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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.

A deck transformation is a homeomorphism $h:X\to X$ satisfying $p\circ h=p$. A connected cover is regular when its deck group acts transitively on each fiber. In the usual connected, locally path-connected covering-space setting, this is equivalent to normality of $p_{\ast }{\pi }_1(X)$ in ${\pi }_1(Y)$.

A two-sheeted cover corresponds to an index-two subgroup, which is normal: its two left and right cosets are the subgroup and its complement. If ${\pi }_1(Y)$ is abelian, every subgroup is normal. These observations prove both regularity assertions.

For examples over the wedge of two circles, label their generators $a,b$ and use three sheets numbered $1,2,3$. A permutation describes the lift of each oriented base edge. For a regular cover take $a=(123)$ and $b=1$: the covering graph has an oriented three-cycle of $a$-edges and a $b$-loop at each vertex. Cyclic rotation of the vertices is a transitive deck group $C_3$. For a nonregular cover take $a=(123)$ and $b=(12)$: the same $a$-cycle is accompanied by a $b$-edge in each direction between 1 and 2, and a $b$-loop at 3. It is connected. A deck permutation must commute with both generators, whose image is all $S_3$; their centralizer is trivial. Thus its deck group is not transitive and the cover is nonregular. Each vertex has exactly one entering and one leaving edge of each label, verifying that these graphs really are three-sheeted covers.

First realize the free isometric action. Choose generators $t_1,\dots ,t_r$ of the finitely presented group $T$. In every dimension $d\ge 3$, the connected sumhas fundamental group the free group $F_r$. Map its generators onto the $t_j$ and take the connected regular covering $N\to B_d$ corresponding to the kernel. Its deck transformation group is $F_r/\ker \pi \cong T$. Lift any Riemannian metric on $B_d$ to $N$. Deck transformations are then Riemannian isometries, and a deck transformation fixing one point is the identity by uniqueness of lifts. This gives the free isometric realization of a finitely generated group. For the trivial group use $S^d$. If $T$ is infinite the covering manifold need not be compact, which is allowed in this first construction. Simple connectivity of $N$ is not being asserted here.

The requested nonhomeomorphism conclusion needs an extra hypothesis. For a counterexample even with distinct subgroups, take $T=S_3$, $U_1=⟨(12)⟩$ and $U_2=⟨(23)⟩$. They are conjugate and therefore Gassman equivalent. For every free isometric action of $T$, the conjugating element induces a Riemannian isometry $U_1\backslash N\to U_2\backslash N$. These quotients cannot be nonhomeomorphic. Identical subgroups give an even simpler obstruction. Thus the stated equivalence condition alone cannot imply the claimed topology.

A sufficient version is: for a finite $T$ with Gassmann equivalent, nonisomorphic subgroups $U_1,U_2$, there are closed isospectral manifolds with fundamental groups $U_1,U_2$, and hence they are not homeomorphic. To prove it, realize $T$ as the fundamental group of a closed smooth manifold $B$ of dimension four. Here is the needed closed-manifold realization of a finitely presented fundamental group. Start with a four-dimensional zero-handle and one one-handle for each generator. Attach two-handles along disjoint embedded boundary circles representing the relators, with arbitrary framings. The boundary is three-dimensional, so finitely many such loops can be chosen disjoint. The resulting compact handle manifold $W$ has ${\pi }_1(W)=T$ by the Seifert-van Kampen theorem.

The inclusion $\partial W\to W$ is surjective on fundamental groups: relative to the boundary the dual handle decomposition uses only handles of indices two, three and four, none of which introduces a fundamental-group generator. Double $W$ along its boundary. In the amalgamated product for the double, both boundary maps are the same surjection onto $T$, so the two copies of $T$ are identified completely and ${\pi }_1(B)=T$. The double is closed, connected and smooth after smoothing its collar. This construction works in any dimension at least four; it does not claim that every finitely presented group is a closed three-manifold group.

