Minimal irrep of $\mathrm{PSL}(2,p) $ $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$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/
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).
 A: $\def\FF{\mathbb{F}}\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$I can do the $p \equiv 1 \bmod 4$ case. It is always defined over $\mathbb{Q}(\sqrt{p})$.
In this case, we can describe the minimal representation as follows: Let $A$ be a two dimensional $\FF_p$-vector space. For $c \in \FF_p$, set $\chi(c) := \left( \tfrac{c}{p} \right)$. Let $U$ be the vector space of all functions $u:A \to \mathbb{Q}$ satisfying
$$u(c \vec{a}) = \chi(c) u(\vec{a}) \ \text{for all}\ c \in \FF_p,\ \vec{a} \in A.$$
The vector space $U$ has dimension $p+1$ since, if we choose one vector on each line in $A$, then the value of $u$ on those vectors determines $u$ everywhere else. Note that $\text{PGL}_2(\FF_p)$ acts on $U$: It is obvious that $\text{GL}_2(\FF_p)$ does, and the condition $p \equiv 1 \bmod 4$ implies that $\chi(-1) = 1$, so the action descends to $\text{PGL}_2(\FF_p)$.
Now, fix a nonzero skew symmetric form $\langle \ , \ \rangle : A \times A \to \FF_p$. Define a linear map $F : U \to U$ as
$$F(u)(\vec{a}) = \frac{1}{p-1} \sum_{\vec{b} \in A} \chi{\big(} \langle \vec{a}, \vec{b} \rangle {\big)} u(\vec{b}).$$
The $\tfrac{1}{p-1}$ factor is for convenience; note that the $p-1$ proportional vectors on any line in $A$ all make the same contribution. Note that this formula is given by a matrix with entries in $\QQ$. (In fact, the matrix entries are in $\ZZ$, because we can just choose one representative vector on each line.)
Note that the operator $F$ is $\text{PSL}_2(\FF_p)$ equivariant since the form $\langle \ , \rangle$ is $\text{SL}_2(\FF_p)$ invariant.
Lemma $F^2 = p \text{Id}$.
Proof:
Choose coordinates on $A$ such that $\langle \ , \ \rangle$ is the standard form, and set $\vec{b}_{\infty} := \left[ \begin{smallmatrix} 1\\0 \end{smallmatrix} \right]$ and $b_j := \left[ \begin{smallmatrix} j\\1 \end{smallmatrix} \right]$ for $0 \leq j \leq p-1$. Then
$$F(u)(\vec{a}) =  \sum_{j} \chi{\big(} \langle \vec{a}, \vec{b}_j \rangle {\big)} u(\vec{b}_j).$$
and
$$F^2(u)(\vec{a}) = \sum_{j,k} \chi{\big(} \langle \vec{a}, \vec{b}_j \rangle {\big)} \chi{\big(} \langle \vec{b}_j, \vec{b}_k \rangle {\big)}  u(\vec{b}_k).$$
It is enough to verify the claim for a spanning set of $U$. Such a spanning set is given by the functions which are supported on the various lines of $A$. By symmetry, it is enough to consider one such function: Let $u(\vec{a}) = \chi(c)$ if $\vec{a} = c \vec{b}_{\infty}$, and $0$ otherwise. Then
$$F(u)(\vec{b}_i) = \begin{cases} 1 & 0 \leq i \leq p-1 \\ 0 & i= \infty \\ \end{cases}.$$
We have
$$F^2(u)(\vec{b}_i) = \begin{cases} \sum_{j=0}^{p-1} \chi(j-i) \\  \sum_{j=0}^{p-1} 1  \end{cases} = \begin{cases} 0 & 0 \leq i \leq p-1 \\ p & i = \infty \end{cases}.$$
In other words, $F^2 = p \text{Id}$ as desired. $\square$
Now, extend scalars to $\QQ(\sqrt{p})$. Then $U$ splits into $\pm \sqrt{p}$ eigenspaces for $F$ and, since $F$ has matrix entries in $\QQ$, its eigenspaces have coordinates in $\QQ(\sqrt{p})$. Since $F$ is $\text{PSL}_2(\FF_p)$-equivariant, the action of $\text{PSL}_2(\FF_p)$ maps each eigenspace to itself. These eigenspaces are the minimal representations.

I clearly have constructed a nontrivial $\tfrac{p+1}{2}$-dimensional representation of $\text{PSL}_2(\FF_p)$, so it must be the minimal representation. How did I come up with this?
Let $B$ be the subgroup $\left[ \begin{smallmatrix} c & d \\ 0 & c^{-1} \end{smallmatrix} \right]$ of $G:=\text{PSL}_2(\FF_p)$. Then we have a character $\psi : B \to \{ \pm 1 \}$ defined by $\psi\left( \left[ \begin{smallmatrix} c & d \\ 0 & c^{-1} \end{smallmatrix} \right] \right) = \chi(c)$. The vector space $U$ is just the induction of $\psi$ from $B$ to $G$, described in the standard way as functions on cosets.
I knew from the general theory that $U = U_+ \oplus U_-$ where $U_+$ and $U_-$ are the two minimal representations, so we should have $\dim \text{Hom}_G(U, U) = 2$. I knew that, if I could find a non-scalar element $F$ of $\text{Hom}_G(U, U)$, then I could recover $U_+$ and $U_-$ as its eigenspaces. The advantage of $\text{Hom}_G(U, U)$ is that I can compute it using matrices with entries in $\QQ$.
The principled way to proceed would then be to compute $\text{Hom}_G(U, U)$ by using Frobenius reciprocity; it should be $\text{Hom}_B(\psi,\ \text{Res}_B^G \text{Ind}_B^G \psi)$. I started trying to do this, but I got confused, so I decided to try guessing instead.
The formula $F^2 = p \text{Id}$ and the presence of $\text{SL}_2$ reminded me of the finite Fourier transform $\Phi(u)(\vec{a})  = \sum_{\vec{b}} \eta(\langle \vec{a}, \vec{b} \rangle) u(\vec{b})$, where $\eta : \FF_p \to \mathbb{C}^{\times}$ is an additive character.  But this is given by a matrix with entries in $\QQ(\zeta_p)$, and I wanted  a matrix with entries in $\QQ$. After some more thought, I realized I could replace the additive character $\eta$ by the multiplicative character which was already present in the definition of $U$, and this worked.
A: $\def\QQ{\mathbb{Q}}\def\FF{\mathbb{F}}\def\Sp{\text{Sp}}\def\SL{\text{SL}}\def\GL{\text{GL}}$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 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 $c_a = \chi(a)$.
