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Use the normalizationThis is ridge regression; the penalty is the squared Euclidean norm. The centering removes the need for an intercept. Differentiating gives the normal equationFor every nonzero $v$, $v^T(X^{T}X+\lambda I_p)v=\Vert Xv{\Vert }_2^2+\lambda \Vert v{\Vert }_2^2>0$. The coefficient matrix is therefore a positive-definite matrix, even if the design matrix has deficient matrix rank. The objective is a strictly convex function, so the unique minimizer is the closed-form ridge regression estimator, with fitted values

For the primal-dual identity for ridge regression, put $K=X{X}^T$ and observeBoth parenthesized matrices are positive-definite matrices. Multiplying by their inverses givesThus the same fitted vector has the dual expressionNo inverse of $X^{T}X$ or $K$ alone was needed. This is also the kernel-ridge hat matrix for the linear kernel.

Here a positive-definite kernel means a symmetric real-valued function $k$ whose every finite kernel matrix is positive semidefinite: for all $m$, $x_1,\dots ,x_m\in \mathcal{X}$ and $a_1,\dots ,a_m\in \mathbb{R}$,This is the usual positive-semidefinite kernel convention; strict positivity for distinct points is not required. In particular $k(x,x)\ge 0$. Write $a=k(x,x)$, $b=k(x^{\prime },x^{\prime })$ and $r=k(x,x^{\prime })$. Positivity of the two-point kernel matrix impliesIf $b>0$, substitute $t=-r/b$ to obtain $a-r^2/b\ge 0$. If $b=0$, the affine expression $a+2tr$ can be nonnegative for every $t$ only if $r=0$. Both cases give the Cauchy-Schwarz inequality for positive-semidefinite kernels:The zero-diagonal case is essential when the positive-definite kernel is not strictly positive.

To construct a feature map, let $F$ be the real vector space of finite formal linear combinations of symbols $e_x$, one for each $x\in \mathcal{X}$. Define a symmetric bilinear form byThe positive-definite kernel property makes this a positive semidefinite bilinear form. It need not yet be an inner product: distinct formal vectors can have zero squared length. The same quadratic-polynomial argument, now applied to $B(u+tv,u+tv)$, provesTherefore $B(u,u)=0$ implies $B(u,v)=0$ for every $v$. The setis the radical of a bilinear form, hence a linear subspace. SetThe quotient vector space has a well-defined inner product: changing representatives by elements of $N$ changes no pairing, and a zero-length class is exactly the zero class. It is therefore an inner product space.

Define $\varphi (x)=e_x+N$. By construction,Every positive-definite kernel thus has a feature representation, including degenerate kernels and the zero kernel. This is its canonical feature space of a positive-semidefinite kernel. The construction already gives the requested inner product space.

A Non-Abelian gauge theory promotes an internal group representation symmetry to a spacetime-dependent transformation and introduces a gauge field so that differentiation transforms covariantly. The symmetry is a gauge redundancy: physically meaningful quantities must be invariant under local changes of field representative. We work in four-dimensional Minkowski spacetime with signature $(+---)$, and use skew-Hermitian matrix generators throughout.

For the compact Simple Lie group $G$, its real Lie algebra has a negative-definite Killing form. Hence $B=-\kappa$ is a positive invariant bilinear form on a Lie algebra. Choose a $B$-orthonormal basis $T_a$, $a=1,\dots ,\dim G$, with $[T_a,T_b]=f_{ab}{}^c{T}_c$. Invariance makes $f_{abc}=B([T_a,T_b],T_c)$ fully antisymmetric. This construction uses the Adjoint representation and works independently of any chosen scalar representation.

Let ${\mathcal{A}}_{\mu }={\mathcal{A}}_{\mu }^a{T}_a$ be a Lie-algebra-valued gauge potential, with the coupling absorbed into this connection. If $R$ is the given finite-dimensional irreducible representation of $G$, its Lie algebra representation $\rho =dR$ definesFor a finite local transformation $U(x)\in G$, require ${\Phi }^{\prime }=R(U)\Phi$ and $D_{\mu }^{\prime }{\Phi }^{\prime }=R(U)D_{\mu }\Phi$. Differentiating the first equation forces the gauge-field transformation lawThus the gauge potential transforms inhomogeneously; an ordinary derivative alone would leave an uncancelled derivative of $U$.

