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On a closed manifold with a Riemannian metric the positive Laplace-Beltrami operator is self-adjoint with compact resolvent. It has a complete smooth orthonormal eigenbasis, finite-dimensional eigenspaces, nonnegative eigenvalues tending to infinity, and polynomial eigenvalue-counting bounds. Elliptic regularity bounds derivatives of its eigenfunctions polynomially in their eigenvalues.

For a nonnegative self-adjoint operator with compact resolvent, this function counts eigenvalues up to a given level with their multiplicities. For the positive Laplace-Beltrami operator on a closed $n$-dimensional Riemannian manifold, $N(\Lambda )\le C(1+\Lambda {)}^{n/2}$. To prove this bound, multiply each vector in the low-eigenvalue subspace by a finite set of chart cutoffs and record its local Fourier coefficients of frequency at most $R$. The total number of retained coefficients is $O(R^n)$, while the sum of discarded squared norms is at most $C{R}^{-2}(1+\Lambda )\Vert u{\Vert }_2^2$ by Parseval identity and the energy bound. For $R$ a sufficiently large multiple of $\sqrt{1+\Lambda }$, vanishing of all recorded coefficients forces $u=0$. The recording map is injective and proves the dimension bound. Inverting this estimate gives ${\lambda }_j\ge c(j+1{)}^{2/n}-1$.

Expanding the matrix determinant from part (b) givesThe finite-interval Airy spectral determinant obeys $\Delta (\alpha k)={\alpha }^2\Delta (k)$. At the origin,This zero comes from coalescence of the three rotated equations; it is removable in the underlying boundary system. Indeed, $v^{\prime \prime \prime }=0$ with $v(0)=v(L)=v^{\prime }(L)=0$ has only the zero quadratic solution.

For $k\ne 0$, a zero of $\Delta$ is equivalent to the existence of a nonzero combination $v(x)={\sum }_{j=0}^2{c}_j{e}^{i{\alpha }^{j}kx}$ satisfying the homogeneous endpoint conditions. This can be checked by expanding the determinant of those three endpoint equations. Such a function satisfies $v^{\prime \prime \prime }=\omega (k)v$, so it is an eigenfunction of $A$ with eigenvalue $-\omega (k)$. Integration by parts yieldsso every nonzero zero has $\mathrm{Re}\omega (k)\ge 0$. None lies on the boundary $\mathrm{Re}\omega =0$: by a cube-root rotation it suffices to take $K=kL$ real. With $u=\sqrt{3}K/2$, the determinant equation becomesThe right-hand side has squared modulus $1+4{\sinh }^{2}u>1$ for $K\ne 0$, whereas the left-hand side has modulus one. Therefore there are no nonremovable spectral poles in $\overline{D}$, and the contour representation has no discrete residue contribution from that domain.

Taken literally as a claim about the operator's discrete spectrum, however, the printed assertion is false. To see this constructively, put $k=-is/L$, $s>0$. ThenThus zeros occur whenBetween successive positive cosine maxima and adjacent negative minima the difference changes sign. There are consequently infinitely many positive roots, producing genuine negative eigenvalues $-s^3/L^3$ of $A$. The third-order resolvent maps into $H^3(0,L)$, whose inclusion into $L^2(0,L)$ is compact, so these are indeed discrete spectrum of an operator with compact resolvent. The valid conclusion is absence of poles in the reconstruction sectors, not absence of operator eigenvalues.

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.

For each of the four standard pairs, integration by parts givesConsequently the eigenvalues are nonnegative. For a positive eigenvalue ${\mu }_n=A{k}_n^4$, the eigenvalue equation is $W_n^{\prime \prime \prime \prime }=k_n^4{W}_n$. Its four characteristic roots are $\pm k_n,\pm i{k}_n$, soHere $C_j$ are coefficients, avoiding a collision with the filament bending modulus $A$. The regular finite-interval self-adjoint operator has compact resolvent; applying the spectral theorem for compact self-adjoint operators to a shifted inverse supplies a complete orthonormal basis of eigenfunctions.

For clamped boundary conditions at zero, a clamped--clamped bending mode can be writtenAt $L$, writing $\beta =kL$, the two remaining boundary conditions areThe determinant is $2(1-\cos \beta \cosh \beta )$. Hence the positive wavenumbers obeyEquivalently, intersect $\cos \beta$ with $\mathrm{sech}\beta$. Numerical root bracketing givesThe entire sequence has the useful large-$n$ descriptionIndeed, put $\beta =(n+\frac{1}{2})\pi +\delta$ in $\cos \beta =\mathrm{sech}\beta$ and use $\cos \beta =(-1{)}^{n+1}\delta +O({\delta }^3)$ and $\mathrm{sech}\beta =2e^{-\beta }+O(e^{-3\beta })$.

The apparent root $\beta =0$ in the determinant equation is spurious for the clamped problem: at zero eigenvalue, $W$ is a cubic polynomial, and its four clamped conditions force $W=0$. For other endpoint choices, zero-energy filament modes must be treated separately from the trigonometric formula. The free-free kernel consists of affine functions, the torqued-torqued kernel consists of constants, and the hinged-hinged kernel is trivial. Including those kernels is necessary for a complete eigenfunction expansion.

For spatially varying tension in filament bending, keep the derivative of the filament tension as well as the curvature term. The first variation isTherefore the Euler-Lagrange equation and fluctuation operator areFor real, sufficiently smooth $\sigma$, the boundary form is the bending boundary form minus $[\sigma (\overline{u}{v}^{\prime }-\overline{u^{\prime }}v){]}_0^L$. Because $\sigma$ vanishes at both ends, the four self-adjoint endpoint conditions for filament bending still apply. The natural endpoint force also reduces there to the bending shear term. Thus $K_{\sigma }$ is a self-adjoint fourth-order scalar differential operator on the same chosen domain, with compact resolvent.

Choose a real orthonormal basis ${\psi }_n$ of eigenfunctions, $K_{\sigma }{\psi }_n={\lambda }_n{\psi }_n$, and write $h={\sum }_n{b}_n{\psi }_n$. Using the endpoint conditions in integration by parts givesThe equipartition theorem now gives, on the strictly positive subspace,This is a formal modal construction; no explicit eigenfunctions are needed. Nonnegative filament tension makes the energy nonnegative. Any surviving zero-energy filament modes must again be fixed. If signed $\sigma$ permits compression, self-adjointness still holds but does not guarantee a canonical ensemble: sufficiently strong compression can create negative eigenvalues and Euler buckling of an elastic filament. For instance, on $A=L=1$, take $\sigma =-Cx(1-x)$ and the clamped trial function $h=x^2(1-x{)}^2$. Thenso the energy is negative when $C>252$, despite $\sigma (0)=\sigma (1)=0$. The equipartition theorem requires a stable positive quadratic energy, not merely a real modal spectrum.

Let $R_z=(A-zI{)}^{-1}$ be compact and let $w$ be another resolvent point. The resolvent identity givesorThe bracket is bounded and the product of a bounded operator with a compact operator is compact. Thus compactness at one resolvent point implies compactness at every resolvent point.

Fix such a $z$. Spectral mapping for the bounded compact operator $R_z$ givesEvery nonzero spectral point of a compact operator is an isolated eigenvalue of finite multiplicity, and zero is its only possible accumulation point. Hence a compact resolvent operator has only isolated eigenvalues of finite multiplicity, with no finite accumulation point. The spectrum is allowed to be empty.