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Define the Lie derivative of a differential form by ${\mathcal{L}}_X\alpha =(d/dt{)}_{t=0}{\varphi }_t^{\ast }\alpha$, using the local flow. The pullback of a differential form respects the wedge product of differential forms and commutes with the exterior derivative. Therefore ${\mathcal{L}}_X$ is a derivation of an algebra of degree zero and commutes with $d$.

Let $D_X={\iota }_{X}d+d{\iota }_X$, where ${\iota }_X$ is the interior product of a differential form. The graded product rules for $d$ and ${\iota }_X$ show that their anticommutator $D_X$ is also a degree-zero derivation of an algebra: the two mixed terms cancel. Since $d^2=0$, it commutes with $d$. On a function $f$, ${\iota }_{X}f=0$ and $D_{X}f={\iota }_{X}df=Xf={\mathcal{L}}_{X}f$. Hence $D_X(df)=d(Xf)={\mathcal{L}}_X(df)$.

In a manifold chart, every differential form is a sum of functions times products $d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. The two derivations agree on its generators $f$ and $d{x}^i$, and therefore agree on every such form. This proves Cartan's magic formula, including the case $p=0$:

Use Cartan's magic formula. The exterior derivative and interior product of a differential form giveThus $d({\iota }_X\alpha )=-2xdx-2ydy+2zdz$, and the first two components cancel:The rotation in the $xy$-plane leaves $ydx-xdy$ invariant, while the dilation in $z$ doubles the infinitesimal weight of $zdz$, providing a geometric check of this Lie derivative.

Put $u=t-z$, $\ell =du=e^0-e^3$, $p=A^{\prime }/A$, $q=B^{\prime }/B$, with primes denoting $d/du$. The exterior derivatives are $d{e}^0=d{e}^3=0$, $d{e}^1=p\ell \wedge e^1$, $d{e}^2=q\ell \wedge e^2$. For a Levi-Civita connection, the connection 1-forms obey both Cartan's first structure equation and metric compatibility. With lower indices ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$, the result isAll other lower-index forms vanish or follow from ${\omega }_{ab}=-{\omega }_{ba}$. Raising the first time index changes its sign: for example ${\omega }^0{}_1={\omega }^1{}_0=p{e}^1$, whereas ${\omega }^1{}_3=-p{e}^1$ and ${\omega }^3{}_1=p{e}^1$.

For direct verification, ${\omega }^1{}_0\wedge e^0+{\omega }^1{}_3\wedge e^3=-p\ell \wedge e^1=-d{e}^1$, and the same holds for index two. For indices zero and three each term wedges $e^1$ or $e^2$ with itself, so it is zero. These forms are metric-compatible, and uniqueness follows from the existence and uniqueness of the Levi-Civita connection. The printed uniqueness statement implicitly includes metric compatibility: the torsion equation alone permits adding nonzero forms $C^a{}_{bc}{e}^c$ with $C^a{}_{bc}=C^a{}_{cb}$, since their wedge with $e^b$ vanishes.

Put $q=r_0^2$ and $\eta =d\psi +\frac{1}{2}\cos \theta d\varphi$. Expanding the metric tensor in the stationary direction gives the useful exact identitiesTaking the exterior derivative givesThe fibered form of the metric tensor has determinant $\det g=-r^6{\sin }^2\theta /16$. Its inverse components needed for the flux areThereforesince $h(1-\alpha \Omega )=1$. This calculation includes the rotational mixed term; dropping it prematurely would miss the exact cancellation.

With the stipulated negative coordinate orientation, ${ϵ}_{tr\psi \theta \varphi }=-r^3\sin \theta /4$. The pullback of the Hodge star operator to the constant-$t,r$ three-dimensional spacelike submanifold is consequentlyThe coordinate ranges and standard sphere identifications giveThus the five-dimensional Komar mass, in the normalization given and $G_5=1$, isThe flux is already independent of $r$ in the vacuum region, so its limit at infinity has the same value. The negative orientation in the question is crucial for the positive sign; reversing that orientation reverses the flux.

If $M$ has complex dimension $n$, its complexified cotangent bundle splits into the $(1,0)$ and $(0,1)$ cotangent bundles. The space of global smooth differential forms of type (p, q) isIn holomorphic coordinates, its elements have the form $\alpha ={\sum }_{I,J}{a}_{I,J}d{z}^I\wedge d{\bar{z}}^J$, where the coefficients are complex-valued smooth functions, $|I|=p$, and $|J|=q$.

