5.1 Even $p$ -rank

Let $L$ be an even lattice of prime level $p$ and genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ with $n>2$ and $n_{p}$ even carrying a holomorphic automorphic product $\unicode[STIX]{x1D6F9}(F)$ of singular weight.

Let $D$ be the discriminant form of $L$ . The oddity formula (see [Reference Conway and SloaneCS99, ch. 15, § 7.7])

$$\begin{eqnarray}e(\operatorname{sign}(D)/8)=\unicode[STIX]{x1D6FE}_{p}(D)\end{eqnarray}$$

implies

$$\begin{eqnarray}e((n-2)/8)=\unicode[STIX]{x1D716}_{p}\biggl(\frac{-1}{p}\biggr)^{n_{p}/2}.\end{eqnarray}$$

Hence, $n=\pm 2\;\text{mod}\;8$ and $k=1+n/2$ is an even integer. Note that $k\geqslant 4$ . Define

$$\begin{eqnarray}\unicode[STIX]{x1D709}=\unicode[STIX]{x1D716}_{p}\biggl(\frac{-1}{p}\biggr)^{n_{p}/2}=-(-1)^{k/2}.\end{eqnarray}$$

Let $E$ be the Eisenstein series of weight $k$ for $\overline{\unicode[STIX]{x1D70C}}_{D}$ corresponding to $0$ . Write

$$\begin{eqnarray}E=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}E_{\unicode[STIX]{x1D6FE}}e^{\unicode[STIX]{x1D6FE}}\end{eqnarray}$$

with

$$\begin{eqnarray}E_{\unicode[STIX]{x1D6FE}}(\unicode[STIX]{x1D70F})=\mathop{\sum }_{m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FE})}a_{\unicode[STIX]{x1D6FE}}(m)q^{m}.\end{eqnarray}$$

Define

$$\begin{eqnarray}c_{k,p,n_{p}}=\unicode[STIX]{x1D709}\frac{2k}{B_{k}}\frac{1}{p^{k}-1}\frac{1}{p^{(n_{p}-2)/2}}.\end{eqnarray}$$

Note that $c_{k,p,n_{p}}$ is positive. By explicit calculation we can derive the following formulas for the Fourier coefficients $a_{\unicode[STIX]{x1D6FE}}(m)$ (see also [Reference ScheithauerSch06, Theorem 7.1]).

Proposition 5.1. Let $\unicode[STIX]{x1D6FE}\in D$ and $m\in q(\unicode[STIX]{x1D6FE})+\mathbb{Z}$ , $m>0$ .

If $q(\unicode[STIX]{x1D6FE})\neq 0\;\text{mod}\;1$ , then

$$\begin{eqnarray}a_{\unicode[STIX]{x1D6FE}}(m)=-c_{k,p,n_{p}}\unicode[STIX]{x1D70E}_{k-1}(pm).\end{eqnarray}$$

Suppose $q(\unicode[STIX]{x1D6FE})=0\;\text{mod}\;1$ . Write $m=p^{\unicode[STIX]{x1D708}}a$ with $(a,p)=1$ . Then

$$\begin{eqnarray}a_{\unicode[STIX]{x1D6FE}}(m)=-c_{k,p,n_{p}}p^{(\unicode[STIX]{x1D708}+1)(k-1)}\unicode[STIX]{x1D70E}_{k-1}(a)\end{eqnarray}$$

if $\unicode[STIX]{x1D6FE}\neq 0$ and

$$\begin{eqnarray}\displaystyle a_{\unicode[STIX]{x1D6FE}}(m)=-c_{k,p,n_{p}}p^{(\unicode[STIX]{x1D708}+1)(k-1)}\unicode[STIX]{x1D70E}_{k-1}(a)+\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{n_{p}/2}\unicode[STIX]{x1D70E}_{k-1}(a)-\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{(n_{p}-2)/2}(p-1)\unicode[STIX]{x1D70E}_{k-1}(m) & & \displaystyle \nonumber\end{eqnarray}$$

if $\unicode[STIX]{x1D6FE}=0$ .

Write

$$\begin{eqnarray}F=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}F_{\unicode[STIX]{x1D6FE}}e^{\unicode[STIX]{x1D6FE}}\end{eqnarray}$$

with

$$\begin{eqnarray}F_{\unicode[STIX]{x1D6FE}}(\unicode[STIX]{x1D70F})=\mathop{\sum }_{m\in \mathbb{Z}-q(\unicode[STIX]{x1D6FE})}c_{\unicode[STIX]{x1D6FE}}(m)q^{m}.\end{eqnarray}$$

Pairing $F$ with the Eisenstein series $E$ (see Proposition 2.3) we obtain

$$\begin{eqnarray}2(k-2)+\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m)=0.\end{eqnarray}$$

In the following, we will often need that $L$ splits a hyperbolic plane $\mathit{II}_{1,1}$ . We give a criterion for this.

Proposition 5.2. The lattice $L$ splits a hyperbolic plane $\mathit{II}_{1,1}$ if and only if

$$\begin{eqnarray}n_{p}=n\quad \text{and}\quad \unicode[STIX]{x1D709}=+1\end{eqnarray}$$

or

$$\begin{eqnarray}n_{p}\leqslant n-2.\end{eqnarray}$$

Proof. Suppose $L$ splits $\mathit{II}_{1,1}$ , i.e.  $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})=\mathit{II}_{n-1,1}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})\oplus \mathit{II}_{1,1}$ . If $n_{p}\leqslant n-2$ , this gives no restriction on $\unicode[STIX]{x1D716}_{p}$ . If $n_{p}=n$ , then the sign rule (see [Reference Conway and SloaneCS99, ch. 15, § 7.7]) applied to $\mathit{II}_{n-1,1}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ implies $\unicode[STIX]{x1D716}_{p}=(-1/p)$ so that

$$\begin{eqnarray}\unicode[STIX]{x1D709}=\unicode[STIX]{x1D716}_{p}\biggl(\frac{-1}{p}\biggr)^{n_{p}/2}=\biggl(\frac{-1}{p}\biggr)^{1+n/2}=+1.\end{eqnarray}$$

The converse is now clear. ◻

Let $d$ be a positive rational number such that $pd$ is integral. We define functions

$$\begin{eqnarray}g_{\unicode[STIX]{x1D6FE}}(d)=\mathop{\sum }_{m>0}c_{m\unicode[STIX]{x1D6FE}}(-m^{2}d),\end{eqnarray}$$

where we assume $m$ to be integral. We have

$$\begin{eqnarray}g_{\unicode[STIX]{x1D6FE}}(d)=g_{0}(p^{2}d)+\mathop{\sum }_{(m,p)=1}c_{m\unicode[STIX]{x1D6FE}}(-m^{2}d).\end{eqnarray}$$

This implies

$$\begin{eqnarray}g_{\unicode[STIX]{x1D6FE}}(d)=g_{0}(p^{2}d)\end{eqnarray}$$

if $q(\unicode[STIX]{x1D6FE})\neq d\;\text{mod}\;1$ .

The divisor of $\unicode[STIX]{x1D6F9}(F)$ is a linear combination of rational quadratic divisors $\unicode[STIX]{x1D6FE}^{\bot }$ , where $\unicode[STIX]{x1D6FE}$ is a primitive vector of positive norm in $L$ . The divisor $\unicode[STIX]{x1D6FE}^{\bot }$ has order $\sum _{m>0}c_{m\unicode[STIX]{x1D6FE}}(-m^{2}\unicode[STIX]{x1D6FE}^{2}/2)$ . Since $\unicode[STIX]{x1D6F9}(F)$ is holomorphic this is a non-negative integer.

Proposition 5.3. Suppose $L$ splits a hyperbolic plane $\mathit{II}_{1,1}$ . Then

$$\begin{eqnarray}g_{\unicode[STIX]{x1D6FE}}(d)\geqslant 0\end{eqnarray}$$

for all $\unicode[STIX]{x1D6FE}\in D$ .

Proof. By the above remark we can assume that $d=q(\unicode[STIX]{x1D6FE})\;\text{mod}\;1$ . Write $L=M\oplus \mathit{II}_{1,1}$ . Choose a representative of $\unicode[STIX]{x1D6FE}$ in $M^{\prime }$ . By adding a primitive element of suitable norm in $\mathit{II}_{1,1}$ we obtain a primitive element $\unicode[STIX]{x1D6FE}\in L^{\prime }$ of norm $\unicode[STIX]{x1D6FE}^{2}/2=d$ . The holomorphicity of $\unicode[STIX]{x1D6F9}(F)$ implies that

$$\begin{eqnarray}g_{\unicode[STIX]{x1D6FE}}(d)=\mathop{\sum }_{m>0}c_{m\unicode[STIX]{x1D6FE}}(-m^{2}d)=\mathop{\sum }_{m>0}c_{m\unicode[STIX]{x1D6FE}}(-m^{2}\unicode[STIX]{x1D6FE}^{2}/2)\geqslant 0.\end{eqnarray}$$

This proves the proposition. ◻

We also define the multiplicative function

$$\begin{eqnarray}h(m)=\mathop{\sum }_{\substack{ d\mid m \\ m/d\;\text{squarefree} \\ (m/d,p)=1}}d^{k-1}.\end{eqnarray}$$

Now we expand the sum $-\sum _{\unicode[STIX]{x1D6FE}\in D}\,\sum _{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m)$ in terms of the non-negative divisor degrees $g_{\unicode[STIX]{x1D6FE}}$ .

Theorem 5.4. Suppose $L$ splits $\mathit{II}_{1,1}$ . Let $c_{p}=1-1/p$ . Then

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m) & {\geqslant} & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D \\ q(\unicode[STIX]{x1D6FE})\neq 0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\,\mathop{\sum }_{m/p=q(\unicode[STIX]{x1D6FE})\hspace{0.2em}{\rm mod}\hspace{0.2em}1}g_{\unicode[STIX]{x1D6FE}}(m/p)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D\setminus \{0\} \\ q(\unicode[STIX]{x1D6FE})=0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\,\mathop{\sum }_{m>0}g_{\unicode[STIX]{x1D6FE}}(m)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{p}c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{m>0}g_{0}(m)h(m).\nonumber\end{eqnarray}$$

Proof. Let $\unicode[STIX]{x1D6FE}\in D$ with $q(\unicode[STIX]{x1D6FE})\neq 0\;\text{mod}\;1$ . Then

