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In this context a nowhere locally homogeneous metric, also called a bumpy metric, is a smooth Riemannian metric for which no two distinct nonempty open subsets are isometric with their induced metrics. Equivalently every local isometry between open subsets is the identity wherever defined: a nonidentity local isometry sends some point $x$ to a distinct point, and restriction to sufficiently small disjoint neighborhoods would violate the first formulation. This is the local-isometry meaning of the terminology here; degeneracy of periodic geodesics is a different use of the word bumpy.

Sunada's local isometry lemma states that on a compact smooth manifold without boundary of dimension $m\ge 2$ the nowhere locally homogeneous metrics contain a residual set in the space of smooth Riemannian metrics with its $C^{\infty }$ topology. In particular they are dense, by the Baire category theorem. A residual set is a countable intersection of open dense sets. The dimension assumption matters: every one-dimensional Riemannian metric is locally $d{s}^2$ in arclength coordinates and has local translations.

The jet bundle of maps $J^k(M,N)$ consists of equivalence classes $j_x^{k}f$ of smooth maps near $x$, where two maps agree to order $k$ at $x$ in coordinate charts. Its projection to $M\times N$ sends $j_x^{k}f$ to $(x,f(x))$. For a multi-index $\alpha$ in $m$ variables there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivatives of order $r$. Thus a coordinate chart consists of the source point, the target point, and $n$ coefficients for every derivative order $1$ through $k$. Summing the counts givesThis includes $J^0(M,N)=M\times N$. The fibre over $(x,y)$ is the space of truncated Taylor maps with fixed constant term $y$, locally modeled onFor $k=1$ its identification with $\mathrm{Hom}(T_{x}M,T_{y}N)$ is canonical. For higher $k$, changes of target coordinates mix derivatives of different orders, so this is a coordinate or connection-dependent description, not a canonical vector-bundle identification. The truncation $J^k\to J^{k-1}$ is an affine bundle modeled on the pullback of ${\mathrm{Sym}}^k(T^{\ast }M)\otimes TN$ over $M\times N$.

For the density assertion put $A={\overline{U}}_i$ and $B={\overline{U}}_j$. Their closures must be distinct. If $A=B$, the identity is an isometry for every Riemannian metric, and the requested complement is empty. Under the intended distinct-domain assumption, the smooth closed-ball hypothesis makes $A$ and $B$ regular closed domains, equal to the closures of their interiors. After interchanging them if necessary, there is a nonempty open set $W$ with compact closure in $\mathrm{int}(A)\setminus B$. Otherwise each interior would be contained in the other closure, forcing $A=B$.

Take any smooth Riemannian metric $g$. If ${\mathrm{vol}}_g(A)\ne {\mathrm{vol}}_g(B)$, it already lies outside ${\mathcal{S}}_{ij}$, because an isometry preserves the Riemannian volume form. If the volumes agree, choose a nonzero nonnegative smooth bump function $\eta$ supported in $W$ and setThese are positive definite Riemannian metrics, they agree with $g$ on $B$, and $g_t\to g$ in $C^{\infty }$ as $t\downarrow 0$, since every derivative of $g_t-g$ is $t$ times a fixed compactly supported smooth tensor. Their Riemannian volume forms satisfyHence ${\mathrm{vol}}_{g_t}(B)={\mathrm{vol}}_g(B)$ whereas ${\mathrm{vol}}_{g_t}(A)>{\mathrm{vol}}_g(A)$ for every $t>0$. In particularThe two domains cannot be isometric for $g_t$. Every $C^{\infty }$ neighborhood of $g$ therefore meets the complement of ${\mathcal{S}}_{ij}$, provingThis localized volume perturbation proves the requested density even when the two domains overlap, and works in every positive dimension. It does not by itself prove the stronger residual local-isometry statement: fixed isometric closures and arbitrary isometric open subsets are different conditions.

Use a radial geometric Kähler potential. Write $s=|z_1{|}^2+|z_2{|}^2$ and $y(s)={\phi }^{\prime }(s)$. The coefficient Hermitian matrix of $i\partial \bar{\partial }\phi (s)$ isIts eigenvalues are $y$ in the complex tangential direction and $y+s{y}^{\prime }$ in the complex radial direction. A Kähler metric is positive precisely when both are positive. The standard form has coefficient matrix $\frac{1}{2}I$, so equality of the Riemannian volume forms is the conditionIndeed, the volume form is ${\omega }^2/2$ and its coefficient is proportional to $\det G$. Multiplying the determinant equation by $2s$ gives $(s^2{y}^2{)}^{\prime }=s/2$, so we can takeAn explicit antiderivative isIt is smooth for $s>0$. Its two eigenvalues areThe form is real and closed because it is $i\partial \bar{\partial }$ of a real function, so it is a Kähler metric on the punctured space. Since $a>0$, neither eigenvalue equals $1/2$, and the metric differs from the Euclidean one. The radial and tangential stretches compensate exactly, preserving the volume form:This is a radial Kähler metric with Euclidean volume in complex dimension two.

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.

