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By the preceding sections, $\eta$ is a nowhere-zero closed differential form and $j^{\ast }\eta =0$. Injectivity of $j^{\ast }:H_{\mathrm{dR}}^1(M)\to H_{\mathrm{dR}}^1(\partial M)$ gives $[\eta ]=0$. The definition of de Rham cohomology then givesfor a smooth $f$, with no critical points. If the boundary were connected, $j^{\ast }df=0$ would make $f$ constant there. A nonconstant function on a compact manifold attains both extrema; a value differing from the common boundary value gives an interior extremum, contradicting $df\ne 0$. If $f$ were constant everywhere there would be the same contradiction. Empty boundary is also impossible, because an extremum would necessarily be interior. Thus $\boxed{\partial M\text{ has at least two connected components}}$, the boundary obstruction for commuting volume-preserving vector fields.

The solid torus $S^1\times D^2$ retracts onto its first circle. Its degree-one de Rham cohomology is generated by that circle's angular one-form, which restricts to a nonzero class on $S^1\times S^1$; hence the restriction map is injective. Its boundary is connected. Extensions of the two specified coordinate vector fields would be tangent there and would satisfy all the forbidden conditions. Therefore no such pair of extensions exists, for any volume form.

The tangent bundle is $TM={⨆}_{p\in M}{T}_{p}M$, with its smooth vector-bundle structure from coordinate charts. A vector field is a smooth section of this bundle. The Lie bracket of vector fields is their commutator as derivations:In coordinates, $[X,Y{]}^k=X^j{\partial }_j{Y}^k-Y^j{\partial }_j{X}^k$.

The four numbered conditions here are hypotheses of one argument, so the following sections assemble its proof. Let $j:\partial M\hookrightarrow M$. Because all $X_i$ are tangent to the boundary, evaluating $\eta ={\iota }_{X_1}\cdots {\iota }_{X_n}\omega$ on a tangent vector $Y$ there evaluates the volume form on $n+1$ vectors in the $n$-dimensional boundary tangent space. Alternation gives $\boxed{j^{\ast }\eta =0}$. This uses the interior product of a differential form and will make the boundary restriction of the de Rham cohomology class vanish.

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.

Write the exact differential form as $\beta =dh$ and take the cotangent fiber translation family ${\rho }_t(x,\xi )=(x,\xi +td{h}_x)$. Its generating vector field is vertical and in local coordinates isWith $\omega ={\sum }_{j}d{x}_j\wedge d{\xi }_j$ and the Hamiltonian vector field convention from Question 3,Thus ${\rho }_t$ is a Hamiltonian isotopy and $f={\rho }_1$ is a Hamiltonian diffeomorphism. The local flow is explicitly defined for every real $t$, regardless of compactness. With the alternative convention ${\iota }_{X_H}\omega =-dH$, the same family is generated by $H=h\circ \pi$.

The definition in the question does not require compact support for the Hamiltonian function; the displayed Hamiltonian diffeomorphism uses that definition.

The Euler-Lagrange field equation obtained by varying the nonminimally coupled scalar field isOne can check stress-energy conservation directly. The divergence of the kinetic contribution is $(\square \Phi ){\nabla }_b\Phi$, because second derivatives of a scalar commute. With $f={\Phi }^2$, the contracted Bianchi identity gives ${\nabla }^a{G}_{ab}=0$, while the Ricci identity givesConsequentlySince ${\nabla }_{b}f=2\Phi {\nabla }_b\Phi$, the full stress-energy tensor satisfies

The general reason is the diffeomorphism Noether identity for a scalar field. Under the Lie derivative along a compactly supported vector field $X$, $\delta g_{ab}=2{\nabla }_{(a}{X}_{b)}$ and $\delta \Phi =X^a{\nabla }_a\Phi$. Invariance of the action under diffeomorphisms gives, after integration by parts,Arbitrariness of $X$ gives the same divergence identity. Conservation requires the scalar equation of motion, with no need to impose the Einstein field equations for the background metric.

The vector field is unchanged by reflection in the $x$-axis, $(x,y)\mapsto (x,-y)$; more generally, linearity of the second equation gives the scaling symmetry $(x,y)\mapsto (x,cy)$. It is also reversible under reflection in the $y$-axis:When $k=0$, $y$ is constant, so arbitrary translations $y\mapsto y+c$ are additional symmetries.

The drift is the linear vector fieldThe eigenvalues of its matrix are $-1\pm i\sqrt{2}$, so the origin is a stable spiral. The field rotates clockwise while contracting toward the origin:

On the two vertical sides of the square, the $x$ component of the vector field satisfiesOn the horizontal sides,Thus the vector field never points outward through the boundary, so the closed square is a positively invariant set and trajectories cannot leave $S$.

At a fixed point, $\dot{y}=0$ gives $x=-y^3$. Substitution into $\dot{x}=0$ givesFor $0<y<1$, both $y^3(2-y^6)$ and $\sin (2y)$ are positive; for $-1<y<0$, both are negative. Hence $y=0$, and then $x=0$, is the unique equilibrium point in $S$.

