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Every smooth left-invariant vector field on a Lie group is a complete vector field. Translate a local integral curve of a vector field through the identity to every other point. The same positive local existence interval works at every initial point, so no maximal integral curve of a vector field can have a finite endpoint. Uniqueness then makes the curve through the identity a one-parameter subgroup.

On ${\mathbb{R}}^2$, the vector fields $X=y{\partial }_x$ and $Y=(x^2/2){\partial }_y$ are complete, since their local flows are $(x+ty,y)$ and $(x,y+t{x}^2/2)$. Their Lie bracket of vector fields is $-(x^2/2){\partial }_x+xy{\partial }_y$. Its integral curve of a vector field starting at $(2,0)$ is $(2/(1+t),0)$, which cannot continue through $t=-1$. The sum has an integral curve of a vector fieldthat cannot continue through $t=2\sqrt{3}$. Thus neither operation preserves complete vector fields on a general noncompact smooth manifold. This contrasts with a compact manifold without boundary, where every smooth vector field is complete.

An integral curve of a vector field $X$ through $p$ is a smooth curve $c:I\to M$, on an interval containing zero, satisfying $c(0)=p$ and $\dot{c}(t)=X_{c(t)}$. The local flow is the smoothly varying family of these integral curves of a vector field:Here $D$ is an open neighborhood of $\{0\}\times M$; for the maximal local flow, its fibre over $p$ is the maximal existence interval of $c_p$. Local existence, uniqueness and smooth dependence for a smooth ordinary differential equation give this family. Uniqueness also gives ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever the relevant curves and compositions are defined, and ${\varphi }_{-t}={\varphi }_t^{-1}$ on their common domains.

Reparametrizing the integral curves of a vector field givesThus the flow of $2X$ is $\boxed{{\tilde{\varphi }}_t={\varphi }_{2t}}$, with domain $\{(t,p):(2t,p)\in D\}$.

Commutation of the two local flows means ${\varphi }_t{\psi }_s={\psi }_s{\varphi }_t$ for all sufficiently small independent parameters $s,t$, on their common local domains. Differentiate this identity in $s$ at zero. The chain rule givesIn a coordinate chart, differentiating again in $t$ at zero gives $D{X}_p{Y}_p=D{Y}_p{X}_p$. But the Lie bracket of vector fields has coordinate expressionso commuting local flows imply $[X,Y]=0$.

For the converse, suppose the Lie bracket of vector fields vanishes. Write $u(t)={\varphi }_t(p)$ in a coordinate chart, and put $J(t)=d{\varphi }_t{|}_p$. Differentiating the local flow equation with respect to the initial point gives the variational ordinary differential equationConsequently $W(t)=Y_{u(t)}-J(t)Y_p$ satisfieswhere the zero Lie bracket of vector fields was used in the second equality. Uniqueness for this linear ordinary differential equation gives $W=0$. This is a local argument and can be continued through successive coordinate charts, proving $d{\varphi }_t(Y)=Y\circ {\varphi }_t$ wherever the local flow is defined.

For fixed $t$, the curve $s\mapsto {\varphi }_t({\psi }_s(p))$ therefore has derivative $Y$ at its current point and starts at ${\varphi }_t(p)$. Uniqueness of the integral curve of a vector field $Y$ identifies it with $s\mapsto {\psi }_s({\varphi }_t(p))$. Hence $\boxed{{\varphi }_t{\psi }_s={\psi }_s{\varphi }_t⟺[X,Y]=0}$ locally. No complete vector field hypothesis is needed.

Using the coordinate formula for the Lie bracket of vector fields, the two components areThus $\boxed{[X,Y]=-(x^2/2){\partial }_x+xy{\partial }_y}$, which is not the zero vector field.

The integral curves of a vector field $X$ solve $\dot{x}=y$, $\dot{y}=0$, whereas those of $Y$ solve $\dot{x}=0$, $\dot{y}=x^2/2$. Their local flows areBoth formulas exist for all real parameters, so both are complete vector fields. Direct composition givesAt $(1,0)$ their first coordinates differ by $ts/2$. Hence these local flows do not commute when $s,t$ are nonzero.

For the Lie bracket of vector fields, start at $(2,0)$. An integral curve of a vector field $[X,Y]$ isIt solves $\dot{x}=-x^2/2$, $\dot{y}=xy$, but $x(t)$ diverges as $t\downarrow -1$. It cannot extend through that finite time as a curve in ${\mathbb{R}}^2$. Complete vector fields are therefore not closed under the Lie bracket.

For the sum $X+Y$, its integral curves of a vector field obey $\dot{x}=y$, $\dot{y}=x^2/2$. Substitution of $x=(1+\alpha t{)}^{-2}$ and $y=\beta (1+\alpha t{)}^{-3}$ requiresChoosing $\alpha =-1/(2\sqrt{3})$ gives the integral curve of a vector fieldIt starts at $(1,1/\sqrt{3})$ and escapes to infinity at the finite time $2\sqrt{3}$. Complete vector fields are not closed under addition either. Both failures use the same pair of complete vector fields.

