Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 19 / 5 / 1 / Solution

For each fixed $k$, the two quotient Riemannian metrics are not isometric. If an isometry $F:N_1\to N_2$ existed, choose a small open set avoiding the finite preimages of cone points and their inverse images under $F$. Composing $F$ with the local inverse of the first base projection and the second base projection gives a local isometry of the base Riemannian metric $h_k$. Local rigidity forces this map to be the identity. Hence $p_{2}F=p_1$ on a dense open set, and by continuity everywhere.

The alleged isometry would therefore be an isomorphism of the two orbifold covers over the same base. Covering-space classification makes $K_1,K_2$ conjugate in $\Gamma$, and applying $\rho$ makes $H_1,H_2$ conjugate in $T$, contrary to the supplied hypothesis. Thus $g_{1k},g_{2k}$ are nonisometric.

For $k\ne \ell$, the surface areas are different:Area is preserved by isometries, so no surface from the $k$th pair is isometric to a surface from the $\ell$th pair. Together these arguments prove the full pairwise nonisometry assertion.