This alone does not prove that $S$ lies in the domain of stability. A compact trapping region containing a unique equilibrium can also contain a periodic orbit or another nontrivial omega-limit set.

The preceding estimate makes $(x_k)$ equicontinuous, andmakes it uniformly bounded. The Arzela-Ascoli theorem therefore supplies a subsequence converging uniformly to a continuous function $x:[0,1]\to {\mathbb{R}}^n$. On every mesh interval,including an endpoint by using the adjacent interval. Consequently $y_k\to x$ uniformly along the same subsequence.

All $x_k(t)$ and $y_k(t)$ lie in one compact ball. The restriction of the continuous vector field $F$ to that ball is uniformly continuous, so $F\circ y_k\to F\circ x$ uniformly. Passing to the limit in the defining equation givesThe fundamental theorem of calculus now makes $x$ differentiable and yields $x^{\prime }(t)=F(x(t))$. This proves the required special case of the Peano existence theorem; uniqueness need not hold because $F$ was assumed continuous but not locally Lipschitz continuous.

The flow equations of the displayed vector field are ${\dot{\psi }}_1=-{\psi }_2$ and ${\dot{\psi }}_2={\psi }_1$. Hence $d\psi /ds=i\psi$, whose solution is $\psi (s)=e^{is}\psi (0)$. The vector field therefore generates the stated phase rotation.

Since $r=(x_j{x}_j{)}^{1/2}$, the chain rule givesFor the radial vector field $g(r)\mathbf{x}$, the product rule and Einstein summation convention give

Choose a Riemannian metric on $X$, using a partition of unity if necessary. For each $p$, letSince $F$ is a submersion, $D_{p}F{|}_{H_p}:H_p\to T_{F(p)}Y$ is an isomorphism. Its inverse depends smoothly on $p$, sodefines a smooth vector field on $X$ satisfying $D_{p}F(v(p))=w(F(p))$. The formula also gives $v(p)=0$ whenever $w(F(p))=0$.

For any vector field $Y$, apply the Leibniz rule to the scalar $\omega (Y)$:Using ${\mathcal{L}}_{V}Y=[V,Y]$ and expanding the Lie bracket in a coordinate chart leavesThe connection terms cancel because the Levi-Civita connection is torsion-free. Applying this formula to each slot of the metric tensor and using metric compatibility givesEquivalently, one may prove both identities at a point in normal coordinates; since both sides are tensors, the result then holds in every coordinate system.

A continuous finite-dimensional vector field has a local solution through every initial point. The theorem asserts existence without uniqueness; local Lipschitz continuity supplies uniqueness through the Picard-Lindelof theorem.

If a periodic vector field has a periodic potential of a conservative vector field, the line integral between points separated by a period vector must vanish, since it is the potential difference. A periodic gradient field need not satisfy this: the constant field $(1,2)$ on the plane has potential $x+2y$, but its integrals over the two unit translations are $1$ and $2$. On the periodic quotient these translation integrals are nonzero periods, obstructing a single-valued potential there.

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.

Let $W$ consist of real-linear maps $\alpha$ from smooth vector fields to smooth functions such that $X_p=0$ implies $\alpha (X)(p)=0$. These are exactly evaluations by smooth differential one-forms. Define ${\omega }_p(v)=\alpha (X)(p)$ using any global field with $X_p=v$; the vanishing condition makes it well-defined. Bump-extended local frame vectors make its coefficients smooth. Also $\alpha (fX)(p)=f(p)\alpha (X)(p)$, so real linearity and pointwise vanishing automatically give $C^{\infty }$-linearity. The evaluation correspondence is a natural module isomorphism.

A vector field $X$ is projectable through a smooth map between manifolds $f:M\to N$ if there is a smooth vector field $Y$ on $N$ with $d{f}_p{X}_p=Y_{f(p)}$ for every $p$. Agreement on fibres is necessary. For a surjective submersion it is also sufficient: smooth local sections of the submersion express $Y$ locally as $df(X)$ and prove its smoothness. An arbitrary non-surjective map may instead give a field only along its image, with extension to $N$ requiring additional choices.

For $V=\xi (x,u){\partial }_x+\eta (x,u){\partial }_u$, its prolongation to derivatives through order $n$ isIt generates a symmetry of $\Delta =0$ exactly when ${\mathrm{pr}}^{(n)}V(\Delta )$ vanishes on the equation manifold.

For a vector field $X$ on $M$, its pushforward by a smooth map between manifolds $f:M\to N$ is the vector field along a map defined by $(f_{\ast }X{)}_p=d{f}_p(X_p)$. Acting on a function $h$ on $N$, it satisfies $(f_{\ast }X{)}_p(h)=X_p(h\circ f)$. For a diffeomorphism this defines a vector field on $N$ by evaluation at $p=f^{-1}(q)$. For a general map there need not be a single vector at each image point: $f(x)=x^2$, $X={\partial }_x$ gives $2x{\partial }_y$, with opposite values over $y>0$.