2. Trace Formula for $T'_m {\circ} W_Q$

For integers $k \geq 2$ , let $p_k(t, m)$ denote the Lucas sequence of the first kind; i.e. $p_k(t, m)$ is the $x^{k-2}$ coefficient in the power series expansion of $(mx^2 - tx + 1)^{-1}$ . Note that in particular, we have by Clayton et al. [ Reference Clayton, D, Ni, Xue and Zummo5 , lemmas 2·2, 2·3] that

(2·1) \begin{align} \begin{cases} p_k(t,m) = (k-1)\,m^{(k - 2)/2} &\text{if } t^2 - 4m = 0, \\[5pt] p_k(t,m) \le \dfrac{2m^{(k - 1)/2}}{\sqrt{\left\lvert{t^2-4m}\right\lvert}} &\text{if } t^2-4m \lt 0. \end{cases} \end{align}

Following Skoruppa–Zagier [ Reference Skoruppa and Zagier20 ], we also define the following class numbers. For integers $N \geq 1$ and $\Delta \leq 0$ , define $H_N(\Delta)$ via

(2·2) \begin{align} H_N(\Delta) = \begin{cases} a^2b \left( \cfrac{\Delta / a^2b^2}{N/a^2b} \right) H(|\Delta / a^2b^2|) & \text{if } a^2b^2 \mid \Delta, \\ 0 & \text{otherwise}, \end{cases}\end{align}

where $a^2b\,:\!=\,(N,\Delta)$ with b squarefree, $\left( \frac{\cdot}{\cdot} \right)$ denotes the Kronecker symbol, and $H(\cdot)$ denotes the Hurwitz–Kronecker class number. Note that in the case of $\Delta = 0$ , $\left({{0}/{1}}\right)=1$ and $H(0) = {-1}/{12}$ , so that $H_N(0) = ({-1}/{12})N$

In the following, we use the notation $Q \| N$ to mean $Q | N$ and $(Q, {N}/{Q}) = 1$ and $\ell = \square$ to mean the integer $\ell$ is a square. In addition, $\sigma_0(N) = \sum\limits_{N' | N} 1$ , $\sigma_1(N) = \sum\limits_{N' | N} N'$ , and B(N) is defined to be the greatest integer $\ell$ such that $\ell^2 | N$ .

Proposition 2·1 (Skoruppa–Zagier [ Reference Skoruppa and Zagier20 ], correction by Assaf [ Reference Assaf1 ]). For $N \ge 1$ coprime to m, $Q \| N$ , and $k \geq 2$ even, we have

\begin{align*} {\mathrm{Tr}}_{S_k(N)} \, T'_m {\circ} W_{Q} &= \sum_{\substack{N' \mid N \\ N/N' \mathrm{ squarefree}}} \mu \left( \frac{Q}{(Q, N')} \right) s'_{k,N'} (m, (Q, N')), \\ {\mathrm{Tr}}_{S_k^{{\mathrm{new}}}(N)} \, T'_m {\circ} W_{Q} &= \sum_{N' \mid N} \alpha\left( \frac{N}{N'} \right) s'_{k,N'} (m, (Q, N')). \end{align*}

Here $\mu$ denotes the Möbius function, $\alpha(N)$ is the nonzero multiplicative function defined on prime powers via

\begin{align*} \alpha(p^r) = \begin{cases} -1 & r = 1 \text{ or } 2 \\ 1 & r = 3 \\ 0 & r \geq 4 \end{cases} \end{align*}

and

(2·3) \begin{align} & s'_{k, N}(m,Q) \nonumber\\ & \quad = -\frac{1}{2 m^{(k-1)/2}} \sum_{Q' \mid Q} \sum_{s} p_{k} \left( \frac{s}{\sqrt{Q'}}, m \right) H_{\frac{N}{Q}} (s^2 - 4m Q') \end{align}

(2·4) \begin{align} &\qquad - \frac{1}{2 m^{(k-1)/2}} \sum_{m' \mid m} \min \left( m', \frac{m}{m'} \right)^{k - 1} \left(B(Q), m' + \frac{m}{m'}\right) \left( B\left( \frac{N}{Q} \right), m' - \frac{m}{m'}\right) \end{align}

(2·5) \begin{align} & \qquad + \frac{1}{m^{(k-1)/2}} {\unicode{x1D7D9}}_{\substack{k = 2 \\ \frac{N}{Q} = \square}} \sigma_0(Q) \sigma_1(m), \end{align}

with the summation over s ranging over all $s \in {\mathbb{Z}}$ such that $s^2 \leq 4mQ'$ , $Q' | s$ , and $( ({s}/{Q'} )^2, {Q}/{Q'})$ is squarefree.

We then estimate the $s'_{k,N}(m,Q)$ terms from this proposition. In the following, and throughout the entire paper, we use big-O notation with respect to both N and k. Note that m here is a fixed constant (so the big-O implied constants will depend on m). And Q is considered to be an arbitrary exact divisor of N (so the big-O bounds will only be stated in terms of N and k, uniform over arbitrary $Q \parallel N$ ).

Lemma 2·2. Fix $m \geq 1$ . Let $N\ge1$ be coprime to m, $Q \| N$ , and $k \geq 2$ be even. Then

\begin{align*} s'_{k,N}(m,Q) = \frac{{\unicode{x1D7D9}}_{m=\square}}{\sqrt{m}} \frac{k - 1}{12} \frac{N}{Q} + O(N^{1/2}\sigma_0(N)\log N), \end{align*}

where $s'_{k,N}(m,Q)$ is as defined in Proposition 2·1.

