On a complex manifold, a divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{m}_{Y}Y$, where $Y$ ranges over irreducible complex analytic hypersurfaces and $m_Y\in \mathbb{Z}$. The local algebra needed here is that the local ring ${\mathcal{O}}_{X,x}$, identified in coordinates with convergent power-series germs, is a regular local ring and a unique factorization domain. In particular, each height-one prime is generated by one irreducible germ, and a nonzero meromorphic germ is a quotient of holomorphic germs. Reduced local equations of the same hypersurface differ by a holomorphic unit. These facts include local branches when a globally irreducible hypersurface has several branches at a singular point.

The order of a meromorphic function along a hypersurface is the exponent of its reduced local equation in the quotient's factorization. Equivalently, at a generic smooth point of $Y$, write $f=t^{m}u$, with $t$ a transverse local equation of $Y$ and $u$ a holomorphic unit; then ${\mathrm{ord}}_Y(f)=m$. Changing a reduced equation multiplies it by a unit and leaves $m$ unchanged. These local orders agree on overlaps and are constant along the generic smooth locus of an irreducible hypersurface. A nonzero global meromorphic function therefore defines its principal divisor on a complex manifoldPositive coefficients record zeros and negative coefficients record poles.

A local defining function $f_i$ for $D$ on $U_i$ is a nonzero meromorphic function whose orders there equal the coefficients of $D$. It can be constructed as the product of local reduced hypersurface equations to the prescribed integer powers. On an overlap, $f_i/f_j$ has no zero or pole along any hypersurface. The unique factorization domain property then makes it a nowhere-zero holomorphic function, after choosing representatives of the germs. The holomorphic line bundle associated to a divisor is obtained by gluing holomorphic frames usingThese ratios satisfy the cocycle condition, and the expressions $f_i{e}_i$ agree on overlaps. They form the canonical meromorphic section of a holomorphic line bundle with divisor $D$.

If $D=\mathrm{div}(f)$ globally, $f_i/f$ is a holomorphic unit on $U_i$, and $(f_i/f)e_i$ is a global nowhere-zero holomorphic section. Hence $[D]$ is holomorphically trivial. Conversely, if $[D]$ has a nowhere-zero global holomorphic frame $e$, write $e_i=a_{i}e$ with $a_i$ holomorphic units. The coefficients $f_i{a}_i$ of the canonical meromorphic section of a holomorphic line bundle agree on overlaps and give a nonzero global meromorphic function $f$. Since multiplication by a unit does not change orders, $\mathrm{div}(f)=D$. Therefore

For a smooth hypersurface $Y\subset X$, the adjunction formula isIndeed, the holomorphic conormal sequence has first term $N_{Y/X}^{\ast }\cong [-Y]{|}_Y$; taking its determinant yields $K_X{|}_Y\cong N_{Y/X}^{\ast }\otimes K_Y$, which gives the formula. Here the canonical bundle is the top exterior power of the holomorphic cotangent bundle.

Apply this first to $W\subset {\mathbb{P}}^4$. Its degree-$d_1$ equation is a section of $\mathcal{O}(d_1)$, so $[W]\cong \mathcal{O}(d_1)$. Since $K_{{\mathbb{P}}^4}\cong \mathcal{O}(-5)$, we obtain $K_W\cong {\mathcal{O}}_W(d_1-5)$. The restricted second equation cuts out $S$ in $W$ and has associated bundle ${\mathcal{O}}_W(d_2)$. A second application of adjunction formula gives the canonical bundle of a regular projective complete intersection

It remains to prove that a nonzero exponent cannot yield a trivial holomorphic line bundle on $S$. Choose a connected component $S_0$ and a hyperplane meeting but not containing it. Such a hyperplane exists: a positive-dimensional compact projective submanifold cannot lie entirely in an affine chart, since all its affine coordinate functions would be global holomorphic functions and hence constant. A hyperplane through one point and not another consequently meets $S_0$ properly. Its equation gives a holomorphic section of ${\mathcal{O}}_{S_0}(1)$ that has both a zero and a nonzero value. For $m>0$, the $m$th power has the same two properties. Were ${\mathcal{O}}_{S_0}(m)$ trivial, this section would be represented by a global holomorphic function, contradicting its constancy on compact connected $S_0$. For $m<0$, triviality would imply triviality of its positive dual power. This proves nontrivial hyperplane powers on a compact projective submanifold and henceThe degrees are positive because the two equations define a hypersurface and a surface of the specified dimensions.

The calculation uses “non-singular” in the usual smooth complete-intersection sense: the equations define reduced smooth divisors, with independent differentials along $S$. If only their reduced zero sets are required to be smooth, polynomial multiplicities cannot determine the canonical bundle. For example, $Q_1=Z_0^m$ and $Q_2=Z_1^4+Z_2^4+Z_3^4+Z_4^4$ have the same smooth quartic surface in the hyperplane $Z_0=0$ for every $m\ge 1$. That surface has trivial canonical bundle, while the displayed polynomial degree sum is $m+4$. Thus unreduced equations must not be substituted into the adjunction formula for the reduced surface.

