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At a point in a normal orthonormal frame, torsion-freeness makes the frame brackets vanish there, hence the dual one-forms have zero exterior derivative there. Differentiating the interior product formula then leaves only ${\sum }_i{E}_i(f_i){\omega }_g$, the divergence of a Riemannian vector field. Since the point is arbitrary, the identity holds globally.

For fixed $p$, the covariant derivative is linear in its tangent-vector input, so $L_p:T_{p}M\to T_{p}M$, $v\mapsto ({\nabla }_{v}X)(p)$, is a well-defined linear map. Its trace defines the divergence of a Riemannian vector field. In the given Riemannian orthonormal frame, write $X={\sum }_j{f}_j{E}_j$. The Levi-Civita connection product rule givesAt $p$, the second sum vanishes by the normal orthonormal frame hypothesis. Thus the $i$th diagonal coefficient of $L_p$ is $E_i(f_i)(p)$, andEquivalently one can take ${\sum }_{i}g({\nabla }_{E_i}X,E_i)$; the trace does not depend on the orthonormal basis chosen.

Use the nonnegative convention for the Laplace-Beltrami operator. For a smooth function $f$ on a Riemannian manifold $(M,g)$, three intrinsic definitions areHere $d$ is the exterior derivative, $d^{\ast }$ its formal $L^2$ adjoint for the Riemannian volume density, $\mathrm{grad}f$ the Riemannian gradient, and $\nabla$ the Levi-Civita connection. The bilinear form $\nabla df$ is the Riemannian Hessian. The equally common nonpositive convention negates all three definitions simultaneously.

In local coordinates let $g^{ij}$ be the inverse metric matrix and $\rho =\sqrt{\det (g_{ij})}$. By the defining relation $g(\mathrm{grad}f,V)=df(V)$,The divergence of a Riemannian vector field can be defined by ${\mathcal{L}}_V(d{V}_g)=(\mathrm{div}V)d{V}_g$. Differentiating the coordinate volume density gives $\mathrm{div}V={\rho }^{-1}{\partial }_i(\rho V^i)$. Thus the second definition becomesFor compactly supported smooth $h$, coordinate integration by parts showsSince this holds for every such $h$, the first and second definitions agree. This argument also works on a nonorientable Riemannian manifold, using its volume density; no global orientation is needed.

For the third definition,The Christoffel symbols satisfy ${\Gamma }_{ik}^i={\partial }_k\log \rho$. This follows by tracing their formula and using ${\partial }_k\log \det g=\mathrm{tr}(g^{-1}{\partial }_{k}g)$. Metric compatibility givesSubstitute both identities into the coordinate divergence expression to obtainThis proves all three definitions equivalent. In geodesic normal coordinates at a point $p$, the common value is $-{\sum }_i{\partial }_i^{2}f(p)$. Finally, on a closed manifold,confirming the chosen positive Laplace-Beltrami operator sign convention.

The integral identity requires compactness of $M$, compact support of $X$, or suitable conditions eliminating flux at infinity. The printed question does not state such a hypothesis. We prove the Riemannian divergence theorem for a compact manifold, and more generally for a compactly supported smooth vector field on an oriented Riemannian manifold with boundary.

Write $j:\partial M\hookrightarrow M$ for the inclusion. Give $\partial M$ the outward-normal-first boundary orientation, and let $N$ be the outward unit normal vector. The Riemannian volume form of the induced Riemannian metric isIndeed, if $(e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis tangent to the boundary, $(N,e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis in $M$. Both sides therefore evaluate to one on the boundary basis.

At a boundary point decompose $X=g(X,N)N+X^{\top }$, where $X^{\top }$ is tangent to $\partial M$. Evaluating on $n-1$ tangent vectors, the term with $X^{\top }$ vanishes: all $n$ arguments of the alternating volume form then lie in the $(n-1)$-dimensional boundary tangent space. Thus the interior product of a differential form satisfiesThe general Stokes theorem for an oriented manifold with boundary states ${\int }_{M}d\eta ={\int }_{\partial M}{j}^{\ast }\eta$ for any smooth differential form $\eta$ of degree $n-1$ with compact support; compactness of $M$ makes the support condition automatic. Applying it to $\eta ={\iota }_X{\omega }_g$ and using the defining equation for the divergence of a Riemannian vector field proves

For a concrete failure without the extra hypothesis, take $M=[0,\infty )$ with Riemannian metric $d{x}^2$ and $X=(\arctan x){\partial }_x$. The divergence of a Riemannian vector field is $\mathrm{div}X=(1+x^2{)}^{-1}$, sobecause $X$ vanishes at the only boundary point. Both integrals are finite. The missing $\pi /2$ is the limiting flux at infinity, so mere integrability of the divergence cannot repair the unrestricted assertion.

The divergence of a Riemannian vector field $X$ is the trace of the covariant derivative map $Y\mapsto {\nabla }_{Y}X$:It is characterized by ${\mathcal{L}}_X{\omega }_g=(\mathrm{div}X){\omega }_g$ in the oriented case. The metric-dual vector field $X_{\theta }={\theta }^{♯}$ is defined by $g(X_{\theta },Y)=\theta (Y)$ for every $Y$, using the musical isomorphism.

On a compact oriented manifold without boundary, the permitted exactness assertion and Stokes theorem imply ${\int }_M\mathrm{div}Y{\omega }_g=0$ for every smooth vector field $Y$. For a smooth function $a$, the product rule givesSince $⟨da,\theta ⟩=X_{\theta }(a)$, the formal-adjoint identity yieldsThis holds for all $a$; taking $a=\delta \theta +\mathrm{div}{X}_{\theta }$ proves the pointwise formulaIt agrees with $\delta \theta =-\ast d\ast \theta$, since $\ast \theta ={\iota }_{X_{\theta }}{\omega }_g$ and $d{\iota }_{X_{\theta }}{\omega }_g={\mathcal{L}}_{X_{\theta }}{\omega }_g$.

The Bochner formula for one-forms, with the positive Hodge Laplacian, isHere $\Delta =d\delta +\delta d$ acts on one-forms. The covariant derivative $\nabla \theta$ is a section of $T^{\ast }M\otimes T^{\ast }M$, and its formal-adjoint composition, the rough Laplacian, isThis expression is independent of the local orthonormal frame. The Ricci term is the zero-order bundle endomorphism specified byThus the identity separates the differential energy of the form from the curvature contribution; the sign of the Ricci term matches positive sectional curvature on the sphere.

Let $\theta$ be a harmonic one-form on a compact connected manifold with nonnegative Ricci curvature. Pair the formula with $\theta$ and integrate:Both integrands are nonnegative and continuous, so each vanishes everywhere. In particular $\nabla \theta =0$, proving that harmonic one-forms are parallel under nonnegative Ricci curvature. This integration uses the Riemannian volume density and formal adjoints, so orientation is not necessary for the final conclusion.

Fix $p\in M$. The evaluation mapis injective: a parallel form with zero value at $p$ stays zero under parallel transport along every path, and connectedness makes every point reachable from $p$. ThereforeThis is the dimension bound for harmonic one-forms under nonnegative Ricci curvature. The bound is sharp: on a flat $n$-torus the $n$ constant coordinate one-forms are parallel and harmonic. If Ricci curvature is positive definite at even one point, the same integral and parallelism argument force every harmonic one-form to vanish.

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 $\pi /2$.