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m>0}c_{j\unicode[STIX]{x1D6FE}}(-m)a_{j\unicode[STIX]{x1D6FE}}(m) & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{(m,p)=1}c_{j\unicode[STIX]{x1D6FE}}(-m/p)\mathop{\sum }_{d\mid m}d^{k-1}\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{(d,p)=1}d^{k-1}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{(t,p)=1}c_{j\unicode[STIX]{x1D6FE}}(-td/p)\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{(d,p)=1}d^{k-1}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{\substack{ t\;\text{squarefree} \\ (t,p)=1}}\,\mathop{\sum }_{(m,p)=1}c_{j\unicode[STIX]{x1D6FE}}(-m^{2}td/p)\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{(d,p)=1}d^{k-1}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{l=1}^{p-1}\,\mathop{\sum }_{\substack{ t\;\text{squarefree} \\ (t,p)=1}}\,\mathop{\sum }_{m=l\hspace{0.2em}{\rm mod}\hspace{0.2em}p}c_{j\unicode[STIX]{x1D6FE}}(-m^{2}td/p)\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{(d,p)=1}d^{k-1}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{l=1}^{p-1}\,\mathop{\sum }_{\substack{ t\;\text{squarefree} \\ (t,p)=1}}\,\mathop{\sum }_{m=l\hspace{0.2em}{\rm mod}\hspace{0.2em}p}c_{lj\unicode[STIX]{x1D6FE}}(-m^{2}td/p)\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{(d,p)=1}d^{k-1}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{\substack{ t\;\text{squarefree} \\ (t,p)=1}}(g_{j\unicode[STIX]{x1D6FE}}(td/p)-g_{0}(tdp))\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{(m,p)=1}(g_{j\unicode[STIX]{x1D6FE}}(m/p)-g_{0}(mp))\mathop{\sum }_{\substack{ d\mid m \\ m/d\;\text{squarefree}}}d^{k-1}\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m/p=q(j\unicode[STIX]{x1D6FE})\hspace{0.2em}{\rm mod}\hspace{0.2em}1}(g_{j\unicode[STIX]{x1D6FE}}(m/p)-g_{0}(mp))h(m).\nonumber\end{eqnarray}$$

For $\unicode[STIX]{x1D6FE}\in D\setminus \{0\}$ with $q(\unicode[STIX]{x1D6FE})=0\;\text{mod}\;1$ we find analogously

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m>0}c_{j\unicode[STIX]{x1D6FE}}(-m)a_{j\unicode[STIX]{x1D6FE}}(m)=c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m>0}(g_{j\unicode[STIX]{x1D6FE}}(m)-g_{0}(mp^{2}))h(m). & & \displaystyle \nonumber\end{eqnarray}$$

For $\unicode[STIX]{x1D6FE}=0$ , we have

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m) & = & \displaystyle c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{d>0}d^{k-1}\,\mathop{\sum }_{(t,p)=1}c_{0}(-td)\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{n_{p}/2}\mathop{\sum }_{(d,p)=1}d^{k-1}\,\mathop{\sum }_{t>0}c_{0}(-td)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{(n_{p}-2)/2}(p-1)\mathop{\sum }_{d>0}d^{k-1}\mathop{\sum }_{t>0}c_{0}(-td)\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{d>0}d^{k-1}\,\mathop{\sum }_{\substack{ t\;\text{squarefree} \\ (t,p)=1}}(g_{0}(td)-g_{0}(tdp^{2}))\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{n_{p}/2}\mathop{\sum }_{(d,p)=1}d^{k-1}\,\mathop{\sum }_{\substack{ t\;\text{squarefree} \\ (t,p)=1}}(g_{0}(td)+g_{0}(tdp))\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{(n_{p}-2)/2}(p-1)\mathop{\sum }_{d>0}d^{k-1}\,\mathop{\sum }_{\substack{ t\;\text{squarefree} \\ (t,p)=1}}(g_{0}(td)+g_{0}(tdp))\nonumber\\ \displaystyle & = & \displaystyle c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{m>0}(g_{0}(m)-g_{0}(mp^{2}))h(m)\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{n_{p}/2}\mathop{\sum }_{(m,p)=1}(g_{0}(m)+g_{0}(mp))h(m)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{(n_{p}-2)/2}(p-1)\mathop{\sum }_{m>0}(g_{0}(m)+g_{0}(mp))h(m).\nonumber\end{eqnarray}$$

Using

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{m>0}g_{0}(m)h(m) & = & \displaystyle \mathop{\sum }_{(m,p)=1}g_{0}(m)h(m)+p^{k-1}\mathop{\sum }_{(m,p)=1}g_{0}(mp)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,p^{2(k-1)}\mathop{\sum }_{m>0}g_{0}(mp^{2})h(m)\nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\mathop{\sum }_{m>0}g_{0}(mp)h(m)=\mathop{\sum }_{(m,p)=1}g_{0}(mp)h(m)+p^{k-1}\mathop{\sum }_{m>0}g_{0}(mp^{2})h(m),\end{eqnarray}$$

we find

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m) & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D \\ q(\unicode[STIX]{x1D6FE})\neq 0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\mathop{\sum }_{m/p=q(\unicode[STIX]{x1D6FE})\hspace{0.2em}{\rm mod}\hspace{0.2em}1}g_{\unicode[STIX]{x1D6FE}}(m/p)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D\setminus \{0\} \\ q(\unicode[STIX]{x1D6FE})=0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\mathop{\sum }_{m>0}g_{\unicode[STIX]{x1D6FE}}(m)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}c_{k,p,n_{p}}^{0}\mathop{\sum }_{(m,p)=1}g_{0}(m)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}\mathop{\sum }_{j=1}^{p-1}c_{k,n_{p},j}^{1}\mathop{\sum }_{m=j\hspace{0.2em}{\rm mod}\hspace{0.2em}p}g_{0}(mp)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}c_{k,p,n_{p}}^{{\geqslant}2}\mathop{\sum }_{m>0}g_{0}(mp^{2})h(m)\nonumber\end{eqnarray}$$

with

$$\begin{eqnarray}\displaystyle c_{k,p,n_{p}}^{0} & = & \displaystyle p^{k-1}-\unicode[STIX]{x1D709}p^{(n_{p}-2)/2},\nonumber\\ \displaystyle c_{k,p,n_{p},j}^{1} & = & \displaystyle p^{n}-a_{k,p,n_{p},j}+\unicode[STIX]{x1D709}p^{(n_{p}-2)/2}((p-1)p^{k-1}-1),\nonumber\\ \displaystyle c_{k,p,n_{p}}^{{\geqslant}2} & = & \displaystyle p^{k-1}(p^{n}-a_{k,p,n_{p},0}+\unicode[STIX]{x1D709}p^{(n_{p}-2)/2}(p-1)(p^{k-1}+1)),\nonumber\end{eqnarray}$$

where $a_{k,p,n_{p},j}$ denotes the number of elements $\unicode[STIX]{x1D6FE}\in D$ of norm $q(\unicode[STIX]{x1D6FE})=j/p\;\text{mod}\;1$ . For $j\neq 0\;\text{mod}\;p$ we have

$$\begin{eqnarray}a_{k,p,n_{p},j}=p^{n_{p}-1}-\unicode[STIX]{x1D709}p^{(n_{p}-2)/2}\end{eqnarray}$$

(see [Reference ScheithauerSch06, Proposition 3.2]) so that

$$\begin{eqnarray}\displaystyle c_{k,p,n_{p},j}^{1} & = & \displaystyle p^{n}-p^{n_{p}-1}+\unicode[STIX]{x1D709}p^{(n+n_{p}-2)/2}(p-1),\nonumber\\ \displaystyle c_{k,p,n_{p}}^{{\geqslant}2} & = & \displaystyle p^{k-1}(p^{n}-p^{n_{p}-1}+\unicode[STIX]{x1D709}p^{(n+n_{p}-2)/2}(p-1)).\nonumber\end{eqnarray}$$

Since $L$ splits $\mathit{II}_{1,1}$ , we obtain the following bounds

$$\begin{eqnarray}\displaystyle c_{k,p,n_{p}}^{0} & {\geqslant} & \displaystyle (1-1/p)p^{k-1},\nonumber\\ \displaystyle c_{k,p,n_{p},j}^{1} & {\geqslant} & \displaystyle (1-1/p)p^{2(k-1)},\nonumber\\ \displaystyle c_{k,p,n_{p}}^{{\geqslant}2} & {\geqslant} & \displaystyle (1-1/p)p^{3(k-1)}.\nonumber\end{eqnarray}$$

Applying the above formula for $\sum _{m>0}g_{0}(m)h(m)$ once more, we obtain

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m) & {\geqslant} & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D \\ q(\unicode[STIX]{x1D6FE})\neq 0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\,\mathop{\sum }_{m/p=q(\unicode[STIX]{x1D6FE})\hspace{0.2em}{\rm mod}\hspace{0.2em}1}g_{\unicode[STIX]{x1D6FE}}(m/p)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D\setminus \{0\} \\ q(\unicode[STIX]{x1D6FE})=0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\,\mathop{\sum }_{m>0}g_{\unicode[STIX]{x1D6FE}}(m)h(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{p}c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{m>0}g_{0}(m)h(m).\nonumber\end{eqnarray}$$

This proves the theorem. ◻

Define $m_{\infty }=\max _{\unicode[STIX]{x1D6FE}\in D}(-\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}}))$ . Note that $m_{\infty }>0$ .

Proposition 5.5. Suppose $L$ splits $\mathit{II}_{1,1}$ . Then

$$\begin{eqnarray}m_{\infty }\geqslant \frac{k-2}{12}.\end{eqnarray}$$

Let $\unicode[STIX]{x1D6FE}\in D$ such that $\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}})=-m_{\infty }$ . Then $c_{\unicode[STIX]{x1D6FE}}(-m_{\infty })$ is a positive integer.

Proof. The function $F_{0}$ is a non-zero modular form for $\unicode[STIX]{x1D6E4}_{0}(p)$ of weight $2-k$ . Applying the Riemann–Roch theorem to $F_{0}$ we obtain

$$\begin{eqnarray}p\unicode[STIX]{x1D708}_{0}(F_{0})+\unicode[STIX]{x1D708}_{\infty }(F_{0})\leqslant -\frac{k-2}{12}(p+1)\end{eqnarray}$$

(see [Reference Hirzebruch, Berger and JungHBJ94, Theorem 4.1]). The formula for the $S$ -transformation (see § 2) implies

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{0}(F_{0})=\unicode[STIX]{x1D708}_{\infty }\biggl(\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}F_{\unicode[STIX]{x1D6FE}}\biggr).\end{eqnarray}$$

Let $\unicode[STIX]{x1D6FE}\in D$ such that $\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}})$ is minimal. Since $L$ splits $\mathit{II}_{1,1}$ , there is a primitive vector $\unicode[STIX]{x1D707}$ in $L^{\prime }$ with $\unicode[STIX]{x1D707}=\unicode[STIX]{x1D6FE}\;\text{mod}\;L$ and $\unicode[STIX]{x1D707}^{2}/2=m_{\infty }$ . Then, the divisor $\unicode[STIX]{x1D707}^{\bot }$ has order $c_{\unicode[STIX]{x1D6FE}}(-m_{\infty })$ which is a positive integer by the holomorphicity of $\unicode[STIX]{x1D6F9}(F)$ . Hence,

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{\infty }\biggl(\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}F_{\unicode[STIX]{x1D6FE}}\biggr)=\min _{\unicode[STIX]{x1D6FE}\in D}\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}}).\end{eqnarray}$$

It follows

$$\begin{eqnarray}p\min _{\unicode[STIX]{x1D6FE}\in D}\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}})+\min _{\unicode[STIX]{x1D6FE}\in D}\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}})\leqslant -\frac{k-2}{12}(p+1).\end{eqnarray}$$

This completes the proof. ◻

We obtain the following inequalities.

