Cube sums of the forms 3p and $3p^2$ II

Abstract

Let $p\equiv 2,5\ \mathrm {mod}\ 9$ be a prime. We prove that both 3p and $3p^2$ are the sums of two rational cubes. We also establish some explicit Gross–Zagier formulae and investigate the 3-part of the full BSD conjecture of the related elliptic curves.

Appendix A: Waldspurger’s local period integral

In this appendix we deal with the explicit local computations that are needed to establish the explicit Gross–Zagier formulae. We shall compute the 3-adic period integral for the 3-adic local newform following [13]. Let ${{\mathbb {F}}}$ be a general p-adic local field with ${{\mathcal {O}}}_{{\mathbb {F}}}$ the ring of integral elements and ${{\mathfrak {p}}}_{{\mathbb {F}}}$ the maximal ideal of ${{\mathcal {O}}}_{{\mathbb {F}}}$. For $\nu$ a multiplicative character on ${{\mathbb {F}}}^\times$, let $c(\nu )$ be the smallest integer such that $\nu$ is trivial on $1+{{\mathfrak {p}}}_{{\mathbb {F}}}^{c(\nu )}$. Similarly, for an additive character $\psi$ on ${{\mathbb {F}}}$, $c(\psi )$ is the smallest integer such that $\psi$ is trivial on ${{\mathfrak {p}}}_{{\mathbb {F}}}^{c(\psi )}$.

Recall $\pi$ is the automorphic representation of ${\mathrm {GL}}_2({{\mathbb {Q}}})$ corresponding to $E_9$ and $\pi _3$ the 3-adic part of $\pi$. Then the level of $\pi _3$ is $3^5$ and we set $c(\pi _3)=5$. Let $p\equiv 2,5\ \mathrm {mod}\ 9$ be an odd prime and let $\chi :{\mathrm {Gal}}({\bar{K}}/K)\rightarrow {{\mathcal {O}}}_K^\times$ be the character given by $\chi (\sigma )=\chi _{3p}(\sigma )=(\root 3 \of {3p})^{\sigma -1}$, resp. $\chi (\sigma )=\chi _{3p^2}(\sigma )=(\root 3 \of {3p^2})^{\sigma -1}$. We also view $\chi$ as a Hecke character on ${{\mathbb {A}}}_K^\times$ by the Artin map and let $\chi _3$ be the 3-adic local component of $\chi$. Assume that $f_3$ is the standard newform of $\pi _3$. We shall compute the following normalized period integral

$$\begin{aligned} \beta ^0_3(f_3, f_3) =\int \limits _{t\in {{\mathbb {Q}}}_3^\times \backslash K_3^\times }\frac{ (\pi (t)f_3,f_3)}{(f_3,f_3)}\chi _3(t)dt \end{aligned}$$

(A.1)

which appears in the proof of the explicit Gross-Zagier formulae.

1.1 A.1. Local arithmetic information

Let $\Theta :K^\times \backslash {{\mathbb {A}}}_K^\times \rightarrow {{\mathbb {C}}}^\times$ be the unitary Hecke character associated to the base-changed CM elliptic curve ${E_9}_{/K}$ coming from the CM theory. Then $\Theta$ has conductor $9{{\mathcal {O}}}_K$. We denote by $\Theta _3$ the 3-adic local component of $\Theta$. Then $\pi _3$ is the local representation of ${\mathrm {GL}}_2({{\mathbb {Q}}}_3)$ corresponding to $\Theta _3$ via the Weil-Deligne construction. Note

$$\begin{aligned}{{\mathcal {O}}}_{K,3}^\times /(1+9{{\mathcal {O}}}_{K,3})\simeq & {} \langle \pm 1\rangle ^{{{\mathbb {Z}}}/2{{\mathbb {Z}}}} \times \langle 1+\sqrt{-3}\rangle ^{{{\mathbb {Z}}}/3{{\mathbb {Z}}}}\times \langle 1-\sqrt{-3}\rangle ^{{{\mathbb {Z}}}/3{{\mathbb {Z}}}}\\&\times \langle 1+3\sqrt{-3}\rangle ^{{{\mathbb {Z}}}/3{{\mathbb {Z}}}}.\end{aligned}$$

