7 Fourier eigenfunctions with double zeroes at lattice points

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

\begin{aligned} F (a) & = a \\ F (b) & = - b \end{aligned}

which double zeroes at all $Λ_{8}$ -vectors of length greater than $\sqrt{2}$ . Recall that each vector of $Λ_{8}$ has length $\sqrt{2 n}$ for some $n \in N_{\geq 0}$ . 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{aligned} ϕ_{- 4} & := \frac{E_{4}^{2}}{Δ} \\ ϕ_{- 2} & := \frac{E_{4} (E_{2} E_{4} - E_{6})}{Δ} \\ ϕ_{0} & := \frac{(E_{2} E_{4} - E_{6})^{2}}{Δ} \end{aligned}

The function $ϕ_{0} (z)$ is not modular; however, it satisfies the following transformation rules:

Lemma 7.2

We have

\begin{aligned} ϕ_{0} (z + 1) & = ϕ_{0} (z) \\ ϕ_{0} (- \frac{1}{z}) & = ϕ_{0} (z) - \frac{12 i}{π} \frac{1}{z} ϕ_{- 2} (z) - \frac{36}{π^{2}} \frac{1}{z^{2}} ϕ_{- 4} (z) . \end{aligned}

Proof ▶

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

\begin{aligned} ϕ_{0} (- \frac{1}{z}) & = \frac{(E_{2} (- 1 / z) E_{4} (- 1 / z) - E_{6} (- 1 / z))^{2}}{Δ (- 1 / z)} \\ = \frac{((z^{2} E_{2} (z) - 6 i z / π) \cdot z^{4} E_{4} (z) - z^{6} E_{6} (z))^{2}}{z^{12} Δ (z)} \\ = \frac{{(E_{2} (z) E_{4} (z) - E_{6} (z) - \frac{6 i}{π z} E_{4} (z))}^{2}}{Δ (z)} \\ = \frac{(E_{2} (z) E_{4} (z) - E_{6} (z))^{2} - \frac{12 i}{π z} (E_{2} (z) E_{4} (z) - E_{6} (z)) E_{4} (z) - \frac{36}{π^{2} z^{2}} E_{4} (z)^{2}}{Δ (z)} \\ = ϕ_{0} (z) - \frac{12 i}{π z} ϕ_{- 2} (z) - \frac{36}{π^{2} z^{2}} ϕ_{- 4} (z) . \end{aligned}

Definition 7.3

✓

For $x \in R^{8}$ we define

\begin{aligned} a (x) := & \int_{- 1}^{i} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} e^{π i ‖ x ‖^{2} z} d z + \int_{1}^{i} ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} e^{π i ‖ x ‖^{2} z} d z \\ - & 2 \int_{0}^{i} ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i ‖ x ‖^{2} z} d z + 2 \int_{i}^{i \infty} ϕ_{0} (z) e^{π i ‖ x ‖^{2} z} d z . \end{aligned}

We observe that the contour integrals in 125 converge absolutely and uniformly for $x \in R^{8}$ . Indeed, $ϕ_{0} (z) = O (e^{2 π i z})$ as $ℑ (z) \to \infty$ . 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

f (z) = \sum_{n \geq n_{0}} c_{f} (n) e^{π i n z}

126

with $c_{f} (n_{0}) \neq 0$ . Assume that $c_{f} (n)$ has a polynomial growth, i.e. $| c_{f} (n) | = O (n^{k})$ for some $k \in N$ . Then there exists a constant $C_{f} > 0$ such that

| \frac{f (z)}{Δ (z)} | \leq C_{f} e^{- π (n_{0} - 2) ℑ z}

127

for all $z$ with $ℑ z > 1 / 2$ .