Let $N$ now be the universal cover of $B$ with the lifted metric. Since $T$ is finite, $N$ is compact and simply connected, with a free isometric deck action of $T$. By Sunada theorem, $M_i=U_i\backslash N$ are isospectral. Because $N$ is their universal cover, ${\pi }_1(M_i)\cong U_i$. Nonisomorphic fundamental groups rule out a homeomorphism. For an infinite $T$ the same qualified argument applies to finite-index subgroups with equal coset characters: divide first by the intersection $K={\mathrm{Core}}_T(U_1)\cap {\mathrm{Core}}_T(U_2)$ of their finite-index subgroup cores, producing a compact intermediate cover and a finite isometry group to which Sunada applies.

For explicit nonhomeomorphic examples, let $p$ be an odd prime. Let $H_p=(\mathbb{Z}/p\mathbb{Z}{)}^3$, and let $K_p$ be the Heisenberg group over a prime field, whose elements are triples with multiplicationBoth have order $p^3$, and every nonidentity element has order $p$. For $K_p$, induction givesso the assertion follows at $k=p$, since $p$ is odd. But $H_p$ is abelian and $K_p$ is not: $(1,0,0)$ and $(0,1,0)$ do not commute.

Embed both groups regularly in $S_{p^3}$. Every nonidentity element in either regular representation has cycle type $p^{p^2}$. Hence both embedded subgroups meet the identity class once, the class of that cycle type $p^3-1$ times, and all other classes zero times. They are nonisomorphic Gassmann equivalent regular subgroups. Taking $p=3$ and using the closed four-manifold construction with $T=S_{27}$ gives an explicit pair of isospectral nonhomeomorphic quotients, completing the intended construction under a sufficient hypothesis.

Finally choose distinct odd primes $p_1,\dots ,p_n$ and takeConjugacy classes in a direct product are products of classes, so their intersection counts factor. Thus all $2^n$ subgroups are Gassmann equivalent. They are pairwise nonisomorphic: the unique Sylow $p_j$-subgroup of $U_{ϵ}$ is its $j$th factor, and whether it is abelian records ${ϵ}_j$. Realize $T_n$ as the fundamental group of a closed four-manifold and quotient its universal cover by these subgroups. The binary family of nonhomeomorphic Sunada quotients givesThis existence statement is valid; the earlier universal nonhomeomorphism assertion from Gassmann equivalence alone is not.

Replace the open knot complement by the compact knot exterior $E_K$; they have the same homotopy type. The abelianizationsends an oriented meridian of a knot to $1$. The connected covering space corresponding to its kernel is the infinite cyclic cover of a knot exterior ${\tilde{E}}_K$. Choose a generator $t$ of its deck transformation group. Thenis the Alexander module of a knot over $\Lambda =\mathbb{Z}[t,t^{-1}]$. It is a finitely presented torsion module, and its order, equivalently the greatest common divisor of the appropriate maximal matrix minors of an Alexander matrix, is the Alexander polynomial of a knot, up to a unit $\pm t^m$.

For a fibered knot, the infinite cyclic cover of a knot exterior is $\Sigma \times \mathbb{R}$. Write the mapping torus with the gluing convention in the question. The deck transformationhas exactly that quotient: $(x,1)$ and $(\varphi (x),0)$ are identified. Choose $t=D$. Its action on $H_1(\Sigma ;\mathbb{Z})$ is ${\varphi }_{\ast }$. If one instead chooses the generator projecting in the positive circle direction, its action is ${\varphi }_{\ast }^{-1}$; this replaces $t$ by $t^{-1}$ and produces an associate for a knot.

Choose an integral basis for $H_1(\Sigma )\cong {\mathbb{Z}}^{2g}$ and let $\Phi$ be the matrix of ${\varphi }_{\ast }$. The Alexander module of a knot has the finite free module presentationThere are no further relations: using $tv=\Phi v$ and the invertibility of $\Phi$, every Laurent combination reduces uniquely to an integral combination of the chosen basis elements. Taking the order of this square presentation gives the monodromy formulaFor a disk fiber both sides are $1$, using the empty determinant convention.

