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The Cartan product formula says that total Steenrod squares and total Steenrod reduced powers preserve the cup product:It is named after Henri Cartan and is distinct from Cartan's magic formula for the Lie derivative of a differential form, named after Élie Cartan.

The Lie derivative of a differential form satisfies ${\mathcal{L}}_X=d{\iota }_X+{\iota }_{X}d$. Both sides agree on functions. On a differential 1-form $\eta$, evaluation at a vector field $Y$ gives $X(\eta (Y))-\eta ([X,Y])$ on both sides, by the definition of the exterior derivative and the Lie bracket of vector fields. Both operators are degree-zero derivations of an algebra on the exterior algebra. Agreement on functions and coordinate differential 1-forms therefore proves agreement on every differential form. Since $d^2=0$, the formula also gives ${\mathcal{L}}_{X}d=d{\mathcal{L}}_X$, so the Lie derivative of a differential form commutes with the exterior derivative.

For a smooth vector field $X$, the Lie derivative is the tensor derivation determined by ${\mathcal{L}}_{X}f=X(f)$ and ${\mathcal{L}}_{X}Y=[X,Y]$. The identity $[X,fY]=f[X,Y]+X(f)Y$ permits its unique extension. On a differential one-form,On differential forms it agrees with the usual Lie derivative of a differential form and Cartan's magic formula.

Let ${\Phi }_t$ be the local flow of a vector field $V$. The Lie derivative of a differential form is ${\mathcal{L}}_V\eta ={\frac{d}{dt}|}_{t=0}{\Phi }_t^{\ast }\eta$. We prove Cartan's magic formulaBoth sides are degree-zero derivations of an algebra on the exterior algebra of differential forms. They agree on a function $f$, since ${\iota }_{V}df=V(f)$. They also agree on $df$: pullback commutes with the exterior derivative, so ${\mathcal{L}}_{V}df=d(Vf)$, while $(d{\iota }_V+{\iota }_{V}d)df=d(Vf)$. Locally every differential form is a sum of products of functions and coordinate differentials, proving the identity in every degree.

For a compactly supported Hamiltonian vector field $V_H$, take the convention ${\iota }_{V_H}\omega =-dH$. Since $d\omega =0$, Cartan's magic formula gives ${\mathcal{L}}_{V_H}\omega =0$, so its complete flow of a vector field consists of symplectomorphisms. Given $p$ and $v\in T_{p}M$, a bump function lets us choose a compactly supported $H$ with $d{H}_p=-{\iota }_v{\omega }_p$, and hence $V_H(p)=v$. Choose such fields for a basis of $T_{p}M$. Their successive small-time flows give a map with invertible derivative at the origin. The inverse function theorem makes the orbit of $p$ open. Every orbit is open; because $M$ is connected, there is only one orbit. Thus the compactly supported symplectomorphisms act transitively on points, by Hamiltonian transitivity on a connected symplectic manifold.

On the unit sphere, the standard symplectic area is $4\pi$. The equator has complementary disks of areas $2\pi ,2\pi$, whereas the latitude at height $1/2$ has complementary disks of areas $\pi ,3\pi$: the cap above height $h$ has area $2\pi (1-h)$. A symplectomorphism must preserve the unordered pair of complementary areas. Therefore no such symplectomorphism exists, by complementary areas obstruct symplectic equivalence of separating curves.

Put ${\xi }_t=\ker {\alpha }_t$, the contact distribution. If $\dim M=2r+1$, the contact form condition is ${\alpha }_t\wedge (d{\alpha }_t{)}^r\ne 0$. It implies that $d{\alpha }_t{|}_{{\xi }_t}$ is a symplectic form on each contact hyperplane: inserting a vector transverse to ${\xi }_t$ in that volume form leaves the nonzero top exterior power $(d{\alpha }_t{|}_{{\xi }_t}{)}^r$. Hence there is a unique smooth time-dependent vector field $Y_t\in {\xi }_t$ satisfyingIts smoothness follows by inverting the smoothly varying nondegenerate matrix of this contact distribution symplectic form. This is the horizontal generator of contact stability.

Let $R_t$ be the Reeb vector field of ${\alpha }_t$. The one-form ${\dot{\alpha }}_t+{\iota }_{Y_t}d{\alpha }_t$ vanishes on ${\xi }_t$, so it is $h_t{\alpha }_t$. Evaluating on $R_t$, using ${\alpha }_t(R_t)=1$ and ${\iota }_{R_t}d{\alpha }_t=0$, identifies the coefficient:By Cartan's magic formula, the Lie derivative of a differential form isbecause $Y_t$ lies in the contact hyperplane.

Let ${\rho }_t$ solve ${\partial }_t{\rho }_t=Y_t\circ {\rho }_t$, with ${\rho }_0={\mathrm{id}}_M$. Since $M$ is compact without boundary and $Y_t$ is smooth on the closed time interval, its solutions cannot escape and exist for the entire interval. Reversing the time-dependent equation supplies a smooth inverse for each ${\rho }_t$. Thus these flow maps give a smooth isotopy. Differentiating the pullback of a differential form along this isotopy yieldsFor each base point this is a scalar linear equation for a covector, initially ${\alpha }_0$. Its solution isBy finite-time flow completeness on a compact manifold, this construction gives the whole time interval. The contact conformal factor along an isotopy is smooth in $(t,x)$ and nowhere zero, proving Gray stability theorem. The precise domains here are ${\rho }_t:M\to M$ and $\rho :M\times [0,1]\to M$; the extra time factor attached to the already indexed ${\rho }_t$ in the source is a notational slip. The proof produces the required family on its whole specified interval. If a parameter domain of all $\mathbb{R}$ is intended, extend the smooth field slightly beyond $[0,1]$, multiply it by a time cutoff, and use compactness to obtain the extended isotopy $\rho :M\times \mathbb{R}\to M$. Its restriction to $[0,1]$ is unchanged.

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 $p=0$:

A local flow of $v$ is a smooth family ${\Phi }^t$ defined near $\{0\}\times X$ such that ${\Phi }^0=\mathrm{id}$, ${\partial }_t{\Phi }^t(x)=v({\Phi }^t(x))$, and ${\Phi }^{s+t}={\Phi }^s\circ {\Phi }^t$ whenever defined. The Lie derivative of a differential form isCartan's magic formula is

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.