The exterior derivative decomposes as $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator $\partial$ and Dolbeault operator $\bar{\partial }$ are its type components:They have bidegrees $(1,0)$ and $(0,1)$ respectively. Keeping the displayed wedge order avoids hiding a sign; moving $d{\bar{z}}_k$ past $d{z}^I$ introduces $(-1{)}^p$. Decomposing $d^2=0$ by type gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

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.

Take the usual convention that the smooth isotopy starts at ${\rho }_0=\mathrm{id}$. The exterior derivative commutes with a pullback of a differential form, soConsequentlyThe right-hand condition says precisely that ${\rho }_t$ is a symplectic isotopy.

Here an exact symplectic manifold is necessarily noncompact in positive dimension: if $\dim M=2n$ and $\omega =-d\lambda$, then ${\omega }^n=-d(\lambda \wedge {\omega }^{n-1})$, contradicting the positive symplectic volume and the Generalized Stokes theorem on a compact smooth manifold without boundary. Thus the paper's introductory word “closed” must here be read as “without boundary”, as its parenthesis suggests, rather than as including compactness.

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.

The Levi-Civita connection of a Riemannian manifold $(M,g)$ is the unique connection on a vector bundle on $TM$ that is torsion-free and compatible with the Riemannian metric. Any such connection must satisfy the Koszul formulaNondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from the right-hand side. Conversely, define $\nabla$ by this formula. Direct substitution shows that it is $C^{\infty }$-linear in $X$, satisfies the Leibniz rule in $Y$, preserves $g$, and obeys ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. It is therefore a torsion-free metric connection. This proves the existence and uniqueness of the Levi-Civita connection.

For the metric vector bundle $E$, choose the stated orthonormal local frame $(e_1,\dots ,e_m)$ and write $d_A{e}_j=A^i{}_j{e}_i$. Since $⟨e_i,e_j⟩={\delta }_{ij}$, metric compatibility givesHence the connection matrix in an orthonormal frame is skew-symmetric:

On an oriented Riemannian $d$-manifold, the Hodge star operator is defined byIt is an orthogonal map $\ast :{\Lambda }^r{T}_x^{\ast }M\to {\Lambda }^{d-r}{T}_x^{\ast }M$ and satisfies ${\ast }^2=(-1{)}^{r(d-r)}$. In dimension $d=2n$ on middle-degree forms, the Adjoint of the Hodge star on middle-degree forms isThus it is self-adjoint when $n$ is even, but it is not always self-adjoint. For $n=1$ on the oriented Euclidean plane,so its matrix in the orthonormal basis $(dx,dy)$ is skew-adjoint.

The Laplace-Beltrami operator on differential forms is the Hodge Laplacianwhere the codifferential $\delta$ is the formal $L^2$ adjoint of the exterior derivative. On a compact manifold without boundary, if $\Delta \omega =\lambda \omega$ for a nonzero differential form $\omega$, then integration by parts givesTherefore the nonnegativity of the Hodge Laplacian yields

On the Euclidean unit ball $D^n$,The outward unit normal along $S^{n-1}$ is the radial vector field $N={\sum }_i{x}^i{\partial }_i$, so part b givesLet $R={\sum }_i{x}^i{\partial }_i$ on the ball and put $\beta ={\iota }_R{\omega }_X$. Direct use of the exterior derivative gives $d\beta =n{\omega }_X$. Therefore the Generalized Stokes theorem yieldsThis proves the volume of a Euclidean unit sphere formula.

An almost complex structure on a smooth manifold $M$ is a smooth bundle endomorphism $J:TM\to TM$ satisfying $J^2=-1$. On the complexified tangent bundle, $J$ has the eigenbundle decompositionwith eigenvalues $i$ and $-i$. A differential form of type (p, q) is a section of ${\bigwedge }^p(T^{1,0}M{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}M{)}^{\ast }$. The operators $\partial$ and $\bar{\partial }$ are the type-$(1,0)$ and type-$(0,1)$ components of the exterior derivative.