The commutator of gauge covariant derivatives determines the gauge field strength:The Lie algebra representation preserves the bracket, so this formula holds for any $R$, including a trivial scalar representation. The gauge field strength itself is defined in the gauge algebra. The transformation of the connection impliesIt is therefore covariant, rather than invariant as a Lie-algebra-valued quantity.

For infinitesimal $U=1+ϵ+O({ϵ}^2)$, the same equations give the infinitesimal gauge transformation with an anti-Hermitian connection:Substitution into $\delta (D_{\mu }\Phi )$ cancels the terms containing ${\partial }_{\mu }ϵ$ and gives $\delta (D_{\mu }\Phi )=\rho (ϵ)D_{\mu }\Phi$, the gauge covariance of a scalar covariant derivative.

The Yang-Mills theory has Lagrangian densityThis is the Killing-form Yang-Mills Lagrangian. Its sign uses a positive internal form $B$ and the stated spacetime signature. In a $B$-orthonormal basis it is $-{\mathcal{F}}_{\mu \nu }^a{\mathcal{F}}^{a\mu \nu }/(4g_{\mathrm{YM}}^2)$. Gauge invariance follows because $B$ is invariant under the adjoint group action. Infinitesimally, the variation is proportional toAfter rescaling ${\mathcal{A}}_{\mu }=g_{\mathrm{YM}}{A}_{\mu }$, the gauge coupling appears in $D_{\mu }={\partial }_{\mu }+g_{\mathrm{YM}}\rho (A_{\mu })$ and in the non-Abelian part of the canonically normalized field strength. Squaring that field strength gives cubic and quartic gauge-field self-interactions. An ordinary local quadratic mass term in the gauge potential is not invariant because its transformation contains ${\partial }_{\mu }ϵ$.

For the scalar field kinetic term, compactness supplies a positive invariant Hermitian form $h$ on its representation space. One can obtain it by averaging any positive form with normalized Haar measure, the unitarization of a compact-group representation. Then $R(U)$ is unitary for $h$, and $\rho (ϵ)$ is anti-Hermitian. The Yang-Mills theory coupled to an arbitrary scalar representation hasHere $V$ is any real gauge-invariant scalar potential, so $V(R(U)\Phi )=V(\Phi )$. For example,is available for every unitary representation, with $\lambda \ge 0$ giving a bounded quartic contribution. Other invariant interactions may exist for particular representations; the potential is not required to depend only on $h(\Phi ,\Phi )$.

The scalar kinetic term is gauge-invariant because both covariant derivatives transform by the same unitary $R(U)$. Infinitesimally its variation vanishes byThe potential is invariant by its defining condition. For a real irreducible representation, use the averaged positive symmetric form instead and write the kinetic term as $\frac{1}{2}h(D_{\mu }\Phi ,D^{\mu }\Phi )$; all transformation and invariance arguments remain valid. Expansion of the covariant kinetic term produces both the linear gauge-scalar interaction and the quadratic term needed by local gauge invariance.

The invariant inner products, gauge covariant derivatives and gauge field strength together construct the full theory for every compact Simple Lie group and every supplied finite-dimensional irreducible representation of the scalar fields. Both finite and infinitesimal transformation laws have been included, and every sign follows the same anti-Hermitian convention.

Use the orientation selected by the volume form, and let ${\mathrm{vol}}_g$ be its normalized Riemannian volume form. The Euclidean metric induces an inner product on the exterior algebra of covectors. The Hodge star operator is the unique map from $p$-forms to $(4-p)$-forms satisfyingFor an oriented orthonormal coframe $e^1,\dots ,e^4$, it sends a basis wedge to the complementary wedge with the sign of the permutation that restores $e^1\wedge \cdots \wedge e^4$. Applying the Hodge star operator twice exchanges blocks of $p$ and $4-p$ covectors, soHere the supplied volume is interpreted in the usual metric-normalized sense. If instead one defines the operator with an arbitrary unnormalized $\mathrm{vol}=f{\mathrm{vol}}_g$, that operator is $f{\ast }_g$ and its square on two-forms is $f^2$. The usual self-duality statements use the metric-normalized Hodge star operator, with the supplied volume specifying orientation.

