2. Siegel’s theory and local densities

For a holomorphic function f on the upper half plane and $\gamma \in \Gamma_0(N)$ we write

\begin{align*}f|[\gamma]_{\frac{m}{2}}(z) \,{:\!=}\, \begin{cases}(cz+d)^{-\frac{m}{2}} f(\gamma z) & \text{if } m \text{ is even} \\ \\[-7pt] \left({\epsilon}_d \big(\frac{c}{d}\big)\right)^{-m} (cz+d)^{-\frac{m}{2}} f(\gamma z) & \text{if } m \text{ is odd and } 4 \mid N ,\end{cases}\end{align*}

where $\big({c}/{d}\big)$ is the extended Kronecker symbol and ${\epsilon}_d = \left({-1}/{d}\right)^\frac{1}{2}$ . Moreover, we denote by $S_{m/2}(N,\chi)$ the space of cusp forms of weight $\frac{m}{2} $ for $\Gamma_0(N)$ and character $\chi$ .

Two positive, quadratic forms Q, Q $^{\prime}$ belong to the same class if $Q' = U^T Q U$ for $U \in \textrm{GL}_m({\mathbb Z})$ . Furthermore, they are in the same genus if they are equivalent over ${\mathbb Z}_p$ for all p, so in particular over ${\mathbb Q}$ . There are only finitely many classes in the genus and the set of automorphs $o(Q) \,{:\!=}\,\{ U \in \textrm{GL}_2({\mathbb Z}) \mid U^T Q U = Q\}$ is finite. We put

\begin{align*}\theta(\textrm{gen}\ Q,z)= \left(\sum_{R \in \textrm{gen}\ Q} \frac{1}{\#o(R)} \right)^{-1} \sum_{R \in \textrm{gen}\ Q} \frac{\theta(R,Z)}{\#o(R)} = \sum_{n} r(\textrm{gen}\ Q,n) e(nz).\end{align*}

Throughout this work, we assume that $q(x) = x^T Qx/2$ is primitive. This implies that $NQ^{-1}$ has level N and that $N/2 \mid \det Q \mid 2 N^{m-1}.$

Recall that $\theta(Q,z) - \theta(\textrm{gen}\ Q,z) \in S_{m/2}(N,\chi)$ . The space $S_{3/2}(N,\chi)$ contains a subspace U generated by theta functions of the form $\sum \psi(n) n e(tn^2z)$ for some real character $\psi$ and $t \mid 4N$ . Their non-vanishing Fourier coefficients are of size $\asymp n^\frac{1}{2}$ which is the order of magnitude of $r(\textrm{gen}\ Q,z)$ , cf. (1). Hence, for $m=3$ the main term needs to be modified.

Two quadratic forms $q_1,q_2$ with matrices $Q_1 = S^T Q_2 S$ and $S \in \textrm{GL}_m({\mathbb Z})$ in the same genus belong to the same spinor genus if $S \in O_{Q}(A_2) \bigcap_{p} O'_{Q_p}(A_2)GL_{m}({\mathbb Z}_p)$ , where $O'_{Q_p}(A)$ is the subgroup of p-adic automorphs $O_{Q_p}(A)$ of determinant and spinor norm 1, cf. [Reference O’Meara21, section 55]. We set

\begin{align*}\theta(\textrm{spn}\ Q,z)= \left(\sum_{R \in \textrm{spn}\ Q} \frac{1}{\#o(R)} \right)^{-1} \sum_{R \in \textrm{spn}\ Q} \frac{\theta(R,Z)}{\#o(R)}.\end{align*}

Then, $\theta(Q,z) - \theta(\textrm{spn}\ Q,z) \in U^{\bot} \subseteq S_{3/2}(N,\chi)$ . For $g(z) = \sum_n b(n) e(nz) \in U^\bot $ , we have by [Reference Waibel30, theorem 1] for $n=tv^2w^2$ with squarefree t and $(w,N)=1$ that

(8) \begin{align} a(n) \ll {g} n^{\frac{1}{4}} \left( {1 + \frac{n^{\frac{3} {14}}}{N^{\frac 1 7}} + \frac{n^{\frac{3} {16}}}{N^{\frac{1} {16}}}+ \frac{\sqrt{v (n,N)}}{\sqrt{N}}} \right) (nN)^{{\epsilon}}.\end{align}

This saving in n is sufficient to obtain an asymptotic formula for r(Q, n), since by a local argument of Blomer, cf. [Reference Blomer2, (1·7)], we can assume that (n, N) is small.

In principle, the Fourier coefficients $ r (\textrm{spn}\ Q,n)$ of $\theta(\textrm{spn}\ Q,z)$ can be computed locally in a similar fashion as $r(\textrm{gen}\ Q,n)$ . However, this process is complicated and tedious. Luckily, in many situations $r (\textrm{spn}\ Q,n)$ and r(Q, n) coincide. For example, this holds if

(9) \begin{align} \begin{split}\textrm{(i)}\ &n \notin \{tm^2 \mid 4t \equiv 0\, (\textrm{mod}\ N), m \in {\mathbb N}\} \text{ or} \\ \\[-10pt] \textrm{(ii)}\ &q \simeq u_1 p^{\nu_1} x_1^2 + u_2 p^{\nu_2} x_2^2 + u_3 p^{\nu_3} x_2^2 \text{ over } {\mathbb Z}_p, \text{at least two } \nu_i \text { are equal for} \\[-3pt] &\text{odd } p \text{ and all three } \nu_i \text{ are equal for } p=2.\end{split}\end{align}

To obtain upper and lower bounds for $r(\textrm{gen}\ Q,n)$ , we need to evaluate the p-adic densities given in (2). If $p \nmid 2nN$ , an easy computation shows that

(10) \begin{align} \beta_p(n,Q) = \begin{cases}1- p^{-\frac{m}{2}}\left(\frac{(\!-1)^\frac{m}{2} \det Q }{p^\frac{m}{2}}\right) & \text{if } m \text{ even}, \\ \\[-9pt] 1 + p^{\frac{1-m}{2}} \left(\frac{(\!-1)^\frac{m-1}{2} n \det Q }{p^\frac{m}{2}}\right) & \text{if } m \text{ odd},\end{cases}\end{align}

cf. [Reference Siegel27, Hilfssatz 12]. For $p \nmid 2N$ , we have by [Reference Siegel27, Hilfssatz 16] that

(11) \begin{align} 1- p^{-r} \leq \beta_p(n,Q) \leq 1 + p^{r},\end{align}

where $r= {m}/{2}$ for even m and $r= {1-m}/{2}$ for odd m. For the remaining p-adic densities, we apply a formula of Yang [Reference Yang32]. For odd p and $\nu_i, u_i$ defined as in (4), we set

\begin{align*}V(l) = \{1 \leq i \leq m \mid \nu_i - l <0 \text{ is odd} \}\end{align*}

and

(12) \begin{align} d(l) = l + \frac{1}{2}\sum_{\nu_i<l} (\nu_i-l), \quad v(l) = \left(\frac{-1}{p}\right)^{\lfloor \frac{\# V(l)}{2} \rfloor}\prod_{i \in V(l)} \left(\frac{u_i}{p}\right).\end{align}

Lemma 6. [Reference Yang32, theorem 3·1] For odd p and $n= p^a t$ with $(t,p)=1$ we have that

\begin{align*} \beta_p(Q,n) = 1 + \left(1+p^{-1}\right) \sum_{\substack{0 < l \leq a \\ \#V(l)\, \text{even}}} v(l) p^{d(l)} + v(a+1) p^{d(a+1)} f(n) \end{align*}

where

\begin{align*} f(n) = \begin{cases} - \frac{1}{p} & \text{if } \#V(a+1) \text{ is even,} \\ \\[-8pt] \left(\frac{t}{p}\right)\frac{1}{\sqrt{p}} & \text{if } \#V(a+1) \text{ is odd}. \end{cases} \end{align*}

To compute the densities for odd $p \mid N, p\nmid n$ , we make a case distinction according to $V(1) = \#\{i \mid \nu_i =0 \}$ . If $V(2) \geq 2$ it follows by the theorem above that

(13) \begin{align} 1 - \frac{1}{p} \leq \beta_p(Q,n) \leq 1 + \frac{1}{p}.\end{align}

For $V(1) =1$ , we obtain $\beta_p(Q,n) = 1 + \Big( {n}/{p} \Big) \prod_{i \in V(1)} \Big({u_i}/{p} \Big)$ .

For the computation of the 2-adic densities Yang provides a slightly more complex formula [Reference Yang32, theorem 4·1]. This gives the following upper bound:

Corollary 7. It holds that

\begin{align*} \prod_p \beta_p(Q,n) \ll (n,N)^{\frac{1}{2}} (nN)^{\epsilon}. \end{align*}

Proof. By (10), (11) and (13) we already know that

\begin{align*}\prod_p \beta_p(Q,n) \ll (nN)^{\epsilon} \prod_{p \mid 2 (n,N)} \beta_p(Q,n).\end{align*}

Let $(n,N) = p^b$ . Then, $d(l) \leq {b}/{2}$ for all l. Hence, Lemma 6 gives for odd p that $\beta_p(Q,n) \ll p^{\frac{b}{2}}$ as $|{v(l)}|=1$ . For $p=2$ we apply [Reference Yang32, theorem 4·1]. The term d(k), defined in [Reference Yang32, (4·3)], satisfies $d(k) \leq 2^{\frac{b+1}{2}}$ which implies that [Reference Yang32, (4·4)] is bounded by $\ll 2^{\frac{b}{2}}$ and hence, also $\beta_2(n,Q)$ .

