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The Cheeger-Gromoll splitting theorem states that a complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line in a Riemannian manifold is isometric to a Riemannian product $N\times \mathbb{R}$ with the Euclidean metric on the line. The factor $N$ is complete and has nonnegative Ricci curvature. Under the isometry one can take the given line to $\{q\}\times \mathbb{R}$, up to translation and reversal. A line means global minimization between every pair of its points, not merely a complete geodesic.

A smooth manifold alone has no distinguished notion of an isometry; one must first choose a Riemannian metric $g$. We therefore interpret the essay as concerning a smooth connected Riemannian manifold $(M,g)$. Its isometry group consists of diffeomorphisms $f$ with $f^{\ast }g=g$. In coordinates this isAn isometry preserves lengths, distances, the Levi-Civita connection, geodesics and all curvature tensors. The Myers-Steenrod theorem ensures that the group is a finite-dimensional Lie group acting smoothly, and that even distance-preserving bijections have the required smoothness.

The crucial rigidity is determination by one first jet. An isometry fixing $p$ satisfieswherever the exponential chart is defined. If also $d{f}_p=I$, it is the identity on a neighborhood. Continuing this equality through overlapping exponential neighborhoods proves it on connected $M$. Therefore the isotropy group at $p$ embeds in $O(T_{p}M,g_p)$. Its dimension is at most $n(n-1)/2$, while an orbit has dimension at most $n$. ConsequentlyOn a compact manifold the group is compact. Indeed distance preservation makes a family of isometries equicontinuous; compactness and the same argument for their inverses give subsequential limits which are inverse distance-preserving bijections. Smoothness and the Lie-group structure then follow from Myers-Steenrod theorem.

Infinitesimal isometries are Killing vector fields. Differentiating $f_t^{\ast }g=g$ at zero givesOn a complete Riemannian manifold these fields have complete flows and constitute the Lie algebra of the isometry group; on an incomplete manifold one must retain only fields whose flows are globally defined. A Killing vector field is fixed by its value and its skew derivative at one point, consistent with the same dimension bound. Along a geodesic, its restriction is a Jacobi field, so local infinitesimal data propagate rigidly.

In dimension three there is no independent Weyl curvature tensor: the Riemann curvature tensor is determined by the Ricci tensor. Large isotropy therefore strongly constrains the geometry. If the full dimension six is attained, the group is transitive with three-dimensional isotropy. That isotropy rotates all tangent two-planes transitively, forcing equal sectional curvatures at each point; transitivity makes the value constant. The simply connected complete models are the round $S^3$, Euclidean ${\mathbb{R}}^3$ and hyperbolic ${\mathbb{H}}^3$. Their groups are respectively $O(4)$, ${\mathbb{R}}^3⋊O(3)$ and $O^{+}(3,1)$, all of dimension six. The orientation-preserving hyperbolic group is $PS{L}_2(\mathbb{C})$.

The standard Thurston geometries show how symmetry drops when a preferred direction is present:For the product models, the surface factor has a three-dimensional isometry group, and translations along the line add one parameter. For Nil and the universal covering of $S{L}_2(\mathbb{R})$ with their standard geometric metrics, a three-dimensional transitive group is enlarged by a circle of isotropy. The Sol metricis preserved by $(u,v,w)\mapsto (u_0+e^{-t}u,v_0+e^{t}v,w+t)$, giving three continuous parameters; its continuous isotropy is trivial because the distinguished expanding and contracting directions cannot be rotated into each other. Additional discrete symmetries may still occur. These dimensions refer to the model metrics, not arbitrary left-invariant perturbations.

A quotient generally has fewer global isometries than its local model. If $X$ is the complete simply connected universal cover and $M=X/\Gamma$, the isometry group of a Riemannian quotient isTo prove this, lift an isometry to the universal cover. Conjugating a deck transformation by the lift again gives a deck transformation, so the lift normalizes $\Gamma$. Conversely a normalizing isometry descends, and two lifts differ by a deck transformation. A flat three-torus retains its three-dimensional translation group, with additional discrete lattice symmetries. A round spherical quotient retains only the normalizer of its finite deck group; the full six-dimensional group does not automatically descend.

