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Let $a\in BV(\Omega )$, $b\in BV(\Omega )\cap L^{\infty }(\Omega )$ and ${\delta }_{t}c=c\circ {\Phi }_t-c$, where ${\Phi }_t$ is the local flow of $\phi e_j$ with nonnegative $\phi \in C_c^{\infty }(\Omega )$. The formula uses common oriented BV traces on a hypersurface; the trace difference of $a$ is zero almost everywhere on $J_b\setminus J_a$. It also holds for both negative-time increments with denominator positive $t$.

The BV slicing theorem reduces the calculation to one dimension. With a right-continuous representative of $a$ and $\mu =Da$, Fubini's theorem yieldsThe inner factor tends to $\phi (s)[b](s)$ and is bounded in absolute value by $2\Vert b{\Vert }_{\infty }\Vert \phi {\Vert }_{\infty }$. Dominated convergence against $|\mu |$ leaves only its measure atoms at the countable slice jumps of $b$. That same bound times $|D{a}_y|$ permits integration over the transverse coordinates. No essential bound on $a$ is used, even if its slice values grow without bound. The formula is useful for opposite-flow fidelity identity for quadratic data with unbounded data and a bounded comparison function.

A smooth vector field is complete if its maximal integral curves of a vector field exist for every real time. Its local flow is then a global one-parameter group of diffeomorphisms; uniqueness of ordinary differential equations proves the group law and the inverse-time identity. On a compact manifold without boundary, every smooth vector field is complete: compactness gives a uniform positive local existence interval through every point, so a curve can be successively extended past any proposed finite endpoint. On $\mathbb{R}$, $x^2{\partial }_x$ is not complete because its solution $x(t)=x_0/(1-t{x}_0)$ has finite-time blowup for $x_0>0$.

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.

The chain rule shows that $\alpha {\varphi }_t{\alpha }^{-1}$ is the local flow of the pushforward of a vector field ${\alpha }_{\ast }X$. If this equals $X$, uniqueness of integral curves of a vector field gives commutation. Conversely, differentiate the commuting identity at $t=0$ to obtain $d{\alpha }_p{X}_p=X_{\alpha (p)}$. All identities hold on their common domains; the vector field need not be complete.

For the local flow ${\varphi }_t$ of a smooth vector field $X$, pull a tensor field $T$ back to the fixed base point before differentiating. Pullback uses $d{\varphi }_t^{-1}$ on vector factors and the dual of $d{\varphi }_t$ on covector factors. The chain rule gives ${\mathcal{L}}_{X}f=Xf$. In coordinates, $d{\varphi }_t^{-1}Y({\varphi }_t(x))=Y+t(dYX-dXY)+O(t^2)$, so ${\mathcal{L}}_{X}Y=[X,Y]$. Differentiation of pullback preserves tensor contractions and obeys the Leibniz rule for tensor products; hence this is the tensor derivation defining the Lie derivative of a tensor field.

An integral curve follows a vector field by solving the displayed ordinary differential equation. For a smooth vector field, integral curves through initial points assemble into its local flow.

Assume $w\in BV\cap L^{\infty }$ minimizes scalar total variation denoising for $g\in BV\cap L^2$. The data $g$ may be unbounded. For the local flow of $\phi e_j$, use $(1-\theta )w+\theta w\circ {\Phi }_{\pm t}$ as two competitors. The total variation under opposite smooth flows and convexity of the total variation seminorm bound the sum of regularizer changes by $O(t^2)$. Minimality and the opposite-flow fidelity identity for quadratic data, evaluated using the BV jump-product limit with one bounded factor, implyLet $\theta \downarrow 0$. The integrands define finite signed Radon measures, since $w$ is bounded and the jump variation of $g$ is finite. Arbitrary nonnegative smooth $\phi$ make their densities nonnegative. Testing every coordinate direction removes the factor $|{\nu }_w\cdot e_j|$ and proves the stated inequality.

For $\xi$ in the Lie algebra of $G$, the left-invariant vector field is $l_{\xi }(g)=(d{L}_g{)}_e\xi$. Its global flow is right translation ${\Phi }^t=R_{\exp (t\xi )}$.

For a smooth function $f$ and a vector field $X$, the Lie derivative is ${\mathcal{L}}_{X}f=X(f)=df(X)$. It is the derivative of $f$ along the local flow of $X$.

For vector fields $X,Y$, the Lie derivative is the Lie bracket of vector fields ${\mathcal{L}}_{X}Y=[X,Y]$. Pulling $Y$ back by the local flow of $X$ and differentiating gives $dY(X)-dX(Y)$. No affine connection is needed.

Let $w\in BV\cap L^{\infty }$ and $g\in BV\cap L^2$ on a bounded domain, $F_g(v)=\frac{1}{2}\Vert v-g{\Vert }_2^2$, and $w_{\pm t}=(1-\theta )w+\theta w\circ {\Phi }_{\pm t}$ for a smooth compactly supported local flow. The displayed formula is exact to an $o(t)$ remainder, and $g$ need not be bounded. Writing $J_t=\det D{\Phi }_t$, its cross-term identity isThe remainder is $o(t)$ because $1-J_{-t}=O(t)$ and weighted $L^1$ convergence follows fromFirst let $t\to 0$, then $K\to \infty$. The mass change of $w^2$ under opposite flows is $O(t^2)\Vert w{\Vert }_2^2$. Combined with the BV jump-product limit with one bounded factor, this evaluates the jump contribution without a bounded fidelity derivative.

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.

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.