1 Introduction
In this paper, we give some exact evaluation of special values of the twisted symmetric square
$L$
-function of elliptic modular forms. The main part of this paper consists of a new construction of a holomorphic kernel function of the twisted symmetric square
$L$
-function and an explicit Fourier expansion of the kernel function. Compared with several previous methods, our approach is rather elementary and direct. Also, it suggests several possible generalizations for studying other modular
$L$
-functions.
Notation.
Some standard notation will be used. Let
$H_{1}:=\{\unicode[STIX]{x1D70F}=u+\text{i}v;v>0\}$
be the upper half-plane. The action of
$\text{SL}_{2}(\mathbf{R})$
on
$H_{1}$
is denoted by
$\big(\!\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\!\big)\unicode[STIX]{x1D70F}:=(a\unicode[STIX]{x1D70F}+b)/(c\unicode[STIX]{x1D70F}+d)$
. The branch of
$z^{\unicode[STIX]{x1D6FC}}$
is taken so that
$-\unicode[STIX]{x1D70B}<\arg z\leqslant \unicode[STIX]{x1D70B}$
. The Petersson inner product is
$$\begin{eqnarray}\langle f,g\rangle _{N}:=\int _{D_{N}}f(\unicode[STIX]{x1D70F})\overline{g(\unicode[STIX]{x1D70F})}v^{k}\,d\unicode[STIX]{x1D707},\quad d\unicode[STIX]{x1D707}:=\frac{du\,dv}{v^{2}},\end{eqnarray}$$
where
$D_{N}:=\unicode[STIX]{x1D6E4}_{0}(N)\setminus H_{1}$
is a fundamental domain,
$\unicode[STIX]{x1D6E4}_{0}(N):=\big\{\big(\!\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\!\big)\in \text{SL}_{2}(\mathbf{Z});c\equiv 0\;(\text{mod}\,N)\big\}$
. We shall use this notation for functions
$f$
and
$g$
whenever the integral is well defined and has a finite value. Let
$\unicode[STIX]{x1D6E4}_{1}:=\text{SL}_{2}(\mathbf{Z})$
be the full modular group. As usual,
$M_{k}(N)$
and
$S_{k}(N)$
are the space of modular forms and cusp forms of weight
$k$
and trivial character on
$\unicode[STIX]{x1D6E4}_{0}(N)$
.
Put
$\text{e}(x):=\text{e}^{2\unicode[STIX]{x1D70B}\text{i}x}$
,
$\text{e}^{\unicode[STIX]{x1D6FC}}(x):=\text{e}(\unicode[STIX]{x1D6FC}x)$
,
$\text{e}_{c}(x):=\text{e}(x/c)$
. The symbol
$\sum _{c(n)}$
indicates the sum over
$c\in \mathbf{Z}/n\mathbf{Z}$
, and the symbol
$\sum _{c(n)^{\ast }}$
is the sum over
$c\in \mathbf{Z}/n\mathbf{Z}$
such that
$(c,n)=1$
. We understand
$\sum _{c(1)^{\ast }}$
as the one term corresponding to
$c=0$
. We put
$\unicode[STIX]{x1D6FF}(A)=1$
when the condition
$A$
is satisfied, and
$\unicode[STIX]{x1D6FF}(A)=0$
otherwise. The automorphic factor for
$\unicode[STIX]{x1D6FE}=\big(\!\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\!\big)\in \text{SL}_{2}(\mathbf{R})$
is denoted by
$j(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D70F}):=c\unicode[STIX]{x1D70F}+d$
. Any product
$\unicode[STIX]{x1D712}\unicode[STIX]{x1D713}$
of two Dirichlet characters is understood as
$\unicode[STIX]{x1D712}\unicode[STIX]{x1D713}(n):=\unicode[STIX]{x1D712}(n)\unicode[STIX]{x1D713}(n)$
, for all
$n\in \mathbf{Z}$
. For integers
$m$
and
$n$
, the symbol
$m\,\mid \,n$
(respectively,
$m\,\nmid \,n$
) means that
$n$
is divisible by
$m$
(respectively,
$n$
is not divisible by
$m$
). For a prime
$p$
, the symbol
$p^{m}\,\Vert \,f$
means that
$p^{m}$
is the exact power of
$p$
dividing
$f$
.◻
For an even integer
$k\geqslant 4$
, let
$$\begin{eqnarray}f(\unicode[STIX]{x1D70F})=\mathop{\sum }_{n=1}^{\infty }b(n)\text{e}(n\unicode[STIX]{x1D70F})\in S_{k}(1)\end{eqnarray}$$
be a cusp form of weight
$k$
and level one. To such an
$f$
and a Dirichlet character
$\unicode[STIX]{x1D712}$
, we associate the twisted symmetric square
$L$
-function
$(\Re s>1)$
$$\begin{eqnarray}L_{2}(s,f,\unicode[STIX]{x1D712}):=L(2s,\unicode[STIX]{x1D712}^{2})\,\cdot \,\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D712}(n)\overline{b(n^{2})}n^{-(s+k-1)}.\end{eqnarray}$$
The series can be continued meromorphically to all
$s\in \mathbf{C}$
. The holomorphy has been established by Shimura, and the following algebraicity result on special values has been obtained by Sturm.
Theorem 1 (Shimura [Reference Shimura30], Sturm [Reference Sturm35, Reference Sturm36]).
Assume that
$\unicode[STIX]{x1D712}$
is a primitive Dirichlet character and satisfies
$\unicode[STIX]{x1D712}(-1)=(-1)^{\unicode[STIX]{x1D708}}$
with
$\unicode[STIX]{x1D708}\in \{0,1\}$
. The series
$L_{2}(s,f,\unicode[STIX]{x1D712})$
for
$f\in S_{k}(1)$
has a meromorphic continuation to
$\mathbf{C}$
, namely, the function
$$\begin{eqnarray}s(s-1)\unicode[STIX]{x1D6E4}((s+k-1)/2)\unicode[STIX]{x1D6E4}((s+k)/2)\unicode[STIX]{x1D6E4}(s+1-\unicode[STIX]{x1D708})\,\cdot \,L_{2}(s,f,\unicode[STIX]{x1D712})\end{eqnarray}$$
is entire. If
$f\in S_{k}(1)$
is a normalized Hecke eigenform, then
$$\begin{eqnarray}L_{2}(l,f,\unicode[STIX]{x1D712})\in \unicode[STIX]{x1D70B}^{k+2l-1}\langle f,f\rangle _{1}\overline{\mathbf{Q}},\end{eqnarray}$$
where
$l$
is any integer with the same parity as
$\unicode[STIX]{x1D708}+1$
such that
$1\leqslant l\leqslant k-1$
.◻
The starting point for the proof is the integral representation
$L_{2}(s,f,\unicode[STIX]{x1D712})=\langle \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}\,\cdot \,E_{s},f\rangle _{4N^{2}}$
. Here
$\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F})$
is a theta series with character
$\unicode[STIX]{x1D712}$
of conductor
$N$
, and
$E_{s}(\unicode[STIX]{x1D70F})$
is a real analytic Eisenstein series of half-integral weight. The function
$\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F})E_{s}(\unicode[STIX]{x1D70F})$
is a real analytic modular form of weight
$k$
on
$\unicode[STIX]{x1D6E4}_{0}(4N^{2})$
. In view of detailed analysis of
$E_{s}(\unicode[STIX]{x1D70F})$
combined with the integral representation, we can obtain the meromorphic continuation of
$L_{2}(s,f,\unicode[STIX]{x1D712})$
. Observing the fact that
$\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F})E_{l}(\unicode[STIX]{x1D70F})$
is a nearly holomorphic modular form, one can deduce
$L_{2}(l,f,\unicode[STIX]{x1D712})=\langle \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}\,\cdot \,E_{l},f\rangle _{4N^{2}}\in \unicode[STIX]{x1D70B}^{k+2l-1}\langle f,f\rangle _{1}\overline{\mathbf{Q}}$
for the indicated special points
$s=l$
. See [Reference Sturm35, Lemma 11, Theorem 4, pp. 240, 235].
On the other hand, in order to compute exact values
$L_{2}(l,f,\unicode[STIX]{x1D712})$
, we need to study the nearly holomorphic modular form
$\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F})E_{l}(\unicode[STIX]{x1D70F})$
of level
$4N^{2}$
. In [Reference Goto11, Reference Hiraoka17], Hida’s identity is employed, which reduces the required computation within
$M_{k}(N^{2})$
. But
$\dim M_{k}(N^{2})$
is usually bigger than
$\dim S_{k}(1)$
. See also [Reference Katsurada18, Reference Katsurada19] for another device, which uses the pull-back of Siegel–Eisenstein series. However, it is desirable to work out needed computations inside the simpler space
$S_{k}(1)$
. This is possible in view of the following main result in this paper.
Theorem 2. Assume that a natural number
$N$
is square-free odd. Let
$\unicode[STIX]{x1D712}$
be a primitive Dirichlet character modulo
$N$
that satisfies
$\unicode[STIX]{x1D712}(-1)=(-1)^{\unicode[STIX]{x1D708}}$
with
$\unicode[STIX]{x1D708}\in \{0,1\}$
. Let
$k\geqslant 4$
be even. Suppose
$1/2<\Re s<k-1-\unicode[STIX]{x1D708}$
,
$s\neq 1$
. Then there exists
${\mathcal{E}}_{s}^{J}\in S_{k}(1)$
satisfying
$$\begin{eqnarray}\langle {\mathcal{E}}_{s}^{J},f\rangle _{1}=\frac{\unicode[STIX]{x1D6E4}(k-1)}{(4\unicode[STIX]{x1D70B})^{k-1}}\,\cdot \,\frac{L_{2}(s,f,\unicode[STIX]{x1D712})}{L(2s,\unicode[STIX]{x1D712}^{2})}\quad \text{for all }f\in S_{k}(1),\end{eqnarray}$$
whose Fourier expansion is given by
$$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{E}}_{s}^{J}(\unicode[STIX]{x1D70F})=\mathop{\sum }_{n=1}^{\infty }a(n,s)\text{e}(n\unicode[STIX]{x1D70F}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle a(n,s)=\unicode[STIX]{x1D6FF}(\exists m\in \mathbf{N},n=m^{2})\,\cdot \,H(s)+\mathop{\sum }_{r\in \mathbf{Z},r^{2}\neq 4N^{2}n}G_{r}(s), & \displaystyle \nonumber\end{eqnarray}$$
where the functions
$H(s)$
and
$G_{r}(s)$
are as follows.
-
(1) If
$4nN^{2}>r^{2}$
, then
$$\begin{eqnarray}\displaystyle G_{r}(s) & = & \displaystyle \text{i}^{-k+\unicode[STIX]{x1D708}}2^{-s-\unicode[STIX]{x1D708}}\unicode[STIX]{x1D70B}^{s+(1/2)}n^{(k-s-1-\unicode[STIX]{x1D708})/2}N^{-2s-\unicode[STIX]{x1D708}}\nonumber\\ \displaystyle & & \displaystyle \times \,(4nN^{2}-r^{2})^{s-(1/2)}\,\cdot \,r^{\unicode[STIX]{x1D708}}A(n,r,s)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\unicode[STIX]{x1D6E4}((k-s+\unicode[STIX]{x1D708})/2)}{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D708}+{\textstyle \frac{1}{2}})\unicode[STIX]{x1D6E4}((s+k-\unicode[STIX]{x1D708})/2)}\nonumber\\ \displaystyle & & \displaystyle \times \,F\biggl(\frac{s+k-1+\unicode[STIX]{x1D708}}{2},\frac{s-k+1+\unicode[STIX]{x1D708}}{2};\unicode[STIX]{x1D708}+\frac{1}{2};\frac{r^{2}}{4nN^{2}}\biggr).\nonumber\end{eqnarray}$$
-
(2) If
$4nN^{2}<r^{2}$
, then
$$\begin{eqnarray}\displaystyle G_{r}(s) & = & \displaystyle \text{i}^{-k+\unicode[STIX]{x1D708}}2^{k-1}\unicode[STIX]{x1D70B}^{s+(1/2)}n^{k-1}N^{-s+k-1}|4nN^{2}-r^{2}|^{s-(1/2)}\nonumber\\ \displaystyle & & \displaystyle \cdot \,r^{-s-k+1}A(n,r,s)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\unicode[STIX]{x1D6E4}((k-s+\unicode[STIX]{x1D708})/2)}{\unicode[STIX]{x1D6E4}(k)\unicode[STIX]{x1D6E4}((s-k+\unicode[STIX]{x1D708}+1)/2)}\nonumber\\ \displaystyle & & \displaystyle \times \,F\biggl(\frac{s+k-1+\unicode[STIX]{x1D708}}{2},\frac{s+k-\unicode[STIX]{x1D708}}{2};k;\frac{4nN^{2}}{r^{2}}\biggr).\nonumber\end{eqnarray}$$
-
(3) Notice that there exists a natural number
$m$
such that
$n=m^{2}$
if and only if
$4nN^{2}=r^{2}$
. Then
$$\begin{eqnarray}H(s)=\unicode[STIX]{x1D712}(m)m^{k-s-1}+G_{2mN}(s)+(-1)^{\unicode[STIX]{x1D708}}G_{-2mN}(s),\end{eqnarray}$$
Here
$$\begin{eqnarray}\displaystyle G_{r}(s) & = & \displaystyle \text{i}^{-k+\unicode[STIX]{x1D708}}2^{s-1}\unicode[STIX]{x1D70B}^{s+(1/2)}m^{k-2+s}N^{-1}\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\unicode[STIX]{x1D6E4}(s-{\textstyle \frac{1}{2}})\unicode[STIX]{x1D6E4}((k-s+\unicode[STIX]{x1D708})/2)}{\unicode[STIX]{x1D6E4}((s+k-1+\unicode[STIX]{x1D708})/2)\unicode[STIX]{x1D6E4}((s+k-\unicode[STIX]{x1D708})/2)\unicode[STIX]{x1D6E4}((s-k+\unicode[STIX]{x1D708}+1)/2)}\nonumber\\ \displaystyle & & \displaystyle \times \,A(n,r,s).\nonumber\end{eqnarray}$$
$F(a,b;c;z)$
is Gauss’ hypergeometric function for
$a,b,c,z\in \mathbf{C}$
with
$|z|<1,c\neq 0,-1,-2,\ldots ,$
$$\begin{eqnarray}\displaystyle & \displaystyle F(a,b;c;z):=\mathop{\sum }_{n=0}^{\infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}n!}z^{n},\qquad (a)_{0}:=1, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (a)_{n}:=\frac{\unicode[STIX]{x1D6E4}(a+n)}{\unicode[STIX]{x1D6E4}(a)}\quad \text{for }n\geqslant 1. & \displaystyle \nonumber\end{eqnarray}$$
The closed forms of
$A(n,r,s)$
in terms of Dirichlet
$L$
-functions and a product of rational functions of
$p^{-s}$
with
$p\,\mid \,N$
are given in Proposition 2 in §2. The infinite sum
$a(n,s)$
is absolutely convergent for
$1/2<\Re s<k-1-\unicode[STIX]{x1D708}$
,
$s\neq 1$
.◻
Let
$\{f_{j}(\unicode[STIX]{x1D70F})\}_{j=1}^{\dim S_{k}(1)}$
be any orthogonal basis of
$S_{k}(1)$
. Suppose that one can find explicit
$c_{\!j}(s)$
satisfying
$$\begin{eqnarray}{\mathcal{E}}_{s}^{J}(\unicode[STIX]{x1D70F})=\mathop{\sum }_{j=1}^{\dim S_{k}(1)}c_{\!j}(s)f_{j}(\unicode[STIX]{x1D70F}).\end{eqnarray}$$
Doing so in
$S_{k}(1)$
is expected to be easier than in the space of higher level. In this case, Theorem 2 implies that
$$\begin{eqnarray}\frac{\unicode[STIX]{x1D6E4}(k-1)}{(4\unicode[STIX]{x1D70B})^{k-1}}\,\cdot \,\frac{L_{2}(s,f_{j},\unicode[STIX]{x1D712})}{L(2s,\unicode[STIX]{x1D712}^{2})}=\langle {\mathcal{E}}_{s}^{J},f_{j}\rangle _{1}=c_{\!j}(s)\langle f_{j},f_{j}\rangle _{1}.\end{eqnarray}$$
For example, let us consider the Kronecker character
$\unicode[STIX]{x1D712}_{5}=(\frac{5}{\cdot })$
and Ramanujan’s
$\unicode[STIX]{x1D6E5}$
. Then
$\unicode[STIX]{x1D708}=0$
,
$k=12$
, and the value (following Stopple’s notation [Reference Stopple33])
$$\begin{eqnarray}4^{12}\unicode[STIX]{x1D70B}^{13}b(1,r)/(5^{2}\,\cdot \,11!):=L_{2}(r,\unicode[STIX]{x1D6E5},\unicode[STIX]{x1D712}_{5})/\langle \unicode[STIX]{x1D6E5},\unicode[STIX]{x1D6E5}\rangle _{1}\end{eqnarray}$$
can be evaluated as (the point
$s=1$
turned out to be a removable singularity)
$$\begin{eqnarray}\displaystyle & \displaystyle b(1,1)=\frac{2^{12}\,\cdot \,3^{6}\,\cdot \,7}{5^{8}},\qquad b(1,3)=\frac{2^{12}\,\cdot \,3^{2}\,\cdot \,7\,\cdot \,2851}{5^{12}\,\cdot \,13}\unicode[STIX]{x1D70B}^{4}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle b(1,5)=\frac{2^{15}\,\cdot \,3\,\cdot \,1511599}{5^{17}\,\cdot \,7\,\cdot \,13}\unicode[STIX]{x1D70B}^{8}. & \displaystyle \nonumber\end{eqnarray}$$
Other examples are given in §4.