The projections $P_{\pm }=(1\pm \ast )/2$ give the Hodge splitting of Euclidean two-forms:Both spaces have dimension three. With $e^4=dt$, bases areThe Hodge star operator is an orthogonal involution on two-forms and therefore is self-adjoint. For a self-dual two-form $H$ and an anti-self-dual two-form $G$,ConsequentlyThis is the wedge orthogonality of opposite-duality two-forms.

For the Yang-Mills action, take an anti-Hermitian special unitary group connection and the fundamental matrix trace, so the positive invariant pairing on its Lie algebra is $⟨X,Y⟩=-\mathrm{tr}(XY)$. Write its gauge curvature as $F=F_{+}+F_{-}$ using the Hodge splitting of Euclidean two-forms, and defineCross terms vanish by the wedge orthogonality of opposite-duality two-forms. ThusWith this anti-Hermitian convention the Second Chern number isFor example, this normalization follows by expanding $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. Interpreting the integral as an integer Second Chern number on ${\mathbb{R}}^4$ assumes the usual decay and gauge behavior that allow extension over the point at infinity. The following norm inequality itself does not require integrality:The Yang-Mills instanton Bogomolny bound is saturated precisely when one component vanishes: $F=\ast F$ or $F=-\ast F$. In the stated convention the self-dual case has $k\le 0$ and the anti-self-dual case has $k\ge 0$. Reversing orientation, or defining topological charge with the opposite sign, reverses this assignment while leaving the absolute-value bound unchanged.

For the self-dual Yang-Mills equations in temporal gauge, choose ${\mathrm{vol}}_g=d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge dt$ and ${ϵ}_{123}=1$. The relevant Hodge star operator identities areWriting $F=F_{ti}dt\wedge d{x}^i+\frac{1}{2}{F}_{jk}d{x}^j\wedge d{x}^k$, the self-dual Yang-Mills equations $F=\ast F$ giveIn temporal gauge, $A_t=0$, so the gauge curvature component is $F_{ti}={\partial }_t{A}_i-{\partial }_i{A}_t+[A_t,A_i]={\partial }_t{A}_i$. HenceThe minus sign follows from placing $dt$ last in the orientation and first in the mixed curvature component.

Write $\mathbf{r}=R\mathbf{\theta }$, $|\mathbf{\theta }|=1$. Uniformly for ${\mathbf{r}}^{\prime }$ in the source ball,For fixed $k$ and $r_0$, the outgoing Green function for the three-dimensional Helmholtz equation therefore has the far-field approximation for an outgoing sourceSubstitution into the source integral yieldsThus the source-to-far-field operator at fixed frequency isIt samples the Fourier transform of the source on the sphere of radius $k$. Its square-integrable kernel makes it a Hilbert-Schmidt operator, hence a compact operator. In factUse complex inner products linear in the first argument: $(f,g)=\int f\overline{g}$. Interchanging the volume and surface integrals givesHence the adjoint source-to-far-field operator isThis is a normalized Herglotz wave function; the change in exponential sign comes from the complex conjugation in the adjoint operator.

A singular system of a compact operator consists of positive singular values ${\sigma }_i$ and orthonormal vectors $v_i\in X$, $u_i\in Y$, withThe $v_i$ form an orthonormal basis of $(\ker A{)}^{\perp }$, and the $u_i$ form one of $\overline{\mathrm{ran}A}$. There are finitely many positive singular values for finite-rank $A$; otherwise ${\sigma }_i\to 0$. Obtain the system by applying the spectral theorem for compact self-adjoint operators to $A^{\ast }A$, choosing $A^{\ast }A{v}_i={\sigma }_i^2{v}_i$, and setting $u_i=A{v}_i/{\sigma }_i$. The inner product identities verify orthonormality and the displayed relations.

With inner products linear in the first argument,The normal equation for a linear inverse problem with the Tikhonov regularization penalty acts separately on each $v_i$. Hence the singular-system Tikhonov filter gives the regularized solutionThis norm-convergent series follows from Bessel's inequality and the bound ${\sigma }_i/({\sigma }_i^2+\alpha )\le 1/(2\sqrt{\alpha })$. Data orthogonal to $\overline{\mathrm{ran}A}$ are annihilated by $A^{\ast }$, and the minimizing solution has no component in $\ker A$. The same formula with $y^{\delta }$ reconstructs noisy data. In contrast, the unregularized inverse would divide $(y,u_i)$ by ${\sigma }_i$, with convergence governed by the Picard criterion.

