Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 20 (Spectral Geometry)

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.

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.

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.

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.

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.