For a closed hyperbolic three-manifold, there are no nonzero Killing vector fields. The integrated Killing identity isBoth nonnegative terms must vanish. The compact isometry group is therefore zero-dimensional and hence finite. The Mostow rigidity theorem additionally says that isomorphisms between the fundamental groups of complete finite-volume hyperbolic three-manifolds are induced by unique hyperbolic isometries; this fixes the complete hyperbolic metric up to isometry and contrasts with the moduli of flat tori. An arbitrary smooth three-manifold need not admit a single homogeneous metric: prime decomposition and torus splitting may lead to several geometric pieces. Thus local homogeneous symmetry, global quotient symmetry and the chosen metric must be distinguished when discussing isometries in dimension three.

Work on a closed compact smooth Riemannian manifold $M$ without boundary, as in the usual discrete spectral setting here. Use the paper's $\Delta ={\mathrm{tr}}_g{\mathrm{Hess}}_g$, and put $P=-\Delta$. The Riemannian heat kernel $H(t,x,y)$ is the smooth kernel, for $t>0$, representing the solution ofThus $({\partial }_t-{\Delta }_x)H=0$ and its distributional initial value in $x$ is the Dirac delta distribution at $y$. The heat trace isThe heat invariants are the coefficients $a_r(M)$ in the short-time heat trace expansion specified below.

To derive its spectral form, take an orthonormal eigenbasis $P{\varphi }_j={\lambda }_j{\varphi }_j$, with $0={\lambda }_0\le {\lambda }_1\le \cdots$, repeating eigenvalues according to multiplicity. For the initial datum ${\varphi }_j$, the function $e^{-t{\lambda }_j}{\varphi }_j$ solves the heat equation. Uniqueness identifies it with the heat-kernel solution. Expanding arbitrary initial data in this basis therefore gives the spectral expansion of the Riemannian heat kernel:Standard elliptic estimates and eigenvalue growth imply smooth convergence for $t$ bounded away from zero: polynomial bounds on derivatives of eigenfunctions are dominated by the exponential factors. On the diagonal all summands are nonnegative, so monotone convergence also justifies integration without needing an interchange of conditionally convergent terms. Since each eigenfunction has unit $L^2$ norm,Equivalently, if the paper's Laplacian eigenvalues are written ${\nu }_j=-{\lambda }_j$, the formula is ${\sum }_j{e}^{t{\nu }_j}$; a growing exponential is not the heat semigroup.

Here are the needed heat kernel expansion facts, stated without proof. On a closed $d$-dimensional smooth Riemannian manifold,with the diagonal remainder uniform in $x$ after any finite truncation. The coefficients are smooth local geometric quantities; $u_1=R/6$. Compactness therefore allows integration term by term:These integrated coefficients are the heat invariants. The uniform leading remainder in particular gives $Z_M(t)=(4\pi t{)}^{-d/2}(\mathrm{Vol}(M)+O(t))$.

Isospectral manifolds have identical heat traces by the derived spectral formula. If their dimensions were $d<e$, multiplying the common trace by $t^{d/2}$ would give a finite positive limit on the $d$-dimensional manifold but divergence on the $e$-dimensional one. Thus the dimensions are equal, and equality of the leading coefficient gives equal Riemannian volume. Explicitly,The absence of a boundary fixes this particular integer-power expansion. With a boundary one must first specify boundary conditions and include the appropriate half-integer boundary terms; the leading dimension-and-volume conclusion remains valid in the usual compact smooth setting.

For $p,q\in M$, define the Riemannian distance byA connected smooth manifold is locally path connected, so any two points are joined by a finite concatenation of coordinate paths. Thus the infimum is finite. It is nonnegative, $d_g(p,p)=0$, and reversing a curve proves symmetry. Concatenating two curves whose lengths approach their respective infima proves the triangle inequality.

For positivity and the topology, choose a coordinate ball about $p$ whose closure is contained in a chart. On its closure there are constants $0<c<C$ withA curve staying in the ball has length at least $c$ times its coordinate displacement. A curve leaving it must first reach its boundary, costing at least $c$ times the positive coordinate distance from $p$ to that boundary. Therefore a distinct $q$ has $d_g(p,q)>0$, whether it lies inside or outside the ball. For $q$ sufficiently close to $p$, the coordinate line segment gives $d_g(p,q)\le C|x(q)-x(p)|$, while the preceding lower bounds prevent a short curve from taking a shortcut outside the chart. These inequalities prove that the Riemannian distance induces the manifold topology. In particular it defines a genuine metric space.

A Riemannian manifold is geodesically complete if every maximal affinely parametrized geodesic exists for all real parameter values. We prove both implications of the Hopf-Rinow theorem rather than assuming global distance minimizers.