Proof ▶

By the product formula 12,

\begin{aligned} | \frac{f (z)}{Δ (z)} | & = | \frac{\sum_{n \geq n_{0}} c_{f} (n) e^{π i n z}}{e^{2 π i z} \prod_{n \geq 1} (1 - e^{2 π i n z})^{24}} | \\ = | e^{π i (n_{0} - 2) z} | \cdot \frac{| \sum_{n \geq n_{0}} c_{f} (n) e^{π i (n - n_{0}) z} |}{\prod_{n \geq 1} | 1 - e^{2 π i n z} |^{24}} \\ \leq e^{- π (n_{0} - 2) ℑ z} \cdot \frac{\sum_{n \geq n_{0}} | c_{f} (n) | e^{- π (n - n_{0}) ℑ z}}{\prod_{n \geq 1} (1 - e^{- 2 π n ℑ z})^{24}} \\ \leq e^{- π (n_{0} - 2) ℑ z} \cdot \frac{\sum_{n \geq n_{0}} | c_{f} (n) | e^{- π (n - n_{0}) / 2}}{\prod_{n \geq 1} (1 - e^{- π n})^{24}} \\ = C_{f} \cdot e^{- π (n_{0} - 2) ℑ z} \end{aligned}

where

C_{f} = \frac{\sum_{n \geq n_{0}} | c_{f} (n) | e^{- π (n - n_{0}) / 2}}{\prod_{n \geq 1} (1 - e^{- π n})^{24}} .

133

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

As corollaries, we have the following bound for $ϕ_{0}$ , $ϕ_{- 2}$ , and $ϕ_{- 4}$ .

Corollary 7.5

There exists a constant $C_{0} > 0$ such that

| ϕ_{0} (z) | \leq C_{0} e^{- 2 π ℑ z}

134

for all $z$ with $ℑ z > 1 / 2$ .

Proof ▶

By Ramanujan’s formula, $E_{2} E_{4} - E_{6} = 3 E_{4}^{'} = 720 \sum_{n \geq 1} n σ_{3} (n) e^{2 π i n z}$ and

(E_{2} (z) E_{4} (z) - E_{6} (z))^{2} = 720^{2} e^{4 π i z} + O (e^{5 π i z}) .

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 $C_{- 2} > 0$ such that

| ϕ_{- 2} (z) | \leq C_{- 2}

135

for all $z$ with $ℑ z > 1 / 2$ .

Corollary 7.7

There exists a constant $C_{- 4} > 0$ such that

| ϕ_{- 4} (z) | \leq C_{- 4} e^{2 π ℑ z}

136

for all $z$ with $ℑ z > 1 / 2$ .

Note that we can take the constants $C_{0}$ , $C_{- 2}$ , and $C_{- 4}$ as

\begin{aligned} C_{0} & = 9 \cdot 240^{2} \cdot e^{π} \cdot \frac{E_{4}^{'} (i / 2)^{2}}{Δ (i / 2)} \\ C_{- 2} & = 3 \cdot \frac{E_{4} (i / 2) E_{4}^{'} (i / 2)}{Δ (i / 2)} \\ C_{- 4} & = e^{- π} \cdot \frac{E_{4} (i / 2)^{2}}{Δ (i / 2)} . \end{aligned}

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{aligned} | \int_{- 1}^{i} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} e^{π i r^{2} z} d z | = | \int_{i \infty}^{- 1 / (i + 1)} ϕ_{0} (z) z^{- 4} e^{π i r^{2} (- 1 / z - 1)} d z | \leq \\ C_{1} \int_{1 / 2}^{\infty} e^{- 2 π t} e^{- π r^{2} / t} d t \leq C_{1} \int_{0}^{\infty} e^{- 2 π t} e^{- π r^{2} / t} d t = C_{2} r K_{1} (2 \sqrt{2} π r) \end{aligned}

where $C_{1}$ and $C_{2}$ are some positive constants and $K_{α} (x)$ 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

| \int_{i}^{i \infty} ϕ_{0} (z) e^{π i r^{2} z} d z | \leq C \int_{1}^{\infty} e^{- 2 π t} e^{- π r^{2} t} d t = C_{3} \frac{e^{π (r^{2} + 2)}}{r^{2} + 2} .

137

Therefore, we arrive at

| a (r) | \leq 4 C_{2} r K_{1} (2 \sqrt{2} π r) + 2 C_{3} \frac{e^{- π (r^{2} + 2)}}{r^{2} + 2} .

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 $\frac{d^{k}}{d r^{k}} a (r)$ .