As a consequence, we obtain by (1) that

(14) \begin{align} r(\textrm{gen}\ Q,n) \ll \frac{n^{\frac{m}{2}-1}(n,N)^{\frac{1}{2}}}{\sqrt{\det Q}} (nN)^{\epsilon}.\end{align}

To obtain lower bounds for $\beta_p(Q,n)$ we follow the approach of Hanke [Reference Hanke15]. For $q \simeq u_1 p^{\nu_1} x_1^2 + \cdots + u_m p^{\nu_m} x_m^2 $ in the form (4) and a primitive solution $x \in ({\mathbb Z}_p)^m$ of $q(x) = n$ , we set

\begin{align*}\nu(x) = \min \{ \nu_i \mid x_i \in {\mathbb Z}_p^\times\}.\end{align*}

We say that x is of good type if $\nu(x)=0$ , of bad type I if $\nu(x)=1$ and of bad type II if $\nu(x) \ge 2$ . If there is a solution of good type, we count solutions modulo p and apply Hensel’s lemma to compute $\beta_p(q,n)$ . As a result, we obtain $\beta_p(q,n) \geq 1- {1}/{p}$ and $\beta_2(q,n) \geq {1}/{4}$ .

In all other cases, we apply reduction maps. Therefore, let $q \equiv q_0 + p q_1 + p^2 q_2$ , where $q_0$ consist of all diagonal terms with $\nu_i=0$ and $pq_1$ of those with $\nu_i=1$ . If there are no good solutions, but at least one of bad type I, we reduce modulo p which gives $\beta_p(q,n) \geq p^{1-\dim q_0} \beta_p(q',{n}/{p})$ for $q' = p q_0 + q_1 + p q_2$ . If there are only bad type II solutions, we reduce modulo $p^2$ which gives $\beta_p(q,n) \geq p^{2-\dim q_0-\dim q_1} \beta_p(q'',{n}/{p^2})$ for $q'' = q_0 + p q_1 + q_2$ . As a consequence, we obtain:

Lemma 8. Assume there is a primitive solution $q(x) = n$ over ${\mathbb Z}_p$ with $\nu \,{:\!=}\, \nu(x)$ as above and $m \geq 4$ . Then,

\begin{align*}r_p(q,n) \geq 1-\frac{1}{p} \quad \text{if } p \equiv 1 \,(\mathrm{mod}\ 4).\end{align*}

Furthermore, we have for $m=4$ that

\begin{align*}r_p(q,n) \geq \frac{1- \frac{1}{p}}{\sqrt{(p^{\nu},n)}} \text{ for } p \equiv 3\,( \mathrm{mod}\ 4) \text{ and } r_2(q,n) \geq \frac{1}{32 \cdot \sqrt{(2^{\nu},n)}}\end{align*}

and for $m \geq 5$ that

\begin{align*}r_p(q,n) \geq \frac{1- \frac{1}{p}}{(p^{\nu},n)} \text{ for } p \equiv 3 \,(\mathrm{mod}\ 4) \text{ and } r_2(q,n) \geq \frac{1}{32 \cdot (2^{\nu},n)}.\end{align*}

Remark. It seems counterintuitive that the bounds for $m=4$ are stronger. One reason is the assumption that n is primitively locally represented which eliminates the worst case scenario for the lower bound, i.e. that $q \sim x_1^2 + x_2^2 + p x_3^2 + p x_4^2$ and $p \equiv 3 \,(\textrm{mod}\ 4)$ .

Proof. The case $m=4$ is essentially proven in [Reference Rouse23, lemma 2]. If $p \equiv 1\, (\textrm{mod}\ 4)$ and $\dim q_0 \geq 2$ , there is always a good solution. Moreover, a bad type solution cannot exist for $q \simeq u_1 x_1^2 + u_2 x_2^2 + u_3 p x_3^2 + u_4 p x_4$ if $p \equiv 3 \,(\textrm{mod}\ 4)$ . Indeed, all solutions of $x_1^2 + x_2^2 + p x_3^2 + p x_4^2 \equiv 0 \,(\textrm{mod}\ p^2)$ satisfy $x \equiv 0 \,(\textrm{mod}\ p)$ for $p \equiv 3 \,(\textrm{mod}\ 4)$ . The cases $m \geq 5$ work analogously to $m=4$ .

Since $\nu \leq v_p(N)$ , it follows for $m=4$ that $\prod_p \beta_p(q,n) \gg (n,N)^{-\frac{1}{2}} (nN)^{-{\epsilon}}$ if n is primitively represented. A form in 5 variables is isotropic over ${\mathbb Z}_p$ and hence if we sort $\nu_i$ by value, there is always a primitive solution x such that $\nu(x) \le \nu_5$ . Hence, we obtain for $m \geq 5$ that

\begin{align*}\prod_p \beta_p(q,n) \gg \prod_{p \mid N} \left( p^{\frac{v_p(\det Q)}{m-4}}, p^{v_p(N)},n\right) ^{-1}(nN)^{-{\epsilon}}.\end{align*}

For a given form, it is often possible to find a primitive solution with $\nu=0$ for all $p \mid (n,N)$ which yields $\prod_p \beta_p(q,n) \gg (nN)^{-{\epsilon}}$ .

3. A uniform bound for the inner product

The aim of this section is to prove Theorem 1, 2 and 3. Recall that the Siegel domain $\mathcal{S}(\delta,\eta)$ consists of all positive matrices M that allow for a decomposition into

\begin{align*}M = V^T D V,\end{align*}

where V is an upper triangular matrix with ones on the diagonal and all other entries $|{v_{i,j}}| \leq \eta$ and D is a diagonal matrix with entries $a_1, \ldots, a_{m} $ that satisfy $0 < a_i \leq \delta a_{i+1}$ .

Let Q denote a positive, integral matrix with even diagonal entries. By [Reference Cassels8, p. 259], the class of Q contains an element $Q' \in \mathcal{S}({4}/{3},{1}/{2})$ . We decompose $Q'=V^T DV$ as above and call $a_1,\ldots,a_m$ the associated diagonal entries of Q. Since $a_1$ equals the upper left entry of Q $^{\prime}$ it is integral and thus, $a_1 \geq 1$ . This construction of $a_1,\ldots,a_m$ with respect to Q is not unique. If we assume that Q $^{\prime}$ is Hermite-reduced, then $a_i$ is of the same size as the i-th successive minimum of Q, cf. [Reference Cassels8, section 12, theorem 3·1].

Lemma 9. Let Q correspond to a positive, integral quadratic form of level N with associated diagonal entries $a_1,\ldots,a_m$ . Then,

(15) \begin{align} a_1 \cdots a_m = \det Q \quad \text{and} \quad \Big(\frac{3}{4}\Big)^{i-1} \le a_i \leq \Big(\frac{4}{3}\Big)^{m-i} N \text{ for every } 1 \le i \le m.\end{align}

Furthermore, it holds that

(16) \begin{align} \sum_{x \leq l} r(Q,x)^2 \ll \left(1 + \frac{l^\frac{1}{2}}{\sqrt{a_1}} + \frac{l}{\sqrt{a_1 a_2}} + \cdots + \frac{l^{m-1}}{\sqrt{a_1 a_2} a_3 \cdots a_m} \right) l^{\epsilon}.\end{align}

More precisely, for $2 \le j \le 2m-2$ the j-th term is given by $\frac{l^{j/2}}{ A_j}$ where

\begin{align*}A_j = \prod_{i=1}^{\lfloor \frac{j+3}{2} \rfloor} \sqrt{a_i} \prod_{k=3}^{\lfloor \frac{j+2}{2} \rfloor} \sqrt{a_k} = \det Q \left(\sqrt{a_1 a_2} \prod_{i= \lfloor \frac{j+5}{2} \rfloor}^{m} \sqrt{a_i} \prod_{k=\lfloor \frac{j+4}{2} \rfloor}^{m} \sqrt{a_k}\right)^{-1}.\end{align*}

Moreover, we have

\begin{align*} \sum_{x \leq l} r(NQ^{-1},x)^2 \ll \left(1 + \frac{ l^{\frac{1}{2}}\sqrt{a_{m}}}{\sqrt{N}} + \sum_{j=2}^{2m-2} \frac{l^\frac{j}{2} \tilde{A}_j }{N^\frac{j}{2} } \right) l^{\epsilon}, \end{align*}

where

\begin{align*}\tilde{A}_j \! = \! \! \prod_{i=m-\lfloor \frac{j+1}{2} \rfloor}^{m} \! \sqrt{a_i} \prod_{k=m-\lfloor \frac{j}{2} \rfloor}^{m-2} \sqrt{a_k}.\end{align*}

Proof. Since $r(Q,n) = r(U^T Q U, n)$ for any $U \in \textrm{GL}_{m}({\mathbb Z})$ , we may assume that $Q \in \mathcal{S}({4}/{3},{1}/{2})$ . If we decompose $Q=V^T D V$ with V and D as above, we obtain that

\begin{align*}q(x) = \frac{1}{2} x^T Q x = \frac{a_1}{2} (x_1+ v_{1,2} x_2 + \cdots + v_{1,m} x_{m})^2 + \dots + \frac{a_{m}}{2} x_{m}^2.\end{align*}

We count the number of solutions of $q(x) =n$ . For $x_3,\ldots,x_{m}$ there are at most

\begin{align*}\left(2 \sqrt{\frac{2n}{a_3}}+1\right) \cdots \left(2 \sqrt{\frac{2n}{a_{m}}}+1\right)\end{align*}

choices. After that, we are left with a binary problem that has $\mathcal{O}(n^{{\epsilon}})$ solutions uniformly in all parameters, cf. [Reference Blomer and Pohl5, lemma 8(a)]. This yields

\begin{align*}r(Q,n) \ll_m \left(1 + \frac{n^{1/2}}{\sqrt{a_3}}+ \cdots + \frac{n^{\frac{m}{2}-1}}{\sqrt{a_3\cdots a_{m}}} \right) n^{\epsilon}.\end{align*}

If we want to count all solutions of $q(x) \leq l$ , we proceed as before, but bound the number of solutions for $x_1$ and $x_2$ by $\big(2\sqrt{{2l}/{a_1}} +1 \big) \text{ and } \big(2\sqrt{{2l}/{a_2}}+1\big).$ Thus, we obtain

\begin{align*}\sum_{x \leq l} r(Q,x) \ll 1 + \max_{1 \leq j \leq m} \frac{l^{\frac{j}{2}}}{ \sqrt{a_1} \ldots \sqrt{a_j}}.\end{align*}

Then, we obtain (16) by applying these bounds and using that $a_1 \cdots a_m = \det Q$ .

