _1)$
, where
$k=k_1+k_2+2e, e\geq 0,$
and
$\chi =\chi _1\chi _2$
. Then,
$$ \begin{align*} \langle f,[g,E^*_{k_2,N}(\cdot;\chi_2)]_e\rangle=\frac{(-1)^e}{e!}\frac{\Gamma(k-1)\Gamma(k_2+e)}{(4\pi)^{k-1}\Gamma(k_2)}\sum_{n=1}^{\infty}\frac{a(n)\overline{b(n)}}{n^{k_1+k_2+e-1}}. \end{align*} $$
To obtain Proposition 5.6, we need to deal with the case
$k_1=k_2$
, which can be done by following Shimura [Reference Shimura19] and Lanphier’s work [Reference Lanphier11]. For
$f(z)=\sum _{n=1}^{\infty }a(n)q^n\in S_{k}(N,\chi )$
and
$g(z)=\sum _{n=0}^{\infty }b(n)q^n\in M_{\ell }(N,\psi )$
, we put
$$\begin{align*}D(s,f,g)=\sum_{n=1}^{\infty}a(n)b(n)n^{-s},\quad \operatorname{\mathrm{Re}}(s)\gg0.\end{align*}$$
We are particularly interested in the following case.
Lemma 5.2 Let
$f=\sum _{n=1}^\infty a(n)q^n\in S_{2\ell }(1)$
be a normalized eigenform with
$\ell =k+2e, e>0,$
and
$k\geq 4$
integers, and let D be an odd fundamental discriminant. Then,
$$ \begin{align*} D(s,f,G_{k,D})=\frac{L(f,s)L(f,D,s-k+1))}{L_{D}(2s-3k-4e+2)},\quad \operatorname{\mathrm{Re}}(s)\gg0. \end{align*} $$
Proof Note that for
$\operatorname {\mathrm {Re}}(s)\gg 0$
, we have
$$ \begin{align*}D(s,f,G_{k,D})=\sum_{n=1}^{\infty}\frac{\sigma_{k-1,\mathbf{1},\chi_D}(n)a(n)}{n^s}, \end{align*} $$
where
$\sigma _{k-1,\mathbf {1},\chi _D}(n)=\sum _{d\mid n}\chi _{D}(d)d^{k-1}$
. A standard computation (see [Reference Westerholt-Raum22, Proposition 4.1]) gives
$$ \begin{align*} \sum_{n=1}^{\infty}\frac{\sigma_{k-1,\mathbf{1},\chi_D}(n)a(n)}{n^s}=\frac{L(f,s)L(f,D,s-(k-1))}{L_{D}(2s-(k-1)+1-(2k+4e))}, \end{align*} $$
as desired.
From Shimura [Reference Shimura19, pp. 786–789],
$D(s,f,G_{k,D})$
has a meromorphic continuation to the whole complex plane and
$D(s,f,G_{k,D})$
is holomorphic at
$s=2k+2e-1$
(see [Reference Shimura19, p. 789].
The Maass–Shimura operators [Reference Shimura19, p. 788, (2.8)] are defined by
$$ \begin{align*} \delta_{\lambda}&=\frac{1}{2\pi i}\left(\frac{\lambda}{2i y}+\frac{\partial}{\partial z}\right),\quad0<\lambda\in\mathbb{Z},\\ \delta_{\lambda}^{(r)}&=\delta_{\lambda+2r-2}\cdots\delta_{\lambda+2}\delta_{\lambda},\quad 0\leq r\in\mathbb{Z}, \end{align*} $$
where we understand that
$\delta ^{(0)}_{\lambda }$
is the identity operator. A relation between Maass–Shimura operators and the Rankin–Cohen bracket is given by
$$ \begin{align} \left(\delta_k^{(n)}f(z)\right) g(z)=\sum_{j=0}^n\frac{(-1)^j\binom{n}{j}\binom{k+n-1}{n-j}}{\binom{k+\ell+2j-2}{j}\binom{k+\ell+n+j-1}{n-j}}\delta^{(n-j)}_{k+\ell+2j}[f,g]_j(z), \end{align} $$
where
$f\in M_{k}(\Gamma )$
and
$g\in M_{\ell }(\Gamma )$
for any congruence subgroup
$\Gamma $
(see [Reference Lanphier11, Theorem 1].
We recall the following two results.
Lemma 5.3 [Reference Shimura19, Lemma 6]
Suppose
$f\in S_{k}(N,\chi ), g\in M_{l}(N,\overline {\chi }),$
and
$k=l+2r$
with a positive integer
$r.$
Then,
$\langle \delta ^{(r)}g,f_{\rho }\rangle =0$
.
Lemma 5.4 [Reference Shimura19, Theorem 2]
Suppose
$f\in S_{2\ell }(|D|)$
with
$\ell =k+2e, e>0,$
and
$k\geq 4$
, and D is a fundamental discriminant. Then,
$$ \begin{align*}D(2k+4e-1-2e,f,G_{k,D})=c\pi^{2k+4e-1}\langle G_{k,D}\delta_{k}^{(2e)}E_{k,|D|}^{\ast}(z;\chi_D),f_{\rho} \rangle,\end{align*} $$
where
$\langle \cdot ,\cdot \rangle $
denotes the non-normalized Petersson inner product defined in (5.1) and
$$ \begin{align*} c&=\frac{\Gamma(2k+4e-k-2(2e))}{\Gamma(2k+4e-1-2e)\Gamma(2k+4e-k-2e)}(-1)^{2e}4^{2k+4e-1}. \\ &=\frac{\Gamma(k)}{\Gamma(2k-1+2e)\Gamma(k+2e)}4^{2k+4e-1} \end{align*} $$
We apply these two results in our situation to obtain the following.