Proposition 7.9

$a (x)$ satisfies 113.

Proof ▶

We recall that the Fourier transform of a Gaussian function is

F (e^{π i ‖ x ‖^{2} z}) (y) = z^{- 4} e^{π i ‖ y ‖^{2} (\frac{- 1}{z})} .

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{aligned} \hat{a} (y) = & \int_{- 1}^{i} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} z^{- 4} e^{π i ‖ y ‖^{2} (\frac{- 1}{z})} d z + \int_{1}^{i} ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} z^{- 4} e^{π i ‖ y ‖^{2} (\frac{- 1}{z})} d z \\ - & 2 \int_{0}^{i} ϕ_{0} (\frac{- 1}{z}) z^{2} z^{- 4} e^{π i ‖ y ‖^{2} (\frac{- 1}{z})} d z + 2 \int_{i}^{i \infty} ϕ_{0} (z) z^{- 4} e^{π i ‖ y ‖^{2} (\frac{- 1}{z})} d z . \end{aligned}

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

\begin{aligned} \hat{a} (y) = & \int_{1}^{i} ϕ_{0} (1 - \frac{1}{w - 1}) (\frac{- 1}{w} + 1)^{2} w^{2} e^{π i ‖ y ‖^{2} w} d w \\ + & \int_{- 1}^{i} ϕ_{0} (1 - \frac{1}{w + 1}) (\frac{- 1}{w} - 1)^{2} w^{2} e^{π i ‖ y ‖^{2} w} d w \\ - & 2 \int_{i \infty}^{i} ϕ_{0} (w) e^{π i ‖ y ‖^{2} w} d w + 2 \int_{i}^{0} ϕ_{0} (\frac{- 1}{w}) w^{2} e^{π i ‖ y ‖^{2} w} d w . \end{aligned}

Since $ϕ_{0}$ is $1$ -periodic we have

\begin{aligned} \hat{a} (y) = & \int_{1}^{i} ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} e^{π i ‖ y ‖^{2} z} d z + \int_{- 1}^{i} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} e^{π i ‖ y ‖^{2} z} d z \\ + & 2 \int_{i}^{i \infty} ϕ_{0} (z) e^{π i ‖ y ‖^{2} z} d z - 2 \int_{0}^{i} ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i ‖ y ‖^{2} z} d z \\ = & a (y) . \end{aligned}

This finishes the proof of the proposition.

Next, we check that $a$ has double zeroes at all $Λ_{8}$ -lattice points of length greater then $\sqrt{2}$ . Using 134, 135, and 136, we can control the behavior of $ϕ_{0}$ near $0$ and $i \infty$ .

Corollary 7.10

We have

\begin{aligned} ϕ_{0} (\frac{i}{t}) & = O (e^{- 2 π / t}) as t \to 0 \\ ϕ_{0} (\frac{i}{t}) & = O (t^{- 2} e^{2 π t}) as t \to \infty . \end{aligned}

Proof ▶

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

| ϕ_{0} (\frac{i}{t}) | \leq | ϕ_{0} (i t) | + \frac{12}{π t} | ϕ_{- 2} (i t) | + \frac{36}{π^{2} t^{2}} | ϕ_{- 4} (i t) | \leq C_{0} e^{- 2 π t} + \frac{12}{π t} \cdot C_{- 2} + \frac{36}{π^{2} t^{2}} \cdot C_{- 4} e^{2 π t} = O (t^{- 2} e^{2 π t}) .

145

Proposition 7.11

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

a (r) = - 4 \sin (π r^{2} / 2)^{2} \int_{0}^{i \infty} ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i r^{2} z} d z .

