7 Fourier eigenfunctions with double zeroes at lattice points

In this section we construct two radial Schwartz functions \(a,b:\mathbb {R}^8\to i\mathbb {R}\) such that

\begin{align} \mathcal{F}(a)& =a\label{eqn:a-fourier}\\ \mathcal{F}(b)& =-b\label{eqn:b-fourier} \end{align}

which double zeroes at all \(\Lambda _8\)-vectors of length greater than \(\sqrt{2}\). Recall that each vector of \(\Lambda _8\) has length \(\sqrt{2n}\) for some \(n\in \mathbb {N}_{\geq 0}\). We define \(a\) and \(b\) so that their values are purely imaginary because this simplifies some of our computations. We will show in Section 8 that an appropriate linear combination of functions \(a\) and \(b\) satisfies conditions 6–8.

First, we will define function \(a\). To this end we consider the following functions:

Definition 7.1

\begin{align} \phi _{-4} & := \frac{E_4^2}{\Delta } \label{eqn: def phi4} \\ \phi _{-2} & := \frac{E_4(E_2 E_4 - E_6)}{\Delta } \label{eqn: def phi2} \\ \phi _{0} & := \frac{(E_2 E_4 - E_6)^2}{\Delta } \label{eqn: def phi0} \end{align}

The function \(\phi _0(z)\) is not modular; however, it satisfies the following transformation rules:

Lemma 7.2

We have

\begin{align} \phi _0(z + 1) & = \phi _0(z) \label{eqn:phi0-trans-T} \\ \phi _0\left(-\frac{1}{z}\right) & = \phi _0(z)-\frac{12i}{\pi }\, \frac{1}{z}\, \phi _{-2}(z)-\frac{36}{\pi ^2}\, \frac{1}{z^2}\, \phi _{-4}(z).\label{eqn:phi0-trans-S} \end{align}

Proof ▶

118 easily follows from periodicity of Eisenstein series and \(\Delta (z)\). For 119,

\begin{align} \phi _{0}\left(-\frac{1}{z}\right) & = \frac{(E_2(-1/z) E_4(-1/z) - E_6(-1/z))^{2}}{\Delta (-1/z)} \\ & = \frac{((z^2 E_2(z) - 6iz / \pi ) \cdot z^4 E_4(z) - z^6 E_6(z))^{2}}{z^{12} \Delta (z)} \\ & = \frac{\left(E_2(z) E_4(z) - E_6(z) - \frac{6i}{\pi z} E_4(z)\right)^2}{\Delta (z)} \\ & = \frac{(E_2(z) E_4(z) - E_6(z))^{2} - \frac{12i}{\pi z}(E_2(z) E_4(z) - E_6(z)) E_4(z) - \frac{36}{\pi ^2 z^2} E_4(z)^{2}}{\Delta (z)} \\ & = \phi _0(z) - \frac{12 i}{\pi z} \phi _{-2}(z) - \frac{36}{\pi ^2 z^2} \phi _{-4}(z). \end{align}

Definition 7.3

✓

For \(x\in \mathbb {R}^8\) we define

\begin{align} \label{eqn:a-definition} a(x):=& \int \limits _{-1}^i\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, e^{\pi i \| x\| ^2 z}\, dz +\int \limits _{1}^i\phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2\, e^{\pi i \| x\| ^2 z}\, dz\\ -& 2\int \limits _{0}^i\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, e^{\pi i \| x\| ^2 z}\, dz +2\int \limits _{i}^{i\infty }\phi _0(z)\, e^{\pi i \| x\| ^2 z}\, dz.\nonumber \end{align}

We observe that the contour integrals in 125 converge absolutely and uniformly for \(x\in \mathbb {R}^8\). Indeed, \(\phi _0(z)=O(e^{2\pi i z})\) as \(\Im (z)\to \infty \). Therefore, \(a(x)\) is well defined. Now we prove that \(a\) satisfies condition 113. The following lemma will be used to prove Schwartzness of \(a\) and \(b\).

Lemma 7.4

✓

Let \(f(z)\) be a holomorphic function with a Fourier expansion

\begin{equation} f(z) = \sum _{n \ge n_0} c_f(n) e^{\pi i n z} \end{equation}

126

with \(c_f(n_0) \ne 0\). Assume that \(c_f(n)\) has a polynomial growth, i.e. \(|c_f(n)| = O(n^k)\) for some \(k \in \mathbb {N}\). Then there exists a constant \(C_f {\gt} 0\) such that

\begin{equation} \left|\frac{f(z)}{\Delta (z)}\right| \le C_f e^{-\pi (n_0 - 2) \Im z} \end{equation}

127

for all \(z\) with \(\Im z {\gt} 1/2\).

