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Put $T=\mathrm{PSL}(3,2)=\mathrm{GL}(3,{\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. It acts on the seven points and seven lines of the Fano plane. Take $U_1$ to be a point stabilizer and $U_2$ to be a line stabilizer; both have index seven and order twenty-four. The fact that point and hyperplane stabilizers are almost conjugate follows by comparing fixed-point counts: for a matrix $g$, the numbers of fixed nonzero vectors and fixed nonzero covectors are bothbecause $g-I$ and its transpose have the same rank, and the dual action $g^{-T}$ has the same fixed space as $g^T$. Over ${\mathbb{F}}_2$, nonzero vectors represent points and nonzero covectors represent lines without any further scalar quotient. The two permutation characters therefore agree. The coset-character formulathen proves Gassmann equivalence. They are not conjugate: a point stabilizer has orbits of sizes one and six on points and hence preserves no three-point line. Its conjugates are again point stabilizers, whereas $U_2$ preserves its line. The point and line stabilizers form the required nonconjugate Gassmann pair.

We construct a family using the Lambert quadrilateral construction of a cone torus. Let $\alpha =\pi /14$. We use the elementary hyperbolic tri-rectangle fact that a quadrilateral with three right angles, remaining angle $\alpha$, and side lengths $x,y$ adjacent to the right-angle vertex opposite $\alpha$ exists uniquely forHere $x,y$ are the sides at the opposite right-angle corner, not the sides incident to $\alpha$. Given any $x>0$, choose $y=\mathrm{arsinh}(\cos \alpha /\sinh x)$. Reflect four copies around the opposite right-angle vertex. The result is a symmetric hyperbolic quadrilateral with all four corner angles $\alpha$ and opposite sides of equal lengths. Pair opposite sides by the axis translations. The quotient is an oriented torus with one cone point, whose total angle is $4\alpha =2\pi /7$. Its two central axis loops $a,d$ are embedded closed geodesics, of lengths $2x,2y$, intersecting once at right angles.

Cut this cone torus along $a$. The result is a hyperbolic pair of pants with two geodesic boundaries of length $\ell =2x$ and the cone point. Reglue those equal boundaries with a Fenchel–Nielsen twist $\tau \in \mathbb{R}$. This gives a two-parameter family $B_{\ell ,\tau }$, with the same cone angle, and marked simple classes $a,d$ meeting once. Near any zero-twist member their geodesic representatives remain embedded, transverse and disjoint from the cone point. The length of $a$ remains $\ell$, and the right-angle matrix product givesFor example, the zero-twist holonomies are $\mathrm{diag}(e^x,e^{-x})$ and $\left(\begin{array}{cc}\cosh y& \sinh y\\ \sinh y& \cosh y\end{array}\right)$; multiply the latter by $\mathrm{diag}(e^{\tau /2},e^{-\tau /2})$ to obtain the twist formula. Their commutator has trace $2-4{\sinh }^{2}x{\sinh }^{2}y=-2\cos (\pi /7)$, confirming exact elliptic order seven. The side-pairing construction supplies discreteness; the trace computation alone would not do so.

The orbifold fundamental group isSend $a$ to the supplied generator $A$ and $d$ to $D$. Their commutator $C=[D,A]$ has order seven, so this defines an epimorphism $\rho :\Gamma \to T$. Its kernel avoids all nontrivial cone stabilizers, which are conjugates of powers of $[d,a]$; therefore it is torsion-free. It defines a smooth closed regular hyperbolic cover $X_{\ell ,\tau }\to B_{\ell ,\tau }$ with deck group $T$. The subgroups $U_i$, of order twenty-four, also avoid every order-seven cone stabilizer. Thusare smooth closed seven-sheeted covers of the cone torus. The orbifold Euler characteristic is ${\chi }_{\mathrm{orb}}(B)=-(1-1/7)=-6/7$, givingEquivalently there is one preimage of the cone point, with local degree seven, and its lifted angle is $2\pi$. The Sunada theorem, in the free-subgroup-action form proved above, makes $S_1(\ell ,\tau )$ and $S_2(\ell ,\tau )$ isospectral for every parameter.