146

Proof ▶

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

\begin{aligned} d (r) = & \int_{- 1}^{i \infty - 1} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} e^{π i r^{2} z} d z - 2 \int_{0}^{i \infty} ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i r^{2} z} d z \\ + & \int_{1}^{i \infty + 1} ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} e^{π i r^{2} z} d z . \end{aligned}

From 119 we deduce that if $r > \sqrt{2}$ then $ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i r^{2} z} \to 0$ as $ℑ (z) \to \infty$ . Therefore, we can deform the paths of integration and rewrite

\begin{aligned} d (r) = & \int_{- 1}^{i} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} e^{π i r^{2} z} d z + \int_{i}^{i \infty} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} e^{π i r^{2} z} d z \\ - 2 & \int_{0}^{i} ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i r^{2} z} d z - 2 \int_{i}^{i \infty} ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i r^{2} z} d z \\ + & \int_{1}^{i} ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} e^{π i r^{2} z} d z + \int_{i}^{i \infty} ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} e^{π i r^{2} z} d z . \end{aligned}

Now from 119 we find

\begin{aligned} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} - 2 ϕ_{0} (\frac{- 1}{z}) z^{2} + ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} = \\ ϕ_{0} (z + 1) (z + 1)^{2} - 2 ϕ_{0} (z) z^{2} + ϕ_{0} (z - 1) (z - 1)^{2} \\ - \frac{12 i}{π} (ϕ_{- 2} (z + 1) (z + 1) - 2 ϕ_{- 2} (z) z + ϕ_{- 2} (z - 1) (z - 1)) \\ - \frac{36}{π^{2}} (ϕ_{- 4} (z + 1) - 2 ϕ_{- 4} (z) + ϕ_{- 4} (z - 1)) = \\ 2 ϕ_{0} (z) . \end{aligned}

Thus, we obtain

\begin{aligned} d (r) = & \int_{- 1}^{i} ϕ_{0} (\frac{- 1}{z + 1}) (z + 1)^{2} e^{π i r^{2} z} d z - 2 \int_{0}^{i} ϕ_{0} (\frac{- 1}{z}) z^{2} e^{π i r^{2} z} d z \\ + & \int_{1}^{i} ϕ_{0} (\frac{- 1}{z - 1}) (z - 1)^{2} e^{π i r^{2} z} d z + 2 \int_{i}^{i \infty} ϕ_{0} (z) e^{π i r^{2} z} d z = a (r) . \end{aligned}

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{aligned} a (r) = & 4 i \sin (π r^{2} / 2)^{2} (\frac{36}{π^{3} (r^{2} - 2)} - \frac{8640}{π^{3} r^{4}} + \frac{18144}{π^{3} r^{2}} \\ + & \int_{0}^{\infty} (t^{2} ϕ_{0} (\frac{i}{t}) - \frac{36}{π^{2}} e^{2 π t} + \frac{8640}{π} t - \frac{18144}{π^{2}}) e^{- π r^{2} t} d t) . \end{aligned}

The integral converges absolutely for all $r \in R_{\geq 0}$ .

Proof ▶

Suppose that $r > \sqrt{2}$ . Then by Proposition 7.11

a (r) = 4 i \sin (π r^{2} / 2)^{2} \int_{0}^{\infty} ϕ_{0} (i / t) t^{2} e^{- π r^{2} t} d t .

From 119 we obtain

ϕ_{0} (i / t) t^{2} = \frac{36}{π^{2}} e^{2 π t} - \frac{8640}{π} t + \frac{18144}{π^{2}} + O (t^{2} e^{- 2 π t}) as t \to \infty .

149

For $r > \sqrt{2}$ we have

\int_{0}^{\infty} (\frac{36}{π^{2}} e^{2 π t} + \frac{8640}{π} t + \frac{18144}{π^{2}}) e^{- π r^{2} t} d t = \frac{36}{π^{3} (r^{2} - 2)} - \frac{8640}{π^{3} r^{4}} + \frac{18144}{π^{3} r^{2}} .

150

Therefore, the identity 148 holds for $r > \sqrt{2}$ .

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 $i R$ for all $r \in R_{\geq 0}$ .

Proposition 7.13

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

Proof ▶

Now we construct function $b$ . To this end we consider the function

Definition 7.14

h (z) := 128 \frac{H_{3} (z) + H_{4} (z)}{H_{2} (z)^{2}} .

151

It is easy to see that $h \in M_{- 2}^{!} (Γ_{0} (2))$ . Indeed, first we check that $h |_{- 2} γ = h$ for all $γ \in Γ_{0} (2)$ . Since the group $Γ_{0} (2)$ is generated by elements $(\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix})$ and $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ 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 $θ_{10}$ 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

h (z) = q^{- 1} + 16 - 132 q + 640 q^{2} - 2550 q^{3} + O (q^{4}) .

