The Exterior Derivative and Cartan Calculus

Replace $X_i$ by $f{X}_i$. For $a\mathrm{\ne }i$, the $a$th derivative term contributes $(-1{)}^a(X_{a}f)\omega ({\widehat{X}}_a)$. If $a<i$, the $(a,i)$ bracket term contributes $(-1{)}^{a+i}(X_{a}f)\omega (X_i,X_0,\dots ,{\widehat{X}}_a,\dots ,{\widehat{X}}_i,\dots ,X_k)=-(-1{)}^a(X_{a}f)\omega ({\widehat{X}}_a)$, because moving $X_i$ to its usual slot takes $i-1$ swaps. If $a>i$, the $(i,a)$ bracket correction from $[f{X}_i,X_a]=f[X_i,X_a]-(X_{a}f)X_i$ contributes $-(-1{)}^{i+a}(-1{)}^i(X_{a}f)\omega ({\widehat{X}}_a)=-(-1{)}^a(X_{a}f)\omega ({\widehat{X}}_a)$. Thus every derivative-of-$f$ term cancels.

All remaining terms are $f$ times the original formula; alternation follows by exchanging adjacent inputs, and smooth coordinate coefficients give a smooth form.

At a point of $U$, extend the prescribed tangent vectors by vector fields on $U$; the invariant formula uses only the values of the form and these fields in $U$.

Applying the same formula to the equal restrictions of $\omega$ and $\eta$ gives equal values at every point of $U$.

For $k=0$ the bracket sum is empty and the invariant formula gives $(df)(X)=Xf$.

This is exactly the defining action of the ordinary differential on a tangent vector.

Evaluate the invariant formula on coordinate vector fields. By [F3] their brackets vanish. On an increasing $(k+1)$-tuple $J=(j_0,\dots ,j_k)$ the resulting value is ${\sum }_{a=0}^k(-1{)}^a{\mathrm{\partial }}_{j_a}{\omega }_{J\setminus j_a}$.

For a function $f$, [F2] in degree zero gives $df({\mathrm{\partial }}_j)={\mathrm{\partial }}_{j}f$, hence $df={\sum }_j{\mathrm{\partial }}_{j}fd{x}^j$. Evaluating ${\sum }_{I}d{\omega }_I\wedge d{x}^I$ on $J$ gives exactly the alternating sum in step 1.1. Uniqueness in [F4] proves the formula. For $k\ge n$ both sides vanish in degree $k+1$; for $k=0$ it is the function formula just established.

In a chart, $df={\sum }_j({\mathrm{\partial }}_{j}f)d{x}^j$, and [F1] expresses $d\omega$ by differentiating each coefficient and adding one coordinate differential. Real linearity of partial differentiation therefore makes $d$ real linear on each degree, and every resulting term has degree one higher. Extending by the finite homogeneous decomposition gives a linear map on ${\mathrm{\Omega }}^{\ast }$.

Write $\alpha ={\sum }_I{a}_{I}d{x}^I$ and $\beta ={\sum }_J{b}_{J}d{x}^J$. The ordinary coefficient product rule gives $d(a_I{b}_J)=b_{J}d{a}_I+a_{I}d{b}_J$. Thus [F1] applied termwise to their wedge product gives $d\alpha \wedge \beta$ from the first summands. In the second summands, [F3] gives $d{b}_J\wedge d{x}^I=(-1{)}^{p}d{x}^I\wedge d{b}_J$, yielding $(-1{)}^p\alpha \wedge d\beta$. Repeated coordinate indices give zero wedges on both sides.

By [F2], these chart identities are the restrictions of the corresponding global forms; equality on a chart cover implies equality on $M$. This globalizes linearity, the degree shift and the product rule, which together are exactly [F4]. The calculation includes degree zero via the empty wedge, and degrees above the dimension give zero.

