text:
\section{Introduction}
In this article we study the number fields generated by the Fourier coefficients of modular forms at the cusps of $X_1(N)$. To do this we study the connections between two actions on spaces of modular forms: the action of $\mathrm{GL}^+_2(\mathbb{Q})$ via the slash-operator and the action of $\Aut(\mathbb{C})$ on the Fourier coefficients of a modular form. A detailed study of these actions was conducted by Shimura in \cite{shimura1975}, where he proved a formula for the action of $\Aut(\mathbb{C})$ on $f|g$ for modular forms of even weight. In Theorem \ref{fg sigma} we give an extension of his result to modular forms of any integral weight and provide two new proofs of it: one using a theorem of Khuri-Makdisi \cite{khuri-makdisi2012} on products of Eisenstein series, and the other using Katz's theory of algebraic modular forms \cite{katz-interpolation}.

We use this theorem to bound the fields generated by the Fourier coefficients of modular forms at the cusps. Let us assume for simplicity that $f$ is a modular form in $M_k(\Gamma_0(N))$, and let $g=\sabcd ABCD \in \mathrm{SL}_2(\mathbb{Z})$. We show in Theorem \ref{thm:number fields general} that the coefficients of $f|g$ lie in the cyclotomic extension $K_f(\zeta_{N'})$, where $K_f$ is the number field generated by the coefficients of $f$, and $N'=N/\gcd(CD,N)$. In the case $f$ has non-trivial Nebentypus, we show in Theorem \ref{thm:characters} that the coefficients of $f|g$ belong to a $1$-dimensional $K_f(\zeta_{N'})$-vector space, which is itself contained in an explicit cyclotomic extension $K_f(\zeta_M)$.

We apply these results in Section \ref{section:al ops} to find number fields that contain the Atkin--Lehner pseudo-eigenvalues of a newform, recovering a result of Cohen in \cite{cohen2018}.

In Section \ref{section:optimizing} we discuss how to choose $g$ among the matrices in $\mathrm{SL}_2(\mathbb{Z})$ that map $\infty$ to a given cusp $\alpha$ so that $N'$ (or $M$) becomes minimal. Assuming that $f\in M_k(\Gamma_0(N))$ is an eigenfunction of the Atkin--Lehner operators, we describe how to further reduce $N'$ by potentially replacing $\alpha$ with its image under a suitable Atkin--Lehner operator. The Fourier expansion of $f|g$ can then easily be obtained from another Fourier expansion $f|g'$ which has coefficients in the field $K_f(\zeta_{\operatorname{gcd}(\delta,N/\delta)})$, where $\delta=\gcd(C,N)$ is the denominator of the cusp $\alpha=A/C$. Note that $\mathbb{Q}(\zeta_{\operatorname{gcd}(\delta,N/\delta)})$ is the field of definition of the cusp $\alpha$ in the canonical model of $X_0(N)$ over $\mathbb{Q}$.

In the last section, we prove in Theorem \ref{thm:exact nf Gamma0} that if $f$ is a newform for $\Gamma_0(N)$ then the number field provided by Theorem \ref{thm:number fields general} is the best possible, in the sense that it is the number field generated by the coefficients of $f|g$.

Recently three algorithms for the computation of the Fourier expansion of $f|g$ have appeared: two algorithms in Sagemath, one by Dan Collins \cite{collins2018} and another by Martin Dickson and the second author \cite{dickson-neururer}. The third algorithm was implemented in PARI/GP by Karim Belabas and Henri Cohen \cite{cohen2018}. While the first algorithm only uses numerical approximations of the Fourier coefficients in order to compute Petersson inner products, the latter two calculate the Fourier coefficients as algebraic numbers. The knowledge of the number field (or vector space) generated by the Fourier coefficients of $f|g$ could provide a significant speed-up for these calculations.

\textbf{Acknowledgements:} We thank Henri Cohen for encouraging us to write this article.

We are grateful to Abhishek Saha for pointing out an alternative approach to the Theorems in Section \ref{section:bounding field}, sketched in Remark \ref{remark:local Whittaker newforms}.

\subsection*{Notations} For any integer $k \in \mathbb{Z}$, we define the weight $k$ action of $\mathrm{GL}_2^+(\mathbb{R})$ on functions $f : \h \to \mathbb{C}$ by
\[
f|_k g(\tau) = \frac{\det(g)^{k/2}}{(c\tau + d)^k} f\Bigl(\frac{a\tau+b}{c\tau+d}\Bigr) \qquad \left(g=\abcd abcd \in \mathrm{GL}_2^+(\mathbb{R})\right).
\]
We will usually omit $k$ from the notation and just write $f|g$ for $f|_k g$.

The automorphism group $\Aut(\mathbb{C})$ acts on spaces of modular forms as follows: for any modular form $f(\tau)=\sum_n a_n e^{2\pi i n\tau/w}$, we let
\[
f^\sigma(\tau) = \sum_n \sigma(a_n) e^{2\pi in\tau/w} \qquad (\sigma \in \Aut(\mathbb{C})).
\]

For any integer $N \geq 1$, we denote $\zeta_N = e^{2\pi i/N} \in \mathbb{C}$.

\section{Eisenstein series}\label{section:eisenstein series}

\subsection{Definitions} We refer the reader to \cite[\S 3]{kato} for more details on Eisenstein series.

For integers $k \geq 1$, $N \geq 1$ and $a,b\in\mathbb{Z}/N\mathbb{Z}$, define the series
\[
E^{(k)}_{a,b}(\tau) = \frac{(k-1)!}{(-2\pi i)^k} \sum_{\substack{\omega \in \mathbb{Z}\tau+\mathbb{Z} \\ \omega \neq - (\tilde{a}\tau+\tilde{b})/N}} \frac{1}{(\omega+\frac{\tilde{a}\tau+\tilde{b}}{N})^k |\omega+\frac{\tilde{a}\tau+\tilde{b}}{N}|^{2s}} \Biggl. \Biggr|_{s=0}
\]
where $\tilde{a},\tilde{b}$ denote any representatives of $a,b$ in $\mathbb{Z}$, and $\cdot |_{s=0}$ denotes analytic continuation to $s=0$ (this is needed only when $k \in \{1,2\}$). It follows from the definition that the weight $k$ action of $\mathrm{SL}_2(\mathbb{Z})$ on these series is given by $E^{(k)}_{a,b} | g = E^{(k)}_{(a,b)g}$ for every matrix $g \in \mathrm{SL}_2(\mathbb{Z})$. In particular, the function $E^{(k)}_{a,b}$ is modular of weight $k$ with respect to the principal congruence subgroup $\Gamma(N)$. If $k \neq 2$, then $E^{(k)}_{a,b}$ is a holomorphic Eisenstein series of weight $k$ with respect to $\Gamma(N)$. If $k=2$, then $\tilde{E}^{(2)}_{a,b} := E^{(2)}_{a,b}-E^{(2)}_{0,0}$ is a holomorphic Eisenstein series of weight $2$ with respect to $\Gamma(N)$.

\subsection{Fourier expansions}\label{subsection:fourier exp} We refer the reader to \cite[\S 3]{katz-interpolation} and \cite[Chap. VII]{schoeneberg} for proofs of the following facts.

If $k \neq 2$, then the Fourier expansion of $E^{(k)}_{a,b}$ is given by
\[
E^{(k)}_{a,b}(\tau) = a_0(E^{(k)}_{a,b}) + \sum_{\substack{m,n\geq 1 \\ m \equiv a (N)}} \zeta_N^{bn} n^{k-1} q^{mn/N} + (-1)^k \sum_{\substack{m,n\geq 1 \\ m \equiv -a (N)}} \zeta_N^{-bn} n^{k-1} q^{mn/N} \qquad (q=e^{2\pi i\tau}).
\]
If $k=2$, then the Fourier expansion of $\tilde{E}^{(2)}_{a,b}$ is given by
\[
\tilde{E}^{(2)}_{a,b}(\tau) = a_0(\tilde{E}^{(2)}_{a,b}) + \sum_{\substack{m,n\geq 1 \\ m \equiv a (N)}} \zeta_N^{bn} n q^{mn/N} + \sum_{\substack{m,n\geq 1 \\ m \equiv -a (N)}} \zeta_N^{-bn} n q^{mn/N} - 2 \sum_{m,n\geq 1} n q^{mn}.
\]
The constant terms $a_0(E^{(k)}_{a,b})$ and $a_0(\tilde{E}^{(2)}_{a,b})$ are elements of $\mathbb{Q}(\zeta_N)$ and are given in \cite[3.10]{kato} and \cite[\S 3]{brunault:regRZ}. We will not need the precise expressions since modularity determines them uniquely.

\begin{proposition}\label{proposition:Galois action on E}
Let $g=\sabcd ABCD\in\mathrm{SL}_2(\mathbb{Z})$ and $\sigma \in \Aut(\mathbb{C})$ such that $\sigma(\zeta_N)=\zeta_N^\lambda$ with $\lambda \in (\mathbb{Z}/N\mathbb{Z})^\times$. If $k \neq 2$, then
\begin{align*}
(E^{(k)}_{a,b} | g)^{\sigma} = (E^{(k)}_{a,b})^\sigma | g_\lambda,
\end{align*}
where $g_\lambda$ is any lift in $\mathrm{SL}_2(\mathbb{Z})$ of the matrix $\sabcd{A}{\lambda B}{\lambda^{-1} C}{D} \in \mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$. If $k=2$, then the same statement holds with $E^{(2)}_{a,b}$ replaced by $\tilde{E}^{(2)}_{a,b}$.
\end{proposition}

\begin{proof}
Note that $(E^{(k)}_{a,b})^\sigma = E^{(k)}_{a,\lambda b}$, so that
\begin{equation*}
(E^{(k)}_{a,b} | g)^{\sigma} = E^{(k)}_{aA+bC,\lambda(aB+bD)} = E^{(k)}_{a,\lambda b} \Bigl| \begin{pmatrix} A & \lambda B \\ \lambda^{-1}C & D \end{pmatrix} = (E^{(k)}_{a,b})^\sigma | g_\lambda.
\end{equation*}
The argument for $\tilde{E}^{(2)}_{a,b}$ is similar.
\end{proof}

\section{The actions of $\mathrm{SL}_2(\mathbb{Z})$ and $\mathrm{Aut}(\mathbb{C})$ on modular forms}\label{section:GL_2(Z/NZ)-action}

In this section we investigate the connection between the natural actions of $\mathrm{SL}_2(\mathbb{Z})$ and $\Aut(\mathbb{C})$ on modular forms.

Let us first recall Khuri-Makdisi's result \cite{khuri-makdisi2012} giving generators of the graded algebra of modular forms. Let $\mathcal{R}_N$ be the subalgebra of $M_*(\Gamma(N))=\bigoplus_{k\geq 0} M_k(\Gamma(N))$ generated by the Eisenstein series $E^{(1)}_{a,b}$ with $a,b\in\mathbb{Z}/N\mathbb{Z}$.

\begin{theorem}\cite{khuri-makdisi2012}\label{thm:khuri-makdisi}
If $N \geq 3$, then $\mathcal{R}_N$ contains all modular forms on $\Gamma(N)$ of weight $2$ and above. In other words, $\mathcal{R}_N$ misses only the cusp forms of weight $1$ on $\Gamma(N)$.
\end{theorem}

\begin{proof}
This follows from combining \cite[Theorem 3.5, Remark 3.14, Theorem 5.1]{khuri-makdisi2012}. The link between our notations and Khuri-Makdisi's notations is
\begin{equation*}
E^{(1)}_{a,b}(\tau) = -\frac{1}{2\pi i} G_1\Bigl(\tau,\frac{a\tau+b}{N}\Bigr) = \frac{1}{2\pi i} \lambda_{(a\tau+b)/N},
\end{equation*}
see \cite[Definition 2.1 and Corollary 3.13]{khuri-makdisi2012}.
\end{proof}

\begin{remark}
If $N=2$, then the algebra $M_*(\Gamma(2))$ is generated by the weight $2$ Eisenstein series $\tilde{E}^{(2)}_{1,0}$, $\tilde{E}^{(2)}_{0,1}$ and $\tilde{E}^{(2)}_{1,1}$, the only relation being $\tilde{E}^{(2)}_{1,0} + \tilde{E}^{(2)}_{0,1} + \tilde{E}^{(2)}_{1,1} = 0$, see \cite[Remark 3.6]{khuri-makdisi2012}.
Of course, if $N=1$ then $M_*(\mathrm{SL}_2(\mathbb{Z}))$ is freely generated by the usual Eisenstein series of weight $4$ and $6$.
\end{remark}

The following theorem is an extension of Shimura's result \cite[Theorem 8]{shimura1975} which deals with the case of modular forms of even weight. We note however, that Shimura works with a much wider class of functions, including Hilbert modular forms and also certain derivatives of them. His methods can be extended to provide a full proof of Theorem \ref{fg sigma}, but we will give two new proofs of it, the first using Khuri-Makdisi's Theorem \ref{thm:khuri-makdisi}. The second proof uses Katz's theory of algebraic modular forms and is given in the appendix along with a brief introduction to the theory of algebraic modular forms.

