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For an oriented smooth surface $S$ with piecewise smooth, consistently oriented boundary, Stokes theorem givesfor a continuously differentiable vector field defined on a neighborhood of $S$. Choose the upward normal vector here. The surface is an annular band of an elliptic paraboloid, parametrized byUsing ${\partial }_{r}r\times {\partial }_{\theta }r$ as the oriented vector area element givesThe sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.

Differentiating the given vector field yieldsThe surface integral is thereforeFor the boundary line integral, the upward normal vector induces counterclockwise traversal of the outer circle and clockwise traversal of the inner circle, as viewed from above. On a circle of radius $R$, parametrized counterclockwise, $z$ is constant andSince each fourth power integrates to $3\pi /4$, the two-circle line integral isconfirming Stokes theorem and the Stokes flux through an annular paraboloid. Reversing the chosen orientation changes both integrals to $-40\pi /27$.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

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-Lindelof 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.

Write the differential one-form as $\omega ={\sum }_i{\omega }_{i}d{x}^i$, and the vector fields as $X=X^i{\partial }_i$, $Y=Y^i{\partial }_i$. The coordinate definition of the exterior derivative givesExpand the right side of the invariant formula:The last line is $\omega ([X,Y])$, by the coordinate formula for the Lie bracket of vector fields. Subtract it and interchange the two indices in the second sum. The remaining terms are exactly $d\omega (X,Y)$. ThereforeThis is the exterior derivative of a one-form evaluated on vector fields; it is valid in any frame, not just a commuting coordinate frame.

Define the pointwise representation of one-forms by vector-field functionals byIt is $\mathbb{R}$-linear, its value is a smooth function, and $X_p=0$ implies $\theta (\omega )(X)(p)=0$. Moreoverso it is a map of $C^{\infty }(M)$-modules. This is the properly typed scalar-multiplication property: the same symbol for a form and its associated functional is understood through $\theta$.

To construct its inverse, let $\alpha \in W$ and $v\in T_{p}M$. Choose a global smooth vector field $X$ with $X_p=v$; a local coordinate field multiplied by a smooth bump function supplies such an extension. SetThis is well-defined: if $X_p=Y_p$, then $X-Y$ vanishes at $p$, so $\alpha (X-Y)(p)=0$. Real linearity of $\alpha$ makes ${\omega }_p$ a covector. In fact the pointwise vanishing assumption also forcesbecause $(f-f(p))X$ vanishes at $p$. Thus $\alpha (fX)=f\alpha (X)$ for every smooth $f$, even though only real linearity was initially imposed.

It remains to prove smoothness, rather than assume continuity of an abstract linear map. Near a fixed point, choose global fields $E_1,\dots ,E_n$ equal to the coordinate basis on a smaller neighborhood, using a bump function equal to one there. On that neighborhood,All coefficients are smooth by the definition of $W$, so $\omega$ is a smooth differential one-form. The construction holds near every point, hence gives $\omega \in {\Omega }^1(M)$ with $\alpha (X)=\omega (X)$ globally.

Every tangent vector can be realized by a global field, so $\theta (\omega )=0$ forces $\omega =0$. The inverse just constructed gives surjectivity. The evaluation map is a natural $C^{\infty }(M)$-module isomorphism ${\Omega }^1(M)\cong W$. No auxiliary metric or connection was chosen.

Let $\omega ={\theta }^{-1}(\alpha )$. The preceding invariant formula for the exterior derivative identifies the expression in the question withIts vanishing for every pair of global vector fields is equivalent to $d\omega =0$: global fields with arbitrary prescribed tangent values are available by bump-function extension. Thus it is exactly the condition that $\omega$ be a closed differential one-form.

By definition, the first de Rham cohomology is closed one-forms modulo exact differential forms. The hypothesis $H_{\mathrm{dR}}^1(M)=0$ makes every closed one-form exact. Hence $d\omega =0$ gives a globally smooth function $g$ with $\omega =dg$, andConversely, if $\alpha (X)=X(g)$, thenby the definition of the Lie bracket of vector fields. Equivalently, $d(dg)=0$. The bracket compatibility condition is necessary and, when $H_{\mathrm{dR}}^1(M)=0$, sufficient for a global potential. The potential is unique up to a constant on each connected component, since the difference of two potentials has zero differential.

A Riemannian metric gives a smooth fibrewise isomorphism $g^{♭}:TM\to T^{\ast }M$, defined by $g^{♭}(v)=g(v,\cdot )$. Its inverse is smooth because the inverse of a smoothly varying invertible matrix is smooth. Define the Riemannian gradient byIt is a smooth vector field satisfying $g(V_{\sigma },Y)=d\sigma (Y)=Y(\sigma )$. Nondegeneracy also gives uniqueness. In coordinates its components are $V_{\sigma }^i={\sum }_j{g}^{ij}{\partial }_j\sigma$.

