Cone-torus construction of genus-four Sunada surfaces ($7(1-1/7)=6=2g-2$)

Take a hyperbolic torus with a single cone point of angle $2\pi /7$. Its orbifold fundamental group is $⟨a,d\mid [d,a{]}^7=1⟩$. An epimorphism to $\mathrm{PSL}(3,2)$ whose commutator has exact order $7$ has torsion-free kernel. The point and line stabilizers have index $7$ and order $24$, hence avoid all nontrivial cone stabilizers. Their quotient covers are smooth surfaces with Euler characteristic $7(-6/7)=-6$, so have genus $4$. Their equal coset permutation characters give equal heat traces by the Sunada theorem. Nonisometry still requires a geometric distinction, such as the intersection test for nonisometric finite covers.

A Lambert quadrilateral with acute angle $\pi /14$ and sides $x,y$ incident to the opposite right-angle vertex satisfies the displayed relation. Reflect four copies around that vertex to make a symmetric hyperbolic quadrilateral whose four corner angles are $\pi /14$. Identifying opposite sides yields a torus with one cone angle $2\pi /7$. The two central axis loops have lengths $2x,2y$ and meet once at right angles. Cutting along one loop gives a pair of pants with two equal geodesic boundaries and the cone point; reglue the boundaries with a Fenchel–Nielsen twist to obtain the two-parameter cone-torus family. Its seven-sheeted torsion-free covers have genus four.