Proof. Note that $\sigma_0(Q) \leq \sigma_0(N) = O(N^\varepsilon)$ and $\sigma_1(m) = O(1)$ , so the third term (2·5) of $s'_{k,N}(m,Q)$ is $O(N^\varepsilon)$ . Also note that since

\begin{align*} \left(B(Q), m' + \frac{m}{m'}\right) \le 2m,\qquad \left( B\left( \frac{N}{Q} \right), m' - \frac{m}{m'}\right) \leq \sqrt{\frac{N}{Q}} \le \sqrt{N}, \end{align*}

the second term (2·4) of $s'_{k,N}(m,Q)$ is of magnitude

\begin{align*} &\leq \frac{1}{2m^{(k-1)/2}} \sum_{m' | m} \min \left( m', \frac{m}{m'} \right)^{k - 1} (2m)\sqrt{N} \leq \frac{(2m)\sqrt{N}}{{2}} \sum_{m' | m} 1 = O(\sqrt N). \end{align*}

Lastly, we consider the first term (2·3) of $s'_{k,N}(m,Q)$ . In the double summation, denote $\Delta \,:\!=\, s^2 - 4mQ'$ , and we break into two cases: when $\Delta \lt 0$ , and when $\Delta = 0$ .

When $\Delta \lt 0$ , we have by (2·1) that

\begin{align*} \left|p_k \left( \frac{s}{\sqrt{Q'}}, m \right) \right| & \leq \frac{2m^{(k-1)/2}}{\sqrt{\left| \frac{s^2}{Q'} - 4m \right|}} = \frac{2m^{(k-1)/2}}{\sqrt{\left\lvert{\Delta}\right\lvert / Q'}}. \end{align*}

Using [ Reference Griffin, Ono and Tsai8 , lemma 2·2], we can bound the Hurwitz–Kronecker class number by

\begin{align*} H(D) \leq \frac{\sqrt{D} (\log D +2)}{\pi}. \end{align*}

So letting $a^2b = \left({N}/{Q}, \Delta \right)$ we have

\begin{align*} \left| H_{\frac{N}{Q}}(\Delta) \right| & \leq \left| a^2b \left( \cfrac{\Delta / a^2b^2}{N/a^2b} \right) H(|\Delta / a^2b^2|) \right| \\[2pt] & \leq \frac{a^2 b \sqrt{|\Delta / a^2b^2|} \left( \log\left| \frac{\Delta}{a^2b^2} \right| + 2\right)}{\pi} \\[2pt] & \leq \frac{a \sqrt{|\Delta|} (\log |\Delta| + 2)}{\pi}. \end{align*}

Combining these,

\begin{align*} \left| p_{k} \left( \frac{s}{\sqrt{Q'}}, m \right) H_{\frac{N}{Q}} (\Delta) \right| & \leq \frac{2m^{(k-1)/2}}{\sqrt{\Delta / Q'}} \frac{a \sqrt{|\Delta|} (\log|\Delta| + 2)}{\pi} \\[2pt] & = \frac{2m^{(k-1)/2} a \sqrt{Q'} (\log|\Delta| + 2)}{\pi} \\[2pt] & \leq \frac{2m^{(k-1)/2} \sqrt{N} (\log(4mN) + 2)}{\pi} \\[2pt] & = m^{(k-1)/2} O(N^{1/2} \log N), \end{align*}

where the last inequality comes from the facts that $a^2 \leq {N}/{Q} \leq {N}/{Q'}$ and $|\Delta| \le 4mQ' \le 4mN$ .

Now, note that in (2·3), the number of terms in the summation over $Q'|Q$ is equal to $\sigma_0(Q) \le \sigma_0(N)$ . Furthermore, in (2·3), the number of terms in the summation over s is $\leq 4\sqrt{m/Q'} = O(1)$ since $Q' | s$ and $|s| \leq 2\sqrt{mQ'}$ . Thus, the contribution of the $\Delta \lt 0$ terms in $s'_{k,N}(m,Q)$ is $O(N^{1/2} \sigma_0(N) \log N)$ .

Finally, we consider the $\Delta = 0$ terms, which will make up the main term of $s'_{k,N}(m,Q)$ . Recall that $(m,Q')=1$ . Then note that $\Delta \,:\!=\, s^2 - 4mQ' =0$ implies that $m=\square$ and $Q'=\square$ . Additionally, $(Q')^2 \,|\, s^2=4mQ'$ implies that $Q' \,|\, 4$ . Hence we must have $Q' = 1 \text{ or }4$ . It is straightforward to verify that the first case occurs precisely when $m=\square$ and $4 \nmid Q$ (for $Q'=1$ , $s=\pm 2 \sqrt m$ ); and the second case occurs precisely when $m=\square$ and $4 | Q$ (for $Q'=4$ , $s=\pm 4 \sqrt m$ ). This means that in all cases, the number of $\Delta = 0$ terms appearing in $s'_{k,N}(m,Q)$ is exactly $2\,{\unicode{x1D7D9}}_{m=\square}$ .

And when $\Delta = 0$ , we have by (2·1) and (2·2) that

\begin{align*} p_k\left({\frac{s}{\sqrt{Q'}}, m}\right) &= (k-1)\,m^{(k-2)/2} \\ \text{and}\quad H_{\frac{N}{Q}} (s^2 - 4m Q') & = H_{\frac{N}{Q}} (0) = \frac{-1}{12} \frac{N}{Q}. \end{align*}

Thus, extracting the $\Delta=0$ terms from $s'_{k, N}(m,Q)$ , we obtain

\begin{align*} &-\frac{1}{2m^{(k-1)/2}} \sum_{Q' | Q} \sum_{\substack{s^2 = 4mQ' \\ Q' | s}} p_{k} \left( \frac{s}{\sqrt{Q'}}, m \right) H_{\frac{N}{Q}} (s^2 - 4m Q') \\ =& -\frac{1}{2m^{(k-1)/2}} (2 {\unicode{x1D7D9}}_{m=\square}) \big( (k-1) m^{(k-2)/2} \big) \Big( \frac{-1}{12} \frac{N}{Q} \Big) \\ = & \frac{{\unicode{x1D7D9}}_{m=\square}}{\sqrt{m}} \frac{k - 1}{12} \frac{N}{Q}, \end{align*}

completing the proof.

