2.1. Notation

The group ${\mathrm{SL}}_2({\mathbb R})$ acts on the upper half-plane $\mathbb H$ by $\gamma z = \frac {az+b}{cz+d}$ , where $\gamma = {\left (\begin {array}{@{}cc@{}}a&b\\c&d\end {array}\right )} $ and $z=x+iy $ . For a positive integer N, let $\Gamma _0(N)$ denote the congruence subgroup consisting of matrices ${\left (\begin {array}{@{}cc@{}}a&b\\c&d\end {array}\right )}$ in ${\mathrm{SL}}_2({\mathbb Z})$ such that N divides c. For a complex number z, let $e(z)$ denote $e^{2\pi i z}$ .

Let $\theta (z) = \sum _{n = -\infty }^\infty e\left (n^2 z\right )$ be the standard theta function on $\mathbb H$ . If $A = {\left (\begin {array}{@{}cc@{}}a&b\\c&d\end {array}\right )} \in \Gamma _0(4)$ , we have $\theta (Az) = j(A, z)\theta (z)$ , where $j(A, z)$ is the so-called $\theta $ -multiplier. For an explicit formula for $j(A, z)$ , see [Reference Soundararajan62, (1.10)]. Let $S_{k+\frac {1}{2}}(4N)$ denote the space of holomorphic cusp forms of weight $k+\frac {1}{2}$ for the group $\Gamma _0(4N)$ . In other words, a function $f : \mathbb H \rightarrow {\mathbb C}$ belongs to $S_{k+\frac {1}{2}}(4N)$ if

1. i) $f(Az) = j(A, z)^{2k +1} f(z)$ for every $A = {\left (\begin {array}{@{}cc@{}}a&b\\c&d\end {array}\right )} \in \Gamma _0(4N)$ ,

2. ii) f is holomorphic,

3. iii) and f vanishes at the cusps.

Any $f \in S_{k+\frac {1}{2}}(4N)$ has the Fourier expansion

$$ \begin{align*} f(z) = \sum_{n> 0} a(f, n)e(nz). \end{align*} $$

We let $c(f,n)$ denote the ‘normalized’ Fourier coefficients, defined by

$$ \begin{align*} c(f,n) = a(f,n)n^{ \frac14-\frac k2}. \end{align*} $$

For $f, g \in S_{k+\frac 12}(4N)$ , we define the Petersson inner product $\langle f, g\rangle $ by

$$ \begin{align*} \langle f, g\rangle = [{\textrm{SL}}_2({\mathbb Z}): \Gamma_0(4N)]^{-1} \int_{\Gamma_0(N) \backslash \mathbb H}f(z) \overline{g(z)} y^{k + \frac12} \frac{dx dy}{y^2}. \end{align*} $$

2.2. The Kohnen plus space and decomposition into old and newspaces

Fix positive integers $k, N$ such that N is odd and square-free. We recall the definition of the Kohnen plus space $S^+_{k+\frac 12}(4N) \subseteq S_{k+\frac 12}(4N).$ The space $S^+_{k+\frac 12}(4N)$ consists of all forms f in $S_{k+\frac 12}(4N)$ for which $a(f,n) = 0$ whenever $n \equiv (-1)^{k+1}$ or $2 \bmod {4}$ . According to the results of [Reference Kohnen33], there exists a canonically defined subspace $S^{+, \mathrm {new}}_{k+\frac 12}(4N) \subset S^+_{k+\frac 12}(4N)$ and a decomposition

(9) $$ \begin{align} S^+_{k+\frac12}(4N) = \bigoplus_{\substack{r, \ell \ge 1 \\ r \ell\mid N}} S^{+, \mathrm{new}}_{k+\frac12}(4\ell)\mid U\left(r^2\right), \end{align} $$

where we define

(10) $$ \begin{align} f\mid U\left(r^2\right) = r^{\frac12 - k}\sum_{n>0 }a\left(f,r^2n\right)e(nz). \end{align} $$

It is known [Reference Soundararajan62, Proposition 1.5] that if $(r, \ell )=1$ , then $U\left (r^2\right )$ takes $S_{k + \frac 12}(4\ell )$ to $S_{k + \frac 12}(4r\ell )$ . It is also useful to note that

(11) $$ \begin{align} c\left(f\mid U\left(r^2\right), n\right) = c\left(f, r^2 n\right). \end{align} $$

2.3. Hecke operators and the Shimura correspondence

For all primes p coprime to N there exist Hecke operators $T\left (p^2\right )$ acting on the space $S_{k+\frac 12}(4N)$ (see [Reference Soundararajan62, Theorem 1.7]). A newform in $S^{+, \mathrm {new}}_{k+\frac 12}(4N)$ is defined to be an element of $S^{+, \mathrm {new}}_{k+\frac 12}(4N)$ that is an eigenfunction of the Hecke operators $T\left (p^2\right )$ for $p\nmid N$ . The newforms are uniquely determined up to multiplication by nonzero complex numbers and are in fact also eigenforms for the operators $U\left (p^2\right )$ for all $p\mid N$ [Reference Kohnen33, Theorem 2]. The space $S^{+, \mathrm {new}}_{k+\frac 12}(4N)$ has an orthogonal basis consisting of newforms.

According to the Shimura lifting [Reference Soundararajan62] as refined by Kohnen in [Reference Kohnen33], there is an isomorphism

(12) $$ \begin{align} S^{+, \mathrm{new}}_{k+\frac12}(4N) \overset{\simeq}{\rightarrow} S_{2k}^{\mathrm{new}}(N) \end{align} $$

as Hecke modules, where $S_{2k}^{\mathrm {new}}(N)$ is the orthogonal complement of the space of cuspidal oldforms of weight $2k$ for $\Gamma _0(N)$ as defined by Atkin and Lehner [Reference Arthur3]. The Shimura lifting takes each newform in $S^{+, \mathrm {new}}_{k+\frac 12}(4N)$ (as already defined) to a newform (in the sense of Atkin and Lehner [Reference Arthur3, Lemma 18]) in $S_{2k}^{\mathrm {new}}(N)$ with the same Hecke eigenvalues. More precisely, if $f \in S^{+, \mathrm {new}}_{k+\frac 12}(4N)$ is a newform and $g \in S_{2k}^{\mathrm {new}}(N)$ is the Shimura lift of f according to isomorphism (12), then for each prime $p \nmid N$ there exists a real number $\lambda _f(p) \in [-2, 2]$ (by Deligne’s bound for the normalized Hecke eigenvalue) such that

$$ \begin{align*} f\mid T\left(p^2\right) = \lambda_f(p) p^{k - \frac12} f, \qquad g\mid T(p) = \lambda_f(p) p^{k - \frac12} g. \end{align*} $$

In view of decomposition (9) and the fact that the operators $U(p)$ with $p\mid N$ commute with $T\left (p^2\right )$ , $p \nmid N$ , a basis of $S^{+}_{k+\frac 12}(4N)$ consisting of eigenforms for $T\left (p^2\right )$ , $p\nmid N$ , is given by

(13) $$ \begin{align} {\mathcal B}_{k+\frac12,4N} = \bigcup_{\substack{r, \ell \ge 1 \\ r \ell\mid N}} \left\{f\mid U\left(r^2\right): f\in {\mathcal B}^{\mathrm{new}}_{k+\frac12,4\ell}\right\}, \end{align} $$

where ${\mathcal B}^{\mathrm {new}}_{k+\frac 12,4\ell }$ is an orthogonal basis of $S^{+, \mathrm {new}}_{k+\frac 12}(4\ell )$ consisting of newforms. Note, however, that it is not necessarily the case that all members of ${\mathcal B}_{k+\frac 12,4N}$ are orthogonal to each other. The following result will be useful for us; recall the definitions of $\Omega (n)$ and $\omega (n)$ from §1.5:

Lemma 2.1. Let $r, \ell $ be positive, odd, square-free integers with $(r, \ell )=1$ and let $f \in S^{+, \mathrm {new}}_{k+\frac 12}(4\ell )$ be a newform. Then for any odd square-free integer n, putting $d=(-1)^kn$ , we have

$$ \begin{align*} c\left(f\mid U\left(r^2\right), n\right) = c\left(f, r^2n\right) = c(f,n) \prod_{p\mid r}\left(\lambda_f(p) - \frac{1}{\sqrt{p}}\left(\frac{d}{p} \right) \right). \end{align*} $$

Additionally, for any odd integer $r \ge 1$ with $(r,\ell )=1$ , we have

$$ \begin{align*} \left\lvert c\left(f,r^2n\right)\right\rvert \le 3^{\Omega(r)} \lvert c(f,n)\rvert. \end{align*} $$

Proof. The first statement follows from [Reference Soundararajan62, Corollary 1.8(i)]. Using Deligne’s bound $\left \lvert \lambda _f(p)\right \rvert \le 2$ and applying [Reference Soundararajan62, Corollary 1.8(ii)], we will establish the second claim by the following simple induction argument. It suffices to show for each $p\nmid 2\ell $ that

$$ \begin{align*} \left\lvert c\left(f,p^{2m} n\right)\right\rvert \le 3^m \lvert c(f,n)\rvert. \end{align*} $$

The case $m=0$ is trivial, and $m=1$ follows from the first claim of the lemma. By [Reference Soundararajan62, Corollary 1.8(ii)], we have, for any $m \ge 1$ ,

$$ \begin{align*} c\left(f,p^{2m+2}n\right) =\lambda_f(p) c\left(f,p^{2m}n\right)-(-1)^{k\left(p^2-1\right)/2} c\left(f,p^{2m-2} n\right). \end{align*} $$

Hence, for $m \ge 1$ we get

$$ \begin{align*} \left\lvert c\left(f,p^{2m+2}n\right)\right\rvert &\le 2 \left\lvert c\left(f,p^{2m}n\right)\right\rvert+\left\lvert c\left(f,p^{2m-2 }n\right)\right\rvert\\ & \le \left(2 \cdot 3^{m}+3^{m-1}\right) \lvert c(f,n)\rvert \le 3^{m+1} \lvert c(f,n)\rvert. \end{align*} $$

2.4. An explicit version of Waldspurger’s formula

A well-known formula of Waldspurger [Reference Waldspurger67] that was refined and made explicit in special cases by Kohnen [Reference Kohnen34] expresses the squares of Fourier coefficients of half-integral-weight eigenforms in terms of central L-values. We state a version of it here for elements of the basis (13).

Proposition 2.2. Let $r, \ell $ be positive, odd, square-free integers with $(r, \ell )=1$ . Let f be a newform in $S^{+, \mathrm {new}}_{k+\frac 12}(4\ell )$ and let $g \in S_{2k}^{\mathrm {new}}(\ell )$ be the Shimura lift of f. Then for any square-free positive integer n with $(n, 4\ell )=1$ , and $d=(-1)^kn$ , we have

$$ \begin{align*} \frac{\left\lvert c\left(f\mid U\left(r^2\right), n\right)\right\rvert^2}{\langle f, f\rangle} = 2^{\omega(\ell)} \frac{L\left(\frac12, g \otimes \chi_d \right)}{\langle g, g \rangle } \frac{(k-1)!}{\pi^k} \left(\prod_{p\mid r}\left(\lambda_f(p) - \frac{1}{\sqrt{p}}\left(\frac{d}{p} \right) \right)\right)^2, \end{align*} $$

provided that

1. i) $d \equiv 1 \pmod {4}$ and

2. ii) for each prime $p\mid \ell $ , we have $\left (\frac {d}{p}\right )=w_p$ , where $w_p$ is the eigenvalue for the Atkin–Lehner operator at p acting on g.

If either of these two conditions is not met, then $c\left (f\mid U\left (r^2\right ), n\right ) = 0$ .

2.5. Estimates on moments of Fourier coefficients

Proposition 2.3. Let $f \in S_{k+\frac 12}^+(4N)$ , where N is odd and square-free. Then there exists $M\ge 2$ such that for all sufficiently large X,

(14) $$ \begin{align} \sum_{\substack{X \leq n \leq MX \\ (n,2N)=1 }} \lvert c(f,n)\rvert^2 \mu^2(n) \asymp_{f, M} X, \end{align} $$

and for any $\varepsilon>0$ ,

(15) $$ \begin{align} \sum_{\substack{X \leq n \leq 2X \\ (n,2N)=1}} \lvert c(f,n)\rvert^4 \mu^2(n) \ll_{f, \varepsilon} X^{1+\varepsilon}. \end{align} $$

Proof. We first prove the estimate (14). For the upper bound we use the fact that $y^{k+1/2}\lvert f(z)\rvert $ is bounded on $\mathbb H$ , and hence we have

(16) $$ \begin{align} \sum_{\substack{X \le n \le 2X \\ (n,2N)=1}} \lvert c(f,n)\rvert^2 \mu^2(n) \le \sum_{X \le n \le 2X} \lvert c(f,n)\rvert^2 & \ll \frac{1}{X^{k-1/2}} \sum_{n} \lvert c(f, n)\rvert^2 n^{k - \frac12} e^{-4\pi n /X}\nonumber\\ & = \frac{1}{X^{k-1/2}} \int_0^1 \lvert f(x+i/X)\rvert^2 dx \ll_f X. \end{align} $$

For the lower bound, we use a result obtained in the proof of [Reference Saha56, Proposition 3.7], which gives, for any $M \ge 1$ ,

$$ \begin{align*} \sum_{(n,2N)=1} \lvert c(f, n)\rvert^2 \mu^2(n) e^{-n/\left(X\sqrt{M}\right)} \gg_f X\sqrt{M}. \end{align*} $$

Using the bound (16) along with partial summation, we can bound the tail end of the sum as follows:

$$ \begin{align*} \sum_{\substack{n \ge M X\\ (n,2N)=1}} \lvert c(f, n)\rvert^2 \mu^2(n) e^{-n/\left(X\sqrt{M}\right)} \ll_f M e^{-\sqrt{M}} X. \end{align*} $$

Combining the two bounds, we have, for M sufficiently large,

$$ \begin{align*} \sum_{\substack{n \le M X\\ (n,2N)=1}} \lvert c(f,n)\rvert^2 \mu^2(n) & \ge \sum_{\substack{n \le MX\\ (n,2N)=1}} \lvert c(f, n)\rvert^2 \mu^2(n) e^{-n/\left(X\sqrt{M}\right)} \\ &= \sum_{\substack{n \ge 1\\ (n,2N)=1}} \lvert c(f, n)\rvert^2 \mu^2(n) e^{-n/\left(X\sqrt{M}\right)} +O_f \!\left(\! M e^{-\sqrt{M}} X \!\right)\! \gg_f X \sqrt{M}. \end{align*} $$

Finally, we note that by the bound (16) the contribution from terms to the left-hand side here with $n \le X$ is $O_f(X)$ , which completes the proof of the lower bound in the estimate (14).