First assume completeness as a metric space. Let a constant-speed geodesic have a finite maximal right endpoint $b$. The bound $d_g(\gamma (t),\gamma (s))\le c|t-s|$ makes $\gamma (t)$ a Cauchy sequence as $t\uparrow b$, hence gives a limit $p\in M$. The distance topology just established puts the final portion inside a relatively compact coordinate ball about $p$. The positive lower bound on the Riemannian metric bounds all coordinate velocity components. The Christoffel symbols are bounded on the closure of that ball, and the geodesic equationthen bounds the coordinate acceleration. The velocity has a limit $v$ as $t\uparrow b$. Local existence and uniqueness for the geodesic equation with data $(p,v)$ continue $\gamma$ through $b$, contradicting maximality. The same argument applies at a finite left endpoint. Hence metric completeness implies geodesic completeness.

Conversely assume geodesic completeness. Fix $p,q$, put $r=d_g(p,q)>0$, and choose $0<ϵ<r$ so small that a closed radius-$2ϵ$ normal ball about $p$ lies inside a convex normal neighborhood. By the Gauss lemma, its radial geodesics minimize; a curve leaving the larger ball cannot improve on a radius-$ϵ$ radial segment. Thus the radius-$ϵ$ geodesic sphere $S_{ϵ}(p)$ is compact and consists of points at distance exactly $ϵ$ from $p$.

The distance splitting through a small geodesic sphere has an elementary proof here. Every path from $p$ to $q$ must cross $S_{ϵ}(p)$, and its initial portion has length at least $ϵ$. HenceThe triangle inequality proves the reverse bound. Continuity of $d_g(\cdot ,q)$ and compactness give a point $x$ achieving the minimum, so $d_g(x,q)=r-ϵ$. Let $\gamma (t)={\exp }_p(tv)$ be the complete unit-speed radial geodesic through $x$ at time $ϵ$.

Consider the closed subsetIt is nonempty since $ϵ\in A$, and has a largest member $t_{\ast }$. Suppose $t_{\ast }<r$. At $x_{\ast }=\gamma (t_{\ast })$, choose a small normal sphere of radius $\eta <r-t_{\ast }$ and repeat the splitting argument. It supplies a radial endpoint $y$ withThe triangle inequality gives $d_g(p,y)\ge t_{\ast }+\eta$, while the broken path following $\gamma$ to $x_{\ast }$ and then the short radial segment to $y$ has exactly that length. It is therefore minimizing. Its two velocities at the joining point must agree: if unit incoming and outgoing velocities are $u,w$, moving the join in direction $w-u$ changes the sum of lengths by $-|u-w{|}^2$ to first order, using the first variation of geodesic energy or the equivalent unit-speed length formula. A corner would strictly shorten it. Uniqueness of the geodesic equation consequently gives $y=\gamma (t_{\ast }+\eta )$, contradicting the maximality of $t_{\ast }$. Thus $t_{\ast }=r$, $\gamma (r)=q$, and $\gamma {|}_{[0,r]}$ is a minimizing geodesic.

This radial continuation proof of Hopf-Rinow also proves compactness of closed balls. For every $R\ge 0$,The inclusion from right to left follows from the length of a radial geodesic; the reverse inclusion uses the minimizing geodesic just constructed. The Riemannian exponential map is defined on all of $T_{p}M$ by geodesic completeness, and the closed tangent ball is compact in a finite-dimensional vector space. Its continuous image is compact. Every Cauchy sequence is bounded, hence has a convergent subsequence in such a ball, and the Cauchy property makes the whole sequence converge. Therefore geodesic completeness implies metric completeness, completing both directions.

For the hyperbolic plane, use the upper half-plane model with $g=(d{x}^2+d{y}^2)/y^2$, $y>0$. Its geodesic equation isEvery nonconstant unit-speed solution is either a vertical curve $x=a$, $y=b{e}^{\pm t}$ or, after translating and possibly reversing $t$,Substitution verifies both equations and $g(\dot{\gamma },\dot{\gamma })=1$. These curves cover every unit initial tangent: a nonvertical tangent at $(x,y)$ determines the circle center $a=x+y{y}^{\prime }/x^{\prime }$ and radius $R=\sqrt{(x-a{)}^2+y^2}$, and its orientation determines the sign of the parameter. Constant solutions and constant-speed rescalings cover arbitrary initial velocities. In every case $y(t)>0$ for all finite $t$, and the solution is defined for every real $t$. Thus the real hyperbolic plane is geodesically complete, directly, and the proved Hopf-Rinow theorem also makes its distance complete.