Proposition 5.5 Let
$f\in S_{2\ell }(1)$
be a normalized eigenform with
$\ell =k+2e, e>0,$
and
$k\geq 4$
. Then,
$$ \begin{align*} \langle[G_{k,D},G_{k,D}]_{2e},f\rangle_{\Gamma_0(|D|)}&=\frac{1}{2}\frac{\Gamma(2k+4e-1)\Gamma(k+2e)}{(2e)!(4\pi)^{2k+4e-1}\Gamma(k)}\frac{L_{D}(1-k)}{L_D(k)}\\&\quad L(f,2k+2e-1)L(f,D,k+2e). \end{align*} $$
Proof Note that
$f_{\rho }=f$
since f is a normalized eigenform. Lemma 5.4 gives
$$ \begin{align*} \langle G_{k,D}\delta_{k}^{(2e)}E_{k,|D|}^{\ast}(z;\chi_D),f\rangle_{\Gamma_0(|D|)}= \frac{\Gamma(2k+2e-1)\Gamma(k+2e)}{(4\pi)^{2k+4e-1}\Gamma(k)}D(2k+2e-1,f,G_{k,D}). \end{align*} $$
$$ \begin{align*}\langle G_{k,D}\delta_{k}^{(2e)}E_{k,|D|}^{\ast}(z;\chi_D),f\rangle_{\Gamma_0(|D|)}=\frac{1}{\binom{2k+4e-2}{2e}}\langle [E_{k,|D|}^{\ast}(z;\chi_D),G_{k,D}]_{2e},f\rangle_{\Gamma_0(|D|)}, \end{align*} $$
which implies that
$$ \begin{align*} \langle [E_{k,|D|}^{\ast}(z;\chi_D),G_{k,D}]_{2e},f\rangle_{\Gamma_0(|D|)}& = \frac{\binom{2k+4e-2}{2e}\Gamma(2k+2e-1)\Gamma(k+2e)}{(4\pi)^{2k+4e-1}\Gamma(k)}\\&\quad D(2k+2e-1,f,G_{k,D}). \end{align*} $$
Since
$G_{k,D}(z)=\frac {L_D(1-k)}{2}E_{k,|D|}^{\ast }(z;\chi _D)$
(4.5) and by Lemma 5.2, we have
$$ \begin{align*} &\langle [G_{k,D},G_{k,D}]_{2e},f\rangle_{\Gamma_0(|D|)}=\frac{L_D(1-k)}{2}\frac{\binom{2k+4e-2}{2e}\Gamma(2k+2e-1)\Gamma(k+2e)}{(4\pi)^{2k+4e-1}\Gamma(k)}\\&\qquad\qquad\qquad\qquad\quad\qquad D(2k+2e-1,f,G_{k,D})\\& \quad =\frac{1}{2}\frac{\Gamma(2k+4e-1)\Gamma(k+2e)}{(2e)!(4\pi)^{2k+4e-1}\Gamma(k)}\frac{L_{D}(1-k)}{L_D(k)} L(f,2k+2e-1)L(f,D,k+2e), \end{align*} $$
as desired.
Now, we prove Propositions 5.6 and 5.7, which generalize [Reference Kohnen and Zagier9, Proposition 1] and [Reference Kohnen and Zagier9, Proposition 2], respectively.
Proposition 5.6 Let
$f=\sum _{n=1}^{\infty } a(n)q^n$
be a normalized eigenform in
$S_{2\ell }(1)$
with
$\ell =k+2e, e>0,$
and
$k\geq 4$
, and let D be an odd fundamental discriminant with
${(-1)^\ell D>0}$
. Then,
$$ \begin{align*} \langle \mathcal{F}_{D,k,e},f\rangle_=\frac{1}{2}\frac{\Gamma(2k+4e-1)\Gamma(k+2e)}{(2e)!(4\pi)^{2k+4e-1}\Gamma(k)}\frac{L_{D}(1-k)}{L_D(k)}L(f,2k+2e-1)L(f,D,k+2e). \end{align*} $$
Proof Recall that (1.11)
$$ \begin{align*} \mathcal{F}_{D,k,e}(z)=\operatorname{\mathrm{Tr}}^D_1[G_{k,D}(z),G_{k,D}(z)]_{2e}. \end{align*} $$
As
$\langle f,g\rangle _{\Gamma _0(M)}=\langle f,\operatorname {\mathrm {Tr}}_N^M g\rangle _{\Gamma _0(N)}$
for
$N\mid M$
, for
$f\in S_k(N),g\in M_{k}(M)$
(see [Reference Gross and Zagier5, p. 271]), we get
$$ \begin{align*} \langle\mathcal{F}_{D,k,e},f\rangle = \langle[G_{k,D}(z),G_{k,D}(z)]_{2e},f\rangle_{\Gamma_0(|D|)}.\end{align*} $$
Then, the result follows from Proposition 5.5.
Proposition 5.7 Let
$g=\sum c_g(n)q^n\in S^{+}_{\ell +1/2}(4)$
be a Hecke eigenform and
$f\in S_{2\ell }(1)$
be the normalized Hecke eigenform corresponding to it by the Shimura correspondence, where
$\ell =k+2e, e>0,$
and
$k\geq 4$
. Let D be an odd fundamental discriminant with
$(-1)^\ell D>0$
. Then,
$$ \begin{align*} \langle g,\mathcal{G}_{D,k,e}\rangle=\frac{3}{2}\frac{\Gamma(k+2e-\frac{1}{2})\Gamma(k+e)}{e!(4\pi)^{k+2e-1/2}\Gamma(k)}\frac{L_D(1-k)}{L_D(k)}|D|^{-k-e+1/2}L(f,2k+2e-1)c_g(|D|), \end{align*} $$
where the Petersson inner product is
$\langle g,\mathcal {G}_{D,k,e}\rangle :=\int _{\Gamma _0(4)\backslash \mathbb {H}}g(z)\overline {\mathcal {G}_{D,k,e}(z)}\operatorname {\mathrm {Im}}(z)^{k+2e+1/2}d\mu $
.