On a chart, write $\omega ={\sum }_I{\omega }_{I}d{x}^I$. Applying the coordinate formula twice gives $d^2\omega ={\sum }_{I,j,\mathrm{\ell }}{\mathrm{\partial }}_{\mathrm{\ell }}{\mathrm{\partial }}_j{\omega }_{I}d{x}^{\mathrm{\ell }}\wedge d{x}^j\wedge d{x}^I\mathrm{.}$ The terms with $j=\mathrm{\ell }$ vanish. For $j\mathrm{\ne }\mathrm{\ell }$, the terms indexed by $(j,\mathrm{\ell })$ and $(\mathrm{\ell },j)$ cancel because mixed partials of the smooth coefficient ${\omega }_I$ agree while $d{x}^{\mathrm{\ell }}\wedge d{x}^j=-d{x}^j\wedge d{x}^{\mathrm{\ell }}$.

The coordinate identity holds on every chart and therefore globally by locality.

The preceding construction exists and has the stated derivation, function, and square-zero properties.

If $D$ has them, then $D(d{x}^i)=D^2{x}^i=0$, and the graded rule determines $D$ on every coordinate expansion; hence $D=d$ locally and globally.

Fix $p\in U$ and tangent vectors $v_0,\dots ,v_k\in T_{p}M=T_{p}U$. Choose a chart around $p$ contained in $U$, extend each vector using constant coordinate coefficients there, multiply by a bump equal to one near $p$, and extend by zero. This gives smooth fields $Y_0,\dots ,Y_k$ on $M$ with $Y_i(p)=v_i$.

Evaluate the invariant formula for $d_M\omega$ on $Y_i$ and the formula for $d_U(\omega {\mathrm{\mid }}_U)$ on $Y_i{\mathrm{\mid }}_U$. Each scalar evaluation of the form restricts to the same smooth function on $U$, so its directional derivatives agree there. Brackets restrict as well: their commutators on smooth functions have identical local expressions. Thus every term in the two formulas has the same value at $p$.

The arbitrary tangent vectors in step 1.1 show equality of the two $(k+1)$-covectors at $p$, and arbitrary $p$ gives $d_U(\omega {\mathrm{\mid }}_U)=(d_M\omega ){\mathrm{\mid }}_U$. Pullback by the open inclusion is restriction because its tangent map is the identity under $T_{p}U=T_{p}M$, proving the claim. The assertion is vacuous if $U$ is empty.

Fix $p\in M$, choose target coordinates $(y^1,\dots ,y^n)$ near $F(p)$, and write $\omega ={\sum }_I{\omega }_{I}d{y}^I$ there. Pullback sends ${\omega }_I$ to ${\omega }_I\circ F$ and $d{y}^a$ to $d(y^a\circ F)$.

On a neighbourhood of $p$, the coordinate formula and the pullback wedge law give $\begin{array}{rl}{\displaystyle d(F^{\ast }\omega )}& {\displaystyle =\sum _{I}d({\omega }_I\circ F)\wedge d(y^{i_1}\circ F)\wedge \cdots \wedge d(y^{i_k}\circ F)}\\ {\displaystyle }& {\displaystyle =F^{\ast }\text{\hspace{0.17em}⁣}(\sum _{I}d{\omega }_I\wedge d{y}^I)=F^{\ast }(d\omega )\mathrm{.}}\end{array}$ Since $p$ was arbitrary, the identity holds on $M$.

Naturality gives $d(F^{\ast }\omega )=F^{\ast }(d\omega )$, so a closed form pulls back to a closed form.

If $\omega =d\eta$, the same equality applied to $\eta$ reads $F^{\ast }\omega =d(F^{\ast }\eta )$, proving exactness preservation.

On the open complement of $\mathrm{supp}\omega$, the form is identically zero.

Locality gives $d\omega =0$ there, so no point of that open set lies in $\mathrm{supp}(d\omega )$.

Two local flows of $X$ have, for each starting point, integral curves with the same initial condition.

Local uniqueness makes the flows equal near $(0,p)$; their pullback curves therefore have the same derivative at zero.