The kernel
${\mathcal{E}}_{s}^{J}$
is a generalization of the kernel functions of Zagier [Reference Zagier39] and of Stopple [Reference Stopple33]. They have treated the cases where
$\unicode[STIX]{x1D712}$
is the principal character or even Legendre’s character
$\unicode[STIX]{x1D712}_{p}=(\cdot /p)$
with any prime
$p\equiv 1\;(\text{mod}\,4)$
. These are applied to obtain: (i) a trace formula of the Hecke operators [Reference Stopple33, Reference Zagier39]; (ii) a new construction of Cohen’s modular forms [Reference Zagier39]; (iii) a simple estimation of Ramanujan’s
$\unicode[STIX]{x1D70F}(n)$
[Reference Hashim and Ram Murty15]; (iv) the non-vanishing of the symmetric square on average [Reference Kohnen and Sengupta21]; (v) an estimation of the central value of the symmetric square on average [Reference Kohnen and Sengupta22]; and (vi) an estimation of the central value of the symmetric square on average in a short interval [Reference Luo24]. Our kernel might be useful for working out similar considerations.
In this paper, we treat the character with square-free odd conductors and modular forms of level one for simplicity. However, it seems possible to treat more general conductors and level. The key idea for constructing the kernel
${\mathcal{E}}_{s}^{J}$
is the restriction of a kind of Jacobi type Eisenstein series, inspired by Heim [Reference Heim16] and Guerzhoy [Reference Guerzhoy13, Reference Guerzhoy14]. This method is elementary and direct compared with several previous researches. What is more, applying this approach to other types of Jacobi–Eisenstein series will give several possible generalizations for studying other modular
$L$
-functions (for example, Hilbert case [Reference Mizumoto26, Reference Takase37], Siegel case and matrix index case [Reference Heim16]). Finally, we mention the papers [Reference Diamantis and O’sullivan4, Reference Diamantis and O’sullivan5, Reference Goldfeld and Zhang10, Reference Nelson27, Reference Panchishkin28], where different types of kernel functions of modular
$L$
-functions have been studied.
2 Non-holomorphic kernel functions and their Fourier expansions
Let
$\unicode[STIX]{x1D712}$
be a primitive Dirichlet character modulo
$N$
(square-free odd), and suppose that
$\unicode[STIX]{x1D708}\in \{0,1\}$
satisfies
$\unicode[STIX]{x1D712}(-1)=(-1)^{\unicode[STIX]{x1D708}}$
. Let
$f\in S_{k}(1)$
be a cusp form of even weight
$k\geqslant 4$
. In this section, we construct a non-holomorphic kernel function of the twisted symmetric square and then compute its Fourier expansion.
For
$2\unicode[STIX]{x1D70E}+k>3+\unicode[STIX]{x1D708}$
(
$\unicode[STIX]{x1D70E}:=\Re (s)$
) and
$(\unicode[STIX]{x1D70F},z)\in H_{1}\times \mathbf{C}$
, we define a Jacobi type Eisenstein series (inspired by [Reference Guerzhoy13, Reference Guerzhoy14, Reference Heim16]) by
$$\begin{eqnarray}\displaystyle E_{k,N^{2}}^{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F},z,s) & := & \displaystyle \frac{v^{s}}{2}\mathop{\sum }_{\substack{ c,d\in \mathbf{Z} \\ (c,d)=1}}\mathop{\sum }_{\unicode[STIX]{x1D706}\in \mathbf{Z}}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706})\unicode[STIX]{x1D706}^{\unicode[STIX]{x1D708}}(c\unicode[STIX]{x1D70F}+d)^{-k}|c\unicode[STIX]{x1D70F}+d|^{-2s}\nonumber\\ \displaystyle & & \displaystyle \times \,\text{e}\biggl(\unicode[STIX]{x1D706}^{2}\frac{a\unicode[STIX]{x1D70F}+b}{c\unicode[STIX]{x1D70F}+d}+2N\unicode[STIX]{x1D706}\frac{z}{c\unicode[STIX]{x1D70F}+d}-N^{2}\frac{cz^{2}}{c\unicode[STIX]{x1D70F}+d}\biggr).\nonumber\end{eqnarray}$$
Here
$a,b$
are any fixed integers so that
$ad-bc=1$
for each pair
$c,d$
. We consider, for
$\unicode[STIX]{x1D70E}>2$
, the specialization
$$\begin{eqnarray}E_{s}^{J}(\unicode[STIX]{x1D70F}):=\frac{\unicode[STIX]{x1D6E4}(k-1)(4\unicode[STIX]{x1D70B})^{(s-k+1+\unicode[STIX]{x1D708})/2}}{2\unicode[STIX]{x1D6E4}((s+k-1+\unicode[STIX]{x1D708})/2)}E_{k,N^{2}}^{\unicode[STIX]{x1D712}}\biggl(\unicode[STIX]{x1D70F},0,\frac{s-k+1+\unicode[STIX]{x1D708}}{2}\biggr).\end{eqnarray}$$
This
$E_{s}^{J}$
is a non-holomorphic kernel of the twisted symmetric square
$L$
-function.
Proposition 1. Let
$\unicode[STIX]{x1D70E}:=\Re (s)$
.
-
(1) For
$2\unicode[STIX]{x1D70E}+k>3+\unicode[STIX]{x1D708}$
, the series defining
$E_{k,N^{2}}^{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F},z,s)$
is absolutely convergent on
$(\unicode[STIX]{x1D70F},z)\in H_{1}\times \mathbf{C}$
. -
(2) The function
$E_{s}^{J}(\unicode[STIX]{x1D70F})$
satisfies
$E_{s}^{J}(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F})=j(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D70F})^{k}\,\cdot \,E_{s}^{J}(\unicode[STIX]{x1D70F})$
for any
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{1}$
, and
$$\begin{eqnarray}\langle E_{s}^{J},f\rangle _{1}=\frac{\unicode[STIX]{x1D6E4}(k-1)}{(4\unicode[STIX]{x1D70B})^{k-1}}\,\cdot \,\frac{L_{2}(s,f,\unicode[STIX]{x1D712})}{L(2s,\unicode[STIX]{x1D712}^{2})}\quad \text{for all }f\in S_{k}(1),\unicode[STIX]{x1D70E}>2.\end{eqnarray}$$
-
(3) The Fourier expansion of
$E_{s}^{J}(\unicode[STIX]{x1D70F})$
is determined from the expansion where
$$\begin{eqnarray}\displaystyle E_{k,N^{2}}^{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F},0,s) & = & \displaystyle v^{s}\,\cdot \,\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F})+\frac{v^{s}}{(2\text{i})^{1/2}N(2N)^{\unicode[STIX]{x1D708}}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n,r\in \mathbf{Z}}\text{e}_{\unicode[STIX]{x1D712}}(n,r,s)r^{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D70F}_{n-(r^{2}/4N^{2})}(v,s+k-1/2-\unicode[STIX]{x1D708},s)\nonumber\\ \displaystyle & & \displaystyle \times \,\text{e}^{-2\unicode[STIX]{x1D70B}(r^{2}/4N^{2})v}\text{e}(nu),\nonumber\end{eqnarray}$$
$2\unicode[STIX]{x1D70E}+k>3+\unicode[STIX]{x1D708}$
and
$$\begin{eqnarray}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F}):=\mathop{\sum }_{\unicode[STIX]{x1D706}\in \mathbf{Z}}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706})\unicode[STIX]{x1D706}^{\unicode[STIX]{x1D708}}\text{e}(\unicode[STIX]{x1D706}^{2}\unicode[STIX]{x1D70F}),\qquad \unicode[STIX]{x1D70F}_{l}(v,\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}):=\int _{-\infty }^{\infty }\text{e}^{-2\unicode[STIX]{x1D70B}\text{i}lu}\unicode[STIX]{x1D70F}^{-\unicode[STIX]{x1D6FC}}\overline{\unicode[STIX]{x1D70F}}^{-\unicode[STIX]{x1D6FD}}\,du,\end{eqnarray}$$
Here
$$\begin{eqnarray}\displaystyle & \displaystyle e_{\unicode[STIX]{x1D712}}(n,r,s):=\mathop{\sum }_{c=1}^{\infty }S_{c}(n,r)c^{-(2s+k-\unicode[STIX]{x1D708})}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle S_{c}(n,r):=\mathop{\sum }_{d(c)^{\ast },\unicode[STIX]{x1D706}(cN)}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706})\text{e}_{c}(\overline{d}\unicode[STIX]{x1D706}^{2}+dn)\text{e}_{cN}(-r\unicode[STIX]{x1D706}). & \displaystyle \nonumber\end{eqnarray}$$
$\overline{d}$
is any fixed integer for
$d$
such that
$d\,\cdot \,\overline{d}\equiv 1\;(\text{mod}\,c)$
if
$c>1$
. If
$c=1$
, the sum over
$d$
is understood as the one term corresponding to
$d=0$
, and then
$\overline{d}$
is chosen by
$\overline{d}=0$
.
Proof of (1).
Since the case
$\unicode[STIX]{x1D708}=0$
is known (for example, see [Reference Choie and Lim2, Lemma 7.6(1), p. 656]), we only prove the case
$\unicode[STIX]{x1D708}=1$
. For all
$\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}\in \mathbf{R}$
and
$z=x+\text{i}y\in \mathbf{C}$
, we know that [Reference Knopp20, proof of Lemma 4, p. 615]
$$\begin{eqnarray}\frac{y^{2}}{1+|z|^{2}}(\unicode[STIX]{x1D6FC}^{2}+\unicode[STIX]{x1D6FD}^{2})\leqslant |\unicode[STIX]{x1D6FC}z+\unicode[STIX]{x1D6FD}|^{2}\leqslant (1+|z|^{2})(\unicode[STIX]{x1D6FC}^{2}+\unicode[STIX]{x1D6FD}^{2}).\end{eqnarray}$$
Hence one has
$|\text{e}(-N^{2}(cz^{2}/(c\unicode[STIX]{x1D70F}+d)))|\leqslant \exp (2\unicode[STIX]{x1D70B}N^{2}(|c|/|c\unicode[STIX]{x1D70F}+d|)|z|^{2})\leqslant \exp (2\unicode[STIX]{x1D70B}N^{2}((\sqrt{1+|\unicode[STIX]{x1D70F}|^{2}})/v)|z|^{2})$
. On the other hand,
$$\begin{eqnarray}\displaystyle \biggl|\text{e}\left(\unicode[STIX]{x1D706}^{2}\frac{a\unicode[STIX]{x1D70F}+b}{c\unicode[STIX]{x1D70F}+d}+2N\unicode[STIX]{x1D706}\frac{z}{c\unicode[STIX]{x1D70F}+d}\right)\biggr| & = & \displaystyle \exp \biggl(-2\unicode[STIX]{x1D70B}\frac{v}{|c\unicode[STIX]{x1D70F}+d|^{2}}\left(\unicode[STIX]{x1D706}+N\frac{I_{c,d,\unicode[STIX]{x1D70F},z}}{v}\right)^{2}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\exp \biggl(2\unicode[STIX]{x1D70B}N^{2}\frac{I_{c,d,\unicode[STIX]{x1D70F},z}^{2}}{|c\unicode[STIX]{x1D70F}+d|^{2}v}\biggr),\nonumber\end{eqnarray}$$
where
$I_{c,d,\unicode[STIX]{x1D70F},z}:=\Im (z(c\overline{\unicode[STIX]{x1D70F}}+d))$
. Since
$|I_{c,d,\unicode[STIX]{x1D70F},z}|\leqslant |z|\,\cdot \,|c\unicode[STIX]{x1D70F}+d|$
, it is sufficient to prove that the series
$$\begin{eqnarray}\mathop{\sum }_{\substack{ c,d\in \mathbf{Z} \\ (c,d)=1}}\mathop{\sum }_{\unicode[STIX]{x1D706}\in \mathbf{Z}}|\unicode[STIX]{x1D706}||c\unicode[STIX]{x1D70F}+d|^{-(2\unicode[STIX]{x1D70E}+k)}\exp \biggl(-2\unicode[STIX]{x1D70B}\frac{v}{|c\unicode[STIX]{x1D70F}+d|^{2}}\left(\unicode[STIX]{x1D706}+N\frac{I_{c,d,\unicode[STIX]{x1D70F},z}}{v}\right)^{2}\biggr)\end{eqnarray}$$
is absolutely convergent on any compact subset in
$H_{1}\times \mathbf{C}$
if
$2\unicode[STIX]{x1D70E}+k>3+1$
.
To prove this statement, for any
$\unicode[STIX]{x1D6FC}>0$
and
$\unicode[STIX]{x1D6FD}\in \mathbf{R}$
, put
$S_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}}:=\sum _{\unicode[STIX]{x1D706}\in \mathbf{Z}}|\unicode[STIX]{x1D706}|\exp (-\unicode[STIX]{x1D6FC}(\unicode[STIX]{x1D706}+\unicode[STIX]{x1D6FD})^{2})$
. It is elementary to see that (see [Reference Choie and Lim2, proof of Lemma 7.6 (1), p. 656])
$$\begin{eqnarray}\displaystyle & \displaystyle S_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}}\leqslant 2\mathop{\sum }_{m\geqslant 1}m\text{e}^{-\unicode[STIX]{x1D6FC}m^{2}}+(2|\unicode[STIX]{x1D6FD}|+3)\mathop{\sum }_{m\geqslant 0}\text{e}^{-\unicode[STIX]{x1D6FC}m^{2}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \mathop{\sum }_{m\geqslant 1}m\text{e}^{-\unicode[STIX]{x1D6FC}m^{2}}\leqslant \frac{1}{2\unicode[STIX]{x1D6FC}}+\sqrt{\frac{2}{\text{e}}}\frac{1}{\sqrt{\unicode[STIX]{x1D6FC}}},\qquad \mathop{\sum }_{m\geqslant 0}\text{e}^{-\unicode[STIX]{x1D6FC}m^{2}}\leqslant 1+\frac{\sqrt{\unicode[STIX]{x1D70B}}}{2}\frac{1}{\sqrt{\unicode[STIX]{x1D6FC}}}. & \displaystyle \nonumber\end{eqnarray}$$
Applying these estimations to the case
$\unicode[STIX]{x1D6FC}=2\unicode[STIX]{x1D70B}v|c\unicode[STIX]{x1D70F}+d|^{-2}$
and
$\unicode[STIX]{x1D6FD}=NI_{c,d,\unicode[STIX]{x1D70F},z}v^{-1}$
, our problem is reduced to the convergence of
$\sum _{(c,d)=1}|c\unicode[STIX]{x1D70F}+d|^{-(2\unicode[STIX]{x1D70E}+k)+j}$
for
$j=0,1,2$
. This implies statement (1).◻
Proof of (2).
The first statement of (2) follows from the expression
$$\begin{eqnarray}\displaystyle & \displaystyle E_{k,N^{2}}^{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F},0,s)=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{\infty }\setminus \unicode[STIX]{x1D6E4}_{1}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F})j(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D70F})^{-k}\Im (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F})^{s}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D6E4}_{\infty }:=\left\{\pm \left(\begin{array}{@{}cc@{}}1 & n\\ 0 & 1\end{array}\right);n\in \mathbf{Z}\right\}. & \displaystyle \nonumber\end{eqnarray}$$
Note that this expression is absolutely convergent for
$\unicode[STIX]{x1D70F}\in H_{1}$
and
$2\unicode[STIX]{x1D70E}+k>3+\unicode[STIX]{x1D708}$
. The second statement of (2) follows from the usual unfolding argument such as
$$\begin{eqnarray}\displaystyle \langle E_{k,N^{2}}^{\unicode[STIX]{x1D712}}(\cdot ,0,s),f\rangle _{1} & = & \displaystyle \int _{0}^{\infty }\int _{0}^{1}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F})\overline{f(\unicode[STIX]{x1D70F})}v^{s+k-2}\,du\,dv\nonumber\\ \displaystyle & = & \displaystyle \frac{2\unicode[STIX]{x1D6E4}(s+k-1)}{(4\unicode[STIX]{x1D70B})^{s+k-1}}\mathop{\sum }_{\unicode[STIX]{x1D706}=1}^{\infty }\frac{\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706})\overline{b(\unicode[STIX]{x1D706}^{2})}}{\unicode[STIX]{x1D706}^{2s+2k-2-\unicode[STIX]{x1D708}}}.\nonumber\end{eqnarray}$$
The computation can be easily justified by
$|f(\unicode[STIX]{x1D70F})|=O(\text{e}^{-2\unicode[STIX]{x1D70B}v})$
(
$v\rightarrow +\infty$
) and
$b(n)=O(n^{k/2})$
combined with the fact that, for
$2\unicode[STIX]{x1D70E}+k>3+\unicode[STIX]{x1D708}$
, there is a constant
$\unicode[STIX]{x1D6FD}>0$
such that
$|E_{k,N^{2}}^{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F},0,s)|=O(v^{\unicode[STIX]{x1D6FD}})$
(
$v\rightarrow +\infty$
).◻
Proof of (3).
