Proof. By Manin’s rationality theorem for periods (see [Reference Kohnen, Zagier and Rankin6, page 202]), there exists a period $\omega ^{-} \in \mathbb {R}$ such that the polynomial $P_K = r_f^{-} / \omega ^{-}$ has coefficients in $K_f$ , the number field generated by the Fourier coefficients of $f$ . Let $\mathcal {O}_f$ be the ring of integers of $K_f$ .

Fix a $3$ -adic embedding $K_f\hookrightarrow \overline {\mathbb Q}_3$ and let $\mathfrak p\mid 3$ be the corresponding prime of $K_f$ . Let $R$ be the ring of integers of the $\mathfrak p$ -adic completion of $K_f$ . This is a complete discrete valuation ring. Let $\pi $ be a uniformiser and normalise $v_{\mathfrak p}$ by $v_{\mathfrak p}(\pi )=1$ . Write ${P_K(x)=\sum _i c_i x^i,}$ and let $m=\min _{c_i\neq 0}v_{\mathfrak p}(c_i).$ Define $P(x)=\pi ^{-m}P_K(x).$ Then, $P\in R[x]$ and at least one coefficient of $P$ is a unit of $R$ . By construction, $P$ is nonzero and satisfies the same relations as $r_f^{-}$ , namely:

$$ \begin{align*} a_2P = \sum_{M\in{\text{Man}}_2} P\mid M, \quad P\mid(1+S)=0, \quad P\mid(1+U+U^2)=0. \end{align*} $$

We now compute the $T_2$ -action modulo the ideal $3R$ . Viewing the integer matrices modulo 3, we readily establish the congruences:

$$ \begin{align*}\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \equiv \varepsilon \pmod 3 \quad\text{and}\quad \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \equiv -\varepsilon \pmod 3.\end{align*} $$

Furthermore,

$$ \begin{align*} \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \equiv -SU\varepsilon \pmod {3} \quad\text{and}\quad \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \equiv SU^2\varepsilon \pmod {3} .\end{align*} $$

Combining these congruences yields

$$ \begin{align*} a_2 P & = P \,\bigg|\, \bigg( \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}+ \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \bigg) \\ & \equiv P \mid ( \varepsilon +(-\varepsilon)+(-SU)\varepsilon+SU^2\varepsilon ) \pmod {3R}, \\ & \equiv P \mid (I_2+(-I_2)+(-SU)+SU^2) \mid \varepsilon \pmod {3R}. \end{align*} $$

Note that $-I_2$ acts trivially because $k$ is even, so $P\mid A=P \mid -A$ for every matrix $A$ . Thus,

$$ \begin{align*}P \mid (I_2+(-I_2)+(-SU)+SU^2) =2P+ P\mid (SU+SU^2),\end{align*} $$

and since $2P \equiv -P \pmod {3R}$ and $-P= P\mid S$ , it follows that

$$ \begin{align*}a_2 P \equiv [ P \mid S \mid (1+U+U^2) ] \mid \varepsilon \equiv 0 \mid \varepsilon \equiv 0 \pmod {3R}.\end{align*} $$

Because some coefficient $u$ of $P$ is a unit in $R$ , the corresponding coefficient of $a_2P$ gives $a_2u\in 3R$ and so $a_2 \in 3R$ . Equivalently, $v_{\mathfrak p}(a_2)\ge v_{\mathfrak p}(3).$ As $\mathfrak {p}\mid 3$ was arbitrary, this inequality holds at every prime of $K_f$ above $3$ . At the other primes, the element 3 is a unit and $a_2$ is already an algebraic integer. Therefore, $a_2/3$ is integral at every finite prime of $K_f$ and we conclude that $a_2/3\in \mathcal {O}_f$ .

Proof. Fix a $3$ -adic embedding $K_f\hookrightarrow \overline {\mathbb Q}_3$ for which the hypothesis $a_2\in 9\overline {\mathbb Z}_3$ holds and let $R$ be the ring of integers of the corresponding completion. By the same normalisation as in Lemma 2.1, choose $P\in R[x]$ proportional to the odd period polynomial, with at least one coefficient a unit.