Proof ▶

By the product formula 12,

\begin{align} \left|\frac{f(z)}{\Delta (z)}\right| & = \left|\frac{\sum _{n \ge n_0} c_f(n) e^{\pi i n z}}{e^{2 \pi i z}\prod _{n \ge 1} (1 - e^{2\pi i n z})^{24}}\right| \\ & = |e^{\pi i (n_0 - 2)z}| \cdot \frac{|\sum _{n \ge n_0} c_f(n) e^{\pi i (n - n_0) z}|}{\prod _{n \ge 1} |1 - e^{2\pi i n z}|^{24}} \\ & \le e^{-\pi (n_0 - 2) \Im z} \cdot \frac{\sum _{n \ge n_0} |c_f(n)| e^{-\pi (n - n_0) \Im z}}{\prod _{n \ge 1} (1 - e^{- 2\pi n \Im z})^{24}} \\ & \le e^{-\pi (n_0 - 2) \Im z} \cdot \frac{\sum _{n \ge n_0} |c_f(n)| e^{-\pi (n - n_0) / 2}}{\prod _{n \ge 1} (1 - e^{-\pi n})^{24}} \\ & = C_f \cdot e^{-\pi (n_0 - 2) \Im z} \end{align}

where

\begin{equation} C_f = \frac{\sum _{n \ge n_0} |c_f(n)| e^{-\pi (n - n_0) / 2}}{\prod _{n \ge 1} (1 - e^{-\pi n})^{24}}. \end{equation}

133

Note that the summation in the numerator converges absolutely because of polynomial growth. The denominator also converges, which is simply \(e^{\pi } \cdot \Delta (i/2)\).

As corollaries, we have the following bound for \(\phi _0\), \(\phi _{-2}\), and \(\phi _{-4}\).

Corollary 7.5

There exists a constant \(C_0 {\gt} 0\) such that

\begin{equation} \label{eqn:phi0-bound} |\phi _0(z)| \le C_0 e^{-2 \pi \Im z} \end{equation}

134

for all \(z\) with \(\Im z {\gt} 1/2\).

Proof ▶

By Ramanujan’s formula, \(E_2 E_4 - E_6 = 3E_4' = 720 \sum _{n \ge 1} n \sigma _3(n) e^{2 \pi i n z}\) and

\begin{equation} (E_2(z) E_4(z) - E_6(z))^{2} = 720^{2} e^{4 \pi i z} + O(e^{5 \pi i z}). \notag \end{equation}

135

Then the result follows from Lemma 7.4 with \(f(z) = (E_2 E_4 - E_6)^2\) and \(n_0 = 4\).

Corollary 7.6

There exists a constant \(C_{-2} {\gt} 0\) such that

\begin{equation} \label{eqn:phi2-bound} |\phi _{-2}(z)| \le C_{-2} \end{equation}

135

for all \(z\) with \(\Im z {\gt} 1/2\).

Corollary 7.7

There exists a constant \(C_{-4} {\gt} 0\) such that

\begin{equation} \label{eqn:phi4-bound} |\phi _{-4}(z)| \le C_{-4} e^{2 \pi \Im z} \end{equation}

136

for all \(z\) with \(\Im z {\gt} 1/2\).

Note that we can take the constants \(C_0\), \(C_{-2}\), and \(C_{-4}\) as

\begin{align} C_0 & = 9 \cdot 240^2 \cdot e^{\pi } \cdot \frac{E_4'(i/2)^{2}}{\Delta (i/2)} \notag \\ C_{-2} & = 3 \cdot \frac{E_4(i/2) E_4'(i/2)}{\Delta (i/2)} \notag \\ C_{-4} & = e^{-\pi } \cdot \frac{E_4(i/2)^{2}}{\Delta (i/2)}. \notag \end{align}

Proposition 7.8

\(a(x)\) is a Schwartz function.

Proof ▶

We estimate the first summand in the right-hand side of 125. By 134, we have

\begin{align} & \left|\int \limits _{-1}^{i}\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, e^{\pi i r^2 z}\, dz\right|=\left|\int \limits _{i\infty }^{-1/(i+1)}\phi _0(z)\, z^{-4}\, e^{\pi i r^2 (-1/z-1)}\, dz\right|\leq \notag \\ & C_1\int \limits _{1/2}^{\infty }e^{-2\pi t}\, e^{-\pi r^2/t}\, dt\leq C_1\int \limits _{0}^{\infty }e^{-2\pi t}\, e^{-\pi r^2/t}\, dt=C_2\, r\, K_1(2\sqrt{2}\, \pi \, r)\notag \end{align}

where \(C_1\) and \(C_2\) are some positive constants and \(K_\alpha (x)\) is the modified Bessel function of the second kind defined as in [ 1 , Section 9.6 ] . This estimate also holds for the second and third summand in 125. For the last summand we have

\begin{equation} \left|\int \limits _{i}^{i\infty }\phi _0(z)\, e^{\pi i r^2 z}\, dz\right|\leq C\, \int \limits _{1}^{\infty } e^{-2\pi t}\, e^{-\pi r^2 t}\, dt=C_3\frac{e^{\pi (r^2+2)}}{r^2+2}. \end{equation}

137

Therefore, we arrive at

\begin{equation} |a(r)|\leq 4C_2\, r\, K_1(2\sqrt{2}\pi r)+2C_3\frac{e^{-\pi (r^2+2)}}{r^2+2}. \end{equation}

138

It is easy to see that the left hand side of this inequality decays faster then any inverse power of \(r\). Analogous estimates can be obtained for all derivatives \(\frac{\mathrm{d}^k}{\mathrm{d}r^k}a(r)\).