To compute the Fourier expansion, we see, as in [Reference Eichler and Zagier7, p. 18] and [Reference Arakawa1], that
$$\begin{eqnarray}\displaystyle & & \displaystyle E_{k,N^{2}}^{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F},0,s)\nonumber\\ \displaystyle & & \displaystyle \quad =v^{s}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D70F})+v^{s}\mathop{\sum }_{c\geqslant 1}\frac{(Nc)^{\unicode[STIX]{x1D708}}}{c^{k+2s}}\mathop{\sum }_{d(c)^{\ast },\unicode[STIX]{x1D706}(Nc)}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706})\text{e}_{c}(\overline{d}\unicode[STIX]{x1D706}^{2})F_{k,N^{2}}\biggl(\unicode[STIX]{x1D70F}+\frac{d}{c},-\frac{\unicode[STIX]{x1D706}}{cN},s\biggr),\nonumber\end{eqnarray}$$
where
$\overline{d}$
is an integer such that
$d\,\cdot \,\overline{d}\equiv 1\;(\text{mod}\,c)$
if
$c>1$
, if
$c=1$
the sum over
$d$
is the one term corresponding to
$d=0$
(and
$\overline{d}=0$
), and
$F_{k,\unicode[STIX]{x1D6FC}}$
is the periodic function defined by
$$\begin{eqnarray}\displaystyle F_{k,\unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D70F},z,s) & := & \displaystyle (-1)^{\unicode[STIX]{x1D708}}\mathop{\sum }_{p,q\in \mathbf{Z}}\frac{\text{e}(-\unicode[STIX]{x1D6FC}((z+q)^{2}/(\unicode[STIX]{x1D70F}+p)))}{(\unicode[STIX]{x1D70F}+p)^{k}|\unicode[STIX]{x1D70F}+p|^{2s}}(z+q)^{\unicode[STIX]{x1D708}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{n,r\in \mathbf{Z}}\unicode[STIX]{x1D6FE}_{n,r}(v,s)\text{e}(nu+rz).\nonumber\end{eqnarray}$$
It follows from the formula
$$\begin{eqnarray}\int _{-\infty }^{\infty }\text{e}\biggl(-\unicode[STIX]{x1D6FC}\frac{z^{2}}{\unicode[STIX]{x1D70F}}-rz\biggr)z^{\unicode[STIX]{x1D708}}\,dx=\biggl(-\frac{r}{2\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D70F}\biggr)^{\unicode[STIX]{x1D708}}\text{e}\biggl(\frac{r^{2}\unicode[STIX]{x1D70F}}{4\unicode[STIX]{x1D6FC}}\biggr)\sqrt{\frac{\unicode[STIX]{x1D70F}}{2\unicode[STIX]{x1D6FC}\text{i}}},\qquad z=x+\text{i}y,\end{eqnarray}$$
which holds for
$\unicode[STIX]{x1D6FC}>0$
,
$\unicode[STIX]{x1D70F}\in H_{1}$
,
$r\in \mathbf{R}$
and
$\unicode[STIX]{x1D708}\in \{0,1\}$
, that the coefficient
$\unicode[STIX]{x1D6FE}_{n,r}(v,s)$
is given by
$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{n,r}(v,s)=\biggl(\frac{r}{2\unicode[STIX]{x1D6FC}}\biggr)^{\unicode[STIX]{x1D708}}\frac{\text{e}^{-2\unicode[STIX]{x1D70B}(r^{2}/4\unicode[STIX]{x1D6FC})v}}{(2\text{i}\unicode[STIX]{x1D6FC})^{1/2}}\,\cdot \,\unicode[STIX]{x1D70F}_{n-(r^{2}/4\unicode[STIX]{x1D6FC})}(v,s+k-1/2-\unicode[STIX]{x1D708},s).\end{eqnarray}$$
This combined with interchanging the summation completes the proof of statement (3). Note that
$|S_{c}(n,r)|\leqslant Nc^{2}$
.◻
Lemma 1. Let
$N$
be the square-free odd conductor of the primitive
$\unicode[STIX]{x1D712}$
and let
$N=\prod _{j=1}^{m}p_{j}$
be the prime factorization. We denote by
$\unicode[STIX]{x1D712}_{j}$
the primitive characters
$\text{mod}\,p_{j}$
so that
$\unicode[STIX]{x1D712}=\prod _{j=1}^{m}\unicode[STIX]{x1D712}_{j}$
, and we define
$\unicode[STIX]{x1D712}_{j}^{\ast }$
by
$\unicode[STIX]{x1D712}_{j}^{\ast }=\prod _{i\neq j}\unicode[STIX]{x1D712}_{i}$
. For any natural number
$c$
with
$c=a\prod _{j=1}^{m}p_{j}^{e_{j}}$
,
$e_{j}\geqslant 0$
,
$(a,N)=1$
, the exponential sum
$S_{c}(n,r)$
in Proposition 1 has the decomposition
$S_{c}(n,r)=S_{c}^{a}(n,r)\prod _{j=1}^{m}S_{p_{j}^{e_{j}}}^{j}(n,r)$
, where
$$\begin{eqnarray}\displaystyle S_{p_{j}^{e_{j}}}^{j}(n,r) & = & \displaystyle \unicode[STIX]{x1D712}_{j}(-1)\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j}^{e_{j}+1})W(\unicode[STIX]{x1D712}_{j})\nonumber\\ \displaystyle & & \displaystyle \cdot \,\left\{\begin{array}{@{}ll@{}}\displaystyle \overline{\unicode[STIX]{x1D712}_{j}}(r) & \displaystyle \text{if }e_{j}=0,\\ \displaystyle 0 & \displaystyle \text{if }e_{j}\geqslant 1\text{ and }p_{j}\,\nmid \,r,\\ \displaystyle p_{j}^{e_{j}}\,\cdot \,\widetilde{H}_{n,r}^{j}(e_{j}) & \displaystyle \text{if }e_{j}\geqslant 1\text{ and }p_{j}\,\mid \,r,\end{array}\right.\nonumber\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \widetilde{H}_{n,r}^{j}(e_{j}):=\mathop{\sum }_{\unicode[STIX]{x1D706}(p_{j}^{e_{j}}),~Q_{j}(\unicode[STIX]{x1D706})\equiv 0\;(\text{mod}\,p_{j}^{e_{j}-1})}\unicode[STIX]{x1D712}_{j}(\unicode[STIX]{x1D706})\overline{\unicode[STIX]{x1D712}_{j}}(Q_{j}(\unicode[STIX]{x1D706})/p_{j}^{e_{j}-1}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle Q_{j}(\unicode[STIX]{x1D706}):=(N^{2}/p_{j}^{2})\unicode[STIX]{x1D706}^{2}+(r/p_{j})\unicode[STIX]{x1D706}+n, & \displaystyle \nonumber\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle S_{c}^{a}(n,r)=\left\{\begin{array}{@{}ll@{}}\displaystyle 1 & \text{if }a=1,\\ \displaystyle \unicode[STIX]{x1D712}(a)\mathop{\sum }_{d(a)^{\ast },\unicode[STIX]{x1D706}(a)}\text{e}_{a}(dQ(\unicode[STIX]{x1D706})) & \text{if }a\geqslant 2,\end{array}\right. & \displaystyle \nonumber\\ \displaystyle & \displaystyle Q(\unicode[STIX]{x1D706}):=N^{2}\unicode[STIX]{x1D706}^{2}+r\unicode[STIX]{x1D706}+n,\qquad W(\unicode[STIX]{x1D712}_{j}):=\mathop{\sum }_{\unicode[STIX]{x1D706}=1}^{p_{j}}\unicode[STIX]{x1D712}_{j}(\unicode[STIX]{x1D706})\text{e}_{p_{j}}(\unicode[STIX]{x1D706}). & \displaystyle \nonumber\end{eqnarray}$$
Proof. We chose the representatives
$\unicode[STIX]{x1D706}(Nc)$
in the form
$\unicode[STIX]{x1D706}=\sum _{j=1}^{m}\unicode[STIX]{x1D706}_{j}\,\cdot \,Nc/p_{j}^{e_{j}+1}+\unicode[STIX]{x1D706}_{m+1}\,\cdot \,Nc/a$
, where
$\unicode[STIX]{x1D706}_{j}(p_{j}^{e_{j}+1})$
,
$\unicode[STIX]{x1D706}_{m+1}(a)$
are any fixed representatives. Similarly, we chose the representatives
$d(c)^{\ast }$
in the form
$d=\sum _{j=1}^{m}d_{j}\,\cdot \,c/p_{j}^{e_{j}}+d_{m+1}\,\cdot \,c/a$
, where
$d_{j}(p_{j}^{e_{j}})^{\ast }$
,
$d_{m+1}(a)^{\ast }$
are any fixed representatives. Note that when
$e_{j}=0$
(respectively,
$a=1$
), we should take
$d_{j}=0$
(respectively,
$d_{m+1}=0$
), which is consistent with our convention
$\sum _{d(1)^{\ast }}g(d)=g(0)$
. We obtain
$S_{c}(n,r)=S_{c}^{a}(n,r)\prod _{j=1}^{m}S_{p_{j}^{e_{j}}}^{j}(n,r)$
by defining
$$\begin{eqnarray}S_{c}^{a}(n,r):=\unicode[STIX]{x1D712}(a)\mathop{\sum }_{d_{m+1}(a)^{\ast },\unicode[STIX]{x1D706}_{m+1}(a)}\text{e}_{a}\biggl(\overline{d}\unicode[STIX]{x1D706}_{m+1}^{2}\frac{cN^{2}}{a}+nd_{m+1}\biggr)\text{e}_{a}(-r\unicode[STIX]{x1D706}_{m+1}),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle S_{p_{j}^{e_{j}}}^{j}(n,r) & := & \displaystyle \unicode[STIX]{x1D712}_{j}^{\ast }(p_{j}^{e_{j}+1})\mathop{\sum }_{d_{j}(p_{j}^{e_{j}})^{\ast },\unicode[STIX]{x1D706}_{j}(p_{j}^{e_{j}+1})}\unicode[STIX]{x1D712}_{j}(\unicode[STIX]{x1D706}_{j})\nonumber\\ \displaystyle & & \displaystyle \times \,\text{e}_{p_{j}^{e_{j}}}\biggl(\overline{d}\unicode[STIX]{x1D706}_{j}^{2}\frac{cN^{2}}{p_{j}^{e_{j}+2}}+nd_{j}\biggr)\text{e}_{p_{j}^{e_{j}+1}}(-r\unicode[STIX]{x1D706}_{j}).\nonumber\end{eqnarray}$$
By definition, the integer
$\overline{d}$
satisfies
$(\sum _{j=1}^{m}d_{j}\,\cdot \,c/p_{j}^{e_{j}}+d_{m+1}\,\cdot \,c/a)\,\cdot \,\overline{d}\,\equiv \,1\;(\text{mod}\,c)$
.
For each fixed
$d_{m+1}$
, the variable change
$\unicode[STIX]{x1D706}_{m+1}$
to
$-d_{m+1}\unicode[STIX]{x1D706}_{m+1}$
in the sum over
$\unicode[STIX]{x1D706}_{m+1}$
implies the statement about
$S_{c}^{a}(n,r)$
. For
$S_{p_{j}^{e_{j}}}^{j}(n,r)$
, the case
$e_{j}=0$
is easy. When
$e_{j}\geqslant 1$
, the variable change
$\unicode[STIX]{x1D706}_{j}$
to
$\unicode[STIX]{x1D706}_{j}+p_{j}^{e_{j}}$
implies that
$S_{p_{j}^{e_{j}}}^{j}(n,r)=\text{e}_{p_{j}}(-r)S_{p_{j}^{e_{j}}}^{j}(n,r)$
, and thus
$S_{p_{j}^{e_{j}}}^{j}(n,r)=0$
for the case
$e_{j}\geqslant 1$
and
$p_{j}\,\nmid \,r$
. Suppose that
$e_{j}\geqslant 1$
and
$p_{j}\,\mid \,r$
. For each fixed
$d_{j}$
, the variable change
$\unicode[STIX]{x1D706}_{j}$
to
$d_{j}\unicode[STIX]{x1D706}_{j}$
in the sum over
$\unicode[STIX]{x1D706}_{j}$
yields
$$\begin{eqnarray}S_{p_{j}^{e_{j}}}^{j}(n,r)=\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j}^{e_{j}+1})\mathop{\sum }_{d_{j}(p_{j}^{e_{j}})^{\ast },\unicode[STIX]{x1D706}_{j}(p_{j}^{e_{j}+1})}\unicode[STIX]{x1D712}_{j}(d_{j})\unicode[STIX]{x1D712}_{j}(\unicode[STIX]{x1D706}_{j})\text{e}_{p_{j}^{e_{j}}}(d_{j}Q_{j}(-\unicode[STIX]{x1D706}_{j})),\end{eqnarray}$$
and then the statement follows by applying Shimura’s lemma [Reference Shimura30, p. 87] to the sum over
$d_{j}$
. Namely, use the identity [Reference Shimura30, (3.8), p. 87]. A similar computation has been found in [Reference Guerzhoy13].◻
By Lemma 1, the arithmetic part
$e_{\unicode[STIX]{x1D712}}(n,r,s)$
of the Fourier coefficients in Proposition 1 has the decomposition
$e_{\unicode[STIX]{x1D712}}(n,r,s)=B(n,r,s)\,\cdot \,\prod _{j=1}^{m}C_{j}(n,r,s)$
by defining
$$\begin{eqnarray}\displaystyle & \displaystyle B(n,r,s):=\mathop{\sum }_{a\geqslant 1,~(a,N)=1}S_{c}^{a}(n,r)a^{-(2s+k-\unicode[STIX]{x1D708})}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle C_{j}(n,r,s):=\mathop{\sum }_{e_{j}\geqslant 0}S_{p_{j}^{e_{j}}}^{j}(n,r)p_{j}^{-e_{j}(2s+k-\unicode[STIX]{x1D708})}. & \displaystyle \nonumber\end{eqnarray}$$
Using
$N$
as odd, we easily have (see [Reference Eichler and Zagier7, p. 21] and [Reference Zagier39, p. 130])
$$\begin{eqnarray}B(n,r,s)=\frac{L_{r^{2}-4N^{2}n}(2s+k-1-\unicode[STIX]{x1D708},\unicode[STIX]{x1D712})}{L(2(2s+k-1-\unicode[STIX]{x1D708}),\unicode[STIX]{x1D712}^{2})},\end{eqnarray}$$
where
$L_{D}(s,\unicode[STIX]{x1D712})$
is the
$L$
-function defined by
$$\begin{eqnarray}\displaystyle & & \displaystyle L_{D}(s,\unicode[STIX]{x1D712})\nonumber\\ \displaystyle & & \displaystyle ~:=\,\left\{\begin{array}{@{}ll@{}}\displaystyle L(2s-1,\unicode[STIX]{x1D712}^{2}) & \text{if }D=0,\\ \displaystyle L(s,\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712})\mathop{\sum }_{a|f}\unicode[STIX]{x1D707}(a)\unicode[STIX]{x1D712}_{D_{K}}(a)\unicode[STIX]{x1D712}(a)a^{-s} & \\ \displaystyle \quad \times \,\unicode[STIX]{x1D70E}_{1-2s,\unicode[STIX]{x1D712}^{2}}(f/a) & \text{if }D\neq 0,D\equiv 0,1\;(\text{mod}\,4),\\ \displaystyle 0 & \text{if }D\equiv 2,3\;(\text{mod}\,4).\end{array}\right.\quad\end{eqnarray}$$
Here
$f$
is defined by
$D=D_{K}f^{2}$
with the discriminant
$D_{K}$
of
$K=\mathbf{Q}(\sqrt{D})$
,
$\unicode[STIX]{x1D712}_{D_{K}}$
is the Kronecker symbol,
$\unicode[STIX]{x1D707}$
is the Möbius function and
$\unicode[STIX]{x1D70E}_{s,\unicode[STIX]{x1D712}}(n)=\sum _{d|n}\unicode[STIX]{x1D712}(d)d^{s}$
.
On the other hand, by Lemma 1,
$$\begin{eqnarray}\displaystyle C_{j}(n,r,s) & = & \displaystyle \unicode[STIX]{x1D712}_{j}(-1)\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j})W(\unicode[STIX]{x1D712}_{j})\nonumber\\ \displaystyle & & \displaystyle \cdot \,\left\{\begin{array}{@{}ll@{}}\displaystyle \overline{\unicode[STIX]{x1D712}_{j}}(r) & \text{for }p_{j}\,\nmid \,r,\\ \displaystyle \mathop{\sum }_{l=1}^{\infty }\widetilde{H}_{n,r}^{j}(l)(\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j})/p^{2s+k-1-\unicode[STIX]{x1D708}})^{l} & \text{for }p_{j}\,\mid \,r.\end{array}\right.\nonumber\end{eqnarray}$$
To compute this
$C_{j}(n,r,s)$
in a closed form, we need the following lemma.
Lemma 2. Assume that
$p_{j}\,\mid \,r$
and put
$r=p_{j}\hat{r}_{j}$
. For
$l\geqslant 2$
, the values of the sum
$\widetilde{H}_{n,r}^{j}(l)$
defined in Lemma 1 are given as follows.