Let $w=k-2$ and write

$$ \begin{align*}P=\sum_{\substack{ n=1\\ n\text{ odd}}}^{k-3} c_n x^n .\end{align*} $$

Using Zagier’s formula, we obtain

$$ \begin{align*} a_2 P & = P \,\bigg|\, \bigg( \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}+ \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \bigg) \\ & = P(2x)+2^w P\bigg( \frac{x}{2} \bigg) +(x+1)^w P \bigg(\frac{2x}{x+1} \bigg) + 2^w P \bigg( \frac{x+1}{2} \bigg). \end{align*} $$

For every odd $n\ge 1$ , we have $2^n=(-1+3)^n\equiv -1+3n \pmod 9$ and since $w$ is even, we also get $2^{w-n}\equiv -1 +3(w-n)\pmod 9$ . Thus, the contribution of the first two terms is

$$ \begin{align*} P(2x)+2^w P\bigg( \frac{x}{2} \bigg) \equiv \sum_{\substack{ n=1\\ n\text{ odd}}}^{k-3} c_n (-2+3w)x^n \equiv (-2+3w)P(x) \pmod {9R[x]}. \end{align*} $$

For the third term, set

$$ \begin{align*}A(x)=\sum_{\substack{ n=1\\ n\text{ odd}}}^{k-3} c_n x^n (x+1)^{w-n} =(x+1)^w P\bigg( \frac{x}{x+1} \bigg).\end{align*} $$

Since

$$ \begin{align*}P\mid U = P\,\bigg|\, \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix} = x^w P\bigg( \frac{x-1}{x} \bigg),\end{align*} $$

we see that $A(x)$ is just $P\mid U$ evaluated at $x+1$ . However, $P\mid U =- P -P\mid U^2$ and

$$ \begin{align*}P\mid U^2 = P\mid ST^{-1}=(P\mid S) \mid T^{-1}=(-P)\mid T^{-1}=-P(x-1),\end{align*} $$

showing that $ A(x)=P(x)-P(x+1). $ Moreover, since

$$ \begin{align*} \sum_{\substack{ n=1\\ n\text{ odd}}}^{k-3} n c_n x^n(x+1)^{w-n} = x(x+1)A'(x)-wxA(x), \end{align*} $$

we obtain, modulo $9R[x]$ ,

$$ \begin{align*} (x+1)^wP\bigg(\frac{2x}{x+1}\bigg) \equiv -(1+3wx)(P(x)-P(x+1)) + 3x(x+1)(P'(x)-P'(x+1)). \end{align*} $$

Similarly,

$$ \begin{align*} 2^wP\bigg(\frac{x+1}{2}\bigg) \equiv (3w-1)P(x+1)-3(x+1)P'(x+1) \pmod{9R[x]}. \end{align*} $$

Combining these four terms gives

$$ \begin{align*} 3bP=a_2P \equiv3D_w(P) \pmod{9R[x]}, \end{align*} $$

where

$$ \begin{align*} \begin{aligned} D_w(P)={}&(w-1)P +x(x+1)(P'(x)-P'(x+1)) \\ &-wx(P(x)-P(x+1)) +wP(x+1) -(x+1)P'(x+1). \end{aligned} \end{align*} $$

Hence,

$$ \begin{align*} bP\equiv D_w(P)\pmod{3R[x]}. \end{align*} $$

Let $\mathfrak m$ be the maximal ideal of $R$ and let $ \kappa =R/\mathfrak m $ be the residue field. Since ${b\in 3R\subseteq \mathfrak m}$ , we have $\overline b=0\in \kappa .$ Also, since $P$ has a unit coefficient, its reduction $\overline P\in \kappa [x]$ is nonzero, which gives $ D_w(\overline P)=0$ in $\kappa [x].$

Let $\delta \in \{0,1,2\}$ be the residue class of $w$ modulo $3$ . Over the field $\kappa $ , define

$$ \begin{align*} H_\delta(Q)=xQ'-\delta Q \quad\text{and}\quad L_\delta(Q)=(x+1)(H_\delta(Q)(x)-H_\delta(Q)(x+1)). \end{align*} $$

Then, for every polynomial $Q\in \kappa [x]$ ,

$$ \begin{align*} D_w(Q)=(2\delta-1)Q+L_\delta(Q). \end{align*} $$

Therefore, $ 0=D_w(\overline P) = (2\delta -1)\overline P+L_\delta (\overline P) $ or, equivalently, $ L_\delta (\overline P)=-(2\delta -1)\overline P. $ If ${k \not \equiv 4 \pmod 6}$ , then $\delta \in \{0,1\}$ , which leads to either $L_0(\overline P) = \overline P$ or $L_1(\overline P) = 2\overline P$ . The next lemma rules out both of these possibilities and completes the proof.

