4
$\begingroup$

Serre's modularity conjecture asserts that a continuous odd irreducible representation $$\overline{\rho} : G_\mathbb{Q} \rightarrow \mathrm{GL}_2(\mathbb{F}_q)$$ must be modular, in the sense that there is some eigenform $f$ so that $\overline{\rho} \sim \overline{\rho}_f$, and moreover gives formulae for the weight and level of $f$.

Is there a version for representations $ \overline{\rho} : G_\mathbb{Q} \rightarrow \mathrm{PGL}_2 (\mathbb{F}_q) $? If so, what is known?

$\endgroup$
2
2
$\begingroup$

The theorem 1.3 in "Lifting Irreducible Galois Representations" by Fakhruddin, Khare and Patrikis applies to this group, it will show that under some refined conditions $\bar{\rho}$ lifts to $\rho:G_{\mathbb{Q}}\rightarrow \text{PGL}_2(W(\mathbb{F}_q))$ which is ramified at finitely many primes and potentially semistable at all primes (and of course is irreducible too). I am not sure if such a representation should come from an odd representation $\tilde{\rho}:G_{\mathbb{Q}}\rightarrow \text{GL}_2(W(\mathbb{F}_q))$, but if it does, then it comes from an eigenform (by the Fontaine-Mazur, theorem of Taylor/Kisin/Emerton).

$\endgroup$
1
  • $\begingroup$ Thanks, that's a really helpful answer! $\endgroup$
    – Siksek
    Nov 9 '18 at 13:11
1
$\begingroup$

View $\overline{\rho}$ as a representation

$$\overline{\rho}: G_{\mathbb{Q}} \rightarrow \mathrm{PGL}_2(\overline{\mathbb{F}}_p).$$

The obstruction to lifting this to a Galois representation with values in $\mathrm{GL}_2(\overline{\mathbb{F}}_p)$ lies in $H^2(\mathbb{Q},\overline{\mathbb{F}}^{\times}_p)$, but this vanishes by a theorem of Tate (here we use the fact that we have extended scalars so that the coefficients are divisible).

Now consider the lifted representation

$$\rho: G_{\mathbb{Q}} \rightarrow \mathrm{GL}_2(\overline{\mathbb{F}}_p).$$

If the image of complex conjugation in $\overline{\rho}$ is non-trivial, then the image in the lift will be non-scalar, and vice versa. Hence Serre's conjecture implies that all such "odd" (complex conjugation is non-trivial) representations come from modular ones.

Some extras: using the fact that the class number of $\mathbb{Q}$ is one, one can twist this representation so that it is unramified at all primes where $\overline{\rho}$ itself is unramified. In fact, examining Tate's theorem more closely, one can even find a lift such that, given any local lift $\varrho$ (a representation of the decomposition group $D_q$ of $\mathbb{Q}$) of $\overline{\rho}|_{D_q}$, one may insist that

$$\rho |_{I_q} = \varrho |_{I_q}.$$

By this result, working out the invariants $N(\rho)$ and $k(\rho)$ associated to $\overline{\rho}$ is a purely local problem. In particular, there is indeed a corresponding projective version of Serre's conjecture (now a theorem, of course!)

Note that, in general, finding a lift will require increasing the coefficient field from $\mathbb{F}_p$ to some finite extension. (If the corresponding class in $H^2(\mathbb{Q},\mathbb{F}^{\times}_p)$ is non-zero, there will never be a lift valued in $\mathrm{GL}_2(\mathbb{F}_p)$. Of course all lifts are twists of each other.)

One warning: for local reasons, it might be the case that any such lift $\varrho$ is more ramified at $q$ than $\overline{\rho}|_{D_q}$ is (this only happens if the latter is ramifed, of course, since unramified representations always have unramified lifts). This happens, for example, if the projective image of $D_q$ has image the Klein $4$-group $K$. (The projective image associated to an abelian representation will be projectively equivalent to $1 \oplus \chi$ and so have cyclic projective image. Hence the lift of any projective faithful representation of $K$ will be non-abelian. However, if the lift came from an extension with no more ramification, it would have to be abelian.)

One reference for some of these remarks is Serre's paper on weight one modular forms which I think is from some point in the '70s. (Serre is working with complex projective representations, but the arguments are pretty much the same.)

(Ah, I now realize this question is from 2014, sorry.)

$\endgroup$
2
  • $\begingroup$ I'm not sure why the action on $\bar{\mathbb{F}}_p^{\times}$ in $H^2$ will be trivial can you explain. $\endgroup$
    – user130124
    Nov 8 '18 at 18:11
  • $\begingroup$ Thanks for a really detailed answer. Yes it's from 2014 but still interesting to me! $\endgroup$
    – Siksek
    Nov 9 '18 at 13:12

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.