Proof Recall that
$\mathcal {G}_{D, k, e}$
is given in (1.12):
$$ \begin{align*} \mathcal{G}_{D, k, e}(z)=\frac{3}{2}\left(1-2^{-k}\left(\frac{D}{2}\right)\right)^{-1}\operatorname{\mathrm{pr}}^{+}\operatorname{\mathrm{Tr}}^{4D}_{4}[G_{k,4D}(z),\theta(|D|z)]_e. \end{align*} $$
Since
$\operatorname {\mathrm {pr}}^{+}$
(1.6) is the projection from
$M_{\ell +1/2}(4)$
to
$M^{+}_{\ell +1/2}(4)$
, we have
$$ \begin{align*} \langle g,\mathcal{G}_{D,k,e}\rangle&=\frac{3}{2}\left(1-2^{-k}\left(\frac{D}{2}\right)\right)^{-1}\langle \operatorname{\mathrm{pr}}^{+}g,\operatorname{\mathrm{Tr}}^{4D}_{4}([G_{k,4D}(z),\theta(|D|z)]_e\rangle\\&= \frac{3}{2}\left(1-2^{-k}\left(\frac{D}{2}\right)\right)^{-1}\langle g,\operatorname{\mathrm{Tr}}^{4D}_{4}([G_{k,4D}(z),\theta(|D|z)]_e\rangle\\& = \frac{3}{2}\left(1-2^{-k}\left(\frac{D}{2}\right)\right)^{-1}\langle g,([G_{k,4D}(z),\theta(|D|z)]_e\rangle_{\Gamma_0(4|D|)} \\&=\frac{3}{4}L_{D}(1-k)\langle g(z),[E_{k,4D}^{\ast}(z;\chi_D),\theta(|D|z)]_e\rangle_{\Gamma_0(4|D|)}\\&=\frac{3(-1)^e}{4}L_{D}(1-k)\langle g(z),[\theta(|D|z),E_{k,4D}^{\ast}(z;\chi_D)]_e\rangle_{\Gamma_0(4|D|)},\end{align*} $$
where we used (4.7) in the second to last equality. Now, Lemma 5.1 gives
$$ \begin{align*} \langle g,\mathcal{G}_{D,k,e}\rangle & =\frac{3(-1)^e}{4}L_{D}(1-k)\frac{(-1)^e}{e!}\frac{\Gamma(k+2e-\frac{1}{2})\Gamma(k+e)}{(4\pi)^{k+2e-1/2}\Gamma(k)}\sum_{n=1}^{\infty}\frac{2c_g(n^2|D|)}{(|D|n^2)^{k+e+1/2-1}}\\ &=\frac{3}{2}\frac{\Gamma(k+2e-\frac{1}{2})\Gamma(k+e)}{e!(4\pi)^{k+2e-1/2}\Gamma(k)}L_D(1-k)|D|^{-(k+e-1/2)}\sum_{n=1}^{\infty}\frac{c_g(n^2|D|)}{n^{2k+2e-1}}. \end{align*} $$
By [Reference Kohnen10, Theorem 1(ii)], we get
$$ \begin{align*} L_{D}(s-(k+2e)+1)\sum_{n=1}^{\infty}\frac{c_g(n^2|D|)}{n^{2k+2e-1}}=c_g(|D|)L(f,s), \end{align*} $$
which implies that
$$ \begin{align*} \langle g,\mathcal{G}_{D,k,e}\rangle=\frac{3}{2}\frac{\Gamma(k+2e-\frac{1}{2})\Gamma(k+e)}{e!(4\pi)^{k+2e-1/2}\Gamma(k)}\frac{L_D(1-k)}{L_D(k)}|D|^{-k-e+1/2}L(f,2k+2e-1)c_g(|D|), \end{align*} $$
as desired.
6 Fourier expansions
In this section, we compute the Fourier coefficients needed for the proof of Theorem 1.1. It is convenient to have explicit formulas for
$G_{k,D}$
and
$\theta $
under the action of certain matrices in
$\operatorname {\mathrm {SL}}_2(\mathbb {Z})$
, which we do in Lemmas 6.1 and 6.2. Propositions 6.3 and 6.4 then give formulas for
$\mathcal F_{D,k,e}$
and
$\mathcal G_{D,k,e}$
, which we use in the final computation of the Fourier coefficients carried out in Lemmas 6.5 and 6.8.
Lemma 6.1 Let
$k\ge 3$
. Suppose D is an odd fundamental discriminant and
$D=D_1D_2$
is a product of two fundamental discriminants. Let
$\gamma =[\begin {smallmatrix} a&b\\c&d \end {smallmatrix}]\in \operatorname {\mathrm {SL}}_2(\mathbb {Z})$
with
$\gcd (c,D)=|D_1|$
. Then,
$$ \begin{align*} G_{k,D}(z)\bigg|_k\begin{bmatrix} a & b\\ c& d\end{bmatrix}=\left(\hspace{-1pt}\frac{D_2}{c}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_1}{d|D_2|}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_2}{|D_1|}\hspace{-1pt}\right)\frac{\varepsilon_{|D_1|}}{\varepsilon_{|D|}}|D_2|^{-1/2}G_{k,D_1,D_2}\left(\frac{z+c^{\ast}d}{|D_2|}\right), \end{align*} $$
where
$c^{\ast }$
is an integer with
$cc^{\ast }\equiv 1\pmod {|D_2|}$
, and
$\varepsilon _n$
is given by
$$ \begin{align} \varepsilon_n=\begin{cases} 1 & n\equiv 1\pmod4, \\ i & n\equiv 3\pmod4. \end{cases} \end{align} $$
Proof We follow the idea in Gross–Zagier [Reference Gross and Zagier5, pp. 273–275]. By equation (4.6), we have
$$ \begin{align*} \frac{2(-2\pi i)^kG(\chi_D)}{(k-1)!|D|^k}G_{k,D}(z)\bigg|_k\begin{bmatrix} a&b\\ c&d\end{bmatrix}&=\sideset{}{'}\sum_{\substack{l,r\in\mathbb{Z}\\D\mid l}}\frac{\chi_D(r)}{(l(az+b)+r(cz+d))^k}\\&=\sideset{}{'}\sum_{\substack{m,n\in{\mathbb{Z}} \\md\equiv nc~\text{mod}~|D|}}\frac{\chi_D(an-bm)}{(mz+n)^k}, \end{align*} $$
where
$(m,n)=(l,r)[\begin {smallmatrix} a&b\\ c&d\end {smallmatrix}]$
. Since
$md\equiv nc\pmod {|D|}$
, we have
$$ \begin{align} d(an-bm)&=adn-bmd\equiv adn-bcn\equiv n\pmod{|D|}, \end{align} $$
$$ \begin{align} c(an-bm)&=anc-bcm\equiv adm-bcm\equiv m\pmod{|D|}. \end{align} $$
Note also that
$\gcd (D_1,D_2)=1$
. Then, (6.2) and (6.3) imply that
$$ \begin{align*} \chi_D(an-bm)&=\chi_{D_1}(an-bm)\chi_{D_2}(an-bm) \\ &=\chi_{D_1}(d)\chi_{D_1}(n)\chi_{D_2}(c)\chi_{D_2}(m). \end{align*} $$