Proposition 7.9

\(a(x)\) satisfies 113.

Proof ▶

We recall that the Fourier transform of a Gaussian function is

\begin{equation} \label{eqn:gaussian Fourier} \mathcal{F}(e^{\pi i \| x\| ^2 z})(y)=z^{-4}\, e^{\pi i \| y\| ^2 \, (\frac{-1}{z}) }. \end{equation}

139

Next, we exchange the contour integration with respect to \(z\) variable and Fourier transform with respect to \(x\) variable in 125. This can be done, since the corresponding double integral converges absolutely. In this way we obtain

\begin{align} \widehat{a}(y)=& \int \limits _{-1}^i\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, z^{-4}\, e^{\pi i \| y\| ^2 \, (\frac{-1}{z})}\, dz +\int \limits _{1}^i\phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2\, z^{-4}\, e^{\pi i \| y\| ^2 \, (\frac{-1}{z})}\, dz\notag \\ -& 2\int \limits _{0}^i\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, z^{-4}\, e^{\pi i \| y\| ^2 \, (\frac{-1}{z})}\, dz +2\int \limits _{i}^{i\infty }\phi _0(z)\, z^{-4}\, e^{\pi i \| y\| ^2 \, (\frac{-1}{z})}\, dz.\notag \end{align}

Now we make a change of variables \(w=\frac{-1}{z}\). We obtain

\begin{align} \widehat{a}(y)=& \int \limits _{1}^i\phi _0\Big(1-\frac{1}{w-1}\Big)\, (\frac{-1}{w}+1)^2\, w^{2}\, e^{\pi i \| y\| ^2 \, w}\, dw\notag \\ +& \int \limits _{-1}^i\phi _0\Big(1-\frac{1}{w+1}\Big)\, (\frac{-1}{w}-1)^2\, w^2\, e^{\pi i \| y\| ^2 \, w}\, dw\\ -& 2\int \limits _{i \infty }^i\phi _0(w)\, e^{\pi i \| y\| ^2 \, w}\, dw +2\int \limits _{i}^{0}\phi _0\Big(\frac{-1}{w}\Big)\, w^{2}\, e^{\pi i \| y\| ^2 \, w}\, dw.\notag \end{align}

Since \(\phi _0\) is \(1\)-periodic we have

\begin{align} \widehat{a}(y)\, =\, & \int \limits _{1}^i\phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2\, e^{\pi i \| y\| ^2 \, z}\, dz +\int \limits _{-1}^i\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, e^{\pi i \| y\| ^2 \, z}\, dz\notag \\ +& 2\int \limits _{i}^{i\infty }\phi _0(z)\, e^{\pi i \| y\| ^2 \, z}\, dz -2\int \limits _{0}^{i}\phi _0\Big(\frac{-1}{z}\Big)\, z^{2}\, e^{\pi i \| y\| ^2 \, z}\, dz\notag \\ \, =\, & a(y). \end{align}

This finishes the proof of the proposition.

Next, we check that \(a\) has double zeroes at all \(\Lambda _8\)-lattice points of length greater then \(\sqrt{2}\). Using 134, 135, and 136, we can control the behavior of \(\phi _0\) near \(0\) and \(i\infty \).

Corollary 7.10

We have

\begin{align} \phi _0\left(\frac{i}{t}\right) & = O(e^{-2 \pi / t}) \quad \text{as } t \to 0 \label{eqn:phi0-near-0} \\ \phi _0\left(\frac{i}{t}\right) & = O(t^{-2}e^{2 \pi t}) \quad \text{as } t \to \infty . \label{eqn:phi0-near-infty} \\ \end{align}

Proof ▶

The first estimate follows from 134 with \(z = i/t\). For the second estimate, by 119, 135, and 136, we have

\begin{equation} \left|\phi _0\left(\frac{i}{t}\right)\right| \le |\phi _0(it)| + \frac{12}{\pi t} |\phi _{-2}(it)| + \frac{36}{\pi ^2 t^2} |\phi _{-4}(it)| \le C_0 e^{-2 \pi t} + \frac{12}{\pi t} \cdot C_{-2} + \frac{36}{\pi ^2 t^2} \cdot C_{-4} e^{2 \pi t} = O(t^{-2}e^{2 \pi t}). \end{equation}

