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Use coordinate charts from the chosen oriented atlas. In each, the matrix $G_x=(g_{ij})$ of the Riemannian metric is smooth and positive-definite, so $\det G_x>0$ and its positive square root is smooth. Define the local Riemannian volume form byIf $y$ is another positive coordinate chart and $J=\partial x/\partial y$, the transformation rule for the metric tensor is $G_y=J^T{G}_{x}J$. Taking determinants givesThe oriented atlas has $\det J>0$, so the wedge product of differential forms transformation from part (a) givesThe local expressions therefore glue to a global smooth top-degree differential form. Its local coefficient never vanishes, so it is a volume form. The global answer is $\boxed{{\omega }_g=\sqrt{\det (g_{ij})}d{x}^1\wedge \cdots \wedge d{x}^n}$ in positive charts. In an arbitrary negatively oriented chart the corresponding expression needs a minus sign; the orientation is essential to the form, although a positive metric volume density exists without it.

Let $M$ be a closed manifold of even dimension $n=2m$, with a Riemannian metric and a spin structure. Let $S=S^{+}\oplus S^{-}$ be its spinor bundle, and let $W$ be a Hermitian vector bundle with a unitary connection of vector-bundle curvature $F^W=({\nabla }^W{)}^2$. We use Clifford multiplication satisfying $c(v)c(w)+c(w)c(v)=-2⟨v,w⟩$, so the Twisted Dirac operatoris self-adjoint and $D_W^{-}=(D_W^{+}{)}^{\ast }$. Here $(e_i)$ is an oriented local orthonormal basis and $(e^i)$ its dual. The Atiyah-Singer index theorem in this setting isThe second factor is the Chern character; the first is the A-hat formThe square root has constant term one. All functions of the vector-bundle curvature matrices mean power series truncated above the dimension of $M$. The brackets select the top-degree differential form. The Bianchi identity shows that these forms are closed, and Chern-Weil connection transgression shows that their de Rham cohomology classes do not depend on the chosen connections on a vector bundle.

The global step is supersymmetric quantum mechanics. On the Hilbert space $L^2(S\otimes W)$, take the grading operator $\Gamma$, the odd supercharge $D_W$, and the nonnegative Hamiltonian $H=D_W^2$. They obey $\Gamma D_W=-D_W\Gamma$ and $D_W^2=H$. The principal symbol of $D_W$ is invertible away from the zero covector, so elliptic regularity and Rellich-Kondrachov compactness theorem give a smooth discrete orthonormal eigenbasis for $H$, finite-dimensional eigenspaces, and a trace-class operator $e^{-tH}$ for $t>0$. Define its supertrace by $\mathrm{Str}A=\mathrm{Tr}(\Gamma A)$.

For every positive eigenvalue $\lambda$ of $H$, the map ${\lambda }^{-1/2}{D}_W$ is an isomorphism between its positive and negative grading eigenspaces, with inverse itself. Their contributions to the supertrace cancel. The zero eigenspaces are $\ker D_W^{+}$ and $\ker D_W^{-}$. ConsequentlyThis also proves independence of $t$ without differentiating an unbounded operator inside a trace. If $K_t$ is the heat kernel, the heat kernel trace formula givesThe remaining task is to compute this local integrand as $t\downarrow 0$.

Fix $p\in M$, use geodesic normal coordinates $x$, and trivialize $S\otimes W$ by radial parallel transport. Expanding $D_W^2$ at $p$ gives the Lichnerowicz formula for a twisted Dirac operator:Indeed the terms with identical indices give the rough Laplacian. The other terms are ${\sum }_{i<j}c(e^i)c(e^j)[{\nabla }_i,{\nabla }_j]$. Decompose this vector-bundle curvature into spin curvature and $F^W$; substituting the spin curvature into the Clifford algebra relations and the algebraic Bianchi identity leaves $\mathrm{Scal}/4$. Thus the calculation retains the curvature of the twisting bundle explicitly.

There is an especially strong cancellation in the spinor supertrace. With $\Gamma =i^{m}c(e^1)\cdots c(e^{2m})$, every ordered product of distinct Clifford generators of degree less than $2m$ has zero supertrace, whileFor an even product of lower degree, conjugate $\Gamma$ times that product by a missing Clifford generator: it changes sign, so its matrix trace is zero. An odd product exchanges the grading summands and also has zero supertrace. For the full product, $(c(e^1)\cdots c(e^{2m}){)}^2=(-1{)}^m$, and $\dim S=2^m$, giving the displayed constant.