A holomorphic line bundle $L\to X$ is a rank-one holomorphic vector bundle: it has local trivializations $L{|}_{U_i}\cong U_i\times \mathbb{C}$ whose fiberwise linear transition functions of a vector bundle are nowhere-zero holomorphic functions. Equivalently, choose local holomorphic frames $e_i$ and write $e_j=g_{ij}{e}_i$ on overlaps, with $g_{ij}\in {\mathcal{O}}^{\ast }(U_i\cap U_j)$ and the cocycle relation on triple overlaps. A holomorphic section over $U$ has coefficients $s_i$ holomorphic on $U\cap U_i$ in $s=s_i{e}_i$; consistency means $s_i=g_{ij}{s}_j$.

The dual bundle has fiber $L_x^{\ast }={\mathrm{Hom}}_{\mathbb{C}}(L_x,\mathbb{C})$. Its dual frames $e_i^{\ast }$ are characterized by $e_i^{\ast }(e_i)=1$, and the frame relation becomes $e_j^{\ast }=g_{ij}^{-1}{e}_i^{\ast }$. Reciprocals of nowhere-zero holomorphic functions are holomorphic, and their cocycle relations persist. This defines the holomorphic dual vector bundle structure on $L^{\ast }$, with holomorphic evaluation pairing $L^{\ast }\otimes L\to {\mathcal{O}}_X$.

Assume now that $X$ is compact and connected, and let $s$ and $t$ be nonzero global holomorphic sections of $L$ and $L^{\ast }$, respectively. The function $t(s)$ is a global holomorphic function. The maximum modulus principle, applied in local complex coordinate lines near a maximum and then propagated on the connected manifold, makes it constant. Moreover it is not identically zero: the zero locus of a nonzero holomorphic section has empty interior by the identity theorem, so the two nonvanishing loci are dense open sets and intersect. Thus $t(s)=c\ne 0$ everywhere.

In particular, $s$ never vanishes. The map $(x,a)\mapsto as(x)$ is a holomorphic bundle isomorphism $X\times \mathbb{C}\to L$; in a local frame its inverse divides by the nowhere-zero holomorphic coefficient of $s$. Conversely, a holomorphic trivialization supplies the constant unit sections of both $L$ and $L^{\ast }$. We have proved the two-section criterion for holomorphic triviality:

Write points of Complex projective space as lines $[Z]\subset {\mathbb{C}}^{n+1}$. The complex tautological line bundle iswith fiber that very line. Its dual is the hyperplane line bundle $\mathcal{O}(1)$. On the affine chart $U_i=\{Z_i\ne 0\}$, the vector $e_i([Z])=Z/Z_i$ is a nowhere-zero holomorphic frame of $\mathcal{O}(-1)$. On $U_i\cap U_j$,Both ratios are nowhere-zero holomorphic functions of the affine coordinates. They exhibit the holomorphic transition functions of a vector bundle for the two bundles explicitly.

Choose a nonzero linear functional $\ell :{\mathbb{C}}^{n+1}\to \mathbb{C}$ whose zero hyperplane is $H$. At $[Z]$, its restriction to $\mathbb{C}Z$ is an element of that line's dual and therefore defines a holomorphic section $s_{\ell }$ of $\mathcal{O}(1)$. In the dual frame above its coefficient isThis expression is holomorphic even along $H$, and its zero locus is exactly $H$. Consequently $s_{\ell }{|}_{{\mathbb{P}}^n\setminus H}$ is the requested nowhere-zero section; its global holomorphic extension vanishes precisely on $H$. The extension need not, and does not, stay nowhere zero on the hyperplane.

The complex structure on a complex manifold splits its complexified cotangent bundle into the $+i$ and $-i$ eigenspaces, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a smooth section of ${\bigwedge }^p(T^{1,0}X{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}X{)}^{\ast }$, so in holomorphic coordinates it has the expressionHere the increasing multi-indices avoid duplicating alternating factors. The exterior derivative has just two type components: the conjugate Dolbeault operator $\partial$ and the Dolbeault operator $\bar{\partial }$. On such a local expression they areHolomorphic changes of coordinates preserve these types, so the definitions are intrinsic. Comparing type components of $d^2=0$ gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

A real (p, p)-form satisfies $\bar{\eta }=\eta$, where complex conjugation acts on both coefficients and differential factors. Conjugation exchanges the two Dolbeault operators, soThus $\bar{\partial }\eta =0$ forces $\partial \eta =0$ as well. Conversely, if $d\eta =0$, its distinct $(p+1,p)$ and $(p,p+1)$ components both vanish. ThereforeThe assertion remains valid in top degree, when the indicated type components are zero automatically.