We prove the three stated conditions are equivalent. If $X,Y$ have type $(1,0)$ and $\alpha$ has type $(0,1)$, the formula for the exterior derivative givesThus closure of $T^{1,0}M$ under the Lie bracket is equivalent to the vanishing of the $(2,0)$ component of $d\alpha$ for every $(0,1)$-form. Complex conjugation of differential-form type gives the corresponding vanishing of the $(0,2)$ component on $(1,0)$-forms. Since every complex one-form is a sum of these two types, this is preciselyThis proves (i)$⟺$(ii).

Under (ii), the $(2,0)$ component of $d^{2}f=0$ is ${\partial }^{2}f$, proving (iii). Conversely, for $X,Y$ of type $(1,0)$,If this vanishes for every smooth complex-valued $f$, then $[X,Y{]}^{0,1}=0$, so (iii) implies (i). These conditions define an integrable almost complex structure.

For a complex manifold, a holomorphic chart $z^j=x^j+i{y}^j$ defines $J_M({\partial }_{x^j})={\partial }_{y^j}$ and $J_M({\partial }_{y^j})=-{\partial }_{x^j}$. The derivative of a holomorphic coordinate change is complex linear, hence commutes with multiplication by $i$; the definitions therefore glue and are independent of coordinates. Locally $T^{1,0}M$ is spanned by the commuting fields $\partial /\partial z^j$, so it is closed under brackets and $J_M$ is integrable. This is the almost complex structure induced by a complex atlas.

We next prove the local Dolbeault-Poincaré lemma. Write a $\bar{\partial }$-closed $(0,q)$-form on a slightly larger polydisc aswhere neither $\alpha$ nor $\beta$ contains $d{\bar{z}}_n$. Apply the supplied one-variable Cauchy-Green operator coefficientwise to $\beta$, obtaining $\gamma$ with $\partial \gamma /\partial {\bar{z}}_n=\beta$. Then $\phi -\bar{\partial }\gamma$ contains no $d{\bar{z}}_n$, and $\bar{\partial }\phi =0$ says that its coefficients are holomorphic in $z_n$ and $\bar{\partial }$-closed in the first $n-1$ variables. Induction on $n$, with the one-variable formula as the base case, makes this remainder $\bar{\partial }$-exact. Hence every $\bar{\partial }$-closed $(0,q)$-form with $q>0$ is $\bar{\partial }$-exact on each bounded polydisc.

Finally, translation by $i$ leaves $d\bar{z}$ unchanged, so this form descends to $\mathbb{C}/(z\sim z+i)$. It is in fact exact on this noncompact complex cylinder, because the invariant function $z+\bar{z}=2\mathrm{Re}z$ descends and satisfies

In local coordinates, forthe exterior derivative isApplying $d$ again gives symmetric second partial derivatives contracted with the antisymmetric wedge $d{x}^k\wedge d{x}^j$, hence $d^2=0$. Expanding coefficients and moving $d\alpha$ past a degree-$r$ form gives the graded Leibniz rule

For a function $f$ and smooth $F:X\to Y$, the chain rule givesEvery differential form is locally a sum of products $f_{0}d{f}_1\wedge \cdots \wedge d{f}_r$. Since pullback preserves products and wedge products, the function case and the graded Leibniz rule implyfor every form.

Let $(z^1,\dots ,z^n)$ be a complex manifold chart, with $z^j=x^j+i{y}^j$. The almost complex structure induced by a complex atlas isThus $J^2=-I$. On an overlap, the derivative of a holomorphic map is complex linear and hence commutes with multiplication by $i$. The two coordinate definitions of $J$ therefore agree, so $J$ is globally well-defined.

The complexified cotangent bundle splits into the $i$- and $-i$-eigenspaces of $J^{\ast }$, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a sumSplitting the exterior derivative according to type definesComplex conjugation sends $d{z}^j$ to $d{\bar{z}}^j$ and conjugates the coefficient derivatives. Term by term this gives the complex conjugation of differential-form type identity

Let $J$ act on a $k$-form byIt acts on a $(p,q)$-form by $i^{p-q}$. Therefore $J^{-1}dJ$ multiplies the $\partial$ component by $-i$ and the $\bar{\partial }$ component by $i$, proving the d c operator formula

A holomorphic vector field is a holomorphic section of $T^{1,0}X$. On the affine chart $U_0\subset {\mathbb{CP}}^n$, put $w_k=z_k/z_0$. The projection isDirect differentiation givesIf $\xi (z)$ is linear and homogeneous, the pushed-forward coefficients are consequently $\xi (1,w)$ when $j\ne 0$, and $-\xi (1,w)w_k$ when $j=0$. They are holomorphic on $U_0$.

