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A Riemannian metric gives a smooth fibrewise isomorphism $g^{♭}:TM\to T^{\ast }M$, defined by $g^{♭}(v)=g(v,\cdot )$. Its inverse is smooth because the inverse of a smoothly varying invertible matrix is smooth. Define the Riemannian gradient byIt is a smooth vector field satisfying $g(V_{\sigma },Y)=d\sigma (Y)=Y(\sigma )$. Nondegeneracy also gives uniqueness. In coordinates its components are $V_{\sigma }^i={\sum }_j{g}^{ij}{\partial }_j\sigma$.

Multiplication of $g$ by the globally defined smooth positive function $e^{2\sigma }$ gives a smooth symmetric two-tensor $\tilde{g}$. For any nonzero tangent vector $v$,Thus $\tilde{g}=e^{2\sigma }g$ is a Riemannian metric. This conformal rescaling of a Riemannian metric changes lengths but preserves angles.

Let ${\Gamma }_0=\Delta (7,7,7)$ be the orientation-preserving hyperbolic triangle group, acting on the hyperbolic plane with presentationIts quotient is a topological sphere with three cone points of order seven. Define the surjective group homomorphismThe stated order conditions make this well defined, and the generation hypothesis makes it surjective. Set $K=\ker \varphi$. Every finite-order element of a hyperbolic triangle group is conjugate to a power of a cone-point generator. None of its nontrivial such powers lies in $K$, since all three images have order seven. Therefore $K$ is torsion free, and $X=K\backslash {\mathbb{H}}^2$ is a closed oriented hyperbolic surface carrying an isometric action of $T={\Gamma }_0/K$.

The orbifold Euler characteristic of the base isThe covering degree is $168$, so $\chi (X)=168(-4/7)=-96$ and $X$ has genus $49$. The subgroups have order $|U_i|=168/7=24$. Point stabilizers for the $T$ action are trivial or cyclic of order seven. Since seven does not divide $24$, each $U_i$ acts freely. Thusare smooth closed hyperbolic surfaces, andEquivalently, each $S_i$ is a seven-sheeted orbifold cover of the base with three branch values, and the Riemann-Hurwitz formula gives the same genus.

Gassmann equivalence means that $U_1,U_2$ meet each conjugacy class of $T$ in the same number of elements. To prove isospectrality directly, let $E_{\lambda }(X)$ be a Laplacian eigenfunction space on $X$, viewed as a representation of $T$. Pullback identifies $E_{\lambda }(S_i)$ with its $U_i$-invariant subspace. Averaging over $U_i$ is the projection onto this subspace, and henceThe character of a representation in this sum is constant on conjugacy classes. The sums are equal by Gassmann equivalence, so the eigenvalues and their multiplicities agree. This is the mechanism of the Sunada theorem. The constructed genus-three hyperbolic surfaces are isospectral.

Nonconjugacy of $U_1,U_2$ in $T$ does not prove nonisometry. For example, an isometry of $X$ outside the given subgroup $T$ might conjugate $U_1$ to $U_2$ and descend to an isometry of the quotients. More generally, an isometry between the quotients lifts to an isometry of ${\mathbb{H}}^2$ conjugating the two groups ${\varphi }^{-1}(U_i)$; there is no requirement that this lift come from an element of $T$. Additional symmetries of the three-cone-point construction are a possible source of this coincidence.

One can prevent this by modifying the common metric, while keeping the $T$ action isometric. Here is a geometric way to make that step precise. On the regular part of the base orbifold $O={\Gamma }_0\backslash {\mathbb{H}}^2$, choose a sufficiently small smooth conformal rescaling of a Riemannian metric of its metric whose local geometry distinguishes base points and local directions. One may use a generic smooth conformal factor with separate curvature features on a countable collection of small disks. Let these disks approach the cone points, with amplitudes decaying sufficiently rapidly that the lifted conformal factor and all its derivatives vanish at their centres. This gives smooth metrics on the covering surface, without retaining an open homogeneous region near a cone point. The purpose is to ensure that a local isometry in the regular part covering a quotient is the identity on the base: the curvature features identify its base point and local frame. Pull the metric back to $X$ and then descend it to both $S_i$. The only points with the cone-point rotational symmetry in their metric germs are the lifts of the cone points; the perturbation removes such symmetries in the regular part. Any quotient isometry must therefore preserve this finite distinguished set.