\begin{theorem}\label{fg sigma}
Let $f\in M_k(\Gamma(N))$ be a modular form of weight $k \geq 1$ and level $\Gamma(N)$. Let $g = \sabcd ABCD \in\mathrm{SL}_2(\mathbb{Z})$ and $\sigma\in\Aut(\mathbb{C})$ such that $\sigma(\zeta_N)=\zeta_N^\lambda$ with $\lambda \in (\mathbb{Z}/N\mathbb{Z})^\times$. Then
\[
(f|g)^{\sigma} = f^{\sigma} | g_\lambda,
\]
where $g_\lambda$ is any lift in $\mathrm{SL}_2(\mathbb{Z})$ of the matrix $\sabcd{A}{\lambda B}{\lambda^{-1} C}{D} \in \mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$.
\end{theorem}

\begin{proof}[First proof]
Let us assume $N \geq 3$, and let $f\in \mathcal{R}_N$. The maps $h \mapsto (h | g)^\sigma$ and $h \mapsto h^\sigma | g_\lambda$ are both $\sigma$-linear ring homomorphisms. We are thus reduced to the case $f=E^{(1)}_{a,b}$, which follows from Proposition \ref{proposition:Galois action on E}. If $f \in S_1(\Gamma(N))$, then $f^2$ and $f^3$ are in $\mathcal{R}_N$, so Theorem \ref{fg sigma} holds for them. Using $f = f^3/f^2$, we get $(f|g)^\sigma = f^\sigma | g_\lambda$. In the case $N=2$, we proceed similarly by applying Proposition \ref{proposition:Galois action on E} to $\tilde{E}^{(2)}_{a,b}$. Finally, the case $N=1$ is trivial.
\end{proof}

\begin{remark}
If we restrict to modular forms on $\Gamma_1(N)$, then Theorem \ref{fg sigma} also follows from the result of Borisov and Gunnells \cite[Thm 5.15]{BG3} that all modular forms of sufficiently large weight are toric.
\end{remark}

By Theorem \ref{fg sigma}, the space $M_k(\Gamma(N);\mathbb{Q}(\zeta_N))$ of modular forms with coefficients in $\mathbb{Q}(\zeta_N)$ is stable under the weight $k$ action of $\mathrm{SL}_2(\mathbb{Z})$. It is thus endowed with a right action of $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$. The Galois group $\Gal(\mathbb{Q}(\zeta_N)/\mathbb{Q})$ also acts on $M_k(\Gamma(N);\mathbb{Q}(\zeta_N))$, by means of the usual action on the Fourier expansion. But more is true: for any $f \in M_k(\Gamma(N);\mathbb{Q}(\zeta_N))$, let us define
\begin{equation*}
f \Bigl| \abcd100\lambda = f^{\sigma_\lambda} \qquad (\lambda \in (\mathbb{Z}/N\mathbb{Z})^\times),
\end{equation*}
where $\sigma_\lambda \in \Gal(\mathbb{Q}(\zeta_N)/\mathbb{Q})$ is the automorphism defined by $\sigma_\lambda(\zeta_N)=\zeta_N^\lambda$. Then the above actions of $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$ and $(\mathbb{Z}/N\mathbb{Z})^\times$ combine to give a right action of $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$ on $M_k(\Gamma(N);\mathbb{Q}(\zeta_N))$.
Indeed
\begin{equation}\label{eq g glambda}
g \abcd 100\lambda = \abcd A{\lambda B}C{\lambda D} = \abcd 100\lambda \abcd A{\lambda B}{\lambda^{-1} C}D = \abcd 100\lambda g_\lambda
\end{equation}
and Theorem \ref{fg sigma} says precisely that both sides of this equality act in the same way on $M_k(\Gamma(N);\mathbb{Q}(\zeta_N))$. Note also that with this definition, the identities $E^{(k)}_{a,b} | g = E^{(k)}_{(a,b)g}$ for $k\neq 2$ and $\tilde{E}^{(2)}_{a,b} |g = \tilde{E}^{(2)}_{(a,b)g}$ are true for any $g \in \mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$.

\begin{remark}
Let $\tilde{Y}(N)$ be the model of $\Gamma(N)\backslash\mathcal{H}$ over $\mathbb{Q}(\zeta_N)$ constructed in \cite[Chapter 6]{shimura1971}. The automorphism group of the $\mathbb{Q}$-scheme $\tilde{Y}(N)$ is $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$. This gives a right action of $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$ on the function field $\mathbb{Q}(\zeta_N)(\tilde{Y}(N))$, which is a subfield of $\mathbb{Q}(\zeta_N)((q^{1/N}))$. The action of $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$ is the slash action of weight $0$, and the action of the diagonal matrix $\sabcd 100\lambda$ with $\lambda \in (\mathbb{Z}/N\mathbb{Z})^\times$ coincides with the natural action of $\sigma_\lambda$. Shimura's original proof of Theorem \ref{fg sigma} relies on this fact.

\end{remark}

\begin{corollary}\label{Kf Qchi}
Let $f \in M_k(\Gamma_0(N),\chi)$ be a nonzero modular form of weight $k \geq 1$, level $N \geq 1$ and Nebentypus character $\chi$. Then the field $K_f$ generated by the Fourier coefficients of $f$ contains the field $\mathbb{Q}(\chi)$ generated by the values of $\chi$.
\end{corollary}

\begin{proof}
We have to prove that every $\sigma \in \Aut(\mathbb{C}/K_f)$ fixes $\mathbb{Q}(\chi)$. Let $g \in \Gamma_0(N)$. Then $f|g=\chi(g) f$. Applying $\sigma$, we get $(f | g)^\sigma = \chi(g)^\sigma f$. But Theorem \ref{fg sigma} implies
\begin{equation*}
(f | g)^\sigma = f^\sigma | g_\lambda = f | g_\lambda = \chi(g_\lambda) f = \chi(g) f,
\end{equation*}
so that $\chi=\chi^\sigma$.
\end{proof}

\begin{remark}
Corollary \ref{Kf Qchi} can also be proved using Katz's theory of algebraic modular forms (see the Appendix), noting that the diamond operators $\langle \delta \rangle$, $\delta \in (\mathbb{Z}/N\mathbb{Z})^\times$ are defined over $\mathbb{Q}$, hence leave stable the space $M_k(\Gamma_1(N);K)$ of modular forms with coefficients in a fixed subfield $K$ of $\mathbb{C}$.
\end{remark}

\section{Bounding the coefficient field of $f | g$}\label{section:bounding field}

\begin{theorem}\label{thm:number fields general}
Let $f \in M_k(\Gamma_1(N))$ be a modular form of integral weight $k \geq 1$ on $\Gamma_1(N)$. Let $K_f$ be the subfield of $\mathbb{C}$ generated by the Fourier coefficients $a_n(f)$, $n \geq 1$. Let $g = \sabcd ABCD \in \mathrm{SL}_2(\mathbb{Z})$.
\begin{enumerate}
\item The modular form $f |_k g$ has coefficients in $K_f(\zeta_M)$ with $M=N/\gcd(C,N)$.
\item If $f \in M_k(\Gamma_0(N))$ then $f |_k g$ has coefficients in $K_f(\zeta_{N'})$ with $N'=N/\gcd(CD,N)$.
\end{enumerate}
\end{theorem}

\begin{proof}
Let $\sigma \in \Aut(\mathbb{C})$. By Theorem \ref{fg sigma}, a sufficient condition for $f|g$ being fixed by $\sigma$ is given by $f^{\sigma} = f = f | g_\lambda g^{-1}$, where $\sigma(\zeta_N)=\zeta_N^\lambda$. We have
\begin{equation}\label{glambda ginv}
g_\lambda g^{-1} \equiv \abcd{AD-\lambda BC}{AB(\lambda-1)}{CD(\lambda^{-1}-1)}{AD-\lambda^{-1}BC} \mod N.
\end{equation}
We see that $g_\lambda g^{-1} \in \Gamma_0(N)$ if and only if $\lambda \equiv 1\mod N'$. If $f \in M_k(\Gamma_0(N))$, then $f|g$ is fixed by every $\sigma \in \Aut(\mathbb{C}/K_f(\zeta_{N'}))$, hence has coefficients in $K_f(\zeta_{N'})$, which proves (2).

Furthermore $AD-\lambda BC = 1+BC(1-\lambda)$ so that $g_\lambda g^{-1}\in\Gamma_1(N)$ if and only if $\lambda \equiv 1\mod N'$ and $\lambda \equiv 1\mod N/\gcd(BC,N)$. Since $B$ and $D$ are coprime, the conjunction of these conditions is equivalent to $\lambda \equiv 1\mod N/\gcd(C,N)$. This proves (1).
\end{proof}

We now turn to modular forms with characters. We will actually bound not only the field of coefficients of $f|g$, but also the \emph{vector space} generated by the coefficients of $f|g$.

In order to state our results, we need some more notation. Let $f \in M_k(\Gamma_0(N),\chi)$, where $\chi$ is a Dirichlet character of conductor $m$ dividing $N$, and let $g = \sabcd ABCD \in \mathrm{SL}_2(\mathbb{Z})$. Put $N'=N/\gcd(CD,N)$, $m'=m/\gcd(BC,m)$ and $M=\lcm(N',m')$. Let $K=K_f(\zeta_{N'})$ and $L=K_f(\zeta_M)$. Since $L=K(\zeta_{m'})$, the extension $L/K$ is abelian and its Galois group $G=\Gal(L/K)$ identifies with a subgroup $G'$ of $(\mathbb{Z}/m'\mathbb{Z})^\times$ by means of the cyclotomic character $\lambda : G \to (\mathbb{Z}/m'\mathbb{Z})^\times$. Since $\Gal(L/K) \cong \Gal(\mathbb{Q}(\zeta_{m'})/K')$ with $K'=K \cap \mathbb{Q}(\zeta_{m'})$, the subgroup $G' \subset (\mathbb{Z}/m'\mathbb{Z})^\times$ corresponds to the subfield $K' \subset \mathbb{Q}(\zeta_{m'})$.

\begin{lemma}\label{lemma chig}
The map $\chi_g : G \to \mathbb{C}^\times$ defined by
\begin{equation*}
\chi_g(\sigma) = \chi(AD-\lambda(\sigma)^{-1} BC) \qquad (\sigma \in G)
\end{equation*}
is a group homomorphism.
\end{lemma}

\begin{proof}
Let $\sigma \in G$. Note that $m'BC$ is divisible by $m$, so that $AD-\lambda(\sigma)^{-1}BC$ is well-defined in $\mathbb{Z}/m\mathbb{Z}$. Let $\lambda_N(\sigma) \in (\mathbb{Z}/N\mathbb{Z})^\times$ denote the cyclotomic character modulo $N$. Since $\lambda_N(\sigma) \equiv 1 \mod N'$, the identity (\ref{glambda ginv}) shows that $g_{\lambda_N(\sigma)} g^{-1}$ is upper-triangular modulo $N$. It follows that $AD-\lambda_N(\sigma)^{-1} BC \in (\mathbb{Z}/N\mathbb{Z})^\times$ and thus $AD-\lambda(\sigma)^{-1} BC \in (\mathbb{Z}/m\mathbb{Z})^\times$. Therefore the map $\chi_g$ is well-defined.

Let us show that $\chi_g$ is a group homomorphism. We may write $\chi_g$ as the composition
\begin{equation*}
G \xrightarrow{\lambda} G' \xrightarrow{\psi} (\mathbb{Z}/m\mathbb{Z})^\times \xrightarrow{\chi} \mathbb{C}^\times
\end{equation*}
where $\psi$ is defined by $\psi(\mu)=AD-\mu^{-1}BC$. Using the relation \eqref{eq g glambda}, we get the following identity in $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$
\begin{equation*}
g_{\mu \mu'} g^{-1} = \abcd100{\mu^{-1}} g_{\mu'} g^{-1} \abcd100\mu g_\mu g^{-1} \qquad (\mu,\mu' \in (\mathbb{Z}/N\mathbb{Z})^\times).
\end{equation*}
Specialising to the case $\mu,\mu' \equiv 1 \mod N'$ and comparing the bottom-right entries, we deduce that $\psi$ is a group homomorphism.
\end{proof}

Note that the character $\chi_g$ takes values in $\mathbb{Q}(\chi)^\times$, which is contained in $K_f^\times$ by Corollary \ref{Kf Qchi}. By the normal basis theorem, $L$ is a free $K[G]$-module of rank 1. Since $K_f$ is contained in $K$, the character $\chi_g$ cuts out a $K$-line $L^{\chi_g}$ in $L$, namely
\begin{equation}
L^{\chi_g} = \{x \in L : \forall \sigma \in G, \sigma(x) = \chi_g(\sigma) x \}.
\end{equation}
We are now ready to state our result.

\begin{theorem}\label{fg Lchig}
The modular form $f | g$ has coefficients in $L ^{\chi_g}$.
\end{theorem}

\begin{proof}
Let $\sigma \in \Aut(\mathbb{C}/K)$ with $\sigma(\zeta_N)=\zeta_N^\lambda$. Since $\lambda \equiv 1 \mod N'$, we have $g_\lambda g^{-1}\in \Gamma_0(N)$. Then
\begin{equation}\label{fg sigma eq1}
(f | g)^\sigma = f^\sigma | g_\lambda = f | g_\lambda g^{-1} g = \chi(AD-\lambda^{-1}BC) f|g = \chi_g(\sigma |_L) f|g.
\end{equation}
In particular $f | g$ is fixed by $\Aut(\mathbb{C}/L)$, hence has coefficients in $L$. Moreover (\ref{fg sigma eq1}) shows that $f|g$ has coefficients in $L^{\chi_g}$.
\end{proof}

We summarise our result and make it slightly more precise as follows.

\begin{theorem}\label{thm:characters}
Let $f \in M_k(\Gamma_0(N),\chi)$, where $\chi$ is a Dirichlet character of conductor $m$ dividing $N$, and let $g = \sabcd ABCD \in \mathrm{SL}_2(\mathbb{Z})$. Put $N'=N/\gcd(CD,N)$, $m'=m/\gcd(BC,m)$ and $M=\lcm(N',m')$. Then $f |_k g$ has coefficients in $K_f(\zeta_M)$.