For the proof of the bound (15), we use equation (13) to reduce to the case $f = f_1\mid U\left (r^2\right )$ , where $f_1 \in S^{+, \mathrm {new}}_{k+\frac 12}(4\ell )$ is a newform with $r\ell \mid N$ . Using Proposition 2.2, it now suffices to prove that

$$ \begin{align*} \sideset{}{^{\flat}}\sum_{\lvert d\rvert\le X} L\left(\tfrac12, g \otimes \chi_d\right)^2 \ll_{g, \varepsilon} X^{1+\varepsilon}, \end{align*} $$

where g is the Shimura lift of $f_1$ and the sum is over fundamental discriminants d. This follows from the approximate functional equation and Heath-Brown’s quadratic large sieve [Reference Henniart25], using a straightforward modification of the proof of [Reference Henniart25, Theorem 2] (see also [Reference Soundararajan and Young65, Corollary 2.5]).

3.1. Statement of main result

Throughout this section, let $k \ge 2$ be an integer and $N \ge 1$ be odd and square-free. The main theorem to be proved in this section is the following:

The main novelty here is that this result holds for all cusp forms $f \in S_{k+1/2}^+(4N)$ , not just Hecke eigenforms, and this is crucial for our later application. Previously it was not apparently even known that there are infinitely many sign changes of $c(f,n)$ as n ranges over square-free integers for $f \in S_{k+\frac 12}^+(4N)$ .

Our proof builds upon the methods developed in [Reference Lester and Radziwiłł39, Reference Matomäki and Radziwiłł42] and relies upon the following two propositions. The first shows that the size of $\lvert c(f,n)\rvert $ is relatively well behaved for most short intervals $[x,x+y]$ .

Proposition 3.2. Let $f \in S_{k+\frac {1}{2}}^+(4N)$ . There exists $M \ge 2$ such that given any $\varepsilon>0$ and $2 \le y \le X/2$ , there are $\gg _{f,M,\varepsilon } X^{1-\frac 32 \varepsilon }$ integers $X \le x \le MX$ such that

$$ \begin{align*} \sum_{\substack{x \leq n \leq x + y \\ (n,2N)=1}} \lvert c(f, n)\rvert \mu^2(n)> \frac{y}{X^{\varepsilon}}. \end{align*} $$

Our other main proposition shows that we can obtain square-root cancelation in sums of $c(f,n)$ over almost all short intervals $[x,x+y]$ .

Proposition 3.3. Let $f \in S_{k+\frac {1}{2}}^+(4N)$ . Then for $1 \leq y \leq X^{\frac {1}{205}}$ , we have

$$ \begin{align*} \sum_{X \le x \le 2X} \left\lvert\sum_{\substack{x \le n \le x+y \\ (n,2N)=1}} c(f,n) \mu^2(n) \right\rvert \ll_f X \sqrt{y}. \end{align*} $$

We will now prove Theorem 3.1 using Propositions 3.2 and 3.3. The proofs of Propositions 3.2 and 3.3 will be given in §§3.2 and 3.3, respectively.

Proof of Theorem 3.1. Observe that if the Fourier coefficients $c(f,n)$ are real and

$$ \begin{align*} \left\lvert \sum_{\substack{x \leq n \leq x + y \\ (n,2N)=1}} c(f,n) \mu^2(n) \right\rvert < \sum_{\substack{x \leq n \leq x + y \\ (n,2N)=1}} \lvert c(f,n)\rvert \mu^2(n), \end{align*} $$

then the interval $[x,x+y]$ must contain a sign change of $c(f,n)$ , where $n\in [x,x+y]$ ranges over odd square-free integers that are coprime to N. We will show that for most integers $X \le x \le MX$ , this inequality holds for intervals of length $y = X^{6 \varepsilon }$ .

By Chebyshev’s inequality, the number of integers $X \le x \le MX$ for which

(17) $$ \begin{align} \left\lvert \sum_{\substack{x \leq n \leq x + y \\ (n,2N)=1}} c(f,n) \mu^2(n) \right\rvert \leq \frac{y}{X^{\varepsilon}} \end{align} $$

does not hold is

$$ \begin{align*} \le \frac{X^{\varepsilon}}{y}\sum_{X \le x \le MX} \left\lvert \sum_{\substack{x \leq n \leq x + y \\ (n,2N)=1}} c(f,n) \mu^2(n) \right\rvert \ll_{f,M} \frac{X^{1+\varepsilon}}{\sqrt{y}}=X^{1-2\varepsilon}, \end{align*} $$

where we have used Proposition 3.3 in the last inequality. By Proposition 3.2, we have

(18) $$ \begin{align} \sum_{\substack{x \leq n \leq x + y \\ (n,2N)=1}} \lvert c(f,n)\rvert \mu^2(n)> \frac{y}{X^{\varepsilon}} \end{align} $$

for all integers $X \le x \le MX$ outside an exceptional set of size $\ll _{f,M, \varepsilon } X^{1 - 3\varepsilon /2}$ . Hence, there exist at least $\gg _{f, M, \varepsilon } X^{1 - 3 \varepsilon /2}$ integers $X \le x \le MX$ such that the bounds (17) and (18) hold. Therefore, we obtain at least $ \gg _{f, M, \varepsilon } \frac {X^{1-3\varepsilon /2}}{y}=X^{1 - 15 \varepsilon /2}$ sign changes of $c(f,n)$ along integers $X \le n \le MX$ that are odd, square-free, and coprime to N.□

3.2. Proof of Proposition 3.2

We first prove the following result, which is an easy consequence of Proposition 2.3:

Lemma 3.4. Let $f \in S_{k+\frac 12}^{+}(4N)$ . Then there exists $M \ge 2$ such that given any $\varepsilon>0$ ,

(19) $$ \begin{align} \sum_{\substack{X \le n \le MX \\ (n,2N)=1 \\ \lvert c(f,n)\rvert \le X^{\varepsilon}}} \lvert c(f,n)\rvert \mu^2(n) \gg_{f,M,\varepsilon} X^{1-\varepsilon/2}. \end{align} $$

Proof. Applying Hölder’s inequality gives

$$ \begin{align*} \sum_{\substack{X \leq n \leq MX \\ (n,2N)=1}} \lvert c(f, n)\rvert^2 \mu^2(n) \le \left( \sum_{\substack{X \leq n \leq MX \\ (n,2N)=1}} \lvert c(f,n)\rvert \mu^2(n) \right)^{2/3} \left( \sum_{\substack{X \leq n \leq MX \\ (n,2N)=1}} \lvert c(f,n)\rvert^4 \mu^2(n) \right)^{1/3}. \end{align*} $$

Hence, using Proposition 2.3 we conclude that

(20) $$ \begin{align} \sum_{\substack{X \le n \le MX \\ (n,2N)=1}} \lvert c(f,n)\rvert \mu^2(n) \gg_{f, M, \varepsilon} X^{1-\varepsilon/2}. \end{align} $$

Also, by the estimate (14) we have

$$ \begin{align*} \sum_{\substack{X \le n \le MX \\ (n,2N)=1 \\ \lvert c(f,n)\rvert> X^{\varepsilon}}} \lvert c(f,n)\rvert \mu^2(n) < X^{-\varepsilon} \sum_{\substack{X \le n \le MX \\ (n,2N)=1}} \lvert c(f,n)\rvert^2 \mu^2(n)\ll_{f,M, \varepsilon} X^{1-\varepsilon}. \end{align*} $$

Combining this with the bound (20) completes the proof.

Proof of Proposition 3.2. Let $C(f,n)=\lvert c(f,n)\rvert \mu ^2(n) 1_{(n,2N)=1} 1_{\lvert c(f,n)\rvert \le X^{\varepsilon }}$ . Applying Lemma 3.4, we see that

$$ \begin{align*} X^{1-\varepsilon/2} \ll_{f,M, \varepsilon} \sum_{\substack{ X+y \le n \le 2MX }} C(f,n) \le \frac{1}{y} \sum_{X \le x \le 2MX} \sum_{\substack{ x \le n \le x+y}} C(f,n) , \end{align*} $$

where the second inequality follows because every term in the sum on the left-hand side is counted $\lfloor y\rfloor +1$ times on the right-hand side. Let $S=\{X \le x \le 2MX : \sum _{x \le n \le x+y} C(f,n) \le y/X^{\varepsilon }\}$ . The contribution to the right-hand side from the integers $x \in S$ is $\ll _{f,M,\varepsilon }\frac {1}{y} \cdot X \frac {y}{X^{\varepsilon }}=X^{1-\varepsilon }$ . Hence we must have

$$ \begin{align*} X^{1-\varepsilon/2} \ll_{f,M,\varepsilon} \frac{1}{y} \sum_{\substack{X \le x \le 2MX \\ x \notin S}}\sum_{\substack{ x \le n \le x+y \\ }} C(f,n) \ll_{f,M, \varepsilon} \frac{1}{y} \cdot y X^{\varepsilon} \sum_{\substack{X \le x \le 2MX \\ x \notin S}} 1, \end{align*} $$

so that $\# \left \{ X \le x \le 2MX : x \notin S\right \} \gg _{f,M, \varepsilon } X^{1-\frac 32 \varepsilon }$ .□

3.3. Proof of Proposition 3.3

Throughout this section, we write

(21) $$ \begin{align} W(u)=u^{\frac{k-1/2}{2}}e^{-2\pi u}. \end{align} $$

The proof of the proposition proceeds directly, beginning with an application of Cauchy–Schwarz. This leads naturally to a shifted convolution sum of Fourier coefficients of f over square-free integers, and to bound this sum we require the following fairly standard result:

Proposition 3.5. Let $f \in S^+_{k+\frac 12}(4N)$ . Then for $1 \leq r\leq X^{\frac {1}{102}}$ , $0 < \lvert h\rvert < X^{\frac {1}{2}}$ , and $v\in \mathbb Z$ with $(v, r) = 1$ , we have, for any given $\varepsilon> 0$ ,

(22) $$ \begin{align} \sum_{n \ge 1} c(f,n) c(f,n+h) e \left ( \frac{n v}{r} \right ) W\left(\frac{n}{X} \right) W\left( \frac{n+h}{X}\right) \ll_{f,\varepsilon} X^{1 - \frac{1}{102} + \varepsilon}. \end{align} $$

Proof. This is an extension of [Reference Lester and Radziwiłł39, Proposition 6.1] to the case of general level, and we will describe how to adapt the arguments given there to this case. The initial step is to use the Fourier expansion of f to express the left-hand side of formula (22) as

$$ \begin{align*} =\frac{1}{X^{k-\frac12}} \int_0^1 f\left(\alpha+\frac{v}{q}+\frac{i}{X} \right) f\left(-\alpha+\frac{i}{X} \right) e(-\alpha h) d\alpha. \end{align*} $$

We now use the circle method following Jutila [Reference Jutila30], as in [Reference Andrianov2 Reference Arthur3, Proposition 2]. An important feature in Jutila’s version of the circle method is that we have freedom over our choice of moduli, which we choose as follows:

$$ \begin{align*} \mathcal Q=\{ Q \le q \le 2Q : q=4Nr p \text{ and } p \equiv 1 \! \! \! \! \pmod 4 \text{ is prime}\}. \end{align*} $$

Write $R=\sum _{q \in \mathcal Q} \varphi (q)$ . Upon applying [Reference Andrianov2 Reference Arthur3, Proposition 2] with $\delta =Q^{-2+\eta }$ , $Q=X^{1/2+2\eta }$ , and $r \le X^{\eta /8}$ , where $\eta>0$ is chosen later, we get that up to a term of size $O_f\left (X^{1-\eta /8+\varepsilon }\right )$ the integral equals

(23) $$ \begin{align} \frac{Q^{2-\eta}}{2 R X^{k - \frac 12}}\sum_{q \in \mathcal{Q}\ } \sideset{}{^*}\sum_{d \! \!\!\! \pmod q} e \!\left(-\tfrac{h d}{q}\right)\! \int_{-\delta}^{\delta} f \left(\tfrac{d}{q}+\tfrac{v}{r}+\alpha+\tfrac{i}{X}\right) f\left(\tfrac{-d}{q}-\alpha+\tfrac{i}{X}\right) e (- \alpha h) d \alpha. \end{align} $$

Notably, to estimate the error term we use the fact that $y^{\frac {k}{2} + \frac 14} \lvert f(z)\rvert $ is bounded on $\mathbb H$ , since f is a cusp form.

Since we have chosen our moduli $q \in \mathcal Q$ such that $4N\mid q$ , we are able to use the modularity of f by applying [Reference Lester and Radziwiłł39, Lemma 6.1], which extends to general level in straightforward way, then once again use the Fourier expansion of f. Consequently, we have transformed the original sum on the left-hand side of formula (22), which is effectively over $n\le X^{1+\varepsilon }$ , to dual sums which are effectively over $m,n \le X^{\varepsilon } Q^2/X $ . The summands in the dual sums include the Fourier coefficients of f twisted by additive characters and factors from the half-integral-weight multiplier system, along with a Kloosterman sum $S(\star , -h;p)$ , where the first argument $\star $ depends on $N,p,m,n,v,r$ . An important observation is that since p is a prime with $0< \lvert h\rvert < p$ , the Weil bound gives $\lvert S(\star , -h;p)\rvert \le 2\sqrt {p}$ for any $\star \in \mathbb Z$ . Using the Weil bound and estimating the dual sums over $m,n$ by applying Cauchy–Schwarz and formula (16) to handle the Fourier coefficients of f, we can show that expression (23) is bounded by

$$ \begin{align*} \ll_{f, \varepsilon} X^{3/4+49\eta/16+\varepsilon}. \end{align*} $$

Recalling our earlier error term of $O\left (X^{1-\eta /8+\varepsilon }\right )$ , which arose from applying Jutila’s circle method, we now take $\eta =4/51$ to complete the proof.