If the perturbed quotients were isometric, their local maps on the dense regular part would preserve the projection to $O$. They would therefore give equivalent connected covers of the punctured base; the correspondence between connected covering spaces and subgroups would conjugate ${\varphi }^{-1}(U_1)$ to ${\varphi }^{-1}(U_2)$ inside ${\Gamma }_0$. Applying $\varphi$ would conjugate $U_1,U_2$ inside $T$, contrary to hypothesis. Thus the perturbed quotients are nonisometric. The same averaging argument still proves isospectrality, because it requires a common $T$-invariant Riemannian metric, not constant curvature. The genus remains three.

This last step produces nonisometric isospectral metrics of variable curvature on genus-three surfaces. An oriented metric defines a Riemann surface through its conformal structure; using a conformal perturbation even preserves the original conformal structures. If “Riemann surface” means specifically its canonical curvature $-1$ metric, the last step should instead be described as producing Riemannian surfaces. The $\Delta (7,7,7)$ orbifold has no hyperbolic deformation parameters: uniformizing the perturbed metrics does not preserve their Laplace-Beltrami operator spectra, and cannot be used to claim nonisometric hyperbolic examples from this argument.

Put $r^2=y_1^2+y_2^2$ and write the Riemannian metric as a conformal rescaling of a Riemannian metric:The base metric is flat, so its Ricci scalar is zero and its Laplacian is $\Delta ={\partial }_1^2+{\partial }_2^2$. The formula for scalar curvature under conformal rescaling simplifies in dimension two toDifferentiate explicitly:Since $e^{-2\omega }=1+r^2$, the required Ricci scalar is

Because the half-plane metric is a conformal rescaling of the Euclidean metric, its angles are Euclidean angles. For each $y>0$, the hyperbolic lines meeting the imaginary axis at a fixed angle $\alpha$ at $iy$ are the Euclidean circleswhere $Y$ denotes the vertical coordinate. These circles have centres on the real axis and are therefore hyperbolic lines; varying $y$ gives the required collection.

To construct a triangle, put one vertex at $i$ and a second at $it$ on ${\ell }_{+}$, with $t>1$. Draw from these vertices the lines making interior angles $\alpha$ and $\beta$ with ${\ell }_{+}$. Their Euclidean centres and radii areIf their other intersection has angle $\gamma$, the law of cosines in the Euclidean triangle formed by the two centres and that intersection givesEquivalently,The right side is greater than one exactly when $\alpha +\beta +\gamma <\pi$. Since $(t+t^{-1})/2$ ranges continuously and strictly increasingly from $1$ to infinity for $t>1$, such a $t$ exists. The three lines therefore bound a hyperbolic triangle with the prescribed angles.

For a conformal rescaling of a Riemannian metric $h=\Omega (d{x}^2+d{y}^2)$, the Taubes equation in flat coordinates isThe first solution $u$ has $\Omega =1$ and vortex number $N$. For $\tilde{g}=e^{u}g$, the second solution therefore obeyswhere we used ${\Delta }_{\tilde{g}}=e^{-u}{\Delta }_g$ and converted the covariant delta distribution to the flat coordinate measure. Adding the equation for $u$ givesThus $u+\tilde{u}$ is again a flat-metric Taubes equation solution. Its vortex divisor is the union of the two divisors, with multiplicities added at coincident zeros, and its total vortex number isThis is vortex composition by conformal rescaling.

In dimension $N$, the conformal rescaling of a Riemannian metric $\tilde{g}=e^{2\omega }g$ changes the Ricci scalar by the displayed formula. In two dimensions the gradient-square term vanishes; for a flat base metric, $\tilde{R}=-2e^{-2\omega }\Delta \omega$.