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Choose an orthonormal basis $e_i$ of the tangent space at $p$ and the unit-speed geodesics with initial velocities $e_i$. Their covariant accelerations vanish, so $(f\circ {\gamma }_i{)}^{\prime \prime }(0)={\mathrm{Hess}}_{g}f(e_i,e_i)$. Taking the metric trace of the Riemannian Hessian proves the formula for $\Delta =\mathrm{div}\nabla$. The positive Laplace-Beltrami operator has the opposite sign.

For a smooth function $f$ and the Levi-Civita connection $\nabla$, the Riemannian Hessian is the covariant two-tensorThe connection rules show that this is linear over smooth functions in both vector-field arguments, so it is a tensor. Torsion-freeness gives its symmetry:Its metric trace defines the paper's Laplace-Beltrami operator, $\Delta ={\mathrm{div}}_g{\mathrm{grad}}_g$; its eigenvalues on a closed manifold are nonpositive. We use $P=-\Delta$ when listing nonnegative eigenvalues or writing $e^{-tP}$.

At $p$, choose an orthonormal basis $e_1,\dots ,e_d$ of the tangent space and unit-speed geodesics ${\gamma }_i(t)={\exp }_p(t{e}_i)$. Differentiating twice along a curve givesThe covariant acceleration vanishes on a geodesic. Taking the trace at $t=0$ proves the geodesic trace formula for the Laplace-Beltrami operator:The last equality is Taylor's formula along the geodesics, and expresses the Laplacian entirely through nearby function values.

For stabilization, let $X_1,X_2$ be the assumed isospectral nonisometric four-dimensional tori, and let $k=n-4>0$. Take the same small flat torus $F_{\epsilon }={\mathbb{R}}^k/\epsilon {\mathbb{Z}}^k$ as an extra factor. On a Riemannian product,If ${\varphi }_j$ and ${\psi }_{\ell }$ are eigenfunctions with eigenvalues ${\lambda }_j$ and ${\mu }_{\ell }$, then ${\varphi }_j{\psi }_{\ell }$ has eigenvalue ${\lambda }_j+{\mu }_{\ell }$. Products of orthonormal eigenbases are complete in the product $L^2$ space, so this lists all eigenvalues with multiplicities, including coincidences between different sums. Consequently the two products are isospectral.

One must also prove that multiplying by a common factor has not accidentally made the two manifolds isometric. ChooseThe injectivity radii are positive because the factors are compact. A closed product geodesic of length less than $\delta$ projects to a closed geodesic of length less than $\delta$ in $X_i$. Such a projected geodesic must be constant: otherwise its initial vector over one period would be a nonzero vector of norm below the injectivity radius mapping back to its starting point under the exponential map. Thus all sufficiently short closed geodesics lie in the $F_{\epsilon }$ directions. Conversely the $k$ coordinate circles, of length $\epsilon$, pass through every point and span those directions.

It follows that the smooth distribution tangent to the small flat-torus factor is characterized intrinsically by the span of tangent vectors to closed geodesics of length below $\delta$. A hypothetical product isometry would preserve this smooth distribution and its orthogonal complement. The latter has leaves $X_i\times \{q\}$; the isometry and its inverse take whole leaves to whole leaves, thereby giving an isometry $X_1\cong X_2$, a contradiction. This proves isospectral stabilization by a small flat torus:This argument works even without assuming that the original four-dimensional torus metrics are flat.

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 $d\ge 4$.

We use $\Delta =\mathrm{div}\nabla$, whose spectrum on a compact manifold is nonpositive; write $P=-\Delta$ for the positive Laplace-Beltrami operator. This convention will also make the heat operator in Question 3 equal to ${\partial }_t-\Delta$.

Let $e_1,\dots ,e_d$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. Since a geodesic has vanishing acceleration for the Levi-Civita connection, differentiation twice along it gives the Riemannian Hessian:Taking its metric trace yields the geodesic trace formula for the Laplace-Beltrami operatorThe positive-sign convention $P$ puts a minus sign in front of this sum.