Proposition 5.6. Suppose $L$ splits $\mathit{II}_{1,1}$ . Then

$$\begin{eqnarray}\biggl(\frac{k-2}{12}\biggr)^{k-1}\leqslant m_{\infty }^{k-1}\leqslant \unicode[STIX]{x1D709}\frac{p^{n_{p}/2}}{c_{p}}\frac{k-2}{k}B_{k}.\end{eqnarray}$$

Proof. Suppose $\unicode[STIX]{x1D708}_{\infty }(F_{0})<\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}})$ for all $\unicode[STIX]{x1D6FE}\in D\setminus \{0\}$ . Then the Eisenstein condition and the estimate in Theorem 5.4 give

$$\begin{eqnarray}\displaystyle 2(k-2) & = & \displaystyle -\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle c_{p}c_{k,p,n_{p}}p^{k-1}g_{0}(m_{\infty })h(m_{\infty })\nonumber\\ \displaystyle & {\geqslant} & \displaystyle c_{p}c_{k,p,n_{p}}p^{k-1}m_{\infty }^{k-1}\nonumber\end{eqnarray}$$

so that

$$\begin{eqnarray}m_{\infty }^{k-1}\leqslant \unicode[STIX]{x1D709}\frac{p^{n_{p}/2}}{c_{p}}\frac{k-2}{k}B_{k}.\end{eqnarray}$$

The assertion now follows from Proposition 5.5. Suppose $\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}})\leqslant \unicode[STIX]{x1D708}_{\infty }(F_{0})$ for some $\unicode[STIX]{x1D6FE}\in D\setminus \{0\}$ . Choose $\unicode[STIX]{x1D6FE}\neq 0$ such that $-\unicode[STIX]{x1D708}_{\infty }(F_{\unicode[STIX]{x1D6FE}})=m_{\infty }$ . Then,

$$\begin{eqnarray}-\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m)\geqslant c_{k,p,n_{p}}p^{k-1}m_{\infty }^{k-1}\end{eqnarray}$$

and the statement follows analogously. ◻

We remark that the first inequality in the proposition is a consequence of the Riemann–Roch theorem and the second is a consequence of the Eisenstein condition.

Theorem 5.7. Let $L$ be an even lattice of level $p$ and genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ with $n>2$ and $n_{p}$ even splitting a hyperbolic plane $\mathit{II}_{1,1}$ . Suppose $L$ carries a holomorphic automorphic product of singular weight. Then for each $c>1/\text{log}\,(\unicode[STIX]{x1D70B}e/6)$ there exists a constant $d$ depending only on $c$ such that

$$\begin{eqnarray}n\leqslant cn_{p}\log (p)+d.\end{eqnarray}$$

Proof. Recall that $k\geqslant 4$ . Using $2\unicode[STIX]{x1D701}(k)=\unicode[STIX]{x1D709}((2\unicode[STIX]{x1D70B})^{k}/k!)B_{k}$ and $k!\leqslant e\sqrt{k}(k/e)^{k}$ we derive from Proposition 5.6 the inequality

$$\begin{eqnarray}1\leqslant e^{2}p^{n_{p}/2}k^{3/2}\biggl(\frac{6}{\unicode[STIX]{x1D70B}e}\biggr)^{k}\end{eqnarray}$$

respectively

$$\begin{eqnarray}0\leqslant 2+\frac{n_{p}}{2}\log (p)+\frac{3}{2}\log (k)-k\log \biggl(\frac{\unicode[STIX]{x1D70B}e}{6}\biggr).\end{eqnarray}$$

If $t$ is a tangent of the real logarithm then $\log (x)\leqslant t(x)$ for all $x>0$ . Thus, $\log (k)\leqslant (k-x)/x+\log (x)$ for all $x>0$ . It follows

$$\begin{eqnarray}0\leqslant -\biggl(\log \biggl(\frac{\unicode[STIX]{x1D70B}e}{6}\biggr)-\frac{3}{2x}\biggr)k+\frac{n_{p}}{2}\log (p)+\frac{3}{2}(\log (x)-1)+2\end{eqnarray}$$

for all $x>0$ . If $x>3/2\log (\unicode[STIX]{x1D70B}e/6)=4.24964\ldots$ this gives an upper bound on $k$ and on $n$ , i.e. 

$$\begin{eqnarray}n\leqslant c(x)n_{p}\log (p)+d(x)\end{eqnarray}$$

with

$$\begin{eqnarray}\displaystyle c(x) & = & \displaystyle \frac{2}{2\log (\unicode[STIX]{x1D70B}e/6)-3/x}\nonumber\\ \displaystyle d(x) & = & \displaystyle (3\log (x)+1)c(x)-2\nonumber\end{eqnarray}$$

in this case. ◻

Note that the proof is constructive. For example, taking $x=20$ gives the bounds $c=3.59750\ldots$ and $d=33.92899\ldots .$

5.2 Odd $p$ -rank

Now let $L$ be an even lattice of prime level $p$ and genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ with $n>2$ and $n_{p}$ odd. Suppose $\unicode[STIX]{x1D6F9}(F)$ is a holomorphic automorphic product of singular weight on $L$ .

Since $n_{p}$ is odd, it follows that $p$ is odd as well.

The oddity formula implies

$$\begin{eqnarray}e((n-2)/8)=\left\{\begin{array}{@{}ll@{}}\displaystyle \unicode[STIX]{x1D716}_{p}\biggl({\displaystyle \frac{2}{p}}\biggr)\quad & \displaystyle \text{if }p=1\hspace{0.2em}{\rm mod}\hspace{0.2em}4,\\ \displaystyle \unicode[STIX]{x1D716}_{p}\biggl({\displaystyle \frac{2}{p}}\biggr)(-1)^{(n_{p}-1)/2}e(1/4)\quad & \displaystyle \text{if }p=3\hspace{0.2em}{\rm mod}\hspace{0.2em}4,\end{array}\right.\end{eqnarray}$$

so that

$$\begin{eqnarray}n=\left\{\begin{array}{@{}ll@{}}\displaystyle \pm 2\hspace{0.2em}{\rm mod}\hspace{0.2em}8\quad & \displaystyle \text{if }p=1\hspace{0.2em}{\rm mod}\hspace{0.2em}4,\\ \displaystyle 0\hspace{0.2em}{\rm mod}\hspace{0.2em}4\quad & \displaystyle \text{if }p=3\hspace{0.2em}{\rm mod}\hspace{0.2em}4.\end{array}\right.\end{eqnarray}$$

Define $k=1+n/2$ and

$$\begin{eqnarray}\unicode[STIX]{x1D709}=\unicode[STIX]{x1D716}_{p}\biggl(\frac{2}{p}\biggr)\biggl(\frac{-1}{p}\biggr)^{(n_{p}-1)/2}.\end{eqnarray}$$

Then

$$\begin{eqnarray}\unicode[STIX]{x1D709}=\left\{\begin{array}{@{}ll@{}}\displaystyle -(-1)^{k/2}\quad & \displaystyle \text{if }p=1\hspace{0.2em}{\rm mod}\hspace{0.2em}4,\\ -(-1)^{(k-1)/2}\quad & \displaystyle \text{if }p=3\hspace{0.2em}{\rm mod}\hspace{0.2em}4.\end{array}\right.\end{eqnarray}$$

Let $\unicode[STIX]{x1D712}(j)=(j/p)$ . Define the twisted divisor sum

$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{l,\unicode[STIX]{x1D712}}(m)=\mathop{\sum }_{d\mid m}\unicode[STIX]{x1D712}(m/d)d^{l}\end{eqnarray}$$

and the generalised Bernoulli numbers $B_{m,\unicode[STIX]{x1D712}}$ by

$$\begin{eqnarray}\mathop{\sum }_{j=1}^{p}\frac{\unicode[STIX]{x1D712}(j)xe^{jx}}{e^{px}-1}=\mathop{\sum }_{m\geqslant 0}B_{m,\unicode[STIX]{x1D712}}\frac{x^{m}}{m!}\end{eqnarray}$$

(see [Reference IwasawaIwa72]). Let

$$\begin{eqnarray}c_{k,p,n_{p}}=\unicode[STIX]{x1D709}\frac{2k}{B_{k,\unicode[STIX]{x1D712}}}\frac{1}{p^{(n_{p}-1)/2}}.\end{eqnarray}$$

The positivity of $L(k,\unicode[STIX]{x1D712})$ implies that $c_{k,p,n_{p}}$ is positive. We describe the Fourier coefficients $a_{\unicode[STIX]{x1D6FE}}(m)$ of the Eisenstein series $E$ .

Proposition 5.8. Let $\unicode[STIX]{x1D6FE}\in D$ and $m\in q(\unicode[STIX]{x1D6FE})+\mathbb{Z}$ , $m>0$ .

If $q(\unicode[STIX]{x1D6FE})\neq 0\;\text{mod}\;1$ , then

$$\begin{eqnarray}a_{\unicode[STIX]{x1D6FE}}(m)=-c_{k,p,n_{p}}\unicode[STIX]{x1D70E}_{k-1,\unicode[STIX]{x1D712}}(pm).\end{eqnarray}$$

Suppose $q(\unicode[STIX]{x1D6FE})=0\;\text{mod}\;1$ . Write $m=p^{\unicode[STIX]{x1D708}}a$ with $(a,p)=1$ . Then

$$\begin{eqnarray}a_{\unicode[STIX]{x1D6FE}}(m)=-c_{k,p,n_{p}}p^{(\unicode[STIX]{x1D708}+1)(k-1)}\unicode[STIX]{x1D70E}_{k-1,\unicode[STIX]{x1D712}}(a)\end{eqnarray}$$

if $\unicode[STIX]{x1D6FE}\neq 0$ and

$$\begin{eqnarray}a_{\unicode[STIX]{x1D6FE}}(m)=-c_{k,p,n_{p}}p^{(\unicode[STIX]{x1D708}+1)(k-1)}\unicode[STIX]{x1D70E}_{k-1,\unicode[STIX]{x1D712}}(a)-\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{(n_{p}-1)/2}\unicode[STIX]{x1D712}(a)\unicode[STIX]{x1D70E}_{k-1,\unicode[STIX]{x1D712}}(a)\end{eqnarray}$$

if $\unicode[STIX]{x1D6FE}=0$ .

We have the following result.

Proposition 5.9. The lattice $L$ splits a hyperbolic plane $\mathit{II}_{1,1}$ if and only if

$$\begin{eqnarray}n_{p}\leqslant n-1.\end{eqnarray}$$

As above, we denote the Fourier coefficients of $F$ by $c_{\unicode[STIX]{x1D6FE}}$ and define the functions $g_{\unicode[STIX]{x1D6FE}}$ . We also define

$$\begin{eqnarray}h_{\unicode[STIX]{x1D712}}(m)=\mathop{\sum }_{\substack{ d\mid m \\ m/d\;\text{squarefree}}}\unicode[STIX]{x1D712}(m/d)d^{k-1}.\end{eqnarray}$$

The function $h_{\unicode[STIX]{x1D712}}$ is bounded below by $h_{\unicode[STIX]{x1D712}}(m)\geqslant (2-\unicode[STIX]{x1D701}(2))m^{k-1}\geqslant m^{k-1}/3$ .

Theorem 5.10. Suppose $L$ splits $\mathit{II}_{1,1}$ . Let $c_{p}=1-1/p$ . Then

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m) & {\geqslant} & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D \\ q(\unicode[STIX]{x1D6FE})\neq 0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\,\mathop{\sum }_{m/p=q(\unicode[STIX]{x1D6FE})\hspace{0.2em}{\rm mod}\hspace{0.2em}1}g_{\unicode[STIX]{x1D6FE}}(m/p)h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D\setminus \{0\} \\ q(\unicode[STIX]{x1D6FE})=0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\,\mathop{\sum }_{m>0}g_{\unicode[STIX]{x1D6FE}}(m)h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{p}c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{m>0}g_{0}(m)h_{\unicode[STIX]{x1D712}}(m).\nonumber\end{eqnarray}$$

Proof. The argument is analogous to the proof of Theorem 5.4. We describe the necessary modifications.