To sum over square-free integers, we will sieve out integers that have a square divisor and require the following estimate for sums of Fourier coefficients:

Lemma 3.6. Let $f \in S_{k+\frac 12}^+(4N)$ . Then

$$ \begin{align*} \sum_{\substack{n \le X \\ (n,2N)=1 \\ d^2 \mid n }} \lvert c(f,n)\rvert \ll_{f,\varepsilon} \frac{X^{1+\varepsilon}}{d^2}. \end{align*} $$

Proof. Just as in the proof of Proposition 2.3, using equation (13) it suffices to consider the case $f=f_1\mid U\left (r^2\right )$ , where $f_1 \in S_{k+\frac 12}^{+,\text {new}}(4\ell )$ is a newform with $r\ell \mid N$ . For $(n,2N)=1$ write $n=s^2m$ , where m is odd and square-free. Also, let $t=rs$ and note $(t,\ell )=1$ , since N is square-free. Applying Lemma 2.1 we have

$$ \begin{align*} c\left(f_1\mid U\left(r^2\right),n\right)=c\left(f_1, t^2 m\right) \ll_\varepsilon t^{\varepsilon} \lvert c(f_1,m)\rvert. \end{align*} $$

Using this bound and then applying Cauchy–Schwarz and the bound (16), we get

$$ \begin{align*} \sum_{\substack{n \le X \\ (n,2N)=1 \\ d^2 \mid n }} \left\lvert c\left(f_1\mid U\left(r^2\right),n\right)\right\rvert \ll_\varepsilon X^{\varepsilon} \sum_{\substack{t \le \sqrt{X} \\ d\mid t}} \sum_{ m \le \frac{X}{t^2} } \lvert c(f_1,m)\rvert \ll_{f, \varepsilon} X^{\varepsilon} \sum_{\substack{t \le \sqrt{X} \\ d\mid t}} \frac{X}{t^2} \ll_\varepsilon \frac{X^{1+\varepsilon}}{d^2}. \end{align*} $$

Proof of Proposition 3.3. To handle the condition that $2n$ be square-free, we first recall that the indicator function of square-free numbers is $\mu ^2(n)=\sum _{d^2\mid n} \mu (d)$ . We then treat the cases of divisors $d\le Y$ and $d>Y$ separately, letting

$$ \begin{align*} \mu^2_{\leq Y}(n) = \sum_{\substack{d^2 \mid n \\ d \leq Y }} \mu(d) , \qquad \mu^2_{> Y}(n) = \sum_{\substack{d^2 \mid n \\ d > Y}} \mu(d). \end{align*} $$

First we consider the large divisors, and get

$$ \begin{align*} \sum_{X \leq x \leq 2X} \left\lvert\sum_{\substack{x \leq n \leq x + y \\ (n,2N)=1}} c(f,n) \mu^2_{>Y }(n) \right\rvert \leq y \sum_{\substack{n \leq 4 X \\ (n,2N)=1}} \left\lvert c(f,n) \mu^2_{> Y}(n)\right\rvert. \end{align*} $$

Using the definition of $\mu ^2_{> Y}(n)$ and applying Lemma 3.6, we see that the right-hand side of this inequality is

(24) $$ \begin{align} \ll y \sum_{\substack{d> Y }} \sum_{\substack{n \leq 4 X \\ d^2 \mid n}} \lvert c(f,n) \rvert \ll_{f, \varepsilon} y X^{1+\varepsilon} \sum_{d >Y} \frac{1}{d^2} \ll y \frac{X^{1+\varepsilon}}{Y}. \end{align} $$

For $ Y \ge \sqrt {y}X^{\varepsilon }$ , this is $\ll X \sqrt {y}$ , as needed.

Next we consider the contribution from the small divisors $d \le Y$ . Let $\widetilde C(f,n)=c(f,n)\mu _{\le Y}^2(n)1_{\left (n,2N\right )=1}$ . Applying Cauchy–Schwarz and using the fact that $W(u)^2 \gg 1$ , for any $u \in [1,2]$ we get

$$ \begin{align*} \sum_{X \le x \le 2 X} \left \lvert \sum_{x \le n \le x+y} \widetilde C(f,n) \right\rvert & \le \sqrt{X} \left( \sum_{X \le x \le 2X} \left \lvert \sum_{x \le n \le x + y} \widetilde C(f,n) \right \rvert^2\right)^{1/2} \\ & \ll \sqrt{X} \left( \sum_{x \ge 1} \left\lvert \sum_{x \le n \le x + y } \widetilde C(f,n) \right \rvert^2 W\left( \frac{x}{X}\right)^2\right)^{1/2}. \end{align*} $$

Assume $y \le X^{1/4}$ . We use the convention that $c(f,n)=0$ if $n \notin \mathbb N$ . To estimate the inner sums on the right-hand side, we expand the square, combine appropriate terms, use the fact that W is a smooth function, and apply the bound (16); the right-hand side then equals

(25) $$ \begin{align} &=\sum_{0 \leq h_1, h_2 \leq y} \sum_{\substack{n \ge 1 }} \widetilde C(f,n+h_1)\widetilde C(f,n+h_2) W\left(\frac{n}{X} \right)^2\nonumber \\ &= \sum_{\substack{\lvert h\rvert \leq y}} \sum_{\substack{n \ge 1}} \widetilde C(f,n) \widetilde C(f,n+h) W\left (\frac{n} {X} \right) W\left ( \frac{n + h}{ X} \right ) \sum_{\substack{0 \le h_1,h_2 \le y \\ h_2-h_1=h}} \left(1+O\left( \frac{y}{n}+\frac{y}{X} \right) \right) + O_f(1) \nonumber \\ &= \sum_{\lvert h\rvert \leq y} (y + 1 - \lvert h\rvert) \sum_{\substack{n \ge 1 }} \widetilde C(f,n) \widetilde C(f,n + h) W \left (\frac{n}{X} \right) W \left ( \frac{n + h}{X} \right ) +O_{f, \varepsilon}\left(y^3 X^{\varepsilon} \right). \end{align} $$

Using the bound (16) once again, we get that the term with $h=0$ in the sum on the right-hand side contributes

(26) $$ \begin{align} \ll y \sum_{n \ge 1} \lvert c(f,n) \rvert^2 W \left ( \frac{n}{X} \right )^2 \ll_{f, \varepsilon} yX. \end{align} $$

We next estimate the contribution from the terms in formula (25) with $h \neq 0$ . Recalling the definition of $\mu ^2_{\leq Y}$ and using the fact that

$$ \begin{align*} 1_{\left(n,2N\right)=1}=\sum_{\substack{d\mid\left(n,2N\right)}} \mu(d)=\sum_{\substack{d\mid 2N \\ d\mid n}} \mu(d), \end{align*} $$

it follows that the contribution to the right-hand side of formula (25) from the terms with $h \neq 0$ is

(27) $$ \begin{align} \ll y \sum_{\substack{0 < \lvert h\rvert \leq y \\ d_1, d_2 \leq Y \\ d_3,d_4 \mid 2N}} \left\lvert \sum_{\substack{d_1^2\mid n, d_2^2 \mid n + h \\ d_3\mid n, d_4\mid n+h}} c(f,n) c(f,n+h) W\left ( \frac{n}{X} \right ) W \left ( \frac{n + h}{X} \right ) \right\rvert. \end{align} $$

For n with $d_1^2 \mid n$ , $d_2^2 \mid n + h$ , $d_3\mid n$ , and $d_4\mid n+h$ , we have $n \equiv a \pmod {r}$ for some $a,r \in \mathbb Z$ with $ r\le 16 N^2 Y^4$ . Using additive characters to detect this congruence, we get that the inner sum in the previous expression is

$$ \begin{align*} & =\sum_{n \equiv a \! \! \! \! \pmod{r}} c(f,n) c(f,n+h) W \left ( \frac{n}{X} \right ) W \left ( \frac{n + h}{X} \right ) \\ &= \frac{1}{r} \sum_{0 \le v < r} e\left(\frac{-av}{r} \right) \sum_{n \ge 1} c(f,n) c(f,n+h)e \left ( \frac{n v}{r} \right ) W \left ( \frac{n}{X} \right ) W \left ( \frac{n + h}{X} \right ). \end{align*} $$

For $v \neq 0$ write $v/r=v'/r'$ , with $(r',v')=1$ , and if $v=0$ , set $r'=1$ . Applying Proposition 3.5, this sum is $\ll _{f, \varepsilon } X^{1-\frac {1}{102}+\varepsilon }$ , provided that $r' \le X^{\frac {1}{102}}$ and $0<\lvert h\rvert < X^{\frac {1}{2}}$ . Hence, by this along with the bound (26) we conclude that for $16N^2 Y^4 \le X^{\frac {1}{102}}$ , the right-hand side of formula (25) is $ \ll _{f, \varepsilon } y X+ y^2 Y^2 X^{1-\frac {1}{102}+\varepsilon }+y^3 X^{\varepsilon }, $ which is $\ll y X$ , as needed, provided that $y \le X^{1/4}$ and

(28) $$ \begin{align} yY^2X^{\varepsilon} \ll X^{\frac{1}{102}}. \end{align} $$

It remains to optimize our parameters. Recall that to handle the contribution of the small divisors, we required $Y \ge \sqrt {y} X^{\varepsilon }$ . We now choose $Y=\sqrt {y} X^{\varepsilon }$ . Taking the constraint (28) into account, the largest we can choose y is $y=X^{\frac {1}{204}-\frac {3}{2}\varepsilon }$ . We conclude by noting that with these choices, we have $16 N^2 Y^4 \ll _N X^{\frac {1}{102}-3\varepsilon } \le X^{\frac {1}{102}}$ , as required for the application of Proposition 3.5.□

4.1. Statement of main result

Theorem 4.1. Let $k \ge 2$ be an integer, $N \ge 1$ be odd and square-free, and $h \in S_{k+\frac 12}^+(4N)$ be a cusp form. Let $\varepsilon>0$ be fixed. Then for all X sufficiently large, there exist at least $X^{1-\varepsilon }$ odd square-free integers n coprime to N such that $X \le n \le 2X$ and

$$ \begin{align*} \lvert c(h,n)\rvert \ge \exp\left(\frac{1}{82} \sqrt{\frac{\log n}{\log \log n}} \right). \end{align*} $$

We will prove this theorem by combining methods of [Reference Gun and Sengupta21] and [Reference Lester and Radziwiłł39, Reference Resnikoff and Saldana50]. Our first job is to reduce the question to bounding central values of L-functions. This is done by using the explicit form of Waldspurger’s formula due to Kohnen (Proposition 2.2).

4.2. Reduction to bounds on L-values

Fix an integer $k \ge 2$ and an odd square-free integer $N\geq 1$ throughout § $4$ . Let h be as in Theorem 4.1. We use the basis (13) to write

$$ \begin{align*} h = \sum_{\substack{r,\ell\geq 1\\ r\ell\mid N}}\sum_{f\in{\mathcal B}_{k+\frac12,4\ell}^{\text{new}}} \alpha_{r,\ell, f} f\mid U\left(r^2\right), \end{align*} $$

where the coefficients $\alpha _{r, \ell , f}$ depend only on $r, \ell , f, h$ . For each odd square-free n, we use Lemma 2.1 to get the following identity for the Fourier coefficients:

$$ \begin{align*} c(h,n)= \sum_{\ell\mid N}\sum_{f\in{\mathcal B}_{k+\frac12,4\ell}^{\text{new}}} c(f,n) \sum_{r\mid \frac{N}{\ell}}\alpha_{r,\ell,f} \prod_{p\mid r}\left(\lambda_f(p) - \frac{1}{\sqrt{p}}\left(\frac{(-1)^kn}{p} \right) \right). \end{align*} $$

We already know that $c(h,n) \neq 0$ for some odd square-free n (this follows from [Reference Schmidt58], for example). So there exist $\ell _0\mid N$ , $f_0\in {\mathcal B}_{k+\frac 12,4\ell _0}^{\mathrm {new}}$ , and a reduced residue class $\eta \bmod 4N$ such that

$$ \begin{align*} \eta \equiv 1 \pmod{4}, \qquad \left(\frac{\eta}p\right)=w_p \quad \text{ for each } p\mid\ell_0, \\ \sum_{r\mid \frac{N}{\ell_0}}\alpha_{r,\ell_0,f_0} \prod_{p\mid r}\left(\lambda_{f_0}(p) - \frac{1}{\sqrt{p}}\left(\frac{\eta}{p} \right) \right) \neq 0. \end{align*} $$

Here, $w_p$ is the eigenvalue of the Atkin–Lehner operator $W_p$ acting on $f_0$ . For brevity, we denote, for each $f \in {\mathcal B}_{k+\frac 12,4\ell }^{\mathrm {new}}$ ,

$$ \begin{align*} \beta_{f} = \sum_{r\mid \frac{N}{\ell}}\alpha_{r,\ell,f} \prod_{p\mid r}\left(\lambda_{f}(p) - \frac{1}{\sqrt{p}}\left(\frac{\eta}{p} \right) \right). \end{align*} $$

We will denote the Shimura lift of $f\in {\mathcal B}_{k+\frac 12,4\ell }^{\mathrm {new}}$ by $g_{f}\in S_{2k}^{\mathrm {new}}(\ell )$ with Fourier coefficients $m^{k - \frac 12}\lambda _{g_f}(m)$ normalized so that $\lambda _{g_f}(1)=1$ . Also, write $g_0$ for $g_{f_0}$ and $m^{k-\frac 12}\lambda _0(m)$ for its Fourier coefficients. For each odd, square-free integer n such that $d=(-1)^k n \equiv \eta \pmod 4N$ , we use the triangle inequality, Cauchy–Schwarz, and Proposition 2.2 to obtain