Lemma A.1

We have $c(\Theta _3)=4$, and $\Theta _3$ is given explicitly by

$$\begin{aligned}\Theta _3(-1)=-1,\quad \Theta _3(1+\sqrt{-3})=\frac{-1-\sqrt{-3}}{2},\ \ \Theta _3(\sqrt{-3})=i,\\\Theta _3(1-\sqrt{-3})=\frac{-1+\sqrt{-3}}{2},\quad \Theta _3(1+3\sqrt{-3})=\frac{-1+\sqrt{-3}}{2}.\\ \end{aligned}$$

Proof

This is [13, Lemma 4.2]. $\square$

The local character $\chi _3$ has conductor ${{\mathbb {Z}}}_3^\times (1+9{{\mathcal {O}}}_{K,3})$, and hence it is in fact a character of the quotient group $K_3^\times /{{\mathbb {Q}}}_3^\times (1+9{{\mathcal {O}}}_{K,3})$. Note that

$$\begin{aligned}K_3^\times /{{\mathbb {Q}}}_3^\times (1+9{{\mathcal {O}}}_{K,3})\simeq \langle \sqrt{-3}\rangle ^{{{\mathbb {Z}}}}\times \langle 1+\sqrt{-3}\rangle ^{{{\mathbb {Z}}}/3{{\mathbb {Z}}}}\times \langle 1+3\sqrt{-3}\rangle ^{{{\mathbb {Z}}}/3{{\mathbb {Z}}}}.\end{aligned}$$

Lemma A.2

We have $c(\chi _3)=4$ and $\chi _3$ is given explicitly by the following tables:

• 1. if $\chi =\chi _{3p}$, then

  $$\begin{aligned} \begin{array}{cccccc} \hline p\ \mathrm {mod}\ 9&{}\qquad \chi _3(1+\sqrt{-3})&{}\qquad \chi _3(1+3\sqrt{-3})&{}\qquad \chi _3(\sqrt{-3})\\ \hline 2&{}\omega ^2&{}\omega &{}1\\ \hline 5&{}\omega &{}\omega &{}1\\ \hline \end{array} \end{aligned}$$

• 2. if $\chi =\chi _{3p^2}$, then

  $$\begin{aligned} \begin{array}{cccccc} \hline p\ \mathrm {mod}\ 9&{}\qquad \chi _3(1+\sqrt{-3})&{}\qquad \chi _3(1+3\sqrt{-3})&{}\qquad \chi _3(\sqrt{-3})\\ \hline 2&{}\omega &{}\omega &{}1\\ \hline 5&{}\omega ^2&{}\omega &{}1\\ \hline \end{array} \end{aligned}$$

Proof

The proof is similar to [13, Lemma 4.3]. Here we just compute one example for the case $\chi =\chi _{3p}$ and $p=2$. Since all the elements in $1+9{{\mathcal {O}}}_{K,3}$ are cubes in $K_3$, for any $t\in K_3^\times$,

$$\begin{aligned}\chi _3(t)=\left( \root 3 \of {3p}\right) ^{\sigma _t-1}=\left( \frac{t,3p}{K_3;3}\right) =\left( \frac{t,6}{K_3;3}\right) ,\end{aligned}$$

where $\sigma _t$ is the image of t under the the Artin map, and $\left( \frac{\cdot ,\cdot }{K_3;3}\right)$ denotes the 3-rd Hilbert symbol over $K_3^\times$. Using the local and global principle,

$$\begin{aligned} \chi _3(1+\sqrt{-3})= & {} \left( \frac{1+\sqrt{-3},3}{K_3;3}\right) \left( \frac{1+\sqrt{-3},2}{K_3;3}\right) \\= & {} \left( \frac{1+\sqrt{-3},3}{K_2;3}\right) ^{-1}\left( \frac{1+\sqrt{-3},2}{K_2;3}\right) ^{-1}=\omega ^2. \end{aligned}$$

$\square$

By Lemma A.1 and A.2, we have the following corollary.