Let $\unicode[STIX]{x1D6FE}\in D$ with $q(\unicode[STIX]{x1D6FE})\neq 0\;\text{mod}\;1$ . Then

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m>0}c_{j\unicode[STIX]{x1D6FE}}(-m)a_{j\unicode[STIX]{x1D6FE}}(m)=c_{k,p,n_{p}}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m/p=q(j\unicode[STIX]{x1D6FE})\hspace{0.2em}{\rm mod}\hspace{0.2em}1}(g_{j\unicode[STIX]{x1D6FE}}(m/p)-g_{0}(mp))h_{\unicode[STIX]{x1D712}}(m). & & \displaystyle \nonumber\end{eqnarray}$$

For $\unicode[STIX]{x1D6FE}\in D\setminus \{0\}$ with $q(\unicode[STIX]{x1D6FE})=0\;\text{mod}\;1$ we find

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m>0}c_{j\unicode[STIX]{x1D6FE}}(-m)a_{j\unicode[STIX]{x1D6FE}}(m)=c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m>0}(g_{j\unicode[STIX]{x1D6FE}}(m)-g_{0}(mp^{2}))h_{\unicode[STIX]{x1D712}}(m). & & \displaystyle \nonumber\end{eqnarray}$$

If $\unicode[STIX]{x1D6FE}=0$ , then

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m) & = & \displaystyle c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{m>0}(g_{0}(m)-g_{0}(mp^{2}))h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D709}c_{k,p,n_{p}}p^{(n_{p}-1)/2}\mathop{\sum }_{(m,p)=1}(g_{0}(m)+g_{0}(mp))\unicode[STIX]{x1D712}(m)h_{\unicode[STIX]{x1D712}}(m).\nonumber\end{eqnarray}$$

Using

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{m>0}g_{0}(m)h_{\unicode[STIX]{x1D712}}(m) & = & \displaystyle \mathop{\sum }_{(m,p)=1}g_{0}(m)h_{\unicode[STIX]{x1D712}}(m)+p^{k-1}\mathop{\sum }_{(m,p)=1}g_{0}(mp)h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,p^{2(k-1)}\mathop{\sum }_{m>0}g_{0}(mp^{2})h_{\unicode[STIX]{x1D712}}(m),\nonumber\end{eqnarray}$$

we obtain

$$\begin{eqnarray}\displaystyle -\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}\,\mathop{\sum }_{m>0}c_{\unicode[STIX]{x1D6FE}}(-m)a_{\unicode[STIX]{x1D6FE}}(m) & = & \displaystyle c_{k,p,n_{p}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D \\ q(\unicode[STIX]{x1D6FE})\neq 0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\mathop{\sum }_{m/p=q(\unicode[STIX]{x1D6FE})\hspace{0.2em}{\rm mod}\hspace{0.2em}1}g_{\unicode[STIX]{x1D6FE}}(m/p)h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}p^{k-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in D\setminus \{0\} \\ q(\unicode[STIX]{x1D6FE})=0\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}\mathop{\sum }_{m>0}g_{\unicode[STIX]{x1D6FE}}(m)h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m=j\hspace{0.2em}{\rm mod}\hspace{0.2em}p}c_{k,p,n_{p},j}^{0}g_{0}(m)h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}\mathop{\sum }_{j=1}^{p-1}\,\mathop{\sum }_{m=j\hspace{0.2em}{\rm mod}\hspace{0.2em}p}c_{k,p,n_{p},j}^{1}g_{0}(mp)h_{\unicode[STIX]{x1D712}}(m)\nonumber\\ \displaystyle & & \displaystyle +\,c_{k,p,n_{p}}c_{k,p,n_{p}}^{{\geqslant}2}\mathop{\sum }_{m>0}g_{0}(mp^{2})h_{\unicode[STIX]{x1D712}}(m)\nonumber\end{eqnarray}$$

with

$$\begin{eqnarray}\displaystyle c_{k,p,n_{p},j}^{0} & = & \displaystyle p^{k-1}+\unicode[STIX]{x1D709}\unicode[STIX]{x1D712}(j)p^{(n_{p}-1)/2},\nonumber\\ \displaystyle c_{k,p,n_{p},j}^{1} & = & \displaystyle p^{2(k-1)}-a_{k,p,n_{p},j}+\unicode[STIX]{x1D709}\unicode[STIX]{x1D712}(j)p^{(n_{p}-1)/2},\nonumber\\ \displaystyle c_{k,p,n_{p}}^{{\geqslant}2} & = & \displaystyle p^{3(k-1)}-p^{k-1}a_{k,p,n_{p},0},\nonumber\end{eqnarray}$$

where $a_{k,p,n_{p},j}$ denotes the number of elements $\unicode[STIX]{x1D6FE}\in D$ of norm $q(\unicode[STIX]{x1D6FE})=j/p\;\text{mod}\;1$ . Since $L$ splits $\mathit{II}_{1,1}$ we have

$$\begin{eqnarray}\displaystyle c_{k,p,n_{p},j}^{0} & {\geqslant} & \displaystyle (1-1/p)p^{k-1},\nonumber\\ \displaystyle c_{k,p,n_{p},j}^{1} & {\geqslant} & \displaystyle (1-1/p^{2})p^{2(k-1)},\nonumber\\ \displaystyle c_{k,p,n_{p},m}^{{\geqslant}2} & {\geqslant} & \displaystyle (1-1/p^{2})p^{3(k-1)}.\nonumber\end{eqnarray}$$

This implies the assertion. ◻

Pairing $F$ with the Eisenstein series $E$ and applying the Riemann–Roch theorem to $F_{0}$ , we obtain the following result.

Proposition 5.11. Suppose $L$ splits $\mathit{II}_{1,1}$ . Then

$$\begin{eqnarray}\biggl(\frac{k-2}{12}\biggr)^{k-1}\leqslant m_{\infty }^{k-1}\leqslant 3\unicode[STIX]{x1D709}\frac{p^{(n_{p}+1)/2}}{c_{p}}\frac{k-2}{k}\frac{B_{k,\unicode[STIX]{x1D712}}}{p^{k}}.\end{eqnarray}$$

We can now derive a bound on $n$ .

Theorem 5.12. Let $L$ be an even lattice of level $p$ and genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ with $n>2$ and $n_{p}$ odd splitting a hyperbolic plane $\mathit{II}_{1,1}$ . Suppose $L$ carries a holomorphic automorphic product of singular weight. Then for each $c>1/\text{log}(\unicode[STIX]{x1D70B}e/6)$ there exists a constant $d$ depending only on $c$ such that

$$\begin{eqnarray}n\leqslant cn_{p}\log (p)+d.\end{eqnarray}$$

Proof. Here we use $2L(k,\unicode[STIX]{x1D712})=\unicode[STIX]{x1D709}\sqrt{p}((2\unicode[STIX]{x1D70B})^{k}/k!)(B_{k,\unicode[STIX]{x1D712}}/p^{k})$ and $L(k,\unicode[STIX]{x1D712})\leqslant \unicode[STIX]{x1D701}(3)$ to obtain

$$\begin{eqnarray}1\leqslant \frac{5}{2}e^{2}p^{n_{p}/2}k^{3/2}\biggl(\frac{6}{\unicode[STIX]{x1D70B}e}\biggr)^{k}.\end{eqnarray}$$

As above, this implies

$$\begin{eqnarray}n\leqslant c(x)n_{p}\log (p)+d(x)\end{eqnarray}$$

with

$$\begin{eqnarray}\displaystyle c(x) & = & \displaystyle \frac{2}{2\log (\unicode[STIX]{x1D70B}e/6)-3/x},\nonumber\\ \displaystyle d(x) & = & \displaystyle (3\log (x)+1+2\log (5/2))c(x)-2\nonumber\end{eqnarray}$$

for $x>3/2\log (\unicode[STIX]{x1D70B}e/6)$ .◻

Note that the constant $d$ is slightly larger here than in Theorem 5.7. Taking $x=20$ we obtain the bounds $c=3.59750\ldots$ and $d=40.52171\ldots .$

5.3 An example

Let $L$ be a lattice of genus $\mathit{II}_{n,2}(2_{\mathit{II}}^{+n_{2}})$ with $n>2$ and $n_{2}=2,~4$ or $6$ carrying a holomorphic automorphic product of singular weight. Then $n\leqslant 34,~42$ respectively $42$ and

$$\begin{eqnarray}\frac{k-2}{12}\leqslant m_{\infty }\leqslant \biggl(2^{(n_{2}+2)/2}\frac{k-2}{k}B_{k}\biggr)^{1/(k-1)}\end{eqnarray}$$

by Theorem 5.7 and Proposition 5.6. The values of the bounds are given in the following table.

Since $m_{\infty }$ is half-integral we obtain the following theorem.

6.1 General results

We derive some general bounds and formulate the Eisenstein condition for reflective modular forms.

Let $L$ be an even lattice of prime level $p$ and genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ with $n>2$ . Let $F=\sum _{\unicode[STIX]{x1D6FE}\in D}F_{\unicode[STIX]{x1D6FE}}e^{\unicode[STIX]{x1D6FE}}$ be a non-zero reflective modular form on $L$ (see § 3). Then $F$ has weight $1-n/2$ ,

$$\begin{eqnarray}F_{0}(\unicode[STIX]{x1D70F})=c_{0}(-1)q^{-1}+\mathop{\sum }_{\substack{ m\in \mathbb{Z} \\ m\geqslant 0}}c_{0}(m)q^{m}\end{eqnarray}$$

with $c_{0}(-1)=0$ or $1$ ,

$$\begin{eqnarray}F_{\unicode[STIX]{x1D6FE}}(\unicode[STIX]{x1D70F})=c_{\unicode[STIX]{x1D6FE}}(-1/p)q^{-1/p}+\mathop{\sum }_{\substack{ m\in \mathbb{Z}-1/p \\ m>0}}c_{\unicode[STIX]{x1D6FE}}(m)q^{m}\end{eqnarray}$$

with $c_{\unicode[STIX]{x1D6FE}}(-1/p)=0$ or $1$ if $q(\unicode[STIX]{x1D6FE})=1/p\;\text{mod}\;1$ and the other components $F_{\unicode[STIX]{x1D6FE}}$ of $F$ are holomorphic at $\infty$ . We define integers $c_{1}=c_{0}(-1)$ and $c_{p}=|\{\unicode[STIX]{x1D6FE}\in D\mid q(\unicode[STIX]{x1D6FE})=1/p\;\text{mod}\;1\text{ and }F_{\unicode[STIX]{x1D6FE}}\text{ singular}\}|$ .

Proof. The conditions imply $F_{0}\neq 0$ . Since $F$ is reflective, the product $F_{0}\unicode[STIX]{x1D6E5}$ is a modular form for $\unicode[STIX]{x1D6E4}_{0}(p)$ of weight $13-n/2$ which is holomorphic on the upper halfplane and at the cusps. Hence, $n\leqslant 26$ . If $n=26$ the function $F_{0}$ must be $1/\unicode[STIX]{x1D6E5}$ . However, as a result, $F$ does not transform correctly under $S$ . This proves the first statement. If $c_{1}=0$ the Riemann–Roch theorem applied to $F_{0}$ gives

$$\begin{eqnarray}-1\leqslant p\unicode[STIX]{x1D708}_{0}(F_{0})+\unicode[STIX]{x1D708}_{\infty }(F_{0})\leqslant \frac{m}{12}(p+1),\end{eqnarray}$$

where $m=1-n/2$ is the weight of $F_{0}$ . This implies the second statement.◻

Pairing $F$ with the Eisenstein series $E$ of weight $k=1+n/2$ we obtain (see Propositions 5.1 and 5.8) the following result.