145

Proposition 7.11

For \(r{\gt}\sqrt{2}\) we can express \(a(r)\) in the following form

\begin{equation} \label{eqn: a double zeroes} a(r)=-4\sin (\pi r^2/2)^2\, \int \limits _{0}^{i\infty }\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, e^{\pi i r^2 \, z}\, dz. \end{equation}

146

Proof ▶

We denote the right hand side of 146 by \(d(r)\). Convergence of the integral for \(r {\gt} \sqrt{2}\) follows from Corollary 7.10. We can write

\begin{align} d(r)=& \int \limits _{-1}^{i\infty -1}\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, e^{\pi i r^2 \, z}\, dz- 2\int \limits _{0}^{i\infty }\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, e^{\pi i r^2 \, z}\, dz\notag \\ +& \int \limits _{1}^{i\infty +1}\phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2\, e^{\pi i r^2 \, z}\, dz.\notag \end{align}

From 119 we deduce that if \(r{\gt}\sqrt{2}\) then \(\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, e^{\pi i r^2 \, z}\to 0\) as \(\Im (z)\to \infty \). Therefore, we can deform the paths of integration and rewrite

\begin{align} d(r)=& \int \limits _{-1}^{i}\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, e^{\pi i r^2 \, z}\, dz +\int \limits _{i}^{i\infty }\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, e^{\pi i r^2 \, z}\, dz\notag \\ -2& \int \limits _{0}^{i}\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, e^{\pi i r^2 \, z}\, dz -2\int \limits _{i}^{i\infty }\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, e^{\pi i r^2 \, z}\, dz\notag \\ +& \int \limits _{1}^{i}\phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2\, e^{\pi i r^2 \, z}\, dz +\int \limits _{i}^{i\infty }\phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2\, e^{\pi i r^2 \, z}\, dz.\notag \end{align}

Now from 119 we find

\begin{align} & \phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2-2\phi _0\Big(\frac{-1}{z}\Big)\, z^2+ \phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2=\notag \\ & \phi _0(z+1)\, (z+1)^2-2\phi _0(z)\, z^2+\phi _0(z-1)\, (z-1)^2\notag \\ & -\frac{12i}{\pi }\, \Big(\phi _{-2}(z+1)\, (z+1)-2\phi _{-2}(z)\, z+\phi _{-2}(z-1)\, (z-1)\Big)\notag \\ & -\frac{36}{\pi ^2}\Big(\phi _{-4}(z+1)-2\phi _{-4}(z)+\phi _{-4}(z-1)\Big)=\notag \\ & 2\phi _0(z). \end{align}

Thus, we obtain

\begin{align} d(r)=& \int \limits _{-1}^{i}\phi _0\Big(\frac{-1}{z+1}\Big)\, (z+1)^2\, e^{\pi i r^2 \, z}\, dz -2\int \limits _{0}^{i}\phi _0\Big(\frac{-1}{z}\Big)\, z^2\, e^{\pi i r^2 \, z}\, dz\notag \\ +& \int \limits _{1}^{i}\phi _0\Big(\frac{-1}{z-1}\Big)\, (z-1)^2\, e^{\pi i r^2 \, z}\, dz +2\int \limits _{i}^{i\infty }\phi _0(z)\, e^{\pi i r^2 \, z}\, dz=a(r).\notag \end{align}

This finishes the proof.

Finally, we find another convenient integral representation for \(a\) and compute values of \(a(r)\) at \(r=0\) and \(r=\sqrt{2}\).

Proposition 7.12

For \(r\geq 0\) we have

\begin{align} \label{eqn:a-another-integral}a(r)=& 4i\, \sin (\pi r^2/2)^2\, \Bigg(\frac{36}{\pi ^3\, (r^2-2)}-\frac{8640}{\pi ^3\, r^4}+\frac{18144}{\pi ^3\, r^2}\\ +& \int \limits _0^\infty \, \left(t^2\, \phi _0\Big(\frac{i}{t}\Big)-\frac{36}{\pi ^2}\, e^{2\pi t}+\frac{8640}{\pi }\, t-\frac{18144}{\pi ^2}\right)\, e^{-\pi r^2 t}\, dt \Bigg) .\notag \end{align}

The integral converges absolutely for all \(r\in \mathbb {R}_{\geq 0}\).

Proof ▶

Suppose that \(r{\gt}\sqrt{2}\). Then by Proposition 7.11

\[ a(r)=4i\, \sin (\pi r^2/2)^2\, \int \limits _{0}^{\infty }\phi _0(i/t)\, t^2\, e^{-\pi r^2 t}\, dt. \]

From 119 we obtain

\begin{equation} \label{eqn: phi asymptotic} \phi _0(i/t)\, t^2=\frac{36}{\pi ^2}\, e^{2 \pi t}-\frac{8640}{\pi }\, t+\frac{18144}{\pi ^2}+O(t^2\, e^{-2\pi t})\quad \mbox{as}\; t\to \infty . \end{equation}

149

For \(r{\gt}\sqrt{2}\) we have

\begin{equation} \int \limits _0^\infty \left(\frac{36}{\pi ^2}\, e^{2 \pi t}+\frac{8640}{\pi }\, t+\frac{18144}{\pi ^2}\right)\, e^{-\pi r^2 t}\, dt =\frac{36}{\pi ^3\, (r^2-2)}-\frac{8640}{\pi ^3\, r^4}+\frac{18144}{\pi ^3\, r^2}.\end{equation}

150

Therefore, the identity 148 holds for \(r{\gt}\sqrt{2}\).

