$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$Let $ p $ be a prime for which $ \PSL(2,p) $ is simple (so $ p \neq 2,3 $).


Is the minimal irrep of $ \PSL(2,p) $ defined over a quadratic extension? In particular I wish to ask:
for $ p $ congruent to $ 1 $ mod $ 4 $ is $ \PSL(2,p) $ always a subgroup of
$$
 \SO\_{d\_{\min}}(\mathbb{Q}(\sqrt{p})) 
$$
and for $ p $ congruent to $ 3 $ mod $ 4 $ if $ \PSL(2,p) $ is always a subgroup of
$$
 \SU\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p})) 
$$


For example, this is true for $ p=5 $ with the "icosahedral" $ A\_5\cong \PSL(2,5) $ subgroup of $ \SO\_3 $. Appropriate generators are given in the section "Coxeter group generators" of <https://en.wikipedia.org/wiki/Icosahedral_symmetry>


Background:


The minimal degree $ d\_{\min} $ of a nontrivial irrep of $ \PSL(2,p) $ is
$$
d\_{\min}=\frac{p+1}{2}
$$
if $ p $ is congruent to $ 1 $ mod $ 4 $ and is
$$
d\_{\min}=\frac{p-1}{2}
$$
if $ p $ is congruent to $ 3 $ mod $ 4 $.


The value of the characters for degree $ d\_{\min} $ irreps are mostly $ 0,1,-1 $s also of course $ d\_{\min} $ and finally either
$$
\frac{1}{2} \pm \frac{\sqrt{p}}{2}
$$
if $ p $ is congruent to $ 1 $ mod $ 4 $ or
$$
-\frac{1}{2} \pm \frac{\sqrt{-p}}{2}
$$
if $ p $ is congruent to $ 3 $ mod $ 4 $. There are always exactly two irreps of degree $ d\_{\min} $ and their characters are related exactly by conjugation in the corresponding quadratic extension. All this information is in the first two nontrivial characters which can be found, for example, here


[http://www2.math.umd.edu/~jda/characters/psl2/](http://www2.math.umd.edu/%7Ejda/characters/psl2/)


Update: At this point I feel confident that, for $ p $ congruent to $ 1 $ mod $ 4 $, there is a $ PSL(2,p) $ subgroup of
$$
 \SO\_{d\_{\min}}(\mathbb{R}) 
$$
and a $ PSL(2,p) $ subgroup of
$$
 \SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{p})) 
$$
I'm still trying to understand if there is a $ PSL(2,p) $ subgroup of
$$
\SO\_{d\_{\min}}(\mathbb{Q}(\sqrt{p})) = \SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{p})) \cap \SO\_{d\_{\min}}(\mathbb{R}) 
$$
Similarly, for $ p $ congruent to $ 1 $ mod $ 4 $, there is a $ PSL(2,p) $ subgroup of
$$
 \SU\_{d\_{\min}}(\mathbb{R}) 
$$
and a $ PSL(2,p) $ subgroup of
$$
 \SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p})) 
$$
I'm still trying to understand if there is a $ PSL(2,p) $ subgroup of
$$
\SU\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p})) = \SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p})) \cap \SU\_{d\_{\min}}
$$


Every representation is a determinant 1 since $ PSL(2,p) $ is simple ( for $ p \neq 2,3 $).


And we can always find a representation defined over the appropriate quadratic extension $ \mathbb{Q}(\sqrt{\pm p}) $ since every character of $ PSL(2,p) $ has Schur index 1 (thanks to @BenjaminSteinberg for this fact that an irrep with Schur index 1 can always be realized over the field generated by its character values for the fact that $ PSL(2,p) $ always has Schur index 1 see part (I) of Theorem 6.1 of <https://link.springer.com/content/pdf/10.1007/BF02762888.pdf>).


