8
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Hi David. What is the LMFDB? $\endgroup$ – Joël Sep 11 '14 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. $\endgroup$ – David Loeffler Sep 11 '14 at 3:40
9
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.