Proposition 6.2. Suppose $F_{0}$ has constant coefficient $n-2$ . Then

$$\begin{eqnarray}\frac{k-2}{k}B_{k}(p^{k}-1)=\unicode[STIX]{x1D709}_{\text{even}}(p^{k-n_{p}/2}c_{1}+p^{1-n_{p}/2}c_{p})-c_{1}\end{eqnarray}$$

with $\unicode[STIX]{x1D709}_{\text{even}}=-(-1)^{k/2}$ if $n_{p}$ is even and

$$\begin{eqnarray}\frac{k-2}{k}B_{k,\unicode[STIX]{x1D712}}=\unicode[STIX]{x1D709}_{\text{odd}}(p^{k-(n_{p}+1)/2}c_{1}+p^{(1-n_{p})/2}c_{p})+c_{1}\end{eqnarray}$$

with

$$\begin{eqnarray}\unicode[STIX]{x1D709}_{\text{odd}}=\left\{\begin{array}{@{}ll@{}}\displaystyle -(-1)^{k/2}\quad & \displaystyle \text{if }p=1\hspace{0.2em}{\rm mod}\hspace{0.2em}4,\\ \displaystyle -(-1)^{(k-1)/2}\quad & \displaystyle \text{if }p=3\hspace{0.2em}{\rm mod}\hspace{0.2em}4,\end{array}\right.\end{eqnarray}$$

if $n_{p}$ is odd.

We will also need the following result.

Proof. Since $\unicode[STIX]{x1D6FE}_{p}(D)$ is a fourth root of unity the oddity formula $e(\operatorname{sign}(D)/8)=\unicode[STIX]{x1D6FE}_{p}(D)$ implies that $n$ is even. Then $n_{p}$ is also even and

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{p}(D)=\unicode[STIX]{x1D716}_{p}\biggl(\frac{-1}{p}\biggr)^{n_{p}/2}.\end{eqnarray}$$

Hence, $n-2=0$ or $4\;\text{mod}\;8$ and $\unicode[STIX]{x1D6FE}_{p}(D)=\unicode[STIX]{x1D716}_{p}$ . The lattice $L$ has determinant $p^{n+2}$ so that $\unicode[STIX]{x1D716}_{1}\unicode[STIX]{x1D716}_{p}=1$ by the sign rule. Now $\unicode[STIX]{x1D716}_{1}=+1$ because $L$ has maximal $p$ -rank and therefore $\unicode[STIX]{x1D716}_{p}=+1$ . Applying the oddity formula again, we obtain $n-2=0\;\text{mod}\;8$ . The second statement follows from the fact that there is only one class in the genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ under the given conditions.◻

6.2 Symmetric forms

Here we classify reflective modular forms that are invariant under $O(D)$ .

Let $L$ be an even lattice of prime level $p$ and genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ with $n>2$ . Then the number of elements $\unicode[STIX]{x1D6FE}$ in $D$ of order $p$ and norm $q(\unicode[STIX]{x1D6FE})=1/p\;\text{mod}\;1$ is given by

$$\begin{eqnarray}p^{n_{p}-1}-\unicode[STIX]{x1D709}_{\text{even}}p^{(n_{p}-2)/2}\end{eqnarray}$$

if $n_{p}$ is even and by

$$\begin{eqnarray}p^{n_{p}-1}+\unicode[STIX]{x1D709}_{\text{odd}}p^{(n_{p}-1)/2}\end{eqnarray}$$

if $n_{p}$ is odd (see [Reference ScheithauerSch06, Proposition 3.2]). Suppose $L$ carries a symmetric reflective modular form $F$ with $[F_{0}](0)=n-2$ . Then the Eisenstein condition takes the form

$$\begin{eqnarray}\frac{k-2}{k}B_{k}(p^{k}-1)=\unicode[STIX]{x1D709}_{\text{even}}(p^{k-n_{p}/2}d_{1}+p^{n_{p}/2}d_{p})-d_{1}-d_{p}\end{eqnarray}$$

if $n_{p}$ is even and

$$\begin{eqnarray}\frac{k-2}{k}B_{k,\unicode[STIX]{x1D712}}=\unicode[STIX]{x1D709}_{\text{odd}}(p^{k-(n_{p}+1)/2}d_{1}+p^{(n_{p}-1)/2}d_{p})+d_{1}+d_{p}\end{eqnarray}$$

if $n_{p}$ is odd. Here $d_{1}$ and $d_{p}$ can be $0$ or $1$ . In the case $n_{p}<n+2$ , the solutions of these equations have been determined in [Reference ScheithauerSch06].

Theorem 6.4. Let $L$ be an even lattice of prime level $p$ and genus $\mathit{II}_{n,2}(p^{\unicode[STIX]{x1D716}_{p}n_{p}})$ with $n>2$ carrying a symmetric reflective modular form $F$ . Suppose $F_{0}$ has constant coefficient $n-2$ . Then $L$ and $F$ are given in the following table.

The $\unicode[STIX]{x1D702}$ -product in the last column is a modular form for $\unicode[STIX]{x1D6E4}_{0}(p)$ whose lift on $0$ gives $F$ .

Conversely, each of these functions is a reflective modular form on $L$ with the above stated properties.

Proof. We only have to consider the case $n_{p}=n+2$ . Then $n=10$ or $18$ and $\unicode[STIX]{x1D709}_{\text{even}}=+1$ by Propositions 6.3 and 6.1. The Eisenstein condition simplifies to

$$\begin{eqnarray}\frac{k-2}{k}B_{k}=d_{p}.\end{eqnarray}$$

Now the left-hand side is $1/63$ for $k=6$ and $2/33$ for $k=10$ . Hence, there are no reflective forms if $n_{p}=n+2$ .◻

6.3 Bounds in the non-symmetric case

In this section, we derive bounds on the signature for reflective modular forms which are not invariant under $O(D)$ .

First, we recall the Riemann–Roch theorem for $\unicode[STIX]{x1D6E4}_{1}(p)$ .

Let $p$ be prime. For $p\geqslant 3$ , the group $\unicode[STIX]{x1D6E4}_{1}(p)$ has $p-1$ classes of cusps which can be represented by $1/c$ with $c=1,\ldots ,(p-1)/2$ of width $p$ and $a/p$ with $a=1,\ldots ,(p-1)/2$ of width $1$ . The cusps of $\unicode[STIX]{x1D6E4}_{1}(2)$ can be represented by $1/2$ of width $1$ and $1/1$ of width $2$ . Let $f\neq 0$ be a meromorphic modular form on $\unicode[STIX]{x1D6E4}_{1}(p)$ of weight $m$ and finite-order character. For $p\geqslant 5$ , there are no torsion points and the Riemann–Roch theorem states

$$\begin{eqnarray}\mathop{\sum }_{c=1}^{(p-1)/2}p\unicode[STIX]{x1D708}_{1/c}(f)+\mathop{\sum }_{a=1}^{(p-1)/2}\unicode[STIX]{x1D708}_{a/p}(f)+\mathop{\sum }_{\unicode[STIX]{x1D70F}\in \unicode[STIX]{x1D6E4}_{1}(p)\setminus H}\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70F}}(f)=\frac{m}{24}(p^{2}-1).\end{eqnarray}$$

For $p=3$ , we have

$$\begin{eqnarray}3\unicode[STIX]{x1D708}_{1/1}(f)+\unicode[STIX]{x1D708}_{1/3}(f)+\frac{1}{3}\unicode[STIX]{x1D708}_{e_{3}}(f)+\mathop{\sum }_{\substack{ \unicode[STIX]{x1D70F}\in \unicode[STIX]{x1D6E4}_{1}(3)\setminus H \\ \unicode[STIX]{x1D70F}\neq e_{3}\hspace{0.2em}{\rm mod}\hspace{0.2em}\unicode[STIX]{x1D6E4}_{1}(3)}}\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70F}}(f)=\frac{m}{3}\end{eqnarray}$$

with $e_{3}=(3+i\sqrt{3})/6$ and

$$\begin{eqnarray}2\unicode[STIX]{x1D708}_{1/1}(f)+\unicode[STIX]{x1D708}_{1/2}(f)+\frac{1}{2}\unicode[STIX]{x1D708}_{e_{2}}(f)+\mathop{\sum }_{\substack{ \unicode[STIX]{x1D70F}\in \unicode[STIX]{x1D6E4}_{1}(2)\setminus H \\ \unicode[STIX]{x1D70F}\neq e_{2}\hspace{0.2em}{\rm mod}\hspace{0.2em}\unicode[STIX]{x1D6E4}_{1}(2)}}\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70F}}(f)=\frac{m}{4}\end{eqnarray}$$

with $e_{2}=(1+i)/2$ if $p=2$ .

Proof. Since $F$ is non-symmetric there are $\unicode[STIX]{x1D6FE}_{1},\unicode[STIX]{x1D6FE}_{2}\in D\setminus \{0\}$ of the same norm such that

$$\begin{eqnarray}f=F_{\unicode[STIX]{x1D6FE}_{1}}-F_{\unicode[STIX]{x1D6FE}_{2}}\neq 0.\end{eqnarray}$$

The function $f$ is a modular form on $\unicode[STIX]{x1D6E4}_{1}(p)$ of weight $m=1-n/2$ and finite-order character.

Let $\unicode[STIX]{x1D6FE}\in D$ and $M=\big(\!\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\!\big)\in \unicode[STIX]{x1D6E4}$ . Then

$$\begin{eqnarray}F_{\unicode[STIX]{x1D6FE}}|_{m,M}=\biggl(\frac{a}{|D|}\biggr)e(-abq(\unicode[STIX]{x1D6FE}))F_{a\unicode[STIX]{x1D6FE}}\end{eqnarray}$$

if $c=0\;\text{mod}\;p$ and

$$\begin{eqnarray}\displaystyle F_{\unicode[STIX]{x1D6FE}}|_{m,M}=\frac{e(-\text{sign}(D)/8)}{\sqrt{|D|}}\biggl(\frac{-c}{|D|}\biggr)\mathop{\sum }_{\unicode[STIX]{x1D707}\in D}e(-c^{-1}dq(\unicode[STIX]{x1D707}))e(-b(\unicode[STIX]{x1D707},\unicode[STIX]{x1D6FE}))e(-abq(\unicode[STIX]{x1D6FE}))F_{a\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707}} & & \displaystyle \nonumber\end{eqnarray}$$

if $c\neq 0\;\text{mod}\;p$ . The coefficient at $F_{0}$ in this sum is

$$\begin{eqnarray}e(-c^{-1}dq(\unicode[STIX]{x1D707}))e(-b(\unicode[STIX]{x1D707},\unicode[STIX]{x1D6FE}))e(-abq(\unicode[STIX]{x1D6FE}))=e(-c^{-1}aq(\unicode[STIX]{x1D6FE})),\end{eqnarray}$$

i.e. only depends on the norm of $\unicode[STIX]{x1D6FE}$ .

