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.)

  • $\begingroup$ Hi David. What is the LMFDB? $\endgroup$
    – Joël
    Sep 11, 2014 at 3:05
  • $\begingroup$ L-Functions and Modular Forms Database (lmfdb.org). They have a searchable database of Artin representations, but unfortunately it doesn't seem to include Joe Buhler's example. $\endgroup$ Sep 11, 2014 at 3:40

2 Answers 2


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.


Here is another proof of the result from Joel's answer, using a little more theory: a two dimensional irrep. of a finite group $G$ corresponds to a simple algebra factor $A$ of the group ring $\mathbb Q[G]$, where $A$ is $4$-dimensional over its centre (which is some number field $F$). Thus $A$ is either $M_2(F)$, or a non-split quat. alg. over $F$. In particular, in the second case $A$ is a division algebra.

If $c \in G$ has order $2$, with image $\bar{c}$ in $A$, then since $c^2 = 1$, we see that $(\bar{c} -1)(\bar{c} + 1) = 0$ in $A$. If $A$ is a division algebra, we conclude that either $\bar{c} = 1$ or $\bar{c} = -1$.

Thus if the image of $c$ is not scalar, we see that $A = M_2(F)$, which is to say that the given irrep. is defined over its trace field.


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