(29) $$ \begin{align} \lvert c(h, n)\rvert &= \left\lvert\sum_{\ell\mid N}\sum_{f\in{\mathcal B}_{k+\frac12,4\ell}^{\text{new}}} c(f,n) \beta_f\right\rvert \nonumber\\ & \ge \left\lvert\beta_{f_0} c_{f_0}(n)\right\rvert -\sqrt{\left(\sum_{\ell\mid N}\sum_{f_0 \ne f\in{\mathcal B}_{k+\frac12,4\ell}^{\text{new}}} \lvert c(f,n) \rvert^2 \right) \left(\sum_{\ell\mid N}\sum_{f_0 \ne f\in{\mathcal B}_{k+\frac12,4\ell}^{\text{new}}} \left\lvert \beta_f\right\rvert^2 \right)} \nonumber \\ & \ge A \sqrt{L\left(\frac12,g_0\otimes\chi_d\right)} - B \sqrt{\sum_{\ell\mid N}\sum_{f_0 \ne f\in{\mathcal B}_{k+\frac12,4\ell}^{\text{new}}} L\left(\frac12,g_f\otimes\chi_d\right)}, \end{align} $$

where $A>0$ and $B>0$ are independent of d. Now Theorem 4.1 follows from the following auxiliary result:

Proposition 4.2. Let $C\ge 0$ be a constant, $\varepsilon>0$ , and let $\eta \pmod {4N}$ be a fixed reduced residue class with $\eta \equiv 1 \pmod 4$ . Given $\ell _0\mid N$ , let $f_0 \in {\mathcal B}_{k+\frac 12,4\ell _0}^{\mathrm {new}}$ be a newform as before with Shimura lift $g_0$ . For sufficiently large X, there are $\ge X^{1-\varepsilon }$ odd square-free integers $n \in [X, 2X]$ , such that $d=(-1)^kn\equiv \eta \pmod {4N}$ and

(30) $$ \begin{align} L\left(\tfrac12,g_0\otimes\chi_d\right)>C\sum_{\ell\mid N}\sum_{\substack{f\in{\mathcal B}_{k+\frac12,4\ell}^{\mathrm{new}}\\ \left(\ell,f\right)\neq\left(\ell_0,f_0\right)}}L\left(\tfrac12,g_f\otimes\chi_d\right)+ \exp\left(\frac1{40}\sqrt{\frac{\log X}{\log\log X}}\right). \end{align} $$

We first explain how Proposition 4.2 implies Theorem 4.1. Put $L_0 = L\left (\frac 12,g_0\otimes \chi _d\right )$ and $L_1 = \sum _{\ell \mid N}\sum _{f_0 \ne f\in {\mathcal B}_{k+\frac 12,4\ell }^{\mathrm {new}}}L\left (\frac 12,g_f\otimes \chi _d\right )$ . Put $C=2B^2/A^2$ . Combining Proposition 4.2 and the estimate (29),

$$ \begin{align*} \lvert c(h, n)\rvert &\ge A \sqrt{L_0} - B \sqrt{L_1}> A \sqrt{CL_1 + \exp\left(\frac1{40}\sqrt{\frac{\log X}{\log\log X}}\right)} - B \sqrt{L_1} \\ & \ge \frac{A}{\sqrt{2}}\left(\sqrt{CL_1} + \exp\left(\frac1{80}\sqrt{\frac{\log X}{\log\log X}}\right)\right) - B \sqrt{L_1}\\ &= \frac{A}{\sqrt{2}} \exp\left(\frac1{80}\sqrt{\frac{\log X}{\log\log X}}\right) \ge \exp\left(\frac{1}{82} \sqrt{\frac{\log n}{\log \log n}} \right) \end{align*} $$

for sufficiently large X. We now proceed with the proof of Proposition 4.2. For $\eta \equiv 1 \pmod 4$ , let

$$ \begin{align*} \mathcal D_{N,\eta}=\left\{d=(-1)^kn : \mu^2(2n)\neq 0,\,(n,N)=1,\, d\equiv \eta\! \! \! \! \pmod {4N}\right\} \end{align*} $$

and

$$ \begin{align*} \mathcal D_{N,\eta}(X)=\mathcal D_{N,\eta}\cap[X,2X]. \end{align*} $$

For each such d we introduce a resonance polynomial

$$ \begin{align*} R(d)=\sum_{m\leq M}r(m)\lambda_0(m)\chi_d(m), \end{align*} $$

where the coefficients $r(m)$ are multiplicative and supported on square-free integers. At primes we set

$$ \begin{align*} r(p)=\begin{cases} \frac L{\sqrt p \log p}&\text{if }L^2\leq p\leq L^4\text{ and }p\nmid N \\ 0&\text{otherwise}, \end{cases} \end{align*} $$

where $L=\frac 18\sqrt {\log M\log \log M}$ , with $M=X^{1/24}$ .

The strategy to prove the estimate (30) is to consider the quantity

$$ \begin{align*} \sum_{d\in\mathcal D_{N,\eta}(X)}L\left(\tfrac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2. \end{align*} $$

Let $\mathcal S$ be the subset of $\mathcal D_{N,\eta }(X)$ for which the estimate (30) holds. Then certainly

$$ \begin{align*} & \sum_{d\in\mathcal D_{N,\eta}(X)} L\left(\frac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2\\ &\quad \leq\sum_{d\in\mathcal D_{N,\eta}(X)}\left(\sum_{\substack{\ell=1\\ \ell\mid N}}^\infty\sum_{\substack{f\in{\mathcal B}_{k+\frac12,4\ell}^{\mathrm{new}}\\ \left(\ell,f\right)\neq\left(\ell_0,f_0\right)}}CL\left(\frac12,g_f\otimes\chi_d\right)+\exp\left(\frac1{40}\sqrt{\frac{\log X}{\log\log X}}\right)\right)\lvert R(d)\rvert^2\\ &\qquad+\sum_{d\in\mathcal S}L\left(\frac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2. \end{align*} $$

Suppose that the following estimates hold:

(31) $$ \begin{align} \sum_{d\in\mathcal D_{N,\eta}(X)} & L\left(\tfrac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2\gg_{k,N} X\cdot\mathcal R\cdot\exp\left(\left(\frac12+o(1)\right)\frac L{\log L}\right), \end{align} $$

(32) $$ \begin{align} \sum_{d\in\mathcal D_{N,\eta}(X)} & L\left(\tfrac12,g_f\otimes\chi_d\right)\lvert R(d)\rvert^2\ll_{k,N} X\cdot\mathcal R\cdot\exp\left(o\left(\frac L{\log L}\right)\right) \qquad \text{for}\ f\neq f_0, \end{align} $$

(33) $$ \begin{align} \sum_{d\in\mathcal D_{N,\eta}(X)} & \lvert R(d)\rvert^2\leq\frac{2X}{\pi^2}\cdot\mathcal R+O(X), \end{align} $$

(34) $$ \begin{align} \sum_{d\in\mathcal D_{N,\eta}(X)} & \lvert R(d)\rvert^6\ll X\cdot\exp\left(O\left(\frac{\log X}{\log\log X}\right)\right), \end{align} $$

where

$$ \begin{align*} \mathcal R=\prod_{L^2\leq p\leq L^4}\left(1+r(p)^2\lambda_0(p)^2\right). \end{align*} $$

Assuming the estimates (31)–(34), the proof of Proposition 4.2 can be finished as follows. We observe that

$$ \begin{align*} X\cdot\mathcal R\cdot\exp\left(\left(\frac12+o(1)\right)\frac L{\log L}\right)&\ll_{k,N}\sum_{d\in\mathcal D_{N,\eta}(X)}L\left(\tfrac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2\\ &\ll_{k,N} X\cdot\mathcal R\cdot\exp\left(o\left(\frac L{\log L}\right)\right)+\sum_{d\in\mathcal S}L\left(\tfrac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2 \end{align*} $$

by using the estimates (31)–(33). Hence

(35) $$ \begin{align} X\cdot\mathcal R\cdot\exp\left(\left(\frac12+o(1)\right)\frac L{\log L}\right)\ll_{k,N} \sum_{d\in\mathcal S}L\left(\tfrac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2. \end{align} $$

On the other hand, the right-hand side can be estimated by Hölder’s inequality and the estimate (34) as

$$ \begin{align*} &\leq \left(\sum_{d\in\mathcal D_{N,\eta}(X)}L\left(\tfrac12,g_0\otimes\chi_d\right)^2\right)^{1/2}\cdot \lvert \mathcal S\rvert^{1/6}\left(\sum_{d\in\mathcal S}\lvert R(d)\rvert^6\right)^{1/3}\\ &\ll_{k,N} X^{1/2+\varepsilon/2}\lvert\mathcal S\rvert^{1/6}\cdot\left(X\cdot\exp\left(O\left(\frac{\log X}{\log\log X}\right)\right)\right)^{1/3}, \end{align*} $$

where, as before, the average of the squares of central L-values is estimated by using the quadratic large sieve of Heath-Brown [Reference Henniart25] (here we can extend the sum to all fundamental discriminants $\le X$ in magnitude by nonnegativity). Combining this with the estimate (35) gives $\lvert \mathcal S\rvert \gg _{k,N} X^{1-\varepsilon /2} \ge X^{1-\varepsilon }$ , as desired.

So it suffices to establish estimates (31)–(34). Notice that the estimates (33) and (34) follow directly from [Reference Gun and Sengupta21, Proposition 3] by simply estimating

$$ \begin{align*} \sum_{d\in\mathcal D_{N,\eta}(X)}\lvert R(d)\rvert^2 & \leq\sum_{\substack{X\leq (-1)^kd\leq 2X\\ d\equiv 1 \! \! \! \! \pmod{4}}}\lvert R(d)\rvert^2\qquad\text{and}\qquad \\ \sum_{d\in\mathcal D_{N,\eta}(X)}\lvert R(d)\rvert^6 &\leq\sum_{\substack{X\leq (-1)^kd\leq 2X\\ d\equiv 1\! \! \! \! \pmod{4}}}\lvert R(d)\rvert^6. \end{align*} $$

The other two estimates are consequences of the following first moment result.

Proposition 4.3. Let $N\geq 1$ be a positive integer and let $g\in S_{2k}^{\mathrm {new}}(\ell )$ for some $\ell \mid N$ . Suppose that u is an odd positive integer coprime to N and write $u=u_1u_2^2$ , with $u_1$ square-free. Let $\Phi $ denote a smooth and compactly supported function in $[1/2,5/2]$ . Then

$$ \begin{align*} &\sum_{d\in \mathcal D_{N,\eta}}L\left(\tfrac12,g\otimes\chi_d\right)\chi_d(u)\Phi\left(\frac{\lvert d\rvert}X\right)\\ &\quad=\frac{X\lambda_g(u_1)}{2Nu_1^{1/2}}\left(\int_0^\infty\Phi(\xi)\mathrm d \xi\right)L_{g,\eta}\left(\tfrac12\right)L\left(1,\mathrm{Sym}^2g\right)\mathcal G(1;u)+O_{k,N, \Phi, \varepsilon}\left(X^{7/8+\varepsilon}u^{3/8+\varepsilon}\right). \end{align*} $$

Here $L_{g,\eta }(1/2)\neq 0$ is the Dirichlet series given in equation (36) and $\mathcal G(1;u)$ is the Euler product defined in equation (41). Furthermore, $\mathcal G(1;\cdot )$ is a multiplicative function satisfying $\mathcal G\left (1;p^k\right )=1+O(1/p)$ at prime powers.

4.3. A twisted first moment asymptotic

In this subsection, we sketch the proof of Proposition 4.3. The starting point is the approximate functional equation (which follows by an easy modification of the proof of [Reference Resnikoff and Saldana50, Lemma 5]) saying that

$$ \begin{align*} L\left(\frac12,g\otimes\chi_d\right)=2\sum_{\substack{m=1\\ \left(m,2N\right)=1}}^\infty\frac{\lambda_g(m)\chi_d(m)}{\sqrt m}W_{g,\eta}\left(\frac m{\lvert d\rvert}\right), \end{align*} $$

where $W_{g,\eta }$ is a smooth weight function defined by, for $c>1/2$ ,

(36) $$ \begin{align} W_{g,\eta}(\xi)=\frac1{2\pi i}\int_{(c)}L_{g,\eta}\!\left( \! s+\frac12 \!\right)\!\frac{\Gamma(s+k)}{\Gamma(k)}\!{\left(\!\frac{\sqrt \ell}{2\pi\xi}\!\right)\!}^s \frac{\mathrm d s}s, \quad L_{g,\eta}(s)=\sum_{\substack{m=1\\ p\mid m\implies p\mid 2N}}^\infty\frac{\lambda_g(m)\chi_d(m)}{m^s}. \end{align} $$

Notice that the value of $L_{g,\eta }(s)$ is the same for each $d\equiv \eta \pmod {4N}$ . The weight function satisfies $W_{g,\eta }(\xi )=L_{g,\eta }(1/2)+O_{k,N,\varepsilon }\left (\xi ^{\frac 12-\varepsilon }\right )$ as $\xi \longrightarrow 0$ and $W_{g,\eta }(\xi )\ll _{k,N,A}\lvert \xi \rvert ^{-A}$ for any $A\geq 1$ as $\xi \longrightarrow \infty $ . Thus the sum we have to evaluate takes the shape

(37) $$ \begin{align} 2\sum_{\substack{m=1\\ \left(m,2N\right)=1}}^\infty\frac{\lambda_g(m)}{\sqrt m}\sum_{d\in\mathcal D_{N,\eta}}\chi_d(mu)W_{g,\eta}\left(\frac m{\lvert d\rvert}\right)\Phi\left(\frac{\lvert d\rvert}X\right). \end{align} $$

Notice that by definition, any $d\in \mathcal D_{N,\eta }$ is square-free and odd. We pick out this property by the identity

(38) $$ \begin{align} \sum_{\substack{\alpha=1\\ \left(\alpha,2N\right)=1\\ \alpha^2\mid d}}^\infty\mu(\alpha)=\begin{cases} 1 & \text{if }d\text{ is square-free},\\ 0 & \text{otherwise}. \end{cases} \end{align} $$

Note that this identity holds without the condition $(\alpha ,2N)=1$ , but this can be added, as by construction $(d,2N)=1$ for $d\in \mathcal D_{N,\eta }$ . Inserting this into the expression (37) that the d-sum is given by

$$ \begin{align*} \sum_{\substack{\alpha=1\\ \left(\alpha,2Nu\right)=1}}^\infty\mu(\alpha)\left(\frac{\alpha^2}{mu}\right)\sum_{r\equiv\eta\overline{\alpha^2}\pmod{4N}}\left(\frac r{mu}\right)W_{g,\eta}\left(\frac m{r\alpha^2}\right)\Phi\left(\frac{r\alpha^2}X\right). \end{align*} $$

We will evaluate the r-sum by applying a version of the Poisson summation formula [Reference Resnikoff and Saldana50, Lemma 7]. The terms where $mu$ is a square will contribute the main term in the zero-frequency term on the dual side, and the rest will give the error term. Using standard arguments [Reference Lester and Radziwiłł39], the contribution of the latter terms can be bounded by $\ll _{k,N, \Phi , \varepsilon } X^{7/8+\varepsilon }u^{3/8+\varepsilon }$ .

