Traceless $\mathrm{SU}(2)$ characters and $\mathbb{Z}/4$ instanton gradings for two-bridge and $(3,n)$-torus knots

We assemble, and where possible independently verify, the representation-theoretic data underlying the pillowcase (symplectic) side of the Atiyah-Floer conjecture for knots, for two-bridge knots and $(3,n)$-torus knots. For a two-bridge knot $b(p,q)$ we give a short self-contained proof that every irreducible traceless $SU(2)$ representation is binary-dihedral; these are the $(p-1)/2$ dihedral characters at meridian angles $\cos(2\pi k/p)$, independent of $q$, and the traceless Riley polynomial is the explicit product $\phi_p(u)=\prod_k (u+4\sin^2(\pi k/p))$, monic of degree $(p-1)/2$ with constant term $\det K$. This gives a transparent account of the Hedden-Herald-Kirk theorem that pillowcase homology equals reduced singular instanton knot homology $I^\natural$ on this family, and of why the figure-eight bubbling and bounding cochains obstructing the general conjecture are structurally inert there. For the $(3,n)$-torus knots we compute the full traceless character variety and prove a dichotomy: exactly $(\det-1)/2$ characters are dihedral, so for $n$ odd every irreducible traceless character is non-dihedral. Passing to the double branched cover $\Sigma(2,3,n)$, we self-compute the $\mathbb{Z}/4$ spectral-flow gradings of the generators from the Fintushel-Stern index and the equivariant $\rho$-invariant, calibrated against the Poudel-Saveliev and Anvari computations; for $n$ odd the gradings split evenly between $1$ and $3$, giving the chain complex $IC^\natural(T(3,n))=(1+a,a,a,a)$, $a=-\sigma/4$. The homology equals this for $n\equiv 1\pmod 6$ (differential zero) but is smaller by $2$ for $n\equiv 5$, where it is nonzero: rank $I^\natural=\sum_i|\Delta_{T(3,n)}|$ throughout, and $T(3,5)=P(-2,3,5)=10_{124}$ has rank $7$, not $9$. We reproduce the first nonzero pillowcase differential, for $8_{19}=T(3,4)$, and identify it as the corner figure-eight bigon absent on two-bridge knots.