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Let $\alpha_0$ be the unique non-trivial character satisfying $\alpha_0^2=1$ of the split torus $\mathrm{T} \subset \mathrm{SL}(2,q)$ and denote by $\mathrm{R}(\alpha_0)$ the character of $\mathrm{SL}(2,q)$ obtained by first extending $\alpha_0$ to the Borel subgroup $\mathrm{B} \subset \mathrm{SL}(2,q)$ and then by inducing to the whole $\mathrm{SL}(2,q)$. It turns out that $\mathrm{R}(\alpha_0)$ is a sum of two irreducible characters $\mathrm{R}_{\pm}(\alpha_0)$. The values of these characters can be explicitly computed and lie in $\mathbb{Q}_p(\sqrt{q})$ (we consider representations over $\overline{\mathbb{Q}}_p$). What can be said about the minimal field of definition of the representations corresponding to these characters?

I am mainly interested in the case $q=p^2$ for odd $p$. In that case, the characters $\mathrm{R}_{\pm}(\alpha_0)$ take values in $\mathbb{Q}_p$ (in fact even in $\mathbb{Z}$). Can the corresponding representations be defined over $\mathbb{Q}_p$? Or at least over some non-ramified extension of $\mathbb{Q}_p$?

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  • $\begingroup$ I thought that the field of definition of an irreducible representation of a finite group was given by the field generated by the values of its character, at least in the semisimple case (char 0, so applies here). $\endgroup$
    – Victor Protsak
    Commented Sep 23, 2016 at 6:04
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    $\begingroup$ @VictorProtsak Not quite. For example, the order $8$ quaternion group has an irreducible $2$-dimensional representation whose character values are integers but which cannot be realized as a real representation. $\endgroup$
    – Tobias Kildetoft
    Commented Sep 23, 2016 at 7:29
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    $\begingroup$ @ML: It would help a lot to include a reference to the character table. I don't understand the claim that the specific characters of these two principal series representations require just $\sqrt{q}$, which makes a reference most important. Also, there is no advantage here in working over a local field or its algebraic closure rather than over $\mathbb{C}$. (Perhaps a tag 'finite-groups' would be more helpful than 'lie-groups'?) $\endgroup$
    – Jim Humphreys
    Commented Sep 27, 2016 at 15:59
  • $\begingroup$ @Victor: What Tobias refers to in his comment is the Schur index. See Chapters 9-10 of the classic text Character Theory of Finite Groups by I.M. Isaacs, for example. $\endgroup$
    – Jim Humphreys
    Commented Sep 27, 2016 at 16:01
  • $\begingroup$ @Jim You are right, it should be either $\sqrt{q}$ or $\sqrt{-q}$, depending on the case. I focus on the case when $q$ is a square of a prime, that is why I forgot about the $\sqrt{-q}$ case. The character table can be found in the book "Representations of $\mathrm{SL}_2(\mathbb{F}_q)$" by C. Bonnafé (Ch. 5). $\endgroup$
    – M L
    Commented Sep 29, 2016 at 0:51

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The answer is yes, the representation is defined over $\mathbb{Q}_p(\sqrt{q})$. We first claim that $R_{+}(\alpha_0)$ and $R_{-}(\alpha_0)$ are not isomorphic. This can easily be seen by using the Bruhat decomposition and concluding that $dim(End(R(\alpha_0)))=2$. Second, let $\psi$ be the character of $R_{+}(\alpha_0)$, and let us write $G=SL(2,q)$ which is a finite group. Then we have the idempotent $$e=\frac{1}{|G|}\sum_{g\in G}\psi(g^{-1})g$$ in the group ring of $G$. This idempotent is already defined over $\mathbb{Q}_p(\sqrt{q})$. One can now easily show that $R_{+}(\alpha_0) = eR(\alpha_0)$. Since $R(\alpha_0)$ and $e$ are defined over $\mathbb{Q}_p(\sqrt{q})$, the same is true for $R_{+}(\alpha_0)$.

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    $\begingroup$ Note that Schur already worked out the character table a century ago, by relying especially on orthogonality relations. Later notions like "Bruhat decomposition" aren't needed. (Also, the actual character value here may involve either $\sqrt{q}$ or $\sqrt{-q}$ depending on congruence mod 4. This complicates matters unless $q$ is an even power of a prime.) $\endgroup$
    – Jim Humphreys
    Commented Sep 28, 2016 at 13:33
  • $\begingroup$ @Ehud Thank you! This is what I was looking for. $\endgroup$
    – M L
    Commented Sep 29, 2016 at 0:53
  • $\begingroup$ Jim is right that we sometimes need $\sqrt{-q}$. A minor remark: shouldn't the whole formula for $e$ be multiplied by the dimension of $R_+(\alpha_0)$ (i.e. $(q+1)/2$)? $\endgroup$
    – M L
    Commented Sep 29, 2016 at 1:11

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