Multiplication of $g$ by the globally defined smooth positive function $e^{2\sigma }$ gives a smooth symmetric two-tensor $\tilde{g}$. For any nonzero tangent vector $v$,Thus $\tilde{g}=e^{2\sigma }g$ is a Riemannian metric. This conformal rescaling of a Riemannian metric changes lengths but preserves angles.

The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincare lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

The topological degree measures an oriented net number of sheets of a map. Its simplest integer-valued setting is a map $f:M\to N$ between connected oriented closed manifolds of the same positive dimension $n$. The hypotheses matter: an integer sign count needs orientations; compactness prevents preimages from escaping; and the same dimension makes a regular fiber discrete. A fundamental class gives the definition for continuous maps:Here $H_n(M;\mathbb{Z})$ and $H_n(N;\mathbb{Z})$ are generated by their chosen fundamental classes. Reversing either orientation changes the sign of the degree of a map between oriented manifolds. Disconnected sources contribute a sum over components; disconnected targets require a degree for each target component rather than one universal integer.

For a smooth map, choose a regular value $y$. The Sard theorem guarantees plentiful choices. Each point of $f^{-1}(y)$ has an invertible derivative, so the inverse function theorem makes it an isolated point. The fiber is finite by compactness of $M$. An oriented coordinate chart defines $\mathrm{sgn}\det d{f}_x$, and the degree as a sum of local degrees becomesAn empty fiber contributes zero. The sign is independent of the chosen oriented charts. Thus the mapping degree counts sheets with signs, not simply the number of preimages: orientation-preserving and orientation-reversing sheets cancel.

A differential-form argument both proves that this count is independent of $y$ and relates it to the topological definition. Around a regular value, take a small neighborhood $V$ whose full preimage is a disjoint union of neighborhoods mapped diffeomorphically onto $V$. Compactness rules out additional preimages approaching $V$ from elsewhere. Choose a smooth top-degree form $\eta$ supported in $V$, with ${\int }_N\eta =1$. The change of variables formula givesAlthough $\eta$ need not be everywhere positive, it represents the same normalized top cohomology class as any normalized volume form. The top de Rham cohomology of a compact connected oriented manifold says that two top forms with equal integrals differ by $d\beta$. Their pullback integrals agree, because the Generalized Stokes theorem gives ${\int }_{M}d(f^{\ast }\beta )=0$. Therefore the degree by integration of a pullback volume form isfor every smooth top form $\eta$. Pairing de Rham cohomology with fundamental classes identifies this integer with the homological definition. This proof also shows that every regular value gives the same signed sum.

The homotopy invariance of mapping degree makes it stable under continuous deformation. Homologically this follows from homotopic maps inducing the same map on homology. For a smooth homotopy $F:M\times [0,1]\to N$, a direct proof uses $d\eta =0$ and the Generalized Stokes theorem on the cylinder to getwith the cylinder orientation chosen to give the indicated boundary signs. Smooth approximation extends the geometric description to continuous maps, while the fundamental class definition already applies without differentiability. The multiplicativity of mapping degree follows from functoriality and givesA constant map has degree zero when $n>0$. If the mapping degree is nonzero, the map is onto: a point outside its image is a regular value with empty preimage. An orientation-preserving diffeomorphism has degree $1$, and an orientation-reversing one has degree $-1$. An orientation-preserving $k$-sheeted covering map has degree $k$; the signs of sheets must be included for other orientation conventions.

Examples show both the strength and the limits of the invariant. On the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, including negative integers and zero; it is the winding number. On the sphere $S^n$, an ambient reflection has degree $-1$, and the antipodal map has degree $(-1{)}^{n+1}$, because the determinant of $-I$ on the ambient ${\mathbb{R}}^{n+1}$ is $(-1{)}^{n+1}$. On the torus, an integer matrix $A$ induces a map of degree $\det A$, obtained by pulling back the constant top form. However, degree does not classify general manifold maps: the identity and an integer shear on the two-torus both have degree one but induce different maps on $H_1$, so are not homotopic.

The mapping degree gives useful obstruction and existence arguments. There is no retraction of a closed ball onto its boundary sphere: such a retraction would make the degree-one identity of the boundary null-homotopic, since it would extend across a contractible ball, contradicting homotopy invariance and the degree-zero constant map. This is the degree obstruction underlying the Brouwer fixed-point theorem. Also an everywhere nonzero tangent vector field on an even-dimensional sphere would, after normalization, produce the homotopy $\cos (\pi t)x+\sin (\pi t)v(x)$ from the identity to the antipodal map. Orthogonality makes every value a unit vector, but the endpoint degrees are $1$ and $-1$, impossible. This proves the corresponding Hairy ball theorem obstruction.