To exploit this, identify the Clifford algebra as a filtered vector space with the exterior algebra by sending an ordered Clifford product to the corresponding wedge product of differential forms. Implement Getzler rescaling: put $x=\epsilon y$, $t={\epsilon }^{2}s$, multiply the degree-$k$ exterior component of the kernel by ${\epsilon }^{-k}$, and multiply the whole kernel by ${\epsilon }^{2m}$. A derivative and a Clifford generator have rescaling degree one; a coordinate has degree minus one. Clifford contractions lower the degree by two relative to the wedge product of differential forms, so the leading product becomes exterior multiplication.

In the radially parallel frame, the linear Taylor term of the connection is determined by its vector-bundle curvature: if $A_i$ is its matrix, then $A_i(x)=\frac{1}{2}{x}^j{F}_{ji}(p)+O(|x{|}^2)$. To check the coefficient, radial gauge means $x^i{A}_i=0$, and integration of $x^j{F}_{ji}$ along the radial segment gives $A_i(x)={\int }_0^{1}s{x}^j{F}_{ji}(sx)ds$. Combining this formula with the spin representation, the symmetries of Riemann curvature tensor, and the Lichnerowicz formula for a twisted Dirac operator, the rescaled square has leading modelHere $\mathcal{R}$ is the skew matrix of tangent-bundle curvature two-forms at $p$; changing the curvature-matrix sign convention changes the sign inside the covariant derivatives but not the determinant below. The coefficients act in $\Lambda T_p^{\ast }M\otimes \mathrm{End}(W_p)$. The scalar-curvature term has degree zero and disappears after multiplication by ${\epsilon }^2$; the linear spin-connection term and the Clifford action of $F^W$ have degree two and survive. Higher Taylor terms have smaller degree and disappear.

For clarity, the analytic meaning of this limiting calculation is the following. Construct the local heat kernel expansion to any fixed order and insert the Taylor expansion of the rescaled coefficients into its transport equations. At each exterior degree, the leading equation is the heat equation of ${\mathcal{H}}_p$; the remaining terms acquire positive powers of $\epsilon$. Equivalently, successive substitutions ingive the same expansion. Polynomial spatial factors are integrable against the local Gaussian kernels, and only finitely many exterior degrees exist. A cutoff outside a fixed normal neighbourhood contributes $O(e^{-c/t})$. Choosing the initial expansion order arbitrarily large bounds its differentiated remainder to any required order. This proves convergence of the rescaled diagonal kernel to the model diagonal kernel, uniformly in $p$ on the compact manifold. Its degree-$2m$ component has no net power of $\epsilon$, precisely the component detected by the spinor supertrace.

We can compute the model kernel rather than invoke a local index formula. Set $F=F_p^W$ and seek its kernel from the origin in the formThe scalar even exterior coefficients of $\mathcal{R}$ commute. On this ansatz the rotational first-order term vanishes because $Q$ is a function of ${\mathcal{R}}^2$. Expanding the model operator gives $-\sum {\partial }_i^2+\frac{1}{2}(\mathcal{R}y)\cdot \nabla +\frac{1}{16}{y}^T{\mathcal{R}}^{2}y+F$, so substitution into the heat equation givesThe Euclidean initial condition $Q(s)\sim s^{-1}I$, $a(s)\sim (4\pi s{)}^{-m}$ selectsThese expressions have well-defined power series at $\mathcal{R}=0$. This is the curvature version of the Gaussian calculation underlying the Mehler formula for Hermite functions.

At $s=1$, $y=0$, multiply the top exterior coefficient by the spinor supertrace constant. Since $(4\pi {)}^{-m}(-2i{)}^m=(2\pi i{)}^{-m}$, and every curvature factor is a two-form, the result isIntegration and the constant supertrace established above prove the boxed Atiyah-Singer index theorem. The mechanism is that odd symmetry pairs the nonzero spectrum globally, while the Clifford filtration makes the surviving local contribution an explicitly solvable curvature oscillator. Odd-dimensional spin Dirac operators have no canonical chiral index of this type; their self-adjoint Fredholm index is zero. For comparison of normalization conventions, Raphaël Ponge's local index calculation writes the same unnormalized curvature model and spinor supertrace constant.