Let $({\theta }^1,\dots ,{\theta }^n)$ be the dual basis of the positively oriented orthonormal basis $(v_1,\dots ,v_n)$. By the definition of the Riemannian volume form,The interior product of a differential form with the outward unit normal isThe vectors $(v_2,\dots ,v_n)$ form a positive orthonormal frame of $T\partial X$ by the outward-normal-first boundary orientation. Consequently the pullback of the last display is the positive unit boundary volume form:

Since $t=u+r$, the hypersurface ${\Sigma }_t$ is an ordinary constant-Minkowski-time slice. Its future unit normal isin the $(u,r,\theta ,\varphi )$ basis. Lowering the index gives $n_a=(-1,-1,0,0)=-{\nabla }_{a}t$. Restricting the metric to $dt=du+dr=0$ gives the Euclidean spherical metricso its induced Riemannian volume form is

The metric on covectors is induced by the inverse matrix $g^{-1}$, and on $p$-forms by the determinant pairingThe Riemannian volume form is the unique positive top form taking value one on every oriented orthonormal frame. The Hodge star operator is uniquely determined byNondegeneracy of the wedge pairing proves existence and uniqueness pointwise, and the smooth metric dependence makes $\ast$ a well-defined smooth bundle map.

On compactly supported forms, Stokes theorem and the graded Leibniz rule givewhere $m=\dim M$; this is the formal $L^2$ adjoint of $d$. The Hodge Laplace-Beltrami operator is

For $g_f=e^{2f}g$, the covector metric scales by $e^{-2f}$, the $p$-form metric by $e^{-2pf}$, and the volume form by $e^{mf}$. ThereforeIf $f$ is constant, the two star factors in the codifferential contribute $e^{-2f}$, so ${\delta }_f=e^{-2f}\delta$. Since $d$ is metric-independent,

The metric and orientation determine the Riemannian volume form: in a positively oriented coordinate chart $(x^1,\dots ,x^n)$,Writing $|g|=\det (g_{ij})$ and $(g^{ij})=(g_{ij}{)}^{-1}$, the Dirichlet energy on a Riemannian manifold with source $f$ is

An orientation selects the positive ordered bases in each tangent space. On an oriented $n$-dimensional Riemannian manifold, the Riemannian volume form ${\omega }_g$ is the unique smooth $n$-form satisfyingfor every positively oriented orthonormal frame. In positively oriented local coordinates,

The metric induces an inner product on the bundle ${\Lambda }^p{T}^{\ast }M$ of $p$-forms. The Hodge star operator is the unique linear mapsuch thatfor all $p$-forms $\alpha ,\beta$. With the codifferential $\delta$, the Laplace-Beltrami operator on differential forms is

The Hodge decomposition theorem says that on a compact oriented Riemannian manifold,an $L^2$-orthogonal direct sum, where ${\mathcal{H}}^p(M)=\ker \Delta$ is the finite-dimensional space of harmonic $p$-forms. Every de Rham cohomology class has exactly one harmonic representative.

Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.

For $s=|z_1{|}^2+|z_2{|}^2>0$ and $a>0$, a geometric Kähler potential with derivative ${\phi }^{\prime }(s)=\sqrt{s^2+a^2}/(2s)$ defines a Kähler metric on ${\mathbb{C}}^2\setminus \{0\}$. Its tangential and radial eigenvalues are $\sqrt{s^2+a^2}/(2s)$ and $s/(2\sqrt{s^2+a^2})$. Their product is $1/4$, giving the same Riemannian volume form as ${\omega }_0=(i/2)\sum d{z}_j\wedge d{\bar{z}}_j$, although the metric is different. This construction makes preservation of volume compatible with anisotropic stretching.

On an oriented Riemannian manifold with boundary, the divergence of a Riemannian vector field obeys the displayed integral identity for smooth compactly supported vector fields $X$, or for arbitrary smooth $X$ if the manifold is compact. Here $N$ is the outward unit normal vector and the boundary has the outward-normal-first boundary orientation. Its induced Riemannian volume form is $j^{\ast }{\iota }_N{\omega }_g$. Splitting $X$ into its normal and tangential components gives $j^{\ast }{\iota }_X{\omega }_g=g(X,N){\omega }_{\tilde{g}}$, since a top form vanishes on $n$ boundary-tangent arguments. Apply the Stokes theorem to $d{\iota }_X{\omega }_g=(\mathrm{div}X){\omega }_g$.

Without a support or decay condition, flux at infinity can invalidate the result even if the integrals converge. On $[0,\infty )$ with Riemannian metric $d{x}^2$, $X=(\arctan x){\partial }_x$ has boundary flux zero but integral of divergence $\pi /2$.

The total Riemannian volume is the integral of the Riemannian volume form, or its density on a nonorientable manifold. In local coordinates the density is $\sqrt{\det g}d{x}^1\cdots d{x}^d$. It is preserved by Riemannian isometries. On a closed manifold it is the leading coefficient among the integrated heat invariants, and on a surface it is the surface area.

A Riemannian metric on a smooth manifold gives a positive section of the density bundle, locally $d{\mu }_g=\sqrt{\det (g_{ij})}|d{x}^1\cdots d{x}^n|$. The absolute Jacobian determinant makes this independent of coordinates without requiring orientability. With an orientation of a smooth manifold, it corresponds to the Riemannian volume form.