-
(a) Suppose
$\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n\neq 0$
. Write as
$r^{2}-4N^{2}n=D_{K}f^{2}$
in terms of the discriminant
$D_{K}$
of
$K=\mathbf{Q}(\sqrt{r^{2}-4N^{2}n})=\mathbf{Q}(\sqrt{\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n})$
. Then
$p_{j}\,\mid \,f=p_{j}f_{j}$
for some
$f_{j}$
and
$\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n=D_{K}f_{j}^{2}$
. Put
$f_{j}=p_{j}^{m_{j}}\hat{f}_{j}$
with
$p_{j}\,\nmid \,\hat{f}_{j}$
.-
(1) If
$p_{j}\,\nmid \,D_{K}$
, then
$\widetilde{H}_{n,r}^{j}(l)=0$
for
$l\geqslant 2m_{j}+2$
. -
(2) If
$p_{j}\,\mid \,D_{K}$
, then
$\widetilde{H}_{n,r}^{j}(l)=0$
for
$l\geqslant 2m_{j}+3$
, and
$\widetilde{H}_{n,r}^{j}(l)=p_{j}^{m_{j}+1}\unicode[STIX]{x1D712}_{j}(-2\hat{r_{j}})\overline{\unicode[STIX]{x1D712}_{j}}(-(D_{K}/p_{j})\hat{f_{j}}^{2})$
for
$l=2m_{j}+2$
. -
(3) Suppose
$2\leqslant l\leqslant 2m_{j}+1$
with
$m_{j}\geqslant 1$
.-
(3-1) Suppose
$l=2e+1$
is odd.If “
$e\leqslant m_{j}-1$
” or “
$e=m_{j}$
and
$p_{j}\,\mid \,D_{K}$
”, we have
$\widetilde{H}_{n,r}^{j}(l)=\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})p_{j}^{e}\,\cdot \,\sum _{c(p_{j})}\overline{\unicode[STIX]{x1D712}_{j}}(c)^{2}$
.If “
$e=m_{j}$
and
$p_{j}\,\nmid \,D_{K}$
”, we have
$\widetilde{H}_{n,r}^{j}(l)=\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})p_{j}^{m_{j}}\,\cdot \,\sum _{c(p_{j})}\overline{\unicode[STIX]{x1D712}_{j}}(c^{2}-D_{K}\hat{f}_{j}^{2})$
. -
(3-2) Suppose
$l=2e$
is even. Then
$\widetilde{H}_{n,r}^{j}(l)=0$
for
$e\leqslant m_{j}$
.
-
-
-
(b) Suppose
$\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n=0$
.-
(1) If
$l=2e+1\geqslant 2$
is odd, then
$\widetilde{H}_{n,r}^{j}(l)=\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})p_{j}^{e}\,\cdot \,\sum _{c(p_{j})}\overline{\unicode[STIX]{x1D712}_{j}}(c)^{2}$
. -
(2) If
$l=2e\geqslant 2$
is even, then
$\widetilde{H}_{n,r}^{j}(l)=0$
.
-
Proof. We can list the solutions
$\unicode[STIX]{x1D706}\;(\text{mod}\,p_{j}^{e_{j}})$
for the quadratic congruence
$Q_{j}(\unicode[STIX]{x1D706})\equiv 0\;(\text{mod}\,p_{j}^{e_{j}-1})$
. The proof is then elementary.◻
Suppose
$p_{j}\,\mid \,r=p_{j}\hat{r}_{j}$
. Using the same notation as in Lemma 2, we consider the power series defined for
$T\in \mathbf{C}$
with
$|p_{j}T|<1$
by
$$\begin{eqnarray}F_{\hat{r}_{j},n}^{j}(T):=\mathop{\sum }_{l=1}^{\infty }\widetilde{H}_{n,r}^{j}(l)T^{l}.\end{eqnarray}$$
By Lemma 2, we get the following lemma.
Lemma 3. Put
$\unicode[STIX]{x1D6FC}_{j}:=\widetilde{H}_{n,p_{j}\hat{r}_{j}}^{j}(1)$
,
$\unicode[STIX]{x1D6FD}_{j}:=\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})\sum _{c(p_{j})}\overline{\unicode[STIX]{x1D712}_{j}}(c^{2})$
and
$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{j}:=\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})\mathop{\sum }_{c(p_{j})}\overline{\unicode[STIX]{x1D712}_{j}}(c^{2}-D_{K}\hat{f}_{j}^{2}),\qquad \unicode[STIX]{x1D6FF}_{j}:=\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})\overline{\unicode[STIX]{x1D712}_{j}}(-(D_{K}/p_{j})\hat{f}_{j}^{2}).\end{eqnarray}$$
The explicit forms of
$F_{\hat{r}_{j},n}^{j}(T)$
defined by (3) are given as follows.
-
(a) Suppose
$\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n\neq 0$
.-
(1) If
$p_{j}\,\nmid \,\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n$
, then
$F_{\hat{r}_{j},n}^{j}(T)=\unicode[STIX]{x1D6FC}_{j}T$
. -
(2) If
$p_{j}\,\mid \,\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n$
, denote
$\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n=D_{K}f_{j}^{2},p_{j}^{m_{j}}\,\Vert \,f_{j}=p_{j}^{m_{j}}\hat{f}_{j}$
.-
(2-1) If
$p_{j}\,\nmid \,D_{K}$
, then
$$\begin{eqnarray}\displaystyle F_{\hat{r}_{j},n}^{j}(T) & = & \displaystyle \frac{T}{1-p_{j}T^{2}}\{\!\unicode[STIX]{x1D6FC}_{j}-(\unicode[STIX]{x1D6FC}_{j}-\unicode[STIX]{x1D6FD}_{j})p_{j}T^{2}\nonumber\\ \displaystyle & & \displaystyle +\,p_{j}^{m_{j}}T^{2m_{j}}(\unicode[STIX]{x1D6FE}_{j}-\unicode[STIX]{x1D6FD}_{j}-p_{j}\unicode[STIX]{x1D6FE}_{j}T^{2})\!\}.\nonumber\end{eqnarray}$$
-
(2-2) If
$p_{j}\,\mid \,D_{K}$
, then
$$\begin{eqnarray}\displaystyle F_{\hat{r}_{j},n}^{j}(T) & = & \displaystyle \frac{T}{1-p_{j}T^{2}}\{\!\unicode[STIX]{x1D6FC}_{j}-(\unicode[STIX]{x1D6FC}_{j}-\unicode[STIX]{x1D6FD}_{j})p_{j}T^{2}\nonumber\\ \displaystyle & & \displaystyle -\,p_{j}^{m_{j}+1}T^{2m_{j}+1}(\unicode[STIX]{x1D6FF}_{j}p_{j}T^{2}+\unicode[STIX]{x1D6FD}_{j}T-\unicode[STIX]{x1D6FF}_{j})\!\}.\nonumber\end{eqnarray}$$
-
-
-
(b) If
$\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n=0$
, then
$$\begin{eqnarray}F_{\hat{r}_{j},n}^{j}(T)=\frac{T}{1-p_{j}T^{2}}\{\unicode[STIX]{x1D6FC}_{j}-(\unicode[STIX]{x1D6FC}_{j}-\unicode[STIX]{x1D6FD}_{j})p_{j}T^{2}\}.\end{eqnarray}$$
Remark.
If
$\unicode[STIX]{x1D712}_{j}$
is quadratic, then we can compute the character sums in rather simple forms. In fact, if
$p_{j}\,\mid \,\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n$
, then
$\unicode[STIX]{x1D6FC}_{j}=-\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})$
,
$\unicode[STIX]{x1D6FD}_{j}=\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})(p_{j}-1)=-\unicode[STIX]{x1D6FC}_{j}(p_{j}-1)$
,
$\unicode[STIX]{x1D6FE}_{j}=-\unicode[STIX]{x1D712}_{j}(-2\hat{r}_{j})=\unicode[STIX]{x1D6FC}_{j}$
,
$\unicode[STIX]{x1D6FF}_{j}=-\unicode[STIX]{x1D6FC}_{j}\unicode[STIX]{x1D712}_{j}(-D_{K}/p_{j})$
. See [Reference Small32, Proposition 4.1, p. 86].
Lemma 3 and Proposition 1 together with the relation
$\unicode[STIX]{x1D70F}_{l}(v/4,\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})=4^{\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}-1}\unicode[STIX]{x1D70F}_{l/4}(v,\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})$
imply the following Fourier expansion of
$E_{s}^{J}(\unicode[STIX]{x1D70F})$
.
Proposition 2. Let
$\unicode[STIX]{x1D70F}=u+\text{i}v\in H_{1}$
and
$\unicode[STIX]{x1D70E}=\Re s>2$
. The Fourier expansion of
$E_{s}^{J}(\unicode[STIX]{x1D70F})$
defined by (1) is given by
$$\begin{eqnarray}\displaystyle E_{s}^{J}(\unicode[STIX]{x1D70F}) & = & \displaystyle \mathop{\sum }_{n\in \mathbf{Z}}c_{n}(v,s)\text{e}(nu),\nonumber\\ \displaystyle c_{n}(v,s) & := & \displaystyle \frac{\unicode[STIX]{x1D6E4}(k-1)(4\unicode[STIX]{x1D70B})^{(s-k+1+\unicode[STIX]{x1D708})/2}}{\unicode[STIX]{x1D6E4}((s+k-1+\unicode[STIX]{x1D708})/2)}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl\{\frac{\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{0},n=0)}{2}v^{(s-k+1+\unicode[STIX]{x1D708})/2}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6FF}(\exists m\in \mathbf{N},n=m^{2})v^{(s-k+1+\unicode[STIX]{x1D708})/2}\unicode[STIX]{x1D712}(m)m^{\unicode[STIX]{x1D708}}\text{e}(m^{2}\text{i}v)\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{r\in \mathbf{Z}}\frac{r^{\unicode[STIX]{x1D708}}}{(2\text{i})^{1/2}2^{2s}}\,\cdot \,\frac{A(n,r,s)}{N(2N)^{\unicode[STIX]{x1D708}}}v^{(s-k+1+\unicode[STIX]{x1D708})/2}\nonumber\\ \displaystyle & & \displaystyle \times \,\unicode[STIX]{x1D70F}_{4n-(r^{2}/N^{2})}\left(\frac{v}{4},\frac{s+k-\unicode[STIX]{x1D708}}{2},\frac{s-k+1+\unicode[STIX]{x1D708}}{2}\right)\text{e}^{-2\unicode[STIX]{x1D70B}(r^{2}/4N^{2})v}\biggr\}.\nonumber\end{eqnarray}$$
Here
$\unicode[STIX]{x1D712}_{0}$
is the principal character, that is,
$\unicode[STIX]{x1D712}_{0}(l)=1$
for all
$l\in \mathbf{Z}$
, and we have defined
$$\begin{eqnarray}\displaystyle & \displaystyle A(n,r,s):=e_{\unicode[STIX]{x1D712}}\biggl(n,r,\frac{s-k+1+\unicode[STIX]{x1D708}}{2}\biggr)=\frac{L_{r^{2}-4N^{2}n}(s,\unicode[STIX]{x1D712})}{L(2s,\unicode[STIX]{x1D712}^{2})}\,\cdot \,\mathop{\prod }_{j=1}^{m}D_{j}(n,r,s), & \displaystyle \nonumber\\ \displaystyle & \displaystyle D_{j}(n,r,s):=\unicode[STIX]{x1D712}_{j}(-1)\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j})W(\unicode[STIX]{x1D712}_{j})\,\cdot \,\left\{\begin{array}{@{}ll@{}}\displaystyle \overline{\unicode[STIX]{x1D712}_{j}}(r) & \text{if }p_{j}\,\nmid \,r,\\ \displaystyle F_{\hat{r}_{j},n}^{j}(\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j})p_{j}^{-s}) & \text{if }p_{j}\,\mid \,r,\end{array}\right. & \displaystyle \nonumber\end{eqnarray}$$
where
$L_{D}(s,\unicode[STIX]{x1D712})$
is as in (2) (see Lemmas 1 and 3 for the notation used in the description of
$D_{j}(n,r,s)$
). This Fourier expansion combined with the estimations in §3 implies a meromorphic continuation of
$E_{s}^{J}(\unicode[STIX]{x1D70F})$
to all
$s\in \mathbf{C}$
.◻
3 Holomorphic kernel functions: proof of Theorem 2
Let us apply the Holomorphic projection lemma to
$E_{s}^{J}(\unicode[STIX]{x1D70F})$
.
Lemma 4 (Zagier [Reference Zagier40, p. 288], Sturm [Reference Sturm34]).
Let
$\unicode[STIX]{x1D6F7}:H_{1}\rightarrow \mathbf{C}$
be a continuous function having the absolutely convergent Fourier expansion
$$\begin{eqnarray}\unicode[STIX]{x1D6F7}(\unicode[STIX]{x1D70F})=\mathop{\sum }_{n=-\infty }^{\infty }c_{n}(v)\text{e}(nu),\qquad \unicode[STIX]{x1D70F}=u+\text{i}v.\end{eqnarray}$$
Suppose that:
-
(1) there exists an even
$k>2$
such that
$\unicode[STIX]{x1D6F7}(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F})=j(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D70F})^{k}\unicode[STIX]{x1D6F7}(\unicode[STIX]{x1D70F})$
for all
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{1}$
; and -
(2) there exists a positive
$\unicode[STIX]{x1D716}>0$
such that
$\unicode[STIX]{x1D6F7}(\unicode[STIX]{x1D70F})=O(v^{-\unicode[STIX]{x1D716}})$
(uniformly with respect to
$u$
) as
$v\rightarrow \infty$
.
For such a function
$\unicode[STIX]{x1D6F7}$
, define
$$\begin{eqnarray}\unicode[STIX]{x1D719}(\unicode[STIX]{x1D70F})=\mathop{\sum }_{n=1}^{\infty }a_{n}\text{e}(n\unicode[STIX]{x1D70F}),\qquad a_{n}=\frac{(4\unicode[STIX]{x1D70B}n)^{k-1}}{\unicode[STIX]{x1D6E4}(k-1)}\int _{0}^{\infty }c_{n}(v)\text{e}^{-2\unicode[STIX]{x1D70B}nv}v^{k-2}\,dv.\end{eqnarray}$$
Then
$\unicode[STIX]{x1D719}\in S_{k}(1)$
and
$\unicode[STIX]{x1D719}$
satisfies
$\langle \unicode[STIX]{x1D719},f\rangle _{1}=\langle \unicode[STIX]{x1D6F7},f\rangle _{1}$
for all
$f\in S_{k}(1)$
. We denote
$\unicode[STIX]{x1D719}=\unicode[STIX]{x1D70B}_{\text{hol}}(\unicode[STIX]{x1D6F7})$
, and call it the holomorphic projection of
$\unicode[STIX]{x1D6F7}$
.◻
Remark.
Assumption (2) can be weakened as
$\unicode[STIX]{x1D6F7}(\unicode[STIX]{x1D70F})=c_{0}+O(v^{-\unicode[STIX]{x1D716}})$
by subtracting an Eisenstein series [Reference Zagier40].
When
$1/2<\Re s<k-1-\unicode[STIX]{x1D708}$
,
$s\neq 1$
, the non-holomorphic kernel
$E_{s}^{J}(\unicode[STIX]{x1D70F})$
satisfies the assumptions in Lemma 4 (in view of
$k\geqslant 4$
, Propositions 1, 2 and the following lemmas). The function
${\mathcal{E}}_{s}^{J}:=\unicode[STIX]{x1D70B}_{\text{hol}}(E_{s}^{J})\in S_{k}(1)$
satisfies
$$\begin{eqnarray}\langle {\mathcal{E}}_{s}^{J},f\rangle _{1}=\frac{\unicode[STIX]{x1D6E4}(k-1)}{(4\unicode[STIX]{x1D70B})^{k-1}}\,\cdot \,\frac{L_{2}(s,f,\unicode[STIX]{x1D712})}{L(2s,\unicode[STIX]{x1D712}^{2})}\quad \text{for all }f\in S_{k}(1),\end{eqnarray}$$
and its Fourier expansion is given by
$$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{E}}_{s}^{J}(\unicode[STIX]{x1D70F})=\mathop{\sum }_{n=1}^{\infty }a(n,s)\text{e}(n\unicode[STIX]{x1D70F}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle \qquad a(n,s)=\frac{(4\unicode[STIX]{x1D70B}n)^{k-1}}{\unicode[STIX]{x1D6E4}(k-1)}\int _{0}^{\infty }c_{n}(v,s)\text{e}^{-2\unicode[STIX]{x1D70B}nv}v^{k-2}\,dv. & \displaystyle \nonumber\end{eqnarray}$$
Here the integrand is the infinite sum
$c_{n}(v,s)$
given in Proposition 2. To compute the integral defining
$a(n,s)$
, the following facts are required for justification of the term-by-term integration of the infinite sum
$c_{n}(v,s)$
.