Next, we consider the quadratic form corresponding to $NQ^{-1}$ . Note that

\begin{align*}N Q^{-1} = N (V^T D V)^{-1} = V^{-1} (N D^{-1}) V^{-T}.\end{align*}

The diagonal entries of $ND^{-1}$ are

\begin{align*}\frac{N}{a_1}, \ldots, \frac{N}{a_{m}} \quad \text{with} \quad \frac{N}{a_i} \ge \frac{3}{4} \frac{N}{a_{i+1}}.\end{align*}

The lowest diagonal entry ${N}/{a_{m}}$ of D equals the lower right entries of $N Q^{-1}$ since $V^{-1}$ is an upper and $V^{-T}$ a lower diagonal matrix with ones on the diagonal. Hence, ${N}{a_{m}} \in {\mathbb Z}$ and thus $a_m \leq N$ . Together with $a_i \geq 1$ and $a_i \leq \frac{4}{3} a_{i+1}$ , this gives (15).

Moreover, it holds that

\begin{align*}\tilde{q}(x) &\,{:\!=}\, \frac{1}{2} x^T N Q^{-1} x =\frac{1}{2} x^T V^{-1} (N D^{-1}) V^{-T} x \\ &= \frac{N}{2 a_1} x_1^2 + \cdots + \frac{N}{2 a_m} ( w_{1,m} x_1 + \cdots + w_{m-1,m} x_{m-1}^2 + x_m)^2,\end{align*}

where $w_{ij}$ are the entries of $V^{-1}$ . Their absolute value $|w_{ij}|$ is bounded from above by a constant. To see this, write $V= I + M$ for the $m\times m$ identity I and a nilpotent matrix M. Then, a simple calculation shows that $V^{-1} = (I+M)^{-1} = I + \sum_{k=1}^{m-1} (\!-1)^k M^k$ .

Next, we count solutions of $\tilde{q}(x) =n$ . For $x_1,\cdots x_{m-2}$ there are

\begin{equation*}\left(2 \sqrt{n\cdot \frac{ 2a_1}{N}}+1\right) \cdots \left(2 \sqrt{n \cdot \frac{ 2a_{m-2}}{N}}+1 \right)\end{equation*}

choices and $\mathcal{O}(n^{\epsilon})$ for $x_{m-1} $ and $x_m$ . It follows that

(17) \begin{align} r(NQ^{-1},n) \ll \left(1 + \frac{n^{1/2} \sqrt{a_{m-2}}}{\sqrt{N}}+ \cdots + \frac{n^{\frac{m}{2}-1}\sqrt{a_1 \cdots a_{m-2} }}{N^{\frac{m}{2}-1}} \right) n^{\epsilon}.\end{align}

Moreover, we obtain by the same argument as above that

\begin{align*}\sum_{x \leq l} r(NQ^{-1},x) \ll 1 + \max_{1 \leq j \leq m} l^{\frac{j}{2}} \cdot \frac{ \sqrt{a_{m-j+1}} \ldots \sqrt{a_m}}{N^{\frac{j}{2}}}.\end{align*}

Combining these bounds yields the last claim.

The level of a form is easily determinable by the local splittings:

Lemma 10. Let $q= x^T Q x/2$ be equivalent over ${\mathbb Z}_p$ to (4) for odd p and to (5) for $p=2$ . Set

\begin{align*}\nu(p) = \max_{1 \leq i \leq m} \nu_i(p) \text{ for odd } p \text{ and } \nu(2) = \max_{1 \leq i \leq m} ( \nu_i(2) + 2 \delta_{ i > 2 r_2}).\end{align*}

Then, the level N of Q is given by $N = \prod_{p} p^{\nu(p)}$ .

Proof. For odd p, choose $U \in \textrm{GL}_m({\mathbb Z}_p)$ such that $\tilde{Q} = U^T Q U$ is diagonal with entries $u_i p^{\nu_i(p)}$ . It follows that $N \tilde{Q}^{-1} = U^{-1} N Q^{-1} U^{-T}$ , thus $p^{\nu_i(p)} \mid N$ for all p. For the other inclusion, we ise that every entry of $p^{\nu(p)} Q^{-1} = U^{-1} p^{\nu(p)} \tilde{Q}^{-1} U^{-T}$ is in ${\mathbb Z}_p \cap {\mathbb Q} = \{{a}/{b} \in {\mathbb Q} \mid p \nmid b \}$ . For $p=2$ , we choose $U \in \textrm{GL}_m({\mathbb Z}_2)$ such that $\tilde{Q} = U^T Q U$ is of the form (5). By

\begin{align*}\begin{pmatrix}2 &\quad 1 \\ 1 &\quad 2\end{pmatrix}^{-1} = \frac{1}{3} \begin{pmatrix}2 &\quad -1 \\ -1 &\quad 2\end{pmatrix}, \quad \begin{pmatrix}0 &\quad 1 \\ 1 &\quad 0\end{pmatrix}^{-1} = \begin{pmatrix}0 &\quad 1 \\ 1 &\quad 0\end{pmatrix},\end{align*}

and the same argument as before, the claim follows

To evaluate the Fourier coefficients of $\theta(Q,z)|\left(\begin{smallmatrix}a & b \\ c & d\end{smallmatrix}\right)$ , we construct a diagonal matrix D that depends on Q and c. To this end, let $c= \tilde{c} 2^t$ for odd $\tilde{c}$ and fix $U \in \textrm{SL}_m({\mathbb Z})$ such that $x^T U^T Q U x/2 \equiv \sum_{i=1}^m q_i x_i^2 \,(\textrm{mod}\ \tilde{c})$ and

\begin{align*}\frac{1}{2}x^T U^T Q U x \equiv \sum_{i=1}^{r} 2^{\tilde{\nu}_i} g_i(x_{2i-1}, x_{2i})+ \sum_{i=2r+1}^m 2^{\nu_i} u_i x_i^2 ~(\textrm{mod}\ 2^t)\end{align*}

for odd $u_i \in {\mathbb Z}$ and $g_i(x_{2i-1}, x_{2i}) = (x_{2i-1}^2 + x_{2i-1} x_{2i} + x_{2i}^2)$ or $x_{2i-1} x_{2i}$ , $\tilde{\nu}_i,\nu_i \le t$ . We set $\tilde{d}_i = (q_i,\tilde{c},N)$ , $t_{2i-1} = t_{2i} = \tilde{\nu}_i$ for $i \leq r$ ,

\begin{align*}t_i = \begin{cases} \nu_i & t \leq \nu_i +1, \\ \\[-7pt] \nu_i +1 & t \geq \nu_i +2, \end{cases} \text{ for } i > 2r \text{ and } D \,{:\!=}\, \textrm{diag}\ (\tilde{d}_1 2^{t_1} ,\ldots,\tilde{d}_m 2^{t_m} ).\end{align*}

Furthermore, we set $\eta = \eta(Q,t) \,{:\!=}\, \#\{ i > 2r \mid t > \nu_i \}$ , $\kern3pt\hat{\kern-3pt d} \,{:\!=}\, d/2$ if $2^2 \mid\!\mid N, 2 \mid\!\mid d$ and $2r<m$ and $\kern3pt\hat{\kern-3pt d}=d$ otherwise. Moreover, let $u_i$ denote the i-th column of U and $\tilde{x}_i \,{:\!=}\, {u_i^T x}/{2^{\nu_i}} \in {\mathbb Z}$ . Then, we set $\delta(x) =0$ if there is $i>2r$ such that $\tilde{x}_i$ is even and $t=\nu_i+1$ or $\tilde{x}_i$ is odd and $t > \nu_i+2$ . In all other cases, we put $\delta(x)=1$ .