For the unit sphere, take $p\in S^d$ and tangent unit vectors $e_i$ orthogonal to $p$. Its great circle geodesics are ${\gamma }_i(t)=p\cos t+e_i\sin t$. For a smooth ambient function $F$ near the sphere, the ordinary chain rule givesThe ambient orthonormal basis is $e_1,\dots ,e_d,p$. Thus, writing $F_r(p)={\partial }_{r}F(rp){|}_{r=1}$ and similarly for $F_{rr}$,This proves the ambient restriction formula for the spherical Laplacian and shows why ambient derivatives normal to the sphere must be subtracted. The equivalent formula in spherical coordinates is

Let ${\mathcal{P}}_{\ell }$ be the vector space of homogeneous polynomials of degree $\ell$ in $d+1$ variables, and ${\mathcal{H}}_{\ell }=\ker \tilde{\Delta }\cap {\mathcal{P}}_{\ell }$ its harmonic polynomials. If $H\in {\mathcal{H}}_{\ell }$, homogeneous polynomial scaling gives $H(r\omega )=r^{\ell }H(\omega )$. Substitution into the spherical coordinates formula givesThe restrictions are the degree-$\ell$ spherical harmonics. A homogeneous polynomial vanishing on the sphere vanishes on every nonzero ray and hence identically, so restriction is injective on ${\mathcal{H}}_{\ell }$.

We prove the harmonic decomposition of homogeneous polynomials that both counts these eigenfunctions and establishes completeness. For an ambient harmonic homogeneous polynomial $H_m$ of degree $m$ and $j\ge 1$, the spherical coordinates formula givesFor $d\ge 1$ its coefficient is nonzero. Inductively assume homogeneous polynomials of degree $\ell -2$ have been decomposed into sums of $r^{2j}{H}_m$. Apply this decomposition to $\tilde{\Delta }Q$ for $Q\in {\mathcal{P}}_{\ell }$, and lift each summand by multiplying by $r^2$ and dividing by the displayed nonzero coefficient. The resulting $R\in r^2{\mathcal{P}}_{\ell -2}$ satisfies $\tilde{\Delta }R=\tilde{\Delta }Q$, so $Q-R\in {\mathcal{H}}_{\ell }$. The same coefficient formula shows that no nonzero element of $r^2{\mathcal{P}}_{\ell -2}$ can be harmonic. Starting at degrees zero and one therefore proves the direct decompositionSince $\dim {\mathcal{P}}_{\ell }=\left(\genfrac{}{}{0ex}{}{\ell +d}{d}\right)$, the multiplicity iswhere the second term is zero for $\ell <2$.

On $r=1$, iterating the decomposition writes every polynomial restriction as a finite sum of spherical harmonics. The Stone-Weierstrass theorem makes these polynomial restrictions dense in $C(S^d)$, since coordinate functions separate points and constants are included; they are consequently dense in $L^2(S^d)$. Integration by parts makes $P$ symmetric on smooth functions and gives orthogonality between distinct eigenvalues. Normalizing within each finite-dimensional eigenspace gives a complete orthonormal eigenbasis $Y_{\ell a}$. To check the full operator spectrum, let $c_{\ell a}=⟨f,Y_{\ell a}⟩$. For smooth $f$, integration by parts gives the coefficients of $Pf$ as ${\lambda }_{\ell }{c}_{\ell a}$. Therefore finite harmonic partial sums approximate both $f$ and $Pf$ in $L^2$, hence approximate in the graph norm. Conversely every coefficient sequence with ${\sum }_{\ell ,a}{\lambda }_{\ell }^2|c_{\ell a}{|}^2<\infty$ is the graph norm limit of its finite harmonic sums. Thus the closure of $P$ has exactly this operator domain and is the real diagonal self-adjoint operator with entries ${\lambda }_{\ell }=\ell (\ell +d-1)$. If $z$ is not one of these entries, the inverse has entries $({\lambda }_{\ell }-z{)}^{-1}$ and is bounded because ${\lambda }_{\ell }\to \infty$. This proves that there are no missing spectral values. The spectrum of the Laplacian on a sphere iswith the above multiplicities; for $\Delta$, negate every eigenvalue. For example $S^1$ has eigenvalues ${\ell }^2$, with multiplicities one at zero and two for $\ell \ge 1$, and $S^2$ has multiplicity $2\ell +1$. The zero-dimensional sphere consists of two isolated points and has only the zero eigenvalue, with multiplicity two.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

The Einstein tensor is $G_{\alpha \beta }=R_{\alpha \beta }-\frac{1}{2}{g}_{\alpha \beta }R$. Taking the metric trace of the Einstein field equations in four dimensions givesand substitution back gives the Trace-reversed Einstein field equations in D dimensions specialized to $D=4$:For linearized gravity, the given equation and $G_{\mu \nu }=8\pi T_{\mu \nu }$ implySince $\bar{h}=-h$, its trace is $\square h=16\pi T$. Using ${\bar{h}}_{\mu \nu }=h_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }h$ then yields

The symmetric covariant two-tensor defined using the Levi-Civita connection by ${\mathrm{Hess}}_{g}f(X,Y)=X(Yf)-({\nabla }_{X}Y)f$. Its negative metric trace is the positive Laplace-Beltrami operator applied to $f$. Along a geodesic it gives the second derivative of $f$.