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In the paired-polygon construction of finite surface covers, a subgroup $U$ acts freely precisely when the displayed intersection is trivial for every vertex word $C_j$ and every $g\in T$. The quotient map $M\to U\backslash M$ is then a normal covering map with deck group $U$ and $\chi (U\backslash M)=\chi (M)/|U|$. Normality of $U$ in $T$ is instead the criterion for the intermediate orbifold cover $U\backslash M\to T\backslash M$ to be normal. These are different covering maps.

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.

Let $p=|a|$, $q=|b|$ and $r=|ab|$ denote element orders. Remove three points from an oriented sphere. Its fundamental group has peripheral generators ${\gamma }_a,{\gamma }_b,{\gamma }_c$ with ${\gamma }_a{\gamma }_b{\gamma }_c=1$. Send them to $a,b,(ab{)}^{-1}$. Since $a,b$ generate $T$, the kernel defines a connected regular covering of the punctured sphere with degree $|T|$ and deck transformation group $T$.

Complete the lifted punctures by disks. Above the three removed points the monodromy cycles have lengths $p,q,r$, respectively, and the disk projection has the local form $z\mapsto z^p$, $z^q$ or $z^r$. This gives a closed connected oriented topological surface $M$ and an orientation-preserving $T$-action. At the three kinds of filled-in points the stabiliser subgroups are conjugates of $⟨a⟩$, $⟨b⟩$ and $⟨ab⟩$, respectively. Every other point has trivial stabiliser subgroup. Orders equal to one simply correspond to unbranched filled points.

The punctured sphere has Euler characteristic $-1$, so its cover has Euler characteristic $-|T|$. Filling the ends adds $|T|/p+|T|/q+|T|/r$ disks. This proves the formula for the branched surface action from two generators:

For a subgroup $U$, its quotient map is always regular as a branched quotient. It is an ordinary unbranched normal covering map exactly when $U$ acts freely, that is,Necessity follows from the listed stabiliser subgroups. Conversely, these conditions eliminate every possible nontrivial point stabiliser. A free finite action gives a covering space, and its deck transformations are $U$, acting transitively on each fibre; hence the cover is normal. In this caseThe condition $U⊴T$ concerns regularity of the different map $U\backslash M\to T\backslash M$; it is not needed for $M\to U\backslash M$ when the action of $U$ is free.

For the specified order-$96$ group, $p=q=3$. Set $c=ab$ and $z=c^3$. Since $z$ has order two, $c^6=1$ and $c^3\ne 1$, so $|c|$ is either two or six. The order-two alternative would make the displayed formula give $\chi (M)=96(1/3+1/3+1/2-1)=16$. That is impossible for a connected closed oriented topological surface, whose Euler characteristic is $2-2g\le 2$. Thus $|c|=6$, and

Each $U_i$ has order eight. By Lagrange's theorem, it contains no element of order three or six, so it meets all conjugates of $⟨a⟩$ and $⟨b⟩$ trivially. In $⟨c⟩$, the only nonidentity element whose order divides eight is $z=c^3$. Since $z$ is central, it is also the unique such element in every conjugate of $⟨c⟩$, and it is excluded from both $U_i$ by hypothesis. Both subgroup actions are therefore free. The quotients are closed oriented surfaces with

The branched-cover construction supplies a smooth structure, with the stabilisers acting by rotations in the disk charts. Average a smooth Riemannian metric over $T$ to make the whole action isometric. The Gassmann equivalence of $U_1,U_2$ then gives isospectral manifolds: for every upstairs eigenspace with character ${\chi }_{\lambda }$, its downstairs multiplicities are $|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\lambda }(u)$, and these are equal because the two subgroups have equal intersection sizes with every conjugacy class. This proves the spectral assertion for any such invariant metric.

To guarantee nonisometry, choose the invariant metric with curvature markers distinguishing finite-cover quotients. Here are the geometric details, since nonconjugacy alone would not guarantee nonisometry for an arbitrary metric. Choose a small disk in the free locus of the $T$-action, with all its translates disjoint. In that disk choose three noncollinear points in a geodesically convex neighbourhood. Supported conformal smooth bump functions can give them three distinct high values as strict Gaussian curvature maxima, occurring nowhere else in the unmodified metric. Arrange these three maxima as one tight cluster, much closer to each other than to any translated cluster, and copy the same modification to every translate. Curvature and its first few derivatives can be adjusted by the second and higher derivatives of the conformal factor; narrow bumps give high distinguished curvature with small change in metric distances. Further small supported perturbations remove accidental maxima of these exact heights. Thus the resulting metric remains smooth and $T$-invariant, and its marked clusters are intrinsic and uniquely labelled by their three curvature values.

Any proposed quotient isometry $F:U_1\backslash M\to U_2\backslash M$ must carry a whole labelled cluster to another. Lift its restriction near that cluster to the corresponding two marked disks in $M$, and identify these disks using the appropriate group element $t\in T$. The resulting local isometry fixes all three marked points. Its differential at the first point fixes the two independent directions of the unique geodesics to the other two; hence it is the identity, and the local isometry itself is the identity near that point. If $p_i:M\to U_i\backslash M$ are the quotient maps, we therefore haveon an open set. Both sides are local isometries on the connected complete Riemannian surface $M$. A local isometry is determined along each geodesic by its value and differential at the initial point, so the equality propagates to all of $M$.

For $u\in U_1$, this equality implies $p_2\circ tu=p_2\circ t$. Evaluating at a free point of the $T$-action shows $tu{t}^{-1}\in U_2$; equivalently these maps differ by a deck transformation of $p_2$. Thus $t{U}_1{t}^{-1}\subseteq U_2$, and equality of subgroup orders gives equality. This contradicts their assumed nonconjugacy. For this marked invariant metric the two quotients are nonisometric, isospectral Riemannian surfaces of genus two. No constant-curvature requirement is imposed or used.

Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\hat{x}\in p^{-1}(x_0)$, the path lifting theorem gives a unique lift of $\gamma$ starting at $\hat{x}$. Send $\hat{x}$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the fibre bijection by path liftingIn particular, all fibres have the same cardinality.

A connected covering is a normal covering map when its deck transformations act transitively on a fibre. Equivalently, for a choice of ${\hat{x}}_0$ above $x_0$,The relevant lifting criterion for a covering space says that a based map $f:(Y,y_0)\to (X,x_0)$ lifts through $p$ exactly whenFor a universal covering, $\hat{X}$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a universal covering map is normal.

Now consider connected finite covers of the closed orientable surface ${\Sigma }_g$. By the classification of connected covering spaces, degree-$n$ connected covers correspond to index-$n$ subgroups of the fundamental group of a closed orientable surface, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

It remains to show that these are the only forced cases. Let $g\ge 2$ and $n\ge 3$. WriteIn the symmetric group $S_n$, putDefine a homomorphism byand send all remaining generators to the identity. The relation is respected becauseThe cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.

LetThe natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\ge 3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the nonnormal finite cover of a higher-genus orientable surface.

Consequently, among connected covers that exist, normality is forced exactly whenwith the qualification that $g=0$ permits only $n=1$, as summarized by the forced normality of finite connected covers of orientable surfaces.

A universal covering has simply connected total space, so its associated subgroup of the base fundamental group is trivial. The trivial subgroup is normal; hence every universal covering map is a normal covering map.