Nonisometry requires further work. A cycle of length $r$ in the monodromy permutation of a simple base geodesic gives a primitive simple lifted geodesic of length $r$ times the base length. From the supplied $A$ action, each cover has one lift of $a$ of length $\ell$ and two of length $3\ell$. From the supplied $D$ action, each has one lift of $d$ of each length ${\ell }_d,2{\ell }_d,4{\ell }_d$. Let ${\beta }_i$ be the unique two-cycle lift of $d$, and let ${\alpha }_{i,1},{\alpha }_{i,2}$ be the two three-cycle lifts of $a$. The fact that cycle intersections count intersections of lifted curves gives their intersection numbers: the seven crossings above the one base crossing are labelled by the sheets, so a crossing belongs to both lifted components exactly when its sheet lies in both cycles. ConsequentlyIndeed, in the first cover the two three-cycles have supports $\{1,2,5\},\{3,6,4\}$ and the two-cycle has support $\{0,3\}$; in the second they have supports $\{1,4,3\},\{2,5,6\}$ and $\{2,5\}$. The lifted curves intersect only over the single base crossing, so these are actual intersection counts of the geodesics, not merely abstract sheet-incidence numbers.

To make this a test against every isometry, we must identify these curves intrinsically, rather than require the isometry to respect the covering map. We use generic isolation of lifted simple geodesics, and give the needed genericity argument. Lengths of closed geodesics are real-analytic functions of the cone-torus length and twist parameters. For a primitive hyperbolic class $w$ and an integer $1\le r\le 7$, consider the equalitiesApart from $w=a,r=3$ or $w=d,r=2$, none is an identity on the full two-parameter family. To see this for the first equality, pinch the simple curve $a$ while keeping the cone angle fixed. The collar lemma makes every class crossing $a$ long. A class disjoint from $a$ lies in the cut pair of pants; its limiting holonomy has positive translation length unless it is a power of one of the new cusp loops, which are conjugate to $a$ in the torus group. Cone-peripheral classes are elliptic and do not represent closed geodesics. Thus the only primitive class with length tending to zero in this degeneration is $a$, up to inversion and conjugacy. This excludes an identity with a positive multiple of $\ell$. Pinching $d$ proves the corresponding assertion for ${\ell }_d$. The argument uses the usual hyperbolic collar and cusp degeneration facts, obtainable by the right-angle quadrilateral formulas; the cone angle $2\pi /7<\pi$ remains fixed. Analytic continuation on the connected length-twist parameter space shows that the same identities cannot hold on a small open neighbourhood of a zero-twist member.

There are countably many class equations, and each nontrivial real-analytic equation has a closed zero set with empty interior. The Baire category theorem therefore gives a parameter $({\ell }_0,{\tau }_0)$ in that small neighbourhood at which both target lengths are attained only by the prescribed lift components. This isolation persists on a smaller open neighbourhood. Indeed, near a fixed compact hyperbolic metric there are uniformly only finitely many primitive geodesic classes below a fixed length bound, by discreteness of the covering group and a compact fundamental region; nearby metrics are uniformly bilipschitz. Only finitely many classes can therefore approach the two fixed finite target lengths. Their strict inequalities persist under a sufficiently small parameter change.

Now keep $\tau ={\tau }_0$ and let $\ell$ vary in a sufficiently small interval $I$ around ${\ell }_0$. For each $\ell \in I$, the only primitive geodesics of length $3\ell$ in either cover are the two $\alpha$ components, and the only primitive geodesic of length $2{\ell }_d$ is $\beta$. An isometry would have to preserve these length classes and their intersection-count multiset, contradicting $\{0,1\}\ne \{0,2\}$. Henceis an interval-parametrized family of isospectral nonisometric hyperbolic Riemann surfaces of genus four. Shrink $I$ again so a small length window around $3{\ell }_0$ contains only those two components for every parameter. Their common length $3\ell$ varies strictly, so members at distinct parameters cannot all be the same surface in different markings. The spectra agree within each pair; this construction does not assert a nontrivial continuously isospectral deformation of one hyperbolic surface.