Let us recall the Whittaker function defined for
$v>0$
and
$(\unicode[STIX]{x1D705},\unicode[STIX]{x1D707})\in \mathbf{C}^{2}$
with
$\Re (\unicode[STIX]{x1D707}-\unicode[STIX]{x1D705}+\frac{1}{2})>0$
by (see [Reference Shimura31, Appendix A.3, pp. 131–136], [Reference Lebedev23, Reference Miyake25, Reference Shimura30])
$$\begin{eqnarray}W_{\unicode[STIX]{x1D705},\unicode[STIX]{x1D707}}(v):=\frac{v^{\unicode[STIX]{x1D707}+(1/2)}\text{e}^{-(v/2)}}{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D707}-\unicode[STIX]{x1D705}+{\textstyle \frac{1}{2}})}\int _{0}^{\infty }\text{e}^{-vt}t^{\unicode[STIX]{x1D707}-\unicode[STIX]{x1D705}-(1/2)}(1+t)^{\unicode[STIX]{x1D707}+\unicode[STIX]{x1D705}-(1/2)}\,dt,\end{eqnarray}$$
which is initially defined for
$\Re (\unicode[STIX]{x1D707}-\unicode[STIX]{x1D705}+\frac{1}{2})>0$
and has a holomorphic continuation to all
$(\unicode[STIX]{x1D705},\unicode[STIX]{x1D707})\in \mathbf{C}^{2}$
. Note that
$W_{\unicode[STIX]{x1D705},\unicode[STIX]{x1D707}}(v)=v^{\unicode[STIX]{x1D707}+(1/2)}\text{e}^{-(v/2)}\unicode[STIX]{x1D6F9}(\unicode[STIX]{x1D707}-\unicode[STIX]{x1D705}+\frac{1}{2},2\unicode[STIX]{x1D707}+1;v)$
using the notation of [Reference Lebedev23, §§9.9–9.12]. In the following, we summarize several known facts.
Lemma 5. The function
$\unicode[STIX]{x1D70F}_{l}(v,\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})$
in Propositions 1, 2 is related to
$W_{\unicode[STIX]{x1D705},\unicode[STIX]{x1D707}}(v)$
by
$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D70F}_{l}(v,\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})=\text{i}^{\unicode[STIX]{x1D6FD}-\unicode[STIX]{x1D6FC}}\nonumber\\ \displaystyle & & \displaystyle \quad \cdot \,\left\{\begin{array}{@{}ll@{}}\displaystyle \frac{\unicode[STIX]{x1D70B}^{(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD})/2}l^{((\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD})/2)-1}}{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})}v^{-(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD})/2}W_{(\unicode[STIX]{x1D6FC}-\unicode[STIX]{x1D6FD})/2,(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}-1)/2}(4\unicode[STIX]{x1D70B}lv) & \text{if }l>0,\\ \displaystyle \frac{\unicode[STIX]{x1D70B}^{(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD})/2}|l|^{((\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD})/2)-1}}{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FD})}v^{-(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD})/2}W_{(\unicode[STIX]{x1D6FD}-\unicode[STIX]{x1D6FC})/2,(\unicode[STIX]{x1D6FD}+\unicode[STIX]{x1D6FC}-1)/2}(4\unicode[STIX]{x1D70B}|l|v) & \text{if }l<0,\\ \displaystyle 2^{2-\unicode[STIX]{x1D6FC}-\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D70B}\frac{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}-1)}{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FD})}v^{1-\unicode[STIX]{x1D6FC}-\unicode[STIX]{x1D6FD}} & \text{if }l=0.\end{array}\right.\nonumber\end{eqnarray}$$
Lemma 6. The function
$W_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D707}}(v)$
satisfies
$$\begin{eqnarray}\displaystyle & \displaystyle W_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D707}}(v)\sim v^{\unicode[STIX]{x1D6FC}}\text{e}^{-(v/2)}\quad \text{as }v\rightarrow \infty , & \displaystyle \nonumber\\ \displaystyle & \displaystyle W_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D707}}(v)=\left\{\begin{array}{@{}ll@{}}\displaystyle O(v^{1/2-|\Re \unicode[STIX]{x1D707}|}) & \text{if }\unicode[STIX]{x1D707}\neq 0,\\ \displaystyle O(v^{1/2}|\log \,v|) & \text{if }\unicode[STIX]{x1D707}=0,\end{array}\right.\quad \text{as }v\rightarrow 0. & \displaystyle \nonumber\end{eqnarray}$$
Moreover, for any compact set
$K$
of
$\mathbf{C}^{2}$
, there exist two positive constants
$A$
,
$B>0$
such that
$$\begin{eqnarray}|W_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D707}}(v)|\leqslant Av^{\Re \unicode[STIX]{x1D6FC}}\text{e}^{-(v/2)}(1+v^{-B})\quad (\forall v>0,\forall (\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D707})\in K).\end{eqnarray}$$
Lemma 7. For
$a>0$
,
$\unicode[STIX]{x1D6FD}>0$
,
$\Re (s)>|\Re (\unicode[STIX]{x1D707})|-1/2$
,
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }\text{e}^{-av}W_{\unicode[STIX]{x1D705},\unicode[STIX]{x1D707}}(\unicode[STIX]{x1D6FD}v)v^{s-1}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\unicode[STIX]{x1D6FD}^{\unicode[STIX]{x1D707}+(1/2)}\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D707}+s+{\textstyle \frac{1}{2}})\unicode[STIX]{x1D6E4}(-\unicode[STIX]{x1D707}+s+{\textstyle \frac{1}{2}})}{\unicode[STIX]{x1D6E4}(s-\unicode[STIX]{x1D705}+1)(a+(\unicode[STIX]{x1D6FD}/2))^{\unicode[STIX]{x1D707}+s+(1/2)}}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,F\biggl(\unicode[STIX]{x1D707}+s+\frac{1}{2},\unicode[STIX]{x1D707}-\unicode[STIX]{x1D705}+\frac{1}{2};s-\unicode[STIX]{x1D705}+1;\frac{2a-\unicode[STIX]{x1D6FD}}{2a+\unicode[STIX]{x1D6FD}}\biggr).\nonumber\end{eqnarray}$$
Lemma 8. We have
$F(a,b;c;z)=(1-z)^{-a}F(a,c-b;c;z/(z-1))$
.
Lemma 7 is taken from [Reference Erdelyi, Magnus, Oberhettinger and Tricomi9, 6.9 (8), p. 337] or [Reference Gradshteyn and Ryzhik12, 7.621.3, p. 822], and Lemma 8 is in (22) [Reference Erdelyi, Magnus, Oberhettinger and Tricomi8, p. 64]. The following Lemma 9 is due to Rademacher [Reference Rademacher29, Theorem 3, pp. 198–199].
Lemma 9. Let
$k>1$
be an integer and let
$\unicode[STIX]{x1D712}$
be a primitive Dirichlet character
$\text{mod}\,k$
.
-
(i) Suppose
$\unicode[STIX]{x1D702}>0$
. One has
$|L(s,\unicode[STIX]{x1D712})|\leqslant \unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})$
for
$s=1+\unicode[STIX]{x1D702}+\text{i}t$
with any
$t\in \mathbf{R}$
. -
(ii) Suppose
$0<\unicode[STIX]{x1D702}\leqslant 1/2$
. Then
$|L(s,\unicode[STIX]{x1D712})|\leqslant (k/2\unicode[STIX]{x1D70B})^{\unicode[STIX]{x1D702}+(1/2)}|1+s|^{\unicode[STIX]{x1D702}+(1/2)}\unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})$
for
$s=-\unicode[STIX]{x1D702}+\text{i}t$
with any
$t\in \mathbf{R}$
. -
(iii) Suppose
$0<\unicode[STIX]{x1D702}\leqslant 1/2$
. Then,
$|L(s,\unicode[STIX]{x1D712})|\leqslant (k|1+s|/2\unicode[STIX]{x1D70B})^{(1+\unicode[STIX]{x1D702}-\unicode[STIX]{x1D70E})/2}\unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})$
for all
$s=\unicode[STIX]{x1D70E}+\text{i}t$
in the vertical strip
$-\unicode[STIX]{x1D702}\leqslant \unicode[STIX]{x1D70E}\leqslant 1+\unicode[STIX]{x1D702}$
,
$t\in \mathbf{R}$
.
For any square-free odd natural number
$Q$
, we denote by
$Q^{\ast }:=(-1)^{(Q-1)/2}\cdot Q$
the associated fundamental discriminant. Recall that the conductor of
$\unicode[STIX]{x1D712}$
is denoted by
$N$
, which is assumed to be a square-free odd natural number. Let
$N=ML$
and
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{M^{\ast }}\unicode[STIX]{x1D712}^{\prime }$
, where
$M,L\geqslant 1$
are square-free odd such that
$\unicode[STIX]{x1D712}_{M^{\ast }}:=(M^{\ast }/\cdot )=(\cdot /M)$
is the quadratic part of
$\unicode[STIX]{x1D712}$
and
$\unicode[STIX]{x1D712}^{\prime }$
is the non-quadratic part of
$\unicode[STIX]{x1D712}$
. So, with the previous notation (see Lemma 1),
$\unicode[STIX]{x1D712}_{M^{\ast }}=\prod _{\unicode[STIX]{x1D712}_{j}^{2}=1_{p_{j}}}\unicode[STIX]{x1D712}_{j}$
, where
$1_{p_{j}}$
is the trivial character modulo
$p_{j}$
. For a given discriminant
$D$
, write as
$D=D_{K}f^{2}$
, where
$D_{K}$
is the discriminant of
$K=\mathbf{Q}(\sqrt{D})$
. Put
$M_{K}:=\text{gcd}(M,D_{K})\geqslant 1$
and
$L_{K}:=\text{gcd}(L,D_{K})\geqslant 1$
(the greatest common divisors). Note that
$\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}\unicode[STIX]{x1D712}^{\prime }$
is a character modulo
$|D_{K}|ML/(M_{K}L_{K})$
. Then
$\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}(q)=0$
for any prime
$q\,\mid \,D_{K}M/M_{K}$
, and
$\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}(q)=((D_{K}/M_{K}^{\ast })/q)(q/(M/M_{K}))$
for any prime
$q\,\nmid \,D_{K}M/M_{K}$
. The character
$\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}:=((D_{K}/M_{K}^{\ast })/\cdot )(\cdot /(M/M_{K}))$
is primitive modulo
$|D_{K}|M/M_{K}^{2}$
, and the character
$\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime }$
is primitive modulo
$|D_{K}|ML/(M_{K}^{2}L_{K})$
. Accordingly, we can apply Rademacher’s estimation (Lemma 9) to the right-hand side of the identity
$$\begin{eqnarray}L(s,\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712})=L(s,\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime })\,\cdot \,\mathop{\prod }_{q\,\mid \,M_{K}}(1-\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}(q)\unicode[STIX]{x1D712}^{\prime }(q)q^{-s}).\end{eqnarray}$$
Note that this identity follows from the fact that
$\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}(q)=\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime }(q)$
for any prime
$q\,\nmid \,M_{K}$
and
$\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}(q)=0\neq \widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime }(q)$
for any prime
$q\,\mid \,M_{K}$
.
For any fundamental discriminant
$D_{K}$
and any natural number
$f$
, we define (see (2) for the notation)
$$\begin{eqnarray}\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f):=\mathop{\sum }_{a|f}\unicode[STIX]{x1D707}(a)\unicode[STIX]{x1D712}_{D_{K}}(a)\unicode[STIX]{x1D712}(a)a^{-s}\unicode[STIX]{x1D70E}_{1-2s,\unicode[STIX]{x1D712}^{2}}(f/a).\end{eqnarray}$$
Lemma 10. Let
$N=\prod _{j=1}^{m}p_{j}$
be square-free odd, which is the conductor of the primitive
$\unicode[STIX]{x1D712}$
. Let
$f$
be a natural number having the decomposition
$f=f_{\ast }\,\cdot \,\prod _{j=1}^{m}p_{j}^{l_{j}}$
with
$(N,f_{\ast })=1$
. Then the following hold.
-
(1) We have the relations
$\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f)=\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f_{\ast })$
and
$\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{1-s}(f)f_{\ast }^{1-s}=\unicode[STIX]{x1D712}(f_{\ast })^{2}\,\cdot \,\unicode[STIX]{x1D6F6}_{D_{K},\overline{\unicode[STIX]{x1D712}}}^{s}(f)f_{\ast }^{s}$
. -
(2) The following estimations hold true.
-
(i) Suppose
$\unicode[STIX]{x1D702}>0$
. One has
$|\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f_{\ast })|\leqslant \unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})\unicode[STIX]{x1D701}(1+2\unicode[STIX]{x1D702})$
for
$s=1+\unicode[STIX]{x1D702}+\text{i}t$
with any
$t\in \mathbf{R}$
. -
(ii) Suppose
$0<\unicode[STIX]{x1D702}\leqslant 1/2$
. Then,
$|\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f_{\ast })|\leqslant f_{\ast }^{1+\unicode[STIX]{x1D702}}\unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})\unicode[STIX]{x1D701}(1+2\unicode[STIX]{x1D702})$
for
$s=-\unicode[STIX]{x1D702}+\text{i}t$
with any
$t\in \mathbf{R}$
. -
(iii) Suppose
$0<\unicode[STIX]{x1D702}\leqslant 1/2$
. Then,
$|\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f_{\ast })|\leqslant f_{\ast }^{1+\unicode[STIX]{x1D702}-\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})\unicode[STIX]{x1D701}(1+2\unicode[STIX]{x1D702})$
for all
$s=\unicode[STIX]{x1D70E}+\text{i}t$
in the vertical strip
$-\unicode[STIX]{x1D702}\leqslant \unicode[STIX]{x1D70E}\leqslant 1+\unicode[STIX]{x1D702}$
,
$t\in \mathbf{R}$
.
-
Proof. For the Dirichlet character
$\unicode[STIX]{x1D712}\text{mod}\,N$
and
$f=\prod _{p\,\mid \,f}p^{m_{p}}$
, we have
$\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f)=\prod _{\text{prime}p\,\mid \,f}\unicode[STIX]{x1D716}_{p,\unicode[STIX]{x1D712}}(s)$
, where
$$\begin{eqnarray}\unicode[STIX]{x1D716}_{p,\unicode[STIX]{x1D712}}(s)=\frac{1-(\unicode[STIX]{x1D712}(p)^{2}p^{1-2s})^{m_{p}+1}-\unicode[STIX]{x1D712}_{D_{K}}(p)\unicode[STIX]{x1D712}(p)p^{-s}(1-(\unicode[STIX]{x1D712}(p)^{2}p^{1-2s})^{m_{p}})}{1-\unicode[STIX]{x1D712}(p)^{2}p^{1-2s}}.\end{eqnarray}$$
Note that if
$q\,\nmid \,f$
, then
$m_{q}=0$
, and
$\unicode[STIX]{x1D716}_{q,\unicode[STIX]{x1D712}}(s)=1$
. Note also that if
$p\,\mid \,f$
and
$p\,\mid \,N$
, then
$\unicode[STIX]{x1D712}(p)=0$
and
$\unicode[STIX]{x1D716}_{p,\unicode[STIX]{x1D712}}(s)=1$
. While if
$p\,\mid \,f$
and
$p\,\nmid \,N$
, then
$\unicode[STIX]{x1D716}_{p,\unicode[STIX]{x1D712}}(1-s)=(\unicode[STIX]{x1D712}(p)^{2}p^{2s-1})^{m_{p}}\unicode[STIX]{x1D716}_{p,\overline{\unicode[STIX]{x1D712}}}(s)$
. These remarks prove statement (1).