Proof. By construction $2 D^{-1} \tilde{Q}$ corresponds to a positive, integral quadratic form. Over ${\mathbb Z}_p$ and p odd, $q= x^T D^{-1} \tilde{Q} x$ is equivalent to $\tilde{u}_1 \frac{p^{\nu_1}}{(p^{\nu_1},\,d)} x_1^2 + \cdots + \tilde{u}_m \frac{p^{\nu_m}}{(p^{\nu_m},\,d)} x_m^2.$ Over ${\mathbb Z}_2$ , q is equivalent to

\begin{align*}\sum_{i=1}^{r_2} 2^{\nu_{i}+1 -t_i} g(x_{2i-1},x_{2i})+ \sum_{i=2r_2+1}^m 2^{\nu_{i}+1 -t_i}u_i x_i^2,\end{align*}

By Lemma 10, this implies that $2 D^{-1} \tilde{Q}$ has level $4{N}/{d}$ if the maximum of $2^{\nu_{i}+1 -t_i} + 2 \delta_{i> 2r}$ is attained for $i>2r$ and $t = \nu_i+1$ and $2{N}/{d}$ otherwise. Hence, ${N}D \tilde{Q}^{-1} D/d$ has even diagonal entries unless $2^2 \mid\!\mid N, 2 \mid\!\mid d$ and $2r < m$ .

For simplicity, we omit the weight ${m}/{2}$ in the slash operator $|[\varrho]_{\frac{m}{2}}$ .

Lemma 12. Let $\varrho = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right) \in \textrm{SL}_2({\mathbb Z})$ , with $\varrho \notin \Gamma_0(N)$ . Moreover, let $D, \eta, \delta(x)$ and $\kern3pt\hat{\kern-3pt d}$ as above Lemma 11 and $\tilde{Q}$ as in Lemma 11. Then,

\begin{align*}\theta(Q,z)|[\varrho] = \frac{\sqrt{2^\eta \det D}}{\sqrt{\det Q}} \sum_{x \in {\mathbb Z}^{m}} \delta(x) \omega e\left(\frac{1}{2} x^T D \tilde{Q}^{-1} D x z\right),\end{align*}

for some $|\omega|=1$ . For $S \,{:\!=}\, \frac{N}{\kern3pt\hat{\kern-3pt d}}D \tilde{Q}^{-1} D$ , it follows that

\begin{align*}\theta(Q,z)|[\varrho] = \sum_{n \geq 0} a(n) e\left(\frac{\kern3pt\hat{\kern-3pt d} nz}{N}\right) \text{ with } |a(n)|\leq 2^{\frac{m}{2}} \frac{\sqrt{\det D }}{\sqrt{\det Q}} r(S,n).\end{align*}

If $d=(c,N)=1$ , we have that $|a(n)| = {r(NQ^{-1})}/{\sqrt{\det Q}}$ .

Proof. By the factorisation $ \varrho = \left(\begin{smallmatrix} 1 & a / c \\ & 1 \end{smallmatrix}\right) \left(\begin{smallmatrix} & - 1 / c \\ c \end{smallmatrix}\right)\left( \begin{smallmatrix} 1 & d / c \\ & 1 \end{smallmatrix}\right)$ and the transformation formula for the generalised theta series, cf. [Reference Siegel27, p. 575], we get that

(18) \begin{align} \theta(Q,z) | [\varrho] = \sum_{x \in {\mathbb Z}^{m}} \alpha(x,Q,\varrho) e\left(\frac{1}{2} x^T Q^{-1} x z\right), \end{align}

where

(19) \begin{align} \alpha(x,Q,\varrho) = \frac{e\left(\frac{3m}{8}\right)}{c^{\frac{m}{2}} \sqrt{\det Q}} e\left(\frac{d}{2 c} x^T Q^{-1} x\right) \sum_{v \in ({\mathbb Z}/c{\mathbb Z})^{m}} e \left(\frac{1}{c} v^T x + \frac{a}{2 c}v^T Q v\right). \end{align}

We write $c = \tilde{c} \, 2^t$ with odd $\tilde{c}$ and decompose this sum into

\begin{align*} \sum_{v \in ({\mathbb Z}/\tilde{c}{\mathbb Z})^{m}} e \left(\frac{a 2^{t} \frac{1}{2} v^T Q v + v^Tx}{\tilde{c}} \right) \sum_{w \in ({\mathbb Z}/2^t{\mathbb Z})^{m}} e \left(\frac{a \tilde{c} \frac{1}{2} w^T Q w + w^Tx}{2^t} \right). \end{align*}

Next, we choose $U \in \textrm{SL}_m({\mathbb Z})$ as in Lemma 11 and substitute v, w by Uv, Uw. Since $ v^TU^TQUv/2 \equiv \sum_{i=1}^m q_i v_i^2 \,(\textrm{mod}\ \tilde{c})$ we obtain for $G(a,b,c) \,{:\!=}\, \sum_{x\textrm{mod}\ c} e\Big(({ax^2+bx})/{c}\Big)$ that

\begin{align*} \sum_{v \in ({\mathbb Z}/\tilde{c}{\mathbb Z})^{m}} e \left(\frac{a 2^{t} \frac{1}{2} v^T Q v + v^Tx}{\tilde{c}} \right) = \prod_{i=1}^m G(q_i a 2^t,u_i^Tx,\tilde{c}),\end{align*}

where $u_i$ denotes the i-th column of U. The i-th sum vanishes unless $d_i \mid u_i^T x$ for $d_i \,{:\!=}\, (\tilde{c},q_i)$ . By completing the square, the previous display equals

(20) \begin{align} \tilde{c}^\frac{m}{2} {\epsilon}_{\tilde{c}}^m \prod_{i=1}^m \sqrt{\tilde{d}_i} \left(\frac{a 2^t \tilde{q}_i }{\tilde{c}_i}\right) e\Big(\!-\frac{\overline{ 2^{t+2} a \tilde{q}_i} \tilde{x}_i^2}{\tilde{c}_i}\Big), \end{align}

where $\tilde{q}_i = q_i / d_i, \tilde{c}_i = \tilde{c}/d_i$ and $\tilde{x}_i=u_i^T x / d_i$ . The absolute value of the term above is $\tilde{c}^\frac{m}{2} \sqrt{\tilde{d}_1} \cdots \sqrt{\tilde{d}_m}$ . Furthermore, we have for U as in Lemma 11 that

\begin{align*} \frac{1}{2}x^TU^TQUx \equiv \sum_{i=1}^{r_1} 2^{\tilde{\nu}_i} g(x_{2i-1},x_{2i}) + \sum_{i=r_1+1}^{r_2} 2^{\tilde{\nu}_i} x_{2i-1} x_{2i} + \sum_{i=2r_2+1}^m 2^{\nu_i} b_i x_i^2 \,(\textrm{mod}\ 2^t), \end{align*}

where $b_i$ are odd integers, $0\leq 2r_1 \leq 2r_2 \leq m$ , $g(x_{2i-1},x_{2i}) = x_{2i-1}^2 + x_{2i-1}x_{2i} + x_{2i}^2$ and $\tilde{\nu_i},\nu_i \leq t$ . For $i > 2r_2$ , we need to evaluate $ G(2^{\nu_i} b_i a \tilde{c}, u_i^T x, 2^t )$ which vanishes unless $u_i^T x \equiv 0\,( \textrm{mod}\ 2^{\nu_i})$ . For $\tilde{x}_i ={u_i^T x}/{2^{\nu_i}}$ , this sum $G(2^{\nu_i} b_i a \tilde{c}, u_i^T x, 2^t )$ equals

\begin{align*} 2^{\nu_i} G(b_ia \tilde{c},\tilde{x_i},2^{t-\nu_i})= \begin{cases} 2{\dfrac{\nu_i+t}{2}} & t= \nu_i, \\ \\[-8pt] 2{\dfrac{\nu_i+t+1}{2}} \delta_{\tilde{x}_i \equiv 1 \,(\textrm{mod}\ 2)} & t = \nu_i+1, \\ \\[-8pt] 2{\dfrac{\nu_i+t+1}{2}} \omega e\Big(\frac{-\overline{b_i a \tilde{c}} (\tilde{x}_i/2)^2}{2^{t-\nu_i}}\Big) \delta_{\tilde{x}_i \,{\equiv}\,0\, (\textrm{mod}\ 2)} & t \geq \nu_i +2, \end{cases} \end{align*}

where $\omega \,{:\!=}\, {\epsilon}_{ b_i a \tilde{c}}^{-1} \left({2^{t-\nu_i}}/{b_i a \tilde{c}}\right) e(1/8)$ . For forms of type $2^{\tilde{\nu}} (x_1^2 + x_1 x_2 +x_2^2)$ , we have that

(21) \begin{align} \sum_{w_1 (\textrm{mod}\ 2^{t})} e\left( \frac{ w_1^2 a \tilde{c} 2^{\tilde{\nu}} + w_1 u_1^T x_1}{2^t} \right) &\sum_{w_2 (\textrm{mod}\ 2^{t})} e\left( \frac{ w_2^2 a \tilde{c} 2^{\tilde{\nu}} + w_2 \left(u_1^T x_1 + a \tilde{c} 2^{\tilde{\nu}} w_1\right)}{2^t} \right) \end{align}

\begin{align} \notag&= 2^{\tilde{\nu}+t} \zeta e\left(\frac{\overline{3a \tilde{c}} \left(\tilde{x}_1^2 - \tilde{x}_1 \tilde{x}_2 +\tilde{x}_2^2 \right)}{2^{t-\tilde{\nu}}}\right) \delta_{ \tilde{x}_1 \in {\mathbb Z}, \tilde{x}_2 \in {\mathbb Z}}\end{align}

where $\tilde{x}_i \,{:\!=}\, {u_i^T x}/{2^{\tilde{\nu}}}$ and $\zeta$ is a root of unity that does not depend on x.