The Sunada theorem is the following. Let $X\to M_0$ be a finite normal Riemannian covering of a connected closed Riemannian manifold, with finite deck transformation group $T$. If $U_1,U_2\le T$ satisfy Gassmann equivalence, meaningthen $U_1\backslash X$ and $U_2\backslash X$ have the same scalar Laplacian spectrum, with multiplicities. Equivalently, an epimorphism ${\pi }_1(M_0)\to T$ yields isospectral covers corresponding to the inverse images of $U_1,U_2$.

Here is the representation-theoretic proof. Every finite-dimensional Laplacian eigenspace $E_{\lambda }$ of $X$ is a group representation of $T$, since isometries commute with the Laplace-Beltrami operator. Eigenfunctions on $U_i\backslash X$ identify with $E_{\lambda }^{U_i}$. Averaging gives the projection $P_i=|U_i{|}^{-1}{\sum }_{u\in U_i}u$, soThe two subgroup orders agree, and the character of a representation ${\chi }_{\lambda }$ is constant on conjugacy classes. The displayed multiplicities are therefore equal for every $\lambda$. This proof also applies when $T$ has fixed points on $X$, provided the two subgroups $U_i$ act freely, so their quotients are smooth manifolds. We shall use this extension for an orbifold base in Question 4.

The precise geometric test for an isometry uses the full isometry group of a common universal cover. Equip the simply connected cover $\tilde{X}$ of $M_0$ with the lifted metric and let ${\Gamma }_i$ be the subgroups of its deck transformation group corresponding to the two covers. ThenIndeed, such an $F$ descends to an isometry of the quotients. Conversely an isometry of the quotients lifts to their simply connected covers; lift its inverse with compatible basepoints to see that the lift is bijective. Since it preserves the lifted metrics locally, it is a global isometry, and equivariance of the covering maps shows that it conjugates the deck subgroups. This proves the criterion, rather than assuming that an isometry must preserve the map to $M_0$. In particular, nonconjugacy inside $T$ alone does not establish nonisometry; the full covering space may have additional isometries. Nonisomorphic fundamental groups are a sufficient topological obstruction, and length-identified geometric configurations provide another.

For a concrete nonisomorphic Gassmann equivalent regular subgroups example, fix an odd prime $p$ and take $T=S_{p^3}$. Embed $E=(C_p{)}^3$ and $B={\mathrm{UT}}_3({\mathbb{F}}_p)$ through their regular permutation actions. Both have order $p^3$. Every nonidentity element of $E$ has order $p$. For $B$, write its elements as $I+N$ with $N^3=0$; thenover ${\mathbb{F}}_p$, including $p=3$, so its nonidentity elements also have order $p$. In a regular action, each nonidentity order-$p$ element has cycle type $p^{p^2}$. Thus each subgroup meets the identity class in one element, the class of that cycle type in $p^3-1$ elements, and every other conjugacy class in zero elements. They are Gassmann equivalent. They are not isomorphic: $E$ is abelian, whereas the Heisenberg group $B$ has $[I+E_{12},I+E_{23}]=I+E_{13}\ne I$.

To obtain arbitrarily many examples in dimension five, choose $r$ distinct odd primes $p_1,\dots ,p_r$ with $2^r\ge n$. For each prime use the preceding pair $E_j,B_j\le S_{p_j^3}$. SetA conjugacy class in a direct product is a product of conjugacy classes, and its intersection count with $U_{\epsilon }$ is the product of the individual counts. Thus all $2^r$ subgroups are Gassmann equivalent. They are pairwise nonisomorphic: the unique Sylow $p_j$-subgroup of $U_{\epsilon }$ is abelian precisely when ${\epsilon }_j=0$, and an abstract group isomorphism preserves this property at each prime.