With this estimate of $s'_{k,N}(m,Q)$ , we can then estimate ${\mathrm{Tr}} \, T'_m {\circ} W_Q$ . In the following, we use the notation $[\![{A}]\!]_{\textit{condition}}$ to denote that the term A only appears when $\textit{condition}$ is satisfied.

Proposition 2·3. Fix $m \geq 1$ . Let $N\ge1$ be coprime to m, $Q \| N$ , and $k \geq 2$ be even. Then

\begin{align*} {\mathrm{Tr}}_{S_k(N)} \, T'_m {\circ} W_Q &= \frac{{\unicode{x1D7D9}}_{m = \square}}{\sqrt{m}} \frac{k - 1}{12} \psi(N) {\unicode{x1D7D9}}_{Q = 1} + O(N^{1/2} \sigma_0(N)^2 \log N), \\ {\mathrm{Tr}}_{S_k^{{\mathrm{new}}}(N)} \, T'_m {\circ} W_Q &= \frac{{\unicode{x1D7D9}}_{m = \square}}{\sqrt{m}} \frac{k - 1}{12} \psi^{{\mathrm{new}}}(N) \eta(Q) + O(N^{1/2} \sigma_0(N)^2 \log N), \end{align*}

where $\psi$ , $\psi^{\mathrm{new}}$ , and $\eta$ are the nonzero multiplicative functions defined on prime powers via

\begin{align*} \psi(p^r) &\,:\!=\, p^r \left({1+\frac 1p}\right), \\ \psi^{{\mathrm{new}}}(p^r) &\,:\!=\, p^r \left({1 - \frac 1p - \left[\!\!\left[{ \frac{1}{p^2} }\right]\!\!\right]_{r\ge 2} + \left[\!\!\left[{ \frac{1}{p^3} }\right]\!\!\right]_{r\ge 3} }\right), \\ \eta(p^r) &\,:\!=\, \frac{-{\unicode{x1D7D9}}_{r=2}}{p^2 - p - 1}. \end{align*}

Note in particular that when N is squarefree, $\eta(Q) = {\unicode{x1D7D9}}_{Q = 1}$ .

Proof. Define the multiplicative function $\theta_M(d) \,:\!=\, {d}/{(M, d)}$ . Observe that $\mu {\circ} \theta_M$ is also multiplicative since $\theta_M(d_1), \theta_M(d_2)$ are coprime whenever $d_1,d_2$ are coprime.

To show the first desired identity, it suffices (by Lemma 2·2 and Proposition 2·1) to show that

(2·6) \begin{align} \sum_{\substack{N' \mid N \\ N/N' \mathrm{ squarefree}}} \mu \left({ \frac{Q}{(Q, N')} }\right) \frac{N'}{(Q, N')} = \psi(N){\unicode{x1D7D9}}_{Q = 1}. \end{align}

Observe that ${\unicode{x1D7D9}}_{N/N' \ \text{squarefree}} = \mu^2(N/N')$ , $\mu ({ {Q}/{(Q, N')} }) = \mu ({{N/N'}/{(N/N', N/Q)} }) = \mu {\circ} \theta_{N/Q}(N/N')$ , and ${N'}/{(Q, N')} = \theta_Q(N')$ . This means that

\begin{align*} \text{LHS of (2.6)} &= [[\mu^2 \cdot\mu {\circ}\theta_{N/Q}] * \theta_Q](N) \\ & = \prod_{p^r \| N} [[\mu^2 \cdot \mu {\circ}\theta_{N/Q}] * \theta_Q](p^r) \\ & = \prod_{p^r \| N} \sum_{i = 0}^r \mu^2(p^i)\, \mu {\circ} \theta_{N/Q}(p^i) \, \theta_Q(p^{r - i}) \\ & = \prod_{p^r \| N} \big( \theta_Q(p^r) + \mu {\circ} \theta_{N/Q}(p) \,\theta_Q(p^{r - 1}) \big) \\ &= \prod_{p^r \| N} \begin{cases} p^r + p^{r-1} & \text{if } p \nmid Q \\ 0 & \text{if } p \mid Q \end{cases} \\ &= \psi(N){\unicode{x1D7D9}}_{Q = 1}. \end{align*}

Note that here, the second-to-last equality comes from the facts that

\begin{align*} \theta_Q(p^s) &= \begin{cases} p^s & \text{if } p \nmid Q \\ 1 & \text{if } p | Q \end{cases} \qquad \text{and}\qquad \mu{\circ}\theta_{N/Q}(p) = \begin{cases} 1 & \text{if } p \nmid Q \\ -1 & \text{if } p | Q. \end{cases} \end{align*}

Similarly, to show the second desired identity, it suffices (by Lemma 2·2 and Proposition 2·1) to show that

(2·7) \begin{align} \sum_{N' | N} \alpha\left( \frac{N}{N'} \right) \frac{N'}{(Q, N')} = \psi^{\mathrm{new}}(N) \eta(Q). \end{align}

And here,

\begin{align*} \text{LHS of (2.7)} &= [\alpha * \theta_Q](N) \\ & = \prod_{p^r \| N} [\alpha * \theta_Q](p^r) \\ & = \prod_{p^r \| N} \sum_{i = 0}^r \alpha(p^{i}) \theta_Q(p^{r-i}) \\ &= \prod_{p^r \parallel N} \big( \theta_Q(p^r) - \theta_Q(p^{r-1}) - [\![{\theta_Q(p^{r-2})}]\!]_{r \ge 2} + [\![{\theta_Q(p^{r-3})}]\!]_{r \ge 3} \big) \\ &= \prod_{p^r \parallel N} \begin{cases} \psi^{\mathrm{new}}(p^r) & \text{if } p \nmid Q \\ - {\unicode{x1D7D9}}_{r=2} & \text{if } p \mid Q \end{cases} \\ &= \psi^{\mathrm{new}}(N) \prod_{p \mid Q} \frac{- {\unicode{x1D7D9}}_{v_p(N)=2} }{ \psi^{\mathrm{new}}(p^2) } \\ &= \psi^{\mathrm{new}}(N) \prod_{p^r \| Q} \frac{- {\unicode{x1D7D9}}_{r=2} }{ p^2-p-1 } \\ &= \psi^{\mathrm{new}}(N) \eta(Q), \end{align*}

as desired.