Use the closed genus $g\ge 2$ version of Teichmüller's uniqueness theorem. Let $f_0:X\to Y$ be a Teichmüller map: for a nonzero holomorphic quadratic differential $q$, normalized by ${\int }_X|q|=1$, and $0<k_0<1$,The value at a zero of $q$ is irrelevant to the Beltrami coefficient, which is defined almost everywhere. For every quasiconformal map $f:X\to Y$ preserving orientation and homotopic to $f_0$, the conclusion isThe normalized $q$ is also uniquely determined when $k_0>0$.

We use the following analytic input, stated with its hypotheses. The Reich–Strebel inequality says that if $f_0$ has the displayed Beltrami coefficient and a competitor $f$ has the same target and homotopy class, then, for $\mu ={\mu }_f$,This is the standard fundamental inequality for an integrable holomorphic quadratic differential; it also holds on a finite-type punctured surface with the homotopy fixing the punctures. Its formulation is given in Gardiner and Hu, §5. We quote this analytic inequality as the lecture result used in the proof.

Put $k=\Vert \mu {\Vert }_{\infty }$. Pointwise, away from the isolated zeros of $q$,Integrating against the probability density function $|q|$ proves $K_0\le K(f)$. If $K(f)=K_0$, all these inequalities are equalities almost everywhere. The strictly increasing last function forces $|\mu |=k_0$ almost everywhere; equality in the triangle inequality then forces $\mu q/|q|=k_0$. ThereforeTwo quasiconformal maps with the same Beltrami coefficient differ by postcomposition with a biholomorphism, by the local chain rule for the Beltrami equation. Consequently $h=f\circ f_0^{-1}$ is a biholomorphism from $Y$ to itself homotopic to the identity.

For completeness, such a biholomorphism is the identity when $g\ge 2$. Apply the uniformization theorem and choose the lift of the homotopy to $\mathbb{H}$ starting at the identity. Its endpoint lift $\tilde{h}$ commutes with every deck transformation. It is a real Möbius transformation. The compact quotient has no parabolic Möbius transformations in its deck group, and freeness excludes elliptic Möbius transformations. Two distinct hyperbolic axes exist: a discrete free group preserving just one axis would be cyclic, contradicting the fundamental group of a closed surface of genus at least two. Commutation with two deck transformations represented by hyperbolic Möbius transformations having distinct axes makes it fix their boundary endpoints individually; there are at least three such endpoints. A Möbius transformation fixing three points is the identity. Hence $h=\mathrm{id}$ and $f=f_0$.

If $q_1$ is another unit-area differential for the same map, then $|q_1|/q_1=|q|/q$, so $q_1/q$ is positive real wherever defined. This meromorphic function is constant by the open mapping theorem, and normalization makes the constant one. Without normalization, positive multiples of $q$ describe the same map. At $k_0=0$, uniqueness of the map still holds in genus at least two, but there is no distinguished $q$. In genus one, translations supply nontrivial biholomorphisms homotopic to the identity, so equality determines the map only up to those biholomorphisms; the genus hypothesis cannot be omitted.

Take $X=S^1\vee S^1$, the wedge of two circles, and label its oriented loops $a$ and $b$. For a fixed $n\ge 3$, letbe permutations of the fibre $\{1,\dots ,n\}$. Construct a labelled graph ${\hat{X}}_n$ with these $n$ vertices, an oriented $a$-edge from $i$ to $\alpha (i)$, and an oriented $b$-edge from $i$ to $\beta (i)$. Map every vertex to the wedge point and each labelled edge homeomorphically onto its corresponding loop. Every vertex has exactly one incoming and one outgoing edge of each label, so this is an $n$-sheeted permutation covering of a wedge of circlesIt is connected because the $n$-cycle $\alpha$ acts transitively on the fibre.

A deck transformation induces a permutation $\gamma$ of the vertices that commutes with both monodromy permutations $\alpha$ and $\beta$. Conversely, such a permutation determines a deck transformation, by the deck transformation group as a monodromy centralizer. The permutations $\alpha$ and $\beta$ generate $S_n$: the conjugates ${\alpha }^j\beta {\alpha }^{-j}$ include the adjacent transpositions, which generate the symmetric group. Hence $\gamma$ centralizes $S_n$ and lies in its center of a group. For $n\ge 3$, the center of $S_n$ is trivial, so $\gamma =1$. A deck transformation fixing one point is the identity by part (a), and therefore

A Wirtinger presentation from a connected knot diagram has one generator per arc and one relator per crossing, with one relator redundant. Its presentation complex is a finite two-dimensional CW complex $X$ with one zero-cell, $n$ one-cells, and $n-1$ two-cells, and the usual diagrammatic construction gives a homotopy equivalence $X\simeq E_K$.