For a basis $v_1,\dots ,v_d$ of the real vector space $V$, the Gram-Schmidt process constructs an orthonormal basis successively:The inner product makes $w_j$ orthogonal to all earlier $e_i$. Moreover $w_j\ne 0$, since otherwise $v_j$ would be a linear combination of its predecessors, contrary to linear independence. Induction gives $\mathrm{span}(e_1,\dots ,e_j)=\mathrm{span}(v_1,\dots ,v_j)$. Starting with a basis of a linear subspace $U$ and extending it to a basis of $V$ gives the same construction with the first $r=\dim U$ vectors spanning $U$.

The orthogonal complement is $U^{\perp }=\{v\in V:⟨v,u⟩=0\text{ for every }u\in U\}$. For every $v\in V$, set $u={\sum }_{i=1}^r⟨v,e_i⟩e_i$. Then $u\in U$ and $v-u\in U^{\perp }$. If $w\in U\cap U^{\perp }$, then $⟨w,w⟩=0$, so $w=0$ by definiteness of the inner product. Thus the direct sum isThis also covers $U=0$ and $U=V$.

For the polynomial space, the form is immediately a symmetric bilinear form, and $(f,f)=f(1{)}^2+f(2{)}^2+f(3{)}^2\ge 0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, so this form is an inner product for $n=1,2$. For $n\ge 3$, the nonzero polynomial $(x-1)(x-2)(x-3)$ has $(f,f)=0$. Therefore, among the stipulated positive integers,

The Riesz-Markov-Kakutani representation theorem identifies the dual of the complex space of continuous functions on a compact space $C(K)$ with finite regular complex Borel measures on $K$: each bounded linear functional has a unique representationHere $|\mu |$ is the variation measure; positive functionals correspond exactly to positive regular measures. This is the measure representation theorem, rather than the Hilbert-space representation of a functional by a vector.

For the spectral construction use the convention that the Hilbert inner product is linear in its first argument. A unital subalgebra of $\mathcal{B}(H)$ is understood to contain $I_H$: an algebra whose abstract identity is a smaller projection could not give a resolution normalized by $P(K)=I_H$. On a compact Hausdorff space, a resolution of the identity is understood to be a normalized regular projection-valued measure: its scalar measures are regular, its values are orthogonal projections, and it is countably additive in the strong operator topology. Regularity is part of the usual convention needed for the uniqueness assertion; the statement is interpreted in this sense.

The Commutative Gelfand--Naimark theorem makes the Gelfand transform an isometric unital star-isomorphism $A\to C(K)$, where $K={\Phi }_A$ is compact Hausdorff. Denote its inverse by $\pi :C(K)\to A\subset \mathcal{B}(H)$. For $x,y\in H$, the bounded linear functional $f\mapsto ⟨\pi (f)x,y⟩$ has norm at most $\Vert x\Vert \Vert y\Vert$. Riesz-Markov-Kakutani gives a regular complex measure ${\mu }_{x,y}$ such thatUniqueness makes these measures sesquilinear in $x,y$. For $f\ge 0$, the identity $\pi (f)=\pi (\sqrt{f})\pi (\sqrt{f}{)}^{\ast }$ shows positivity, so ${\mu }_{x,x}$ is positive with total mass $\Vert x{\Vert }^2$.

For every Borel set $B$, the bounded sesquilinear form $(x,y)\mapsto {\mu }_{x,y}(B)$ determines an operator $P(B)$ by Hilbert-space Riesz representation theorem, withIn particular $0\le P(B)\le I$ and $P(B)$ is self-adjoint. We now verify the projection identity rather than assume it.

For continuous $f$, uniqueness of the representing measure applied to continuous test functions givesThese identities imply $P(B)\pi (f)=\pi (f)P(B)$. Testing once more against a continuous $g$ givesThe restriction $1_B{\mu }_{x,y}$ of a finite regular Borel measure is still regular, so uniqueness in Riesz-Markov-Kakutani yields ${\mu }_{P(B)x,y}=1_B{\mu }_{x,y}$. Consequently, for any Borel $C$,Taking $C=B$ proves that $P(B)$ is an orthogonal projection. Also $P(K)=I$ and $P(\varnothing )=0$. For disjoint $B_j$, these projections are orthogonal, and scalar countable additivity gives weak countable additivity. If $B={\bigcup }_j{B}_j$, then $R_N=P(B)-{\sum }_{j\le N}P(B_j)$ is the projection of the remaining union andThus countable additivity holds in the strong operator topology. The scalar measures are the regular measures already constructed. This completes the scalar-measure construction of a projection-valued measure.