More precisely, let $G'$ be the subgroup of $(\mathbb{Z}/m'\mathbb{Z})^\times$ corresponding to the abelian number field $K_f(\zeta_{N'}) \cap \mathbb{Q}(\zeta_{m'})$, and let $\zeta$ be any $m'$-th root of unity such that
\begin{equation*}
c_{\chi,g} := \sum_{\mu \in G'} \chi(AD-\mu BC) \zeta^\mu
\end{equation*}
is nonzero (such a $\zeta$ always exists). Then $f |_k g$ has coefficient in $c_{\chi,g} \cdot K_f(\zeta_{N'})$.
\end{theorem}

\begin{proof}
The fact that $f | g$ has coefficients in $L=K_f(\zeta_M)$ was proved in Theorem \ref{fg Lchig}. Let $K=K_f(\zeta_{N'})$ and let $\pi_{\chi_g} : L \to L^{\chi_g}$ be the $K$-linear projector associated to the linear character $\chi_g : G \to K^\times$. It is given explicitly by
\[
\pi_{\chi_g}(x) = \frac{1}{|G|}\sum_{\tau\in G} \chi_g(\tau^{-1})\tau(x) \qquad (x \in L).
\]
Since $L$ is generated as a $K$-vector space by the $m'$-th roots of unity, there exists $\zeta \in \mu_{m'}$ such that $\pi_{\chi_g}(\zeta) \neq 0$. We set $c_{\chi,g}=|G|\cdot \pi_{\chi_g}(\zeta)$, so that
\begin{equation*}
c_{\chi,g} = \sum_{\tau \in G} \chi_g(\tau^{-1}) \tau(\zeta) = \sum_{\tau \in G} \chi_g(\tau^{-1}) \zeta^{\lambda(\tau)} = \sum_{\mu \in G'} \chi(AD-\mu BC) \zeta^\mu.
\end{equation*}
By Theorem \ref{fg Lchig}, the modular form $f|g$ has coefficients in $L^{\chi_g} = c_{\chi,g} \cdot K$.
\end{proof}

\begin{remark}
The choice $\zeta=\zeta_{m'}$ does not always work. For example, take the newform $f$ of weight $k=3$, level $N=9$ and character $\chi$ of conductor $m=9$, with $\chi(4)=\zeta_3$ and $\chi(-1)=-1$. We have $K_f=\mathbb{Q}(\chi)=\mathbb{Q}(\zeta_3)$. Taking $g=\sabcd0{-1}13$, we get $N'=3$, $m'=9$ and $G'=\{1,4,7\}$, so that $c_{\chi,g}=0$ for $\zeta=\zeta_9$. On the other hand, for $\zeta=\zeta_9^2$ we get $c_{\chi,g}=3\zeta_9^2$ and $f | g$ indeed has coefficients in $\zeta_9^2 \cdot \mathbb{Q}(\zeta_3) = \langle \zeta_9^2, \zeta_9^5 \rangle_{\mathbb{Q}}$.
\end{remark}

\begin{remark}
Theorem \ref{thm:characters} says that the coefficients of $f|g$ belong to a vector space which has the same dimension as in the $\Gamma_0(N)$ case.
\end{remark}

\begin{remark}\label{rk:composite field}
Theorem \ref{fg Lchig} also shows that the coefficients of $f|g$ lie in the fixed field $L^{\ker \chi_g}$. Let $\chi'_g : G' \to \mathbb{C}^\times$ be the character defined by $\chi'_g(\mu)=\chi(AD-\mu^{-1}BC)$ (using the notations of the proof of Lemma \ref{lemma chig}, we have $\chi'_g = \chi \circ \psi$ and $\chi_g = \chi'_g \circ \lambda$). Then the field $L^{\ker \chi_g}$ is equal to the composite $F \cdot K_f(\zeta_{N'})$, where $F$ is the subfield of $\mathbb{Q}(\zeta_{m'})$ corresponding to the kernel of $\chi'_g$.
\end{remark}

\begin{remark}\label{remark:local Whittaker newforms}
An alternative approach towards proving Theorems \ref{thm:number fields general} and \ref{thm:characters} would be to use local Whittaker newforms as in \cite{corbett-saha}. In particular Proposition 3.3 in \textit{loc.}\ \textit{cit.}\ gives an explicit formula for the Fourier coefficients of $f|g$ in terms of Whittaker newforms and the Galois action on such newforms is described in the proof of Proposition 2.17.
\end{remark}

\section{Atkin-Lehner operators}\label{section:al ops}

For a divisor $Q$ of $N$ with $\gcd(Q,N/Q)=1$ we define the Atkin-Lehner operator on $M_k(\Gamma_1(N))$ as follows. Choose $x,y,z,w\in\mathbb{Z}$ with $x\equiv 1\mod N/Q$ and $y\equiv 1\mod Q$ such that the matrix $W_Q = \sabcd{Qx}y{Nz}{Qw}$ has determinant $Q$. Note that $W_Q = h_Q \sabcd Q001$ with $h_Q=\sabcd{x}{y}{\frac{N}{Q}z}{Qw} \in \mathrm{SL}_2(\mathbb{Z})$. For a modular form $f\in M_k(\Gamma_1(N))$ we have
\[
f|_k W_Q(\tau) = Q^{k/2} \left(f|_k
h_Q\right)(Q\tau).
\]
Therefore we can apply our previous results to find a module that contains the coefficients of $f|h_Q$, or equivalently the coefficients of $Q^{-k/2}f|W_Q$, and reprove a theorem of Cohen.

\begin{corollary}[Theorem 2.6 in \cite{cohen2018}]\label{cor:number field of atkin-lehner}
Let $Q$ be a maximal divisor of $N$ and let $f\in M_k(\Gamma_1(N))$ and $K_f$ be the subfield of $\mathbb{C}$ generated by its Fourier coefficients. Then
\begin{enumerate}
\item\label{item:atkin-lehner 1} The modular form $f|_k W_Q$ has coefficients in $Q^{k/2} \cdot K_f(\zeta_Q)$.
\item\label{item:atkin-lehner 2} If $f\in M_k(\Gamma_0(N),\chi)$ for a character $\chi$ of conductor $m$, then $f|_k W_Q$ has coefficients in $Q^{k/2}G'(\chi_Q) \cdot K_f$, where $G'(\chi_Q)$ is the Gauss sum of the primitive character associated to the $Q$-part of $\chi$.
\end{enumerate}
\end{corollary}
\begin{proof}
The first statement follows directly from Theorem \ref{thm:number fields general}.

Now let $f\in M_k(\Gamma_0(N),\chi)$. We will prove that the coefficients of $f|h_Q$ lie in $G'(\chi_Q)\cdot K_f$, which is equivalent to the second statement. First we determine the character $\chi_{h_Q}$ from the previous section. Splitting $\chi$ as a product of its $Q$-part and its $N/Q$-part we observe
\[
\chi_{h_Q}(\sigma) = \chi_Q(-\frac{N}{Q}zy\lambda(\sigma)^{-1})\chi_{N/Q}(Qwx) = \chi_Q(\lambda(\sigma)^{-1}) = \overline{\chi_Q(\sigma)}.
\]
The conclusion now follows from Theorem \ref{thm:characters}. Since $N' = 1$, $m' = m_Q$, and for $\zeta=\zeta_{m'}$ we have $c_{\chi,h_Q}=G'(\chi_Q)$.
\end{proof}

Let $f\in S_k(\Gamma_0(N),\chi)$ be a newform. Then according to \cite[\S 1]{AtkinLi78} there exists a newform $\tilde{f} \in S_k(\Gamma_0(N),\overline{\chi_Q}\chi_{N/Q})$ and an algebraic number $\lambda_Q(f)$ of absolute value $1$ such that
\begin{equation}\label{fWQ atkin-li}
f|W_Q = \lambda_Q(f) \tilde{f}.
\end{equation}
The number $\lambda_Q(f)$ is called the pseudo-eigenvalue of $f$ at $Q$. By looking at the first Fourier coefficient in (\ref{fWQ atkin-li}), we get the following result.
\begin{corollary}
If $f\in S_k(\Gamma_0(N),\chi)$ is a newform, then the pseudo-eigenvalue $\lambda_Q(f)$ is in $Q^{k/2}G'(\chi_Q) \cdot K_f$ and $\tilde{f}$ has coefficients in $K_f$.
\end{corollary}

This should be compared to the following theorem of Atkin-Li, where an explicit formula for $\lambda_Q(f)$ is derived in a special case.
\begin{theorem}\cite[Theorem 2.1]{AtkinLi78}
Let $f=\sum_n a_n e^{2\pi in\tau} \in S_k(\Gamma_0(N),\chi)$ be a newform, $q$ be a prime dividing $N$ and $Q = N_q$. If $a_q \neq 0$, then
\[
\lambda_Q(f) = Q^{k/2-1} \frac{G(\chi_Q)}{a_Q},
\]
where $G(\chi_Q) = \sum_{u \in (\mathbb{Z}/Q\mathbb{Z})^\times} \chi_Q(u) e^{2\pi iu/Q}$ is the Gauss sum of $\chi_Q$.
\end{theorem}

\section{Optimising the coefficient field}\label{section:optimizing}

We may reduce the number fields provided by Theorems \ref{thm:number fields general} and \ref{thm:characters} as follows. Let $\alpha \in \mathbb{P}^1(\mathbb{Q})$ be a cusp, and let $g = \sabcd ABCD \in \mathrm{SL}_2(\mathbb{Z})$ such that $g \infty = \alpha$. In order to compute the Fourier expansion of $f$ at $\alpha$, we may replace $g$ by $g T^u$ with $T=\sabcd1101$ and $u \in \mathbb{Z}$. Then $f | gT^u$ depends only on the class of $u$ modulo $w$, where $w$ is the width of $\alpha$ on the appropriate modular curve. The following proposition gives the minimal value of the integer $M$ from Theorem \ref{thm:characters} when $u$ varies in $\mathbb{Z}/w\mathbb{Z}$.

\begin{proposition}\label{lem:minimal number field - fixed cusp}
In the notation of Theorem \ref{thm:characters}, let $M'$ be the minimal value of $M$ for $gT^u$ as $u$ varies. Then
\[
M' = \frac{N_C}{\gcd(C,N)} \cdot m_{\overline{C}},
\]
where $N_C=\prod_{p|C}p^{v_p(N)}$ is the $C$-part of $N$ and $m_{\overline{C}} = m/m_C$ is the prime to $C$ part of $m$. Moreover, $M'$ is attained for any $u$ such that $N/N_C$ divides $uC+D$.
\end{proposition}
\begin{proof}
Replacing $g$ with $gT^u$ changes $D$ to $uC+D$ and $B$ to $uA+B$. Varying $u$, we need to determine the minimal value of
\[
M_u = \lcm\left(\frac{N}{\gcd(C(uC+D),N)},\frac{m}{\gcd(C(uA+B),m)}\right).
\]
Since $C$ and $D$ are coprime we have $\gcd(C(uC+D),N)=\gcd(C,N)\gcd(uC+D,N)$, so $N/\gcd(C(uC+D),N)$ is divisible by $N_C/\gcd(C,N)$. Therefore $N_C/\gcd(C,N)$ divides $M_u$.