Corollary A.3

If $p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9$, and $\chi =\chi _{3p},\text { resp. }\chi _{3p^2}$, then the local character $\Theta _3\overline{\chi }_3$ is given explicitly by

$$\begin{aligned}&\Theta _3\overline{\chi }_3(-1)=-1,\quad \Theta _3\overline{\chi }_3(1+\sqrt{-3})=1,\\&\Theta _3\overline{\chi }_3(1-\sqrt{-3})=1,\quad \Theta _3\overline{\chi }_3(1+3\sqrt{-3})=1,\quad \Theta _3\overline{\chi }_3(\sqrt{-3})=i.\end{aligned}$$

If $p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9$, and $\chi =\chi _{3p^2},\text { resp. }\chi _{3p}$, the local character $\Theta _3\overline{\chi }_3$ is given explicitly by

$$\begin{aligned}&\Theta _3\overline{\chi }_3(-1)=-1,\quad \Theta _3\overline{\chi }_3(1+\sqrt{-3})=\omega ,\\&\Theta _3\overline{\chi }_3(1-\sqrt{-3})=\omega ^2,\quad \Theta _3\overline{\chi }_3(1+3\sqrt{-3})=1,\quad \Theta _3\overline{\chi }_3(\sqrt{-3})=i.\end{aligned}$$

1.2 A.2. Local period integrals

In [11] and [10], a new kind of test vectors (the so-called minimal vectors) for Waldspurger’s local period integral of supercuspidal representations is investigated in detail. The minimal vectors arise naturally from the compact-induction theory. Their matrix coefficients and Whittaker functionals have better properties than the new forms. For example, their matrix coefficients are almost multiplicative on the support. In particular, the local period integrals for minimal vectors can be evaluated explicitly in many cases. Using the explicit Kirillov model, the newform can be expressed as a sum of minimal vectors. The main idea in [13] is to evaluate the local period integrals of new forms by the local period integrals of minimal vectors.

For the convenience of the readers to track the results we used in [13], we will use the notation there and give the setup in our case. Let ${{\mathbb {F}}}={{\mathbb {Q}}}_3$ be the base field, $\varpi =3$ the uniformizer, ${{\mathfrak {p}}}=(3)$ the maximal ideal and $q=3$ the cardinality of the residue field of ${{\mathbb {F}}}$. Then ${{\mathbb {E}}}=K_3={{\mathbb {Q}}}_3(\sqrt{-3})$ is the quadratic extension of ${{\mathbb {F}}}$ with discriminant $D=-3$. Besides, there is another quadratic extension ${{\mathbb {L}}}$ of ${{\mathbb {F}}}$ over which the supercuspidal representation $\pi _3$ is parametrized by the character $\theta _3$ via the compact-induction as in [13, Section 2.2]. In our case, ${{\mathbb {L}}}$ is also $K_3$, and we are in the situation that ${{\mathbb {E}}}={{\mathbb {L}}}$ both ramified over ${{\mathbb {F}}}$.

Let $(\cdot ,\cdot )$ be the nondegenerate invariant pairing on $\pi _3$. For two vectors $\varphi , \varphi '\in \pi _3$, we follow the notation in [13] to denote the Waldspurger’s local period integral as

$$\begin{aligned} {\left\{ \varphi ,\varphi ' \right\} }:=\int \limits _{t\in {{\mathbb {Q}}}_3^\times \backslash K_3^\times }(\pi (t)\varphi ,\varphi ')\chi _3(t)dt, \end{aligned}$$

(A.2)

where $K_3$ is embedded in $\mathrm {M}_2({{\mathbb {Q}}}_3)$ as in Section 2.2. A vector $\varphi \in \pi _3$ is called a test vector for $(\pi _3,\chi _3)$ if the local period integral ${\left\{ \varphi ,\varphi \right\} }\ne 0$.

Proposition A.4

• (1) Suppose ${\left\{ \varphi ,\varphi \right\} }=0$, then ${\left\{ \varphi ,\varphi ' \right\} }={\left\{ \varphi ,\varphi ' \right\} }=0$ for any $\varphi '$.

• (2) Suppose $|{\left\{ \varphi ,\varphi \right\} }|=|{\left\{ \varphi ',\varphi ' \right\} }|$, then $|{\left\{ \varphi ',\varphi \right\} }|= |{\left\{ \varphi ,\varphi ' \right\} }|$.