By the integral theorem permitted in the question, continuous $f$ satisfies $⟨(\int fdP)x,y⟩=\int fd{\mu }_{x,y}=⟨\pi (f)x,y⟩$, and hence $\int fdP=\pi (f)$. For $T\in A$, this proves the spectral theorem for a commutative operator algebra:Any other regular resolution giving these integrals has the same scalar integrals on all of $C(K)$; Riesz-Markov-Kakutani uniqueness forces the same scalar measures and therefore the same projections on every Borel set.

For nonempty open $U\subset K$, compact Hausdorff normality supplies a nonzero continuous function $f$ supported in $U$. If $P(U)=0$, the stated squared-norm identity for spectral integrals gives $\pi (f)=\int fdP=0$, contradicting the isometry of $\pi$. This proves the full support of a faithful spectral measure propertyFaithfulness of the representation is essential here.

The spectral theorem for normal operators says that a bounded normal operator $T$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on $\sigma (T)$ such that $T=\int zdE(z)$. Its support is all of $\sigma (T)$, and bounded Borel functions have the associated Borel functional calculus for a normal operator.

For the proof sketch, $A=C^{\ast }(I,T)$ is commutative because $T$ commutes with $T^{\ast }$; polynomials in these two operators commute, as do their norm limits. The map ${\Phi }_A\to \sigma (T)$, $\phi \mapsto \phi (T)$, is onto by the character of an algebra formula and spectral permanence for C-star algebras. It is one-to-one because a character of an algebra preserves the star operation and its values on $T,T^{\ast }$ determine it on their dense polynomial algebra. It is therefore a homeomorphism from compact ${\Phi }_A$ to the Hausdorff spectrum. Transport the resolution just constructed through this homeomorphism. It gives the formula for $T$ and full support; conversely a regular resolution for $T$ gives the same integrals for polynomials in $T,T^{\ast }$, hence by density the same continuous functional calculus and the same resolution. This argument works without separability of $H$.

Finally choose disjoint nonempty relatively open sets $U,V\subset \sigma (T)$ around two distinct spectral points, and put $Q=E(U)$. Full support gives $Q\ne 0$ and $E(V)\ne 0$, while $QE(V)=0$, so $Q\ne I$. The spectral integral, or multiplicativity of its Borel calculus with $1_U$, gives $QT=TQ$ and also $Q{T}^{\ast }=T^{\ast }Q$. Thus $Y=QH$ is closed, nonzero and proper, and $T(Qx)=Q(Tx)\in Y$. The spectral projection gives a reducing subspace conclusion isIn fact it is a reducing subspace, since it is also invariant under $T^{\ast }$.

Under the standard smooth manifold hypotheses, the first answer is yes, although the isomorphism requires a choice. A partition of unity combines Euclidean inner products in manifold charts to produce a Riemannian metric $g$. Nonnegative local weights summing to one preserve positivity, and local finiteness preserves smoothness. The musical isomorphism isIt is smooth and fiberwise invertible, with inverse having local matrix $(g_{ij}{)}^{-1}$. It is therefore a vector bundle isomorphism. There is no preferred such map before a metric or equivalent additional structure is chosen.

The second answer is no. On $S^1=\mathbb{R}/2\pi \mathbb{Z}$, compare the trivial vector bundle $S^1\times \mathbb{R}$ with the Möbius line bundleBoth have rank one. A continuous section of a vector bundle of $L$ is represented on $[0,2\pi ]$ by a continuous function $s$ satisfying $s(2\pi )=-s(0)$. If it were nowhere zero, the intermediate value theorem would prevent its endpoint signs from being opposite. Thus $L$ has no nowhere-zero section. The trivial bundle has the section $1$, and a bundle isomorphism would carry that section to a nowhere-zero section of $L$, a contradiction. Hence