Let $p\nmid C$. Since $C(uA+B) = A(uC+D)-1$, we have
\begin{align*}
v_p\left(\frac{N}{\gcd(C(uC+D),N)}\right) &= v_p(N) - \min(v_p(uC+D),v_p(N)),\\
v_p\left(\frac{m}{\gcd(C(uA+B),m)}\right) &= v_p(m) - \min(v_p(A(uC+D)-1),v_p(m)).
\end{align*}
If $v_p(uC+D)\neq 0$, then $v_p\left(m/\gcd(C(uA+B),m)\right)=v_p(m)$. On the other hand, if $v_p(uC+D)=0$, then $v_p\left(N/\gcd(C(uC+D),N)\right)=v_p(N)\geq v_p(m)$. In all cases, we have $v_p(M_u) \geq v_p(m)$, which proves that $m_{\overline{C}}$ divides $M_u$. It follows that $M_u$ is always divisible by $N_C/\gcd(C,N) \cdot m_{\overline{C}}$.

Now choose $u$ such that $N_{\overline{C}} = N/N_C$ divides $uC+D$. This is possible because $C$ and $N_{\overline{C}}$ are coprime. Then $\gcd(uC+D,N)=N_{\overline{C}}$ so that
\begin{equation*}
\frac{N}{\gcd(C(uC+D),N)} = \frac{N}{\gcd(C,N) N_{\overline{C}}} = \frac{N_C}{\gcd(C,N)}.
\end{equation*}
Moreover $\gcd(C(uA+B),m)=\gcd(A(uC+D)-1,m)$ is coprime to $N_{\overline{C}}$ and thus divides $m_C$. It follows that
\begin{equation*}
m_{\overline{C}} \mid \frac{m}{\gcd(C(uA+B),m)}
\end{equation*}
On the other hand, if $p\mid C$, then
\begin{equation*}
v_p\left(\frac{m}{\gcd(C(uA+B),m)}\right) \leq v_p\left(\frac{m}{\gcd(C,m)}\right) \leq v_p\left(\frac{N_C}{\gcd(C,N)}\right).
\end{equation*}
Hence $M_u = (N_C/\gcd(C,N)) \cdot m_{\overline{C}}$.
\end{proof}

In practice, we may further reduce the field of coefficients as follows. Let $f \in M_k(\Gamma_0(N))$ be an eigenvector of the Atkin--Lehner operators and $g = \sabcd ABCD \in \mathrm{SL}_2(\mathbb{Z})$. The denominator of the cusp $\alpha=g\infty = A/C$ of $X_0(N)$ is $\delta := \gcd(C,N)$.

Now let $Q$ be a maximal divisor of $N$, and let $W_Q$ be the associated Atkin-Lehner involution of $X_0(N)$. Using the notations of Section \ref{section:al ops}, if $f$ is an eigenvector of $W_Q$ with eigenvalue $\lambda_Q(f) \in \{\pm 1\}$, then we may write
\begin{align*}
f | g &= \lambda_Q(f) f | W_Q g = \lambda_Q(f) f | h_Q\abcd Q001 \abcd ABCD\\
&= \lambda_Q(f) f| h_Q \abcd {\frac{AQ}{\gcd(C,Q)}}s{\frac{C}{\gcd(C,Q)}}{r}\abcd {\gcd(C,Q)}{rBQ-sD}{0}{\frac{Q}{\gcd(C,Q)}}
,
\end{align*}
where $r,s$ are chosen, so that $r\frac{AQ}{\gcd(C,Q)} - s\frac{C}{\gcd(C,Q)}=1$.

The action of the upper triangular matrix $\sabcd {\gcd(C,Q)}{rBQ-sD}{0}{\frac{Q}{\gcd(C,Q)}}$ on Fourier expansions is easily calculated. We now try to find $Q$ such that $f|g'$ has coefficients in the minimal possible number field, where $g'=h_Q\sabcd {\frac{AQ}{\gcd(C,Q)}}s{\frac{C}{\gcd(C,Q)}}{r}\in\mathrm{SL}_2(\mathbb{Z})$.

 Let $\delta = \delta_Q \delta_{\overline{Q}}$ where $\delta_Q$ is the $Q$-part of $\delta$. Then the cusp $\alpha'=g'\infty = W_Q \alpha$ has denominator $\delta' = \frac{Q}{\delta_Q} \delta_{\overline{Q}}$ and we may choose $Q$ such that $M':=N_{\delta'}/\delta'$ is minimal. Explicitly, the choice
 \begin{equation*}
 Q=\prod_{\substack{p | N \\ 0<v_p(\delta)\leq v_p(N)/2}} p^{v_p(N)}
 \end{equation*}
gives the minimal value $M'=\gcd(\delta,N/\delta)$. By Proposition \ref{lem:minimal number field - fixed cusp}, there exists $v \in \mathbb{Z}$ such that the form $f | g' T^v$ has coefficients in $K_f(\zeta_{M'})$.
Thus the problem of finding the Fourier expansion of $f|g$ reduces to finding the the eigenvalue $\lambda_Q(f)\in\{\pm1\}$, calculating the Fourier expansion of $f|g'T^v$ which is over a potentially much smaller number field than that of $f|g$, and finally applying an upper triangular matrix to $f|g'T^v$. Some information, such as the vanishing order of $f|g$ or the absolute value of its Fourier coefficients, can be extracted directly from $f|g'T^v$ without further calculation.

\section{Determining the exact number field}\label{section:determining number field}
Our final goal is to determine the exact coefficient field of $f|g$ when $f$ is a newform. In this section, we assume that $f$ is a newform of (even) weight $k \geq 2$ on the group $\Gamma_0(N)$.

We will need the following theorems of Newman \cite{newman1955,newman} on the congruence subgroup $\Gamma_0(N)$, where $N$ is a fixed integer $\geq 1$.

\begin{theorem}\cite[Theorem 3]{newman1955}\label{newman1}
Every intermediate subgroup between $\Gamma_0(N)$ and $\mathrm{SL}_2(\mathbb{Z})$ is of the form $\Gamma_0(M)$ for some positive divisor $M$ of $N$.
\end{theorem}

In the following, we denote by $R$ the matrix $\sabcd1011$.

\begin{corollary}\label{Gamma0M RM}
Let $M$ be a positive divisor of $N$. The group $\Gamma_0(M)$ is generated by $\Gamma_0(N)$ and $R^M=\sabcd10M1$.
\end{corollary}

\begin{proof}
Let $\Gamma$ be the group generated by $\Gamma_0(N)$ and $R^M$. By Theorem \ref{newman1}, we have $\Gamma=\Gamma_0(M')$ for some $M'$ dividing $N$. Since $R^M \in \Gamma_0(M')$, the integer $M'$ divides $M$. Moreover $\Gamma$ is contained in $\Gamma_0(M)$, so that $M$ divides $M'$. It follows that $\Gamma=\Gamma_0(M)$.
\end{proof}

\begin{theorem}\cite{newman}\label{newman2}
The normaliser of $\Gamma_0(N)$ in $\mathrm{SL}_2(\mathbb{Z})$ is equal to $\Gamma_0(N/s)$, where $s$ is the largest divisor of $24$ such that $s^2$ divides $N$. Moreover, the quotient group $\Gamma_0(N/s)/\Gamma_0(N)$ is cyclic of order $s$, generated by the class of $R^{N/s} = \sabcd10{N/s}1$.
\end{theorem}

\begin{proof}
The first assertion follows from \cite[Theorem 1]{newman}. By Corollary \ref{Gamma0M RM}, the group $\Gamma_0(N/s)$ is generated by $\Gamma_0(N)$ and $R^{N/s}$. It follows that the quotient group $\Gamma_0(N/s)/\Gamma_0(N)$ is generated by the class of $R^{N/s}$, and it is easy to see that this class has order $s$.
\end{proof}

\begin{proposition}\label{Stab f}
Let $F$ be a nonzero element of the new subspace $S_k^{\textrm{new}}(\Gamma_0(N))$. Then its stabiliser
\begin{equation*}
\mathrm{Stab}(F) = \{g \in \mathrm{SL}_2(\mathbb{Z}) : F | g = F\}
\end{equation*}
is equal to $\Gamma_0(N)$.
\end{proposition}

\begin{proof}
Since $\mathrm{Stab}(F)$ contains $\Gamma_0(N)$, Theorem \ref{newman1} implies that $\mathrm{Stab}(F)=\Gamma_0(M)$ for some positive divisor $M$ of $N$. But $F$ belongs to the new subspace, so we must have $M=N$.
\end{proof}

\begin{proposition}\label{fg fsigma}
Let $f$ be a newform of weight $k \geq 2$ on $\Gamma_0(N)$. Let $g \in \mathrm{SL}_2(\mathbb{Z})$ and $\sigma \in \Aut(\mathbb{C})$ such that $f | g = f^\sigma$. Then we have $f^\sigma=f$ and $g \in \Gamma_0(N)$.
\end{proposition}

\begin{proof}
By Proposition \ref{Stab f}, the stabilisers of $f$ and $f^\sigma$ are both equal to $\Gamma_0(N)$. On the other hand $\mathrm{Stab}(f | g) = g^{-1} \mathrm{Stab}(f) g$, so that $g$ normalises $\Gamma_0(N)$. By Theorem \ref{newman2}, we have $g \in \Gamma_0(N/s)$, and there exists an integer $m \in \mathbb{Z}$ such that $\Gamma_0(N) g = \Gamma_0(N) R^{mN/s}$. Hence $f | g = f | R^{mN/s}$.

We now make use of the Atkin-Lehner involution $W_N = \sabcd 0{-1}N0$. Let $w \in \{\pm 1\}$ be the root number of $f$, defined by $f | W_N = w f$. Since $W_N$ is defined over $\mathbb{Q}$, we also have $f^\sigma | W_N = w f^\sigma$. Applying $W_N$ on both sides of the equality $f | g = f^\sigma$, we get
\begin{equation*}
w f^\sigma = f^\sigma | W_N = f|gW_N = f | R^{mN/s} W_N = f | W_N \left(W_N^{-1} R^{mN/s} W_N \right) = w f | R'
\end{equation*}
with $R'=W_N^{-1} R^{mN/s} W_N = \sabcd1{-m/s}01$. The Fourier expansion of $f | R'$ is given by
\begin{equation*}
f | R'(z) = f(z-m/s) = \sum_{n \geq 1} a_n(f) e^{-2\pi imn/s} e^{2\pi inz}.
\end{equation*}
Comparing the first term of the Fourier expansions, we get $e^{-2\pi im/s}=1$. This implies that $s$ divides $m$, $f^\sigma = f$ and $g \in \Gamma_0(N)$.
\end{proof}

We are now in the position to determine the exact number field of $f | g$. This refines Theorem \ref{thm:number fields general}(2) for $\Gamma_0(N)$ newforms.

\begin{theorem}\label{thm:exact nf Gamma0}
Let $f$ be a newform of weight $k \geq 2$ on $\Gamma_0(N)$. Let $g=\sabcd ABCD \in \mathrm{SL}_2(\mathbb{Z})$. Then the field generated by the Fourier coefficients of $f|_k g$ is equal to $K_f(\zeta_{N'})$ with $N'=N/\gcd(CD,N)$.
\end{theorem}

\begin{proof}
We have to show that every automorphism of $\mathbb{C}$ fixing $f|g$ also fixes $K_f(\zeta_{N'})$. Let $\sigma \in \Aut(\mathbb{C})$ such that $(f | g)^\sigma = f |g$. Define $\lambda \in (\mathbb{Z}/N\mathbb{Z})^\times$ by $\sigma(\zeta_N) = \zeta_N^\lambda$. By Theorem \ref{fg sigma}, we have $f^\sigma | g_\lambda = f | g$, so that $f^\sigma = f | g g_\lambda^{-1}$. By Proposition \ref{fg fsigma}, we have $f^\sigma = f$ and $g g_\lambda^{-1} \in \Gamma_0(N)$. It follows that $\sigma$ fixes $K_f$, and we have already seen in the proof of Theorem \ref{thm:number fields general} that the condition $g g_\lambda^{-1} \in \Gamma_0(N)$ is equivalent to $\lambda \equiv 1 \mod{N'}$. Therefore $\sigma$ fixes $K_f(\zeta_{N'})$.
\end{proof}
\begin{remark}
An inspection of the proofs shows that Proposition \ref{fg fsigma} and Theorem \ref{thm:exact nf Gamma0} are valid for elements $f=\sum a_n q^n$ of the new subspace of $S_k(\Gamma_0(N))$ that are eigenfunctions of $W_N$ and satisfy the following condition: if $s$ denotes the largest divisor of $6$ whose square divides $N$, then there exists $n\in\mathbb{N}$ which is coprime to $s$ such that $a_n$ is a non-zero rational number. One family of such forms is given by the traces $\sum_\sigma f^\sigma$ of newforms $f$, where the sum is over all embeddings of $K_f$ into $\mathbb{C}$.
\end{remark}
\begin{question}
What is the $\mathbb{Q}$-vector space (or $K_f$-vector space) generated by the Fourier coefficients of $f | g$? Furthermore, can we bound effectively the denominators of these coefficients? Note that the $q$-expansion principle implies that the Fourier expansion of $f|g$ lies in $\mathbb{Z}[[q^{1/N}]] \otimes K_f(\zeta_{N'})$, so that the denominators of $f|g$ are indeed bounded. In fact, by Remark \ref{rem SL2 action}, we know that the denominators of $f|g$ divide some fixed power of $N$.
\end{question}

\begin{question}
It would be interesting to generalise Theorem \ref{thm:exact nf Gamma0} to newforms with non-trivial character. Is the number field provided by Remark \ref{rk:composite field} best possible?
\end{question}