This implies that for all $M\in \unicode[STIX]{x1D6E4}$ , the function $f|_{m,M}$ is a linear combination of functions $F_{\unicode[STIX]{x1D6FE}}$ with $\unicode[STIX]{x1D6FE}\neq 0$ . Hence,

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{s}(f)\geqslant -1/p\end{eqnarray}$$

for all cusps $s$ of $\unicode[STIX]{x1D6E4}_{1}(p)$ . It follows that

$$\begin{eqnarray}-\frac{p-1}{2}\biggl(1+\frac{1}{p}\biggr)\leqslant \frac{m}{24}(p^{2}-1).\end{eqnarray}$$

This proves the theorem. ◻

Note that the bounds do not hold in the symmetric case.

Using Theorem 6.5, we can determine the non-symmetric forms on lattices of prime level by analysing the obstructions in a finite number of cases. For $p=3$ , which is the most complicated case, we describe this explicitly in the next section. The other cases are analogous.

6.4 Level $3$

In this section we determine the reflective forms on lattices of level $3$ and signature $(n,2)$ where $n=4,~6,~8$ or $10$ .

Let $L$ be a lattice of genus $\mathit{II}_{10,2}(3^{\unicode[STIX]{x1D716}_{3}n_{3}})$ and $F$ a reflective form on $L$ . Suppose $F_{0}$ has constant coefficient $[F_{0}](0)=8$ . Then $c_{1}=1$ (see Proposition 6.1) and the Eisenstein condition gives the following value for $c_{3}$ (see Proposition 6.2).

Since $c_{3}$ should be a non-negative integer, this already excludes the first three cases.

The space $S_{6}(\unicode[STIX]{x1D6E4}(3))$ has dimension $3$ and is spanned by the functions $\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{A_{2}}^{2}$ , $\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{A_{2}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}}$ and $\unicode[STIX]{x1D702}_{1^{6}3^{6}}$ . The liftings of these functions generate the obstruction space $S_{\overline{\unicode[STIX]{x1D70C}}_{D},6}$ .

Pairing $F$ with the lift $F_{\unicode[STIX]{x1D702}_{1^{6}3^{6}},0}$ of the $\unicode[STIX]{x1D702}$ -product $\unicode[STIX]{x1D702}_{1^{6}3^{6}}(\unicode[STIX]{x1D70F})=\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})^{6}\unicode[STIX]{x1D702}(3\unicode[STIX]{x1D70F})^{6}$ we obtain

$$\begin{eqnarray}1-\frac{1}{3^{n_{3}/2}}-\frac{c_{3}}{3^{(n_{3}+4)/2}}=0.\end{eqnarray}$$

Next, we consider the case $n=8$ . Let $L$ be a lattice of genus $\mathit{II}_{8,2}(3^{\unicode[STIX]{x1D716}_{3}n_{3}})$ and $F$ a reflective modular form on $L$ with $[F_{0}](0)=6$ . Then we obtain the following for $c_{3}$ .

The discriminant form of type $3^{+1}$ contains no elements $\unicode[STIX]{x1D6FE}$ of norm $q(\unicode[STIX]{x1D6FE})=1/3\;\text{mod}\;1$ . Hence, this case can be excluded.

The space $S_{5}(\unicode[STIX]{x1D6E4}(3))$ has dimension $2$ and is spanned by the functions $\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{A_{2}}$ and $\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}}$ . The liftings of these functions generate the obstruction space $S_{\overline{\unicode[STIX]{x1D70C}}_{D},5}$ .

The lattice $A_{2}$ has genus $\mathit{II}_{2,0}(3^{-1})$ and is isomorphic to its rescaled dual $A_{2}^{\prime }(3)$ . The theta functions of $A_{2}$ can be written as

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}_{A_{2}} & = & \displaystyle \frac{\unicode[STIX]{x1D702}_{1^{3}}+9\unicode[STIX]{x1D702}_{9^{3}}}{\unicode[STIX]{x1D702}_{3^{1}}},\nonumber\\ \displaystyle \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}} & = & \displaystyle {\textstyle \frac{1}{2}}(\unicode[STIX]{x1D703}_{A_{2}^{\prime }}-\unicode[STIX]{x1D703}_{A_{2}}).\nonumber\end{eqnarray}$$

They transform under $S=\big(\!\begin{smallmatrix}0 & -1\\ 1 & 0\end{smallmatrix}\!\big)$ as

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}_{A_{2}}|_{1,S} & = & \displaystyle \frac{e(-1/4)}{\sqrt{3}}(\unicode[STIX]{x1D703}_{A_{2}}+2\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}})=\frac{e(-1/4)}{\sqrt{3}}\unicode[STIX]{x1D703}_{A_{2}^{\prime }},\nonumber\\ \displaystyle \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}}|_{1,S} & = & \displaystyle \frac{e(-1/4)}{\sqrt{3}}(\unicode[STIX]{x1D703}_{A_{2}}-\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}}).\nonumber\end{eqnarray}$$

Let $\unicode[STIX]{x1D6FE}\in D$ be of norm $q(\unicode[STIX]{x1D6FE})=1/3\;\text{mod}\;1$ . Then the lift of $\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{A_{2}}$ with respect to the dual Weil representation $\overline{\unicode[STIX]{x1D70C}}_{D}$ on $\unicode[STIX]{x1D6FE}$ is given by

$$\begin{eqnarray}F_{\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{A_{2}},\unicode[STIX]{x1D6FE}}=F_{1/3}+F_{1/1}\end{eqnarray}$$

with

$$\begin{eqnarray}F_{1/3}=\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{A_{2}}(e^{\unicode[STIX]{x1D6FE}}+e^{-\unicode[STIX]{x1D6FE}})\end{eqnarray}$$

and

$$\begin{eqnarray}F_{1/1}=-\frac{1}{3^{(n_{3}-1)/2}}\mathop{\sum }_{\unicode[STIX]{x1D707}\in D}e((\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707}))g_{j_{\unicode[STIX]{x1D707}}}(e^{\unicode[STIX]{x1D707}}+e^{-\unicode[STIX]{x1D707}}),\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D702}_{1^{8}}(\unicode[STIX]{x1D703}_{A_{2}}+2\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}})=g_{0}+g_{1}+g_{2}\end{eqnarray}$$

and $g_{j}|_{5,T}=e(j/3)g_{j}$ . Note that $g_{0}=0$ . We obtain an analogous result for the lift of $\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}}$ with respect to $\overline{\unicode[STIX]{x1D70C}}_{D}$ on an element $\unicode[STIX]{x1D6FE}\in D$ of norm $q(\unicode[STIX]{x1D6FE})=2/3\;\text{mod}\;1$ .

Let

$$\begin{eqnarray}M=\{\unicode[STIX]{x1D6FE}\in D\mid q(\unicode[STIX]{x1D6FE})=1/3\hspace{0.2em}{\rm mod}\hspace{0.2em}1\text{ and }F_{\unicode[STIX]{x1D6FE}}\text{ singular}\}.\end{eqnarray}$$

We assume now that $M$ is non-empty. Then $|M|=c_{3}=2\cdot 3^{(n_{3}-1)/2}$ and $M=-M$ because $F_{\unicode[STIX]{x1D6FE}}=F_{-\unicode[STIX]{x1D6FE}}$ . The crucial result to determine the structure of $M$ is the following proposition.

Proposition 6.7. Let $\unicode[STIX]{x1D6FE}\in D$ be of norm $q(\unicode[STIX]{x1D6FE})\neq 0\hspace{0.2em}{\rm mod}\hspace{0.2em}1$ . Then

$$\begin{eqnarray}|M\cap \unicode[STIX]{x1D6FE}^{\bot }|=\left\{\begin{array}{@{}ll@{}}2|M|/3\quad & \text{if }\unicode[STIX]{x1D6FE}\in M,\\ |M|/3\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

Proof. Let $\unicode[STIX]{x1D6FE}\in D$ be of norm $q(\unicode[STIX]{x1D6FE})=1/3\hspace{0.2em}{\rm mod}\hspace{0.2em}1$ . Suppose $\unicode[STIX]{x1D6FE}\in M$ . Then pairing $F$ with $F_{\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{A_{2}},\unicode[STIX]{x1D6FE}}$ gives

$$\begin{eqnarray}2-\frac{1}{3^{(n_{3}-1)/2}}\mathop{\sum }_{\unicode[STIX]{x1D707}\in M}(e((\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707}))+e(-(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707})))=0\end{eqnarray}$$

so that

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D707}\in M}(e((\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707}))+e(-(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707})))=|M|.\end{eqnarray}$$

This implies

$$\begin{eqnarray}|M\cap \unicode[STIX]{x1D6FE}^{\bot }|=2|M|/3.\end{eqnarray}$$

If $\unicode[STIX]{x1D6FE}\notin M$ the same argument shows $|M\cap \unicode[STIX]{x1D6FE}^{\bot }|=|M|/3$ . In case $q(\unicode[STIX]{x1D6FE})=2/3\hspace{0.2em}{\rm mod}\hspace{0.2em}1$ the statement follows from pairing $F$ with $F_{\unicode[STIX]{x1D702}_{1^{8}}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D708}+A_{2}},\unicode[STIX]{x1D6FE}}$ .◻

The proposition implies that $M^{\bot }$ is an isotropic subgroup of $D$ . Let $\unicode[STIX]{x1D6FE}\in M$ and $\unicode[STIX]{x1D707}\in M^{\bot }$ . Then $M\cap \unicode[STIX]{x1D6FE}^{\bot }=M\cap (\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707})^{\bot }$ . Hence, the group $M^{\bot }$ acts on $M$ by translations.

Proof. The sets $M\cap \unicode[STIX]{x1D6FE}^{\bot }$ and $M\cap \unicode[STIX]{x1D707}^{\bot }$ are both subsets of $M\setminus \{\pm \unicode[STIX]{x1D6FE}\}$ . Hence,

$$\begin{eqnarray}|(M\cap \unicode[STIX]{x1D6FE}^{\bot })\cap (M\cap \unicode[STIX]{x1D707}^{\bot })|\geqslant 4|M|/3-(|M|-2)=|M|/3+2.\end{eqnarray}$$

Since $(M\cap \unicode[STIX]{x1D6FE}^{\bot })\cap (M\cap \unicode[STIX]{x1D707}^{\bot })\subset (M\cap (\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707})^{\bot })$ this implies $|M\cap (\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707})^{\bot }|=2|M|/3$ and $\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707}\in M$ .◻

Proposition 6.9. Let $\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707}\in M$ such that $(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707})=0\;\text{mod}\;1$ . Then

$$\begin{eqnarray}(M\cap \unicode[STIX]{x1D6FE}^{\bot })\cap (M\cap \unicode[STIX]{x1D707}^{\bot })=M\cap (\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707})^{\bot }.\end{eqnarray}$$

Proof. We have $|M\cap \unicode[STIX]{x1D6FE}^{\bot }|=|M\cap \unicode[STIX]{x1D707}^{\bot }|=2|M|/3$ so that

$$\begin{eqnarray}|(M\cap \unicode[STIX]{x1D6FE}^{\bot })\cap (M\cap \unicode[STIX]{x1D707}^{\bot })|\geqslant 4|M|/3-|M|=|M|/3.\end{eqnarray}$$

On the other hand, $(M\cap \unicode[STIX]{x1D6FE}^{\bot })\cap (M\cap \unicode[STIX]{x1D707}^{\bot })\subset (M\cap (\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707})^{\bot })$ and $|M\cap (\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707})^{\bot }|=|M|/3$ because $q(\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D707})=2/3\;\text{mod}\;1$ . This implies the statement.◻

Proposition 6.10. Let $\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707},\unicode[STIX]{x1D708}\in M$ such that

$$\begin{eqnarray}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707})=(\unicode[STIX]{x1D707},\unicode[STIX]{x1D708})=2/3\quad \text{mod}\;1.\end{eqnarray}$$

Then

$$\begin{eqnarray}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D708})=2/3\quad \text{mod}\;1.\end{eqnarray}$$

A consequence of this result is the following proposition.