The statements in (2) follow from elementary considerations apart from (iii). Statement (2)(iii) follows from Rademacher’s Phragmen–Lindelöf theorem [Reference Rademacher29, Theorem 2, p. 195]. ◻
Proposition 3. Let
$r$
,
$n$
be integers and write as
$r^{2}-4nN^{2}=D_{K}f^{2}$
, where
$D_{K}$
is the discriminant of
$K=\mathbf{Q}(\sqrt{r^{2}-4nN^{2}})$
and
$f$
is a natural number. Let
$f=f_{\ast }\,\cdot \,\prod _{j=1}^{m}p_{j}^{l_{j}}$
with
$(N,f_{\ast })=1$
be as in Lemma 10. Put
$N=ML$
as in the above discussion after Lemma 9. For any fixed decompositions
$M=M_{1}M_{2}$
and
$L=L_{1}L_{2}\;(M_{i},L_{i}\in \mathbf{N},i=1,2)$
, assume that
$r$
satisfies
$\text{gcd}(M,D_{K})=M_{1}$
and
$\text{gcd}(L,D_{K})=L_{1}$
. We define
$$\begin{eqnarray}{\mathcal{L}}_{n,r}(s):=L(s,\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime })\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f).\end{eqnarray}$$
-
(1) Suppose that the primitive
$\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime }$
is a non-trivial character modulo
$|D_{K}|ML/(M_{1}^{2}L_{1})$
.-
(i) Let
$\unicode[STIX]{x1D702}>0$
. For all
$s=1+\unicode[STIX]{x1D702}+\text{i}t$
with any
$t\in \mathbf{R}$
,
$$\begin{eqnarray}|{\mathcal{L}}_{n,r}(s)|\leqslant \unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})^{2}\unicode[STIX]{x1D701}(1+2\unicode[STIX]{x1D702}).\end{eqnarray}$$
-
(ii) Let
$0<\unicode[STIX]{x1D702}\leqslant 1/2$
. For all
$s=-\unicode[STIX]{x1D702}+\text{i}t$
with any
$t\in \mathbf{R}$
,
$$\begin{eqnarray}\displaystyle |{\mathcal{L}}_{n,r}(s)| & {\leqslant} & \displaystyle \biggl(\frac{|D_{K}|ML}{2\unicode[STIX]{x1D70B}M_{1}^{2}L_{1}}\biggr)^{\unicode[STIX]{x1D702}+(1/2)}|1+s|^{\unicode[STIX]{x1D702}+(1/2)}\nonumber\\ \displaystyle & & \displaystyle \cdot \,f_{\ast }^{1+\unicode[STIX]{x1D702}}\unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})^{2}\unicode[STIX]{x1D701}(1+2\unicode[STIX]{x1D702}).\nonumber\end{eqnarray}$$
-
(iii) Let
$0<\unicode[STIX]{x1D702}\leqslant 1/2$
. For all
$s=\unicode[STIX]{x1D70E}+\text{i}t$
in the vertical strip
$-\unicode[STIX]{x1D702}\leqslant \unicode[STIX]{x1D70E}\leqslant 1+\unicode[STIX]{x1D702}$
,
$t\in \mathbf{R}$
,
$$\begin{eqnarray}\displaystyle |{\mathcal{L}}_{n,r}(s)| & {\leqslant} & \displaystyle \biggl(\frac{|D_{K}|ML|1+s|}{2\unicode[STIX]{x1D70B}M_{1}^{2}L_{1}}\biggr)^{(1+\unicode[STIX]{x1D702}-\unicode[STIX]{x1D70E})/2}\nonumber\\ \displaystyle & & \displaystyle \cdot \,f_{\ast }^{1+\unicode[STIX]{x1D702}-\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})^{2}\unicode[STIX]{x1D701}(1+2\unicode[STIX]{x1D702}).\nonumber\end{eqnarray}$$
-
-
(2) Suppose that the primitive
$\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime }$
is the trivial character modulo
$|D_{K}|ML/(M_{1}^{2}L_{1})$
. In this case, where
$$\begin{eqnarray}\displaystyle L_{r^{2}-4N^{2}n}(s,\unicode[STIX]{x1D712}) & = & \displaystyle L(s,\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712})\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D712}}^{s}(f)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D701}(s)\cdot \mathop{\prod }_{q\mid M_{1}}(1-q^{-s})\cdot \unicode[STIX]{x1D6F6}_{M_{1}^{\ast },\unicode[STIX]{x1D712}_{M_{1}^{\ast }}}^{s}(f),\nonumber\end{eqnarray}$$
$L_{r^{2}-4N^{2}n}(s,\unicode[STIX]{x1D712})$
is the
$L$
-function defined by (2). By Lemma 10, we can obtain the similar estimations for this
$L_{r^{2}-4N^{2}n}(s,\unicode[STIX]{x1D712})$
under an additional assumption that
$s\neq 1$
. ◻
Proof. Statement (1) follows from Lemmas 9 and 10. To see the identity in statement (2), note that if
$\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime }$
is trivial, then
$|D_{K}|ML/(M_{1}^{2}L_{1})=1$
,
$L=1$
,
$\unicode[STIX]{x1D712}^{\prime }=\unicode[STIX]{x1D712}_{0}$
is principal and
$|D_{K}|=M$
,
$M_{1}=|D_{K}|=M$
. Then apply the identity (**).◻
Lemma 11. Suppose that
$\hat{r}_{j}^{2}-4(N^{2}/p_{j}^{2})n\neq 0$
, where we use the same notation as in Lemmas 2 and 3. If
$\Re (s)=\unicode[STIX]{x1D70E}>1/2$
, then
$|F_{\hat{r}_{j},n}^{j}(\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j})p_{j}^{-s})|\leqslant 1+p_{j}^{1-\unicode[STIX]{x1D70E}}\,\cdot \,(1-p_{j}^{1-2\unicode[STIX]{x1D70E}})^{-1}$
.
Proof. We can deduce the lemma similarly to Lemma 3 by using Lemma 2, since
$0<p_{j}^{1-2\unicode[STIX]{x1D70E}}<1$
for
$\unicode[STIX]{x1D70E}>1/2$
.◻
Fix any
$n\geqslant 1$
. Let
$r$
be any integer satisfying
$r^{2}>4nN^{2}$
. The infinite sum in
$c_{n}(v,s)$
occurs from such integers
$r$
. Using Lemma 5, we evaluate
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }v^{(s-k+1+\unicode[STIX]{x1D708})/2}\biggl|\unicode[STIX]{x1D70F}_{4n-(r^{2}/N^{2})}\biggl(\frac{v}{4},\frac{s+k-\unicode[STIX]{x1D708}}{2},\frac{s-k+1+\unicode[STIX]{x1D708}}{2}\biggr)\biggr|\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\text{e}^{-2\unicode[STIX]{x1D70B}(r^{2}/4N^{2})v}\text{e}^{-2\unicode[STIX]{x1D70B}nv}v^{k-2}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad =K_{1}\biggl|4n-\frac{r^{2}}{N^{2}}\biggr|^{\unicode[STIX]{x1D70E}/2-(3/4)}\int _{0}^{\infty }v^{(k+\unicode[STIX]{x1D708})/2-(3/4)-1}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\biggl|W_{1/4-(k-\unicode[STIX]{x1D708})/2,s/2-(1/4)}\biggl(\unicode[STIX]{x1D70B}\left|4n-\frac{r^{2}}{N^{2}}\right|v\biggr)\biggr|\text{e}^{-\unicode[STIX]{x1D70B}(v/2)(4n+(r^{2}/N^{2}))}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad =K_{1}\biggl|4n-\frac{r^{2}}{N^{2}}\biggr|^{(\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D708}-k)/2}\unicode[STIX]{x1D70B}^{-(k+\unicode[STIX]{x1D708})/2+(3/4)}\int _{0}^{\infty }v^{(k+\unicode[STIX]{x1D708})/2-(3/4)-1}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,|W_{1/4-(k-\unicode[STIX]{x1D708})/2,s/2-(1/4)}(v)|\text{e}^{-\unicode[STIX]{x1D70B}(v/2)(4n+(r^{2}/N^{2}))(r^{2}/N^{2}-4n)^{-1}}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \,K_{1}\biggl|4n-\frac{r^{2}}{N^{2}}\biggr|^{(\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D708}-k)/2}\unicode[STIX]{x1D70B}^{-(k+\unicode[STIX]{x1D708})/2+(3/4)}\int _{0}^{\infty }v^{(k+\unicode[STIX]{x1D708})/2-(3/4)-1}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,|W_{1/4-(k-\unicode[STIX]{x1D708})/2,s/2-(1/4)}(v)|\text{e}^{-\unicode[STIX]{x1D70B}(v/2)}\,dv,\nonumber\end{eqnarray}$$
where
$K_{1}:=2^{\unicode[STIX]{x1D70E}-(1/2)}|\unicode[STIX]{x1D6E4}((s-k+\unicode[STIX]{x1D708}+1)/2)|^{-1}$
. By Lemma 6, the last integral is convergent if
$1/2<\unicode[STIX]{x1D70E}<k+\unicode[STIX]{x1D708}$
, and its value is independent of
$r$
. So, in order to justify the term-by-term integration of the infinite sum
$c_{n}(v,s)$
, we need to check the convergence of
$$\begin{eqnarray}S:=\mathop{\sum }_{r\in \mathbf{Z},r^{2}>4nN^{2}}|r|^{\unicode[STIX]{x1D708}}|A(n,r,s)|(r^{2}-4nN^{2})^{(\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D708}-k)/2}.\end{eqnarray}$$
Let
$r^{2}>4nN^{2}$
and
$r^{2}-4nN^{2}=D_{K}f^{2}$
, as in Proposition 3. Fix any
$N_{1}\,\mid \,N=N_{1}N_{2}$
, any
$M_{1}\,\mid \,M=M_{1}M_{2}$
and any
$L_{1}\,\mid \,L=L_{1}L_{2}$
. It is sufficient to consider each of the partial sums
$S_{N_{1},M_{1},L_{1}}$
of
$S$
over
$r\in \mathbf{Z}$
satisfying
$r^{2}>4nN^{2}$
,
$N_{1}\mid r$
,
$N_{2}\nmid r$
,
$\text{gcd}(M,D_{K})=M_{1}$
and
$\text{gcd}(L,D_{K})=L_{1}$
. We denote this condition for
$r\in \mathbf{Z}$
by
$(C)$
. For such an
$r$
, the identity (**) after Lemma 9 and the explicit form of
$A(n,r,s)$
in Proposition 2 imply that
$$\begin{eqnarray}\displaystyle |A(n,r,s)| & = & \displaystyle \frac{|L_{r^{2}-4N^{2}n}(s,\unicode[STIX]{x1D712})|}{|L(2s,\unicode[STIX]{x1D712}^{2})|}\,\cdot \,\mathop{\prod }_{j=1}^{m}|D_{j}(n,r,s)|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{|{\mathcal{L}}_{n,r}(s)|}{|L(2s,\unicode[STIX]{x1D712}^{2})|}\,\cdot \,\mathop{\prod }_{q\,\mid \,M_{1}}(1+q^{-\unicode[STIX]{x1D70E}})\times N^{1/2}\mathop{\prod }_{p_{j}\,\mid \,N_{1}}|F_{\hat{r}_{j},n}^{j}(\unicode[STIX]{x1D712}_{j}^{\ast }(p_{j})p_{j}^{-s})|.\nonumber\end{eqnarray}$$
Let us prove that the partial series
$S_{N_{1},M_{1},L_{1}}$
defined above is convergent if
$k\geqslant 4$
and
$1/2<\unicode[STIX]{x1D70E}<k-1-\unicode[STIX]{x1D708}$
,
$s\neq 1$
. Since
$1/2<\unicode[STIX]{x1D70E}$
, Lemma 11 yields
$$\begin{eqnarray}|A(n,r,s)|\leqslant \frac{|{\mathcal{L}}_{n,r}(s)|}{|L(2s,\unicode[STIX]{x1D712}^{2})|}\,\cdot \,\!\mathop{\prod }_{q\,\mid \,M_{1}}(1+q^{-\unicode[STIX]{x1D70E}})\times N^{1/2}\!\mathop{\prod }_{p_{j}\,\mid \,N_{1}}\!(1+p_{j}^{1-\unicode[STIX]{x1D70E}}\!\,\cdot \,(1-p_{j}^{1-2\unicode[STIX]{x1D70E}})^{-1}).\end{eqnarray}$$
Hence, our problem is reduced to the convergence of
$$\begin{eqnarray}T_{N_{1},M_{1},L_{1}}:=\mathop{\sum }_{r\in \mathbf{Z}\text{ with }(C)}|r|^{\unicode[STIX]{x1D708}}|{\mathcal{L}}_{n,r}(s)|(r^{2}-4nN^{2})^{(\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D708}-k)/2}.\end{eqnarray}$$
Here the sum is taken over
$r\in \mathbf{Z}$
satisfying the condition
$(C)$
mentioned above. Now decompose
$T_{N_{1},M_{1},L_{1}}=T_{N_{1},M_{1},L_{1}}^{1}+T_{N_{1},M_{1},L_{1}}^{2}$
, where
$T_{N_{1},M_{1},L_{1}}^{1}$
(respectively,
$T_{N_{1},M_{1},L_{1}}^{2}$
) is the partial sum such that
$\widetilde{\unicode[STIX]{x1D712}_{D_{K}}\unicode[STIX]{x1D712}_{M^{\ast }}}\unicode[STIX]{x1D712}^{\prime }$
is non-trivial (respectively, trivial). When
$\unicode[STIX]{x1D70E}>1$
, statement (1)(i) of Proposition 3 combined with
$k-\unicode[STIX]{x1D70E}>1+\unicode[STIX]{x1D708}$
implies that
$$\begin{eqnarray}\displaystyle T_{N_{1},M_{1},L_{1}}^{1} & {\leqslant} & \displaystyle C_{1}\mathop{\sum }_{r\in \mathbf{Z}\text{ with }(C)}|r|^{\unicode[STIX]{x1D708}}(r^{2}-4nN^{2})^{(\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D708}-k)/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2C_{1}\unicode[STIX]{x1D6FC}^{(k+\unicode[STIX]{x1D708}-\unicode[STIX]{x1D70E})/2}\unicode[STIX]{x1D701}(k-\unicode[STIX]{x1D70E})<+\infty ,\nonumber\end{eqnarray}$$
where we put
$C_{1}:=\unicode[STIX]{x1D701}(\unicode[STIX]{x1D70E})^{2}\unicode[STIX]{x1D701}(2\unicode[STIX]{x1D70E}-1)$
, and we take a positive constant
$\unicode[STIX]{x1D6FC}:=\unicode[STIX]{x1D6FC}(n,N)$
such that
$\unicode[STIX]{x1D6FC}(r^{2}-4nN^{2})>r^{2}$
for all
$r\in \mathbf{Z}$
such that
$r^{2}>4nN^{2}$
. When
$-\unicode[STIX]{x1D702}<1/2<\unicode[STIX]{x1D70E}\leqslant 1+\unicode[STIX]{x1D702}$
with any fixed
$0<\unicode[STIX]{x1D702}\leqslant 1/2$
, it follows from statement (1)(iii) of Proposition 3 that
$$\begin{eqnarray}\displaystyle T_{N_{1},M_{1},L_{1}}^{1} & {\leqslant} & \displaystyle C_{2}\mathop{\sum }_{r\in \mathbf{Z}\text{ with }(C)}|r|^{\unicode[STIX]{x1D708}}(D_{K}f_{\ast }^{2})^{(1+\unicode[STIX]{x1D702}-\unicode[STIX]{x1D70E})/2}(r^{2}-4nN^{2})^{(\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D708}-k)/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2C_{2}\unicode[STIX]{x1D6FC}^{\unicode[STIX]{x1D708}/2}\mathop{\sum }_{r\in \mathbf{N}\text{ with }(C)}(D_{K}f_{\ast }^{2})^{(1+\unicode[STIX]{x1D702}-\unicode[STIX]{x1D70E})/2}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(D_{K}f_{\ast }^{2}\,\cdot \,\mathop{\prod }_{j=1}^{m}p_{j}^{2l_{j}}\biggr)^{(\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D708}-k)/2+(\unicode[STIX]{x1D708}/2)}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2C_{2}\unicode[STIX]{x1D6FC}^{\unicode[STIX]{x1D708}/2}\biggl(\mathop{\sum }_{\substack{ D_{K}>0 \\ \text{fundamental discriminant}}}\;D_{K}^{(1+\unicode[STIX]{x1D702}-k)/2}\biggr)\biggl(\mathop{\sum }_{f_{\ast }\geqslant 1}f_{\ast }^{1+\unicode[STIX]{x1D702}-k}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\prod }_{j=1}^{m}\mathop{\sum }_{l_{j}\geqslant 0}p_{j}^{(\unicode[STIX]{x1D70E}-k)l_{j}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2C_{2}\unicode[STIX]{x1D6FC}^{\unicode[STIX]{x1D708}/2}\,\cdot \,\unicode[STIX]{x1D701}((k-\unicode[STIX]{x1D702}-1)/2)\unicode[STIX]{x1D701}(k-\unicode[STIX]{x1D702}-1)\mathop{\prod }_{j=1}^{m}(1-p_{j}^{\unicode[STIX]{x1D70E}-k})^{-1}<+\infty .\nonumber\end{eqnarray}$$
Here we put
$C_{2}:=(ML|1+s|/(2\unicode[STIX]{x1D70B}M_{1}^{2}L_{1}))^{(1+\unicode[STIX]{x1D702}-\unicode[STIX]{x1D70E})/2}\unicode[STIX]{x1D701}(1+\unicode[STIX]{x1D702})^{2}\unicode[STIX]{x1D701}(1+2\unicode[STIX]{x1D702})$
and recall that
$k\geqslant 4$
.
Similar consideration is applicable for
$T_{N_{1},M_{1},L_{1}}^{2}$
whenever
$1/2<\unicode[STIX]{x1D70E}<k-1-\unicode[STIX]{x1D708}$
,
$s\neq 1$
. The term-by-term integration of the infinite sum involved in
$c_{n}(v,s)$
defining
$a(n,s)$
is now justified, and we complete the proof of Theorem 2 using Lemmas 5, 7 and 8.
4 Examples for the special values of
$L$
-functions
4.1 Special functions
First of all, we give preparatory lemmas to compute the Fourier coefficients
$a(n,s)$
for some special points
$s$
in exact closed forms.