There are appropriate extensions rather than an unrestricted integer degree for every pair of manifolds. For a proper map between connected oriented manifolds without boundary of equal dimension, the regular-value count remains finite and is invariant; compactly supported top forms or locally finite homology replace the compact fundamental-class description. Without orientations, degree modulo two counts preimages modulo two and uses mod-two fundamental classes. For manifolds with boundary one uses relative mapping degree for maps of pairs, keeping the boundary away from the chosen target value. Maps of unequal dimensions, or nonproper maps where preimages can escape to infinity, do not in general admit this same signed sheet-count invariant. The central principle is an integer signed count stable under homotopy, with its hypotheses explicitly preserved.

Differentiate the circle group action at angle zero. The resulting real vector field isIn real coordinates $z=x+iy$, this is $K=-y{\partial }_x+x{\partial }_y$; the complex coefficients are conjugates, so it is indeed real. Its integral curves satisfy $\dot{z}=iz$ and hence $z(t)=e^{it}z(0)$, with period $2\pi$ away from the fixed points.

The affine coordinate does not cover the whole complex projective line, so check the other chart. With $w=1/z$, the same vector field is $K=-iw{\partial }_w+i\bar{w}{\partial }_{\bar{w}}$, smooth at $w=0$. It vanishes at $z=0$ and $w=0$, the two fixed poles. It therefore generates the stated global rotation of $S^2$, not merely a flow in one chart.

Set $s=|z{|}^2$. Contraction of the symplectic form with the rotation vector field givesWith the convention fixed in the unheaded solution, ${\iota }_K\omega =-dH$, so the Hamiltonian function isIts derivative is $dH=ds/(1+s{)}^2$. In the chart $w=1/z$, $H=1/(1+|w{|}^2)+C$, so it extends smoothly across infinity; the additive constant is the kernel freedom already identified. Thus $K=X_H$ globally, proving that the action is a Hamiltonian group action. With $C=0$, the moment map for rotation of the complex projective line takes values in $[0,1]$, reaching its endpoints at the two fixed poles.

One can check the normalization in polar coordinates: $\omega =2r(1+r^2{)}^{-2}dr\wedge d\varphi$ has total symplectic area $2\pi$, not $4\pi$, and $H=r^2/(1+r^2)$. Equivalently, in a polar angle with $r=\tan (\vartheta /2)$, $H=(1-\cos \vartheta )/2$. This explains the half-height factor that would be lost by replacing the printed form with the unscaled round-sphere area form. Under the opposite convention ${\iota }_{X_H}\omega =dH$, the Hamiltonian is the negative of this function, up to a constant.

A covariant derivative on the tangent bundle is an affine connection: it assigns a vector field ${\nabla }_{X}Y$ to vector fields $X,Y$, is linear over smooth functions in $X$, is real-linear in $Y$, and obeys the Leibniz rule ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$. Thus it differentiates a vector field while taking account of how its local basis changes. It extends to tensor fields by the Leibniz rule, compatibility with tensor contraction, and ${\nabla }_{X}a=X(a)$ for a scalar.

In a local basis $e_{\mu }$, the connection components are defined by ${\nabla }_{e_{\rho }}{e}_{\nu }={\Gamma }^{\mu }{}_{\nu \rho }{e}_{\mu }$. In a coordinate basis, $e_{\rho }={\partial }_{\rho }$, soA covariant derivative of a tensor field has tensorial transformation properties, but the connection components themselves do not: changing the coordinate basis introduces second derivatives of the coordinate change. They describe the connection in a chosen basis, rather than components of a tensor.

Literally, the printed hypothesis with positive $C$ is impossible for pairs at distance less than one, even when $U$ is constant. Use the corrected log-Lipschitz modulus $\mu$ from part (e). Local existence for the continuous finite-dimensional vector field follows from the Peano existence theorem. Boundedness gives $|X(t)-X(s)|\le \Vert U{\Vert }_{\infty }|t-s|$. Thus a trajectory cannot escape to infinity in finite time; at a finite endpoint it has a limit, and local existence there extends it. Solutions exist globally.

For uniqueness, let $X,Y$ have the same initial point and set $\delta (t)=|X(t)-Y(t)|$. This absolutely continuous function satisfies, almost everywhere,This is the Osgood uniqueness criterion, rather than the usual linear Gronwall inequality, because the velocity need not be Lipschitz continuous. To see the argument directly, put $w_{\epsilon }=\delta +\epsilon$. As long as $w_{\epsilon }\le 1$, monotonicity of $\mu$ givesFor $z_{\epsilon }=1-\log w_{\epsilon }$, this becomes $z_{\epsilon }^{\prime }\ge -C{z}_{\epsilon }$. Integration givesFor any fixed finite interval, this upper bound tends uniformly to zero as $\epsilon \downarrow 0$. Taking $\epsilon$ small prevents a first exit through $w_{\epsilon }=1$, so the bound remains valid throughout that interval. Thus $\delta =0$. Backward time has the same proof. Every initial point has exactly one global trajectory.

The decisive fact is ${\int }_0^{1}dr/\mu (r)=\infty$. For different nearby initial points, the same comparison givesuntil the separation reaches one. This quantitative continuous dependence makes the flow a homeomorphism, although it need not be a smooth diffeomorphism.