Use the curvature convention $\mathcal{R}$ of Solution 1, and put $T=\dot{\gamma }$. On continuous piecewise smooth perpendicular vector fields with zero endpoint values, the Riemannian index form is the symmetric bilinear formA Jacobi field is a smooth field $J$ along $\gamma$ satisfying $D_t^{2}J+\mathcal{R}(J,T)T=0$. The point $\gamma (t_0)$ is a conjugate point to $\gamma (0)$ along this geodesic when there is a nonzero Jacobi field with $J(0)=J(t_0)=0$. Such a field is perpendicular to $T$, because $⟨J,T⟩$ is affine and vanishes at both endpoints.

The conjugate-point criterion for the Riemannian index form, stated without proof, isAt a first conjugate endpoint the form is nonnegative but has a nonzero nullspace, consisting of endpoint-vanishing Jacobi fields. A conjugate point strictly inside gives a negative direction. This explicitly distinguishes the non-strict and strict senses of positivity in the question; in terminology where “positive definite” already means strict positivity, the second line is the positive-definiteness condition.

Choose $n-1$ perpendicular orthonormal fields $E_1,\dots ,E_{n-1}$ by parallel transport along $\gamma$, and put $V_i(t)=\sin (\pi t/L)E_i(t)$. They are admissible andThe tangent has unit length. The assumed Ricci curvature bound and the two elementary sine-square and cosine-square integrals give the sine index-form bound for positive Ricci curvature:Since $n\ge 2$, at least one of the admissible $V_i$ therefore has $I(V_i,V_i)<0$.

If the Riemannian manifold is complete, the Hopf-Rinow theorem supplies a minimizing geodesic between any two points. Its Riemannian index form is nonnegative: any smooth fixed-endpoint variation has nonnegative second variation of geodesic energy, since a length minimizer of constant speed also minimizes energy on a fixed parameter interval by the Cauchy-Schwarz inequality. The same conclusion for piecewise smooth fields follows by smooth approximation, or by piecewise variations. The negative direction above excludes a minimizing segment longer than $\pi /\sqrt{\kappa }$. Taking the supremum over pairs yields the diameter conclusion of the Bonnet-Myers theorem:

The completeness assumption is essential. An explicit incomplete positively curved strip with infinite diameter isThe map $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the unit sphere, so $K=1$ and, in dimension two, $\mathrm{Ric}=g$. Thus the required Ricci curvature bound holds with $n=2$, $\kappa =1$. The unit-speed meridian $t\mapsto (t,0)$ reaches the missing boundary $u=\pi /4$ in finite time, so the strip is geodesically incomplete. On the other hand every curve joining $(0,0)$ to $(0,A)$ hassince $\cos u\ge 1/\sqrt{2}$ throughout the strip. Therefore its intrinsic diameter is infinite, in particular larger than $\pi$. This is the unwrapped strip with $v\in \mathbb{R}$, not a strip with longitude identified modulo $2\pi$; that distinction prevents spherical shortcuts from invalidating the example.

Take the nonnegative scalar Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$ on a closed compact Riemannian manifold, and include every eigenvalue with its multiplicity. The spectrum contains substantial global geometric information, but generally does not determine the Riemannian metric up to isometry. Specifying the operator and its domain matters: the spectrum on functions is different information from the spectra of the Hodge Laplacian on differential forms, and a boundary requires fixed boundary conditions.

The Weyl law makes dimension and volume audible. If $N(\lambda )$ counts eigenvalues at most $\lambda$, thenwhere ${\omega }_d$ is the volume of the Euclidean unit ball. Thus the growth exponent recovers $d$, and its leading coefficient recovers $\mathrm{Vol}(M)$. The multiplicity of zero gives the number of connected components: a function in the kernel has $\int |\nabla f{|}^2=0$ and hence is constant on each component.

More detailed invariants come from the heat trace, which is determined directly by the spectrum:The first heat invariants for a closed manifold areHere $R$ is scalar curvature, $\mathrm{Ric}$ the Ricci curvature, and $\mathrm{Rm}$ the Riemann curvature tensor. A divergence term in the local coefficient integrates to zero on a closed manifold. The expansion supplies further integrated curvature expressions, rather than a pointwise curvature map.