Proof. Suppose that $P$ has degree $d$ and that $P(x) = a_d x^d + a_{d-2} x^{d-2} + O(x^{d-4})$ with $a_d\neq 0$ . Then,

$$ \begin{align*} H_\delta(x) = (d-\delta)a_d x^d + (d-2-\delta)a_{d-2} x^{d-2} + O(x^{d-4})\end{align*} $$

and

$$ \begin{align*}H_\delta(x) - H_\delta(x+1) = -(d-\delta)a_d \bigg[ d x^{d-1} + \binom{d}{2} x^{d-2} \bigg] + O(x^{d-3}),\end{align*} $$

which shows that

$$ \begin{align*}L_\delta(P)(x) = -d(d-\delta)a_d x^d - \binom{d+1}{2}(d-\delta)a_d x^{d-1} + O(x^{d-2}).\end{align*} $$

If $d\equiv 0\pmod 3$ , then clearly $\deg (L_{\delta })<d$ , so $L_\delta (P)$ cannot equal any nonzero scalar multiple of $P(x)$ .

If $d\equiv 1\pmod 3$ , then $L_\delta (P)(x) = (\delta -1)a_d x^d + (\delta -1)a_d x^{d-1} + O(x^{d-2})$ . For $L_\delta (P)$ to be a scalar multiple of $P(x)$ , it must mirror the structure of $P$ . Because $P(x)$ is odd, it has no $x^{d-1}$ term, meaning $L_\delta (P)$ must also have a zero coefficient for $x^{d-1}$ , that is,

$$ \begin{align*}(\delta-1)a_d = 0.\end{align*} $$

Since $a_d \neq 0$ , we must have $\delta = 1$ . However, if $\delta = 1$ , the $x^d$ coefficient $(\delta -1)a_d$ also becomes $0$ . This means $\deg (L_1(P)) < d$ , so $L_1(P)\neq 2P$ . For $\delta \in \{0, 2\}$ , the required $x^{d-1}$ term fails to vanish.

If $d\equiv -1\pmod 3$ , then the leading term of $L_\delta (P)$ is $-(\delta +1)a_d x^d$ and the $x^{d-1}$ term vanishes. We consider three cases.

1. (1) Let $\delta =0$ . The leading coefficient of $L_0(P)$ evaluates to $-a_d$ , whereas that of $P$ is $a_d$ . Thus, $L_0(P) \neq P$ .

2. (2) Let $\delta =1$ . This time, the leading coefficient of $L_1(P)$ is $-2a_d\equiv a_d\pmod 3$ , whereas that of $2P$ is $2a_d\equiv -a_d\pmod 3$ . Hence, $L_1(P) \neq 2P$ .

3. (3) Let $\delta =2$ . Then, the leading coefficient of $L_2(P)$ is $0\pmod 3$ . Since ${\deg (L_2(P)) < d}$ , we have $L_2(P) \neq \lambda P$ for all $\lambda \neq 0$ .

Proof of Corollary 1.3.

We use the notation from the proof of Theorem 1.2. The condition that $3$ is unramified in $K_f$ implies that $3R=\mathfrak m$ , the maximal ideal of $R$ .

First, suppose $k\equiv 4\pmod 6$ . Then, $\delta =2$ , so $\overline b\,\overline P=L_2(\overline P).$ By Lemma 3.1, we must have $\overline b=0$ , giving $a_2\equiv 0\pmod {9\overline {\mathbb Z}_3}.$

Now, suppose $k\not \equiv 4\pmod 6$ , so $\delta \in \{ 0,1 \}$ . The computation in the proof of Theorem 1.2 shows that

$$ \begin{align*} bP\equiv D_w(P)\pmod{3R[x]}, \end{align*} $$

where $b=a_2/3\in R$ . We emphasise that this part does not require that $a_2\in 9R$ ; indeed, this assumption was only used in the theorem to conclude that $b\in 3R$ . Then,

$$ \begin{align*} \overline b\,\overline P = (2\delta-1)\overline P+L_\delta(\overline P). \end{align*} $$

If $\overline {P}$ has degree $N$ and leading term $u\neq 0$ , then the coefficient of $x^N$ in $L_\delta (\overline P)$ is ${-N(N-\delta )u.}$ This yields

$$ \begin{align*} \overline b=(2\delta-1)-N(N-\delta). \end{align*} $$

It is easy to see that for $\delta \in \{0,1\}$ , we have $\overline {b}\in \{1,2\}$ . Thus, in either case, ${b\equiv 1\text { or }2\pmod {3R}.}$ Equivalently, $a_2\equiv 3\text { or }6\pmod {9\overline {\mathbb Z}_3}.$