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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.

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 ${\varphi }_s$ be the local flow of the smooth vector field $X$, with ${\varphi }_0=\mathrm{id}$ and $\frac{d}{ds}{\varphi }_s(p)=X_{{\varphi }_s(p)}$. The flow definition of the Lie derivative of a tensor field isFor a tensor field of type $(r,s)$, the mixed tensor pullback transports the tensor at ${\varphi }_s(p)$ back to $p$: it applies $(d{\varphi }_s{)}_p^{-1}$ to every contravariant factor and $(d{\varphi }_s{)}_p^{\ast }$ to every covariant factor. Thus every difference quotient belongs to the same tensor space at $p$. This definition gives another tensor field of the same type and uses no choice of affine connection. The tensor Lie derivative is defined wherever the local flow exists, including points where $X$ vanishes.

The literal assumptions suffice for the scalar quadratic fidelity problem. One can avoid regularity of level-set boundaries by moving only a clipped part of the minimizer. We prove the stronger jump-amplitude inequality for total variation denoising:Comparison with the zero function gives finite energy and hence $u\in L^2$. Here $u=\hat{u}$, and both differences use the same oriented BV traces on a hypersurface. Reversing the normal reverses both differences and leaves the inequality unchanged. Outside $J_f$, the two traces of $f$ agree, so the inequality would read $-[u{]}^2\ge 0$ at a jump of $u$. It therefore gives the requested no-new-jumps property of total variation denoising in every dimension.

First use residual-preserving clipping of an ROF minimizer. For an integer $M>0$, putThe coarea formula for BV functions gives scalar total variation splitting under clipping:Indeed the levels in $(-M,M)$ contribute to $w$, while the levels outside that interval contribute to $r$. This uses the full signed coarea formula. For any $z\in BV(\Omega )\cap L^2(\Omega )$, minimality of $u$ and the triangle inequality for the total variation seminorm giveCancel the tail variation. Thus $w$ is a bounded ROF denoising minimizer for $g\in BV(\Omega )\cap L^2(\Omega )$. The data $g$ may still be unbounded. This is an exact reduction preserving $w-g=u-f$; it does not truncate the data and then pass to a limit of different reconstructions.

We next prove the jump-amplitude inequality for a bounded ROF minimizer, allowing its data to be unbounded. Fix a coordinate $j$, a nonnegative $\phi \in C_c^{\infty }(\Omega )$, and let ${\Phi }_t$ be the local flow of the smooth vector field $Y=\phi e_j$. The flow is the identity near the domain boundary, preserves each line parallel to $e_j$, and satisfies ${\Phi }_{-t}={\Phi }_t^{-1}$. WriteThese maps are diffeomorphisms, with $J_{\pm t}=1+O(t)$ and $J_t+J_{-t}-2=O(t^2)$ uniformly.

The useful trace calculation is the BV jump-product limit with one bounded factor. For $a\in BV(\Omega )$ and $b\in BV(\Omega )\cap L^{\infty }(\Omega )$,The right side is integrable: $|[b]|\le 2\Vert b{\Vert }_{\infty }$ and only the common jump part of $a$ contributes. The same limit holds with both increments replaced by their negative-time increments, still dividing by positive $t$.

Here is why this calculation needs only one bounded factor. By the BV slicing theorem, almost every coordinate slice of $a$ and $b$ has one-sided representatives. On one such interval let $\mu =Da$ and use its right-continuous representative. Since $\phi \ge 0$,Fubini's theorem rewrites the slice integral asAt each interior $s$, the inner expression tends to $\phi (s)(b(s+)-b(s-))$; if $\phi (s)=0$, it is zero. Its absolute value is at most $2\Vert b{\Vert }_{\infty }\Vert \phi {\Vert }_{\infty }$. Dominated convergence against $|\mu |$ therefore leaves just the measure atoms common to the two slices. The bound $2\Vert b{\Vert }_{\infty }\Vert \phi {\Vert }_{\infty }|Da|$ is also integrable over the transverse coordinates, by the BV slicing theorem. Integrating the slice jump sums gives the surface integral and its factor $|{\nu }_b\cdot e_j|$. Negative-time increments give the same product, because both slice differences reverse sign. At no point is a uniform bound on $a$ across the slices required.

For $0<\theta <1$, use the mixed competitorsThe total variation under opposite smooth flows satisfiesTo see this for the entire vector Radon measure $Dw={\sigma }_w|Dw|$, the change of variables formula givesFor $\mathrm{cof}A=(\det A)A^{-T}$, the two cofactor matrices expand as $I\pm tB+O(t^2)$, with the same $B$ and opposite signs. Their norm expansions on $|{\sigma }_w|=1$ have cancelling linear terms. Integrating proves the estimate, including the absolutely continuous, jump and Cantor parts. The BV transformation formula is also given in Lemma 4.2 on differentiable regularizers; that lemma does not assume bounded data. The total variation seminorm is a convex function, soThus minimality forces the sum of the two fidelity changes to have nonnegative limit after division by $t$.

It remains to evaluate that sum without bounding $g$. Put $F_g(v)=\frac{1}{2}\Vert v-g{\Vert }_2^2$. Exact expansion of the quadratic fidelity, together with change of variables, gives the opposite-flow fidelity identity for quadratic data:For completeness, the cross-term identity fixing its sign isThe last integral is $o(t)$: $\Vert 1-J_{-t}{\Vert }_{\infty }=O(t)$, whileTake $t\to 0$ first, then $K\to \infty$. Here $g\in L^1$ and the local flow is strongly continuous in $L^1$. The remaining Jacobian mass term is $O(t^2)\Vert w{\Vert }_2^2$. This argument avoids multiplying an unbounded fidelity derivative by an uncontrolled derivative measure.