For $t= \tilde{\nu}$ the claim is obvious. If $t = \tilde{\nu}+1$ , the latter sum vanishes unless $\tilde{x}_2 + a \tilde{c} w_1$ is odd and (21) equals

\begin{align*}2^{\tilde{\nu}+t} e\left(\frac{\overline{a \tilde{c}} (\tilde{x}_1^2 - \tilde{x}_1 \tilde{x}_2 + \tilde{x}_2^2)}{2}\right).\end{align*}

If $t \geq \tilde{\nu}+2$ , the second sum vanishes unless $\tilde{x}_1 + a \tilde{c} w_1$ is even. In this case (21) equals

\begin{align*}2^{\frac{3\tilde{\nu}+t+1}{2}} &\omega \sum_{w_1 (\textrm{mod}\ 2^{t-\tilde{\nu}-1})} e\left(\frac{(2 w_1 - \overline{a \tilde{c}} \tilde{x}_2)^2 a \tilde{c} + (2 w_1 - \overline{a \tilde{c}} ) \tilde{x}_1 - a v_1^2}{2^{t-\tilde{\nu}}} \right) \\[3pt] &= 2^{\frac{3\tilde{\nu}+t-1}{2}} \omega e\left(\frac{\overline{a \tilde{c}} \left(\!-\tilde{x}_2^2+ \tilde{x}_1 \tilde{x}_2\right)}{2^{t-\tilde{\nu}}}\right) G(3a\tilde{c},2(2\tilde{x}_2-\tilde{x}_1),2^{t-\nu}) \\[3pt] &= 2^{\tilde{\nu}+t} \zeta e\left(\frac{\overline{3a \tilde{c}} \left(\tilde{x}_1^2 - \tilde{x}_1 \tilde{x}_2 +\tilde{x}_2^2 \right)}{2^{t-\tilde{\nu}}}\right).\end{align*}

For forms of type $2^{\tilde{\nu}} x_{1} x_{2}$ we obtain by a simple computation

(22) \begin{align} &\sum_{w_{1}(\textrm{mod}\ 2^t)} e\left(\frac{w_1 u_1^T x_{1}}{2^t} \right) \sum_{w_{2} (\textrm{mod}\ 2^t)} e\left(\frac{w_2 \left(u_2^T x_{2} + a w_1 2^{\tilde{\nu}}\right)}{2^{t}} \right) = 2^{t+\tilde{\nu}} e\left(\frac{\overline{a \tilde{c}} \tilde{x}_1 \tilde{x}_2}{2^{t-\nu}}\right)\end{align}

provided that $\tilde{x}_1, \tilde{x}_2 \in {\mathbb Z}$ .

It follows that $\alpha(x,Q,\varrho)$ vanishes unless $D^{-1} U^T x \in {\mathbb Z}^m$ . If this holds, we have that

\begin{align*}\alpha(x,Q,\varrho) = \omega \, \delta(D^{-1} U^T x) \frac{\sqrt{2^\eta \det D}}{\sqrt{\det Q}}\end{align*}

for some $\omega$ with $|{\omega}|=1$ . Substituting x by $U^{-T} D x$ in (18) yields the first claim of the lemma.

The absolute value of the n-th Fourier coefficient of $\theta|[\varrho](z) = \sum_n a(n) e(\kern3pt\hat{\kern-3pt d} nz/N)$ is given by

\begin{align*}|a(n)| = \left|\sum_{\frac{1}{2}x^T S x=n } \alpha(x,Q,\varrho)\right| \leq 2^{\frac{m}{2}}\frac{\sqrt{ \det D} }{\sqrt{\det Q}} r(S,n),\end{align*}

where $S\,{:\!=}\, \frac{N}{\kern3pt\hat{\kern-3pt d}} D \tilde{Q}^{-1}D$ . For $d=(N,c)=1$ , we have that $\eta= \delta(x)=1$ for all x. To prove $|a(n)| = r(NQ^{-1}) / \sqrt{\det Q}$ , it remains to show that $\alpha(x,Q,\varrho)$ only depends on n and not on the particular choice of x. By (19) it suffices to consider the Gauss sum modulo c. By (20), the dependence on x modulo $\tilde{c}$ is given by

\begin{align*} e\left(\!-\frac{\overline{2^{t+2}a} \sum_{i=1}^m \overline{q_i} \tilde{x}_i^2}{\tilde{c}} \right) = e\left(\!-\frac{n\overline{2^{2t+2}aN} }{\tilde{c}} \right). \end{align*}

For the last step, we use that $d_i=1$ and hence, $U^TQU/2 = \textrm{diag}(q_1,\ldots,q_m)$ is invertible modulo $\tilde{c}$ which implies that $n = x^T N Q^{-1} x/2 \equiv N \sum_{i=1}^m \overline{q_i} \tilde{x}_i^2 \,(\textrm{mod}\ \tilde{c})$ . If N is even, then c must be odd. If N is odd, then m must be even and there are no diagonal terms in the splitting over ${\mathbb Z}_2$ . Since the inverse of $(\begin{smallmatrix} 2 & 1 \\ 1 & 2 \end{smallmatrix})$ modulo $2^t$ is given by $\overline{3} (\begin{smallmatrix} 2 & -1 \\ -1 & 2 \end{smallmatrix})$ we obtain that

\begin{align*} n = \frac{1}{2} x^T N (U^T Q U)^{-1} x = N \sum_{i=1}^{r_1} \overline{3} \left(x_{2i-1}^2 - x_{2i-1} x_{2i} + x_{2i}^2\right) + N \sum_{i=1}^{r_1} x_{2i-1} x_{2i} \,(\textrm{mod}\ 2^t). \end{align*}

It follows by (21) and (22) that the dependence on x of the Gauss sum modulo $2^t$ is given by $e\Big(\!-{n\overline{\tilde{c}aN}}/{2^t} \Big)$ .

This provides the necessary tools to prove our main result:

Proof of Theorem 1 and 2. Both f and g are cusp forms since the value of $\theta(Q,z)$ at each cusp depends only on a genus-invariant term. Set $\mu = [\textrm{SL}_2({\mathbb Z}) : \Gamma_{0}(N)]$ , $\tilde{\mu} = [\textrm{SL}_2({\mathbb Z}): \Gamma(N)]$ and recall that

\begin{align*} \frac{\tilde{\mu}}{\mu} = N^2 \prod_{p \mid N } \left( {1- \frac{1}{p}} \right). \end{align*}

We obtain

\begin{align*} \langle g,g \rangle = \int_{\Gamma_{0}(N) \backslash {\mathbb H}} |{g(z)}|^2 y^{\frac{m}{2}-2} dx\,dy = \frac{\mu}{\tilde{\mu}} \sum_{j=1}^{\tilde{\mu}} \int_{\textrm{SL}_2({\mathbb Z}) \backslash {\mathbb H}} \big|g|[\tau_j])(z)\big|^2 y^{\frac{m}{2}-2} dx\,dy \end{align*}

for a set of coset representatives $\tau_i$ of $\Gamma(N) \backslash \textrm{SL}_2({\mathbb Z})$ . On this set, we define an equivalence relation

\begin{align*} \tau_i \sim \tau_j \Leftrightarrow \tau_i \in \Gamma(N) \tau_j T \quad \text{for } T= \left. \left\{ \begin{pmatrix} 1 &\quad x \\ &\quad 1 \end{pmatrix} \, \right| \, 0 \leq x \leq N-1 \right\}. \end{align*}

For the remaining matrices, a set of representatives is given by

\begin{equation*}\left\{ \varrho_{a,c} = \begin{pmatrix} a &\quad * \\ c &\quad * \end{pmatrix} \in \textrm{SL}_2({\mathbb Z}) \mid a, c\, (\textrm{mod}\ N)\right\}.\end{equation*}

For simplicity, we set $\varrho \,{:\!=}\, \varrho_{a,c}$ . It follows that

(23) \begin{align} \langle g,g \rangle = \frac{\mu}{\tilde{\mu}} \sum_{c\, (\textrm{mod}\ N)} \sum_{\substack{a (\textrm{mod}\ N) \\ (a,c,N)=1}} \bigcup_{t \in T} \int_{t F} \big|g|[\varrho](z)\big|^2 y^{\frac{m}{2}-2} dx\,dy, \end{align}

where $F = \{x+iy \mid - {1}/{2} \le x \le {1}/{2}, |x+iy| \geq 1 \}$ is a fundamental domain of $\textrm{SL}_2({\mathbb Z}) \backslash {\mathbb H}$ . We put $d=(c,N)$ and write $g|[\varrho] = \sum_n a_{\varrho}(n) e\big({n \kern3pt\hat{\kern-3pt d} z}/{N} \big)$ for $\kern3pt\hat{\kern-3pt d}$ either $d/2$ or d. By the previous display, it follows that

\begin{align} \notag \langle g, g \rangle &\le N \frac{\mu }{\tilde{\mu}} \sum_{\substack{ c (\textrm{mod}\ N) \\ (c,N)=d}} \sum_{\substack{a (\textrm{mod}\ N) \\ (a,d)=1}} \int_{\frac{1}{2}}^{\infty} \sum_{n=1}^\infty |a_{\varrho}(n)|^2 \textrm{exp}\left(\!- \frac{2 \pi y n d}{N}\right) y^{\frac{m}{2}-2} dy \\[4pt] \notag &\ll N \frac{\mu }{\tilde{\mu}} \sum_{\substack{ c\, (\textrm{mod}\ N) \\ (c,N)=d}} \sum_{\substack{a (\textrm{mod}\ N) \\ (a,d)=1}} \Big(\frac{N}{d}\Big)^{\frac{m}{2}-1} \sum_{n \leq (N/d)^{1+{\epsilon}}} \frac{|a_{\varrho}(n)|^2}{n^{\frac{m}{2}-1}} + \mathcal{O}(N^{-100}). \end{align}