Proposition 6.11. Let $\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707}\in M$ such that $(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707})\neq 0\;\text{mod}\;1$ . Then

$$\begin{eqnarray}M\cap \unicode[STIX]{x1D6FE}^{\bot }=M\cap \unicode[STIX]{x1D707}^{\bot }.\end{eqnarray}$$

Proof. Let $\unicode[STIX]{x1D6FE}\in M$ . Then the elements $\unicode[STIX]{x1D707}\in M$ with $(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707})\neq 0\;\text{mod}\;1$ are in $\pm \unicode[STIX]{x1D6FE}+M^{\bot }$ . Hence, $M$ decomposes as

$$\begin{eqnarray}M=(\unicode[STIX]{x1D6FE}+M^{\bot })\cup (-\unicode[STIX]{x1D6FE}+M^{\bot })\cup (M\cap \unicode[STIX]{x1D6FE}^{\bot })\end{eqnarray}$$

so that

$$\begin{eqnarray}|M|=2|M^{\bot }|+2|M|/3.\end{eqnarray}$$

This implies the statement. ◻

Proposition 6.14. The set $M$ is of the form

$$\begin{eqnarray}M=\mathop{\bigcup }_{i=1}^{3}(\unicode[STIX]{x1D6FE}_{i}+M^{\bot })\cup \mathop{\bigcup }_{i=1}^{3}(-\unicode[STIX]{x1D6FE}_{i}+M^{\bot })\end{eqnarray}$$

with $\unicode[STIX]{x1D6FE}_{i}\in M$ and $(\unicode[STIX]{x1D6FE}_{i},\unicode[STIX]{x1D6FE}_{j})=0\hspace{0.2em}{\rm mod}\hspace{0.2em}1$ for $i\neq j$ .

Let $H$ be an isotropic subgroup of $D$ of order $|H|=3^{(n_{3}-3)/2}$ . Then the lift of $9\unicode[STIX]{x1D702}_{1^{-9}3^{3}}$ , with respect to $\unicode[STIX]{x1D70C}_{D}$ on $H$ , is given by

$$\begin{eqnarray}F_{9\unicode[STIX]{x1D702}_{1^{-9}3^{3}},H}=F_{1/3}+F_{1/1}\end{eqnarray}$$

with

$$\begin{eqnarray}F_{1/3}=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in H}9\unicode[STIX]{x1D702}_{1^{-9}3^{3}}e^{\unicode[STIX]{x1D6FE}}\end{eqnarray}$$

and

$$\begin{eqnarray}F_{1/1}=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in H^{\bot }}g_{j_{\unicode[STIX]{x1D6FE}}}e^{\unicode[STIX]{x1D6FE}}\end{eqnarray}$$

where $\unicode[STIX]{x1D702}_{1^{3}3^{-9}}(\unicode[STIX]{x1D70F}/3)=g_{0}(\unicode[STIX]{x1D70F})+g_{1}(\unicode[STIX]{x1D70F})+g_{2}(\unicode[STIX]{x1D70F})$ and $g_{j}|_{-3,T}=e(j/3)g_{j}$ . Note that

$$\begin{eqnarray}\displaystyle g_{0} & = & \displaystyle -3\unicode[STIX]{x1D702}_{1^{-9}3^{3}},\nonumber\\ \displaystyle g_{1} & = & \displaystyle 0.\nonumber\end{eqnarray}$$

The function $F_{9\unicode[STIX]{x1D702}_{1^{-9}3^{3}},H}$ has $0$ -component $F_{0}=6\unicode[STIX]{x1D702}_{1^{-9}3^{3}}$ and is reflective. The singular components are the $F_{\unicode[STIX]{x1D6FE}}$ with $\unicode[STIX]{x1D6FE}\in H^{\bot }$ and $q(\unicode[STIX]{x1D6FE})=1/3\hspace{0.6em}{\rm mod}\hspace{0.2em}1$ . The discriminant form $H^{\bot }/H$ is of type $3^{-3}$ . It is generated by elements $\{\unicode[STIX]{x1D6FE}_{1},\unicode[STIX]{x1D6FE}_{2},\unicode[STIX]{x1D6FE}_{3}\}$ with $q(\unicode[STIX]{x1D6FE}_{i})=1/3\;\text{mod}\;1$ and $(\unicode[STIX]{x1D6FE}_{i},\unicode[STIX]{x1D6FE}_{j})=0\;\text{mod}\;1$ for $i\neq j$ . We obtain the following result (see Theorem 2.1 and Proposition 2.2).

We can decompose $L=K\oplus \mathit{II}_{1,1}(3)$ , where $K$ has genus $\mathit{II}_{7,1}(3^{-\unicode[STIX]{x1D716}_{3}(n_{3}-2)})$ and assume that $H$ is a maximal isotropic subgroup of the discriminant form of $K$ . Then the embedding $K\subset K_{H}$ gives an embedding $L\subset L_{H}$ and identifies the corresponding domains ${\mathcal{H}}_{L}$ and ${\mathcal{H}}_{L_{H}}$ . Proposition 3.4 implies the following proposition.

We calculate the product expansions of the automorphic product $\unicode[STIX]{x1D6F9}$ corresponding to $F_{9\unicode[STIX]{x1D702}_{1^{-9}3^{3}},0}$ on $L$ of genus $\mathit{II}_{8,2}(3^{-3})$ .

First, we decompose $L=K\oplus \mathit{II}_{1,1}(3)$ . Then $K=E_{6}\oplus \mathit{II}_{1,1}$ . We choose a primitive norm $0$ vector $z$ in $\mathit{II}_{1,1}(3)$ .

Proposition 6.17. The expansion of $\unicode[STIX]{x1D6F9}$ at the cusp corresponding to $z$ is given by

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{\unicode[STIX]{x1D6FC}\in K^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[9\unicode[STIX]{x1D702}_{1^{-9}3^{3}}](-\unicode[STIX]{x1D6FC}^{2}/2)}\mathop{\prod }_{\unicode[STIX]{x1D6FC}\in (3K^{\prime })^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[-3\unicode[STIX]{x1D702}_{1^{-9}3^{3}}](-\unicode[STIX]{x1D6FC}^{2}/6)}\nonumber\\ \displaystyle & & \displaystyle \quad =1+\sum c(\unicode[STIX]{x1D706})e((\unicode[STIX]{x1D706},Z))\nonumber\end{eqnarray}$$

where $c(\unicode[STIX]{x1D706})$ is the coefficient at $q^{n}$ in $\unicode[STIX]{x1D702}_{1^{9}3^{-3}}$ if $\unicode[STIX]{x1D706}$ is $n$ times a primitive norm $0$ vector in $K^{+}$ and $0$ otherwise.

Proof. The product expansion of $\unicode[STIX]{x1D6F9}$ at the cusp corresponding to $z$ is

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{\unicode[STIX]{x1D6FC}\in {K^{\prime }}^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[F_{\unicode[STIX]{x1D6FC}}](-\unicode[STIX]{x1D6FC}^{2}/2)}(1-e(1/3)e((\unicode[STIX]{x1D6FC},Z)))^{[F_{\unicode[STIX]{x1D6FC}+z/3}](-\unicode[STIX]{x1D6FC}^{2}/2)}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,(1-e(2/3)e((\unicode[STIX]{x1D6FC},Z)))^{[F_{\unicode[STIX]{x1D6FC}+2z/3}](-\unicode[STIX]{x1D6FC}^{2}/2)}.\nonumber\end{eqnarray}$$

By the above formulas for the components of $F_{9\unicode[STIX]{x1D702}_{1^{-9}3^{3}},0}$ , this product is equal to

$$\begin{eqnarray}\displaystyle \mathop{\prod }_{\unicode[STIX]{x1D6FC}\in K^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[9\unicode[STIX]{x1D702}_{1^{-9}3^{3}}](-\unicode[STIX]{x1D6FC}^{2}/2)}\mathop{\prod }_{\unicode[STIX]{x1D6FC}\in (3K^{\prime })^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[-3\unicode[STIX]{x1D702}_{1^{-9}3^{3}}](-\unicode[STIX]{x1D6FC}^{2}/6)}. & & \displaystyle \nonumber\end{eqnarray}$$

Since $\unicode[STIX]{x1D6F9}$ has singular weight, the Fourier expansion of $\unicode[STIX]{x1D6F9}_{z}$ is supported only on norm $0$ vectors of $K^{\prime }$ . Hence, $\unicode[STIX]{x1D6F9}_{z}$ has the stated sum expansion.◻

This is the twisted denominator identity of the fake monster superalgebra [Reference ScheithauerSch00] corresponding to an element of class $3A$ in $O(E_{8})$ (see [Reference ScheithauerSch01, Proposition 6.1]).

Now, we decompose $L=K\oplus \mathit{II}_{1,1}$ with $K=E_{6}\oplus \mathit{II}_{1,1}(3)$ and choose a primitive norm $0$ vector $z$ in $\mathit{II}_{1,1}$ . Then we have the following result.

Proposition 6.18. The expansion of $\unicode[STIX]{x1D6F9}$ at the cusp corresponding to $z$ is given by

$$\begin{eqnarray}\displaystyle & & \displaystyle e((\unicode[STIX]{x1D70C},Z))\mathop{\prod }_{\unicode[STIX]{x1D6FC}\in {K^{\prime }}^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[\unicode[STIX]{x1D702}_{1^{3}3^{-9}}](-3\unicode[STIX]{x1D6FC}^{2}/2)}\mathop{\prod }_{\unicode[STIX]{x1D6FC}\in K^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[9\unicode[STIX]{x1D702}_{1^{-9}3^{3}}](-\unicode[STIX]{x1D6FC}^{2}/2)}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{w\in W}\det (w)\unicode[STIX]{x1D702}_{1^{-3}3^{9}}((w\unicode[STIX]{x1D70C},Z)),\nonumber\end{eqnarray}$$

where $W$ is the reflection group of $K^{\prime }$ generated by the roots of norm $\unicode[STIX]{x1D6FC}^{2}=2/3$ .

This is the twisted denominator identity of the fake monster algebra corresponding to an element of class $3C$ in $\mathit{Co}_{0}$ (see [Reference ScheithauerSch04, Proposition 10.7]).

Again, let $L$ be a lattice of genus $\mathit{II}_{8,2}(3^{\unicode[STIX]{x1D716}_{3}n_{3}})$ carrying a reflective modular form $F$ .