On a closed surface, $R=2K$, $|\mathrm{Ric}{|}^2=2K^2$ and $|\mathrm{Rm}{|}^2=4K^2$, so the same coefficients becomeThe Gauss-Bonnet theorem gives $\int KdA=2\pi \chi (M)$; hence the constant term of the heat trace is $\chi (M)/6$. In particular the Euler characteristic, and for a connected orientable surface its genus, are spectrally determined. There is even spectral determination of hyperbolic curvature on a surface: if this spectrum agrees with that of a curvature-$-1$ surface, the coefficients give $\int K=-A$ and $\int K^2=A$, whenceThus $K=-1$ everywhere. This determines local curvature, while leaving room for globally nonisometric hyperbolic surfaces.

Geodesic information is also accessible in important settings. For closed hyperbolic surfaces, the Selberg trace formula makes the scalar spectrum equivalent to the unmarked length spectrum with its appropriate multiplicities. This does not supply a marked length spectrum: it does not identify which group element or intersecting curve has a given length. For general manifolds, one must not simply assert this equivalence; the singularities of the wave trace can be affected by degeneracies and cancellations.

The historic development separates these positive results from the inverse problem. Weyl's 1911 eigenvalue asymptotics established the leading geometric information. The Minakshisundaram-Pleijel heat expansion of 1949 brought local curvature into the short-time analysis. In 1964 Milnor exhibited nonisometric flat tori in dimension sixteen with identical scalar spectra, showing that all eigenvalues together need not recover the metric. Kac's 1966 question about a vibrating drum popularized the corresponding inverse problem for planar domains. Vignéras's 1980 hyperbolic examples and the Sunada theorem of 1985 extended the supply of counterexamples; the latter turns Gassmann equivalence of finite subgroups into a geometric construction. The four-dimensional tetracode length-isospectral lattice construction and the planar examples of Gordon, Webb and Wolpert in 1992 made nonuniqueness explicit in smaller dimensions and in the original drum setting.

Nor is scalar isospectrality a universal topological invariant strong enough to determine the manifold. For example, Tetra and Didi are scalar-isospectral closed orientable flat three-manifolds with first Betti numbers one and zero. Their topology therefore differs. The spectra of all Hodge Laplacians would detect these Betti numbers through the dimensions of their zero eigenspaces, illustrating the extra information obtained by changing the operator. Dimension, volume and many integrated invariants are determined; the full metric and, in general, the topology are not.

Let $M$ be a compact Riemannian manifold without boundary. Let $\delta =d^{\ast }$ be the formal $L^2$ adjoint operator of the exterior derivative, and let ${\Delta }_H=d\delta +\delta d$ be the Hodge Laplacian. Write ${\mathcal{H}}^r(M)=\ker {\Delta }_H\cap {\Omega }^r(M)$ for the smooth harmonic differential forms. The Hodge decomposition theorem states that this space is finite-dimensional andan $L^2$-orthogonal direct sum. Thus every smooth $r$-form is uniquely the sum of a harmonic, an exact and a coexact form; its exact and coexact components are unique, though the primitives used to write those components need not be unique. AlsoThe isomorphism sends a harmonic form to its de Rham cohomology class. For example, a closed coexact component $\delta \gamma$ must vanish because $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$; an exact harmonic form also vanishes by orthogonality. These facts explain uniqueness of the harmonic representative of a cohomology class.

On an oriented $n$-manifold the codifferential is $\delta =(-1{)}^{n(r+1)+1}\ast d\ast$ on $r$-forms. The general formulation above also holds on nonorientable Riemannian manifolds: use the globally defined Riemannian volume density for the $L^2$ inner product and its formal adjoint, rather than require a global ordinary volume form. The closed-manifold hypothesis is essential unless suitable boundary conditions are added.

Use the sign convention in the supplied scalar Laplace-Beltrami operator identity, and let $X={\omega }^{♯}$ be the metric-dual vector field. Integrate that Bochner formula for one-forms over the closed Riemannian manifold, with its volume density $d{V}_g$. The integral of a scalar Laplacian is zero: on an oriented manifold this is the Riemannian divergence theorem applied to its gradient, and the same density formula is valid without orientability. HenceBoth terms are nonnegative, because $\mathrm{Ric}\ge 0$. The smooth nonnegative integrand must therefore vanish everywhere: a strictly positive value would remain positive on an open neighborhood of positive volume and make its integral positive. In particular $|\nabla \omega {|}^2=0$ at every point, provingThus harmonic one-forms are parallel under nonnegative Ricci curvature. The same argument also gives $\mathrm{Ric}(X,X)=0$ pointwise. No orientation or connectedness assumption is needed for this step.