Since
$D_1,D_2\mid (md-nc)$
,
$(d,D_1)=1, (c,D_2)=1,$
and
$(c,D)=|D_1|$
, we must have
$D_1\mid m$
, and
$n\equiv c^{\ast }md\pmod {|D_2|}$
. By the Chinese Remainder Theorem, we can choose
$c^{\ast }$
such that
$D_1\mid c^{\ast }$
. Now, we write
$n=c^{\ast }md+l|D_2|$
. It follows that
$$ \begin{align} \frac{2(-2\pi i)^kG(\chi_D)}{(k-1)!|D|^k}G_{k,D}(z)\bigg|_k\begin{bmatrix} a&b\\ c&d\end{bmatrix}&=\sideset{}{'}\sum_{\substack{m,l\in\mathbb{Z} \nonumber\\ D_1\mid m}}\frac{\chi_{D_1}(d)\chi_{D_1}(c^{\ast}md+l|D_2|)\chi_{D_2}(c)\chi_{D_2}(m)}{(mz+mc^{\ast}d+l|D_2|)^k}\\&=\chi_{D_2}(c)\chi_{D_1}(d|D_2|)\sideset{}{'}\sum_{\substack{m,l\in\mathbb{Z}\\ D_1\mid m}}\frac{\chi_{D_2}(m)\chi_{D_1}(l)}{(mz+mc^{\ast}d+l|D_2|)^k} \nonumber\\ &=\chi_{D_2}(c)\chi_{D_1}(d|D_2|)|D_2|^{-k}\sideset{}{'}\sum_{\substack{m,l\in\mathbb{Z}\\ D_1\mid m}}\frac{\chi_{D_2}(m)\chi_{D_1}(l)}{\left(m\frac{z+c^{\ast}d}{|D_2|}+l\right)^k}. \end{align} $$
Note that (4.4) implies that
$$ \begin{align} \sideset{}{'}\sum_{\substack{m,l\in\mathbb{Z}\\ D_1\mid m}}\frac{\chi_{D_2}(m)\chi_{D_1}(l)}{\left(m\frac{z+c^{\ast}d}{|D_2|}+l\right)^k}=\frac{2(-2\pi i)^kG(\chi_{D_1})}{|D_1|^k(k-1)!}\chi_{D_2}(|D_1|)G_{k,D_1,D_2}\left(\frac{z+c^{\ast}d}{|D_2|}\right). \end{align} $$
Plugging (6.5) into (6.4) gives
$$ \begin{align} G_{k,D}(z)\bigg|_k\begin{bmatrix} a & b\\ c& d\end{bmatrix}&=\chi_{D_2}(c)\chi_{D_1}(d|D_2|)\chi_{D_2}(|D_1|)\frac{G(\chi_{D_1})}{G(\chi_D)}G_{k,D_1,D_2}\left(\frac{z+c^{\ast}d}{|D_2|}\right). \end{align} $$
From [Reference Cohen3, Proposition 2.2.24, p. 49] we know that
$$ \begin{align*} G(\chi_{D_1})=\varepsilon_{|D_1|}|D_1|^{1/2}\quad\mathrm{and}\quad G(\chi_{D})=\varepsilon_{|D|}|D|^{1/2}, \end{align*} $$
which implies that
$$ \begin{align*} G_{k,D}(z)\bigg|_k\begin{bmatrix} a & b\\ c& d\end{bmatrix}=\left(\hspace{-1pt}\frac{D_2}{c}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_1}{d|D_2|}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_2}{|D_1|}\hspace{-1pt}\right)\frac{\varepsilon_{|D_1|}}{\varepsilon_{|D|}}|D_2|^{-1/2}G_{k,D_1,D_2}\left(\frac{z+c^{\ast}d}{|D_2|}\right), \end{align*} $$
as desired.
Lemma 6.2 Let D be an odd fundamental discriminant and
$D=D_1D_2$
be a product of two fundamental discriminants. Then,
$$ \begin{align*} \theta(z)\bigg|_{\frac{1}{2}}\begin{bmatrix} |D|&0\\4|D_1|&1 \end{bmatrix}=\varepsilon_{|D_2|}^{-1}|D|^{1/4}|D_2|^{-1/2}\theta\left(\frac{|D_1|z+4^{\ast}}{|D_2|}\right), \end{align*} $$
where
$4^{\ast }$
is an integer such that
$44^{\ast }\equiv 1\pmod {|D_2|}$
.
Proof Since
$(4,D_2)=1$
, there exist
$n,m\in \mathbb {Z}$
such that
$n|D_2|+4m=1$
and
$$ \begin{align*} \begin{bmatrix} |D|&0\\4|D_1|&1 \end{bmatrix}=\begin{bmatrix} |D_2|&-m\\4&n \end{bmatrix}\begin{bmatrix} |D_1|&m\\0&|D_2| \end{bmatrix}. \end{align*} $$
It follows that
$$ \begin{align*} \theta(z)\bigg|_{\frac{1}{2}}\begin{bmatrix} |D|&0\\4|D_1|&1 \end{bmatrix}=\theta(z)\bigg|_{\frac{1}{2}}\begin{bmatrix} |D_2|&-m\\4&n \end{bmatrix}\begin{bmatrix} |D_1|&m\\0&|D_2| \end{bmatrix}. \end{align*} $$
Recall that the transformation law for
$\theta $
(see, e.g., [Reference Koblitz8, p. 148]) gives
$$ \begin{align*} \theta(z)\bigg|_{\frac{1}{2}}\begin{bmatrix} |D_2|&-m\\4&n \end{bmatrix}=\left(\frac{4}{n}\right)\varepsilon_{n}^{-1}\theta(z), \end{align*} $$
where
$\varepsilon _n$
is as in (6.1). Since
$n|D_2|+4m=1$
and
$D_2\equiv 1\pmod 4$
, we have
$\varepsilon _{n}=\varepsilon _{|D_2|}$
. Hence,
$$ \begin{align*}\theta(z)\bigg|_{\frac{1}{2}}\begin{bmatrix} |D|&0\\4|D_1|&1 \end{bmatrix}&= \varepsilon_{|D_2|}\theta(z)\bigg|_{1/2}\begin{bmatrix} |D_1|&m\\0&|D_2| \end{bmatrix}\\&=\varepsilon_{|D_2|}|D|^{1/4}|D_2|^{-1/2}\theta\left(\frac{|D_1|z+4^{\ast}}{|D_2|}\right), \end{align*} $$
which gives the desired result.
Next, we give some computations generalizing the lemma in [Reference Kohnen and Zagier9, p. 193].
Proposition 6.3 Let
$k\ge 4$
and
$e>0$
with
$\ell =k+2e$
and let D be an odd fundamental discriminant with
$(-1)^{\ell }D>0$
. Then,
$$ \begin{align*} \mathcal{F}_{D,k,e}(z)=\sum_{D=D_1D_2}\left(\hspace{-1pt}\frac{D_2}{-1}\hspace{-1pt}\right)|D_{2}|^{-2e}U_{|D_2|}([G_{k,D_1,D_2}(z),G_{k,D_1,D_2}(z)]_{2e}), \end{align*} $$
where the summation is over all decompositions of D as a product of two fundamental discriminants, and
$U_{|D_2|}$
is the operator defined in (3.1).