On the other hand, from the definition 125 we see that \(a(r)\) is analytic in some neighborhood of \([0,\infty )\). The asymptotic expansion 149 implies that the right hand side of 148 is also analytic in some neighborhood of \([0,\infty )\). Hence, the identity 148 holds on the whole interval \([0,\infty )\). This finishes the proof of the proposition.

From the identity 148 we see that the values \(a(r)\) are in \(i\mathbb {R}\) for all \(r\in \mathbb {R}_{\geq 0}\).

Proposition 7.13

We have \(a(0) = -\frac{i}{8640}\).

Proof ▶

These identities follow immediately from the previous proposition.

Now we construct function \(b\). To this end we consider the function

Definition 7.14

\begin{equation} \label{eqn: h define} h(z) := 128 \frac{H_3(z) + H_4(z)}{H_2(z)^2}. \end{equation}

151

It is easy to see that \(h\in M^!_{-2}(\Gamma _0(2))\). Indeed, first we check that \(h|_{-2}\gamma =h\) for all \(\gamma \in \Gamma _0(2)\). Since the group \(\Gamma _0(2)\) is generated by elements \(\left(\begin{smallmatrix} 1 & 0 \\ 2 & 1 \end{smallmatrix}\right)\) and \(\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)\) it suffices to check that \(h\) is invariant under their action. This follows immediately from 27–29 and 151. Next we analyze the poles of \(h\). It is known [ 8 , Chapter I Lemma 4.1 ] that \(\theta _{10}\) has no zeros in the upper-half plane and hence \(h\) has poles only at the cusps. At the cusp \(i\infty \) this modular form has the Fourier expansion

\begin{equation} h(z)\, =\, q^{-1} + 16 - 132 q + 640 q^2 - 2550 q^3+O(q^4).\notag \end{equation}

152

Let \(I=\left(\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix}\right)\), \(T=\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)\), and \(S=\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)\) be elements of \(\Gamma _1\).

Definition 7.15

We define the following three functions

\begin{align} \psi _I\, :=\, & h-h|_{-2}ST \label{eqn:psiI-define}\\ \psi _T\, :=\, & \psi _I|_{-2}T \label{eqn:psiT-define}\\ \psi _S\, :=\, & \psi _I|_{-2}S. \label{eqn:psiS-define} \end{align}

Lemma 7.16

More explicitly, we have

\begin{align} \psi _I(z) & = 128 \frac{H_3(z) + H_4(z)}{H_2(z)^2} + 128 \frac{H_4(z) - H_2(z)}{H_3(z)^2} \label{eqn:psiI-explicit} \\ \psi _T(z) & = 128 \frac{H_3(z) + H_4(z)}{H_2(z)^2} + 128 \frac{H_2(z) + H_3(z)}{H_4(z)^2} \label{eqn:psiT-explicit} \\ \psi _S(z) & = 128 \frac{H_2(z) + H_3(z)}{H_4(z)^2} - 128 \frac{H_2(z) - H_4(z)}{H_3(z)^2} \label{eqn:psiS-explicit} \end{align}

Lemma 7.17

The Fourier expansions of these functions are

\begin{align} \psi _I(z)\, =\, & q^{-1} + 144 - 5120 q^{1/2} + 70524 q - 626688 q^{3/2} + 4265600 q^2 + O(q^{5/2}) \label{eqn: psi fourier I}\\ \psi _T(z)\, =\, & q^{-1} + 144 + 5120 q^{1/2} + 70524 q + 626688 q^{3/2} + 4265600 q^2 + O(q^{5/2}) \label{eqn: psi fourier T}\\ \psi _S(z)\, =\, & -10240 q^{1/2} - 1253376 q^{3/2} - 48328704 q^{5/2} - 1059078144 q^{7/2}+O(q^{9/2}).\label{eqn: psi fourier S} \end{align}

Definition 7.18

For \(x\in \mathbb {R}^8\) define

\begin{align} \label{eqn:b-definition} b(x):= & \int \limits _{-1}^{i}\psi _T(z)\, e^{\pi i \| x\| ^2 z}\, dz + \int \limits _{1}^{i}\psi _T(z)\, e^{\pi i \| x\| ^2 z}\, dz \\ -& 2\, \int \limits _{0}^{i}\psi _I(z)\, e^{\pi i \| x\| ^2 z}\, dz - 2\, \int \limits _{i}^{i\infty }\psi _S(z)\, e^{\pi i \| x\| ^2 z}\, dz \nonumber . \end{align}

Now we prove that \(b\) is a Schwartz function and satisfies condition 114.