Use the nonnegative convention for the Laplace-Beltrami operator. For a smooth function $f$ on a Riemannian manifold $(M,g)$, three intrinsic definitions areHere $d$ is the exterior derivative, $d^{\ast }$ its formal $L^2$ adjoint for the Riemannian volume density, $\mathrm{grad}f$ the Riemannian gradient, and $\nabla$ the Levi-Civita connection. The bilinear form $\nabla df$ is the Riemannian Hessian. The equally common nonpositive convention negates all three definitions simultaneously.

In local coordinates let $g^{ij}$ be the inverse metric matrix and $\rho =\sqrt{\det (g_{ij})}$. By the defining relation $g(\mathrm{grad}f,V)=df(V)$,The divergence of a Riemannian vector field can be defined by ${\mathcal{L}}_V(d{V}_g)=(\mathrm{div}V)d{V}_g$. Differentiating the coordinate volume density gives $\mathrm{div}V={\rho }^{-1}{\partial }_i(\rho V^i)$. Thus the second definition becomesFor compactly supported smooth $h$, coordinate integration by parts showsSince this holds for every such $h$, the first and second definitions agree. This argument also works on a nonorientable Riemannian manifold, using its volume density; no global orientation is needed.

For the third definition,The Christoffel symbols satisfy ${\Gamma }_{ik}^i={\partial }_k\log \rho$. This follows by tracing their formula and using ${\partial }_k\log \det g=\mathrm{tr}(g^{-1}{\partial }_{k}g)$. Metric compatibility givesSubstitute both identities into the coordinate divergence expression to obtainThis proves all three definitions equivalent. In geodesic normal coordinates at a point $p$, the common value is $-{\sum }_i{\partial }_i^{2}f(p)$. Finally, on a closed manifold,confirming the chosen positive Laplace-Beltrami operator sign convention.

On a closed compact Riemannian manifold, the heat kernel for the positive Laplace-Beltrami operator is the smooth kernel $K_M(t,x,y)$ of $e^{-t{\Delta }_M}$ for $t>0$. It solves $({\partial }_t+{\Delta }_x)K_M=0$ and tends to the delta kernel as $t\downarrow 0$, so thatFor an orthonormal eigenbasis with eigenvalues repeated according to multiplicity,The exponential factors and elliptic regularity ensure smooth convergence for positive time. If a boundary is present, the same construction uses a specified boundary condition preserved by the covering; the closed-manifold case is the one needed below.

Let $p:N\to M=N/U$ be the finite normal locally isometric covering map. The heat equation for $f\circ p$ on $N$ is the lift of the one for $f$ on $M$. Integrate the lifted solution over a fundamental domain for the deck transformation group $U$, splitting the integral over $N$ into its translates. This givesThere is no averaging factor in this kernel formula: each translate accounts for one lift of the integration variable. To verify it directly, the sum solves the lifted heat equation and its integral against $f$ converges to $f(px)$; uniqueness of the heat solution identifies the kernel. Replacing either lift $x$ or $y$ only permutes the sum, since isometries preserve the heat kernel.

Integrating the diagonal and using the covering degree gives the heat trace formulaSuppose $U$ is contained in a finite group $T$ of isometries of $N$. For $s\in T$, change variables $x=sy$ and use invariance of both volume and the heat kernel. ThenThus $F_t$ is a class function on $T$, and for its conjugacy classes $\mathcal{C}$,

Consequently, if $U_1,U_2\le T$ are Gassmann equivalent and act freely, their orders are equal and the displayed class-by-class sums give $Z_{N/U_1}(t)=Z_{N/U_2}(t)$ for every $t>0$. Equal heat traces determine equal eigenvalues with multiplicities: the least eigenvalue at which the multiplicities differed would give a nonzero leading exponential in their trace difference as $t\to \infty$, a contradiction. Hence the two smooth quotient manifolds are isospectral. This proves the Sunada theorem, with the free-action hypothesis ensuring that the quotients are manifolds rather than orbifolds. The theorem alone makes no nonisometry claim.

The musical isomorphisms are the pointwise maps induced by the Riemannian metric:In local coordinates they are $(X^{♭}{)}_i=g_{ij}{X}^j$ and $({\alpha }^{♯}{)}^i=g^{ij}{\alpha }_j$. The inverse metric makes these inverse bundle maps, hence inverse maps between smooth vector fields and smooth differential one-forms.