Proof We consider the following system of representatives (Lemma 3.1) of
$\Gamma _0(|D|)\backslash {\mathrm {SL}_{ 2 }}({\mathbb {Z}})$
,
$$ \begin{align*} \left\{\begin{bmatrix} 1 & 0 \\ |D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix} \qquad \text{where } D=D_1D_2, \quad \mu \text{ mod }|D_2|\right\} \end{align*} $$
and
$D_1,D_2$
are fundamental discriminants. By (6.6), we have
$$ \begin{align*} G_{k,D}(z)\bigg|_{k}\begin{bmatrix} 1 & 0 \\ |D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix} =\left(\hspace{-1pt}\frac{D_2}{|D_1|}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_1}{|D_2|}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_2}{|D_1|}\hspace{-1pt}\right)\frac{G(\chi_{D_1})}{G(\chi_D)}G_{k,D_1,D_2}\!\left(\!\frac{z+\mu+|D_1|^*}{|D_2|}\!\right), \end{align*} $$
where
$|D_1|^*|D_1|=1\ \mod |D_2|$
. We then compute
$\mathcal {F}_{D,k,e}(z)$
, which is
$$ \begin{align*} &\operatorname{\mathrm{Tr}}_1^D([G_{k,D}(z),G_{k,D}(z)]_{2e}) \\ =&\sum_{D_1D_2=D}\sum_{\mu \text{ mod } |D_2|} \left[G_{k,D}(z),G_{k,D}(z)\right]_{2e}\bigg|_{2k+4e}\begin{bmatrix} 1 & 0 \\ |D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix} \\ =&\sum_{D_1D_2=D}\sum_{\mu \text{ mod } |D_2|}\left[G_{k,D}(z)\bigg|_{k}\begin{bmatrix} 1 & 0 \\ |D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix},G_{k,D}(z)\bigg|_k\begin{bmatrix} 1 & 0 \\ |D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix}\right]_{2e} \\ =&\sum_{D_1D_2=D}\sum_{\mu \text{ mod } |D_2|}\left(\hspace{-1pt}\frac{D_2}{-1}\hspace{-1pt}\right)|D_2|^{-1}\left[G_{k,D_1,D_2}\!\left(\!\frac{z+\mu+|D_1|^*}{|D_2|}\!\right),G_{k,D_1,D_2}\!\left(\!\frac{z+\mu+|D_1|^*}{|D_2|}\!\right)\!\right]_{2e}\!, \end{align*} $$
where we used the well-known fact
$G(\chi _{D_1})^2=\left (\hspace {-1pt}\frac {D_1}{-1}\hspace {-1pt}\right )|D_1|$
and
$G(\chi _D)^2=\left (\hspace {-1pt}\frac {D}{-1}\hspace {-1pt}\right )|D|$
in the last equality (see, e.g., [Reference Cohen3, Corollary 2.1.47 on p. 33]). On the other hand, we have by our equivalent definition (3.2) of the U operator that
$$ \begin{align*} &U_{|D_2|}([G_{k,D_1,D_2}(z),G_{k,D_1,D_2}(z)]_{2e}) \\=&\sum_{v~\text{mod}~|D_2|}|D_2|^{(2k+4e)/2-1}[G_{k,D_1,D_2}(z),G_{k,D_1,D_2}(z)]_{2e}\bigg|_{2k+4e}\begin{bmatrix} 1 & v\\ 0 & |D_2| \end{bmatrix} \\=&\sum_{v~\text{mod}~|D_2|}|D_2|^{(2k+4e)/2-1}\left[|D_2|^{-k/2}G_{k,D_1,D_2}\left(\frac{z+v}{|D_2|}\right),|D_2|^{-k/2}G_{k,D_1,D_2}\left(\frac{z+v}{|D_2|}\right)\right]_{2e}\\=&\sum_{v~\text{mod}~|D_2|}|D_2|^{2e-1}\left[G_{k,D_1,D_2}\left(\frac{z+v}{|D_2|}\right),G_{k,D_1,D_2}\left(\frac{z+v}{|D_2|}\right)\right]_{2e}. \end{align*} $$
It follows that
$$ \begin{align*} &\quad \,\sum_{D=D_1D_2}\left(\frac{D_2}{-1}\right)|D_{2}|^{-2e}U_{|D_2|}([G_{k,D_1,D_2}(z),G_{k,D_1,D_2}(z)]_{2e})\\&=\sum_{D_1D_2=D}\sum_{v~\text{mod}~|D_2|}\left(\frac{D_2}{-1}\right)|D_2|^{-1}\left[G_{k,D_1,D_2}\left(\frac{z+v}{|D_2|}\right),G_{k,D_1,D_2}\left(\frac{z+v}{|D_2|}\right)\right]_{2e}\\&=\operatorname{\mathrm{Tr}}_1^D([G_{k,D}(z),G_{k,D}(z)]_{2e}), \end{align*} $$
as desired.
Proposition 6.4 Let
$k\geq 4$
and
$e>0$
and let D be an odd fundamental discriminant with
$(-1)^kD>0$
. Then,
$$ \begin{align*} \mathcal{G}_{D,k,e}(z)=\sum_{D=D_1D_2}\left(\hspace{-1pt}\frac{D_2}{-|D_1|}\hspace{-1pt}\right)|D_2|^{-e}U_{|D_2|}([G_{k,D_1,D_2}(4z),\theta(|D_1|z)]_e), \end{align*} $$
where the summation is over all decompositions of D as a product of two fundamental discriminants, and
$U_{|D_2|}$
is the map defined in (3.1).
Proof The proof follows a similar outline to Proposition 6.3. From Proposition 3.9, we know that
$\mathcal {G}_{D, k,e}(z)=\operatorname {\mathrm {Tr}}^{4D}_4\left [G_{k,D}(4z),\theta (|D|z)\right ]_e$
. We use the coset representatives (Lemma 3.1) for
$\Gamma _0(4|D|)\backslash \Gamma _0(4)$
,
$$ \begin{align*} \left\{\gamma_{D_1,\mu}=\begin{bmatrix} 1 & 0 \\ 4|D_1| & 1 \end{bmatrix}\begin{bmatrix} 1 & \mu \\ 0 & 1 \end{bmatrix}: \qquad \text{where } D=D_1D_2,\quad \mu \pmod {|D_2|}\right\}, \end{align*} $$
where
$D=D_1D_2$
is a product of fundamental discriminants. By a simple casework, we have
$$ \begin{align}\frac{\varepsilon_{|D_1|}}{\varepsilon_{|D|}\cdot\varepsilon_{|D_2|}}\left(\hspace{-1pt}\frac{D_1}{|D_2|}\hspace{-1pt}\right)=\left(\hspace{-1pt}\frac{D_2}{-|D_1|}\hspace{-1pt}\right).\end{align} $$
Now, Lemmas 6.1 and 6.2 together with (6.7) imply that
$$ \begin{align*} &\quad \,\sum_{D_1D_2=D}\sum_{\mu~\text{mod}~|D_2|}[G_{k,D}(4z),\theta(|D|z)]_e\big|_{k+2e+\frac{1}{2}}\gamma_{D_1,\mu} \\ &=\sum_{D_1D_2=D}\sum_{\mu~\text{mod}~|D_2|} \frac{\varepsilon_{|D_1|}}{\varepsilon_{|D|}\cdot \varepsilon_{|D_2|}}\left(\hspace{-1pt}\frac{D_1}{|D_2|}\hspace{-1pt}\right)|D_2|^{-1}\\ &\qquad \times\left[G_{k,D_1,D_2}\left(\frac{4z+|D_1|^*+4\mu}{|D_2|}\right),\theta\left(\frac{|D_1|z+4^*+|D_1|\mu}{|D_2|}\right)\right]_e \\ &=\sum_{D_1D_2=D}\sum_{\mu~\text{mod}~|D_2|}\left(\hspace{-1pt}\frac{D_2}{-|D_1|}\hspace{-1pt}\right)|D_2|^{-1} \\ &\qquad \times \left[G_{k,D_1,D_2}\left(\frac{4(z+4^{\ast}|D_1|^*+\mu)}{|D_2|}\right),\theta\left(\frac{|D_1|(z+4^*|D_1|^*+\mu)}{|D_2|}\right)\right]_e \\ &=\sum_{D_1D_2=D} \left(\hspace{-1pt}\frac{D_2}{-|D_1|}\hspace{-1pt}\right)|D_2|^{-e}U_{D_2}[G_{k,D_1,D_2}(4z),\theta(|D_1|z)]_e, \end{align*} $$
as desired.
