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Take the orientation induced by the future-directed unit normal and positive lapse function $\alpha$. In coordinates adapted to the 3+1 decomposition of spacetime, any component ${ϵ}_{\ell ijk}$ with all four indices spatial vanishes: only three distinct spatial indices are available. The shift vector therefore contributes nothing, andHere $[ijk]$ is the Levi-Civita symbol, and the last equality uses the metric determinant in a 3+1 decomposition, $|g|={\alpha }^2\gamma$. The right side is precisely the spatial metric volume tensor, with component $\sqrt{\gamma }$ at $ijk=123$, the appropriate permutation sign, and zero for repeated indices. HenceMore invariantly, the spatial volume form is the restriction to the hypersurface of the contraction of the spacetime volume form with $n$.

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.

For the spherical coordinates in the question, the coordinate tangent vectors are mutually orthogonal, with lengths $1,r,r\sin \theta$. Their metric tensor is therefore $\mathrm{diag}(1,r^2,r^2{\sin }^2\theta )$ and the Jacobian determinant is $r^2\sin \theta$. Applying the divergence formula to the gradient givesFor example, the formula follows directly from $\Delta u=|\det h{|}^{-1/2}{\partial }_i(|\det h{|}^{1/2}{h}^{ij}{\partial }_{j}u)$ for this diagonal metric. These coordinates are valid away from the polar axis and angular seam; the Laplacian is a smooth geometric operator there as well, described by other charts. The coordinate volume form integrates over $1/2<r<2$, $0<\theta <\pi$, $0\le \varphi <2\pi$, with the usual periodic identification.

An orientation of a smooth $n$-dimensional manifold can be specified by a nowhere-vanishing smooth $n$-form $\omega$: an ordered tangent basis is positive precisely when $\omega$ evaluates positively on it. Multiplication by a positive smooth function preserves the orientation. This is the differential-form formulation of the equivalent formulations of orientability of a smooth manifold.

Use the standard convention that the compact connected manifold is nonempty and has no boundary. Choose a positive volume form $\omega$ for its orientation. It is a closed differential form, because there are no nonzero $(n+1)$-forms. For $n>0$, compactness makes its integral finite, and positivity on a coordinate neighborhood together with nonnegativity elsewhere givesIf $\omega =d\eta$, the Generalized Stokes theorem would give ${\int }_M\omega ={\int }_{\partial M}\eta =0$, a contradiction. ThereforeFor $n=0$, a nonempty connected manifold is a point, and $H_{\mathrm{dR}}^0(M)=\mathbb{R}$ directly. The boundary hypothesis matters: a closed interval is an oriented compact connected manifold with boundary with $H_{\mathrm{dR}}^1=0$. Thus the result is about ordinary boundaryless manifolds, not an unrestricted version including boundary.

Use the double covering map $\pi :S^n\to {\mathbb{RP}}^n$ and its deck involution, the antipodal map $a(x)=-x$. The given sphere cohomology vanishes outside degrees $0,n$, so only those degrees can survive as invariant classes. The degree-zero action is the identity because the sphere is connected for $n\ge 1$.

For the top degree, orient $S^n$ as the boundary of the unit ball. A positive tangent basis $(v_1,\dots ,v_n)$ at $x$ means that $(x,v_1,\dots ,v_n)$ has positive determinant in ${\mathbb{R}}^{n+1}$. Under the antipodal map every one of these $n+1$ vectors changes sign, so the orientation sign is $(-1{)}^{n+1}$. Equivalently, the sphere volume form isIts cohomology class is nonzero by the integral argument and therefore generates the one-dimensional top de Rham cohomology. Thus the top class is invariant precisely when $n$ is an odd integer. Applying the quotient result yieldsIn particular ${\mathbb{RP}}^1$ has one copy of $\mathbb{R}$ in each of degrees zero and one. These are real cohomology groups; integral torsion is not detected by de Rham cohomology.