For a covariant two-tensor $\beta$, its metric trace is ${\mathrm{tr}}_g\beta =g^{ij}{\beta }_{ij}={\sum }_i\beta (e_i,e_i)$ in a Riemannian orthonormal frame. The contraction is independent of the chosen frame. There is a terminology distinction here: if $\beta$ is literally an alternating differential two-form, this trace is zero. The object needed below is the covariant two-tensor $D(X^{♭})$, generally not alternating. Interpreting that tensor as an alternating form would make the stated divergence construction impossible.

Let $D$ be the Levi-Civita connection. The Riemannian gradient, divergence and positive Laplace-Beltrami operator areThus $g({\mathrm{grad}}_{g}f,Y)=Yf$, and in coordinatesThis choice makes the Laplacian eigenvalues nonnegative on a closed Riemannian manifold. The alternative convention $+\mathrm{div}\mathrm{grad}$ reverses all eigenvalue signs, but none of the comparisons below.

For the Riemannian submersion, let $F=f\circ \pi$ and choose orthonormal horizontal lifts $E_i$ of a base frame and an orthonormal vertical frame $V_{\alpha }$. The horizontal part of $D_{E_i}{E}_i$ projects to the base Levi-Civita connection; this follows directly from its Koszul formula applied to basic horizontal fields, because their inner products and projected brackets are pulled back from the base. Therefore the horizontal trace of the Riemannian Hessian of $F$ is the pulled-back base Hessian trace. Since $V_{\alpha }F=0$, its vertical terms areWriting $H={\sum }_{\alpha }(D_{V_{\alpha }}{V}_{\alpha }{)}^{\mathrm{hor}}$ for the unnormalised fibre mean curvature vector givesThe totally geodesic submanifold condition on the fibres makes $H=0$. Consequently eigenfunction pullback by a Riemannian submersion givesHere the Riemannian submersion is onto, so the map is injective. Its image is exactly the members of the upstairs eigenspace constant on each whole fibre: such a function descends smoothly by local sections of the submersion, and the displayed identity forces the descended function to have the same eigenvalue. In particular, whenever these eigenspaces are finite dimensional, the downstairs multiplicity is no larger than the upstairs multiplicity. The upstairs eigenspace may also contain functions varying along fibres.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with the metric induced by the Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its dual lattice is ${\Lambda }^{\ast }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. The characters $e_w(x)=e^{2\pi i⟨w,x⟩}$ descend to the quotient and form an orthogonal Fourier basis of $L^2$ on it. Direct differentiation gives the spectrum of a flat torus:with multiplicity equal to the number of dual lattice vectors of each length. In particular zero has multiplicity one.

Let the given four-dimensional isospectral manifolds be ${\mathbb{R}}^4/{\Lambda }_1$ and ${\mathbb{R}}^4/{\Lambda }_2$. For $d=4+k$, take their products with the same $k$ circles of length $\epsilon$, equivalently the lattices ${\Lambda }_i\oplus \epsilon {\mathbb{Z}}^k$. The dual lattices are ${\Lambda }_i^{\ast }\oplus {\epsilon }^{-1}{\mathbb{Z}}^k$, so their Laplacian eigenvalues areThe given equality of base spectra with multiplicities makes these product spectra identical.

Choose $\epsilon$ smaller than the shortest nonzero vector length in either original Euclidean lattice. By short-vector cancellation for flat tori, the shortest nonzero vectors in the product lattice are exactly $\pm \epsilon e_j$ in its new factor. To spell out the nonisometry argument, any isometry of the product flat tori lifts to an isometry of their universal Euclidean covers. This is affine, and its orthogonal linear part maps one product lattice onto the other. It must map the span of their shortest vectors, namely the new $k$-dimensional factor, onto itself; orthogonality then preserves the original four-dimensional factor and maps ${\Lambda }_1$ onto ${\Lambda }_2$. That would give an isometry of the original flat tori, contrary to the assumption. Isospectral nonisometric flat tori therefore exist in every dimension $d\ge 4$.

Use the positive Laplace-Beltrami operator on a compact Riemannian manifold without boundary. Its heat kernel is the smooth integral kernel $K(t,x,y)$, $t>0$, of the solution operators for the heat equation:The last condition holds, in particular, in $L^2$ for smooth initial functions. The heat trace isThe heat invariants are the coefficients in the small-time heat kernel expansion of this trace:Equivalently, the local diagonal coefficients $u_j(x,x)$ integrate to $a_j$; $R_g$ is the scalar curvature. Some normalizations absorb $(4\pi {)}^{-d/2}$ into the coefficients. These definitions here use the displayed normalization consistently.

