Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-20/4/solution
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 20 4 Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-14
Let . Since and the scalar centre is trivial,Choose to stabilize a nonzero vector and to stabilize a two-dimensional subspace. There are seven choices of either object, so both subgroups have index and order .
Their coset permutation characters agree. The number of nonzero vectors fixed by is . A two-dimensional subspace is the kernel of a unique nonzero covector, so its stabilizer action is the dual action . Its fixed-point count isusing equality of the matrix ranks of a matrix and its transpose. For a subgroup , counting the representatives of fixed cosets givesThus the equal characters and subgroup orders give equal intersection counts with every conjugacy class. This proves are Gassmann equivalent.
They are not conjugate. For example, the stabilizer of contains every matrix with and . An invariant plane containing would give an invariant line in , impossible for the full action. An invariant plane not containing would be a complement to that line, but the arbitrary shears do not preserve any such complement. Hence a point stabilizer preserves no plane, whereas every conjugate of does. The two subgroups are nonconjugate.
We now use a cone-torus construction of genus-four Sunada surfaces. Let be a hyperbolic torus with one cone point of angle . Its orbifold fundamental group has presentationMap to the supplied generators of . Their commutator has exact order , so this gives a surjection with torsion-free kernel : every finite-order element of is conjugate to a power of the cone generator, and no nonidentity such power lies in . Thus is a smooth closed hyperbolic surface with a -action. Its quotients by are smooth too, since order- subgroups contain no element of order and therefore meet no cone stabilizer.
The orbifold Euler characteristic of is . Each quotient cover has degree , soThe Sunada theorem proved in Question 3 makes isospectral for every choice of the base hyperbolic metric.
Here is an explicit supply of base metrics. A hyperbolic trirectangle has three right angles and a fourth angle . If its sides adjacent to the opposite right-angle vertex have lengths , the trirectangle identity is . Thus varies freely, with . Reflecting four copies about their two perpendicular centre lines gives a quadrilateral with four angles and opposite sides of equal length. Identifying opposite sides produces a cone torus: all four corners give total angle . Its two centre-line geodesics intersect once and have lengths . Cut along the shorter chosen geodesic and reglue with a small twist ; this retains the cone angle and allows an additional metric parameter. The remaining cut surface has two equal-length geodesic boundaries and the cone point.
We give a concrete intersection test for nonisometric finite covers, rather than relying on subgroup nonconjugacy alone. Choose sufficiently small. The trirectangle geometry gives an embedded collar of width about . The shortest perpendicular joining the two boundary components after cutting has length . At zero twist it closes to , the unique shortest closed geodesic crossing once. Uniqueness follows from uniqueness of the shortest perpendicular; all other once-crossing classes have a positive length gap, since only finitely many closed geodesics have bounded length. For sufficiently small twists, the same marked class remains the unique shortest once-crossing geodesic. A geodesic crossing at least twice traverses the collar at least twice, so for small its length is greater than .
Geodesics disjoint from remain entirely in the cut surface and have twist-independent lengths. In a bounded window around they supply only finitely many possible length values, including lengths of iterates. Meanwhile varies nontrivially with by the hyperbolic seam identity: its hyperbolic cosine has a positive multiple of . Choose a small twist away from those finitely many coincidences. Then the only base geodesic of length is , up to orientation. By taking smaller if necessary, the only base geodesics of length at most are powers of : those crossing are long, and the nonperipheral geodesics of the limiting two-cusped cone pair of pants have lengths bounded away from zero.
In each cover, primitive lift components of a base geodesic are indexed by cycles of its monodromy permutation, with length equal to the cycle size times its base length. The permutation of has a unique singleton in each action, so each has a uniquely length-identified primitive lift of length . The permutation of has one two-cycle: it is for and for . Hence each has a unique primitive lift of length ; the iterate of the singleton lift is not primitive. At the single base intersection of , intersections of their lifted components are exactly their common sheet labels. ThereforeAn isometry would have to preserve the uniquely identified primitive lengths and their geometric intersection numbers. This contradiction proves that and are not isometric, even allowing orientation reversal.
Finally, let range over a sufficiently small positive interval, choosing a permissible twist for each value. The singleton cycle of makes the hyperbolic systole of each cover equal to . Distinct values therefore give distinct isometry classes. We obtain uncountably many pairs of nonisometric, isospectral genus-four Riemann surfaces.
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