For a holomorphic function $f$, $\bar{\partial }f=0$ and $\partial \bar{f}=0$. Writing its real part as $u=(f+\bar{f})/2$ givesHence the real part of a holomorphic function is a pluriharmonic function.

For the local exactness assertion, work on a smaller polydisc whose closure lies in the original one. The allowed one-variable integral is the Cauchy-Green operator $T_j$ in coordinate $z_j$. Applied coefficientwise, it satisfies ${\partial }_{{\bar{z}}_j}{T}_{j}a=a$ on a smaller disc and commutes with differentiation in all other coordinates; the smooth parameter dependence follows by applying the integral to a smooth compactly supported extension in the integration variable. This extension can be made with a cutoff equal to one on the smaller disc.

Begin with a Dolbeault operator closed form $\phi$ and decompose it uniquely as $\phi =d{\bar{z}}^1\wedge \alpha +\beta$, where neither $\alpha$ nor $\beta$ contains $d{\bar{z}}^1$. Set ${\psi }_1=T_1\alpha$. The $d{\bar{z}}^1$ component of $\bar{\partial }{\psi }_1$ is exactly $d{\bar{z}}^1\wedge \alpha$, sohas no $d{\bar{z}}^1$ factor. It is still $\bar{\partial }$-closed. The $d{\bar{z}}^1$ component of $\bar{\partial }{\phi }_1=0$ now shows that every coefficient of ${\phi }_1$ is holomorphic in $z_1$.

Apply the same construction in $z_2$ to remove its antiholomorphic differential factor. The operator $T_2$ preserves holomorphicity in $z_1$, since it commutes with ${\partial }_{{\bar{z}}_1}$. Consequently the second subtraction introduces no previously removed factor. Continue through all $n$ coordinates, shrinking the polydisc when necessary. The final remainder has positive antiholomorphic degree but contains none of the $d{\bar{z}}^j$, and hence is zero. This coordinate elimination for local Dolbeault primitives constructsIf $q>n$, the original form is already zero, so one can take $\psi =0$.

Finally let $\phi$ be a smooth $(0,1)$-form on the complex projective line. Choose $0<r<R$ and the two coordinate neighborhoods $U=\{|z|<R\}$ and $V=\{|z|>r\}\cup \{\infty \}$, the latter expressed in $w=1/z$. They cover the sphere. The one-variable construction, integrating over slightly larger closed discs in each coordinate chart, gives smooth functions $a$ on $U$ and $b$ on $V$ with $\bar{\partial }a=\phi {|}_U$ and $\bar{\partial }b=\phi {|}_V$.

Their difference $h=a-b$ is a holomorphic function on the annulus $r<|z|<R$. Its Laurent series splits asThe first series extends holomorphically throughout $U$, while the second extends throughout $V$, including infinity, where it has value zero. Thus $a-h_{+}$ and $b+h_{-}$ agree on the overlap and glue to a global smooth function $F$. Both corrections are holomorphic, soThis Laurent gluing of Dolbeault primitives on the projective line uses no Hodge theorem.

A Riemann surface is a connected one-complex-dimensional complex manifold: it has local complex coordinates with holomorphic transition maps. These coordinates induce an orientation and a conformal structure. Fix a closed oriented topological surface $S_g$ of genus $g$. A point of Teichmüller space is a marked Riemann surface $(X,f)$, where $f:S_g\to X$ is an orientation-preserving marking. Two pairs are equivalent if there is a biholomorphism $h:X\to X^{\prime }$ with $h\circ f$ isotopic to $f^{\prime }$. Thus the marking remembers how curves on the reference surface sit in $X$; forgetting it gives the moduli space of Riemann surfaces, obtained by quotienting by the mapping class group.

For $g\ge 2$, the uniformization theorem supplies a unique curvature-$-1$ hyperbolic metric in each conformal class. A pants decomposition has $3g-3$ cuffs, and their positive lengths together with their real twists give Fenchel–Nielsen coordinates. HenceFor completeness, ${\mathcal{T}}_0$ is a point, whereas ${\mathcal{T}}_1$ is the upper half-plane, parametrizing marked complex tori $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$ with $\mathrm{Im}\tau >0$.

The Wolpert generic spectral rigidity theorem says that, for fixed $g\ge 2$, there is a closed proper real-analytic exceptional subset of ${\mathcal{T}}_g$ such that a hyperbolic surface outside that subset is determined up to isometry by its unmarked length spectrum, equivalently its Laplacian spectrum. This is an unoriented isometry statement: reversing orientation cannot be detected by either spectrum.

The Huber spectral equivalence theorem says that two closed curvature-$-1$ hyperbolic surfaces have the same Laplacian spectrum, with multiplicities, if and only if they have the same unmarked length spectrum, with multiplicities. One may consistently count unoriented closed geodesics including their iterates; alternatively, the primitive multiset determines the multiset with iterates and conversely. This equivalence comes from the Selberg trace formula.