We use the closed-manifold realization of a finitely presented fundamental group in dimension five. Its construction here is short. From a finite presentation of $T$ with $s$ generators, start with the connected sum of $s$ copies of $S^1\times S^4$. Its fundamental group is free on those generators. Represent the finitely many relators by disjoint embedded circles, which is possible by general position for curves in a manifold in dimension five. Their oriented normal bundles are trivial. Perform framed circle surgery, removing $S^1\times D^4$ and inserting $D^2\times S^3$. The Seifert-van Kampen theorem shows that each surgery kills exactly the normal closure of the corresponding relator: the attaching circle bounds the new disk, and the new $S^3$ factor contributes no fundamental group. The resulting closed connected five-manifold $M_0$ has ${\pi }_1(M_0)=T$.

Choose any smooth Riemannian metric on $M_0$ and lift it to its universal cover $X$. Since $T$ is finite, $X$ is a compact simply connected five-manifold on which $T$ acts freely and isometrically. The Sunada theorem makes all $U_{\epsilon }\backslash X$ isospectral, whileshows that they are pairwise nonhomeomorphic, hence nonisometric. Taking any $n$ members proves for every positive integer $n$, there are $n$ pairwise nonisometric closed five-manifolds with the same Laplacian spectrum.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$. For the global discrete spectral statements, take $M$ to be a closed manifold with a smooth Riemannian metric. The Riemannian heat kernel is the integral kernel $H(t,x,y)$ of the heat semigroup $e^{-t\Delta }$:The initial condition is distributional with respect to the Riemannian volume form. Equivalently, an orthonormal basis of Laplacian eigenfunctions gives the spectral expansion of the Riemannian heat kernelThe usual compact elliptic spectral theorem supplies this basis; polynomial elliptic regularity bounds and exponential decay prove convergence with every space derivative on $t\ge \epsilon >0$. These analytic results may be used as clearly stated external ingredients. We will also construct the kernel locally and prove its short-time expansion.

On a sufficiently small neighbourhood of the diagonal, let $r=d(x,y)$ and write $x={\exp }_y\xi$ in geodesic normal coordinates, with $d{V}_x=J(y,\xi )d\xi$. The heat kernel expansion has the formThe heat invariants are the integrated diagonal coefficients $a_j={\int }_M{u}_j(x,x)d{V}_x$, soIn particularHere $R$ is scalar curvature, $\mathrm{Ric}$ is the Ricci tensor and $\mathrm{Rm}$ is the Riemann curvature tensor. The last expression is the integrated second scalar heat coefficient; a local divergence term integrates to zero on a closed manifold. The exponent of the leading heat-trace term determines dimension, and these coefficients determine volume, total scalar curvature, and the displayed integrated curvature combination. On a surface $R=2K$, where $K$ is Gaussian curvature, so $a_1=\frac{1}{3}\int KdA=\frac{2\pi }{3}\chi (M)$ by the Gauss-Bonnet theorem, and $a_2=\frac{1}{15}\int K^{2}dA$. Thus the Euler characteristic, and the genus of a connected orientable closed surface, are determined. The first three coefficients also detect constant Gaussian curvature, since they determine $\int (K-\overline{K}{)}^{2}dA$. These conclusions do not claim that the pointwise curvature function or the metric is determined.

Here is the local construction. Put $g_t=(4\pi t{)}^{-m/2}{e}^{-r^2/(4t)}$. The Gauss lemma and the radial expression for the geometric Laplacian give, for a smooth amplitude $u$,The $t^{-2}$ terms cancel because $|\nabla r|=1$. Define $D=r{\partial }_r+(r/2){\partial }_r\log J$. The singular term disappears with $u_0=J^{-1/2}$, normalized by $u_0(y,y)=1$. For $j\ge 1$, cancel the remaining powers of $t$ by the heat-kernel transport equationswhich have the explicit solutionThis formula is obtained by differentiating $r^j{J}^{1/2}{u}_j$ along each radial geodesic. It proves smoothness jointly in $x,y$, including the diagonal, by induction: the integrand is smooth on a fixed compact integration interval. A homogeneous solution of $(D+j)u=0$ has $r^j{J}^{1/2}u$ constant along a ray, and smoothness at $r=0$ forces this constant to vanish for $j>0$. Thus the smooth transport coefficients are unique. In particular $u_j(y,y)=-({\Delta }_x{u}_{j-1})(y,y)/j$. Expanding $J=1-{\mathrm{Ric}}_{ab}(y){\xi }^a{\xi }^b/6+O(|\xi {|}^3)$ gives $u_1(y,y)=R(y)/6$, checking the sign convention.