Using equation (36), the zero-frequency contribution is given by

(39) $$ \begin{align} & \frac{X}{2N}\cdot\frac1{2\pi i}\int_{(c)}\left(\int_0^\infty\Phi(\xi)\xi^s\mathrm d \xi\right)L_{g,\eta}\left(s+\tfrac12\right)\frac{\Gamma(s+k)}{\Gamma(k)}\left(\frac{\sqrt \ell X}{2\pi}\right)^s \nonumber\\ &\quad \times \sum_{\substack{\alpha=1\\ \left(\alpha,2Nu\right)=1}}^\infty\frac{\mu(\alpha)}{\alpha^2}\sum_{\substack{m=1\\ \left(m,2N\alpha\right)=1\\ mu \text{ is a square}}}^\infty\frac{\lambda_g(m)}{m^{s+1/2}}\frac{\varphi(mu)}{mu}\frac{\mathrm d s}s. \end{align} $$

A simple computation shows that for $(m,2N)=1$ ,

$$ \begin{align*} \sum_{\substack{\alpha=1\\ \left(\alpha,2Num\right)=1}}^\infty\frac{\mu(\alpha)}{\alpha^2}\cdot\frac{\varphi(mu)}{mu}=\prod_{p\nmid 2N}\left(1-\frac1{p^2}\right)\prod_{p\mid um}\left(1+\frac1p\right)^{-1}, \end{align*} $$

leading to

(40) $$ \begin{align} & \sum_{\substack{\alpha=1\\ \left(\alpha,2Nu\right)=1}}^\infty\frac{\mu(\alpha)}{\alpha^2}\sum_{\substack{m=1\\ \left(m,2N\alpha\right)=1\\ mu \text{ is a square}}}^\infty\frac{\lambda_g(m)}{m^{s+1/2}}\cdot\frac{\varphi(mu)}{mu} \nonumber \\ &\quad =\prod_{p\nmid 2N}\left(1-\frac1{p^2}\right)\sum_{\substack{m=1\\ \left(m,2N\right)=1\\ mu \text{ is a square}}}^\infty\frac{\lambda_g(m)}{m^{s+1/2}}\prod_{p|um}\left(1+\frac1p\right)^{-1}. \end{align} $$

For $mu\text { a square}$ and $u=u_1u_2^2$ with $u_1$ square-free, it follows that $m=u_1\ell ^2$ for some $\ell \in \mathbb Z$ . Hence the right-hand side of the previous display is

$$ \begin{align*} &\prod_{p\nmid 2N} \left(1-\frac1{p^2}\right)\sum_{\substack{\ell=1\\ \left(\ell,2N\right)=1}}^\infty\frac{\lambda_g\left(u_1\ell^2\right)}{u_1^{s+1/2}\ell^{2s+1}}\prod_{p\mid u\ell}\left(1+\frac1p\right)^{-1}\\ &=\frac1{u_1^{s+1/2}}\prod_{p\nmid 2N} \!\left( \! 1-\frac1{p^2} \!\right)\!\prod_{p\nmid 2Nu}\!\left(\! 1+\sum_{k=1}^\infty\frac{\lambda_g \!\left(\! p^{2k} \!\right)}{p^{k(2s+1)}}\!\left(\!1+\frac1p\right)^{-1}\!\right) \\ &\qquad \times \prod_{p\mid u}\!\left(\sum_{k=0}^\infty\frac{\lambda_g\left(p^{2k+\text{ord}_p\left(u_1\right)}\right)}{p^{k(2s+1)}}\!\right)\!\left(\!1+\frac1p \!\right)^{-1}\\ &=\frac1{u_1^{s+1/2}}L\left(2s+1,\text{Sym}^2g\right)\prod_{p\mid 2N}L_p\left(2s+1,\text{Sym}^2h\right)^{-1}\prod_{p\mid u}\left(1-\frac{1}p\right)\left(\sum_{k=0}^\infty\frac{\lambda_g\left(p^{2k+\text{ord}_p\left(u_1\right)}\right)}{p^{k(2s+1)}}\right)\\ &\quad\cdot L_p\left(2s+1,\text{Sym}^2g\right)^{-1}\prod_{p\nmid 2Nu}\!\left(\! 1-\frac1{p^2}\!\right)\!\left(\! 1+\sum_{k=1}^\infty\frac{\lambda_g(p^{2k})}{p^{k(2s+1)}}{\!\left(\! 1+\frac1p \!\right)\!}^{-1} \!\right)\! L_p\left(2s+1,\text{Sym}^2g\right)^{-1}. \end{align*} $$

From this it is easy to see, by using the Euler product expression of the symmetric square L-function, that for $p\nmid 2Nu$ the corresponding Euler factor is

$$ \begin{align*} =\left(1-\frac1p\right)\left(1-\frac1{p^{2s+1}}\right)\left(\frac1p\left(1-\frac{\alpha_p^2}{p^{2s+1}}\right)\left(1-\frac{\beta_p^2}{p^{2s+1}}\right)+1+\frac1{p^{2s+1}}\right), \end{align*} $$

where $\left \{\alpha _p,\beta _p\right \}$ are the Satake parameters of the cusp form g at p.

Similarly, for $p\mid u, p\nmid u_1$ , the corresponding Euler factor is $\left (1-\frac {1}{p}\right )\left (1-\frac 1{p^{4s+2}}\right )$ . For $p\mid u_1$ , the corresponding Euler factor is $\left (1-\frac {1}{p}\right )\left (1-\frac 1{p^{2s+1}}\right )\lambda _g(p)$ by using the relations $\lambda _g\left (p^j\right )=\left (\alpha _p^{j+1}-\beta _p^{j+1}\right )/\left (\alpha _p-\beta _p\right )$ for $j\geq 0$ and $\alpha _p\beta _p=1$ . And finally, for $p\mid 2N$ the corresponding Euler factor is clearly $L_p\left (2s+1,\text {Sym}^2g\right )^{-1}$ .

To summarize, the right-hand side of equation (40) equals

$$ \begin{align*} \frac{\lambda_g(u_1)}{u_1^{s+1/2}}L\left(2s+1,\text{Sym}^2g\right)\mathcal G(2s+1;u), \end{align*} $$

where $\mathcal G(2s+1;u)=\prod _p \mathcal G_p(2s+1;u)$ is the Euler product locally given by

(41) $$ \begin{align} &\mathcal G_p(2s+1;u)\nonumber\\&\quad =\begin{cases} L_p\left(2s+1,\text{Sym}^2 g\right)^{-1}&\text{if }p\mid 2N,\\ \left(1-\frac1{p}\right)\left(1-\frac1{p^{2s+1}}\right)&\text{if }p\mid u_1,\\ \left(1-\frac1{p}\right)\left(1-\frac1{p^{4s+2}}\right)&\text{if }p\mid u_2,\ p\nmid u_1,\\ \left(1-\frac1p\right)\left(1-\frac1{p^{2s+1}}\right)\left(1+\frac1p\left(1-\frac{\alpha_p^2}{p^{2s+1}}\right)\left(1-\frac{\beta_p^2}{p^{2s+1}}\right)+\frac1{p^{2s+1}}\right)&\text{if }p\nmid 2Nu. \end{cases} \end{align} $$

By estimating trivially, it follows that $\mathcal G(2s+1;u)$ extends analytically to the domain $\mathrm {Re}(s)>-1/4$ and is bounded there by

$$ \begin{align*} \ll\prod_{p\nmid 2Nu}\left(1+\frac{O(1)}{\sqrt p}\right)\ll_{N,\varepsilon} u^\varepsilon. \end{align*} $$

Consequently, the s-integrand in the expression (39) extends to an analytic function the domain $\mathrm {Re}(s)>-1/4$ (apart from a simple pole at $s=0$ ). Thus moving the line of integration in formula (39) to the line $\mathrm {Re}(s)=-1/4+\varepsilon $ shows that the expression equals

$$ \begin{align*} \frac{X\lambda_g(u_1)}{2Nu_1^{1/2}}\left(\int_0^\infty\Phi(\xi)\mathrm d \xi\right)L_{g,\eta}\left(\frac12\right)L\left(1,\text{Sym}^2g\right)\mathcal G(1;u)+O_{k,N, \Phi, \varepsilon}\left(X^{3/4+\varepsilon}\right), \end{align*} $$

where the main term comes from the residue at $s=0$ and the error term from the contour shift. It follows immediately from the definition of $\mathcal G(s;u)$ that $\mathcal G(1;\cdot )$ is multiplicative and that $\mathcal G\left (1;p^k\right )=1+O(1/p)$ at prime powers. This concludes the sketch of the proof of Proposition 4.3.

4.4. Proofs of the estimates

We are now ready to prove the estimates (31) and (32). As the arguments are similar to [Reference Gun and Sengupta21], we will be brief. Let us denote

$$ \begin{align*} A_{g,\eta}(\Phi)=\frac1{2N}\left(\int_0^\infty\Phi(\xi)\mathrm d \xi\right)L_{g,\eta}\left(\frac12\right)L\left(1,\text{Sym}^2g\right) \end{align*} $$

and

$$ \begin{align*} B_g(u)=\frac{\lambda_g\left(u_1\right)}{u_1^{1/2}}\mathcal G(1;u) \end{align*} $$

for $u=u_1u_2^2$ with $u_1$ square-free.

Let $\Phi $ be a compactly supported smooth weight function. Our aim is now to evaluate

(42) $$ \begin{align} \sum_{d\in\mathcal D_{N,\eta}}L\left(\tfrac12,g\otimes\chi_d\right)\lvert R(d)\rvert^2\Phi\left(\frac{\lvert d\rvert}X\right) \end{align} $$

for $g=g_f\in S_{2k}^{\mathrm {new}}(\ell )$ , where $f\in {\mathcal B}_{k+\frac 12,4\ell }^{\mathrm {new}}$ with $\ell \mid N$ , by choosing $\Phi $ appropriately for given f.

By opening the definition of $R(d)$ and using Proposition 4.3, the sum (42) equals

$$ \begin{align*} \sum_{d\in\mathcal D_{N,\eta}} & L\left(\tfrac12,g\otimes\chi_d\right)\Phi\left(\frac{\lvert d\rvert}X\right)\sum_{n_1,n_2\leq M}r(n_1)r(n_2)\lambda_0(n_1)\lambda_0(n_2)\chi_d(n_1n_2)\\ &=X\cdot A_{g,\eta}(\Phi)\sum_{n_1,n_2\leq M}r(n_1)r(n_2)\lambda_0(n_1)\lambda_0(n_2)B_g\left(n_1n_2\right)\\ &\quad +O_{k,N,\Phi,\varepsilon}\left(X^{7/8+\varepsilon}\sum_{n_1,n_2\leq M}r(n_1)r(n_2)\lvert\lambda_0(n_1)\lambda_0(n_2)\rvert(n_1n_2)^{3/8+\varepsilon}\right) \end{align*} $$

as $r(n)$ vanishes, unless $(n,N)=1$ . Using Deligne’s bound, we obtain as in [Reference Gun and Sengupta21, §6] that the error term is $\ll _{k,N,\Phi } X^{99/100}$ .

By making use of the fact that

$$ \begin{align*} n_1n_2=\frac{n_1n_2}{(n_1,n_2)^2}\cdot(n_1,n_2)^2, \end{align*} $$

with the first factor on the right-hand side square-free in our case, as the function $r(n)$ is supported only on square-free integers, the main term can be written as

$$ \begin{align*} X\cdot A_{g,\eta}(\Phi)\sum_{n_1,n_2\leq M}r(n_1)r(n_2)\lambda_0(n_1)\lambda_0(n_2)\frac{\lambda_g\left(\frac{n_1n_2}{(n_1,n_2)^2}\right)}{\left(\frac{n_1n_2}{(n_1,n_2)^2}\right)^{1/2}}\cdot\mathcal G(1;n_1n_2). \end{align*} $$

Our aim is to use multiplicativity, and so we need to extend the sum over all integers. To do so we must show that the terms with max( $n_1,n_2)>M$ can be added with a tolerable error. Using Rankin’s trick, these terms contribute

$$ \begin{align*} &\ll_{k,N} X\sum_{\max\left(n_1,n_2\right)>M}r(n_1)r(n_2)\lvert\lambda_0(n_1)\lambda_0(n_2)\rvert\cdot\lvert\mathcal G(1;n_1n_2)\rvert\cdot\left\lvert\frac{\lambda_g\left(\frac{n_1n_2}{(n_1,n_2)^2}\right)}{\left(\frac{n_1n_2}{(n_1,n_2)^2}\right)^{1/2}}\right\rvert\\ &\ll_{k,N} X\sum_{n_1,n_2=1}^\infty r(n_1)r(n_2)\lvert\lambda_0(n_1)\lambda_0(n_2)\rvert\cdot\lvert\mathcal G(1;n_1n_2)\rvert\cdot\left\lvert\frac{\lambda_g\left(\frac{n_1n_2}{(n_1,n_2)^2}\right)}{\left(\frac{n_1n_2}{(n_1,n_2)^2}\right)^{1/2}}\right\rvert\left(\frac{n_1n_2}M\right)^\alpha \end{align*} $$

for any $\alpha>0$ , which is chosen optimally later.