\section{Appendix: Algebraic modular forms}
\label{section:appendix}

Here we recall the theory of algebraic modular forms, in order to give a second proof of Theorem \ref{fg sigma}. For more details on this theory, see \cite[Chap. II]{katz-interpolation} and the references therein.

\begin{definition} Let $R$ be an arbitrary commutative ring, and let $N \geq 1$ be an integer. A \emph{test object} of level $N$ over $R$ is a triple $T=(E,\omega,\beta)$ where $E/R$ is an elliptic curve, $\omega \in \Omega^1(E/R)$ is a nowhere vanishing invariant differential, and $\beta$ is a \emph{level $N$ structure} on $E/R$, that is an isomorphism of $R$-group schemes
\begin{equation*}
\beta : (\mu_N)_R \times (\mathbb{Z}/N\mathbb{Z})_R \xrightarrow{\cong} E[N]
\end{equation*}
satisfying $e_N(\beta(\zeta,0),\beta(1,1)) = \zeta$ for every $\zeta \in (\mu_N)_R$. Here $\mu_N = \operatorname{Spec} \mathbb{Z}[t]/(t^N-1)$ is the scheme of $N$-th roots of unity, and $e_N$ is the Weil pairing on $E[N]$ \footnote{Our definition of the Weil pairing is the reciprocal of Silverman's definition \cite[III.8]{silverman:AEC}. With our definition, we have $e_N(1/N,\tau/N)=e^{2\pi i/N}$ on the elliptic curve $\mathbb{C}/(\mathbb{Z}+\tau\mathbb{Z})$ with $\mathrm{Im}(\tau)>0$.}.
\end{definition}
If $\phi : R \to R'$ is a ring morphism, we denote by $T_{R'} = (E_{R'},\omega_{R'},\beta_{R'})$ the base change of $T$ to $R'$ along $\phi$.

The isomorphism classes of test objects over $\mathbb{C}$ are in bijection with the set of lattices $L$ in $\mathbb{C}$ endowed with a symplectic basis of $\frac1N L/L$ \cite[2.4]{katz-interpolation}. Another example is given by the Tate curve $\mathrm{Tate}(q) = \mathbb{G}_m/q^\mathbb{Z}$ \cite[\S 8]{deligne:formulaire}. It is an elliptic curve over $\mathbb{Z}((q))$ endowed with the canonical differential $\omega_\mathrm{can}=dx/x$ and the level $N$ structure $\beta_\mathrm{can}(\zeta,n)= \zeta q^{n/N} \mod{q^\mathbb{Z}}$. The test object $(\mathrm{Tate}(q),\omega_\mathrm{can},\beta_\mathrm{can})$ is defined over $\mathbb{Z}((q^{1/N}))$.

\begin{definition} An \emph{algebraic modular form} of weight $k \in \mathbb{Z}$ and level $N$ over $R$ is the data, for each $R$-algebra $R'$, of a function
\begin{equation*}
F = F_{R'} : \{\textrm{isomorphism classes of test objects of level $N$ over $R'$}\} \to R'
\end{equation*}
satisfying the following properties:
\begin{enumerate}
\item \label{AlgMF cond 1} $F(E,\lambda^{-1} \omega,\beta) = \lambda^k F(E,\omega,\beta)$ for every $\lambda \in (R')^\times$;
\item \label{AlgMF cond 2} $F$ is compatible with base change: for every morphism of $R$-algebras $\psi : R' \to R''$ and for every test object $T$ of level $N$ over $R'$, we have $F_{R''}(T_{R''}) = \psi(F_{R'}(T))$.
\end{enumerate}
We denote by $M_k^{\mathrm{alg}}(\Gamma(N);R)$ the $R$-module of algebraic modular forms of weight $k$ and level $N$ over $R$.
\end{definition}
 Evaluating at the Tate curve provides an injective $R$-linear map
\begin{equation*}
M_k^{\mathrm{alg}}(\Gamma(N);R) \hookrightarrow \mathbb{Z}((q^{1/N})) \otimes_\mathbb{Z} R
\end{equation*}
called the $q$-expansion map. The \emph{$q$-expansion principle} states that if $R'$ is a subring of $R$, then an algebraic modular form $F \in M_k^\mathrm{alg}(\Gamma(N);R)$ belongs to $M_k^\mathrm{alg}(\Gamma(N);R')$ if and only if the $q$-expansion of $F$ has coefficients in $R'$.

Algebraic modular forms are related to classical modular forms as follows. To any algebraic modular form $F \in M_k^\mathrm{alg}(\Gamma(N);\mathbb{C})$, we associate the function $F^\mathrm{an} : \h \to \mathbb{C}$ defined by
\begin{equation*}
F^\mathrm{an}(\tau) = F\Bigl(\frac{\mathbb{C}}{2\pi i\mathbb{Z}+2\pi i\tau\mathbb{Z}},dz,\beta_\tau \Bigr)
\end{equation*}
with $\beta_\tau(\zeta_N^m,n):=[2\pi i(m+n\tau)/N]$.
\begin{proposition}
The map $F \mapsto F^\mathrm{an}$ induces an isomorphism between $M_k^\mathrm{alg}(\Gamma(N);\mathbb{C})$ and the space $M_k^!(\Gamma(N))$ of weakly holomorphic modular forms on $\Gamma(N)$ (that is, holomorphic on $\h$ and meromorphic at the cusps). Moreover, the $q$-expansion of $F$ coincides with that of $F^\mathrm{an}$.
\end{proposition}

We now interpret the action of $\mathrm{SL}_2(\mathbb{Z})$ on modular forms in algebraic terms. Let $F \in M_k^\mathrm{alg}(\Gamma(N);\mathbb{C})$ with $f=F^\mathrm{an}$, and let $g = \sabcd abcd \in \mathrm{SL}_2(\mathbb{Z})$. A simple computation shows that
\begin{equation}\label{fg F}
(f |_k g)(\tau) = F\Bigl(\frac{\mathbb{C}}{2\pi i(\mathbb{Z}+\tau \mathbb{Z})},dz,\beta'_\tau\Bigr)
\end{equation}
where the level $N$ structure $\beta'_\tau$ is given by
\begin{equation}\label{beta' beta}
\beta'_\tau(\zeta_N^m,n) = \beta_\tau(\zeta_N^{md+nb},mc+na).
\end{equation}
Let $\psi : (\mathbb{Z}/N\mathbb{Z})^2 \to \mu_N(\mathbb{C}) \times \mathbb{Z}/N\mathbb{Z}$ be the isomorphism defined by $\psi(a,b)=(\zeta_N^b,a)$. Let us identify the level structure $\beta_\tau$ (resp. $\beta'_\tau$) with the map $\alpha_\tau = \beta_\tau \circ \psi$ (resp. $\alpha'_\tau = \beta'_\tau \circ \psi$). Then (\ref{beta' beta}) shows that
\begin{equation}
\alpha'_\tau(a,b)=\alpha_\tau((a,b)g).
\end{equation}
What we have here is the right action of $\mathrm{SL}_2(\mathbb{Z})$ on the row space $(\mathbb{Z}/N\mathbb{Z})^2$, which induces a left action on the set of level $N$ structures. As we will see, all this makes sense algebraically. For any $\mathbb{Z}[\zeta_N]$-algebra $R$, we denote by $\zeta_{N,R}$ the image of $\zeta_N=e^{2\pi i/N}$ under the structural morphism $\mathbb{Z}[\zeta_N] \to R$.

\begin{lemma}
If $R$ is a $\mathbb{Z}[\zeta_N,1/N]$-algebra, then there is an isomorphism of $R$-group schemes $(\mathbb{Z}/N\mathbb{Z})_R \xrightarrow{\cong} (\mu_N)_R$ sending $1$ to $\zeta_{N,R}$.
\end{lemma}

\begin{proof}
Note that $(\mu_N)_R = \operatorname{Spec} R[t]/(t^N-1) = \operatorname{Spec} R[\mathbb{Z}/N\mathbb{Z}]$ and $(\mathbb{Z}/N\mathbb{Z})_R = \operatorname{Spec} R^{\mathbb{Z}/N\mathbb{Z}}$. If $R=\mathbb{C}$, then $\mathbb{C}[\mathbb{Z}/N\mathbb{Z}] \cong \mathbb{C}^{\mathbb{Z}/N\mathbb{Z}}$ because all irreducible representations of $\mathbb{Z}/N\mathbb{Z}$ have dimension 1. This isomorphism $\mathcal{F}_\mathbb{C}$ is given by the Fourier transform, and both $\mathcal{F}_\mathbb{C}$ and $\mathcal{F}_\mathbb{C}^{-1}$ have coefficients in $\mathbb{Z}[\zeta_N,1/N]$ with respect to the natural bases. It follows that in general $R[\mathbb{Z}/N\mathbb{Z}] \cong R^{\mathbb{Z}/N\mathbb{Z}}$ and this isomorphism sends $[1]$ to $(\zeta_{N,R}^a)_{a \in \mathbb{Z}/N\mathbb{Z}}$.
\end{proof}

Let $R$ be a $\mathbb{Z}[\zeta_N,1/N]$-algebra. We have an isomorphism of $R$-group schemes
\begin{equation*}
\psi_R : (\mathbb{Z}/N\mathbb{Z})^2_R \to (\mu_N)_R \times (\mathbb{Z}/N\mathbb{Z})_R
\end{equation*}
given by $\psi_R(a,b)=(\zeta_{N,R}^b,a)$. The group $\mathrm{SL}_2(\mathbb{Z})$ acts from the right on the row space $(\mathbb{Z}/N\mathbb{Z})^2_R$ by $R$-automorphisms, and for $\alpha : (\mathbb{Z}/N\mathbb{Z})^2_R \xrightarrow{\cong} E[N]$ we define
\begin{equation}\label{action g alpha}
(g \cdot \alpha)(a,b)=\alpha((a,b)g) \qquad ((a,b) \in (\mathbb{Z}/N\mathbb{Z})^2).
\end{equation}
Using $\psi_R$, we transport this to a left action of $\mathrm{SL}_2(\mathbb{Z})$ on the set of level $N$ structures of an elliptic curve over $R$. Given a test object $T=(E,\omega,\beta)$ over $R$, we define $g \cdot T := (E,\omega,g\cdot \beta)$. For any $F \in M_k^\mathrm{alg}(\Gamma(N);R)$, we define $F|g \in M_k^\mathrm{alg}(\Gamma(N);R)$ by the rule $(F|g)(T) = F(g \cdot T)$ for any test object $T$ over any $R$-algebra $R'$. The computation (\ref{fg F}) then shows that the right action of $\mathrm{SL}_2(\mathbb{Z})$ on $M_k^{\mathrm{alg}}(\Gamma(N);\mathbb{C})$ corresponds to the usual slash action on $M_k^!(\Gamma(N))$.

\begin{remark}\label{rem SL2 action}
The action of $\mathrm{SL}_2(\mathbb{Z})$ on algebraic modular forms over $\mathbb{Z}[\zeta_N,1/N]$-algebras has the following consequence: if a classical modular form $f \in M_k^!(\Gamma(N))$ has Fourier coefficients in some subring $A$ of $\mathbb{C}$, then for any $g \in \mathrm{SL}_2(\mathbb{Z})$, the Fourier expansion of $f|g$ lies in $\mathbb{Z}[[q^{1/N}]] \otimes A[\zeta_N,1/N]$.
\end{remark}

We now interpret the action of $\Aut(\mathbb{C})$ in algebraic terms (see \cite[p. 88]{ohta}). Let $\sigma \in \Aut(\mathbb{C})$. For any $\mathbb{C}$-algebra $R$, we define $R^\sigma := R \otimes_{\mathbb{C},\sigma^{-1}} \mathbb{C}$, which means that $(ax) \otimes 1 = x \otimes \sigma^{-1}(a)$ for all $a \in \mathbb{C}$, $x \in R$. We endow $R^\sigma$ with the structure of a $\mathbb{C}$-algebra using the map $a \in \mathbb{C} \mapsto 1 \otimes a \in R^\sigma$. We denote by $\phi_\sigma : R \to R^\sigma$ the map defined by $\phi_\sigma(x)=x \otimes 1$. The map $\phi_\sigma$ is a ring isomorphism, but one should be careful that $\phi_\sigma$ is not a morphism of $\mathbb{C}$-algebras, as it is only $\sigma^{-1}$-linear. For any test object $T$ over $R$, we denote by $T^\sigma$ its base change to $R^\sigma$ using the ring morphism $\phi_\sigma$.