Lemma 7.19

\(\psi _S(z)\) can be written as

\begin{equation} \label{eqn:psiS-new} \psi _S(z) = -\frac{H_2^3 (2 H_2^3 + 5 H_2 H_4 + 5 H_4^2)}{2 \Delta }. \end{equation}

162

Proof ▶

Using 157 and 59 gives

\begin{align} \psi _S & = -128 \frac{H_3 + H_2}{H_4^2} - 128 \frac{H_2 - H_4}{H_3^2} \\ & = -128 \frac{H_3^2 (H_2 - H_4) + H_4^2 (H_2 - H_4)}{H_3^2 H_4^2} \\ & = -128 \frac{(H_2 + H_4)^2 (2H_2 + H_4) + H_4^2(H_2 + H_4)}{H_3^2 H_4^2} \\ & = -128 \frac{H_2 (2H_2^2 + 5 H_2 H_4 + 5 H_4^2)}{H_3^2 H_4^2} \\ & = -128 \frac{H_2^3 (2H_2^2 + 5 H_2 H_4 + 5 H_4^2)}{H_2^2 H_3^2 H_4^2} \\ & = - \frac{1}{2} \frac{H_2^3 (2H_2^2 + 5 H_2 H_4 + 5 H_4^2)}{\Delta }. \end{align}

Lemma 7.20

There exists a constant \(C_S {\gt} 0\) such that

\begin{equation} \label{eqn:psiS-bound} |\psi _S(z)| \le C_S e^{- \pi \Im z} \end{equation}

169

for all \(z\) with \(\Im z {\gt} 1/2\).

Proof ▶

Proof is similar to that of Lemma 7.5. By Proposition 6.38, 6.39 and 6.40, we can write Fourier expansion of the numerator of \(\psi _S\) as

\begin{equation} H_2(z)^3 (2 H_2(z)^2 + 5 H_2(z) H_4(z) + 5 H_4(z)^2) = \sum _{n \ge 3} a_n e^{\pi i n z} \end{equation}

170

with \(a_3 = 16^3 \cdot 5 = 20480\) and \(a_n = O(n^k)\) for some \(k {\gt} 0\). Now the result follows from Lemma 7.4.

Proposition 7.21

\(b(x)\) is a Schwartz function.

Proof ▶

We have

\begin{align} & \int \limits _{-1}^{i}\psi _T(z)\, e^{\pi i r^2 z}\, dz=\int \limits _{0}^{i+1}\psi _I(z)\, e^{\pi i r^2 (z-1)}\, dz=\notag \\ & \int \limits _{i\infty }^{-1/(i+1)}\psi _I\Big(\frac{-1}{z}\Big)\, e^{\pi i r^2 (-1/z-1)}\, z^{-2}\, dz=\int \limits _{i\infty }^{-1/(i+1)}\psi _S(z)\, z^{-4}\, e^{\pi i r^2 (-1/z-1)}\, dz.\notag \end{align}

Using 169, we can estimate the first summand in the left-hand side of 161

\begin{equation} \left|\int \limits _{-1}^i \psi _T(z)\, e^{\pi i r^2 z}\, dz \right|\leq C_1\, r\, K_1(2\pi r). \end{equation}

171

We combine this inequality with analogous estimates for the other three summands and obtain

\begin{equation} |b(r)|\leq C_2\, r\, K_1(2\pi r)+C_3\, \frac{e^{-\pi (r^2+1)}}{r^2+1}. \end{equation}

172

Here \(C_1\), \(C_2\), and \(C_3\) are some positive constants. Similar estimates hold for all derivatives \(\frac{\mathrm{d}^k}{\mathrm{d}^k r} b(r)\).

Proposition 7.22

\(b(x)\) satisfies 114.

Proof ▶

Here, we repeat the arguments used in the proof of Proposition 7.9. We use identity 139 and change contour integration in \(z\) and Fourier transform in \(x\). Thus we obtain

\begin{align} \mathcal{F}(b)(x)= & \int \limits _{-1}^{i}\psi _T(z)\, z^{-4}\, e^{\pi i \| x\| ^2 (\frac{-1}{z})}\, dz + \int \limits _{1}^{i}\psi _T(z)\, z^{-4}\, e^{\pi i \| x\| ^2 (\frac{-1}{z})}\, dz \notag \\ -& 2\, \int \limits _{0}^{i}\psi _I(z)\, z^{-4}\, e^{\pi i \| x\| ^2 (\frac{-1}{z})}\, dz - 2\, \int \limits _{i}^{i\infty }\psi _S(z)\, z^{-4}\, e^{\pi i \| x\| ^2 (\frac{-1}{z})}\, dz. \notag \end{align}

