Let $f$ be a weight 1 modular form (let's say cuspidal, new, normalized, and a Hecke eigenform). Then there's an associated Artin representation $\rho\_f: \operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q}) \to GL\_2(\mathbf{C})$.


The character of this representation is determined by $f$, so it takes values in the number field $L = \mathbf{Q}(a\_n(f): n \ge 1)$. Are there examples where $\rho\_f$ is not definable over $L$, i.e. isn't conjugate to a morphism $\operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q}) \to GL\_2(L)$? 


(There are lots of examples of *even* representations in the LMFDB where this happens, coming from even Artin representations with image $Q\_8$ or $S\_4$, but I can't find any odd ones.)



Field of definition of Galois representations of weight 1 modular forms
No. The fact that $\rho(c)$ has eigenvalues +1 and -1 implies that $\rho$ is defined over $L$. Here is a low-tech, longish argument (there are shorter ones using slightly more technology), which is essentially from an old paper of Wiles.


Work in a basis $(e\_1,e\_2)$ where $\rho(c)$ is the diagonal matrix $(1,-1)$. Now let $g \in G:= Gal( \overline {\bf Q}/\bf Q)$ and write $$\rho(g)=\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$$ in the same basis. Writing that $tr\ \rho(g)$ is in $L$ gives $a+d \in L$; writing that $tr \rho(gc)$ is in $L$ gives $a-d \in L$. Hence $a,d$ are in $L$ (for all $g \in G$).


Next takes a second element $g' \in G$, and write $$\rho(g')=\left(\begin{matrix} a' & b' \\ c' & d' \end{matrix}\right),$$
so that 
$$\rho(gg') = \left(\begin{matrix} aa'+bc' & \* \\ \* & \* \end{matrix}\right).$$
By the above paragraph applied to $g$, $g'$ and $gg'$, we know that $a$, $a'$ and $aa'+bc'$ are all in $L$. Thus $bc'$ is in $L$ (for all $g, g' \in G$, that is). Now we can certainly choose a $g$ such that $b \neq 0$ (otherwise $\rho$ would be reducible), and by changing the basis $(e\_1,e\_2)$ into $(be\_1,e\_2)$ (which does not affect the matrix of $\rho(c)$, hence does not affect the preceding conclusions), we may assume that $b=1$ for that $g$. We deduce that $c' \in L$ for all $g' \in G$. Next we can find a $g'\in G$ such that $c' \neq 0$, and then $bc' \in L$ implies $b \in L$. So we have shown that for all $g \in G$, all four coefficients of $\rho(g)$ are in $L$. QED.