3. Trace Formula for $T'_m$ over the sign pattern spaces

In [ Reference Martin11 , proposition 3·2], Martin derived a formula for $\dim S_k^{{\mathrm{new}},\sigma}(N)$ in terms of the traces ${\mathrm{Tr}}_{S_k^{{\mathrm{new}}}(N)}W_Q$ . Here, we prove a similar formula for ${\mathrm{Tr}}_{S_k^{\sigma}(N)}T'_m$ and ${\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)}T'_m$ in terms of the traces ${\mathrm{Tr}}_{S_k(N)} T'_m {\circ} W_Q$ and ${\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)} T'_m {\circ} W_Q$ , respectively.

We note that the identity (3·1) underlying this lemma can also be interpreted as a special case of orthogonality for characters over a finite group. (The set of all exact divisors of N forms a group via the binary operation $Q_1 * Q_2 \,:\!=\, {Q_1 Q_2}/{(Q_1,Q_2)^2}$ . And sign patterns for N make up the characters on this group.)

Lemma 3·1. For $N\geq1$ ,

\begin{align*} {\mathrm{Tr}}_{S_k^{\sigma}(N)}T'_m &= \frac{1}{2^{\omega(N)}}\sum_{Q\mid\mid N}\sigma(Q){\mathrm{Tr}}_{S_k(N)} \, T'_m {\circ} W_Q, \\ {\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)}T'_m &= \frac{1}{2^{\omega(N)}}\sum_{Q\mid\mid N}\sigma(Q){\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)} \, T'_m {\circ} W_Q.\end{align*}

Proof. We show this result just for the full space $S_k(N)$ , and the argument for the newspace $S_k^{\mathrm{new}}(N)$ is identical.

First, observe that

(3·1) \begin{align} \sum_{Q\mid\mid N}\sigma(Q)\sigma '(Q) &= \prod_{p^r \| N} \sum_{Q \| p^r} \sigma(Q) \sigma'(Q) = \prod_{p^r \| N} \left({ 1 + \sigma(p^r) \sigma'(p^r) }\right)\nonumber\\ &= \prod_{p^r \| N} 2 {\unicode{x1D7D9}}_{\sigma(p^r) = \sigma'(p^r)} = 2^{\omega(N)}{\unicode{x1D7D9}}_{\sigma =\sigma '}. \end{align}

Then, proceeding by splitting ${\mathrm{Tr}}_{S_k(N)}T'_m{\circ}W_Q$ into a sum of ${\mathrm{Tr}}_{S_k^{\sigma '}(N)}T'_m{\circ}W_Q$ for all sign patterns $\sigma'$ of N:

\begin{align*} \frac{1}{2^{\omega(N)}}\sum_{Q\mid\mid N}\sigma(Q){\mathrm{Tr}}_{S_k(N)} \, T'_m{\circ}W_Q &= \frac{1}{2^{\omega(N)}}\sum_{Q \mid\mid N}\sigma(Q)\sum_{\sigma '} {\mathrm{Tr}}_{S_k^{\sigma '}(N)}T'_m {\circ}W_Q \\ &= \frac{1}{2^{\omega(N)}}\sum_{Q \mid\mid N}\sum_{\sigma '}\sigma(Q)\sigma '(Q){\mathrm{Tr}}_{S_k^{\sigma '}(N)}T'_m \\&= \frac{1}{2^{\omega(N)}}\sum_{\sigma'} 2^{\omega(N)} {\unicode{x1D7D9}}_{\sigma = \sigma'}{\mathrm{Tr}}_{S_k^{\sigma'}(N)}T_m' \\&= {\mathrm{Tr}}_{S_k^{\sigma}(N)}T'_m, \end{align*}

as desired.

We can then estimate ${\mathrm{Tr}}_{S_k^\sigma(N)} T'_m$ and ${\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)} T'_m$ by applying Proposition 2·3 to Lemma 3·1.

Corollary 3·2. Fix $m \geq 1$ . Then for $N \ge 1$ coprime to m and $k \geq 2$ even,

(3·2) \begin{align} {\mathrm{Tr}}_{S_k^\sigma(N)} \, T'_m &= \frac{{\unicode{x1D7D9}}_{m = \square}}{\sqrt{m}} \frac{k - 1}{12} \frac{\psi(N)}{2^{\omega(N)}} + O(N^{1/2}\sigma_0(N)^2\log N), \end{align}

(3·3) \begin{align} {\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)} \, T'_m &= \frac{{\unicode{x1D7D9}}_{m = \square}}{\sqrt{m}} \frac{k - 1}{12} \frac{\psi^{{\mathrm{new}}}(N)}{2^{\omega(N)}} \prod_{p^r||N} \left({ 1 + \sigma(p^r) \frac{- {\unicode{x1D7D9}}_{r=2}}{p^2 - p -1}}\right) + O(N^{1/2} \sigma_0(N)^2 \log N). \qquad \end{align}

