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Continuation around one positively oriented loop induces an invertible linear map $M$ on the two-dimensional solution space. Its inverse is continuation around the reversed loop. Over $\mathbb{C}$, choose an eigenvector with eigenvalue $\lambda \ne 0$, and let $w$ be the corresponding nonzero solution. After continuation it becomes $\lambda w$.

Choose $\sigma$ with $e^{2\pi i\sigma }=\lambda$. The product $h(z)=z^{-\sigma }w(z)$ has trivial monodromy, so it is single-valued and holomorphic on the punctured disc. Its Laurent series, convergent on compact subannuli, gives the Monodromy eigenfunction Laurent representationNo assumption of a regular singular point has been made, so the negative part need not terminate.

For the second-order Fuchsian equation represented by the Papperitz symbol, the coefficient of $w^{\prime }$ isThis follows from the sums of local exponents at its two finite regular singular points. Abel's Wronskian identity, obtained by differentiating $w_1{w}_2^{\prime }-w_1^{\prime }{w}_2$ and using the differential equation, gives $W^{\prime }=-pW$. Thus, on every branch domain avoiding zero and one,The normalized expression is locally constant. Its analytic continuation around either singularity is unchanged: the power factors cancel the determinant of the local monodromy, equivalently the integral of $-p$ around that singularity. It therefore defines a single-valued holomorphic function on the punctured plane, locally constant near both punctures. Those punctures are removable, so the normalized expression extends to an entire function. This argument also covers resonant exponents and avoids assuming two pure power solutions in that case.

For a first-order system, it is important to specify the frame in which its coefficients are written. In the simple-pole convention intended here, a Fuchsian linear differential system has, at a finite singular point $a$,The residue is the matrix $R_a={\lim }_{z\to a}(z-a)A(z)$, computed entry by entry. Such a point is a regular singular point: on a sector, the coefficient bound $\Vert A(z)\Vert \le C/|z-a|$ and integration along rays give at most power growth of solutions. Intrinsically, a regular singular point means moderate power growth of all solutions on sectors, allowing logarithmic factors. This intrinsic definition is more general than requiring a simple coefficient pole in an arbitrary meromorphic frame.

For $A(z)=R/z$, a fundamental matrix on a chosen complex logarithm branch isDifferentiating the matrix exponential verifies ${\Phi }^{\prime }=R\Phi /z$. A counterclockwise circuit replaces $\log z$ by $\log z+2\pi i$, soThis is the residue and monodromy of a constant Fuchsian system. A change of solution basis conjugates $M$. Its eigenvalues are $e^{2\pi i\lambda }$ for the residue eigenvalues $\lambda$; a nontrivial Jordan block supplies the corresponding logarithmic terms in $z^R$. The monodromy does not uniquely recover $R$, since integer shifts of residue eigenvalues can leave their exponentials unchanged.

The coordinate-independent coefficient is the matrix-valued one-form $A(z)dz$. With $t=1/z$, its coefficient isBeing Fuchsian at infinity means $A(z)=O(z^{-1})$, while being ordinary there means $A(z)=O(z^{-2})$. If the finite simple poles are $a_1,\dots ,a_s$, subtract their principal parts:Each entry of $H$ is entire and tends to zero at infinity. Liouville's theorem therefore gives $H=0$. Expanding this partial fraction decomposition at infinity yieldsThus the sum of all residues, including the residue at infinity of the one-form, is zero. If infinity is ordinary, $R_{\infty }=0$ and the finite residues alone sum to zero. Treating $A$ merely as a scalar function of $z$, without transforming $dz$, would give the wrong infinity convention.

Put $\omega =e^{2\pi i/3}$. The possible finite singular points of $(B+zC)/(1-z^3)$ are $1,\omega ,{\omega }^2$, andA point $\zeta$ is removable exactly when $B+\zeta C=0$ as a matrix. At infinity,which is holomorphic at $t=0$. Consequently the actual set is $S=\{\zeta :{\zeta }^3=1,B+\zeta C\ne 0\}$, with $\{1,\omega ,{\omega }^2\}$ the generic set. The finite residues sum to zero, as follows also from ${\sum }_{{\zeta }^3=1}{\zeta }^{-1}={\sum }_{{\zeta }^3=1}{\zeta }^{-2}=0$.