Let $F \in M_k^{\mathrm{alg}}(\Gamma(N);\mathbb{C})$ be an algebraic modular form. For any $\mathbb{C}$-algebra $R$, we define
\begin{align*}
F^\sigma_R : \{\textrm{isomorphism classes of test objects of level $N$ over $R$}\} & \to R \\
T & \mapsto \phi_\sigma^{-1}\bigl(F_{R^\sigma}(T^\sigma)\bigr).
\end{align*}
One may check that the collection of functions $F^\sigma_R$ satisfies the conditions (\ref{AlgMF cond 1}) and (\ref{AlgMF cond 2}) above, hence defines an algebraic modular form $F^\sigma \in M_k^\mathrm{alg}(\Gamma(N);\mathbb{C})$. Moreover, since the Tate curve is defined over $\mathbb{Z}((q))$, one may check that the map $F \mapsto F^\sigma$ corresponds to the usual action of $\Aut(\mathbb{C})$ on the Fourier expansions of modular forms: for every $F \in M_k^\mathrm{alg}(\Gamma(N);\mathbb{C})$ and every $\sigma \in \Aut(\mathbb{C})$, we have $(F^\sigma)^\mathrm{an} = (F^\mathrm{an})^\sigma$.

We finally come to the second proof of Theorem \ref{fg sigma}.

\begin{proof}
Let $f \in M_k(\Gamma(N))$ with corresponding algebraic modular form $F \in M_k^{\mathrm{alg}}(\Gamma(N);\mathbb{C})$. Let $g \in \mathrm{SL}_2(\mathbb{Z})$ and $\sigma \in \Aut(\mathbb{C})$. We take as test object $T=(\mathrm{Tate}(q),\omega_\mathrm{can},\beta_{\mathrm{can}})$ over $R=\mathbb{Z}((q^{1/N})) \otimes \mathbb{C}$. Since a modular form is determined by its Fourier expansion, and unravelling the definitions of $F|g$ and $F^\sigma$, it suffices to check that the test objects $g \cdot T^\sigma$ and $(g_\lambda \cdot T)^\sigma$ over $R^\sigma$ are isomorphic. Since $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$ acts only on the level structures of the test objects, we have to show that
\begin{equation}\label{to show}
g \cdot \beta_\mathrm{can}^\sigma \cong (g_\lambda \cdot \beta_\mathrm{can})^\sigma.
\end{equation}
For any scheme $X$ over $R$, let $X^\sigma$ denote its base change to $R^\sigma$ along $\phi_\sigma$. Since $\phi_\sigma$ is a ring isomorphism, the canonical projection map $X^\sigma \to X$ is an isomorphism of schemes, and we also denote by $\phi_\sigma : X \to X^\sigma$ the inverse map.

Put $E=\mathrm{Tate}(q)$ and $\beta = \beta_\mathrm{can}$. Let $\alpha = \beta \circ \psi_R : (\mathbb{Z}/N\mathbb{Z})^2_R \xrightarrow{\cong} E[N]$. By functoriality, the level structure $\beta^\sigma$ is given by the following commutative diagram
\begin{equation}\label{big diagram}
\begin{tikzcd}
(\mathbb{Z}/N\mathbb{Z})^2_R \arrow[dotted]{d}[swap]{\gamma} \arrow{r}{\psi_R} \arrow[bend left=20]{rr}{\alpha} & (\mu_N)_R \times (\mathbb{Z}/N\mathbb{Z})_R \arrow{r}{\beta} \arrow{d}{\phi_\sigma}[swap]{\cong} & E[N] \arrow{d}{\phi_\sigma}[swap]{\cong} \\
(\mathbb{Z}/N\mathbb{Z})^2_{R^\sigma} \arrow{r}{\psi_{R^\sigma}} \arrow[bend right=20]{rr}{\alpha^\sigma} & (\mu_N)_{R^\sigma} \times (\mathbb{Z}/N\mathbb{Z})_{R^\sigma} \arrow{r}{\beta^\sigma} & E^\sigma[N].
\end{tikzcd}
\end{equation}
Let us compute the dotted arrow $\gamma$. Since $\phi_\sigma$ is $\sigma^{-1}$-linear, we have $\phi_\sigma(\zeta_{N,R}) = \zeta_{N,R^\sigma}^{\lambda^{-1}}$. It follows that
\begin{equation}
\phi_\sigma(\psi_R(a,b)) = \phi_\sigma(\zeta_{N,R}^b,a) = (\zeta_{N,R^\sigma}^{\lambda^{-1}b},a) = \psi_{R^\sigma}(a,\lambda^{-1}b)
\end{equation}
so that $\gamma(a,b)=(a,\lambda^{-1}b)$. We may thus express $\alpha^\sigma$ in terms of $\alpha$ by
\begin{equation}\label{alpha sigma}
\alpha^\sigma(a,b) = \phi_\sigma \circ \alpha \circ \gamma^{-1}(a,b) = \phi_\sigma \circ \alpha(a,\lambda b) = \phi_\sigma \circ \alpha \biggl((a,b) \abcd 100\lambda \biggr).
\end{equation}
Let us make explicit both sides of (\ref{to show}). By (\ref{action g alpha}) and (\ref{alpha sigma}), the left hand side is given by
\begin{equation}
(g \cdot \alpha^\sigma)(a,b) = \alpha^\sigma((a,b) g) = \phi_\sigma \circ \alpha \biggl((a,b) g \abcd 100\lambda\biggr).
\end{equation}
Let us now turn to the right hand side of (\ref{to show}). By (\ref{action g alpha}), we have $(g_\lambda \cdot \alpha)(a,b) = \alpha((a,b)g_\lambda)$. Applying the commutative diagram (\ref{big diagram}) with $\alpha$ replaced by $g_\lambda \cdot \alpha$, we get
\begin{equation}
(g_\lambda \cdot \alpha)^\sigma(a,b) = \phi_\sigma \circ (g_\lambda \cdot \alpha) \biggl((a,b) \abcd 100\lambda \biggr) = \phi_\sigma \circ \alpha \biggl((a,b) \abcd 100\lambda g_\lambda \biggr).
\end{equation}
Finally, we recall equation \eqref{eq g glambda}, which states $g \sabcd 100\lambda  = \sabcd 100\lambda g_\lambda$.

\end{proof}
\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


If you are a blogger like me and you have an iPhone then these iPhone apps are for you. Now a day most of the bloggers operate their blogs from their mobile and if you are one of the mobile blogger then have a look to the following list which I have compiled specially for the bloggers.These apps will keep you updated about every information about your blogs/websites right from your iPhone/iPod touch.
Here is my definitive list of the top must have apps that every blogger must know or have on their iPhone and/or iPod Touch.
A free RSS reader application without ads. Simply add your feeds, then manage and read them. Feed caching support for offline reading and unread items indicator.
Evernote for iPhone lets you create notes, snap photos, and record voice memos that you can then access any time from your iPhone, computer, or the web.
Colorpedia is a tool that will help you make good color decisions. It uses color names from the Color Naming Experiment , making Colorpedia a rich and accurate source of the most commonly used color names.
Analytics App is the premier app for Google Analytics . Upto 55 reports, showing data up to 100 rows and more reports and data than any other Analytics app also Supports multiple logins (new!) and multiple accounts.
Dropbox is the easiest way to sync and share your files online and across computers.
TwitBird is a fast, flexible, full-featured twitter client for iPhone and iPod Touch. It has a lot of cool features that you see in other twitter apps but rarely together. It offers everything you'd expect for a good twitter client, from your timeline to search.
Send money to your friends, manage your account, and more, with the PayPal app. It's free, secure and more convenient than going to the ATM, writing checks, or sending gifts the traditional way.
People are using Simplenote to keep notes, lists, ideas and more. Your notes automatically synchronize with your computer and all your devices. It's extremely easy to use. For those who want more power, you can also use tags, pins, versions, and sharing.
It's easy to manage your WordPress blog or site from your iOS device. With WordPress for iOS, you can moderate comments, create or edit posts and pages, and add images or videos with ease.
FREE 2-way video calls, voice calls and live chat for your iPhone and iPod touch. Make 2 way Video calls to other fringsters on any advanced Smartphone (iPhone, Android, Nokia or other) on 3G and WiFi. Connect with your fring buddies and other popular social networks: Google Talk, MSN Messenger, Yahoo!, Facebook, Twitter & SIP!
If you know any other iPhone app that you think which will be useful for bloggers then do share with us via comments. I'm sure that there are others I've missed.
.-= Novacep´s last blog ..Turkcell T10 =-.
Its so informative and helpful to know about the mobile apps. Tanx for such a great info.


Hats off to Linda Lamb, partner within the family law team, who has boldly undertaken to do a sky dive for the charity Help for Heroes.
Linda's challenge takes place on Sunday April 18 when she will jump 12,000 metres from a small plane with just a parachute for company.
Linda will be accompanied in her endeavour by a team of teachers and pupils from Eastbourne College. They are hoping to collectively raise £15,000 for Help for Heroes. All funds will go directly to the charity which provides practical direct support to those servicemen and women wounded in the line of duty in the current conflicts in Iraq and Afghanistan.
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text:
import {
  resolve as resolvePath,
} from "path";

import {
  default as webpack,
} from "webpack";

import {
  default as ExtractTextPlugin,
} from "extract-text-webpack-plugin";

let FILENAME_FORMAT;
let BABEL_PLUGINS;
let PRODUCTION_PLUGINS;

if (process.env.NODE_ENV === `production`) {
  FILENAME_FORMAT = `[name]-[chunkhash].js`;
  BABEL_PLUGINS = [];
  PRODUCTION_PLUGINS = [
    // Same effect as webpack -p
    new webpack.optimize.UglifyJsPlugin(),
    new webpack.optimize.OccurenceOrderPlugin(),
  ];
} else {
  // When HMR is enabled, chunkhash cannot be used.
  FILENAME_FORMAT = `[name].js`;
  BABEL_PLUGINS = [
    [
      `react-transform`,
      {
        transforms: [
          {
            transform: `react-transform-hmr`,
            imports: [`react`],
            locals: [`module`],
          }, {
            transform: `react-transform-catch-errors`,
            imports: [`react`, `redbox-react`],
          },
        ],
      },
    ],
  ];
  PRODUCTION_PLUGINS = [];
}

export default {
  devServer: {
    port: 8080,
    host: `localhost`,
    contentBase: resolvePath(__dirname, `../../public`),
    publicPath: `/assets/`,
    hot: true,
    stats: { colors: true },
  },
  output: {
    path: resolvePath(__dirname, `../../public/assets`),
    pathinfo: process.env.NODE_ENV !== `production`,
    publicPath: `assets/`,
    filename: FILENAME_FORMAT,
  },
  module: {
    loaders: [
      {
        test: /\.woff(2)?(\?v=[0-9]\.[0-9]\.[0-9])?$/,
        loader: "url?limit=10000&minetype=application/font-woff"
      },
      {
        test: /\.(ttf|eot|svg)(\?v=[0-9]\.[0-9]\.[0-9])?$/,
        loader: "file"
      },
      {
        test: /\.scss$/,
        loader: ExtractTextPlugin.extract(`style`, `css!sass`, {
          publicPath: ``,
        }),
      },
      {
        test: /\.js(x?)$/,
        exclude: /node_modules/,
        loader: `babel`,
        query: {
          plugins: BABEL_PLUGINS,
        },
      },
    ],
  },
  plugins: [
    new webpack.EnvironmentPlugin(`NODE_ENV`),
    new ExtractTextPlugin(`[name]-[chunkhash].css`, {
      disable: process.env.NODE_ENV !== `production`,
    }),
    ...PRODUCTION_PLUGINS,
  ],
};

redpajama_set_name: RedPajamaGithub


Allergens are substances that can provoke allergic reactions. An allergic reaction is caused when the immune system of the body reacts in a hypersensitive way to certain substances (allergens). Indoor allergens can come from amongst others dust mites, pets, mould and pollen.
Dust mites are located in (older) mattresses, upholstered furniture, carpets and on mouldy walls. The excrements of dust mites contain allergens.
An allergy to animal hair is an allergic reaction that is caused by the contact with an animal. Pets spread hair and skin scales which are so light that they can hang in the air for days. Dog and cat hair allergies are the most common allergies against animal hair, but guinea pigs, rabbits and birds feathers can also cause them.
An allergy against mould can be detected through different ailments that are caused by the inhalation of mould spores. The symptoms of such an allergy are sneezing attacks, a runny nose, coughing and asthma.
Pollen can be spread by the wind and represent the main problem for allergic persons since anemophilous plants often free big.
amounts of pollen into the air which is then disseminated across long distances through the wind and can be blown from the outside inside housings. Hence the pollen concentration can periodically be very high and can strongly impede the respiratory system of people suffering from this allergy.
Each year, hay fever season causes unpleasant ailments such as sneezing, a running nose and itchy eyes. Up to 20% of people suffer from allergic rhinitis whose most common form is called hay fever and starts at a very young age and represents an increasing trend.
Up to 80% of people which suffer from asthma also suffer from allergies such as hay fever. Unfortunately for asthma patients, these are not the only ailments they suffer from. Because asthma and hay fever are closely connected, asthma patients do not only suffer from hay fever but experience an increase in asthma symptoms. The nose and the lungs form our respiratory system. When our upper respiratory tract is irritated by pollen, it can influence the lungs and aggravate asthma problems. The combination of asthma and allergies is often underestimated.
At the same time spring means the kick-off of hay fever season and the first pollen (birch, oak, hornbeam, hazel tree and alder) fly around. People with a pollen allergy know about these problems and avoid to go outside if their work permits it.
Clean Air Optima air purifiers free indoor air from pollen!
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Galexia is undertaking a Second Privacy Impact Assessment (PIA) for the Digital Transformation Agency (DTA) on the proposal to establish the Trusted Digital Identity Framework (TDIF).
The purpose of this PIA is to assist in identifying and managing privacy issues that are raised by the establishment of the TDIF.
Individual PIAs for other accredited TDIF Participants (such as the Identity Exchange, Attribute Providers and Credential Providers) (as required).
This PIA is the second public PIA undertaken in relation to the TDIF. Many issues were the subject of findings and recommendations in the first PIA.
This PIA considers compliance with privacy legislation and relevant privacy measures contained in the TDIF documentation. The PIA also briefly considers issues around overall privacy management and governance. The currency of this PIA is as at end of September 2018.