Proof. Using Lemma 3·1 and then Proposition 2·3, we obtain

\begin{align*} {\mathrm{Tr}}_{S_k^\sigma(N)}T_{m}' &=\frac{1}{2^{\omega(N)}}\sum_{Q\mid\mid N}\sigma(Q){\mathrm{Tr}}_{S_k(N)}T_m'{\circ} W_Q \\ &=\frac{1}{2^{\omega(N)}} \sum_{Q \| N} \sigma(Q) \frac{{\unicode{x1D7D9}}_{m=\square}}{\sqrt{m}}\frac{k-1}{12}\psi(N) {\unicode{x1D7D9}}_{Q=1} +O(N^{1/2}\sigma_0(N)^2\log N)\\ &=\frac{{\unicode{x1D7D9}}_{m=\square}}{\sqrt{m}}\frac{k-1}{12}\frac{\psi(N)}{2^{\omega(N)}}+O(N^{1/2}\sigma_0(N)^2\log N),\\ {\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)}T_{m}'&=\sum_{Q\mid\mid N}\sigma(Q){\mathrm{Tr}}_{S_k^{{\mathrm{new}}}(N)}T_m'{\circ} W_Q \\&=\frac{1}{2^{\omega(N)}} \sum_{Q\|N} \sigma(Q) \frac{{\unicode{x1D7D9}}_{m=\square}}{\sqrt{m}}\frac{k-1}{12}\psi^{{\mathrm{new}}}(N) \eta(Q)+O(N^{1/2}\sigma_0(N)^2\log N) \\ &=\frac{{\unicode{x1D7D9}}_{m=\square}}{\sqrt{m}}\frac{k-1}{12}\frac{\psi^{{\mathrm{new}}}(N)}{2^{\omega(N)}}\prod_{p^r||N} \left({ 1 + \sigma(p^r) \frac{- {\unicode{x1D7D9}}_{r=2}}{p^2 - p -1}}\right)+O(N^{1/2}\sigma_0(N)^2\log N),\end{align*}

as desired.

We point out here that the main term of (3·2) always grows faster than the error term $O(N^{1/2} \sigma_0(N)^2\log N) = O(N^{1/2+\varepsilon})$ . Furthermore, the main term of (3·3) grows faster than the error term, assuming it is not the case that $4 \| N$ with $\sigma(4)=+1$ (because after excluding this one particular case, the $\prod_{p^r||N} ({ 1 + \sigma(p^r) ({- {\unicode{x1D7D9}}_{r=2}}/({p^2 - p -1}))})$ factor in (3·3) is $\ge 1/2$ by (1·5)). In other words, the main terms of this corollary grow faster than the corresponding error terms for all admissible sign patterns $\sigma$ .

4. Proportion of Sign Patterns

In this section, we show Theorems 1·1 and 1·2.

Theorem 1·1. Consider $N \ge 1$ and $k \ge 2$ even. Then as $N+k \to \infty$ , the proportion of cusp forms with global sign $+1$ is given by

\begin{align*} \frac{\dim S_k^+(N)}{\dim S_k(N)} &\sim \frac12 + \frac{{\unicode{x1D7D9}}_{N=1}}{2}, \\[3pt] \frac{\dim S_k^{{\mathrm{new}},+}(N)}{\dim S_k^{\mathrm{new}}(N)} &\sim \frac12+\frac{ {\unicode{x1D7D9}}_{N=\mathrm{cubefree\ square}}}{2} \prod_{p|N} \frac{-1}{p^2-p-1}. \end{align*}

Moreover, consider sign patterns $\sigma$ for N. Then as $N+k \to \infty$ , the proportion of cusp forms with sign pattern $\sigma$ is given by

(4·1) \begin{align} \frac{\dim S_k^\sigma(N)}{\dim S_k(N)} &\sim \frac{1}{2^{\omega(N)}},\\[-10pt]\nonumber \end{align}

(4·2) \begin{align} \frac{\dim S_k^{{\mathrm{new}},\sigma}(N)}{\dim S_k^{\mathrm{new}}(N)} &\sim \frac{1}{2^{\omega(N)}} \prod_{p^r||N} \left({ 1 + \sigma(p^r) \frac{- {\unicode{x1D7D9}}_{r=2}}{p^2 - p -1}}\right). \end{align}

Proof. First, substitute $m=1$ and $Q=1$ into Proposition 2·3 to obtain the dimensions of the entire spaces:

(4·3) \begin{align} \dim S_k(N) = \frac{k-1}{12} \psi(N) + O(N^{1/2+\varepsilon}), \end{align}

(4·4) \begin{align} \dim S_k^{\mathrm{new}}(N) = \frac{k-1}{12} \psi^{\mathrm{new}}(N) + O(N^{1/2+\varepsilon}).\end{align}

Now, for the first two claims, we use Proposition 2·3 at $m=1$ and $Q=1,N$ to obtain

\begin{align*} \dim S_k^+(N) &= \frac12\left({ {\mathrm{Tr}}_{S_k(N)}W_1+{\mathrm{Tr}}_{S_k(N)}W_N}\right) \\ &= \frac{k-1}{12} \psi(N) \frac{ 1 + {\unicode{x1D7D9}}_{N=1} }{2} + O(N^{1/2+\varepsilon}), \\ \dim S_k^{{\mathrm{new}},+}(N) &= \frac12\left({ {\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)}W_1+{\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)}W_N}\right) \\ &= \frac{k-1}{12} \psi^{\mathrm{new}}(N) \frac{ 1 + \eta(N) }{2} + O(N^{1/2+\varepsilon}) \\ &= \frac{k-1}{12} \psi^{\mathrm{new}}(N) \frac{1+{\unicode{x1D7D9}}_{N = \text{cubefree square}} \prod_{p|N} \frac{-1}{p^2-p-1} }{2} + O(N^{1/2+\varepsilon}).\end{align*}

Comparing these estimates with (4·3) and (4·4) then yields the desired result.

For the last two claims, we substitute $m=1$ into Corollary 3·2 to obtain

\begin{align*} \dim S_k^\sigma(N)&=\frac{1}{2^{\omega(N)}}\frac{k-1}{12}\psi(N)+O(N^{1/2+\varepsilon}), \\[4pt] \dim S_k^{{\mathrm{new}},\sigma}(N)&=\frac{1}{2^{\omega(N)}}\frac{k-1}{12}\psi^{{\mathrm{new}}}(N)\prod_{p^r||N}\left({1+\sigma(p^r)\frac{-{\unicode{x1D7D9}}_{r=2}}{p^2-p-1}}\right)+O(N^{1/2+\varepsilon}).\end{align*}

Comparing these estimates with (4·3) and (4·4) similarly yields the desired result.