Conversely, suppose a Fuchsian linear differential system has its finite singular points among these three roots and has no singularity at infinity. Then $Q(z)=(1-z^3)A(z)$ extends to an entire matrix function, and ordinariness at infinity gives $Q(z)=O(z)$. By the Cauchy estimate, each entry has zero derivatives of order two or higher, so $Q(z)=B+zC$. The same argument applies to any subset $S$, with additional cancellations at the omitted roots. This proves the requested normal form.

There is a genuine convention limitation: the normal form is not true under intrinsic regular singularity alone in an unrestricted frame. For example, if $N\ne 0$ and $N^2=0$, then $A(z)=N/(z-1{)}^2$ has fundamental matrix $I-N/(z-1)$. Its solutions have only power growth, so $1$ is intrinsically a regular singular point, and infinity is ordinary. Nevertheless its coefficient has a double pole and cannot equal $(B+zC)/(1-z^3)$. The simple-pole convention makes the preceding residue and normal-form assertions correct.

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.

Near infinity, $g(z{)}^2=(z-2)(z^2-1)\sim z^3$, so $g(z)\sim z^{3/2}$. One positive circuit around infinity changes $z^{3/2}$ by the monodromy factor $e^{3\pi i}=-1$. Thus analytic continuation does not return to the original value, and $\infty$ is a branch point.

A commutator contour around two branch points cancels the total monodromy of its integrand. For the contour circling $1,0$ clockwise and then $1,0$ anticlockwise, with initial positive-real branches, the Beta function obeysThe contour integral is entire in $z$, giving a meromorphic continuation of the Beta function.

A fundamental matrix for $F^{\prime }=RF/z$ on a chosen logarithm branch is $\exp (R\log z)$. A counterclockwise circuit changes the logarithm by $2\pi i$, giving the displayed monodromy matrix and the cyclic group it generates. Changing the solution basis conjugates the matrix. The residue eigenvalues exponentiate to the monodromy eigenvalues; the latter do not determine the former uniquely.

Fix ordered branch positions and conjugacy classes $C_i$ in a finite group $G$. The generating product-one tuples $(g_1,\dots ,g_r)\in \prod C_i$ form the inner Nielsen set. The type is rigid when this set is nonempty and a single simultaneous-conjugation orbit. Over $\mathbb{C}(T)$ this fixes the connected Galois cover up to isomorphism: tuples prescribe the monodromy epimorphism of the punctured sphere, and conjugate epimorphisms give isomorphic covers; the local completions at punctures are uniquely determined by their cyclic monodromy.

Over a cone point of order $m$, a cycle of length $\ell$ in the monodromy on sheets has ramification degree $\ell$ and residual cone angle $2\pi \ell /m$. The lifted inherited hyperbolic metric is smooth precisely when every such cycle has length $m$. Equivalently the subgroup defining the cover meets no conjugate of a nonidentity power of a cone generator.

For a Serre fibration $F\to E\to S^n$ with $n\ge 2$, the homological Serre spectral sequence has only columns $0,n$, and its only potentially nonzero differential isUsing an oriented fundamental class of $S^n$ identifies the source and target with $H_i(F)$ and $H_{i+n-1}(F)$. This gives the Wang homomorphism $\Delta$, and the resulting kernel/cokernel filtration pieces splice into the Wang sequence over a sphere.

For a fibration over the circle, the degree-preserving boundary map is instead the monodromy action minus the identity, $T_{\ast }-1:H_i(F)\to H_i(F)$, up to the chosen orientation sign. It arises from the cellular boundary with the fibre-homology local coefficient system. It is already incorporated in the $E_2$ page; there is no additional $d_1$ differential on an $E_2$-starting Serre spectral sequence. The distinction is visible for a product over $S^1$: its monodromy is the identity, so this boundary map is zero. The fundamental group action in the general construction is essential here.^[ref]