• (3) Let $\varphi =\sum _{i\in I}\varphi _i$ with I a finite index set. If ${\left\{ \varphi _i,\varphi _i \right\} }\ne 0$ for only one vector say $\varphi _j$, then

  $$\begin{aligned}{\left\{ \varphi ,\varphi \right\} }={\left\{ \varphi _j,\varphi _j \right\} }.\end{aligned}$$

Proof

For any nontrivial functional ${\mathcal {F}}\in {\mathrm {Hom}}_{K_3^\times }(\pi _3\otimes \chi _3,{{\mathbb {C}}})$, we have

$$\begin{aligned} {\left\{ \varphi _1,\varphi _2 \right\} }=C {\mathcal {F}}(\varphi _1)\overline{{\mathcal {F}}(\varphi _2)} \end{aligned}$$

for some non-zero constant C independent of the test vectors, as $\dim {\mathrm {Hom}}_{K_3^\times }(\pi _3\otimes \chi _3,{{\mathbb {C}}})=1$ by Proposition 4.1 and the theorem of Tunnell–Saito ([27, 21]). Then

$$\begin{aligned}&|{\left\{ \varphi ,\varphi \right\} }|=|C{\mathcal {F}}(\varphi )^2|, \\&|{\left\{ \varphi ',\varphi ' \right\} }|=|C{\mathcal {F}}(\varphi ')^2|, \\&|{\left\{ \varphi ,\varphi ' \right\} }|=|C{\mathcal {F}}(\varphi )\overline{{\mathcal {F}}(\varphi ')}|. \end{aligned}$$

Now the results are clear. $\square$

Recall $\theta _3$ is the 3-adic character which parametrizes the supercuspidal representation $\pi _3$ via the compact-induction construction. The test vector issue for Waldspurger’s local period integral is closely related to $c(\theta _3\overline{\chi }_3)$ or $c(\theta _3\chi _3)$. We can work out these by using Lemma A.1, A.2 and Corollary A.3, and the relation between $\theta _3$ and $\Theta _3$ in [13, Theorem 2.10].

Let $\psi$ be the additive character such that $\psi (x)=e^{2\pi i \iota (x)}$ where $\iota :{{\mathbb {Q}}}_3\rightarrow {{\mathbb {Q}}}_3/{{\mathbb {Z}}}_3 \subset {{\mathbb {Q}}}/{{\mathbb {Z}}}$ is the map given by $x\mapsto -x\ \mathrm {mod}\ {{\mathbb {Z}}}_3$. Recall that by [13, Lemma 2.1], for a multiplicative character $\nu$ over ${{\mathbb {Q}}}_3$ with $c(\nu )\ge 2$, there exists $\alpha _\nu \in {{\mathbb {Q}}}_3^\times$ with $v(\alpha _\nu )=-c(\nu )$ such that

$$\begin{aligned} \nu (1+u)=\psi (\alpha _\nu u) \end{aligned}$$

(A.3)

for any $u\in {{\mathfrak {p}}}^{\lceil c(\nu )/2\rceil }$, where $\lceil \cdot \rceil$ means the smallest integer not less than the given number. Moreover, $\alpha _\nu$ is uniquely determined $\ \mathrm {mod}\ {{\mathfrak {p}}}^{-\lceil c(\nu )/2\rceil }$. Also recall that for any additive character $\xi$ of ${{\mathbb {Q}}}_3$, the Langlands $\lambda$-function of the extension $K_3/{{\mathbb {Q}}}_3$ is $\lambda _{K_3/{{\mathbb {Q}}}_3}(\xi ):=W({\mathrm {Ind}}_{G_{{{\mathbb {Q}}}_3}}^{G_{K_3}}(1_{K_3}), \xi )$ where $1_{K_3}$ is trivial representation of the absolute Galois group $G_{K_3}$ and W means the local $\varepsilon$-factor, see [17]. Now we prove the following key lemma.

Lemma A.5

If $p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9$, and $\chi =\chi _{3p},\text { resp. }\chi _{3p^2}$, we have $\theta _3\overline{\chi }_3=1$. If $p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9$, and $\chi =\chi _{3p^2},\text { resp. }\chi _{3p}$, we have $c(\theta _3\overline{\chi }_3)=2$ and $\alpha _{\theta _3\overline{\chi }_3}=\frac{1}{3\sqrt{-3}}$. Moreover, in all cases, $c(\theta _3\overline{\chi }_3)\le c(\theta _3\chi _3)$.