An integral curve of a vector field $X$ is a smooth curve $\gamma :I\to M$ satisfying $\dot{\gamma }(t)=X_{\gamma (t)}$ for every $t$ in its interval of definition. Passing through $p$ at time zero means $\gamma (0)=p$. The tangent-vector equality is intrinsic; in a coordinate chart $x:U\to {\mathbb{R}}^n$ it becomesSmoothness makes the coordinate right side locally Lipschitz continuous. The Picard-Lindelöf theorem supplies a locally unique solution through each coordinate initial value.

In fact, two integral curves of $X$ with the same initial point agree throughout their common time interval containing zero. To pass from local to global uniqueness, the set of coincidence times is nonempty and closed by continuity and the Hausdorff property of the smooth manifold. At a coincidence time, put both curves in one coordinate chart; local ordinary differential equation uniqueness makes them agree on a neighborhood of that time. Thus the coincidence set is open as well as closed in the connected common interval, so it is the whole interval. This argument applies around any prescribed time, not only zero.

By joining all extensions that agree on overlaps, one obtains the unique maximal integral curve of a vector field through $p$. Its maximal interval need not be all of $\mathbb{R}$: completeness is an additional property of the vector field.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

Choose local coordinates $y^1,\dots ,y^{D-1}$ on the hypersurface $\Sigma$, and let ${\Phi }_s$ be the local flow of the vector field $X$. The map $F(s,y)={\Phi }_s(p(y))$ follows the integral curves of a vector field starting at $p(y)\in \Sigma$. At $s=0$, its differential maps the $s$ direction to $X$ and the remaining directions to a basis of $T_p\Sigma$. Transversality makes these vectors linearly independent, so the inverse function theorem makes $F$ a local manifold chart.

Set $x^0=s$ and transport the $y^i$ unchanged along each integral curve of a vector field. Then the adapted coordinates satisfyThe flow-box theorem gives this construction locally. The parameter is the flow parameter, and need not be arc length; a global manifold chart would require additional hypotheses.

A smooth vector field on a smooth manifold $M$ is a smooth section of the tangent bundle: it assigns $v(p)\in T_{p}M$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v={\sum }_i{a}_i\partial /\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.

For a Lie group $G$, write $L_a(h)=ah$ for Left translation on a Lie group. A left-invariant vector field satisfiesThus it is determined by its value at the identity. Given the printed $X\in T_{g}G$, first translate it to the identity:The unique left-invariant vector field with value $X$ at $g$ isThe smoothness of multiplication makes this a smooth vector field, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.

We prove the completeness of left-invariant vector fields: this left-invariant vector field is a complete vector field. The Picard-Lindelöf theorem, applied in a local chart, gives a unique local integral curve of a vector field $u:(-\epsilon ,\epsilon )\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an integral curve of a vector field through $h$, becauseCrucially, the same interval $(-\epsilon ,\epsilon )$ works for every initial point.

Let $\gamma$ be the maximal integral curve of a vector field through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in (a,b)$ with $b-t_0<\epsilon /2$. The curveexists on $(t_0-\epsilon ,t_0+\epsilon )$ and agrees with $\gamma$ on the overlap by uniqueness of the local ordinary differential equation. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb{R}$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb{R}$.

After establishing completeness, uniqueness also gives the one-parameter subgroup law for the global curve through $e$:Both sides, as curves in $t$, are integral curves of a vector field through $u(s)$ at $t=0$. Defining the Exponential map of a Lie group by $u(t)=\exp (t\xi )$, the answer is

The identity component $G^{\circ }$ is an open normal subgroup, and every open identity neighbourhood generates it. Let $G^{\circ }$ be the connected component of $e$. The product of connected spaces is connected, so the image of $G^{\circ }\times G^{\circ }$ under multiplication is connected and contains $e$. It is therefore contained in $G^{\circ }$. Inversion has the same property. Thus $G^{\circ }$ is a subgroup.

A smooth manifold is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^{\circ }$. Its translates $hO$, $h\in G^{\circ }$, are open and lie in $G^{\circ }$, and cover $G^{\circ }$. Therefore the identity component of a Lie group is open in $G$. Conjugation by any $a\in G$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^{\circ }$ is a normal subgroup.

If $U$ is an open neighbourhood of $e$ in $G^{\circ }$, let $H=⟨U⟩$, allowing inverses in the meaning of generated subgroup. For each $h\in H$, $hU$ is open and lies in $H$, so $H={\bigcup }_{h\in H}hU$ is open in $G^{\circ }$. Every other left coset of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^{\circ }$. It is nonempty, so $H=G^{\circ }$.

A quadratic form on $V={\mathbb{R}}^n$ is a function $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Equivalently, $Q(tv)=t^{2}Q(v)$ and the polarizationis a bilinear map. In coordinates there is a unique real symmetric matrix $A$ with $Q(v)=v^{T}Av$. No assumption of nondegeneracy or positive definiteness is needed.

