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A geodesic polygon in the hyperbolic plane is bounded by finitely many geodesic segments. For curvature $-1$, its area is its angle defect, as given by hyperbolic polygon area. Pairing equally long edges produces hyperbolic surfaces, with smoothness at glued vertices determined by their total angles.

Use an oriented geodesic hyperbolic $2s$-gon $D$ with paired edges of equal length. Identify each pair by a length-preserving map reversing the boundary directions, to make an oriented base surface $M_0$ with possibly singular vertices. Assign to the $s$ edge pairs generators $A_1,\dots ,A_s$ of $T$, with inverse labels for reversed crossings. For each $g\in T$, take a copy $D_g$ and glue its positively labelled edge of type $k$ to the corresponding mate in $D_{g{A}_k}$. The generators make the glued surface $M$ connected. Left multiplication $D_g\mapsto D_{hg}$ commutes with these right-labelled gluings and gives the $T$ action with quotient $M_0$.

Existence for an arbitrary finite group can even be arranged with no cone singularities. If $T$ has $d$ generators, take $h\ge \max \{2,d\}$ and a regular hyperbolic polygon with $4h$ edges, all angles $\pi /(2h)$, paired with side word ${\prod }_{i=1}^h{a}_i{b}_i{a}_i^{-1}{b}_i^{-1}$. Its vertices form one cycle of angle $2\pi$, giving a smooth base of genus $h$. Label the $a_i$ by the generators of $T$, padded by identities, and every $b_i$ by the identity. The single vertex word is ${\prod }_i[A_i,1]=1$, so every lifted vertex also has angle $2\pi$. The resulting connected ordinary normal covering map has degree $|T|$ and deck group $T$.

This is the paired-polygon construction of finite surface covers. Interior and open-edge points are regular hyperbolic points; possible singularities occur only at vertices. When the base has cone points, the map is an orbifold covering and may be branched as a map between the underlying topological surfaces. It should not be called an unbranched manifold covering at those branch points.

A vertex cycle of a paired polygon is the collection of vertices identified to one point on $M_0$. Let $r$ be the number of cycles, ${\theta }_j$ the sum of the polygon angles in cycle $j$, and $C_j\in T$ the gluing word recorded on going around it. If $q_j=|C_j|$ is the order of that element, a lift closes after $q_j$ turns. It has angle $q_j{\theta }_j$, and there are $|T|/q_j$ such lifted vertices. The base and lifted cone orders, when they are integer orbifold orders, areAn orbifold homomorphism requires $q_j\mid m_j$. The lifted point is smooth exactly when $q_j{\theta }_j=2\pi$, equivalently $q_j=m_j$ in that case. For arbitrary cone metrics the angle formulas remain valid even when an angle does not correspond to an integer orbifold order.

Counting faces, edges and vertices gives the Euler characteristic of a paired-polygon surface cover:These are ordinary topological Euler characteristics, rather than orbifold Euler characteristics. If the lifted surface is smooth, the equivalent orbifold formula isFor a closed oriented smooth surface, $\chi =2-2g$ defines its genus. These formulas also follow from the Riemann-Hurwitz formula by subtracting the ramification deficits $|T|-|T|/q_j$ over each base vertex.

For $U\le T$, the quotient map $M\to M_1=U\backslash M$ is an ordinary normal covering if and only if $U$ acts freely. All possible point stabilizers in this construction are conjugates of $⟨C_j⟩$, so the concrete necessary and sufficient condition isThis is the freeness criterion for a subgroup of a polygon-cover deck group. Under it the deck group is $U$, the degree is $|U|$, andThere is no requirement $U◃T$ for this map. That different condition characterizes normality of the intermediate cover $M_1\to M_0$. If fixed points are retained and one instead speaks of an orbifold covering, the quotient is regular as an orbifold cover, but ordinary Euler characteristic need not divide by $|U|$; orbifold Euler characteristic does.

For the low-genus examples takeLet $U_1$ be a nonzero-vector stabilizer and $U_2$ a plane stabilizer. Their actions have seven points, so $|U_1|=|U_2|=24$ and $[T:U_i]=7$. They are almost conjugate subgroups: the numbers of fixed nonzero vectors for a matrix $A$ and its dual action $A^{-T}$ are both $2^{\dim \ker (A-I)}-1$. Thus the two coset permutation characters agree, which is equivalent to Gassmann equivalence. They are not conjugate: a point stabilizer has a common fixed nonzero vector, whereas a plane stabilizer fixes no nonzero vector globally. In particular, since $7\nmid 24$, neither subgroup contains any nonidentity element of a cyclic subgroup of order seven. All cone monodromies of exact order seven therefore act freely on the intermediate covers.

For genus three, start with a hyperbolic triangle whose three angles are $\pi /7$ and double it across one edge. The resulting quadrilateral $D$ has anglesThe side pairings produce three base vertex cycles, each of total angle $2\pi /7$, and an underlying sphere. Choose generators $A,B$ with $|A|=|B|=|AB|=7$; the three vertex words are $A,B,(AB{)}^{-1}$. Hence $s=2$, $r=3$ and all $q_j=7$. The full cover is smooth, andThe hyperbolic polygon area of $D$ is $8\pi /7$, so the seven-sheeted surfaces have area $8\pi$, agreeing with Gauss-Bonnet. Their equal spectra follow from the Sunada theorem. For complete numerical generator data, one possible pair isThese generate all of $T$ and have the required orders. This construction establishes the requested isospectrality; it does not establish nonisometry. In the constant-curvature triangle example a lifted reflection can identify the point and plane covers, as in reflection intertwining of triangle-cover coset actions. Breaking that symmetry with a generic metric gives nonisometric genus-three surfaces, but then the metric need not have constant curvature. Those are distinct assertions.