152

Let $I = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ , $T = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ , and $S = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ be elements of $Γ_{1}$ .

Definition 7.15

We define the following three functions

\begin{aligned} ψ_{I} := & h - h |_{- 2} S T \\ ψ_{T} := & ψ_{I} |_{- 2} T \\ ψ_{S} := & ψ_{I} |_{- 2} S . \end{aligned}

Lemma 7.16

More explicitly, we have

\begin{aligned} ψ_{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}} \\ ψ_{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}} \\ ψ_{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}} \end{aligned}

Lemma 7.17

The Fourier expansions of these functions are

\begin{aligned} ψ_{I} (z) = & q^{- 1} + 144 - 5120 q^{1 / 2} + 70524 q - 626688 q^{3 / 2} + 4265600 q^{2} + O (q^{5 / 2}) \\ ψ_{T} (z) = & q^{- 1} + 144 + 5120 q^{1 / 2} + 70524 q + 626688 q^{3 / 2} + 4265600 q^{2} + O (q^{5 / 2}) \\ ψ_{S} (z) = & - 10240 q^{1 / 2} - 1253376 q^{3 / 2} - 48328704 q^{5 / 2} - 1059078144 q^{7 / 2} + O (q^{9 / 2}) . \end{aligned}

Definition 7.18

For $x \in R^{8}$ define

\begin{aligned} b (x) := & \int_{- 1}^{i} ψ_{T} (z) e^{π i ‖ x ‖^{2} z} d z + \int_{1}^{i} ψ_{T} (z) e^{π i ‖ x ‖^{2} z} d z \\ - & 2 \int_{0}^{i} ψ_{I} (z) e^{π i ‖ x ‖^{2} z} d z - 2 \int_{i}^{i \infty} ψ_{S} (z) e^{π i ‖ x ‖^{2} z} d z . \end{aligned}

Now we prove that $b$ is a Schwartz function and satisfies condition 114.

Lemma 7.19

$ψ_{S} (z)$ can be written as

ψ_{S} (z) = - \frac{H_{2}^{3} (2 H_{2}^{3} + 5 H_{2} H_{4} + 5 H_{4}^{2})}{2 Δ} .

162

Proof ▶

Using 157 and 59 gives

\begin{aligned} ψ_{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} (2 H_{2} + H_{4}) + H_{4}^{2} (H_{2} + H_{4})}{H_{3}^{2} H_{4}^{2}} \\ = - 128 \frac{H_{2} (2 H_{2}^{2} + 5 H_{2} H_{4} + 5 H_{4}^{2})}{H_{3}^{2} H_{4}^{2}} \\ = - 128 \frac{H_{2}^{3} (2 H_{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} (2 H_{2}^{2} + 5 H_{2} H_{4} + 5 H_{4}^{2})}{Δ} . \end{aligned}

Lemma 7.20

There exists a constant $C_{S} > 0$ such that

| ψ_{S} (z) | \leq C_{S} e^{- π ℑ z}

169

for all $z$ with $ℑ z > 1 / 2$ .

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 $ψ_{S}$ as

H_{2} (z)^{3} (2 H_{2} (z)^{2} + 5 H_{2} (z) H_{4} (z) + 5 H_{4} (z)^{2}) = \sum_{n \geq 3} a_{n} e^{π i n z}

170

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

Proposition 7.21

$b (x)$ is a Schwartz function.

Proof ▶

We have

\begin{aligned} \int_{- 1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z = \int_{0}^{i + 1} ψ_{I} (z) e^{π i r^{2} (z - 1)} d z = \\ \int_{i \infty}^{- 1 / (i + 1)} ψ_{I} (\frac{- 1}{z}) e^{π i r^{2} (- 1 / z - 1)} z^{- 2} d z = \int_{i \infty}^{- 1 / (i + 1)} ψ_{S} (z) z^{- 4} e^{π i r^{2} (- 1 / z - 1)} d z . \end{aligned}

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

| \int_{- 1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z | \leq C_{1} r K_{1} (2 π r) .