We are now ready to compute the Fourier expansions of
$\mathcal {F}_{D,k,e}$
and
$\mathcal {S}_D(\mathcal {G}_{D,k,e})$
.
Lemma 6.5 Let
$k\ge 4$
,
$e>0$
and let D be an odd fundamental discriminant with
$(-1)^{k}D>0$
. Then, we have the Fourier expansion
$$ \begin{align*} \mathcal{F}_{D,k,e}(z)= \sum_{n\ge 1}f_{D,k,e}(n)q^n, \end{align*} $$
where
$$ \begin{align*} f_{D, k, e}(n)&= \!\sum_{D=D_1D_2}\left(\hspace{-1pt}\frac{D_2}{-1}\hspace{-1pt}\right)|D_2|^{-2e}\!\!\sum_{\substack{a_1,a_2\ge 0\\ a_1+a_2=n|D_2|}}\sum_{d|(a_1, a_2)}\left(\hspace{-2.5pt}\frac{D}{d}\hspace{-1pt}\right)d^{k-1}\sigma_{k-1, D_1, D_2}\left(\!\frac{a_1a_2}{d^2}\!\right) C_{e, a_1, a_2},\\C_{e,a_1,a_2}&=\sum_{r=0}^{2e}(-1)^ra_1^ra_2^{2e-r}\binom{2e+k-1}{2e-r}\binom{2e+k-1}{r}. \end{align*} $$
Proof By Proposition 6.3, we have
$$ \begin{align*} f_{D,k,e}(n)=\sum_{D=D_1D_2}\left(\hspace{-1pt}\frac{D_2}{-1}\hspace{-1pt}\right)|D_{2}|^{-2e}F_{D_1, D_2, e}(n), \end{align*} $$
where
$F_{D_1, D_2, e}(n)$
is the
$n|D_2|$
-th Fourier coefficient of
$[G_{k,D_1,D_2}(z),G_{k,D_1,D_2}(z)]_{2e}.$
Note that
$$ \begin{align*} G_{k,D_1,D_2}(z)^{(r)} = \sum_{n\ge 0} n^r \sigma_{k-1, D_1, D_2}(n)q^n, \end{align*} $$
which implies that the
$n|D_2|$
-th Fourier coefficient of
$G_{k,D_1,D_2}^{(r)}(z)G_{k,D_1,D_2}^{(2e-r)}(z)$
is
$$ \begin{align*} \sum_{\substack{a_1, a_2\geq 0\\a_1+a_2=n|D_2|}} a_1^r \sigma_{k-1, D_1, D_2}(a_1)a_2^{2e-r} \sigma_{k-1, D_1, D_2}(a_2). \end{align*} $$
It follows that
$F_{D_1, D_2, e}(n)=$
$$ \begin{align*} &\hspace{10pt} \sum_{r=0}^{2e}(-1)^r\binom{2e+k-1}{2e-r}\binom{2e+k-1}{r} \sum_{\substack{a_1, a_2\geq 0\\a_1+a_2=n|D_2|}} a_1^r \sigma_{k-1, D_1, D_2}(a_1)a_2^{2e-r} \sigma_{k-1, D_1, D_2}(a_2)\\ &=\sum_{\substack{a_1, a_2\geq 0\\a_1+a_2=n|D_2|}} \sigma_{k-1, D_1, D_2}(a_1)\sigma_{k-1, D_1, D_2}(a_2)\sum_{r=0}^{2e}a_1^ra_2^{2e-r}(-1)^r\binom{2e+k-1}{2e-r}\binom{2e+k-1}{r}\\ &=\sum_{\substack{a_1, a_2\geq 0\\a_1+a_2=n|D_2|}} \sigma_{k-1, D_1, D_2}(a_1)\sigma_{k-1, D_1, D_2}(a_2)C_{e, a_1, a_2}\\ &=\sum_{\substack{a_1,a_2\ge 0\\ a_1+a_2=n|D_2|}}\sum_{d|(a_1, a_2)}\left(\frac{D}{d}\right)d^{k-1}\sigma_{k-1, D_1, D_2}\left(\frac{a_1a_2}{d^2}\right) C_{e, a_1, a_2}, \end{align*} $$
where the last equality is given by the Hecke multiplicative relation [Reference Kohnen and Zagier9, p. 194]
$$ \begin{align*} \sigma_{k-1, D_1, D_2}(a_1)\sigma_{k-1, D_1, D_2}(a_2) =\sum_{d|(a_1,a_2)}\left(\frac{D}{d}\right)d^{k-1}\sigma_{k-1, D_1, D_2}\left(\frac{a_1a_2}{d^2}\right). \end{align*} $$
This finishes the proof.