The Jacobian determinant of ${\varphi }_t$ is $e^t$, while every coordinate and every coordinate differential is multiplied by $e^t$ under ${\psi }_t$. Hence the pullback of a differential form operation givesTo compare directly with Cartan's magic formula, note $d\mu =0$ and computeThe first two differentiated terms vanish because their coordinate differentials repeat; the last contributes $dz\wedge dx\wedge dy=\mu$. AlsoTheir sum is $2\alpha$, as required. Equivalently the divergence of $X$ relative to the standard volume form is one.

At any point, the linearly independent vectors $X_1,\dots ,X_n$ can be completed by $Y$ to a basis of $T_{p}M$. A nowhere-zero volume form is nonzero on every basis, soThus $\boxed{\eta \text{ is nowhere zero}}$. In particular, if it is an exact differential form $df$, its potential has no critical points. This is the role of pointwise independence in the boundary obstruction for commuting volume-preserving vector fields.

The tangent bundle is $TM={⨆}_{p\in M}{T}_{p}M$, with its smooth vector-bundle structure from coordinate charts. A vector field is a smooth section of this bundle. The Lie bracket of vector fields is their commutator as derivations:In coordinates, $[X,Y{]}^k=X^j{\partial }_j{Y}^k-Y^j{\partial }_j{X}^k$.

The four numbered conditions here are hypotheses of one argument, so the following sections assemble its proof. Let $j:\partial M\hookrightarrow M$. Because all $X_i$ are tangent to the boundary, evaluating $\eta ={\iota }_{X_1}\cdots {\iota }_{X_n}\omega$ on a tangent vector $Y$ there evaluates the volume form on $n+1$ vectors in the $n$-dimensional boundary tangent space. Alternation gives $\boxed{j^{\ast }\eta =0}$. This uses the interior product of a differential form and will make the boundary restriction of the de Rham cohomology class vanish.

A symplectomorphism is a diffeomorphism, so it preserves the genus of a surface. It also preserves the orientation determined by the symplectic form and, by integration of a pullback of a differential form, preserves the symplectic area.

Conversely, equal genus of a surface gives a diffeomorphism $f:{\Sigma }_0\to {\Sigma }_1$. We can choose $f$ to preserve the orientations determined by ${\omega }_0,{\omega }_1$: if necessary, compose with an orientation-reversing diffeomorphism of the standard closed orientable surface of that genus of a surface, obtained by reflection of a symmetric handle model. On ${\Sigma }_0$ put $\eta =f^{\ast }{\omega }_1$. Both $\eta$ and ${\omega }_0$ are positive volume forms for the same orientation, and their integrals agree. By top-degree de Rham cohomology, $[\eta ]=[{\omega }_0]$.

The family ${\omega }_t=(1-t){\omega }_0+t\eta$ consists of positive volume forms, hence symplectic forms in dimension two, with constant de Rham cohomology class. Moser theorem supplies $\psi$ with ${\psi }^{\ast }\eta ={\omega }_0$. ConsequentlyTherefore the genus of a surface and symplectic area classify compact connected symplectic surfaces.

There is an orientation convention in this statement: the two volume forms must be positive for one fixed orientation, as is usual in Moser theorem for volume forms. Without that convention, the stated necessity is false. For example, on an oriented two-sphere, take a positive volume form ${\beta }_0$ and an orientation-reversing diffeomorphism $r$, and set ${\beta }_1=r^{\ast }{\beta }_0$. Then $(r^{-1}{)}^{\ast }{\beta }_1={\beta }_0$ but $\int {\beta }_1=-\int {\beta }_0$. We prove the intended result with a common orientation.

If ${\varphi }^{\ast }{\beta }_1={\beta }_0$, positivity forces $\varphi$ to preserve the orientation, and integration of the pullback of a differential form gives equality of the integrals.