For $v$ in the domain of the exponential map at $p$, let ${\gamma }_v$ be the unique geodesic with initial data $(p,v)$ and define ${\exp }_p(v)={\gamma }_v(1)$. The domain consists of initial velocities whose geodesics exist on $[0,1]$; it is open, contains zero, and need not be the whole tangent space. The ordinary differential equation theorem used here says that a smooth vector field has unique maximal integral curves, with an open flow domain and smooth dependence on initial point and time. Applied to the geodesic equation on $TM$, it gives smoothness of the exponential map. Rescaling the parameter gives ${\exp }_p(tv)={\gamma }_v(t)$ and hence $(d{\exp }_p{)}_0(v)=v$. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighbourhood of zero to one of $p$. Coordinates in an orthonormal tangent basis transported by this map are geodesic normal coordinates.

The geodesic sphere of radius $r$ is $S_r(p)=\{x:d(p,x)=r\}$. For small $r>0$, it is ${\exp }_p(\{v:\Vert v\Vert =r\})$, a smooth compact hypersurface; larger distance spheres need not be smooth. The Gauss lemma statesIn particular radial and angular directions are orthogonal and the radial coordinate measures arc length.

For a piecewise smooth path $c$, put $L(c)=\int \Vert \dot{c}(t)\Vert dt$. The Riemannian distance is $d(x,y)=\inf L(c)$ over such paths from $x$ to $y$. Connectedness makes it finite. Reversal and concatenation give symmetry and the triangle inequality; positivity and compatibility with the manifold topology follow locally from geodesic normal coordinates and the Gauss lemma.

Choose $r_0>0$ so that ${\exp }_p$ is a diffeomorphism on a neighbourhood of the closed tangent ball of radius $2r_0$. For $r<r_0$, radial geodesics realize the distance from $p$ on that ball. Indeed, Gauss lemma bounds any path remaining in the normal ball below by its radial change, while a path leaving the radius-$2r_0$ ball already has length at least $2r_0$. Thus the radius-$r$ sphere is the compact exponential image described above.

Given $q\ne p$, take $0<\delta <\min \{r_0,d(p,q)\}$. The continuous function $z\mapsto d(z,q)$ attains its minimum at some $p_0\in S_{\delta }(p)$. Every path from $p$ to $q$ crosses that sphere, since the distance from $p$ is continuous. Splitting at a crossing gives length at least $\delta +{\min }_{S_{\delta }(p)}d(\cdot ,q)$. Taking the infimum over paths, and using the triangle inequality in the other direction, proves the distance splitting through a small geodesic sphere equalityThis compact local argument does not assume completeness or the existence of a globally minimizing geodesic to $q$.

For the Lie-group clause, a bi-invariant Riemannian metric has adjoint-invariant identity inner product. Differentiating this invariance gives $⟨[X,Y],Z⟩+⟨Y,[X,Z]⟩=0$. The Koszul formula for left-invariant vector fields has no metric-derivative terms, and this skew-adjointness reduces it toThis is the Levi-Civita connection of a bi-invariant metric.

Let $\sigma$ be the integral curve through the identity of the left-invariant field with identity value $v$. Uniqueness and left translation show $\sigma (s+t)=\sigma (s)\sigma (t)$ wherever the local curves are defined. The field is complete: a fixed local existence interval about the identity translates to an equally long interval at every point of the group. Restarting the curve before any proposed finite endpoint extends it beyond that endpoint. Thus $\sigma$ exists on $\mathbb{R}$, and uniqueness now gives the group law for every $s,t$.

Since ${\nabla }_{X}X=0$, this integral curve is a geodesic. Conversely any geodesic starting at the identity has the same initial data as one of these curves and agrees with it by uniqueness. Therefore the geodesics of a bi-invariant metric are one-parameter subgroups conclusion isThe zero velocity gives the constant, trivial subgroup. Here ${\exp }_G$ is the Exponential map of a Lie group; for this metric it agrees at the identity with the Riemannian exponential.

At each point, use the inverse metric on the cotangent space. It induces an inner product on $p$-forms byEquivalently, wedges of distinct members of an orthonormal coframe form an orthonormal basis. Choose an orientation to define the ordinary global Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩d{\mathrm{vol}}_g$. It maps $p$-forms to $(n-p)$-forms and obeys ${\ast }^2=(-1{)}^{p(n-p)}$. We use real forms; complex forms are obtained by complex-linear extension. Without an orientation an ordinary global star is not available without twisting the target, so orientation is implicit in this first part.