Lemma 6.6 Let
$k\geq 4$
,
$e>0$
and let D be an odd fundamental discriminant with
$(-1)^k D>0$
. Then, we have the Fourier expansion
$$ \begin{align*}\mathcal{S}_D\left(\mathcal{G}_{D,k,e}(z))\right) = \sum_{n \geq 1} g_{D,k,e}(n)q^n,\end{align*} $$
where
$$ \begin{align*} g_{D,k,e}(n)&= |D|^{e}\sum\limits_{D=D_{1}D_{2}}\left(\frac{D_{2}}{-1}\right)|D_{2}|^{-2e}\sum_{\substack{a_{1},a_{2}\geq0\\ a_{1}+a_{2}=n|D_{2}| } }\sum_{d|(a_{1},a_{2})}\left(\frac{D}{d}\right)d^{k-1}\\&\quad\sigma_{k-1,D_1, D_2}\left(\frac{a_{1}a_{2}}{d^{2}}\right)E(a_1, a_2), \end{align*} $$
$$ \begin{align*} E(a_1, a_2)=\sum\limits_{r=0}^{e}(-1)^{r}\binom{e+k-1}{e-r}\binom{e-1/2}{r}4^{r}\left(a_{1}a_{2}\right)^{r}(a_{2}-a_{1})^{2(e-r)}. \end{align*} $$
Proof By Proposition 6.4 and (1.9), we have
$\mathcal {S}_D(\mathcal {G}_{D,k,e}(z))=$
$$ \begin{align*} &\sum\limits_{D_{1}D_{2}=D}\left(\frac{D_{2}}{-|D_{1}|}\right)|D_{2}|^{-e} \sum\limits_{r=0}^{e}(-1)^{r}\binom{e+k-1}{e-r}\binom{e-1/2}{r} \mathcal{S}_{D}\left[U_{|D_{2}|}(G_{k,D_{1},D_{2}}(4z)^{(r)}\theta(|D_{1}|z)^{(e-r)})\right]\hspace{-1pt}, \end{align*} $$
where we abuse notation to move the Shimura operator
$\mathcal {S}_D$
into the sums. Note that
$$\begin{align*}G_{k,D_{1},D_{2}}(4z)^{(r)}&=\sum_{n\geq0}(4n)^{r}\sigma_{k-1,D_{1},D_{2}}(n)q^{4n},\\\theta(|D_{1}|z)^{(e-r)}&=\sum_{n\in\mathbb{Z}}(n^{2}|D_{1}|)^{e-r}q^{n^{2}|D_{1}|}.\end{align*}$$
This allows us to rewrite the product
$$ \begin{align*} &G_{k,D_{1},D_{2}}(4z)^{(r)}\theta(|D_{1}|z)^{(e-r)} =\sum_{n\geq0}c_{r}(n)q^{n},\\&\quad c_{r}(n):=\sum_{m\equiv n\quad \mod 2}\left(n-m^{2}|D_{1}|\right)^{r}\sigma_{k-1,D_{1},D_{2}}\left(\frac{n-m^{2}|D_{1}|}{4}\right)(m^{2}|D_{1}|)^{e-r}, \end{align*} $$
where we take the convention that
$\sigma _{k-1,D_{1},D_{2}}(x)=0$
if
$x\notin \mathbb {\mathbb {Z}}$
or
$x<0$
. It follows that
$$ \begin{align}U_{|D_2|}\left(G_{k,D_{1},D_{2}}(4z)^{(r)}\theta(|D_{1}|z)^{(e-r)}\right)=U_{|D_2|}\left(\sum_{n\geq1}c_{r}(n)q^{n}\right)=\sum_{n\geq1}c_{r}(n|D_2|)q^{n}.\end{align} $$
Now, we compute the D-th Shimura lift of (6.9). If we write
$$ \begin{align*} \mathcal{S}_D\left(\sum_{n\geq1}c_{r}(n|D_2|)q^{n}\right)=\sum_{n\geq1}a_{r,D_2}(n)q^{n}\end{align*} $$
for some
$a_{r,D_2}(n)$
, then by the definition of
$\mathcal {S}_D$
(1.1), we have
$a_{r,{D_2}}(n)=$
$$ \begin{align*} \sum_{d|n}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)d^{k+2e-1}\hspace{-2pt}\sum_{m\in\mathbb{Z}}\hspace{-2pt}\left(\!|D_{2}||D|\frac{n^{2}}{d^{2}}-|D_{1}|m^{2}\!\right)^{r} \hspace{-2pt}\!\left(m^{2}|D_{1}|\right)^{e-r}\!\sigma_{k-1,D_1, D_2}\hspace{-2pt}\left(\!\frac{|D_{2}||D|\frac{n^{2}}{d^{2}}-m^{2}|D_{1}|}{4}\!\right). \end{align*} $$
Note that we can write
$$ \begin{align*}\frac{|D_{2}||D|\frac{n^{2}}{d^{2}}-m^{2}|D_{1}|}{4}=|D_{1}|a_1a_2,\quad\mathrm{ where}~a_{1}=\frac{|D_{2}|\frac{n}{d}+m}{2}~\mathrm{and}~a_{2}=\frac{|D_{2}|\frac{n}{d}-m}{2}.\end{align*} $$
It follows that
$a_{r,D_2}(n)=$
$$ \begin{align} &\hspace{10pt}\sum_{d|n}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)d^{k+2e-1}\sum_{\substack{a_{1},a_{2}\geq0\\ a_{1}+a_{2}=\frac{n}{d}|D_{2}| } }\left(4|D_{1}|a_{1}a_{2}\right)^{r}(a_{2}-a_{1})^{2(e-r)}|D_{1}|^{e-r}\sigma_{k-1,D_1, D_2}(|D_{1}|a_{1}a_{2}) \nonumber\\ & =\sum_{\substack{a_{1},a_{2}\geq0\\ a_{1}+a_{2}=n|D_{2}| } }\sum_{d|(a_{1},a_{2})}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)d^{k+2e-1}\left(4|D_{1}|\frac{a_{1}a_{2}}{d^2}\right)^{r}\left(\frac{a_{2}-a_{1}}{d}\right)^{2(e-r)}\nonumber\\ &\quad|D_{1}|^{e-r}\sigma_{k-1,D_1, D_2}\left(|D_{1}|\frac{a_{1}a_{2}}{d^{2}}\right) \nonumber\\ &=\sum_{\substack{a_{1},a_{2}\geq0\\ a_{1}+a_{2}=n|D_{2}| } }\sum_{d|(a_{1},a_{2})}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_{2}}{|D_{1}|}\hspace{-1pt}\right)d^{k-1}|D_{1}|^{e}\nonumber\\ &\quad \left(4a_{1}a_{2}\right)^{r}(a_{2}-a_{1})^{2(e-r)}\sigma_{k-1,D_1, D_2}\left(\frac{a_{1}a_{2}}{d^{2}}\right).\nonumber\\ \end{align} $$
Now, we substitute (6.9) back into our equation for
$\mathcal {S}_D(\mathcal {G}_{D,k,e}(z))$
. Let
$$ \begin{align*}\mathcal{S}_D(\mathcal{G}_{D,k,e}(z)) = \sum_{n\geq 1} g_{D,k,e}(n)q^n.\end{align*} $$
Then,
$g_{D,k,e}(n)=$