Apply the BV jump-product limit with one bounded factor with $(a,b)=(g,w)$ and $(w,w)$, and use minimality. For every nonnegative $\phi$ and every coordinate $j$,Let $\theta \downarrow 0$. These are inequalities for finite signed Radon measures, so arbitrary nonnegative smooth tests imply nonnegativity of their densities. Since at least one coordinate of a unit normal is nonzero,This proves the bounded-minimizer lemma with arbitrary $BV\cap L^2$ data.

Finally return to $w=T_{M}u$ and $g=f-u+w$. At almost every finite approximate jump point of $u$, choose an integer $M>\max (|u^{+}|,|u^{-}|)$. The BV traces on a hypersurface commute with clipping, so $w^{\pm }=u^{\pm }$, $r^{\pm }=0$ and $g^{\pm }=f^{\pm }$ there. It is consequently a jump point of $w$, and its inequality is exactly $[u]([f]-[u])\ge 0$. A countable union over $M$ removes all exceptional surface-null sets. Outside $J_f$, the BV traces on a hypersurface of $f$ agree almost everywhere. We concludeThe mechanism is exact scalar coarea splitting, paired smooth-flow variations, a one-bounded-factor BV trace limit, and localization through integer clipping levels. It works in every dimension under the printed hypotheses, without essential boundedness of $f$ or $u$ and without regularity of their level-set boundaries.

In total variation calibration notation, $\mathrm{div}z=(u-f)/\alpha$. Since this divergence is itself in the BV space, the proved inequality equivalently readsThus an upward output jump forces the appropriate nonpositive jump of the calibrated divergence. This is a consequence of the variational argument above, with common oriented BV traces on a hypersurface; no curvature of the rectifiable interface or differentiability of its normal is assumed.

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 is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.

Now put ${\psi }_t=\alpha \circ {\varphi }_t\circ {\alpha }^{-1}$. The chain rule givesThus ${\psi }_t$ is the local flow of ${\alpha }_{\ast }X$. In the case $M=N$, if ${\alpha }_{\ast }X=X$, uniqueness of integral curves of a vector field makes ${\psi }_t={\varphi }_t$ on their common domains, or $\alpha \circ {\varphi }_t={\varphi }_t\circ \alpha$. Conversely, differentiating this commuting identity at zero gives $d{\alpha }_p{X}_p=X_{\alpha (p)}$, hence ${\alpha }_{\ast }X=X$. This proves that diffeomorphism invariance of a vector field is equivalent to commuting with its local flow, with every identity understood on the domain where its compositions exist.

The Lie derivative of a function is differentiation along the flow, while the Lie derivative of a vector field compares a vector with its pullback by that flow. Their index-free expressions areHere $X(f)$ is the directional derivative of the smooth function, and $[X,Y]$ is the Lie bracket of vector fields, characterized by $[X,Y](f)=X(Y(f))-Y(X(f))$. To see the second expression directly, if ${\varphi }_t$ is the local flow of $X$, differentiating $(d{\varphi }_t{)}^{-1}Y\circ {\varphi }_t$ at $t=0$ gives $dY(X)-dX(Y)=[X,Y]$. This uses the flow definition of the Lie derivative of a tensor field.

For a smooth vector field $X$, solve the initial-value problem $d{\varphi }_t(p)/dt=X({\varphi }_t(p))$, ${\varphi }_0(p)=p$. Local existence, uniqueness and smooth dependence for ordinary differential equations give a local flow. Uniqueness yields ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever both sides are defined, and ${\varphi }_{-t}$ is the inverse of ${\varphi }_t$ on the corresponding domains. Thus each local flow map is a diffeomorphism between open subsets.

A group of diffeomorphisms of the entire manifold for every real $t$ requires a complete vector field. This follows, for example, on a compact manifold without boundary: local existence times can be made uniform on a finite cover, and repeated continuation prevents finite-time escape. Without completeness, the source's group is only local. The vector field $x^2{\partial }_x$ on $\mathbb{R}$ has flow $x\mapsto x/(1-tx)$ and blows up at $t=1/x$ for $x>0$, so smoothness alone cannot imply a global one-parameter group.

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 $p=0$:

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.

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 index order in $M_{\mu }{}^{\nu }{x}^{\mu }$ fixes the rotation sign. With the entries specified in the paper, the integral curves of a vector field satisfyShift to $u=x^1-A^2$, $v=x^2+A^1$. Then $\dot{u}=v$, $\dot{v}=-u$, so $u^2+v^2$ is constant. If $x(0)=x_0$, put $u_0=x_0^1-A^2$ and $v_0=x_0^2+A^1$. Solving this linear ordinary differential equation gives the full local flow:The planar motion is a circle centred at $(A^2,-A^1)$, traversed clockwise as $\lambda$ increases. A nonzero vector $(A^3,\dots ,A^n)$ produces a helical orbit of a Euclidean Killing vector; zero drift gives a circle. At zero radius the curves are straight lines in the remaining directions, or stationary points if that drift also vanishes. These formulas are defined for every real $\lambda$.