For $c=0$ , we have $d=N$ and $a_\rho(n) = r(Q,n) - r(Q',n)$ . Hence, the contribution of this case is bounded by $\mathcal{O}(N^{\epsilon})$ . For $c \neq 0$ we apply Lemma 12. This gives

\begin{align*} |a_\varrho(n)|^2 \le \frac{\det D_\varrho}{\det Q} (r(S_\varrho,n)^2 + r(\tilde{S}_\varrho,n)^2), \end{align*}

for matrices $D_\varrho, S_\varrho,\tilde{S}_\varrho$ given as in Lemma 12. Let $a'_1,\ldots,a'_m$ denote the associated diagonal entries of $S_\varrho$ . By Lemma 9 we infer that

\begin{align*} \sum_{n \le (N/d)^{1+{\epsilon}}} \frac{r(S_\varrho,n)^2}{n^{\frac{m}{2}-1}} &\ll \left( 1+ \sum_{j=m-1}^{2m-2} \Big(\frac{N}{d}\Big)^{\frac{j-m}{2}+1} \frac{\sqrt{a'_1 a'_2}}{\det S_d} \prod_{i=\lfloor \frac{j+5}{2}\rfloor}^m \sqrt{a'_i} \prod_{k=\lfloor \frac{j+4}{2}\rfloor}^m \sqrt{a'_k}\right) N^{\epsilon}. \end{align*}

To estimate this sum, we use $a'_1 a'_2 \ll (\det S_\varrho)^{\frac{2}{m}}$ and $a'_i \ll N$ for $3 \le i \le m$ and $\det S_\varrho = (N/d)^m (\det D_\varrho)^2 (\det Q)^{-1}$ . This yields that the j-th term is bounded by

\begin{align*} N^{-\frac{m}{2}+1} d^{\frac{3m-j}{2}-2} (\det Q)^{1-\frac{1}{m}} (\det D)^{\frac{2}{m}-2}. \end{align*}

If m is even, we use that $a_i \gg 1$ and bound the $j=m-1$ term by

\begin{equation*}\frac{\sqrt{N}}{\sqrt{d}\sqrt{\det S_d}} \prod_{i=\frac{m}{2}+2}^{m} \sqrt{a'_i} \ll \frac{d^{\frac{m}{2}} N^{-\frac{m}{4}} \sqrt{\det Q}}{\det D_\varrho}.\end{equation*}

By construction, $\det D_\varrho$ only depends on $d=(c,N)$ . Since there are $\phi({N}/{d})$ choices for $(c,N)=d$ and ${\phi(d)}/{d}N$ for a, it follows by the previous five displays that

(24) \begin{align} \langle g,g \rangle \ll N^{\epsilon} + \frac{N^\frac{m}{2}}{\det Q}\sum_{d \mid N} \frac{\det D_\varrho}{d^\frac{m}{2}} + \frac{N^{\frac{m}{4}}}{\sqrt{\det Q}}+\frac{N}{ \det Q^\frac{1}{m}} \sum_{d \mid N} \frac{d^{\lfloor\frac{m-3}{2}\rfloor}}{(\det D_\varrho)^{1-\frac{2}{m}}}.\end{align}

For the first d-sum, we use that

(25) \begin{align} \frac{\det D_\varrho }{\det Q} \leq \frac{F(D_\varrho,\frac{m}{2})}{F(Q,\frac{m}{2}) } \le \frac{d^{\frac{m}{2}}}{F(Q,\frac{m}{2})}.\end{align}

This bound is sharp for at least one d which can be constructed as follows. Let $v_p(j)$ defined as after (5), choose $k_p$ maximal such that $v_{p}(k_p) \geq {m}/{2}$ and set $d= \prod_{p \mid N} p^{k_p}$ .

By (24) and (25) we obtain the bound for $m=3$ and $m=4$ . For $m \ge 5$ , we estimate

\begin{equation*}\frac{d^{\frac{m}{2}-\frac{3}{2}}}{(\det D_\varrho)^{1-\frac{2}{m}}} \le \frac{d^{\frac{m}{2}-\frac{3}{2}}}{F(D_\varrho,\frac{m-1}{2}-\frac{1}{m})^{1-\frac{2}{m}}} \le \frac{N^{\frac{m}{2}-1}}{F (Q,\frac{m-1}{2}-\frac{1}{m})^{1-\frac{1}{m}}}.\end{equation*}

By definition, the same bounds hold for $\langle \,f,f \rangle$ . If Q is diagonal, we can do slightly better. The Fourier coefficients $a_\varrho(n)$ of $(f|[\varrho])(z)$ are bounded by

\begin{equation*}\frac{\det D_\varrho}{\det Q} (r(S_\varrho,n) + r(\textrm{gen}\ S_\varrho,n)).\end{equation*}

For the latter term, we make use of (14):

\begin{align*} \frac{\det D_\varrho}{\det Q} \! \sum_{n \le (N/d)^{1+{\epsilon}}} \! \frac{r(\textrm{gen}\ S_\varrho,n)^2}{n^{\frac{m}{2}-1}} \ll \frac{d^m}{N^m \det D_\varrho } \! \sum_{n \le (N/d)^{1+{\epsilon}}} \! n^{\frac{m}{2}-1} (n,N) \ll \Big(\frac{N}{d}\Big)^{-\frac{m}{2}} N^{{\epsilon}}.\end{align*}

As a result, the contribution of $r(\textrm{gen}\ S_\varrho,n)$ is bounded by $\mathcal{O}(N^{{\epsilon}})$ .

For $Q = \textrm{diag}(a_1,\ldots,a_n)$ , the diagonal entries of $S_\varrho$ are given by

\begin{equation*}\frac{N (a_1,d)^2}{ d \, a_1}, \ldots, \frac{N (a_m,d)^2}{ d \, a_m}. \end{equation*}

We sort these entries by value and denote them by $a'_1,\ldots,a'_m$ . For $3 \le l_1,l_2 \le m+1$ , we estimate

\begin{align*}\sqrt{a'_1 a'_2} \prod_{i=l_1}^m \sqrt{a'_i} \prod_{j=l_2}^m \sqrt{a'_j} \le\Big(\frac{N}{d}\Big)^{\frac{2m+4-l_1-l_2}{2}} \frac{\det D_\varrho}{\sqrt{a_m a_{m-1}}}. \end{align*}

Thus, we get

\begin{align*}\sum_{d \mid N} \Big(\frac{N}{d}\Big)^{\frac{m}{2}}\frac{\det D_\varrho}{\det Q} \sum_{n \le (N/d)^{1+{\epsilon}}} \frac{r(S_\varrho,n)^2}{n^{\frac{m}{2}-1}} \ll \left( \frac{N^{\frac{m}{2}}}{\det Q} \sum_{d \mid N} \frac{\det D_\varrho}{d^{\frac{m}{2}}} + \frac{N}{\sqrt{a_m a_{m-1}}} \right) N^{\epsilon}.\end{align*}

Applying (25) gives the second upper bound of Theorem 1.

It remains to determine a lower bound:

Proof of Theorem 3. By (23), we have that

(26) \begin{align} \langle \,f,f \rangle \geq \frac{\mu}{\tilde{\mu}} N \sum_{\substack{c\,(\textrm{mod}\ N) \\ d=(c,N)}} \sum_{\substack{a (\textrm{mod}\ N) \\ (a,d)=1}} \sum_{n=1}^\infty |a_{\varrho}(n)|^2 \int_{1}^{\infty} \textrm{exp}\left(\!- \frac{4 \pi y n d}{N}\right) y^{\frac{m}{2}-2} dy,\end{align}

where $\varrho= (\begin{smallmatrix}a & * \\ c & *\end{smallmatrix})$ . For $t>0$ , it holds by substitution and [13, 2·33·5] that

\begin{equation*}\int_{1}^{\infty} \textrm{exp}(\!- t y) y^{\frac{m}{2}-2} dy \gg t^{1-\frac{m}{2}} \textrm{exp}(\!- t).\end{equation*}

We drop all terms with $(c,N) > 1$ . For the remaining c and a, we have by Lemma 12 that $|a_\varrho(n)|^2 = (\det Q)^{-1} |r(NQ^{-1},n) - r(\textrm{gen}\ NQ^{-1},n)|^2$ . Since $\frac{\mu}{\tilde{\mu}} N \varphi(N) = 1$ it follows that

\begin{align*}\langle \,f,f \rangle \gg \frac{N^{\frac{m}{2}}}{\det Q} \sum_{n=1}^\infty \frac{r(NQ^{-1},n)^2}{n^{\frac{m}{2}-1}} e^{- \frac{4 \pi n}{N}} - \frac{N^{\frac{m}{2}}}{\det Q}\sum_{n=1}^\infty \frac{r(\textrm{gen }NQ^{-1},n)^2}{n^{\frac{m}{2}-1}} e^{- \frac{4 \pi n}{N}}.\end{align*}

By (14) the latter term is bounded by $\mathcal{O}(N^{\epsilon})$ . For the former term, we apply Hermite’s theorem. This gives

\begin{equation*}\min N Q^{-1} \leq (4/3)^\frac{m-1}{2} N (\det Q)^{-\frac{1}{m}}.\end{equation*}

As a result, we obtain

\begin{align*} \sum_{n=1}^\infty \frac{r(NQ^{-1},n)^2}{n^{\frac{m}{2}-1}} \exp\Big(\!- \frac{ 2 \pi n}{N}\Big) \gg (\min N Q^{-1})^{1-\frac{m}{2}}. \end{align*}

This yields the claim.