Proof

Let $\psi _{K_3}(x)=\psi \circ \mathrm {Tr}_{K_3/{{\mathbb {Q}}}_3}(x)$, be the additive character of $K_3$. By the definition of $\psi _{K_3}$ and Lemma A.1, we know that $\alpha _{\Theta _3}=\frac{1}{9\sqrt{-3}}$. Now let $\eta$ be the quadratic character associated to the quadratic field extension $K_3/{{\mathbb {Q}}}_3$. Then by [1, Proposition 34.3],

$$\begin{aligned} \lambda _{K_3/{{\mathbb {Q}}}_3}(\psi ')=\tau (\eta ,\psi ')/\sqrt{3}=-i, \end{aligned}$$

here $\tau (\eta ,\psi ')$ is the Gauss sum and $\psi '(x)=\psi (\frac{x}{3})$ is the additive character of level one. By [17, Lemma 5.1],

$$\begin{aligned} \lambda _{K_3/{{\mathbb {Q}}}_3}(\psi )=\eta (3)\lambda _{K_3/{{\mathbb {Q}}}_3}(\psi ')=-i. \end{aligned}$$

Then define $\Delta _{\theta _3}$ to be the unique level one character of $K_3$ such that $\Delta _{\theta _3}|_{{{\mathbb {Z}}}_3^\times }=\eta$ and

$$\begin{aligned}\Delta _{\theta _3}(\sqrt{-3})=\eta ((\sqrt{-3})^3\alpha _{\Theta _3})\lambda _{K_3/{{\mathbb {Q}}}_3}(\psi )^{3}=-i.\end{aligned}$$

Then by [13, Theorem 2.10], $\theta _3=\Theta _3\Delta _{\theta _3}$. By Corollary A.3 we can easily check that:

• (1) If $p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9$, and $\chi =\chi _{3p},\text { resp. }\chi _{3p^2}$, $\theta _3\overline{\chi }_3$ is the trivial character.

• (2) If $p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9$, and $\chi =\chi _{3p^2},\text { resp. }\chi _{3p}$, $\theta _3\overline{\chi }_3$ is of level 2 and by definition we can choose $\alpha _{\theta _3\overline{\chi }_3}=\frac{1}{3\sqrt{-3}}$.

Since $\chi _3$ is a cubic character, $\theta _3\chi _3=\theta _3\overline{\chi }_3^2$. Since $c(\chi _3)=c(\overline{\chi }_3)=4$, $c(\theta _3\chi _3)=4$ and the last assertion follows. $\square$

Recall $c(\theta _3)=c(\Theta _3)=c(\chi _3)=4$ and $c(\pi _3)=5$, let $n=(c(\pi _3)-1)/2=2$. For the supercuspidal representation $\pi _3$ of ${\mathrm {GL}}_2({{\mathbb {Q}}}_3)$, the Kirillov model ${\mathscr {K}}(\pi _3,\psi )$ is the unique realization of $\pi _3$ on the Schwartz function space ${{\mathcal {S}}}({{\mathbb {Q}}}_3^\times )$ such that

$$\begin{aligned} \pi _3\left( \begin{pmatrix} a &{} b \\ 0 &{} 1 \end{pmatrix} \right) \varphi (y)=\psi ( by)\varphi (ay),\quad \varphi \in {{\mathcal {S}}}({{\mathbb {Q}}}_3^\times ). \end{aligned}$$

(A.4)

By [13, Lemma 2.11], we have the minimal vector $\varphi _0={\mathrm {Char}}(\varpi ^{-2}U_{{\mathbb {F}}}(1))$ in the Kirillov model where $U_{{{\mathbb {F}}}}(1)=1+{{\mathfrak {p}}}$. For more details on minimal vectors we refer to [13, 2.5]. Recall K is embedded into $\mathrm {M}_2({{\mathbb {Q}}})$ as in Section 2.2 which linearly extends the following map:

$$\begin{aligned} \sqrt{-3}\mapsto \begin{pmatrix} a &{} 3^{-2}b \\ 3^3c &{} -a \end{pmatrix} :={\left\{ \begin{array}{ll} \begin{pmatrix} 3 &{} -2p/9 \\ 54/p &{} -3 \end{pmatrix} , &{}\text { if }K \text { is embedded under }\rho _1;\\ \begin{pmatrix} 9 &{} -2p/9 \\ 374/p &{} -9 \end{pmatrix} , &{}\text { if }K \text { is embedded under }\rho _2;\\ \begin{pmatrix} 0 &{} -p/18 \\ 54/p &{} 0 \end{pmatrix} ,&\text { if } K \text {is embedded under } \rho _3; \end{array}\right. } \end{aligned}$$

(A.5)

where $b, c\in {{\mathbb {Q}}}\cap {{\mathbb {Z}}}_3^\times$.

