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My question is that:

For any prime $p>7$, is there a mod $ p $ representation $ \bar{\rho}:G_{\mathbb{Q}}\to \operatorname{GL}_{2}(\mathbb{F}_{q}) $ of the absolute Galois group $ G_{\mathbb{Q}} $ of the rational number field $ \mathbb{Q} $ where $ \mathbb{F}_{q} $ is a finite field of characteristic $p$ such that

  1. $ \bar{\rho} $ is unramified at $ p $;
  2. $ \bar{\rho} $ is even, i.e. the determinant of the image of a complex conjugation $ c $ is $ 1 $;
  3. the image $ \operatorname{Im}(\bar{\rho}) $ of $ \bar{\rho} $ contains $ \operatorname{SL}_{2}(\mathbb{F}_{q}) $?

My motivation for asking this question comes from Theorem 1 in Allen and Calegari - Finiteness of unramified deformation rings. The paragraph following Theorem 1 in the above paper states that if the mod $p$ representation $ \bar{\rho} $ contains $ \operatorname{SL}_n(\mathbb{F}_q)$ , then …. I asked myself, does such $ \bar{\rho} $ really exist (especially in even case)?

If we drop conditions 1 and 2, then it is a special case of inverse Galois problem for $\mathbb{Q}$, and it seems that it has an affirmative answer, cf. Wiese - On projective linear groups over finite fields as Galois groups over the rational numbers.

If we drop condition 2 and the condition 1 is weakened in the sense that we only consider the tame representation (but not unramified at $p$), then it's called the Tame Inverse Galois Problem, and it seems that it has an affirmative answer, cf. Sara Arias de Reyna's Ph. D. thesis "Galois Representations and Tame Galois realizations".

I think the most serious assumption here is 1.

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  • $\begingroup$ Maybe you can lift this projective representation? lmfdb.org/NumberField/11.11.31376518243389673201.1 $\endgroup$
    – Aurel
    Commented May 31, 2022 at 21:04
  • $\begingroup$ Oh I reread the question and I guess you wanted an answer for every $p$, not an example? $\endgroup$
    – Aurel
    Commented May 31, 2022 at 21:08
  • $\begingroup$ @Aurel Yes, I wanted an answer for every $p$. $\endgroup$
    – Nobody
    Commented Jun 1, 2022 at 20:51
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    $\begingroup$ @LSpice: I edited to add the author and correct on-line address of the cited Ph. D. Thesis. $\endgroup$
    – F Zaldivar
    Commented Jun 3, 2022 at 2:36
  • $\begingroup$ @FZaldivar, thanks! I have deleted my comment, since it is no longer relevant (and had a typo). $\endgroup$
    – LSpice
    Commented Jun 3, 2022 at 3:03

1 Answer 1

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Either you want an example for one $p$ which is given in the comments or you are asking about an open case of the inverse Galois problem. Even if you drop the requirement that the representation is unramified at $p$ it’s not know if there exist such representations for any sufficiently large prime $p$. (Your comment on that point is wrong since the linked paper only constructs odd representations when $p>2$.)

version 2 of the question: If you want to find such representations after you restrict to totally real fields then it is elementary: $S_n$ is the Galois group of a totally real extension of $\mathbb Q$ for any $n$ (take a random polynomial over $\mathbb Q$ with n real roots) so $\operatorname{SL}_n(\mathbb F_p)$ is the Galois group of a totally real extension of some totally real field for any $n$ and $p$. By making a further extension you can ensure the representation is also unramified at all primes. That seems to suggest there are many examples in the linked paper over totally real fields. For $\mathbb Q$ it seems like an open problem.

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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jun 1, 2022 at 16:26
  • $\begingroup$ @YufanLuo p=11 is not “all p” and 11 is not “sufficiently large” so what is the point of your comment? $\endgroup$
    – user483372
    Commented Jun 2, 2022 at 17:48
  • $\begingroup$ Thanks for your answer, I have re-edited my description for the question to make it clear. $\endgroup$
    – Nobody
    Commented Jun 2, 2022 at 18:27
  • $\begingroup$ @user483372 please ignore it. $\endgroup$
    – Nobody
    Commented Jun 2, 2022 at 18:32

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