4. Estimates for the error term

We start by deriving uniform estimates for the ternary case:

Lemma 13. Consider a positive, integral, ternary quadratic form $q(x) = x^T Q x/2$ . Then,

\begin{align*}|r(Q,n) - r(\mathrm{spn}\ Q,n)| \ll \frac{\sqrt{N}}{(\det Q)^\frac{1}{6}} \left(\frac{n^\frac{13}{28}}{N^{\frac{1}{7}}} + \frac{n^\frac{7}{16}}{N^{\frac{1}{16}}} + n^\frac{1}{4} \frac{\sqrt{(\tilde{n},N^\infty)} v^{\frac{1}{4}} \sqrt{(n,N)}}{\sqrt{N} }\right) (nN)^{\epsilon},\end{align*}

where $\tilde{n}$ is the largest divisor of n such that $(\tilde{n},N^\infty)$ is squarefree and

\begin{align*}v= \prod_{q\,{\rm{anisotropic}} \,{\rm{over}}\,p} (p^{\infty},N).\end{align*}

Proof. By a local argument of Blomer [Reference Blomer2, section 1·3], it is possible to write

\begin{align*} r(Q,n) - r(\textrm{spn}\ Q,n) = \sum_{j} \gamma_j r(Q_j,m_j v_j w_j) - r(Q'_j,m_j v_j w_j) \end{align*}

for $Q_j, Q'_j$ in the same spinor genus and of level dividing N, $\sum_j |\gamma_j| \ll (nN)^{\epsilon}$ , $m_j v_j w_j\mid n$ for all j, $(m_j,N)=1$ , $w_j$ squarefree, $v_j \mid N^2$ and q anisotropic over all prime divisors of $v_j$ . The claim now follows by applying Theorem 1 and (8).

For $m \geq 4$ , we determine an effective lower bound for n with respect to Q. To this end, recall the definition of F(Q, s) given in (6).

Lemma 14. Let Q correspond to a positive, integral, quadratic form in m variables. Then, we have for $m=4$ that

\begin{align*} |r(Q,n) - r(\mathrm{gen}\ Q,n)| &\ll n^{\frac{m}{4}-\frac{1}{2}} \left(\frac{N}{\sqrt{F(Q,2)}} + \frac{\sqrt{N}}{(\det Q)^\frac{1}{8}}\right) \\ &\times \min \left(\sqrt{N}, 1 + \frac{n^\frac{1}{4} (n,N)^\frac{1}{4} }{\sqrt{N}}\right) (nN)^{\epsilon}.\end{align*}

Let $\prod_p \beta_p(q,n) \gg \beta^{-1} (nN)^{-{\epsilon}}$ . Then, $r(Q,n) \geq 1$ holds if

\begin{align*}n \gg \beta^4 (n,N) \Big(N \frac{\det Q}{F(Q)} + (\det Q)^\frac{3}{4} \Big)^2 \quad \text{or} \quad n \gg \beta^2 N^2 \Big(N \frac{\det Q}{F(Q)} + (\det Q)^\frac{3}{4} \Big).\end{align*}

For $m \geq 5$ , we have $r(Q,n) \geq 1$ if

\begin{align*}n \gg \left( \beta^2 \sqrt{(n,N)} \left(N^{\frac{m}{2}-1} \frac{\det Q}{F\left(Q,\frac{m-1}{2}-\frac{1}{m}\right)} + (\det Q)^{1-\frac{2}{m}} \right) \right)^{\frac{2}{m-3}}.\end{align*}

Moreover, we have that

\begin{align*}\beta \leq \begin{cases}\sqrt{(n,N)} & \text{if } m=4 \text{ and } n \text{ primitively locally represented}, \\ \\[-7pt] \min( (n,N), (n,(\det Q)^{\frac{1}{m-4}})) & \text{if } m\geq 5 \text{ and } n \text{ locally represented}.\end{cases}\end{align*}

Proof. We apply (3) to bound the Fourier coefficients of $f(z) =\theta(Q,z)- \theta(\textrm{gen}\ Q,z)$ . For even m this is a result from [Reference Iwaniec and Kowalski18, corollary 14·24]. Since the Weil bound also holds for Kloosterman sums twisted by a quadratic character, cf. [Reference Waibel30, lemma 4], this result can be extended to odd m. In addition, we have for even m that

\begin{align*}a(n) \ll {f} n^{\frac{m}{4}-\frac{1}{2}} N^\frac{1}{2} (nN)^{\epsilon}\end{align*}

by applying Deligne’s bound, cf. [Reference Schulze–Pillot and Yenirce26, theorem 11].

For $m=4$ , this result implies a substantial saving compared to previous lower bounds such as [Reference Rouse23, theorem 1]. For example if $(n,N) =1$ and $\det Q = F(Q,2)$ , a lower bound is given by $n \gg (N^2 + (\det Q)^{4/3})N^{\epsilon}$ . For $m=6$ , we obtain

\begin{align*}n \gg (n^2,\det Q)^\frac{2}{3} (n,N)^\frac{1}{3} \left( N^\frac{2}{3} (\det Q)^\frac{2}{3} + \det Q\right) N^{\epsilon}\end{align*}

which improves the lower bound $n \gg (\det Q)^{\frac{12}{5}}$ of Hsia and Icaza [Reference Hsia and Icaza16].

5. Applications

We start with the problem of representing an integer by three squares of almost primes. To this end, we want to sieve the sequence

\begin{align*}\mathcal{A} = \{x_1 x_2 x_3 \mid x_1^2 +x_2^2 + x_3^2 = n, x_i \in {\mathbb Z}_{>0}.\},\end{align*}

where $n \equiv 3\,(\textrm{mod}\ 24)$ , $5 \nmid n$ . Let d denote a squarefree integer and $\mathcal{A}_d$ the subset of all $a \in \mathcal{A}$ that are divisible by d. Moreover, let $\textbf{l} = (l_1,l_2,l_3)$ , $\mu(\textbf{l}) = \mu(l_1) \mu(l_2) \mu(l_3)$ and $q_\textbf{l}(x) = l_1^2 x_1^2 + l_2^2 x_2^2 + l_3^2 x_3^2$ with corresponding matrix $Q_\textbf{l}$ . An application of the inclusion-exclusion principle gives

\begin{align*}|\mathcal{A}_d| = \mu(d) \sum_{\textbf{l} \in {\mathbb N}^3, [l_1,l_2,l_3] =d } \mu(\textbf{l}) r(Q_\textbf{l},n).\end{align*}

We decompose $r(Q_\textbf{l},n) = r(\textrm{gen}\ Q_\textbf{l},n) + a(n)$ . For the main term, we have

\begin{equation*}r(\textrm{gen}\ Q_\textbf{l},n) = \frac{\pi n^\frac{1}{2}}{4 l_1 l_2 l_3} \prod_p \beta_p(Q_\textbf{l},n).\end{equation*}

For $\textbf{1}=(1,1,1)$ , we have $\beta_p(Q_\textbf{1},n) \gg n^{-{\epsilon}}$ for $n \equiv 3\, (\textrm{mod}\ 8)$ . Hence, we may set

\begin{equation*}\omega(\textbf{l},n) \,{:\!=}\, \prod_p \beta_p(Q_\textbf{l},n) \Big( \prod_p\beta_p(Q_\textbf{1},n)\Big)^{-1}\end{equation*}

and

\begin{equation*} \frac{\Omega(d)}{d} = \prod_{p \mid d} \frac{\Omega(p)}{p} \,{:\!=}\, \mu(d) \sum_{\substack{l \in {\mathbb N}^3, [l_1,l_2,l_3]=d}} \mu(l_1) \mu(l_2) \mu(l_3) \frac{\omega(\textbf{l},n)}{l_1 l_2 l_3}.\end{equation*}

The functions $\omega(\textbf{l},n)$ are explicitly computed in [Reference Blomer and Brüdern3, lemma 3·1 and 3·2]. This yields for $n \equiv 3 \,(\textrm{mod}\ 24)$ , $5 \nmid n$ that $0 \leq \Omega(p) <p$ . Furthermore, there is constant $A \ge 2$ such that

\begin{align*}\prod_{z_1 \le p < z} \left(1- \frac{\Omega(p)}{p} \right)^{-1} \le \left(\frac{\log z}{\log z_1}\right)^3 \left(1+ \frac{A}{{\log z_1}}\right) \text{ for } 2 \le z_1 \le z.\end{align*}

For $X \,{:\!=}\, \frac{\pi}{4} n^\frac{1}{2} \prod_p \beta_p(Q_\textbf{1},n)$ , we obtain that $|\mathcal{A}_d| = \frac{\Omega(d)}{d} X + R_d(\mathcal{A})$ , where

\begin{align*}R_d(\mathcal{A}) = \mu(d)\sum_{\substack{l \in {\mathbb N}^3, [l_1,l_2,l_3]=d}} \mu(\textbf{l}) (r(Q_\textbf{l},n) - r(\textrm{gen}\ Q_\textbf{l},n)).\end{align*}

