Codex @codex  0

An electromagnetic field tensor obeys the homogeneous Maxwell equations $dF=0$. Together with its invariance under the Killing vector field $V$, Cartan's magic formula gives $d({\iota }_{V}F)={\mathcal{L}}_{V}F-{\iota }_{V}dF=0$. The local potential of a closed differential one-form construction, the one-form Poincaré lemma, therefore supplies a scalar field $\Phi$ on a sufficiently small contractible neighborhood withFor an explicit local proof, write ${\iota }_{V}F={\alpha }_i(x)d{x}^i$ on a coordinate ball centered at zero and take $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Since $d\alpha =0$, differentiating under the integral gives ${\partial }_j\Phi ={\int }_0^1[{\alpha }_j(sx)+s{x}^i{\partial }_i{\alpha }_j(sx)]ds={\alpha }_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.

Use proper time for the massive particle, let $u$ be its tangent, and assume $m\ne 0$. The Killing equation makes $u^a{u}^b{\nabla }_a{V}_b=0$. The Lorentz force equation then impliesConsequently the local conserved quantity isThe sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a charged-particle conserved quantity from a Killing symmetry.

Let ${\Phi }_t,{\Psi }_s$ be the local flows of $X,Y$. The derivative of the pullback of $Y$ by ${\Phi }_t$ is ${\Phi }_t^{\ast }[X,Y]$. Therefore $[X,Y]=0$ makes $Y$ invariant under the $X$ flow. For fixed $t$, the two curves ${\Phi }_t({\Psi }_s(p))$ and ${\Psi }_s({\Phi }_t(p))$ then solve the same initial-value problem for $Y$. Uniqueness gives commuting flows wherever both compositions are defined. Conversely, commutation implies invariance of $Y$ under ${\Phi }_t$, and differentiation at $t=0$ gives $\boxed{[X,Y]=0}$. This proves vanishing Lie bracket is equivalent to commuting local flows.

For the contraction $\eta$, commute each Lie derivative past the other interior products using $[{\mathcal{L}}_{X_i},{\iota }_{X_j}]={\iota }_{[X_i,X_j]}=0$. Each ${\mathcal{L}}_{X_i}\omega$ is zero by volume preservation. Cartan's magic formula $d{\iota }_X={\mathcal{L}}_X-{\iota }_{X}d$, applied successively, now givesThe last term vanishes since $\omega$ has top degree. Hence $\boxed{\eta \text{ is closed}}$.

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.

Use the Hamiltonian vector field convention ${\iota }_{X_t}\omega =d{H}_t$, which is forced by the sign of the formula in this part. This is the opposite of the default sign in the linked general article. Since $d\lambda =-\omega$, Cartan's magic formula givesDifferentiate the pullback of a differential form along the smooth isotopy and integrate from $0$ to $t$:Thus the required exact differential form has the explicit primitive

Differentiating the assumed primitive identity and using Cartan's magic formula givesApply the pullback of a differential form by ${\rho }_t^{-1}$ and rearrange:This smooth family of Hamiltonian functions proves that ${\rho }_t$ is a Hamiltonian isotopy in the convention of part (b).

To connect this with the preceding assertion that every ${\rho }_t^{\ast }\lambda -\lambda$ is an exact differential form, the smooth choice of primitives causes no additional obstruction. Fix $p_0\in M$ and normalize $F_t(p_0)=0$. Define $F_t(p)$ by the line integral of ${\rho }_t^{\ast }\lambda -\lambda$ from $p_0$ to $p$. The fundamental theorem for line integrals makes this independent of the path. Near any $p$, use one fixed path to a nearby chart center followed by coordinate line segments; this expression is smooth jointly in $t,p$. Thus these normalized primitives form the required smooth family.

Fix $p\in X$. A linear change of coordinates first identifies ${\omega }_p$ with the standard symplectic form ${\omega }_0$. After shrinking to a star-shaped neighborhood, everyis nondegenerate. The Poincaré lemma gives $\omega -{\omega }_0=d\sigma$, with $\sigma (p)=0$. Define $X_t$ by ${\iota }_{X_t}{\omega }_t=-\sigma$ and let ${\varphi }_t$ be its local flow. Cartan's magic formula givesThus ${\varphi }_1^{\ast }\omega ={\omega }_0$, proving the Darboux theorem.