For genus four, use the cone-torus construction of genus-four Sunada surfaces. Take a regular hyperbolic quadrilateral with four anglesand pair opposite sides in the usual torus pattern $a,b,a^{-1},b^{-1}$. All four vertices belong to one cycle, whose total angle is $2\pi /7$; the underlying base is a torus with one cone point. Choose $A,B$ of order four generating $T$, with commutator $[A,B]$ of order seven. For example,Only the commutator is a cone monodromy, so the order-four side labels do not create cone points of order four. Here $s=2$, $r=1$, $q_1=7$, givingThe quadrilateral has area $12\pi /7$, and each seven-sheeted smooth quotient has area $12\pi$. Again the order-seven cone stabilizer meets neither $U_i$ nontrivially, so the metrics are smooth and the Sunada theorem gives identical spectra. The required group and subgroup orders are therefore 168 and 24 in both constructions; the genus-three side-generator orders are 7 and 7, and the displayed genus-four side-generator orders are 4 and 4, with cone-word order 7 in both cases.

In curvature-minus-one polar coordinates the hyperbolic metric is $d{s}^2=d{r}^2+{\sinh }^{2}rd{\theta }^2$. Thus the area element is $\sinh rdrd\theta$, givingIt is asymptotic to $\pi e^r$.

For a usual locally finite tessellation with compact congruent tiles of nonempty interior, let their diameter be at most $D$ and choose in each a congruently placed interior ball of radius $r_0>0$. These balls have disjoint interiors. The number of tiles meeting a geodesic segment of length one is uniformly bounded: their ball centres lie in a ball of radius $D+1$, and the disjoint $r_0$-balls lie in the larger ball of radius $D+1+r_0$. Comparing areas supplies a constant $K$. A path of length $R$, divided into at most $\lceil R\rceil$ such segments, therefore crosses at most $K(\lceil R\rceil +1)$ tiles. A small generic perturbation avoids vertices if steps are defined by crossing sides.

Consequently the union of tiles at most $n$ steps from the initial tile contains a metric ball of radius $cn-C$ for fixed $c>0,C$. If $A$ is tile area, their number $N(n)$ is at least $\mathrm{Area}B(cn-C)/A$, an exponential lower bound. Conversely any chain of $n$ adjacent tiles lies in $B(D(n+1))$ about a point in the initial tile; the interiors are disjoint, so $AN(n)\le \mathrm{Area}B(D(n+1))$. Thus $N(n)$ is bounded above and below by positive exponential functions of $n$. No precise common growth exponent is claimed.

An explicit example is the regular right-angled pentagon tiling $\{5,4\}$: reflect a regular hyperbolic pentagon with all angles $\pi /2$ in its sides. Four pentagons meet at each vertex, and their interiors tile the plane. Its tiles have area $3\pi -5\pi /2=\pi$ by the hyperbolic polygon area formula. The compactness and ordinary local finiteness assumptions exclude pathological decompositions with degenerate tiles.

In the curvature $-1$ normalization, the hyperbolic triangle area is its angle defect:Join one vertex of a convex geodesic polygon to its nonadjacent vertices. This gives $n-2$ hyperbolic triangles with disjoint interiors. Their angles at every polygon vertex sum to that vertex's interior angle. Adding the hyperbolic triangle areas therefore gives the hyperbolic polygon areaFor a regular hyperbolic polygon with prescribed area, place $n$ equally spaced vertices on a hyperbolic circle of radius $R>0$ and join consecutive vertices by geodesics. Rotations through $2\pi /n$ and reflections in radial lines show that this is a convex regular polygon. The triangle from its centre to a vertex and the midpoint of an adjacent side is right angled, with angles $\pi /n$, $\alpha (R)/2$, and $\pi /2$, and hypotenuse $R$. The angle form of the hyperbolic law of cosines givesConsequently $A(R)=(n-2)\pi -n\alpha (R)$ is continuous and strictly increasing. As $R\downarrow 0$, $\alpha (R)\to \pi -2\pi /n$, so $A(R)\to 0$; as $R\to \infty$, $\alpha (R)\to 0$, so $A(R)\to (n-2)\pi$. The intermediate value theorem proves existence for every permitted $A$. Indeed the required radius isThe argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.

After the $2n$ sides are identified in pairs, the polygonal-schema Euler count gives one face, $n$ edges and some number $V\ge 1$ of vertices. The Euler characteristic of a closed orientable genus-$g$ surface is $2-2g$, so

For genus two, take a regular octagon in the Poincare disc model and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling isThe quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequentlyEquivalently, the hyperbolic polygon area formula shows that the regular hyperbolic octagon fundamental polygon has areaas required by the Gauss-Bonnet theorem for a genus-two surface of curvature $-1$.

In curvature $-1$, a regular $n$-gon of circumradius $R>0$ has interior angle $\alpha (R)=2\arctan (\cot (\pi /n)/\cosh R)$, by the right-triangle angle form of the hyperbolic law of cosines. Thus its hyperbolic polygon area is $(n-2)\pi -n\alpha (R)$, continuously and strictly increasing from $0$ to $(n-2)\pi$. Every area in that open interval is therefore attained at exactly one circumradius.