Products below combine matrix multiplication with the wedge product of differential forms of differential forms. Differentiate the projection identity $p^2=p$ to getMultiplying by $p$ on both sides also gives $pdpp=0$; applying the same argument to $1-p$ gives $(1-p)dp(1-p)=0$. Thus at each point $dp$ is off-diagonal relative to $\mathrm{im}p\oplus \ker p$, whereas $(dp{)}^2$ is diagonal and commutes with $p$. For $\omega =p(dp{)}^2$,because $d(dp)=0$.

The graded Leibniz rule and cyclicity of the matrix trace give, since $\omega$ has even degree,Here $(d\omega ){\omega }^{k-1}$ is off-diagonal: $(dp{)}^3$ is off-diagonal and every power of $\omega$ is diagonal. Its matrix trace is zero. Hence $\boxed{d\mathrm{tr}({\omega }^k)=0}$ for every $k\ge 1$; for $k=0$ the trace is the constant matrix size.

For the determinant, work over the commutative algebra of even differential forms. If $f(0)=1$, write $f(\omega )=I+N$, where $N$ has strictly positive degree and is nilpotent for dimensional reasons. The finite power series identityholds over this algebra, as follows either from the formal determinant identity over characteristic-zero commutative rings or by differentiating $\det (I+sN)$ in the scalar variable $s$. Each $N^j$ is a linear combination of positive powers of $\omega$. The matrix trace of the logarithm is therefore a closed differential form, and its finite exponential is closed by the graded Leibniz rule. Thus $\boxed{d\det f(\omega )=0}$. The projection connection interprets $\omega$ as the curvature of $pd$ on $\mathrm{im}p$, but the proof above directly establishes both closedness claims.

The space of smooth differential forms of degree $p$ isThus a form assigns smoothly to each point an alternating $p$-linear function on its tangent space. Here ${\Omega }^0(M)=C^{\infty }(M)$, and forms of degree greater than $\dim M$ vanish. The wedge product of differential forms gives the graded exterior algebra of forms.

The exterior derivative is an $\mathbb{R}$-linear map of degree one satisfying the following axioms: $df(X)=X(f)$ for functions; for a $p$-form $\alpha$,and $d^2=0$. The sign is the graded Leibniz rule.

To construct it, write in a coordinate chartThis formula is linear, agrees with differentiation on functions and obeys the graded Leibniz rule. Its square vanishes because symmetric second partial derivatives are contracted against antisymmetric wedge products.

The construction is independent of the chart. For coordinates $y^j=y^j(x)$, the chain rule gives $d{y}^j={\sum }_i({\partial }_i{y}^j)d{x}^i$, and the coordinate formula gives $d(d{y}^j)=0$ by equality of mixed partial derivatives. Consequently, expressing a form as ${\sum }_I{b}_{I}d{y}^I$ and applying the $x$-coordinate formula gives exactly ${\sum }_{I}d{b}_I\wedge d{y}^I$, the $y$-coordinate formula. The local constructions agree on overlaps and therefore define a global exterior derivative.

For uniqueness, first derive locality from the axioms. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ supported in that neighborhood and equal to one near $p$. Then $\chi \alpha =0$, andimplies $(d\alpha {)}_p=0$. Thus the value depends only on the local germ of $\alpha$. Local coordinate functions can be extended using bump functions, so the same axioms apply to their germs. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forces precisely the coordinate formula above. The exterior derivative exists and is unique, as expressed by uniqueness of the exterior derivative from its axioms.

The connection difference as an endomorphism-valued one-form follows because the graded Leibniz rule terms cancel:Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the endomorphism bundle connection, requireThe right side is $C^{\infty }$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\varphi$ by $f\varphi$ gives $df\otimes \varphi +f\tilde{D}\varphi$. In a frame with connection matrix $A$, its extension to an endomorphism-valued $k$-form $b$ isThe endomorphism-valued exterior product combines the wedge of differential forms with composition in $\mathrm{End}E$. In particular,Expanding $d(A+a)+(A+a)\wedge (A+a)$ proves the curvature difference formula:

The Chern connection is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. Take Hermitian inner products linear in the first variable. In a holomorphic local frame write $⟨eu,ev⟩=v^{†}Hu$, with $H$ a positive Hermitian matrix. The second condition forces $A^{0,1}=0$, and metric compatibility isIts $(1,0)$ component determines $A$ uniquely:Conversely this formula and its conjugate transpose satisfy the metric equation. Under a holomorphic local frame change, $H^{\prime }=g^{†}Hg$ and $\partial g^{†}=0$, so the displayed formula transforms by $A^{\prime }=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.