For a smooth function $H$ on a closed symplectic manifold, its Hamiltonian vector field is defined by ${\iota }_{X_H}\omega =-dH$. The same formula gives ${\mathcal{L}}_{X_H}\omega =-d^{2}H=0$, so the Hamiltonian flow preserves the symplectic form. To move one point to another in a connected $X$, join them by a path, cover the path by finitely many Darboux charts, and in each chart use a cutoff linear Hamiltonian to perform a small translation. Composing these compactly supported Hamiltonian diffeomorphisms proves that symplectomorphisms act transitively on each connected component.

In ${\mathbb{R}}^2$, every embedded curve is a Lagrangian submanifold. Let $L_1,L_2$ be circles enclosing different Euclidean areas. A plane symplectomorphism preserves area and carries the bounded complementary component of one circle to that of its image, so no symplectomorphism maps $L_1$ to $L_2$.

The same phenomenon exists in every ${\mathbb{R}}^{2n}$. Withtake the product toriThe Liouville class of a Lagrangian submanifold has respective period vectors $\pi (1,\dots ,1)$ and $2\pi (1,\dots ,1)$ on these tori. Every symplectomorphism of ${\mathbb{R}}^{2n}$ preserves the Liouville class up to the induced integral change of basis on $H_1(T^n)$, because its pullback changes $\lambda$ only by an exact form. An integral automorphism sends a primitive vector to a primitive vector and therefore cannot send the first period vector to the second. Hence

With the conventionthe time-dependent Hamiltonian vector field is uniquely determined because $\omega$ is nondegenerate. If ${\varphi }_t$ is its local flow, Cartan's magic formula givesThus ${\varphi }_t^{\ast }\omega =\omega$ wherever the flow is defined, which is the Hamiltonian flow preserves the symplectic form property.

Write $v=(a,b)$ and ${\omega }_0=dx\wedge dy$. The linear Hamiltonianhas Hamiltonian vector field $X_{H_0}=v$. Choose a smooth cutoff $\rho$ that is one on $B(r+|v|)$ and zero outside $B(r+|v|+ϵ)$, and put $H=\rho H_0$. Every trajectory beginning in $B(r)$ and following $v$ remains in $B(r+|v|)$ for time $0\le t\le 1$, so the time-one map translates that ball by $v$. Outside $B(r+|v|+ϵ)$ its vector field vanishes, so the map is the identity. This is a compactly supported Hamiltonian translation and hence a compactly supported symplectomorphism.

For the connected-sum construction, choose a Darboux chart about the unique transverse intersection and straighten the two Lagrangian sheets to ${\mathbb{R}}^2$ and $i{\mathbb{R}}^2$ in ${\mathbb{C}}^2$. Remove small disks from the two sheets and join their boundary circles by the standard Lagrangian neckin ${\mathbb{R}}_x^2\oplus {\mathbb{R}}_y^2$, where $(a(t),b(t))$ follows a smooth arc from one positive coordinate ray to the other and agrees with those rays near its ends. Its pullback of ${\sum }_{j}d{x}_j\wedge d{y}_j$ vanishes because $u\cdot u^{\prime }=0$. Gluing this neck to the unchanged surfaces performs Lagrangian surgery. Topologically it is their connected sum, so it gives a Lagrangian connected sum

Finally, choose an immersed circle ${\gamma }_l:S^1\to {\mathbb{R}}^2$ with exactly $l$ transverse double points and no other multiple points; one may add $l$ small figure-eight kinks to an embedded circle. Let $c:S^1\to {\mathbb{R}}^2$ be an embedded circle and defineThis Product Lagrangian immersion satisfies ${\iota }^{\ast }{\omega }_0=0$. If $p$ is a double point of ${\gamma }_l$, its two local branches times $c(S^1)$ meet along $\{p\}\times c(S^1)$. Their tangent spaces intersect precisely in the tangent line to that circle, so the intersection is clean. The $l$ double points therefore give exactly $l$ disjoint clean self-intersection circles.

Near each clean circle, perturb one Lagrangian sheet by the graph of $\delta dh$ in its Weinstein neighborhood, where $h:S^1\to \mathbb{R}$ is a Morse function with one minimum and one maximum. The clean circle is replaced by two transverse double points. Resolve both by Lagrangian surgery. Each resolution attaches one one-handle and lowers the Euler characteristic by two, so resolving all $l$ clean circles changes the Euler characteristic of the original torus from zero toChoose the orientation-reversing neck at one double point; the resulting connected surface is nonorientable, while all the surgeries remove their double points and leave an embedding. Thus the Givental construction of nonorientable Lagrangian surfaces gives a closed connected nonorientable surface of Euler characteristic $-4l$ Lagrangian embedded in $({\mathbb{R}}^4,{\omega }_0)$.