171

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

| b (r) | \leq C_{2} r K_{1} (2 π r) + C_{3} \frac{e^{- π (r^{2} + 1)}}{r^{2} + 1} .

172

Here $C_{1}$ , $C_{2}$ , and $C_{3}$ are some positive constants. Similar estimates hold for all derivatives $\frac{d^{k}}{d^{k} r} b (r)$ .

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{aligned} F (b) (x) = & \int_{- 1}^{i} ψ_{T} (z) z^{- 4} e^{π i ‖ x ‖^{2} (\frac{- 1}{z})} d z + \int_{1}^{i} ψ_{T} (z) z^{- 4} e^{π i ‖ x ‖^{2} (\frac{- 1}{z})} d z \\ - & 2 \int_{0}^{i} ψ_{I} (z) z^{- 4} e^{π i ‖ x ‖^{2} (\frac{- 1}{z})} d z - 2 \int_{i}^{i \infty} ψ_{S} (z) z^{- 4} e^{π i ‖ x ‖^{2} (\frac{- 1}{z})} d z . \end{aligned}

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

\begin{aligned} F (b) (x) = & \int_{1}^{i} ψ_{T} (\frac{- 1}{w}) w^{2} e^{π i ‖ x ‖^{2} w} d w + \int_{- 1}^{i} ψ_{T} (\frac{- 1}{w}) w^{2} e^{π i ‖ x ‖^{2} w} d w \\ - & 2 \int_{i \infty}^{i} ψ_{I} (\frac{- 1}{w}) w^{2} e^{π i ‖ x ‖^{2} w} d w - 2 \int_{i}^{0} ψ_{S} (\frac{- 1}{w}) w^{2} e^{π i ‖ x ‖^{2} w} d w . \end{aligned}

Now we observe that the definitions 152–154 imply

\begin{aligned} ψ_{T} |_{- 2} S = & - ψ_{T} \\ ψ_{I} |_{- 2} S = & ψ_{S} \\ ψ_{S} |_{- 2} S = & ψ_{I} . \end{aligned}

Therefore, we arrive at

\begin{aligned} F (b) (x) = & \int_{1}^{i} - ψ_{T} (z) e^{π i ‖ x ‖^{2} z} d z + \int_{- 1}^{i} - ψ_{T} (z) e^{π i ‖ x ‖^{2} z} d z \\ + & 2 \int_{i}^{i \infty} ψ_{S} (z) e^{π i ‖ x ‖^{2} z} d z + 2 \int_{0}^{i} ψ_{I} (z) e^{π i ‖ x ‖^{2} w} d w . \end{aligned}

Now from 161 we see that

F (b) (x) = - b (x) .

Now we regard the radial function $b$ as a function on $R_{\geq 0}$ . We check that $b$ has double roots at $Λ_{8}$ -points.

Lemma 7.23

There exists a constant $C_{I} > 0$ such that

| ψ_{I} (z) | \leq C_{I} e^{2 π ℑ z}

173

for all $z$ with $ℑ z > 1 / 2$ .

Proof ▶

By 162, 154, 27, and 29,

ψ_{I} (z) = \frac{H_{4}^{3} (2 H_{4}^{2} + 5 H_{4} H_{2} + 5 H_{2}^{2})}{2 Δ} .

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{aligned} ψ_{I} (i t) & = O (t^{2} e^{π / t}) as t \to 0 \\ ψ_{I} (i t) & = O (e^{2 π t}) as t \to \infty . \end{aligned}

Proof ▶

By 154, we have

ψ_{I} (i t) = (i t)^{- 2} ψ_{S} (\frac{- 1}{i t}) = - t^{- 2} ψ_{S} (\frac{i}{t}) .

177

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

Proposition 7.25

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

b (r) = - 4 \sin (π r^{2} / 2)^{2} \int_{0}^{i \infty} ψ_{I} (z) e^{π i r^{2} z} d z .

