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 ▶

115 easily follows from periodicity of Eisenstein series and \(\Delta (z)\). For 116,

\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 122 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 110. 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}

123

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}

124

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

red Note that the assumption on the polynomial growth holds when \(f\) is a holomorphic modular form, where the proof can be found in [ 11 , p. 94 ] for the case of level 1 modular forms. But we just add this for simplicity, and we can prove it for “specific” \(f\) such as Eisenstein series, theta functions, and their combinations.

Proof ▶

By the product formula ??,

\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}

130

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

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}

131

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}

132

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}

132

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}

133

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 122. By 131, 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 122. 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}

134

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}

135

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 110.

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}

136

Next, we exchange the contour integration with respect to \(z\) variable and Fourier transform with respect to \(x\) variable in 122. 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 131, 132, and 133, 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 131 with \(z = i/t\). For the second estimate, by 116, 132, and 133, 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}

142

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}

143

Proof ▶

We denote the right hand side of 143 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 116 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 116 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 ??–116 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}

146

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}

147

Therefore, the identity 145 holds for \(r{\gt}\sqrt{2}\).

On the other hand, from the definition 122 we see that \(a(r)\) is analytic in some neighborhood of \([0,\infty )\). The asymptotic expansion 146 implies that the right hand side of 145 is also analytic in some neighborhood of \([0,\infty )\). Hence, the identity 145 holds on the whole interval \([0,\infty )\). This finishes the proof of the proposition.

From the identity 145 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}

148

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 28–30 and 148. 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}

149

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 111.

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}

159

Proof ▶

Using 154 and 60 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}

166

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

Proof ▶

Proof is similar to that of Lemma 7.5. By Proposition 6.37, 6.38 and 6.39, 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}

167

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 166, we can estimate the first summand in the left-hand side of 158

\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}

168

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}

169

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 111.

Proof ▶

Here, we repeat the arguments used in the proof of Proposition 7.9. We use identity 136 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 149–151 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 158 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}

170

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

Proof ▶

By 159, 151, 28, and 30,

\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}

171

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 151, 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}

174

and combined with 166 we get 172. 173 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}

175

Proof ▶

We denote the right hand side of 175 by \(c(r)\). By Corollary 7.24, the integral in 175 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 155 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}

179

Indeed, from definitions 149-151 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 178 and 179 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}

180

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 155 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}

181

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}

182

Therefore, the identity 180 holds for \(r{\gt}\sqrt{2}\).

On the other hand, from the definition 158 we see that \(b(r)\) is analytic in some neighborhood of \([0,\infty )\). The asymptotic expansion 181 implies that the right hand side of 180 is also analytic in some neighborhood of \([0,\infty )\). Hence, the identity 180 holds on the whole interval \([0,\infty )\). This finishes the proof of the proposition.

We see from 180 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 \(b(0) = 0\).