An element $g\in {\mathrm{GL}}_n(\mathbb{R})$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the polarization identity, this is equivalent to preserving $B$, or in coordinates toThis is a closed Matrix Lie group. The Lie algebra of a quadratic-form stabilizer isTo prove necessity, differentiate $g(t{)}^{T}Ag(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g^{\prime }(0)=X$. The derivative at zero is $X^{T}A+AX$.

To prove sufficiency, suppose $X^{T}A+AX=0$ and consider the matrix exponential $g(t)=e^{tX}$. ThenIts value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed tangent space description even for a degenerate quadratic form.

Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate quadratic form it is the corresponding Special orthogonal Lie algebra. To see the degenerate case explicitly, choose a basis withWriting $X=\left(\begin{array}{cc}P& R\\ S& T\end{array}\right)$, the condition becomeswhile $S,T$ are arbitrary. Thus the nullspace of the quadratic form is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q={\mathrm{GL}}_n(\mathbb{R})$ and $T_e{G}_Q=M_n(\mathbb{R})$.

The directional derivative along the specified vector field givesUse the magnitude formula for the curvature of an integral curve of a vector field:It vanishes on the plane $z=0$ and on the $z$ axis away from the origin, consistent with straight field lines there. At the origin the vector field vanishes, so it fails the nowhere-zero hypothesis and this field-line curvature is not defined.

As an independent geometric check, an integral curve of a vector field satisfies $\dot{x}=x$, $\dot{y}=y$, $\dot{z}=-z$, giving $(x,y,z)=(a{e}^u,b{e}^u,c{e}^{-u})$. Its velocity is $(x,y,-z)$ and acceleration $(x,y,z)$; the ordinary curvature of a space curve formula $|r^{\prime }\times r^{\prime \prime }|/|r^{\prime }{|}^3$ gives the same expression.

The relevant Penrose singularity theorem states: a connected time-oriented four-dimensional globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a nonempty compact orientable boundaryless trapped surface, and the null convergence condition is future null-geodesically incomplete. Under the Einstein field equations, the null energy condition supplies the curvature hypothesis. A trapped surface here is spacelike and has both future normal null expansions strictly negative. The conclusion is an incomplete null geodesic, not necessarily a divergent curvature invariant at an identifiable point.

Suppose, for contradiction, that every future null geodesic is complete. Write $S$ for the trapped surface andThe equality follows because the causal future of a compact set is closed in a globally hyperbolic spacetime. This future horismos is a closed achronal boundary. It is nonempty: the restriction of a Cauchy time function to compact $S$ has a minimum, and a point at that minimum cannot be chronologically preceded by another point of $S$.

Normalize the two future null normal directions along $S$ against a smooth future timelike field, fixing the affine scale continuously. Their initial null expansions are continuous and strictly negative. Compactness of the normalized normal bundle gives a uniform $c>0$ with ${\theta }_{\pm }\le -c$ everywhere on $S$. The preceding null focusing theorem forces a conjugate point to a spacelike surface along each normal generator within affine length $L=2/c$.

By the supplied boundary-generator result, every point of $E^{+}(S)$ is reached by an orthogonal future null geodesic which has no earlier conjugate point. No such boundary generator can remain on the boundary beyond its first focal point. Every point of $E^{+}(S)$ therefore lies inThe parameter set is compact. Future null completeness makes its geodesic flow defined throughout this common finite interval, and smooth dependence on initial data makes its image $K$ compact. Since $E^{+}(S)$ is closed and contained in $K$, it is compact. This is the compactness of the future horismos of a trapped surface step; a merely pointwise finite bound would not suffice without compactness and uniform normalization.

Choose a smooth complete timelike vector field, obtained if necessary by positive rescaling against a complete auxiliary Riemannian metric. Its inextendible integral curves of a vector field meet a chosen smooth Cauchy hypersurface $C$ exactly once. Projection along those curves defines a continuous map ${\pi }_C:E^{+}(S)\to C$. The achronal set property makes this map an injective function: two boundary points on the same timelike integral curve of a vector field would be timelike related.

An achronal boundary is a topological hypersurface without boundary, even at nonsmooth generator junctions; the submanifold property allowed in the question supplies this fact. Thus domain and codomain are both three-dimensional topological manifolds without boundary. Invariance of domain makes ${\pi }_C(E^{+}(S))$ open in $C$. Compactness makes it closed in the Hausdorff space $C$, and it is nonempty. Connectedness of the spacetime gives connectedness of $C$, so this image is all of $C$. That would make $C$ compact, contradicting the noncompact Cauchy hypersurface hypothesis. The completeness assumption is false, proving the theorem.

The noncompactness hypothesis and strict trapping are essential to this version. Global hyperbolicity cannot simply be omitted in the compactness and projection steps, and nonpositive initial expansion with zeros does not supply the uniform focusing bound used here.

An integral curve of a vector field in phase space, representing the positions and momenta of a moving system. A periodic one-dimensional Hamiltonian orbit encloses the area used to define the action variable.