Recall that we defined the notion of admissible sign patterns, where all sign patterns are admissible except for the case of $4 \| N$ with $\sigma(4) = +1$ over the newspace. We introduced this definition because the proportions (4·1) and (4·2) are always nonzero except in this one exceptional case.

In fact, it turns out that $W_4$ always has Atkin–Lehner sign $-1$ over $S_k^{\mathrm{new}}(N)$ . This means that for inadmissible sign patterns $\sigma$ , there are no forms at all in $S_k^{\mathrm{new}}(N)$ with sign pattern $\sigma$ .

Proof. It suffices to show that ${\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)} W_1 + {\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)} W_4 = 0$ . Using Proposition 2·1, we compute

\begin{align*} & {\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)} W_1 + {\mathrm{Tr}}_{S_k^{\mathrm{new}}(N)} W_4 \\ & \quad = \sum_{N' | N} \alpha\left( \frac{N}{N'} \right) \big[{ s'_{k, N'}(1, (1, N')) + s'_{k, N'}(1,(4, N')) }\big] \\ & \quad = \sum_{L | \frac{N}{4}} \sum_{N' \in \{L, 2L, 4L\}} \alpha\left( \frac{N}{N'} \right) \big[{ s'_{k, N'}(1, (1, N')) + s'_{k, N'}(1,(4, N')) }\big] \\ & \quad = \sum_{L | \frac{N}{4}} \big[ \alpha\left( \frac{N}{L} \right) \big[ s'_{k, L}(1,1) + s'_{k, L}(1,1) \big] + \alpha\left( \frac{N}{2L} \right) \big[ s'_{k, 2L}(1,1) + s'_{k, 2L} (1,2) \big] \\ & \qquad + \alpha \left( \frac{N}{4L} \right) \big[ s'_{k, 4L}(1,1) + s'_{k, 4L}(1,4) \big] \big] \\ & \quad = \sum_{L | \frac{N}{4}} \alpha \left( \frac{N}{4L} \right) \big[{ -2 s'_{k, L}(1,1) - s'_{k, 2L}(1,1) - s'_{k, 2L}(1,2) + s'_{k, 4L}(1,1) + s_{k, 4L}(1,4) }\big]\\ & \quad =: \sum_{L | \frac{N}{4}} \alpha \left( \frac{N}{4L} \right) [{ * }]. \end{align*}

Now, we show that $[{*}] = 0$ . Computing each of the terms appearing in $[{*}]$ (see (2·3)-(2·5) for the definition of $s'_{k,L}$ ) yields

\begin{align*} s'_{k,L}(1,1) & = -\frac{1}{2} \Big[{p_k(0,1)H_L({-}4) + 2p_k(1,1)H_L({-}3) + 2p_k(2,1)H_L(0) + B(L) - 2{\unicode{x1D7D9}}_{\substack{k = 2 \\ L = \square}}}\Big], \\ s'_{k, 2L}(1,1) & = -\frac{1}{2} \Big[{p_k(0,1)H_{2L}({-}4) + 2p_k(1,1)H_{2L}({-}3) + 2p_k(2,1)H_{2L}(0) + B(L)}\Big], \\ s'_{k, 2L}(1,2) & = -\frac{1}{2} \Big[ p_k(0,1)H_L({-}4) + 2p_k(1,1) H_L({-}3) + 2p_k(2,1) H_L(0) + p_k(0,1) H_L({-}8) \\ & \quad + 2p_k(\sqrt{2}, 1) H_L({-}4) + B(L) - 4 {\unicode{x1D7D9}}_{\substack{k = 2 \\ L = \square}} \Big], \\ s'_{k, 4L}(1,1) & = -\frac{1}{2} \Big[p_k(0,1)H_{4L}({-}4) + 2p_k(1,1) H_{4L}({-}3) + 2p_k(2,1) H_{4L}(0)\\ & \quad + 2B(L) - 2{\unicode{x1D7D9}}_{\substack{k = 2 \\ L = \square}}\Big], \\ s'_{k, 4L}(1,4) & = -\frac{1}{2} \Big[2p_k(1,1) H_L({-}3) + p_k(0,1) H_L({-}8) + 2p_k(\sqrt{2}, 1) H_L({-}4)\\ & \quad + p_k(0,1)H_L({-}16) + 2p_k(2,1)H_L(0) + 2B(L) - 6 {\unicode{x1D7D9}}_{\substack{k = 2 \\ L = \square}} \Big]. \end{align*}

Then grouping all the terms of $[{*}]$ by the $p_k(t,m)$ yields

\begin{align*} [{*}] =& \; \frac{1}{2}p_k(0,1) [{3H_L({-}4) + H_{2L}({-}4) - H_{4L}({-}4) - H_L({-}16)}] \\[1pt] & + p_k(1,1) [{ 2H_L({-}3) + H_{2L}({-}3) - H_{4L}({-}3) }] \\[1pt] & + p_k(2,1) [{ 2H_L(0) + H_{2L}(0) - H_{4L}(0) }] \\[1pt] = & \; \frac{1}{2}p_k(0,1) \left[ \frac{3}{2} \left( \frac{-4}{L} \right) - \frac{3}{2} \left( \frac{-16}{L} \right) \right] \\[1pt]& + p_k(1,1) \frac{{\unicode{x1D7D9}}_{3 {\nmid} L}}{3} \left[ 2 \left( \frac{-3}{L} \right) + \left( \frac{-3}{2L} \right) - \left( \frac{-3}{4L} \right) \right] \\[1pt]& + p_k(2,1) \frac{-1}{12} [{ 2L + 2L - 4L }] \\[1pt] = & \; 0, \end{align*}

as desired.

5. $\mu_p$ -equidistribution of Hecke eigenvalues

In this section, we prove Theorem 1·3. For a fixed prime p, let $\mu_p$ denote the p-adic Plancherel measure $\displaystyle \mu_p(x) = (({p+1})/{\pi})({(1-x^2/4)^{1/2}}/({(p^{1/2}+p^{-1/2})^2-x^2}))dx$ . Then over the spaces $S_k^\sigma(N)$ and $S_k^{{\mathrm{new}},\sigma}(N)$ (for admissible sign patterns $\sigma$ ), we show that the Hecke eigenvalues for $T_p'$ are $\mu_p$ -equidistributed as $N+k \to \infty$ .