Differentiating $H^{-1}$ givesTherefore the $(2,0)$ vector-bundle curvature is zero. There is no $(0,2)$ component, andThe last equality is ${\bar{\partial }}^{2}A=0$ in holomorphic local frames, so is intrinsic. This is Dolbeault closedness of Chern curvature. The induced endomorphism bundle connection is itself the Chern connection for the Hilbert-Schmidt metric on $\mathrm{End}E$: the tensor product of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic Dolbeault operator. These two properties and uniqueness suffice; no further matrix calculation is needed.

For two metrics, both Chern connections have the same $(0,1)$ part, so $a\in A^{1,0}(\mathrm{End}E)$. The $(1,1)$ part of the curvature difference formula is just ${\bar{\partial }}_{\mathrm{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both vector-bundle curvatures have type $(1,1)$. ThusBy Dolbeault theorem, this proves that the curvature Dolbeault classis independent of the metric.

Endomorphism composition and wedge product of differential forms define the powers in $A^{k,k}(\mathrm{End}E)$, rather than in a tensor power of the endomorphism bundle. The Dolbeault operator is a derivation, so $\bar{\partial }({\Theta }^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statementAll powers of vector-bundle curvature have even total degree, so the derivation introduces no extra signs before $a$. Consequentlyis metric independent. For $k>n$, the form space and the resulting class are zero. This proves metric independence of curvature Dolbeault powers without assuming that endomorphism-valued forms commute.

Taking the exterior derivative of the left-invariant differential forms yieldsFor example, $d({\rho }^{-1}d{x}^1)=-{\rho }^{-2}d\rho \wedge d{x}^1$. With the antisymmetric structure constants of a Lie algebra found above, the two terms with indices $(0,1)$ and $(1,0)$ combine asThe same computation holds for superscript $2$, and all superscript-$0$ terms vanish. Hence the Maurer-Cartan equation isThe factor $1/2$ avoids counting each antisymmetric pair twice. If coefficients were not required to be antisymmetric in $j,k$, their symmetric parts would be invisible to the wedge product of differential forms; the standard structure constants of a Lie algebra fix this ambiguity.

The Hamiltonian flow of $H$ solvesCompactness and the absence of a boundary make this smooth vector field complete, so the flow exists for all real $t$. The defining equation and Cartan's magic formula giveTherefore the pullback of a differential form differentiation rule yieldsso $({\varphi }_H^t{)}^{\ast }\omega =\omega$. Pullback commutes with the wedge product of differential forms, and consequentlyThus the flow preserves the symplectic volume, in fact the entire symplectic form.

Use real differential forms and the usual convention that a smooth manifold has no boundary unless one is specified. On a noncompact manifold the formal-adjoint identity below uses compactly supported forms. On a compact manifold it has no boundary contribution; if a boundary were allowed, extra boundary conditions would be necessary.

The Riemannian metric gives the dual inner product on $T_x^{\ast }M$ and hence the inner product on exterior powers of a cotangent space:Extend this bilinearly. Equivalently, the increasing wedge products of an orthonormal coframe are an orthonormal basis of ${\Lambda }^p{T}_x^{\ast }M$. The Hodge star operator is the unique map $\ast :{\Lambda }^p{T}_x^{\ast }M\to {\Lambda }^{n-p}{T}_x^{\ast }M$ satisfying $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$. If $e^1,\dots ,e^n$ is a positively oriented orthonormal coframe, $\ast e^I=\epsilon (I,I^c)e^{I^c}$. Moving the $p$ factors of $I$ across the $n-p$ factors of its complement changes the sign by $(-1{)}^{p(n-p)}$, soThe basis formula also proves that $\ast$ is an isometry.

For $\beta \in {\Omega }^{p-1}(M)$ and $\alpha \in {\Omega }^p(M)$ with compactly supported product, the Stokes theorem and the Leibniz rule for the exterior derivative giveThus $(d\beta ,\alpha {)}_{L^2}=(-1{)}^p{\int }_M\beta \wedge d\ast \alpha$. The formal adjoint operator $\delta$ satisfies $\ast \delta \alpha =(-1{)}^{p}d\ast \alpha$. Applying the inverse star and its square on degree $p-1$ gives the codifferentialThe exponent is congruent to $p+(p-1)(n-p+1)$ modulo two, so this convention is consistent in every degree. Define the Hodge Laplacian by $\Delta =d\delta +\delta d$ and a harmonic differential form by $\Delta \alpha =0$. On a compact manifold,Both nonnegative terms vanish for a harmonic form, giving $d\alpha =\delta \alpha =0$. Conversely these two equations imply harmonicity.