Proposition A.6

Suppose ${\mathrm {Vol}}({{\mathbb {Z}}}_3^\times \backslash {{\mathcal {O}}}_{K,3}^\times )=1$ so that ${\mathrm {Vol}}({{\mathbb {Q}}}_3^\times \backslash K_3^\times )=2$. For $f_3$ being the newform corresponding to $\pi _3$, K being embedded in $\mathrm {M}_2({{\mathbb {Q}}})$ as in (A.5), we have

$$\begin{aligned} \beta ^0_3(f_3,f_3)={\left\{ \begin{array}{ll} 1, &{}\text { if }p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9, \chi =\chi _{3p},\text { resp. }\chi _{3p^2}\\ &{}\qquad \text { and }K \text { is embedded under }\rho _2;\\ 1, &{}\text { if }p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9,\,\chi =\chi _{3p},\text { resp. }\chi _{3p^2}\\ &{}\qquad \text { and } K \text { is embedded under }\rho _3;\\ 1/2, &{}\text { if }p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9, \chi =\chi _{3p^2},\text { resp. }\chi _{3p}\\ &{}\qquad \text { and } K \text { is }\text { embedded under }\rho _1.\\ \end{array}\right. } \end{aligned}$$

Proof

We may assume $f_3$ to be $L^2$-normalized. To evaluate $f_3$ for the embedding in (A.5) is equivalent to use the standard embedding

$$\begin{aligned} \sqrt{-3}\mapsto \begin{pmatrix} 0 &{} 1 \\ -3 &{} 0 \end{pmatrix} \end{aligned}$$

(A.6)

of $K_3$ for the corresponding translate of the newform. In particular the embedding in (A.5) is conjugate to the standard embedding by the following

$$\begin{aligned} \begin{pmatrix} a &{} 3^{-2}b \\ 3^3c &{} -a \end{pmatrix} =\begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} ^{-1}\begin{pmatrix} 0 &{} 1 \\ -3 &{} 0 \end{pmatrix} \begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} , \end{aligned}$$

(A.7)

where we have used the fact that $\mathrm {Nm}(\sqrt{-3})=-a^2-3bc=3$. Thus we have

$$\begin{aligned} \beta ^0_3(f_3,f_3)&=\int \limits _{{{\mathbb {Q}}}_3^\times \backslash K_3^\times }\left( \pi _3\left( \begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} ^{-1}t\begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} \right) f_3,f_3\right) \chi _3(t)dt \end{aligned}$$

(A.8)

$$\begin{aligned}&=\int \limits _{{{\mathbb {Q}}}_3^\times \backslash K_3^\times }\left( \pi _3\left( t\begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} \right) f_3,\pi _3\left( \begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} f_3\right) \right) \chi _3(t)dt, \end{aligned}$$

(A.9)

where $K_3$ is embedded in $\mathrm {M}_2({{\mathbb {Q}}}_3)$ as in (A.6). By definition, the integral in (A.9) is just

$$\begin{aligned} {\left\{ \pi _3\left( \begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} \right) f_3,\pi _3\left( \begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} \right) f_3 \right\} } \end{aligned}$$

for the standard embedding. Note that by [13, Corollary 2.12],

$$\begin{aligned} \pi _3\left( \begin{pmatrix} -9c &{} a/3 \\ 0 &{} 1 \end{pmatrix} \right) f_3= & {} \frac{1}{\sqrt{(q-1)q^{\lceil \frac{c(\theta _3)}{2 e}\rceil -1}}}\nonumber \\&\sum \limits _{x\in ({{\mathbb {Z}}}_3/\varpi ^{\lceil \frac{c(\theta _3)}{2 e}\rceil } {{\mathbb {Z}}}_3)^\times }\pi _3\left( \begin{pmatrix} 1 &{} a/3 \\ 0 &{} 1 \end{pmatrix} \begin{pmatrix} x &{} 0 \\ 0 &{} 1 \end{pmatrix} \right) \varphi _0 \end{aligned}$$