Lemma 12. Suppose that
$\unicode[STIX]{x1D708}\in \{0,1\}$
,
$k\geqslant 4$
is even. Let
$1\leqslant s<k-1-\unicode[STIX]{x1D708}$
be a natural number such that
$s\equiv \unicode[STIX]{x1D708}+1\;(\text{mod}\,2)$
. If
$4nN^{2}>r^{2}$
, then
$G_{r}(s)$
in Theorem 2 has the form
$$\begin{eqnarray}\displaystyle G_{r}(s) & = & \displaystyle \text{i}^{\unicode[STIX]{x1D708}}(-1)^{(s+1+\unicode[STIX]{x1D708})/2}2^{s-1}\unicode[STIX]{x1D70B}^{s}n^{(k-s-1)/2}N^{-2s}(4nN^{2}-r^{2})^{s-(1/2)}A(n,r,s)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\unicode[STIX]{x1D6E4}(k-s)\unicode[STIX]{x1D6E4}(s)}{\unicode[STIX]{x1D6E4}(k+s-1)}C_{k-s-1}^{s}\biggl(\frac{r}{2N\sqrt{n}}\biggr).\nonumber\end{eqnarray}$$
Here
$C_{m}^{s}(x)$
is the Gegenbauer polynomial defined by [Reference Temme38, (6.37), p. 152]
$$\begin{eqnarray}C_{m}^{s}(x):=\mathop{\sum }_{j=0}^{\lfloor m/2\rfloor }\frac{(-1)^{j}(s)_{m-j}}{j!(m-2j)!}(2x)^{m-2j},\end{eqnarray}$$
where
$\lfloor \unicode[STIX]{x1D6FC}\rfloor$
denotes the integer such that
$\lfloor \unicode[STIX]{x1D6FC}\rfloor \leqslant \unicode[STIX]{x1D6FC}<\lfloor \unicode[STIX]{x1D6FC}\rfloor +1$
and
$(a)_{0}:=1$
,
$(a)_{n}:=\unicode[STIX]{x1D6E4}(a+n)/\unicode[STIX]{x1D6E4}(a)$
for
$n\geqslant 1$
.
Proof. We have
$$\begin{eqnarray}\displaystyle & & \displaystyle F\biggl(\frac{s+k-1+\unicode[STIX]{x1D708}}{2},\frac{s-k+1+\unicode[STIX]{x1D708}}{2};\unicode[STIX]{x1D708}+\frac{1}{2};\frac{r^{2}}{4nN^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{((k-s-1-\unicode[STIX]{x1D708})/2)!}{(\unicode[STIX]{x1D708}+{\textstyle \frac{1}{2}})_{(k-s-1-\unicode[STIX]{x1D708})/2}}P_{(k-s-1-\unicode[STIX]{x1D708})/2}^{(\unicode[STIX]{x1D708}-(1/2),s-(1/2))}\biggl(1-\frac{r^{2}}{2nN^{2}}\biggr).\nonumber\end{eqnarray}$$
Here
$P_{n}^{(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})}(x)$
is the Jacobi polynomial defined for
$-1<x<1$
,
$\unicode[STIX]{x1D6FC}>-1$
,
$\unicode[STIX]{x1D6FD}>-1$
and a natural number
$n$
by (see [Reference Temme38, (6.35), p. 151])
$$\begin{eqnarray}P_{n}^{(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})}(x):=\frac{(\unicode[STIX]{x1D6FC}+1)_{n}}{n!}F(-n,1+\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}+n;\unicode[STIX]{x1D6FC}+1;(1-x)/2).\end{eqnarray}$$
Lemma 12 follows from the relations (see [Reference Temme38, 6.12, p. 169], [Reference Temme38, (6.37), p. 152], [Reference Temme38, (6.19), p. 143])
$$\begin{eqnarray}\displaystyle P_{2n+\unicode[STIX]{x1D708}}^{(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FC})}(x) & = & \displaystyle (-1)^{n}\frac{\unicode[STIX]{x1D6E4}(2n+\unicode[STIX]{x1D6FC}+1+\unicode[STIX]{x1D708})\unicode[STIX]{x1D6E4}(n+1)}{\unicode[STIX]{x1D6E4}(n+\unicode[STIX]{x1D6FC}+1)\unicode[STIX]{x1D6E4}(2n+1+\unicode[STIX]{x1D708})}x^{\unicode[STIX]{x1D708}}\nonumber\\ \displaystyle & & \displaystyle \times \,P_{n}^{(\unicode[STIX]{x1D708}-(1/2),\unicode[STIX]{x1D6FC})}(1-2x^{2}),\quad \unicode[STIX]{x1D708}=0,1,\nonumber\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle C_{n}^{\unicode[STIX]{x1D706}}(x)=\frac{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D706}+{\textstyle \frac{1}{2}})\unicode[STIX]{x1D6E4}(n+2\unicode[STIX]{x1D706})}{\unicode[STIX]{x1D6E4}(2\unicode[STIX]{x1D706})\unicode[STIX]{x1D6E4}(n+\unicode[STIX]{x1D706}+{\textstyle \frac{1}{2}})}P_{n}^{(\unicode[STIX]{x1D706}-(1/2),\unicode[STIX]{x1D706}-(1/2))}(x), & \displaystyle \nonumber\\ \displaystyle & \displaystyle P_{n}^{(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})}(-x)=(-1)^{n}P_{n}^{(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})}(x) & \displaystyle \nonumber\end{eqnarray}$$
and Legendre’s formula
$2^{2s-1}\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D6E4}(s+1/2)=\unicode[STIX]{x1D70B}^{1/2}\unicode[STIX]{x1D6E4}(2s)$
(see [Reference Temme38, (3.4), p. 46]).◻
Recall that the Gegenbauer polynomial and Jacobi polynomial can be determined from (see [Reference Temme38, (6.45), p. 155], [Reference Temme38, (6.18), p. 143])
$$\begin{eqnarray}\displaystyle & \displaystyle (1-2xt+t^{2})^{-s}=\mathop{\sum }_{m=0}^{\infty }C_{m}^{s}(x)t^{m}\quad (-1<x<1,~|t|<1), & \displaystyle \nonumber\\ \displaystyle & \displaystyle P_{n}^{(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})}(x)=\frac{(-1)^{n}}{2^{n}n!}(1-x)^{-\unicode[STIX]{x1D6FC}}(1+x)^{-\unicode[STIX]{x1D6FD}}\frac{d^{n}}{dx^{n}}\{(1-x)^{\unicode[STIX]{x1D6FC}}(1+x)^{\unicode[STIX]{x1D6FD}}(1-x^{2})^{n}\}. & \displaystyle \nonumber\end{eqnarray}$$
Lemma 13. Let
$D\equiv 0,1\;(\text{mod}\,4)$
and let
$D=D_{K}f^{2}$
, where
$D_{K}$
is the discriminant of
$K=\mathbf{Q}(\sqrt{D})$
and
$f$
is a natural number. For any odd prime
$p$
, put
$p^{\ast }:=(-1)^{(p-1)/2}p$
and
$\unicode[STIX]{x1D712}_{p^{\ast }}:=(\frac{p^{\ast }}{\cdot })$
the Kronecker symbol. Denote by
$m_{p}$
the highest power of
$p$
dividing
$f$
, and by
$\unicode[STIX]{x1D712}_{0}$
the principal character, that is,
$\unicode[STIX]{x1D712}_{0}(m)=1$
for all integers
$m$
. Then
$L_{D}(s,\unicode[STIX]{x1D712})$
defined in (2) satisfies the following statements.
-
(1) If
$p\,\nmid \,D_{K}$
, then
$$\begin{eqnarray}L_{p^{\ast }D}(s,\unicode[STIX]{x1D712}_{0})=L_{D}(s,\unicode[STIX]{x1D712}_{p^{\ast }})\frac{1-p^{(1-2s)(m_{p}+1)}}{1-p^{1-2s}}.\end{eqnarray}$$
If, moreover,
$p\,\mid \,f$
, then
$$\begin{eqnarray}L_{D}(s,\unicode[STIX]{x1D712}_{p^{\ast }})=L_{D/p^{\ast }}(s,\unicode[STIX]{x1D712}_{0})\frac{1-p^{1-2s}}{1-p^{(1-2s)m_{p}}}.\end{eqnarray}$$
-
(2) If
$p\,\mid \,D_{K}$
, then
$$\begin{eqnarray}\displaystyle & & \displaystyle L_{p^{\ast }D}(s,\unicode[STIX]{x1D712}_{0})=L_{D}(s,\unicode[STIX]{x1D712}_{p^{\ast }})\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{1-p^{(1-2s)(m_{p}+2)}-\unicode[STIX]{x1D712}_{D_{K}/p^{\ast }}(p)p^{-s}(1-p^{(1-2s)(m_{p}+1)})}{(1-p^{1-2s})(1-\unicode[STIX]{x1D712}_{D_{K}/p^{\ast }}(p)p^{-s})}.\nonumber\end{eqnarray}$$
Proof. For any quadratic character
$\unicode[STIX]{x1D70C}\;\text{mod}\,M_{\unicode[STIX]{x1D70C}}$
and
$f=\prod _{q|f}q^{m_{q}}$
, we obtain
$$\begin{eqnarray}\unicode[STIX]{x1D6F6}_{D_{K},\unicode[STIX]{x1D70C}}^{s}(f)=\mathop{\prod }_{\substack{ q|f \\ (q,M_{\unicode[STIX]{x1D70C}})=1}}\frac{1-q^{(1-2s)(m_{q}+1)}-\unicode[STIX]{x1D712}_{D_{K}}(q)\unicode[STIX]{x1D70C}(q)q^{-s}(1-q^{(1-2s)m_{q}})}{1-q^{1-2s}}.\end{eqnarray}$$
If
$p\,\nmid \,D_{K}$
, then the number
$p^{\ast }D_{K}$
is a fundamental discriminant,
$p^{\ast }D=(p^{\ast }D_{K})f^{2}$
,
$\unicode[STIX]{x1D712}_{p^{\ast }D_{K}}(q)=\unicode[STIX]{x1D712}_{D_{K}}(q)\unicode[STIX]{x1D712}_{p^{\ast }}(q)$
for any prime
$q$
. If, moreover,
$p\,\mid \,f$
, then
$m_{p}\geqslant 1$
,
$D/p^{\ast }=(D_{K}p^{\ast })(f/p)^{2}$
. While, if
$p\,\mid \,D_{K}$
, the number
$D_{K}/p^{\ast }$
is a fundamental discriminant and
$p^{\ast }D=(D_{K}/p^{\ast })(pf)^{2}$
,
$\unicode[STIX]{x1D712}_{D_{K}/p^{\ast }}(q)=\unicode[STIX]{x1D712}_{D_{K}}(q)\unicode[STIX]{x1D712}_{p^{\ast }}(q)$
for any prime
$q\neq p$
. These remarks and the first equation of statement (1) of Lemma 10 yield Lemma 13.◻
4.2
$N=5$
,
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{5}$
Using Lemmas 12 and 13, we give an exact closed form of the first few of the Fourier coefficients
$a(n,s)$
given in Theorem 2 for special
$\unicode[STIX]{x1D712}$
and
$s$
. We use Cohen’s function
$H(s,N)$
defined in [Reference Cohen3, p. 273]. It is defined for any natural numbers
$N$
and
$s$
such that
$(-1)^{s}N\equiv 0,1\;(\text{mod}\,4)$
by
$$\begin{eqnarray}\displaystyle H(s,N) & := & \displaystyle L(1-s,\unicode[STIX]{x1D712}_{D_{K}})\mathop{\sum }_{a|f}\unicode[STIX]{x1D707}(a)\unicode[STIX]{x1D712}_{D_{K}}(a)a^{s-1}\unicode[STIX]{x1D70E}_{2s-1}(f/a),\nonumber\\ \displaystyle & & \displaystyle \quad (-1)^{s}N=D_{K}f^{2},\qquad K=\mathbf{Q}(\sqrt{(-1)^{s}N})\nonumber\end{eqnarray}$$
$$\begin{eqnarray}L_{(-1)^{s}N}(s,\unicode[STIX]{x1D712}_{0})=(-1)^{\lfloor s/2\rfloor }\frac{2^{s-1}\unicode[STIX]{x1D70B}^{s}}{\unicode[STIX]{x1D6E4}(s)N^{s-(1/2)}}H(s,N),\end{eqnarray}$$
where
$\lfloor \unicode[STIX]{x1D6FC}\rfloor$
denotes the integer such that
$\lfloor \unicode[STIX]{x1D6FC}\rfloor \leqslant \unicode[STIX]{x1D6FC}<\lfloor \unicode[STIX]{x1D6FC}\rfloor +1$
and
$\unicode[STIX]{x1D712}_{0}$
is the principal character.
Proposition 4. Let
$k\geqslant 4$
be even, let
$N=5$
and let
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{5}=(\cdot /5)$
be the even Legendre character. For any odd natural number
$s$
such that
$1\leqslant s\leqslant k-3$
,
$$\begin{eqnarray}\displaystyle a(1,s) & = & \displaystyle 1+2c_{k,s}\,\cdot \,\mathop{\sum }_{r=1}^{9}\unicode[STIX]{x1D712}_{5}(r)C_{k-s-1}^{s}(r/10)H(s,5(100-r^{2}))\nonumber\\ \displaystyle & & \displaystyle +\,2c_{k,s}\,\cdot \,5^{s-1}\,\cdot \,\{-C_{k-s-1}^{s}(0)H(s,20)+2C_{k-s-1}^{s}(1/2)H(s,15)\},\nonumber\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle a(2,s) & = & \displaystyle 2d_{k,s}\,\cdot \,\mathop{\sum }_{r=1}^{14}\unicode[STIX]{x1D712}_{5}(r)C_{k-s-1}^{s}(r/(2\sqrt{50}))H(s,5(200-r^{2}))\nonumber\\ \displaystyle & & \displaystyle +\,4d_{k,s}\,\cdot \,5^{s-1}\,\cdot \,\{-C_{k-s-1}^{s}(0)H(s,40)+C_{k-s-1}^{s}(5/\sqrt{50})H(s,20)\},\nonumber\end{eqnarray}$$
where the constants
$c_{k,s}$
,
$d_{k,s}$
are defined by
$$\begin{eqnarray}c_{k,s}:=(-1)^{s}\frac{2^{2s-2}\unicode[STIX]{x1D70B}^{2s}}{\unicode[STIX]{x1D701}(2s)}\,\cdot \,\frac{5^{-s+1}}{5^{2s}-1}\,\cdot \,\frac{\unicode[STIX]{x1D6E4}(k-s)}{\unicode[STIX]{x1D6E4}(s+k-1)},\qquad d_{k,s}:=2^{(k-s-1)/2}\,\cdot \,c_{k,s}.\end{eqnarray}$$
Proof. First, note that
$s=1$
can be included, and therefore, for any odd
$s$
such that
$1\leqslant s\leqslant k-3$
,
$$\begin{eqnarray}a(1,s)=1+G_{10}(s)+G_{-10}(s)+\mathop{\sum }_{|r|\leqslant 9,5\,\nmid \,r}G_{r}(s)+\mathop{\sum }_{r=0,\pm 5}G_{r}(s).\end{eqnarray}$$
In fact, for
$s\in \mathbf{C}$
with
$1/2<\Re (s)<k-1$
,
$s\neq 1$
, one has
$F_{\pm 2,1}^{1}(5^{-s})=5^{-s}(-1+5^{2-2s})/(1-5^{1-2s})$
by Lemma 3(b). By Proposition 2 combined with
$L(s,\unicode[STIX]{x1D712}_{5}^{2})=(1-5^{-s})\unicode[STIX]{x1D701}(s)$
, we obtain
$$\begin{eqnarray}A(1,\pm 10,s)=\frac{5^{-s+(1/2)}\unicode[STIX]{x1D701}(2s-1)}{(1-5^{-2s})\unicode[STIX]{x1D701}(2s)}(-1+5^{2-2s}).\end{eqnarray}$$
It is then easy to see that
$A(1,r,s)$
is holomorphic at
$s=1$
for any
$r$
. It follows from the formula for
$a(1,s)$
and
$G_{r}(s)$
in Theorem 2 that the point
$s=1$
turned out to be a removable singularity. In view of the factor
$\unicode[STIX]{x1D6E4}((s-k+1)/2)^{-1}$
, we have
$G_{\pm 10}(s)=0$
and
$G_{r}(s)=0$
for
$|r|>10$
when
$s$
is odd and
$1\leqslant s\leqslant k-3$
. We apply Lemma 12 and use the formula of
$A(1,r,s)$
in Proposition 2 to compute
$G_{r}(s)$
for
$|r|<9$
. Note that
$L_{r^{2}-100}(s,\unicode[STIX]{x1D712}_{5})=L_{5(r^{2}-100)}(s,\unicode[STIX]{x1D712}_{0})$
for
$5\,\nmid \,r$
and
$L_{r^{2}-100}(s,\unicode[STIX]{x1D712}_{5})=L_{(r^{2}-100)/5}(s,\unicode[STIX]{x1D712}_{0})$
for
$r=0,\pm 5$
by Lemma 13. Note also that these can be described by Cohen’s function by the identity (4). Using the values
$W(\unicode[STIX]{x1D712}_{5})=\sqrt{5}$
and
$F_{\pm 1,1}^{1}(5^{-s})=2\,\cdot \,5^{-s}$
and
$F_{0,1}^{1}(5^{-s})=-2\,\cdot \,5^{-s}$
from Lemma 3(a)(1), the formulas of
$G_{r}(s)$
for any odd
$s$
such that
$1\leqslant s\leqslant k-3$
are given by
$$\begin{eqnarray}G_{r}(s)=c_{k,s}\,\cdot \,\left\{\begin{array}{@{}ll@{}}\displaystyle \unicode[STIX]{x1D712}_{5}(r)C_{k-s-1}^{s}(r/10)H(s,5(100-r^{2})) & \text{if }|r|\leqslant 9,5\,\nmid \,r,\\ \displaystyle 2\,\cdot \,5^{s-1}C_{k-s-1}^{s}(1/2)H(s,15) & \text{if }r=\pm 5,\\ \displaystyle -2\,\cdot \,5^{s-1}C_{k-s-1}^{s}(0)H(s,20) & \text{if }r=0.\end{array}\right.\end{eqnarray}$$
This proves the claim about
$a(1,s)$
. Similarly, we see that
$$\begin{eqnarray}a(2,s)=\mathop{\sum }_{|r|\leqslant 14,5\,\nmid \,r}G_{r}(s)+\mathop{\sum }_{r=0,\pm 5,\pm 10}G_{r}(s).\end{eqnarray}$$
The required values in the case
$5\,\mid \,r$
are
$F_{0,2}^{1}(5^{-s})=-4\,\cdot \,5^{-s}$
,
$F_{\pm 1,2}^{1}(5^{-s})=0$
and
$F_{\pm 2,2}^{1}(5^{-s})=2\,\cdot \,5^{-s}$
, which follows from Lemma 3(a)
$(1)$
. Analogous to the case
$n=1$
, one has
$$\begin{eqnarray}G_{r}(s)=d_{k,s}\,\cdot \,\left\{\begin{array}{@{}ll@{}}\displaystyle \unicode[STIX]{x1D712}_{5}(r)C_{k-s-1}^{s}(r/(2\sqrt{50}))H(s,5(200-r^{2})) & \text{if }|r|\leqslant 14,5\,\nmid \,r,\\ \displaystyle 2\,\cdot \,5^{s-1}C_{k-s-1}^{s}(5/\sqrt{50})H(s,20) & \text{if }r=\pm 10,\\ \displaystyle 0 & \text{if }r=\pm 5,\\ \displaystyle -4\,\cdot \,5^{s-1}C_{k-s-1}^{s}(0)H(s,40) & \text{if }r=0.\end{array}\right.\end{eqnarray}$$
This completes the proof of the statement about
$a(2,s)$
.◻
Remark.