The Buser finite length spectrum theorem is the finite-determination ingredient: for each $g\ge 2$ and $\epsilon >0$ there is a finite number $L(g,\epsilon )$ such that, for any two closed hyperbolic surfaces of genus $g$ with hyperbolic systole at least $\epsilon$, equality of their length multisets through $L(g,\epsilon )$ implies equality of their entire length spectra. The finite-length statement is used in Buser's chapter on Wolpert's theorem. It determines the full spectrum, not necessarily the isometry class.

Here is a proof of the Buser finite length spectrum theorem through finite spectral matching by polynomial invariants. We use the following subsidiary results, stating the required content. First, Mumford's compactness theorem makes the part of the genus-$g$ moduli space of Riemann surfaces with hyperbolic systole at least $\epsilon$ compact. Second, marked hyperbolic holonomy representations of a closed oriented surface lift from ${\mathrm{PSL}}_2(\mathbb{R})$ to ${\mathrm{SL}}_2(\mathbb{R})$, and normalized representatives can be chosen continuously on sufficiently small parameter charts. Their hyperbolic translation length satisfiesThird, the length spectrum of a compact hyperbolic surface is locally finite, and metrics on compact marked parameter charts are uniformly bilipschitz equivalent. Finally, the Hilbert basis theorem implies that a descending sequence of algebraic sets in a finite-dimensional affine space stabilizes. These are compactness, holonomy, elementary geometric finiteness and algebraic finiteness results; none is the finite unmarked-spectrum assertion we are proving.

By compactness, finitely many relatively compact marked charts cover the thick moduli space of Riemann surfaces. Choose normalized lifted hyperbolic holonomy representations on these charts and take their compact closures. Their finite union $K$ is a compact set of determinant-one matrix tuples representing every surface in the thick part, possibly more than once. Each tuple records the images of the $2g$ generators of the fundamental group. Thus $K$ sits in a fixed finite-dimensional real matrix space. Every group word is a polynomial in these matrix entries: multiplication is polynomial, and for a determinant-one two-by-two matrix its inverse is its adjugate, also polynomial. Consequently each $q_{\gamma }$ is a polynomial on the ambient space. Squared traces remove the sign ambiguity of the lift, and $4{\cosh }^2(\ell /2)$ is strictly increasing for positive $\ell$.

Enumerate the unoriented nontrivial free homotopy classes as ${\gamma }_1,{\gamma }_2,\dots$, including the iterates, and write ${\ell }_j(x)$ for their geodesic lengths at $x\in K$. The uniform finiteness of short geodesic classes follows here directly: on each of the finitely many compact marked charts, choose a fixed reference metric $h$ and a uniform bilipschitz constant $C$. A class with ${\ell }_j(x)\le R$ somewhere on that chart has reference length at most $CR$. Only finitely many classes have that reference length, and a finite union of these finite collections is finite. In particular, for every $j$ the candidate setis finite. If two parameters in $K$ have equal lengths for classes $j,k$, then $k\in F_j$. These candidate sets depend on the fixed compact set $K$ and the source label $j$, not on the number of labels currently being matched.

For every $N$, define $E_N$ in the full ambient space of pairs of matrix tuples $(x,y)$ as follows. Require an injective choice $\sigma :\{1,\dots ,N\}\to \mathbb{N}$ with $\sigma (j)\in F_j$ andand independently require an injective choice $\tau :\{1,\dots ,N\}\to \mathbb{N}$ with $\tau (j)\in F_j$ andThere are finitely many possible pairs of choices $(\sigma ,\tau )$ because every $F_j$ is finite. For each choice these are finitely many polynomial equations. Therefore $E_N$ is a finite union of algebraic sets, hence an algebraic set. Restricting an injective matching to its first $N$ entries provesThe Hilbert basis theorem now supplies $N_0$ such that $E_N=E_{N_0}$ for all $N\ge N_0$. This step takes place in a fixed affine space and uses polynomial equations, so it gives stabilization of the actual sets, not just stabilization of germs at individual parameters.

For $(x,y)\in E_{N_0}\cap (K\times K)$, stabilization gives injective partial length matchings in both directions for every $N$. Fix any length $a>0$. Only finitely many labels have length $a$ at $x$ or at $y$, by local finiteness. Take $N$ large enough to contain every such label at $x$. Its injective matching into the labels at $y$ preserves $q_j$, and hence preserves length, so the multiplicity of $a$ at $x$ is at most its multiplicity at $y$. The reverse matching gives the opposite inequality. Thus the two multiplicities are equal for every $a$, proving equality of the entire unmarked length spectra. No assumed global matching or unjustified exchange of infinitely many neighborhoods is needed.