$$ \begin{align*} &\sum_{D=D_1D_2}\left(\hspace{-1pt}\frac{D_2}{-|D_1|}\hspace{-1pt}\right)|D_2|^{-e}\sum_{r=0}^e(-1)^r\binom{e+k-1}{e-r}\binom{e-1/2}{r}a_{r,D_2}(n)\\& \quad = \sum\limits_{D=D_{1}D_{2}}\left(\hspace{-1pt}\frac{D_{2}}{-|D_{1}|}\hspace{-1pt}\right)|D_{2}|^{-e}\sum\limits_{r=0}^{e}(-1)^{r}\binom{e+k-1}{e-r}\binom{e-1/2}{r}\\ &\qquad \times\sum_{\substack{a_{1},a_{2}\geq0\\a_{1}+a_{2}=n|D_{2}|}}\sum_{d|(a_{1},a_{2})}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)\left(\hspace{-1pt}\frac{D_{2}}{|D_{1}|}\hspace{-1pt}\right)d^{k-1}|D_{1}|^{e} \left(4a_{1}a_{2}\right)^{r}(a_{2}-a_{1})^{2(e-r)}\sigma_{k-1,D_1, D_2}\left(\frac{a_{1}a_{2}}{d^{2}}\right) \\ &\quad =|D|^{e}\sum\limits_{D=D_{1}D_{2}}\left(\hspace{-1pt}\frac{D_{2}}{-1}\hspace{-1pt}\right)|D_{2}|^{-2e}\sum_{\substack{a_{1},a_{2}\geq0\\a_{1}+a_{2}=n|D_{2}|}}\sum_{d|(a_{1},a_{2})}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)d^{k-1}\sigma_{k-1,D_1, D_2}\left(\frac{a_{1}a_{2}}{d^{2}}\right) \\&\qquad \times\sum\limits_{r=0}^{e}(-1)^{r}\binom{e+k-1}{e-r}\binom{e-1/2}{r}4^{r}\left(a_{1}a_{2}\right)^{r}(a_{2}-a_{1})^{2(e-r)}\\& \quad =|D|^{e}\sum\limits_{D=D_{1}D_{2}}\left(\hspace{-1pt}\frac{D_{2}}{-1}\hspace{-1pt}\right)|D_{2}|^{-2e}\left(\sum_{\substack{a_{1},a_{2}\geq0\\ a_{1}+a_{2}=n|D_{2}| } }\sum_{d|(a_{1},a_{2})}\left(\hspace{-1pt}\frac{D}{d}\hspace{-1pt}\right)d^{k-1}\sigma_{k-1,D_1, D_2}\left(\frac{a_{1}a_{2}}{d^{2}}\right)E(a_1, a_2)\right), \end{align*} $$
as desired.
7 Discussion
It is a folklore conjecture that
$S_{2\ell }^{0,D}(1)=S_{2\ell }(1)$
. Luo [Reference Luo13] showed that for
$\ell $
sufficiently large one has
$\dim S_{2\ell }^{0,1}(1)\gg \ell $
. Our Theorem 1.2 (the case
$D=1$
was proved earlier by Xue [Reference Xue24, Proposition 3.5]) provides a possible different approach to the conjecture. By studying the linear independence of
$\mathcal {G}_{D,k,e}$
or
$\mathcal {F}_{D,k,e}$
, one could obtain lower bounds on the dimension of
$S_{2\ell }^{0,D}(1)$
.
Conjecture 7.1 For
$\ell $
even, D a positive fundamental discriminant, the set
$\{\mathcal {G}_{D, k,e}~|~k+2e=\ell , 1 \leq e \leq \lfloor \frac {\ell }{6}\rfloor \}$
is linearly independent.
We checked this conjecture computationally in the
$D=1$
case up to
$\ell =1000$
and for prime D less than 50 up to
$\ell =100$
, using code written in Pari/GP [Reference Kayath7]. In particular, we computationally verified that the matrix
$$\begin{align*}\begin{bmatrix} g_{D, \ell-2,1}(4) & g_{D, \ell -2,1}(8) & \ldots & g_{D, \ell -2,1}(4\lfloor \frac{\ell}{6}\rfloor) \\ g_{D, \ell-4,2}(4) & g_{D, \ell-4,2}(8) & \ldots & g_{D, \ell -4,2}(4\lfloor \frac{\ell}{6}\rfloor)\\ \vdots & \vdots & \ddots & \vdots \\ g_{D,\ell-2\lfloor \frac{\ell}{6}\rfloor, \lfloor \frac{\ell}{6}\rfloor}(4) & g_{D,\ell-2\lfloor \frac{\ell}{6}\rfloor, \lfloor \frac{\ell}{6}\rfloor}(8) & \ldots & g_{D,\ell -2\lfloor \frac{\ell}{6}\rfloor,\lfloor \frac{\ell}{6}\rfloor}(4\lfloor \frac{\ell}{6}\rfloor)\\ \end{bmatrix},\end{align*}$$
where
$\mathcal {G}_{D,k,e} = \sum _{n \geq 1} g_{D,k,e}(n)q^n$
for
$1\le e\le \lfloor \frac {\ell }{6}\rfloor $
, has nonzero determinant. Further work in this area should try to prove that this determinant is nonzero in general.
The conjecture would have several interesting consequences. Using the isomorphism between
$S^{0,D}_{\ell +1/2}(4)$
and
$S^{0,D}_{2\ell }(1)$
given by the D-th Shimura lift, we find that the dimension of
$S^{0,D}_{2\ell }$
would be at least
$\lfloor \frac {\ell }{6}\rfloor $
. Since the dimension of
$S_{2\ell }(1)$
for even
$\ell $
is
$\lfloor \frac {2\ell }{12} \rfloor = \lfloor \frac {\ell }{6} \rfloor $
and
$S^{0,D}_{2\ell }(1) \subseteq S_{2\ell }(1)$
, we would conclude that
$S^{0,D}_{2\ell }(1)=S_{2\ell }(1)$
, settling the conjecture on the nonvanishing of twisted central L-values for Hecke eigenforms.
This would then imply that
$S^{0,D}_{\ell +1/2}(4) = S^{+}_{\ell +1/2}(4)$
, so the Kohnen plus space for k even is generated by Hecke eigenforms whose D-th coefficients are nonzero for all fundamental discriminants D. Further, we would conclude that
$\{\mathcal {G}_{D, k,e}\}_{k+2e=\ell ,1\leq e \leq \lfloor \frac {\ell }{6} \rfloor }$
is a basis for
$S^{+}_{\ell +1/2}(4)$
, and the set
$\{\mathcal {G}_{D, k,e}\}_{k+2e=\ell , 0\leq e \leq \lfloor \frac {\ell }{6} \rfloor }$
is a basis for
$M^{+}_{\ell +1/2}(4)$
(since the
$0$
-th Rankin–Cohen bracket produces a modular form which is non-cuspidal but still in the Kohnen plus space). To the best of our knowledge, a similar basis was first mentioned by Henri Cohen in a MathOverflow post.
Acknowledgements
We thank the anonymous referees for the detailed comments and insightful advice that have greatly improved the exposition of this article.
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