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The affine germ surface over the plane with the eight roots of unity removed is compactified by adding the eight simple branch points above those roots and two unbranched points above infinity. The degree-two map to the Riemann sphere has total ramification eight, so the Riemann-Hurwitz formula gives genus three.

If two smooth plane conics over $\mathbb{C}$ meet at four distinct points, their tangent incidence curve is a smooth projective genus one curve. Projection to $C_2$ is a double cover branched at those four intersections. The tangent equation becomes a quadratic whose discriminant cuts out $C_1\cap C_2$; its four simple zeros make the cover smooth and connected. The Riemann-Hurwitz formula then gives genus one.

Let $m=\mathrm{ord}(A)$, $n=\mathrm{ord}(B)$ and $r=\mathrm{ord}(AB)$. Cut the Riemann sphere along arcs joining three marked points, say $0,1,\infty$, so the complement is simply connected. Take copies labeled by the elements of $T$ and glue the slit banks with monodromies given by left multiplication by $A,B,(AB{)}^{-1}$. The product relation ensures consistency. The surface is connected because $A,B$ generate $T$. At a puncture with monodromy order $m$, fill each cycle using the local complex coordinate $z=w^m$; use the analogous charts at the other two punctures. This constructs the three-branch-point regular surface cover $N\to {\mathbb{P}}^1$.

Right multiplication by inverses defines a left action of $T$, commutes with the monodromy gluing and extends holomorphically over the filled points. Average a compatible metric over this finite group to obtain a metric for which $T$ acts by Riemannian isometries. When the genus is at least two, the unique compatible curvature-minus-one metric from the uniformization theorem is an invariant choice. Away from the three marked fibers the action is free. The nontrivial stabilizer subgroups are the conjugates of $⟨A⟩$, $⟨B⟩$ and $⟨AB⟩$, of orders $m,n,r$, omitting order-one groups.

There are $|T|/m$, $|T|/n$, $|T|/r$ points over the corresponding branch points. The Riemann-Hurwitz formula givesThis construction also covers the spherical and toroidal cases. If one monodromy has order one it is simply an unbranched marked fiber, with no nontrivial stabilizer.

For the supplied coset actions, faithfulness gives $m=n=3$. Compose permutations with the rightmost factor acting first. The two products $AB$ have cycle decompositionsThus $r=6$. The index is twelve and each subgroup has order eight, so $|T|=96$. ConsequentlyAll cycles of $A,B,AB$ on each coset set have the full orders $3,3,6$. No nontrivial power of any of these branch monodromies fixes a coset. The fixed-coset criterion means that each $U_i$ has trivial intersection with every conjugate of each cyclic branch stabilizer. Since these are all stabilizers on $N$, both $U_i$ act freely. Their quotient surfaces are therefore smooth closed surfaces, not cone-point orbifolds. Sunada theorem supplies isospectrality, and the unbranched Euler characteristic formula givesChoose the hyperbolic metric for these surfaces and their common genus-nine cover.

In this natural construction the two genus-two surfaces are isometric, by an orientation-reversing map. Define the sheet permutationDirect composition verifiesComplex conjugation of the base sphere fixes its three real branch points and reverses the two chosen generating loops, so their monodromies become the inverses. These identities supply a lift, after relabeling sheets by $P$, to an anticonformal map between the two branched covers. It extends across the branch points by the local power charts. An anticonformal map between closed hyperbolic Riemann surfaces is a Riemannian isometry, by uniqueness of the compatible hyperbolic metric.

Equivalently, take the hyperbolic triangle group of signature $(3,3,6)$. Reflection in the side joining the two order-three vertices conjugates its two vertex rotations to their inverses. The displayed sheet permutation identifies the reflected subgroup for the first covering with the subgroup for the second, up to an irrelevant change of base sheet. This is the orientation-reversing equivalence of branched-cover monodromy. The exhibited isometry is anticonformal, so the conclusion is about Riemannian isometry, without asserting that this map is holomorphic. Merely knowing the subgroups were almost conjugate would not have settled isometry; the explicit inverse-monodromy symmetry does. Nor does the data force this symmetry for every arbitrarily chosen invariant deformation of the cover's metric: it is present for the constructed hyperbolic metric.

To analyze the genus one tangent incidence curve of two conics, choose projective coordinates in which $C_1$ has equation $XZ-Y^2=0$. Its points and tangent lines are parametrized by $[a:b]\in {\mathbb{P}}^1$ asFor $x=[X:Y:Z]\in C_2$, the incidence equation is a nonzero homogeneous quadratic in $a,b$, so the projection $E\to C_2$ is finite of degree two. Its discriminant vanishes exactly when $Y^2-XZ=0$, that is, at $C_1\cap C_2$. There are four such points; Bézout theorem says their intersection multiplicities sum to four, so all four intersections are transverse.

Locally on $C_2$, after choosing an affine tangent-parameter chart and completing the square, the incidence equation is $w^2=u(s)$, where $s$ is a local parameter and $u$ has a simple zero at each intersection. This is smooth, with ramification index two there; away from those points the roots are distinct and the cover is étale. The analogous chart covers a root at infinity. Thus $E$ is smooth everywhere, and the cover has exactly four ramification points. It is connected: the discriminant has odd valuation at each of its four zeros, so it cannot be a square in the function field of $C_2$. The associated quadratic extension is therefore a field, not two separate sheets.

A smooth plane conic over $\mathbb{C}$ is isomorphic to the projective line. Apply the Riemann-Hurwitz formula to this connected degree-two cover:Consequently $\boxed{g(E)=1}$, and $E$ is a smooth projective genus one curve.