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The normalized Killing one-form $\eta =F^{-1}{V}_{\mu }d{x}^{\mu }$ is a closed differential form. The Poincaré lemma gives a local scalar potential $\varphi$ with $\eta =d\varphi$, hence $V_{\mu }=F{\partial }_{\mu }\varphi$. Choose the adapted chart of part (i), so $V={\partial }_0$. ThenThe closed differential form condition supplies the integrability of these spatial derivatives. For a fixed $g$ and $V$, $f$ is determined locally up to an additive constant; its derivatives cannot be chosen freely despite the wording in the paper.

Use ${\bar{x}}^0=\varphi$, ${\bar{x}}^i=x^i$. The Jacobian determinant is one, so this is a local coordinate change. Holding ${\bar{x}}^i$ fixed gives $V={\partial }_{\bar{0}}$, while its metric-dual one-form is $v=Fd{\bar{x}}^0$. Thus ${\bar{g}}_{00}=F<0$ and ${\bar{g}}_{0i}=0$. The Killing equation still holds, so ${\partial }_{\bar{0}}{\bar{g}}_{\mu \nu }=0$. The metric is locally static:A closed differential form need not be globally exact; neither a global potential nor a global static chart follows without additional topological assumptions.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

The formal adjoint of $\mathcal{L}=\Delta -k^2$ is itself. With the bilinear convention used in Green second identity, take any complex-valued $v$ satisfying $\mathcal{L}v=0$. The product rule for divergence gives the familyEquivalently, the differential one-formis a closed differential form. Arbitrary linear combinations of adjoint solutions give further members of this family. One can also write it as $\nabla \cdot [v^2\nabla (u/v)]=0$ wherever $v\ne 0$, but the first formula has no division and remains valid at zeros of $v$. For a sesquilinear convention replace $v$ by $\overline{v}$; no such conjugation is built into the bilinear identity used below.

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.

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.

The given $\gamma$ is a simple closed curve. Assume it is essential; for a contractible class both infimal lengths are already zero. Write ${\ell }_n={\ell }_{X_n}(\gamma )$. The collar lemma supplies an embedded annulus with coordinates $r\in (-w_n,w_n)$ and $t\in \mathbb{R}/\mathbb{Z}$, and metricSet $y(r)={\ell }_n^{-1}{\int }_0^r\mathrm{sech}sds$. This changes the metric to a positive scalar multiple of $d{t}^2+d{y}^2$, so the conformal modulus of an annulus isThe extremal length of its core curves is $1/M_n$. Allowing all curves homotopic to $\gamma$ in $X_n$ can only decrease infimal lengths, while the area of a metric on all of $X_n$ is at least its area on the collar. Thus $\lambda (\gamma ,X_n)\le 1/M_n$. Use the particular conformal metric $|q_n{|}^{1/2}$, whose area of a quadratic differential is one:This proves the required implication uniformly over all the area-one holomorphic quadratic differentials on these surfaces.

The converse is false. Here is an explicit slit connected sum of translation tori. Start with square flat copies of a torus $T_1=\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ and $T_{\delta }=\mathbb{C}/(\delta \mathbb{Z}+i\delta \mathbb{Z})$, where $\delta \to 0$. Cut a horizontal slit of physical length ${\delta }^2$ in each, centred in an interior coordinate disk, and cross-glue the banks by translation. The resulting surface has genus two. Its two slit endpoints have cone angle $4\pi$, so the locally defined $dz$ extends to a holomorphic one-form ${\omega }_{\delta }$ with two simple zeros. Its flat area is $1+{\delta }^2$. Setwhich is a holomorphic quadratic differential of area one. Let $\gamma$ be a horizontal generator in the small torus, taken away from the slit and fixed by the marking of this small handle. Then

To verify that its hyperbolic length does not tend to zero, construct a uniform lower bound on extremal length. On the unit square torus choose a disk $D$ about the eventual slit centre and a smooth cutoff function $\chi$ equal to one on a smaller disk and supported in $D$. Let $x$ denote a local real coordinate on $D$; in the first term below, $dx$ is the globally defined torus one-form, while $\chi x$ is extended by zero outside $D$. The real closed differential formis globally defined, vanishes on the smaller disk, and has period one on the horizontal generator. Pull it to $T_{\delta }$ by the rescaling map $z\mapsto z/\delta$, and extend it by zero across the slit and over the other torus. For sufficiently small $\delta$, the slit is inside the region where the form vanishes. Hence this extension ${\alpha }_{\delta }$ is a smooth closed differential form on the connected sum, with ${\int }_{\gamma }{\alpha }_{\delta }=1$.