Finally choose a numberSuppose the two length multisets agree through $L$. For each of the first $N_0$ labels at $x$, select a distinct equal-length label at $y$, which is possible precisely because the truncated multisets agree with multiplicity. Each selected label belongs to the corresponding $F_j$. Do the same in the reverse direction. Then $(x,y)\in E_{N_0}$, and the previous paragraph proves equality of the full length spectra. Since $K$ was selected using only the fixed thick genus-$g$ parameter space, the resulting finite $L$ depends only on $g,\epsilon$. This proveswith each intersection interpreted as a multiset. By the Huber spectral equivalence theorem, the corresponding entire Laplacian spectra also agree. This is the finite matching reduction needed before the analytic and geometric rigidity steps in the Wolpert generic spectral rigidity theorem.

Subgroups $U,V$ of a finite group $T$ have Gassmann equivalence whenSumming over the conjugacy classes gives $|U|=|V|$. Equivalently, their coset permutation representations have the same character of a representation. In particular, almost conjugate subgroups need not be conjugate.

Put $p=\mathrm{ord}A$, $q=\mathrm{ord}B$, and $r=\mathrm{ord}(AB)$. Begin with the sphere punctured at three points. Its fundamental group is $⟨a,b,c\mid abc=1⟩$, and the epimorphism sending $a,b,c$ to $A,B,(AB{)}^{-1}$ defines a connected regular $|T|$-sheeted covering space. Around the punctures the monodromies have cycles of lengths $p,q,r$. Fill each lifted puncture by a disk, using the local branched covering models $z\mapsto z^p$, $z\mapsto z^q$ and $z\mapsto z^r$. This produces a closed oriented surface $M$, with $T$ acting by homeomorphisms that are smooth in these local models. This construction works even if one of $p,q,r$ is one.

Triangulate the base sphere with the three marked points as vertices and two faces. Lifting this triangulation givesso the Euler characteristic isIn the hyperbolic case $1/p+1/q+1/r<1$, this is equivalently the triangle cover construction for Sunada surfaces. Take the hyperbolic triangle groupand its epimorphism $\theta :\Delta \to T$ with the same images. All finite-order elements of $\Delta$ are conjugate to powers of its three elliptic generators. Because their images have the same exact orders, $K=\ker \theta$ is torsion-free and $M=K\backslash {\mathbb{H}}^2$. It can be assembled from $2|T|$ hyperbolic triangles of angles $\pi /p,\pi /q,\pi /r$. Both examples below belong to this hyperbolic case.

For $M_1=U\backslash M$ and $M_2=V\backslash M$ to be smooth surfaces normally covered by $M$, the subgroup actions must be free. The precise condition is that neither subgroup meet any conjugate in $T$ of a nonidentity power of $A$, $B$ or $AB$. These are exactly the possible point stabilizers. Then $M\to M_1$ and $M\to M_2$ are regular coverings, with deck transformation groups $U,V$ respectively. No normality of $U,V$ in $T$ is needed: normality would concern the further covering of the base orbifold, not these coverings from $M$. For a free subgroup $W$,

Choose any $T$-invariant smooth Riemannian metric on $M$, for example by averaging an arbitrary metric. It descends to both quotients. On functions let $R_t$ be the unitary action of $t\in T$, and let $E=e^{-s{\Delta }_M}$ for $s>0$. The averaging operator $P_W=|W{|}^{-1}{\sum }_{w\in W}{R}_w$ projects orthogonally onto the $W$-invariant functions, which identify with functions on $W\backslash M$. The trace formula underlying the Sunada theorem isThe expression $\mathrm{Tr}(R_w{e}^{-s{\Delta }_M})$ is a class function, because ${\Delta }_M$ commutes with $T$. Thus Gassmann equivalence makes the two sums identical. Alternatively, apply the same averaging formula to each finite-dimensional Laplacian eigenspace: its invariant dimension is $|W{|}^{-1}{\sum }_{w\in W}{\chi }_{\lambda }(w)$. This proves equality of every eigenvalue multiplicity and hence isospectrality of the two quotient metrics.

If $V=tU{t}^{-1}$ for some $t\in T$, the action of $t$ gives an isometry between the quotients for every such metric. Nonconjugacy is therefore necessary. To make it sufficient, choose the quotient orbifold metric generically so that its regular locus has no nonidentity local isometries; this is the equivariant, or orbifold, version of the Sunada local isometry lemma. Smooth perturbations on regular coordinate balls provide this genericity while their lifts remain smooth on $M$. Any isometry between $M_1$ and $M_2$, when composed locally with their projections, would then identify the same points of the base regular locus. By continuation it is an isomorphism of their coverings of that locus, and the classification of connected covering spaces forces $U,V$ to be conjugate in $T$. Thus nonconjugate subgroups give non-isometric isospectral quotients for a suitable generic invariant metric. The genericity belongs to the base regular locus; the covering metric on $M$ itself necessarily has the prescribed deck symmetries.