Common Tags, Elements, Attributes, Formatting, Links, Head, Images, Tables, Lists, Layout, Blocks, Forms, Iframe, URL Encode.
Convas, Drag/Drop, Geolocation, Audio/Video, Web Storage, Form elements, Form attributes.
Dimension, Display, Positioning, Align, Navigational Bar, Image Gallary, Media Types.
Introduction , Functions, Conditions, Loops, JS Objects, JS Windows, JS Libraries.
Introduction, Selectors, Events, Hide/Show, Slide, Animate, Stop(), Set, Get, Add, Remove, JQuery Traversing, JQuery AJAX.
*	Material Design colors *	Material design layouts * New concepts in material design .
*	Accelerometer *	Orientation *	Camera .
Deploying JAVA Beans in a JSP Page. Introduction to struts framework.
SELECT, WHERE, AND & OR, ORDER BY, INSERT INTO, UPDATE , DELETE etc.
LIKE, IN, BETWEEN, JOIN, INNER JOIN, LEFT JOIN , RIGHT JOIN, FULL JOIN.
Constraints, Unique, Primary Key, Foreign Key, Delete Drop Alter, View, Date etc.
COUNT(), MAX(), MIN(), FIRST(), LAST(), GROUPBY, HAVING, UCASE(), LCASE(), MID(), LEN(), NOW() and more.
Introduction of Creating DLL in VB.
For Final Year BBA/MBA Student Live Project Training/Internship with all concept Learning in Retail, Finance etc.


text:
The Restoration of In re Winship: A Comment on Burdens of Persuasion in Criminal Cases After Patterson v. New York
Ronald J. Allen, State University of New York at BuffaloFollow
At the conclusion of its last term, the Supreme Court rendered what should have been a most unremarkable decision. In Patterson v. New York, the Court upheld New York's affirmative defense of extreme emotional disturbance, which requires a defendant who seeks to reduce his offense from murder to manslaughter to prove by a preponderance of the evidence that he acted under extreme emotional disturbance. Had the case come before the Court seven years earlier, it could have been swiftly dispatched with a brief opinion upholding the New York statute on the grounds that the issue of extreme emotional disturbance does not arise until the state proves beyond a reasonable doubt the "essential elements" of the crime intent and causation--and that "extreme emotional disturbance" is but a slightly modified version of the defense of provocation, for which many states had long placed the burden of proof upon the defendant.
Ronald J. Allen, The Restoration of In re Winship: A Comment on Burdens of Persuasion in Criminal Cases After Patterson v. New York, 76 Mich. L. Rev. 30 (1977).
Criminal Law Commons, Jurisprudence Commons, Supreme Court of the United States Commons

redpajama_set_name: RedPajamaCommonCrawl


Inspired by sportswear, this varsity jacket-style cardigan is distinguished by its colour scheme. KENZO's must-have emblem, the Tiger head, adorns the chest giving this piece a bold graphic look.
77% cotton, 14% polyester, 9% polyamide.


These enlargers are used and may have minor cosmetic marks from previous use, but have been reconditioned to factory specifications and are in perfect working condition. Supplied with 120V 75W opal enlarging lamp, condenser lenses, filter drawer, red safety filter, and instruction manual. Lenses and negative carriers to suit your requirements can be purchased separately.


Create galleries by searching on Flickr.
It's fully automatic and brings relevant, fresh images to your site.
Justified Image Grid allows you to create a gallery based on a predefined Flickr search. Various factors are available to narrow your results and ensure that the content is relevant. A useful choice is to only search among your photos. By using PHP template tags one could connect to the WordPress search results to let visitors perform Flickr searches in their online portfolio. Match individual pictures... or photos by tag - from your Photosteam or through a global Flickr search. It's a fun and quick way to populate galleries.
This gallery looks for photos taken at Blizzcon, but only those that are allowed to be shown by their licensing. They are sorted by relevance.
Onlookers admire the original conceptual art from several Blizzard games.
Taxis await outside the Anaheim Convention Center.
You can't go wrong with this one, since you manage your own portfolio and all the results will be just what you expect. Flickr allows tagging your photos and this is where you can take advantage of that. See the same thing on Flickr: Lake by Pircho.
By searching for an image you can pick individual or a small group of photos from Flickr. This photo is called Fiery.
How does global Flickr search work and what are the options?
Just enter your Flickr search query and set up narrowing options. Your gallery will update itself as results change.
Perform a Flickr search by arbirtary text or just by tags.
Set tags matching mode to ALL or ANY.
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Only show those that are geotagged or not.
Restrict results by only looking for content with a certain license.
Really great plugin, worth the money. Ever since PhotoSwipe 4 was added as lightbox, JIG is my favourite masonry gallery plugin. Thanks a lot for the great work. Keep it up!
This plugin not only solves the issue of badly cropped gallery images, it gives you an enormous amount of control over how the gallery looks and functions. It's worth every cent for that alone. Add to that the awesome, prompt and cheerful support from Firsh and you really have a 6 star product.... but I'm limited to 5.


Unique ANCSL Club - A mystery club?
Lovely beech headed scare neck driver. Shaft also stamped J&D Clark , Musselburgh. Lovely club with character, great for display.
This is a rare split-hosel mashie by Gourlay. All original condition. Speaks for itself really! Few still extant in the world.


Are we watching Frank's dream? Or did he really push Cathy Durant down the steps, and did Claire really poison Tom Yates, and did they each coerce Doug into copping to homicide over dinner, and did Frank really go full monologist in entrance of the Senate Judiciary committee before saying his resignation? Cards is hyperreality, in fact, but much of this hour felt like a power fantasy unfurling within the head of the embattled commander-in-chief. And the filmmakerly execution here (episode director: Robin Wright) gave each Stunning Development a hint of surrealism. I want to promote it as a result of I slipped on ice and broke my leg the day after New Years. It is all I can do to get up the stairs on crutches simply to go to the physician, and I've to have assist doing that. That is no place for the geriatric set. Mr. B wants to maintain it if we go to the difficulty to fix it. The problem is, after his knee replacement, he was able to stroll when was released from the hospital. I nonetheless can't stroll without assist. It is vitally frustrating once I look at all these stairs just to get to my automotive.
I began with the web. This was 3 years ago and the research was lengthy. I started on the lookout for information on house coaching a duck, what to feed a duck, duck as pets, and far more. It was very laborious to search out information on Domestic Duck Care. Now, over the three years quite a bit has changed. Pet Duck Care and Duck Diapers are simple to find in the major search engines. I suppose that more individuals are elevating Geese for Pets. But, take my recommendation; there are all sorts of various opinions and recommendation for Home Duck and Goose Care. I did some learning of my own on these sites and providers that at the moment are simpler to seek out. I must say that, I solely belief the care and advice from the source that I first discovered.
Despite the fact that Camille only contributes to the show part-time, she still holds our curiosity. Her greatest claim to fame was her marriage to Kelsey Grammar, the star of widespread, long-working sitcoms Cheers and Frasier. Their subsequent divorce, and not using a pre-nup, left her with quite a fortune: $50 million! Before she met Kelsey (who's value $a hundred and twenty million), Camille was an actress and dancer. Throughout their marriage she worked as a author and producer for their company, Grammnet Productions. Now, it seems, she is content to raise her children and be a Real Housewife.
On the entrance, a gate offers a view into the courtyard, although a wall to the appropriate prevents the gaze penetrating any further. The edge is defined by a lintel that slips over the top of the wall. This lintel reveals a twin aspect, a technical certainly one of taking water to the swimming pool on the opposite side of the wall, and a metaphorical one in every of framing the gaze so as to present a hint of something else beyond. As one passes the entrance, tiny openings cut by means of the front wall of the home and sign the presence of the unseen occupants.
Hello… I've a male (neutered) and a feminine cat and so they have been brilliant…. at all times use the litter tray, out of the blue although, certainly one of them has determined to start out peeing on or in a bag I use for work which is stuffed with clothes….. I discovered it and washed the bag and clothes and while it was drying I got one other bvag to make use of as an alternative… I have just gone to fill it with garments and found pee all over it…… this bag will have to be destroyed because it has wheels and is unwashable…. please help… its not like its pee on the floor or in a room… it appears to be on bags!!!