Write $n=n_1n_2$ to express the double sum as a single sum over n, and note that by the fact that $r(\cdot )$ is supported only on square-free integers coprime to N, the only integers n which contribute to the sum over n satisfy $p^3 \nmid n$ and $(n,N)=1$ . Hence, by the multiplicativity of $r(\cdot )$ we can express the sum over n as an Euler product, and the expression on the right-hand side of the previous display equals

$$ \begin{align*} &= XM^{-\alpha}\prod_{L^2\leq p\leq L^4}\left(1+2r(p)p^{\alpha-1/2}\lvert\mathcal G(1;p)\rvert\cdot\frac{\left\lvert\lambda_0(p)\lambda_g(p)\right\rvert}{\sqrt p}+r(p)^2p^{2\alpha}\left\lvert\mathcal G\left(1;p^2\right)\right\rvert\cdot\lvert\lambda_0(p)\rvert^2\right)\\ &\ll XM^{-\alpha}\exp\left(\sum_{L^2\leq p\leq L^4} \left(\frac{8Lp^{\alpha-1}}{\log p}+\frac{4L^2p^{2\alpha-1}}{(\log p)^2}\right) \left( 1+O\left( \frac1p\right)\right)\right), \end{align*} $$

where the last estimate follows from Deligne’s bound $\left \lvert \lambda _g(p)\right \rvert \leq 2$ , the fact that $\mathcal G\left (1;p^k\right )=1+O(1/p)$ , and the definition of $r(n)$ .

Let us now choose $\alpha =1/(8\log L)$ . Then by the prime number theorem and partial summation, the previous is

$$ \begin{align*} \ll X\cdot\exp\left(-\frac{\log M}{8\log L}+\frac{8L}{\log L}+\frac{4L^2}{(\log L)^2}\right)\ll X \end{align*} $$

by the choices $L=\frac 18\sqrt {\log M\log \log M}$ and $M=X^{1/24}$ . From the foregoing arguments we deduce that

(43) $$ \begin{align} &\sum_{d\in\mathcal D_{N,\eta}} L\left(\frac12,g\otimes\chi_d\right)\lvert R(d)\rvert^2\Phi\left(\frac{\lvert d\rvert}X\right)\\ &\quad = A_{g,\eta}(\Phi)\cdot X\cdot\prod_{L^2\leq p\leq L^4}\!\left(\! 1+2r(p)\mathcal G(1;p)\cdot\frac{\lambda_0(p)\lambda_g(p)}{\sqrt p}+r(p)^2\mathcal G(1;p^2)\cdot \lambda_0(p)^2 \!\right) \nonumber \\ &\qquad +O_{k,N,\Phi}(X) \nonumber \\ &\quad =A_{g,\eta}(\Phi)\cdot X\cdot\mathcal R\cdot\exp \!\left(\! 2\sum_{L^2\leq p\leq L^4}\frac{r(p)\lambda_0(p)\lambda_g(p)}{\sqrt p}+O_{k,N,\Phi} \!\left(\frac L{(\log L)^3} \!\right)\!\right)\! + O_{k,N,\Phi}(X), \nonumber \end{align} $$

where the last step follows exactly as in [Reference Gun and Sengupta21, §6].

We now apply this result with $g=g_f$ , where $f\in {\mathcal B}_{k+\frac 12,4\ell }^{\mathrm {new}}$ with $\ell \mid N$ . For the estimate (31) we choose $\Phi $ to be supported on the interval $[1,2]$ such that $\Phi (t)=1$ for $t\in [11/10,19/10]$ and satisfying $0\leq \Phi (t)\leq 1$ . For the estimate (32) we choose $\Phi $ to be supported in $[1/2, 5/2]$ with $\Phi (t)=1$ for $t\in [1,2]$ and again satisfying $0\leq \Phi (t)\leq 1$ . These choices lead to

$$ \begin{align*} & \sum_{d\in\mathcal D_{N,\eta}(X)} L\left(\frac12,g_0\otimes\chi_d\right)\lvert R(d)\rvert^2\\ &\quad \geq\frac{4X\mathcal R}{5N}\cdot L_{g_0,\eta}\left(\frac12\right)L\left(1,\text{Sym}^2g_0\right)\cdot\exp\left(2\sum_{L^2\leq p\leq L^4}\frac{r(p)\lambda_0(p)^2}{\sqrt p}+O\left(\frac L{(\log L)^3}\right)\right)\\ &\qquad +O_{k,N}(X) \end{align*} $$

for $f=f_0$ and

$$ \begin{align*} &\sum_{d\in\mathcal D_{N,\eta}(X)} L\left(\frac12,g_f\otimes\chi_d\right)\lvert R(d)\rvert^2\\ &\quad \leq\frac{X\mathcal R}N\cdot L_{g_f,\eta}\left(\frac12\right)L\left(1,\text{Sym}^2g_f\right)\exp\left(2\sum_{L^2\leq p\leq L^4}\frac{r(p)\lambda_0(p)\lambda_{g_f}(p)}{\sqrt p}+O\left(\frac L{(\log L)^3}\right)\right)\\ &\qquad +O_{k,N}(X) \end{align*} $$

for $f\neq f_0$ .

To obtain the estimates (31) and (32) we use partial summation and the following estimate, which is a consequence of the Rankin–Selberg theory [Reference Wu and Ye69, Theorem 3]:

Lemma 4.4. With the notation as given, we have

$$ \begin{align*} \sum_{p\leq x}\lambda_0(p)\lambda_{g_f}(p)\log p=x\cdot 1_{f=f_0}+o(x). \end{align*} $$

5.1. Preliminaries

Denote by J the $4 \times 4$ matrix given by $ J = \left (\begin {smallmatrix} 0 & I_2\\ -I_2 & 0\\ \end {smallmatrix}\right )$ , where $I_2$ is the identity matrix of size $2$ . Define the algebraic groups ${\mathrm{GSp}}_4$ and $\mathrm {Sp}_4$ over ${\mathbb Z}$ by

$$ \begin{gather*} {\textrm{GSp}}_4(R) = \left\{g \in {\textrm{GL}}_4(R) \mid {}^t\!{g}Jg = \mu_2(g)J, \ \mu_2(g)\in R^{\times}\right\}, \\ \mathrm{Sp}_4(R) = \{g \in {\textrm{GSp}}_4(R) \mid \mu_2(g)=1\}, \end{gather*} $$

for any commutative ring R. The Siegel upper half-space $\mathbb H_2$ of degree $2$ is defined by

$$ \begin{align*} \mathbb H_2 = \left\{ Z \in \mathrm{Mat}_{2\times 2}({\mathbb C}) \mid Z ={}^t\!{Z},\ \mathrm{Im}(Z) \text{ is positive definite}\right\}. \end{align*} $$

Let k and N be positive integers. Let $\Gamma ^{(2)}_0(N) \subseteq \mathrm {Sp}_4({\mathbb Z})$ denote the Siegel congruence subgroup of level N – that is,

(44) $$ \begin{align} \Gamma^{(2)}_0(N) = \mathrm{Sp}_4({\mathbb Z}) \cap \left(\begin{smallmatrix}{\mathbb Z}& {\mathbb Z}&{\mathbb Z}&{\mathbb Z}\\{\mathbb Z}& {\mathbb Z}&{\mathbb Z}&{\mathbb Z}\\N{\mathbb Z}& N{\mathbb Z}&{\mathbb Z}&{\mathbb Z}\\N{\mathbb Z}&N {\mathbb Z}&{\mathbb Z}&{\mathbb Z}\\\end{smallmatrix}\right). \end{align} $$

We define

$$ \begin{align*} g \langle Z\rangle = (AZ+B)(CZ+D)^{-1}\qquad\text{for } g=\left(\begin{smallmatrix} A&B\\ C&D \end{smallmatrix}\right) \in \mathrm{Sp}_4({\mathbb R}),\ Z\in \mathbb H_2. \end{align*} $$

We let $J(g,Z) = CZ + D$ . Let $S_k(\Gamma ^{(2)}_0(N))$ denote the space of Siegel cusp forms of weight k and level N; precisely, they consist of the holomorphic functions F on $\mathbb H_2$ which satisfy the relation

(45) $$ \begin{align} F(\gamma \langle Z\rangle) = \det(J(\gamma,Z))^k F(Z) \end{align} $$

for $\gamma \in \Gamma _0^{(2)}(N)$ , $Z \in \mathbb H_2$ , and vanish at all the cusps. Any F in $S_k(\Gamma ^{(2)}_0(N))$ has a Fourier expansion

(46) $$ \begin{align}F(Z) =\sum_{S \in \Lambda_2} a(F, S) e^{2 \pi i \mathrm{Tr}(SZ)}, \end{align} $$

with $\Lambda _2$ defined in equation (2). We have the relation

(47) $$ \begin{align} a(F, T) = \det(A)^ka\left(F, {}^t\!{A}TA\right) \end{align} $$

for $A \in {\mathrm{GL}}_2({\mathbb Z})$ . In particular, the Fourier coefficient $a(F, T)$ depends only on the ${\mathrm{SL}}_2({\mathbb Z})$ -equivalence class of T. We say that a matrix $S \in \Lambda _2$ is fundamental if ${\mathrm{disc}}(S)= -4 \det (S)$ is a fundamental discriminant. Given a fundamental discriminant $d<0$ and $K= {\mathbb Q}(\sqrt {d})$ , let $\operatorname {\mathrm {Cl}}_K$ denote the ideal class group of K. It is well known that the ${\mathrm{SL}}_2({\mathbb Z})$ -equivalence classes of matrices in $\Lambda _2$ of discriminant d can be identified with $\operatorname {\mathrm {Cl}}_K$ ; so the expression $\sum _{S \in \operatorname {\mathrm {Cl}}_K}a(F, S) \Lambda (S)$ makes sense for each $\Lambda \in \widehat {\operatorname {\mathrm {Cl}}_K}$ .

5.2. Constructing half-integral-weight forms

Each F in $S_k(\Gamma ^{(2)}_0(N))$ has a Fourier–Jacobi expansion $F(Z) = \sum _{m> 0} \phi _m(\tau , z) e^{2 \pi i m \tau '}$ , where we write $Z= \begin {pmatrix}\tau &z\\z&\tau ' \end {pmatrix}$ and for each $m>0$ ,

(48) $$ \begin{align} \phi_m(\tau, z) = \sum_{\substack{n,r \in {\mathbb Z} \\ 4nm> r^2}}a \left(F, {\left(\begin{array}{@{}cc@{}}n&r/2\\r/2&m\end{array}\right)}\right) e^{2\pi i (n \tau + r z)} \in J_{k,m}^{\text{cusp}}(N). \end{align} $$

Here $J_{k,m}^{\text {cusp}}(N)$ denotes the space of Jacobi cusp forms of weight k, level N, and index m.

Given a primitive matrix $S = {\left (\begin {array}{@{}cc@{}}a&b/2\\b/2&c\end {array}\right )} \in \Lambda _2$ (i.e., $\gcd (a,b,c)=1$ ), we let ${\mathcal P}(S)$ denote the set of primes of the form $ax^2+bxy+cy^2$ . The set ${\mathcal P}(S)$ is infinite; indeed, by [Reference Iwaniec28, Theorem 1(i)],

(49) $$ \begin{align} \lvert\{p \in {\mathcal P}(S): p \le X\}\rvert \gg_S \frac{X}{\log X}. \end{align} $$

For each prime p dividing N, define the operator $U(p)$ acting on the space $S_k(\Gamma ^{(2)}_0(N))$ via

(50) $$ \begin{align} a(U(p) F, S)= a(F, pS). \end{align} $$

Proof. The claim that there exists $S_0 = {\left (\begin {array}{@{}cc@{}}a&b/2\\b/2&c\end {array}\right )} \in \Lambda _2$ such that $a(F, S_0)\neq 0$ and $d_0 = b^2-4ac$ is odd and square-free follows from [Reference Schmidt58, Theorem 2.2].

Now let $p \in {\mathcal P}(S_0)$ , $p \nmid 2Nd_0$ . The fact that $h_p \in S_{k-\frac {1}{2}}(4pN)$ follows from $\phi _{p} \in J_{k,p}^{\text {cusp}}(N) $ and [Reference Manickam and Ramakrishnan41, Theorem 4.8]; by definition, $h_p$ belongs to the Kohnen plus space. Next we prove equation (51). Let $\gcd (m, 4p)=1$ . Observe that the sum $\sum _{\substack {0 \le \mu \le 2p-1 \\ \mu ^2 \equiv -m \pmod {4p}}}$ is nonempty if and only if $-m$ is a square modulo $4p$ , in which case the sum has exactly two terms. Indeed, we get

(52) $$ \begin{align} a(m) = a\left(F, {\left(\begin{array}{@{}cc@{}}\frac{m+\mu_0^2}{4p}&\frac{\mu_0}{2}\\\frac{\mu_0}{2}&p\end{array}\right)}\right) + a\left(F, {\left(\begin{array}{@{}cc@{}}\frac{m+\mu_1^2}{4p}&\frac{\mu_1}{2}\\\frac{\mu_1}{2}&p\end{array}\right)}\right), \end{align} $$

where $0 \le \mu _0 \le 2p-1$ satisfies $\mu _0^2 \equiv -m \pmod {4p}$ and $\mu _1 = 2p-\mu _0$ . Using equations (47) and (52) and the identity ${\left (\begin {array}{@{}cc@{}}1&-1\\0&-1\end {array}\right )} {\left (\begin {array}{@{}cc@{}}\frac {m+\mu _0^2}{4p}&\frac {\mu _0}{2}\\\frac {\mu _0}{2}&p\end {array}\right )}{\left (\begin {array}{@{}cc@{}}1&0\\-1&-1\end {array}\right )} = {\left (\begin {array}{@{}cc@{}}\frac {m+\mu _1^2}{4p}&\frac {\mu _1}{2}\\\frac {\mu _1}{2}&p\end {array}\right )}$ , we obtain equation (51).