The abelianization of the knot group is $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$, generated by a meridian of a knot. Every homomorphism to the cyclic group $\mathbb{Z}/2$ factors through this abelianization, and reduction modulo two is its unique surjection. Thus the requested map $\alpha$ is unique.

Its kernel determines a two-sheeted covering space $\hat{X}$. The nontrivial deck transformation acts on cellular chains and homology, giving them module structures over the group ringLift one copy of every cell of $X$; its two deck translates form a free $\hat{R}$-basis. Hence $C_0\cong \hat{R}$, $C_1\cong {\hat{R}}^n$, and $C_2\cong {\hat{R}}^{n-1}$. If $\tilde{X}$ is the infinite cyclic cover, its cellular chains are free over $R=\mathbb{Z}[t^{\pm 1}]$, and imposing $t^2=1$ gives

For an odd prime $p$, the two idempotents $e_{+}=(1+t)/2$ and $e_{-}=(1-t)/2$ split the group algebraThe plus summand is the cellular chain complex of $X$ with ${\mathbb{F}}_p$ coefficients, while the minus summand is $C_{-}=C_{\ast }^{\mathrm{cell}}(\tilde{X}){\otimes }_R{R}_{-}$. Therefore

On the minus summand, the boundary $t-1:C_1\to C_0$ becomes multiplication by $-2$, which is invertible in ${\mathbb{F}}_p$, so $H_0(C_{-})=0$. After the corresponding cancellation, the remaining square boundary matrix is an Alexander matrix specialized at $t=-1$. It is singular over ${\mathbb{F}}_p$ exactly whenor equivalently when $p$ divides the knot determinant $\det K$. Thus $H_{\ast }(C_{-})$ is nonzero exactly in that case.

Finally, $\det K=|{\Delta }_K(-1)|$ is a nonzero odd integer, so the minus complex is acyclic over $\mathbb{Q}$. Since a knot exterior has the rational homology of a circle,

Send each orientation-preserving isometry to its isotopy class. This is a group homomorphismFor $g\ge 2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the hyperbolic plane: after composing with a deck transformation, its lift commutes with the surface fundamental group. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the injection of a finite hyperbolic isometry group into a mapping class group.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the unit sphere is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat torus, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

Write ${\mathbb{F}}_2=\mathbb{Z}/2$. A double covering has a deck transformation $\tau$ that exchanges the two points in every fibre. Every singular simplex $\sigma :{\Delta }^q\to Y$ has exactly two lifts $\tilde{\sigma }$ and $\tau \tilde{\sigma }$. Define the mod-two transfer chain map of a double covering byand let $f_{\#}$ send a simplex of $X$ to its composite with $f$. Uniqueness of lifted faces shows that $T$ commutes with the boundary operator, so both maps are chain maps.

For each base simplex, its two lifts span a copy of ${\mathbb{F}}_2^2$. On this summand, $T$ is $a\mapsto (a,a)$ and $f_{\#}$ is $(a,b)\mapsto a+b$. Hence the sequence is exact in every degree:

A connected Riemann surface whose universal cover is the unit disc inherits the complete curvature-minus-one metric displayed on its cover. Möbius transformations preserving the disc are isometries, so deck transformations preserve the metric and it descends to the surface. This uniformization notion must be distinguished from the potential-theoretic existence of a Green function. Holomorphic maps between such surfaces contract their metric by lifting to Schwarz-Pick theorem on the covers.

A covering map between Riemannian manifolds that is a local isometry. The metric on the covering manifold is the pullback of the base metric; its deck transformations are isometries. This local metric agreement makes the Laplace-Beltrami operator commute with pullback of smooth functions.