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Two isospectral manifolds remain isospectral after taking a Riemannian product with the same closed factor: eigenvalues add, with multiplicities given by the product Laplacian eigenspace decomposition. For nonisometric closed connected factors $M_1,M_2$, choose $\epsilon <\delta <{\min }_i\mathrm{inj}(M_i)$. Every closed product geodesic of length less than $\delta$ has constant projection to $M_i$, since a nonconstant closed geodesic shorter than the injectivity radius is impossible. The short coordinate geodesics of the small flat torus span exactly its tangent factor at every point. Any product Riemannian isometry must preserve that intrinsically defined smooth distribution and its orthogonal complement, hence restrict to an isometry of the original factors. Thus these products are still nonisometric. Taking $M_i$ to be four-dimensional tori proves the higher-dimensional stabilization statement without assuming a cancellation theorem for arbitrary products.

Use the intended trace bound $\mathrm{Ric}\ge (n-1)kg$ and write $D=\pi /\sqrt{k}$. We state precisely the allowed Bishop volume comparison with a positive-curvature model. Setwhere ${\omega }_{n-1}$ is the area of the unit $(n-1)$-sphere. For $V_p(r)=\mathrm{Vol}B(p,r)$, the ratio $V_p(r)/v_k(r)$ is nonincreasing on $0<r<D$, tends to one at zero, and is at most one. Before the cut time, the radial volume density $J(r,\theta )$ in geodesic polar coordinates is at most $s_k(r{)}^{n-1}$.

By the Bonnet-Myers theorem, $\mathrm{diam}M\le D$. Therefore $V_p(D)=\mathrm{Vol}M$. If distance-$D$ endpoints exist, they do not affect this volume equality: the distance sphere is contained in the smooth image ${\exp }_p(D{S}_p^{n-1})$, which has zero $n$-dimensional Riemannian volume. Taking limits in the comparison ratio at $D$ is legitimate. The assumed equality of total volume gives $V_p(D)/v_k(D)=1$, so monotonicity forcesfor every $p$.

We now extract the equality geometry, rather than merely naming a rigidity theorem. On a small normal neighborhood of $p$, the continuous nonnegative difference $s_k(r{)}^{n-1}-J(r,\theta )$ has integral zero. It is therefore identically zero. Let $A(X)={\nabla }_X{\partial }_r$ on a distance sphere and $h=\mathrm{tr}A={\partial }_r\log J$. Equality of the density gives $h=(n-1)s_k^{\prime }/s_k$. The traced radial Riccati equation for distance spheres readsSince $s_k^{\prime \prime }=-k{s}_k$, substituting $h$ makes the sum of the following two nonnegative quantities zero:Equality in the Cauchy-Schwarz inequality for the eigenvalues of the self-adjoint operator $A$ forces $A=(s_k^{\prime }/s_k)I$. The angular metric $g_r$ in geodesic polar coordinates consequently satisfies ${\partial }_r{g}_r=2(s_k^{\prime }/s_k)g_r$. Smoothness at the center fixes its integration constant, since $r^{-2}{g}_r\to g_{S^{n-1}}$. Henceon the normal ball: the Riemannian metric is locally the round curvature-$k$ metric. As $p$ was arbitrary, $M$ has constant sectional curvature $k$.

The classification of complete positive constant-curvature manifolds now gives $M=S_k^n/\Gamma$ with $\Gamma$ a finite freely acting group of isometries. Covering degree gives $\mathrm{Vol}M=\mathrm{Vol}{S}_k^n/|\Gamma |$. Volume equality forces $|\Gamma |=1$, provingThis establishes spherical rigidity of maximal total volume under the correct normalization.

Under the literal trace bound, the assertion is false in dimension four. Take the Riemannian product $M=S^2(r)\times S^2(r)$ with $r^2=1/(\sqrt{6}k)$. Its Ricci curvature is $r^{-2}g=\sqrt{6}kg\ge kg$, andBut a mixed tangent two-plane has sectional curvature zero, whereas a factor plane has curvature $\sqrt{6}k$. Thus this manifold is not the round sphere despite having the literal hypotheses.

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.

Suppose, for a contradiction, that $X=M\times \mathbb{R}$ has a complete metric with zero Ricci curvature. Since $M$ is closed and connected, the complement of $M\times [-A,A]$ has two unbounded components in any complete metric on $X$: their closures are noncompact, while closed bounded sets are compact by the Hopf-Rinow theorem. Thus $X$ is disconnected at infinity, and part (a) supplies a line. The Cheeger-Gromoll splitting theorem gives an isometryThe Riemannian product curvature formula makes $N$ complete and Ricci-flat. A three-dimensional Ricci-flat Riemannian manifold is flat: for any orthonormal triple, the three trace equations are ${\mathrm{Ric}}_{11}=K_{12}+K_{13}$ and their cyclic companions. Solving gives $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2=0$. Every plane can occur in such a triple, so every sectional curvature vanishes, and hence so does the Riemann curvature tensor.

The complete simply connected universal cover of $N$ is therefore Euclidean ${\mathbb{R}}^3$, and that of $X$ is Euclidean ${\mathbb{R}}^4$. But the topological universal cover of $X$ is $\tilde{M}\times \mathbb{R}$. Its deformation retraction onto $\tilde{M}\times \{0\}$ shows that contractibility of this cover would imply contractibility of $\tilde{M}$, contrary to the hypothesis. Therefore no complete Ricci-flat metric exists on $M\times \mathbb{R}$. Notice that no product assumption was made about the hypothetical original metric.

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.

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.

Give $S^2\times \mathbb{R}$ the Riemannian product of the unit round metric and the Euclidean metric. It is complete and has infinite diameter. The round sphere has scalar curvature $2$, while the line has scalar curvature $0$; scalar curvature is additive under Riemannian products, soThus this manifold has a strictly positive uniform lower bound on scalar curvature but violates the conclusion of the Bonnet-Myers theorem. Its Ricci curvature vanishes in the $\mathbb{R}$ direction, showing precisely why a scalar-curvature bound is insufficient.

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

The Hadamard-Cartan theorem states that if a complete simply connected Riemannian manifold has nonpositive sectional curvature, then for every point $p$ its exponential mapis a diffeomorphism. In particular, the manifold is diffeomorphic to Euclidean space and is contractible.