Now let $n=4$. The printed eigenvalue relation omits the subscript on its right-hand side; the intended equation is $\ast {\alpha }_{\pm }=\pm {\alpha }_{\pm }$. On two-forms ${\ast }^2=1$, soThese are respectively self-dual two-forms and anti-self-dual two-forms, and sum to $\alpha$. Their uniqueness follows because the $+1$ and $-1$ eigenspaces have zero intersection. Star is an orthogonal involution here, hence is self-adjoint and its opposite eigenspaces are orthogonal.

For closed two-forms define $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$. Adding $d\xi$ to $\alpha$ changes the integral by ${\int }_{N}d(\xi \wedge \beta )=0$; adding an exact form to $\beta$ has the same effect. This is the real de Rham intersection form in dimension four. The wedge product of differential forms in degrees two and two is symmetric because $(-1{)}^{2\cdot 2}=1$; hence $Q$ is a symmetric bilinear form on $H_{\mathrm{dR}}^2(N;\mathbb{R})$.

In four dimensions $\delta =-\ast d\ast$ in every degree. If $h$ is a harmonic two-form, $\delta h=0$ gives $d\ast h=0$, and $dh=0$ gives $\delta \ast h=-\ast dh=0$. Thus $\ast h$ is harmonic, and ${\mathcal{H}}^2={\mathcal{H}}^{+}\oplus {\mathcal{H}}^{-}$ is the harmonic star-eigenspace decomposition. Use the permitted unique harmonic representative for each de Rham cohomology class. For harmonic $h,k$,The mixed terms vanish by orthogonality. The first restriction is positive definite and the second negative definite. In particular a nonzero harmonic $h$ has $Q(h,\ast h)=\Vert h{\Vert }_{L^2}^2>0$, proving nondegeneracy. Hence the signature of the intersection form from harmonic duality isThe finite dimensions and the harmonic description also follow from the Hodge decomposition theorem.

For completeness, the Hodge decomposition theorem on a compact oriented Riemannian manifold without boundary states the finite-dimensional harmonic space and the $L^2$-orthogonal direct sumThe three summand components are unique, although their potentials need not be. Given an exact three-form $\beta =d\omega$ on $N$, decompose its smooth two-form potential as $\omega =h+d\xi +\delta \psi$, with $\psi \in {\Omega }^3$. Since $dh=0$ and $d^2=0$, $\beta =d\delta \psi$. Put $a=\delta \psi$. In dimension four, $a=-\ast d\ast \psi$, and ${\ast }^2=1$ on two-forms gives $\ast a=-d\ast \psi$, so $d\ast a=0$. ConsequentlyThis constructs a self-dual primitive of an exact three-form. The sum contains no factor $1/2$: it is twice the self-dual projection of the coexact potential. Using the unscaled projection alone would give $d\sigma =\beta /2$. This proves the final unheaded request.

The complex structure splits the complexified cotangent bundle into its $(1,0)$ and $(0,1)$ parts. Thus the space of smooth differential forms of type (p, q) isHere $\Gamma$ denotes smooth sections of a vector bundle; exterior multiplication identifies the displayed bundle with a subbundle of the complexified exterior algebra. In holomorphic coordinates, such a differential form of type (p, q) has the expressionwhere the coefficients are smooth and both multi-indices are increasing. Negative bidegrees, or bidegrees exceeding the complex dimension, give the zero space.

On a complex manifold, the exterior derivative has just two type components, $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator and the Dolbeault operator are, respectively,Keeping the differentiating factor at the front fixes the signs. These definitions are intrinsic because holomorphic coordinate changes preserve type. The two operators are the type projections of the exterior derivative, raising $p$ and $q$, respectively.

For a holomorphic map $f:M\to N$, its differential is a complex-linear map, so the pullback of a differential form sends $(1,0)$ covectors to $(1,0)$ covectors and $(0,1)$ covectors to $(0,1)$ covectors. In coordinates $w$ on $N$,The pullback of a differential form commutes with the wedge product of differential forms, so it preserves the counts of these factors. ConsequentlyIt also commutes separately with the Dolbeault operator and the conjugate Dolbeault operator, because it commutes with $d$ and preserves type.