We can use the command in Mathematica such as NumberFieldDiscriminant to get
$D_{K}$
and
$f=\sqrt{D/D_{K}}$
, NumberFieldClassNumber and DivisorSum to get
$H(s,N)$
, KroneckerSymbol and JacobiSymbol to compute the character sums in
$F_{\hat{r},n}^{j}(T)$
, and FactorInteger, IntegerExponent etc.
Example 1 (
$k=22$
,
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{5}$
,
$s=1$
).
Suppose
$\dim S_{k}(1)=1$
and
$S_{k}(1)=\mathbf{C}f$
, where
$f$
is the normalized Hecke eigenform. We have
${\mathcal{E}}_{s}^{J}=a(1,s)f$
for
$1/2<\Re s<k-1-\unicode[STIX]{x1D708}$
,
$s\neq 1$
, and thus
$$\begin{eqnarray}\frac{L_{2}(s,f,\unicode[STIX]{x1D712})}{\langle f,f\rangle _{1}}=\frac{(4\unicode[STIX]{x1D70B})^{k-1}}{\unicode[STIX]{x1D6E4}(k-1)}L(2s,\unicode[STIX]{x1D712}^{2})\,\cdot \,a(1,s).\end{eqnarray}$$
As mentioned in the proof of Proposition 4, we can put
$s=1$
when
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{5}$
. Note that
$$\begin{eqnarray}C_{10}^{1}(x/2)=-1+15x^{2}-35x^{4}+28x^{6}-9x^{8}+x^{10}.\end{eqnarray}$$
For
$g(r):=H(1,5(100-r^{2}))$
, we have
$g(1)=20$
,
$g(2)=12$
,
$g(3)=20$
,
$g(4)=8$
,
$g(6)=14$
,
$g(7)=12$
,
$g(8)=6$
,
$g(9)=8$
, and
$H(1,20)=2$
,
$H(1,15)=2$
. By Proposition 4, the exact
$L$
-value at
$s=1$
for the unique normalized Hecke eigenform
$f=f_{22}\in S_{22}(1)=\mathbf{C}f_{22}$
is given by
$$\begin{eqnarray}\frac{L_{2}(1,f,\unicode[STIX]{x1D712}_{5})}{\unicode[STIX]{x1D70B}^{23}\langle f,f\rangle _{1}}=\frac{2^{39}\,\cdot \,71}{3^{3}\,\cdot \,5^{22}\,\cdot \,7\,\cdot \,11\,\cdot \,19}.\end{eqnarray}$$
In addition to the data given above,
$h(r):=H(1,5(200-r^{2}))$
takes the values
$h(1)=8$
,
$h(2)=14$
,
$h(3)=4$
,
$h(4)=20$
,
$h(6)=8$
,
$h(7)=12$
,
$h(8)=12$
,
$h(9)=4$
,
$h(11)=8$
,
$h(12)=4$
,
$h(13)=4$
,
$h(14)=2$
, and
$H(1,20)=2$
,
$H(1,40)=2$
. Hence
$a(1,2)=-288a(1,1)$
. This is consistent with the fact that
$S_{22}(1)=\mathbf{C}f_{22}$
and the Fourier expansion
$$\begin{eqnarray}f_{22}=(240G_{4})(-504G_{6})\unicode[STIX]{x1D6E5}=q-288q^{2}-128844q^{3}+\cdots \,.\end{eqnarray}$$
Here
$G_{k}$
is the normalized Eisenstein series of weight
$k$
, and
$\unicode[STIX]{x1D6E5}$
is Ramanujan’s
$\unicode[STIX]{x1D6E5}$
[Reference Zagier40, (4), p. 241].
Example 2 (
$k=24$
,
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{5}$
,
$s=1$
).
It is known that
$\dim S_{24}(1)=2$
. A pair of basis are given by
$$\begin{eqnarray}\displaystyle & \displaystyle h_{1}:=(240G_{4})^{3}\unicode[STIX]{x1D6E5}=q+696q^{2}+162252q^{3}+\cdots \,, & \displaystyle \nonumber\\ \displaystyle & \displaystyle h_{2}:=\unicode[STIX]{x1D6E5}^{2}=0q+q^{2}-48q^{3}+\cdots \,, & \displaystyle \nonumber\end{eqnarray}$$
and the Hecke eigenbasis for
$S_{24}(1)$
are given by (see [Reference Zagier40, Example, p. 258])
$$\begin{eqnarray}f_{1}:=h_{1}+(-156+12\sqrt{144169})h_{2},\qquad f_{2}:=h_{1}+(-156-12\sqrt{144169})h_{2},\end{eqnarray}$$
and thus
$$\begin{eqnarray}\displaystyle & \displaystyle f_{1}-f_{2}=24\sqrt{144169}h_{2},\qquad f_{1}+f_{2}=2h_{1}-312h_{2}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle f_{1}+f_{2}+312h_{2}=2h_{1}. & \displaystyle \nonumber\end{eqnarray}$$
Suppose that
$1/2<\Re s<k-1-\unicode[STIX]{x1D708}$
,
$s\neq 1$
. Consider the expressions
$$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{E}}_{s}^{J}=b_{1}(s)h_{1}+b_{2}(s)h_{2}=c_{1}(s)f_{1}+c_{2}(s)f_{2}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \frac{\unicode[STIX]{x1D6E4}(23)}{(4\unicode[STIX]{x1D70B})^{23}}\,\cdot \,\frac{L_{2}(s,f_{j},\unicode[STIX]{x1D712})}{L(2s,\unicode[STIX]{x1D712}^{2})}=\langle {\mathcal{E}}_{s}^{J},f_{j}\rangle _{1}=c_{\!j}(s)\langle f_{j},f_{j}\rangle _{1}. & \displaystyle \nonumber\end{eqnarray}$$
Then the coefficients
$b_{j}(s)$
can be determined from the relations
$$\begin{eqnarray}a(1,s)=b_{1}(s),\qquad a(2,s)=696b_{1}(s)+b_{2}(s)=696a(1,s)+b_{2}(s),\end{eqnarray}$$
and the coefficients
$c_{\!j}(s)$
are written as
$$\begin{eqnarray}c_{1}(s)=\frac{b_{1}(s)}{2}+\frac{156b_{1}(s)+b_{2}(s)}{24\sqrt{144169}},\qquad c_{2}(s)=\frac{b_{1}(s)}{2}-\frac{156b_{1}(s)+b_{2}(s)}{24\sqrt{144169}}.\end{eqnarray}$$
These quantities have explicit forms by means of Proposition 4. In addition to the data given in Example 1,
$$\begin{eqnarray}\displaystyle C_{22}^{1}(x/2) & = & \displaystyle -1+66x^{2}-715x^{4}+3003x^{6}-6435x^{8}+8008x^{10}-6188x^{12}\nonumber\\ \displaystyle & & \displaystyle +\,3060x^{14}-969x^{16}+190x^{18}-21x^{20}+x^{22}.\nonumber\end{eqnarray}$$
Hence, in the case
$k=24$
,
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{5}$
, we can put
$s=1$
and the exact
$L$
-values for the Hecke eigenbasis
$f_{1},f_{2}$
are given by
$$\begin{eqnarray}\displaystyle & \displaystyle c_{1}(1)=\frac{340568118825984}{2193450927734375}+\frac{21189748684833792}{2193450927734375\sqrt{144169}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \frac{L_{2}(1,f_{1},\unicode[STIX]{x1D712}_{5})}{\unicode[STIX]{x1D70B}^{25}\langle f_{1},f_{1}\rangle _{1}}\,\cdot \,\frac{L_{2}(1,f_{2},\unicode[STIX]{x1D712}_{5})}{\unicode[STIX]{x1D70B}^{25}\langle f_{2},f_{2}\rangle _{1}}=\frac{2^{83}\,\cdot \,109\,\cdot \,54449}{3^{7}\,\cdot \,5^{47}\,\cdot \,7^{2}\,\cdot \,13\,\cdot \,17\,\cdot \,19\,\cdot \,23^{2}\,\cdot \,144169}. & \displaystyle \nonumber\end{eqnarray}$$
In both examples, the prime divisors
$71$
,
$109$
and
$54449$
of the discriminants of the Hecke fields appear in the numerators [Reference Doi, Hida and Ishii6, Reference Goto11, Reference Hiraoka17].
4.3
$N=3$
,
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{-3}$
We give here an analogous result for an odd character case.
Proposition 5. Let
$k\geqslant 6$
be even, let
$N=3$
and let
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{-3}=(\cdot /3)$
be the odd Legendre character. For any even natural number
$s$
such that
$2\leqslant s\leqslant k-4$
,
$$\begin{eqnarray}\displaystyle a(1,s) & = & \displaystyle 1-2c_{k,s}\,\cdot \,\mathop{\sum }_{r=1}^{5}\unicode[STIX]{x1D712}_{-3}(r)C_{k-s-1}^{s}(r/6)H(s,3(36-r^{2}))\nonumber\\ \displaystyle & & \displaystyle -\,2c_{k,s}\,\cdot \,(-1)^{s/2}2^{1-s}\frac{3^{s}-1}{3^{2s-1}-1}3^{s-1}(-3^{3s-2}+3^{2s-1}+3^{s-1}-1)\nonumber\\ \displaystyle & & \displaystyle \cdot \,C_{k-s-1}^{s}(1/2)\frac{\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s)}{\unicode[STIX]{x1D70B}^{s}},\nonumber\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle a(2,s) & = & \displaystyle -2d_{k,s}\,\cdot \,\mathop{\sum }_{r=1}^{8}\unicode[STIX]{x1D712}_{-3}(r)C_{k-s-1}^{s}(\sqrt{2}r/12)H(s,3(72-r^{2}))\nonumber\\ \displaystyle & & \displaystyle +\,2d_{k,s}\,\cdot \,3^{s-1}\,\cdot \,\{\!-2C_{k-s-1}^{s}(\sqrt{2}/4)H(s,21)\nonumber\\ \displaystyle & & \displaystyle +\,2C_{k-s-1}^{s}(1/\sqrt{2})H(s,12)\!\},\nonumber\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle a(3,s) & = & \displaystyle -2e_{k,s}\,\cdot \,\mathop{\sum }_{r=1}^{10}\unicode[STIX]{x1D712}_{-3}(r)C_{k-s-1}^{s}(\sqrt{3}r/18)H(s,3(108-r^{2}))\nonumber\\ \displaystyle & & \displaystyle -\,2e_{k,s}\,\cdot \,3^{s-1}\,\cdot \,\{\!-C_{k-s-1}^{s}(\sqrt{3}/6)H(s,33)\nonumber\\ \displaystyle & & \displaystyle +\,C_{k-s-1}^{s}(1/\sqrt{3})H(s,24)\!\},\nonumber\end{eqnarray}$$
where the constants
$c_{k,s}$
,
$d_{k,s}$
,
$e_{k,s}$
are defined by
$$\begin{eqnarray}\displaystyle & \displaystyle c_{k,s}:=\frac{2^{2s-2}\unicode[STIX]{x1D70B}^{2s}}{\unicode[STIX]{x1D701}(2s)}\,\cdot \,\frac{3^{-s+1}}{3^{2s}-1}\,\cdot \,\frac{\unicode[STIX]{x1D6E4}(k-s)}{\unicode[STIX]{x1D6E4}(s+k-1)},\qquad d_{k,s}:=2^{(k-s-1)/2}\,\cdot \,c_{k,s}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle e_{k,s}:=3^{(k-s-1)/2}\,\cdot \,c_{k,s}. & \displaystyle \nonumber\end{eqnarray}$$
Example 3 (
$k=12$
,
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{-3}$
,
$s=2$
).
The polynomial
$C_{9}^{2}(x/2)$
has the form
$$\begin{eqnarray}C_{9}^{2}(x/2)=30x-140x^{3}+168x^{5}-72x^{7}+10x^{9},\end{eqnarray}$$
and
$g(r):=-120H(2,3(36-r^{2}))$
takes the values
$g(1)=8640$
,
$g(2)=6480$
,
$g(4)=2880$
,
$g(5)=1440$
. In addition,
$h(r):=-120H(2,3(72-r^{2}))$
takes the values
$h(1)=14400$
,
$h(2)=18720$
,
$h(4)=12960$
,
$h(5)=8640$
,
$h(7)=2880$
,
$h(8)=720$
,
$-120H(2,21)=480$
,
$-120H(2,12)=240$
, and
$k(r):=-120H(2,3(108-r^{2}))$
takes the values
$k(1)=47520$
,
$k(2)=33120$
,
$k(4)=31680$
,
$k(5)=33120$
,
$k(7)=18720$
,
$k(8)=10080$
,
$k(10)=720$
, and
$-120H(2,33)=1440$
,
$-120H(2,24)=720$
. By Proposition 5, the Fourier expansion of
${\mathcal{E}}_{2}^{J}$
is
$$\begin{eqnarray}{\mathcal{E}}_{2}^{J}=\frac{2^{8}\,\cdot \,5\,\cdot \,7}{3^{5}\,\cdot \,11}(q-24q^{2}+252q^{3}+\cdots \,)\in S_{12}(1).\end{eqnarray}$$
This tells us the value
$$\begin{eqnarray}\frac{L_{2}(2,f,\unicode[STIX]{x1D712}_{-3})}{\unicode[STIX]{x1D70B}^{15}\langle f,f\rangle _{1}}=\frac{2^{17}}{3^{7}\,\cdot \,5^{3}\,\cdot \,11^{2}},\qquad f=\unicode[STIX]{x1D6E5}\in S_{12}(1)=\mathbf{C}\unicode[STIX]{x1D6E5}.\end{eqnarray}$$
Example 4 (
$k=22$
,
$\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{-3}$
,
$s=2$
).
Similarly, we obtain
$$\begin{eqnarray}\displaystyle C_{19}^{2}(x/2) & = & \displaystyle -110x+1980x^{3}-10296x^{5}+24024x^{7}-30030x^{9}+21840x^{11}\nonumber\\ \displaystyle & & \displaystyle -\,9520x^{13}+2448x^{15}-342x^{17}+20x^{19},\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}{\mathcal{E}}_{2}^{J}=\frac{2^{9}\,\cdot \,13\,\cdot \,17\,\cdot \,67}{3^{12}\,\cdot \,11}(q-288q^{2}-128844q^{3}+\cdots \,)\in S_{22}(1),\end{eqnarray}$$
which tells us the exact value
$$\begin{eqnarray}\frac{L_{2}(2,f,\unicode[STIX]{x1D712}_{-3})}{\unicode[STIX]{x1D70B}^{25}\langle f,f\rangle _{1}}=\frac{2^{28}\,\cdot \,67}{3^{19}\,\cdot \,5^{6}\,\cdot \,7^{3}\,\cdot \,11^{2}\,\cdot \,19},\quad f=f_{22}\in S_{22}(1)=\mathbf{C}f_{22}.\end{eqnarray}$$
We refer the reader to [Reference Doi, Hida and Ishii6, Reference Goto11, Reference Hiraoka17–Reference Katsurada19, Reference Stopple33, Reference Zagier39] for motivation and other examples of exact computations of the special values of
$L$
-functions.
Acknowledgements
The author would like to thank Professor Shin-ichiro Mizumoto for discussions about his paper [Reference Mizumoto26], Professor Kohji Matsumoto for suggestions about estimating
$L(s,\unicode[STIX]{x1D712})$
, Professor Hidenori Katsurada for his preprint [Reference Katsurada19] and the referee for detailed reading and indicating misprints. The second-named author is supported by JSPS Grant-in-Aid for Young Scientists (B) 25800021 and JSPS Grant-in-Aid for Scientific Research (C) 17K05175.