(A.10)

where $\lceil \cdot \rceil$ means the smallest integer not less than the given number and $e=2$ is the ramification index of $K_3/{{\mathbb {Q}}}_3$. Denote

$$\begin{aligned} \varphi _{a, x}=\pi _3\left( \begin{pmatrix} 1 &{} a/3 \\ 0 &{} 1 \end{pmatrix} \begin{pmatrix} x &{} 0 \\ 0 &{} 1 \end{pmatrix} \right) \varphi _0. \end{aligned}$$

In order to compute $\beta ^0_3(f_3,f_3)$, we just need to consider $\{\varphi _{a, x'},\varphi _{a, x''}\}$ for $x',x''\in ({{\mathbb {Z}}}_3/\varpi {{\mathbb {Z}}}_3)^\times$.

If $p\equiv 2$, resp. $5 \ \mathrm {mod}\ 9$ and $\chi =\chi _{3p}$, resp. $\chi _{3p^2}$, we embed K into $\mathrm {M}_2({{\mathbb {Q}}})$ under $\rho _2$ or $\rho _3$. In this case, $c(\theta _3\overline{\chi }_3)=0$ and $a=0$ or 9. By (A.4), $\varphi _{a, x}(y)=\psi ((a/3)y)\varphi _0(xy)$. Since $x\in {{\mathbb {Z}}}_3^\times$, $\psi ((a/3)y)=1$ or $e^{2\pi i x^{-1}/3}$ depending on $a=0$ or 9 on the support of $\varphi _0(xy)$. Then $\{\varphi _{a, x'},\varphi _{a, x''}\}=\omega ^i\{\varphi _{0, x'},\varphi _{0, x''}\}$ for some $i=0, 1, 2$ and $i=0$ if $x'=x''$ since we take a dual pair. By the $l:=c(\theta _3\overline{\chi _3})/2=0$ case in [13, Section 2.4], we have a unique $x\ \mathrm {mod}\ 3$ for which ${\left\{ \varphi _{0,x},\varphi _{0,x} \right\} }={\mathrm {Vol}}({{\mathbb {Q}}}_3^\times \backslash K_3^\times )=2$ is nonvanishing. By Proposition A.4, we have

$$\begin{aligned} \beta ^0_3\left( f_3,f_3 \right) =\frac{1}{(q-1)q^{\lceil \frac{c(\theta _3)}{2 e_{{\mathbb {L}}}}\rceil -1}} {\left\{ \varphi _{0,x},\varphi _{0,x} \right\} }=\frac{1}{2}\cdot 2=1. \end{aligned}$$

(A.11)

If $p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9$, and $\chi =\chi _{3p^2},\text { resp. }\chi _{3p}$, we embed K into $\mathrm {M}_2({{\mathbb {Q}}})$ under $\rho _1$. In this case, we have $c(\theta _3\overline{\chi }_3)=2l=2$ and $a=3$. This is the case $l=1$ and $n-l=1$ as in [13, Section 2.4]. Noting $\alpha _{\theta _3}=\alpha _{\Theta _3}=\frac{1}{9\sqrt{-3}}$, we set

$$\begin{aligned} D'=\frac{1}{\alpha _{\theta _3}^2\varpi _{{\mathbb {L}}}^{2c(\theta _3)}}=-3. \end{aligned}$$

According to [13, Section 2.4], [10, Section B.2.3], for $u\in {{\mathcal {O}}}_{{\mathbb {F}}}$ and $v\in {{\mathcal {O}}}_{{\mathbb {F}}}^\times$,

$$\begin{aligned} \pi _3\left( \begin{pmatrix} 1 &{} u \\ 0 &{} 1 \end{pmatrix} \begin{pmatrix} v &{} 0 \\ 0 &{} 1 \end{pmatrix} \right) \varphi _0 \end{aligned}$$

is a test vector for $(\pi _3,\chi _3)$ if and only if (u, v) is a solution of the equation

$$\begin{aligned} \frac{D}{D'}v^2-\left( 2\varpi ^n\alpha _{\theta _3{{\overline{\chi }}_3}}\sqrt{D}-2\sqrt{\frac{D}{D'}}\right) v+(1-Du^2)\equiv 0\ \mathrm {mod}\ \varpi ^{n-\lfloor \frac{l}{2}\rfloor }, \end{aligned}$$