Proof. We just state the argument here for $S_k^{{\mathrm{new}},\sigma}(N)$ , and an identical argument works for the full space $S_k^{\sigma}(N)$ .

To show asymptotic $\mu_p$ -equidistribution of the Hecke eigenvalues, it suffices (e.g. by Kuipers-Niederreiter [ Reference Kuipers and Niederreiter10 , theorem 1·1], Serre [ Reference Serre19 , proposition 2]) to show that

(5·1) \begin{equation} \frac{1}{\dim S_k^{{\mathrm{new}},\sigma}(N)}\sum_{\lambda \in \text{eigv}\big({T'_p|_{S_k^{{\mathrm{new}},\sigma}(N)}}\big)} X_n(\lambda)\longrightarrow\int_{-2}^{2} X_n(x) \mu_p(x) \quad\text{as } N+k\to\infty\end{equation}

for the Chebyshev polynomials $X_n(x) \,:\!=\, U_n(x/2)$ for $n \ge 0$ , since they span the space of all polynomials, which is a dense subspace of all continuous functions on $[-2,2]$ .

The integral on the right hand side of (5·1) has value $p^{-n/2} {\unicode{x1D7D9}}_{{\textit{n} even}}$ , by Serre [ Reference Serre19 , equation (20)].

Additionally, it is well known (see [ Reference Serre19 , lemme 1], for example) that

\begin{equation*}T_{p^n}' = X_n(T_p').\end{equation*}

This means that

\begin{align*} \text{LHS of (5.1)} &= \frac{1}{\dim S_k^{{\mathrm{new}},\sigma}(N)}\sum_{\lambda \in \text{eigv}\big({T'_p|_{S_k^{{\mathrm{new}},\sigma}(N)}}\big)} X_n(\lambda) \\ &= \frac{ {\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)} X_n(T'_{p}) }{ {\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)} T'_{1} } \\ &= \frac{ {\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)} T'_{p^n} }{ {\mathrm{Tr}}_{S_k^{{\mathrm{new}},\sigma}(N)} T'_{1} } \\ &\to p^{-n/2} {\unicode{x1D7D9}}_{{\textit{n}\ \mathrm{even}}} \quad \text{as } N+k \to \infty,\end{align*}

where for the last step, we used Corollary 3·2 at $m=p^n$ and $m=1$ (since $\sigma$ is an admissible sign pattern). This verifies (5.1), completing the proof.

6. Applications

Here, we state some applications of Theorem 1·3 analogous to those in Serre [ Reference Serre19 , section 6].

For each $N\geq1$ and $\sigma$ admissible sign patterns for N, choose an eigenbasis $\{f_i\}_{i=1}^{\dim S_k^\sigma(N)}$ of $S_k^\sigma(N)$ with respect to the Hecke operators $T_m$ satisfying $(m,N)=1$ . For each m coprime to N, let $x_{i,m}$ denote the $T_m$ eigenvalue of $f_i$ . Then for each i, let $K_i^\sigma \,:\!=\, {\mathbb{Q}}(\{x_{i,m}\}_{(m,N) = 1})$ denote the totally real field generated by the $x_{i,m}$ (see Cohen–Strömberg [ Reference Cohen and Strömberg6 , corollary 10·3·7(c), remark 10·6·3(b)]). Then let $s(N,k,\sigma)_r$ be the number of i such that $[K_i^\sigma\,:\,\mathbb{Q}]=r$ (which is well defined as the $K_i^\sigma$ are independent of the eigenbasis chosen, up to permutation (cf. Serre, section 6.1 [ Reference Serre19 ])).

Similarly for $S_k^{{\mathrm{new}}, \sigma}(N)$ , we have the newform basis $\{f^{\mathrm{new}}_i\}_{i = 1}^{\dim S_k^{{\mathrm{new}}, \sigma}(N)}$ , and we define $K_i^{{\mathrm{new}}, \sigma}$ and $s(N,k,\sigma)^{\mathrm{new}}_r$ analogously. Then the following improves a result of Serre [ Reference Serre19 , théorème 5] (and Binder [ Reference Binder3 , corollary 10·0·5]), by extending to the sign pattern spaces.

Corollary 6·1. Fix $k\geq 2$ even, p prime, and $r \geq 1$ . Then for N coprime to p and $\sigma$ admissible sign patterns for N,

\begin{align*} \frac{s(N, k, \sigma)_r}{\dim S_k^\sigma(N)} &\to 0 \qquad \text{as } N \to \infty,\\[5pt] \frac{s(N, k, \sigma)^{\mathrm{new}}_r}{\dim S_k^{{\mathrm{new}}, \sigma}(N)} &\to 0 \qquad \text{as } N \to \infty. \end{align*}

Proof. We show the proof only over the full space, as the proof over the newspace is identical. For N coprime to p, let $s(N,k,\sigma,p)_r$ denote the number of indices i such that $[\mathbb{Q}(x_{i,p})\,:\,\mathbb{Q}] \leq r$ . Note that $s(N,k,\sigma,p)_r \geq s(N,k,\sigma)_r$ , so it suffices to show

\begin{equation*}\frac{s(N,k,\sigma,p)_r}{\dim S_k^\sigma(N)}\to 0\end{equation*}

as $N\to \infty$ while $(N,p)=1$ . Note that if $x=x_{i,p}$ satisfies $[\mathbb{Q}(x_{i,p})\,:\,\mathbb{Q}]\leq r,$ then x also satisfies:

1. 1. x is a totally real algebraic integer of degree $\leq r$ ;

2. 2. for $\tau \in \text{Gal}(\overline{{\mathbb{Q}}}/{\mathbb{Q}})$ , the Galois conjugate $x^\tau$ of x satisfies $|x^\tau|\leq 2p^{(k-1)/2}$ .