Define a nonnegative conformal metric density by the pointwise norm of ${\alpha }_{\delta }$ relative to the flat metric. Two-dimensional scale invariance givesindependently of $\delta$. For every representative $\tilde{\gamma }$ homotopic to $\gamma$,because the period of a closed differential form is unchanged by homotopy. This is the extremal length lower bound from a closed one-form; therefore $\lambda (\gamma ,X_{\delta })\ge 1/C>0$. If ${\ell }_{X_{\delta }}(\gamma )$ tended to zero along any subsequence, the collar estimate would force $\lambda (\gamma ,X_{\delta })\to 0$, a contradiction. In fact its hyperbolic lengths are uniformly bounded away from zero. ThusFor every fixed $g>2$, replace $T_1$ by the area-one translation surface $X_{2g-2}$ constructed in the polygon argument; its genus is $g-1$. Cut its slit inside a nonsingular flat coordinate disk. The new connected sum has genus $g$, and the area normalization, small-handle length bound and closed-one-form energy argument are unchanged. Thus the converse fails at every fixed genus $g\ge 2$.

By the preceding sections, $\eta$ is a nowhere-zero closed differential form and $j^{\ast }\eta =0$. Injectivity of $j^{\ast }:H_{\mathrm{dR}}^1(M)\to H_{\mathrm{dR}}^1(\partial M)$ gives $[\eta ]=0$. The definition of de Rham cohomology then givesfor a smooth $f$, with no critical points. If the boundary were connected, $j^{\ast }df=0$ would make $f$ constant there. A nonconstant function on a compact manifold attains both extrema; a value differing from the common boundary value gives an interior extremum, contradicting $df\ne 0$. If $f$ were constant everywhere there would be the same contradiction. Empty boundary is also impossible, because an extremum would necessarily be interior. Thus $\boxed{\partial M\text{ has at least two connected components}}$, the boundary obstruction for commuting volume-preserving vector fields.

The solid torus $S^1\times D^2$ retracts onto its first circle. Its degree-one de Rham cohomology is generated by that circle's angular one-form, which restricts to a nonzero class on $S^1\times S^1$; hence the restriction map is injective. Its boundary is connected. Extensions of the two specified coordinate vector fields would be tangent there and would satisfy all the forbidden conditions. Therefore no such pair of extensions exists, for any volume form.

Let $\omega$ be the Kähler form associated with the metric. The Lefschetz operator of a Kähler manifold and adjoint Lefschetz operator areHere the adjoint is taken with respect to the pointwise Hermitian inner product on complexified forms, or equivalently the global $L^2$ inner product. It lowers degree by two and bidegree by $(1,1)$. In terms of the Hodge star operator, $\Lambda ={\ast }^{-1}L\ast$.

Since $d\omega =0$, $dL=Ld$, so $L$ takes closed differential forms to closed differential forms and exact differential forms to exact differential forms. This already gives its action on de Rham cohomology. For both operators together, use the Kähler identities: $L$ commutes with the Hodge Laplacian, and taking adjoints shows that $\Lambda$ does too. Thus both preserve harmonic differential forms.

When $M$ is compact, the Hodge decomposition theorem identifies $H^k(M,\mathbb{C})$ with its space of harmonic $k$-forms. If $h_{\alpha }$ is the unique harmonic differential form representing $[\alpha ]$, defineThe resulting forms are harmonic and therefore closed, and uniqueness of $h_{\alpha }$ makes these definitions independent of the initial representative. The cohomological definition of $\Lambda$ uses harmonic differential forms as representatives: $\Lambda$ need not take an arbitrary closed form to a closed form.

The cotangent fiber translation $f$ is a diffeomorphism, with inverse $(x,\xi )\mapsto (x,\xi -{\beta }_x)$. Because $\pi \circ f=\pi$, the definition of the canonical one-form on a cotangent bundle givesTaking the exterior derivative yieldsIf $\beta$ is a closed differential form, this is $f^{\ast }\omega =\omega$, proving that the cotangent fiber translation is a symplectomorphism.

The exterior derivative is the real-linear degree-one map $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ characterized by $df(X)=Xf$ on smooth functions, the graded Leibniz ruleand $d^2=0$. In local coordinates $(x^1,\dots ,x^m)$, write $\alpha ={\sum }_I{\alpha }_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$ using multi-index notation. Since $d{x}^i=d(x^i)$ and hence $d(d{x}^i)=0$, the defining rules forceThis proves local uniqueness, and the coordinate formulas agree on overlaps because the same rules are preserved by the pullback of a differential form. They also directly define an operator satisfying all three rules, proving existence.

An exact differential form is a form $\omega =d\eta$. Let $\pi :S^{2n}\to {\mathbb{RP}}^{2n}$ be the antipodal double covering map and let $a(x)=-x$. For a $2n$-form $\omega$ on real projective space, $a^{\ast }{\pi }^{\ast }\omega ={\pi }^{\ast }\omega$, while the mapping degree of the antipodal map is $-1$. ThereforeBy the stated criterion, ${\pi }^{\ast }\omega =d\eta$. The invariant primitive under a finite group actionstill satisfies $d\bar{\eta }={\pi }^{\ast }\omega$ and descends to a form $\beta$ on ${\mathbb{RP}}^{2n}$. Since pullback through a covering is injective on differential forms, ${\pi }^{\ast }(d\beta -\omega )=0$ implies $d\beta =\omega$. Thus every $2n$-form on ${\mathbb{RP}}^{2n}$ is exact; this is the top-degree differential forms on even-dimensional real projective space are exact result.

