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Expansion along the first column gives the characteristic polynomialThus the eigenvalue $2$ has algebraic multiplicity two. Solving $(M-2I)(x,y,z{)}^T=0$ gives $y=0$ and then $z=0$, leaving $x$ arbitrary. Its eigenspace is $\mathrm{span}\{(1,0,0{)}^T\}$ and has geometric multiplicity one. A diagonalisable matrix needs an eigenvector basis, so $M$ is not diagonalisable.

The symmetry between the last two coordinates gives the eigenvector $(0,1,-1{)}^T$ with eigenvalue $2$. On vectors $(s,t,t{)}^T$, the remaining eigenvalue equation iswhose characteristic polynomial is ${\lambda }^2-5\lambda +4=(\lambda -1)(\lambda -4)$. Corresponding eigenvectors are $(1,-1,-1{)}^T$ and $(2,1,1{)}^T$. Thus all eigenvalues and eigenspaces areNonzero scalar multiples give all eigenvectors in each eigenspace. The three displayed vectors are mutually orthogonal and have squared norms $3,2,6$, respectively.

An eigenvalue can be complex even though $P$ is real, so use the complex Hermitian inner product. For a nonzero complex eigenvector $\mathbf{v}$, the real orthogonal matrix also satisfies $P^{\ast }P=P^{T}P=I$. Henceand $\boxed{|\lambda |=1}$. Taking complex conjugates in $P\mathbf{v}=\lambda \mathbf{v}$ gives $P\overline{\mathbf{v}}=\overline{\lambda }\overline{\mathbf{v}}$ because $P$ is real. The conjugate vector is nonzero, so $\boxed{{\lambda }^{\ast }=\overline{\lambda }\text{ is also an eigenvalue}}$.

For the unheaded geometrical conclusion, the third eigenvalue of $Q$ must be real: nonreal eigenvalues of a real matrix occur in conjugate pairs. Its modulus is one, so it is $1$ or $-1$. Exact algebraic multiplicity two for $1$ excludes a third $1$, leaving $-1$. By the supplied diagonalisability fact, the $1$-eigenspace is a two-dimensional plane and the $-1$-eigenspace is a line. They are orthogonal: preservation of the inner product between a fixed vector and a negated vector gives $\mathbf{u}\cdot \mathbf{v}=-\mathbf{u}\cdot \mathbf{v}$. Therefore $Q$ is the orthogonal reflection in its fixed plane. It fixes the two tangential directions and reverses the normal direction; with unit normal $\mathbf{n}$, $Q=I-2\mathbf{n}{\mathbf{n}}^T$.

Consider two finite assemblies of the same number $n$ of congruent bounded Euclidean tiles. Label their matching faces so the same prescribed face identifications are used in both assemblies. For each face type $c$, encode the first assembly by a symmetric involution $A_c$: a paired face exchanges its two tile indices, an exterior Dirichlet boundary condition contributes diagonal $-1$, and an exterior Neumann boundary condition contributes diagonal $+1$. Let $B_c$ encode the second assembly. Face coordinates are pulled back to the same reference face; if identifications have additional face isometries, these pullbacks must be included in the intertwining operators.

The transplantation theorem says that a constant matrix $C$ with $C{A}_c=B_{c}C$ for every face type sends the vector of tile restrictions of a Laplacian eigenfunction on the first assembly to one on the second, by $g=Cf$. If $C$ is invertible, it is a bijection of each eigenspace and the assemblies are isospectral with multiplicities.

To prove it, each component solves the same interior equation $\Delta f_j=\lambda f_j$, and a constant linear combination solves that equation too. On a face, let $u$ be the vector of boundary values and $v$ the vector of outward normal derivatives. All matching and exterior conditions are exactlyFor an internal face these equations say that values agree and the two outward derivatives sum to zero. For diagonal $-1$ they impose zero value, and for diagonal $+1$ zero normal derivative. The intertwining identity gives $B_c(Cu)=Cu$ and $B_c(Cv)=-Cv$, so the transplanted functions match and satisfy the correct boundary conditions. Applying $C^{-1}$ proves the eigenspace bijection. At corners the same reasoning is interpreted in the finite-energy weak domain: no value jump and cancellation of normal flux prevent an extra distributional source. It gives the usual self-adjoint Dirichlet/mixed Laplacians on polygonal assemblies.

Here is an explicit transplantation by reflection parity. Take a rectangle of width $2a$ and height $b$, with the Dirichlet boundary condition on every exterior side. Cut it at its vertical midline into two congruent half-rectangles, pulling the right-hand half back by reflection. The midline gluing matrix isReplace the connected assembly by the disjoint union of two width-$a$, height-$b$ rectangles: the first has Dirichlet conditions on all sides; the second has Dirichlet on three sides and Neumann on the side representing the cut. Its corresponding matrix is $D=\mathrm{diag}(-1,1)$. The invertible, indeed orthogonal, matrixintertwines the midline conditions. Every other face has matrix $-I$ on both assemblies, which automatically intertwines. The transplantation theorem proves equality of the complete spectra, although one surface is connected and the other has two components with the required uniform and mixed conditions.