We make the change of variables \(w=\frac{-1}{z}\) and arrive at

\begin{align} \mathcal{F}(b)(x)= & \int \limits _{1}^{i}\psi _T\Big(\frac{-1}{w}\Big)\, w^{2}\, e^{\pi i \| x\| ^2 w}\, dw + \int \limits _{-1}^{i}\psi _T\Big(\frac{-1}{w}\Big)\, w^{2}\, e^{\pi i \| x\| ^2 w}\, dw \notag \\ -& 2\, \int \limits _{i\infty }^{i}\psi _I\Big(\frac{-1}{w}\Big)\, w^{2}\, e^{\pi i \| x\| ^2 w}\, dw - 2\, \int \limits _{i}^{0}\psi _S\Big(\frac{-1}{w}\Big)\, w^{2}\, e^{\pi i \| x\| ^2 w}\, dw.\notag \end{align}

Now we observe that the definitions 152–154 imply

\begin{align} \psi _T|_{-2}S=& -\psi _T \notag \\ \psi _I|_{-2}S=& \psi _S \notag \\ \psi _S|_{-2}S=& \psi _I. \notag \end{align}

Therefore, we arrive at

\begin{align} \mathcal{F}(b)(x)= & \int \limits _{1}^{i}-\psi _T(z)\, e^{\pi i \| x\| ^2 z}\, dz + \int \limits _{-1}^{i}-\psi _T(z)\, e^{\pi i \| x\| ^2 z}\, dz \notag \\ +& 2\, \int \limits _{i}^{i\infty }\psi _S(z)\, e^{\pi i \| x\| ^2 z}\, dz + 2\, \int \limits _{0}^{i}\psi _I(z)\, e^{\pi i \| x\| ^2 w}\, dw.\notag \end{align}

Now from 161 we see that

\[ \mathcal{F}(b)(x)=-b(x). \]

Now we regard the radial function \(b\) as a function on \(\mathbb {R}_{\geq 0}\). We check that \(b\) has double roots at \(\Lambda _8\)-points.

Lemma 7.23

There exists a constant \(C_I {\gt} 0\) such that

\begin{equation} |\psi _I(z)| \le C_I e^{2 \pi \Im z} \end{equation}

173

for all \(z\) with \(\Im z {\gt} 1/2\).

Proof ▶

By 162, 154, 27, and 29,

\begin{equation} \psi _I(z) = \frac{H_4^3(2 H_4^2 + 5 H_4 H_2 + 5 H_2^2)}{2 \Delta }. \end{equation}

174

The denominator is not a cusp form (i.e. has a nonzero constant term), hence Lemma 7.4 concludes the proof with \(n_0 = 0\).

Corollary 7.24

We have

\begin{align} \psi _I(it) & = O(t^2 e^{\pi /t}) \quad \text{as } t \to 0 \label{eqn:psiI-near-0} \\ \psi _I(it) & = O(e^{2 \pi t}) \quad \text{as } t \to \infty . \label{eqn:psiI-near-infty} \end{align}

Proof ▶

By 154, we have

\begin{equation} \psi _I(it) = (it)^{-2} \psi _S\left(\frac{-1}{it}\right) = -t^{-2} \psi _S\left(\frac{i}{t}\right). \end{equation}

177

and combined with 169 we get 175. 176 follows from Lemma 7.23.

Proposition 7.25

For \(r{\gt}\sqrt{2}\) function \(b(r)\) can be expressed as

\begin{equation} \label{eqn: b double zeroes} b(r)=-4\sin (\pi r^2/2)^2\, \int \limits _{0}^{i\infty }\psi _I(z)\, e^{\pi i r^2 \, z}\, dz. \end{equation}

178

Proof ▶

We denote the right hand side of 178 by \(c(r)\). By Corollary 7.24, the integral in 178 converges for \(r{\gt}\sqrt{2}\). Then we rewrite it in the following way:

\[ c(r)=\int \limits _{-1}^{i\infty -1}\psi _I(z+1)\, e^{\pi i r^2 \, z}\, dz-2\int \limits _{0}^{i\infty }\psi _I(z)\, e^{\pi i r^2 \, z}\, dz+ \int \limits _{1}^{i\infty +1}\psi _I(z-1)\, e^{\pi i r^2 \, z}\, dz. \]

From the Fourier expansion 158 we know that \(\psi _I(z)=e^{-2\pi i z}+O(1)\) as \(\Im (z)\to \infty \). By assumption \(r^2{\gt}2\), hence we can deform the path of integration and write

\begin{align} \label{eqn: inside proof 1} \int \limits _{-1}^{i\infty -1}\psi _I(z+1)\, e^{\pi i r^2 \, z}\, dz=& \int \limits _{-1}^{i}\psi _T(z)\, e^{\pi i r^2 \, z}\, dz+\int \limits _{i}^{i\infty }\psi _T(z)\, e^{\pi i r^2 \, z}\, dz\\ \int \limits _{1}^{i\infty +1}\psi _I(z-1)\, e^{\pi i r^2 \, z}\, dz=& \int \limits _{-1}^{i}\psi _T(z)\, e^{\pi i r^2 \, z}\, dz+\int \limits _{i}^{i\infty }\psi _T(z)\, e^{\pi i r^2 \, z}\, dz. \end{align}