Choose a smooth cutoff $\eta (x,y)$ that is one near the diagonal and supported where these normal coordinates exist, and form the heat parametrixIt tends to the identity kernel as $t\downarrow 0$: in normal coordinates the Gaussian is an approximate identity, and $u_{0}J=J^{1/2}\to 1$ at the centre. Its residual $R_N=({\partial }_t+{\Delta }_x)P_N$ is $\eta g_t{t}^N{\Delta }_x{u}_N$ plus cutoff-derivative terms supported a positive distance from the diagonal. Those latter terms are smaller than every power of $t$. Hence $R_N=O(t^{N-m/2})$ uniformly. Every prescribed finite number of derivatives has the same kind of estimate after increasing $N$; differentiating a Gaussian costs only finitely many powers of $t$.

For $N$ large enough that the residual is continuous and bounded up to time zero, use the Volterra parametrix correction. For kernels defineSet $Q=-R_N+R_N\ast R_N-R_N\ast R_N\ast R_N+\cdots$. A bounded residual with bound $C$ givesso this series converges and $Q+R_N+R_N\ast Q=0$. Because $P_N$ has identity initial kernel, differentiating the convolution gives $({\partial }_t+{\Delta }_x)(P_N\ast Q)=Q+R_N\ast Q$. Thusis an exact kernel with the correct initial condition. Taking $N$ sufficiently large gives any required finite regularity; uniqueness of the heat equation makes the kernels obtained from different $N$ equal. For uniqueness, the difference applied to smooth initial data has zero initial value, and the energy identity $\frac{d}{dt}\Vert v{\Vert }_2^2=-2\Vert \nabla v{\Vert }_2^2$ forces $v=0$. Parabolic regularity then gives smoothness for positive time.

The Gaussian part of $P_N$ has uniformly bounded spatial $L^1$ norm, so the correction is $O(t^{N+1-m/2})$. This proves, uniformly near the diagonal,with corresponding derivative estimates on increasing the truncation order. In particular on every parabolic neighbourhood $r\le B\sqrt{t}$ it is the stated Gaussian expansion with relative remainder $O(t^{N+1})$, and on the diagonal it yields the complete heat trace expansion. The standard off-diagonal Gaussian version on compact subsets of a sufficiently small convex normal neighbourhood follows from the same convolution estimates with the Gaussian factor retained: the inequalitylocalizes the convolution to the unique joining geodesic; its transverse Hessian is positive, and Gaussian integration gives the successive transport orders. The additive estimate above already specifies an appropriate uniform near-diagonal remainder without asserting a relative estimate far from the diagonal on the basis of a crude bound.

If $M$ is noncompact, use the minimal Friedrichs extension heat semigroup; if there is a boundary, specify boundary conditions. The same construction is local in the interior. One may extend a relatively compact coordinate neighbourhood to a closed auxiliary manifold; localization errors are supported a positive distance away and are exponentially small near the smaller diagonal neighbourhood. Thus the local coefficients remain the same. Finite integrated heat invariants and a discrete spectrum require the global hypotheses used above.

Let $M={\mathbb{R}}^d/\Lambda$ be a flat torus, where $\Lambda$ is a full-rank Euclidean lattice, and use the nonnegative Laplace-Beltrami operator $\Delta =-{\sum }_{k=1}^d{\partial }_k^2$. The dual lattice isFor $w\in {\Lambda }^{\ast }$, the function ${\varphi }_w(x)=\mathrm{vol}(M{)}^{-1/2}{e}^{2\pi i⟨w,x⟩}$ is periodic under $\Lambda$, so it defines a function on $M$, and direct differentiation gives $\Delta {\varphi }_w=4{\pi }^2|w{|}^2{\varphi }_w$. Integrating characters on a fundamental parallelepiped proves that these functions are orthonormal. After choosing a lattice basis, the ordinary Fourier series theorem on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ proves completeness, so there are no additional eigenvalues. Thus the spectrum of a flat torus iswith multiplicity equal to the number of dual vectors of the given length; the zero eigenvalue occurs once.