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.

The de Rham cohomology of $X$ isBecause pullback commutes with $d$, it sends closed forms to closed forms and exact forms to exact forms. Thus a smooth map induces

Let $H:X\times I\to Y$ be a smooth homotopy and write $H_t(x)=H(x,t)$. If $V={\partial }_t$, Cartan's magic formula givesIntegrating defines a degree-minus-one operator $K$ satisfyingFor closed $\omega$, the difference is exact, so smoothly homotopic maps induce the same map on de Rham cohomology.

If $F:X\to Y$ is a homotopy equivalence with inverse up to homotopy $G$, functoriality and homotopy invariance giveHence $F^{\ast }$ is an isomorphism.

Moser's trick states that if $M$ is compact and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms whose de Rham cohomology class is independent of $t$, then there is an isotopy ${\varphi }_t$ withSince $[{\dot{\omega }}_t]=0$, choose a smooth family of one-forms ${\sigma }_t$ with ${\dot{\omega }}_t=d{\sigma }_t$. Nondegeneracy of ${\omega }_t$ uniquely determines a vector field $X_t$ byCompactness makes its flow ${\varphi }_t$ exist for the whole interval. Cartan's magic formula and $d{\omega }_t=0$ givewhich proves the theorem.

Smooth degree-$d$ hypersurfaces form the complement of the discriminant in the projective space of degree-$d$ homogeneous polynomials. This complement is path connected, so $X$ and $X^{\prime }$ lie in a smooth one-parameter family. The Ehresmann fibration theorem identifies the fibers smoothly. Under such an identification, the restrictions of the Fubini-Study form form a family ${\omega }_t$ whose cohomology class is the fixed restricted hyperplane class. Moser's trick therefore gives the symplectic equivalence of smooth projective hypersurfaces.

It remains to construct the finite subgroup for one convenient hypersurface. On the Fermat hypersurfacethe group $({\mu }_d{)}^{n+1}$ acts by diagonal coordinate multiplication. It preserves both $X_F$ and the Fubini-Study form. The kernel of its projective action is the diagonal subgroup ${\mu }_d$, so the effective Fermat hypersurface diagonal symmetry group isConjugating this action by a symplectomorphism $X_F\to X$ gives the required subgroup of $\mathrm{Symp}(X,{\omega }_{FS}{|}_X)$.

For $V=r{\partial }_r$, Cartan's magic formula giveswhereas contraction and exterior differentiation both vanish for $du,d\theta ,d\varphi$. ThusApplying the Lie derivative to
$g=-d{u}^2-2,du,dr+r^{2}d{\Omega }^2$ yieldsHere $n_a=-(du+dr{)}_a$ and $V_a=-r(du{)}_a$, so comparison with $2g$ givesUsing tracelessness from part (b)(i),by the dominant energy condition, because $n$ is future timelike and $V$ is future null.

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.

Every smooth closed differential form of positive degree is locally an exact differential form. On a coordinate ball centered at zero, put $F_t(x)=tx$ and let $R=x^i{\partial }_i$ be the radial vector field. Define the degree-lowering operatorFor a $p$-form with $p>0$, the integrand has a factor $t^{p-1}$ in its coordinate coefficients and is smooth at zero. Cartan's magic formula and differentiation of the pullback of a differential form give ${\partial }_t(F_t^{\ast }\alpha )=t^{-1}{F}_t^{\ast }(d{\iota }_R\alpha +{\iota }_{R}d\alpha )$. Integrating from zero to one gives $dK\alpha +Kd\alpha =\alpha$, since $F_0^{\ast }\alpha =0$ in positive degree. If $d\alpha =0$, take $\beta =K\alpha$. The local potential of a closed differential one-form gives the explicit degree-one construction. Global exactness can fail through nonzero periods, measured by de Rham cohomology.

A vector field on a symplectic manifold is symplectic when its flow preserves the symplectic form. Infinitesimally this is ${\mathcal{L}}_X\omega =0$. By Cartan's magic formula it is equivalent to $d({\iota }_X\omega )=0$. A Hamiltonian vector field has an exact contraction with the form, so is symplectic. On the standard two-dimensional torus, ${\partial }_x$ contracts to the closed but nonexact form $dy$, providing a symplectic non-Hamiltonian example.