For genus two, an explicit choice is the following permutation group on twelve labels, with permutation products composed from right to left:These parameters are those in Kang's genus-two construction. They admit a direct finite verification. Starting with $T_0=\{1\}$ and iterating $T_{j+1}=T_j\cup T_{j}A\cup T_{j}B$, the successive sizes are $1,3,7,13,23,39,61,81,93,96$, after which the set is unchanged. Thus $|T|=96$, and direct multiplication gives $(p,q,r)=(3,3,6)$ and $|U|=|V|=8$. The conjugacy-class triples $(|C|,|C\cap U|,|C\cap V|)$, obtained as $C=\{tx{t}^{-1}:t\in T\}$, areThis proves Gassmann equivalence. The subgroup $V$ fixes the labels $5,11$, whereas $U$ has no common fixed label, so they cannot be conjugate even in the full symmetric group. Both are groups of order eight, so they meet no nonidentity order-three element. The only order-two power of $AB$ iswhich is central in $T$ and lies in neither subgroup. This also excludes its conjugates and proves freeness. The quotient Euler characteristics and genera are thereforeThe covering surface $M$ has genus nine.

For genus three, use $T={\mathrm{GL}}_3({\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. Let $U$ stabilize a nonzero vector and let $V$ stabilize a two-dimensional subspace. Both have index seven and order twenty-four. An element $t$ fixes $2^{\dim \ker (t-I)}-1$ nonzero vectors. Hyperplanes correspond to nonzero dual vectors over ${\mathbb{F}}_2$, so the number fixed in the dual action is $2^{\dim \ker (t^T-I)}-1$, the same number. Equality of these permutation characters proves that the point and hyperplane stabilizers are almost conjugate.

They are not conjugate. In coordinates where $U$ fixes $e_1$, its elements include all matrices $\left(\begin{array}{cc}1& \alpha \\ 0& C\end{array}\right)$ with arbitrary row $\alpha$ and $C\in {\mathrm{GL}}_2({\mathbb{F}}_2)$. An invariant plane containing $e_1$ would project to a line invariant under every $C$, which is impossible. A plane not containing $e_1$ would be the graph of a linear functional on the quotient, and arbitrary $\alpha$ would move that graph. Thus $U$ stabilizes no plane, whereas every conjugate of $V$ does.

Choose the concrete generatorsOver ${\mathbb{F}}_2$, each of $A,B,AB$ has order seven. The same closure iteration has sizes $1,3,7,15,31,59,103,144,168$, verifying that $A,B$ generate $T$. Since seven does not divide $24$, neither stabilizer contains a nonidentity power of these generators, or any conjugate of such a power. We may therefore use $(p,q,r)=(7,7,7)$ in the triangle cover construction for Sunada surfaces. The quotient calculation isHere the covering surface $M$ has genus forty-nine. Choosing the generic invariant metrics described above completes both requested pairs.

Here “bumpy” means a nowhere locally homogeneous metric, not the different convention concerning nondegenerate closed geodesics. A smooth Riemannian metric has this property if every local isometry between connected open subsets is the identity on its domain; in particular, no two distinct open subsets are isometric. The Sunada local isometry lemma states that, on a compact smooth manifold without boundary of dimension at least two, such Riemannian metrics form a residual set and are dense in the $C^{\infty }$ topology. The dimension restriction matters: in dimension one, arclength coordinates always give nontrivial local isometries.

An element $j_x^{k}f$ of the jet bundle of maps $J^k(M,N)$ records the source point $x$, target value $f(x)$ and all derivatives through order $k$. If $\dim M=m$ and $\dim N=n$, there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivative multi-indices of order $r$, and $n$ components for each one. ThusFor the projection $j_x^{k}f\mapsto (x,f(x))$ onto $M\times N$, the fiber has dimension $n((\genfrac{}{}{0ex}{}{m+k}{k})-1)$. In source and target coordinates its Taylor coefficients identify it withThis is a coordinate-dependent identification for $k>1$, since a change of coordinates mixes different derivative orders. For $k=1$ the fiber is canonically $\mathrm{Hom}(T_{x}M,T_{y}N)$; for higher orders, truncation to the previous jet bundle of maps is an affine bundle modeled on ${\mathrm{Sym}}^k(T_x^{\ast }M)\otimes T_{y}N$. For $k=0$ the fiber is a point.

The density assertion in the printed question requires distinct closed domains. Literally, for $i=k$, the identity is an isometry for every Riemannian metric, so ${\mathcal{S}}_{ii}$ is the whole metric space and its complement is empty. Merely indexing the same closed ball twice has the same defect. The intended assertion is for distinct embedded closed balls $K_i={\overline{U}}_i$ and $K_k={\overline{U}}_k$, with an isometry meaning an onto diffeomorphism preserving the metric. The proof below establishes precisely this corrected assertion.