It remains to show that a two-dimensional Euclidean lattice is determined up to an orthogonal transformation by its vector-length multiset. This is the two-dimensional lattice reconstruction from vector lengths. Let $v$ be a shortest nonzero vector, of length $a$. It is primitive: $v=ku$ with integer $|k|>1$ would give a shorter nonzero vector. The lattice vectors on its line are therefore precisely the integer multiples of $v$. Subtract two occurrences of each length $ka$, $k=1,2,\dots$, from the length multiset. This removes exactly one primitive lattice line, even when other vectors happen to have the same lengths. The smallest remaining length $b$ is the length of a shortest vector $w$ independent of $v$.

These two vectors form a basis of the lattice. Otherwise a nonzero coset of $\mathbb{Z}v+\mathbb{Z}w$ has a lattice representative $u=sv+tw$ with $|s|,|t|\le 1/2$. It cannot have $t=0$, since primitivity of $v$ would then force $u=0$. It is therefore independent of $v$, whileThe inequality is strict: if both coefficients are nonzero, the vectors are not parallel and the triangle inequality is strict; if $s=0$, then $|u|\le b/2$. This contradicts the choice of $w$.

The covolume $A$ is also determined by the length multiset. If $N(R)$ counts lattice vectors in a radius-$R$ disk, thenTo justify the estimate, tile the plane by a bounded fundamental parallelogram. Only cells meeting a fixed-width neighbourhood of the disk boundary contribute an error; their total number is $O(R)$. The lattice basis $v,w$ has Gram matrixThus $|c|=\sqrt{a^2{b}^2-A^2}$ is determined. Replacing $w$ by $-w$ makes $c\ge 0$, so the Gram matrix is determined entirely. Equal Gram matrices give an orthogonal map between the lattice bases and hence between the lattices.

Two isospectral two-dimensional flat tori consequently have orthogonally equivalent dual lattices. Taking duals shows that their original lattices are orthogonally equivalent too, and the orthogonal map descends to an isometry of the quotient tori. Two isospectral flat tori of dimension two are isometric.

For an oriented three-manifold M with prescribed framed boundary endpoints P, the relative Kauffman bracket skein module is the complex vector space generated by properly embedded framed tangles, modulo framed isotopy and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted circle multiplies by $\delta =-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a circle has value $\delta$. For a nonempty uncolored link this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the quantum dimensions. Gluing compatible boundary pieces composes skeins; disjoint union gives tensor products. With no endpoints this is the Kauffman bracket skein module. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed isotopy.

For $A\ne 0$, the Temperley-Lieb algebra ${\mathrm{TL}}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed circle removed with factor $\delta$. There are $\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands givesEvery crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.

Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum ${\sum }_{s=0}^{m-1}{A}^{2(m-1-2s)}$ if its denominator vanishes. Set ${\Delta }_j=(-1{)}^j[j+1]$. These quantum integers satisfy ${\Delta }_0=1$, ${\Delta }_1=\delta$, and ${\Delta }_j=\delta {\Delta }_{j-1}-{\Delta }_{j-2}$. The Jones-Wenzl idempotent $f_n$ is characterized by identity-matching coefficient one andStarting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes 1$ and $C=F{e}_{n-1}F$. Whenever the denominators are nonzero the recursion isHere is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $({\Delta }_{n-1}/{\Delta }_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu ={\Delta }_{n-1}/{\Delta }_{n-2}$. Also $FC=CF=C$. Expanding $(F-{\mu }^{-1}C{)}^2$ proves idempotence. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $C{e}_{n-1}=\mu F{e}_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.