text:
Car Rental South Africa
Car Hire Vereeniging
Car Hire in Vereeniging Town Centre
Vereeniging Town Centre car rental offers the best way to explore the part of Southern Gauteng known as the Vaal Triangle. Lying 60km or so from Johannesburg, this triangle includes the towns of Vanderbijlpark and Sasolburg too. As an important hub within the industrial sector, Vereeniging manufactures cables, bricks, tiles, processed lime and farm equipment. The town also boats some rather fascinating landmarks, along with a decent number of accommodation choices and facilities. There are quite a few things to see and do after collecting your car rental in Vereeniging Town Centre. Highlights in the town and close surrounds include the Treaty of Vereeniging Memorial, City Library, stone age artefacts at Klip River Terrace, Suikerbosrand Nature Reserve, Vereeniging Drive-In (open all year except for winter between June and August), Jabulani Butterfly Garden, Orange Farm Township Park, Treehaven Waterfowl Trust and De Kraal Antiques and Collectables to name but a few.
Car Rental in Vereeniging Town Centre
Southern Gauteng may not be as famous as other South African areas for tourism, but that doesn't mean that it isn't a rewarding destination. Business travellers may travel to this part of the world for its industrial plants or to be within fairly easy reach of Johannesburg. Leisure travellers on the other hand will be able to enjoy the picturesque surrounds, rich history, unique culture and laid-back atmosphere of the town. And what better way to explore your surrounds than self-drive! Drive South Africa works with a network of leading rental companies operating within Southern Africa, to bring you the best deals on short and long term rentals across a wide range of late model vehicles. From standard to luxury; 4x4 to van and plenty of options in between, planning your trip has never been simpler. Book your Vereeniging Town Centre car rental from Drive South Africa today, for the best service at the best rates.
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Discover how Grape Seed Extract can prevent skin aging and why you should include them in your diet.
Grapes are most often relished as your 5 p.m. companion (in the form of a nice, woodsy glass of Merlot) but these humble fruits can make you happy in other ways: specifically, by keeping your skin youthful and vibrant. The seeds of grapes, in particular, have been shown to work wonders for skin when they're consumed. After all, they are the fuel that grows entire vines! Let's explore what the fuss over grape seed extract is about—and why you should consider adding them to your daily regimen.
What Exactly Is Grape Seed Extract?
Surely you've heard of grape seed oil (since it's quite often used in topical skincare treatments for its moisturizing abilities), but grape seed extract (GSE) is another incredible ingredient you should consider.
How much can the humble grape really help against the never-ending battle against the telltale signs of aging? It turns out, a lot.
First of all, grape seeds are powerful antioxidants that serve as your frontline soldiers in the fight against free radicals. In short, free radicals are unstable atoms that damage cells in the body, including skin cells. They're caused by pollution, smoking, stress and sun damage, and they're even created out of the blue, too, just because you're alive.
When free radicals damage skin cells, it can lead to premature aging, meaning you're likely to get a lot more wrinkles and sagging unless you provide your body with enough antioxidants to neutralize these effects. Without antioxidants to balance them out, free radicals would run rampant throughout your body, causing irreparable damage.
So remember, antioxidants and free radicals are natural enemies, and antioxidants are definitely your friend.
Among the most powerful antioxidants in grape seed extract are proanthocyanidins and catechins. These are believed to be significantly more potent than the antioxidants in vitamin E and C—and this is a big deal considering the two of the most powerful and oft-studied skincare ingredients ever.
Moreover, the polyphenols or OPCs (oligomeric procyanidins) found in grape seeds are also extremely nutritious for the body, thanks to their free radical-scavenging abilities.
Polyphenols are also known for their job of binding collagen fibers. Collagen, the most abundant protein in the body, begins to deteriorate as we age. Grape seed extract prevents collagen from being damaged. This further allows for more elasticity in your skin, keeping it firm and perky, and preventing the dreaded sag and deepening of wrinkles. It even allows for better movement of joints and other areas of connective tissue.
UV damage also prevents skin from retaining that bounce-back that makes skin snap back into place after being stretched. If everyone really understood how badly the sun damages their skin, the whole world would probably never be caught dead sunbathing again! Alas, educating people on the dangers of sun damage still hasn't quite convinced everybody to use sunblock properly and efficiently. But you, dear reader, are now armed with the knowledge that sun exposure is the very devil, and you'd do well to steer clear of it, or at least try to undo some damage from your youth by embracing substances like grape seed extract.
How are grape seed extract manufactured?
Grape seed extracts are seed blends made primarily from grapes. Polyphenols are extracted from grape seeds using hot water. The extract is then purified, concentrated and spray dried into powder.
A hundred years ago, seeds were pressed in a manual wedge press to extract the precious oil from grape seeds. Today, pharmaceutical-grade extracts often are made by combining high heat and solvents to produce grape seed extract that can be used as a supplement, or grape seed oil that can be added to our diet.
Although it may be tempting to think every part of a grape is the same, that's simply not true. GSE shouldn't be confused with resveratrol, which comes from the skin (not the seeds) of grapes used to make red wine. Resveratrol is more abundant in red wine than white because the fermentation process for red wine takes longer.
One caveat: resveratrol is chemically related to estrogen, and can either promote estrogen activity or block it. You might want to steer clear of it if you have breast, ovarian or uterine cancer, or if you're trying to become pregnant or taking an oral contraceptive. Resveratrol may also interact badly with blood thinners or drugs like ibuprofen. Talk to your doctor if this concerns you.
GSE also shouldn't be confused with French pine bark. French pine bark comes from trees growing along the southwest coast of France. French pine bark contains procyanidins, also found in red wine, which are antioxidants that fight against oxidative stress.
French Pine Bark OPC offer a variety of amazing benefits for your body, skin and overall health. Besides the known anti-oxidant benefits, what makes this miracle bark so amazing (and better than the OPC's found in grape seeds), are the additional healing and protective properties: anti-bacterial, anti-viral, anti-aging, anti-inflammatory, and anti-allergic properties. In addition, it has been discovered that French Pine Bark can help reduce muscle soreness and improve conditions related to poor circulation, high blood pressure, osteoarthritis, diabetes, aging skin, eye disease and stamina.
At this point it should be abundantly clear why you should incorporate grape seed extract into your anti-aging endeavors, especially if you want to maintain youthful skin for as long and possible.
This is one of the reason why we have recently upgraded our best seller and most popular Taut Premium Collagen Replenishment Formula with grape seed extract. Each glass bottle contains a good amount of grape seed extract and provides all these wonderful benefits that translate into visible results—namely firmer, uplifted, brighter skin and the reduced appearance of discoloration and dark spots that occur as a result of sun damage, acne scarring or the regular aging process.
Our Taut Premium Collagen Advanced formula combines grape seed extract with other potent and proven ingredients that synergistically work to preserve and protect youthful skin. Along with its antioxidant, collagen-supporting and brightening powers, grape seed extract works with 13,000 milligrams of hydrolyzed marine collagen (potent concentration to trigger your body's natural wound repair response to start producing collagen), elastin, hyaluronic acid, vitamins B6, C & E ,ceramides, and DNA, all of which create a glowing, taut, soft and uplifted complexion.
Discover our advanced formula and how this little bottle can help reverse signs of aging, promote youthful radiant skin and protect your skin from free-radical damages.


The foundation of the ABALOBI app suite, where fishers co-produce knowledge. A personal logbook with analytics and sharing options. Personal accounting. Safety-at-sea integrations via Weatherdock.
Fishers record various oceanic, atmospheric and fisheries parameters as part of their daily logbook, and utilise dashboards to visualise their data.
Fishers communicate within their community and across fishing communities via an integrated chat platform.
Fishers own their data but can share with various fisheries stakeholders.
Overview of income & expenditure within dashboards.
Navigational, weather prediction and SOS functionalities using mobile phone and VMS integration.
Latest fishing regulations, notices and management meeting outcomes.
Digitised community catch monitoring at the landing site and along the shoreline.
Rendering community catch monitoring more effective using tablet-based technology.
Dashboards of data collected, per fisher and landing site.
Communication with fishers, other monitors and fisheries managers via an integrated chat platform.
Real-time fishery data and communications for co-management.
Dashboards of community catch monitoring data for government managers and fisher data for local co-management committee members, as selectively shared by fishers.
Communication with fishers, monitors and other fisheries managers via an integrated chat platform.
Registers of fishers, vessels and co-operatives.
Posting of latest fishing regulations, notices and management meeting outcomes for fisher and monitor audiences.
Co-operative member and fleet management. Transparent collective accounting. Catch value-adding.
Collective accounting package for payment of members, and fleet income and expenditure calculations.
Transparent transactions and member benefits.
Supporting women's equitable participation in fisheries. Post-harvest fish processors log product value-adding and expenses. As these processors are predominantly women, this renders the role of these community members more visible and promotes gender equity in the co-operative.
Traceable slowfish. By linking with fisher data from ABALOBI FISHER, sales of catches can be linked to origin.
Skipper, crew member, fleet and processing establishment licensing and auditing are kept up to date, with renewal alerts.
Integration with online banking and invoicing platforms. Possibility for co-operatives to leverage collective insurance schemes.
Fish with an ecological and social 'story'. Towards community-supported fisheries. Traceability. Empowerment in the value chain.
Live catch-of-the-day posted by fishers and co-operatives. Real-time spatial and catch data.
Transparent product pricing and trends, based on posts and sales, available to fishers and retailers.
Slowfish with a 'story'. Fair trade. Public profile of fisher co-operative based on various ecological and social criteria.
Consumer, chef and other retailers log their details and seafood requests.


text:
Q: How can i make a search box to search by custom field Hy i am trying to make a search box for a week to search for post that have a specific custom field.
I have a post_type - produse with 2 custom field

*

*
CF1- culoare  / alb,maro,verde

*
CF2- marime   / 120x200, 200x300

Ok i want to make a checkbox form and select, after i select wath i want and push search i want to retrive the post that i have atached the custom field.
This i have made this from search around web.. but this is not working.. it load the cpu on my web hosting..
<?php
$field = get_field_object('culoare');
$values = explode (',', $_GET['culoare']);
?>

<form>
    <fieldset>
        <legend>Cauta dupa culoare:</legend>
        <?php if (!empty($field['choices'])) : $i = 0; foreach( $field['choices'] as $choice_value => $choice_label ): ?>
        <?php $culoareIndex = strtolower($choice_value); $i ++; ?>
        <input type="checkbox" name="culoare" value="<?=$choice_value?>" id="checkbox-1-<?=$i?>" class="regular-checkbox-<?=$culoareIndex?>"
        <?php if( in_array($choice_value, $values) ): ?> checked="checked" <?php endif; ?> />
        <label for="checkbox-1-<?=$i?>"></label>
         <?php endforeach; endif; ?>
        <input type="submit" value="Search" />
    </fieldset>
</form>

Functions.php
add_action('pre_get_posts', 'my_pre_get_posts');

function my_pre_get_posts( $query ) {
    if( is_admin() ) { return; }
    $meta_query = $query->get('meta_query'); // get original meta query

    // validate type
    if( (empty($_GET['culoare'])) )
    {
        return;
    }

    // get types as an array
    // - use explode to get an array of values from type=a|b|c
    $culori = explode('|', $_GET['culoare']);

    // set compare for the meta_query
    // - http://codex.wordpress.org/Class_Reference/WP_Query
    $meta_query['relation'] = 'OR';

    foreach( $culori as $culoare )
    {
        $meta_query[] = array(
            'key'       => 'culoare',
            'value'     => '"' . $culoare . '"',
            'compare'   => 'LIKE',
        );
    }

    $query->set('meta_query', $meta_query); // update the meta query args

    return; // always return
}

Can somebody show me how is done,please, and point me in the right direction, i'm not so good with php and i want this to be made.
Thanks

A: The generated SQL for that query is going to look something like:
SELECT SQL_CALC_FOUND_ROWS wp_posts.ID
FROM wp_posts
INNER JOIN wp_postmeta ON (wp_posts.ID = wp_postmeta.post_id)
INNER JOIN wp_postmeta AS mt1 ON (wp_posts.ID = mt1.post_id)
INNER JOIN wp_postmeta AS mt2 ON (wp_posts.ID = mt2.post_id)
WHERE 1=1
  AND wp_posts.post_type = 'post'
  AND (wp_posts.post_status = 'publish')
  AND ((wp_postmeta.meta_key = 'culoare'
        AND CAST(wp_postmeta.meta_value AS CHAR) LIKE '%\"a\"%')
       OR (mt1.meta_key = 'culoare'
           AND CAST(mt1.meta_value AS CHAR) LIKE '%\"b\"%')
       OR (mt2.meta_key = 'culoare'
           AND CAST(mt2.meta_value AS CHAR) LIKE '%\"c\"%'))
GROUP BY wp_posts.ID
ORDER BY wp_posts.post_date DESC LIMIT 0,10

There are three things you should notice. The first is that your query is generating an incorrect LIKE clause-- '%\"c\"%'. You do not want to quote your values when you create the meta_query.  Do simply:
'value'     => $culoare,

As you have it written, the query will search every row in the table but come up with nothing.
Secondly, LIKE queries, especially with wildcards-- the %s-- are not very efficient. The effects of that inefficiency are going to become more pronounced the larger your $wpdb->postmeta table becomes and the more conditions you try to match against.
Third, LIKE queries with an OR connector are not efficient. The query has to look at every row in the table to work out the result. Again, the effects of that inefficiency are going to become more pronounced the larger your $wpdb->postmeta table becomes and the more conditions you try to match against.
I suspect that those three problems are why this "load the cpu on my web hosting", by which I think you mean the server slows to a crawl and times out or crashes.
There is no simple solution.

*

*Try to add limiting conditions-- post dates, post authors, something
to limit the scope of the search.

*Do you need the LIKE operator? An = or IN search would probably
be quicker.

*Is there any way you can cache some of this information rather than
run a search every time? If there is, you should pursue that avenue.

redpajama_set_name: RedPajamaStackExchange


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text:
Arsenal will analyse CCTV footage in order to catch the fan who threw a missile at Dele Alli in League Cup defeat
Thursday 20 December 2018, 10:12am
Dele Alli scored a sumptuous goal to knock Arsenal out of the League Cup. Credit: PA
Arsenal are analysing CCTV footage in a bid to identify the supporter who hit Dele Alli on the head with a plastic bottle during Tottenham's 2-0 Carabao Cup quarter-final win at the Emirates Stadium.
Alli, who put on a masterclass to guide Spurs through to a last-four tie with Chelsea, was struck by a missile in the second half having scored a sublime goal and set up Son Heung-min's opener.
It earned revenge for Tottenham's 4-2 Premier League defeat to the Gunners just 17 days earlier, when a Spurs fan was arrested for throwing a banana skin at Pierre Emerick-Aubameyang.
The England international said the unsavoury incident made his goal even sweeter.
"It is what it is. It made the goal a little bit sweeter and the win," he told Sky Sports.
Pochettino, who on Tuesday declined the chance to distance himself from rumours linking him to Manchester United, hopes the culprit can be caught but was keen not to tarnish the whole Arsenal fanbase.
"It was lucky that it wasn't a big issue," the Argentinian said. "For me, okay we play a derby and we hate each other in a good way but that is in a bad way when you go to damage some people or an opponent.
"I cannot talk in general because only one stupid person made a mistake. I respect the Arsenal fans, like I respect the professionals from Arsenal and I respect our fans.
Dele Alli put in a typically combative display against arch rivals Arsenal. Credit: PA
"For one you cannot talk about everyone. I hope that Arsenal take responsibility and through the camera they can identify because it wasn't a problem but it can happen to be a big problem."
Arsenal had their chances as Aaron Ramsey and Alexandre Lacazette both hit the woodwork, but they were second best.
Mesut Ozil's non-selection was again the main talking point after the game and boss Unai Emery was particularly evasive when asked about his future.
Ozil was not included in the matchday squad and his future at the Emirates Stadium looks bleak.
Emery said: "It was a tactical decision, I thought the players today that are with us, were for us the best for this match. We have 24 or 25 players and when they are playing in one or not playing in another it is for me the decision and today it is tactical.
"The focus is to analyse this match and for Saturday, it is a very important match. Tomorrow we are going to train with every player.
"I am thinking about the match on Saturday and not thinking about another situation. Every player is important, I decided he didn't play and it was only a tactical decision."