The $k$th de Rham cohomology isthe closed differential forms modulo the exact ones. For the product, let $p:M\times S^1\to M$ be projection and choose a closed one-form $\nu$ on the circle with ${\int }_{S^1}\nu =1$. Averaging differential forms over the circle is cochain-homotopic to the identity, so every class has a rotation-invariant representative; if such a representative is exact, averaging a primitive gives an invariant primitive. Every invariant $k$-form has a unique decomposition $p^{\ast }\alpha +p^{\ast }\beta \wedge \nu$, andClosedness and exactness are therefore componentwise. Hence the mapis a well-defined bijection $H_{\mathrm{dR}}^k(M)\oplus H_{\mathrm{dR}}^{k-1}(M)\to H_{\mathrm{dR}}^k(M\times S^1)$, proving the de Rham cohomology of a product with a circle formula.

Suppose first that the left-invariant 1-form $\alpha$ is closed. The scalar function $\alpha (l_{\xi })$ is constant for every $\xi$, because both the form and vector field are left-invariant. Cartan's magic formula therefore givesPart b says that $g\mapsto R_g^{\ast }\alpha$ is locally constant. Since $G$ is connected, it is constant, and its value at the identity is $\alpha$. Hence $R_g^{\ast }\alpha =\alpha$ for every $g$, so $\alpha$ is bi-invariant.

Conversely, if $\alpha$ is bi-invariant, then all these Lie derivatives vanish. Cartan's formula and the constancy of $\alpha (l_{\xi })$ give ${\iota }_{l_{\xi }}d\alpha =0$. The left-invariant vector fields span every tangent space, so $d\alpha =0$.

Connectedness cannot be omitted. The orthogonal group $O(2)$ has an abelian Lie algebra, so every left-invariant 1-form is closed by the Maurer-Cartan equation in a Lie-algebra basis. Conjugation by a reflection acts as $-1$ on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the closed left-invariant 1-form criterion.

The de Rham cohomology of $X$ isBecause pullback commutes with $d$, it sends closed forms to closed forms and exact forms to exact forms. Thus a smooth map induces

Let $H:X\times I\to Y$ be a smooth homotopy and write $H_t(x)=H(x,t)$. If $V={\partial }_t$, Cartan's magic formula givesIntegrating defines a degree-minus-one operator $K$ satisfyingFor closed $\omega$, the difference is exact, so smoothly homotopic maps induce the same map on de Rham cohomology.

If $F:X\to Y$ is a homotopy equivalence with inverse up to homotopy $G$, functoriality and homotopy invariance giveHence $F^{\ast }$ is an isomorphism.

Every smooth closed differential form of positive degree is locally an exact differential form. On a coordinate ball centered at zero, put $F_t(x)=tx$ and let $R=x^i{\partial }_i$ be the radial vector field. Define the degree-lowering operatorFor a $p$-form with $p>0$, the integrand has a factor $t^{p-1}$ in its coordinate coefficients and is smooth at zero. Cartan's magic formula and differentiation of the pullback of a differential form give ${\partial }_t(F_t^{\ast }\alpha )=t^{-1}{F}_t^{\ast }(d{\iota }_R\alpha +{\iota }_{R}d\alpha )$. Integrating from zero to one gives $dK\alpha +Kd\alpha =\alpha$, since $F_0^{\ast }\alpha =0$ in positive degree. If $d\alpha =0$, take $\beta =K\alpha$. The local potential of a closed differential one-form gives the explicit degree-one construction. Global exactness can fail through nonzero periods, measured by de Rham cohomology.

On a fiberwise star-shaped neighborhood, let $H_s$ scale fibers by $s$ and let $E$ be the radial vector field. For a closed differential form $\delta$ with zero pullback to the zero section, the homotopy primitive $\eta ={\int }_0^1{s}^{-1}{H}_s^{\ast }({\iota }_E\delta )ds$ satisfies $d\eta =\delta$. For a two-form, it vanishes as an ambient covector along that section, providing the fixed-submanifold condition in the relative Moser theorem.

Let a neighborhood of a submanifold retract onto it by fiberwise radial contraction in a tubular neighborhood. If a closed differential form $\beta$ pulls back to zero on the submanifold, the radial homotopy formula gives $\beta =d\alpha$, with $\alpha$ vanishing along the zero section. Applying the homotopy operator to $\beta$ gives the primitive directly. This is the relative ingredient in the relative Moser theorem.