For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta -{\Delta }_{n-2}/{\Delta }_{n-1})f_{n-1}=({\Delta }_n/{\Delta }_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\mathrm{tr}(f_n)={\Delta }_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.

A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}{e}_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted Jones-Wenzl twist eigenvalue identity, including its exponent:A negative framing twist acts by the inverse scalar.

Choose $A=e^{\pi i/(2r)}$, with an integer $r\ge 3$. Then $[r]=0$ but $[1],\dots ,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the reduced Temperley-Lieb skein theory by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\dots ,r-2$, with dimensions and twistsTheir dimensions are nonzero. The recurrence decomposes $f_j\otimes 1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup idempotent is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all matrix blocks split. Applying the recurrence to the decompositions proves the fusion of Jones-Wenzl colors ruleOne can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.

We next prove the Kirby color handle-slide identity, rather than assume it. In the annular skein space let ${\hat{f}}_b$ be the closure of color b and define the Kirby color $\Omega ={\sum }_b{d}_b{\hat{f}}_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_c{i}_c=1_c$ and ${\sum }_c{i}_c{p}_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtainingClosing the resulting diagrams and straightening the bent strand gives $\mathrm{tr}(V_c^b{U}_c^b)=\mathrm{tr}(p_c{i}_c)=d_c$. Since b is simple, $V_c^b{U}_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. ThereforeTo slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution ${\sum }_c{i}_c{p}_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed handle slide.

We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed Hopf link with colors i,j. On a fusion channel k, its double braiding has eigenvalue ${\theta }_k/({\theta }_i{\theta }_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identityFor $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey ${\hat{f}}_j=X{\hat{f}}_{j-1}-{\hat{f}}_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the Hopf pairing of Jones-Wenzl colorsIndeed the recurrence is the sine identity $\sin ((j+1)x)=2\cos x\sin (jx)-\sin ((j-1)x)$ with $x=(i+1)\pi /r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now givesFor example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is ${\sum }_{k=1}^{r-1}\sin (ik\pi /r)\sin (jk\pi /r)=\frac{r}{2}{\delta }_{ij}$. Thus no degeneracy remains in this pairing.

The Gauss sums of Jones-Wenzl colors are $p_{\pm }={\sum }_j{d}_j^2{\theta }_j^{\pm 1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion proveIn the last step ${\sum }_i{d}_i{N}_{ij}^k=d_j{d}_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_{+}{T}^{-1}d$, with $T=\mathrm{diag}({\theta }_j)$. The matrix H and vector d are real, and each twist has modulus one, so conjugation gives $H{T}^{-1}d=p_{-}Td$. Apply H again and use its orthogonality to obtain $\boxed{p_{+}{p}_{-}={\mathcal{D}}^2}$. Hence neither factor vanishes.

We may use the surgery existence theorem and Kirby calculus: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by handle slides and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery link L let $\sigma (L)$ be the signature of its symmetric surgery linking matrix, put $\kappa =p_{+}/\mathcal{D}$, and define the Jones polynomial surgery invariantA handle slide preserves the colored bracket by the proved identity and changes the linking matrix by congruence, preserving its signature. Adding a disjoint plus-one-framed unknot multiplies the bracket by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so the net factor is $p_{+}/(\mathcal{D}\kappa )=1$. For a minus-one-framed unknot it is $\kappa p_{-}/\mathcal{D}=p_{+}{p}_{-}/{\mathcal{D}}^2=1$. Kirby calculus therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has ${\tau }_r(S^3)=1$; reversing orientation conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the Jones polynomial.

For real projective 3-space the surgery diagram is a single unknot of framing two. Thus $m=1$, $\sigma =1$, and its colored evaluation is ${\sum }_j{d}_j^2{\theta }_j^2$. Since $\mathcal{D}\kappa =p_{+}$, the explicit answer isThe denominator is nonzero by the proved Gauss-sum identity. For instance at $r=3$ the twists are $1,-i$, the squared dimensions are both one, and the numerator vanishes, so this normalized invariant is zero for ${\mathbb{RP}}^3$ at that level.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.