We have

\begin{align} \label{eqn: c1}c(r)=& \int \limits _{-1}^{i}\psi _T(z)\, e^{\pi i r^2 \, z}\, dz+\int \limits _{1}^{i}\psi _T(z)\, e^{\pi i r^2 \, z}\, dz -2\int \limits _{0}^{i}\psi _I(z)\, e^{\pi i r^2 \, z}\, dz\\ & +2\int \limits _{i}^{i\infty }(\psi _T(z)-\psi _I(z))\, e^{\pi i r^2 \, z}\, dz.\nonumber \end{align}

Next, we check that the functions \(\psi _I,\psi _T\), and \(\psi _S\) satisfy the following identity:

\begin{equation} \label{eqn: c2}\psi _T+\psi _S=\psi _I.\end{equation}

182

Indeed, from definitions 152-154 we get

\begin{align} \psi _T+\psi _S=& (h-h|_{-2}ST)|_{-2}T+(h-h|_{-2}ST)|_{-2}S\notag \\ =& h|_{-2}T-h|_{-2}ST^2+h|_{-2}S-h|_{-2}STS.\notag \end{align}

Note that \(ST^2S\) belongs to \(\Gamma _0(2)\). Thus, since \(h\in M^!_{-2}\Gamma _0(2)\) we get

\[ \psi _T+\psi _S=h|_{-2}T-h|_{-2}STS. \]

Now we observe that \(T\) and \(STS(ST)^{-1}\) are also in \(\Gamma _0(2)\). Therefore,

\[ \psi _T+\psi _S=h|_{-2}T-h|_{-2}STS=h|_{-2}-h|ST=\psi _I. \]

Combining 181 and 182 we find

\begin{align} c(r)=& \int \limits _{-1}^{i}\psi _T(z)\, e^{\pi i r^2 \, z}\, dz+\int \limits _{1}^{i}\psi _T(z)\, e^{\pi i r^2 \, z}\, dz -2\int \limits _{0}^{i}\psi _I(z)\, e^{\pi i r^2 \, z}\, dz\notag \\ & -2\int \limits _{i}^{i\infty }\psi _S(z)\, e^{\pi i r^2 \, z}\, dz\notag \\ =& b(r).\notag \end{align}

At the end of this section we find another integral representation of \(b(r)\) for \(r\in \mathbb {R}_{\geq 0}\) and compute special values of \(b\).

Proposition 7.26

For \(r\geq 0\) we have

\begin{equation} \label{eqn:b-another-integral}b(r)=4i\, \sin (\pi r^2/2)^2\, \left(\frac{144}{\pi \, r^2}+\frac{1}{\pi \, (r^2-2)}+\int \limits _0^\infty \, \left(\psi _I(it)-144-e^{2\pi t}\right)\, e^{-\pi r^2 t}\, dt\right).\end{equation}

183

The integral converges absolutely for all \(r\in \mathbb {R}_{\geq 0}\).

Proof ▶

The proof is analogous to the proof of Proposition 7.12. First, suppose that \(r{\gt}\sqrt{2}\). Then by Proposition 7.25

\[ b(r)=4i\, \sin (\pi r^2/2)^2\, \int \limits _{0}^{\infty }\psi _I(it)\, e^{-\pi r^2 t}\, dt. \]

From 158 we obtain

\begin{equation} \label{eqn: psi asymptotic} \psi _I(it)=e^{2\pi t}+144+O(e^{-\pi t})\quad \mbox{as}\; t\to \infty . \end{equation}

184

For \(r{\gt}\sqrt{2}\) we have

\begin{equation} \int \limits _0^\infty \left(e^{2\pi t}+144\right)\, e^{-\pi r^2 t}\, dt =\frac{1}{\pi \, (r^2-2)}+\frac{144}{\pi \, r^2}.\end{equation}

185

Therefore, the identity 183 holds for \(r{\gt}\sqrt{2}\).

On the other hand, from the definition 161 we see that \(b(r)\) is analytic in some neighborhood of \([0,\infty )\). The asymptotic expansion 184 implies that the right hand side of 183 is also analytic in some neighborhood of \([0,\infty )\). Hence, the identity 183 holds on the whole interval \([0,\infty )\). This finishes the proof of the proposition.

We see from 183 that \(b(r)\in i\mathbb {R}\) far all \(r\in \mathbb {R}_\geq {0}\). Another immediate corollary of this proposition is

Proposition 7.27

We have \(b(0) = 0\).

Proof ▶

These identities follow immediately from the previous proposition.