$G=\operatorname{SL}(2,q)$; for $n$ not listed explicitly below, $\delta_n(G) = \delta^{\natural}_n(G)=1$.

For $G = \operatorname{SL}(2,q)$ with $q$ arbitrary or for $G = \operatorname{PSL}(2,q)$ with $q$ even, $$\begin{aligned}
\delta_n(G) = \delta^{\natural}_n (G)  \
     \left\{ \begin{array}{cll}
= \frac{2}{q}& \text{if }n = \frac{q \pm 1}{2} \text{ and $q$ is odd},& \\
\ge \frac{1}{6}& \text{if }n = q - 1,&\\
     \end{array} \right.\end{aligned}$$ and $\delta_{q+1} (G) \ge \frac{1}{6}$ whereas $\delta^{\natural}_{q+1} (G) =  1$.

For $G = \operatorname{PSL}(2,q)$ and $q$ odd, $$\begin{aligned}
\delta_n(G) = \delta^{\natural}_n (G)  \  \left\{ \begin{array}{cll}
= \frac{2}{q}&\text{if } n = \frac{q - 1}{2} \text{ and $q \equiv 3 \bmod 4$},& \\
= \frac{2}{q}&\text{if } n = \frac{q + 1}{2} \text{ and $q \equiv 1 \bmod 4$},& \\
\ge \frac{1}{6}&\text{if } n = q - 1,&\\
     \end{array} \right.\end{aligned}$$ and $\delta_{q+1} (G) \ge  \frac{1}{6}$ whereas $\delta^{\natural}_{q+1} (G) =  1$.

We remark that we completely determine $\delta_{q \pm 1}(G)$ for $G=\operatorname{SL}(2,q)$ and $\operatorname{PSL}(2,q)$ in Section \[sec:sl2q\], but the exact formulas are a bit complicated and depend on divisibility conditions of $q \mp 1$. In particular, $\delta_{q \pm 1}(\operatorname{SL}(2,q)) = \frac 16$ if and only if $12 | (q \mp 1)$, and $\delta_{q \pm 1}(\operatorname{PSL}(2,q)) = \frac 16$ if and only if $24 | (q \mp 1)$.