The lift of $\unicode[STIX]{x1D702}_{1^{3}3^{-9}}(\unicode[STIX]{x1D70F})=\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})^{3}\unicode[STIX]{x1D702}(3\unicode[STIX]{x1D70F})^{-9}$ with respect to $\unicode[STIX]{x1D70C}_{D}$ on $0$ is given by

$$\begin{eqnarray}F_{\unicode[STIX]{x1D702}_{1^{3}3^{-9}},0}=F_{1/3}+F_{1/1}\end{eqnarray}$$

with

$$\begin{eqnarray}F_{1/3}=\unicode[STIX]{x1D702}_{1^{3}3^{-9}}e^{0}\end{eqnarray}$$

and

$$\begin{eqnarray}F_{1/1}=3^{(11-n_{3})/2}\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in D}g_{j_{\unicode[STIX]{x1D6FE}}}e^{\unicode[STIX]{x1D6FE}},\end{eqnarray}$$

where $\unicode[STIX]{x1D702}_{1^{-9}3^{3}}(\unicode[STIX]{x1D70F}/3)=g_{0}(\unicode[STIX]{x1D70F})+g_{1}(\unicode[STIX]{x1D70F})+g_{2}(\unicode[STIX]{x1D70F})$ and $g_{j}|_{-3,T}=e(j/3)g_{j}$ . The modular form $F_{\unicode[STIX]{x1D702}_{1^{3}3^{-9}},0}$ is reflective and has $0$ -component

$$\begin{eqnarray}F_{0}(\unicode[STIX]{x1D70F})=q^{-1}+(3^{(11-n_{3})/2}-3)+\cdots \,.\end{eqnarray}$$

Suppose $L$ has genus $\mathit{II}_{8,2}(3^{-7})$ . Then the level $1$ expansion of the theta lift of $F_{\unicode[STIX]{x1D702}_{1^{3}3^{-9}},0}$ on $L$ is the twisted denominator identity of the fake monster superalgebra corresponding to an element in $O(E_{8})$ of class $3A$ and the level $3$ expansion gives the twisted denominator identity of the fake monster algebra corresponding to an element in $\mathit{Co}_{0}$ of class $3C$ .

The case $n_{3}=9$ has already been described above because we have the following result.

Proof. We decompose $L=K\oplus \mathit{II}_{1,1}(3)$ , where $K$ has genus $\mathit{II}_{7,1}(3^{-7})$ and choose a primitive norm $0$ vector $z$ in $\mathit{II}_{1,1}(3)$ . Then the product expansion of the theta lift $\unicode[STIX]{x1D6F9}$ of $F_{\unicode[STIX]{x1D702}_{1^{3}3^{-9}},0}$ at the cusp corresponding to $z$ is given by

$$\begin{eqnarray}\displaystyle e((\unicode[STIX]{x1D70C},Z))\mathop{\prod }_{\unicode[STIX]{x1D6FC}\in K^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[\unicode[STIX]{x1D702}_{1^{3}3^{-9}}](-\unicode[STIX]{x1D6FC}^{2}/2)}\mathop{\prod }_{\unicode[STIX]{x1D6FC}\in (3K^{\prime })^{+}}(1-e((\unicode[STIX]{x1D6FC},Z)))^{[3\unicode[STIX]{x1D702}_{1^{-9}3^{3}}](-\unicode[STIX]{x1D6FC}^{2}/6)}. & & \displaystyle \nonumber\end{eqnarray}$$

The Fourier coefficients $[\unicode[STIX]{x1D702}_{1^{3}3^{-9}}](n)$ vanish for $n=1\;\text{mod}\;3$ and $K=E_{6}^{\prime }(3)\oplus \mathit{II}_{1,1}(3)$ contains no elements $\unicode[STIX]{x1D6FC}$ of norm $-\unicode[STIX]{x1D6FC}^{2}/2=2\;\text{mod}\;3$ . This implies that the first product extends only over the elements $\unicode[STIX]{x1D6FC}\in K$ satisfying $\unicode[STIX]{x1D6FC}^{2}/2=0\;\text{mod}\;3$ , i.e.,  $\unicode[STIX]{x1D6FC}\in 3K^{\prime }$ . Now, $[\unicode[STIX]{x1D702}_{1^{3}3^{-9}}](3n)=-[3\unicode[STIX]{x1D702}_{1^{-9}3^{3}}](n)$ so that the product is constant. This finishes the proof.◻

Now we consider the case $n=6$ . Let $L$ be a lattice of genus $\mathit{II}_{6,2}(3^{\unicode[STIX]{x1D716}_{3}n_{3}})$ and $F$ a reflective form on $L$ with $[F_{0}](0)=4$ . We find the following value for $c_{3}$ .

Hence, we can assume that $F_{0}$ is holomorphic at $\infty$ and $n_{3}=4$ or $6$ .

The space $S_{4}(\unicode[STIX]{x1D6E4}(3))$ has dimension $1$ and is spanned by the function $\unicode[STIX]{x1D702}_{1^{8}}$ . The liftings of this function generate the obstruction space $S_{\overline{\unicode[STIX]{x1D70C}}_{D},4}$ .

Let $\unicode[STIX]{x1D6FE}\in D$ be of norm $q(\unicode[STIX]{x1D6FE})=1/3\;\text{mod}\;1$ . Then the lift of $\unicode[STIX]{x1D702}_{1^{8}}(\unicode[STIX]{x1D70F})=\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})^{8}$ with respect to the dual Weil representation $\overline{\unicode[STIX]{x1D70C}}_{D}$ on $\unicode[STIX]{x1D6FE}$ is given by

$$\begin{eqnarray}F_{\unicode[STIX]{x1D702}_{1^{8}},\unicode[STIX]{x1D6FE}}=F_{1/3}+F_{1/1}\end{eqnarray}$$

with

$$\begin{eqnarray}F_{1/3}=\unicode[STIX]{x1D702}_{1^{8}}(e^{\unicode[STIX]{x1D6FE}}+e^{-\unicode[STIX]{x1D6FE}})\end{eqnarray}$$

and

$$\begin{eqnarray}F_{1/1}=-\frac{1}{3^{(n_{3}-2)/2}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D707}\in D \\ q(\unicode[STIX]{x1D707})=1/3\hspace{0.2em}{\rm mod}\hspace{0.2em}1}}e((\unicode[STIX]{x1D707},\unicode[STIX]{x1D6FE}))\unicode[STIX]{x1D702}_{1^{8}}(e^{\unicode[STIX]{x1D707}}+e^{-\unicode[STIX]{x1D707}}).\end{eqnarray}$$

As above, we define

$$\begin{eqnarray}M=\{\unicode[STIX]{x1D6FE}\in D\mid q(\unicode[STIX]{x1D6FE})=1/3\hspace{0.2em}{\rm mod}\hspace{0.2em}1\text{ and }F_{\unicode[STIX]{x1D6FE}}\text{ singular}\}.\end{eqnarray}$$

Then $|M|=4\cdot 3^{(n_{3}-4)/2}$ and $M=-M$ . Pairing $F$ with $F_{\unicode[STIX]{x1D702}_{1^{8}},\unicode[STIX]{x1D6FE}}$ we obtain the following result.

Proposition 6.21. Let $\unicode[STIX]{x1D6FE}\in D$ be of norm $q(\unicode[STIX]{x1D6FE})=1/3\;\text{mod}\;1$ . Then

$$\begin{eqnarray}|M\cap \unicode[STIX]{x1D6FE}^{\bot }|=\left\{\begin{array}{@{}ll@{}}\displaystyle 5|M|/6\quad & \displaystyle \text{if }\unicode[STIX]{x1D6FE}\in M,\\ \displaystyle |M|/3\quad & \displaystyle \text{otherwise.}\end{array}\right.\end{eqnarray}$$

This excludes the case $n_{3}=4$ . We assume now that $n_{3}=6$ . Then the proposition shows that $M$ must be of the form

$$\begin{eqnarray}M=\{\pm \unicode[STIX]{x1D6FE}_{1},\ldots ,\pm \unicode[STIX]{x1D6FE}_{6}\}\end{eqnarray}$$

with $(\unicode[STIX]{x1D6FE}_{i},\unicode[STIX]{x1D6FE}_{j})=0\;\text{mod}\;1$ for $i\neq j$ . In particular, $M^{+}=\{\unicode[STIX]{x1D6FE}_{1},\ldots ,\unicode[STIX]{x1D6FE}_{6}\}$ is a basis of $D$ . Let $\unicode[STIX]{x1D6FE}\in D$ be of norm $q(\unicode[STIX]{x1D6FE})=1/3\;\text{mod}\;1$ with $\unicode[STIX]{x1D6FE}\notin M$ . Then $\unicode[STIX]{x1D6FE}$ is a linear combination of four of the $\unicode[STIX]{x1D6FE}_{i}$ so that $|M\cap \unicode[STIX]{x1D6FE}^{\bot }|=4$ . Hence, the principal part of $F$ satisfies all obstructions coming from $S_{\overline{\unicode[STIX]{x1D70C}}_{D},4}$ . This implies that a reflective modular form with constant coefficient $4$ on $\mathit{II}_{6,2}(3^{+6})$ exists.

We give an explicit construction. Let $\unicode[STIX]{x1D6FE}\in D$ be of norm $q(\unicode[STIX]{x1D6FE})=1/3\;\text{mod}\;1$ . Then the lift of $\unicode[STIX]{x1D703}_{A_{2}}^{2}/\unicode[STIX]{x1D702}_{1^{8}}$ on $\unicode[STIX]{x1D6FE}$ with respect to $\unicode[STIX]{x1D70C}_{D}$ is given by

$$\begin{eqnarray}F_{\unicode[STIX]{x1D703}_{A_{2}}^{2}/\unicode[STIX]{x1D702}_{1^{8}},\unicode[STIX]{x1D6FE}}=F_{1/3}+F_{1/1}\end{eqnarray}$$

with

$$\begin{eqnarray}F_{1/3}=\frac{\unicode[STIX]{x1D703}_{A_{2}}^{2}}{\unicode[STIX]{x1D702}_{1^{8}}}(e^{\unicode[STIX]{x1D6FE}}+e^{-\unicode[STIX]{x1D6FE}})\end{eqnarray}$$

and

$$\begin{eqnarray}F_{1/1}=\frac{1}{3^{3}}\mathop{\sum }_{\unicode[STIX]{x1D707}\in D}e(-(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D707}))g_{j_{\unicode[STIX]{x1D707}}}(e^{\unicode[STIX]{x1D707}}+e^{-\unicode[STIX]{x1D707}})\end{eqnarray}$$

where $\unicode[STIX]{x1D703}_{A_{2}^{\prime }}^{2}/\unicode[STIX]{x1D702}_{1^{8}}=g_{0}+g_{1}+g_{2}$ and $g_{j}|_{-2,T}=e(j/3)g_{j}$ . Note that $g_{2}=\unicode[STIX]{x1D703}_{A_{2}}^{2}/\unicode[STIX]{x1D702}_{1^{8}}$ .

The function $\unicode[STIX]{x1D702}_{(1/3)^{-3}1^{2}3^{-3}}(\unicode[STIX]{x1D70F})=\unicode[STIX]{x1D702}_{1^{-3}3^{2}9^{-3}}(\unicode[STIX]{x1D70F}/3)$ is a modular form for $\unicode[STIX]{x1D6E4}(3)$ of weight $-2$ . If we decompose $\unicode[STIX]{x1D702}_{(1/3)^{-3}1^{2}3^{-3}}=h_{0}+h_{1}+h_{2}$ with $h_{j}|_{-2,T}=e(j/3)h_{j}$ , then $g_{2}=h_{2}$ , $g_{1}=4h_{1}$ and $g_{0}=4h_{0}$ . It follows