178

Proof ▶

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

c (r) = \int_{- 1}^{i \infty - 1} ψ_{I} (z + 1) e^{π i r^{2} z} d z - 2 \int_{0}^{i \infty} ψ_{I} (z) e^{π i r^{2} z} d z + \int_{1}^{i \infty + 1} ψ_{I} (z - 1) e^{π i r^{2} z} d z .

From the Fourier expansion 158 we know that $ψ_{I} (z) = e^{- 2 π i z} + O (1)$ as $ℑ (z) \to \infty$ . By assumption $r^{2} > 2$ , hence we can deform the path of integration and write

\begin{aligned} \int_{- 1}^{i \infty - 1} ψ_{I} (z + 1) e^{π i r^{2} z} d z = & \int_{- 1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z + \int_{i}^{i \infty} ψ_{T} (z) e^{π i r^{2} z} d z \\ \int_{1}^{i \infty + 1} ψ_{I} (z - 1) e^{π i r^{2} z} d z = & \int_{- 1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z + \int_{i}^{i \infty} ψ_{T} (z) e^{π i r^{2} z} d z . \end{aligned}

We have

\begin{aligned} c (r) = & \int_{- 1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z + \int_{1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z - 2 \int_{0}^{i} ψ_{I} (z) e^{π i r^{2} z} d z \\ + 2 \int_{i}^{i \infty} (ψ_{T} (z) - ψ_{I} (z)) e^{π i r^{2} z} d z . \end{aligned}

Next, we check that the functions $ψ_{I}, ψ_{T}$ , and $ψ_{S}$ satisfy the following identity:

ψ_{T} + ψ_{S} = ψ_{I} .

182

Indeed, from definitions 152-154 we get

\begin{aligned} ψ_{T} + ψ_{S} = & (h - h |_{- 2} S T) |_{- 2} T + (h - h |_{- 2} S T) |_{- 2} S \\ = & h |_{- 2} T - h |_{- 2} S T^{2} + h |_{- 2} S - h |_{- 2} S T S . \end{aligned}

Note that $S T^{2} S$ belongs to $Γ_{0} (2)$ . Thus, since $h \in M_{- 2}^{!} Γ_{0} (2)$ we get

ψ_{T} + ψ_{S} = h |_{- 2} T - h |_{- 2} S T S .

Now we observe that $T$ and $S T S (S T)^{- 1}$ are also in $Γ_{0} (2)$ . Therefore,

ψ_{T} + ψ_{S} = h |_{- 2} T - h |_{- 2} S T S = h |_{- 2} - h | S T = ψ_{I} .

Combining 181 and 182 we find

\begin{aligned} c (r) = & \int_{- 1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z + \int_{1}^{i} ψ_{T} (z) e^{π i r^{2} z} d z - 2 \int_{0}^{i} ψ_{I} (z) e^{π i r^{2} z} d z \\ - 2 \int_{i}^{i \infty} ψ_{S} (z) e^{π i r^{2} z} d z \\ = & b (r) . \end{aligned}

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

Proposition 7.26

For $r \geq 0$ we have

b (r) = 4 i \sin (π r^{2} / 2)^{2} (\frac{144}{π r^{2}} + \frac{1}{π (r^{2} - 2)} + \int_{0}^{\infty} (ψ_{I} (i t) - 144 - e^{2 π t}) e^{- π r^{2} t} d t) .

183

The integral converges absolutely for all $r \in R_{\geq 0}$ .

Proof ▶

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

b (r) = 4 i \sin (π r^{2} / 2)^{2} \int_{0}^{\infty} ψ_{I} (i t) e^{- π r^{2} t} d t .

From 158 we obtain

ψ_{I} (i t) = e^{2 π t} + 144 + O (e^{- π t}) as t \to \infty .

184

For $r > \sqrt{2}$ we have

\int_{0}^{\infty} (e^{2 π t} + 144) e^{- π r^{2} t} d t = \frac{1}{π (r^{2} - 2)} + \frac{144}{π r^{2}} .

185

Therefore, the identity 183 holds for $r > \sqrt{2}$ .

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 $b (r) \in i R$ far all $r \in R_{\geq} 0$ . Another immediate corollary of this proposition is

Proposition 7.27

We have $b (0) = 0$ .

Proof ▶