Equivalently, odd reflection eigenfunctions vanish on the cut, while even reflection eigenfunctions have zero normal derivative there. A direct separation-of-variables check gives the connected rectangle's eigenvaluesEven $m=2j$ gives the fully Dirichlet half-rectangle spectrum, and odd $m=2j+1$, $j\ge 0$, gives the mixed half-rectangle spectrum. The split preserves every multiplicity, including coincidences between different pairs of indices.

Take $N$ to be a closed compact Riemannian manifold and use the positive Laplace-Beltrami operator. Its Riemannian heat kernel $K_N(t,x,y)$ is the integral kernel of $e^{-t{\Delta }_N}$:It solves $({\partial }_t+{\Delta }_x)K_N=0$ for $t>0$ and tends to the identity kernel as $t\downarrow 0$. For an orthonormal eigenbasis, the spectral expansion of the Riemannian heat kernel isThe bar is required for a complex basis. With boundary, a specified invariant boundary condition must also be included in this definition.

For the quotient formula use the covering-space interpretation: $U$ acts freely and properly discontinuously by Riemannian isometries. On compact $N$ such a discrete group is finite. Freeness alone, without this covering hypothesis, does not justify the image sum: an infinite dense subgroup of circle rotations, for example, acts freely but has no manifold quotient. Positive-dimensional group actions require a different quotient analysis.

For lifts $x,y$ of $\bar{x},\bar{y}\in M$, the heat kernel on a finite isometric quotient isInvariance of $K_N$ under simultaneous Riemannian isometries and reindexing the sum show that this is independent of both lifts. A local isometry commutes with the Laplacian, so the sum satisfies the quotient heat equation. For its initial condition, integrate over a fundamental domain $F\subset N$ against a lifted function. The terms combine into the integral over all of $N$, whose initial limit is $f(x)$. Uniqueness of the heat evolution proves the formula. There is no factor $1/|U|$ in this kernel formula.

On the diagonal the sum is $U$-invariant. Its integral over $F$ is therefore $1/|U|$ times its integral over $N$, giving the heat traceThus the normalization factor appears in the trace, not the kernel. Put $F_t(g)={\int }_N{K}_N(t,x,gx)dV(x)$. For any finite isometry group $T$, changing variables $x=hy$ proves $F_t(hg{h}^{-1})=F_t(g)$: it is a class function on $T$.

For Gassmann equivalent subgroups $U_1,U_2\le T$, their intersections with each conjugacy class have equal size, and their orders are equal. If both act freely, the heat traces of their quotients satisfyso they agree for every $t>0$. Since $Z_M(t)={\sum }_{\lambda }m(\lambda )e^{-t\lambda }$, equality determines the spectrum with multiplicities: take $t\to \infty$ to recover the smallest eigenvalue and its multiplicity, subtract that term, and repeat. This proves Sunada theorem. Equivalently, projection onto $U_i$-invariant functions averages the group action, and the quotient eigenvalue multiplicity is $|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\lambda }(u)$, with ${\chi }_{\lambda }$ the eigenspace character of a representation. Almost conjugacy equalizes these averages.

Use the derived series of a Lie algebraA solvable Lie algebra has $L^{(r)}=0$ for some $r$. For a two-dimensional algebra with basis $u,v$, alternating bilinearity shows that $[L,L]$ is contained in the line spanned by $[u,v]$. A one-dimensional Lie algebra is abelian, so $[[L,L],[L,L]]=0$. Every two-dimensional complex Lie algebra is solvable, with derived length at most two. This dimension argument actually works over any field with the alternating definition of a Lie bracket.

There is a genuine qualification in the last request: the assertion about all Irreducible Lie algebra representations is true for finite-dimensional representations. Without that restriction, the infinite-dimensional simple module for the two-dimensional affine Lie algebra is a counterexample. Take $L=\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, and let it act on $V=\mathbb{C}[t]$ byThen $[x,y]f=tf(t-1)-(t-1)f(t-1)=yf$, so this is a Lie algebra representation. Any nonzero submodule is stable under multiplication by $t$, hence is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ implies $p(t)\mid p(t-1)$. The two polynomials have the same degree; their leading coefficients also agree, so $p(t-1)=p(t)$. In characteristic zero this forces $p$ to be constant. Thus $V$ is infinite-dimensional and irreducible, while $L$ is solvable. The literal unrestricted assertion is false.