To prove uniqueness, let $K_1,K_2$ satisfy the heat kernel conditions and let $w$ be the difference of their solutions with the same smooth initial function. It solves $w_t=-\Delta w$ and tends to zero in $L^2$ as $t\downarrow 0$. The heat equation energy identity, using integration by parts on the closed manifold, gives for $t>0$Integrating from $s>0$ to $t$ gives $\Vert w(t){\Vert }_{L^2}\le \Vert w(s){\Vert }_{L^2}$; sending $s\downarrow 0$ proves $w(t)=0$. Thus for every smooth test function $f$,For fixed positive $t$ and $x$, smoothness makes this difference a smooth function of $y$. Taking its conjugate as test function shows its squared norm is zero. The heat kernel is unique.

The compact elliptic spectral theorem for the self-adjoint elliptic Laplace-Beltrami operator gives a complete orthonormal basis of smooth eigenfunctions ${\varphi }_j$, with discrete eigenvalues $0\le {\lambda }_0\le {\lambda }_1\le \cdots$ counted with multiplicity. Solving the heat equation for initial data ${\varphi }_j$ gives $e^{-t{\lambda }_j}{\varphi }_j$, so the unique heat kernel has its eigenfunction expansion and its heat trace isThe series converges for every $t>0$. This proves that the spectrum with multiplicities determines the heat trace.

Conversely, its large-time behaviour recovers the distinct eigenvalues and their multiplicities successively. The multiplicity of zero is $m_0={\lim }_{t\to \infty }{Z}_M(t)$. After subtracting the already determined terms, letbe the positive remaining sum, and let $\mu$ be its smallest remaining eigenvalue. ThenIndeed, after factoring $e^{-t\mu }$, the next distinct eigenvalue is separated from $\mu$ by a positive gap. The remaining tail tends to zero: for fixed $t_0>0$ its absolute sum is bounded by a constant times $e^{-({\mu }_{\mathrm{next}}-\mu )(t-t_0)}$, using finiteness of $Z_M(t_0)$. Induction recovers every spectral term. Thus the heat trace and the spectrum with multiplicities determine one another.

Finally the leading coefficient among the heat invariants is positive, soThe common heat trace of isospectral manifolds therefore forces them to have the same dimension. Once the dimension is recovered, its leading coefficient also recovers their common volume.

Take a finite presentation $T=⟨x_1,\dots ,x_r\mid R_1,\dots ,R_s⟩$. For a closed-manifold realization of a finitely presented fundamental group, begin with a five-dimensional ball and attach $r$ one-handles, giving a compact oriented five-manifold $W_1$ with fundamental group the free group on the $x_i$. Represent the relators by disjoint embedded circles in its four-dimensional boundary. General position allows them to be embedded and disjoint, and their oriented normal bundles admit framings. Attach a two-handle along each such circle. The resulting compact five-manifold $W$ hasby the Seifert-van Kampen theorem. Now set $B=\partial W$. It is a compact connected smooth four-manifold without boundary. Turning this handle decomposition upside down gives a relative handle decomposition of $(W,B)$ with handles of indices three, four and five only. Attaching such handles does not change the fundamental group: there are no new one-handles or two-handle relations. Thus ${\pi }_1(B)\to {\pi }_1(W)$ is an isomorphism andThese are the differential-topological handle facts used in the construction; the dimension-five thickening ensures the boundary retains the prescribed fundamental group.

Now assume $T$ finite. Choose any smooth Riemannian metric on $B$ and lift it to its universal cover $\tilde{B}$. This cover is compact, since its degree is $|T|$, and simply connected. Its deck transformations give a free isometric $T$-action. For the two Gassmann equivalent subgroups putEach is a closed Riemannian manifold, and $\tilde{B}$ is its simply connected universal cover, with deck transformation group $U_i$. Consequently ${\pi }_1(M_i)\cong U_i$.

Here is the complete spectral argument behind the Sunada theorem in this situation. Each finite-dimensional upstairs eigenspace $E_{\lambda }$ is a representation of $T$, because the action is isometric and commutes with the Laplace-Beltrami operator. Functions downstairs correspond exactly to invariant functions upstairs, so their multiplicities are $\dim E_{\lambda }^{U_i}$. Averaging the representation matrices over a subgroup is a projection onto its invariant subspace. Taking its trace givesThe character ${\chi }_{\lambda }$ is constant on each conjugacy class of $T$. Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every such class $C$; summing these equalities also gives $|U_1|=|U_2|$. The two averages are therefore equal for every $\lambda$. The quotients are isospectral, with ${\pi }_1(M_i)\cong U_i$.