Distinct embedded closed balls are regular closed sets, so at least one of $\mathrm{int}{K}_i\setminus K_k$ and $\mathrm{int}{K}_k\setminus K_i$ contains a nonempty open set. Indeed, if both were empty, taking closures would give both inclusions and hence equality. Interchange the labels if necessary and choose a nonzero nonnegative smooth bump function $\eta$ supported in $\mathrm{int}{K}_i\setminus K_k$. For any given Riemannian metric $g$, make the localized volume perturbationThe Riemannian volume form satisfies $d{V}_{g_t}=e^{mt\eta }d{V}_g$. ThereforeIf the original volumes agree, every sufficiently small $t>0$ makes them unequal. If they already disagree, continuity preserves that inequality for sufficiently small $t$. An onto isometry must preserve volume, so each of these perturbed Riemannian metrics belongs to $\mathcal{C}{\mathcal{S}}_{ik}$. Finally $g_t\to g$ in $C^{\infty }$: all derivatives of $e^{2t\eta }-1$ tend to zero uniformly on compact sets. This proves density of $\mathcal{C}{\mathcal{S}}_{ik}$ for distinct closed balls, in fact in every positive dimension. The localized volume perturbation settles this particular density claim directly. It also does not imply the stronger Sunada local isometry lemma by itself, since a general local isometry need not send one preselected basis ball onto another.

Use the following skew-symmetric matrix, whose columns reduce modulo three to four of the specified nonzero tetracode words:Set $B_{+}=3I+S$, $B_{-}=-3I+S$, and define the full Euclidean lattices $L_{\pm }^0=B_{\pm }{\mathbb{Z}}^4$. Their determinants have absolute value $144$, sinceFor arbitrary positive real numbers $a_1,\dots ,a_4$, put $D=\mathrm{diag}(\sqrt{a_1},\dots ,\sqrt{a_4})$ and $L_{\pm }=D{L}_{\pm }^0$. This is the tetracode length-isospectral lattice construction. We prove equality of the length multisets by an explicit bijection, not just equality of the shortest lengths.

Modulo three, both lattices have image $\mathrm{im}S=T$. The matrix $S$ has rank two over ${\mathbb{F}}_3$; its four column lines are precisely the four lines in $T$. Also $S^2=0$ modulo three, so $\ker S=\mathrm{im}S=T$. Each nonzero tetracode word has exactly one zero coordinate. Every $v\in L_{\pm }^0$ therefore has some coordinate divisible by three. Moreover, all four coordinates of $v$ have the same parity, because every entry of either $B_{\pm }$ is odd.

For $v\in L_{+}^0$, choose the least index $i$ with $3\mid v_i$ and set $w=v-2v_i{e}_i$. The coordinate reflection bijection for tetracode lattices will show that $w\in L_{-}^0$. First, $w\equiv v(\mathrm{mod}3)$, so $Sw\equiv 0(\mathrm{mod}3)$. Second, positive-lattice membership implies $Sv\equiv 3v\equiv -v(\mathrm{mod}4)$. HenceThe expression in parentheses has $i$th coordinate zero and every other coordinate $v_j+v_i{S}_{ji}$ even, by the common parity and $S_{ji}=\pm 1$. Thus $(I-S)w\equiv 0(\mathrm{mod}4)$. Together the congruences modulo three and four say $(-3I-S)w\equiv 0(\mathrm{mod}12)$, which is precisely $B_{-}^{-1}w\in {\mathbb{Z}}^4$.

Conversely, if $v\in L_{-}^0$, the same reflection has unchanged reduction modulo three, and now $Sv\equiv v(\mathrm{mod}4)$. ThusIts $i$th coordinate is $4v_i$ and its other coordinates are even multiples of two, so it vanishes modulo four; it vanishes modulo three as well. The reflected vector is in $L_{+}^0$. The chosen index is unchanged by reflection, including for vectors reducing to zero, so this map is an involution between the two lattices. It preserves every $|v_j|$ and therefore preserves $\Vert Dv{\Vert }^2={\sum }_j{a}_j{v}_j^2$. The two lattices have the same length spectrum, including every multiplicity, for every positive choice of weights.

To prove a genuine failure of isometry, chooseThese four numbers are linearly independent over $\mathbb{Q}$: changing the signs of the independent square roots in the multiquadratic field $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ isolates each coefficient of a rational relation. If an orthogonal transformation $O$ maps $L_{+}$ to $L_{-}$, thenand $P^T\mathrm{diag}(a_j)P=\mathrm{diag}(a_j)$. Comparing rational coefficients of the independent $a_j$ in the diagonal entries gives $P_{ji}^2={\delta }_{ji}$. Consequently $P$ is diagonal with entries $\pm 1$.

Such a coordinate sign change must preserve $T$ modulo three. For each zero coordinate $i$, there are exactly two opposite nonzero tetracode words. Preserving their line requires the other three signs to be equal. Taking two different zero coordinates forces all four signs to be equal. Thus $P=I$ or $P=-I$, either of which would imply $L_{+}^0=L_{-}^0$. ButThis contradiction proves $L_{+}$ and $L_{-}$ are not isometric. Allowing affine isometries changes nothing: translating the image of zero back to zero reduces them to the orthogonal case.