For $ p $ congruent to $ 1 $ mod $ 4 $ we can always find an orthogonal representation since the Frobenius-Schur indicator is $ 1 $ (indeed every irrep of $ PSL(2,p) $ for $ p $ congruent to $ 1 $ mod $ 4 $ has FS indicator $ 1 $, this is related to the fact that for such $ p $ every element of $ PSL(2,p) $ is strongly real, meaning conjugate to its inverse by an involution, and thus the group as a whole is totally orthogonal, see <https://arxiv.org/abs/1811.05343> for details).



Minimal irrep of $\mathrm{PSL}(2,p) $
$\def\QQ{\mathbb{Q}}\def\FF{\mathbb{F}}\def\Sp{\text{Sp}}\def\SL{\text{SL}}\def\GL{\text{GL}}$Okay, time to do both cases using the Weil representation. This is going to rely on some Key Facts which I can check from my [explicit matrices for the Weil representation](https://mathoverflow.net/questions/432407) and which I do not know how to prove abstractly.


**Notation:** Let $\zeta$ be a primitive $p$-th root of unity. Let $L = \QQ(\zeta)$. Let $p^{\ast} = (-1)^{(p-1)/2} p$ and let $K = \QQ(\sqrt{p^{\ast}})$, this is the unique quadratic subfield of $L$. Let $\chi(a)$ be the quadratic residue symbol $\left( \tfrac{a}{p} \right)$.
For $a \in \FF\_p^{\times}$, let $\sigma\_a$ be the element of $\text{Gal}(L/\QQ)$ with $\sigma\_a(\zeta) = \zeta^a$.


For $a \in \FF\_p^{\times}$, let $t(a)$ be the matrix $\left[ \begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix} \right]$ in $\SL\_{2}(\FF\_p)$. I'll write $\gamma : \SL\_{2}(\FF\_p) \to \SL\_{p}(\mathbb{C})$ for the Weil representation. (This argument will also extend to the Weil representation of $\Sp\_{2k}(\FF\_p)$ for $k>1$.)


Here are the key facts:


**Key Facts** We can choose the map $\gamma$ such that all of the following are simultaneously true:


(1) All the matrices $\gamma(g)$ are in $\SL\_{p}(L)$.


(2) For $a \in \FF\_p^{\times}$, we have $\sigma\_{a^2}(\gamma(g)) = \gamma(t(a)^{-1} g t(a))$.


(3) The matrices $t(\gamma(a))$ are $\pm 1$ times the permutation matrices with permute $\FF\_p$ by multiplication by $a$. Specifically, the sign is $\chi(a)$.


**Writing down the minimal representation:**


The involution $t(-1)$ is central in $\SL\_2(\FF\_p)$. Using Key Fact (3) above, it acts on the Weil representation with eigenvalues $1$ and $-1$, having multiplicities $\tfrac{p + \chi(-1)}{2}$ and $\tfrac{p-\chi(-1)}{2}$ respectively; the $1$-eigenspace is the representation which factors through $\text{PSL}\_2(\FF\_p)$. I'll write $\gamma\_+(g)$ for the action of $g \in \SL\_2(\FF\_p)$ on this $1$-eigenspace. We will need to understand this more explicitly:


When $p \equiv 1 \bmod 4$, the matrix $\gamma(t(-1))$ is the permutation matrix $b \mapsto -b$ on $\FF\_p$. So we can think of the $1$-eigenspace as even functions on $\FF\_p$, or in other words as functions on $\FF\_p/\{ \pm 1 \}$. We can think of $\gamma\_+(t(a))$ as the permutation $a$ acting on $\FF\_p/\{\pm 1\}$, times $\chi(a)$.


When $p \equiv 3 \bmod 4$, the matrix $\gamma(t(-1))$ is $-1$ times the permutation matrix $b \mapsto -b$ on $\FF\_p$. So we can think of the $1$-eigenspace as odd functions on $\FF\_p$. Write $S$ for the group of quadratic residues in $\FF\_p^{\times}$. For any $a \in \FF\_p^{\times}$, exactly one of $a$ and $-a$ is in $S$, so an odd function is determined by its value on $S$, so we can think of the $1$-eigenspace as functions on $S$. We can think of $\gamma\_+(t(a))$ as the permutation $\chi(a) a$ acting on $S$.