(A.12)

where $\lfloor \cdot \rfloor$ means the largest integer not bigger than the given number. In our case $u=1$ and by Lemma A.5, $\alpha _{\theta _3\overline{\chi }_3}=\frac{1}{3\sqrt{-3}}$, we have the discriminant of (A.12) (as an equation of v)

$$\begin{aligned} \Delta&= 4\varpi ^{n}\alpha _{\theta _3\overline{\chi }_3}\sqrt{D}\left( \varpi ^{n}\alpha _{\theta _3\overline{\chi }_3}\sqrt{D}-2\sqrt{\frac{D}{D'}}\right) +4\frac{D}{D'}D \nonumber \\&\equiv 4\cdot 9\cdot \frac{1}{3\sqrt{-3}} \cdot \sqrt{-3}\cdot (-2)+4 \cdot (-3)\ \mathrm {mod}\ {\varpi ^2}\nonumber \\&\equiv -8\cdot 3-4\cdot 3 \ \mathrm {mod}\ {\varpi ^2}\nonumber \\&\equiv 0 \ \mathrm {mod}\ {\varpi ^2}. \end{aligned}$$

(A.13)

Then we can get a unique solution $x\ \mathrm {mod}\ \varpi$ of (A.12), which means there is only one $\varphi _{a,x}$ in the expression (A.10) such that $\{\varphi _{a,x},\varphi _{a,x}\}=1/q^{\lfloor l/2\rfloor }$ is nonvanishing. Again, by Proposition A.4,

$$\begin{aligned} \beta ^0_3\left( f_3,f_3\right) =\frac{1}{(q-1)q^{\lceil \frac{c(\theta _3)}{2 e_{{\mathbb {L}}}}\rceil -1}}\frac{1}{q^{\lfloor l/2\rfloor }}=\frac{1}{2}. \end{aligned}$$

(A.14)

$\square$

Let $f'$ be the admissible test vector of $(\pi , \chi )$ as defined in [3, Definition 1.4], which is the test vector used in [3] to establish the general explicit Gross–Zagier formula while we use the new form f. So we need to compare the local period integrals of $f'$ and f in order to get our explicit Gross–Zagier formulae as in [3, Theorem 1.6]. We just remark here that in our case, $f'$ is different from f only at the place 3 and the 3-adic part $f_3'$ is a $\chi _3^{-1}$-eigenvector under the action of $K_3^\times$. For more details of admissible test vectors, we refer to [3, Definition 1.4]. The following corollary compares the local period integrals of $f'$ and f and is what we need to establish the explicit Gross–Zagier formulae in Theorem 4.2.

Corollary A.7

For the admissible test vector $f'_3$ and the newform $f_3$ we have

$$\begin{aligned}\frac{\beta _3^0(f'_3,f'_3)}{\beta _3^0(f_3,f_3)}={\left\{ \begin{array}{ll} 2, &{}\text { if }p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9, \chi =\chi _{3p},\text { resp. }\chi _{3p^2} \\ &{}\qquad \text { and } K \text { is embedded under }\rho _2,\\ 2, &{}\text { if }p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9, \chi =\chi _{3p},\text { resp. }\chi _{3p^2}\\ &{}\qquad \text { and }K \text { is embedded under }\rho _3,\\ 4, &{}\text { if }p\equiv 2, \text { resp. } 5 \ \mathrm {mod}\ 9, \chi =\chi _{3p^2},\text { resp. }\chi _{3p} \\ &{}\qquad \text { and }K \text { is embedded under }\rho _1. \end{array}\right. }\end{aligned}$$

Proof

Keep the normalization of the volumes in Proposition A.6. Since $f_3'$ is a $\chi _3^{-1}$-eigenvector under the action of $K_3^\times$, we have $\beta _3^0(f'_3,f'_3)={\mathrm {Vol}}({{\mathbb {Q}}}_3^\times \backslash K_3^\times )=2$. Then the corollary follows from Proposition A.6. $\square$