For the flat tori, take period lattices equal to the dual lattices:The standard spectrum of a flat torus lemma states that its characters of a real torus are indexed by the dual of its period lattice. Here they are $e^{2\pi i⟨v,x⟩}$ for $v\in L_{\pm }$, with Laplacian eigenvalue $4{\pi }^2\Vert v{\Vert }^2$. These characters give a complete orthogonal basis, so our length-preserving bijection proves that $X_{+}$ and $X_{-}$ are isospectral manifolds. The second standard lemma is that an isometry of flat tori lifts to an affine isometry of their Euclidean universal covers. Its orthogonal part maps the period lattices to one another, and therefore maps their dual lattices to one another. This would give the impossible isometry $L_{+}\cong L_{-}$. Hence $X_{+}$ and $X_{-}$ are isospectral non-isometric flat four-dimensional tori.

Use the nonnegative Laplacian convention $\tilde{\Delta }=-{\sum }_{j=1}^{d+1}{\partial }_j^2$ and $\Delta =-\mathrm{div}\mathrm{grad}$ on the unit sphere. In polar coordinates the Euclidean Riemannian metric is $d{r}^2+r^2{g}_{S^d}$, its volume form is $r^{d}drd{V}_{S^d}$, and therefore the Laplacian in polar coordinates isConsequently the required restriction identity isFor the opposite sign convention all eigenvalues below change sign, and the restriction identity changes accordingly.

Let ${\mathcal{P}}_n$ be the vector space of degree-$n$ homogeneous polynomials in $d+1$ variables, and let ${\mathcal{H}}_n$ be its subspace of harmonic polynomials. Write $H(r\omega )=r^{n}Y(\omega )$ for $H\in {\mathcal{H}}_n$. Substitution in the Laplacian identity givesThus each restricted harmonic polynomial is a spherical harmonic with eigenvalue ${\lambda }_n=n(n+d-1)$.

To count these spherical harmonics, give polynomials the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_j$ is adjoint to ${\partial }_j$, so multiplication by $r^2$ is adjoint to the ordinary signed Laplacian ${\sum }_j{\partial }_j^2$. In these finite-dimensional inner product spaces, the orthogonal complement of $r^2{\mathcal{P}}_{n-2}$ is exactly ${\mathcal{H}}_n$. Hence the Fischer decomposition isMultiplication by $r^2$ is injective, while restriction of a homogeneous polynomial to the sphere is injective: homogeneity recovers its value at every nonzero point. Counting monomials now gives the eigenvalue multiplicitywhere the second term is zero for $n<2$.

It remains to establish that these are all the eigenvalues and their full multiplicities. Iterating the Fischer decomposition expresses every restricted polynomial as a sum of the restricted harmonic polynomials. Restricted polynomials separate points and contain the constants, so the Stone-Weierstrass theorem makes their span dense in continuous functions on the sphere, hence dense in its L2 space. The Laplace-Beltrami operator is self-adjoint, and the distinct numbers ${\lambda }_n$ strictly increase for $d\ge 1$. Its different eigenspaces are therefore orthogonal. Any additional eigenfunction, or any additional vector in one of these eigenspaces orthogonal to its known spherical harmonics, would be orthogonal to a dense subspace and hence zero. The standard compact elliptic spectral theorem then shows that the spectrum of the round sphere is precisely ${\lambda }_n=n(n+d-1)$ with multiplicity $m_n$. For $S^1$ this gives multiplicity one at zero and two at $n^2$ for $n\ge 1$. If the degenerate case $d=0$ is included, $S^0$ has two points and its Laplacian is zero on a two-dimensional function space.

A subgroup $H\le G$ is a benign subgroup when there are an embedding of $G$ into a finitely presented group $E$ and a finitely generated subgroup $L\le E$ with $G\cap L=H$. Finite generation is required of $L$, not of $H$ itself. We use Britton's lemma, the normal form theorem for an amalgamated free product, and, for the final quotient assertion, the Higman embedding theorem.

Choose witnesses $E_H,L_H$ and $E_K,L_K$ for $H,K$. The amalgamated free product $P=E_H{\ast }_G{E}_K$ embeds both witnesses and their common copy of $G$. It is finitely presented: take their two finite presentations and add the finitely many equalities identifying the images of a finite generating set of $G$. These equalities identify all elements of $G$, so they present precisely the amalgam. The original intersections $G\cap L_H=H$ and $G\cap L_K=K$ are preserved by the embeddings.

Now form the finitely presented groupOnly finitely many commutation relators, one for each of the chosen subgroup generators, are needed. The subgroup $Q=⟨G,t,s⟩$ is exactly the multiple HNN extension of $G$ centralizing $H$ and $K$. To prove this claim, take a word in $G,t,s$ reduced for those associated subgroups. It remains reduced in $T$: a $t$-pinch with a coefficient in $G$ occurs precisely when that coefficient lies in $L_H\cap G=H$, and an $s$-pinch precisely when it lies in $K$. Britton's lemma therefore prevents any extra relation, and it also embeds $G$.