For functions, ${\mathrm{\Phi }}_t^{\ast }f=f\circ {\mathrm{\Phi }}_t$, whose derivative at zero is $Xf$.

For vector fields, differentiating the pullback uses the inverse-time pushforward convention and yields the established bracket $[X,Y]$.

Pullback by each local diffeomorphism preserves tensor products and commutes with contraction.

Differentiate these identities at $t=0$ and use the ordinary product rule for the tensor-product identity.

Evaluate the tensor derivation formula on ${\mathrm{\partial }}_{i_1},\dots ,{\mathrm{\partial }}_{i_k}$ and differentiate the component function along $X$.

Since ${\mathcal{L}}_X{\mathrm{\partial }}_i=[X,{\mathrm{\partial }}_i]=-({\mathrm{\partial }}_i{X}^j){\mathrm{\partial }}_j$, moving the input terms to the other side gives the displayed plus-sign formula.

Apply the tensor derivation law to the coordinate tensor expansion of $T$.

The coordinate bracket formula and ${\mathcal{L}}_X{\mathrm{\partial }}_i=[X,{\mathrm{\partial }}_i]$ give ${\mathcal{L}}_X{\mathrm{\partial }}_i=-({\mathrm{\partial }}_i{X}^j){\mathrm{\partial }}_j$. The coefficient derivative is $X^j{\mathrm{\partial }}_j{T}^{i_1\cdots i_k}$, and applying the preceding identity in each tensor slot supplies the displayed minus terms.

If ${\mathrm{\Phi }}_t^{\ast }T=T$, differentiating at zero gives ${\mathcal{L}}_{X}T=0$.

Conversely the derivative of ${\mathrm{\Phi }}_t^{\ast }T$ is ${\mathrm{\Phi }}_t^{\ast }({\mathcal{L}}_{X}T)$; it vanishes when ${\mathcal{L}}_{X}T=0$, so the pullback curve is constant on each common domain.

The wedge of alternating forms is the alternating restriction of their tensor product.

Restricting the tensor product rule for ${\mathcal{L}}_X$ to that alternating product yields the degree-zero graded rule.

For a function $f$, the right side is ${\iota }_{X}df=Xf={\mathcal{L}}_{X}f$; for a one-form $\alpha$, evaluate both sides on $Y$ and use $d\alpha (X,Y)=X\alpha (Y)-Y\alpha (X)-\alpha ([X,Y])$.

Both ${\mathcal{L}}_X$ and $d{\iota }_X+{\iota }_{X}d$ are degree-zero derivations, so equality on functions and exact coordinate one-forms extends to every local coordinate expression and hence globally.

Apply $d$ to ${\mathcal{L}}_X=d{\iota }_X+{\iota }_{X}d$; $d^2=0$ leaves $d{\mathcal{L}}_X=d{\iota }_{X}d$.

Applying Cartan's formula to $d\omega$ gives ${\mathcal{L}}_{X}d\omega =d{\iota }_{X}d\omega$, the same expression.

Both $[{\mathcal{L}}_X,{\iota }_Y]$ and ${\iota }_{[X,Y]}$ are degree-$-1$ graded derivations, so it suffices to compare them on functions and one-forms. They vanish on functions. For a one-form $\alpha$, the tensor derivation rule gives $({\mathcal{L}}_X\alpha )(Y)=X(\alpha (Y))-\alpha ({\mathcal{L}}_{X}Y)=X(\alpha (Y))-\alpha ([X,Y])$. Hence $[{\mathcal{L}}_X,{\iota }_Y]\alpha ={\mathcal{L}}_X(\alpha (Y))-({\mathcal{L}}_X\alpha )(Y)=\alpha ([X,Y])={\iota }_{[X,Y]}\alpha$. The equality extends to all forms by the derivation rules.