Let $F_3(s)$ and $f_3(s)$ denote the three dimensional sieve functions given in [Reference Diamond and Halberstam10, theorem 1]. Then, it holds by [Reference Diamond and Halberstam10, theorem 1] for $\kappa =3$ that:

Lemma 15. Let $\mathcal{A}, X, \Omega$ as above and write v(d) for the number of prime factors of d. Furthermore, let

\begin{align*}\sum_{d \leq X^\tau} \mu^2(d) 4^{v(d)} |R_d(\mathcal{A})| \ll \frac{X}{(\log X)^4} \end{align*}

for $\tau \in {\mathbb R}$ with $0< \tau <1$ . Then, it holds for any $u,v \in {\mathbb R}$ with $1/\tau <u <v$ and $\tau v > \beta_3 \,{:\!=}\, 6.6408$ that

\begin{align*}|\{P_r \mid P_r \in \mathcal{A}\}| \gg X \prod_{p < X^\frac{1}{v}} \Big(1 - \frac{\Omega(p)}{p}\Big)\end{align*}

provided only that

(27) \begin{align} r > 3 u -1 + \frac{3}{f_3(\tau v)} \int_{1}^{v/u} F_3(\tau v-s) \Big(1- \frac{u}{v}s \Big) \frac{ds}{s}.\end{align}

We prove the following improvement of [Reference Lü20, lemma 4·1]:

Lemma 16. Let $n \equiv 3\, (\mathrm{mod}\ 24), 5 \nmid n$ and $\tau < {3}/{58}$ . Then,

\begin{align*}\sum_{d \leq n^\frac{\tau}{2}} \mu^2(d) 4^{v(d)} |R_d(\mathcal{A})| \ll n^{\frac{1}{2}-{\epsilon}}.\end{align*}

Proof. We only consider combinations of $l_1,l_2,l_3$ such that either $r(Q_\textbf{l},n) \neq 0$ or $r(\textrm{gen}\ Q_\textbf{l},n) \neq 0$ . The remaining forms are isotropic over all odd primes and satisfy $2 \nmid l_1l_2l_3$ . It follows by (9) that $r(\textrm{gen}\ Q_\textbf{l},n) = r(\textrm{spn}\ Q_\textbf{l},n)$ . Since the level of $Q_\textbf{l}$ is given by 4d we infer by Lemma 13 that

\begin{align*}&\sum_{d \leq n^{\tau/2}} \mu^2(d) \sum_{ [l_1,l_2,l_3]=d} \mu^2(\textbf{l}) 4^{v(d)} |r(Q_\textbf{l},n) - r(\textrm{gen}\ Q_\textbf{l},n)| \\[4pt] &\ll \sum_{d \leq n^{\tau/2}} d^{\frac{2}{3}} \left( \frac{n^{\frac{13}{28}}}{d^{\frac{2}{7}}} + \frac{n^{\frac{7}{16}}}{d^{\frac{1}{8}}} + n^{\frac{1}{4}} \frac{\sqrt{(n,d)(n,d^2)}}{d} \right) d^{\epsilon},\end{align*}

where we estimated $\#\{ \textbf{l} \in {\mathbb N}^3 \mid [l_1,l_2,l_3] =d\} \le \tau(d)^3 \ll d^{\epsilon}$ in the second step. For $\tau < {3}/{58}$ the display above is bounded by $\ll n^{\frac{1}{2}-{\epsilon}}$ .

To avoid the computation of $F_3(s)$ and $f_3(s)$ , we apply a well known trick. If we set $\tau u = 1 + \zeta - \zeta/ \beta_3$ and $\tau v = \beta_3 / \zeta + \beta_k -1$ , it follows by (10·1·10), (10·2·4) and (10·2·7) in [Reference Halberstam and Richert14] that

\begin{align*}\frac{3}{f_k(\tau v)} \int_{1}^{v/u} F_k(\tau v-s) \Big(1- \frac{u}{v}s \Big) \frac{ds}{s} \le (3+ \zeta) \log \frac{\beta_3}{\zeta} -3 + \zeta \frac{3}{\beta_3}.\end{align*}

This gives the following slightly weaker version of (27):

\begin{align*}r > \frac 3 \tau (1+ \zeta) -1 + (3+\zeta) \log \frac{\beta_3}{\zeta} -3 - \zeta \frac{3 ( \frac 1 \tau -1)}{\beta_3} \,{:\!=}\, m(\zeta).\end{align*}

For $\tau = 3/58$ it follows that

\begin{align*}r > \min_{0 < \zeta < \beta_3} m(\zeta) = m(0.0560831\ldots) \approx 71.3875.\end{align*}

This proves Corollary 4.

Even for four squares of primes, current technology is not sufficient to prove that every $n \equiv 4\,(\textrm{mod}\ 24)$ is represented. By means of sieving and Theorem 1, we can address this problem for almost primes. Here, the main input is:

Lemma 17. Let $n \equiv 4\, (\mathrm{mod}\ 24)$ , $\tau < {1}/{2}$ , $q_\textbf{l}(x) = l_1^2 x_1^2 + l_2^2x_2^2+l_3^2 x_3^2 + x_4^2 l_4^2$ . Then,

\begin{align*} \sum_{d \leq n^\frac{\tau}{2}} \mu(d) \sum_{\substack{ l \in {\mathbb N}^4 \\ d=[l_1,l_2,l_3,l_4]}} \mu(\textbf{l}) 4^{v(d)} |r(Q_\textbf{l},n) - r(\mathrm{gen}\ Q_\textbf{l},d)| \ll n^{\frac{1}{2}-{\epsilon}}. \end{align*}

By applying the weighted four dimensional sieve from [Reference Diamond and Halberstam10] it follows that every $n\equiv 4\, (\textrm{mod}\ 24)$ is represented by $x_1^2 + x_2^2 + x_3^2 + x_4^2$ with $x_1 x_2 x_3 x_4 \in P_{20}$ . However, a recent approach by Tsang, Zhao [Reference Tsang and Zhao29] and Ching [Reference Ching9] yields much better results. Their idea is to choose one of the $x_i$ ’s as a prime and then to combine Chen’s switching result with a three dimensional sieve. This shows that $n = p^2 + x_1^2 + x_2^2 + x_3^2$ for a prime p and $x_1x_2x_3 \in P_{12}$ .

To tackle the problem of representing an integer by three squares of smooth numbers, we follow the approach in [Reference Blomer, Brüdern and Dietmann4]. The underlying idea is to choose integers $d_1,d_2,d_3 \in [n^{\eta}, 2 n^{\eta}]$ with $\eta$ as large as possible such that

\begin{align*}q(x) = d_1^2 x_1^2 + d_2^2 x_2^2 + d_3^2 x_3^2 = n\end{align*}

is soluble. To obtain a lower bound for $r(\textrm{spn}\ Q,n)$ , we choose $d_1,d_2,d_3$ to be distinct primes that are $\equiv 1 \,(\textrm{mod}\ 4)$ and coprime to n. This implies that

(28) \begin{align} r(\textrm{spn}\ Q,n) = r(\textrm{gen}\ Q,n) \gg \frac{n^{\frac{1}{2}}}{{d_1 d_2 d_3}} (nN)^{-{\epsilon}} \gg n^{\frac 1 2-3\eta - {\epsilon}}.\end{align}

Furthermore, we obtain by Lemma 13 that

(29) \begin{align} | r(Q,n) - r(\textrm{spn}\ Q,n)| &\ll (d_1d_2d_3)^\frac{2}{3} \left(\frac{n^{\frac{13}{28}}}{(d_1d_2d_3)^{2/7}} + \frac{n^{\frac{7}{16}}}{(d_1d_2d_3)^{1/8}} \right) n^{\epsilon} \end{align}

\begin{align} \notag &\ll \Big(n^{\frac{13}{28}+ \frac{8}{7} \eta }+ n^{\frac{7}{16}+ \frac{13}{8} \eta }\Big) n^{ {\epsilon}}.\end{align}

Hence, for $\eta = {1}/{116}-{\epsilon}$ , equation (29) is dominated by (28) and it follows that $r(Q,n) \geq 1$ for n sufficiently large. This proves the first claim of Corollary 5.

To derive results for sums of four squares, we make use of the distribution of smooth numbers in short intervals. For every $n \in {\mathbb N}$ it is possible to choose x with largest prime factor not exceeding $n^\frac{1}{4}$ such that $0 \leq n -x^2 \leq n^\frac{5}{8}$ and ${n -x^2 \not\equiv 0,1,4,7\,(\textrm{mod}\ 8)}$ , see [Reference Blomer, Brüdern and Dietmann4, section 5] for details. By our previous result for three squares, this implies that every sufficiently large n is represented by the sum of four squares whose largest prime divisors do not exceed $n^\frac{\theta}{2}$ where $\theta = {285}/{464}$ .

The bounds for three smooth squares can be improved, if we choose $d_1 = e_1 e_2$ , ${d_2 = e_1 e_3}$ and $d_3 = e_2 e_3$ for $e_1,e_2,e_3$ mutually coprime and $(e_1e_2e_3,n)=1$ . However, this only works, if $\big({p}/{n} \big) =1$ for every prime divisor p of $e_1 e_2 e_3$ , since then $\beta_p(n,Q) = 1 + \big({p}/{n} \big) =2$ . If this latter condition is satisfied, we can choose $\eta = {1}/{80}- {\epsilon}$ .