Here is a proof of the intended finite-dimensional assertion, including the essential Lie theorem argument. We prove that a nonzero finite-dimensional representation $\rho :L\to \mathrm{End}(V)$ of a complex solvable algebra has a common eigenvector, by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise $[L,L]\ne L$; choose a hyperplane $K$ containing $[L,L]$. It is a solvable ideal, and $L=K\oplus \mathbb{C}x$. Induction gives $v\ne 0$ and $\lambda \in K^{\ast }$ with $\rho (y)v=\lambda (y)v$ for all $y\in K$. Write $X=\rho (x)$ and suppress $\rho$ on elements of $K$.

Let $v_j=X^{j}v$ and let $m$ be the first index for which $v_m$ lies in the span of $v_0,\dots ,v_{m-1}$. The space $W=\mathrm{span}(v_0,\dots ,v_{m-1})$ is $X$-invariant. Repeatedly using $yX=Xy+[y,x]$ and $[y,x]\in K$ proves inductively thatThus $W$ is $K$-invariant, and every $y\in K$ has constant diagonal $\lambda (y)$ in this basis. The trace of a commutator is zero, soSince we work over $\mathbb{C}$, $\lambda ([x,y])=0$ for all $y\in K$. Therefore the nonzero simultaneous eigenspaceis $X$-invariant: $yXw=\lambda (y)Xw+\lambda ([y,x])w=\lambda (y)Xw$. A complex linear operator on a nonzero finite-dimensional space has an eigenvector. An eigenvector of $X{|}_{V_{\lambda }}$ is consequently a common eigenvector for $L$.

Its span is a one-dimensional submodule. If $V$ is irreducible, that span must be all of $V$. Hence every finite-dimensional irreducible representation of a finite-dimensional complex solvable Lie algebra has dimension one. Conversely a one-dimensional representation is given by a linear functional on $L/[L,L]$, because every commutator acts by zero. Passing repeatedly to quotients also gives the simultaneous triangularization of a Lie algebra representation. Neither the eigenvector step nor the trace argument extends to the infinite-dimensional counterexample above.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

We first prove the needed Fredholm alternative for a compact operator without assuming self-adjointness. If $C$ is compact, the finite-rank approximation theorem for compact operators on a Hilbert space follows directly by covering the compact closure of $C$'s unit-ball image by finitely many balls of radius $\epsilon$, spanning their centers by a finite-dimensional space $E$, and using the orthogonal projection $P_E$. Then $\Vert C-P_{E}C\Vert \le \epsilon$.

Choose a finite-rank $F$ with $\Vert C-F\Vert <1$, put $E_0=C-F$, and factorThe first factor is invertible by the Neumann series, and $G$ has finite rank. With $V=\mathrm{ran}G$ and $H=V\oplus V^{\perp }$, the second factor has the triangular formIt is injective precisely when the finite-dimensional block $I_V-G{|}_V$ is injective, and onto precisely when that block is onto. These properties are equivalent in finite dimension. This proves that $I-C$ is injective if and only if it is onto; in that case its inverse is bounded.

For $\lambda \ne 0$, apply this result to $C=K/\lambda$. If $\lambda$ lies in the spectrum of a bounded operator $K$, then $K-\lambda I$ is not invertible, hence is not injective. Therefore every nonzero spectral point of $K$ is an eigenvalue. On its eigenspace $E_{\lambda }$, the operator $K$ equals $\lambda I$. Compactness of $K$ would make the closed unit ball of $E_{\lambda }$ compact after dividing by $\lambda$; part (a) forces $\dim E_{\lambda }<\infty$.

It remains to rule out infinitely many distinct eigenvalues away from zero. Suppose $|{\lambda }_n|\ge \epsilon >0$ are distinct, with respective eigenvectors $v_n$. Distinct eigenvalues give linearly independent eigenvectors. Let $V_n=\mathrm{span}(v_1,\dots ,v_n)$ and choose a unit vector $h_n\in V_n\cap V_{n-1}^{\perp }$. Since $V_n$ is invariant and $(K-{\lambda }_{n}I)V_n\subset V_{n-1}$,For $m<n$, $K{h}_m\in V_{n-1}$, and orthogonality gives $\Vert K{h}_n-K{h}_m\Vert \ge |{\lambda }_n|\ge \epsilon$. This contradicts compactness. Thus for each $\epsilon >0$ there are only finitely many distinct spectral points with modulus at least $\epsilon$.

Taking $\epsilon =1/j$ shows that the nonzero spectrum is countable; if it is infinite, its distinct elements can be enumerated as a sequence tending to zero. Since the spectrum is closed, zero then also belongs to it. In an infinite-dimensional space zero belongs to the spectrum even if there are only finitely many nonzero spectral points: an invertible compact $K$ would make $I=K^{-1}K$ compact, contradicting part (a). The only possible accumulation point is zero, and every nonzero eigenspace is finite-dimensional. Zero need not be an eigenvalue. The argument uses invariant spans, not an orthogonal eigenbasis, which is unavailable for a general non-self-adjoint compact operator.

A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.

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.