The commutator $[{\mathcal{L}}_X,{\mathcal{L}}_Y]$ is a degree-zero derivation. On a function $f$ it is $X(Yf)-Y(Xf)=[X,Y]f$. On a one-form $\alpha$, evaluate the commutator at a vector field $Z$ and use the one-form formula from step 1.1 twice. The terms where $X$ and $Y$ differentiate $\alpha (Z)$ combine to $[X,Y](\alpha (Z))$; the remaining terms combine by the Jacobi identity to $-\alpha ([[X,Y],Z])$. This is $({\mathcal{L}}_{[X,Y]}\alpha )(Z)$, proving the second identity on all forms. Finally, for any alternating form $\omega$, $({\iota }_X{\iota }_Y\omega )(Z_1,\dots )=\omega (Y,X,Z_1,\dots )=-\omega (X,Y,Z_1,\dots )=-({\iota }_Y{\iota }_X\omega )(Z_1,\dots )$. This proves the third identity, including degrees zero and one.

Relatedness and the definition of pullback give ${\iota }_X(F^{\ast }\omega )=F^{\ast }({\iota }_Y\omega )$; [F1] also gives $d(F^{\ast }\omega )=F^{\ast }(d\omega )$.

Apply [F2] to both sides of the first identity and use step 1.1: ${\mathcal{L}}_X(F^{\ast }\omega )=d(F^{\ast }({\iota }_Y\omega ))+F^{\ast }({\iota }_{Y}d\omega )=F^{\ast }({\mathcal{L}}_Y\omega )\mathrm{.}$

The cocycle writes ${\mathrm{\Phi }}_{t+h,s}^{\ast }{\omega }_{t+h}={\mathrm{\Phi }}_{t,s}^{\ast }({\mathrm{\Phi }}_{t+h,t}^{\ast }{\omega }_{t+h})$. Divide the increment by $h$.

As $h\to 0$, variation of ${\omega }_t$ gives ${\dot{\omega }}_t$ and the short evolution gives ${\mathcal{L}}_{X_t}{\omega }_t$, proving the formula.

Cartan's formula gives ${\mathcal{L}}_X\omega =d({\iota }_X\omega )+{\iota }_X(d\omega )=0$.

The flow-invariance criterion now makes $\omega$ invariant on every local flow domain.

At a point, complete a local annihilator frame ${\theta }^1,\dots ,{\theta }^r$ to a coframe. A form vanishes whenever all inputs lie in $\mathcal{D}$ exactly when every wedge monomial contains some ${\theta }^a$.

Hence those forms constitute the ideal generated by the frame, a description independent of the frame chosen.

For tangent fields $X,Y$ in $\mathcal{D}$, $d{\theta }^a(X,Y)=-{\theta }^a([X,Y])$; thus involutivity forces each $d{\theta }^a$ to vanish on $\mathcal{D}$ and so to have the displayed coframe decomposition.

Conversely that decomposition vanishes on pairs from $\mathcal{D}$, so every ${\theta }^a([X,Y])=0$ and $[X,Y]\in \mathcal{D}$; the frame-independent ideal statement is the same condition.

Extend the nowhere-zero $\alpha$ to a local coframe. The Pfaffian condition is $d\alpha =\eta \wedge \alpha$.

In that coframe, $\alpha \wedge d\alpha =0$ is equivalent to the absence of every component of $d\alpha$ not containing $\alpha$, hence to $d\alpha \in (\alpha )$. Apply [F1] and then [F2] in both directions.

A nowhere-zero one-form has constant rank one and defines a smooth hyperplane distribution.

Since $d\alpha =0$, one has $\alpha \wedge d\alpha =0$, so the codimension-one criterion gives integrability.

The graded Leibniz rule gives $d(f\omega )=df\wedge \omega +fd\omega$.

Take $M=\mathbb{R}$, $f=x$, and $\omega =1$; then $d(f\omega )=dx$ whereas $fd\omega =0$.

Cartan's formula and ${\iota }_{fX}=f{\iota }_X$ give ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +df\wedge {\iota }_X\omega$.

On ${\mathbb{R}}^2$, take $f=x$, $X={\mathrm{\partial }}_y$, and $\omega =dy$; the correction is $dx\mathrm{\ne }0$.