It remains to show $h_p \neq 0$ , for which we will show that $a(-d_0)\neq 0$ . Let $x_0, y_0$ be integers such that $cx_0^2 + bx_0y_0 + ay_0^2=p$ . Since $\gcd (x_0, y_0) = 1$ , we may pick integers $x_1, y_1$ such that $A= {\left (\begin {array}{@{}cc@{}}y_1&y_0\\x_1&x_0\end {array}\right )} \in {\mathrm{SL}}_2({\mathbb Z}).$ Then $S' = {}^t\!{A}SA$ is ${\mathrm{SL}}_2({\mathbb Z})$ -equivalent to $S_0$ and has the form $S'={\left (\begin {array}{@{}cc@{}}\frac {-d_0+\mu _0^2}{4p}&\frac {\mu _0}{2}\\\frac {\mu _0}{2}&p\end {array}\right )} \in \Lambda _2$ . By equation (47), $a(F,S') \neq 0$ since $a(F, S_0) \neq 0$ . Hence by equation (51), $a(-d_0) = 2 a(F,S') \neq 0$ .

5.3. Proofs of Theorems A and B

We are now ready to prove Theorems A and B in slightly stronger forms.

5.4. L-functions of Siegel cusp forms

For the rest of this section, we assume that $k>2$ .

Given an irreducible cuspidal automorphic representation $\pi $ of ${\mathrm{GSp}}_4({\mathbb A})$ , we let $L(s, \pi )$ denote the associated degree $4\ L$ -function (known as the ‘spin’ L-function). Furthermore, we let $L(s, \mathrm {std}(\pi ))$ denote the associated degree $5\ L$ -function (the ‘standard’ L-function) and $L(s, \mathrm {ad}(\pi ))$ denote the associated degree $10\ L$ -function (the ‘adjoint’ L-function). Each of these L-functions is defined as an Euler product with the local L-factor at each prime (including the ramified primes) constructed via the associated representation of the dual group ${\mathrm{GSp}}_4({\mathbb C})$ using the local Langlands correspondence (which is known for ${\mathrm{GSp}}_4$ by the work of Gan and Takeda [Reference Gun, Kohnen and Soundararajan19]). More precisely, for $n=4,5,10$ , let $\rho _n$ denote the irreducible n-dimensional representation of ${\mathrm{GSp}}_4({\mathbb C})$ given as follows: $\rho _4$ is the inclusion ${\mathrm{GSp}}_4({\mathbb C}) \hookrightarrow {\mathrm{GL}}_4({\mathbb C})$ , $\rho _5$ is the map defined in [Reference Radziwiłł and Soundararajan52, A.7], and $\rho _{10}$ is the adjoint representation of ${\mathrm{GSp}}_4({\mathbb C})$ on the Lie algebra of $\mathrm {Sp}_4({\mathbb C})$ . Then $L(s, \pi )$ , $L(s, \mathrm {std}(\pi ))$ , and $L(s, \mathrm {ad}(\pi ))$ correspond to the representations $\rho _4$ , $\rho _5$ , and $\rho _{10}$ , respectively.

We say that an element F of $S_k(\Gamma ^{(2)}_0(N))$ gives rise to an irreducible representation if its adelization (in the sense of [Reference Böcherer and Das5 Reference Bump and Ginzburg7, §3]) generates an irreducible cuspidal automorphic representation of ${\mathrm{GSp}}_4({\mathbb A})$ . The automorphic representation associated to any such F is of trivial central character and hence may be viewed as an automorphic representation of ${\mathrm{PGSp}}_4({\mathbb A}) \simeq {\mathrm{SO}}_5({\mathbb A})$ . It can be checked [Reference Böcherer and Das5 Reference Bump and Ginzburg7, Proposition 3.11] that if F gives rise to an irreducible representation, then F is an eigenform of the local Hecke algebras at all primes not dividing N. In addition, we say that such an F is factorizable if its adelization corresponds to a factorizable vector in the representation generated.

We say that an irreducible cuspidal automorphic representation $\pi $ of ${\mathrm{GSp}}_4({\mathbb A})$ arises from $S_k(\Gamma ^{(2)}_0(N))$ if there exists $F \in S_k(\Gamma ^{(2)}_0(N))$ whose adelization generates $V_\pi $ (in which case, by definition, F gives rise to the irreducible representation $\pi $ , which therefore must be of trivial central character by the previous comments). We say that an irreducible cuspidal automorphic representation $\pi $ of ${\mathrm{GSp}}_4({\mathbb A})$ is ‘CAP’ if it is nearly equivalent to a constituent of a global induced representation of a proper parabolic subgroup of ${\mathrm{GSp}}_4({\mathbb A})$ . If a CAP $\pi $ arises from $S_k(\Gamma ^{(2)}_0(N))$ , then by [Reference Pitale and Schmidt47, Corollary 4.5] it is associated to the Siegel parabolic (i.e., it is of Saito–Kurokawa type). Such a $\pi $ is nontempered at almost all primes (in particular, it violates the Ramanujan conjecture). On the other hand, if $\pi $ arises from $S_k(\Gamma ^{(2)}_0(N))$ and is not CAP, then $\pi $ satisfies the Ramanujan conjecture by a famous result of Weissauer [Reference Weissauer68, Theorem 3.3].

Thus, the space $S_k(\Gamma ^{(2)}_0(N))$ has a natural decomposition into orthogonal subspaces

$$ \begin{align*} S_k\left(\Gamma^{(2)}_0(N)\right)=S_k\left(\Gamma^{(2)}_0(N)\right)^{\textrm{CAP}} \oplus S_k\left(\Gamma^{(2)}_0(N)\right)^{ \rm T}, \end{align*} $$

where $S_k(\Gamma ^{(2)}_0(N))^{\mathrm{CAP}}$ is spanned by forms F which give rise to irreducible representations of Saito–Kurokawa type, and $S_k(\Gamma ^{(2)}_0(N))^{ \rm T}$ is its orthogonal complement, spanned by forms F which give rise to irreducible representations that are not of Saito–Kurokawa type. Furthermore, one can get a basis of each of the spaces $S_k(\Gamma ^{(2)}_0(N))^{\mathrm{CAP}}$ and $S_k(\Gamma ^{(2)}_0(N))^{ \rm T}$ in terms of factorizable forms. We refer the reader to [Reference Dickson, Pitale, Saha and Schmidt15, §§3.1 and 3.2] for further comments related to the foregoing discussion.

If $\pi $ arises from $S_k(\Gamma ^{(2)}_0(N))$ and is of Saito–Kurokawa type, then there exist a representation $\pi _0$ of ${\mathrm{GL}}_2({\mathbb A})$ and a Dirichlet character $\chi _0$ satisfying $\chi _0^2=1$ such that $L^N(s, \pi ) = L^N(s, \pi _0) L^N(s+1/2, \chi _0) L^N(s-1/2, \chi _0)$ . Additionally, if N is square-free, then only $\chi _0=1$ is possible, by a well-known result of Borel [Reference Böcherer8], and so in this case we have $L^N(s, \pi ) = L^N(s, \pi _0) \zeta ^N(s+1/2) \zeta ^N(s-1/2)$ . There exists another typical situation where the L-function factors: we say that a $\pi $ arising from $S_k(\Gamma ^{(2)}_0(N))$ is of Yoshida type if there are representations $\pi _1$ and $\pi _2$ of ${\mathrm{GL}}_2({\mathbb A})$ such that $L(s, \pi ) = L(s, \pi _1) L(s, \pi _2)$ ; in this case (after possibly swapping $\pi _1$ and $\pi _2$ ), $\pi _1$ arises from a classical holomorphic newform of weight $2$ and $\pi _2$ arises from a classical holomorphic newform of weight $2k-2$ (see [Reference Böcherer and Das5 Reference Bump and Ginzburg7, §4] for more details). We let $S_k(\Gamma ^{(2)}_0(N))^{ \rm Y}$ denote the subspace of $S_k(\Gamma ^{(2)}_0(N))^{ \rm T}$ spanned by forms which give rise to an irreducible representation of Yoshida type, and let $S_k(\Gamma ^{(2)}_0(N))^{ \rm G}$ – which represents the general type – denote the orthogonal complement of $S_k(\Gamma ^{(2)}_0(N))^{ \rm Y}$ in $S_k(\Gamma ^{(2)}_0(N))^{ \rm T}$ . So we get the following key orthogonal decomposition into subspaces:

(53) $$ \begin{align} S_k\left(\Gamma^{(2)}_0(N)\right)=S_k\left(\Gamma^{(2)}_0(N)\right)^{\textrm{CAP}} \oplus S_k\left(\Gamma^{(2)}_0(N)\right)^{ \rm Y} \oplus S_k\left(\Gamma^{(2)}_0(N)\right)^{ \rm G}. \end{align} $$

In the notation of [Reference Shahidi59], the three subspaces on the right-hand side correspond to the global Arthur packets of type (P), (Y), and (G), respectively. In the sequel we will be mostly concerned with the space $S_k(\Gamma ^{(2)}_0(N))^{ \rm G}$ , because the other two spaces are easier to handle. The following proposition collects together some relevant facts about the associated L-functions that follow from the work of Arthur [Reference Andrianov2]:

Proof. Since F corresponds to the ‘general’ Arthur parameter, Arthur’s work [Reference Andrianov2] (see also [Reference Shahidi59, §1.1]) shows that $\pi $ has a strong lifting to a cuspidal automorphic representation $\Pi _4$ of ${\mathrm{GL}}_4$ ; clearly $\Pi _4$ has trivial central character, since $\pi $ does. It is known from [Reference Andrianov2, Theorem 1.5.3] that $\Pi _4$ is self-dual and symplectic.Footnote ⁷ The fact that the lifting is ‘strong’ – that is, corresponds to a local lift at all places – implicitly uses the fact that the local Arthur parameters coincide with the local Langlands parameters of Gan and Takeda [Reference Gun, Kohnen and Soundararajan19]; this consistency of local parameters follows from [Reference Chan and Gan10]. The required properties of $L(s, \pi \times \sigma )$ now follow from the Rankin–Selberg theory of ${\mathrm{GL}}_n$ .

For the second assertion, let $\wedge ^2$ denote the exterior square, and note that $\wedge ^2\rho _4=\mathbf {1}+\rho _5$ ; hence

(54) $$ \begin{align} L\left(s,\Pi_4,\wedge^2\right)=L(s, \mathrm{std}(\pi))\zeta(s). \end{align} $$

Recall that $L(s, \Pi _4 \times \Pi _4) = L\left (s,\Pi _4,\mathrm {Sym}^2\right ) L\left (s,\Pi _4,\wedge ^2\right )$ . Since $\Pi _4$ is symplectic, $L\left (s,\Pi _4,\wedge ^2\right )$ has a simple pole at $s=1$ and $L\left (s,\Pi _4,\mathrm {Sym}^2\right )$ is an entire function by [Reference Borel9, Theorem 7.5]. It follows from equation (54) that $L(s, \mathrm {std}(\pi ))$ is holomorphic and nonzero at $s=1$ . Together with [Reference Gritsenko20, Theorem 2], we obtain that $L(s, \mathrm {std}(\pi ))$ has no poles on $\mathrm {Re}(s)=1$ . On the other hand, by [Reference Kohnen32, Theorem A] and [Reference Huang and Lester26], $L\left (s,\Pi _4,\wedge ^2\right )$ is the L-function of an automorphic representation of ${\mathrm{GL}}_6({\mathbb A})$ of the form ${\mathrm{Ind}}(\tau _1\otimes \dotsb \otimes \tau _m) ,$ where $\tau _1,\dotsc ,\tau _m$ are unitary, cuspidal, automorphic representations of ${\mathrm{GL}}_{n_i}({\mathbb A})$ , $n_1+\dotsb +n_m=6$ . Since $L\left (s,\Pi _4,\wedge ^2\right )$ has a simple pole at $s=1$ , it follows that exactly one of the $\tau _i$ – say $\tau _m$ – is the trivial representation of ${\mathrm{GL}}_1({\mathbb A})$ . Canceling out one $\zeta $ factor, we see that

(55) $$ \begin{align} L(s, \mathrm{std}(\pi))=L(s,\tau_1)\dotsm L(s,\tau_{m-1}). \end{align} $$

So we take $\Pi _5={\mathrm{Ind}}(\tau _1\otimes \dotsb \otimes \tau _{m-1}) $ ; the foregoing discussion shows that $\Pi _5$ is a (strong) lifting of $\pi $ from ${\mathrm{GSp}}_4({\mathbb A})$ to ${\mathrm{GL}}_5({\mathbb A})$ with respect to the map $\rho _5$ of dual groups ${\mathrm{GSp}}_4({\mathbb C}) \rightarrow {\mathrm{SO}}_5({\mathbb C}) \subset {\mathrm{GL}}_5({\mathbb C})$ . Observe that each $\tau _i$ is unitary, cuspidal, and unramified outside primes dividing N. By Rankin–Selberg theory, $L(s, \Pi _5 \times \sigma )=\prod _{i=1}^{m-1}L(s, \tau _i \times \sigma ) $ satisfies the usual properties of an L-function. To complete the proof, it suffices to show that if $\sigma $ is an irreducible cuspidal representation of ${\mathrm{GL}}_1({\mathbb A})$ or ${\mathrm{GL}}_2({\mathbb A})$ such that the set of ramification primes for $\sigma $ either is empty or contains at least one prime outside N, then $\tau _i \not \simeq \hat {\sigma }$ for each i. First, consider the case where $\sigma $ is a character, in which case we are reduced to $n_i=1$ . In this case, since $\tau _i$ is unramified outside N, the assumption on $\sigma $ means that the situation $\tau _i \simeq \hat {\sigma }$ will force $\tau _i$ to be unramified everywhere, in which case the right-hand side of equation (55) would have a pole on $\mathrm {Re}(s)=1$ . This contradicts the observation from before that $L(s, \mathrm {std}(\pi ))$ has no poles on $\mathrm {Re}(s)=1$ . Next consider the case where $\sigma $ is a cuspidal representation of ${\mathrm{GL}}_2({\mathbb A})$ . An identical argument to the foregoing reduces us to the case where $\tau _i \simeq \hat {\sigma }$ is unramified at all finite primes. The easy relation $\wedge ^2\rho _5 = {\mathrm{Sym}}^2\rho _4=\rho _{10}$ implies that

(56) $$ \begin{align} L\left(s,\Pi_5,\wedge^2\right)= L(s, \mathrm{ad}(\pi)) = L\left(s,\Pi_4,\mathrm{Sym}^2\right), \end{align} $$

which we know is an entire function. On the other hand, if some $\tau _i$ in equation (55) has $n_i=2$ and is unramified at all finite primes, then $L\left (s,\Pi _5,\wedge ^2\right )$ will contain a factor of $L(s, \omega _{\tau _i})$ and so will have a pole on $\mathrm {Re}(s)=1$ . This contradiction completes the proof that each $L(s,\tau _i \times \sigma )$ and hence $L(s,\Pi _5\times \sigma )$ is entire.