**Conjugating into $\SL(K)$**


Let $\delta$ be a matrix in $\GL(L)$. We'd like to have $\delta \gamma\_+(g) \delta^{-1} \in \SL(K)$ for all $g$ in $\SL\_2(\FF\_p)$. Now, $K$ is the fixed field of the subgroup $\{ \sigma\_{a^2}: a \in \FF\_p \}$, so we want to have
$$\sigma\_{a^2}(\delta \gamma\_+(g) \delta^{-1}) = \delta \gamma\_+(g) \delta^{-1}.$$
We can rewrite the LHS as
$$\sigma\_{a^2}(\delta) \sigma\_{a^2}(\gamma\_+(g)) \sigma\_{a^2}(\delta)^{-1} =
\sigma\_{a^2}(\delta) \gamma\_+(t(a))^{-1} \gamma\_+(g) \gamma\_+(t(a)) \sigma\_{a^2}(\delta)^{-1}.$$
So we want
$$\sigma\_{a^2}(\delta) \gamma\_+(t(a))^{-1} \gamma\_+(g) \gamma\_+(t(a)) \sigma\_{a^2}(\delta)^{-1} = \delta \gamma\_+(g) \delta^{-1}$$
or, in other words,
we want
$$\delta^{-1} \sigma\_{a^2}(\delta) \gamma\_+(t(a))^{-1}$$
to commute with all $\gamma\_+(g)$.
Since the representation $\gamma\_+$ is irreducible, by Schur's lemma, this will happen if and only if
$$\sigma\_{a^2}(\delta) = c\_a \delta \gamma\_+(t(a)) \qquad (\ast)$$
for all $a \in \FF\_p^{\times}$, for some scalar $c\_a \in L^{\times}$.
Our goal, therefore, is to write down an invertible matrix $\delta$ with entries in $L$, satisfying $(\ast)$.


**Finding $\delta$**


In the case that $p = 3 \bmod 4$, we think of the rows and columns of $\delta$ as indexed by $S$. We define the matrix $\delta$ by $\delta\_{yx} = \zeta^{y^2 x^2}$. Since $x \mapsto x^2$ is a permutation of $S$, this is a Vandermonde matrix and hence invertible. (Really, $y^2$ could be replaced by any bijection $S \to S$, but the $x^2$ term is important, and I thought it was most elegant to use the same formula for both.)


For $a \in S$, we have $\sigma\_{a^2}(\zeta^{y^2 x^2}) = \zeta^{a^2 (y^2x^2)} = \zeta^{y^2 (ax)^2}$. The latter is the $(y,x)$ entry in $\delta \gamma\_+(t(a))$. This checks $(\ast)$ for $a \in S$; if $-a \in S$ instead, then note that $\sigma\_{a^2} = \sigma\_{(-a)^2}$ and $\gamma\_+(t(a)) = \gamma\_+(t(-a))$. So we have checked $(\ast)$ with $c\_a = 1$.


Now, let's do the $p \equiv 1 \bmod 4$ case. In this case, we think of the rows and columns of $\delta$ as indexed by $\FF\_p/ \{ \pm 1 \}$. Once again, we define $\delta$ by $\delta\_{yx} = \zeta^{y^2 x^2}$. Once again, this is a Vandermonde matrix, and thus invertible. If $\gamma\_+(t(a))$ acted by the permutation of multiplication by $a$ on $\FF\_p/ \{ \pm 1 \}$, we would once again verify $(\ast)$ with $c\_a = 1$. Since, in fact, $\gamma\_+(t\_(a))$ is $\chi(a)$ times this permutation matrix, we still win, but with $c\_a = \chi(a)$.