More intrinsically, $\xi (z){\partial }_{z_j}$ is a linear vector field $Az$ on ${\mathbb{C}}^{n+1}$. Its flow $e^{tA}$ preserves complex lines and induces the projective transformationsDifferentiating gives a globally defined projectivization of a linear vector field, whose expression on $U_0$ is the field just computed. This proves extension across the other affine charts.

Finally choose distinct complex numbers ${\lambda }_0,\dots ,{\lambda }_n$ and the diagonal fieldIts projectivization vanishes at $[z]$ precisely when $({\lambda }_0{z}_0,\dots ,{\lambda }_n{z}_n)$ is proportional to $z$, so $[z]$ is an eigenline. The distinct eigenvalues leave exactly the $n+1$ coordinate points. Hence ${\mathbb{CP}}^n$ has a holomorphic vector field with finitely many zeroes.

For any differential form $\omega$, Cartan's magic formula givesUsing $d^2=0$,Thus the Lie derivative commutes with the exterior derivative on every differential form.

On $p$-forms in dimension $n$, the defining identity for the Hodge star operator givesFor $n=4$ and $p=2$, therefore, ${\ast }^2=1$. For every $\alpha \in {\Omega }^2(N)$ defineThen $\ast {\alpha }_{+}={\alpha }_{+}$, $\ast {\alpha }_{-}=-{\alpha }_{-}$, and $\alpha ={\alpha }_{+}+{\alpha }_{-}$. The two eigenspaces of the involution $\ast$ have zero intersection, which proves uniqueness. They are respectively the spaces of self-dual and anti-self-dual two-forms.

Now suppose $N$ is compact and let $\beta$ be an exact three-form, say $\beta =d\theta$. Apply the Hodge decomposition theorem to the two-form $\theta$:Set $a=\delta \psi$. Then $da=d\theta =\beta$ and $\delta a={\delta }^2\psi =0$. For a two-form in dimension four, $\delta =-\ast d\ast$, so $d\ast a=0$. The self-dual formsatisfiesThus every exact three-form is the exterior derivative of a self-dual two-form.

A diffeomorphism $\varphi :M\to M$ is a smooth bijection with smooth inverse. It acts covariantly on a differential form by the pullback of a differential form:For vector fields $X_0,\dots ,X_p$,The coordinate definition of the exterior derivative, or its characterization as the unique natural graded derivation extending the differential of functions, gives

A smooth map between manifolds $f:M\to N$ pulls a function back by composition and pulls a covariant tensor back by applying its differential to each vector argument. The pullback of a differential form is the alternating-tensor case and commutes with the exterior derivative.

A smooth map $\varphi :M\to N$ pulls a differential form on $N$ back to one on $M$ by applying its differential $d\varphi$ to each tangent vector argument. In particular, a parametrization of a curve pulls a one-form back to an ordinary integrand on an interval.

For $\alpha =xdy$ and the parabola $\varphi (t)=(t,t^2)$, $0\le t\le 1$, substitute $x=t$ and $dy=2tdt$ to obtainThe same endpoints gave $0$ and $1$ along the two square-edge paths in the introduction, so this third answer is consistent with path dependence. By contrast, ${\varphi }^{\ast }d(xy)=d(t^3)=3t^{2}dt$ gives integral $1$, the same endpoint difference as before.

The pullback of a differential form commutes with the wedge product of differential forms and the exterior derivative: ${\varphi }^{\ast }(\alpha \wedge \beta )={\varphi }^{\ast }\alpha \wedge {\varphi }^{\ast }\beta$ and ${\varphi }^{\ast }(d\alpha )=d({\varphi }^{\ast }\alpha )$. These identities make changes of coordinates and parametrizations compatible with differentiation and integration.

It is the alternating-tensor case of pullback.

The four-dimensional membrane Fierz rearrangement identity $(C{\gamma }_{mn}{)}_{(\alpha \beta }(C{\gamma }^n{)}_{\gamma \delta )}=0$ makes the displayed supertranslation-invariant four-form closed. It is locally the exterior derivative of a three-form potential, which can change by an exact form under target supersymmetry. Pulling the potential back to a closed worldvolume gives an invariant Wess-Zumino brane coupling. The charge magnitude and normalization of this form are fixed together by cancellation of the area variation under kappa symmetry.