Finally, the assertion concerning the degree $10\ L$ -function $L(s, \mathrm {ad}(\pi ))$ follows from the identity (56) and the fact that $L\left (s,\Pi _4,\mathrm {Sym}^2\right )$ represents a holomorphic L-function. Since symmetric square L-functions are accessible via the Langlands–Shahidi method, the nonvanishing on $\mathrm {Re}(s)=1$ follows (see [Reference Shahidi60, §5] and [Reference Shimura61, Theorem 1.1]).

For our future applications, we will also need the facts that $L(s, \mathrm {ad}(\pi ) \times \sigma )$ has the properties of an L-function and is entire for certain special automorphic representations $\sigma $ of ${\mathrm{GL}}_1({\mathbb A})$ or ${\mathrm{GL}}_2({\mathbb A})$ . The next two lemmas achieve this for $\sigma $ a quadratic character or $\sigma $ of the form ${\mathcal {AI}}\left (\Lambda ^2\right )$ , where $\Lambda $ is a character of ${\mathbb A}_K^\times $ , K is a quadratic field, and ${\mathcal {AI}}$ denotes automorphic induction.

Proof. Let $\pi _K$ and $\Pi _{4,K}$ be as in the statement of the lemma. Since $\Pi _4$ is the lifting of $\pi $ , it follows from the definition of base change that $\Pi _{4,K}$ is the lifting of $\pi _K$ . Let $\sigma $ be an arbitrary cuspidal automorphic representation of ${\mathrm{GL}}_1({\mathbb A}_K)$ or ${\mathrm{GL}}_2({\mathbb A}_K)$ . By the definition of lifting, we have

$$ \begin{align*} L(s, \pi_K \times \sigma) = L(s, \Pi_{4,K} \times \sigma), \end{align*} $$

and so to prove that $\pi _K$ and $\Pi _{4,K}$ are cuspidal it suffices to show that $L(s, \Pi _{4,K} \times \sigma )$ has no poles. By the adjointness formula [Reference Pitale, Saha and Schmidt48, Proposition 3.1], we have $L\left (s, \Pi _{4,K} \times \sigma \right ) = L(s, \Pi _4 \times {\mathcal {AI}}(\sigma ))$ . Note that ${\mathcal {AI}}(\sigma )$ is an automorphic representation of ${\mathrm{GL}}_2({\mathbb A})$ if $\sigma $ is a character of ${\mathbb A}_K^\times $ , and ${\mathcal {AI}}(\sigma )$ is an automorphic representation of ${\mathrm{GL}}_4({\mathbb A})$ if $\sigma $ is a cuspidal representation of ${\mathrm{GL}}_2({\mathbb A}_K)$ . By Proposition 5.4, $L(s, \Pi _4 \times {\mathcal {AI}}(\sigma ))$ is entire when $\sigma $ is a character. This reduces us to the case where $\sigma $ is a cuspidal representation of ${\mathrm{GL}}_2({\mathbb A}_K)$ ; in this case, $L\left (s, \Pi _{4,K} \times \sigma \right ) = L(s, \Pi _4 \times {\mathcal {AI}}(\sigma ))$ has a pole if and only if $\Pi _4 \simeq {\mathcal {AI}}(\sigma )$ (recall that $\Pi _4$ is self-dual). However, by looking at a prime p which ramifies in K, we see that this is impossible: at any such prime, the local Langlands parameter of $\Pi _{4,p}$ is the local lifting of a representation of ${\mathrm{GSp}}_4\left ({\mathbb Q}_p\right )$ that has a vector fixed by the local Siegel congruence subgroup of level p (since N is square-free), but an inspection of [Reference Andrianov2 Reference Borel9, Table 1] tells us that the local parameter of ${\mathcal {AI}}\left (\sigma _p\right )$ can never equal one of those. Hence we have completed the proof that $L\left (s, \Pi _{4,K} \times \sigma \right )$ has no poles.

Proof. It can be verified (as a formal identity that holds at each place) that

$$ \begin{align*} L\left(s, \mathrm{Sym}^2\left(\pi_{4,K} \times \Lambda\right)\right) = L\left(s, \mathrm{ad}(\pi) \times {\mathcal{AI}}\left(\Lambda^2\right)\right). \end{align*} $$

By Lemma 5.5, $\pi _{4,K} \times \Lambda $ is a cuspidal representation of ${\mathrm{GL}}_4({\mathbb A}_K)$ . Furthermore, if $\pi _{4,K} \times \Lambda $ is self-dual, then it must be symplectic, since $\Pi _{4,K}\times \Lambda $ is the lift of $\pi _K\times \Lambda $ from ${\mathrm{GSp}}_4({\mathbb A}_K)$ to ${\mathrm{GL}}_4({\mathbb A}_K)$ . So by [Reference Borel9, Theorem 7.5], $ L\left (s, \mathrm {ad}(\pi ) \times {\mathcal {AI}}\left (\Lambda ^2\right )\right )$ has the properties of an L-function and is entire.

To show the same fact for $L(s, \mathrm {ad}(\pi )\times \chi _K) = L\left (s, \mathrm {Sym}^2(\Pi _4)\times \chi _K\right )$ , we appeal to [Reference Takeda66, Theorem, p. 104], which shows that the only possible poles of $L\left (s, \mathrm {Sym}^2(\Pi _4)\times \chi _K\right )$ are at $s=0, 1$ (these two possible poles are related by the functional equation). However, by what we have already proved, $L(s,\mathrm {ad}(\pi )\times {\mathcal {AI}}(1)) =L( s, \mathrm {ad}(\pi ))L(s, \mathrm {ad}(\pi )\times \chi _K)$ is entire. So a pole at $s=1$ for $L(s, \mathrm {ad}(\pi )\times \chi _K)$ would imply that $L(1, \mathrm {ad}(\pi ))=0$ . This contradicts the last part of Proposition 5.4.

5.5. A consequence of the refined Gan–Gross–Prasad identity

Let $K = {\mathbb Q}(\sqrt {d})$ , where $d<0$ is a fundamental discriminant. Recall from the introduction that for any character $\Lambda $ of the finite group $\operatorname {\mathrm {Cl}}_K$ , we define $B(F, \Lambda ) = \sum _{S \in \operatorname {\mathrm {Cl}}_K}a(F, S) \Lambda (S)$ . The refined Gan–Gross–Prasad conjecture in the context of Bessel periods of ${\mathrm{PGSp}}_4 \simeq {\mathrm{SO}}_5$ (and more generally, for Bessel periods of orthogonal groups) was formulated in [Reference Lester and Radziwiłł40, equation (1.1)] and made more explicit in the context of Siegel cusp forms in [Reference Dickson, Pitale, Saha and Schmidt15]. This conjecture implies an identity expressing the square of $\lvert B(F, \Lambda )\rvert $ as a ratio of L-values, up to some local integrals. For the purpose of this paper, we only need a relatively weak consequence of this identity, which we formulate explicitly as Hypothesis G. In this subsection we do not assume that N is square-free.

Suppose that $F \in S_k(\Gamma ^{(2)}_0(N))^{ \rm T}$ gives rise to an irreducible representation $\pi $ . Then F is said to satisfy Hypothesis G if there exists a constant $C_F$ such that for each imaginary quadratic field $K={\mathbb Q}(\sqrt {d})$ (with $d<0$ a fundamental discriminant) and each character $\Lambda $ of $\operatorname {\mathrm {Cl}}_K$ , we have

(57) $$ \begin{align} \lvert B(F, \Lambda)\rvert^2 \le C_F \lvert d\rvert^{k-1} L\left(\tfrac12, \pi \times {\mathcal{AI}}(\Lambda)\right). \end{align} $$

Proof. Note that the subspace of $S_k(\Gamma ^{(2)}_0(N))^{ \rm T}$ generated by forms that give rise to a fixed irreducible representation $\pi $ has a basis consisting of factorizable forms. So for the purpose of verifying Hypothesis G, it suffices to prove the bound (57) for factorizable F whose adelization $\phi = \otimes \phi _v$ generates $\pi $ . Assuming the truth of [Reference Dickson, Pitale, Saha and Schmidt15, Conjecture 1.2] for $(\phi , \Lambda )$ , and combining it with the explicit calculations performed in [Reference Dickson, Pitale, Saha and Schmidt15, §2.2, §§3.3–3.5], we see that

$$ \begin{align*} \vert B(F, \Lambda)\rvert^2 = A_F \lvert d\rvert^{k-1} L\left(\tfrac12, \pi \times {\mathcal{AI}}(\Lambda)\right) \prod_{p\mid N_F}J_p\left(\phi_p, \Lambda_p\right), \end{align*} $$

where $A_F$ depends on F, $N_F\mid N$ is the smallest integer such that $F \in S_k(\Gamma ^{(2)}_0(N_F))^{\mathrm{T}}$ , and the normalized local factors $J_p\left (\phi _p, \Lambda _p\right )$ (which depend on p, $\phi _p$ , $\Lambda _p$ , and d) are defined in [Reference Dickson, Pitale, Saha and Schmidt15, equation (30)]. To complete the proof, it suffices to show that $J_p\left (\phi _p, \Lambda _p\right ) \ll _{p,\phi _p} 1$ (i.e., is bounded by some constant that depends on p and $\phi _p$ but not on d or $\Lambda _p$ ). It suffices to show this for the unnormalized local factors $J_{0,p}\left (\phi _p, \Lambda _p\right )$ defined in [Reference Dickson, Pitale, Saha and Schmidt15, equation (29)], since the normalized local L-factors $J_p\left (\phi _p, \Lambda _p\right )$ differ from these only by certain absolutely bounded L-factors appearing in [Reference Dickson, Pitale, Saha and Schmidt15, equation (30)].

To show this, we move to a purely local setup. Let p be a prime dividing $N_F$ . Let $F ={\mathbb Q}_p$ . We fix a set M of coset representatives of $F^\times / (F^\times )^2$ such that all elements of M are taken from ${\mathbb Z}_p$ , and each $r \in M$ generates the discriminant ideal of $F\left (\sqrt {r}\right )/F$ . We let $r_p$ equal the unique representative in M that corresponds to d. The assumptions imply that

$$ \begin{align*} r_p = d u_p^2, \qquad \text{for some }u \in \left({\mathbb Z}_p^\times\right)^2. \end{align*} $$

Let K equal $F\times F$ (the ‘split case’) if $r_p \in F^2$ and $K=F\left (\sqrt {r_p}\right )= F(\sqrt {d})$ (the ‘field case’) if $r_p \notin F^2$ . Fix the matrix $S=S_d$ as in [Reference Dickson, Pitale, Saha and Schmidt15, (75)], so that $S_d$ has discriminant d. Let $T(F)=T_S(F) \simeq K^\times $ be the associated subgroup of ${\mathrm{GSp}}_4(F)$ and let $N(F)\subset {\mathrm{GSp}}_4(F)$ denote the unipotent for the Siegel parabolic and $\theta _S$ the character on $N(F)$ given by $\theta _S\left ({\left (\begin {array}{@{}cc@{}}I_2&X\\0&I_2\end {array}\right )}\right ) = \psi (\mathrm {tr}(SX))$ , where $\psi $ is a fixed unramified additive character. Then we need to show that the integral

$$ \begin{align*} J_{0,p}\left(\phi_p, \Lambda_p\right) =\int_{F^\times \backslash T(F)}\int^{{\textrm{St}}}_{N(F)}\left\langle \pi_p\left(t_pn_p\right) \phi_p , \phi_p \right\rangle \Lambda_p^{-1}\left(t_p\right) \theta_S^{-1}\left(n_p\right)dn_p dt_p \end{align*} $$

is bounded by some quantity that does not depend on d or $\Lambda _p$ . Note that the superscript ‘ ${\mathrm{St}}$ ’ denotes a stable integral as in [Reference Lester and Radziwiłł40]; this means that the integral can be replaced by any sufficiently large compact subgroup (as we will do later). Put $S'={\left (\begin {array}{@{}cc@{}}-\frac {r_p}4&0\\0&1\end {array}\right )}, A' = {\left (\begin {array}{@{}cc@{}}u_p&0\\0&1\end {array}\right )}$ if $d \equiv 0 \pmod {4}$ . In the case where $d \equiv 1 \pmod {4}$ , we put $S' = {\left (\begin {array}{@{}cc@{}}\frac {1-r_p}4&\frac 12\\\frac 12&1\end {array}\right )}, A' = {\left (\begin {array}{@{}cc@{}}u_p&0\\\frac {1-u_p}{2}&1\end {array}\right )}$ . In either case, $A' \in {\mathrm{GL}}_2\left ({\mathbb Z}_p\right )$ and $S'={}^t\!{A}SA$ . To show that $J_p$ is bounded independently of d, we use a simple change of variables $\left (n_p \mapsto An_p{}^t\!{A}, \ t_p \mapsto At_p'A^{-1}\right )$ to see that the integral $J_{0,p}\left (\phi _p, \Lambda _p\right )$ remains unchanged when the matrix S is replaced by $S'$ . This shows that $J_{0,p}\left (\phi _p, \Lambda _p\right )$ depends not on the actual value of d but only on the class of d in $F^\times / (F^\times )^2$ together with d modulo $4$ , of which there are only finitely many possibilities. To show that the resulting integral is absolutely bounded independently of $\Lambda $ , we replace the integral $\int ^{{\mathrm{St}}}_{N(F)}$ by $\int _{p^{-m_p}N\left ({\mathbb Z}_p\right )}$ , where $m_p$ depends only on the level N (as follows from the argument of [Reference Dickson, Pitale, Saha and Schmidt15, Proposition 2.14]). The resulting integral is absolutely convergent [Reference Lester and Radziwiłł40, Theorem 2.1(i)] and hence bounded independently of $\Lambda $ .