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.

Using ${\mathcal{L}}_X=[d,{\iota }_X]$ and $d^2=0$, expand the graded commutator with ${\iota }_Y$; the antiderivation signs give $[{\mathcal{L}}_X,{\iota }_Y]={\iota }_{[X,Y]}$.

The same expansion together with the bracket characterization of ${\mathcal{L}}_X$ gives $[{\mathcal{L}}_X,{\mathcal{L}}_Y]={\mathcal{L}}_{[X,Y]}$; alternating insertion gives ${\iota }_X{\iota }_Y+{\iota }_Y{\iota }_X=0$.

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 $dx\mathrm{\ne }0$.

The invariant formula uses only the form, vector fields, their action on functions, and their Lie brackets.

No metric, connection, or inner product occurs in that construction, so the asserted dependence is false.

For $\omega =(-ydx+xdy)\mathrm{/}(x^2+y^2)$, writing $r^2=x^2+y^2$ gives ${\mathrm{\partial }}_x(x\mathrm{/}{r}^2)-{\mathrm{\partial }}_y(-y\mathrm{/}{r}^2)=0$; therefore [F1] gives $d\omega =0$ on ${\mathbb{R}}^2\setminus \{0\}$.