For sufficiency, top-degree de Rham cohomology gives a differential form $\alpha$ of degree $n-1$ with ${\beta }_1-{\beta }_0=d\alpha$. Put ${\beta }_t=(1-t){\beta }_0+t{\beta }_1$. Positivity makes every ${\beta }_t$ a volume form. The interior product of a differential form gives an linear isomorphismThus there is a unique smooth time-dependent vector field $X_t$ satisfying ${\iota }_{X_t}{\beta }_t=-\alpha$. Since $M$ is compact, its local flow exists throughout $0\le t\le 1$ and defines a smooth isotopy ${\varphi }_t$ starting at the identity. By Cartan's magic formula, and since a top-degree differential form has zero exterior derivative,HenceThe zero-dimensional case, a single point, is immediate from equality of the integrals.

Put $h_{ab}=\delta g_{ab}$, $h=g^{ab}{h}_{ab}$, and $h^{ab}=g^{ac}{g}^{bd}{h}_{cd}$. Throughout this part use the Riemann curvature tensor convention printed in the paper. Varying $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives the inverse-metric variation. The determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ gives the volume form variation. Together these areIn Lorentzian coordinates the latter reads $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}h$.

For the connection, vary metric compatibility ${\nabla }_a{g}_{bc}=0$:Add the equations with derivatives $a,b$, subtract that with derivative $c$, and use symmetry of the lower indices of the Levi-Civita connection. This givesThe difference of two connections is a tensor, so this formula is covariant.

Varying the coordinate curvature formula yields the Palatini identityThe contractions of the connection variation areFinally, $\delta R=(\delta g^{ab})R_{ab}+g^{ab}\delta R_{ab}$ gives the metric variation of scalar curvature:The double divergence may equivalently be written ${\nabla }_a{\nabla }_b{h}^{ab}$: relabeling the contracted dummy indices and using symmetry of $h^{ab}$ gives the same expression. The derivative terms are a covariant divergence, useful when varying a curvature-dependent action.

Define the stress-energy tensor by metric variation with the scalar held fixed:Use compactly supported variations to remove boundary terms. The first covariant derivative of a scalar field is an ordinary derivative, so it has no connection variation. Varying the kinetic term and the volume form gives

For the curvature term set $f={\Phi }^2$. Using part (a), its variation isApplying integration by parts twice moves each pair of derivatives from the metric variation onto $f$. The result iswhere $G_{ab}=R_{ab}-R{g}_{ab}/2$ is the Einstein tensor and $\square ={\nabla }^c{\nabla }_c$ is the d'Alembert operator. Comparing with the defining variation gives the stress-energy tensor of a nonminimally coupled scalar field:The factor $2\xi$ follows from the action's coupling $-\xi R{\Phi }^2$; a convention using $-\xi R{\Phi }^2/2$ would instead give $\xi$ in that bracket. At $\xi =0$ this reduces to the massless Klein-Gordon scalar stress-energy tensor.

The determinant identity $\delta \log |\det g|=g^{ab}\delta g_{ab}$ impliesHence the metric volume form varies as

The volume form in cylindrical coordinates is $dV=rdrd\varphi dz$. Integrating around the axis and over the height gives cylindrical-shell volume $2\pi hrdr$. Hence the volume integral isFor unit density, this is the kinetic energy of the rotational motion.

For $w=x^1+i{x}^2$, $z=x^3+i{x}^4$ and the standard oriented Euclidean metric, the real and imaginary parts of $dw\wedge dz$, together with $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. The printed complex four-form $dw\wedge dz\wedge d\bar{w}\wedge d\bar{z}$ is four times the metric volume form and gives the same orientation.

Under SU(2) as the three-sphere, a Skyrme model field with $U\to I$ at infinity defines $S^3\to S^3$. Choose $T_i=-i{\tau }_i$ and positive left volume form ${\theta }^1\wedge {\theta }^2\wedge {\theta }^3$. The normalized form is $-(24{\pi }^2{)}^{-1}\mathrm{tr}(U^{-1}dU{)}^3$, so the topological baryon number in the Skyrme model equals $\mathrm{deg}U$. Generator and orientation conventions fix the sign.