Let A(p,k,r) be the set of all x satisfying conditions (1) and (2). Such a set must be finite, as the characteristic polynomials of its elements x must have degree $\leq r$ and bounded integer coefficients. Let $A'\subseteq [-2,2]$ be the dilation of A(p,k,r) under the mapping $x\mapsto {x}/{p^{(k-1)/2}}.$ The set A’ is finite, so it has $\mu_p$ measure 0, and it is independent of N and $\sigma$ . Then since the ${x_{i,p}}/{p^{(k-1)/2}}$ are $\mu_p$ -equidistributed as $N\to\infty$ by Theorem 1·3, the proportion of i such that ${x_{i,p}}/{p^{(k-1)/2}}\in A'$ (and hence $x_{i,p}\in A$ ) tends to 0 as $N\to\infty$ (e.g. by Serre [ Reference Serre19 , proposition 1]). This is equivalent to saying that ${s(N,k,\sigma,p)_r}/{\dim S_k^\sigma(N)}\to0$ as $N\to\infty,$ completing the proof.

We note here that over the sign pattern newspaces $S_k^{{\mathrm{new}},\sigma}(N)$ , we have $K_i^\sigma = {\mathbb{Q}}(f_i)$ , where $\{f_i\}_{i = 1}^{\dim S_k^{{\mathrm{new}},\sigma}(N)}$ is the newform basis for $\dim S_k^{{\mathrm{new}},\sigma}(N)$ . From this perspective, Corollary 6·1 can be interpreted as a weak form of the Maeda Conjecture for general level.

Recall that the classical Maeda Conjecture states that $\deg {\mathbb{Q}}(f) = \dim S_k(1)$ (i.e. is maximum possible) for each normalised Hecke eigenform $f \in S_k(1)$ . For squarefree levels N, a corresponding conjecture has also been stated for the newspaces. Since the Hecke operators restrict to the $S_k^{{\mathrm{new}},\sigma}(N)$ ; for newforms $f \in S_k^{\mathrm{new}}(N)$ , the maximum possible degree of ${\mathbb{Q}}(f)$ is $\dim S_k^{{\mathrm{new}},\sigma_f}(N)$ , where $\sigma_f$ denotes the sign pattern for f. In particular, Tsaknias and Chow-Ghitza posited a version of the Maeda Conjecture for squarefree level: that for newforms $f \in S_k^{\mathrm{new}}(N)$ , $\deg {\mathbb{Q}}(f) = \dim S_k^{{\mathrm{new}},\sigma_f}(N)$ for sufficiently large weight k (see [ Reference Tsaknias21 , conjecture 2·4] and [ Reference Chow and Ghitza4 , conjecture 5·2]).

For general level N, however, the picture is not as clear. In this case, there seem to be plenty of examples where $\deg {\mathbb{Q}}(f)$ is not maximal. However, weaker forms of the Maeda Conjecture have still been investigated. For example, Roberts conjectured that for any fixed weight $k \ge 6$ , $\deg {\mathbb{Q}}(f) = 1$ for only finitely many newforms f [ Reference Roberts16 , conjecture 1·1]. Later, Martin made the same conjecture for arbitrary fixed degree r [ Reference Martin12 , question 3]. Here, Corollary 6·1 can be interpreted as a density 1 version of these conjectures. In particular, for any fixed $r \ge 1$ , Corollary 6·1 shows that the proportion of newforms $f \in S_k^{{\mathrm{new}},\sigma}(N)$ with $\deg {\mathbb{Q}}(f) \le r$ tends to 0 as $N \to \infty$ (along N coprime to p).

We can also use Corollary 6·1 to obtain information about the largest degree of $K_i^\sigma$ and $K_i^{{\mathrm{new}},\sigma}$ . The following result follows immediately from Corollary 6·1.

Corollary 6·2 (cf. Serre, Théorème 6 [ Reference Serre19 ]). Fix $k \geq 2$ even and p prime. Let $r(N,k, \sigma)$ be the maximum degree among the $K^\sigma_i$ associated with $(N,k,\sigma)$ , where $\sigma$ is an admissible sign pattern for N, and similarly define $r(N, k, \sigma)^{\mathrm{new}}$ . Then for N coprime to p and $\sigma$ admissible sign patterns for N,

\begin{align*} r(N,k, \sigma) &\to \infty \qquad \text{as } N \to \infty, \\ r(N,k, \sigma)^{\mathrm{new}} &\to \infty \qquad \text{as } N \to \infty. \end{align*}

Now, let $J_0(N)$ denote the Jacobian of the modular curve $X_0(N)$ . Note that the Atkin–Lehner involutions act naturally on $X_0(N)$ , and thus also act on $J_0(N)$ by functoriality. Up to ${\mathbb{Q}}$ -isogeny, we have the decomposition:

\begin{align*} J_0(N)\cong \prod_{\sigma} J^{\sigma}_0(N), \end{align*}

where $W_Q=\sigma(Q)$ on $J^{\sigma}_0(N)$ for all $Q||N$ . Now, each $J_0^{\sigma}(N)$ can be decomposed into a product of ${\mathbb{Q}}$ -simple factors:

\begin{align*} J_0^{\sigma}(N)\cong \prod A^\sigma_j. \end{align*}

Following the same argument as Serre (and Ribet [ Reference Ribet15 ]), we see that the number of factors $A^\sigma_j$ with dimension r is exactly $({1}/{r})s(N,2,\sigma)_r$ for each $r\ge1$ . Now, Corollary 6·2 implies the following.

Corollary 6·3 (cf. Serre [ Reference Serre19 , théorème 7]). Fix p prime. Let $r(N,\sigma)$ be the largest dimension of the simple ${\mathbb{Q}}$ -factors of $J_0^\sigma(N)$ , where $\sigma$ is an admissible sign pattern for N. Then for N coprime to p and $\sigma$ admissible sign patterns for N,

\begin{align*} r(N, \sigma) \to \infty \qquad \text{as } N \to \infty. \end{align*}