On ${\mathbb{C}}^{n+1}\setminus \{0\}$, let $\Omega =d{z}_0\wedge \cdots \wedge d{z}_n$ and let $E={\sum }_i{z}_i{\partial }_{z_i}$ be the Euler vector field. The numerator is its interior product of a differential form, soBoth numerator and denominator have homogeneous scaling degree $n+1$. Thus this meromorphic $n$-form is invariant under constant nonzero scalings, and ${\iota }_E\alpha =0$ makes it horizontal for the quotient to Complex projective space.

To check descent using the hint, take a local holomorphic section $Z$ of the quotient and another section $Z^{\prime }=\lambda Z$, where $\lambda$ is a nowhere-zero holomorphic function. In the evaluationevery term involving $d\lambda$ has two collinear arguments and vanishes. The remaining term is ${\lambda }^{n+1}{Z}^{\ast }({\iota }_E\Omega )$, while $f(\lambda Z)={\lambda }^{n+1}f(Z)$. Hence $Z^{\prime \ast }\alpha =Z^{\ast }\alpha$. The forms therefore glue to a meromorphic section of the canonical bundle of Complex projective space.

On the affine chart $z_j\ne 0$, use the section $z_j=1$ and list the remaining coordinates as $t_1,\dots ,t_n$ in increasing original-index order. If $F$ is the resulting polynomial, thenAt a point of $Y$, some derivative $F_{t_r}$ is nonzero. Otherwise all the homogeneous derivatives except possibly $f_{z_j}$ would vanish; the Euler homogeneous function theorem gives ${\sum }_i{z}_i{f}_{z_i}=(n+1)f=0$ there, forcing $f_{z_j}=0$ too. This contradicts the permitted smoothness criterion. The holomorphic inverse function theorem now makes $F$ a transverse coordinate, so the pole is simple. On a neighbourhood with $F_{t_r}\ne 0$, rearranging the wedge product of differential forms givesBy part (a), its Poincaré residue is consequentlyThe remaining $t$ coordinates form a holomorphic coordinate system on $Y$, and $F_{t_r}$ is nonzero in this neighbourhood. The displayed top form is therefore nowhere zero. These neighbourhoods cover $Y$ and part (a) guarantees agreement on their overlaps. The residue is a nowhere-vanishing global holomorphic section:This is the explicit residue trivialization for a degree n+1 projective hypersurface, consistent with the adjunction formula $K_Y\cong {\mathcal{O}}_Y((n+1)-n-1)$, but obtained here directly from the indicated form.

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 $p=0$:

Let $\sigma (z)=\overline{z}$, componentwise complex conjugation. Interpret the printed $|z^2|$ in the potential as $|z{|}^2={\sum }_j|z_j{|}^2$. With $D=1+|z{|}^2$, the local Fubini-Study form isIts pullback of a differential form under $\sigma$ iswhere the last equality swaps $j,k$ and uses antisymmetry of the wedge product of differential forms. Since ${\sigma }^2=I$, it is an anti-symplectic involution, with fixed-point set ${\mathbb{R}}^n$. Part (c) gives

The wedge product of differential forms is the alternating tensor productFor nonzero $\eta \in {\Lambda }^{n-1}(({\mathbb{R}}^n{)}^{\ast })$, choose a volume form $\mu$. There is a nonzero vector $v$ with $\eta ={\iota }_v\mu$. Extend $v$ to a basis and use its dual coframe; then $\eta$ is a scalar multiple of $e^2\wedge \cdots \wedge e^n$, hence equals $\xi \wedge {\eta }_0$. The zero form is immediate.

To integrate a top form on a compact oriented $n$-manifold, choose a finite oriented atlas and a subordinate partition of unity; integrate each compactly supported coordinate expression and sum. A smooth map pulls forms back byIf $F^{\ast }\omega =\omega$ for a nowhere-zero top form, $F$ preserves its orientation. The change-of-variables theorem gives

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 Hodge star operator is self-adjoint on real two-forms in four Euclidean dimensions, so its opposite eigenspaces are orthogonal. Since $H\wedge G=⟨H,\ast G⟩{\mathrm{vol}}_g$, a self-dual two-form and an anti-self-dual two-form also have zero wedge product of differential forms. Applied to Lie-algebra components with an invariant pairing, this identity eliminates mixed terms from the Yang-Mills action.