With the codifferential ${\delta }_p=(-1{)}^{n(p+1)+1}\ast d\ast$ for $p>0$ and ${\delta }_0=0$, define the Hodge Laplacian $\Delta =d\delta +\delta d$. On functions it is the nonnegative-sign Laplace-Beltrami operator, $\Delta f=-\mathrm{div}\mathrm{grad}f$. A harmonic differential form satisfies $\Delta \alpha =0$.

For $\alpha$ of degree $p$, the displayed sign formula and star square giveApplying the identities again, at degrees $p-1$ and $p+1$, proves the Hodge star commutes with the Hodge Laplacian relationSince star is invertible, $\alpha$ is harmonic if and only if $\ast \alpha$ is harmonic. This part does not need compactness; zero-degree and top-degree terms are interpreted as zero when their degrees are outside the range.

For the remaining cohomological statements assume a compact oriented manifold without boundary. The Hodge decomposition theorem gives the $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^p=\ker \Delta$ is finite-dimensional and consists of smooth forms. Formal adjointness gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic differential forms are closed differential forms and coclosed differential forms.

If $\Delta f\ge 0$, formal adjointness and $\Delta 1=0$ give ${\int }_M\Delta f=0$. A continuous nonnegative function with zero integral vanishes, so $\Delta f=0$. Then $\Vert df{\Vert }_{L^2}^2=⟨\Delta f,f⟩=0$. Thus the a function with nonnegative Laplacian on a closed manifold is locally constant assertion isIt is globally constant when $M$ is connected. The PDF does not explicitly include connectedness here; on two disjoint circles one may take different constants, so literal global constancy needs that qualification. Boundaryless is the standard manifold convention in this argument; otherwise boundary conditions are necessary.

For a closed differential form $\alpha$, write its Hodge decomposition as $\alpha =h+d\beta +\delta \gamma$. Since $d\alpha =dh=0$, we have $d\delta \gamma =0$. Formal adjointness yields $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$, so $\alpha =h+d\beta$. Thus $h$ represents its de Rham cohomology class. If two harmonic representatives differ by $d\beta$, their difference $h$ has $\Vert h{\Vert }^2=⟨d\beta ,h⟩=⟨\beta ,\delta h⟩=0$. Therefore

Finally let $h\in G$. A connected Lie group admits a smooth path $h_t$ from the identity to $h$. The maps $R_{h_t}$ give a smooth homotopy between the identity and $R_h$. The homotopy invariance of de Rham cohomology says their pullbacks agree on cohomology. The maps $R_{h_t}$ preserve orientation, since their Jacobian signs cannot change from the identity along the path. As an orientation-preserving isometry, $R_h$ commutes on forms with star, exterior differentiation, codifferentiation and consequently the Hodge Laplacian. Hence $R_h^{\ast }\alpha$ is harmonic and represents the same cohomology class as a harmonic $\alpha$. Uniqueness now proves the harmonic forms are fixed by a connected isometric group action conclusionThe argument also applies componentwise when $M$ is disconnected.

Take the real inner productThe stationary distribution is strictly positive for an irreducible Markov chain. Detailed balance gives $⟨Pf,g{⟩}_{\pi }=⟨f,Pg{⟩}_{\pi }$, so $P$ is a self-adjoint operator. Choose a real orthonormal eigenbasis $f_1,\dots ,f_m$, where $m=|S|$, $P{f}_j={\lambda }_j{f}_j$, $f_1\equiv 1$, ${\lambda }_1=1$, and $\pi (f_j)=0$ for $j\ge 2$.

For fixed $y$, apply the spectral decomposition to $g_y(z)=1_{\{y\}}(z)/\pi (y)$. Its coefficient against $f_j$ is $f_j(y)$, whereas $(P^t{g}_y)(x)=P^t(x,y)/\pi (y)$. Thus the spectral kernel expansion of a reversible chain isThis holds for all integers $t\ge 0$; at $t=0$ each multiplier is one, including when ${\lambda }_j=0$. Negative eigenvalues are allowed, and retain their signs for odd $t$.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.